Problem Books in Mathematic
I I
Jpl Springer

Problem Books in Mathematics
Edited by P.R. Halmos
Springer
New York
Berlin
Heidelberg
Barcelona
Budapest
Hong Kong
London
Milan
Paris
Santa Clara
Singapore
Tokyo

Problem Books in Mathematics
ijtia
Series Editor: P.R. Halmos
Polynomials
by Edward J. Barbeau
Problems in Geometry
by Marcel Berger, Pierre Ращи, Jean-Pic Berry, and Xavier Saint-Raymond
Problem Book for First Year Calculus
by George W. Bluman
Exercises in Probability
by T. Cacoullos
I
An Introduction to Hubert Space and Quantum Logic
by David W. Cohen
Unsolved Problems in Geo]
by Hallard T. Croft, Kenneth \J. Falconer, and Richard K. Guy
metry
h\J Fa
Problems in Analysis
by Bernard R. Gelbaum
Problems in Real and Complex Analysis
by Bernard R. Gelbaum
Theorems and Counterexamples in Mathematics
by Bernard R. Gelbaum and John M.H. Olmsted
Exercises in Integration
by Claude George
Algebraic Logic
by S.G. Gindikin
Unsolved Problems in Number Theory (2nd ed.)
by Richard K. Guy
An Outline of Set Theory
by James M. Henle
(continued after index)

Gabor J. Szekely
Editor
Contests in Higher Mathematics
Miklos Schweitzer Competitions 1962-1991
With 39 Illustrations
Springer

Gabor J. Szekely
Department of Mathematics .
Eotvos Lorand and Technical University
Muzeum krt. 6-8
1088 Budapest, Hungary
Series Editor:
Paul R. Halmos
Department of Mathematics
Santa Clara University
Santa Clara, CA 95053
USA
Mathematics Subject Classification (1991): 15-06, 05-06, 26-06, 51-06
Library of Congress Cataloging-in-Publication Data
Contests in higher mathematics rMiklos Schweitzer competitions,
1962-1991 / Gabor J. Szekely (ed.)
p. cm. — (Problem books in mathematics)
Includes bibliographical references and index.
ISBN 0-387-94588-1
1. Mathematics — Competitions -[■ Hungary. 2.
exercises, etc. I. Schweitzer, Miklos, 1923-1945.
Gabor J., 1947- . III. Series. I
QA99.C62 1995
510'.76-dc20
Mathematics — Problems,
II. Szekely,
95-25361
Printed on acid-free paper.
© 1996 Springer-Verlag New York, Inc.
All rights reserved. This work may not be translated or copied in whole or in part without the
written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New
York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly
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adaptation, computer software, or by similar or dissimilar methodology now known or
hereafter developed is forbidden.
The use of general descriptive names, trade names, trademarks, etc., in this publication, even
if the former are not especially identified, is not to be taken as a sign that such names, as
understood by the Trade Marks and
by anyone.
Production managed by Hal Henglei
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Printed and bound by R.R. Donnelley
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manufacturing supervised by Jeffrey Taub.
editor's LaTeX files.
& Sons, Harrisonburg, VA.
ISBN 0-387-94588-1 Springer-Verlag New York Berlin Heidelberg

Preface
"I had the opportunity to speak with Leo Szildrd about the
contests of the Mathematical and Physical Society, and about
the fact that the winners of these contests turned out later to
be almost identical with the set of mathematicians and
physicists who became outstanding ..."
(J. Neumann, in a letter to L. Fejer, Berlin, Dec. 7, 1929)
The solutions to deep scientific problems rarely come to us easily. Thus, it
is important to motivate students to begin efforts on these kinds of
problems. Scientific competition has proved to be an effective stimulant toward
intellectual efforts. Successful examples include the "Concours" for
admission to the "Grandes Ecoles" in France, and the "Mathematical Tripos" in
Cambridge, England. At the turn of the century, mathematical contests
helped Hungary become one of the strongholds of the mathematical world.
With the revolution in 1848 and the Compromise in 1867, Hungary broke
free from many centuries of rule by the Turks and then the Hapsburgs, and
became a nation on equal footing with her neighbor, Austria. By the end
of the 19th century, Hungary entered a period of cultural and economic
progress. In 1891, Baron Lorand Eotvos, an outstanding Hungarian
physicist, founded the Mathematical and Physical Society. In turn, the Society
founded two journals: the Mathematical and Physical Journal in 1892 and
the Mathematical Journal for Secondary Schools in 1893. This latter
journal offered a rich variety of elementary problems for high school students.
One of the first editors of the journal, Laszlo Ratz, later became the teacher
of John Neumann and Eugene Wigner (a Nobel prize winner in physics).
In 1894, the Society introduced a mathematical competition for high school
students. Among the winners there were Lipot Fejer, Alfred Haar, Todor
Karman, Marcel Riesz, Gabor Szego, Tibor Rado, Ede Teller, and many
others who became world-famous scientists.
The success of high school competitions led the Mathematical Society to
found a college-level contest. The first contest of this kind was organized in
1949 and named after Miklos Schweitzer, a young mathematician who died
in the Second World War. Schweitzer placed second in the High School
Contest in 1941, but the statutes of the fascist regime of that time
prevented his admission to college. Schweitzer Contest problems are proposed
and selected by the most prominent Hungarian mathematicians. Thus,

PREFACE
Schweitzer problems reflect thfe interest of these mathematicians and some
aspects of the mainstream of Hungarian mathematics. The universities of
Budapest, Debrecen, and Szeged have alternately been designated by the
Society Presidium to conductj the Schweitzer Contests. The jury is
chosen by the mathematics departments of the universities in question from
among the mathematicians working in the host city. The jury sends out
requests to leading Hungarian mathematicians to submit problems suitable
for the contest. The list of problems selected by the jury is posted on the
bulletin boards of mathematics departments and of local branches of the
Mathematical Society (copies are available to anyone interested). Students
may use any materials available in libraries or in their homes to solve the
contest problems. In ten days the solutions are due, with the student's
name, faculty, course, year, aijid university or high school recorded on the
solution set.
The Schweitzer competition is one of the most unique in the world.
Winners of the contests have gone on to become world-class scientists. Thus,
the Schweitzer Contests are of interest to both math historians and
mathematicians of all ages. They serve as reflections of Hungarian mathematical
trends and as starting points for many interesting research problems in
mathematics. The Schweitzer problems between 1949 and 1961 were
previously published under the title Contests in Higher Mathematics, 1949-1961
(Akademiai Kiado, Budapest, 1968; Chapter 4 of this book summarizes the
mathematical work of M. Schweitzer). Our book is a continuation of that
volume.
We hope that this collection of Schweitzer problems will serve as a guide
for many young mathematicians and math majors. The large variety of
research-level problems may spark the interest of seasoned mathematicians
and historians of mathematics.
I wish to close by acknowledging the outstanding work of Dr. Marianna
Bolla as Managing Editor. In addition, without the constant assistance of
Dr. Dezso Miklos as Technical Editor, we could not have this book.
Bowling Green, OH
August 26, 1995
Gabor J. Szekely

Contents
Preface
Chapter 1.
Chapter 2.
Chapter 3.
Problems of the Contests
Results of the Contests
Solutions to the Problems
3.1 Algebra (Jozsef Pelikan)
3.2 Combinatorics (Ervin Gyori)
3.3 Theory of Functions (Janos Bognar and Vilmos
Totik)
3.4 Geometry (Balazs Csikos)
3.5 Measure Theory (Janos Bognar)
3.6 Number Theory (Imre Z. Ruzsa)
3.7 Operators (Janos Bognar)
3.8 Probability Theory (Gabriella Szep)
3.9 Sequences and Series (Jeno Torocsik)
3.10 Topology (Gabor Moussong)
3.11 Set Theory (Peter Komjath)
Index of Names
ν
1
49
55
55
121
150
244
330
362
388
404
465
517
552
565

1. Problems of the Contests
The letter in parentheses after the text of a problem refers to the section
in Chapter 3 containing its solution. The topics include these areas of
mathematics:
A: Algebra
C: Combinatorics
F: Theory of Functions
G: Geometry
M: Measure Theory
N: Number Theory
O: Operators
P: Probability Theory
S: Sequences and Series
T: Topology
N: Set Theory
Thus, for example, P.3 refers to problem in "Probability Theory"
section.
When available, the names of proposers are in brackets at the very end
of each problem.
l

2
1. PROBLEMS OF THE CONTESTS
1962
1
F = G holds if and only if
numbers α φ 0 and b. (N.lj
Let / and g be polynomials with rational coefficients, and let F and G
denote the sets of values cjf / and g at rational numbers. Prove that
f(x) = g(ax + b) for some suitable rational
) [E. Fried]
Determine the roots of uijity in the field of p-adic numbers. (A.l)
[L. Puchs]
Let A and В be two Abeliitn groups, and define the sum of two homo-
morphisms η and χ from ^4 to В by
α(ν + x) = Щ + aX f°r aU a € A-
With this addition, the se^; of homomorphisms from A to В forms an
Abelian group H. Suppose now that A is a p-group (p a prime
number). Prove that in this сафе Н becomes a topological group under the
topology defined by taking the subgroups pfcif(fc=l,2,...)asa
neighborhood base of 0. Prove tjhat Η is complete in this topology and that
every connected component of Η consists of a single element. When is
Η compact in this topology? (A.2) [L. Puchs]
4. Show that
π
1<ж<з/<^
Л
+ 2/2)Ξ(-1)[£ΐ1] (mod p)
for every prime ρ = 3 (mo^ 4). ([.] is integer part.) (N.2) [J. Suranyi]
5. Let / be a finite real function of one variable. Let Df and D_f be its
upper and lower derivatives, respectively, that is,
Df(x) = \imsup
h,k-+0
h,k>0
h+k>0
f(x+h)-\f(x-k)
h+\k
Df(x)--
Ε with the exception of a
[J. Czipszer]
Prove that the function
= lim inf
h,k->0
h,k>0
h+k>0
f(x+h)-f(x-k)
h+k
Borel-measurable functions. (M.l) [A. Csa-
Show that Df and D_ f are)'.
szar]
Let Ε be a bounded subset of the real line, and let Ω be a system of
(nondegenerate) closed intervals such that for each χ G Ε there exists an
/ G Ω with left endpoint x.\ Show that for every ε > 0 there exist a finite
number of pairwise nonoverlapping intervals belonging to Ω that cover
subset of outer measure less than ε. (Μ.2)
'<*>-/'
dx
y/{x2-l){l-№x2)
(where the positive value of the square root is taken) is monotonically
decreasing in the interval 0 < ϋ < 1. (F.l) [P. Turan]

1. PROBLEMS OF THE CONTESTS 3
8. Denote by M(r,f) the maximum modulus on the circle \z\ = r of the
transcendent entire function f(z), and by Mn(r, f) that of the
nth partial sum of the power series of f(z). Prove the existence of an
entire function fo{z) and a corresponding sequence of positive numbers
Π < Γ2 < · · · —> +oo such that
r Afn(rn,/0)
hmsup — — = +oo.
n_oo M(rn,/0)
(F.2) [P. Turan]
9. Find the minimum possible sum of lengths of edges of a prism all of
whose edges are tangent to a unit sphere. (G.l) [Muller-Pfeiffer]
10. From a given triangle of unit area, we choose two points independently
with uniform distribution. The straight line connecting these points
divides the triangle, with probability one, into a triangle and a
quadrilateral. Calculate the expected values of the areas of these two regions.
(P.l) [A. Renyi]
1963
1. Show that the perimeter of an arbitrary planar section of a tetrahedron
is less than the perimeter of one of the faces of the tetrahedron. (G.2)
[Gy. Hajos]
2. Show that the center of gravity of a convex region in the plane halves
at least three chords of the region. (G.3) [Gy. Hajos]
3. Let R = Ri ®R2 be the direct sum of the rings R\ and R2, and let N2 be
the annihilator ideal of R2 (in R2). Prove that R\ will be an ideal in
every ring R containing R as an ideal if and only if the only homomorphism
from R\ to N2 is the zero homomorphism. (A.3) [Gy. Pollak]
4. Call a polynomial positive reducible if it can be written as a product
of two nonconstant polynomials with positive real coefficients. Let f(x)
be a polynomial with /(0) φ 0 such that f(xn) is positive reducible
for some natural number n. Prove that f(x) itself is positive reducible.
(A.4) [L. Redei]
5. Let Я be a set of real numbers that does not consist of 0 alone and
is closed under addition. Further, let f(x) be a real-valued function
defined on Η and satisfying the following conditions:
f(x)<f(y) ifx<y and f(x + y) = f(x) + f(y) (x,y e H).
Prove that f(x) = ex on Я, where с is a nonnegative number. (F.3)
[M. Hosszu, R. Borges]
6. Show that if f(x) is a real-valued, continuous function on the half-line
0 < χ < oo, and
oo
f2(x)dx < oo,
then the function
g(x) = f(x)-2e-x Г e*f(t)dt
Jo
L

4 1. PROBLEMS OF THE CONTESTS
satisfies
/»oo /»oo
/ rf{x)dx= / f(x)da
Jo Jo
(F.4) [B. Szokefalvi-Nagy]
7. Prove that for every convex] function f(x) defined on the interval — 1 <
χ < 1 and having absolute value at most 1, there is a linear function
h(x) such that
ι
\f(\r)-h{x)\dx<4-Vs.
1
/:
(F.5) [L. Fejes-Toth]
8. Let the Fourier series
— + //"(O'k cos kx + bk sin kx)
k>l
of a function f(x) be absolutely convergent, and let
°1 + %>4+ι+%+ι (fe=l,2,...).
Show that
τ / (f(x + l/l) - f(x - h)f dx (h > o)
is uniformly bounded in h. (S.l) [K. Tandori]
9. Let f(t) be a continuous function on the interval 0 < t < 1, and define
the two sets of points [
At = {(t,0):t e [j),l]}, Bt = {(/(*), 1): * G [0,1]}.
Show that the union of all segments AtBt is Lebesgue-measurable, and
find the minimum of its measure with respect to all functions /. (M.3)
[A. Csaszar]
10. Select η points on a circle independently with uniform distribution. Let
Pn be the probability that the center of the circle is in the interior of
the convex hull of these η points. Calculate the probabilities P3 and P4.
(P.2) [A. Renyi]
1964
1. Among all possible representations of the positive integer η as η =
Σί=ιαΐ w*tn positive integers k, a\ < 0,2 < · · · < a^, when will the
product Πΐ=ι αί ^e maximum? (C.l)
2. Let ρ be a prime and let
h(x,y) = akx + bky (k = l,...,p2),

1. PROBLEMS OF THE CONTESTS 5
be homogeneous linear polynomials with integral coefficients. Suppose
that for every pair (£, η) of integers, not both divisible by p, the values
/*;(£, 77), 1 < к < ρ2, represent every residue class mod ρ exactly ρ times.
Prove that the set of pairs {(α^,δ^) : 1 < к < ρ2} is identical mod ρ
with the set {(ra, n) : 0 < ra, η < ρ — 1}. (N.3)
3. Prove that the intersection of all maximal left ideals of a ring is a (two-
sided) ideal. (A.5)
4. Let Αχ, Α2,..., An be the vertices of a closed convex n-gon К numbered
consecutively. Show that at least n — 3 vertices Ai have the property
that the reflection of Ai with respect to the midpoint of Ai-\Ai+i is
contained in K. (Indices are meant mod n.) (G.4)
5. Is it true that on any surface homeomorphic to an open disc there exist
two congruent curves homeomorphic to a circle? (G.5)
6. Let y\{x) be an arbitrary, continuous, positive function on [0, A], where
A is an arbitrary positive number. Let
fiX
yn+1(x) = 2 y/^)dt (n=l,2,...).
Jo
Prove that the functions yn (x) converge to the function у = x2 uniformly
on [0,4]. (S.2)
7. Find all linear homogeneous differential equations with continuous
coefficients (on the whole real line) such that for any solution f(t) and any
real number c, f(t + c) is also a solution. (F.6)
8. Let F be a closed set in the η-dimensional Euclidean space. Construct a
function that is 0 on F, positive outside F, and whose partial derivatives
all exist. (F.7)
9. Let Ε be the set of all real functions on / = [0,1]. Prove that one cannot
define a topology on Ε in which /n —► / holds if and only if fn converges
to / almost everywhere. (S.3)
10. Let £i,£2j · ·· j£2n De independent random variables such that P{si =
1) = p{si = -1) = 1/2 for all i, and define Sk = Σ*=1 eu l<k< 2n.
Let 7V2n denote the number of integers к £ [2, 2n] such that either
Sk > 0, or Sk = 0 and Sk-i > 0. Compute the variance of Α^η· (Ρ·3)
1965
1. Let ρ be a prime, n a natural number, and S a set of cardinality pn. Let
Ρ be a family of partitions of S into nonempty parts of sizes divisible
by ρ such that the intersection of any two parts that occur in any of the
partitions has at most one element. How large can |P| be? (N.4)
2. Let R be a finite commutative ring. Prove that R has a multiplicative
identity element (1) if and only if the annihilator of R is 0 (that is,
aR = 0, ae R imply a = 0). (А.6)
3. Let a, 6o? &ъ · · · ? &n-i be complex numbers, A a complex square matrix of
order p, and Ε the unit matrix of order p. Assuming that the eigenvalues

6 1. PROBLEMS OF THE CONTESTS
of A are given, determine the eigenvalues of the matrix
/b0E Μ b2A2 ■■■ bn-iA»-^
abn^A"-1 b0E Μ ··· bn-2An-2
abn_2An-2 a6„_1An-1 b0E ■■■ 6n_3An-3
\abiA ab2A2 ab3A3 ··· b0E )
(O.l)
4. The plane is divided into domains by η straight lines in general position,
where η > 3. Determine the maximum and minimum possible number
of angular domains among them. (We say that η lines are in general
position if no two are parallel and no three are concurrent.) (G.6)
5. Let A — A1A2A3A4 be a tetrahedron, and suppose that for each j φ к,
[Aj,Ajk] is a segment of length ρ extending from Aj in the direction
of Ak- Let pj be the intersection line of the planes [AjkAjiAjm] and
[j4fcj4ii4m]. Show that there are infinitely many straight lines that
intersect the straight lines pi,P2,P3,P4 simultaneously. (G.7)
6. Consider the radii of normal curvature of a surface at one of its points
Pq in two conjugate directions (with respect to the Dupin indicatrix).
Show that their sum does not depend on the choice of the conjugate
directions. (We exclude the choice of asymptotic directions in the case
of a hyperbolic point.) (G.8)
7. Prove that any uncountable subset of the Euclidean η-space contains
an uncountable subset with the property that the distances between
different pairs of points are different (that is, for any points P\ φ Ρ<ι
and Qi φ Qi of this subset, P\P<i = Q1Q2 implies either Pi = Qi and
Pi = Q2, or Pi = Q2 and P2 = Qi). Show that a similar statement is
not valid if the Euclidean η-space is replaced with a (separable) Hubert
space. (T.l)
8. Let the continuous functions fn(x), η = 1,2,3,..., be defined on the
interval [a, b] such that every point of [a, b] is a root of fn(x) = /m(^)
for some η φ πι. Prove that there exists a subinterval of [a, b] on which
two of the functions are equal. (S.4)
9. Let / be a continuous, nonconstant, real function, and assume the
existence of an F such that f(x + y) = F[f(x), f(y)] for all real χ and y.
Prove that / is strictly monotone. (F.8)
10. A gambler plays the following coin-tossing game. He can bet an arbitrary
positive amount of money. Then a fair coin is tossed, and the gambler
wins or loses the amount he bet depending on the outcome. Our gambler,
who starts playing with χ forints, where 0 < χ < 2C, uses the following
strategy: if at a given time his capital is у < С, he risks all of it; and
if he has у > С, he only bets 2C — y. If he has exactly 2C forints, he

1. PROBLEMS OF THE CONTESTS
7
stops playing. Let f(x) be the probability that he reaches 2C (before
going bankrupt). Determine the value of f(x). (P.4)
1966
1. Show that a segment of length h can go through or be tangent to at
most 2[ft/\/2] + 2 nonoverlapping unit spheres. ([.] is integer part.)
(G.9) [L. Fejes-Toth, A. Heppes]
2. Characterize those configurations of η coplanar straight lines for which
the sum of angles between all pairs of lines is maximum. (G.10) [L. Fejes-
Toth, A. Heppes]
3. Let /(n) denote the maximum possible number of right triangles
determined by η coplanar points. Show that
limM = oo and lim Щ = 0.
η—>oo Τι η—>οο Τι
(G.ll) [P. Erdos]
4. Let / be an ideal of the ring of all polynomials with integer coefficients
such that
(a) the elements of / do not have a common divisor of degree greater
than 0, and
(b) / contains a polynomial with constant term 1.
Prove that / contains the polynomial 1 + χ + χ2 + (- xr~l for some
natural number r. (A.7) [Gy. Szekeres]
5. A "letter T" erected at point A of the rr-axis in the rry-plane is the
union of a segment AB in the upper half-plane perpendicular to the x-
axis and a segment CD containing В in its interior and parallel to the
rr-axis. Show that it is impossible to erect a letter Τ at every point of
the rr-axis so that the union of those erected at rational points is disjoint
from the union of those erected at irrational points. (M.4) [A. Csaszar]
6. A sentence of the following type is often heard in Hungarian weather
reports: "Last night's minimum temperatures took all values between
—3 degrees and +5 degrees." Show that it would suffice to say, "Both —3
degrees and +5 degrees occurred among last night's minimum
temperatures." (Assume that temperature as a two-variable function of place
and time is continuous.) (T.2) [A. Csaszar]
7. Does there exist a function f(x,y) of two real variables that takes natural
numbers as its values and for which f(x, y) = /(y, z) implies χ = у = ζΊ
(N.l) [A. Hajnal]
8. Prove that in a Euclidean ring R the quotient and remainder are always
uniquely determined if and only if R is a polynomial ring over some field
and the value of the norm is a strictly monotone function of the degree
of the polynomial. (To be precise, there are two more trivial cases: R
can also be a field or the null ring.) (A.8) [E. Fried]
9· If Sm^-oo lam| < oo, then what can be said about the following
expression?
1 ~t^xj
fim 7; ""Γ / . \0"m-n + Я-m-n+l + * * * + a>m+n\
ι—юо /rt. -\- I ^—-J

8 1. PROBLEMS OF THE CONTESTS
(S.5) [P. Turin]
10. For a real number χ in the interval (0,1) with decimal representation
0.αι(χ)α2(χ)...αη(χ)...,
denote by η (χ) the smallest nonnegative integer such that
fln(x) + lftn(x)+2fln(x)+3fln(x)+4 = 1966.
Determine J*0 n(x)dx. (abed denotes the decimal number with digits
a, 6,c, d.) (P.5) [A. Renyi]
1967
1. Let
f(x) = a0 + aix + a2x2 + ai0x10 + ацх11 + αι2ζ12 + a^x13 (ai3 ^ 0)
and
g(x) = b0 + hx + 62^2 + M3 + ^ii^U + b12x12 + 6i3^13 (63 7^ 0)
be polynomials over the same field. Prove that the degree of their
greatest common divisor is at most 6. (A.9) [L. Redei]
2. Let К be a subset of a group G that is not a union of left cosets of a
proper subgroup. Prove that if G is a torsion group or if К is a finite
set, then the subset
кек
consists of the identity alone. (A. 10) [L. Redei]
3. Prove that if an infinite, noncommutative group G contains a proper
normal subgroup with a commutative factor group, then G also contains
an infinite proper normal subgroup. (A.ll) [B. Csakany]
4. Let αχ, α2,..., адг be positive real numbers whose sum equals 1. For a
natural number г, let щ denote the number of a^ for which 21_г > a^ >
2_< holds. Prove that
00
^V/^2Z7<4+v/log2iV.
i=l
(A.12) [L. Leindler]
5. Let / be a continuous function on the unit interval [0,1]. Show that
n^l-l f(Xl + "n+Xn)dXl...dXn = f(£)
rind
lim / ··· / /(tyxi... xn)dxi... dxn = f
n-*°° Jo Jo
e·

1. PROBLEMS OF THE CONTESTS
9
(P.6)
6. Let A be a family of proper closed subspaces of the Hubert space Η =
I2 totally ordered with respect to inclusion (that is, if Ζα,Ζ/2 £ A
then either L\ С 1/2 or L2 С L\). Prove that there exists a vector
χ € Η not contained in any of the subspaces L belonging to Α. (Τ.3)
[B. Szokefalvi-Nagy]
7. Let U be an η χ η orthogonal matrix. Prove that for any η χ η matrix
A, the matrices
1 m
Am = ^—J2u-jAUj
771+ 1 Г-ί
converge entrywise as m —> 00. (0.2) [I. Kovacs]
8. Suppose that a bounded subset 5 of the plane is a union of
congruent, homothetic, closed triangles. Show that the boundary of S can be
covered by a finite number of rectifiable arcs. (G.12) [L. Geher]
9. Let F be a surface of nonzero curvature that can be represented around
one of its points Ρ by a power series and is symmetric around the normal
planes parallel to the principal directions at P. Show that the derivative
with respect to the arc length of the curvature of an arbitrary normal
section at Ρ vanishes at P. Is it possible to replace the above symmetry
condition by a weaker one? (G.13) [A. Moor]
10. Let σ(5η, к) denote the sum of the kth powers of the lengths of the sides
of the convex τι-gon оп inscribed in a unit circle. Show that for any
natural number greater than 2 there exists a real number fe0 between 1
and 2 such that a(Sn,ko) attains its maximum for the regular n-gon.
(G.14) [L. Fejes-Toth]
1968
1. Consider the endomorphism ring of an Abelian torsion-free (resp.
torsion) group G. Prove that this ring is Neumann-regular if and only if
G is a discrete direct sum of groups isomorphic to the additive group
of the rationale (resp., a discrete direct sum of cyclic groups of prime
order). (A ring R is called Neumann-regular if for every a £ R there
exists a β e R such that αβα = α.) (Α.13) [Ε. Fried]
2. Let αϊ, α2,..., αη be nonnegative real numbers. Prove that
(S.6) [J. Suranyi]
3. Let К be a compact topological group, and let Ρ be a set of continuous
functions defined on К that has cardinality greater than continuum.
Prove that there exist xq £ К and / φ g £ F such that
/(so) = g(xo) = max f(x) = тзхд(х).
(T.4) [I. Juhasz]

10
1. PROBLEMS OF THE CONTESTS
4. Let / be a complex-valued, completely multiplicative, arithmetical
function. Assume that there exists an infinite increasing sequence Nk of
natural numbers such that
f(n) = Акф О provided Nk<n<Nk + ±y/N~k.
Prove that / is identically 1. (N.5) [I. Katai]
5. Let /с be a positive integer, ζ a complex number, and ε < 1/2 a positive
number. Prove that the following inequality holds for infinitely many
positive integers n:
>(|-ε)-.
(F.9) [P. Turan]
6. Let 21 = {A\...) be an arbitrary, countable algebraic structure (that is,
21 can have an arbitrary number of finitary operations and relations).
Prove that 21 has as many as continuum automorphisms if and only if
for any finite subset A' of A there is an automorphism έα> of 21 different
from the identity automorphism and such that
(x)tta' — x
for every χ eA'. (A. 14) [M. Makkai]
7. For every natural number r, the set of r-tuples of natural numbers is
partitioned into finitely many classes. Show that if /(r) is a function
such that f(r) > 1 and ϋπν-^ /(г) = +oo, then there exists an infinite
set of natural numbers that, for all r, contains r-tuples from at most f(r)
classes. Show that if f(r) -f* +oo, then there is a family of partitions
such that no such infinite set exists. (C.2) [P. Erdos, A. Hajnal]
8. Let η and к be given natural numbers, and let A be a set such that
W <- *$.
For i = 1,2,..., η + 1, let Ai be sets of size η such that
\АгПАа\<к (i^j),
n+l
A = \jAi.
2 = 1
Determine the cardinality of A. (C.3) [K. Corradi]
9. Let f(x) be a real function such that
lim M = 1
x—++oo ex
Σ
1°<'<*тт
fcV

1. PROBLEMS OF THE CONTESTS
11
and \f"{x)\ < c\f'(x)\ for all sufficiently large x. Prove that
x—>+oo ex
(F.10) [P. Erdos]
10. Let h be a triangle of perimeter 1, and let Я be a triangle of perimeter
λ homothetic to h. Let /ii, /12,... be translates of h such that, for all
г, hi is different from hi+2 and touches Η and /ii+i (that is, intersects
without overlapping). For which values of λ can these triangles be chosen
so that the sequence /ii, /12,... is periodic? If A > 1 is such a value, then
determine the number of different triangles in a periodic chain h\, /12,...
and also the number of times such a chain goes around the triangle H.
(G.15) [L. Fejes-Toth]
11. Let Αι,..., An be arbitrary events in a probability field. Denote by С к
the event that at least к of A\,..., An occur. Prove that
ЦР(Ск)<ЦР(Ак).
k=l k=l
(P.7) [A. Renyi]
1969
1. Let G be an infinite group generated by nilpotent normal subgroups.
Prove that every maximal Abelian normal subgroup of G is infinite.
(We call an Abelian normal subgroup maximal if it is not contained in
another Abelian normal subgroup.) (A.15) [J. Erdos]
2. Let ρ > 7 be a prime number, ζ a primitive pth root of unity, с с
rational number. Prove that in the additive group generated by the
numbers 1,C?C2? C3 + C-3 there are only finitely many elements whose
norm is equal to c. (The norm is in the pth cyclotomic field.) (A. 16)
[K. Gyory]
3. Let f(x) > 0 be a nonzero, bounded, real function on an Abelian group
G, 9i,-..,gk are given elements of G and λχ,..., λ* are real numbers.
Prove that if
к
X^/((fc:r)>0
holds for all χ G G, then
=1
к
$]λ, >0.
г=1
(S.7) [A. Mate]
Show that the following inequality holds for all к > 1, real numbers
αχ, d2,..., α*;, and positive numbers xi,X2,...,Xk-
к к
7 X% У &%Χ% in Χί
In -!=i < '^L·^
г=1 г=1

12
1. PROBLEMS OF THE CONTESTS
(S.8) [L. Losonczi]
5. Find all continuous real functions /, g and h defined on the set of positive
real numbers and satisfying the relation
f(x + y)+ g(xy) = h(x) + h(y)
for all χ > 0 and у > 0. (F.ll) [Z. Daroczy]
6. Let #ο be a fixed real number, and let / be a regular complex function
in the half-plane Re ζ > xo for which there exists a nonnegative function
F e Li(—oo, oo) satisfying \f(a + ίβ)\ < F(0) whenever a > rr0, — °° <
β < +oo. Prove that
/»а+гоо
/ f(z)dz = 0.
J a —too
(F.12) [L. Czach]
7. Prove that if a sequence of Mikusinski operators of the form με~λ8 (λ and
μ nonnegative real numbers, s the differentiation operator) is convergent
in the sense of Mikusinski, then its limit is also of this form. (О.З)
[E. Gesztelyi]
8. Let / and g be continuous positive functions defined on the interval
[0, oo), and let Ε С [0, oo) be a set of positive measure. Prove that the
range of the function defined on Ε χ Ε by the relation
F(x,y)= [ f(t)dt+ [ g(t)dt
Jo Jo
has a nonvoid interior. (M.5) [L. Losonczi]
9. In η-dimensional Euclidean space, the union of any set of closed balls (of
positive radii) is measurable in the sense of Lebesgue. (M.6) [A. Csa-
szar]
10. In η-dimensional Euclidean space, the square of the two-dimensional
Lebesgue measure of a bounded, closed, (two-dimensional) planar set
is equal to the sum of the squares of the measures of the orthogonal
projections of the given set on the η-coordinate hyperplanes. (M.7)
[L. Tamassy]
11. Let Αχ, Α2,... be a sequence of infinite sets such that \Ai Π Aj\ < 2 for
г ф j. Show that the sequence of indices can be divided into two disjoint
sequences i\ < %2 < ... and j\ < J2 < ... in such a way that, for some
sets Ε and F, \Ain Π E\ = 1 and \Ajn Π F\ = 1 for η = 1,2,.... (C.4)
[P. Erdos]
12. Let A and В be nonsingular matrices of order p, and let ξ and η be
independent random vectors of dimension p. Show that if ξ, η and ξ A +
Ύ]Β have the same distribution, if their first and second moments exist,
and if their covariance matrix is the identity matrix, then these random
vectors are normally distributed. (P.8) [B. Gyires]

1. PROBLEMS OF THE CONTESTS
13
1970
1. We have 2n + 1 elements in the commutative ring R:
α,αι,...,αη,£ι,...,£η.
Let us define the elements
n
ak = ka + y^ajQi ·
г=1
Prove that the ideal (σο, σι,..., σ&,...) can be finitely generated.
(A.17) [L. Redei]
2. Let G and Η be countable Abelian p-groups (p an arbitrary prime).
Suppose that for every positive integer n,
pnG^pn+lG.
Prove that Я is a homomorphic image of G. (A. 18) [M. Makkai]
3. The traffic rules in a regular triangle allow one to move only along
segments parallel to one of the altitudes of the triangle. We define the
distance between two points of the triangle to be the length of the shortest
such path between them. Put (nJX) points into the triangle in such
a way that the minimum distance between pairs of points is maximal.
(G.16) [L. Fejes-Toth]
4. If с is a positive integer and ρ is an odd prime, what is the smallest
residue (in absolute value) of
£(2;V (modP)?
n=0 ^ '
(N.6) [J. Suranyi]
5. Prove that two points in a compact metric space can be joined with a
rectifiable arc if and only if there exists a positive number К such that,
for any ε > 0, these points can be connected with an ε-chain not longer
than К. (Т.5) [M. Bognar]
6. Let a neighborhood basis of a point χ of the real line consist of all
Lebesgue-measurable sets containing χ whose density at χ equals 1.
Show that this requirement defines a topology that is regular but not
normal. (T.6) [A. Csaszar]
7. Let us use the word iV-measure for nonnegative, finitely additive set
functions defined on all subsets of the positive integers, equal to 0 on
finite sets, and equal to 1 on the whole set. We say that the system 21 of
sets determines the iV-measure μ if any iV-measure coinciding with μ on
all elements of 21 is necessarily identical with μ. Prove the existence of
an TV-measure μ that cannot be determined by a system of cardinality
less than continuum. (M.8) [I. Juhasz]

14
1. PROBLEMS OF THE CONTESTS
8. Let πη (χ) be a polynomial of degree not exceeding η with real coefficients
such that
kn(s)| < y/l-x2 for -1<ж<1.
Then
|<(x)|<2(n-l).
(F.13) [P. Turan]
9. Construct a continuous function /(#), periodic with period 2π, such that
the Fourier series of f(x) is divergent at χ = 0, but the Fourier series of
f2{x) is uniformly convergent on [0,2π]. (S.9) [P. Turan]
10. Prove that for every ϋ, 0 < ϋ < 1, there exist a sequence λη of positive
integers and a series X^LX αη such that
(i) λη+ι — λη > (λη) ,
(ii) lim У^ anrXn exists,
n=l
(iii) Y^ an is divergent.
n=l
(S.10) [P. Turan]
11. Let ξι, &j · · · be independent random variables such that Εξη = m > 0
and Var(£n) = σ2 < oo (n = 1,2,...). Let {an} be a sequence of positive
numbers such that an —> 0 and Σ™=ι an — °°· Prove that
Ρ I lim Vafc^ = oo J = 1.
\ k=l /
(P.9) [P. Revesz]
12. Let #i,..., ϋη be independent, uniformly distributed, random variables
in the unit interval [0,1]. Define
h(x) = - ф{к : i?fc < x}.
Prove that the probability that there is an x0 G (0,1) such that h(xo) =
x0, is equal to 1 - £. (P.10) [G. Tusnady]
1971
1. Let G be an infinite compact topological group with a Hausdorff
topology. Prove that G contains an element 9 φ I such that the set of all
powers of g is either everywhere dense in G or nowhere dense in G.
(A.I9) [J. Erdos]
2. Prove that there exists an ordered set in which every uncountable subset
contains an uncountable, well-ordered subset and that cannot be
represented as a union of a countable family of well-ordered subsets. (N.2)
[A. Hajnal]

1. PROBLEMS OF THE CONTESTS
15
3. Let 0 < ak < 1 for к = 1,2, Give a necessary and sufficient condition
for the existence, for every 0 < χ < 1, of a permutation πχ of the positive
integers such that
k=l
(S.ll) [P. Erdos]
4. Suppose that V is a locally compact topological space that admits no
countable covering with compact sets. Let С denote the set of all
compact subsets of the space V and U the set of open subsets that are not
contained in any compact set. Let / be a function from U to С such
that /(17) С U for all U GU. Prove that either
(i) there exists a nonempty compact set С such that f(U) is not a proper
subset of С whenever С С U GU,
(ii) or for some compact set C, the set
r\C) = \J{U&V:f(U)CC}
is an element of U, that is, / 1{C) is not contained in any compact
set.
(T.7) [A. Mate]
5. Let λι < λ2 < ... be a positive sequence and let К be a constant such
that
n-l
Σ\1<Κ\2η (η = 1,2,...)·
k=l
Prove that there exists a constant K' such that
n-l
Y^\k <K'\n (n=l,2,...).
k=l
(S.12) [L. Leindler]
6. Let a(x) and r(x) be positive continuous functions defined on the interval
[0, oo), and let
liminf(x — r(x)) > 0.
x—+oo
Assume that y(x) is a continuous function on the whole real line, that
it is difFerentiable on [0, oo), and that it satisfies
y\x) = a(x)y(x - r(x))
on [0, oo). Prove that the limit
lim y(x)exp
< — / a(u)du >
exists and is finite. (F.14) [I. Gyori]

16
1. PROBLEMS OF THE CONTESTS
7. Let η > 2 be an integer, let S be a set of η elements, and let Ai, 1 <
г < га, be distinct subsets of S of size at least 2 such that
Ai Π Aj фФ, А{ПАкф 0, Aj ПЛ^0 imply Α{ Π Α, Π Afc φ 0.
Show that m < 2n~1 - 1. (C.5) [P. Erdos]
8. Show that the edges of a strongly connected bipolar graph can be
oriented in such a way that for any edge e there is a simple directed path
from pole ρ to pole q containing e. (A strongly connected bipolar graph
is a finite connected graph with two special vertices ρ and q having the
property that there are no points x, у, χ φ у, such that all paths from
χ to ρ as well as all paths from χ to q contain у.) (С.6) [A. Adam]
9. Given a positive, monotone function F(x) on (0, oo) such that F{x)/x is
monotone nondecreasing and F(x)/x1+d is monotone nonincreasing for
some positive d, let λη > 0 and an > 0, η > 1. Prove that if
f^\nF[an^2^) <oo,
n=l \ k = l nJ
or
Σλ^(Σα*ΊΓΐ <00'
n=l \jfe=l AnJ
then Y^=1 an is convergent. (S.13) [L. Leindler]
10. Let {φη(χ)} be a sequence of functions belonging to L2(0,1) and having
norm less than 1 such that for any subsequence {фПк(х)} ^ne measure
of the set
1 N
{x e (0,1): Ι^=ΣФпк(х)\>У}
tends to 0 as у and N tend to infinity. Prove that фп tends to 0 weakly
in the function space L2(0,1). (M.9) [F. Moricz]
11. Let С be a simple arc with monotone curvature such that С is congruent
to its evolute. Show that under appropriate differentiability conditions,
С is a part of a cycloid or a logarithmic spiral with polar equation
r = ae*. (G.17) [J. Szenthe]
1972
1. Let Τ be a nonempty family of sets with the following properties:
(a) If X e T, then there are some Υ eT and Ζ eT such that Υ Π Ζ = 0
and Υ U Ζ = Χ.
(b) If X e Τ, and Υ U Ζ = Χ, Υ Π Ζ = 0, then either Υ Ε Τ or Ζ Ε Τ.
Show that there is a decreasing sequence X0 D Χλ D X2 D ... of sets
Xn Ε Τ such that
oo
f)Xn = Q>.
n=0

1. PROBLEMS OF THE CONTESTS 17
(C.7) [F. Galvin]
2. Let < be a reflexive, antisymmetric relation on a finite set A. Show
that this relation can be extended to an appropriate finite superset В
of A such that < on В remains reflexive, antisymmetric, and any two
elements of В have a least upper bound as well as a greatest lower bound.
(The relation < is extended to В if for x, у G A, χ < у holds in A if and
only if it holds in В.) (К.З) [E. Fried]
3. Let Xi (i = 1,2,...) be a sequence of distinct positive numbers tending
to infinity. Consider the set of all numbers representable in the form
oo
μ = У^гцХг,
2=1
where щ > 0 are integers and all but finitely many щ are 0. Let
L(x) = Σ ι and M(x) =Σ1'
Xi<x μ<χ
(In the latter sum, each μ occurs as many times as its number of
representations in the above form.)
Prove that if
hm = 1,
ж-»оо L(X)
then
r M{x+1) 1
hm = 1.
z-»oo M\X)
(F.15) [G. Halasz]
4. Let G be a solvable torsion group in which every Abelian subgroup is
finitely generated. Prove that G is finite. (A.20) [J. Pelikan]
5. We say that the real-valued function f(x) defined on the interval (0,1)
is approximately continuous on (0,1) if for any x0 G (0,1) and ε > 0 the
point #ο is a point of interior density 1 of the set
H = {x: \f(x) - f(x0)\ < ε}.
Let F С (0,1) be a countable closed set, and g(x) a real-valued
function defined on F. Prove the existence of an approximately continuous
function f(x) defined on (0,1) such that
f(x) = g(x) for all χ e F.
(M.10) [M. Laczkovich, Gy. Petruska]
6. Let P(z) be a polynomial of degree η with complex coefficients,
P(0) = 1, and \P{z)\ < Μ for \z\ < 1.

18
1. PROBLEMS OF THE CONTESTS
Prove that every root of P(z) in the closed unit disc has multiplicity at
most Cy/n, where с = c(M) > 0 is a constant depending only on M.
(F.16) [G. Halasz]
7. Let f(x, y, z) be a nonnegative harmonic function in the unit ball of R3
for which the inequality f(xo, 0,0) < ε2 holds for some 0 < ^o < 1 and
0 < ε < (1 — xq)2. Prove that /(я, y, ζ) < ε in the ball with center at
the origin and radius (1 - 3ε1/4). (F.17) [P. Turan]
8. Given four points Αι, A2, As, A± in the plane in such a way that A± is the
centroid of the AA1A2A3, find a point A$ in the plane that maximizes
the ratio
mini<j<j<fc<5 T(AjAjAk)
maxi<i<j<fc<5 T(AiAjAk)
(T(ABC) denotes the area of the triangle A ABC.) (G.18) [J. Suranyi]
9. Let К be a compact convex body in the η-dimensional Euclidean space.
Let Pi, P2,..., Pn+i be the vertices of a simplex having maximal volume
among all simplices inscribed in K. Define the points Pn+2, Pn+3? · · ·
successively so that Pk (k > η + 1) is a point of К for which the volume
of the convex hull of Pi,..., Pk is maximal. Denote this volume by Vk.
Decide, for different values of n, about the truth of the statement "the
sequence V^+i, Vn+2,... is concave." (G.19) [L. Fejes-Toth, E. Makai]
10. Let T\ and T% be second-countable topologies on the set E. We would
like to find a real function σ defined on Ε χ Ε such that
0 < σ(χ, у) < +oo, σ(χ, χ) = 0,
σ{χ, ζ) < σ{χ, у) + а(у, ζ) (χ, y,z e E),
and, for any ρ Ε Ε, the sets
Vi (ρ, ε) = {χ: σ{χ, ρ) < ε} (ε > 0)
form a neighborhood base of ρ with respect to 71, and the sets
V2(p,ε) = {χ : σ(ρ,χ) < ε} (ε > 0)
form a neighborhood base of ρ with respect to T2. Prove that such a
function σ exists if and only if, for any ρ e Ε and 7^-open set G Э ρ
(г = 1,2), there exist a 7^-open set G' and a T^-closed set F with
peGf CFCG. (T.8) [A. Csaszar]
11. We throw N balls into η urns, one by one, independently and uniformly.
Let Xi = Xi{N, n) be the total number of balls in the zth urn. Consider
the random variable
N
y(N,n)= min \Xi-—\.
1<г<п 71
Verify the following three statements:

1. PROBLEMS OF THE CONTESTS
19
(a) If η —> oo and N/n3 —> oo, then
p(m^<x)^l_e-xV^ forallx>0.
п У п
(b) If η —> oo and iV/n3 < Κ (Κ constant), then for any ε > 0 there is
an A > 0 such that
P(y(N,n)< A)> 1-е.
(c) If η —> oo and iV/n3 —> 0 then
P(y(JV,n) <1)->1.
(P.ll) [P. Revesz]
1973
1. We say that the rank of a group G is at most r if every subgroup of
G can be generated by at most r elements. Prove that there exists an
integer s such that for every finite group G of rank 2 the commutator
series of G has length less than s. (A.21) [J. Erdos]
2. Let R be an Artinian ring with unity. Suppose that every idempotent
element of R commutes with every element of R whose square is 0.
Suppose R is the sum of the ideals A and B. Prove that AB = В A.
(A.22) [A. Kertesz]
3. Find a constant с > 1 with the property that, for arbitrary positive
integers η and к such that η > ck, the number of distinct prime factors
of (I) is at least k. (N.7) [P. Erdos]
4. Let f(n) be the largest integer к such that nk divides n!, and let F(n) =
max2<m<n/W· Show that
lim ψ^ = 1.
n-»oo Π log log Π
(N.8) [P. Erdos]
5. Verify that for every χ > 0,
T(x + 1) > g
(F.18) [P. Medgyessy]
6. If / is a nonnegative, continuous, concave function on the closed interval
[0,1] such that /(0) = 1, then
/ xf(x)dx < - \ f(x)d:
Jo J L/o
I2
lx\ ,

20
1. PROBLEMS OF THE CONTESTS
(F.19) [Z. Daroczy]
7. Let us connect consecutive vertices of a regular heptagon inscribed in a
unit circle by connected subsets (of the plane of the circle) of diameter
less than 1. Show that every continuum (in the plane of the circle) of
diameter greater than 4, containing the center of the circle, intersects
one of these connected sets. (G.20) [M. Bognar]
8. What is the radius of the largest disc that can be covered by a finite
number of closed discs of radius 1 in such a way that each disc intersects
at most three others? (G.21) [L. Fejes-Toth]
9. Determine the value of
sup [\ogE£-E\og£],
1<£<2
where ξ is a random variable and Ε denotes expectation. (P. 12)
[Z. Daroczy]
10. Find the limit distribution of the sequence ηη of random variables with
distribution
„f , 2(27-1)я\\ 1 ,. ,
Ρ Ι ηη = arccos(cos — ) I = - (j = 1,,..., n).
(arccos(.) denotes the main value.) (P.13) [B. Gyires]
1974
1. Let Τ be a family of subsets of a ground set X such that \Jf£fF = X,
and
(a) if А, В G T, then A U В С С for some С G T\
(b) if An G Τ (η = 0,1,...), В G Τ, and А0 С Αι С ..., then, for some
к > 0, Ап Π Β = Ак П В for all n > к.
Show that there exist pairwise disjoint sets ΧΊ {η e Γ), with X =
U{Xy : 7 G Γ}, such that every ΧΊ is contained in some member of T,
and every element of Τ is contained in the union of finitely many Xy's.
(K.4) [A. Hajnal]
2. Let G be a 2-connected nonbipartite graph on 2n vertices. Show that the
vertex set of G can be split into two classes of n elements each such that
the edges joining the two classes form a connected, spanning subgraph.
(C.8) [L. Lovasz]
3. Prove that a necessary and sufficient condition for the existence of a
set S С {1,..., n} with the property that the integers 0,1,..., n — 1 all
have an odd number of representations in the form χ — у, ж, у G 5, is
that (2n — 1) has a multiple of the form 2 · 4* — 1. (N.9) [L. Lovasz,
J. Pelikan]
4. Let R be an infinite ring such that every subring of R different from {0}
has a finite index in R. (By the index of a subring, we mean the index of
its additive group in the additive group of R.) Prove that the additive
group of R is cyclic. (A.23) [L. Lovasz, J. Pelikan]

1. PROBLEMS OF THE CONTESTS 21
5. Let {fn}^=o De a uniformly bounded sequence of real-valued measurable
functions defined on [0,1] satisfying
JO
Further, let {cn} be a sequence of real numbers with
oo
Σ 4 = +oo.
71=0
Prove that some re-arrangement of the series Σ™=0 cnfn is divergent on
a set of positive measure. (M.ll) [J. Komlos]
6. Let f(x) = Σ™=1 an/(x + n2), (x > 0), where Σ™=1 \an\n~a < oo for
some a > 2. Let us assume that for some β > Ι/α, we have f(x) =
as χ —> oo. Prove that an is identically 0. (S.14) [G. Halasz]
7. Given a positive integer m and 0 < δ < π, construct a trigonometric
polynomial f(x) = a0 + Y^™=1(ancosnx + bnsmnx) of degree m such
that /(0) = 1, f6<\x\<v \f(x)\dx < c/m, and πΐ3Χ-π<χ<π \f'(x)\ < c/δ,
for some universal constant c. (S.15) [G. Halasz]
8. Prove that there exists a topological space Τ containing the real line
as a subset, such that the Lebesgue-measurable functions, and only
those, extend continuously over T. Show that the real line cannot be an
everywhere-dense subset of such a space Т. (Т.9) [A. Csaszar]
9. Let A be a closed and bounded set in the plane, and let С denote the
set of points at a unit distance from A. Let ρ G C, and assume that the
intersection of A with the unit circle К centered at ρ can be covered by
an arc shorter than a semicircle of K. Prove that the intersection of С
with a suitable neighborhood of ρ is a simple arc of which ρ is not an
endpoint. (T.10) [M. Bognar]
10. Let μ and ν be two probability measures on the Borel sets of the plane.
Prove that there are random variables £ъ£2???ь??2 sucn ^na^
(a) the distribution of (ξι, &) is β and tne distribution of (7/1,7/2) is v,
(b) ξι < 77i, £2 < 772 almost everywhere, if and only if /i(G) > i/(G) for
all sets of the form G = υ^=1(—00, Xi) x (-00, yi). (P.14) [P. Major]
1975
1. Show that there exists a tournament (T, —>) of cardinality Hi containing
no transitive subtournament of size Νχ. (A structure (T, —>) is a
tournament if —> is a binary, irreflexive, asymmetric, and trichotomic relation.
The tournament (T, —>) is transitive if —> is transitive, that is, if it orders
Т.) (К.5) [A. Hajnal]
2. Let An denote the set of all mappings / : {1,2,..., n} —> {1,2,..., n)
such that /_1(г) := {к : f(k) = i} φ 0 implies /_1(j) Φ 0, 3 Ξ
{1,2,...,г}. Prove
fc=0

22
1. PROBLEMS OF THE CONTESTS
(C.9) [L. Lovasz]
3. Let S be a semigroup without proper two-sided ideals, and suppose that
for every a, b G S at least one of the products ab and ba is equal to one
of the elements a, b. Prove that either ab = a for all a, 6 G S or ab = b
for all a,be S. (A.24) [L. Megyesi]
4. Prove that the set of rational-valued, multiplicative arithmetical
functions and the set of complex rational-valued, multiplicative arithmetical
functions form isomorphic groups with the convolution operation fog
defined by
d\n
(We call a complex number complex rational, if its real and imaginary
parts are both rational.) (N.10) [B. Csakany]
5. Let {fn} be a sequence of Lebesgue-integrable functions on [0,1] such
that for any Lebesgue-measurable subset Ε of [0,1] the sequence JE fn
is convergent. Assume also that limn fn — f exists almost everywhere.
Prove that / is integrable and fEf = limn fE fn. Is the assertion also
true if Ε runs only over intervals but we also assume fn > 0? What
happens if [0,1] is replaced by [0, oo) ? (M.12) [J. Szucs]
6. Let / be a differentiable real function, and let Μ be a positive real
number. Prove that if
\f(x + t)- 2f(x) + f(x-t)\<Mt2 for all χ and t,
then
\f'(x + t)-f'(x)\<M\t\.
(F.20) [J. Szabados]
7. Let a < a' < b < b' be real numbers, and let the real function / be
continuous on the interval [a, b'] and differentiable in its interior. Prove
that there exist с е (a,b), с' е (a',bf) such that
№-f(a) = f'(c)(b-a),
f(bf)-f{a') = f'{c'){b'-a'),
and с < с'. (F.21) [В. Szokefalvi-Nagy]
8. Prove that if
m
Y^an< Nam (m = 1,2,...)
n=l
holds for a sequence {an} of nonnegative real numbers with some positive
integer N, then cti+p > pcti for г, ρ = 1,2,..., where
iN
с*г= Σ αη (г=1,2,...)-
п=(г-1)ЛГ+1
(S.16) [L. Leindler]

1. PROBLEMS OF THE CONTESTS
23
9. Let /o?c, a,g be positive constants, and let x(t) be the solution of the
differential equation
{[lo + cta]2x')'+ g[l0 + cta] sin χ = 0, t>0, ~ <x< |,
satisfying the initial conditions χ (to) = xq, xf(to) = 0. (This is the
equation of the mathematical pendulum whose length changes according
to the law Ζ = lo + cta.) Prove that x(t) is defined on the interval [to, oo);
furthermore, if a > 2 then for every xo φ 0 there exists a to such that
liminf \x(t)\ > 0.
t—>oo
(F.22) [L. Hatvani]
10. Prove that an idempotent linear operator of a Hilbert space is self-adjoint
if and only if it has norm 0 or 1. (0.4) [J. Szucs]
11. Let Χι, X2,. · ·, Xn be (not necessary independent) discrete random
variables. Prove that there exist at least n2/2 pairs (г, j) such that
Н(Х< + ХА>\ mm {H(Xk)},
О 1<к<п
where H(X) denotes the Shannon entropy of X. (P. 15) [Gy. Katona]
12. Assume that a face of a convex polyhedron Ρ has a common edge with
every other face. Show that there exists a simple closed polygon that
consists of edges of Ρ and passes through all vertices. (G.22) [L. Lovasz]
1976
1. Assume that R, a recursive, binary relation on N (the set of natural
numbers), orders N into type ω. Show that if f(n) is the nth element of
this order, then / is not necessarily recursive. (N.6) [L. Posa]
2. Let G be an infinite graph such that for any countably infinite vertex set
A there is a vertex ρ joined to infinitely many elements of A. Show that
G has a countably infinite vertex set A such that G contains uncountably
infinitely many vertices ρ joined to infinitely many elements of A. (CIO)
[P. Erdos, A. Hajnal]
3. Let Η denote the set of those natural numbers for which r(n) divides
n, where r(n) is the number of divisors of n. Show that
(a) n\ e Η for all sufficiently large n,
(b) Η has density 0.
(N.ll) [P. Erdos]
4. Let Ζ be the ring of rational integers. Construct an integral domain J
satisfying the following conditions:
(a) Z С j;
(b) no element of / \ Ζ is algebraic over Ζ (that is, not a root of a
polynomial with coefficients in Z);
(c) J only has trivial endomorphisms.
(A.25) [E. Fried]

24
1. PROBLEMS OF THE CONTESTS
5. Let Su = Σ*=ι. bjZj (y = 0, ±1, ±2,...), where the bj are arbitrary and
the Zj are nonzero complex numbers. Prove that
\Sq\ < η max ISU.
0<H<n
(S.17) [G. Halasz]
6. Let 0 < с < 1, and let η denote the order type of the set of rational
numbers. Assume that with every rational number r we associate a
Lebesgue-measurable subset Hr of measure с of the interval [0,1]. Prove
the existence of a Lebesgue-measurable set if С [0,1] of measure с such
that for every χ G Η the set
{r:xe Hr}
contains a subset of type η. (Μ.13) [Μ. Laczkovich]
7. Let /i, /2,.. ·, fn be regular functions on a domain of the complex plane,
linearly independent over the complex field. Prove that the functions
fifki 1 < г, /с < n, are also linearly independent. (F.23) [L. Lempert]
8. Prove that the set of all linear combinations (with real coefficients) of
the system of polynomials {xn + xn }£i0 is dense in C[0,1]. (F.24)
[J. Szabados]
9. Let D be a convex subset of the η-dimensional space, and suppose that
D' is obtained from D by applying a positive central dilatation and then
a translation. Suppose also that the sum of the volumes of D and D'
is 1, and D Π D' φ 0. Determine the supremum of the volume of the
convex hull of D U D' taken for all such pairs of sets D,D'. (G.23)
[L. Fejes-Toth, E. Makai]
10. Suppose that r is a metrizable topology on a set X of cardinality less
than or equal to continuum. Prove that there exists a separable and
metrizable topology on X that is coarser than r. (T.ll) [I. Juhasz]
11. Let £1,^2,... be independent, identically distributed random variables
with distribution
p(fr = -i) = p(fr = i) = i.
Write Sn = ft +& + ■■■+ Cn (n = 1,2,...), So = 0, and
Tn = —= max Sk .
yjn 0<k<n
Prove that lim inf (log n) Tn = 0 with probability one. (P. 16) [P. Re-
n—+00
vesz]

1. PROBLEMS OF THE CONTESTS
25
1977
1. Consider the intersection of an ellipsoid with a plane σ passing through
its center O. On the line through the point О perpendicular to σ, mark
the two points at a distance from О equal to the area of the intersection.
Determine the loci of the marked points as σ runs through all such
planes. (G.24) [L. Tamassy]
2. Construct on the real projective plane a continuous curve, consisting
of simple points, which is not a straight line and is intersected in a
single point by every tangent and every secant of a given conic. (G.25)
[F. Karteszi]
3. Prove that if a, x, у are p-adic integers different from 0 and p\x, pa\xy,
then
1 (!+«)'-lsb*(l+«) (moda).
у χ χ
(N.12) [L. Redei]
4. Let ρ > 5 be a prime number. Prove that every algebraic integer of
the pth cyclotomic field can be represented as a sum of (finitely many)
distinct units of the ring of algebraic integers of the field. (A.26)
[K. Gyory]
5. Suppose that the automorphism group of the finite undirected graph
X = (P,E) is isomorphic to the quaternion group (of order 8). Prove
that the adjacency matrix of X has an eigenvalue of multiplicity at least
4.
(P = {1,2,..., n} is the set of vertices of the graph X. The set of edges
Ε is a subset of the set of all unordered pairs of elements of P. The
group of automorphisms of X consists of those permutations of Ρ that
map edges to edges. The adjacency matrix Μ = [rriij] is the η χ η
matrix defined by m^ = 1 if {г, j} e Ε and m%j = 0 otherwise.) (A.27)
[L. Babai]
6. Let / be a real function defined on the positive half-axis for which
f(xy) = xf{y) + yf(x) and f(x + 1) < f(x) hold for every positive
χ and y. Show that if /(1/2) = 1/2, then
f(x) + /(1 - x) > -x log2 χ - (1 - x) log2(l - x)
for every χ e (0,1). (F.25) [Z. Daroczy, Gy. Maksa]
7. Let G be a locally compact solvable group, let ci,...,cn be complex
numbers, and assume that the complex-valued functions / and g on G
satisfy
η
Σ ckf(xyk) = f(x)9(y) for all x,yeG.
k=l
Prove that if / is a bounded function and
inf Ref{x)X(x)>0
for some continuous (complex) character χ of G, then g is continuous.
(F.26) [L. Szekelyhidi]

26
1. PROBLEMS OF THE CONTESTS
8. Let ρ > 1 be a real number and R+ = (0, oo). For which continuous
functions g : R+ —> R+ are the following functions all convex?
x = (ж1,...,жп+1) еШ1+\ п= 1,2,...
(S.18) [L. Losonczi]
9. Suppose that the components of the vector u = (щ,..., un) are real
functions defined on the closed interval [a, b] with the property that
every nontrivial linear combination of them has at most η zeros in [a, b].
Prove that if σ is an increasing function on [a, b] and the rank of the
operator
A(f)= I u(x)f(x)da(x), /6 CM,
J a
is r < n, then σ has exactly r points of increase. (F.27) [E. Gesztelyi]
10. Let the sequence of random variables {Xm, m > 0}, Xq = 0, be an
infinite random walk on the set of nonnegative integers with transition
probabilities
Pi = P(Xm+i = i + 11 Xrn = i) > 0, i > 0,
qi = P{Xm+1 = г - 1 \Xm = i) > 0, i > 0.
Prove that for arbitrary к > 0 there is an a^ > 1 such that
Pn(k) = P[ max Xj = к
satisfies the limit relation
1 L
lim -У]Рп(А;К<оо.
L—+oo L· £-^
n=l
(P.17) [J. Tomko]
1978
1. Let Η be a family of finite subsets of an infinite set X such that every
finite subset of X can be represented as the union of two disjoint sets
from H. Prove that for every positive integer к there is a subset of X
that can be represented in at least к different ways as the union of two
disjoint sets from H. (C.ll) [P. Erdos]
2. For a distributive lattice L, consider the following two statements:
(A) Every ideal of L is the kernel of at least two different homomor-
phisms.
(B) L contains no maximal ideal.
Mn(x) =

1. PROBLEMS OF THE CONTESTS
27
Which one of these statements implies the other?
(Every homomorphism φ of L induces an equivalence relation on L:
a rsj b if and only if αφ = bip. We do not consider two homomorphisms
different if they imply the same equivalence relation.) (A.28) [J. Varlet,
E. Fried]
3. Let 1 < a\ < α<ι < · · · < an < χ be positive integers such that J^™ Va* —
1. Let у denote the number of positive integers smaller than χ not
divisible by any of the a^. Prove that
ex
logx
with a suitable positive constant с (independent of χ and the numbers
α<). (Ν.13) [I. Z. Ruzsa]
4. Let Q and R be the set of rational numbers and the set of real numbers,
respectively, and let / : Q —> R be a function with the following property.
For every h e Q, x0 G R,
/(ж + Л)-/(ж)->0
as χ G Q tends to xq. Does it follow that / is bounded on some interval ?
(F.28) [M. Laczkovich]
5. Suppose that R(z) = 4^l^>-_00 an^n converges in a neighborhood of the
unit circle {z : \z\ = 1} in the complex plane, and R(z) = P(z)/Q(z) is a
rational function in this neighborhood, where Ρ and Q are polynomials
of degree at most k. Prove that there is a constant с independent of к
such that
oo
У^ \o>nI ^c^2 max \R(z) I·
|z|=l
n=—oo
(S.19) [H. S. Shapiro, G. Somorjai]
6. Suppose that the function g : (0,1) —> R can be uniformly approximated
by polynomials with nonnegative coefficients. Prove that g must be
analytic. Is the statement also true for the interval (—1,0) instead of
(0,1)? (F.29) [J. Kalina, L. Lempert]
7. Let Τ be a surjective mapping of the hyperbolic plane onto itself which
maps collinear points into collinear points. Prove that Τ must be an
isometry. (G.26) [M. Bognar]
8. Let Xi,...,Xn be η points in the unit square (n > 1). Let Ti be
the distance of Xi from the nearest point (other than Xi). Prove the
inequality
r\ + ---+r2n <4.
(G.27) [L. Fejes-Toth, E. Szemeredi]
9. Suppose that all subspaces of cardinality at most Hi of a topological
space are second-countable. Prove that the whole space is second-
countable. (T.12) [A. Hajnal, I. Juhasz]

28
1. PROBLEMS OF THE CONTESTS
10. Let Yn be a binomial random variable with parameters η and p. Assume
that a certain set Η of positive integers has a density and that this
density is equal to d. Prove the following statements:
(a) \\mn^00 P(Yn G H) = d if Η is an arithmetic progression.
(b) The previous limit relation is not valid for arbitrary H.
(c) If Η is such that P(Yn G H) is convergent, then the limit must be
equal to d.
(P.18) [L. Posa]
1979
1. Let the operation / of к variables defined on the set {1,2, ...,n} be
called friendly toward the binary relation ρ defined on the same set if
/(αϊ, α2, ...,ak) ρ f(h,b2,..., 6*)
implies α* ρ bi for at least one г, 1 < г < к. Show that if the operation
/ is friendly toward the relations "equal to" and "less than," then it is
friendly toward all binary relations. (C.12) [B. Csakany]
2. Let V be a variety of monoids such that not all monoids of V are groups.
Prove that if A G V and В is a submonoid of A, there exist monoids
S G V and С and epimorphisms φ : S —> A , φι : S —> С such that
{{ε)φϊ1)φ = В (е is the identity element of С). (А.29) [L. Marki]
3. Let g(n, k) denote the number of strongly connected, simple directed
graphs with η vertices and к edges. (Simple means no loops or multiple
edges.) Show that
Yf(-l)fcfl(n,*) = (n-l)!.
k=n
(C.13) [A. A. Schrijver]
4. For what values of η does the group SO(n) of all orthogonal
transformations of determinant 1 of the η-dimensional Euclidean space possess
a closed regular subgroup? (G < SO(n) is called regular if for any
elements x, у of the unit sphere there exists a unique φ G G such that
φ{χ) = у.) (А.30) [Ζ. Szabo]
5. Give an example of ten different noncoplanar points Ρχ,.,.,Ρδ, Qi,
..., Q5 in 3-space such that connecting each Pi to each Qj by a rigid
rod results in a rigid system. (G.28) [L. Lovasz]
6. Let us define a pseudo-Riemannian metric on the set of points of the
Euclidean space E3 not lying on the z-axis by the metric tensor
/100 \
0 1 0^ ,
\0 0 -y/x2 + y2J
where (x,y,z) is a Cartesian coordinate system in E3. Show that the
orthogonal projections of the geodesic curves of this Riemannian space

1. PROBLEMS OF THE CONTESTS
29
onto the (#, y)-plane axe straight lines or conic sections with focus at the
origin. (G.29) [P. Nagy]
7. Let Τ be a triangulation of an η-dimensional sphere, and to each vertex
of Τ let us assign a nonzero vector of a linear space V. Show that if Τ has
an η-dimensional simplex such that the vectors assigned to the vertices
of this simplex are linearly independent, then another such simplex must
also exist. (C.14) [L. Lovasz]
8. Let Kn(n = 1,2,...) be periodical continuous functions of period 2π,
and write
/»2π
M/;*) = / f(t)Kn(x - t)dt.
Jo
Prove that the following statements are equivalent:
(i) /02π IM/; x) - f(x)\ <te -► 0 (n -► oo) for all /eli [0,2π].
(ii) fcn(/;0) —► /(0) for all continuous, 27r-periodic functions /.
(S.20) [V. Totik]
oo
9. Let us assume that the series of holomorphic functions У^Л(^) is ab-
k=l
solutely convergent for all ζ G С Let ii С С be the set of those points
where the above sum function is not regular. Prove that Η is nowhere
dense but not necessarily countable. (S.21) [L. Kerchy]
10. Prove that if ai (i = 1,2,3,4) are positive constants, a2 — a^ > 2, and
CL1CL3 — CL2 > 2, then the solution (x(t), y(t)) of the system of differential
equations
χ = a\ — a<ix + а$ху,
у = а4х-у- а3ху (х,у е R)
with the initial conditions x(0) = 0, y(0) > a\ is such that the function
x(i) has exactly one strict local maximum on the interval [0,00). (F.30)
[L. Pinter, L. Hatvani]
11. Let {£fcj}fc°j=i be a double sequence of random variables such that
^г^ы = 0((log(2|i-fc|+2)log(2|j-i|+2))-2) (i, j,k,l = 1,2,.. .)·
Prove that with probability one,
^ m η
Σ^Σλ^ ~* ^ ^ max(m? П) ~* °°·
mn ,
k=l 1=1
(P. 19) [F. Moricz]

30
1. PROBLEMS OF THE CONTESTS
1980
1. For a real number x, let ||ж|| denote the distance between χ and the
closest integer. Let 0 < xn < 1 (n = 1,2,...), and let ε > 0. Show
that there exist infinitely many pairs (n, m) of indices such that η φ πι
and
• ( l \
\\Xn ~ Xm\\ < mm ε, г .
\ 2\n-m\J
(C.15) [V. T. Sos]
2. Let Η be the class of all graphs with at most 2H° vertices not containing
a complete subgraph of size Νχ. Show that there is no graph Η e Η
such that every graph in Η is a subgraph of Η. (Ν.7) [F. Galvin]
3. In a lattice, connect the elements a/\b and a\/b by an edge whenever a and
b are incomparable. Prove that in the obtained graph every connected
component is a sublattice. (A.31) [M. Ajtai]
4. Let Τ e SX(n, Z), let G be a nonsingular η χ η matrix with integer
elements, and put S = G~lTG. Prove that there is a natural number к
such that Sk G SX(n,Z). (N.14) [Gy. Szekeres]
5. Let G be a transitive subgroup of the symmetric group £25 different from
525 and A25. Prove that the order of G is not divisible by 23. (A.32)
[J. Pelikan]
6. Let us call a continuous function / : [a, b] —> R2 reducible if it has a
double arc (that is, if there are a<a<P<^<6<b such that
there exists a strictly monotone and continuous h : [α, β] —> [7, δ] for
which f(t) = f(h(t)) is satisfied for every a < t < /3); otherwise / is
irreducible. Construct irreducible / : [a, b] —> R2 and g : [c, d] —> R2
such that /([a, b\) = g([c, d]) and
(a) both / and g are rectifiable but their lengths are different;
(b) / is rectifiable but g is not.
(F.31) [A. Csaszar]
7. Let η > 2 be a natural number and p(x) a real polynomial of degree at
most η for which
max |p(x)|<l, p(-l) =p(l) = 0.
—1<ж<1
Prove that then
ι // μ ncos^- / 1 1 \
\p'(x)\ < , \= F- <x< 5Γ ·
1/П Л" ,/1-zW^ V cos£ cos^y
(F.32) [J. Szabados]
8. Let f(x) be a nonnegative, integrable function on (0,2π) whose Fourier
series is f(x) = ao + ^2^=1 а>к cos(rikx), where none of the positive
integers rik divides another. Prove that |a&| < ao- (S.22) [G. Halasz]
9. Let us divide by straight lines a quadrangle of unit area into η subpoly-
gons and draw a circle into each subpolygon. Show that the sum of the
perimeters of the circles is at most π>/η (the lines are not allowed to cut
the interior of a subpolygon). (G.30) [G. and L. Fejes-Toth]

1. PROBLEMS OF THE CONTESTS
31
10. Suppose that the Тз-space Χ has no isolated points and that in X any
family of pairwise disjoint, nonempty, open sets is countable. Prove
that X can be covered by at most continuum many nowhere-dense sets.
(T.13) [I. Juhasz]
1981
1. We are given an infinite sequence of l's and 2's with the following
properties:
(1) The first element of the sequence is 1.
(2) There are no two consecutive 2's or three consecutive l's.
(3) If we replace consecutive l's by a single 2, leave the single l's alone,
and delete the original 2's, then we recover the original sequence.
How many 2's are there among the first η elements of the sequence?
(S.23) [P. P. Palfy]
2. Consider the lattice L of the contractions of a simple graph G (as sets
of vertex pairs) with respect to inclusion. Let η > 1 be an arbitrary
integer. Show that the identity
/- \ - ( ( λΝ
χΛ V* = V ΙχΛ| V »
\i=0 / j=Q
. \ 0<i<n ,
holds if and only if G has no cycle of size at least n+2. (C.16) [A. Huhn]
3. Construct an uncountable Hausdorff space in which the complement of
the closure of any nonempty, open set is countable. (T.14) [A. Hajnal,
I. Juhasz]
4. Let G be a finite group and /C a conjugacy class of G that generates G.
Prove that the following two statements are equivalent:
(1) There exists a positive integer m such that every element of G can be
written as a product of m (not necessarily distinct) elements of /C.
(2) G is equal to its own commutator subgroup.
(A.33) [J. Denes]
5. Let К be a convex cone in the η-dimensional real vector space Rn, and
consider the sets A = К U (-K) and В = (Rn \ A) U {0} (0 is the
origin). Show that one can find two subspaces in Rn such that together
they span Rn, and one of them lies in A and the other lies in B. (G.31)
[J. Szucs]
6. Let / be a strictly increasing, continuous function mapping 7 = [0,1]
onto itself. Prove that the following inequality holds for all pairs x, у G I:
l-cos{xy)< f f(t)sin(tf(t))dt+ [ /-^sin^/-1^))^.
Jo Jo
(F.33) [Zs. Pales]
7. Let U be a real normed space such that, for any finite-dimensional,
real normed space X, U contains a subspace isometrically isomorphic to

32
1. PROBLEMS OF THE CONTESTS
X. Prove that every (not necessarily closed) subspace V of U of finite
codimension has the same property. (We call V of finite codimension if
there exists a finite-dimensional subspace N of U such that V + N = U.)
(F.34) [A. Bosznay]
8. Let W be a dense, open subset of the real line R. Show that the following
two statements are equivalent:
(1) Every function / : R —> R continuous at all points of R \ W and non-
decreasing on every open interval contained in W is nondecreasing on
the whole R.
(2) R \ W is countable.
(T.15) [E. Gesztelyi]
9. Let η > 2 be an integer, and let X be a connected Hausdorff space
such that every point of X has a neighborhood homeomorphic to the
Euclidean space Rn. Suppose that any discrete (not necessarily closed)
subspace D of X can be covered by a family of pairwise disjoint, open
sets of X so that each of these open sets contains precisely one element
of D. Prove that X is a union of at most Ni compact subspaces. (T.16)
[Z. Balogh]
10. Let Ρ be a probability distribution defined on the Borel sets of the real
line. Suppose that Ρ is symmetric with respect to the origin, absolutely
continous with respect to the Lebesgue measure, and its density function
ρ is zero outside the interval [—1,1] and inside this interval it is between
the positive numbers с and d (c < d). Prove that there is no distribution
whose convolution square equals P. (P.20) [T. F. Mori, G. J. Szekely]
1982
1. A map F : P(X) —> P(X), where P(X) denotes the set of all subsets
of X, is called a closure operation on X if for arbitrary А, В С X, the
following conditions hold:
(i) ACF(A);
(и) А с В => F(A) с F(B);
(iii) F(F(A)) = F(A).
The cardinal number min{|A| : А С X, F(A) = X} is called the density
of F and is denoted by d(F). A set Η С X is called discrete with respect
to F if и $. F(H — {u}) holds for all и е Н. Prove that if the density
of the closure operation F is a singular cardinal number, then for any
nonnegative integer n, there exists a set of size η that is discrete with
respect to F. Show that the statement is not true when the existence
of an infinite discrete subset is required, even if F is the closure
operation of a topological space satisfying the T\ separation axiom. (T.17)
[A. Hajnal]
2. Consider the lattice of all algebraically closed subfields of the complex
field С whose transcendency degree (over Q) is finite. Prove that this
lattice is not modular. (A.34) [L. Babai]
3. Let G(V,E) be a connected graph, and let dc{x,y) denote the length
of the shortest path joining χ and у in G. Let rc(x) = max{dc(x,y) :
у e V} for χ e V, and let r(G) = min{rc(^) · x £ V}. Show that if

1. PROBLEMS OF THE CONTESTS
33
r{G) > 2, then G contains a path of length 2r(G) — 2 as an induced
subgraph. (C.17) [V. T. Sos]
4. Let
№ = Σ ρ"·
P\n
ра<п<ра + 1
Prove that
log log η
hmsup f(n) =1.
n_»oo Π log Π
(N.15) [P. Erdos]
5. Find a perfect set Η с [0,1] of positive measure and a continuous
function / defined on [0,1] such that for any twice differentiable function
g defined on [0,1], the set {x G Η : f(x) = g(x)} is finite. (M.14)
[M. Laczkovich]
6. For every positive a, natural number n, and at most an points Xi,
construct a trigonometric polynomial Ρ (χ) of degree at most η for which
/»2π
P(xi) < 1, / P(x)dx = 0, and maxP(x) > en,
Jo
where the constant с depends only on a. (F.35) [G. Halasz]
7. Let V be a bounded, closed, convex set in Rn, and denote by r the
radius of its circumscribed sphere (that is, the radius of the smallest
sphere that contains V). Show that r is the only real number with the
following property: for any finite number of points in V, there exists a
point in V such that the arithmetic mean of its distances from the other
points is equal to r. (G.32) [Gy. Szekeres]
8. Show that for any natural number η and any real number
d > 3n/(3n — 1), one can find a covering of the unit square with η
homothetic triangles with area of the union less than d. (G.33)
9. Suppose that К is a compact Hausdorff space and К — U^_Q-An, where
An is metrizable and An с Am for η < т. Prove that К is metrizable.
(T.18) [Z. Balogh]
10. Let ρο,ρι,... be a probability distribution on the set of nonnegative
integers. Select a number according to this distribution and repeat the
selection independently until either a zero or an already selected number
is obtained. Write the selected numbers in a row in order of selection
without the last one. Below this line, write the numbers again in
increasing order. Let Ai denote the event that the number г has been selected
and that it is in the same place in both lines. Prove that the events
Ai (i = 1,2,...) are mutually independent, and P(Ai) = pit (P.21)
[T. F. Mori]

34
1. PROBLEMS OF THE CONTESTS
1983
1. Given η points in a line so that any distance occurs at most twice, show
that the number of distances occurring exactly once is at least [n/2].
(C.18) [V. T. Sos, L. Szekely]
2. Let / be an ideal of the ring R and / a nonidentity permutation of the
set {1,2,..., k} for some k. Suppose that for every 0 φ a G i2, αϊ φ Ο
and Ια φΟ hold; furthermore, for any elements Xi, x2, · · ·, %k £ I,
XlX2 •••Xk= X\fX2f ' · · Xkf
holds. Prove that R is commutative. (A.35) [R. Wiegandt]
3. Let / : R —> R be a twice differentiable, 27r-periodic even function. Prove
that if
holds for every x, then / is π/2-periodic. (F.36) [Z. Szabo, J. Terjeki]
4. For which cardinalities к do antimetric spaces of cardinality к exist?
(Χ, ρ) is called an antimetric space if X is a nonempty set, ρ : X2 —>
[0, oo) is a symmetric map, ρ(χ, у) = 0 holds iff χ = у, and for any
three-element subset {01,02,^3} οι Χ
ρ(αι/, α2/) + #(^2/, аз/) < gfai/, аз/)
holds for some permutation / of {1,2,3}. (K.8) [V. Totik]
5. Let # : R —> R be a continuous function such that χ + g(x) is strictly
monotone (increasing or decreasing), and let и : [0, oo) —> R be a
bounded and continuous function such that
u(t) +
/ 0(u(e))de
is constant on [l,oo). Prove that the limit \imt-^oo u(t) exists. (F.37)
[T. Krisztin]
Let Τ be a bounded linear operator on a Hubert space H, and assume
that ||Tn|| < 1 for some natural number n. Prove the existence of
an invertible linear operator A on Η such that ||ATA_1|| < 1. (0.5)
[E. Druszt]
Prove that if the function / : R2 —> [0,1] is continuous and its average
on every circle of radius 1 equals the function value at the center of the
circle, then / is constant. (F.38) [V. Totik]
Prove that any identity that holds for every finite η-distributive lattice
also holds for the lattice of all convex subsets of the (n — l)-dimensional
Euclidean space. (For convex subsets, the lattice operations are the set-
theoretic intersection and the convex hull of the set-theoretic union. We
call a lattice n-distributive if
η η
*Л(\/и)= \/(хА( \/ и))
2=0 7=0 0<г<п

1. PROBLEMS OF THE CONTESTS
35
holds for all elements of the lattice.) (A.36) [A. Huhn]
9. Prove that if Ε С R is a bounded set of positive Lebesgue measure, then
for every и < 1/2, a point χ = x(u) can be found so that
\(x-h,x + h)nE\ >uh
and
\(x-h,x + h)C)(R\E)\ >uh
for all sufficiently small positive values of h. (M.15) [K. I. Koljada]
10. Let R be a bounded domain of area t in the plane, and let С be its
center of gravity. Denoting by Tab the circle drawn with the diameter
AB, let К be a circle that contains each of the circles Tab (А, В е R).
Is it true in general that К contains the circle of area 2t centered at C?
(G.34) [J. Szucs]
11. Let Mn С Rn+1 be a complete, connected hypersurface embedded into
the Euclidean space. Show that Mn as a Riemannian manifold
decomposes to a nontrivial global metric direct product if and only if it is a real
cylinder, that is, Mn can be decomposed to a direct product of the form
Mn = Mk χ Rn~k (k < n) as well, where Mk is a hypersurface in some
(k + l)-dimensional subspace Ek+1 С Rn+1, Rn~k is the orthogonal
complement of Ek+1. (G.35) [Z. Szabo]
12. Let Χι,..., Xn be independent, identically distributed, nonnegative
random variables with a common continuous distribution function F.
Suppose in addition that the inverse of F, the quantile function Q, is also
continuous and Q(0) = 0. Let 0 = Xq:ti < Χ\:η < · · · < Χη·.η be the
ordered sample from the above random variables. Prove that if EX\ is
finite, then the random variable
Δ = sup
0<3/<l
- Τ (*+l-0№:n-*t-l:n)- / i1 ~ u)dQ(u)
tends to zero with probability one as η —> oo. (P.22) [S. Csorgo,
L. Horvath]
1984
1. Let к be an arbitrary cardinality. Show that there exists a tournament
TK = (VK,EK) such that for any coloring / : EK —> к of the edge set EK,
there are three different vertices χ0,χι,χ2 G VK such that
XQX\,X\X2,X2XQ G Ek
and
|{/(^o^i),/(xix2),/(^2a?o)}| < 2·
(A tournament is a directed graph such that for any vertices x,y G
VK, χ Φ у exactly one of the relations xy G Ек, у χ G EK holds.) (C.19)
[A. Hajnal]

36
1. PROBLEMS OF THE CONTESTS
2. Show that there exist a compact set К С R and a set А С R of type Fa
such that the set
{xeR:K + xcA}
is not Borel-measurable (here K + x = {y + x:ye К}). (М.16)
[M. Laczkovich]
3. Let a and b be positive integers such that when dividing them by any
prime p, the remainder of a is always less than or equal to the remainder
of b. Prove that a = b. (N.16) [P. Erdos, P. P. Palfy]
4. Let #i, #2,2/1,2/2,21,^2 be transcendental numbers. Suppose that any
3 of them are algebraically independent, and among the 15 four-tuples
only {яьz2,?/i, 2/2}, {^b^2,^i,z2}, and {2/1,2/2,^1,^2} are algebraically
dependent. Prove that there exists a transcendental number t that
depends algebraically on each of the pairs {^1,^2}, {2/1,2/2}, and {^1,^2}.
(A.37) [L. Lovasz]
5. Let αο, αϊ,... be nonnegative real numbers such that
00
]P an = 00 .
n=0
For arbitrary с > 0, let
ί k
l г=о
nj(c) = min < к : с · j < ]P аЛ , j = 1,2,... .
Prove that if Σ°10αϊ < oo, then there exists а с > 0 for which
Yl7Lianj(c) < °°, and if Σί^οαί = °°j ^en tnere exists а с > 0 for
which X^ anj(c) = 00. (S.24) [P. Erdos, I. Joo, L. Szekely]
6. For which Lebesgue-measurable subsets Ε of the real line does a positive
constant с exist for which
sup / eitxf(x)dx\ <c sup / einxf(x)da
—oo<t<oo\jE I n=0,±l,... \Je
for all integrable functions / on Ε? (Μ.17) [G. Halasz]
7. Let V be a finite-dimensional subspace of C[0,1] such that every nonzero
/ G V attains positive value at some point. Prove that there exists a
polynomial Ρ that is strictly positive on [0,1] and orthogonal to V, that
is, for every / G V,
rl
f(x)P(x)dx = 0.
/
./0
(F.39) [A. Pinkus, V. Totik]
8. Among all point lattices on the plane intersecting every closed convex
region of unit width, which one's fundamental parallelogram has the
largest area? (G.36) [L. Fejes-Toth]

1. PROBLEMS OF THE CONTESTS
37
9. Let Xq,Xi,... be independent, identically distributed, nondegenerate
random variables, and let 0 < a < 1 be a real number. Assume that the
series
ХУ**
k=0
is convergent with probability one. Prove that the distribution function
of the sum is continuous. (P.23) [T. F. Mori]
10. Let Χι, Χ2,... be independent random variables with the same
distribution:
P{Xi = l) = P{Xi = -l)=l- (i= 1,2,...)-
Define
So = 0, Sn = X1+X2 + ---+Xn (n=l,2,...),
ξ(χ, η) = I {к : 0 < к < η, Sk = x}\ (χ = 0, ±1, ±2,...),
and
а(п) — Ι {χ: ζ(χιη) = ι} Ι (η = ο, ι,...).
Prove that
P(liminf α(η) = 0) = 1
and that there is a number 0 < с < 00 such that P(limsup a(n) /log η =
с) = 1. (P.24) [P. Revesz]
1985
1. Some proper partitions P1?..., Pn of a finite set S (that is, partitions
containing at least two parts) are called independent if no matter how
we choose one class from each partition, the intersection of the chosen
classes is nonempty. Show that if the inequality
ψ<\Ρΐ\-\Ρη\ (*)
holds for some independent partitions, then Pi,..., Pn is maximal in
the sense that there is no partition Ρ such that P, P\,..., Pn are
independent. On the other hand, show that inequality (*) is not necessary
for this maximality. (C.20) [E. Gesztelyi]
2. Let S be a given finite set of hyperplanes in Rn, and let О be a point.
Show that there exists a compact set К СЖп containing О such that
the orthogonal projection of any point of К onto any hyperplane in S
is also in K. (G.37) [Gy. Pap]
3. Let к and К be concentric circles on the plane, and let к be contained
inside K. Assume that к is covered by a finite system of convex angular
domains with vertices on K. Prove that the sum of the angles of the

38
1. PROBLEMS OF THE CONTESTS
domains is not less than the angle under which к can be seen from a
point of K. (G.38) [Zs. Pales]
4. Call a subset S of the set {Ι,.,.,η} exceptional if any pair of distinct
elements of S are coprime. Consider an exceptional set with a maximal
sum of elements (among all exceptional sets for a fixed n). Prove that
if η is sufficiently large, then each element of S has at most two distinct
prime divisors. (N.17) [P. Erdos]
5. Let F(x, y) and G(x, y) be relatively prime homogeneous polynomials of
degree at least one having integer coefficients. Prove that there exists a
number с depending only on the degrees and the maximum of the
absolute values of the coefficients of F and G such that F(x, у) ф G(x, y) for
any integers χ and у that are relatively prime and satisfy max{|x|, |y|} >
с. (А.38) [К. Gyory]
6. Determine all finite groups G that have an automorphism / such that
Η % f(H) for all proper subgroups Η of G. (A.39) [B. Kovacs]
7. Let pi and p2 be positive real numbers. Prove that there exist functions
fi : R —> R such that the smallest positive period of fa is pi (i = 1,2),
and /i — /2 is also periodic. (A.40) [J. Riman]
8. Let 2/(\/5+l) < ρ < 1, and let the real sequence {an} have the following
property: for every sequence {en} of O's and ±l's for which Σ™=1 enpn =
0, we also have Σ™=1 enan = 0. Prove that there is a number с such
that an = cpn for all n. (S.25) [Z. Daroczy, I. Katai]
9. Let D = {z e C: \z\ < 1} and D = {w e C: \w\ = 1}. Prove that if for
a function / : D χ Β —> С the equality
,{az + b aw + b\ (h aw + b\
\bz + a bw + aj \a bw + a/
holds for all ζ G D, w e В and a, b G C, |a|2 = 1 + |6|2, then there is а
function L : ]0, oo[—> С satisfying
L(pq) = Цр) + L(q) for all p, q > 0
such that / can be represented as
/ 1 - \z\2 \
f(z,w)=L[- Цтг for all zeD,weB.
\\w — z\z J
(F.40) [Gy. Maksa]
10. Show that any two intervals А, В С R of positive lengths can be count-
ably disected into each other, that is, they can be written as countable
unions A = A\ U A% U... and В = B\ UB2 U... of pairwise disjoint sets,
where Ai and Bi are congruent for every г G Ν. (Μ.18) [Gy. Szabo]
11. Let ξ(Ε,π,Β) (π : Ε —> В) be a real vector bundle of finite rank, and
let
TE = V£(B Ηξ (*)
be the tangent bundle of E, where V£ = Ker άπ is the vertical subbundle
of te- Let us denote the projection operators corresponding to the

1. PROBLEMS OF THE CONTESTS
39
splitting (*) by υ and h. Construct a linear connection V on V£ such
that
Vx V Υ - Vy V X = υ[Χ, Υ] - v[hX, hY].
(X and Υ are vector fields on E, [.,.] is the Lie bracket, and all data are
of class C°°). (G.39) [J. Szilasi]
12. Let (П,Д, Ρ) be a probability space, and let {Xn,Tn) be an adapted
sequence in (Ω, Д, P) (that is, for the σ-algebras Tn, we have T\ С
^2 ί'·· ί A and for all n, Xn is an ^-measurable and integrable
random variable). Assume that
E{Xn+i\Fn) = \Xn + \Xn-i (n = 2,3...)·
Prove that supn B|Xn| < oo implies that Xn converges with probability
one as η —> oo. (P.25) [I. Fazekas]
1986
1. If (Д <) is a partially ordered set, its dimension, dim (Д <), is the least
cardinal к such that there exist к total orderings {<a: a < к} on A
with <= ΓΊα<κ <a. Show that if dim(A, <) > No, then there exist
disjoint Α), Αλ<ΖΑ with dim(A0, <), dim(Ab <) > Ν0· (Ν.9) [D. Kelly,
A. Hajnal, B. Weiss]
2. Show that if k < η/2 and Τ is a family οϊ k χ k submatrices of an η χ η
matrix such that any two intersect then
(C.21) [Gy. Katona]
3. (a) Prove that for every natural number /c, there are positive integers
a\ < a2 < · · · < ak such that a^ — ay divides a^ for all 1 < г, j < k,
(b) Show that there is an absolute constant С > 0 such that αχ > kCk for
every sequence αχ,..., α& of numbers that satisfy the above divisibility
condition. (N.18) [A. Balogh, I. Z. Ruzsa]
4. Determine all real numbers χ for which the following statement is true:
the field С of complex numbers contains a proper subfield F such that
adjoining χ to F we get С. (А.41) [Μ. Laczkovich]
5. Prove the existence of a constant с with the following property: for every
composite integer n, there exists a group whose order is divisible by η
and is less than nc, and that contains no element of order n. (A.42)
[P. P. Palfy]
6. Let U denote the set {/ G C[0,1] : |/(ж)| < 1 for all χ e [0,1]}. Prove
that there is no topology on C[0,1] that, together with the linear
structure of C[0,1], makes C[0,1] into a topological vector space in which the
set U is compact. (T.19) [V. Totik]

40
1. PROBLEMS OF THE CONTESTS
7. Prove that the series ]T cpf(px), where the summation is over all primes,
unconditionally converges in L2[0,1] for every 1-periodic function /
whose restriction to [0,1] is in L2[0,1] if and only if ^ \cp\ < oo.
(Unconditional convergence means convergence for all rearrangements.)
(F.41) [G. Halasz]
8. Let ao = 0, αχ,..., α^ and bo = 0, &i,..., bk be arbitrary real numbers,
(i) Show that for all sufficiently large η there exist polynomials pn of
degree at most η for which
р«(-1) = «ц, p«(l) = bi, » = 0, !,...,*, (1)
and
тах|рп(ж)| < —, (2)
|ж|<1 ηΔ
where the constant с depends only on the numbers a^, b{.
(ii) Prove that, in general, (2) cannot be replaced by the relation
lim n2 · max |pn(#)| — 0· (3)
n-»oo |яг|<1
(F.42) [J. Szabados]
9. Consider a latticelike packing of translates of a convex region K. Let
t be the area of the fundamental parallelogram of the lattice defining
the packing, and let tm{n(K) denote the minimal value of t taken for
all latticelike packings. Is there a natural number N such that for any
η > N and for any К different from a parallelogram, ntm[n(K) is smaller
than the area of any convex domain in which η translates of К can be
placed without overlapping ? (By a latticelike packing of К we mean a
set of nonoverlapping translates of К obtained from К by translations
with all vectors of a lattice.) (G.40) [G. and L. Fejes-Toth]
10. Let Χι, Χ2,... be independent, identically distributed random variables
such that Xi > 0 for all i. Let EXi = m, Vai(Xi) = σ2 < 00. Show
that, for all 0 < a < 1,
lim η Var
n—+00
Χλ + · · · + Xn
*λ α2σ2
) m2(1"«
(P.26) [Gy. Michaletzki]
1987
1. Let us color the integers 1,2,..., N with three colors so that each color
is given to more than N/4 integers. Show that the equation χ = y+z has
a solution in which x, y, ζ are of distinct colors. (C.22) [Gy. Szekeres]
2. A binary relation -< is called a quasi-order if it is reflexive and transitive.
The infimum of the quasi-order (Q, -<) is the greatest subset J С Q such
that
(i) for every В e Q there is an A G J with A -< B, and
(ii) A -< Я, А, В e J imply В -< A.

1. PROBLEMS OF THE CONTESTS
41
Let X be a finite, nonempty alphabet, let X* be the set of all finite words
from X, and let V be the set of infinite subsets of X*. For A,BeV, let
A -< В if every element of A is a (connected) subword of some element
of B. Show that (V, -<) has an infimum, and characterize its elements.
(N.10) [Gy. Pollak]
Let A be a finite simple groupoid such that every proper subgroupoid
of A has cardinality one, the number of one-element subgroupoids is at
least three, and the group of automorphisms of A has no fixed points.
Prove that in the variety generated by A, every finitely generated free
algebra is isomorphic to some direct power of Α. (Α.43) [A. Szendrei]
Let the finite projective geometry Ρ (that is, a finite, complemented,
modular lattice) be a sublattice of the finite modular lattice L. Prove
that Ρ can be embedded in a projective geometry Q, which is a cover-
preserving sublattice of L (that is, whenever an element of Q covers in
Q another element of Q, then it also covers that element in L). (A.44)
[E. Fried]
Let / and g be continuous real functions, and let g ψ 0 be of compact
support. Prove that there is a sequence of linear combinations of
translates of g that converges to / uniformly on compact subsets of R. (F.43)
[Sz. Gy. Revesz, V. Totik]
Is it true that if A and В are unitarily equivalent, self-adjoint operators
in the complex Hilbert space W, and A < B, then A+ < Β+Ί (Here A+
stands for the positive part of А.) (О.б) [L. Kerchy]
Let χ : [0, oo) —> R be a differentiable function satisfying the identity
x'(t) = —2x(t) sin21 + (2 — | cos t\ + cos t) / x(s) sin2 s ds
Jt-i
on [1, oo). Prove that χ is bounded on [0, oo) and that lim^oo x(t) = 0.
Does the conclusion remain true for functions satisfying the identity
t) f x(s)i
Jt-1
x'(t) =-2x(t)t + (2-\co8t\+coet) / x(s)sds?
(F.44) [L. Hatvani]
8. Let с > 0, с Φ 1 be a real number, and for χ e (0,1) let us define the
function
oo
f(x)=l[(l+cx2k).
k=0
Prove that the limit
Urn «
s->l-0 f(x)
does not exist. (F.45) [V. Totik]
9. Show that there exists a constant Ck such that for any finite subset V
of the Ar-dimensional unit sphere there is a connected graph G such that
the set of vertices of G coincides with V, the edges of G are straight line

42
1. PROBLEMS OF THE CONTESTS
segments, and the sum of the kth powers of the lengths of the edges is
less than ck. (G.41) [V. Totik]
10. Let F be a probability distribution function symmetric with respect to
the origin such that F(x) = 1 — x~lK{x) for χ > 5, where
K(x) = {
1 ifxe[5,cx))\U~=6(n!,4n!),
f, ifrr G (n!,2n!], n>5,
3 — 2^r ifa?e(2n!,4n!), η > 5.
Construct a subsequence {n^} of natural numbers such that if
Xi,X2,... are independent, identically distributed random variables
with distribution function F. then for all real numbers χ
lim Ρ I -Vl?· <πχ }
1 1
= - Η—arctanrr.
2 7Γ
(P.27) [S. Csorgo]
1988
1. Define a partial order on all functions / : R —> R by the relation / -< g if
/(ж) < #(:r) for all χ G R. Show that this partially ordered set contains a
totally ordered subset of size greater than 2H° but that the latter subset
cannot be well-ordered. (N.ll) [P. Komjath]
2. Suppose that a graph G is the union of three trees. Is it true that G can
be covered by two planar graphs? (C.23) [L. Pyber]
3. Let G be a finite Abelian group and x,y e G. Suppose that the factor
group of G with respect to the subgroup generated by χ and the factor
group of G with respect to the subgroup generated by у are isomorphic.
Prove that G has an automorphism that maps χ to y. (A.45) [E. Lukacs]
4. Let Φ be a family of real functions defined on a set X such that к о h G Φ
whenever fa G Φ (г G /) and h : X —> R7 is defined by the formula
h(x)i = fi(x), and
(1) k: h(X) —> R is continuous with respect to the topology inherited
from the product topology of R7. Show that / = sup{(?j : j G J,Qj G
Ф} = inf {hm : m G M, /im G Ф} implies / G Ф. Does this statement
remain true if (1) is replaced with the following condition?
(2) k: h(X) —> R is continuous on the closure of h(X) in the product
topology.
(T.20) [A. Csaszar]
5. Let us draw a circular disc of radius r around every integer point in the
plane different from the origin. Let Er be the union of these discs, and
denote by dr the length of the longest segment starting from the origin
and not intersecting Er. Show that
lim(dr ) = 0.
r-»(T Г

1. PROBLEMS OF THE CONTESTS
43
(G.42) [M. Laczkovich]
6. Let Η С R be a bounded, measurable set of positive Lebesgue measure.
Prove that
.. . eX((H + t)\H) n
hminf — 7-r-^—- > 0,
t-»o \t\
where H + t = {x + t:xe H} and λ is the Lebesgue measure. (M.19)
[M. Laczkovich]
7. Let S be the set of real numbers q such that there is exactly one 0-1
sequence {an} satisfying
oo
n=l
Prove that the cardinality of S is 2*°. (S.26) [P. Erdos, I. Joo]
8. Let / and g be holomorphic functions on the open unit disc D, and
suppose that |/|2 + \g\2 G Lipl. Prove that then f,g G Lip|.
A function h : D —> С is in the Lipa class if there is a constant К such
that
|Л(*)-Л(и;)| <K\z-w\a
for every z,w G D. (F.46) [L. Lempert]
9. We say that the point (αι,α2,α3) is above (below) the point (&ь&2,&з)
if αχ = bi, a2 = b2 and a3 > 63 (a3 < 63). Let ei, e2,..., e2k {k > 2) be
pairwise skew lines not parallel with the z-axis, and assume that among
their orthogonal projections to the (x, y)-plane no two are parallel and no
three are concurrent. Is it possible that going along any of the lines the
points that are below or above a point of some other line e^ alternately
follow one another? (G.43) [J. Pach]
10. Let a G C, \a\ < 1. Find all values of b G С for which there exist
probability measures with characteristic function φ satisfying ф(2) = а
and 0(1) = b. (P.28) [T. F. Mori]
1989
1. Let ρ be an arbitrary prime number. In the ring G of Gaussian integers,
consider the subrings
An = {pa+pnbi : a,b e Z}, n= 1,2,... .
Let R С G be a subring of G that contains An+\ as an ideal for some n.
Prove that this implies that one of the following statements must hold:
R = An+1; R = An;orleR. (A.46) [R. Wiegandt]
2. Let η > 2 be an integer, and let Ωη denote the semigroup of all mappings
g : {0, l}n —> {0, l}n. Consider the mappings / G Ωη, which have the
following property: there exist mappings <& : {0, l}2 —> {0,1} (г =
1,2,..., η) such that for all (0,1,0,2,..., an) G {0, l}n,
/(αϊ, α2,..., αη) = (#ι(αη, αχ), ^(^ι, &2), · · ·, 9η(<ΐη-ι> αη)) ·

44
1. PROBLEMS OF THE CONTESTS
Let Δη denote the subsemigroup of Ωη generated by these /'s. Prove
that Δη contains a subsemigroup Γη such that the complete
transformation semigroup of degree η is a homomorphic image of Γη. (A.47)
[P. Domosi]
3. Let πι < Π2 < ... be an infinite sequence of natural numbers such that
nk' tends to infinity monotone increasingly. Prove that ΣΤ=ι Vnfc
is irrational. Show that this statement is best possible in a sense by
giving, for every с > 0, an example of a sequence n\ < n^ < ... such
that п\/2к > с for all к but ΣΤ=ι l/nk is rational. (N.19) [P. Erdos]
4. Cancelled.
5. Characterize the sets А С R for which
A + B = {a + b:aeA, b e B}
is nowhere-dense whenever В с R is a nowhere dense set. (T.21)
[M. Laczkovich]
6. Find all functions / : R3 —> R that satisfy the parallelogram rule
f(x + y)+f(x-y) = 2/(x) +2/(y), x,ye R3,
and that are constant on the unit sphere of R3. (F.47) [Gy. Szabo]
7. Let К be a compact subset of the infinite-dimensional, real, normed
linear space (X, || · ||). Prove that К can be obtained as the set of
all left limit points at 1 of a continuous function g : [0,1[—> X, that
is, χ belongs to К if and only if there exists a sequence tn G [0,1[
(n = 1,2,...) satisfying Итп_оо tn = 1 and Итп_оо \\g{tn) - x\\ = 0.
(0.7) [B. Garay]
8. For any fixed positive integer n, find all infinitely differentiable
functions / : Rn —> R satisfying the following system of partial differential
equations:
£a?*/ = o, fc = i,2,....
(F.48) [L. Szekelyhidi]
9. Suppose that HTM is a direct complement to a vertical bundle VTM
over the total space of the tangent bundle Τ Μ of the manifold M.
Let ν and h denote the projections corresponding to the decomposition
TTM = VTM θ HTM. Construct a bundle involution Ρ : TTM ->
TTM such that Ρ oh = voP and prove that, for any pseudo-Riemannian
metric given on the bundle VTM, there exists a unique metric
connection V such that
VXPY - VYPX = Po h[X, Y]
if X and Υ are sections of the bundle HTM, and
VXY-VYX = [X,Y]

1. PROBLEMS OF THE CONTESTS
45
if X and Υ are sections of the bundle VTM. (G.44) [J. Szilasi]
10. Let Y(k), к = 1,2,... be an m-dimensional stationary Gauss-Markov
process with zero expectation, that is, suppose that
Y(k + 1) = A Y(k) + e{k + 1), к = 1,2,...
Let Hi denote the hypothesis A = A^ and let РДО) be the a priori
probability of Я^, г = 0,1,2. The a posteriori probability P\{k) =
Ρ(Ηι \Υ(ί),..., Y(k)) of hypothesis Hi is calculated using the
assumptions Pi(0) > 0, P2(0) > 0, Pi(0) + P2(0) = 1.
Characterize all matrices A0 such that P{limfc_>oo Pi(fc) = 1} = 1 if Щ
holds. (P.29) [I. Fazekas]
1990
1. Let A be a finite set of points in the Euclidean space of dimension d > 2.
For j = 1,2,..., d, let Bj denote the orthogonal projection of A onto
the (d — l)-dimensional subspace given by the equation Xj = 0. Prove
that
fti^i^i^r
\d-l
(G.45) [I. Z. Ruzsa]
Prove that for every positive number K, there are infinitely many
positive integers m and N such that there are at least KN/\ogN primes
among the integers m + 1, m + 4,..., m + Ν2. (Ν.20) [I. Z. Ruzsa]
Let n= pk (pa, prime number, к > 1), and let G be a transitive subgroup
of the symmetric group Sn. Prove that the order of the normalizer of G
in Sn is at most \G\k+\ (A.48) [L. Pyber]
Let Ρ be a polynomial with all real roots that satisfies the condition
P(0) > 0. Prove that if m is a positive odd integer, then
k=0
for all real numbers x, where / = P~m. (F.49) [J. Szabados]
We say that the real numbers χ and у can be connected by a ί-chain of
length к (where δ : R —> (0, oo) is a given function) if there exist real
numbers xq, xi, - - -, Xk such that xq = x, Xk = У, and
Ч-Ц
fXj-l +Xj\
< ^^pi г = 1,...Л
Prove that for every function δ : R —> (0, oo) there is an interval in which
any two elements can be connected by a ί-chain of length 4. Also, prove
that we cannot always find an interval in which any two elements could
be connected by a ί-chain of length 2. (F.50) [M. Laczkovich]

46
1. PROBLEMS OF THE CONTESTS
6. Find meromorphic functions φ and ψ in the unit disc such that, for any
function / regular in the unit disc, at least one of the functions f — φ
and / — ψ has a root. (F.51) [G. Halasz]
7. Denote by B[0,1] and C[0,1] the Banach space of all bounded functions
and all continuous functions, respectively, on the interval [0,1] with the
supremum norm. Is there a bounded linear operator
T:£[0,1]-+C[0,1]
such that Tf = f for all / G C[0,1]? (0.8) [G. Halasz]
8. Let A\ ,..., An be a sequence of η > 3 points in the Euclidean plane
R2. Define the sequence A^ , ...,Αη (i = 1,2,...) by induction as
follows: let Aj be the midpoint of the segment Aj M^ , where
A„+i = Ai . Show that, with the exception of a set of zero Lebesgue
measure, for every initial sequence (A\ ,..., An ') G (R2)n, there exists
a natural number N such that the points A\ ,..., An * are consecutive
vertices of a convex n-gon. (G.46) [B. Csikos]
9. Prove that if all subspaces of a Hausdorff space X are σ-compact, then
X is countable. (T.22) [I. Juhasz]
10. Let X and Υ be independent identically distributed, real-valued random
variables with finite expectation. Prove that
E\X+Y\ >E\X-Y\.
(P.30) [T. F. Mori]
1991
1. To divide a heritage, η brothers turn to an impartial judge (that is, if
not bribed, the judge decides correctly, so each brother receives (l/n)th
of the heritage). However, in order to make the decision more favorable
for himself, each brother wants to influence the judge by offering an
amount of money. The heritage of an individual brother will then be
described by a continuous function of η variables strictly monotone in
the following sense: it is a monotone increasing function of the amount
offered by him and a monotone decreasing function of the amount offered
by any of the remaining brothers. Prove that if the eldest brother does
not offer the judge too much, then the others can choose their bribes so
that the decision will be correct. (F.52) [V. Totik]
2. Suppose that η points are given on the unit circle so that the product of
the distances of any point of the circle from these points is not greater
than two. Prove that the points are the vertices of a regular n-gon.
(G.47) [L. I. Szabo]
3. Prove that if a finite group G is an extension of an Abelian group of
exponent 3 with an Abelian group of exponent 2, then G can be
embedded in some finite direct power of the symmetric group £3. (A.49)
[G. Czedli, B. Csakany]

1. PROBLEMS OF THE CONTESTS 47
4. Let η > 2 be an integer, and consider the groupoid G = (Zn U {oo}, o),
where
( x + 1 if χ = y e Zn,
: о у = <
{ oo otherwise.
(Zn denotes the ring of the integers modulo n.) Prove that G is the only
subdirectly irreducible algebra in the variety generated by G. (A.50)
[A. Szendrei]
5. Construct an infinite set Η С C[0,1] such that the linear hull of any
infinite subset of Η is dense in C[0,1]. (F.53) [V. Totik]
6. Let α > 0 be irrational.
(a) Prove that there exist real numbers 01,02,03,04 such that the
function / : R -> R,
f(x) = ex[a\ + θ2 sin χ + аз cos χ + 04 cos(arr)]
is positive for all sufficiently large #, and
liminf/(x) =0.
x—>+oo
(b) Is the above statement true if a2 = 0?
(F.54) [T. Krisztin]
7. Given an > an+i > 0 and a natural number μ, such that
hmsup < μ,
П A/in
prove that for all ε > 0 there exist natural numbers AT and no such that,
for all η > щ the following inequality holds:
η Νη
k=l k=l
(S.27) [L. Leindler]
8. Prove that if {a^} is a sequence of real numbers such that
00 00 / 2n-l \ X/2
^|afc|/A: = oo and $1 I $1 k(ak - ak+1)2 \ < 00,
fc=l n=l 4^=2"-! /
then
dx = 00.
/
Jo
\y^aksm(kx)\
\k=i I
(F.55) [F. Moricz]
9. Let h : [0,00) —> [0,00) be a measurable, locally integrable function, and
write
H(t):= f h(s)ds (t>0).
Vo

48 1. PROBLEMS OF THE CONTESTS
Prove that if there is a constant В with H(t) < Bt2 for all t, then
/»oo pi
/ е~я« / eH^dudt
Jo Jo
t
00.
(F.56) [L. Hatvani, V. Totik]
10. Consider the equation f'{x) = f(x + l). Prove that
(a) each solution / : [0, oo) —> (0, oo) has an exponential order of growth,
that is, there exist numbers a > 0, b > 0 satisfying \f(x)\ < aebx,
x>0;
(b) there are solutions / : [0, oo) —> (—oo, oo) of nonexponential order of
growth.
(F.57) [T. Krisztin]
11. Does there exist a bounded linear operator Τ on a Hubert space Η such
that
oo oo
ПТ"(Я) = {0} but ПГ(Я)^{0},
n=l n=l
where ~ denotes closure? (0.9) [L. Kerchy]
12. Let Χι, Χ2, · · · be independent, identically distributed random variables
such that, for some constant 0 < a < 1,
Ρΐχλ = 2k'a\ = 2~k, к = 1,2,...
Determine, by giving their characteristic functions or any other way, a
sequence of infinitely divisible, nondegenerate distribution'functions Gn
such that
?P. Π n^ <x\-Gn{x)
—оо<ж<оо I ^ '* ' )
(P.31) [S. Csorgo]
0 as η —► oo.

2. Results of the Contests
SUMMARY
Year of Number of Number of Number of
contest problems competitors correct solutions
1962
1963
1964
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
10
10
10
10
10
10
11
12
12
11
11
10
10
12
11
10
10
11
10
10
10
12
10
12
10
10
10
10
10
12
23
56
34
52
43
34
37
33
27
26
18
35
14
63
17
30
25
34
19
42
29
28
17
34
20
21
28
40
17
17
50
402
188
321
333
258
169
191
161
159
95
184
47
441
104
138
74
84
64
137
56
138
41
183
62
84
128
224
46
74
49

50
2. RESULTS OF THE CONTESTS
List of Prize Winners and Honorably Mentioned Competitors
(Notations: 1. First prize, 2. Second prize, 3. Third prize,
H. Honorably mentioned.)
1962
1. Gabor Halasz
2. -
3. -
H. Bela Bollobas, Arpad Elbert, Istvan Juhasz, Gyorgy Petruska,
Domokos Szasz
1963
1. Arpad Elbert
2. Gyula Katona, Domokos Szasz
3. -
H. Gabor Halasz, Janos Komlos, Miklos Simonovits, Jozsef Szucs
1964
1. Bela Bollobas, Peter Vamos
2. Gabor Halasz, Istvan Juhasz, Gerzson Kery, Miklos Simonovits,
Domokos Szasz
3. -
1965
1. Bela Bollobas, Jozsef Fritz, Miklos Simonovits
2. Gerzson Kery, Attila Mate, Jozsef Pelikan
3. -
H. Laszlo Gerencser, Elod Knuth, Laszlo Lovasz, Lajos Posa, Gyorgy
Vesztergombi
1966
1. Bela Bollobas, Endre Makai, Miklos Simonovits
2. Laszlo Lovasz, Lajos Posa
3. Laszlo Gerencser, Miklos Laczkovich, Jozsef Pelikan, Ferenc Szigeti
1967
1. Laszlo Lovasz, Attila Mate
2. Laszlo Gerencser, Eors Mate
3. -
H. Laszlo Babai, Robert Freud, Miklos Laczkovich, Miklos Simonovits,
Ferenc Szigeti
1968
1. Peter Gacs, Laszlo Lovasz, Endre Makai
2. Miklos Laczkovich, Lajos Posa
3. Laszlo Babai
1969
1. Peter Gacs, Laszlo Lovasz
2. Miklos Laczkovich
3. Laszlo Babai, Endre Makai, Jozsef Pelikan, Lajos Posa, Imre Z.
Ruzsa

2. RESULTS OF THE CONTESTS
51
1970
1. Laszlo Lovasz
2. Laszlo Babai, Miklos Laczkovich
3. Peter Gacs, Endre Makai, Jozsef Pelikan, Lajos Posa, Imre Z. Ruzsa
1971
1. Zsigmond Nagy, Laszlo Babai
2. Lajos Posa
3. Jozsef Pelikan, Imre Z. Ruzsa, Jeno Deak
1972
1. Imre Z. Ruzsa
2. Zsigmond Nagy
3. Laszlo Babai, Peter Frankl, Janos Pintz
1973
1. Laszlo Babai, Peter Komjath
2. Janos Pintz
3. Peter Frankl, Zsigmond Nagy, Imre Z. Ruzsa
H. Ervin Bajmoczi, Zoltan Balogh, Jozsef Beck, Ervin Gyori, Emil Kiss,
Tamas Mori, Zsolt Tuza, Endre Boros, Laszlo Lempert, Janos
Revizcky
1974
1. Laszlo Lempert, Imre Z. Ruzsa
2. Peter Frankl
3. Ervin Gyori
1975
1. Imre Z. Ruzsa
2. Peter Komjath, Laszlo Lempert, Vilmos Totik
3. Zoltan Ftiredi, Gabor Somorjai
H. Ervin Bajmoczi, Ervin Gyori, Tamas Mori, Peter Pal Palfy, Maria
Szendrei
1976
1. Imre Z. Ruzsa
2. Peter Komjath
3. -
H. Vilmos Totik, Ervin Bajmoczi, Zoltan Ftiredi, Mihaly Gereb, Ervin
Gyori, Emil Kiss, Janos Kollar
1977
1. Ferenc Gondocs, Vilmos Totik
2. Zoltan Ftiredi
3. Gabor Czedli, Peter Pal Palfy
H. Tamas Bara, Vilmos Komornik, Andras Sebo, Nandor Simanyi,
Sandor Veres
1978
1. Vilmos Totik
2. Zoltan Ftiredi, Janos Kollar, Nandor Simanyi
3. Emil Kiss
H. Tibor Krisztin, Zoltan Magyar

52
2. RESULTS OF THE CONTESTS
1979
1. Janos Kollar, Peter Pal Palfy
2. -
3. Mihaly Gereb, Gabor Ivanyos, Akos Seress
H. Emil Kiss, Tibor Krisztin, Zsolt Pales, Nandor Simanyi
1980
1. Janos Kollar
2. Akos Seress
3. Nandor Simanyi
H. Gabor Ivanyos, Zoltan Magyar
1981
1. Gabor Ivanyos, Zoltan Magyar
2. Balazs Csikos, Akos Seress
3. Peter Hajnal, Dezso Miklos, Mario Szegedy
H. Gabor Moussong, Zalan Bodo, Andras Zempleni
1982
1. Zoltan Magyar, Gabor Tardos
2. Akos Seress
3. Balazs Csikos, Andras Szenes
H. Peter Hajnal, Dezso Miklos, Mario Szegedy, Andras Zempleni
1983
1. Zoltan Magyar
2. Balazs Csikos, Gabor Tardos
3. Mario Szegedy, Jozsef Varga
H. Peter Hajnal, Zoltan Buczolich, Andras Szenes, Akos Seress
1984
1. Gabor Tardos
2. Mario Szegedy
3. Zoltan Buczolich
H. Balazs Csikos, Andras Szenes
1985
1. Gabor Tardos
2. Andras Szenes
3. Geza Bohus, Gabor Elek, Gyula Karolyi
H. Ferenc Beleznay, Laszlo Erdos, Miklos Mocsy, Tibor Odor, Endre
Szabo, Laszlo Szabo, Zoltan Szabo
1986
1. Gabor Tardos
2. Gabor Elek, Endre Szabo, Zoltan Szabo
3. Laszlo Erdos, Jeno Torocsik
Я. Gyula Karolyi, Akos Magyar
1987
1. Akos Magyar
2. Istvan Sigray
3. Gyula Karolyi
H. Laszlo Erdos, Gabor Hetyei, Sandor Kovacs, Jeno Torocsik

2. RESULTS OF THE CONTESTS
53
1988
1. Geza Kos, Zoltan Szabo
2. Laszlo Erdos
3. Istvan Sigray, Jeno Torocsik
H. Sandor Kovacs, Akos Magyar, Andras Benczur, Tamas Keleti,
Miklos Mocsy
1989
1/ Laszlo Erdos
2. Sandor Kovacs
3. -
H. Andras Benczur, Gyorgy Birkas, Andras Biro, Gabor Drasny, Geza
Kos, Akos Magyar, Laszlo Majoros, Miklos Mocsy, Tibor Szabo,
Zoltan Szabo
1990
1. Andras Benczur
2. Gabor Drasny
3. Andras Biro
H. Tamas Hausel, Geza Makay
1991
1. Andras Biro
2. -
3. Tamas Fleiner, Geza Kos
H. Matyas Domokos, Gabor Hajdii, Gergely Harcos, Tamas Keleti, Vu
Ha Van

3. Solutions to the Problems
3.1 ALGEBRA
Problem A.l. Determine the roots of unity in the field ofp-adic
numbers.
Solution. The p-adic number a_mp~m Η (- a_ip-1 + ao + a\p -\ +
Q>nVn + * * * (0 ^ ai < p) is а p-adic integer iff all the coefficients with
negative index are equal to 0, and it is a unit in the ring of p-adic integers
iff furthermore uq ^ 0. It is clear that every p-adic number a can be
written in the form a = /3pr, where /3 is a p-adic unit and r an integer.
Since the product of p-adic units is again a p-adic unit, we deduce that a
p-adic number can be a root of unity only in the case when it is a p-adic
unit.
Every root of unity can be written in the form ε = a + αρΓ, where r is
a positive integer, α is a p-adic unit, and 0 < a < p. Now if εη = 1, then
an = 1 (mod p) holds true. So the exponent of α is a divisor of n.
Consider the case ρ φ 2. Let exp(a) = k\ we are going to show к = n.
Let β = ek, and suppose β φ I. Then β is of the form /3=1+ 7ps, where
7 is a p-adic unit. If к φ η were the case, then some power of /3 would
be 1. To show the impossibility of this, it is enough to show that a power
with prime exponent of a number of the form /3 = 1 +7ps (7 a p-adic unit)
cannot be 1. Let q be a prime number; using the binomial theorem, we get
/39 = 1 + q^p3 + 6p2s = 1 + φρ3 if q ф р;
while in the case q = p, using ρ > 2 we get
βΡ = 1 + 7ps+1 + 6p2s+1 = 1 + φρ3+\
where 7 is a p-adic integer, whereas φ, as we can easily see, is a p-adic unit.
This guarantees that β4 is different from 1. We have proved that in the
field ofp-adic numbers every root of unity is necessarily a (p — l)th root of
unity. Now we show that such roots of unity indeed exist.
Let go be a primitive root of unity mod p. We are going to prove that it
is possible to determine numbers 0 ^ αι < ρ in such a way that the p-adic
number
£ = go + aip-\ h anpn Η
55

56
3. SOLUTIONS TO THE PROBLEMS
should be a (p — l)th primitive root of unity. It is clear that any power
of ε with exponent less than ρ — 1 is different from 1. So it is enough
to show that for a suitable choice of the numbers dj, the natural numbers
9j — <7ο + αιρΗ bujP7 satisfy g?~l = 1 (mod pJ+1). This will be proved
by induction. For j = 0 the statement is true, as go is a primitive root.
Suppose that the statement holds for some j, that is,
Let aJ+i be the solution of the congruence g0x = Cj (mod p). Then
#;ί ξ (»+%+ι^+1)ρ_1 = ι+с,у+1 - ft<w+1
ξ 1 + (Cj. - 9jaj+1 V+1 ξ 1 (mod ^+2).
Therefore ε and its powers are different (p — l)th roots of unity; their
number is ρ — 1. There can be no other roots of unity since the polynomial
xp-i _ ι can nave at most ρ — I roots in a commutative field.
Let us now deal with the case ρ = 2. Similarly to the case of odd prime
numbers, we can prove that a power with an odd exponent of a dyadic
number of the form 1 + · · · can be 1 only in case the number itself is 1.
So if a dyadic number is a primitive n-th root of unity then η can have no
odd prime divisors. There are two second roots of unity, 1 and the number
—1 = 1 + 2 +22 Η 2rH as well. There can be no more second roots
of unity, because the polynomial x2 — 1 can have at most two roots in a
commutative field.
We show that there are no other roots of unity in the field of dyadic
numbers. Suppose there were another root of unity; then this would be
necessarily a primitive root of unity belonging to some power of 2, so there
would surely exist a primitive further root of unity, that is, a dyadic number
η with η2 = — 1. Clearly, η must have the form η = 1 + 2r -\ , where
r > 1. This gives
-1 = (1 + 2r)2 = 1 + 2r+1 + 22r (mod 2r+2).
Therefore,
-2 ξ 2Γ+1 + 22r (mod 4), that is, 2r + 22r~1 = 1 (mod 2),
which is clearly impossible. This concludes the proof. D
Problem A.2. Let A and В be two Abelian groups, and define the
sum of two homomorphisms η and χ from A to В by
α(ν + Χ) — Щ + aX f°r a^ a e A.
With this addition, the set of homomorphisms from A to В forms an
Abelian group H. Suppose now that A is a p-group (p a prime number).

3.1 ALGEBRA
57
Prove that in this case Η becomes a topological group under the topology
defined by taking the subgroups pkH (k = 1,2,...) as a neighborhood base
of 0. Prove that Η is complete in this topology and that every connected
component of Η consists of a single element. When is Η compact in this
topology?
Solution. Я is clearly a commutative group whose 0 element is the ho-
momorphism mapping every element of A to 0 G Я, and for η G Я, —η is
the mapping defined by α(—η) = —αη.
To prove that Я is a topological group, we have to check the following:
(i) The intersection of the neighborhoods of 0 is 0 alone. Suppose namely
that η G pkH(k = 1, 2,...). As A is a p-group, the order of any α G A
is of the form pn for some n depending on a. Because η G pnH, we have
V — PnX f°r some χ G Я, and so
Щ = α{ρηχ) = {pna)X = Οχ = 0,
proving η = 0.
(ii) To any neighborhood U of 0 there is a neighborhood V of 0 such
that V + (—V) С С/, a suitable choice being V = U.
(iii) If a is contained in some neighborhood U of 0, then there exists a
neighborhood V of 0 such that a + V CU. Again the choice V = U will
do.
(iv) The intersection of any two neighborhoods of 0 contains a
neighborhood of 0, because for any two neighborhoods, one of them contains the
other.
We now come to the proof of completeness. We have to prove that if
Vi, Щ, · · · ? Ήη, · · · is a Cauchy sequence, then it converges to some element
of Я. By repeating members of the sequence, if necessary, we can assume
that 77ΐ+ι — Tfr G ргН, that is, ηί+ι = щ + ρτΰχ. Furthermore, define 770 as
0.
Now define mappings Xi(i = 0,1,2,...) in the following way:
If the order of a G A is pk then let
axi = a(ui +pi?i+i + · · · +/-1tfz+fc-i).
It is easy to check that χι is indeed a homomorphism from A to В and
Xi =ui+PXi+i-
We will show that the limit of the sequence 770,771,..., ryn,... is χο· For
this it is enough to show χο — Vi £ РгЯ, actually χ0 —η{ = ρχχ^ which we
prove by induction. For г = 0, we have χο — ^0 = Xo — 0 = P°Xo- Using
our previous observations and the induction hypothesis, we have
Xo-^г-ы =Xo~Vi-plui =plXi~plui =pl(Xi-ui) =pl+lXi+u
proving the completeness of Я.

58
3. SOLUTIONS TO THE PROBLEMS
We now prove that the connected component of 0 consists of 0 alone; to
prove this, it is enough to show that the intersection of all open-closed sets
containing 0 is 0. As pkH is an open subgroup, it is closed as well, and as
the intersection of all of them (for к = 1,2,...) is already 0 alone, we get
the desired result.
For compactness we are going to prove the following result: Η is compact
if and only if the index of pkH in Η is finite for every k.
Necessity of this condition is easy to establish: the cosets χ + pkH(x G
H) cover H. If Η is compact, then already a finite number of them have
to cover if, which is precisely what the condition says.
Suppose now that all the subgroups pkH have finite index. We are going
to show that if a family of closed subsets of Η has the property that any
finite number of them have a nonempty intersection, then the whole family
has a nonempty intersection.
First, we show the following: let U\(\ G Λ) be subsets of the set
Η such that the intersection of any finite number of them is nonempty.
Suppose that Η is the disjoint union of two subsets S and T. Then at
least one of the families S Π U\(X G Λ) and Τ Π U\(\ G Λ) inherits
the same intersection property. (In particular, either S or Τ meets all
the Uχ.) Suppose that neither family inherited the property. Then we
would have values Ai, Аг,..., Ar and Ar+i, Ar+2, · · ·, Ar+S such that the
intersection of the S Π U\. and also of the Τ Π U\r+j would be empty
(г = 1,2,..., r, j = 1,2,..., s). In other words, the sets
(.rU^)nS and (Д^+^ПГ
are empty, or equivalently
(nUXi)cT and (fiU^)QS.
Consequently,
(2*^) СГП5 = 0,
a contradiction. We then get the same type of statement for any
decomposition of Η into a finite number of pairwise disjoint subsets.
Let us now consider closed subsets U\(\ G Λ) of our topological group Η
satisfying the finite intersection condition. Repeatedly applying the above
procedure, we see that there exists a sequence щ, 772,..., щ»· · · of elements
of Η for which
4i +PH 2 r?2 +Р2Я 2 · · · 2 Vk +pkH D... (1)
and such that for any к the intersection of the U\(X G Л) with щ + ркН
also satisfies the finite intersection condition; in particular, any U\ and
ηΐι+ρ1*!! have a common element Xk,\-

3.1 ALGEBRA
59
In view of (1) and the completeness of H, the sequence 771,772,..., rfc,...
has a limit 77 for which щ + pkH = η +pkH. For any λ we have χ^Α G
η + pkH\ thus the limit of the sequence χι,Α, X2,\, · · ·, Xk,\, · · · 1S V· Since
U\ is closed and \k,\ G U\, we get that 77 is contained in every U\, which
completes the proof of the compactness of Η. Π
Problem A.3. Let R = Ri 0 R2 be the direct sum of the rings Ri and
i?2, and let iV2 oe ^ne annihilator ideal of R2 (in R2). Prove that R\ will
be an ideal in every ring R containing R as an ideal if and only if the only
homomorphism from R\ to N2 is the zero homomorphism.
Solution. First suppose there is a ring R containing R as an ideal and
such that Ri is not an ideal of R. Since R is an ideal in Д, we have R\f С R
and rR\ С R for every f G R. Suppose for example, that fR\ С Ri for
some f G R. (The case R\f С Ri can be treated similarly.) Using the
element f we will define a nonzero homomorphism from R\ to N2. The
direct sum property implies that for every r\ G R\ the element fv\ G R can
be uniquely decomposed in the form fv\ = g{r\) + A(ri)j where д(г\) G R\
and h{r\) eR2.f1 will be the desired homomorphism. It is clearly defined
for all elements of R\. Furthermore,
(i) h(ri) G N2 for every r\ G R\.
Namely, for every r2 G R2 we have
h{ri)r2 = g(ri)r2 + h{ri)r2 = (gin) + ft(ri))r2 = (m)r2 = f(rir2) = 0
and
r2h(ri) = r2(fri) = (r2f)ri G ARi С Д1?
but on the other hand, r2/i(ri) G i22? proving r2/i(ri) = 0, which gives (i).
(ii) h(ri +r[) = h(ri) + Mri) f°r everv ri? ri G i?i.
Since
r{ri + r^) = #(ri + r^) + /i(ri + r[) = fri + rri
= 0(ri) + Mn) + g(r[) + h{r[) = g(ri) + g(r[) + h(n) + h(r[),
where g(r{) +g(r[) G R\ and h(ri)-\-h(r[) G i22? we get the validity of (ii).
(iii) ft(rir£) = Mri)Mri) for every ri,r[ G i?i.
We will show that both sides of the equality are equal to zero. For
the right-hand side, this is clear because h(r\) is an annihilator of i?2 and
h(r[) G i?2 · For the left-hand side, we have to consider the product f(rir[):
f(riri) = (frijrj = (0(1*1) + ΜΓι))Γί = #(ηΚ G Дь
so ft(riri) = 0.
This establishes the first part of the statement.
For the reverse statement, we use the same idea. We assume that there
is a nonzero homomorphism h from R\ to iV2, and we construct the element
f in such a way that it should produce h.

60
3. SOLUTIONS TO THE PROBLEMS
We denote by Rq the zero ring over the additive group of the integers.
The element of Ro corresponding ton gZ will be denoted by η. The
additive group of the ring R is defined by R+ = Rq Θ Ri θ RJ, and for
the elements of R
r = η + r\ + Г2, rf = nf + r[+ r2,
we define the multiplication by
rrf = т\г'х + т^т'ч + nh(r[) + n'h{r\).
It is routine to check that R is a ring and R is a subring of it. In fact,
R is even an ideal of R in view of RR С R.
To verify that Ri is not an ideal of Д, take an η e R\ with h(ri) φ 0.
Then \t\ = h{r\) φ R\ as h(r\) G i?2, and this shows that R\ is not an
ideal of R. D
Problem A.4. Call a polynomial positive reducible if it can be written
as a product of two nonconstant polynomials with positive real coefficients.
Let f(x) be a polynomial with /(0) φ 0 such that f(xn) is positive reducible
for some natural number n. Prove that f(x) itself is positive reducible.
Solution. Consider the product representation of f(xn):
s
7(*n) = ГЫ*)' Μ
where the gj(x) are polynomials with positive coefficients. Suppose that
some of the polynomials Qj{x) contain a term с^хк with nonvanishing Cjk
such that η \ k. Then the product on the right-hand side of (1) contains
the term Cjk Yli^3-,9г(0)хк. The coefficient of this term does not vanish,
since Πί=ι 9i(P) = f(9) Φ 0· Since the polynomials gj(x) all have positive
coefficients, the right-hand side of (1) will contain the term xk with a
nonvanishing coefficient, which is a contradiction in view of η \ k.
This means that all the polynomials gj(x) have the form
h
9j(x) = Σ*>3ΐχίη (bji > 0,1 = 0,1,..., /,·),
2=0
that is, gj(x) is actually a polynomial of xn: gj(x) = gj{xn). So upon
substituting xn = y, we get
m = ΐ[-9Μ,
proving the statement of the problem. D

3.1 ALGEBRA
61
Problem A.5. Prove that the intersection of all maximal left ideals of
a ring is a (two-sided) ideal
Solution. Let us denote by В the intersection of all maximal left ideals of
the ring R (if there are no maximal left ideals, the statement is void). В
is clearly a left ideal, so we only have to prove that b e В and r e R imply
br G B, or equivalently br φ Β implies b φ Β. Since an element is not in
В if and only if there is a maximal left ideal not containing it, we actually
have to prove the following statement:
If for some elements b G B, r G R there is a maximal left ideal Μ with
br φ Μ, then there exists a maximal left ideal N with b φ N. Let
N = {x G R | XT G M}.
We are going to prove that N has the desired properties.
1. AT is a left ideal: if s G R and χ G iV, then xr G Μ implies sxr G Μ (Μ
being a left ideal), which in turn implies sx G N. Also χ G N, у G N
clearly imply χ — у G N.
2. b(£N asbr <£ M.
3. N is maximal. To prove this, we have to show that for any element
α of R with α ^ N the only left ideal of R containing both α and N
is R itself. We have ar £ Μ since a £ N. As Μ was maximal, every
element of R is contained in the left ideal generated by ar and M; in
particular, for every с G R, cr can be written in the form
cr = yar + nar + m,
where у G R, m G M, and η is an integer. This implies
(c — ya — na)r = m G M,
so
d = с — ya — na G AT,
but this shows that с = ya+na+d is contained in the left ideal generated
by a and N, which was to be proved. D
Remark. If R has an identity element, then В is of course the Jacobson
radical.
Problem A.6. Let R be a Gnite commutative ring. Prove that R has
a multiplicative identity element (1) if and only if the annihilator of R is 0
(that is, aR = 0, α G R imply α = 0).
Solution 1. If R has an identity element e, then aR = 0 (a G R) implies
0 = ae = a, so the annihilator of R is indeed 0.

62
3. SOLUTIONS TO THE PROBLEMS
Conversely, suppose R is a finite commutative ring whose annihilator is
0. This implies that to any element a of R different from 0 there is an
element b of R such that ab φ 0. If R = (0), then R surely has an identity
element. So suppose R φ (0) and ao is an arbitrary element of R different
from 0. Then the remark made above shows that for any natural number
η we can find an element an G R such that the relations
clocli φ 0, αοαια,2 φ 0,..., α0αι... an φ 0,...
hold true. Since R is finite, we have numbers m and η (0 ^ m < n) such
that
α0αι... аш = (α0αι... am)(am+i... an). (1)
This means that the set Ε of elements e φ 0 of the ring R for which
there exists a 0 φ d G R such that
de = d (2)
is nonempty. Choose an element e from Ε so that the number of elements
0 Φ d G R corresponding to e that satisfy (2) should be maximal. (Such
an e exists because R is finite.) We will show that e is an identity element
of Д.
Suppose, on the contrary, that for some a G R we have ae — α φ 0. Then
in view of (1) there are elements r,s G R such that
0 φ (ae — a)r = (ae — a)rs (3)
holds. (We allow r to be an empty product.)
Equation (3) implies
ar(e — es + s) = ar φ 0 (4)
consequently, 0 φ e — es + s e Ε. If de = d, then
d(e — es + s) = de — des + ds = d — ds + ds = d\ (5)
furthermore,
are — ar = (ae — a)r φ 0. (6)
Equations (4), (5) and (6) together show that the element e — es-\-s G Ε
satisfies condition (2) for more d G R than e. This contradiction shows the
existence of an identity element in R. D
Solution 2. Suppose the commutative ring R with n(> 2) elements has
annihilator 0. Then R cannot be nilpotent. If R were nilpotent, there
would exist an integer к ^ 2 such that Rk = 0, Rk~l φ 0 would be
satisfied, but this would imply that Rk~l is some nonzero annihilator of R,
a contradiction.

3.1 ALGEBRA
63
Next, we show that there is an element a of R such that no power of
a is equal to 0. Suppose that to every element a^ (г = 1,2,..., η) of R
there exists a natural number U with a> = 0. Let Ζ be the maximum of the
li (i = 1,2,..., n). If a product in R does not vanish, then every ai can
occur in it at most (I — 1) times. So every product with n(l — 1) +1 factors
vanishes; in other words, Дп(/_1)+1 = 0, contrary to the nonnilpotency
of Д.
So suppose a G R is such that all the elements a·7 (j = 1,2,...) are
nonzero. Since R is finite, these powers cannot be all different, say ak =
ak+l (I > 0). This implies ak = ak+l = ak+21 = ak+sl = ... . Choosing m
big enough to satisfy ml > k, we have
( „ml\2 „ml „ml „k-\-ml „ml—к „к „ml—к „ml
[a ) = a -a = a -a = a · a = α ,
so am/ is a nonzero idempotent of R.
Let us write am/ = e for simplicity. We can write R as a direct sum of
the ideals A and B:
R = A®B, (7)
where A is the set of all elements of the form re (r G R) and В is the set
of all elements of the form r — re (r G R). Clearly, A and В are ideals of
R. Any element r of R can be written in the form r = re + (r — re), thus
R = A + B. Furthermore, ea = a for all a in A, whereas eb = 0 for all b in
B. This implies А П В = 0, so (7) holds indeed. If В = 0 then R = A, so e
is the desired identity element of Д. li Β φ 0, then in view of 0 Φ e G A, R
is a direct sum of two rings, each of which contains fewer than η elements.
Consider this case and suppose by induction that all commutative rings
with fewer than η elements and with annihilator 0 have an identity element.
(The one-element ring clearly has an identity element.) If ζ G A is such that
ζ A = 0, then in view of AB = 0 we have zR = z(A + Β) = ζ A + zB = 0,
implying ζ = 0. Similarly, from w G B, wB = 0 we get w = 0. This
means that the annihilators of both A and В are 0, so by induction A has
an identity element e\ and В has an identity element e<i. But then e\ + e2
is an identity element of the ring R = Α Θ Β. Π
Remark. The statement of the problem is true for commutative Artinian
rings, too (see R. Baer, Inverses and zerodivisors, Bull. Amer. Math. Soc,
48 (Щ2), 630-638).
Problem A.7. Let I be an ideal of the ring of all polynomials with
integer coefficients such that
(a) the elements of I do not have a common divisor of degree greater
than 0, and
(b) I contains a polynomial with constant term 1.
Prove that I contains the polynomial 1 + χ + χ2 -\ (- xr~l for some
natural number r.
Solution. According to assumption (b), for a suitable f e Z[x] we have
l + x/6/.

64 3. SOLUTIONS TO THE PROBLEMS
Because of
(l + */)*r + *r+1(-/)=*r
for any r > 0, we have
xr G(l+x/,xr+1).
Repeatedly applying this observation, we get
(1 + xf,xr+1) D (l+xf,xr) D..-D (1 + xf,x) Э 1,
consequently,
1,ж,...,жг G(l+^/,^r+1),
which in turn means that we can find gr, hr G Ζ [χ] for which
1 + χ + · · · + xr = (1 + xf)gr + rrr+1/ir (1)
holds true.
We claim that gr and hr can be chosen in such a way that
deghr^degf (2)
holds. To prove this, observe that the polynomial gr figuring in (1) can be
written as
gr =xr+1q + p
where p, q G Z[x] and degp ^ r. From (1) we get
1 + χ + · · · + xr = (1 + xf)p + xr+1 [(1 + xf)q + Лг],
and choosing ρ (resp., (1 +#/)<? + /ir) as the new gr (resp., hr) we clearly
get an hr satisfying (2) because of degp ^ r.
Suppose s > r and
1 + χ + · · · + xs = (1 + xf)g3 + xs+1hs, (3)
where deghs ^ deg/. Subtracting from (3) xs~r times (1), we get
1+X+...+ xs-r~l = (1 + ж/)(0в - ^rxs-r) + xs+l(hs - hr).
This means that if there are indices s > r such that
hs — hr G /
holds, then
l + rr + .-'+rr*-7*-1 G/
is also true.

3.1 ALGEBRA
65
We prove that the ideal contains a nonzero constant polynomial. Take
a ρ G / such that p^O and degp is minimal. (Such a p exists in view of
1 + xf φ 0.) For any q G J let
q = pu + v,
where u, ν G Q[x] and degv < degp. Multiplying by a suitable nonzero
integer a, we get
aq = pui +vi,
where щ, v\ G Щх] and of course deg^i < degp still. This gives
v\ = aq — pu\ G /,
so by the choice of ρ and in view of deg^i < degp, we have v\ — 0.
This means that to an arbitrary q e I we can find а и G Q[x] for which
q = pu. Let ρ = φρ\, where p\ is a primitive polynomial and φ G Z. Then
q = ρι(φη), which in view of Gauss's lemma implies φη G Щх]. This in
turn implies that for any q G / we have p\ \ q. In view of condition (a),
this is only possible if p\ = 1; consequently φ G /, thus (φ) С /.
Finally, we show that there are values s > r such that
hs-hr G (y>).
Indeed, any coefficient of a polynomial can take on at most φ values
ταοάφ\ furthermore, deg/ir ^ deg/(r = 1,2,...), so there exist values
r < s(^ (^sZ+i -\-1) for which all coefficients of hs — hr are divisible by
φ. This proves the statement. D
Remark. The following, more general, statement can be proved:
Let R be a unique factorization domain whose proper factor rings are all
finite. Let / be an ideal of the polynomial ring R[x] satisfying the following
two conditions:
(a) The elements of / do not have a common divisor of degree greater
than 0.
(b) / contains a polynomial with constant term 1.
Then to every natural number N there exists a natural number r ^ N
for which N | r + 1 and
1 + χ + · · · + xr G J.
Problem A.8. Prove that in a Euclidean ring R the quotient and
remainder are always uniquely determined if and only if R is a polynomial
ring over some Reld and the value of the norm is a strictly monotone
function of the degree of the polynomial. (To be precise, there are two more
trivial cases: R can also be a Reld or the null ring.)

66
3. SOLUTIONS TO THE PROBLEMS
Solution. We shall work with the following definition of a Euclidean ring:
let R be a ring and N the set of nonnegative integers. R is called a Euclidean
ring if there exists a map φ: R —> N having the following properties:
i) φ[α) = 0 <=ϊ a = 0.
ii) For any a,b G R (b ф 0) there exist q,r e R such that
a = bq + r and φ(τ) < ip(b).
iii) The unicity of the quotient and remainder means that if a = bq + r =
bqi + r\ and φ(τ) < <^(6), φ(ν{) < <^(6), then r = r\ and q = q\.
With these definitions, we will prove that if R is a (commutative)
Euclidean ring whose quotient and remainder are unique, then R is the null
ring (which consists of the single element 0), a (commutative) field, or a
polynomial ring over a (commutative) field.
Note that although commutativity of R is usually included in the
definition of a Euclidean ring, we did not assume it, and in the following 16-step
proof the first 15 steps never use the commutativity of R. In the remarks,
following the proof we shall devote some space to the noncommutative case.
Before embarking on the proof, let us make a simplification. If the values
actually taken on by φ are 0 = no < n\ < ··· < Пк < ..·, then instead
of the value φ (a) = Пк take the value φ'(α) = k. φ' is equivalent to φ in
the sense that φ (a) < ^(6) if and only if φ'(a) < 4>'(b) and φ' also satisfies
conditions i), ii), and iii). So from now on we can assume rik = k.
And now to the proof.
1. R has no zerodivisors.
If ab = 0, а ф 0, b ф 0 held, then the decomposition 0 = α·0+0 = ab+0
would contradict iii).
2. If с ф 0, then φ(αο) > φ(α).
Suppose φ (ас) < φ (a), then ас = ac + 0 = α·0 + ас would contradict
ш).
Introduce the following notation:
Ti = {a: aeR, φ (a) < г} (г = 0,1,2,...).
Clearly,
T0 = {0}, T0 С Γι С · · · С Тк С ... and R = U Т{.
г=0
3. If R = Tq, then R is the null ring.
This is clear.
In the following, we assume RфTo.
4. R has an identity 1 and 1 G Т\.
Since R φ Tq, there is an α φ 0 with a G T\. If a = aq + r, we
have φ(ν) < φ(α) = 1, so r = 0, a = aq. Multiplying by some b and
using step 1, we get ab = aqb, b = qb, so q is a left-sided identity.
Proceeding in the same way with an arbitrary c: cb = cqb, с = cq, so
q is a two-sided identity; let us denote it by 1. If φ(α) < φ(1), then
α = 1·α + 0 = 1·0 + α, so by unicity a = 0; consequently <^(1) = 1.

3.1 ALGEBRA
67
5. φ(α — b) ^ πιαχ{φ(α), 4>(b)}.
Suppose max{<^(a),<£(6)} < φ(α — b). Then a = (a — b) · 1 + b =
(a — b) · 0 + a, together with 1^0, contradicts iii).
As a consequence, we have
6. Ti is an additive group.
7. φ{—α) = φ(α).
φ(—α) = φ(0 — α) ^ ΤΆ3χ{φ(ΰ),φ(α)} = φ(α). In a similar way, we
have φ(α) ^ φ{—α), so φ(α) = φ(—α).
8. If (f(b) < φ(α), then φ{α — b) = φ{α).
Step 5 gives (p(a—b) ^ φ(α). On the other hand, φ(α) = (f(b—(b—a)) ^
max{(^(6),(^(6 — a)} = max{(p(b),(p(a — 6)}, so because <^(6) < φ(α)
we have φ(α) ^ φ(α — 6), and thus φ(α — b)= φ{α).
9. An element a of R is a (two-sided) unit if and only if φ(α) = 1
(equivalently, a G Ti and а ф 0).
Let φ(α) = 1, and let b be an arbitrary element of R. We have b =
aq + r with <£>(r) < φ{α) = 1, so r = 0, 6 = aq. If, in particular,
6=1, then 1 = σαι, which also implies a\ = αχααι, 1 = αια, and thus
a\ = a"1. Conversely, if ab = 1, then step 1 gives а Φ 0, b ф 0 and
step 2 gives y?(a) ^ φ(α6) = φ(1) = 1, thus φ(α) = 1.
Steps 6 and 9 together imply
10. T\ is a (skew) field.
If R = T\, then we are again finished (and we see that actually every
skew field — not only the commutative ones — satisfies i), ii), and iii)
with φ(α) = 1, whenever а ф 0). Suppose now that R^Ti. We will
prove that R is the polynomial ring over T\.
11. If <p(b) = 1, then ip(ab) = φ(α).
Step 2 implies namely φ{α6) ^ φ{α) = φ((α6) · b~l) ^ φ{αβ), so
(p(ab) = φ(α).
The next statement is the converse of this.
12. If for some αφΰ ip(ab) = φ(α), then <^(6) = 1.
Let a = (ab)q+r, where (p(r) < ip(ab). Then r = a(l—bq). If <^(6) ф 1,
then 1 — bq φ 0, so in view of step 2, <^(r) ^ φ(α), so <^(α) < φ{α6), а
contradiction.
13. //у>(ж) = 2, then <p(xk) = k+ 1.
We prove the statement by induction on k. For к = 1 the statement is
clearly true. Suppose that φ(χ]ς~1) = к. Then, using steps 2 and 12,
we have <p(xk) = φ(χ]ς~1-χ) > φ(χ]ς~1), so φ(χ]ς) ^ fc+1. On the other
hand, let φ(α) = k+ 1 and α = xk~l -b + r where (j>{r) < φ(χ1ς~1) = к.
Then (р(хк~гЬ) = φ(α — г) = φ(α) = к + 1 > к = φ(χ1ς~1). Using
step 12, we get b & T\. So if b = же + s with <^(s) < φ(χ) = 2, then
(^(жс) = (p(b — s) = ip(b) because <^(6) ^ 2 > <^(s). Thus φ(χβ) ^ 2,
and consequently с ф 0. By substitution, we get α = xkc + xk~ls + r.
Here φ{α) = k+1; in case <^(s) = 1, we have ip(xk~ls) = φ(χ1ς~1) = к;
in case <£>(s) = 0, that is, s = 0, we have φ(χ1ς~1 · s) = 0, finally, we
know <^(r) < /c. All in all φ(α) > maxf^^"1^),^)}, so using step
8, ip(xk) ^ ip(xkc) = φ(α - xk~ls - r) = φ(α) = k+l.
14. // φ{α) = к + 1, then a can be written in the form a = ao + xot\ +

68
3. SOLUTIONS TO THE PROBLEMS
h xkOLk, where αζ G T\ (г = 0,1,..., к) and α& φ 0.
If к = 0, the statement is clear. Suppose it is true for к — 1. For
к > 0, let a = xhb + r, where φ(τ) < φ(χ]ς) = к + 1, so ip(xkb) =
φ{α — г) = φ(α) = к + 1 = (р(хк), thus <£>(&) = 1. This means b G Т\,
b ф0. Because <£>(r) ^ /c, we have r = xk~lotk-\ Η l· xot\ + c*o with
15. ж is transcendental over T\.
Suppose xkctk + · · · + #αι + ao = 0 (a^ G Ti). Then in view of
ip(xkctk) = φ^-1**]^-! + · · · + xol\ + ao) ^ /с, a^ = 0 follows. By
repetition of the argument, we get α^_ι = 0,..., a\ = 0, ao = 0.
16. If R is commutative, then R = Ti[x], and the Euclidean value of the
norm is a strictly monotone function of the degree of the polynomial.
This follows from steps 10, 14, and 15.
Finally, we remark that in a polynomial ring over a commutative field
quotient and remainder are indeed unique — the only possible quotient and
remainder are the ones that we get using the usual division algorithm. D
Remarks.
1. The noncommutative case is tricky. The point is that although as a set
R is the same as T\ [χ], R is not actually isomorphic to T\ [x] equipped
with the usual multiplication. If we define the product of two
polynomials in the usual way, that is, the product of two polynomials
η m
2=0 2=0
is defined as the polynomial
m+n
/ ; x cit
2=0
where сг is defined by the equation
Ci = aoh + aibi-ι -\ h аг6о
(with α_ι = a_2 = · · · = δ-ι = b-2 = · · · = 0), the statement of the
problem does not remain true. But by suitably defining the product
of two polynomials, the statement remains true. For arbitrary α G R
there exist ασ and aT which satisfy
ax = χασ + aT where φ(ατ) < φ(χ) = 2,
so aT GTi. Clearly,
(a + by = ασ +ba and (a + b)T = aT + bT.
Furthermore,
(ab)G = aaba and (ab)T = aTbT.

3.1 ALGEBRA
69
So a —> ασ is an endomorphism and σ : Τ\ —> Ti is an automorphism,
because for α G T\ (α φ 0) there exists an inverse of a, and since lx = xl
implies 1σ = 1 (so Tf ^ 0), we get 1 = (αα~ι)σ = ασ(α_1)σ, which
shows ασ e T\ and T\ being a field, each endomorphism of it is actually
an automorphism. If we now take a suitable automorphism a —> ασ
of T\ and a suitable derivation a —> ατ (with respect to σ; by this we
understand a map of T\ into itself, satisfying (a + b)T = ar + bT and
(ab)T = aTba + αδτ) then, defining in the polynomial ring the product
of two polynomials by ax = χασ + ατ, we can see that Τι[:τ]σ)Τ has the
required properties.
2. In the formulation of the problem the words "strictly monotone"
function of the degree are not superfluous. If, for instance, we have
polynomials / and g with deg / = deg g, but ip(f) > <p(g), say, then R is still
Euclidean, but with a suitable constant a we have deg (a/ + g) = deg/
and af + g = fq + r, where deg r < deg /, but also af + g = fa + g
where (p{g) < <£>(/), so in view of r φ g the remainder is unique. (We
have in mind a φ with φ(τ) < <£>(/), of course.)
Problem A.9. Let
f(x) = a0 + a\x + a2x2 + αιο^10 + anx11 + αι2ζ12 + αΐ3^13 (^13 Φ 0)
and
g{x) = b0 + bix + 62z2 + M3 + Ьцх11 + 6i2z12 + 6i3z13 (63 Φ 0)
be polynomials over the same field. Prove that the degree of their greatest
common divisor is at most 6.
Solution. Let
fi(x) = a0 + a\x + a2x2, /2(^) = aio + aii# + a>i2x2 + ai3^3
gi(x) =b0 + b\x + &2^2 + b3x3, g2(x) = 610 + Ьцх + &i2^2 + 6i3#3
(the problem says 6ю = О, but we will not use this). This implies
f{x) = /iW+AW, g(x) = 9i(x)+x1092(x),
so
f{x)92{x) ~ 9{x)f2{x) = fi(x)92(x) ~ f2{x)gi{x).
The right-hand side is divisible by the greatest common divisor of f{x)
and g(x), since the left-hand side is divisible by it. But the polynomial on
the right-hand side has degree 6, because fi(x)g2(x) has degree at most 5
but f2(x)gi(x) has degree precisely equal to 6 since a\s · 63 Φ 0. So the
greatest common divisor of f(x) and g(x) divides a polynomial of degree
6; therefore its degree can be at most 6. D

70
3. SOLUTIONS TO THE PROBLEMS
Remarks.
1. The upper bound of 6 for the degree of the greatest common divisor
cannot be improved as shown by the polynomials x10+x13 and x3+x12,
whose greatest common divisor is x3 + xQ.
2. It is possible to prove in a similar manner the following generalization:
if f(x) and g(x) are polynomials with coefficients in the same field,
grad / = n, grad g ^ n, and g{x) has a term of degree fc, but the
terms of degree fc + l,...,fc + r — 1 are missing from both polynomials
(k + r ^ n), then the greatest common divisor of / and g has degree at
most η — r.
Problem A. 10. Let К Ъе а, subset of a group G that is not a union of
left cosets of a proper subgroup. Prove that ifG is a torsion group or if К
is a unite set, then the subset
consists of the identity alone.
Solution. Suppose there is an element а е G, different from the identity
element such that α e Π k~lK. Then for every к G К we have к α G К,
кек
that is, Κα С К.
First we observe that if we have equality here, then К is the union of
some left cosets of the subgroup Η generated by a. Indeed, Κα = К
implies Каг = К for all integers г, so Κ Η = К, which in turn means
К = UkeKkH.
Now we only have to prove that in the cases mentioned in the problem
we indeed have Κα = К. This is clear if К is finite since Κα has the same
cardinality as К and is a subset of it. On the other hand, if α has finite
order n, then К D Κα implies
KDKaDKa2D..-D Kan - K,
again proving К a = K. D
Remark. If we assume that G is an Abelian group, then the assumption
that G is a torsion group can be replaced by the weaker assumption that
every element of К has finite order.
Problem A. 11. Prove that if an infinite, noncommutative group G
contains a proper normal subgroup with a commutative factor group, then
G also contains an infinite proper normal subgroup.
Solution. First we prove a lemma:

3.1 ALGEBRA
71
Lemma. If the infinite group G has a proper normal subgroup N
satisfying N <£ Z(G) (Z(G) denotes the centre of the group), then G has an
infinite proper normal subgroup.
Proof. To prove the lemma, we can assume that N is finite. For every
g e G, the mapping
φ9 : η ι—► g~1ng (η Ε Ν)
is clearly an automorphism of N. Furthermore, the mapping g \-> φ9 is
a homomorphism of G to some subgroup Φ of the full group of
automorphisms of TV, and Φ is clearly finite in view of the finiteness of N. Because
N Q. Z(G), Φ contains nonidentity automorphisms so the kernel F of the
homomorphism g \-> φ9 is a proper normal subgroup of G that is infinite
in view of G/F ~ Φ.
Now let if be a proper normal subgroup of the infinite, noncommutative
group G such that G/H is commutative. If Η <£ Z(G), then we can apply
the lemma. If, on the other hand, Η С Z(G) and h is any element of G
not contained in Z(G), then using the commutativity of G/H we see that
(H,h) is a normal subgroup of G. Clearly, (H,h) <£ Z{G)\ on the other
hand, (if, h) Φ G since G is noncommutative. Applying the lemma with
N — (H, h) completes the proof. D
Remark. The condition of noncommutativity cannot be dropped from
the statement of the problem, as shown by the group Zjpoo. Conversely, the
alternating group of countably infinite degree shows that the condition of
the existence of a proper normal subgroup with commutative factor group
cannot be dropped either.
Problem A. 12. Let αϊ, α2,..., αχ be positive real numbers whose sum
equals 1. For a natural number i, let щ denote the number ofak for which
21-' >ak> 2~l holds. Prove that
oo
^^2^<4+ν/ϊο^ΑΓ.
г = 1
Solution. We know that
oo N oo
Σ^<Σα^ = 1 and ]Cn* = iv>
г = 1 j=l г=1

72 3. SOLUTIONS TO THE PROBLEMS
so, by applying Cauchy's inequality, we get the following estimate:
oo ι [logJV] , oo ι
2=1 V 2 = 1 V 2 V
z=[logAT] + l
[logiV] ν 1/2 /[logAT] N i/2 , oo ч 1/2 / oo 1 χ
Σ') (Σ5) + (?*) (Σ^)
Λ ,2,
t=[logJV] + l i=[logJV] + l
^VlogiV+ViV—^ (Σ?) ^VlogiV + ^.
/|bgiVl+l\ \*=1 /
2^ 2 /
This proves a somewhat stronger form of the statement with y/2 in place
of 4. D
Remarks.
1. We can get the stronger estimate
± yf^V^v+o
log log AT
in a similar way if we start with the decomposition
oo ι Κ ι oo ι
Σ Hi — X^ Hi + V^ Hi
г=1 v г=1 V г=К+1 V
2 2 2s
where К = [log N + log log N].
£. If we use Holder's inequality instead of Cauchy's inequality, we get a
similar estimate:
ёфг^г+од.
where ρ > 1, l/p+ l/q = 1, and C(p) is a positive constant depending
only on p.
Problem A. 13. Consider the endomorphism ring of an Abelian torsion-
free (resp. torsion) group G. Prove that this ring is Neumann-regular if
and only ifG is a discrete direct sum of groups isomorphic to the additive
group of the rationals (resp., a discrete direct sum of cyclic groups of prime

3.1 ALGEBRA
73
order). (A ring R is called Neumann-regular if for every a G R there exists
а β e R such that αβα = α.)
Solution. In the following, "group" will always mean "commutative
group", where the operation is written as addition. Instead of Neumann-
regularity, we shall just write regularity.
First, we prove the necessity of the condition. If the group G is torsion-
free, call it Go, if it is a torsion group, then it can be written as a (discrete)
direct sum: G = G\ θ · · · θ Gi θ ..., where in the group Gi all elements
have order a power of pi (pi denotes the zth prime number).
Let ρ be a prime number, and let Φρ be the map of G defined by Φρ: a —>
pa (a e G). Φρ is clearly an endomorphism. Because of the regularity, there
is an endomorphism Ψρ with ΦρΨρΦρ = Φρ. Then for any a G G,
pa=p29p(a). (1)
It follows from (1) and the fact that Φρ is an endomorphism that if p2a = 0,
then pa = 0. Therefore every element {φ 0) of Gi (i > 0) has order pi.
In case G = Go, we can cancel (1) by p, which shows that every element a
of Go is divisible by every prime, thus Go is a divisible group. Furthermore
— as Go is torsion-free — the quotient is uniquely determined. Let po = 0,
and denote by Kq the field of rational numbers, whereas for г = 1,2,...,
denote by Ki the field with pi elements. From our previous observations
we see that Gi is a vector space over the field Ki (г = 0,1,2,...). Invoking
Zorn's lemma, we see that Gi has a basis that is just equivalent to the
direct sum decomposition formulated in the problem.
Sufficiency will follow if we prove the following stronger statement: The
endomorphism ring of the direct sum G = Go θ G\ θ · · · θ Gi Θ... is regular
if Gi is a vector space over Ki (г = 0,1,2,...). Here "Θ" can mean either
discrete or complete direct sum. First, we prove that the endomorphism
ring of each Gi is regular.
Let a be an endomorphism of Gi. Then α is a linear transformation of
Gi as a vector space. Since in a vector space every subspace is a direct
summand, there exist subspaces {/*, Vi such that
Gi = Ker a0 0i=VS0lma.
It is clear that a maps the elements of Ui onto Im α in a 1-to-l way. Let us
define β: Gi -^ Gi by the conditions /3(a) = 0 for a G Vi and /3(a) = b for
a G Im a, where b is the unique element of Ui with a(b) = α. β is clearly
an endomorphism of Gi with αβα = a. This proves the regularity of the
endomorphism rings of each of the Gi (г = 0,1,2,...).
Now let a be an endomorphism of G. Since for every prime p, if α G G
has order p, then a{a) again has order ρ or it is 0, a maps Gi to itself
if г ^ 1. Now let α G Go, and suppose a{a) G Gi (г ^ 1). We know
Pia(a) = 0. As Go is a vector space over Ко, there is an χ G Go with
PiX = a. So p2a(x) = 0, thus a(x) G G^, so Pi(x(x) = a(a) = 0. Applying

74
3. SOLUTIONS TO THE PROBLEMS
this argument not only to a itself, but to a and subsequent projections to Gi
(г > 1), we see that each component of a(a) belonging to some Gi (г > 1)
is zero, so a maps Go to itself.
These observations imply that every a uniquely determines a vector
(ao, ai, c*2, · ·.) where a^ is an endomorphism of G^, and conversely every
such vector uniquely determines an endomorphism a of G from which we
get back the original vector. We clearly have (αβ)ϊ = осф{. Now let α be an
endomorphism of G, (ao, ai, α2, ·. ·) the corresponding vector. We have
already proved the existence of endomorphism βι of Gi such that α^β^αι =
ai (i = 0,1,2,...). The endomorphism β belonging to (Аь/Зъ/?2, · · ·)
clearly satisfies αβα = α, so the endomorphism ring of G is regular. D
Problem A. 14. Let 21 = (A;...) be an arbitrary, countable algebraic
structure (that is, 21 can have an arbitrary number of unitary operations
and relations). Prove that 21 has as many as continuum automorphisms if
and only if for any unite subset A' of A there is an automorphism πΑ' of
21 different from the identity automorphism and such that
(χ)πΑ' = x
for every χ G A!.
Solution. Suppose first that the condition mentioned in the problem is
not satisfied. Let A! be a finite subset of A with the property that every
automorphism of 21 fixing A! pointwise is necessarily the identity. Then
for any two automorphisms πι and π2 that satisfy (χ)-κ\ = (#)π2 for all
χ G Α', πχπ^1 is the identity, so πι = π2. The number of automorphisms
of 21 is therefore less than or equal to the number of mappings of A' to A,
and so the number of automorphisms is countable.
Suppose now that the condition mentioned in the problem is satisfied.
We can assume that A = {1,2,...}. We define recursively an increasing
sequence of finite subsets of A: Ao С А\ С · · · С Ап С ... and a sequence
of automorphisms of 21: πι,..., πη,... . Let A0 = 0. Now let η > 0, and
suppose we have already defined An, which is finite, and π^ for all г < n.
Let πη be an automorphism of 21 that fixes An pointwise and is not the
identity, and let
An+1 = {1,..., n} U (J (Α^τή1 ... πελη U {min{fc | (*)πη φ к}} .
(ει,...,εη)
β<€{0,1}
Then Αη+ι is again finite, and by this procedure we have defined An
and πη for all n. The sequence so defined has the following properties:
α) πη is the identity on An but is different from the identity on An+\.
β) If Si G {0,1} (г = 1,..., η), then
(An)^1 ...πεηη αη+1.

3.1 ALGEBRA
75
7) \Л£=1Ап = A and all the An are finite.
Let (ει,..., εη,...) be an arbitrary infinite sequence with en £ {0,1}
(n = 1,2,...). Let us define the product Jl^Li πη — π m the following
way. If к е A, let
(k)n = (k^...nZk, (1)
where rik denotes the smallest number for which we have к £ АПк. Because
of 7), π is indeed a map of A into itself. For any η > η&, we have
(k)n = (ftK1 .·.<"= ((ft)*?1 ... π£* )π^+Υ ...<»= (*K1 ... π£*
(2)
because in view of /3) we have (k)^1.. .7rn^fc £ ^4nfc+i and in view of a)
7rnfc+i,... ,πη are all equal to the identity map on Ank+i. Taking into
consideration that 21 contains only finitary operations and relations, we see
that in view of (2) π preserves operations and relations. For every Ζ G A
there exists an η with Ζ € An+i. Thus, there exists а к with (k)^1 ... π^η =
I. Because of a) we have for any m > η (k)^1 ... π^1 = I so in view of (2)
we have (k)n = I. Thus π is an automorphism. We are going to show that
ίοΓ(ει,...,εη,...)^(ε,1,...,ε^,...) π = Π~=ι<η Φ *' = Ш°=1 ^ is
satisfied.
Let n be the smallest number for which εη φ ε'η. We may assume
εη = 0,ε'η = 1. Then π^1 .. . π^_Υ = π^1 .. . π^γ. Let Ζ G Αη+ι be an
element with (Ζ)πη φ I. Let /с be the number for which (k)^1 ... π^ΐγ = /.
In view of a) and (2) we have (k)n = (k)^1 ... π£η = (ft)*"?1 ... π^_Υ = /,
so {к)-к' = (1)πη φ (k)n = I. This means that the number of different
automorphisms of 21 is at least as much as the number of infinite sequences
of 0's and l's. Taking into consideration the fact that A is a countable
set, we see that the cardinality of the set of automorphisms of 21 is indeed
continuum. D
Remark. The following example shows that the statement of the
problem is no longer true if we allow relations with infinitely many variables.
Let 21 = {A,R), where A is the set of natural numbers and the relation
R(ai,..., αη,...) with countably many variables is true if and only if we
have an = η with finitely many exceptions. Then for any automorphism
π of 21, (η)π = η holds with finitely many exceptions. Thus 21 has only
countably many automorphisms although the condition for finite subsets
formulated in the problem is clearly satisfied by this 21.
Problem A. 15. Let G be an infinite group generated by nilpotent
normal subgroups. Prove that every maximal Abelian normal subgroup
ofG is infinite. (We call an Abelian normal subgroup maximal if it is not
contained in another Abelian normal subgroup.)
Solution. Let the group G be generated by nilpotent normal subgroups,
and suppose it has a maximal Abelian subgroup A that is finite.

76
3. SOLUTIONS TO THE PROBLEMS
The centralizer С of A in G is a normal subgroup of G that has finite
index in G. Indeed, if we map every element g of G to the mapping χ ι—►
g~lxg of A, then this is a homomorphism from G to the automorphism
group of A and the kernel of this homomorphism is C. Therefore G/C is
finite.
Suppose В is an Abelian normal subgroup in G. Then we have \B\ ^ n,
where η = \A\ · \G : C\. Indeed, we have С Π Β = Α Π Β, because otherwise
the Abelian normal subgroup A(C Π Β) of G would strictly contain A.
Therefore \B : Α Π Β\ ζ \G : C\ because В/А С\В = В/С Γι Β ~ CB/C ζ
G/C, which gives |B| = \A П B\ · \B : Α Π Β\ ζ \A\ · \G : C|.
Let if be a nilpotent normal subgroup of G. We claim that the nilpo-
tency class of Η is at most 2n — 2. Consider namely the lower central series
ΟΪΗ:
H = H1>H2>->Hk> Hk+1 = 1.
It is well known that for the members of the lower central series, [Hi, Hj] ^
Hi+j holds, so as soon as r > ft/2 we have [Hr,Hr] ^ Щг = Нк+\ = 1,
that is, the characteristic subgroup Hr of Η is Abelian. By what we said
previously, this implies \Hr\ ^ n. This clearly implies that there can be
at most η Hr such that r falls in the range к + 1 ^ r > fe/2, that is,
к + 1 ^ 2n - 1 holds.
The group G is nilpotent. Indeed, for any elements g\, g2,..., g<in-\ £ G,
[•••[[5,ι,5,2],5,3],···,5,2η-ι] = 1
is true, because g\, g2,..., #2n-i belong to some subgroup of G whose nilpo-
tency class is at most 2n — 2.
It is true that С = A. If this were not true, then the nilpotent group G/A
would contain the nonidentity normal subgroup С/A. By a well-known
theorem, this implies that the center of G/A has a nonidentity intersection
with С/A; in other words, there is an element g G С \ A for which Ag
is contained in the center of G/A thus the subgroup (A,g) would be an
Abelian normal subgroup properly containing A.
All this would imply that G is a finite group. Indeed, the subgroup С
being identical with A is finite and we know that it has finite index in G,
so \G\ = \C\ · \G : C\ is finite. D
Problem A. 16. Let ρ > 7 be a prime number, ζ a primitive pth root of
unity, с a rational number. Prove that in the additive group generated by
the numbers 1, ζ, ζ2, ζ3 + ζ~3 there are only Rnitely many elements whose
norm is equal to с (The norm is in the pth cyclotomic Reld.)
Solution. Let
L<*>(*) = Xl + ckx2 + c2kx3 + (c3fc + C3/>4 (ft = l,... ,p - l).
It is enough to show that the inequality
iL^Or)...^-1^)! <|c| (1)

3.1 ALGEBRA
77
has only a finite number of rational integer solutions x\, x2, £3, £4.
For any such solutions
p-l p-l
2p~1\c\ > 2ρ~λ Yl I L(k)(x) |> 2р-х Π I Im L(fc)(s) |
p-l
= niKfc-rfc)^ + (c2fc-r2fc)^i
fc=l
p—1 p—1
= J](c* _ r*) 1 Д(Я2 + (c* + г*)жз) 1
fc=l fc=l
= p\F(x2,x3)\.
But for any A: with ρ \ к we know that £fc + £_fc is an algebraic number
of degree (p — l)/2 ^ 3, so by applying Thue's theorem we see that the
homogeneous polynomial F(x2,xs) with rational integer coefficients takes
on rational integer values only at finitely many places, so there are only a
finite number of possible values for x2 and x%.
Therefore, it is enough to show that for any fixed x2 and rr3, the number
of possible values of x\ and X4 is also finite. For x2 = x3 = 0, the proof
of this statement is analogous to the previous proof. If one of x2 and
xs is different from 0, then the absolute value of each of the L^k\x) is
bounded from below in view of |l/fc)(:r)| ^ |Im l№{x)\ by a positive bound
independent of x\ and #4, and as the product of them is bounded from
above in view of (1), their absolute value is also bounded from above by a
bound independent of x\ and X4. But then the system of equations
xi + (C3 + C3)*4 = L(1)(*) " (C*2 + С2*з),
xi + (С6 + C6)*4 = L^\x) - (ζ2χ2 + С4*з)
guarantees that \x\\ and \χή\ are bounded from above, so the number of
possible pairs χι, Χ4 is indeed finite. D
Problem A. 17. We have 2n + 1 elements in the commutative ring R:
α,αι,...,αη,0ι,...,#η.
Let us define the elements
η
2=1
Prove that the ideal (σο, σι,..., σ&,...) can be finitely generated.
Solution. First, we are going to show that we can assume that R is a ring
with identity. There is a standard way to embed an arbitrary ring R as

78
3. SOLUTIONS TO THE PROBLEMS
an ideal into a ring with identity R+: as a set R+ consists of the ordered
pairs (a, n) with a £ R and η an integer, and the operations are defined in
the following way:
(a, n) + (6, m) = (a + b,n + ra), (a, n)(6, ra) = (ab + ma + n&, nra).
R can be identified with the ideal of R+ formed by the elements of the form
(α,Ο), and a subset of R generates the same ideal in R and R+. To verify
this statement, take a subset Η of R and denote the ideals generated by
Η in R and R+ by / and J, respectively. Clearly, / С R Π J С J. On the
other hand, / consists of elements of the form (a, 0), and multiplying these
elements by elements of the form (6, ra), we again get elements οι I so I is
an ideal of i?+, too. This proves J С I.
This shows that the ideal of R mentioned in the text of the problem is
an ideal of R+, too, and if it is finitely generated as an ideal of R+, then
the same elements clearly generate it as an ideal of i2, too. Therefore, from
now on we can safely assume that R is a ring with identity.
We are going to prove the following generalization of the problem:
Theorem. Suppose we have polynomials fi(x) (i = 0,..., n) with
coefficients in the ring (with identity) R. Using these polynomials and fixed
elements ρο,..., Qn of R, we form the elements
η
** = 5>(*)е? (fe = o,i,2,...).
г=0
Then the ideal (σο, σι,..., σ&,...) is equal to the ideal (σο,..., σθ) where
* = E(deg/* + l).
We get the original problem as the special case where the polynomials
axe f0{x) = x, fi{x) = а» (г = 1,... ,n), and ρ0 = 1.
Proof. Let ri be the degree of fc (x). If
η s—1
F(x) = l[(x - Qi)ri+1 =x°- YfiiX>
г=0 j=0
and D is the operator of taking the derivative with respect to χ and then
multiplying by x, then Drn(xkF(x)) vanishes at the place Qi if m ^ r».
(We can see this if we take into consideration the fact that a root with
multiplicity t of a polynomial Ρ (χ) is also a root with multiplicity at least
(t - 1) of the polynomial D(P(x)).) As Вш(хк) = кшхк, we have
s-l
Drn(xkF{x)) = {s + к)шх3+к - 5j3j(j + к)шх>+к,
3=0
and consequently
s-l

3.1 ALGEBRA
79
Let
Ji\%) — / j1fm,ix
m=0
Then
as+k = JTfii* + b)Qt+k = EE^(S + №'+"
i=0 i=Om=0
i=0m=0j=0 j=0 i=Om=0
= Σ>ί>^'+*W+* = ΣΜ+* ·
j=0 г=0 j=0
This means that σθ+*. is contained in the ideal generated by at with
smaller indices, and this proves the theorem. D
Problem A. 18. Let G and Η be countable Abelian p-groups (p an
arbitrary prime). Suppose that for every positive integer n,
pnG^pn+1G.
Prove that Η is a homomorphic image of G.
Solution. We say that an element д £ G (д ф 0) is of infinite height if
for every natural number η there is an χ G G that satisfies pnx = g. The
elements of G that are of infinite height together with 0 form a subgroup A
of G. Let G* = G/A be the factor group of G with respect to A. Clearly, G*
is a finite or countably infinite Abelian p-group. G* contains no elements
of infinite height. To see this, pick a g* G G* such that pnx* = g* has
a solution x* £ G* for every natural number n. Take an element g £ G
which is mapped to g* by the natural homomorphism G -+ G* and an
χ e G which is mapped to x* (for some fixed n). We see that pnx — g is
mapped to 0, that is, pn — χ e A, so pnx — g = pny for some у е G, which
gives g = pn(x — y), and this in turn implies g £ A, that is, g* = 0.
Now a well-known theorem of Prufer gives that G* is a direct sum of
cyclic groups (in the case where G* is finite, this follows more easily from
the fundamental theorem of finite Abelian groups).
Now we show that for every natural number n,
pnG* ^pn+1G*.
We know that pnG φ pn+1G, so for some h £ G the equation pnh = ρη+ιχ
has no solutions in G. We want to show that pnh* = pn+ly* has no
solutions y* £ G* (h* denotes the image of h in G* with the natural
homomorphism). Otherwise, for a suitable у £ G, pnh — pn+1y would be

80
3. SOLUTIONS TO THE PROBLEMS
mapped to 0 G G*, so pnh — pn+ly = pn+lz would hold for some ζ G G,
giving the contradiction pnh = pn+1{y-\-z). (This, incidentally, shows that
G* cannot be finite.)
Suppose the direct sum decomposition of G* is
oo
i=l
where Ci is a cyclic group of order pki. Then the set {k\, fo,... } cannot be
bounded. This follows from the fact that if, for some n, all Ci were of order
at most pn, then pnG* = pn+1G* = 0 would hold, contrary to our previous
observation. Suppose с* is an element generating Ci. Then every element
g* £ G* can be written uniquely in the form g* = Σ^ι α%°%^ where α* is
an integer taken modulo pki and, with finitely many exceptions, all the щ
are equal to 0.
Now take a finite or countably infinite Abelian p-group H, the elements
of which are 0, hi, /12, · ·. · Choose a series cix ,ci2,... in such a way that the
order of cia is greater than the order of hi. Then we have a homomorphism
of Ci0 onto the cyclic subgroup generated by hj. We map all the other
С ι to 0. Since G* is a direct sum of these cyclic subgroups, there is a
homomorphism of G* to Η extending the above mentioned homomorphisms
of the cyclic subgroups. As every hj is the image of some element of
Cij, this extended homomorphism is clearly onto. The composition of the
natural homomorphism G —> G* and our homomorphism is the required
homomorphism of G onto Η. Π
Problem A.19. Let G be an infinite compact topological group with
a Hausdorff topology. Prove that G contains an element д φ 1 such that
the set of all powers of д is either everywhere dense in G or nowhere dense
inG.
Solution. Suppose G is an infinite compact group with the property that
the closure of every cyclic subgroup (φ 1) has an inner point, that is, every
closed subgroup (φ 1) of G is open. We are going to prove that G has a
dense cyclic subgroup.
First, let G be commutative. The closure of a cyclic subgroup Α φ 1 is
open so it has a finite index η in G; therefore, the continuous isomorphism
φ : χ —> xn (χ € G) maps G to this subgroup. Gn Φ 1 is a compact, thus
closed, thus open subgroup of G, therefore Gn Π A is dense in Gn. So the
cyclic subgroup <^-1(Gn Π A) is dense in G.
Now we show that the index of every open subgroup of G is a power
of the same prime number. If ρ is a prime divisor of n, then because
Gp ф G the open subgroups Gv% (i = 0,1,2,...) form a strictly decreasing
chain, so their intersection is a closed subgroup of infinite index, and so
this intersection is the identity. The topology of G that we get by choosing
the subgroups Gp as a base for neighborhoods of 1 is coarser than the

3.1 ALGEBRA
81
original compact topology of G, so it is equal to it. Consequently, every
open subgroup of G has index a power of p, since it contains a subgroup
some Gpt.
In the following, we do not suppose that G is commutative. The center
Ζ of G is an open subgroup. Indeed, the closures of the cyclic subgroups
(Φ 1) of G are open subgroups and cover G, so finitely many of them cover
G already, the intersection of these is a subset of Z, and therefore Ζ is an
open subgroup.
The group G/Z is a p-group. Indeed, if every open subgroup of Ζ has
index a power of p, and Η ^ G is a subgroup that contains Ζ as a subgroup
of index q (q is prime, therefore Η is commutative), then all elements {φ 1)
2
of the factor group Hq/Hq (φ 1) formed using open subgroups of Ζ have
order q\ thus ρ = q and consequently the finite group G/Z is a p-group.
Finally, we prove that G is actually commutative. Let V be a subgroup
with Ζ ^ V ^ G such that the center of G/Z is V/Z. Then for every
g £ G, taking the endomorphism of V defined by χ —> [x,g] (χ (Ξ V), the
image of V will be finite (as the kernel of the endomorphism contains a
subgroup Ζ that itself has finite index) so this image is the identity, thus
V = Ζ and thus G = Z, because G/Z is a finite p-group. D
Remarks.
1. Paragraphs 3, 5 and 6 of the elementary solution described above can
be omitted if we use a theorem of Baer which states that if the factor
group with respect to the center of a (discrete) group G is finite, then
the commutator subgroup of G is also finite.
2. Jozsef Pelikan noticed the following. If we use the fact that every infinite
compact group has a proper closed subgroup, then the statement of the
problem can be reformulated in the following way:
Statement If G is an infinite, compact, topological group, every closed
subgroup (φ 1) of which is open, then G is topologically isomorphic to the
topological group of the p-adic integers (p prime).
Problem A.20. Let G be a solvable torsion group in which every
Abelian subgroup is finitely generated. Prove that G is finite.
Solution. A finitely generated Abelian torsion group is finite. Therefore,
the problem is equivalent to the following one: If in a solvable torsion group
every Abelian subgroup is finite, then G itself is finite. We are going to
prove the following stronger statement: If in a solvable torsion group every
Abelian normal subgroup is finite, then G itself is finite.
First we prove a lemma: If a group G has a commutative normal
subgroup with a commutative factor group, and furthermore every Abelian
normal subgroup of G is finite, then G itself is finite.
Let Η be such a normal subgroup. Using Zorn's lemma, we see that
the partially ordered set of all Abelian normal subgroups of G containing

82
3. SOLUTIONS TO THE PROBLEMS
Η has a maximal element F. If an element g is permutable with every
element of F, then F and g generate a commutative normal subgroup (it is
normal because G/H is commutative), so by the maximality of F we have
g € F, that is, F is its own centralizer. Consequently, G/F is isomorphic
to a subgroup of Aut (F) that is itself finite in view of the finiteness of F.
Thus G is also finite.
Now we prove the statement using induction on the length of the derived
series of the group G. If this length is 1, there is nothing to prove. If
the length is 2, the statement follows from the previous lemma. In the
general case, consider P, the last term of the derived series, different from
the identity. Ρ is a commutative normal subgroup (hence finite) and the
length of the derived series of G/P is one less than the length of G. We
only have to check that every commutative normal subgroup of G/P is
finite. But if N/P is a commutative normal subgroup of G/P, then N
satisfies the conditions of the lemma, so N is finite (hence N/P is also
finite), completing the proof. D
Remark. In the problem, the condition on the solvability (or some
weakened form of it) cannot be dropped: Novikov and Adian constructed
examples of infinite groups where the order of the elements is finite (in fact
bounded) but every Abelian subgroup is finite (actually, cyclic).
Problem A.21. We say that the rank of a group G is at most r if
every subgroup of G can be generated by at most r elements. Prove that
there exists an integer s such that for every unite group G of rank 2 the
commutator series of G has length less than s.
Solution. We start by showing that if G is a finite group of rank 2, N a
normal subgroup of it, and N* the intersection of the normal subgroups of
N that have prime index, then the following statement is true:
Statement. If in every rank-2 automorphism group of N/N* the length
of the commutator series is at most n(> 0), then h(G), the length of the
commutator series of G, is at most η + 2.
Proof. The condition implies that the group obtained by restricting the
inner automorphism of G/N* to N/N* has a commutator series whose
length is at most n. In other words, taking the member G^ of the
commutator series G ^ G' > G" > ... and forming the commutator group
[TV, G(n)], this will be a subgroup of N*. Now in case N < G^n\ applying
this observation to N' in place of N we see that the group Ν'/Ν'* is cyclic,
for if a and b generate N, then the commutators [a, b] and [Nf ,N] generate
TV7, so Ν'/Ν" is cyclic. Thus Ν'/Ν"* is commutative, because — again
by the first observation — N"/Nff* is a subgroup of the center of N'/N"*,
which gives N"* = N", and thus Nm = N". Therefore, h{G) ^ η + 2.
We now show that for every subgroup A of the group L = GL<2(p) of
2x2 invertible matrices over the field of ρ elements (p prime), we have
h(A) ^ 5 (this bound is not sharp).

3.1 ALGEBRA
83
In the group L of order (p2 — l)p(p — 1), the matrices of determinant 1
form a normal subgroup S = SL2(p) and the factor group L/S is
commutative. If с is a generating element of the multiplicative group of the field
with ρ elements, then the diagonal matrix whose diagonal entries are с and
c~l is an element of order ρ — 1 in S. By multiplying with a generator
element of the multiplicative group of the field with p2 elements, we get
an automorphism of order p2 — 1 of the same field; this implies that S
has an element of order ρ + 1. As a consequence, we see that every Sylow
subgroup of odd order of S is cyclic, since any odd prime power dividing
(p — l)p(p+ 1) divides precisely one of the three factors. Therefore, if A is
a subgroup of rank 2 of L, then for every normal subgroup N of Α Π 5, any
automorphism group of N/N* has a commutator series of length at most
2 (because N/N* is a direct product of a cyclic group of odd order and at
most two groups of order 2); therefore, by the statement first proved we
have h(A Π S) < 4, and consequently h(A) < 5.
Now for every normal subgroup iVof a finite group G of rank 2, it is true
that every Sylow subgroup of N/N* is either of prime order or the direct
product of two groups of prime order. Therefore every automorphism group
В of rank 2 of N/N* is isomorphic to a subgroup of the direct product of
groups A of the abovementioned type, which implies h(B) < 5, and then,
once more using the first proved statement, we get h(G) < 7 (which is
again not the exact bound). D
Remarks.
1. Instead of the straightforward proof described above, we could have
obtained the statement of the problem by applying the theorem of
Blackburn on finite p-groups of rank 2 and the theorem of Zassenhaus on
solvable matrix groups.
2. The statement of the problem does not remain true if we consider finite
groups of rank 3 instead of finite groups of rank 2. A counterexample
is provided by the p-Sylow subgroups of the automorphism group of
the direct product of two cyclic groups of order pn (p > 2 prime,
n=l,2,...).
Problem A.22. Let R be an Artinian ring with unity. Suppose that
every idempotent element of R commutes with every element of R whose
square is 0. Suppose R is the sum of the ideals A and B. Prove that
AB = В A.
Solution. We will use two lemmas.
Lemma 1. Every idempotent e of R lies in the center of R.
Proof. Let r be an arbitrary element of R. Then (er — ere)2 = 0, and
consequently er — ere is permutable with e. Therefore,
er — ere = e(er — ere) — (er — ere)e = 0, er = ere.

84
3. SOLUTIONS TO THE PROBLEMS
We get re = ere in a similar fashion. Thus er = re, proving the lemma.
Lemma 2. If A is a nonnilpotent right ideal of a (right) Artinian ring
R, then A can be written as a direct sum A = eR Θ Ν, where e is an
idempotent element of A and iV is a suitable nilpotent right ideal of R.
Proof. The proof of this lemma can be found, for instance, in A. Kertesz,
Vorlesungen uber artinsche Ringe, Akademiai Kiado, Budapest, 1968;
VEB, Deutsche Verlag, Berlin, 1975, Theorem 6.23, p.155.
Turning to the proof of the statement, we see that it is enough to prove
that ab £ Β A for any a e A and b £ B. Since R is the sum of the ideals A
and B, we have elements αο (Ξ A and bo £ В such that 1 = ao + &o· This
gives
ab = 1 · ab = (a0 + bo)kab = a^ab + Σδ^ζ (bi G Β,αι G A).
If A is nilpotent, then for к sufficiently large we have a^ab = 0, that is,
ab £ Β A. If A is not nilpotent, then using Lemma 2, we have A = eR®N;
in particular ao = er + η (r e R,n € Ν). Using Lemma 1, we get
a§ = exk + nfc for suitable elements Xk of Д.
Since η is nilpotent, we have nk = 0 for sufficiently large fc, that is,
a§ = err*;, and consequently
ab = exfca6 + Σδ^ = 6'e + Σ^α* G В A (bf e B). D
Problem A.23. Let R be an infinite ring such that every subring of R
different from {0} has a finite index in R. (By the index of a subring, we
mean the index of its additive group in the additive group of R.) Prove
that the additive group of R is cyclic.
Solution. Choose an element а £ R such that the order of α in the additive
group R+ is either infinite or a prime number p. Depending on these two
cases, let Μ denote the ring of the integers or the finite field of ρ elements.
Then the subring Ra generated by a consists of the elements of the form
/(a), where / is a polynomial with coefficients from Μ whose constant
term is 0.
Suppose f(a) φ 0 for every such / φ 0. Then the elements of the form
/(a2) form a subring that has infinite index in i?a, so also in R. Hence,
there exists a polynomial f φ 0 over Μ with constant term 0 such that
f(a) = 0. Take such an / whose degree is the least possible. We have
f(x) = x(c+h(x)),
where h φ 0 (because for с G M, с ф 0 we know ca φ 0) and h is also
a polynomial over Μ with zero constant term. Let b = h(a). Since the
degree of / was minimal, we have b φ 0. Then
b2 = h(a) · h(a) = h(a)(c + h(a)) — с · h(a) = —c · h(a) = —c-b.

3.1 ALGEBRA
85
This implies that the subring Rb generated by b consists of the elements of
the form с · b (c £ M) alone; therefore, the additive group R^ is cyclic.
By the assumption of the problem R£ has finite index in R+, so R+ is
finitely generated and, as R+ is infinite, R£ is also infinite. So a could not
have finite order in i2+, and consequently R+ is torsion-free. By the
fundamental theorem of finitely generated Abelian groups, R+ can be written
as a direct sum of infinite cyclic groups. But R+ contains an infinite cyclic
group of finite index, namely R£, so the rank of R+ is 1. Thus R+ is an
infinite cyclic group. D
Problem A.24. Let S Ъе а semigroup without proper two-sided ideals,
and suppose that for every a, 6 £ S at least one of the products ab and ba
is equal to one of the elements a, b. Prove that either ab = a for all a, 6 £ S
or ab = b for all a, b £ S.
Solution 1. S is clearly an idempotent semigroup. Applying a theorem
of McLean (see A. H. Clifford, and G. B. Preston, The Algebraic Theory
of Semigroups, vol. L, AMS, Providence, R.L, 1961, p. 129), we get that
there is a congruence relation θ on S such that S/θ is a semilattice and
each class of the congruence relation θ is a rectangular band. It can be seen
easily that S/θ itself has no proper two-sided ideals, so it consists of one
element; in other words, S is a rectangular band. Recalling the definition,
this means that there are sets X and Υ such that S is isomorphic to the
semigroup (Χ χ Υ\ ·), where multiplication is defined by the rule
(si,2/i)(^2,2/2) = (si,2/2) Ы eX.VieY).
But then the condition formulated in the problem can hold only in the
case where either X or Υ consists of a single element, and this proves the
statement. D
Solution 2. From the second and first conditions of the problem, we see
in turn that
(1) S is idempotent,
(2) for any elements a, b £ 5, there exist elements x,y £ S such that
b = xay.
Let us define relations L, R on S in the following way:
aLb <£=> ab = a,
aRb 44 ab = b.
L is clearly transitive, and in view of (1) it is also reflexive. Suppose for
some a, b € S we have aLb. Then using (1) and (2), we see
a = ab = axay — axay2 = (axay)y = ay,
b = xay = x(ay)(ay) = (xay)(ay) = ba,

86
3. SOLUTIONS TO THE PROBLEMS
therefore bLa. Thus, we have proved that L (and similarly R) is an
equivalence relation. The definition of L and R and the second condition of the
problem give in turn that
(3) aLb and aRb are both satisfied only in case a = 6,
(4) for any elements a, b G 5, either aLb or aRb is satisfied.
But for equivalence relations L, R, (3) and (4) can be true simultaneously
only in the case when one of L and R is the full relation 5x5. D
Problem A.25. Let Ζ be the ring of rational integers. Construct an
integral domain I satisfying the following conditions:
(a) Z С /,.
(b) no element of I \Z is algebraic over Ζ (that is, not a root of a
polynomial with coefficients in Z);
(c) I only has trivial endomorphisms.
Solution 1. Choose a transcendental real number a, and let A be the set
of numbers of the form f{a)/g(a), where /, g G Z[x) and the polynomial g
is primitive, that is, the greatest common divisor of its coefficients is 1. A
is an integral domain with identity that satisfies
a) A \ Ζ contains no algebraic number,
b) to every a e А (афО) there is a β G Α (β φ 0) such that αβ G Z.
Let Μ be the set of all integral domains / with Ζ < I < R that satisfy
a) and b). Μ satisfies the conditions of Zorn's lemma, so it contains a
maximal element J. We are going to show that
c) for every a G J (a > 0) there exist elements /3i,/32, · · · ,/3fc £ J and
7i? 72, · · ·, 7k £ J such that
к п
г=1 j=l
Indeed, let α £ J, α > 0 be arbitrary. If α is an integer or y/a G J, then
the statement holds trivially. Now let α £ J \ Ζ, α φ β2 {β G J), and look
at the integral domain Jy/a\ this satisfies b). If β + 7^7 φ 0, (/3,7 G J),
then there is a δ G J (δ φ 0) such that <$(/32 — 72a) G Ζ holds (we can
assume β2 - η2α φ 0) and so (/3 + ^^α)(δβ - δ^/α) G Ζ.
Since J ^ «/[\/5], the maximality of J implies that J[\/a] cannot satisfy
a). So there are /3,7 G J with 7^0 such that /3 + 7\/^ is algebraic, that
is, there is a polynomial f(x) = Σΐ=οαίχ1 ^ ^M w^^ f(@ + 7>/**) = 0·
We have
N
/(/3 + 7v^) = Σ α*(/3 + 7у^Г = С + D^t,
г=0
where С and Д being polynomials with integer coefficients of /3, 7 and a,
belong to J. If D = 0, then С = 0, so /(/3 - 7^) = С - £>л/« = 0, that
is, /3 — jy/a is also algebraic. This implies that jy/ά, consequently 72a,
is algebraic, so by a) 72a = η is an integer, and thus the statement of c)

3.1 ALGEBRA
87
holds for a. If, on the other hand, D φ 0, then С + Dyfa = 0 implies
aD2 = C2, so c) is satisfied again.
Now we show that J satisfies the requirements of the problem. Let φ
be a nontrivial endomorphism of J. Clearly, 0(1) = 1, and so 0(n) = η
for every η G Z. If a € J (α φ 0), then b) implies φ(α) φ 0. If α > 0,
then it is easy to deduce from c) that 0(a) > 0, which means that φ is
order-preserving. But then φ can only be the identity that we easily verify
if we extend in an operation-preserving way to the quotient field of J and
observe that in this way φ becomes an order-preserving map, fixing all
rational numbers. D
Solution 2. We prove that the integral domain
τ π, Γ ι ι il
J = Ъ \x, —, -, —
[ χ χ + 3 χ + 10 J
satisfies the conditions of the problem. The elements of J \ Ζ are clearly
transcendental, so we only have to show that J has no nontrivial
endomorphism. J is evidently the set of rational functions of the form
f(x)/(xk(x + 3)n(x + 10)m), where f(x) e Z[x] and fc,n,ra > 0. This
shows that the invertible elements of J are precisely the elements of the
form exk(x+3)n(x+ 10)m, where ε = ±1 and к, п, т are arbitrary integers.
Let φ be an endomorphism of J. If φ is not identically zero, then 0(1) = 1,
so the image of an invertible element is invertible. This implies
ф(х) = егхк1 (x + 3)ni (x + 10)mi, (1)
φ(χ + 3) = e2xk2 (x + 3)n2 (x + 10)m2 = φ(χ) + 3, and (2)
φ(χ + 10) = е3хкз (х + 3)Пз (х + 10)тз = ф(х) + 10, (3)
where ε* = ±1 and к^п^т^ are suitable integers (г = 1,2,3). If k\ < 0,
then (1) and (2) imply that k% = k\ (the right-hand side of (2) has a pole
of order — fei in 0, thus the left-hand side too), which gives
e2xk2 (x + 3)n2 (x + 10)m2 = ει (χ + 3)ηι (χ + 10)mi + 3x~kl . (4)
Substituting χ = 0, we get
ε23η210τη2 =ei3ni10mi,
which gives e\ — e^ n\ = пг, т\ — m^. But this is impossible in view
of (4); therefore k\ > 0. We get similarly ki > 0, щ > 0, and га; ^ 0
(г-1,2,3).
If ni > 0, then substituting in (3) χ = — 3 we get щ = 0 and
ε3 · (—3)fc3 · 7тз = 0, which is clearly impossible. Therefore, щ = 0 and
similarly mi = 0. We can't have k\ = 0, because then φ(χ) = ε\ would give
ф(х + 3) = 2 or 4, contradicting (2). Therefore k\ > 0, and substituting
χ = 0 in (2), we get fc2 = 0 and ε2 · 3П2 · 10m2 = 3. This gives ε2 = 1,
n2 = 1, and m2 = 0; therefore
φ(χ + 3) = χ + 3 = φ(χ) + 3, that is, ф(х) = ж.
Therefore, 0(a) = a holds for all a G J, which was to be proved. D

88
3. SOLUTIONS TO THE PROBLEMS
Problem A.26. Let ρ > 5 be a prime number. Prove that every
algebraic integer of the pth cyclotomic field can be represented as a sum of
(finitely many) distinct units of the ring of algebraic integers of the field.
Solution. Put ζ = е27гг/р, and let us denote the pth cyclomatic field by
Kp = Q(C). It is well known that 1 + C, 1 + C2> ·. ·, 1 + Cp_1 and ζ are
units in Kp. So ει = (ζ + С-1)2 and ει = ~(C2 + C~2) are also units.
Furthermore,
ει+ε2 = 2. (Ι)
We show that ε ι and ε2 are independent, that is, there are no rational
integers α and 6, at least one of them is nonzero, and such that
e?4 = l. (2)
Suppose the contrary. With the notation 4a = a', —2b = &', we deduce
from (2)
(С + гУ-^ + Г2)6'. (3)
Let d denote the smallest positive integer such that 2d = 1 (mod p).
Clearly, 3 ζ d ζ p— 1. It is well known that Kp is a normal extension of Q
and the automorphisms of Kp are determined by ζ —> ζ1 (г = 1,2,..., ρ— 1).
Repeatedly applying the automorphism ζ —> ζ2 to (3), we get
к2+с2г' = (с22+с2У
22\α' _ ,v23 , /--23\Ь'
(С2 +CT =(C +Г2)
(с2""1 + с2'"1)0' = (c2d+ζ~2Ύ = (с+ζ-Г
(4)
_/d i/d
(3) and (4) impliy
If α' φ V , then ζ + ζ-1 is a root of unity. But ζ + ζ-1 is a real number,
so the only way it can be a root of unity is for it to be equal to 1 or —1.
But this would imply that ζ is a sixth or third root of unity, which is not
possible. So o! = bf and consequently a' = ±b' φ 0. But then (3) implies
[K + CW + r2)]"'^ (5)
or
(A±£LY =
U2+c-v
1. (6)
But the left-hand side of both (5) and (6) is a power with exponent α' φ 0
of a real number that can be equal to 1 only if the numbers themselves are
equal to 1 or —1. But this would imply that ζ is a root of a polynomial of
degree at most 6 from Z[x], which is different from the seventh cyclomatic
polynomial. Since this is impossible, ε ι and ε2 are indeed independent.

3.1 ALGEBRA 89
As is well known, ζι = 1, ζ2 = С, · · ·, Cp-ι — Cp~2 f°rm an integer basis of
Kp, which means that every algebraic integer of Kp can be represented as a
linear combination of these units with integer coefficients. Therefore, every
algebraic integer of Kp can be written as a sum of units of the form ±ζ^\^2
(1 ^ к ζ ρ— 1;г, j G Z), and it follows from the previous observations that
these units are pairwise different.
Now let α be an arbitrary algebraic integer in Kp, and suppose that
m q p— 1
α = Σ Σ Σ αϋ^ε{ε\ , (7)
χ—I j=nk=l
where α^, Ζ, га, η, <? G Ζ. We can assume that some α^ φ 0 and that (7)
is a representation for which the value of
Μ = Σ IM (8)
ij,k
is minimal.
Using induction on M, we prove that a has a representation of the form
r s p—l
i=l j=nk=l
where r, s G Ζ and afijk = — 1, 0, or 1. This is trivially true for Μ = 1.
Suppose Μ ^ 2 and that the statement is true for all a such that
^2\aijk\ <M.
Let α be an algebraic integer in Kp that can be represented in the form (7)
with property (8) (if such an element exists at all). In view of (1), clearly
%Αε\ = Ck4+14+αεί4+1 · (9)
Applying (9) repeatedly to (7), any a^ with |α^| > 2 can finally be
reduced to —1, 0, or 1. Furthermore, by the minimality of M, J2ij,k \aijk\
remains unchanged during the repeated applications of (9). So, after a
finite number of steps, we arrive at
t p—1 и ν p—l
α = ΣΣα>4^ε{4 + Σ ΣΣ^ΑΑ, (10)
j=nk — l г=/+1 j=n k=l
where bijk,t,u,v G Ζ and a'ijk = — 1, 0, or 1. In addition,
j,k i,j,k
and
so we can apply the induction hypothesis to the second term of (10), and
this concludes the proof. D

90
3. SOLUTIONS TO THE PROBLEMS
Problem A.27. Suppose that the automorphism group of the unite
undirected graph X = (Ρ, Ε) is isomorphic to the quaternion group (of
order 8). Prove that the adjacency matrix of X has an eigenvalue of
multiplicity at least 4. (P = {1,2,..., n} is the set of vertices of the graph X.
The set of edges Ε is a subset of the set of all unordered pairs of elements
of P. The group of automorphisms of X consists of those permutations of
Ρ that map edges to edges. The adjacency matrix Μ = [т^] is the η χ η
matrix defined by rriij = 1 if {i,j} G Ε and rriij = 0 otherwise.)
Solution. Let π be a permutation of the set P, and let Απ be the
corresponding permutation matrix, that is, the matrix that has 1 in the ith row
and jth column if ίπ = j, and 0 otherwise. We see that Απ is an orthogonal
matrix: A% = A~l. We also see easily that the element in the ith row and
jth column of Α~λΜΑπ is just m^j^. This means that
(1) π G Aut(X) <=> Α~λΜΑπ = Μ.
We are going to use the following well-known (and trivial) fact:
(2) If AB = В A (A and В are η χ η matrices), then the subspace
belonging to some eigenvalue of В ("eigensubspace") is an invariant subspace
of A
Let the eigensubspaces of Μ (in Rn) be Vi,...,V8 and dimV^ = щ.
The subspaces Vi are pairwise orthogonal (with respect to the usual scalar
multiplication), and their direct sum is Rn. (This follows from the fact
that Μ is a real symmetric matrix.) So, applying (2) we see that the group
of the orthogonal matrices that are permutable with Μ is a subgroup of
the group 0(Vi) χ · · · x 0(VS) (here χ denotes direct product, 0(V) the
group of orthogonal transformations of the Euclidean space V; later 0(k)
will denote the group οι к х к real orthogonal matrices). In view of (1)
Aut(X) is isomorphic to a subgroup of the group Ο(ηι) χ · · · χ 0(ns). Since
0(1) < 0(2) < 0(3), it will be enough to prove the following:
(3) The quaternion group is not isomorphic to any subgroup of 0(3) χ
• · · x 0(3). To prove this, suppose that the quaternion group Η = {±1, ±г,
±j, ±k} has an embedding
/ : Η -> 0(3) χ···χΟ(3),
and let us denote by pr the rth projection of the group 0(3) χ · · · χ 0(3)
to 0(3). Now fr = f opr : Η —> 0(3) is a homomorphism. We claim the
following:
(4) The kernel of any homomorphism h : Η —> 0(3) contains — 1 G H.
This implies (3) as it gives — 1 G ker/r for all r, thus — 1 G ker/, proving
that / cannot be an injection. To prove (4), let us recall the full list of
finite subgroups of 0(3):
(5) A finite subgroup of 0(3) is isomorphic to one of the following:
A5 χ Z2,54 x Z2, A4 x Z2, Zn χ Z2, Dn χ Z2, A5, S4, A4, Dn, Zn,

3.1 ALGEBRA
91
where At denotes the alternating group of degree t, St the symmetric group
of degree t, Zt the cyclic group of order t, and Dt the dihedral group of
degree t (thus order 2t) (see H. S. M. Coxeter, Introduction to Geometry,
Wiley, New York, 1969, Table III).
(6) Cosequence: 0(3) has no subgroup isomorphic to H. So h : Η —>
0(3) cannot be an embedding. Therefore кег/ι contains am χ e Η different
from the identity. But then either χ or x2 is equal to —1, so necessarily
-1 Gker/i.
This concludes the proof. D
Problem A.28. For a distributive lattice L, consider the following two
statements:
(A) Every ideal of L is the kernel of at least two different homomor-
phisms.
(B) L contains no maximal ideal.
Which one of these statements implies the other?
(Every homomorphism φ of L induces an equivalence relation on L: а ~ b
if and only if αφ = b(p. We do not consider two homomorphisms different
if they imply the same equivalence relation.)
Solution.
(A) IMPLIES (B). Assume, by contradiction, that L contains a maximal
ideal M. Suppose we have a homomorphism φ whose kernel is equal to M.
We will show that the image of L \ Μ by φ is a single point. *Since we also
know that under the equivalence relation induced by φ all elements of Μ
are equivalent and no element of L \ Μ is equivalent to an element of M,
we get that there is at most one homomorphism whose kernel is M. Let
α and b be two elem ents of L \ M. The following description of the ideal
(Μ, α) generated by Μ and α is well known:
(Μ, α) = {x e L\3me Μ : χ ^ α V га} .
In view of the maximality of M, we have (M,a) = L. So for some
element ra G Μ we have b ^ a V ra, therefore bip ^ αφ\/ πιφ = αφ. Similarly,
αφ ^ bφ consequently, αφ = bφ, as we stated.
(B) DOES not IMPLY (A). We construct a counterexample. Let
L = {V | V С R finite} U {(-oo, x]UF|xGR,FcR finite} .
The operations on L should be the set-theoretic union and intersection. In
this way, L is a sublattice of the lattice of subsets of R and therefore L
is distributive. We claim that it has no maximal ideal, but on the other
hand, the ideal {0} is the kernel of a single homomorphism.
Let J be a proper ideal of L. Not all half-lines of the form (—oo, x] can
occur in J because otherwise all finite subsets of R would be elements of J
forcing J — L. So there is an χ G R such that (—oo, x] £ J which of course
implies (—oo, y] £ J for all y^x. So the ideal
J* = {(-oo, z]\JV\z^x,V СЖ finite} U {V \ V С R finite}

92
3. SOLUTIONS TO THE PROBLEMS
is a proper ideal properly containing J. Consequently, J is not a maximal
ideal.
To prove the other claim, let φ be a homomorphism whose kernel is {0}.
We show that φ is a monomorphism, which means that the equivalence
relation induced by φ is such that each element is equivalent only to itself,
and this contradicts (A). Assume, on the contrary, that a, 6 G L, а ф b but
αφ = bip. The symmetric difference of a and b is nonempty, so there is a
ρ G R with, say, pea and ρ £ b. This implies
{ρ}φ = ({ρ} Λ ά)φ = {ρ}φ Λ αφ = {ρ}φ Λ 6<£ = ({ρ} Λ Ъ)ч> = ®φ = 0,
in contradiction with ker<^ = {0}. Thus φ is indeed a monomorphism as
claimed. D
Problem A.29. Let V be a variety of monoids such that not all monoids
of V are groups. Prove that if A e V and В is a submonoid of A, there
exist monoids S G V and С and epimorphisms φ : S —> A , φι : S —> С
such that ((β)φϊ1)φ = В (е is the identity element ofC).
Solution. Take DeV, which is not a group. Then there exists an element
χ G D that is not invertible. The powers of χ with positive integer
exponents and the identity element form a submonoid of D, and the dichotomy
underlying the previous description of this submonoid is a congruence
relation of it. Taking the natural homomorphism, we see that the variety V
contains the monoid Ε = {/, д} (/ ф д) with operation defined by /·/ = /,
f '9 = 9' f = 9,&ъад'д = д.
Consider the set
S={(a,g)\aeA}U{(bJ)\beB}.
On the direct product Α χ Ε of the monoids A and E, we have
{аъд)(а2,д) = (ага2,д), (δι, /)(62, /) = (6ι*>2, /),
(αι,#)(6ι,/) = (агЬид) (α* G A,bi G B),
so S is actually a submonoid οι Α χ Ε proving S G V. We see at the
same time that the two sets defining S give rise to a congruence relation on
S. Let φι be the natural homomorphism with respect to this congruence
relation and denote the image of S by C. Clearly, the set of the elements
of S that are mapped to the identity element of С equals {(6, /) | b G B}.
If we denote by φ the projection of an ordered pair to its first component,
we see that φ : S —> A is an epimorphism, and we have {(b, f) \ b e Β} φ =
В, which concludes the proof. D

3.1 ALGEBRA
93
Problem A.30. For what values of η does the group SO(n) of all
orthogonal transformations of determinant 1 of the η-dimensional Euclidean
space possess a closed regular subgroup? (G < SO(n) is called regular if
for any elements x,y of the unit sphere there exists a unique φ G G such
that φ{χ) = у.)
Solution. The answer is that such a subgroup exists precisely for η = 2 and
η = 4, namely for η = 2 the group SO(2) itself and for η = 4 the symplectic
group denoted by Sp(l), which we get by taking the multiplications with
quaternions whose absolute value is 1.
We show that there are no other solutions of the problem. For odd n,
SO(n) certainly does not contain a regular subgroup since in this case every
transformation φ G SO(n) has a fixed vector.
Suppose therefore that η is even, η = 2m. In this case, for every φ G G
(G is subgroup in question) the space R2m decomposes as the direct sum
of m pairwise orthonal two-dimensional subspaces, invariant with respect
to φ: R2m = fti θ · · · θ ftm. Clearly, on every subspace /ι^, φ induces a
rotation ψ{ with an angle a*.
Now we prove two lemmas:
Lemma 1. For every φ G G there is a Jordan decomposition R2m =
h\ 0 · · · Θ hm such that ot\ = · · · = am.
Proof. There is a one-dimensional subgroup Φ of G such that φ G Φ
(this is well known). It is also well known that there exists an element
g G Φ whose powers are dense in Φ. Let R2m = hi 0 · · · θ /im be a Jordan
decomposition with respect to g, and let the restriction of g to hi be <ft.
Then gi is a rotation with angle Д on hi. At least one of the βΐ$ is an
irrational multiple of 2π, and consequently every other /3j is an irrational
multiple of 2π; otherwise for some /с, дк ф id would fix an element of hj.
This gives that the restriction Φ; of Φ to hi is the group SO(2) for all i.
Let us consider the group homomorphism λ: Φ^ —> Φ^, which maps every
element g* G Φ* (where g* Ε Φ) to ^* G Φ^. Thus λ is continuous and
1-to-l in view of the regularity and compactness of G, so it produces an
automorphism of the topological group SO(2). But there are just two such
automorphisms: the identity and (identifying SO(2) with the unit circle of
C) the conjugation. These map a rotation with angle α to a rotation with
angle a, so — using \(<Pi) = φ^ — we get a^ = aJ? which was to be proved.
This lemma immediately implies
Lemma 2. If some φ G G maps a unit vector to a vector orthogonal to
it, then φ2 = —id and φ maps any other vector у to a vector orthogonal
to it; furthermore, the vectors y, ip(y) span a plane that is invariant with
respect to φ.
Proof. Let R2m = hi 0 · · · 0 /im be the decomposition whose existence is
assured by Lemma 1. Suppose that χ is orthogonal to φ(χ) and χ = χι +
\-Xm, ψ(χ) = φ(χι)~\ \-φ(Χπι) the Jordan decomposition with respect

94
3. SOLUTIONS TO THE PROBLEMS
to φ and consider the inner product (χ,φ{χ)) = Y^Li(xi^(xi)) = 0.
Lemma 1 guarantees that either (χί,φ(χ{)) ^ 0 for all terms or (χι,φ(χι))
for all terms, therefore necessarily (χί,φ(χΐ)) = 0 for all terms. Thus,
Lemma 1 implies that φ rotates with an angle π/2 in every subspace hi,
and from this the statements of the present lemma follow immediately.
Now the question formulated in the problem is answered by the following
Proposition. If the group SO(2m) (2m > 4) has a closed, regular
subgroup G, then 2m = 4 (furthermore, it is true that G is the universal
covering group of SO(3), that is, Sp(l)).
Proof. Let the unit vectors eo, ei be arbitrary, but orthogonal, and let
φι G G be the group element with <^i(eo) = e\. Let e<i be a unit vector
that is orthogonal to the plane spanned by eo, ei, and let ез = <£i(e2)· By
Lemma 2, the vectors eo, ei, e2, ез are pairwise orthogonal (the subspace
orthogonal to an invariant subspace is itself invariant) and denoting by
ψχ G G the group element with ipi{eo) = e^, then the following formulas
are already true:
φ\ = -id, φ\ = -id, φ\ = -id,
The formulas φ%φ^ — —ψ^ψχ m the last row follow immediately from the
relation {ψΐψ^)2 = —id. In the case 2m = 4, the proof is finished.
It remains to show that the case 2m > 4 cannot occur. Suppose 2m > 4,
and let К be the subspace spanned by eo, ei, e2, ез. If we choose a unit
vector e4 that is orthogonal to К and define e$ = <^i(e4), ее = (^2(^4),
e7 = <^з(е4), then the vectors e4, es, eg, e7 are pairwise orthogonal and
span a subspace K*, orthogonal to K, consequently eo,ei,...,e7 is an
orthonormal system of vectors. Denoting by φι G G the group element
with (fi(eo) = e» (i = 0,1,..., 7), the transformations ψι have the following
multiplication table:
φο
φι
ψ2
Ψ3
ψ4
<Рь
ψ6
φι
φο
id
φι
φι
φζ
φι
φ$
φβ
φι
φι
φι
-id
-φ?>
φι
-φ$
φ4
-φι
φβ
φι
φι
φζ
-id
-φι
-φβ
φ7
φ4
-φ$
φ3
φ3
-φι
φι
-id
-φ7
-φβ
φ$
φ4
φ4
φ4
φζ
φβ
φι
-id
-φι
-φι
-φ*
φ$
φ$
-φ*
-φι
φβ
φι
-id
-φ*
φι
φβ
φβ
φι
-φ*
-φ*
φι
φζ
-id
_-ψι
φ7
φ7
-φβ
φъ
-φ4
φ3
-φι
φι
-id
From this multiplication table, we deduce the following identities:
(^4^5)^6 = {φ^φιφ^φβ = {-φ4φ*φι)φβ = φιφβ
= ч>\{ч>2Ч>а) = (^1^2)^4 = ψζψα = φι
φ^φβφβ) = φ^φιφιφιφ*) = φ^-φιφιφιφ*)
= ^4(^1^2) = φ*φ3 = -φζφ* = -φι ·

3.1 ALGEBRA
95
This contradicts the associativity of the multiplication, so the case 2ra > 4
cannot occur. D
Problem A.31. in a lattice, connect the elements а A b and α V b by
an edge whenever α and b are incomparable. Prove that in the obtained
graph every connected component is a sublattice.
Solution. Let us denote the fact that two elements α and b belong to
the same connected component of the graph by а ~ b. The statement
to be proved is that whenever b\,..., bn is a path in the graph, we have
b\ ~ b\ V&n, which in turn reduces to b\ V6n_i ~ b\\Jbn. We will denote
the fact that two elements χ and у are incomparable by χ || y.
Lemma. Let χ \\ y,a = x\/y,b = χ Ay, ζ arbitrary. Then one of the
following two cases holds:
(a) a V ζ ~ b V z\
(b) one of a, 6, ζ is ~ a V ζ and one of a, 6, ζ is ~ b V z.
This will indeed prove the statement of the problem. In the case 6n_i =
x\/y and bn = χ Ay, we can apply the lemma with a = bn — 1, b = 6n,
ζ = b\. In case (a), the proof is finished immediately, and in case (b) we get
that one of b\,bn-\,bn is ~ b\ V6n, which suffices to conclude the proof.
In the case 6n_i = χ А у and bn = x\/y, we can proceed similarly.
We will prove the lemma in six steps. First, we deal with that part of
the statement that refers to a V z.
1. If a ^ z, then a V ζ = α, ζ so the first part of (b) holds.
2. Suppose a £ z\/y. This gives ж || zVy since χ ^ z\/y would
imply χ ^ y, whereas in case χ ^ z\/y we would have a = x\/y ^ z\/y.
Therefore a\Jζ = xVyVζ > xA(y\/ζ). In view of xAa = x, we can
further deduce χ Л(у V ζ) = χ Λ(α л(у V г)) < χ ν(α Л(у V ζ)) as soon as we
establish χ \\ aA(zWy). To prove the latter statement, suppose first χ ^
aA(zWy). Then in view of у ^ aA(zWy), we have a = x\/y ^ aA(z\/y),
contrary to the initial assumption. On the other hand, if χ ^ аЛ(г Vy),
then у ^ α Λ(ζ V у) implies у ^ χ, which is not the case.
But x\/(aA(y\/z)) = a since x,aA(y\/z) ^ α and aA(zWy) ^ y, thus
χ ν(αЛ(у Vг)) > x\Jу = a. We arrived at the conclusion a ~ a\/z.
The case a ^ z\/χ can be treated similarly.
3. So we can assume a ^ z\/x, a ^ z\/y, and α || z. Thus χ ^ £Vy,
which implies z\Ja = z\Jy. Similarly z\Ja = z\Jχ consequently z\Jχ =
z\Jy = z\/a. We can further assume χ \\ ζ and у || z. (Suppose namely
χ ^ z. This gives aV£ = yVζ ~ a. The case у ^ ζ is similar, and in case
χ ^ ζ от у ^ ζ we get α ^ ζ.)
Suppose now ζ \\ (z\/b) A a. This implies
z\/a ~ ζ Aa = ((z\/b) Az) Aa = ((z\/b) Aa) Az
< {{z\Jb)Aa)\Jz = zWb.

96
3. SOLUTIONS TO THE PROBLEMS
As to the last equality, ^ is clear and we get the other direction from
b^ (ζνδ)Λα.
What if ζ and (z V b) A a are comparable? ζ ^ (z V b) Α α ^ a is excluded,
so this can happen only if ζ > (ζ V b) Α α ^ 6, and consequently ζ = ζ V b.
4. Thus, we can assume ζ = z\/b, ζ Vχ = z\Jу = ζ Va. If we have
г Л χ = 6, we get z\/a = z\/x>zAx = b, giving ζ V a ~ 6, and similarly
in the case ζ Ay = b.
5. In the remaining case let t = ((χ Α ζ) v(y Л г)). Then z\/a = zVx >
ζ Ax = χ At since ζ ^ £ and zAx^t. Certainly, χ \\ t since χ ^ t ^ ζ
would imply χ ^ ζ and ж > £ implies χ Ay ^ у Az > b, which is impossible.
So χ At < x\/t = x\/(yAz). We have χ \\ у Az since χ ^ yAz ^ у is
impossible and χ ^ у A z gives χ Ay ^ у Az > b, a contradiction.
So, finally, ж V(y Лг)>ж Л(у Λ ζ) = b since 6 ^ ζ. This gives α V ζ ~ 6,
proving the part of the lemma that refers to α V ζ.
6. Now let us deal with the statement concerning b V z. In case b ^ z,
we have b\/ζ = b,z. Otherwise, b\/ ζ > bAz, and applying the part of the
statement that has been already proved to the dual lattice, we get that
b Α ζ ~ to one of a, 6, ζ or b A z ~ a A z. In case α ^ z, we have aAz = a,z,
whereas in case a \\ ζ using a\Jζ > aAz we get a\Jζ ~ 6Vz, and this
finally concludes the proof of the lemma. D
Problem A.32. Let G be a transitive subgroup of the symmetric group
S25 different from S25 and A25. Prove that the order of G is not divisible
by 23.
Solution. We assume that 23 | |G|, and we try to reach a contradiction
from that assumption. Denote the stabilizer of a point χ by Stab (x).
By the transitivity and our assumption, all the Stab (x) are isomorphic
permutation groups containing a 23-cycle. Suppose that Stab (x) is not
transitive. Then taking the fixed point у of Stab (#), we see that the fixed
point of Stab (y) is x. That would give a pairing of the 25 points, which is
impossible. So G is 2-transitive, and as Stab (x, y) contains a 23-cycle G is
actually 3-transitive. We have seen that the 23-cycles of G act transitively
so we can take G to be the group generated by them. We know then that
G < A25 and G is 3-transitive.
By a well-known theorem (see H. Wielandt, Finite Permutation Groups,
Academic Press, New York, 1964, 13.10), the order of G cannot be
divisible by 11. We know that Stab (x, y) has prime degree and is
transitive. According to Burnside's theorem (Wielandt, 7.3) it is either 2-
transitive or isomorphic to a subgroup of the group of all affine mappings:
ζ ^ az + b (a,be GF(23)).
In the first case, G is 4-transitive, so its order is divisible by 25-24-23-22,
therefore by 11, which is excluded. So Stab (я, у) is a subgroup of a group
of order 2-11-23. Again 11 is excluded. An element of order 2 would
be of the form ζ ι—► — ζ + 6, which is a product of 11 transpositions and
is thus not contained in A25. Therefore, the order of Stab (я, у) is 23, G

3.1 ALGEBRA
97
is sharply 3-transitive, but then by the theorem of Zassenhaus (Wielandt,
20.5), 25 — 1 = 24 has to be a prime power, which is not the case. D
Problem A.33. Let G be a finite group and /C a conjugacy class of G
that generates G. Prove that the following two statements are equivalent:
(1) There exists a positive integer m such that every element of G can
be written as a product of m (not necessarily distinct) elements of
1С.
(2) G is equal to its own commutator subgroup.
Solution. Suppose first that (1) is satisfied, and consider the natural
homomorphism φ : G —► G/G'. Since for any h,k G /C there exists а д е G
such that к = g~lhg we have ip(h~lk) = (p(h~1g~1hg) = 1. Thus /C is
contained in the coset KG' = φ(Κ) (h e 1С).
Now let g = hi· · -hm and g' = h[· · · Н'ш (hi·· · /im, h[ · · · Ыш e /С) be
any two elements of G. By our previous remarks,
ψ(9) = ν(Λι) · · · V(M = <f(h)m = ν(Λί) · · · φ(Κη) = φ(9').
In other words G/G' has only one element, that is, G = G'.
Conversely, suppose (2) is satisfied and let us denote \G\ by n. Since
G = G', every element g of G can be written as a product of commutators:
g = a^b^aibi · · · a;lb~larbr = a^b^aibi · · · a^-lb^-larbr. (1)
Since /C generates G and the inverse of an element of /C can be replaced
by its (n — l)th power, it is true that every a^, bi (i = 1,... ,r) can be
written as a product of elements of /C. Substituting these representations
in (1), we can deduce the existence of a natural number kg such that g
can be written as a product of kg · η elements of /C. But then for any
к ^ kg, we can also write g as a product of к · η elements of /C by simply
adding a suitable number of factors of the type hn = 1 (h e 1С). Let
/ = max{fcg | g e G}. Then every element of G can be written as a product
of m = η · I elements of /C. D
Remark. In the proof of (2) => (1) the only property of /C that we used
was the fact that /C is a system of generators of G.
Problem A.34. Consider the lattice of all algebraically closed subfields
of the complex field С whose transcendency degree (over Q) is finite. Prove
that this lattice is not modular.
Solution. For any set of numbers ЯСС, let us denote by A(H) the
smallest algebraically closed subfield of С containing H. Let us consider
a system of numbers α,/3,7ξ (ξ < ωι) that are algebraically independent
(over Q). Since the algebraic closure of a countable set of numbers is
countable, such a system does exist. We are going to show that for a

98
3. SOLUTIONS TO THE PROBLEMS
Α(\α,β\)
Figure A.l.
suitable ξ < ω\ the following diagram is actually the Hasse diagram of a
sublattice of the lattice of algebraically closed subfields of C, and this will
prove the statement of the problem.
The containments indicated by the lines are evident. Furthermore,
Α(Α({α})υΑ({Ίξ,α + βΊξ})) = Α({α,β,Ίξ})
since the right-hand side clearly contains the left-hand side, while the left-
hand side contains each of the numbers a, /3, 7$. We have Α({α}) φ
Α({α,β}) in view of the original assumption, so all that remains to be
proved is to show that for some ξ we have
Α({α,β}) Π Α({Ίξ,α + βΊξ}) = А{Щ .
For this, it is enough to prove
A({a,β}) П (Α({Ίξ,α + βΊξ}) \ АЩ = 0. (1)
Suppose ξι φ £2· Then the transcendency degree of
A(A({76,a + /376})UA({7^2,a + /37^})) = ^({7^,7^,a,/3})
is equal to 4; therefore, the transcendency degree of
is equal to 0, which means
(Α({Ίξι,a + βΊξι}) \ A(9)) П (Α({Ίζ2,α + βΊξ2}) \ А(Щ) = 0.
Now the set Α({α,β}) is countable and the sets A({^,a + /37ξ}) \
А(Щ are pairwise disjoint, so there exists a ξ for which (1) holds, and this
concludes the proof. D
Α(\Ύ,α+β7.\)

3.1 ALGEBRA
99
Problem A.35. Let I be an ideal of the ring R and f a nonidentity
permutation of the set {1,2,..., k} for some k. Suppose that for every
0 φ α Ε R, αϊ φ 0 and Ια φ 0 hold; furthermore, for any elements
xi,x2,...,xk Ε /,
X\X2 •••Xk= X\fX2f ' · · Xkf
holds. Prove that R is commutative.
Solution. Let raE{l,2,...,fc} the smallest number that is not fixed by /.
Clearly, m < mf. We then have x\x2-·-xk = x\ — - Xm-iXmf · · · xkf, that
is, χι · · · rrm_i(:rm · · · xk - xmf · · · xkf) = 0 for any xu ..., xk Ε /. Fixing
the elements x2,..., xk we see that x2 · · · хш-\{хш · · · xk — Xmf · · · %kf)
annihilates / so has to be equal to 0. Repeating the argument, we get
Xm'"Xk = Xmf ' ' ' Xkf for any Хш, . . . , Xk Ε /.
Let η = mf. Then for every α e R and хш · · · xk Ε /, we get
Ο^τη^τη+Ι ' ' ' %k = ^^mf^(m-\-l)f ' ' ' %kf = %m%m+l ' ' ' %n—lQ'%n%n+l ' ' ' %k·
(We get the second equality by using ахш^ Ε /.) Rearranging,
\αΧγη ' ' ' %n—l ^m ' ' ' *^n—l^/^n " " " ^Ck == ^
for any α Ε R and rrm · · · xk Ε /. Again applying our previous argument,
we get
αΧγγχ ' ' ' Χγχ—\ — %m ' ' ' ^n—l^
for any α Ε R and rrm · · · χη-ι Ε /.
Now let a, 6 Ε R and хш · · · xn-\ Ε / be arbitrary elements. Then we
have
{аЪ)хш · · · Χη-ι = Xm'" xn-i(a>b) = (хш · · · xn-ia)b
= (axm · · · xn-i)b = (axm)xm+i · · · xn-\b
= 6(ажт)жт+1 · · · χη-ι = {ba)xrn · · · xn-\.
Applying once more our previous argument to the resulting identity
[ah - ba)xrn · · · xn_i =0 (a, b Ε R, Хщ · · · xn-i £ /),
we get ab — ba = 0 or ab = 6a, which was to be proved. D
Remark. The proof did not use the fact that / was a two-sided ideal
of the ring R, only the fact that / was a left ideal in the multiplicative
semigroup of the ring R.

100
3. SOLUTIONS TO THE PROBLEMS
Problem A.36. Prove that any identity that holds for every unite n-
distributive lattice also holds for the lattice of all convex subsets of the (n —
l)-dimensional Euclidean space. (For convex subsets, the lattice operations
are the set-theoretic intersection and the convex hull of the set-theoretic
union. We call a lattice n-distributive if
*A(\Zyo = \Λ*Λ( V и))
0<t<n
i*3
holds for all elements of the lattice.)
Solution. Let us introduce some notation. Let En~l denote the Euclidean
space of η — 1 dimensions; for X С Еп~г, let C(X) denote the convex hull
of X; and let L be the lattice of convex subsets of En~l. For an arbitrary
Я С En~l, denote by L(H) the partially ordered set ({X П Я: X e L}, C).
L(H) is a complete lattice since it is closed with respect to forming
arbitrary intersections and it has a greatest element. If we denote the
lattice operations in L(H) by V# and Ля, respectively, then we see that for
X,Y e L(H) we have X AH Υ = Χ Π Υ and X VH Υ = С(X U Υ) Π Η.
For a finite Я, L(H) is also finite since \L(H)\ < 2'я1.
First, we show that for a finite Я С Е71-1, L(H) is η-distributive. Let
X,Yo,...,Yn e L(H) be arbitrary elements of L(H). By Caratheodory's
theorem for an arbitrary subset U of Ε71-1,
C(U)= U C(V).
vcu
\V\<n
Therefore
also holds and so
η In
XAH V Yi=XnHnC[ uYi)=HnXnHn U С U Yt
i=o \t=o J j=o I t=q
= ЯП U \XnHnC(uYi)
j=0 V ft?
That means that L(H) is n-distributive.
Let ρ be a lattice-theoretic term in к variables, let Я С Εη~λ, and
denote by ρ (resp., рн) the term function induced by the term ρ on L

3.1 ALGEBRA
101
(resp., L(H)). For arbitrary Хъ...,Хк G L, рн(Х\ ПН,...,ХкПН) is
a monotone function of Я, that is, Hi С Я2 implies
pHl (Χλ Π Ηλ,..., Xk Π Ηλ) С рн2 (Χλ Π Я2,..., Хк Π Я2).
This follows easily from the monotonicity of U, П, and C, using induction
on the length of the term p.
We claim that for arbitrary X\,..., Xk G L and h G p(^i,..., Xk) there
is a finite Я С En~l such that /ι G Я and /ι G pH{X\ Π Я,..., Xk Π Я).
This statement will be proved by induction on the length of p. If ρ is
just a variable, then we can choose Я = {h}. Suppose that the statement
is proved already for terms whose length is less than the length of pn,
and let ρ = p'/\p". Then there are finite sets Hi and Я2 such that h G
p'Hl{Xi ПНи...,ХкП Ях), h G р'Ь2(Х1 Π Я2,... ,Xk Π Я2), and /ι G Яь
/ι G Я2. Let Η = Ηλϋ Я2. Then we have
h G р7Я1 (Χι Π Яь ... ,Xk Π Нг) Пр^2(Хг ПН2,...,ХкП Я2)
С р'н (Хг Π Я,..., Хк П Я) Π р£ (Хх Π Я,..., Хк Π Я)
= №ЙПЯ,...ДкПЯ).
Now let p = p'\/p". Then
/iGp(X1,...,Xfc)CC(p,(^b...,^)u/№,...,^)).
By Caratheodory's theorem, there exist finitely many points h[,..., h'u G
р'(Хъ ...,Xk) and ft'/, ...,^Ε ρ"(ΑΊ, .-,**) such that
Λ G С ({fti,..., tiu, ft7/,..., ft"}) and u + t; < n.
(Actually, we could obtain и ^ 1, ν ^ 1, but we don't need this now.) By
the induction hypothesis there exist finite sets H\,... ,HU,G\,... ,GV С
En~l such that ft< G p7Hi№ n^,...,Ifcn Я;) and Щ G ρ7- (Χι Π
Gi?..., Xk Π Gj) (1 < г < η, 1 < j < ν). Let Я = {/г} U Ηλ ϋ · · · U
ffttUGiU-'-UGt,. Now
ЛбС({л; л;,л? л;})пя
СС(р'н(Х1ПН,...,ХкПН)иР'11(Х1ПН,...,ХкПН))ПН
= ря№пя,...д,пя).
Now let p(xi,...,xk) = q(xi,... ,xk) be an identity of lattices that is
satisfied by every η-distributive lattice, and let X\,..., Xk G L be
arbitrary. We have to prove p(X\,..., Xk) = q{X\,..., Xk). To prove the
inclusion p{X\,..., Xk) С q(Xi, ...,Xk), take an arbitrary ft G p{Xi,...,
Χ*). Then for a suitable finite Я С En~l we have ft G рн{Х\С\Н, ...,ХкП
Я). Since 1/(Я) is finite and η-distributive we have
h G qH(Xi Π Я,..., Jffc Π Я) С ten-i(Jfi Π £η"\ ...,Χ* Π Я71'1)
This establishes the inclusion p(X\,..., Xk) С q(X\,..., Xk), and the
reverse inclusion can be proved in the same way. D

102
3. SOLUTIONS TO THE PROBLEMS
Problem A.37. Let #i, #2,2/1,2/2,21,2:2 be transcendental numbers.
Suppose that any 3 of them are algebraically independent, and among the
15 four-tuples only {zi,£2,2/1,2/2}* {^1,^2,^1,^2}, and {2/1,2/2,21,2:2} are
algebraically dependent Prove that there exists a transcendental number
t that depends algebraically on each of the pairs {^1,^2}, {2/1,2/2}, and
{21,22}·
Solution. Take (nonzero) polynomials p, q, r with rational coefficients such
that
p(zi,Z2,2/i,2/2) =0,
^(2/1,2/2,21,22) =0.
Since the polynomials in one variable,
Pi(X) =^№^2,2/1,2/2)
and
Qi(X) = q(X,X2,zi,z2),
have a common root, their resultant is zero:
R(puQi)=0.
But it is well known that R(pi, <?i) is a polynomial with integer coefficients
of the coefficients of p\ and q\. Therefore, it is a polynomial with rational
coefficients of the numbers #2,2/1,2/2,21, z^
R(Pi,Qi) = 7(^2,2/1,2/2,21,2:2).
By the conditions, we know that X2 does not depend algebraically on the
numbers 2/1,2/2,21 whereas Z2 does depend on them; consequently, X2 does
not depend algebraically on 2/1,2/2,21,2:2. Therefore, /(#2,2/1,2/2,21, Z2) = 0
implies that /(X, 2/1,2/2,21, Z2) is the (identically) zero polynomial of X.
Now take an arbitrary rational number u. Then we have
7(^,2/1,2/2, 21, 2:2) =0,
and consequently the polynomials
p2(X) =р(Х,щу1,у2) and q2(X) = q(X,u, zu z2)
have a common root t (since the resultant of the polynomials P2(X) and
q2(X) is equal to /(и, 2/1,2/2, 21, 2:2) ).
Define now polynomials ρ*, q* in three variables as follows:
p*(X,Y1,Y2)=p(X,u,YuY2),
q'(X,Y1,Y2)=q(X,u,Y1,Y2).

3.1 ALGEBRA
103
It is clear that we can choose и in such a way that none of the polynomials
is the zero polynomial (if, for every rational number u, at least one of them
were the zero polynomial, then for every complex number и the same would
hold, in particular for и = x2 which is certainly not the case).
Then we have
P*(t,yuy2) = 0,
so t, 2/i, 2/2 are algebraically dependent. But 2/1,2/2 are algebraically
independent, so ί is a transcendental number and depends algebraically on the
pair {2/1,2/2}· Similarly, t depends algebraically on the pair {2:1,2:2}.
To prove that t depends algebraically on the pair {^1,^2}, suppose
the contrary. Then {x\,X2,t} would be a transcendency base of the
algebraically closed field generated by {x\, rr2,2/1,2/2}, and similarly it would
be a transcendency base of the algebraically closed field generated by
{^1,^2,2:1,2:2}. But these two fields cannot be isomorphic, since for
example {#i,#2,2/1? ^1} are algebraically independent. D
Problem A.38. Let F(x, y) and G(x, y) be relatively prime
homogeneous polynomials of degree at least one having integer coefficients. Prove
that there exists a number с depending only on the degrees and the
maximum of the absolute values of the coefficients of F and G such that
F(x, y) φ G(x, y) for any integers χ and у that are relatively prime and
satisfy max{|x|, \y\} > c.
Solution. In the course of the proof, we will denote by c, ci, C2, C3 numbers
that depend only on the degrees and the maximum of the absolute values
of the coefficients of F and G.
Let χ and у be arbitrary relatively prime integers that are solutions of
the equation
F(X,Y) = G(X,Y). (1)
We will show that max{|x|, \y\} < с for a suitable constant c, and this is
just the statement of the problem. We can suppose xy φ 0 since otherwise
in view of (x,y) = 1 we have max{|x|, \y\} < 1. Let us denote the degrees
of F and G by m and n, respectively, and let us assume m ^ n, say.
Let f(X) = F(X, 1) and g(X) = G(X, 1). By assumption, F and G are
relatively prime homogeneous polynomials so at least one of f(X) and g(X)
is a nonconstant polynomial. Actually, we can assume that none of them
is a constant. If f(X) were a constant, then denoting the coefficient of Xй
in G by a, we would get α φ 0. Now (1) together with (x,y) = 1 implies
2/ I a, and since F(X, a) — G(X, a) is not the identically zero polynomial,
we get тах{|ж|, \y\} < c\.
/ and g are relatively prime because F and G are relatively prime.
Thus, denoting by R the resultant of / and g, we have R φ 0. Furthermore
|Д| < C2- By a well-known theorem (see, for example, L. Redei, Algebra,
Akademiai Kiado, Budapest, 1954, Thm. 196, p. 376), there exist
polynomials with integer coefficients A(X) and B(X) such that deg A < degg ^ n,

104
3. SOLUTIONS TO THE PROBLEMS
deg В < deg / < ra, and
A(X)f(X) + B(X)g{X) = R.
This implies that there exist homogeneous polynomials with integer
coefficients Αλ(Χ,Υ), Вг(Х,У) and A2(X,Y), B2(X,Y) such that
Аг(Х,Y)F(X, Y) + Вг(Х,Y)G(X, Y) = R- χ™+η~λ (2)
and
A2(X,Y)F(X,Y) + B2{X,Y)G{X,Y) = R- у™*""1. (3)
But then, using (1), (2), (3), and the fact that (x,y) = 1, we see that
F(x,y) and G(x,y) are different from 0 and both are divisors of R. So,
in case F(x,y) = G(x,y) = d we have d \ R, and therefore \d\ < c2. This
implies that x/y is a root of the polynomial with integer coefficients
h(t) = cT-nfn(t)-gm(t),
which is not identically zero since / and g are relatively prime.
But then by Rolle's theorem we have тах{|ж|, \y\} < C3, and then with
the choice с = max{ci,c3} we get the statement we wanted to prove. D
Remark. It is not hard to get explicit expressions for ci, c2, C3, and с
using the degrees of F and G and the maximum of the absolute values of
their coefficients.
Problem A.39. Determine all unite groups G that have an
automorphism f such that Η % f(H) for all proper subgroups Η ofG.
Solution. If φ has a fixed point α e G, а ф 1, then φ((α)) = (α) so
G = (α), and as every subgroup of (a) is also fixed, G = (a) cannot have
any proper subgroups, consequently it is a cyclic group of prime order.
Suppose now that φ is fixed-point-free. By a well-known result (see, for
example, D. Gorenstein, Finite Groups, Harper & Row, New York, 1968,
Theorem 10.1.2), for every prime ρ there exists a p-Sylow subgroup Ρ of
G with φ{Ρ) = Ρ, so G has to be a p-group. Furthermore, as the center
is a characteristic subgroup and p-groups have nontrivial center, G has to
be Abelian. Also, in an Abelian p-group, the elements of order ρ form a
characteristic subgroup so every element of G (Φ 1) must have order p,
that is, G is an elementary Abelian group:
G = ®ZP. (1)
2=1
We now prove that groups of the form (1) indeed possess an
automorphism φ with the required property. G can be considered as a vector space

3.1 ALGEBRA
105
of dimension η over the finite field GF(p), and automorphisms can be
identified with the invertible linear transformations of this space. Furthermore,
subgroups are the same as subspaces. Now for a subgroup Η the condition
Η ^ φ{Η) means that Η is not an invariant subspace of φ. It is well known
that a linear transformation with an irreducible characteristic polynomial
can have no invariant subspace. It is also well known that for every η there
exists a polynomial of degree η over GF(p) that is irreducible over GF(p).
Finally, it is an elementary result that to every polynomial there is a
linear transformation with the given polynomial as characteristic polynomial.
Combining these observations, we see that a φ with the required properties
exists indeed.
So the groups with the required property are the elementary Abelian
groups (including the one-element group). D
Problem A.40. Let p\ and p2 be positive real numbers. Prove that
there exist functions fi : R —> R such that the smallest positive period of
fi is pi (i = 1,2), and Д — /2 is also periodic.
Solution. If pi/p2 is rational, there exist positive integers m,n such that
npi = mp2- Let us take for г = 1,2 the functions
f 1, if x = kpi, keZ,
\ 0, otherwise.
The smallest positive period of fi is pi. But /1, /2, and consequently
/1 — /2 are also periodic with period np\ — mp2 = p.
Suppose now that P1/P2 is irrational. Define for г, j = 1,2
f aj? if x = a1p1 +
\ 1, otherwise.
a2p2, ai,a2GZ, гф$,
fi is well defined, because χ = ctipi + OL2P2 — os[pi + ^2^2 ((Xi,&i £
Z, cti Φ α'{) would imply that p\/p2 is rational.
Now fi is periodic with period pi, since
/.<*+*>-{?_7
aj = /<(ж), if x = aipi + OL2P2,
(x), otherwise.
It is clear from the definition that fi{x) = 0 ^=^ χ = otiPi, so the smallest
positive period of fi is p^ On the other hand, f\ — /2 is periodic with
period p\+P2, since
(4 /W α. α. ϊ ί("2 + 1)-(*ι + 1) =
tl-l = (/i-/2)(a:),
(α2 + 1)-(αι + 1) = (/ι-/2)(χ), if x=a1p1-\-a2p2,
otherwise. D

106
3. SOLUTIONS TO THE PROBLEMS
Problem A.41. Determine all real numbers χ for which the following
statement is true: the field С of complex numbers contains a proper subfield
F such that adjoining χ to F we get С
Solution. We will prove that a field F with the desired properties exists if
and only if either χ is transcendental or χ is algebraic but some conjugate
of χ is nonreal. First, if χ belongs to one of the classes described above,
then there is an automorphism φ of С for which χ £ φ(Κ). Then F = <£>(R)
will be the suitable proper subfield: С = F(x).
Conversely, we show that if χ is algebraic and all conjugates of χ are
real, then no subfield has the required property. Suppose on the contrary
that С = L(x) for some proper subfield L of С Then χ is also algebraic
over L, so С is a finite extension of L. We use the following well-known
theorem.
Theorem. If К is an algebraically closed field, char К = 0, and К is a
finite extension of some proper subfield L, then | К : L \= 2.
In view of this theorem, we have | С : L |= 2. Next we prove that
г ф L. Suppose we have x2 + ex + d = 0 with c, d G L and г G L. This
implies χ = с' + \fd' for some c',d' G L, and consequently С = L(d').
But this in turn implies \fd = а + by/d, with suitable a, b G L. Thus
\[d = a2 + b2d + 2aby/d, so a2 + b2d = 0 and 2ab = 1, and therefore
4a4 + 4a262d = 0 and 4a262 = 1. This gives 4a4 + d = 0, d = -4a4,
so \[d = ±i2a2, showing \fd G L. This leads to L = L(\/d), and this
contradiction proves г £ L.
So С = L(z), which implies that L is real closed, thus L can be ordered.
Let us consider A = {y G L | у algebraic}. Then A can be ordered, too.
But a well-known theorem states that every ordering of any subfield of
the field of all algebraic numbers is Archimedean, so A is a field with an
Archimedean ordering, thus there exists an embedding φ : A —► R. But
then there exists an extension φ' of φ such that φ' is an embedding of all
algebraic numbers to С Now L(i) = С implies that A(i) is the set of all
algebraic numbers. Clearly, φ'{ϊ) = ±г. It is also clear that φ' permutes
the conjugates of χ among themselves, so φ'{χ) G R. Now we surely have
φ'(χ) G 4>'{A{i)), so φ'{χ) G φ'(Α(%)) Π R = φ'(A). Thus φ'(χ) G φ'(A),
that is, χ G A, which is a contradiction, proving our statement. D
Problem A.42. Prove the existence of a constant с with the following
property: for every composite integer n, there exists a group whose order
is divisible by η and is less than nc, and that contains no element of order
n.
Solution. If η = pk(k > 2), then G = Z£ is suitable (Zp denotes the cyclic
group of order p). Suppose η = Ρι1ρ22 · · 'Prr (r ^ 2). We take
G = PSL2(q) xL·,
4 ' P1P2

3.1 ALGEBRA
107
where g is a prime satisfying q= I (mod p{) and q = — 1 (mod p2)- Then
we have
(g-l)g(g+l) η q-l g+1 1
|G| = = · q · n,
^ P1P2 Pi Pi 2,
which is divisible by η and \G\ < {q3n/piP2)- If there were an element of
order η in G, then PSL2(tf) would contain an element of order pip2- But
a well-known theorem of Dickson (see, for example B. Huppert, Endliche
Gruppen I, Springer, Berlin, 1967, Hauptsatz II.8.27., p. 213) states that
the order of any cyclic subgroup of PSL^o) divides one of g, (g — l)/2 and
(g + l)/2. As these numbers are pairwise relatively prime and pi\(q— l)/2
and P2\(q + l)/2, we see that PSL^o) contains no element of order p\P2\
consequently, G contains no element of order n.
In our construction, g was an arbitrary prime number in a prescribed
residue class mod p\P2- A well-known theorem of Linnik (see, for example,
K. Prachar, Pnmzahlverteilung, Springer, Berlin, 1957, Satz X.J^.l., p.
364) states that the smallest such prime number satisfies g < (p\P2)C with
some constant C. Then we have
\G\<(piP2)3C— = (p1P2)3C-1n<n3C. Ώ
PlP2
Problem A.43. Let A be a unite simple groupoid such that every
proper subgroupoid of A has cardinality one, the number of one-element
subgroupoids is at least three, and the group of automorphisms of A has
no fixed points. Prove that in the variety generated by A, every finitely
generated free algebra is isomorphic to some direct power of A.
Solution. We will show that
Every subalgebra of a finite direct power of A
(*)
is isomorphic to some direct power of A.
For an arbitrary natural number к = {1,..., к} and for any / С к,
denote by pr/ the projection
Ak -► A1: (αϊ,..., ак) -► (oi)ieI.
It is clear that if В is a subgroupoid of Ak then pr/5 is a subgroupoid of
A1.
Denote by U the set of those elements и of A for which {u} is a
subgroupoid of A. Furthermore, we shall use the following notation: if В С Ак,
1 ^ г < к and α^+ι,..., a^ G A, then
B(xi,...Xi,at+i,...,ajfe)
= {(xu...,Xi) e Аг \(хъ...хиаг+1,...,ак) е В} .

108
3. SOLUTIONS TO THE PROBLEMS
It is clear that if В is a subgroupoid of Ak and ai+i, · · · ,<*>к £ U, then
B(x\,... Xi, di+i,..., a^ is a subgroupoid of A%. We shall denote the
operation in the groupoid A by "o".
Our first observation is that о is surjective. It cannot be constant
because then A would have a single one-element subgroupoid. On the other
hand, the set of the elements that we get as results of the operation о is a
subgroupoid, so it can only be A itself.
Now we prove two lemmas.
Lemma 1. For every natural number η ^ 1, if В is a subgroupoid of An
with Un С В, then В = Ап.
Proof. Suppose for some η there exists a subgroupoid В of An for which
Un С В С Ап holds, and choose the minimal value of n, for which this can
happen. Clearly, η > 2. Now for an arbitrary и G U, B(xi,...,xn-i,u) is
a subgroupoid of Ап~г, containing Un~l. So, by the minimality of n, we
have B(xi,..., xn-i,u) = An~l. So U is a subset of the set
S={ae A\An~1x{a}CB},
which implies \S\ > 3. On the other hand, in view of β С An, we have
S С A. Finally, using the surjectivity of o, we can easily check that S is a
subgroupoid. The contradiction proves Lemma 1.
Lemma 2. A2 has only the following subgroupoids:
(a) subgroupoids with one element;
(b) {u} χ A, respectively, A x {и} (и е U)\
(c) automorphisms of A\
(d) A\
Proof. Suppose С is a subgroupoid of A2 not in the list above. Clearly,
pTiC = A (i = 1,2); furthermore, for every element и G С/, С(х, и) (and
similarly C(u,x)) is one of the sets Α, {υ} (υ G U). By Lemma 1, U2 % C,
so by the previous observations only the following two cases are possible:
С = ({a} x A) U (A x {b}) U С for
some C'C(A\ U)2 and a,b G U
or
С = {(щиа) \и eU}UC' for some
С С (A\U)2 and some permutation (2)
σοΐυ.
In case (1), let Τ = {a G A \ (a, a) G C}. Clearly, Τ is a subgroupoid of
A. As Τ Π С/ = {α, 6}, Τ is a proper subgroupoid, so \T\ = 1 and a = b.
By our assumptions, there is an automorphism π of A for which α φ απ.
Consider the following set:
С = {χ G A\3 у G A such that (ж,у), (χπ_1,2/π_1 G С} .

3.1 ALGEBRA
109
We can easily see that С is a subgroupoid of A and С Π U = {α, απ}. Thus
2 ^ |С| < \A\, which is impossible.
In case (2), let
D = {(x, y) G A2 | 3 ζ G A such that (x, z), (у, ζ) e С} ,
and denote by Δ the diagonal of A2, that is, Δ = {(α, α) | a G A}. It is
easy to verify that D is a subgroupoid of A2 and D C\U2 С Δ. Since,
priC = рггС = A and С itself is not a permutation, we have Δ С D.
So the transitive closure of D is a nontrivial congruence of A, which is
impossible.
Now we turn to the proof of (*). We prove by induction on η that if В
is a subgroupoid of An, then В is isomorphic to some direct power of A.
The case η = 1 being trivial, suppose η > 2. If В = An, there is nothing to
prove, and if for some index г G η we have |рг»В| = 1, then В ~ prn_{i}i?
and by induction we are ready. So we can suppose Β φ An and рг»В = А
for every г G n. Choose a minimal index set I С η such that ρτιΒ φ Α1,
and let к = \I\, I = {i\ < · · · < 4}, and С = ρτιΒ. Clearly, к > 1 and
prk_{j}C = pr/_{<i}B = Ak~l for every j'Gk. (3)
Consider the subgroupoid of A2 of the form
C(tii,...,tijfe_2,a;,2/) (1*1,.. .,и*-2 Ξ 17).
Because of (3), С(щ,..., г^_2, #, у) has the property that both projections
of it is A, so in view of Lemma 2 — it is either A2 or an automorphism
of A. Since Lemma 1 — and our assumption on С imply that Uk % C,
there exists (u[,..., г^_2) G i7/c-2 for which C(u[,..., wj._2, x, y) is an
automorphism. Suppose that there exists also some (u'{,..., u^_2) G Uk~2
for which C(u'{,..., г^'_2, ж, у) = A2. Then in the sequence
C«,..., <, u-+1,... г4_2, x, у) (г = 0,..., к - 2)
there are two consecutive terms such that the first is an automorphism of
A and the second is A2. So we can assume u'2 = u2', ... ,u'k_2 = u^_2.
Look at the subgroupoid С = C(x,u2,... ,ufk_2,y,u[) of A2. Clearly
C'(ui,y) = 1, C'(u'{,y) = A. But by Lemma 2 we see that this cannot
happen. So we know that
С(щ,..., Uk-2, x, y) is an automorphism
(4)
of A for every ui,...,Uk-2 £ U.
Now let
D ={(x,y) G A2 I 3 zi,..., Zfc_i G A, such that
(zi,...,zfc_i,x), (zi,...,Zfc_i,y) G C}.

110
3. SOLUTIONS TO THE PROBLEMS
It is easy to check that D is a subgroupoid of A2 and Δ С D (to prove
this we use the following consequence of (3): pi>C = A). So by Lemma 2
either D = A or D = A2.
Suppose that D = A2, and fix arbitrary elements u,v G U, и φ v. Since
(u, v) G D, there exist elements αχ,..., α^_χ G A such that
(αϊ,...,α*_ι,u), (ab ...,afc_bi;)eC.
So for the subgroupoid C* of Ak~l defined by
C* = C(xu ·,Xk-uu) Π (χι,...,Xfc_i, v),
we have (αχ,..., α^_χ) G С*. On the other hand, because of (4) we have
C* Π Uk~l = 0, (5)
which implies that not all of αχ,..., α^_χ are elements of U. For example,
let αχ e A\U. Since αχ is contained in the subgroupoid prxC* of A, we
have prxC* = A. So for every щ G t/, C*(iix, #χ,..., Xk-2) φ 0 holds. Let
Ζ ^ к — 2 be the greatest index such that there exist elements щ,... ,щ G U
with the property that the set
E = C*(ui,...,ui,xu..., Xk-i-i)
is nonempty. In the case Ζ < к — 2, the maximality of Z, while in the case
I = k — 2 (5) shows that (ргхЕ)ПС/ = 0. On the other hand ρΓχΕ ^ 0 since
Ε φ Φ. Thus ρτιΕ is a proper subgroupoid of A which is different from the
one-element subgroupoids {u} (u eU). This contradiction proves D = Δ.
The equality D = Δ shows that the projection С —► prk_^}C (= A^-1)
is injective, so it is an isomorphism. Consequently, the same is true for the
projection В —► prn_{ifc}#. Using the induction hypothesis, we
immediately get our claim for B. This concludes the proof of (*).
Since in the variety generated by a finite algebra each finitely generated
free algebra is isomorphic to a subalgebra of some finite direct power of the
original algebra, (*) proves the statement of the problem. D
Problem A.44. Let the finite projective geometry Ρ (that is, a finite,
complemented, modular lattice) be a sublattice of the finite modular lattice
L. Prove that Ρ can be embedded in a projective geometry Q, which is a
cover-preserving sublattice of L (that is, whenever an element of Q covers
in Q another element ofQ, then it also covers that element in L).
Solution. Let us denote by 0 the smallest element of P. Let L' = {a G
L | 0 < a}. Then L' is a (clearly modular) cover-preserving sublattice of L.
Evidently, it is enough to solve the problem for L' in place of L. Let Q be
the sublattice of L' generated by those atoms of V which are less than \p
(lp denotes the greatest element of P). Let δ be the dimension function

3.1 ALGEBRA
111
on Q. Take a q G Q. Then there is a finite set αϊ,..., αη G Q of atoms of
Z/ such that 1q = <? V αϊ V · · · V αη and such that the value of η is minimal.
Then
δ(αι V··· Van) = η
(Q being modular) and 6(1q) = 6(q) + n. Using the well-known relation
6{x V y) + 6{x Л у) = δ(χ) + «(у), (1)
we get <$(<7 Λ(αι V · · · V αη)) = 0, that is, a\ V · · · V αη is a complement of
q, and therefore Q is a projective geometry. Now let /3 G Q be an atom of
Q. Then there exists an atom α G Z/ of Z/ with α ^ /3(^ lp). So α G Q,
and since /3 was an atom, /3 = a, and thus /3 is an atom in Z/, too.
Now, if <7i,<?2 G Q and <?i -<q #2 then there is an atom β G Q such
that q2 = qi V/3 (using the fact that Q is a projective geometry), but then
the modularity of Z/ implies that q\ -<l> q% also holds. So Q is indeed a
cover-preserving sublattice of Z/.
We are going to embed Ρ in Q. For an element ρ G Ρ let /(ρ) = \/{a G
Z/ I a ^ p; a is an atom}. The map / is injective. If namely for two
elements pi,p2 G Ρ we have p\ ^p<i, then there is an element τ G Ρ such
that r ^ pi, and г Лр2 = 0. Then there is an atom α G Z/ with α ^ г and
thus α < /(pi) and αΛ/(ρ2) < гЛр2 = 0, proving /(pi) ^ /(p2).
Next we prove 7(piVp2) = 7(pi)V/(p2). The definition of / gives
immediately 7(piVp2) ^ 7(pi)V/(p2). To prove the inequality in the
other direction, let p2 < p2 be an atom of Ρ such that pi Vp2 — Pi V.P2
and pi Λ Pa = 0. Clearly, /(p2) < /(p2) and /(pi Vp2) = /(pi Vp2), which
means that for proving the inequality in the other direction we can assume
Pi Λρ2 = 0. Let а < pi VP2 be an atom of Q. We have to prove а <
/(pi)V/(p2). In case α ^ p\ or а ^ Р2, this is clear. Suppose αΛρι =
аЛрг = 0. Let αϊ = ριΛ(ανρ2) and a2 = P2A(aVpi). Using the
modularity of Z/, we see that αϊ and a2 are atoms and α Vp2 = αϊ VP2
and α Vpi = a2 Vpi are true. Again using the relation (1) (where δ denotes
the dimension function of V) for the elements χ = p\ V α and у = p^ V а,
we get that the dimension in V of the element (pi Va)A(p2 Va) is two.
Since αϊ φ ot^ and αι,α2 ^ (pi Va) л(р2 Va), we see that ot\\l a^ =
(pi V а) Л(р2 V a). But this implies a < ai V a2, thus a < /(pi) V /(P2) ·
It remains to be proved that /(ρι)Λ/(ρ2) = f{p\/\P2)· It is evident
from the definition of / that /(Ρ1ΛΡ2) ^ 7(Ρι)Λ/(ρ2). Since Q is a
projective geometry, it is enough to prove that if β ^ /(pi) Л/(рг) is an
atom, then /3 ^ /(pi Лрг). But if for an atom /3 we have /3 ^ f(pi) and
/3 < /(P2), then /3 < pi,/3 < P2, and consequently /3 < Pi Лр2, therefore
/3 < /(pi ЛР2), concluding the proof. D
Problem A.45. Let G be a finite AbeJian group and x,y e G. Suppose
that the factor group of G with respect to the subgroup generated by χ
and the factor group ofG with respect to the subgroup generated by у are
isomorphic. Prove that G has an automorphism that maps χ toy.

112
3. SOLUTIONS TO THE PROBLEMS
Solution 1. It is enough to deal with the case where G is a p-group.
Denote the subgroup generated by χ (resp., y) by X (resp., Y). Suppose
αχ,..., an is a base of G. Then every element of G can be written uniquely
in the form k\a\ -\ (- knan, where 0 ^ ki < o(aJ. For the given element
#, choose a base so that the number of nonzero coefficients ki should be
minimal. By multiplying with suitable integers relatively prime to p, we
can achieve that all these nonzero coefficients are powers of p; let us arrange
them in decreasing order:
x=peia1 + ---+pekak, (1)
where e\ > e2 ^ · · · ^ ek ^ 0 (k < n). Let us denote the order of a^ by pdi
if г > к and by pei+f^ where fi > 0 if г ^ к.
We show that e\ > e2 > ··· > ek. Suppose, on the contrary that
ei = e^+i, and let us say fi ^ fi+\ holds. Then, by replacing a^ by
di +ai+i, we would get a base giving fewer nonzero terms in representation
(1). Next we show /i > /2 > ··· > Л- Otherwise, assuming fi ^ /z+i,
replacing ai+\ by p€i~€i+1ai + a^+i we would get a base giving less nonzero
terms in the representation (1).
Now we determine the invariants of G/X (that is, the orders of the cyclic
groups whose direct sum is isomorphic to G/X). Take
a'i = Pei~eia>i + ·'' +pei-1-eiai-i + a* (2)
for г ^ к and o!i = di for г > к. Then α'1? α2,..., α!η is a set of generators.
Furthermore, p^+f^a'i = p^i+1x G X, so the order of the image of o!{ in the
factor group is at most ρβ*+^+ι (we define fk+i to be 0). Since o(x) = pf1,
we get that the images of the elements a[ form a base of G/X and their
respective orders are
„ei+.fc „ег+Уз „ек-i+fk „ek „dk+ι γΛη
У ι У j · · · j У ? У j У ? · · · j У
Therefore, if we know the isomorphism type of G and G/X, we can
uniquely determine the exponents /1, ei,..., Д, e^, namely, we omit the
common invariants and we use the fact that
ei + /1 > ei + /2 > e2 + /2 > e2 + /3 > · · · > e*_i + fk > ek + fk ^ 0 .
Consequently, if G/X and G/У are isomorphic, then in a suitable base
with o(bi) = o(di) we have
y=peib1 + ---+pekbk
with the same exponents ei,..., ek as in (1).
This means that the automorphism of G mapping a^ to bi maps χ
toy. D
Solution 2. We assume again that G is a p-group and proceed by
induction. We distinguish two cases according to whether X is contained in pG
or not.

3.1 ALGEBRA
113
In the first case, we have p{G/X) = pG/X and {G/X)/p{G/X) ~
G/pG. In the second case, p{G/X) = (pG + X)/X and {G/X)/p{G/X) ~
G/(pG + X) has an order smaller than G/pG; furthermore, pG/pX =
pG/{XC\pG) ~ (pG + X)/X = p(G/X). Since G/X ~ G/Y, either the
first case or the second case applies to both of them.
In the first case, pG/X = p{G/X) ~ p{G/Y) = pG/Y, so by the
induction hypothesis there is an automorphism of pG mapping χ to y.
It is easy to see that every automorphism of pG can be extended to an
automorphism of G so in this case the proof is finished.
In the second case, pG/pX ~ pG/pY so by the previous statements we
can assume px = py. It is easy to see that in the elementary Abelian group
G/pG (which is actually a vector space over the field with ρ elements) there
is a maximal subgroup Μ containing the image of neither χ nor y. We have
\G : M\ = ρ, Μ Π Χ = pX = pY = Μ Π У, and G = Μ + Χ = Μ + Υ.
Now it is clear that G has an automorphism that acts as the identity map
on Μ and maps χ to у. П
Remark. For p-groups a third possible way to solve the problem is to
prove that the automorphism exists iff h(pkx) = h(pky) for each fc, where
h(g) = h means that there is an element ζ of the group such that phz = g
and there is no ζ such that ph+1z = g (h is the "height" of g). An integer d
is called a divisor of the element g of an Abelian group if there is an element
ζ of the group such that dz = g. It is easy to see that for finite Abelian
groups the above statement means that the automorphism exists iff the
divisors of χ and the divisors of у are the same. One of the contestants,
Balazs Montagh, later proved the following generalization: If G is a finitely
generated Abelian group and x\, x^ - - -, xn, УъУч, · · · ? 2M £ G, then there
exists an automorphism f οι G such that f(xi) = у ι for each г iff the
divisors of a\X\ + (- anxn and the divisors of o.\y\ Л V anyn are the
same.
Problem A.46. Let ρ be ал arbitrary prime number. In the ring G of
Gaussian integers, consider the subrings
An = {pa + pnbi : a, b G Ζ}, η = 1,2,... .
Let R С G be a subring of G that contains An+i as an ideal for some
n. Prove that this implies that one of the following statements must hold:
R = An+1; R = An; or 1 G R.
Solution. Since ρ G Αη+\ and Αη+\ is an ideal in R, we have for any
χ = α + Ы G R, pa + phi = p(a + Ы) G Αη+\. This implies pb = pn+16/,
that is, b = pnb' for a suitable integer b'. Therefore, all elements of R are
of the form χ = a + pnbi.
Now we distinguish two cases.
First, let us assume that for all elements χ = a +pnbi of R we have ρ \ a.
Then a = pa' for a suitable integer a', that is, χ = pa' +pnbi G An. Since

114
3. SOLUTIONS TO THE PROBLEMS
x is arbitrary, this shows R ^ An. Clearly, An+i is an ideal in An, and the
factor ring An/An+i has ρ elements. Thus An+i ^ R ^ An implies either
R = An+i or R = An.
Now let us assume that there exists an element χ = a + pnbi in R such
that ρ \ a. Then for suitable integers и and ν we have
pu + av = 1.
This implies
xv = av + pnbvi = 1 — pu + pnbvi.
In view of ρ G Αη+ι, this gives
1 + pnbvi = xv + pu e R + An+i = Д .
With the notation ζ = xv + pu, we have
pnbvi = ζ — 1,
and therefore
Since η ^ 1 and ζ G ϋ, we finally get
1 = -p(p2"-ibV) - z2 + 2z G An+l + R + R = R ,
that is, 1 e R. D
Problem A.47. Let η > 2 be an integer, and let Ωη denote the
semigroup of all mappings д : {0,1}η —> {0, l}n. Consider the
mappings f G Ωη, which have the following property: there exist mappings
Qi : {0, l}2 —> {0,1} (г = 1,2,..., η) such that for all (αϊ, α2,..., αη) G
{0,1}η,
/(аь α2,..., αη) = (#ι(αη, αϊ), д2(а>1,а2),..., #η(αη_ι, αη)).
Let Δη denote the subsemigroup ofVin generated by these f's. Prove that
Δη contains a subsemigroup Tn such that the complete transformation
semigroup of degree η is a homomorphic image of Yn.
Solution. We prove a generalization of the problem, namely the one when
{0,1} is replaced by any group (G, +) of order at least two. (The set
{0,1} with addition mod 2 is such a group.) This generalization causes no
additional complication.
So let Ωη be the semigroup of all mappings Gn —> Gn with the operation
f,g G Ωη, αι,...,αη G G: (fg)(au · · · ,an) = g (/(ab ... ,an)).

3.1 ALGEBRA
115
Let Η denote the set of those mappings / G Ωη for which there exist
mappings #1,..., gn : G2 —> G such that for every ai,.. .,an e G,
/(ab ...,an) = (#i(an,ai),g2(ai,a2),... ,#n(an-i,an)) ·
Let Δη denote the subsemigroup of Ωη generated by H. Finally, let Tn
denote the semigroup of transformations p: {1,... ,n} —> {1,... ,n} with
the operation
p,qeTn, l^i^n: (pq)(i) = q(p(i)).
Let us consider the mapping ψ: Tn —> Ωη defined by
peTn, ai,...,an eG : ψ(ρ)(αι,... ,an) = (ap(1),... ,ap(n)).
We shall prove that ^ has the following properties:
(1) φ is a homomorphism;
(2) ψ is injective;
(3) ψ(Τη) С Δη.
Then ^(^n) = Γη will be a subsemigroup of Δη, and in view of the
above-mentioned three properties the inverse of φ maps ψ(Τη) homomor-
phically onto Tn, which was the statement of the problem.
Let us now prove properties (1), (2), and (3).
(1) Let p, q e Tn, ab ..., an e G. Then
ip{pq){ai,...,an) = (α(ρς)(ΐ), · · · ,α(ρς)(η))
= K(p(i))> · · ·' ая(р(п))) = ^W(a(p(i), · · ·, ap(n))
= ^Ы^ЫС^ь · · ·, an)] = [V>(pM?)](ai, · · ·, an),
which shows that φ is a homomorphism.
(2) Let ρ and q be different elements of Tn. We show that ψ{ρ) φ il>(q); in
other words, φ is injective. ρ φ q means that there is an г (1 < г < η)
for which ρ(ι) φ q(i). Since G has at least two elements, it has elements
αϊ,..., an such that ap^ φ aq^y But then
φ(ρ)(α1,...,αη) = (α(ρ(1),...,αρ(η))
φ (α(σ(1),..., aq{n)) = x/;(q)(au ..., αη).
(3) The proof that, for ρ eTn, ψ(ρ) G Δη holds will be broken down into
the following steps (a) toy (e).
(a) First, we prove the statement for the case when ρ is a transposition
interchanging two (cyclically) neighboring element, that is, one of
the transpositions (1,2), (2,3), ... ,(n — l,n), (n, 1). For instance,
let ρ = (1,2). For the other cases, the proof is similar. We represent
the mapping
ψ(ρ){αι,α2,α3,...,αη) = (α2,αι,α3,... ,αη)

116
3. SOLUTIONS TO THE PROBLEMS
as a product of elements of H. In what follows, '—►' always denotes
a mapping in H.
(αχ, α2, аз, α^,α^,... ,αη) —>
—► (-αη,αι,α3 -\-α2,α4 -α3,α5 - α4,...,αη - αη_ι) —►
—> (-αη,αι,α3 + α2,α± + а2,аъ - α4,... ,αη - αη_ι) —>...—>
—> (-αη,αι,α3 + α2,α4 + а2,аъ + α2,... ,αη + α2) —►
—► (^2,^1,^3 + ^2,^4 + α2,α5 + α2,...,αη + α2) —>
—> (α2,α2 + αϊ,аз + ^2 + αι,α4 — аз,а$ — α4,... ,αη — αη_ι) —>
—> (α2,αι,α3,α4 — a3,as — α4,... ,αη — αη_ι) —>
—> (α2,αι,α3,α4,α5 — α4,... ,αη — αη_ι) —>...—>
—> (α2,αι,α3,α4,α5,... ,αη).
This procedure is correct for η > 5, and for η < 5 we can use a
similar but simpler procedure,
(b) If ρ is an arbitrary permutation, then it can be represented as a
product of transpositions interchanging neighboring elements. Since
^ is a homomorphism and the images of these transpositions are
products of elements in Я, the same is true for ψ(ρ).
c) Suppose that ρ G Tn is such that p(i) φ % for some г, but p(j) = j
for every j φ г. Without loss of generality, we can assume that
p(i) = 1. Let us consider the following representation:
(αι,α2,α3,... ,α»_ι,αΐ,αΐ+ι,... ,αη) —>
—> (αχ,α2 + αι,α3 — α2,... ,α^-ι — α;_2, —α^ι,α^ι,... ,αη) —>
—> (αι,α2 +αχ,α3 + αι,... , а^_1 — a;_2, — а»_1,а»+1,... ,αη) —>
—► (аьа2 + аьа3 + ab...,ai_i + αϊ, -α»_ι,αί+ι,... ,αη) —►
—► (αι,α2,α3 — α2,... , a^_i — αΐ_2,αι,α»+ι,... , αη) —>
—> (αι,α2,α3,... ,α^-ι — αΐ_2,αι,α»+ι,... ,αη) —>···—>
—> (α1,α2,^3, · · · ,^г-1,а1,аг+1, · · · , ^η) ·
The case г = 2,3 can again be handled in a similar but simpler way.
d) Now let ρ be a transformation that is identical on its image set:
p(p(i)) = р(г) (г = 1,..., η). Let pk,i denote the following
transformation of type (c):
pkti(i) = fc, РкЛз) = h if JV * ·
Then this ρ can be represented in the following form:
Ρ=Ρρ(1),1·.·Ρρ(η),η.
To prove this, observe that on the right-hand side the first
transformation from the left that changes г is рр{%)^ and it maps г to p(i).

3.1 ALGEBRA
117
If some later pp(k),k {к > г) changes this element, then necessarily
к = p(i). But then pp(k),k maps к = p(i) to p(k) = p{p{i)) = p(i)\
thus, nevertheless, it actually fixes p(i).
Now in view of (c), each of the pp(k),k can be represented as a
product of elements of if, so ρ has such a representation as well,
e) To conclude the proof, let ρ be an arbitrary transformation from Tn.
We shall show that there is a permutation q such that r = qp is the
identical map on its image set. This will establish the statement,
since in this case ρ = q~lr (a permutation has an inverse), ψ(ρ) =
/0(ς,_1)ν,(Γ)? and (b) and (d) show that il>{q~l) and tp(r) belong to
Δη, so the same is true for ψ(ρ).
To construct the suitable q to the given p, we use induction. Let
Po = ρ and q0 be the identical permutation. Suppose that we have
already constructed pi and <ft in such a way that qi is a permutation
and if к ^ г belongs to the image of p, then Pi(k) = k. Now if
г + 1 does not belong to the image of p, let ρι+\ = pi, <ft+i = <ft.
However, if г + 1 does belong to the image of ρ then it also belongs
to the image of pi, so Pi(j) = г + 1 for some j and in this case
let pi+i = (i + lj)pi, qi+ι = (i + 1, j)ft> where (г + 1, j) denotes
the transposition of г + 1 and j. Then qi+\ is again a permutation,
Pi+i(i + 1) = г + 1, and if for some element к < г + 1 of the image
of ρ it was true that pi(k) = k, then pi+i(k) = к also holds. The
reason for this is that к < г + 1 trivially implies к ф г + 1 and к = j
cannot hold either since Pi{k) = к < г + 1 = Pi(j). Finally, we
choose q = qn. Then q is a permutation, and pn = qnp = qp is the
identical map on its image set. D
Problem A.48. Let η = pk (pa prime number, к > 1), and let G be
a transitive subgroup of the symmetric group Sn. Prove that the order of
the normalizer ofG in Sn is at most \G\k+1.
Solution. First, we show that if Τ is a minimal transitive subgroup of
G, then Τ can be generated by к elements. Since Τ is transitive, we have
pk | \T\. Let Ρ be a Sylow p-subgroup of T; we are going to show that Ρ
is transitive, thereby proving that Τ is necessarily a p-group.
Denoting the stabilizer of a point α in the group G by Ga, we have
| G : Ga | = pk, so
pk | | G : Ga | · | Ga : Ga Π Ρ \ = | G : Ρ | · | Ρ : Ga Π Ρ | .
Since (| G : Ρ \ ,p) = 1, we get pk \ \ Ρ : Ga Π Ρ |, that is, pk \ \ Ρ : Pa |,
proving that Ρ is indeed transitive.
Now let Μ be a maximal subgroup of T. Then Μ is not transitive, so
MTa cannot be transitive either. Therefore MTa = M, that is, Ta < M.
This shows that Ta is contained in the intersection of all maximal subgroups
of T, the Frattini subgroup of T. Therefore | Τ : Φ (Τ) | < | Τ : Ta | = pk,

118
3. SOLUTIONS TO THE PROBLEMS
so by a well-known theorem of Burnside, Τ can indeed be generated by к
elements: Τ = (gu ... ,gk).
We next observe that the elements of N(G), the normalizer of G in £n,
act by way of conjugation on the elements of G, permuting the elements
of G among themselves. The number of possible images of the k-tuple
(<7ъ · · · j 9k) is at most \G\ . Conjugation by the elements n,m G N(G) has
the same effect on (#i,..., g^) if and only if n~1m G C(T), the centralizer of
Τ in Sn. Thus \N(G)\ < \G\k \C(T)\. But the group C(T) is easily shown to
be semiregular. Suppose that s G C(T) fixes the point a and for any other
point β choose a t G Τ with at = β. Then β = at = ^sts'1 = (/3)s~\
that is, /3s = /3, so s is the identity, proving semiregularity. This shows
|C(T))| < pk, consequently
|iV(G)|<|G|fc./<|G|fc+1 . D
Problem A.49. Prove that if a unite group G is an extension of an
Abelian group of exponent 3 with an Abelian group of exponent 2, then G
can be embedded in some finite direct power of the symmetric group S3.
Solution. Applying the Schur-Zassenhaus theorem, we see that the
extension is actually a split extension, that is, a semidirect product. Using the
fundamental theorem of finite Abelian groups, we get that G actually has
the form G = ZJ1 χ ZJ, where ξ is a homomorphism ξ : Z£ -> Aut(Z^) =
GLm(3). (GLm(3) is the group of automorphisms of Z31, considered as a
vector space over the field of three elements.) Applying Maschke's theorem
to the representation ξ, we see that there is a basis 61,62, - - -, &m of the
vector space Z31, the elements of which are eigenvectors of all elements of
£(Z2). Let αϊ, a2,..., an be a basis in Z^, then bj is an eigenvector of ξ(αι)
with eigenvalue c^· G {1,2}, say. Denote the elements (1,2,3) and (1,2) of
£3? by σ and r, respectively. For an arbitrary b = Σ™=λ \bi G Z31, let
6=(aAl,...,aA-,l,...,l)G53m+n,
and let
а< = (гс'1-1>гед-1>...,гс^-1>1,..., ?,i,...,i)GS3™+n.
Then В = {b I b G Ζψ} is a normal subgroup of 5^+n, isomorphic to Z£\
The mapping щ \-> hi can be extended to a homomorphism Щ —> S^"1"71,
and then A = {a \ a G Ζξ} is a subgroup of S^"1"71, isomorphic to Zg.
Let us consider the homomorphism 77 : A —> Aut(£) mapping each ж G A
to the conjugation by ж. Then for the subgroup В A of SJ1"1"71 we have
В А ~ В y\ A. On the other hand, for any aeZJ and b G Z31 we have

3.1 ALGEBRA
119
(it is enough to check this for basis elements b{ and dj). This shows that
indeed G ~ В A. D
Remark. Instead of referring to the Schur-Zassenhaus theorem, we can
simply consider a Sylow 2-subgroup.
Problem A.50. Let η > 2 be ал integer, and consider the groupoid
G = (Zn U {oo}, o), where
_ J x + 1 if χ = у е Zn,
\ oo otherwise.
(Zn denotes the ring of the integers modulo n.) Prove that G is the only
subdirectly irreducible algebra in the variety generated by G.
Solution. (Q will denote the groupoid G considered as a member of the
variety of the problem.)
The unary operation taking the constant value oo is a polynomial
expression in С}, for example,
oo = (xoi)oi (0)
(This unary operation will also be denoted by oo.)
Furthermore, it is easy to check that for the polynomial expressions
f(x)=xox and x\J У = fn~l{x°y) (1)
of Q the following hold:
V is a semilattice operation and oo is (2)
the greatest element with respect to it,
/ is an automorphism with respect to W, (3)
and the following identities hold:
fn{x)=x, fk{x)\Jfl{x) = oc if кф\ (0<ft,Z<n). (4)
It is also true that the original operation о can be expressed using V and
/ since the identity
xoy = f(xVy) (5)
holds true in Q.
Now take any algebra С = (С; о) in the variety generated by £?, and
consider the algebra С = (С; V, /, oo) whose basic operations are the
polynomial expressions of С defined according to (0) and (1). As all the properties
discussed above are described by identities, they hold simultaneously for С
(resp. C). Identity (5) (together with the definitions included in (0) and

120
3. SOLUTIONS TO THE PROBLEMS
(1)) ensures that a map φ : С —> Zn U {oo} produces a homomorphism
φ : С —> Q if and only if φ : С —> £' is a homomorphism. So instead of Q
and C, we can work with the algebras Q' and C, which can be handled more
comfortably. Also denote by < the natural partial order of the semilattice
(C;V).
Let a be an arbitrary element of the set C, different from oo, and consider
the following subsets of C:
а = {сеС\с^Г(а)} (iezn), C00 = C\|JC,i.
iezn
We will show that {Co,..., Cn_i, C^} is a partition of С. Clearly, the
union of these sets is C, and none of them is empty, since fl(a) e Ci (г е
Zn) and oo G COo · If с G Ci Π Cj for some 0 < i ^ j < n, that is, с < /г(а)
and с ^ p (a), then using (2) and (3) we get
/'"'(с) Vе < Р~ЧГ(а))\/Г(а) = fj(*)\/fj(a) = fj(a) < oo,
which, in view of (4), implies j — г = 0. So Co,. ·., Cn_i are pairwise
disjoint, and C^ is clearly disjoint from each of them.
So the following map φ : С —> 5' is well defined:
c<pa = i if ceCi.
Clearly, y?a is surjective, it is permutable with the operation /, and
oo<^a = oo. For arbitrary c,d Ε С and Ζ G Zn, we have
(c\Jd)(pa = l <=> c\/d^fl(a) & c,d^fl(a) & οφα = 1,άφα=1.
This implies that φα is permutable with V too, so φα : С —> Q' is a
surjective homomorphism.
As αφa = 0, we have for every element b G С the following: α<^α =
6<^α <Φ=> b ^ a. So if a, 6 are different elements of the algebra С (say
а ф oo), then in case ft^awe have αφα ф b(pa, whereas in case b < a (this
implies b ф oo) we have αψι φ Ь<ръ. Consequently, the homomorphisms
(^?α(α G C\{oo}) give a representation of С as a subdirect power of Q'. This
shows that С can be subdirectly irreducible only in case it is isomorphic
tog7.
It is easy to check that the algebra Q' is simple, so necessarily
subdirectly irreducible. Namely, if for some congruence relation σ of Q' we have
α σ b,a фЬ (say α φ oo), then a = a\/ α σ a\/ b = oo so ΐοτ any 0 ^ к < η
we have fk(a) σ fk{oo) = oo. D

3.2 COMBINATORICS
121
3.2 COMBINATORICS
Problem C.l. Among all possible representations of the positive
integer η as η = Σί=1 di with positive integers k, αχ < α2 < · · · < α^, when
will the product Π*=ι αΐ oe maximum ?
Solution. First, we are going to investigate the properties of the extremal
integer sets. Suppose that An — {αϊ, α2,..., α^} is extremal.
Claim 1. There are no integers i and j such that a\ < i < j < a^ and
г ψ An·) J ψ Αη.
If it is not the case, then there are some elements ar,as e An such that
ar + 1 ^ An, as — 1 £ An, and ar + 3 < as. Then
(ar + 1)(cls — 1) = aras -\- (as — ar) — 1 > aras -\- 2 > aras,
so replacing the elements ar, as of An by ar + 1 and as — 1, the sum of the
elements of An does not change, but their product increases, a contradiction
to the extremality of An.
Claim 2. 2 < ax if к > 1.
If ai = 1, then replacing the elements αι,ακ of An by a^ + 1 gives us а
"better" system, a contradiction to the choice of An.
Claim 3. a\ = 2 or a\ = 3 if η > 5.
If αϊ > 4, then replacing the element αϊ of An by 2 and αϊ — 2, the sum
of the elements of An does not change, but their product increases since
2(αι - 2) = 2ai - 4 = ai + (ax - 4) > ab
a contradiction to the extremality of An.
If αϊ = 4, then к > 1 by η > 5, and replacing the elements αϊ,α2 of An
by 2, αϊ — 1, and a2 — 1, the sum of the elements of An does not change,
but their product increases, since
2(αι - l)(a2 - 1) = 2axa2 - 2(ax + a2) + 2 = αλα2 + (аг - 2)(a2 - 2) - 2
> aia2 + 6 — 2 = aia2 + 4 > aia2,
a contradiction to the extremality of An.
Claim 4- If ai — 3 and г is an integer such that αϊ < г < α^, г £ Ап
then г = аь — 1.
If not, then i + 2 e An Ъу claim 1, and replacing г + 2 by 2 and г, the
sum of the elements of An does not change, but their product increases,
since
2г>г + а1 + 1 = г + 4>г + 2,
a contradiction to the extremality of Д

122
3. SOLUTIONS TO THE PROBLEMS
Now, we are ready to determine the extremal sets An. Let A(i,j, I)
denote the set {г,г + 1,...,/- 1,/ + 1,... ,г +j - 1,г + j}, and let s(i,j,l)
denote the sum of the elements of A(i, j, I). For к = 2,3,..., let 5(2, к +
2,fc + 2) =2 + 3+··· + Α; + (Α;+1) = (^-l = Lk. By claims 1-4, An has
at least two elements if η > 5, and if к > 2 then the possible extremal sets
are as follows: A(2, fc + 2, fc + 2), A(2, fc + 2, fc +1),..., A(2, fc + 2,3), A(2, fc +
2,2),A(3,fc + 3,fc + l). Obviously, 5(2, к + 2,fc + 2) = L*,s(2,fc + 2, fc +
1) = Lfc + l,...,s(2,fc + 2,3) = Lfc + (fe-l),e(2,fe + 2,2) = L* + к and
5(3, A: + 3, к + 1) = Lk + (fc + 1) = £fc+i — 1, so the sums of the elements of
the above sets are the integers in the interval [Lk,Lk+i), and each integer
appears exactly once. Thus, for any η > 5, we have exactly one set of
type above and this is the extremal set An for this n. If 1 < η < 4, then
obviously An = {n} is the only extremal set. D
Remark. Consider the following generalization of the problem: Let f(x)
be an arbitrary function that is strictly concave from below in the interval
(0,+00) (f(x) = log χ in the problem above). Let η = Σί=1α», where
α,ι < a,2 < · · · < a>k are positive integers, к is not fixed. When will the
sum Σ%=ι f(ak) be maximum for a given n? It can be proved that also in
this case every extremal set can be obtained from a sequence of consecutive
natural numbers deleting at most one element.
Problem C.2. For every natural number r, the set ofr-tuples of
natural numbers is partitioned into finitely many classes. Show that if f(r) is
a function such that f(r) > 1 and linv^oo f(r) = +00, then there exists
an infinite set of natural numbers that, for all r, contains r-tuples from
at most f(r) classes. Show that if f(r) -/+ _|_оо^ then there is a family of
partitions such that no such infinite set exists.
Solution. Let N denote the set of natural numbers. We will use the
following well-known theorem of Ramsey. If the r-tuples of natural numbers
are divided into finitely many classes, then there is an infinite subset N' of
N such that every r-tuple in N' is contained in the very same class.
To prove the first statement, we define a sequence No 2 ' · · 2 Nr D ...
of subsets and a sequence χι,..., xr,... of natural numbers by induction
on r. Let No = N. Suppose that r > 0 and that Nr and Xi (г < г)
are defined, Nr is infinite. Let xr be an arbitrary element of Nr and let
Nfr = Nr — {#1,..., xr}. For any set А' С {#!,..., xr}, divide the (r — s)-
tuples of N'r into finitely many classes where s = |A'\. Put two (r — s)-tuples
into the same class if and only if adding the elements of A' to them the
resulting r-tuples are in the same class of the original partition. Applying
Ramsey's theorem 2r times, we get that there is a subset Nr+i of the set
N'r such that
\A\ = \B\ = r, (1)
А, В С {ж1,...,жг}иМг+1, AC\{xi,...,xr) = В П {жь ... ,xr} (2)
imply that A and В are in the very same class.

3.2 COMBINATORICS
123
So, we defined the sequences No 2 * * * 2 Nr D ...; xi,...,ir, Let
X = {#1,..., xr,... }. Now, (1) and (2) imply that X has the following
properties.
(*) Suppose that \A\ = |B| = г, А,В С X, AC\ {a?i,... ,xr} = В П
{#ι,... ,#r} for any r. Then A and Б are in the very same class.
Since f(r) > 1 and f(r) —> +oo as г —> +oo, thus there exists a
monotone subsequence xTk such that
\{xrk: rk<r}\ <log2/(r).
Let X' = {x ri, хГ2,... }. This set X' is obviously infinite and meets the
requirements of the problem by (*).
Now, we are to prove a statement that is stronger than the second part of
the problem. We show that a set of cardinality continuum has the desired
partition, as well. Let S = [0,1]. We divide the set of r-element subsets
of S into r classes for any r. Let X = {xo,... ,xr-i} be an r element
subset of S with xq < · · · < xr_\. We put X into the zth class if exactly
г of the intervals (xj,Xj+i) (j = 0,1,..., r - 2) are longer than 1/r. If
S' is an arbitrary infinite subset of 5, then using the fact that S' has an
accumulation point, it is easy to see that for any г there is a number tq
such that for r > r$ the set S' contains a set out of the zth class of the
r-tuples, and so our proof is complete. D
Problem C.3. Let η and к be given natural numbers, and let A be a
set such that
w <- m?-
For i = 1,2,...,η+ 1, let Ai be sets of size η such that
n+1
A - (J A.
г=1
Determine the cardinality of A.
Solution 1. Let фх denote the number of sets Ai containing the point x.
Obviously,
n+l
J2 Φχ = ^{AiHAjl -η + ^|ΛΠ^·|,
xEAj i = l ίφϊ
and so
Σ Фх<п + пк = n(k+ 1). (1)
xeAj
Add these up for all j. On the left-hand side, we get
j=l xEAj xEAxEAj xEA

124
3. SOLUTIONS TO THE PROBLEMS
Estimating it by the inequality between arithmetic and quadratic means,
we get
x€A
Adding up the right-hand sides of (1) for all j, we get n(n + l)(k + 1).
Thus,
n2(^1)2<n(n + l)(fc+l),
that is,
n(n + 1)
\л\>
fc+1
Since the opposite inequality was supposed, it implies the equality above.
Naturally, к + l\n(n + 1) is needed for the existence of such a family of
sets. D
Solution 2. We will use the following theorem: Let Fi,..., Fn be
arbitrary polynomials of some events A\,..., An, and let c\,..., cn be arbitrary
real numbers. Then, the inequality
N
^2ckp(Fk)>0
k=l
holds in any probability space and for any events Ai,...,An provided that
it holds in the trivial probability space.
Thus, it is sufficient to verify that the inequality
ρ{Αιυ··-υΑ^1)>^^^ρ(Αί)--^ψ Σ P(AinAj)
holds when Ai = 0 or Ω. If Ai = 0 for all г, then we get 0 > 0, which holds.
If exactly / of the A^s are Ω (I > 1), then we have to prove that
ι>^±4ι
-©■
(fc+1)2 (fc+l);
that is,
I2-I 2fc+l л -(A iV>n
(fc + 1)* (fc + i)2z + 1~U + i 7
which holds as well. Now, let Ai (i = 1,..., n+1) be the sets given in the
problem, and consider the probability space such that Ω = A = U^1^

3.2 COMBINATORICS
125
and the probability of each point is l/\A\. So, even in this space,
nr/n = i> 2fc+1 ψ№ 2 ν l^n^l
2fc + 1 (η + 1)η 2 /n+l\ fc
-(*+ΐ)*~~μΓ~~(*+ϊ)4 2 Jr·
Reducing this, we obtain that \A\ > n(n + l)/(^ + 1)· We supposed that
\A\ < n(n + l)/(fe + 1), so \A\ = n(n + l)/(fe + 1). D
Problem C.4. Let Αι, Α2,... be a sequence of infinite sets such that
\Ai Π Aj\ < 2 for г φ j. Show that the sequence of indices can be divided
into two disjoint sequences i\ < i% < ... and j\ < J2 < · · · in such a
way that, for some sets Ε and F, \Ain Π E\ = 1 and \Ajn Π F\ = 1 for
n=l,2,....
Solution. Suppose that for к > 3, there are some finite disjoint sets Ek
and Fk such that
either \A{ nBfc| = l or \A{ Π Fk\ = 1, (1)
if г > fc then |A< ПЕк\<1 or |A< Π Ffc| < 1. (2)
For к = 3, it is easy to construct such sets E$ and F3. If we show that
there are some finite disjoint sets Ek+\ 2 Ek and Fk+i 2 -Fife satisfying (1)
and (2) for Afc+i, then the sets Ε = U™=1Ek F = U™=1Fk have the desired
properties.
First, suppose that Ak+ι meets both Ek and Fk· According to (2), it is
possible only if, say, \Ak+\ Π Ek\ = 1, and then Ek+i = £k and i^+i = F*
satisfy the conditions (1) and (2).
Now, suppose that Ak+ι does not meet at least one of the sets Ek and
Fk, say, Ek. Consider all the sets Αι that meet Ek U Fk in at least three
elements. Since the given three elements can be contained in finitely many
sets Ai only, the number of these sets is finite. Let us denote them by
Aix,..., Aim. Let efc+i denote an arbitrary element of the infinite set
m к
Afc+1-(U^jU((j^)|j^
r=l г=1
and let Ek+i = Ek U {ε^+ι}, Fk+i = Fk. These sets satisfy (1) obviously.
Furthermore, if г > к + 1 and e^+i ^ A», then (2) holds as well, and if
ek+i G Ai then A» differs from the sets A^,... ,Aim by the choice of e^+i,
and so \Ai n(EfcUFfc)|<2, which yields (2). D

126
3. SOLUTIONS TO THE PROBLEMS
Problem C.5. Let η > 2 be an integer, let S be a set of η elements,
and let Ai, 1 < г < га, be distinct subsets of S of size at least 2 such that
Ai Π А0 ф$,А{Г\Акф 0, Aj Г\Акф$ imply Ai Π Ad Г\Ак^$.
Show that m < 2n"1 - 1.
Solution 1. We will prove the statement by induction on n. It obviously
holds for η = 2. Assume that η > 2, and let A\ φ She a maximal element
of the set
К = {Ai : 1 < i < fe},
that is, if A\ С Ai, then either i = 1 or Ai = S. Choose an arbitrary
element χ of the set S — Αχ, and let
K1 = {AiGK:x^Ai},
К2 = {А{еК:хе АиA1nAi = 0},
K3 = {AieK :xe Au Ax с AJ,
К4 = {А{еК :xe Au AxnAi^ 0, Αλ % Α{}.
Obviously,
κ = κλυκ2υκ3υκ4. (i)
We will estimate the cardinality of the set K. The inductional hypothesis
implies that
\Κχ\ <2n"2-l. (2)
Let I = \Аг\. Then
\K2\ <2n"/"1-l (3)
since every element of К is a set of at least two elements. By the maximality
of Αχ, the only element that may be contained in K2 is 5, that is,
l*s| < 1. (4)
Finally, the elements of the set K4 can meet the sets S — A\ and A\ in
at most 2n~l~1 and 2l — 2 distinct sets, respectively (they cannot meet A\
in A\ or 0). Pairing these intersections in all possible ways, we get
\KA\ < 2n"/"1(2/ - 2) = 2n~l - 2n~l.
Actually, this estimate can be sharpened. If X G If4, then
Υ = (X-A1)U(A1-X)£K4
since the sets A\, Χ, Υ do not satisfy the intersection conditions for the sets
Ai (the sets A\ Π Χ, Α\ Π У, Χ Π Υ are not empty, but the set Αχ Π Χ Π Υ
is empty). Since the mapping
X-+Y = (X-A1)U(A1-X)

3.2 COMBINATORICS
127
is one-to-one for a fixed set A\ — and if χ G Χ, Αχ Π Χ φ 0, and Αι С X,
then the same holds for the set Υ so the cardinality of K4 is at most half
of the estimate above, that is,
\K4\ <2^-2_2n-z-1. (5)
Summarizing (1) to (5), we get the desired inequality, \K\ < 2n_1 — 1. D
Solution 2. We will prove the statement by induction on n. It obviously
holds for η = 2. Assume that η > 2. We distinguish two cases.
Case 1. There are no г and j such that A{ U A3; = S and Αχ Π Aj is of
one element.
In this case, it is not difficult to show the statement. Considering an
arbitrary element of the set 5, the number of sets A{ not containing χ is at
most 2n_2 — 1 by the inductional hypothesis. The number of sets X С S
containing χ is 2n_1. At most half of them, that is, at most 2n~2, appear
as a set Ai, since if X = Ai, then according to the assumption above, there
is no j such that Aj = (S — X) U {x}. Thus, the number of sets Ai is at
most 2n"1 - 1.
Case 2. There is an element χ e S such that A\ U A2 = S and A\ Π Α2 =
Let г and s denote the cardinality of the sets Αχ and A2, respectively.
Clearly, r + s = η + 1. The number of the sets A» such that А» С Αι is at
most 2r_1 — 1 by the inductional hypothesis. Similarly, the number of the
sets Ai such that Ai С A2 is at most 2S_1 — 1. If A» is not a subset of Ai
or A2, then Αι Π Α» φ 0, A2 Π Α» φ 0, and, of course, ΑλΓ\Α2φ 0. Thus
Αι Π Α2 Π Ai ^ 0 by the conditions of the problem, that is, χ G A^ Now,
Ai = {ж} U {Ai — Αι) U (Ai — A2), and since the nonempty sets Ai — A\
and Ai — A2 can be chosen in 2S_1 — 1 and 2r_1 — 1 ways, respectively, the
number of these sets Ai is at most (2S_1 — l)(2r_1 — 1). Adding up the
partial results obtained, we get that the number of all sets Ai is at most
2n"1 - 1. D
Remark. It can be shown that if к = 2n_1 — 1, then, necessarily, the sets
Ai are exactly the sets of at least two elements containing a given element
χ G S. It can be shown by induction, say, like we proceeded in the first
solution.
Problem C.6. Show that the edges of a strongly connected bipolar
graph can be oriented in such a way that for any edge e there is a simple
directed path from pole ρ to pole q containing e. (A strongly connected
bipolar graph is a finite connected graph with two special vertices ρ and
q having the property that there are no points x,y, χ φ у, such that all
paths from χ to ρ as well as all paths from χ to q contain y.)
Solution. First, we prove that there is a real-valued function / defined on
the vertices such that

128
3. SOLUTIONS TO THE PROBLEMS
(i) f(p) = l,f(q)=0,
(ii) if f(x) = f(y), then x = y,
(iii) if χ φ ρ, then there is an edge xy such that f(y) > f(x),
(iv) if χ φ q, then there is an edge xy such that f(y) < /(#).
The graph G is connected so there is a path ρ = x0,..., rrm = q from ρ
to q. In this subset, f(xk) = 1 — k/m is a desired function.
Now, suppose that / is a desired function on a proper subset V\ С V(G).
We show that / can be extended for some bigger set. Let ζ G V2 =
V{G) — V\. The graph G is connected so there is a path from ζ to p. Let
x\ be the last point of this path belonging to V2, and let rr0 be the next
point of this path. The graph G is strongly connected so there is an x\ — ρ
от χι — q path not containing x0. Let xi,X2,... be such a path, and let
η > 1 be the smallest index i such that Xi G V\ (there is such a point since
p, q are in Vi). The points χη,χο are distinct elements of V\ so the open
interval f{xo),f{xn) is nonempty by (ii) and it contains infinitely many
elements even if we delete the values of / in Vi. Thus, we can assign a
strictly monotone subsequence of η — 1 members to the points χι,...,χη-ι
so that the sequence f(xo),f(xi),... ,f(xn) is strictly monotone. Now,
(i)-(iv) hold in the extended domain as well: (i) and (ii) hold obviously, as
well, as (iii) and (iv) for χ G V\. Finally, if χ = Xi (1 < г < η — 1), then
Xi-\ and Xi+\ is an appropriate choice for y, respectively.
Since G is finite, it implies that there is a function / satisfying (i)-(iv).
Now, let us orient the edges of G as follows. An edge joining χ and
у should be oriented from χ to у if and only if f(x) > f(y). We show
that this orientation has the desired properties. Let e be an edge from
some x\ to some y\. For η > 1, let xn+i be a neighbor of xn such that
/(#n+i) > f(xn) if xn Φ V· (Such a point exists by (iii).) Similarly, let
yn+i be a neighbor of yn such that /(yn+i) < f(yn) if yn Φ Q- (Such a
point exists by (iv).) Since G is finite, after a while we get that xr = p,
ys = q for some r,s > 1. Then xr,xr_u ... ,ге2,яъ2/ъ2/2, · · · ,у3-\,Уз is a
desired path through e. D
Problem C.7. Let Τ he a, nonempty family of sets with the following
properties:
(a) IfX eJ7, then there are some Υ G Τ and Ζ G Τ such that ΥΠΖ = 0
and Υ U Ζ = Χ.
(b) ИХ eT,andYOZ = X,YC\Z = $, then either Υ e Τ or Ζ eT.
Show that there is a decreasing sequence Xo 2 X\ 3 X2 2 · · · of sets
Xn G Τ such that
00
Π χη = 0.
n=0
Solution. We will show that the statement of the problem holds even if
the condition (a) is replaced by the following weaker condition:
(a') If X G T, then there are some Υ G Τ and Ζ G Τ such that YC\Z = 0,
YUZCX.

3.2 COMBINATORICS
129
We will prove it by contradiction. Suppose that the statement does
not hold, that is, if Xq D Χχ D X2 2 · · · for some sets Xn G T, then
n%L0Xn Φ 0.
1. For any set A G T, there is a set A' G Τ such that А' С A and
A' = U^-^Ai where the sets Ai are pairwise disjoint elements of T.
By condition (a'), there exist sets Ai,A[ G Τ such that A\ U A[ С
Α, ΑιΠΑί = 0. Similarly, there exist sets A2,A'2 G Τ such that A2UA2 С
Αι, A2 Π A2 = 0, and so on. We are done if Ug^ G T. If it is not
the case then by condition b), the sets A0 = A — U^Ai G Τ and
A' = A = и^0Аг have the desired properties.
2. For any set A G Τ there are some sets B,C e J7 such that В П С = 0,
£uCCAandif£CPC£uC then Pef.
Consider the sets Ai existing by 1. We try to define a sequence of sets
Ni as follows. Let Ni = A', and if A^-i is defined, then let Ni be an
arbitrary set satisfying the following four conditions:
NieT,
oo
Ni С \jAj,
j=i
Ni OAjeJ7 for infinitely many j.
According to the third condition, Γ\^=1Νΐ = 0 if the sequence is infinite,
so there is an η such that Nn is defined but iVn+i cannot be defined. Let
Β = ΝηΠΑη and С = NnCiAk for some index к > η such that ΝηΓ\Α^ G J7.
Suppose that there is a set Ρ such that В С Ρ С В U С but Ρ g J7. Then
Nn — Ρ G T, and it is easy to see that iVn+i = Nn — Ρ would be an
appropriate choice. (B G Τ by the choice Ρ = B.)
Thus, for A e J7, there are some sets B\,C\ G Τ such that B\ UCi Ci,
BiC\C\ = 0. Similarly, for C\ G .F, there are some sets B2,C2 G Τ such that
£2 UC2 С Ci, В2ПС2 = 0, and so forth. Let P{ = U^Bj for i = 1,2,....
Then ^ D Pi+1, fl^^i = 0, and P{ G F because ДС^СДи Сь а
contradiction. D
Remark. The problem was motivated by the following set theoretical
game. There are two players: Black and White. Let X\ be a given set.
White partitions it into two sets, Y\ and Z\, Then Black chooses one of
the sets ΥΊ,Ζι, let us denote this set by X2, and partitions it into some
sets Y2 and Z2. Then White chooses one of these sets Y2 and Z2 (let us
denote it by X3) and partitions it into some sets У3 and Z3, and so on.
Thus, the players construct a count ably infinite sequence Xi,X2,X%,...
of sets. White wins if n^Xi φ 0 and Black wins if n^Xi = 0. If the
statement of the problem were false, that is, if there were a family Τ of sets
satisfying (a) and (b) such that П^Хг Ф 0 for any decreasing sequence
of sets Xi, then White would have a winning strategy for any set X\ G T\
White would partition X^+x so that >2fc+b ^2fc+i £ J7 and, from among

130
3. SOLUTIONS TO THE PROBLEMS
>2fc and Zzk, would choose a set contained in T. Finding this kind of set
families is motivated by the fact that we do not know who has a winning
strategy in sets X\ of different cardinalities.
Problem C.8. Let G be a 2-connected nonbipartite graph on 2n
vertices. Show that the vertex set of G can be split into two classes of η
elements each such that the edges joining the two classes form a connected,
spanning subgraph.
Solution. Let F be a spanning tree of the graph G. The vertices of F
can be colored with red and blue so that if two vertices are of the same
color then they are not adjacent. Actually, F has exactly two colorings like
this, which can be obtained from each other by interchanging the colors
red and blue. Notice that the statement of the problem is equivalent to the
following one: There exists α spanning tree F in G such that the numbers
of the red and blue vertices are the same.
Let Fi and F2 be two spanning trees in G, and let χ be a vertex that
is a leaf (a vertex of degree one) in both spanning trees. We say that the
spanning trees F\ and F2 are neighboring at χ if F\ — χ = F2 — x. We will
prove the following lemma, which implies the statement above, as we will
see.
Lemma. Let F and F' be two spanning trees of G, and let Τ be a
common subtree of them. Then there is a sequence F$ = F, Fi,..., Fk = Ff
of spanning trees such that Τ С Fi and Fi and Fi+i are neighboring at some
vertex not in T.
Proof. We prove the lemma by "backward" induction on the number of
vertices in T. If Τ has 2n — 1 vertices, then F and F' are neighboring.
Suppose now that Τ has 2n — I vertices (I > 2). Let xy and uv be an edge
of F and F' leaving T, respectively, such that x,u G T. If xy = uv then
T' = T+xy is a common subtree and there is a desired sequence of spanning
trees by the inductional hypothesis. If у φ ν, then let F" be a spanning tree
containing the tree T+xy+uv as a subgraph. Then T+xy С FnF", and so
by the inductional hypothesis, there is a sequence F = Fo, F1?..., Fm = F"
of spanning trees such that Fi and Fi+i are neighboring (г = 0,1,..., m—1)
and Τ С T+xy С Fi. Similarly, T+uv С F'ClF" and so by the inductional
hypothesis, there is a sequence F" = Fm, Fm+i,..., Fk = F' of spanning
trees such that Fi and Fi+i are neighboring (г = m, m + 1,..., к — 1) and
Τ С Τ + uv С Fi, and F = F0, Fi,..., Fk = F' is the desired sequence of
spanning trees.
Finally, suppose that xy φ uv but у — v. Since G — у is connected, it
has a vertex ζ £ Τ joined to a vertex w e T. Now, let F" and F"f each
be a spanning tree containing Τ + xy + wz and Τ + uv + wz, respectively.
Then, as above, there is a desired sequence of spanning trees from F to
F", from F" to F'", and from F'" to F", and the union of these sequences
is a desired sequence from F to F'.

3.2 COMBINATORICS
131
Since G is nonbipartite, it contains a cycle С of odd length. Let χ be a
vertex of C, and let у and ζ be its neighbors in C. Let T0 be a spanning
tree of G — χ containing С — χ. Let F = To + xy and Ff = To + xz. Color
F so that χ is red.
According to the lemma, there is a sequence F = ίο, ίι,..., F^ = i"
of spanning trees such that i1» and Fi+\ are neighboring at some vertex
Xi φ χ. Color Fi with colors red and blue so that the colorings of Fi and
ii+i differ from each other at most in the color of xit It defines some
colorings of the spanning trees F\,..., Fk = F1 starting with the coloring
of Fq = F such that χ is red in every coloring. Clearly, every vertex got
different colors in the colorings of F and F', except x. Let a^ denote the
number of red vertices in the coloring of Fi. Then
\αί+ι — a»| < 1,
ao + &k = 2n + 1.
Thus,
either clq <n < a^ or a^ < η < ao,
so there is an index 0 < г < к such that di = n, and this is what we wanted
to prove. D
Problem C.9. Let An denote the set of all mappings f : {1,2,..., n} —>
{1,2,..., n} such that f-1 (<) := {k : f(k) = i} φ 0 implies f~\j) φ 0, j G
{1,2,...,г}. Prove
k=0
Solution 1. Let an denote the cardinality of the set An- Then the number
of mappings f e An such that f(i) = l holds for exactly / integers г is equal
to (^)an_j. Since Ζ is positive for any function /, we have
n-l
Tl / \ 71—1
ι=ι V£/ z=o
for η > 1.
Using the notation 6n = an/n\, we have 6q = 1 and
n—1 .J
6» = Σ т^гпА
z=o
Clearly, 6n < 2n, so if |z| < 1/2, then the series X^0 ^n^n is absolutely

132 3. SOLUTIONS TO THE PROBLEMS
convergent. For the sum f(z) of this series, we have
oo n—1
n=0 n=l !=0 {П l>·
1=0 \n=Z+l v ' J
oo
= 1 + Yy - \)blZl = 1 + (e2 - l)f(z).
1=0
It implies that
1 1 ^ ekz v^v^ 1 kn
^ = 2 ' 1-е*/2 = ^ 2*+ϊ = ^ ^ ri¥+iZn
1 k=0 k=0n=0
n=o \ k=o /
Thus,
k=0
which we wanted to prove. D
Solution 2. Let sn^ denote the number of surjective mappings of the set
{1,2,..., n} onto the set {1,2,..., г}. Then the cardinality of the set An
is
η
2_^sn,i-
г=1
Counting the mappings of the set {1,2,..., n} onto the set {1,2,..., &},
we get the equality
fcn _ z_> (i )Sn*'
Now, we have to prove that
oo к (к\
2_j Sn^ ~Δ^Δ^ 2^+1 Sn'u
г=1 к=1г=1
that is, using sn,n+i = sn,n+2 = · · · = 0 and supposing that the
rearrangement does not change the sum, we have to prove that
η η / oo /k\ \
i=l г=1 \k=i J

3.2 COMBINATORICS
133
In the parentheses, we have the sum of the terms of a negative binomial
distribution ((J/2fe+1 is the probability of the event that flipping a coin,
we get the (г + l)st head in the (k + l)st flip), and so the coefEcient of
every term sn^ on the right is 1. Since the series in the parentheses are
absolutely convergent, the rearrangement did not change the sum, and the
proof is complete. D
Problem CIO. Let G be an infinite graph such that for any count ably
infinite vertex set A there is a vertex ρ joined to infinitely many elements of
A. Show that G has a countably infinite vertex set A such that G contains
uncountably infinitely many vertices ρ joined to infinitely many elements
of A.
Solution. We prove the statement by contradiction. Assume that G is a
graph with vertex set V such that
(1) for any countably infinite vertex set А С V, the set Ha of vertices
ρ e V — A joined to infinitely many elements of A is a nonempty,
countable set.
The set Ha is obviously infinite since if not, then for the set В = AuHa,
there is no vertex ρ e V — В joined to infinitely many elements of B. It
implies, that deleting finitely many vertices from G, the resulting graph
has property (1).
Notice that the vertex set V is clearly uncountable since (1) holds for
A = V as well.
Now, we show that we can delete finitely many vertices from G so that
the resulting graph G\ has no finite vertex set X covering V with the
exception of a countable set. (A vertex set X covers the union of the
sets of neighbors of the elements of X.) Suppose it is not the case. Then
there is a finite set X$ covering all elements of V except a countable set.
Deleting Xq, the resulting graph still must have a finite set X\ with the
same property, and so on. So if we do not get a desired graph after finitely
many steps then we can define the pairwise disjoint finite sets Xn covering
V with the exception of some countable sets. Let A = U^0Xn. Then
every element of V (except the elements of a countable set) is joined to
infinitely many elements of A, that is, V — Ha is countable. Since Ha is
countable by (1), it implies that V is countable, a contradiction.
As we have seen, G\ has property (1) as well, so we may assume that
Gi=G.
Let А С V be an arbitrary countably infinite set. We show that there is
a countably infinite set F(A) С V — A such that На С F(A) and for any
finite subset {αϊ,..., αη} οι A, the set F(A) has infinitely many elements
not joined to any vertex а* (г = 1,..., η). Let A = {ai, a2,... } and for
each n, put infinitely many elements of V — A not joined to any element
of {αχ,..., αη} into F(A). It can be done, as we have seen. Then take the
union of Ha and the set of the elements obtained.

134
3. SOLUTIONS TO THE PROBLEMS
Now, let Aq С V be an arbitrary countably infinite set, and let
Ai = F{A0 U Ax U · · · U Ai-г) (i = 1,2,...),
au = f(\JaA.
For any element ρ G Αω and for any index г, ρ is joined to finitely many
elements of Ai. Really, ρ e Αω С V — U?Z0Ai С V — Ai+i, and so ρ £
F(i4oUiliU..-Ui4<_i) эЯАг
Let U^0Ai = {αϊ, α2,... }, Aj = {&i, &2, · · · }· Let us define an infinite
set С = {ci, c2,... } such that if С С U^0Ai and η < ra, then cm is joined
to neither an nor bn. It yields the desired contradiction since ρ £ U^Ai
for ρ e He (if ρ = an, then it is joined to at most η elements cm), and
so, ρ e He С H\j°±0Ai С F (υ^0Ai) = Αω. However, it is a contradiction
again, since if ρ = 6n, then it is joined to at most the elements ci, c2,..., cn
of С
Thus, let'ci be an arbitrary element of Aq, and suppose that c\, c2,..., cn
are given so that {αϊ, α2,..., αη, ci, c2,..., cn} С A0 U ii U · · · U Адт for
some N. Then, Α^+ι has an infinite subset Τ such that the elements of Τ
are not joined to any of the elements αϊ, α2,..., αη. As we have seen above,
each of the elements 6i, 62,..., bn is joined to finitely many elements of T,
so by choosing cn+i out of the rest of T, we meet the requirements. D
Remark. Assuming the continuum hypothesis, it is possible to construct
a graph G satisfying the conditions of the problem in which every countably
infinite set A has a countably infinite subset В such that Нв is countable.
Problem C.ll. Let Η be a family of finite subsets of an infinite set
X such that every finite subset of X can be represented as the union of
two disjoint sets from H. Prove that for every positive integer к there is
a subset of X that can be represented in at least к different ways as the
union of two disjoint sets from H.
Solution. The solution is based on the following theorem of Ramsey.
Theorem. Let Υ be an arbitrary infinite set and let η be an arbitrary
natural number. Subdivide the family of all subsets of η elements of Υ
into two classes. Then, there exists an infinite set Yf CY such that every
subset of η elements of the set Y1 belongs to the very same class.
For a set A, let [A]n denote the set of subsets of η elements of the set
A. For a given natural number k, we show that there is a natural number
m > к and an infinite set Ζ С У such that [Z]m С Н. It implies the
statement of the problem, since then any subset of 2m elements of the set
Ζ can be obtained as the union of two disjoint members of Η in at least
(2^) > к different ways.
We find a desired set Ζ as follows. Subdivide the set [X]2 into two
classes depending on whether or not the elements are in H. By Ramsey's

3.2 COMBINATORICS
135
theorem above, there exists an infinite set X2 С X such that
either [X2]2 С Н or [X2]2 Π Η = 0.
Then classify the subsets of three elements of X2 based on whether or
not they belong to H. Again by Ramsey's theorem above, there exists an
infinite set X% С Х2 such that
either [X3]3 с Η or [X3]3 Π Η = 0.
Proceeding further, we obtain a sequence X D X2 D · · · D X2k of infinite
subsets of X with the following property:
either [Im]mcH or pfm]mn« = 0
for 2 < m < 2k. For such an m, we naturally have
either [X2k]m С И or [Х2к]т Π Η = 0
as well. We are done if we can prove that the second possibility cannot be
the case. Actually, choose a subset A of 2k elements of the set X2k- By the
conditions of the problem (which are used only at this point), A = В U С
for some sets B,C GH. One of these sets, say B, has at least к elements.
Then for m = |B|, we have В е [Хм]™ П И, that is, [X2fc]m Π Η φ 0, and
so [X2fc]m С W, which we wanted to prove. D
Remarks.
1. We did not use the fact that the finite subsets of X can be obtained as
the union of two disjoint sets in H; this part of the condition can be
omitted. Furthermore, it is sufficient to assume that for a fixed integer
c, the finite subsets of X can be obtained as the union of at most с
members of H.
2. Considering natural numbers instead of sets and sum instead of union,
we get the following analogous problem: let if be an arbitrary set of
natural numbers. Suppose that every natural number is the sum of two
elements of H. Does it imply that for any natural number k, there is
a number that can be obtained as the sum of two elements of Η in at
least к different ways? This problem of S. Sidon is open. However, the
multiplicative version of the problem (where we say "product" instead
of "sum" in all cases) has been answered affirmatively. It follows from
the statement of the original problem, as follows: let X be the set of
all prime numbers, and assign the set of prime divisors to any natural
number in H. Thus, we obtain a family Η meeting the conditions of the
original problem. Thus, there exists a set А С X that can be obtained
as the union of two disjoint members of Η in at least к different ways.
Then, the product of the prime divisors in A can be obtained as the
product of two elements of Η in at least к different ways.

136
3. SOLUTIONS TO THE PROBLEMS
Problem C.12. Let the operation f of к variables denned on the set
{1,2,..., n} be called friendly toward the binary relation ρ defined on the
same set if
/(αϊ, α2, ...,ак) р /(Ьь 62, · · ·, h)
implies a,i p bi for at least one г, 1 < г < к. Show that if the operation f is
friendly toward the relations "equal to" and "less than," then it is friendly
toward all binary relations.
Solution. Since / is friendly with the relations "equal to" and "less than,"
we have
/(1,...,1)</(2,...,2)<...</(п,...,п)
and thus /(г,..., г) = г for all г. Hence, the following property holds:
(*) If /(αϊ,...,α*) = a, then there is an г such that di = α (since
/(αϊ,..., α*.) = a = /(a,...,a) and / is friendly with the relation
"equal to").
It is sufficient to prove that if /(αϊ,..., α^) = α and /(&ι,..., bk) = b
then there is a subscript г such that ai = α and bi = b. Let 7 = {г | a^ = a}
and J = {j \ bj = b}. Property (*) shows that 7, J φ 0; however, what we
need is that 7 П J Φ 0. This is obviously true for η — 1. Now, let η > 2,
and assume that 7 Π J = 0. We distinguish two cases:
Case 1. α φ b. Let a = b for г G 7 and q = α otherwise.
Furthermore, let di = a if г G J and di = b otherwise. Then /(ci,...,Cfc) ^
/(αχ,...,α*) = a (since q = 6 ^ a = a^ if г G 7 and Ci = α φ α^ Ίΐ
г φ 7), and similarly /(db ..., dk) φ /(6Ь ..., bk) = b. Property (*)
implies that /(ci,..., ck) and /(di,..., dk) could only be α or b and thus
/(ci...,Cfc) = 6 and f(di...,dk) = a. Without loss of generality, we
may assume that α < 6, that is, /(di,..., dk) < /(ci,... ,c^). Since / is
friendly with the relation "less than" we must have an г such that di < q.
On the other hand, however, if г G 7 or г G J then a = d^ otherwise,
d = α < b = di, a contradiction.
Case 2. a = b. Let a = с(ф a) if г G 7 and q = α otherwise;
similarly di = с if г G J and di = a otherwise; and finally e» = α if
г e J and бг = с otherwise. Similarly to the previous case, we have
/(ci,...,Cfc) ^ /(ab...,afc) = a, /(di,...,dfc) ^ /(&i,... A) = b = a
and /(ei,...,efc) ^ /(di,... ,d^). However, all the values /(ci,... ,c^),
/(di,...,dfc) and /(ei,...,efc) must be either α or c, and thus
/(ci,...,Cfc) = /(di,...,dfc) = c, and so /(eb...,efc) = a. Again, it is
enough to see the case when α < 6, that is, /(ei,..., e^) < /(ci,..., c^),
which implies the existence of an index г such that e^ < c^. However, we
have q = бг = с if г G 7, Сг = e^ = α if г G J, and a = a < с = ei otherwise.
This is a contradiction again, which makes the proof complete. D

3.2 COMBINATORICS
137
Problem С.13. Let g(n, k) denote the number of strongly connected,
simple directed graphs with η vertices and к edges. (Simple means no loops
or multiple edges.) Show that
η — η
"£(-1)кд(п,к) = (п-1)\.
к=п
Solution 1. We prove the statement by induction on n. For η = 2, the
statement is obviously true. Let us write the left-hand side in the form
Ем)'*™. (i)
G
where G runs over the graphs described above. For every G in the
summation (1), consider the edges of G having one end vertex at n. We distinguish
two cases:
Case 1. G has exactly two edges incident to η and the other endpoint
of both of them is the same vertex p.
In this case, G\{n} is strictly connected as well, and so
(_1)|£(σ\{η})| = (_i)|£(G)l.
Thus, for a fixed p, the sum of these terms is (n — 2)! according to the
induction assumption and summing them up for ρ = 1,..., (η — 1), we get
exactly (n — 1)!.
Case 2. There are more than two edges of G incident to n, or there are
two such edges of it having different other end vert ices.
There must be two vertices ρφ q, 1 < p, q <n — 1, such that pn G E(G)
and nq G E{G). In this case, the graph having the same edge set as G,
except that it has the edge pq if G does not have it, or it does not have this
edge if G does have it, is strongly connected if and only if G is strongly
connected; thus we can pair these graphs such that every such graph G not
containing the edge pq has its pair G-\-pq. For every graph, we have a pair,
and the sum in (1) is 0 for such a pair; thus, the sum is 0 for the graphs in
this case.
Summing up (—1)Ι£(σ)Ι for the graphs in Cases 1 and 2, we get that (1)
is(n-l)!. D
Solution 2. If Q is a set of graphs satisfying the properties of the problem,
let us define
μ^-Σί-1)1*™
GeG
For an arbitrary graph G, assign the permutation {αϊ = 1, α<ι,..., an} of
the vertex set such that if α& is already chosen we pick α^+ι as the smallest

138
3. SOLUTIONS TO THE PROBLEMS
number having an edge from the set {αϊ, α2,..., α^} to it. (We always have
such an edge and vertex since the graph G is strongly connected.)
For any given permutation π, let ζ}π denote the set of graphs having π
as the assigned permutation. Since we have
E(-d|e(G)I= Σ MO,
G πΕ5η-ι
and since the number of permutations of the set {2,..., n} is (n — 1)!, it
suffices to prove that μ(ζ}π) = 1 for all π.
Let us color the edges of these graphs with blue and red as follows: let
the edges going along the permutation π be colored blue and the other ones
colored red (that is, the blue edges are of form a^a^ for г < j). Clearly,
(*) for every vertex ν φ 1, there is a blue edge pointing to v, but there is
no blue edge акЩ if щ < ctj for some к < j < I.
If Qa is the set of graphs corresponding to a fixed set α of blue edges,
that is, having exactly those edges blue that belong to α (provided we have
the fixed permutation π), then we have
Μ£π) = Σ Μ£α),
Gacg„
Let now us consider these sums μ(£α) now. If in any graph in Qa there is
a red edge aidj such that г — j > 1, then pick the edge having the smallest
j and among them that having the biggest i. Let Q%J be the subset of Qa
in which the graphs have aidj as the red edge picked, and let G G Q1^. The
way we chose j ensures that the edges aiCLj, ajdj-i,..., ύ^αι are edges in G
and furthermore — because of the way we constructed π — we can reach
di-i from the vertex a\ = 1 through blue edges only. The graph, having
the same edge set as G with the exception of the edge a^-i, is strongly
connected if and only if G is strongly connected. Thus, we can pair the
graphs of Q)£ so that the pairs differ only in the edge didi-i. Such a pair
contributes 0 to the sum; thus, μ(ΰ^) = 0 and Σμ^1^) = 0.
hj
There is exactly one set of red edges not containing any red edge to pick
up, namely {didi-i \ г = 2,..., η}, and so
μ(0α) = (_i)l°l(-i)"-i = (_i)M-(n-i).
It is easy to see that any set of edges satisfying property (*) and having
all edges going along the permutation can be chosen as the set a of blue
edges. Thus, we have
м&) = Σ (-ΐ)|α|-(η-1}
Gacg„
P2 P3 Pn / \ Π Pj / ч
■SE-EftK'-nB-rfr,
Z2 = 1Z3 = 1 in = 1\3/ j=2lj=l XJ/

3.2 COMBINATORICS
139
where Pk denotes the maximal in-degree of vertex k. Since we have
J>'>"-ft) - <-» (I·-»"(I;) - ft))=<-'><°-ч -·
we get μ(£π) = 1, and topthe proof is complete. D
Problem C.14. Let Τ be a triangulation of an η-dimensional sphere,
and to each vertex of Τ let us assign a nonzero vector of a linear space V.
Show that if Τ has an η-dimensional simplex such that the vectors assigned
to the vertices of this simplex are linearly independent, then another such
simplex must also exist.
Solution 1. Let A = (ao,..., an) be a simplex in Τ such that the vectors
v(ao),..., v(an) assigned to the vertices of A are linearly independent. We
will call an (n — l)-dimensional simplex in Τ red if for any index 0 < г <
η — 1, only one vector in the set
Xi = (v(a0),..., v(di)) - (v(a0),..., ^-i))
is assigned to the vertices of it. (Here, (v\,... ,г^) denotes the subspace
generated by the vectors v\,..., Vk·) Obviously, (a0,..., an) is red.
Lemma. If the vectors assigned to the vertices of an η-dimensional
simplex in Τ are not linearly independent, then the number of its red faces is
Oor 2.
Proof. If the set ν (S) of vectors assigned to the vertices of S does not
contain any vector in one of the sets Xo? · · · ? ^n-i? then obviously S has no
red face. If υ(S) contains an element of each of the sets Xq? · · · ? Xn-i and
a vector not in X0 U · · · UXn-\ then the vectors assigned to the vertices of
S are linearly independent, a contradiction. So, we may assume that υ(S)
contains an element of each set Xj (j = 0,1,..., n— 1) and that it contains
exactly two elements of, say, Xi and exactly one element of the other X/s.
Then S has two red faces by which can be obtained from 5, deleting one
of the vertices with vectors in Xi assigned to it. The proof of the lemma
is complete.
Now, consider the graph G whose vertices are the η-dimensional sim-
plices of Τ and where two vertices are joined by an edge if they share a
common red face. In this graph, the degree of the simplex A is 1. Thus, G
contains one more vertex of odd degree. By the above lemma, the vectors
assigned to the vertices of the corresponding simplex are linearly
independent which we wanted to prove. D
Solution 2. We prove the following, slightly more general, statement: If
К is an arbitrary η-dimensional complex with boundary 0 (that is, К is
a cycle mod 2), and if we assign nonzero vectors of the linear space V to

140
3. SOLUTIONS TO THE PROBLEMS
the 0-dimensional simplices (that is, the vertices) of K, then the number of
η-dimensional simplices with linearly independent assigned vectors cannot
be 1.
We prove the statement by induction on n. It obviously holds for η = 1.
Suppose that η > 2. Suppose that dimv(A) = η Η- 1 for A e K. We have
to show that К contains one more such simplex. Let β be an (n - 1)-
dimensional face of A. If С φ A is a simplex such that В is a face of it,
then v(C) % v(B) implies that dimv(C) > dimv(B) = n, that is, С has
the desired properties. So, we may assume that if С φ A and В is a face
of A, then υ(C) Qv(B).
Let K0 = {S e K\v(S) С v(B)} and let Kx = 6{K0), where δ denotes
the boundary. Obviously, δ{Κι) = 6(6(Kq)) = 0, and since A £ Ко and
δ(Κ) = 0, В is thus the face of an odd number of simplices in Kq, that is,
В G K\. By the induction hypothesis, there exists a simplex B\ φ Β in K\
such that dimi>(i?i) = dimi;(i?), and so v{B\) = v(B). Since B\ e δ(Κ0)
but B\ $. δ(Κ), there is a thus simplex A\ in К such that B\ is a face
of A\ but v{A\) 2 v(B), and so dimv(yli) > dimv(Bi) = n. So, the
vectors assigned to the vertices of A\ are linearly independent. To make
the proof complete, notice that it is impossible that A = A\ because if it
is the case then A has two faces В and B\ such that v(B) = v{B\) and so
dim υ (A) = dimv(B) = n, a contradiction. D
Problem C.15. For a real number x, let \\x\\ denote the distance
between χ and the closest integer. Let 0<жп<1 (η =1,2,...), and let
ε > 0. Show that there exist innnitely many pairs (n, m) of indices such
that η φ πι and
\\Xn -XrnW < П11П
\6'2\n-m\)'
Solution. Let us fix ε > 0. We show that if η is large enough, 0 < X{ <
1 (г = 1,..., n), then there exists a pair (z,j) of indices such that \\xi —
Xj\\ < πηη(ε, l/over2\i—j\). It suffices since it implies that by subdividing
the infinite sequence :ri, #2,... into blocks of η terms, we can find a desired
pair of indices in each block and the difference \i — j\ does not change if
the indices in the block are replaced by the indices in the whole sequence.
Let / : {1,..., n} —> {1,..., n} be a permutation such that
0<x/(i) < <x/(n) < 1-
We may assume that
ll*/G) - */(*fi)ll < min (ε, 2m_1f{i + 1)\)
for г = 1,..., η since if it is not the case then we are done. Let A denote
the set of the indices 1 < г < η such that \\xf(i) — xf(i+i)\\ > £· Since the

3.2 COMBINATORICS
141
cyclically taken intervals [z/(i),£/(2)), [s/(2),s/(3))> · · ·, [^/(n),^/(i)) cover
the cyclic interval [0,1)
1-§l|S,W-iB/(4+1)ll-SC + S2|/(i)-1/(i + l)l'
and \A\ < Ι/ε. That is,
2^Ι/(0-/(< + ΐ)Γ
Applying the inequality between the arithmetic and harmonic means, we
get the inequality
In the sum Σ™=1 \f(i) — f(i + 1)|, every integer j G {1,... ,n} appears
exactly twice with positive or negative signs, so that the total sum of the
signs is 0. Obviously, the sum is maximum when the large numbers have
coefficient +2 and the small numbers have coefficient —2. Thus,
Σΐ/(0-/(ί + ΐ)Ι<ΐτ
when η is even, and
2
Σ 1/(0-/(< +i)l <^
when η is odd. Combining this fact with the preceding inequality, we
obtain 1 — \Α\ε > (1 — |A|/n)2, which is a contradiction if \A\ > 1 and
η is sufficienly large. If A = 0, then we get 1 = 1, which means that
η is even and the numbers |/(г) — /(г + 1)| are equal (mean inequality).
However, if /(г) > n/2, then /(г) appears twice with positive sign, that is,
/(г) > /(г - 1), /(г) > /(г + 1), implying that /(г - 1) = /(г + 1), which
is possible only if η < 2. But it is easy to see that equality cannot hold in
this case either, and we got the desired contradiction in all cases. D
Remark. The existence of a pair (г, j) of indices such that \\xi — Xj\\ <
πιίη(ε, l/c\i — j\) can be proved if с < y/b (the proof is much more
complicated). The statement is false if с > y/b since it would imply that for an
irrational number a, there are infinitely many rational numbers p/q such
that |a — p/q\ < l/cq2.

142
3. SOLUTIONS TO THE PROBLEMS
Problem C.16. Consider the lattice L of the contractions of a simple
graph G (as sets of vertex pairs) with respect to inclusion. Let η > 1 be
an arbitrary integer. Show that the identity
ι
*Λ V* =V
\г=0 / j=0
*л
ν
V y*\
. 0<i<n I
/
holds if and only ifG has no cycle of size at least η + 2.
Solution. It can be seen that a, b G vJLqZ» holds if and only if there
is a natural number I and a path а = u0,ui,...,ui = b such that for
every 0 < j < Ζ, (щ,щ+г) G Zi for some 0 < г < к. Furthermore,
(α, b) G z\ Λ ζ2 holds if and only if there is a natural number I and a path
а = u0, iii,..., щ = b such that (г^, щ+ι) G z\ and (щ,щ+1) G z2 for every
0 < г < /. The inequality > always holds in equality (1), so what we have
to prove is that the inequality < holds if and only if G has no cycle of at
least η + 2 vertices. Suppose that vo,...,Vk is a cycle with к > η + 1. If
χ contracts (vo,Vk), y% contracts (ι^,ι^+ι) (0 < г < η), and yn contracts
the vertices vn,i;n+i,..., г;*, then the left-hand side of (1) is χ and the
right-hand side of it is not since (г>о, Vk) is not contracted in the right-hand
side.
Conversely, suppose that G has no cycle of at least n+2 vertices, and let
(a, b) G χ Λ Vi=0yi. We show that (a, b) is an element of the right-hand side
of (1) as well. The assumption (a, b) G ^AV^=0yi implies that there is a path
а = v0,vi,... ,Vk = b in G such that (vt,Vt+i) G χ and (vuvt+i) G Л"=0уг
for 0 < t < к. Let us fix t. There is a path vt = щ, щ,.. .,щ = vt+i such
that for every 0 < s < I, we have (u3,u3+i) G ym for some 0 << n. Since
the vertices uo,ui,... ,щ constitute a cycle, / + 1 < η + 2, that is, / < η
and some of the y's do not appear among these ym. Thus,
(vUVt+i)e \j yi
i=0,...,j —l,j"+l,...n
for some j, and hence (a, 6) is contained in the right-hand side of (1). D
Problem C.17. Let G(V,E) be a connected graph, and let dc(x,y)
denote the length of the shortest path joining χ and у in G. Let rc{x) =
тах{ас(х,у) : у G V} for χ G V, and let r(G) = min{rc(^) : x G V}.
Show that if r(G) > 2, then G contains a path of length 2r(G) — 2 as an
induced subgraph.
Solution 1. Let η = r(G), and let χ be a vertex of G such that rc{x) = η
and the cardinality of the set {y : d(x, y) = n} is minimum. Let xn be a
vertex such that d(x,xn) = n, and let χ = xq, χι, у = x2, · · ·, xn be a path

3.2 COMBINATORICS
143
joining χ and xn. We show that if G does not contain a path of length 2n—2
as an induced subgraph then rc(y) = η and that d(x, z) < n(z e V) implies
that d(y, z) < n, which contradicts the choice of x, and d(y, xn) = η — 2.
rG(y) = η follows from the other statement, so it is sufficient to prove
that one.
Since d(y, z) < d(x, z) + 2, it is sufficient to consider the cases d(x, z) =
n — 2 and d(x,z) = n— 1. Let ж = zq,zi, ... ,ζ* — ζ (к = η — 2 or η— 1) be
a path of length d(x, z) between χ and z. By the minimality of the paths,
if X{ = yj (i > 0,j > 0) then г = j, but then the length of the path
у = X2,.. ·, x% — Zi,... ,Zk is at most к and we are done. Similarly, if X{Zj
is an edge of the graph, then г — 1 < j. If this is an edge of G and г > 2,
then the length of the path у = X2,..., Xi, Zj,..., Zk is at most к (< η) and
if г = 1 then the length of the path y, x\, Zj,..., z^ is к — j' + 2. We are done
if either j > 2 or к = η — 2, so we may assume that j = 1 and A: = η — 1
and then xn,..., хг, zi,..., Zk is an induced path of length 2n — 2. If G
does not contain any edge X{Zj, then жп,..., x\, xq, z\, ..., z^ is an induced
path of length 2n — 2 or 2n — 1, which we wanted to prove. D
Solution 2. (sketch) Every graph G with r(G) = η contains a connected,
induced subgraph G' such that r(G') = η and r(G") < η for every
connected, induced, proper subgraph G" of G', and it is sufficient to take such
a graph G'.
1. If С is a path, then its length is η — 1.
2. If G' — {x} is not connected, then it has two components and one of
them is a path.
3. Let С be the set of vertices χ such that there is no vertex у such that
G — {y} is not connected and the component of G — {y} containing χ is
a path. Then С induces a 2-connected subgraph.
4. For every vertex χ G C, we get a path of length к if we delete the vertices
of the component of G — {x} containing С — {χ}.
5. For every vertex χ G C, there is exactly one vertex у Ε С such that
d(x,y) = n — k.
6. For such a pair of vertices {x, y}, the graph induced by С — {щ, у} is not
connected, that is, joining χ and у in two components, we get a cycle of
length at least 2(n — k).
7. For some 0 < к < η — 2, the graph G' is a cycle of length 2(n — k) with
a pending path of length к at each vertex of it. D
Remarks.
1. The second solution shows that the statement is sharp; it is false for
path of length 2n — 1.
2. There is an infinite graph G such that r(G) = 3, and every induced path
in G is of length at most 3.
Problem C.18. Given η points in a line so that any distance occurs
at most twice, show that the number of distances occurring exactly once is
at least [n/2].

144
3. SOLUTIONS TO THE PROBLEMS
Solution. Suppose that the points are on the real line at the numbers
Pi < V2 < "' < Vw For апУ index 1 < г < n, let us take the set Ai of
lengths of segments from pi to the right, that is, Ai = {pj — Pi : г < j}.
Obviously, \Ai\ — n — i.
We prove (by contradiction) that \A{ Π Aj\ < 1 for г < j. Suppose that
u, υ e Ai Π Aj, и φ υ, and, say, u = pkl-pi = pk2 - pj and υ = pmi ~Pi =
Pm2 ~ Pj- But then, the distance pj - Pi = Pk2 - pkl = Pm2 ~ pmi occurs
three times, a contradiction.
Now we estimate the number of distances occurring from below. As we
have seen,
\Ai - (Αλ U · · · U i4<_i)| = \Ai - ((Ai Π Αλ) U · · · U (Ai Π А^г))\
= \Ai\ - \(Ai Π Αι) U · · · U (Ai Π i4i_i)|
> η — г — (г — 1) = η — 2г + 1.
Hence,
liliU-.-Uilnl
= \Аг U (А2 - Αλ) U (А3 - (Αλ U A2)) U · · · U (Ап - (Аг U · · · U Αη_χ))|
>η-1 + η-3 + η-5 + ··· + η- 2[η/2] + 1,
that is,
f n2/4 if η is even,
У (пг - l)/4 if η is odd.
Now let d\ and б?2 denote the number of distances occurring once and
twice, respectively. Obviously,
d\ + 2б?2
and
©
di + d2 >
Г
/4 if η is even,
2-l)/4 if η is odd.
Subtracting the equality from the double of the inequality, we obtain the
desired inequality, d\ > [n/2]. D
Problem C.19. Let к be an arbitrary cardinality. Show that there
exists a tournament TK = (VK, EK) such that for any coloring f : EK —> к
of the edge set EKy there are three different vertices χ0,χι,χ2 Ε VK such
that
XqXI,X\X2,X2X0 € EK
and
\{f(XQXl)J(xiX2)J(X2Xo)}\ < 2·

3.2 COMBINATORICS
145
(A tournament is a directed graph such that for any vertices x, у G VK, χ ^
у exactly one of the relations xy G Ек, у χ G EK holds.)
Solution. By a famous theorem of Erdos and Rado, there exists a triangle-
free graph G = (V,F) with chromatic number 2*. Order the vertex set V
of this graph G in an arbitrary way, and let < denote this ordering. Now
we define a tournament on V as follows. For x, у G V, χ < у, let xy G Ε if
xy G F and let yx G Ε if xy £ F.
We show that the resulting tournament (V, E) meets the requirements of
the problem. Let / : Ε —> к be an arbitrary coloring of E. For every χ G V,
consider the set Аж = {f(yx) : у < x,yx e E} С к. Let us fix a vertex
χ G V. The cardinality of the set {y G V : у < ж, ух G F or ж < у, :ry G F}
is greater than 2K, so there is a vertex xf such that Ax = Ax* and x'x Ε Ε Ίΐ
Ju ^^ Ju , JuJu Ε Ε Ίΐ χ < x'. Let ^i and #2 denote the smaller and the bigger
element from among χ and ж', respectively. Then f(x\X2) G A^ = AX2,
that is, there exists a vertex xq G V such that
#0 < XU χβχ1 € Я> and /(S0Sl) = f(xiX2J-
However, by the definition of E,
#o#i ? #i#2 G i1,
and so
and
that is, the vertices жо?#ъ#2 have the desired properties. D
Remark. It is easy to see that we cannot replace 2 by 1 in the statement.
For any tournament Τ = (V,E) and for any ordering < of V, define a
coloring / of Ε as follows: if χ < у, then let f(xy) = 0 Ίϊ xy Ε Ε and let
f(yx) = I'nyx Ε Ε. Obviously, there are no three vertices x,y,z eV such
that xy, yz,zx G Ε and / is constant on these three edges.
Problem C.20. Some proper partitions Pi,..., Pn of a unite set S
(that is, partitions containing at least two parts) are called independent if
no matter how we choose one class from each partition, the intersection of
the chosen classes is nonempty. Show that if the inequality
1-γ<\Ρι\--\Ρη\ (*)
holds for some independent partitions, then Pi,..., Pn is maximal in the
sense that there is no partition Ρ such that P,Pi,...,Pn are independent.

146
3. SOLUTIONS TO THE PROBLEMS
On the other hand, show that inequality (*) is not necessary for this max-
imality
Solution. We prove the statement by contradiction. Suppose that (1)
holds and that there is a proper partition P0 such that Po, Pl, · · · ? Pn are
independent. Let us choose one of the classes of each partition and take
the intersection of them. For the sake of brevity, let us call these
intersections class intersections. Obviously, the total number of class intersections
is |Pq||-PiI · · · \Pn\ and апУ two class intersections are disjoint since among
the classes constituting any two class intersections there are two classes
belonging to the same partition, and so they are disjoint. On the other
hand, all the class intersections are nonempty since Pq, Pi, ..., Pn are
independent. Thus, we have
|ΡοΙ|ΡιΙ·.·Ι^η|<|5|. (2)
Combining (1) and (2) and using the fact that \S\ Φ 0, we obtain that
| Po| < 2, which is a contradiction to the assumption that Pq is a proper
partition. So, we proved that (1) implies the maximality.
On the other hand, the next example shows that (1) is not necessary for
the maximality. Let S = {αχ,..., α8}. Let
Pi = {Sii, £12}, Pi — {S21, £22}
be the partitions where
S11 = {^1,^2,^3,^4},
#12 = {^5,^6,^7,^8},
S21 = {αϊ, as},
^22 = {α2,α3,α4,α6,α7,α8}.
Then
SnnS2i = {a1}, (3)
and the other class intersections are not empty either, that is, P\, Ρ2 are
independent. Equation (1) does not hold since |Pi||P2| = 4 = |5|/2, and we
show that {Pi,P2} is maximal. Suppose that Pq,Pi,P2 are independent
for some proper partition P0 = {S01, ^02, · · · }· Then the intersection of
S01 with the set (3) is nonempty and so а\ е S0i· We similarly obtain
αχ e So2- But then Sqi Π 5ο2 φ 0, a contradiction to the assumption that
Pq is a partition. D
Problem C.21. Show that if к < η/2 and Τ is a family of к х к
submatrices of an η χ η matrix such that any two intersect then
Mr;)2

3.2 COMBINATORICS
147
Solution. For any к х к submatrix Μ G F, let Rm and См denote
the Ar-tuple of its rows and columns, respectively. Obviously, Rm and См
determine Μ in a unique way. The condition of the problem says that
Rm, Π Rm2 ^ 0 and CMl П Сма ^ 0
for any two matrices Mi, M2 G T. Let us take the families
П = {RM :M eJ7} and С = {CM : Μ G J7}.
Then TZ and С are families of subsets of к elements of a set of η elements
such that any two members of К and С have nonempty intersection,
respectively. Thus,
ι*·· и *(;:;)
by the famous Erdos-Ko-Rado theorem, and so
Obviously, the bound (^ij) is sharp since this is exactly the number
οι к χ к submatrices containing a given element. D
Problem C.22. Let us color the integers 1,2,..., N with three colors
so that each color is given to more than iV/4 integers. Show that the
equation χ = у + ζ has a solution in which x,y,z are of distinct colors.
Solution. We prove the statement by contradiction. Suppose that we can
color the integers 1,2,..., N with red, green, and blue so that each color is
given to more than N/4 integers and there are no ж, у, ζ of distinct colors,
where χ = у + ζ. We may assume that 1 is red. Then by our hypothesis,
there are no green and blue integers such that their difference is 1. We will
call a nonempty set S С {1,..., Ν} an interval if it consists of consecutive
integers, that is, if there are some integers 1 < a < b < N such that
S = {s : a < s < b}. If we delete the red integers, then the remaining
integers can be partitioned into intervals. Furthermore, each interval is
monochromatic by the observation above, since if an interval contains some
integers χ < у of distinct colors, then there are two consecutive integers of
distinct colors among x,x + l,...,y.
Suppose that there exist intervals of at least two integers in both colors
green and blue. Let A and В denote the longest green and blue interval,
respectively. If a G A and b G B, then \a — b\(= a — b or b — a) is not
red. Hence, the set С = {\a — b\ : a G A,b G B} does not contain any red
integer. If A = [αϊ, аг], В = [61,62], then
Г [61 - а2,62 - αϊ] if а2 < 6Ь
Ι [αϊ -62,а2 -6i] if 62 < аь

148
3. SOLUTIONS TO THE PROBLEMS
so it is an interval of |Α| + |i?| — 1 > |A|, |£| integers. The set С does not
contain any red integers, so it is monochromatic, as we have seen above.
But it is a contradiction to the maximal choice of the intervals A and B.
On the other hand, at least one of the colors green and blue contains two
consecutive integers, since otherwise the number of red integers would be
greater than or equal to the number of green or blue integers. In that case,
the number of blue or green integers would be at most iV/2, a contradiction
to the assumption that each color is given to more than N/4 integers.
Suppose now that there are no two consecutive green integers but there
are two consecutive blue integers, that is, \B\ > 2 for the longest blue
interval B. If 1 < s < N is green, then s > 2 and s — 1 and s + 1 are
red. Suppose that the distance between any two green integers is at least 3.
Then the intervals [s — 1, s + 1] are pairwise disjoint for the green integers
s and each of them contains two red integers. If N is not green, then it
implies that the number of red integers is at least double the number of
the green integers, which is impossible if each color is given to more than
N/4 integers. If N is green, then it implies that nr > 2ng — 1, where
nr and rig are the number of red and green integers, respectively. If 2 is
not green, then 1 is not counted in the intervals above, so nr > 2ng, a
contradiction again. If 2 is green, then take a blue interval В = [61,62]
such that \B\ > 2,62 < N. Now, 62 — 1 is blue, 62 -l· 1 is red, 2 is green,
and they constitute a solution to the equation χ = у + ζ such that χ, у, ζ
are of distinct colors, a contradiction.
Thus, we may assume that there is a green integer s such that s + 2 is
green as well. Either 61 > s + 2 or 62 < «s, since \B\ > 2 and BC\{s, s + 2} =
0. We may assume that 61 > s + 2. Consider the set
С = {b - s : b e B} U {b - s - 2 : b e B}.
Since 61 < 62 so С is an interval such that \C\ = \B\ + 2. The interval С
does not contain any red integers, so it is monochromatic. But С is not
green because \C\ > 2 and not blue because of the maximal choice of В, а
contradiction. D
Problem C.23. Suppose that a graph G is the union of three trees. Is
it true that G can be covered by two planar graphs?
Solution. The answer is no. We construct a graph that is the union of
three acyclic graphs but that cannot be covered by two planar graphs. Any
acyclic graph can be extended to a tree, so it implies the statement.
Let A and В be a set of η and (3) elements, respectively, where the
value of η will be determined later. Let A U В denote the vertex set of
the graph. Let us choose three elements of A in all possible ways and join
each of these triples to an element of В so that distinct triples are joined to
distinct vertices in B. Let G denote the resulting bipartite graph. It can
be obtained as the union of three acyclic graphs in the following way: for
any vertex b G B, the three edges incident to b are put into three distinct

3.2 COMBINATORICS
149
subgraphs. Any vertex b G В is of degree one in each of the three resulting
subgraphs, so these subgraphs do not contain any cycle.
Suppose that G is the union of two planar graphs G\ and G2. We may
assume that G\ and G% have no common edge.
For every vertex b G B, let us proceed as follows. Delete one of the edges
of G incident to b so that the remaining two edges belong to the same graph
G{. (It can be done since two of the edges incident to В always belong to
the very same subgraph.) Now, replace the vertex b and the remaining two
edges incident to b by one edge joining the end vertices of these two edges
in A. If G\ and G% are planar, then the resulting graphs are as well. The
resulting graphs are defined on A. The number of replacements is (3), and
an edge is obtained in at most η — 2 different ways since a couple can be
extended to a triple in at most this many ways. So, the resulting graph Η
has at least
(3) =n(n-l)
n-2 6
edges. It is known that a planar graph of η vertices has at most 3n—6 edges
(if η > 3). Thus, Η has at most 6n — 12 vertices. From these estimates,
we obtain
^<6n-12,
that is,
n2 - 37n + 72 < 0.
But this inequality does not hold if η > 35. Thus, G cannot be covered by
two planar graphs if η > 35. D

150 3. SOLUTIONS TO THE PROBLEMS
3.3 THEORY OF FUNCTIONS
Problem F.l. Prove that the function
1
dx
/«=/'
^/(x2-i)(i-u2x2)
(where the positive value of the square root is taken) is monotonically
decreasing in the interval 0 < ϋ < 1.
Solution 1. Substitute
t =
1 - ϋ2χ2 '
While χ increases in the interval I 1, — ], the value t increases in (0, +oo).
Differentiating the relation
ϋψχ2 - t2 + x2 - 1 = 0,
we obtain
dx _ t{l - ϋ2χ2)
dt " χ(1 + ϋ42)
and, consequently,
dx
y/{x2 - 1)(1 -"^i5)
dx Г+ос dx dt
Π dx _ Γ°° dx
Λ ί(1-ΰ2χ2)~70 dt
ί(1 - №x2)
_ f+°° dt _ [+0° dt
Now, for increasing ϋ the integrand decreases. Since the limits of
integration are independent of ϋ, the integral is also monotonically decreasing. D
Solution 2. Map the first quadrant of the z-plane (excluding the points
2=1 and ζ = Ι/ΰ by semicircles open from below) to the ги-plane with
the help of the function
<K
w —
Jo x/TT1
уЧ1-С2)(1-02<2)
(taking the value of the square root that is positive on the positive half-
axis). If, starting from 0, ζ = χ + iy runs through the segment 0 < ζ < 1,
then starting from 0, w = и + iv obviously runs through the segment

3.3 THEORY OF FUNCTIONS
151
If 1 < ζ < —, then it is clear that w runs through the segment
ϋ
u = A, 0<v < ,, = B.
Jo
dx
y/(x2-l)(l-u2x2)
If ζ runs through the segment 1/ΰ < ζ < oo, then w runs from the point
A + Bi along the horizontal line ν = Β to the point (A — C) + Bi, where
C: / + " *"
/
^ W
VV - l){u2x2 - 1)
On the other hand, if, starting from 0, ζ runs through the positive part of
the imaginary axis, then, starting from 0, w runs through the segment
и = 0, 0 < υ < Д
where
/»+oo
Jo
dy
V(i + y2)(i + i»V)
Since the mapping is domain preserving, we have the relations
A = С, Б = Д
the latter of which implies the statement. D
Problem F.2. Denote by M(r, f) the maximum modulus on the circle
\z\ = r of the transcendent entire function f{z), and by Mn(r,f) that of
the nth partial sum of the power series of f(z). Prove the existence of
an entire function fo(z) and a corresponding sequence of positive numbers
т\ < Г2 < · - - —> +oo such that
v Mn(rn,/o)
hmsup —— — = +oo.
n-^oo M(rn,/o)
Solution. By a theorem of Fejer, there exists a power series
oo
/(ζ) = "£αηζη,
71=0
which defines a regular function in the disc \z\ < 1 such that \f(z)\ < 1
and such that the sequence of the partial sums
Sr{z) = ^2,anzn
n=0
is unbounded at the point ζ = 1.

152
3. SOLUTIONS TO THE PROBLEMS
We shall use this function f(z) for constructing the function fo(z) that
meets the requirements of the problem.
We define sequences of numbers nk, mk, and ck as follows. Let no =
mo = Co = 0. Then suppose that n^_i, m^_i, and ck-i have already been
defined.
We begin by defining mk. Since limsupsn(l) = oo, there is an mk with
η—>+οο
k-1
]
1*1=*'
(1)| > к у max|sni(z)|;
in addition, we may assume that mk > nk-\.
We now define ck. Since smk(z) is continuous, for ck sufficiently close to
1 we also have
k-l
(ck)\ > к у max|sni(z)|.
l^olzl=k
We additionally assume that ck is real, further
ck>ck-1 and ck>l--^.
Finally, we define nk in the following way. Since sn(z) tends to f(z) in
the disc \z\ < 1, for sufficiently large nk we have
max IsnjbWI < 1·
\z\=ck
On the other hand, the absolute convergence of the power series of f(z)
implies the convergence of Y^=0 |an|c£; thus, for sufficiently large nk,
Σ ia»ic£ < L
П>Пк
In addition to these two inequalities, we require that nk satisfy nk > mk.
Now consider the power series
ni П2 η
Σα«2"+ Σ α"(|) + ···
η=1 η=η\+1
ΟΟ Пк п ОО
Σ Σ °»Ш =Σ{-»*©—»*-t©}·
k=l n=nfc_i+l /c=l
We show that the function /o(^) defined by this power series has the desired
property.
Prescribing an arbitrarily large positive R, for к > 2R and \z\ < R we
have \an(z/k)n\ < |an|(l/2n). Since ΣΖ=ο |an|(l/2n) is convergent, the

3.3 THEORY OF FUNCTIONS
153
power series of fo(z) is absolutely convergent in the disc \z\ < R. Thus
/0(2:) is a transcendental entire function.
We give an upper estimate for fo(z) on the circle \z\ = kck. In view of
our previous remarks,
i/oMi<EMfb".-.(f)b~-.©
u=i I
OO Tlfc
+КШ1+ Σ Σ 41)
Z=/e+l η=Πί_ι+1
к—1 оо Пк /τ \ η
-2Σ,1^ ismWi+livfxisnfc(2)i+ Σ Σ κι(-γ)
/=0l*l-*c* |z|-c /=H1 n=«,.1+l \ * /
k-l
<2Vmax|s„,(2)| + max |s„fc(z)| + V |an|c£
*—' \z\=k \z\=cu *—'
/=0'
fc-1
|z|=cfc
n>rifc
Setting ΣΪ~β max|z|=fc |sniO*)| = Tk, we obtain
М(Агсь/0)<2Г*+2.
Next we give a lower estimate for the corresponding partial sum on the
circle \z\ = kck; since we are concerned with the maximum, it is sufficient
to do this at the point ζ = kck.
The modulus of the m^th partial sum =
I k-l
= \smk (I) --»»_, (|)+ΣΚ (f) --».-t (I)}
/=i
fc-l
that is,
> |smfc(c*)| - 2 ^ max |βηι(ζ)| > (k - 2)Tk;
Mmk(kckJ0) > (к - 2)Tk.
Now, since limsup|snz(l)| = +00, the numbers
Z->+oo
k-l
tend to +00. Thus, for sufficiently large к we have Tk > 1 and,
consequently,
Mmk(kck,f0) (k-2)Tk fc-2
M(kck,f0) ~ 2(Tfc + l) 4 ·

154
3. SOLUTIONS TO THE PROBLEMS
Therefore, taking rmk = kck, and for η different from the rrik choosing rn
with the consideration of monotonicity but otherwise arbitrarily, we obtain
r Afn(rn,/0) ,.. Afmfc(rmfc,/0) ^r fc-2
hmsup —— r-f > hmsup * — > hmsup —-— = +00 . D
n-^+00 M(rn,/o) fc-^+oo M(rmfc,/o) fc-^+oo 4
Problem F.3. Let Η be a, set of real numbers that does not consist
of 0 alone and is closed under addition. Further, let f(x) be a real-valued
function defined on Η and satisfying the following conditions:
f(x)<f(y) ifx<y andf(x + y)=f(x)+f(y) (x,y G H).
Prove that f(x) = ex on H, where с is a nonnegative number.
Solution. Let xq be an element of Η other than 0, and let с = f(xo)/xo.
It can be seen by induction that f(nx) —nf{x) (n = 1,2,...) for every χ
in H. Therefore
f(nx0) = cnxQ. (1)
Let у be an arbitrary element of H. Then there is a positive integer no
such that {y-\-tiqXq)xq > ®' ^ n 1S a sufficiently large positive integer, then
there exist two positive integers mn and μη such that
mnxo < n(y + щхо) < μηΧο and \mn - μη\ = 1. (2)
In view of the monotonicity of the function /(rr), we have
f(mnx0) < f (n(y + n0xo)) < ί(μηΧο),
so by (1),
cmnxo < nf(y + n0xo) < c/inx0.
It follows that
c—xq < f(y + n0xo) < c—xq. (3)
η η
According to (2),
mn μη
—x0 ->y + n0xoi —xo -* У + rco^o·
η η
Thus, from (3) we obtain that f(y + noxo) = c(y + tiqXq). Consequently,
f(y) = /(y+^o^o)-/(^o^o) = c(y+n0xo)-cn0xo = cy. The monotonicity
of the function f(x) yields с > 0. D

3.3 THEORY OF FUNCTIONS
155
Problem F.4. Show that if f(x) is a real-valued, continuous function
on the half-line 0 < χ < oo, and
then the function
satisfies
poo
/ f2(x)dx < oo,
Jo
g(x) = f(x)-2e-x Γ elf{t)dt
Jo
poo poo
/ g2{x)dx= I f2(x)dx.
Jo Jo
Solution. We assume that the function f(x) is square integrable in the
Lebesgue sense on the half-line (0, oo). The relation
f(x)-g(x) = 2e-' /%*/(*)* (1)
Jo
implies that
(f(x) — д{х)У = f{x) + g(x) almost everywhere. (2)
Using the Schwarz inequality, we obtain
β~ω [ e*f(t) dt\ <е-ш\[ e*f (t) dt\ + β~ω \[ e%f(t) dt
\J0 I L/O \Ju>/2
<e~
iJo
•ω/2 \ 2 / ρω/2 >
e2tdt\ / f2{t)dt
+ e~
e2tdt
/ω/2
/ω/2
f(t)dt
pOO pOO
<е~ш'2 / f2(x)dx+ / f2(t)dt,
JO Ju/2
whence it follows that
lim е~ш ί e*f(t) dt = 0.
ω-οο J0
(3)
Relation (1) assures that f(x) —g(x), and therefore (1/2) (f(x) — g{x))2, is
absolutely continuous on each bounded subinterval of (0, oo). By (1) and
(2),
j\nx)-94x))dx=[{m^)dx
U{x)-g{x)?
= 2e
-2ω
JO
(£e'mdti-
Making use of (3), the statement follows. D

156
3. SOLUTIONS TO THE PROBLEMS
Problem F.5. Prove that for every convex function f(x) denned on
the interval — 1 < χ < 1 and having absolute value at most 1, there is a
linear function h{x) such that
ί \f(x)-h(x)\dx<A-VS.
Solution. We prove more than stated. We establish the existence of a
constant к such that
/.
ι
\f(x)-k\dx<A-Vs. (1)
ι
Without loss of generality, we may assume that f(x) is continuous even
at the endpoints of the interval [—1,1] and that /(—1) > /(1). Since
f(x) is continuous and convex on [—1,1], there is a largest interval [ci,C2]
(—1 < c\ < С2 < 1) on which f(x) is minimal. Introduce the notation
min f(x) = p, max f(x) = q.
z€[-l,l] *€[-l,l]
Let φι (у) be the inverse of the restriction to [— l,ci] of the function f(x),
and let
( , (ГЧУ) if P<V<№,
<hW |χ .f /(1)<у<д>
where f~1(y) denotes the inverse of the restriction to [c2, 1] of the function
f(x). Obviously, the function ф(у) = ф2(у) — ф\{у) is continuous and
strictly increasing on the interval \p,q\; further φ(ρ) = φ2(ρ) — φι(ρ) =
02 — Ci and φ(ς) = φ2(θ) — Ф\{ч) — 2. We distinguish between two cases.
a. If c2 — c\ < 1, then by the above properties of the function ф(у) there
is one and only one number к (G \p,q\) with ф{к) = 1. It can be shown
that к satisfies (1). Put ф\(к) = d, ф2(к) = e, D = (d,k), Ε = (e,fc),
G = (-1, fe), Η = (1, fe), A = (-1,1), and В = (1,1). The lines AD and
BE intersect at a point F. By the convexity of f(x) and the relation
|/(#)| < 1? ^ is obvious that the graphs of the restrictions of f(x) to the
segments [— l,d], [d, e], and [e, 1] lie in the triangles AGD, DFE, and
EHB, respectively. Therefore,
/ \f(x)-k\dx= [ \f(x)-k\dx+f \f(x)-k\dx+ [ \f(x)-k\dx
J-l J-l Jd Je
< t(AGDA)+t(DFEA)+t(EHBA),
where t stands for area. If F lies above the line у = — 1, then — DE
being the mid-parallel of the triangle ABF — the altitudes perpendicular

3.3 THEORY OF FUNCTIONS 157
to GH of the triangles in question are equal, so the sum of the areas of
the triangles is
-mGD+-m'DE+-mEH = -m{GD + DE + EH) = m < 1, (2)
Ζ Ζ Ζ Ζί
where m = BH. If F lies below the line у = — 1, then (denoting by J
the point of intersection of the line у = — 1 and the segment AF, and
further denoting by К the point of intersection of the line у = — 1 and
the segment BF) it is obvious that the graph of the restriction to [d, e]
of f(x) remains in the trapezoid Τ determined by the vertices D, J, Κ, Ε
and, consequently,
/
ι
\f{x) -k\dx< t(AGDA) + t(T) + t(EHBA),
/-1
while m = Β Η > 1. Then
t(AGDA) + t(EHBA) = \mGD+\m'EH = \m.
Zi Li Li
Since J К = (2m — 2)m_1, we have
,m. 1 /2m-2 Л /л
ί(Τ) = 2(—+ 1J(2-m)·
It follows that
/
1 ι ,/ χ , ■ , 4m - 2 - m2 A (2
\f(x) -k\dx< = 4 - — + m
m \m
1/2
/2 \ '
< 4 —2 ί — m) =4-V§.
(3)
Relations (2) and (3) imply (1).
b. If c2 - ci > 1, then let fe = p. Setting A = (-1,1), В = (1,1), D =
(ci,p), £ = (c2,p), G = (—l,p), and if = (l,p), it follows as before that
/
ι
\f(x) -p\dx< t{AGDA) + t{EHBA)
-1
< -m · GO + -m · EH = -m (1 - (c2 - ci)) < 1,
2 2 2
which completes the proof. D
Remarks.
1. Several participants have noted that the estimate cannot be improved
in general. For instance, in the case of the function
fix) = i~l if 0 < χ < 1 - V2/2,
1 ll + V§(a?-l) if 1-\/2/2<ж<1,
the estimate is the best possible.

158
3. SOLUTIONS TO THE PROBLEMS
2. Paul Turan called the attention of the organizing committee to the
fact that S. Bernstein (Doklady Akad. Nauk SSSR, 1927, 405-407) had
proved the following theorem:
Theorem. Let f(x) be an η +1 times differentiable function on the
interval [—1,1] satisfying the condition f(n+1\x) > 0 on [—1,1]. The expression
j \f{x)-Rn{x)\dx,
where Rn(x) denotes a polynomial of degree n, is minimal when Rn(x) is
the Lagrange interpolation polynomial of degree η that coincides with f(x)
at the points cos (hn/(n + 2)) (h = 1,2,... η + 1).
Problem F.6. Find all linear homogeneous differential equations with
continuous coefficients (on the whole real line) such that for any solution
f(t) and any real number c, f(t + c) is also a solution.
Solution. As usual in the literature, we restrict attention to differential
equations with the coefficient of the term of highest order identical to 1.
Let the differential equation
y{n)(x) + /i(x)y(n_1)(«) + · · · + Ux)y{x) = о (l)
have the desired properties, and let φ(χ) be a solution. Let с be an arbitrary
real number. Obviously,
—*^- = {-^)t-x+c (г = 1'2'-'п)' (2)
since together with ф{х) ф(х + с) also satisfies (1), it follows that
that is,
фЩ) + h(t - c)^"-1^) + · · · + fn(t - с)фЦ) = 0,
and this is true for any real constant с and all real values of t. This means
that all solutions of (1), so for example, η linearly independent solutions
of (1), satisfy the differential equation
У(п)(х) + /i(x - с)^-^*) + ··· + /„(*- c)y{x) = 0; (3)
hence — as is well known — it follows that the coefficient functions of the
differential equations (1) and (3) coincide:
fi(x) = fi(x-c) (г = 1,2,...,п). (4)

3.3 THEORY OF FUNCTIONS
159
Choosing χ = 0, we obtain /i(0) = /*(—c) for every с (г = 1,2,..., η). Thus
fi(x) = /г(0), and so the differential equation (1) has constant coefficients.
Conversely, if (1) has constant coefficients, then from (2) it follows easily
that together with any solution ф(х) ф(х + с) is also a solution. D
Remarks.
1. Assuming the conditions for a single с only, from (4) we obtain that the
coefficient functions are periodic with period c.
2. If, for any solution ф(х), ф(х + C\) and ф(х + c2) are also solutions,
then together with ф(х) obviously ф((х + c{) + c2) = ф(х + C\ + c2)
is also a solution. Consequently, assuming the conditions for a (finite
or infinite) set {ca}, the condition will also be fulfilled by the elements
of the smallest additive semigroup that contains the numbers ca. In
view of the previous remark, this implies that it is sufficient to assume
the conditions only for values of с that generate an additive semigroup
containing a sequence that tends to zero.
Problem F.7. Let F be a closed set in the η-dimensional Euclidean
space. Construct a function that is 0 on F, positive outside F, and whose
partial derivatives all exist.
Solution. Define a function фг{у) in the following way:
ФЛу) =
_ J ev-r2 if |y| < r2,
0 if \y\>r2.
Obviously, фг(у) is positive if |y| < r2, and infinitely difFerentiable in the
intervals (—oo,r2) and (r2,+oo). Further, it is clear that (dkфг(у)/dyk)
has the form Як(у)Фг(у), where Rk(y) is a rational fractional function; so
lim ί^ω lim ^l = o.
x->r2+o dyk x->r2-o dyk
Therefore (акфг(у)/аук) is continuous in (—oo,r2) and (r2,+oo) and has
right-hand and left-hand limits at the point у = r2, whence it follows by
induction that фг(у) has continuous derivatives of any order at у = r2 and,
consequently, at all points y.
Denote by En the η-dimensional Euclidean space. For α G En, χ G En,
а — (αι,α2,...,αη), χ = {χι,χ2,... ,xn) put
ya(x) = (:ri - αϊ)2 + (x2 - a2)2 + · · · + (xn ~ a>n)2·
Obviously, for every x, all partial derivatives of ya(x) exist.
Finally, let G С En be an open ball of center a and radius r chosen
arbitrarily, and let
fG(x) = фг{Уа(х))'

160
3. SOLUTIONS TO THE PROBLEMS
Since фг(у) is infinitely differentiable for any у and all partial derivatives
of ya(x) exist for any x, all partial derivatives of /g(#) exist for any x.
Moreover, it is evident that if χ e G then 0 < ya(x) < r2, so fc{x) > 0,
while if χ £ G then /g(#) = 0. Finally, each partial derivative of /g(#) is
bounded since it is continuous everywhere and zero outside G.
The complement F of F relative to En is an open set, so it can be
represented in the form F = U^=1G^, where Gi, G2,... are open balls.
Since each partial derivative of the functions fc1(x), fG2(x)i ··· 1S
bounded, there exist positive constants ck {k = 1,2,...) such that all
partial derivatives of order not higher than к of the function fck (x)
(including fGk{x) itself) have absolute value less than (l/c^)(l/2/c) for every
x. Then, obviously, the function series
00
J2ckfGk{x)
is absolutely convergent; we set
00
F(x) = ^ckfGk(x).
k=l
Forming a partial derivative of order г of the terms of the series, the
definition of ck ensures that, beginning with the ith term, the absolute
value of each term can be majorized by 1/2*, so the sum of the partial
derivatives is absolutely convergent. It follows that all partial derivatives
of F(x) exist. Finally, it is clear that F(x) = 0 for χ e F and F(x) > 0 for
χ $. F, thus the function F(x) has the required properties. D
Remark. Let Fi and F2 be disjoint closed sets in the n-dimensional
Euclidean space. By a similar method, one can construct an infinitely
differentiable function that is equal to 0 on F\, equal to 1 on F2, and
positive and less than 1 outside F\ and F2.
Problem F.8. Let f be a continuous, nonconstsnt, real function, and
assume the existence of an F such that f{x + y) = F[f(x), f(y)] for all real
χ and y. Prove that f is strictly monotone.
Solution. Suppose that / is not strictly monotone. Then by the continuity
of / there exist real numbers si < s2 with /(si) = /(s2). If ε > 0 is
arbitrary, then by the continuity of / there are values t\ < t2 in the closed
interval [si, s2] such that t2 — £1 < ε and /(£i) = /(^2)· Then, however,
f[t + (t2 - to] = f[(t - «o + *2] = F[f{t - to, /(*2)]
= F[f(t - to, /(*01 = № - h) + h] = №
for all real values t. Thus r = t2 — t\ is a period of /. Since τ < ε where
ε > 0 is arbitrary, it follows that the continuous function / has arbitrarily
small periods. Hence / is constant, contrary to the assumption. D

3.3 THEORY OF FUNCTIONS
161
Problem F.9. Let к be a positive integer, ζ a complex number, and
ε < 1/2 a positive number. Prove that the following inequality holds for
inRnitely many positive integers n:
>(\-еГ.
Solution. Put
αη(ζ) = αη= 2^ I £ )z ' η =1,2,....
We have to prove that limsup y/\an\ > 1/2.
If \w + wk+1z\ < 1, then
Therefore, the series 1 + X)^Li fln^n is the power series of the function
1/(1 — (w + г^/с+12;)) at the point u? = 0. According to the Cauchy-
Hadamard criterion, it is sufficient to establish that the radius of
convergence of the power series is not greater than 2.
For this purpose, it is sufficient to show that some zero of the polynomial
1 — w — wk+1z (k > 1) has absolute value not greater than 2. If \l/z\ <
2/c+1, then this follows from the fact that the product of the zeros of the
polynomial equals 1/z. If \l/z\ > 2/c+1, then |гу/с+12;| < 1 < |1 — w\ on the
circle \w\ = 2, so by Rouche's theorem the polynomial considered and the
polynomial 1 — w have the same number of zeros inside this circle.
In the case ζ = —1/4, к = 1, using relation (1), it is easy to see that
an(—1/4) = (n + l)/2n. This shows that the assertion concerning the
limsup cannot be improved. D
Remarks.
1. One participant proved the following, stronger, statement:
limsup |anI1/71 > fc/(fc + l), and this is sharp for every k;
limsup\an(z)\l/n assumes its minimum for ζ = (—kk/(k+ l)/c+1). (We
omit the proof, which is rather long.)
2. Several participants proved that the an satisfy the following recursive
definition:
ao = · · · = dk = 1, αη+ι = 0"n + a>n-kZ if η > fe.

162 3. SOLUTIONS TO THE PROBLEMS
Problem F.10. Let f(x) be a real function such that
x—>+oo ex
and \f"(x)\ < c\f'{x)\ for all sufficiently large x. Prove that
urn Ш _ ,.
x—>+oo еж
Solution 1. Suppose that \f"(x)\ < c\f'(x)\ for χ > x0. We first show-
that χ > xq and t < l/c imply
Ι/'(* + ί)<-^Ι/'(*)Ι· (!)
We may assume that \f'(x +1)\ > \f'(x)\, since otherwise (1) is trivially
fulfilled. Put
to = min{*': t' > 0, |/(:r+ 01 = l/(* + *)|}·
Since the function |/'(£)l1S continuous and does not intersect the horizontal
line of ordinate |/'(# + *) Ι f°r £ € [x,x + t0) while it remains under this line
at the point x, therefore |/'(0I ^ If'(x+t)\ on the whole interval [x, x+to].
Prom the Lagrange mean value theorem,
\f(x + t)\ - |/'(χ)| < № + *>)1 - Ι/'ΟΌΙ = ίο|/"(01
<Μ/'(0Ι< W(* + *)I>
which gives (1).
Prom (1) we see that if f'{x) = 0 for some χ > xo, then \f'(x')\ = 0
for all x' > x; but this is impossible since then e~xf(x) would tend to
0. Consequently, f'{x) is either positive for all χ > xq or negative for
all χ > xo- However, because e~xf(x) —► 1, the function f(x) cannot be
monotone decreasing. Thus
/'(*) > 0 (2)
for all χ > xo-
Prom (1) and (2), we obtain that in case χ > x0 + l/c and t < l/c,
(1 - ct)f(x + u)< f'(x) < —^/'(* - «)
for 0 < и < t. Hence, by integration,
(l-ct)[f(x + t)-f(x)]<tf'(x) < Y±-j\f{x) -f(x-t)},

3.3 THEORY OF FUNCTIONS
163
that is,
l-ct
f{x + t)c% f{x)
pX + t
<
/'(*)
<
/(*) _ /(s-*)c-t
t(l - ct)
We first consider the left-hand side of (3). For fixed t,
x—>oo у e^+t g# /
(3)
Hence
and therefore
/'(*)
lim inf
x—юо еа
> (1-ci)-
liminf^^> lim
(i-rf)-
1
*
1.
x—>oo e^ t—>+o
From the right-hand side of (3), it similarly follows that
1 1-е"*'
/'0*0
lim sup < lim
я^оо еж t->+o
which proves the statement. D
1-е* ί
= 1,
Solution 2. We shall make use of the following theorem, which is well
known and easy to prove.
Theorem. Let g be a two times differentiable function such that
Итж_+00 g(x) exists and is finite, whereas Ιρ'Όε)! < С if χ > xq for suitable
real numbers С and x$. Then Цт^^+оо g'{x) = 0.
Now let g(x) = (f(x)/ex). Then
jM _ /'(*)-/(*) - χ _ ffl-2/;(x) + /W
» ^ ~ ex ' 9 W ~ ex
By assumption, \imx^+00 g(x) = 1. From the other assumption, it
follows that for sufficiently large χ the derivative f'{x) is of constant sign.
Since limx_+00 f(x) = +oo, it must be positive. Thus, if x\ is sufficiently
large and χ > x\, then
\f\x) - f(Xl)\ < Γ \f"(t)\dt < с Г f'(t)dt = /(*) - /On).
JX\ JXl
Therefore |/"0e)I < c'f(x) and 1/4^)1 ^ c'f(x) with a suitable constant d
and, consequently, ^"(ж)! < (3c' + l)^(^) if ж > #ι. It follows that, for
some С and rr0, Ь'Ч^)! < С Ίΐ χ > xq.

164 3. SOLUTIONS TO THE PROBLEMS
By the theorem stated at the beginning,
lim g'(x)= lim IM^IM. = 0,
x—>+oo x—>+oo еж
whence \imx^+00(ff (x)ex) = 1. D
Remark. In some sense, it is necessary to assume that \f" / f\ and |(/'|
are bounded. To show this, let
9(x) = 1 +
sin ж2
ж
Then
and
- sin x2
g'[x) = (- 2 cos χ
liminf g'(x) = — 2 < 2 = lim sup </ (ж).
ж->+оо ж->+оо
Problem F.ll. Find aJi continuous real functions f,g and h defined
on the set of positive real numbers and satisfying the relation
f(x + y)+ g{xy) = h(x) + h(y)
for all χ > 0 and у > 0.
Solution. Choose у = 1 in the equation, and then
^х) = Л(*0-/(* + !) +Ml) (1)
for χ > 0. Substituting (1) into the original equation, we obtain
h{x) + h(y) - h(xy) = f{x + y) - f(xy + 1) + Ml)· (2)
Put
H(x,y) = h{x) + h(y) - h(xy).
Then
H(xy, z) + Я(я, у) = Я(я, yz) + Я(у, ζ) (3)
for any triple of positive numbers x, y, z. By relation (2),
H(x, y) = f(x + y)~ f(xy + 1) + Λ(1);
putting this into (3), we find that
f(xy + z)- f(xy + 1) + f(yz + 1) = f(x + yz) + f(y + z) - f(x + y)
(x,y,z>0). (4)

3.3 THEORY OF FUNCTIONS 165
Since / is continuous on the set of positive numbers, passing to the limit
ζ —> 0 (ζ > 0), from (4) it follows that
f{xy) - f(xy + 1) + /(1) = f(x) + f(y) - f(x + y). (5)
Introduce the notations
f*(t) = f(t)-f(t + i) + /(i) (*>o)
and
F(x,y) = f(x) + f(y)-f(x + y).
Then
F(x + y,z)+F(x,y) = F(x,y + z) + F(y,z) (x,y,z>0) (6)
and, by (5),
F(x,y) = r(xy). (7)
Putting (7) into (6), we obtain
/* (xz + yz) + Г (ху) = Г (xy + χζ) + Γ Ы (ж, у, ζ > 0). (8)
Hence, choosing ζ = l/у and writing
и = -, v = xy, (9)
У
it follows that
Π« + ΐ) + /» = Γ(« + «) + πΐ) (ίο)
for all positive values of и and г>, since for positive и and г> the system of
equations (9) can be satisfied by suitable positive values χ and y. Prom
(10), interchanging и and v,
/*(u + l) + /»=/> + l)+/*(«),
whence, taking г> = 1,
/•(и+1)=/» + Г(2)-Л1).
Substituting this into (10),
/* (« + «) +Г (i) = /*(«) +/'(«) + /· (2)-r(i),
whence, by the continuity of /*,
f*(t) = at + p,

166
3. SOLUTIONS TO THE PROBLEMS
where α and /3 are constants. Then in view of (5),
axy + β = f(x) + f(y) - f(x + y),
and therefore, with the notation f(x) = f(x) = (a/2)x2 — /3, we find that
f(x + y) = f(x) + f(y). Thus f(x) = ΊΧ and
f(x) = -^xi+7X + p. (11)
Putting (11) into (2), we see that the function
h(x) = h(x) + ^x2 -Ίχ-δ (<5=^-7-/3 + /ι(1))
Δ Δ
satisfies the equation
h{x) + h(y) = h(xy) (x, у > 0),
which yields h(x) = к\пх. Consequently,
h(x) = -^x2 + Ίχ + κλτιχ - δ. (12)
Finally, from (1), using (11) and (12), it follows that
g(x) = /dnrr + ax - 26 - β. (13)
We have shown that the solutions of the equation can only be the
functions of the forms (11), (12), and (13). On the other hand, it is easy to see
that the functions (11), (12), and (13) are solutions of the equation for any
choice of constants a, /3, 7, 6, and к. D
Remark. Most participants first show that /, g, and h are two times
continuously difFerentiable functions and then reduce the problem to a
differential equation. Several of them note that when using this method it
is sufficient to assume the integrability of /, g, and h on every bounded,
closed subinterval of the set of positive numbers.
Problem F.12. Let xo be a fixed real number, and let f be a regular
complex function in the half-plane Re ζ > xq for which there exists a non-
negative function F e Zq(—00,00) satisfying \f(a + ιβ)\ < ^(/3) whenever
a > xo, —00 < /3 < +00. Prove that
l
J ot
a+ioc
f(z)dz = 0.
ot—ioo
Solution. Let xo < a\ < a^. Let {/3n} and {7n} be sequences of real
numbers tending to +00 such that ^(/3η) —► 0 and F{—ηη) —> 0 . By the
Cauchy integral theorem,
ραι+ιβη ρα2+ΐβη pot2-i~fn pai-i~fn
/ f(z)dz+ f(z)dz+ f(z)dz+ f(z)dz = 0
J OL\—i~1n Joti+ίβη Jot2+iPn J Ci2 — i~1n

3.3 THEORY OF FUNCTIONS
167
(the path of integration is always the connecting segment). Since
pOt-2
1+ίβη
and similarly
therefore
f(z)dz
pa2
\ f(a + ίβη
)da
<(a2-a1)F(Pn)^0,
I/
Q!i-i7n
f(z)dz
< (a2 - ai)F(-7„) -> 0,
/ f(z)dz= f(z)dz,
J αι—гоо J a.2 — ioo
(1)
which means that the integral in question is independent of a. (Here the
improper integrals exist since the integrand admits an integrable majorant.)
Denote by A the common value of the integrals appearing in (1).
Apply our result to the function f(z)/z (which is analytic, for instance,
in the half-plane Re ζ > 1). Then
В = Г™ Mdz = ir f±±*&άβ, α > max{1,,o}
Ja-ioc z J-oc α + Φ
is independent of a. Let a —► oo. Since
I /(a + г/3)
a + г/З
<F(/3) (a>l),
integration and transition to the limit can be interchanged by the theorem
of Lebesgue, and we obtain В = 0. Then
A = A-aB= / f(z)h--) dz = - / /(a+ </?)-
β
+ ίβ
άβ.
Here, again, F(P) is a common majorant of the integrands, and for each
fixed value of β the integrand tends to 0 as a —* oo. Consequently, by the
theorem of Lebesgue, the integral tends to 0, whence A = 0. D
Remark. All participants first show that the integral under consideration
is independent of a. Then some of them choose the function f(z)/z and
the way described above, while others work with the function e~tzf(z),
where t > 0 is a real number. They establish similarly (using the fact that
this function also satisfies the conditions of the problem and tends to 0 as
Re ζ —* oo) that
/
а+гоо
e-tzf(z)dz = 0;
hence, letting t —* 0, the desired equation follows.

168
3. SOLUTIONS TO THE PROBLEMS
Problem F.13. Let πη{χ) be a polynomial of degree not exceeding η
with real coefficients such that
Κ (χ) Ι < у/1-х2 for -1<ж<1.
Then
\π'η(χ)\<2(η-1).
Solution. The background of the problem is provided by the Markov
inequality: If the polynomial P{x) of degree η has absolute value less than
1 through the interval (—1,1), then its derivative has absolute value less
than n2 on the same interval. Of similar type is Bernstein's theorem: If a
trigonometric polynomial of order η has absolute value not greater than 1,
then its derivative has absolute value not greater than n.
We have strengthened the hypothesis of Markov's theorem and wish to
prove an estimate much sharper than that of the Markov theorem. To this
end, we shall need Bernstein's theorem cited above as well as the following
theorem:
Theorem. If a polynomial Q(x) of degree A; satisfies
IQ(:r)'<Trhp for -i<x<l>
then
\Q(x)\ < к + 1 for the same x.
For both theorems, see for example, /. P. Natanson, Constructive
Function Theory 1-3, 1964~65 (translated from the Russian), sections V.l and
VI. 6.
Now we prove the statement of the problem. We may assume that
πη(χ) is nonconstant. Prom the relations πη(±1) = 0, it follows that
πη(χ) = (1 — x2)f(x), where f(x) is a polynomial of degree not exceeding
η — 2 and
\f(x)\<—L= for -1<я<1,
VT=
2
whence \f{x)\ < η — 1. Let χ = costf. It is well known that in this case
f(x) = /(costf) = F(u) is a trigonometric polynomial of order not
exceeding η - 2. Write G(i?) = F(tf)sintf. Since G(tf) arises from f(x)y/l-x2
by the substitution χ = costf, therefore |G(#)| < 1. The trigonometric
polynomial F(#) has order not greater than η — 2, so G(u) has order not
greater than η — 1, that is,
|C(tf)|<n-l. (1)
We calculate the derivative of 7rn(cos#) with respect to ϋ in two ways.
On the one hand, we have
—7rn(costf) = <(costf)(-sintf). (2)
air

3.3 THEORY OF FUNCTIONS
169
On the other hand,
J I
—7rn(costf) = — (G(tf)sintf) = G'(tf)sintf+ G(tf)costf
αϊ/ <ш
= G' (tf) sin i? + F (ϋ) sin 0 cos ϋ. (3)
Comparing (2) and (3),
-<(cos ι?) = G'(ti) + F(u) cos ι?, (4)
that is,
|<(x)| < (n - 1) + (n - l)|z| = (1 + \x\)(n - 1) < 2(n - 1). (5)
The proof is complete. D
Remark. It should be noted that for sin ϋ = 0, we cannot divide by sin #,
but the continuity of π'η(χ) ensures that the assertion of the problem is true
also in this case. Equality can only hold if πη{χ) = (1 — x2)Qn-2{x) or
the negative of this, where Qn-2(x) stands for the Chebyshev polynomial
of the second kind of degree η — 2. This case can also be characterized by
the relation G(#) = ±sin(n — 1)#. Then in the Bernstein inequality, for
the values ϋ corresponding to \x\ = 1, we have equality, and in the triangle
inequality applied in (5) the terms have equal sign, so equality really holds
for these πη.
Problem F.14. Let a(x) ала r{x) be positive continuous functions
denned on the interval [0, oo), and let
liminf(x — r(x)) > 0.
X—ЮО
Assume that y(x) is a continuous function on the whole real line, that it is
differentiable on [0, oo), and that it satisfies
y'(x) = a(x)y(x - r(x))
on [0, oo). Prove that the limit
lim y(x) exp < — / a(u)du >
x^°° I Jo J
exists and is unite.
Solution. Integrating (2), we obtain
y(x) = y{u) + f a(t)y(t - r(t)) dt {x>u> 0). (3)
Ju

170
3. SOLUTIONS TO THE PROBLEMS
If y(x) is any continuous function on the interval (—oo,0], then it is easy
to see that y(x) can be uniquely extended to the whole real line so that
(3) is valid. In fact, suppose that xq is the supremum of those values up
to which unique extension is possible. Then in the neighborhood of x$ of
some radius 6, the function r(t) is greater than some positive ε. Denote
by η the smaller of the numbers ε and δ. Then, by (3), the values of y(x)
taken for χ < xq — η/2 uniquely determine the values of у(х) in the interval
(#o — ^?/2, #o + v/fy- This contradicts the choice of xq.
A similar reasoning shows that if y(x) is positive on (—oo,0] then it is
positive on the whole real line. In this case, with the help of (2) we obtain
that y(x) is monotonically increasing on [0, oo). Put
z{x)=y(x)e-fox"(t)dt (x>0). (4)
Differentiating and using (2) we obtain
z\x) = a{x)e~ So "W dt [y(x - r(x)) - y(x)]. (5)
To prove the statement of the problem, first suppose that y(x) is
positive on (—oo,0]. Then, as we have seen, y(x) is positive everywhere and
increasing for χ > 0. Thus, ζ (χ) is positive for all χ > 0; further, ζ (χ)
is monotone decreasing for sufficiently large x. Equation (1) implies that
χ — r(x) > 0 if χ is large; so from (5) by the monotonicity of y(x) it follows
that z'{x) is negative. Since z(x) is positive and decreasing, Цт^-юо z(x)
exists.
To prove the general case, represent the function y(x) on (—oo, 0] in the
form
y{x) = У1(х)-у2{х), (6)
where у ι (χ) and y2 (χ) are positive, continuous functions. As we have seen,
2/i (x) and 2/2 (#) can be extended to the whole real line so that they satisfy
the differential equation (2) for χ > 0. By the uniqueness of the solution of
(2) mentioned above, it is also clear that (6) remains valid on the whole real
line. Defining the functions z\(x) and Z2{x) in analogy with (4), it follows
as above that the limits lmxc-^o z\(x) and Ит^-юо z2(x) exist. This implies
the existence of the limit
lim Z(X) = lim (z\{x) — Z2{x))>
X—ЮО V X—ЮО
The solution of the problem is complete. D
Problem F.15. Let \i (i = 1, 2,...) be a sequence of distinct positive
numbers tending to infinity. Consider the set of all numbers representable
in the form
oo

3.3 THEORY OF FUNCTIONS
171
where щ > 0 are integers and all but finitely many щ are 0. Let
L(x) = Σ l and Μ(χ) = Σ L
(In the latter sum, each μ occurs as many times as its number of
representations in the above form.) Prove that if
x—>oo L\X)
then
lim МШ = 1.
x—юо
M(x)
Solution. (When in the solution we speak of a μ, we mean not only its
value but also its representation by a fixed sequence {r^}, keeping in mine-
that a number can possibly be represented in several ways.)
If A*i = ΣηΙ ^г and μ2 = Ση! λίι then kt μι|μ2 mean that щ ' <
nf] (<= 1,2,...)· Then
μ =
Σ λ<
η>1 integer ,λ{
ηλί|μ
and
Σ"=Σ Σ λ-
μ<χ μ<χη>1,λί
ηλί|μ
In the inner sum, μ' = μ — ηλι is also a μ-number; let us sum with respect
to this. Then
Σ/-Σ Σ *.
μ<χ μ'<χ η>ι,λί
n\i<x—μ'
With the notation
Σ λ< = £(г/)
η>1,λί
n\i<y
(replacing μ' by μ) we have
Σμ=Σ£(χ-μ). (1)
μ<χ μ<Χ
Apply this to χ + 1 and subtract the two relations from each other:
Σ μ=Σ{£{χ+1-μ)-£{χ-μ))+ Σ £(*+1-*0·
χ<μ<χ+1 μ<Χ χ<μ<χ+1

172 3. SOLUTIONS TO THE PROBLEMS
The left-hand side is at least x(M(x+ 1) — M(x)), and the second sum
on the right-hand side is at most £(1) (M(x + 1) — M(x)) (since C(y) is
increasing). Consequently,
{x - £(1)) (M(x + 1) - M(x)) < Σ (C(x +1-μ)~£{χ- μ)).
μ<χ
Неге С(у) is expressed through the sequence λ^ alone, and from the
hypothesis relating to the latter we shall deduce that
whence for sufficiently large K, C(y+1)—C(y) < eC(y) if у > К. Therefore,
decomposing the sum based on whether μ < χ — К от μ > χ — Κ, and
using (1) again,
Σ {€{χ+1-μ)-ε{χ-μ))<ε £ C{x - μ)
μ<χ — Κ μ<χ — Κ
< ε Υ] C(x — μ) = ε Υ] μ < εχΜ(χ)
μ<χ μ<χ
and
]Γ (Цх +! - μ) - Цх - μ)) < ЦК + ΐ)Μ(χ).
χ—Κ<μ<χ
We have thus obtained
{χ - £(1)) (Μ (χ + 1) - Μ (χ)) < εχΜ(χ) + C{K + l)M(rr),
Μ(χ + 1) - Μ(χ) εχ + CjK+l)
М{х) ~ х-С{1) '
М(х+1)-М(х)
hmsup -η— < ε,
ж^оо Μ (Χ)
and since ε > 0 is arbitrary,
Μ(χ+Ϊ)-Μ(χ) Λ
hmsup — тт7~\ = 0·
It remains to prove (2). We may write
Α*)=Σλ< Σ ^ Σλ^ + Σ λ<
>|L(|) + |(L(a:)-L(|)) = lxL(x). (3)
Now
С(х+1)-С(х)= Σ λ<·
η>1,λί
X<n\i<X+l

3.3 THEORY OF FUNCTIONS
173
For a fixed ε > 0, decompose the sum according to the cases λ^ < ε (χ + 1)
and Xi > ε(χ + Ι):
^ ^ mmXi ^
\ι<ε{χ+1) * <η<*±1 \i<e{x+l)
Αί — Αί
(since the interval corresponding to η has length not greater than
l/minAi= constant). Further, using (3),
ттАЧ,<^+1) ттА<
< -Щ-Ь(х) < -^-£(z).
mm Xi mm A^
In the other part, η < (χ + 1)/ε(χ + 1) = Ι/ε, so
l<n<J f <λί<^ l<n<J \ \ / /
The inner sum consists of finitely many terms (for fixed ε > 0) and, by the
assumption, each of them satisfies the relation
Ч^)-*®51(: + 1Н(1) "('(;))-№:
))
as χ —► oo. Therefore, again by (3), the entire sum is o((x+ l)L(x))
o(C(x)). As a result,
C(x + 1) - C(x) < -4^Цх) + о (C(x)),
minAi
whence
C(x + 1) - C(x)
C(x)
The proof is complete. D
0.
Remark. The solution above is based on relation (1). Application of this
relation is motivated by the following argument.
If Xi = log pi, where Pi is the ith prime, then the numbers μ are the log η
where η > 1 are integers, each appearing once. Although the condition on
L(x) does not hold in this case, namely L(x + 1)/L(x) —> e, it is natural
to start from the formula applied in prime number theory (for example, in
the proof of Chebysev's theorem), the formula corresponding to the prime
factorization of n\.

174
3. SOLUTIONS TO THE PROBLEMS
Problem F.16. Let P(z) be a polynomial of degree η with complex
coefficients,
P(0) = 1, and \P(z)\<M for \z\ < 1.
Prove that every root of P(z) in the closed unit disc has multiplicity at
most Cy/n, where с = c(M) > 0 is a constant depending only on M.
Solution 1. It is sufficient to examine the multiplicity of number 1. In fact,
if we prove something for 1 then we may apply the result to the polynomial
p(z) = P(ctz) with \ct\ < 1, and in this way we obtain the same estimate
for all roots lying in the unit disc.
The idea of the solution is the following. We consider the integral
Γ2π
F(P)= / log|P(e^)|#
Jo
and show that it exists and is nonnegative. Then we estimate it from above,
once in the neighborhood of 1 with the aid of the multiplicity of 1 and the
degree of P, and once at other points using the condition \P(z)\ < M.
It is sufficient to prove the existence of the integral for polynomials of
the form ζ — zq. If
then
The existence of
P(z) = cf[(z-Zi),
i=l
η
log \P(z)\ = log \c\ + J2lo& \z - ζλ
i=l
ρ2π
/ \οΕ\β^-ζ0\άφ
Jo
is evident if |zo| φ 1. Next let |zo| = 1. Without loss of generality, we may
assume that zq = 1 (a substitution φ = η + φο takes them into each other).
Then
logH-l| = 2sin^,
and
Γ2π
I log M2 sin |j άφ
really exists and is equal to 0.
Next, compute the integral
ρ2π
/(«) = / ι
Jo
log
1- —
α
άφ (α φ 0)

3.3 THEORY OF FUNCTIONS
175
for α φ 1. Obviously, its value depends on the absolute value of α only
(again, a substitution as above may be applied), so it is the same for the
numbers αεί, ole^ - · · ,αεη, where the Ej are the nth unit roots. Therefore,
4/(α) = £/(«*) =/ ΙοβΠίΐ-—)
cty
/»2π
/*ζπ I pin</> I
= / Η1_^Η#·
./0 I a I
Now, if |a| > 1, then for η —> oo the integral on the right-hand side tends
to zero since 1 — ein<t>/an —> 1 uniformly; thus f(a) = 0. On the other
hand, if \a\ < 1 then
n(f(a) + 2nlog\a\) = / log \an - ein<t>\ άφ -> 0
since 1 - \a\n < \an - ein(t>\ < 1 + |a|n; so in this case f(a) = -2tflog \a\
is only possible. In each of the three cases, we have f(a) > 0.
In our case, the relation P(0) = 1 implies
P{z)
ЙН)·
whence
(1)
*хрн£/ы<о.
Now let P(z) = {z- l)kQ{z) = a0 + агг + · · · + αη2η, where Q(l) ^ 0,
We estimate F(P) with the help of k, n, and M. Let
/»27Γ /»ε /»2π—ε
F(P)= / = / +/ = Ή + *2·
Then
ρ2π
F2< log Μ # = 2tflog Af.
7o
We split Fi again into two parts:
Fi = У" log|(z - 1)к\аф + J* log|Q(e^)| άφ = F3 + F4.
Clearly,
F3 = k f log i2sin Μ J аф<2к ί ]χ%φάφ
= 2fes(log£-l).
(2)
(3)
(4)

176 3. SOLUTIONS TO THE PROBLEMS
For estimating F4, we need an estimate of Q, which we obtain from the
coefficients of the expansion of Q about 1. Let Q(l + z) = R(z). We
calculate the coefficients of R from those of Ρ using the formula R(z) =
P{z + l)/zk\
η j
Ρ(*+1) = έβί(*+1)^=έέβ/^);
j=0 j=0 m=0 ^ '
η η / · \
rrt=k i=m. N '
m=k j=m
since for m < к the coefficient of z™ is 0 by our assumption. Thus
R(z) = n£bmzm, bm= J2 °j(mJ+fc)·
m=0 j=m+k ^ '
Further, by the Cauchy inequalities, \a,j\ = \p(j\0)\/j\ < M.
Putting this into (5), we find
К\<м ± ( j \=m( n\l \
If \z\ = δ and δ(η - fc)/(fc + 2) < 1, then in view of (6),
1ВД1
(5)
(6)
Μ
oo
<
y6m( n + i \
n + \\ (л , гп — к ш г2п — кп — к — 1
<
k + lj \l + 6k + 2 +δ fc + 2 fc + 3 + "
Since \е*ф-1\ =2|sin(0/2)| < |0|, therefore if ε < ((fc + 2)/2(n-fc)), then
for |0| < ε we have
1^)1=ие--.)|<м(^;)1-^<2м(^;).
If к > 2, then using the relation f! > t'e_t we obtain
η + 1\ _ (n + l)n(n - 1)... (n - fc + 1)
fc+1/ (fc+1)!
_ (n2 - l)n(n - 2)... (n - fc + 1) nk+1 ( en \fc+1
(fc + 1)! < (fc+1)! < \k + l) '

3.3 THEORY OF FUNCTIONS
177
Thus
log \Q(e^)\ < log(2M) + (fe + 1) log
en
and, consequently,
F4 <2ε
log(2M)+ (*;+!) log
en
k + 1
Collecting everything, by (1), (2), (3), and (4)
2β7ϊ ЕП
(π + ε) log Μ + ε log , ι Λ + ek log , . . > 0.
к + 1
к + 1
(7)
Now if η = к2/2c, let ε = с/к (this fulfills the condition ε <((k + 2)/2(n -
A;))). Then (7) becomes
(. + f)logM + |log^- + clog-A->0.
We only make things worse if we also replace к + 1 by k. Moreover, c/k =
k/2n < 1/2 gives π + с/к < 4, c/k\og(ek/c) < (с/к) · (ek/c) = e < 3, so
finally
41ogM + 3>clog2,
c<81ogM + 6. ^
Since /с = \/2cn, relation (8) means that we have proved the assertion of
the problem with
c{M) = v/l61ogM + 12. D
Solution 2. It is sufficient to study the multiplicity of 1 (see the previous
solution). We establish the following lemma:
Lemma. Assume that the polynomial ωη(ζ) of degree η satisfies
|^n(^)| < Μ if |ζ| < 1. Let ζ = 1 be a root of multiplicity ί for ωη(ζ).
Then there exists an absolute constant c\ (c\ — (4ε/π) + ε) such that for
z = е^,\ф\< £π/2η we have
{z - iy \ i ) '
Proof. Setting LJn(z) = U2n{z), the assertion takes the form
U2n{z)
(z-l)
11
<
Μ*(ψγι,
in which form we shall prove it for all polynomials of degree 2n. Let the
roots of the equation z2n + 1 = 0 be
Zk
-* {k= l,2,...,2n).

178
3. SOLUTIONS TO THE PROBLEMS
Proposition. There exist complex numbers α& (к = ί, ί+1,..., 2η—£+1)
such that
/ ν 2n-i+l / 1 \ 2n-i+\
зфк,2п+1-к
Suppose that ω2η(ζι/) φΟ [ί < ν < 2η — ί+1). (Such v obviously exists
since ω2η has at most 2n — 2£ roots different from 1.) For ί <i <2n —ί+1,
i φ ν, put
1
ai~(Zi-iru2n-^ {«-*,)'
j^i,2n-i+l
Then for ζ = ζι (i φ ν) the left-hand side is equal to the right-hand side.
Indeed, the terms of the sum appearing on the right-hand side are zero
except for the term with к = г; Ίϊ к φ 2η — г + 1, then this is true because
of the factor
2n-£+l
Ц (zi-zj)=0,
зфк,2п+\-к
while if к = 2n — г + 1, then because of the factor
*—^- + 1 = 0.
For к = г, the two sides are equal because of the definition of a;. Now
choose the remaining av so that the coefficients of ζ2η~2£+1 and 1/z are
0. This can be done because the two coefficients are the negative of each
other since
2n-£+l
Π (-*>■) =l
3=t
зфк,2п+1-к
for all k\ together with each — Zj appearing in the product, its reciprocal
—Z2n+i-j also appears.
Thus, the condition
1 2n-£+l
- ]P akU2n{Zk) = 0
Z k=£
is to be fulfilled and, as α;2η(^ι/) φ 0, this is possible if we choose av properly.
But then both the left-hand and the right-hand members are polynomials
of degree 2n — 2£ which coincide at 2n — 2£ + 1 points (the points ζ = Zi,
ί < г < 2n — 2ί + 1, г φ ν), so they are equal identically. (Of course, hence

3.3 THEORY OF FUNCTIONS
179
it follows that they are equal for ζ = ζν and therefore, similar to the case
1
3=1
ϊφν,Ίη—ν+λ
also holds.)
Since \zf -l\ = \Zi- l/zi\ = \z{ - z2n+i-i\ and
2n
H(zi-zj) = \(z2« + l)'z=Zi\=2n,
3=1
зФ*
we have
2n
(z2-l) J] (zi-zj)
3 = 1
j^i,2n+l—г
= 2n.
Relying on this, and setting φ = π/η, we obtain the following estimate of
\a,i\ (for ί <i < n, but also for г > η because of the symmetry):
Ш =
2n-£+l
(Zi-l)^ Π (^-zi)
j^i,2n+l —г
Г*-1 2n
Lj = 1 j=2n-£+2
(z,-l)M
(*?-i)
2n
<
£-1 £-2
ro*-j>]I№+j>]·2
j=i j=o J_
' 2n
[<* - i) I]
2£-l
< A 2У a/ . < 22e .
(ί-ΙΫ'-'-φ^-ΐ-η
(We have made use of the relation τ/2 < sinr < r, τ < ττ/2.)

180
3. SOLUTIONS TO THE PROBLEMS
Now we have to give an upper estimate of the expression
2n-e+i
Π (*-**)
<4
2n-M-l
Π (*-**)
for each ζ = егг^, \ψ\ < (£π/2η). Considering pairs and setting ζ = егг^,
Zj = ег1?, we find that
\(ζ - Zj)(z - z2n-j+i)\ = 4sin -у- sin —γ-
= 2(cos^ - cosu) < 2(1 - costf);
that is, in the interval \ψ\ < (ίπ/2ή) the expression attains its maximum
at the point ζ = 1. On the other hand,
2n-i+l
Π ι1"**)
= 4-
n5i(i-«i)
<
4(l2n + 1)
. 21-2
<
Γβ=ϊα-**)| [n3=iu - i)!
8 /cpn\
|(^-2)! (f)
Since z — e1^, where |^| < (^/2) · (π/η), we have
2Imz
21mzi
< 1
for all i > ί, and therefore
^2n(^)
(z-l)
2i
^ с 2<2i *,2 /СоП\2^-2 ^o/CiTlX2^
<n.2.T.M».(^) <M^(-) ,
which proves the lemma.
To prove the assertion of the problem, let
9(ΰ) = \ωη(β^)\2,
un{z) = c Y[ (z-zv) Y[ (ζ-ζμ),
\ζν\<1 \ζμ\>1
<(*)=C Π ί1""-) Π (*"*μ)·
\ζν\<1
Ι*μΙ>1
Then by Example 43 in Gy. Polya, and G. Szego, Problems and Theorems
in Analysis, Springer, Berlin, 1976, vol 2, p. 82, \ωη{ζ)\ = |ω*(ζ)| for
\z\ = 1 (this is evident), and
ίϋιΜ- π ι,κι

3.3 THEORY OF FUNCTIONS
181
whence
μ;(ο)| > κ(ο)| = ι,
and by Example 53 in the same vol 2, p. 84,
/»2π
2π
1 Γ
- J \ogg{u) άϋ = log K(0)|2 > log 1 = 0.
On the other hand, g(u) < M2 for all tf, and for \ΰ\ < (£π/2η)
g(*) = μ*(β«)| < μ2 (ψ\ζ -1\)2£ < м2 (ψή2£,
so in this interval
bg g{$) < 2 log Μ + 2ί log °ψ ϋ.
We use this relation only for \ϋ\ < (ί/nci) (< {i/ri) · (π/2)). At the
remaining points, we apply log^(^) < 2 log Μ to obtain
0 < / log#(tf) du<2n-2\ogM + 2-2£ "C1 log (^^) A?
ί Ζ*1 ί2
27r-21ogM + 2-2i / logrdr = ^logM + 4 (-1),
л " Jo
С\П Jq С\П
that is,
ft
4 < AnlogM,
С\П
ft < ncilogM -n,
i < y/i^yflogM^ = с(М)л/п.
The proof is complete. D
Remark. If η is large, then we may choose c\ = 4ε/π + ε.
Problem F.17. Let f(x,y,z) be a nonnegative harmonic function in
the unit ball of R3 for which the inequality f(xo, 0,0) < ε2 holds for some
0 < #ο < 1 &nd 0 < ε < (1 — #ο)2· Prove that f(x,y,z) < ε in the ball
with center at the origin and radius (1 — 3ε1/4).
Solution. We regard / only in the interior of the unit ball. We write xo
instead of (xo, 0,0) and χ instead of (x, y, z), and let \x\ denote the length of
the vector (ж, y, z) in R3. Let 0 < A < 1, 0 < В < 1, max(A, B) < R < 1,
and suppose that \x\ < A and |жо| < B. By Poisson's formula (see, for
example, the reference in Remark 1 later) the values of / inside the sphere

182
3. SOLUTIONS TO THE PROBLEMS
5(0, R) with center at the origin and of radius R are given by the values
on 5(0, R):
R2-\A2 f /(0
f( л я3-Μ3 f
rdSe
S(0,R)
χ-ξ\*
Д2-№ Г (\χο-ξ\\3 f(0 dC,
4πβ ]3<QtR)\\x-(i\) '\*ο-ξ\3 ξ
/5(0,Д)
where S% denotes the surface element on 5(0, R).
Since \xq — ζ\ < R + \xq|, \x — ζ\ > R — \x\, we obtain from the nonneg-
ativity of / that
/Д+|хо|\3 Д2 - |*|2 / /(Q
/Д+|хо|\3 Д2-|х|2 ., λ
(д+ы)2 д+игЛжо)</ы(*+*)2 д+^
Д-|ж0| {R-\x\)2JK OJ~JK OJ R-B {R-A)2'
This holds for all R < 1, hence with the notations 1 — A = a and 1 — Β = β
we get by letting R tend to 1 — 0,
fM < fU Л! + Д)2 ! + Л ,, ,(2-/3)2(2-α) 8
/(х) < /Ы 1-β (1_л)2 = /Ы ^ < /Ы^·
Thus, for α2β > 8ε we have f(x) < (1/ε) · f(xo)- In the problem ε <
(1 — #ο)2, hence |жо| < 1 — ^1^2 and \x\ < 1 — 3ε1/4, so we can choose
a = 3ε1/4 and β = ε1/2, because then α2β = 9ε > 8ε. Then from the
preceding estimate
/(ζ)<^·ε2 = ε. □
Remarks.
1. The above solution is virtually the same as the usual proof for Harnack's
inequality (see, for example, Theorem 1.18 in the book W. K. Hayman,
and P. B. Kennedy, Subharmonic Functions, Math. Soc. Monographs,
9, London, Academic Press, 1976) according to which if / is a nonneg-
ative harmonic function in the unit ball of Rm, then for |£| < ρ < 1
(Т^г/(°) */«> * (Γ^Μ°>·
2. The statement of the theorem is a certain strengthening of the maximum
principle for harmonic functions. In fact, the maximum principle asserts
that if / is nonnegative and harmonic in the unit ball, then f(xo) =
0 implies that / vanishes identically. The problem yields that if / is
nonnegative and harmonic in the unit ball and f(xo) is "small," then
in a disk of radius "almost 1" around the origin / is also "small." In
this form, the statement is valid on an arbitrary, bounded, connected
domain (this version is also often called Harnack's theorem), which can
be seen by a standard argument using chains of overlapping disks.

3.3 THEORY OF FUNCTIONS
183
Problem F.18. Verify that for every χ > 0,
Γ'(χ+1)
Г(х + 1)
> log χ.
Solution. It is known (see, for example, 1.7.(3) in Erdelyi et al., Bateman
Manuscript Project, McGraw-Hill, New York, 1953) that if χ > 0, then
Г(х) > О and
P(x)
Г(х)
χ ^—' \v x + v I
v=l x '
where С is Euler's constant. From this, we get
Г(х)
Г(х)
= Σ (ж+и)2 >0;
that is,
v=0
Γ(χ)
Γ(χ)
is strictly increasing. On the other hand,
/ T7^dt = logT(x + 1) - logr(x) = logrr,
Γ(ί)
where we used Г(ж + 1) = хГ(х) for all x. Hence, by the mean value
theorem,
Г+1 Γ(ί) _, Γ (ξ)
for some χ < ξ < χ + 1,
and so
Hx + i)>r4j) =
Γ(χ + ΐ) >г(о logx·
Problem F.19. If f is a nonnegative, continuous, concave function on
the closed interval [0,1] such that /(0) = I, then
f1 2 Г f1 Ϋ
/ xf(x)dx < -\ f(x)dx\ .
Solution. Let A = /0 f(x)dx and В = JQ xf(x)dx. Integrating by parts,
we obtain
B = A-f (J f(t)dt) dx.
(1)

184 3. SOLUTIONS TO THE PROBLEMS
Since / is concave, its curve lies above the chord joining the points (0, /(0))
and (x, /(#)), that is, for 0 < t < x,
f{t)>M^lt + li (2)
χ
where we have also taken into account that /(0) = 1. If we integrate (2),
we obtain
г/ ч , f(x) - I x2 1 ., ч 1 , .
0 Χ Δ Δ Δ
Using (1) and (2), we arrive at
that is,
"4(л-\). (4)
From the inequality 0 < (2A-1)2 = AA2 -4A+1, it follows that A-1/4 <
A2. Substituting this into (4), we finally arrive at
L
°*\Ы)Ф-·
(5)
and this is what we had to prove. D
Remarks.
1. The proof does not use the positivity of /.
2. The equality
В = |Л2 (6)
occurs if and only if
f(x) = l-x. (7)
Clearly, (7) implies (6). Conversely, if (6) holds, then by (5)
A=\. (8)
Furthermore, because of the continuity of /, we must have equality in
(2) for all 0 < t < x. But then
fit) - 1 fix) - 1
—^ = —— = c = constant,
t χ
and so / is of the form f(x) = ex + 1. Using (8), we can conclude
1 Г1 с
2=A = J f(x)dx=- + l,
from which с = — 1 follows. Thus, f(x) = 1 — x, as we have claimed.
3. The statement is a special case of the following theorem that can be
found in the book 7. M. Jaglom, and V. I. Boltjainskii, Convex Figures
(in Russian), Gostehizdat, Moscow, 1951: If the boundary of a convex
domain contains a segment of length 1, then the distance of the weight
point of the domain from that segment is at most 2/3 times the area.

3.3 THEORY OF FUNCTIONS
185
Problem F.20. Let f be a differentiable real function, and let Μ be a
positive real number. Prove that if
then
\f(x + t) - 2f(x) + f{x-t)\<Mt2 for all χ and t,
\f'(x + t)-f'(x)\<M\t\.
Solution. We shall prove more; namely, we shall not assume the
differentiability of / in advance. It will be enough to assume its continuity.
The inequality
\f(x + t) - 2f(x) + f{x -t)\<M-t2
means that for all χ and t we have
f{x + t)-2f(x) + f(x-t)<M.t2
and
(1)
(2)
f(x + t) - 2f(x) + f{x -t)>-M· t2.
Let hi(x) = f(x) — (M/2)x2. For this function, we obtain from (1)
hi(x + t) - 2hi(x) + hi(x - t) = f(x + t) - 2f(x) + f(x -t)- Mt2 < 0.
This means that h\ is a continuous function with nonnegative second-order
symmetric differences. We know that then h\ is concave. The same
argument based on (2) yields the convexity of h\(x) = f(x) + (M/2)x2. It is
well known that h\ and h<i then have one-sided derivatives for which
hf >(x) > h\+,(x),
h{2-\x) < h?\x).
It follows that f^ and /(+) also exist at every point and satisfy
f(-Xx) = h[-\x)+Mx,
f{+)(x) = h{+)(x) + Mx,
f(~\x) = h{z\x) - Mx,
f(+\x) = h^\x)-Mx.
(3)
(4)
(5)
(6)
(7)
(8)
By (3), (5), and (6) we have /(") > /(+), while (4), (7), and (8) yield
/^ < f^- Thus, f(~^ = f(+\ which means that / is differentiable, and
so both hi and /i2 are also differentiable.

186 3. SOLUTIONS TO THE PROBLEMS
Since the derivative of a differentiable concave (convex) function is
decreasing (increasing), we can conclude for χ < у that
h[(y) - h[(x) = f'(y) - /'(*) - Μ ■ (у - χ) < 0 (9)
and
h'2(y) ~ h'2(x) = f'(y) - f{x) + Μ · (у - χ) > 0. (10)
Equations (9) and (10) can be summarized as
\f'(y)-f'(x)\<M\y-x\,
and this is equivalent to the statement. D
Remark. The statement can be generalized as follows. Let
Δ№) = E(-i)fcQ/ (* + (£- *)*)
be the rth symmetric difference of /. If
|ΔΓ(χ)| < Mf
for all χ and £, then
|ΔΡ7ω(*)| < Mf-j, ι < j < r.
This can be derived from the formula
Δ?/(*)= / ·· / f(k) \x-^t + ui + --- + uk)du1---duk,
provided the (к — l)th derivative of / is absolutely continuous.
Problem F.21. Let a < a' < b < У be real numbers, and let the
real function f be continuous on the interval [a, b'] and differentiable in its
interior. Prove that there exist ce (a,&), d G (α',δ') such that
/(6)-/(a) = /'(c)(6-a),
/(&') - /(a') = f(c')(b' - a'),
and с < с'.
Solution 1. First, we verify the following statement.
Statement. Let p,q,q' be real numbers such that ρ < q < qf. Suppose
that the real function / is continuous on the interval [p, qf] and differentiable
in its interior. Then for every r G (p, q) for which
f(q)-f{p)=f'(r)(q-p)

3.3 THEORY OF FUNCTIONS
187
holds, there exists an r1 e (p, qf) with
ftf)-№ = fVM-p)
and r1 > r.
Proof. To prove this, we may assume /(p) = f(q') = 0, since otherwise
we can work with the function
q -p
If f(q) — 0? then the statement is obvious, so we can also assume that
f(q) > 0. Let us choose the point r e (p, q) so that
f(q)=f(q)-№=f'(r)(q-p).
If now f(r) < 0, then in the interval [r, q) there is a point p' such that
f(pr) — 0· Therefore in the interval (p;, q') there is an r' with f(rf) = 0,
and this r' satisfies the requirements.
We have to examine the case when f(r) > 0. Since
m = /(g) ~ /Cp) > o,
q-p
there is a p' G (r, ^') with the property that the ratio
f(p')~f(r)
p' — r
is bigger than 0, that is, f(p') > f(r). But then there is a point q" in the
interval (p;, q') for which f(q") = /(r), and so for a suitable point r' of the
interval (r, q") we have f(r') = 0, and with this the statement is verified.
Now we turn to the solution of the problem. Let us choose the point
d e {af,b) so that
f(b)-f(a') = f'(d)(b-a').
On applying the above statement, we get ac'G (a', br) with the property
f(b')-f(a') = f'(c')(b'-a')
and d > d, and it similarly follows that there isacE (a,b) for which
f(b)-f(a)=f'(c)(b-a)
and d > c. This proves the assertion of the problem. D
Remark. It can be seen from this proof that the assertion is valid in the
cases a < a' < b < b' or a < a' < b <b' as well.

188
3. SOLUTIONS TO THE PROBLEMS
Solution 2. Let
_ f(b) -f(a) _ f(V)-f(a')
b-a ' " b'-a' '
r = inf{c G (a, 6) : /'(c) = £>}, r' = sup{c' G (a', 6') : /V) = D'}.
Our aim is to prove that r < r'. Suppose, on the contrary, that r > rf.
Then
a<a' <r' <r <b<b'.
Because of the Darboux property of derivative functions, f'(x) lies on one
side of D for all χ G (a, r), and we may suppose that for all such χ we have
f'(x) > D. Likewise, in (r, bf)(C (r',6')) the derivative f'(x) lies on one
side of D'. We distinguish two cases.
Case 1. f(x) > D' for all χ G (r, b'). By the mean value theorem, we
have
f(r) - f(a) > D(r - a), f(b') - f(r) > D'(b' - r).
Since by definition
f(b) - f(a) = D(b - a), f(b') - f(a') = D'(b' - a'),
we get by subtraction
f(b)-m<D(b-r), f(r)-f(a')<D'(r-a'). (1)
On the other hand, again using the mean value theorem
f(b)-f(r)>D'(b-r), f(r)-f(a')>D(r-a'). (2)
Since b — r > 0 and r — a1 > 0, we can conclude D > D' from the left
sides of (1) and (2), while from their right sides it follows that Df > D.
This contradiction proves our claim in Case 1.
Case 2. f'(x) < Df for all χ G (r, b'). Similarly as before, we get
D'(b - a!) < f(b) - f(a') < D(b - a'),
and from here /'(6) < D' < D < f'{a'). Let Τ G (D1\ D) such that Τ φ
ff(r). Because of the Darboux property, there must be a place t G (a', b)
where f'(t) = T. This, however, can be neither in (a', r) (because there
f'(x) > D >T) nor in (r,6) (because there f'(x) < Df < T). Neither
can t be equal to r (because of the choice of T), and we have arrived
again at a contradiction. This proves the claim in Case 2. D

3.3 THEORY OF FUNCTIONS
189
Problem F.22. Let /q, с, а, д be positive constants, and let x(t) be the
solution of the differential equation
([Io + cta]2x'y + g[l0 + cta] sin χ = 0, t > 0, -^ < ж < |,
satisfying the initial conditions x(to) = xo, x'{to) = 0. (This is the equation
of the mathematical pendulum whose length changes according to the law
I = lo + cta.) Prove that x(t) is denned on the interval [to, oo); furthermore,
if a > 2 then for every xq φ 0 there exists a to such that
liminf \x(t)\ > 0.
t—юо
Solution. With the notation у = [lo + ct"]2^, equation (1) transforms to
/= У
[lo + ct<*\*
y' = -g[lo + cta] sinrr (t > 0, |x| < |, у G r) .
Applying the Picard-Lindelof theorem to the latter system, we get that the
solution satisfying the given initial conditions exists in a right neighborhood
of to. From the theorem of "continuation up to the boundary" (see the
book L. Sz. Pontriagin, Ordinary Differential Equations, Addison-Wesley,
London, 1962), we can conclude that if x(t) exists on the interval [to,T)
but it cannot be continued beyond T, then we must have \x(t)\ —> π/2
as t —> Τ — 0, since it follows from equation (1) that x'{t) is bounded on
bounded intervals.
Let us introduce the notation l(t) = lo + cta and consider the function
V(t,x,x') = ^(^)2 + 2(l -cosж),
9
which is nothing else than the mechanical energy of the pendulum
divided by l(t)/2. From equation (1), we get by differentiation that v(t) =
V{t,x{t),x'{t)) is a nonincreasing function. Therefore,
2 > v(t0) = 2(1 - cosrr0) > v(t) > 2(1 - cosx(t)),
which makes \x(t)\ —> π/2, as t —> T — 0 is impossible. Thus, x(t) is defined
on the whole interval [to, oo).
To prove the second statement, let us start from the equation
x(t)=x0-g τ]— -to / [ίο + ста] sinx(r)drds,
J to llo+csa\2 J to
which follows from (1) by integrating twice and taking into account the
initial conditions. If a > 2, then
Е%^?1Ус!"]М1
< oo,

190
3. SOLUTIONS TO THE PROBLEMS
and so for large to
\x(t)\ > Ы - g f _} f[lo + cra]drds > ψ
for all t on the interval [£q, oo), and this is what we needed to prove. D
Remark. It can be proved that if 0 < a < 2, then for any solution of (1)
that is defined on the whole interval [ίο,οο), we have
lim x(t) = 0
t—>oo
(see L. Hatvani, On absence of asymptotic stability with respect to a part
of the vanables, J. Anal Math. Mech., 40 (1976), 223-225).
Problem F.23. Let /i, /2, · · ·, /n be regular functions on a domain of
the complex plane, linearly independent over the complex field. Prove that
the functions fifk, 1 < i,k < n, are also linearly independent.
Solution. First, we prove the following lemma:
Lemma. Let /г, #г, 1 < i < η be regular functions in the domain D, and
suppose that not every gi vanishes identically. If Σ™=1 fi9i — 0? then the
/i's are linearly dependent.
Proof. Indeed, for every h φ 0 and ζ e D
Q = y> fi(z + h)gj(z + h) - fi(z)gj(z)
^ h
i=l
Έ^-/- , ^fi(z + h)-fi(z) , V^//^ (9i(z + h)-9i(z)\ h
= "L9i(z + h) + Σ,Μ*)[ 1 )h'
г=1 г=1 \ /
Letting h tend to zero first on the real and then on the imaginary axis, we
obtain
and
E/K*)^-i>(*№) = o
2=1 2=1
for all ζ e D, from which Σ™=1 flJJi = 0 follows. Repeating this argument
we get Σ"=1 fi9i = 0 for every m.
Without loss of generality, we can assume that 0 e D. Let
00 00
Λ(ζ) = Σο5'ν and φ) = ^ψζ>.
j=0 j=0

3.3 THEORY OF FUNCTIONS
191
These series converge in some disk {z : \z\ < δ}. We may also assume that
6q 7^ 0 for at least one г (in the opposite case, we may factor out an
appropriate power of ζ from every <ft). For every ra, the function Σ7=ι f\ 9i
takes the value
at ζ = 0, and so Σ7=ι b0 fa = 0 because every coefficient in its power series
vanishes. This proves our lemma.
Now we prove that if, besides the assumption of the theorem, the
functions #j, 1 < j < m are also regular and linearly independent on D, then
the same is true of the system /^, 1 < г < η, 1 < j < га. This is obviously
stronger than what the problem asked for.
In fact, suppose that Σ7=ι Σ™=ι Cijfi9j = 0. Then
ΣΓ=ι ίΐ(ΣΤ=ι fyjdj) — 0, and so it follows from our lemma that
ΣΤ=ι Cij9j — 0 for all i. From this, we get c^· = 0 for every г and j because
of the assumed linear independence of the functions #j, and this is exactly
what we wanted to prove. D
Problem F.24. Prove that the set of all linear combinations (with real
coefficients) of the system of polynomials {xn + xn }^0 is dense in C[0,1].
Solution. Let Ρ be the set of the linear combinations. Since the set
of polynomials is dense in C[0,1] (Weierstrass's theorem), it is enough to
prove that every power xn, η = 0,1,2,..., can be uniformly approximated
by polynomials from P.
For η = 0,1, we have xn e P. If η > 1, then let
m— 1 .
г=0
1 ТП г 1
= хп + — Y(-l)i+1xn2t := хп + — · А{х).
m *—? m
г=1
Obviously, /^ G P. The sum defining A(x) consists of terms that alternate
in sign and decrease in absolute value, and the first term lies in between 0
and 1. Hence, 0 < A(x) < 1. As a consequence,
\хп-Ш*)\<±
for all χ e [0,1], which proves our claim. D
Remark. The problem can be generalized as follows. Let g(n) be a
strictly increasing function defined on and taking values from the set of
nonnegative integers. Then the linear combinations of the system {xn +
x9(n)} form a dense set in C[0,1]. The proof goes along the lines presented
above.

192
3. SOLUTIONS TO THE PROBLEMS
Problem F.25. Let f be a real function denned on the positive half-
axis for which f(xy) = xf(y) + yf(x) and f(x + 1) < f(x) hold for every
positive χ and y. Show that if f (1/2) = 1/2, then
f(x) + f(l-x) > -xlog2x- (l-x)log2(l-x)
for every χ G (0,1).
Solution. Let φ(η) = f(n)/n. Then φ(ηπί) = φ(η) + φ(πι) and (η +
1)φ(η + 1) < ηφ(η) hold for all η, га G N. Hence, there exists ace
R, for which φ(η) = clog2n for every η G N (see p. 19 in J. Aczel,
and Z. Daroczy, On Measures of Information and Their Characterizations,
Academic Press, New York, 1975). From here, because /(1/2) = 1/2 and
0 = /(l) = i/(2) + 2/
2) 2
(/(2)+ 2),
we get
f(n) = —n · log2 η (η G
Since for every positive χ and у
f(x + y)- f(x) - f(y) = у
f
H-'(!)
<o,
we have
f(x + y)<f(x) + f(y).
Let η G N and 0 < χ < 1. Then using what we have obtained so far, we
can write
0 = /(!) =/[(* + !-*)"]
= /
± Qzfc(l - *)-*] < ±f [(^(1 - *)-*
ς(Ι)^(Ι)·^-)-^ς(Ι)/(^-γ
fe=0
/ + //.
Prom the functional equation on /, we obtain
f(xk) = kxk-xf{x) and /((1 - x)k) = fc(l - x)k~lf{l ~ *),

3.3 THEORY OF FUNCTIONS 193
hence the sum representing II can be seen to be equal to nf(x) +nf(l — x),
which yields
f{x) + /(1 - x) > I £ (fc)3^1 - χ)η~" loS2 (fc) =: S^)- (!)
Now let
tt\x) = (fyxk(l-x)n-k, k = 0,l,.-.,n.
Since these (for a fixed x) take their maximum for either к = [nx] or
к = [nx] + 1, we easily obtain from Stirling's formula for the factorials
appearing in the binomial coefficients that Ншп-юо p^ (x) = 0 uniformly
in k. Using this and 1ш1г_>о+о t log21 = 0, we get
lim ±Y,pin\x)\og2p(i:\x)=0. (2)
k=0
On the other hand,
1 n
-Σ^η)(^)1^2Ρΐη)(^)
fc=0
= \ Σ (^^ί1 " *)""* · [loS2 (fc) + *l°g2 * + (* - *0 bg2(l - x)]
= Sn(:r)+:rlog2:r + (l -x)log2(x- 1),
from where, with the aid of (2), we can conclude that
lim Sn(x) = — rr log2 χ — (1 — χ) log2(l — χ).
η—>oo
Taking (1) into account, we get the statement. D
Problem F.26. Let G be a locally compact solvable group, let c\,...,
cn be complex numbers, and assume that the complex-valued functions f
and д on G satisfy
η
Σ °kf(xyk) = f(x)9(y) ^r all x,yeG.
Prove that if f is a bounded function and
inf Ref(x)x(x) >0
xeG v y v '
for some continuous (complex) character χ of G, then д is continuous.

194
3. SOLUTIONS TO THE PROBLEMS
We will give two solutions. Neither of them will use the local compactness,
and the second solution will not use the solvability of G either. Hence, the
claim is true on any topological group.
Solution 1. It is known (see for, example, F. R. Greenleaf, Invariant
Means on Topological Groups, Van Nostrand, Princeton, 1969) that
because of the solvability of G there exists a right-invariant mean m on the
space of complex-valued, bounded functions on G. That is, m has the
following properties: m is (complex) linear; ra(l) = 1; m(f) = m(/); if / > 0,
then m(f) > 0; furthermore, m(fx) = m(f) for every bounded function /
defined on G, where fx(t) = f(tx), x,t e G.
Let /, g, and χ be as in the problem. Since Re /χ > inf^GiRe
f{x)x(x) > 0, we have
m(Re/x) > m(MRef(x)x(x)) = m£Bef(x)x(x) > 0,
xeG xeG
and from this
to(/x) = m(Re/x) + im(lmfx) φ 0.
Now let x, у G G be arbitrary, and let us multiply the equality
η
J2ckf(xyk)=f(x)g(y)
k=l
by χ(χ):
η
Y^ckf{xyk)x{x) = f(x)x(x)g(y).
k=l
On the left-hand side, we make use of the identity
x(x) = x(xyk)x(y~k) = x(xyk)xk(y),
by which we obtain
η
J2ckf(xyk)x(xyk)xk(y) = f{x)x(x)g{y). (1)
k=l
This holds for all x,y G G. Apply now to both sides as functions of χ the
mean m, and make use of
rnx{f(xyk)x{xyk)) = m(fx) φ 0
(here mx denotes that the argument has to be considered as a function of
x). It follows that
η
Т,скхк{у) = д{у), (2)
k=l
from which the continuity of g is obvious. D

3.3 THEORY OF FUNCTIONS
195
Remark. One can prove with the same method the generalization that
one obtains by allowing different functions Д on the left of the assumed
equality.
Solution 2. In the second solution, we will not use the existence of the
mean m; otherwise, the argument that follows is similar to the one above.
We start from the identity
η
J2ckf(xyk)x(xyk)xk(y) = f(x)x(x)g(y)
k=l
that we obtained in the first solution in (1). Applying this with χ =
у-1,..., χ = у~ш and adding the resulting equalities together, we obtain
with
sm(y) = ^2f(xy J)x(xy J)
the identity
'k-l
Σc* sm(y) + Σ " Σ I №)х№) I t{y) = Sm(y)g(y).
k=l \ \j=0 j=m-k + li
By the assumption,
ReSm(y) > mc -^oo as m —> oo,
(3)
where с denotes a positive lower bound on the real parts of the products
f(x)x(x). Hence, the numbers |5m(y)| tend to infinity as m —> oo. Now if
we divide (3) by 5m(y) and let / tend to infinity, then we again arrive at
(2) in view of the boundedness of /. This completes the proof. D
Problem F.27. Suppose that the components of the vector
u = (uo,...,un) are real functions defined on the closed interval [a,b]
with the property that every nontrivial linear combination of them has at
most η zeros in [a, b]. Prove that ifa is an increasing function on [a, b] and
the rank of the operator
A(f) = f u(x)f(x)da(x), f 6 C[a,b],
J a
is r < n, then σ has exactly r points of increase.
Solution. By the assumption that there isaO^cGR71"1"1 orthogonal onto
the range of A, that is, (c, A(f)) = 0 for every / G C[a, b]. This means that
/
J a
ь
(c,u(t))/(t)da(t) = 0

196
3. SOLUTIONS TO THE PROBLEMS
holds for every / continuous on [a, b]. In particular,
I
J a
(c,u(t))2da(t) = 0,
from which it follows that every point of increase of σ is a zero of (c, u(t)).
Since c^ 0, the assumption in the problem implies that σ has at most
η points of increase. Let these be χ ι < · · · < rrm, πι < η, and let the
corresponding jumps be Δσχ,..., Дат. Then
m
i=l
Because of the assumption made on u, the vectors u(xi),... ,и{хш) are
easily seen to be linearly independent — furthermore Δσ^ > 0 — for every
г = 1,..., га. Hence the range of A is the subspace spanned by the vectors
u(xi),..., и(хш). Then its rank is ra, and this was to be proved. D
Problem F.28. Let Q and R be the set of rational numbers and the
set of real numbers, respectively, and let f : Q —> R be a function with the
following property. For every h e Q, xo £ R,
f(x + h)-f(x)-+0
as χ G Q tends to xo. Does it follow that f is bounded on some interval ?
Solution. The answer is no. Consider the following function:
f(?) =loglog29,
where p/q is a rational number written in the form where the denominator
is positive and the fraction cannot be simplified. This is not bounded
on any interval. On the other hand, if χ = p/q and h = k/m, then
χ + h = (pm + qk)/qm, where the last fraction may be simplified. In any
case, f(x + h) < loglog2^m, and so
t( , u\ t( \ ^ ι bg 2^ +log m
/(* +JO-/(*)< log log2g ■
Now if χ —> #0, then q —> oo, and so
limsup/(x + h) — f(x) < 0.
χ—>a?o
On applying this with h and ^o replaced by — h and χ ο + /ι, respectively,
we obtain the opposite inequality,
liminf/(x + ft) - f(x) > 0.
X—*Xq
Thus, this / is a counterexample. D

3.3 THEORY OF FUNCTIONS
197
Problem F.29. Suppose that the function g : (0,1) —> R can be
uniformly approximated by polynomials with nonnegative coefficients. Prove
that д must be analytic. Is the statement also true for the interval (—1,0)
instead of (0,1)?
Solution. Let Pi,P2? · · · be polynomials with nonnegative coefficients, and
assume that limn_00pn(x) = g{x) pointwise on (0,1). We shall prove that
g is analytic. This is stronger than what the problem asked for, since we
will not use the assumption that the limit is uniform.
We claim that the polynomials pn are uniformly bounded on every disc
{z : \z\ < p} for every ρ < 1. In fact, let pn{z) = Σ ак*к- Then for \z\ < p,
\Pn(z)\ < Y^ak\z\k < Y^akpk =pn{p) <K = K(p),
since the sequence {pn(p)} is convergent, and hence it is bounded.
We also know that the sequence {pn} is convergent on a segment of
the unit disk, hence we can invoke Vitaly's theorem to conclude that {pn}
is convergent inside the unit disc and that the convergence is uniform on
every compact subset of the open unit disk. This implies that G(z) =
1ш1п_юоpn{z) = g(z) is analytic in the unit disk, and so its restriction to
(0,1) also has this property. This proves the first part of the problem.
The answer for the second part is negative. In fact, we shall prove much
more, namely that the following result holds:
Result. A continuous function g : [—1,0] —> R can be uniformly
approximated by polynomials with nonnegative coefficients if and only if #(0) > 0.
The necessity of the condition is obvious, hence we only have to deal with
its sufficiency. Let G be the set of all continuous functions on [—1,0] that
are the uniform limit of some sequence of polynomials with nonnegative
coefficients. Obviously, G is closed for addition, multiplication and forming
uniform limits; furthermore, G contains every polynomial with nonnegative
coefficients.
First, we prove that the function —x is in G. This immediately follows
from the fact that the polynomials x{\ + x)n — χ have nonnegative
coefficients and \x(l + x)n\ < (1/n) for every — 1 < χ < 0 (this can be easily
verified by differentiation). However, then — xn = {—x)xn~l also belongs
to G for every η > 1, and so G contains every polynomial whose constant
term is nonnegative.
Now let g be an arbitrary continuous function on [—1,0] with #(0) > 0.
By Weierstrass's theorem, there is a sequence of polynomials pn converging
uniformly to / on [—1,0]. But then the same is true of the sequence with
terms
Pn{x)=Pn{x)-Pn{0) + g(0),
and these polynomials are from G. Hence, g also belongs to G, which
proves the sufficiency part in our claim. D

198
3. SOLUTIONS TO THE PROBLEMS
Problem F.30. Prove that if di (i = 1,2,3,4) are positive constants,
(i2 — fl4 > 2, and ахаз — a2 > 2, then the solution (x(t), y(t)) of the system
of differential equations
χ = a\ — a^x + a^xy,
у = а±х-у- a3xy (x, у G R)
with the initial conditions x(0) = 0, y(0) > αχ is such that the function
x(t) has exactly one strict local maximum on the interval [0,00).
Solution, χ strictly increases in a neighborhood of t = 0, hence it is
enough to prove that χ changes sign exactly once.
Let us draw the pieces of the hyperbolas χ = 0 and у = 0 lying in the
first quadrant of the phase plane (x,y) (see Figure F.l):
χ = 0 : азх [y I + αχ = 0
V «з/
у = 0 : -α3 (χ + — J (у - — J - — = 0.
V аз/ V аз/ аз
Figure F.l.
Denote (xo,yo) the intersection of these. The trajectory of the solution
cannot leave the region / bounded by the curves χ = 0, у = 0, у > ус
Therefore, it is enough to prove that y(t) > yo? t > 0, and the trajectory
intersects the curve χ = 0. If the trajectory intersecting the half-line χ =
#съ У > Уо leaves the region i7 bounded by the lines χ = 0, χ = xo,
e = (04 — d2)x — y + di =0, then it must intersect the curve χ = 0, namely
in the opposite case
y(t) i zi, x(t) + y(t) i const > 0, and so x(t) / x\ (t —> 00).

3.3 THEORY OF FUNCTIONS
199
From this, it follows that x\ = xo and y\ = yo because the point (xi,yi)
must lie on the curves χ = О, у = 0. But this is a contradiction for x(t) > xq
for large t.
Next, we show that the trajectory must indeed intersect the half-line
x = #0, у > yo- Write the function y/x in the form
y_ _ -a$xy + a±x-y _ (a4 - a2)x - у + a\
χ asxy — a2x + cl\ ^з^У ~ a2# + cl\
From this, it follows that the values of y/x on the line e would be equal to
— 1, which is bigger than a4 — a2. Hence the trajectory does not intersect
the line e. We can also see that the function y/x is a decreasing function
of χ in the region IV = {(#, y) e II : у < a2/as} along the lines (a4 —
a>2)x — у + С = 0 (αχ < С = const). Since
it follows that
_ -азхру + Q4ffo - У
азхоу - a2x0 + cl\
Xq
_ -a3x0(y - yo) ~ (y - Уо) = _ | x
о>зхо(у-уо) \ азхо
a-(i + -L).
(x,y)eir
Now we show that
1 Η < a2 — a4.
a2xo
Simple calculation gives the formula
(aids - a2) + \/(ага3 — a2)2 + 4αια3(α2 - α4)
#ο
2аз(а2 — а4)
from which аз#о > (αια2 — ^2)/(^2 — ^4) follows. Thus, to conclude the
required inequality, it is enough to verify
aids — a2>
1-
CL2~ 0,4
which is obvious from our assumptions.
Now let e' be a line through the point P, the tangent of which lies in
between —{a2 — a4) and — (1 + l/аз^о). Since
(х,у)ее'П11' V аз*о7
the trajectory coming from the region II cannot intersect e1 from above.
If it does not intersect the line χ = xo either, then (x(t),y(t)) —> (x\,yi) φ
(#o?2/o)? (t —> 00) would follow, which is a contradiction.
With this, the proof is done. D

200
3. SOLUTIONS TO THE PROBLEMS
Problem F.31. Let us call a continuous function f : [a, b] —> R2
reducible if it has a double arc (that is, if there are a<ct<P<j<6<b
such that there exists a strictly monotone and continuous h : [α, β] —> [7, δ]
for which f(t) = f(h(t)) is satisfied for every а < t < β); otherwise f is
irreducible. Construct irreducible f : [a, b] —> R2 and g : [c, d] —> R2 such
that /([a, b\) = g([c, d]) and
(a) both f and g are rectinable but their lengths are different;
(b) f is rectifiable but д is not.
Solution. In both a) and b) the curve / will be the following: Let А С [0,1]
be Cantor's ternary set, and let £ be the unit interval [0,1] considered as
a curve on the plane. It is known that there is a continuous and increasing
surjection φ : A —> £ (write χ G A in the ternary form χ = 0.ειε2 · ·.
where each Ej is 0 or 2, and then let φ{χ) be defined by the binary form
0.(ει/2)(52/2)...). Let φ be the restriction of / to A. In the
complementary intervals of A of lengths 3_n, let / run through circles of radius 3_n in
a continuous fashion. It can be easily seen that / is continuous, irreducible,
and rectifiable.
In case (a), let д : [-1,1] -> R2 be defined on [-1,0] by g(t) = £(-£),
and on [0,1] let g coincide with /. Then g is irreducible, but its length is
longer than that of / by 1.
In case (b), let us divide the set of dyadic rational numbers of [—1,1],
from which there exists a monotone correspondence with the
complementary intervals of A and hence with the circles that / runs through, into
the sets So, Si,..., each of which is dense in [—1,1]. Let us also divide the
parameter interval of g : [— 1,1] —> R2 into the consecutive intervals In of
length 2~n, η = 0,1,... . On Д, let g run from E(0) to £(1/2) running
also through the circles corresponding to the points of Si Π Д. On /2 ? let
g run from Ε (I/2) to -E(O) running also through the circles corresponding
to the points of S2 Π 1<ι. On /3, let g run from E(0) to £(1/3) and so on.
Finally, on /0 let us define g so that it goes from Ε (1) to £(0), and besides
the circles corresponding to the points of So, it should also run through
every other circle through which it has not yet run. Obviously, the continuity
of g has to be checked only at the point 1, which can be easily done. Since
each set Si is dense, g is irreducible. But it is not rectifiable, because for
every η we can write into it a polygon (with vertices £(0), £(1/2), £(0),
£(1/3), ..., £(l/n), £(0)) of length 2 (1/2 + 1/3 + · · · + 1/n). D
Remark. If we like, we can also assume that [a, b] = [c, d].
Problem F.32. Let η > 2 be a natural number and p(x) a real
polynomial of degree at most η for which
max |р(я)|<1, p(-l) = p(l) = 0.
Prove that then

3.3 THEORY OF FUNCTIONS
201
Solution. Let с = cos(n/2n). It is enough to prove that for \x\ < l/c we
have \p(x)\ < 1, because then by applying the classical Bernstein inequality
on the interval [—l/c, l/c], it follows that
Let r, |r| > 1 be the number outside [—1,1] with the smallest absolute
value for which \ρ(τ)\ = 1. We have to show that τ > l/c. If we apply a
linear transformation, this amounts to the same as proving that if \p{x) | < 1
for —1 < χ < 1 and \p(l)\ = 1, then for с < χ < 1 the polynomial p(x)
cannot vanish, that is, p{x) φ 0.
Suppose that this is not true, and let Tn{x) = cos(narccosx) be the
Chebyshev polynomial of degree n. We know that for some sequence 1 =
xo > x\ > · · · > xn = —1, we have Tn(xi) = (—1)\ Furthermore, Tn(c) =
0, and Tn is positive to the right of c. We shall arrive at a contradiction
by proving that the polynomial ρ — Tn has η + 1 zeros. One zero is χ = 1.
If г > 0, then there is a zero in the interval [rr^z+i]· If this happens
to be one of the endpoints of this interval, then it is a double zero, and
hence there are at least η — 1 zeros in [— 1,#ι] (counting the possible root
x\ only once). By our assumption, there is an x\ < χ < 1 for which
0 = p(x) < Tn(x), so ρ — Tn must vanish in at least one point of [χι,χ]
(recall that p(x\) — Tn(xi) > 0). Thus, ρ — Tn has at least η + 1 zeros.
But since its degree is at most n, this can only happen if ρ = Τη, which
is not possible, because p(x) < Tn(x). The obtained contradiction proves
our claim. D
Remark. It can be proven that equality occurs if and only if p{x) =
±Tn(x/c).
Problem F.33. Let / be a strictly increasing, continuous function
mapping I = [0,1] onto itself. Prove that the following inequality holds for all
pairs x,y e I:
pX py
l-cos(xy)< f(t)an(tf(t))dt+ /-^tjsinit/-1^))*.
Jo Jo
Solution. Let
Γι = {(и, v):0<u<x, 0 < v < у},
Г2 = {(и,v):0<u<x, 0<υ < f(u)},
Τ3 = {(u,v):0<v<y, 0<u< Γ\ν)}.
Then Γι С Т2 иГ3, Г2 ПГ3 = 0, and each of these sets is a Borel set, hence
for every nonnegative and Lebesgue-integrable function д on [0,1] x [0,1],
we have
ry rx ρχ rf(u) ry rf'1^)
/ / g(u, v)dudv < / g(u, v)dvdu + / / g(u, v)dudv.
Jo Jo Jo Jo Jo Jo

202
3. SOLUTIONS TO THE PROBLEMS
Thus, with the choice
, ч д2(1 — cos(xy))
g(x, y) = —- = sin xy + xy cos xy > 0,
oyox
we get
l — cos(xy)< / f(u)sin(u - f(u))du+ I f~1(v)sin(v · f~1(v))dv . D
Jo Jo
Problem F.34. Let U be a real normed space such that, for any finite-
dimensional, real normed space X, U contains a subspace isometrically
isomorphic to X. Prove that every (not necessarily closed) subspace V
of U of finite codimension has the same property. (We call V of finite
codimension if there exists a finite-dimensional subspace NofU such that
V + N = U.)
Solution. By applying induction, we may assume that V has codimension
1. Let W be a finite-dimensional normed space, and consider W θ W with
the norm ||(u?i,u?2)|| = max{||u?i||, ||w2||}· Let W\ θ W<i be the isometric
isomorphic image of W θ W in U. It is enough to show that V Π (W\ Θ W^)
contains a subspace that is isometrically isomorphic to W.
HVf) (Wx θ W2) = Wx θ W2, then we are done. If V Π (Wi θ W2)
has codimension 1 in W\ Θ W2, then there is an 0 ^ / : W\ Θ W2 —> R
linear functional with kernel V Π {W\ θ W2). With the notations f\{w\) =
f(w\ Θ 0), f2(w2) = f(w2 θ 0) we get two linear functionals on W\ and
W2, respectively, and
f(wi θ w2) = /i(wi) + /2(w2)·
Without loss of generality, we can assume that ||/i|| > Ц/2Ц and 0 <
||/i||. Consider the linear functional
λ/ι--λ||/2||.
By the Hahn-Banach theorem, there is a φ G Wf* for which \\φ\\ =
H/all/ll/ill and φ{/1) = -||/2||. If ||/2|| ф 0, then let B(g) = <p(g) · /2/||/2||
(g e WZ), while in the case ||/2|| = 0 let В = 0. This В : W? -+ W£ is a
linear mapping of norm at most 1. Since our spaces are of finite dimensions,
we have В = A* for some linear operator A : W2 —> W\.
Since ЦАЦ = \\A*\\ = \\B\\ < 1, the mapping w2 —> Aw2 θ w2 is an
isometry of W2 into ^ШИ^. Furthermore, the range of this mapping is
part of V Π (Wi θ W2) because
fi(Aw2) + f2(w2) = (i4*/i)(^2) + /2W = -/2W + /2(^2) = 0.
Thus, we have found an isometry from W2 into Vf) (W\ Θ W2), by which
we verified that V Π {W\ θ W2) contains a subspace that is isometrically
isomorphic to W. D

3.3 THEORY OF FUNCTIONS
203
Problem F.35. For every positive a, natural number n, and at most
an points Xi, construct a trigonometric polynomial P{x) of degree at most
η for which
Ρ2π
P{x%) < 1, / P(x)dx = 0, and maxP(x) > en,
Jo
where the constant с depends only on a.
Solution. We shall construct a P of degree 2n instead of n, but this is the
same as if we solved the problem with 2a instead of a.
Without loss of generality, we can assume that the midpoint of the
largest contiguous interval determined by the points Xi is 0 (if this was not
the case, then we would have to translate the trigonometric polynomial
to be constructed below). This means that there is no xi in the interval
(—π /αη, π /an). Let
, χ 1 x-^ sin(n + h) χ
pi(x) = -+Σ0Ο8νχ= Ln* = Dn{x)
Li Li Sin c\
v=l Δ
be the nth Dirichlet kernel. For this Pi(0) = n + (1/2). Furthermore, P\
monotonically decreases on (0, π/η), for π > \x\ > π/η, its absolute value
is smaller than 1/28ΐη(π/2η) ~ η/π, and at the point π /an its value is
Mn+l)^ sing
2sin^ I '
where the coefficient of η is smaller than 1 and where A ~ В means that
the ratio А/В tends to 1 as η —> oo. Hence, there exists а с < 1 such that
P\{x) — en is negative at every xi but P\(0) — en > n(l — c) is positive.
Let
be the Fejer kernel. We have
P2(0) = ^-^ and - / P2(x)dx = 1.
2 π Jo
Therefore, if P3(x) = P2(x)(Pi{x) - en), then
P3(0) > c'n2, P3(xi) < 0,
and
2
1 ί^π I 1 ί^π
- / Ps(x)dx\ < c"n- I P2(x)dx = c"n.
wr Jo I 7г Уо

204
3. SOLUTIONS TO THE PROBLEMS
Finally, let
Then
/·2π
/ P(x)dx = 0, P(xi) < 1,
Jo
because of the previous estimate, and
P(0)>£7„-1. D
Problem F.36. Let f : R —> R be a twice differentiable, 2n-periodic
even function. Prove that if
/"(*) + /(*)
/(* + 3π/2)
hoids for every ж, then / is π/2-periodic.
Solution. Since l/f(x + 3π/2) is defined and / is continuous for all x, f
must have the same sign on R. From the parity of /, we get f"{—x) =
f"(x)- On applying the identity of the problem at —x, we get
f"(-x) + f(x)= l
/(-х + Зтг/2)'
therefore
„. З7Г. .. З7Г. ,. З7Г.
/(* + γ) = f(-x + γ) = f(* - γ)
holds for every x. Thus, / is 37r-periodic, and since it is also 27r-periodic,
it follows that π is a period of /. Hence, the condition in the problem can
be written in the form
/(χ + π/2)'
Let us consider the function g(x) = f(x + π/2). This is also even:
g(-x) = f(-x + |) = f{x - |) = f(x + |) = g(x).
In view of g'(x) = f'(x + π/2) and g"(x) = f"(x + π/2), we can write
}"{x)+f{x)=m (1)
д"(Х)+д(Х) = Л-у (2)

3.3 THEORY OF FUNCTIONS
205
Multiplying here by g and / and subtracting the second equality from the
first, we get
о = /"<?-/</' = (/'<?-/</)',
and so the function с = f'g — fgf is constant. Furthermore, since the
derivative of an even function is odd, we find that с must be odd, hence
с = 0. Since g does not have a zero, we can conclude that {f/g)' = c/g2 = 0,
and therefore f/g is constant.
/ is continuous and periodic, so it takes its maximum and minimum at
some points x\ and xq. Hence g(xo) — f(xo + яг/2) > f(xo) and g{x\) =
f{x\ + π/2) < f{x\). These inequalities and the fact that the ratio f/g is
constant show that this constant is equal to 1, which means that f{x) =
g(x) for every x, and this is exactly what we had to prove. D
Problem F.37. Let g : R —> R be a continuous function such that
x+g{x) is strictly monotone (increasing or decreasing), and let и : [0, oo) —>
R be a bounded and continuous function such that
u(t) + / 9(u(s))ds
Jt-i
is constant on [l,oo). Prove that the limit lim^oo u(t) exists.
Solution. It is enough to prove the existence of limt_00(ii(i) + g(u(t)))
because χ + g(x) is continuous and strictly increasing.
Consider the function v(t) = Jtl g(u(s))ds on the interval [l,oo). For
this,
v'(t)=g{u(t))-g{u(t-l)).
Since и is bounded, say и : [0, oo) —> [a, 6], we get that v' is also bounded.
Hence, ν is uniformly continuous on [0, oo). But we have assumed that
u{i) + v{t) is constant, hence the uniform continuity of и also follows. On
the other hand, g is uniformly continuous on [a, 6], which yields the uniform
continuity of v' via the preceding formula.
Next we show that v' has limit zero at infinity. To this end, it is enough
to show that the integral $™(v'(s))2ds is finite, because the integrand is
nonnegative and uniformly continuous. We can write
J\v'(s))2ds = f (g{u{s)) - g(u(s - 1)))^
= 2 J g(u(s))(g(u(s))-g(u(s-l)))ds
-J g(u(s))2ds + j g(u(s-l))2ds
= 2 / g(u(s))v'(s)ds - / g(u(s))2ds + / g(u(s))2ds.
Jl Jt-i Jo

206
3. SOLUTIONS TO THE PROBLEMS
Since g is bounded, the second and third terms are bounded with a bound
independent of t. The same is true of the first term, namely, using v' = — v!
(which follows from the assumption that u(t) + v(t) is constant), we can
get
rt nt ru(t)
2 / g(u(s))v'(s)ds = -2 / g(u(s))u'(s)ds = -2 g{x)dx,
J\ Λ Ju(i)
and g is bounded in [a,b]. With this, we have verified that the above
integral is finite, furthermore
lim u'(t) = - lim v'(t) = 0.
t—юо t—юо
Let us now notice that
u(t) + g(u(t)) = (u(t) + v(t)) + (g(u(t)) - [ g(u(s))ds)
Jt-i
= (u(t) + v(t)) + [ (g(u(t)) - g(u(s)))ds.
Jt-i
By our assumption, the first term is constant. For the second term, we get
from the mean value theorem that for every s G [t — 1, t) there is a χ G (s, t)
with
|u(t)-«(e)| = |(t-*)«'(χ)| <|«'(χ)|.
Therefore,
lim sup \u(t) — u(s)\ = 0.
t->°°e€[t-l,t]
Hence, the integrand in the preceding formula, and together with it also the
second term, tends to zero as t —> oo. Thus, the limit limi_00(iA(i)+^(ii(i)))
exists, and the proof is complete. D
Problem F.38. Prove that if the function f : R2 —> [0,1] is continuous
and its average on every circle of radius 1 equals the function value at the
center of the circle, then f is constant.
Solution. It is enough to prove that for every 1 > а > 0 the function
-i pa pa
9a(z) = g(z) = ^2 / / f(z + x + iy)dxdy
is constant, since then we get the constancy of / for a —> 0. Note that g
also satisfies the assumptions of the problem, namely by Fubini's theorem,
h Г9{z+eit)dt=hfjb £ £f{z+eit+x+iy)dxdydt
-ι pa pa -ι p2x
-i pa pa
= Г~2 / / f(z + x + iy)dxdy = g(z).
(1)

3.3 THEORY OF FUNCTIONS
207
Obviously, g is uniformly continuous. Hence there is a positive function
δ such that
lim δ(ε) = 0,
ε—0 Ч '
and for every z, z' e R2 = С
\9(ζ)-9(ζ')\<δ(\ζ-ζ'\). (2)
Now let Q be the set of all g : R2 —> [0,1] that satisfy the mean value
property (1) and the smoothness property (2) (with the above δ, which
we consider to be fixed). Q consists of uniformly bounded and uniformly
equicontinuous functions, hence Q is compact with respect to the uniform
norm. Thus, the functional g(l)—g(0) attains its supremum a on Q for some
go. Since Q is clearly translation and rotation invariant, we can conclude
that for every g in Q and for every z, z' with \z — z'\ = 1 the inequality
\g(z)-g(z')\<a (3)
holds. But then, on applying the mean value property (1) for go, we get
that
-ι /»27Γ -ι /»27Γ
ос = 9o(l) - 9ο(0) = ^J (9o(l + eil) - go{eil))dt <—J adt = α,
which, in view of the continuity of the functions involved, is only possible
if we have equality everywhere under the integral sign, that is,
до(1 + еи)-д0(еи)=а.
In particular, #o(2) — <7o(l) = ol.
Applying the same procedure to this equality, we can conclude #o(3) —
#o(2) = a, and in general go(k + 1) — go(k) = ol for every к = 1,2,
Adding these together, we arrive at go{n) — #o(0) = not, which, taking into
account the boundedness of go, is only possible if a = 0.
Return now to (3). This says that for every g in Q we have g(z) = g{z')
provided \z — z'\ = 1. But every two points on the plane can be joined by
a chain of points with consecutive distance 1; hence, we can conclude that
every function in Q is constant. Thus, the function ga from the beginning
of the proof is also constant because it is a member of Q, and this proves
our claim. D
Remarks.
1. The conclusion holds if we assume only the nonnegativity of /. This is
a result of H. Shockey, and J. Deny.
2. The statement is a certain strenghtening of the fact that a bounded
and harmonic function on the plane is constant; namely, the harmonic
functions are exactly the functions that have the mean value property
on every circle (not just on those with radius one).
3. The statement is the continuous variant of the following well-known
problem, which can be solved along the lines discussed above: If we
write numbers from [0,1] into the lattice points of the plane in such a
way that every number equals the average of the four neighboring ones,
then all the numbers are the same.

208
3. SOLUTIONS TO THE PROBLEMS
Problem F.39. Let V be a finite-dimensional subspace of C[0,1] such
that every nonzero f G V attains positive value at some point. Prove that
there exists a polynomial Ρ that is strictly positive on [0,1] and orthogonal
to V, that is, for every f G V,
f f(x)P(x)dx = 0.
Jo
Solution. Let Д,...,Д be a basis in V, and let us define a mapping
/:[0,1]->К*Ьу/(х) = (/1(х),...,Л(х)). If λ = (λ1?... ,λ*), then
XTf(x) = J2\ifi(x)eV,
and hence there is an χ G [0,1] for which
XTf(x) > 0.
That is, if L is any half-space containing the origin, then f(x) is an inner
point of L for some x. Hence, the origin is contained in the convex hull of
{/(*)|*€ [o,i]}.
But then there exists an ε > 0 such that this set contains the ball with
center at the origin and of radius ε. Let us consider the points of the form
(eei,...,eejfe),
where e; = ±1. For any choice of the values ei,..., e^, there are numbers
ci > 0,..., Cfc+i > 0, Σ а = 1, χι,..., Xk+ι € [0,1] for which
Approximating first the distribution Σj cj$Xj by positive functions, and
then these functions by polynomials, we get the existence of a strictly
positive polynomial ρ for which
Y^Cjfi(xj)- J fi(x)p(x)da
ε
<2-
Now let J С {1,..., к}. Let us choose the numbers ej as follows:
ei = 1 if г G J; e» = —1 if г $. J,
and let pj denote the polynomial ρ corresponding to this choice. Then
/
PJJi > 2' г е J'

3.3 THEORY OF FUNCTIONS
209
and
Hence, in every quadrant of the space Rk, there is a point of the form
( / Pj/ь···, / Pjfk) ·
Thus, the origin belongs to the convex hull of these points, that is, with
some dj > 0,
/ (]CdjPJ) Λ = 0' * = 1>···Λ
and at the same time the polynomial Y^jdjpj is strictly positive on
[0,1]. D
Remark. For more information see A. Pinkus, and V. Totik, One-sided
L1-approximation, Can. Bull. Math., 25 (1986), 84-90.
Problem F.40. Let D = {z e C: \z\ < 1} and D = {w eC: \w\ = 1}.
Prove that if, for a function f : D χ В —> С, the equality
„(αζ + b aw + b\ ,, ч ,/b aw + b\
\02 + а tmi + a/ \a o^ + a/
holds for all ζ e D, w e B, and a, b G C, |a|2 = 1 + |6|2, then there is a
function L : ]0, oo[—> С satisfying
L(pq) = L(p) + L(tf), for all p, q > 0,
such that / can be represented as
/1- U|2 \
f(z,w) = L Ί Цтг), foraJJ ζ e D.w e В.
\\w — zr J
Solution. First, we prove that if / satisfies equality (1), then there exists
a φ : D —> С such that
f(z,w) = <p(zw), (z,w) eDx B, (2)
'(V.)^-?)-""'+^ ''s€D· (3)
Let a2 = w hold for (z,w) e Dx B, b = 0, and α G C. Then, from equality
(1) it follows that
f(zw,l)=f(z,w)+f (0,1). (4)

210
3. SOLUTIONS TO THE PROBLEMS
By substituting ζ = 0, w = 1, we get /(0,1) = 0, and so, by (4), (2)
holds for the following function φ : D —> C:
0(z) = /(z,l), zeD.
From (1) by (2), we get
(az + b aw-\-b\ (b aw + b\
(z,w)eDxB, |a|2 = l + |6|2. (5)
Let t,s e D. If we substitute
1 , 1-s
a = ,— , b = as, w = ζ, ζ = tw
чЛ"
\s
2'
for (5), then (3) follows from the fact (a · w + b)/(bw + a) = 1.
Let A = {и е С : Re и > 0} and
^(u) = φ(^) ueA. (6)
г/ + 1
Substituting for u,v e A: t = (u— l)/(u + 1), s = (v — l)/(v + 1), from
(3), we obtain
ψ(ηΚβυ + i Imt;) = ψ(ν) +ψ(υ), u,v e A. (7)
We are going to show that
ф(и)=ф(Кеи), иеА. (8)
Let χ e (0, +oo), у еШ. Then from (7) it follows that
ψ{1 + 2iy) = ψ((1 + iy)l + гу) = ^(1 + iy) + ^(1 + гу) = 2^(1 + iy)
and
^(1 + 2iy) = ф(2) + ψ(1 + 2iy) - ψ{2) = ф(2 + 2гу) - ψ{2)
= ф((1 + гу)2) - ф(2) = ф(1 + гу) + ф(2) - ф(2) = ф(1 + iy).
Hence, ψ(1 + iy) = 0, and from (7) we get
ψ(χ + iy) = ψ(χ · 1 + iy) = ф(х) + ^(1 + iy) = ψ{χ).
Finally, let L(p) = ψ(ρ), ρ e (0, +oo). Then from (6), it obviously
follows that
L{pq) = L(p) + L(q), p,qe (0, oo).
On the other hand, from (2), (6), and (8), we obtain
f(z,w) = φ(ζιυ) = ψ(-—^Ч = ^(Re- ^ J
VI — zw/ \ l — zw/
\\w — zr / \ \w — zr /
^4)=l(^
\w — zr' \\w — ζ
for all (z, w)eDxB. D

3.3 THEORY OF FUNCTIONS
211
Problem F.41. Prove that the series Σρ°ρί(Ρχ)> where the
summation is over all primes, unconditionally converges in L2[0,1] for every 1-
periodic function f whose restriction to [0,1] is in L2 [0,1] if and only if
Σ» \cp\ < °°· (Unconditional convergence means convergence for all
rearrangements.)
Solution. Let fp(x) = f(px)· Suppose first that the sum in question is
finite. Since we have ||/p|| = ||/||, where || · || denotes the L2[0, l]-norm, we
get
Σ h<WpII ^ Σ \срШ\ = ιι/ιι Σ i°pi < °°·
Hence the completeness of L2[0,1] implies the convergence of J^cpfp in
any rearrangement.
Conversely, suppose that Σ \cp\ 1S divergent. Let Ap denote the norm-
preserving operation / —> fp. If δ is an arbitrary number, there is a
finite subset p», i 6 /, of the primes such that |X^e/cPil > δ- Let
si = EieicPiAPi and H = ifliejPi · J С I}. If η € Я, then let
Jn = {i€l · Pi\n}, η = Y[{pi :ie I\ Jn}, vn = £{Cp : ρ G Jn}·
Consider the functions e2nint e L2[0,1], η e H. These form an or-
thonormal system, and Ap(e2nint) = e2ninpt. The function Епея β2πίη<
has norm 2'я'/2. Apply the operator 5/ to this function. Then for η G Я,
the coefficient of e2ntnt in the resulting function is
Σ^
Pi\n
if Ιρι = η for some г G /, and 0 otherwise. Hence, the coefficient is exactly
vn. Thus,
ιι^(Σε2πίηί)ΐι2^Σ^ = ^Σ(^+^)
пен пен пен
>^Σκ+^)2 = 2ΐ*'-2(Σ^)2.
пен nei
that is,
Ш > \f>-
Since here δ is arbitrary, we can form a sequence of operators 5/fc with
some increasing sequence I\ С h С · · · of subsets of the natural numbers,
the union of which contains every natural number such that the norms
||5/fc || tend to infinity as к —> oo. Hence by the Banach-Steinhaus theorem,
there is an / for which the sequence S7fc(/), к = 1,2,... is not convergent
in L2[0,1], and this proves the necessity of the condition. D

212
3. SOLUTIONS TO THE PROBLEMS
Problem F.42. Let ao = 0, αχ,..., a^ and bo = 0,6i,..., b^ be
arbitrary real numbers.
(i) Show that for all sufficiently large η there exist polynomials pn of
degree at most η for which
р<Р(-1) = <ц, p<p{l) = bit i = 0,l,...,k, (1)
and
max\pn(x)\ < —, (2)
с
|ж|<1lx '"x " ~ ri
where the constant с depends only on the numbers a», 6».
(ii) Prove that, in general, (2) cannot be replaced by the relation
lim n2 · max |pn(#)| = 0· (3)
n-»oo N<1
(4)
Solution. Let us search J
pn in the form
. 2k+l
Pn(x) = ~2 Z2 С3)ПГ(2т+1)5(ж),
П s=0
where
772 —
η 1"
4fc + 2 2
(n>4fe-l),
(5)
and Tj (x) = cos(j arccos x) is the Chebyshev polynomial of degree j. Then
the degree of pn is at most n, and conditions (1) lead to the system of
equations
2k+l 2k+l
Σ c..»igL+i).(-l) = °*η2. Σ c-.n^+i).(!) = 6«"2' < = 0, · · ·, fc.
(6)
We are going to prove that for large enough η this system can be uniquely
solved.
First of all, because TJ°(-1) = (-l)^Tf }(1), we get from (6) by
addition that
Σ <W2$L+1).(D = ^o^V i = 0,..., *. (7)
If we differentiate the well-known differential equation
(1 -x2)Tj(x) -χΤ'ό(χ) + j2Tj{x) = 0
for the Chebyshev polynomials (г — 1) times and substitute χ = 1, we arrive
at
0m _ 32 -(г-1)2^-р,1Ч _ _ j2(J2 - 1) · · · О'2 - (< - I)2)
Ti (!) = 2t-l ^ (1)
(2* — l)!l
j2^+0(j2i-2), i=l,...,fe;j->oo>
(2»-l)!!

3.3 THEORY OF FUNCTIONS
213
whence from the above equations and from (5) the equations in (7) take
the form
I>,n{*240(n-2)} = £1^+1{Я-Щпл, i-0 fc. (8)
3=0
22i(2m+l)
Now, if η tends to infinity, then in the limit this takes the form (recall that
a0 = b0 = 0)
к b _
Y^c2ss2i = ±-^{2k+l)6a, i = 0,...,fe,
s=o Δ
which has a unique solution, for its determinant is the Vandermonde
determinant formed of the elements l2,22,..., A;2, and so it is different from zero.
But then (8) has a unique solution for sufficiently large n, and C2S,n = 0(1),
s = 0,..., k. Similarly, it follows from (6) that C2e+i,n = O(l),s = 0, ...,fc.
Hence, (4) satisfies (2) whenever we choose с so large that с > ^s=0 |св)П|
is satisfied.
The fact that in general (3) cannot hold follows from Markov's
inequality:
max \p'\ < n2 max \pn\.
*€[-l,l]' ~ s€[-l,l]'
Indeed, using this we can see that (3) would imply
lim max\pfn{x)\ = 0,
n-»oo |ж|<1
which contradicts (1) unless a\ = 0 and b\ = 0. D
Problem F.43. Let f and g be continuous real functions, and let g ^ 0
be of compact support Prove that there is a sequence of linear
combinations of translates of g that converges to f uniformly on compact subsets
ofR.
Solution. If {gn} is a sequence of linear combinations of translates of g
for which \gn(x) — f(x)\ < I/n for every χ in the interval [—n,n], then this
sequence uniformly converges to / on every compact subset of the real line.
Hence, it is enough to verify that any / can be arbitrarily well approximated
on any finite closed interval I by linear combinations of translates of g.
Let us suppose for an indirect proof that this is not the case, and I is
such an interval, that the linear hull of translates of g is not dense in С (I).
Then, by the Hahn-Banach theorem, there is a linear functional L G C*(I)
that vanishes on every translate of g. By the Riesz representation theorem,
L can be identified by integration with respect to a signed measure μ with
support in /. Let h(x) = g(—x). Then
/oo /»oo /»n
h(t — x)dp(x)= / g(x — t)dp(x)= / g(x — t)dp(x) = 0.
-oo J— oo J—n

214 3. SOLUTIONS TO THE PROBLEMS
Taking the Fourier transform here (which is possible because h and μ
have compact support), we find F{h) · Τ{μ) = 0. However, the functions
!F{h) and Τ{μ) are analytic, hence one of them must be identically 0. But
this is a contradiction since neither h nor μ is identically zero. This proves
the statement of the problem. D
Problem F.44. Let χ : [0, oo) —> R be a differentiable function
satisfying the identity
xf(t) = —2x(t) sin21 + (2 — | cos t\ + cos t) / x(s) sin2 s ds
Jt-i
on [l,oo). Prove that χ is bounded on [0, oo) and that lim^oo x(t) = 0.
Does the conclusion remain true for functions satisfying the identity
x'(t) = -2x(t)t + (2 - | cos 11 + cos t) / x(s)sds ?
Λ-1
Solution. Let us consider the equation
x'(t) = -2a(t)x(t) + (2- |cost| + cost) / a(s)x(s)ds,
Jt-i
where a(t) > 0 is a continuous function. If x(t) is a solution, then let us
estimate the upper right derivative £)+|x(i)| of \x(i)\:
D+\x(t)\ < -2a(t)\x(t)\ + (2- |cost\ + cost) / a(s)\x(s)\ds
Jt-i
= -2a(t)\x(t)\ + 2 / a(s)\x(s)\ds
Jt-i
— (|cos11 — cost) / a(s)\x(s)\ds
Jt-l
= -2[ a(t)\x(t)\ds - a(t + s)\x(t + s)\dsj
— (|cosί| — cost) / a(s)\x(s)\ds
Jt-l
d f° fl
= —2— / / a(u)\x(u)\duds
at y_! Jt+S
—(|cosί| — cost) I a(s)\x(s)\ds.
Jt-l
Hence,
/0 pt pt
/ a(u)\x(u)\duds) <-(\cost\-cost) / a(s)\x(s)\ds.
-1 Jt+s Jt-l
(i)

3.3 THEORY OF FUNCTIONS
215
Since the right-hand side of (1) is nonpositive, we get that the function
\x(t)\+2 / a(u)\x(u)\duds (2)
J-l Jt+s
is decreasing, and so \x(t)\ is bounded.
For к = 0,1,..., let Hk = [(2k + 1)π - 1, (2k + 1)π + 1]. We show that
max/ a(s)\x(s)\ds —> 0 (A: —> oo). (3)
teHk Jt_1
If we suppose that on the contrary, (3) does not hold, then there is an
a > 0 and a sequence {tn} such that tn e U£L0#fc, tn —> oo, and
/
a(s)|x(s)|ds > a.
/tn-i
Then at least one of the equalities
ptn-l/2 ptn
/ a(s)|x(s)|ds>a/2, / a(s)|x(s)|ds > a/2
Λη-1 Λη-1/2
holds. Since | cos ί | — cos ί > 0 on the interval H^ = [(2k + 1)π — 3/2, (2k +
1)π + 3/2], there exists a β > 0 such that at every point of an interval of
length 1/2 and containing tn, we have
(| cos£| — cost) / a(s)\x(s)\ds > β.
Jt-i
However, this and (1) imply
\x(t)\ + 2 / / a(u)\x(u)\duds —> —oo (t —> oo),
-/-1 Λ+s
which is impossible. Hence, (3) holds.
From (3) it follows that
max/ / a(u)\x(u)\duds —> 0 (A: —> oo). (4)
«ея^ 7_x JtJrS
Using (4) and the monotonicity and nonnegativity of the function in (2), we
can see that to prove the existence of the limit of x(t) at infinity it is enough
to show that there is a sequence {tn} of points of the set U^=0Hk for which
tn —> oo, x(tn) —> 0 as η —> oo. If there was not such a sequence, then there
would be a fco G N and a 7 > 0 with the property that \x(t)\ > 7 for every
t e Hk and к > ко. This, however, contradicts (3), whether a(t) = sin2 t
or a(t) = t.
This proves the claim of the problem, and we have also obtained that
the same conclusion holds if we use the equation in the second half of the
problem. D

216
3. SOLUTIONS TO THE PROBLEMS
Problem F.45. Let с > 0, с φ 1 be a real number, and for χ G (0,1)
let us define the function
Prove that the limit
does not exist.
f(x) = Y[(l + cx*k).
k=0
hm ———
s->i-o f(x)
Solution. We argue indirectly, hence let us assume that the above limit
exists. In general, let S be the set of all positive real numbers q for which
the limit
lim l£-l=:g(q)
*->i-o f(x) yy4J
exists. Thus, the indirect assumption is that 3 G S. First, we are going to
show that this implies S = R+, and we also get an explicit representation
on the function g.
Note that 2 G S because
lim f-^l= lim —— = -!-. (1)
*—1-0 f(x) *—l-Ol+СЖ 1 + C
Furthermore, by substitution it is easy to verify that S is a multiplicative
subgroup of the real field, and
0(01)0(02) = 0(0102) for qu q2 e S. (2)
The numbers 2m3*, m,/ = 0,±1,... form a subgroup £1 of S that is
dense on the positive real line (use the fact that by the prime factorization
theorem the number log 2/ log 3 is irrational, hence numbers of the form
m log 2 + Ζ log 3 form a dense set on R).
Using the monotonicity of /, we can see that g(q) < 1 if q > 1 and
0(0) > 1 if 0 < 1. This implies g{q) = q1 for q G S\ for some 7 < 0. In fact,
let #(2) = a, g(S) = 6, and let us choose 7 to satisfy а = 27. If b φ 37, say
b < 37, then by choosing a number q = 2m3*, Ζ < 0, in the interval [1,2]
such that (b/S1)1 > 1, we get a q G S\ with q > 1 and g(q) > 1, which is
not possible. Hence, g{q) = q1 for q = 2,3, from which the same follows
for all q G S\ by the group property (2).
Now it is easy to show that S = R+ and that for every q G R+ we have
9(q) = <77· (3)
In fact, if q G R+ is arbitrary, then for every ε > 0 there are 91,92 G S\
such that
0i<0<02, 1<^\<1 + ε.
0(02)

3.3 THEORY OF FUNCTIONS
217
But the monotonicity of / implies that
f(xq) f(xq)
g(qi) > limsup ^ > liminf J-j—>- > g(q2),
x_i_o }[x) *—1-0 f(x)
and here the left- and right-hand sides can be arbitrarily close to q1 if ε is
chosen sufficiently small. This verifies (3) for all q. We will not use it, but
it is clear from (1) that 7 = — log2(l + c).
Since
lim -7- ^— = σ7,
x-i-o {1-х)"*
we can conclude from (3) that / is of the form
f(x) = (l-x)-"L(x) (4)
(with ρ = —7), where L is slowly varying inthe sense that
Цхя) _ 1
for every q > 0.
Let
lim
Ж—1-0 L(x)
f{x) = ^2,akxk, 0<x < 1,
fc=0
and 5n = X^=0 flfc- By a well-known Tauberian theorem (see G. H. Hardy,
Divergent Series, Clarendon Press, Oxford, 1989, Ch. VII, Theorem 108),
we can derive from (4) with any fixed q G (0,1) and some ρ > 0 that
*n~p/(<71/n),
where an ~ bn means that the ratio an/bn tends to 1. However,
m—1
«2™ = П(1 + С)+С=(1+С)т + С
and
from which
S2m+2m-l = (1+ С)"* + C(l + C)m_1 + C2,
С 52n,+2m-! /(gl/(2"-13))
1 + C~ S2m ~ /(ς1/(2—ΐ·2))
1 +
(l_gl/(2·"-^))-,,
~ (1-ς1/(2*"-1·2))-ρ
follows if m —»· 00. Hence,
/2.2m-i\-p /34
\3 · 2"1-1/ ~ U/
ρ
1 +
lie {2) '

218
3. SOLUTIONS TO THE PROBLEMS
and by (1) and (4),
1 + с = 2p.
Thus, 1 + 2c = 3P, and ρ must satisfy the equation
3P - 2 · 2P + 1 = 0,
which certainly holds for ρ = 0 and ρ = 1, but does not hold for any other
p, because the function 3* — 2 · 2* + 1 is convex on [0, oo). If ρ = 0, then
с = 0; while if ρ = 1, then с = 1, and these values for с were not allowed.
The obtained contradiction proves the claim. D
Remark. Note that if с = 1, then f(x) = 2/(1 — я), and the limit
*->l-0 /(ж)
exists for every g > 0.
Problem F.46. Let / and g be holomorphic functions on the open unit
disc D, and suppose that \f\2 + \g\2 G Lipl. Prove that then f,g G Lip^.
A function h : D —> С is in the Lipa ciass if there is a constant К such
that
\h(z)-h(w)\<K\z-w\a
for every z,w e D.
Solution. We shall use the following Hardy-Littlewood theorem: if a
function h : D —> С satisfies
№)l < ^=п Ш
with some constant M, then h G Lip^. In fact, in order to verify the
Hardy-Littlewood theorem, it is enough to show that if \z — z'\ < δ < 1/2,
then
\f(z)-f(z')\<CVE. (2)
But if w = (1 — δ)ζ, w' = (1 — δ)ζ', then (1) easily implies that
rl*l Μ
/•1*1 Μ
\№-f(w)\< -==dt
J(i-6)\z\ vi — t
(i-«)M
= 2M(v/H-Vl(i-*)H) < /Ί ?f „, , < mV«.
v1 —С1 —*)l«l
A similar argument shows that
\f(z') - f(w')\ < СуД and \f(w) - f(w')\ < CVS,

3.3 THEORY OF FUNCTIONS
219
from which (2) follows.
We shall prove that
С1(~\\Ъ ι iJ/^12 К
l/W + lsW< —
z\.
where К is the Lipschitz constant corresponding to |/|2 + \g\2. From here
our statement follows by the Hardy-Littlewood theorem.
Let ζ e D and r > 0 be such that the circle with radius r and center at
ζ is contained in D. Let us apply Parseval's identity on this circle:
f [Z)i r2«>\f(z)\2 + \f'(z)\2r2,
1 Γ _ -
-W |/(z + e«r)|ai» = E
2π Jo ^o
2π
or in rearranged form,
m
lj\\f(z^eier)\2-\f(z)\2)de>\ff(.
z)\2r2.
Applying the same inequality with / replaced by g, and adding the two
together we, get
h Γ((l/("+Λ)|2 + w+ei<?r)|2)"(l/(2)|2 + l5(z)|2)) M
>(|/'(г)|2 + |9'(г)|2)г2.
But because |/|2 + \g\2 e Lipl, the modulus of the integrand on the left-
hand side is at most if |ez^r| = Kr, hence the integral is at most 2nKr.
Thus,
±2nKr = Kr>(\f'(z)\2 + \g'(z)\2)r2,
that is,
|/'(*)|2 + |5'(*)|2<V·
r
Since r can be any number smaller than 1 — |z|, by letting r tend to 1 — \z\
we obtain
\f\z)\>+g>{z)?<-^-
ι — \z\
and this is what we had to prove. D
Remark. The example / = 0, g{z) = (1 — z)1/2 shows that the conclusion
is sharp.

220 3. SOLUTIONS TO THE PROBLEMS
Problem F.47. Find all functions f : R3 —> R that satisfy the
parallelogram rule
f(x + У) + f(* -У)= 2/(x) + 2/(y), я, у e R3,
and that are constant on the unit sphere of R3.
Solution. Suppose that the function / : R3 —> R satisfies the functional
equation
/(* + У) + f(* -У)= 2/(ж) + 2/(y), x, у € R3, (1)
and is constant on the unit sphere
f(u) =c, ue R3, ||u|| = 1. (2)
We shall prove that, with the help of an additive function а : R —> R, /
can be written in the following form:
f(x) = a(\\x\\2), xeM3. (3)
We begin by showing that / takes equal values on vectors of equal norms.
So first let x,y e R3, \\x\\ = \\y\\ < 1. Then there is a vector ζ G R3 such
that \\x\\2 + ||ζ||2 = 1 = ||y||2 + ||z||2 and z±x, z±.y. By the Pythagorean
theorem, \\x ± z\\ = 1 = \\y ± z\\, and therefore by (l)-(2),
2f{x) = f(x + z) + f{x -z)- 2f(z) =c + c- 2f(z)
= f(v + z)+f(y-z)-2f(z)=2f(y),
that is, f(x) = f(y).
On the other hand, setting у = 0 in (1),
2f(x) = f{x + 0) 4- f{x - 0) = 2f(x) + 2/(0),
whence /(0) = 0. Thus the substitution у = χ leads to the relation
f(2x) = f{x + x) + fix -x) = 2fix) + 2fix) = 4/(ж), xeM3.
Consequently, if for some positive number r and all x, у G R3 satisfying
\\x\\ — \\у\\ < r we have fix) = /(y), then for any xf,yf G R3 satisfying
ΙΙ^ΊΙ = \\у'\\ < 2r, with the notation χ = χ'/2, у = у'/2, we obtain ||ж|| =
11 y 11 < r and therefore also
fix') = /(2x) = 4/(x) = 4/(г/) = /(2у) = /(у')·
So, by induction, fix) = /(y) whenever x,у e R3 and ||ζ|| = ||y||.

3.3 THEORY OF FUNCTIONS
221
By what we have proved, f(x) depends only on ||ж|| or equivalently,
same, on \\x\\2', in other words, there exists a function a : R —> R such that
f(x) = a(\\x\\2), xe R3.
It remains to verify the additiveness of the function a. To this end, fix
λ, μ G R+ arbitrarily, and choose vectors x, у G R3 so that λ = \\x\\2,
μ = ||y||2, and x±.y. Then, in view of (1) and the Pythagorean theorem,
2α(λ) + 2α(μ) = 2α (\\χ\\2) + 2a (\\y\\2) = 2f(x) + 2f(y)
= f(x + y) + f(x-y) = a (\\x + y\\2) + a (\\x - y\\2)
= a (\\x\\2 + \\y\\2) + a (\\x\\2 + \\y\\2) = 2α(λ + μ\
Thus a is additive on R+ and, as we may choose its value on R_ arbitrarily,
setting a{—λ) = — α(λ), λ G R+, we get the additive function desired.
Conversely, simple substitution shows that the functions of the form (3)
are solutions of the problem (l)-(2). D
Remarks.
1. We obtain the same solution if we assume (1) only for у satisfying \\y\\ =
1 and, in a more general way, consider vectors of a real inner product
space of dimension at least three. Of course, in this case we encounter
further, essential difficulties.
2. The two-dimensional case has proved to be still harder. It is an open
question whether in this case the problem (l)-(2) admits solutions
different from (3).
Problem F.48. For any fixed positive integer n, find all infinitely dif-
ferentiable functions f : Rn —> R satisfying the following system of partial
differential equations:
η
£9f/ = 0, k = l,2,....
Solution. Let η G N, and consider the system of partial differential
equations
η
Σ#"7 = 0, fe=l,2,... (1/n)
for the infinitely difFerentiable unknown function / : Rn —> R. Obviously,
linear combinations of any partial derivatives of solutions are solutions
again. We show that there is a universal solution in the sense that all
solutions are linear combinations of partial derivatives of this universal
solution. We begin with an important observation.
Lemma 1. If / is a solution of system (1/n), then d2nf = 0 (г =
1,2,..., η), so / is a polynomial of degree not greater than 2n — 1 in each
variable.

222
3. SOLUTIONS TO THE PROBLEMS
Proof. For the commuting partial differential operators df, д\, д^,
consider the power sums
Pk = %k+%k + -~+%k, fe=l,2,...,
and the elementary symmetric polynomials
s2 = dldl + dldl + --- + dl_ldl
s3 = dldldl + д\д%д\ + ■■■ + dl_2dl_xdl
sn = d\d\...dl.
Then the differential equations can be written in the form
Pfc/ = 0, fc = l,2,....
Further, by the Newton formulas,
Pi - Si = 0,
P2-S1P1 +252 = 0,
P„ - 5!Pn_! + · · · + (-l)"-1^-^! + (-l)nn5n = 0.
Hence, it follows that
Sif = Pi/ = 0,
S2f = \(S1P1f-P2f)=0,
Snf = -(Sn-iPif - Sn-2P2/ + · · · + (-l)n_1Pn/) = 0.
η
Now, for fixed г G {1,2,..., n}, consider the differential operator
p(%) = (%-%)(%-%)...(%-eft
= d?n - s^-2 + s2d*n-4 - · · · + (-i)nsn.
We see that P(df) = 0 and, therefore,
dfnf = Si0?n"2/ - 52^n"4/ + · · · + (-l)n+15n/ = 0.
I be defined by
Lemma 2. Let η G N, and let the function Qn : Rn
the following determinant:
/X
,2n-l Jln-Z
Qn(^b^2,-..,^n) = det
r2n-l „2n-3
Wr!
,2n-l ~2n-3

3.3 THEORY OF FUNCTIONS
223
Then / = Qn is a solution of the system (1/n).
Proof. We proceed by induction for n.
If η = 1, then the assertion is obvious:
Qi{xi)=xi, Q?(*i)=0·
Suppose the assertion is true for η — 1, that is,
n-l
^2dikQn-i(xuX2, - " ,Xn-i) = 0, fc= 1,2,... .
г=1
Expand the determinant Qn according to the first column:
η
Qn{xuX2,..., xn) = 2j(-l)J'~1^n~1Qn-i(^ij · · · jSj-ijSj+ij · · · j Sn)·
It follows that
n
η η
= X)X)(-l)j-1^fc[a;f"1Qn-i(xi,...,Xi-i,xi4.i,...,Xn)]
d2fe
»=1 j=l
η
Σ
η
+Σ
3=1
-IV'"1
(-1)
dxf j
Xj I Ww—l^lj · · · j *£j — 1? ^j + Ij · · · j xn)
{-l)j-lxf l^d1kQn-l{xi, · · · , ^-1,^ + 1, . . . , Xn)
ϊφά
The second sum vanishes by the induction hypothesis. Obviously, the first
sum also vanishes if к > η, whereas for к < η it is the expansion according
to the first column of the following determinant:
(x\
2n-2k-l Jln-Ъ
(2n - l)(2n - 2)... (2n - 2k) · det
2n-2k-l r2n-3
v«
2n-2fc-l ~2n-3
x\ \
X2
This, too, is zero since the first and (k + l)th columns of the determinant
coincide.
Theorem. The function / : Rn —> R satisfies the system of partial
differential equations (1/n) if and only if it is a linear combination of partial
derivatives of Qn.

224
3. SOLUTIONS TO THE PROBLEMS
Proof. Lemma 2 shows that Qn satisfies the system (1/n). Therefore,
each of its partial derivatives and all linear combinations of them will also
be solutions.
The converse can be proved by induction for n.
For η = 1, it is obvious, since /" = 0 implies
f(xi) = axi +b = aQi(xi) + bdiQi(xi).
Suppose that the assertion is true for all natural numbers not greater
than n. Let / be a solution of the system (1/n+l). By Lemma 1, d^+12/ =
0, and so d^i1/ 1S independent of xn+i'-
) = Φθ(Χΐ,Χ2,---,Χη)-
Since together with any solution its partial derivatives are also solutions, φο
satisfies (1/n + 1) and, since it does not depend on :rn+i, it satisfies (1/n)
as well. Thus, by the induction hypothesis, we have the representation
ФоЫ, x2, ---,Χη) = Σ С<0дТд22 · · · d%nQn(xi,X2, · · · , Χη)
α
with suitable constants Cq G R (α Ε Щ) among which only finitely many
are different from 0. On the other hand, we note that
(—l)n
Qn(?l, x2, · · · , Xn) = (Ο -μ ι\|^η+ΐ1(3η+ΐ(^1^2, · · · , #n+l),
which can be seen from the expansion of Qn+i according to its last row:
η
Qn+i(xi^2,...^n+i) = ^^++11(-l)fc])fc(n).
k=0
Here Dk(n) is the respective minor and, obviously, Dn(n) =
Comparing the above results it follows that
*> = Σ co (^I)!^1^2 · · · dZ»dln+YQn+1.
Next, define the function g0 : Rn+1 —> R by the relation
*> = Σ°0(2(η + 1)!^2α2 · · · d^d°n+1Qn+1.
Clearly, g0 is a solution of (1/n + 1), and d^[1(f — go) = φο — Φο = 0.
Thus /ι = / -g0 is a solution of (1/n + l) and d^jVi = 0, that is, d^/i
does not depend on xn+\:
0η+ΐ/ΐ(χ1>χ2,··.,Εη+ι) = φΐ{Χ\,Χ2,'-,Χη)'

3.3 THEORY OF FUNCTIONS
225
Pursuing the process, we obtain the functions 0o, φι,..., 02n, <7o, <7ъ · · · ?
g2ru and /1, /2, · · ·, Лп+ъ Here each of the functions g0,9u · · · > 9in is a
linear combination of partial derivatives of Qn+i- So with the help of Lemma
2, we see by induction that for к = 1,2,..., 2n the function Д+ι = fk~9k is
a solution of the system (l/n+1), and 0j£l*+1/fc+i = «ηΐΙ*+1(Λ -0fe) =
Фк — Фк = 0. Consequently, /2n+i does not already depend on :rn+i, that
is,
/2n+l (Xl, Ж2j · · · j ^n, Sn+i) = 02n+l (#1, #2, · · · , ^n)
where, similar to the arguments above, it can be proved that 02n+i is a
linear combination of partial derivatives of Qn+i- Finally, the
representation
/ = 90 + 9l Η l· 9<ln + /2n+l
proves the theorem. D
Problem F.49. Let Ρ be a polynomial with all real roots that satisfies
the condition P(0) > 0. Prove that ifm is a positive odd integer, then
^/(fc)(o) fc n
Σ π^χ > °
k=0
for all real numbers x, where f = P_m.
Solution. Let
k=o K'
and consider the polynomial Q(x) = PTn(x) · F(x). Denoting by η the
degree of P, from the factor Pm (counting multiplicities) we obtain η · πι
real roots of Q. Assume, contrary to the assertion of the problem, that F
also has real roots. We distinguish between two cases:
1. F has degree m — 1; then F has at least two real roots since its degree
is even.
2. F has degree not greater than m — 2.
According to Rolle's theorem, we can count at least mn + 1 roots (with
multiplicities) of Q' in the first case, and at least mn roots in the second
case. We note that because of the relations P(0) φ 0 and F(0) φ 0, the
root 0 could be counted at most once. Zero is actually a root of Q', of
multiplicity not less than m — 1 at that, since for j = 1,2,..., m — 1 we
have
qu)(o) = Σ (Λ [(ρ-)θ(ο)]. [fu-')(o)1
1=0 ^ '
= Σ ft) [(^m)(0(o)] · [/ϋ-ι)(ο)] = (Pm · /)ω(ο) = о,

226
3. SOLUTIONS TO THE PROBLEMS
the function Pm · / being equal to the constant 1.
Consequently, at least m — 2 roots can be joined to those counted so far.
Thus Q' has at least nm + m — 1 roots in case 1, and at least nm + ra — 2
roots in case 2. Since, however, the degree of Q' is equal to nm + m — 2
in case 1 and is not greater than nm + m — 3 in case 2, we obtain that
Q' is identically 0. Therefore, Q is a constant polynomial, and hence it is
identically 0, contrary to the relation Q(0) Φ 0.
Thus, F has no real roots and, since F(0) = P~m(0) > 0, we have
F(x) > 0 for all real values of x. D
Problem F.50. We say that the real numbers χ and у can be connected
by a δ-chain of length к (where δ : R —> (0, oo) is a given function) if there
exist real numbers x0, хг,..., Xk such that x0 = x, Xk = y, and
\xi - Xi-i\ <δ( -^ J , г = 1,..., к.
Prove that for every function δ : R —> (0, oo) there is an interval in which
any two elements can be connected by a δ-chain of length 4. Also, prove
that we cannot always find an interval in which any two elements could be
connected by a δ-chain of length 2.
Solution. To prove the first assertion, using the Baire category
theorem, choose a positive integer η and an interval 7 such that the set
Hn = {χ: δ(χ) > I/n} is dense in 7; we may also assume that the length
|7| < l/10n. Let К be the middle third of 7; we show that in it, any two
elements can be connected by a ί-chain of length 4. So let x.y G K. If
с G КГ\Нп and b = c+ (y — x)/2, then reflection in b followed by reflection
in с give translation by χ — у. If we insert this between two reflections
in a, we obtain a translation by у — χ, which carries χ into y. In order
that in this way we obtain a ί-chain connecting χ and y, we have to care
only about the following two things: first, α G Hn should be satisfied and,
second, 2α —χ should fall in the <$(6)/2-neighborhood of b. Since, however,
Hn is dense in 7, this can easily be achieved.
Proving the second half of the problem requires a bit longer argument.
Let В be the set of all real numbers that can be written using a finite
number of binary digits. We specify certain pairs of points of В so that for
suitable δ they cannot be connected by a ί-chain of length at most 2, and
every interval contain such a pair of points. To this end, let Ρ = U^P»,
where
22i ' 2^ ) ''"' \ 22i ' 2?ϊ) J '
We first define δ on elements of В in the following obvious manner: if
χ G В ends in 1/2г, then let δ(χ) = 1/2г+1 (if χ is an integer, we take
г = 0). Then, evidently, pairs of points of Ρ cannot be connected by 6-
chains of length 1, nor by those of length 2 and passing through an element
ofR
И

3.3 THEORY OF FUNCTIONS
227
We extend δ to R in the following way. В is an additive subgroup of
R, so R is a disjoint union of residue classes of the form a + B. If a £ B,
we define δ on a + В so that for b e a + Bi the value 6(b) be smaller than
min{2|x - b\,2\y - b\} for all (x,y) e U)=1Pj, where В = и£0В<? and B{
consists of those numbers whose fractional part can be written using exactly
г digits. It is easy to see that δ may be defined in this way. To finish the
proof, we verify that for any (x, y) G Ρ and ζ G a + B the numbers x, 2z, у
cannot form a ί-chain. Suppose that (x,y) G P%. Then (y — x)/2 = 1/2*.
If x, 2z, у form a <$-chain, then
δ (| + z) > \x - 2z\ = 2 |ж - (| + z)
whence by the definition of ί for j > г we obtain ж/2 + ζ ^ α + Bj. We
similarly obtain у/2 + ζ £ a + Bj. Then, however, the fractional part of
(y — x)/2 can be written using at most г — 1 digits, which contradicts the
relation (y - x)/2 = 1/2*. D
Remark. It is not known what happens if we require the existence of
ί-chains of length not greater than 3.
Problem F.51. Find meromorphic functions φ and ψ in the unit disc
such that, for any function f regular in the unit disc, at least one of the
functions f — φ and f — φ has a root.
Solution. Introduce the notation
κ(ζ) = φ(ζ) - φ(ζ) φ 0,
9{ζ) = ί{ζ)-φ{ζ),
9(ζ)
h(z) =
ФУ
Examine the functions that satisfy the following conditions:
1. / = Ηκ + φ is regular;
2. hn is nowhere 0;
3. (h — 1)k is nowhere 0.
We claim that there exist functions φ and к, meromorphic in the unit
disc, with κ{ζ) Φ 0 if \z\ < 1 and such that the above set of conditions
cannot be satisfied by any h. This immediately gives the assertion of the
problem.
If we choose φ and к so that the locations and orders of their poles
coincide, then the fulfillment of condition 1 implies the regularity of h. If
all three conditions are satisfied, then the values of h(z) can be 1 only
at the poles of к, and even this can be executed by choosing к and φ
properly. Assuming first-order poles, we only have to take care that,
denoting by A(zo) and Β(ζ$) the residues of к and φ at the pole zo,
respectively, the value B{zq)/A{zq) is different from —1. Actually, then
^(^o) = — B{zq)/A{zq) Φ 1 by condition 1.

228
3. SOLUTIONS TO THE PROBLEMS
Thus, the regular function h does not take on values 0 and 1 in the unit
disc. By Schottky's theorem, it follows that \h{z)\ remains below a bound
depending on |/i(0)| and \z\ only. We show that this is impossible. Consider
the following sequences of functions:
. 1 1
Φη = <t>= - Η г»
ζ ζ - %
Calculate the residues of the two poles:
An(0) = -2, Bn(0) = l, An(^j=2-n+\ Bn(£)=l.
Hence, it is easy to see that the functions meet the requirements stated so
far. If, however, the three conditions above are fulfilled, then hn(0) = 1/2
and /in (1/2) = —2n_1, contrary to the conclusion of the Schottky theorem.
Therefore, if η is sufficiently large, the statement of the problem holds for
^n? φη and the function ψη = κη + φη obtained from them. D
Problem F.52. To divide a heritage, η brothers turn to an impartial
judge (that is, if not bribed, the judge decides correctly, so each brother
receives (l/n)th of the heritage). However, in order to make the decision
more favorable for himself, each brother wants to influence the judge by
offering an amount of money. The heritage of an individual brother will
then be described by a continuous function of η variables strictly monotone
in the following sense: it is a monotone increasing function of the amount
offered by him and a monotone decreasing function of the amount offered
by any of the remaining brothers. Prove that if the eldest brother does not
offer the judge too much, then the others can choose their bribes so that
the decision will be correct.
Solution 1. In terms of functions, the problem can be expressed in the
following way. We are given the continuous functions
9l(Xl, · · · ? xn), · · · j02(Sb · · · j xn), · · · j0n(Sb · · · , Sn)
defined on [0, oo)n (here gj(x\,..., xn) is the deviation of the heritage of
the jth brother from l/n times the whole heritage, provided that the judge
is offered x\ units of currency by the first brother, X2 units by the second,
etc.), the sum of the functions being 0, and the function gj(xi,... ,xn)
being strictly increasing in the variable Xj but strictly decreasing in all
other variables Xk, к Ф j. Further, we know that the judge is originally
impartial, that is, <7j(0,..., 0) = 0 for all j. We have to show that there
exists an > 0 such that for any 0 < xn < an there are values x\ = xi(xn),
... ,xn-i = xn-i(xn) with gj(x\,...,xn) — 0 for all j. We prove more (to

3.3 THEORY OF FUNCTIONS
229
be exact, we must prove more in order that the proof below remain valid),
namely, that there even exist x\ = xi(xn), ... ,xn-i = Xn-i(xn), which,
besides satisfying the relations just mentioned, tend to 0 as xn —> 0.
We apply induction for n. If η = 1, there is nothing to prove. Assume
that the assertion holds for n— 1 functions. We claim that there is a number
b > 0 such that if 0 < x2, · · ·, xn < b are arbitrary, then there exists one
and only one у = y(x2, · · ·, xn) satisfying gi(y, X2, · · ·, xn) = 0, whereas
у is a continuous, strictly increasing function of the variables ι2,...,α;η.
Really, the uniqueness of у follows from g\ being strictly increasing in its
first variable. If X2 = хз = · · · = xn = Q, then we may choose у = 0,
while for other values the existence of у can be seen as follows. Since
<7i(l, 0,..., 0) > 0, therefore the continuity assumed ensures the existence
of b > 0 such that for 0 < ж2,..., жп < 6 we have
gi(l,X2,...,xn) > 0.
On the other hand,
#ΐ(0,Ζ2,···,Ζη) < 0.
So, again by continuity, the у above must exist. If x'3 > Xj, j φ 1, then
9i(2/(^2, · · ·, x'jj · · ·, xn)j ^2, · · ·, x'j, · · ·, xn)
= 0 = 9\{y(x2, · · ·, xn), X2, · · ·, Xn)
> 91 (^(^2, · · · , Xn), X2, · · · , Xj, . · · , Xn),
and this shows that y(x2,.. ·, жп) is an increasing function of Xj. We next
verify the continuity of y. Since
gi(y(x2,-· ,xn) -ε,Χ2,···,Χη) < 0 < gi(y(x2,--.,xn) +ε,Χ2,·..,χη),
there is a δ > 0 such that for \xj — x'3\ < δ, j = 2,3,..., n, we have
9i(y(x2,---,xn) -ε,Χ2,--·,χ'η) < ° < 9i(y(x2,---,xn) +ε,χ2, · · · ,x'n)-
By the foregoing, this proves the inequalities
y(x2, ...,χη)-ε< y(x'2, ...,x'n)< y(x2, · · ·, xn) + ε.
After these preparations, consider the η — 1 functions
hj(x2, --·,χη)= 9j{y(x2, · · ·, xn),X2, ···, Xn), j = 2,..., n.
By the properties of y, they are continuous, have sum 0, and satisfy the
relations hj (0,..., 0) =0 for all j. We claim that they are strictly monotone
in the required sense. Actually, if к Φ j and Xk increases, then y(x2, · · ·, xn)
also increases and therefore hj decreases. Thus, if x2 increases, then all
hk, к φ j, will decrease, so hj must increase since the sum of the hi is
zero. Consequently, the η — I functions hj satisfy the hypotheses of the

230
3. SOLUTIONS TO THE PROBLEMS
problem. So, by the induction hypothesis, there exists a!n > 0 such that for
any 0 < xn < a'n there are x2 — #2(#n), · · · ,#n-i = Xn-i(xn) satisfying
hj(x2,..., xn) = 0 for all j > 2, and here each Xj(xn) tends to 0 as xn —> 0.
Hence, there exists min{a^,6} > an > 0, where b is the constant used
during the preparation, such that if 0 < xn < an, then 0 < Xj(xn) < b for
j = 2,...,n — 1. Then, however, with the values
χι = xi(xn) := yOE2(zn),...,zn_i(£n),zn),
^2 = ж2(жп)? · · · ixn-l — xn-l\xn)'>
the relation gj(xi,..., #n) = 0 is valid for all j (for j = 1 this follows from
the definition of y), and here each Xj(xn) tends to 0 as xn —> 0 (we again
make use of the continuity of y). Thus, we have proved the statement also
for η functions. D
Solution 2. Let <^ (г = 1,2,..., η) be the functions defined in
Solution 1.
Let ei = (0,..., 0,1,0,..., 0) be the zth unit vector in Rn (the ith
coordinate is 1, the others are 0). By the assumptions, g\{ei) < 0 if г > 1. In
view of the continuity of gi, there is a number e% > 0 such that 0 < χ ι < ε χ
implies g\{x\e\ + ег) < 0. Set ε = min^: 2 < г < η}. We show that if
0 < x\ < ε, then there exist nonnegative numbers x^ £3,..., xn such that
gi(xi, X2, - - · j Xn) = 0 for 1 < г < п. То this end, it is sufficient to prove
that gi{xi, X2, · · · 5 Xn) = 0 for г > 2 (recall that the sum of all functions <&
is zero).
Let 0 < x\ < ε be an arbitrary number, which will be fixed in what
follows. Set
G(rr2,...,Zn) = max gi(x1,x2,...,xn)
2<i<n
and
Я = {(ж2,...,жп) GRn_1 >0: G(xux2,...,xn) < 0}.
Then
a. Я С [0, l]n_1, since if Xi > 1 for some г > 2 then 0 > gi(x\ei + ег) >
gi(xi,x2, · · ·, xn). Therefore £\ ^ = 0 gives G(x2,..., a?n) > 0.
b. Я is closed, being a level set of a continuous function.
c. Я is nonempty, since (0,0,..., 0) G Я. It follows that #i(xi, #2? · · ·, Xn)
takes its minimum on Я at some point (ж2, Xg,..., x^).
We prove that
0»(Я1,Ж2>ж31--чжп) = ° fora11 *>2·
By the definition of Я, it is clear that
&(яъЖ2,...,а£) <0 (г = 2,...,п).
Suppose that for some г > 2 we have
0t(si,a&...,a£) <0.

3.3 THEORY OF FUNCTIONS
231
Then making x® a little larger, we remain in H, but the value of g\ becomes
smaller, which contradicts the choice of (ж§,... ,ж£). This contradiction
proves the statement. D
Remarks.
1. It is easy to show by examples that the assertion becomes false if we
replace the condition of strict monotonicity by monotonicity in the wide
sense.
2. Here is a simple example that if a > 0 is arbitrary then a correct decision
cannot already be achieved in case the eldest (say, the nth) brother offers
the judge at least a units of currency: let
9j(xu ... ,жп) = (и - 2)xj Η j— -χι Xj-i - Xj+ι xn
Xj + ι
if j = 1,2,..., η — 1, and let
αχι αχ2 αχη-ι
gn(xi, ...,xn) = (n- l)xn -
χι + 1 x2 + 1 xn-i + 1
Problem F.53. Construct an infinite set Η С С[0,1] such that the
linear hull of any infinite subset of Η is dense in C[0,1].
Solution. Let {#fc}£Li be dense in C[0,1] (gk Φ 0). We show that the set
Η = {M~=2, where
^"SillftiJn* '
fr[ \1Ы1ос
meets the requirements.
If L is any bounded linear functional on C[0,1], set
Σ( L9k \
fc=1VIW<J
M^Ehrrl·
Μ
Then hi, is analytic in the unit disc, and Lhn = Hl (l/n)· Let
ti = \Γίη\ 1 Ι^Π2 1 ^Пз J · · · J
be an infinite subset of Я, and let L G C[0,1]* be any functional that
annihilates H'. Then Hl (l/nm) = LhUm = 0 for all m, so the analyticity
of tiL yields Hl = 0. Hence Lgk = 0 (k = 1,2,...), that is, L = 0. This,
however, means exactly that the linear hull of H' is dense in C[0,1]. D
Remark. Of course, the proof works in any separable Banach space.

232
3. SOLUTIONS TO THE PROBLEMS
Problem F.54. Let a > 0 be irrational.
(a) Prove that there exist real numbers αϊ, α2, аз, а4 such that the
function f : R -> R,
f(x) = ex[a\ + ^2 sin ж + аз cos ж + а4 cos(arr)]
is positive for all sufficiently large x, and
liminf/(x) = 0.
(b) Is the above statement true if a2 = 0?
Solution.
(a) Put
f{x) = ex{2 — cos (x — 2πα) — cosarr),
where a is to be defined later. Then f(x) > 0, while f(x) = 0 if χ = 2ктт and
χ — 2πα = 2ηπ for some integers к and n. Hence a(n + a) = k. Assuming
that also f(xf) = 0 for some χ' φ χ, we obtain a(nf + а) = к' for some
integers nf φ η and к' ф к. The relations a{n + a) = к, а{п' + ά) = к' yield
а = (к — к')/(п — n'), a contradiction. Thus f(x) > 0 for all sufficiently
large values of x.
Choose α G [0,1) so that, for some sequence {rik} of natural numbers,
Щ_
a
<
0-пкъ/а
Пк
(mod 1).
The existence of an α of this kind can be established as follows. Let К be
the circle of perimeter 1, and po G K. Starting from po, lay 1/a on К (in
a given direction) η times to obtain pn. Denote by In the interval
0—ηπ/α
0-ηπ/α
Pn ~
П
■ > Pn + "
on K. Let no = 1, and suppose that {n^}^0 are given. Define nm+i
so that nm+i > nm and 1Пт+1 С 1Пт. There is an nm+i of this kind
since {PnJn^Lk ls dense in К and e-mr/a/n —> 0 (n —> oo). We have
n^=07nfc. Then
Jno Э Jni Э ..., and |Jn| —► 0 (n —► oo). Let а
/ (^Л = e2nfcx/« Γ2 _ cos n^L _ 2πΛ _ со82пкЖ
= e2nfc-/« [ι _ cos27r (?* - a - mfc)] ,
where m^ is an integer satisfying the relation
e-nkn/oc
a
a-ruk
<
Пк

3.3 THEORY OF FUNCTIONS
233
Therefore, using the inequality cos и > 1 — и2/2, we obtain
D (k -► oo).
2тг2
Hence, liminf:c__(_00 /(ж) = 0.
(b) We show that in the linear hull of the functions еж, ех cos x, ex cos ax
there is a function with the required properties if and only if for any ε > 0
the inequality
m
a
-πη/2
<ε-
n
m
(*)
holds for infinitely many rational numbers — (m,n G N). Let
η
g{x) = ex{a\ + аз cos χ + сц cos ax)
be such a function. Then, necessarily, a\ = \а$\ + |а4|, and а^ ^ 0 for
г = 1,3,4. Let {xk} be a sequence that tends to oo and satisfies g(xk) —> 0
as к —> oo. Since αϊ = |аз| + |а4|, we have
xk = тгпк + #ь a^fc = ^ra^ + Δ*,
where n^ and m^ are natural numbers, n^ —> oo, mk —> oo, ^ —> 0 and
Δ*. —> 0 as A: —> oo. Let β € (0,1) and b = тт{|аз|, |а4|}. If к is sufficiently
large, then
β > g(xk) = e™k+6k (αϊ + α3 cos^nfc + Sk) + a4 cos(nmk + Ak))
= e™k+6k (αϊ - |α3| cos6k - |α4| cos Δ*)
= e™k+8k (*i " Ы + l^slf +o(*2) - Ы + M^ + ο(Δ2))
> e™k~'\ (\a3\6l + |α4|Δ2) > e™k Α (*,2 + Δ2) ,
since cos и = 1 — u2/2 + o(u2) for и —> 0. Hence, for sufficiently large /c,
№1, |Δ*| < )/ψβ-^2
and, therefore,
Ι πτη^ + Δ^ га J
а —
mk
nk
nnk + 5fc nfc
Δ* " ^**
(*+&)
ftfc
< |Afc|+2a|gfc| < 2a_+l ^ \ъфе~™к12
(π - l)nfc
^fc

234
3. SOLUTIONS TO THE PROBLEMS
Since β (Ξ (0,1) is arbitrary, we obtain that for any ε > 0 the relation (*)
is valid for infinitely many rational numbers ^.
Next let ε > 0 be fixed, and suppose that (*) has infinitely many rational
solutions rafc/rifc, mk £ N, nk £ N, к = 1,2,... . Passing to subsequences
if necessary, we may assume that all the nk have the same parity, all the
mk have the same parity, and
a
mk
Пк
р-жпк/2
<ε (fc=l,2,...).
n>k
Let a\ = 2, and furthermore,
аз
and
CL4
\-l if
-ί1 "
1-1 if
if the members of {nk } are odd,
the members of {nk} are even
1 if the members of {m^} are odd,
the members of {m^} are even.
Then g(x) = ex{a\ + аз cos χ + α± cos ax) > 0 for χ > 0, and
д(пк7г) = enfc7r [2 + аз cos пкж + 04 cos апкк]
_ en*^ ^ _ COS7r(arik _ mfc)]
-n2(ank - mk)2 + о ((anfc - mk)2)
nfc7r 2 2 -7rnfc _ J2J2
ε π
< enk7Vn^ze
for sufficiently large k. Consequently,
liminf #(:e) < ε2π2.
x—>+οο
Thus, if for any ε > 0 the relation (*) has infinitely many solutions, then
\immfg(x) = 0.
χ—>+οο
If α is an algebraic number then, by the Thue-Siegel-Roth theorem, for
any δ > 0 there are only finitely many m/n such that
m
a
η
<
,2+6 *
So, if (*) for any ε > 0 has infinitely many rational solutions m/n, then
α must be transcendent. There is an α of this kind. For instance, let
N > e71"/2 be an integer, n\ = TV, and nk+i = 7Vnfc, к = 1,2,..., and
further let α = Σΐίι Vn»· Then with the notation

3.3 THEORY OF FUNCTIONS
235
we have
a - ctk <
and, therefore,
α
ГПк
пк+1 Νη><
2пк
-ттпк/2
Since
]упк (iVe-7r/2)nfc пк
2пк
(ЛГе-тг/г)^
it follows for any ε > 0 that
О, к —> oo ,
a —
ГПк
пк
о-кпк/2
< ε-
пк
for all sufficiently large indices к. П
Problem F.55. Prove that if {ак} is a sequence of real numbers such
that
oo / 2n-l
1/2
^2\ак\/к = oo and ^ I ^ к(ак - ак+г)2 \ < oo,
k = l
n=l \k=2n~1
then
Г
Jo
}aksm(kx)
k = l
dx = oo.
Solution. Although the condition
lim ak = 0
к—>oo
(i)
does not appear in the statement of the problem, by the well-known Cantor-
Lebesgue theorem, this follows from the fact that the series
yj OLk sin kx
(2)
k=l
is convergent almost everywhere. We note that for the application of this
theorem it would be sufficient if the series (2) were convergent on a set of
positive measure. To make reference easier, we list the remaining
conditions:
flfc
k = l
00,
(3)

236 3. SOLUTIONS TO THE PROBLEMS
oo / 2n-l \ 1/2
n=l \fc=2"-1 /
where
Aak := α^ ~ flfc+i (fc = 1,2,...).
Prom (4) it follows that the sequence {ak} has bounded variation, that
is,
oo
]Γ|Δα*|<οο. (5)
Really, by the Cauchy inequality,
oo oo 2n—1
£>afc| = £ £ \Аак\
к=1 n=lfc=2n-1
oo / 2n-l \ 1/2
<Σ r-1 Σ ιδ-*ιί
n=l \ fc = 2»-1 /
oo / 2n-l \ 1/2
<Σ Σ *ι**π ·
n=l \fc=2n-1 /
Consider the nth partial sum of (2). By Abel's rearrangement, we obtain
η η
Σак sinкх = ΣDk(x)Aak + an+1Dn(x), (6)
fc = l fc = l
where Dn(x) is the conjugate Dirichlet kernel:
a / ч v^ . , cosf-cos (n+^)x
Dn(x):=^sinA:x= 2 9 .Д ^- (n=l,2,...)·
Introduce the notation
_ , ч cos (п+\)х,
Ό^··=—^ψ- (« = 0,1,···).
Then
ДДх) = L>n(x) ~ D0(x) (n = 0,1,... ; D0(x) = 0),
and from (6) we derive
У^ a* sinfex = ]P Dk(x)Aak - D0(x) ]P Δα^ + an+i^nW - αη+ι Д)(я)
k=l k=l k=\
η
= ]P Dk(x)Aak - aiD0(x) + an+i 1)п(ж)
η
= ]P Dk(x)Aak + an+i 1>п(ж),
fc=0

3.3 THEORY OF FUNCTIONS
237
where the convention α$ := 0 and its consequence Δαο = —αϊ are used. It
follows that the series (2) is convergent:
^2aksinkx = ^2Dk(x)Aak =: f(x)
(7)
k=l
k=0
for every χ with the possible exception of the case χ = 0 (mod27r).
In the following, we make use of an inequality of the Sidon type: for any
integer η > 2 and numerical sequence {bk},
f
J τ In
2n-l
Σ №(*)
k=n
dx < С [ Σ kti]
2n-l \ V2
I
k=n
(8)
where С is a positive constant. To see this, we first apply the Cauchy-
Schwarz inequality and then exploit the orthogonality of the system
{cos (k+ ^) x}:
f
<
2n-l
Σ bkDk(x)
k=n
dx
J π/η
2n-l
k=n
COS
(*+3l·
2sinf
dx
dx
π/η (2 Sin f)2
V2 / /2n-l
/ ( Σь*cos lk+
ν /e=n
a·'
2 \ 1/2
(2n-l \ X/2 /2n-l \ X/2
/e=n / \ k=n /
Let s > 1 be an integer. According to (7),
rr/U+1)
/ \№\άχ>Σ
2S-1 r±
j'-i _
J2Dk(x)&ak
k=0
dx
j=l J+i
J2Dk(x)&ak
k=j
dx := h - h · (9)
Using the inequality
Dk(x) + -
χ
we obtain that
t/(j+1)
/i>
< fc + 1 (0<χ<π; fc = 0,1,...),
2s-1 f*/j j-i
Σ/ ΣΔα^τ-Σ/ Bfc+1)iAa*id*
7=ί Α/ϋ+i) |^ί |x i=i Α/ϋ+ΐ) fc=0
:= Λι — Λ2 ·

238
Since
by the foregoing
3. SOLUTIONS TO THE PROBLEMS
,„(i + I)>i- ' ,
Ju - h It - w+T)) ■
It is easy to see that
2S-1
ί —ι II с» - σο
У^ .,. L· <max|a,|V ——- < VlAaJ.
whence
2S-1
Similarly,
2s-lj-l
fe+1
^ = ^ΣΣτ77τπΐΔ^ι^7ΓΣΐΔ^ι^27ΓΣΐΔ^ΐ·
j=l k=0J^ ' k=0 k=l
2S-1
Therefore
Z —± | | OO
7ι>^1Μ_(ι + 2π)^|Δα,|.
We turn to estimating 72:
2 ~3 f*/j
rr/U+l)
-ΣΣ/
g 2—1 «тг/j
<ΣΣ
/=1 ,=2«-ι ·/π/ϋ·
2*-l oo 2n-l
Σ + Σ Σ Ι β*(*)Δα*
fc=j n=/+lfc=2n"
dx
2'-l
Σ Дь(*)А<Чь
2*-l oo
fc=J
π/j
dx
+
β * —χ u*j /•π/7
Σ Σ Σ /„■♦,,
2η-1
Σ -Dfc(a?)Aajb
fc=2"
dx =: /21 + /22 ·
Making use of the elementary inequality
2
sin χ > —χ (0 < χ < — ) ,
~ π V ~ ~ 2/
we obtain
s /·τ2-'+1 „ 2'-1 ^l^O2""1
sE/, έ Σ ιδ*ι*-=|ϊΣι**ι·

3.3 THEORY OF FUNCTIONS
239
We now apply (8):
^22
ο© -π2"
-ΣΣί,
oo η—Ι /.π2-ζ +
n=2 i=l 7π2"'
2n-l
5^ Dk(x)Aak
ife=2»-i
2n-l
5^ Dk(x)Aak
к=2п~г
2n-l
5^ Dk(x)Aak
k=2n~1
ч 1/2
drr
dx
drr
oo / 2n-l
n=2 \fc=2n
Consequently,
2S-1
oo / 2n-l
1/2
^^Σι^ι+^Σ Σ *iw
fe=l n=2 \fe=2"-1
Relying on (9), (10), and (11), we find that
2 -1 ы
(И)
Γ
2^ &k sin kx
k = l
*>Σ?
k = l
- 1 + 2π +
π1η2
oo / 2n-l
1/2
ΣΐΔα*ι-σΣ Σ *ΐΔα*ι2
k = l
n=2 U=2n"1
In view of conditions (3), (4), and (5), this already yields the desired
conclusion. D
Problem F.56. Let h : [0, oo) —> [0, oo) be a measurable, locally inte-
grable function, and write
H(t) := / h(s)ds (t > 0).
Jo
Prove that if there is a constant В with H(t) < Bt2 for all t, then
poo ft
/ e-*<«> / e*<«>dudt = oo.
Jo Jo
Solution 1. Interchanging the order of integration, we obtain
/»oo ft /»oo ft
/ e-ff(«) / eH^dudt= / / el"»(*)-*(«)] du A
Jo ./o Jo Jo
/»oo /»oo
= / / е[-я(*)-я«]ЙЫи
«/о Л
poo poo г /»t
= / / exp -/ h(s)
Jo Ju L Jw
dtdu.

240
3. SOLUTIONS TO THE PROBLEMS
It is sufficient to prove the existence of a constant с > 0 such that
r2T ru+l/T
i{t):=Jt L exp[-Juh^
for large T. We have
r2T ru+\/T
ds
dtdu> с
I{T)>-L L
=4
exp
ru+l/T
— / h(s) ds
Ju
dtdu
ru+l/T
■ / h(s)ds
Ju
du.
Introduce the notation
Qt
( ru+l/T
:=lu€[T,2T]: J
h(s) ds > 10B
!■
and find an upper estimate for the Lebesgue measure μ{(ίτ) of the set Qt-
r2T ru+l/T r2T+l/T rs
-1/T
whence
p'Zl pU+L/1 fZl+L/l pS
10Βμ(Τ) < / h(s)dsdu< / h(s)duds
J Τ J и J Τ J s-l/Τ
1 /-2T+1/T ! / t\2
= TjT М-)Л<гВ(2Г+-) ,
Therefore,
1 J[Tt\
β-ΙΟΒ
2T]\Q7
е"11ШЛ|>-=_(r-M(Qr))
> e
-10B
ЫУ
c>0,
provided that Τ > 1. D
Solution 2. We first show that if #: [1, oo) —> (0, oo) is measurable and
locally integrable, further
/'
g(s) ds < Bit2 (Bi = constant)
for every t, then
f°° 1
ds = oo.

3.3 THEORY OF FUNCTIONS
241
Indeed, let Τ > 1 be any real number. By the Bunyakovski-Schwarz
inequality
Ч/»ЧГ
IT
(g(j)—==dt
g(t)
So
/•2Γ
*4 °m)\L жй|£В,-4Г2
/.2T
/*2T ι J
It W)dU
/"" 1 1
/ -r- dt > — (T > 0),
which yields the assertion.
Let
№ ■■-
PH(t)
J* eff(«) du
(ί>1).
According to the problem, we have to prove that
Г 1 J
/ —ττ dt = oo.
To this end, by the above remark, it is sufficient to verify the inequality
I g(s)ds<B1t2 (t> 1)
with a suitable B\. This is simple:
ft rt eH{s)
\ q(s) = / —: ds
Л Л ЦенЫаи
= In / ея« du - In [ eH^ du.
Jo Jo
Since Я is monotone increasing,
/ #(s) < In \teH{tA = lnt + Я(«)
<£ί2 + 1ηί <#ι*2
with a suitable #i. D

242
3. SOLUTIONS TO THE PROBLEMS
Problem F.57. Consider the equation f'(x) = f(x + 1). Prove that
(a) each solution f : [0, oo) —> (0, oo) has an exponential order of growth,
that is, there exist numbers а > 0, b > 0 satisfying \f(x)\ < аеЪх,
x>0;
(b) there are solutions f : [0, oo) —> (—oo, oo) of nonexponential order of
growth.
Solution. In the statement of the problem, a constant с has been omitted.
The statement is correctly, f'(x) = cf(x + 1), where 0 < с < 1/e.
Then there exists a λ > 0 such that λ = cex and so eXx is a positive
solution. On the other hand, the equation f'{x) = f(x + 1) has no
positive solution /. If it had one, then / would be strictly increasing, and the
Lagrange mean value theorem would give /(1) > /(1) — /(0) = /'(ξ) =
/(£+1)>/(1),а contradiction. The equation f'(x) = cf(x+1) has a
positive solution on [0, oo) if and only if с < 1/e (see T. Knsztin, Exponential
bound for positive solutions of functional differential equations, unpublished
manuscript).
(a) Proof of the exponential order of growth. If /: [0, oo) —> (0, oo)
satisfies the equation f'(x) = cf{x-\-1), put a(x) = f(x)/f(x). Then
/(*) = /(0)exp
(/««*).
and
that is,
a(x) = cexp ί / a(s) ds\ > 0;
In
a(x) Г+ / ч J
—— = / a(s)ds, x>0.
с Jx
Choose к so that к > α(0) and к > \п(к/с). Begin to define the
sequence {xn}£L0 in the following way:
xo = 0, X\= max{x G (0,1]: a(x) < k} .
Since
I
a(s)ds = \n^-<\n-<k,
Jo с с
Χι is well defined. If a?o,..., a?n axe given, put
xn+i = max{x G (xn,Xn + 1]: ol(x) < k} .
Since
J X
Xn+1a(s)ds = ln^^-<\n-<k,
~ С С

3.3 THEORY OF FUNCTIONS 243
xn+i is well defined. Since a(x) > к on (xn+i,xn + 1], it follows that
xn+2 > Xn + 1· Hence
In
[0,7l] С (J[a?Z_i,a?Z].
1 = 1
Therefore, if χ G [n — 1, n), then
/ a(s) ds < a(s) ds < Y^ / a(s) ds
JO JO l = 1 Jxi-i
1=1 -*χι-ι 1=1 C
к
< 2nln - < 2nk < 2xk + 2k.
с
Consequently,
/(*) = /(0) exp (Г a(s) ds] < f(0)e2ke2kx (x > 0).
(b) There is a function φ G C°°[0,1], not identically zero, such that
фМ(0) = 0(n)(l) = 0; n = 0,1,2,... . For example, ф{х) =
exp (l(x(x - 1))) if χ G (0,1), and /(0) = /(1) = 0. Then the formula
f(x + n) = ^n\x), n = 0,l,...; *G[0,1],
defines a solution of the equation f(x) = cf(x + 1). Let χ G (0,1) be
such that φ(χ) ^ 0. By Taylor's theorem, there exists an η G (0, x)
such that
1=0
Hence
which increases faster than any a · ebn as n —> oo. Thus / has a
nonexponential order of growth. D

244
3. SOLUTIONS TO THE PROBLEMS
3.4 GEOMETRY
Problem G.l. Find the minimum possible sum of lengths of edges of
a prism all of whose edges are tangent to a unit sphere.
Solution. If the edges of a prism are tangent to a sphere, then each face
of the prism intersects the sphere in a circle, tangent to the sides of the
face. Hence, each face is a circumscribed polygon.
The translation along the generators of the prism that takes one base
polygon to the other moves the inscribed circle of the first polygon to
the inscribed circle of the second. These circles are congruent and have
parallel planes since the bases of the prism are congruent and parallel.
Since the translation that moves one of two congruent and parallel circles
on the sphere to the other is perpendicular to the plane of the circles, the
generators of the prism are perpendicular to the base, thus, the prism in
question is a right prism.
Sides of a right prism are rectangles. Since they are also circumscribed,
and only squares are circumscribed among rectangles, each side of the prism
is a square. Consequently, the edges of the bases and the generators have
the same length, the bases are regular, and the sum of the lengths of the
edges of the prism is three times the perimeter of the base.
Let us cut the prism and the sphere by a plane through the center of the
sphere, parallel to the base. The prism is cut in a polygon congruent to the
base polygons; the sphere is cut in a great circle. The great circle contains
the point of contact of every tangents of the sphere orthogonal to the plane
of the great circle. Therefore, it contains a point from each generator.
We get that the base polygon is inscribed in a unit circle. Since the base
polygon has equal sides, it must be a regular polygon. We conclude that
the prisms that satisfy the prescribed condition are regular prisms, with
the length of the generators equal to the length of the sides of the base
polygon scaled in such a way that the circumcircle of the base polygon has
radius 1.
Conversely, every prism of this type satisfies the condition of the
problem, since the center of such a prism is located at distance 1 from each
edge. Indeed, the distance to the center from a generator is clearly equal
to 1. On the other hand, the orthogonal projection of the center onto a
lateral face, which is a square, is the center of the face. Thus the distances
to the center from the sides of this square are equal.
To get the sum of the lengths of the edges of such a prism, we have
to multiply by 3 the perimeter of a regular polygon inscribed in a unit
circle. It remains to decide which of these regular polygons has the shortest
perimeter.
The perimeter of a regular n-gon inscribed in a unit circle is equal to
η α

3.4 GEOMETRY
245
where α = π/η. Since sin x is concave on the interval (Ο,π), the slope of
the chord connecting the origin to the point (x,sinx) is decreasing as χ is
increasing. Therefore, sin a/a is minimal when α = π/η is maximal, that
is, η is minimal. Thus, the minimum is attained by a regular triangle.
As for the prisms, the minimum of the sum of the edge lengths is
obtained for a regular prism over a triangular base, with lateral faces being
squares, and the value of the minimum is 3 · 6 · sin 60° = 9>/3- □
Problem G.2. Show that the perimeter of an arbitrary planar
section of a tetrahedron is less than the perimeter of one of the faces of the
tetrahedron.
Solution. Obviously, planar sections of a tetrahedron are triangles or
quadrangles, since each face of the tetrahedron contains at most one edge
of the intersection provided that the intersection does not coincide with
the face.
The case of quadrangles can be reduced to the case of triangles. Namely,
we show that if we translate the plane of a quadrangle intersection in both
directions until it passes through the nearest vertex, then one of the
translated planes intersects the tetrahedron in a triangle (possibly degenerated
to an edge), the perimeter of which is not shorter than the perimeter of
the quadrangle.
If both translated planes intersect the tetrahedron in a triangle, then
introduce the notation shown in Figure G.l and express the sides of the
quadrangle with the help of the sides of the two triangles. Since the planes
of the three sections are parallel, the ratio of the segments of a straight line
cut off by these planes is the same for all straight lines. Set
λ = BP : AB = BQ : ВТ = FR : UF = GS : AG.
Then for the corresponding sides we have
ρ = λαι, q = 6i + (1 - λ)(62 - h) = \Ьг + (1 - λ)62 ,
г = с2 + A(ci - с2) = Xci + (1 - Л)с2, s = (1 - λ)α2 .
From these, p + q + r + s = λ(αι + Ъ\ + C\) + (1 — λ)(α2 + 62 + c2), that is,
К = ΧΚι + (1 — X)K2, where К = ρ + q + r + s, K\ — a\ + h\ + ci, and
K2 = a2 + 62 + c2. This implies
К <шах{КиК2}.
If one of the triangles AAUT, or ABGF, or both degeneratse to the edges
AD or ВС, respectively, then this inequality remains valid, putting Κ ι =
2AD or К2 = 2BC, respectively. Then we get that the perimeter of one of

246
3. SOLUTIONS TO THE PROBLEMS
Figure G.l.
the faces of the tetrahedron is greater than the perimeter of the intersection
quadrangle.
Hence, it is enough to consider the case of triangle intersections. We
may suppose that the intersection triangle has a vertex in common with
the tetrahedron. Otherwise we may translate the plane of the triangle
to reach this situation (see Figure G.2). The translated plane meets the
tetrahedron in a triangle similar to the original one, with ratio of similarity
greater than 1 and it passes through that vertex of the nonintersected face
of the tetrahedron that is nearest to the intersecting plane.
If the intersecting plane is parallel to the face not met by it, the assertion
is trivial. Otherwise, we show that the perimeter of AKBM is smaller than

3.4 GEOMETRY
247
the greater of the perimeters of AKBC and AKBD. For this purpose,
consider the smallest ellipsoid of revolution with foci К and В that contains
both С and D. Since Μ is an internal point of it, we get
~KM + ЖВ < max{FC + UB,WD + Ш?} .
But then, either кКвм < kKBD < kABD or кКвм < кКвс, where kPQR
denotes the perimeter of APQR. In the first case, the assertion is proved,
the second case can be finished by a repeated application of the previous
arguments. D
Problem G.3. Show that the center of gravity of a convex region in
the plane halves at least three chords of the region.
Solution. Let us denote the disc by Τ and its barycenter by S. If X is
a point of the boundary curve G of T, then denote by Y(X) the second
intersection point of the straight line XS and the curve G. Let f(X) =
~XS - Y{X)S. We have f{X) = -f(Y(X)) for any X e G. Since f{X)
changes continuously as X runs over the arc it attains any value
between f(X) and —f(X). This implies the existence of a point Χι e
G such that X\S = Y(Xi)S. If X\S were the only straight line whose
intersection with Τ is halved by 5, then / would be positive along one of
the arcs Χ{Ϋ(Χ\). Reflecting this arc through 5, the reflected arc together
with the other arc Λ\Υ(Χι) of G would bound a domain Ti, the barycenter
of which is different from 5, since T\ lies in a half-plane bounded by a
straight line passing through S. Therefore, the barycenter of the union
T<i = T\ U Τ should be different from S. On the other hand, T2 is centrally
symmetric with respect to 5, so its barycenter should be invariant under
the reflection in 5, which means that its barycenter should be 5, which is
a contradiction.
The same arguments show that if / has finite zeros, then it can not be
nonnegative on a half-arc of G. Suppose that there are only two straight
lines X\S and X2S for which the segment of the straight line cut off by
the figure is halved by S (X\,X2 G G). By the previous remark, / must
be negative (or positive) on the arcs X2Y(X\) and XiY(X2), which is
a contradiction, since the sign of / is opposite on arcs opposite to one
another. We conclude that at least three chords of the figure are halved by
S. The existence of four such chords cannot be proved in general, as it is
shown by the case of an arbitrary triangle. D
Problem G.4. Let Αι, A2,..., An be the vertices of a closed convex
n-gon К numbered consecutively. Show that at least η — 3 vertices Ai
have the property that the reflection of Ai with respect to the midpoint of
Ai^iAi+ι is contained in K. (Indices are meant mod n.)
Solution. We shall call a vertex Ρ of К reflectible with respect to К if its
reflection in the midpoint of the segment connecting the two neighbors of
Ρ belongs to K.

248
3. SOLUTIONS TO THE PROBLEMS
1. Let us begin with the first nontrivial case, with the case of quadrangles.
For both pairs of opposite sides, the sum of angles lying on one of the two
sides is at least π; let us denote by A a common vertex of two such sides
and denote the other vertices in a cyclic order by В, C, D (s§e Figure
G.3). We show that A is reflectible with respect to the quadrangle. We
can construct the reflection A' of A by taking a parallel to AB through
D, which intersects side ВС in a point Ε because <ADC + <BAD > π,
and then taking the parallel to AD through B, which intersects the
segment DE at Af because <DAB + <ABC > π. Therefore, A' belongs
to the quadrangle ABCD.
Figure G.3. Figure G.4.
2. If we show that among any four vertices of the convex polygon K, at least
one is reflectible with respect to K, then the assertion of the problem is
proved.
We shall say that a vertex Ρ of the polygon precedes the neighboring
vertex R if going around К starting from Ρ in the direction of R is a
movement of positive orientation. Consider four arbitrary vertices of К
and apply for them the previous notation, that is, let <ADC + <BAD > π,
<DAB + <ABC > π (see Figure G.4). If the vertex Ε of К that follows
A is not B, then Ε is an internal point of the intersection S of three half-
planes Si, 52, S3, where Si is bounded by the straight line AD and contains
ABCD, S2 is bounded by ВС and contains ABCD, and S3 is bounded by
AB and does not contain ABCD. Similarly, if the vertex F that precedes
A is not D, then F is an internal point of the domain Τ = T\ П Т2 П T3,
where the half-plane T\ is bounded by the straight line AB and contains
ABCD, T2 is bounded by DC and contains ABCD, and T3 is bounded
by AD and does not contain ABCD. It is enough to prove that A is
reflectible with respect to the quadrangle AECF, since then it is obviously
reflectible with respect to К as well. For this purpose, it suffices to show
that <EAF + <AFC > π and <FAE + <EAC > π. It is enough to see

3.4 GEOMETRY
249
the first inequality; the second can be shown in a similar way. But we have
<EAF + <AFC = <EAC + (<CAF + <AFC)
= <EAC + π - <ACF > <BAC + π - <ACD
= <BAC + (<CAD + <ADC) = <BAD + <ADC > π,
which proves the assertion of the problem. D
Remarks.
1. The assertion cannot be improved; that is, there exists for all η a convex
n-gon having exactly η — 3 reflectible vertices. Indeed, consider a convex
quadrangle ABCD with no parallel sides. It is easy to see that such a
quadrangle has only one reflectible vertex; denote it by A. Now consider
a convex n-gon К such that B, C, and D are vertices of К while the
other vertices of К lie in a neighborhood of radius ε of A (obviously,
such a convex n-gon exists for any n). It can be easily seen that if ε is
sufficiently small, then В, С, and D are not reflectible with respect to
K.
2. The assertion does not hold for concave polygons.
Problem G.5. Is it true that on any surface homeomorphic to an open
disc there exist two congruent curves homeomorphic to a circle ?
Solution. The answer is no: There exists a surface homeomorphic to a
disc which does not contain two congruent curves homeomorphic to a circle.
We show this by giving an example.
We start from a minimal surface, that is, from a surface for which
Η = 1/2(#1 + д2) = 0. (Неге Н denotes the Minkowski curvature, and
gi and g2 are the principal curvatures of the surface.) Although
unnecessary, we explain why we are looking for counterexamples among minimal
surfaces. If two congruent copies of a surface intersect one another in a
closed curve, then by covering one copy with the other, the two positions
of the intersection curve give two congruent curves on the surface, which
are different in general. If two copies are tangent to one another at an
isolated common point, then by a slight movement of one of them, they
can be made to intersect one another along a curve. The choice of minimal
surfaces is justified by the fact that if Η > 0 at a point of a surface, then
we can always find a congruent surface so that this point is an isolated
common point of them. We can get such a surface by reflecting the
original surface in the tangent plane at the point in question and then rotating
it 90° about the normal of the surface. This follows from Euler's theorem,
since by the inequality
gi cos2 φ + g<i sin2 φ > —g2 cos2 φ — g\ sin2 φ,
the curvatures of the normal sections of the original surface are always
greater than the curvatures of the corresponding normal sections of the

250
3. SOLUTIONS TO THE PROBLEMS
transformed surface, and therefore the first surface is above the second one
in a small neighborhood (looking at them from the direction of the normal
vector).
Now we show that in a sufficiently small domain of a minimal
surface, any two congruent closed curves bound congruent domains of the
surface. For this purpose, take two copies of the surface and move one of
them so that the congruent curves cover one another. Suppose that the
piece of the minimal surface is small enough to ensure that both surfaces
can be obtained as the graph of the single-valued functions ζ = fi(x,y)
and ζ = /2(^,2/)· The areas of the domains bounded by the closed curve
are obtained as integrals of the lengths of the vectors rai(—l,pi,qi) and
77i2(—1,^25^2)? respectively, over the same domain. (Here pi and qi denote
the partial derivatives of fi.)
We use the fact that the variation of the surface area of a minimal
surface is 0, that is, if a minimal surface is embedded into a one-parameter
family of surfaces all having the same boundary, then the derivative of
the area of these surfaces with respect to the parameter is zero at the
minimal surface. Consider the one-parameter family of surfaces defined by
ζ = λ/ι + (1 — λ)/2. The area of a member of this family is obtained as the
integral of the magnitude of the vector Ami + (1 — λ)ττΐ2. This magnitude
is a strictly convex function of the parameter λ if mi 7^ 777,2 and does not
depend on λ if mi = m2. Consequently, the area, obtained by integrating
convex functions of λ, is itself a convex function of λ. Its derivatives at
λ = 0 and 1 can vanish only if it is a constant function, and this happens
only if mi = Ш2 at any point of the domain, that is, if the two pieces of
surfaces coincide.
According to this observation, it is enough to find a minimal surface
that does not contain different congruent pieces. Since two irreducible
algebraic surfaces that have a domain in common coincide, it is enough to
present an algebraic minimal surface that has no or only a finite number of
automorphisms different from the identity. (An automorphism of a surface
is a bijective mapping of the surface onto itself that can be extended to a
congruence of the space.) If the surface has finite automorphisms, then a
small neighborhood of a point different from its automorphic images has
no automorphisms different from the identity.
Actually, any algebraic minimal surface would do, provided it is not a
surface of revolution. However, we need to find only one of them. A simple
computation shows that for the surface
χ = u3 — 3uv2 + 3u ,
у = ν3 — Su2v + Sv,
ζ = 6uv,
Η = 0 and К = -\{u2 + v2 + l)2 (K is the Gauss curvature). The latter
attains its minimum only at и = ν = 0; thus this point and the Dupin
indicatrix at this point must be fixed by any automorphism of the surface.

3.4 GEOMETRY
251
There may be only a finite number of such automorphisms, however, since
the Dupin indicatrix is a hyperbola by К < 0. D
Problem G.6. The plane is divided into domains by η straight lines
in general position, where η > 3. Determine the maximum and minimum
possible number of angular domains among them. (We say that η lines are
in general position if no two are parallel and no three are concurrent.)
Solution. The minimal number of angular domains is three. Indeed,
the convex hull of the intersection points of the straight lines is a convex
polygon having at least three vertices. Each vertex is an intersection point
of two straight lines and those half-lines of these lines, which go outside the
convex hull, bound an angular domain of the considered subdivision, since
there are no intersection points on them.
On the other hand, one can always position η > 3 straight lines in
the plane so that the number of angular domains is exactly 3. Such a
construction is given by η tangents to a quadrant (see Figure G.5).
Figure G.5.
The maximal number of angular domains is
(_l)n+i _ x
that is, η if η is odd and η — 1 if η is even. Indeed, each straight line
is divided by the others into two half-lines and some segments. Angular
domains can be bounded only by these half-lines, and each half-line bounds
at most one angular domain (otherwise there would be a point that lies on
three of the straight lines). It follows that 2n half-lines can bound no more
than η angular domains.
For odd n, we can present an (essentially unique) construction with η
straight lines and η angular domains: Consider the longest diagonals of a
regular n-gon. These η straight lines are in general position. There is an

252
3. SOLUTIONS TO THE PROBLEMS
Figure G.6.
angular domain at each vertex of the polygon. Thus the number of angular
domains is η (see Figure G.6).
We prove that if η straight lines in general position bound η angular
domains, then η is odd. Removing the two half-lines considered above
from each straight line gives η segments. Any endpoint of such a segment
is shared by exactly two segments. If two of the segments have different
endpoints, then they cross one another since the intersection point of their
straight lines can lie neither outside nor at the end of the segments. Thus,
the union of the segments yields a (number of) self-intersecting closed
broken line(s). Fix an orientation on each of these broken lines, and consider
three consecutive segments a, 6, and с on one of them. Since a and с
intersect, they lie on the same side of b. Omitting the endpoints of a and b
(that is, three points), we can couple the remaining vertices, saying that
the points P, Q form a pair if Ρ is on the same side of b as the segments
a and c, and Q is the neighboring vertex that comes after Ρ according to
the fixed orientation on the broken line containing P. Hence, the number,
n, of vertices is odd.
In the case of even n, we can attain the maximal (n— 1) angular domains.
This follows directly from Figure G.6; if we remove one straight line, the
number of angular domains decreases by 2. D
Problem G.7. Let A = A1A2A3A4 be a tetrahedron, and suppose that
for each j ^ k, [Aj,Ajk] is a segment of length ρ extending from Aj in the
direction of Ak> Let pj be the intersection line of the planes [AjkAjiAjm]
and [AkAiAm]. Show that there are infinitely many straight lines that
intersect the straight lines Pi,P2,P3,P4 simultaneously
Solution. It is natural to assume that we are in the projective space
obtained from the Euclidean space by joining ideal elements.

3.4 GEOMETRY
253
Exclude first the singular cases, and assume the tetrahedron has no
edges of length p. Let {j, k, Z,ra} = {1,2,3,4}. Denote the intersection
point of the straight lines Α^Αχ and Aj^Aji by В^ш. Obviously, Bjm is
the intersection point of the plane 5m = [Aj A^ A/] opposite to Am and the
straight line pj. Consider now the plane Sj (see Figure G.7). The points
Bkj, Bij, Bmj are adjacent to a straight line eJ? as is easy to see by a
multiple application of Menelaos' theorem. This implies that ej intersects
Pi? P2? Рз? P4 simultaneously. (Indeed, the points Bkj, Bij, Bmj belong to
Pk·, Ph Pm, respectively, while pj and ej are coplanar.) The line ej is not
an edge of the tetrahedron, since it would result, the excluded case. The
lines ei, e2, ез, е± are all different since they lie in different planes of the
tetrahedron, but none of them coincides with an edge.
Figure G.7.
Therefore, the lines Pi, P2? Рз? P4 are intersected by four different straight
lines simultaneously. If two of the lines pi (i = 1,2,3,4) are intersecting,
then their point in common must be contained in the plane spanned by the
two other straight lines; hence we can easily find infinitely many straight
lines intersecting all four of them.
If Pi? P2? Рз? Р4 are mutually skew, then take pi, ^2? Рз- It is known that
the straight lines that intersect all of these three lines sweep out a doubly
ruled second-order surface. One family of straight lines on this surface is
given by the straight lines intersecting pi, ^2? Рз? while pi, ^2? Рз belong
to the other family. The line p± has more than two points in common with
the surface, hence it is also a generator of it and intersects every straight
line in the first family.
Now let us discuss the singular cases.
(a) If the three edges starting from one of the vertices Ak have length
p, then ρ^ is not properly defined, while the three other lines are
adjacent to A^. As pk is not determined, the statement cannot be
applied to this case. If, however, we define pk as an arbitrary straight

254
3. SOLUTIONS TO THE PROBLEMS
line in the two coinciding planes that should define it, the statement
is clearly true.
(b) If there are two edges of length ρ starting from a vertex Ak, then
denote by Αι the fourth vertex, not lying on these edges. It is easy
to see that in this case pk is the intersection of the planes Sk and Si,
pi С Si, while the two other lines, pj, рш, go through A^. This means
that any straight line e such that A^ £ e С Si intersects all the lines
P%-
(c) Assume now that the tetrahedron has exactly one edge of length p.
In this case, one of the straight lines e^ coincides with this edge, and
therefore we can only say that there are three different straight lines
among ei, б2, ез, е±. However, to apply the arguments used in the
generic case, it suffices to have three straight lines that intersect all of
Ръ Ί>2·> Рз? P4- Thus, pi, p2, рз, Р4 are intersected by infinitely many
straight lines simultaneously.
в*
Figure G.8.
(d) Assume, finally, that the tetrahedron has two edges of length p,
opposite to one another, say A1A3 = A2A4 = ρ = a. Introduce the
notation A\A2 = 6, A2A3 = c, A3A4 = d, A1A4 = f (see Figure
G.8). Applying Menelaos' theorem, we obtain the following divided
ratios:
(АгАзВн) = —, (АгАьВъ) = ^ ,
а- с а- а
/4 4,-.ч c — a /44^4 b — a
(A2A4B31) = -, (A2A4B13 ) = -.
a — a a — j
Turning to cross ratios,
{АгАгВпВи) = ,(* ~ °J,(<1 ~ **! = (^^4^13^31),
{a-c){f-a)

3.4 GEOMETRY
255
but then
{A1A3B24B42) = {B31B13A4A2).
Let us take into consideration that A\ and B$\ are on p$; As and B^
are on pi; B2± and A4 are on p2\ В\2 and A2 are on p4. The equality
we have just obtained means that pi, p2, p$, p± cut the skew lines
A1A3 and A2A^ in two 4-tuples having the same cross ratio. Thus
these lines belong to the same family of generators on a doubly ruled
second-order surface. This implies the statement of the problem. D
Problem G.8. Consider the radii of normal curvature of a surface at
one of its points P$ in two conjugate directions (with respect to the Dupin
indicatrix). Show that their sum does not depend on the choice of the
conjugate directions. (We exclude the choice of asymptotic directions in
the case of a hyperbolic point.)
Solution. If Pq is a parabolic point, then the indicatrix is a couple of
parallel lines, and a direction not parallel to them is conjugate only to
their (that is, the asymptotic) direction. The radius of normal curvature is
infinite in the asymptotic direction and finite in any other direction, so the
sum of the radii of normal curvature in two conjugate directions is always
infinite.
It is known that the absolute value of the radius of normal curvature in
a given direction is the square of the length of the segment from Pq to the
indicatrix point in the given direction. According to this, we need only to
show that the sum (for the elliptic case) or the difference (for the hyperbolic
case) of the square of the length of conjugate half-diameters of the Dupin
indicatrix does not depend on the choice of the conjugate diameters.
A theorem of Apollonius states that the indicatrix is an ellipse at an
elliptic point.
Now suppose that Pq is a hyperbolic point, that is, the indicatrix is a
pair of conjugate hyperbolas whose equation is of the form
in a properly chosen coordinate system. Let us denote by si and s2 the
common asymptotes of them (see Figure G.9). Let S be a point on one of
the hyperbolas and e\ be the tangent of the hyperbola at S.
This tangent intersects s\ at the point L, s2 at the point M. Draw a
tangent (different from s2) from Μ to the other hyperbola. This tangent
touches the hyperbola at the point Τ and crosses si at N. It is known
that the area of the triangle enclosed by the asymptotes and a tangent
of the hyperbolas is equal to ab. Thus, APoLM and APqMN are of
the same area, which, due to their common altitude MM*, yields PqL =
PoN. It is also known that S halves LM and Τ halves MN. Hence,
APqST is the median triangle of ALMN; consequently, PqS \\ e2 and

256
3. SOLUTIONS TO THE PROBLEMS
Figure G.9.
PqT II e\. This means, however, that PqS and PqT are conjugate half-
diameters. Applying the cosine law for the triangles APqLM and APqMN,
and using the equality PqL = PqN shown above, we get
4 (PoT2 - PoS2} =Ш2 - Ш?
=Щ2 + P0M2 - 2LP0P0M cos a-
NP0 +P0M -2NP0P0Mcos(7t-a)
= - ALPqPqM cos α =
-16— — cos a,
where LPq/2 and PqM/2 are the contravariant coordinates of the vector
P^ with respect to a basis consisting of unit vectors pointing in the
directions of the asymptotes. The product of them (as it is known) does
not depend on the choice of the point S of the hyperbola. In such a way,
ί PoT — PoS) does not depend on 5, that is, on the choice of the
conjugate pair of diameters. D
Problem G.9. Show that a segment of length h can go through or be
tangent to at most 2[h/y/2] + 2 nonoverlapping unit spheres. ([.] is integer
part.)
Solution. Let us denote by e the supporting straight line of the given
segment of length /ι, project the centers of the spheres onto this straight
line, and consider the open intervals of length \[2 on e centered at the
projections of the centers. We claim that no point of e belongs to more
than two such intervals. If not, there would exist a cylinder of radius 1 and
altitude m < \[2 containing three points А, Б, and С (the centers of the
spheres) that are at least 2 units away from one another. We shall show
that this is impossible.

3.4 GEOMETRY 257
Figure G.10.
In the following, we shall use the term "the plane of a point" for the
plane that is orthogonal to e and passes through the point. The plane of
A intersects the cylinder in a circle centered at K. Let us project В and С
onto the plane of A. Denoting the projections by B' and C", respectively
(see Figure G.10), we have
m2+M2>22,
that is,
FI2>22-m2>2,
hence, AB'KA is obtuse angled at K, since the radius of the circle equals
1. It follows that if we move A to the perimeter of the circle along the
radius К A, ABf and hence AB increase (and similarly, so do AC and
hence AC). The same is true for the other points, so we may suppose
that A, B, and С are on the surface of the cylinder, at distance 1 from e.
Since we required only m < y/2 for the height of the cylinder, we may also
suppose that one of the points, say A, lies in the upper base of the cylinder
and one of the points, say C, lies in the lower base.

258
3. SOLUTIONS TO THE PROBLEMS
Consider the plane of B, and let Β ι be the reflection of В in e, A' and С
be the projections of A and С onto the plane of B, respectively (see Figure
G.ll). Since the distances between the planes of А, В and С are smaller
than \/2, while the distances between the points are not less than 2, the
distances between the projections of these points are greater than y/2. This
implies that the endpoints A' and С lie on different arcs determined by В
and B\. If we move both A' and C" toward В along the arc, and move A
and С in a corresponding way, we can manage to have AB = С В = 2. But
in this case Α'Βχ is just the distance of the planes of A and B, and C'B\
is just the distance of the planes of С and B. Therefore
Consequently,
Thus,
А'Вг+С'Вх <т<\/2.
~Ήσ <~ArB[ + TJrB[<ypi.
~AC2 = m2 + ArCj2 <2 + 2,
and so AC < 2. This is a contradiction that proves the
proposition.
Let us elongate the segment of length h in both directions by a segment
of length \/2· Since the segment goes through or is tangent to the spheres,
the intervals assigned to the spheres are all contained in this enlarged
segment. Since every point of the enlarged segment is covered by at most two
intervals, we get that the number of spheres is at most
/i + 2—
VI
= 2
A
72.
+ 2.
To see this, we put the intervals in an increasing order by their left end-
points (this ordering is not always unique). Then the intervals standing at
odd (or even) positions are necessarily disjoint. Thus, the number of all
intervals is at most
A
V2\
+ 1 +
A
V2\
+ 1,
It is easy to see that the given estimation is exact. D
Problem G.10. Characterize those configurations of η coplanar
straight lines for which the sum of angles between all pairs of lines is
maximum.
Solution. We may suppose that all straight lines go through one point.
Starting from an arbitrary straight line, let us number the straight lines
in counterclockwise orientation (the order of coinciding straight lines is

3.4 GEOMETRY
259
arbitrary): ei,..., en. For 1 < г < [n/2], we say that e/ is the zth neighbor
of ek if
/ = к + г (mod η).
We shall denote the zth neighbor of ek by e\. Let us denote by (e,/) the
angle between the straight lines e and / and denote by < e, / > the angle
of the counterclockwise rotation that takes e to /. For г fixed,
(ei,ei) + — + (en,ejl)<<ei,ei>+- + <en,ejl>
= z(<ei,e2> + <e2,e3>H \-<en,ei >)=ίπ,
and equality holds if and only if
<efc,ei><^ (1)
for all k. Thus, for odd n, the sum of angles between all pairs of straight
lines can be estimated as follows:
n-l n-1
Σ[(β^ί)+···+(«»>«»)]^Σίπ=!^Ι' (2)
i=l i=l
and equality holds if and only if (1) holds for all к = 1,2,... ,n and г =
1,2,..., (η — 1)/2. For this, it is enough to require that
<ек,еГ1)/2><^ (3)
hold for к = 1,2,... ,n.
For even n, the sum of angles is
Σ [(ei, ei) + · · · + (en, ejj] + ^ [(eb e?/2) + · · · + (e„, e^2)]
H_1
2 9
г=1
and equality holds if and only if
n/2 ^ . К
<ek,ek/ ><-
holds for к = 1,2,..., п. For even n, these inequalities imply the requested
characterization, since the inequalities
n/2 ^ . 7Г , n/2 ^ . 7Г
< efc, efc' >< - and < efc+n/2, efc;n/2 >< -
can be satisfied simultaneously only if
n/2 ^ 7Г
2
<ek,ek' >=-,

260
3. SOLUTIONS TO THE PROBLEMS
that is the (n/2)th neighbors are perpendicular to one another. To
summarize, if η is even, the maximum is attained if the family of straight lines
is composed of n/2 orthogonal pairs and any family of orthogonal pairs of
straight lines yields maximal sum of angles.
Returning to the case of odd n, let us observe that orthogonal couples
of straight lines can be removed from the family, since the omission of an
orthogonal pair there decreases the sum of angles by (n — 1)^ and
n2 (n - 2)2
4 4
Suppose that after the removal of orthogonal pairs there remain 2m + 1
straight lines. These straight lines must be different, because if e were a
double line then the mth neighbor of the mth neighbor / of e would be
e, which would mean that e and / are perpendicular. If e is an arbitrary
member of the 2m + 1 straight lines, then conditions (3) imply that each
quadrant of the plane determined by e and the straight line perpendicular
to e must meet exactly m straight lines. Adding to such a family of an odd
number of straight lines an arbitrary number of orthogonal pairs, we get
all the configurations that yield the maximum, and the maximum is equal
to [(n2 - l)/4] · (π/2) by (2).
Finally, if we do not suppose that the straight lines go through one point,
then taking a configuration characterized above and replacing each straight
line by a parallel one, we can get all the requested configurations. D
Problem G.ll. Let f(n) denote the maximum possible number of
right triangles determined by η coplanar points. Show that
limM = 00 and шЩ- = о.
η—>oo ΤΙό
Solution.
Lower estimation. We shall show the following sharper result. There
exists a constant с > 0 such that
f(n) > en2 log η (η> 3).
Suppose first that η = (ЗА; + l)2, where /с is a natural number. Consider
those points of the lattice of points with integer coordinates, which lie in
the square spanned by the vertices (0,0), (0,3/c), (ЗА;, ЗА;), (3/c, 0). The
number of these points is obviously (ЗА; + l)2. We shall count only those
right triangles for which the right-angled vertex (q, r) satisfies к < q < 2A;,
к <r <2k and the two other vertices of which lie in the square of sides 2A;
centered at (<?, r). Obviously, all these right triangles are good for us.
We have to determine the number of right triangles spanned by the
lattice points of the square (—A;,—A;), (—A;, A;), (A;, A;), (A;,—A;) so that the
right-angled vertex lies in the origin (0,0).

3.4 GEOMETRY
261
The perpendicular sides of such a right-angled triangle lie on straight
lines given by equations of the form ix = jy and jx = —iy, where (г, j) = 1,
—k < i,j < к. Obviously, we estimate the number of right triangles from
below if we take into consideration only those straight lines that satisfy
0 < г < j as well. In this case, each of the straight lines ix = jy and
jx = —iy contains 2 [к/j] lattice points different from (0,0), so the number
of right-angled triangles determined by them is
4
For a fixed j, we find at least
fel2 £
к2
right triangles (φ denotes Euler's phi function), which gives
i=i J
right-angled triangles altogether. Now returning to the original problem,
the (3k + l)2 points we constructed yield at least
(2fc + l)a*2 Σ ^
i=i J
right triangles.
Introducing the function
j<x
and summing up partially, we get
/((3fc + l)2)>(2A; + l)2fc2^^^
Combining this with the well-known inequality
Ф(х) > cix2, (1)
we get
k-l 1
/((ЗА: + l)2) > c2(2k + l)2k2 У> c3k4 log к.
Η3

262
3. SOLUTIONS TO THE PROBLEMS
Now let η be an arbitrary number. Set к = [{y/n — l)/3] > y/n/A. Then
f(n) > /((ЗА: + l)2) > c3k4 log к > en2 log η.
Upper estimation. We shall prove a sharper result again: using the "de-
scente infinie" method we show that
/(n)<n2v^. (2)
The statement is trivial for η = 1,2,3,4,5 (for these values n2y/n > (3)).
If (2) were not true, then there would be a smallest natural number η such
that
f(n) > n2y/n.
Let us take η points Pi,..., Pn on the plane that span f(n) right-angled
triangles. We claim that there must be a straight line that contains from the
η points at least 2\Jn ones. Consider all possible ordered couples (P^, P7),
and assign to each of them the number of right triangles PiPjP^ such that
the right angle is at Pi. This number is the number of points different
from Pi on the straight line perpendicular to PiPj at Pi. The sum of these
numbers is twice the number of right triangles and so it is at least 2n2y/n.
Since the number of summands is n(n — 1)(< n2), one of the summands is
at least 2y/n, and thus one of the perpendicular lines contains at least 2y/n
points.
Let us remove the points of this straight line and examine how many
right triangles are destroyed by this at most. We divide the triangles
destroyed into two groups:
(A) triangles for which the right-angled vertex is omitted;
(B) triangles with right-angled vertex lying off and one of the other
vertices lying on the critical straight line.
We can give an upper bound for the number of triangles of type A in
such a way that we have n(n — l)/2 choices for the hypotenuse of such a
triangle, and a segment PiPj may serve as hypotenuse for at most two right
triangles of type A (Thales' circle!), so the number of triangles of type A
is at most n(n — 1) < n2.
In the case of triangles of type B, we have n(n — l)/2 choices for those
vertices that are not required to lie on the critical straight line, and we
can choose at most two points on the critical straight line which form a
triangle of type В together with a fixed couple of points P%Pj so that the
right-angled vertex is either at Pi or at Pj. It follows that the number of
triangles of type В is also at most n(n — 1) < n2.
In such a way, the removal of the points of the critical straight line
destroys at most 2n2 right-angled triangles; consequently, the remaining,
at most, η — 2 points span at least n2y/n — 2n2 right triangles. By the
induction hypothesis,
n2y/n-2n2 < (n-2y/n)2Jn-2y/n< (n - 2y/n)2 y/n,

3.4 GEOMETRY
263
from which
2n2 < Апу/п,
that is,
n<4,
for which cases we have already seen that the assertion is true. D
Remarks.
1. We could have avoided the use of (1) using only the elementary fact that
ψ{ρ) = Ρ — 1 f°r prime numbers. However, this would have given us only
the estimation
f(n) > c4n2 log log η,
where c± is a suitable positive constant.
2. Laszlo Lovasz remarks that there is a positive constant C5 such that the
number of right triangles in the lattice construction described in the first
part of the solution is at most
c5n2 log η.
3. Bela Bollobas proves the statement of the problem in the following more
general form: Let us denote by /(n, k) the maximal number of right-
angled triangles spanned by η points of the k{> 2)-dimensional space,
where the maximum is taken for all possible configurations of the points.
Then
en2 log η < /(η, к) < dkn3~2 ,
where с and d^ are positive constants. The lower bound comes from
our previous arguments because of f(n,k) > /(n, 2). We prove the
upper bound for /с = 3. Denote by R a collection of η points in the
three-dimensional space, and denote by ρ the number of those straight
lines that contain more than Зу/ΰ points of R. Let ei, e2,..., ep stand
for these straight lines. e\ contains at least Зу/ΰ points of R, e\ and
e2 together contain at least Зу^Зу^— 1) points, because two different
straight lines may have at most one common point, and so on, ei,..., ep>
cover at least
p'-i
^(Зу^-1) >pf3yfii-p'2
2 = 1
points of R for every pf < p. Consequently, ρ < y/n. We prove that if
we remove at most y/rt "big" straight lines, the number of right triangles
decreases at most by 4n5/2. Let e be a "big" straight line. Those right
triangles that have at least one vertex on e can be divided into two
classes:
i. Triangles that have two vertices on e. The third vertex of such a
triangle can be chosen out of at most η points. This point, together
with a point on e, leaves only two possibilities, and so the number of
right triangles of this class is at most 2n2.

264
3. SOLUTIONS TO THE PROBLEMS
ii. Triangles that have only one vertex on e. We have n(n — l)/2 < n2/2
choices for the two other vertices, which form a right-angled triangle
together with at most 4 points on e, thus the number of right triangles
of this type is less than 4n2/2 = 2n2. Since we have at most y/n
"big" straight lines, their removal causes the vanishing of not more
than y/nAn2 = 4n5/2 right triangles. After the removal, each straight
line contains at most Зу/n points. Let us denote by q the number
of planes passing through the point Ρ and least 6n3/4 points of the
remaining family. By the idea used above, we get
?'6n3/4 - qf23y/n < η
for all q' < q, since two planes have at most Sy/n common points.
Prom here, q < n1/4. There are less than n1/43y^ triangles satisfying
that its right-angled vertex is at Ρ, one of its perpendicular sides
lies in one of these "big" planes, while the other is perpendicular to
the plane. The number of right triangles with right-angled vertex
at P, not satisfying these conditions, can be estimated by the sum
Σΐ=ι /*s*j where I denotes the number of orthogonal pairs of a plane
and a straight line passing through Ρ such that both the plane and
the straight line contains at least one point different from Ρ and /;,
Si denote the number of points different from Ρ lying on the plane
and the straight line of the ith pair, respectively.
But then
EM<6n3/4EHfe7/4·
Thus, the number of right triangles in R is less than
4n5/2 + 6n7/4n<10nn/4.
We remark that the case к > 3 requires the consideration of not only
straight lines and I (I < k)-dimensional linear subspaces but also that
of circles and spheres.
Problem G.12. Suppose that a bounded subset S of the plane is a
union of congruent, homothetic, closed triangles. Show that the boundary
of S can be covered by a finite number of rectinable arcs.
Solution 1. Let us denote the boundary of S by Η and let Ρ e Η be an
arbitrary point. Then one can find a sequence of points {Pn} converging to
Ρ such that each point Pn lies on the boundary of a triangle Δη constituting
S.
We may suppose that each point Pn lies on the boundary of Δη at the
same position (that is, the translation that takes Anto Лт takes Pn to
Рщ). (This can be shown by our assumptions. Translate the triangles Δη
to a fixed triangle Δ; denote by Qn the point corresponding to Pn. The
sequence {Qn} is bounded, and it thus contains a convergent subsequence

3.4 GEOMETRY
265
{Qnk}, which tends to Q. Translating Δ back to Anfc, Q corresponds to
P'nk · Obviously, P'Uk —> Ρ and the points Р'Пк lie at the same position on
the boundary of Anfc). Obviously, each triangle Δη contains a homothetic
triangle Δ^ of half size such that Pn is a vertex of Δ^ lying at the same
position for all n. These triangles converge to a homothetic triangle Δ ρ
with vertex Ρ such that the interior of Δ ρ contains no point of H.
Let us number the three vertex positions of the triangles homothetic
to Δη. This gives rise to a natural ordering of the vertices of triangles
homothetic to Δη. Let us define the sets Hi С Η (г = 1,2,3) in such a
way that a point Ρ G Η belongs to Hi if and only if Ρ is the ith vertex of
Δ'ρ. Obviously, Η = Hi U H2 U Я3.
Let us study H\. Denote by α the angle lying at the first vertex of Δρ.
Let us introduce a coordinate system on the plane so that the y-axis shows
in the direction of the bisectrix of the angle a, the direction of the rr-axis
coincides with the direction of the supplementary angle of a. Let us divide
the plane into bands of equal width parallel to the rr-axis. If the width of
the bands is chosen to be small enough (for example, if it is smaller than
the bisectrix of Δ ρ that passes through P), then the intersection of H\
and one of the closed bands is the graph of a function у = /(#), where / is
defined on a bounded subset of the rr-axis and satisfies Lipschitz condition
with constant к = cot (α/2), which gives the possibility to extend it to
the closure of its original domain. The extension of / is also Ar-Lipschitz.
After this, by extending / to the components of the open complement of its
domain in a linear way, we can obtain a Ar-Lipschitz extension of /, defined
on a sufficiently large bounded interval. The graph of this extension is a
rectifiable curve that covers the intersection of Hi and the considered band.
Since S is bounded, it crosses only a finite number of bands; thus, H\ can
be covered by a finite number of rectifiable arcs. We can proceed similarly
for Hi and Щ, so the theorem is proved. D
Solution 2. Let R2 be the two-dimensional Euclidean vector space, and
let Я С R2 be a nonsingular closed triangle whose angles are the regions
Αι, A2, A3, that is,
Н = А1ПА2ПА3. (1)
The statement of the problem can be reformulated as follows.
Statement. If Μ (с R2) is bounded, than the boundary of Μ + Η can
be covered by a finite number of rectifiable arcs. (Prom now on, we use
operations on complexes.)
Since Μ is bounded, there exists a finite set V С R2 such that Μ С
V + \H. Setting Μ (υ) = (υ + \Η) Π Я, we get
M= U Μ (υ); (2)
vEV
hence,
M + H= U (Μ(υ) + Η). (3)
vev

266
3. SOLUTIONS TO THE PROBLEMS
First we show that
M(v) + Η = η (Μ(υ) + Ai) (4)
г=1
for all υ G V. It is clear from (1) that
M(v) + Η С η (Μ(ϋ) + А;).
г=1
Suppose now that
ρ G Π (Μ(г;) + АО .
г=1
Then p G Μ (г;) + Α{ (г = 1,2,3), but
(ρ- ΜΗ) Π (АД Я) С (ρ-ν - Iff) П (АЛЯ) = 0
must be satisfied at least for one i. Thus
(р-М(и))ПЯ^;
and this means that
pe M(v) +Я.
Hence,
M(v) + Η D η (Μ(ϋ) + АО
г=1
holds as well, and this proves (3).
Let us denote by fr (X) the boundary of a set X. Using that V is finite,
(3) and (4) yield
fr (M + Я) С U U fr (Μ(υ) + А{).
vev i=i
Now it is enough to show the following proposition.
Proposition. Let A be the angular domain of angle 2a (0 < 2a < π),
the vertex of which is the origin and the bisectrix of which is the negative
y-axis. Let Μ С R2 be a bounded set; furthermore,
T = {(x,y) GR2: 0<x< 1}.
Then
G = fr (М + А)ПТ
is a rectifiable curve.
Proof. To show this, let (£, #(£)) denote for 0 < £ < 1 the point of G the
whose abscissa is £, that is, set
#(£) = sup{y — \x — t\ cot α: (ж, у) G Μ} .
Since g satisfies the Lipschitz condition
\g{t)-g{t')\<\t-t'\cota,
the latter proposition is trivial. D

3.4 GEOMETRY
267
Problem G.13. Let F be a surface of nonzero curvature that can be
represented around one of its points Ρ by a power series and is symmetric
around the normal planes parallel to the principal directions at P. Show
that the derivative with respect to the arc length of the curvature of an
arbitrary normal section at Ρ vanishes at P. Is it possible to replace the
above symmetry condition by a weaker one ?
Solution 1. Since the normal planes parallel to the principal directions
are perpendicular, the symmetry of F around them implies the symmetry
around their intersection, that is, around the normal of F at P. Since
the plane of normal sections contains the normal line, normal sections at
Ρ are also symmetric around the normal. The analyticity of the surface
ensures that the normal sections have arc length and curvature at each
point, which is a differentiable function of the arc length. Let g(s) be the
curvature expressed as a function of the arc length. If Ρ corresponds to the
parameter so, then the symmetry of the normal section around the normal
at Ρ gives the identity g(so + s) = g(so — s). Differentiating with respect
to s at s = 0, we get g'(so) = —g'(so), that is, g'(so) = 0, which proves the
proposition.
As we see, symmetry around the normal planes parallel to the principal
directions can be substituted by the symmetry around the normal of F
at P. Π
Solution 2. By our assumption, introducing a coordinate system with
origin at P, the surface is given by an equation of the form
oo
Ζ = Σ агкХ1ук ,
i,k=0
where αοο = 0. We may also suppose that the x- and y-axes show in the
principal directions of F. In this case, aw = αοι = 0. With such a choice
of coordinate system, according to our assumptions, z(x,y) = z(—x,y) =
z(—x,—y) = z(x,—y), which can be the case only if a^ = 0 whenever
either г or /с is odd. Then the equation of the surface has the form
oo
ζ = J2 amkX2iy2k ·
i,k=0
A normal section at Ρ can be parameterized as follows:
r = r(£) : χ = cit, у = c2t,
oo oo
■ z= Σ a2i<2k(cit)2i(c2t)2k ==J2bnt2n,
i,k=0 n=l
where the constants c\, c2 {c\ +c\ φϋ) depend on the plane of the normal
section, and the coefficients bn are functions of a2i^k·

268
3. SOLUTIONS TO THE PROBLEMS
The curvature of the curve r = r(t) can be computed by the formula
^_ у/|г'|2|г"|2-(г'-г")2
9"' ~ |r'|3/2
By our assumption, #(0) φ 0, and instead of needing to prove the equation
dg/ds = 0, it is enough to show the relation gf(0) = 0, since
dg dg dt dg 1
ds dt ds dt |r'|
Taking into consideration that
r' = ic1,c2,^2n6nt2n-1),
Γ" = ί 0,0, ]T 2n(2n - 1) bnt
\ n=l
r'" = | o, 0, JT 2n(2n - l)(2n - 2) bni
±2n-2
the relation </(0) = 0 is obtained by direct computation.
The above solution shows that in order to weaken the symmetry
condition, it is enough to require that the equation ζ = Y^k=oaikXlyk of
the surface in the coordinate system attached to the normal and principal
directions does not contain terms of order three. D
Problem G.14. Let σ(5η, к) denote the sum of the kth powers of the
lengths of the sides of the convex n-gon Sn inscribed in a unit circle. Show
that for any natural number greater than 2 there exists a real number ко
between 1 and 2 such that σ(5η, ко) attains its maximum for the regular
n-gon.
Solution. We shall show the following sharper result.
If 1 < к < (ίΜΐ(π/η))/(π/η), then a(Sn,k) < σ(5*,Α:) for any convex
n-gon inscribed in a unit circle, where 5* is the regular n-gon inscribed in
a unit circle.
Proof. To show this statement, it is enough to consider only those
polygons Sn that contain the center of the circle on the boundary or inside.
Indeed, let Sn be a polygon not satisfying this condition, and let A^ ..., An
be the vertices of Sn. Suppose that A\A2 is the closest side of Sn to the
center of the circle. Denote by A'2 the antipodal pair of Αι, and consider
the polygon S'n with vertices A^ A'2, ^3, · · ·, An. Then ΑχΑ'2 > ΑλΑ2,
and А'2Аз > А2А$ (since the angle opposite to the side A$A2 in ΔΑ2Α2Α3
is obtuse). For к > 1, obviously σ(5^, к) > σ(5η, к).

3.4 GEOMETRY
269
Now let Sn be a convex n-gon inscribed in a unit circle, such that Sn
contains the center of the circle on the boundary or inside. Let xi,...,xn
denote half of the central angles corresponding to the sides of Sn. Then
η η
σ(5η, к) = 2к y^smkXj, where 0 < χι < —, У^#г = ?τ·
г=1 г=1
Our proposition is well known for к = 1, so it suffices to deal with the case
к > 1. Let us fix the exponent 1 < к < (ί&η(π/η)/(π/η). Since the function
(tan x)/x is continuous and strictly increasing in the interval (0, π/2,
furthermore, Y\vnx_+Q(tanx)/χ = 1, there exists a unique real number a = a(k)
such that 0 < a < π/η and к = (tan a)/a.
Consider the function f(x) = sin^ x. Since
f"(x) = k(k - 1) sin*-2 χ cos2 χ - к sin* χ = к sin*-2 x(k cos2 χ - 1),
one can easily check that f(x) is convex on the interval
[ 0, arc cos (1/y/k) J and concave on the interval ί arc cos (1/y/k), π/2 J.
Therefore, it is concave on the interval (α, π/2) because
,/7/ ч tana . k_2 /tana 2 Л л
/"(a) = sin* 2 a ( cos2 a - 1 I < 0.
Set
"i cos ax, if 0 < χ < a,
. . f fcshr xa(
t sin x,
if a < χ < π/2.
Obviously, ^ is continuous in the interval [0, π/2]. One can see easily by
a study of the derivative that the function (sin^ x) /x is strictly increasing
on the interval [0,a] and strictly decreasing on the interval [α,π/2]. One
can derive from this that g is convex on [0, π/2] and also that f(x) < g(x)
holds for χ G [0,π/2]. These facts, together with the inequality
η
^2g(xi)<ng(^ ,
i=l
obtained by an application of Jensen's inequality, yield
η
a{Sn,k) = 2kYjf(xi)<n2kg(^.
2 = 1
Since α < π/η, we have /(π/η) = ^(π/η), and thus
π
σ(5η, к) =< n2* sin* - = σ(5;, к).
η
The proposition is proved. D

270
3. SOLUTIONS TO THE PROBLEMS
Remark. Many contestants observed that if 3 < η' < η and 1 < к <
(tan(n/n))/(n/n), then we also have the inequality a(Sn',k) < a(S*,fc).
Indeed, considering the derivative of the function у = я sin*(π/я), we see
that
, . h π kn . b_i π π . к_л π { . π кж π\
у = sin* sin* - cos — = sin* - sin cos — I > 0
xx xx χ \ χ χ χ J
provided that χ > 2 and к < (tan(n / χ)) / (π / χ) . Using the relations
π/η' > π/η and
tan -77 tan;:
rv_ ^ η
JL TL '
η' η
we obtain
σ(5^, fc) = 2*nsin* - > 2knf sin* ^ = a(S^ k) > σ(5η, fc).
Problem G.15. Let h be a triangle of perimeter 1, and let Η be a
triangle of perimeter λ homothetic to h. Let /ΐχ,/ΐ2,... be translates of
h such that, for all i, hi is different from hi+2 and touches Η and /ц+ι
(that is, intersects without overlapping). For which values of λ can these
triangles be chosen so that the sequence Λχ, /12,... is periodic? If λ > 1 is
such a value, then determine the number of different triangles in a periodic
chain Λχ, /12,... and also the number of times such a chain goes around the
triangle H.
Solution. We may restrict ourselves to the case of regular triangles since
any triangle can be transformed into a regular one by an affine
transformation and affinities preserve homothetic and touching position of triangles
and also the ratio of perimeters of homothetic triangles.
Let А, В, С be the vertices of the triangle Я; let Pi,Qi, Ri be the vertices
of the triangle /i^; and let a, 6, с and 'РгЛи T% be the sides of the triangles Η
and hi opposite to the vertices А, В, С and Pi,Qi,Ri, respectively. These
sides are considered to be half-open segments excluding the endpoint С
from side a, the endpoint Ri from side pi, etc. It is assumed that the
vertices A, B, and С correspond homothetically to the vertices Pi, Qi, and
Ri, respectively (see Figure G.12).
There are two ways a triangle hi can touch the triangle H: either a
vertex of Η lies on the side of hi opposite to the homothetically corresponding
vertex of hi, or a vertex of hi lies on the side of Η opposite to the
homothetically corresponding vertex. According to our assumptions, we always
have exactly one of the two cases. In the first case, we say that hi touches
a vertex of Я; in the second, we say that hi touches a side of H. The
common point of the two triangles will be called the point of contact. We
fix the orientations (Pi, Qi, Ri) and (A, B, C) of the triangles hi and H.

3.4 GEOMETRY
271
С
\ h, /
Ρ
ι
A
Figure G.12.
If hi touches a side of H, then denote by f(hi) the distance between the
point of contact and the vertex of Η that comes next to the contact point
according to the fixed orientation.
If hi touches a vertex of H, then denote by f(hi) the distance between
the point of contact and the vertex of hi that comes next to the contact
point according to the orientation opposite to the fixed one. Obviously,
0 < f(hi) < λ/З in the first case, 0 < f(hi) < 1/3 in the second (see Figure
G.13).
Figure G.13.
It is clear that the contact points of hi and /ι^+ι are contained in one
(closed) side of H. Thus, the fixed orientation of Η gives rise to an ordering
of the triangles hi and /4+1-
We may establish the following facts.
Let h' denote an arbitrary translate of h touching H. We want to describe
the translates h" of h that touch both h! and Η and come after h' (according
to the fixed orientation). (See Figures G.14.a-e).
(a) If h' touches a side of Η and f(hr) > 1/3, then h" is unique, it touches
the same side of H, and the distance of contact points is 1/3 (thus
f{h") = f(h') - 1/3)) (Figure G.14.a).
(b) If h! touches a side of Η and f(hf) < 1/3, then h" is unique, it touches
the vertex of Η next to the contact point of h! with respect to the

3. SOLUTIONS TO THE PROBLEMS
Figure G.14.a
Figure G.14.b
fixed orientation, and f{h") = 1/3 - f{h') (Figure G.14.b).
(c) If Ы touches a vertex of Η and f{hr) < λ/З, then h" is unique,
it touches the side of Η that contains the contact point of hf, and
f(h") = λ/З - f(ti) (Figure G.14.c).
(d) If h! touches a vertex of Η and f(hf) = λ/З, then we can choose for
h" any triangle that touches the vertex of Η next to the contact point
of ti satisfying f{h") < (1 - λ)/3 (Figure G.14.d).
(e) Finally, if h! touches a vertex of Η and f{hf) > λ/З, then h" is unique
again, it touches the vertex of Η next to the contact point of Λ/, and
it satisfies f{h") = (1 - λ)/3 (Figure G.14.e).
I. Suppose first that λ < 1. We show that in this case one can always
find a periodic chain of triangles.
Let h\ touch the vertex A of Я so that f(hf) = 1 — λ. Set к =
— (—λ/(1 — λ)). Starting from hi and advancing according to the fixed
orientation, one can construct the sequence /i2,..., /i2fc+i uniquely by the

3.4 GEOMETRY
273
f(h')<£
f(h")=$-f(h')
Figure G.14.C
oa(h'k-b^
h"
Figure G.U.d
previous observations (a)-(e), so that the even triangles /i2,..., h^k touch
a side of Я, Αχ,..., Аг^+х touch a vertex of Я, and f(h2k+i) > λ. Then
we can choose /12^+2 according to (d) so that /(Агл+г) = 1 — λ is fulfilled.
The triangle /i2fc+2 can be obtained from Αχ by a rotation of angle 2πί/3
about the center of Я. Therefore, continuing the construction from /i2fc+2?
choosing f(ti4k+3) = 1 — λ, we can close the chain by setting /i6fc+4 = Αχ,
which finishes the proof.

274
3. SOLUTIONS TO THE PROBLEMS
Figure G.14.e
II. Now let us examine the case λ > 1. Then only the cases (a), (b),
and (c) are possible, and each h" is uniquely determined by h!. The same
would be true if we changed the orientation; hence, there are exactly two
translates of h that touch Η and a given Ы. Consequently, given h\ and h2
that touches h\ and Я, the sequence /ΐχ, h2,... can be continued uniquely.
Suppose, we are given a periodic chain /ΐχ, /ι2,... with period η(/ιη+χ =
h\) winding around Η к times. The phrase "the chain winds around Η к
times" makes sense since the contact points of the triangles hi follow one
another according to a fixed orientation of the boundary of H. We may
suppose that this orientation coincides with the orientation (А, В, С).
Observations (a), (b), and (c) make clear that the contact points of the
triangles hi that touch a side of Η form a cyclic sequence of η — 3k elements
running around Η к times such that the peripheral distance between two
consecutive contact points is 1/3. Thus,
-(n-3fc) = Xk,
о
that is,
λ=^Τ' n = 3fc(A + 1)· ί1)
One obtains from this equation that in the case λ > 1, the existence of a
periodic chain implies the rationality of λ. In this case, the number of times

3.4 GEOMETRY
275
this chain goes around the triangle Η is the denominator in the reduced
simple fraction form of 3λ, and the number of triangles contained in the
chain is obtained as the product of this winding number and 3(λ + 1) (of
course, there is a periodic chain for any integer multiple of η and к with
period η and winding number k).
We still have to show that if (1) holds for some natural numbers к and
n, then there exists a periodic chain with period n. Then it will also prove
that the rationality of λ is sufficient for the existence of a periodic chain.
In addition to this proposition, we show that no matter where we put
the starting triangle /ΐχ, the sequence of /i^s will get closed at the nth step
after к rounds around H. We may restrict ourselves to the case when /ΐχ
touches a side of H, since either hi or /12 touches a side of H, and if the
chain starting from /i2 gets closed after the nth step, then so does the chain
starting from /ΐχ because of the unique reconstructability of the sequence
in both directions.
Let us construct the beginning part /ΐχ,..., /im+x of the sequence until
it goes around Η к times and the distance between the contact points of
/im+x and h\ is less than 1/3. This situation is surely arrived at sooner or
later, because at most λ consecutive elements of the chain can touch the
same side of H. Then, as above, we have
0<
m — Sk
k\
4 <2>
Expressing λ with the help of (1) and multiplying by 3, we get
0 < \n-m\ < 1.
The integers η and m can satisfy the latter inequality and the equivalent
relation (2) only if equality holds on the left, that is, η = m and /in+i —
hi. Π
Problem G.16. The traffic rules in a regular triangle allow one to
move only along segments parallel to one of the altitudes of the triangle.
We define the distance between two points of the triangle to be the length of
the shortest such path between them. Put ^J1) points into the triangle in
such a way that the minimum distance between pairs of points is maximal.
Solution. In the course of the solution, "distance" will refer to the ordinary
distance of points while the italic "distance" will mean distance in the new
metric. Let ABC be the given triangle, AB = 1. In order not to lose
the idea of the solution among the precise verification of simple geometric
facts, we shall omit technical details.
(a) Construction for lower bound. Let us divide each side of the triangle
into η — 1 equal parts, connect each node of ВС to a corresponding node
of С A by a segment parallel to AB, and divide these segments into η —
2, η — 3,..., 2,1 equal parts, respectively, in the order of their distances

276
3. SOLUTIONS TO THE PROBLEMS
from the side AB. We obtain the same set of points if we take through
all nodes on the sides AB, ВС, and С A straight lines parallel to the two
other sides and consider all possible intersections of these straight lines.
One may expect that this system of (nJX) points is optimal. Let us denote
by rn the minimum of distances for this system.
(b) Theorem. Let Q) < Ν < (η^). If we are given N points in AABC,
then one can always find two among them with distance <rn. If N = (nJ )
and the distance of any two points is at least rn, then the system of points
coincides with the one constructed in (a).
(c) For the proof, we need the proposition that a ball of radius r with
center R in the metric space described in the problem is a regular hexagon
centered at R, the vertices of which are obtained by moving R off to distance
r parallel with one of the altitudes. The proof of this proposition is left to
the reader.
(d) We show here that if the distance of any two points of a certain subset
of the AABC is greater than rn = 2c = \/3/n — 1, then the subset contains
at most (2) points. Consider triangles of the triangular lattice constructed
in (a). Triangles homothetic to AABC with ratio l/(n — 1) will be called
recumbent, and those homothetic to ABC with ratio — l/(n — 1) will be
called standing. There are (™) recumbent triangles. Let us place a ball of
radius с that is a regular hexagon of side с around the centers of recumbent
triangles. They cover the recumbent triangles and the standing ones as well,
since each standing triangle is surrounded by three recumbent ones and
the hexagons put around the centers of the neighboring triangles cover the
standing triangle obviously. Thus, the given Q) balls cover AABC, that
is, every point of the considered subset belongs to one of these hexagons.
Because the diameter of such a hexagon is 2c = rn, a system of points for
which the distance between any two points is greater than rn meets each
hexagon in at most one point; hence, it contains at most (£) points. This
proves the first part of the theorem.
(e) We can show the uniqueness for N = C^J1) points in the following
way. If q < 1 and we shrink AABC from A with ratio q, then we get a
triangle AB'C that contains at most (2) points. If q —> 1, we obtain that
side ВС contains at least η points. Of course, side ВС cannot contain
more than η points, and the points have to divide side ВС into η — 1
equal parts. Let AB+C+ be the dilatation of ABC with respect to A with
ratio (1 — l/(n — 1)). The hexagons of side с around the points lying on
ВС cover the difference of the triangles ABC and AB+C+. Consequently,
AAB+C+ contains (2) points, so that the distance of any two points is at
least rn = rn-i (1 — l/(n — 1)) = rn_iAB+. Since the theorem is trivial
for η = 2, we may complete the proof by induction. D
Problem G.17. Let С be a simple arc with monotone curvature such
that С is congruent to its evolute. Show that under appropriate
differentiability conditions, С is a part of a cycloid or a logarithmic spiral with
polar equation r = ae0'.

3.4 GEOMETRY
277
Solution. Let rx = r(s), s G [0,£] be the parameterization of С by arc
length for which the radius of curvature p(s) is a monotone increasing
function. The evolute Ε is parameterized then by Г2 = r(s) + p(s)n(s),
where η is the principal normal vector. The length of the arc of the evolute
bounded by r2(0) and r2(s) is a(s) = p(s) — po, where po = p(0). Let Φ be
the congruence that superimposes С on E. There are two possibilities:
1. Φ takes r(0) to r2(0);
2. Φ takes r(0) to r2(£).
The radius of curvature of the evolute at the point r2(s) is p(p(s) — po) in
the first case and p(£ — p(s)-\-po) in the second. Since, by Frenet equations,
dv2 _ d2v2 _ 1
d^~n' ~d^~~p(s)p'(s) '
we have p(p(s) — po) = p(s)p'(s) in the first case and p(£ — p(s) + po) =
p(s)pf(s) in the second.
We show that s = p(s) — po for all s G [0,ί\ in the first case. Otherwise,
we could find an interval [a, b] by the suitable differentiability conditions
such that a = p(a) — p0? b = p(b) — p0? and s > p(s) — po or s < p(s) — po
holds everywhere inside [a, b]. However, in this case, we would have
b _ a = p(b) - p(a) = t p'(s)ds = t p{p{s}~Po)ds фЬ-а
Ja Ja PiS)
since p(s) — po > s implies p(p(s) — po) > p(s) and p(s) — p0 < s implies
p(p(s) — po) < p{s). We conclude that the solutions belonging to the first
case have a natural equation of the form s = p(s) — po- These curves are
logarithmic spirals the polar, equation of which has the form r = αβΰ', and
one easily checks that any arc of such a spiral is a solution.
In the second case, £ — s = p(£ — p(s) + po) holds for all s G [0,i]. To
see this, we have to show first that Φ is an orientation-reversing
transformation of the plane. We may suppose without loss of generality that
while r(s) is running along the curve C, the unit tangent vector t(s) is
rotating counterclockwise. Meanwhile r2(s) is running along the evolute
E, and its unit tangent vector n(s) is also rotating counterclockwise. Since
Φ takes r(0) to r2(£), the point Φ(Γ(δ)) is running along Ε in the opposite
direction and its unit tangent vector is rotating clockwise. Therefore, Φ
reverses orientation. Orientation-reversing isometries of the plane are glide
reflections, so the evolute F of the evolute of С is a translate of C. Let
Гз(в) denote the center of curvature of Ε at r2(s). The length of the arc
of F lying between τ%(£) and ^(s) is equal to p(£ — p(s) + po) — po- Mono-
tonicity of the curvature implies that the direction of the tangent vector
uniquely characterizes the point at which the vector is tangent to the curve.
As a consequence, we get that the translation above takes r(s) to r3(s).
For this reason, p(£ — p(s) + po) = £ — s holds, as we wanted to show.
From here, we obtain p(s)p'(s) = £ — s + po by our previous considerations.
Thus, in the second case, the solutions are given by arcs of the curves

278 3. SOLUTIONS TO THE PROBLEMS
with natural equation p2(s) + (£ — s + p0)2 = c. These curves are cycloids
given by χ = a(u — sin-u), у = α(1 — cos ία), where a = \\/ρ2 + (ί + Po)2·
We may conclude that, in the second case, the solutions are those sub-
arcs P\p2 of the arc corresponding to the parameter domain и G [0, π] for
which a = у ρ2(Ρ\) + [P1P2 + p(Pi)]2 (for example, the whole arc fits the
requirement). D
Remark. We may pose the more general problem of finding curves that
are similar to their evolute. The solution of this problem involves epi- and
hypocycloids and logarithmic spirals of the general polar equation r = aecl9
as well.
Problem G.18. Given four points Αι,Α2, Α3, Α4 in the plane in such
a way that A4 is the centroid of the Α Αχ Α2 As, find a point Аъ in the plane
that maximizes the ratio
nrnii<z<j<fc<5 T(AjAjAk)
maxi<i<j<fc<5 T(AiAj Ak)
(T(ABC) denotes the area of the triangle AABC.)
Solution. The medians, which go through the centroid of the triangle,
divide the plane into six angular regions. We claim that there is exactly
one point in each of these regions for which the ratio in question attains its
maximal value, namely 1/4. Let F be the midpoint of A2A3, let Ε denote
the reflection of A in F, let R denote the trisecting point of the segment
A3E that is closer to A3, and finally, let A$ denote the point on the half-
line A4R for which A4A$ = 3/2 · A4R. A$ is the point in the angular region
A3A4E, that yields the maximum (see Figure G.15).
Figure G.15.
Suppose for a while that A$ is an arbitrary point in the angle A3A4E.
If this point is not in AA3A4E, then using the notation Т(А±АЪЕ) =

3.4 GEOMETRY
279
2T(A1A5A4) = 26, T(A3A5A4) = а, Т(АгА2А4) = с, Т(АгА2Аъ) = d, we
сап express the area of the hatched domain as
a + 2b + c = d, (1)
since this area is the sum of the area of three triangles each having a side
of length Αχ A2 = A3E, and the sum of the corresponding altitudes equals
the altitude of AAiA2A$ corresponding to the side A\A2. Formula (1)
remains true also when A$ is inside ΑΑ<$Α±Ε, only we have to take signed
areas. In both cases, the areas with no sign satisfy
a + 26 + с < d.
From this,
min(a,6, c) < -d. (2)
Therefore, the ratio in question is < 1/4.
The necessary and sufficient condition of having equality in (2) is a =
b = с = 1/4. On one hand, this means that A$ is four times farther
from the straight line A\A2 than A4 is; on the other hand, it implies that
Τ(Α3Α4Α5) = T(AiA4A5), that is, А±АЪ _L A1A3. These two conditions
together are satisfied only by the point A$ constructed above.
One can show by a direct computation that in this case AA\A2A$ has
the largest area among all AAiAjAk and that T(AiAjAk) > d/4 for all
i,j,k. D
Problem G.19. Let К Ъе а, compact convex body in the n-dimensional
Euclidean space. Let Pi, P2,..., Pn+\ be the vertices of a simplex having
maximal volume among all simplices inscribed in K. Define the points
Рп+2, ^п+з, · · · successively so that Pk {к > η + 1) is a point of К for
which the volume of the convex hull of Pi,..., Pk is maximal. Denote this
volume by Vk- Decide, for different values of n, about the truth of the
statement "the sequence Vn+i, Vn+2,... is concave."
Solution. The statement makes no sense for η = 0 and is obviously true
for η = 1. Consider the case η = 2.
It is clear that the vertices of the starting simplex lie on the boundary
of K, otherwise we could move the point lying inside К farther from the
opposite hyperplane. We show by general induction for the two-dimensional
case that further points are also chosen from the boundary.
The base clue has already been proved. Suppose that the first η points
lie on the boundary. In this case, they are vertices of a convex n-gon, each
side of which cuts off a (possibly trivial) part of the convex figure. The
new point is chosen from one of these parts, which results in the polygon
"growing" by a triangle. Because of maximality, the new point must come
from the boundary.

280
3. SOLUTIONS TO THE PROBLEMS
To prove concavity, we have to show that when taking two consecutive
differences of the sequence, the second does not exceed the first. If the
new points come from arcs belonging to different sides, then this is
obvious, since otherwise we would have to add these points to the sequence
in the opposite order. If two consecutive points are chosen from the same
arc over a side of the convex hull of preceding points, then let A and В
denote the endpoints of this side and С and D be the points added to
the sequence. Since С is taken first, the area of AABC is greater than
or equal to the area of AABD. Denoting by Ε the intersections of the
diagonals of the quadrangle spanned by the points ABCD, the statement
follows from the inequality T(EAB) > T(ECD), since the first difference
is Τ {ABC), while the second is T(ABD) + T(ECD) - T(EAB). Observe
that the supporting half-lines of the convex figure taken at A and В in the
half-plane bounded by AB containing С (see Figure G.16) are parallel or
intersect each other; otherwise the one from A and В that was chosen later
into the sequence would not have the maximality property. Let us draw
a straight line through С parallel to AB. This intersects the straight line
BD at a point D' lying outside the segment BD, so it suffices to prove
the inequality with D' instead of D. The triangles ABE' and CD'E' are
similar, and by the above observation, the segment AB is not shorter than
the segment CD', and thus the statement is implied.
Now we are going to prove that the sequence is not necessarily concave
for η > 3. For η = 3, the regular octahedron serves as a counterexample.
Let us choose four arbitrary, noncoplanar vertices of the octahedron for
the vertices of the first inscribed simplex. The volume of this simplex is
maximal. Indeed, suppose that ABC Ό is an inscribed simplex of maximal
volume having as many common vertices with the octahedron as possible.
If one of the vertices, say A, is not a vertex of the octahedron, then consider
the plane through A parallel to the plane BCD. By the maximality
assumption, this plane may not contain an internal point of the octahedron,
but it has a nonempty intersection with the octahedron and thus contains
at least one vertex of the octahedron, which could have been chosen
instead of A. The contradiction proves our proposition. One can similarly

3.4 GEOMETRY
281
show, that the fifth point can also be taken from the vertices of the
octahedron. The point is that opposite faces of the octahedron are parallel, so
the points lying at maximal distance from two neighboring faces form an
edge of the octahedron. The sixth point then must be the sixth vertex of
the octahedron. The second difference of this sequence is twice as much as
the first one; hence the sequence is not concave.
Turning to the general case, consider a prism over the octahedron, that
is, the convex hull of a three-dimensional octahedron and an (n — 3)-
dimensional simplex lying in a complementary subspace of the octahedron.
We can show that this is a suitable counterexample in the same way as
above, since due to maximality, the starting simplex must contain all the
vertices not belonging to the octahedron and the further points must be
chosen from the three-dimensional plane generated by the octahedron. D
Problem G.20. Let us connect consecutive vertices of a regular
heptagon inscribed in a unit circle by connected subsets (of the plane of the
circle) of diameter less than 1. Show that every continuum (in the plane of
the circle) of diameter greater than 4, containing the center of the circle,
intersects one of these connected sets.
Solution. Let Αχ, A2,..., A? be the vertices of the heptagon, let О be the
center of the circumscribed circle, and let Hi, H2,..., Η γ denote the
connected sets that connects the pairs Ai,A2, A2, A$, ... ,ΑγΑι, respectively.
Denote by С a continuum in question, by D the circle of radius 2 centered
at O.
By the assumption а(Щ) < 1 (г = 1,2,..., 7), Hi is inside D and
О £ Щ. On the other hand, conditions d{C) > 4 and 0 G С imply that
there exists a point Q G С which is outside D. Suppose to the contrary
that UiHi and С are disjoint. In this case, the distance δ(χ) of a point
χ G Hi from the closed set С U D not covering χ is positive. Consider
the open disc U(x,6(x)) of radius δ(χ) centered at χ for all χ G Hi. The
union Гг = UX£HiU(x,6(x)) of these discs is open and connected, and the
same holds for the union Г = U^IY The set Γ is inside D; furthermore,
Γ Π (С U D) = 0. Since Г is open and connected, one can draw a closed
broken line T* inside Г that connects the points Ai,A2,..., Αγ. Let Τ be
the boundary of the connected component of R2 \ T* containing Ο. Τ is a
simple, closed, broken line inside D such that Τ and С are disjoint, О G С
is inside T, and Q G С is outside T.
We are going to show that the existence of such a T contradicts the fact
that С is a continuum. Denote by В the set of points that lie inside T, by
К the set of points lying outside T. By С П D = 0, we have
С = (С Π В) U (С Π К),
where
(a) both COB and С Π К are open-closed in the subspace topology of
C; furthermore,

282
3. SOLUTIONS TO THE PROBLEMS
b) none of them is empty because О еСПВ and Q e С ПК. However,
these two facts together contradict the connectedness of the
continuum C. Π
Remark. One can give a counterexample for the proposition of the
problem if С is required only to be connected, dropping the assumption that
it is also closed. Indeed, draw two disjoint, infinite, broken lines into a
square EFLJ so that one of them starts from E, the other starts from F,
and both have closure that cover the segment LJ. Thus, M\ = T\ U J and
M2 = T<i U J are two disjoint connected sets. Let us shrink and move the
square and the figures contained in it in such a way that the points E, F
that correspond to the vertices Ε and F get onto the segment ΑγΑχ in the
order A7, E, F, A\ while L gets inside the heptagon. (We denote the image
of a figure under this similarity by putting a bar over the corresponding
sign.) Let Q* be a point on the half-line OF whose distance from О is
greater than 4. Then the claim of the proposition does not hold for the
connected sets Hj = AjAj+i (j = 1,2,..., 6), H7 = E7~E{\J~M{\JLAU and
the bounded and connected set С = О J U M2 U FQ*. (This remark is due
to Laszlo Babai.)
Problem G.21. What is the radius of the largest disc that can be
covered by a unite number of closed discs of radius 1 in such a way that
each disc intersects at most three others?
Solution. Draw a closed disc of radius 1 around each vertex of a square
of side y/2. These four discs cover the disc of radius y/2 drawn around the
center of the square.
We are going to show that a circle of radius greater than \/2 does not
have a covering having the prescribed properties.
In the following, we shall denote by' the boundary of a set (the boundary
of A is A') and "disc" will always mean a closed disc.
Let К be a disc of radius greater than \/2, and suppose that it has
a covering with the prescribed properties. Let us consider a subsystem
Αι, Α2,..., Am of the covering in such a way that
(i) the system Αχ, A^ ..., Am covers K'',
(ii) if any of the discs Αχ, A2,..., Am are removed, the remaining discs fail
to cover K'. Such a subsystem can be produced obviously by omitting
unnecessary discs one by one. There are no single points and empty
sets among the intersections Κ'Γ\Α^ since the discs different from A{
cover a closed subset of K'', so if Κ' Π Α{ is empty or consists of one
point, then Ai can be omitted.
We claim that m > 5. Obviously, Κ' Π Ai is an arc with diameter not
greater than 2; however, the central angle of such an arc is less than π/2
in a circle of radius greater than y/2. Hence one needs at least five discs to
cover K'.
Now let X\ and X2 be the endpoints of the arc Κ'πΑχ. We may suppose
without loss of generality that A% covers X\ and As covers X2 since if both

3.4 GEOMETRY
283
endpoints of an arc Κ' Π Αι were contained in the disc Aj{i φ j), then the
whole arc would lie in Aj and Αχ could be omitted.
Now if we suppose that ((A[ Π Κ) \ A2) \ A3 is not empty, then it is
an arc of positive length. Observe that this arc must be covered by one
disc, otherwise A\ would intersect more then three discs. Denote by A this
disc, and consider the arc Af3 Π К. This arc must be covered by A and
Αχ together. Otherwise, A would also have a point in common with the
disc that covers the remaining part, that is, A would intersect at least four
other discs. Therefore, Auii D A'3 Π К. But then A covers the endpoint
of Κ' Π A3 not contained in Αχ, and since four circles can not cover Kf,
A must intersect at least one more disc, which would be the fourth disc
intersected by A, and this is impossible.
We conclude that ((A[ ΠΚ)\ A2) \ A3 = 0, that is, A2 U A3 э A[ Π К.
Similarly, we can find further two discs, A4 and A5, such that Αχ U A4 э
A'3 Π Κ, Αχ U Аъ э А2 Π К, and Αι,.,.,Α^ are different. However, this is
a contradiction, since in this case A\ would intersect more than four other
circles.
Thus, a disc of radius \/2 admits a covering with the prescribed
properties, but larger discs do not. On the basis of the above proof, we can also
see that the covering of the disc of radius y/2 is "essentially" unique (that
is, unique up to rotations). D
Problem G.22. Assume that a face of a convex polyhedron Ρ has a
common edge with every other face. Show that there exists a simple closed
polygon that consists of edges of Ρ and passes through all vertices.
Solution 1. Let S be the distinguished face of P, and suppose that Ρ
has η vertices off S. The proof goes by general induction with respect to
n. The base clause is obviously true for η = 1; now let η > 1. It is easy
to see that by omitting the edges of S from the graph of edges of P, we
obtain a tree. (Indeed, imagine that Ρ is a planet and its edges are dams,
S is filled with water, while the other faces are empty basins and explode
the dams that bound S.) For this reason, Ρ has a vertex В not belonging
to S such that the edges starting from В end on S with the exception of
one edge. Let us denote by С the endpoint of the exceptional edge and by
A\, A2,..., Am the endpoints of the other edges ordered in correspondence
with an orientation of S. Finally, let Aq be the vertex of S that precedes the
vertex Ai^andletAm+i be the vertex that follows the vertex Am according
to the above orientation of S. (Figure G.17 shows a case with m = 3.)
Now if the half-line С В intersects the plane of S in a point A, then
construct a polyhedron P' by gluing the prism AA\... АШВ to P. P'
is obviously convex itself and satisfies the conditions of the proposition.
However, В is not a vertex of P', and thus P' has only η — 1 vertices off
the plane of S. By the induction hypothesis, the edge skeleton of P'
contains a simple closed polygon that passes through each vertex of P'. This
polygon goes through A in one of the orders j4oAAm+i, AqAC, CAAm+i-

284
3. SOLUTIONS TO THE PROBLEMS
Figure G.17.
According to these three cases, replace vertex A of this polygon by the
paths Αχ... Am-iBAm, ΑχΑ2 ... AmB, ΒΑχΑ2 ... Am, respectively. The
simple closed polygon we obtain this way passes through each vertex of P.
If CB is parallel to the plane of 5, then choose a plane 7 such that
7 crosses the half-line BU but has empty intersection with P. Now if
we take the projective augmentation of the Euclidean space by adding a
plane ω of points at infinity and apply a projectivity Π that takes 7 to ω,
then the polyhedron UP will completely consist of proper points, and the
intersection point of the half-line (TIC) (TIB) and the plane of Π5 will also
be proper; consequently, we may apply the previous consideration.
We can proceed similarly if the half-line intersects the plane of S. Π
Solution 2. Let S be the distinguished face, and let G be the graph
obtained by removing the edges of S from the graph formed by the vertices
and edges of the polyhedron. G is a tree, and the degree of a vertex of G
not belonging to S is at least three.
Let us choose an arbitrary face, B\, of P, and color the edges of G that
lie on the boundary of B\. Suppose that we have already chosen the faces
#1, #2, ·, Д· in such a way that
(1) Bi and Bj have no vertex in common if г φ j, (i,j = 1,2,..., r), and
(2) for all г (г = 1,2,..., r), there exists an edge of G that connects Bi
to one of the faces £1, £2, · · ·, #ζ-ι·
Using that G is connected, we get that if Ρ has a vertex not belonging
to S that is not covered by the faces i?i, i?2, ·. ·, #r? then there is a vertex
χ among these that is connected to some Bi (i G {1,2,..., r}) by an edge
xy. Since the degree of χ is at least three, χ is covered by a face Br+\
such that xy is not an edge of Д-+1- If some vertices of Br+\ were covered
by the faces Bj (j = 1,2,..., r), then one of these vertices, say z, could
be connected to χ along edges of G without passing through points of the
faces Bj. On the other hand, according to the induction hypothesis, we

3.4 GEOMETRY
285
can join χ and ζ by a path in G in such a way that all the vertices we go
through belong to some Bj, which contradicts the fact that G is a tree.
Consequently, the choice of faces Bi, B2, · · ·, #r+i meets requirements (1)
and (2). Now color the edges of G that lie on the boundary of £r+i.
At the end of this process also color the edges of S that do not belong to
any of the faces Bj. Thus, we obtain a system of colored edges that cover
all vertices of Ρ and form a polygon with the required properties. D
Problem G.23. Let D be a convex subset of the η-dimensional space,
and suppose that D' is obtained from D by applying a positive central
dilatation and then a translation. Suppose also that the sum of the volumes
of D and D' is 1, and Df)Df φ$. Determine the supremum of the volume
of the convex hull of DU D' taken for all such pairs of sets D, Df.
Solution. The conditions imply that D and D' are bounded convex sets
with nonempty interior. The problem is equivalent to the determination
of the supremum of the quantity V(co(D U D'))//(V(D) + V(D')) , where
besides, having the properties just mentioned, D and D' have nonempty
intersection and D' is obtained from D by a positive central dilatation and
a translation. Let the ratio of the dilatation be a fixed positive number
λ. We show that in the case of a fixed λ, the supremum in question is
(1 + λ + · · · + λη)/(1 + λη) = /(λ), and this supremum is attained. The
case η = 1 is trivial; let η > 1 in the following.
First, we show that the case λ = 1 can be derived from the case λ φ Ι.
Indeed, if λ = 1, then stretch D' from a point ρ G DC\D' with ratio 1 + ε.
Since D and the obtained D meet each other at p, we have
V(co (D U £>')) < ^(co (P u D'e)) < fiX + e)(V(D) + V(D'e)).
Taking the limit ε \ 0, the right-hand side tends to f(l)(V(D) + V(D')).
Assume now that λ φ 1. Changing the role of D and D' if necessary,
we may also suppose that 0 < λ < 1, since /(λ) = /(λ-1). Then D' can
be obtained from D by one central dilatation, the center О of which will
be chosen for the origin.
V(D) = supVr(P), where supremum is taken for (closed) polyhedrons
Ρ С D. It is easy to see that V(P) tends to V(D) if and only if r(P, D) =
suPx€D d{x, P) tends to 0, where d(x, P) denotes the distance between χ
and P. It is also not difficult to see that if r(P,D) < ε, then r(co(P U
P'),co(D U D')) < ε (Ρ' denotes the image of Ρ in D'). Suppose that Ρ
contains a common point of D and D' and its preimage with respect to the
similarity; then Ρ ΠΡ' φ 0. Taking such polyhedrons P, if V(P) -> V(D),
then У (со (PUP')) —> У (со (DUD')), and thus to prove that the supremum
is < /(λ), we may restrict ourselves to polyhedrons.
Now let D be a (closed) polyhedron, D' = AD, ρ e D Π D'. We may
assume that 0 ^ D. It is easy to see that со (D U XD) = υχ<μ<ιμΌ. Set
Ε = (υ0<μ<ιμΌ) \ D. Then
(J μΌ = Όυ(Ε\ΧΕ),
λ<μ<1

286
3. SOLUTIONS TO THE PROBLEMS
and thus,
V\ (J дД =ПО) + (1-А")У(£).
\λ<μ<1 /
We show that V(E) < V(D) · λ/(1 - λ) . Let Fu F2,..., Fr be those
(n — 1)-dimensional faces of D whose hyperplane separates О strictly from
D. Then the segment that connects О to ρ intersects the hyperplane щ at
a point pi. Therefore, using the notation q = (1/λ) · ρ, we have
/^\ /^\ \
d(0,Ki) = =d(q,ni) < -z^d(q^i) = -d(q,Ki).
qpi qp 1-Х
Since Ε is the union of pyramids over the faces Fi with vertex O, and
since the pyramids over the faces Fi with vertex q are contained in D, the
previous inequality gives
V{E) = -yjd{0^i)Vn.l{Fi)
г=1
(V^_i denotes the (n — l)-dimensional volume). Prom this,
v( U μϋ]
\λ<μ<1 J
V(co(DUD')) _ ул<м<1^ ) _ V(D) + (1 - Xn)V(E)
V(D) + V(D') ~ V(D)(1 + An) ~ V(D)(1 + Xn)
1 + (Λ + Λ2 + ··· + Λ") _
^ TT> "П)'
Now we show that /(λ) is an exact maximum. Let D be a simplex with
vertices pi,P2, · · · ,Pn+i; the corresponding vertices of D' are p'^p'^ . ·. ,Κι+ъ
and let ρ'χ = p'2. (For λ = 1, we choose for D' the translate of D for
which pi = p2·) In this case, co(D U D') is made up of the simplex
D' and a truncated pyramid (or a prism if λ = 1), so its volume is
XnV(D) + (1 + λ + · · · + X^ViD) = f{X){V{D) + V(D')).
It remains to determine supA>0/(A). We claim that /(λ) < /(1) =
(η + l)/2, that is,
(n + 1)(1 + λη) - 2((1 + λ + · · 4- λη) > 0.
However, this follows from summing up the inequalities
1 + Лп-Л*-Лп-* = (1-Л*)(1-Лп-*)>0 for 0<г<п. D

3.4 GEOMETRY
287
Problem G.24. Consider the intersection of an ellipsoid with a plane
σ passing through its center O. On the line through the point О
perpendicular to σ, mark the two points at a distance from О equal to the area of
the intersection. Determine the loci of the marked points as σ runs through
all such planes.
Solution. Let В be an arbitrary body, О a point of it. Denote by
Τ(σ\Β,0) the pair of points P = {P+,P~} that lie on the straight line
perpendicular to σ at О and whose distance from О is equal to the area of
the intersection of σ and B: Area (σΠ B). Let Φ(Β) stand for the figure
formed by the points Ρ as σ is varied:
Φ(Β) = υ^(σ;Β,0).
σ
Let Σ and t be a plane and a straight line intersecting one another
orthogonally at O, U the stretching in the direction of t with ratio λ, and U*
the stretching in the direction of Σ with ratio λ. Obviously, U and W are
affinities.
Lemma. К*{Ф{В)) = Ф(Ы(В)).
Proof. Set
σ'=ΖΥ(σ), Ρ = Τ{σ,Β,0),
pf = Τ{σ'-Μ{Β), Ο) = F(U(a);U{B)M{0)).
The planes σ and σ' intersect each other in a straight line m, lying in the
plane Σ. Thus P' belongs to the plane Λ spanned by t and P. (Figure
G.18 shows the trace of the mentioned figures in the plane Λ viewed from
the direction of m.) Consider the points U e σ and U' G σ' that lie over a
given point N e ΣΠΛ (that is, UN, U'N _L Σ). They satisfy Ш7 = X~NU,
and by elementary properties of affinities,
OP7 _ Area (σ' ПК{В)) OU7
~OP ~ Area (σ Π Β) ~ ΌΠ '
This equation shows that AOPP' can be obtained from AOUU' by a
dilatation and a right-angled rotation about О in the plane Л. This similarity
takes ΣπΛ to t, the point N to the intersection Μ of t and PP', and since
similarities preserve angles and the ratio between corresponding segments,
we get that PP' _L t and ¥F = MP, which means that U*{P) = P', and
this proves the lemma.
Now let a, 6, с be the half-axes of the ellipsoid Ε (Ε(α, 6, c)). The
equation of Ε takes the canonical form
a2 "I" b2 ■»" C2

288
3. SOLUTIONS TO THE PROBLEMS
Figure G.18.
in a suitable coordinate system. Consider the sphere S2(a) of radius a
around the origin. Then Φ(52(α)) is also a sphere: 52(α2π). Now let U\ be
the stretching with ratio λ = b/a in the direction of the y-axis. Ui{S2{a))
is the ellipsoid Ei(a,6,a), and by the lemma,
Φ(Ει) = Φ(%(52(α))) = %(Φ(52(α)))
= Щ(S2(a2n)) = E2(abn, α2π, аЪж).
Now let Ы2 be the stretching with ratio c/a in the direction of the z-axis.
Then U2(Ei) is the given ellipsoid E(a, 6, c), and a repeated application of
the lemma gives
Ф(Е) = Ф(и2(Ег)) = и;(Ф(Е1)) = Щ(Е2) = E3(bcn,acn,abn),
which was to be determined. D
Remarks.
1. We remark that the proposition of the lemma is true also for n-dimen-
sional bodies В and (n — l)-dimensional hyperplanes. Thus, an
application of the lemma gives an immediate answer to the n-dimensional
version of the problem as well.
2. We also remark that the lemma, which lies in the base of the solution,
concerns an arbitrary body В and an arbitrary point О of it. Hence, it
is applicable to a much wider class of problems.
Problem G.25. Construct on the real projective plane a continuous
curve, consisting of simple points, which is not a straight line and is
intersected in a single point by every tangent and every secant of a given
conic.
Solution. Removing a straight line from the projective plane we obtain a
Euclidean plane. Let us introduce Cartesian coordinates (x, y) on it and

3.4 GEOMETRY
289
denote by К the circle of radius 1 centered at (0, —2). К is a conic of the
projective plane. All conies of the projective plane are equivalent, that is,
one can, find a projective transformation V for any conic С such that V
takes С to K. Therefore, if Г is a solution of the problem with respect
to the circle K, then V~X(T) is a solution with respect to С since the
projectivity V and its inverse V~l are bijections that take straight lines
into straight lines, and secants and tangents of a conic into a secant or
tangent of the image conic, respectively. Thus if Г meets the requirements
of the problem with the conic K, then so does /Р~1(Г) with the conic С
First, we construct on the Euclidean plane a continuous curve Г
consisting solely of simple points that is not a straight line and is intersected in
a single point by every secant and tangent of К that is not parallel to the
rr-axis. Let у = f(x) be a function defined on the whole x-axis such that
(a) / is continuous and differentiable,
(b) 0 < /(*) < 1,
(c) the absolute value of the derivative of / is less than 1,
(d) / is increasing on the left of χ = 0 and decreasing on the right,
(e) the limits linx^oo f(x) and Ηηχ^-οο f(x) exist.
We show that secants of the graph Γ of / do not intersect K. Indeed,
suppose that £(x) is a linear function such that
i(Xl) = /On), £{x2) = /Ы; χι < X2 · (i)
The graph J of £(x) is a secant of Γ. We have
\e'(x)\ = |tana| < 1
because of (c), where a is the direction angle of 1. If a = 0, then ΐ either
coincides with the x-axis or lies above it because of (b); hence it does not
meet K. If a > 0, then the zero xq of ί satisfies xq < x\ < x2 by (b)
and (1). In this case, the assumption 0 < xq together with α > 0 and
(1) implies f(x\) < /(#2), 0 < #1 < #2, which is in contradiction with
(d). If, however, x0 < 0 and |tana| < 1, then J does not meet K. We
can prove similarly that 1 has empty intersection with К also in the case
—1 < tana < 0. We conclude that the secants of Г do not meet K, or,
in other words, the tangents and secants of К have at most one point in
common with Г. Those secants and tangents of К that are parallel to the
rr-axis (denote the set of them by χ) obviously have empty intersection
with Γ. Other tangents and secants with nonzero direction angle intersect
Γ since Γ divides the plane into two connected parts, one of which contains
the half-plane у < 0 while the other contains the half-plane у > 1. A
straight line with nonzero direction angle has points in both half-planes,
hence it must cross the graph Г of /. Thus, the curve Г of the Euclidean
plane has the required properties.
We augment the Euclidean plane to a projective plane by adding an
ideal point to each family of parallel lines. Attaching the ideal point P^
of the rr-axis to Г, we get a closed curve Г on the projective plane, which

290
3. SOLUTIONS TO THE PROBLEMS
is continuous by (e) and is intersected in a single point, namely in P^, by
the straight lines of χ. The points of Γ are simple since the graph Γ of the
continuous function / consists of simple points and we have closed Γ by
adding one single boundary point.
Finally, we have to present a nonlinear function that has properties (a)-
(e). For example, у = е~х /2. This function gives a solution according to
the previous considerations. But one can easily construct other suitable
functions. The simple, though not differentiable, function whose graph
is obtained from the rr-axis by replacing the segment between the points
A(-2,0), C(2,0) by the broken line ABC, where В has coordinates (0,1),
also does the job. D
Problem G.26. Let Τ be a surjective mapping of the hyperbolic plane
onto itself which maps collinear points into collinear points. Prove that Τ
must be an isometry.
Solution.
(A) First we show that Τ is injective. The proof (based on the idea of
Nandor Simanyi) consists of three steps.
1. We claim that the inverse image of a point is convex. For this
purpose, we need to show that if Τ (A) = T(B), then every
point С of the segment AB is mapped to Τ (A). Suppose it is
not so: T(C) φ Τ (A) for some С G AB. Let e and / be the
straight lines perpendicular to T(A)T(C) at Τ (A) and T(C),
respectively. Let us choose the points Ρ and Q in such a way
that their images are points of e and /, respectively, different
from Τ (A) and T(C) (see Figure G.19). There are no three
collinear points among A,B,P,Q since there are obviously no
three collinear points among their images Τ (A), T(B), T(P),
T(Q). According to this, the straight line CQ intersects side AB
of AABP in an inner point and thus must cross one of the two
other sides. The image of this second intersection point must lie
on both e and /, which is a contradiction since e and / do not
intersect each other.
ρ Τ(Ρ) e T(A)=T(B)
T(Q) T(C)
Figure G.19.

3.4 GEOMETRY
291
2. Now we show that the inverse image of a point X contains no
interior point. Suppose to the contrary that the interior of T~l(X)
is not empty. Draw a straight line / through X. We claim that
int (Γ_1(/ \ {Χ})) φ 0. Indeed, consider a point Ζ of /
different from X and denote by Ρ a point of its pre-image (see Figure
G.20). Then the images of straight lines that go through Ρ and
intersect T~l(X) lie in /. In other words, T_1(/) contains a pair of
opposite angular domains with common vertex P. Since T~X(X)
is completely contained in one of these angles, it is disjoint from
the other; therefore, int (T_1(/ \ {Χ})) φ 0, as we claimed. Now
rotate / about X and consider the sets T_1(/ \ {X}). They form
an uncountable family of disjoint subsets of the hyperbolic plane
such that each set has an interior point, which is a contradiction.
T_1(X)
X
Z=T(P)
Figure G.20.
3. Now we are in the position to prove that the preimage
T~l(X) of every point X consists of one single point. Again
we use the indirect method. If T~l{X) is not a point, then it is
a segment, half-line, or a straight line according to the previous
considerations. In any case, T~l{X) is contained in a straight
line, which will be denoted by e in the following. Let A and В
be two points of e that are mapped to X (see Figure G.21). It is
- easy to see that T(e) is a part of a straight line д. Draw a straight
line / through X other than д. One can show as above that the
interior of the set T_1(/ \ {X}) is not empty. Rotating / about
X, we can again get a contradiction.
g^T(e)
Z=T(P)
fX=T(A)=T(B)
Figure G.21.

292 3. SOLUTIONS TO THE PROBLEMS
(B) The major difficulties of the proof are over. As a next step, we show
that the images of noncollinear points are noncollinear. Indeed, if the
noncollinear points P, Q, and R were mapped to a straight line e,
then the images of the straight lines PQ, QR, and RP would also be
contained in e. Consequently, every straight line that twice crosses the
boundary of APQR twice would be mapped into e. However, such a
straight line passes through any point of the plane; therefore, the image
of the whole plane should be contained in e, which contradicts the fact
that Τ is surjective.
(C) From this it follows that T_1 also takes collinear points to collinear
points; the image of a straight line (under T) is a straight line, and
the images of intersecting straight lines are intersecting, the images of
nonintersecting straight lines are nonintersecting.
(D) Τ preserves the ordering on straight lines. Indeed, if A, B, and С are
three different points of a straight line e, then, as is known, С does not
separate A and В if and only if one can find straight lines a, 6, and
с through A, B, and C, respectively, such that a and b intersect each
other but с intersects neither a nor b (see Figure G.22). Τ preserves
the latter property by (C), hence preserves ordering.
Figure G.22.
(E) Therefore, Τ preserves half-planes, half-lines, and segments.
Furthermore, it preserves the ordering of pencils. Thus, the images of
asymptotic half-lines are asymptotic.
(F) Now we show the existence of a length d such that Τ maps segments
of length d onto segments of length d. Let us perform the following
construction (see Figure G.23.a).
We denoted the two half-lines of e determined by Ρ by / and д. Let us
take a point X not lying on e and draw the half-lines h and к starting
from X and asymptotic to / and g, respectively. Take a point Υ on the
elongation of к beyond X; let the segment Υ Ρ intersect h at U. Draw
an asymptotic half-line to g from U, and suppose that it intersects the
segment XP in V. Denote by Q the intersection point of YV and g.

3.4 GEOMETRY
293
Figure G.23.8L
We claim that Q does not depend on the choice of X and Y.
It is very easy to check this statement in the Cayley-Klein model
of hyperbolic geometry (and it is also sufficient, since we know that
if a proposition is true in every Cayley-Klein model, then it is a
true theorem of hyperbolic geometry). The above construction is
a construction of a complete quadrangle in the model (see Figure
G.23.b); consequently, denoting by / and J the horizontal points
of / and g, respectively, the pairs J, Ρ and /, Q are conjugate, and
the points P, J, and / determine the fourth harmonic Q uniquely.
Figure G.23.b
It is clear that the length d of the segment PQ does not depend on
the choice of Ρ either. Since Τ transforms Figure G.23.a into a similar
one, it moves segments of length d into segments of length d, as we
stated.
Of course, the segments of length nd are also invariant under T, as

294
3. SOLUTIONS TO THE PROBLEMS
are the angles of regular triangles of side nd. Because the angle of
these triangles tends to zero as η tends to infinity, there exist
arbitrarily small invariant angles. As the sum of invariant angles is also
invariant, a dense set of angles will be invariant under T. Taking into
consideration that Τ preserves the ordering on pencils, it follows that
Τ preserves angles. However, in hyperbolic geometry, a triangle is
determined up to isometries by its angles, so the image of every triangle
under Τ is a triangle congruent to the original one, hence Τ is an
isometry. D
Problem G.27. Let Χχ,..., Xn be η points in the unit square (n > 1).
Let T{ be the distance of Xi from the nearest point (other than Xi). Prove
the inequality
rl + ---+rl <4.
Solution. We shall prove by general induction. The base clause is trivial
for η = 2. Let η > 2, and suppose that the proposition is true for any
system of points consisting of less than η points. This assumption means
also that if the points Y\,..., Y^ are enclosed in a square of side a and qj
denotes the distance of Yj from the other points, then for 1 < к < η we
have^=l9j2<4a2.
Let us divide the square into four congruent squares by the medians.
Denote the small squares by iVi,..., N± (see Figure G.24). We shall
separate some cases according to the possible distributions of the points Xi
in the small squares and then study each case individually. If a point Xi
lies on the common boundary of two squares, then decide which square it
belongs to; it does not matter how.
Nt
N4
N,
N3
Figure G.24.
Case A. None of the small squares contains exactly one point. Here we
may distinguish two further subcases: Either all points are in one small

3.4 GEOMETRY
295
square or the points are distributed in more than one square. In the first
case, applying the induction hypothesis for Χχ,..., Χη-ι, we get
Since r\ < 2, the proposition of the problem is true in this case.
In the second subcase, however, each small square contains less than η
points (and if it contains a point at all, then it certainly contains more than
one). Therefore, by the induction hypothesis,
XiZN» W
for 1 < ν < 4, from which the proposition of the problem follows.
Case B. There is exactly one small square, say N\, that contains one
point. For example, let X\ G N\. Inequality (1) holds also in this case for
ν = 2,3,4. Therefore, if r\ < 1, then we are ready. We may suppose for
this reason that r\ > 1. Of course, r\ < 2, and this estimation would be
enough if one of the small squares were empty. Thus, we may suppose in
the following that each small square contains a point; from this it follows
that r\ < 5/4 because of the point in N4.
• Xi
II
N„
If
N43
Figure G.25.
Let us divide N4 into four smaller squares with the help of its medians,
and we will denote these squares by AT41, AT42, AT43, AT44 (see Figure G.25).
As ri > 1, the shaded domain contains no point. We shall show that
Ε гг2<^ 0 = 3,4). (2)
XieNAj
This follows from the induction hypothesis — applied for AT4j — if the
number of points contained in iV4j is not exactly one. If, however, N±j

296
3. SOLUTIONS TO THE PROBLEMS
contains exactly one point, then its distance from another point in N4 is
at most л/5/4, that is, (2) holds in any case. Then
f>t2 = r?+ Σ <·!+ Σ »?+ Σ 'Ι
г=1 Χι£Ν2 Χι£Νζ Χ«€ΛΓ4
5 5
^4 + 1 + 1 + 2Ϊ6<4·
Case С. The number of points is equal to one in exactly two small
squares. In this case, the squares can be coupled in such a way that squares
in a couple are neighbors and one of the squares in a couple contains
exactly one point while the number of the points in the other is not one (see
Figure G.26). Assume, for example, that Νχ and N± are in one couple and
Νχ is the one that contains exactly one point. Obviously, it suffices to show
that
Σ ^2·
Xi£Nx\jNA
This can be done in the same way as in case B.
1 point-
1
1 point
1 point
1 point
Figure G.26.
Case D. Exactly three squares, say Νχ, N2, and N3, contain one point.
Call these points Χχ, Χ2, and X3, respectively. Let us divide N4 into
four small squares, as shown in Figure G.27. First of all, we remark that
r\ < 5/4.
We distinguish three subcases. Assume first that the rectangles iV^uA^
and ΛΓ43 U A^42 both contain a point. Then
^6)ЧУ
13
16'
Consequently,
X>2 = r2 + r22+r32 + £ г? <^| + | + ^| + К4.
г=1 XiGN4

3.4 GEOMETRY
297
N2
N3
Figure G.27.
Second, suppose that, for example, iV4i U N42 contains no point but ЛГ43
and iV44 do contain a point. In that case, r§ < 13/16, r\ < 1 + 1/16, and
furthermore, (2) is now valid. Hence,
J\f = r? + r2 + r§+ Σ r^1U + l + Y6 + 2ie<4·
i=l XiGNt
Third, it remains to study the case when neither iV41 nor AT42 contains
a point and one of N43 and iV44 is also empty. Applying the induction
hypothesis to the other (nonempty) square we get
1>г2 = г? + г2 + гз2+ £ г2<Ё + Ё + Ё+1=4.
i=l Xi£NA
Case Ε. The last case that remains is when each of the small squares
contains one point. Let X{ GiVj. We prove that
d(Xu X2)2 + а(Х2,Хз)2 + d(X3,X4)2 + d(X4, X1)2 < 4. (3)
((d(P, Q) denotes the distance between the points Ρ and Q.) This already
implies the inequality of the problem.
Let us think of (3) as a strictly convex function defined on the set
x^=i^i/ С R8. Such a function can attain its maximum only at the
extremal points of the boundary of its domain. An extremal point
corresponds to such a configuration of the points Χι in which every Xi is
positioned at a vertex of the corresponding square Ni. Since for these
cases (3) can be easily verified, (3) holds in general. D
Remark. Contestants B. Brindza, V. Komornik, P.P. Palfy, V. Totik,
and Zs. Tuza described those configurations of the points for which the
sum of squares in question is equal to 4. They found that there are only
• X,
N41
N44
N42
N43
,X2
·χ3

298
3. SOLUTIONS TO THE PROBLEMS
two possibilities: η = 2 and the points are positioned at opposite vertices
of the square, or η = 4 and the points are put to the vertices of the square.
Problem G.28. Give an example of ten different noncoplanar points
Pi,..., P5, Qi,..., Q5 in 3-space such that connecting each Pi to each Qj
by a rigid rod results in a rigid system.
Solution. First, we describe two constructions.
I. Let д be a rigid structure. Take a further point P, and connect it by
a rigid rod to three points Qi, Q2, Q3 of g, for which P, Qi, Q2? Q3 are
not coplanar. Then a rigid system is obtained.
II. Let g be a rigid system and PQ a rod in it. Take a point Τ on this
rod and connect it to two further points R, S of g, for which P,Q,R,S are
not coplanar. The system obtained will also be rigid.
Putting it the other way around, these two statements mean that to get
a rigid realization of a graph g, it is enough to have
(a) a rigid realization of a graph obtained from g by deleting a point of
degree 3, or
(b) a rigid realization of a graph obtained from g by deleting a point of
degree four and connecting two of its neighbors by an edge.
In our case, the graph to be realized is as shown in Figure G.28.
Qi Q2 Q3 Q4 Q5
Figure G.28.
We omit the edge P\Q\. Without drawing edges between P^'s and Q/s
it is sufficient to find a realization of the following graphs:
by(b), P2 P3 P4 P5
0000
OOOO О
Qi Q2 Q3 Qa Qb
again by (b), P2 P3 P4 Рь
О О О О
о о о о
Q2 Q3 Q4 Яъ
again by (b), P3 P4 P5
о о о

3.4 GEOMETRY
299
by (a),
о
Q2
О -
Q3
Q4
-O
Qb
P3
О
P4
by (a),
o-
Q3
Pa
— О
by (a)
O —
Qz
Qi
Pa
O-
— О
Qn
Ръ
- О
О-
Q4
Q5
The last graph is a tetrahedron, which is obviously rigid. D
Problem G.29. Let us define a pseudo-Riemannian metric on the set
of points of the Euclidean space E3 not lying on the z-axis by the metric
tensor
10 0
0 1 0
0 0 -V^^Ty2,
where (x,y,z) is a Cartesian coordinate system in E3. Show that the
orthogonal projections of the geodesic curves of this Riemannian space
onto the (x, y)-plane are straight lines or conic sections with focus at the
origin.
Solution 1. Let a, /3, and 7 run through the indices 1,2, and denote by
(χλ,χ2,χ3) the coordinates (#,y, z). We compute the differential equation
of the geodesies. We can write the metric tensor and its inverse (дгк) in
the following form
д= -^(χψ + 0*2)2, ff
33
vV)2 + (*2)2
Using the formula
г* = l V4 nih fd9jh 1 d9hk dfljfc }
jk 2z^y \ dxk "·" dxj dxh (
/1=1

300
3. SOLUTIONS TO THE PROBLEMS
for the Christoffel symbols, we obtain
1 αβ — L αβ — L 3β — L аЗ ~ U J
1 ΤΊ
г7 = - ·
33 2 ^1)2 + (Ж2)2 '
г3 = г3 = * ж7
73 37 2(х1)2 + (х2)2'
By these, the equation of the geodesies,
xj" +J2rijkxj'xk' = 0
can be written in the form
J' + \ , X =z,2 = 0; (1)
у" + \-7==2*,2 = *'> (2)
^ y/ x1 + y1
z" + -^—2x'z' + -^—2y'z' = 0. (3)
x1 + уг хг +уг
This system is satisfied by the curves
χ = as + b, у = as + b, ζ = с,
the projections of which onto the (я, y)-plane are straight lines.
Furthermore, looking at the equations, we see that if z' = 0 at the initial point of
an integral curve, then z' = 0 along the curve.
Assume ζ' φ 0. Then the third equation can be written in the form
_-5 Μ^+Λ.
Therefore,
ln*' = ln
y/x2 + y2' V^2 + У2
(where с is an arbitrary positive constant).
Substituting this into equations (1) and (2), we get
^ 2 (χ2 + 2/2)3/2 U'
» ■ c2 У n
У 2 (x2+t/2)3/2 U-

3.4 GEOMETRY
301
This system is non other than the well-known system of differential
equations
„ д (с* 1 \
дх \2 y/x*+y*J '
д (с* 1 \
У =
ду 12 Jx^+ψ
of the Kepler problem, the solutions of which are conic sections with focus
at the origin. D
Solution 2. The problem of finding geodesies of the given space is
equivalent to the problem of finding stationary curves of the variational problem
corresponding to the integral
[ \/х2+У2- Vx2 + y2*2dt.
Since the variational function is positively homogeneous of degree one in
the variables x, y, i, it is well known that those stationary curves of it that
satisfy the condition
x2 + y2 - \Jx2 + y2z2 = 1 (4)
coincide with the stationary curves of the problem corresponding to the
integral
' b2 + y2-yfi^tfz2dt.
l·
Changing over to the coordinate system (r, φ, ζ) with the coordinate
transformation χ = r cos φ, у = r sin φ, ζ = ζ, we obtain the variational problem
/<
(r2 + Γζφζ +rzz)dt.
The integrand function is independent of φ and z. Therefore, the Euler-
Lagrange equations
δφ dt δφ' dz dt dz
imply that
dF A OF „
-—■ = A, —— = В (A and В are constant),
οφ dz
that is, we obtain the first integrals
2г2ф = A, -2rz = В.

302
3. SOLUTIONS TO THE PROBLEMS
We introduce the new unknown function u(tp) = 1/τ(φ). For this, we
have
1 . 1 du . 1 du ( A \ _ Adu
r = —u = —2l~4) = —o~
u ir αφ
Therefore, by the relation (4),
= f + ?
2 -2 -2
^ φ —ΤΖ
A2 (du\2 , 1 (А2 Л 1 (В2 Л
_А2
4
Х%Ь«\
в2
и.
4
Let us differentiate with respect to ψ:
(A*_ [fu
\ 2 [dtp2
+ u
B2\ du n
The equation du/άφ = 0 characterizes circles centered at the origin.
Dividing by du/άφ, we get the equality
£u _ B^
άφ2+4~ 2Α2'
the general solution of which can be written in the form
B2
ν>{φ) = TjTtf (l + e cos(^ + ω))'
that is,
r{<f)
os(y> + ω) V
l + ecos((^ + o;) V^ 2A2 J '
and this is the focal equation of a conic section with focus at the origin. D
Remark. Analogously, the trajectory of a point moving in a conservative
force field can be obtained as the spacelike projection of a geodesic curve
of a suitable Riemannian metric on the four-dimensional space-time plane.
Problem G.30. Let us divide by straight lines a quadrangle of unit
area into η subpolygons and draw a circle into each subpolygon. Show
that the sum of the perimeters of the circles is at most π^/η (the lines are
not allowed to cut the interior of a subpolygon).
Solution. It is known that the area of an n-gon circumscribed about a
circle of radius R is at least i22ntan(7r/n) . Cutting the quadrangle by
one straight line after the other, we determine the number of the domains
produced and the sum of the angles of the domains. When we draw a

3.4 GEOMETRY
303
new straight line, the number of polygons increases by k, and the sum
of the angles increases by at most 2kn, depending on how many nodes
of the previous decomposition the new straight line goes through. This
way, if the number of sides of the polygons in the resulting decomposition:
ki,k2, · · · ,fcn, then
η
5>,-4)<о.
2 = 1
Denote by Тг the area of the ith domain and by Ki the perimeter of the
inscribed circle. We know that
Kf <
4π2
h tan £
π -* * '
Then
Σ***Σ·
— VTi<n,
i=l
Υ—*-
\ f-f ki tan ?. \ .
V г=1 г ^г V г=1
Σι
■f-' \/ fcj tan £.
by the Cauchy-Schwarz inequality. Therefore, it suffices to show that
4
Υτ^-ιτ>η·
If ki = 4, then the corresponding term in the sum is just equal to 1. If
ki > 5, then by £](&» ~ 4) < 0, this term comes together with (ki — 4)
triangles, and for this reason, it is enough to see that for к > 5,
4 7 + (к-4)^^<к-3.
к tan
'3tanf ~
Since
3tan|
- < 0.7699,
it suffices to show that for к > 5,
4
Artan^
< 0.23A: + 0.079.
The left-hand side is less than 4/π, and this is enough for к > 6. When
к = 5, the left-hand side is less than 1.11 and the right-hand side is greater
than 1.2. D

304
3. SOLUTIONS TO THE PROBLEMS
Problem G.31. Let К be a convex cone in the η-dimensional real
vector space Rn, and consider the sets A = KU{-K) and В = (Rn\A)U{0}
(0 is the origin). Show that one can find two subspaces in Rn such that
together they span Rn, and one of them lies in A and the other lies in B.
Solution. First, we show that if χ G intK then χ G К. Take a simplex
with vertices in К that has χ in its interior. The vertices can obviously be
moved to К by a small perturbation in such a way that χ remains in the
simplex; thus χ G K.
Now we show that if χ £ К, then there exists и G Rn such that (u, x) <
0, but (u, y) > 0 for every у G K.
Let к be the point of К nearest to x. The perpendicular bisector hy-
perplane of the segment connecting χ to к separates χ from K, that is,
there exist и G Rn and b G R such that (u, y) > b for every у G K, but
(u, x) <b. Since О G K, b < 0. If we had (u, y) = ε < 0 for some у G K,
(u, Xy) = Χε < b would hold with a suitable λ > 0. Thus, (и, у) > О for
every у G K.
We prove the proposition of the problem using induction. The statement
is obvious for η = 0. Assume that η > 1 and the statement is true for Rs
provided that 0 < s < n. If the maximal number of linearly independent
elements of К is s and s < n, then К is contained in an (n — l)-dimensional
subspace H. In this case, let E0 С К U {-К) and F0 С (Я \ A) U {O} be
subspaces with which the proposition holds in H.
Taking Eo for Ε and the span of F0 and a vector / G Rn \ Η for F, we
find that the proposition is true. We may thus assume that К contains η
linearly independent vectors and, as a consequence, it has an interior point.
Set _
N = {u: (u,x) > 0 for every χ G K} .
If N = {O}, then Έ = Rn, and thus К = Rn. Therefore, we may assume
that Ν φ {Ο}. Choose a maximal linearly independent system ui,...,Uk
consisting of vectors from N. Let Η be the hyperplane
Η = {χ: (щ-\ \-Uk,x) = 0}.
If Η Π Κ is a cone,
{HC\K)U {-{H ПК)) = НГ\А, (Я \ (Я Π A)) U {О} = Я П Я;
therefore, the proposition can be applied to Η Π A and Η ΠΒ. We obtain
subspaces Eo С Я Π A and F0 С Η Π Β, which span Я. Let us observe
that
E0 С {χ: (tii,ж) = (ti2,s) = *·· = (м*,ж) = 0}.
Let /i,..., lm be a basis of E0, and let Д,...,// be a basis of Fo- We may
obviously assume that Ζχ,..., lm G If; furthermore, m + / > η — 1 holds.
(K \ H) has at least one interior point, call any of them e, and let Ε be

3.4 GEOMETRY
305
the subspace generated by Eq and e. If we show that Ε С A, then we are
done with the choice F = Fq.
First, we show that Eq С К. If χ G Eq and и G AT, then и = \\Ui +
(- Afciifc, and thus
(ti, ж) = λι(ϋι, ж) + · · · + Afc(tifc, χ),
from which χ € K. Now, if lo € Eo, then lo € К and hence lo + eGX;
moreover, /0 + e £ int If С К. Prom this, A(/0 + e) € A for every A G R,
that is, Ε С A. D
Problem G.32. Let V be a bounded, closed, convex set in Rn, and
denote by r the radius of its circumscribed sphere (that is, the radius of
the smallest sphere that contains V). Show that r is the only real number
with the following property: for any finite number of points in V, there
exists a point in V such that the arithmetic mean of its distances from the
other points is equal to r.
Solution.
(a) Since V is bounded, it is easy to see that there exists a minimal sphere
of radius r that contains V. Call this sphere G, and denote its boundary
by 5, its center by О (the origin). We show that О е co(S f)V). Suppose
that this is not true. Then there exists a hyperplane Η that separates О
from S Π V, since S Π V is compact. We may assume that
Η = {(#1, · · ·, xn) : Xn = c}, where 0 < с < r.
Since r is minimal, the spheres of radius less then r centered at Ok =
(0,..., 0,1/fe) do not cover V. Therefore, we can find points a^ G V such
that \dk — 0\ > r. The sequence {a^} possesses a convergent subsequence
Q>ki —► a*· Obviously, a* G V and |a*| = r; therefore, a* G S C\V and
thus a* = (al,..., α*), where a* > c. Then, for г > г0, aki = (a\,..., агп),
where агп > с and Σ"=1(α})2 < τ2 since a\ G V. From this, we get
f л - 1/2
' n—1
K-o*J= ΣΚ·)2 + Κ-ι/^)2
<(r2-2c/A:i + l/A:2)1/2<r,
but this is impossible.
This shows that О G со (Sn V), and therefore О G со (V) = V as well. If
αχ,..., am G V, then |а^| < г, and thus the average of the distance between
О and the points a^ is < r. On the other hand, s = (1/m) · Σ™=ι сц £ V',
and by the minimality of r we can find a point у in V such that \s — y\ > r.
Thus,
- ς \ai - y\ > \- Σ(α* - у) =|β"y| - Γ·
m *-^ \m

306
3. SOLUTIONS TO THE PROBLEMS
By the convexity of V, the segment [О, у] lies in V, and one can obviously
find a point ζ on this segment for which (1/ra) · Σ \ai ~~ A — r-
(b) If и > r, then for αχ = О, there is no such point χ G V for which
\d\ — x\ = U.
(c) Now let и < r. Since О G со (S Π У), there exist points αϊ,..., α& G
5 Π У and numbers £i,...,£fc such that for 0 < U, Σί=ι^ = 1 anc^
Σΐ=ι ^аг = 0. Set ε = (г — и)/кг > 0, and choose the natural numbers q
and pi in such a way that
к
Σρ* = ?' and \Pi/Q ~U\<e (i = 1,..., k).
г=1
Let the system of points bi,...,bq consist of the points ai, so that ai is
taken pi times for each i. Then \bj\ = r (since bj G S) and
1 ς 1 k
~У2ЬЧ = \~y^Piai
«ti
г=1
^2(Pi/Q ~ U)a>i
г=1
< £/2, la*l= ^Γε·
г=1
We show that for any χ e V, (l/q) · Σ?=ι |Ь* — ж| > гг. Indeed, denoting
by (α, b) the dot product, for χ G V, we have
...1(ф
> г г
qr
Σ*
г=1
> г — кг ε > и. D
Problem G.33. Show that for any natural number η and any real
number d > 3n/(3n — 1), one can find a covering of the unit square with η
homothetic triangles with area of the union less than d.
Solution. Let us circumscribe congruent homothetic triangles about the
faces of a regular hexagonal tessellation such that the area of the triangles
is 3/2 times larger than the area of the hexagonal face. These triangles
cover the plane with density 3/2.
Let the radius of the circle inscribed in one of the triangles be equal
to 1. Let us translate the sides of every triangle toward its center by
ε. The density of the new triangles is equal to (3/2) · (1 — ε)2. There are
holes produced in the covering: upside-down triangular holes circumscribed
about a circle of radius ε. The relative number of the holes is twice the
number of the triangles, so the density of the holes is (3/2) · 2ε2 = 3ε2.
Let us cover each hole with some density d > 1. By this we mean that
we cover each hole with some plates, whose total area is equal to d times
the area of the hole. The joint density of the triangles and the plates is

3.4 GEOMETRY
307
D = (3/2) · (1 -ε)2 + 3<fe2. D is minimal at ε = l/(2d-1) . The minimum
isD = 3d/(2d+l) .
Let Dn denote the infimum of the densities of the coverings with
nomothetic triangles of η different sizes. As we have seen, D\ < 3/2 . Since the
holes in the above construction can be covered by homothetic triangles of
(n — 1) different sizes with density arbitrarily close toDn_i, we have
B,< «°L_lBlS №
that is,
2D! + 1' ° - 2D2 + 1'
on
Dn < - . D
n - 3n - 1
Problem G.34. Let R be a bounded domain of area t in the plane,
and let С be its center of gravity. Denoting by Tab the circle drawn with
the diameter AB, let К be a circle that contains each of the circles Tab
(A,BeR). Is it true in general that К contains the circle of area 2t
centered at C?
Solution. We shall show that the answer to the question is negative. Let
R be the semicircle in the plane with Cartesian coordinates (я, у) defined
by the inequalities x2 + y2 < l,x > 0. The area of R is t = π/2, and
the radius of the circle of area 2t is equal to 1. Assume that Ρ e Tab for
some pair of points А, В G R. Denote the position vectors of the points
Ρ, Α, Β by ρ, α, 6, respectively. The condition Ρ e Tab is equivalent to the
inequality
|p— (a —6)/2| < \a-b\/2.
Since
we have
|p - (a + 6)/2| > |p| - |(a + 6)/2|,
\p\ < \a + b\ /2 + \a - b\ /2 < ((\a + b\2 + \a- b\2)/2)1/2
= ((H2 + |6|2 + 2(a, b) + |a|2 + |6|2 - 2(a, 6))/2)1/2
= (|a|2 + |6|2)1/2<V2.
Therefore, if we choose for К the circle of radius y/2 centered at the origin,
then К will contain all the circles Tab (А, В е R). Let us compute the
coordinates (хс,Ус) οι the center of gravity С of the figure R. By the
symmetry of the figure, у с = 0. To compute xc, we use the well-known
method
Uo t I 3 J
2 4
xc = - I x2y/l — x'dx= T |--(i-arj"'-| = — = — .

308
3. SOLUTIONS TO THE PROBLEMS
We want to show that the circle of radius 1 centered at (4/3π, 0) does
not contain the circle K. For this purpose, we have only to show that
4/3π+ 1 > V2- Using π < 3.2, 2.4-0.416 < 1, and (1.416)2 > 2, we obtain
έ+1>3^2 + 1 = ά + 1>1·416>^·
Therefore, R is indeed a counterexample to the proposition in question. D
Remark. Competitors disproved the proposition with many different
counterexamples. Most constructions resembled the set-theoretic union
of the circle x2 + y2 < 1 and the ellipse x2/(l + ε)2 + у2/ε2 < 1 for some
suitably small positive ε.
Problem G.35. Let Mn С Rn+1 be a complete, connected hypersur-
fa.ce embedded into the Euclidean space. Show that Mn as a Riemannian
manifold decomposes to a nontrivial global metric direct product if and
only if it is a real cylinder, that is, Mn can be decomposed to a direct
product of the form Mn = Mk χ Rn~k (k < n) as well, where Mk is a
hypersurface in some (k+ 1)-dimensional subspace Ek+1 с Rn+1, Rn~k is
the orthogonal complement of Ek+1.
Solution. The solution rests upon the two fundamental equations of
classical surface theory, namely, the Gauss equation
R(X, Y)Z = g(Z, A(X))A(Y) - g(Z, A(Y))A(X)
and the Codazzi-Mainardi equation
(4XA)(Y) = (4YA)(X).
In these formulas, Χ, Υ, Ζ are vector fields tangential to the hypersurface,
g(X, Y) is the first fundamental form, A(X) = άχτα is the Weingarten
map (m is the unit normal vector field of the hypersurface, άχ is the
directional derivative in Rn+1), VχΥ is the covariant derivative, and, finally,
R(X, Y)Z is the Riemannian curvature tensor.
A(X) is a self-adjoint linear mapping on each tangent space. It is also
called the extrinsic geometrical fundamental form, since with its help one
can describe the shape of the surface in the space. For example, the
following statements are known and can simply be derived from the Codazzi-
Mainardi equations.
• Let W® be the kernel of Ap at the point ρ € Mn, and assume dim W%
= constant = к on some open subset U С Мп. Then the distribution of
the subspaces W% is integrable, and the integral manifolds are open subsets
of a subspace Шк С Rn+1.
• The Riemannian curvature of the hypersurface Mn is 0 if and only if
the rank of A is at most 1 at any point. For such a surface let U denote the
open set of those points at which Α φ 0, and let η be a unit vector field on

3.4 GEOMETRY
309
U consisting of eigenvectors of A such that A(n) = λη holds with λ φ 0.
Then, by the previous proposition, integral manifolds perpendicular to η
are open subsets of some subspace Rn-1 in Rn+1. Let c(s) be a curve on
such an integral manifold parameterized by arc length, and consider the
function X(s) = X(c(s)). Codazzi-Mainardi equations yield by a simple
computation that λ' = <^(s)A, where <£>(s) := — g(c(s), Vnn), and thus
A(s) = A(0)e/o^dt.
Consequently, if λ φ 0 at one of the points of the integral manifold
perpendicular to n, then λ φ 0 along the whole integral manifold and also at
its boundary points. For this reason, if Mn is complete, then the maximal
integral manifolds perpendicular to η must fill the whole subspace Rn-1.
Consider a bending of such a complete hypersurface having Riemannian
curvature 0 onto Rn. (If Mn is not simply connected, then we bend its
universal covering space.) The above integral manifolds will be bent onto
parallel hyperplanes of Rn; Thus the distance between the points of an
integral manifold and another integral manifold is constant. This means, that
the integral manifolds themselves are parallel, (n— 1)-dimensional subspaces
Rn_1 in Rn+1, since otherwise the above distance would not be constant.
The orthogonal complement R2 of the subspaces Rn-1 cuts Mn in a curve
M1, and we obviously have the cylinder decomposition Mn = Μ1 χ Rn_1.
To sum up, we have the following propositions.
Proposition 1. A complete hypersurface of Riemannian curvature 0 is
always a cylinder of the form Mn = Μ1 χ Rn_1, and M1 is a curve in the
orthogonal complement R2 of Rn-1.
Proposition 2. If a complete hypersurface Mn decomposes into a metric
direct product in the form Mn = Mk χ Mn~k, then either Mn has
Riemannian curvature 0, or the Weingarten map vanishes at each point on the
tangent spaces of one of the manifolds.
Proof. Denote by Tp the tangent space of Mn at ρ G Mn and by
Tp = Tp χ Tp the decomposition of the tangent space corresponding to the
direct product decomposition of Mn. Since for a metric direct product the
Riemannian curvatures are also multiplied in a proper way, ЩТ^Т^Х = О
holds. Combining this with the Gauss equation we get
g(X, Α(Τ}))Α(Ί$) = g{X,A{Tl))A{Tl).
This identity can hold for a self-adjoint mapping A only if A has rank 1 or
A vanishes along one of the subspaces X£.
We have to show that if the Riemannian curvature of the surface Mn is
not identically equal to zero, then A vanishes along Тг at each point of the
manifold.
Let ρ € Mn be a point at which R(X, Y)Z φ 0. Then, as we have seen
above, A vanishes on one of the subspaces X£. Suppose that A{T^) = 0
holds. By Gauss's equation, the Riemannian curvature R(2\X, Y)Z of the

310
3. SOLUTIONS TO THE PROBLEMS
manifold Mn~k vanishes, while the curvature R(1\X,Y)Z of the manifold
Mk is different from 0 at p. Since the tensor R^(X, Y)Z is constant along
each copy of Mn~k in the direct product, A(T2) = 0 along the copy of
Mn~k that goes through p. This means that the Riemannian curvature
of Mn~k is equal to 0 identically. It follows also that A(T2) = 0 in every
point q = (<7i, £2) £ Mk χ Mn~k such that the tensor RW is different from
0 at (ft.
It remains to show that A(T2) = 0 also at points q = (91,92) f°r which
Let Vk С Mk be a maximal connected open subset such that R^ = 0
at each point of Vk. The Riemannian curvature vanishes on the part U =
yk χ мп~к с Мп of the hypersurface, and for this reason, the rank of
A is at most 1 here. We have to show that the only eigenvector η of A
corresponding to the nonzero eigenvalue lies in the tangent space T1.
Suppose to the contrary that η ^T1 at some point p. Let N be the
maximal integral manifold perpendicular to η and going through p. Since
the submanifolds AT, Vk, and Mn~k are subspaces (totally geodesic sub-
manifolds with 0 curvature) of the locally Euclidean space [/, the angle
between η and the subspace T1 is constant and nonzero along N. Since
the eigenvalue λ of η does not vanish on the boundary of N either, Α φ 0
on the boundary of N and η ^Τ1 holds at these points as well. In any
neighborhood of such a boundary point, one can find a point q = (91,92)
such that i2^|9l φ 0, and thus the image A(Tq) lies in the subspace T^ ,
while at the boundary points A(Tp) is not contained in j£. For this
reason, the existence of such a boundary point contradicts the continuity of A.
However, such a boundary point does exist obviously, otherwise N would
be complete and its orthogonal projection onto the manifold Vk would give
a complete open subset of Vk. This is impossible because of the definition
of Vk. This proves Proposition 2.
The theorem follows immediately from the two propositions. Indeed, it is
clear that cylinders decompose into a metric direct product, and conversely,
if Mn decomposes into a metric direct product, then either R(X, Y)Z = 0,
and then by Proposition 1 Mn is a cylinder, or A(T2) = 0, and then the
copies of Mn~k are parallel (n — k)-dimensional subspaces. Mn is obviously
a cylinder in the latter case as well. D
Problem G.36. Among all point lattices on the plane intersecting
every closed convex region of unit width, which one's fundamental
parallelogram has the largest area?
Solution. Let r* be the lattice generated by the vertices of a regular
triangle of altitude \. The area of the fundamental parallelogram is equal
to l/(2>/3).
Let r be another lattice such that r is not congruent to r*, but the area
of the fundamental parallelogram of r is also 1/(2л/3). We shall show that

3.4 GEOMETRY
311
in this case, there exists a regular triangle of altitude 1 which shares no
point in common with r. Let P*,Q* and P, Q be the two nearest points in
the lattices r* and r, respectively. Obviously,
1
a = PQ <a = P*Q* = _ .
Let the straight line PQ be horizontal. Place a regular triangle with
altitude 1 onto the plane in such a way that one of its sides lies horizontally
below PQ, while the other two sides go through Ρ and Q, respectively. The
distance of the straight line PQ from the upper vertex of the triangle is
(a\/3/2), and its distance from the base side is (1 — ay/3/2). The distance
between two horizontal ranges of points in r is equal to l/(2\/3a). Since
л ay/3 1 алД л 1
1--2-<^<-^' °<а<^'
the triangle contains no lattice points other than Ρ and Q. Translating the
triangle a little downward, we obtain a triangle of unit width that contains
no lattice points.
It remains to show that r* has a point on every closed convex figure к
of width 1. For this purpose, let us remark that the radius of the maximal
inscribed circle of a convex figure of width 1 is at least 1/3. This can be
proved from the facts that every convex figure is contained in a triangle
or a band circumscribed about the inscribed circle of the figure and that
among the triangles circumscribed about a given circle, the width of the
regular triangle is maximal. As a consequence, the width of к is at most
three times the radius of its inscribed circle. Now we need only remark
that every circle of radius 1/3 contains a lattice point from r*. Therefore,
r* has a point in common not only with every convex figure of unit width,
but also with the inscribed circle of such a figure. D
Problem G.37. Let S be a given unite set ofhyperplanes in Rn, and let
О be a point. Show that there exists a compact set К С Rn containing О
such that the orthogonal projection of any point of К onto any hyperplane
in S is also in K.
Solution. We may assume without loss of generality that О is the origin
of the space Rn and that the normal vectors of the hyperplanes in S
generate Rn (if this latter condition is not fulfilled, then we may add further
hyperplanes to S until we reach the desired property).
Denote by V\ the set of orthogonal projections onto the (n —
^-dimensional linear subspaces of Rn parallel to a hyperplane in S. For г = 2,..., η,
let Vi be the collection of all projection operators, the image of which is
(n — z)-dimensional and can be obtained as the intersection of the images of
some operators from V\. Then, by the finiteness of S and the assumption we
made for the normal vectors of the hyperplanes, Vi is finite and nonempty

312
3. SOLUTIONS TO THE PROBLEMS
for every 1 < г < п. Thus, Vn has one element, Vn = {Pn}, where Pn is the
zero operator. For the sake of unified notation, set Vo = {Po}, where Pq is
the identical transformation. If Pi £ Vi and Pj £ Vj, and the image of Pi
contains the image of Pj, or equivalently, if the kernel of Pi is contained in
the kernel of Pj, then we write Pi > Pj.
The orthogonal projection onto a plane s from S can be given in the form
χ ι—► Psx + ps, where Ps £ V\ is the operator of the orthogonal projection
onto the (n — l)-dimensional linear subspace parallel to s, and ps is the
vector drawn from О perpendicular to s. Obviously, P3ps = 0 for every
seS.
Lemma. One can find a sequence of numbers 0 = Ro < R\ < · · · < Rn
such that for any 1 < j < n, Pj £ Vj, and R> Rj, the set
G{Pj, R) := {xGRn: Ράχ = 0 and ||P^||2 < R2 - R2
for every 0 < г < j - 1, P3 < Pi e Vi}
is mapped into itself by any projection χ \-> Psx + ps onto a plane s € S
for which P3 > Pj.
Since Pn = 0, applying the lemma for j = η and R = Rn, we obtain that
the closed set К = G(0, Rn) is mapped into itself by any projection onto
a plane in S. On the other hand, this set is bounded, for it is contained
in the sphere of radius Rn centered at the origin. Thus, it satisfies the
requirements formulated in the problem.
Proof. We prove the lemma by induction on j. For j = 1, let P\ £ V\
be an arbitrary projection. Then
G(PUR) = {xeRn: P1x = 0,\\x\\ < R}.
If, for some s £ 5, Ps > Pi, then, since the images of the operators Pi
and Ps are (n — l)-dimensional, Ps = Pi. Therefore, Psxq + ps = p3 for
any xo £ G(Pi,R). Thus, if we take supsG5 \\p3\\ for i?i, the claim of the
lemma holds.
Now assume that the lemma has been proved for the values
1,2,..., j — 1 and that the numbers Ro < Ri < · · · < Rj-ι have
already been constructed. Let Pj £ Vj be an arbitrary projection and s £ S
such that Ps > Pj.
If PjXo = 0, then Pj(P3xq + p3) = PjXo + PjPs = 0. Thus, projection
onto s preserves the first defining property of the elements of G(Pj,R).
Second, we show that if R > Rj-ι and ^o £ G(Pj, R), then
\\Pi(PsXo+Ps)\\2<R2-R2 (l)
holds for 0 < г < j - 2 and Pj < Pi £ Vi.
If Ps > Pi, then Pi(Psxo + ps) = Д#о? and there is nothing to prove.
If Ps > Pi, then let us denote by Pi+ι the projection operator whose
image is the intersection of the images of Ps and Pi. Consider the set
G(Pi+1, R*) = G(Pi+1, y/E? - HPi-ixoll2) ·

3.4 GEOMETRY
313
Then
R* = R — ||Pi+i:ro|| > R — {R — Ri+i) = ^i+i'
and therefore, R* > i2»+i. Thus, г + 1 < j — 1, and by the induction
hypothesis, the projection χ ι—► Psx+ps maps G(P;+i, R*) into itself (since
P3 > Pi+1). However, (P0 - Pi+1)x0 e G(Pi+1,R*), since for P<+i(P0 -
Pi+i)xo = 0 and Pi+i <Pk^Vk (k = 0,..., г), we have
||P*(P0 - Рг+ιΜ2 = ||(Pjb - Рг+ιΜ2 = llPfc^o II2 -ЦРг+i^oll2
< it — Лд. — ||i^^_ixo|| = R — Rfc .
Therefore, Ps(Po — Ρ,+ι)χο + Ps = (-fa — -FVh^o + Ps is also contained in
G(Pi+1,R*). Thus,
\\Pi[{P. - PJ+i)xo + Po]||2 < Д*2 - Rf,
that is,
\\Pi(PsXo+Pa) - Pi+ixo\\2 <R2- ЦРг+i^oll2 - R2,
from which we get (1).
To prove the Lemma we have only to show that if R > Rj-ι is sufficiently
large, then (1) holds for г = j — 1 and any xo £ G(Pj,R) and any Pj-i £
Vj-ι such that Pj-ι > Pj.
If P3 > Pj-i, then we are done because P7_i(Ps^o + p3) = Pj-\Xq.
If Ps > Pj-i, then the intersection of the images of Ps and Pj-\ is
(n — j)-dimensional and consequently coincides with the image of Pj.
We show that Pj-iPs is a contraction on the kernel of the operator
Pj. For this purpose, it suffices to show that if χ φ 0, PjX = 0, then
||Pj_iPe:r|| < ||ж||. The assumption that the two sides of the latter
inequality are equal yields ||Рвж|| = ||ж||, from which we obtain Psx = x. Repeating
the same argumentat yields Pj-ix = x- Therefore, χ is in the image of Ps
and P/-1, that is, in the image of Pj. However, this contradicts χ φ 0 and
PjX = 0. This means that Pj-iP3 is indeed a contraction, that is, there
exists a number 0 < q < 1 such that PjX = 0 implies ||Pj_iPsx|| < q\\x\\.
Now let us choose the numbers Tj = Tj(Ps, Pj-i, Pj) in such a way that
qR + R1 < yjRt-R'*^, when R>Tj. (2)
(This is possible since if both sides are divided by R, the left-hand side
tends to q and the right-hand side tends to 1 as R —> oo.)
Now, if R > Tj and xq G G(P/, Д), then
||Ρ,·_ι(Ρ^ο +ρθ)|| < 9||x0|| + \\Pa\\ <qR + Ri;
therefore, by (2), (1) holds for г = j — 1. Now let Rj be the maximum of
the numbers Tj(Ps,Pj-i,Pj) for all Ps,Pj-i,Pj. Then we obtain that (1)
holds for R> Rj, with г = j — 1, if
PjeVj, x0eG(Pj,R), Pj < Pj_i eVj-U
seS and P3 > Pj.
This completes the proof of the lemma. D

314
3. SOLUTIONS TO THE PROBLEMS
Problem G.38. Let к and К be concentric circles on the plane, and
let к be contained inside K. Assume that к is covered by a unite system
of convex angular domains with vertices on K. Prove that the sum of the
angles of the domains is not less than the angle under which к can be seen
from a point of K.
Solution. Let О be the common center of the circles, and r and R the
radius of к and K, respectively. Let us define a function F on the interior
of /с as follows. If the distance from О to Ρ is p, then set
If Τ is a measurable set in the interior of k, then let
m(T)= I F{P)dP.
The function m is obviously a measure on the interior of k. We show that
m has the following property:
If д and h are half-lines starting from a point A of К such that д is
tangent to k, h intersects k, and the angle between h and д is a, then the
angular domain bounded by д and h cuts the interior of /с in a domain T,
the measure m of which is a.
Figure G.29.
Let us compute ra(T) by successive integration. Let e be the half-line
starting from О perpendicular to h. Let us introduce a polar coordinate
system on the plane, with polar axis e. Then a point P(p, φ) belongs to Τ

3.4 GEOMETRY
315
if and only if d < ρ < r, — ϋ < φ < ϋ, where d = pcosu = Rsin(e — a) is
the distance of О from /ι, ε is the angle between the half-lines АО, and g
(see Figure G.29). Based on these, we get
/» /»r /»arccos(d/p)
m(T)= / F(P)dP= / / ί(ρ)ράψάρ
JT Jd 7-arccos(d/p)
= / 2/(p)parccos(-)dp
Jd Ρ
Γ ο« ^ /^8ΐη(ε-α)\
= / 2/(p)parccos I I dp.
jRsm(e-a) \ Ρ J
Therefore, we have to show that, for 0 < α < 2ε, we have
Г «г/ ч /R-sin(e-a)\ .
= / 2/(p)parccos I I dp = a.
JRsm(e-a) \ Ρ )
Substituting R · sin (ε — α) = ί, this means that
/
2/(p)parccos(-) dp — ε — arcsin( —)
t Ρ R
if — r < t < r. This relationship is obviously true for t = r. Thus, it is
enough to show that the derivatives of the two sides with respect to t are
the same:
J KHJH : dp = . -Γ<ί<Γ
у/(?=1* VR2-t2'
i;
[rVMLdp= f2-^-2l2 R\:^dP
1 , /i?2-r2p2-i2
π ^Д2 - ί2 V Д2 - ί2 г2 - Ρ2
р=Г
1
Vi?2 - t2 '
p=t
The proposition proved, combined with the additivity of the measure m,
yields immediately that if both half-lines bounding a convex angular
domain of angular measure α intersect k, then the m measure of the
intersection of the angle and к is equal to a (since the intersection can be obtained
as the difference of two segments of k).
Now consider a covering of к with a finite number of convex angular
domains. Let us denote by αϊ,..., an the angular measure of the angles,
and by Ti,... ,Tn the intersections of the domains with k. Then, by the
proposition just proved, a* > m(Ti), г = 1,... ,n. (Equality holds here if
and only if the both boundary lines of the ith angle have a point in common
with k.) Therefore, by the subadditivity of measures,
oti + · · · + an > m(Ti) + · · · + m(Tn) > m U T{ ) = 2ε
СИ -
(since υΓ=ι -^ ^^s ^ne interior of к and the m measure of к is exactly 2ε).
This completes the proof. D

316 3. SOLUTIONS TO THE PROBLEMS
Problem G.39. Let ξ(Ε, π, Β) (π : Ε —> Β) be a real vector bundle of
unite rank, and let
ΤΕ = νξθΗξ (*)
be the tangent bundle of E, where νξ = Ker dn is the vertical subbundle
of те- Let us denote the projection operators corresponding to the splitting
(*) by υ and h. Construct a linear connection V on νξ such that
VXWY-VY\/X = v[X, Υ] - v[hX, hY].
(X and Υ are vector fields on E, [.,.] is the Lie bracket, and all data are
of class C°°).
Solution. Let C°°(E) be the ring of differentiable (of class C°°) functions
Ε -> R, and denote by X(E) and XV(E) the C°°(£)-module of all vector
fields on Ε and the submodule of vertical vector fields, respectively. A
mapping
D: XV(E) x XV(E) -> XV{E), (Z,W) н-> DZW
will be called a pseudoconnection on νξ if it has the formal properties of
linear connections, that is, it is C°°(i£)-linear in Ζ and an R-linear
derivation in W (DzfW = (Zf)W + fDzW, f e C°°{E)). Assume that D is a
pseudoconnection such that
DZW - DwZ = [Z, W] for every Z, W G XV{E\
and define the mapping
V: X(E) x XV(E) -+ XV(E), (X,Y) ~ VXY
by the formula
VxY:=DvXY + v[hX,Y].
Then V is a linear connection. It can be seen immediately that V is R-
linear in X and Υ and that for / G C°°(E) we have
VxfY = fDvXY + (vX)fY + fv[hX, Y] + v(hX)fY
= fVxY + (vX + hX)fY = fVxY + (Xf)Y
(applying linearity at each point and vY = Y). Finally, a similarly simple
computation shows that VjxY = fVxY. We show that V satisfies the
required relationship. Indeed,
VxvY - VYvX = DvXvY - DvYvX + v[hX, vY] - v[hY, vX]
= [vX, vY] + v[hX, vY] - v[hY, vX]
= υ([υΧ, vY] + [hX, vY] - [hY, vX])
= v([X, Y] - [hX, hY]) = v[X, Y] - v[hX, hY].

3.4 GEOMETRY
317
It remains to show that an "auxiliary" pseudoconnection D does exist.
We shall prove this by a simple local construction. Assume dim Β = η,
rank ξ = r. By the local triviality of the vector bundle, each point of
В has a neighborhood U such that π-1([7) —> U is diffeomorphic to the
trivial bundle U χ Rr -> Rr. If φ: π-1 (С/) -> С/ х Rn is a vector bundle
isomorphism, (иг)™=1 is a local coordinate system on [/, and (Za)£=1 is the
dual basis of the canonical basis of Rr, then
хг := иг ο π (1 < г < η),
уа := la о рг2 о φ (1 < α < г)
is a local coordinate system on π~ *([/), and a direct computation shows
that the vector fields д/дуа form a local basis of Xy(E). Let us define the
pseudoconnection D by the requirement
D д =0 (Ka,/3<r).
(It follows from standard methods of the theory of linear connections that
D can be defined this way.) Then for any two vector fields
over 7г-1([/), we have
aw? д
DzW = Za-
дуа дуР '
and consequently, DzW — DyyZ = [Z, W], which completes the proof. D
Problem G.40. Consider a latticelike packing of translates of a convex
region K. Let t be the area of the fundamental parallelogram of the lattice
denning the packing, and let tm-m(K) denote the minimal value oft taken
for all latticelike packings. Is there a natural number N such that for any
η > N and for any К different from a parallelogram, ntm-m(K) is smaller
than the area of any convex domain in which η translates of К can be
placed without overlapping ? (By a latticelike packing of К we mean a set
of nonoverlapping translates of К obtained from К by translations with all
vectors of a lattice.)
Solution. We show that such an N does not exist. Let h be the branch of
the hyperbola defined by the equation xy = 1 that lies in the first quadrant
of the plane. Let A be the intersection point of the rr-axis and the tangent
of h at (a, 1/a), and let В be the intersection point of the y-axis and the
tangent of h at (1/a, a). Let us replace the arcs of the hyperbola "beyond"
(a, 1/a) and (1/a, a) with the segments of the tangents between A, (a, 1/a)

318
3. SOLUTIONS TO THE PROBLEMS
and Β, (1/α, a), respectively. Denote by g the curve consisting of the two
segments and the arc of the hyperbola between them. Then the area of
any triangle bounded by the coordinate axes and a tangent of h equals 2.
On the other hand, if χ —> oo, then the area of the domain between the
coordinate axes and g tends to infinity as well. Therefore, we can choose a
in such a way that the ratio of the two areas is equal to an arbitrarily small
ε. Let us round two opposite corners of the unit square with the help of
two suitably scaled copies of the arc g cutting off the square two domains
of area δ. The area of the rounded square q equals 1 — 26. Since q is central
symmetric, tm[n(q) is the area of the circumscribed hexagon of q having
minimal area, that is, tm-m(q) = 1 — 2εδ. On the other hand, η translates
of q can be packed into a domain of area η — 26. Thus, for any n, we can
choose an ε > 0 such that en < 1, and then ntm[n(q) = η — 2εδ >η — 2δ. Π
Problem G.41. Show that there exists a constant c& such that for any
unite subset V of the k-dimensional unit sphere there is a connected graph
G such that the set of vertices of G coincides with V, the edges of G are
straight line segments, and the sum of the kth powers of the lengths of the
edges is less than c^.
Solution 1. Let G be a connected graph such that the set of vertices of G
is V, the edges of G are straight line segments, and the sum of the length
of the edges is minimal. By the characterization of G, G is a tree. Take an
open ball about the midpoint of every edge of G with radius r(y/S — l)/4,
where r is the length of the edge. We prove that these spheres do not
intersect one another. Assume, to the contrary, that the segments AB and
CD are edges in G, p(A, B) = R, p(G, D) = r, and
S((A + B)/2, R{VS - l)/4) Π S((C + D)/2, r{Vs - l)/4) φ 0,
where ρ is the distance and 5(a, r) is the open ball of radius r centered at
a. By symmetry, we may assume R> r. Then
p(A, (C + D)/2) < р(Д (A + B)/2) + p((A + B)/2, (G + D)/2)
< R/2 + R(y/S - l)/2 = RVS/2.
Since at least one of the angles (A, (G + D)/2, C) and (A, (G + D)/2, D) is
not obtuse if Α φ (G + £>)/2, we get
min{p(A G), p(A, D)} < (p2(A, (G + D)/2) + r2/4)1/2
<(m2/± + R2/£)iri = R.
If the edge AB is removed from the graph G, then the subgraph is the
union of two trees. By symmetry, we may assume that В is contained in
the same component as G (and D as well). But then removing from G the
edge AB and adding to it the shorter of AC and AD, the length of which is
less then R, we obtain a connected graph G satisfying all the requirements
for G, but the sum of the length of its edges is less then the sum for G,
and this is a contradiction.

3.4 GEOMETRY
319
Since the balls we constructed are disjoint and are contained in a sphere
of radius 2, their total volume is less than the volume of the ball of radius
2. Thus,
А В is an edge in G
where шк is the volume of the Ar-dimensional unit ball. Therefore,
Σ рк{А'в)^{тЬ\)2к=Ск'
AB is an edge in G ^ * '
as was to be proved. D
Solution 2. It suffices to show the following claim:
Claim. If Τ is a box such that the ratio of any two edges of Τ is less
than or equal to 3, and V С Τ is a finite set of points, then there exists a
connected graph G = G(V) set of vertices V such that the sum of the kth
power of the edges of G satisfies
S(G) <ck Vol (T),
where ck is a constant depending only on k, Vol (T) is the volume of T.
Using induction on the number of points in V, we shall show that one
may choose ck = 3k+1kk/2. Let |V| = n. The statement is true for η = 1,2,
and assume it has already been proved for 1,2,..., η — 1. Let us shrink T,
that is, move the faces of Τ inside as long as the assumptions on Τ are
not violated and Τ contains V. This way, we make the inequality to prove
sharper. If Τ cannot be shrunk further and e is one of the longest edges of
T, and furthermore, Qi and Q2 are the faces perpendicular to e, then V
has a point on both Qi and Q<i (otherwise, we could shrink Τ further). Let
Ti, T2, and T3 be the consecutive boxes obtained by cutting Τ into three
equal parts by hyperplanes perpendicular to e. We distinguish two cases
depending on whether T2 contains a point of V or not.
Case 1. V has no point in T2. Let Vi = V Π T2, V3 = V Π Τ3. Since
Qi^V ψ 0 for г = 1,2, we have \Vi\ < n. Applying the induction hypothesis
for Ti,Vi and T3,V3, we obtain connected graphs G\ = G{V\) and G3 =
G(V3), for which S(Gi) < cfcVol(Ti) = (l/3)Vol(T), г = 1,3. Connecting a
point of G\ to a point of G3, we obtain a connected graph G such that
S(G) < Sid) + S{G3) + {Vk)k\e\k
< ^cfcVol(T) + (Vk)k3kVo\(T) < cfcVol(T)
by the choice of ck.
Case 2. V has a point in T2. Choose χ £ V П T2 and cut Τ into two
closed boxes Tx* and X^ by a hyperplane passing through χ perpendicular
v^-l
Uk,

320
3. SOLUTIONS TO THE PROBLEMS
to e. Then T* also has the property that the ratio of any two of its edges
is not more than 3. Thus, the induction hypothesis can be applied again to
T\*, T\* Π V and T2*, Τ2* Π V since the sets T\* Π V and T2* Π V have less than
η points. Therefore, one obtains the connected graphs G\ = G(V Π Τ\*)
and G*2 = G(V Π Τ2*) such that
5(G;)<cfcVol(7;*) (i = 1,2).
Setting G* = Gj U G2 yields a connected graph on the set of vertices V
and
5(G*) = S(G;) + 5(G3) < cfcVol(Ti) + cfcVol(T2*) = cfcVol(T). D
Remarks.
1. We may choose for G a Hamiltonian circuit as well, but the proof of this
fact is more difficult.
2. One can prove that c2 = 4 is the smallest suitable constant and that
it is good for the Hamiltonian circuit problem also. It is an interesting
problem to find the exact constants for к > 2.
Problem G.42. Let us draw a circular disc of radius r around every
integer point in the plane different from the origin. Let Er be the union
of these discs, and denote by dr the length of the longest segment starting
from the origin and not intersecting Er. Show that
lim(dr ) = 0.
г—►О Г
Solution. Let the second endpoint (ж, у) of one of the longest segments be
on the circumference of the disc centered at (fc, n). Reflecting the segment
in the point (fc/2, n/2), we obtain that the segment connecting the points
(k — x,n — y) to (fc, n) has a point in common only with the circle centered
at (fc, n). Since the points (0,0), (k — χ, η — у), (χ,у), and (к,η) are
vertices of a parallelogram, the two other sides of which have length r,
this parallelogram contains no lattice points in its interior. The segment
connecting the origin to (fc, n) has a point in common only with the disc
about its endpoint. The length of this segment differs from dr by at most
r.
Let Dr be the length of the longest segment connecting the origin to a
lattice point and having a point in common only with the disc about its
endpoint. As we observed, | Dr — dr |< r. According to this, it suffices to
show that
lim (Dr--) = 0.
For this, we shall apply the well-known theorem of lattice geometry saying
that a triangle that has lattice points for its vertices but contains no further

3.4 GEOMETRY
321
lattice points in its interior and on the boundary (except for the vertices)
has area 1/2.
Let V = (n, k) be the endpoint of a segment of length Dr, and let
Η be the domain containing those points of the plane whose orthogonal
projection onto the straight line OV lies in the closed segment OV. Since
translates of Η with integer multiples of the vector (n, k) cover the whole
plane, the distance of any lattice point Ρ £ Η from the straight line OV
is at least r if Ρ is different from О and V. Let Q be a lattice point in Η
different from О and V such that the distance of Q from the straight line
OV is minimal. Then there are no further lattice points on the segments
OQ and OV, nor in the interior of triangle OVQ, because these points
are closer to OV than Q. Furthermore, there are no lattice points in the
interior of the segment OQ since every disc of radius r centered at a lattice
point different from О and V is disjoint from the segment OV. Therefore,
the area of triangle OVQ is equal to 1/2. Consequently, the distance of
Q from the straight line OV, that is, the altitude of triangle OVQ, is
1/ | OV |. We find that the distance of Q from OV is at least r if and only
if | OV |< 1/r.
This yields another characterization of Dr: Dr is the length of the
longest segment connecting О to a lattice point without going through
further lattice points (that is, the coordinates of the endpoint are relative
primes) and having length at most 1/r.
By this, Dr < 1/r, of course.
It is well known that for any ε > 0, if η is large enough, then there is a
prime number between η and n + en.
Let 0 < ε < \/2 — 1, and choose a prime ρ satisfying
1 1 л
<p< 1.
(l + e)r
Such a prime exists by the theorem mentioned above, provided that r is
small enough.
Let q be the largest nonnegative integer such that
Р2+Я2<\-
Then q < ρ since if we had q > p, then by
< (Λ ,1 λ <Ρ, p2 + q2>2p2>^
V2r (l + e)r " У ч ~ y r2
would hold. On the other hand, q > 1, as for r < 1, p2+l2 < (1/r—1)2+1 <
1/r2. Thus, 1 < q < p, and since ρ is a prime, ρ and q are relative primes.
This implies that the segment connecting the origin to (p, q) is suitable.
Therefore
2
D2 > p2 + q2 > p2 +

322
3. SOLUTIONS TO THE PROBLEMS
that is,
We conclude that for any 0 < ε < \/2 — 1, and for any sufficiently small r,
we have
-- Wl- (Λ , χ2 <Dr <-■
Since
the proposition follows immediately. D
Problem G.43. We say that the point (01,02,^3) is above (below)
the point (&1,&2,&з) if οί = &i, fl2 = &2 and аз > 6з (аз < &3). Let
ei, б2, · · ·, в2к (к > 2) be pairwise skew lines not parallel with the ζ-axis,
and assume that among their orthogonal projections to the (x, y) -plane no
two are parallel and no three are concurrent. Is it possible that going along
any of the lines the points that are below or above a point of some other
line ei alternately follow one another?
Solution. Assume that it is possible. From this, we derive a contradiction.
Let P{j be the point of e*, which is above or below the line e3 ; (1 <
h J < 2/c, г φ j). In the first case, we call Pij an upper point, and in the
latter case we call it a lower point. In the text below, by a point of ei we
mean one of the points Pij. Those two points of e^ that have all points of
ei on one side will be called the "ends" of e^. Two points of a line will be
called neighbors if there is no point of the line between them.
Lemma. Every straight line contains at least two points that are above
or below an end of another line.
Proof. Obviously, it is enough to consider the straight line e\. We may
also assume that e\ is horizontal (that is, parallel with the (x, y)-plane),
since this can always be managed applying a transformation of the form
(x, y, z) 1—► (x, y,z + ay + βζ), which leaves the location of upper and lower
points unchanged. Let S be a plane perpendicular to ei, and denote by
/2, /3? · · ·, /2Α; the orthogonal projections of the lines ег, ез,..., e2k onto
S (see Figure G.30). Let us introduce a coordinate system on S having
horizontal and vertical axes and take from among /2, /3,..., j^k the ones,
say /2 and /3, whose slope is minimal and maximal, respectively. We claim
that if e2 is below (above) ei, then each point of it lies on the negative
(positive) side of e\ with respect to the positive direction of the horizontal
coordinate axis. Assume that this is not so and, for example, e<i is below
e\ and has a point on the positive side of e\ (the other cases can be treated
analogously). Let P%Q^ be the nearest to Р2Д among these points. This

3.4 GEOMETRY
323
is an upper point since it is a neighbor of a lower point. The point Р2)г0
below it lies on the straight line e%Q. It is easy to see that one can find a
sequence of indices zi,..., in such that
1. in = 1;
2. the points Ρ%ά^ά_λ and Pij,ij+1 are neighbors on e^. (j = 1,2,..., η — 1);
3. Pij,ij+1 is in the negative direction from Pijiij_1 (j = 1,2,..., η — 1).
Figure G.30.
It is also easy to see, however, that in this case, the slope of one of
the straight lines /ix,..., fin_1 is less than the slope of /2, and this is a
contradiction.
This way, we proved that one of the ends of e2 and ез is above or below
e\. Therefore, the lemma holds.
We remark that there cannot be more than two ends above or below
a straight line, because it would mean that the 2k straight lines would
have more than 4k ends. Therefore, there are exactly two points on every
straight line that are above or below an end of another line, and these latter
lines are the ones giving the minimal and maximal slopes after the above
transformation and projection.
Let the ends of e\ be above or below e2 and ез, respectively. Since the
straight line contains 2k — 1, that is, an odd number of points, either both
e2 and e3 are above ei or both of them are below e\. We may assume
that they are above e\. We also suppose that e2 and e3 are horizontal
since this can be reached by applying a suitable transformation of the form
(:r, у, ζ) ι—► (χ, у, ζ + αχ + Py). Let S be a plane perpendicular to e2, and
introduce a coordinate system on S as shown in Figure G.31.
By the previous remark, among the orthogonal projections of the straight
lines ej(j φ 2) onto 5, the projection of e\ has maximal slope, e\ has
maximal slope, since e\ goes below e2 and has a point in the positive direction
from Pi>2, and since the projection of ез is horizontal, that of e\ has
positive slope. This implies that P\^ is higher than P\2 (its z-coordinate is

324
3. SOLUTIONS TO THE PROBLEMS
Figure G.31.
greater). Repeating the same argument but changing the role of e2 and ез,
we obtain that P1)2 is higher than P13. However, these two implications
contradict one another.
We conclude that the indirect assumption is false: upper and lower
points cannot follow one another alternating on each straight line. D
Problem G.44. Suppose that HTM is a direct complement to a
vertical bundle VTM over the total space of the tangent bundle Τ Μ of the
manifold M. Let υ and h denote the projections corresponding to the
decomposition TTM = VTM Θ HTM. Construct a bundle involution
Ρ : TTM —> TTM such that Poh = voP and prove that, for any pseudo-
Riemannian metric given on the bundle VTM, there exists a unique metric
connection V such that
VXPY - VYPX = Po h[X, Y]
if X and Υ are sections of the bundle HTM, and
VXY-VYX = [X,Y]
if X and Υ are sections of the bundle VTM.
Solution. As usual, call the sections of the bundles VTM —> TM and
HTM —> Τ Μ vertical and horizontal vector fields on TM, and denote
their modules by XyTM and XjjTM, respectively. If X(TM) denotes the
module of vector fields in TTM —> TM, then
X(TM) = XVTM Θ XHTM,
and, for simplicity, the projections corresponding to this decomposition are
also denoted by υ and h.

3.4 GEOMETRY 325
Let (С/, (и1,..., un)) be a chart for Μ, and consider the induced chart
(π-1(ί/),(χ1,·..,χ";2/1,.·.,2/η))
for TM, where π : TM —> Μ is the projection of the tangent bundle, and
xi = uio π, y>) = v{ul) (v e TM, 1 < i < n).
When this chart is fixed, there exist unique smooth functions
N* :k-1(U)-*R (l<t,fe<n)
such that the vector fields
form a local basis in XjjTM. Then
д δ
0 -τ,-^τ ) (1<г<п)
is a local basis in jC(TM), and on the coordinate patch π 1(C/) every vector
field X e X(TM) can be written in the form
X — X{ — |- Yi
δχ* ду{'
where the first term is horizontal and the second is vertical.
Define the map Ρ by the formula
6хг дуг 6хг дуг
An easy calculation using the above formulas shows that this map is
independent of the choice of the chart. From the formula, it is immediate that
P2 = 1 and Ρ ο h = υ ο P. It is also clear that Ρ is a bundle involution.
Since X(TM) = XVTM θ ХнТМ, and a connection V : X(TM) χ
XyTM —> XyTM is linear over the function ring in its first variable, it
suffices to define V on the summands XyTM χ XyTM and ХнТМ х
XyTM.
Case 1. Consider the map F : (XyTM)3 -> XyTM defined by
F(X, У, Ζ) =Χ(Υ, Ζ) + Y(Z, X) - Z(X, Y)
-(X,[y,Z]) + (y,[Z,X]) + (Z,[X,y]),
where (., .) denotes the pseudo-Riemannian metric given on the vertical
bundle. For fixed X and Υ, the map Ζ \—> F(X, Υ, Ζ) is linear over the

326
3. SOLUTIONS TO THE PROBLEMS
function ring, and, since the form (., .) is nondegenerate, there exists a
unique VXY G XyTM such that
(2VXY,Z) = F(X,Y,Z)
for all Ζ G XyTM. An argument analogous to the proof of the theorem
on Levi-Civita connections shows that the map (X,Y) ·—► VχΥ satisfies
all requirements.
Case 2. The restriction of the connection V to XjjTM χ XyTM is
defined by the requirement that
(2Ус/У, Z) = U(Y, Z) + PY(Z, PU) - PZ(PU, Y)
- {PU, Ρ ο h[PY, Ρ Ζ]) + (У, Р о h[PZ, U])
+ (Z,Poh[U,PY])
hold for all U G ХнТМ, Υ, Ζ G XyTM. A simple but lengthy calculation
shows that this restriction of the connection also has all required properties.
In order to prove its uniqueness, suppose that the map
V : XHTM χ XyTM -> XyTM
also fulfills the requirements of the problem. Then, for all U G ХнТМ,
Υ, Ζ G XyTM, we have
u(y, z) + py(z, pu) - pz(pu, y) (=} (v^y, z) + (у, vvz)
+ (vpyz, pu) + (z, vpypu)
- (Vpzpu,y) - (pu,vpzy)
=2(V'UY, Z) + (VPYPU - V'uY, Z)
+ (V'uZ - VPZPU, Y) + (VPYZ - VPZY, PU)
{^2(VuY,Z)-(Poh[U,PY],Z)
- (Ρ ο h[PZ, C/], Υ) + (Ρ ο h[PY, PZ\ PU),
where at (1) compatibility of V with the metric is used, and at (2) the
assumption VXPY - VYPX = Po h[X,Y] is used. Therefore, it follows
that
(VUY,Z) = (VUY,Z),
that is,
{VVY - Vc/У, Z) = 0 for all Ζ G XyTM.
Since the form (., .) is nondegenerate, this implies that V' = V, which
completes the proof. D

3.4 GEOMETRY
327
Problem G.45. Let A be a finite set of points in the Euclidean space
of dimension d > 2. For j = 1,2, ...,d, let Bj denote the orthogonal
projection of A onto the (d— l)-dimensional subspace given by the equation
Xj =0. Prove that
f[\Bj\>\A\d-\
i=i
Solution. We use induction on the cardinality of the set A. If \A\ = 0 or
1, then the statement is obvious.
If | A\ > 1, then the projection of A onto a suitable coordinate axis
contains at least two distinct points. Without loss of generality, we can assume
that this coordinate is the dth. This means that a suitable hyperplane with
equation xd = α divides A into two disjoint nonempty subsets: A = XUY.
Let Cj and Dj denote the projections of X and У, respectively, onto the
hyperplane Xj = 0; then \Cj| + \Dj\ = \Bj\ for each j e {1,..., d — 1}, and
\Cd\ < \Bd\ and |Ai| < \Bd\.
By the induction hypothesis, we have
d-l
Μ'-^ΐΒϋΐ-Πι^ι and
d-l
Using the inequality
{αϊ ak)i + (fti bk)i < j J^(a, + 6,)
which is an equivalent form of the usual inequality between arithmetic and
geometric means, we have
Π IB/^-i) = |Д,|1/«-Ч · Π (\Cj\ 4- l^l)1^
3=1 3=1
>ι^ι1/^-1)· (Πic.i1^-1) + Π|х?,-|А
\3=1 3=1
>\X\ + \Y\ = \A\,
that is,
which is what we wanted to prove. D

328
3. SOLUTIONS TO THE PROBLEMS
Problem G.46. Let A[',..., An be a sequence of η > 3 points in the
Euclidean plane R2. Define the sequence A^\ ..., An > (i = 1,2,...) by
induction as follows: let Aj be the midpoint of the segment A^~ Α^~λ \
where ^+1 = ^i · Show that, with the exception of a set of zero
Lebesgue measure, for every initial sequence (A[ \ ..., An ) £ (R2)n, there
exists a natural number N such that the points A ι \ ..., An ^ are
consecutive vertices of a convex n-gon.
Solution. If R2 is naturally identified with the complex plane C, then our
sequences can be considered as vectors in the space Cn, and the induction
step corresponds to the linear transformation
Д*1,-.*п)=^—2~ '···'—2~ )■
In Cn there exists a basis consisting of eigenvectors of L: we have
L(ej) = Xjej (j = l,2,...,n)
for the vectors ej = (Ι,ε·7, ε2·7,... ,e'n_1^) and scalars Xj = (1 + Sj)/2
(j = 1,..., n), where ε = е27гг/п is a primitive nth root of unity.
All vectors ν in Cn can be written in the form ν = Σΐΐ=ιαίβ3- Since
en = (1,1,...,1), adding the vector anen means a translation of our system
of points in the original plane R2 by a fixed vector. Therefore, it suffices to
consider the problem restricted to the subspace generated by the vectors
ei,..., en_i. Suppose that
Applying the induction step N times, we obtain
An elementary calculation shows that |Aj| < |λι| for 1 < j < η — 1, and
|λη-ι| = |λι|. So, for 1 < j < η - 1, we have Ιλ^/ΙλιΙ^ —> 0 when
N —> 0. Therefore, it suffices to show that for almost all υ the vector
a\e\ +an_ien_i corresponds to successive vertices of a convex n-gon. This
is indeed the case if |αχ| Φ |an-i|j because e\ is a vector corresponding
to successive vertices of a regular n-gon, and the vector a\e\ + an-ien-i
corresponds to the sequence of points obtained from successive vertices of
this regular n-gon by applying the linear map ζ \-> a\z + an-\z in R2. This
transformation is nondegenerate, since a\z + an-\z — 0 can only hold when
z = 0or|ai| = |an_i|, and thus it takes convex n-gons into convex n-gons.
To summarize the property required in the problem is true for all vectors
Σ?=ι о»з^з of Cn except when |αχ| = |an-i|? and this exceptional set does
indeed have zero Lebesgue measure. D

3.4 GEOMETRY
329
Problem G.47. Suppose that η points are given on the unit circle so
that the product of the distances of any point of the circle from these points
is not greater than two. Prove that the points are the vertices of a regular
n-gon.
Solution. Let the circle be the unit circle around the origin in the complex
plane. Let zi, z<i,..., zn be the complex numbers that represent the points.
By a suitable rotation, we may assume that z\ · Z2 zn = 1.
Consider the following polynomial:
P(w) = (w- zi){w -z2)-"(w- zn)
= wn + ащ)71'1 + · · · + an-iw + 1 = wn + Q(w) + 1.
Then \P(z)\ is the product of the distances of the point represented by the
complex number ζ from the given points. So, if ζ is a complex number of
absolute value 1, then \P(z)\ < 2.
Let wi,W2,-..,wn denote the nth roots of unity. It is well known that
Wi + w\ + · · · + w% = 0 for all к = 1,2,..., η — 1. This implies that
Q(wi) Η l· Q(wn) = 0. If Q(w) is not identically zero, then, for some j,
the complex number Q(wj) is different from zero and has nonnegative real
part. Then |P(u?j)| = |2 + Q(wj)\ > 2, which contradicts the assumption.
Therefore, the polynomial Q is identically zero, that is, P(z) = zn+l. Then
the roots ζι, Ζ2, · · ·, zn of the polynomial P(z) form a regular n-gon. D

330
3. SOLUTIONS TO THE PROBLEMS
3.5 MEASURE THEORY
Problem M.l. Let f be a finite real function of one variable. Let Df
and D_f be its upper and lower derivatives, respectively, that is,
■τΛΐ./ ч ,. f(x+h)-f(x-k) „„ ч ,. . f(x+h)-f(x-k)
Df (x) = \imsup^ {—^ '-, D/(g)=Hminf/v ! Ί -.
h,k>0 h,k>0
h+k>0 h+k>0
Show that Df and D_ f are Borel-measurable functions.
Solution. It is obviously sufficient to prove the assertion for Df. It is also
clear that
{x : ~Df{x) > с} = и П Aik
i=lk=l
for all real values c, where Aik denotes the union of intervals [a, b] shorter
than 1/fc satisfying (f(b) — f(a))/(b — a) > c+ l/i. We show that Aik is
a set of type FG which implies the assertion.
More generally let {/7} (7 Ε Γ) be a system of open intervals, and let
A = и7£г^7 (where 77 stands for the closure of ΙΊ). Obviously,
A = BUCUD,
where Β = υΊ^γΙΊ, С is the set of all points that are left-hand endpoints
of at least one ΙΊ but are not interior points of any of them, and D is the
set of those points that are right-hand endpoints of at least one ΙΊ but do
not lie in the interior of any ΙΊ. Clearly, В is an open set. If χ e C, let rx
be a rational point in the interior of the interval ΙΊ that has the left-hand
endpoint x. Points rx corresponding to distinct points χ are distinct since
in the case χ < у, rx = ry, the point у would lie in the interior of the
interval ΙΊ starting at x. Consequently, C, and similarly D, is countable.
So each of the sets В, С, Д and therefore A as well is of type FG. D
Remark. It is interesting that for the left-hand and right-hand
derivatives a similar theorem does not hold. For instance, if Ε is a nonmeasurable
set that along with its complementary set is dense, and f(x) is the
characteristic function of E, then D f(x) = 0 or +00 if χ e Ε or χ φ Ε,
respectively, and so D f(x) is nonmeasurable in this case.
It should be noted that in the book by S. Saks (Theory of the Integral,
Hafner, New York, 1937, Vitali's covering theorem, pp. 112-113), the
author proves only Lebesgue measurability of D f and D_ f with much more
powerful tools, although generalized to arbitrary dimension.

3.5 MEASURE THEORY
331
Problem M.2. Let Ε be a bounded subset of the real line, and let Ω
be a system of (nondegenerate) closed intervals such that for each χ G Ε
there exists an I G Ω with left endpoint x. Show that for every ε > 0 there
exist a unite number of pairwise nonoverlapping intervals belonging to Ω
that cover Ε with the exception of a subset of outer measure less than ε.
Solution. Let Δ be a bounded open interval containing E. Denote by Sn
the set of all points of Δ having neighborhoods not containing points of Ε
that are starting points of intervals belonging to Ω and longer than 1/n.
Obviously Sn is an open set, Sn+i С Sn and, furthermore, (C\™=lSn)C\E = 0
since at each point of Ε there begins at least one nondegenerate interval
belonging to Ω and if — is already less than the length of the latter, then
η
the point cannot belong to Sn. It is a well-known property of the outer
measure that for any set A the outer measure λ* (Χ Π A) as a function of
X is a measure on the Lebesgue-measurable sets. Thus, taking A = Ε and
using the relation £n+i С Sn as well as the finiteness of the outer measure
of E, we obtain
lim X*(Sn ΠΕ) = λ*(( η Sn) ΠΕ) = λ*(0) = 0.
η—>·+οο η=1
Fix n0 so that λ*(5ηο Π Ε) < ε/2. If the set Ελ = Ε - (Sno Π Ε) is
nonempty, which we may assume, then in each neighborhoood of any of
its points there is a point of Ε that is also a point of E\ and at which an
interval longer than 1/no begins. There is a point а\ е E\ such that
А>((-°°,а1)ПД1)<2(яоЛ(д) + 1),
and by the former arguments we may assume at once that some interval
[αι,δι] starting from a\ and belonging to Ω is longer than 1/no- Then we
choose (if still possible) a point α>2 G E\ such that а>2 > Ь\,
А-Цб^ПЦ) <2(ηοΛ(£Δ) + 1),
and
[u2j 62] £ Ω, b2 — α2 > —,
По
and so on. The procedure ends after a finite number of steps, namely
we can construct at most ηολ(Δ) + 1 intervals [α^,δ*] in this way (since
bk—ak> l/no)· What is left out from Ε by these intervals is covered by
SnonE and the not more than ηολ(Δ) + 1 sets (6fc_i, α^)Γ\Ει (bo = —00) of
outer measure not exceeding ε/(2(ηολ(Δ) +1)). Their total outer measure
is less than ε, which completes the proof. D

332
3. SOLUTIONS TO THE PROBLEMS
Remarks.
1. In addition to being nonoverlapping, the intervals we have chosen are
disjoint.
2. The assertion cannot be replaced by the stronger statement that it is
possible to choose an at most countable number of nonoverlapping
intervals such that the part not covered by them has measure 0. This is shown
by the following simple example: Ε = (0,1), Ω = {[χ, 1] : χ e (0,1)}.
Problem M.3. Let f(t) be a continuous function on the interval 0 <
t < 1, and define the two sets of points
At = {(t, 0): t e [0,1]}, Bt = {(/(*), 1): t e [0,1]}.
Show that the union of all segments AtBt is Lebesgue-measurable, and find
the minimum of its measure with respect to all functions f.
Solution. We first show that the set A = Uo<t<iAtBt is closed, hence
Lebesgue-measurable. Let Pn be a convergent sequence of points in A,
Pn —> Pq. By the definition of A, to every η there is a tn such that Pn e
AtnBtn. According to the Bolzano-Weierstrass theorem, the sequence tn
contains a convergent subsequence: tnk —> to. The distance from the point
Pnk to the segment AtoBto is not greater than max{|/(tnfc) — f{to)\, \tnk —
£o|}, which tends to 0 by the continuity of f(t). It follows that Pq g
At0Bto С A. Since A contains the limit point of any convergent sequence
of its points, A is closed.
A simple calculation shows that the point of the segment AtBt that
lies on the straight line у = с has abscissa (1 — c)t + cf(t), which, by the
continuity of f(t), is a continuous function of t. Consequently, if two points
of the set A lie on the line у = с, then A contains the segment that joins
these points.
Now we can determine the minimum of the measure of A. If f(t) is
constant, then A is a triangle of unit base and unit altitude, so it has
measure 1/2. If the segments AoBo and A\B\ do not intersect, then the
trapezoid with vertices Ao, Bo, Αι, and B\ is a subset of A, so the measure
of A is not less than 1/2. If the segments AqBo and A\B\ intersect at some
point C, then the triangles AqCA\ and BqCB\ are subsets of A, so the
measure of A is not smaller than
where d denotes the distance from Д) to B\. By a simple calculation, we
obtain that the minimum of t(d) on the positive half-axis is \/2-l. Thus,
the measure of A cannot be less than л/2 — 1. On the other hand, for
f(t) = (y/2 — 1)(1 — £), the measure of A is exactly \/2 — 1. D
Remark. I. N. Berstein (Doklady Acad. Nauk. SSSR Ц6 (1962), 11-
13) refers to the result of the problem but he states erroneously that the
minimum is ^.

3.5 MEASURE THEORY
333
Problem M.4. A "letter T" erected at point A of the x-axis in the xy-
plane is the union of a segment AB in the upper half-plane perpendicular
to the x-axis and a segment CD containing В in its interior and parallel to
the x-axis. Show that it is impossible to erect a letter Τ at every point of
the x-axis so that the union of those erected at rational points is disjoint
from the union of those erected at irrational points.
Solution 1. We call width of the letter Τ erected at A the minimum of
the lengths CB and BD. We devide the irrational points into countably
many classes Hi,..., H^,... . The class H^ consists of all irrational points
for which the widths of the letters Τ erected at them are greater than
l/k. Baire's theorem implies that the set of all irrational points cannot be
represented as the countable union of nowhere-dense sets, so some Hk is
dense in a suitable interval /. Selecting an arbitrary rational point A of
the interval /, denote by δ the width of the letter Τ erected at it. The class
Η^ contains an element A\ whose distance from A is less than min{<$, l/k}.
The letters Τ erected at A and A\ obviously intersect each other since they
both have widths greater than the distance from A to A\. The proof is
complete. D
Solution 2. Project on the rr-axis the segment CD parallel to the rr-axis
of each letter T, and denote the projection by Ix. The projection Ix is an
interval containing χ in its interior. We prove the existence of an irrational
α and a rational r such that Ia and Ir contain r and a, respectively, in
their interior. This already implies that the letters Τ erected at a and r
intersect.
For each rational number r = p/q (we shall write all rational numbers in
irreducible form), we choose a closed subinterval Jr of Ir containing r in its
interior and having length not greater than 1/q2. Consider a sequence of
rational numbers where the denominators are strictly monotone increasing
and each number lies in the intervals Jr corresponding to the previous
terms:
Pi P2
r\ = —,r2 = —,..., rn
Qi Q2
qi <Я2 <·" <qn< -"
ΐγι t *Jf\ 1 1 "Г2 ' 1***1 1 "Tjx
_ Pn
Qn
)
-1 *
Such a sequence obviously exists and, moreover, it is a Cauchy sequence, so
it has a limit Нтп^оо rn = a. Since JTn contains the points rn+i, rn+2, · · ·
and is closed, it follows that α G JTn and, in particular,
ι , 1
\OL-rn\ < -τ-.
Qn
From the theory of numbers it is well known that in this case a can only be
irrational. Ia contains a in its interior, so for sufficiently large η we have

334
3. SOLUTIONS TO THE PROBLEMS
rn G Ia. On the other hand, a G JTn С lTn. Thus, for sufficiently large n,
the letters Τ erected at the points α and rn intersect each other. D
Solution 3. Assume that the assertion is false, that is, we have succeeded
in erecting a letter Τ at each point of the real line so that none of the
letters Τ erected at rational points intersects a T erected at an irrational
point. We may assume that the letters Τ are symmetric, that is, В is the
midpoint of the segment CD (let С always have the smaller abscissa among
С and D).
We define a monotonically increasing sequence of numbers αι,α2,...,
αη,... by recursion. Let a\ be any real number. If an is already defined
as a rational number, then let αη+ι be irrational, while if an is irrational,
then let αη+ι be rational, and choose αη+ι(> an) so that αη+ι — an is
smaller than one-quarter of the length of BnDn. Denote the abscissa of Dn
by δη. Making use of the fact that the letters Τ erected at an and αη+ι
do not intersect, we obtain the following sequence of inequalities:
θίη+ι < On+i < ocn + 2(αη+ι - an) < an + 2 = < bn.
It follows that {an} is bounded, and so a = liman exists, and also that
a < δη for every n. Then, however, for sufficiently large n, the letter Τ
erected at an intersects the Τ erected at a, which contradicts the initial
assumption. D
Remarks.
1. In addition to the assertion of the problem, a contestant has proved
that in any interval there is a continuum of irrational numbers a such
that the letter Τ erected at a intersects a T erected at a rational point.
Really, from the reasoning of Solution 2 it turns out that if r\, r2,...
is a sequence of rational numbers such that Jri D Jr2 D ... and q\ <
q2 < ..., then {rn} is obviously convergent and its limit is a "suitable"
irrational number. Let r = p/q be a rational number belonging to the
interval (a, b) such that Jr С (α, b). Let tq = po/tfo and r\ = ρι/qi be
rational numbers satisfying q0 > q, q\ > q, Jro С Jr, Jri С Jr, and
Jro Π Jri = 0. The existence of r$ and r\ of this kind is evident. Once
we have defined the rational numbers nb...,ifc (h = 0,1;...; i^ = 0,1),
let nlv..,ifc,o and nlv..,ifc,i be two rational numbers such that
<Ζϊι,...,ύ,0 > 9ii,...,ifc? *7zi,...,Zfc,l > 9<i,...,tfcj
Jrilf...,ifc,0 С Jrilt...tik, Jrilt...,iktl С Jrilf...,ifc>
*H „O^H,.,,,!^.
With each number 0 < χ < 1, we associate an irrational number a G
(a, b) such that the letter Τ erected at a surely intersects a T erected at
some rational point. If the infinite dyadic form of χ is
χ = 0.iii2 ·..,

3.5 MEASURE THEORY
335
we associate with χ the limit of the sequence
' i\ ·> ι гi,Z2 ? · · · ? ' ii,Z2,...,Zfc ? · · · ·
This will be a "suitable" irrational number, and the construction
guarantees that αϊ and c*2 associated with distinct numbers χ ι and X2 are
distinct.
2. The union of a segment AB in the upper half-plane that is erected
perpendicularly at point A of the rr-axis and a segment ВС that is
parallel to the rr-axis will be called a [ora] depending on whether the
abscissa of С is greater than or less than that of B. Two contestants
proved that if we erect a |~ at each rational point and a ] at each irrational
point, then the union of the [ intersects the union of the ]. This assertion
is obviously sharper than that of the problem. Two other contestants
showed by examples that it is possible to erect a |~ at each point so that
the [ are pairwise disjoint. Indeed, if the points В of the [ lie on the
graph of a strictly monotone decreasing positive function and the length
of ВС is arbitrary, then clearly any two [ are disjoint.
3. It is easy to see that the next theorem generalizes the statement of the
problem: Let X be a complete metric space of second category, and let
X = PUQ, where Ρ and Q are disjoint dense sets. Then it is
impossible to define a real function / on X such that each point ρ e Ρ has a
neighborhood Vp for which χ G VpC\Q implies f(x) < f(p) and such that
each point q G Q has a neighborhood Vq for which χ G Vqf) Ρ implies
This was proved by some contestants for complete metric spaces and
compact metric spaces, respectively. It was also shown that the
hypothesis of completeness cannot be dropped.
Problem M.5. Let f and g be continuous positive functions denned
on the interval [0, oo), and let Ε С [Ο,οο) be a set of positive measure.
Prove that the range of the function denned on Ε χ Ε by the relation
F(x,y)= [ f(t)dt+ [ g(t)dt
Jo Jo
has a nonvoid interior.
Solution. Put
px px
φ(χ) = / f(t)dt, φ{χ) = / g(t)dt.
Jo Jo
The function φ and its inverse φ~ι (which obviously exists) are absolutely
continuous, so they map sets of measure 0 onto sets of measure 0.
Therefore, the image φ(Ε) = A of the set Ε of positive measure is measurable
and has positive measure. Similarly, the set ψ(Ε) = В has positive
measure. Since the range of the function F(x, y) considered on Ε χ Ε is exactly

336
3. SOLUTIONS TO THE PROBLEMS
the set A + В = {a + b : a G A, b G B}, it is sufficient to prove that for sets
A and В of positive measure, the sum A + В contains some interval.
Let и and ν be points of density 1 of A and B, respectively, and let ε > 0
be chosen so that the relations 0 < <$ < 2ε, 0 < <$' < 2ε, <$ + <$'>() imply
μ(Αη [u-m+ «'])> £(« + «')» μ(Β η [ν-«, ν+ «'])> £(« + «')»
where μ stands for Lebesgue measure. We show that (u + υ — e,u + v-\-e) С
A + B. Let t e (u + v — e,u + v + ε) and
A* = t- A = {t-a: aeA}.
Then A* f][t — u,v + ε] is congruent with Α Π [t — г> — ε, ία], so
μ(Α* Π [t — u,v + ε]) > -{u + v + ε — ί) and
μ(ΒΓ\ [ί-η,υ + ε]) > -(η + υ + ε-t),
from which it follows that А* ПВ ^ ®. Let χ G Α* Π Б; then t-ieA,
χ e B, and, consequently, i = (i-a;) + a;ei + R D
itemarfcs.
i. Two contestants showed that it is sufficient to assume that / and g are
positive functions integrable in every finite interval.
2. A contestant remarked that if we know of Ε only that it has positive
outer measure, then the statement becomes false.
3. Denote by A\ and B\ the set of all points of density 1 of A and B,
respectively. If и G A\ and ν G Bi, then by the density theorem of
Lebesgue и and ν have density 1 also in A\ and B\, respectively, so by
the solution above, A\ + B\ contains some neighborhood of и + v. This
means exactly that A\ + B\ is an open set (see J. В. Н. Kemperman, A
general functional equation, Trans. Am. Math. Soc. 86 (1957), 28-56,
Theorem 2.2).
Problem M.6. In η-dimensional Euclidean space, the union of any set
of closed balls (of positive radii) is measurable in the sense of Lebesgue.
Solution. Let {Ga} (a G A) be an arbitrary set of closed balls of positive
radii. Put Η = Uae^Ga. Denote by G the set of all closed balls (of positive
radii) that are contained in some Ga. Clearly, the elements of G constitute
a cover of Η in the sense of Vitali. Therefore, by Vitali's covering theorem
(see for example S. Saks, Theory of the Integral, Hafner, New York, 1937,
p. 109), we can choose an at most countable number of pairwise disjoint
balls Si, #2, · · · belonging to G such that H\UiSi has Lebesgue measure
0. Since Si С Н, the relation
н= (н\изг)и(изг)
implies that Η is Lebesgue-measurable. D

3.5 MEASURE THEORY
337
Remarks.
1. Most contestants solved the problem with the help of Vitali's covering
theorem.
2. Several contestants remarked that Η can be obtained as the union of at
most count ably many Jordan measurable sets.
3. One contestant proved the following generalization of the problem: The
union of any set of convex sets with nonvoid interiors is measurable in
the sense of Lebesgue.
Problem M.7. In η-dimensional Euclidean space, the square of the
two-dimensional Lebesgue measure of a bounded, closed, (two-dimensional)
planar set is equal to the sum of the squares of the measures of the
orthogonal projections of the given set on the η-coordinate hyperplanes.
Solution 1. We solve the problem in the generalized form where a
bounded, closed subset of Η or an r-dimensional subspace L is projected
to the r-dimensional coordinate subspaces.
Η is the difference of two bounded open sets N2 G iVi. Each of them is
the union of an at most countably infinite number of disjoint parallelepipeds
tu and t2j. Their images under a linear transformation Τ (the orthogonal
projection) are parallelepipeds whose volume (Lebesgue measures, denoted
by λ) satisfies
X(Ttu) = cX(tu) and \{Tt2j) = c\(t2j),
where с is a constant depending on the position of the two subspaces and
the direction of projecting only. Further, since a projection takes the union
and the difference of sets into the union and the difference, respectively, of
the image sets, from the additivity of the measure it follows that TH is
measurable and X(TH) = cX(H).
lfTilmm,ir is the orthogonal projection to the coordinate subspace [x^,...,
xir] and Cilmm.ir denotes the corresponding constant, then the sum of the
squared measures of the projections is
Σ Х2(Тп..лгН) = £ Cil...irX\H) = λ2(#) Σ ch...ir, (1)
ii,...,ir ii,...,ir ii,...,ir
where the summation is extended to all combinations i\,..., ir of order r
of the numbers 1,2,..., n.
If ei,..., en are unit vectors of a Cartesian system of coordinates in the
η-dimensional Euclidean space and Η is the r-dimensional cube spanned
by the orthogonal unit vectors
η
uk = ^а[к)ег (k = l,...,r)
i=l

338
3. SOLUTIONS TO THE PROBLEMS
belonging to L, then X(H) = 1, and the value c^...^ = \(TilmmmirH) is equal
to the determinant \Ailmmmir | formed of the columns ii,..., ir of the matrix
A = {a\ '}. Then, however, by the Binet-Cauchy formula,
Σ cl-v= Σ \Ail...ir\2 = \AA*\ = l,
ii,...,ir ii,...,ir
which, owing to (1), verifies our assertion. D
Solution 2. We prove the assertion in the previous generalized form
where a (not necessarily compact) Lebesgue-measurable set Η lying in
the r-dimensional subspace L is projected to the r-dimensional coordinate
subspaces.
Let
ei? · · · j en be the orthonormal basis vectors of a Cartesian system
of coordinates in the η-dimensional Euclidean space En, and щ,..., ur be
the orthonormal basis vectors of a system of coordinates in the subspace
L. We may assume that L contains the origin since this can be achieved
by translation, and translation means translation of the projections, so
the measure of Η and the measures of its projections remain unchanged.
Consequently, we may choose
η
uk = ^2а\к)е{ (fc = l,...,r),
2 = 1
where the matrix A = {a\ '} satisfies \AA*\ = 1.
Denote by Ailmmmir the operator of orthogonal projection from L to the
coordinate subspace [x^,... ,x%r] of En. The operator Ailmmmir sends the
basis vectors Uk to the vectors of the latter subspace with components
α\λ ,..., ol\J. Let (-4^...^) be the matrix of this projection operator (that
is, the square matrix formed of the columns ii,..., ir of A), and let \Ailmm,ir \
be the determinant of this matrix.
The Lebesgue measure of Η is given by
μ = / Xh d\,
where χπ denotes the characteristic function of H. By the formula for
transformation of integrals, the measure of А^.^Н is
/ XAil...irHd\= \Ailt..ir\ / χΗάλ= ^...J/i.
J[xi1,...,xir] JL
(This is valid also when l-A^...^ | = 0, since in this case A^.^H lies in an
at most (r — l)-dimensional subspace, that is, a subspace of measure 0 with
respect to λ.) Summing for all combinations ii,... ,ir of order r, by the
Binet-Cauchy formula we obtain
Σ { jxA4.,rHd\Y = μ2 Σ l*i-U2 = μ2\ΑΑ*\ = ν2,
*lv»*r г,,. „. 1 *lv»*r
[Хг1,...,Х1г\
which proves the assertion. D

3.5 MEASURE THEORY
339
Problem M.8. Let us use the word N-measure for nonnegative, finitely
additive set functions defined on all subsets of the positive integers, equal
to 0 on finite sets, and equal to 1 on the whole set. We say that the system
21 of sets determines the N-measure μ if any N-measure coinciding with μ
on all elements of 21 is necessarily identical with μ. Prove the existence of
an N-measure μ that cannot be determined by a system of cardinality less
than continuum.
Solution. We shall need a definition and a lemma.
Definition. We say that the system A consisting of certain subsets of
a set is independent, if for any distinct members Xi,...,Xj, ΥΊ,..., Y^ of
A, the set
Хг Π · · · Π Xj Π ΪΤ Π · · · Π Yk
is infinite (here U denotes the complement of U).
Lemma. Any countable set has an independent system of subsets that
is of power continuum.
Proof. Indeed, take the following set for the countable basic set. Let
Μ be the set of all sets of real numbers that can be obtained as finite
unions of intervals belonging to any type of closedness but having rational
endpoints. Clearly, there is only a countable number of such sets. Denote
by Μa the set of those sets of real numbers described above that contain
the real number a. Then the system {Ma} of subsets of M, for a running
through the set of all real numbers, has power continuum. It is easy to
see that {Ma} is independent. Really, for given real numbers αϊ,... ,aJ?
6i,.. .,6fc of A, the set
Μαι Π · · · Π Ma. П Mbl Π · · · П Mbk
consists of those elements of M, that is, those sets of real numbers described
above, that contain αχ,... ,a,j but do not contain &i,..., b^. The set of these
sets, however, is countably infinite.
Next we construct an iV-measure that takes the values 0 and 1 only. To
this end, we shall need Zorn's lemma, so our proof is not purely
constructive.
A system of subsets of the set N of all real numbers is called a filter
if it is closed for taking finite intersections and if together with any set it
contains all larger sets as well. To exclude the filter consisting of all subsets,
we also require that a filter should not contain the empty set. Since the
union of an increasing chain of filters is again a filter, to any filter with
the help of Zorn's lemma we can find a maximal filter containing it. This
is called an ultrafilter. If a filter contains neither the set A nor A, then
it can be extended by A. If it contained both, then it would also contain
their intersection, that is, the empty set. Therefore, an ultrafilter contains
exactly one of the sets A and A. If we now associate with an ultrafilter

340
3. SOLUTIONS TO THE PROBLEMS
Μ the N-measure μ that takes the value 1 on elements of Μ and 0 on
their complements, then we obtain a set function with values 0 and 1 that
is defined on all subsets of N and is additive. Additiveness could fail only
if neither of the sets has measure 0, but this is impossible: as the empty
set does not belong to our ultrafilter, two disjoint sets of measure 1 do
not exist. We shall prove for a measure of this kind that it cannot be
determined by less than a continuum number of sets.
From the subsets of N, in the manner described in the lemma, we
construct an independent system of power continuum, and denote its elements
by Na. If с is an infinite sequence consisting of distinct real numbers, we
denote by Sc the union of the complements of the corresponding sets Na.
The filter generated by the Na, the 5C, and the complements of finite sets
consists of all sets containing finite intersections of these. To see that the
latter is really a filter, we have to show that it does not contain the empty
set. In the opposite case, we could find αχ,..., am and ci,..., cn so that
the set
X = Ναι Π · · · Π Nam Π 5C1 Π · · · Π 5Cn
is finite. Неге с* is an infinite sequence of distinct real numbers, and
consequently we can choose real numbers &i,..., 6n from the elements of
ci,..., cn so that αχ,..., am, and &i,..., bn are distinct. Independence
implies that the set
Ναι Π · · · Π Nam ΠΪν^η · · · Π N^ = Υ
is infinite. On the other hand, the definition of the Sc yields Υ С X. This
contradicts our assumption. It follows that the system of sets considered
does not contain the empty set; hence it is a filter. We extend this filter to
an ultrafilter Μ and define the iV-measure μ with the help of the latter.
We have to prove that μ cannot be determined by a system 21 of power
less than continuum. If 21 is a determining system containing, among
others, sets of measure 0, then these can be replaced by their complements:
it is sufficient to consider determining systems 21 all of whose elements
belong to M. Taking a 21 of this kind, we assume that it contains less than
a continuum of sets.
If we consider the finite intersections of the sets in 21, their power is
still less than continuum. Therefore, we may assume from the beginning
that 21 is closed for taking finite intersections. If there were a Z G Μ that
contained no element of 21, then Ζ would not be disjoint from any set in 21
and, consequently, the sets containing sets of the form Ζ Π A (A G 21) would
constitute a filter. This filter could be extended to an ultrafilter, and the
latter would define an iV-measure v. Then μ and ν would coincide on the
elements of 21, whereas μ(Ζ) = 0 and v{Z) = 1. This is impossible since
21 is a determining system. So each element of Μ contains a set from 21.
Then at least one element W of 21 is contained in infinitely many elements
of {Na}; let с be a countable sequence of the corresponding subscripts.
The set Na is disjoint from W for each element of c, from which it follows
that Sc = UNa is also disjoint from W. But this means that Μ contains

3.5 MEASURE THEORY
341
two disjoint sets, which contradicts our assumption. The contradiction has
been caused by the assumption that μ can be determined by less than a
continuum of sets. For the completion of the proof, it remains to show that
μ vanishes on every finite set. This is evident since μ{Ό) = 1 whenever D
is the complement of a finite set. D
Remarks.
1. If the continuum hypothesis is assumed, the problem becomes trivial.
In fact, it is easy to prove that a countable set cannot determine an
iV-measure.
2. It would be a sharpening of the problem to prove that no iV-measure can
be determined by less than a continuum number of sets. This, however,
is undecidable within the usual Zermelo-Fraenkel system of axioms for
set theory.
3. The problem can be generalized to other cardinal numbers. For any
cardinal m, there exists a finitely additive 0-1 measure that vanishes on
all sets of power less than m and cannot be generated by less than 2m
subsets. To prove this, it is sufficient to give an adequate generalization
of the lemma. One contestant proved this generalization.
4. One may raise the question of whether there are relatively many or
relatively few iV-measures for the determination of which 2m subsets are
needed. There are altogether 22™ iV-measures, and there are equally so
many iV-measures that cannot be determined by less than 2m sets. In
the original case where the basic set is countable, one contestant proved
this.
Problem M.9. Let {φη(χ)} be a sequence of functions belonging to
L2(0,1) and having norm less than 1 such that for any subsequence
{фПк(х)} tne measure of the set
1 N
tends to 0 as у and N tend to inRnity Prove that фп tends to 0 weakly in
the function space I/2(0,1).
Solution. Suppose there is a function / e L2(0,1) such that
(x)dx
does not tend to 0 as η —> oo. Then in view of the relations
/ Φη(χ)ί(*
Jo
I / Фп(х)П
\Jo
x)dx
< UnWWfW < 11/11 (n = i,2,...)

342
3. SOLUTIONS TO THE PROBLEMS
(replacing, if necessary, f(x) by βιϋ f{x) with a suitable ϋ), there is an
a > 0 and a subsequence {0nfc (#)}£! ι such that
Re
/ ΦηΛΧ)ί(
JO
x)dx > a
for every к (||/|| stands for the norm of the function f(x) in L2(0,1)).
Obviously, to every ε > 0 there is δ > 0 such that the relations m(A) < <$,
А С (0,1) imply ||хл/||2 < £ (here m(A) denotes the measure of the set
A, and Xa(x) denotes the characteristic function of A). Let ε be a positive
number less than a, let δ be a positive value corresponding to ε in the
sense just indicated, and choose 2/(ε) and Ν(ε) so that for Ν > Ν(ε) the
measure of the set
Ε
^ΣΧ(*)
>2/(ε)
XG(0,1):
is smaller than δ. Then, for Ν > Ν(ε), it follows that
Na < У2 Re / ΐ(χ)ΦηΛχ)dx
k=l J°
r Ν Ν Γ
= Re /_ f(x) Σ Фпк (x)dx + Re Σ / f^^k (x)dx
J Ε ι ι ι ι «/ .C/
fc=l
< VN [_\f(x)\ \^=ΣΦηΜ dx + Σί \Пх)Фпк{*)\<Ь
Je Ι*-™ fe=i I k=ijE
< VNy(e)\\f\\2 + N\\XEf\\2 < VNy(e)\\f\\2 + Νε,
that is, y/N(a — ε)< 2/(ε)||/||2, which is impossible for sufficiently large N
since a — ε > 0. Consequently,
/
Jo
cj)n(x)f(x) dx —> 0 (n —> oo)
for all functions / G L2(0,1), as stated. D
Remarks. Essentially all acceptable solutions have followed this way.
1. Two contestants noted that it is sufficient to require the following: for
any subsequence {фПк{х)} an(^ апУ number с > 0, the measure of the
set
χ e (0,1) :
N
ΣΦηΛχ)
k=l
>cNl
tends to 0 as N —> oo.
2. One contestant showed by a counterexample that without boundedness
of the sequence {фп(х)} in the norm of L2(0,1) the assertion is not

3.5 MEASURE THEORY
343
always true. He also mentioned that the hypotheses of the problem do
not imply strong convergence.
3. Another contestant pointed out that it is sufficient to make the following
assumption: to any subsequence {n^} of the natural numbers, there is a
function f(N) t^ 0 for which f(N)/N —> 0 as N —> oo and the measure
of the set
1 N I
x в (0,1) :
№)
k=l
>y
tends to 0 as у —> oo, N —> oo. He also noted that the sequence {фп(х)}
does not necessarily converge either in measure or in mean.
4- A contestant observed that the hypothesis may be weakened as follows:
from any subsequence of the sequence {фп(х)} one can choose a
subsequence {фПк(х)} for which the measure of set (1) tends to 0 as у —> oo,
N —> oo.
5. One contestant called attention to the fact that if one assumes bounded-
ness of the sequence {фп(х)} in the norm of Lp(0,1), where 1 < ρ < oo,
then it can be shown that, under the hypothesis of the problem,
I
ι
φη(χ)/(χ) dx —> 0 (n —> oo)
follows for all functions f(x) e Lq(0,1), where 1/p+l/q = 1.
6. Four contestants treated the problem in the complex case.
Problem M.10. We say that the real-valued function f(x) defined on
the interval (0,1) is approximately continuous on (0,1) if for any x0 e (0,1)
and ε > 0 the point xq is a point of interior density 1 of the set
H = {x: \f(x)-f(x0)\<e}.
Let F С (0,1) be a countable closed set, and g(x) a real-valued function
denned on F. Prove the existence of an approximately continuous function
f(x) denned on (0,1) such that
f(x) = g(x) for all χ G F.
Solution 1. Let F = {ci,C2,... }. Denote by I(x,n) the closed interval
of center χ and radius 1/n. By the "central p-multiple" (0 < ρ < 1) of the
interval (a, b) we shall mean the closed interval
α+1(6-α)(1-ρ),6-^(6-α)(1-ρ)
We shall define a system Jn of intervals by recursion so that it has the
following properties. Jn consists of countably many disjoint closed intervals,
none of which intersects F. Further, cn is the only accumulation point of

344
3. SOLUTIONS TO THE PROBLEMS
the set of endpoints of the intervals forming Jn and, finally, 7fm and Kn are
disjoint if m φ η, where Kn stands for the union of the intervals forming
«Λι·
Suppose that for η' < η we have already defined Jn> so that they have
the properties required. Then the closure of the set U£~J7ffc = Ln is
Ln U{ci,...,cn_i},
and it does not contain cn. Therefore, it does not intersect a suitable
neighborhood 7(cn, rn) of cn either. We may assume that rn > n. For each
j (j = 1,2,...), consider the interior of the set 7(cn, rn + j) \ 7(cn, rn +
j + 1)\F. This is the union of a (finite or) countable number of open
intervals. Obviously, the total length of these open intervals is equal to the
measure of the set 7(cn,rn+j)\7(cn,rn+j + l) = M(n,j), so the union of
a finite number of them has at least (1 — 1/j) times this measure. Take the
central (1 — l/j)-multiples of these finitely many intervals, and let these
new intervals constitute the system Jnj. Put Jn = U^j Jnj.
Clearly, the systems Jn of intervals so obtained possess the required
properties (in particular, the recursion is correct). Further, the intervals
forming Jnj fill at least (1 — 1/j)2 times the measure of M(n,j) and,
therefore, Kn, the union of the intervals forming Jn, has density 1 at cn.
Define / in the following way. Let f(x) = g(cn) if χ G Kn or χ = cn. If
we prove that the set U™=1Kn UF is closed, then / can be defined at points
not belonging to this set so that on the countable number of disjoint open
intervals forming the complement of this closed set, / is linear, and at the
endpoints of the intervals it takes the values already fixed. It is obvious that
the function / defined in this way is approximately continuous in points
of F and continuous in all other points. To see the latter, observe that for
χ φ F the distance from χ to F is positive, say greater than 1/m, where
mis a positive integer. Thus, by the assumption rn > n, Kn does not
intersect the l/2m-neighborhood of χ if η > 2m. Consequently,
7(ж,2ш)П ( U KnUF) =7(ж,2ш)П С™ Кп).
The right-hand side, however, is the union of a finite number of closed
intervals. This proves, on the one hand, that \J^=lKn U F is closed and,
on the other hand, that the restriction to I(x, 2m) of the function /
constructed with the help of this set is a continuous polygon function. D
Solution 2. Let F = {n, r2,... }. We start from the idea that the closer
χ is to Гг, the more f(x) should "feel" the value taken at n. We try to
achieve this by a definition of the following type:
Σ°° 9(Гг)
г=1 и(г)\х—Гг\ *с л т^ (л\
х) = ^оо ι 1) if х i F. (1)

3.5 MEASURE THEORY
345
We now impose various conditions on the order of u(i) to ensure
convergence of this series and approximate continuity of the function obtained,
and finally we show that these conditions can be fulfilled.
The series will always be convergent if, for instance,
u(i) > i2\g(ri)\ and u(i) > г2 (г = 1,2,...)· (2)
Really, for χ £ F, the closedness of F gives пищ \х — n\ > 0.
/ will be continuous in all points of (0,1) \ F (in a small neighborhood,
both the numerator and the denominator are sums of uniformly convergent
series of continuous functions), hence it is also approximatively continuous
there.
Consider therefore an r*. Let d = minj<i \n — rj|, δ < d/2. We show
that in the interval (r* — δ, r» + δ), for δ sufficiently small, the numerator
in (1) will generally be around g{ri)/{u{i)\x — n|), and the denominator
around l/(u(i)\x — Г{\). In any case,
_ 9iTj)
3<i v*
and
ΣϊδΡ^-ад w
Σ-τττ r = 0(1)
(since \x — rj\ > d — δ/2 > d/2). We divide the values χ into two classes.
One consists of those χ that satisfy the relation
i + |gfo)l ^ ι
<T2 (4)
u(j)\x-rj\ j2
for all j > i. For these x, by (3) and (4),
fM = mfei+QW = 9(η) + 0(\χ-η\)
/u щ£=я + 0{1) ι + θ(\χ-η\) '
which lies in the ε-neighborhood of g(ri) if δ is sufficiently small. Thus, we
have to achieve that the measure of such χ in (r^ — δ, r* + δ) be (2 + o(l))<$.
If, however, (4) does not hold, then
The measure of such "bad" χ in (r* — 5, r» + δ) is
kt-rj|<6+sj
If now Ej <\ri — rj\2/j2, then this measure is
<2 Σ ΐ!^^φ)*£ΐ = 0(*2) = ο(ί)
|г4-г,|«+Ь^
as <$ —> 0 (since |г» — rj\2/j2 < |г» — rj|/4 and so (3/4) · |r» — rj| < δ).
Consequently, together with (2), we have only a finite number of conditions
for each j, so they can be realized. D

346
3. SOLUTIONS TO THE PROBLEMS
Problem M.ll. Let {/n}i?Lo oe a uniformly bounded sequence of real-
valued measurable functions deuned on [0,1] satisfying
/1/»a = i.
Further, let {cn} be a sequence of real numbers with
oo
Σ>« = +0°·
n=0
Prove that some re-arrangement of the series $^L0 cnfn is divergent on a
set of positive measure.
Solution. We first show that
oo
71=0
is divergent on a set of positive measure. Suppose that (1) yields a function
д that is finite a.e. By Egorov's theorem there is a set А С [0,1] on which
the series (1) is uniformly convergent and for which X(A) > 1 — (/2if),
where К is a common upper bound of the functions f%. Then, by the
uniform convergence, the limiting function g(t) is integrable on A, and
Y^clf fn{t)dt= f g(t)dt <+oo. (2)
n=0 ·>* JA
On the other hand,
/ fi(t)dt= f mdt- f fi(t)dt>\-K± = \,
J A JO ·Λθ,1]\Α ΖΚ Ζ
which, in view of (2), yields
1 °°
n=0
a contradiction.
As a second step we show that the series Y^™=0cnfn has a subseries
divergent on a set of positive measure. Let ει,..., εη,... be independent
random variables on some field (Ω, Α, Ρ) taking the values 0, 1 with
probability 1/2, and let 0 < t < 1 be a number for which (1) is divergent. Then
for the random variables encnfn(t) we have
oo oo
^Var (encnfn(t)) = Y^c2Jl(t) = +<x
71=0 71=0

3.5 MEASURE THEORY
347
and therefore, by Kolmogorov's three-series theorem, X]^L0£ncn/n(t) 1S
divergent with positive probability. Thus, the set of all elements (a;, t) in
the probability field (Ω, Α, Ρ) χ [0,1] for which
oo
Σ>η(ωΚ/η(') (3)
n=0
is divergent has positive measure. Consequently, there exists ω G Ω for
which the series (3) is divergent on a set of positive measure.
It follows that there is a set D of positive measure and an infinite
sequence πι < Π2 < ... such that
oo
Y,cnkfnk{t) (4)
71 = 0
is divergent for alii G D\ so there exists 6(t) > 0 such that for arbitrarily
large indices the partial sums of (4) have oscillation not less than 6(t).
Define next the numbers Ni = 1 < N2 < ... as follows. Let the number
Nk+i be such that the measure of the set
Dk = < f G D : max
ΝΗ<ν<μ<ΝΗ+1
Σ^/η,φ
>Ш
is greater than (1 — (l/2/c+1)) · X(D). Such an Ν^+ι exists since for
sufficiently large Njc+i, all values t e D belong to Dk. The set
D* = η Dk
k=l
has a positive measure, and
max
Nfc<^<M<Nfc+i
Y^cnifni(t)
>
- 2
if t G D* and к > 1. Therefore, any rearrangement of the original series
Σ^=ο Cnfni for which the sums X^=fc^ Cmfm consist of consecutive terms,
is divergent at all points t G D*. D
Problem M.12. Let {fn} be a sequence of Lebesgue^integrable
functions on [0,1] such that for any Lebesgue-measurable subset Ε of [0,1] the
sequence JE fn is convergent. Assume also that limn fn = f exists almost
everywhere. Prove that f is integrable and JEf = Hmn JE fn. Is the
assertion also true if Ε runs only over intervals but we also assume fn > 0?
What happens if [0,1] is replaced by [0, oo) ?
Solution. Let Ε = [0,1] or Ε = [0, oo), and Ε = U™=1Ek, where Ek is a
measurable set of finite measure and on Ek the convergence of the sequence

348
3. SOLUTIONS TO THE PROBLEMS
{/n} is uniform. By Egorov's theorem, such a sequence {E^} obviously
exists. It is clear that fE f exists for every k, and fFf = limn fF fn if F
is a measurable subset of Ek.
The formula μ{β) = limn fG fn defines a finitely additive signed measure
on measurable subsets G of E. The theorem of Beppo-Levi shows that
the existence of / / together with the relation μ((?) = / / for each
Je Jg
measurable subset G of Ε are equivalent to the σ-additivity of μ. But
the latter follows from the well-known fact that the limit of a pointwise
convergent sequence of finite signed measures is a finite signed measure (in
particular, it is σ-additive; see P. R. Halmos, Measure Theory, Springer,
New York, 1974, p- 170, relation (14)). It can also be seen that / |/n —
JE
/I —> 0 (n —> 00). Thus, we have answered the first and third questions of
the problem. The answer to the second question is negative, as shown by
the example fn{x) = η if 0 < χ < —, and fn{x) = 0 if — < χ <1. Π
η η
Problem M.13. Let 0 < с < 1, and let η denote the order type of the
set of rational numbers. Assume that with every rational number r we
associate a Lebesgue-measurable subset Hr of measure с of the interval [0,1].
Prove the existence of a Lebesgue-measurable set Η С [0,1] of measure с
such that for every χ G Η the set
{r :xe Hr}
contains a subset of type η.
Solution. We give two solutions. Both make use of the following simple
lemma:
Lemma. Let Μ be a system of sets that consists of certain subsets of
the interval (0,1), and suppose that for each Α Ε Μ there is an χ G A such
that Α Π (0, χ) G Μ and Α Π (χ, 1) G Μ. Then every element of Μ has a
subset of type η.
Proof. Arrange the rational points of (0,1) in a sequence 7*1, Г2,... . Let
A G Μ be arbitrary. We shall define by recursion a sequence of points
xn e A having the following properties:
1. It is ordered in the same way as the sequence {rn};
2. An (0, xn) e M, An (xn, 1) G Μ for every n, and Α Π (χι, Xj) G Μ for
every pair Xi < Xj.
By assumption, there is an x\ G A such that Α Π (0, #ι) G Μ and
An(xi, 1) G M. Next suppose that χι, x%,.. .,xn have already been chosen,
and let the immediate neighbors of rn+i from among ri, Г2,..., rn be r* and
rj, ri < Гу By the induction hypothesis, we have X{ < Xj and An(xi, Xj) G
Μ , so there is an xn+\ Ε An (xi,Xj) such that Α Π (xi,xn+i) G Μ and
An(жп+1,Xj) G M. A similar choice of xn+\ is possible in the cases where
rn+i > ri (i < n) or rn+i < Г{ (i <n). The proof of the lemma is complete.

3.5 MEASURE THEORY
349
Solution 1. We need the following auxiliary theorem:
Auxiliary theorem. Let the sets Hn G [0,1] be measurable and have
measure not less than с (η = 1,2,...). Then the set Η of all points χ
for which the sequence {η : χ G Hn} has positive upper density is also
measurable and has measure not less than c.
Proof. Denote by fn the characteristic function of Hn, and put
0n = -(/i+/2 + "- + /n)·
Obviously, χ G Η if and only if gn{x) does not tend to 0 as η —> oo, which
yields the measurability of H. Since 0 < gn(x) < 1 for every n, we have
X(H)> / gn(x)dx= / gn(x)dx- / gn(x)dx
Jh Jo «Лод]\я
= c~ gn{x)dx.
J[0,1]\H
Now ж G [0,1] \ Я implies limn_+00 <7n(:r) = 0, so by Lebesgue's theorem on
the passage to the limit under the integral sign,
lim / gn(x)dx = 0.
п-*°°./[0Д]\Я
Thus, from the previous inequality, we obtain X(H) > c, which proves the
auxiliary theorem.
Let {rn} be an arrangement of the rational points of (0,1) in a sequence
of uniform distribution, that is, for any subinterval (a, b) G (0,1), the
density of the sequence {n : rn G (a, b)} is b — a. (A simple example
of an arrangement of this kind is the following. We first fix an arbitrary
arrangement in a sequence {sn}, then from {sn} we consecutively choose
the elements of minimal index lying respectively in the intervals
[0,1],
'4
2'
η
1 2
η' η
n-\
η
meanwhile, we should make sure that each element is selected only once.)
Let A G (0,1), and assume that the sequence {n : rn G A} has positive
upper measure. We show that A contains a subset of type η. By the lemma
we have presented in advance, it is sufficient to prove the existence of an
χ G A such that both {n : rn G Α Π (0, χ)} and {n : rn G Α Π (χ, 1)} have
positive upper density.
Denote by d the upper density of the sequence {n : rn G A}, and let
0 = #0 < χι < ''' < xm — 1 be a subdivision of [0,1] finer than d/3.
Since {rn} is uniformly distributed, the upper densities of the sequences
{n : rn G ΑΠ (xi_i,Xi)} are smaller than d/3 for every i. Consequently,

350
3. SOLUTIONS TO THE PROBLEMS
denoting by di the upper density of the sequence {n : rn G Α Π (0, X{)}, we
have 0 < di — di-i < d/3 for г = 1,2,..., m. Since dm = d, there is an г
with 0 < di-ι < d{ < d. Then d^-i < d» implies that the set Α Π (ггг-ъ я^)
is infinite, and it is easy to see that for any element χ G An(^_i, Xi)}, the
sequences {n : rn G Α Π (0, ж)} and {η : rn G А П (ж, 1)} also have positive
upper measures.
Now consider the sets Hr appearing in the statement, and let Hi be
the set of those χ for which the sequence {n : χ G #Γη} has positive
upper measure. By the previous observation, if χ G Hi, then {r : χ G
Яг} contains a subset of type 77. Applying our auxiliary theorem to the
sequence of the sets НГп, it follows that Hi is measurable and λ(Ηι) > c,
as required. D
Solution 2. If the closure of a set A has positive measure, then A contains
a subset of type η since the system Μ = {А С (0,1) : X(A) > 0} obviously
satisfies the condition of the lemma. It is therefore sufficient to prove that
the set
H2 = {x: \({r : χ G Hr}) > 0}
is measurable and λ(#2) > с.
Consider the function f(x) = \({r : χ G Hr}). Clearly, 0 < f(x) < 1;
if we show that f(x) is Lebesgue-measurable and /0 f(x) dx > c, then we
shall be done.
For any A G [0,1], put
An= U
2n ' 2n
Obviously, AiD A2D ... and C\™=lAn = A\ hence limn_+00 \{An) = λ(Α).
So let /n(^) = A({r : χ G #r}n); then /n \ /, and therefore we need only
to show that fn(x) is Lebesgue-integrable with /0 fn(x) dx > c.
We have the relation
ι 2П
^ <=ι
where дпЛ^) = 1 or 0 depending on whether A has or does not have a
point in [(< - 1)/2п,г/2п]. Clearly,
/n,iW := 9nA{r : x e Hr}) = sup < kr : r G ^-, ^ > ,
where kr is the characteristic function of Hr. The function kr is integrable
and /0 kr(x) dx = c, so fn^ is integrable and /0 fn,i{x) dx > с Since fn =
(l/2n) · Σί=1 /η,ή it follows that fn is integrable and JQ fn(x) dx > с D
Remark. Denote by Щ the set of those χ for which the set {r : χ G Hr}
contains a subset of type η. Let {rn} be a fixed arrangement of the rational

3.5 MEASURE THEORY
351
points of (0,1) in a uniformly distributed sequence, and denote by Hi the
set of those χ for which the sequence {η : χ G НГп} has positive upper
density. Let H2 be the set of those χ for which the closure of the set
{r : χ e Hr} has positive measure. Finally, let Щ be the set of those χ
for which {r : χ G Hr} is dense in a subinterval of (0,1). It is easy to see
that for any set A of rational numbers, the upper density of the sequence
{n : rn G A} is at most X(A). Therefore, Я3 С Hi С H2 С Я0 holds for
any system Hr (thus Solution 1 proves a deeper assertion than Solution 2,
since it shows from a narrower set that its measure is at least c). On the
other hand, the statement Х(Щ) > с is false: it can be shown that, for
any 0 < с < 1, there exists a system Hr satisfying the conditions of the
problem and such that the set {r : χ G Hr} is nowhere dense for every
χ G [0,1], in particular, Щ = 0.
We also note that if the sets Hr are measurable, then so is Hq. Actually,
it can be proved that if the Hr are Borel (or, more generally, analytic) sets,
then Hq is analytic, hence measurable. From this it follows easily that Hq
is also measurable in this general case.
Problem M.14. Find a perfect set Η С [0,1] of positive measure and a
continuous function f defined on [0,1] such that for any twice differentiable
function g deRned on [0,1], the set {x G Η : f(x) = g(x)} is unite.
Solution. Delete an interval of length 1/6 from the center of [0,1],
intervals of length 1/18 from the centers of the remaining two intervals, and,
following the procedure, intervals of length 1/2 · 3n+1 from the centers of
the 2n intervals remaining after the nth step. This is possible because the
length of the remaining intervals is
-(-IV-d^-Hgd)')
= 2"n II — —I = 2"n_1 > —-—.
\ 2) 2-3"*1
Denote by /n,i (n = 0,1,...; г = 1,..., 2n) the open intervals of length
1/2 ·3n+1 deleted in the nth step. Put Η = [0,1] \Uln4. Then Η is perfect
and has measure 1/2. Denote by h the function that vanishes on Η and
whose graph on In^ is an isosceles triangle of altitude 1/n for η = 0,1,...
and г = 1,..., 2n. Then h is continuous on [0,1], and so the function
/(ж)= Г h(t)dt
Jo
is continuously differentiable on [0,1] and satisfies the relation f'(x) = 0
for all χ G H. Let g be twice differentiable on [0,1]; we establish that the
set A = {x G Η : f(x) = g(x)} is finite. Suppose A is infinite. Then

352
3. SOLUTIONS TO THE PROBLEMS
there is a convergent sequence Xk —> xo m A. Since f'(xo) = 0, we have
9*(χο) = 0; hence by l'Hopital's rule,
lim g(f)-gfe> = lim ^
x-*xo (x — Xq)2 x-*xo 2(x — Xq)
g'(x)-g'(xo) _ 9"Ы)
It follows that
lim
= lim . ч
x-+xo 2[x — Xq)
f(xk)-f(xo) g"(xo)
k^oo (Xk — Xq)2 2
The contradiction will arise from the fact that
lim
X—>Xq,X£H
№-f(*o)
(x-Xq)2
= 00.
Indeed, let χ G Η, χ φ xq, be arbitrary. Then there is an interval ln^ that
separates χ and xq. Let η be the smallest subscript for which there exists
an In,i of this kind. It is easy to see that \x — x0\ < 2~n. On the other
hand, if, say, χ > xq, then
f(x) - f(x0) = [ h{t) dt> [ h(t) dt
Jxo JIn,i
1
1
Hence
№-f(*o)
1
2 2·3η+* η 12·3η·τΓ
> ι = (*ϊη·-±-.
(Χ - Xq)2
If χ —> xq, then η —> οο and (4/3)η · (1/η) —> οο, which proves the
assertion. D
Problem M.15. Prove that if Ε с R is a bounded set of positive
Lebesgue measure, then for every и < 1/2, a point χ = χ (и) can be found
so that
\(x-h,x + h)r\E\ >uh
and
\(x-h,x + h)C\(R\E)\ >uh
for all sufficiently small positive values of h.
Solution. We shall prove a stronger statement, namely, we verify the
conclusion for и = 1/2. Let
F(x) = га((-оо,ж)ПЯ),
where m denotes the Lebesgue measure. Then F is continuous, increasing,
and satisfies \F(x) — F(y)\ < \x — y\ for all я,у С R. By Lebesgue's

3.5 MEASURE THEORY
353
density theorem, there exists χ G R, which is a point of density 1 of J5,
and hence there is a positive fe G R with F(x + fe) — F(x) > fe/2. Since
for sufficiently small χ we have F(x + fe) — F(x) = 0, continuity ensures
the existence of an x0 satisfying F{xq + fe) — F(xq) = fe/2. Consider the
function G(x) = F(x) - F(x0) -{x- xq)/2. Then G(x0) = G(x0 + fe) = 0.
As almost every point of R is a point of density 1 of either Ε or R \ £,
we have |Gr/(rc)| = 1/2 a.e. Thus, G is not identically 0 on [xq,xq + fe].
Consequently,
either max G(t) > 0 or min G(i) < 0.
te[xo,xo+k] te[x0,x0+k]
It is sufficient to consider the first possibility since the second is similar to
it. Let x\ G (жо, #o + fe) be a point at which G assumes its maximum just
mentioned. Then for every
0 < h < min{^i — xq, x0 + fe — #ι},
wehaveG(^i-/i) < G(xi) > G(xi+h), that is, F(xi-h)+h/2 < F(xi) >
F{xl + h)-h/2. Thus, F(x1 + h)-F(x1-h) < F(xi)+h/2-(F(xl)-h) =
3Λ/2 and F(xi + ft) - F(a?i - ft) > F{x{) - (F{x{) - ft/2) = ft/2. From
these two inequalities, we obtain
m ((χι - ft, χι + ft) Π Ε) > -
and
m ((χι - ft, a?i + ft) П (R \ E)) > -.
The proof is complete. D
Remark. One contestant did not make use of the boundedness of Ε and
for и = 1/2 proved the assertion in the case where Ε is measurable and
neither Ε nor R \ Ε is a null set, and also in the case where Ε С R,
m*(E) φ 0, and m*(E) φ 0 (here m* and m* denote the exterior and
interior Lebesgue measures, respectively).
Problem M.16. Show that there exist a compact set К С R and a set
А С R of type Fa such that the set
{xeR:K + xCA}
is not Borel-measurable (here K + x = {y + x:ye K}). Show that there
exist a compact set К С R and a set А с R of type Fa such that the set
{xeR:K + xcA}
is not Borel-measurable (here K + x = {y + x:ye K}).
Solution. Let Ρ and К be bounded perfect sets such that the sums χ + у
(χ G Ρ,у G К) are pairwise distinct, that is, the relations X\,X2 G P,

354
3. SOLUTIONS TO THE PROBLEMS
Уъ 2/2 € Κ, χι + yi = X2 + 2/2 imply rri = x2 and yi = y2. It is easy to see
that the sets
p = iEw; ai = 0'1; < = 1>2,..
and
10*
„ 2=1
* = Ew; αί = 0'2; i = 1'2·
li10 J
are of this kind. It is well known that there exists a set U С Ρ χ Κ of type
G$ such that the set
В = {χ G Ρ : there exists у е К with (ж, у) G {/}
(which is the projection of U onto P) is not Borel-measurable.
Put V = (Ρ χ K) \ U and A = {x + у : (ж, у) G V}. The mapping
0(ж, у) = χ + у (ж G Ρ, у G К) is continuous and, by the choice of Ρ
and K, one-to-one. Therefore, φ is a homeomorphism of the compact set
Ρ χ К onto {χ + у : χ e P,y e К}. Since V is of type Ρσ, it follows that
Л = ф(у) is an iVtype subset of φ(Ρ χ К) = Ρ + К. Since the latter set
is closed in R, the set A is a subset of type Fa of R as well.
It is easy to see that К + χ С A if and only if φ~ι(Κ + χ) С φ~λ(Α),
that is, if χ G Ρ and
({ж} χ Κ) Π С/ = 0,
(that is, if xE P\£). So
{rr GR: K + iC A} = P\£,
which is not Borel-measurable. D
Problem M.17. For which Lebesgue-measurable subsets Ε of the real
line does a positive constant с exist for which
sup / eitxf(x)dx\ <c sup / einxf(x)dx
-oo<t<oo \Je I n=0,±l,... \JE
for all integrable functions f on Ε ?
Solution. We show that (*) holds for those and only those Lebesgue-
measurable sets that — apart from a subset of measure 0 — can be covered
by a set consisting of the finite union of closed intervals and containing no
pair of congruent points modulo 2π.
Denote by £ the system of all sets satisfying the requirements of the
problem and by Τ the system of all sets having the properties just
described.

3.5 MEASURE THEORY
355
If Ε G £, then those points of Ε for which there is a congruent point in
Ε (mod 2π) form a set of measure 0. Indeed, if, for example, the set
E' = Ε Π {Ε + 2jn) Π (2for, 2(k + 1)π)
had positive measure (j φ 0 integer, Ε + u denotes the translate by и of
E), then with
1 if χ G E',
-1 iixeE' -2j,
I 0 otherwise,
№ = {
it would follow that
/ elnxf(x) dx = 0 for every n,
Je
while, as / φ 0 a.e.,
/ eitxf(
\Je
sup / eitxf(x) dx
>0.
The same property trivially holds also for every Ε G T.
Denote by * the reduction modulo 2 with range space (—π,π]. That is,
u* is the (signed) deviation of и from the nearest multiple of 2π (if there
are two such multiples, then u* = π). Furthermore, put
E* = К : и G Ε} ,
h*(x) = h(u), where χ G Ε*, χ = и*, и G Ε.
By the property above, the latter definition is correct for a.e. χ G E* since
there is only one such u. Therefore, we may write
φ(ί)= [ eitxf(x)dx= [ (eitx)*f*(x)dx,
JE JE
φ(η)= ί einxf(x)dx= [ (einx)*f*(x)dx.
JE JE
Here f*(x) G £i(E*), and every element of Ci(E*) occurs among the f*(x),
where f(x) G Ci(E).
We show that the relation
/ a(x)f*(x)dx\ < cmax|0(n)| = с max / elnxf*(x)dx
\JE* I n JE*
is valid for every function /* G C\(E*) if and only if there exists a function
Kx) — Σαη€ιηΧι ΣΙαη| < +oo? such that b(x) = a{x) for a.e. point
χ G E*. Then the choice с = Σ^^^ \αη\ is possible.

356
3. SOLUTIONS TO THE PROBLEMS
Suppose first that there exists a function b(x) of this kind. Then
/ a(x)f*(x)dx\ = \[ b(x)f*(x)dx\
\Je* I \Je* I
oo r I
Τ an / einxr(x)dx\
^^^ J E*
n= — oo
OO
Σ ο,ηφ(ή)\< Σ |an|max|0(n)|.
ι= — oo I n= — oo
Conversely, consider the sequences {jn}n^-oo that tend to 0 as \n\ —> oo;
they form a Banach space with respect to the norm max|,;n|. The set
of all sequences of the form {φ(η)} is a subspace of this space, and the
functional fE+ a(x)f* (x) dx is linear and, by our assumption, bounded on it.
According to the Hahn-Banach theorem, this functional can be boundedly
extended to all sequences {jn}. But the bounded linear functionals on the
latter have the form Σ anjn, where Σ \an\ < +°o. Consequently, restricted
to the sequences {φ(η)},
/ a(x)f*(x)dx = ΣαηΦ(η) = У2 an einxf*(x)dx
= [ b(x)f*(x)dx
JE*
(where b(x) = ^апегпж), for every /* e C\(E*). Hence, a(x) = b(x) for
a.e. χ e E*.
Apply the proposition just proved to the function a(x) = eltx with fixed
t. Let Ε e T. If the finite number of intervals covering Ε is somewhat
augmented, then the set obtained will not contain any pair of points
congruent modulo 2π either. Let k(x) be a "sufficiently smooth" function that
equals 1 on the original intervals, and 0 outside the augmented intervals.
Finally, put
oo
b(x)= Σ fc(x + 2j)e'*(x+?iir).
j=—oo
Since at each point at most one term of the series is different from 0, the
definition of b(x) makes sense. Furthermore,
b(x) = a{x)* = {eitxy ΊϊχβΕ*.
The function b(x) is periodic with period 2π and Fourier coefficients
2W_
OO
oo
b(x) dx
oo
Σ Hx + 2j)e'*(x+!yir) dx
-- — oo
к(х) dx.
3-гпх
J = -oo
Atx—inx

3.5 MEASURE THEORY
357
Then
1 f00
\bn\ < — / \k(x)\dx< ci,
or, after integrating by parts twice,
1 f°°
C2
i)2
(for a suitable choice of k(x)). Hence,
..- „(n-<)2
Applying the statement we have proved previously, it follows that Ε e S.
Conversely, suppose that Ε G £. We establish the existence of δ > 0 such
that for any two Lebesgue density points u, υ of E, the relation \u — v\ > π
implies \(u — v)*\ > δ. Otherwise, indeed, there are sequences {un}, {vn}
of density points with \un — vn\ > π but (un — vn)* —> 0. It can be shown
that there is a number t such that [t(un — vn)]* -f* 0. To see this, first
let un — vn = wn be a bounded sequence; then t = 1/sup \un — vn\ is an
appropriate value. If wn is unbounded, then we may assume that wn —> oo
and, what is more, wn+i/wn —> oo. We form a sequence tn as follows:
* η * i (tn-iwn)*-n/2
to = u, tn = tn_i , η = ι, ζ,... .
Here tn — tn-\ = 0(1/гуп), and by the fast increase of wn the limit of the
sequence tn exists. Put
t = \imtn = ^2(tn - tn_i).
n=l
Then
t-tn = 0 (^—) ·
Consequently,
( W \ 7Г
twn = (t - tn)wn + tnwn = О —— + tn-iwn - (tn-iwn)* + -,
and, since tn-iwn — (tn-iwn)* is a multiple of 2π,
{tWny = o(-^) +| = |+0(1) ,4 0.
So, this t is suitable.

358
3. SOLUTIONS TO THE PROBLEMS
Using this t, we form the function
a(x) = (eitx)*, xeE\
and apply the statement proved earlier. Omitting a set of measure 0, a{x)
can be extended to a 27r-periodic continuous function (even a function with
absolutely convergent Fourier series), to be denoted by b(x). Since
[un ,tin + - ) ΠΒ,
\ η η J
as well as its *, has positive measure, and similar statements hold for vn,
after the omission of the set of measure 0 mentioned above, there remain
some points in these sets. Let u'n, v'n be such points. Then u'n — un —> 0,
vn~vn -* 0, hence «-<)* -+ 0, and a (V*) = b (V*), a (/*) = b (/*).
Therefore, by the continuity of b(x), it follows that a (V*) — a (г/*) —> 0.
But
α «)=**·«, α(τ/;)=β«««,
whence
0 = lim[t« - v'n)]* = lim[t(tin - t;n)]*,
and this contradicts the construction of t. Thus δ exists.
For each density point of E, consider its neighborhood of radius Й/4.
Their union S is open and almost covers E. For any two points u\ and
v\ of 5, the relation |(iai — v\)*\ < δ/A. implies \u\ —v\\< Й/4; otherwise
(assuming that δ < π), |τχι— vi | > (3/2)π would follow, and by the
construction of S there would exist density points и and ν in Ε with \u — u\ \ < δ/A,
\v — vi\ < δ/4, hence \u — v\ > π, while \(u — v)*\ < \(u\ — νι)*|+2·5/4 < Й,
which contradicts the property of i.
Thus, the closure of S cannot have two congruent points modulo 2π.
Hence, it also follows that the measure of S does not exceed 2π, and since
each component of the open set S has length greater than or equal to δ/2,
the number of the components must be finite, that is, the closure of S is a
finite union of closed intervals. Consequently, Ε G T. D
Problem M.18. Show that any two intervals А, В С R of positive
lengths can be countably disected into each other, that is, they can be
written as countable unions A = A\ U A% U ... and В = B\ U B% U ... of
pairwise disjoint sets, where Ai and Bi are congruent for every г G N.
Solution 1. In R, consider the equivalence relation
~={(*,2/)eR2: x-yeQ},
which induces the disjoint classification R = U7GrQ7· Obviously, each
equivalence class Q7 is dense in R, so Α Π Q7 and В C\Q1 are countably

3.5 MEASURE THEORY
359
infinite sets. Consequently, assuming the axiom of choice, there exists a
bijection
φΊ : Α Π Q7 —► Β Π Q7
between them for each 7. Then φΊ(χ) ~ ж for every 7 eT, χ e AnQy.
Let Q = {<7i, #2,... } be a numbering of the rationale, and consider the
sets
Ai = U {χ G Α Π Q7 : φΊ(χ) - x = qi) ,
В» = U {φΊ(χ) : x e ΑΠ Q7, 07(ж) - χ = qi} ,
7£Г
for г G N. Prom the definitions, we see that
00 л oo
A= U Ai and JB = U B»
г=1 г=1
are partitions, and Д = A» + q^ for every г G N. D
Solution 2. Denote by σ the relation that holds between the subsets of R
if and only if they can be countably dissected into each other. It is easy to
verify that σ is a translation-invariant equivalence relation satisfying
( U Sn) σ ( υ Γη)
\n=l / \n=l /
if Sn and Tn, respectively, are pairwise disjoint sets (n G N), and SnaTn
for every n.
We now observe that any interval is the disjoint union of a countably
infinite number of open intervals and a countably infinite set. So, by the
above remarks, we may assume that A = ]0, a[, Β = ]0, b[, 0 < a < b.
In R, we now consider the equivalence relation
~= {(^,y)GR2: x-yeQ}.
Its equivalence classes are dense in R, so — assuming the axiom of choice
— there exists a set X c]0, a/3[ containing exactly one element of each
class. Put
Qa = Qn]o,^[, QB = Qn\o,b-^[
and
A*= U (X + q), B*= U (X + q),
qeQA Q^Qb
where X + q stands for the set X translated by the number q G R.
Obviously, A* and B* are countably infinite, disjoint unions of sets congruent
with X, so Α*σΒ*. One also easily verifies the inclusions
a 2a
3'T
с А* с А
[3,6-3] С Б* С В.

360
3. SOLUTIONS TO THE PROBLEMS
Put
A-=]0^[\A\ A+=]^,a[\A\
в-=]о,|[\в*> в+=]ь-|,ь[\в*.
A simple calculation shows that B~ = A~ and B+ = A+ + (b — a).
Consequently, the partitions
A = A~UA*UA+ and В = B~ U B* U B+
prove the statement. D
Remarks.
1. A direct consequence of the statement is the significant fact that there
is no σ-additive, translation-invariant measure defined on all subsets of
R for which the unit interval has measure 1.
2. The first proof can immediately be applied to subsets with nonvoid
interior of nondiscrete, separable, Hausdorff topological Abelian groups.
3. By the usual tools of measure theory, the statement can be extended
to Lebesgue-measurable sets as follows. Any sets А, В С R of positive
Lebesgue measures can almost be countably dissected into each other,
that is, they have partitions
oo oo
A = U Ai and B= U Bi
2=0 2=0
such that Aq and Bo have measure 0, while Ai is congruent with Bi, for
every г e N.
4. The assertion of the previous remark cannot be sharpened by deleting
the sets of measure 0, since — as noticed by a contestant — a set of first
Baire category cannot be countably dissected into a set of second Baire
category; however, among sets of positive Lebesgue measure, there are
examples of this kind.
Problem M.19. Let Η с R be a bounded, measurable set of positive
Lebesgue measure. Prove that
1# . p\((H + t)\H)
liminf -^—-ff^—- > 0,
t->o |t|
where H + t={x + t:xe H} and λ is the Lebesgue measure.
Solution. We show that
л. . \((H + t)\H) л
liminf-^—-f^—- > 1·
t—o \t\ ~

3.5 MEASURE THEORY
361
Lemma. For any 0 < ε < 1, there is an interval [a, b) such that X(H Π
[a,b))>(l-e)(b-a).
Proof. Assume that the conclusion of the lemma is false. Let us cover
Η Ъу а, countably infinite number of intervals (closed from the left), so that
the sum of the lengths of the intervals is less than Л(Я)/(1 — ε). By the
definition of the Lebesgue measure, this is possible. Let the zth interval be
[di,bi). Then by the indirect assumption,
\(НП[<ь,Ь))<(1-е)(Ь-<ь)
and therefore
oo oo
X(H) <^А(ЯП [aub{)) < (1 - ε) £(Ь ~ ai) < *(Я),
г=1 г=1
which is a contradiction. The proof of the lemma is complete.
Let 0 < ε < 1, and let the interval [a, b) be such that
Л(ЯП[а,6)) >(1-ε)(6-α).
Let 0 < t < b - a and η = [(6 - a)/t] + 1. Then
η
(1 - ε)(6 - а) < Л(Я П [а, Ь)) <^А(ЯП[а + (Ь 1)ί, α + kt)).
k=l
Hence, there exists an integer k, 1 < к < η, such that
\{H Π [a + {k - l)t, α + kt)) > ^~ε)^~α)
>(l-g)(b-fl) = (l-g)t
" ^ + 1 1 + *V
Put
Ai = Я Π [α + (fc + г - l)t, а + (fc + i)t) (i = 0,1,2,...).
Since, Η being bounded, A» is empty if г is sufficiently large, we have
oo oo
λ ((Я + ί) \ Я) > Σ λ ((Л, + ί) \ Λί+1) > J] (λ(^) - λ(^+1))
г=0 г=0
= λ(Λ)) = λ (Я Π [а + (к - 1)ί, α + fet)) > j1"^.
1 + b-a
Quite similarly,
А((Я-*)\Я)>£^.
+ 6-a
It follows that
,;„(М(Й + ()\Я)Ч1:„ 1-ε
liminf , , —- > lim з— = 1 — ε.
t-o |t| ~ t-o 1 + glj
But this is true for every 0 < ε < 1. Thus,
,. . r\((H + t)\H) ,
liminf—^ т-г^—'- > 1. D
*-o \t\

362
3. SOLUTIONS TO THE PROBLEMS
3.6 NUMBER THEORY
Problem N.l. Let f and g be polynomials with rational coefficients,
and let F and G denote the sets of values of f and g at rational numbers.
Prove that F = G holds if and only if f{x) = g(ax + b) for some suitable
rational numbers α φ 0 and b.
Solution. By a "polynomial" we shall always mean a polynomial with
rational coefficients, and by the "range" of a polynomial we shall mean the
set of its values assumed at rational numbers.
We use two classifications of polynomials. We write / ~ g if the ranges
of these polynomials coincide, and / « g if f(x) = g(ax + b) for all χ with
suitable rational numbers α φ 0, b. These are clearly equivalence relations,
and they are compatible with multiplications by a constant, that is, if / ~ g
or / ~ 9, then cf ~ eg or cf « eg, respectively (с Ф 0 rational). Our aim
is to show that these relations are identical. One implication is clear: if
f ~ g, then / ~ g holds obviously.
The converse will be proved in several steps.
1. It is sufficient to prove the statement for polynomials with integral
coefficients. Indeed, assume f ~ g, and let с be an integer such that the
coefficients of cf and eg are all integers. We have cf ~ eg, thus cf « eg,
assuming the integer case, which yields / « g as wanted.
2. If / ~ g are polynomials with integral coefficients, we show that they
must be of equal degree. Indeed, assume that
f(x) = axn + ..., g(x) = bxk + ...,
where α φ 0, b φ 0, and η < к. Choose a prime pj(ab. We have
f(l/p) = c/pn, where pj(c. By assumption, g also takes on this value
at some number u/v with (u, v) = 1. Now if pj(v, then ρ cannot divide
the denominator of g(u/v); consequently, ρ must divide v. Now, by
(p,b) = 1, the exponent of ρ in the denominator of b(u/v)k is higher
than in any other term of g(u/v), thus the denominator (in the reduced
form) of g(u/v) contains ρ with an exponent > k. This yields η > k,
hence η = к as asserted.
3. It is sufficient to prove the statement for polynomials with integral coffi-
cients where at least one of the leading cefficients is 1. Indeed, let f ~ g,
f(x) = axn + ... . Then we have
a""1/ (f) ~ a"-7(*) ~ a^gix),
and here an~1f(x/a) still has integral coefficients while its leading
coefficient became 1. By assumption, we conclude an~1f(x/a) « an~1g(x),
which yields f ~ g.
4. Let / ~ g be polynomials with integral coefficients, f(x) = xn + ...,
g(x) = bxn+... . We claim that every prime divisor of b has an exponent

3.6 NUMBER THEORY
363
> п. То show this, take a p\b and consider the polynomials
AW = Ρ"'""1/ (p^r) -!>"<"->/ (p) ~ S.W
We know that all coefficients of /i as well as all but the leading coefficient
of g\ are integral, and the leading coefficient of /1 is 1. Hence every value
of /i has the following property: if (in the reduced form) its denominator
is divisible by p, it is divisible by pn. By /ι ~ g\ the same must hold for
g\. Now #i(l) = (b/pn) + с with an integer c, and by p|6 the exponent
of ρ in the denominator is strictly less than n, thus it must be 0, that
is, pn\b.
5. It is sufficient to prove the statement for polynomials with integral cof-
ficients where both leading cefficients are 1. Let / ~ g be polynomials
with integral coefficients, f(x) = xn + ..., g{x) = bxn + ... . Let cn be
the largest nth power dividing b. Consider the polynomials
/1(x) = c"("-1)/(^T)=x" + ...J
9l(X) = c^9(-^)=±x» + ....
We have /ι ~ g\. By step 4, the exponent of any prime in the leading
coefficient of g\ must be either 0 or > n, and by the definition of с this
coefficient must be 1 or —1. We can exclude the case —1 for even n, since
then one domain would be bounded from below and the other from above.
Finally, if η is odd, then replacing с by — с we can change a —1 into 1. So if
the statement is true under this restriction, then we obtain /i « g\, which
yields / « g.
6. We can also assume that the coefficient of xn~l is 0 in both polynomials.
Indeed, if
f(x) =xn + axn~x + g(x) = xn + bxn~l + ...,
then the polynomials
Л(х) = nnf (^ - a) ~ 9l(x) = nng (^ - a)
have this property, and /i « g\ again implies f ~ g.
7. Combining the previous arguments, we find that to solve the problem it
is sufficient to prove the following statement:
Statement If / ~ g are polynomials of the form
f(x) =xn + axn~2 + ..., g{x) = xn + bxn~2 + ...
with integral coefficients, then f ~ g.
Proof. Observe that, since the leading coefficient is 1, the only rational
numbers where / and g assume integral values are the integers.

364
3. SOLUTIONS TO THE PROBLEMS
Since f(x + l)— g(x) = nxn~1 + ..., we have g(x) < f(x + l) for large x,
and similarly we obtain g{x) > f(x — 1). Thus, for a large positive integer
x, the only positive rational у that can satisfy f(y) = g(x) is у = χ. If
η is even, then there can also be a negative y, and similarly we find that
its only possible value is у = —χ. This means that one of the equations
f(x) = g(x) and f(—x) = g(x) has infinitely many solutions; hence it must
be an identity. This concludes the proof. D
Problem N.2. Show that
Ц (х2 + у2) = (-1)[Ч1} (mod ρ)
1<х<у<^
for every prime ρ = 3 (mod 4). ([.] is integer part.)
Solution 1. Write ρ = 4k + 3. Consider all the sums x2 + y2, 1 < x, у <
2k +1. Let r denote the number of those pairs x, у for which this sum ξ 1.
First, we show that the number of those pairs for which x2 + y2 = —1
(mod p) is r +1. Since every quadratic residue has a unique representation
in the form x2, 1 < χ < 2k + 1, while every nonresidue has a unique
representation in the form —y2, the solutions of x2 + y2 = x2 — (—y2) = 1
count the consecutive numbers in the form (nonresidue, residue) within
the sequence 1,2,... ,p — 2, ρ — 1. Similarly, the pairs with x2 + y2 = —1
correspond to pairs of type (residue, nonresidue). Since this sequence starts
with a residue and (recall that ρ = — 1 (mod 4)) ends with a nonresidue,
the second case must happen r + 1 times.
Next, we prove that the number of solutions of x2 + y2 = a2 (mod p),
1 < x, у < 2fc + 1, is r for every a. Indeed, the mapping χ = ±αχι,
у = ±ayi, where the signs are chosen so that x, x\,y, y\ are all in [1,2fc + l],
provides a one-to-one correspondence between solutions of x2 + y2 ξ a2
and x\ + y\ = 1. Similarly, we obtain that the number of solutions of
x2 + y2 ξ -1 is τ + 1.
The above observations mean that these sums x2 + y2 represent every
quadratic residue τ times and every nonresidue r +1 times. Since the total
number of these pairs is ((p — l)/2)2 while the number of residues and the
number of nonresidues are both equal to (p—1)/2, we find r = (p—3)/4 = k.
Now, the product we want to compute is not over all these pairs but
only over those with χ < у. Consider a quadratic residue a. We know that
the total number of solutions of x2 + y2 = a is k. If among them there are
и with χ < y, υ with χ = у, and w with χ > у, then и = w by symmetry
and ν is 0 or 1, hence и = [к/2]. Analogously, for a nonresidue the number
of solutions with χ < у is [(к + l)/2].
The product of all quadratic residues is

3.6 NUMBER THEORY
365
and the product of nonresidues is
^.„.......(-(^Jm-.)^^).)'
ξ —1 (mod p).
Hence, the original product is
Solution 2. Denote this product by P. In the product, all sums of pairs of
quadratic residues are multiplied. Let g be a primitive root modp, and put
h = g2. The numbers 1, /ι, /ι2,... ,h2k represent every quadratic residue
once; thus we have
P = Y[ (/i* + /iJ') (modp).
0<i<j<2k
This implies
Ρ- Y[ (ti-hi)= Y[ (h2i-h2j) (modp).
0<i<j<2k 0<i<j<2k
Here each product is the value of a nonzero Vandermonde determinant.
Thus, this equation can be rewritten as
Ρ · 1/(1, /ι, /ι2,..., h2k) ξξ 1/(1, /ι2,..., hAk) (mod p).
Since h2k+1 = gAk+2 = ι (mod ρ), the generators of the second
Vandermonde determinant are congruent to
l,h2,...,h2k,h,h\...,h2k-\
Thus, the second determinant can be obtained from the first by k+(k — 1) +
Ь 2 + 1 = k(k + l)/2 transpositions of rows. By the familiar properties
of determinants, this means
Ρ S(-1)^S (_1)№]S (-!)№] (modp). D
Problem N.3. Let ρ be a prime and let
hfay) = акх + Ьку (к = Ι,.,.,ρ2),
be homogeneous linear polynomials with integral coefficients. Suppose
that for every pair (£, η) of integers, not both divisible by p, the values
Iki^v), 1 < к < ρ2, represent every residue class mod ρ exactly ρ times.

366
3. SOLUTIONS TO THE PROBLEMS
Prove that the set of pairs {(flfc,&fc) : 1 < к < p2} is identical mod ρ with
the set {(m, n) : 0 < m, η < ρ — 1}.
Solution 1. Assume that the statement does not hold. Then there are
numbers г ф j such that α* ξ clj and bi = bj (every congruence is meant
modp). Consider the number of those triplets (k,x,y), (x,y) φ (0,0) that
satisfy
lk(x,y) = li(x,y).
By the assumption, for every fixed pair (x, у) ф (0,0), the number of
solutions in к is p\ thus the total number of these triplets is p(p2 — 1).
Now consider the solutions in x, у for a fixed k. If к = г or j, this is an
identity, which means 2(p2 — 1) solutions. For any other к it is easy to see
that the number of solutions is at least ρ — 1. This means that the total
number of solutions is at least
(p2 - 2)(p - 1) + 2(p2 -1)= p(p2 - 1) + p(p - 1) > p(p2 - 1),
a contradiction. D
Solution 2. It is sufficient to prove that for every и and ν there is exactly
one к with a& = u, bk = v.
The uniformity of representations implies that for every (ξ, η) Φ (0,0),
we have
ρ2
27гг /
k=l
We multiply both sides by e(2™/p)-«+*"/) to get
p2
k=l
for (£,77) ^ (0,0). For ξ = η = 0, the same sum obviously gives p2. Now
summing these sums for all possible values of ξ and 77, including (0,0), we
obtain
£=0r/=0fc=l
Changing the order of the summations, we get
5 = \^ \^ V^ g^iiafc-ti^+ibfc-t;)!,)
fc=l£=0 77=0
Л>-1
2
uA(j2^{bk~v)1')) =p2·

3.6 NUMBER THEORY
367
Now, a typical factor in this product vanishes, unless ak = и (for the first)
or bk = ν (for the second). Thus, the whole product is 0, unless a^ = и
and bk = v, in which case it is equal to p2. Since the sum of these products
is p2, this case must happen for exactly one value of к. D
Remarks.
1. The method of the second solution can be applied to prove the following
generalization of the problem: Let ρ be a prime, r a positive integer,
and the Z^'s homogeneous linear polynomials in η variables of the form
lk(xu · · ·, xn) = ο,ιχι + ·- + aknxn (k = 1,..., rpn).
Assume that for every Ar-tuple (£i,..., ξη) of integers, not all divisible by
p, the values of Zfc(£i,..., ξη) represent every residue class mod ρ exactly
rpn~l times. Then every η-tuple (mi,...,mn) is represented mod ρ
among the n-tuples (a\,..., α£) of coefficients for exactly r values of k.
2. If we relax the requirement of uniform representation to those pairs (£, η)
where neither ξ nor η is divisible by p, then the following can be asserted:
Assertion. Let f(m,n) denote the number of those pairs (α^,Μ that
satisfy dk = m, bk = n. Then the values /&(£, 77) will be uniformly
distributed for all of the p2 — 2p + 1 admissible pairs (£, η) if and only if
/(m, n) can be represented as /(m, n) = g(m) + h(n).
Problem N.4. Let ρ be a prime, n a natural number, and S a set of
cardinality pn. Let Ρ be a family of partitions of S into nonempty parts of
sizes divisible by ρ such that the intersection of any two parts that occur
in any of the partitions has at most one element. How large can |P| be?
Solution 1. This maximum is (pn — l)/(p— 1). Let Η be the set to be
partitioned, and let Ci,..., Ck be its partitions into sets whose cardinalities
are multiples of p, such that any two classes may have at most one common
element. Consider an h G H, and let Ci(h) be the class of Ci that contains
h. By the assumptions, the sets
σι(Λ)\{Λ},..·,σ*(Λ)\{Λ}
are pairwise disjoint, and each has at least ρ — 1 elements. Consequently,
we have k(p - 1) < pn - 1, that is, к < (pn - l)(p - 1).
Now we prove the corresponding lower estimate. Consider the n-dimen-
sional projective space Pn over a finite field К of ρ elements, a hyperplane σ
in it, and the affine space P'n = Ρη\σ. Observe that Pn has (pn+1—l)/(p— 1)
points, σ has (pn — l)/(p — 1) and P'n has pn.
For an arbitrary point Pea, consider those lines of Pn that contain
Ρ but do not lie in σ. If we omit the point Ρ from each such line, the
remaining affine lines (as sets of points) form a partition Ср of the affine
space P'n into sets of cardinality p. The number of these partitions is the
same as the number of points of σ, that is, (pn — l)/{p — 1). □

368
3. SOLUTIONS TO THE PROBLEMS
Solution 2. We keep the upper estimation from the first proof, and present
a different construction to show the lower bound.
For Η we take the set of all η digit numbers in base p. This set has pn
elements. The partitions will be given in forms of integer-valued functions;
f(x) will mean the number of the class containing a given χ G H. The
functions will be indexed by pairs (j, a), where j = 0,..., η - 1, and for a
given value of j, the possible values of a are α = 0,..., p·7 — 1. The number
of possibilities for a is p·7; thus the total number of these functions is
9 „ ι Pn ~ 1
p-1
as wanted.
We define the function fja(x) as follows. Let the representations of χ
and a to base ρ be
Χ = ξο+ξιΡ + ·-+ξη-ιΡη~\
a = a0 + Οίχρ Λ V OLj-ip*~l.
Let [m] denote the (smallest nonnegative) residue of an integer m mod p.
Now we put
fja(x) = [ξο + ototj] + [ξι + a^j]p + · · · + fo_i + α,-ι^ν-1
+ ^·+ι^ + ··· + £η-ιΡη~2.
In the case j = 0, a = 0, we interpret this as
f00{x) =ξΐ+ξ2Ρ+'"+ ξη-ΐρη~2.
We show that every class contains ρ elements. Indeed, given the value
of
fja(x) = у = 77o + · · · + Vn-2PU~2,
we have ξι^ = Щ-ι for к = j + 1,..., n, we can choose the value of ξj
that will give ρ possibilities, and after fixing £j we can uniquely determine
ξο,. · ·, Cj-i from the congruences
ξι + ol^j = щ (modp) (i = 0,...,j-l).
Next, we have to show that the intersection of two classes has at most
one element, that is, the system of equations
fja(x) = 2/, fj'a*(x) = У
has at most one solution. Assume first that j φ j', say f < j. Then we
obtain £j = 77J from the second equation, and we already know that fja(x)
and £j determine χ uniquely.
Finally, consider the case j = f. We must have α φ α', say α& φ a'k for
some digit к < j — 1. Then the congruences
£k+c*k£j =Vk (modp),
6 + a'kt>3 = rfk (m<>d P)
determine £j, which together with faj (x) determines χ uniquely. D

3.6 NUMBER THEORY
369
Problem N.5. Let f be a complex-valued, completely multiplicative,
arithmetical function. Assume that there exists an innnite increasing
sequence Nk of natural numbers such that
f(n) = АкфО provided Nk<n<Nk + 4a/A^.
Prove that f is identically 1.
Solution. We show a slightly stronger statement, where the 4 is replaced
by 2 + ε. Assume that / is (nonzero and) constant on the intervals Ik =
[Nk, Nk + Mfc], where Mk = (2 + е)у/Щ.
First, we prove that f(n) φ 0 for any n. Indeed, if Mk > n, then
nx e Ik for some x\ hence f(n)f(x) = f(nx) φ 0. Next, we use an interval
of constancy to create another. Let / be constant on / = [Ν, Ν + M].
If for an η we can find an integer χ such that nx,(n + l)x G /, then
f(n)f(x) = f(nx) = /((n + l)x) = f(n + 1)/(ж), hence /(n) = f(n + 1).
Now this condition can be reformulated as
Ν Ν + Μ
— <χ< —.
η η + 1
Such an integer χ can be found if
N + M N
η + 1 η
>1,
or, after rearranging, n2+n(l—M)+N < 0. This means that η lies between
the roots of the corresponding quadratic equation, that is, (M — l)2 > AN
(which shows that our method does not work with Μ = (2 — e)y/N) and
M-l-y/(M-l)2-4N Μ-1 + λ/(Μ-1)2-4ΑΓ
2 -П- 2 '
If Μ = (2 + e)y/N, then the above interval includes an interval of the form
Γ = [ciM, C2M], where 0 < c\ < c<i are constants depending on ε.
Now we repeat this argument for the interval Γ. The result is an interval
I" = [сз,С4М] of constancy. As I runs over all the intervals Ik, these
intervals I" cover the whole half-line [сз,оо]; thus /(n) is constant for
η > C3. Now take an arbitrary positive integer m, and select an η > C3.
We have /(n) = f(mn), hence f(m) = 1 as asserted. D
Remark. The statement will not hold if the Ay/Nk is reduced to exp
(c\/log Nk log log log Nk) with a suitably small positive с
To see this, consider a sequence Ik = [Nk,Nk + Mk] of intervals, Nk >
Mk > Nk-i + Mk-ъ w^h the property that every number η e Ik has a
prime factor ρ > Mk. Choose the numbers Ak φ 0 arbitrarily. We claim
that there is a completely multiplicative function / that is identically Ak
on Ik; if not, every Ak is 1, and then our function will not be identically 1.

370
3. SOLUTIONS TO THE PROBLEMS
We construct our function recursively. Assume that f(p) is fixed for
every prime ρ < Nk-i + Mk-i\ we define f(p) up to Nk + M^. For every
η £ Ik, let pn > Mk be the largest prime divisor of n. These primes are
distinct, since for η φ η' we have (n,m) < \nf — n\ < Mk- Choose f(p)
arbitrarily for every prime that does not occur among the pn's, and then
set f(pn) to achieve /(n) = A^.
Now we have to find such a sequence of intervals. Assume that /χ, ... ,
Ik-1 are given. Take a large N, say N > exp Α^_χ; we try to find an Ik in
[N/2,N].
Let R(x, y) denote the number of those integers η < χ whose prime
factors are all less than y. If Μ is such that M(R(N, M) + 1) < N/2,
then the interval [N/2, N] contains a subinterval of length Μ that is free of
these numbers. This interval can serve as our Ik. The feasibility of taking
Μ = exp(c\/log Nk log log log Nk) follows from Rankin's inequality:
/ log log у \
R(x, y) < χ exp I j log χ + 0(log log y) 1 .
Problem N.6. If с is a positive integer and ρ is an odd prime, what is
the smallest residue (in absolute value) of
2=1
W2;V (modp)?
ΎΊ—П ^ /
n=0
Solution. Let q = (p—1)/2. With this notation, we have 2j+l ξ —2(q—j)
for every j. Consequently, we have
2n\ _ (2n)! _ 2n · n! · 1 · 3 ·... · (2n - 1)
η j n!2 n!2
2n ■ 1 · 3 ■... - (2n - 1)
n!
2η·(-2Γ·9·(9-1)·...·(9-η+1)
(-4)
"(I)·
n!
Hence,
n=0 x 7 n=0 x 7
This is congruent to 0 if 1 — 4c is divisible by p. Otherwise, it is 1 if 1 — 4c
is a quadratic residue mod ρ and —1 if it is a nonresidue. D

3.6 NUMBER THEORY
371
Problem N.7. Find a constant с > 1 with the property that, for
arbitrary positive integers η and к such that η > ck, the number of distinct
prime factors of (£) is at least k.
Solution. Write t = [1,2,..., n]. We shall show that the number of prime
factors of (^) is at least к for η > t + k.
Take a prime ρ < k, and let ps < к < pk+1. The exponent of ρ in the
decomposition of k\ is Σΐ=ι^/ρ% and in t it is s. Among the numbers n,
... , n — k + 1, there are at least [fc/p2] multiples of рг. For г < s, these also
enter the decomposition of the corresponding (n — j, £); thus the exponent
of ρ in the product (n, t) · (n — 1, t) ·... · (n — /с + 1, t) is at least as high as
in k\. Since this holds for every p, we conclude that
fc!|(n, t) · (n - 1, t) ·... · (n - A: + 1, t).
This implies that
k-l
Пт^
г
г=0
(η — г,£)
The factors on the left side are pairwise coprime, since (n — г, η—j)|(i—j)\t
for i^j, and if η > ί + /с, they are all larger than 1. Taking a prime factor
of each, we find к different prime factors of (^).
Since t = exp(k + o(k)) by the prime number theorem, every с > е
satisfies the requirement of the problem for large k. If we want to exhibit a
concrete value of с that works for every k, we can argue as follows. Applying
Chebyshev's estimate,
П?<4Х'
p<x
we obtain
Therefore,
Π ps<Uk U p< kV~k~4k
p»<fe<p»+1 p<y/k Vk<p<k
\ к
t + k< k^Ak = Uk1/Vk) < 9k
since the function x1!^ assumes its maximum at e2, and its value is e2/e <
2.1. D
Remark. The value с = 9 we gave can be further reduced with a little
extra effort; one of the contestants gave с = 4.3. It seems to be difficult to
decide the improvability of с = e + ε for large k.
If the "prime Ar-tuple conjecture" is true, then there are arbitrary large
values of η for which (^) has exactly к prime factors.

372
3. SOLUTIONS TO THE PROBLEMS
Problem N.8. Let f(n) be the largest integer к such that nk divides
n!, and let F(n) = max2<m<n 7(n)· Show that
lim 4^*n = L
n-*oo П log log П
Solution. First, we find an upper estimate for f(n). Let η = Пг=1^Г* be
the prime factorization of n. Since η·^η)|η!, we have
aj(n)<£
η
я.
η
t is,
η log^
^zlogPz < 77-^r r,
/(n) p< - 1
all i. Summing these inequalities, we obtain
loff η = \ rv- loff n.
<
η ^ log ^
Let ςΊ < ^2 < · · · be the sequence of all primes. Since {\ogp)/{p — 1) is
decreasing, we have
Combining the last two inequalities, we get
/W * (i + «.»!££
On the other hand,
n=pV...plh>qi...qk>2k.
Consequently, к <C logn, log к < loglogn+0(l), and the previous estimate
of /(n) yields
д„)й(1 + 0(1))!~
as wanted.
Now we construct a number m <n with a large value of /(m).
Define mi by (mi+1)! < η < (mi+2)!. We have mi ~ (logn)/(loglogn)
as η —> oo. Let ρ be the largest prime up to (n/mi!)1/3, and put m =
p3mi!. We have m < η obviously, and, since n/mi! > mi —> 00, we have
ρ ~ (n/mi!)1/3, that is, m ~ n.

3.6 NUMBER THEORY 373
Since (mi!)m/mi \m\, the exponent of any prime other than ρ is at least
m/mi times as high in m! as in m. We have to compare the exponents of
p. The exponent of ρ in mi! is Σί=ι [mi/Pl] < mi/{p ~ 1)? hence in m
it is less than mi/(p — 1) + 3. On the other hand, the exponent of ρ in
m! is Σίΐι [т/Рг] > [m/p] > im/p) ~ 1· Therefore, the quotient of these
exponents is at least
m -i
J ~ 1 m
A+3 Ш1'
because ρ —> oo, but
Ρ < (n/mi!)1/3 < ((mi + l)(mi + 2))1/3 = o(m).
Consequently, we have
F(n) > f(m) > (1 + 0(1))^ = (1 + o(l))^- = (1 + o(l))^^,
mi mi log η
as asserted. D
Remark. It can be proved that
n(loglogn-logloglogn)
logn
\\ogn) '
Problem N.9. Prove that a necessary and sufficient condition for the
existence of a set S С {Ι,.,.,η} with the property that the integers
0,1,..., η — 1 all have an odd number of representations in the form χ —
у, х, у е 5, is that (2n — 1) has a multiple of the form 2 · 4* — 1.
Solution. Let S be a set with this property, and write
f(x) = J2xa-\ 9(χ) = Σχη~α,
aeS aeS
which we regard as polynomials over GF(2). Then we have
f(x) = x-'si;), (1)
X
and
2n-2
f(x)g(x) =J2XJ Σ l = l + x + χ2 + ''' + *2П~2' (2)
3=0 a,beS
a—l+n—b=j
by our assumptions on S. Conversely, if / and g are polynomials over
GF(2) that satisfy (1) and (2), then the set
S = {a : the coefficient of xa~l in / is 1}
has the desired property.

374
3. SOLUTIONS TO THE PROBLEMS
Now put
1 + χ + x2 + · · · + x2n~2 = pi{x). ..pi(x), (3)
where the factors p\, ... , p\ are irreducible over GF(2). With a suitable
choice of the suffices, we have
f =Pi ...pr, g =pr+1...pi
by (2).
Observe that the roots of 1 + χ + χ2 + · · · + x2n 2 are the (2n — l)th
roots of unity except 1. Furthermore, we have
(1 + χ + x2 + · · · + x2n~2)f = 1 + x2 + · · · + x2n~*
and
(1 + χ + x2 + · · · + x2n~2) - (x + ж2)(1 + ж2 + · · · + x2n~*) = 1,
and hence this polynomial has no multiple roots. Since the roots of / are
the reciprocals of the roots of g by (1), no pi can have two roots that are
reciprocals of each other. Conversely, if no pi has reciprocal roots, then
the polynomials p\, ... , p\ can be coupled so that the roots of any pi are
the reciprocals of the roots of its mate. We now multiply one element from
each pair to get the polynomial / and the others to get g\ these obviously
satisfy (1) and (2). Thus, such a set S exists if and only if for every root a
ofl + x + x2-\ b£2n~2, ol and 1 /a belong to different irreducible factors.
We construct this decomposition (3).
Lemma. Consider the permutation of the roots of the equation x2n~l —
1 = 0 given by the operation of squaring. Let C\, ... , C\ be the cycles of
this permutation. The irreducible factors of x2n~l — 1 are the polynomials
Ц(х-0, i = l,...,Z.
We prove this lemma at the end of the solution; see also E. R. Berlekamp,
Algebraic Coding Theory, McGraw-Hill, New York, 1968, Chapter 6.
Hence, our condition for the existence of the set S is satisfied if and only
if the roots a and I/a cannot be tranformed into each other by repeated
squarings for any α φ 1, that is,
a2"+1 φ 1
for all к > 0 and (2n— l)th roots of unity α φ 1. This can be reformulated
as
(χ2η-1-1,χ2*+1-1)=χ-1
for all k. Since
{xu-l,xv-l) = x^u^-l,
the set S exists if and only if
(2n-l,2* + l) = l (4)

3.6 NUMBER THEORY 375
for all k. Now let d be the smallest exponent for which 2n — l\2d — 1. If d
is even, then
2n-l|(2d/2-l)(2d/2 + l),
but 2n - l/(2d/2 - 1), and thus (2n - 1,2d/2 + 1) > 1, which contradicts
(4). Conversely, if d is odd, then we have
(2n-l,2* + l) = (2d-l,2* + l) = 1.
Consequently, S exists if and only if d is odd, which is clearly equivalent
to the condition given in the problem.
Proof of the lemma. Let
N
k=0
be a polynomial over any field of characteristic 2. We have
(</>(z))2 = f^z2*.
k=0
Thus, the equation
(φ(χ))2 = φ(χ2)
holds identically if and only if a\ — a^ for all k, that is, every a^ is 0
or 1. On the other hand, the above equality is clearly equivalent to the
assumption that for every root α of 0, a2 is a root as well. D
Problem N.10. Prove that the set of rational-valued, multiplicative
arithmetical functions and the set of complex rational-valued,
multiplicative arithmetical functions form isomorphic groups with the convolution
operation fog denned by
(/°5)(п) = £/(%ф.
d\n
(We call a complex number complex rational, if its real and imaginary parts
are both rational.)
Solution. Let Q be either the field of rational numbers or the field of
complex rational numbers, and let Gq denote the group of Q-valued
multiplicative functions. It is well known that Gq is an Abelian group. Now
we prove that for every natural number η and every / G Gq,the equation
9 ° 9 ° · · · ° 9 = 9
y ν '
η times
is solvable and the solution is unique.

376 3. SOLUTIONS TO THE PROBLEMS
Indeed, we have #(1) = 1 by multiplicativity. For every prime power pk,
we have
f(Pk)= Σ 9(pkl)...g(pk")
feiH hk„=k
Σ g(pkl)...g(pkn) + ng{Pk).
ki-\ \-kn=k
ki<n
This yields a recursion for f(pk), which shows the unicity. To show the
existence, consider the function g whose values are defined by the above
recursion at prime powers and that is extended multiplicatively to the other
numbers, g о · · · о д is a multiplicative function that coincides with / at
prime powers; thus they must be identical.
Thus, both groups are divisible, torsion-free Abelian groups. By the
fundamental theorem of divisible Abelian groups, both are isomorphic to
some discrete direct power of the additive group of rational numbers.
Since we can prescribe the values of a multiplicative function at prime
powers arbitrarily, we conclude that the cardinality of Gq is 2^° for both
choices of Q. This implies that the direct power must have 2^° factors in
both cases. This proves the isomorphy of the two groups. D
Problem N.ll. Let Η denote the set of those natural numbers for
which r(n) divides n, where r(n) is the number of divisors ofn. Show that
(a) n! G Η for all sufficiently large n,
(b) Η has density 0.
Solution.
(a) We show that r(n!)|n! for all η φ 3,5. We have
oo
«!=ПЛ αρ = Σ\Φ%
р<п г=1
and consequently,
τ(η!) = JI(op + l).
p<n
To prove r(n!)|n!, it is sufficient to find, for every prime ρ < η, a
natural number h(p) < η such that ap + l\h(p) and h(p) φ h(q)
whenever ρ φ q. If ρ < y/n, put h(p) = ap + 1. We have
oo oo
h(p) = 1 +ap < 1 + 5}n/2'] < 1 + ]T n/2* = 1 + n,
2=1 2=1
hence h(p) < n. Also, if ρ < q < y/n, then
η η (q — p)n ^ η
= Λ^—L·^ > _ > i;
ρ q pq pq

3.6 NUMBER THEORY
377
consequently, [n/p] > [n/q]. Since also \p/jP\ > [n/qj], for all j, we
can infer that h(p) > h{q). For the primes y/n < ρ < η, we define
h(p) by recursion. Assume that h(q) is already defined for every prime
q < p. We need to find a multiple of ap + 1 that is not among the
already distributed numbers h(q), q < p. Observe that [n/jP\ = 0,
for j > 2; thus ap = [n/p]. Hence, the number of multiples of ap + 1
up to η is
η
_ap + 1
=
η
l + [n/p}_
= (p-i)
n + p
>
>
1 + [n/p]
p-l
n/p\ > η - (n/p)
- 1+{n/p)
with strict inequality unless ρ = n. The number of already occupied
values is π(ρ) — 1 < (ρ — l)/2, with strict inequality unless ρ = 3,
5, or 7. Thus there is a free number for h(p) except possibly in the
cases η = ρ = 3, 5, 7. In these cases, the divisibility r(n!)|n! holds
for η = 7 and does not hold for η = 3 or 5.
(b) We split Η into two parts. Choose a parameter К > 1. We put those
numbers in which the exponent of every prime is < К into H\ and
the rest into #2· Consider first an η G H\. If η = ΓΠ=ι Pj3 j аз < ^
then r(n) = Π(α.? + 1) consists exclusively of primes < K, and by
r(n) \n, the exponent of every prime is less than K. Consequently,
-«
(Up)
к
<K
кг
On the other hand, we have r(n) > 2s, and these inequalities together
imply
s<[K2\og2K] = r.
It is well known that the sequence of numbers that have at most r
prime factors has density 0, hence so does H\.
Every element of Щ is divisible by the if th power of a prime. Hence,
the number of elements of H2 up to χ is at most
00 /»oo
Hence, the density of Η is at most l/(K — 1). Since К was arbitrary,
Η must have density 0. D
Remark. Let H(x) denote the number of elements of Η up to x. It can
be shown that
^logi)"'1/2^' <C H(x) <C x(\ogx)~1/2,
for χ > χο(ε). We outline a proof of these inequalities.

378
3. SOLUTIONS TO THE PROBLEMS
We start with the upper estimate. Write η in the form
n = 2lp1p2...pkm2,
where p\, ... , pk are odd primes and m is an odd integer.
The number of possibilities for m is < y/x, for Ζ it is О (log x). Hence, the
number of those η for which p\...pk < x1^3 is 0(xb^\ogx) =
o(x(\ogx)~1^2). In what follows, we assume that pi...pk > x1^3.
The exponent of each pj in η is odd, hence 2к\т(п), so к < I if η G H.
For fixed I and m we have
τ
Pi-"Pk<y =
2<m2'
and у > xx 1Ъ by the previous assumption. Now by a classical theorem of
Hardy and Ramanujan, the number of integers < у that are products of к
distinct primes is
У (с + log log у)*"1
logy (*-1)!
with an absolute constant с Since log χ > logy > (1/3) log χ in our case,
this is
у (с + log log x)k~l
logx (k — 1)!
We have to sum this for all possible values of / and m. Write / = к + j; we
know that j > 0. With this substitution, our sum is
^ fe-?- 2 x (c + loglogrr)
^ logx (fe-1)!
k,j,m
Here the sum over m and j contributes only a constant factor, while the
sum over к is just the power series expansion of
expc + loglog* = c,(log;r)1/2
This concludes the proof of the upper estimate.
The lower estimate can be obtained by considering numbers of the form
n = 2q~1qpi...pk,
where q, pi, ... , pk are distinct odd primes. This number has r(n) = 2h+1q
divisors, thus η e Η if к < q — 2. For a fixed q, q < (1 — ε) log log x, the
number of such η < χ is approximately
ι q χ (log log гг)**-3
4 log* (q-3)\ '
The lower estimate follows by taking q ~ \ log log x.

3.6 NUMBER THEORY
379
Problem N.12. Prove that if а, x, у are p-adic integers different from
0 and p\x, pa\xy, then
1 (i + «)'-isbg(i + aQ (mode).
У X X
Solution. Every p-adic number a can be uniquely represented in the form
a = ρ^^ε, where ε is a unit and v(a) is an integer; p-adic integers are
characterized by v{a) > 0. Hence, a divisibility α\β is equivalent to u(a) <
ν{β). In particular, the condition pa\xy means 1 + v(a) < v{x) + v(y), and
the statement to be proved means ν {A) > и (a), where
1= l(l + s)*-l log(l + aQ
у χ χ
By the previous observation, it is sufficient to show that ν {A) > v{x) +
"M -1·
The series
and
χ2 ■ . -x»-l*"
log(l + χ) = χ - — + · · · + (-Ι)""1 — + ...
λ Π
are convergent since ρ\χ, (^) is a p-adic integer, and η — и (η) —> oo.
Substituting these expansions into A, we obtain
oo
A=J2Bm
n=2
where
= {(у-1)(у-2)...(у-(п-1))-(-1Г-1(п-1)!}:""1
П!
Here the first factor is a polynomial in у whose constant term vanishes;
thus it is a multiple of y. Consequently, Bn = cyxn~l/n\ with a p-adic
integer c, thus
v{Bn) > v(y) + (n - 1)|/(ж) - ι/(η!).
To prove that v(A) > v{x) + v(y) — 1, it is sufficient to show that v(Bn) >
v(x) + v(y) — 1 for every n; by the previous inequality, this would follow
from v(n\) < (n — 2)v{x) + 1. Since v(x) > 1 by assumption, we need only
to show v(n\) < η — 1. But the exponent of ρ in n! is
OO r -ι OO
v(n\)=y \ — \< > -7 = -<n.
Ρ3
j=1P> Ρ
j=i L'
Consequently, z/(n!) < η — 1 since it must be an integer. This concludes
the proof. D

380
3. SOLUTIONS TO THE PROBLEMS
Problem N.13. Let 1 < a\ < a2 < · · · < an < χ be positive integers
such that ΣιΊ/α» — 1· Let У denote the number of positive integers
smaller than χ not divisible by any of the di. Prove that
ex
У>
logx
with a suitable positive constant с (independent of χ and the numbers a^.
Solution. The number of multiples of dj up to χ is [x/clj]. Hence, the
total number of multiples is
<ri-l<r-<*·
We have to improve on this trivial bound.
Choose an xo such that the number of primes between χ/2 and χ is at
least x/(S\ogx) for all χ > xo', such an xq exists by the prime number
theorem. For χ < xo, we use the trivial estimate у > 1.
For χ > xo, we distinguish two cases.
If the number of dj > χ/2 is less than x/(6\ogx), then there are at
least x/(6logx) primes χ/2 < ρ < χ that are not contained among the dj.
These primes cannot be divisible by any ay, thus in this case we obtain
У > x/{§\ogx).
If the number of dj > χ/2 is at least x/ (6 log x), then we have
Ε 1^1- Σ -<i--^r^ = i-^—.
а&,2а* ,W2U' X6l0gX 61°gX
Hence, the number w of integers up to x/2 that are divisible by some aj
satisfies
This implies
w< £
CLj<x/2
x/2
X X
2 121ogx'
w > — 1. D
~ 12 log χ
Problem N.14. Let Τ e SL(n, Z), Jet G be a nonsingular nxn matrix
with integer elements, and put S = G~lTG. Prove that there is a natural
number к such that Sk e 5L(n, Z).
Solution. Let d be the absolute value of the determinant of G, a
positive integer by the assumption. Every element of G~l is a fraction with
denominator d. Since we have
gk = G-lTkQ = Q-Цгрк _ j)G + /?

3.6 NUMBER THEORY
379
Problem N.12. Prove that if а, х, у are p-adic integers different from
0 and p\x, pa\xy, then
l(l+*)'-ls "*(! + *) (mode).
у χ χ
Solution. Every p-adic number α can be uniquely represented in the form
a = ρ^α)ε, where ε is a unit and v{a) is an integer; p-adic integers are
characterized by v(a) > 0. Hence, a divisibility α\β is equivalent to v(a) <
ν(β). In particular, the condition pa\xy means 1 + u(a) < u(x) + v{y), and
the statement to be proved means v(A) > v(a), where
1(1+д?)у-1 log(l + a;)
у χ χ
By the previous observation, it is sufficient to show that v{A) > v{x) +
u(y) - 1.
The series
(l + x)" = l+Qx+...Qx4...
and
log(l + *)=x-^ + -.. + (-l)"-1—+ ...
2 η
are convergent since p\x, (J[) is a p-adic integer, and η — v{n) —> oo.
Substituting these expansions into A, we obtain
oo
A=J2Bm
n=2
where
= {(у-1)(у-2)...(у-(п-1))-(-1Г-1(п-1)!}^.
Here the first factor is a polynomial in у whose constant term vanishes;
thus it is a multiple of y. Consequently, Bn = cyxn~l/n\ with a p-adic
integer c, thus
u(Bn) > i/(») + (n - l)v(x) - ι/(η!).
To prove that z/(A) > z/(x) + v(y) — 1, it is sufficient to show that v(Bn) >
v{x) + v{y) — 1 for every n; by the previous inequality, this would follow
from v{n\) < (n — 2)v(x) + 1. Since v(x) > 1 by assumption, we need only
to show v(n\) < η — 1. But the exponent of ρ in n! is
OO r -ι OO
u(n\) = > — < > — = < n.
3 = 1 L^J J = l^ ^
Consequently, v{n\) < η — 1 since it must be an integer. This concludes
the proof. D

380
3. SOLUTIONS TO THE PROBLEMS
Problem N.13. Let 1 < a\ < a2 < · · · < an < χ be positive integers
such that Σ™1/α,ΐ < 1. Let у denote the number of positive integers
smaller than χ not divisible by any of the a^. Prove that
ex
У>
logx
with a suitable positive constant с (independent of χ and the numbers di).
Solution. The number of multiples of dj up to χ is [x/clj]. Hence, the
total number of multiples is
< W-l < Y-<x.
We have to improve on this trivial bound.
Choose an xq such that the number of primes between x/2 and χ is at
least rr/(31og:r) for all χ > χ$; such an x$ exists by the prime number
theorem. For χ < xo, we use the trivial estimate у > 1.
For χ > xo, we distinguish two cases.
If the number of dj > x/2 is less than rr/(61og:r), then there are at
least rr/(61og:r) primes x/2 < ρ < χ that are not contained among the dj.
These primes cannot be divisible by any a^; thus in this case we obtain
У > x/(6\ogx).
If the number of dj > x/2 is at least rr/(61og:r), then we have
1
^ α, ^ α7
л, — ix7- x 6 log χ 6 log χ
αό<χ/2 J clj>x/2 J ° °
Hence, the number w of integers up to x/2 that are divisible by some dj
satisfies
αά<χ/2
This implies
Tx/2
[ аз .
w > -
<-l·
X
12 log x'
y> \-] -w> — 1. D
y ~ Y2\ ~ 121og:r
Problem N.14. Let Τ e SL(n, Z), Jet G be a nonsingular η χ η matrix
with integer elements, and put S = G~lTG. Prove that there is a natural
number к such that Sk G SL(n, Z).
Solution. Let d be the absolute value of the determinant of G, a
positive integer by the assumption. Every element of G~l is a fraction with
denominator d. Since we have
Sk = G-iTkG = G-i(T* _ /)G + /,

3.6 NUMBER THEORY
381
2
it is sufficient to prove that there is a positive integer к for which every
element of Tk — I is a multiple of d (the determinant of Sk is
automatically 1).
By the box principle, there are two matrices among T°, T1, ... , Td"
say Tl and Tm, Ζ > m, such that every element of Tl is congruent to the
corresponding element of Tm. This means that every element of Tl — Tm =
jrni^jii-m _ /) is a multiple of d. This property is preserved if we multiply
it by the matrix T~m, which has integer elements by Τ e 5L(n, Z). D
Problem N.15. Let
f(n) = £ pa.
p\n
pa<n<pa + 1
Prove that
log log η
limsup /(n) =1.
n_*oo П lOg П
Solution. It is well known that the number ω(η) of different prime divisors
of η is at most (1 + o(l)) · (log n)/(log log n). Since obviously f(n) < ηω(η),
we conclude immediately that
loglogn
hmsup/(n)— < 1.
n_*oo Π log Π
To prove the other inequality, we select a parameter к and try to compose
an η from primes below k. The number of these primes is ~ k/ log к by
the prime number theorem. Now take another number m, which we shall
later specify in terms of k. Every prime ρ < к has a power between m
and km. We cover the interval [ra, Arm] with intervals of the type [m(l +
ε)ι_1,7τι(1 + ε)2], where г = 1, ... , [clogfe], с depends only on ε. For a
suitable г, there are at least Ζ = [c\k/'(log k)2] primes ρ < к that have
a power in this interval. We select / of them; let Ρ be the product of
these / primes. If Ρ < επι, then there is a multiple of Ρ in the interval
[m(l + ε)1, m(l + ε)ζ+1]; let η be such a multiple. We have
f(n) > im(l + ε)*-1 > in(l - ε)2.
We have to estimate η in terms of /. We have Ρ < kl anyway; thus the
choice m = kl+l guarantees Ρ < επι \ik> l/ε. Further, we have
η < m(l + e)i+l < fem(l + ε)2 < fc/+2,
hence / > (log n)/(log /c) — 2. Also η <kk from the above inequality. Thus
/с ^> (logn)/loglogn, and we obtain
,>(1_£)j2ifL· /(η)>(1-ε)3η1°^
log log η' log log η'
for large k. D
Remark. It can be shown that
nlogn
max/n ~ -—■ .
n<x loglogn

382
3. SOLUTIONS TO THE PROBLEMS
Problem N.16. Let a and b be positive integers such that when
dividing them by any prime p, the remainder of α is always less than or equal
to the remainder ofb. Prove that a = b.
Solution. By taking a ρ > max(a,6), we immediately see that a < b.
Assume that a < 6, and write
к = min(a, b — a).
By a theorem of Sylvester and Schur, there is a prime ρ > к that divides
(k)' We show that this prime contradicts the assumption.
We know
p|6(6-l)...(6-fc+l),
and thus the residue of b is a number r < к — I. Now if a < 6/2, then
к = a < p, and thus the residue of a is a = к > r, as claimed.
If a > 6/2, then we have к = b — a. Write 6 = qp + r. We have
a — (q— l)p = (6 — k) — (q — l)p = p + r — k>r,
while p + r — к < ρ — 1, and thus this is the residue of a and it was larger
than r. D
Problem N.17. Call a subset S of the set {1,..., n} exceptional if any
pair of distinct elements of S are coprime. Consider an exceptional set
with a maximal sum of elements (among all exceptional sets for a fixed n).
Prove that if η is sufficiently large, then each element of S has at most two
distinct prime divisors.
Solution. We prove the following slightly more general statement.
Statement. We call a subset S of the set {1,..., n) Ar-exceptional if any
к distinct elements of S are coprime. Consider a Ar-exceptional set with a
maximal sum of elements (among all exceptional sets for a fixed n). If η
is sufficiently large, then each element of S has at most two distinct prime
divisors.
By maximality, every prime ρ < η must have a multiple in S. We divide
the primes ρ < η into two classes: into Ρ we put those for which there is
an m > 1 such that pm G 5, and into Q those for which the only multiple
of ρ in S is ρ itself.
We can estimate the number of ρ e Ρ, ρ > χ, as follows. For every
such p, take an m > 1, pm G 5, and then a prime q\m. These primes
satisfy q < n/x and any prime is represented at most к — I times among
them; thus the number of our p's is at most (k — 1)π(η/χ). Consequently,
if 1 < χ < у < η, then the number of primes q G Q, χ < q < y, is at least
n(y)-n(x)-(k-l)n-. (1)

3.6 NUMBER THEORY
383
Now assume that there is a t G S that has at least three distinct prime
divisors, say t = uav^w^r, where и < υ < w are primes, and a, /3, 7, and r
are positive integers. Our plan is to find two suitable primes p,q G Q and
replace t,p, q by up and vq. We must have ρ < η/и, q < η/υ; thus the loss
is
η η η η 11
t + p + q<n-\ ь-<п+- + - = —п,
и ν 2 3ο
If we can achieve up > en and vq > en, where с = 11/12, then the inclusion
of up and vq offsets this loss. If there are at least two elements of Q in both
(cn/u,n/u] and (cn/v,n/v], then this is possible. (We need two elements
to guarantee that ρ φ q.) By (1), for this it is sufficient that
η en ,, . и
π π— > (к- 1)тг- + 2
и и с
and
η сп ,, „ч ν
π π— > (к- 1)π- +2.
ν ν с
By the prime number theorem, these inequalities hold if ν < 6y/n with a
suitable constant δ depending on к (say, δ = 1/(3A;) is a good choice).
If ν > δ<^η, so that the previous procedure fails, we argue as follows.
We have
η η I
u- — < "2 < 72·
vw νΔ οΔ
We try to find a prime ρ G Q and replace t and ρ by u^p and vw with a
suitable exponent j. The new number is admissible if ρ < njuK The loss
is at most ρ + η, while the sum of the new terms is pu^ + vw > pu^ + δ2η.
Thus the process yields a profit if p(u^ — 1) > (1 — δ2)η, that is, we need
to find a p G Q satisfying
1-δ2 η
-η < ρ <
ν? - 1 ν?
By (1), such a p exists if
(2)
Recall that и < 1 /δ2, hence there is a power of и in the interval (<$~4, δ~6].
With this as iaj, the left side of (2) is of order n/logn, while the right side
stays bounded, so (2) will hold for every sufficiently large n. D
Problem N.18.
(a) Prove that for every natural number k, there are positive integers
αϊ < a>2 < · · · < a>k such that a\ — dj divides ai for all 1 < г, j < k,
ΐφ3·

384
3. SOLUTIONS TO THE PROBLEMS
(b) Show that there is an absolute constant С > 0 such that a\ > kck for
every sequence αχ,..., α& of numbers that satisfy the above divisibility
condition.
Solution.
(a) For к = 1, we can put αχ = 1. Now we show the inductive step from
к to к + 1. Assume that 0 < αχ < · · · < α^ is such a set. Write
b = Π*=ι аг· Then the numbers b<b + ai<---<b + ak form a
suitable collection of к + 1 numbers. Indeed, we have
(6 + a*) — (b + dj) = di — aj\a,i\b + a,i,
for 1 < i,j < к, г φ j, and also (b + ai) — b = сц\Ь.
(b) Consider any prime ρ < к. If ai = aj (mod p), then a^ = 0 (mod p)
by the divisibility condition. Consequently, there are at most ρ — I
numbers a^ that are not divisible by p. Consider now the divisible
ones. Dividing them by p, we again get an admissible system. Thus
there can be at most ρ — 1 numbers not divisible by ρ among them.
Repeating this argument, we find that the exponent of ρ in A =
Πί=ι аг 1S at least
(fc-(p-l)) + (fc-2(p-l))+...+ (fe-[^_](p-l))
fc2
> —
- 3p
lp-1
if ρ < у/к. Consequently, we have (with с = 1/3)
A > Yl pck2/p,
p<Vk
and hence
a*; = max aj > A1/k > Y[ pck/p
p<Vk
= exp ck У^ > exp c'k log к = kck
p<Vk
for к > 4. From the relation α& — αι|αι, we infer a\ > a^/2 > 2, hence
ai = Wa? > y/2a[ > y/a^ > kCk
with С = d/2. We have proved the inequality a\ > kck for к > 4.
For A: = 3 it is easy to see that the minimal possible value of αχ is 2,
hence it holds if С < (log 2)/(3 log 3). For к = 1 and 2, it fails by the
examples {1} and {1,2}. D

3.6 NUMBER THEORY
385
Problem N.19. Let щ < n^ < ... be an infinite sequence of natural
l/2fc
numbers such that nk tends to infinity monotone increasingly. Prove
that Σ™=1 l/rik is irrational Show that this statement is best possible in
a sense by giving, for every с > 0, an example of a sequence щ < ri2 < ...
such that nk > с for all к but Σ™=11/щ is rational.
Solution. Assume indirectly that Σ Vnfc = p/q> where ρ and q are
positive integers. Then, for arbitrary k, we have
к
Σ1* Σ 1 = -·
Multiplying both sides of this equality by qri\... n^, we see that
oo 1
qni...nk Σ —
t=fc+l Щ
is a positive integer for all k. Now we show that
oo 1
Πχ...Пк У^ >0 (fc—>oo),
i=k+l l
and this contradiction will disprove the indirect assumption.
By the monotonicity condition, we have
2-(fc+i) 2-fc 2-<k-V ^ ^ 2-1
пк+1 >nk >пк_г >--->пг ,
which yields
о — 1 ι _1_Q — ^
ni...nk <nk+{'"^ .
On the other hand, by applying the inequalities
2-(k+i) 2-(fc+2) ^ 2~(fc+3) ^
nk+1 < nk+2 < nk+3 <...
and obvious estimations, we obtain
111 111
+ + + ...< + ^—+ ^- + ...
Пк+ι nk+2 rifc+з rik+i nk+1 n£+1
1 1 1
+ ^— + ^— + ■
Wfc+i nfc+1 nk+1
1
These two estimations together imply
°° 1 ^1-2"'
Σ1 ^ nfc+l ™fc+l / _2-(fc+1)V
_^ щ nk+1 - ι nfc+i -1 ν k+l )
i=k+l

386
3. SOLUTIONS TO THE PROBLEMS
Here the first term tends to 1 and the second to 0 as к —> oo; thus the
whole expression tends to 0.
To solve the second part of the problem, we are going to construct a
sequence such that
™k
oo 1
> с (к = 1,2,... )and V^ — is rational
* ^ 71.1.
k=ink
Take an arbitrary integer щ > с2 + 1, and let rik+i = n\ — пк + 1 for
к = 1,2,... . The definition implies
rifc+i - 1 > (nk - l)2 > · · · > (ni - lfk > c2k+\
Consequently, n2 > с for all k. On the other hand, an induction shows
that
к
^l_ _ _1 1
^ щ ni - 1 nk(nk
i=1 --. --χ - n^(^ - 1)'
Thus, the sum of the series is the rational number l/(ni — 1). D
Problem N.20. Prove that for every positive number K, there are
infinitely many positive integers m and N such that there are at least
KN/ log N primes among the integers m + 1, m + 4,..., m + N2.
Solution. The basic idea of the proof is to average for several values of m
while excluding divisibility by small primes.
Let Q = 3 · 5 · ... · pi, the product of the first / odd primes, and let с
be a natural number < Q such that — с is a quadratic nonresidue for the
moduli 3,5,... ,p/; such а с exists by the Chinese remainder theorem. We
try to find m in the form m = c+ kQ, where 1 < к < Ν2.
Consider an 1 < г < TV; by the choice of c, we have (г2 + c,Q) = 1.
By the prime number theorem for arithmetical progressions, the number
of primes among the first N2 elements of this progression is asymptotically
equal to
1 N2Q
0(Q)logAT2Q;
thus for sufficiently large N (for a fixed value of Q), it exceeds
Q N2
3ct>(Q) log N'
Thus, the total number of primes (counted with multiplicity) among the
integers
i2 + c+kQ, l<i<N, l<k<N2
is at least
Q N3
30(Q)logiV

3.6 NUMBER THEORY
387
Consequently, for a suitable value of m = с + kQ, there are at least
Q N
30(Q)logiV
primes among the integes m + l,m + 4, ...,ra + N2.
Since
and the sum J] 1/p is divergent, we can select Q so that Q/30(Q) > if.
Then the previous arguments yield that for every sufficently large N, the
integer m can be chosen so that there are more than Κ · (Ν/ log Ν) primes
in the sequence m + 1, m + 4,..., m + AT2. Π

388
3. SOLUTIONS TO THE PROBLEMS
3.7 OPERATORS
Problem O.l. Let a, bo, &i, · · ·, 6n-i be complex numbers, A a complex
square matrix of order p, and Ε the unit matrix of order p. Assuming that
the eigenvalues of A are given, determine the eigenvalues of the matrix
В =
/b0E ЬгА b2A2
abn-iA11-1 b0E biA
abn-2An-2 abn-iA"-1 b0E
bn^A"-1 \
bn-2An-2
bn-zAn~*
\abiA
ab2A2
ab3A3
b0E
/
Solution. We show that if the eigenvalues of A are λι,..., λρ then the
eigenvalues of В will be the numbers
ф(акХ^) (k = l,...,n; j = l,...,p),
where αϊ,..., an denote the roots of the equation zn — a = 0, and
φ(χ) = b0 + bix + · · · + hn-ixn~l.
In fact, if Ao, -Ai,..., An-\ are quadratic matrices of order ρ having
complex entries, a is an arbitrary complex number and
/Aq Аг А2 ··· Αη_λ\
aAn-i A0 Αι ··· An_2
С =
then
where
\αΑλ αΑ2 aA3 · · · A0 )
det С = det Μ(αλ)... det M(an),
(1)
M{x) = A0 + Axx + · · · + Αη_ιχη_1.
For a = 0 the assertion of this theorem is trivial. For α Φ 0, the validity of
the theorem follows from the simple fact that
/мы ··· (о) \
с = w
\(0)
И'-1,
■·· M(an)J

3.7 OPERATORS
389
where W is the Kronecker product with Ε of the Vandermonde matrix
built from the roots of the equation zn = a, that is,
W=V®E=
(E
αλΕ
,n-l
Ε \
OLnE
\<- lE ··· c%-lE/
Apply (1) to the matrix В — λ£ (where £ is the unit matrix of order
np):
η
det(B - X£) = Yl det (ф(акА) - XE).
k = l
On the other hand it is well known that the eigenvalues of the matrix
ф(акА) = b0E + hakA + · · · + b^a^A"-1
are
0(α^λι),...,0(α^λρ).
This proves the assertion. D
Remark. Another group of solutions is based on the theorem stated, but
not proved, by one participant in the following general form:
Theorem. If the eigenvalues of the quadratic matrix A of order ρ are
λι,..., λρ, further maxj \Xj\ = λ, and fij(z) (i,j = 1,...,n) are regular
functions on the disc \z\ < λ, then the eigenvalues of the matrix
(hM) ■■■ fln(A)\
\/ηΐ(Λ) ··· fnn(A)/
are given by the eigenvalues of the matrices
/7ll(A<) '·· /ΐη(λί)\
(г = 1,...,р).
\/ηΐ(λζ) ··· fnn{\)/
One participant proved this theorem in the less general case where the
functions fij(z) are polynomials; this proof can be applied in the more
general case as well.

390
3. SOLUTIONS TO THE PROBLEMS
Problem O.2. Let U be an η χ η orthogonal matrix. Prove that for
any η χ η matrix A, the matrices
m
771+ 1 t-ί
j=0
converge entry wise asm->oo.
Solution 1. For an arbitrary η χ η matrix A, let \\A\\ denote the norm of
A, that is,
Pll = sup ||Ar||,
IM=i
where χ varies in the η-dimensional Euclidean space. If U is orthogonal,
then U~l is also orthogonal, and ||£M|| = \\AU\\ = \\A\\.
Entrywise convergence and convergence in norm of a sequence of
matrices are equivalent. Thus, by the Bolzano-Weierstrass theorem, from
a norm-convergent sequence of matrices it is possible to select a norm-
convergent subsequence.
For a given orthogonal matrix C/, natural number m, and arbitrary
matrix B, put
Bn
1 in
771+1 ^
г=0
We shall need the following lemma:
Lemma. For any natural number m,
lim ||(Am)p-Ap||=0.
ρ—кзо
Proof. Let ρ > т. Then
1 1
Ρ 771
г=0.7=0
^ ^ Ρ+™
= η+ΐηΛΐ Σ s(k)U~kAUk,
Ό + 1 771 + 1 f—'
k=0
where
s(k) = Σ 1< TTi+1 (fe = 0,...,p + m);
0<i<p,0<j<m
if 771 < к < ρ, then s(k) = m + 1. Hence
^ ^ p+™ ρ
-Ц- —— Υ s(k)U-kAUk — Υ U~kAUk
+ 1771+1 f^L ' p+lf-^
Ρ
m—l
k=0
\\\Am)p Ap\\ -
1 1
Ρ + 1 771+ 1
p+1
as ρ —> oo. This proves the lemma.
p+m
k=0
J2(s(k)-m-l)U-kAUk+ J2 s(k)U~kAUk
k=p+l
k=0

3.7 OPERATORS
391
Now the assertion of the problem can be proved as follows. Since \\Am \\ <
|| A|| for all natural number ra, the sequence {Am} contains a subsequence
{Amfc} that is convergent in norm to a matrix H. Then for к —> oo we
have
\\H - U^HUW < \\H - Amk || + \\Amk - U-xAmkU\\
+ \\U-\Amk-H)U\\
<2||Я-ЛтЛ -b—L-piHO,
mk + 1
whence Η = U~lHU. Thus, Hp = Η for every natural number p.
We prove that limp-^ || Ap—H\\ = 0. Actually, for every natural number
к we have
\\AP - H\\ = \\AP - Hp\\ < \\AP - (Amfc)p|| + ||(Amfc)p - Hp\\
<PP-(AmJp|| + pmfc-tf||,
whence by the lemma we obtain
limsup \\Ap - H\\ < \\Amk - A\\.
p—+00
For к —> oo, the assertion follows. D
Solution 2. Consider the natural embedding of the η-dimensional real
space in the η-dimensional complex space. In the latter, the η by η real
orthogonal matrices are unitary matrices. Thus, we prove more than
required if in the statement of the problem we replace the orthogonal matrix
by a unitary matrix, and real matrices A by matrices with complex entries.
So let U = (Uij) be a unitary matrix in the η-dimensional complex
space. Then, as we know, in a suitable basis the matrix U has the form
(Uij) = (eidij), where 6ij is the Kronecker symbol and |εχ| = · · · = |εη| = 1.
Obviously,
(U3)ik=e{6ik, (U-j)ik=e7j6ik.
Now, if A = (a,ik) is an arbitrary matrix, then
(U~3AU3)ik = Y^e^36irarse3s6sk = e^3aike3k = e^3e3kaik.
r,s
Therefore,
^ τη m
{Am)ik = ^Τϊ Σ^α* = -^ £еГЧ
If Si = ek, then ej3e3k = 1, so lim^oo (Am)ik = aik.

392
3. SOLUTIONS TO THE PROBLEMS
On the other hand, if ε% Φ ε*, then
τη /— \m+l _
(*»)* = ^TI Σ № = m + l ε7ε.-1
as m —> oo, since
[SiSk) -
£%£k — 1
\e%£k ~ 1
We see that in this case limm_+00 (Am)ik = 0.
It remains to note that if for a sequence of matrices [α\™ J we
V / m=l
know that for each pair of indices г, /с the numbers α\™' tend to some a,ik
as m —> oo then the sequence of matrices, evidently, is entry wise convergent
to the matrix (a^). D
Solution 3. As a generalization of the problem, we prove the following
theorem.
Theorem. Let Я be a complex Hilbert space, and U a unitary operator
in H. Then, for any compact operator A of Я, the sequence of operators
1 m
<l>m(A) = —-J£U-jAW (m = 0,l,...)
771+1 ^—ί
is weakly convergent, that is, for every pair of elements /, g in Η
lim ((Фтп(А)-Фп(А))^д) = 0,
m,n—+ oo
where (/, #) denotes the inner product of the elements /, g G H.
(This statement obviously contains the statement of the problem.
Indeed, the η by η real orthogonal matrix appearing in the problem induces
a unitary operator of the η-dimensional complex Hilbert space; in this n-
dimensional space all linear operators, in particular those induced by η by
η real matrices, are compact, and in finite-dimensional spaces weak and
entrywise — so-called strong — convergences of operators coincide.)
Proof. For proving the theorem, we first observe that the mappings A —>
Фгп(А) (т = 1,2,...), defined for all bounded operators A of Я, have the
following properties:
a. For any pair of bounded linear operators А, В of Я and any pair of
complex numbers a and /3,
Фт(сь4 + βΒ) = аФт(А) + βΦ^Β) (т = 0,1,...).
b. For any bounded linear operator A of Я,
||Фт(Л)||<Р|| (m = 0,l,···),
where \\A\\ stands for the norm of A.

3.7 OPERATORS
393
с. If the sequence {Ak}^L1 of bounded linear operators is uniformly
convergent to a bounded linear operator A, that is, \\Ak — A\\ —> 0 as
к —> oo, and the sequence {Фгп(Ак)}^=0 is weakly convergent for each
/c, then the sequence {Фгп(А)}^=0 is also weakly convergent.
From these three properties only с needs to be verified. So let {Аь}<£=1
be a sequence of operators with the properties above, and let / and g be
two elements of the Hilbert space H. Without loss of generality, we may
assume that ||/|| = ||#|| = 1. Then
\((Фгп(А)-Фп(АМд)\
= \([Фт(А) - Фт(Ак) + Фт{Ак) - Фп(Ак) + Фп(Ак) - Фп(А)\ f,g)\
< \(Фт(А - Ak)f,g)\ + \((Фт(Ак) - Фп(Ак)) f,g)\ + \(Фп(Ак - A)f,g)\
< 2\\А - Ак\\ + \((Фт(Ак) - Фп(Ак)) f,g)\
(we have made use of properties a and b and applied the Schwarz
inequality).
Now let ε > 0 be arbitrary. Then, by one of the assumptions on the
sequence {Α^})^, there is an index к = к(е) such that
\\ΑΗε)-Α\\<^.
Next, let N = N(k(e),e) be a positive integer satisfying
|((Фт(л(£))-Фп(л(£)))/,^)|<|
for m, η > N. By the assumptions, such N exists. Then, in view of the
foregoing, we obtain
\((Фт(Л)-Фп(А))/,д)\<е
for m,n> N. This proves property с
It is well known that every compact operator is the uniform limit of
finite rank operators, further every finite rank operator is a finite linear
combination of rank 1 operators. Consequently, by properties a. and c, it
is sufficient to prove our assertion for operators of rank 1. So let A be an
operator of rank 1. Then there are two elements φ and φ in the space Η
such that
for all / in H. Now, if h and g are any two elements of H, then
^ τη ^ m
(*m(A)h, 5) = —^ ^ (AWh, Uig) = ^ΓΓιΣ (Ujh, φ) (CTty, <?) ·
i=o j=o
Denote by Η (8) Η the tensor product of Η with itself. (The definition
of the Hilbert space Η (8) Η is the following. Denote by Щ the algebraic

394
3. SOLUTIONS TO THE PROBLEMS
tensor product of Η with itself. As is well known, this is the free module
generated by the symbols x®y(x,yeH) and having the complex field for
operator domain. By the inner product of two elements
η m
г=1 j=l
of Я0, we mean the number
П 771
<^> = ΣΣ(^4)(Μ)· (ΐ)
г=1 j=l
Two elements ψ, ψ' G Яо are considered to be identical if (ψ —ψ'',ψ — ψ') =
0. On the factor space Я0 that arises after this identification, (1) defines a
norm. Completing Я0 in the metric induced by this norm, we obtain the
Hilbert space Η = Η ® Я.)
Let Uq be the operator of Я satisfying the relation
U0(x <8> y) = (Ux (8) U~ly) (x, у е Н).
C/o is uniquely determined in this way and is unitary in H. On the other
hand it is clear that
1 m
j=o
By a classic theorem of ergodic theory (see for example F. Riesz and B.
Sz. Nagy, Functional Analysis, Blackie, London, 1956, §144), the right-
hand side is convergent for every pair of elements ψ, ψ' G Η and so, in
particular, for the pair ψ = Η®ψ, ψ' = φ® g. The proof is complete. D
Problem O.3. Prove that if a sequence of Mikusinski operators of
the form με~Χ3 (\ and μ nonnegative real numbers, s the differentiation
operator) is convergent in the sense of Mikusinski, then its limit is also of
this form.
Solution. Let /ine~AnS be the convergent sequence of operators considered.
We may assume that μη —> μ and λη —> λ (0 < μ, λ < +οο) as η —> οο; in
fact, passing to a suitable subsequence, this can always be achieved without
any influence on convergence and limiting operator.
By the definition of convergence, there exists an operator {p}/{q} ({p},
{q} G C[0, oo), {p}, {q} φ {0}) such that
Ш
(1)

3.7 OPERATORS
395
is an almost uniformly (that is, uniformly on each finite interval)
convergent sequence of functions in C[0, oo). We may assume that p(0) = 0,
since otherwise {p}/{q} in (1) can be replaced by the equal expression
({p}{l})/({tf}{l})· Define ρ to be zero on the negative half-line; then ρ
will be a continuous function on the whole real line and not identically zero
on the positive half-line. From (1) it follows that the sequence of functions
μηβ-Χη3{ρ} = {μηΡ^-\η)} (2)
is almost uniformly convergent.
If λ = +oo, then we see that (2) is almost uniformly convergent to the
zero function; thus
μηε-Χη3 -+ 0 = 0 · e~s (n -+ oo).
If λ < +oo, then we show that μ is also finite. Let to be a point with
p(t0 - λ) φ 0. Then p(t0 - λη) —► p(t0 - λ) (η —► oo), so μηρ^ο ~ Κ) can
only be convergent if μ < +oo. Then, however, μηρ(ί — λη) tends almost
uniformly to the function {μρ(ί — λ)} = με~Χ3{ρ}. Consequently,
μηε"ληθ —► μβ~λ3 (η —► oo). D
Problem 0.4. Prove that an idempotent linear operator of a Hilbert
space is self-adjoint if and only if it has norm 0 or 1.
Solution 1. If Τ is self-adjoint, then it is an orthogonal projection
operator; it is well known that such operators have norm 1 or 0. Conversely, let
\\T\\ < 1. For any element χ of the space and any number μ, we have
Τ (μΤχ -{χ-Τχ))= μΤχ
and therefore
\μ\*\\Τχ\\*<\\μΤχ-(χ-Τχψ
= -2Re \μ(Τχ, χ - Τχ)} + |μ|2||Τ^||2 + ||χ - Τχ\\2.
Consequently,
Re [μ{Τχ, χ - Τχ)] < - \χ - Τχ\\2.
But this is possible for every μ only if
(Trr, χ - Tx) = 0.
Hence, it follows in a simple and well-known manner that T* = T.
The proof applies to real as well as complex spaces. D

396
3. SOLUTIONS TO THE PROBLEMS
Solution 2. The kernel (null space) of T, in view of the property T2 = T,
coincides with the range οι Ι — Τ, that is, with the set of all elements of the
form (I — T)x where χ runs through all elements of the space. It is a well-
known fact, and easy to verify, that the orthogonal complement of the range
of I—T is equal to the kernel of I—T*. By a well-known result of Nagy (see,
for example, B. Sz. Nagy and C. Foia§, Harmonic Analysis of Operators
on Hilbert Space, Akademiai Kiado and North-Holland Publ. Co., 1970),
from the inequality \\T\\ < 1 it follows that the invariant elements of Τ and
T* are the same. So, the kernel of / — T* coincides with the kernel οι Ι — Τ,
which, in turn, coincides with the range of Τ (the latter fact follows from
the property (I -T)2 = 1 - T). Thus
(Tx, x-Tx)=0
for every x, which implies the statement. D
Problem O.5. Let Τ be a bounded linear operator on a Hilbert space
H, and assume that \\Tn\\ < 1 for some natural number n. Prove the
existence of an invertible linear operator A on Η such that \\ΑΤΑ~λ\\ < 1.
Solution. Let (.,.) denote scalar multiplication in H. It is easy to verify
that the formula
[х,у] = ^2(Тх,Гу)
i=0
defines a new scalar product on Я. Let Η denote the space equipped with
the scalar product [.,.]. Obviously,
(χ,χ)<[χ,χ}<(Ψ\\Τ\Α(χ,χ)
for all χ G Я, that is, the norm in Η is equivalent to that in Я.
Consequently, Η is also a Hilbert space.
Since the dimension of a Hilbert space is the minimal power of complete
sets, that is, sets whose closed linear span is the whole space, the
dimensions of Η and Η are equal. But Hilbert spaces of equal dimension are
isomorphic. So, consider a Hilbert space isomorphism A : Η —> Я. Since
the underlying vector space for Я and Я is the same, A is an invertible
linear operator on Я. Let χ G Я be arbitrary, and put у = Α~λχ. Then
(χ, χ) - (ATA~lx, ATA~lx) = (Ay, Ay) - (ATy, ATy) = [у, у] - [Ту, Ту]
= Σ{Τν,Τν) - П^{Т^у,Т^у) = (у, у) - {Т"у,Т"у),
г=0 г=0
which is nonnegative since ||Tn|| < 1. Thus, we have obtained
(x,x) >(ATA-lx,ATA~lx)

3.7 OPERATORS
397
for all χ e if, that is, ЦАТА г\\ < 1. Therefore, the operator A satisfies
the conditions of the problem. D
Remark. One contestant showed that the positive square root of the
positive, bounded, self-adjoint, linear operator Σ£Γ0 Т*гТг can be chosen
for A.
Problem O.6. Is it true that if A and В are unitarily equivalent,
self-adjoint operators in the complex Hilbert space H, and A < B, then
A+ < B+? (Here A+ stands for the positive part of A.)
Solution. We first show that there are self-adjoint operators A and В in
C2 with eigenvalues λι,λ2 and Лз,Л4, respectively, such that A < В but
A+ ^ B+. For instance, let A and В have matrices
(_°i ij) «* (J о)-
respectively. Then for (x,y) G C2, we have
(B(x, y), (x, y)) - (A(x, y), (ж, y)) = \x\2 + xy + xy + \y\2 = \x + y\2 > 0,
that is, A < B. From the matrices, it is clear that A and В are self-
adjoint, В positive semidefinite, A indefinite. Thus B+ is equal to B, and
A+ is positive semidefinite with A+(0,1) φ (0,0), since (0,1) is not an
eigenvector of A. Therefore,
(i+(0,l),(0,l)) >0 whereas (£+(0,1), (0,1)) = 0.
SoA+£B+.
Now let Η = h (complex). For simplicity, we write each element of Η
as a series Σ™=ι αη^η convergent in l^ where
n.
en = (0,...,0,i;0,...).
Let
(oo \ oo
Antonym
n=l / n=3
where λη = λ3 if η = 4k + s, 1 < s < 4, and λι, A2, λ3, λ4 are described
above. Here e\ and e2 are identified with (1,0) and (0,1). Further, let
(oo \ oo
/J ο-η^η J = -Β(αι, α2) + Σ ^ηΟ-η^η·
n=l / n=3

398 3. SOLUTIONS TO THE PROBLEMS
Both A and В are sums of two self-adjoint operators, so they are self-
adjoint. Since В — A = В — A, we have
/ oo oo \
/ (B- A)^2anen,^2anen \ = ^(£ - i)(aba2), (aba2)^ > 0,
\ n=i n=i /
that is, A < B. Similarly, B+ — A+ = B+ — A+ and, consequently,
A+ £ B+.
We now prove that A and В are unitarily equivalent. Let
(oo \ oo
71=1 / 71=3
where U\ denotes the unitary transformation of coordinates that reduces
the matrix of В to diagonal form:
(U1BUr1)(a,b) = (\3a,\4b).
U\ is obviously unitary, and
(oo \ oo
Σ αηβη Ι = λ3αιβι + A4a2e2 + Σ ληαηβη ·
η=1 / η=3
Let
(oo \ oo
Σαηβη)= Σα/(η)βη'
η=1 / η=1
where / denotes the following permutation of N: f(n) = n—4 for η = 4/с+З
or η = 4k + 4 with к > 1. Further, /(3) = 1, /(4) = 2, and f(n) = η + 4
ϊοτ η = 4k + 1 οτ η = 4k+ 2 with к > 0. Then
(oo \ oo oo
/i an^n ι, ΣЬпвп^= Σ afw^n
n=l / n=l n=l
oo oo / oo \
= Σ unh-Hn)= (Σαηβη' υϊι ΙΣbnen J)'
n=l n=l \n=l /
which implies that C/2 is also unitary. Further, we have
(oo \ oo
Σα"βΑ = Σχ^α^^
n=l / n=l
Finally, let
(oo \ oo
Σα^^Ι= ^3(^1^2) + Σαηβη>
n=l / n=3
where (С/^~МС/з)(а,Ь) = (λια, λ26), and U3 is unitary. With this third
unitary transformation, we have
(U3U2U1BUr1U21U^1) (f>nen j = A I f>nen) ,
and so A and В are unitarily equivalent.
Consequently, the assertion is false. D

3.7 OPERATORS
399
Problem 0.7. Let К be a compact subset of the infinite-dimensional,
real, normed linear space (X, || · ||). Prove that К can be obtained as the set
of all left limit points at 1 of a continuous function g : [0,1[—> X, that is, χ
belongs to К if and only if there exists a sequence tn G [0,1[ (n = 1,2,...)
satisfying lim^oo tn = 1 and lim^oo \\g{tn) - x\\ = 0.
Solution. The compact set К С X has, for every ε — \jn (n G N), a finite
ε-net zi(n),Z2(n),.. -,Zjn(n) G K, that is, if χ G K, then \\x — Zi(n)\\ < ε =
1/n for some 1 < г < jn. Denote by xo, χι, X2,... the sequence obtained by
writing the finite (l/n)-nets zi(l),Z2(l),...,2;J-1(l),zi(2),Z2(2),...,Zj2(2),
zi(3),... one after the other. It would already be possible to define a
piecewise linear function g: [0,1[—> X such that #(1 — 1/fe) = Xk (к G iV),
and therefore each point of К is a left-hand limit point at 1 of the function
g. The problem is that g may have further limit points that do not belong to
K. So, we have to improve the procedure by interpolating a new sequence
of function values 2/o ? 2/i, 2/2, · · · · To define the latter, we shall need the
following proposition (which is essentially the same as the well-known Riesz
lemma on almost orthogonal vectors):
If L С X is a linear subspace of finite dimension, then there exists a
vector у G X\L such that ||y|| = 1 and dist(y,L) = 1, where
dist(A,B) = inf{||α -Ъ\\:аеА,Ье В}
denotes the distance between the subsets А, В С X, and the singleton {a}
is replaced by a. Indeed, since X is infinite dimensional, it contains a vector
υ G X \ L, and the subset
D = {и: и G L, \\v — u\\ < \\v\\}
of the finite-dimensional subspace L is bounded and closed. Thus D is
compact, so it contains a vector г^о G Д which lies closest to v. Then,
setting ^o = ν — щ,
\\vo\\ = min{||i; — u\\: и G D} = dist(i;,L) = dist^o,!/).
Consequently, the vector у = г>о/||г>о|| has the required properties.
Next, we define a sequence in X by induction. Let Lq = lin{^o,^i},
and using the proposition just proved, choose a vector у о G X \ Lq such
that
112/o 11 = 1 =dist(2/0>bo).
If 2/oj 2/ij ··· j 2/fc-i (fe £ N) are already defined, then set
Lk = lin{xo,2/o?^i?2/i?---?^fc-i?2/fc-i?^?^+i}?
and using the preceding proposition again, choose the vector yk G X \Lk
so that
\\ук\\ = 1 = dist(j/jfe,Ljfe).

400
3. SOLUTIONS TO THE PROBLEMS
For simplicity, we first define a continuous function φ: [0, oo[—> X that
has К for the set of limit points at oo. For к G No = NU{0} and 0 < r < 1,
put
ф(3к + r) = (1 - r)xk + r(xk + yk),
ф(3к + 1 + r) = (1 - г)(жл + у*) + г(жл+1 + yk),
0(3fc + 2 + r) = (1 - г)(хк+г + yk) + rxk+1.
Let us verify the desired properties.
a. By the construction of the sequence xq,xi,X2,..., for any χ G К there
is a subsequence xkn —> x\ setting rn = 3/cn, we have
lim ф(тп) = lim xkn = x.
n—>oo n—>oo
b. On the other hand, we show that no χ $. К can be a limit point of φ
at oo. In fact, the closedness of К yields
δ = min{dist (χ, K), 1} > 0,
and if for some To G [3&o, 3ko + 3] (&o G No) we have dist (ж, ф(то)) <
δ/3, then for every τ > Зко + 3 the relation dist (ж, </>(т)) > δ/З is valid.
Now there exists a natural number ко < к G N such that
bl. either ЗА;+ (25/3) < r < 3fc + 3- (25/3), in which case the relation
0(ro) G Lfc and the definition of φ imply
25
dist(0(ro),0(r)) > y
and therefore
с
dist (ж, ф(т)) > dist (0(ro), 0(r)) - dist (ж, 0(ro)) > -,
b2. or |3fc — t\ < 25/3, in which case for xk G K, by the definition of
φ, we have
9Л
dist(xfc,0(r)) < y,
and so
с
dist (ж, 0(r)) > dist (rr, if) - dist (К, ф{т)) > -.
о
Thus, there is indeed no sequence rn —> oo with 0(rn) —> ж. Finally,
the required function g: [0,1[—> X is provided by the transformation
ί#) = 0 (т^-г)' *G[0,1[. D

3.7 OPERATORS
401
Problem 0.8. Denote by B[0,1] and C[0,1] the Banach space of all
bounded functions and all continuous functions, respectively, on the
interval [0,1] with the supremum norm. Is there a bounded linear operator
T:B[0,1]-+C[0,1]
such that Tf = f for all f G C[0,1] ?
Solution. Suppose the existence of such an operator T.
For any 0 < а < 1, consider the function
f 0 if χ < a,
I 1 it χ > a.
The function ha = fa — Tfa is, apart from the jump of size 1 at point a,
continuous and satisfies T(ha) = T(fa — Tfa) = 0.
Replacing ha by — ha is necessary. We may assume that ha > 1/3 in a
small, punctured, right-hand or left-hand neighborhood of a.
Let a\ = 1/2, and define the sequence αι,α2,... so that сц G (0,1)
and di belongs to a corresponding small neighborhood of а^_1 for all г > 2.
Then for the function gn = Σ™=1 hai, we have gn G B[0,1], gn is continuous
apart from the jumps of size 1 at the points αχ,..., αη, and gn > n/3 in a
small, one-sided, punctured neighborhood of an. It is easy to construct a
continuous function fn satisfying the relation
||/n - 9n\\ sup \fn(x) - gn{x)\ < -.
*€[0,1] *
(To speak exactly, the sign of equality is valid.) Then, for the norm of the
operator T, we obtain the estimate
ЦГЦ > llTn[fn~9n^ > 2\\Tfn - Tgn\\ = 2||/n|| > \n - 1,
\\jn-9n\\ 3
valid for any natural number n. Consequently, Τ cannot be a bounded
operator. It follows that there is no operator Τ satisfying the requirements
of the problem. D
Problem O.9. Does there exist a bounded linear operator Τ on a
Hilbert space Η such that
f)T"(#) = {0} but f) Т»(ЯГ ?έ {0},
n=l n=l
where denotes closure?

402
3. SOLUTIONS TO THE PROBLEMS
Solution. There exists an operator Τ of this kind. Let Я be a Hubert
space of countably infinite dimension, and let {en}'^)=1 be an orthonormal
basis in H. We define Τ on the basis vectors first. Let
0 if n = 1,
Ten = { en+i if η > 2 is not a square,
otjei + ^еп+1 if η = j2 for an integer j > 2.
For the sequence {otj}^L2i we assume that 0 < \ctj\ < 1 for each j, and
oo
a = Y] \aj\2 < oo.
i=2
We next extend the definition of Τ to all of H. For an arbitrary vector
χ e Я, put
oo
Tx := ]P £nTen, where ξη = (χ, en).
n=l
First, we have to prove that the series that defines Τ χ is convergent, that
is, the sequence of the partial sums is a Cauchy sequence. Let к < I be
positive integers. Then
ΣξηΤβη
n=k
< Σ κ»ι2 +
n=k
к<р<1
since, substituting the defining expressions of the vectors Ten in the sum
J2n=k £>nTen, in the linear combination of basis vectors so obtained the
coefficients of the vectors en with k + 1 < η <l + l have absolute value less
than or equal to 1, the coefficient of e\ is J2k<j2<i аз£з2> an(^ ^ne remaining
basis vectors do not appear in the linear combination. Using the Cauchy
inequality, it follows that
/ j Sri-* en
n=k
<Σι^ι2+ Σ κ
n=k \k<j2<l
I
<(1 + α)ΣΚη|2·
n=k
Since Y^=1 |£n|2 = INI2 < oo? the series Y^=1 ζηΤεη is convergent. The
linearity of Τ is now obvious, and if in the preceding inequality we write
к = 1 and let / tend to infinity, then we obtain the boundedness of T,
namely, ||T|| < 1 + a.
We show that e\ eTl(H) for every positive integer Z. In fact,
rp2j-2
eo-i)2+i for j > 2

3.7 OPERATORS
403
and therefore
ei + \ep+i = ±TeP = Τ*'1 (1-еи_1)а+Л € TV-\H).
J aj \aj J
Thus,
ei + ieia+1 еТ2Н(Я)сГ!(Я) if 2j - 1 > i,
and letting j tend to infinity, we obtain
ei = lim ί ei + -ej2+1 J G Tl(H)~.
It remains to show that П£гТ1(Н) = {0}. Suppose ж G П^Т^Я).
From the definition of Τ it is clear that, for any Z, the vectors en with
2 < η < I + 1 are orthogonal to Tl(H). Thus (ж, en) = 0 for all integers
η > 2. On the other hand, if Ту Φ 0 for some у е Н, then there necessarily
exists an integer η > 2 with (y, en) ^ 0. Then, however, (Ту, en+i) ^ 0
and, by the foregoing, ж = TO = 0. D

404
3. SOLUTIONS TO THE PROBLEMS
3.8 PROBABILITY THEORY
Problem P.l. From a given triangle of unit area, we choose two points
independently with uniform distribution. The straight line connecting
these points divides the triangle, with probability one, into a triangle and
a quadrilateral. Calculate the expected values of the areas of these two
regions.
Solution. First, observe that an affine transformation does not change
the ratio of the areas, therefore the expectation of the ratio does not change.
We will use this fact by choosing the type of the triangle in the most
convenient way for our purposes.
Denote the vertices of the triangle by A\, A2, As, and the chosen points
by X and Y. The probability that the line e connecting the points X and
Υ crosses one of the points A{ (i = 1,2,3) is equal to zero because of the
independence of X and Y. So the probability that we will get a triangle
and a quadrilateral by dividing the triangle by the line e is equal to one.
Denote by Αχ the event that the point Ai is a vertex of the small triangle.
The events Ai (i = 1,2,3) form a complete system of events, so by the
theorem of "complete expectations,"
Е(^)=^Е(^\А)Р(А),
where t denotes the area of the small triangle, and Τ denotes the area of
the original triangle.
If the original triangle is equilateral, then E(t/T\Ai) is independent of
г (so does P(Ai) for г = 1,2,3), hence
Д(|) = ЕфЛг) £>(А) = Ефлг).
г=1
Suppose now that we have a right-angled isosceles triangle, Ai is the
right-angled vertex, and the equal sides have unit length. Denote by F(a, b)
the (conditional) probability, that for the line segments ξ and η of the sides
cut by e, we have ξ < a and η < b, assuming that ξ < 1 and η < 1. That
is,
F(a,ft) = P(£<a,q<ft|£<l,4<l).
Since a < 1 and b < 1,
_ P(i<a,V<b)
ΡΜ-Ρ(ξ<ι,η<ιγ
If X = (x\,X2) and Υ = (2/1,2/2), then
Ρ(ξ < α,η <b) = с / dxidx2dyidy2,
JA{a,b)

3.8 PROBABILITY THEORY
405
where с is a constant, independent of a and 6, and Δ(α, b) is the domain
of the four-dimensional space such that, for the corresponding line, ξ < a
and η < b. By substitutions and yi = byl, the domain Δ (α, b) is
transformed to Δ(1,1), and therefore
Ρ(ξ < α,η <b) = ca2b2 / dx^dx^dy^dy^
= α42Ρ(ξ <1,77<1),
that is, F(a, b) = a2b2. So the density function is
„ 1N d2F(a,b) Λ ι
and thus the ratio of the areas of the triangles is given by
\ab
9{a,b)= ι =ab.
2 ' L ' L
The expectation of the ratio is
Εφ = Ε g(a,b) = J J g(a,b)f(a,b) da db
1 Ί 4
4a2b2 dadb=-.
II
Jo Jo
So the expectations of the areas of the triangle and the quadrilateral are
4/9 and 5/9, respectively. D
Problem P.2. Select n points on a circle independently with uniform
distribution. Let Pn be the probability that the center of the circle is in
the interior of the convex hull of these n points. Calculate the probabilities
P3 andP4.
Solution 1. Assume that the circumference of the circle is one. Fix a
point on the circle, and introduce the arclength parameter. Let ri,..., rn
be the parameter values associated with the chosen points. Denote by An
the event that the center of the circle is not inside the convex hull of the
points. Furthermore, denote by Bi the event that there is no point on
the arc (r^, r; + 1/2). By the independence and uniform distribution of the
points, it is obvious, that, except for an event of probability zero, the events
Bi are mutually disjoint and An = B\ + · · · + Bn, P(Bi) = l/2n_1, and
therefore Pn = 1 - P(An) = 1 - η/2η"1. Hence P3 = 1/4, P4 = 1/2. D
Solution 2. Denote by TiTj the length of the shorter arc with endpoints
Ti and Tj and by Bij (1 <i < j <n) the event that all of the points are

406
3. SOLUTIONS TO THE PROBLEMS
on the shorter arc with endpoints Ti and Tj. (If r^ and Tj are the endpoints
of a half-circle, then Bij denotes the event that all of the points are on
one of the half-circles defined by them.) It is obvious that An = J^ij Bij-
If the points are distinct, the events B^ are mutually disjoint. Since the
points are independent and uniformly distributed, they do not coincide with
probability one, and so P(An) = J^ . P(Bij). Furthermore, Pfcrj < x) =
2x, so {P{7ytj < x)Y = 2 (0 < χ < 1/2) and Ρ(Β^\τγτά < χ) = xn~2.
Because of the theorem of total probability,
,1/2
Hence,
ft-i-PW-i-Q^.i-jii. □
Problem P.3. Let ει, ε2,..., ε2η be independent random variables such
that P{ei = 1) = P(ei = -1) = 1/2 for all i, and define Sk = £*=1 eu 1 <
к < 2n. Let iV2n denote the number of integers к G [2, 2n] such that either
Sk > 0, or Sk = 0 and Sk-i > 0. Compute the variance of iV2n.
Solution. It is known (see, for example, W., Feller, An Introduction to
Probability Theory and Its Applications, Wiley, New York, 1957, Vol. I,
p. 77) that the distribution of iV2n is given as
(2k\ (2(n-k)\
P(N2n = 2k)=ykn22nn-k } (fc = 0,l,...,n), (1)
2n = 2k + 1) = 0 (k = 0,1,..., η - 1). (2)
The symmetry of the distribution implies that £(iV2n) = n, and from
(1) and (2) we conclude that
η (2k\ (2{n-k)\
E(Nl) = £(»)' U" ·
k=0
Assume that \x\ < 1, and consider the functions
oo /2k\
F(x) = ^2(2k)2^-x2k and G(x) = (l-x2)~1/2.
It is easy to see that
oo /2k\
ОД = Е^ж2*· (3)
k=0

3.8 PROBABILITY THEORY 407
The generating function of Е(Щп) is F(x)G(x), that is,
oo
F(x)G(x)=-£E(Nl)x2k.
k=0
Using (3), it is easy to verify that
F(x) = x(xG\x))',
which implies that
F(x) = x(x2(l - χ2)"3/2)' = 2x2(l - χ2)"3/2 + Sx\l - χ2)"5/2.
Therefore,
F(x)G(x) = 2x2(l - x2)-2 + 3x4(l - x2)-3. (4)
Simple calculations show that
oo
(l-*2)-2 = X>x2fe-2
k=l
and
So from (4), we get
oo oo о oo /ο ι \
F(x)G(x) = £2fcz2* + Σ 5*(* - ^ = Σ ( ψ + ок) *"·
к = 1 к = 1 к=1 ^ '
Therefore,
E(Nl) = \п2 + \п .
Hence,
Var (ΛΓ2η) = Е{Щп) - Ε*{Ν2η) = 0η2 + ±») - η2 = ^^±il . D
Problem P.4. A gambler plays the following coin-tossing game. He can
bet an arbitrary positive amount of money Then a fair coin is tossed, and
the gambler wins or loses the amount he bet depending on the outcome.
Our gambler, who starts playing with χ forints, where 0 < χ < 2C, uses
the following strategy: if at a given time his capital is у < С, he risks all of
it; and if he has у > C, he only bets 2C — y. If he has exactly 2C forints,

408
3. SOLUTIONS TO THE PROBLEMS
he stops playing. Let f(x) be the probability that he reaches 2C (before
going bankrupt). Determine the value of f(x).
Solution 1. We are going to prove that f(x) = x/2C (0 < χ < 2C).
First, we show that f(x) is nondecreasing. Let 0 < x\ < X2 < 2C, and
suppose there is a sequence of tossings such that he can reach 2C from χ ι.
We show that by the same sequence of tossings he can reach 2C from X2
(maybe earlier), so /(#2) > f(xi) ·
Let us see how the amount of the player's money changes after one toss:
(a) if x\ < X2 < C, then 2x\ < 2x2 for a win, and 0 < 0 for a loss;
(b) if x\ < С < Χ2·> then 2x\ < 2C for a win, and 0 < 2(x2 — C) for a loss;
(c) if С < χι < x2, then 2C < 2C for a win, and 2(m - C) < 2(x2 - C)
for a loss.
Thus if x\ < X2, then, for the new amounts x[ and xf2, we have x[ < x'2.
Define χ = (a/2n) 2C (1 < a < 2n -1 , a is odd, η = 1,2,...). We will
prove by induction that for such an ж, f(x) = a/2n = x/2C. For η = 1,
/((1/2) · 2C) = f(C) = 1/2, since starting from С the player will have
2C with probability 1/2. Suppose the formula holds for η = к (к > 1),
that is, f{x) = f{{a/2k) · 2C) = a/2*. If χ = (a/2*+1) · 2C and a < 2k
(that is, χ < C), then the player will win 2x forints or lose everything with
probabilities 1/2, 1/2, respectively. So
/m = !/(£*?) + 1/(0) = ^ = ^.
If χ = (a/2fc+1) · 1С ала 2k < a < 2k+1 (α φ 2k since it is odd), then the
inequality χ > С implies
,, x 1«„^ч lrfa-2k \ 1 la-2k a x
f(x) = -f(2C) + -2f (-p- ~ 2CJ = - + -2-ΊΓ = ^ = w
Suppose now xo φ (^/2n) · 2C (1 < a < 2n — 1, a is odd), and suppose
f(x0) φ xq/2C, but, for example, f(xo) > xq/2C .Then there should exist
a number having the form a/2n such that
ё<|г</Ы, (i)
and it contradicts the fact that f(x) is nondecreasing. Namely, because of
(1), #ο < (a/2n) · 2C, and, on the other hand,
£ = /(£*?)</(*>)·
We can obtain a similar contradiction for f(xo) < xq/2C, therefore
f(x)=x/2C(0<x<2C). Π

3.8 PROBABILITY THEORY
409
Solution 2. Introduce the notation g(x) = f(2Cx) (0 < χ < 1). Since
/(0) = 0 and f(2C) = 1, we get g(0) = 0 and g(l) = 1. It is easy to verify
that
fi9(2x) ifxe[0,i],
9[X) \I + I9(2x-1) ifx€(|,l]. W
To see this, observe that if χ G (0,1/2] , the gambler wins or loses 2Cx
forints with probability 1/2, so
g(x) = ±g(2x) + ±g(0) = ±g(2x).
If χ G (1/2,1], the gambler will have 2C or (2x — l)2C forints with equal
probabilities 1/2 , so
g{*) = \g(i) + \a(2x -1) = \ + \g(2x -1).
We will next show that the only bounded solution of the functional equation
(2) is the identity function g(x) = χ (from which we get the same solution
as before).
Put h(x) = g(x) — x. Then h(x) is bounded and
h(x) = { \
\h{2x) if are [0, |],
5Λ(2χ-1) ifxe(i,l].
(3)
We are going to show that h(x) = 0. Since h(x) is bounded, both Μ =
suPx€[o,i] h(x) and m = 1п^€[0д] h(x) are finite, and
h(x)
^ 9
Щ2х) < f if^G[0,|],
zh(2x-l)<f ifzG(§,l].
The definition of Μ implies that Μ < M/2, that is, Μ < 0. A similar
argument shows that m > 0, that is, h(x) = 0. D
Problem P.5. For a real number χ in the interval (0,1) with decimal
representation
0.αι(χ)α2(χ) ...αη(χ)...,
denote by n(x) the smallest nonnegative integer such that
an(x)+lfln(x)+2fln(x)+3an(x)+4 = 1966.
Determine f0 n(x)dx. (abed denotes the decimal number with digits a, 6,
c,d.)
Solution 1. The integral is understood as a Lebesgue integral. Introduce
the following notation:
Ln = {χ : χ G (0,1), n(x) = n},
An = {x : χ G (0,1), αη+ι(χ)αη+2(χ)αη+ζ(χ)αη+Α(χ) = 1966}.

410
3. SOLUTIONS TO THE PROBLEMS
λ is the Lebesgue measure, λη = A(Ln), an = X(An), where (n = 0,
1,2,...). Observe that Ln and An are the unions of finitely many intervals,
so they are measurable sets.
First, we show that Σ™=ο ^η = 1, that is, the integrand is defined almost
everywhere. If
H = {x: xe (0,1),
ак+1{х)ак+2(х)ак+ъ(х)ак+А{х) φ 1966, к = 1,2,... }
and
Cn = {x: xe (0,1),
а4к+1(х)а4к+2(х)а4к+з(х)а4к+4(х) Φ 1966, к = 1,2,... },
then Η С Сп (η = 0,1,2,...) and
Therefore, limn-^ Л(СП) = 0, λ(Η) = 0. The lengths of the intervals in An
are equal to 10~(n+4), the number of these intervals is 10n (we can choose
the first η decimal digits in 10n different ways), and therefore an = 10~4.
Obviously, ΑηΠ Ln = Ln (n = 0,1,...), and if η — 3 < к < η — 1, then
An Π Lk = 0 since two sequences 1966 cannot overlap. If 0 < к < η — 4,
then
X(AnnLk) = X(An_k_4)X(Lk) = 10"4ЛЬ
since the assumption that χ is in Lk restricts only the first (k + 4) decimal
digits of x. So
η η—4
ΙΟ"4 = λ(Αη) = Σ λ(Λι Π Lfc) = Σ 10"4λ* + λ-
fc=0 k=0
that is,
η—4
104An = l-^Afc.
k=0
By the equality Σ£10 ^k = h we get
oo
Σ xk = io4xn+4. (i)
For the integral of n(x),
«1 oo oo oo
/ n(x)dx = ^ηλη = ^ ^ Afc.
^° n=0 n=0fc=n+l

3.8 PROBABILITY THEORY 411
Hence, by (1),
«1 oo oo
/ n(x)dx = Σ 104λη+4 = Ю4 Σ χη
J° η=0 η=4
= 104(1-λ0-λι-λ2-λ3),
and since λ0 = λι = λ2 = λ3 = ΙΟ-4, we clearly get
/ n(x)dx = ΙΟ4 - 4 = 9996. D
Jo
Solution 2. We solve the problem in a more general form. Let σ =
si,S2? · · · ?Sfc be an integer given in the decimal system, and for x G (0,1)
define the function n(x) as follows: let n(x) be the smallest nonnegative
integer such that
an(x)+lan(x)+2 · · · 0>n(x)+k = SlS2 . . . Sk.
We are going to calculate the value of /0 n(x)dx. Let (Ω,Α,Ρ) be the
probability space of the Lebesgue-measurable subsets of the interval (0,1)
endowed with the Lebesgue measure. Then on the set Ω = (0,1), the digits
αϊ, α2,..., α*;,... are independent and
РЫ=г) = ^ (к =1,2,...; г = 0,1,...,9).
On this space, n{x) is a random variable, and its expectation is
/ n(x) dx.
Jo
Put Pn = P{n{x) = n) and Qn = Ρ(αη+ιαη+2 · · -Q>n+k = σ) = 1/10*
(n = 0,l,...).
Let / be the smallest natural number such that shifting σ by / positions,
the digits on the same positions coincide. (That is Si = si+i for all г =
1,2,..., к — I). Obviously, 1 < I < к . Rewrite к in the form к = pi + r,
where 0 < p, 1 < r < I. The σ consists of ρ blocks r = s\S2 ... si of
length Z, and the first r digits of this block r. Notice that
Qn = {PoQn-k + PiQn-k-i + -" + Pn-kQo)
+ (Pn_fc+rl(T(/c-r) + Pn-k+r+iW-{k-r-l) + · · · + Pn-il0-1 + Pn)
(n = fe,fe + l,...) (2)
since the event
An+lfln+2·· · an+k — V
can occur in two different cases, as follows:

412
3. SOLUTIONS TO THE PROBLEMS
(a) Block σ appears only from the (г + l)th index (0 < г < η — /с), and
after η — к — г digits it occurs again. (The probability of this case is
-*гЦ;п—/с—z*J
(b) Block σ appears from the (n — k + r + z/)th index (г = 0,1,...,ρ), thus
an-k+r+il+lQ"n-k+r+il+2 · · · an+r+il — &,
and it appears again starting from the (n + l)th index. Note that
besides the previous assumption, it is also necessary that the values of
the undefined (k — r — il) digits are given. The probability of this case
is Ρη_£_|_Γ_|_ΐ/10
Assume that Pn = Qn = 0 if η = —1, —2, Then (2) holds also for
0 < η < к. Multiply (2) by zn, add up the resulting equations for all values
of n, and introduce the notation
Ρ{ζ) = Σρη*η, QW = £)Qn^
n=0 n=0
to get
ρ
Q{z) = zkP{z)Q(z) +J2zk~r~ilpWl0~{k~r~il)·
г=0
Using the obvious formula
00 1 1
Ж*) = Σ Ток*" = 10*(l-z)'
we have
P(z) =
zk + 10fc(! _ z) £P=o zk-r-UW-(k-r-U) '
From P(l) = 1 we can see that n(x) is defined almost everywhere, and
Г1 p
/ n(x)dx = P'(l) = -k + J2l0r+il .
This last equality shows that the integral is always an integer. If σ = 1966,
then к = I = r = A, p = 0, and thus
/ n(x)dx =
Jo
104 - 4 = 9996.
If σ = 1961, then к = 4, / = 3, ρ = г = 1, and thus
10 + 104 - 4 = 10,006.
/ n(x)dx =
Jo

3.8 PROBABILITY THEORY 413
If σ = 1919, then к = 4, / = r = 2, ρ = 1, thus
/ =102 + 104-4= 10, (
Jo
,096,
and, finally, if σ = 1111, then к = 4, Ζ = г = 1, ρ = 3, thus
10 + ΙΟ2 + ΙΟ3 + ΙΟ4 - 4 = 11,106. D
/ n(x)dx =
Jo
Remarks.
1. The problem can obviously be generalized for any number system not
just the decimal one.
2. One can reduce the problem to compute the expectation of the first
return time of a recurrent sequence of events. In this case, the value of
the integral can be determined using the theorem by Erdos, Feller, and
Pollard.
Problem P.6. Let f be a continuous function on the unit interval [0,1].
Show that
^ΐ!-Ί!Κχ1+'η+Χη)άχι·-·άχη=ί(^
lim / · · · / /( Цх\ ...xn)dxi... dxn =/(-].
n-*°°Jo Jo Vе/
Solution 1. Let к be an arbitrary positive integer and n> k. Consider the
multinomial expansion of (Σ^=1 Xi)k- The number of terms that contain
all variables with powers not exceeding one is
n(n - 1)... (n - к + 1) = nk + 0{nk~l).
The integrals of these terms on the unit cube
Rn = {{xi,...,xn): 0<Яг<1} (г = 1,...,п)
are equal to 1/2*4 The number of terms containing at least one Xi with a
power higher than 1 is not greater than η · nk~2 = nk~1. The integrals of
these terms on Rn are not greater than 1. From this observation, we get
as η —> oo.

414
3. SOLUTIONS TO THE PROBLEMS
An easy calculation shows that
as η —> oo.
Hence, we have shown that the statements are valid for f(x) = xk.
It is also obvious that the statements hold for constant functions, and if
they are valid for two continuous functions, they are valid for their linear
combinations as well. So we have proved the theorem for polynomials.
If the function / is continuous on [0,1], then by the Weierstrass
approximation theorem, for any positive ε there is a polynomial p(x) such
that \f(x) —p(x)\ < ε/3 (0 < χ < 1). Since the statements are valid for
polynomials, there is a K such that
/ · · · / p{ ψχι...χη)(1χι
\Jo Jo
for any n> K. Therefore, for any n> K,
.dx
.dx
»-"Q)
<
<
|Р(г1±^)*-"*-'Ш
< С t\f (Xi+'"+xn\ (xi+—hx"N\
'к "Jo I V η ) P\ η )
l!-l!p(xl+"n+Xn)dxi-dxn-p^)
'G)-'G)
dx\ ...dxn
and similarly,
\Jo Jo
x\ ...xn)dx\...dx
*-'G)
< ε.
Now the proof is completed. D
Solution 2. Let ξι,..., £n,... be mutually independent, uniformly
distributed random variables on the interval (0,1). Then the variables ηη =
l°g£n (n = 1,2,...) are also mutually independent, and they have the
same distribution. Since Ε(ξη) = 1/2 and Ε(ηη) = — 1 (η = 1,2,...), a
theorem by Kolmogorov implies that
Ψη
£ι + ··· + £η
η

3.8 PROBABILITY THEORY
415
and
η
with probability 1.
By the assumptions the function / is bounded on the interval [0,1] and
continuous at χ = 1/2 , and the function g(x) = f(ex) is also bounded on
(—oo,0] and continuous at χ = — 1. Using the Lebesgue theorem, we get
я(/Ы) - я(/ф) = /φ,
ад^)) - вд-i)) = я(/ф) = f(-j,
е е
and, finally, the definitions of φη, φ'η, and g imply that
/®-w<*>>-jf-j£/(s±;^)<*...*.
and
/ Q J = ВД<^)) = /···/ /(^ι···^)ώι · · · dxn. U
Remark. The statements also hold in the more general case when / is
bounded, integrable, and continuous at χ = 1/2 and χ = 1/e, respectively.
In the first statement we can also drop the condition of boundedness (as
was pointed out by Laszlo Lovasz).
Problem P.7. Let Ai,...,An be arbitrary events in a probability field.
Denote by Ck the event that at least к of Αι,..., An occur. Prove that
flP(Ck)<f[P(Ak).
k=l k=l
Solution. We start with three remarks:
Comment A. The statement is true for η = 2, that is,
P(A + B)P(AB) < P(A)P(B).
To see this, use the notation AB = x, AB = y, and AB = z. Then our
inequality becomes
(P(x) + P(y) + P{z))P{x) < (P(x) + P(y)){P{x) + P(z)),
which obviously holds.
Comment B. If A\ D · · · D An, then Ck = Аь.

416 3. SOLUTIONS TO THE PROBLEMS
Comment C. Ck is identical to the event that at least к occur from the
events A\ + A2, A\A2, A3,..., An, that is, from the events Αι + A2, A\A2,
A3,..., An, exactly as many occur as from the events Αι, A2,..., An. This
statement is trivial.
Our proof is based on the above comments. Put A® = Ai (i = 1,..., n),
and assume that the events Αμ, ... ,A% are already defined. Next, define
the events Αμ+1, ... ,A^+1 as follows: choose a pair A?, Αμ of events, г < j,
for which A? 2 А?, Щ 2 Αμ (that is, A? and Αμ· are incomparable) and
define Αμ+1 = A? + Αμ, Αμ+ι = ΑμΑμ and Αμ+1 = Αμ for к φ г J.
Continue this procedure until there exists at least one pair of incomparable
events. This procedure will terminate in finitely many steps, since in each
step the number of incomparable pairs decreases. This observation follows
from the facts that Αμ+1 and Αμ+ι become comparable, and any event is
comparable to at least as many of the events Αμ +Α**, ΑμΑμ, as of Αμ and
By comment A,
Пр(^+1)<Прю· ί1)
к=1 к=1
Observe that comment С implies that the events Ck remain unchanged, and
because of comment В they are identical to the final events Αμ. Therefore,
(1) implies the original statement.
From the proof of comment A, we see that if Π£=ι P(A%) φ 0 and
Ρ(ΑμΑμ)Ρ(ΑμΑμ) φ 0, then strict inequality holds in (1). Hence, equality
holds iff either Π£=ι ^(^fc) = 0 or the original system of events {A^} are
already "ordered." D
Remarks.
1. If multiplication is replaced by addition, then in the statement we
always have equality. That is,
г=1 г=1
2. Let Sk(x\,..., xn) be the Arth elementary symmetric polynomial of the
variables xi,...,xn. Then
Sib(P(Ci),..., P(Cn)) < 5Л(Р(Л1),..., P(An)).
Slight modifications of the above proof can be applied to show the
validity of this more general statement.
Problem P.8. Let А and В be nonsingulai matrices of order p, and
let ξ and η be independent random vectors of dimension p. Show that
if £, η and ξ Α + ηΒ have the same distribution, if their first and second

3.8 PROBABILITY THEORY
417
moments exist, and if their covariance matrix is the identity matrix, then
these random vectors are normally distributed.
Solution. Put ζ = ξΑ + ηΒ. Without loss of generality, we can suppose
that Ε(ξ) = Ε(η) = Ε(ζ) = 0. Since ξ and η are independent,
Var (ξ) = Ε[{ξΑ + ηΒ)*(ξΑ + ηΒ)} = A* Var (ξ)Α + В* Var (η)Β,
where Var (£), Var (77), and Var (ζ) denote the covariance matrices of £, η,
and ζ, respectively. They are equal to the identity matrix, so
A*A + B*B = I.
Thus, for arbitrary vector t,
(tA\tA*) + (tB\tB*) = (t,t),
and by the regularity of A and B,
\tA*\<\t\ \tB*\<\t\ (i^O), (1)
where |. | denotes the Euclidean length of vectors.
Since ξ and η are independent,
E(ei(t,M+nB)) = Е(е«*л*'Я)Е(е«*в*«)).
If φ{ί) denotes the common characteristic function of ξ, η, and ζ, then
4>{t) = <p{tA')<p{tB·). (2)
We are going to show that φ(ί) ф 0. Assume that φ(ί) has real roots. Then
its continuity implies that there is a root to with smallest absolute value.
Since φ(0) = 1, to ф 0. Thus, (1) implies that
φ(ί0Α*)φΟ, φ(ίοΒ*)φΟ,
and this contradicts equation (2).
Consider the function
and choose the branch of the logarithmic function for which ^(0) = 0.
The assumptions imply the existence of vectors
φ'it) = ivW, v'(o) = ίΕ(ξ) = о

418
3. SOLUTIONS TO THE PROBLEMS
and matrices
Therefore,
and hence
ψ" it) = ^φ(ί), φ"(0) = -Var (ξ) = -Ε.
>
= 0,
t=o
£**>
= 0,
t=o
m = o((t,t)) if |t|-o.
(3)
We are going to show that ψ(ί) = 0. Suppose ψ(ί) ф 0. Then there exists
a positive ε and a t Φ 0 such that \i/>(t)\ > e(t, t). Because of the continuity
of <p(t), there exists such a t\ with minimal absolute value , and (3) implies
that t\ Φ 0. From inequality (1), we conclude that
and
From (2),
№(M*)|<e(M*,M*)
|^ιΒ*)|<ε(*ιΒ*,*ιΒ*).
so
|V(*i)| < m1A*)\ + \tP(t1B*)\<e[(t1A*,t1A*) + (t1B*,t1B*)]=e(t1,t1).
This inequality contradicts the definition of t\. Consequently, ψ(ί) = 0,
and so
Problem P.9.. Let £i,£2,··· oe independent random variables such
that Εξη = m > 0 and Var(£n) = σ2 < oo (n = 1,2,...). Let {an} be а
sequence of positive numbers such that an —> 0 and Σ™=ι αη = oo. Prove
that
Ρ [ lim Σα*& = oo = 1.
k=l
Solution. We can solve the problem using the Kolmogorov inequality, but
in the solution given below we only use the Chebyshev inequality.
Instead of an —> 0 and Var (ξη) = σ2, it is sufficient to assume that {an}
is bounded and Var(£n) = σ2. For instance, suppose 0 < an < 1. Put
Sn = Σ£=1 аь an(^ define the sequence {rn} by Srn < η2 < SVn+i· Then
n2 -1 < Srn < n2.
(1)

3.8 PROBABILITY THEORY
419
If limn Σ£=1 ак^к = oo does not hold, then there is a K such that
η
5>Λ&<ΙΓ (2)
fc=l
occurs infinitely many times.
It is sufficient to show that for any fixed K, this event has probability 0
since the union of these events for η = 1,2,... includes all event sequences
for which limn Σ ak£,k = °o does not hold.
Let Ak be the event that there is an η between τ к and гь+ι such that
(2) holds. Note that (2) holds infinitely many times if infinitely many Ak
occur. Denote the probability of Ak by P^. We have to show that with
probability one only finitely many Ak occur. We prove this by verifying
tnat ΣΤ=ι pk is convergent. In this case, J2k>rn pk —► 0, and X^>m pk is
greater than the probability that infinitely many A^ occur. Therefore, this
latest probability is equal to 0. (Actually, here we have proved and applied
the Borel-Cantelli lemma.)
Put щ = Σ<=ια<& and Фк = sr=ri+ia*l&l- If (2) holds for some
Tk < ζ < rk+1 with ζ replacing n, then either
Гк
αϊξί < к —
i=l Z
or
CLi£i < -k y,
i=rk + l
supposing k2m> 6K.
In the last case, ^ = ΣΓ=γ£+ι αΐ\ζί\ > k2m/3, so at least one of the
inequalities фь > k2m/S and щ < к2т/2 is valid. Thus,
Ъ<р(щ<Щ+р(ь>*РУ (3)
We have to show that on the right side both terms are small.
It is obvious that Е(щ) = mSrk and Var(^) < o2Srk since a^ < 1
implies that
Var (щ) <σ2Σ "Ι < °* Σ α* = σ2^ *
г=1 г=1
Furthermore,
ВД) < Е{\ + ξ*) = 1 + Я2(&) + Var β) = 1 + m2 + σ2 = A
and
Var(|fc|) = E{&) - Я2(|6|) < Я(£2) < Α.

420
3. SOLUTIONS TO THE PROBLEMS
Since the absolute values of independent random variables are independent,
ЕШ < (Srk+1 - Srk) max Ε(\ξ{\) < (Srk+1 - Srk)A = 0(k)
гк<г<Гк+1
and
nt+i
Var(^)< У, ai m^ Var№i\)<(Srk+1-Srk)A = 0(k).
. *~^, л гк<г<гк+1
г=гк + 1
Now we can use the Chebyshev inequality and (1) to see that
E(Vk) ~ k2m, Var (щ) ~ k2a,
P(m <^)< P(|% - E(Vk)\ > ^) = оф,
Р(фк > ^) < Р{\Фы - ЕШ\ >^) = θφ
if к is large enough. From these relations, the convergence of J] P*.
follows. D
RemarL·.
1. One can sharpen the proof to see that J2k<na^k tends to infinity as
fast as J2k<nak w^n probability one.
2. The condition an = 0(Sn) is weaker than boundedness, and it does
not guarantee the convergence to infinity with probability one, but the
slightly stronger condition
(which is still weaker than boundedness) does.
P.10. Let #i,...,#n be independent, uniformly distributed, random
variables in the unit interval [0,1]. Define
h(x) = -#{k:uk <x}.
η
Prove that the probability that there is an xo G (0,1) such that h(xo) = xo,
is equal to 1 — ^.
Solution 1. Let us denote by Д the interval ((k — l)/n, к/ή). Consider
the elementary events that for every г, ϋι belongs to 1^., where ki is given
for every i. So we decomposed the original sample space to the union of
nn disjoint events of probability l/nn (events with probability zero are
omitted). These events will be called atoms.

3.8 PROBABILITY THEORY
421
Since the value (1/n) · Σ·&·<χ 1 = Mx) 1S an integer multiple of 1/n,
in order to check whether h(xo) = xo holds for some x0? it is sufficient to
know which intervals 1^. contain ui, that is, which atom will be the result
of the experiment series. They have the same probability; therefore, it is
sufficient to determine the number of atoms for which there is no к such
that exactly к ϋι occur until к (1 < к < η). Denote by An (1 < к < η)
the number of atoms for which there is no к such that the first к intervals
contain exactly к ϋι values.
Let Bk denote the number of atoms for which there are exactly к ύι
values in Uj<klj but for any t < к, Uj<tlj does not contain exactly t ϋι
values. So every atom that has exactly к ϋι values in the first к intervals
belongs to exactly one Bk- Next, we determine the value of Bk- We can
choose the tf^s belonging to Uj<klj in (£) different ways. We can choose
the first к intervals 1^. containing these ϋι values in Ak different ways,
and we can arrange the remaining η — к ϋι values in the remaining η — к
intervals in (n — к)п~к different ways. So we have the recursion
пп-Ап = ^/(^\Ак(п-к)п-к, Аг = 1. (1)
We are going to verify that An = nn_1 by mathematical induction. It is
easy to check that Αι = 1° and A2 = 21 or As = 32. Suppose that we have
already verified this relation up to η — 1. In order to show this equation
for n, we have to show that
пп-пп-1 = ^(П\к-1(п-к)п-к. (2)
fe=l ^ '
It is easy to see by interchanging к and n—k that this equation is equivalent
to
nn-i(n - 1) = J] (Л k\n - k)kk-\n - k)n~k-\ (3)
k=i ^ ^
We will prove this equality by a well-known theorem of Cayley. Consider
the trees having the numbers 1,2,..., η as vertices. (A tree is a connected
graph not containing a circle.) Two graphs are different if they do not
have the same (z,j) pairs as edges. By the Cayley theorem, there are nn~2
different such trees. Now choose in every tree in all possible different ways
one vertex and one edge. We will call these graphs vertex-edge-signed trees,
and consider these graphs different if they are different before our vertex-
edge selection, or the selected vertex-edge pair is different. Since a tree has
η — 1 edges, we have nn~1(n — 1) different vertex-edge-signed trees, and
this number is exactly the left-hand side of (3). Let us count the number
of these vertex-edge-signed trees in another way. By omitting the signed
edge, we get a tree with к vertices with one signed vertex and a normal tree
with (n — k) vertices (k = 1,2,..., η — 1). On the other hand, if we choose

422
3. SOLUTIONS TO THE PROBLEMS
к vertices out of n, define a tree on them, select a vertex and define a tree
from the remaining η — к vertices, and connect the two trees by an edge,
which will be called the signed edge, then we get a vertex-edge-signed tree.
Thus, the two procedures are the reverse of each other. In this way, we get
n-l
Σ (fe) k * kk~2(n ~ k)n~k~2k(n - k)
different graphs, which is the right-hand side of (3). D
Solution 2. The first part of the solution is the same, but here we give an
algebraic proof of (2).
Lemma 1. If 0 < j < m, then
-l)kkj = 0.
Proof. Denote by D the operation of differentiation followed by a
multiplication by x. Thus Drn(xk) = krnxk. It is easy to see by induction
that if #0 is a ro°t of the polynomial p(x) with multiplicity mo, then xq
will also be a root of DTn(p(x)) with multiplicity at least mo — m. Apply
this observation to the polynomial p(x) = (1 — х)ш and the operator D3,
j < m. Then we get that χ = 1 is a zero of the polynomial
^((i-*r) = f;(™)(-i)W4
This completes the proof of Lemma 1.
Lemma 2.
n-i /#rA
\n — к „„η-1 „η—1
η —ι / \
(z-k)n-k = nzn-1-nn
Proof.
n-l
η—1 / \ η—к / Ί \
ς(ι>>-ς(„:;!><-*γ->-
3 =
η —1 η—к
^^fc!j!(n-fc-j)!

3.8 PROBABILITY THEORY
423
η—1 η—к
η— in — κ / \ / ·\
= Σ Σ Q (п^")^(-1)"-^*-^ -η-1
= Σ Σ (J) (η ^(-ΐ)-*-'*-1-' - η-1
= Σ Σ (η) fnr,;V'(-1)n"fc"iiJfen"1"i''nn"1 +ηζ"~1
= Σ (")(-1)и-^(Е (η I'Vi)**"-'-1) - "η_1+^η_1
j=o \Э / к=1 \ /
= ηζη-1-ηη~1
since, by Lemma 1, the inner sum is equal to zero.
Substitute ζ = η to get (2). D
Problem P.ll. We throw N balls into η urns, one by one,
independently and uniformly Let Χι = Χι(Ν,η) be the total number of balls in
the ith urn. Consider the random variable
N
y(N,n)= mm \Xi--Y
1<ι<η Π
Verify the following three statements:
(a) If η —> oo and N/n3 —> oo, then
p I 2/C^ < χ ι j _ e-xV2/^ foraJJx>0.
Π λ/ n
(b) If η -^ oo and Ν/η3 < Κ (Κ constant), then for any ε > 0 there is
an A > 0 such that
P(y(N,n) <A)>\-e.
(c) If η —> oo and AT/n3 —> 0 then
P(y(JV,n) <1)->1.
Solution. The basic idea of the proof is the following: Xi is a B(p, N)
(binomial distribution with parameters p, N), where ρ = 1/n. If the X^s
were independent, the statements could be proved easily. But of course

424
3. SOLUTIONS TO THE PROBLEMS
they are not, since X\ -\ + Xn = N. However, we will show that к of
them can be considered independent for sufficiently small values of k.
Suppose that the X[s are independent and their distribution is B(p, N).
Let αϊ,... a\z be natural numbers in the interval [0, N],
a = P(Xi = a^, г = 1,... к), οι = Ρ{Χ[ = α^, г = 1,..., к).
Introduce some additional notations:
a = ^aii di=pN + bi, Ь = ^Ьг .
г=1 г=1
We will show that a ~ af under the conditions к = 0(1), bi = o(y/~N), and
N>n.
Obviously,
and
so
/V!
a = —r : pa{\ - kp)N~a
(Il!Li *!)(*-α)!
а'=(й0)р"{1-рГ"·
a _ a> _ N^^jN - a)!(l - p)fcjV-°
a (Пы^-а()!)(1-М"-«
The value of /3 will be estimated by the Stirling formula:
The error of log£, that is, the difference of the logarithms of the true and
estimated β values can be determined from the 0(l/t) terms as
since
Oi=pN + o(y/N), a = kpN + o(v/iV), ρ = ► 0
η
and Ν > η implies N —> oo . The terms e~l and л/27т are cancelled, and
the value of the \Д terms can be given as
'Nb-ijN-a) _ J l-§
Illi(N-ai) Vnti(l-t)
= l+o(l).

3.8 PROBABILITY THEORY 425
So, finally,
NN(<k-V(N-a)N-a (l-p)kN~a
Ilki=i(N - ai)n~ai (ΐ-ΜΝ"α
= (i-j^)N-a.n!Li(i-^)N-a*
(1 -kp)N~a ' (l-p)kN-a
( b \N-a к , , ч N-di
= у" ТфТкр)) '■ Π у1 - щЬр))
Since each term converges to 1, we can apply the formula log(l + ε) =
се + Ο (ε2). The value of the second-order term is
The value of the first-order term is
-b(N - kpN - b) ^bi(N-pN-bj)
N(l-kp) +^ N(l-p)
h2 k л2
N(l-kp) f^N(l-p)
which implies that log/3 = o(l).
Solution to part (a). Put
ak = P(\Xi - pN\ < x\/ —; i = 1,..., fe),
V n°
c = Ply(Ar,n)<^W-3 I .
Recall that if Αι,..., An are arbitrary events and
bj — / v -P (^ii? · · · ? Aij),
l<ii<--'<ij<n
then
2s+l 2s
J2(-l)Sj <P(A[...A-n) < Σ(-1)181
j=0 j=0
(see, for example, K. Bogname, J. Mogyorodi, A. Prekopa, A. Renyi,
D. Szasz, Exercises in Probability Theory (in Hungarian), Tankonyvkiado,
Budapest, 1971, Problem 1.2.11.c). So
2s+l у ч 2s у ч
7=0 V/ 7=0 4*//

426 3. SOLUTIONS TO THE PROBLEMS
Note that
ak ~ a'k = Ρ I \X[ - pN\ < x^ -j; г = 1,..., к I ,
since we can apply the previous asymptotic relation for any system Xi —
pN = hi, when |6;| < XyjN/n3. Since a'k = a,k, with
ot = p(\X[-PN\<xJ^,
>---Ъ-»ЪУ+о((к1у>)+о(±(;у
The first sum is the first к terms of (1 — a')n, and substituting them by
(1 - a')n itself, the error will be only 0((fc^1)a//c+1). From the local De
Moivre-Laplace theorem we conclude that
a ~ \ —
V 7ГП
(here we use the fact that N/n3 —> oo).
Select an arbitrary, small ε > 0. If к is sufficiently large, then the first
error term is less than ε, the second term is o((l + a')n) = o(l), and finally,
log(l
- a') = -a' + 0{a'2), η log(l - a') = -J-x + o(l),
which proves the assertion.
Solution to part (c). We are going to prove for an arbitrary, small ε > 0
that there exist δ and no such that for η > щ and N/n3 < 0, we have
P(y(N,n) <1) >l-e.
If we wish to apply the previous reasoning, we face the difficulty that
(fc+i)a' +1 1S not smaU enough, and (1 + a')n is not bounded if a' is too
large. We can, however, overcome this difficulty since these quantities can
be replaced by
([^')(7Г - ('WT*·
respectively. Consider the probability that for any of the first [А л/N/n]
urns, \Xi — N/n\ < 1. (If δ is small enough, then Ay/N/n < ή). Here

3.8 PROBABILITY THEORY
427
where с remains between positive bounds for N >n. (We can assume this
since otherwise the assertion is obvious.) Then the proof can be completed
as it was demonstrated in the proof of (a).)
Solution to part (b). We can assume that with the δ given in the previous
part Ν/η3 > δ, otherwise, the previous result implies the desired inequality.
Now a' = cA/n, where с is between positive bounds (they depend on δ and
K). The proof then can be completed with this a' in the same way as was
shown at the end of the solution to part (a). D
Problem P.12. Determine the value of
sup [log Εξ- Ε log £],
1<£<2
where ξ is a random variable and Ε denotes expectation.
Solution. Since log ξ is concave in £, and 1 < ξ < 2,
log£ = log[(2 - ξ) + (ξ - 1)2] > (ξ - l)log2,
which implies that £(log£) > [Ε(ξ) - l]log2. Note that 1 < Ε(ξ) < 2,
and therefore
log Ε(ξ) - E(\og ξ) < log Ε(ξ) - [Ε(ξ) - 1] log 2
< max [log t- (£-l)log2l
~ l<t<2
= log*0-(*o-l)log2 = tf ,
where £o = 1/I°g2 G [1,2]. We will next show that К is the least upper
bound. Consider the random variable £o defined by Ρ(ξο = 1) = 2 —to and
Ρ{ξ0 = 2) = t0-I. Then log£(£0) = log*o and £(log£0) = (*o - l)bg2,
therefore, К = - log log 2 - 1 + log 2. D
Remark. A similar proof shows that if φ(ϊ) is a concave and continuous
function on the closed interval [a, b], then
sup {φ(Ε(ξ)) - Ε(φ(ξ))} = max [p(t) - ^φ(α) - j^fv(b)].
a<£<6 a<t<b О — α О — a
A further generalization is also possible for continuous but not necessarily
concave functions φ.
Problem P. 13. Find the limit distribution of the sequence ηη of
random variables with distribution
Ρ (ηη = arccos(cos2 J ~^ )π)\ = - (j = 1,,..., n).
(arccos(.) denotes the main value.)

428
3. SOLUTIONS TO THE PROBLEMS
Solution. Let ξη be defined as
P^" = ^) = ^ (j = 1,2,-··,«)·
λη η
Then ηη = arccos (cos2 π£η). Since ξη e (0,1) (n = 1,2,...), and the
function arccos(cos2 πχ) maps the interval (0,1) into (0, π/2), Ρ(ηη < χ) =
0 for χ < 0 and Ρ(ηη < χ) = 1 for χ > π/2. Suppose χ e [0, π/2].
Since cos χ is strictly decreasing in the interval [0, π], arccosя is strictly
decreasing in the interval [—1,1]. The distribution of ξη is symmetric with
respect to 1/2, therefore,
Ρ(ηη < x) = P(arccos(cos2 πξη) < χ) = P(cos2^£n) > cos x)
= P(cos^£n) > Vcosx) + P(cos^£n) < — д/cos x)
= Ρ(πζη > arccos(—Vcosx)) + Ρ(π£η < arccos(v/cosx))
лт^/ ^ / ч лт^/^ arccos л/cosχч
= 2Ρ(π£η < arccos Vcosx) = 2P(£n < - ).
π
For xe(0,n/2), arccos -^/cosχ e (0, π/2); thus only the value of Ρ(ξη < у)
for ye(0,1/2) is to be determined.
Let к be the largest integer such that k/n < y. Then
П. * * 71. 71. 71.
к + 1
η ^—' η η η η
2 j - 1 ^ к
2η ^ η
Therefore, Итп_*оо Ρ(ξη <у) = У, and consequently,
Γθ if χ < 0,
lim Ρ(τ7η < я) = < I arccos ^/cosx if ж G (0, f),
n—+00 Ι π ^
I 1 ifs> f. D
Remarks.
1. The random variable ηη may take the same value twice.
2. The numbers Xj = cos((2j — 1)/2η)π (j = 1,2,..., n) are the zeros of
the Chebyshev polynomial Tn = cos(n arccos x), and the corresponding
Cotes numbers are Anj = π/η. Using the characteristic function of ηη,
one can easily prove the following.
Theorem. Let / be a continuous function in [—1,1], and let ξη be a
random variable defined by
Ρ(ξη = /(λ,)) = i (j = 1,... ,n), (n = 1,2,...).
Assume that ξ is a random variable with density function 1/(π\/ΐ — x2) in
the interval (—1,1) and zero otherwise. Then the sequence {ξη} converges
weakly to the random variable /(ξ).

3.8 PROBABILITY THEORY
429
In the problem, select f(x) = arccosrr2 so that the limit distribution of
the sequence {ηη} is the distribution of η = arccos£2. Simple calculations
show that the density function of η is
/θ 1 ΤΓ
= if xe(0, —) and 0 otherwise.
π ^/l-tan2! 2
Problem P. 14. Let μ and ν be two probability measures on the Borel
sets of the plane. Prove that there are random variables £1, £2, 771, Щ such
that
(a) the distribution of (£1,^2) is μ and the distribution of (771,772) is z/,
(b) £1 < 771, £2 < 772 almost everywhere, if and only if μ{ϋ) > v{G) for
all sets of the form G = υ*=1(—oo, Xi) x (—00, yi).
Solution. The necessity part of the assertion is obvious, since if ξ = (£ι, ξ2)
and 77 = (771,772), then ηεϋ implies £eG, so
v(G) = P(VeG) < P&G) = μ(β).
First, we prove the sufficiency part of the assertion for the special case
where μ and ν are concentrated to the finite set χι, ... ,rrm and yi, ... ,yn,
respectively. Define the graph Q with vertices x\,..., rrm, yi,..., yn and
where the vertices x^ and % are connected by an edge if and only if Xi < yj
((a, b) < (c, d) means α <c and b <d). Define a random variable (£, 77) that
takes the values (xi,yj) with probability a^·. These numbers aiyj satisfy
the following relations:
(a) aiyj > 0;
(b) IT=iabJ = K%);
(c) Σ^=ιαΜ = Μ^);
(d) If dij > 0, then there is an edge between x^ and yj.
By the Konig-Egervary theorem, such numbers a^j exist if and only if
for any set Υ С {yi,..., yn} the μ measure of the set X consisting of all
points Xi connected with the elements of Υ is at least v(Y). Assume that
У={уи---,Ук} with yi = (yl y'l)
and
G = U?=i(-oo,i4)x(-oo,j4')·
Then a point x» is connected to Υ if and only if x» G G. Therefore,
Σ МЫ = β(Χ П G) = μ(β) > v(G) > ν(Υ).
XicX
Consequently, there are numbers aij having properties (a)-(d), and any
vector variable (£, η) that has the values (xi,yj) with probabilities aiyj
satisfies the requirements of the theorem.

430
3. SOLUTIONS TO THE PROBLEMS
Now consider the general case. Denote by F(x, y) the distribution
function of the underlying variable (£, η), and put
F^x', x") = μ((-οο, χ') χ (-00, χ")),
F„(y',y") = v((-oo,yf)x(-oo,y")).
The existence of a distribution function F satisfying
(i) F(x,oo) = F^x)
(ii) F(oo,y) = F„(y)
(iii) F{y,y) = Fv{y)
solves the problem, since for a random variable (£, η) having distribution
F, ξ < η holds with probability one because F(y, y) = F(oo, y).
Divide the square [—n, n] x [—n, n] into n4 small squares of sides 1/n.
Define μη as follows: concentrate the μ measure of every point into the
closest node. (More precisely, consider the set of all points of the plane
that are closest to a given vertex, and concentrate the μ measure of this
set to the given vertex. If a point is in the same distance from two or
more vertices, then select the one that has the smallest abscissa among the
vertices with smallest ordinate.)
Let vn be defined analogously. Then (μη, vn) satisfies the requirements of
the theorem, since they are measures concentrated in finitely many points.
Therefore, there is a variable (£n?f?n) such that ξη < ηη , the distribution
of ξη is μη, and the distribution of ηη is vn. Let Fn be the distribution
function of (£n?f?n)· Then
Fn(y,y) = Fn(oo,y),
furthermore,
Fn(xf, x", oo, oo) = μη((-οο, χ') x (-oo, x")),
Fn(oo,oo,y',y") = vn((-oo,y) x (-00,2/")).
Let S be a countable and everywhere-dense set of the continuity points of
Fv and Ρμ. Select a subsequence щ such that F(a,b) = lim^oo FUi (a, b)
exists for all a, b G S. Extend the definition of F by the relation
F(x, y) = sup{ lim Fni (a, b) : a <x, b <y, a,b e S}.
i—>oo
It is easy to verify that F is a distribution function and that it satisfies (i),
(ii), (iii). Thus, the proof is completed. D

3.8 PROBABILITY THEORY
431
Problem P. 15. Let Χι, X2, · · ·, Xn be (not necessarily independent)
discrete random variables. Prove that there exist at least n2/2 pairs (г, j)
such that
H(Xi+Xj)>\ пап {ЩХк)},
О 1<к<п
where H(X) denotes the Shannon entropy of X.
Solution. Consider the graph G with vertices {1,2,..., n}, and assume
that г and j are connected with an edge if and only if the inequality
H{Xi+Xj)>\mm H{Xk) (1)
does not hold. There is no loop in G, since for all г
H(Xi +Х{) = H(Xi) > \ min H(Xk).
О 1<к<п
We are going to show that there is no triangle in G. Let г, j, and Ζ be
different elements of {1,2,..., ή]. Obviously,
Xi = \{[Хг + Xj] + [Xt + Χι) - [Xj + Xt]),
that is,
H(Xi) = H([Xi + Xj] + [Xi + Xt] - [Xj + Xt])
< H(Xt + Xj) + H(Xi + Xt) + H(Xj + Xt)
< 3 тах{Я№ + Xj), Н(Х{ + Xt), H(Xj + Xt)} .
So
\ min Н(Хк)<\н(Х<)
О 1<к<п О
< тах{Н(Х{ + Xj), Н(Х{ + Χ/), Η(Χά + Xt)}.
Therefore, there are two vertices among z,j, and / that are not connected.
By a well-known theorem of Pal Turan (see, Mat Fiz. Lapok 48, (1941),
pp. 436-452), the number of edges of such a graph is not larger than n2/4.
Consequently, the number of pairs (г, j) such that (1) holds is at least
n2 - 2 · (n2/4) = n2/2. D
Remark. The bound n2/2 is the best possible. Let X be a random
variable such that H(X) φ 0, and define
Xi = X2 = ··· = X[n] = X,
^[§ ]+i — · · · — Xn = —X-
If either 1 < г < n/2 < j < η or 1 < j < n/2 < i < n, then H(Xi+Xj) = 0.
The number of such pairs (г, j) is equal to [n2/2].

432
3. SOLUTIONS TO THE PROBLEMS
Problem P. 16. Let £i,£2, · · · be independent, identically distributed
random variables with distribution
P(fr = -l) = P(fr = l) = I.
Write Sn = £i + ξ2 + · · · + ξη (η = 1,2,...), So = 0, and
Tn = —= max Sk .
у/П 0<k<n
Prove that lim inf (log n) Tn = 0 with probability one.
Solution. Denote maxa<k<bSk by (Sb - Sa)*, and max0<KnSfc by S*.
First, suppose that we have two sequences g(n) and f(n) such that
oo
n=l
and
for all ε > 0. If such sequences g(n), f(n) are found, then by the Borel-
Cantelli lemma with probability one, only finitely many of the events
S*nn) >g(f(n))
and infinitely many of the events
(S/(n+i)-S/(n)r + V7Wg(/(n))
y/f(n + l) Mog/(n + l)
occur. Consequently, infinitely many of the events
<
y/f(n + 1) " y/f(n + 1) bg /(n + 1)
will occur with probability one. Hence, it is sufficient to find sequences
g(n), f(n) satisfying (a) and (b).
We are going to apply the approximation
"(%<')"·№£·-**'*·

3.8 PROBABILITY THEORY 433
(see, for example, A. Renyi, Foundation of Probability, Holden Day, San
Francisco, 1970). This approximation holds if χ is replaced by a sequence
xn =о(пз).
So
ρ ( (g/(n+l) ~ S/(n))* + V7(n)g(/(")) ^
V7(n + 1) log/(n + l)
>p[ (g/("+i)~g/(n))* ^ g y7(")g(/(")) \
- \v7(n + l)-/(n) log/(n + l) vTF+I) J
where
л_ ^ у7Ш/(п))
log/(n + l) >//("+!)
is small enough. Therefore, we have to find sequences f(n) and g(n) such
that
(n) Sv"^+ir
00
< 00
< 00.
"^ /.OO
(iii) V / е"^2^
η=ίΛ(/(η))
It is easy to see that, for example, g(f(n)) = n3 and f(n + 1) = (n!)14
satisfy these relations. D
Problem P.17. Let the sequence of random variables {Xm, m > 0},
Xo = 0, be an infinite random walk on the set of nonnegative integers with
transition probabilities
Pi = P(Xm+i = i + 11 Xrn = i) > 0, i > 0,
q{ = P(Xm+i = i-l\Xm = i)>0, i > 0.
Prove that for arbitrary к > 0 there is anajt >1 such that
Pn(fc) = P\ max X, = fc ]
\0<j<n J
satisfies the limit relation
1 L
lim -У2рп(к)о% <°° ·
L—>oo L
n=l

434
3. SOLUTIONS TO THE PROBLEMS
Solution. Introduce the notation
Pn4fc) = P( max X,-< ft),
0<j<n
P*n(k) = P( max Xj < к and Xn = i) {0<i<k).
Obviously, Pn(k) < P*{k) = T,i=oPi,n(k)· lt is еа£5У to see that the
following inequality holds:
к
Σ^η(*)»»+ι.. .ft+fc+i+p:+k+1(k) < p:(k).
i=0
Define
Ek = 3% {PiPi+1 ' ' -Vi+k+l}-
0<г<к
Then 0 < Ek < 1 and
^n+*+i(*)<(l-e*)^n(*)·
Using mathematical induction, one can easily show that for all positive
integers n,
Pn*(*0<(i-^)1/(fe+lb2·
1 — £t
If
К α* < [— л
■ε*
then
p„*(fcK<(i-efc)2"-2,
and this inequality implies the statement. D
Problem P.18. Let Yn be a binomial random variable with parameters
η and p. Assume that a certain set Η of positive integers has a density
and that this density is equal to d. Prove the following statements:
(a) linin^oo P(Yn G H) = d if Η is an arithmetic progression.
(b) The previous limit relation is not valid for arbitrary H.
(c) If Η is such that P(Yn G H) is convergent, then the limit must be
equal to d.
Solution.
(a) By the assumptions of the problem,
Р(У„ = ft) = QpV-* (i = l-p).
For any given k, liuin^oo P(Yn = k) = 0; consequently, limn^oo P(Yn G
H) does not change if finitely many elements of Η are changed. We

3.8 PROBABILITY THEORY
435
can therefore assume that Η is a, complete residue class with respect to
a module D:
Η = {k: k = amodD}.
Then
-k
P{YneH)= £ QpV"
(i)
k=a{D)
Denote by ε a primitive Dth unit root. It is known that
D-l
We can write (1) in the following form:
oo D-l
г — к Лк—ά)ν
oo и — ι Ί / \
fc=0 i/=0 ^ W
^ v=0 fe=0 W
!/ = 0
since in the last sum, for ν > 0, the absolute values of the corresponding
terms are less than one, so they converge to zero as η —> oo.
(b) We construct a sequence Η with zero density such that
limsupn_+00P(l^l G H) = 1. The expectation and variance of Yn are
np and npq, respectively, so the Chebyshev inequality implies that
P(\Yn-np\>Xy/rm)<^.
If λη is a sequence such that linin^oo λη = oo, then
lim P(\Yn-np\ <ХпУ/п) = 1.
n—>oo
Let rik be an arbitrary increasing sequence of natural numbers, and let
H=(J(nfc-[nj|/4],n* + [n*/4]).
k = l
By the previous observation,
lim P(Ynk 6 Я) = 1,
k—>oo
and so
limsupP(rn 6ff) = l.

436
3. SOLUTIONS TO THE PROBLEMS
Finally, if rik increases sufficiently fast (for example, Пк = №к), then
Η has zero density,
(c) The majority of the contestants verified that if Η has density, then
lim P(Yn G H) = d in the sense of Cesaro, that is,
η η
From this property, the original statement follows, since for arbitrary
convergent sequence, the Cesaro limit is the same as the usual limit.
One can also prove that we do not need to assume the existence of the
density of Я, since if P(Yn G H) is convergent, then Η automatically
has a density that equals linin^oo P(Yn G H).
A more general result can also be verified: for an arbitrary set Η of
natural numbers, define h{x) = \{n G Η : η < x}\. Then
lim^_Mnp)=0
n—+00 77, Tip
Thus Η has a density if and only if P(Yn G H) is convergent in the
sense of Cesaro.
After these remarks, we can start to prove part (c). We are going
to prove that if the density of Η is d, then pn = P(Yn G H) —► d in
the sense of Abel, that is, Σ^0(ρη —Pn-i) = d in the sense of Abel,
namely, Ηπι^ι.ο Σ™=ο(Ρ,"<~ Ρη-ι)χΗ = d. If we prove this result, then
we are ready, since Нп^-^ P(Yn G H) = с implies P(Yn G H) —> с in
the sense of Abel, that is, с = d.
For |ж| < 1,
oo 00
G(x) = Σ(ρη -pn-i)xn = (1 -χ)ΣρηΧη
71=0 71=0
oo / \
n=o хен v /
= (i-»)Ew*E'1(n"1),«('l"t+1)(prt
кен n=o
Substitute ζ = рт/(1 — 9ж). If x —> 1 — 0, then ζ —> 1 — 0, and
furthermore (1 — x)I(1 — qx) ~ 1 — ζ. So
lim G(x)= lim (1 - z) V zk (2)
if this last limit exists.
Let an be the indicator sequence of Я, that is, an equals 1 if η G Я,

3.8 PROBABILITY THEORY
437
and equals 0 if η £ Η. The limit of an in the sense of Cesaro is the
density of Я, which is d. Then, by a theorem of Frobenius, the limit of
the sequence an in the sense of Abel also exists and equals d. This Abel
limit is the right-hand side of (2). Consequently, \imx^i-0G(x) = d,
which was to be proved. D
Problem P.19. Let {&ι}™ι=ι oe a double sequence of random variables
such that
Εξφι = 0((log(2|t - k\ + 2) log(2|j -1\ + 2))~2) (i,j, k, I = 1,2,...).
Prove that with probability one,
1 771 П
V^ V^£ы —> 0 as maxim,n) —> oo.
ran ^ ^ v '
k = l1 = 1
Solution. Let
b+m c+n
S(b,c\m,n) = У2 ^2 ^kl wnere ^?c^0? and ra, n>l.
k=b+ll=c+l
Using assumption (1) and the identity thickmuskip=1.4mu
b+m c+n r b+m c+n — lc+n — l
E{S2(b,c,m,n)) = Σ Σ Я(&)+2| Σ Σ Σ £(&i6M+i)
k=b+ll=c+l ^k=b+ll=c+l j = l
c+n b+m—lb+m—k χ
+ Σ Σ Σ E(^k+i,i)
1=с+1к=Ь+1 г = 1 '
b+m—lc+n — lb+m—kc+n—l
+2 Σ Σ Σ Σ (^««&w+j)
k=b+l l=c+l i = l j=l
+-E(&,/+j&+t,/)J,
it is easy to see that
τη—1 η —1
^ftw.))-acggwi + a0^1 + wy) (2)
= CM ran——-—гтгт;—-—-г l· (ra,n = 1,2,...).
1 (log2ra)2(log2n)2/ v ' ' ' J
Put
M(6,c;ra,n) = max max |5(6,c,fc,Z)|,
1</с<тп 1<Ζ<η

438 3. SOLUTIONS TO THE PROBLEMS
and use the following result (Theorem 4 in F. Moricz, Momemt inequalities
for the maximum of partial sums of random fields, Acta. Sci. Math. 39
(1977), pp. 353-366).
Assume that the nonnegative function /(&, c; ra, n) (6, c,m,n G N)
satisfies the following inequalities:
/(6, c; ft, n) + f(b + ft, c; m - ft, n) < /(6, c, m, n),
/(6, c, m, n) + /(6, с + г; m, η - г) < /(6, с; m, n),
for 6 > 0, с > 0, and 1 < ft < га, 1 < г < п. Let x(m) and λ(η) be two
nondecreasing sequences, and define К and Л as follows:
Κ(1) = χ(1), Α(1) = λ(1).
For m > 2 and η > 2,
*(»»») = χ(Λ) + *(Λ-1), Л=[|(т + 2)],
A(n) = A(t) + A(i-l), < = [^(n + 2)]
([. ] denotes the integer part of real numbers). Assume that for some 7 > 1,
and for all 6, с > 0 and ra, η > 1,
E(\S(b, c; ra, n)|7) < tf7(ra)A7(n) /(6, c; m, n).
Then
£(M7(6, c; m, n)) < tf7(ra)A7(n) /(6, c; m, n)
for all 6, с > 0 and ra, η > 1.
In our case, choose /(&, c; ra,n) = mn and x(m) = y/m/\og2m, λ(η) =
y'n/ log 2n, 7 = 2. Note that /(6, c; m, n) is an additive set function on the
rectangle [b + 1,b + ra] x [c + 1, с + η]. Furthermore, for 2P < ra < 2P+1,
Ρ Ρ ofc/2
tf(ro) < *(2"+1 - 1) = Σ X(2P) = Σ ΤΓΓΤ
\P+1J \\og2mJ
Consequently,
E(M\b,c;m,n) = 0{mn^^^}. (3)
In the case b = с = 0, define S(ra,n) = 5(0,0;m,n) = ΣΓ=ιΣΓ=ι£μ·
Since (2) holds,
oo oo
22&22/
< 00 .
22fc22/ ν- ν » " w^^22*22'(fe+l)2(Z + l)2

3.8 PROBABILITY THEORY 439
A theorem of B. Levi implies that with probability one,
^5(2*2')-0 as max(M) - oo. (4)
To complete the proof, we have to show that
-T-T max max \S(m,n) - S(2k,2l)\ -»0 (5)
with probability one, as max(fc, I) —► oo. Consider the following
decomposition:
S(m, n) = S(2k, 2l) + S(2k, 0;m-2k, 2l)
+ 5(0,2l;2k,n-2l) + S(2k,2l;m-2k,n- 2l),
from which we see that the left-hand side of (5) is not greater than
-^-j{M(2fc, 0; 2fc2') + M(0,2l; 2k, 2l) + M{2k, 2l; 2k, 21)}.
By (3),
OO OO
22/c22/
and the theorem of B. Levi again implies that
oo oo 1
k=01=0
^М(2к,21;2к,21)^0 (6)
with probability one, as max(fc, I) —> oo. Notice that
/-1
М(2*,0;2*,2')< ^ M(2k,2q\2k,2q),
q = -l
where, by definition, M(2k,2~l;2k ,2~λ) = M(2*,0;2*,l). A Toeplitz
lemma and (6) imply that with probability one
-LM(2k,0;2\2l)< Σ ^^(S^^.SVO
q=-l
as max(fc, Z) —> oo. A similar argument shows that
1 М{0,#;2к,&)-+0,
2*2'
as max(fc, I) —> oo, with probability one.
Combining (4) and (5), we complete the proof. D

440
3. SOLUTIONS TO THE PROBLEMS
Problem P.20. Let Ρ be a probability distribution defined on the Borel
sets of the real line. Suppose that Ρ is symmetric with respect to the origin,
absolutely continous with respect to the Lebesgue measure, and its density
function ρ is zero outside the interval [—1,1] and inside this interval it is
between the positive numbers с and d (c < d). Prove that there is no
distribution whose convolution square equals P.
Solution. Assume indirectly that there exists such a distribution Q. Then
Q([—1/2,1/2]) = 1, and its moments satisfy the relations
/ xkdQ(x)\ = / xkdQ(x)
J-oo I «/-1/2
< 1.
Thus,
limsup \ L-—i = 0.
*->οο V fc!
Therefore, the characteristic function ifQ of Q is analytic on the real line.
Since Ρ is symmetric, φ ρ is real. The equality φρ(0) = 1 and the continuity
of φ ρ imply that, for any χ with sufficiently small absolute value, ψρ{χ) >
0. For such χ values ψρ{χ) = {(Pq{x))2, so <Pq{x) is also real. Since ifQ
is analytic, it is real everywhere, which implies that φρ{χ) > 0 for all x.
By a well-known theorem (see for example E. Heivitt, and K. Stromberg,
Real and Abstract Analysis, Springer, 1965, p. 409), we know that if a
density function is bounded and its characteristic function is nonnegative,
then its characteristic function is integrable. Therefore, φρ is integrable,
so the density function ρ should be equal to a continuous function almost
everywhere, but because of its "jumps" at —1 and 1, this is impossible. D
Remark. For many related negative and positive decomposition results
see 7. Z. Ruzsa and G. J. Szekely, Algebraic Probability Theory, Wiley,
New York, 1988 and G. J. Szekely, Paradoxes in Probability Theory and
Mathematical Statistics, Reidel (Kluwer), Dordrecht, 1986.
Problem P.21. Let ρο,Ρι,... be a probability distribution on the set
of nonnegative integers. Select a number according to this distribution
and repeat the selection independently until either a zero or an already
selected number is obtained. Write the selected numbers in a row in order
of selection without the last one. Below this line, write the numbers again
in increasing order. Let Ai denote the event that the number г has been
selected and that it is in the same place in both lines. Prove that the events
Ai (i = 1,2,...) are mutually independent, and P(Ai) = pi.
Solution. We will show that for any к (к = 1,2,...), and any sequence
1 < h < h < -" < ifc,
P{Ai1 Ai2... A{k) = Pi1Pi2 · · · Pik ·

3.8 PROBABILITY THEORY
441
Modify the experiment in such a way that in addition to zero and a
repetition of a number, we also stop at selecting any of the numbers ii, Z2,..., %k -
Consider now a (finite) outcome of the original experiment in which A^A^
... Aik occurs. (The probability of infinite outcome is equal to zero.) From
the original sequence, omit the values ii,Z2,..., г*. In this way, an
arbitrary outcome of the modified experiment can be uniquely obtained. On
the other hand, it is easy to see that there is only one way to insert the
numbers ii, г2,..., ik in such way that event AixAi7_ ... Aik occurs, that is, the
numbers ii, г2,..., ik are placed in their increasing locations. This one-to-
one correspondence implies that the probability of the event A^A^ ... Aik
is equal to p^p^ ...pik-l. □
Problem P.22. Let Xi,...,Xn be independent, identically distributed,
nonnegative random variables with a common continuous distribution
function F. Suppose in addition that the inverse of F, the quantile function
Q, is also continuous and Q(0) = 0. Let 0 = Xq:u < X\:n < · · · < Xn-.n be
the ordered sample from the above random variables. Prove that if EX\
is finite, then the random variable
Δ = sup
0<y<l
ι Μ + 1 py
- Έ (п+1-*)№:п-*г-1:п)- / (1 - ti)dQ(ti)
n ri Jo
i=l
tends to zero with probability one as η —> oo.
Solution. Introduce the notation
[ny] + l
Hn{y) =~ J2 {Π + 1 - l){Xi:n ~ Xi-l:n)
and
HF(y)= f\l-u)dQ(u).
Jo
Thus, HF(1) = Е(Хг) and
Hn(l)=]imHn(y) = -Txu (1)
3/Ti η 2—'
г=1
that is, the problem generalizes the strong law of large numbers for non-
negative random variables.
Because of the continuity of F, the independent random variables Yi =
F(Xi), 1 < г < η, are uniformly distributed on the interval (0,1). Denote
by En(y) = n~l χ {k : 1 < к < η , Yk < у} the empirical distribution
function of Υί, У2, · · ·, Yn, and let
I Yn:n if у = 1

442
3. SOLUTIONS TO THE PROBLEMS
be their empirical quantile function. Further, let
Fn(x) =n~1 χ {k : l<k<n, Xk < x} (0 < χ < oo)
and
I Xn:n ity = l
be the empirical distribution and quantile function of the original sample,
respectively.
Consider the random function
rUniy)
Gn(x)= / (l-En(u))dQ(u).
Jo
By the continuity of F and Q,
rQ(Un(y)) rQniy)
Gn(y) = (1 - En(F(x)))dx = (1 - Fn(
Jo Jo
x)dx
with probability one, where the exceptional set (where the original sample
elements coincide) is independent of y. If (k — I)/n < у < к/п for some
integer 1 < к < η, then this last integral is the following:
/ kn(l-Fn(x))dx = f2 [ "" (1-Fn(x))dx,
JO i = 1 JXi-l:n
where
к . _ 1
V(l - )(Xi:n - Xi-l:n) = Hn(y).
*-^ Π
г=1
Since Gn(l) = Hn(l) with the common value given in (1), we conclude
that for η = 1,2,...,
P{ sup \Hn(y)-Gn(y)\=0} = l.
0<y<l
It is sufficient to show that for η —> oo,
Δ;= sup \Gn(y)-HF(y)\-+0
0<y<l
with probability one. Obviously,
Δ;< sup \Gn(y)-HF(U(y))\+ sup \HF(Un(y)) - HF(y)\
0<у<1 0<у<1
= Δΐ1> + Δ(ι2). (2)

3.8 PROBABILITY THEORY
443
Notice that for 0 < у < 1, Un(y) has only η different values from the
interval [0,1]; therefore,
ΔΡ<Δ^= sup | f\l-En{u))dQ{u)- [\l-u)dQ(u)\.
0<y<l JO JO
Let 0 < ε < 1 be an arbitrary value. Then
Δΐ3)</ (l-En(u))dQ(u) + [ (l-u)dQ(u)
+ Q(l- ε) sup \y-En(y)\.
Since Q(l — ε) < oo, the third term tends to zero with probability one
(Glivenko-Cantelli theorem, see, for example, A. Renyi, Probability
Theory, Akademiai Kiado, Budapest, 1970, VII. §8). The first term (1(A)
denotes the indicator function of the event A) can be written as
/ (l-En(u))dQ(u) = -T f (l-I({Ui<u}))dQ(u),
and because of the strong law of large numbers, this quantity tends to
/le(l — u)dQ{u) with probability one.
In summary,
POimsupA^ < 2 / (1 - u)dQ{u)} = 1,
n—>oo Jl—ε
where the upper bound can be made arbitrarily small by choosing a
sufficiently small ε since E(X\) < oo. Thus, we have proved that the first term
of (2) tends to zero (Δη —> 0) with probability one as η —> oo.
It is known (see, for example, M. Csorgo and P. Revesz, Strong
Approximations in Probability and Statistics, Akademiai Kiado, Budapest, 1981,
p. 162) that the Glivenko-Cantelli theorem is also valid for the quantile
function of a uniformly distributed sample. Therefore,
sup \Un(y)-y\ ->0
0<y<l
with probability one, as η —> oo. Using the fact that the integral J^(l —
u)dQ(u) is a continuous function of y, for the second term in (2) we get
P{ lim Δΐ2) = 0} = 1,
η—+oo
which completes the proof. D
Remark. Suppose η machines start working at time t = 0, and let
X\-.n < ^2:n ^ · · · be the failure times for these machines. Then nHn(y)
is the total time until the ([ny] + l)th failure. The best published
result (AT. A. Langberg, R. V. Leon, and F. Proschan, Characterization of
nonparametric classes of life distribution, Annals of Probability 8(1980),
pp. 1163-1170, Theorem 3.2) shows only pointwise convergence, that is,
for all fixed 0 < у < 1, Hn(y) —> Hp(y) with probability one, as η —> oo.

444
3. SOLUTIONS TO THE PROBLEMS
Problem P.23. Let Xo, -ΧΊ,... be independent, identically distributed,
nondegenerate random variables, and let 0 < а < 1 be a real number.
Assume that the series
oo
k=0
is convergent with probability one. Prove that the distribution function of
the sum is continuous.
Solution. Define
oo oo
Z = Y^ak-lXk and Y = ^akXk.
k=l k=0
Then Υ = Xq + aZ, where Υ and Ζ have the same distribution, and Xq
and Ζ are independent.
Assume that the distribution function of Y\s not continuous. Define
p = max P(Y = a),
a
and denote by αχ,..., αη the points for which
P(Y = aj)=p.
Then
ρ = P(Y = aj) = P(X0 +aZ = aj)
= ς ρ(χ°=x^z=^)
^—' a
P(Xo=x)>0
< Υ P(X0 = x)maxP(Z=^^)<p.
*-^ г a
P(Xo=x)>0
Therefore, we have equality everywhere. This implies ^2xeR P(Xo = x) =
1, that is, Xq has a discrete distribution, and
P(Z = £lZ£) =p if ρ(χ0 = x) > o.
α
Put
Κ = {χ : P(X0 = x)> 0},
Pa = {x: P{Z=^—^)=p].
a
Then
Kcf]Paj.

3.8 PROBABILITY THEORY
445
Since \Pa\ = η and Pa = P0 + a,
К с Po+aj,
that is,
K-djCPo , j = l,...,n.
This relation can hold only if \K\ = 1, that is, if Xq is degenerate. D
Remark. For another proof and some related results see 7. Z. Ruzsa
and G. J. Szekely, Algebraic Probability Theory, Wiley, New York, 1988,
Section 5.5.
Problem P.24. Let Χι, X2,... be independent random variables with
the same distribution:
P(Xi = l) = P(Xi = -l)=l- (г = 1,2,...).
Define
So = 0, Sn = Хг + X2 + · · · + Xn (n = 1,2,...),
ξ(χ, η) = I {к : 0 < к < η, 5fc = ж} | (ж = 0, ±1, ±2,...),
and
а(п) = Ι {χ : ξ(χ, η) = 1} Ι (η = 0,1,...).
Prove that
P(liminf α(η) = 0) = 1
and that there is a number 0 < с < 00 such that P(limsup a(n)/log η =
c) = l.
Solution. Represent the values of £0, S\,... in a two-dimensional
coordinate system as follows: starting from the origin, make a step to the right,
and move up or down if Xn = +1 or Xn = — 1. Note that a(n) counts the
points whose second coordinates are visited exactly once by this random
walk.
First, we show that
P(liminf a(n) = 0) = 1,
which means that the events An = {a(n) = 0} occur infinitely many times
with probability one. In the moments when the random walk returns to
zero, there are infinitely many such points with probability one, the value
of a(n) is at most one, since only the extreme values can occur once.
However, the probability that between two returns the random walk visits
the extreme values at least twice is positive. Therefore, among the infinitely
many instants, there will be one, and thus there will be infinitely many ones,
when the value of a(n) is zero.

446
3. SOLUTIONS TO THE PROBLEMS
Let us consider the second statement. We will proceed via several
lemmas. Denote by An the set of second coordinates visited only once, that
is, An = {x : there is a k, 1 < к < η, such that Sk = x, and for j ^ /c,
Sj ^x}. Then a(n) = |Ai|.
Lemma 1. Нтп_оо P(Sj > 0, 0 < j < n, Sn - S3 > 0, 0 < j < n) =
c* >0.
Proof. Define the following conditional distribution:
/in(dx,dy) = Ρ I —=5n = dx, —■= sup Sk=dy |Sj>0 1 <j <n) .
\y/n Vno<k<n J
If η = 2m + 1, then
P(Sj > 0, 0 < j < n, Sn - Sj > 0, 0 < j < n)
= P(Sj > 0, 0 < j < m, Sm > sup (Sn - Sk) - (Sn - 5m),
m<k<n
Sn - Sj > 0, m < j < n, Sn-Sm> sup (Sk - Sm))
0<k<m
= P(Sj > 0, 0 < j < m) P{Sn -Sj>0,m<j< n)/im * /im(A),
where
A = {(ж1,у1,ж2,у2) : x\ >У2 -^2, X2 > 2/1 ~^i}.
We know that the sequence μη of measures tends weakly to a measure
μ* without atoms and has a positive value on the open subsets of the set
{(x,y) : x > 0, у > 0}. On the other hand,
9 1
P(Sj > 0, 0 < j < m)P(Sn -Sj>0,m<j<n)~- ,
so the statement of the lemma holds for integers of the form 2m + 1. For
integers of the form 2m, the proof is similar.
Lemma 2. Suppose к ~ alogn, ак(п) = ( fc ). Then for arbitrary
η > 0, there is an no = no(k, η) such that for any n > no,
[(c* -77) logn]* < Eak(n) < [(c* +i7)logn]\
Proof. Put
C{r,t) = {Sr<Sj<St,r<j<t},
D1(t) = {Sj<St, 0<j<t},
D2(t) = {Sj > St, t<j<n}.
Then
P(C(r, t)) = P(C(0, t-r)) = ^(l + o(l)),
P(D2(t)) = -£=(1 + 0(1)).
V n — t

3.8 PROBABILITY THEORY
447
Put
ΑΪ=ΑηΠ{ζ:ζ>0},
*+(п) = И+|, at(n)=^n)y
and for 0 < ji < J2 < * * * < jk < n, use the notation
Bju...jb = Di(h)C(JuJ2) · · · C(jk.1Jk)D2(jk).
Then
0<ji<-<jk<n
since the event -Bjlv..,jfc means that
(ji,···,^) с Д+
and the event a(n) = I contains exactly (£) such events. Put
U{j,l)= Σ P{C(ji,J2))-...-P{C(jk-UJk)) = U{0,l-j).
j=ji<—<jk=l
So
E<*t(n) = J2u(0,r)J2P(D1(j))P(D2(r + j))
r=0
3=0
1 1
< const · J] ^(0, г) Σ -^ ^^7^7 " C°nSt' Σ ^°'r
< const
On the other hand,
j=0
k-1
<[(c* + n)logn]'
Ea+(n)> Σ P(DiUi)C(ji,J2)...CUk-1,jk)Da(jk))
0<3\<n/3
n/3k>jt-jt_1>0
n/3
COnSt ν—ν n/rx , ...
n/3
J = l
Ί k-1
> const [(c* - |) log ^]K_1 > [(c* - τ,) logn]'
since к ~ α logn, so [(c* - n/2)/(c* - η)]0'08™ > logn if n is large enough.

448
3. SOLUTIONS TO THE PROBLEMS
We can get the same result for An in a similar way. So
f At plus maybe one more point, if Sn > 0,
An =i \
\ A~ plus maybe one more point, if Sn < 0.
So the statement of the lemma is valid for a(n).
Lemma 3. For all К > 0 and ε > 0, there is an no = щ(К, ε) such that
for all η > no,
n-(*/0-« < Ρ(α(η) < K\og2n) < n-(K/c*)+£.
Proof. Put A: = (K log n)/c*. Then
P (Q(n, > *,<**„) = Р(ад(„, > (**"")) < ^ .
If η is large enough, then we can use Lemma 2 to get
P(a(n) > K\og2Tl) < n-(^/c*)log((c*e)/(c"+,)) < η-(Κ/ο")+ε.
For the proof of the lower bound, put qm = P(a(n) = m). Then
Eak(n) = Y^qm\™\.
m ^ '
Put A:' = (K/c*) · (1 + ε2) log η. Then for any sufficiently small ε,
m>(/c+e)log2n ^ '
and
m<Klog2 n
because if fe/x = ((K + e)/c*) · logn, then k" > kf and
Eim\ _ ^ /m\ (™)
i>(K+e)log2n V 7 m ч 7 U"/
m>(K+e)\og
/(K+e)log2n\
/(Κ+ε) log2 n\
/(K+e)log2n\
^y^'")·

3.8 PROBABILITY THEORY 449
Again using Lemma 2 with η = ε3, we get
Σ **(£) < (c*logn)*'exp{(-^ +0(e2))logn}
m>(K+e)log2n
< ^£Jafe<H
if η is sufficiently large. On the other hand,
Γ°/η)
Σ Ц£)*71п5Ц Σ ЦТ)·
/tflog'nN
m<Klog2 π χ 7 \ к ) m<K log2 n
Using Lemma 2 again, we get
m<Klog2n V /
These inequalities imply
К log2 n<m<(K+e) log2 η
So
Σ ?m(™J >^afc,(n)
Klog2n<m<(K+e)log2n V 7 I fc' /
1 E(ak,(n)) Ι-ίΐι^^^ΐι+^ΐ
- 3^(κ+εμοδ2^ - 3"
This last expression tends to 1 for ε > 0 and 77 > 0, so we have checked the
lower bound. Using the above lemmas, we are able to prove that
log n J
Ρ limsup —о— = с I = 1.
V log2 '
Put
Mn = inf{j : Sj > n}.
Then
Mn
lim —£ = 0 and M(n+1)2 - Mn2 > 2n +1.
Consider the instant when the random walk reaches the height k2 for the
first time. Start a Ar-step walk from this point, and denote it by Uk , that
is,
Uk = {0, 5,Mfc2+i - SWfc2 j · · · j SMk2+k - £Mfc2 }·

450
3. SOLUTIONS TO THE PROBLEMS
These walks are independent from each other. Put
Γ the walk Uk visits exactly once at least Ί
[ (c* — έ) log к points with positive coordinates J
Then P(Bk) > fc-1+£/2 if к is large enough. So Y,P(Bk) = oo, hence
P(limsup Afc) = 1. But on the event Лк ot(Mk2 +k) > (c* — ε) log2 /c, and
since Mfc2 + A: < ks + fe for sufficiently large values of /c,
a(n) c* — ε
log2 η 64
for infinitely many η with probability one.
The upper bound of Lemma 3 with К = с* + ε shows
Σ ρ(αΗ > (c* +ε) W2 n) < °°>
that is,
α(η) ^ *
hmsup—ο~ < с + ε,
log n
for large values of η with probability one. Recall that the Kolmogorov
0-1 law implies that lim sup a(n)/ log2 η is constant with probability one;
therefore,
1 log2 η J
log'
with a suitable c*/64 <c<c*. D
Remark. Unfortunately, the original formulation of the second
statement of the problem was incorrect. It contained a log divisor instead of
log2. The proof of the correct statement appeared in P. Major, On the
set visited once by a random walk, Probab. Th. Rel. Fields 77, (1988),
pp. 117-128. The present proof follows this work.
Problem P.25. Let (Ω, А, Р) be a probability space, and let (Xn, Tn)
be an adapted sequence in (Ω, Д, P) (that is, for the σ-algebras Tn, we have
T\ С J^2 Q ''' Q A, and for all n, Xn is an Tn-measurable and integrable
random variable). Assume that
E(Xn+1\Fn) = ±Xn + ±Xn-1 (n = 2,3...).
Prove that supnE\Xn\ < oo implies that Xn converges with probability
one as η —> oo.
Solution. Put Yn = Xn + (1/2) · Xn_i (n = 2,3,...). Then Yn is an
.^-measurable and integrable random variable. Furthermore,
Ε{Υη+ι\Τη) = Ε I Xn+i + -zXnl^n J = ^Xn + o^n_1 ~*~ o^n = ^n
for η > 2. That is, (Y^T^ η = 2,3...) is a martingale.

3.8 PROBABILITY THEORY
451
Since supn£'(|yrn|) < (3/2) · supn25(|Xn|) < oo, the martingale
convergence theorem implies that Yn is convergent with probability one, and
lim Υη(ω) = Υ (ω)
η—+οο
if ω G Ω', where Ρ(Ω') = 1. Неге У is a random variable that is finite with
probability one, and therefore we may assume that it is finite for ω G Ω'.
We will show that for ω G Ω', the sequence Χη(ω) is also convergent.
Let ω G Ω' be given, and an = Χη(ω). We will prove that if the sequence
bn = αη + (1/2)*αη-ι converges to c, then an converges to (2/3)·с Assume
first that с = 0. Then for any ε > 0, there is an N such that |6n| < ε/2 as
η > N. So
ι ι υ. 1 ι ^ υ. ι . 1ι ι ^ ε+ \αη\
|βη+ι| = l&n+i - 2a"l b l&n+il + -\an\ <
for all n> N. Using this inequality for η = Ν,..., Ν + к — 1, we get
ι ι ^ ε ε ε ΙαΛτ| Wn\
|α„+*|<- + ϊ + ··· + ^ + ^<ε + ^<2ε
if /с is large enough. However, for all η > N + к, \ап\ < 2ε. That is, the
sequence converges to zero.
If с t^ 0, then the same proof applies for the sequence a'n = an — (2/3) ·
с D
Problem P.26. Let Χλ, X2,... be independent, identically distributed
random variables such that X{>0 for all i. Let EXi = m, Var(Xi) = σ2 <
oo. Show that, for all 0 < α < 1,
lim η Var (
П—+00 V
Jf!+...+ Jfn
Π
α\ α2σ2
J ~ m2(l-a)·
Solution. Put Xn = (Xx + · · · + Xn)/n and Sn = Xx + · · · + Xn. Notice
first that
nVar
p^r
= n£(^ - ma)2 - п(Д(Х^ - m«))2.
First we show that the second term tends to zero as η —> oo. The quadratic
Taylor polynomial of (l + z)a with remainder term implies that for ζ > — 1,
1 — (1 + ζ)α < (1 — α)ζ2 — αζ. From the Jensen inequality and this relation,
we get
0 < ma - E(X°) = maE 1
m
<m°
<i-«)*(^-i)'-eB(^-1)
πι«σ2
πι2 η

452
3. SOLUTIONS TO THE PROBLEMS
which implies the assertion. Consider next the first term. Introduce the
notation Zi = (Xi/m)-\ and Zn = (Ζι + Ζ2 + · · - + Zn)/n. Then Z{ > -1,
E(Zi) = 0, Var (Zi) = σ2/га2, furthermore for arbitrary 0 < ε < 1,
n£(X" - raa)2 = m2anE((l + Zn)a - l)2
= m2e[n£?(((l + Zn)e-l)2J^n|<e)
+ n£;(((i + znr-i)27^n|>£)].
In the next step, we will use the following inequalities. In the first term,
a\z\/(l + |г|) < |(1 + z)a - 1| < a|z|/(l - |г|) for \z\ < 1. In the second
term, |(l+z)a —1| < \z\a for ζ > —1. Observe, furthermore, that if \z\ < ε,
then \z\/(l + |г|) > \z\/(l + ε) and |z|/(l - |zj) < \z\/(l - ε), therefore,
1 <nE(X"ma)2
m'
2a
n^-^E(Z2nLzn{^)
(1 + ε)2
< га
2a
(V о
(1-е)'
Ζη\<ε
c + nE{\Zn\2aI^n>e)
Therefore, it is sufficient to show that
ηΕ(ΖηΙ\ζη\<ε) ^ ^2 ^d ηΕ(ΖηΙ\ζη1>ε)^0.
Let φ be the standard normal distribution function. Then for large n,
^=nS(zb>nS(Z^n|<J = ^S
/\/nZn
1, ./KZ-
\ aim ) (^)<ψ
(VnZn
since s/nZnliajm) —► φ in probability. In addition,
nE(\Zn\'aIlZnl>s) = егапЕ
< ηε2αΕ
Zn
la
1-Е.
>1
Zn
7|%Ч>1
ηε
2α-2
E(Z2J\zn{>e)-0-
Since
we get
^ = nE(zl) = nE(zll^n{^) + ηΕ(Ζ2ηΙιτη{>ε),
ηΕ{ΖηΙ^η\<ε)
m*
as η —> oo. D

3.8 PROBABILITY THEORY
453
Problem P.27. Let F be a probability distribution function symmetric
with respect to the origin such that F(x) = 1 — x~lK(x) for χ > b, where
ifxG[5,cx))\U~=5(n!,4n!),
ifx e (n!,2n!], η > 5,
3—2^τ ifz E(2n!,4n!), n>5.
Construct a subsequence {rik} of natural numbers such that if
Χι, X2,... are independent, identically distributed random variables with
distribution function F, then for all real numbers χ
\ ι Jl* 1 ι ι
limP — > X~ < πχ > = - + -
*->oo I nk ^ J I 2 π
arctan x.
Solution. Observe first that the independence of Χχ, X2,... implies that
for any rik > 1 and t ^ 0 the characteristic function of the random variable
*nk j=1
^rifc = У ν ^
can be given as
Пк
The assumed symmetry implies that, for any s^O,
Л (5) = 1^(1_д(е^*1) = ^ Г fl-cos—x)dF(x)
\s\ \s\J-oo\ nk )
=R£(i-cos^x)dir(x)-2£(i-cos2/)d(^fenfc))·
nfc
The above form of the characteristic function is useful since by a suitable
choice of subsequence {rik} we can guarantee that these functions converge
to e-'*'. Note that this is the characteristic function of the Cauchy
distribution, which is the limit distribution given in the problem. (See A. Renyi,
Probability Theory, Akademiai Kiado, Budapest, 1970, IV. §10, VI. §2.)
The first term of the above representation of hnk tends to zero if for
к —> oo, Пк —> oo. It is easy to see that if A" = 1, then the second term
(with the choice Пк = к) tends to the limit
ft f°° 1-cosy,
2 / z-^-dy = π.
Jo У2

454
3. SOLUTIONS TO THE PROBLEMS
Thus, in this case Пк = к would be a suitable choice. Therefore, it is
sufficient to choose the sequence {пк} so that for any у > 0 and s^O,
K((y/\s\)' nk) —► 1 as к —► oo.
Let {dk } be an arbitrary sequence of positive integers such that a^ —> oo
and a,k/k -+ 0 as к -+ oo. Then for arbitrary χ > 0 and sufficiently large
к, Ак\ < akklx < (к + 1)! and for any such k, K(akk\x) = 1. That is,
if we choose Пк = а,кк\, к = 1,2,..., then the continuity theorem of P.
Levy, implies that hnk(s) —> π for к —> oo. Therefore, E(ettYnk) —> e-'*' for
arbitrary t φ 0 as к —> oo, and the repeated application of the continuity
theorem completes the proof. D
Problem P.28. Let a G C, \a\ < 1. Find aJJ values ofbeC for which
there exist probability measures with characteristic function φ satisfying
0(2) = α and 0(1) = b.
Solution. If φ is a characteristic function with the required properties,
then 0(0) = 1, φ(—1) = 6, and φ(—2) = α, and the self-adjoint matrix
0(0) 0(1) 0(2) \ /1 6 o\
0(-l) 0(0) 0(1) =6 1 6
0(-2) 0(-l) 0(0) У \а b 1/
is positive semidefinite. So its determinant is real and nonnegative:
1 + ab2 + a62 - 266 - aa > 0. (1)
If а = u + vi and b = χ + yi, then (1) can be rewritten as
(2 - 2u)x2 - Avxy + (2 + 2u)y2 < 1 - u2 - v2. (2)
Let a' be a complex number such that its square equals a. It is easy to see
that if \a\ = 1, then (1) holds for the points of the interval between a' and
—a' (since |6| < 1, it does not hold for the other points of the connecting
line). If \a\ < 1, then (1) holds for the points in the interior and on the
curve of the ellipse with major axis >/2 + 2\a\ and foci a' and —a'. These
statements are shown next. Assume first that \a\ = 1. Let the random
variable X be defined as arga'. Then φχ(1) = af and φχ(2) = a'2 = a,
so 6 = a' is suitable. A similar proof shows that 6 = —a' also satisfies the
required properties.
Next assume that \a\ < 1, and let 6 be a boundary point of the ellipse.
In this case, we have equality in (1). If 62 — a = re2m, then (1) implies
that
r2 = (62 - a)(62 - a) = |6|4 - af - ab2 + \a\2
= |6|4-2|6|2 + 1 = (|6|2-1)2,
that is, r = 1 - |6|2.

3.8 PROBABILITY THEORY
455
Put с = Ree m6, ωχ = α+arccosc, and ω2 = α—arccosc. Furthermore,
let
1 lme~aib
Pi = о +
V2
2 2^ПГ^2'
1 Ime"ai6
2 2^ПГ^2
Since |b| < 1, pi and p2 are positive. Define X as follows:
P(X = cji) =pi and Ρ(=ω2)=Ρ2-
Then the characteristic function φ οι Χ satisfies
0(1) = pie«i< ч-рае*** = eai(piearccosc +p2earccosc)
= ea*((pi + p2) cos(arccos c) + i(pi —P2) sin(arccos c))
= ea<((Pi +P2)c + 2t(pi -p2)v/T^)
= eai((Ree"ai6) + i(Ime-ai6)) = b
and
0(2) = Ple2uJli + p2e2uJ2i = e2ai(pie2arccosc +p2e"2arccosc)
= e2al((pi + P2) cos(2 arccos c) + i(pi — p2) sin(2 arccos c))
= β2α<((Ρι +P2)(2c2 - 1) + i(pi -р2)(2сУП^))
= e2a*(2(Ree"ai6)2 - 1 + 2i(Ree-aib)(Ime-aib))
= e2ai((e~aib)2 + \e~aib\2 -l) = b- e2ai(l - \b\2)
= b2-re2ai = b2-(b2-a)=a.
Hence 0(1) = 6 and 0(2) = a. Finally, we show that the set of the suitable
b values is convex, and consequently, the interior points are also good. Let
61 and 62 be two suitable points, and let Q\ and Q2 be two probability
distributions such that 0qx(2) = </>q2(2) = α, φζ}!^) = &i, and </>q2(1) =
62. Define Qs = XQi + (1 — A)Q2 with 0 < λ < 1; then Q3 is also a
probability measure, </>q3(2) = a, and </>q3(1) = Xb\ + (1 — λ)62. D
Problem P.29. Let Y(k), к = 1,2,... be an m-dimensionalstationary
Gauss-Markov process with zero expectation, that is, suppose that
Y{k + 1) = A Y{k) + e{k + 1), к = 1,2,...
Let Hi denote the hypothesis A — Ai, and let P»(0) be the a priori
probability of Щ, г = 0,1,2. The a posteriori probability Px(k) = P(Hi\Y(l),
...,У(к)) of hypothesis H\ is calculated using the assumptions Pi(0) >
0, P2(0)>0, Pi(0) + P2(0) = l.

456
3. SOLUTIONS TO THE PROBLEMS
Characterize all matrices A0 such that P{\imk-+oo Pi{k) = 1} = 1 if H0
holds.
Solution. Let Υ (I) be an m-dimensional, normally distributed random
vector with zero expectation and covariance matrix D. The so-called "white
noise" e(k), fc = 1,2,..., is independent from the past of the process Y(k),
that is the e(A;)'s are m-dimensional, normally distributed random vectors
independent from Υ (I) and from each other, and they have zero expectation
and covariance matrix Q.
The background of the problem is as follows: from hypothesis Η χ and
#2, we accept the one for which the posterior probability converges to 1.
We examine the robustness of our procedure, that is, if H0 holds, then our
decision must result in the hypothesis that is "closer" to Hq.
Suppose that Q is regular. Since the process is stationary, each Y(k)
has the same covariance matrix D, к = 1,2,... . From (1) we obtain the
following matrix equation:
D = E{Y{k + l)Y*(k + 1)) = E([A Y(k) + e(k + 1)][A Y{k) + e{k + 1)]*)
= E(A Y(k)Y*(k)A*) + E{e(k + l)e*(k + 1)) = ADA* + Q.
Hence Q = D — ADA*, therefore, D is also regular. Since Q is positive
definite and D is at least semidefinite, for any χ G Rn such that Dx = 0,
0 < x*Qx = x*Dx - x*ADA*x = -(A*x)*D(A*x) < 0,
that is, χ = 0.
Since Pi(0) + P2(0) = 1, Рг(к) + P2(k) = 1 for all к = 1,2,..., and
therefore Hindoo Pi(fe) = 1 if and only if Hindoo Pi(k)/P2(k) = oo. The
Bayes theorem implies that, for г = 1,2,
р(я,|У(1) = yu..., Y(k) = ») = Ру^пщиЛт,...^)
where py(i),...,y(fc) (or Ру(1),...,у(&)|я») denotes the probability density
function of variables Y(l),..., Υ (к) (or the corresponding conditional density
function under Щ). This relation implies that
PjH^Yjl) = yi,..., Y(k) = yk) _ Py(i),...,y(t)|Hl(yi, · · · ,yk)Pi(0) ()
P(H2\Y(l)=yi,...,Y(k)=yk) ру(1),...,т|я2(2/ь···,!fc)ft(0) ' ^ )
Under the hypothesis Hi, the random variables
Y(l) « iV(0, D), Y(k + 1) - AiY(k) и iV(0, D)
are independent. (Here N(M,S) denotes the normal distribution with
expectation Μ and covariance matrix S. In the following, we will denote
the density function of N(M, S) by Pn(m,s)-)

3.8 PROBABILITY THEORY 457
In this case,
РУ(1),...,У(к)\нЛУи-чУк)
= Pn{o,d)(2/1 )Pn{o,q)(2/2 - Myi) · · ·Pn(o,q){Ук - А{ук_г)
= 1 e-hvlD^yi . l p-%(y2-Aiy1yQ-1(y2-Aiy1)
y/(2n)m\D\ VWr\Q\
χ . χ fi-\iyk-Aiyk-{)*Q-x{yk-Aiyk-1)
V(2*)mIQI
V(2^^|D||Q|*-i
Substituting this relation into (2) and using the variables У(1),..., Υ (к),
we get the following ratio of the posterior probabilities:
к
]Щ = Щ «p(^ Σ>ω ~ МГЦ - WQ^inj) - A2Y(j -1)]
J—2
- J ί>ω - ^(i - WQ-1™ - MY{j -1)]).
J=2
That is, if the hypothesis Я0 holds, that is, the process is based on the
matrix Aq. Then with the random variable
L(Y(l),...,Y(k))
k-1
= J^Uo - A2)Y(j) + e(j + l)]*Q-1[(^o - A2)Y(j) + e(j + 1)]-
k-l
- J^Uo - AJYU) + eU + WQ-'Uo - AJYV) + e(j + 1)]
we get
PlW_Pl(o)ex __jy(fc))
(±L(Y(l),...,Y(k))y
P2(fc) p2(o)
The question to be answered is as follows: when does the limit relation
lim L(Y(l),...,Y(k)) = oo
k—>oo
hold with probability one?
The process (Y(k), e(k +1) fe = 1,2,...) is an ergodic Gauss-Markov
process. By the ergod theorem,
lim h(Y(l),...,Y(k))
к—+ос К
= E([(A0 - A2)Y(\) + e(2)]*Q-1[(AQ - A2)Y(1) + e(2)])-
- E([(A0 - Л0У(1) + e(2)]*Q-1[(A0 - A)y(l) + ε(2)])
= tr [(Λ) - A2YQ-1{A0 - A2)D] - tr [(Ao - AtfQ-^Ao - AJD] = L

458
3. SOLUTIONS TO THE PROBLEMS
with probability one. If L > 0, then Hindoo P\(k)/P2(k) = oo with
probability 1, and this is the same as liim^oo P\(k) = 1. D
Remark. For L < 0, we get lim^oo Рг(^) = 1 with probability 1. One
can also prove that for L = 0, neither P\(k) nor Рг(^) tends to 1 with
probability one, that is, no appropriate decision between Щ and H2 can
be made.
Problem P.30. Let X and Υ be independent identically distributed,
real-valued random variables with finite expectation. Prove that
E\X + Y\>E\X-Y\.
Solution 1. For any real numbers α and /3,
\α + β\ - \a - β\ = 2sign(a/3) min{|a|, \β\}.
Therefore,
E(\X + Y\)-E(\X-Y\)=E(\X + Y\-\X-Y\)
=2E(mm{\X\,\Y\}sign(XY))
/»oo
=2/ {P(\X\>t,\Y\>t,XY>0)
Jo
-P(\X\ > t, \Y\ >t,XY< 0)}dt.
Since X ала Υ are independent, identically distributed variables, and
P(Z >t) — P(Z > t) almost everywhere,
E(\X + Y\-\X-Y\)
/»oo
= 2 / {P(X > t)]2 + [P(X < -t)]2 - 2 P(X > t)P(X < -t)}dt
Jo
/»oo
= 2 / [P(X > t) - P(X < -t)]2dt > 0.
hence,
E(\X + Y\)>E(\X-Y\). Π
Solution 2. This problem has a simple solution, if we use the Fourier
transform of generalized functions. We may assume that X has a nice
density function f(x), say an infinitely many times differentiable function
with finite support. Then the statement of Problem P.30 can be rewritten
as
J\x\(f*f)(x)dx>J\x\(f*f-)(x)dx,

3.8 PROBABILITY THEORY
459
where * denotes convolution, and f~(x) = /(—#)· This formula can be
rewritten by means of the Plancherel formula if \x\ is considered as a
generalized function. The Fourier transform of \x\, when it is considered as a
generalized function, equals —2σ~2. (See, for example, /. M. Gelfand, G.
E. Shilow, Verallgemeinerte Funtionen (Distributionen) I. VEB Deutscher
Verlag der Wissenschaften, Berlin (1960) Vol. 1). The Plancherel formula,
which is actually the definition of the Fourier transform of generalized
functions, and the definition of σ~2 as a generalized function (see formula (5)
on page 60 of Gelfand and Shilow's book) state that the last formula can
be rewritten as
" 2i°° h Ψ{σ)+f2{~a) ~2/2(0)]da
> -2 Γ -i [/>)/"- (σ) + /(-σ)/-(-σ) - 2/(0)/- (0)1 da,
Jo σ L J
where ~ denotes the Fourier transform. A simple calculation shows that the
last formula is equivalent to the relation
/
Jo
1
/(σ)-/(-σ)
da <0.
This relation clearly holds, since /(σ) — /(—σ) is a purely imaginary
number. D
Remark. The second solution was suggested by Peter Major.
Problem P.31. Let X\, X2,... be independent, identically distributed
random variables such that, for some constant 0 < а < 1,
ρ{χλ =2к/а\ = 2~к, А: = 1,2,.
Determine, by giving their characteristic functions or any other way, a
sequence of inRnitely divisible, nondegenerate distribution functions Gn
such that
sup
-оо<ж<оо
r{Xl+nya+Xni*}-Gn(X)
0 as η
00.
Solution 1. Put Sn = X\ + · · · + Xn, η = 1,2,... . The characteristic
function
oo 1
^)=Я(е'«0=ЕеЙЯ№/в)а · i€R·
fc=i г
Introduce the sequence 7„ = n/2^lognl, η = 1,2,..., where \u\ = min{n G
N : и < η}. Obviously, 1/2 < jn < 1 for all n. The independence

460
3. SOLUTIONS TO THE PROBLEMS
assumption implies that for all t the value of the characteristic function of
the random variable Sn/n1^ is
Mt) = E{e^s"'^) = φη(1/η^) =(l + £y«<"">V»"· - Ι)±λ
_ / 1 yv «2('/-)(*-гь«-1)/71/- _ ,ν 1 \
r/ahk/a - i)llL
'2n
у г=— [logn]+l
For a given 1/2 < 7 < 1, define
00
mo = Σ (?nTl~hlla -1) ^ «^
r=—00
and put
Since for all t e R,
7i-i/«
IM*)I < 27^2^ +71-1/Ί*ΐΣ2(1/β-1^ < 27+3^1=17^1*1.
r=l r=0
the above definition is correct, and £7(.) is the characteristic function of
the random variable
00 cyrja.
zi = Σ ^w.
r=—00 '
where the Yr(jyYs are independent Poisson-distributed random variables
with expectation E(Yr(/y)) = 7/2r, r = 0, ±1,±2,... . For the sake of
simplicity, ΖΊ can be defined as the limit in distribution of the partial
sums
n or/a
Σ Ж*(7).
r=—n
By the continuity theorem of P. Levy it is easy to see that this is possible.
Obviously, Zr is an infinitely divisible random variable.
Observe that, for any fixed t e R, £7(£) is continuous as the function of
7 on the interval [1/2,1]. (£7(£) as a bivariable function is continuous on
the domain [1/2] χ R).
We are going to prove that the distribution function ΗΊ of Z7, which is
uniquely defined by the relation
/00
еНхаЩ(х), t € R,
-OO

3.8 PROBABILITY THEORY
461
is continuous on the whole real line for all 7 G [1/2,1]. Simple calculations
show that
-l/a poo
fj^)\dt=1-f
\ r=-oo
= j™ exp ί-7 Σ (l - cos^/")) 1J de
< /00expi-7^(1-cos(s2-fc/a))2a)ds
S / exp I —7
(1)
where for arbitrary s > 0, k(s) denotes the smallest integer к > 1 such
that log(s/7r) < A: - 1. Since k(s) < 2 + log(s^) and s2_/c < π/2, for all
к > k(s), s2~k/a < π/2. Thus, using the inequality
1 — cos χ > —zx
(os.sf).
we get
00 .00 . 00 ..
Σ (l-cos(S2-fe/a))2fc>As2 £ 2(1-2/«)«=>±s2 £ i
/e=/t(s)
^5_ 2*;
2
к=к(з)
π
(2)
Hence
/00 /»o
|i7(t)|di < 271/a /
-00 JO
e-27/™ds < ^
The continuity of ΗΊ follows from the inversion formula of characteristic
functions. (See, for example, A. Renyi, Probability Theory, Akademiai
Kiado, Budapest, 1970.)
Consider the random variables Zln with distribution functions Gn{x) =
ΗΊη(χ), χ e R , and characteristic functions Ψη(£) = £y„(*)j t e R, η —
1,2,....
Since for η —> oo
oo
4.(0- Σ (eit2^/e-i)|-o
r=— [logn] + l
for arbitrary £ e R, it is easy to see that
φη{ΐ) - 9n(t) -> 0 as η -+00. (3)

462
3. SOLUTIONS TO THE PROBLEMS
Finally, introduce
A(Fn,Gn) = sup \Fn(x) - Gn(x)\,
— оо<ж<оо
where
F^x)=p\-§k^xY xeR>
and consider a sequence {n'} of positive integers tending to oo. Since
1/2 < 7n/ < 1, the Bolzano-Weierstrass theorem implies that there is a
subsequence {n"} С {η'} such that for some 7 G [1/2,1], ηηη —> 7 as
n" —> 00. In this case,
Л(^,СП.)<А(^,Я7) + Л(СП.,Я7)
and Ψη//(ί) = £7n„ (t) -> £7(£) for allt G R as n" -> 00.
From the continuity theorem of Levy, we know that Gn"(x) =
ΗΊη„(χ) —> ΗΊ(χ) at any continuity point of the limit function. Since
the limit function is continuous everywhere, it converges uniformly by a
well-known theorem of Polya. Hence
Л(СП»,Я7) —> 0 as n" —> 00.
The relation (3) and a similar argument imply that
A(Fn//,ff7)->0 as n/7->cx).
Since the sequence {n'} was arbitrary,
A(Fn,Gn) ->0 as n-^00. D
Solution 2. Define the sequence ΖΊη of random variables in the same way
as above:
00
Ίη = 2гет> L" = Hognl, ΖΊη = η-1/" j; 2(L-+I-)/«yP(7„)>
r= — 00
where the l^(7n)'s are independent Poisson-distributed random variables
with expectation 7n/2r = n2~Ln~r . The "three-series theorem" implies
that the sum defining ΖΊη is convergent. We will use estimations (1) and
(2), which guarantee that the absolute value of the characteristic function
£n(£) of the distribution function Gn(x) of the random variable ΖΊη is
integrable, and this integral remains bounded independently of n. The
distribution functions Gn(x) are obviously infinitely divisible, and we will
show that they satisfy the requirements of the problem. This assertion will
follow from the following two statements.

3.8 PROBABILITY THEORY 463
Put Sn — η 'α Χ^_ι -^k ' an<^ ^ Fn(x) denote its distribution
function.
(a) There exists a sequence en —> 0 of positive numbers such that for all
χ G (—00,00),
£n < Fn(x) < Gn(x + εη) + εη.
(b) For all n, the density function <7n(:r) of Gn(x) exists, and there is a real
Μ that is independent of n, and for all χ e (—00,00), gn(x) < M.
First, we show that the requirements of the problem follow from (a) and
(b). Indeed,
Fn(x)-Gn(x) = Fn(x)-Gn(x+en)+Gn(x+en)-Gn(x) < εη+Μ εη, (4)
Gn(x)-Fn(x) = Gn(x-en)-Fn(x)+Gn(x)-Gn(x-en) < εη+Μ εη. (5)
Since εη —> 0, from (4) and (5) we have the limit sup^. \Gn(x) — Fn(x)\ —> 0,
which was to be proved.
Statement (b) follows from estimations (1) and (2).
Proof of statement (a). Let dn be a sequence of real numbers that tends
to infinity slowly enough. Put
r= — 00
00
Z(2)=n-l/« £ 2(Ln+r)/ayr(7n)|
r=Ln+dn
η
Sn] = n1/a^2XkI(Xk<2^n-dn)/a)Ykbn),
k = l
η
Sn = ^~1/a 22 X^(Xk>2Ln+dn)/^)Yk{7n),
k=l
S(3) = sn-s(i)-s(2),
η η η η '
where Ια is the indicator function of the set A.
The relations P{Sn2) φ 0) -> 0, P(z£2) ^ 0) -> 0, ^(^1}) -> 0,
E(Zn ) —> 0, and simple calculations imply that
Sn-SP^0 and Z7n - Z<3) =► 0, (6)
where => denotes the stochastic convergence.
(3)
Observe that Sn J can be written in the form
Ln+dn
S(3) = £ 2l/aVin-l/a^
l=Ln—dn

464 3. SOLUTIONS TO THE PROBLEMS
where v\ = ${j, 1 < j < η ,Xj = 2//a}, I = 1,2,..., and $A denotes
the cardinality of the set A. Denote the distribution function of Sn by
Fn (#), and that of Ζ η by Gn(x)· We are going to prove that
Var(Fi3>(*),Gf(*))-+0, (7)
where Var ({Pi}, {qi}) denotes the distance Σι \Ρ* ""9*1 °f the discrete
distributions {pi} and {qi}.
From relations (6) and (7), statement (a) follows immediately. We only
have to prove the convergence in (7).
Denote by distr {Yy(7n), \j\ < dn} and distr {vi+Ln, \l\ < dn} the joint
distributions of the random variables {^(7n)} and {^/+Ln}? respectively,
(\j\ < dn, \l\ < dn). The limit relation (7) follows from the convergence
Var (distr {У,Ы, |j| < <U, distr {z//+Ln}, \l\ < dn). (8)
This last statement can be proved as follows. Since dn tends to infinity
slowly enough, by restricting the range {pi , |/| < dn} of the random
variables appearing in (8) by log2n, the resulting error does not affect the
validity of the following estimation:
P{vi+Ln=Pi , \j\<dn) =
Ut-dnpiin-EtdnPiy
dn / dn \n-Et-dnPi
π 2-p»(Ln+/). i_ y* 2(-L-+/) ]
l=-dn \ l=-dn J
dn /. _q—Ln— l\pi
= (ΐ+θ№^\^Ρ(Υη(Ίη)=Ρι , \l\<dn).
= 11+0(^)1 Μ Ρ{Υι{Ίη)=Ρι)
The proof is complete. D

3.9 SEQUENCES AND SERIES 465
3.9 SEQUENCES AND SERIES
Problem S.l. Let the Fourier series
2
— + /^(afc cos kx + bk sin kx)
к>1
of a function f(x) be absolutely convergent, and let
4 + ь1>а2к+1 + ь2к+1 (k = i,2,...).
Show that
1 ί2π
- / (f(x + h) - f(x - h))2 dx (h > 0)
is uniformly bounded in h.
Solution. By the Denjoy-Lusin theorem, X^xdafcl + \bk\) < oo. Hence
X><°° (Qk = y/a2k+Pk; fc = l,2>...). (1)
k = l
So we obtain ΣΤ=ι #1 < °°· By tne Riesz-Fischer theorem and the
completeness of trigonometric functions, it follows that (f(x))2 and thereby
(f(x + h)-f(x-h))2 is integrable. Easy calculation shows that
oo
f(x + h) — f(x — h)~2 2^ {bk sin kh cos kx — ak sin kh sin kx).
k = l
The Parseval formula implies that
/oo
{f(x + h) - f(x - h))2 dx = 4n^2 Ok sin2 kh· (2)
о k=1
By the condition ρ\ > ρ|+1 (к = 1,2,...), we obtain кдь < Σ|=1 Qi (^ =
1,2,...). So by (1), kQk = O(l). Using sin2 χ < |ж|, by (1) we obtain
oo oo
^2Q2kSm2kh<\h\J2kgl = 0(h).
k=l k=\
By this and (2), we conclude the theorem. D
Remark. Instead of the absolute convergence of the Fourier series and
Qk > Qk+i (k = 1,2,...), it is enough to suppose that ΣΤ=ι ^Qk < °°·
Gabor Halasz and Tibor Nemetz remarked that by the conditions of the
theorem (Ι/Λ) · /02π(/(α + h) - f(x - h))2 dx = o(h) holds.

466
3. SOLUTIONS TO THE PROBLEMS
Problem S.2. Let yi(x) be an arbitrary, continuous, positive function
on [0, A], where A is an arbitrary positive number. Let
yn+1(x) = 2 [ vCT)* (n=l,2,...
./o
)·
Prove that the functions yn{x) converge to the function у = x2 uniformly
on [0,A].
Solution. We will not only prove the theorem but also approximate the
rate of the convergence.
Let y{(x) and yl*(x) be given continuous functions on [0,A] such that
0 < y{{x) < yl*{x), for all 0 < χ < A. Denote by y{(x),y2*(a;),...,y*(x),...
and by у I * (χ), уз * (χ) ? · · · ? У η* (χ) j · · · the functions obtained from y{ (x) and
yl* (x) by the above iteration. Obviously, у*(я) < у** (χ) for all 0 < χ < Α.
So it is enough to prove that the claim of the problem holds for a constant
function of value С > 0, since if we choose С to be the maximum (minimum)
of yi(x), then the nth element of the function series received from the
constant function will be larger (smaller) than yn(x) at each x.
So let Yx(x) = C, where С > 0, and let Y2{x), У3(ж),..., Yn(x) be the
functions obtained by the iteration. We can show by induction that
^n\x) = CnX ι
where cn is defined by the recursion
cn+i = _ _i_ (n = 1,2,...; ci = C). (1)
Take the 2n+1th power of each side of (1):
J2n+1 _ 2n !
w._li — ^n
The multiplier of C2n is positive and converges to a finite positive value (to
e2) as η goes to infinity. So the quotient of the neighboring elements of the
series c\,c\ ,... c2 ,... remains between two positive constants. Therefore,
by appropriate constants m and M,
mn<< <Mn (n = l,2,...).
Hence, by
it follows that
m
n/2n
<cn<Mn/2n (n = l,2,...),
cn = e°W2n). (2)

3.9 SEQUENCES AND SERIES
467
Obviously,
Yn(x) -x2 = sMVa-1) . (Cn _ !) + (^-d/2"-1) _ x2).
By (2), for all 0 < χ < A,
\x>-W^-(cn-l)\<A>\cn-l\=0(^).
It is easy to show by derivation that the function achieves
its maximum on the interval (0,1) at χ ο = (1 — ^г)2 , and
г-и/г""1) 2 _ г-а/г»"1)
с0 х0 — х0
For А > 1, К ж < А
(ι-41/2η_1))
<
1)
2П/
1
2П"
, -^ _Г *«> _Г ^Х,
ι 2 ±
д. 2П-
я2|=яГ 2*=т\1-х2*=т\<А2\1-А2*=т\ = 0[ — \.
2п
Thus,
ВД-2 = о(^)
Hence, it follows that, starting from the function yi(x), which satisfies the
conditions of the problem, we get
yn(x)-x2 = θ(—J ,
where the constant included in the sign О depends only on A, and on
the maximum and minimum of y\{x). So the claim of the problem is
proved. D
Remarks.
1. Gabor Halasz proved the following generalization of the problem. If
the function G(y) is monotone increasing and continuous for all у > 0,
G(0) = 0, G(y) > 0 for у > 0, and
У GG
G(y)
dy < +oo,
but
Jw)dy = +0°'
then obtaining the functions yn(x) by the recursion
X
Уп+iW = / G(yn(t))dt,

468
3. SOLUTIONS TO THE PROBLEMS
it follows that the functions yn(x) uniformly converge to the unique
solution of the differential equation y'(x) = G(y), which goes through
(0,0) and does not agree with the constant function (x = 0) in any
neighborhood of the origin.
2. If we suppose that y\ (χ) is nonnegative instead of positive, we can state
the following:
Let δ be the upper bound of those χ values for which yi(u) = 0 for all
0 < и < χ, and let /(#,<$) be the function such that /(#,<$) = 0 for
0 < χ < δ and f(x, δ) = (χ — δ)2 for δ < χ < A; then the functions
yn{x) uniformly converge on [0,A] to the function /(#,<$).
Problem S.3. Let Ε be the set of all real functions on I = [0,1]. Prove
that one cannot define a topology on Ε in which /n —► / holds if and only
if fn converges to f almost everywhere.
Solution. Let
~fc-l к
I<
2m+k —
Orn ' 2m
(m = 0,l,2,...;fe=l,2,...2m),
and denote by fn the characteristic function of In. Then {/n} will nowhere
converge to 0; but if {fnk} is an arbitrary subsequence and хПк G /η^, where
{xnkl} converges to a number xG/, then obviously {fnkl} —> 0 everywhere
but at ж, so in particular, almost everywhere. So if Τ is a topological space
and xn -f+ G T, then there exist, a neighborhood of χ such
that for some subsequence хПк, хПк £ V, and hence for any subsequence
Remarks.
1. The solution shows that the same statement is true if in the definition of
almost-everywhere convergence we give the role of the zero-measurable
sets to a set system 3?, where for any χ e I, {x} G 3? but 7^3?.
2. With a slight modification of the solution, one can show that the
statement holds when I is the set of real numbers or a closed interval.
Problem S.4. Let the continuous functions fn(x), η = 1,2,3,..., be
denned on the interval [a, b] such that every point of [a, b] is a root of
fn(x) = fm(x) for some η φ πι. Prove that there exists a subinterval of
[a, b] on which two of the functions are equal.
Solution 1. Denote by Enrn the zero set of /n(#) — fm{x) for η φ т.
These sets сап be enumerated. Let
Mi,M2,M3,...,il4,...
be an enumeration of these sets. If M\ is a nowhere-dense set on [a, 6], then
there exists a subinterval [αχ, b\] of [a, b] that is disjoint from M\. If M2 is a

3.9 SEQUENCES AND SERIES
469
nowhere-dense set on [a, 6], then there exists a subinterval [аг, Ьг] °f [аъ &ι]
that is disjoint from M2. If all of the Mk are nowhere dense on [a, 6], then
we can continue this method. The intersection of all the intervals
[a, b] D [ab 61] D · · · D [afc, 6fc] 2 ...
is not empty and does not contain a root of fn(x) = fm(x) for any пфт.
This is a contradiction. So there is an M^ that is dense on some [c, d] part
of [a, 6], that is, the roots of the corresponding equality fn(x) = fm(x)
are dense on [c, d]. By the continuity of the functions, it follows that
/n(s) = frn(x) on [c, d]. D
Solution 2. The union of the sets Enrn of the previous solution is the
interval [a, b]. If none of the sets Enrn is dense on any subinterval of [c, d],
then their union cannot be an interval, since an interval cannot be a union
of countably many nowhere dense sets. So there is a set Егш that is
dense on some [c, d] subinterval of [a, &]; that is, the set of the roots of
the corresponding fn(x) — /m (#) is dense on [c, d]. Since the functions are
continuous, each point of [a, b] is a root. D
Remark. Many participants used the Baire category theorem to
generalize the problem.
Problem S.5. IfJ2t^-oo \am\ < °°, then what can be said about the
following expression?
1 +00
hni -z——г /\ \o-m-n + am_n+i "I Ь ап
n->oo 2n + 1 z—'
Solution. Since Y^1_00 \am\ = σ < oo, Σ^Ξ-00 α™ = S exists* Let
+00
-J Tbv
= 9 1 1 / у |am—η Τ" flm-n+1 τ * * * Τ" flrn+n|·
m=— 00
We will show that Ишп-юо Cn = |S|.
Let ε be an arbitrary positive number. Then there is a natural number
Μ such that
У^ |am| < e·
|m|>M
Hence, for η > M,
(2n + i)cn= Σ |...|+ Σ 1-1+ Σ 1-1-
\m\>n+M η+Μ>|τη|>η—Μ \m\<n—M

470
3. SOLUTIONS TO THE PROBLEMS
So
Lm+n\
<
У^ am-n Η \-a,
m+n
/ ^ Щт—п "τ * * * τ Q"n
\т\>п+М \т\>п+М
< (2n+l) J2 Ια™1 - (2п+1)^
|т|>М
and
Ι^τη-η Η l· а
'm+n I -^
σ < 4Μσ.
n+M>|m|>n—M n+M>|m|>n—M
Since m — η < —Μ < Μ < τη + η if |m| < η — Μ, it follows that if
|m| < η — Μ, then
||ат-п + * * * + a>m+n\ — \S\\ < 2_^ lam| < ^?
\m\>M
so
2 |am_n + ... + am+n| - (2n - 2M + 1)|S|
|m|<n—M|
< 51 ||flm-n + --- + aro+n|-|5||<(2n-2M + l)e.
|m|<n —M
Therefore,
|(2n + l)Cn - (2n - 2m + 1)|S|| < (2n + 1)ε + 4Μσ + (2n - 2M + 1)ε.
Dividing by (2n + 1) and taking the limit as η —> oo,
limsup |Cn - |S|| < ε + ε = 2ε.
η—>οο
Since ε is an arbitrary positive number, we conclude that
lim Cn = |S|. D
n—>oo
Remark. Let (Μ, σ,μ) be a commutative measurable group. Namely,
Μ is an additive group, σ is an invariant σ-algebra of the subsets of Μ
under addition and μ is an invariant measure, where μ{Μ) > 0. Suppose
that the function S:MxM-+MxM defined by S(x, y) = (x,x + y)
maps measurable sets of Μ χ Μ into measurable sets ofMxM.
Let Αχ, Α2,... (Ai Ε σ) be a sequence of sets symmetric about 0 (that
is, Ai = —Ai) such that
UfAi = M; Ai С Ai+l\ 0 < μ(Αί) < oo;
μ(Αί+ϊ -Ai) = ο(μ(Αί)); A{ + А, С Ai+i.
Cte'ra. If f(x) G Ζα(Μ,μ), then
п^~й(А~) J \j Κχ + ^ά^>\ άμχ = \l ^άμ
Μ \At
\M
This can be proved by similar methods. Miklos Simonovits gave this
generalization.

3.9 SEQUENCES AND SERIES 471
Problem S.6. Let αχ, α2,..., an be nonnegative real numbers. Prove
that
\г=1 / \г=1 / г=1 г=1
Solution l. If one of the a* is zero, then the statement follows immediately
from the known inequality between the nth power mean and lower power
means. It also follows that equality holds iff all the other clj numbers are
equal. So we can suppose a* > 0 for г = 1,2,..., п.
Denote by Σ^ the sum of the ftth power of the numbers αχ, α2,..., αη,
and by Sk their ftth elementary symmetrical polynomial. We generalize
the inequality:
Пк _ l\ ΣχΣ*_ι < nSk + (ft - 1) (fj Efc,
l<ft<n. (1)
The problem is the special case of inequality (1) for ft = n.
For η = 1,2, ft = 1, equality holds. So we can suppose η > 3, ft > 2.
We introduce the following notation. If αχ,..., αη > 0 are real numbers,
then denote by [αχ,..., αη] the sum (1/n!) · Σ α7ι · · · αΓηη j where we sum
over all permutations ΐχ,..., in of 1,..., n.
Lemma. If η > 3 and v, a±, · · · > 0, δ > 0, are real numbers, then
[ι/ + 2ί,0,0,α4,...]-2[ι/ + ί,ί,0,α4,...] + Κ«,«,α4,---] >0. (2)
Equality holds iff all of the ai are equal.
Proof. We cut the sum (2) into (з)(п — 3)! terms, so it is enough to
prove that for arbitrary real numbers 6χ, &2, &з > О,
Ъ\+2в + Ъ»2+2в + Ь^2в - {Щ+вЪв2 + Щ+8Ь{ + Ь\+вЬ1 + К+8Ъ{
+ ь^вь1 + ь»+6ь62) + ь\ъв2ь1 + ъв№1 + ь№щ > о,
where equality holds iff Ь\ = Ъ2 = Ъ$. This is straightforward by the
following obvious inequality:
χμ(χ - y)(x -z) + у»(у - x)(y -z) + ζμ(ζ - x)(z - y) > 0,
for x,y,z > 0,/i > 0. Equality holds iff χ = у = ζ. We conclude the
lemma by changing x, y, z, and μ into af, a2, a%, and и/δ, respectively. The
ideas in the proof of the lemma appears in G. H. Hardy, J. E. Littlewood,
Gy. Poly a, Inequalities, Cambridge University Press, Cambridge, 1959.
To prove inequality (1), write it in the following form:
;:;)ς«^'*·*+((*-4;)-(;:;))*. <»

472 3. SOLUTIONS TO THE PROBLEMS
The following equalities hold by the definition of [αχ,..., an]
n(n — 1)
i#i
n-2
п. *—J s_^ —/
n-1
/c n—к
(4)
If we substitute the inequalities (4) for (3), and making use of the relation
(fc) = Έ ' (fc-i)' we can WTl^e (3) ш the following form:
(n- l)-[fe-1,1,0,... 0] < [1,... 1,0,... 0] + | к - 1 - - ) [fe, 0,... 0]. (5)
η—2 /с η—к η—1
By (η — 1) · (fe/n) = к — 1 — (fe/n) + 1, we can write (5) as
[k - 1,1,0,. ^] - [V·. 1,0,. ^]
n-2
< (fe - 1 - -)[fe, 0,. .M - [k - 1,1,0,. ^]. (6)
n-1
n-2
We introduce the following notation:
Δ* = [t + 1,1,1^, ^0 ] - [t, Ι,^Λ, ^0 ]
k-t-1 n-k+t-l k-t n-k+t-1
n>3, k>2, £ = 0, l,...,fc-1.
By substituting i/ = £—1,5=1 into the lemma, it is straightforward
that
ο = δ5<δϊ<···<δϊ_1. (7)
Equality holds iff all of the a^s are equal.
We can write (6) in the following form:
Σ>**(*-ι-£)δΪ-ι.
This inequality holds, since by (7),
k-2
J2At <(k-2)Akk_x and fe-2<fe-l--.
t=o

3.9 SEQUENCES AND SERIES 473
It is straightforward that if η > 3 and к > 2, then equality holds iff
к — n and all of the a^s are equal. D
Remark. Inequality (1) was first proved by Laszlo Lovasz. He also
proved it for the case when some of the a^s are zero, but he did not
examine the cases of equality. The previous proof is the generalization of Peter
Gacs's proof. He used this method for proving the case к = n, that is, for
proving Problem S.6.
Solution 2. If one of the a^s is zero, then the statement follows
immediately from the Chebyshev inequality for uniformly ordered sequences. It
also follows that equality holds iff all the other numbers clj are equal. So
we can suppose щ > 0 for г = 1,2,..., п. Let
η η η η
F(au ... ,αη) = ηY[a,i + (η - 1) ^< - ^α* J^aJT1.
г=1 г=1 г=1 г=1
By the homogeneity of F, it is enough to prove that
F(ai,..., an) > 0 if αχ Η h an = n. (1)
F(ai,..., an) = 0 holds for η = 1,2. Let us suppose η > 3. It is enough
to prove (1) in those cases when F has a local minimum at αχ Η \-an = η,
that is, when
1 dF n
_a =(n_i)(an_an-i) + TTa. = λα. (<=1ι...ιη) (2)
П dCLi A *
г=1
for some real λ.
By the Descartes sign rule and Л™=1аг > 0, the polynomial (n — l)x —
(n — l)xn~l + \x + ΠΓ=ι ai cann°t have more than two positive roots. So
by (2) it is enough to prove that
F{x,...x,y,...,y) > 0 for x,y>0.
к п—к
By the homogeneity, it is enough to prove that g(x) > 0, where
0(ж) = ^(ж,...ж,1,...,1), 0<ж<1. 0(1) = 1.
к п—к
If we show g(x) > 0 for 0 < χ < 1, /с ^ 0, /с ^ 1, then it also follows
that in case η > 3, α^ > 0, equality holds iff all of the a^s are equal. Let
0<ж<1, 0 < к < п. By rearranging, we get
g(x) = fe(n - k)(xn - xn~l - χ + 1) - kxn + nxk - η + fe.

474
3. SOLUTIONS TO THE PROBLEMS
Hence,
1^1 = k(n - k)(l - x71-1) - n(l + · · · + xk~l) + fe(l + · · · + xn~l)
1-х
= k(n -k)-(n- fe)(l + · · · + xk~l) + k(xk + · · · + xn~l)
-k{n-k)xn-1 >0,
since
k(n-k)>(n-k)(l + ---+xk-1) and k(xk + · · · + ζη_1) > k(n-k)xn-1.
Equality holds in both inequalities if fc = 1, к = n— 1, that is, iff η = 2. D
Solution 3.
η η η ι η η
ηΣα*-Σα*-Σα'_1 = 2ΣΣ(αί-αί)2κί_2+«!_4·+···+αΓ2)
2 = 1 2=1 2=1 2=1 J = l
for arbitrary t, since
-α,)2(α<
for i,j = 1,2,.. .n.
(a< - a?)2(a-"2 + · · · + a*"2) = a■ + a) - a^r1 - a^r1 (1)
If
η
0 < an+i < an < · · · < αχ and Y^ аг = 1,
2=1
then
η η η
2 = 1 2=1 2 = 1
= <+1(ΐ-η) + <;}. (2)
We shall use induction. Equality holds for η = 1,2. Suppose that η > 2,
Gn+i < fln < · · · < ^ь and that the inequality of the problem holds for n.
We are going to prove it for η + 1. By the homogeneity, we can suppose
that Σ™=ι аг = 1· By the induction,
η η
(п-^^аГ + пД^-^^"1^0· (3)
2=1 2 = 1 2 = 1
We have to prove
n+l n+1 n+1 n+1
"Σα"+1 + (η+1)Πα'-Σα*Σα"^0· (4)
2=1 2=1 2 = 1 2=1

3.9 SEQUENCES AND SERIES
475
The left side of (4) can be written in the following form:
η η η
η Σ <+1 + ηα^ΧΙ + ηαη+ι JJ α{ + αη+ι JJ α{
г=1 2=1
η
-(α + αη+1)£>?+<+1). (5)
г=1 г=1 2=1
2 = 1
By (3), it is enough to prove that (5) is at least as large as αη+ι times
the left side of (3). So, by rearrangement, we have to prove that
2 = 1 2 = 1 2 = 1 2 = 1
Π
+ ап+1(Паг-(1-пК+1-<;})>0. (6)
2 = 1
By (1) and аг > αη+1 (г = 1,2,... ,η), the difference of the first two
terms is
\ Σ ί> - ^)2κ_1+α?~4· + · - ·+aV
- αη+1(α?-2 + <"Ч + · · · + a;"2)] > 0. (7)
2
2=1 J = l
By (2), the third term is also nonnegative.
By (7), if η > 2, then equality holds iff αλ = a2 = · · · = an. If an+1 > 0
and ai = a2 = · · · = an, then the third term of (6) can be zero only in the
case аг — αη+1 = 0; otherwise, strict equality would hold in (2). D
Problem S.7. Let f(x) > 0 be a nonzero, bounded, real function on
an Abelian group G, gx,..., g^ are given elements of G and X\,..., Xk are
real numbers. Prove that if
^Xif(9ix)>0
2=1
holds for all χ G G, then
к
Σλί>ο.
2 = 1
Solution 1. We can suppose that f(gi) > 0. Denote by An the set of those
elements that can be written in the form g*1,..., g*j*k, where the maximum

476
3. SOLUTIONS TO THE PROBLEMS
absolute value of the numbers a\,..., α& is n, where η > 0 is an integer.
Denote by S(H) the sum J2xeH f(x)> wnere Η is a finite set.
.n{S(An+1)-S(An_l)=0
n>o S(An)
holds, since if for some ε > 0 and for all η > 0,
S(An+1) - SiAn-г)
S(An)
> ε
would hold, then
S(An+1) > S(iln_i) + eS(An) > (1 + s)S(i4n-i)
and so
5(Л2п+1)>(1 + е)"5(Л1)
would hold, which is a contradiction, since
By (i),
S(An)< sup f(x)(2n+l)k
xeG
s^w^,slwMi°'
that is,
hence
and therefore
г=1 г=1
S(An) - S(9iAn)
S(An)
S(An) - S(9iAn)
S(An)
S{An+1) - S(i4„_i)
<
S(An)
Σλ^~Σ\χ*
i=l
i=l
S(An+1) - S(An_!)
5(ЛП)
It follows that
к
S(An+1) - 5μη_χ)
έί έί ">0 5^
0. D
Solution 2. Let us again suppose f(gi) > 0. It is enough to consider
the subgroup G' generated by the elements gi,...,gk· We are going to

3.9 SEQUENCES AND SERIES 477
define a discrete translation-invariant measure on G'. Let ε > 0, and let
a\...ak be integers. For χ = g^1 ...^fc, denote by με(χ) the number
(1 + ε)-(Ι«ιΙ+-+Ια*Ι). The measure of G is finite since
oo oo
£ με(χ) < 2* £ ■ · ■ £ (1 + ε)-<αι+-+β*>
xGG αι=0 afc=0
oo oo
= 2* £ (1+ε)-*ι ···£(! +ε)-*,
ai=0 Q!fc=0
and so
Furthermore,
and so
/ ί(χ)άμε{χ)
0 < S = f(x) άμ,ε(χ) < oo.
G'
με(9ΐχ) = (1 + ε)±1με(χ),
IM&s) - με(χ)\ < εμ^χ).
Integrating inequality (1) on G', we get
к
χ) άμε (χ)
С г=1 г=1 С
г=1 qi г=1 г=1
So by ε —► О,
S^A^>0
г=1
it follows that
к
5]Аг>0. D
г=1
iteraarfcs.
1. Peter Gacs and Andras Simonovits remarked that the statement of the
problem holds for an arbitrary group if к = 2.
2. Peter Gacs and Laszlo Lovasz showed that the statement generally does
not hold for a noncommutative group.
3. Laszlo Lovasz, Endre Makai, and Imre Ruzsa showed that f(x) has to
be bounded.
4- Laszlo Lovasz discussed situations where the converse theorem holds.

478
3. SOLUTIONS TO THE PROBLEMS
Problem S.8. Show that the following inequality holds for all к > 1,
real numbers αχ, α2,..., α&, and positive numbers χχ, χ2,..., Xk ·
fc к
У %i У а^Х^ 1П 3^2
In _Si < i=L·
Σ^α< Σ**
г=1 г=1
Solution 1. By the known inequality relating weighted arithmetic and
geometric means,
^Тк1Х'Ш> \f{yXA^— (xi>0,tfi>0>i = l>2J...fe;fc>l).
2-,г=1 Жг \г=1 / Ζ-α=1 Ж*
If we substitute уг· = x~ai for г = 1,..., к and take —1 times the logarithm
of each side, then we get the claim of the problem. D
Solution 2. We are going to prove the following generalization of the
problem.
Generalization. Let (X, S, μ) be a measure space, where μ(χ) > 0. Let
x(s) > 0 and let a(s) be measureable functions on X, and x(s), x(s)1~a^s\
a(s)x(s)\n(x(s)) are integrable functions on X. In this case,
/ χ(β)άμ J a(s)x(s) In χ(β)άμ
1П /*(β)ΐ-«(')£ίμ " /d(s)d/i ' (1)
Equality holds iff x(s)~a^ is a constant for almost all s e X.
Proof. The following inequality holds for all positive d:
d-l>\nd. (2)
Put
f a(s)x(s) In χ(β)άμ
и = exp - r .
X
By substituting d = u-x(s)~a^s\ multiplying the inequality by x(s), and
integrating each side on X, we get the following equality:
и fгф)1"0^ άμ - ί χ(8)άμ>0. (3)
χ χ
This is equivalent to the claim of the generalization. Equality holds in
(2) iff d = 1. So equality holds in (3) (that is, in (1)) iff и · ф)"а^ is a
constant for almost all s e X, that is, iff x(s)~a^ is a constant for almost
all seX. Π

3.9 SEQUENCES AND SERIES
479
Problem S.9. Construct a continuous function f(x), periodic with
period 2π, such that the Fourier series of f(x) is divergent at χ = 0, but the
Fourier series of f2(x) is uniformly convergent on [0,2π].
The background of the problem. The following statement is a special case
of the Wiener-Levy theorem. If a positive function f(x) is periodic by 2π
and the Fourier series of f2(x) is absolutely convergent, then the Fourier
series of f(x) is also absolutely convergent. The following question arose.
Can the absolute convergence be replaced by uniform convergence? The
problem above states that it cannot be replaced without assuming the pos-
itivity. (R. Salem constructed an f(x) whose Fourier series is uniformly
convergent, but the Fourier series of f2(x) is not.) There are more ways
to solve the problem. Laszo Lovasz slightly modified Lipot Fejer's known
construction: By adding the Fejer polynomials under a relatively strong
gap condition, he received a continuous function f(x), whose Fourier
series is divergent at χ = 0, and the Fourier series of f2(x) is uniformly
convergent. It is also nontrivial to prove the uniform convergence of the
Fourier series of f2(x). Lajos Posa and Peter Gacs solved the problem in
an essentially similar way: They considered the function f(x), instead of
its Fourier series and used the observation that if the following inequality
holds for a continuous, 27r-periodic function g(x), then the Fourier series
of g(x) is uniformly convergent:
\g(x)-g(v)\<
l+e
(See the Dini-Lipschitz condition in /. P. Natanson, Constructive Function
Theory 1-3, 1964-65, VII. S§.)
Now we prove the claim.
Solution 1. Denote by sn(f\x) the nth partial sum of the Fourier series
of f(x), where f(x) is periodic by 2π and integrable, that is,
π
/,.ч 1 f ,./~n4sin(2n + 1)ϋ 1Λ
,„(/;*) = -У /(MJ-L-J-d*. (0)
If, on [0,2π],
then
. (2n + l)x
/п(ж)=81П^ —^-, (1)
π/2
2/°"±lU 1 r sin2(2n+l)tf JQ
• απ
/Jf ^ 1 /,sin2(2n+l)i? Ift If
sn(fn\0) = - / \ J dv > - /
π J sin ν π J
sintf
о о
π/2 η ηπ
1 / sin2(2n+Ш Ift 1 /,sin^i?jn A л
0 0 \Z)

480
3. SOLUTIONS TO THE PROBLEMS
where A\ and later A2, As,... are positive constants. Note that if \g(x)\ <
1 is integrable, then for all x,
\sn(g\x)\ <A2\ogn. (3)
The function we are going to construct will have the form
F(x)= ΣCv sin —Η—x> ^
ι/=1 Δ
where cv, nv are still undefined. Outside [0,2π] we will get the function by
periodic continuation. The restriction
\cv\<4Tv (у = 1,2,...) (5)
and F(0) = F(2n) = 0 will guarantee the continuity of F(x) and the
absolute convergence of (4). Let c\ = 1, c2 = 1/4, щ = 1, and suppose
that Ci, C2,..., θμ-ι,ομ and Πι,n2,...,τιμ-ι are already defined.
Since the functions fn(x) — defined in (1) — are twice differentiable
and their Fourier series are uniformly convergent, the following inequality
holds at each x, for m > As
\sm(fnu;x)\<2, μ = 1,2,...,ι/-1. (6)
Denote by n„ the smallest positive integer for which
JL
A,
ομ1οβημ>-£- (7)
and
ημ > 2 + max{A3,3ημ_ι} . (8)
If
CM+1
= min^ -f, - >,
[ 4 logr^J
(9)
then (5) is straightforward. So i^rr) is defined.
We will use the fact that the following numbers are pairwise different:
rii + rik + l and nj-nm (1 < г < к, l<m<j). (10)
By πι = 1 and by (8), the numbers above are different if 1 < г < к <
2, 1 < πι < j < 2. If they are different for щ, n^, rij, nm, where г, k,j,m
are less then μ, then by joining ημ to щ, n^, rij, nm, we get the following
sums and differences:
ημ + ημ + 1 > ημ + ημ_ι + 1>···>ημ + ηι + 1
> ημ - η2 > · · · > ημ - ημ_ι.

3.9 SEQUENCES AND SERIES 481
Even the smallest of them, ημ — ημ_ι, is larger than the largest of the
previous ones, ημ_ι + ημ_ι +1. So the numbers at (10) are indeed pairwise
different.
By (4) and (5),
F2(n) = o^2 βμ°ν{0Ο8(ημ ~ Uv)X ~ COS(?V + nv + l)x}
μ,ν
1 °°
г=1 \<ν<μ
- - ^2 cl cos(2ni/ + 1)ж - J^ ομαν ο,ο&(ημ + n„ + l)x.
v=\ \<ν<μ
Since the numbers at (10) are pairwise different, this gives the Fourier
series of F2(x). By (5), this is absolutely convergent. We still have to prove
that the Fourier series of F(x) is divergent at χ = 0. To see this, divide
the series into three parts:
μ —1 oo
sn(F;0) = cMsn(/n;0) + ^CjSn(/n/,0) + ^ <^ημ(/ημ;0)
J=l J=M+1
= C/i + C/2 + f/3. (11)
By (2) and (7),
Ut > μ. (12)
By (6) and (8),
Finally, by (3) and (5),
μ-1
|[/2|<2^с,<|. (13)
\U3\ < Α2^ημ ]Γ cji Α4ομ+1^ημ < A4. (14)
So by (11), (12), (13), and (14), the Fourier series of F(x) is divergent
at χ = 0. D
Solution 2. We will define the function f(x) instead of its Fourier series,
such that (0) will hold for f2(x), /(0) = 0, and
π
limsup f E±0*f(t)dt > 0. (15)
Π—+00 J *
о
f(x) will be an even function. Its nth subsum 5n(0) at 0 thus has the
following form:

482
3. SOLUTIONS TO THE PROBLEMS
π
Sn(0) = ljf(2t)
sin(2n + l)t
sint
dt.
If a continuous function is convergent, then its Fourier series and the
Fejer means will converge to the same value, to the value of the function.
So to see the divergence of Sn(0), it is enough to prove that Sn(0) -f* 0. To
prove this, we modify the kernel:
sin(2n + l)t . rt л sin2nt . л , - .
— — =sin2n£cot£+cos2n£ = bsin2n£ | cott — - ] +cos2nt.
suit
t
(cott-i).
cott — (1/t) can be modified to be continuous at 0, so by applying the
Riemann lemma twice, we get
Г „/л 4sin(2n + l)t , ,. Γ „/Λ 4sin2nt ,
imsup / f(2t)—Ц J-dt = hmsup / f(2t) dt
n—KX) J Sin t n—юо J *
0 0
7Γ
2
/si
/(2t)-
k sin 2nt
hmsup / f{t) —
n—юо J
t
sin nt
t
dt >0.
Thus Sn(0) -h 0, that is, Sn(0) is divergent if (15) holds. Since /(n) is
even, it is enough to construct it on [Ο,π].
Let
: же n , δη=η , ρη
(n>0)
and
Λ» = [οη,αη_ι], 9η(χ) = tinSinpnX.
Define gn(x) = <7п(ж) between the second and the last roots of g*(x) on
/n, and gn(x) = 0 elsewhere. We will define f(x) as a sum of a subsequence
of the sequence {gn(x)}·
Construct a sequence щ < n^ < ... of natural numbers such that
π
/
smpnk+1x
fk(x)dx
<ρ if fk =^gni,
2=1
>-fc
< ρ2~* for all j < fc,
(16)
(17)
where ρ is a sufficiently small constant, say ρ = 10~5. To see that such a
sequence exists, observe that, on the one hand, by pn —> oo, by the
continuity of gn(x)/x, and by the Riemann lemma, all terms of the integral in (16)

3.9 SEQUENCES AND SERIES
483
converge to zero; on the other hand, by the continuity of (sinQnx)/x, the
integral in (17) converges to 0 for any fixed j. Define f(x) = Y^kLo9nk(x)·
Obviously, f(x) is everywhere continuous, even at 0. We will show that the
Fourier series of f(x) does not converge to 0 at 0, that is, it is divergent. For
the reader's convenience, denote by Sn « Tn if Sn/Tn —► 1. By (16) and
(17), the difference of J*0 (sinpnkt)/tdt and J^nfc_1 9nk(t) · (sinpnkt)/tdt is
a maximum of 2ρ.
anfc-i πε-(η*-1)3
/ 9nM^Y^dt^ j 6Пк*т2е<Ыг
π6^-(η*-ι)3 ^ 2
6nu I du
и
I
π
*».( Σ ^)/sin2u
\ m=l / n
re3n2-3nfc-l
du
37Γ _2 о ЗТГ
«γ»* »* = γ·
So the Fourier series of f(x) is divergent at 0.
Now we are going to show that
i/2<i)-/2<»)|=o(G«))2)· <i8)
that is, the Fourier series of f2(x) is uniformly convergent. Let χ > у and
x £ In, У £ Im for some n> m. We move у to the sine curve pn sinpn£ of
Im such that у lies on the same quarter period of the sine curve as #, but
f2(x) remains the same. |rr — y| does not increase if η = m or η > га. So
we have to prove (18) only in the case when χ and у are on the same In
and \x — y\ < π/ρη. By the Lagrange mean value theorem,
I/2(x) - f(y)\ = «nl s[nPlx ~ sinPn2/l < ^lPn\x - y\
0S2„ \x - y\(log\x - y\)2 nch
= 28nPn dog A)2 · (19)
Since /i(log h)2 monotonically increases if h < e-2, by (19) we have to prove
(18) only in the case |rr — y| = π/ρη; but this case is trivial. This finishes
the solution. D
Remark. Solution 1 proves more, as it proves that the Fourier series of
f2(x) is absolutely convergent.

484
3. SOLUTIONS TO THE PROBLEMS
Problem S.10. Prove that for every ϋ, Ο < ϋ < 1, there exist a
sequence Xn of positive integers and a series Σ™=ι αη such that
(i) λη+ι - λη > (λη)*,
oo
(ii) lim > anrXn exists,
n=l
oo
(Hi) ^2 an JS divergent.
n=l
The background of the problem. Andras Simonovits noted that λη+ι >
c\n (c > 1) and the (C,l) summability together guarantee the convergence.
Moreover, if we substitute χ = e~y in G. H. Hardy, Divergent Series,
Clarendon Press, Oxford, 1949, §7.13, then Theorem 114 can be stated as
follows. Given с > 1 constant and a sequence of natural numbers, such
that λη+ι > c\n (c > 1), Σ™=ι o>nrXn convergent in the unit circle and its
limit exists at r —> 1 + 0, then Σ™=1 αη is convergent.
So this theorem, conjectured by Littlewood and proved by Hardy and
Littlewood, states that the Abel summability and the convergence are
equivalent notions in the case of sequences satisfying the Hadamard gap
condition λη+ι > c\n (c > 1). Our problem states that the Hadamard gap
condition cannot be substituted with a much weaker condition. The
participants gave two kinds of generalization for the problem. Some of them
substitute the Abel summability with the stronger (C,l) summability;
others showed that for any sequence λη satisfying liminf(An+i — λη)/λη = 0,
there exists a sequence Σ™=λ Q>n that satisfies the conditions of the
problem. This latter statement is also a generalization of the theorem, since if
An = e^, then
λη+1 - λη = e^1 -e^ = (e^+ϊ-^ - l)e^
= e^(ev^+U+r _ χ) = (χ + (0(i)) e
2у^'
and therefore,
Xn+1 Xn л Xn+1 Xn r q . t
► 0, -^ ► oo for ν < 1.
Xn Xn
Solution. We will prove the following statement, which is slightly weaker
than the previous two generalizations.
Theorem. For any series {λ^} that satisfies
liminfAfe+rAfe=0>
Xk
there exists a divergent series J2nG=1 an that is (C,l) summable and an = 0
only if η is not in {λ^}.

3.9 SEQUENCES AND SERIES
485
Proof. Let 1 < mi < m2 < m3 < ... be a subsequence of λχ,..., λ&,...
such that
Ik — 1
/c > mn < rri2k and < -.
П=1
Choose amk to be ( —l)/c+1, the others to be zero. Since Σ?=1 flj is 0
infinitely many times and 1 infinitely many times, Σ™=1 clj is divergent.
Denote Σ™=1 %· by sn. Since
o<tw^Sl + S2 + --- + s"<i,
Π
£n is monotone decreasing in [ra2bm2fc+i) and monotone increasing in
[m2fc-i,77i2fc)· Therefore, to prove tm —► 0, it is enough to show that
W -> 0:
_ (m2 - mi) + (Ш4 ~ тз) + · · · + (m2fe - m2k-i) 11 n
7Tl2fc л AC
Hence, tm —► 0, that is, Σ™=1 flj is (C,l) summable. D
Problem S.ll. Let 0 < a^ < 1 for к = 1,2, Give a necessary and
sufficient condition for the existence, for every 0 < χ < 1, of a permutation
πχ of the positive integers such that
*-Σ
k=l
Solution. Obviously, the following condition is sufficient:
inf a,k = 0, supafc = 1.
к к
We are going to prove that this condition is also necessary. Let χ G (0,1)
be fixed. For a permutation π, let απ(^) = bk and
dk = 2k
. h 62 . bk*
(do = s).
If we choose the permutation π such that 0 < dk < 1 holds for all /c,
then obviously
¥'
k=l
Let us suppose that π(1),π(2),... ,n(k) are already defined such that
0 < do, di,..., dfc < 1. We want to define π (к + 1) such that 0 < d^+i < 1.
This is equivalent to
2dfc-K5fc+i <2<fc, (1)

486
3. SOLUTIONS TO THE PROBLEMS
and also equivalent to
2(1 - dk) - К 1 - bk+1 < 2(1 - dk). (2)
We have to take care that ak will be used exactly once in the
construction.
Let щ be the element with the smallest index such that a/ ^ {b\, 62? · · · ,
bk]. If 2dk — 1 < сц < 2dfc, then we can choose bk+\ = сц. If сц > 2dk,
then let η be the smallest natural number such that 2n+1dk > a/; then
2n+ldk <2сц < l + a/? so
2n+4 -Ka/< 2n+4.
Since inffc ak = 0, &fc+i, ^fc+2? · · · ? ^fc+n can be chosen sufficiently small,
such that by dJ+i = 2dj — fy+i, rfj+n will be close enough to 2nd^. So we
can achieve
2dk+n -I <ai < 2dk+n.
Thus, we can choose 6fc+n+i = сц.
If a/ < 2dk — 1, or equivalently 1 — сц > 2(1 — dk), then we repeat the
above process by substituting 1 — щ, 1 — a3·,, 1 — dj in the role of щ, clj , dj,
respectively, and by using (2) instead of (1); and in the meantime, we use
the equality 1 — dJ+i = 2(1 — dj) — (1 — fy+i). By supfc a^ — 1, we can
choose the numbers 1 — bj to be sufficiently small, that is, bj is sufficiently
close to 1.
Repeating the process above, we obtain the required permutation. D
Problem S.12. Let λι < λ2 < ... be a positive sequence and let К be
a constant such that
n-l
Σ\\<Κ\1 (n=l,2,...)· (1)
k=l
Prove that there exists a constant K' such that
n-l
Y^\k <K'\n (n=l,2,...). (2)
k=l
Solution 1. We can suppose that К is an integer. We are going to prove
by induction that (2) holds if K' = SK. The theorem is straightforward if
η > SK, since К > 1. Let η > SK.
We observe that by the monotonicity of {λη} and by (1),
n-l
4Κ\ΐ_4Κ<Σ\ΐ<κ\2η,
k=l

3.9 SEQUENCES AND SERIES 487
so λη-4κ > λη/2. By this, by the induction, and by the monotonicity of
{An},
n—1 П—4К—1 n—1
Σλ* = Σ λ*+ Σ Xk
k=l к=1 к=п-4К
n-1
< 8Κλη_4Κ + J2 λ* ^ ±K\n + 4#λη = 8Κλη.
k=n-4K
Thus, the solution is complete. D
Solution 2. The monotonicity of {λη} is superfluous. We are going to
prove the following. If μϊ > 0, then the necessary and sufEcient condition
of
n-l
^2μί<Κμη, η =1,2,... (*)
t=i
is that there exist с > 0 real and r natural numbers such that, for each n,
μη+ι > ομη and (a)
μη+Γ > 2μη. (b)
This implies the statement of the theorem, since if λ^+1 > cX^ and
\2n+r > 2\2n, then λη+ι > y/c\n and \2n+2r > \2n+r > 4λ£, so An+2r > 2λη.
(a) is necessary: by (*) and by μι > 0, trivially μη-\ < Κμη, so (a)
holds if с = I/K.
(b) is necessary: by (*) again,
1
r2 > ~f?^n·, · · · ?/^п+г-1 >
Adding them together, we get
βη+Ι > 17/^5 βη+2 > ~7?№п-> ••-,μη+Γ-Ι > ~τ?1*·η-
βη+Ι Η Ь Vn+r-l >
so by (*)
A*n+1 + · · · + μη+r-l > —JF~^ni
1 / Ν Γ_1
A*n+r > -^(μη+1 Η l· /in+r-l) > Κ2 μη-
Thus, (b) holds if r > 2K2 + 1.
(a) and (b) are sufficient: define μο = μ~\ = · · · = 0.
η—1 г оо г оо
Σ^ = Σ Σ **»-«-*■ < ΣΣ^τ
t=l t=l j=0 i=l j=0
г=1 г=1 \ г=1 /
In the last step, we used that, by (a), /im+/ > c^m holds.
This finishes the solution. Moreover, a similar argument shows that
the theorem remains true if the exponent 2 is replaced by any exponent
a >0. D

488
3. SOLUTIONS TO THE PROBLEMS
Problem S.13. Given a positive, monotone function F(x) on (0, oo)
such that F(x)/x is monotone nondecreasing and F(x)/x1+d is monotone
nonincreasing for some positive d, let λη > 0 and an > 0, η > 1. Prove
that if
f>nWan£^)<oo, (1)
n=l V fe=l n)
or
Σ^ρ(ς**τ)<Ο0> (2)
n=l \jfe=l AnJ
then Σ™=1 αη is convergent.
Solution 1. If χ > 1, then F(x)/x > F(l). If χ < 1, then F(x)/x1+d >
F(l). We can suppose that F(l) = 1.
F{x) > χ if χ > 1, (3)
and
F(x) > xl+d if χ < 1. (4)
First, we prove the statement of the problem when (1) holds. Denote by
Ση the sum over those numbers η that satisfy
k=i n
Denote by J^ the sum over the other numbers n. Then, by (1) and (3),
(n \ η
o>n Σхк/к 1 > Σαη ΣXk - Σαηλι·
/c=l / η fc=l η
Since λι > 0,
Σ αη < οο. (5)
η
By (1) and (4),
οο > Y^'XnF ίαη £ Xk/Xn) > £'an+V ( Σ λ* ) · (6)
η \ k=l ) η \k=l )
Divide the sum Ση m^° two parts:
Σ II ^-^yll X~^"
= L· ω + L· (2)'
η η

3.9 SEQUENCES AND SERIES 489
where the first sum runs over those numbers η that satisfy
/ n ν (l + d)/d
o-n < Κ/ Ι Σ ^k
\k=l /
and the second runs over the other numbers n.
Then by one of the theorems of Abel and Dini, and by (1 + d)/d > 1,
/ η \ (!+<*)/<*
Σ"(ΐ)αη < Σλ™/ (Σ хч <ο°· w
η η=1 \k=l )
By (6),
00 > Σ wanadnKd Σλ*
Kk=l /
\fc=l / ) k=l n
Comparing this with (5) and (7), we conclude that Σ™=ι αη 1S
convergent.
Now, we prove the statement of the problem when (2) holds. If an = 0
for all but finitely many n, then the theorem is trivial; otherwise, we can
delete those λη, αη pairs for which an = 0. So we can suppose that an > 0.
Let X = {n : an > 1}. If η G X, then
EQfeAfe QnAn
k=i λη λη
and so by (2) and (3),
oo > Σλη Σ ~\^ = Σ Σ α*λ* - Σ
nex k=i n nexk=i nex
where no is the smallest element of X. So X is a finite set and the sequence
an is bounded. Let μη = anXn. Then (2) can be written in the following
way:
oo / η \
< OO.
S£4*££.
By the boundedness of the sequence an, (1) is satisfied with the substitution
λη = βη and, as we proved before, from this it follows that Σ™=ι CLn is
convergent. D

490 3. SOLUTIONS TO THE PROBLEMS
Solution 2. We prove the statement of the problem when (2) holds, but
we will not use (1). Solution 2 is the same as Solution 1 through formula
(4). Denote by J]n the sum over those numbers η that satisfy
±<ψ>1. (5)
k=i An
By (2) and (3), we obtain
ι ( n \ ι η
00 > Σ XnF 1 Σα*λ*7λη I > 5Z Σα*λ*
η \k=l J η k=l
^ / J ап0Лп0 = αη0λη0 / J 1-
П<71() 71<По
By choosing no such that ano φ 0, it follows that (5) holds only for
finitely many n. So, by (2) and (4),
oo / η \ \1+d °° / n \1+d
»>ΣΣτ *„ = £ап-« Σβ*λ* · (β)
n=l U=l n / n=l U=l /
Hence, assuming that not all of the an's are zero, it follows that
oo
n=l
and therefore
λη -> oo. (7)
Let ης be the largest natural number η such that 2q < Xn < 2ς+1; if
such an η does not exist, then nq can be chosen arbitrarily. We will see
that in the latter case the multiplier of Xnq is zero. Continuing (6),
l+d
°°>Σν(Σα*λ*) ^ΣνΙ Σ α*λ*'
η=1 \k=l ) q=l \к:2я<\к<2я+1
oo
> y^2_d^+1)29^1+^S'1+d = 2~dy^29S'1+d
q=l q=l
where
Sq = ]P α*.
fc:29<Afc<29+1
So there is a positive К for which
oo
j224si+d<K.
q=l

3.9 SEQUENCES AND SERIES
491
Hence, for any natural number q,
Therefore
which trivially implies
2qSl+d < K
Sq<K2~q^1+d\
У^ Sq < 00.
q=l
By the definition of Sq and by (7), there is an no such that Σΐ£=ι &q —
Σ™=η 0"η· Therefore, J^Li αη < °o, which finishes the solution. D
Remarks.
1. Originally, Laszlo Leindler placed part (1) of this problem at the
competition committee's disposal. Atilla Mate noticed that the statement
also follows from (2).
2. Laszlo Babai states a similar theorem for integrals: Let F(x) be a
positive monotone function on (Ο,οο) such that F(x)/x is monotone non-
decreasing and there is a positive d for which F(x)/x1+d is monotone
nonincreasing. Furthermore, let A(t) and A(t) be absolutely continuous
functions on (Ο,οο), A(t) monotone nondecreasing, A'(t) > 0 almost
everywhere, and lim^o A(t) = lim^o Λ(ί) = 0; then any of
and
oo
|A'(i)F(^A(i))di<oo
OO t
< 00
implies lim^oo A(t) < oo. (By the monotonicity of F(x)/x, the
functions after the / signs are measurable.)
Problem S.14. Let f(x) = Σ™=ι a>nl(x + n2), (x > 0), where
Σ™=ι Ιαη|^_α < °o for some a > 2. Let us assume that for some β > 1/α,
we havef(x) = 0(e-x) as χ —> oo. Prove that an is identically 0.
Solution. Obviously, the series defining f(x) is convergent at each χ φ
—ηα (η = 1,2,...), so f(x) is meromorphic on the whole plane. Let
oo
Γ(χ)=Π(ΐ+^). h(x) = F(x)f(x).
71=1

492
3. SOLUTIONS TO THE PROBLEMS
Since α > 2, this product is convergent for all x. Easy calculation shows
that F(x) is an entire function, so for any ε > 0,
/(х)=0(еИ1/а+£) .
The function h(x) is also entire, and there is a sufficiently small ε such
that, for χ > 0,
|M*)| = |/(x)| · \F(x)\ = 0(e-*S)0{e-*Ua+') = o(l). (1)
Furthermore, for any x,
00 I I / I I \ / °° I I \
n=l τηφη x 7 \n=l /
So if Μ(r) = max \F{x)\, then
/ 1/α+ε\
log M(r) _^(rl'a+^
0(1). (2)
So by a Phragmen-Lendelof type of theorem (Gy. Polya and G. Szego,
Problems and Theorems in Analysis, Springer, Berlin, 1976, vol. /,
III.6.332), (1) and (2) can hold for an entire function h(x) only if h(x)
is constant. So by (1), h{x) = 0. Hence f{x) = 0. Therefore, if for some
η, αη φ 0, then f(z) has a pole at ζ = —na. But this is a contradiction, so
a\ = a2 = · · · = 0. D
Problem S.15. Given a positive integer m and 0 < δ < π, construct a
trigonometric polynomial f(x) = α0+Σ™=1(αη cos nx+bn sin nx) of degree
m such that /(0) = 1, f6<\x\<n \f{x)\dx < c/m, and ταοχ-π<χ<π \f(x)\ <
c/δ, for some universal constant c.
Solution 1. Let ϋ = max{<$/2,1/m}. Extend the following function
periodically to all real numbers:
κ,,./'-ιΐι· "Ms*
ΨΚ J \ 0, if 0< |ж| <π.
This nearly satisfies our purpose, it just is not a trigonometric
polynomial. Let us examine the Fejer kernel of its Fourier series. Let
f(x) =c0 φ(χ - t)Km(t) dt,

3.9 SEQUENCES AND SERIES 493
where
sin2 2^11
Km{t) = 2(m+l)sin2§
and cq is chosen such that /(0) = 1. Obviously,
•d/2 l/2m
-tf/2 -l/2m
1 = /(0)>со^ У Km(t)dt>^ J Km(t)
trivially if |f| < l/2m. Then
Hence,
l^mWI > ^m.
l = /(0)>§, co<20.
Since y? is absolutely continuous,
π
f'(x) = co f <p'(x-t)Km(t)dt.
—π
So
χ+ϋ π
l/'COl < ^ j Km{t) < I y'Km(t)dt < *£.
Ж-1?
Estimate J6<\X\<7Z \f(x)\dx in cases # = δ/2 and # = 1/ra. If # = 1/m,
then
π ж+6
A \f(x)\dx = 2 f\f(x)\dx<2 l I Km(t)dtdx
δ 6 χ—δ
π+ϋ π
W j Km(t) dt < W j Km(t) dt = ^.
δ<\χ\<π δ δ χ-δ
<
6-0
If ϋ = δ/2, then
π+0 π
/ 1/0*01 ^ < 4ι? / /fm(t)d* = 2ί / +^lfm(t)dt.
6<|ζ|<π 6-0 6/2
Using elementary estimates on the interval 0 < t < π + <5/2 (< (3/2)π),

494
3. SOLUTIONS TO THE PROBLEMS
and so
π+δ/2
2δ ί Km(t) dt<26 ί ^dt =
240
m
δ/2
This finishes the solution. D
Solution 2. Start from the function
δ/2
h(:
X) = {
^(Δ2-*2)2, if*<A,
where the parameter Δ will be defined later.
Let
h(x) = Σ Cne%
and
s(x)= Y^cneinx.
n=—l
We will show that Δ can be chosen in such way that f(x) = s2(x)/s2(0)
satisfies the conditions of the problem. Obviously, f(x) is a trigonometric
polynomial of degree not more than ra, and /(0) = 1. We will estimate the
difference of s(x) and h(x). The Cauchy-Schwartz inequality yields
\S(x)-h(x)\< Σ |СпП|т-Т<л1
^—' \n\ Л
\n\>l ' ' \
Σ iCnTli2
\п\<1
Ν
Σ ;Ы Σ <«·>>#
\η\>1 \ η=-οο
By Parseval's formula,
-Δ
hence
Similarly,
So if
then
\s(x) -h(x)\ <
\s\x)-h\x)\<J±.
/Δ>9,
(1)

3.9 SEQUENCES AND SERIES 495
Moreover,
so
\П*)\-щ№)\\*Ы < ξ
suffices if
Δ > δ. (2)
This inequality and (1) also hold if we choose Δ = max{<$, 9//}. The
integral of \f(x)\ can be estimated in two ways. If Δ = <$, then
j \f(x)\dx=^ J \s\x)\dx=J^ I (s(x)-h(x))2dx
δ<\χ\<π δ<\χ\<π δ<\χ\<π
π
-iW)I{s{x)-h{x))2dx=im Σ|c"12
\η\>1
27Г ν—^ о о 27Г 1
W Еп2с2п<АР<Т
If Δ = 9/1 ± δ, then
_ι/_ _π η=-ί
00 /· 72
<2тг^|сп|2 = 4 / h2(x)dx<SA= i-.
δ<\χ\<π -π
οο
This finishes the solution. D
Problem S.16. Prove that if
m
Y^an< Nam (m = 1,2,...)
n=l
holds for a sequence {an} of nonnegative real numbers with some positive
integer N, then oti+p > pcti for i,p = 1,2,..., where
iN
αι= Σ an (г = 1,2,...).
п=(г-1)ЛГ+1

496
3. SOLUTIONS TO THE PROBLEMS
Solution 1. Since
/(t+p-l)JV+l {i+p)N
<*i+p = QL(i+p-i)N+i Η *~α(ί+ρ)Ν>—\ 22 αη_| >" /J Qn
(г+р-1)ЛГ ЛГ 2ЛГ
JV
Ν
N ·
n=l n=l
> —AT ^ αη = ]>^αη + ^ αη + ··· +
n=l n=l n=iV+l
(г+р-1)ЛГ
+ Σ αη = αι + α2-\ \-ai+p-U
п=(г+р-2)ЛГ+1
it follows that
(Xi+p > Οίι + &2 Η h ai+p_i.
Hence ai+i > a*, and therefore
Яг+р > oti + ai+i Η Ь ai+p_i > pa^. D
Solution 2. Similar to the previous solution,
г-1
аг>^а^. (1)
k=l
We are going to prove that
г
а,+р>2^"1^а, (2)
k=l
for any г and p.
Indeed, (2) is the same as (1) for ρ = 1, and if (2) holds for p, then by
г+1 / г \ г
α,+ρ+ι > 2*"1 ^ ак = 2Ρ-1 ai+1 + ^ α J > 2*>"12 ^ a*
fc=l V k=l J k=l
it also holds for p+1. Since a* > 0, from (2) it follows that ai+p > 2p~1ai,
which finishes the solution. D
Remark. We can similarly prove the following, more general statement.
Let {an} be a sequence of nonnegative real numbers such that for some c,
n-l
Q>n > cy aj.
i=l
Define oti as we did in the problem. What is the largest constant dp such
that for all such sequences the following holds?
&i-\-p :_ CLp&i

г=1
3.9 SEQUENCES AND SERIES 497
In the problem, с = 1 /(Ν — 1). First, we will find a cp such that
η
Q"n+p ^ ^p / ; ai·
e\ = с suffices for ρ = 1. Since
an+p+i > ep Σ ai > ep(l + c) ^ α<?
г=1 г=1
it is not a bad idea to choose ep+i = ep(l + c), that is, ep = c(l + c)p_1.
Then
n-l
(г+р)ЛГ
ЛГ
гЛГ
ЛГ
п=(г+р-1)ЛГ+1 j=l n=l \j=l J
So we can choose dp such that
ΛΓ
άρ = ΣC(X + c)^"1^^'-1 = (1 + cjk-^Kl + c)N - 1].
i=i
It is easy to see that if we define a\ = 1, an = c(l + c)n~2 (n > 2), then for
any N and ρ, αρ+ι = dpai holds, so the inequality cannot be sharpened.
As stated, in the problem c= l/(N — 1), so
N
N
(ρ-Ι)ΛΓ
but even for с = 1/iV it is true that dp > 2P_1.
*>=Ш
>#-\e-l\
Problem S.17. Let S„ = Σ"=ι bjZ% (y = 0, ±1, ±2,...), where the bd
are arbitrary and the Zj are nonzero complex numbers. Prove that
\Sq\ < η max |S„|.
0<|i/|<n
Solution. Let Π"=ι(* - Zj) = ^Ук=оакгк and тахо>*>п |a*| = \an
Obviously, \am\ > 1. Then
k=0
k=0
k=0
Hence
l-Sbl =
Σ (-?№
k=0\k^m
< У^ \Sk-m\<n max ISJ. Π
, ΓΤ', 0<|i/|<n
Remark. One can show that equality holds iff b\ = 62 = · · · = bn and
the set of numbers z^ is the same as {a · е^27Гг^/^п+1^^ (j = 1,2,... ,n)},
where a is an arbitrary constant of absolute value 1.

498
3. SOLUTIONS TO THE PROBLEMS
Problem S.18. Let ρ > 1 be a real number and R+ = (0, oo). For
which continuous functions g : R+ —> R+ are the following functions all
convex?
Mn(x) =
ж = (xu ..., жп+1) G RJ+1, η = 1, 2,...
Solution. We prove that Mn (n = 2,3,...) are all convex iff д is a constant
function. If д is a constant function, then by Minkovsky's inequality,
Mn(x + y)<Mn(x) + Mn(y) feyelf), (1)
and by Mn(ax) = ctMn(x) (a G R+,:r G R++1), it follows that Mn is
convex.
Conversely, if all of the Mn are convex, then we will show the following.
If M2(x, tx, 1) is a convex function of χ for all и G (1 — δ, 1 + δ), δ > 0, then
the function д is a constant.
Taking each side of the inequality
Μ2(αχι + (1 -а)х2,щ1) < aM2(xi,u,l) + (1 - а)М2(х2,и,1)
(xi,x2eR+, не (1-М + £),ае (0,1))
to the power p and using the inequality
at + (1 - a)s < (atp + (1 - a)sp)1/p (t, s > 0, a G (0,1))
between the arithmetic mean and the n-th power mean, we obtain
Μξ(αχι + (1 - а)ж2, Щ 1) < aAff (жь и, 1) + (1 - α)Μξ(χ2, и, 1).
Deleting υΡ = αυΡ + (1 — α)υΡ from both sides, we get
д(и)(1 - vP) ^ ag(u)(l-uP) + _ a)g(v)(l - vP)
forxi,x2 G R+,uG (1 -5, l + i),QG (0,1).
If г/ < 1, then divide by g(u)(l — vP) and take the limit и —> 1 — 0. Since
# is continuous, we get
< a——г τ— + (1 - α)-
g{ctxi + (l-ct)x2)+90) ~ g{xi)+g(l) Jg(x2)+g(l)'
Using the same method for и < 1, we get the opposite inequality.
Therefore, equality holds. So the function / = l/(g(x) + #(1)) satisfies the
following Jensen's equality:
f(ax! + (1 - a)x2) = af(xi) + (1 - a)f(x2) (xu x2 G R+, ae (0,1)).

3.9 SEQUENCES AND SERIES
499
Since / is continuous and positive, f(x) = С х + D, where С and D are
nonnegative constants. So
Cx + D = * <-1т=2/(1) = 2(С + Д).
This does not hold for large values of χ when С > 0. Therefore, С = 0 and
# is constant. D
Remark. Some of the participants remarked that we do not have to
suppose the continuity of g, since it follows from the convexity of Mn.
Moreover, g is continuous if M2(x,u, 1) is a convex function of χ for и G
(1-«,! + «).
Problem S.19. Suppose that R(z) = Y^=_00 Q>nzn converges in a
neighborhood of the unit circle {z : \z\ = 1} in the complex plane, and
R(z) = P(z)/Q(z) is a rational function in this neighborhood, where Ρ
and Q are polynomials of degree at most k. Prove that there is a constant
с independent of к such that
/ \o>n\ < ck2 max \R(z

absolution. Denote by D the unit disc of the complex plane D = {z :
|ζ| < 1}, and by dD its boundary. We can suppose that \R(z)\ < 1 on the
perimeter of the unit circle. So we have to prove that J2<^>=_00 \an\ < ck2.
If we prove
-1
Σ \ап\<ск2,
n=— inf
then we can apply it to R(l/z)/z, so that we obtain Σ™=0 \αη\ < ck2,
which finishes the solution.
We know that
г
where we can integrate along dD, or along such a curve Г, such that R(z)
does not have a pole in the region bounded by dD and Г. Now we will try
to choose skillfully Г С D. By the previous results,
Σ \αη\ <^j> \R{z)\(l + \z\ + · · · + |*|* + · · ·) \dz\

500
3. SOLUTIONS TO THE PROBLEMS
We are looking for а Г for which the integrand is not too big in the
rightmost integral. Because the multiplier is 1/(1 — |z|), we will have to
bring Γ as far from the unit circle as possible. The only problem is that
we do not know anything about the behavior of R{z) far away from dD.
Denote the poles of R in the unit disc by <?i, q2,..., qn, where the points
are included in the sequence with appropriate multiplicity. Obviously, h <
k. The Blaschke product
4u
"w-IIrfi
i/=l
qvz
is holomorphic in D, vanishes at the points qv, and has absolute value
not more than 1 in the points of the unit circle. So S(z) = R(z)Q(z) is
holomorphic in D and has absolute value not more than 1 in the boundary
of the unit circle. Thus, by the maximum modulus principle, \S(z)\ < 1,
for ζ e D, and so
1ВД1 <
Π Λ z-qv
(N<i).
(2)
This is the approximation we were seeking.
Let
T = dizeD :
If гбГ, then by (2),
z-qv
1 -qvz
> 1-
fe+1
1 <v<hz
Ί·
«•^(ГГ^-^^У
< e.
(3)
The curve defined above is not necessarily a Jordan curve, but it consists
of finitely many closed Jordan curves. Nonetheless, R(z) will be
holomorphic in the region bordered by Г and dD. So all of the consequences of (1)
will hold.
Obviously, Г consists of some subcurves of the curves
\ = U :
■Чи
\-qvz
fc+l/
If qv = 0, then Г^ is a circular arc. But it is also a circular curve when
qv Φ 0, since Γ^ is obtained from the circular arc \w\ = k/(k + 1) by the
transformation
w=~,—=;, (4)
1 -qvz
and such a transformation maps a circle onto a circle. (Of course, qv
is not the centerpoint of IV Moreover, if we are considering Q as the

3.9 SEQUENCES AND SERIES
501
: |z| = li
Figure S.l.
Poincare model of the hyperbolic geometry, then the transformation (4) is
a conformal transformation of the hyperbolic space, so qv will be the non-
Euclidean center of IV The non-Euclidean radius of Γ^ is approximately
log/с, see Figure S.l.)
Since Г С ujT,,, by (3),
Σε / \dz\ e v-^ f \dz\
n=-oc r v=lTv
The latter sum can be approximated by substituting (4) and some
calculations as
/ \dz\ ^ ft Г \dz\ ft Г \dw\ ^ 2 ft Ω 7
/1-2 J l-\z\2 J l-\w\2 i_(J^Y
И=т#г
(Here we had to do "some calculations" to prove the equality above.
The equality also shows that curve |dz|/(l — \z\2) is invariant under the
conformal transformation of the hyperbolic plane.) So
-1
2_. \an\ < 4efc2,
n=—oo
and as we noted above it follows that
oo
Σ \an\ < 8efc2. D
Remark. The inequality above is not sharp. Gabor Somorjai showed
with a better curve Γ that
y] \an\ < cklogk.

502
3. SOLUTIONS TO THE PROBLEMS
Problem S.20. Let Kn(n = 1,2,...) be periodical continuous
functions of period 2π, and write
kn(f\x) = / .
Jo
f(t)Kn(x-t)dt.
Prove that the following statements are equivalent:
(*) JT IM/'»*) - /0*01 dx -> 0 (n -> oo) for all f e Li[0,2tt].
(ii) kn(f;0) —> /(0) for all continuous, 2n-periodic functions f.
Solution 1.
(i) —>(ii): Denote by C[0; 2π] the space of the continuous functions under
the supremum norm. Let us examine the sequence of operators kn(·; ·) :
||fen(.;.)||= sup ||M/;-)IUi
ll/IUi<i
(1)
2π
2π
2π
< sup / |/(t)| dt / \Kn(x - t)\ dt< \Kn(t)\ dt.
Миг<1{ { {
On the other hand, the functions
h ifxG[0;«],
χ β (<5;2π),
by the continuity of Kn(t), satisfy
2π
Λ0
\0, if:
IM/i;OIUi-*/l«n(i)l* (*->o),
and so
2π
ΙΜ·;·)ΙΙ = /ΐ^η(«)|Λ.
By applying the Banach-Steinhaus theorem to the sequence {fcn(·; ·)},
2π
we obtain f \Kn(t)\ dt < C, where С is the same constant for all n.
о
If / is continuous, then
\kn(f;x)-kn(f-x-\-h)\ =
2π
f(№-f(t-h))Kn(x-t)dt
<Csup\f(t)-f(t-h)\,
and so the functions kn{f;x) are equicontinuous.

3.9 SEQUENCES AND SERIES 503
Let us suppose that (ii) does not hold. Then there is a continuous /, an
ε > 0, and a sequence {vn} such that kVn (/; 0) > /(0) + ε for η = 1,2,... .
By the uniform continuity of {fc^ (/;#)}, there is a δ such that, for all n,
fc„n (/; x) > f(x) + | if 0 < χ < 6П(п = 1,2,...). Hence,
2π 6
У'|ft„n(/;*)-/(*)|<fc>у'(Μ/;*)-/(*№>*! (n = i,2,...),
о о
but this contradicts (i).
(ii)—>(i): It is well known that the norm of the linear functional kn(-; 0) :
С[0;2тг] —> R is /0π |ifn(t)|dt. So by the Banach-Steinhaus theorem,
JQ π |ifn(£)|d£ < С (n = 1,2,...), where С is a constant. Since
Γ2π
left
kn(f, x0) = / f(t)kn(x0 - t) <
Jo
ρ2π
= / /(s0 + t)Kn(-t) dt = kn(f(x0 + t); 0)
Jo
is continuous for all /, kn(f;x) —> /(#) holds everywhere. But kn(f;x) <
Csup(/), so by Lebesgue's theorem,
L
2π
|fen(/,a?)-/(s)|ds->0 (n->oo, /eC[0,2tt]).
Let / G Li[0,2π] and let ε > 0 be arbitrary. Choose a continuous
function g in such a way that \\g — f\\b1 < ε holds. Then, by (1),
/ \kn(f,x)-f(x)\dx< / \kn(f,x)-kn(g;x)\dx
Jo Jo
η2π
+ / \kn(g\x)-g(x)\dx,
Jo
ρ2π
/ \g(x)-f(x)\dx<C\\g-f\\Ll
Jo
r2n
+ / IM0;z)-0(z)|dx+||0-/||i,i.
Jo
By the previous argument, the middle term above goes to 0 if η —> oo, so
/»2π
/ \kn(f,x)-f(x)\dx
Jo
becomes arbitrarily small for sufficiently large n. This finishes the
solution. D
Solution 2. We are going to use the following version of the Banach-
Steinhaus theorem. Let X and Υ be Banach spaces, and let A, An : X —> Υ

504
3. SOLUTIONS TO THE PROBLEMS
(n = 1,2,...) be linear bounded operators. The sequence {An} pointwise
converges to A iff there is a constant С and a closed system S С X such
that
Pn||<C (n=l,2,...), (1)
Anx —> Ax (n —> oo) (2)
hold for all χ e S. (S С X is called a closed system if the linear hull of S
is dense in X.)
Let A\ : L\ —> Li be the operator fcn(·;·), A1 be an identity on Li,
A2n : С[0,2тг] -> R be the functional Α:η(·;0), and A2 : С[0,2тг] -> R be
the mapping / —> /(0). As we showed in the previous solution, ||-A^|| =
\\А„\\ = J0n \Kn(t)\dt, so (1) holds for {A\} and {A^} at the same time.
Let S = {em = etrnx}m=i- Since S is closed in L\ and in C[0,2π], and
/»2π /»2π
(^em)(x) = / eimtKn(x -t)dt= eim^x+t^Kn(-t) dt
Jo Jo
Ρ2π
пгтх
/ eimtKn(-t)dt = em(x)A2nern,
Jo
it follows that
/»2π
||^nem —-^ em||Li — / \em\x)Anem — em(^J| uX
Jo
ρ2π
= \An€"rn ~ A em\ I \em\x)\dX
Jo
= 2n\\A2nem-A2ern\\R,
so (2) holds for {A\} and A1. Furthermore, (2) also holds for {A^} and A2
simultaneously. So we proved the equivalence of the pointwise convergences
A\ -> A1 and A\ -> A2. D
Problem S.21. Let us assume that the series of holomorphic functions
oo
У^Л(я) is absolutely convergent for all ζ e С Let Η С С be the set of
k=l
those points where the above sum function is not regular. Prove that Η is
nowhere dense but not necessarily countable.
Solution. Let gn(z) = Σ%=ι Mz) and #(*) = ΣΤ=ι Μζ)· Then 9n(z) is
holomorphic and gn(z) —> g(z) for all ζ G С. То prove the first part of the
statement of the problem, we have to show that inside each circle К of C,
there is a circle к in which g(z) is holomorphic. Let φ(ζ) = supn |<7п(г)|·
This exists by the convergence and is finite for all ζ e С Let Ak = {z e
K\k — 1 < φ(ζ) < к} (к = 1,2,...). By Baire's theorem, there exists
an N such that An is dense inside some circle к С К. If φ(ζο) > N for

3.9 SEQUENCES AND SERIES
505
some z0 e k, then |pn(z0)| > N for some n, and by the continuity of gn(z)
there is a neighborhood of zq where also φ(ζ) > TV, which contradicts the
fact that An is dense in k. So φ(ζ) < N if ζ e k, and so 1^(^)1 < Ν Ίΐ
ζ e k. By the Vitali-Montel theorem, there is a subsequence of {gn{z)}
that is equiconvergent inside k. Since gn{z) is convergent, this subsequence
converges to g{z). Hence, by Weierstrass's theorem, g(z) is holomorphic
inside k. So we proved that Η is nowhere dense. D
RemarL·.
1. In the solution, we did not use the fact that the series of functions
ΣΤ=ι fk(z) 1S absolutely convergent. To prove the second part of the
statement of the problem, we will need the following result.
Mergelyan's theorem. Let К be a compact set of C, such that its
complement is connected, and let / be is a continuous complex function
on K, such that it is holomorphic inside K. Then for any ε > 0 there
exists a polynomial P, for which \f(z) — P(z)\ < ε for all ζ e K. (See
W. Rudin, Real and Complex Analysis, McGraw-Hill, London, 1970.)
Let A = {z\lm ζ < 0}, В = {z\lm ζ > 0}, С = {ζ: ζ is real}. There
are series of compact sets {An}, {Bn}, and {Cn} (n = 1,2,...), such
that
An С An_|_i, Bn С #n+i, Cn С Cn+i,
C\(An U Bn U Cn) is connected for η = 1,2,...
and
\JAn = A, \J~Bn=B, \J~Cn = C.
1
So An U Bn U Cn is compact, its complement is connected, and
(z)= ( 1, ifzeAnUBn,
Hence, by Mergelyan's theorem, there is a polynomial Р£уП for any ε > 0
and n, such that |1 — Pe>n| < ε, Ίϊ ζ e AnU Bn, and |Pe>n(z)| < ε, if
zeCn (n= 1,2,...). Let
fi(z) = Qi(z), f2(z) = Q2(z)-Qi(z),..., fk(z) = Qk(z)-Qk-i(z),
where Qn(z) = P(i/2n)-n(z) (n = 1,2,...) is a series of polynomials.
Then Σ2=ι/*(*) = <?»(*), so
^> ( 1, if ζ is not real,
!>(*) = {0> if2isreal.
Here if is the set of real numbers, and is hence uncountable. On the
other hand, for any ζ eC there exists an N such that ζ e AnUBnUCn
if η > Ν, so in this point z,
oo oo oo 1
Σ \f*+k(z)\ = Σ \QN+k(z) ~ QN+k-l(z)\ < Σ 22N+k-l < °° 5
k=l k=l k=l
hence, ΣΤ=ι fk(z) 1S convergent at z.

506
3. SOLUTIONS TO THE PROBLEMS
2. We can prove a stronger statement: the measure of Η can be positive.
Let R be the real axis, / the imaginary axis. It is enough to show that if
S is a nowhere-dense subset of /, then there is a sequence {/(^)}^Li of
holomorphic functions that satisfies the conditions of the problem and
R χ S contained in the Η set obtained from this sequence {f(n)}™=1.
So let S be a nowhere-dense, open set of /. Then I\S~ is a dense set
of /, so it is a union of countably many disjoint intervals; let these be
ik = (a*., bk) (к = 1,2,...). (We can suppose that ak and bk are finite.)
Let {r\ }^i be monotone decreasing, {r\ }fZ\ monotone increasing,
such that 7*i < l[ and r\ —> ak, s\ ' —> bk if / —> oo. Let {an}™=1 be а
sequence of natural numbers, such that it contains each natural number
infinitely many times. Let η be an arbitrary number. Let an = к,
denote by I the cardinality of {m: m < n, am = к}, and let
Kn = {z
An = {z
Bn = {z
Ln = {z
Mn = {z
The set Pn = Kn U An U Bn U Ln U Mn is compact, and its complement
is connected in C~. So by Runge's theorem, there is a polynomial fn
such that
Ι/η(*)Ι<2ί if zeKnULnUMn
and
\fn(z)-n\<— if zeAnUBn.
So
(i) The functions fn are polynomials, and hence holomorphic on С
(ii) For any ζ G C, there exists an N0 such that ζ G Kn U Ln U
Mn for any η > Nq. So |/n(^)| < l/2n f°r any η > щ, and
Σ~=ιΙ/η(ζ)Ι<°°·
(iii) Let z0 G R x S arbitrary. Let ε > 0 and Ue = {z\\z — z0\ <
ε}. There is an index ко such that ί7ε/2 Π (R x гко) φ 0; hence,
Ue Γ\{ζ: Imz = ako} φ 0 or Ue C\{z\ Imz = bko} φ 0.
Let us suppose that Ue Π {ζ: Imz = ako} φ 0 and z\ G U£ Π {ζ: Imζ =
α*;0}. (The other case can be discussed similarly.) Let n0 > max(|Re2:o| +
ε, llm^ol + ε) such that if η > щ and an = ко, then Kn Π f/e Φ 0. Let
#n0 — Ση1ι#η· Since #ηο is continuous at zi, there is a 0 < <$ < ε such
that for all ζ G Vg = {z: \z — z\\ < <$}, \gno{z) — gni\ < 1· Let Πι > no such
that ani = k0, m > 5 + 2|#no(zi)| and Κηι Π Vs φ 0. Let λχ G Αηι Π t/$
and λ2 G dKni Π C/^, where 9i^ni denotes the boundary of Kni.
r2l-l < ^mZ — S2/-l' \R<ez\ < nL
1шг = Г2| ' |Rez|<n},
lmz = S2l ' |Re2;|<n},
r2/-i — ^mz — mm(afc — n, —n)? |Rez| < nL
s2/-i — ^m2; — т^х(^ + n,n), |Rez|<n}.

3.9 SEQUENCES AND SERIES 507
If η > n0 andn^m, then |/η(λι)| < l/2n and |/ηι(λι) - щ\ < 1/αηι.
So
|/(λι) - дп(гг) - щ\ < \gno(\i) - 9η0(ζι)\ + l/m(Ai) - гц|
oo 1
+ Σ /„(λ!)<ι+ Σ ^<2.
η>ηο,η^πι η=ηο + 1
On the other hand, |/η(λ2)| < l/2n if η > no, so
oo
Ι/(λ2) - ft,0(*i)| < Ι9η0(λ2) - 9η(ζι)\ + Σ Ι/»(λ2)Ι ^ 2·
η=ηο + 1
Hence,
|/(λι)| > \gno(zi) + ηι| - |/(λι) - 9ηο(Ζι) - m| > щ - IWi)! - 2
and
|/(λ2)| < |/(λ2) - gno(Zl)\ + \gno(Zl)\ <2+\gno(Zl)\.
So
|/(λι)|-|/(λ2)|>η1-4-2Λΐ0(ζ1)>1.
Therefore, the total variation of |/| is more than 1 in Ue. Since ε > 0 is
arbitrary, |/| is not continuous at zq. So |/| is not continuous in the points
of R χ S. That is, {/n}^=i satisfies the conditions of the problem, and the
set H, obtained from {f(n)}^>=1, contains R χ S. By this, the statement
of the remark is proved.
Problem S.22. Let f(x) be a nonnegative, integrable function on
(0,2π) whose Fourier series is f(x) = α$ + Σ™=ι dk cos(rikx), where none
of the positive integers nk divides another. Prove that |α*| < a0.
Solution. Let Kn(x) = 1/2 + (1 — ^-j-) cos:r +... be the nth Fejerkernel.
It is known that Kn{x) > 0. By integrating, we get 0 < /0 π f(x)Kn(rikx) =
π(α0 + (1 - l/(n+ l))anfc), so if η —> oo, then аПк < a0. Kn(x - π) =
1/2- (1- 1/(п+1))со8ж+..., so
2π
0 < / f(x)Kn(nkx - π) = π ί α0 - Μ — J аПк \ .
о
Therefore, if η —> oo, then anfc < —ao- □

508
3. SOLUTIONS TO THE PROBLEMS
Problem S.23. We are given an infinite sequence of Vs and 2's with
the following properties:
(1) The first element of the sequence is 1.
(2) There are no two consecutive 2's or three consecutive 1 's.
(3) If we replace consecutive Vs by a single 2, leave the single 1 's alone,
and delete the original 2's, then we recover the original sequence.
How many 2's are there among the first η elements of the sequence?
Solution. Since, by the above conditions, the first η — 1 elements of the
sequence determine the nth one, there can be only one sequence. It is
easy to see that the following sequence of numbers 1 and 2 satisfies the
conditions of the problem: in this sequence, the index of the nth 2 is larger
by 2 or 3 than the index of the (n — l)th 2 iff the nth element of the
sequence is 1 or 2, respectively.
The following condition is equivalent to (3):
(3') If we change all l's into 12 and all 2's into 112, then we obtain the
original sequence.
Denote by /(n) the number of 2's among the first η elements. We are
going to prove that, for all n,
/(/(n) + 2n) = n, (a)
/(/(n)+2n-l)=n-l, (b)
/(/(n) + 2n + 1) = n. (c)
To prove this, change the first η elements of the sequence as described
in (3'). Hence, we obtain 3 · f(n) + 2 · (n — /(n)) = /(n) — 2n elements,
such that the number of 2's is n, so we proved (a). By (3'), this /(n) — 2n
ends with 12, and by (2), the (/(n) — 2n + l)th element is 1, so we obtain
(b) and (c). Obviously,
/(1) = 0 and /(2) = 1. (d)
Now we are going to prove that if function д : N —> N satisfies the
conditions (a)-(d), then д = f.
f(n) = g(n) follows from (d) for η = 1,2. Let us suppose that we proved
f(k) = g(k) for all к < η. Since ft(fc) = f(k) + 2fc, h(k + 1) - h(k) < 3, and
so there is а к < η natural number for which η = f(k) + 2k + εη, where
εη = 0, ±1. Then, by (a), (b), or (c) and the induction,
/(n) = /(/(fe) + 2fc + en) = |*'_i
Moreover,
9(n) = g(f(k) + 2k + εη) = g(g(k) + 2k + en) =
( fe, if en = 0,1,
\ k- 1, if εη = -1.
Hence, /(n) = g(n).

3.9 SEQUENCES AND SERIES
509
Finally, we are going to show that g(n) = [(\/2 — l)n + 1 — 1/у/Щ
satisfies the conditions (a)-(d), so f(n) = [(л/2 — l)n + 1 — l/y/Щ holds.
Obviously, #(1) = 0 and #(2) = 1. To verify (a), write g(n) in the form
g(n) = (y/2-l)n+l-l/y/2]-e(n), where ε(ή) G [0,1]. (However, e(ri) = 0
cannot hold, but we do not need this fact.) Hence,
g(g(n) + 2n) = [(VS - 1) (VS - l)n + 1 - ^= - e(n) + 2n) + 1 - -J=]
= [n + (l-e(n))(V2-l)]=n.
We can similarly verify (b) and (c). D
Remark. After writing down the first few elements of /(n), it becomes
plausible that f(n) is the integer part of some linear function. If we
substitute the function f(n) = [an + b] into (a) and (b), we obtain that (a)
and (b) hold only for the pair a = y/2 — 1, b = 1 — l/y/2 obtained in the
solution.
Problem S.24. Let αο, αχ,... be nonnegative real numbers such that
oo
Σ αη = 00 .
n=0
For arbitrary с > 0, let
п0(с) = min < к : с · j < ^ аЛ , j = 1,2,... .
Prove that if Σ°10 α? < oo, then there exists aoOfor which Σ°°=ι ап0(с) <
oo, and if Σ°^0of = oo, then there exists a с > 0 for which Y^JLi anj(c) —
00.
Solution. Let
к
= Σα*'
Sk
i=0
and denote by Xkj the characteristic function of the interval Ikj =
[(l/j).ak-u(l/j)'ak]. Let
oo
/(c) = Σ <Ч (c); ^: (°' °°) "" (°' +0°] *
i=i
Since Xfc,j(c) = 1 iff rij(c) = /c, the function / = J2kj UkXkj is Lebesgue-
measurable. So
/ f(c)dc= J2 akX(IktjU[a,b]).
J I A
k=oo ,j=p

510
3. SOLUTIONS TO THE PROBLEMS
We are going to give a lower and upper estimate for the value of the integral.
Lower estimate. We sum over only those pairs k, j for which
Ik j С [α, 6],
so if
or, equivalently,
Note that
So
b
Jm
a
dc>
1 1
α < -Sk-i < -Sk < o,
J J
sk . sk-i
b ~J - a
Wkj) = у ·
Sk-i/a
oo л oo /» ι
fc=° ' fc=° (.*/*) + !
Va?in Sfc~l/a =Va?in 6gfc-1
f0 Wb + i ifcs <"* + «*■
Upper estimate. We sum over only those pairs /c, j for which
4,,· Π [α, 6] ^0,
so if
1 . . Sk
-Sk-i < b and a < —,
J J
or, equivalently,
Sk-l . . . Sk
~ΊΓ <·? - T'
ο α
then
//(c)dc<f;a2fcj)^f;e2(1 + | +
a /с—О у J J k—0
In- 6Sfe
ask-i + aft
If limsupan = a > 0, then an > a/2 for infinitely many n. So J] a£ =
oo. Since infinitely many terms of the sum Σ°°=ι o>nj(a/2) are larger than
an, /(a/2) =oo.
So let us suppose liman = 0. Then sn/sn-\ —> 1. Therefore,
limln :L— = In - > 0,
ask + ab a
lim In = In -.
ask-i + ab a

3.9 SEQUENCES AND SERIES 511
So there exist real numbers A> О, В > 0, С > 0, and D such that
So if Σ al < °°? then JT f(c)d>c < oo· Therefore, /(c) < oo almost
everywhere, and if J] a£ = oo, then /6° /(c) dc = oo. So for any a < 6, the
set
oo
is a dense set for an any ra.
However, the function an.(c) is a left-upper-semicontinuous function of
c, so the following sum is also left-upper-semicontinuous:
oo
Therefore, the set {anj(c) > ™1 is dense and open. So the set {/ = oo}
is of second category and therefore is not empty. D
Problem S.25. Let 2/(y/b +1) < ρ < 1, and let the real sequence {an}
have the following property: for every sequence {en} ofO's and ±l's for
which Σ™=ι enPn = 0, we also have Σ™=ι епап = 0. Prove that there is a
number с such that an = cpn for all n.
Solution. We start with a definition.
Definition. Let L denote the set of all positive and strictly decreasing
sequences {ln} for which L = Σ™=ι ^η < oo. We call such a sequence
interval filling if for every χ G [0, L] there is a sequence εη G {0,1} with
ΣΟΟ j
n=l εηΙ"η-
We need three lemmas.
Lemma 1. If a sequence {ln} G L satisfies
oo
λ„< Σ Xi (« = 1.2,·..),
i=n+l
then it is interval filling.
Proof. For χ G [0, L] we define the numbers en inductively as follows:
{*> if YJi=i £ixi + λη < s;
0, if E?=i4Ai>x.

512
3. SOLUTIONS TO THE PROBLEMS
For every η for which εη = 0,
oo n—1
0 < X - ^2 £i^i < X ~ Σ, £i^i < K,
г=1 г=1
hence if there are infinitely many such n, then χ = Σ*£=1 ε* limn_»0 k· If,
however, there are only finitely many such n's, then for the largest one
n—1 oo oo
x ~ / J £гЛ{ < An < у ^ Xi = у ^ 6i\i,
i=l i=n+l i=n+l
from which
oo
n=l
so our claim holds even in this case.
Lemma 2. For 2/(>/5 + 1) < ρ < 1, the sequence
p, p2, p3, ...
is interval filling, and for any natural number N, the sequence
is interval filling too.
Proof. By Lemma 1, it is enough to verify that
oo
pn < Σ pi for a11 neN
i=n+l
and
oo
pn_1 < Σ pi fora11 n€N
i=n+l
are satisfied. The first of these is equivalent to 1 < p/(l — p) and the second
one to 1 < p2/(l — p), hence each of these is true if 2/(л/5 + 1) < ρ < 1.
Lemma 3. Let A and В be nonempty disjoint sets such that their union
is the set of the natural numbers, and let 2/(\/5 + 1) < ρ < 1. Then there
exist nonempty sets А' С А, В' С В with the property
Σρί = Σ^ (i)
t£A' i€B'
Proof. Let L = p/(l — p) and χ = Σί£Αρι- Let us choose an N G A
for which χ < L — pN. If A is finite, then N can be the largest element in

3.9 SEQUENCES AND SERIES 513
A; otherwise, we can choose any element in A with sufficiently large index.
Since Lemma 2 says that the sequence
is interval filling, there is a set С С Ν \ {Ν} for which
г£С г€А
Let A' = А \ С, В' = С \ A. Then (1) clearly holds, and because N e A',
the set A' is not empty. From this and (1), it follows that B' is not empty,
either.
After these preparations, we turn to the proof of the statement in the
problem. For every с Ε R, the sequence a'n = an — cpn also satisfies the
assumptions of the theorem. Suppose, on the contrary, that the claim is
not true. Then we can choose a real с for which there is an η with a!n > 0
and also another one with a!n < 0. Let
A = {n e N : a'n > 0}, В = {η e N : a'n < 0}.
On applying Lemma 3, we get two sets A' and B' with the properties stated
there, and with the aid of these we define
Г 1, if η e A!\
en = < -1, if η e Bf-,
0, otherwise.
Then
but
г=1 i£A' i€B'
г=1 i£A' i£B'
and this contradicts the hypothesis of the problem. The contradiction
obtained proves the claim. D
Remark. The result is true for any 0 < ρ < 1 (see Z. Daroczi, I. Katai,
T. Szabo, On Completely Additive Functions Related to Interval-filling
Sequences, Arch. Math. Ц (1990), 173-179).
Problem S.26. Let S be the set of real numbers q such that there is
exactly one 0-1 sequence {an} satisfying
71 = 1
Prove that the cardinality of S is 2**°.

514
3. SOLUTIONS TO THE PROBLEMS
Solution. Let {bn} be a sequence such that 61 = 62 = 1, &3n = 1, &3n+i —
0, and 6зп+2 = 0 or 1 (n = 1,2,...). There is exactly one number q > 1
such that
Such a q exists, since the series Σ™=ι bnxn is convergent if \x\ < 1,
continuous, it takes the value 0 at χ = 0, and goes to infinity if χ —> 1—.
We will prove that if there is a sequence {an} with elements 0 or 1 and
00
Σαη _ ,
n=l 4
then an = bn for all n.
Let us suppose that this is false. Let к be the smallest number such that
ak^bk. Then
00 ,
On
nn /^ nn '
Since
qn *-^ qn
n=k n=k
Qn ~~ Q Q2 Q3n qb — q2 '
n=l n=l
that is, <?5 - <?4 - <?3 - 2g2 + q + 1 > 0 holds,
<?2 + <? = 1 g3-g2-g-i=1 (g3-g2-g-i)(g3-i)
g3-l 93-l (93-l)2
_t g(g5-g4-g3-2g2 + g+l) + l
(<?3-i)
So
oo 1 1
n=l ч
Suppose that bk = 0 and a^ = 1. If η > 1, then at least one of 6n, 6n+i,
and 6n+2 is zero, so
bn frn+i ^n+2 < J_ 1
Hence
^n on+l дП+2 ^n on+l
y>6n= у- ^<\V I + 1 "I
L·^ an Δ^ί an — L·^ \ Qk+3m+l Qk+3m+2 J
п=кЧ п=к+1Ч тп=0 ^Ч Ч '
jk L·^ \ q3m-2 ^ дЗт-l J gk — L^t qU '
τη—ί η^=κ
9
and this is a contradiction

3.9 SEQUENCES AND SERIES
515
Otherwise, suppose that bk = 1 and α& = 0. Since at least one of 6n,
6n+i, and 6n+2 is 1,
bn frn+i frn+2 > 1
qti on+l nn-\-2 qn+2 '
SO
ν^>_! v_l_-i.ii у l )
>^ > лП q\z / > лП+Зтп л/с ι / > β3τη Ι
n=fc v v m=l v ч \ m=l v /
ι f^( ι M'fiLifi
*fc I Z-/ l дЗтп-2 + дЗтп-l J """ Z-/ дЗтп I gk 2^ дП
>
qK
Ου ou
n=fc+l Qn n=k Qn '
and this is also a contradiction. So only the sequence {bn} satisfies the
conditions.
The cardinality of different sequences {bn} is 2**°. If two {bn} sequences
are different, then the corresponding numbers q are also different. So the
cardinality of S is maximum 2**°. But it cannot be more than this, since
the cardinality of the set of all real numbers is 2**°. D
Problem S.27. Given an > αη+ι > 0 and a natural number μ, such
that
hm sup < μ,
η βμη
prove that for all ε > 0 there exist natural numbers N and щ such that,
for all η > no the following inequality holds:
η Νη
k=l k=l
Solution. We have to find a natural number no such that for any ε > 0
there exists a natural number N such that
£ Efc=l a*
holds for all η > no, where condition (1) holds.
Choose a real number ν in the interval (limsupn(an/an/Lt),/i) . Define а
natural number Μ such that αη/αημ < и, for all η > Μ. Therefore, for all
natural numbers I > 0 and n> M, the inequality αη/νι < αημι holds.

516
3. SOLUTIONS TO THE PROBLEMS
First, we show that the series Σ^=1 ak 1S divergent. Obviously,
Μμι Μμι Μμι .
k=l k=l k=l
Here Μ and ам are fixed, so when Ζ —> oo, then from ν < μ it follows that
the right side of the inequality also goes to infinity, that is, we estimated the
series Σ£=1 ak from below by a divergent series. This divergence provides
us with a number K, for which J2k=1 a>k < Σ^=μ+ι ak\ so f°r any n> K,
T,k=iak ^2Efc=M+iflfe·
For given ε, choose / such that 1/ε < (μ/и)1/2. Therefore,
ε 2 2Ζ^=Μ+ια* 22^=Μ+ια*
Σημ — 1 / ν-^ημζ— 1 n;y
fe=MM^Q(Lfe/^J+i)/i < 1^к=м^ак <Efc=iQfe n

3.10 TOPOLOGY
517
3.10 TOPOLOGY
Problem T.l. Prove that any uncountable subset of the Euclidean n-
space contains an uncountable subset with the property that the distances
between different pairs of points are different (that is, for any points Pi φ
P2 and Qi Φ Q2 of this subset, P\P2 = Q1Q2 implies either Pi = Q\ and
P2 — Q2, or P\ = Q2 and P2 = Q\). Show that a similar statement is not
valid if the Euclidean η-space is replaced with a (separable) Hilbert space.
Solution. For the proof of the first statement of the problem, we say
that a subset of the Euclidean η-space En has property Τ if the distances
between all pairs of points of this subset are different. By induction on
n, we prove that if all subsets of a set Я(С En) that have property Τ
are countable, then Я is countable. The case η = 0 is trivial. Suppose
that the statement is true for η — 1(> 0). Let Я(С En) be a set such
that all subsets with property Τ are countable. By Tukey's lemma, there
exists a maximal subset Μ of Я with property Т. By maximality of M,
any point of Η \ Μ either has equal distance from two distinct points of
Μ or has a distance from a point of Μ that equals the distance between
some pair of points in M. Therefore, Я is covered by the perpendicular
bisector hyperplanes of pairs of points of M, together with the spheres
centered at and passing through points of M. The set of these hyperplanes
Fi and spheres Si is countable (г = 1,2,...), since Μ is countable. Each
Я Π Sj is a countable set. Indeed, the preceding argument can be applied
to Я Π Sj instead of Я (since Я П Sj satisfies all assumptions made on Я);
therefore, the set of S^'s and F/'s that replace the preceding S^'s and F^s
and cover Я Π Sj is countable; by the induction hypothesis Η Π F[ and
(since Sj Π S[ is contained in a hyperplane of En) HC\SjC\S[ are countable
sets. Thus, Η Π Sj is countable since it can be covered by a countable
family of countable sets. Further, by the induction hypothesis, each Η Π Fj
is a countable set. Therefore, Я is a countable set since it is covered by a
countable family of countable sets.
In order to prove the second statement of the problem, it suffices to
construct an uncountable subset of the Hilbert space spanned by the or-
thonormal basis e* (г = 1,2,...), such that the set of distances between
all pairs of points in this subset is countable. To this end, consider an
uncountable set 21 of infinite sets of natural numbers with pairwise finite
intersection. It is well known that such a set 21 exists. (For instance, the
set of bounded, monotone sequences of rational numbers has this
property.) Each set A(e 21) determines a vector XXi(V2') *
ea, in the Hilbert
space, where a* runs through A in increasing order. The set К of these
vectors is uncountable. On the other hand, the distance between any pair
of elements of К is the square root of a rational number, for if the vectors
α = Σ°^!(1/2Ζ) · eQi and α' = Σ^ι(1/2Ζ) * e'a. are determined by the sets
A and A! (e 21), then

518 3. SOLUTIONS TO THE PROBLEMS
\\a - a'f = \\a\\2 = \\a'\\2 - 2(a,a') = \ - 2 У J-,
where in the last term we have a finite sum since Α Π A' is finite.
Remark. Istvan Juhasz and Bela Bollobas pointed out that a similar
argument proves the following generalization of the first statement of the
problem: If τη is a regular cardinal number, then any subset of
cardinality m of the Euclidean η-space has a subset of cardinality τη in which all
distances between pairs of points are different; further, they considered
generalizations of the statement to other metric spaces.
Problem T.2. A sentence of the following type is often heard in
Hungarian weather reports: "Last night's minimum temperatures took all
values between —3 degrees and +5 degrees." Show that it would suffice to
say, "Both —3 degrees and +5 degrees occurred among last night's
minimum temperatures." (Assume that temperature as a two-variable function
of place and time is continuous.)
Remark. The formulation of the problem allows for various models. The
proof is simplest when the country is assumed to be compact; not assuming
compactness requires a different proof and yields a more general theorem.
The argument needs some modification if the time interval is replaced with
a (not necessarily metrizable) compact space. In the following solutions,
the space is connected and the time is compact. The proof is carried out
for metric spaces in the first one and for arbitrary topological spaces in the
second one.
Solution 1. Since any continuous image of a connected space is connected,
and in the real line only the intervals are connected sets, it is sufficient to
prove that the function that assigns to each point the minimum
temperature attained there during the night is continuous.
Let Ε be a connected metric space, / be a compact metric space, R
be the real line, and / : / —> R be a continuous function. For any fixed
χ e E, the function /(#, t) is continuous on the compact space /; therefore,
g(x) = mmt£i f(x,t) exists. We show that g(x) is continuous. If this
were not the case, then there would exist an ε > 0 and a sequence {xk}
(xk G E, к = 1,2,...) such that x^ —> #, but
\g(xk)-g(x)\ >ε.
We consider two cases:
(a) There is a subsequence of {xn} such that
g(xnk) <9{x)~e.
If f(xnk,tk) — 9(xnk), then the sequence {^} has an accumulation
point t e /, and we can assume that tk —> t. Then
f(?nk,tk) = 9(xnk) <g{x)-e<f{x,t) -ε,
and / cannot be continuous.

3.10 TOPOLOGY
519
(b) There is a subsequence of {xn} such that
g(xnk) >g(x) + e.
If /(#, t) = g(x), then
f(xnk,t) > д(хПк) > g(x) +ε> f(x,t) + ε,
and / cannot be continuous.
Since at least one of the cases must hold, we have a contradiction, which
proves the statement. D
Solution 2. Now let Ε be a connected topological space and / be a
compact topological space. We keep other notations of Solution 1.
We prove again that g(x) is continuous on E. For ε > О, и G Ε, and
υ G /, let U(u, v) and V(u, v) be respective neighborhoods of и and ν such
that for all (я, t) G U(u, ν) χ V(u, v), we have
|/(tx,t;)-/(^,t)| <ε.
For и fixed, the neighborhoods V(u, v) (v G /) cover /. Then, by its
compactness, / is covered by a finite number of them:
V(% ^i), V(u, v2),..., V(u, vk).
Then U = Di=1U(u, Vi) is a neighborhood of u, and if χ G C/, then \g(x) —
g(u)\ < ε, for if t G /, then t G V(u, Vj) for some j, and then, since
we have
(x, t), (u, t) G U(u, Vj) χ V(u, Vj),
\f(x,t)-f(u,t)\<e. D
Remarks.
1. Assuming space and time to be compact topological spaces, Attila
Mate considered temperature to take values in a metric space and
proved, under these conditions, that the function g(x) is continuous.
Compactness of space is not essential in this model.
2. Gyorgy Vesztergombi proved that the statement of the problem
remains true if the definition of night depends on place (astronomical
night), provided that the beginning and the end of the night are
continuous functions of place.
3. Miklos Simonovits gave an example to show that it is essential to
assume compactness of night. (If the night is not compact, then, of
course, we have to speak of local infimum instead of local minimum.)
Let Ε = [-1. + 1], / = (-oo, +oo), f(x, t) = e"'2*2. Then
g(x) =inf/(x,t) = I
for χ = 0,
Ге/',ч~'"/ [0 for χ ^0,
and g(x) is not continuous.

520
3. SOLUTIONS TO THE PROBLEMS
4- Juhasz and Posa gave examples that show that it is not sufficient to
assume partial continuity of f(x,t). Juhasz's example is the following:
put Ε = I = [-1,+1], and
f(x,t) = {
In this case,
1 if χ < 0, or χ > 0 and t > 0,
1 + 1 if χ > 0 and -x < t < 0,
1 + f if χ > 0 and -1 < t < -x.
( 1 for χ < 0,
0 for χ > 0
is not continuous, and the statement of the problem is not true.
Problem T.3. Let A be a family of proper closed subspaces of the
Hilbert space Η = I2 totally ordered with respect to inclusion (that is, if
Li,L2 G A, then either L\ С L2 or L2 С L\), Prove that there exists a
vector χ G Η not contained in any of the subspaces L belonging to A.
Solution 1. More generally, we prove the statement for separable Banach
spaces. Let В be a separable Banach space, and let R be a system of sub-
spaces L of В that satisfies the requirements of the problem. Suppose that
Ul£rL = B. Consider a countable, everywhere-dense subset {xi,x2,... }
of B. By recursion, define a countable increasing sequence of elements of R
as follows. Let L\ be an element of В that contains x\. Suppose that the
subspaces Li,..., Ln (g R) have already been defined. Let Ζ,(η+1) be an
element of R that contains xn+\. (Such an element exists by the
assumptions.) One of the subspaces Li,..., Ln, Z/n+1) contains all others; let this
one be denoted by Ln+\. Then Ln С Ln+i and xn G Ln (n = 1,2,...).
Now, let L be an arbitrary element of R. Since L is a proper closed subspace
of B, it cannot contain all elements of the dense set {xi,x2, · · · }· Suppose,
say, that Xk £ L. By definition, x^ G L^, so L С Lk since the system В is
ordered with respect to inclusion. This means that U^L; = B.
Now, for all natural numbers n, let fn be a continuous linear functional
on В for which ||/n|| = η and fn(x) =0Ίΐ χ e Ln. (Such functionals exist.
For example, take an arbitrary element yn in the complement of Ln. Since
Ln is closed, the distance d of yn from the subspace Ln is positive. Consider
now the linear subspace [yn] +Ln, where [yn] denotes the one-dimensional
subspace generated by yn, and let
1 Λ is ;
[yn] + Ln and fn(x) = 0 for χ G Ln, and
where χ G Ln and λ is a complex number, fn ) is obviously linear on
\n\d\ nd nd
SUP τ\ и = sup и м = ^Ί Π μ = η'
\\лУп ~ χ\\ хеьп \\Уп ~ x\\ inf*ez,n \\yn - x\\

3.10 TOPOLOGY
521
By the Hahn-Banach theorem, /4 is extendable to a functional fn defined
on the whole В that has the required properties.) Now, if χ € В, then
liniy^oo fn(x) = 0, since, by Ug^Li = Β, χ is contained in some 1^, and
the sequence {Ln} of subspaces is increasing with respect to inclusion.
So, the sequence {/n} of functionals is pointwise convergent; therefore, by
the Banach-Steinhaus theorem, supn ||/n|| < oo. But this contradicts the
choice of fn. D
Solution 2. We prove the following generalization of the problem.
Generalization. Let Μ be a separable topological space of second Baire
category, and let 21 be a system of nowhere-dense closed subsets of Μ
ordered by inclusion. Then Ol^^L Φ Μ.
(It is obvious that the space I2 and the family of subspaces in the problem
and, more generally, separable Hilbert and Banach spaces with a similar
family of subspaces all satisfy the hypotheses of this statement; indeed,
these spaces equipped with the norm topology are separable topological
spaces of second Baire category and, in any of these spaces, a proper closed
subspace is nowhere dense.)
Proof. Suppose that, to the contrary, Ui,eaL = M, and consider a
countable, everywhere-dense set R = {xi,X2,···} in M. Put
Mk=\J{LeK:xkeL} (fe = l,2,...).
Then Mk is a nonempty, nowhere-dense subset of Μ for all k. Therefore,
U^Mfc is of first category, and
oo
U Мкф Μ.
k=l
On the other hand, if L € 21, then L φ Μ. L cannot contain R, that is,
there is a natural number к such that xk £ L. Since 21 is ordered with
respect to inclusion, L С Мк follows from the definition of Mk. Thus,
Ul£%L С U^Mfc, a contradiction. D
Remark. Several contestants remarked that the statement of the
problem is not true for nonseparable Hilbert spaces. Attila Mate gave the
following example. Let if be a nonseparable Hilbert space. Then Η is
isomorphic to a Hilbert space X0L2(a;i), where K is a suitably chosen
Hilbert space, and ω\ is the set of countable ordinals considered as the
discrete measure space in which the measure of all singletons is 1. Then
(J (ίθΙ2(ί))=ίΦί2(ωι).

522
3. SOLUTIONS TO THE PROBLEMS
Problem T.4. Let К be a compact topological group, and let F be a
set of continuous functions defined on К that has cardinality greater than
continuum. Prove that there exist xq g К and f φ д G F such that
/(so) = дЫ) = max/(ж) = тзхд(х).
хек хек
Solution. In the proof, we use the following theorem by Paul Erdos.
Theorem. If the edges of a complete graph of cardinality greater than
continuum are labeled with natural numbers, then there exists an
uncountable complete subgraph with all edges labeled with the same number. See,
for example, P. Erdos, A. Hajnal, and R. Rado, Partition Relations for
Cardinal Numbers, Acta Math. Acad. Sci. Hung. 16 (1965), Theorem 1.
If Fx denotes the system of functions in F whose maximum is x, then it
is clear that the cardinality of Fx is greater than continuum for some real
number x. Suppose that f=x Π g=x = 0 for all pairs of functions / φ g G
Fx. Consider the complete graph with the functions in Fx as vertices. If
/ Φ g G Fx, then there exists a natural number η such that f>x-(i/n) Π
9>x-(i/n) = 0; otherwise, the closed sets {/>ж-(1/п),#>я-(1/п)}п=1>2>>>
would form a family with the finite intersection property whose intersection,
by compactness, would be nonempty, but this intersection is contained
in the set f=x Π g=x. By the theorem quoted above, there is an η and
an uncountable F' С Fx such that all edges of F' are labeled with n.
Thus, the sets fx>i/n, /GF' are nonempty, open, and pairwise disjoint.
This contradicts the well-known fact that К admits a finite Haar measure,
and therefore any family of pairwise disjoint, nonempty, open sets in К
is countable. So, the statement of the problem is true for some pair of
functions / φ g G F. D
In fact, we proved the following theorem.
Theorem. The statement of the problem holds for a compact space
К if К satisfies the following condition: any family of pairwise disjoint,
nonempty, open sets in К is countable.
Problem T.5. Prove that two points in a compact metric space can
be joined with a rectifiable arc if and only if there exists a positive number
К such that, for any ε > 0, these points can be connected with an ε-chain
not longer than K.
Solution. If A and В are two points of the metric space, then let t(A, B)
denote their distance. By a rectifiable arc joining A and В we mean a
homeomorphic image of the real interval [a, b], where a and b are mapped
to A and B, and for any subdivision α = to < h < t2 < · · · < tn=i <tn=b,
denoting the image of U by T{, we have
n-l
Σί(7ί,Τ4+1) < К
г=0

3.10 TOPOLOGY
523
for some fixed K. The infimum of such numbers К is called the length of
the arc. Therefore, the half of the problem stating that if two points can
be joined with a rectifiable arc, then for any ε > 0 they can be joined with
an ε-chain of length at most K, is obvious.
Now suppose that, for any ε > 0, A and В are joinable with an ε-chain
not longer than K. Let L = {Ho,..., Hn} be a sequence of points in the
metric space. We use the following notation:
q(L) = max £(Яг_1,Яг)
l<i<n
and
η
K(L) = Ytt(Hi-1,Hi).
2 = 1
For 0 < h < 1, let L(h) denote the Hk for which
к k+1
^ЦЩ-иНд <h.K(L) < £*(#<_!,#<),
г=1 г=1
and let L(l) = Hn. Then, obviously,
ЦЦНОЦНг)) < |/ц - h2\ ■ K(L) + q(L).
Now choose an indefinitely refining sequence SO of chains connecting A and
B, where the lengths of these chains do not exceed a fixed constant K. The
hypotheses in the problem guarantee the existence of such a sequence of
chains. Arrange the rational points of the interval [0,1] into a sequence.
Then let Sk be an indefinitely refining sequence SO of chains connecting A
and В such that for к = 1,2,...,
(a) Sk is a subsequence of 5^_i, and
(b) L(h) is convergent in the first к rational points when the chains are
taken from Sk and к is kept fixed.
(More precisely, instead of L(h), we should write Ln>^(/i), where this
denotes L(h) for the nth chain of Sk- In (b), к and h are fixed while
η —> oo.) The sequence of chains Sk can be defined for all к by compactness
of the space. Now define the function f(h) on the set of rational numbers
in [0,1] by the formula
f(h) = lim Lktn(h)
n—>oo
if h is one of the first к rational numbers. This definition is correct by (a).
Let hi and h2 be rational numbers in [0,1]. Choose a sufficiently large к
so that hi and h2 occur among the first к rational numbers. Then
*(/(Λι),/(Λ2))<^·|Λι-Λ2|.
Since a uniformly continuous function defined on a dense subset of the
interval [0,1] can always be continuously extended over the whole interval

524
3. SOLUTIONS TO THE PROBLEMS
[0,1], provided the target space is compact, the function f(h) is extendable,
and the last estimate remains valid. The image of the interval [0,1] is a
continuous curve connecting the points A and B, and from the estimate it
immediately follows that all approximating chains have length at most K.
Therefore, the curve is rectifiable of length < K.
In general, it is not true that we obtain an arc. But if we choose the
above К to be the smallest constant such that, for any ε > 0, A and В are
joinable with an ε-chain not longer than K, then we get an arc. Indeed,
otherwise there would exist и and υ in [0,1] (u Φ ν), with f(u) = f(v).
Then we could delete the image of the open interval (u, v) from the curve
constructed above, and we would get a rectifiable curve from A to В whose
length K+ is less than K. Then, by our initial observations, for any ε > 0,
A and В would be joinable by an ε-chain shorter than K+. This contradicts
the minimality of K, proving the statement of the problem. D
Remarks.
1. Laszlo Lovasz proved that, instead of compactness, assuming only local
compactness and completeness of the space, the statement remains valid.
By deleting a chord from a disk in the plane, we obtain a locally compact
space in which two points on different sides of the chord cannot be joined
with an arc although they are joinable with an arbitrarily fine ε-chain.
Therefore, it does not suffice to assume only local compactness of the
space. Completeness alone is not sufficient either. To show this, Lovasz
gave the following example.
Example. Let eo, ei,..., en,... be an orthonormal basis in the Hubert
space I2. Consider the segments connecting 0 with the endpoints of the
vectors ei, б2,..., and similarly the segments connecting eo with the points
ei +eo,e2 + eo, ... . Divide the segment between en and en + eo into η equal
parts, that is, consider the points en + (i/n) · eo when г = 1,2, ...,n — 1.
The segments and points just defined form a closed set in the Hilbert
space; therefore, they define a complete metric space. In this space the
points 0 and eo for any ε > 0 are joinable with an ε-chain not longer than
3, but they cannot be joined with an arc. This space can be made into
a connected counterexample by connecting the points en + ((г — 1)/η) · eo
and en + (i/n) · eo with suitably chosen, nonrectifiable arcs.
2. A sketch of the proof of Lovasz's generalization is the following.
Proof. We call a point accessible if, for some £, В and the point are
joinable with an arc of length t and, for any ε > 0, the point and A are
joinable with an ε-chain not longer than К — t, where К is the minimal
constant used above. Of the accessible points χ and у we say that χ is
finer than у if, in the definition of accessibility of x, the arc from В can be
chosen through y. Then, using Zorn's lemma, we define a "finest" point xo,
which we show to be necessarily A. Assuming the contrary, take a compact
neighborhood U of xo that does not contain A. Then define a point ζ on
the boundary of U that is a limit point of certain points of chains from xq
to A in the definition of accessibility of xq. We show that ζ is accessible

ЗЛО TOPOLOGY
525
and finer than x0. To this end, it suffices to see that xq and ζ are joinable
with a sufficiently short arc. But we can apply the result already obtained
for compact spaces to the points x$ and z, and this proves the statement.
Problem T.6. Let a neighborhood basis of a point χ of the real line
consist of all Lebesgue-measurable sets containing χ whose density at χ
equals 1. Show that this requirement defines a topology that is regular but
not normal.
Solution.
(1) Let m(A) denote the measure of the Lebesgue-measurable set A. Let A
and В be basis neighborhoods of the point x, we show that Α Π В is as
well. Since Α Π Β is measurable and χ G Α Π Β, we only have to show
that its density at the point χ is 1, that is,
т((АпВ)Ш)
m(I)
when the interval / shrinks to x:
I m(A ПВО I)
I W)
_ \m((I-A)U(I-B))\
~ I m(I) I
m(/ - A) m(I - B)
~ m(I) + m(I)
The last expression converges to 0 when / shrinks to x. So Α Π Β is
indeed a neighborhood of x. In order to obtain a topology, we show
that any neighborhood A of χ contains a neighborhood В of χ that is a
neighborhood of all of its points. Let В be the set of points in A where
the density of A is 1. Obviously, χ G B, and by the Lebesgue density
theorem, A — В has measure 0. Therefore, В is also measurable, and
at points of В, В has the same density as A, that is, 1. So, we have a
topology indeed. Let Ε denote this new topology, and call its open sets
Ε-open while keeping the adjective "open" for open sets in the usual
topology. Similarly, we distinguish closed and Ε-closed sets. It is not
obvious but it is true that the Ε-open sets are measurable. We can prove
this as follows.
Let Η be an Ε-open set. We can assume that Η is bounded; then
the outer measure m(H) of Η is finite. Now put
α = sup{m(C) : С С Я, С is measurable},
and for all η choose Cn so that Cn С Η and m(Cn) > а — (1/n). Then
S = UnCn is a subset of Η with the property that every measurable
m{I)-m{I-{AC\B))
m(I)

526
3. SOLUTIONS TO THE PROBLEMS
subset of Η—S has measure 0. (5 is usually called the measurable core of
H.) If V С Η is measurable and Q = H-S, then, by (Vf)Q)U(Vr\S) =
V, V Π Q is measurable, and therefore it has measure 0. So,
m(V)=m{Vr\S).
Q decomposes as the union of two sets Ζ and X, where the former has
measure 0, and the density of Q at all points of the latter is 1. In order
to prove that Η is measurable, we show that Q has measure 0, and to
this end, we need X to be empty. This will imply measurability of H.
Arguing by contradiction, we suppose that ρ G X. Let / denote an
interval shrinking to p. By definition of X, m(Q Π /) = (1 + o(l))m(I).
Now let U be an ^-neighborhood of ρ contained in H. Then
m(U Π / Π S) = m(U Π /) = (1 + o(l))m(J),
so U Π S has density 1 at p. Therefore, by / - (U Π S) D Q Π /, we have
m(Q П J) < m(I - (C/ Π S)) = o(m(I)),
which contradicts our assumption.
(2) We show that the topology Ε is regular. Suppose that К is an E-closed
set, and χ £ К. Then χ has a neighborhood A that does not meet K.
So, πι(ΑΠΐ) = (1 +o(l))ra(7), where the interval / shrinks to x. Thus,
m(A Π I) = o(m(I)). (Here A denotes the complement of the set A.)
К С A, so m(K ΠΙ) < m(A Π /), and therefore
m(Knl)
m(I)
Put xn = χ — (1/n) and yn = χ + (1/n), η = 1, 2,... . It is well known
that, for any measurable set X and ε > 0, there exists an open set G
such that X С G and m{G) < m(X) + ε. Choose the open set Gn in
the interval [xn-i, xn+2] so that Κ Π [χη, χη+ι] Q Gn and
m{Gn) < m{K Π [xn, xn+i]) + ^
hold. For η = 1, we require G\ С (—oo, #2]. Choose the sets iJn in a
similar way for the intervals [yn+i,yn]. Put
£ = (-00, χλ) U Gi U G2 U · · · U Ηλ U Я2 U · · · U (уь оо);
then В is open and contains K. Since open sets are Ε-open (because they
have density 1 at every point), it will suffice to prove that the density of
В at χ is 0. Then the complement of the Ε-closure of В and В will be
disjoint ^-neighborhoods of χ and K, respectively. It is enough to show
that m(I Π B)/m(I) —> 0 for / = [x,x + к] (к —> 0), because a similar
argument shows the corresponding estimate for the left-hand intervals,

3.10 TOPOLOGY
527
and these two together imply the same for arbitrary intervals / shrinking
to x. Assume Ут+ι < x + к < уш. Then m(I) = к > l/(ra + 1). We
have m(B Π I) < Y^Lrn_2 ™>(Hi), since Hi Π [χ, ym] = 0 for г < m — 2.
Therefore
oo oo 1
т(ВП1)< Σ m(Kn[yi + l,yi})+ £ -.
i=m—2 i=m—2
1
and so
m(I)
= т{КП[х,уп-2]) + 1^,
т(КГ\[х,уТп_2\) ^
- 1 "·" 1
m(BC\I) ^ ra(if n[rr,ym_2]) , 2^
m+l m+1
If / shrinks to x, then m —> oo, so both terms in the right-hand side of the
last inequality converge to 0. This is obvious for the second term; for the
first one, use the already established convergence m(K Π I)/m(I) —> 0
and the fact that (m + l)/(m — 2) < 4 if m > 3. So, В indeed has
density 0 at x, and the space is regular.
(3) We show that the space is not normal; namely, that an Ε-closed set
K\ of second Baire category and an Ε-closed everywhere-dense set K2
cannot be separated by -Ε-open sets. Since a countable set has measure
0, and thus is Ε-closed, the existence of such a set K2 is clear. The
existence of such a K\ is also well known: for every n, there exists a
nowhere-dense set of measure greater than 1 — 1/n in [0,1]; the union
of these is a first category set of measure 1; and the complement of
this in [0,1] is of second category and has measure 0 and is therefore
Ε-closed. Since K\ is chosen to have measure 0, K2 can be chosen in
the complement of Κι, then Κ ι and K2 are disjoint.
Suppose that there exist two disjoint Ε-open sets C\ and C2 with
K\ С C\ and K2 С C2. Then, for any χ G Ci, there is a neighborhood
А С d of χ for which
^4^Ul, andthus "^Ul
m{I) ™{I)
when / shrinks to x. Therefore, by Κ ι С Ci, for any χ G ΑΊ, there
exists an no such that for η > no,
m(Cin[s-(l/n),a? + (l/n)]) 1
2/n ~ 2'
that is,
mid Π
1 1
χ ,x-\—
η η
> - (η > n0).
η
If we divide the elements of K\ into a countable number of subsets
depending on whether the last inequality holds from a certain index,

528
3. SOLUTIONS TO THE PROBLEMS
) ~ An
Η
then at least one of these subsets is dense in an interval [a, b]. Thus,
there exists an no such that the last inequality holds from this no on an
everywhere-dense subset of the interval (a, b).
Since K2 is everywhere dense, we can choose an element у of K2 in
(a, b). Since у G C2, there is an η > no with
m(C2n[y-(l/n),y + (l/n)]) >3
2/n ~ 4'
that is,
/^ Γ 1 1
m С2П h/ ,y+-
\ [ η η
if ι is dense in (a, 6), so there exists an χ in the interval (у, у + (l/8n))
for which
/^ Г 1 1",
πι Gi Π ж ,aH—
\ [ η nj
holds. The length of the interval I = [y — (l/n),x + (1/n)] is at most
(2/n) + (l/8n). Ci and C2 are disjoint and
1 3
m(Ci Π Л > - and ra(C2 П J) > —.
η 2η
Therefore,
2 + ^ - > m(I) > mid Π I) + m(C2 Π /) > (1 + ^ -,
о J η \ λ J η
a contradiction, which proves that the topology Ε is not normal. D
Remarks.
1. In order to solve the problem, it is not necessary to prove that E-open
sets are measurable, but this simplifies the proof at various points.
2. Lajos Posa proved the nonnormality statement using a cardinality
argument. The sketch of his proof is as follows.
Proof. There exists a set of cardinality с = 2^° and of measure 0. All
subsets of this set are Ε-closed; therefore it can be partitioned into two
disjoint Ε-closed sets in 2C different ways. If the space were normal, then for
each such partition we could find two disjoint Ε-open sets that separate the
two parts. It can be proved that the pairs consisting of the Ε-closures of the
separating Ε-open sets corresponding to different partitions are different.
It is also easy to see that the Ε-closure of every Ε-open set equals the
Ε-closure of an Εσ-subset, which only differs by a set of measure 0. Since
the cardinality of the set of i^-sets is only c, the cardinality of the set of
all pairs of closures of i^-sets is also only с Therefore, the cardinality of
the set of pairs of separating Ε-open sets is only c, a contradiction.
3. Laszlo Lovasz proved complete regularity (which is stronger than
regularity). He noticed that the above technique yields that if A and В
are disjoint Ε-closed sets, then they can be separated by an open set
containing В and an Ε-open set containing A. Then the proof of the
Urysohn lemma applies.

3.10 TOPOLOGY
529
Problem Т.7. Suppose that V is a locally compact topological space
that admits no countable covering with compact sets. Let С denote the
set of all compact subsets of the space V and U the set of open subsets
that are not contained in any compact set. Let f be a function from U to
С such that f(U) С U for all U GU. Prove that either
(i) there exists a nonempty compact set С such that f(U) is not a proper
subset of С whenever С С U eU,
(ii) or for some compact set C, the set
Г\С) = \J{U e U : f(U) С С}
is an element of U, that is, /_1(C) is not contained in any compact
set.
Solution. The statement is trivial. Indeed, if /_1(0) = V, then (ii) holds.
If not, then for an arbitrary χ G V — /_1(0), (i) holds with С = {χ}.
(We do not use that V is not σ-compact, only noncompactness of V is
necessary.) D
Remark. Laszlo Babai proves that if, under the hypotheses of the
problem, (ii) is not true, then (i) holds with a set С of two elements. The proof
of this is fairly difficult.
Problem T.8. Let T\ and T2 be second-countable topologies on the set
E. We would like to find a real function σ defined on Ε χ Ε such that
0 < σ(χ, у) < +oo, σ(χ, χ) = 0,
σ(χ, ζ) < σ(χ, у) + а(у, ζ) (χ, y,ze Ε),
and, for any ρ G Ε, the sets
νί(ρ,ε) = {χ : σ(χ,ρ) < ε} (ε > 0)
form a neighborhood base of ρ with respect to 71, and the sets
V2{p, ε) = {χ : σ(ρ, χ) < ε} (ε > 0)
form a neighborhood base of ρ with respect to T2. Prove that such a
function σ exists if and only if, for any ρ G Ε and %-open set G Э ρ
(г = 1,2), there exist a %-open set G' and a %-i-closed set F with ρ G
σ cfcg.
Solution. Suppose that there exists a function σ(χ, у) with the required
properties. Let ρ G E, G be open in 7^, ρ G G. Then there exists ε > 0
with Vi(p,e) С G. Let 0 < δ < ε, G = Vi(p,6), and let F be the Ъ-%-
closure of G. Using the triangle inequality, it easily follows that G, F, G

530
3. SOLUTIONS TO THE PROBLEMS
indeed satisfy the hypotheses. (The assumption on second countability is
not needed in this direction.)
In what follows, G always denotes an open set and F a closed set. Super-
(2)
scripts indicate which topology this means. For example, G^ denotes
the 72-closure of the 7i-open set G^. We prove that the hypothesis given
in the problem is sufficient for the existence of σ. The proof is a
modification of the well known-proof of the Urysohn metrization theorem.
Tikhonov lemma. If F^nF™ = 0, then there exist G(1> and G(2> such
that G^ d /t(2)i G(2) d F(i)? and G(i) π G(2) = 0e
Proof. Let ρ e FW. Then pe Ε - F(3_i), this set is 7^-open, so by
the assumption there exists a G' = G'^ ~ (p) such that ρ G G'^ (p) and
С/(з-г)(р)(г) cE_ F(3-<)i that is, G,(3-i}(p)(l) Π F*3"*) = 0.
To each such G,{j\p), assign an element U^ of the countable basis in
Tj·. ρ e U^ С G^3\p). We have countably many sets U^\ these can be
arranged as Un\ η = 1,2,..., (j = 1,2). We also have
00 (3 η
F(3-«) c у [/W> {/W nF(i)=0. (1)
n=l
Put
fc = l
These Unl are 7^-open, they also satisfy (1), and obviously Uni C\Uin* =
0. Therefore, the sets
oo
G«=(Jt/« (г = 1,2)
n = l
clearly satisfy the requirements of the lemma.
Urysohn lemma. If F^ С G^2\ then there exists a real function p(x, y)
on Ε χ Ε such that
0 <p(x,y) < 1, p(x,x) = 0,
р(ж, г) < р(ж, у) + p(y, ζ) (ж, у, ζ е Ε); (2)
the sets
ν^(ρ,ε) = {χ:ρ(χ,ρ)<ε} (ε > 0) (3)
are 7i-open; the sets
ν^(ρ,ε) = {χ:ρ(ρ,χ)<ε} (ε > 0) (4)
axe 72-open; and
p(x,y) = l if arefW, t/^G^. (5)

3.10 TOPOLOGY
531
Proof. Introduce the notations g£2) = 0, F0(1) = F^\ G{?] = G&\
F[ = E. Let Dbea countable dense subset of the closed interval [0,1]:
D = {r0, Γι, r2,... }, where r0 = 0, τλ = 1.
By recursion on n, we define the sets Gn and Fn so that Gi Cfi
and, for rn < rm, Fn С Gin hold. These inclusions are true in the
cases already defined (n, m = 0,1). If the sets with subscripts less than η
(η > 2) have already been defined, then find the neighbors of rn among
Γο, Γι,..., rn-i- Let them be r^ and r/: r^ < rn < ri, 0 < k, I < η — 1.
Apply the Tychonoff lemma to the sets F^ ' and Ε — G\j : by r^ < r\ and
the induction hypothesis, F^ Π (Ε - GJ2)) = 0, so by the the Tychonoff
lemma, there exist Gn and Gn* with Gn Π Gn* = 0 and
that is, using the notation Fn — Ε — Gn*,
and G^2) Π GlV = 0 means that
G(2) c F(l)_
It is clear that the validity of the induction hypothesis is inherited to these
sets, that is, the recursive definition is correct.
For χ e E, put
g(x) = mi({rn:xeG^}U{l}\
f(x) = sup({rn:x^F^}U{0}),
(0 < f(x),g(x) < 1). It is clear that
/|F<1} = g\F™ = 0, f\E - G(2) = g\E - G(2) ξ 1.
Further, if η > 2, then
g(x)<rn =» xeGn2) =» xe^ =» /(ж) < rn,
so /(ж) < g(x) for all χ e E. But if /(ж) < #(#) for some χ e E, then
there would exist гп,Гт G -D such that
/(ж) < rn < rm < ^(ж).
Then xefi and χ £ Gin , but Fn С G™ , a contradiction. Therefore,
f(x) = g(x) on E.

532
3. SOLUTIONS TO THE PROBLEMS
For x,y G E, put
fO, if/(*) >/(*,),
P(x,y) I/M-/(*), Xf(x)<f(v).
Obviously, 0 < p(x,y) < 1, p(x,x) — 0, and (2) and (5) hold. To prove (3)
and (4), by (2) it suffices to show that ρ G mt^V^\p,e) (p G Ε, ε > 0)
(< = 1,2).
For г = 1:
Let ε > 0, ρ G Ε. Then
Vp(1)(p,e) = {* : p(*,p) < ε} - {* : f(p) - f(x) < ε} D Ε - F^ Э ρ
if η is chosen so that f(p) — ε < rn < f(p). In case f(p) = 0, Fp (ρ, ε)
Эр. E-l
For г = 2:
ЕЭр. Ε - Frl^ and Ε are Τχ-open.
Vp<2>(p,e) = {χ : ρ(χ,ρ) < ε} = {χ : /(χ) - f(p) < ε}
= {χ : g(x) - g(p) < ε} D G™ э р
if g(p) <rn< g{p) + ε. In case #(p) = 1, v£2)(p, ε) = Ε.
This proves the Urysohn lemma.
We turn now to the proof of the theorem.
Let Ρ be the set of all pairs (U^\ V^), where U^ and V^ are elements
of the countable basis for the topology % and
7(3-0
UM cV^ (for г = 1,2).
Then Ρ is countable:
P = {(tfi,VU(C2,V2),...}.
Now for к = 1,2,... we define a function р^(ж, у).
Put (t/fc, 14) = (C/^}, V^). If г = 2, then apply the Urysohn lemma to
the pair of sets FW = C/f} , G(2) - Vfe(2), and call pfe the function ρ given
(2)
by the lemma. If г = 1, then put F<2> = U^ , G™ = V^\ and apply
the version of the Urysohn lemma where the superscripts (1) and (2) are
interchanged and p(a, b) is replaced with p(6, a) everywhere. (Then (3) and
(4) simply interchange and (2) remains the same.) Call pk the function ρ
thus obtained.
Put
oo 1
°&У) = Σ^Ρ*(ζ,2/).
k=l

3.10 TOPOLOGY
533
It is clear that 0 < σ(χ, у) < 1, σ(χ, χ) = 0 and, by (2), σ(χ, ζ) < σ(χ, у) +
а(у, ζ) holds for all χ, у G Ε.
We prove that for ε > 0 and ρ e Ε
реЫ^Уг(р,е) (г = 1,2).
Let N be sufficiently large so that Σ™=Ν -^ < f · Then
fc=l
But, by (3) and (4), the right-hand side is a 7^-open set containing p.
It only remains to show that if ρ G int^C/, then ρ G Vi(p, ε) С U for
some ε > 0. Since ρ G int^C/, there is a G^%\ then, by the assumptions,
there is a G'^ in the countable basis for %, such that
PeG,{i\ РМсС^с[/.
Therefore, (G,{i),G^) = (U^\v^) G Ρ for some A:. So, using ρ G G/(0
and (5),
^(ft^)cVW(p,l)cG« D
Problem T.9. Prove that there exists a topological space Τ containing
the real line as a subset, such that the Lebesgue-measurable functions, and
only those, extend continuously over T. Show that the real line cannot be
an everywhere-dense subset of such a space T.
Solution.
1. Let {fi : г G /} be the set of all Lebesgue-measurable functions, Xi be
a copy of the real line with the usual topology (г G /), and
If h : R —> Τ is the map defined by
Pi(h(x)) = fi(x),
where pi is the projection onto Xi, then h is injective and h(R) can be
considered as a copy of the real line. We prove that Τ is as required.
Let / be a measurable function, say, f = fi- Then / = pi\h(R) and
a continuous extension of this over Τ is pi. Conversely, if к : Τ —> R is
continuous, then A;|/i(R) is measurable. Indeed, by a well-known theorem
(see, for example, R. Engelking, Outline of General Topology, PWN-
Polish Sci. Publ, Warsaw, 1968, p. 98, Problem R), к only depends on

534
3. SOLUTIONS TO THE PROBLEMS
countably many coordinates, that is, к = к' ο ρ, where ρ : Τ —> Τ' is the
projection onto the product V = Y[ieI, X%, where Г С I is countable
and k' : T' —> R is continuous. Now,
{ж : k(h(x)) >c} = {x : k'(p(h(x))) >c} = {x: p(h(x)) G t/c}
where Uc is open in T'. The sets of the form
i€l'
where Y* = Xi with a finite number of exceptions when Yi is an interval
with rational endpoints, form a countable basis in T'. Therefore, the
set {x : k(h(x)) > c} can be written as a countable union of sets of the
form {x : p(h(x)) G Пге/' **}· But
| χ : p(h(x)) еЦУг\={х: /<(x) G Y{ (г G /)},
being a union of countably many measurable sets, is measurable. Thus,
{x : k(h(x)) > c} is measurable.
2. Suppose that Τ has the required properties and EcTis dense. Since
all measurable functions are continuous, it is obvious that the topology
of Τ induces the discrete topology on R. Let ρ eT. Let /* denote the
extension of the function f(x) = χ over Τ (this is unique since R is dense
in T). Further, consider the function
, ,_ / 0, if* = /*(p),
and let #* be its continuous extension over T. Put
Up = {q: \9*{P) ~9*(θ)\ < 1, I/*(p) " Г (β) I < ^Τϊ} '
Let ж G C/p Π R (since R is dense, such an χ exists); we show that
χ = f*(p). Indeed, if χ φ /*(ρ), then
lff(*)l
>lfl*(p)| + i,
к-/*Ы1
which contradicts that |^*(p) -^*(ж)| < 1. So Up Π R = {/*(p)}. From
this it also follows that f*(q) = f*(p) for every q G Up\ indeed, UqC\UpC\R
is nonempty (since R is everywhere dense), but its only element must
equal both /*(<?) and /*(p). So /*(p) = /*(9).
Now for an arbitrary, nonmeasurable, real function φ, the function
φ* = Ψ о /*
is a continuous extension of φ, because this function, being constant on
the neighborhood Up of any point ρ G T, is continuous. Thus,
nonmeasurable functions also have continuous extensions over T, and this is a
contradiction. D

3.10 TOPOLOGY
535
Problem Т. 10. Let A be a closed and bounded set in the plane, and
let С denote the set of points at a unit distance from A. Let ρ G C, and
assume that the intersection of A with the unit circle К centered at ρ
can be covered by an arc shorter than a semicircle of K. Prove that the
intersection of С with a suitable neighborhood of ρ is a simple arc of which
ρ is not an endpoint.
Solution. Let ab be the minimal arc of К containing Κ Π A (possibly
a = b). Introduce Cartesian coordinates in the plane so that ρ is the
origin, a and b lie in the left half-plane symmetrically with respect to the
horizontal axis, and in case a ^ 6, b lies in the upper half-plane.
We claim that with a suitable δ > 0, the intersection C\ of С with the
upper half-plane and with the disc of radius δ centered at ρ is a simple arc
starting from ρ that only meets the horizontal axis at p. Then a similar
statement is true for the lower half-plane, and this proves the theorem. In
order to prove our claim, it suffices to show that, for any r < δ, the set С
meets the upper semicircle KT of radius r centered at ρ in a single interior
point. In this case, assigning to points of C\ their distance from ρ is a
topological map from the compact set C\ onto the interval [0,5], since it is
continuous and bijective. Let Τ be a convex plane sector, containing a and
b in its interior, with vertex at p, and with the horizontal coordinate axis
as axis of symmetry. Then, obviously, ρ(ρ, A — T) = 1 + ει > 1. Further,
there is an e% > 0 such that if χ G T, say, in the upper half-plane, and
ρ(χ,ρ) < ε2, then the triangle pxb has an obtuse angle at p; Therefore,
Q(x,b) < Q(p,b) = 1, and so ρ(χ,Α) < 1.
Now let δ < πιίη(ει, ε2). If г < 5, then Kr Π C\ is nonempty. Indeed, let
с and d be the endpoints of Kr in the left and right half-planes, respectively.
Then ρ(ά, A) < 1 by the above. On the other hand, if χ G A — T, then
Q(x,c) > ρ(ρ,χ) — ρ(ρ,с) >1-\-ε— δ > 1\
and if χ G Α Π Τ, then the triangle xpc has an obtuse angle at p, so
Qfax) > ρ(ρ,χ) > 1.
Thus, g(c, A) > 1. By continuity of ρ, there is a point у on the arc Kr for
which g(y, A) = 1, that is, у e C\. It also follows that c,d £ C\.
On the other hand, suppose that 2/1,2/2 £ C\ Π Kr. By the above,
2/1,2/2 ^ T. Let 2/1 be the one of 2/1,2/2 that is closer to d, and let t G A be
a point such that ρ(2/2? t) = 1. Then t eT, because
eM < 0(P,ifc) + e(ifc.t) < 1 + « < 1 + ei.
Since ρ^ι,ί) > 1 = ρ(2/2? £), the point t is on the same side of the
perpendicular bisector of the segment 2/12/2 (which goes through p) as 2/2· Therefore,
the angle yiptL, measured in positive orientation, is less than 180 degrees
and obviously is greater than 90 degrees. Then the triangle y<ipt has an
obtuse angle at p, and so 0(2/2, £) > Q(p,t) > 1, a contradiction. D

536
3. SOLUTIONS TO THE PROBLEMS
Problem Т.Н. Suppose that τ is a metrizable topology on a set X
of cardinality less than or equal to continuum. Prove that there exists a
separable and metrizable topology on X that is coarser than т.
Solution 1. We say that a set A is separated by a family of subsets if for
any a, 6 G Α, α ^ 6, there exists a set in the family that contains α but
does not contain b. If the cardinality of A is at most continuum, then A is
always separated by a countable family of subsets. Indeed, embed A into
R and take the subsets corresponding to intervals with rational endpoints.
First, we show that there exists a sequence of closed subsets in X that
separates X. It is well known that every metrizable space admits a σ-
discrete basis. Let В = U™=1Bn be a basis in X such that Bn consists of
pairwise disjoint, nonempty sets. Then the cardinality of Bn is less than or
equal to continuum, so we can take a family {Вгп : г = 1,2,... } (a family
of families of sets) that separates Bn. Let U%n be the union of the elements
of Bln\ then XJ%n is an open set with respect to the topology r.
The family {U^} separates X. Indeed, let x, у G Χ, χ ^ y. Since В is a
basis, there is an η and V G Bn such that χ e V and у ^ V. There may
or may not exist a V G Bn with у G V, but there is at most one such V',
since Bn is a disjoint family. Choose а Вгп such that V G Вгп and V £ Вгп
(or, an arbitrary Вгп containing V if there is no V'). Then, obviously,
χ G Uxn and у £ Uxn. As the family {Uln} separates X, so does the family
consisting of the complements of the sets Uxn. Arrange these complements
into a sequence Fi, F2,... . Then {Fn} is a family that separates X and
consists of closed sets.
Let d be a bounded metric on X that induces the topology r. Consider
the following map / : X —> I1 :
t( \ - (d^x) d^x) \
JW)-y 2i , 22 ,··)·
Then / is injective, because if χ ^ у there would exist an Fn with χ G Fn
and у £ Fn, and then d(Fn,x) = 0 ^ d(Fn,y). The map / is continuous,
for if D denotes the distance in I1, then
00 00
D(f(x),f(y)) = J22-n\d(Fn,x)-d(Fn,y)\ < ^~nd(x,y)=d(x,y).
71=1 71 = 1
Consider now the topology on X determined by / as the inverse image
topology. Since I1 is a separable metric space (and separability is a
hereditary property of metric spaces) and / is injective, this topology is separable
and metrizable and, since / is continuous, it is coarser than r. D
Solution 2. For any cardinal number m > 0, we denote by J(m) the
following metric space: the points of J(tn) are the pairs (г, χ), where г G /

3.10 TOPOLOGY
537
(/ is an arbitrary set of cardinality m) and χ G [0,1], with the points of the
form (г,0) identified. The metric d of J(m) is defined as
t \x-yu и г = J.
In the proof, we use the theorem that asserts that every metric space of
weight m is topologically embeddable into the topological product of count-
ably many copies of the space J(m). (See, for example, R. Engelking,
Outline of General Topology, PWN-Polish Sci. Publ, Warsaw, 1968, p. 197,
Theorem 7.)
Let Ρ denote the following property of topological spaces (X, r): there
exists on X a separable metrizable topology coarser than т. It is obvious
that all subspaces of a space with property Ρ and topological products of
countable families of spaces with property Ρ also have property P. It is
also clear that metric spaces satisfying the hypotheses of the problem are
of weight less than or equal to с (continuum). Since, by the theorem cited
above, all metric spaces of weight < с are embeddable into the topological
product of countably many copies of J(c), it suffices to show that J(c) has
property P.
It is obvious that J(c) is homeomorphic to the metric space whose
underlying set is the unit disc and in which the distance between two points
on the same radius is their usual distance, and the distance between two
points on different radii is the sum of the norms of the points. The usual
topology of the unit disc is coarser than this topology and is separable and
metrizable. D
Problem T.12. Suppose that all subspaces of cardinality at most Ni
of a topological space are second-countable. Prove that the whole space is
second-countable.
Solution. We argue by contradiction. Suppose that the topological space
X is not second-countable but that all subspaces of cardinality < Ni
are. We construct a certain subspace Υ of cardinality < Ni, and second-
countability of Υ will lead to a contradiction.
The subspace Υ is the union of subspaces Υα (α < ω ι) to be defined
by transfinite recursion. Together with the Уа, we simultaneously define
a family Qa of open sets in X so that 5а|^а is a basis for Ya. (If Η is a
family of subsets in X, and Ζ С X, then U\Z = {Η Π Ζ : Η G Η] is the
trace of H in Ζ.) The recursion is as follows: Yq = Qo = 0. If the countable
Υξ and Q% have already been defined for ξ < α < ωι, then define Υα and
Qa the following way. The family U^<a^, being countable, is not a basis
for X, therefore there exists a point ρ G X and a neighborhood V of ρ such
that, for any element G of U$<a(?£ containing p, we have G (f. V. Choose
a point pc G G \ V for each such G, and put
Y<*= \jY^U{p}ulpG:PSG€ (j£4
ξ<α Ι ξ<α

538
3. SOLUTIONS TO THE PROBLEMS
It is clear that Ya is countable; therefore, by the assumptions, it is second-
countable. Let Qa be a countable family of open sets in X such that Ga\Ya
is a basis in Ya.
Consider the subspace Υ = Ua<UJlYa. It obviously has cardinality < N1,
so it is second-countable. The crucial observation now is that υα<ωι0α\Υ
is a basis for Υ. Let us postpone the proof of this statement and show first
how to complete the solution, assuming that this observation is valid.
Since Υ is second-countable, the basis Ua<UJlGa\Y contains a countable
basis. Therefore, for a sufficiently large a, the family υξ<αζ}ξ\Υ is a basis
for Y. Then, with this a, U^<aQ^\Ya is a basis for Ya. But this is a
contradiction since U^<aG^\Ya does not contain a basis of neighborhoods
for the point ρ used in the definition of Ya.
It only remains to show that υα<ωιGa \Y is a basis for Y. This statement
is immediate from the following lemma.
Let У be a second-countable topological space, and let Υ = υα<ωιΥα,
where Ya С Υβ for a < β. For a < ω\, let Qa be a family of open sets in
Υ such that ζ}α\Υα is a basis for Ya. Then Ua<UJlGa is a basis for Y.
The proof is again by contradiction. Suppose that υα<ωι0α is not a
basis for Y. We construct a sequence {Va : a < ω{\ of open sets with the
property λίβ \ υα>βνα 7^ 0 (β < ω\), but such a sequence cannot exist in a
second-countable space.
The sets Va, together with a sequence of points ra G Va, are defined
by transfinite recursion as follows. Since υα<ωι0α is not a basis for У,
there exists a point q e Υ and a neighborhood U of q such that G (jL U
whenever q G G G υα<ωι0α. Put ro = q and Vq = Y. If 1^ and r% have
already been defined for ξ < a < ωχ, then first choose a ρ < ω\ with
YPD {Γξ : ξ < a}. Since QP\YP is a basis for Yp, there exists a set Va G Qp
such that q G Yp Π Va С U. On the other hand, Va <jt U; therefore we can
choose a point ra eVq\ U. Thus, the sequences {Va}, {ra} axe defined.
Notice that if β < α, then τ β φ Va. Indeed, τβ eYp\U while Va Π Yp С
U. Thus, Τβ Ε V/з \ υα>βνα (β < ω\). But this contradicts that Υ admits
a countable basis {JE?i, JE?2,...}. Indeed, for each α choose а ВПа with
To. Ε ВПа С V. Since there are only countably many JE?n, there exists
a > β with na = Πβ = п. But then τ β G Bn and τ β φ Va D Bn, which is
impossible. D
Remark. Unfortunately, the phrase "at most" was missing from the text
of the problem when it was posed for the competition. The statement is
not true in that version; a counterexample is provided by the set of natural
numbers endowed with the topology in which the nonempty, open sets are
the sequences of density 1. This is not a second-countable space, but the
hypothesis is vacuously true. The contestants usually noticed this defect
and considered the correctly modified version of the problem.
Seven solutions were submitted. Best results were obtained by Emil
Kiss. He proved the statement for Тз-spaces among other arguments that
contained all the ideas necessary to prove the statement in full. Nandor
Simanyi proved the statement assuming the continuum hypothesis and reg-

3.10 TOPOLOGY
539
ularity of the space. Vilmos Totik used the hypothesis that every point has
a basis of neighborhoods of cardinality N1, and Zoltan Szabo assumed first-
countability of the space.
Problem T.13. Suppose that the T^-space X has no isolated points
and that in X any family of pairwise disjoint, nonempty, open sets is
countable. Prove that X can be covered by at most continuum many
nowhere-dense sets.
Solution 1. By transfinite recursion on α < ωι, we define a family of open
sets in X indexed by sequences of length α consisting of natural numbers,
that is, by functions / : α —> ω. For α = 0, the only such function is
the empty set; put Gq = X. If α is a limit ordinal and / : α —> ω,
then put Gf = int (r\gGg), where g ranges over restrictions of / to all
smaller ordinals. If α has the form α = β + 1 and Gf = 0 for some
/ : β —> ω, then we define Gg = 0 for all extensions g : β —> ω of /. If
Gf φ 0, then we can define a sequence {Hi,H2,...} of open sets such
that Hn С Gf, Hn Π Hm = 0, Hn φ 0 for all η φ m, and Ηι,Η2,... is a
maximal system of nonempty, open sets in Gf. (Here we used regularity of
X and the property formulated in the problem. It is easy to avoid that this
system is finite.) Let the sets Η ι, H2,... be indexed with the extensions
9i,92,- · · of / over /3+1; the cardinality of these is also countably infinite:
{Hi,H2,... } = {G9l,G92,...}. So the sets Gf are defined for all functions
/ : α —> ω (α < ωι). It is obvious that Gf f)Gg = 0 if f,g : а —> ω and
f φ g, and that Gf D Gg if / С д.
First, we prove that there is no ρ G X that is contained in Gf with
some / : а —> ω for all α < ωι. Indeed, let us suppose the contrary.
Then, by our preceding remarks, for all α < ωι there exists precisely one
function /α:α-+ω with ρ G Gfa, and these functions are restrictions of
a single function F : ωι —> ω. Then none of the sets Gfa is empty, and
if we define the functions ga : a + 1 :—> ω by ga{Q = ία{ζ) {ζ < οι) and
<7α(α) = /α+ι(α) + 1, then the sets G9a are pairwise disjoint, nonempty,
open sets, which contradicts the assumptions.
Thus, each point ρ "dies out" at a certain level below ωι. Therefore, the
space X can be covered as follows:
ьии к- и 3
α<ωι f:a—Kjj
I g-.a+Ι^ω .
V 90f )
UU U ГК-С/
limit V/*,/
It is immediate that this is a covering by 2^° sets. It remains to show that
the summands are nowhere-dense sets. The set
Gf- (J G*
79
p:a+l-*
9Df

540
3. SOLUTIONS TO THE PROBLEMS
is nowhere dense because otherwise it would have a nonempty interior, and
the family of the sets Gg would not be maximal. For the other type of
summands, denoting the restriction of / to β by gp for β < α, we have
Π G„ С Π Ggg+1 cf]G90,
β<α β+Kot β<α
that is, the summand is a closed set minus its interior, which, being the
boundary of a closed set, is nowhere dense. D
Solution 2. Let {Ga : a < λ} be a well-ordering of the set of nonempty,
open sets of X. Since there are no isolated points, for each a < λ there
exists a pair Ga , Ga of nonempty, open sets with Ga r\Ga = 0 and
Got UGa Я Ga. Let ρ e X be arbitrary, and define the set' H(a,p) by
transfinite recursion on α as follows. Suppose that the sets Η(β,ρ) are
already defined for β < a. If (υβ<αΗ(β,ρ)) П Ga ^ 0, put H(a,p) = 0.
If (υβ<αΗ(β,ρ)) Π Ga = 0 and ρ G gL°\ put H(a,p) = Go)]', otherwise
put H(a,p) = Ga . Finally, let Fp = X — Ua<\H(a,p). It is obvious that
Fp is a closed set containing p\ moreover, it is nowhere dense, because if it
contained a nonempty, open set, then it would contain another nonempty,
open set, say, Ga, which would not contain p, but then, by the construction,
Ga would not be a subset of Fp.
It remains to prove that the number of different Fp's cannot be greater
than continuum.
For each ρ there can only be countably many a with H(a,p) φ 0, since
the sets H(a,p) are pairwise disjoint. Let the set of these be {&ξ(ρ) :
ξ < φ}, where φ = φ(ρ) < ω\. We claim that if ρ and q are points
with φ(ρ) = φ(ς) such that, for every ξ < φ(ρ), Η(α^(ρ),ρ) = G^(p)
implies H(a^(q),q) = G^K , (that is, for each ξ < φ(ρ), the points ρ
and q "ramify in the same direction"), then Fp = Fq. Taking this
statement for granted, we can complete the proof by the observation that
the cardinality of 0-1 sequences indexed by countable ordinals is
continuum; therefore the number of different Fp's is at most continuum. It
remains to prove our claim. It will suffice to show, by transfinite
induction, that &ξ(ρ) = &z(q)· If &ζ(ρ) — <2ζ(<ζ) holds for all ζ < ξ,
then obviously Η(αζ(ρ),ρ) = H^(q),q), and а$(р) is the first ordinal
a > supja^ : ζ < ξ} with Ga Π (и^<^Я(а^,р)) = 0; similarly, a^(q) is the
first α with Ga Π (и^<^Я(а^, q)) =0, and therefore а$(р) = ot^(q). D
Remark. Janos Kollar generalized the statement of the problem; by a
modified version of Solution 1, he proved the statement for T2-spaces. In
fact, Solution 2 does not use regularity, but this was not noticed by the
author.
Problem T.14. Construct an uncountable Hausdorff space in which
the complement of the closure of any nonempty, open set is countable.

3.10 TOPOLOGY
541
Solution 1. Let X be an arbitrary uncountable set. It is easy to see that
there exists a topology on X with the required properties if and only if there
exists a family A of subsets of X that satisfies the following conditions:
(1) \A\ = ω or A = 0 for all A G Д;
(2) A is closed under finite intersections;
(3) for every pair a,b e Α, α φ b, there exist A,B e A such that a e A,
b G В and Α Π Β φ 0;
(4) if А! С A and A e A \ {0}, then (иД') Π A = 0 implies | U A!\ < ω.
Indeed, if A has properties (l)-(4), then X endowed with the topology
generated by the basis A satisfies the requirements of the problem.
Conversely, if r is a topology required by the problem, then the family A of all
countable r-open sets satisfies the conditions (l)-(4).
Now choose X to be the set ω\ of countable ordinals. Let {(ba,ca) :
a G ω{\ be a sequence of all pairs of elements of ω\ with ba φ ca. By
transfinite recursion, we construct a sequence of families Aa · ot G ω\ of
subsets of ω\ so that for all α G ω\, we have
(la) \A\ = ω or A - 0 for all A G Да,
(2α) ^β<αΛβ С Да,
(За) Аа is closed under finite intersections,
(4a) |A*| <<*>,
(5a) there exist В,СбЛ with 6a € B, ca e C, and Б П С = 0,
(6а) if A G Да \ Цд<аД/з and A' G и)9<аД)д, then \А Π Д'| = ω.
Assuming that we have constructed such a sequence, we show that A =
Ua<u>i Да satisfies conditions (l)-(4). (1), (2), and (3) are obvious. In
order to check (4), let Л' С A and A G A \ {0}, with (uAf) Π A = 0. Let
a G ω\ be the smallest ordinal such that Д G Да- We shall prove that
UА* С иДа. To this end, it is sufficient to see that for each A' e Af, there
exists В G Aa with А' С В. Let Д' G Д' be arbitrary, and let β G ω\
denote the smallest ordinal В G Αβ such that В э A' and В П Д = 0.
(The definition of /3 is correct since there exists В G A with BDi' and
В П Д = 0; for example, Б = Д' is such.) Then β > a would contradict
(6a), so β < a, that is, ВеД^С Да.
In order to construct the sequence {Aa : α G o;i}, we need the following
simple lemma.
Lemma. If \E| = ω and ε is a nonempty, countable family of subsets
of E, then there exist subsets В and С of Ε such that BnC = 0 and
\B Π A\ = \C Π Д| = ω for all Д G ε.
Proof. Indeed, put ε = {An : η e ω} (possibly with multiple
appearances), and for η G ω, let An = и^еа;Дп^ be a partition of An into pairwise
disjoint infinite sets. Arrange the sets An\z (n,k G ω) into a sequence
{A'n : η G ω}. Since each A'n is infinite, by recursion we can define the
elements 6n, cn G ω such that 6n, cn e Afn\ ({bk : к <n}U{ck : к < η}) for
all η Εω. Then Б = {bn : η G ω} and С = {cn : η e ω} are as required.
Turning now to the construction of the sequence {Aa '· α < α;ι}, let
Во, Со С ω\ be two arbitrary countably infinite sets such that bo G Bo,

542
3. SOLUTIONS TO THE PROBLEMS
cq G Co and Во Г\ Co = 0. Then Aq = {i?o?CO,0} satisfies conditions
(lo)-(6o). Suppose that for some ordinal аЕц, q>0, we already have
defined the sequence {Αβ : β < a} so that conditions (Ιβ)-(6β) are fulfilled
for all β < a. Then, by applying the lemma to Ε = υβ<α(υΛβ) and
ε = ϋβ<αΛβ, there exist countably infinite sets B,C С Όβ<(χ(ΌΑβ) such
that \ВГ\А\ = \СГ\А\ = ω for all A G ^β<αΑβ. HAa is defined to consist of
finite intersections from the family υβ<αΑβυ{Β U {ba},С U {ca}}, then it
is routine to check that conditions (1α)-(6α) are fulfilled, which completes
the proof. D
Solution 2. Let Dbea countable, dense subspace of the product space
2ωι, which exists by the Marczewski-Pondiczery theorem. Let F = {fa -
a G ωχ} С 2ωι \ D such that
/β|(ωι\7)^//3ΐ(^ι\7) and /|(a;i\7)^/e|(wi\7)
for all a, /3,7 G α;ι, α φ /3, and f e D. Choose £) U F to be the underlying
set of the space X to be defined; let D be open in X, and let its subspace
topology coincide with the topology inherited from 2ωι.
It remains to define basis neighborhoods for fa (а Ец). То this end,
let В = {Ba : a G ω\] be the family of elementary open sets in the product
space 2ωι, let the finite sets Ta (а Ец) be the supports of the Ba, and
put λα = sup(U^<aT^) + 1. Then let the neighborhoods of /a be the sets
of the form {/Q}U(BnD), where В ranges over the elements of В that
contain fa and are supported in ω ι \ λ.
It is easy to check that we have defined a Hausdorff topology. By the
construction, for all Ba G β, we have
c\x(Ba Πϋ)Ό{/β:ββω1\(α + 1)},
and thus the closure of any nonempty, open set in X has a countable
complement. D
Remark. It can be shown that the maximum cardinality of spaces
satisfying the hypotheses of the problem is 2ω.
Problem T.15. Let W be a dense, open subset of the real line R. Show
that the following two statements are equivalent:
(1) Every function f : R —> R continuous at all points of R \ W and
nondecreasing on every open interval contained in W is nondecreasing
on the whole R.
(2) R\W is countable.
Solution. Suppose first that F = R\W is uncountable. Then we construct
a continuous function / : R —> R that is constant on every subinterval of
W and that is nonconstant and decreasing on the whole R.
We may assume that F has no isolated points (otherwise we pass to the
set of its points of condensation). Let W = UV, where V is a countable

3.10 TOPOLOGY
543
family of pairwise disjoint, open intervals. Consider the natural ordering
of V inherited from R. Since F is nowhere dense and perfect, this ordering
of V is dense (that is, between any two elements of V, there exists a further
element of V); therefore, there exists an order-preserving map q^Vq from
the set Q of rational points of the interval [0,1] into V. For xeR, put
f(x) = 1 - sup{q G Q : (-oo, x]C\Vq^ 0}.
The function / is obviously constant on elements of V, /(x) = 1 for χ G Vq,
f(x) =0 for χ G Vi, and / is monotone nonincreasing. Moreover, / takes
all values t G [0,1], since f(x) = t for χ = supU{V^ : q G Q Π [0,1 - £]}.
Therefore, / has no jumps and / is continuous.
Suppose now that R \ W = {an : η G N} is countable. Let ε > 0 be
arbitrary; we show that for any pair x, у G R, χ Φ у, we have f(x) <
f{y)+e.
Since / is continuous at the points an (n G N), for each η G N there
exists an open neighborhood Un of an such that for all u, ν G Un the
inequality f(u)—f(v) < ε/2η+1 holds. Let W = UV, where V is a countable
family of open intervals such that the restriction of / to any element of V
is nondecreasing. The family U = V U {C/n : η Ε iV} is an open cover of the
real line R; consider a finite subfamily U' = V U {Un : η < n0} of U such
that V'cV and W covers the interval [x, y]. Let L be the set of points y'
in [x, y] for which there exists a finite sequence χ = z$ < z\ < · · · < Zk = y'
such that [zi-ι, zi] С f/W with some U^ G W for all 1 < г < к. It is easy
to see that L is an open-and-closed subset of [x, y] containing x, and so
L = [x,y]. Therefore, there exists such a sequence zo,...,Zk with Zk = y-
Then
к no
f(x) - f(y) = Σ Uizi-ύ - f(zi)) < Σ ^r < ε. Π
i=l n=0
Remark. Statement (2)=>(1) of the problem can also be proved by trans-
finite induction on isolated points of R \ W. In connection with a few
incorrect solutions, it seems worth noting that such an induction does not
necessarily terminate at the first infinite ordinal.
Problem T.16. Let η > 2 be an integer, and let X be a connected
Hausdorff space such that every point ofX has a neighborhood homeomor-
phic to the Euclidean space Rn. Suppose that any discrete (not necessarily
closed) subspace D of X can be covered by a family of pairwise disjoint,
open sets ofX so that each of these open sets contains precisely one element
of D. Prove that X is a union of at most Hi compact subspaces.
Solution. Let X be a space that satisfies the hypotheses of the problem.
First, we prove the following.

544
3. SOLUTIONS TO THE PROBLEMS
Statement (*). If F is an arbitrary subspace of X, then there exists a
family Q of open subsets of X homeomorphic to Rn such that (UQ) Π F
is dense in the subspace F, and the family Q is σ-disjoint, that is, Q is a
union of countably many families consisting of pairwise disjoint sets.
If F = 0, then this statement is obvious. If F Φ 0, then consider a
maximal family U of pairwise disjoint, nonempty, separable, relative open
subsets of the subspace F. (Such a family exists by Zorn's lemma.) For
each U e U, let S(U) = {xn(U) : η = 0,1,...} be a countable, dense
subset of U. For each n, the set Sn = {xn(U) : U G U} is a discrete
subspace of F and, thus, of X. By the hypothesis of the problem, there
exists a family Qn = {Gn(U) : U G U} of pairwise disjoint, open subsets of
X, such that xn{U) G Gn(U) for all U G U. Since X is locally Euclidean,
we may assume that Qn consists of sets homeomorphic to Rn. Then the
family Q = U£L0(7n satisfies the requirements in (*), since the closure of
(U(?) Π F D U™=0Sn in the subspace F contains the closure of UK in F,
and, by maximality of U, the set UU is dense in F.
We define a sequence {(Fa,Qa) : α G ц} by transfinite recursion as
follows. (As usual, ω\ denotes the set of all countable ordinals.) Let Fo =
X and Go be a maximal family of pairwise disjoint, open subsets of X
homeomorphic to Rn. If a G ωχ, a > 0 and {(i^, Οβ) '· β < ol\ has already
been defined, then put Fa = X \ U^<a(U^), and let Qa be a σ-disjoint
family of open sets of X homeomorphic to Rn such that (U(?a) C\Fa is dense
in Fa. (The existence of such a family Qa is guaranteed by (*).)
We show that X = UaeuJl (U(?a). Arguing by contradiction, assume that
there exists a point χ G X \ UaGu;i(Ui/a), and consider a neighborhood
V of χ that is homeomorphic to Rn. Since (UQa) Π Fa is dense in Fa =
Χ \ υβ<α(υΟβ), it follows that the sets Va = {υβ<α (U^)) C\V (a G ωλ)
form a strictly monotone increasing sequence of order type ω\ of open sets
in V; this is impossible in a space homeomorphic to Rn.
Therefore, the family Q* = UaeuJlQa is a covering of X by open sets
homeomorphic to Rn, and Q* is a union of at most Hi disjoint families.
Since a subspace homeomorphic to Rn can only meet countably many
disjoint open sets, each G G G* meets at most Hi members of Q*.
Finally, consider the equivalence relation on Q* in which G, G' G G* are
equivalent if and only if there exists a finite sequence Go,..., Gk E5* such
that Go = G,Gk = G', and G{ Π Gi+1 φ 0 (г = 0,1,..., к - 1). Using the
result of the last paragraph, induction on к shows that every equivalence
class contains at most Hi sets from Q*. On the other hand, if Q С G* is an
equivalence class, then UQ is open and closed in X, and then Q = Q* by
connectedness of X. Therefore, the cardinality of Q* is at most Hi.
Thus, X is a union of at most Hi subsets homeomorphic to Rn, and
therefore it is a union of at most Hi compact sets. D

3.10 TOPOLOGY
545
Remarks.
1. The most well-known nonmetrizable topological manifold, the long line,
is an example of a space that satisfies the hypotheses of the problem but
is not a union of count ably many compact subspaces.
2. Several contestants proved the statement of the problem using the
assumption that all separable subspaces of X are second-countable. Gabor
Moussong showed that there exists a model of ZFC where this
assumption holds. Note, however, that by a construction of Μ. Ε. Rudin and
P. H. Zenor in some other models of ZFC there exist separable and not
second-countable topological manifolds that satisfy the hypotheses of
the problem.
3. Gabor Moussong noticed that the statement of the problem follows from
the continuum hypothesis, since the cardinality of a connected
topological manifold of dimension > 1 is continuum.
Problem T.17. A map F : P{X) -+ P{X), where P(X) denotes the
set of all subsets of X, is called a closure operation on X if for arbitrary
A,BcX, the following conditions hold:
(0 ACF(A);
(ii) AcB^· F(A) с F(B);
(Hi) F(F(A)) = F(A).
The cardinal number min{|A| : А С X, ^(^4) = X} is called the density
ofF and is denoted by d(F). A set Η С X is called discrete with respect
to F if и φ F(H — {и}) holds for all и G H. Prove that if the density
of the closure operation F is a singular cardinal number, then for any
nonnegative integer n, there exists a set of size η that is discrete with
respect to F. Show that the statement is not true when the existence of
an infinite discrete subset is required, even if F is the closure operation
of a topological space satisfying the T\ separation axiom.
Solution.
(a) We prove the statement by induction on n. For η = 0, the statement
is obvious; assume that it is true for n. Let the density of the closure
operation F on X be λ with cf(A) = κ < λ. Let \A\ = λ, with
F{A) = X, and well-order A in order type λ. By recursion, define a
sequence В = {x$ : ξ < λ} by x$ = min (A — F ({χη : η < ξ})). Then
F(B) d F(A) = X. Let С С В such that \C\ = к and С is cofinal in
£, and for Υ с В put F'(Y) = F(Y UC)nB.
It is easy to see that F' is a closure operation on В and that d(Ff) =
λ. By the induction hypothesis, there exists a set {x^ : г < n} that is
discrete with respect to F'. Let £n > ξι (i < n) be such that x$n G C.
Then the set {x%. : г < n} is discrete since x%n φ F ({х& : г < η}) С
F ({χη : η < £η}), and χξί φ F ({χξ. :i^j,j< η}) С
F' ({χ^ : г Φ J, j < η}) for г < п.
(b) Let λ be a singular, strong-limit cardinal number, and let Η С Ρ(λ)
be a maximal, almost disjoint family of countable sets. (Such a family

546
3. SOLUTIONS TO THE PROBLEMS
exists by Zorn's lemma.) Call F С λ closed if, for each Я G W, \F Π
H\ = ω implies F D Η. It is easy to see that by this a 7\ topological
space is defined in which there are no infinite discrete subsets. It
remains to prove that the density is λ. Let A$ С λ, with \Aq\ < λ.
For ξ < ωχ, we define a sequence Αξ by the following recursion: if
ξ = η + 1, then put Αξ = U{H e Η : \Η Π Αη\ = ω}, and if ξ is a limit
ordinal, then put Αξ = υ{Αη : η < ξ}. Since l-A^+il < \Αη\ω, it is easy
to see that \Αξ\ < \Αο\ω < λ for ξ < ωχ. On the other hand, Αωι is
obviously closed, and therefore Aq is not dense. D
Problem T.18. Suppose that К is a compact Hausdorff space and
QAn, where An is metrizable and An С Am for n < m. Prove
that К is metrizable.
Solution. First, we prove the following lemma.
Lemma. If a subspace Я of a metric space A is not separable, then there
exists an uncountable, discrete, closed subspace Ζ С Я.
Proof. Let Zn be a maximal subset of A such that d(x, y) > l/n for all
x, у € Zn, χ φ у. If a € A, then d(a,x) < l/2n can hold for at most one
point of Zn, so Zn is discrete. On the other hand, by maximality of Zn,
for each h £ Я there exists χ £ Zn such that d(h, x) < l/n, so U™=1Zn is
dense. If Я is not separable, then this implies that Zn is uncountable for
some n.
Now we prove that An is separable for all n. Indeed, assuming that it is
not separable, there is an uncountable discrete closed subset Zn in An; this
set is dicrete in An+i and is not separable, so there exists an uncountable,
discrete, closed subset Zn+1 in Zn. Thus, we obtain a sequence Zm such
that Zmi D Zm2 for mi <m2, and Zm is an uncountable, discrete, closed
subset of Am. Since К is compact, the set Z'm of accumulation points of
Zm is nonempty; therefore n^=nZ^ φ 0. On the other hand, Z^dAm = 0
since Zm is discrete and closed in Am, and therefore r\^=nZmr\U^=nAm =
0, and which is a contradiction.
Fix к and consider the set An nijt (k < n), which is a separable metric
space and therefore has a countable basis {Яп,т · m £ ω}· ^et Gn,m be
an open set in Ak such that Яп>т = Gn,m Π An. The family {Gn,m : n >
k, m (Ξ a;} is a subbasis for a Hausdorff topology on A^, which is coarser
than the compact topology of Ak; therefore, these two topologies are equal.
This means that Ak is a second-countable, compact topological space. Let
{fn,k : η G a;} be a family of continuous functions on Ak that separates the
points. Let Fn^ be a continuous function on К for which Fn,k\Ak = fn,k-
(Such functions Fn^k exist by the Tietze theorem.) The functions Fn^
separate the points of K; therefore the metric defined by these functions
induces the topology of K. D

3.10 TOPOLOGY
547
Problem T.19. Let U denote the set {/ G C[0,1] : \f(x)\ < 1 for all
χ G [0,1]}. Prove that there is no topology on C[0,1] that, together with
the linear structure of C[0,1], makes C[0,1] into a topological vector space
in which the set U is compact.
Solution. We assume that topological vector spaces are Hausdorff spaces.
Arguing by contradiction, assume that C[0,1] is a topological vector
space in which U is compact. Then U is closed since the topology is
Hausdorff. For each / G C[0,1] and δ > 0, the set
Ufi6 = {дв C[0,1] : \g(x) - f(x)\ < δ for all χ G [0,1]}
is also compact and closed.
Define the functions fn,gn G C[0,1] for each natural number η > 2 as
fn(x) = <
9n(x) = {
f 0 ifO<x<I-l,
nx+l-f ifl-l<x<l,
1 if \ < χ < 1,
0 if0<x<£,
nx-f if I < ж i+l,
1 if \ + \ < x < 1.
Put Kn = {g G C[0,1] : gn(x) < g(x) < fn(x) for all χ G C[0,1]} (n > 2).
Then K2 D Ks D · · ·, and Kn is compact and closed since Kn = С//п_1д П
C/Pn_i,i. So η^°=2ϋΓη ^ 0. Let / G Π^=2ϋΓη. Then f(x) = 0 for χ < 1/2,
and /(x) = 1 for χ > 1/2, which is impossible, since / is continuous. This
contradiction shows that U cannot be compact. D
Problem T.20. Let Φ be a family of real functions defined on a set X
such that к о h G Φ whenever fc e Φ (i e I) and h : X —> R7 is defined by
the formula h(x)i = fi(x), and
(1) k: h(X) —> R is continuous with respect to the topology inherited
from the product topology of R7. Show that f = sup{#j : j G J, gj G
Φ} = inf{/im : m G M, hm G Φ} implies f G Φ. Does this statement
remain true if (1) is replaced with the following condition?
(2) k: h(X) —> R is continuous on the closure of h(X) in the product
topology.
Solution. We may obviously assume that J and Μ are disjoint.
Define ρ : X —> RJuM by p(x)j = <7j(#), f°r 3 ^ «A and p{x)m = hm(x),
for m e M. Then, for all у G p{X), we have sup{yj : j G J} = inf{ym : m G
M}. Define к : p(X) -> R by fc(y) = sup{% : j G J} = inf{ym : m G M}.
Then / = к op, so, in view of (1), it suffices to show that к is continuous
onp(X).
Let у G p(X) and ε > 0. We shall define a neighborhood S of у in
p(X) such that \k(z) — k(y)\ < ε for all ζ G S. Choose j0 £ J with

548
3. SOLUTIONS TO THE PROBLEMS
9j0(y) > Ну) ~ (ε/2) = sup{^(y) : j e J} - (ε/2), and choose m0 e Μ
with hmo(y) < k(y) + (ε/2) = inf{/im(y) : m G Μ} + (ε/2). Such j0 and
ra0 exist by the definition of supremum and infimum. Put
S= {ze p(X) : \zjo - yjo\ < - and \zmo - ymo\ < - J .
Then, for ζ G S, we have
k{z) = sup{zj :jeJ}> zjo > yjo - - > k(y) - ε
and
k(z) = M{zm :me M} < zmo < ymo + - < k(y) + ε,
that is,
\k(z)-k(y)\<e.
Therefore, for every у G p(X) and ε > 0, there exists a neighborhood S
of у such that \k(z) — k(y)\ < ε for all ζ G S, that is, к is continuous on
p(X). This solves the first part of the problem.
If condition (1) is replaced with (2), then the statement is false.
Indeed, let X = N be the set of natural numbers, and let Φ consist of
all functions / : N —> R such that the sequence {/(n)}^! is convergent.
We claim that this is a counterexample.
Suppose that fa e Φ (г G /), that h : N —> R is defined by h(n)i = fi(n)
(i G /), and that к : h(X) —> R is continuous. Let b be the element of
R7 such that bi = lim/i. Then, by the definition of h, h(n)i —> fy for all
г G /; therefore, /i(n) —> b in the product topology. This implies that b is
contained in the closure of h(X) since it is the limit of a sequence in h(X).
By continuity of к on /i(X), we have that k(h(n)) —> /c(lim/i(n)) = Ar(6),
that is, the sequence k(h(n)) is convergent, and therefore к о h G Ф. This
shows that the hypotheses of the problem are satisfied.
Let J = Μ = Ν; let gj(n) = 1, for η < j and η even, gj(n) = 0 otherwise;
and let hm(n) = 0, for η < m and η odd, /im(n) = 1 otherwise. These
functions are indeed in Φ, since gj(n) —> 0 and hm(n) —> 1 for all j G J and
m e M. On the other hand, sup{^j(n) : j G J} = inf{/im(n) : m G M}
equals 1 if η is even and equals 0 if η is odd. This sequence is not convergent;
therefore / = sup{(?j(n) : j G J} = inf{/im(n) : m G M} does not belong
к)Ф. D
Problem T.21. Characterize the sets А С R for which
А + В = {а + Ь:аеА, b e В}
is nowhere-dense whenever В с R is a nowhere dense set.
Solution. We prove that the sets with this property are the bounded sets
with countable closure. More precisely, we show that for a set А С R the
following are equivalent:

3.10 TOPOLOGY
549
(i) If В С R is nowhere dense, then so is A + B;
(ii) For any sequence an (n = 1,2,...) of positive numbers, there exists
a finite family of intervals Ij (j = 1,..., k) that covers A and such
that the length of Ij is \Ij \ = a j (j = 1,2,..., k);
(iii) A is bounded and its closure is countable.
(i) => (ii): Suppose that (ii) does not hold. Then there exists a sequence
consisting of positive numbers an (n = 1,2,...) such that for any finite
family of intervals Ij, \Ij\ = clj (j = 1,2, . ..,&) we have A <jL U^=1/j.
Using this, we shall construct a nowhere-dense set В such that A + В
contains all rational numbers, which contradicts (i).
Arrange the rational numbers into a sequence rn (n = 1,2,...). Now we
define the numbers xn and yn by induction on n. Let X\ G A be arbitrary,
and yi = η — x\. Suppose that к > 1 and the numbers yi, 2/2, · · ·, 2/fc-i
have already been defined. Then, by our assumption, the intervals
Ij = {гь-Уз~Ц ι ^ - % + у) , j = 1,2,..., fe - 1
do not cover A. Therefore, we can choose an element хь е A\ O^Z^Ij.
Finally, put yk = Гк~ Xk- Then, with В = {ук : к G N}, the set A + В
contains all rational numbers since τ к = Xk + У к £ A + В. We show that
every point of В is isolated. Indeed, for j < k, we have Xj £ (r^ — yj —
CLj/2,rk - yj + clj/2). Therefore, yk = П - xk <£ {yj ~ aj/^Vj + а//2).
This means that the set В contains at most j points in the neighborhood
of radius clj/2 of the point yJ? and thus yj is an isolated point of B. This
implies that В is nowhere dense.
(ii) => (iu): If (ϋ) is true, then A is obviously bounded. Arguing by
contradiction, suppose that (iii) does not hold. Then the closure of A is
uncountable and, by the Cantor-Bendixson theorem, contains a nonempty,
bounded, and perfect set P. Let us call a sequence (αι,α2,...) of
positive numbers insufficient if for any family of closed intervals Ij, \Ij\ = dj
(j = 1,2,..., /c), we have Ρ <£ Okj=1Ij. In order to arrive at a
contradiction, by induction we shall define an infinite sequence of positive numbers
(αχ,α2,...) whose all finite initial segments are insufficient. Let a\ be a
positive number that is smaller than the diameter of P; then (αϊ) is
obviously insufficient. Suppose that η > 1 and we have already defined an
insufficient sequence (αϊ, α2,..., an). We show that there exists a number
αη+ι such that (αϊ, аг,..., αη+ι) is insufficient. Suppose this is not true;
then for every к G N and αη+ι = l/к there exists a sequence of intervals
Jj\Jfe> ΙΌ,*I = % (j = 1,2,..., η + 1) such that Ρ С U^1/^ holds. We may
assume that the intervals Ij^ all meet the set P, and thus they are all
contained in a bounded set. By passing to a suitable subsequence, we may also
assume that for each fixed j = l,2,...,n + l, the sequence of intervals Ij^
converges to an interval Ij when к —> oo. Then \Ij\ = dj, for j = 1,2,..., n,
and /n+i consists of a single point. Since (αϊ, α2,..., αη) is insufficient, the
set Ρ is not contained in U^=1/j. Moreover, being perfect, Ρ has
infinitely many points in the complement of \Jj=lIj. So we can choose a point

550
3. SOLUTIONS TO THE PROBLEMS
x G P\ Wj+flj. But then, foi sufficiently large k, χ £ U^1/^, which
contradicts the choice of the intervals Ij^. This proves that the sequence
(αϊ, α2,..., 1/fc) is insufficient, completing the induction.
(iii) => (i): Suppose that, to the contrary of (i), with a nowhere-dense
set B, the set A + В is dense in some interval /. Let A0 denote the closure
of A; then Aq + В is also dense in /. Define the sets Aa for all countable
ordinals as follows: If β > 0 is an ordinal such that Aa has already been
defined for all a < /3, then let Α β = Πα<βΑα if β is a limit ordinal, and
let Αβ be the set of accumulation points of Aa if β = a + 1. Since A$
is countable, there exists a countable ordinal 7 such that ΑΊ = 0. Then
obviously ΑΊ + Β = 0 and ΑΊ + Β is not dense in /. Let β denote the
smallest ordinal for which Α β + Β is not dense in /; then β > 0. Let J
be a subinterval of J with (Αβ + Β) Π J = 0, and let 0 < ε < | J|/3. Let
C/(iJ, <$) denote the neighborhood of radius δ of the set H. It is easy to see
that [/(Я, i) + Bc U(H + £, <5) holds for every set Η and δ > 0. Thus,
υ(Αβ, ε) + В С υ(Αβ + Б, ε), and therefore υ(Αβ, ε) + В does not meet
the middle third К of J.
Suppose that β is a limit ordinal. Since {Aa: a < β} is a nested sequence
of compact sets whose intersection is Αβ, there exists a < β with Aa С
[/(Α^,ε). Then Aa + В does not meet if. But this is impossible, since
a < β and Aa + В is dense in /.
Finally, suppose that β = a + 1. Then V = Aa\ υ(Αβ,ε) is a finite
set, since all accumulation points of Aa are in Αβ. Then the set V + В is
nowhere dense. Therefore, from (Aa + В) С) К = (V + В) С) К, it follows
that (Aa +В)Г\ К is nowhere dense. But again this contradicts that a < β
and that Aa + В is dense in /. This completes the proof. D
Problem T.22. Prove that if all subspaces of a Hausdorff space X are
σ-compacty then X is countable.
Solution. Since X itself is a union of countably many compact sets, it is
sufficient to prove the statement for compact spaces X.
Suppose that X has no subset that is dense in itself. Then there is a
well-ordering on X in which all initial segments are open. Indeed, suppose
that the points ρβ G X have already been defined for all β < α and that
Χα = {ρβ : β < α} φ Χ. Then there exists a point pa G X \ Xa that
is not an accumulation point of Χ \ Χα, and so pa G Ga Q Xa U {Pa}
with a suitable open neighborhood Ga of pa. This recursion defines a well-
ordering of X, and υβ<αΰβ С Ха = υβ<α{ρβ} Q ^β<αΟβ shows that
Χα = ^β<αΟβ is an open set for each a. An initial segment of order type
ω\ in X cannot be σ-compact since a compact subset in X must have a
greatest element. Therefore, X is countable.
Thus, it suffices to prove that X does not contain a subset dense in
itself. Arguing by contradiction, assume that there exists such a subset
Υ. Then Υ is dense in itself and closed. We define a continuous function
/ : Υ —> [0,1] as follows. Since У is a compact Hausdorff space, for any
two distinct points ρ and q there exist open sets Gp, Gq and closed sets Fp,

3.10 TOPOLOGY
551
Fq such that ρ G Gp С Fp, q G Gq С Fg, and Fp Π Fg = 0. So, for all finite
0-1 sequences {ε^}, we can define closed sets -Aei,...,en С У so that
Λ1ν..,εη,0 Π Αει,...,εη,1 = 0 and Αει,...,εη,0 U Αει,...,εη,1 Q Л^,...^
hold for all natural numbers η and ει,..., εη G {0,1}. For any infinite 0-1
sequence ει, ε2,..., the set
oo
^(ει,ε2,...) = p| Λει,ε2,...,εη
n=\
is nonempty and closed. If χ G Α (ει, ε2,...), then put
г=1
The function / maps Α(ει,ε2, · · ·)? which is a closed set since it is an
intersection of closed sets, continuously onto the interval [0,1]. Then /
extends continuously over Υ with Im / = [0,1]. The interval [0,1] contains
2^° σ-compact sets (that is, i^-sets), since the number of closed sets in
[0,1] is 2*V Therefore, there exists a set Ζ С [0,1] that is not σ-compact;
then its inverse image /_1(Z) С Υ cannot be σ-compact, which contradicts
the hypothesis of the problem. This proves that X contains no subset that
is dense in itself. D

552
3. SOLUTIONS TO THE PROBLEMS
3.11 SET THEORY
Problem N.l. Does there exist a function /(x, y) of two real variables
that takes natural numbers as its values and for which /(x, y) = f(y,z)
implies χ = у = ζ?
Solution 1. First, we show that it is sufficient to construct for each real
number c, two sets Ac and Bc of natural numbers that satisfy the following
conditions:
(1) 1 φ Ac, 1 φ Bc for all c,
(2) Ac Π Bc = 0 for all c,
(3) Ac Π Bd φ 0 for с φ d.
Indeed, put f(c,c) = 1, and for с ф d, let /(c, d) be an arbitrary element
of the nonempty set Ac Π Bd- Now, if
ffay) = f(y,z) =тф\,
then
m e АХПВУ and m e AyC\ Bz,
that is,
m e Ax Π Βν Π Ay Π Bz С Ay Π By = 0,
which is a contradiction. But m = 1 is only possible when χ = у = ζ.
In order to construct Ac and £c, map the set R of rational numbers
bijectively onto the set N = {2,3,... }; let n(r) denote the image of r e R.
Assign to each real number с a sequence {гк{с)}™=1 that converges to с
and consists of rational numbers different from c. Let
4 = {n(rfc(c)):fe = l,2,...}cJV,
BC = N- Ac.
It is easy to see that these sets Ac and £c satisfy conditions (1), (2), and
(3). Thus, we have constructed a function with the required property. D
Solution 2. Arrange the set of rational numbers into a sequence ri, Г2,
.... Define the function /(x, y) the following way:
If x = y, put /(x,y) = 1;
If χ < y, put /(x, у) = 2г, where г is smallest possible with χ < η < у;
If χ > у, put /(χ, у) = 2г+1, where г is smallest possible with χ > r^ > y.
We show that /(x, y) is as required. It is obviously defined on the reals,
and its range is the set of natural numbers.
Suppose that
ffay) = f(y>z) = m·
If m = 1, then χ = у = ζ is the only possibility. If m > 1 and m = 2г, then
x <ri <y and у <ri < z,

3.11 SET THEORY 553
which is a contradiction. Similarly, if m > 1 and m = 2% + 1, then
x> Гг> у and у > u> ζ,
which is again a contradiction. Therefore, χ = у = ζ follows. D
Solution 3. We will decompose the plane into a countable family of pair-
wise disjoint subsets with the following property: if one of the lines parallel
to the coordinate axes through the point (у, у) intersects one of the subsets,
then the other does not, except when the subset is the line χ = у, which
will be one of the subsets. It suffices to construct such a partition of the
open half-plane below the line χ = у, because by reflecting in the line χ = у
and adding the line χ = у, we obtain the required partition of the whole
plane.
Put
A> =
Al-
A°-k-
Akn-
An-k-
= {(*,
= {(*,
= {(*,
= {(*
= {(*
У)
У)
У)
,y)
>У)
where fe, η = 1,2,... . It is obvious that the union of these sets is the open
half-plane below the line x = y.
This defines a partition of the whole plane, as we said above. Label the
sets of this partition by the natural numbers, and assign to the point (x, y)
the label of the set containing (x, y) as the value of the function /(x, y).
If x, y, ζ are real numbers with /(x, y) = /(y, 2), then the points (x, y)
and (y, z) are in the same subsets, but then (x, y) (and (y, z)) is on the
line through the point (y,y) parallel to the x-axis (y-axis, respectively),
and any of the subsets can only meet one of these, except for the subset
{(x, y) : χ = y}, but then χ = у = ζ. Π
Remarks.
1. Obviously, the existence of such a function is a question of cardinality.
In general, it makes sense to ask whether there exists a function /(x, y)
defined on a set A of cardinality m that takes its values in a set В of
cardinality η (both m and η are infinite) and for which /(x, y) = /(y, z)
implies χ = у = ζ. J. Gerlits, L. Lovasz, L. Posa, and M. Simonovits
proved that such a function exists if and only if m < 2n. Here we present
the proof by Posa.
x> 0,
χ > k,
—k<x
2k-1
2n
У
к
<
<0|,
-1 <y
-fc + 1,
2k
<x < τ-,
— 2"'
-2k + 1
<
С т. < —
<k},
y<-
2k-
2n
2k+ 2
-*},
2
— <
У .
■2k
- 2n J
in
2n
<У<
-2k + 1 Ί
2n J'

554
3. SOLUTIONS TO THE PROBLEMS
Let m < 2n. It suffices to give the construction for a set A of cardinality
2n. Let a be the smallest ordinal of cardinality n. We may assume that
A is the set of sequences of the numbers 0 and 1 of order type a. Define
the function /(x, y) as follows:
If χ = y, put /(x, y) = 0;
If χ φ у, then let /3 be the smallest ordinal where the elements in
χ and у are different, let j be the /3th element of χ (j = 0 or 1), and
put /(x, y) = (/3, j). The range of / is the set of the symbol 0 and the
ordered pairs (/3, j), where β < a and j = 0 or 1. Therefore, it has
cardinality η and can be mapped bijectively onto B.
/(x, y) = /(y, ζ) = (β^) means that all elements before /3 agree both
in χ and y, and in у and z, and the /3th element is j in χ and y. But
then the /3th elements of χ and у are equal, which is a contradiction.
Therefore, /(x, y) = f(y,z) = 0, which implies that χ = у = ζ. (This
part of the statement is also proved by Bollobas and Juhasz.)
Suppose that m > 2n, and there exists a function required in the
problem. Consider the range of /(x, y) when χ is kept fixed. This is a
subset of B, and since the cardinality of all possible x's is greater than
the cardinality of the set of all subsets of B, for some х\ф хъ the range
of /(xi, y) equals that of /(x2, y)> So, the value /(xi, x2) is contained in
the range of /(x2,y), that is, for some yo? /(#ъ#2) = /(#2?Уо)? which
is a contradiction.
2. Bela Bollobas and Miklos Simonovits notice that there exists a function
F(xi,... ,xn) defined on the real numbers that takes its values in the
natural numbers, and for which
F(a,x2,...,xn) = F(yi,a,y3,...,yn) = ··· = F(m,... ,un-i,a)
implies a = Xi = yi = · · · = щ (г = 1,2,..., η). For example,
F(xu...,xn)= flp{}Xi'Xi)
is such a function, where the pij (г, j = 1,..., n) are different primes,
and /(x, y) is the function defined in either solution of the problem.
Problem N.2. Prove that there exists an ordered set in which every
uncountable subset contains an uncountable, well-ordered subset and that
cannot be represented as a union of a countable family of well-ordered
subsets.
Solution. Let R be the set of all limit ordinals less than ωχ, and for а е R,
let /a (ft) be a monotone increasing sequence (of order type ω) of ordinals
converging to a. Order R as follows: for a, /3 G i2, a -< /3 if /a(ft) < ίβ{η)
holds for the first η with /a(n) φ fiAn)·

3.11 SET THEORY
555
Any uncountable subset X of R contains a well-ordered subset of order
type ω\ with respect to the ordering -<. Indeed, let
X\n = {(αο, αϊ,..., αη) : there exists a e X, such that /a(m) = a(ra)
for all m < n}.
For η sufficiently large, X\n is uncountable, since for large enough η the
ordinal
7n = sup{/a(n) :aeX}
equals ωι, since, by /a(n) —> a,
sup{7n : η < a;} = sup{a : a G X} = ωι.
Order X\n lexicographically, that is,
(α0,...,αη) -<' (/30,...,/3n)
if am < Pm for the smallest m with am φ /3m. Then X|n is well-ordered
and, if η is large enough for X\n to be uncountable, it contains a well-
ordered subset of order type ω\. Let X' be a subset of X such that for
any
(α0,...,αη) G X|n,
there exists a unique a e X' with
/a(0) = a0, . . . , /a(™) = OLn.
Then the ordered set (Xf, -<) is isoir. Orphic to (X|n, -<;). Since we have
seen that the latter contains a well-ordered subset of order type ωι, the
same is true of the former. This is what we wanted to prove.
Now we show that R is not a union of a countable family of well-ordered
subsets. First we make a digression.
A function that maps a set of ordinals into the ordinals is called
regressive if /(ξ) < ξ holds for all ordinals ξ φ 0 in the domain of /. If the
domain of / is a subset of ωι and the set
tf:/(0<M}
is not cofinal with ωι for any μ < ωι, then / is called divergent. X С ωι
is called thin if there exists a regressive divergent function defined on X;
otherwise, it is called stationary.
The set ωι is stationary. Indeed, assuming that / is a regressive function
on ωι, the sequence
e,/(0,/(/(0),···,
being a descending sequence of ordinals, can contain only a finite number
of elements different from zero. Therefore,
ui= \J Xn

556
3. SOLUTIONS TO THE PROBLEMS
where
Xo = {0},
and
Xn+i = {ξ : /(0 € Xn}.
Assuming further that / is divergent, induction on η shows that supXn <
ω\ for all n, which obviously is a contradiction.
From this it easily follows that the set R of limit ordinals less than ω\
is also stationary.
It quickly follows from the definition that the union of a countable family
of thin sets is also thin. Further, if X is stationary and / is a regressive
function on X, then for some μ < ω\ the set
Χμ = {£ ·· № = μ}
is stationary (this is a special case of Fodor's theorem). Indeed, assume
that all Χμ are thin, and let /μ be a regressive divergent function defined
on Χμ. Then the function g defined on X by
g(£) = max(/i, /μ(£)) (£ e Χμ, μ < ωλ)
is regressive. It is divergent, too, because for all ν < ω\ the set
is not cofinal with ωχ, since neither one of the summands in the right-
hand side is, because the functions /μ are divergent. This contradicts the
assumption that X is stationary, which proves our statement.
Let X С R be stationary. We show that X is not well ordered with
respect to the ordering -<. Put Yli = X and
Yn = {ae Yn-i : fain) = <5n},
where δη is the smallest ordinal for which Yn is stationary (such δη exists
by the Fodor theorem proved above).
Put
where
Y^ = {ae Гп_1 : fa(n) < U·
By the choice of <$n, Y^ is not stationary, and neither is X — X'.
Put
δ = sup{<$n : η < ω}.
It is obvious that δ <ω\. Let α G X1 be greater than δ. (Χ' is stationary
and so it is cofinal with ω\.) Since fa(n) —> a, for some η < ω we have
fain) > δ > δη,

3.11 SET THEORY
557
so /a(m) = 6m cannot hold for all m. It follows from a G X' that $а{т) >
6m for the smallest m with $а(т) φ 6m. Thus, for this ra, α is greater
than all elements of Ym with respect to the ordering -<.
Put Xo = X, cto = a, and X\ = Ym. Since Ym is stationary, we can
repeat the above argument with X\ instead of Xq. We obtain a\ and X2,
then a2 and X3, and so on. Since с*о >- ot\ >- a2 >- ..., X is not well-
ordered with respect to the ordering -<, and this is what we wanted to
prove.
Now, if R = \Jn<UJRni then some Rn is stationary, and this Rn is not
well-ordered with respect to -<. This proves the second statement of the
problem. D
Remark. Laszlo Babai showed that for each cardinal number κ > ω,
there exists an ordered set В of cardinality к such that all subsets of
cardinality к contain a well-ordered subset of cardinality к and В is not a union
of a family of cardinality less than к of well-ordered subsets.
Problem N.3. Let < be a reflexive, antisymmetric relation on a unite
set A. Show that this relation can be extended to an appropriate finite
superset В of A such that < on В remains reflexive, antisymmetric, and
any two elements of В have a least upper bound as well as a greatest lower
bound. (The relation < is extended to В if for χ, у e A, x < у holds in A
if and only if it holds in B.)
Solution. В will consist of all the elements and subsets of A without
identifying the elements with the corresponding singletons. In what follows,
lowercase letters denote elements, whiie capitals mean subsets.
Define < as follows:
α < 6, as given,
a<P, if α G Ρ,
Ρ<α, ifa&P,
Ρ < Q, if Ρ С Q.
This is obviously a reflexive and antisymmetric relation. To show the other
property, it suffices to treat incomparable elements, that is, two
incomparable elements in A, or two subsets of A, none containing the other, as a
subset.
If a, b G A are incomparable, {a, 6}, A \ {a, b} are the least upper (resp.
greatest lower) bounds. If P,Q С A, then PUQ and Pf)Q will serve as
the least upper and greatest lower bounds. D
Problem N.4. Let J7 be a famiiy of subsets of a ground set X such that
UFeJrF = X, and
(a) if А, В G T, then AUBCC for some С G J7;
(b) if An G Τ (η = 0,1,...), В G Ту and А0 С Αι с ..., then, for some
k>0,AnnB = AknBforalln> k.

558
3. SOLUTIONS TO THE PROBLEMS
Show that there exist pairwise disjoint sets I7 (7 G Γ), with X =
U{Xy : 7 G Γ}, such that every ΧΊ is contained in some member of
T, and every element of Τ is contained in the union of finitely many
X/s.
Solution. Let < be a well-ordering of F. Let Γ be the set of all finite
subsets of F. Well order Γ as follows:
7b 72 ^Γ, 7i ^72, 7i = {Αι, ..., An}, 72 = {Bu..., Bm},
An < · · · < A\, Bm < ... B\, m < η.
Put 71 > 72 if Ai = Bi for 1 < г < т. If, on the other hand, г < т is
the least number with Ai φ Bi then put 71 < 72 or 71 > 72 if Ai < Bi
or Ai > Bi. A straightforward checking shows that this well orders Γ. We
notice that 71 < 72 implies 71 < 72.
For 7 G Γ, we define ΥΊ G F as follows. If 7 = {A}, put ΥΊ = A. Assume
that Yg is defined for δ < 7. Then let ΥΊ be a member of F, covering every
Υβ {β < 7). This is possible by (a).
Now put Χγ = ΥΊ \ υδ<ΊΥδ- The only nontrivial thing to prove is that
every A G F is covered by finitely many Χβ. We prove by transfinite
induction on 7 that ΥΊ is covered by finitely many X$'s. If this first fails
for 7, there exists β\ < β2 < · · · < 7 such that ΥΊ Π Χβη φ 0. Let
ΥΊ = {Αι,..., Ak}, Αι > · · · > Ak. For every η there is a 1 < j < к such
that
Au...,Aj-i G/3n, Aj £βη.
We can assume, by shrinking, that j is the same for every n. Put r\n =
max(/3i,... ,/3n) < 7. As 7 is minimal, l^n meets finitely many Χβ. As
Ym С 1^2 С ..., by (b) there is an m such that YVm ΠΥΊ = 1^TO+1 Γ\ΥΊ = ...,
so У„т D У„т+< ПУ7 2 ^m+i ПУ7 ^ 0 (t = 0,1,...) that is, Υηψη meets
XVm, ^r/m+i 5 · · · 5 a contradiction. D
Problem N.5. Show that there exists a tournament (T, —>) of
cardinality Hi containing no transitive subtournament of size Hi. (Ά structure
(T, —>) is a tournament if —> is a binary, irreflexive, asymmetric, and tri-
chotomic relation. The tournament (T, —>) is transitive if —> is transitive,
that is, if it orders T.)
Solution. We use the existence of a Specker type, that is, an ordered set
(A, <) of size Hi, not containing subsets similar to ωι, ω{, any
uncountable subset of the reals. See (P. Erdos, A. Hajnal, A. Mate, R. Rado,
Combinatorial set theory: Partition Relations for Cardinals, Akademiai
Kiado, Budapest, 1984, P- 326.) Enumerate A as {a(a): a < ωι}, and let
{x(a): a < ωι} be different real numbers in [0,1]. If a < β, put a <— β iff
either x(a) < χ(β) and a(a) < α(β) or x(a) > χ(β) and a(a) > α(β).

3.11 SET THEORY
559
Let В С Abe uncountable. Let X = {a < ω\: a(a) G B}.
Claim 1. There is an α G X such that
{peX: a(a) < α(β) and x(a) < χ(β)}
is uncountable.
Proof. For a e B, let /(a) be the least £ G [0,1] such that χ(β) < t
for all but countably many β G Χ, α(β) > a. Since / is a non-increasing
real-valued function on (A, <), it can only have countably many different
values; otherwise, there would be an uncountable subset of A, similar to a
set of reals. Hence / is constant on an uncountable Во С В. Define Xo
analogously to X. We can choose a0 G Xo such that {β e Xo'· χ(β) >
x(cto)} is uncountable; otherwise (A, <) would contain a subset of type ω{.
Now we can choose a e Xo such that a(a) > a(cto) and x(a) < x(cto).
Since g(a(a)) = g(a(cto)) > #(α), there are uncountably many β G X such
that α(β) > a(a) and χ(β) > x(a).
Claim 2. There are uncountable Χο?^ι !Ξ Χ such that if α G Xo,
β G Χι, then α(α) < α{β) and χ(α) < x(/3).
Proof. Let C/ be the set of those α G X such that
{/3GI: α(α) < α{β) and χ(α) < χ(β)}
is countable, and let L be the set of those a e X such that
{/3 G X: α(β) < a(a) and x(/3) < x(a)}
is countable. If U or L is uncountable, we get a contradiction by Claim 1,
so we can select a G X\U\L, and put
X0 = {β е X: a(/3) < a(a) and χ{β) < x{a)} ,
Χλ = {β^Χ: a(a) < α(β) and x{a) < χ(β)} .
By an application of Claim 2 to Xo and the Specker type (A, >), one
can find Υο,Υι ζζ Xo such that a G Yo, /3 G Y\ imply a(a) < a(/3), and
x(a) > χ(β). Now select a e Yo, β e Χι, Ύ e Yi with α < /3 < 7. Then
a <— /3 <— 7 and 7 <— a, so our tournament is not transitive on5. D
Remark. The result in the problem was first proved by R. Laver.
Problem N.6. Assume that R, a recursive, binary relation on N (the
set of natural numbers), orders N into type ω. Show that if f(n) is the nth
element of this order, then f is not necessarily recursive.
Solution. We use the following well-known facts. There is a set А С N
that is recursively enumerable, that is, the range of a recursive function, but

560
3. SOLUTIONS TO THE PROBLEMS
is not recursive, namely, its characteristic function is not recursive. For
every infinite т.е. set A there is a recursive function enumerating A's
elements in a one-to-one manner. An infinite set possesses an increasing such
enumeration if and only if it is recursive.
Take a (necessarily infinite) т.е. but not recursive set A, and a function
g enumerating A without repetition. Put aRb iS g(a) < g(b). Clearly, R is
recursive and orders N into type ω. If /(n)=the nth element by this order,
and / is recursive, then the recursive h(n) = g(f(n)) would enumerate A's
elements in increasing order, which is impossible. D
Problem N.7. Let Η be the class of all graphs with at most 2*° vertices
not containing a complete subgraph of size Ni. Show that there is no graph
Η eH such that every graph in Η is a subgraph of H.
Solution. We have to show that if (V,G), a graph with |V| = 2^°, does
not contain a complete graph on Hi vertices, then there is a graph (W,H)
with these properties such that (W, H) cannot be embedded into (V, G).
Let W be the set of those functions injecting a countable ordinal
(possibly Ο Φ 0) into a complete subgraph of G. To define Я, join two such
functions if one extends the other. Clearly,
\W\ = ^ |a->V|<»i.(2^)No=2«°.
α<ωι
First, we show that (W, H) does not contain a complete Hi-gon. Assume
that {fa: a < ω\} are pairwise joined, and that the ordinals Dom (/a) are
in increasing order. By the construction of Я, for β < α, fa extends f$, so
the union of the /a's gives a function /: ω\ —> V onto a complete Hi-gon
in (V, G), a contradiction.
Next we prove that (W, H) may not be embedded into (V, G). Assume
that h: W —> V is such an embedding. By transfinite recursion on α < ω\,
we define fa: a —> V, fa G W, and xa = h(fa) G V in such a way that fa
extends f$ for β < a. Put /o = 0, /o G W. If //3 (β < a) are defined, then
{χβ: β<α} = Η{/β: β < a}
is a complete subgraph, so we can define /α(/3) = Χβ- This clearly satisfies
the requirements. But then {xa: a < ω{\ is a complete subgraph in G, a
contradiction. D
Remark. This is an unpublished result of R. Laver.
Problem N.8. For which cardinalities к do antimetric spaces of
cardinality к exist?
(Χ, ρ) is called an antimetric space if X is a nonempty set, ρ : X2 —> [0, oo)
is a symmetric map, ρ(χ, у) = 0 holds iff χ = у, and for any three-element
subset {01,02,^3} of X
Q(o>if,Q>2f) + Qfaf, ^3/) < Qfaif, ^3/)
holds for some permutation f of {1,2,3}.

3.11 SET THEORY
561
Solution. For 0 < к < 24 First, if X С R is nonempty, then (Χ, ρ) is
antimetric, where g(x,y) = (x — y)2 and clearly \X\ can take any value in
the given interval.
For the other direction, let (Χ,ρ) be antimetric and \X\ > 2*V Color
the complete graph on X by the integers as follows. Let {x, y} get color k
iff 2* < ρ(χ, у) < 2/c+1. By the Erdos-Rado theorem (see P. Erdos, A. Haj-
nal, A. Mate, R. Rado, Combinatorial Set Theory: Partition Relations for
Cardinals, Akademiai Kiado, Budapest, 1984, P- 98), there exist x, y, ζ
such that {x, y}, {x, z}, {y, z} get the same color. But then, x, y, ζ do not
meet the requirement on antimetricity. D
Problem N.9. If (A, <) is a partially ordered set, its dimension,
dim (A, <), is the least cardinal к such that there exist к total orderings
{<a: a < к} on A with < = Па<к <a. Show that if dim(A, <) > No, then
there exist disjoint Aq, Αχ С A with dim(Ao, <), dim(Ai, <) > Hq.
Solution. Assume indirectly that (A, <) is a counterexample of minimal
cardinality λ. Without loss of generality, A = λ. Put
7 = {£CA:dim(£,<)<N0}.
Clearly, if В has \B\ < λ, then Bel.
Claim 1. If Bn e I (n = 0,1,...) and В is such that every pair of В is
covered by some Bn, then Bel.
Proof. Straightforward from the definition.
Claim 2. There are Β0,Βλ el such that B0 Π Βλ = 0 and £0 U i?i g 7.
Proc/. If not, then 7 is a σ-complete primideal containing every subset
of cardinality < λ. For a < λ, let <a,n be a total order on a establishing
dim (a, <) < N0. If, for /3 < 7 < λ we put /3 <n 7 if {a < λ: /3 <a,n 7} g J,
then <n establish that dim (λ, <) < Hq, a contradiction.
We can assume that Bo,B\ as above are selected with к = \B\\ minimal.
Put
h = {С С Вг: Bo U С Ε /} .
Then #1 ^ 7i and every subset of Bi of size < я is in Д. Repeating
the argument of Claim 2, we get that there exist Co, С ι e 11 such that
Co Π Ci = 0 and #0 U Co, 50 U Ci ^ 7. Repeating the above argument for
Co, Bo this time, one gets Do,D\ С 7?0, Α) Π D\ = 0 such that Do U Co,
£>i U Co e 7, £>o U D\ U Co ^ 7. As every pair of D0 U D\ U Ci is contained
either in 7?o or in Do U Ci or in Di U Ci, by Claim 1, either Do U Ci or
£>i U Ci is not in 7. In the former case, Do U C\ and D\ U Co are two
disjoint sets not in 7, and we conclude similarly in the other case, too. D

562
3. SOLUTIONS TO THE PROBLEMS
Problem N.10. A binary relation -< is called a quasi-order if it is
reflexive and transitive. The infimum of the quasi-order (Q, -<) is the greatest
subset J CQ such that
(i) for every В e Q there is an A G J with A<B, and
(ii) A<B,A,BeJ imply В -< A.
Let X be a finite, nonempty alphabet, let X* be the set of all finite
words from X, and let V be the set of infinite subsets of X*. For
А, В eV, let A -< В if every element of A is a (connected) subword of
some element ofB. Show that (V, -<) has an infimum, and characterize
its elements.
Solution. As follows, subword will always mean connected subword. Let
α <b denote that α is a subword of b. Call A eV minimal, if for all w G A,
all but finitely many subwords of the elements of A contain w, as subword.
Claim 1. If A is minimal, В -< A, then A -< B.
Proof. Assume that w G A and к is so large that every subword in A
longer than k, as В -< A, is a subword, so w < b.
Claim 2. If A G V is not minimal, then there exists а В G V such that
В -< A but A ^ B.
Proof. Let w be a subword of A such that
Η = {h: there is an a G A, h < a, w ^ h}
is infinite. Then Η -< A, but A -ft Η is established by w.
CZazra 5. For every A eV there is a minimal В -< A.
Proof. Such а Б is given by the following algorithm. Step 1: Let X =
{αϊ,..., an}. If, among the subwords of A, infinitely many do not contain
αϊ, let Hi be their set. Hi eV, Hi ^ A, and it does not contain a\. Then
get successively Я2,Яз,... using а2,аз, If Hn-i is defined, it may
only contain an, so it is minimal. We can therefore assume the existence
of Щ-1 such that all but finitely many subwords of Ηι-ι contain a*.
Put bi = ai, Bi = Hi-ι -< A, and there follows Step 2. Here Step k:
If 6fc_i, Bk-ι are given, and bi < · · · < 6fc_i, Bk-ι -< · · · -< i?i -< A are
given, and all but finitely many subwords of Bt contain bt, consider the
word bk-ιαι. If infinitely many subwords of Bk-ι omit it, let H*~l be this
set. frfc-iui therefore is not a subword of H^-1. Repeating this to bk-ia2,
we get #2~\ etc. If we can reach H%z\, then there exists a natural number
m, such that in every subword of length 2m a bk-ι can be found in the first
m positions, but it must be followed by an. So we can select H^~l as Bk
and bk-iai as bk- By this process, we get bi < b<i < ..., Bi >- B2 >-
Put L = {bi, b2,... }. Then L -< A and L is minimal. In fact, for bi G L ,
6i_|_i, 6i_|_2? * * * £ Bi, so only finitely many subwords of them omit bi.
We get that the minimal members constitute the infimum. Claim 3 gives
(i), Claim 1 (ii), and the maximality holds as if A is nonminimal, В -< А

3.11 SET THEORY
563
is such that A -ft В, С is minimal, С -< В, then С -< A, A -ft C, so the
addition of С would violate (ii). D
Problem N.ll. Define a partial order on all functions f : R —> R by
the relation f -< g if f(x) < g(x) for all χ e R. Show that this partially
ordered set contains a totally ordered subset of size greater than 2H° but
that the latter subset cannot be well-ordered.
Solution. For the first part of the problem, it suffices to find a family of
more than 2^° subsets of R totally ordered by C, as then the characteristic
function will do the job. In order to show that if к is an infinite cardinal,
then there are к+ subsets of a set of size к, ordered by C, let λ be the least
cardinal such that 2λ > к. By Cantor's theorem, λ exists and is < к. If
we find a totally ordered set of size κ+, with a dense subset of cardinality
< к, we are done by the Dedekind cuts. For such a set, take the λ —> {0,1}
functions, ordered lexicographically. A dense subset is formed by those
functions that take 0 from a place onward. Their number is
^2'al <\-K = K.
ot<\
For the second part, assume that fa < /β for α < β < (2**°) . Let
xa G R satisfy fa(xa) < /α+ι(#α)? for (2^°) many a, xa = x, and so the
fa(x) are (2^°) different elements of R, which is impossible. D

Index of Names
A
Abel, N. EL, 489
Aczel, J., 192
Adam, Α., 16
Adian, S. I., 82
Ajtai, M., 30
В
Babai, L., 25, 32, 50, 51, 282, 491,
529, 557
Baer, R., 63, 81
Bajmoczi, E., 51
Balogh, Α., 39
Balogh, Z., 32, 33, 51
Вага, Т., 51
Beck, J., 51
Beleznay, F., 52
Benczur, Α., 53
Berlekamp, E. R., 374
Bernstein, I. N., 332
Bernstein, S., 158
Birkas, Gy., 53
Biro, Α., 53
Bodo, Z., 52
Bognar, J., vii
Bognar, M., 13, 20, 21, 27
Bognarne, K., 425
Bohus, G., 52
Bolla, M., vi
Bollobas, В., 50, 263, 518, 554
Boltjainskii, V. I., 184
Borges, R., 3
Boros, E., 51
Bosznay, Α., 32
Brindza, В., 298
Buczolich, Z., 52
Burnside, W., 96, 118
С
Cayley, Α., 421
Clifford, A. H., 85
Corradi, K., 10
Coxeter, H. S. M., 91
Csakany, В., 8, 22, 28, 46
Csaszar, Α., 2, 4, 7, 12, 13, 18, 21,
30,42
Csikos, В., vii, 46, 52
Csorgo, M., 443
Csorgo, S., 35, 42, 48
Czach, L., 12
Czedli, G., 46, 51
Czipszer, J., 2
D
Daroczy, Z., 12, 20, 25, 38, 192,
513
Deak, J., 51
Denes, J., 33
Deny, J., 207
Dini, U., 489
Domokos, M., 53
Domosi, P., 44
Drasny, G., 53
Druszt, E., 34
Ε
Elbert, Α., 50
Elek, G., 52
Engelking, R., 533, 537
Eotvos L., ν
Erdelyi, Α., 183
Erdos, J., 11, 14, 19
Erdos, L., 52, 53

566
INDEX OF NAMES
Erdos, P., 7, 11, 12, 15, 16, 19, 23,
26, 33, 36, 38, 43, 44, 145,
522, 558, 561
F
Fazekas, I., 39, 45
Fejer, L., v, 151, 479
Fejes-Toth, G., 30, 40
Fejes-Toth, L., 4, 7, 9, 11, 13, 18,
20, 24, 27, 30, 36, 40
Feller, W., 406
Fischer, E., 471
Fleiner, Т., 53
Foia§, C, 396
Frankl, P., 51
Freud, R., 50
Fried, E., 2, 7, 9, 17, 23, 27, 41
Fritz, J., 50
Fuchs, L., 2
Furedi, Z., 51
G
Gacs, P., 50, 51, 473, 477, 479
Galvin, F., 17, 30
Garay, В., 44
Geher, L., 9
Gelfand, I. M., 459
Gereb, M., 51, 52
Gerencser, L., 50
Gerlits, J., 553
Gesztelyi, E., 12, 26, 32, 37
Gondocs, F., 51
Gorenstein, D., 104
Greenleaf, F. R., 194
Gyires, В., 12, 20
Gyori, E., vii, 51
Gyori, I., 15
Gyory, K, 11, 25, 38
Η
Haar, Α., ν
Hajdu, G., 53
Hajnal, Α., 7, 10, 14, 20, 21, 23,
27, 32, 35, 39, 522, 558, 561
Hajnal, P., 52
Hajos, Gy., 3
Halasz, G., 18, 21, 24, 30, 33, 36,
40, 46, 50, 465, 467
Halmos, P. R., 348
Harcos, G., 53
Hardy, G. H., 217, 470, 484
Hatvani, L., 23, 29, 41, 48, 190
Hausel, Т., 53
Hayman, W. K., 182
Heppes, Α., 7
Hetyei, G., 52
Hewitt, E., 440
Horvath, L., 35
Hosszu, M., 3
Huhn, Α., 31, 35
Huppert, В., 107
I
Ivanyos, G., 52
J
Jaglom, I. M., 184
Joo, I., 36, 43
Juhasz, I., 9, 13, 24, 27, 31, 46,
50, 518, 520, 554
К
Kalina, J., 27
Karman, Т., ν
Karolyi, Gy., 52
Karteszi, F., 25
Katai, I., 10, 38, 513
Katona, Gy., 23, 39, 50
Keleti, Т., 53
Kelly, D., 39
Kemperman, J. В. Н., 336
Kennedy, P. В., 182
Kerchy, L., 29, 41, 48
Kertesz, Α., 19, 84
Kery, G., 50
Kiss, E., 51, 52, 538
Knuth, E., 50
Koljada, K. I., 35
Kollar, J., 51, 56, 540
Komjath, P., vii, 42, 51, 52

INDEX OF NAMES
567
Komlos, J., 21, 50
Komornik, V., 51, 297
Kos, G., 53
Kovacs, В., 38
Kovacs, I., 9
Kovacs, S., 52, 53
Krisztin, Т., 34, 47, 48, 51, 52,
242
L
Laczkovich, M., 17, 24, 27, 33, 36,
39, 43, 44, 45, 50, 51
Langberg, Ν. Α., 443
Laver, R., 559, 560
Leindler, L., 8, 15, 16, 22, 47, 491
Lempert, L., 24, 27, 43, 51
Leon, R. V., 443
Levi, В., 439
Linnik, Yu. V., 107
Littlewood, J. E., 470
Losonczi, L., 12, 26
Lovasz, L., 20, 22, 23, 28, 29, 36,
50, 51, 263, 415, 473, 477,
479, 524, 553
Lukacs, E., 42
Μ
Magyar, Α., 52, 53
Magyar, Z., 51, 52
Major, P., 21, 450, 459
Majoros, L., 53
Makai, E., 18, 24, 50, 51, 477
Makay, G., 53
Makkai, M., 10, 13
Maksa, Gy., 25, 38
Marki, L., 28
Mate, Α., 11, 15, 50, 491, 519,
521, 558, 561
Mate, E., 50
McLean, R. P., 85
Medgyessy, P., 19
Megyesi, L., 22
Michaletzki, Gy., 40
Miklos, D., vi, 52
Mocsy, M., 52, 53
Mogyorodi, J., 425
Montagh, В., 113
Moor, Α., 9
Mori, T. F., 32, 33, 37, 43, 46, 51
Moricz, F., 16, 29, 47, 438
Moussong, G., vii, 52, 545
N
Nagy, P, 29
Nagy, Zs., 51
Neumann, J., ν
Novikov, S. P., 82
Natanson, I. P., 168, 479
Nemetz, Т., 465
О
Odor, Т., 52
Ρ
Pach, J., 43
Pales, Zs., 31, 38, 52
Palfy, P. P., 31, 36, 39, 51, 52, 297
Pap, Gy., 37
Pelikan, J., vii, 17, 20, 30, 50, 51,
81
Petruska, Gy., 17, 50
Pinkus, Α., 36, 209
Pinter, L., 29
Pintz, J., 51
Pollak, Gy., 3, 41
Polya, Gy., 180, 462, 470, 492
Pontriagin, L. S., 189
Posa, L., 23, 28, 50, 51, 479, 520,
528, 553
Prachar, K., 107
Prekopa, Α., 425
Preston, G. В., 85
Proschan, F., 443
Pyber, L., 42, 45
R
Rado, R., 145, 522, 558, 561
Rado, Т., ν
Ramsey, F. P., 122, 134
Ratz, L., ν
Redei, L., 3, 8, 13, 25, 103

568
INDEX OF NAMES
Renyi, Α., 3, 4, 8, 11, 425, 433,
443, 453, 461
Revesz, P., 14, 19, 24, 37, 443
Revesz, Sz. Gy., 41
Reviczky, J., 51
Riesz, F., 394
Riesz, Μ., ν
Riman, J., 38
Rudin, Μ. Ε., 545
Rudin, W., 505
Ruzsa, I. Z., vii, 27, 39, 45, 50, 51,
440, 445, 477
S
Saks, S., 330, 336
Salem, R., 479
Schrijver, Α. Α., 28
Schottky, W., 228
Schur, I., 382
Schweitzer, M., v, vi
Sebo, Α., 51
Seress, Α., 52
Shapiro, H. S., 27
Shilow, G. E., 459
Shockey, H., 207
Sidon, S., 135
Sigray, I., 52, 53
Simanyi, N., 51, 52, 290, 538
Simonovits, Α., 477, 484
Simonovits, M., 50, 470, 519, 553,
554
Somorjai, G., 27, 51, 501
Sos, V. Т., 30, 33, 34
Stromberg, K., 440
Suranyi, J., 2, 9, 13, 18
Sylvester, J. J., 382
Szabados, J., 22, 24, 30, 40, 45
Szabo, E., 52
Szabo, Gy., 38, 44
Szabo, L. I., 46, 52
Szabo, Т., 53, 513
Szabo, Z., 28, 34, 35, 52, 53, 539
Szasz, D., 50, 425
Szegedy, M., 52
Szego, G., v, 180, 492
Szekely, G. J., vi, 32, 440, 445
Szekely, L., 34, 36
Szekelyhidi, L., 25, 44
Szekeres, Gy., 7, 30, 33, 40
Szemeredi, E., 27
Szendrei, Α., 41, 47
Szendrei, M., 51
Szenes, Α., 52
Szenthe, J., 16
Szep, G., vii
Szigeti, F., 50
Szilard, L., ν
Szilasi, J., 39, 45
Szokefalvi-Nagy, В., 4, 9, 22, 394,
396
Szucs, J., 22, 23, 31, 35, 50
Τ
Tamassy, L., 12, 25
Tandori, K., 4
Tardos, G., 52
Teller, Ε., ν
Terjeki, J., 34
Tomko, J., 26
Torocsik, J., vii, 52, 53
Totik, V., vii, 29, 34, 36, 39, 41,
42, 46, 47, 48, 51, 209, 297,
539
Turan, P., 2, 3, 8, 10, 14, 18, 158,
431
Tusnady, G., 14
Tuza, Zs., 51, 297
V
Vamos, P., 50
Varga, J., 52
Varlet, J., 27
Veres, S., 51
Vesztergombi, Gy., 50, 519
Vitali, G., 336, 510
Vu, H. V., 53
W
Weiss, В., 39
Wiegandt, R., 34, 43
Wielandt, H., 96, 97

INDEX OF NAMES
569
Wigner, Ε., ν
Ζ
Zempleni, Α., 52
Zenor, P. H., 545
Zorn, M., 86, 345

Problem Books in Mathematics (continued)
Demography Through Problems
by Nathan Keyfitz and John A. Beekman
Theorems and Problems in Functional Analysis
by A. A. Kirillov and A.D. Gvishiani
Exercises in Classical Ring Theory
by T.Y. Lam
Problem-Solving Through Problems
by Loren С Larson
A Problem Seminar
by Donald J. Newman
Exercises in Number Theory
by D.P. Parent
Contests in Higher Mathematics:
Miklos Schweitzer Competitions 1962-1991
by Gabor J. Szekely (editor)

One of the most effective ways to stimulate students to enjoy
intellectual efforts is the scientific competition In 1894, the Hungarian
Mathematical and Physical Society introduced a mathematical
competition for high school students. Among the winners were Lipot Fejer.
Alfred Haar, Todor Karman, Marcel Riesz. Gabor Szego, and many
others who became world-famous scientists. The success of high
school competitions led the Mathematical Society to found a college
level contest, named after Miklos Schweitzer. The problems of the
Schweitzer contests are proposed and selected by the most
prominent Hungarian mathematicians This book collects the problems
posed in contests between 1962 and 1991, which range from
algebra, combinatorics, theory of functions, geometry, measure theory,
number theory, operator theory, probability theory, and topology to set
theory Solutions are included.
The Schweitzer competition is one of the most unique in the world.
Experience shows that this competition helps identify research
talents This collection of problems and solutions in several fields in
mathematics can serve as a guide for many undergraduates and young
mathematicians The large variety of research-level problems should
interest more mature mathematicians and historians of mathematics
as well.
ISBN Q-3fi?-445fifi-l
ISBN 0-387-94588 1
9 ,,780387,l94588