Major American Universities Ph.D.
Qualifying Questions and Solutions
PROBLEMS and
SOLUTIONS in
MATHEMATICS
Compiled by:
Chen Ji-Xiu, Jiang Guo-Ying,
Pan Yang-Lian, Qin Tie-Hu,
Tong Yu-Sun, Wu Quan-Shui
and Xu Sheng-Zhi
Edited by:
LI TA-TSIEN
World Scientific

PROBLEMS and
SOLUTIONS in
MATHEMATICS

Major American Universities Ph.D.
Qualifying Questions and Solutions
PROBLEMS and
SOLUTIONS in
MATHEMATICS
Compiled by:
Chen Ji-Xiu, Jiang Guo-Ying,
Pan Yang-Lian, Qin Tie-Hu,
Tong Yu-Sun, Wu Quan-Shui
and Xu Sheng-Zhi
Edited by:
LI TA-TSIEN
Fudan University
O World Scientific
U New Jersey • London • Singapore • Hong Kong

Published by
World Scientific Publishing Co. Pte. Ltd.
5 Toh Tuck Link, Singapore 596224
USA office: Suite 202, 1060 Main Street, River Edge, NJ 07661
UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE
Library of Congress Cataloging-in-Publication Data
Problems and solutions in mathematics / [edited by] Li Ta Tsien.
p. cm.
Includes bibliographical references.
ISBN 9810234791 -ISBN 9810234805 (pbk)
1. Mathematics — Problems, exercises, etc. I. Li, Ta-ch'en.
QA43.P754 1998
510'.76~dc21 98-22020
CIP
British Library Cataloguing-in-Publication Data
A catalogue record for this book is available from the British Library.
First published 1998
Reprinted 2002, 2003
Copyright © 1998 by World Scientific Publishing Co. Pte. Ltd.
All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,
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This book is printed on acid-free paper.
Printed in Singapore by Uto-Print

V
PREFACE
This book covers six aspects of graduate school mathematics: Algebra,
Topology, Differential Geometry, Real Analysis, Complex Analysis and
Partial Differential Equations. It contains a selection of more than 500 problems
and solutions based on the Ph.D. qualifying test papers of a decade of
influential universities in North America. The mathematical problems under
discussion are kept within the scope of the textbooks for graduate students.
Finding solutions to these problems, however, involves a deep understanding
of mathematical principles as well as an acquisition of skills in analysis and
computation. As a supplement to textbooks, this book may prove to be of
some help to the students in taking relevant courses. It may also serve as a
reference book for the teachers concerned.
It has to be pointed out that this book should not be regarded as an all-
purpose troubleshooter. Nor is it advisable to take the book as an exemplary
text and commit to memory all the problems and solutions and make an
indiscriminate use of them. Instead, the students are expected to make a selective
survey of the problems, take a do-it-yourself approach and arrive at their own
solutions which they may check against those listed in the book. It would be
gratifying to see that the students can work out the problems on their own and
come up with better solutions than those provided by the book. If the students
fail to do so or their solutions may turn out to be incomplete, it may reveal
the inadequacy of their knowledge or approach, thus spurring them to greater
efforts to promote their skills. The very purpose of the authors in writing the
book is just to help the students to discover the truth by trial and error.
This book was inspired by Professor K. K. Phua's proposals. We are
particularly grateful to him for his support. We also wish to thank Dr. Xu Pei-
jun, Professors Zhang Yin-nan, Hong Jia-xing and Chen Xiao-man for their
painstaking efforts to collect test-oriented data. For selecting problems and
providing solutions, we wish to acknowledge the following professors
respectively: Wu Quan-shui (Part I), Pan Yang-lian (Part II), Jiang Guo-ying (Part

vi
III), Tong Yu-sun, Xu Sheng-zhi (Part IV), Chen Ji-xiu (Part V) and Qin
Tie-hu (Part VI). We are also indebted to Professor Guo Yu-tao for carefully
reading and correcting the manuscript. Finally, we pay tribute to Dr. Cai
Zhi-jie for printing out the manuscript.
Li Ta-tsien
Department of Mathematics
Fudan University
Shanghai 200433
China

vii
CONTENTS
Preface , (v)
Part I. Algebra (1)
1. Linear Algebra (3)
2. Group Theory (26)
3. Ring Theory (44)
4. Field and Galois Theory (59)
Part II. Topology (81)
1. Point Set Topology (83)
2. Homotopy Theory (99)
3. Homology Theory (118)
Part III. Differential Geometry (151)
1. Differential Geometry of Curves (153)
2. Differential Geometry of Surfaces (171)
3. Differential Geometry of Manifold (194)
Part IV. Real Analysis (229)
1. Measurablity and Measure (231)
2. Integral (256)
3. Space of Integrable Functions (283)
4. Differential (302)
5. Miscellaneous Problems (322)
Part V. Complex Analysis (333)
1. Analytic and Harmonic Functions (335)
2. Geometry of Analytic Functions (360)
3. Complex Integration (377)
4. The Maximum Modulus and Argument Principles (413)
5. Series and Normal Families (433)
Part VI. Partial Differential Equations (455)
1. General Theory (457)
2. Elliptic Equations (472)
3. Parabolic Equations (496)
4. Hyperbolic Equations (513)
Abbreviations of Universities in This Book (539)

Parti
Algebra

3
Section 1 Linear Algebra
1101
Let V be a real vector space of dimension at least 3 and let T £ Endji(V).
Prove that there is a non-zero subspace W of V, W ^ V, such that T(W) C W.
(Indiana)
Solution.
Make V into an iR[A]-module by denning A • v = T(v) for all v £ V. Thus
for £ a,A; € JR[A] andugy
i=0
(n \ n
«=o / «=o
It is clear that a subspace W of V is an iR[A]-submodule of V if and only if
T(W) C W.
Now suppose V is a simple iR[A]-module. Then V ~ JR[A]/7 for some
maximal ideal of JR[A]. Since iR[A] is a P.I.D., there exists an irreducible
polynomial /(A) of JR[A] such that I = (/(A)). So
3 < dim^F) = dim«ffl[A]/(/(A)) = deg/(A).
This implies that we have an irreducible polynomial /(A) with degree > 3 in
M[X]. This is a contradiction. Hence V is not a simple iR(A)-module, that is,
there is a non-zero subspace W of V, W ^ V, such that T(W) C W.
1102
Let V be a finite dimensional vector space over a field K.
Let 5 be a linear transformation of V into itself. Let W be an invariant
subspace of V (that is, SW C W). Let m(2), mi(i), and rri2(i) be the minimal
polynomial of 5 as linear transformation of V, W and V/W respectively.
(a) Prove that m(i) divides m\(t) ■ 1712(^)-

4
(b) Prove that if mi(t) and m2(t) are relatively prime, then
m(t) = mi{t) ■ m2(t).
(c) Give an example of a case in which m(t) ^ mi(t) • m2(t).
(Indiana)
Solution.
As usual, V can be viewed as a i£"[i]-module via the linear transformation
S. Since W is an S-invariant subspace of V, W is a if^j-submodule of V.
Then it is clear that (m(t)) = Ann^jV, (m,i(t)) = Ann^jW and (m2(t)) =
AnnK[t]V/W.
(a) Since
mi(<) • m2(t) ■ V C mi(<) • W = 0,
miW • w2(<) € Ann^jF = (m(t)).
Hence m(t) divides mi(t) ■ m2(t).
(b) Since
m(t) € Annif^V C Ann^jW = (m^i)),
m^i) divides m(i). Similarly, m2(i) divides m(i). Since mi(<) and m2(t) are
relatively prime, mi(t)-m2(t) divides m(i). Then we have m(t) = m!(i)-m2(i),
since m(t), m1(i) and m2(i) are all monic polynomials.
(c) Let W be a 2-dimensional vector space over the field Q of rational
numbers and S : W —► W be a linear transformation with minimal polynomial
i2 + 1. Let V = W © W and S : V -> V be the natural extenaion of S to V.
Then it is clear that m(t) = m^t) = m2(t) = t2 + 1. So m(t) rfc. m\(t) ■ m2(t)
in this example.
1103
Let V be a finite dimensional vector space over 1R and T : V —> V be a
linear transformation such that (a) the minimal polynomial of T is irreducible
and (b) there exists a vector w G V such that {T'w | i > 0} spans V. Show
that V doesn't have proper T-invariant subspace.
(Indiana)
Solution.
V can be viewed as a module over the polynomial ring R[\] via /(A) • x —
f(T) ■ (x), for any /(A) £ JR[A] and a; € V. Then we have V = JR[A] • w, a
cyclic module, since {T!w | i > 0} spans V by (b). Let m(A) be the minimal

5
polynomial of the linear transformation T : V —* V. Then m(A) £ Annjj[A](w).
Since m(A) is irreducible, we have
2R[A]/(m(A)) ~ 2R[A] • v = V
(we may assume that V ^ 0). So V is an irreducible IR[A]-module. Thus, V
does not have proper T-invariant subspace.
1104
Let A be an n x n matrix with entries in (F. Show that A has n distinct
eigenvalues in (F if and only if A commutes with no nonzero nilpotent matrix.
(Indiana)
Solution.
Necessity. Suppose that A has n distinct eigenvalues A1,A2,---,A„ in (F.
Then there exists an invertible n x n matrix P such that
PAP-1 =diag{Ai,"-,A„}.
If A commutes with some nilpotent matrix B, we have to show B = 0. Since
Ai, A2, • • •, A„ are distinct and
P.4P-1 = diag{A!, ...,AB}
commutes with P~XBP, P~1BP is a diagonal matrix. But the nilpontency of
B implies that P~1BP is nilpotent. Hence we have P~1BP = 0. So B = 0.
Sufficiency. Suppose that the characteristic polynomial of A has multiple
roots. We have to show that A commutes with some nonzero nilpotent matrix.
Let diag(Ji, J2, • • • ,Jt) be the Jordan canonical form of A and
Pi4P-1=diag(J1,J2, ••-,/«).
where P is an n x n invertible matrix and J,- is a Jordan block of order e*.
Without loss of generality, we may assume that ex > 1 (If all the e, = 1, then
it is easy to see that A commutes with some nonzero nilpotent matrix).
Let Bi be the Jordan block of order ei, with 0 on the diagonal. Then
J\B\ = B\3\ and Si(^ 0) is nilpotent. Let
B' = diag(Bi,B2,---,B«)
where B,(i > 2) = 0 € Me.(W). Then
B' -PAP'1 =PAP~1 -B'.
Taking B = P~1B'P, we have B ^ 0, which is nilpotent, and AB = BA.

6
1105
Suppose V is a finite dimensional vector space over a field K, T : V —>
V a linear map such that the minimal polynomial of T coincides with the
characteristic polynomial, which is the square of an irreducible polynomial in
K[T). Show that if T?, U* and u? are any three non-zero vectors in V, then at
least two of the three subspaces spaned by the sets {T'lf}i>0, {T*lf },>o and
{T! «/};>o coincide.
(Stanford)
Solution.
V can be viewed as a module over the polynomial ring F[X] simply by
f(X)-x = f(T) -(x) for any x G V, /(A) £ F[X\. Let {ux,u2, ■■ ■ ,un} be abase
of V over F, A — (aij)nXn be the matrix of T relative to the base. In general,
a normal form for AI — A in Mn(F[X]) has the form
diag{l,--.,l,d1(A),---,d.(A)}
where the d;(A) axe monic of positive degree and d,(A)|dj(A) if i < j. By
the structure theory of finite generated modules over P.I.D., there exist Zi(i =
1,2, ---,3) € V such that V = F[\]-z1®F[\]-z2®- • -®F[\]-z, where Ann(z;) =
(di(\)). Here, according to the assumptions, the minimal polynomial ra(A) of
Tis det(\I -A), so
m(A) = d,(X) = det(\I-A).
Hence s = 1 and
V = F[\]-z1^F[X]/(m(X))
is cyclic. Since m(A) is the square of some irreducible polynomial, V = .F[A]-,zi
has exactly two non-zero submodules. Obviously, the three subspaces
generated by the sets {T'"u*}i>o, {T* k },->o and {T* u?},->0 are non-zero submodules
of V over .F[A]. So at least two of them coincide.
1106
Let V be a finite dimensional vector space over (T with basis {vx, •■■,«„}.
Let a be a permutation on {i^, • • •, i>n} and thus induce a linear transformation
A on V. Show that j4 is diagonalizable.
(Harvard)

7
Solution.
By re-ordering the elements «i, «2, • • •, vn, we assume that
a = (^1--1^)(^,-, + 1---¾) --(vi. + i-•«»), (1 < «i < i2 <•••<**< n),
when <r is expressed as the product of disjoint cycles (This decomposition may
have 1-cycles). Let Wj be the subspace of V generated by {Vi._1 + 1, • • •, vt.}
for j = 1, 2, • • •, s + 1 (i0 — 0, is+i = n). Then the Wj are invariant subspaces
of A and V = Wi © W2 © • • • © Ws+1. Let M,- be the matrix of A\Wj :
Wj —> Wj- relative to the base {^-., + 1,---,¾} of Wj over (T. Then M =
diag{Mi, • • •, Ms+i} is the matrix of A relative to the base {i>i, v2, ■ ■ •, i>„}. So
it suffices to prove that every Mj is diagonalizable.
Hence, without loss of generality, we may assume that <r is the n-cycle
(«i, «2, • • •, vn). The matrix of A relative to the base {«i, «2, • • •, vn} is
M
It is easy to see that the minimal polynomial of M is A™ — 1, and thus M is
diagonalizable.
This completes the proof that A is diagonalizable.
1107
Let V be a finite dimensional vector space over the field of rational numbers.
Suppose T is a non-singular linear transformation of V such that T"1 = T2+T.
Prove that 3 divides the dimension of V, and prove that if dim V = 3, then all
such T's are similar.
(Harvard)
Solution.
Since T_1 = T2 + T, T is annihilated by the polynomial A3 + A2 - 1.
Obviously, A3 + A2 — 1 is irreducible over the field Q of rational numbers. Thus
A3 + A2 — 1 is the minimal polynomial m(A) of T.
Now let n be the dimension of V over Q, A be the matrix of T relative to
( °
0
V i
l
0
0
0 •
1 •
0 •
• 0
• 0
0 \
0
1
o /
some base of V, diag{ 1, - - •, 1, ^i (A), • • •, d3 (A)} be the normal form for AI — A

8
where the c£;(A) are monic of positive degree and dj(A)|dj(A) if i < j. By the
irreduciblity of da(X) = m(A) = A3 + A2 — 1, we have
di(X) = d2(A) = • • • = ds(X) = A3 + A2 - 1.
Since det(AJ - A) = di(A) • • • ds(X),
3 • s = deg(det(AJ - ,4)) = n.
Thus we have proved that 3 divides the dimension of V.
If dimF = 3, then XI - A is equivalent to diag{l, 1, A3 + A2 - 1}. The
/01 0
rational canonical form for A (or T) is 0 0 1 I . It follows that all the
V 1 0 -1
T's are similar when dimF = 3.
1108
Let Fq be a finite field with q = pn elements, where p is a prime. Let
II : Fq —> Fq be the Probenius automorphism II(a;) = xp. Prove that II
considered as a linear map over Fp is diagonalizable if and only if n divides
pn — 1. (Here is a misprint. It should be "n divides p — 1".)
(Harvard)
Solution.
It is wellknown that Fp C Fq is a Galois extension with
Ga.\(Fq/Fp) = {i,n,n2,- ■■ ,11^1}.
By the Normal Base Theorem, there exists a u G Fq such that {u, II(u), II2(u),
• • •, II"_1(u)} is a base for Fq over Fp. When II is considered as a linear map of
Fq over Fp, the matrix of II relative to the base {u, II(u), II2(u), • • •, nn_1(u)}
/ 0 1 0 ••• 0 \
0 0 1
M -
0
\ 1
0
1
o )
The normal form for XE — M is diag{l, • • •, 1, A™ — 1} and the minimal
polynomial m(A) of M is A" - 1 = det(A£ - M).
Suppose that II is diagonalizable as a linear map over Fp. Then m(A) =
A™ — 1 has no multiple root, and all the roots of A™ — 1 are in Fp. On the other

9
hand, all the root of A™ - 1 forms a subgroup of F* = .Fp\{0}. Thus n divides
p-l.
Conversely, if n divides p — 1, A™ — 1 has no multiple root and
An - 1 = (A - 1) • (A - ad)(X - a2d) • • • (A - a^'1^)
where d = ^~ and a is the generator of the group F*. Hence M is similar
to diag{l,a ,---,a^n~1'd} in Mn(Fp). So II is diagonalizable as a linear map
over Fp.
1109
Let A(t) be a non-singular matrix elements are differentiable
functions of real variable t. Let A'(t) denote the matrix formed by the derivatives
of the elements. Show that the derivative of the determinant det A satisfies
d
dt
(det ,4) = det A ■ trace^' -.4-1).
Solution.
Let
Then
A[t)
A\t) =
I au(t) a12(t)
a2i(t) 0,22(t)
\ anl(t) an2(t)
I a'n(*) a'12(t)
a'21(t) a!22(t)
ai„(t)
a2n{t)
Qnn XJ')
ai„(<) \
«2»(*)
and
/ ^11 A2X ■■• Anl \
A12 A22 ■
A'1 = (det 4)-1
\ Ain A2n
where Aij is the algebraic cofactor of ay(i). Hence
A„2
(Harvard)
n n ii
det ,4 • traced"1) = ^ a^Ay + E a'2j(t)A2j + ■ ■ ■ + ^ a'nj(t)A
= EE4(')^ = e(E4(W)-
i j j \ i )

10
For 1 < k < n, let
/ On(<)
«*i(*)
\ anl(t)
ai2(t) ■
o*2(*) •
a„2^) •
•• aln(t) \
- <n{t)
■ ■ ann(t) /
So,
det A ■ tva.ce(A'A x) = det Ax + det A2 H h det ^.
On the other hand, by definition,
d(detA) = -§( £ (-1)T(P1-Pn)alpi(t)a2p2(t)...anpn(t)
dt
\(Pl--Prt)
= E (-i)r(pi"-p")(Ea^w---a^w---a^w
(Pl—Pn) \k=l /
= El E (-1)^-^^(^)---^(^)---^^(^)
k = l \(Pl-pn)
~ det j4i + det A2 + h det .4,,.
Hence we have proved that
d
dt
(det A) = det A • trace(,4' -.4-1).
1110
Let V be the vector space of polynomials p(x) — a + bx + ex2 with real
coefficients a, b, and c. Define an inner product on V by
1 f1
Om) = 2 / P(x)l(x)dx.
(a) Find an orthonormal basis for V consisting of polynomials <j>o(x), (j>i(x),
and <j>2{x), having degree 0,1, and 2, respectively.
(b) Use the answer to (a) to find the second degree polynomial that solves
the minimization problem
min / (p(x) — x )2dx.
(Courant Inst.)

11
Solution.
(a) Since l,x, x2 is a base of V, we can orthonormalization l,x,x2 to get
an orthonormal basis
<p0(x) = l, $!(x) = VZx, 4>2{x) = —2~ + ~o~x2
of V with degree 0, 1 and 2 respectively as usual.
(b) Let
p0(x) - c04o(x) + ci<^i(a;) + c2<p2(x) G V (c,- € SI)
and
qo(x) = x3 -po(x)
such that qo(x) is orthogonal to V, that is
(x3-po(x),ti(x)) = 0 (t = 0,1,2).
Then for any p(x) £ V,
f {p{x)-x3)2dx = 2|x3-p(x)|2
= 2\x3 - p0(x) + p0(x) -p{x)\2
= 2\q0{x) + p0{x) - p(x)\2.
Since (q0(x),po(x) -p(«)) = 0,
J (p(x) - x3)2dx = 2\q0{x)\2 + 2\p0(x) - p(x)\2.
It follows that
min I (p(x)~x3)2dx = 2\q0(x)\2.
ptv J-i
Since (50(^)1 <Pi(x)) := 0 if and only if
(a;3, <j>i{x)) = (po(x),fc(x)) = ct (i = 0,1, 2).
By an easy calculation, we get Co = 0, Ci = ^ and c2 = 0. Hence Po(x) — |a;
and qo(z) = a;3 — fa;. Obviously,
j\x3-\x)2dx=^.

12
Thus, when p(x) = |a;,
is minimal, and
mm
pe
f {p{x)-x3fdx
in I (p(x) — x3)2dx —
175
1111
Suppose that A is an n x n matrix and that
xi \ / 2/1 \
y
x„ }
, V* = (3/l,"-)3/n).
2/n /
Suppose that all the entries of A, x, and t/ are real.
(a) Show that there exist numbers a and b so that
det(j4 + sxy*) = a + bs.
(b) Show that if det(A) ^ 0 then a = det(A) and b = det(A) ■ y*A"1x.
(c) Is it true that a = 0 if det (4) = 0?
(Courant Inst.)
Solution.
(a) Directly,
det(j4 + sxy*)
= det
= det
+s
an +sxxyx
a„i + sxny1
an • • • aiv
aln + sx-iyn
&nn "T S%nyn
O-nl
det
/ E12/1
021
ZlJ/n
02n
+ det
/ an
E22/1 •
031
\ a„i
ai„ >
• a;2j/n
0-3n
O-nn J

Oil
a In
+ h det
cn-i,i
anyi
&n— l,n
Hence det(-4 + sxy*) = a + bs for some a and b, for any s.
(b) Since for any s, det(j4 + sxy*) = a + bs as in (a), we have
and
b = det
a-ln
x2yn
- X]xiyjAv + X]x2yjA2j h h X]^¾^
j=l j=l j=l
n / n \
where j4;j is the algebraic cofactor of a^. If det .A ^ 0,
-4ii •• Anl
n]
A~l =
1
det(j4)
l In
Thus
b = Ylxi Ewifj
/^11 ••• Anl
= y*-[
- y*(det(A)-A~1)x
- det(A)-y*A~1x.
(c) If det(^) = 0, a = det(^) = 0.

14
1112
Let A be an n x n real matrix with distinct (possibly complex) eigenvalues,
Ai, A2, • • •, A„, and corresponding eigenvector vi, v2, ■• ■ ,vn. Assume that Ai =
1 and that |A,| < 1 for 2 < j < n. Prove that lim Anv exists. Define
n—>oo
T : <En -► <En by T(v) = lim Anv. Find the dimensions of the kernel and
image of T and give basis for both.
(Courant Inst.)
Solution.
Let P = («i, V2, ■ ■ ■, vn). Since Ai, A2, • • •, An are distinct, P is an invertible
matrix in M„((T) and P~XAP — diag{Ai, A2,- ••, An}.
For any v G(Tn,
lim j4"w = lim PdiagjA?, A^, • • •, A£} • P-1 • u
n—»oo n—»oo
= P-diag{l,0, ---,0}^-1 -v
(Since lim (XnYn) — lim X„ • lim Yn). Let (ei,e2,- • • ,en) be the standard
n—>oo n—>oo n—>oo
orthonormal basis of (F™. Then the matrix of T with respect to this basis is
P'-1diag(l, 0,---,0) -P'. Let
/i \
: =p'
/n /
Then (/i, /2, ■ ■ •, /„) is a basis of (F™ and
diag(l, 0, • • •, 0) = P' • P'-1 • diag(l, 0,---, 0)P'P'"1
is the matrix of T with respect to the basis (A,/2, • • •,/n)- Hence {/i} is
a basis of Im(T), {/2,---,/n} is a basis of ker(T), and dim(Im(T)) = 1,
dim(ker(T)) = n - 1.
1113
Let A be a matrix. Define
sin^) = ^-^43 + ^5-""
For
A-l 7 "3
A " 4 l -3 7

express sin(-A) in closed form.
Solution.
Denote
Then B is similar to
and
Obviously,
Hence
1 1
4 0
l -l I \ o io M l -l
7 -3
-3 7
4 0
0 10
1 1
1 1
1 -1
-1 ( I I
= ? 2i
15
(Courant Inst.)
7 -3
-3 7
sin(j4)
1 1
1 -1
+
1 /7r\5
(!)"
f /4 0
4 ' V 0 10
45 0
0 105
l l
K(i)'-(
5! V5
1 /7T\3 ^43 0
0 103
■ l
1 1
1 -1
where
E(-i)*
;t=i
(2fc-l)!
0
fc=l
\
/
So,
sin(A) = ( ! _!
1 1
1 -1
1/1 -1
2 V -1 1
sm7r u
0 sin^
0 0
0 1
\ \
\-\

16
1114
Let A be the 9x9 tridiagonal matrix
/-2 1
1 -2 1
,4 =
V
All entries not shown are zero.
(a) Show that
1
-2 J
xTAx = -(x\ + (x2 - xx)2 + --- + (2:9- x8)2 + x\).
(b) Use part (a) to show that A has all negative eigenvalues.
(c) Let B be the following 10 x 10 tridiagonal matrix, which agrees with A
except in the first row and column:
B
( 1 -3
-3 -2 1
1 -2 1
\
1
\
/
Let Amax(B) be the largest eigenvalue of B. Show that Amax(jB) > 1.
(d) Show that B has 9 negative eigenvalues.
(Courant Inst.)
Solution.
(a) A direct verification shows that (a) is true.
(b) Since xTAx < 0 for all x £ M9 and xT Ax = 0 if and only if x = 0, all
the eigenvalues of A are negative.
(c) Obviously, the symmetric matrix \maxI — B is semi-definite positive.
Then for any Y £ M10, YT(\maJ - B)Y > 0, this is, YTBY < AmaxyTy.
Taking YT = (1,-1,0,---,0), we have YT BY = 5< Amax • 2. Thus Amax > 1.
(d) Let Fi = {(0,0=1,---,0:9) I x{ € M} C 1R10. Then for any Y £ Vi,
YTBY < 0 and YTBY = 0 if and only if Y - 0. If B has more than two non-
negative eigenvalues. Let Ai and A2 be two of them, we have 1/1,1/2 € JRW such

17
that 3/1 ±3/2, yjyi = V2V2 = 1 and By{ = A j 3/,- (i = 1,2). Let V2 = (2/1,2/2),
the subspace spaned by 2/1 and y2. Then for any 2/ = 043/1 + a22/2 € V2,
3/T£3/ = a?Ai + af-A2 > 0.
Since
dim Vi + dim V2 = 9 + 2 = 11 > 10,
there exists 0 ^ 2/ € Vi f~l V2. Then we have yT By < 0 (y ^ Q, y e Vi) and
yTBy > 0 (3/ G V2), a contradiction. Thus B has 9 negative eigenvalues.
1115
Let T be a real symmetric, positive-definite, n x n matrix with distinct
eigenvalues Ai > A2 > • • • > An. Show that
A2 = max mm -.—r,
v xev-{0} I as I
where V ranges over all two dimensional subspaces of lRn and |x| is the
Euclidean norm of x.
Hint. Show = by showing < and > separately. You may wish to express T
in a basis of eigenvectors.
(Courant Inst.)
Solution.
By assumption, there exists an nxn orthogonal matrix P such that P'TP =
diag(Ai, A2,-• •, An). Since for any orthogonal matrix H and any 3/ G Mn,
\Hy\ — M, we have
\Tx\ . |P'-1diag(A1,A2,---,An)-P-1a;|
mm -j—r — mm
xgv-{o} |a;| xev-{o} |a;|
|diag(Ai,---,A„) • P"1x\
min
xev-{o} |P_1a;|
If V is a 2-dimensional subspace, P~XV is also two-dimensional. Hence
\Tx\ . |diag(Ai,---,A„) ■ x\
max mm -r~r = max mm
v rgv-{o} |a;| v xev-{o} \x\
when V ranges over all two dimensional subspaces of IRn.
Now let {e!,e2, • • • ,en} be the standard orthonormal basis of Mn, and
V = (eue2). Then
|diag(Ai, •• •, A„)a;|
mm j—:
xgV-{0} \x\

18
- min < ——\ _ -
I y/a\ + a2
"2
o^ (0.^0,2) em2
= A2 (Ai > a2 > 0).
It follows that
\Tx\ ,
max mm > A2-
v xev-{o} I as | ~~
On the other hand, for any two dimensional subspace V of Mn. we can
find an orthogonal basis {/1,/2,---,/11} of M1 such that V = (/1, /2). Let
(/1,/2,---,/.1)3, = <3(ei,e2,---,e„)T.
Then Q = (qij)nxn is an orthogonal matrix, and if 0 ^ as = ai/i + a2/2 € V.
|diag(Ai,---,A„) • x\
I Z) (Aiaiqrifc + A2a252fc)2
fc=i
A/ai + a
y fc=i y fc=i
So,
diag(Ai,---,A„)a;
mm —r—: —
l£V-{0} \x\
< min < —— :—-
[ y/al + a\
0 # (aia2) € iR2 \ < A2,
and
Thus
\Tx\ ,
max min -p-T- < A2.
v a;ev-{o} I as I ~~
• \Tx\
max min ——7- = M
v xev-{o} |as|
when V ranges over all two dimensional subspace of Mn.

19
1116
For any n x n matrix P, consider the sum
oo
R(P)=J2PK
k=o
oo
(a) Prove that if £ ||Pfc|| < oo, then (J-P)"1 exist and R(P) = (I-Py1.
(b) Assume that ||P|| < 1 in a matrix norm induced by a vector norm.
Prove that \\(I- P)'1]] < j^.
(c) Use part (a) to compute the inverse of
/ 1 1 0 0 \
0 110
0 0 11
\ 0 0 0 1 /
(Courant Inst.)
Solution.
(a) First, we claim that the norms of all eigenvalues of P are < 1. Let
Ai, A2, • • •, An be the n eigenvalues of P and
/ Ai *
Q-XPQ =
V 0 An
be the Jordan form of P. Denote
<j>m(x) = l + x + ---+xm.
Then $m(\i), <l>m{^2), • •• ,$m(K) are the n eigenvalues of
Sm = E + P + P2 + ■ ■ ■ + Pm
and
Q-1SmQ =
I ^m(Al)
V 0
</>m(A„)
Since J2 \\Pk\\ < 00 and the norms over vector space are all equivalent,
fc=o
R(P) = VPJ= lim Sn
*■—rf m—>oo
fc = 0

20
is convergent. Since
lim Q^SnQ = Q-1 ( lim Sm) Q,
Ti—*oo \m—*oo /
lim <j>m(Xi) exists, that is, V A™ is convergent (i = 1,2,---,71). Hence
ra^°° m = 0
|A;| < 1 (1 < i < n).
Thus all the eigenvalues of I — P are non-zero, and (I — P) is invertible.
Since
Sm(I-P)
lim S„
m—*oo
Thus
= I-Pm+1,
= lim (I-P""^^-^)-1
= (l- lim Pra+1)(J-P)-1
\ n—*oo /
= (/-0)(7-^-1
= (i-py1-
R(P) = Y^Pk = (I-P)-1.
k=0
(b) By definition,
sup
(\\Pv\
I INI
v non-zero vector
}■
Since ||P|| < 1, for any non-zero vector u, \\Pu\\ < \\P\\ ■ \\u\\ < ||u|| for the
corresponding vector norm.
Now suppose (I — P) -u = 0 for some vector u. Then ||u|| = ||J-u|| = ||Pw||.
We must have u = 0. The matrix I — P is therefore invertible, and we can
write (I - P)_1 =1+ P(I- P)_1. Then we have
11(/-P)"1!!^
\p\\-\\(i-p)-l\\ = i
(i-p)-1!!.
Thus ||(J - P)"1!! < irfpji. since ||P|| < 1.
(c) Denote
/0-10 0 \
0 0-10
0 0 0-1
\ 0 0 0 0 )

21
Then A = I - N. So
A'1 = (I-N)'1 = I + N+ N2+ N3
/1-1 1 -1 \
0 1-11
0 0 1-1
\ 0 0 0 1 /
■• = 1+ N + N2
NJ
1117
Let G be a p-group and G x V —> V be a linear action of G on a finite
vector space over a finite field 5"p». Using Sylow theory, prove that there exist
a basis of V in which all the transformation v —> <r • v(a £ G) are unipotent
matrices.
(Columbia)
Solution.
Let
G' = {V ->V,v -KT-v\a eG}.
Then G' is a p-group, since there is a surjective homomorphism G ^> G'
(< Endf(V)). We fix a base of V over 2^». Then G' is isomorphic to a
subgroup If of GL(m, J-pn), where m = dim.Fp„ V.
It is well known that
\GL{m,Tpn)\ = (p»™-l)G>"»-p»)...(p'
and
= P
n(m-l)
r /1
u = <
l\o
nm n(m —1)\
*\1
1/
is a Sylow p-subgroup of GL(m, J-P")- By Sylow's Theorem, there exists some
P € GLn{m,Tpn) such that PHP'1 C C/. Thus, if we change the base of V
via P, the corresponding matrices of the linear transformations in G' are in U,
which are unipotent matrices.

22
1118
Let S be an endomorphism of a finite dimensional vector space V over a
field F whose characteristic polynomial is not equal to its minimal polynomial.
Show that there is an endomorphism T of V so that T commutes with S but T is
not a polynomial in S (T is a polynomial in S if T = aoI+aiS+a2S2-i haj,5fc
for some k and a8 G F.)
(Stanford)
Solution.
Let
F[S] = {/(S) : F -> F | /(A) G 2?[A]}.
Then as F-vector space, dirnp F[S] = degm(A), where m(A) is the minimal
polynomial of S. Since the minimal polynomial of S is not equal to its
characteristic polynomial, dirnp F[S] < dim^ V.
On the other hand, let di(X), ck(A), • • • i ^s(A) be the invariant factors ^ 1
of S and let n; = degdi(X), then by Frobenius theorem, the dimension of the
vector spcae over F of matrices commutative with the matrix of S is
s
i=i
Obviously,
s
N > Yl ni = dimF v-
i=i
So there is a linear transformation T of V such that T commutes with S but
T is not a polynomial in S.
1119
Let A be an n x n real matrix, all of whose (complex) eigenvalues are real
and positive. Show that for any integer m > 1, there exists at least one n x n
real matrix B with Bm = A.
Hint. Make use of the Jordan form S + N of a conjugate of A, where S is
diagonal and N is strictly triangular.
(Stanford)

23
Solution.
Suppose first that
/ a 1
0 a
0 0
\ 0 0
o\
1
a /
be a Jordan block with a real and positive. For any integer m > 1, let b = y/a
and
5 =
/ 6 1
0 b
0 0
\ 0 0
1
o\
1
6/
It is easy to see that
I b"
Bm =
cimbm-i
bm
C2mbm-1
clb™-1
\ o
fjn — lum—n + l \
bm
)
Obviously, XE—A and XE—Bm have the same invariant divisors {1, ■
a)n}, A and Bm are similar. Thus
J4 = <3-1JBra<3 = (<3-1JB<3)™
,1,(A-
for some invertible n x n real matrix Q.
Now let A be an n x n real matrix, all of whose eigenvalues are real and
positive. Then there exists an n x n invertible real matrix P such that
p-1AP = d\&g(J1,J2, •••,•/*)
where Ji (1 < i < k) are Jordan blocks with diagonal elements real and
positive. As proved in the above, for any integer m > 1, any 1 < i < k, there
exists real matrix S,- such that 5™ = Ji (1 <i < k). Hence
P-XAP = (diag(JB1,JB2,---,JBfc))r

24
and
where
/
I BX
P ■
\ V o
0 \
Bk J
Bm,
B = F-1-diag(B1,B2,---,Bk) P.
1120
Let F be a finite dimensional vector space over an algebraically closed field
K, and let T : V —> F be a linear transformation.
(a) Show that there are T-invariant subspaces Vi C V and elements of
a; £ K such that F is the direct sum of the F and (T — a,-/) : F; —► F is
nilpotent. (A transformation AT is nilpotent if iV™ = 0, for some n > 1.)
(b) show that there are polynomials S(T),N(T) £ K[T] such that T =
S(T) + N(T), S(T) : V -> V is diagonalizable, and N(T) : V -> F is nilpotent.
Hint. Use the chinese Remainder Theorem.
(Stanford)
Solution.
Viewing F as a module over the polynomial ring K[X] via T. Let (A —
ai)eu,..., (A-ai)e-i; (A-a2)e-, • • •, (A-a2)e^; (A-an)e-, • • •, (A-an)e—
be the elementary divisors of Vk[\], where 0:1,0:2, • • •, an are distinct and
0 < en < ei2 < • •• < elri;- ■ ■; 0 < enl < en2 < • • • < enTn.
By the structure theorem of finitely generated modules over PID, we may write
F = K[\]w11®---®K[X]wlri ®K[X]w2i @---@K[X]w2r3
© • • • © K[X]wni © • • • © K[X]w„rn,
where all the Wy £ F and
Ann(wij) = ((A - o;)6'-').
Denote Fj = K[X]wij. Then
F = Fu © • • • © Firi © • • • © F„i © • • • © Vnrn.

25
The Vij's are T-invariant subspace and (T — a{I) : 1¾ —> 1¾ is nilpotent. Thus
(a) is proved.
Since
(A - ai)e^ , (A - a2)e-=, • • • (A - any—
are pairwise relatively prime, there exists some S(X) G K[X] such that
S(X) = ai( mod (A - ai)eir<), (1 < i < n),
by Chinese Remainder Theorem. Then
S(T) • Wij = a; • Wij
for all Wij and it is easy to see that S(T) : V —> V is diagonalizable. Let
jV(A) = A - S(X). Obviously, T = S(T) + N(T) and X - at \ N(X) for any
1 < i < n. So the minimal polynomial m(A) | N(X)h for some positive integer
h. Thus N(T) : V -> V is nilpotent. Thus (b) is proved.

26
SECTION 2
GROUP THEORY
1201
Let G be the group of real 2x2 matrices, of determinant one. Describe
the set of conjugacy classes of elements of G.
(Harvard)
Solution.
Let g G G, Ai and A2 be the eigenvalues of g viewed in M2((T). Then
Ai-A2 = 1.
i) If Ai = A2, then Ai = A2 = 1 or Ai = A2 = —1. g is similar to
a) ( I ? ) °r b) ( "o1 -1 ) or c) ( 0 i ) °r d) ( "o1 -1 )
in GL2(M).
In case a), g = I Q 1 j ;
-1 0
In case b) g = ,
In case c), let A be an invertible 2x2 matrix with real entries such that
AgA-
,-i /11
0 1
We take N = d~ = • A if d = det A > 0, or
»-ho-m-,1 !
if d = det A < 0, then N eG and
1 1
NgN-* - , Q 1
tff/tf-1 = f
1 -1
0 1

27
Hence g is conjugate to f i ) or ( n 1 I in C But I „ J is not
conjugate to ( „ 1 J in G, for if
'■<Slk'= J?
where A = [ ) £ G, then we have 021 = 0, an = —a22 and
\ 0-21 a22 J
det.4 = — a\x = 1 which is a contradiction. Thus in case c), we have two
conjugacy classes
1 1
0 1
and
1 -1
0 1
-1 1
In case d), g is similar to I _ , J in GI/2(IR)- As in case c), g i
-1 1
0 -1
-1 -1
in G and
-1 1
0 -1
similar (conjugate) to
( -1 -1 \ •
is not conjugate to I n J in G. Thus in case d), we have two conjugacy
dasses
-1 1
0 -1
and
-1 -1
0 -1
ii) If Ai ^ A2 and A; G M (i = 1,2), then g is similar to f ^1
GL2{M).
There exists M G GL2(1R) such that
in
""-1=(o*:)
We take N = d~ = • M if d = det M > 0 or
-1 0
* = i-*)-\ 0 1
if det M = d < 0. Then N £ G and
M
">"-> - (A; I)
Thus g is conjugate to f . ) in G. So in this case, we have the conjugacy
classes
{[(
A 0
0 1/A
A G -ZRand A ^ ±1
}■

28
iii) If Ai ^ A2 and A; € @\1R (i = 1, 2), we denote Ai = cos# + is'm8 and
A2 = cos6 — is'm6 (since Ai • A2 = 1 and trace (g) = Ai + A2 € JR), where
|0| < rr, 0 ^ 0. 0 is similar to ( cos6 + is'me ° . . ) in GI2(<P),
1 ' ' 7- » ^ g cos ^-«sin ^ y v y'
and also similar to ( . „ „ | in GL2((P)-
\ -sin# cos6 J v '
As it is well known, that two matrices A, B £ M2{M) are similar in M2((F)
implies that A and B are similar in M2{M). So there exists M £ GL2{JR) such
that
>«- »*•-1 / cos# sin#
\_ — sin a cos 0
As in the above, we conclude that g is conjugate to
cos 6 sin 6
— sin 6 cos #
in G. But
and
are not conjugate in G, for if
A =
such that
cos 6 — sin 6
sin # cos 6
cos # sin #
— sin 6 cos 0
cos 6 — sin 0
sin 6 cos #
/ an a12 A
V a2i a22 /
GG
cos # sin 6
)A-i _ f cos 6 — sin 6 \
~ V sintf cos6> J '
— sin 6 cos #
then 012 = a.21, an = —022 and
det.4 = -(au + a221) = 1,
which is a contradiction. So in this case, we have the conjugacy classes
cos 6 sin 6
— sin 6 cos #

29
and
cos 6 — sin 6
sin 6 cos 6
To sum up, the set of conjugate classes of elements of G is
U
0
A2
{[(»■
Ai, A2 € M and Ax • A2
■'}
)]M
A = ±l
cos 6 sin #
— sin 6 cos 0
)]|o<|0|<tJ.
1202
Let G — GLn(Fq), the group of invertible n x n matrices over the finite
field Fq with q = pr, p a prime, C/ = Un(Fq), the subgroup of upper triangular
matrices with l's on the diagonal
a) Calculate the orders of G and U. Deduce that U is a Sylow p-subgroup
ofG.
b) Deduce that every p-subgroup of G is conjugate to a subgroup of U.
c) Determine the number of G-conjugates of U.
d) Show that g G G has p-power order iff g = I + N with Nn = 0.
e) Show that G contains elements of order qn — 1
Hint. Make use of Fqn.
(Columbia)
Solution.
a) When forming a matrix in GLn(Fq), we may choose the first row in
qn — 1 ways (a row of zeros not being allowed), the second row in qn — q ways
(no multiple of the first row being allowed), the third row in qn — q2 ways (no
linear combination of the first two rows being allowed), and so on. Thus we
can conclude that the order of GLn(Fq) is
n
(qn-l)(qn-q)(qn-q2)--(qn~qn-1) = ¢^^ ■ Rtf ~ 1)
i=l
= p^-n^'-i).
i=i

30
When forming a matrix in U„(Fq), we may choose the first row in q™_1
ways, the second row in qn~2 ways: the third rows in qn~3 ways and so on.
Hence the order of Un(Fq) is
„n—1 „n —2
q -q ---q
"("-1)
™t"-P
« 2 =P
So U = U„(Fq) is a p-subgroup of G = GLn(Fq). Since p j [G : t/] =
n
J! (p" - 1): U is a Sylow p-subgroup of G.
b) It follows directly from Sylow's Theorem.
c) By Sylow's Theorem, the number of G-conjugates of U is equal to [G :
Ng(U)] where Ng(U) is the normalizer of U in G.
Suppose g = (gij)nxn € NG(U), or gllg-1 = U. Then for any 1 < i < j <
n, there exists some
( 1 U12
0 1 U23
0 0 1
\ 0 0
0
"In \
1 )
eu
such that g(I + Etj) = ug where Etj is the matrix with a lone 1 in the (i,j)
place, 0's elsewhere. Checking the last row of g(I+ Eij) and ug, we get </nl = 0.
So gni = 0 if i < n. Then checking the (n — l)th row, we get gn-i,i = 0 if
i < n — 1. By this way, we get gtj = 0 if i > j. Hence g £ T = Tn(Fq), the set
of matrices (dij) £ G with a,j = 0 (0 < j < i < n).
Conversely, if g £ T. It is easy to see that gUg~l C U (by matrix
multiplication). So g £ NG(U) and we have NG(U) = T.
We can define a map 6 : T —> F* x ■ ■ ■ x F* by mapping a matrix onto its
principal diagonal, matrix multiplication shows that 8 is a group epimorphism
whose kernel is precisely U = Un(Fq). So the order of T is
rn(n-l)
G>r-i)"-P*-v
Hence the number of G-conjugates of U is J\ (l' — 1)/(¾ — !)"•
i=i
d) Suppose g 6 G has p-power order. Then (</) is a p-subgroup and (g)
is conjugate to a subgroup of C/ by b). There exists an h £ G such that
hgh~l £ U. Denote hghT1 = I + M where M is a matrix with zero on and
below the diagonal. Obviously Mn = 0 and g = I + h_1Mh = I + N where
N = h~1Mh and Nn = h~1Mnh = 0.

31
Conversely, if g = I + N with Nn = 0, the Jordan canonical form for N is
a matrix with zero on and below the diagenal. Thus g is similar (conjugate)
to some matrix in U. So the order of g is p-power.
e) Consider the finite field Fqn as an extension of the field Fq and as a vector
space over Fq. For any a(^ 0) G Fg*, we have a linear map a; : Fqn —> Fqn,
ai(x) = ax (over Fq) which is invertible. We fix a base of Fqn over Fq, then
we can obtain a group homomorphism a : F*n —> G = Gn(Fq), <j(a) — a;.
Obviously, a is injective. It is well known that F*n has an element with order
qn — 1. It follows that G contains element of order qn — 1.
1203
(a) Let G be a finite group and H a subgroup. Prove that if G = (J gHg-1,
g€G
then H = G.
(b) Recall that a subgroup M of a group G is said to be maximal if the
only subgroups H satisfying M C H C G are H = M and H = G. Let G be a
finite group with the property that all of its maximal subgroups are conjugate.
Prove that G is cyclic of prime power order.
{Indiana)
Solution.
(a) Let S be the set of all subgroups of G. Then we have an action of G on
the set S defined to be g ■ K = gKg-1 for any g £ G and K G S. Obviously,
the stabilizer of H under this action is the normalizer N{H) of H in G, and
the length of the orbit defined by H is [G : N(H)].
IfG= (J gHg'1, then
g€G
\G\ =
U 9*g-1
g€G
< [G : A-(JST)] - (|J5T| -1) + 1.
If If is not normal in G, then [G : N(H)] > 1. Hence
|G| < [G
< [G
< [G
N(H)]-\H\-[G:N(H)] + l
N(H)]-\H\
H]-\H\ = \G\.
This is a contradition. So H is normal in G. Hence
G = (J gHg-1 = H.
g€G

32
(b) First we claim that G is cyclic. Suppose that G is not cyclic. Then for
any a £ G, (a) < G. So (a) is contained in some maximal subgroup of G (since
G is finite). Let H be a maximal subgroup of G. Then we have G = \J gHg-1
g€G
by assumption. By (a) G = H. This contradicts the maximality of H. Hence
G is cyclic. Now, it is easy to see that G is cyclic of prime power order since
all of its maximal subgroup are conjugate.
1204
Suppose that H and K are normal subgroups of a finite group G and that
G/H ~ K.
(a) Give an example to show that G/K need not be isomorphic to H.
(b) Show that if H is simple, then G/K ~ H.
(Columbia)
Solution.
(a) Let G = Qs = {±1, ±i, ±j, ±k} be the Hamilton's Quaternion group
and H = (i) = {±l,±i}, K = (-1) = {±1} be the subgroups of G generated
by i and — 1 respectively. Obviously, we have H <\G, K <\G and G/H is cyclic
of order 2, so it is isomorphic to K. But G/K = {1, i,j, k} ~ K$ (Klein four
group), K is not isomorphic to H.
(b) Let {1} = Kq < K\ < • • • < Kn-\ < Kn = K be a composition series of
ii". Since G/if ~ K, there exist subgroups Hi such that If,- 3 H, Hi/H ~ .K";
(i = 0,1,-- -,n) and If; < If,+ i (i = 0, • • • ,n — 1).
Suppose if is simple, then
is a composition series of G with composition factors H and Hi/Hi-i ~
Ki/Ki-i (i = 1, 2, • • •, n). On the other hand,
{1} = iT0 < iTi < ■-■ < #„_! < if < G
is a normal series of G with factors K{/Ki-i (i = 1,2, • • • ,n) and G/K. By
the Jordan-Holder Theorem, we have the isomorphism G/K ~ If.
1205
Prove that if G is a finitely generated infinite group then for each positive
integer n, G has only finitely many subgroups of index n.

33
Hint. Let H < G and define HG = f] g~lHg. Show that if H < G
g€G
(finite index) then there exists a homomorphism of G/HG into S[G:H], where
Sn denotes the symmetric group on n letters.
( Columbia)
Solution.
Let H be any subgroup of G such that [G : H] = n. Then GxG/H -> G/H,
g ■ (xH) = (gx) ■ H defines an action of G on G/H, where G/H denotes the
set of all left cosets xH(x £ G). The kernel of this action is HG = f] g~lHg.
g€G
Hence the group G/HG is isomorphic to a subgroup of Sn, the symmetric group
on n letters. This induces a homomorphism G —> Sn with HG as its kernel.
Since G is finite generated, there are only finite number of homomorphisms
from G to Sn. It follows that the set
{HG\H <G,[G:H] = n}
is finite. Since G/HG is finite, G/HG has only finitely many subgroups of
index n. Thus G has only finitely many subgroups of index n.
1206
Let
G={(o a"1 ) ael''ieJR} CGL(2,M).
(Columbia)
Show that G is solvable but not nilpotent.
Solution.
Let 6 be the map G —> 1R* x 1R* mapping I _ x I to its diagonal
(a,a-1). Matrix multiplication shows that 8 is a homomorphism, and
*..= {( J })|M*}.
Obviously, N = ker# (~ (-ZR, +)) is an abelian subgroup of G, and G/N ~
M* x M* is also abelian. Hence G is a solvable group.
Suppose
(; „-.)e c(G)-

34
the center of G. Then for any I :, Ug,
a b \ ( x V \ _ f x y \ ( a b
0 a"1 J \ 0 x-1 J ~ \ 0 I"1 j \ 0 a"1
So ay + bx~l = xb + ya~l for any x ^ 0, 3/ G M. It follows that 6 = 0, a = ±1.
Thus
A direct discussion as above shows that C(G/C(G)) = {1}. Hence
Ci(G) = C2(G) = • • • = Cn(G) = • • • g G
where Ci+1(G)/Ci(G) = C(G/d(G)) (C0(G) = 1). So G is not nilpotent.
1207
Let p be a prime and let V be an n-dimensional vector space over Fp. Let
G = GLFp(V). Recall that |G| = (pn - l)(pn -p) ■ ■ ■ (pn - p""1).
(a) Recall that a linear transformation is called semisimple if its minimal
polynomial is separable. Prove that a transformation T £ G is semisimple if
and only if Tp _1 = 1 for some positive integer m.
(b) Let If be a subgroup of G of order a power of p. Show that H can be
simultaneously upper triangularized, that is, there is a basis of V with respect
to which all of the elements of H are upper triangular.
Hint. Find a Sylow p-subgroup of G.
(Indiana)
Solution.
(a) Let T £ G and /(A) be the minimal polynomial of T over Fp. Then A
| /(A). If T is semisimple, that is, /(A) is separable (remark: here separability
means that /(A) has no multiple roots), then /(A) = /i(A)/2(A) • • • /jt(A) for
some distinct irreducible polynomials /; (1 < i < k). Denote n; = deg/;(A)
and m = ni • 7¾ • • • n^. Let E be a splitting subfield of Xp — A over Fp and
Ei = .Fp[A]/(/i(A)) (1 < i < k). Then \E\ = pm and E{ is a field of order
pn\ Since njlm, -E; is isomorphic to a subfield of E. So -E contains an element
vTn — 1
n whose minimal polynomial over Fp is /,(¾). Since r; ± 0 and rj" "x = 1,
/,(A)|(APm-1-l), (1 < t < fe). Hence /(A) = A(A)-/2(A) • • •/fc(A)|(A^-1-l).
Thus TP"1'1 = 1.

35
Next suppose Tp™ * = 1 for some positive integer m. Then /(A)|(Apm_1 —
1). Since (XP"1'1 - 1)' = -\Pm~2 ± 0, /(A) has no multiple roots. Thus T is
semis imple.
(b) Let U = Uf„(V), the subgroup of upper triangular matrices with l's
on the diagonal. Then
IttI n-1 n-2 n(n-i)
\U\ = p ■ p ■ ■■ p = p 2
(when forming a matrix in U, we may choose the first row in pn_1 ways, the
second row in pn~2 ways and so on). Since
\G\ = (p»-i)(p»_p)...(p»_p»-i)
= 1^^-(^-1)-(^-1)---(^-1),
U is a Sylow p-subgroup of G.
Now let If be a subgroup of G of order a power of p. By Sylow's Theorem,
H is contained in some Sylow p-subgroup K of G and K is conjugate to U.
Hence there exists an element g in G such that gHg-1 C gKg-1 = C/. Thus
we have proved that H can be simultaneously upper triangularized.
1208
Let p and q be distinct prime numbers. Let G be a group of order p3 • q such
that its commutator subgroup K is of order q. Let If be a p-Sylow subgroup
of G.
(a) Show that H is abelian and G = HK.
(b) Show that there are elements h £ H and k E K such that hfc ^ feh.
(c) From (b) show that p divides q — 1.
(J?irfia7ia)
Solution.
(a) Since ii" is the commutator subgroup of G. K is normal in G and G/K
is abelian. So If • K = iflf is a subgroup of G. Since \K\ = q, \H\ = p3,
If n if = {e} and |J2\K"| = p3 • q = |G|. Hence G = Ifif and If ~ H/(e) =
H/H C\K~ HK/K ~ G/isT is abelian.
(b) Suppose that for any elements h £ If and fe G K holds hfc = kh. Since
X is abelian and If is abelian, G = HK is abelian. This contradicts the fact
that the commutator subgroup K of G is of order q. This proves (b).
(c) First we claim that If is not normal in G. Otherwise, for any h £ If
and k e K, hkh~lk~l £ H D K = {e} and so hk = kh. By Sylow Theorem,

36
the number of p-Sylow subgroups of G is greater than 1 and divides q. So it is
q. Again by Sylow Theorem, p\q — 1.
1209
(a) Let p, q be primes, p > q > 2. Let G be a group of order pq2. Show G
has a subgroup of order pq.
(b) What can you say if q = 2 (and p > q)?
(Indiana)
Solution.
(a) Let rp be the number of Sylow p-subgroups of G. Then, by Sylow's
Theorem, rp\q2 and p\rp — 1. Since p, q are primes and p > q > 2, it is easy to
see that rp = 1. So G has only one Sylow p-subgroup H, which is normal in
G and of order p. By Cauchy's Theorem, G also contains an element of order
q. So G has a subgroup K of order q. Thus H • K is a subgroup of order p • q
since if is normal in G.
(b) If 5 = 2, G may not contain a subgroup of order p-q. For example, -44,
the alternating group of degree 4, is a group of order 3-22. _A4 does not have
a subgroup of order 6.
1210
Let A be the abelian group on generators e, /, and </, subject to the relations
9e + 3/ + 6g = 0,
3e + 3/ = 0,
3e - 3/ + 6g = 0.
Give a decomposition of .4 as a direct sum of cyclic groups of prime order or
infinite order.
(Stanford)
Solution.
Let F be the free abelian group Ze\ ®Ze2 ®Ze3, K be the subgroup of F
generated by /i = 9ei + 3e2 + 6e3, /2 = 3ei + 3e2 and /3 = 3ei - 3e2 + 6e3.
Obviously A ~ F/K. Denote
/ 9 3 6 \
M = 3 3 0 € M3(Z).

37
It is easy to get the normal form diag{3, 6,0} of M in M3(Z) and to find
P :
and
Q =
such that QMP = diag{3, 6,0} (P and Q are invertible matrices in M3(Z)).
Let
(e'1,e'2,e^)' = p-1(e1,e2,e3)'
and
Then
and
So
(/{,/2,/3)' = Q-(/i,/2,/3)'.
F =Ze[®Ze2®Ze3
K = ZfY +Zf2 +Zf3 = 3Ze\ © 6Ze'2.
A = F/K = Ze[ ®Ze'2 ®Ze'3/ZZe[ © 6Ze2
~ Z/(3)@Z/(6)@Z
~ ^2 ©^3 ©^3©^.
1211
Let M be an n x n matrix of integers. Suppose that M is invertible when
viewed as a matrix of rational numbers, i.e., that there exists an n x n matrix
N with rational entries so that MN equals the n x n identity matrix. View
M as an endomorphism of Zn.
a) Show that Zn/MZn is finite.
b) Show that the order of Zn /MZn is equal to the absolute value of the
determinant of M.
(Stanford)

38
Solution.
a) It follows directly from b). It is also obvious from the facts that the
map g : Zn/MZn -> Zn/\M\Zn, S(X) - M*X is injective, where \M\ is
the determinant of M, M* is the adjoint of M, and the fact \Zn/\M\Zn\ =
(|detM|)n.
b) Let D = diagjdi, d2, ■ ■ ■, dn } be the normal form of M in Mn (Z), where
the di ^ 0 and d,-|d,-+i (i = 1, 2, • • • , n — 1), and P,Q £ Mn(Z) be invertible
matrices in Mn(Z) such that D = QMP. Obviously
det D = di ■ d2 ■ ■ ■ dn = det Q • det M • det P = det M or - det M.
Let {ei, e2, • ••,€„} be a base for the free module Zn. Denote
(e'!,e'2,-••,<)' = P-1(ei,e2,---,e„)'.
Then {61,62,---,6(,} is another base oiZ". Let
(h, f2,- ■ ■, fn)' = M ■ (ei,e2,- ■ ■ ,en)' = MP ■ (e[,e2,- ■ ■ ,e'n)'.
Then MZn is generated by {/1, /2, • • •, /„}• Since Q is invertible in M„(JT),
MZ™ can be generated by {/{, /2, • • •, f'n} where
(/(, /2, •••,&)'= Q- (/1,/2, ■••,/»)'•
Obviously,
(/{,/2, ■■-,/»)' = Q-(/1,/2, ■■-,/»)'
= QMP-(e'1,e2,---,e;)'
= diag{di,d2,---,^} • (e'i,e2, • • • ,e(J'.
Hence
Zn/MZn = Ze\®Ze'2®---®Ze'n/{dle'l,d2e'2,---,dne'n)
=< ZI{di)®Z/(d2)®---®ZI(dn).
It follows that the order of^n/MJrn is 1^| • |d2| K| = | det M|.
1212
Let D = ■£[£] with £ = ~1+2^- Calculate the order of the additive group
G = D2/K where if is the D-submodule of D2 generated by (2£ + 1,£ - 1),
(£ + 2,£ - 4) and (21, 21). Then express Casa product of cyclic groups.
(Harvard)

39
Solution.
Since £ is a root of the irreducible polynomial x2 + x+l, D — Z[£] = 7©7£
(as additive group).
Let {ei,e2} be a base of the free £)-module D2. Then
D2 = Dei © De2 =Zer ®Z£ex ®Ze2 ®Z£e2,
{ei,£ei,e2,£e2} is a base of D2 as7-module, and {(1 + 2£)ei + (-1 + £)e2, (2 +
£)ei + (~4 + £)' e2, 21 • ei + 21 • e2} is a generating subset of the D-submodule
K of D2. For any a + b£eD,
(a+&0[(l + 20ei + (-l+Oe2]
= a(ei + 2^ei - e2 + £e2) + 6(-2ei - £ei - e2 - 2£e2),
(a+&0[(2 + Oei + (-4 + Oe2]
= a(2ei + £ei - 4e2 + £e2) + 6(-ei + £ei - e2 - 5£e2),
(a+^)(21ei+21e2)
= a(21 -61+0-^1 + 21-62 + 0-^2)
+6(0 • ei + 21 • £ei + 0 • e2 + 21 • £e2).
It is easy to see that {ei + 2£ei — e2 + £e2, — 2ei — £ei - e2 - 2£e2, 2ei + £ei -
4e2 + £e2,-ei + £ei - e2 - 5£e2, 21ei + 21e2, 21£ei + 21£e2} is a generating
subset of K as 7-module (additive group).
Denote
/ 1
-2
2
-1
21
^ 0
2
-1
1
1
0
21
-1
-1
-4
-1
21
0
It is routine to get the normal form
/10 0 0 \
0 10 0
0 0 3 0
0 0 0 21
0 0 0 0
\ 0 0 0 0 /
for A in Msxi(Z) and to find invertible matrices P G M6(Z) and Q £ M${Z)
such that PAQ = N. It follows that D2/K ~ 7/(3) ©7/(21) and the order of
D2/K is 63.

40
1213
Prove that SL(2,Z) is generated by I „ 1 J and I 1 „ I where
SL(2,Z) is the group of 2x2 matrices with integral coefficients and determinant
= 1.
(Stanford)
Solution.
Suppose that M = ( a d j € SL(2,Z).
0 1 \ / 0 -1 W 1 i
If a = Oor b = 0, then M has the form . 1 j i>\ i j /Mn 1
and ( n ,_i I where 6,d G.2T.
If both a and c are nonzero, we claim that I , ) is a prodct of the
\ c d J
matrices in SL(2,Z) with the form
V 0 1 J ' \ c' 1 J ' \ 1 d'J
Since det M = ad— 6c = 1, a and c are relatively prime. It is easy to know
that if a = 1,
a b\_ ( 1 0 \ / 1 b
xc d J ~ \ c lJV° l
and if c = 1,
a b\ _ ( I o\/0 -1
c d J ~ \0 1 J \ 1 d
If a > c > 1, there exist some q, ai £Z such that 1 < ai < cand a = qc+a\.
Hence
a b \ __ ( 1 q \ ( ai b'
^ c d ) ~ \0 1 ) \ c d
for some b' EZ.
Similarly, if c > a > 1, let c = ha + C\, 1 < C\ < a, then
a b\ _ ( I 0\ ( a b \
c d ) ~\h i;Vci d' J
for some d' (£Z.

41
According to the above discussions, it can be derived that M = ( , )
\ c d J
is a product of the matrices in SL(2,Z) with the form
( 0 1 J ' ( c' 1 J ' ( 1 d> )
for some b', c', d' £ 2.
So to prove that SL(2,Z) =11 /Mi n ))' ^e subgr°up
generated by ( I and I ), it suffices to prove that all
1 V \ / 1 0 \ / 0 -1 \ / 0 l\ /-1 V
0 1 J ' \ c' 1 ) ' \ 1 <f J ' V -1 d! J ' V 0 -1
. // 1 ±1 \ / 0 -1 \\ , ,, , ,, _
are in I I cc I , I „ II where b , c ,a £Z.
It is easy to check this by the property of the elementary matrices. This
completes the proof.
1214
Let G be a finite group, K a normal subgroup of G and P a Sylow subgroup
of K. If N is the normalizer of P in G (i.e., JV = {# e G \ gPg~l ~ P} then
show that G= K ■ N.
[Indiana)
Solution.
For any g £ G, we have gPg~l C gKg~l C ii" since ii" is normal in G.
Hence gPg~l is also a Sylow subgroup of i£\ By Sylow Theorem, there exists
he K such that gPg~l = hPh~l. Hence h-lgP(h~lg)-1 = P and h"^ £ JV.
So we have g = h- h~1g £ K ■ N. Thus we have G = K • N.
1215
Let P be a Sylow subgroup of a finite group G. Let JV = {x £ G | xPx-1 =
P). Let # be asubgroup of G, H D N. Prove: If y £ G such that yHy~l = H,
then y € H.
(Indiana)

42
Solution.
Obviously, P is a Sylow subgroup of H. Since yPy~l C yHy~l C H,
yPy~l is also a Sylow subgroup of H. So there exists an h G H such that
yPjT1 = hPh.-1. Hence /i_1yP • (/i_1y)_1 = P and h-1?/ C N. So
1216
Let G be a finite group. The probability of G to be commutative is defined
by
\{(a,b)eGxG\ab=ba}\
P{G) ~ \G^G\ •
(a) If G is not commutative, prove that P(G) < 5/8.
(b) Give an example such that P(G) is exactly 5/8.
(Stanford)
Solution.
(a) Let C(G) be the center of G and Cg{o) = {b £ G | ab = 6a} be the
centralizer of a in G for any a 6 G. If G is not commutative, then, obviously,
\G/C(G)\ > 4.
Since
|{(a,6)GGxG|afc=6a}l = E |C6(a)|
a6Cf
= \C(G)\-\G\ + ]T |CG(a)|,
aeG-C(G)
|C(G)| , „ |Cc(«)|
W> = ^+ £
|G| ' ^, |G|2
1 ' a£G-C(G) ' '
MlciG)l+ £^
agG-C(G)
< ^[|C(G)|+^(|G|-|C(G)|]
2 vi+ igi ;
1 /, In 5
* 2"<1+4>=8

43
(b) Let G be the dihedral group D4 = {x{yi | 0 < i < 1, 0 < j < 3,x2 =
y4 = l,xy = y3x}. Then \G\ = 8 and C(G) = {l,y}. For any a G G - C(G),
obviously, 3 < |Cc(a)| < 8, and so |Cc(a)| = 4. It is easy to see that P(G) =
5/8.

44
SECTION 3
RING THEORY
1301
a) Prove the ringZ[yf^2\ is Euclidean.
b) Using a), find all integer solutions to the equation y2 + 2 = x3.
(Harvard)
Solution.
a) It is readily known that Z[\/—2\ is a subring of <T, hence an integral
domain. For any a = m + n\f^2 £Z[\^2\, we define 6(a) = m2 + 2n2. So we
have a map 6 :Z[yf^2\* —> IN, a t—> 6(a).
Suppose that o,i/0£ Z[yf^2\. Then ab~x = /j, + v\f^2 where fj, and v
are rational numbers. We can find integers u and v such that \u — fj,\ < 1/2
and \v -is\ < 1/2. Then
a = b(n + v\f—2)
= b[(u + n-u) + (v + v- v)V^2]
= bq + r
where q = u + v\/—2 is in Z[\/—2\ and
r = a — bq = b(n — u) + b(v — v)y—2.
Obviously r G Z[y/^2] and
«5(r) = <5(6) - (1/^ — w|2 + 2|^ — ^|2)
< 6(b)- Q +2 -l^j<6(b).
Hence Z[\^2] is Euclidean.
b) By a), Z[\/^2] is a unique factorization domain. {1,-1} is the set
of units of Z[\^2]. Suppose (x0,y0) be an integer solution to the equation
y2 + 2 = x3. In the ring ^[^/^2], we have (y0 + \/=:2)(yo - \/-2) = Zq- Since
the integral divisor of y0 + V^2 or yo — V—2 can only be ±1, z0 is not a
prime element in Z[y/^2]. If a = m + ny/^2 is an irreducible divisor of z0,
a — m — ny/^2 is also an irreducible divisor of z0, and if a|yo + \/—2, then

45
a\yo — V—2 from which it follows that a3\y0 + \^2,a3\y0 - \/-2~- So without
loss of generality, yo + ^/^2 is of the form (m + n^/^2)3, m,n £Z. Hence
J y0 = m3 — 6mn2
\ 1 = 3m2n - 2n3 = (3m2 - 2n2) • n,
Thus it is easy to conclude that the integer solution to the equation y2 + 2 = x3
are
{^=5„ and (^=-5
(^ £ = 3 ^3: = 3
1302
Let p be a prime. Show that for any element a £ Z/pZ, there exist i,c£
Z/pZ such that a = b2 + c2.
(Harvard)
Solution.
When p = 2, it is trivial.
Assume that p ^ 2. Let 5 be the set {62 | b £ Z/pZ}. Then 5 =
{0,1,2 , •••, (^^-)2} has exactly ^|— elements. On one hand for any i ^ j,
o < i < ^, o < j < ^, ?2 - j2 =r+7 • r^j ^ 0 (0 < i + j < p - 1,
0 < i - j < ^ if i > j), {0, T, 2, • • •, (2^1)2} are ^±L elements in S. On the
other hand, for any n2 £ S where ^=- <n<p, 0<p—n< ^=- and n2 =
^ = (n - p)2 = ^=^2. Thus S' = {b2 \ b £Z/pZ} = {0,T,22,...,(HI)2}
has exactly ^|— elements.
Now for any element a £ Z/pZ, the set a — S := {a — b2 \ b £ Z/pZ} has
exactly ^^ elements. Since a — S C .2T/p£ and Z/pZ has only p elements,
(o-S)nS/U. Hence there exist b,c £ Z/pZ such that a - b2 = c2, that is
a = b2 + c2.
1303
For a ring -R, R* denotes its multiplicative group of units, Mn(R) the ring
ofnxn matrices over R, and GLn(R) = Mn(R)*.
(a) If a £ R is nilpotent (a™ = 0 for some m) then 1 - a G R*.
(b) Let J be a nilpotent ideal of R (Jm = 0 for some m).

46
Show that GLn(R) —> GLn(R/J) is surjective, with kernel
I + Mn(J) = {A e Mn(R) \ A = I mod Mn(J)}.
a,b,c G Tq > is a p-Sylow subgroup
(c) Let Tq denote the finite field with q = pr elements (p prime). What is
the order of GL3(Fq)l
((lab
(d) Show that U = I 0 1 c
of GL3{Tq).
(e) Show that GL3(Tq) contains an element of order q3 — 1.
(f) Find the order of GLs(Z/25Z), and describe a 5-Sylow subgroup.
(Columbia)
Solution.
(a) Since (l-a)(l + a+---+01^) = (l + a+- • ■ + am-1)(l-a) = l-ara = 1,
1-aG R*.
(b) Let A = (aij)nxn G GLn(R/J) and B = (J>jj)n-x.n = A_ 1 where
A = (ay)m and £ = (6y)»x» G Mn(#). ^B = BA = J = diag(T, T, • • •, 1)
implies that 7 — AB G Mn(J) and J — BA G Mn(J). So there exist integers
mi and m2 such that (J - AB)mi = (/- B^)™2 = 0. Then it is easy to find
some X and Y G Mn(R) such that 4X = I and y.4 = I. Thus ,4 G GLn(R)
and GLn(R) —> GL„(R/J) is surjective.
Obviously, the kernel of this map is
{J4=(a„)|l = diag{T,T,.-.,l}}
= {4 € M„(ii) I A = I mod Mn(J)}
= I + Mn(J).
(c) By calculating the possibility of the row vectors of A in GL3(Tq), it is
easy to see that
|Gi3(^,)| = (q3 - i)(?3 -q)(q3 - q2) = q3(q3 - i)(?2 - i)(« -1)-
(d) \U\ = q3. Since U is a subgroup of GL3(Tq) with order p3r and
|GI3(^)| = P3r(p3r - l)(p2r - 1)(/ - 1),
U is a p-Sylow subgroup of GL3(Tq).
(e) We consider the field extension Tq C :Fg». There exsits a G Tqt. Such
that :F*3 = (a). Obviously, Ta : Tq> —* Tq>, x >-+ az is an invertible linear

47
transformation over Tq. Let A G GL3(Tq) be the matrix of Ta relative to
some base for Tqs/Tq. Since o(a) = q3 — 1, the order of A in GL3(Tq) is 53 — 1.
(f) Let R = X/25X and J = 5Z/25Z < R. Then J2 = 0 and R = R/J ~
X/5X. By (b), the kernel of the surjective homomorphism
GL3(X/25X) -» GL3(X/5X)
is I + M3(J). Obviously
|J + M3(J)| = 59
and
|GI3(^/5Z)| = 53(53 - 1)(52 - 1)(5 - 1) = 27 • 3 • 53 • 31.
So
|G£3(^/25Z)| = 27-3-512-31.
Let
P={A= (oijOaxa € GL3{X/2hX) \ a(A) eUC GL3{X/hX)}
where <x is the homomorphism GL3(X/25X) -> GL3(Z/5Z). Then P/kercr ~ U
and |P| = I ker<x| . \U\ - 512. Thus P is a 5-Sylow subgroup of GL3(X/25X).
1304
Describe an infinite set of integral solutions (a, b) of the Pell's equation
x2 - 2y2 = 1 (i.e., a, 6 € X such that a2 - 262 = 1).
(Indiana)
Solution.
Suppose that (a, b) is an integral solution of z2 — 2y2 = 1. Then a must be
odd and b even. Let a = 2a' + 1 and b = 2b'. Then we have b'2 = a'(a'2+1K So
a '°2 ' must be a square number. Hence either both y and a' + 1 are square
numbers or a' and S-J^- are square numbers.
If ^ and a' + 1 are square numbers, say a' = 2b2,, a' + 1 = a2, for some
integers a0, 60, then b' = b% • a2,, (a0,&o) is a solution of x2 — 2y2 = 1, and
a = 2a2, — 1, b = ±2ao&o-
Prom the above consideration, it is easy to see that if (ao, &o) is a solution of
x2 — 2y2 = 1, then (ai = 2oq — l,b= 2ao&o) is another solution of x2 — 2y2 = 1.
Obviously, (3, 2) is an integral solution of x2 — 2y2 = 1. So if we take
ao = 3, 60 = 2 and a,- = 2a?_1 - 1, 6,- = 2a,--i&;_i (i > 1), we get an infinite
set of integral solutions {(a,-, 6,) | i > 0} of the pell's equation x2 — 2y2 = 1.

48
Remark. If a' and s-^ are square numbers, say a' = Oq, a'+l = 26q, then
(ao, 60) is an integer solution of x2 — 2y2 = -1 and a = 2oq + 1, 6 = ±2ao&o-
Conversely, if (ao,b0) is an solution of x2 — 2y2 = — 1, then (a = 2oq + 1,6 =
±2ao&o) is an solution x2 — 2y2 = 1. For example, from the solution (7, 5) of
x2 - 2y2 = -1, we get a solution (99, 70) of x2 - 2y2 = 1.
1305
Let p be a prime. Let iJbea commutative ring in which ap = a for all
a£ R.
(a) If R is an integral domain, prove that R is isomorphic (as a ring) to
Z/pZ.
(b) In the general case (i.e., R not necessarily an integral domain), prove
that R is isomorphic (as a ring) to a subring of a direct product (not necessarily
finite) of rings each of which is isomorphic to Z/pZ.
(Indiana)
Solution.
(a) Suppose that R is an integral domain. For any a (^ 0) G R, since
aP — a = 0.(0^-1 — 1) = 0, ap_1 = 1. Hence R is a field, and for any a 6 R, a is
a root of the polynomial xp — x = 0. It follows that iJ has at most p elements.
Since p is a prime, Char(-R) = p and iJ ~ Z/pZ.
(b) Since for any a E R, ap = a, the nil radical of R is 0, that is, the
intersection of all prime ideals of R is 0. Hence
R^-> Yl R/P.
all prime p
For any prime ideal P of R, R/P is an integral domain and for any a G R/P,
aP = aP — a. By (a), R/P ~Z/pZ. Thus we have proved that R is isomorphic
to a subring of a direct product of rings, each of which is isomorphic to Z/pZ.
1306
Let R be a commutative ring with 1. If R satisfies the a.c.c. (ascending
chain condition) for finitely generated ideals then show that R satisfies the
a.c.c. for all ideals. Give example (without proof) of such a ring which is an
integral domain but not a p.i.d.
(Indiana)

49
Solution.
If R satisfies the a.c.c. for finitely generated ideals, then for any ideal I of
R, I is finitely generated. For otherwise, we can construct a strictly ascending
chain of finitely generated subideals of I. Hence R is a Noetherian ring, i.e.,
R satisfies the a.c.c. for all ideals.
Z[x], the polynomial ring over Z, satisfies the a.c.c. for all ideals, but not a
p.i.d.
1307
Let R be a commutative ring. Let A be an ideal of R.
(a) Show that S={l + a|a£j4.}isa multiplicatively closed subset of R.
(b) Show there is a one-to-one correspondence between the prime ideals of
S~XR and those prime ideals P of R such that P + A ^ R.
(c) If A is contained in every maximal ideal of R, what is S~1R1
(d) If A is a maximal ideal of R, what can you say about the structure of
S_1JJ?
(Indiana)
Solution.
(a) Obviously, 1 = 1 + 0 £ 5. For any 1 + a and 1 + b, a, b G A,
(1 + a)(l + 6) = 1 + (a+ b+ab) £ S.
It follows that S is a multiplicatively closed subset of R.
(b) It is well known that there is an one-to-one correspondence between
the prime ideas of S_1R and those prime ideals P of R such that P <1 S = 0.
Obviously, P n S ^ 0 if and only if P + A ^ R here. This is what we need to
prove (b).
(c) If A is contained in every maximal ideal of R, then A C J(R), the
Jacobson radical of R. So every elements in S is invertible in R. Hence
(d) If A is a maximal ideal of R, then S~XA is the unique maximal ideal
of S~XR by (b). So S_1 R is a local ring, and it is easy to see that S_1R is
isomorphic to R^, the localization of R at the maximal ideal A.
1308
Let I be a nilpotent ideal in a ring R, let M and N be iJ-modules, and let
/ : M —* N be an iJ-homomorphism. Show that if the induced map

50
/ : M/IM -» N/IN
is surjective, then / is surjective.
(Harvard)
Solution.
Since
/ : M/IM -> JV/IiV
is surjective, f(M) + IN = N. It follows that
I ■ N/f(M) = IN + f(M)/f(M) = N/f(M).
Hence
N/f(M) = I ■ N/f(M) = I2 ■ N/f(M) = ••. = /". N/f(M) = • • • = 0,
because I is nilpotent. So N = f(M) and / is surjective.
1309
Let Fbea finite field containing 5 elements. Let t be transcendental over
F. Explicitly construct one non-archimedean absolute value | | on F(t). If f(t)
is the completion of F(t) with respect to | |, show that the set of a(t) £ F(t)
satisfying \a(t)\ < 1 is a local ring.
( Columbia)
Solution.
Let p = p(t) = t - 1 € F[t]. We define Vp : F(t) -> 1R by Vp(0) = oo and
Vp(f(t)) = k if 0 ± f(t) = p(t)k ■ b(t)/c(t), where b(t), c(t) G F[t] and
(Hi), Pi*)) = 1 = (c(<),P«)-
Obviously, we have
i) Vp(f(t)) = oo if and only if f(t) = 0,
ii) Vp(f(t) ■ g(t)) = Vp(f(t)) + Vp(g(t)) and
iii) Vp(f(t) + g(t)) > mm(Vp(f(t)),Vp(g(t))).
Then, by denning |/(2)|p = 2~v^f^\ we get an absolute value | \p (called
p-adic absolute value) on F(t). Obviously, | \p is non-archimedean (\f(t) +
g(t)\p <max{\f(t)\p,\9(t)\p}).
Suppose F(t) is the completion with respect to | \p. Let R = {a(t) | a(t) £
~F(tj, \a(t)\p < 1} and M = {a(t) G R \ \a(t)\ < 1}. Since
l«W+W)lp<max{l«(<)lp.l^Wlp}

51
and
\a(t)P(t)\p = M*)\pW)\p,
R is a subring of the field F(t) and M. is an ideal of R.
For any a(t) £ R\M., we have |a(2)|p = 1 and |a(tf)_1|p = 1 where a(t)_1 €
F(t) is the inverse of a(t). So, a(t)~x € R and a(t) is invertible in R. Thus R
is a local ring with maximal ideal M..
1310
Prove that the ideal generated by X\, ■ ■ • ,Xn in the polynomial ring(T[Xi,
• • •, Xn] cannot be generated by fewer than n elements.
(Stanford)
Solution.
Suppose that {Yi, Y2, •••, Ym} is another generating subset of the ideal
(Jli, X2, • •• ,Xn) of the ring(T[Xi,X2, •• • ,Xn]. We have to show that m>n.
We can write
Xi = a,iYi + h aimYm + /;
for any X{ where a;i,a;2, •• • ,a;m G <T, and /,■ is a sum of monomials in
Yi, • • •, Ym of degree two or greater. In the same way, we also have
Yj = bjiXi + bj2X2 + h bjnXn + gj
for j = 1, 2, • • •, m, where 6,1, • • •, bjn G (T, and gj is a sum of monomials in
Xi, X2, • • •, Xn of degree two or greater.
So
x, = XXy, +/•■
u=i
= 5Z "'J ( 5Z 6i*^ +9k\ + fi
j=l \k=l I
n I m
— Vj I z2i a>jbjk I ■X'fc + (terms in X\, • • • Xn of degree > 2)
k=i \j=i J
(i=l, 2,---, n).
Since Xi, X2, • • •, Xn are algebraically independent over (T,
m
53a''i6ifc = (5ifc (*)fe= 1)2,---,71).

52
It follows that
I an
Hence m
021
\ a„i
> n.
ai2 •
a.22 ■
an2 ■
■ aim \ / &11
• 0.2m
Onm /
&21
\ bmi
b12 •
b22 •
bm2 •
•■ hn \
• b2n
Umn J
— In xn
1311
Let A be a commutative ring, and let I and J be two (proper) ideals such
that every prime ideal of .4 contains either I ox J but no prime ideal contains
both I and J. Prove that
A ~ Ax x A2
for some (nontrivial) rings Ai and A2.
(Stanford)
Solution.
Since every prime ideal of A contains either I ox J but no prime ideal
contains both I and J, we have IJ C N(A) (the nil radical of A) and I+J = A.
There exist a 6 I, b G J such that a+b = 1, and since ab £ IJ C iV(.A), there
exists an integer n such that (ab)n = a™ • 6™ = 0. Let Ji = (an) and J2 = (6n).
Then I\ and /2 are proper and I\ + I2 = -4, since
1 = (a + &)2n € (an) + (bn) = h + I2.
By Chinese Remainder Theorem, we have
h n /2 = /1 • h = (an) ■ (bn) = (anbn) = 0
and thus A ~ A/Ii x A/I2 where Ajl\ and A/I2 nontrivial.
1312
Define / : [0,1) -► [0,1) by
/(*) =
2z,
if 0 < 2z < 1,
Find all z such that
2x - 1, if 1 < 2¾ < 2.
/(/(/(/(/(/(*))))))=*•
(Indiana)

53
Solution.
For any real number a, [a] denotes the largest integer < a (Gauss function).
We denote {a} = a — [a] here. Then it is clear that
f(x) = 2x- [2x] = {2x}
for any x £ [0,1). Hence
/(/(/(/(/(/(*) • • •) = {2 • {2 • {2 • {2 • {2 • {2x} ■ • •}.
Since for any real number a and positive integer n, we have
{n • {a}} = n{a] — [n ■ {a}] = n(a — [a]) — [n(a — [a])]
= na — n • [a] — [na — n[a]] = na — n[a] — ([na] — n[a])
= na— [na] = {na},
/(/(/(/(/(/(^)))))) = {26*} = {64*}.
Notice that x = {64¾} = 64¾ — [64 • x] if and only if 63¾ = [64¾]. Since
x £ [0,1), 63z = [64¾] = [63¾ + x] (to be an integer) if and only if
/n _! — *!B\
Xe{ '63'63'*"'63j*
Thus /(/(/(/(/(/(¾)) • • •) = x if and only if x = ^, 0 < k < 62.
1313
Let T be the ring of all real trigonometric polynomials
N
f(x) = ao + \_] an cosnx + &n sinnx-
n = l
Define deg f(x) = N where ajv or bpf ^ 0. Show that deg / -g = deg / + deg #.
Use this result to prove that T is an integral domain which is not a unique
factorization domain.
(Columbia)
Solution.
By using the orthogonality of the set {l,cosn:c,sinn:c | n £ IN}, it is easy
to see that
N
f(x) = ao + /, o.n cos nx + bn sin nx = 0
n=l

54
if and only if all the coefficients of f(x), i.e., ao, ai, • • •, ajv and bo, 61, • • •, 6/v
are zero.
Let
jv
/(¾) = oq + /^ an cos nz + &n sin nx
and
n=l
M
3(3;) = co + 2_j c™ cos m:c + dm sin mi,
Suppose that deg/(z) = N and Aegg{x) = M. Since
(an cos nx + bn sin nx) ■ (cm cos rai + dm sin mi)
6nc
cos(n + m)x -)
sin(n + m)x
+ J
f(x)-g(x)
N+M
cos(n — m)x
Ken
■ sin(n — m)x,
cos kx
n — m=k
- sin kx
m—n=k /
The coefficients ofcos(N + M)x and sm(N + M)x in f(x)-g(x) are ajvCM~^dM
and aivdM+V^M respectively. If both of them are zero, then
a-N -cm • d,M = bpf ■ dM = —bpfCM,
and so bN{c2M + d2M) = 0. Since c2M + d2M ^ 0, bpf = 0. Then we have
o-nCm = aN<lM — 0- Hence a/v = 0, which is contrary to a/v or 6/v 7^ 0.
Thus we have proved that
deg(/(a;) • g(x)) - N + M = deg/(a;) + degg(x).
Now /(a;) • </(:c) = 0 can happen only if either f(x) = 0 or g(x) = 0 by the
above degree relation. So T is an integral domain.
If f(x) ■ g{x) = 1, then the degree relation implies that deg/(a;) = 0 =
degg(x). Hence the units of T are {±1}.

55
Obviously, in T we have
cos 2a; = (cos x + sin a;)(cos x — sin a;)
= (1 +V^sina;)(l - v^sina;).
Again by the degree relation, all the factors cos a; ± sin a;, 1 ± \/2sina; are
irreducible, and cos a; ±sina; and 1 ± \/2sina; are not associates. Thus cos 2a;
in T does not have an essentially unique factorization into irreducible elements.
Hence T is not a unique factorization domain.
1314
Let A be a Noetherian integral domain integrally closed in its field of
fractions K. Let I be a finite separable field extension of K. If B is the integral
closure of A in I, outline a proof that B is Noetherian.
(Stanford)
Solution.
First, we claim that the trace function TiLfK ^ 0. Let p = Char(.K')
and n = [Ir : K]. If p = 0 or (p,n) = 1, it is easy to see that Tr^/j^ ^ 0
(TrL/K(b) = nb for any b £ K). Now, suppose that p rfc. 0 and p \ n. Since
I D K is a finite separable extension, L — K[a] for some a E L. Denote
a = «1,0:2, ■•• ,an be the set of the conjugates of a. For any f(a) £ I,
Tri/A:(/(a)) = £ /(«,). Let x» - dx""1 + c2a;"-2 + (-1)" - c„ be the
1=1
minimal polynomial of a over K and j be the minimal positive integer such
that p \ j and Cj ^ 0. Then by using Newton's identities on the elementary
symmetric polynomials, we get
^/^) = (-l)B+''+1- j-CjtO.
Thus Tri/if £ 0.
Let ui, U2, • • •, un be a basis of I over K contained in B. The bilinear
function (x,y) -+ TiL/K(xy) is non-degenerate since Tr^/j^ ^ 0. Let «i, v2> • • •> vn
be the dual basis of ui, u2, • • •, un (the elements in I satisfying ^^,^(¾¾) =
(5y for all i, j). Then, for any x £ 5, x has the form fci«i + k2v2 + • • • + k„v„
(k{ £ B). Since a;u,- £ B, fe,- = Tr^/j^ani,) £ A for any i. Hence B C
j4wi + j4w2 + 1- Avn. Since A is Noetherian, 5 is Noetherian.

56
1315
If A is a commutative ring and A[[x]] is the ring of formal power series over
oo
A, show that if / = J2 onxn is nilpotent, then all the elements a, £ A are
n = 0
nilpotent. If A is Noetherian, prove the converse, i.e., that if all the elements
a; are nilpotent, then / is nilpotent.
(Stanford)
Solution.
oo
Suppose that f = J2 anXn is nilpotent. It is easy to see that fm = 0
n=0
implies that a™ = 0. So, first we get that ao is nilpotent. Assume that
c.0,0,1, ■ ■ • ,a/t is nilpotent. Since A is commutative, ao + a\x + ■ ■ ■ + OkXk is
nilpotent and
oo / oo \
f-(a0 + aix + --- + akxk) = J2 anxn = xh+1{ ^ a^"-*"1 j
oo
is nilpotent. Hence J2 onxn~k+1 is nilpotent and so ajt+i is nilpotent. By
n = k+l
induction, all the elements a; are nilpotent.
Conversely, suppose that A is Noetherian and all the elements a,- are
nilpotent. Let I be the ideal generated by {a* | 0 < i < oo}. Since A is
commutative Noetherian, I is finitely generated and nilpotent. If Im — 0, that is,
&1&2 • • • bm = 0 for any b\,62, • • •,bm £ I, then it is easy to see that fm = 0.
Hence / is nilpotent.
1316
Let F be a field, and for n > 1 let Rn be the subring of the ring of
polynomials F[x] consisting of polynomials
/0 + fix + f2x2 + ---+ fkxh G F[x]
such that /1 = /2 = • • • = /„ = 0.
(a) Show that Rn is a Noetherian ring.
(b) Show that the field of fractions of Rn is F(a;), and determine the integral
closure of Rn in its field of fractions.
(Stanford)

57
Solution.
(a) Since F[x] is Noetherian and
F[x] = Rn+Rnx + Rnx2 + --- + Rnxn
is finitely generated as iE„-module, Rn is a Noetherian ring by Artin-Tate
Lemma (or Eakin's Theorem).
(b) Observe that
a(x) _ xn+1a(x)
b(x) ~ x^+^b(x)
for any a(x)/b(x) G F(x). It is clear that the field of fractions of R„ is F(x).
Obviously, x is integral over R„, since £ is a root of Xn+1 - xn+1 G R„[X].
Hence F[x] is integral over R„. On the other hand, if a(x)/b(x) G F(x) is
integral over Rn, a(x)/b(x) is integral over F[x\. Since F[x] is an integrally
closed domian, a(x)/b(x) £ F[x]. Hence F[x] is the integral closure of Rn in
its field of fractions.
1317
Describe all subrings of Q. (A subring contains 1 by definition.)
(Stanford)
Solution.
Let R be a subrings of<5. Then R D Z by definition. Let
S={Q^p£Z\ -£R}.
p
Obviously, S is a multiplicatively closed subset of Z containing 1. Let Zs be
the localization of Z at the multiplicatively closed subset S. For any q/p £Z$
(qeZ,p&S), % = q- ±£R. Hence Zs C R.
On the other hand, for any q/p G R, we may assume that (p, q) = 1 and
pi + qk = 1 for some I, k £Z. Then
p p p
So p G S and q/p G Zs. Hence RCZS. It follows that R=Zs where
S = {Q^peZ\-£R}.

58
Thus
{Zs | S is a multiplicatively closed subset of Z}
is the set of all subrings of Q (S can be choosed as the complement in Z of the
union of some prime ideals of Z).

59
SECTION 4
FIELD AND GALOIS THEORY
1401
1) Let X be a finite set and G a subgroup of the group of permutations
of X. Define a relation ~ on X by requiring x ~ y if either x = y or the
transposition (x,y) (which interchanges x,y £ X and leaves all other elements
fixed) is an element of G. Show the following.
(a) ~ is an equivalence relation.
(b) If G acts transitively, then all equivalence classes are distinct and
contain the same number of elements.
(c) If Card(X) is a prime number and if G acts transitively and contains
at least one transposition then G must be the whole permutation group of X.
2) Suppose / G(?[a;] is irreducible and has degree p, a prime number. If /
has exactly p—2 real roots and 2 complex roots, show that the Galois group of
/ over Q is the symmetric group Sp on p symbols. Show that the polynomial
(x2 + 4) • x ■ (x2 - 4)(3:2 -16)-2
is irreducible and determine its Galois group over Q.
Solution.
1) (a) By the defintion of ~, ~ is reflexive and symmetric
(z.yXy^X^y)"1 = (x,z), (x^ y,y^z),
it is easy to see that ~ is transitive. Hence ~ is an equivalence relation.
(b) Let {xi, x2, • • •, xn} and {2/1,2/21'' • ij/m} be two equivalence classes of
~ determined by x = Xi and y = yi respectively. Since G acts transitively on
X, there exists a G G such that <r(x) = y. For any 1 < i < n, we have
(0-(3:1),0-(3:,-)) = 0-(3=1, 3;i)o-_1 € G.
Thus {0-(3:1), 0-(3:2), • • •, 0-(3:,,)} are n distinct elements belonging to the
equivalence class determined by yi = 0-(3:1). Hence m > n. Similarly, we have
n > m. Thus m = n. So all equivalence classes contain the same number of
elements.
(Columbia)
Since

60
(c) Suppose G acts transitively and contains at least one transposition.
Then, by (b), all the equivalence classes contain the same number of elements,
say n, and n > 1. So Card(X) = m ■ n, where m is the number of distinct
equivalence classes determined by ~. Since Card(X) is a prime, m = 1 and
n = Card(X). Thus for any x, y £ X, x ~ y, i.e., (x, y) G G. It follows
that G is the whole permutation group of X, which is generated by all the
transpositions.
2) Suppose
f(x) = f[(x - n)
in (P[x\. So E = (?(ri, • • •, rp) is a splitting field of f(x) over Q contained in (F.
Identify G = Gal(S/Q) with a permutation group of the set X = {n, ••-,?>}
of the (distinct) roots by r\ —» t;|x. For any r,-, r^ € X, since /(e) is irreducible
and /(rj) = 0 = f(rj), there exists an isomorphism of Q(rt)/Q intoQ(rj)/Q by
sending r; to r^. Since E = Q(ri, ••• ,rp) is a splitting field of f(x) over (?(r,),
and also over Q(rj), this isomorphism can be extended to an automorphism r\
of E/Q. Then 7y G Gal(S/<5) and 77(7-,-) = 7-j, which shows G acts transitively
onX.
Consider the conjugation automorphism on(T. This maps f(x) to itself. Let
7*1 and 7*2 be the two non-real roots of f(x). Thus the conjugation interchanges
n and 7-2 = fi and fixed all other real roots. Hence the restriction of the
conjugation to E is an element of G and it is a transposition. Thus the Galois
group G of f(x) over Q is the symmetric group Sp on {ri, • • •, rp}.
By Eisenstein criterion,
f(x) = (x2 + 4)x(x2 - A)(x2 -16)-2
is irreducible over Q. Let
g(x) = (x2 + A)x(x2 - 4)(x2 - 16).
The real roots of g(x) are 0, ±2, ±4, and the graph of y = g(x) has the form
Fig.1.1

61
Since \g(n)\ > 2 for any odd integer n, it is easy to see that f(x) = g(x) —2
has five real roots and two non-real roots. Hence the Galois group of f(x) over
Q is S7.
1402
Let p be an odd prime. Let (p be a primitive pth root of unity and g a
primitive root ( mod p) (i.e., g is a generator for (Z/pZ)*). Fix e, a divisor of
p — 1 and put / = (p — l)/e. Define
v,=ff:«Pfs+i
j=o
Show that, for any i, ^ generates a subfield of (?(Cp) of degree e over ¢.
Hint. Use the generator a : (p —+ (CP)9 of the Galois group oiQ((p) overt?.
(Columbia)
Solution.
Since g is a primitive root ( mod p), a : (p i—> (Cp)9 is an automorphism of
Q(CP) over <? and
G = Gal(Q((p)/Q) =< <r >= {cr,cr\- ■■ ,0^2^1 = 1}.
Obviously, {cr((p),cr2(Cp),- ■ ■ ,^-1^) = CP} is a base for Q((p)/Q.
Let H =< cre >< G. Then \H\ = f,
®((p) ■■ Invff] = |J5T| = /
and [Inv(If) :(?] = e. Since H is normal in G, lnv(H)/Q is a Galois extension
and Gal(Inv(H)/Q) ~ G/H.
Now for any i,
/-1
, = 0

62
and
= X>bW+1)+'k,o
i=o
/-i
= E ^+!'(cP) + ^+!'(cP)
i=i
/-1
= E ^+i(cP)+^'(cP)
j=i
/-i
= E^KiO
It follows that r\i e Inv(If) and {770,771,.- • ,»7e-i} is the orbit Gal(Inv(If)/(?)
(¾) of r/i for any i. Since {cr((p),cr2((p), ■■■ ,<Tp_1(Cp)} is a base for ®((P)/Q,
vo = ^¢0 + ^(0 + -- + ^-¾).
^1 = ^2(CP) + ^+1(CP) +--- + ^-^(^),
e-1
are distinct. Hence JJ (2-77,) is the minimal polynomial for 77; overt? (for any
3=0
i). It follows that 77* is a primitive element for lnv(H)/Q, i.e., Inv(If) = ¢(71,)
for any i. Thus 7¾ generates a subfield of Q((p) of degree e.
1403
Let K be a field and x be an element of an extension of K such that
x is transcendental over K. Put G = Aut(K(x)/K), and let If denote the
subgroup of G consisting of the substitutions x —► x + b with b £ K.
(a) Let A.flG #[X], AB ¢ #, gcd(J4,S) = 1, and put y = A(x)/B(x).
Show in succession that the polynomial A — yB £ ^(y)^] is not 0, that x is

63
algebraic over K(y), that y is transendental over K, that A—yB is irreducible
in K(y)[X], and that
[K(x) : K(y)] = max{deg ,4, deg B).
(b) Show that the elements of G are given by the fractional linear
substitutions x —► ax + b/cx + d with a,b,c,d G K and ad — be ^ 0.
(c) Show that when if is infinite, then G is infinite and the fixed field of G
is K. Find the fixed field of H.
(d) Show that when K = Wq and put
z = (z«2 - xy+1/(xq - x)«2+1,
then ordG = q3 — q and the fixed field of G is Wq{z). Conclude that Wq(x)
is a Galois extension of a given field L with Wq C. L C. lFq(x) if and only if
I D Wq(z). Find the fixed field of H.
(Columbia)
Solution.
(a) Since gcd(,4, B) = 1, there exist 5,Te #[X] such that AS + BT = 1
in #[X]. Suppose A - yB = 0 in #(y)[X]. Then
y = ,4Sy+ ^ = ,4^ + 71).
It follows that deg.A = 0 and degS = deg(j4) = 0, which is contrary to
AB ¢ K. Thus A-yB ¢0.
Since A(x) — yB(x) = 0, x is algebraic over K{y). If y is algebraic over if,
£ must be algebraic over K, which is contrary to the assumption. Hence y is
transendental over K.
Again, since gcd(.A, B) = 1 in K[X] (C K(y)[X]), A — yB is irreducible in
^M^] = -KWM- Thus A~yB is irreducible in K(y)[X}. Thus the minimal
polynomial of x over K(y) is the monic polynomial which is a multiple in K(y)
of .4 — yS and
[#(£) : #(y)] = deg(,4 - yB) = maxjdeg A, deg B).
(b) By (a), y = 4(^)/5(0:) is a generator of #(£)/# (i.e., K(y) = K(x)) if
and only if maxjdeg A, deg B} = 1, or if and only if y has the form ax+b/cx+d,
ad — be ^ 0.
(c) By (b), it is easy to see that G ~ GL2(K)/K* ■ I2, where
iT-/2 = {diag(a,a) \a&K*}.

64
If K is infinite, so is G.
Obviously, K C Inv(G), the fixed subfield of G. On the other hand, if
there exists some y = f(x)/g(x) £ lnv(G)\K, then by (a), K(x)/K(y) is a
finite dimensional simple extension. Hence AutK^y\K(x) is finite, which is
contrary to the facts that G C A\itjc^K(x) and G is infinite. Thus we have
Inv(G) = K.
Similarly, we have Inv(If) = K, since H is also infinite.
(d) Suppose K = Fg. Then
ordG
and
GL2(Fq)
Ft ■ h
(q2-l)(q2-q) _q3 q
q-1
[Fq(x):lnv(G)}=q3-q.
For any
a : x i—► ax + b/cx + d, (ad — be ^ 0)
in G, it is routine to check
<r(z) = <r((xi2 - xy+1/(x« - x)q2+1) = z.
Hence F(z) C Inv(G). On the other hand, we have
(Xq2 - X)q+1 /(Xq - x)q2+1
(Xq -X)\ 1
Denote
and
y(XV-X)} (X«-X)«2-«
(l + xi-1 + x^-1)2 + • ■ ■ + x^-1)^1
(X«-X)«2-« '
A = (1 + Xi-1 + X^"1)2 + • • • + x(q~Vq)q+1
B = (Xq -X)q2~q.
Then gcd(,4, B) = 1 and
maxjdeg A, deg B} = q — q.
Hence we have [Fq(x) : Fq(z)\ = q3 - q. Thus Inv(G) = Fq(z) and Fq(x)
is a Galois extension of a given field L with Fq C i C Fq(x) if and only if
L D F(z).

65
Similarly, we have [Fq(x) : Inv(fi')] = q since \H\ = q. Let A = Xq - X
and B = 1 in Fq[X] and let z' = x« - x £ Fq{x). Then z' € Inv(fi') and by
(a), [Fq(x) : Fq(z')} = q. Thus
Inv(fi-) = Fq(z') = Fq(x« - x).
1404
Let E = <F(Y) with Y an indeterminate, F = (E(Z) with Z = Yn + Y~n,
and C = e2"/n.
(a) Show that there are unique automorphisms a and t of #/(£ such that
cr(Y) = (Y and t(Y) = Y_1, and that the subgroup G of Aut(£) which these
generate is isomorphic to Dn, the Dihedral group of order 2n.
(b) Show that Y is a root of a polynomial of degree In with coefficients in
F.
(c) Show that EjF is a Galois extension with Galois group G.
(Columbia)
Solution.
(a) Since Y is a generator of E over (F, there are unique homomorphisms
a and t of E to £ over <E such that <r(Y) = (Y and t(Y) = Y"1. Obviously,
crn = 1 = t2. Hence <r and t are automorphisms of E/W. Since ord(<r) = n,
ord(r) = 2 and ra = <7™-1t in G, G =< <t,t >~ Dn, the Dihedral group of
order 2n.
(b) Since
X2n-£-X + l = X2n -(Yn + Y_n) + 1
= (Xn -Yn)(Xn-Y-n) e F[X],
Y is a root of X2n -ZX + 1.
(c) Obviously, E/lnv(G) is a Galois extension with Galois group G and
[E : Inv(G)] = 2n. Since <r(£) = <r(Yn + Y~n) = £ and t(Z) = Z, we have
Z € Inv(G) and F =W(Z) C Inv(G). It is readily verified that X2n-£X + 1 is
irreducible over F (i? is transendental over(F and X2n — i?X + 1 is irreducible
in W[Z][X],iox example, using 1403). £ is a splitting field of X2n -ZX + 1
since
X2n-ZX + l = (X - Y)(X - (Y) ■ ■ ■ (X - Cn_1Y)
(X - Y"X)(X - (Y-1)... (X - C-iy-1)

66
in E. It follows that E/F is Galois and [E : F] = 2n. Hence we have
F = Inv(G) and E/F is Galois with Galois group G.
1405
Let F = Q(x) be the field of rational polynomials in one variable x over
Q (i.e., the quotient field of the polynomial ring(?[a;]). Consider the elements
<r, t in Aut<$(F) (i.e., field automorphisms of F) given by cr(x) = 2 — x and
(a) Find the subgroup G (of Au^(F)) generated by a and t.
(b) If K is fixed field of G in F, find a finite subset S of F such that
#=<?(S).
(c) How many subfields lie strictly between K and Fl How many of these
are Galois over K? Justify your answers.
(Indiana)
Solution.
(a) Since cr2(x) = x, t2(x) = x and tct(x) = |5| = ctt(x), we have <r2 = 1,
t2 = 1 and <tt = tct. Hence
G=(<t,t)~K4,
where K4 is the Klein 4-group.
(b) Obviously a(x — 1) = 1 — x and t(x — 1) = —~. It is easy to see that
(a; - 1)2 + ^jp- € K, the fixed field of G in F. Denote rj = (x - 1)2 + (^riw•
Then ¢(77) C .K" C F. Let
Then f(t) is irreducible in<5(7y)[i] and F is a splitting field of f(t) since
/W^-* + l)(* + z-l)-(*-^)(*+^)
in F[t]. HenceQ(rj) CFisa Galois extension and [F : Q(r])] = 4. On the other
hand, K C F is a. Galois extension and [F : K] — \G\ = 4. It follows that
K=Q(V)=q((x-1)2 ' X
(z-1)2
(c) By the fundamental theorem of Galois theory, there are exactly 3 sub-
fields lying strictly between K and F which correspond to the three proper

67
subgroups of G ~ K4. Since all the subgroups of K4 are normal, all these 3
subfields are normal over K (finite dimensional and separable). Thus all these
3 subfields are Galois over K.
1406
Consider Q(t), the field of quotients of the polynomial ring ¢[2]. Let a and
t be elements of Aut<j(<5(2)) given by cr(t) = j^~ and r(t) = —t. Let G be the
subgroup of Aut<j((5(i)) generated by cr and t.
(a) Identify G.
(b) Let H be the subgroup of G generated by a2 and t. Find a £ Q(t) such
that Q(a) is the fixed field of H (Justify your answer).
(Indiana)
Solution.
(a) Obviously, r2(t) = t, a2(t) = a (f=±) = -±, <r3(i) = -£±± and <r4(i) =
i. It is easy to see that t2 = 1, <r4 = 1 and to- = <t3t. It follows that G is
isomorphic to the Dihedral group D4.
(b) Since
i2 , M - ,2 I ^2 1 M _ -(2 1 1 j2 . 1
the fixed field of H. Hence,
Q(W^) CInvHCQ(t)
and obviously
HJ(i) • Invff] = |H| = 4.
On the other hand,
f(x) = z4 - (> + 1) . z2 + 1
= (*» -1») (*a - i
is irreducible over Q (t2 + £) and Q(i) is a splitting field over Q (t2 + £) of
separable polynomial f(x). So
Q(t) 2QU2 + ^

68
is a Galois extension and
It follows that
Q(t)-Q[t
InvH :Q\t2
that is,
InvJT =Q\t
h).
= 4.
= 1,
t2
1407
Let Q denote the field of rational numbers, let K = (?(a), where a is a root
of f(x) = x3 - 3a; + 1.
a) Prove that f(x) is irreducible over Q.
b) Prove that K/Q is Galois.
Hint. Consider a2 — 2.
c) Find a generator of the Galois group Gal(K/Q).
(Indiana)
Solution.
a) If f(x) — x3 — 3a; + 1 is reducible over Q, then f(x) has a factor with
degree one in Q[x], that is, f(x) has a root in Q. Since f(x) = x3 — 3a; + 1 is
monic, the rational roots must be integral factors of 1. But ±1 are not roots
of f(x). This is a contradiction. Thus f(x) is irreducible over Q.
b) Let /3,7 be the other two roots of f(x) = x3 — 3a; + 1 in a splitting field
of f(x) over K. Obviously, a + /3 + 7 = 0, a/37 — ~1 and *ne discriminant A
of f(x) is 34. Hence /3 + 7 = -a and
/3-7 =
(a -/3)(/3- 7)(7 -a)
(a -/3)(7 -a)
Va
-a2+ (/3+ 7)0:-,87
v/8T-a
3(1 - 2a)
v/8T
3(1-a2)"

69
It follows that a, j^ = —a2 — a + 2 and — yr^ — a — a2 — 2 are the roots of
f(x). So K is a splitting field of f(x). Hence K/Q is Galois since Char(<5) = 0.
(We may check a2 — 2 is a root of f(x) directly by using the hint).
c) From a) and b), it is easy to see that the Galois group Gal(K/Q) ~ A3.
a i-+ a — 2
a: K=Q(a)->K
is a generator of Gal(if/<5).
1408
Let K be a field and x an indeterminate.
(a) Show that the rational functions /„ = ^- (a G if) are linearly
independent over if.
(b) As if-modules, K[x] and if (a;) have what dimensions?
(c) Let G denote the additive group of if, acting on K(x) by a G G sending
x to x + a. Assume that if is infinite. Let / G if (a;). Show that the G-orbit
of / spans a finite dimensional if-module if and only if / G K[x].
(Columbia)
Solution.
(a) Suppose that {/a = ~^ \ a G if} is linearly dependent over if.
There exist some non-zero elements ai,---,an G K and some distinct
elements ai, a.2, • • •, an G K such that
n
^«t/a; = 0.
t=l
Hence
« \i& j
and
(z-ai) JJ(a;-aj),
which is not true. Thus {/a = ^^ | a G if} is linearly independent.
(b) dimjc K[x] = dimjc K{x) = oo.
(c) Suppose
f(x) = a„xn + an^ixn~l H h axx + a0 G if [a;].

70
We claim that for any n+ 1 distinct elements ai,a2,- ■ ■ ,an+i in K (note
that K is infinite), f(x + «i),/(x + 02), •• -,f(% + On+i) spans the subspace
generated by the G-orbit of /.
For any a £ G, we take (Ai> A2> • • •, A1+1) € Kn+1 to be the solution of the
equation system
/ 1 1
«1 «2
\ a? ccn2
Then it is clear that
(x + a)n = p^x + oi)" + /32(x + a2)n
and also
(x + a)* = fa(x + a{f + f32(x + a2y -
for 1 < i < n. Hence
/(x + a) = Ai/(z + «1) + M(x + a2) ■
... 1 \
••• «n+l
• • • n,2
■■ < + l 1
( ^ \
X2
x3
\ xn /
=
( X \
a
a2
\ an J
-Pn+i(x + an+i)n
■ Pn+l(x + «n+l)'
-/5n+l/(a: + «n+l).
This shows that {f(x + a;) | 1 < i < n + 1} spans the subspace generated by
the G-orbit of /.
On the other hand, suppose f(x) £ K(x), and the G-orbit of/, {f(x + a) \
a £ K} spans a finite dimensional if-module. We write f(x) = g(x)/h(x)
where g(x), h(x) £ K[x] and (g(x), h(x)) = 1. Let f(x + ai),f(x + a2), ■ • •, f(x
+an) (oj £ K) span the subspace < {f(x+a) \ a £ K} >. Then for any a £ K,
there exist (3\, (32> • • •, Aj in if such that
/i(a; + a)
n
= ]TA/(z + a»)
«=i
£a
g(^ + o«)
'-^^l h(x + OLi)
n
E A#(z + at) n Ma; + a,)
«=1 £*»
n
f] h(a; + a,-)
i = l

71
Since (g(x + a), h(x + a)) = 1, we have
n
h(x + a)\Y[h(x + ai)
t=i
for any a £ F. Since K is infinite, an easy discussion in a splitting field of h(x)
over F will lead to degh(x) = 0. Thus we have
1409
Let £ be a finite Galois extension of F and let f(x) be an irreducible
polynomial in F[x]. Show that all the irreducible factors of f(x) over E are of
the same degree.
(Columbia)
Solution.
For any a £ G = Ga.\(E/F), we still denote a to be the isomorphism
E[x] —► E[x] which extends a on E and maps x to a;. Since /(a;) £ F[x],
cr(f(x)) — f(x). Let e(x) be a monic irreducible factor of f(x) over E. Then,
for any a £ G, <r(e(x)) is an irreducible factor of f(x) over E.
We prove in the following that all monic irreducible factors of f(x) over
E arise in this way. Thus all the irreducible factors of f(x) over E are of the
same degree.
Suppose e'(x) is another monic irreducible factor over E. Let a and a'
be roots of e{x) and e'(x) in some extension field of E. Then, a and a' are
roots of f(x), which is irreducible over F. Hence we have an isomorphism
j] : F(a) —► F(a') which sends a to a (a £ F) and a to a'.
Since S/F is Galois, we can write E = F(f3), where /3 is a root of g(x),
a separable irreducible polynomial over F. Obviously E is a splitting field of
g(x). Then £(a) = F(a)(/3) and £(«') = F(a')(/3) are splitting fields of ff(a;)
over F(a) and ^(a') respectively. Hence rj : F(a) —► F(a') can be extended
to an isomorphism r\ of -E(a) onto E(a'). Note that ^(/3) may not be /3, but
77(/3) is a root of rj(g(x)) = </(:c). It follows that t)\e : E —* E is in G.
Now, since a and a' are roots of the monic irreducible polynomial e(x) and
e'(x) over E? respectively, r\ : -E(a) —» E(a') is an isomorphism and rj(a) = a',
we must have (r]\E)(e(x)) = e'(x) and dege(a;) = dege'(a;).

72
1410
(a) Let K be a field of characteristic p > 0. Show that the polynomial
tp —t — c m K[t] is either irreducible or splits completely into p linear factors
over K.
Hint. If u is a root of tp — t — c then so is u + 1.
(b) Let F be the splitting field of the polynomial t62 — 1 over Z5. Show
that [F : Z5] = 3.
Hint. First prove that the zeroes of t62 — 1 form a cyclic group G of order
62.
(Indiana)
Solution.
Suppose that tp — t — c = f(t) ■ g(t) in K[t] where f(t) is a monic polynomial
of degree n, 1 < n < p— 1. Let S be a splitting field of tp — t — c and let u G S
be a root of this polynomial. Then for any m G Zp, the prime field of K,
(u + m)p — (u + m) — c = up + mp — u — m— c = up — u — c = 0.
Hence we have
/«•</(*)= n (*-«-™)
and there exist ij, 12, • • •, i„ GZp such that
/(i) = (t-u- ii)(t -u~i2)---(t-u- in).
Comparing the coefficients of the term of degree n — 1, we obtain n • u + i\ +
«2 H h *n € iiT. So we have n-u G if. Since p- u = 0 and there exist integers
v and w such that w • n + wp — 1, u = (v ■ n + wp) ■ u = v(n ■ u) £ K. Thus we
have
tp -t-c= Y[ (* -u -m)
in K\t).
(b) Let G be all the zeroes of t62 — 1 in F. Obviously, G is a subgroup of
F* and G is cyclic of order 62 since
(t62 - 1)' = 62t61 = 2t61 £ 0.
It follows that
[F:Zb}>3.

73
Let E be a, extension field of Z5 such that [E : Z5] = 3. Then E* is a cyclic
group of order 53 — 1 = 124. Let G' be its unique subgroup of order 62. Then
all the elements of G' satisfies t62 — 1. So t62 — 1 splits in E and E is a splitting
field of t62 - 1. Thus we have E ~ F and [F : Z5] = 3.
1411
(a) Suppose you are given a field L, Q C L C (F, such that L/Q is algebraic
and every finite field extension K/L, K C <E is of even degree. Show that every
finite field extension of L must in fact have degree equal to a power of 2.
(b) Show that such a field L actually exists.
(Indiana)
Solution.
(a) Let K/L (K C (F) be a finite field extension. We have to show [K : L]
is a power of 2. For this purpose, we may assume that K is Galois over L.
Let G - QsXK/L and \G\ = 2" • m where m is odd. By Sylow's Theorem,
G has a subgroup H of order 2™. If K' is the corresponding subfield of K/L,
then [if : K'] = 2™ and [K' : L] = m. Since L has no proper odd dimensional
extension field, we must have m = 1, and so K' ~ L and [K : L] = 2™.
(b) Let i be the field of real algebraic numbers, that is, the subfield of M
of numbers which are algebraic oveiQ. Then<# ClCff and L/Q is algebraic.
Now for any finite field extension K/L, K C (F, there exists some element
a £ K such that if = L(a) by Primitive Element Theorem. Let f(x) be
the minimal polynomial of a over £. Then f(x) has no real root since f(x)
is irreducible in L[x]. So, /(a;), when decomposed in JR[a;], is a product of
irreducible polynomials of degree 2. Hence,
[K:L]=[L(a):L] = degf(x)
is even. By (a), it is in fact a power of 2.
1412
Let K be a field of characteristic p ^ 0. The set {a;p | x £ if} is a subfield
of K that is denoted by Kp (no proof required).
(a) Let L be an intermediate field between Kp and K. If [i : Kp] is finite,
prove that it is a power of p.

74
(b) A subset B of K is called p-independent if for any finite set 61, 62, • • •, 6m
of distinct elements of B
[KP(b1,b2,---,bm):K''}=pm.
Prove that if Kp ^ K, then K contains a maximal p-independent subset B.
(c) Prove that the set B of part (b) satisfies KP(B) = K.
[Indiana)
Solution.
(a) If [L : Kp] is finite, there exists a finite set of elements {61, 62, • • •, 6„}
such that
#p(&i,&2,---,M = £.
Without loss of generality, we can assume that 6; ¢ Kp(bi ■ ■ ■ bi-i) for any
1 < i < n (Kp(bu • • •, 6;_i) = Kp when i = 1). Since
6f £J^C K*(bu ■•-,*-!)
and
bi ¢^(61,---,6,-1),
2P — 6f is irreducible in Kp(bi, ■ ■ ■, 6,--1)[*]- Hence
[^(61, ---,6,) : ^(6i,---,6,_i)] = p.
It follows that
[L:KP] = [^(6i,62,---,6n):i]
n
= J][^(61, ---,6,) : ^(61,---,6,-1)]
i = l
= P".
(b) If Kp ^ if, there exist p-independent subsets. For example, if 6 £
K\KP, B = {6} is a p-independent subset of K. Now suppose that {5,- | i £ J}
is a chain of p-independent subsets of if. Let 5 = (J 5,-. Then for any
finite set {61, • • •, 6m} of distinct elements of B, there exists some i, such that
{61,62, • • •, 6m} C 5,-. Since Bi is a p-independent subset,
[K»(b1,b2,---,bm):Kp)=pm.
It follows that B is p-independent. By Zorn's Lemma, K contains a maximal
p-independent subset.

75
(c) Let B be a maximal p-independent subset of K. Suppose KP(B) C K.
Let b £ K\KP(B). Then B U {6} is p-independent. The reason is that, for any
distinct elements 6i,62, • • • ,6ra in B U {6}, if 6 ¢ {6i, b2, ■ ■ ■, bm}, then
{6l,62,-",6m}CB
and
[#p(6i,...,6ra):l^]=pra,
and if b £ {6i, b2, ■ • • ,bm}, say, b — bm, then {6i, •• • ,6m_i} C 5 and we still
have
[K"(h,---^):^]
= [1^(61, •••,6m) :1^(61,---,6^1)1-^(61,---,6^0 :1^]
Since B C BU {6}, the independency of 5 U {6} contradicts the maximality of
B. Thus we have KP(B) - K.
1413
Let K be a finite field with pr elements (p a prime) and n be a positive
integer. If m is an integer which divides n and f(t) G Iffi] is an irreducible
polynomial of degree m, show that / divides ip —t.
(Indiana)
Solution.
Let E be a splitting field off'"" -t over #. Then |£| = prn and [S : K] = n.
Let a be a root of the irreducible polynomial f(t) in some extension field of
K. Then \K{a)\ = prm and [K(a) : K] = m.
Since m\n, E contains a subfield L such that K C L C E and i ~ 1^((¾).
Hence there exists an element {3 E L C E such that /(/3) = 0, that is, f(t) is
the minimal polynomial of/3. Since f3p — /3 = 0, we have /(2) | (tp — t).
1414
Let K/F be a finite extension of fields and let L and E be intermediate
fields, with E/F Galois and [K : L] = p, a prime. Prove that if p does not
divide [E : F] then £CI.
(jTitfema)

76
Solution.
Let f(x) be a separable irreducible polynomial over F such that E is its
splitting field. Let
E-L = E(L) = L(E)
be the composite of E and L in K. Then E ■ L = L(E) is a splitting field of
f(x) over L. Hence E ■ L is Galois over L. Let a £ E be a root of f(x). Then
£ = .F(a) and E ■ L = L(E) = L(o). For any
a £ Gal(E • L/L) = Ga\{L(a)/L),
it is clear that cr\E £ Aut(E/L D E). And further, we have
Gal(£ • X/i) ~ Gal(£/I n E).
Now suppose E <£ L. Then E ■ L = K, since L C E ■ L and [if : i] = p, a
prime. It follows that
p= \Gal(E-L/L)\ = \Ga\(E/L(lE)\
dividers \Gal(E/F)\ = [E : F], contrary to the assumption. Thus we have
ECL.
1415
Let Ki be the subfields of (T defined as follows: Ko = Q- If i > 0, Ki+i is
the smallest subfield of (T containing the set
{6 £ W | 6n £ Ki for some n > 0}.
Let
OO
K=\jKt.
«=o
(1) Prove K is a field.
(2) Let f(x) £ K[x] be irreducible. Prove that deg(f) > 5.
(Indiana)
Solution.
(1) Since
KoCKiC-'-CKiC Ki+1 C • • •
OO
is a chain of subfields of <E. It is clear that |J K{ is a subfield of (T.

77
(2) (Remark: deg/(a;) may be 1).
Let f(x) G K[x] be an irreducible polynomial with deg/(a;) > 1. There
exsits some i such that f(x) G iiT,-[a;]. Suppose deg/(a;) < 4. By the formulas
for the roots of quadratic, cubic and quartic equations and
Ki+1 = {6 eW | 6n G Ki for some n > 0},
f(x) splits in Ji"j+3[a;], hence in K[x\. Contradicts the irreducibility of f(x).
Hence deg/(a;) > 5.
1416
Let K be an extension field of Fp, the field with p elements. Let a be an
algebraic element in K. Prove that [Fp(a) : Fp] is the smallest positive integer
m such that a9(m) G Fp, where g(m) = vjrr~
[Indiana)
Solution.
Let n = [Fp(a) : Fp]. Then \Fp(a)\ = pn and a^""1 = 1. Since (a^"))1'-1 =
aP"'1 = 1, a»W G i^.
On the other hand, if a3^"1' G Fp, for some positive integer m, then
a""-1 = (0^))^ = 1,
so a is a root of xp — x. Let E be a splitting field of xp — x over Fp and
a G -E. Then [£ : Fp] = m and fp C fp(a) C £. Hence
n=[Fp(a):Fp]\[E:Fp} = m.
Thus [.Fp(a) : F] is the smallest positive integer m such that a9^"1' G Fp.
1417
Let f D K be a field extension of finite degree m. Let / G -K"[i] be an
irreducible polynomial of degree n. If m and n are coprime then show that /
remains irreducible in F[t].
(Indiana)
Solution.
Suppose that f(t) is reducible in F[t] and let f(t) = g(t) ■ h(t) in F[t] where
g(t) is a irreducible polynomial in F[t] of degree fe, 1 < k < n. Let E = -F^a),

78
where a is a root of g(t) in some extension field of F. Then [E : F] = fe since
g(t) is irreducible in F[t]. So
[E:K] = [E : F][F : K] = k ■ m.
On the other hand,
[E:K] = [E: K(a)} ■ [K(a) : K] = [E : K(a)] ■ n,
since a is a root of f(t) and f(t) is irreducible in K[t]. It follows that n | fe • m,
which contradicts (m,n) = 1 and 1 < fe < n. Thus f(t) is irreducible in F[t],
1418
Find a Galois extension E over Q with Gal(E/Q) cyclic of order 16.
(Stanford)
Solution.
For any positive integer n, the cyclotomic field Q(zn) over Q is a Galois
extension and K?(zn) : (?] = ¢(11-) where zn is an n-th primitive root of the
unit, 4>(n) is the Euler (^-function. It is easy to see that |Gal(<5(zj)/(9)| = 4>(n)
and Gal(<5(z„)/(?) ~ Aut(G) where G is the cyclic group of order n. When n
is prime, Aut(G) is cyclic of order n — 1.
So if we take n = 17, E = ¢(217), then E is a Galois extension of Q with
Gal(£y<?) cyclic of order 16.
1419
Find a Galois extension E over (# with Gal(E/Q) cyclic of order 32.
(StaTi/ond)
Solution.
As in 1418, for any positive integer n, the cyclotomic field Q(zn) over Q is
a Galois extension and [(?(zn) : <9] = </>(n) where zn is an n-th primitive root of
the unit, <j>(n) is the Euler (/>-fui -,tion. It is easy to see that |Gal(<5(zj)/(?)| =
(j>(n) and Gal((?(z„)/(?) ~ Aut( *) where G is the cyclic group of order n.
When n = 2™ and m > 3, it is well known that Aut(G) ~ Z2 ®Z2™-i-
By the Fundamental Theorem of Galois Theory, if we take E = Inv(JT2),
then(? C£isa Galois extension and Gal(E/Q) ~Z2™-?-
Taking m = 7, then Q C E is a cyclic extension of order 32.

79
1420
Let E/F be a finite Galois extension, G = Ga\(E/F) and a £ E. Consider
the F-linear map Ma : E —► E, Ma(x) = ax. Show that its trace is given by
Trjr(Ma) = J](T(a) where a varies over G.
(Columbia)
Solution.
Let z be a primitive element of E/F, [E : F] = n and
G=GA\{E/F) = {aua2,---,an}.
Then
n
= xn+a1xn~1+ ■■■+an
is the minimal polynomial of z over F, and <ri(z), ^2(2:),---, an(z) are distinct.
Now, for any a £ E, a has the form
a0 + aiZ-\ ha„_izn_1 («,•£?),
since {1, z, • • •, z™-1} is a base for E/F. To prove that
n
TrF(Ma) = 2
1=1
it suffices to prove that
n
TrF(M2k) = ^^(^)
i = l
for any 1 < k < n — 1.
Obviously, /(a;) is also the minimal polynomial of the F-linear map Mz :
E —► E, Mz(x) = z ■ x. Since f(x) has distinct roots ori(z),or2(z), •• • ,<r„(z) in
E, the matrix of M2 (G Mn(.F)), say, relative to the base {1, z, • • • ,zn_1}, is
similar to diag{<ri(z), • • • ,<r„(z)} in Mn(E). Anyway, we have
n
TrJr(M2) = ^<Ti(z)
1 = 1

80
and for any 1 < i < n — 1,
TrF(M2>) = TrF((M2)fc)
= ai(z)k + a2(z)k +■ ■■ + an(z)h
i=i
This completes the proof.

Part II
Topology

83
SECTION 1
POINT SET TOPOLOGY
2101
Let A and B be connected subspaces of a topological space X, such that
A fl B ^ 0. Prove that A U B is connected. If j4 and B are path connected,
need iUB be path connected?
(Indiana)
Solution.
Let / be any continuous map from AU B to S° = {-1,1}- Since A is
connected, /|^ must be constant. Without loss of generality, we may assume
that A C /_1( —1)- By the same reason, f\g is also constant. Let Xq G A fl B.
We have f(xo) = —1. Since / is continuous, there exists a neighborhood of Xo,
say C/, such that U C /_1(—1). But since x0 G 5, there is a point of B which
belongs to U. Therefore we have B C /_1(—1). Hence / is not surjective. It
means that A U B is connected.
The following example shows that if A and B are path connected then
A U S needs not to be path connected. Let
,4= {(0,0)} C R2
and
B = {(a;,sin-) | 0 < x < 1}.
Then A and 5 are path connected and -4 ns =-4, but A Li B is not path
connected.
2102
Suppose that A and 5 are compact subspaces of spaces X and Y
respectively, and that N is an open neighborhood of
AxB CX xY.
Prove that there are open sets U C X and V C Y such that

84
Ax B CU xV CN.
(Indiana)
Solution.
Let' xbea point of A. For any y £ B, since (x, y) belongs to Ax B and N
is an open neighborhood oiAxBcXxY, there exist open sets Uy(x) C X
and Vy(x) C Y such that
(x,y) &Uy(x) x Vy{x)c N.
Therefore the family of open sets {Vy(x), y £ B} covers B. Since B is compact,
p
there is a subcover {Vyi(x), i — 1, • • • ,p} such that B C \J Vyi(x). Let U(x) =
i=i
p p
p| Uyi(x) and V(x) = (J V^z)- It is obvious that U(x) and V(x) are open
i=l i-\
sets of X and Y respectively and that
{x} x B C t/(z) x V{x) C iV.
On the other hand, {U(x),x E .A} is an open cover of A. Since A is also
compact, there exists a subcover {U(xi),j = 1, • • •, q} such that A C [J U{xj).
y=i
Let C/ = (J U(xj) and V = |~| V(a;j). It is easy to see that U and V are open
j=\ j=i
sets of X and y respectively and that AxBcUxVcN.
2103
Let X be a locally compact HausdorfF space. Let A and B be disjoint
subsets of X, with .4 compact and B closed. Does there exist a continuous
function / : X -► [0,1] such that /jA = 0 and /|B = 1?
(Cincinnati)
Solution.
If X is compact, then X is normal and the existence of / is obvious. Hence
we may assume that X is noncompact. We denote by X* the one-point com-
pactification of X. Since X is locally compact and HausdorfF, X* is compact
and HausdorfF, and consequently is also a normal space. Let J* = Ju{oo}.
It is easy to see that A and F = Sujoo} are two disjoint closed subsets of X*.
Then by the Urysohn Lemma there exists a continuous function / : X* —► [0,1]

85
such that /\a = 0 and f\p = 1. Therefore, the restriction of f on X, f, satisfies
the requirements that /|^ = 0 and /|b = 1-
2104
No proofs, only the correct answers to the question asked, are required for
this problem.
If X and Y are topological spaces, the join of X and Y is the quotient
space
X*Y = (X xY x [0,1])/ ~,
where
{x = x' and t = t' = 0
or
y = y' and t = t' = 1.
(a) S° * S° and S1 * S° are homeomorphic to familiar spaces. What space
are they?
(b) Describe X * S° for a general space X.
(Indiana)
Solution.
(a) S° * S° is homeomorphic to the unit circle S1, and S1 * S° is
homeomorphic to the unit sphere S2.
(b) Generally, X * S° is homeomorphic to the quotient space X x [0,1]/ ~
obtained from the cylinder X x [0,1] by collapsing X x {0} and X x {1} to
two points p and q respectively.
2105
(a) Define quotient map.
(b) Show that if X is compact, Y is HausdorfF and / : X —► Y is continuous
and onto, then / is a closed map.
(c) Show that if / satisfies the condition of (b) then / is a quotient map.
(Indiana)
Solution.
(a) Let X be a topological space and ~ be an equivalence relation on X.
Define by X/ ~ the space of equivalence classes under ~. By the quotient
map 7r : X —► X/ ~ we mean the map which assigns to x G X the equivalence

86
class containing x. The quotient space X/ ~ may be topologized by defining
a subset U C X/ ~ to be open if and only if 7r_1(t/) is open in X. Under
this topology, 7r becomes a continuous map. More generally, if / : X —► Y is
a continuous map, there is naturally associated an equivalence relation on X
such that X\ ~ £2 if and only if f{x\) = /(0:2)- If Y is homeomorphic to X/ ~
under the map i : X/ ~—» Y and f = ioir then we call / a quotient map.
(b) Let A be a closed subset of X. Since X is compact, A is compact too.
It follows from the continuity of / that f(A) is a compact subset of Y. Since
Y is Hausdorff, /(.4.) is closed in Y. Hence / is a closed map.
(c) Let ~ be the equivalence relation on X associated to the map /. Denote
by [a;] the equivalence class containing x. Then we define a map i : X/ ~—» Y
by i([x]) = f(x). Since / is onto, i is obviously a 1 — 1 map. By the result of
(b), the quotient space X/ ~ is compact. Hence i is a continuous 1 — 1 map
from the compact space X/ ~ to the Hausdorff space Y, and, therefore, is a
homeomorphism. It is clear that / = i 0 -k. So / is a quotient map.
2106
Let / : X —► Y be a continuous function from a space X to a Hausdorff
space Y. Let C be a closed subspace of Y, and let U be an open neighborhood
of f-l(C) inX.
(a) Prove that if X is compact then there is an open neighborhood V of C
in Y such that /_1(V) C C.
(b) Give a counterexample to show that if X is not compact, then there
need not be such a neighborhood V.
[Indiana)
Solution.
(a) Let W = X — U, then W is closed in X. Since X is compact, W
is compact too. Since / is a continuous function, f(W) is a compact set of
Y, and consequently is a closed set of Y because Y is a Hausdorff space.
Let V = Y — f(W). Then V is an open neighborhood of C in Y. Since
W C f-l(f(W)), we see that
/_1(V) = X - f~l(f(W)) CX -W = U.
(b) Let X = .R and Y = S1. / : X —► Y is the continuous function denned
by f(t) = e2Kit for < € -R. Take C = {1} € 51. It is obvious that
f-l(C) = {n\neZ}.

87
Let U = (J Un be the open neighborhood of / X(C) in X, where
£/„ = ( \-n,n+ -).
n n
Since
lim \Un\=- = Q,
n—>oo n
one can easily prove that there does not exist such a neighborhood V.
2107
Let X be a normal topological space and A C X a, closed subspace.
(a) Show that the quotient space Y obtained by collapsing A to a point is
normal.
(b) Does this result hold if normality is replaced with regularity?
(Indiana)
Solution.
(a) Let 7r : X —► Y be the identification map and yo £ Y be the point
which A collapses to. Let U and V be two nonempty closed sets in Y such
that U n V = 0. Then 7r_1(C/) and 7r_1(V) are two nonempty disjoint closed
sets in X. If yo £ U LI V, then, by the normality of X, it is clear that there
exist two disjoint open sets W\ and W2 in X containing 7r_1(C/) and 7r_1(V)
respectively such that Win.A = 0 for i equal to 1 and 2. Thus we see that -k(W\)
and 7r(VP2) are two disjoint open sets in Y and contain U and V respectively.
If yo £ U (or V), we only need to take the sets Wi and VP2 without the
restrictions that Wt 0 A = 0.
(b) Let X be a regular space which is not a normal space. It means that
there exist two disjoint closed sets A and B in X such that they cannot be
separated by disjoint open sets in X. Then the quotient space Y obtained by
collapsing A to a point y0 is not regular, because one can prove that the point
y0 and the closed set ir(B) in Y cannot be separated by disjoint open sets in
Y.
2108
Let p : E —► B be a covering map with E locally path connected and simply
connected. Let X be a connected space, let / : X —► B be a continuous map,

88
and /i, /2 : X —► E be two lifts of /. Prove that there is a deck transformation
g : E —► £ such that /2 = 3/1.
(Indiana)
Solution.
Take a point xo £ X and let /i(a;o) = eo G E. Then eo G p_1(6o), where
&o = p(eo)- Let /2(2=0) = ei. Since pfi = pf2 = /, we see that ei G p_1(60).
By the assumptions p ; E ^ B is the universal covering map. Thus there exists
a deck transformation g such that </(eo) = e\. Therefore, gfi(xo) = /2(2:0)-
Let
A={x€X\gf1(x) = f2(x)}.
It is obvious that A is not empty.
It is not difficult to prove that A is both-open-and-closed in X. Thus, by
the connectedness of X, we see that A = X, which means /2 = gfi : X —► E.
2109
Let p : X —► X be an n-sheeted covering projection, n < 00. Suppose that
X is compact. Prove that X is compact.
(J7i<&a7ia)
Solution.
Suppose that U = {U\,\ G A} is an open covering of X. For any point
x G X, let p_1(x) = {xi,- • • ,xn}. Let W(a;) be an elementary neighborhood,
i.e., W(x) is an path-connected open neighborhood of x such that each path
component of p~1(W(x)) is mapped topologically onto W(x) by p. Let
n
p-\W(x))=\JWi(x),
1=1
where Wi(a;) is a path component of p~1(W(x)) such that S; G Wi(x). Choose
a P, £ W such that x.{ G £/,-. Let ^(a;) be the path component of Wi(x) n Hi
containing X{. It is obvious that each Vj(x) is open and Vi(x) (1 Vj(x) = 0 for
n ^,
i ^ j. Since p is an open map, |~| p(Vi(x)) is an open neighborhood of x in X.
Choose another elementary neighborhood V(x) of a; such that
n
F(x)cf|P(W)-
i=l

89
Let
n
p-\V(x)) = (J vXx),
1=1
where V-(x) is the path component of p~l(V(x)) for any i. Then it is easy to see
that V/(x) C Vi(x) C Ui for i = 1,- •• ,n. It follows that p_1(V(a;)) C (J *7,-.
That is, we have proved that for any point x G X there exists an elementary
neighborhood V(x) of x such that p~1(V(x)) can be covered by a finite number
of sets in U. Since X is compact, X can be covered by a finite number of V(x{),
i = 1, • • •, m, and consequently X can be covered by a finite subcover of U.
2110
Let T and U be two different topology on X such that X is compact and
Hausdorff with respect to both. Prove that T (£U. (Recall that T C U means
that every set in the topology T is contained in U.)
[Indiana)
Solution.
We use the reduction to absurdity. Suppose that T Cli. Let (X,T) and
(X,U) denote the topological spaces of X with respect to T and U respectively.
h : (X,U) —► (X,T) is the identity map of X. Then h is a 1 — 1 map from
the compact space (X,U) to the Hausdorff space (X, T). We claim that h is a
continuous map. For any point xq G X. Let U be any open neighborhood of
xo in (X,T). Since T C U, U is also an open neighborhood of £o in (X,U).
It is obvious that h(xo) — xq and h(U) = £/. Hence h is continuous at xq.
Thus h is a homeomorphism from (X,U) to (X, T), which means U = T. This
contradicts the assumption.
2111
Let X be a topological space and let A C X. Show that if C is a connected
subset of X that intersects both A and X — A, then C intersects BdA. (Recall
that BdA = I n (X-4).)
(Indiana)
Solution.
We use the reductio ad absurdum. Suppose that C <1 BdA = 0. Take
U = C DA and V = CC) (X - A). Since C D A C U and C n (X - A) C V,

90
from the assumption, we see that both U and V are nonempty subset of C.
It is clear that C = U U V and both U and V are closed subsets of C, and,
consequently, that both U and V are open sets of C. But
unv = cnAn(X-A) = cnBdA = ®,
which is a contradiction to the connectedness of C.
2112
Let (X,d) be a metric space. For any subspace A C X and real number
e > 0, let
Oe{A) = {x £ A : d(x,a) < e for some a £ A},
Ce(A) = {x £ A : d(a;,a) < £ for some a £ A}.
(a) Prove that Oe(A) is an open subspace of X.
(b) If A is compact, show that Ce(A) is closed in X. Must Ce(A) be closed
for a general subspace .4 of X?
Solution.
(a) Let xo be any point of Oe(A). By the definition of Oe(A), there exists
a point a £ A such that <2(a;o, a) < e, i.e., <5 = e — d(xo, a) > 0. Then, for any
x £ O6/4(x0),
c
d(x, a) < d(x, x0) + d(x0, a) < - + (e — S) < e,
which means that 0^/4(20) C Oe(A). Thus Oe{A) is an open subspace of X.
(b) Let xo be any cluster point of Ce(A). Then there exists a sequence
{xn} in Ce(.A) such that xn ^ xo for any n and £n —► xo as n —► 00. By the
definition of Ce(A), for each xn there exists an an £ A such that d(xn, an) < e.
Since A is compact, without loss of generality, we may assume that an —► a for
some a £ A. Thus we have
d(x0,a) = lim d(xn,an) < e,
n—>oo
which means a;0 £ Ce(A). So C£(j4) is closed in X. If .A is not compact, we
give a counterexample as follows. Take X = [0,2] C -R and j4 = [0,1). Then
Ci(-A) = [0, |) is not closed in X.

91
2113
Suppose that X is a dense subspace of a topological space Y. Prove or give
counterexamples to the following assertions:
(a) If X is HausdorfF, then Y is HausdorfF.
(b) If X is connected, then Y is connected.
(Indiana)
Solution.
a) This assertion is not correct. We give a counterexample as follows. Let
Y = {a, b, c}. The topology on Y is determined by the family of open sets
T={{a,6,c},{a,c},{6,c},{c},0}.
Since any neighborhood of the point a always contains the point c, Y is not a
HausdorfF space. But it is easy to see that the subspace {c} is dense in Y and
is HausdorfF.
b) This assertion is true. We give a proof to it as follows. If Y were not
connected, then there would exist a nonempty proper subset U of Y which
is both-open-and-closed in Y. Let A = X (1 U. Since X is dense in Y and
U is open in Y, A would be a nonempty open set of X. On the other hand,
X — A = X (1 (Y — U) would be an open set of X, because Y — U is open in
Y. Thus A would be a nonempty subset of X, which is both-open-and-closed
in X. Since X is dense in Y and Y — U is open in Y, X — A is nonempty. It
means A ^ X. This contracts the connectedness of X.
2114
Let X and Y be topological spaces, X = U U V, and / : X —> Y be a
function so that f\u and /|y are continuous.
a) If U and V are open in X, show that / is continuous.
b) Give an example where U and V are not open in X and / is not
continuous.
(Indiana)
Solution.
a) Let N be an open set of Y. Since f\u and /y are continuous, /-1 |c/(JV)
and /-1|y(.W) are open in, respectively, U and V. But C/ and V are open in
X, therefore, f~1\u(N) and /_1|y(iV) are also open in X. Thus
r1(N) = ri\u(N)urlw(N)

92
is open in X, and consequently, / is continuous.
b) Let X = [0, 2], U = [0,1) and V = [1, 2]. / : X -» fl is denned by
Then f\u and /|y are continuous, but / is not continuous.
2115
Let q : X —> Y be a quotient space projection from a topological space X
to a connected topological space Y. Assume that q~1(y) is connected for each
yeY.
a) Show that X is connected.
b) Is X necessarily connected if the map q is not assumed to be a quotient
mapping? Justify your assertion.
(Columbia)
Solution.
a) Suppose that X is not connected. Thus, there exist two disjoint
nonempty open subsets oiX,U and V such that X = UUV. Then q(U)C\q(V) = 0.
For otherwise, let y £ q(U) <1 q(V). Therefore,
q-1(y) = (q-\y)riu)\j(q-\y)nv),
and it is obvious that q~1(y) H U and 5-1(2;) <1 V are both nonempty open
subsets of q~1(y), which contradicts the connectedness of q~1(y). Since q is
a quotient mapping, V = q~1(q(V)) and U = q~1(q(U)), q(U) and q(V) are
disjoint nonempty open subsets such that Y = q{U) Uq(V), which contradicts
the connectedness of Y. Thus X must be connected.
b) The following example shows that the assumption that q is a quotient
mapping is necessary for X to be connected. Let X = U U V, where
U = {(¢,0) G R2 |0< x< 1}
and
V={(l,y)£R2\l<y<2}.
The topology of X is induced from the topology of R2. Let Y = [0,1], the unit
interval of R, and q : X —> Y be the map denned by q(x, 0) = x for (a;, 0) € U
and qr(l, y) = 1 for (1,3/) £ V. Then q is continuous but is not a quotient
mapping. For each y £ Y, q~1(y) is connected. But X is not connected.

93
2116
Let X and Y be topological spaces, with Y compact. Let p : X x Y —> X
be the usual projection onto the first factor. Show that p is a closed map.
(Cincinnati)
Solution.
Let U C X xY be a, closed subset and Xo € X — p(U). Then, for any y £Y,
(xo,y) ¢ U. Since U is closed, there exist an open set WXo(y) of X and open set
Vy(x0) of Y such that (x0,y) £ WXo(y) x Vy(x0) and (WXo(y) x Vy(x0))DU = 0.
Since Y is compact, there must exist a finite number of Vyi(xo), ■ • •, Vyn(xo)
such that
n
r = U ^,-^0).
f=i
Let
n
W(x0) = f]WXQ(yi).
i=i
Then W(a;o) is an open neighborhood of xo in X. Since (W(zo) x Y) <1 U = 0,
we see that W(x0)C)p(U) = 0, i.e., W(x0) C X - p(U). Thus X - p(U) is an
open set of X, and consequently, p(C/) is closed in X, which means that p is a
closed map.
2117
Let Y be a connected subset of the topological space, and let Z be a set
such that y is a subset of Z and ^ is a subset of the closure of Y. Prove that
Z is connected.
(Minnesota)
Solution.
According to the assumptions, we have Y C Z C Y. Let f : Z ^> S° be any
continuous map, where S° — {—1,1} is the O-dimensional sphere with discrete
topology. Since Y is connected, without loss of generality, we may assume that
f(Y) = {1}, i.e., y C /_1(1). Taking closures of these two sets with respect to
Z and noting that /_1(1) is a closed subset of Z, we get (Y)z c /_1(1). But
as well-known, (Y)z = Y C\Z = Z. Thus we have Z c /_1(1), and it follows
that / is not surjective. It means that there does not exist any continuous
surjective map from Z to S° and therefore Z is connected.

94
2118
Let A be a connected subspace of a connected set X. If C is a component
of X\A, show that X\C is connected.
(Cincinnati)
Solution.
We first prove that if X is a connected space, U is a connected subset of X,
and if V is a both-open-and-closed subset with respect to X\U, then U U V is
connected. Suppose that C/ U V is not connected. Then let U UV = W\ U W2
where W\ and W2 are two disjoint nonempty both-open-and-closed subsets of
U UV. Since U is connected, without loss of generality, we may assume that
U C W2. Thus W\ is a both-open-and-closed subset of V, and, consequently,
a both-open-and-closed subset of X\U. Hence W\ is a nonempty both-open-
and-closed subset of (U U V) U (X\U) = X, which contradicts the assumption
that X is connected. So the above statement is proved.
Now suppose that X\C is not connected. Let X\C — U U V where U
and V are two disjoint nonempty both-open-and-closed subsets of X\C. Since
A C X\C and A is connected, we may assume that A C V. By the above
fact, C U U is connected because C is connected. Since Cut/ C X\A and
C(1U = 0, we see that CUU is a connected subset of X\A containing C, which
contradicts that C is a component of X\A. Hence X\C must be connected.
2119
1) A metric space X has property S if for every e > 0 there is a cover of
X by connected sets each of which has diameter < e.
a) Prove a metric space X has property S if X has a dense subset with
property S.
b) Suppose X is a subset of a metric space. Suppose the closure of X has
property S. Must X have property S?
(Cincinnati)
Solution.
a) Suppose that X has a dense subset j4 which has property S. Then,
for any e > 0, there is a cover {F0,a G T} of ^4 by connected sets such that
each Va has diameter < e/2. Since each V« is connected, Va is also connected

95
and obviously has diameter < e. Since A = X and A = [j Va, we see that
{Va, a € r} is a cover of X by connected sets each of which has diameter < e.
Thus X has property S.
b) We give a counterexample as follows. Let R denote the euclidean real
line, X be the set of all rational numbers. Then X = R has property S, but it
is easy to see that X does not have property S.
2120
Let X be the topologist's sine curve denned by
X = {(x,smir/x)\Q <x <l}(j{(Q,y)\-l<y <2}
U{(s, 2) | 0 < x < 1} U {(l,y), 0<y<2}cR2.
(i) Sketch X.
(ii) Let / : X —> X be continuous.
Show that either /(X) = X or else there exists S > 0 such that
/(X) n {(s,i/) | 0< s < «, -^<y<^} = 0.
(Toronto)
Solution.
(i) X is shown in the Figure below.
<0,2>
<0,1>
0
<-l,0>
ao>
Fig.2.1
A = f(X) n {(z,sin7r/:c) | 0 < x < 1}
(5 = inf{a; | (a;,sin7r/a;) £ A}.
(ii) Let
and

96
If S = 0, we claim that f(X) = X. To prove it, we first note that since X is
path connected, f(X) is also path connected. Hence in this case there exists
a So, 0 < <5o < 1, such that
{(^sinTr/z) | 0 < x < S0} C f(X).
Therefore it is easy to prove that the set
{(0,y)|-l<y<l}c/(X).
Once again, using the fact that f(X) is path connected, we see that f(X) = X.
If S > 0, then it is obvious that
f(X)n{(x,y)\0<x<6,-l<y<l} = 1b.
2121
Let T = S1 x S1 denote the torus.
(i) Show that T can be covered by 3 contractible open subsets.
(ii) Show that T cannot be covered by 2 contractible open subsets.
( Toronto)
Solution.
(i) It is well-known that the torus T can be identified to the quotient space
of a square X obtained by identifying opposite sides of the square X according
to the directions indicated by the arrows as shown in the Figure below.
Fig.2.2
Thus let C/x = X - {a U b}, U2 = X - I and U3 = X - II. (See the Figure
above.) Then U\, U2 and U3 are contractible open subsets and T = C/1UC/2UC/3.

97
(ii) Suppose that T could be covered by 2 contractible open subsets. From
the Van Kampen theorem it would follows that the fundamental group 7i"i(T) =
0. It contradicts the known fact that Tti(T) pzZ ®Z.
2122
Let U = {Ua}a€j be an open cover of the space X.
a) Give the definition for "W is locally finite".
b) If U is locally finite show that, for any subset K c J, [j Up is closed.
( Toronto)
Solution.
a) By definition, U is said to be locally finite, if each point p £ X has a
neighborhood which intersects only a finite number of Ua.
b) For any point p ¢. (J Up, since U is locally finite, there is an open
f*€K
neighborhood W of p which intersects only a finite number of sets in U,
particularly, only a finite number of Up for (3 £ K, say, Up1, ■ ■ ■, Upn. Since p ¢ Upi
for each /3,-, there is an open neighborhood Vi of p such that V; D Up{ = 0.
Then let V = W 0 Vi 0 ■ ■ ■ 0 Vn. It is clear that V is an open neighborhood of
p such that V (1 ( (J Up) — 0. Hence (J Up is closed.
2123
Let S be a set and let F be a family of real valued functions on S such
that f(si) = /(s2) for all / £ F implies Si = s2. Prove that there exists
a weakest topology in S amongst all those for which all members of F are
continuous. Show further that the resulting topological space satisfies the
Hausdorff separation axiom.
{Harvard)
Solution.
Let
U = {/_1((a, b)) | (a, b) is any open interval of R and / £ F}.
Then there exists a unique topology T on the set S of which U is the topology
subbase. It is easy to see that T is the weakest topology on S amongst all
those for which all members of F are continuous. Suppose that Si and s2 are

98
two distinct points of S. By the assumption there exists at least an / G F
such that f(si) ^ /(^2)- Hence we may take two open intervals (01,61) and
(a2,62) such that /(a,-) £ (a;, 6,) for i = 1,2 and (ai,6i) fl (02, 62) = 0. Then
£/,• = /_1((ai,6i)) and t/2 = f~l((a2,b2)) are two disjoint open sets of (S,T)
such that S{ £ C; for i = 1,2. So (S, T) is HausdorfF.

99
SECTION 2
HOMOTOPY THEORY
2201
a) Give generators and relations for the fundamental groups of the torus
and of the oriented surface of genus 2.
b) Compute the fundamental group of the figure 8 and draw a piece of its
universal covering space.
(Harvard)
Solution.
a)
(a) {by
Fig.2.3
As is well-known, we can present the torus T as the space obtained by
identifying the opposite sides of a square, as shown in Fig.2.3 (b). Under the
identification the sides a and b each become circles which intersect in the point
xo. Let y be the center point of the square, and let U = T — {y}. Let V
be the image of the interior of the square under the identification. Since V is
simply connected, by the Van Kampen theorem, we conclude that 7Ti(T, x\)
is isomorphic to iri(U, xi) modulo the smallest normal subgroup of tti(U, x\)
containing the image <p*(iri(UC]V, £i)), where <f>t is the homomorphism induced

100
by the inclusion map <fi : UC\ V —> U. It is easily seen that aUb is a deformation
retract of U. Hence ~K\{U, xq) is a free group on two generators a and /3, where
a and /3 are presented by circles a and 6, respectively. It is also clear that
jti(!7,xi) is a free group on two generators a' = 6~1a8 and /3' — 6~l(36,
where S is the equivalence class of a path d from £o to X\. (See Fig.2.3 (b).)
On the other hand, it is easy to see that -K\{U D V,X\) is an infinite cyclic
group generated by 7, the equivalence class of a closed path c which circles
around the point y once, and, consequently, that ¢,,,(7) = a'[3'a.'~lf3'~l. The
smallest normal subgroup of iri(U, xi) containing <j>*(iri(U n V, x\)) is just the
commutator subgroup of iri(U, xi). Thus 7Ti(T, Xi) is a free abelian group on
two generators a' and /3'. Changing to the base point x0, we see that 7i"i(T, x0)
is a free abelian group on two generators a and /3, which are presented by
circles a and b, respectively.
In a similar way, we can see that the fundamental group of the oriented
surface of genus 2 is a free group on four generators ai,/3i, a2,/32 with the single
relation [ai,/3i][o!2,/32], where a 1,(3\, a.2,P2 are presented by circles 01,61,02,¾)
respectively, and [a,-,/3,-] denotes the commutor a,/3,¢2^/3^1. (See Fig.2.3 (a).)
b)
Fig.2.4
Let X denote the figure 8 space, as shown in Fig.2.4. Let U = X — {q},V =
X — {p}. By the Van Kampen theorem we can prove that 7Ti(X, xo) is a free
group on two generators a, /3, where a, /3 are presented by circles a and b,
respectively.
The following picture is a piece of its universal covering space.
Fig.2.5
Under the covering map 7r, each level segment is mapped on the circle a

101
according to the direction indicated by the arrow, and each vertical segment
is mapped on the circle b according to the direction indicated by the double
arrows.
2202
Let A be a connected, closed subspace of a compact Hausdorff space X,
and suppose / : A —> A is a continuous map. For each positive integer n let
f"(A) = fof0...0f(A).
n times
oo
(i) Show that B = f] fn(A) is connected.
n=l
ii) Suppose 7 : S1 —> X — B is a nullhomotopic map. Show that there exists
a positive integer n such that 7(^1) C X — fn(A) and such that the induced
map 7' : S1 —> X — fn(A) (~y'(s) = -y(s) for s £ S1) is also nullhomotopic.
(Indiana)
Solution.
i) Since X is compact and A is closed, A is also compact. Thus it is easy
to see that /"(-4) is compact, closed and connected. Noting that fn+1(A) C
00
f"(A) C A for any n, and that A is compact, we see that B = f] fn(A)
n = l
is a nonempty closed subset of A. Suppose that B is not connected. Then
B = U U V, where U and V are disjoint nonempty closed subsets of B. It is
obvious that U and V are also closed subsets of A. Since A is obviously compact
and Hausdorff, A is a normal space. Therefore, there exist disjoint open subsets
of A, Wx and W2 such that U C Wi and V C W2- Thus it follows that
2? C Wi U W2 — W, which is an open subset of A. We claim that there exists
a positive integer N such that fN(A) C W. For, otherwise, there would be a
squence in A, {xn}, such that xn G /"(-4) but xn ¢. W for every n. It means
that xn £ A — W for every n. Since j4 — W is closed in A and, consequently, is
a compact subset, there exists at least a limit point xq of the sequence {£„}.
Noting that /"(-4) is compact for every n and fn+1(A) C fn(A), we can see
that Xo £ B c W, which is a contradiction. Thus fN(A) C W, and therefore,
fN(A) = (fN(A) n wt) u (fN(A) n w2),
which contradicts the fact that fN(A) is connected.
ii) Let F : S1 x I —> X — B be a homotopy between 7 and the constant
map. Since ^(51 x J) is compact and X — B is open, ^(51 x J) is also closed in

102
X. Since X is normal, there exists open sets U and V such that F(SX x I) c U
and B C V. From the proof for i), it follows that there exists a positive N
such that fN(A) C V. Therefore,
F(Sl x I) c U C X - fN(A).
Particularly, 7(^1) C X — fN(A). The remainder of ii) is obvious.
2203
Let RPn be the real projective n-space and Tm be the m-torus S1 x • • • x S1
(m factors). Prove that any continuous map RPn —> Tm is null-homotopic.
(Indiana)
Solution.
It is well-known that Rm is a universal covering space of Tm. Let P :
Rm —> Tm be the universal covering map. It is also well-known that
tt^T"1) &Z® ■■■<$! (m times)
and that iri(RPn) m Z2. Therefore, for any continuous map / : RPn —> Tra,
the induced homomorphism /„ : -K\(RPn) —> 7Ti(Tm) is trivial. Thus there
exists a lifting of /, say /, such that pf — f. Let po be a fixed point of Rm.
Define # : RPn x J -> Rm by
fT(x,t) = (1-*)/(»;)+ <P0-
Then E is a homotopy between / and the constant map po- So / is null-
homotopic and consequently / is also null-homotopic.
2204
Let X be the quotient space obtained by collapsing {pt.} x S1 C S1 x S1
to a point

103
Compute 7Ti(X) and H*(X).
(Columbia)
Solution.
Fig.2.6
X may be identified with the space shown in Fig.2.6, which is obtained by
identifying the sides of a 2-gon. Let y be the center of the 2-gon, U — X — {y},
and V be the interior of the 2-gon. Then, U and V are open subsets, U,
V, and U 0 V are path connected, and V is simply connected. Thus, by
the Van Kampen theorem, tti(X, xi) is isomorphic to the quotient group of
7Ti(C/, X\) modulo the smallest normal subgroup of tti(U, x\) containing the
image </>*(7ri(£/ D V, Xi)), where </>, is the homomorphism induced by the
inclusion map <f> : U f) V —> U. It is easy to see that a is a deformation ratract of U.
Therefore, iri(U,Xi) is an infinite cyclic group on the generator 6~la6, where
S is the equivalence class of a path d connecting Xq and X\. (See Fig.2.6) It
is also clear that -K\(U n V, Xi) is an infinite cyclic group on the generator 7,
where 7 is the equivalence class of a closed path c which goes once round the
point y. It is easy to see that ¢^(7) = lKl(u,x1)- Therefore we conclude that
■Ki.{X)=Z.
To compute H*(X), we may apply the Mayer-Vietoris sequence to the pair
(U, V). The conclusion is that Hj(X) is an infinite cyclic group for i equal to
0,1, 2 and is zero otherwise.
2205
(i) Suppose n > 2. Does there exist a continuous map / : Sn —> S1 which
is not homo topic to a constant?
(ii) Suppose n > 2. Does there exist a continuous map / : RP" —> S1
which is not homotopic to a constant?
(iii) Let T = S1 x S1 be the torus. Does there exist a continuous map
f :T—> S1 which is not homotopic to a constant?
( Toronto)

104
Solution.
(i) Let -k : R —> S1 be the universal covering map denned by 7r(<) = e2™',
and / : Sn —> S1 be a continuous map. Since 7ri(Sn) = 0 for n > 2, we see that
there is a lifting of f, f : Sn —> R such that 7r/ =: /. Since R is contractible,
/ must be homotopic to a constant map C : Sn —> R, hence / is homotopic to
■k o C = C, which is also a constant map from Sn to S1, i.e., there does not
exist any continuous map / : 5™ -> 51 which is not homotopic to a constant
map.
(ii) Since -Ki(RPn) = Z2, any continuous map / : RPn —> S1 induces
a trivial homomorphism /* : iri(RPn) —> tt^^1), and, consequently, has a
lifting / : RPn —> iJ such that ir o f = f. By the same argument as in (i), /
must be homotopic to a constant.
(iii) Denote T = Sl x S1 by T = (e2,rrtl,e2,r,'t2), 0 < <i, t2 < 1. Define
/ : T - 51 by
f(e2Kitl,e2Kit2) = e2Kitl £ S1.
It is easy to see that the induced homomorphism /* : -ffi(T) —> Hi(Sl) maps
one generator of Hi(T) to the generator of Hi(Sl), and another generator
to zero. Hence /„ is not trivial, which means that / is not homotopic to a
constant.
2206
A continuous map of topological spaces: p : E —> B is called a fibration if it
has the homotopy lifting property — that is, for any pair of continuous maps
I—fG:XxI—t3 and h : X x {0} —> E such that ph = G|xx{o} there exists
a continuous map H : X x I —> E such that H|xx{o} = ^ and pH — G. Let
p : E —> 2? be a fibration, 60 £ B be a base point, F = p-1(&o) (the "fiber"),
and eo £ F. Let i : F —> E denote the inclusion map.
(i) If F is path connected, prove that p# : iri(E,eo) —* 7Ti(-B, b0) is surjec-
tive.
(ii) Prove that in general the 3-term sequence
*i(F,e0) -> 7ri(jE;,eo) -> 7Ti(B,60)
in which the homomorphisms are i# and p#, respectively, is exact.
(Indiana)
Solution.
(i) Let a = [/] G -Ki(B,bo), where / : J —> 2? is a closed path at &o
which represents a. Let G : I x I —> B be a continuous map denned by

105
G(t,s) = f(st) for (s,t) £ I x I, and h : I x {0} —> E be the constant map
such that h(t,0) = eo- It is obvious that ph = G|jx{o}- Therefore there
exists a continuous map H : I x J —> jE such that .ff|jx{o} = ^ and P-ff = G.
Particularly we have pH(t, 1) = /(<)• Let C : J —> jE be a path in E denned
by c(t) = H(t, 1). Then pc = /. Let c(0) = ei and c(l) = e2. It is clear that
ei and e2 belong to F. Since F is path-connected, we may choose two paths
01 andjf2 in F such that #i(0) = e2, <?i(l) = ei, <?2(0) = ex and g2(l) = e^
Thus / = </2 * c * </i * ^2-1 is a dosed path at ei, where g2X is the inverse path
of g2. Noting that p</i, pg2 and p*/^-1 are ^1 constant path at bo, we have
P#[f] = b/] = \P92] ■ \pc] ■ \pgi] • [P02-1] = \pc} = [f] = «,
which means that p# is surjective.
(ii) Let a = [/] € 7i"i(.F,eo). It is obvious that p# • i#(a) = [p/] = [bo],
where bo is the constant path at bo. Hence im# C kerp#. On the other hand,
suppose that a — [f] 6 kerp#. Then there exists a homotopy G : I x J —> 2?
between p/ and the constant path &o such that G(t, 0) = (pf)(t), G(t, 0) = bo,
and G(0, s) = G(l, s) = bo for any s. By the homotopy lifting property, there
exists a continuous map H : I x I —> jE such that Plf = G and H(t, 0) = /(<).
Let ci = fi'|{0}xJ) c2 = #|ix{i} and c3 = #|{i}xJ. Then ci, c2 and c3 are
paths in F and cx(0) = e0, ci(l) = c2(0), c2(l) = c,j~ 1(0) and c3"1(l) = e0.
Hence c = ci x c2 x cj1 is a closed path in F with base point eo. It is easy to
see that / is homotopic to C. (See Fig.2.7.) Therefore, «=[/] = [c] = «'#[c].
i.e., kerp# C imi#. Hence the sequence mentioned above is exact.
(0,1)
(0,0) J (1,0)
Fig.2.7
2207
Is the canonical map q : S2 —> RP2 (which identifies antipodal points of
S ) nullhomotopic? Why or why not?
(Indiana)

106
Solution.
It is well-known that q is also the universal covering map from S2 to RP2.
Suppose that q is nullhomotopic. Then q is homotopic to a constant map
c : S2 —> RP2, and therefore q has a lifting q : S2 —> 52 which is also homotopic
to the lifting of c, a constant map c : S2 —> 52. But it is obvious that q is
the identity map or the antipodal map from S2 to S2. Hence deg<? = ±1. On
the other hand, we have deg<? = degc = 0. This is a contradiction. Thus we
conclude that q is not nullhomotopic.
2208
Let p : E —> B be an universal cover with E and B path-connected and
locally path-connected. Let T : B —> B be a map so that T" = Id and so that
T(b) = b for some b <E B. (Here Tn = ToTo-■-oT, n times.) Show that there
is a map f : £ -> E so that poT = Top and f" = 7d.
(/TirfiaTia)
Solution.
Choose eo € P_1(^) an^ consider the map Top: E —> B. We have
T o p(e0) = T(6) = b. Since ~K\{E) is trivial, there is a lifting oi T o p, T :
(E,e0) —> (jE,e0) such that poT = T op. Since
pof" = (poT) of"-1 = To(pf"-1) = ... = T" op
and
f "(e0) = T""1^ (e0)) = T"-^) = • • • = T(e0) = e0,
we see that T" is a lifting of T" op at the base point e0. On the other hand,
since T" = Id, it follows that identity map Id : (E, eo) —> (E,eo) is obviously
a lifting of I^p at the base point eo- Therefore, by the uniqueness of lifting,
we conclude that T" = Id.
2209
Let T2 = S1 x S1, and let I C T2 be the subset S1 x {1} U{1} x 51. Prove
that there is no retraction of T2 to X.
{Indiana)
Solution.
We use the reduction to absurdity. Suppose that there is a retraction of T2
to X, denoted by r. Let i : X —> T2 be the inclusion map. Then roi : X —> X is

107
the identity map. Hence r*oit : ~K\(X) —> ~Ki{X) is the identity homomorphism.
It is well-known that -K\(T2) is abelian and that ~K\{X) is an non-abelian free
group on two generators denoted by a and b. Hence we have ab ^ ba and
i*(a)i*(b) = i*(b)it(a). Therefore we have
ab = rtit(ab) = rt(it(a)i*(b)) — r*(it(b)it(a)) = rtit(b)rti*(a) = ba,
which is a contradiction.
2210
Let A C R3 be the union of the x and y-axis,
A = {(x,y,z)\(y2 + z2)(x2 + z2) = ()},
and let p = (0,0,1).
a) Compute H^R3 - A).
b) Prove that tti(R3 — A,p) is not abelian.
(Indiana)
Solution.
Let X = {(x,y,z) | x2 + z2 = 1} be a circular cylindrical surface, pi and
P2 denote the points (1,0,0) and (—1,0,0) respectively. Then it is easy to see
that X — {pi,P2) is a deformations retract of R3 — A. It is also clear that
X — {pi,P2) is homotopically equivalent to the space Y as shown in Fig.2.8.
r -A~r=p
Fig.2.8
a) Y has a structure of a graph with 4 vertices and 6 edges. The Euler
Characteristic K(Y) = 4 — 6 = —2. Hence the rank of H\(Y) is equal to
1 _ (_2) = 3. Therefore we conclude that .ffi(.R3 - A) & H^Y) fvZ <$Z <$Z.
b) Let e be a point on the arc ab different from a and b. Take U = Y — {c}
and V = Y — {e}. Then we have Y = U U V. It is easy to see that U has the

108
same homotopy type as Sl and that V has the same homotopy type as the
"eight figure space". Since U (1V is contractible, by the Van Kampen theorem
we see that tti(Y) is a free group generated by ~K\(U) and -K\(V). Therefore we
conclude that tti(R3 — A,p) (¾ Tti(Y) is a free group on three generators and,
consequently, that 7Ti(-R3 — A,p) is not abelian.
2211
Let p : (Y,y) —> (Y,y) be a regular covering space; that is, p*(iri(Y ,y))
is a normal subgroup of ir\(Y,y). Suppose that / : X —> Y is a continuous
function from the path-connected space X to Y with f(xo) = /(^ l) = j/, and
that there is a lifting of /, /o : (X,xo) —> (Y,y). Show that there is a second
lifting, /i : (X, xx) —> (y, i/). (/i need not be distinct from /o.)
(j7i<Ha7ia)
Solution.
Let fo{xi) = y'. Since
P(S0 = pfo(xi) = f(xi) = y, y' € p'^y).
Since p is a regular covering space, there is a deck transformation 7 such that
f(y') = J/. Then let
/1 = 7 0 /0 : JT - ?•
Therefore
7i(»i) = 7(/0(^1)) = 7(iT) = i/.
Hence /1 is the lifting of / we want.
2212
Let X be the identification space obtained from a unit 2-disk by identifying
points on its boundary if the arc distance between them on the boundary circle
is ^-. Compute the fundamental group of X.
(Indiana)

Solution.
109
avA//;y//A//Xa
¾v¾¾¾y
Fig.2.9
Take a point y in the open disk. (See Fig.2.9.) Let U — X — {y} and let
V be the open disk. Then both U and V are path connected subsets of X
and X = U UV. Since V is simply connected, by the Van Kampen theorem,
7Ti(X) is isomorphic to the quotient group of tti(U) with respect to the least
normal subgroup containing ¢^(^(¢/ n V)), where <j> : (U n V) —> -K\(U) is
the homomorphism induced by the inclusion <j> : U <1 V —> U. Take a point
Xi £ U (IV as the base point. (See Fig.2.9.) It is clear that -K\{U O V,Xi) is
an infinite cyclic group generated by 70 the closed path class of a closed path
c which circles around the point y once. Since the circle a is a deformation
retract of U, it is clear that iri(U, Xo) is an infinite cyclic group generated by
a' the closed path class of a. Therefore iti(U, x\) is an infinite cyclic group on
generator a = 7_1a'7, where 7 is the path class of a path d from xq to x\.
It is also clear that </>*(7c) = 3a. Hence the least normal subgroup containing
<I>*{-k\{U n V)) is isomorphic to ZX. It follows that -ki(X) &Z/ZZ =Z3.
2213
Sketch a proof of the Fundamental Theorem of Algebra (every nonconstant
polynomial with complex coefficients has a complex zero) using techniques of
algebraic topology.
(Indiana)
Solution.
Let (F denote the complex plane and f(z) be a polynomial of positive degree
with complex coefficients. We may consider / to be a continuous nonconstant
map / : (T —> (T. Note that \f(z)\ —> 00 as \z\ —> 00; hence, we may extend /

110
to a map of the one-point compactification of <E
f:S2^S2
by setting /(00) = 00, where 00 denotes the north pole. Then we may first
prove that if /(z) = zfc, k > 0, then the degree of the extension / : S2 —> S2 is
equal to k. Furthermore, we may prove that if / is any polynomial of degree
k > 0 then the degree of the extension / : S2 —> S2 is still equal to k. Noting
the fact that if a continuous map / : S2 —> S2 is not surjective then the degree
of / is zero, we may prove the Fundamental Theorem of Algebra by means of
the reduction to absurdity.
2214
Let X denote the subspace of R3 that is the union of the unit sphere S2,
the unit disk D2 in the x-y plane, and the portion, call it A, of the z axis lying
within S2.
(a) Compute the fundamental group of X.
(b) Compute the integral homology groups of X.
(Indiana)
Solution.
(a)
Fig.2.10
It is clear that X has the homotopy type of the one point union Xi V X2
where X\ and X2 are each homeomorphic to the union of the unjt sphere
and the portion of the z axis lying within S2. (See Fig.2.10.) To compute
tti(Xi V X2), we take U = Xi V X2 - {p} and V = Xx V X2 - {q}. Then
we have U UV = Xi V X2- Since U 0 V is contractible, by the Van Kampen,
7Ti(Xi V X2) is a free product of the groups Tti(U) and 7Ti(V) with respect to
the homomorphisms induced by the inclusion maps. It is obvious that
n(U) « n(V) « n(Xi) « 7n(X2) « ^(s1).

111
Therefore, iti{X) is a free product generated by two generators. We can take
as generators the closed path classes which are determined by the closed paths
omp and omq respectively. (See Fig.2.10.)
(b) Applying the Mayer-Vietoris sequence to the pair (U, V), we see that
Hi(X) « Hi{U) ® Hi(V) w Hi(X{) $ Hi(X2).
Noting that Hi(Xi) « Hi(X2) which are infinite cyclic for i equal to 0,1, 2 and
are zero otherwise, we conclude that
{Z®Z, i = l,2,
Z, i = 0,
0, otherwise.
2215
Let X be a path-connected space, / : X —> Y a continuous function, and
Boi^i € X. Suppose that the induced homomorphism
/* : -Ki(X,x0) -► iri(Y,f(x0))
is surjective. Show that
/* : tti(X, a;!) -> 7Ti(y, /(a=i))
is also surjective.
(jTirfiaTia)
Solution.
Since X is path-connected, there exists a path C : [0,1] —> X such that
c(0) = x0 and c(l) = Ki. Then c = / oc is a path connecting /(zo) and /(zi)
in y. For any [a] £ 7Ti(y/(^i)), where a is a closed path at /(^l), cac-1 is a
closed path at /(½). By the assumption, there is a closed path h at £o such
that f*([h]) = \cac~1]. It means that f o h and cac-1 are homotopic. Thus
/ o (c_1/ic) and a are homotopic, and, consequently, f*([c~lhc]) = [a]. Hence
f*:irl(X,x1)^ir1(YJ(x1))
is surjective.

112
2216
Let B denote the "figure eight space". Let p : X —> B and q : Y —> B be
2-fold covering maps, where both X and Y are connected. Prove that X and
Y are homotopy equivalent, but not necessarily homeomorphic.
(Indiana)
Solution.
V< /,
Fig.2.11
For any 2-fold covering map p : X —> B, let p~1(xo) = {eo,ei}. (See
Fig.2.11.) Since the automorphism group A(X,p) « Z2 and X is connected, it
is not difficult to see, by considering the liftings of the circles a and b in X, that,
in substance, X has only two different types as shown in Fig.2.11. Then it is
easy to see that they are homotopy equivalent to an 3-leaved rose G3. But the
spaces X and Y shown in Fig.2.11 are not homeomorphic. Otherwise, suppose
that / : X —> Y is a homeomorphism. Then X — {eo} is homeomorphic to
Y — {/(eo)}. Since X — {eo} is contractible and Y — {/(eo)} is obviously not
contractible, we come to a contradiction.
2217
Calculate the fundamental group of the space RP2 x S2.
(Indiana)
Solution.
By the formula
we see that
Tn(RPl x S2) x -ki(RP2) ® 7n(S2) k,12 $ {0} »Z2.

113
2218
Let Z denote the figure 8 space, Z = X V Y, X and Y circles. Let x, y £
iri(Z,*) be the elements in Z defined by X, Y, where * denotes the vertex
Fig.2.12
(a) Let h : tti(Z, *) —> Z/6Z be the homomorphism satisfying h(x) — 2 and
h(y) = 3, and let p : Z —> Z denote the covering space corresponding to the
kernel of h. (p*(-Ki(Z,*)) = ker(h).) If A is a path component of p~1(X) and
B is a path component of p~1(Y), how many intersection points of A and B
are there? (i.e., what is the cardinality of the sat A 0 B?)
(b) If G is a finite group, h : tti(Z, *) —> G a surjection, and p : Z —> Z the
corresponding cover, prove that the number of intersection points of a path
component A of p~x(X) with a path component B of p~x(Y) divides the order
of G.
(Indiana)
Solution.
(a) Since h(x) = 2 and h(y) = 3 and iri(Z,*) is generated by x and y, it
follows that the homomorphism h is surjective. Thus tti(Z, *)/ker h is
isomorphic to the group Z/6Z = Z&. Hence the covering space p : Z —> Z is a 6-fold
cover. We denote p_1(*) by
P_1(*) = {eo,ei,e2,e3,e4,e5}.
Then, from h(x) = 2, we see that the path component of p~1(X) contains
exactly the points {e0,e2,e4} of p_1(*) which corresponds to the elements
{0,2,4} of Ze respectively. By the same reason, we see that the path component
oip~1(Y) contains exactly the points {e0, 63} of p-1(*). Since
p-1(X)nP-1(Y)=p-\*),
we conclude that A D B = {eo}.

114
(b) Since h is a surjection and G is a finite group, the corresponding cover
p : Z —> Z is a finite fold cover. Suppose that h{x) = r and h(y) = s. Then
r generates a subgroup Hi of G and s generates a subgroup of H2 of G. In a
similar way as in (a), we see that the number of intersection points of A and
B is equal to the number of elements in Hi fl H2. By Lagrange's theorem, it
divides the order of G.
2219
Let X be the result of attaching a 2-cell D2 to the circle S1 by the map
/ : S1 —► S1 given in terms of complex numbers by z —► z6.
(a) Compute, with proof, the fundamental group of X.
(b) Compute, with proof, the homology of the universal cover of X.
[Indiana)
Solution.
(a)
Fig.2.13
Represent X as the space obtained by identifying the edges of a hexagon, as
shown in Fig.2.13. Under the identification the edges a become a circle through
the point Xq. Let y be the center point of the hexagon, and let U = X — {y}.
Let V be the image of the interior of the hexagon under the identification.
Then, U and V are open subsets, U, V, and U 0 V are arcwise connected, and
V is simply connected. Let xi be a point in UC\ V. It is clear that U(l V has the
same homotopy type with S1, and that x1(r/n V, xi) is an infinite cyclic group
generated by 7, the homotopy class of a closed path c which circles around the
point y once (see Fig.2.13.)
Applying the Van Kampen theorem, we conclude that
V>i :iri(U,xi) -^-ki(X,Xi)
is an epimorphism and its kernel is the smallest normal subgroup containing
the image of the homomorphism
4>i :irl(UnV,xl) -nri(tf,a;i),

115
where ipi and <pi are homomorphism induced by inclusion maps.
It is obvious that the circle a is a deformation retract of U. Thus x^C/, a;0)
is an infinite cyclic group generated by a and, consequently, tti(U,Xi) is an
infinite cyclic group generated by a' = 7_1a7, where 7 is the homotopy class
of a path d from Xq to Xi.
It is obvious that ¢1(7) = a'6. Hence the smallest normal subgroup
containing <j>i(ni(U n V, xi)) is isomorphic to 6Z. Thus we conclude that
xi(JT)»Z/6Z=.Z6.
(b) Since X is a finite 2-dimensional CW complex and ^i(X) = Z&, its
universal covering space X is a 6-fold covering space and is also a 2-dimensional
CW complex. It is well-known that the Euler characteristic X(X) = QX(X).
But it is easy to see that X(X) = 1, hence X(X) = 6. From H0(X) « Z,
#i(X) = 0 and H2(X) « H2(X, X1), where X1 is the 1-skeleton of X, we see
that H2(X) is a free Abelian group of rank 5. So we conclude that
0, i > 3,
Z ®Z ®Z ®Z ®Z, i = 2,
0, i = 1,
Z, i = 0.
2220
If X is any topological space and S1 denotes the unit circle in the complex
plane with its usual topology as a topological group with multiplication given
by the multiplication of complex numbers, then it is known that the set [-X\ S1]
of homotopy classes of maps from X to S1 inherits a natural group structure.
(a) Define this group operation explicitly and indicate the group identity
and how inverses are formed. You do not need to prove your assertions.
(b) Compute this group explicitly for X = point, S1, S2, and T2 = S1 x 51.
(Indiana)
Solution.
(a) We denote the homotopy class of a map / : X —► S1 by [/] and write
S1 as
Sx = {e2'ie G <F I 0 < 6 < 1}.
Then the multiplication of [-^,51] is defined by [/] • [g] — [f • g], where the
map f ■ g : X —► S1 is denned by (/ • g)(x) — f(x) • g(x) for any x £ X. Here
H.-(X) =

116
f(x) ■ g(x) is defined by the multiplication of complex numbers. The identity of
this multiplication is [e], where the map e : X —► Sl is defined by e(x) = e2*'
for any x £ X. The inverse of [/] is the homotopy class of a map /, which is
defined by f(x) = l/f(x) for any x E X.
(b) If X = point, then [X, S1] obviously has only one element. So the group
[X, S1] is trivial. For the case of X = S1, one can easily prove that
For the case of X = S2, it is easy to see that each homotopy class [/] can
be ^presented by a map / : S2 —► S1 which maps the northpole N of S2 to
the point po = e2** of S1. Then by the facts that S2 is simply connected and
the universal covering space of S1, R is contractible, one can easily prove that
[S2, Sl] « ^(S1) = 0. Now we discuss the case of X = T2 = Sl x Sl. For any
map / from S1 x S1 —► 51 we define two maps /i and /2 from 51 —► S1 by
/x(0) = f(e2'ie,p0) and /2(6>) = /(po,e2"ie) for any e2"B £ S1, respectively.
Then let ^ : [X^1] ^1 ®Z is denned by ¢([/]) = (deg/i,deg/2). We have
H[f]-[9}) = ^([/•ff]) = (deg(/.ff)i,deg(/.ff)2)
= (deg(/i •ffi),deg(/2-flf2))
= (deg/1+deg<?i,deg/2 + deg<?2)
= (deg/1,deg/2) + (deg<?1,deg<?2)
= Hif}) + <K\g])-
Therefore, <j> is a homomorphism from [X, 51] toZ (&Z. Note that /j can be
extended to a map from X to S1, still denoted by /1, by
which is homotopic to / under the homotopy map F : X x J —► S1 denned by
We2iri0 e2iriV ^) _ f(e2xie e2irit+2iri(l-t)V y
Thus one can easily prove that <j> is a monomorphism. It is clear that (/1
is a epimorphism, and cosequently <j> is an isomorphism. We conclude that
[X,Sl]ttZ®Z.

117
2221
If X is a path-connected space whose universal cover is compact, show that
xi(X, xo) is finite.
(Indiana)
Solution.
Let x : X —► X be the universal cover of X. If k\(X, xq) were not finite,
then ir~l(xo) would be a closed set of infinite points in X. Since X is compact,
7r-1(a;o) must have at least a limit point, say x, such that ir(x) — Xq. Thus it is
easy to see that x is not a local homeomorphism at x, which is a contradiction.
2222
Prove that if X is locally path connected and simply connected then every
map X —► Sl is homotopic to a constant. What can you say if we just assume
that X is path connected, locally path connected and the fundamental group
of X is finite?
(Indiana)
Solution.
Let exp : R —► Sl denote the exponential covering map, i.e., the universal
covering space of 51. Since X is locally path connected and simply connected,
■Ki(X) = 0, and /*(xi(X)) = 0 for any map / : X —► 51. Hence there exists a
lifting of /, / : X —► R such that exp(/) = /. Since iJ is simply connected, /
is homotopic to a constant map c. Denote the homotopy between / and c by
If. Then exp(If) is the homotopy between / and c, i.e., / is homotopic to a
constant map.
Suppose that 7Ti(X) is finite. Since 7^(51) k, Z and ft(-Ki(X)) is a finite
subgroup of iTi(Sl), we see that f*(iti(X)) must be trivial. So the above
argument still works in this case, and the same conclusion holds.

118
SECTION 3
HOMOLOGY THEORY
2301
Prove the 3x3 Lemma.
Consider the following commutative diagram of abelian groups
0
1
A3
hi
B3
Sii
c3
i
0
0
1
2i A2
£2 i
£i R
—► u2
eil
3 C2
I
0
0
1
^ A1
C2I
£i R
—* #1
Cil
^ d
1
0
If all 3 columns and the first two rows are short exact, then the last row is also
short exact.
(Harvard)
Solution.
To prove the exactness at C3, we show that 72 is injective. Let c £ C3 and
72(c) = 0. Since 61 is surjective, there is a b £ B3 such that c = 6\(b). By
the commutativity we see ^(b) £ kercj. Hence there is an 02 € -A 2 such that
£2(0-2) — faity- Then since
C2(«i(a2)) = /3i(£2(a2)) = /?i(/32(6)) = 0
and (2 is injective, we have «1(0.2) = 0, i.e., 02 € kerai = ima2. Thus there is
an a £ A3 such that a2(a) = 02- Since
HH°) ~b) = /W«) - /32(6) = e2a2(a) - /32(b) = e2(a2) - /32(6) = 0
and /32 is injective, we see 62(a) — b. Therefore
¢ = ^(4) = 4^(0) = 0.
Hence 72 is injective.
0
0
0

119
Now we prove that ker7i C im72- For any c G ker7i, since £1 is surjective,
there is an b G #2 such that c = £\(b). Thus it is easy to see that /3i(6) G ker^i,
and consequently that there is an ai G -4i such that (2(0-1) = /3i(6). Since
c*i is surjective, there is an 02 G A2 such that 0:1(02) = a\. Thus by the
commutativity, we have b — £2(02) G ker/3i. Therefore there is a 63 G S3 such
that f32(b3) = b- £2(a2). Then
72(^1(¾)) = £1/32(¾) = £i(b) - £i£2(a2) = c.
Hence c G im72. In a similar way we may prove that im72 C ker7i. Thus the
exactness at C2 is proved.
We leave the proof of the exactness at C\ to the reader.
2302
Prove that if M is a compact manifold of odd dimension, then X(M) = 0.
Show examples of compact 4-manifolds with X = 0,1,2,3,4.
(Columbia)
Solution.
Since M is compact, M is ^2-orientable. Suppose that dimM = 2m + 1.
Therefore,
2m+l
X(M) = Y^ (-!)*' dim^i(M,^2).
i = 0
By the Poincare duality theorem,
dimHi(M,Z2) = dimH2m+1_i(M,Z2)
for any i. Thus, since i and 2m + 1 — i have different parity, they appear in
the sum with opposite signs. Therefore X(M) = 0.
Let X0 = T2 x T2, Xj = RP2 x RP2. Denote by Uh the connected sum of h
projective planes. Then it is well-known that X(Uu) — 2—h. Let X2 = U3 x t/4,
X3 = U3 x t/5, and X4 = S2 x 52. Using the fact that
X(M1 x M2) = X(M1) x A*(M2),
we see that X(X{) — i for i equal to 0,1, 2, 3 and 4.

120
2303
Let X be a topological space. The suspension EX of X is defined to be
the identification space obtained from X x [—1,1] by identifying X x {—1}
to a point and X x {1} to another point. For example the sphere Sn is the
suspension of Sn_1 with the north and south poles corresponding to the two
identification points.
Compute the homology of EX in terms of the homology of X.
(Illinois)
Solution.
Let p\ and p2 be the identification point respectively. Set U = EX — {pi}
and V = EX - {p2}. Then U and V are open sets of EX, and EX = U U V. It
is obviously to see that U and V are contractible spaces and X is a deformation
retractor of U <1 V. By Mayer-Vietoris sequence, we have
#g(EX)
I*,
(*).
«>o,
¢ = 0,
where Hq-\(X) denotes the reduced homology of X.
2304
{Cn\n>Q} Cn=Z3
Cn-\ denned by
(s — i,0, 0) n even
(Q,s + t,s + t) n odd.
Consider the chain complex
C =
On '• l-'n —*
(r,s,t) -»
(r,s,t) -»
Compute Hn(C) for all n.
Solution.
When n is even, (r, s, t) G ^„(c) if and only s = t. So ^n(c) = {(r, s, s) G
C„}, i.e., Zn(c) is isomorphic toZ (&Z.
Noting that im<9n+i = {(0,t,t) £ C„} for even n, we have Hn(c) = Z for
even n.

121
When n is odd, (r, s, t) G Zn(c) if and only if s + t — 0. So Zn(c) is
isomorphic to Z 0 Z, and noting im<9„+i = {(r, 0,0) G C„} « JT, we have
fl-„(c)=Z.
Therefore, Hn{c) k,Z for any n.
2305
Construct a CW complex which has the following ^-homology groups:
H0(X) = Z,
Hi(X) = Z&Z/2Z,
H2(X) = Z173,,
H3(X) = Z,
Hn(X) = 0, ifn>4.
(Columbia)
Solution.
Denote by X1 the "figure eight space" as shown below,
Let X2 be the space obtained by attaching two 2-cells to X1, one by the
map /i : S1 —► xq and the other by the map /2 : Sl —► a such that /2(2) = z2.
Therefore, the image of one 2-cell under /1 is homeomorphic to S2, the 2-
sphere. It is well-known that Hi(Xl) =Z (&Z, which has two generators a and
b, consider the following diagram
0^H2(X2)±H2(X2,Xl) * ^(X1)-^^*2)-^*2,*1)
T fi* T/iU1*
The square is commutative and the level rows are exact. Since
H2(X2,X1) = imfu(&\mf2* &Z®Z,

122
im/ils1* = 0 and im^ls1* = 2Z> we may see that im<9* « 1Z and ker<9* « Z.
Since H\{X2,Xl) = 0, we have Hi(X2) Pa Z ®Z2. It is also easy to see that
H2(X2)=Z.
Now let X be the space obtained by attaching two 3-cells to X2, one by the
map </i : S2 —► £o and the other by a map </2 : S2 —► 52 such that deg</2 = 3.
Then in a similar way as above, we may conclude that the space X satisfies
the requirements in the problem.
2306
Compute Hp(Sn V Sn V • • • V Sn), n > 0, for all p.
(Columbia)
Solution.
Denote by Sn(q) the space Sn V S" V ■ ■ ■ V 5",. When n = 0, 5°(«) has
5 times
5 + 1 points. Then Hp(S°(q)) is a free abelian group of rank q for p equal to
0 and is zero otherwise. In fact
Sn(q+l) = Sn(q)VSn.
Let a e Sn(q) and b £ Sn. It is easy to see that U = Sn(q +1)- {a} has
the homotopy type of Sn(q) and V = Sn(q + 1) — {b} also has the homotopy
type of Sn(q). It is also clear that Sn(q + 1) = U U V and U 0 V has the
homotopy type of Sn(g — 1). Thus, when n > 0, by induction on q and the
Mayer-Vietoris sequence of the pair (U, V), we may prove that Hp(Sn(q)) is a
free abelian group of rank q for p equal to n and is zero otherwise, where Hp
is the reduced homology group.
2307
Build a CW complex X by adding two 2-cells to S1, one by the map z —► z4
and the other by the map z —► z6. What is the homology of this space?
(Indiana)

123
Solution.
Fig.2.14
H2{Kl)
H2(K2)^H2(K2,K1) ^
T/.
X is a 2-dimensional CW complex with 2-skeleton K2 = X, 1-skeleton
K1 — S1 and 0-skeleton K° = {p}. K2 is obtained from K1 by attaching
two 2-cells via the maps / and g as indicated in Fig.2.14. Kl is obtained
from K° by attaching one 1-cell. Let Kn = K2 for n > 3. Then we have
a chain complex K = {C„(if),rf„}, where Cn(K) = Hn(Kn, Kn~l) and d„ :
Cn{K) —► Cn-i^) is denned to be the composition of homomorphisms,
Hn(Kn,Kn~l) % frB_1(jr-1)^1 frI,_1(jr*-1,jirB-2),
where 5* is the boundary operator of the pair (if™,if™-1) and j„_i is the
homomorphism induced by the inclusion map. It is well-known that Hn(X) (¾
Hn(K). It is obvious that Hn(X) = Hn(K) = 0 for n > 3. To compute #2(fc),
consider the following diagram:
T(/|sO*
The square is commutative and it is well known that /* is a monomorphism
and d't is an isomorphism. Noting /|Si is the map z —► z4, we see that the
image of (/Is1)* *s ^Z- ^n *^e same wav we see that the image of ({/Is1)* is 6Z-
Since
H2(K2tK1)Mimft®ungt,
it follows that im<9* is isomorphic to the subgroup 1Z of Hi (if1), and that
ker<9* is isomorphic to Z ®Z2. Thus, since j2 is injective, we see that
H2(K2) « imj2 « kerd* «Z $Z2.
It is clear that ji is an isomorphism. Therefore, \md2 ~ 2Z. Noting that
ker^ « fl^i^if0) « Z, we have #i(iiQ « ^/2Z « JT2- Thus we conclude
that
' ^0^2, i = 2,
^2, i = 1,
Z, i = 0,
0, otherwise.
¢(1) = {

124
2308
Let
X = {(x,y,z)£R3 \xyz = 0}.
a) Compute H*(X,X - {0}).
b) Prove that any homeomorphism h : X —► X must leave the origin fixed.
(Indiana)
Solution.
a) Let U = D3 <1 X, where
D3 = {(x, y, z)£R3\x2 + y2 + z2 < 2}.
It is easy to see that U is contractible. Thus
Hq(X,X - {0}) « Hq(U, U - {0}) « Hg-^U - {0})
for any q. Let 5 = 52 O X, where S2 is the unit 2-sphere. Then S is a
deformation retract of U - {0}. So #*(*/ - {0}) « ff*(5). To compute #*(S),
let
Wi = S-{(0,0,1),(0,0,-1)}
and
W2 = S-{(-1,0,0),(1,0,0)}.
Applying the Mayer-Vietoris sequence of the pair (W\, W2), we see that
¢ = 2,
otherwise.
ifg(A,A - |U)j - u
b) Let x be any point of X different from the origin. Similarly we obtain
*,<*,*-<.»={;,•*•*• 1^;
in a coordinate axis,
otherwise.
Let h : X —► X be any homeomorphism. Since the local homology groups are
invariant under homeomorphism,
H2(X, X - {0}) « H2(X,X- {/(0)}).
From the above results, it follows that /(0) = 0.

125
2309
a) Write down the Mayer-Vietoris sequence in reduced homology which
relates the spaces Sl x R1, S1 x Rl — {p}, and a disk D in S1 x R1 about the
point p. (S1 = 1 — sphere)
b) Use a) to calculate the homology of S1 x R1 — {p}.
(Indiana)
Solution.
a) The Mayer-Vietoris sequence is
► Hq(D - {p}) t Hq(Sl xR1- {p}) ® Hq(D)
t H^S1 x R1) A Hq^{D - {p}) - • • •.
b) Since S1 x R1 and D — {p} have the same homotopy type with S1, it
follows that
Hq(Sl x R1) « Hq(D - {p}) « HgiS1).
Since D is contractible, the nontrivial part of the Mayer-Vietoris sequence is
0^Ifi(Z?-{p}) -t H^S1 x R1 - {p}) ^ JSf^S1 x ii1)-» 0,
which is split exact. Thus we have
Z®Z, i=l,
H^S1 x R1 - {p}) = { Z, f = 0,
0, otherwise.
2310
Let the real projective plane RP2 be embedded in the standard way in the
real projective 5-space RP5. Compute Hq (RP5/RP2), where RP5/RP2 is
the space obtained from RP5 by identifying RP2 to a point.
(Indiana)
Solution.
Since RP5 is compact HausdorfF and RP2 is a strong deformation retract
of a compact neighborhood of RP2 in RP5, we see that
Hq (RP5/RP2) « Hq(RP5,RP2) for any q.

126
It is well-known that
r z, ¢ = 0,5, r I, q = 0,
Hq(RP5)=lz2, ¢ = 1,3, and Hq(RP2) = I Z2, q = 1,
(^ 0, otherwise, (^ 0, otherwise.
Thus, from the exact sequence of the pair (RP5,RP2), it follows that
( Z, ¢ = 0,5,
Hq {RP5/RP2) = < Z2, 5=3,
\ 0, otherwise.
2311
Let
Y = {(xuxt) G R? | 2:10:2 = 0 and x2 > 0}.
Prove that Y x R is not homeomorphic to R2.
(Indiana)
Solution.
Let
Y x R= {(xi,x2,x3) <ER3 | xxx2 = 0,x2> 0}.
Then the point O = (0, 0,0) € Y x R. To compute the local homology groups
Hi(Y x R,Y x R-{0}), i = 0,1,2,---,
we take an open neighborhood of the point O, U = D3 D (Y" x R), where D3
is an open ball centered at O. Therefore
Hi(Y xR,Y xR-{0})fnHi(UtU-{0})taHi.1(U-{0})1
because U is contractible. Let X be the space as shown in the following figure

127
It is easy to see that X is a deformation retract of U — {0} and that
ffi(X) ?*Z®Z. Thus
H2(Y xR,Y x R-{Q})kZ®Z.
It is well-known that for any point p £ R2 the local homology group
H2(R2,R2-{p})*Z.
Since the local homology groups are isomorphic under homeomorphism, it
follows that Y x R is not homeomorphic to R2.
2312
Let A be a nonempty subset of X and X U CA be the union of X with the
cone of A; that is, X U CA is obtained from the subset X x {0} U A x [0,1] of
X x [0,1] by identifying A x {1} to a point.
Prove that the reduced homology group Hn(X U CA) is isomorphic to
Hn(X, A), for every n.
(Jn<Kana)
Solution.
Let Y = X x {0} U A x [0,1] and B = A x {1}. Then XUCA = Y/B.
Let x : y —► y/S be the identification map and yo denote the point ir{B) in
Y/B. Let U = A x [~,l] c Y. Then y0 is a strong deformation retract of
n(U). Thus, in the exact sequence of the triple (Y/B,ir(U),yo),
■ ■ ■ ^ Hn(w(U),y0) ^ Hn(Y/B,y0)
-» Hn(y/B,ir(U)) - fl-„-i(x(CT),jto) -» • • •
it follows that Ht(-K(U),yo) = 0. Hence, the inclusion map of pairs induces an
isomorphism
H*(Y/B,y0)nH*(Y/B,Tr(U)).
Let V = Ax [|, 1]. By the excision property, we have an isomorphism
H.(Y,U) t* H.(Y - V,U - V).
Since 5 is a strong deformation retract of U, it follows from the exact sequence
that
H*(Y, B) n H*(Y,U).

128
Thus, we have
Ht(Y,B)t*Ht(Y-V1U-V).
In a similar way the set x( V) may be excised from the pair (Y/B, t(U)) to
give an isomorphism
Ht(Y/B,y0) « Ht(Y/B,w(U)) « #*(y/5 - w(V),w(U) - w(V)).
Now the restriction of the map x gives a homeomorphism of pairs
tt:(Y-V,U-V)^ (Y/B - tt(V), tt(U) - tt(V)),
and so an isomorphism of their homology groups. All of these combine to give
an isomorphism
H*(Y/B,y0)&H*(Y,B).
Since Y admits obviously a strong deformation retraction onto X x {0}
which maps B onto A x {0}, we have
H.(Y,B)kiH*(X,A).
On the other hand, from the exact sequence of the pair (Y/B, yo), it follows
that
H*{Y/B,y0) « &*{Y/B) = H*(XUCA).
Therefore, Hn(X U CA) is isomorphic to Hn(X, A) for every n.
2313
For any topological space X let SX denote its (unreduced) suspension.
(SX is the quotient space (X x [0,1]/ ~, where ~ denotes the equivalence
relation generated by requiring that (x,t) ~ (y, s) if s = i = 0 or s = t = 1.)
If / : X —► y is a map, let S/ : SX —► Sy be the map of suspensions induced
by the map (x,t) -» (f(x),t) of X x [0,1].
(i) Prove that if / : X —► Y is a homotopy equivalence then so is S/.
(ii) Using only the Eilenberg-Steenrod axioms (Homotopy, Exactness,
Excision, Dimension) for a homology theory prove that 5j(EX) is naturally
isomorphic to If,_i(X). (Here Hi denotes reduced singular homology.)
(Indiana)

129
Solution.
(i) Denote by Ilx :^x/-» EX the quotient map, and by po, Pi the
equivalence classes of (x, 0) and (x, 1), x £ X, respectively. Let / : X —► Y be
a continuous map. By the definition we see that
£/ = ny o(/ x M2)on^\
where id : J —► J is the identity map. Suppose that g : X —► Y is another
continuous map which is homotopic to / by a homotopy G : X x J —► Y
such that G(x,0) = f(x) and G(x, 1) = </(:c) for any a; £ X. Then define
G : (X x J) x I -> £Y by
G((x,*),s) = ny(G(jr,s),t).
Let If : £X x I —> £Y be a map denned by H(x,s) = G(U^l(x), s) for any
(x, s) £ £-^ x I, where, if x = po or pi, II^1(J) means any point (x, 0) or (x, 1).
It is easy to check that H is well-defined and continuous. It is also easy to see
that H(x, 0) = £/(£) and H(x, 1) = £</(£)• Therefore £/ is homotopic to £</,
and consequently it follows that if / : X —► Y is a homotopy equivalence then
so is £/.
(ii) Let
cr = {nx(x,*)|*>i}
and
V={Ux(x,t)\t< -}.
It is obvious that U and V are both open sets of £X and that U and V
are both contractible. Hence H{(U) = H{(V) = 0 for all i. It follows from
the homology sequence of the pair (£X, U) that Ifj(£X) ai Hi{EX,U). Let
W = {n;c(:c,i) | i > 2/3}. Then W C 17. By the excision property we have
Hi(EX, U) » ffi(£X -W,U-W) = Hi(V,U - W).
It is easy to see that U — W has the same homotopy type of X. On the other
hand, from the homolopy sequence of the pair [V, U — W) we see that
Hi(V, U-W)fn &i-i(U - W).
Thus we conclude that ^,(£X) « ^_i(X).

130
2314
Consider the following commutative diagram of abelian groups in which
the row and column are exact sequences.
At
j-3
Suppose that 7 and /¾ are surjective. Prove that 03 is injective.
(Indiana)
Solution.
Let 03 € A3 such that 03(03) = 0. Then from the exactness there is an
a.2 £ -4.2 such that 02(02) = 03. Since 7 is surjective, there is an ai £ ij such
that 7(01) = /^2(02)- Hence by the commutativity we have
7(01) = /32(ai(ai)) = /32(a2),
so oi(ai) — 02 € ker/32, and there is a 61 £ B\ such that /3i(&i) = a(a\) — a2.
Since /¾ is surjective, there is a 60 € Bq such that /3o(&o) = &i- Once again by
the exactness we have 0 = /3i/3o(&o) = /3i(&i), which means that oi(ai) — a.2 =
0. Hence 03 = 02(02) = 0201(0.1) = 0. It means that kero3 = 0, i.e., 03 is
injective.
2315
Suppose that a topological space X is expressed as the union U U V of two
open path-connected subspaces such that U(~)V is path connected, H\(U) = 0,
and the inclusion UC\V —* V induces asurjective homomorphism Hi(U(~)V) —►
Hi(V). (Here H denotes singular homology.)
a) Prove that H^X) = 0.

131
b) Give an example to show that the analogous assertion becomes false in
general if Hi is replaced everywhere by H2.
(Indiana)
Solution.
a) Consider the Mayer-Vietoris sequence of the pair (U, V)
► H2(X) ^H^UOV)^ Hi(U) ® JTi(V) ^ H^X) ± H0(U n V).
Since U <1 V is path-connected, H0(U (1 V) = 0. It means Hi(X) = inn/)*. By
the hypothesis, <j>* is surjective so that
keri/>, = im<^ = H^U) 0 J2"i(V).
Therefore inn/)* = 0, i.e., H\(X) ~ 0.
b) Let X = S2, the unit 2-sphere,
and
F={(i,y^)e52U< ^}.
Then X = U U V. It is obvious that {7, V and U (1 V are path-connected
and that H2(U) = H2(V) = 0. So the homomorphism H2(U (IV) ^ H2(V)
induced by the inclusion map is surjective. But H2(X) &Z.
2316
Let
Dn = {x £ Rn | \x\ < 1}
denote the standard unit ball in Euclidean space Rn, and
Sn~1 = {x£Rn | |a:| = l}
denote the standard (n — l)-sphere. Suppose that / : Dn —> D" is a continuous
map such that the restriction of / to Sn~x is a homeomorphism from Sn~l to
Sn_1. Prove that / is surjective.
(Indiana)
Solution.
We use the reduction to absurdity. Suppose that there is a point x0 € D"
such that x0 ¢ f(Dn). By the assumption, we see that xo € D" — S™-1.

132
Therefore Sn~l is a deformation retract of Dn — {xq}. Let r : Dn — {xo} —►
S11"1 be a retraction. Then g = r o / is a continuous map from D™ to Sn_1
and </|s"-i •' S™-1 —► S™-1 is just the restriction of / to S™-1, which is
a homeomorphism. Let i : Sn~l —► Dn be the inclusion map. Therefore
g o i : Sn~l —► Sn~l is a homeomorphism. Hence
(</ o i), : Hn^S"'1) -» fTn.^S—1)
is an isomorphism. But (</ o i)„ = </„ o i„ and i* : Hn-i(Sn~1) —► If„_i(.Dn) is
obviously a trivial homomorphism because of Hn-i{Dn) = 0, and consequently
(g o t)„ is trivial. This is a contradiction.
2317
Let X be a connected CW complex with two 0-cells, three 1-cells, three
2-cells, and no higher-dimensional cells. Assume H\{X) « Z (&Z/Z. Compute
the Euler characteristic of X and determine (with proof) all possibilities for
H2{X).
[Indiana)
Solution.
The Euler characteristic of X, X{X) = 2 — 3 + 3 = 2. On the other hand,
X(X) = ra.nkH0{X) - rantff^X) + rank.ff2(X).
Therefore rank.H^-X') = 2. Let Xk denote the fe-skeleton of X. By the
assumption, we see that H3(X3,X2) = 0, and H2(X2,X1) =Z®Z®Z. Therefore
H2(X) « ker{d* : H^X^X1) -» H^X^X0)},
where d* is the composition of homomorphisms
H2{X2,XX) h H^X1) h H^X^X0).
Thus because H2{X2 ,Xl) is free abelian, H2(X) is also a free abelian group
of rank 2, i.e., H2(X) &Z&Z.

133
2318
Let X be a CW complex with exactly one fe + 1-cell. Prove that if Hk(X;Z)
is a nontrivial finite group then Hk+i(X;Z) = 0.
(Indiana)
Solution.
Denote by Xk the fe-skeleton of the CW complex X, and by dk the
composition of homomorphisms of pairs
Hk{Xh,Xh-1) h H^X*-1) h fffc.!^*"1, JTfc-2).
It is well-known that
Hk(X;Z)&kerdk/imdk+1
and
Hk+1(X;Z) ss keidk+1/imdk+2.
By the hypothesis,
Hk+1(Xk+1,Xk)^Z.
We choose a generator a in Hk+i(Xk+1,Xk). We claim that dk+1(a) ^ 0.
Otherwise we would have imdk+1 = 0. Therefore
Hk(X;Z)Kkeidk c Hk(Xk,Xk~1).
But it is well-known that Hk(Xk,Xk~1) is a free abelian group, and,
consequently, kerdfc is trivial or free abelian. It means that Hk(X;Z) is not a non-
trivial finite group and contradicts the hypothesis. Thus we have dk+1(a) ^ 0.
It follows that kerdfc+1 = 0 and, therefore, Hk+i(X;Z) = 0.
2319
Let
A = {(3:1,3:2,2:3,2:4) G -K4 I xi = 0}
and
4
B = {(3:1,3:2,3:3,3:4) G R I 3:4 = 0}

134
Let X = AU B. Compute the relative homology groups Hi(X,X — (0,0,0,0))
for all i.
[Indiana)
Solution.
Let U be the unit open ball centered at the point (0,0,0,0) in R4. Then
U = U fl X is an open neighborhood of (0,0,0,0) in X. By the excision
property we have
fri(A-,A--(o,o,o,o))«fri(cr,cr-(o,o,o,o))
for all i. It is obvious that U is contractible and that U — (0,0,0,0) has a
deformation retract CU D, where
C = {(xi,x2, x3, x4) £ R4 | xi = 0, x\ + x\ + x\ = 1}
and
D = {(xi, X2, X3, X4) £ R4 I X4 = 0, x\ + x\ + x\ = 1}.
(See Fig.2.15)
Fig.2.15
Let W! = CUD - {pi,p2} and W2 = C U D - {p3,P4}- (See Fig.2.15.)
Then Wi and W2 are open subsets of C U D and Wi U W2 = C U D. It is clear
that both C and D are homeomorphic to the unit sphere S2 and that D and
C are deformation retracts of Wi and W2 respectively. It is also clear that S1
is an deformation retract of Wi <1 W2. Therefore, applying the Mayer-Vietoris
sequence to the pair (Wi, W2), we see that
{Z®Z®Z, i = 2,
Z, i = 0,
0, otherwise.
From the homology sequence of the pair (U, U — (0, 0,0,0)), we see that
Hi+1(U, ¢/-(0,0,0,0))« ff<(t/-(0,0,0,0))

135
for all i > 0. Hence we conclude that
{Z®Z®Z, i = 3,
Z, i = 0,1
0, otherwise.
2320
Let X be a path connected and locally path connected space and let x £ X.
Let Y = S x S1 x S2. Show that if iri(X,x) is finite, then any continuous
map / : X —► Y induces the trivial map /* : Hi(X,Z) —► Hi(Y,Z) for all i
different from 0 and 2.
[Indiana)
Solution.
It is easy to see that R2 x S2 is the universal covering space of Y. By the
Kiinneth theorem, we have
Hi(R2 xS2) = Yl HJ(R2) ® ^(S2) = { o' *
j+k=i
It is clear that
0,2,
otherwise.
miS1 xS1 x S1) = miT2) ® miS2) = Z ®Z.
Since tti(X,x) is finite, the homomorphism /„ : iri(X, x) —► xi(y, f(x)) must
be trivial. Therefore, we may lift / to a map / : X —► iJ2 x S2 such that
7r o / = /, where x : iJ2 x S2 —► Y is the covering map. Hence we have
/. = x. o /* : ^(X,^) -> Hi{Y,Z)
for any i. But it is obvious that
7 = Ht(X,Z) -> #,(722 x S2,Z)
is trivial for i different from 0 and 2, so is
f* : Hi(X,Z) ^ Hi(Y,Z).

136
2321
Call a commutative diagram of abelian groups
A A B
/31 17
C -» D
6
"exact" if the sequence of groups and homomorphisms
0 _> A (aJ] B $ C 7-^* D -» 0 is exact.
It is an interesting fact that if
/31 17
C -» D
and
7l le
D -► F
«'
are exact, then so is
A a^ E
/31 ie
C -» 2?
Prove exactness of
(7n<2tana)
Solution.
By the assumptions, we have Sf3 = ya and ea' = S'y. Therefore, we have
(«5'«5 - e){/3,a'a) = S'S/3 - ea'a = 6'ja - 6'<ya = 0.

137
It follows that im(/3, a'a) c ker(<5'<5 - e). Let (c,e) G ker(<5'<5 - e), i.e., 6'6(c) -
e(e) — 0. By the assumptions, we have im(a,/3) = ker(7 - 6) and im(a',7) =
ker(e — 6'). Thus, noting that
(e,%))Gker(£-<5'),
we see that there exists a b G B such that a'(b) = e and y(b) = 6(c). Hence we
have (b, c) G ker(7 — 6). Therefore, there exists ana£i such that a(a) = b
and /3(a) = c, and, consequently,
(c,e) = (f3,a'a)(a) G im(/3, a'a).
It means that
ker((5'(5 - e) C im(/3, a'a).
The exactness at C ® E is proved.
2322
Let X be a non-empty compact HausdorfF space and / : X —► X be a
continuous map. Prove that there exists a non-empty closed subset A of X
such that /(-4) = A. Give an example to show that compactness is essential
for this assertion.
(Indiana)
Solution.
Let F\ = f(X). Then Fi is a non-empty closed subset of X. We define
Fn+i = f(Fn) inductively for n G 2+. It is clear that {Fn,n E Z+} is a
sequence of non-empty closed subsets in X and that
F1DF2D---DFnD Fn+1 0---.
oo
Since X is compact, we see that the subset A = f] F„ is a non-empty closed
n = l
subset of X. We claim that /(-4) = .4 holds.
We give an example to show that compactness is essential for this assertion.
Let X = (0,1] and / : X —► X is denned by f(x) = \x for x G X. Suppose
that a non-empty closed A of X satisfies /(-4) = A. Then there would exist an
x0 G -4 such that a; < a;0 for any x G -4. In fact a;0 = sup a;. Since /(-4) = .4
and Xo G -4, there would exists an xi G -4 such that /(a;i) = £0, i-e., £0 = §£1.
Therefore Xq > xi = 2xq, which is a contradiction.

138
2323
Recall that
H3(S3;Z)&H3(RP3;Z)nZ.
(RPn is n-dimensional real projective space.) Prove that there is no function
/ : S3 —► RP3 inducing an isomorphism on the third homology.
(Indiana)
Solution.
It is well-known that S3 is the 2-fold universal covering space of RP3.
Let x : S3 —► -RP3 denote the universal covering map. Then it is clear that
the degree of x is equal to 2. Let / be a function from S3 to RP3. Since
xi(S3) = {0}, there is a lifting of /, /: S3 -» 53 such that x/= /. Denote
by a and 6 the generators of Hs(S3) and Hz(RP3) respectively. Then we have
/* (a) = x* (/* (a)) = deg / • x* (a) = 2 • deg / • 6,
because deg/ E.2T, it is obvious that /„ is not an isomorphism.
2324
Let
X = {(x,y,z)\xy = 0).
(a) Compute HX(X - (0,0,0)).
(b) Using part a, show that X is not homeomorphic to R2.
(c) Prove or disprove: X is homotopy equivalent to R2.
(Indiana)
Solution.
(a) In fact, X = xi U X2 where xi is the plane y = 0 and X2 is the plane
a; = 0. Denote by A and 5 the unit circle in xx and x2 respectively. Then
it is easy to see that AU B is a deformation retract of X — (0,0,0). Take
U = AUB-(0,0,1) and V = ,40 5-(0,0,-1). Therefore, U and V are both
contractible, U LiV = AU B and U (IV has four path components which are
all contractible. Applying the Mayer-Vietoris sequence to the pair (U, V), we
see that
fli(j:-(0,0,0))!sfli(4UB) =z®z®z.
(b) Suppose that there exists a homeomorphism f : X —* R2. Then
f\x-(o,o,o) '■ X — (0,0,0) —► R2 — /(0,0,0) is also a homeomorphism, which

139
would induce an isomorphism from H1(X - (0,0,0)) to Hi(R2 - /(0,0,0)).
But it is clear that
Hi (ii2-/(0,0,0))« Hi(5^=^.
It is a contradiction.
(c) Let F : X x I —* X be & map defined as follows.
F((x vz)t)-l (:C'y'Z)' ifa; = 0'
*((s,y,z),i)-| {tXiytZ)t if x#0.
Then F is a deformation retraction of X onto the plane X2. Hence X is
homotopy equivalent to R2.
2325
Let (B",5n_1) be the standard ball and sphere pair in Rn, n > 1. Suppose
that / : (jB™,^™"1) —► (Jl,j4) is a continuous map and /Is"-1 : S™"1 —► ^4 is a
homeomorphism. Show that if Hn(X) = 0 then H„(X,j4) =Z.
(Indiana)
Solution.
Consider the following diagram
Hn(X)±Hn(X,A) ^ Hn-!(A) ± Hn-!(X)
U* # T/i* # TA
in which the level rows are exact and the squares are commutative. By the
assumption, /i* is an isomorphism, and therefore
Hn.1(A)^Hn.1(Sn-1)=Z.
It is well-known that 5J, is an isomorphism and Hn-\(Bn) = 0. Thus we have
keri* = H„-i(A) —1.
Since Hn(X) ~ 0, <9* is a monomorphism. Hence we have
Ifn (X, ^4) rj imd* « ker i* = JT.

140
2326
(a) Describe a CW structure on S2 x S5 and use it to compute the homology
of S2 xS5.
(b) Compute the homology of S2 x S5 with 2 points removed.
(Indiana)
Solution.
(a) Let x0 £ S2, xi £ S5. Then S2 is obtained by attaching a 2-cell to
xq, and S5 is obtained by attaching a 5-cell to x\. Denote by S2 V S5 the one
point union of S2 and S5, which can be considered as the space obtained by
attaching a 2-cell and a 5-cell to the point (xq,xi) G S2 x S5. It is easy to see
that S2 x S5 is homeomorphic to the space obtained by attaching a 7-cell to
S2VS5. Therefore the CW structure on S2 x S5 has a 0-cell, a 2-cell, a 5-cell
and a 7-cell. Hence it is obvious that Hi(S2 x S5) is an infinite cyclic group
for i equal to 0, 2, 5, and 7, and is zero otherwise.
(b) Let X = S2 x S5 - {pi,P2>, C = S2 x 55 - {pi} and F = 52x55- {p2}-
Then Z7, V are both open sets of S2 x 55 and UDV = X, and UUV = S2 x 55.
It is easy to see that S2VS5 is a deformation retract of both U and V. Applying
the Mayer-Vietoris sequence to the pair (U, V) and using the fact that
Hi(S2 V S5) » 5,(52) ® 5i(55),
we conclude that
r Z, i = 0,2,5,6,
y 0, otherwise.
2327
Let Abe a subspace of S2 x S2 homeomorphic to the 2-sphere S2. State
what the homology groups Hq(S2 x S2) are (no proof is required). What can
you say about the possibilities for the relative homology groups Hq(S2 x S2, A)?
(Indiana)
Solution.
By the Kiinneth formula we conclude that
( Z, 5 = 0,4,
Hq(S2 xS2)= < Z®Z, ¢ = 2,
{ 0, otherwise.

141
From the homology sequence of the pair (S2 x S2,A), using the above result
and the fact that Hi{A) « Hi(S2),>yre see that
Hi(S2 x S2, A) & Hi(S2 x S2)
for i > 4 and that the nontrivial parts of the sequence are
0 -» H3(S2 x S2, A) h H2(A) -^ H2(S2 x 52) ^ # 2(S2 x S2, A) -» 0
and
0 -» # i(S2 x S2, il) * J5f0(i4) -» H0(S2 x 52) -» #0(S2 x S2, A) -» 0.
Since 2?o(-A) = 0, .ffi(S2 x S2, A) = 0 for i equal to 0 and 1. By the exactness,
we have #3(^2 x S2,A) « keri* and ^2(^2 x S2,A) k,1 ®Z/\mit, where u
is the homomorphism induced by the inclusion map i : A —► S2 x 52.
2328
Compute the homology groups of the space X obtained as the union of the
2-sphere S2 and the z-axis in R3.
(Indiana)
Solution.
Let
X* = X — {the open interval (—1,1) in the z-axis}.
Then the space X may be viewed as a space obtained from X* by attaching
a 1-cell. It is clear that H{(X, X*) is infinite cyclic for i equal to 1 and is zero
otherwise. It is easy to see that X* has the homotopy type of S2. Thus from
the homology sequence of the pair (X, X*) we conclude that Hi(X) is infinite
cyclic for i equal to 0,1,2 and is zero otherwise.
2329
Define the "unreduced suspension" SX of a space X to be the quotient
space of I x X obtained by identifying {0} x X to one point and {1} x X
to one point. (This is the union of two "cones" on X.) Show that there is a
natural isomorphism ax ■ H{(X) —» #,-+1 (£X), for all i > 0.
(Indiana)

142
Solution.
Let {0}'x X and {1} x X be identified to the points xq and xi respectively.
Take U — EX — {x0} and V = EX - {a;i}. Then U and V are open sets
of EX, and U U V = EX. Consider the Mayer-Victories sequence of the pair
(u,vy.
► Hi+1(U f)V)^ Hi+i(U) ® Hi+i(V) ^ Hi+i(2X) -½ Ht(U n V) -»• • •.
It is clear that U and V are contractible and that UC\ V has the homotopy type
of X. Thus the homomorphism A : if,+i(SX) —► Hi(UCiV) is an isomorphism
and we may obtain an isomorphism A* : 5,+i(EX) —► Hi(X). The inverse of
A*, ax : Hi(X) —► .ffj+^SX) is just what we are looking for. The naturality
of crx can be derived from the naturality of the Mayer-Vietories sequence.
2330
Denote by X the union of the torus S1 x S1 with the disc D2, where D2
is attached to T2 by identifying 3D2 with a meridian curve S1 x {xo} in the
torus, where xo £51. (See below.)
Fig.2.16
(a) Calculate Hn(X) for all n > 0.
(b) Is T2 a retract of X? Why or why not?
{Indiana)
Solution.
(a) Consider the homology sequence of the pair (X,T2). It is well-known
that Hi(X,T2) = 0 for i± 2 and H2(X,T2) » U{H2{D2,dD2)) » Z, where
/* is the homomorphism induced by the adjunction map / : £)2 —► X. Thus
the nontrivial part of the sequence is as follows.
0^H2(T2)±H2(X)±H2(X,T2) 5 fi'1(T2)^fi'1(X)^0
~T /* T A*
0^#2(D2,dD2) ^ #i(d£>2)^0

143
It is easy to see that im<9* = im/i* and ker d't = ker/i*. We know that
H1(T2) = Z ®Z is generated by a and b. (See Fig.2.16) It is clear that
/i*(.ffi(<9D2)) = Z ® {0}. Thus we see that #i(X) k, Z ® Z/Z & Z im&
H2{X) « H2(T2)&Z. Hence we conclude that
ff ry> _ J 0, n > 3,
^"W-\ Z, n = 0,l,2.
(b) We claim that T2 is not a retract of X. Otherwise, there would exist
a retraction map r : X —► T2 such that r|T2 = ic£T2. Let i : T2 —► X be
the inclusion map. Then rot = idj2 : T2 —► T2, and, consequently, we
have r» ojt = id : H\(T2) —► Hi(T2). In other words, we have the following
commutative diagram
zoz —z — zoz
because of Hi(T2) = JT ®Z and Hi(X) =Z. This is a contradiction.
2331
Given homomorphism h : A —► B and g : C —> B, the pull back of h via </
is the group
g*(A) = {(c,a)£CxA\g(c) = h(a)}.
(a) Let g*(h) : g*(A) —► C be the homomorphism obtained by restricting
the projection onto the first factor C x A —► C to </*(.A). Prove that the kernel
of g* (A) is isomorphic to the kernel of h.
9* (A) A
l9*(h) [h
C -» B
9
(b) Let ^4 = Z/4Z, B = .2T/2Z, and let h : A —► B be the surjection denned
by h(l) = 1. Let C = Z/8Z and let g : C —► B be the surjection defined by
</(l) = 1. Identify the group g*(A), with explanation.
(Indiana)

144
Solution.
(a) Suppose that (c,a) £ ker g*(h). Then g*(h)(c,a) = lc, and, therefore,
C = lc, the identity of C. By the definition of g*(A), we see that
keig*(h) = {(lc,a) | a G ker h}.
Let F : keig*(h) —► ker h be defined by F(lc,a) = a. It is easy to see that F
is an isomorphism.
(b) In this case, we have /i-1(0) = {0,2}, h_1(l) = {1,3}, 0_1(O) =
{0,2,4,6} and g~l(l) = {1,3,5,7}. Thus it is clear that the group g*(A)
consists of the following 16 elements: (0,0), (0,2), (2,0), (2,2), (4,0), (4,2),
(6,0), (6,2), (1,1), (1,3), (3,1), (3,3), (5,1), (5,3), (7,1), (7,3).
2332
Recall that if C is a homeomorphic copy of the circle in S3, then Hi(S3 — C)
is infinite cyclic for i equal to 0 or 1 and is zero otherwise. Assuming this fact
compute
(a) The homology of R3 — C, when C is a homeomorphic copy of the circle
in R3.
(b) The homology of Y = R3 — X, where X C -R3 is a homeomorphic copy
of the "figure-eight space" (i.e., the one-point union of two circles.)
{Indiana)
Solution.
(a) Denote S3 = R3 U {oo}. Let A = S3 - C, B = S3 - {oo}. Then A and
B are open subsets in S3, and A U B = S3 and A (1 B = R3 — C. We have the
following Mayer-Vietoris sequence.
> Hi+1(S3) ^ Hi(R3 -C)t Hi(S3 - C) $ Hi{S3 - {oo})
*; £,(^)-^-1(^-(7)--..
Noting that
for any i and that
and
Hi(S3-{oo}) = Hi(R3)=Q
H(S3) = { 2' * = 3,
'^ ' \ 0, otherwise,
HCi3 — C) = < % =
'^ \ 0, otherwise,

145
we can see that
'^ ' | 0, otherwise,
(b)
Fig.2.17
Represent X as shown in Fig.2.17. Let U — X — {p\} and V = X — {p2}.
Then we see that U and V have homotopy type of Sl and that U <1 V is
contractible. Since U U V — X, we have
Y = R3-X = (R3-U)n(R3- V)
and
(R3 -U)U (R3 -V) = R3 -(Un V).
From the result of (a), we see that Hi(R3-U) and JT,-(.R3-V) is infinite cyclic
for i equal to 1 or 2 and is zero otherwise. It is easy to see that
Hi(R3 -(Uf) V)) » Hi(S2).
Due to the above facts, the nontrivial part of the Mayer-Vietoris sequence
of the pair (R3 -U,R3- V) is
0 -» H2(Y) h H2(R3 -U)® H2(R3 - V) ^ ^2(^3 -U(1V)
k,X®Z &Z
A H1(Y)^H1(R3-U)®H1(R3-V)^0.
K.Z ®Z
We claim that the homomorphism rj>* is an epimorphism. Take a sufficiently
large r > 0 such that the sphere S2(r) belongs to R3 — X. Then S2(r) is a
deformation retract of R3 — U <1 V. Hence we can consider the generator of
H2(S2(r)), [c], as the generator of H2(R3 — U C\ V). Since the representative
chain c of [c] can also be considered as a chain of (R3 — U) (1 (R3 — V), by
the definition of if>* we have [c] = ip*([c], 0). The claim is proved, which means
imA = 0 too. Thus we get Hi(Y) —Z ®Z and a split exact sequence
Q^H2(Y)tz®Z^Z -»0.

146
It follows that H2(Y) — Z. So we conclude that
{0, i > 3,
Z®Z, \Zl]
Z, i = 0.
2333
It is known that if X C S3 is homeomorphic to S1 then H*(S3 — X) «
If* (S1). Use this fact to compute the homology of S3 — Y where Y is a subspace
of S3 homeomorphic to the disjoint union of two copies of S1.
(Indiana)
Solution.
Denote Y by Y = A U B, where A (1 B = 0 and both A and B are
homeomorphic to S1. Therefore,
S3-Y = (S3-A)n(S3-B).
Noting S3 — A and S3 — B are open sets of S3 and
(S3 -A)U (S3 -B) = S3,
in the Mayer-Vietoris sequence we have
► Hq+1(S3) ± Hq(S3 -Y)*S Hq(S3 -A)® Hq(S3 - B)
hHq(S3)^Hq^(S3-Y)t...
Using the fact that
Hq(S3 -A)k Hq(S3 - B) « Hq{Sl) = | *'
q = 3,
«#3,
¢ = 1,
«#1,
and that
we can easily see that
#g(S3 - Y) = {
( 0, <z>3,
Z, ¢ = 2,
jre.2r, ¢ = 1,
z, q = o.

147
2334
(a) Sketch pictures of the universal covering of the one point union S1 V S2
and of the connected 2-fold covering (no proofs required).
(b) Compute the homology of the connected 2-fold covering space of S^'VS2.
(Indiana)
Solution.
(a)
S5 S5 s* S'v.S2
P> Vi vi ()s'
Fig.2.18
The universal covering of the one point union S^\IS2 is shown as in Fig.2.18.
The covering map x, restricted on each S2, is the identity map, and, restricted
on R, is the exponential map: R —► 51. The connected 2-fold covering of
S1 V S2 is shown as follows.
Fig.2.19
The covering map x, restricted on each S2, is the identity map, and,
restricted on S1, is the 2-fold covering map: z —► z2 from S1 to S1.
b) To compute the homology of the connected 2-fold covering space of
S1 V S2, i.e., the homology of S2 V S1 V S2, we take
U = S2V 51 V S2 - {the antipodal point of px}
and
V = S2 V S1 V S2 - {the antipodal point of p2}
(see the above figure). Then it is easy to see that S1 is a deformation retract
of U fl V, and that 51 V S2 is a deformation retract of C/ and V.
Thus, it is clear that

148
and
Hq(U) « Hq(V) « Hq(S2) ® 5,(51) = { *' J = ^ 2;
Therefore, the nontrivial part of the Mayer-Vietoris sequence is
h H1(U)®H1(V)^H1(S2VS1VS2)^0.
It is easy to see that the inclusion maps k : Uf)V —► 17 and /: 17 DV —► V induce
injective homomorphisms fe* : H\{U C\ V) —► Hi(U) and /* : Hi(U <1 V) —►
■ffi(^)) respectively. Hence the homomorphism
^: i?i(t/ n v) -»#i(tf) ® ffi(v)
is injective, i.e., ker^>* = 0. Thus imA = 0. It means that
H2(S2 V S1 V 52) » ker A » iim/>, » If2(*7) $ J5f2(V) =^r©-2r.
It is clear that
Hr(S2 V Sl V 52) » Hi(CT) ® Hi(V)/im^« »Z.
Hence, we have
{Z®Z, ¢ = 2,
Z, ¢ = 1,
0, otherwise.
2335
Compute the homology of S1 x 51-point.
(Indiana)
Solution.
Fig.2.20
It is easy to see that the space X = A U B is a deformation retract of the
space H = S1 x 51-point, where A and 5 are each homeomorphic to S1 and

149
A n B = {xo} as shown in Fig.2.20 Choose points a £ A and b £ B such
that a ^ xo and 6 ^ xq. Let U = X — {b}, and let V = X — {a}. It is
clear that A and 5 are deformation retracts of U and V, respectively, and that
U C\V — X — {a, b) is contractible. Applying Mayer-Vietoris sequence to U
and V, we have
r 0, n> 2,
fl^S1 x 51 - point) = Hn(X) = Hn{A) $ #„(5) =jze2, n = 1,
[ Z, n = 0.
2336
Suppose the following diagram is commutative, the rows are exact, and -y„
is an isomorphism for all n.
. 4 !i n £^ r £i 4 , _> ...
* sin ' -^n * ^n * -^-n—1 —'
i«n IAi J.7n i«n-l
i' }' «'
"~* An ^ -°n -1 ^n ^ An-1 "*
Construct an exact sequence
• " ' —► -An * An © -On —► -On —► A.n — l *
Write out the proof of exactness at B'n.
(Indiana)
Solution.
The following sequence is exact:
>A (a2i^A>ffiB ^ B' 6"°^oj-A ,_>...
' jt-ji * JT-n W -On ► -On * JT-n —1 ► •
where A is defined by
A(a',b) = i'n(a,)-f3n(b)
for any (a', b) £ A'n ® Bn. We give the proof of exactness at B'n as follows.
Let u £ kev(6n 07" * oj'n). Then 7,71 o j'n(u) £ kei6n. Due to the exactness
at Cn, there exists a, b £ Bn such that jn(&) = 7,71 °in('u)) and consequently,
7n • jn(b) = j'n(b). From the commutativity, we have jn o(3n(b) = i{,(u). Thus
/3„(6) — u£ kerjn and there exists an a' £ j4{, such that i'n(a') = /3n(&) — «• It
means that
u = /3n(6)-^(a') = A(-a',-6).

150
So ker((5„ o -yn *• o j'n) c imA.
Now we prove
imA C ker((5„ o 7^1 o j^).
Suppose that u £ imA, i.e., u = i'n(a') — f3n(b) for some a' £ A'n and b £ Bn.
Since j^ 0 i'n = 0 and j^ 0 /3n = 7„ 0 j„, it is easy to see that
<5n ° In1 ° in(U) = -<*n ° Jn(&) = 0.
Thus the exactness at B'n is proved.
2337
Suppose that X is a space and f '. X —► Y, g : X —► Z are two maps of
X into contractible spaces Y and Z. Let M be the mapping cylinder of / and
g, that is, M is the identification space obtained from the disjoint union of Y,
X x I and Z by identifying each (a;,0) with f(x), and each (a;, 1) with g(x).
Prove that HqM = Hq-iX.
(Indiana)
Solution.
Let U = X x [0,3/4] = Y/ ~, where X x [0,3/4] 1 Y denotes the disjoint
union of X x [0,3/4] and Y, the equivalence relation ~ is determined by (a;, 0) ~
f(x). In the same way, let V = X x (|, 1] = Z/ ~', where the equivalence
relation ~' is determined by (a;, 1) ~ g(x). Then we have M = U U V and
£/■ n V = X x (1/2,3/4). Noting U, V and *7 n V are open sets of M, by
Mayer-Vietoris sequence, we have the following exact sequence:
► Hq(U f)V)^ Hq(U) ® Hq{V) -» H,(M) -^ irg-i(c/- n F)
-^H,-!^)®^!^)-*... (1)
Since X x {0} = Y/ ~ is homeomorphic to Y and is a deformation retract of
U, from the assumption that Y is contractible, we have Hq(U) = Hq(Y) = 0.
In the same way, Hq(V) = ffg(^) = 0.
It is obvious that U C\ V has the same homotopy type with X. So Hq(U D
V) ~ Hq{X). Thus, from (1), Hq{M) = 5g_i(X).

Part III
Differential Geometry

153
SECTION 1
DIFFERENTIAL GEOMETRY OF CURVES
3101
Let a(s) be a closed plane curve. Define the diameter da of a(s) to be
da = sup ||a(s)-a(i)||.
t,l£M
Now assume that the curvature k(s) > 1 for all s.
i) For any N £ Z+ sketch an example of such an a(s) with da > N.
ii) Assume further that a is a simple closed curve. Prove that da < 2 (or
some other constant independent of a; 2 is the best possible such constant).
(Indiana)
Solution.
i) The following is an example of such an a(s) with k(s) > 1 and da > N.
Fig.3.1
ii) From the hypothesis that k(s) > 1 for all s, we know that the simple
closed curve a is an oval.
For every oval, by Blaschke, if we take the origin O as shown in the figure
and denote by p(6) the distance from 0 to the tangent I at the point (x,y)
of the oval, where the oval is counterclockwise orientated and 6 denotes the
oriented angle from the z-axis to 1, then the oval can be parameterized by 6
as follows
f x(6) = p(0) sin 0+p'(6) cos 0,
\ 2,(0) = -p(0) cos 6 + p'(8) sin 6.

154
p(6) is called the support function of the oval. From this we can conclude that,
by direct computation, the relative curvature of the oval is kT(6) = (p(0) +
p"{6))-\
Fig.3.2
Now we can prove a more general result of Blaschke:
Let two ovals C and C\ in a plane be internally tangent at a point O.
Suppose that, at every pair of points P and Pi where C and Ci have the
same tangent orientation, the curvatures of C and C\ satisfy the inequality
fci(-Pi) < fc(-P)- Then the domain encircled by C\ must contain the domain
encircled by C.
Fig.3.3
In fact, take the tangent point O of C and C\ as the origin, and their
common tangent line as the z-axis. Let p{0) and P\{0) be the support functions
of C and C\, respectively. From the above, we can say that the support
function p(6) must be the solution to the following initial value problem of
ODE
f P"(e)+p(6) = 1^-v
\ p(0) = i/(0) = 0,
because p'{6) is exactly the distance from 0 to the normal line of C at point
(x(6),y(6)). Hence, p(6) can be uniquely determined by k(6) = kr(6) of C,
p{8)
f
Jo
i(0 - ¢)
kW
d<j>.

155
Analogously, for the curve Ci, we also have the similar expression of pi(6).
Thus,
»<"-*" = /^72 "■<•-'>*
By the hypothesis, we know that
k{<j>) - fci(^)
k(t)ki(t)
>0.
As for the sign of sin(0 — <f>), firstly, if 0 < 6 < ir, then owing to 0 < $ < 6,
we have sin(6> - <j>) > 0. Therefore, pi{0) - p{6) > 0. Secondly, if tt < 6 < 2tt,
we make the reflections of the oval C and C\ with respect to their common
tangent line at O and reverse the orientation of the z-axis. Then we can get
P\(8) — p(0) > 0, where 6 and its corresponding original 6 satisfy 6 + 6 = 2-k.
Hence, we always have P\{6) > p{6). Noticing that every oval is the envelope of
all its tangent lines, we see that the domain encircled by C must be contained
in the domain encircled by C\. The assertion of Blaschke is proved.
If we take a circle with radius 1 and centered at cc{sq) + N(so) as Ci, and
take a as C, then ii) follows immediately from the above assertion.
3102
Let a be a regular C°° curve in M3 with nonvanishing curvature. Suppose
the normal vector N(t) is proportional to the position vector; that is, N(t) =
c(t)a(t) for all t, where c is a smooth function. Determine all such curves.
(Indiana)
Solution.
For convenience, we assume that the parameter t is the arc length of the
curve a. Differentiating both sides of the equality N(t) = c(t)a(t) with respect
to t, we have, by the Frenet formula,
-k(t)T(t) - T(t)B(t) = c'(t)a(t) + c(t)T(t).
Thus we can immediately obtain k(t) = constant, r(t) = 0 and c(t) = —k.
Therefore, by the fundamental theorem of the theory of curves, we know that
the curve a must be a circle (or a part of it).

156
3103
A surface S C M3 is called triply ruled if at every p £ S we can find
three open line segments Li, L2, L3 lying in S such that L\ n £2 H £3 = {p}.
Determine all triply ruled surfaces.
Solution.
If 5 is a triply ruled surface, then, by the hypothesis, there are three
different asymptotic directions at every point of S. Observe that every asymptotic
direction (du,dv) satisfies
L(u, v)du2 + 2M(u, v)dudv + N(u, v)dv2 - 0,
where L(u, v), M(u, v), N(u, v) are the coefficients of the second fundamental
form of S at point p(u, v). Noticing that the above equation is of 2nd order
with respect to du : dv and it has two roots, we obtain L(u, v) = M(u, v) =
N(u,v) = 0 for all (u,v). In other words, every point of S is a planar point.
Therefore, S must be a plane (or a part of it).
3104
Let 7 : (a, b) —► M3 be smooth with |7'[ = 1 and curvature k and torsion
t, both nonvanishing. Denote the Frenet frame by {T, N, B}. Assume there
exists a unit vector a €. M3 with
T ■ a — constant = cos a.
a) Show that a circular helix is an example of such a curve
b) Show that N ■ a = 0.
c) Show that k/r = constant = ± tana.
Solution.
a) Let a circular helix be parameterized as follows
(Indiana)
-y(s) = (rcosws,rsin ws, hus),

157
where r, h, w = (r2 + h2) ? are all constants. Then it is easy to verify that s
is the arc length parameter. Hence
T(s) = (-rw s'w. u>s, ru; cos ws, haj).
If we take a = (0, 0,1), then T(s) • a = hu = constant.
b) Differentiating T-a = constant with respect to the arc length parameter
s, we obtain k(s)N(s) • a = 0, from which follows N(s) ■ a = 0 for all s (E (a, b).
c) By the property of b), we may assume that the constant vector
a = cos a • T(s) ± sin a ■ B(s).
Differentiating the above equality with respect to s, we have
0 = (cos a • k(s) ± sin a • t(s))N(s).
Thus, for all s g (a,b), k(s)/r(s) = ±tana.
3105
Show that if 7 is a geodesic on the cone z = \/x2 + y2, (x, y) € iR2\{0,0},
then 7 intersects itself at most a finite number of times.
(Indiana)
Solution.
If we cut the cone along a generator I and develop it into a plane, then the
cone becomes an infinite sector without the vertex, and the geodesic 7 becomes
a straight line on the developed infinite sector. Noticing that the central angle
of the sector is \/27r, an obtuse angle, and the image of 7 on the sector must
be one of the following three cases:
a generator,
a straight line never intersecting the generator I,
two rays which start from the generator 1,
then we can conclude that 7 never intersects itself.
3106
Let 7 : (a, b) —► M3 be a C°° curve parameterized by arc length, with
curvature and torsion k(s) and t(s). Assume k(s) ^ 0, t(s) ^ 0 for all
s € (a, 6), and let T and N denote the unit tangent and normal vectors to

158
7- The curve 7 is called a Bertrand curve if there exists a regular curve
7 : (a, b) —► ttt3 such that for each s g (a, 6) the normal lines of 7 and 7 at s
are equal. In this case, 7 is called the Bertrand mate of 7 and we can write
7~(s) = 7(3) + rN(s) for some r = r(s) g 1R. (Note that s might not be an arc
length parameter for 7.)
(a) Prove that r is constant.
(b) Prove that if 7 is a Bertrand curve (with r as above), then there exists
a constant C such that rk(s) + Ct(s) = 1 for all s g (a, b).
(c) Prove that if 7 has more than one Bertrand mate, then 7 is a circular
helix.
(Indiana)
Solution.
(a) Prom y(s) — -y(s) + r(s)N(s) it follows by differentiation that
— ds
T(s)— = (1- r(s)k(s))T(s) + r'(s)N(s) - r(s)r(s)S(s).
Taking inner products at both sides with N(s) = ±N(s), we have that r'(s) =
0, namely r = const.
(b) From (a), now we have
rj Q ff a
T(s) = -(1 - rk(s))T(s) - -rr(s)B(s).
If we denote
T(s) = a(s)T(s) + b(s)B(s),
then by differentiating with respect to s, we obtain
_ ds
k(s)N(s)— = a'(s)T(s) + (a(s)k(s) + b(s)T(s))N(s) + b'(s)B(s).
Hence, from N(s) = ±N(s) we know that a'(s) = b'(s) = 0, namely
ds ds
—(1 — rk(s)) = const, -^rr(s) = const.
dsy w/ ds v '
Therefore, there exists a constant C such that rk(s) + Ct(s) — 1 for all s g
(a, b).
(c) Suppose that 71 and 72 are the Bertrand mates of 7. Then, by (b),
there exist constants r\, r2, Ci, C2 such that for all s g (a,6),
f nfe(s) + Cit(s) = 1,
\ r2k(s) + C2t(s) = 1.

159
Because the non-zero constants rlt T2 are not equal, the above system of linear
algebraic equations has solution k(s) = const, t(s) = const. Hence 7 must be
a circular helix.
3107
Let x(s) be a curve in M3 parameterized by arc-length. Assume that t(s) ^
0 and fc'(s) ^ 0 for all s. Show that a necessary and sufficient condition for
x(s) to lie on a sphere is that
1 , 1 k'2(s)
- — ' — - • — constant.
k2(s) ' T2(S) k4(s)
(Indiana)
Solution.
Suppose that x(s) lies on a sphere centered at the origin. Then we may
assume that x(s) = a(s)T(s) + b(s)N(s) + c(s)B(s), where {T(s), N(s), B(s)}
is the Frenet frame field along x(s), and a(s), b(s), c(s) are suitable functions
to be ascertained later. Differentiating (x(s),x(s)) = R2 with respect to s, we
have (x(s),T(s)) — 0, from which it follows that a(s) ~ 0, Vs. Differentiating
(x(s),T(s)) = 0 with respect to s again, we obtain
l+k(s)(x(s),N(s)) = 0,
which means that b(s) = —jr?j- At last, still differentiating (x(s),N(s)) =
~kTT) wi*h respect to s, we obtain
<z(s),5(s)> = - k'(S)
namely,
Therefore,
<*) = -;
k2(s)T(S)'
fe'(s)
k2(s)r(s)
R2 = {x(s),x(s)) =
k2(s) t2{s) fe4(s) '
Conversely, differentiating
k(syH } k2(s)r(s)
*(*) + if^W + i5?£bW.

160
with respect to s, we have
X{S} + W)N^ + 1¾)
B(s)\ =
r(£)
k(s)
k2(s)T(s)
that vanishes identically because k'(s) ^ 0 and
1 1 k'2(sY
0
d
ds
= -2
Then
k2(s)
k'(s)
k3(s)
2k'(s)
k2(s)T(s)
x(s) +
T2(s) k4(s)
2 fc,(g) ( fc,(g)
fc2(s)r(s) Vfe2(s)r(8)
T(£) , /^ fc'(S)
1
■N{s)
k2(s)T(s)
k'(s)
B(s)
is a constant vector, denoted by m. Hence (x(s) — m, a;(s) — m)
namely, x(s) lies on a sphere centered at m.
N(s)
constant,
3108
Let M C St3 be the torus obtained by rotating the circle {(0, y, z) : (y —
2)2 + z2 = 1} around the z-axis, and let c{t) — (2cosi,2sin<, 1) ("top circle").
Is this curve a geodesic on Ml Explain without long computations.
[Indiana)
Solution.
Observe that the geodesic curvature of a curve on M can be computed as
follows
kg = ±fc(<)sin0(<),
where k(t) is the curvature of the curve, and 6{t) is the angle between the
normal of M and the principal normal of the curve at the point corresponding
to the parameter t.
Then, the curve c(t) is not a geodesic on M, because neither the top circle
is a straight line, nor its principal normal, which is orthogonal to the z-axis, is
parallel to the normal of M along c(t), which is parallel to the z-axis.

161
3109
Let 7 : [0,1] -► St3 be a C°° curve with |-y| = 1 and nonvanishing curvature.
Assume the torsion t = 0.
(a) Show that 7 lies in a plane.
(b) What happens if the curvature is allowed to vanish at a point?
(Indiana)
Solution.
(a) I7I = 1 implies that the curve is parameterized by arclength s. From
the Frenet formula we know that the binormal vector field of 7, denoted by B,
is constant. Thus, ^(7(3)!?) = 0, which means that j(s)B = constant, that
is, 7 lies in a plane.
(b) The vanishing curvature at point so means 7(^0) = 0, i.e., so is a
stationary point of 7's tangent vector field. If the point is the strict extreme
value point of the tangent vector field, then it is an inflection point of the curve
7-
3110
Let / : M —► M be positive and smooth. Let M be the surface in M3
obtained by rotating the graph {(x, f(x)) : x £ M} of / in the xz plane about
the x axis. Characterize in terms of / the set of x such that ±yh) is a principal
curvature of M at (x,0,f(x)).
Hint. Local coordinate computations are not necessary.
(Indiana)
Solution.
At P = (xp, 0, f(xp)), in the direction of the circle of latitude, the
corresponding normal curvature
k„ = kcos6 = ——-cos0(a;p),
f(xp)
where 0(xp) is the angle between the normal of M and the principal normal
of the circle of latitude at P. Since every circle of latitude on M is a line of
curvature, then the corresponding normal curvature kn is a principal curvature.
Thus, kn = ± *,l ) implies that cos 0(¾) = ±1, that is, the curve {(x,0,f(x)) :
x g M} has a tangent parallel to the x axis at P. Namely, f'(xp) = 0.

162
Besides, the meridian passing P is also a line of curvature. Its curvature
k at P is just the other principal curvature of M. Thus, the second possible
case is
, \f"M\ = 1
(l + /'2K))3/2 f(Xpy
Therefore, we conclude that the set of a; such that ±t^t is a principal curvature
of Mat (x,QJ(x)) is
3111
Let a(s) C M3 be a smooth curve parameterized by arclength. Assume
that the position vector a(s) is always a linear combination of the binormal
and normal vector B(s), N(s) of a(s). Show that a(s) does not pass through
oem3.
(Indiana)
Solution.
If the position vector a(s) is always a linear combination of the binormal
and normal vectors, then (T(s),a(s)) = 0. Integrating the obtained equality,
we have (a(s),a(s)} = const. Therefore, if a(s) passes through the origin, we
will get a(s) = 0, the trivial case.
3112
Let M2 C -ffi3 be the cylinder x2 + y2 = 1. Suppose the curve a(s) g M2
is parameterized by arclength.
i) If k(s) > 0 and t(s) = 0, show that a(s) is a closed curve. (Here fe, t
are curvature and torsion of a in St3.)
ii) If kg(s) = 1, show that a(s) is a closed curve (kg is the geodesic curvature
inM).
(Indiana)
Solution.
i) The hypothesis of torsion t(s) = 0 implies that the curve a is a plane
curve, whereas the hypothesis of curvature k(s) > 0 implies that the plane 7r

163
where the curve a lies does not parallel the generating line of the cylinder M.
Therefore, the curve acMfli must be closed.
ii) If one develops M into a plane 7r, then the corresponding plane curve
of a, denoted by a, has the same geodesic curvature kg(s) = 1. Thus the
curvature of a is k = \Jkj + k2 = 1 which means that it is a circle with radius
1 in 7r. So, a must be closed.
3113
Let T be a two dimensional distribution in M3 defined by
T(w) = span j-,- + ^j
i) Show that T is not involutive.
ii) Given a C°° curve a(s) e M2 = {(x,y,Q) : x,y € M} C M3 and
a(so), show that there exists a unique C°° curve /3(s) g iR3 such that /3'(s) g
?>(»),/?(so) = a(s0) and tt(/3(s)) = a(s), where 7r(a;, ;y,z) = (as,y,0).
iii) Show that if a(s) is a simple closed curve of length L bounding the
region 0. in M2 then /3(s0 + £) - /3(so) = (0,0, ±A) where A = area of £1.
(Indiana)
Solution.
i) That [&> % + *<&] = £ and { J^4 + *£> ^} Me linearly
^dependent shows that T is not involutive.
ii) Let a(s) = (x(s),y(s),0). By 7r(/3(s)) = a(s), we may assume that
/3(s) = (x(s),y(s),z(s)) where z(s) is unknown. Then /3'(s) (E ^/3(») implies
that
^) = a(s}l+b(s)[ly+x(s)l)
for suitable functions a(s) and 6(s), i.e.,
'&(*) dM M£\__ {a{s)Ms),b{s)x{s}}.
ds ' ds ds
Hence
«., = !*£>, ,,, = ^.
Thus the problem of finding /3( s) reduces to solving the following initial value
problem

164
When the given plane curve a(s) is C°°, the unknown function z(s) can be
uniquely determined by
dy{s)ds
Z(s) = £ ,(,) dg
and so is the space curve (3(s), i.e.,
/3(5)= (x(s),y(s), J' x(s)^-ds
iii) If a(s) is simple and closed, then we easily have
(3(s0 + L) -/3(s0) = (x(s0 + L)-x(s0),y(s0 + L) - y(s0), j> xdy\
= (0,0, ±A),
where the sign is determined by the orientation of the curve a.
3114
Call a normal vector field v along a space curve 7 parallel if v is always
tangent to 7.
i) Show that the angle through which a parallel normal vector field v turns
relative to the principal normal N along 7 is given by the total torsion of 7,
i
L
r(s)ds.
/0
Here s and L denote arclength and the length of 7, respectively.
ii) Show that the total torsion of any closed curve 7 which lies on a sphere
in M3 must vanish.
(Indiana)
Solution.
i) The hypothesis that v is always tangent to 7 means that (v, i/) — 0, and
hence, the normal vector field v has constant norm. Therefore, without loss of
generality, we may assume that \v\ = 1, and v(s) = cos#(s) • N(s) + sin#(s) •
S(s), where 6(s) is a smooth function globally defined on 7, which measures
the angle between v(s) and N(s). Thus we have
v = (-sine ■ N+ cos6 ■ B)6+cos6-(-kT-TB)+sinO-tN
= - cos e ■ kT - (e - r) sin 6» • N + (6 - r) cos 6 ■ B.

165
Noting that v is parallel to T, we see that the above equality implies that
8(s) = t(s), and hence
0(L) - 0(0) - f T(s)ds.
Jo
ii) For any closed curve 7 which lies on a sphere in M3, the unit normal
vector field v of the sphere along 7 satisfies the above mentioned hypothesis.
Therefore, we have the relation
6(s)= J T(s)ds + 6(Q).
Jo
On the other hand, the smooth function 0(s), s g [0,i] can be regarded as a
lift of a certain differentiable map / : [0, L] —► S1 into M1. The fact that 7 is
closed means that
I r(s)ds = 0(L) - 0(0) = 2n7r,
Jo
where the integer n is just the degree of the map /, i.e., n = deg /. Let po £ 7
and 7 contract to po smoothly on S2. Thus we get a family of curves {7^},
t £ [0,1]. Furthermore, we may assume that for every t g [0,1), 7t overlaps
with 7 = 70 about Po = 7i- Also, for every 7^, we have the corresponding map
ft such that /0 = /, /1 = the constant map into po- Because the degree of a
map is homotopically invariant, finally we have
L
r(s)ds = 2-k • deg /1 = 0.
3115
Let M be a surface in St3 and let P be a plane. Suppose M and P intersect
orthogonally. Show that the intersection curve (parameterized by arclength)
is a geodesic on M.
(Indiana)
Solution.
Let k(s) be the curvature of the intersection curve C = M n P. If k = 0,
then C is naturally a geodesic on M. Otherwise, along the segment where
k(s) ^ 0, the normal of M is parallel to the principal normal of C, hence the
segment is also a geodesic one.
L

166
3116
Let a(s) be a C2 curve in M3 parameterized by arclength. Suppose that
for some function f(s), a"(s) = f(s)a(s). What can you deduce about f(s),
a(s)?
(Indiana)
Solution.
From
f(s)a(s)a'(s) = \d{a\s)f = 0
we have
f(s)d(a(s))2 = 0.
If f(s) = 0, then a"(s) = 0, i.e., a(s) is a straight line.
If /(s) ^ 0, then (a(s))2 = const, i.e., a(s) is a spherical curve.
Furthermore, differentiating
a"(s) = k(s)N(s) = f(s)a(s)
with respect to s, we have
k'(s)N(s) + k(s)(-k{s)T(s) - t(s)B(s)) = f'(s)a(s) + f(s)T(s)
which implies that
f(s) = -k2(s), t(s) = 0, *'(*) = 0.
Hence k(s) = const, and a(s) is a circle or part of it with radius I—.
3117
Suppose 7(f) parameterizes a space curve with curvature function «;(<).
Define a new curve 7 by setting, for each t € St, j(t) := cy(t/c), where c (E lit
is an arbitrary fixed constant. Derive the curvature function for this new curve.
(Indiana)
Solution.
The hypothesis is
W(t) x 7"(*)l
"(*)
if+\\3
|7'(*)I

167
Therefore, from ~j'(t) — -y'(t/c) and y"(t) = 7"(</c)/c we have
|7'(*)X7"(*)I "(I)
«(*) =
|7'WI3
3118
Let a : (a, 6) —► iR3 be a smooth curve with nonvanishing curvature.
i) Show that if the torsion of a vanishes identically then there is a plane
■k C M3 containing a, that is, a(t) g 7r for all t g (a, 6). Is 7r unique?
ii) Is the conclusion of i) still true if the curvature is allowed to vanish at
a single point c g (a, 6)? [Of course, the torsion is not denned at c, but is
assumed to be zero at all other points.]
[Indiana)
Solution.
i) Let a be reparameterized by arclength s\ namely, suppose a = a(t(s)),
s € (si>s2) with t(si) = a, t(s2) = b. The condition k ^ 0 means that we can
define the Frenet frame field {T,N,B} along a. Then the hypothesis t = 0
implies that B is a constant vector field. Hence, j^(a,B) = 0, from which
follows
(a(t(s)), B) = const = (a(t(s0)), B).
Therefore, the plane ir : (p — a(t(so)), B) = 0 contains the curve a. Obviously,
the connectedness of a and the smoothness of the Frenet frame field imply
that 7r is unique.
ii) If the curvature k is allowed to vanish at a single point c g (a, b), then,
according to the above discussion, a(t), t g (a, c) must be on a plane 7iv, and
a(t), t g (c, 6) must be on another plane 7r2. Of course, maybe, -K\ ^ 7T2. A
counterexample is as follows. Let
f (<,/(<), 0) if *<0,
a(t)= <^ (0,0,0) if t = 0,
I (t, 0,/(t)) if *>0,
where the function / is denned by
/W_1 0 if * = 0.

168
3119
Let a and /3 be two regular curves in M3. The curve /3 is called an involute
of a if for all t, f3(t) lies on the line tangent to a at a(t) and («'(<),/3'(i)) = 0.
Show that every involute of a generalized helix a is a plane curve. (Recall that
a is a generalized helix if for some constant vector u ^ 0, (u,a'(s)) = const,
where s is arclength for a.)
(Indiana)
Solution.
For convenience, let a and /3 be parameterized by their arclength s and
si respectively, and let s, s\ represent their corresponding points. Then we
may assume /3(si) = a(s) + X(s)T(s). Differentiate the equation with respect
to s. Using the Frenet formulas and the hypothesis (T(s),Ti(si)) = 0, we
can ascertain that /3(«i) = a(s) + («o — s)^(s)- Differentiate the obtained
expression for /3 successively. Noting that a being a generalized hrelix implies
k t
k' t'
= 0,
we obtain by straightforward calculation
d „/ x d2 „, x d3
^(-1),5^(.1),5
X^i^^Xa^))^'
i.e., the torsion of /3 vanishes everywhere. Therefore, /3 is a plane curve.
3120
Suppose that the unit normal vector to a surface M c St3 is constant along
a regular curve a C M. Deduce that in this case, a is an asymptotic (Note: A
curve in a surface is called asymptotic if its acceleration is everywhere tangent
to that surface.) curve, that a plane contains it, and that the Gauss curvature
of M vanishes at each point of a.
(Indiana)
Solution.
Let a be parameterized by arclength s, and M be locally parameterized by
^(u1,^2). Firstly, by the Weingarten formula, we have, along a,
dn v~* idv? dX TTr (dX\ TTr. ..
0=- = - Vw-——^ = -W -7-= -W(a').
ds ^ J ds du' \ds J v '

169
Further, by the Gauss formula, we immediately deduce that, along a
£
d2u{
~ds>
£r-
dv? duk \ dX
~jk ds ds ) du*
j,k=l I j,k = l
7 + £ ni*
dv? duk
ds ds
= kgn x a' + (W(a'),a')n = kgn x a',
where kg is geodesic curvature. Hence a" is everywhere tangent to the surface.
Secondly, along a, noting that n is constant, we have d(a(s),n) = 0.
Therefore,
(a(s), n) = const = (a(s0), n),
namely, (a(s) —a(so),n) — 0, which means that the curve a is contained by a
plane (p — a(s0),n) = 0.
Thirdly, from III - 2.ffII + Kl = 0 it follows that along a
K = -
dn
ds
2H{W(a'),a) = 0.
3121
Sketch the closed regular plane curve /3 : [—7r, tt] —► M2 having /3(0) = (0, 0)
and/3'(0) = (1,0), if/3's curvature function fe(s) (s = aiclength) is odd, satisfies
I k(s)ds ft* 371-/2,
./o
and has the following graph. (Include an explanation with your sketch.)
(Indiana)
Fig.3.4

170
Solution.
The image of the plane curve /3 is a "figure eight", as shown in Fig.3.5, which
has z-axis as its tangent line at (0, 0). And j/-axis is almost its "tangent" line
at (0, 0), too. This assertion follows from
J' k(s)ds = J* —da = 0«) - 0(0) « y,
where 6(s) is the oriented angle formed by /3'(s) and /3'(0).
y
Cj
p
(0,0)
Fig.3.5

171
SECTION 2
DIFFERENTIAL GEOMETRY OF SURFACES
3201
i) Show that there exists a metric on the plane so that some geodesies are
simple closed curves.
ii) Let a(s) be a simple closed geodesic as described above and K(x) be
the Gaussian curvature of the above metric at x £ St?. Compute
interior of a
Here the integral is with respect to the Riemannian volume induced by the
metric.
(Indiana)
Solution.
i) Let S2 = {(x1,x2,x3) e St3 : x\ + x\ + x\ = 1}, and M2 be the
xi ° x2 plane. For every p (E St2, its coordinates with respect to the Xi and
x2 axes are denoted by (u, v). Suppose that 7r : S2\{iV} —► M2 is the stereo-
graphic projection from the north pole of S2 into the plane St2, which maps
(xi,x2,X3) g S2\{N} to (u,v) g St2. By direct calculation, we obtain
_ 2u __ 2v __ u2 + v2 - 1
Xl ~ u2 + v2 + l' X2 ~ u2 + v2 + V X3 ~ u2 + v2 + l
Then the metric of S2\{N} can be expressed by
2_ Hdu2 + dv2)
S ~ (u2 + v2 + l)2'
Now, using the pull back of ds2 by (7r-1)*, we can obtain a 2-dimensional
Riemannian manifold (St2,(ir~1)*ds2). Thus, the map 7r : (S2\{N},ds2) —>
(St2,(ir~l)*ds2) is an isometry. Since all great circles which do not pass
through the north pole are closed geodesies on S2\{i\T}, then their images
are simple closed geodesies on (St2, (ir~1)*ds2).

172
ii) Denote by D the simply connected domain encircled by the simple closed
geodesic a. Then, using the famous Gauss-Bonnet formula, we immediately
have
K(x) = 2irX(D) = 2tt.
interior of a
3202
Let M2 C M3 be a surface containing x = 0. Assume that J— and ■£— are
tangent to M at x = 0, and that in that basis the Weingarten map
Let a be the curve (near x = 0) obtained by intersecting M with the 0=10:3
plane.
i) What is the normal curvature of a at x = 0?
ii) What is the geodesic curvature of a at x = 0?
iii) What is the Gaussian curvature of M2 at x = 0?
iv) Sketch the surface near x = 0. You may assume that the normal to the
surface at x = 0 is (0, 0,1).
(Indiana)
Solution.
i) By the Meusnier theorem, the normal curvature of a at x = 0 is
ii) Since a is obtained by intersecting M with 0=11:3 plane, i.e., a is a normal
section, then its curvature at x = 0 is fe(0) = |fen(0, =^-)1 = 4. By the relation
among fe(0),fen(0, gf-) and the geodesic curvature fe3(0, g|-)
we immediately obtain fe3(0, g|-) = 0.
iii) Because the eigenvalues of the Weingarten map L are ±5, we know that
the Gaussian curvature of M2 at x = 0 is K(0) = —25; or more directly,
K(Q)^det(l _34) =-25.
/

173
iv) From iii), we know that x = 0 is a hyperbolic point of M2. Noting that
the normal to the surface at x = 0 is (0, 0,1) and fcn(0, §|-) = 4 > 0, we sketch
the surface near x = 0 as follows.
Fig.3.6
3203
Let M ~ {(x,y,z) €. M3 : z = 6 — (x2 + y2)} (a paraboloid of revolution).
D = {(x, y, z) g M : z > 2}, and w = yz2dx + xzdy + x2y2dz. Orient M and
evaluate
I zdxAdy + du.
JD
[Indiana)
Solution.
Choose the unit outward pointing normal as the orientation of M. Let
P = {(x,y,z) e M3 : x2 + y2 < A,z = 2} and take (0,0,-1) as its normal
vector. Then, by the Stokes' formula, we have that
dw = 0.
Hence,
L
/^ = -/
JD JP
dw.
Therefore,
I zdx A dy + du = / [6 - (x2 + y2)]dx Ady- do.
JD JD JP
Firstly, we have
/ [6 - (x2 + y2)]dx Ady= f dO f (6 - r2)rdr = 16tt.
Jd Jo Jo

174
Secondly, noticing that
dw = z dy Adx + 2yzdz A dx + zdx A dy
+xdz A dy + 2xy2dx Adz + 2x2ydy A dz,
we have
— I du = — / 4<fy A da; + 2<2a; A dy = —2 I dy Adx = — 87r.
'p Jp
Hence,
zcfa; Ady + dw = 871".
X
3204
Let 5 be a surface difFeomorphic to the ordinary 2-sphere. Suppose there is
a C°° metric of positive curvature on S for which there exist two simple closed
geodesies 71 and 72. Show that 71 and 72 must intersect.
(Indiana)
Solution.
Here we assume the Jordan curve theorem.
If 71 and 72 do not intersect, then 71 and 72 encircle a domain D such
that the Euler characteristic x(D) = 0 and dD = 71 U72. Applying the global
Gauss-Bonnet formula to the domain D C S and noting that 71 and 72 are
geodesies of S, we have
I Kda = 2irX(D) = 0,
Jd
which contradicts the assumption of positive curvature.
3205
Let D be the disk {(x, y) e St2 : x2 + y2 < f} and let O = (0, 0) € D. Let
(21,^2) = (r, 6) be the usual polar coordinates and define a metric on D\{0}
by
1 0
(9ij)- \ 0 ^0)
where h(r, 6) = r2[l - 2r2 + r4 sin2 6}2

175
(a) Prove that this metric extends to a smooth metric on D.
Hint. Use a smooth coordinate system.
(b) Show that line segments in D which pass through the origin (suitably
parameterized) are geodesies.
[Indiana)
Solution.
(a) Set
{x = r cos 0,
y = r sin 6.
Then flv '*< = r. So, on D\{0}, we can choose (x,y) as the coordinates. From
{dx = cos 6dr — r sin 6d6
dy = sin Odr + r cos 6d6
follows
{dr = cos dx + sin 6dy
d6 = -(cosOdy — r sin #cfa;).
Hence the metric of D\{0} may be rewritten as
ds2 = dr2 + h(r, 6)d02
= [cos2 6 + (1 - 2r2 + 4r4 sin2 6>)2 sin2 6}dx2
+2 cos 6» sin 6[1 -(1- 2r2 + 4r4 sin2 6>)2]da;^
+[sin2 6» + cos2 6>(1 - 2r2 + 4r4 sin2 6)2]dy2
= gndx2 + 2g12dxdy + g22dy2.
When r —» 0, we see that gu = 1 + o(r), g\2 — °{r)i 922 = 1 + °{r), and
-J^J- = o(l), -^- = o(l). Therefore, this metric can extend to a smooth metric
on D.
(b) For any line segment I in D, we can express f\{0} as the union of
Ji = {(r,0) : 0 < r < \,6 = 60} and l2 = {(r,6) : 0 < r < \, 6 = 60 + tt}.
Using ds2 = dr2 + h(r,6)d62, we know that both /i and l2 are geodesies in
D\{0}. If we use the extended metric, the geodesic curvature kg of I must be
a continuous function of the coordinates. Because /i and l2 are geodesies by
either metric, we know that kg\o = 0. Hence, as a whole line segment, I is a
geodesic, too.

176
3206
i) Let M be a surface (without boundary) in M3. Suppose M is inside a
ball of radius R and suppose a point p g M is on the boundary sphere. Show
that the Gauss curvature of M at p is > -gj•
ii) Show that there is no closed minimal surface in M3.
(Indiana)
Solution.
i) The surface M and the sphere S2, the boundary of the ball, have the
same tangent plane at p. Let x be a normal plane of M and S2 at p. Then,
using the local canonical forms of the normal section curves Mfli and S2 n7r
at p, one can easily show that, at p, the curvature of M n 7r is not less than
that of S2Dir = S1, and p is an elliptic point of M. Hence the Gauss curvature
of M at p is greater than or equal to -^.
ii) Suppose that M is a closed minimal surface in M3. Then there is a
family of spheres that contain the surface M inside and have a fixed center.
Ler R be the infimum of their radii. Then, there must be at least one common
point of M and the sphere with radius R and centered at the fixed point. Using
the result of i), one concludes that the Gauss curvature of M at p, K(p) > -^.
But it contradicts the fact that , for minimal surfaces,
K = kxk2 = ~h\ < 0.
3207
Let H2 = {(x,y) € St2,y > 0} be the upper half-plane in M2 and let H2
have the Riemannian metric g such that
9(x, y) = ~^(x,y) for (x,y) € H2,
where (x, y) is the usual inner product on M2.
(a) Compute the components of the Levi-Civita connection (i.e., the Chris-
toffel symbols).
(b) Let V(0) = (0,1) be a tangent vector at the point (0,1) G H2. Let
V(t) = (a(t),b(t)) be the parallel transport of V(0) along the curve x = t,

177
y = 1. Show that V(t) makes an angle t with the direction of the y-axis in the
clockwise direction.
Hint. Write a(t) = cosO(t), b(t) = sm6(t) where 6(t) is the angle V(t)
makes with the z-axis.
(Indiana)
Solution.
(a) From the hypothesis we have E = G = -\, F = 0 and hence
r1 -El-n r1 -El--- r1 --9l
U~2E~^ 12~2E~ y' i22~ IE
= 0,
r2 - El - - r2 _ Gl _ n p2 _ G2 _ 1
111 ~ 2G~j/' ll2~2G-U' l22~2G~~y-
(b) For convenience, denote the curve (x,y) = (^,1) = (u1(<),u2(<)) and
the parallel transport vector field V(t) = (a(t),b(t)) = (vx(t), v2(t)). Then
V(t) must be the solution to the following initial value problem
^+4^(0^ = 0, i=l,2
(^(0),^(0)) = (0,1).
Noticing the above expression of r'fc, along the curve, we see that the problem
is equivalent to
r *p = mi «1 = _a(<)
1 (a(0),6(0)) = (0,1).
Therefore, writing a(t) = cos6(t), b(t) = sin0(i), we have immediately 6 = —t.
3208
Consider the hyperboloid S of one sheet x2 + y2 — z2 = 4.
(a) Define the Gauss curvature K of S.
(b) Use the definition from part (a) to compute K at (0,2,0).
(Indiana)
Solution.
(a) The Gauss curvature K of S at P is denned hy K — fci&2, where fcl7 &2
are the two principal curvatures of S at P.
(b) At P = (0,2,0), in the direction of the circle of latitude, the normal
section is a circle x2 + y2 = 4. Thus, by taking the outer unit normal vector
field as the orientation of S, the corresponding principal curvature ki = — |;

178
whereas in the direction of the meridian, the normal section is a component of
l
2"
the hyperbola y2 — z2 = 4, thus the corresponding principal curvature k? — 1
Therefore, the Gauss curvature K = fci&2 = — 4
■r
3209
Calculate the total geodesic curvature of a circle of radius r on a sphere of
radius R> r.
(Indiana)
Solution.
By the Gauss-Bonnet formula, the total geodesic curvature is given by
I kgds = - I I Kdcr + 2irx(ty
Jc J Jil
—^•2-kR(R- y/R2-r2) + 2ir
= 2tt
VR2 - r2
R '
where C is the given circle, and fi is the spherical cap encircled by C.
3210
Let M be a compact surface embedded in M3, with smooth unit normal
vector field v.
Show that the mapping Pc : M —► M3 denned by Pc(x) = x + ev(x) will
immerse M provided |e| is the reciprocal of neither principal curvature fei or
&2 at x g M. In this case, express the principal curvatures of M£ := PC(M) at
Pe(x) in terms of fei and fe2.
(Indiana)
Solution.
If, locally, we take the lines of curvature as the parametric lines of M, then
by the Rodriques equation we have on Pe(M)
d „ , , ,. , , dx
Thus,
-^) = (1-^)^, i=l,2.
—P.(x) x ^P£(x) = (1 - ctl)(l - ek2)^ x ^ # 0

179
provided that |e| is the reciprocal of neither ki or &2 at x (E M, which shows that
the mapping P£ immerses M into M3. Furthermore, again by the Rodriques
formula, we see that the u1 and u2 parametric lines are the lines of curvature
in Pe(M), and the corresponding principal curvatures are eki — 1 and £&2 — 1,
respectively.
3211
Let A : M —► St be a smooth function with compact support, and consider
the Riemannian surface obtained by equipping M2 with a metric g of the form
</(v,w):=eA«(v,w)
for v,w g T(Xty)M2. Assuming the Gauss curvature K of this metric is
everywhere non-negative, use the Gauss-Bonnet Theorem to deduce that in fact,
the surface is flat.
{Indiana)
Solution.
Using the Descartes coordinates (x,y) in ttt2, we can express the metric
of the Riemannian surface fit2 by ds2 = ex^y\dx2 + dy2), which is obviously
conformal to that of the Euclidean plane ttt2. Take a positive number a and
consider a square domain V in St2, whose vertices are -4(-a, —a), B(a,—a),
C(a, —a), D(—a,a) respectively. Suppose that, in M2, the corresponding part
of V is V, and the corresponding points of A, B, C, D are A, B, C, D,
respectively.
Then, by the Liouville formula, we can see that the geodesic curvatures of
the segments of coordinate curves AB and CD are
Kg
d6 1 dlnE 1 <91nG . „
~~ — i 7= —^— cos t) -i — —-— sin 0 ,
ab \ds 2VG dy 2VE Ox J e=o,y=-a
1 dX'
Kg
2Ve^ dy J y=_a '
dJ9 1 d\aE n 1 31nG . „
~~ — i -: 7= —7.— cos 6 H = —-— sin 0 ,
cd \ds 2VG dy 2^E dx J e=K,y=a
1 d\'
2\f^dy)y=a'
whereas both BC and DA are geodesies of Et2. Applying the Gauss-Bonnet

180
formula to V C fit2, we have
J „kgds+ J J Kda = 0,
JdV J Jv
where the Gauss curvature
K=- — AlneA = - — —
eA ex dy2
Noting the area element da = exdxdy and the line element ds2 = ex^dx2
along AB and CD, from the above integral equality we can obtain
'dX\ _ /dX\
,dy)y=-a \dy)y=a
Because the number a is arbitrarily chosen, letting a —► 0, it leads to
\dy)y=o~
Besides, by the hypothesis K > 0, we have
$*»
namely, -£ is a decreasing function. Thus,
\dy)y=-a~ ~ \dyJy=a'
which combining the above equality shows that -£ = 0 in M1. Therefore,
K = 0, i.e., St is flat.
3212
Let M2 C St3 be a smooth compact surface such that M2 C {(x, y, z) : z >
0}. Assume that M2C\{(x, y,z) : z = 0} is asmooth curve a(s), parameterized
by arclength.
i) Show that a(s) is an asymptotic curve on M2.
ii) Show that a'(s) is always a principal direction.

181
iii) Let's now drop the assumption that M2 n {(x, y, z) : z = 0} is a curve.
What kind of a set could M2 n {(a;, y, z) : z = 0} be?
[Indiana)
Solution.
i) The hypotheses imply that along the curve a(s), the plane {(x,y,z) :
z = 0} is a tangent plane of the surface M2. Thus, the unit normal vector field
n(s) of M2 is constant along a(s). Therefore, along a(s), nj'' = 0, which
means that the tangent vector of a(s) for every s is an asymptotic direction.
Hence, a(s) is an asymptotic curve on M2.
ii) Noticing the above fact, we see that along a(s), n£*' = —0 • a'(s), from
which the Rodriques equation says that a'(s) is always a principal direction
with the principal curvature 0.
iii) If we drop the assumption that M2 n {(a;, y, z) : z = 0} is a curve, then
the set M2 n {(x,y,z) : z = 0} consists of elliptic or parabolic points of M2,
at which the plane {(x,y,z) : z = 0} is tangent to M2.
3213
(a) Construct an example of a non-compact C°° surface in ttt3 with a
sequence of closed geodesies {<r,} such that length (<r,-) —► 0.
(b) Show that this is not possible if the surface is compact.
(Indiana)
Solution.
(a) Consider the following surface S of revolution generated by a C°°
vibrating curve C, illustrated in Fig.3.7, rotating around the z-axis which is
the asymptote of the curve. Let Pi denote the points where C has horizontal
tangents. Then, on S, Pi draw closed geodesies <r,-, and length (<r;) -+ 0.
Fig.3.7
(b) Suppose S is a compact surface. If there exist closed geodesies <r,- such

182
that length (en) —► 0, then by the Gauss-Bonnet formula we have
Kdcr = 2rX(Oi),
J Jiii
where £), denotes the domain encircled by <r,. Because S is compact, when
i is sufficiently large, we may suppose £),■ is simply connected. Thus, setting
i —► oo in the above equality, we come to a contradictory result 0 = 2ir.
3214
Let (r(s),0,z(s)) be a unit speed curve in Et3 with r(s) > 0. Consider
the surface of revolution (s, 6) —► (r(s) cos 6, r(s)sm 6, z(s)). On this surface
compute the covariant derivatives V_e_ Jj and V a_-§g in terms of J^ and Jj.
[Indiana)
Solution.
The direct computation gives
I = ds2 + r2(s)d62 = Eds2 + GdO2.
Denoting (s,6) = (it1,!*2), we have
V7 d pi d a r2 d Gi d G2 d , d
v£do = T22rs+V22de = -2EdS + 2Gde = -r{s}r{s}d-s'
d_ _ rld_.r2d__ E1d_,G1d__r'(s)d
T*ds 21ds 21d0 2Eds 2G 80 r(s) 80'
where r'fc denotes the ChristofFel symbol (i,j,k = 1,2).
3215
Let M2 C St3 be an embedded compact surface of genus > 1. Show that
the Gauss curvature of M must vanish somewhere on M.
(Indiana)
Solution.
By the Gauss-Bonnet formula, we have
J J&
Kda = 2ttX(Mz) = 4tt(1 - g) < 0.
M2

183
Firstly, we claim that because of the compactness of M2, there must be a
point P with K{P) > 0; hence, by the continuity, there exists a domain U of
P such that K\u > 0. Secondly, we show that there exists another point Q
with K(Q) < 0. Otherwise, we would have
0 > J I Kda > I I Kdcr > 0,
a contradiction. Finally, by the connectedness, the continuous function K must
vanish somewhere on M2.
3216
Consider the torus-of-revolution T obtained by rotating the circle (x — a)2 +
z2 = r2 around the z-axis:
T = {(x, y, z) : (x2 + y2 + z2 + a2 - r2)2 - Aa2{x2 + y2) = 0}.
Parameterize this torus, compute its Gauss curvature function K, and verify
that JT KdA = 0 by explicit calculation.
[Indiana)
Solution.
Thus obtained torus-of-revolution T can be parameterized by
X(u, v) = ((a + r cosu) cos v, (a + r cost*) sin v, rsintt), 0 < u, v < 2ir.
A straightforward computation gives the coefficients of its first and second
fundamental forms
E = r2, F = Q, G= (a+rcosuf;
L = r, M = 0, N = cosu-(a +rcosu).
Therefore, its Gauss curvature function
_ LN — M2 _ cos u
~ EG- F2 ~ r(a + rcosu)'
Noting that the area element on T is
dA = VEG — F2dudv = r(a + r cos u)dudv,
we have immediately
* y»2ir *2ir
/ KdA = I dv I cos udu = 0.
Jt Jo Jo

184
3217
Let
x(t)= |icos(<),isin(<),^J , 0<*<2tt
be a curve on S2 C M3. Let X0=^£T,in ^S2- Compute the parallel
2 12 ' ' 2 f
translation of Xq along x(t).
(Indiana)
Solution.
Consider the cone that is tangent to the unit sphere S2 along the curve
x. This cone minus one generator is isometric to an open set D C JR2 given
in polar coordinates by 0 < p < +oo, 0 < 6 < \/Z-k. Because the parallel
translation in the plane coincides with the normal Euclidean one, we obtain
the result that, for a displacement t of a moving point p along x starting from
(|,0, ^) (corresponding to the central angle 6 = *^t in the domain D), the
oriented angle formed by the tangent vector x'(t) with the parallel translation
vector X(t) of X0 is given by 2ir - 6 = 2ir - ^-t.
3218
Show that for any Riemannian metric on S2 with \K\ < 1, where K is the
Gauss curvature, the area of S2 is not less than 4ir.
(Indiana)
Solution.
By the Gauss-Bonnet formula, for any Riemannian metric on S2, we have
Thus, if \K| < 1, then
/ Kdv = 2irX(S2) = 4.71-.
Js2
4t < / \K\dv < Area (S2).
3219
Define "geodesies", and characterize (with proof):
i) All geodesies on the unit sphere S2 C M3.

185
ii) All geodesies on the surface {(x,y,z) €. M3 : x2 + y2 = 1}.
(Indiana)
Solution.
Let a(s) be a curve on a surface M parameterized by arclength. Along a,
we have
E dv? duk
a" (a) = k(s)N(s)
y-^ 1 d2u' y-^ ■ dui duk
~ ^L, \ ds2 + 2s jk ds ds
= kgn x a + knn,
dX
ds ds
where ul,u2 are local coordinates on M, n is the unit normal vector field of
M along a, and kg is the geodesic curvature of a. Then, a is a geodesic of M
if and only if along a, kg = 0 or
d2u' v—* „• dui duk
i) On the unit sphere, every great circle is a geodesic, because along a,
the principal normal vector N is parallel to n. On the contrary, owing to the
uniqueness of the initial value problem
-2^- + 2-,^-^-5- = 0. «=1,2
~ U0> ds
"0>
every geodesic on M, that starts from a given point and is tangent to a given
direction, must be a great circle.
ii) Since geodesies are intrinsic objects, then if we develop the cylinder to
a plane, every geodesic must become a straight line. Therefore, every geodesic
on the cylinder must be a helix, or a circle of latitude, or a straight generating
line.
3220
Give an example (e.g., draw a picture) to show that a connected surface
can have two points which are not jointed by any geodesic. What is the usual
topological hypothesis that prevents this problem?
(Indiana)

186
Solution.
Let 7r be a plane and p, q two points in 7r. Let r be an interior point of the
line segment pq. Then the surface 7<"\{Y} is a case in point. "Completeness" is
the usual topological hypothesis that prevents this problem.
3221
Define "minimal surface in iR3", and prove that the eaten oid, obtained by
rotating the graph of y = cosh(a;) around the z-axis in M3, is minimal.
(Indiana)
Solution.
A minimal surface is a surface with mean curvature
1 EN - 2FM + GL _
~ 2 EG-F2 ~ '
The straightforward calculation shows that the catenoid
X(x,0) = (z,cosh(i)cos(6>),cosh(i)sin(6>))
is a minimal surface in St3.
3222
The "Clifford" torus in S3 can be parameterized using charts of the form
X(u,v) = —-=(cosu, sinu, cos v,sinv),
V2
where u and v are constrained to lie within intervals length < 2ir.
i) Compute the metric [g,y] on the Clifford torus for a coordinate chart of
the indicated type.
ii) Figure out the Clifford torus' Gauss curvature function.
Hint. Calculation is not necessary here.
iii) Deduce from the result of ii) that the Euler characteristic of a torus is
zero.
(Indiana)
Solution.
i) From
Xn = —p(— sin u, cost*, 0,0), Xv = —,= (0,0, — sin v, cos v)
V2 v2

187
we easily have
9n = 922 = 2, 9i2 = 921 = 0.
ii) The Gauss equation implies that the Clifford torus' Gauss curvature
function K = 0.
iii) Using the Gauss-Bonnet formula, we see that the Euler characteristic
of a torus T
iiT) = hjjT
Kdcr = 0.
3223
Let / : M2 —► M be a smooth function with a critical point (e.g., a minimum
or maximum) at the origin X\ = x2 = 0.
i) Show that the principal curvatures of the graph
z = f(xi,x2)
at (0,0,/(0,0)) € M3 are the same as the eigenvalues of the Hessian matrix
[d2f/dxidxj] at (0,0).
ii) Show that M3 contains no compact embedded surface with strictly
negative Gauss curvature at all its points.
Hint. Look at the "lowest" point on the surface.
(Indiana)
Solution.
i) The hypothesis means that J£- and |£- vanish at (0,0). Then, by
straightforward calculation, we have
( 1 0 \ I 0
[9ij]o,o - ( 0 1 J , [%](o,o) = I %)
d2f
dx\
d3}
dx^dx?
Therefore, the principal curvatures of the graph
z = f(x1,x2)
at (0, 0, /(0, 0)) € St3 are the roots of the following equation
(0,0)
det[n,j - \gij](0,o) =
a*? A
dxidx? dx\
dx^dx-2
»~2 «
0.

188
ii) Let p(xo,yo,zo) De the "lowest" point on the surface. (The existence of
such a point follows from the compactness of the surface.) Since the surface is
above the plane 7r : z = zq and p is the common point of -k and the surface, then
■k is the tangent plane of the surface at p. By observing the normal section at
p, it is easy to conclude that any normal curvature of the surface at p is greater
than or equal to zero, if we take (0,0,1) as the unit normal of the surface at
p. Thus the Gauss curvature of the surface at p is not less than zero.
3224
Let M be a 2-dimensional manifold smoothly embedded in M3 with unit
normal n.
a) Prove that for each p £ M there exist an open neighborhood Up of p in
M3 and a smooth function F : Up —► M such that .F-1(0) = UPC\M.
b) Find F if M is the graph of a smooth function / : IR? —► IR.
(Indiana)
Solution.
Since the inclusion map M —► M3 is an embedding, for each coordinate
neighborhood V c M of p £ M, there exists a neighborhood U of p in IB?,
such that V = U <1 M. Using the local coordinates (u,v) for V and (x, y,z)
for U, we can express this embedding as (u, v) —► (x(u, v), y(u, v),z(u, v)).
Noticing that
'd(x,y,zy
ran
k
d(u, v)
we know from the implicit function theorem that there exists a neighborhood
Vp C V of p, such that on Vp,
{x = x(u, v)
y = y(u,v)
has smooth inverse
u = u(x,y)
v = v(x,y).
Then, we can find a neighborhood Up C U of p in M3, such that Vp = Up n M,
and in terms of the local coordinates (x,y), the embedding can be expressed
by (a;, y) -» (a;, y, /(a;, y)), where /(a;, y) = z(u(x, y), v(x, y)).
Therefore, setting F : Up —► St3 by F(x,y,z) = z — f(x,y), we have
jr-i(0) = [7pnM.

189
b) If M is the graph of a smooth function / : Et2 —► ttt, then for each p g M,
we can take Up - M3, and F : M3 —► M denned by F(x, y,z) = z — f(x, y).
3225
Let M be a 2-dimensional manifold smoothly embedded in M3 with unit
normal n. Assume that M is the boundary of a bounded convex open set.
Assume that n is the exterior normal and that the Gauss curvature K of M
is everywhere positive. [Recall that S C M3 is denned to be convex if for each
two points of S the line segment joining these points lies in S. You may use
without proof the fact that M lies on one side of each of its tangent planes.]
a) Define the Gauss (or sphere) map
V: M - S2 = {(x,y,z) :x2 + y2 +z2 = 1}.
b) Prove that rj is one-to-one.
c) Assuming that rj is one-to-one, prove that 77 is a diffeomorphism onto
S2.
d) Show that
J K(p)dp=4ir.
Jm
(You may assume b) and c) if you wish.)
(Indiana)
Solution.
a) For each p £ M, define r)(p) as the end point of the unit exterior normal
n(p) after parallel translating it to the origin of M3.
b) and c). We first prove that rj : M —> S2 is a local diffeomorphism.
For each p G M, there exist a coordinate neighborhood £ C M of p and a
coordinate map h : £ —+ U C St2, which is a diffeomorphism, such that on U
the Gauss map has the following expression
■q 0 h~ (u, v) = (a(u, v), /3(u, v), 7(1*, v)) = n(u, v),
where n(u, v) is the unit normal at X(u, v) £ £ C M.
Since n„ x n, = -ff-X„ x Xv and if > 0, Xu x I, ^ 0, the rank of the
Jacobi matrix
(au «» \
A, A =2.
7« 7* /

190
Thus, the implicit function theorem says that 77 o hr1 has a smooth inverse on
a neighborhood of h(p) (which may be smaller than U); hence 770/1-1 is a local
difFeomorphism. Therefore, in the neighborhood of h(p), 77 = (77 0 h-1) 0 h is a
difFeomorphism. Thus, on M, 77 is a local difFeomorphism.
Next, we show that 77 : M —> S2 is surjective. Since 77 : M —> S2 is a local
difFeomorphism, 77 is an open map. Besides, because M is compact and S2 is
a Hausdorff space, 77 is also a closed map. Thus f)(M) is an open, closed and
non-empty set of S2. The connectedness of S2 implies that 77(M) = S2.
Now we prove that 77 : M —> S2 is globally one-to-one by leading to a
contradiction. Suppose that there are two distinct points P,Q £ M such that
77(F) = r)(Q). From the above argument we know that there are neighborhoods
Ep, Sq of P,Q respectively, such that 77^ : SP —> ?7(Ep),?7|s : Sq —>
77(¾) are difFeomorphisms. Because P / Q, in If we can choose £p,£q so
small that £p HSq = 0. Now take the inverse images of ?7(£p) HtjCEq) under
77^ and 77|sQ, respectively. Namely, set
U = (r7|Sj3)-1(r7(Sp)n77(SQ)),
v = (^)^(^)^(¾)).
Thus, UDV = 0 and 77(^/) = 77(F), which implies r)(M\U) = S2. On the other
hand, it is easy to show that M is compact, connected and oriented. Then the
Gauss-Bonnet formula gives
/ da =/ Kda = 4x, (by noting if > 0),
where da, da have local expressions
da = \XU x XB|chicfo;, da = |n„ x n»|(iu(iw.
Hence
4x = / #<2<r = / #d<r + / #d<7
./M JM\U JU
> J da+ J Kda = 4x +/ #^7,
Since J^ ifrfcr > 0, we arrive at a contradiction.
In the end, noticing that differentiability is a local property, we see that
the globally one-to-one, surjective local difFeomorphism is naturally a global
difFeomorphism.

191
Remark. In fact, this problem is the famous Hadamard theorem. Using
the theory of covering map, we can simplify its proof as follows.
Firstly, as the above, show that 77: M —> S2 is a local difFeomorphism.
Since M is compact and S2 is connected, then the local difFeomorphism
r): M —+ S2 is also a covering map.
Further, because S2 is simply connected, and M C St3 is connected and
hence path connected, we know that the covering map rj must be a homeomor-
phism and hence a global difFeomorphism.
3226
Let M be a 2-dimensional manifold smoothly embedded in St3 with unit
normal n.
a) Explain what is meant by intrinsic and extrinsic quantities on M.
b) Are the principal curvatures intrinsic?
c) Discuss why the covariant derivative on M, denned using the covariant
derivative on St3, is intrinsic.
d) Assuming c), discuss why the Gauss curvature of M is intrinsic.
(Indiana)
Solution.
a) The terminology "intrinsic quantities" means those geometrical
quantities that are definable only by the first fundamental form of M and its
derivatives. Otherwise, they are called "extrinsic quantities". In other words,
intrinsic quantities are those that are invariant under isometric correspondence, but
extrinsic ones are not.
b) The principal curvatures are not intrinsic; they are extrinsic. For
example, consider a plane and a cylinder.
c) Although the covariant derivative on M is defined by using the covariant
derivative on St3, the last local expression of the covariant derivative of M
involves the tangent vector field and the ChristofFel symbols of M. Therefore,
it is intrinsic.
d) For each p £ M, let C be a simple closed curve encircling a simply
connected domain D where p lies. Let Aw denote the angle variance caused by
a tangent vector after parallel translating it around C once. Using the Hopf's
rotation index theorem and the Gauss-Bonnet formula, we can show that the
Gauss curvature of M at p can be expressed by
Aw
K(p) = lim
D^SSDdcr

192
Since parallel translation is intrinsic, then the Gauss curvature is intrinsic, too.
3227
By revolving the curve 7 sketched below around the a;-axis, we get a surface
of revolution M2 C ffi3. Compute JM KdA, where K is the Gauss curvature
and dA the area form, on M. (Make sure to justify your answer.)
(Indiana)
Fig.3.8
Solution.
The surface M2 of revolution generated by the curve 7 is homeomorphic
to a section of a cylindrical surface. Thus the Euler characteristic number
x(M2) = 0. Besides, the boundary of M2 consists of two circles which are just
geodesies of M2, because along these circles the normal vector of M2 is parallel
to the principal normal vector of the two circles respectively. Therefore, by
the famous Gauss-Bonnet formula, we have immediately
f KdA = 0.
JM
3228
A surface S2 immersed in a Riemannian manifold N is said to be ruled if
it can be parameterized near any point by a mapping X : (0,1)2 —> N such
that for each fixed v0 £ (0,1), the it-parameter curve X(u, vo) is a geodesic in
N.
a) When n = 3, show that the Gauss curvature of a ruled surface in Mn is
nowhere positive.
b) Show it for arbitrary n.
(Indiana)

193
Solution.
a) Since every geodesic in N = M3 is a straight line, then S2 can be
characterized locally by X(u, v) = a(v) + (u— |)/(v), where l(v) is the direction
of the geodesic corresponding to v G (0,1). Thus, through computation, we
can easily deduce that the first coefficient of the second fundamental form of
S2
fl2y
L=<»,-^) = 0,
which shows that the Gauss curvature of the surface
LN-M2 M2
~ EG-F2 ~ ~ EG-F2 ~
b) Similar to the above, suppose S2 can be characterized locally by
X(u, v) = a(y) + ul(v),
where, for convenience, we may assume that v G (0,1) is the arclength of the
curve a(v), \l(v)\ = 1, and (a'(u), l(v)} = 0. Then, by a routine work, we see
that the first fundamental form of S2 is
ds2 = du2 + (1 + 2u(a'(v), /'(«)) + «2|/'(«)|2)d«2.
Since the Gauss equation says
XT du2
it suffices to show that 9g^ > 0. However, we have
dVG {a'(v), l'(v)) + u\l'(v)
du y/i + 2u{a'(v), /'(«)) + u2\l'(v)\2 '
d2VG \l'(v)\2-{a'(v),l'(v))2
du2 (1 + 2u{a'(v), /'(«)) + u21/'(«)|2)3/2 '
Noting that
\l'(v)\2 - {a'(v),l'(v))2 = \a'(v) x l'(v)\2 > 0,
we obtain the desired result.

194
SECTION 3
DIFFERENTIAL GEOMETRY OF MANIFOLD
3301
Let Mn be a Riemannian manifold and x a 2-plane in T*Af.
i) Let {X,Y} be an orthonormal basis of x. Use this basis to define the
sectional curvature K(tt) and show that it depends only on x and not the
particular basis chosen.
ii) Define the Riemann curvature tensor in terms of covariant
differentiation. Explain why it is a tensor.
iii) Recall that if Mn C 2Rn+1 is a hypersurface, then
R(X,Y)Z = (LY,Z)LX - (LX,Z)LY,
where L is the Weingarten map. Using this or any other valid method, compute
all sectional curvatures of the sphere {x £ ttt4 : \x\ = 3}.
(Indiana)
Solution.
i) The sectional curvature is defined by K(~k) = ~R(X,Y,X,Y), where
R( ) is the Riemannian curvature tensor field of Mn. If {X,Y} is another
orthonormal basis of x, then X = aX + bY, Y = cX + dY. Noticing that
I , I is an orthogonal matrix, and R( ) is a 4th order covariant tensor
field, we can easily have R(X,Y,X,Y) = R(X,Y, X, Y), which proves that
K(ir) depends only on x.
ii) For any vector fields X, Y, Z, W on M™, define the Riemannian curvature
operator by R(X, Y)Z = Vx Vy^ — Vy VXZ — ^?[x,y]Z and the Riemannian
curvature tensor field by R(X, Y, Z, W) = (R(X,Y)Z, W), respectively. Then
for any C°° function / on Mn, using the properties of covariant differentiation
and of the inner product, we can conclude from straighforward computation
that
R(fX,Y,Z,W) = R(X,fY,Z,W) = R(X,Y,fZ,W)
= R(X,Y,Z,fW)^fR(X,Y,Z,W).
Thus, in terms of local coordinate frame field, one can show that Vp G M",
R(X,Y,Z, W)\p is dependent only on Xp,Yp,Zp,Wp E Tp{Mn). Therefore,

195
R( )\p is a well-defined 4th order multilinear function. Besides, the inner
product and covariant differentiation are all C°°. Hence, R( ) thus denned is
a C°° tensor field.
iii) For the sphere M" = {x £ -ffi4,|a;| = 3}, we can regard the position
vector field x as the normal vector field of Mn. Let {e,-} be a local orthonormal
frame field about i (as a point) of Mn. Then by the original definition of
covariant differentiation, one can easily show that Ve;x = e,-, where V denotes
the covariant differentiation in ttt4. Hence, the Weingarten map is L : X >—►
L(X) = — Vx f = —X/Z. Therefore, if {X,y} is any orthonormal basis of an
arbitrary 2-plane x in Tp(Mn),
K(-k) = -{R(X,Y)X,Y) = (LX,X)(LY,Y) - {LY,X){LX,Y) = 1/9.
3302
Let C = {x G St3 : 0 < X{ < 1} be the unit cube in M3. Suppose
F : St3 —> M4 is a 1-1 C°° immersion in some neighborhood of C. The image
F(C) is then a compact Riemannian submanifold of M4 with boundary and
therefore has a volume. Justify the following formula:
Also, evaluate the integrand if
F(x) = ((1 + zi)2, (1 + x2)2, (1 + x3)2, (2 + ^)3).
dxid,X2d.X3.
(Indiana)
Solution.
Consider the following four points in F(C)
Pi = F(xx, x2,x3), P2 = F(x1 + Ax1,x2,x3),
P3 = F(xi,x2 + Ax2,x3), P4 = F(x!,x2,x3 + Ax3),
where |Axi|, |Aa;2|, |Aa;3| are sufficiently small. Construct three vectors as
follows
P2P[ = F(xi+Ax1,x2,x3)-F(x1,x2,x3)=-—Azi H ,
UX\
QP
P3P[ = F(x1,x2 + Ax2,x3)-F(x1,x2,x3)=-—Ax2+---,
dF
P4Pi = F(xx,x2,x3+Ax3)-F(x1,x2,x3) = -—Ax3 +
ox3

196
Let Act be the volume of the parallelopiped spanned by the vectors P2P1, P3P1
and P4P1 in M4. Then
Act = \\P2Pl AP3P1AP4P1I
F* (£) AF* (£) AF* (^) ^Az2A*3 + -
is an infinitesimal when Ao=i —> 0, Ao=2 —> 0, Ao=3 —> 0. We take the principal
part of the infinitesimal Acr as the volume element, namely, the volume element
of F(C) is
da =
* \dxi) * \dx2) * \dx3)
dxid,X2dx3.
Hence the desired formula follows immediately.
Furthermore, if F(x) = ((1 + o=i)2, (1 + x2)2, (1 + x3)2, (2 + x3f), then, by
denoting F(x) = F(x1,x2,x3) = (2/1,2/2,2/3,2/4), we have that
- (£) =
2(1 + ^2)^-,
<?2/2
M^l) = 2(1 + ^)/- + 3(2 +^/-
#2/3
#2/4
Since ^7.
a a
and
2{l + x3)— + Z{2 + x3)2 —
#2/3 #2/4
form an orthonormal frame, we obtain that
4(1 + 0=3)2+9(2 + 0=3)4
dxi) * \dx2) * \dx3
= 4(1 + o;i)(l + x2)>/4(l + x3)2 + 9(2 + 0:3)4.
3303
There is no submersion from S3 into M2.
(Indiana)

197
Solution.
Let S3 C M4 be defined by {(x\ x2, x3, x4) G M4; (x1)2 + (x2)2 + (x3)2 +
(a;4)2 = 1}. Observe the following three vector fields
2<9 i<9 4 <9 , d
ox1 ox2 ox3 dx4
4 d , d , <9 , d
x* = x jzi + x in-* in-xVi-
ox1 ox1 ox* ox*
■w 3^ *d id 7 d
ox1 ox1 oxi ox*
where (a;1 4) G S3. It is easy to see that they are nowhere-vanishing
tangent vector fields on S3. Since (a;1)2 + (a;2)2 + (a;3)2 + (a;4)2 = 1, then,
without loss of generality, we may assume a;4 ^ 0. Hence, from the fact that
a;2
a;4
-a;3
-x1
X3
x4
X4
-a;2
X1
det I a;4 a;3 -a;2 I = (a;4)3 + a;4(a;3)2 + a;4(a;2)2 + a;4(a;1)2 = a;4 ^ 0,
V -a;3 a;4 x1 J
we know that X\, X2, X3 are three linearly independent vector fields.
Furthermore, by direct calculation, we obain
[Xlt X2] = 2X3, [X2, X3] = 2Xlt [X3, Xx] = 2X2.
Now we suppose that there is such a submersion x : S3 —+ M2. Then,
x* : Tp(S3) —> TF(p)(iR2) is a surjection for every point p G S3. Thus, we may
assume that, for example, irtX2,irtX3 are linearly independent, and x*Xi =
airtX2 + bTTtX3 at p. Then, on the one hand,
[x*Xi,x*X2] = 6[x*X3,x*X2] = birt[X3,X2]
= — 2birtXi = — 2abwtX2 — 262x*X3;
on the other hand,
[x*Zi,x*X2] = x*[Xi,X2] = 2x*X3.
Therefore,
tt»X3 = —a6x*X2 — b ^*X3
from which we get ab = 0 and -b2 = 1. Similarly, from [^3,^] = 2X2 we
can deduce that —a2 = 1 and ab = 0. They are all contradictory. So, there is
no submersion from S3 into ttt .

198
3304
Let M" and Nk be Riemannian manifolds. Then M x N is naturally a
Riemannian manifold with the product metric. If x\, ■ ■ •, xn are local
coordinates on M and yi, • • •, yjt are local coordinates on N, the product metric in
local coordinates (xi, • • •, xn, y\, • • •, yu) looks like
g= ( Sij(x)
0
0
9pg(y)
i) Let X be a vector field onMxJV "along" M (i.e., in local coordinates no
yp and -£— are present) and Y be a vector field along N. Show that DxY = 0.
ii) Show that at z £ M x N, some sectional curvatures always vanish, (e.g.,
product manifolds in the product metric never have strictly positive curvature.)
(Indiana)
Solution.
i) Using the properties of the Riemannian connection on M x N, we only
need to verify D a g|- = 0. In fact, observing that the inverse matrix of g is
of the form
g-i = ( 9'j{x) 0
0
gpq(y)
o.
and the first class of ChristofFel symbols of M x N satisfy
r - 1 (dgth dgph _ dgip\ _
l'p'h ~ 2\dyp + dx, dxj-^
Y- = 1 (dg" d9pr d&p\ - Q
,p'r 2 V dyp dx{ dxr)
we easily conclude that
""BVp j dxi g dVi "X dxi it dyq
Obviously, we also have D^g_g§- = 0.
ii) By the definition of the curvature operator, using the result of i), we
have
r(— — ^- -o r(——}— = o
\ dxi' dyp J dyh ' \ dxi' dyp J dyq
This means that R (gfr, ^-) = 0, Hence, for arbitrary X along M and Y
along N, using the C°°(M x iV)-linearity of the curvature operator, we have
R(X,Y,X,Y) = 0.

199
Therefore, the sectional curvature determined by X and Y always vanishes.
Obviously, here X may involve yp, and Y may involve X{.
3305
Let F : M —> N be smooth, X and Y be smooth vector fields on M and
AT, respectively, and assume FtX = Y; that is, that
F.p(X(p)) = Y(F(p)) for p G M.
a) Let w be a smooth 1-form on N. Define the Lie derivative iyw of w
with respect to Y. (If you use local coordinates, you must verify independence
of choice of coordinates.)
b) Prove that
F*(LYu) = Lx(F*u).
c) Let Z be a smooth vector field on M such that FtZ = W, where W is
a smooth vector field on N. Show that LyW = Ft(LxZ).
(Indiana)
Solution.
a) iyw is denned by iyw = d(Y [w)+ Y [du, where the symbol "[" denotes
the interior product of a vector field with a form, for example, if fi is a p-form,
then Y[Q is a (p — l)-form denned by
(Y^)(Y1,---,Yp_1) = Q(Y,Y1,---,Yp_1).
Thus, Ly w is a well-defined 1-form on N.
b) Let Z be a smooth vector field on M such that FtZ — W, where W is
a smooth vector field on N. Then we have
F*(LYw)(Z) = (F*(d(YH+ 11^))(^)
= (dF*(YLu))(Z) + (F*(Y[du))(Z)
= (dF*(w(Y)))(Z) + (YldwX^)
= (du(F,X))(Z) + du(F.X,F*Z)
= (d(F*w)(X))(Z) + (F*du,)(X,Z)
= (d(X[F*w))(Z) + dF*w(X,Z)
= (d(X 1^) + ^1.^^)(^)
= (IXF*W)(Z).

200
c) Noticing that LYW = [Y,W] and [FtX,FtZ] = Ft[X,Z], we
immediately obtain LYW = Ft(LxZ).
3306
Let r, 6, z be the usual cylindrical coordinates in St3. Let
w — [2rz sin 6+3z r cos 6>+5r2sin26>]d6>+[-2r cos#-f6.zr sm6]dz+[4zr2 sin6]dr.
Let the curve 7 be given in rectangular coordinates by
y(t) = (a;(i), y(t), z(t)) = (cosi,sini,4sin5i + sin2icos8i), 0 < t < 2x.
Evaluate Jr w.
(Indiana)
Solution.
Let a function / be denned by
f(r,6,z) = -2rzcos0 + 3z2rsin6>, (r, 0, z) G iR3.
Then / is a C°° function in iR3 and we have
df = [2r2sin6+3z rcos6]d6+[—2rcos6+Qzrsm6]dz + [—2zcos 0+3z2 sin 6]dr.
It is easy to see that /|7 is also a C°° function, and by the invariance of the
form of first order differentiation, the above expression of df is an exact 1-form
on the closed curve 7. Noticing that 7 is a closed curve, we have, by using the
Stokes' theorem,
J w = \{w-df)=. I 5r2 sin2 6d6 + (4zr2 sin 6 + 2z cos 6 - 3z2 sin 6>)dr
Without loss of generality, we may assume 0 < 6 < 2x and r = 1. Then
/■ /■2,r , f2* 1 - cos 26»
/ w = / 5sin26><26> = 5 / d0 = 5x.
J-, Jo Jo 2

201
3307
Let M be a C°° Riemannian manifold. Assume the theorem that there is
a unique C°° mapping V : X(M) x X(M) -> X(M) denoted by V : (X, Y) -*
Vj^y which has the following linearity properties: For all /, g £ C°°(M) and
X, X', Y, Y' E X(M), we have
Vfx+gx'Y = f(VXY)+g(Vx.Y),
Vx(fY+gY') = fVxY +gVxY' + (Xf)Y + (Xg)Y',
[X,Y] = VxY-VyX,
X(Y,Y') = (VXY,Y') + (Y,VXY').
a) Suppose a : [a, b] —+ M is a smooth curve. Define what it means for a
vector field Y on a to be parallel along a. Derive the differential equations
that must be satisfied if Y is parallel along a.
b) Let X, Y £ X(M). Let p G M and let a : [0, b] —> M be an integral
curve of X such that a(0) = p, da/dt = X(a(i)). Show that
(V* r)(p) = jt[P-ltY(a(t))}t=0,
where Pa-,o,t '■ Ta^M —► Ta^M is the parallel transport along a.
(Indiana)
Solution.
a) As known, thus defined mapping V is the Riemannian connection. Using
the properties of V, we can prove that V±aY is completely determined by a(t)
and Y(a(t)). So Vdo, Y is well-defined for every vector field along a.
Now we give a definition that a vector field Y on a is parallel along a, if
and only if Via. Y = 0, \/t £ [a, 6]. In order to derive the differential equations,
we choose a coordinate neighborhood with local coordinates {x'}. Then
Denoting
\7 A-Vr'i-
k

202
we have, by the properties of V,
V = E^(yiw^(«w))
E
i
E
dYJ(t) 3
dt dx~i
(«(*))+ ^WV^—(«(*))
*m+v^a)^YHt)
_d_
dxi
*(«(*))•
Therefore, the desired differential equations are
dYk(t)
dt
+ E(^-)^^W = o, v*.
•J
d<
b) Choose a basis {ei, •••,en} of Tp(M), where n = dimM. Let
e,-(i) = Po,-o,t(ei)-
Since Po;o,t is a parallel isomorphism for every t, then {ei(i), • • •, en(t)} is a
basis of T0(t)(M). Hence we can denote
*
Then, by the properties of the connection V and by the fact that e,-(i) is
parallel along a(t), i = 1, • • •, n, we immediately have
On the other hand, because
is equivalent to
we can write

203
Therefore,
A /7V'
3308
Let M be a Riemannian manifold with the property that given any two
points p,q £ M the parallel transport of a vector from TpM to TqM is
independent of the curve joining p and q. Prove that the curvature of M is
identically zero, i.e., R(X, Y)Z = 0 for all X, Y, Z £ X(M).
(JncfoaTia)
Solution.
For an arbitrary point p £ M, take a coordinate neighborhood D oip with
local coordinates a;1, • • •, xn. Let
i
be n linearly independent vectors in TpM. Using the hypothesis that the
parallel transport of a vector from TpM to TqM is independent of the curve
joining p and q, then for q £ D, we can transport every Vpi from 3],M to
TqM. Thus, we can define n linearly independent vector fields Vi,---,Vn
in D. Obviously, all Vt's are well-defined, and if we transport along special
coordinate curves, then
• = **<» = £ (¾^+E^.w) stw^ E^wstw.
fc \ / / fc
namely
Now, by the Ricci identity, we have
<jm(z> - <mjw = -2^)4--.00 = o.
i
Noting the linear independence of VJ's, we immediately have, in D,
R^jm = 0 k,l,j,m= 1, ---,71,
i.e., the curvature of M is identically zero.

204
3309
Let M be an n-dimensional Riemannian manifold. Suppose there are
n orthonormal vector fields X\,- • • ,Xn that commute with each other (i.e.,
[X{, Xj] = 0; i, j = 1, • • •, n), show that the sectional curvature of M is
identically zero.
(Indiana)
Solution.
Set
k
Then, from (Xi,Xj) = 6;j it follows that
(v^x,-, ^) + (:^^) = 0,
i.e., T}k. +T'k. = 0; i,j,k = 1,---,71. On the other hand, from [Xi,Xj] — 0 it
follows that
Vx.Xj~Vx.Xi = ^,^1 = 0,
i.e., Tkj - Tj, = 0; i, j, k = 1, •••,"■• Namely, the ChristofFel symbols T^-'s
are antisymmetric with respect to k, j, and symmetric with respect to i,j.
Therefore,
pfc _yj —V? F' T"" F^ F*
1 ij ~~ L ik ~~ ki ~~ Lkj — L jk ~ j* ~ U'
which means r*,- = 0; i, j, k = 1, • • •, n. Hence the sectional curvature of M is
identically zero.
3310
Let X be a smooth vector field on a Riemannian manifold M. The
divergence of X, denoted by div(X), is denned by the function trace (VX).
i) If M is closed (i.e., compact without boundary), show that
JM
div(X)dv = 0.
IM
ii) If M is compact with boundary dM, show that
f div(X)dv = f (X, N)ds,
JM JdM

205
where N is the outer normal vector field of dM.
Hint. Consider w £ A1(M) denned by u>(Y) = (X, Y), and try to use the
Stokes' Theorem.
(Indiana)
Solution.
In fact, this problem is the famous Green theorem. The outline of the proof
is as follows.
Firstly, one can show that, for example, by means of the normal coordinate
system about p £ M, thus denned div(X) satisfies div(X)flM = d(i(X)£lM),
where Qm ls the volume form of M and i( ) is the interior product operator.
Next, for p £ dM, choose an oriented orthonomal frame field about p
{ei, • ■ • ,en} such that, at p, e\ = Np. Let {u1, ■• ■ ,wn} be the dual frame field
of {ei, •••,en}. Then the volume elements of M and dM are respectively
dv = Ojif(p) = uj1 A ••• Aw™, ds = £ldM{p) = w2 A ■ ■ ■ Aw™.
Observing that, at p,
i(X)Clm = i(X)(w1A---Awn)
= w1(X)w2 A'"Aw" + (terms involve w1)
= (X, N)£Iqm + (terms involve w1),
and along dM w1 = 0, one can obtain, by Stokes' theorem,
f d\v(X)dv= f d(i(X)nM)= f (X,N)CldM= f (X,N)ds.
JM JM JdM JdM
If M is compact without boundary, the right hand side of the above formula
vanishes naturally.
3311
Let w1, • • •, wk be one-forms. Show that {w' }f=1 are linearly independent
if and only if to1 A w2 A ■ ■ • A wk ^ 0.
(Indiana)
Solution.
Let w1, • • •, wk be denned on an n-dimensional manifold with n > k.
Suppose w1 A • ■ ■ A wk ^ 0. If w1, ■ ■ ■, iu* are linearly dependent, then
without loss of generality, we may suppose that
wk = axw1 H h ajt-iiu*-1

206
with suitable functions ai, ■ ■ ■ ,a,k-i- Thus we have
ffi'A-Au^ij'A-A wk~1 A (aiw1 + •••+ ak-iwk~1) = 0,
which contradicts the above hypothesis.
On the contrary, if w1, • • •, w are linearly independent, then we can extend
them to a basis {w1, • ■ ■ ,wk,wk+1, • • •, wn}. Thus
implies that
u'A'-Ab'a wk+1 A • ■ ■ A iufc ^ 0
w1 A ■ ■ • A wk ^ 0.
3312
Let M be a Riemannian manifold. Let p £ M.
(a) Show that there exists 6 > 0 such that
exPp :B6(Q)cTpM -+M
is a diffeomorphism onto its image.
(b) Show that there exists e > 0 such that expp(jBe(0)) is a convex set.
Hint, let d(x) = distance from x to p. Show that d2 is convex in a
neighborhood of p.
(Indiana)
Solution.
(a) This is just the existence of the normal neighborhood of p. Use the fact
that deup is nonsingular at p, and then the implicit function theorem.
(b) The existence of convex neighborhoods is a classic result due to J. H.
C. Whitehead. Refer to every standard textbook on differential geometry.
3313
Let
A = < I , J : a, b, c, d G ffi, ad — be = 1 >
Show that
(i) A is a differentiable manifold.

207
(ii) A is a Lie group with the standard matrix multiplication as a product.
(Indiana)
Solution.
Define a map F : Gl(2,M) -* Gl(l,M) by F(X) = detX. Then F is a
smooth homomorphism between Lie groups, and the rank of F is constant.
Therefore, the kernel of F
baF = F-1(l) = A
is a closed regular submanifold of Gl(2, M) and thus a Lie group.
3314
(a) Let / be a smooth function on a Riemannian manifold M. Let grad/
be the vector field defined by the equation
(grad/, v)p = dpfv, v G TpM.
Let (a;1, ■ ■ ■, xn) be local coordinates around p. Find the expression for gradf
in terms of x1, • ■ ■, xn.
(b) For a vector field X define the divergence of X, div(X) as the trace
of the operator Y —► DyX where D is the Levi-Civita connection. Find the
expression for the divergence of X in a local coordinate system (a;1, ■ ■ •, xn).
(c) Use (a) and (b) to find the expression for the Laplacian A in local
coordinates, where A acting on a smooth function / is defined by A/ = div(V/).
(Indiana)
Solution.
For convenience, we omit the suffix p.
(a) Let
grad/ =£<*'£/•
Then, from
it follows that
<grad/'a^> = £aWfK^
9 ^ . V^.I_ ._ * ( d \ _ d/
dxh
a -^ ft?"

208
Thus
dxk dx
grad/^^^I^A
k,l
(b) Denote
* = E**#--
I-*1 dx'
i
Then, by the definition of divergence, we have
k,l i \ k
(<0
A/ = div(grad/) = ^
a? (X>,fc^Fj + Er^fc,^r
i,k \ m
3315
Let M be a compact connected Riemannian manifold without boundary.
Let / be a smooth function satisfying A/ = 0. Show that / = const.
Hint. Use the definitions in Problem to show that div(/V/) = | V/|2 + /A/.
(Indiana)
Solution.
For any function /, by straight calculation, we have
div(/V/) = |V/|2 + /A/.
Now, noticing the hypothesis of this problem, by Green's theorem we have
/ |V/|2^ = 0,
Jm
which implies df = 0 everywhere, that is, / = const.

209
3316
Suppose F : M —► N is a smooth map between difFerentiable manifolds, and
is homotopically trivial. Show that in this case, F*uj will be exact whenever
w is a closed l-form on N. (Note: F is homotopically trivial if it extends to a
smooth mapping F : M x [0,1] —> N such that F(x, 0) = F(x), for all x G M,
while -F(:c, 1) = q G N (q constant) for all x.)
[Indiana)
Solution.
Consider the map G : M —► {q} and the inclusion i : {q} —> N. Then the
map F : M —> N is homotopic to the composition ioG. Furthermore, we know
that F* and (ioG)*=G*oi* induce the same homomorphism
F** = (ioG)** :H1(N,d)-^H1(M,d).
Since i*(Z1(N,d)) c ^({¢),(2) = 0, the induced homomorphism F** = (i o
G)** is a zero homomorphism, i.e., F*(H1(N,d)) = 0. In other words, for
every w G Z1(N,d), F*w G S1(M, d), namely, F*w is exact.
3317
Regard St9 as the space of all 3 x 3 matrices with real entries. Does the
subset
S = {4 G M9 : det(A) = 0}
form a smooth submanifold of iR9?
(Indiana)
Solution.
Observe that {A G St9 : rank.A < 1}, the union of all axis in ttt9, is
closed. Then M := M9\{A G St9 : rank.A < 1} is an open submanifold of
M9. Define a map F : M -> M1 by .^(4) = det(,4), A G M. Noting that
dF = (^11,^12,^13,^21,^22,^23,^31,^32,^33) where Aij is the algebraic
complement of the corresponding entry a,j of j4, we see that i&nkdF = 1
on M. Therefore, F_1(0) = {.4 G M9 : det(A) = 0} is a closed regular
submanifold of M. Hence, it is also a submanifold of M9.

210
3318
Let
w = X\dx2 A dxz A dx± + x2dx\ A dx3 A dx± + x3dxi A dx2 A dx4.
Compute /s3 w, where S3 = {x £ ]R4 : \x\ — 1}, oriented as the boundary of
the unit ball (assume standard orientation on HZ1).
(Indiana)
Solution.
Denote D4 = {x £ M4 : |x| < 1}. Then, by the Stokes theorem we have
I w = / w = I dw
Js* JdD* Jd*
= / (dxi A dx2 A ^0:3 A dx4 -f ^2 A dxi A (¾ A ^4
Jd*
+dx3 A cfo;i A dx2 A ^0:4)
/ (¾ A dx2 A ^3 A dx4 = vol(D4) =
x
2
3319
Prove that
{(^2,^4): rank (( £ ^))=1}
is a three-dimensional submanifold of M4.
Solution.
(Indiana)
Define a map F from iR4\{(0, 0, 0, 0)} into M1 by
F(x1,x2,x3,x4) =
£1 0=2
Z3 ¢4
= 0=10:4 - 0:22:3.
Then i*1 is a C°° map with (£4 —0:3 -0=2 £1) as its Jacobi matrix which has
constant rank 1 on iR4\{(0, 0, 0, 0)}. Thus
F-\0) ={(Xl,x2,x3,x4): rank (( ** ** ))=1}

211
is a regular submanifold of iR4\{(0, 0, 0, 0)} with dimension 3. Noting that
iR4\{(0,0, 0, 0)} is an open submanifold of M4, we see that
{(*!, x2, x3, x4): rank ((^ ^))=1}
is also a three-dimensional submanifold of M4.
3320
Let vector fields Xi,X2 on M4 be denned by
y d _l d v d , d
OX2 OX3 OXi OX4
i) Is there a 2-dimensional submanifold M2 of iR4 such that for each p G
M2,X1(p),X2(p)eTpM2?
ii) Is there a nonconstant function / in the neighborhood of 0 G M4 such
that XJ = 0 and X2f = 0?
(jTirfiaTia)
Solution.
i) No, there is not. The reason is that the bracket [Xi,.^] = -£- -£—
does not satisfy the Frobenius condition.
ii) Yes, for example, we can set / = X\x2 — (x3 + £4).
3321
Let M = {(¢,1/) : x,y G M3, \\x\\ = 1, \\y\\ = 1, {x,y) = 0}.
i) Show that M is a smooth compact embedded submanifold of M6 and
explain how M can be identified with the unit tangent bundle of S2.
ii) Show that M is orientable.
(Indiana)
Solution.
i) Identify M6 with {(x, y) :x,y G M3} and define a map F : M6 -+ M3 by
F(x,y) = (f1,f2,h) = (\\x\\2,\\y\\2,(x,y)).
It is easy to verify that the Jacobi matrix of the C°° map F
fitf f f\ \ 1 2xi 2x2 2x3 0 0 0
0(fi,j2,l3) \_| Q Q Q 2j/i 2^2 2^3
ato.sa.sa.I/i.Ifc.ttO, x yi y2 y3 Xl X2 X3

212
has constant rank three when /i = ||a;||2 = 1, /2 = \\y\\2 = 1 and /3 = {x,y) =
0. Therefore, F'1 (1,1,0) = M is a closed embedded submanifold of Et6.
Besides, since ||(£,2/)|| = (|M|2 + ||2/||2) = = V% which means M is bounded in
M6, M must be compact.
Naturally, for every (x, y) £ M, if we regard x £ S2 and y £ TX(S2) with
\\y\\ = 1, then we can identify M with the unit tangent bundle of S2.
ii) Because S2 is orientable, S2 has a covering {(Ua,<f>a)} of coherently
oriented coordinate neighborhoods. By using identification of M with the
unit tangent bundle of S2, for every (x,y) £ M, x has the local coordinates
(ua,u2) £ Ua, and y is uniquely determined by the oriented angle 6a at x
from g^j- to the unit tangent vector y £ TX(S2). Thus, (C/0 x Ia,<j>a x V'a) is
a coordinate neighborhood of (a;, y) £ M, where J0 = (6a — e, 6a -f e) with e
being a suitable positive real number and ij>a is the map from (x, y) to 0O.
If (Ua x J0, <^0 x V>a) H (C//3 x 1/), (j)/} x ^/3) ^ 0, then the transition function
has the following Jacobian
d(ulul,6a)
3(1^,11^,6/3)
* 1
d«,<)
which means that {(Ua x Ia,$a x ij>a)} form a covering of coherently oriented
neighborhoods. Hence, M is orientable.
3322
Let F : M —+ N be a local isometry between connected Riemannian
manifolds M and N. Show that if M is complete, so is N and F is a covering
map.
(Indiana)
Solution.
Because F is a local isometry, F(M) is open in iV. If y is a limit point of
F(M) in JV, then there is a point a; £ M such that there exists a geodesic in
i\T connecting F(x) and j/. The local isometry of F and the completeness of M
imply that the above geodesic can be uniquely lifted to a geodesic in M starting
from x, and the image of its end point under F must be y. Therefore, F(M) is
also closed in N. Thus, the connectedness of N implies that F(M) = N, i.e.,
F is surjective. Besides, since F maps every geodesic of M into a geodesic of
N, the Hopf-Rinow theorem means that N is complete, too.
Next, we show that F is a covering map. For every x' £ N, take S > 0
so small that exp^., : B'(6) —> B'6 is a diffeomorphism, where £'(($) = {^ £

213
TX<(N) : \v\ < 6} and B'6 = {y' G N : d(y',x') < 6}. Since F is a local
isometry, F~1(x') is discrete. Denote F~1(x') = {xa} C M and set Ba(S) =
{v G TIo(M) : |i/| < 6} and 5° = {y G M : d(i/,a;Q) < 6}. Then, we claim
that B'6 is an admissible neighborhood of x' and i*1 is a covering map.
Firstly, we claim that F-1(B'6) = [j Bf. In fact, if z G f_1(BJ), then
there is a unique geodesic 7 : [0,1] —> jB£ such that 7(0) = F(z) and 7(1) = x'.
Since i*1 is a local isometry, there exists a geodesic 7 : [0,1] —+ M such that
F(j(t)) = 7(2), \/t. Hence ^(7(1)) = x' and 7(1) = xa for some a. Besides,
L(j) = L(f) < S means that z = 7(0) G Bf, i.e., F~'L(B'6) C \jBf. On the
a
other hand, (J B" C ir_1(JB^) is obvious. Thus the claim is proved.
a
Secondly, we say that for any a, F : B" —> B'6 is a difFeomorphism. Since
M is complete, we have the following commutative diagram
B"(S) °i—~B\&-)
exp exp
! I
S° -Bl
and expj. is surjective. Besides, we know that dF, expx, are diffeomorphisms.
Therefore, F 0 exp,,. = exPa;' °dF is a difFeomorphism and hence exp^. is
an immersion. So, exp^. is also a difFeomorphism. Hence F = exp^./ odF 0
(expj. )_1 is a difFeomorphism.
Thirdly, we claim that if a ^ /3, then Bf <1 B% = 0. Otherwise, if there
is 2: G S" n 5^, then there exist unique geodesies ja, 7/3 connecting z with
a;0, xp respectively. Let 7 be the unique geodesic in B'6 connecting F(z) and
x'. Because both F : B" —> B'6 and F : B^ —+ B'6 are isometric, we have
F(ja) = 7 = F(f/}). In other words, 70,7/3 are the lifts of 7 through 2. The
uniqueness implies ya = 7/3. In particular, xa = 7o(l) = 7/3(1) = a;/3, which
contradicts a ^ (3.
3323
Let F : M —> M be an isometry of a Riemannian manifold M.
i) Show that each component of X = {x G M : F(x) = x) is an embedded
totally geodesic submanifold of M.

214
Hint. Use exponential coordinates.
ii) Give an example in which the components of X have different
dimensions.
(Indiana)
Solution.
i) First, we show that X has submanifold structures. For x G X C M,
set B(S) = {v G TX(M) : |i/| < 6} and B6 = {y G M : d(x,y) < 6}, where
S is so small that expx : B(S) —+ B{ is a diffeomorphism. Define V = \y G
TX(M) : dF(v) = iv}- Thus, V is a subspace of TX(M). Then we claim that
XC\B( = expa.(VnJB((5)). If this is proved, because expx(Vn B(8)) is obviously
a submanifold of M, we can assert that X has submanifold structures. In
order to prove the claim, we first assume that y = X 0 B( and v G B(S)
such that expx v — y. Let 7 : [0,1] —+ M be the unique shortest geodesic
7(i) = expx(tv) connecting x and j/. Since x,y E X, and i*1 is an isometry,
^(7) is also a shortest geodesic connecting F(x) = a; and F(y) = j/. Thus
the uniqueness implies ^(7) = 7. In particular, dF(j(0)) = 7(0), namely,
dF(is) = v. Therefore, v G V which means that y G expx(V <1 B(6)), i.e.,
XDB( C expx(V n B(6)). On the other hand, suppose that v G V(1 B(S) and
j/ = expx v. Let the geodesic 7 : [0,1] —+ M be denned by 7(2) = expx(tis).
From dF(v) = 1/ follows ^(7(0)) = 7(0). Then, that F is an isometry implies
F(-y) = 7. In particular, F(y) = F(-y(l)) = 7(1) = y, which means that
y£X<lB6, i.e., expx(V <1 B(S)) CXC\B6.
Next, we show that every geodesic 7 : (a, b) —+ X parameterized by ar-
clength is also a geodesic of M. For any so G (a, 6), let £(s) be a geodesic
of M such that ((s0) = 7(so), C(so) = 7(so)- Since F(((s0j) = ((s0),
dF(C(so)) = C(so) and ^ is an isometry, then .F(C) and ( are two geodesies of
M which satisfy the same initial conditions. Therefore, ^(¢) = ¢, i.e., £ lies
in X. Besides, ( is naturally a geodesic of X. Thus, in a neighborhood of so,
( = 7. Because So is arbitrarily chosen. 7 is a geodesic of M. Hence, X is
totally geodesic.
ii) Let
M = {(x, y, z) G M3 : y > 0, z = 0} U {(a;, y, 2:) G M3 : x = 0, y < 0},
and F be a reflection with respect to the plane z = 0, i.e., F(x,y,z) =
(0:,1/,-2:). Then M is a 2-dimensional manifold, F is an isometry, and
X = {(x,y,z) EM3 :y>Q,z = 0}Ll{(x,y,z) £ M3 : x = Q,y < 0,2: = 0}.

215
3324
Compute the de Rham cohomology groups of the circle S1. Do so directly;
i.e., without citing the de Rham Theorem.
(Indiana)
Solution.
since B°(S1, d) = 0 and S1 is connected, then
H°(S\d) = Z°(S\d) = {/ G C0O(S1,iR1) | df = 0} = M1.
For Hk(S1,d),k > 1, since there are no non-vanishing fe-forms on S1, we
have Zk(S1, d) = 0 and hence #fc(S\ d) = 0.
Besides, observe that
Z1(S1,d) = C0O(S1,A1(S1)), B\S\d) = {df\fe C0O(S1,iR1)}.
Let 0 be the polar coordinate characterizing S1. Then ^ is a non-vanishing
vector field on S1. Let d6 be its dual non-vanishing 1-form on S1(Caution:
Here d0 is only a formal symbol, because 6 in usual sense is not a globally
well-defined function on S1), and it is not exact. For every w = g(6)d0 G
C0O(S1,A1(S1)), define a function
Because £1(0) = fl(2ir), Cl is globally well-defined on S1. Hence, denoting
1 f2K
2t Jo
we see that u> — Cd6 = dfi, i.e., w — Cd# is exact. Therefore,
Hx(S^,d) = .^(S1, (2)/51 (S\d) = {CdB \ C G iR1} = M1.
3325
Let X denote a submanifold of Euclidean space TEn, and set
UeX := {a;+ ^ : a; G X,i/ G iV^X, |iv| < e},
B(X,e) := {yG En : \y - x\ < e for some x G X}.

216
Show that UeX C B(X,e) for all e. Show that the two are not generally
equal. (Consider examples of 1-dimensional submanifold in IE2.) Can you
give conditions which imply equality?
(Indiana)
Solution.
Setting y = x -f v immediately implies that UeX C B(X,e).
Let X be an open line segment in IE2. Then considering the boundary of
X, i.e., the end points, can show that UeX is a proper subset of B(X, e).
If X has no boundary, e.g., either compact or complete, then UeX =
B(X,e).
3326
Consider a Riemannian manifold (M, </). Call a vector field Z on M a
killing vector field if Z generates a 1-parameter group of isometries of M.
i) Show-that when Z is Killing, we have Lzg = 0, i.e.,
Z(g(X,Y)) = g(LzX,Y) + g(X,LzY) (**)
for all vector fields X and Y on M. Here Lz denotes the Lie derivative along
Z.
ii) Show that the expression (**) above is equivalent to
g(VxZ,Y) = -g(VYZ,X),
where V denotes the Levi-Civita connection for (M, g).
(Indiana)
Solution.
In local coordinates (a;1, • • •, xm) of (M,g), let the 1-parameter group of
m
isometries generated by a vector field Z = ^ z' ^- be expressed by i1 =
j = i
^(x1, ■■ ■ ,xm;t) := x*(x,t) such that
m m
^2 gij(x)dx'dxJ = ]P gk,(x)dxkdxl,
ij=l k,l=l
where gij(x) = g (^-, gfj), and ^ = ^(x^) satisfies
dx*
x,(x,Q) = x', —r|t=0=.z*.

217
Thus we have
m c—k a—l
i s v^ i—\Oxr ox
k,l=l
Differentiating the obtained equality with respect to t and then setting t = 0,
we obtain
m /5<7" dzk dzk\
K = l x '
which can be written as
9kjz* + gikZj = 0,
or equivalently
9 fo + Z, ~) + g (^-, V + z) = 0.
\ «*■ ox* J \ox' b*j J
From this follows what we desire in ii).
Noting that the Levi-Civita connection V satisfies
Z(g(X, Y)) = g(VzX, Y) + g(X, VZY)
and the Lie derivative satisfies LzX = [Z,X], we easily obtain (**).
3327
Let M2 be a connected Riemannian manifold and X, Y complete vector
fields on M. Assume that the flows Xt and Yt are isometries of M for all t.
i) Show that the integral curves of X are curves of constant geodesic
curvature.
ii) Assume that X and Y are linearly independent at all x £ M2 and that
their flows commute Xt o Y„ = Y„ o Xt. Conclude that (X,X), (X,Y) and
{Y, Y) are constant on M.
(Indiana)
Solution.
i) For every p £ M, take a local coordinate neighborhood about p such
that the coordinate curves are the integral curves of X and their orthogonal
trajectories. Namely, we may assume that X = X1-^ and </i2 = (-g^r, g^j) =
0. Because the flow Xt of X for every t is an isometry, the vector field X
should satisfy the Killing equation X,j + Xjj = 0; i, j = 1, 2, or equivalently,
^fvkdgij dXk dXk\ n . . , „

218
Taking i = 1, j = 2, we obtain gu^gr = 0, i.e., X1 = X1(x1). Now, make the
following coordinate transformation
d(x1,x2)
x-2 _ ^2 a/„l _5\ 71 u-
If we still adopt the original notations, then the vector field X = ^fr- Hence
from the corresponding killing equation, taking i = j = 1 and i = j = 2, we
obtain
dgn _ n %22 _
da;1 ~ ' da;1 ~ '
which means that the metric of M2 about p can be written as
ds2 = ffn^Xds1)2 + 022^)(^2)2.
Then, using the Liouville formula, we can compute the geodesic curvature of
the a; ^coordinate curve as follows:
d6 1 dlngn 1 d\ng22 . „
kg = 3---.— a , cos^1 + — . sing
ds 2y^22 da;2 2y^ii da;1
- ' dln9ll=kg(X%
2Jg^i dx2
where # = 0. Therefore, along every integral curve of X, -^ = 0, i.e., kg ~
const.
ii) Analogously, about p, take the integral curves of X and Y as the a;1
and x2 coordinate curves, respectively. Then, we have locally X = X1 ~^r,
Y = Y^g^y. Because their flows are commutative Xt oY, — Ys o Xt, that is
equivalent to
rxyi-x1^ d Y2dXl J--o
[A,rj~X dxiW~Y ~Wdxi-0,
we immediately have
dY2 _ d_x^ _
a?" ~ ' 'dx2 ~ °-
Therefore, we can make a suitable coordinate transformation and then, if
adopting the original notations, X = g^r, Y — -^. Again, by the
corresponding Killing equations, we can obtain -^fc = 0; i,j,k = 1,2, i.e., all </,j's
are constants. Noting the expressions of X and Y, we obtain
(X,X) = 3n = const, (X,Y) = ^12 = const, (Y,Y) = g22 = const.

219
3328
Let M be a compact Riemannian manifold without boundary. For any
/ G C°°(M), define V/ G X{M) and A/ G C°°(M) as follows:
At any p G M, choose an orthonormal frame field {ei, •••,e„} around p
and then define
(v/)(p) = Ete/x*) md (A^)(p) = - Em*/) - (v««e.-)/]-
* i
Verify first that V/ and A/ are well denned (i.e., they do not depend on the
choice of orthonormal frame) and then show that
y*/A/=y*nv/n2.
(Indiana)
Solution.
Let {e{,- •• ,e„} be another orthonormal frame field around p. Then we
may suppose that
i
for suitable functions a\, i,j= 1, • • • ,n. Noting
(e;,e,) = (e*,e*j) = 6ij,
we have
E ai al = **« E ai4 = s'k •
3 J
Using these equalities and the properties of Riemannian connection, we can
easily obtain
E>.-/) = E(e.w<
i i
Em*/) - (v^.-)/] = E[e.*(e.*/)-(v«re.*)/]-
i i
Hence V/ and A/ are all well denned.
Using the definition of the Laplacian here, through direct claculation, we
have div(/V/) = ||V/||2 - /A/. Then Green's theorem implies
y*/A/=y*||V/||2.

220
3329
Let p(xi, x2, x3, x4) = (x\ + x\)(x3 + x\) and 5(0:1, x2,x3, x4) = x\ + x\ -f
x3 + x\. Define
Sa,b = {x £ M4 I p(x) = a and 5(0;) = b}.
For what a, 6 > 0, is Sa,b a manifold? Explain.
(Indiana)
Solution.
Denote a = x\ + x\ and f3 = x3 + x\. Then a/3 = a and a + /3 = 6
means that a and /3 are two roots of the equation A2 — bX + a = 0. Therefore,
62 — 4a > 0 is the prerequisite condition.
1. If a = b = 0, then £1 = £2 = £3 = X4 = 0. Thus So,o is a 0-dimensional
manifold.
2. If a = 0, b ^ 0, then
<So,& = {x £ -K4 I £1 = £2 = 0, x\ + £4 = 6 or x\ + £2 = b, x3 = x4 = 0}.
Using the theorem of closed regular submanifolds proved by rank theorem, we
can easily show that So,b is a 1-dimensional submanifold of ttt4.
3. If a ^ 0, b ^ 0, then when b2 - 4a > 0
■Sa.fc = < x e M
1-
2 2 _ ^ + vP—4a 2 2 _ ^ — \/&2 — 4a
, , 6 — \fb2 — 4a 2 2 6 + V&2 — 4a
or x{+x^= , x3 + x4 =
and when b2 — 4a = 0
Sa,6 = {a; G iR4 I x\ + x\ = x\ + £4 = -}.
b,
Analogously, we can show that they are all 2-dimensional submanifolds of M4.
3330
Let F : M —> N be a C°° map between two C°° manifolds. Assume that
F is onto. Let X be a smooth vector field on M.
(i) Show by an example that dF(X) may not be a vector field on N.

221
(ii) Suppose Y = dF(X) is a smooth vector field on N. Show that F takes
integral curves of X into integral curves of Y.
(iii) Suppose X\, Y\, and X2, Y2 are related as X and Y in (ii) above. Show
tha.tdF([X1,X2}) = [Y1,Y2}.
(Indiana)
Solution.
(i) Let M = {(¢,1/) : x,y G M], N = {x : x G ffi}, X = (x2 + y2)-^,
and i*1 : M —> AT be denned by F(x,y) = a;. Then, if y\ ^ y2, we have
dF(X(x,y1)) zfi dF(X(x,y2)). Therefore, as a vector field, dF(X) is not well
defined in every point of N.
(ii) Let a(s) be an arbitrary integral curve of X. Then, along a, we have
da (Jj) = X. Hence,
which means that F takes integral curves of X into integral curves of Y.
(iii) Using local coordinates, by direct computation, one will obtain the
desired equality.
3331
Let Mn be a Riemannian manifold. Show that whenever / : M —+ ttt is a
smooth function, there is a unique vector field V/ (called the gradient of / on
M) such that
<V/(p),7(«)> = |/(7(<))
whenever 7 is a smooth curve in M with 7(2) = p.
(Indiana)
Solution.
Let (a;1, ■ ■ •, xn) be the local coordinates about p and set p : (xq, ■ • ■, x%).
If y(t) is the i-th coordinate curve that passes p, i.e., 7(2) has the following
expression
x'=t, xj = xJ0 (j^i),
then, by
<V/(P),7(0> = |/(7(0) =j£

222
we have the expression
where g'^'s are the components of the matrix [g,j]-1 = [(^7, afj)]-1- ^ is
easy to verify that the above expression of V/(p) is independent of the choice
of local coordinates.
3332
The space MnXn of n x n real matrices forms an n2 dimensional Euclidean
space, in which the dot product between A = [a,j] and B = [6,j] is given by
n n
(A,B):=J212a'Jb'J-
i=ij=i
Let Sn_1 be the unit sphere in Mn, and define a : Sn_1 -+ MnXn as the map
sending x = (xi, • • ■, xn) in Sn_1 to the symmetric matrix <r(x) = -75^¾].
i) Show that a maps Sn~1 into the sphere of radius 4= centered at the
origin in MnXn.
ii) Prove that a is a local isometry (i.e., the pull-back via a of the dotpro-
duct metric defined above on MnXn is the standard one on S™"-1).
(Indiana)
Solution.
n
i) Let x = (si, • • •, xn) £ S""1. Then £ a;? = 1. Hence
1 " 1
(<r(x),<7(x)) = -5^ z,?z2 = 5,
ij=l
which means that <r(x) is on the sphere of radius -y= centered at the origin in
Mnxn.
ii) Let i : Sn_1 -+ Mn be the standard inclusion map, and t : Mn -+ HUlxn
be denned by (xi,--• ,xn) —> [ytj] = -fy?ixj]- Suppose that
^ dz/ ^ dx'
•=i «=i

223
belong to Tx(Sn_1). Then we have
n n
^TVz,- = ^w'xi = o.
i=l
Therefore
dcr(v) = dr o di I V^ wfc -—
\k=i dXh)
a
Hence we have
1 _"_ n
(do-(v),do-(w)) = - 2^ ("'^J +^^.)(¾1¾ 4-ii/a;,-) = ^^v'w' = (v,w).
3333
Let M be the Riemannian manifold obtained by equipping Mn with a
metric conformal to its usual one; i.e., a metric of the form \gij] = e2^[<5,j],
where / : Mn —► .22 is a smooth function, and (5^- is the Kronecker delta. Let
d = ^7 denote the standard coordinate basis vector fields.
i) Show that for arbitrary indices i, j, k £ {1,2, • ■ ■, n}
ii) Show that when n = 2, the sectional curvature of M along an ei,e2
plane at p is given by
_--2/(p)
5a;i <9a;|
^2+^3- 0>)-
(Indiana)

224
Solution.
i) Set
/=1
From gij = e2f8ij it follows that
<Ve,.e,,efc> = ^,,^ = 2^-^ + ^7
ii) The sectional curvature of M along an ei,e2 plane at p is just the Gauss
curvature of M at p. Noting that E = G — e2f, F = 0, from the Gauss
equation we obtain
K{p) = l
EG
((VEh\ , f(VG)i\
[{ Vg )2 + { Ve)x
—(g+gw
3334
Let X, Y be complete vector fields on a manifold M and let Xt, Yt be the
flows induced by them.
i) Show that X, oYt = Yt o Xs ioi ail s,t £ St implies [X, Y] = 0.
ii) Prove the converse to i).
(Indiana)
Solution.
Observe that for a diffeomorphism F : M —► M, the complete vector field
Y is F-invariant, i.e., FtY = Y, if and only if F o Yt = Yt o F, Vt G iR.
i) Suppose that I, o 7( = Yt o X, for all s,t G -ffi. Since Xs is a
diffeomorphism for each s, then Y is X,-invariant. Thus
[X, Y] = Ix Y = lim V - X„Y] = 0.
«—>o s
ii) Now suppose that [X, Y] = 0. Then
0 = X„[X,Y] = [X„X,X„Y] = [X,XstY] = LX(X,.Y).

225
Hence for each p G M and any / G C°°(p), we have
0 = (Lx(X„Y))pf = Kmoj-[(Xs.Y)pf -(XAst(XstY))pf]
= i!m0 -h[ix"Y)pf ~ (Xe+*>>Y^ = -fs{x**Y)pf
for all s£ffi. Therefore,
(XstY)pf = (X0tY)pf = Ypf.
Since / is arbitrary, we obtain XstY = y, namely, y is X,-invariant. Hence
X,oYt = Yto Xs for all s,t £ M.
3335
If <j> '■ Gi —► Gi is a Lie group homomorhism, show that for all v G LG\,
we have
<^(exp(w)) = exp(<^„v).
(Indiana)
Solution.
Note that exptv, t G M1 is a 1-parameter subgroup of G\ generated by
v G LGi. Since <^ is a Lie group homomorphism, then <j)(exptv) is also
a 1-parameter subgroup of (¾ generated by a suitable w G £(j2, namely,
<^(expiv) = exptw.
Jjet \X , • • • , X m) and (y1, • • •, yn) be local coordinates of Gi and (¾ about
the identities e± and e2, respectively. Locally, <^ can be expressed by
<f,(x1,---,xm) = (<f,1(x1,..-,xm),---,r(x\...,xm));
and exptv and expiu; can be denoted by
(a;1 (t),-,xm(t))
and
(<f>i(x\t),---,xm(t)),...,r^1(t),---^mm
respectively. Assuming that
m „ n „
1=1 0=1

226
we have
V dt Jt=0 \ dt /t=Q
Therefore,
i=i a=i \ox / x=x{o) °lr
m n
_ ^-^ ^-a J dx' \ ( d(j)a
ti^Adt ) «=o V dx* Jx=x{0) dya
Furthermore, by left translation, we have d<f>(v) = w, i.e., <f>tv = w. Hence
<j)(exptv) — expt<f>tw. Taking t = 1 completes the proof.
3336
Consider the linearly independent vector fields r and v on U :~ iR4\0
whose values at x = (xi, X2, x3, £4) G -ffi4 are given by
r^ : = (0:1,0:2,2:3,0:4),
v* : = (-2=2,2:1,-2:4,2:3)
a) Is the rank-2 distribution defined by these two vector fields in U
completely integrable?
b) Is the rank-2 distribution orthogonal to these two vector fields completely
integrable?
(Indiana)
Solution.
a) Direct calculation gives [r, v] = 0. Thus the Frobenius theorem
guarantees that the rank-2 distribution denned by r and v in U is completely
integrable.
b) Construct the following linear algebraic equations about 3/1,3/2,3/3 and
f 2;iyi + 2^2 + x3y3 + x4y4 = 0,
\ -2=2yi + 2;iy2 - 2;4y3 + 2;3y4 = 0,
which are equivalent to
2:1 x2 \ ( yi \ I x3 x4 \ I y3 I _ Q
-2:2 2:1 I \ y2 / V -2:4 x3 I \ y4

227
Because (2^22,23,24) € -ffi4\0 = U, without loss of generality, we may
assume 21 ^ 0. Therefore, we obtain
Vi \ _ -1 ( xxx3-\-x2x4 xxx4-x2x3 \ ( y3
Setting
and
3/2 ) X\ + X\ \ X2X3 - X\X4 x2x4 + XxX3 J \ y4
Vs \ _ ( -(*i+si)
V4 ~\ 0
0
-(xl + x22)
respectively, we obtain the following two linearly independent vector fields a
and /3 which define a rank-2 distribution orthogonal to the above one and
whose values at x are given by
a = (xix3 + 22^4, x2x3-xix4, -(xl+xl), 0),
/3 = (xix2-x2x3, x2x4 + xix3, 0, -(21+22)).
By a long but straightforward calculation, we have [a, /3] = -2(2^ + 2^)v
which does not belong to the distribution defined by a and /3. Thus the
Probenius theorem tells us that this distribution is not completely integrable.
3337
Let G be a Riemannian manifold with a global frame-field {ej}™=1.
a) Show that any connection on G is competely determined by its effect on
the frame field, i.e., by the vector fields Veiej, i, j = 1, • • •, n.
b) Show that when G is a Lie group with a bi-invariant metric ( , ), and
the frame-field is left-invariant, we characterize the Levi-Civita connection on
(G, ( , )) by setting,
for all i, j — 1, • •• ,n.
(Indiana)
Solution.
a) For arbitrary smooth vector fields X and Y on G, we have
x = YJxiei, Y = j2yiei-
i = l

228
Then, motivated by the properties of connection, we can well define
Thus defined operator V certainly satisfies all properties of a linear connection
on G.
b) For every left invariant vector field X on G, let g(t) denote the unique
1-parameter subgroup of G such that </(0) = e, gjt ' = Xe, where e is the unit
element of G. Noting that g(t) is a geodesic of G and Xg^ = -¾^ we have
obviously
VxX = Rlm =E(*m =0.
<& t=o <# \ <# ) t=0
Besides, because the metric of G is bi-invariant, we know that VxX = 0 is
valid everywhere. Especially, if X = e; +Cj, then Vei+ej(e,- +ej) = 0 implies
that
Ve,ej + V6j.ei = 0.
On the other hand, the Levi-Civita connection V satisfies
Ve;ej - Vejei - [e,-,ej].
Thus, we obtain

Part IV
Real Analysis

231
SECTION 1
MEASURABLITY AND MEASURE
4101
Let S C [0,1] be the set denned by the property that x £ S if and only if
in the decimal representation of x the first appearance of the digit 2 precedes
the first appearance of the digit 3. Prove that S is Lebesgue measurable and
find its measure.
(Stanford)
Solution.
For any x £ [0,1) there is a unique sequence {^(^))^=1 of integers
satisfying the following properties
(1) 0 < pn(x) < 9, (2) Vn, 3m>n: pm(x) < 8 and (3) x = £ To^-
71 = 1
Then
S = {1}U{Z6 [0,1) |Vn,p„(a;)#2andft,(a;)^3}
\J{x e [0,1) | 3n,pn(x) = 2, Vi < n,Pi(x) ^ 2 and Pi(x) £ 3}.
Let
A = {x G [0,1) | Vn,pn(x) £ 2 and pn(x) £ 3}
and
B = {xe [0,1) | 3n,pn(x) = 2, Vi < n,Pi(x) # 2 and p,{x) £ 3}.
Then
[0,1)\J4=(J |J
hi i fc„-l 2 fcj fcn-1 <-
10 10"-1 10»'10 10"-1 10"
and
n-M II \h< i fc"~i i 2 fcl i i ^-1 i 3 ^
^-U U io+ " io"-1 10»' io io"-1 io»;
It follows that both A and B are measurable and therefore is S. Since
_2_
10"
™([0,i)V4) = E7^-8"~1 = 1
n=l

232
and
oo
m<B) = Y — .Sn-1 = -,
y ' ^ 10™ 2
n=l
we have
m(S) = m(A) + m(B) = -.
4102
Define 51 = M/Z endowed with the natural Lebesgue measure. Consider
on 51 the equivalence relation: [x] ~ [y] ■& x — y £ Q, for a;, y £ IR and [ ]
denoting the class in 51. For each £ £ S1/ ~, choose a representative £ £ 51,
and let £ C 51 be the set of these points, i.e.,
£ = {£|£e£,V£eSV~}.
Show that E? is not measurable.
(Stanford)
Solution.
51 is an abelian group under the binary operation ([a;], [y]) >—> [x + y] and
the inverse operation [a;] >—► [—x].
For any [a;] £ S1 there is by the construction of£a unique element [y] £ E
such that x — y £ ¢. Then [a;] = [y] + [r], where r £ [0,1) n<5 is such that
r = x — y mod JT. So we conclude that S1 = (J (E? + [r]).
re[o,i)n<?
If r,s £ [0,1)n(? are such that (E + [r])(l(E+ [s]) ^ 0 there are [a;], [y] £ E
such that [a;] + [r] = [y] + [s]. It follows that x = y mod<5 and therefore
[a;] = [y] by the construction of E. Thus [r] = [s], which then implies that
r = s since —l<r — s<l.
It follows that {E + [r]\re [0,1) DQ} is a partition of S1.
Let m denote the natural Lebesgue measure on S1 such that 772(51) = 1.
If E is measurable then
m(S1)= J2 rn(E + [r})= ^ m(E)'
re[o,i)m? re[o,i)m?
If m(E) = 0 then 771(51) = 0, a contradiction; Otherwise 771(51) = 00, a
contradiction, too.

233
4103
Let X be a set and V C V(X), V closed under finite intersection. Denote
by TZ the ring generated by V. Puthermore, let 7r be the smallest system,
V C 7r C V(X) such that 7r is closed under the following operations:
(i) finite disjoint unions.
(ii) differences A\B, B C A.
Prove that 7r = H.
(Iowa)
Solution.
Obviously, it CH. We have
{AeK\AnBe-K, Bev}=n (l)
since the former is a ring containing V. Also we have
{B G % | AC) B£ir,A£ 11)=11, (2)
since by the equality (1) the former is a ring containing V. By equality (2),
% = -k since for any A G %, A = A n A.
4104
Let (J,* be the Lebesgue outer measure on M and A, B subsets of St such
that
in{{\x-y\ \x £A,y£B} >Q.
Prove or disprove that
H*(A\JB)=n*(A)+n*(B).
(Iowa)
Solution.
We will show the equality. Let r = d(A, B) > 0. Let
U = {x G M | d(x, A) < -}
and
V = {x G M | d(x, B) < T-}.

234
Then U and V are disjoint open sets containing A and B respectively. We
have
(J,*(AUB) - inf{//(W) | Wopen,,4u£ C W}
= M{fi(W)\W open, AUB CW CU\JV}
= inf {/*( Wi) + /i(W2) I Wi open, ,4 C Wx C U, B C W2 C V}
{~ inf {/i(Wi) | Wi open, iC^CP}
+ inf{/i(P72) | W2 open, 5 C P72 C V}
= //(,4) + //(5),
where the equality (1) follows from the following equality
inf (X + Y) = inf X + inf Y, X, Y C iR.
4105
a) Consider a measurable space (X, fi) with a finite, positive, finitely
additive measure //. Finite additivity means that whenever {5,-} is a finite
collection of mutually disjoint measurable sets, then /i(UB;) = ^2 fJ.(Bi). Prove that
// is countably additive if and only if it satisfies the following condition:
If An is a decreasing sequence of sets with empty intersection then
lim fi(An) =0.
r—*oo
b) Now that suppose X is a locally compact HausdorfF space, that B is the
Borel a-algebra, and that // is a finite, positive, finitely additive measure on
S. Suppose moreover that // is regular, that is for each B £ B,
//(5) — sup{//(i£") \ K C B and K is compact}.
Prove that // is countably additive.
(Iowa)
Solution.
a) The sufficiency. Let {Bn} be countably many measurable sets which are
oo oo
mutually disjoint. Let An = (J 5,-. Then f] An = 0. We have
!=n+l n=l
(n \ oo
i = l / !=1

235
Therefore /i is a measure. The necessity is obvious.
b) If fi is not countably additive, by a) there is a decreasing sequence {An}
of measurable sets with empty intersection such that
lim n{An) = m{fj,(An) > 0.
n—*oo n
For each n there is a compact Kn contained in An such that
fi(An) < /i(Kn) + ^1^/^(^)-
( " \ " 1
^k\flK' < Em^^o < 2 inf m^o,
V i=i / i=i
which implies that
/iff) ^* J ^°
Then
and therefore
n ^ # 0.
Thus { p| if,- | n G -W} is a decreasing sequence of nonempty compact subsets
! = 1
oo
in the compact space K\. So f] Kn ^ 0, which contradicts the fact that
n=l
oo
n An = 0.
n=l
4106
For / : [0,1] -> M, let £c{x| /'(a;) exists}. If m(E) = 0, show that
m(f(E)) = 0.
(Indiana-Purdue)
Solution.
Denote
F = {x\xeE,\f'(x)\<M},
where M is any positive number. It suffices to prove m(f(F)) = 0. Set
Fn = {x | x G F, |/(y) - /(s)| < M\y - x\ if |y - x\ < -}.

236
Then
Ft CF2 C---,771(2^) = 0,
and
f(F) C Uf(Fn) = lim f(Fn).
n—*oo
For any e > 0, take a sequence {in,fc} of open intervals such that
F„ C \Jl„,k,rn(Intk) < -, ^m(Jn,fc) < e.
k k
For i,yefnn J„,fc, we have
l/W-/(s/)l<Mm(Jnil).
Therefore
m*(f(Fn)) = m* (f(Fnn (\JInA\ )
< ^m*(/(irnnjn,fc))
< M^m(In,k)<Me.
Thus m*(/(Fn)) = 0, which implies that m(f(F)) ~ 0.
4107
Suppose A C M is Lebesgue measurable and assume that
,nn 6 — a
m(Af)(a,b))< -^-
for any a, 6 G St, a < b. Prove that m(A) = 0.
(Jotwa)
Solution.
If m(A) ^ 0 there is an n such that m(A D(n,n + 1)) ^ 0. There is an
open subset U in (n,n + 1) such that
A n (n, n + 1) C U C (n, n + 1)
and
m(C/) < 772.(4 (l(n,n + 1)) + £,

237
where e < m(A f) (n, n + 1)).
There are at most countably many disjoint intervals (aj,6j)'s such that
f/ = U(ai,6j).
i
Then
A<l(n,n+l)=[JAn(aj,bj).
i
We have
i
< y^ h ~ ai
- L_j 2
i
= 2m(f/)< 2(m(j4n(n'n+1))+£)
which deduces that
m(A(l (n,n + 1)) < £,
a contradiction.
4108
Choose 0 < A < 1 and construct the Cantor set Kx as follows: Remove
from [0,1] its middle part of length A; we are left with two intervals Ii and Ij,.
Remove from each of them their middle parts of lengths A|Jij|, etc. and keep
doing this ad infinitum. We are left with the set Kx- Prove that the set Kx
has Lebesgue measure zero.
(Stanford)
Solution.
Claim. For any n £ M, the total length of intervals removed in the n-th
step is A(l- A)""1.
The claim holds for n = 1. Assume that it holds for any n < k. Then the
total length of intervals removed in the k + 1-th step is
A^l-^TA^-ArM^l-A)*

238
By induction the claim holds for any n £ ffl.
It follows that the Lebesgue measure of K\ is
oo
l-^A(l-A)"-1 =0.
4109
Suppose // is a positive Borel measure on Et such that
(i) //([0,1]) = 1, and (ii) fi(E) = n(x + E) for any Borel set E of M and
every x £ M. Does this imply that // is the Lebesgue measure? Justify your
answer.
(Iowa)
Solution.
Yes, // is the Lebesgue measure. For any x £ M define
^ _ / M(M). * >o,
»W-\ -/.((^, 0]), x<0.
Then g : ttt —► St is nondecreasing and right-continuous. Moreover, for any
x,y £ Et with x < y, fi(x,y] — g(y) — g(x). It follows that the measure // is
induced by g.
For any £, y £ iR, from either
fi(x, x + y] =//(0, y], y>0
or
/i(z + y,z] = //(y,0], y<0
we have
0(z + y) = ff(z) + ff(y).
From the right-continuity of g, we conclude that g(x) = £#(1). However,
ff(i) = //((0,1]) = //([o,i])-//({o})
= 1- lim //((--,0])
n-»oo n
= 1- lim(ff(0)-ff(-i)) = l.
n—>oo n
It follows that g(x) = a; and therefore // is the Lebesgue measure.

239
4110
Let U be a a-algebra of subsets of a set X and fin : U —* M be signed
measures such that fi(E) = lim fJ.„(E) exists for every E £ U. Prove that /i
n—»00
is a signed measure.
(Jotwa)
Solution.
Let, for each E Ell,
w ==2^1
1 |/*nl(£)
.=1-- - + i^iw
Then /I is a finite measure and /in's are absolutely continuous with respect to
/Z. Given any mutually disjoint measurable sets E?„'s, by the Vitali-Hahn-Saks
Theorem one has
lim sup
Mm I (J -Efc
\k=n
= o.
Then
U=i
= lim lim /im II Ek ) + fim [
n—*oo m—>oo \ \ ^-^ / \
\ \fc=l / \
(n / oo
5^/i(^) + /im ( (J £fc
fc = l
oo
u ^
fc=n+l y
^fc=n+l
fc=l
which shows that fi is a generalized measure and therefore a signed measure.
4111
Let A be Lebesgue measure on Et. Show that for any Lebesgue measurable
set E C M with X(E) = 1, there is a Lebesgue measurable set A C E with
A(^) = |.
(Iowa)

240
Solution.
Define the function / : M —> [0,1] by
f(x) = \(ED(-oo,x]), xeM.
It is continuous by the following inequality
\f(x)-f(v)\<\*-y\, x,yeM.
Since lim fix) = 0 and lim f(x) = 1, there is a point x0 £ 1R such that
X —* — OO X —*oo
/(a;0) = |. Put j4 = £ n (—oo,a;o]> which is required.
4112
Let fi be a complex Borel measure on [0,oo). Show that if
oo
e-nxdn(x) = 0
for all n = 0,1, 2, • • •, then fi = 0.
(Jotwa)
Solution.
Let
S = spanje-"* |n£»U{0}}
then S is a self-adjoint subalgebra of C([0, oo]) separating the points of [0, oo].
It follows from the Stone-Weierstrass Theorem, S is dense in C([0, oo]) under
the supremum norm topology. Therefore for any / £ C([0,oo]),
/ fd/i = 0,
./[0,oo]
which then implies that fi = 0.
4113
Let A C [0,1] be a measurable set of positive measure. Show that there
exist two points x' ^ x" in A with a;' — x" rational.
(Indiana-Purdue)
Solution.
Denote all the rational numbers in [—1,1] by ri,r2, • • • ,rn, ■ ■ -. Denote
A„ - {x + rn | x£ A}.
Then m(An) = m(A) > 0. An C [-1, 2]. Thus
I

241
(J 4, C [-1,2].
n=l
Suppose that An (1 Am = 0 if n ^ m. Then
oo
Y,rn(An) < m([-l,2}) = 3,
n=l
which contradicts m(A) > 0. Therefore there must be some m, n such that
An n Am ^ 0. Take z G An n Am- Then we can find x', x" G A such that
Thus
z = x' + rn = x" + rn
x' -x" =rm- rn.
4114
Let E
rove that E is Lebesgue measurable with
Lebesgue measure 0. Also show that E + E = [0,1].
Solution.
Obviously, E C [0,1). We will show next that
(Iowa)
[0,l)\£ = u{
Xl
Xn-1 , 2 Xi
... - -» Xn-1 + 1\
Ly + • • • + ^rrr + ^ y + • • • + 3„_i -J
0 <»!,•••,a;„-i < l,n£w}.
(1)
Indeed, for any x G [0,1)\-E, i = J] |j (where 0 < xn < 2 and for any n
n = l
there is an m > n such that a;m < 1), there is at least one n such that xn = 2.
Let n = min{fe | x^ = 2} then 0 < £i, • • •, a;„_i < 1 and therefore
x G
3 +•••+ 3„_i + 3„. 3 3„_i
The converse inclusion is obvious.
By equality (1), E is measurable and since
00 1 °° 2n_1
^([0,1)^) = 2 £ ^Ey = 1'
n = 10<Ii,"',In-l<l
n=l

242
m(E) = 0. Since Ec[Q,\],E + EC [0,1]. For any x G [0,1) say x = £ p-
n=l
(where 0 < a;n < 2, and for any n, there is an m > n such that xm ^ 2).
Define
f x Q<x <1, ^^
1, Xn — Z,
and
OO , OO //
Then a;'- ^ ^ and xrr — ^ |^ belong to i£ so that z'-f z" — z. Obviously,
n=l n=l
This completes the proof.
> , 3" / \ ^ 3"
\n=l / \n=l
4115
Let / : M —► iR+ be measurable, and let e > 0. Show that there exists
g : M —► iR+ measurable such that (i) ||/ — </||oo < e (ii) for every r G M,
\{x\g(x)=r}\ = 0.
(Indiana-Purdue)
Solution.
Take {rn} such that 0 < n < r2 < • • • < rn < • • •, limrn = -foe, rn+\ —
rn < £, for all n. Set fi(x) = /(a;) + arcctgz, Denote
En = {x \rn_! < f^x) <rn}.
Set 0i = £rnX£„. Then
gi(x) > arcctga;, ||/i - ffi||oc < £•
Set #(:c) = <?i(a;) - arcctgz. We have g > 0 and ||/ - ^H^ = ||/i - 0i||oo < e.
For every r E St,
{x\g(x) = r} = {a; | 0i(a;) - arcctgu = r}
- [J{x | x £ -E„,arcctga; = rn - r}.
n
Since \{x \ arcctgz = rn - r}j = 0, |{a; | g{x) - r}\ - 0.

243
4116
Let / be the function on [0,1] denned as follows f(x) = 0 if x is a point
on the Cantor ternary set and f(x) = ^ if x is in one of the complementary
intervals of length 3-p.
a) Prove that / is measurable.
b) Evaluate I f(x)dx.
Jo
(Stanford)
Solution.
Let K denote the Cantor ternary set. Then
[0,1]\#=|J |J (0-a1---op_1l,0-a1a2---ap_12).
p>l af=0,2
By the definition, we have
J — 2_j 2-^ ~X(0-a1---ap_1l,0-a1a2---ap_12),
p>la;=0,2^
which then is measurable, and
f1 f(x)dx = y y --- = y-—= inV3.
4117
Let Bbea Lebesgue measurable subset of M with m(E) < oo and let
f(x) = m((E + x)(lE).
Show that
(a) f(x) is a continuous function on M.
(b) lim f(x) = 0.
X—*+oo
(Illinois)
Solution.
We will show first that for any Lebesgue measurable set E,
lim rn((E + h) n E) - m(E).
h—*0

244
n
If E is of the form U (a;,/3^), where (ai,/3i)'s are finitely many mutually
i= 1
disjoint open intervals of finite length, then
n
m{E) = £>,■-«,)
i = l
n
= lim VV/3; A (fa + h)-aiW (ai + h))
h—»00 *—'
• =1
n
= lim^m((a,-,/3,-)n(a,-+fc,A + h))
= limm I (J(aJ7/3i)n(ai + h,/3; + h) I
< ljmm(En(E + h))
h->0
< hmm(En(E + h))<m(E).
h—*0
So
lim m((E + h)C\E) = m(E).
h—*0
If E is a compact set, then there is, for any e > 0, an open set E?' of the above
form such that E C E' and m(E'\E) < e. Let F = £'\£ one has
m(E) < m(E') = lim m((E' + h) n £')
= lim m(((.E + ii)n£) U((£ + fc) n F) <J((F + h)DE) <J((F + h) 0 F))
h—*Q
< lim m((E + h) C\ E) + 3e < Em m((E + h)r\E) + 3e < m(E) + 3e.
fc->0
fc^O
It follows that
m(E) = lim m((E + h) 0 E).
ft->0
In general, there is an increasing sequence {En} of compact sets such that
lim m(En) ~ m(E).
En C E and
Then
i(E) = lim m(En) = lim lim m((En + h) D En)
v ' n->oo n->oofc-»0
< lim lim m((E + h) 0 E) = lim ?ti((£ + fc) n £).
»->ooh-»0 h-»0

245
So
lim m((E + h) n E) = m(E).
(a) For any y,x £ ffi
l/(y) - /(=5)1 = \rn(((E + y)\(E + x))nE)-m((E + x)\(E + y))\
< m((E + y)\(E + x))+M((E + x)\(E + y))
= m(E + (y - x)\E) + m(E + (x - y)\E)
= m(E) - m(E + (y - x)) n E) + m(E)
-mp+(j;-y))n£)-tO, as y -> x.
(b) If E? is compact, then there is an r > 0 such that for any x > r,
(i + £)n£ = |. The claim follows.
In general there is an increasing sequence {En} of compact sets such that
En C E and lim m(En)= m(E). Then
n—»00
lim m((E + x)C\E)
= ^lirn^ m((£ + i)fl£)- m((En + x) n £„) + m((.En + a;) n £„)
= lim lim mp-fi)ni;)\((i;„ + i)n£n)) + m((£„ + i)n£„)
n-tiKl-t+oo // w / /
< lim (m((E + a;)\(K + x)) + m(£\£n)) = 0.
4118
Let A C St be a. set of positive Lebesgue measure. Prove that
<p(x) = / XA(tx)XA(t)dt
is continuous at x = 1. Use this result to prove that there exists an e > 0
such that for any m E M with |m — 1| < e, the line y = mx has a non-void
intersection with A x A.
(Iowa)
Solution.
If B is a bounded open set, say (J (a,, 6j), where 2 < m < oo and (a,, 6,)'s
are mutually disjoint open intervals, then for any n < m and any i>0we
have
(J(a;ai, xbi) n (a,-, h) CxBflB.

246
We have
//(5) = sup 2_^(bi — a,i)
n<m
= sup lim y]fi((xaj,xbi) Pi (0,,6,))
n<m
,= 1
< lim fJ,(xB n 5)
< Kmn(xBr\B) </i(B),
(1)
where // is the Lebesgue measure. If K is a compact set, there is a decreasing
00
sequence {Bn} of bounded open sets such that K = p) Sn. Since
n=l
/i(iB. n Bn) - /i(xjr DJT) < n{x{Bn\K)) + n(Bn\K)
we have by (1)
/i(i£") = lim fj,(Bn) = lim lim//(a;!?,, Pi 5n)
< lim lim(fj,(xK C\ K) + fi(x(Bn\K)) + fi(Bn\K))
n-»oo a:—► 1
= lim ( lim //(a: if D K) + 2//( £„ Pi if))
n-»oo \:c-»l /
= lim fijxK n JT)< Jim p.(xK D K) < n(K).
x—l x^1
(2)
In general, there is an increasing sequence {Kn} of compact sets such that
oo oo
(J K„ C A and fi(A\ (J ifn) = 0. By (2) we have
n=l
//(,4) = lim u,(Kn)= lim lim//(a;.ff„ D iJT„)
x n-»oo n->Ml-»l
< lim //(a;,4 Pi ,4) < Iim n(xA C\ A) < //(,4).
(3)
Since
Hx)-^(i)| =
/ + 0O
(XlA(t)XA(t)-XA(t))dt
-OO
= //(,4)-//-,4 0,4
we have by (3) lim <p(x) = ^(1).
x—»1

247
If there is no e > 0 with the property mentioned in the question, then for
any n £ IN there is an mn £ M such that \mn — 1| < £ and the line y = mnx
has a void intersection with Ax A. However, this implies that mn(Af) A) = 0,
and therefore
H(A)= lim n(mn(AnA)) = 0,
n—*oo
a contradiction.
4119
Let /i be a countably additive measure on a set S with fi(S) < +oo that
is without atoms, i.e., if A is a measurable set with fJ.(A) > 0, then there is a
measurable set B C -A such that 0 < /i(B) < (J,(A). Prove that the range of fi
is the closed interval [0,/i(S)].
(Courant Inst.)
Solution.
If there is a t0 £ (0, /i(5)) not in the range of fi. Let
V = {A\ A measurable and p(A) < t0}/ ~,
where ~ is an equivalence relation: A ~ B if /i(j4\jB) = //(B\j4) = 0. Then P
is a partially ordered set: [A] < [B] if fi(A\B) = 0. Given a totally ordered set
Q of V, let /3 = sup{/i(j4) | [A] £ 2}. Then (3 <t0 and there is an increasing
oo
sequence {[.A„]} of Q with {(j,(An)} increasing to /3. Let A = (J j4„, then
n=l
/3 < fi(A) = lim /x (J Ai
\i=l
= nlim I /x I |J i4i\i4„ ) + /i(j4„) J
= lim /^(^) =/3 < i0.
n—>oo
So /i(,4) = /3 < <0 and therefore [,4] £ V. For any [B] £ S, if [B] < [An] for
some n, then [B] < [A]; otherwise, [B] = [A] since
lx(A\B)<J2»(An\B) = 0
n=l
H(B\A) = n(B) - fi(A n B) = /3 - /3 = 0.
n=l
and

248
It follows that [A] is an upper bound of Q in V. There is by Zorn's Lemma
a maximal element [A0] of V. It follows that fi(B) > t0 - fi(A0) whenever
B C S\A0 and fi(B) > 0. (e.g. fi(S\A) > t0 - K^o)). Let
^ = {B|BC5\i4o,MB)>0}/~,
where ~ is denned as above. Equip TZ with the partial order < as above. Given
a totally ordered set S of TZ. Let
a = inf{/i(B) | [B] G 5},
then a > t0 — fi(A) and there is a decreasing sequence {[Bn]} of S with {/i(Bn)}
decreasing to a. Let
oo
5=n *».
n = l
then
< lim ( J2 KBn\Bi) + /J f)Bi = /*(B) < a.
n—*oo
So 11(B) = a > 0. Therefore /i(B) ><0 - /*(A))- [5] G ft. For any [C] G 5, if
[Bn] < [C] for some n, then [B] < [C]; otherwise [B] = [C] since
oo
/x(C\B) < £ »(C\Bn) = 0
n=l
and
/i(B\C) = /i(B) -/i(BflC) = «-a = 0,
It follows that [B] is a lower bound of S in TZ. There is by Zorn's Lemma a
minimal element [Bo] of TZ. However, by the assumption that there is a subset
Co of B0 with 0 < fi(Co) < fJ-(Bo), [Co] < [B0] and [Co] ^ [B0], a contradiction.
4120
Let X be a compact HausdorfF space. Consider the <r-algebra B generated
by the compact G( sets. Show that any positive measure /jonB which is finite
on compact sets is automatically regular.
(Iowa)

249
Solution.
Recall that sets in B are called Baire sets and each compact Baire set K is
a G( set. Also recall that E is outer regular if
fj,(E) = inf {//(V) | E C V, V open and V G B}
and E? is inner regular if
H(E) = sup{fi(K) \K CE,K compact and K G B).
Let K and O denote the classes of compact sets and open sets in B,
respectively. Let
rc ={.0(iirA£0 I #«,£<££},
» = 1
where the symbol U means disjoint union. Then % is a ring such that B —
S(K).
By definition, each set in K is outer regular. Let us show each set U in O
is inner regular. For any e > 0, there is a V in O such that X\U C V and
fj,(V) < fi(X\U) + e. Then Jf\V C f/ and fj,(U) < /i(X\V) + e.
Let us preceed in five steps to show the regularity of /i.
Step 1. For any pair K, L in K, K\L is regular. For any e > 0 there are
a B G O and anMGK such that K C B and (j,(B\K) < e and such that
M C B\L and fi(B\L) < (i(M) + e. Then we have
K\L C B\I, /x(B\I) < ^K\L) + e;
ifnMCif\i, fJ.(K\L) <n{K C\M) + e.
Thus ii"\i is regular.
Step 2. If (¾ | 1 < i < n} is a finite class of mutually disjoint, regular
n
sets, then U Ei is regular.
Obviously
(i( OBfj <inf|/i(F)|p^CFGol. (1)
For any e > 0, there are B\, ■ ■ ■, Bn G O such that Ei C S,- with
fJ,(Bi\Ei) < -, i=l, •••,n.
v n
Consequently,
n n
• = 1 i=l

250
and
n
which shows the outer regularity of (J Ei. The inner regularity follows from
!=1
the following inequalities
^ U EiJ = Z) sup{/W \Ei2KieK)
= sup j^/^i) I^D^G/ci
< supi/i(if)| Q Cif G/ci.
Step 3. If {.En | n G IN} is an increasing sequence of regular sets then
oo
(J En is regular.
71 = 1
oo
Let £ = U £„. Obviously
n = l
/i(£) <inf{/i(F) |£Cl^eO}.
For any e > 0 there are Vi, ■ ■ ■, Vn, ■ ■ ■ G O such that En C Fn and /i(Vn) <
oo
fi(En) + £-. Let V = UKe O, then
71=1
£CF, /x(V) < n{E) + £,
which shows the outer regularity of E, while the inner regularity follows from
the following inequalities
fi(E) = sup/i(£n)
n
= supsup{/i(if) I K C En,K G K}
n
< sup{n(K) | £ 2 -K" G £}•
oo
Step 4. If {.£„} is a decreasing sequence of regular sets, then f] En is
71=1
regular.

251
Let E = f] En. For any e > 0, n G M, there is a if„ £ H such that
n=l
K„ C En and /*(£„) < (i(Kn) + -^, then
OO
EDp\Kn
n = l
and
/*(^?) < ^ f n is-™ j+c
which shows the inner regularity of E. The outer regularity follows from the
following inequalities,
inf{//(V) | £ C V G 0} < inf inf{fi(V) \ En C V £ O}
n
= inf/i(£„)=/i(£).
n
Step 5. Let
S = {E £B\E is regular }.
By steps 1 and 2, <S contains TZ. By steps 3 and 4, <S is a monotone class. It
follows that S — B.
4121
Suppose that \i is a nonnegative Borel measure on iR™. Let
f(r) = sup{/i(S(a;,r)) | x G Mn}.
Assume /(r) is finite for all r > 0 and assume
r~^0 rn
Prove that /i is identically 0.
(Indiana)
Solution.
Let
C(x,r) = {y£Mn \~r < y,- - a;,- < r, i = 1,---,n}
and let
g(r)=sup{ti(C(x,r))\x£lRn}.

252
Since C(x,r) C B(x,-*/n + Ir),
lim^=0.
r-0 rn
Given a compact set K of ffl1, let s > 0 be such that K C C(0, s). For any
e > 0 there exists an r > 0 such that #(r) < (2r)ne. There exist finitely many
x1,--- ,xk £ Mn such that
k
K C U^»CC(0,2s)
•=i
and C(x',rys are mutually disjoint. Then
k k
H{K) < £/.((7(^,7-))^(27-)^
•'=1 i=i
k
= J2HC(x\r))e<\(C(0,2s))e.
!=1
Letting £ —> 0, we get that /i(i£") = 0 and therefore /i is identically 0.
4122
Let fi be a finite Borel measure defined on iR™, and define a function / on
-ZR™ by /(a;) = fi(B(x, 1)) where S(a;, 1) denotes the open ball centered at x of
radius 1. Prove that / attains its minimum on each compact set of Mn.
(Indiana)
Solution.
Let K be any nonempty compact set of Stn and let a = inf/(if). There
exists an sequence (¾} of K such that
lim /(¾) = a.
k—*oo
Assume without loss of generality that xk —> x in K. It is easy to show that
B(x,l) C lim B(xk,l).
k—*oo
Hence
H(B(x, 1)) </i( lim 5(^,1) ) < ljmfi(B(xk,l)) = a,
\fc->oo / fc->0

253
which shows that
Hence a is finite and / attains its minimum on K.
4123
Let E be the set of all numbers in [0,1] which can be written in a decimal
expansion with no sevens appearing. Thus,
1 27 28
-=0.333..,- = 0.2699..,-=0.2800-..6^.
(i) Compute the Lebesgue measure of E.
(ii) Determine whether E is a Borel set.
(Indiana)
Solution.
For any x £ [0, l]\E, write x = 0.aia2 • • • an • • •. Let
n = min{fe > 1 | a* = 7}.
If a* = 0 for all k > n then
x = 0.aia2 • • • an_i7 = 0.aia2 • • • an_i699 • • • £ E,
a contradiction. It follows that
0.aia2 • • • an_i7 < x < 0.aia2 • • • an_i8.
and therefore
[0,1]\E C U{(0.aia2 • • • a„_i7,0.aia2 • • • an_i8) | ai, • • , an_i ^ 7,
n = l,2,-..}.
The reverse inclusion is obvious. So E is Borel measurable and
1
10^
oo x
1(S) = l-^9-1x—= 1-1 = 0.
71=1
4124
Let / : M —> iR™ be a function such that for all x,y £ M
l/(x) - /(y)|" < el*l+l»l |x - y|. (*)
Show that if E C St is a measurable set with mi(E) = 0 then mn(f(E)) = 0.
(Indiana)

254
Solution.
Assume without loss of generality that E is bounded. Let E C (—r,r).
For any e > 0 there exists an open set U such that E C U C (—r,r) and
mi(U\E) < e. Write C/ = (J(a,-,&,), where (a,-, 6,-)'s are mutually disjoint.
Then condition (*) implies that
f((aiA)) C B(f(ai),(e2r\bi - ai\)±).
It follows that
mn{f{U)) < J2Cne2r\bi -¾| < Cne2r£,
i
which implies that m*n{f{E)) = 0 and therefore f(E) is measurable. Hence
m„(/(£))=0.
4125
Let / : iRn —> iR be an arbitrary function having the property that for
each £ > 0, there is an open set U with X(U) < e such that / is continuous on
Mn\U (in the relative topology). Prove that / is measurable.
(Indiana)
Solution.
Let Uk be an open set such that A(£4) < £ and / is continuous on Mn\Uk.
Let /¾ = fxm.n\uh, *nen /fc is measurable. For any e > 0,
™*({* I l/fc - /l("0 > e}) = m*{{x G r/fc | |/(x)| > e}) < i.
It follows that {/¾} converges to / in measure. Since the Lebesgue measure is
complete / is measurable.
4126
Let A C St be a Lebesgue measurable set and let rA = {rA \ x G A} where
r is a real number. Assume that rA = A for every nonzero rational number r.
Prove that either A or M\A has Lebesgue measure zero.
(Indiana)

255
Solution.
Let M* = M\{0} and B = A\{0}. Then rB = B and r(M*\B) = M*\B
for every nonzero rational number r. If m(IR*\B) = m(]R\A) > 0 there
exists a compact subset K of M*\B with positive Lebesgue measure. For any
compact subset L of B, define function / : M* —> [0, oo) by
/(*)=/ Xi^XL-^-1*)-, zGiR*.
./jr' y
Then / is continuous and for any nonzero rational number r,
f(r)= f Xk{y)XrL{y)^- < I x*'\b(v)xb(v)^l=0.
Jm.' y Jm* y
Hence f(x) = 0 for any x £ M*. Since
/ »(,)*/ «-(»)* = / /wf = o,
yjR- y jr' y Jm.' x
we conclude that m(i_1) = 0 and therefore m(i) = 0. It follows that m(B) =
0 and therefore m(A) = 0.
4127
Let /i be a <r-finite measure on the measure space (X, m). Prove that there
exists a probability measure v on (X, m) (v(x) = 1) such that fJ. is absolutely
continuous with respect to i/, and v is absolutely continuous with respect to /i.
(Indiana)
Solution.
There exists a sequence {-En}^i of mutually disjoint measurable sets of
oo
finite and positive measure such that X = (J En. For each measurable set E
n=l
let
then iv is the desired probability measure.

256
SECTION 2
INTEGRAL
4201
Prove or disprove that the composition of any two Lebesgue integrable
functions with compact support f,g : IR —* IR is still integrable.
(Stanford)
Solution.
It is not true. For example, let
f(x) = X{o}(x) and g(x) = X{o,i}(*)-
Then / and g are integrable functions with compact support. However, since
g o f(x) = 1, the function g o / is not integrable.
4202
Let / G Li(0,1). Assume that for any x G (0,1) and every e > 0, there is
an open interval Jx C (0,1) such that
x G Jx, rn(Jx) < £, and / fdm = 0.
Prove that for every open interval I C (0,1)
J fdm = 0.
(Illinois)
Solution.
There is a measurable set E of measure zero such that any x G (0,1)\E is
a Lebesgue point of /, i.e.,
lim —!— / f(t)dt = f(x). (1)
/3-o>0
3-a->0

257
For any x £ (0,1)\E and for any n there is an open interval Jn C (0,1) such
that x £ Jn, rn{Jn) < £ and
I f(t)dt = 0.
It follows by equality (1) that f(x) = 0, i.e., f(x) = 0 a.e.. Therefore
J fdm=Q.
4203
Let (X, M., fi) be a positive measure space with f^(X) < oo. Show that a
measurable function / : X —> [0,oo) is integrable (i.e., one has Jx fdfi < oo)
if and only if the series
oo
2>({* i/(*) >»»
n=0
converges.
(Jotwa)
Solution.
Suppose that / is integrable. Then
oo
!>({* i/(*) >«»
n=0
= S E^1 |m</(x)<m + l})
oo m
m=0n=0
oo
= ^(m + l)/j({i|m</(i)<m+l})
m=0
oo oo
= ^T mn({x | m < f(x) <m + l})+Y^ Kix I m < f(x) <m + l})
m=0 m=0
oo .
< £ / f(x)dp(x) + p(X)
m=0J{x\m<f{x)<m+l}
= /(/ + l)d/i <
Jx

258
Conversely,
* oo .
I fdfi = y, I M
JX m=0J{x\m<f{x)<m + l}
< 5Z(m + l)/i({s|m</(x)<m + l})
m=0
oo
which shows that / is integrable.
4204
(a) Is there a Borel measure fi (positive or complex) on ttt with the property
that
Jm
fdfi = /(0)
for all continuous / : M —>(F of compact support? Justify.
(b) Is there a Borel measure /i (positive or complex) on M with the property
that
f fd/i = /'(0)
Jm.
for all continuously differentiate / : M —>(F of compact support? Justify.
(Jotwa)
Solution.
(a) Yes. Let fi(E) = Xb(0) for any Borel set E.
(b) No. If there were such a Borel measure, let ip > 0 be a continuously
different iable function of compact support, taking value one on [—1,1]. Then
a contradiction occurs from the following limits
lim f <p(t)e«dt = f <p(t)dt > 0
n^°° Jm Jjr
and
1 € n
lim (<j>(t)e*y \t=o= lim —
n—>oo n—»00 n
= 0.
4205
Let £ be a Banach space, (X,ir,(j,) a probability space, and f : X —> E
such that g o / is /i-integrable for every # G £'•

259
Define
L-.E'^M, L(g) = f g o fdfi.
Does L G E"? Justify your answer.
(Iowa)
Solution.
It is ture that L E E". Define the linear operator
T : E' — LX(X), ip^ipof.
Assume that ipn —* <p in E' and T(ipn) —* h in LX(X). It follows that ipn o /
converges to <p o f everywhere and to h in measure and therefore h = <p o f in
i1(X). By the closed graph theorem, T is bounded. We have
\L(g)\< [ \gof\d»< [ ||r||||ff||d/x<||r||/i(jr)||ff||.
Jx Jx
So L is bounded.
4206
Let (X, Ai,fi) be a positive measure space with n(X) < oo, and let / and
g be real-valued measurable functions with
[ fdn= [
Jx Jx
gdfi.
X
Show that either (a) / = g a.e., or (b) there exists a.n E £ M such that
/ /d/i > / gdfi.
Je Je
(Iowa)
Solution.
If (b) does not hold, then for any E G M,
/ /^ < / fl^M-
Since
[ fdfi= f
Jx Jx
gdfi,
x

260
for any E £ M,
/ fdfi= / gdfi.
Je Jb
For any e > 0 the sets
E+ = {x\f(x)>g(x)+e}
and
E-={x\ f(x) < g(x) - e}
are measurable. From the equalities
/ fdfJ. - / gdfi
Je+ Je+
and
/ fdfJ.= gdfi,
JE_ JE_
we conclude that fJ.(E+) = fJ.(E-) = 0. It follows that
M{* I /(*) # »(*)}) = /M U <* I \9(x) ~ f(x)\ > \) ) = °-
Therefore (a) holds.
4207
Let (X, M., /i) be a positive measure space, and S a closed set in(F. Suppose
that
<y»es
KE) JE
whenever E £ M. and /i(-E) > 0. Show that {x E X \ f(x) ¢ S} has measure
0.
(Jotwa)
Solution.
Since (F is second countable, one finds that (T\S is the union of countably
many closed balls {z £ (D \ \z — \n\ < £„}'s. If
/x({s G X | /(a;) £ S}) # 0,

261
there is at least an n such that
»({x£X\\f(x)-\n\< €^)^0.
But then -Arr JE fdfj, belongs to {z G (T | \z — An| < £„}, not to 5, where
S = {x G -X" | \f(x) — An| < £„}), a contradiction.
4208
Let / : [0,1] -> (0, oo) and let 0 < a < 1. Show that
where inf is extended over all measurable E C [0,1] with m(E) > a.
(Indiana -Purdue)
Solution.
Obviously,
[0,l] = (/>l>u(0(i>/>;^)).
Thus
™«/> D)+ E-((£>/> ^))=1-
n = l 7 7
Take an N such that
£>((;>'^)H
Suppose that E C [0,1], m(£) > a. Denote Ex = E (f > ^), £2 = -E\-Ei-
Then m(£2) < f, so m(£i) > f. Therefore
Thus
inf (//}> « >0.

262
4209
Let {/„} be a sequence of real-valued functions in i1(iR) and suppose that
for some / G ^(M),
/■+0° 1
/ \fn(t) - f(t)\dt < -^, n>l.
Prove that /„ —> / almost everywhere with respect to Lebesgue measure.
(Illinois)
Solution.
Since
/n °° ( 1 1
< oo,
there is, by Levi's Lemma, a measurable set E of measure zero such that for
any t G M\E,
n
sup^l/fc+i -fk\(t) < oo.
n fc=i
Therefore for any i G M\E,
/„(*)=/i(*) + £(/fc-/*-i)(*)
k=2
converges. It follows that fn —> / almost everywhere.
4210
Let /i be a finite measure on ttt, and define
'w=1rpsw)- *effi-
Show that f(x) is finite a.e. with respect to the Lebesgue measure on M.
(Indiana)
Solution.
Let
g(x) = [ R^' X * °'
y{ ' \ 0, x = 0,

263
then g £ i1(iR, di/), where
dx
dv(x) =
1 + z2'
and
f{x) = I g(x-t)dn(t).
Jm
Since
J (J \g(x-t)\<h,(t)\dn(t)
[([ \ln\x-t\\ dx f \\n\x-t\\ dx \
Jx\J\*-t\<i U-A1'2 i + ^2 4-t,>i 1» -*l1/2 i + *2J m
ill WWr^W)
Jm \J\x\<\ M1'2 Jir1+x2 J
/ix-«i<i \x-ir- i-i-x- j\x_t]
<
< +00,
by Pubini's Theorem, the function
x 1—> I g(x — t)dfi(t)
Jm
is finite a.e. with respect to the measure v. The conclusion follows from that
the measure v and the Lebesgue measure are equivalent.
4211
Let (X, M,fi) be a positive measure space, fn : X —> [0, 00] a sequence of
integrable functions, and / : X —> [0,00] an integrable function. Suppose that
fn —> / a-e- [fAi and that
/ fndfJ.^ fdfi.
Jx Jx
Prove that
I |/-/„|d/i-0.
(Jotwa)
Solution.
Assume, without loss of generality, that /i is totally <r-finite, i.e.,

264
X = (J Xn,
n=l
where Xn's are mutually disjoint measurable sets of finite and positive measure.
Let
, . v> 1 Xxn{x)
Then g is integrable. Let dv{x) = g(x)dfj,(x). Then v is a finite measure on
X, equivalent to /i. Let
dXn(x) = f„(x)dfi(x), n E JV.
Then A„ < i/.
For any j4 G M, {/A fndfJ.} is bounded and therefore admits a convergent
subsequence. If {/. fnkdfi} is any such subsequence, then by Fatou's Lemma
one has
/ fdfi = / fd(i+ / /d/i
•/.x J a Jx\a
= / lim fnkd(j,+ / lim /„kd/i
JAk^oo Jx\Ak^°°
< lim / fnkdfi+ lim / /„fcefyt
k-'oo./A fc — oo Jx\A
lim / fnkdfJ.= / /d/i.
fc—*oo
It follows that
lim / f„kdfi= / /d/i
and therefore
lim / fndfj,= / /d/i.
"^°° 7a 7a
For any £ > 0 there is a (5 > 0 such that An(S) < e whenever v(E) < (5,
by the Vitali-Hahn-Saks Theorem. There is by EgrofF's Theorem an E with
v(E) < S such that \^-\ converges to £ uniformly on X\E. Then
0 < lim f \fn - f\d(i < lim" f \fn - f\dn
= lim" If \fn-f\dfi+ f \fn-f\dA
n~>° \Jx\E JE J

265
m lim I /
< lim
IX\E
9 9
dv + 2e] = 0.
4212
Let fi be a <r-finite, positive measure on a <r-algebra M. in a set X.
(a) Show that there exists W G L1^) which takes its values in the open
interval (0,1). Show also that
Jl(E)= f WdfJ,
J E
is a positive, finite measure on Ai, and that
J1(E) = 0 <£> n{E) = 0
for £GM.
(b) Show that if / is a complex function on X which is measurable with
respect to M, then
[ fdji= f fwdfi.
Jx Jx
(Iowa)
Solution.
oo
(a) Let {Xn} be a disjoint sequence of M. such that X = (J Xn and
/i(jr„) > 0. Let
00 .
w = T- xx-
^2-l+fi(Xn)
as desired.
The set function /i is of course a positive, finite measure onA<. If fi(E) = 0
then ]1(E) — 0. Conversely,
1 Wdfi = 0
Je
implies that
H({x \x£ E,W(x) ^0)) = 0,
i.e., n(E) = 0.

266
(b) If / is a simple function, i.e.,
n
where fJ,(Ei) < +00, i = 1,---,71, then both jx fdfi and Jx fWdfJ, equal to
n
J2 at]l(Ei).
i=l
In general, if / is integrable with respect to the measure /i, there is a
sequence {/„} of simple functions integrable with respect to fi, such that
lim / |/„ - /|<# = 0.
n^°° Jx
Then /nW's are integrable with respect to /i, and since
lim f \fnW - fmW\dn = lim f \fn - fm \d]l = 0,
lim fnW = fW,
n—*oo
we see that fW is integrable with respect to fi. We have
and therefore
lim I fnWdfj,= f fWdfi
n^°°Jx Jx
I fdji = I fWdfj,.
Jx Jx
Conversely, if fW is integrable with respect to /i, assuming without loss
of generality that / is positive-valued, there is a sequence of simple functions
{/„} such that \fn\ < f and lim /„ = /. Then fnW is integrable with respect
n—*oo
to /i and hence /„ is integrable with respect to /i. Since
lim / \fm - f„\d]l = Q
m,n—*oo J y
and lim /„ = /,/ is integrable with respect to Jl. Accordingly,
n—»oo
[ fdji= [ fwd^.
Jx Jx

267
4213
Let m denote the Lebesgue measure on [0,1] and let (/„) be a sequence in
L1(m) and h a non-negative element of L1(m). Suppose that
(i) J fnQdm —+ 0 for each g 6 C([0,1]) and
(ii) l/nl < h for all n.
Show that
r
f„dm —* 0
L
for each Borel subset A C [0,1].
Solution.
For any e > 0, there is a S > 0 such that
(Iowa)
Je
hdm < e
whenever m(E) < 8. For such a 6 there are a compact set K and an open
set U such that (1) K C A C U and (2) m(U\K) < 6. There is a continuous
function g : [0, 1}-* M such that (3) 0 < £ < 1, (4) g = 1 on K and (5) ff = 0
outside U. Then we have
lim
n—»oo
/ /„dm = lim / f„XAdm
J a n"°° \Jo
< lim
n—»oo
<
<£.
I /" I I f1
\Ja I KO
Em ( / /„ffdra + / /iXc/\A-rfm)
It follows that
lim /
/„dm = 0.
4214
Let {/n} be a sequence of non-negative measurable functions in LP(R) for
some 1 < p < oo. Show that /„ -» /(7^) if and only if f% -» /^(11).
(Indiana-Purdue)

268
Solution.
Suppose that ||/n - f\\p -» 0. Then ||/n||p -» ||/||p, i.e.,
J n-Jr.
Denote
/„ = min(/„, /), 7» = max(/„, /).
Then /„ < / and
l/n - /I < l/n - /I
which implies that ||/n — f\\p —> 0. Just as above, we have
Since
7p+T„^fp + fp,
so
J\K-fp\ = J(rn-fz)-> Jfp-Jfp = o.
Conversely. Suppose that
/l/p-/pl-o.
Then
/»"/, />0.
Therefore
Since
\J(fp-fp)\<J\fp-fp\->o,
it follows that
For any e > 0, take JVi such that
|/(/„p-/P) <£P/2

269
for n > Nx
and any
m(A) < +00 and
Then
for n > Nx.
if m(e) < 6.
Take 6 >
Thus
measurable set E.
[ fP<
J A"
/ Pn
J A"
• 0 such that
Take A,
: ep/2.
<ep
/'<?
a measurable set,
such that
I
fPn<ZP
if m(e) < 6 and n > Ni. Denote 77 = e/m{A)r. Take N such that N > Ni
and m(|/n - f\> rj) < 8 li n > N. We have
O"-^"'
VP / . \ 1/P
< (/ l/n-/N +(/ |/„"/|
\-/|/n-/l>»J / VA(\fn-f\<V)
+ {J..W'-,r^'
Vp / . \ i/p , . , i/p
<
/ fn) +([ n +(/ /«p)
f/ fP)lP+\ [ \fn-f\P)
\Ja° J \->M\fn-f\«l) J
< 4e+ (rfm(A))v = 5e,
for any n > N.
4215
Let 1 < p < 00. All parts refer to Lebesgue measure on R.
(a) Give an example where {/„} converges to / pointwise, ||/„||p < M, Vn
and \\fn - f\\p J* 0.

270
(b) If {/„} converges to / pointwise and ||/„||p —+ M < +00, what can you
conclude about ||/||p? Justify this conclusion.
(c) Show that if {/„} converges to / pointwise and ||/„||p —* \\f\\P, then
ll/n " /Up - 0
(Indiana-Purdue)
Solution.
(a) Consider L[Q, I]. Set /„ = nxto,±)> Then
lim /n(z) = /(z) = 0, ¢€[0,1]
n—»00
and
ll/n-/|| = ||/n|| = l.
(b) We conclude that ||/||p < M. By the Fatou's lemma,
J \f\pdx < lim I \fn\pdx = Mp.
(c) Set
<?„=2P(|/np> + |/n-|/«-/lP-
Then gn > 0, and
lim gn(x) = 2p+1 \f\p
n—»00
pointwise. Using the Fatou lemma, we have
2P+1 f \f\pdx < lim f g„(x)dx = 2P+1 f \f\»dx - lhH f \f„ - f\pdx.
Therefore
lim f \f„ - f\pdx = 0.
n—>oo
4216
Suppose /„ is a sequence of measurable functions on [0,1] with
/ \fn(x)\2dx < 10
Jo
and /„ —* 0 a.e. on [0,1]. Prove
/ |/„|<k-0.
Jo

271
Hint. Use Egorov and Cauchy-Schwartz.
(Stanford)
Solution.
For any e > 0 there is by Egorov's Theorem a measurable set E C [0,1]
such that (1) m([0,1] | E) < e and (2) /„ converges to 0 uniformly on [0, l]\E.
We have by the Cauchy-Schwartz Inequality
f \fn\dx = i |/„|(te+ I \fn\dx
JO J[0,1]\E JE
[ \f„\2dx) /i([0,l]\£)*+ I \fn\dx
J[0,1]\E / JE
/[0,l]\i
<
J[OA]\E
and therefore
lim [ \fn\dx < VT0/i([0,1]\#)* < \/l0e-
Let e—+0, and we have
ix-0.
lim / j/njdi
4217
Let (X, M., fJ,) be a probability space (a positive measure space with (J.(X)
1). Show that if / is an integrable function with values in [1, oo), then
(Iowa)
Solution.
It suffices to show that
lim '°S(JW<M , ,ogM,. (1)
X
pj.0 p
If / = 1 a.e., equality (1) holds; Otherwise, for any 0 < p < 1 and any x £ X
0 < 7(X)P ~ 1 = /(z)?p log /(z) < /(z), (0 < £ < 1).
P

272
By the Dominated Convergence Theorem we have
lim l°g(/x/P"^) = nm log(/x 1 + (fp" - l)d/i)
n->oo pn n->oo pn
= / log/d/i.
where {pn} is any sequence of (0,1) decreasing to 0.
It follows that
log(Jx /Pd/i) _
lim-
pJ.o p
/ log fdfJ,.
Jx
4218
Let {an}n>2 be a sequence of real numbers with \an\ < logn. Consider
oo
y^ ann~x, V2 < x < oo.
n=2
a) Prove that this series converges in Lr[2, oo)
b) Prove that
°° yOO yOO °°
y^ / ann~xdx = / yj ann~xdx,
71 = 2^2 ^2 n = 2
where the sum on the right is the pointwise limit (you need not prove that this
pointwise limit exists).
(Stanford)
Solution.
a) Since
El \an\n~xdx - Y^ -^-n-2 < 'V — < oo.
7, ^-1 logn ^ n2
n=2'/2 n=2 6 n = 2
oo
The series ^ ann~x converges in ^[2,00).
n = 2
b) Let
n
/n(z) = ^2,a-kk~x, x>2.
k = 2

273
Then
OO CO
(i) lim fn(x) = J2 akk~x and (ii) |/„(z)| < £ \an\n~*.
n^x k=2 k=2
oo
By a), x h-> 52 lajtlfc-* is integrable, and by the Dominated Convergence
k=2
Theorem,
00 yOO .00
Y^ / ann~xdx = lim / fn(x)dx
^2 »^°°J2
,00
= / lim fn(x)dx
J2 n->oo
yOO OO
= / yj ann~xdx.
4219
Let S be a bounded Lebesgue measurable set in M, and let {cn} be a
sequence in JR. Show that
lim I cos2(nt + cn)d\(t) = -MS),
>^°°Js 2
is
where A is Lebesgue measure.
Solution.
By the Riemann-Lebesgue Lemma, we have
(Iowa)
lim / cos2(nt + c„)d\(t)
n^oojs
f l + cos(2nt + 2cn)
= lim / « Xs(t)dX(t)
(1 cos 2c f
«A(S)+—TT^ / cos2ntxs(t)dt
2 % Jm
-^£2. J sm2ntXs(t)dt\
\KS).

274
4220
(a) Is the function
f(x,y)={ 0?W' (^)^(°>°),
A I 0, (x,y) = (0,0).
Lebesgue integrable on the unit square 0<£<l,0<y<l?
(b) Compute the repeated integrals in the two orders.
(c) Does the integral
J IJ \f(x,y)\dx) dy
exist? (If you use a theorem, be explicit!)
Solution.
(a) The function / is not integrable on the unit square, for
/ / \f(x,y)\dxdy > / \f(x,y)\dxdy
J0 J0 J{(x,y)\x,y>0,x^ + y^<l}
- ff "
Jo Jo
(Iowa)
cos 20
—s—rdrdd
> / / drd6
r1 r* cos i
Jo Jo r
r1 dry/2*
Jo V~2~4
(b) We have
-dy dx
/(0,1] 1/(0,1] (x2 + V2)2 ,
X2(\-igt) 2
./(0,1] J(o,arctgi] sec4t
J(o,x] J(o,
•Ao.ll 1
sec tdtdx
x2 cos 26dtdx
arctgi]
x3 , l-ln2
rffl =

275
and
J(0,1] 1./(0,
x2-V2 J \ , ln2-l
i] (^TWd£J dy = ~^'
(c) If/0(/0
|/(a;,2/)|da;J */ did exist, then /(0?1]x(0?1] |/(z, y)|d(:c, ?/) would
exist by Fubini Theorem, which contradicts (a). So /0 f/0 \f(x, y)\dx) dy does
not exist.
4221
Let (X,Ai,fi) be a measure space and / £ £(/*)• Evaluate
l/l
2>
lim / n In ( 1 + I — I Id//.
n^°°Jx \ \n J J
(Iowa)
Solution.
It is easy to show that for any x > 0
ln(l + z2) < x.
I /I2
It follows that for any n £ IN the function nln(l + [J-^-) is dominated by |/|
and therefore integrable. By the Dominated Convergence Theorem
lim / n
n->°° Jx
1 lim nln
L
In j 1+( !£)) dM
1+,M)>
Od/i = 0.
4222
Suppose that / is a measurable real valued function on M such that t
etxf(t) is in I^JR) for all x £ (-1,1). Define the function
/+00
etxf(t)dt
-CO

276
for all 16 (—1,1). Prove that <p is difFerentiable on (—1,1).
Solution.
Since for any x G (-1,1),
\etxtf(t)\ < etx-
tx 2 I'lO-H), .,,,.
-re * \f(t)\
l-\x\
t(x+sigati=p-)
\f(t)l
(Iowa)
and
x + signi -
1
x ±
l-ld
<i,
we conclude that t h-» etxtf(t) is also in X1(iR), x G (—1,1)-
Next we show that
. y + oo »+oo
f- I etxf(t)dt = / etxtf(t)dt.
&£ J —CO J —CO
For any fixed x G ( — 1,1), for any y > x,
etvf(t) _ etxf(t) ^
etxtf(t)
y- x
\(et2-etx)tf(t)\, (x<z<y)
\eu"t2f{t){z - x)\ (x<z'< z)
*{
\y-x\etl-¥t2f(t), t > 0 and x < y < ^,
\y - x\etxt2f(t), t < 0 and x < y < 2 .
By the same method as above, we see that t i—» etxt2f(t) is integrable. It
follows that
W-oo V
lim I I " Jy"' ' Jy"' - etxtf(t) )dt = 0
yW-oo V y~x
and
Urn A" (•"*'»-f'">-."«/(«) | « = 0.
1 ./-00 V
»T
y-x
4223
Evaluate
lim / fl + -)"e-2:rds,
n-»oo Jq \ n/

277
justifying any interchange of limits you use.
Hint. First show that (l + f)" < ex for x > 0.
Solution.
Since for any n £ IN and x > 0,
(Stanford)
i=o v y
n
i=0
n! k*
(n — i)!n* i!
1+s(i-;)6-i)-f' -ni
1=1 v ' x '
rri\ n+l)\ n+lj V n + lj i\
x
i=l
n+1
n+ 1
1+£( * JU+t)=(1 + ^+t
and
lim 1 +
n)
have
l+*)"<e*.
v n)
By the Dominated Convergence Theorem
lim I (l + -)ne-2xdx
n-»oo ^q \ n/
f°° I x\n
lim / X[o,n](a;) (1+-) e~2xdx
/ lim x[0n](x) 1+- e 2xdz
/
Jo
e~xdx = 1.

278
4224
Let (X, A, //) be a probability space and / : X —> [1, oo) a measurable
function. Prove or disprove:
I f In fdn> J fdfi J la fdfi.
(Iowa)
Solution.
Under the condition that /In/ is integrable we conclude that both / and
In/ axe integrable, and
J /d/i f ln/d/i< I /ln/d/i.
Indeed, since
0 < / < { e' f(x) -e'
— — 1^ /In/, otherwise
and 0 < In / < / In / we see that both / and In / are integrable. Since
Int < tint for any t > 0,
/ /d// / In/d// — I finfdfi
Jx Jx Jx
= \f (f(x)lnf(y) + f(y)lnf(x))dfz(x)dfz(y)
z JXxX
-\ I (f(x)lnf(x) + f(y)ln(y))dfz(x)d^y)
z JXxX
ZJxxx \ f(x) f(x) f(x)J
< 0.
4225
For i = 1,2, let Xt = W (the natural numbers), Let Af,- = 2W (the er-
algebras of all subsets of JV) and let //; be the counting measure. For the
function / : Xi x X2 —* -K denned by

279
f -2-, j = i,
f(i,j)= I 2-', j = i+l,
[ 0, otherwise.
Compute the iterated integrals
JXi \Jx2
fdfj.2 dfii
and
/ ( / /<*Mi ) dM2-
How do you reconcile your answers with Fubini's Theorem?
Solution.
For any i £ X\, j \—» /(1,.7) is integrable and
/ /(t,j')dM2(j) = -2-' + 2-' = 0.
Therefore
1(1 /d/^Wi = 0.
«, \Jx2 /
For any jel2i»H f(hJ) ls integrable and
1
'2'
We have
f f(i,l)d^(i) =
JXi
f f{i,j)d^{i) = 2-J (j>2).
JXi
(Iowa)
Since / is integrable ( £) 1/1(¾) j) = 2), the two iterated integrals exist
and coincide by the Fubini's Theorem.
4226
Let (fi, /4) be any measure space. For / S L1(fi, /i), and for A > 0 define
p(A) = n({x e n | /(¾) > A})

280
and
i>(X) = li({x€il\f(x)>-\}).
Show that the functions ip and V> are Borel measurable and that
ll/lli= /°°(^A) + V>(A))dA.
Jo
Hint. As usual, it may be helpful to consider / positive first.
(Iowa)
Solution.
The measurablity of <p and V> follows from that they are monotone.
The function (x, A) t—» X(o,oo)(\f\(x) ~ ^) *s measurable. On one hand,
J (Jx(o,oo)(\f\(x)-Wn(*))d\
- J il X{x\\f(x)\>x}(x)dfj,(x)j dX
= / rt{x\\f(x)\>\})d\
Jo
= /0O(^(A)+V(A))dA;
Jo
but on the other,
f ( f0 X(o,oo)(l/l(aO - WX) dfz(x) = // dXdfz(x)
Jn \Jo / Jn Jo
= f \f\(x)d^(x) = \\f\W.
Jn
By Fubini's Theorem we have
ll/lli= /°°(^A) + V-(A))dA,
Jo
provided that either side exists.
4227
Let / G I2(0,1). Set
F(x) = f f(t)dt.
Jo

281
Show that
f fl-h
vJo
F(x + h) - F(x)
h
2 y
dx < ell
for each h, 0 < h < 1 where c is a positive number independent of /.
(UC, Irvine)
Solution.
We have
' yl-/l
Jo
F(x + h)~ F(x)
h
2 5
dx
1 I y»l —ft fX-\-h
' V
dx
rl-7i rx+h
\2dt ■ hdx
*i-h rt
4f f f dt f\f{t)\2dx + f "dt f \f(t)\2dx
Vn \Jo Jo Jh Jt-h
1 I /■i_" rx+n
i
+ f dt f \f(t)\2dx)
Jl-h Jt-h J
-J= ( fh \f(t)\Hdt + f h \f(t)\2hdt + f \f{t)\2{\ - t)dt
V« \Jo Jh Jl-h )
-L ( t \f(t)\2hdt+ /1 h \f(t)\2hdt+ f \f(t)\2hdt)
\h \Jo Jh Jl-h I
4228
Let /, g G X1(0,1) and assume that f(x)g(y) = f(y)g(x) for all x,y £ [0,1].
Show that
/ / f(x)g(y)dxdy = 2 / f(x)g(y)dA,
Jo Jo Ja

282
where
& = {(x,y)\Q<x<y< 1}
and dA denotes the planar Lebesgue measure.
Solution.
We have
(Iowa)
/ / f(x)g(y)dxdy = / f(x)g(y)dA
J0 J0 J[0,l]x[0,l]
= I f(x)g(y)dxdy+ f f(x)g(y)dA
Ja J{(^,y)\o<x=zy<i}
+ [ f(x)g(y)dA
•>{(x,y)\o<y<x<i]
= / f(x)g(y)dxdy+ / f(x)g(y)dA
Ja J{(x,y)\o<x<y<i}
= 2 f f(x)g(y)dA.
Ja

283
SECTION 3
SPACE OF INTEGRABLE FUNCTIONS
4301
Let X be a positive measure space of total measure 1. Show that for any
[0, oo)-valued measurable function / on X,
is a nondecreasing function of p G (l,oo). Under what circumstances is I(p)
strictly increasing as a function of p?
(Iowa)
Solution.
For any p, q G (1, oo) with p < q, let a = £ and /3 = ^ then £ + | = 1.
By Holder's inequality, we have
which shows that I(p) is a nondecreasing function of p G (1, oo). The function
I is strictly increasing if and only if / is not almost everywhere equal to the
constant function.
4302
Let (X,M.,fJ.) be a fixed measure space, let {pi}"=1 be positive numbers
such that pi > 1, i = 1,..., n, p > 1 and
t- = --
f^iPi P
If, for i G {1, ■ ■ ■, n}, /,- G CPi (//), must it be the case that
/i/2"-/„G£p(M)?
Justify.
(Iowa)

284
Solution.
Yes, /1 • ■ • /„ G Cp(fJ,). Moreover
ll/i-"/«llp<IIMU---||/«lk- (i)
Pi P2
In case n = 2, |/i|p G L~* (fJ,) and |/2|p G C^(fj.), and by Holder's inequality,
Assume that the conclusion and the inequality (1) hold for n = k. Then for
n = k + 1,
1 111 1
0<—+ •■■+ —= = pPfe+1 < 1,
which shows that pPkJrlp > 1- By inductive assumption /i • • ■/¾ belongs to
£pi.+i-p(At) and
f(/i-"/fc)"-+1-'J ^11/11^---11/.11^-
Then by the case n = 2 we conclude that (/i ■ ■ ■ /jt)/jt+i G £p and
||(/l---/n)/k + l||p < ll/i---/fc||j^*±i_||/fc + ilU + i
< II/iIIpi '•■ll/fcllptll/fc+illpi.+i)
which completes the proof.
4303
Let X — {a, b, c}, let M. be the er-algebra of all subsets of X, and let fi be
the measure determined by
A*({a}) = 0, fJ,({b}) = 1, and m({c}) = oo.
What are the dimensions of the vector spaces L1^), L2(fi), and Lx(fi)?
Justify your answers.
(Iowa)

285
Solution.
We have dim L1(fz) = dim L2 (//) = 1 and dim L°° (fi) = 2. Since each
function / of L1^) or L2(fi) vanishes at c and two functions taking the same value
at b coincide in L1(fj,) or L2(fi), we conclude that dim X1 (//) = dim L2(fi) = 1.
Since two functions are equal in L°°(fi) if and only if they are identical on
{b, c} we have dim L°°((j,) — 2.
4304
Letf,geL2([Q,l]),
/•i
I fdm = 0.
./o
Show that
( J fgdmj <(f f2dm) I f g2dm -( f gdmj
(Indiana-Purdue)
Solution.
Denoting
we have
Jo
gdm,
2
fgdmj - I f(g - a)dm
< ( f f2dm) ( I (g- a)2dm
= I / f2dm ) ( / g2dm - 2a / gdm + a
= (i'H {lls'dm~ (/.'•*■)')
4305
Let / be a non-negative function in LP (IE) for some 1 < p < oo and let
r + s = p, r > 0, s > 0. Also let /j,(:c) = f(x + h).

286
(i) Show that ftf € L\R).
(ii) Investigate what happens to ||/ft/*||i as \h\ —* oo.
{Indiana-Purdue)
Solution.
(i) Obviously fh G LP(R). From /£ G L$, fs G £*, and | + | = 1, it
follows that
fkf € i1^).
(ii) We claim that
lim \\rhf\\i = o.
\h\—»00
For any e > 0, take N such that /£c fp < £ where £7 = [-JV, TV]. Now if
\h\ > 2N, then x + h £ E whenever x G E. Therefore
Wflf'Wi = I \flfs\ + I l/fc/'l
* (M *(/,'")*+(W(lmf
4306
Let f : R—> R+ be measurable, and let 0 < r < 00. Show that
ijf/i/-(l'l //v
in
(Indiana-Purdue)
Solution.
Set p = 1 + i. Then p > 1 and ± + ^ = 1. Since
„1 = jf/Jr*< (//*-)*(//--")
,^(/,)(/rf,
we have

287
which implies
A-AU'-')
4307
Let / be a bounded measurable function on (0,1). Prove that
lim ( f \f(x)\pdx] ' = ess sup|/|,
p^°° \Jo J
where
ess supl/l = ll/Hoc = inf{< | m({|/| > <}) = 0}.
(Illinois)
Solution.
For any e > 0,
m({* | |/(*)| >||/||oo-e})>0.
Then
= lim lim (H/IU - e)m({x \ \f(x)\ > ||/|U - e})*
e—»0p—»00
< lim lim f / |/(z)|pdz )
e^Op^oo \/{*ll/(*)l>ll/ll=o-<r} /
< lim lim ( f \f(x)\pdx
£—► cop—>oo \Jo
= lim (/ |/(z)|pdz
< Em ( I \f(x)\pdx
<
Therefore
lim ( [ \f(x)\pdx
p^°° \Jo

288
4308
Let / be a nonnegative measurable function on [0,1] satisfying
m({x | f(x) > t}) < T-^J, i>0.
Determine those values of p, 1 < p < oo for which f £ LP and find the
minimum value of p for which / may fail to be in LP.
(Illinois)
Solution.
If 1 < p < 2, then / 6 LP. The minimum value of p for which / may fail
to be in LP is 2.
Indeed, for any p S [1,2),
oo oo
52m({x\f*(x)>n}) = X>({z|/(z)>n?})
n=l n=l
oo . oo .
< £ ——t < £ — < oo,
n=l1 + n" n = ln"
it follows that fp is integrable. Let f(x) =-^--1. Then /2 is not integrable.
However for any t > 0
™({z I/fa) > <}) = mfiaj-r-l^ilj
({"M(iw})
= m
1 1
<
(i + ty i + t2
4309
With Lebesgue measure on [0,1], prove that
5= {/€C[0,1] | ll/Hoo < 1}
is not compact in ^[0,1].
(Iowa)
Solution.
It suffices to show that S is not closed in i1([0,1]). For each n S IN, let

289
1, 0<z<|,
-4nz + (2n+l), §<z<|+^-
Then /„ G S and lim /„ = x\o ii in ^[0, !]• Since X\o ii ¢ 5, 5 is not closed.
n—»0O L ' 2J L ' 2J
4310
Let Ji be Hilbert space X2(0,2ir), with inner product defined by
(u,v) = / u(x)v(x)dx, u,tieW.
./o
Consider the elements u„ £ W, n = 1,2,---, defined by m„(k) = sin(ns) for
x G (0,2ir). Show that
(a) the set {un}^=1 is closed and bounded, but not compact, in the strong
(i.e., norm) topology of H.
(b) un —> 0 as n —+ oo in the weak topology of H, i.e., for every »gW.
lim (u„,v) - 0.
n—»oo
(Sian/onZ)
Solution.
(a) Obviously, the set {un}^L1 is bounded. For any m, n > 1 with m ^ n
\\um — ttnlb = I / (sinms — sinns)2da; ) = Vzir,
which shows that {um \ m G -EV} is closed. Since it admits no convergent
subsequence, it is not compact.
(b) For any v G H, v G X1(0,2ir) by Holder's inequality. By Riemann-
Lebesgue Lemma,
y2)T
lim(tin,v)= lim / v(x)sinnxdx = 0.
n—*oo n—*oo Jq
4311
Let Hi be the Sobolev's space on the unit interval [0,1], i.e., the Hilbert
space consisting of functions / G L2[0,1] such that

290
+ 0O
2
where
1 = E (l + n2)l/(n)|2<°o,
1 f1
f(n)=7T~ / f(x)exp(—2irinx)(h
2*" Jo
are Fourier coeffents of /. Show that there exists constant C > 0 such that
II/IIl- < cil/lli-
(5ian/or«Z)
Solution.
Since for any £ S [0,1]
+ 0O + 0O i + 0O .
E i/(«)«2"'nxi< ( E »2i/(")i2)5( £ ^)+i/(0)i. (1)
n = —oo
n = —oo n= —oo
+ oo
the series ^ /(^)e27rmx converges to f(x) both in #1 and in L°°. It follows
n = —oo
from (1) that
II/IU-<c||/||i.
4312
Let / be a periodic function on M with period 2ir such that /|[o,2tt] belongs
to L2(0,2ir). Suppose
+ 0O
f(X)= e a^inx-
n = — oo
For each h £ M define the function /¾ by fh(x) = f(x — h).
(i) Give the Fourier expansion of //,.
(ii) Find the L2-novra \\fh — f\\2 in terms of the a„ and h.
(iii) Prove that
hm^i>0,
h-o \h\
unless / is constant almost everywhere.
(Stanford)

291
Solution.
(i) We have
+00
h(x) = E *nein(x-h)
n = —00
+ 00
= Yl "ne-inhein-
(ii) We have
(+00
J2 2ir\ane-inh - an\
n = -oo
/ +00
E 8*
. 2 nh. 2
sin — |a„|
(iii) If / is constant almost everywhere, then a„ — 0, n ^ 0. It follows that
fh — f and ||/h — /H2 = 0- If / is not constant almost everywhere, there is an
n ^ 0 such that an ^ 0. Then
h^o \h\ ~ h^o \h\
= Vzir|nan| > 0.
4313
Let fn(x) be an orthonormal family of functions in the Hilbert space
L2(0,1). Prove that
E / f»^dt
7. = 1^°
< X,
for all x G [0,1], and that this inequality is sharp (equality) if and only if
{/„ I n = 1,---} span a dense subspace of £2(0,1).
(Iowa)
Solution.
By Bessel's inequality
00 I *x
E \ f»^dt
n=ll"/o
00 I /-1
= E / X(0,x] (<)/«(<)
n = ll"/°
OO
= EKX(0,*]./»)|2<IIX(0,x]ll2 = *- (1)
n=l

292
If {/„ | n G IN} span a dense subspace of L2(0,1), then {/„} is an orthonomal
basis of L2(0,1) and therefore (1) is sharp. Conversely, since
span{x(0,x] I 0 < x < 1} = span{x(o?6] | 0 < a < b < 1}
is a dense subspace of L2(0,1) while
|/G I2(0,l)|||/||2 = £|(/,/n)|2|
is a closed subspace of L2(0,1), containing span{x(o,x] | 0 < x < 1}, it follows
that
j/ g i2(o, i)|ii/n2 = f; k/, /n)i2| = i2(o, l)
and therefore, {/„} span a dense subspace of L2(0,1).
4314
If / is a function on M, let ft be the translate ft(x) — f(t + x). Prove that
if / is square integrable with respect to Lebesgue measure, then
lim||/t-/||L3(H)=0.
(Stanford)
Solution.
Suppose that / : 2R —> (C be a continuous function with compact support.
Then ft converges to / uniformly. We have
Jim ||/t-/||2 = Hm(7 \f(t + x) - f(x)\2dx
«->o
lim(/ \f(t + x)-f(x)\2dx] =0,
*-° \Jif-[-i,i] /
where if = supp(/) and
K - [-1,1} = {x - y \ x £ K,y ¢[-1,1}}
is compact. In general, there is a sequence {/„} of continuous functions with
compact support converging to / in L [M). Then
fim||/t-/||2 < jm(ll/t-/nt||2 + ||/nt-/n||2 + ||/„-/||2)
< jim(2||/„ - /||2 + \\fnt ~ fnh) = 2||/„ - /||2-

293
Letting n —* oo, we have
lim||/«-/||2 = 0.
4315
A trigonometric polynomial is a function p on M of the form
N
P(6)= J2 c^in\
n=-N
for some C„ G W, N > 0. Suppose / is continuous and 2ir-periodic on M, and
let
r2)r
Show that for every e > 0, there exists a trigonometric polynomial
N
n=-N
such that
m= E c»eine
n=-JV
Il/-Pl|oo <£
and
|Cnl < lanl f°r aU n-
Hint. You may wish to think about harmonic functions on the unit disk.
(Stanford)
Solution.
Regard / as a continuous function on the unit circle S1 of (F. Then
On = -^J m^aidz)
where <r(dz) is the Harr measure on S1 such that c(Sr) = 2ir. Let u be the
harmonic extension of / to the unit disc D = {z \ \z\ < 1}. Then
lim sup \u(rz) — f(z)\ = 0.
Moreover
+ 0O
L(rZ)= J2 «nr|n|z", zeS1.

294
There is an N such that
J2 |a„|rl"l<£/2
|n|>JV
where r is fixed so that
sup \u(rz) — f(z)\ < e/2.
Put
N
p(z)= J2 «nr|n|zn, zeS1,
n=-N
which is required.
4316
Prove or disprove the following statements:
(i) the set of continuous functions on the interval [0,1] is dense in £°°([0,1]).
(ii) L°°([0,1]) is a separable metric space.
(Stanford)
Solution.
(i) False. Since C([0,1]) is separable while I°°([0,1]) is not, C([0,1]) is not
dense in I°°([0,1]).
(ii) False. Let
S = {/ £ L°°([0,1]) | f(x) = 0 or 1, -i- < x < -, n G M, /(0) = 0}.
Then as a subset of X°°([0,1]), S is of cardinal N. However, since the distance
between any two elements in S is 1, S is not separable.
4317
Let {/n}^=i be a sequence in 1^((0,1]) (with Lebesgue measure) and
{9n}n=i a sequence in Li([0,l]), where 1+1 = 1. If lim /„ = / in LP
and lim gn = g in Lq, is it true that fngn —> fg in measure? Justify.
n—*oo
(Iowa)
Solution.
Yes, fngn —* fg in measure. Indeed, since

295
ll/nffn - /fflll < ll/n ~ /llpllffnllg + ll/llpllffn ~ ffllg -» 0 as n -» OO,
we have for any e > 0
A({» I |/ngn(z)-/g(z)| >£})
< - / l/nffn -/ff|(a;)da;
£ ^{^Il/n9n-/9|(x)>e}
. ll/ngn ~ /gill n
< * 0 as n —+ oo,
where A is the Lebesgue measure.
4318
Let X be a measure space with measure fi and suppose that fJ,(X) < oo.
Let
S = {(equivalence class) of measurable complex functions on X}.
(Here, as usual, two measurable complex functions are equivalent if they agree
a.e.) For / £ S, define
p(f)=L
i/i
+i/i
d/i.
Show that d(f,g) — p(f — g) is a metric on S, and that /„ —* f in this metric
if and only if /„ —* f in measure.
(Stanford)
Solution.
Obviously, d(f,g) = d(g,f) > 0 and d(f,g) = 0 iff ^^ = 0 a.e., iff
/ = g in S. For /, g, /i £ <S, we have
*/,!
+ \f-g\ + \g-h\ p
= d(f,g) + d(g,h).

296
Therefore d is a metric on S.
If /n —* / in measure, then i+i}~_/| —+ 0 in measure by the equality
f€XlTT07l *'} = {*eX\U--'l2zh} (0<E<1)'
By the Dominated Convergence Theorem d(/„,/) —* 0.
Ifd(/n,/)-0, then
p({x G X | \fn - f\(x) > e}) = m({zG*
l/n - /!(*)
l + \fn~f\(x)
\fn~f\(x)
>
&})
dx
e i^l + |/,-/IW
d(fn,f)—*0 as n -+ oo.
4319
Let 3 be a measurable function such that /^ |/<?| < oo for every / G I^(2R)
(fixed p > 1). Prove that there is a constant M such that
Jm.
l/ffl < ^I/Ip
all / G Z*(2R).
(Stanford)
Solution.
Define for each n £ IN a measurable function gn
ffn(z)
J ff(z). if
I 0, el
\9(x)\ < n and \x\ < n>
else.
Then gn G Lq(M) and for any / G IP(2R).
Iffi/l<lff2/l<-"<lffn/l<"-<lff/|.
Moreover, lim gnf — gf. It follows that the sequence of bounded linear
n—*oo
functional / h-» J]Rf(x)gn(x)dx on LP(M) converges to / >-* J]Rf(x)g(x)dx,
pointwise. By the Banach-Steinhaus Theorem, f ^> Jm f(x)g(x)dx is
continuous on LP(IR) and therefore there is an h G Lq(M) such that
/ /(3)(/(3) = / /(sWaOds, / G If
Jm Jm
a.e.
which then implies that g =' h G Lq(M)

297
4320
Let S1 denote the unit circle (the set of complex numbers with modulus
one, or the real numbers modulo 2ir). The convolution of two functions on S1
is
2*
f * g(a) = ~ J ' f(6)g(a - 9)dB.
Suppose that / is an element of L2(S1) with the property that its Fourier
coefficient
*>* Jo
2*
n0d6
is non-zero for all n £Z. Show that the linear space {/ * k \ k G L2(Sr)} is
dense in L2(S1).
(Iowa)
Solution.
Let for any n £ Z, Xn(0) = e'n0 then {Xn}nez is an orthonormal basis of
I^S1). We have
(f*Xn)(9) = ~ I"f(a)Xn(6 - a)da
2ir JQ
Jo
p.n0 r2*
. /:_■:.-'""da
2ir
= f(n)em°.
Therefore {/ * k \ k G ^(S1)} contains {%„ | n E.Z} and therefore is dense in
L2(S').
4321
For functions f,g£ L2(M), define the convolution / * g by
r+oo
/+oo
f(y)g(x - 2/)dy,
-oo
where the integral is with respect to Lebesgue measure.
a) Show that / * g G L°°(IR). Do not neglect to check the measurablifcy of
f*9-

298
b) Suppose that / has the property that for all g G L2(M) the convolution
f*g is also an element of L2(M). Define Tf : L2(M) -» L2(M) by Tf(g) = f*g.
It is evident that Tf is a lineaj operator; you need not check this. Show that
Tf is a bounded linear operator.
Hint, closed graph theorem.
(Iowa)
Solution.
Since by Holder's inequality
/+0O / y + OO \ 2 / y + OO \ 2
^ \f(y)g(x-y)\dy < (j_ \f(y)\2dyj (J \g(x-y)\2dy)
< II/II2IMI2, (1)
/ * g(x) is well defined.
a) We will show that / * g is uniformly continuous. Indeed,
Urn nl(/*ff)(si)-(/*ff)(32)|
|x2-Xi|-»0
/+OO
f(y)(g(xi -2/)- 9(22 - 2/))cty
■00
< , Urn JfhfJ \g(x1-y)-g(x2-y)\2dy
k3-xi|->o
r+oc - *
lim ||/||2(7 °° |ff(^i -^2 + 2/)-^(2/)12^
-*i|-»0 V-oo /
|x2-»l|-»0
= 0.
By (1), f * g is bounded and therefore belongs to L°°(1R).
b) Let {gn} be a sequence of L2(M) such that
gn -* g and Tfg„ -* h in L2(M).
Since
1(/ * ffn " / * ff)(s)l < ll/lbllffn " <?l|2 - 0, X G M
we see that h= f * g = Tfg. By the Closed Graph Theorem, Tf is bounded.
4322
If f eL^M), define

299
/-t-oo
f(x)e^dx.
-OO
Prove that, for any / G ^(M), /(0 -* 0 as |£| -* oo.
Solution.
If / is simple, say
then
lim |/(0I = , lim
i
n
J2ak
(Stanford)
(1)
e>?/3fc _ eii°>k
Jfc=l
0.
In general, there is a sequence {/„} of functions of the form (1) such that
lim ||/„ - /||! = 0.
n—►oo
Then
lim" |/(0I < ,Iim" (|/(0 -/„(01+ 1/-(01)
l?l — oo l?l —oo
< lim (||/„ - /IK + |/„(0|) = ||/n - /111-
1^1 — oo
Letting n —> +oo, we have
lim 1/(01 = 0.
1^1—oo
4323
Let / : [0,1] —+ [0, oo) be an essentially bounded function, ||/||oo > 0. Show
that
Jim (^/(^^)/(1/(,)-^
(Indiana)
Solution.
For any a with 0 < a < ||/||oo) let
Ea = {x£ [0,1] | f(x) > a}

300
and Fa - [0, l]\Ea then X(Ea) > 0, where A is the Lebesgue measure. For
any k 6 IN, by the Dominated Convergence Theorem,
nl™{jj^n+kdX)/{jEJ^ndX
< lim —-- / ^-M ttfUtdx
n-> oo \(E,
0.
L_ f (M\
Hence
Jjsa (j[ /Wn+1^) / (/1 f(x)nd
> lim (a / /(z)"dz + / f(x)n+1dx]
/(J f(x)ndx + J f(x)ndx
a.
Letting a | ||/||oo, we get that
}™M!f(x)n+ldx) / (J!f{x)ndx
4324
Let (X,At,//) be a measure space for which (j,(X) < oo. Let 1 < p <
oo. Suppose that {/¾} is a sequence in LP(X) such that sup 11/¾||p < oo and
k
lim fk(x) = f(x) exists for fi-a.e. x. Prove that
n—*oo
lim ||/fc-/||1 = 0.
K—»0O
(Indiana)
Solution.
If on the contrary lim ||/jt — /||i > 0 there exists a subsequence of {fk}, also
k—*oo
denoted {fk}, such that lim \\fk — /||i > 0. Since {|/jt|} is bounded in LP(X)
fc—»00
and LP(X) is reflexive there exists a subsequence {l/jtj} of {1/¾| | k 6 IN},
converging to |/| weakly in LP(X).

301
By the Vitali-Hahn-Saks Theorem, for any e > 0 there exists a 6 > 0 such
that for k = 1,2,---,
j \h>W + f \fW<e
Je Je
whenever (J,(E) < S. For such 6 > 0 there exists by Egroff's Theorem a set E
such that fJ,(E) < 6 and {/¾} converges to / uniformly on X\E. Then
Rm" / |/fcl - /|d/i < IS / |/fcl - /|d/i+ IS / (|/fc,| + |/|)d/i < e.
l^°°Jx '^°°Jx\E >-°°Je
Letting e —+ 0, we get that
lim / |/fcl - /|d/i = 0,
l^°° Jx
a contradiction.

302
SECTION 4
DIFFERENTIAL
4401
Let / : [0,1] -* M and g : [0,1] x [0,1] -* M satisfy the inequality
f(y)-f(x)<g(y,x)(y-x) ( for all x, y G (0, 1)). (*)
Assume also that g is nondecreasing in each variable, i.e., u < x, v < y =>
g(u,v) < g(x,y). Show that lim g(x,y) = 4>(x) exists except in a countable
set and that for 0 < x < y < 1 we have
f(y)-f(x)= [%(t)dt.
Jx
Hint. Observe that (*) is equivalent to
f(y) - f(x)
g(x,x) <g(y,x) < < g(x,y) < g(y,y), 0 < x < y < 1. (**)
y x
(Iowa)
Solution.
Let 4>(x) = g(x, x) then <f> is nondecreasing and
f 4>(x) < g(y,x) < fiy)HZfJx) < g(x,y) < 4>(y), x<y,
\ 4>{v) < g(x, y) < WyZfJx) < g(y, x) < 4>{x), y<x. U
By (1) we have an inclusion
{x | lim g(x,y) ^ $(x)} C {x \ lim <j>(y) ^ <f>(x)}
y-*x y->x
while the latter set is at most countable. So lim g(x, y) ± 4>(x) exists except
y->x
in a countable set. Again by (1) / is Lipschitz and therefore absolutely
continuous. However, since by (1), f'(x) = 4>(x) whenever <j> is continuous at x,
we have
f(y)-f(x)= f f'(t)dt= I* 4>{t)dt.
Jx J X

303
4402
Prove that a function / : [0,1] —* M is Lipschitz, with
\f(x)-f(y)\<M\x-y\
for all x,y £ [0,1], if and only if there is a sequence of continuously differen-
tiable functions f„ : [0,1] —* M such that
(I) \f'n(x)\ <M for all x£ [0,1];
(II) fn(x) -> f(x) iov all x £ [0,1}.
Hint. There are several different ways to do this problem; one is to use the
Fundamental Theorem of Calculus.
(Stanford)
Solution.
If / is Lipschitz, then / is differentiable almost everywhere and moreover
for any x £ [0,1]
fix)=/(o) + r f(x)dx.
Jo
Since |/'(k)| < M, x £ [0,1] there is a sequence {<?„} of continuous functions
such that
|<?„(z)|<M, a; €[0,1]
and
lim / \gn(x)-f'(x)\dx = 0.
For any n £ IN, define
/»W = /(0)+ / g„(t)dt, a; €[0,1],
./o
which is required.
Conversely, by the Mean Value Theorem
\f(x)-f(y)\ = Jim \f„(x) - fn(y)\
= lim \K(t„)\\x - y\
n—>co
< M\x — y\.
This completes the proof.

304
4403
Let {/n} be a sequence of absolutely continuous real-valued functions on
[0,1] such that
oo
(a) f(x) = J2 f„(x) converges for every x E [0,1].
(b)/„ ( £ l/iOOlW <+«>.
Show that / is absolutely continuous on [0, !]•
Solution.
Let
(Illinois)
so*) = £/;(*)
n=l
in ^[0,1]. Then
OO OO
/(*) = £(/«(*)" /«(«)) + £/« W
n=l n=l
oo oo »x
= £/»(<>) + £ / f'niX)dx
= /(0) + / ff(t)dt.
Jo
It follows that / is absolutely continuous.
4404
Assume that / G 4C(7) for every I C R. If both / and /' are in L\R)
show that (i) JR f = 0, (ii) f(x) -> 0 as \x\ -» oo.
(Indiana-Purdue)
Solution.
Since / G 4(7(/), for every I C R,
[' f'(t)dt
Jo
/(z)-/(0).

305
Since /' G L1, lim f_ f'(t)dt exists, which means lim fix) exists. Since
X—» + cx> U X—» + oo
/Gl1, we must have lim fix) — 0. Therefore
X—» + oo
/• + 0O
/ /'(z)dz = lim /(s) -/(0) =-/(0).
In the same way, we have lim fix) = 0 and
X—» — oo
/ /'(z)dz =/(0).
«/ — OO
a yO y + oo
/ f'(x)dx = I f'(x)dx + / /'(z)dz = 0.
Thus
4405
Let {/„} C AC([0,1]), /„(0) = 0 for every n. If {/;} is Cauchy (I1), show
that there is / G AC([Q, 1]) such that /„ —+ f uniformly on [0,1].
{Indiana-Purdue)
Solution.
Since /„ G AC([0,1]),
f,
Thus
.(x)= f f'n{t)dt.
Jo
\fn(x) - fm(x)\ < f \f'n{t) - f'm{t)\dt - 0.
JO
So there exists an / G C([0,1]) such that /„ —* f uniformly on [0,1].
Moreover, there exists g G L1 such that f'n —* g. Then
\f'\fn(t)-9(t)]dt\< I |/;(t)-ff(t)|cft-»0.
Mo I >/o
Therefore
/(¾) = lim / /;(t)dt = / ff(t)cft
n->°° 7o ./o
which implies / G 4C([0,1]).

306
4406
(a) Assume that / G AC (I) and /' G L°°(I). Show that / is Lipschitz.
(b) Show that the following two statements are equivalent.
(1) f : I —> R is Lipschitz
(2) e > 0 =>• 36 - 6(e) > 0 such that {/,- = [a,-, by]} C I with
£ 141 <«=►£!/(&,•)-/(«>)!<*•
(Indiana -Purdue)
Solution.
(a) For any xux2 G /,
l/(z2)-/(zi)|= / /'(z)cfe
/ /'(
<||/'||ockl-^2|.
(b) (1) =>• (2) obviously.
(2) =>• (1). Take 6 > 0 such that £ |/(bj)-/(ay)| < e for any Ij = [aj,bj] C
/> X) 1-¾ I — ^- ^or anv ^i) ^2) € I, Xi < X2, we give the following fact.
Suppose that x2-xi> 6. Take N such that f < x'~Xl < 5. Take {cy}
such that Ki = Co < Ci < • • • < cjv = »2- c* — c,_i < 5. Then
\f(x2) ~ /(a:i)| < £ 1/(¾) - /(c.-OI < TV < 2-\x2 - Xl\.
Suppose that x2 — x1 < 6. Take N > 0, such that | < -/^(^2 — ^l) < ^-
Then N\f(x1) - f(x^)\ < 1. So we have
\f(x2)-f(xi)\<^<-^\x2-xl\.
Therefore we find a Lipschitz constant |.
4407
Let G„, n = 1,2, • • • be open subsets of [0,1] such that
Let
/» = £|Gnn[o,z]|.
n>l

307
Show that
(i) / G AC{% 1]).
(ii) \f(x') - f(x")\ < M\x' -x"\, x', x" G [0,1] iff there exists no such that
Gn = 0, n > no-
(Indiana -Purdue)
Solution.
oo
(i) For any e > 0, take N such that £ ^ < §. Set S = ^. If {(a,-, 6,)}
JV+l
is a sequence of disjoint open intervals in [0,1] and XX^> — o,i) < 6, then
N
< E^ + E
J] |G„n [0,-,6/.
N+l 1
< l + ^E^-^x | + | = e'
which means / G -AC([0,1]).
(ii) Suppose that there exists no such that Gn = 0 for n > n0. Then for
z" > z',
!/(*") - /(*')! = £ |G„ n [a:', z"]| < n0|z" - z'|-
l
Conversely, suppose that for any natural number K, there is k' > K, Gk' ^ 0-
Then GK ^ 0- Take x G G^, 6 > 0 such that (ai-iS,i + S)cGif. Take a,6,
a < b,
K
a,b £ (x - 6,x + 6) C GK = f]Gt.
1 = 1
oo
i/w-/(«)i = EiG»nM
n=l
K
> ^|G„n[a,6]|
Then
n=l
K
Emi
K|6-a|

308
4408
For what values 'a' and '&' is the function
■^ ~ ( |s|asi"l-"l6
x = 0,
sin|x|", x ^ 0
(Iowa)
(i) of bounded variation in (—1,1); (ii) defferentiable at '0'.
Solution.
The function / is of bounded variation if and only if
ml';:, - <">{^0, - <*>{U2>.. <i>
To show (1), let us first establish the following equality
V(/) = ^p(W)I + / \f'(*)\d*
fl \a^- + bcosxb\ \
sup I |easine6| +
l>e>0 \
provided that either side is finite.
Indeed, since / is continuously differentiable on (0,1], for any e > 0, / is
of bounded variation on [e, 1] and
'(x)\dx.
Then
V(/)=/V(
sup f|/(e)|+ / |/'(*)|<te
l>e>0 \ Je
1
SUP (!/(£)- /(0)| + \/(/))
l>e>0
< suP(V(/) + V(/)) = V(/)
l>e>0
0 e 0
sup (|/(e) - /(0)| + \f(Xl) - /(e)| + • • • + |/(1) - /Oc-i)!)
0<e<xi<-<xn = l
< sup (|/(e)| + / |/'(a:)|da
0<e<l \ Je

309
Case I. b = 0. Then sup |/(e)| < oo iff a > 0, while a > 0 implies that
0<e<l
f11 /•// \i 7 f1 I a sin 11 ,
SUP / I/ (2J)|ckc = sup / ———ax <
<e<lJc 0<e<lj£ X1 a
OO.
0<
Case II. b > 0. Then sup |/(e)| < oo iff a + b > 0, for
0<E<1
1/00
sine
-<»+*> I
Since a + b = 0 and a + 6 > 0 imply respectively that
and
x_0 j;26-l 3
x^OXa+b-1
the function /' is integrable on (0,1].
Case III. b < 0. If a + b < 0, there is a 6 > 0 such that 0 < x < 6 implies
that
-n/3 ... \a\ ^ \b\
T^-t^ir-
LetN=\^
+ 1. Then
I \f\x)\dx > j
Jo Jo
6 "-'Icosa;6! - \a\x~b
«.1 — a — b
dx
°° /.(2Jfc7r)b iii
E /
= <
2
~ 13((2^+0-(2^+¾)1^) „,L.ft
Z^ 2(^+6) ' «+°<U>
~ |t|ln(l+-^)
Z^ -2b '
a + 6= 0.
If a + 6 < 0, since
£tU»* + s) '-<»*>■♦*/*<
(2lT)^TT(a+b)
66
and
E*
t=i
b — oo,

310
if a + b = 0, since
and
we have
If a + b > 0, then
+oo;
f1\f'(x)\dx =
Jo
lira In (l + 4--) /4r = 1
oo 1
^121
jt=i
/ |/'(*)|<fe =
Jo
12* = °°'
it=i
-foo.
sup |/(e)| < -foo
0<E<1
and
l/'WI = 1¾^^)^)^1
and therefore /' is integrable.
(ii) / is differentiable at 0 if and only if
f b>0 ( b<0
\ a + 6> 1 °r \ a> 1.
4409
Prove or disprove
a) Let / be a real function on [0, 2ir] satisfying \f(x) — f(y)\ < \x — y\, all
x,y £ [0, 2ir] and f'(x) = 0 a.e. on [0,2ir]. Then / must be constant, b) Let /
be a real valued function of bounded variation on [0, 2ir] with f'(x) = 0 a.e..
Then / must be constant.
(Stanford)
Solution.
a) The conclusion is ture. Since by the condition that
\f(x)-f(y)\<\x-y\, x,y£[0,2ir}
f is absolutely continuous, / is constant.

311
b) is false. Let
/(z) = 0 on [0,71-) and f(x) = 1 on[7r,27r].
Then / is of bounded variation with f'(x) = 0 a.e., but / is not constant.
4410
Let f,g e ^(M). Show
a)
b) if
then
c) if
l^o hjt f{x)dx = /W' a"e-
i ra+h
lim - / f(x)dx = c,
,. r+t f{x + h)-f{x)J
lim / ^ f—^-^dK = f(a + t)-c a.e.
/ \f(x + h)-f(x)
lim / ; g(x)
'm.
then there are constants a, c such that
dx = 0
/(a+ 2)= / 3(s)dK + c a.e.
Can you deduce that f'(x) — g(x) a.e.?
State explicitly the theorems you use.
(Stanford)
Solution.
Assume, without loss of generality, that / and g are real-valued. Since the
function
I f(t)dt= f f+(t)dt- i /_(t)dt
J —CO J—CO J— CO
is of bounded variation, it is difFerentiable almost everywhere by Lebesgue's
Theorem.
a) For any e > 0 there is a continuous function h : M —+ M such that
r+°°
e.
\f(x) - h(x)\dx <
-CO

312
We have
<
<
J -co I a,t J-co
/ b/ (f(t)-Ht))dt + h(x)-f(
J -oo \adj J -co
/+oo I j fi I y + oo
-/ (/(*)-*(*))<« + / 1^)-/(.
■ OO "^ J — CO \J — CO
x)
r+oo,^
Ids
r+0OU
| cfa;
r+oo
:c)|d:c
dx + e
/+OO I J / [X [X \
/+oo f A tx A tx \
.. (s/..(/-Xw,) + £/.tf-X>-«"*)* +
/ + oo ____
|/-ft|(a;)(ic + e < 2e.
-oo
Letting e —> 0, we see that
- + 0O
and therefore
, + 0O I . ,*
/ b/ w*-**)
«/ — oo I ""^ «/ — oo
r fit)dta^f(x)
J — oo
cte = 0,
_d_
etc
b) We have by a)
f{x + h)- f(x)
lim f
*-»°\./« h Ja h J
dx
a+t
lim if
ill
a+h+t
h
dx
lim
h
a+t+h f(T\ fa+h
~Tdx~ I h
+t n .L n
i
= f(a + t)-c a.e. t G M.
c) By a), there is an a S M and a c such that
I ra+h
lim — / fix)dx = c.
h^o h Ja

By b) and by the assumption, we have
ra+t
f(a + t)— / g(x)dx — c
Ja
Therefore
i.e.,
0, a.e. < G M.
ra+t
f(a + t)= g(x)dx + c a.e. t £ M,
Ja
/(z) = / g(t)dt + c a.e.xeM.
Ja
It follows that f'(x) = </(a;) a.e..
4411
Let F : (a,b) —* M be measurable.
(1) Prove that the following two statements are equivalent:
(a) There is an / S L2(a, b) such that
F(x) = / fdm (ra the Lebesgue measure);
Ja
(b) There is an M > 0 such that
y.|f(st)-.F(st-l)|2 <Jtf
for any finite partition £o < a?i < • • • < xn of (a, 6).
(2) Show that the smallest constant M in (b) is equal to \\f\\\.

314
Solution.
(1) (a) =>• (b). By the Cauchy-Schwarz inequality
£1^)--^-1)12
(=i
%i %i-l
f(x)dx
<
i I rxi
E—— \ ■
ij;^Mf-.w,,|,*)(£>)
ii/iii. (i)
(b) =>• (a) Obviously, for any mutually disjoint intervals (an,/3n) of (a, b),
n G IN, we have
£|fQ3„)-f(«„)|2 <M
n=l
/3n - «n
(2)
Since
^|F(A)-F(aO
«=i
" |F(fl.)-F(a,-)| /s—-
u'=l
A - a.
1.1=1
<
\
U^ifii-a,),
1=1
F is absolutely continuous, and therefore there is an integrable function / :
(a, 6) —+ M such that
F(z)= J f(t)dt + c, x G (a, b).
J a
We will show that if E1, ■ ■ ■, En are any mutually disjoint measurable sets,
then for any e > 0
y —nk—I f f(x)d2
^m(Ei) + e\jEiJK
< M.
(3)
If E{'s are open sets of (a,b), say .E,- = U (a,-,-,/3,-,-) (A, at most countable),

315
then by inequality (2)
■0U r\2
I V riJ f\
El Jej J I _ V^ JGA, u
m( F,A A- r ~ 2-^
i€A; V'J a,J u
£
<
ra(.E;) + e
i€A;' J J jeA,-
E ^1./^/I'.£&%-«.>>
£
.•=li€A,^~"a<>
< M.
ra(#i) + e
r;/i2
If i?,-'s are compact sets there is for each i a decreasing sequence {E{j}JL1 of
oo
open sets of (a,b) such that Ei - f] Eij and Etlr\Ekl = 0 (i ^ fc). Applying
i=i
(3) to the mutually disjoint open sets Eij, ■ ■ ■, Enj and passing to the limits
we see that (3) holds for Ei,- ■■ ,En.
In general, there is for each i an increasing sequence {Eij}Ji1 of compact
oo
sets such that I) Eti C Ei and lim fJ-(Eij) = m(E{). Applying (3) to the
mutually disjoint compact sets Eij, ■■ ■, Enj and passing to the limits we see
that (3) holds in general.
To show that / is square integrable, it suffices to show that
oo
y^ nm(En) < oo,
n=l
where En = {x\n< \f{x)\2 < n + 1}. Let
E^ = {x\y/n< f(x) < Vn+l}
and
E~ = {x\ -Vn+l < f(x) < -y/n}.
Then
^jnm(En)

316
= £(nm(tf+) + nm(tf-))
n=l
< lim V lim , ,, / /
= lim lim > —-—x- / /
+
+
1
m(tffc ) + e
1
m(£-)+e
Je~
Jet
< M.
(2) Define, for each / G L2(a,b),
sup
£
a=xo<",<arn =6 \ ._- *^i *^i—1
2\ 2
then || ■ || is a norm on L2(a,b) by (3). Moreover ||/|| < ||/||2 by (1). For any
/ G C[a,b], there is £, G (zj,-1,:^) (¾ = a + £(& - a)) such that
)■
Then
A™Ei^«)i2(<-<_1)
i=l
limV U*-, .'
n—»00 ^-^ v* — vi.
<
It follows that 11/11 = H/H2 for any / G C[a,6]. For any / G I2(a,6) there is a
sequence {/n}£Li of C[a,b] such that ||/n — /||2 —+ 0, so
We have
lim ||/„ - /|| < lim ||/n - /||2 = 0.
n—►oo n—>oo
/||= lim ||/n||= lim ||/„||2 = ||/||2.
This completes the proof.

317
4412
Let X = [0,1], M the cr-algebra of Borel subsets of X. Let a(t) = 22,
/3(2) = 23 and define measures /4 and <f> on M by
fi(E) = J Ida and <f>(E) = [ ld/3.
J E Je
Does j£ exist? Does ^ exist? Compute the value of the Radon-Nikodym
derivatives that exist. Justify.
(Iowa)
Solution.
For any
and
Borel subset E of X,
KE)
4>(E) =
Je
*
2tdt
-- \ 322d2
Je
We have
fi(E) = 0 O 22 = 0 a.e. 2G Eo X(E) = 0
O 322 = 0 a.e. teEo X(E) = 0
«• 4>{E) = 0.
It follows that /4 ~ 4> ~ A, where A the Lebesgue measure. We have
djJL_djJL /dj>_ _ , 2 _ _2_
d<!>~ dXl dX ~ '6 ~ 32
and
d<t> _ 32
djc ~ ~2'
4413
Let A be the Lebesgue measure on M and /4 and v the Borel measures on
M defined by:
oo 1
^A) = E2^A((n'2n)n^'
n=l
oo 1 „
3"
n=l

318
Is /4 <C v or v <C /*? Find the corresponding derivaties if they exist.
(Iowa)
Solution.
It is true that v <C /* and it is false that /4 <C v- Because fJ,(A) = 0 implies
for all n G ff, A((n,2n) D -4) = 0 and therefore fi((n, §n) D -4) = 0, i.e.,
zv(.4) = 0; however, let A = (§, 2] and then v(A) = 0 but /i(.4) = \.
We show next that
Since
and
we have
and
dv
dfi
f 0, x < 0,
oo
^ , a; > 0.
X) FTX(„,2„](^)
n = l
/" °° 1
M(^) = / X) ^TX(„,2n](z)dz
^An = l
/" °° 1
^n = l J
dfi _ y^ 1
^-2^ 2S"X(».2»1
oo 1
Zj Qn^(».f"]'
n = l
(1)
Therefore (1) holds.
4414
Let /4 be the Lebesgue measure on [0,oo]. Define
oo 1
^ n3 JEn[n,n + l]
to(E)
JEr\[l,cc] x
dfi
for any Lebesgue measurable subset E of [0, oo]. Is fi <C /*2 or/and fi2 <C Mi?
If so, find the corresponding derivatives.
(Iowa)

319
Solution.
Since
/" °° 1
M#) = / J2 ZSXln,n+i)(x)dx
jEn=in
f*2(E) = / X[i,oo)(a;)^<ic,
we have
oo 1
fj.i(E) = 0 & ^2 -^X[n,n+i)XE(x) = 0, a.e. x G [0, oo)
n=l
O \/n<EM, XEn[n,n+i)(x) = 0, a.e. z e [0, oo)
O Vn G JZV, ^(£7 n [n, n + 1)) = 0
O /i(S (~1 [1, oo)) = 0
O XEn[i,oo){x)— - 0
O fj,2{E) = 0.
It follows that both /ix <C /^2 and /i2 <C /ii- Moreover
d/x2 0
d/i2
d/ii
(z) = {
1
X2 ( E prXIn.n + ljW
0 < a; < 1,
, z> 1,
0, 0 < x < 1.
From /i2([0,1)) = 0 and /i([0,1)) = 1, /i <C ^2 does not hold.
4415
Let 4> : M —* M be a bounded, continuously differentiable function with a
bounded derivative, and assume g 6 L(M). Define
z))dz, i e JR.
a) What additional assumption on <^> will insure that / is well defined? (i.e.,
that 4>(tg(-)) G I(2R) for all < e 2R).

320
b) Under the additional assumption in a) above, show that / is differen-
tiable, i.e., f'(t) exists for alii £ St.
(Indiana)
Solution.
a. Assume <f>(0) = 0 then / is well defined. Indeed, for any t, x there exists
an s with 0 < s < 1 such that
Since
$(tg(xj) = tg(x)$'(y) \y=stg{x) ■
sup \4>Xy)\ < oo 4(tg(.)) e L(M).
y£lR
b. We will show that
/'
'(<) = / *'(*</(
Jm.
x)g(x)dx.
Indeed, for any s,t £ 2R (s ^ t),
$(sg(x)) - $(tg(x))
s-t
= \(<t>'(rg(x)) - 4>'(tg(x))g(x
< 2 sup \4>'(y)\\g(x)\
ytm.
- 4>'(tg(x))g(x)
and
lim
s->t
4>(sg(x)) - 4>(tg(x))
s-t
- 4>'(tg(x))g(x)
0.
The conclusion follows from the Dominated Convergence Theorem.
4416
Let {fk} denote a sequence of nondecreasing functions defined on (0,1)
with the property that lim fk(x) = 1 for almost every x £ (0,1). Prove that
k—*oo
lim fk(x) = 0 for almost every x £ (0,1).
k—*oo
(Indiana)
Solution.
There are sequences {an} and {&„} such that 0 < an < bn < 1,
lim (bn -an)=l

321
and
lim fk(an) = lim fk(bn) = 1
k—*oo k—*oo
for any n £ .EV. By Fatou's Lemma, for any n £ IN
/ lim fl(x)dx < Hm / fk(x)dx
< lim(/fc(6„)-/fc(a„)) = 0.
it—»00
It follows that lim /^ is integrable and
fc—*oo
I
1
lim fk(x)dx = 0.
0 fc—»00
Hence lim f'k{x) — 0 for almost every x £ (0,1).
k—*oo

322
SECTION 5
MISCELLANEOUS PROBLEMS
4501
Let S C R be a set of real numbers with the property that |SH \-Sn \ < 1
for every finite subset {Si, ■ ■ ■, Sn} C S. Show that S is countable.
(Indiana -Purdue)
Solution.
Since S n (£, n) and S n (—n, — £) must be finite and
S = Sn(\J(dn)U(-n,-h)U{0}\,
\n = l /
S is countable.
4502
Let {Ia}, a £ I\ be a collection of closed intervals in R. Show that
U M u '°
is countable.
(Indiana-Purdue)
Solution.
Obviously U /° is an open set. We have
aer
oo
Ul°=U(an,/3n),
a€T n=l
where an,/3ne [J /°. For any x £ U^AU^ there must be a such that
agr a a
x £ /a, but z can not be in 1°. Take n such that
/°C(a0,/3„).

323
If Ia C (an,/3n), then x ¢ Ia, which is a contradiction. Thus we must have
x = an or /3n.
Therefore
UMU^CUK./?"}-
a a n
which implies U-^AU-^* *s countable.
a a
4503
Show that any infinite set of non-empty, mutually disjoint, open sets in a
separable metric space X is countable.
(Stanford)
Solution.
Let {Ui | i G 1} be such an infinite set. For each i G I there is an a* G D
such that a,- G £/";, where D is a countable dense subset of X. Then a; ^ a/
whenever i ^ j. It follows that the map, i >—» a* of / to D is injective and
therefore I is countable.
4504
Prove that for almost every x G [0,2ir]
lim sin(ns) = 1.
n—»00
(5ian/o7t2)
Solution.
Let
X
A — {x G (0, 2ir) | — is irrational}.
Then A is a measurable set of measure 2ir. Moreover, for any xGi,
lim sin(ns) = 1.
n—*oo
Indeed for any xei, since {fc^ — 2/ | k, I G Z} is a dense subgroup of M,
there are sequences {kn} and {/„} of if such that
x 1
lim (&„ 2ln) = -.
n—*oo 7f Z

324
Since 5 ¢ {k^ — 2l \ k,l £Z}, {kn} admits a subsequence {k'n} either increasing
to +00 or decreasing to —00. If lim k'n = +00 then
n—»00
lim sin(k'nx) = lim sm(k'nx — 2/^ir) = 1;
n—*oo n—*oo
Otherwise lim (—3k'n) = +00 and
n—.00
lim((-3*;)J + 2(3Z; + l))=i
and therefore
lim sin(-3&;z) = lim sin(-3*4:c + 2(3l'n + l)ir) = 1.
4505
Let / G C(-f)- Show that there exists a sequence of polynomials {pn} such
that p„ —+ f uniformly on I and Pi(x) < ^2(2) < • • ■ for every i£/.
(Indiana-Purdue)
Solution.
For any n, take a polynomial pn such that
lf»W-[/W-^]l<^. *e/.
Then
Obviously, Pi(x) < P2(x) <••• and p„ —+ / uniformly on I.
4506
Let / G C([0, 1]). Show that there is a sequence of odd polynomials pn(x)
with pn~* f uniformly on [0,1] iff /(0) = 0.
(Indiana-Purdue)
Solution.
Suppose that there is a sequence of odd polynomials pn with pn —+ f. Then
/(0) = lim p„(0) = 0.

325
Conversely, suppose that /(0) = 0. Set
f(x) = -f(-x), ie[-i,o].
Then / is a continuous function on [—1,1]. Take a sequence of polynomials
pn(x) with pn —+ f uniformly on [-1,1]. Set
Pn(x)= -\p„(x) - Pn(-X)}.
Then Pn is an odd polynomial for any n. We have
lim Pn(x) = lim -\p„(x) - p„(-x)}
n—>oo 2*
= \\f(*) -f(-x)} = f(x)
uniformly on [0,1].
4507
a) Show that the mapping I: C[0,1] -* C[0,1]
(If)(*) = *' + \£ WW
is a contracting mapping on C[0,1], with the supremum norm.
b) Show that there exists one and only one smooth function / on [0,1]
satisfying the conditions:
/ £f(x) = e* + xf(x%
I /(0) = i.
(Stanford)
Solution.
a) For any / and g in C([0,1]),
||I/ - Ig\\ = max \ f \f(t) - g(t)\dt < h\f - ff||.
x€[o,i] 2 JQ 2
b) By the Banach's Theorem, there is only one function / E C[0,1] such
that 1(f) - /, i.e.,
f(x) = e*+±JX f(t)dt.
(1)

326
By (1) / is smooth and satisfies the conditions:
{
/(0) = 1.
(2)
If g is another smooth function satisfying (2), then
2
1 f
g(x) = ex + -J g(x2)dx,
i.e., 1(g) = g, by the uniqueness of /, g = f.
4508
Given / : R —> R bounded and uniformly continuous and {Kn}, n =
1,2, 3, ■■■,#„ E L1 such that
(i) \\Kn\\i < M<oo, n = 1,2,3,---.
(ii) J Kn —> 1 as n —> oo.
(i") /{x:|x|>«} l-K"n| -* 0 as n -+ oo for all 6 > 0.
Show Kn * f —* f uniformly.
(!7C, Irvine)
Solution.
Take Mx > 0 such that |/(z)| < Mx for all x <E R. For any e > 0, by (ii)
there is an Ni such that
If n> .ZVi. Thus
|y"xn(y)dy-
\J Kn(y)f(x)dy - f(x)
<
2Mi
< Mi-
e e
2Mi 2
(1)
Take 5 > 0 such that
\f(x - y) - f(x)\ <
4M
holds for all x £ R any |y| < 5. For the above 5, take N2 such that
/ \Kn(y)\dy<
J{y-\y\X}
8M1'

327
if n > N2 ■ Therefore
(Kn*f)(x)-J Kn(y)f(x)dy
< J\Kn(y)\\f(x-y)-f(x)\dy
= I \Kn(y)\\f(x - y) - f(x)\dy
J{r-\y\>t]
+ [ \Kn(y)\\f(x-y)-f(x)\dy
J{y:\y\<6}
< 2Mx / \Kn(y)\dy+ [ \Kn(y)\dy
J\v\>6 J\v\<6
£
AM
< 2^1-8^ + 11^111-4^
< i + M.-^=£-.
4 AM 2
Set N = max(Ni,N2). It follows from (1), (2) that
\(Kn *f)(x) -f(x)\ < \(Kn*f)(x)-J Kn(y)f(x)dy
+
\jKn(y)f(
x)dy - f(x)
holds for all x <E R if n > N.
£ £
< 2+2=£
4509
(2)
Let f : R —> (—00,00) be upper semicontinuous and define m : R —>
[-00,00) by m(x) — liminf f(y). Let S = {x \ f(x) — m(x) > 1}.
y->x
(a) Show S is closed.
(b) Show: HI is an open interval contained in S, then m(x) — —00 on I.
(c) Show that S is nowhere dense.
(UC, Irvine)
Solution.
(a) For any xq £ Sc, there exists e > 0 such that
f(x0) — m(xo) < 1 — e < 1.

328
Take 6 > 0 such that
f(x) < /(a*) + |, f(x) > m(x0) - |
for £ G 0(xq,S). Therefore
m(x) = liminf f(y) > m(xQ) - -
y-*x I
holds for x G O(xo,6). Thus
/(a;) - m{x) < /(z0) + r - m(x0) + -<l-e + e = l
if x G O(xo,6), i-e., x G Sc. So Sc is open.
(b) Suppose that there is xo G I such that m(xo) > —oo. As in (a), we can
find S > 0 such that m(s) > m(xo) — \ for x G O(zo; £) C /■ By the definition
ofra(i), there must be x G O(xo,6) such that /(5) < m(xo) + |. Therefore
/(z) — m(x) < 1, i.e., x^S which contradicts z G I C 5.
(c) Suppose that S is not nowhere dense. Since S is closed, there is an open
interval I C 5. From (b), m(s) = —oo for i£/. Denote
4i = {ie/,/(i)<-i}.
j4i is a non-empty open set. Take an open interval I\ such that 7i C A\. In the
same way, we can take an open interval /2 satisfying I2 C /1 and /(k) < —2
for x G /2- By induction, we obtain a sequence of open intervals {/„} such
that In C /„-1) /(z) < —n f°r z G /„. Obviously f(x) = -00 for z G C\In,
which is a contradiction.
4510
Prove that any topological metric space is homeomorphic to a bounded
metric space.
(Stanford)
Solution.
Suppose that (X,d) is a metric space. Define another metric d on X by
d(x,y) = v , x,y<EX.
It is easy to show that the identity mapping of X is a homeomorphism of (X, d)
to(X,d).

329
4511
Let M be a metric space with distance function d, suppose A : M —* M is
a distance nonincreasing periodic map of order 3, i.e.,
AoAoA = Id and d(Ax,Ay)<d(x,y).
(i) Show that A is a continuous bijective isometry.
(ii) Give an example of a complete metric space M and an isometry A on
M, periodic of order 3, which has exactly two fixed points.
(Stanford)
Solution.
(i) Since A-1 = A2, A-1 is distance nonincreasing, too. Therefore A is an
isometry and therefore continuous.
(ii) Let M be the 2-dimensional sphere
M = {(z, x) G <T x M | \z\2 + x2 = 1}.
The isometry A: M —* M defined by
A(z,x) = (e*T'z,x), (z,x)<EM,
is a periodic map of order 3, having exactly two fixed points (0,1) and (0,-1).
4512
Let if be a Hilbert space and let / : H —> M be a continuous convex
function such that f(xn) —> oo whenever \\xn\\ —> oo. Prove that / attains a
minimum.
(SUNY, Stony Brook)
Solution.
Let a = inf/(if). There is a sequence {xn} of H such that
lim f(xn) = a.
n—*oo
If
sup||a;n|| = +oo,
n
there is a subsequence {ainj^j such that \\xnk\\ —* +oo, then
a = lim f(xnk) - +oo,

330
a contradiction. So sup ||a;„ || < +00. By the weak compactness of the closed
n
ball of H, there is a subsequence {xnk} converging weakly to a point x £ H.
For any /3 > a there is an N £ IN such that for any k > N, f(x„k) < /3.
Since
W y/ 11.11
x £ {xnk I k > N} C cov{xnk I k > N} = cov{z„k | k > N}
and for any y £ cov{snk | k > N}, f(y) < /3, one has f(x) < /3. Therefore
a < f(x) < lini/3 = a.
/3J.a
It follows that a is finite and / attains its mininum at x.
4513
A is the subset of ^(M) consisting of all functions / satistying \f(x)\ < 1
a.e. on M. Prove that A is closed in the norm topology of X1(2R).
(Stanford)
Solution.
If {fn} is a sequence of A such that / = lim /„ exists in L1(2R). Then
{/n} converges to / in measure and therefore by F.Riesz Theorem there is a
subsequence {/„,,} converging to / almost everywhere. It follows that \f(x)\ <
1 a.e. on M. So A is closed.
4514
Let A be a bounded linear operator on Hilbert space H. Recall that the
adjoint A* is the unique bounded linear operater on H such that (Ax,y) —
(x, A*y) for all x,y £ H.
(a) Show that ||j4*|| = ||j4||, where ||.A|| is the norm of A.
(b) Show that AA* — A* A cannot be the identity on H. (You may wish to
use (a) prove this.)
(Stanford)
Solution.
(a) Indeed, since
11^*11 = SUP 11^*2/11 = SUP SUP \(x,A*y)\
113/11 < 1 I|9||<1||*II<1
= sup sup \(Ax,y)\ < \\A\\
l|9ll<l|N<l

331
and a fortiori \\A\\ — ||-4*||.
(b) First, we will show that
\\A*A\\ = ||4|2 = sup (A* Ax, x). (1)
Indeed equalities (1) follow from the following inequalities
||,4||2 = sup (A*Ax,x) < sup \\A*
Nl<i Nl<i
= ll^^l < 11^111141 = 11^12.
If for some bounded linear operator A on H, AA* - A* A = I then
||A4*|| = sup (AA*x,x)
H*||=i
= sup (A* Ax + x,x)
= sup (A*Ax, x) + 1
11*11=1
= \\A*A\\ + 1.
It follows that ||j4||2 = ||j4||2 + 1, a contradiction.
4515
Suppose that A,BCM are Lebesgue measurable, with m(A) > 0, m(B) >
0. Show that
A + B = {xeM\x = a + b, aE A,b ^ B}
contains an interval of positive Lebesgue measure.
(Indiana)
Solution.
Assume without less of generality that A and B are compact sets. Since
/ m((x - B) (~1 A)dx = /(/ X(x~B)nA(y)dy) dx
Jm. Jir \Jm. J
= /(/ Xb(x - y)xA(y)dy) dx
= m(A)m(B) > 0,

332
there exists an xq £ M such that m((xo — B)^A) > 0. Since x i—» m((x — B)C\A)
is continuous, the set {x \ m((x — B) O A > 0} is nonempty and open. Then
the conclusion follows from the following inclusion
{x | m((x - B) 0 4) > 0} C A + B.
4516
Let X be a compact Hausdorff topological space and let /i be a finite regular
Borel measure on X. Is it true that if / : X —► M is /i-measurable then
there exists a sequence {/n}^Li of continuous real-valued functions such that
lim fn = f a.e. [//j? Justify.
n—»0
(Iowa)
Solution.
Yes. For any n £ IN there is a compact subset Fn of X such that
n
fi(X\Fn) < £ and / is continuous on Fn. Let Xn = [J Fi. Then (1)
i = l
fi(X\Xn) < ^-, (2) / is continuous on Xn, and (3) fi I X\ [J Xn I = 0. There
V n>l )
is for each n £ IN a real-valued continuous function /„ : X —► M such that
/„ = / on Xn. Then lim /„ = / a.e. \p].
n—*oo

Part V
Complex Analysis

335
SECTION 1
ANALYTIC AND HARMONIC FUNCTIONS
5101
True-False. If the assertion is true, quote a relevant theorem or reason; if
false, give a counterexample or other justification.
(a) if /(z) = u + iv is continuous at z = 0, and the partials ux, uy, vx, vy
exist at z = 0 with ux = vy and Uy — —vx at z = 0, then /'(0) exists.
(b) if /(z) is analytic in fi and has infinitely many zeros in fi, then / = 0.
(c) if / and g are analytic in fi and /(z) • g(z) = 0 in fi, then either / = 0
or g = 0.
(d) if /(z) is analytic in fi = {z; Rez > 0}, continuous on fi with \f(iy)\ < 1
(-00 < y < +oo), then |/(z)| < 1 (z £ ft).
(e) if ~Y^anzn has radius of convergence exactly R, then 'YJn3anzn has
radius of convergence exactly R.
(f) sin y^z is an entire function.
(Indiana -Purdue)
Solution.
(a) False. A counterexample is f(x,y) = •y'^cyl. / satisfies Cauchy-
Riemann equations at z = 0, but /'(0) doesn't exist.
(b) False. A counterexample is /(z) = sin^^i- / is analytic in fi = {z :
\z\ < 1}, and has zeros z = 1 — ^, n = 1, 2, • • •. But / is not identically zero
in fi.
(c) True. If neither of / and g is identically zero in fi, then both / and 3
have at most countably many zeros in fi, and the zeros have no limit point in
fi. Then /(z) • g(z) is not identically zero in fi.
(d) False. A counterexample is /(z) = ez, which is analytic in fi, and
continuous on fi with \f(iy)\ = 1. But /(z) is not bounded in fi.
(e) True. Because lim vn? = 1, it follows from
n—»oo

336
(f) False, sin y/z is not analytic at z — 0. Actually, z = 0 is a branch point
of sin ^fz.
5102
(a) Let /(z) be a complex-valued function of a complex variable. If both
/(z) and zf(z) are harmonic in a domain fi, prove that / is analytic there.
(b) Suppose that / is analytic with |/(z)| < 1 in \z\ < 1 and that /(±a) = 0
where a is a complex number with 0 < \a\ < 1. Show that |/(0)| < a2. What
can you conclude if this holds with equality.
(c) Determine all entire function / that |/'(z)| < I/(2) I-
Solution.
(a) It is well known that the Laplacian can be written as
d2 d2 A d2
A = 1 = 4 -.
dx2 dy2 dzdz
Because
it follows from
^L(,A,), = ^/,,, + ,^/(,,,
=/W = o
and
that
dzdz
d2
dzdz
(zf(z)) = 0
dz
which implies that /(z) is analytic in fi.
(b) Define
„/ , ,, , 1 - az 1 + az
F(z) = f(z) —,
z — a z + a
then F(z) is analytic in {\z\ < 1}. When \z\ = 1,
1 — az 1 + az
z — a z + a
1,
hence
lim \F(z)\<l,
(Stanford)

337
which implies that |.F(2)| < 1 for |z| < 1. Take z = 0, we obtain
1/(0)1 < H2.
When it holds with equality, we have F(z) = e'e, which is equivalent to
2 + a
m = ei9Y
az 1 + az'
(c) From |/'(2)l < l/(2)l> we know that / has no zero in (F, which implies
that =¾¾ is also an entire function. It follows from 4-f^f < 1 that tW = c,
T\z) i\z) J\z)
\c\ < 1. Integrating on both sides, we obtain log/(2) = cz + d. Hence /(2) =
c'ecz, where c and c' are constants and \c\ < 1.
5103
Let G be a region in <F and suppose u : G —* M is a harmonic function.
(a) Show that |^ — i^- is an analytic function on G.
(b) Show that w has a harmonic conjugate on G if and only if ||- — i|^- has
a primitive (anti-derivative) on G.
(/nrfiana)
Solution.
(a) Let
-P(z, 2/) = ^-» Q(x,y) = --*-■
ox ay
Because u is a harmonic function, we have
dP__dQ^_ (Pu d2u _
dx dy dx2 dy2
We also have
dP_ 5Q _ d2u d2u
dy dx dxdy dxdy
So P(x,y) and Q(x,y) satisfy the Cauchy-Riemann equations, hence
... du .du
P + iQ=—-x —
ox ay
is analytic on G.
(b) If |f- — if2- has a primtive, then for any closed curve c C G, the integral
f (du .du\ , f Idu , du \ . ( du , du , \
JArx-ld-y)dz-JArxdx + Vxdy)+i{-Vydx+o-xdy)=0-

338
It follows that
f du du ,
Jc dy dx
holds for any closed curve c C G. Hence we can define a single-valued function
v(x,y) on G:
v(x,y)= / -—dx + — dy,
Jzo dy ox
where z0, z £ G, and the integral is taken along any curve connecting z0 and
z in G. Because
dv _ du dv du
dx dy' dy dx'
we know that v(x, y) is a harmonic conjugate of u(x, y) on G.
On the contrary, if u has a harmonic conjugate v on G, then
, dv . dv , du , du ,
dv = ^-az + —dy = - —cte + -^-dy.
dx dy dy dx
For any closed curve c C G, we have
l\Tx-lTy)dZ = JAo-xdX+d-ydy)+l{-d-ydX+d-xdy)
t(du, , du, \ /dv dv, \
JATxdx+d-ydy)+id-x + rydy)
d(u + iv) — 0.
Hence £ - ^% has a Primitive //. (if - ^) <** on G.
5104
Suppose that u and v are real valued harmonic functions on a domain fi
such that u and v satisfy the Cauchy-Riemann equations on a subset S of fi
which has a limit point in fi. Prove that u + iv must be analytic on fi.
(Indiana-Purdue)
Solution.
Because u and t> are harmonic functions, /i = ^- — i'■-£ and /2 = |^ — i-^-
are analytic functions on fi. The reason lies on the fact that the real and
imaginary parts of /i and /2 satisfy the Cauchy-Riemann equations respectively
(see 5103 (a)).
I:

339
By the assumption of the problem,
du _ dv du dv
dx dy' dy dx
when z G S C ft- Hence
_ du ,d
dx
.du _ _ . /dv .dv\
dy ~ 2~l \dx ldy)
when 2 G S. Because the subset S has a limit point in ft, by the uniqueness
theorem of analytic functions, we know that /i = i/2 holds for all 2 G ft. It
follows from /1 = i/2 for 2 G ft that
du _ dv du dv
dx dy' dy dx
for 2 G ft, which implies that u + iv is analytic on ft.
5105
Let Q = [0, 1] x [0,1] C<T be the unit square, and let / be holomorphic in
a neighborhood of Q. Suppose that
/(2 + 1)- /(2) is real and > 0 for 2 G [0, i]
/(2 + i) - /(2) is real and > 0 for 2 G [0, 1].
Show that / is constant.
(Indiana)
Solution.
Because / is holomorphic on the closed unit square Q, by Cauchy integral
theorem, we have
I f(z)dz = I f(x)dx+ I f(l + yi)idy- [ f(x + i)dx
JdQ JO JO Jo
- / f(yi)idy
Jo
= I {fix) - f{x + i))dx + i f (/(1 + yi) - f{yi))dy
Jo Jo
= 0.
As
fix) -f{x + i)<0

340
for 0 < x < 1 and
/(1 + yi) - /(j/t) > 0
for 0 < y < 1, by comparing the real and imaginary parts in the above identity,
we obtain that f(x+i) = f(x) for 0 < x < 1 and f(l+yi) = f(yi) for 0 < y < 1.
Hence /(z) can be analytically extended to a double-periodic function by
/(z) = /(z+l) = /(z + i),
which is holomorphic in (F and satisfies
|/(z)|<max{|/(z)|}<+oo.
This shows that /(z) must be a constant.
5106
Let / be continuous on the closure S of the unit square
S = {z = x + iy eW : 0 < x < 1,0 < y < 1},
and let / be analytic on S. If R/ = 0 on S (~1 ({y = 0} U {y = 1}), and if I/ = 0
on S fl ({k = 0} U {x = 1}), prove that / = 0 everywhere on S.
(Indiana)
Solution.
Define F(z) — JQ f(z)dz, where the integral is taken along any curve in S
which has endpoints 0 and z. Then F(z) is analytic in S and continuous on
S. For z G OS, we choose the integral path on dS and consider the differential
form f(z)dz in the integral. Let / = u + iv, then
f(z)dz = (udx — vdy) + i(vdx + udy).
On 5 (~1 ({y = 0} U {y = 1}) we have u = 0 and dy = 0, and on 5 (~1 ({x =
0} U {x = 1}) we have v = 0 and ds = 0. Hence we obtain Re(/(z)dz) = 0 on
dS which implies ReF(z) = 0 when z G dS.
Let G{z) = eF(-z\ Then G(z) is analytic in S and \G(z)\ = 1 when z G dS.
Because G(z) has no zeros in S, so 1/G(z) is also analytic in S and |l/G(z)| = 1
when z G 5S. Apply the maximum modulus principle to both G(z) and
1/G(z), we obtain |G(z)| = 1 for z G S, which implies that G(z) is a constant
of modulus 1. It follows from G(z) = eF^ that F(z) is also a constant. Hence
/(z) = F'(z) = 0.

341
5107
(a) Find the constant c such that the function
1 c
/(*) =
Z4 + Z3 + Z2 + Z - 4 2—1
is holomorphic in a neighborhood of z = 1.
(b) Show that the function / is holomorphic on an open set containing the
closed disk {z : \z\ < 1}.
(Iowa)
Solution.
(a) As
lim(z- 1)
z—»1
= lim
*-iv ' z4 + z3 + z2 + z
1
I (z* + z3 + z2 + z- 4)'
1
z1™ 4z3 + 3z2 + 2z + 1
1
~ 10'
we know that z = 1 is a simple pole of ;4 .„_.;,_. _4 with residue equal to -^.
Hence when c = ^, /(z) is holomorphic in a neighborhood of z = 1.
(b) When \z\ < 1, we have
|z4 + z3 + z2 + z - 4| > 4 - |z4 + z3 + z2 + z| > 4 - |z|4 - |z|3 - |z|2 - \z\ > 0,
and the equalities hold if and only if z = 1, which shows that z = 1 is the only
zero of z4 + z3 + z2 + z — 4 in {z : \z\ < 1}. By (a), we obtain that /(z) has
no singular point in {z : \z\ < 1}, hence /(z) is holomorphic on an open set
containing {z : \z\ < 1}.
5108
Let P(z) be a polynomial of degree d with simple roots z%, Z2, ■ • •, z<j. A
"partial fractions" expression of -p has the form:
1 d c
v ' n=l

342
(a) Give a direct formula for cn in terms of P.
(b) Show that pj-p: really has a representation of the form (*).
(c) Give a formula similar to (*) that works when Z\ — 22 but all other
roots are simple.
(Courant Inst.)
Solution.
(a)
which is the residue of -pr^r at z — zn.
(b) Let
1 d c
^) = ^-1^7=^--
Then /(2) is analytic on(F and lim /(2) — 0. By Liouville's theorem, /(2) is
z—* 00
identically equal to zero, hence
P(z) ~ 2-.\z-zn
v ' n=l
(c) Denote the Taylor expansion of P{z) at z — z\{ — z%) by
00
P(z)=J2^n(z-Z1)n.
n = 2
Then the Laurent expansion of -pl^y at 2 = z\ is
^+ 7=^: +EM*-*0".
P(z) (z-zx)2 *"*i n=0
where ,„ _ s
, _ 1 _ 2 a3 2P,"(21)
Cl_a2- P"(2!)' °2 a\ 3P"(zi)2"
Hence pK has the form:
P{z) ~ (2 - 2i)2 2-21 ~^ Z _ Z"

343
5109
Suppose / is meromorphic in a neighborhood of D (D = {\z\ < 1}) whose
only pole is a simple one at z = a £ D. If f(dD) C M, show that there is a
complex constant A and a real constant B such that
,, , Az2 + Bz + A
f(z) = -7
(z — a)(l — az)
Solution.
Assume that the residue of / at z = a is A\. Define
g(z) = f(z)- Al AlZ
(Indiana)
z — a 1 — az
It is obvious that g(z) is analytic on D and g(dD) C M. By the reflection
principle, g(z) can be extended to an analytic function on the Riemann sphere
<F, hence g(z) must be a constant. Suppose g(z) = Bi, then Bi is real and
/w = ^ + T^+5l
z — a 1 — az
Az2 + Bz + A
(z — a)(l — az)'
where 4 = Zi - aBi, B = -{aAi + aA~i) + Bi(l + \a\2) £ M.
5110
Let K\, K2, ■ ■ ■, Kn be pairwise disjoint disks in(T, and let / be an analytic
n
function in <F\ (J ifj. Show that there exist functions /1, /2, ■ ■ ■, /„ such that
3 = 1
(a) /j is analytic in (E\Kj, and
(b)/(*)=£ £(*) for* e<F\U*i ■
(/nrfiana)
Solution.
Assume i^x = {z; |z — zx| < rx}. Choose ex > 0 sufficiently small, such that
n
Si = {z;n< Iz-zi^n + eJCffXlJif,-.

344
In Si, /(2) has the Laurent expansion
+00
/(*)= E a^C*-*!)*-
k= — 00
Set
/1(2)= E 4^(^-^-
fc=—00
/1(2) is analytic in W\K 1. Because /(2) — /1(2) has an analytic continuation
n
to K\, /(2) — /1(2) is analytic in <F\ [J ify.
i=2
Assume X2 = {2; \z — z2\ < r2}- Choose e2 > 0 sufficiently small, such
n
that E2 = {z;r2 < \z-z2\ < r2 + e2} C <F\ |J if,. In E2, /(2)-/1(2) has the
i=2
Laurent expansion
+00
/w-/iW= E 42)(*-*2)fc.
^=-00
0 .
Set /2(2) = ^2 a), \z — z2)h. /2(2) is analytic in W\K2. Because /(2) —
k=— oo
/1(2)-/2(2) has an analytic continuation to X2, /(2)-/1(2)-/2(2) is analytic
in<F\ (J #y-
J = 3
Repeat the above procedure n — 1 times, we get a function /(2) — /1(2) —
/2(2) -----/,,-1(2), which is analytic in W\Kn. Set
/„(2) = /(2) - /i(2) - /2(2) /n_i(2).
Then we have
n
/(2) = E/A2)*
j=i
n
where fj(z) is analytic mW\Kj, and the above identity holds for z£ff\ [J ifj.
5111
Recall that a divisor Df of a rational function /(2) on (T is a set {p £
<F U {00}}, consisting of zeros and poles p of /(2) (including the point 00),

345
counted with their multiplicities np 6 Z. Let / and g be two rational functions
with disjoint divisors. Prove that
n s(p)np = n /(<z)ng-
P€D, q£Dg
(SUNY, Stony Brook)
Solution.
Let pi (i = 1, 2, ■ ■ •, n) be all the zeros and poles of f(z) with multiplicities
nPi respectively. It should be noted that p,- is a zero of / when nPi > 0 and
a pole of / when nPi < 0. By the property of rational functions, we have
n
^2 nPi = 0. Similarly, let qj (j = 1, 2, ■ ■ ■, m) be all the zeros and poles of g(z)
i=i
m
with multiplicities mqj respectively, then we have ^ mqj = 0.
i=i
First we assume that the point oo is not a zero or a pole of / or g, then /
and g can be represented by
n
/(z)=^n(z-P')npi
i=l
and
m
ff(z) = 5^2 -%■)"*■
Then
n n n m
n ^)np = 11^)""=n5npi ■nnte-*)B"m"
p€-D/ 1 = 1 i=l i = 1J = 1
n m
and
m m m n
n /(9)"- = n f^)mqi=n ^ n 11(¾ - ^nr
q€Da j=l j = l j=li = l
n m
= nn(^«i)",ira,j'
. = 1.7 = 1

346
In case the point oo is a zero or a pole of / or g, we may assume pn = oo
without loss of generality. Then
n-l
f(z) = A'[[(z-pi)n>!
i=l
and
m
g(z) = BY[(z-qj)m^.
i=i
m
Since ^ m9j. = 0, we may assume that g(pn) = g(oo) = B. Hence
i=i
n-l
n ^p)np = n s(Pi)npi ■ Bn>«
p£Df i=l
n —1 n—1 m
= (11^)-5-- -n n^-^)B,,m'j
i=l i=lj=l
n— 1 m
.-=1 j=i
and
m mm n — 1
n /(«)"' = n /fe-r^=n ^ n n (¾ - ^r^-
g€i>g j=i j=i j=i»=i
n— 1 m
•=ij=i
which completes the proof of the problem.
5112
Let f(z) be the "branch" of log 2 defined off the negative real axis so that
/(1) = 0.
(a) Find the Taylor polynomial of / of degree 2 at —4+ 3i, simplifying the
coefficients.
(b) Find the radius of convergence R of the Taylor series T(z) of /(2) at
-4 + 3t.

347
(c) Identify on a picture any points z where T{z) converges but T(z) ^
/(z), and describe the relationship between / and T at such points. If there
are no such points, is this something special to this example, or a general
impossibility? Explain and/or give examples.
{Minnesota)
Solution.
(a) When z is in the neighborhood of zq = —4 + 3i, we have
/(z) = logz = log[(-4 + 3i) + (z + 4 - 3t)]
z + 4 - 3i
log(-4 + 3i) + log
1 +
-4 + 3i
, r •/ • 3X 2 + 4-3i
= log 5 + i(ir — arcsin -) -\
v 5; -4 + 3i
_1 /z + 4-3A2
~2\ -4 + 3i ) +'"'
Hence the Taylor polynomial of / of degree 2 at —4 + 3i is
c0 + cx(z + 4 - 3i) + c2(z + 4 - 3i)2,
where
3
Co = log 5 + i(ir — arcsin -),
5
c\
and
4 + 3i
25 '
25 + 24z
Co = .
2 1250
(b) Denote the Taylor series of /(z) at —4 + 3i by T(z). Because logz has
only z = 0 and z = oo as its branch points, and has no other singular point, the
radius of convergence R of T(z) is equal to the distance between z = —4 + 3i
and z = 0. Hence R = 5.
(c) Denote the shaded domain shown in Fig.5.1 by fi. When z £ fi =
{z : \z + 4 - 3i| < 5,Imz < 0}, T(z) ± /(z). It is because T(z) in ft is the
continuation of logz at —4 + 3i in the disk {z : |z + 4 — 3i| < 5}, while /(z)
in fi is the continuation of log z at — 4 + 3i in the slit plane <F\(—oo, 0]. Hence

348
the difference is 2iri, i.e., T(z) = f(z) + 2iri.
Fig.5.1
5113
Let / be the analytic function defined in the disk A = {z : \z — 4| < 4} so
that /(z) = z»(z + l)s in A and f(x) is positive for 0 < x < 8. An analytic
function g in A is obtained from / by analytic continuation along the path
starting and ending at z = 4 (see Fig.5.2). Express g in terms of /.
(Indiana)
Fig.5.2
Solution.
Denote the closed path in Fig.5.2 by T, and denote the change of 4>{z) when
z goes along T from the start point to the end point by Ar<^(z). Then
g(z) = |ff(z)|e''arg^^ = |/(2)|e'(arg/(,)+Ararg/(,))_
We have
Ararg/(z) = - Arargz + -Ararg(z + 1) = -(2tt) + -(-2t)
it
Hence
g(z) = e-Vf(z).

349
5114
Define
/(2) = sinV? ■
(a) Where is / single-valued and analytic?
(b) Classify the singularities of /.
(c) Evaluate Jjj|=25/(z)dz.
(Indiana)
Solution.
(a) It is known that z = 0 and z = oo are the branch points of function y/z.
Let r = {z : \z\ = r}, and when z goes along T once in the counterclockwise
sense, yfz is changed to —yfz, while /(z) is changed to
p~V* _ eV* pV* _ p~V*
sin(—yfz) s'my/z
which is still /(z). Hence z = 0 and z = oo are no longer the branch points of
When z is in the small neighborhood of z = 0, /(z) can be represented by
Eli v-^ (-1) -
n^22~Eo^^22
/(z) = C n=0 n=0
sin^ V^ (-1)- 2n+j
n=0
2 E (2n + l)!Z"
_ n=0
2^ (2n+l)!Z
n=0 v
which implies that z = 0 is a removable singular point of /(z). It is obvious
that z = n2ir2 (n = 1,2, ■ ■ ■) are poles of /(z). Hence /(z) is single-valued and
analytic in <F\{z = n2ir2 : n = 1,2, ■ ■ ■}.
(b) We have
lim — ^- = f^— = 2nir(-l)n(en* - e""),
^^n=7r= (sin^/z)' cos(n7r) '
v v ' 2n;r
which shows that z = n2ir2 are simple poles of /(z) with residues
2mr(-l)n(en^-e-n^).

350
As to z = oo, it is the limit point of the poles of /(z), and hence is a
non-isolated singular point of /(z).
(c) /(z) has only one pole z = ir2 in the disk {z : \z\ < 25}. Hence
I
f(z)dz = 2iriRes(/, ir2)
z\<25
= -4ir2i(eK - e-*).
5115
Let fi be the plane with the segment {— 1 < x < l,y = 0} deleted. For
which of the multi-valued functions
(*) /(*) = t£?'
0>) »(*) = 71b*
can we choose single-valued branches which are holomorphic in fi. Which of
these branches are (is) the derivative of a single-valued holomorphic function
in ft. Why?
(Indiana-Purdue)
Solution.
Let T be an arbitrary simple closed curve in ft, and denote by Ar</>(z) *ne
change of (f>(z) when z goes continuously along T counterclockwise once. It is
known that / and g can be represented by
Vl-z
f(z) - Z = eO°S '- \ log(l+*)- i log(l-z)}
ei[arg^- iarg(i+*)- larg(i-z)]
VT^I
and
ff(z) = 1 -P{-^oS(l+^)-|log(l-^)}
ei[-iarg(i+^)-iarg(i-^)]_
Vl-2
1
VTT
z'
Because
and
Ar[argz - -arg(l + z) - -arg(l - z)] = 0
Arr ^1 + ^--8^1-^1--( ° {-1 <*< 1,!/= 0} not inside T
Arh-arg(l + z)- 2arg(l-z)j - j _2x {_i < -,. < i,j, = 0} inside V

351
we have Ar/(z) = 0 and Ar<?(z) = 0. Hence both /(z) and </(z) have single-
valued branches which are holomorphic in fi, and each of / and g has two
single-valued branches.
In order to know which of / and g has a single-valued primitive in fi, we
consider the integrals frf(z)dz and frg(z)dz. If the segment {—1 < x <
1,2/ = 0} is not inside T, it is obvious that /r f(z)dz = 0 and /r g(z)dz = 0.
If the segment { — 1 < x < l,j/ = 0} is inside T, we consider the Laurent
expansion of / and g about z = oo:
It follows that /r f(z)dz = 0 and /r g(z)dz = ±2ir. Hence we obtain that
both of the single-valued branches of / are the derivatives of single-valued
holomorphic functions in fi, and the primitives are /^ f(z)dz + c, where the
integral is taken along any curve connecting zq and z in fi. But neither of the
branches of g is the derivative of a single-valued holomorphic function in fi.
5116
(a) Let D C ff be the complement of the simply connected closed set
{ee+ie | 6 G M} U {0}. Let log be a branch of the logarithm on D such that
loge = 1. Find loge15. Justify your answer.
(b) Let 7 denote the unit circle, oriented counterclockwise. By lifting the
integration to an appropriate covering space, give a precise meaning to the
integral / (log z)2dz and find all possible values which can be assigned to it.
(Harvard)
Solution.
(a) The set {e0+'e \ 9 £ M} U {0} is a spiral which intersects the positive
real axis at {e2nx : n = 0,±1, ±2,-■ ■}. The single-valued branch of logz is
denned by loge = 1. Hence loge15 = loge + Aplogz, where T is a continuous
curve connecting z = e and z = e15 in D and Aplogz is the change of logz
when z goes continuously along T from z = e to z = e15. It follows that
Aplogz = Arlog|z| + iApargz, and Arlog|z| = 15 — 1 = 14. Because
e £ (e°,e2K),e15 £ (e4K,e6K), we know that when T connects e and e15 in D,
Apargz must be 4ir. Hence log e15 = 1 + (14 + 4iri) = 15 + 4tu.
(b) Define the lift mapping by w = log z which lifts the unit circle 7 one-to-
one onto a segment with length 2ir on the imaginary axis of to-plane. Because

352
both the starting point of 7 and the single-valued branch of log z on 7 can be
arbitrarily chosen, the segment on to-plane can be denoted by [it, i(t + 2ir)),
where t can be any real number. Hence we have
ri(t+2x) . ,i(t+2x)
(\ogzfdz = I w2ewdw = (w2ew)tit+2x)-2 wewd
J y Jit Jit
fi(t+2x)
= e!'*(-4iri - 4ir2) - (2wew)\'^t+2x) + 2 / ewdw
Jit
= -4ir(< + ir + i)e{t = 4ir(t + ir + i)ei(-t+*\
which implies that the set of values being assigned to the integral J (log z)2dz
is a spiral {4ir(s + i)e" : s £ M}.
hill
Find the most general harmonic function of the form /(|z|), z £ (F\0. Which
of these /(|z|) have a single valued harmonic conjugate?
(Indiana)
Solution.
Because /(|z|) is harmonic, we have reason to assume that the function /
(with real variable t) has continuous derivatives f'(t) and f"(t). Note that the
Laplacian
02 d2 . d2
and
A = 1 = 4
dx2 dy2 dzdz'
5>» = BA^ = i/-(|.|)-Vf.
e=f(\*\) = 7/"(M)+7^t/'(|z|).
we obtian
dzdzu " V Vl u 4|z|
where t = \z\. This differential equation is easy to solve, and the solution is
f(t) = a log 2 + /3, where a, /3 are two real constants. Hence the most general
harmonic function of the form f(\z\) in(F\0 is alog|z| +/3.
Since log \z\ has no single-valued harmonic conjugate in(F\0, we know that
when f(\z\) has a single-valued harmonic conjugate in(F\0, it must be a
constant.

353
5118
Consider the regular pentagram centered at the origin in the complex plane.
Let u be the harmonic function in the interior of the pentagram which has
boundary values 1 on the two segments shown and 0 on the rest of the
boundary. What is the value of u at the origin? Justify your claim.
(Stanford)
Solution.
Denote the interior domain of the pentagram shown in Fig.5.3 by D, and the
ten segments of the boundary by l\, l2, • • •, ho, put in order of counterclockwise.
Fig.5.3
Then denote the harmonic function on D with boundary values 1 on 4 and
0 on the rest of the boundary by ujt(z), k = 1,2, ■ ■ ■, 10. By the symmetry of
domain D, we have
u2{z) = ui(-z),
u3(z) = ui(e~^'z),
u4(z) = u2(e~~^'z),
u5(z) = ui(e--^'z),
um(z) = u2(e = 'z).
It follows from
10
u(z) = ^2uk(z) = l
Jt=i
10
and Ml(0) = w2(0) = ■ ■ ■ = wlo(0) that uk(0) = ^ for k = 1, 2, ■ ■ ■, 10. Hence
«(0) = «i(0) + M5(0) = ^.

354
5119
Suppose G is a region in <F, [0,1] C G, and h : G -+ M is continuous.
/i|g\[o,i] is harmonic, does this implies that h is harmonic on G?
(lotos)
Solution.
The answer is No.
A counterexample is h(z) — Re-v/z(z - 1), where the single-valued branch
of \Jz(z - 1) is chosen by \/z(z - l)|z=2 = a/2- Since \Jz(z - 1) is analytic
in(F\[0,1], h(z) is harmonic there. When 0 < x < 1,
lim \/z(z — 1) = \A(1 - z)i,
y>o
lim \/z(z - 1) = --y/a;(l - s)i.
z=x-\-yi—*x
y<o
Hence h(z) = 0 when z = x,0 < x <l, and h(z) is continuous on(T. But h(z)
is not harmonic on(T, because z = 0 and z = 1 are branch points of \/z(z — 1).
Remark. If the problem is changed to h : G —* W is continuous and
/i|g\[o,i] is holomorphic, then h must be holomorphic on G.
5120
Let 7 be an arc of the unit circle. Suppose that u and v are harmonic in
D = {z : \z\ < 1} and continuously differentiable on D U 7. If the boundary
values satisfy u — v on 7 and the radial derivatives satisfy §jf- = §f on 7> prove
that m = b in D.
(/nrfiana)
Solution.
Let m* be a conjugate harmonic function of u in D and v* be a conjugate
harmonic function of v in £). We know that a variation of Cauchy-Riemann
equations for / = u + iu* and g = v + iv* are
du du* du du*

355
and
dv _ dv* dv _ dv*
r!fr = ~d6) de~~r~bT'
It follows from the continuous differentiability of u and v on D U 7 that u*
and v* can be continuously extended to D U 7 and then are also continuously
differentiable on D U 7. Let z0 be a fixed point on 7, and for z £ 7 denote
the subarc of 7 from zq to z by jz. Without loss of generality, we may assume
that u*(z0) = v*(zo) — 0. Then for z £ 7,
<9w* , <9w* ,„ f du* ,„ /" <9u
■d0
or
[ du* du* ,. f du* Jn f
uiz) = L-8vdr+wde=lwde=L
Hence we obtain two functions / = u + iu* and g — v + iv* which are analytic
in D and continuous on D U7, such that / = g on 7. Let F = f — g. Then by
the reflection principle, F can be analytically extended to an analytic function
on D U 7 U £>*, where D* = {z : |z| > 1}. Since F = 0 on 7, we obtain F = 0
on £) U 7 U £>*, which implies u — v in D.
5121
Use conformal mapping to find a harmonic function U(z) defined on the
unit disc {\z\ < 1} such that
.. TU ;<k /+1 for 0 < 9 < ir
lim Ulre ) = < , . . „
r-a- v ; \ -1 for ir < 6 < 2ir.
Give the correct determination of any multiple-valued functions appearing in
your answer.
(Courant Inst.)
Solution.
It is easy to know that w = — if^y is a conformal mapping of the unit
disc D = {z : \z\ < 1} onto the upper half plane H = {w : Imw > 0}. The
boundary correspondence is that the negative real axis {w : —00 < w < 0}
corresponds to the arc rx = {z = e,e : 0 < 6 < ir} and the positive real axis
{w : 0 < w < +00} corresponds to the arc 1^ = {z = e'0 : ir < 6 < 2ir}.
It is well known that u(w) = -argw — 1 is a harmonic function in H and
assume +1 on the negative real axis and —1 on the positive real axis. Hence
tti \ 1 -z+ l\ 2 /z+ l\ 0
U(z) = u(-i- -) = -arg(- -) - 2,
Z — 1 TV Z — 1

356
where the single-valued branch of arg(f^y) is defined by arg(|^)|j=o = ir, is
a harmonic function in D = {z : \z\ < 1} with the boundary values +1 on Ti
and —1 on ^.
Remark. This problem can be solved directly from the Poisson formula
as follows:
™ J\(\-\ Q - z K
2^7ri \C-zJ iC 2Wr2 \C-z) <
* Jr, \t-z) <
= ^Imjy d(21og(C-z)-logOJ-l
= -Ari{2arg(C-z)-argC}-l
TV
2 z+l
= —arg z.
5122
Determine all continuous functions on {z £ (F : 0 < |z| < 1} which are
harmonic on {z : 0 < \z\ < 1} and which are identically 0 on {z G<F : |z| = 1}.
(Minnesota)
Solution.
Suppose u(z) is a continuous function on {0 < \z\ < 1} which is harmonic
on {0 < \z\ < 1} and identically zero on {\z\ = 1}. Let *du = —uydx + uxdy
and A — /,, *du, where A is a real number not necessarily zero, Denote
v(z) = Jz *du, then v(z) is the conjugate harmonic function of w(z), but may
be not single-valued. Define
/(z) = (w(z) + iv(z)) - — log z,

357
then /(2) is a single-valued analytic function on {0 < \z\ < 1} and Re/(2) is
identically zero on {\z\ = 1}.
00
Let /(2) = Yl anZn be the Laurent expansion of /(2) on {0 < \z\ <
1}, then lim ?/|a_n| = 0 and lim y/\a^~\ < 1. Define g(z) = V bnzn,
n-00 ^ n-00 n=-oo
satisfying b_n = —bn for n = 0,1, 2, • • •, and &_„ = a_n for n = 1,2, ■ ■ ■. Then
3(2) is an analytic function on {0 < \z\ < +00}. When \z\ = 1, it follows from
Re&o = 0 and
— 1 00 00 00
Re J^ 6"z" = Re J2 b-"z~n = Re J2 -^z_" = -Re J2 6"z"
n= —00 n = l n—\ n = l
00
that Reg(z) = 0. Then /(2) — 3(2) = ^ cnzn is an analytic function in
n=0
{|2| < 1} and Re(/(2) — g(zj) is identically zero on {\z\ = 1}. Consider
F(z) = ef^~9^ which is analytic and does not assume zero in {\z\ < 1}, and
|F(2)| = 1 on {\z\ = 1}, by the maximum and minimum modulus principles,
we have F(z) = e'a, hence /(2) = g(z) + ia.
From the above discussion, we finally obtain
+00 A
u(z) = Re V bnzn + — log \z\,
nzz — 00
where 6_„ = — bn and lim a/|6„| = 0.
n—*oo
5123
(a) Let /(2) be a holomorphic function in the disc \z\ < r whose zeros in
this disc are given by aj., 02, ■ ■ •, an counted with multiplicity. Suppose further
that |aj| < r for all j = 1,2, ■ ■ ■, n, and |/(0)| = 1. Jensen's formula states
that
^/o"iogi/(^)id^|:iog(^).
Prove this.
(b) With the hypotheses and notations of (a), let n(t) be the number of a,j
(j = 1, 2, • • ■, n) such that \a.j\ < t. Using Jensen's formula, show that
r2w
[ n{t)j=h r i°gif{rei9)ide-

358
(c) For r < R deduce an estimate on n(r) in terms of max log \f(Rel9)\.
0<9<2k
(d) What can be said about the zeros of bounded holomorphic functions in
the unit disc?
(Harvard)
Solution.
(a) Let
n i —
r — a,- 2
'W = A"j5^tj.
then F(z) is holomorphic and has no zero in the disc {\z\ < r}, which
implies that log|F(z)| is harmonic in {\z\ < r}. By the mean value theorem of
harmonic functions,
2-K
dd.
Noting that
and
we obtain that
log\F(0)\ = ±J Xlog|F(re*
1^)1=1/(0)161^=6]¾
\F(reie)\=\f(re'%
J2\og(f-) = ±- i* \og\f(reie)\d6.
fr[ \ai\ 21T Jo
(b) It is obvious that log r^r = J7 , y. By the definition of the function
n(t) we have
which shows that the identity holds.
(c) Apply the identity in (b), we have
-Jo log\f(ReiS)\de = Jo «MT>/r n(i)T>n(r) log:
Denote max log \f(Ret9)\ by M(R), we obtain
0<B<2w
n(r) < M(R)/log -.
r

359
(d) Let /(z) be a bounded holomorphic function in {z : \z\ < 1}. We know
that /(z) can have countably many zeros. Suppose z = 0 is a zero of /(z) of
multiplicity m > 0 with ' m? ' = a, and let the other zeros be ordered by
0 < lai| < la2| < ■■■■ Obviously \an\ —> 1. Apply Jensen's formula in (a) to
F(z) = ^¾ with 0 < r < 1 such that there is no zero of / on {|z| = r}, we
have
±- i * log\f(reie)\d6 = J2 log(,^)+log(|a|rm).
Since /(2) is bounded, we assume
2ir
p / Xlog|/(rew)|d0<M.
For any n, we can choose r such that r > |an|, and hence
n
Elog(rn)^ E log(r^)<M-log(|a|rm).
J = l ' U |«j|<r ' Jl
Let r —+ 1, we obtain
n
JJ|aj|> |a|e~M> 0,
i=i
00
which implies that the series ^2 (1 — \a,j |) is convergent.

360
SECTION 2
GEOMETRY OF ANALYTIC FUNCTIONS
5201
Find a one-to-one holomorphic map from the unit disk {\z\ < 1} ont
slit disk {\w\ < 1} - {[0,1)}.
(SUNY, Stony E
Solution.
We construct the map by the following steps:
z-f 1
zi = <f>i{z) = i : {z : \z\ < 1} -+ {zi : Imzi < 0};
z — 1
22 = <M*i) = y^i - 1 + 2i (\/zf - 1 __ =\/2i):
{zi : Imzi < 0} -+ {z2 : |z2| < 1 and Imz2 > 0};
w = <f>3(z2) = z\ : {z2 : |z2| < 1 and Imz2 > 0} -+
{w : \w\ < l}\{w : Imw = 0,0 < Rew < 1}.
Then w = ¢3 0 ¢2 ° <f>i(z) — f{z) is a one-to-one holomorphic map from the
unit disk {\z\ < 1} onto the slit disk {\w\ < 1}\{[0,1)}.
5202
(a) Find a function / that conformally maps the region {z : |argz| < 1}
one-to-one onto the region {w : \w\ < 1}. Show that the function you have
found satisfies the required conditions.
(b) Is it possible to require that /(1) = 0 and /(2) = |? If yes, give an
explicit map; if No, explain why not.
(Illinois)
Solution.
(a) C = /i(z) — z% — e2los* (logl = 0) is a conformal map of {z :
|argz| < 1} onto {C : R^ > 0}, and w — /2(C) = fey is a conformal map of
{C : Re(," > 0} onto {w : \w\ < 1}. Hence
w = f(z) = /20 /x(z) = — -
Z2 + 1

361
is a conformal map of {z : |argz| < 1} onto {w : \w\ < 1} with /(1) = 0 and
/(2)
22
27 + 1 ^
(b) Suppose w — /(z) is an arbitrary conformal map of {z : |argz| < 1} onto
{w : \w\ < 1} with /(1) = 0. Then w = F(w) — f o /_1({o) is a conformal map
of {w : \w\ < 1} onto {w : \w\ < 1} with ^(0) = 0, and w = F(w) = f°f~1(w)
is a conformal map of {w : \w\ < 1} onto {w : \w\ < 1} with F(Q) = 0. By
Schwarz's lemma, we have both 1^(^)1 < \w\ and 1-^(^)1 < \w\, which implies
that |/(z)| = 1/(^)1 for every z S {z : |argz| < 1}. Since
we cannot require that /(2) = ^.
5203
(1) Find one 1-1 onto conformal map / that sends the open quadrant
{(x,y) : x > 0 and y > 0} onto the open lower half disc {(x,y) : x2 + y2 <
1 and y < 0}.
(2) Find all such /.
(Toronto)
Solution.
(1) Let ¢ = <^i(z) = z2. It is a conformal map of {z = x + iy : x >
0 and y > 0} onto {£ = £ + it) : T) > 0}.
Let w = <f>2(C) = a/C2 ~ 1+C> where ^/C2 — 1 = — \/2i- It is a conformal
<=•
map of {C = £ + Z77 : 77 > 0} onto {w = u+ iv : u2 + v2 < 1 and t> < 0}.
Then w = ¢2 ° ^1(2) = \/z4 — 1 + z2, where \/z4 — 1 = — i/2i is a
required conformal map.
(2) If/ is an arbitrary conformal map satisfying the condition of (1), then
¢2 1 0/0 <f>i1(Q is a conformal map of the upper half plane onto itself, which
can be represented by ^(¢) = ffxj, where a, b,c,d£ M, ad—bc> 0. Hence /
can be written as ¢2 ° ip ° <f>i(z)-
5204
Map the disk {|z| < 1} with slits along the segments [a,I], [—1,-6] (0 <
a < 1, 0 < b < 1) conformally on the full disk {|u>| < 1} by means of a function

362
w = /(z) with /(0) = 0, /'(0) > 0. Compute /'(0) and the lengths of the arcs
corresponding to the slits.
(Harvard)
Solution.
We construct the conformal mapping by the following steps.
(i) Zl = Mz) = z + I : {|z| < \}\{[a, l]u[-l,-6]} -ff\{[-6- \,a+%}.
It has the point correspondences ¢1(0) = oo, <f>i(a) = a + ^, 4>\(b) = — b — A
¢1(1) = 2 and tfi(--l) = -2.
(ii) z2 = 4>2{zi) = Xila+l) =g\{[-fc-F»°+ B — <T\[0,+cx>). It has
the point correspondences
¢2(00) = -1, ^2(-2)= (? )2
77 + va
and
^)=(¾)2.
Va v
and it is easy to know that
(¢2 0 ^i)'(O) = -(a+^ + b+^)<0.
(iii) z3 = ¢3^2) = ^/ii : <F\[0,+oo) —► {z3 : Imz3 > 0}. It has the point
correspondences
*,(-!) = *, fc((£f£)')=±$z£)
and
*((^» = ±(£9-
4= — a/«' ' ^4=-^/^
Va v V<»
For the convenience of computation, let
Vb ± + V~b
JL
zr + VS' ^-\/«'
v^
We also know that <j>'3( — 1) = — |.
(iv) to = <f>4(z3) = -^4 : {z3 : Imz3 > 0} —► {io : |iu| < 1}. It is obvious
that ¢4(2) = 0 and <f>'4(i) = — §.

363
Now we define w — /(2) = ¢4 0 (f>3 0 ¢2 ° <f>i(z)- From the above discussion,
we know that / maps the unit disk with slits [—1,-6] and [a, 1] conformally
onto the unit disk with /(0) = 0 and
/'(0) = ^(i) • &(-l) • (¢2 0 ^)'(O) = I(a + -a + b + -b) > 0.
What correspond to the slits are the arc with endpoints j^| and ^H
containing point 2 = — 1 and the arc with endpoints g^4 and ^4 containing
point 2=1. The lengths of the two arcs are
. A-i A + i Th'^1
<i = arg— : - arg— : = 4arctgJA = 4arctg-y-
>A + i &A-i 6 ^ + v^'
and
h = arg- \ - arg——4 = 4arctg- = 4arctg^ -=.
B - 1 B + i B -i=.fV&
vb
5205
Let 0 < e < tt, let 7e denote the arc {e!* : e < t < 2ir — e} and let fie be
the complement of 7e in the Riemann sphere. If/ is the conformal map of the
unit disk onto fie, /(0) = 0, /'(0) > 0, describe the part of the unit disc that
/ maps onto {\z\ > 1}.
(Stanford)
Solution.
We are going to find the map./ by the following steps:
2 — e~ie
2l = <t>r(z) = e'e ■ -j~r ■ {z ■ |*| < 1} — K = Imzi < 0},
with ^x(0) = e~ie, arg^i(O) = -f - e.
22 = ^2(21) = y/zl: {21 : Im2i < 0} —» {22 : Re22 > 0, Im22 < 0},
with 4>2{e-ie) = e"f% arg^e"") = -§.
— -t
C = (^,3(22) = ef•' • Z2~6 ' : {22 : Re22 > 0, Im22 < 0} - £>i
22 - e 2 *

364
(shown in Fig.5.4), with ^3(e~ = ') = 0, arg^e't') = f + |, where Di is a
domain bounded by {£ = e'0, § < 6 < 2ir — |} and an circular arc l£ which is
orthogonal to {|C| = 1} and connects points e^' and e~i' in {|<^| < 1}.
Fig.5.4
Let $(z) = ¢3 0 ¢2 ° 4>i{z), then $ maps {z : \z\ < 1} conformally onto D\
with ¢(0) = 0, $'(0) > 0. After considering the boundary correspondence, we
know that l£ corresponds to the arc {z = e1' : \t\ < e} under the map $. Since
the symmetric domain of {\z\ < 1} with respect to arc {z = e%t : \t\ < e} is
{\z\ > 1}, and the symmetric domain of Dx with respect to le is D2 — {|C| <
1}\£>1, by the reflection principle, <&(z) can be extended to a conformal map
of fi£ onto {C : |<^| < 1}. Hence the conformal map / in the problem is nothing
but the inverse of ¢, and the domain / maps onto {\z\ > 1} is D2, which is
bounded by circular arcs le and j^ = e'e : \8\ < |}.
5206
Suppose that w = /(z) maps a simply conncted region G one-to-one and
conformally onto a circular disk Dr with center w — 0, radius r, such that
/(a) = 0 and |/'(a)l = 1 f°r some point a S G.
(1) Prove that the radius r = r(G, a) of Dr is uniquely determined by G
and a.
(2) Determine r(G, a) if G is the region between the hyperbola xy = 1
(x > 0, y > 0) and the positive axes, and if a = 1 + |.
(Indiana)
Solution.
(l)Suppose f = g(z) is another conformal map of G onto a circular disk
Dri with center ¢ = 0 and radius n, such that g(a) — 0 and |<?'(a)| = 1, then
w = F(Q — /°<?_1(C) is a conformal map of {C : |<^| < r\\ onto {w : \w\ < r}

365
with F(0) = 0 and \F'(0)\ = ^| = 1. Apply Schwarz's lemma to F(Q and
we have |.F'(0)| < ^-, hence r% < r. For the same reason, apply Schwarz's
lemma to F~1(w) and we have r < n, which implies r\ = r. In other words,
r is uniquely determined by G and a.
(2) We construct a conformal map of G onto a circular disk Dr in the
following steps:
2x = 4>i(z) = z2 : G —» {zi : 0 < Im2i < 2},
with <^(i+i) = i+i, i^i(i+i)j = ^5.
22 = ^2(21) = e?Zl : {21 : 0 < Im2i < 2} -» {22 : Im22 > 0},
with <£2(f + i) = te^f, |^2(| + *)| = f e^.
4 22 — ic&K 4
w = ^3(22) = —e rr~ : iZ2: Imz2 > 0} -► {w : \w\ < -7=-},
Voir Z2 + ie*T v5t
with ^(tei') = 0, ^(iei*)! = —^.
V57re »
Define /(2) = <f>3 ■ ¢2 ° <f>i(z)> then u> = /(2) : G —► {if : |iu| < -7?-}, with
/(«) = 0, |/'(a)| = 1. Hence r(G,a) = ^U.
5207
Let T(;z) = jj^| be a Mobius transformation.
(a) Assume that z\, 22 £<T are two distinct fixed points for T, i.e., T(zi) =
Zi, i = 1,2. Show that there exists a constant c such that
T(z) — z\ 2 — z\
T{z)-z2 =Cz-z2'
(b) Use (a) to find an expression for Tn(z), n= 1,2, 3,---, if
T<*> = 7=3-
(/owa)
Solution.
(a) Let a £ (F be a point different from 21, 22. Because the cross ratio is
invariant under Mobius transformations, we have
(T(z),Zi,T(a),z2) = (2,21,0,22),

366
which is
Denoting
we obtain
T(z) — Z\ T(a) — zx z — Zi a — Z\
T(z) - Z2 ' T(a) — z2 z - Z2 ' a — Z2
T(a) — Z\ a~ zi
T(a) - z2 ' a - z2
T(z) — Z\ _ z — Z\
T(z) - z2 z - z2'
(b) Since Tn(z) - T(Tn~1(z)), it is easy to have
Tn(z) - Zj. _ Tn~l{z) - Zl = _2T"-2(z) - Zt = _ ,n 2 - 21
T"(;z)-^ "CT"-1(2)-z2 °T"-2(z)-z2 -'"_Cz-z2'
When T(z) = ^-ry-, by solving the equation ^ry = z, we obtain that z = ±1
are two fixed points of T. Choose a = 2, then T(a) = 5, hence c = |+j : |^j —
2.
It follows from
T"(*)-l=2n*-l
T"(z)+1 z+1
that
^ ^ (2»+ 1)-(2--l)z"
5208
(a) Justify the statement that "the curves
X2 +tA = 1
a2 + A 62 + A
form a family of confocal conies".
(b) Prove that such confocal conies intersect orthogonally, if at all.
(c) Show that the transformation w — § (2+i0 carries straight lines through
the origin and circles centered at the origin into a family of confocal conies.
(Harvard)
Solution.
(a) Without loss of generality, we assume a > b > 0. When — a2 < A < — b2,
the curves form a family of hyperbolas, while when A > — b2, the curves form

367
a family of ellipses. Suppose the focuses of the conies are (±c(A),0). When
-a2 < A < -b2,
c(A) = V(^ + A) + [-(62 + A)] = y/a2 - b2.
When A > -62,
c(A) = J{a2 + A) - (62 + A) = ^/a2 - 62.
Hence the curves
*' +tA = 1
a2 + A ' 62 + A
form a family of confocal conies.
(b) Suppose (xQ, y0) is the intersection point of
x2 y2
1 : a2 + Ai + b2 + Ai ~
and
z2 y2
' a2 + A2 62 + A
where Ai ^ A2. It followa-from
^o , Vo _ i
a2 + Ax 62 + Ai ~
and 2 2
x° i y° = 1
a2 + A2 &2 + A2
that
+77__4__^=0.
(a2 + A!)(a2 + A2) (&2 + A^fr2 + A2)
Noting that the tangent vector of Li at (zoi^o) is r*i = (^2^x7, 55¾¾^)) and
the tangent vector of L2 at (zo,yo) is ~r*2 = ('a*+\3' P+^t)' we ^ave
_> _> _ ^0 . Vo n
Tl' T2~ (a2 + Ai)(a2 + A2) (b2 + Ai)(62 + A2)
which implies that the confocal conies intersect orthogonally, if at all.
(c) Let z = re'e, and
1/ In 1/ 1 X /1 */ l\-d
w — u-\-tv= -(z +-)= ;r(r + -) cos 6 + -(r ) sin 0.

368
The image of straight lines through the origin is
2 2
u v
cos2 8 sin2 6
which are hyperbolas in to-plane. Because
cos2 6 + sin2 8 = 1,
the focuses of the hyperbolas are (±1,0).
The image of circles centered at the origin is
u2 v2
+ tt rrr = l.
\(r+±)2 "i(r-i)2
which are ellipses in to-plane. Because |(r + -)2 — |(r — -)2 = 1, the focuses
of the ellipses are (±1,0). Hence the transformation
carries straight lines through the origin and circles centered at the origin into
a family of confocal conies.
5209
If / : £)(0,1) = {z : \z\ < 1} —»<F is an analytic function which satisfies
/(0) = 0, and if
|Re/(z)| < 1 for all z G £)(0,1),
prove that
1/(0)1 < z-
TV
(Indiana)
Solution.
It is easy to know that
e^-1
w = g(Q
e^t + 1
is a conformal mapping of the domain {£ : |Re<^| < 1} onto the unit disk
{w : \w\ < 1} with g(0) = 0. Hence w — F(z) = g o f(z) is analytic in

369
£)(0,1) and satisfies .F(O) = 0 and |.F(z)| < 1. By Schwarz's lemma, we have
|F'(0)| < 1. Because
F'(z) = 9'(f(z)).f(z) = re*'" ■ TO
w y VM ;; J K ' (e^/^) + 1)2 '
it follows from /(0) = 0 that
l/'(0)| < -.
TV
5210
Let ft = {z S <F; — 1 < Imz < 1}, and let T be the family of all analytic
functions / : ft -»<F such that |/| < 1 on ft and /(0) = 0. Find
Solution.
It is obvious that
sup |/(1)|.
/6^
e*z - 1
C = /o(*) = -n
(Indiana)
ei2 + 1
is a conformal mapping of ft onto the unit disk with the origin fixed. For
any analytic function w = /(z) : ft —►<F such that |/| < 1 and /(0) = 0, we
consider the composite function w = F(Q — f o /J" (¢). ^(¢) is analytic in
the unit disk such that |.F(C)| < 1 and F(0) = 0. By Schwarz's lemma,
\f(0\ < ICI-
Choose Co = ^-»—-) we have
S el + l
^(¢0)1 = 1/(1)1 <Kol = fJ-^.
e 2 + 1
The equality holds if and only if F(Q = e!0£, which implies
sup |/(l)| = il^,
and the supremum is attained by /(z) = e!0/o(z), where 6 is a real number.

370
5211
Let / be an analytic function on D — {z; \z\ < 1} such that /(0) = —1, and
suppose that |1 + f(z)\ < 1 + |/(z)| whenever \z\ < 1. Prove that |/'(0)| < 4.
(Indiana)
Solution.
Let fi = W\{w = u + iv : u > 0 and v = 0}. It follows from |1 + f(z)\ <
l + |/(z)| that /(D) Cfi.
Set g(w) = v^-~^, (-^A" = i)- Then 3 0 /(z) is an analytic function
^ w? = — 1
on D with 3 0 /(0) = 0 and \g 0 f(z)\ < 1. By Schwarz's lemma,
Since
Kff 0 /)'(o)| < 1.
g'(w) =
we have g'( — 1) = —4- From
we obtain
'w(^/w + l)2 '
(ffo/)'(0) = ff'(-l)/'(0),
l/'(0)| < 4.
5212
Let P be the set of holomorphic function / on the open unit disc so that
(i) Both the real and imaginary parts of /(z) are positive for \z\ < 1, (ii)
/(0) -l + i. Let E - {/(|) : / G P}. Describe E explicitly.
(Minnesota)
Solution.
Let f £ P and define
C - F(z) = /2(2> ~ 2i
C l) /2(2) + 2i"
Then F is a holomorphic function on the unit disc with F(0) = 0 and |.F(z)| <
1. By Schwarz's lemma, we have \F(z)\ < \z\, which implies |.F(!)| < |. It
should be noted that when / changes in P, F(|) can take any value in the
disc {C : |C| < |}. Because w = ^¾^ (that is the inverse of C = ^ff) is a

371
conformal mapping of {£ : |C| < |} onto {w : \w — ^i| < |}, we obtain that
the set {/2(§) : / G P} is equal to
{u> : |u> — i\ < -} = {w = pe"^ : \<f> — — | < arcsin -,p2 —— psin<f> + 4 < 0}.
o o Zi 0 o
Hence
1 ir 1 4 90
£ = {/(-) : / G P} = {reie : \0 - -| < - arcsin -, r4 - yr2 sin 20 + 4 < 0}.
If we denote the two roots of p2 — 2£-ps'm$ + 4 = 0 by pi{<j>), P2{<j>) where
Pi(4>) < P2(<f>) and |<^> — j| < arcsin |, the set E can also be represented by
lreie : |0 _ I| < I arcsin |, v7^) < r < VM20)\ ■
5213
Let
fi = {to = u + iv : -^- + -j > 1).
If T is the family of all analytic function on fi such that |/| < 1 in fi and
lim f(w) = 0, find sup |/(8)|. Your answer should be an explicit number,
w—>oo fcT
and you should prove your assertion.
{Indiana)
Solution.
Define w = 4>(z) = 2(|+ -), it is easy to know that w = 4>(z) is a conformal
map of D = {z : \z\ < 1} onto fi with ¢(0) = oo and ¢(4 - -\/l2) = 8.
Then F(z) = /o<£(z) = /(2( § + })) is analytic in D and satisfies F(0) = 0
and |.F(z)| < 1. By Schwarz's lemma,
\n*)\ < \*\-
Hence
|/(8)| = |F(4 - v/12)| < 4 - y/n.
This upper bound can be reached if we let / = ^-1 which belongs to family T
and satisfies <^>_1(8) = 4— vT2. So we obtain
sup |/(8)| = 4-V12.
/6^

372
5214
Let D be the upper-half and let f ^ id be a conformal map of D onto itself
such that f o f — id. Prove that / has a unique fixed point inside D.
(SUNY, Stony Brook)
Solution.
Since / is a conformal map of D onto itself, it can be written as /(2) = ff+3,
where a,b,c,d £ M and ad — be > 0. Then
(a2 + bc)z + b(a + d)
3 Jy ' c(a+d)z + d2 + bc '
It follows from / 0 / = id that 6( a + d) = c(a + d) = 0 and a2 -\- be —
d2 + bc^ 0.
If a + d ■£■ 0, then 6 = c = 0. Hence ad— be > 0 and a2 + 6c = d2 + bc impies
/ = id, which contradicts the condition / -£ id. Thus we have a + d = 0 and
the inequality ad — be > 0 can be written as be + a2 < 0.
Now we consider the equation /(2) = ^±| = 2, which is equivalent to
cz2 + (d - 0)2 - 6 = 0. Since A = (d - a)2 + 46c is equal to 46c + 4a2 < 0, we
know that /(2) = 2 has two conjugate roots, one in the upper-half plane and
the other in the lower-half plane. So / has a unique fixed point inside D.
5215
Let fi be a convex, open subset of W and let / : fi —» W be an analytic
function satisfying Re/'(2) > 0, 2 £ fi. Prove that / is one-to-one in fi (i.e., /
is injective).
{Indiana)
Solution.
Let 2X ^ 22 be two arbitrary points in fi. L : 2(2) = 21+2(22—21), 2 S [0,1]
is the line segment connecting 21 and 22. Since fi is convex, L C fi, we have
/(¾) - f(Zl) = [ f'(z)dz = I f'(z(t))(z2 - 2i)d*.
J L Jo
Hence
z2 — zl Jo
Since Re/'(2) > 0 for 2 6 fi, we know that J f'(z(t))dt ^ 0, which implies
/(21) ^. /(22) whenever 21 ^ z2-

373
5216
Show that if the polynomial P(z) = anzn + a„_iz"-1 + ■ ■ ■ + a,\z + ao,
n > 1, is one-to-one in the unit disk \z\ < 1 and a\ = 1, then \nan\ < 1.
(SUNY, Stony Brook)
Solution.
It follows from the univalence of P(z) in {\z\ < 1} that P'(z) = nanzn~l +
(n- l)a„_1zn_2H \-2a,2Z + ai ^ 0 for all z 6 {\z\ < 1}. In other words, the
roots of P'(z) are all situated outside the open unit disk. Let z\, Z2, ■ ■ ■ ,zn_i
be the roots of P'(z), then \zj\ > 1 for j — 1, 2, ■ ■ ■, n — 1. Because P'(z) can
also be written as nan(z-zi)(z-Z2) ■ ■ ■ (z- zn_x), by comparing the constant
terms, we have
n-l
(-l)n-1nan Y[z3 =ai.
Since a,i = 1, we obtain
i i lail
lna"l = ^I <1-
n i^i
5217
Let P(z) be a polynomial on the complex plane, not identically zero; let
H = {z : Rez > 0}.
(a) If all roots of P{z) lie in H, show that the same is true for the roots of
dP/dz.
(b) For any non-vanishing polynomial P(z), use the result in (a) to show
that the convex hull of the roots of P(z) contains the roots of dP/dz.
(Courant Inst.)
Solution.
(a) Let zi, Z2, ■ ■ ■, zn be the zeros of P(z). By assumption,
Rez,- > 0 (j = 1,2,---, n),
and P(z) = a(z — z{)(z - z2) ■ ■ ■ (z - zn). It follows that
, , P'(z) 1 1 1
log P(z))' = —if = + + ■ ■ ■ + .
v ° " P(z) Z - Zi z- z2 z- z„

374
When z 6 {z : Rez < 0}, then | < arg(z — Zj) < ^-, or equivalently, Rejrjr <
n ,. .
0. Hence Re ^2 —j: < 0, which shows -pjfi can not be zero on {z : Rez < 0}.
(b) Let zi,Z2, • • ■ ,zn be the zeros of P(z), and I is a directed straight line
passing through two zeros Z]. and z/ such that the other zeros are on the right
side of I (including on I). Denote the intersectional angle from the positive
direction of the imaginary axis to I by 6. When z is on the left side of I, we
have Re{e_i0(z- Zj)} < 0. Hence
I P(')l fri'-'i
which shows that the zeros of P'(z) do not lie on the left side of I. After
considering all the directed straight lines passing through two of the zeros of
P(z) such that the other zeros are on the right side of the line, we obtain that
the zeros of P'(z) lie on the convex hull of the zeros of P(z).
5218
Let /(z) be a Laurent series centered at 0, convergent in(F\{0}, with residue
b at z = 0.
(a) Show that there exists ¢ on {z 6<T : \z\ = 1} with
l/(0-C_1l > l&- i|,
(b) Characterize those functions with
max |/(C)- ¢-^ = 16-11.
(Minnesota)
Solution.
(a) Let /(z) = Y, Kzn, then
n= —oo
H
2** J\C\ = l

375
If 1/(0 - C_1| < \f> - 1| holds for all C with |C| = 1, then
Ifr-^^ffi^lrtO-C1!- I |dCI<|fc-i|,
which is a contradiction. Hence there exists ¢ with |C| = 1 such that
l/(0-C_1l>|fc-i|-
(b) If max |/(C) - C"1! = \b - 1|, it follows from
\b-i\<±-f \f(Q-Cl\\<K\
that
\f(Q-Cl\ = \b-i\
holds for all C with |C| = 1.
Let /(0 - ¢-1 = (b - l)eW), where < = e*e and ¢(9) is a continuous
real-valued function. It follows from
'-'-is/*,..*"0-<"'>«
that
_L I'* eW)+')M = it
2Wo
which implies that ¢(6) = —9, and hence
/(0-c1 = ^
holds on {£ : |<^| = 1}. Apply the discreteness of zeros for analytic functions
to f(z) - f, we obtain f(z) = f, z G<F\{0}.
5219
Assume / is analytic in a neighborhood of D, f maps D into D, and /
maps 5D into dD, where D = {z : \z\ < 1}.
(a) Show that Vz G d£>, /'(z) ^ 0.
(b) Show that fg[avgf(eie)] > 0 for 9 in JR.

376
(c) Assume that /(0) = /'(0) = 0 and f\gr, is a two-to-one map from dD
onto dD. Show that f(z) ^ 0 whenever 0 < \z\ < 1.
(Indiana-Purdue)
Solution.
(a) Assume f'(z0) = 0, where z0 6 dD. Let f(z0) = w0 G dD. Then
f(z)-w0 = (z~z0)ng(z),
where n > 2,
3(2) = 60 + &i(z - 20) + &2(z - 2o)2 H ,
with 60 i=- 0. Let T be an arc in D denned by T — {z S D : \z — z0\ = r},
and denote by Ar<^>(2) the change of 4>(z) when 2 goes along the arc T in
the counterclockwise sense. It is demanded that r is sufficiently small such
that Ararg(2- z0) > ^ and \g(z) - b0\ < ^ when zGT. It follows from
/(2) - ibo = (* - 2o)"ff(2) (n > 2), that
Ararg(/(2) - w0) = nArarg(2 - 20) + Arargff(2) > — - - > ir,
which implies that /(2) assumes values outside the disk D when zGT. It is
a contradiction to the fact that / maps D into D. Hence /'(2) ^ 0 for all
2Gd£>.
(b) Let 2 = re'0, and w = /(2) = iJe8*. A variation of the Cauchy-Riemann
equations for analytic function w = /(2) is
dR ndtp dR M
or 06 06 or
Since / maps dD into dD, we know that §f (eie) = 0. If §£(e*e) = 0, then
at point eie, §£ = §f = §£=§£ = 0, which implies that ff(e^) = f'(eie) =
0. But from (a) it is impossible. If %(ei0) < 0, it follows from r§£ = i?§£
that f^(e!'e) < 0. Since R = 1 when r = 1, §£(e*'e) < 0 implies that i? > 1
when r < 1. This is also impossible. Hence we obtain
£<•"> = ![■*««")] ><>.
(c) Because /|au is a two-to-one map from dD onto dD, ^A|j| = 1arg/(2) =
2, which implies that /(2) has two zeros (counted by multiplicity) in D. Since
/(0) = /'(0) = 0, 2 = 0 is a zero of / of multiplicity m = 2. Hence /(2) has
no zero in {0 < \z\ < 1}.

377
SECTION 3
COMPLEX INTEGRATION
5301
Evaluate the integral
J\z\
(Indiana)
ee'dz.
'1*1=2
Solution.
Function ee' is analytic in {z : 0 < \z\ < +oo}, and its Laurent expansion
around z = 0 is:
j.
j.
1 2 . 1 «
eez = l + et + ^-e? + h -^e" +
2! n!
= l + {l+-z + h-^ + -} + h{l+l + h^)2 + -}
1 f n 1 /n\2 1
+-- + ^{1+z + 2y(z) + -} + -■
The coefficient of the term j in the above development is
1 + 1 + 5J + -+(^TTJ!+- = e-
By the residue theorem, we obtain
j eeidz= 2iriRes (ee^,o) = 2irei.
5302
Evaluate
/
dz
where 7 is the positively oriented circle {\z\ — 1}.
(Indiana)

378
Solution.
It is obvious that ■ \ is analytic in{z:0<|z|<l}, and with z = 0 as a
pole. The Laurent expansion of . *a z around z = 0 can be obtained as follows:
1 1
sin3z (2_i23+^25_...)3
1
= ?{1+3(r2-r4+-)+6(r2-r4+-^+-
Hence the coefficient of the term j in the above development is |. By the
residue theorem, we have
dz _ .„ ( 1
2iriRes ( —5—,0 = iri.
77 sin z \sin z
5303
For what value of a is the function
f{Z) = l (}+^)C0SZdZ
single-valued?
(Indiana)
Solution.
Function F(z) = (j + -^) cos z is analytic in {z : 0 < \z\ < +00}, and its
Laurent expansion around z = 0 is:
a t a\ 1 ( a 1\
The necessary and sufficient condition for /(z) to be single-valued is that
the residue of F(z) at z = 0 is zero, i.e., the coefficient of the term j in the
above development is zero. Hence we obtain a = 2.

379
5304
Define
yoo
/1(2)=/ (l + zte-ty1e-tcos(t2)dt.
Jo
What is the largest possible P so that h(z) is analytic for \z\ < P?
(Indiana-Purdue)
Solution.
When 2 = —e,
V ' Jo e*-et
It is easy to see that when i-tl,
cos(i2) A
e*-et (i-1)2'
where A = |cosl, which implies that the integral is divergent. Hence P can
not be larger than e.
For any r < e, let \z\ < r. Consider the integral
^-1 s
""CI a.
+ 2i
It follows from |e* + 2i| > e* — ri and the convergence of the integral
cos(i2)|
e* — rt
that
f°° lcc
/^ cos(i2)
Jo
dt
dt
/0 e* + 2i
is uniformly convergent in any compact subset of {2 : \z\ < e}. By Weierstrass
theorem, we know that h(z) is analytic in {2 : \z\ < e}. Hence the largest
possible P is equal to e.
5305
Let /(2) be analytic in S = {2 G<F; |jz| < 2}. Show that

380
- f * f(eH) cos2 Ut = 2/(0) + /'(0).
(Iowa)
Solution.
It is easy to see that
/(0) = 7^-.1 l^1dz=± T fie^dt,
2ttiJ{z{ = 1 z 2irJ0
/'(0) = ~[ ^z = J- /^ /(e'V'dt.
Note that
It follows from the above three equalities that
2/(0)+ /'(0) = i.y X/(e")(2 + e" + e-«)<ft
= -/ /(e")cos2|di.
5306
Suppose that the real-valued function u is harmonic in the disk {\z\ < 2},
v is its harmonic conjugate and u(0) = t>(0) = 0. Show that
£u\zy(z)^=l-£(u\z)+v\z))^,
where j(t) = e2wit, t e [0,1].
(SUNY, Stony Brook)
Solution.
Let /(z) = u(z) + iv(z). Then /(z) is analytic in {z : |z| < 2}, and we have
= 2iri/4(0) = 0,
IM -

381
It follows from
and
u(z) =
v(z) =
(/"™G))
/(2) + /(*)
2
2i
that
fA>)A>)* = --jjf(/V>+ 7^)-21/(1)14) *
which implies that
Ju\z)v\z)^ = \j{u\z)+v
\*))di-
5307
Let / be an analytic function on an open set containing D(0,1) = {z; \z\ <
!}■
(a) Prove that
eT//m n!
dz
n ' * Jo
e-n,e[Ref(e,e)]d6.
(b) If /(0) = 1, and if Ref(z) > 0 for all points z G D(0,1), prove that
dzn
(0)
<2(n!).
(Indiana)

382
Solution.
(a) Assume that
f(z) = J2*kZn,
Jfc=0
have
e-(„+*)iedd _ Q_
By Cauchy Integral Formula,
^(0)--^/ l«±d<-^ Trie*
dz"
0\ -ni0
)e-"!0de.
Hence
dn/
dzn
(0)
ni /,2'r „ .„ „1 r2*
= f I f(e>°)e-«>°de+^ /(e«)e-»«
= -/ %-"i0[Re/(e''e)]de.
^ Jo
d0
(b) Because Re/(z) is harmonic on D(0,1), by the mean-value formula of
harmonic functions,
^- f Ref(ei0)d9 = Re/(0) = 1.
2ir 7o
Noting that Re/(e!'e) > 0, we have
g(o)| = I^V-iimO]*
< — / |e-"i0|[Re/(e!'e)]de
*" Jo
= -/ Ref(e'e)d6
* JO
= 2(n!).

383
5308
If / is analytic in the unit disk and its derivative satisfies
\f'(z)\<(l-\z\)-\
show that the coefficients in the expansion
f(z) = J2anz"
satisfy \an\ < e for n > 1, where e is the base of natural logarithms.
(Stanford)
Solution.
It follows from
that
where
na,
f(z) = J2^nZn
n-0
oo
/'(z) = 5>anz"-\
n=l
= J_ f f-^*±dz, (0 < r < 1).
It is obvious that
l«i| = l/'(0)| < 1 <e.
For n > 1, we choose r = 1 — -,
1
2irn
f £(£
2)
dz
<
i (i)
iw
—-■ n/lN -2t(1--)
27rn (1-i)" n;
(1 + rf-^e.
n — 1
5309
Let / = u + it) be an entire function.

384
(a) Show that if u2(z) > v2(z) for all z £(£, then / must be a constant.
(b) Show that if |/(z)| < .4+5^171 for all z 6 <F with some positive numbers
A,B,h, then /(z) is a polynomial of degree bounded by h.
(Stanford)
Solution.
(a) Let
F(z) = e~f2^ = e-(»*(*)-»3(*))-x»(*M*)m
Then F(z) is an entire function with
\F(z)\ = e-(»2(*)-»2(*)) < 1.
By Liouville's theorem, F(z) must be a constant, which implies that /(z) is a
constant.
(b) Let
OO
f(z) = ^>nz".
n=0
Then
1 f /Wj
GL, = / , , dZ.
27TZ 7u| = ij z"+i
'1*1
For any integer n > h,
,4 + 5^
Letting R —► +oo, we obtain that a„ = 0, which implies that /(z) is a
polynomial of degree bounded by h.
5310
Let / be an entire function that satisfies |Re{/(z)}| < |z|" for all z, where
n is a positive integer. Show that / is a polynomial of degree at most n.
(Indiana)
Solution.
Let R be an arbitrary positive number. Then it follows from Schwarz's
theorem that when \z\ < R,
f^ = h\ Re^/(c)} ■ r4 ■ v + iIm^(°)>-
2i" J\a=R C - z <

385
Especially when \z\ = y,
1/(2)1 < ^- • 3iT ■ 2* + |lm{/(0)}| = 3U" + |lm{/(0)}|,
which implies that there exist constants A, B such that
|/(z)|<il|z|"+B
holds for all z G <F.
Let
OO
Jfc = 0
where
1 / /(2)^
2tti Ju[ = r z*+i
'1*1
Hence when k > n,
1 fc| - 2ir ,/M=r Izl^H"1' ' - rk y h
which shows that /(z) is a polynomial of degree at most n.
5311
Compute the double integral
/ / cos zdxdy
where D is the disk given by {z = x + iy G<F : x2 + y2 < 1}.
Solution.
First we have the following complex forms of Green's formula:
/ / wzdxdy = // -(wx — iwy)dxdy
= -ttt / u>(cte - idy) = -xt / u>dz,
<" JdD Zl JdD
/ / wjdxdy = / / -(% + iwy)dxdy
= — / u)(dx + idy) = —- / Wz.
2z 7aD 2z yai>
(Iowa)

386
The problem can be solved directly by either one of the above two forms:
/ / cos zdxdy = — I z cos zdz = — / dz = ir;
J Jd 2* J\z\=\ 2z j,z]=1 z
/ / cos zdxdy = — —- / sin zdz = / sinzd(-)
7 7p 2ty,,,=1 2z'7|,,=1 V
1 f sin z
2iL = i
, -dz = ir.
z1
5312
Let
/(z) = 5>nz"
n=0
be analytic in £)= {|z| < 1} and assume that the integral
A= J j \f'(z)\2dxdy
is finite.
(a) Express A in terms of the coefficients an.
(b) Prove that
\f(z) - /(0)| < J± log —L-p
y ir l — |z|
for 2g]).
Solution.
(a) By
OO
n=l
we have
(Indiana)
A = / / |/'(z)|2dxdy= / rdr / X(/V))(/'(re"))<»
= f rdr r (f^na^e^-'A (f^na^-'e-^-'A 60.

387
Noting that
we obtain that
£■>--'»"={;,
k = L
A = [ [ \f'(z)\2dxdy= [ rdr f *f>2|an|V
J Jd Jo Jo n=1
»1 oo oo
= 2» / x)n2ia»i2r2n"idr=,rz;nia»i2-
Jo n=l n=l
(b) By Cauchy's inequality, we have
oo oo , . .
n = l n=l V /
d0
1/(^)-/(0)1 =
OO °° 1
1 n = l n=l
|2" = J^log —
1- Izl2'
5313
Let / be analytic in {0 < |z| < 1} and in L2 with respect to planar Lebesque
measure. Is 0 a removable singularity? Proof or counterexample.
(Stanford)
Solution.
The answer to the problem is Yes.
Let the Laurent expansion of / in {z : 0 < |z| < 1} be
/(*)= Yl a«2"'
where
Frc
an=^Lr<Mdz> (»=°.=^.-■■)■
^ |/(re^'
,, i r
'-de,

388
we have
r2x \'2, f2x
\a lVn+
1 < (^ J J 1/(^)1^) < i- J2J \f(reiS)\2rd6.
Let e < 1 be a small positive number, and then
[ \an\2r2n+ldr < J- / ' / |/(rei(,)|2rdrd0 < i- / / |/(z)|2d*dy.
A 27r JO 7e ^ 7 J0<\z\<l
Then an must be zero when n < — 1. Otherwise, let e —* +0, the left side of
the above inequality will tend to infinity, while the right side of the inequality
is finite, which leads to a contradiction. Hence
oo
/(z) = 5>nz",
n = 0
which shows that z = 0 is a removable singularity of /.
5314
Evaluate the integral
/ 1 12' a M1 p-
J\z\=p F-a|2
[Indiana)
Solution.
Let z = peie, a = re'*
r \dz\ _ r2K pde
/M=p|z-a|2 ~ J0 p2 + r2-pr^(0-^ +eW-0))
pde
-
Jo
J\z\ = p
_ f pidz
Ju
p2 + r2 - pr(eie + e~ie)
pdz/(iz)
p2 + r2 — rz — p2r/z
l\z\=p rz2 - (/92 + r2)z + p2r

389
When r < p,
\dz\ f pidz
f \dz\ t
J\z\=p \z - a\2 J\z\=Pr
\z\=p\z-a\* J\z\=Pr(z- ^)(z -r)
= 2in ■ —Res
r V(2-1r)(z-r)
2irp
p2 — r2
When r > p,
pidz
r \dz\ = r
J\z\=P\z-a\* L
/|,|=P^-«I2 J\Z\=P r(z - ^)(z - r)
2iri ■ —Res I = , ^-
r \(z-^-)(z-r) rt
2irp
r2 — p2
Evaluate
Jo a-
5315
dx . .
—> lal > !.
+ sin x
by the method of residues.
Solution.
Denote
dx
(Columbia)
sin2 x
JQ a + sir
It is obvious that 1(a) is an analytic function in {a : \a\ > 1}. Then we have
,. , /"= dx /"= 2dx
M01) = / 9— = / ^
Jo a + sinJ a; y0 2a + 1
r dx -~ r -
J0 2a + 1 - cos x 2 J_ K 2a + 1
cos 2¾
dx
— cos a;

390
Let z = e'x, then
dx =
cos a; =
f
dz
iz
Z+ 2_1
2 '
idz
and
/(a) = X, = 1 22- 2(2a+ l)2+r
Denote the two roots of 22 — 2(2a+ 1)2+ 1 = 0 by 2X and 22. Since 2X -22 = 1,
we may assume that \zi\ > 1, |22| < 1. By the residue theorem we have
/ idz 2ir
1(a) = /
\z\ = i (z-zi)(z-z2) Zi-Z2
2ir ir
%/(zi + 22)2 - 42i22 2y/aJa~+T)'
It should be noted that —, w is also analytic in {a : \a\ > 1}, and the
2-^/0(0+1)
branch of ^/a(a + 1) should be chosen by argiv/a(a + 1) |a>i= 0.
5316
Consider the function
g(*»g) = ,_,_ * ■ B-
1 + 2 sin 0
(a) Use the residue theorem to find an explicit formula for
r2w
/(2)= f g(z,6)de
Jo
when \z\ < 1.
(b) Integrate the Taylor expansion
00
g{z,6) = YJ9n{0)zn
n = 0
term by term to find the coefficients in the Taylor expansion
00
/(*) = £/„*».
n=0

391
(c) Verify directly that (a) and (b) agree when \z\ < 1.
(Courant Inst.)
Solution.
(a) Let < = eie. Then
C2-i
° 2* ~ 2iC '
and
where ¢1 = j(\/l — z2 — 1), ¢2 = |( —\/l — z2 — 1), and the single-valued branch
of Vl — z2 in {|z| < 1} is denned by \/l — z2 \z=o= 1- Because |Cx • C21 = 1,
we know that ¢1. G {|C| < 1} and ¢2 G {|C| > 1}. Hence
,, x „ ■ 2 1 2ir
/(z) = 2iri •
Ci - ¢2 VT^z2 "
(b) It follows from |sin#| < 1 and \z\ < 1 that
00
ff(z,fl) = 5](-l)fc8bfcfl-Zfc, (|Z|<1).
Jfc=0
Since the series converges uniformly for all 6 G [0, 2ir], the integration with
respect to 6 can be taken term by term, and
.2t /00 \ 00
/(*) = / £(-l)fc sin* 0 ■ zk ]d6 = J^ akz",
where
r2x
ak= (-l)fcsinfc 6dd.
Jo
It is easy to obtain that a,2n-i = 0 and
^=4^.^^ = ^^.2..
(c) In order to verify that (a) and (b) agree when \z\ < 1, we develop the
function /(z) in (a) into a power series:
f(z) = -jJL= = 2*(1 - z2)-* = 2x^(-l)"C^z2"
n = 0

392
Since
(-1) c_i-(-1) ^ -"(^rjir'
we know that the results in (a) and (b) agree when \z\ < 1.
5317
If a is real, show that
/R
■R
exists and is independent of a.
Solution.
First we have
( UC, Irvine)
e -(*+<«)
It follows from the existence of
a „ — x
< e° -e
lim
JJ->oo
that
fR
/ e~x:>dx
J-R
rR
lim / e-^+ia^dx
R^oo J_R
exists.
Define /(z) ■= e~z and choose the contour of integration T = Ti U T2 U
r3 U T4 as shown in Fig.5.5.
y
r3 at
'•1
-fi 0
jr.
r, fi
Fig.5.5
As /(z) is analytic inside T, by Cauchy integral theorem,
[ f(z)dz = [ f(z)dz+ [ f(z)dz+ f f(z)dz+ f f(z)dz
Jr JTi Jr2 Jra Jr4

393
/* e-'*dx + ie~R2 f e^-Wdy - /" e^'+^'dx
J-R JO J-R
~ie-R2 /V
Jo
' + 2Ryi
= 0.
Letting R —► oo, it follows from the facts that e R —> 0 (R —* oo) and
I r eV'±2RVidy\ < re»adl/
I ./0 I ./0
that
/R [R
e-(*+*»)'da, = lim / e—'ds = V^-
i? R^°°J-r
5318
Let n > 2 be an integer. Compute
r°° l
7o i+»n
da;.
Solution.
(Iowa)
Fig.5.6
Let /(z) = yxj-j-, and select the integral contour T as shown in Fig.5.6.
/(z) has one simple pole z = e»! inside T. By the residue theorem, we have
I f(z)dz = 2iriRes(/,e»!).
°e^dx
+ x"
= (!-«*')/ T^T + / " iBe»f(Re»)d6.
Jo L + x Jo

394
It is obvious that
and
j:
lim j iRe'°f(Re,v)d6 = 0,
Res(/,e»!)= -
Letting R —► oo, we obtain
t°° dz 2iri
Jo l + *n~ ne^*i (i .
ne
2iri
LA n(e»'—e~»') rising'
5319
Evaluate
with full justification.
Solution.
i»0O
I cos(x2)dx
Jo
(Minnesota)
Define
/(*) = e~z ,
and choose the contour of integration r = £2 T^ as shown in Fig.5.7. Because
/(z) = e~z is analytic on T and inside T, by Cauchy integral theorem, we
have
[ f(z)dz =J2f f(z)dz = 0.
Jr trj ^

For the integral of /(z) on T2, we make a change of variable by w = z , then
dw
e ■
where
I f(z)dz = f
Jr2 Jy,
6 'V~z>
ZU>2
72 = {w : \w\ = R2,0 < axgw < —}.
By Jordan's lemma, we have
lim I f(z)dz = 0.
For the integral of /(z) on T3, we have
fR
j f(z)dz = -/ e-'^et'dx
Jr* Jo
f ^-1 2 , ■ 2x. ■ f V%, 2 • 2xj
= — / -^-(cosa; +smi )0¾ — 1 I -—(cosa; -sini )dx.
Jo 2 y0 2
It is well known that
~2~
/ f(z)dz = f e~x2dx
Jrt Jo
when R —► 00. Hence we obtain by letting i? —► co that
f°° f°° v/2ir
I (cos x2 + sin x2)dx + i I (cosx2 — s'mx2)dx = ,
./0 ./0 2
which implies
/2ir
/ cos x dx — I sin a; da; =
./0 »/0
5320
Evaluate
r°° sin2 z ,
—7.—dx.
f
Jo
[Iowa

396
Solution.
Define
/(*)
l-e
2iz
and select the integral contour T as shown in Fig.5.8. Because /(z) is analytic
inside T, by Cauchy integral theorem,
J f(z)d
!z = 0,
where
r£ 1 - e2ix
/ f(z)dz = / —dx+ iReief(Rei0)d6 + ~-dx
Jv Je x Jo J-R x
r°
+ / ieeief(eeie)d8
/R O _ p2ix _ p-2ix i"T
— dx + / iRei0f(Rei0)d6
+ / ieei9f(eeie)d9
j 4sin_x&+ j* iReief{Rei0)d6 + f ieeief(eew)d6.
Je x Jo J-k
It is easy to see that
and
lim ( iReief(Reie)d6 = Q
•R->°° Jo
r°
lim / ieeie/(ee!'e)^ = -7riRes(/,0).
J 7T

397
Since the Laurent expansion of / about z = 0 is
oo
f(z) = J2a"zn>
n = -l
where a_i = — 2i, we know that Res(/, 0) = — 2i.
Letting £-*0 and R —» oo, we obtain
/•oo • 2
r sin a; , ir
-ax
7o »=2 "~ 2'
5321
Let /(z) be holomorphic in the unit disk \z\ < 1. Prove that
/ f(x)dx=— f(z)\ogzdz
Jo *t™J\z\=i
1*1=
where respective integration goes along the straight line from 0 to 1 and along
the positively oriented unit circle starting from the point 2=1. The branch
of log is chosen to be real for positive z.
(SUNY, Stony Brook)
Solution.
Fig.5.9
Let the contour of integration T be shown as in Fig.5.9, and the single-
valued branch of logz be chosen by axgz\z=_i = ir. Since /(z)logz is
holomorphic inside the contour T, by Cauchy integral theorem,
J f(z)\ogzdz = Q,

398
where
I f(z)logzdz = / f(x)logxdx + / f(z)logzdz
Jr Je J\z\ = l
+ / /(x)(logx + 2iri)da;+ / f(eeie)log(eeie)ieeiBdB
./1 ./2*
= -2iri / f(x)dx + / /(2) log zdz
- / f(eeie) \og(eei0)ieei0d6.
Jo
It is easy to see that
r2x
lim f f(eeie) \og(eeie)ieeied9 = 0.
Letting e —► 0, we obtain
/ f(x)dx = — I /(z) log zdz,
./0 2l™ ^Ul=i
where the integration contour |z| = 1 has starting point and end point z = 1,
and the value of log z at the starting point z = 1 is defined as 0.
5322
Find the value of
,2w
/ log|a + 6e^|c^
Jo
where a and b are complex constants, not both equal to zero.
(Harvard)
Solution.
First we assume \a\ > \b\, and then the multi-valued analytic function
log(a + bz) has single-valued branch on {z : \z\ < 1}. Take e"^ = z, then
d<}>= ^, and
I log|a + 6e^|^ = Re{/ \og(a + bei4,)dA
= Jr ^±^4
= Re{2irloga} = 2irlog \a\.

399
When \a\ < |b|, we have
/ log|a + 6e'*|cty = / log|6 + ae'*|d0
Jo Jo
= 2irlog|6| = 2irlog|6|.
In the case \a\ = \b\, let b = ae'a. Then
/ log|a + 6e'*|cty = / (log \a\ -flog |1 + e'(*+°>|)cty
Jo Jo
= 2irlog\a\+ J log|l + e**|cty.
J — )T
i(n-e)
i(-7T4-£
Fig.5.10
In order to evaluate the integral
f log|l+e'*|d0,
J—T
we define
/w =
log(l-fz)
where the single-valued branch is defined by log(l + z) \z=o= 0. Choose a
contour of integration r = V£ \J^e as shown in Fig.5.10. Since /(z) is analytic
on r and inside T, by Cauchy integral theorem, Jr f(z)dz = 0. Because
II.
f(z)dz
<
10B I 4. E
1_£ 2 -^-^0 (e-»0),
we have
/ log|l + e*'*|cty = Re f log(l + e*'*)cty
«/ — 7T ^-7T

400
limRe{£log(l + z)§}
limRe {^/(^ = 0.
Hence we obtain
/
Jo
log|a + be'^ldtf) = 2irmax{log |a|, log \b\}.
Evaluate
Solution.
I
5323
log a;
o (1+*)3
;d,X.
(Iowa)
Let
Fig.5.11
/w = a
log2 2
(1+2)3
and select the integral path T as shown in Fig.5.11. The single-valued branch
of log z is chosen by argz|J=_i = ir. By the residue theorem, we have
I
/(z)dz = 2iriRes(/,-l),

401
where
rR l^«2 „, f2w
+ / ieei9 f(eei9)d9
+ / ieei9f(eei9)d9.
J2x
It is obvious that
p2jt
and
lim / iRei9f(Rei9)d6 = 0
R^°° Jo
r°
lim / ieet9f(eei9)d6 = 0.
«> 27T
'2;i
In order to find Res(/, —1), we consider the Laurent expansion of / about
z= -1:
f(z\ = l°g2[(* + l)-l]_(^ + log[l-(*+l)])2
^ ' (2+1)3 (2+1)3
(^-(2+1)-1(2 + 1)2-.-.)2
(2+1)3
oo
£ an(2+l)",
n = —3
where a_i = 1 — iri. Hence
2iriRes(/, -1) = 2iri + 2ir2.
As e —► 0 and R —► oo, it turns out that
+ 4tt2
j0 (i + xy
Comparing the imaginary parts on the two sides of the above identity, we
obtain
f —
Jo (1
logx d* = -L
+ x)
3

402
5324
Evaluate the following integrals:
(a) /-/r (*'-4)to8(*+T) (the integration is over the imaginary axis),
(b) /0°° ^rr[dx for a in the range -1 < a < 2.
(Courant Inst.)
Solution.
(a)
Fig.5.12
Define
/(*)
(22 - 4) l0g(2 + 1) '
The single-valued branch for log(2-f 1) is chosen by log(2+ l)|J=o = 0, and
the contour T of integration is shown in Fig.5.12. As /(2) is analytic on and
inside T except a simple pole at 2 = 2, we have
J f(z)dz~2iriRes(f,2),
J f(z)dz = f2 f(Rei0)iReied9- [' f(eei0)ieeil
f(z)dz- / f(z)dz,
■iR Jie
where
d6
and
Because
Res(/, 2) = lim
(2-2)
2(22 -4)log(2 + l) 4log 3
lim [' f(Reie)iReied9 = 0

403
and
lim J 2 f(eei0)ieei0d6 = iriRes(/, 0) = -^,
by letting e —» 0 and R —► oo, we obtain
f+i°° * = ™(1--A_)
7-ioc (z2-4)log(z+l) 4l log 3^
(b)
Define
/(*) =
z3 + l'
The single-valued branch for za is chosen by argz|J=:r>0 = 0, and the contour
T of integration is shown in Fig.5.13. As /(z) is analytic on and inside T except
a simple pole at z = e■»', we have
I f(z)dz = 2iriRes(/, e**'),
where
[ f(z)dz = f f(x)dx + j f(Rei0)iRei0d6- f e^if(x)-e^idx
Jr Je Jo Je
- / f(ee'0)ie
Jo
)iee>"de,
Jo
and
Re8(/,«7') = Hm (z - e »!')/(2)
2-»e J
e »
3e »!
]__
3ei(2-^K'

404
Because
when a < 2 and
,3s.
lim I * f(Rei9)iRei9d6 = 0
lim I
f(ee*v)iee,vd6 = 0
when a > —1, by letting e —► 0 and i? —► oo, we obtain
t°° ■*»
/•°° a:"
/o *3 + l
da;
3sin(^7r)"
5325
Show that
L
xa , ir(l-a)
ax =
/o (1 + x2)2 4cos(^-)'
for —1 < a < 3, a ^ 1. What happens if a = 1?
(.ffarwurd)
Solution.
Let
za
/(Z) = (1 + Z2)2'
where (argza)j=x>o = 0, and select the integral path T as shown in Fig.5.14.
By the residue theorem, we have
L
f(z)dz = 2iriRes(/(z),i),
where

405
r°
+ / ieeief{eeie)d8
= (1 + e«-) jR j^f^dx + £ iReief(Reie)de
+ / ieei9f(eei9)d6,
J T
and
Res(/(z),i) = lim
z—*i
It follows from a < 3 that
and from a > — 1 that
(z + i)2\ Ai
lim f iReief(Reie)d6 = Q,
R^ooJ0
lim / ieeief(eeie)d8 = 0.
1 - a , ™
e =
Letting e —> 0 and R —> oo, we obtain
r°° -^o
(1 + e"°»i (TTW*
ir(l - a) ,™
-^-: -e 2
When a^l,
when a = 1.
/0° x^_
Jo (1 + *
/
Jo
da;
ir(l — a) ir(l — a)
~4cos(^)'
t 4.pr?)
x , ,. ir(l — a) 1
zdx = lim —
(1 + ¾2)2 a-i4cos(^) 2
5326
(a) Prove that
converges if 0 < a < 1.
i»0O
/ eixx~adx
Jo

406
(b) Use complex integration to show that
I -n , . TO „, ,.
I x cosxdx = sin — • r( — a + 1).
Jo 2
Solution.
(a)
/ e'xx adx = I I x a cos xdx + I x a cos xdx I +i I x
It follows from a < 1 that
I x~a cosxdx
Jo
is convergent. It is also obvious that
(Harvard)
' s'mxdx.
/ cosxdx < 2, /
sin xdx
<2,
i "is monotonic decreasing and
lim x a = 0 for a > 0.
X—» + oo
By Dirichlet's criterion, we know that J1 x a cos xdx and /0°° a; ° sin xdx are
also convergent. Hence JQ e'xx~adx is convergent when 0 < a < 1.
(b)
Let /(z) = z ae z, and the contour of integration T is chosen as shown in
Fig.5.15. The single-valued branch of /(z) on T is definde by z~a\z=x>0 > 0.

407
By Cauchy integral theorem,
x ae = 'e %xidx
R
f f(z)dz = f x-ae-xdx+ /2 iReief(Reie)d6+ f
Jr Je Jo JR
r°
+ / ieeief{eeie)dB = 0.
It follows from a < 1 that
t°
lim / ieeief(eeie)d6 = 0,
and from a > 0 and Jordan's lemma that
lim /2 iRei0f(Rei0)d6 = 0.
Letting e —* 0 and i? —> 00, we have
i»0O *0O
r(-a + l)= / x-ae-xdx = ie-^i x-ae~ixdx.
Jo Jo
Multiplying both sides by e~*~*, and comparing the imaginary parts, we obtain
f°° -a j • ™*™
/ a; cosxaa; = sin — r(—a + 1).
Jo 2
5327
Use a change of contour to show that
cos(ax)^ _ [°° te-°^
t2 + l
f°° cos(ax) _ r
J0 x+/3 X~ J0
dt,
provided that a and /3 are positive. Define the left side as a limit of proper
integral and show that the limit exists.
(Courant Inst.)
Solution.
Since
fA
I cos(ax)dx
Jo
2
< ~,
a

408
—j-3 is monotonic with respect to x and
x+0
lim
x-»+oo X + p
the convergence of the integral
t°° cos(aa;)
0,
dx
follows from Dirichlet's criterion.
Define
m
z + pV
and choose the contour of integration r = Ti U T2 U T3 as shown in Fig.5.16.
By Cauchy integral theorem, we have
I f(z)dz = f ^r.dx + [ f(z)dz - f
Jr Jo x+pi Jr J0
x+p
R
-dx
r*e-a'(z-pi), f t( ., f
-2 +/32
It follows from Jordan's lemma that
x+p
dx = 0.
R->co
lim / f(z)dz = 0.
Letting R —* 00 and considering the real part in the above identity, we obtain
,oc te-a„t
dx =/ —z——at.
f°° cos(ax) , f°° xe~ax , f
Jo ^WdX = h ^W*dX = h
t2 + l

409
5328
(a) Let c be the unit circle in the complex plane, and let / be a continuous
(F-valued function on c. Show that
*'W = /
/(C)
<K
is a holomorphic function of z in the interior of the unit disk.
(b) Find a continuous / on c which is not identically zero, but so that the
associated function F is identically zero.
{Minnesota)
Solution.
(a) Let zq be an arbitrary point in the unit disk. Then 1 — |z0| = p > 0.
Choosing 8 > 0 such that 5 < p, we prove that
/«)
'w = /
c
-dC
has a power series expansion in {\z — Zq\ < 5}.
It is clear that
20
C
< -< 1
P
when |z — zo| < 5 and ¢ £ c. We can also assume |/(C)| < M because / is
continuous on c. Thus
/(C) =
C-z
/(C)
/(C)
1
(C - z0) - (z - z0) C-zo 1- fEfS"
/(Q ~ /z-z0\"
C-zo^M-zo)
n=0
= £
n=0
/(C)
C - 20
/z-Zp\"
vc-20y '
As
/(C)
C -2o
/z-Z0\"
vc-20;
M
< —
P
oo / \n oo / -, n
and 52 ~ (p) is convergent, the series £) /^ (f^fo) converges
uniformly for all C S c. Hence termwise integration is permissible, and we obtain
*>(*)= //^-^ = f>„(z-z0)",
7c C "2 "^
n=0

410
where \z — zq\ < 5 and
Since zo is arbitrarily chosen in the unit disk, Ff(z) is holomorphic in {\z\ < 1}.
(b) Take /(0 = J (K| = 1). Then
In fact, /(<^) can be taken as ,,_',, for any positive integer n and fixed
z0 G {z : \z\ < 1}. When (Gc,
L{®*=L(i- 4 - «.>■ 4»=I a - .<)(i- .«)• *=°
5329
Let [a, b] be a finite interval in 1R and define, for z in D = <F — [a, 6],
J a
fb dt
t-Z
Show that /(z) is analytic in D. Given c, a < c < 6, calculate the limit of /(z)
as z tends to c from the upper half plane and as z tends to c from the lower
half plane.
(UC, Irvine)
Solution.
For any zq G D, choose S > 0 sufficiently small such that {z : \z — zo\ <
6} (~1 {z = x + iy : y = 0, a < x < b} = 0. When \z - z0\ < 6, a < t < b, we have
1 1 _ 1 1 _A (*-*o)n
= £
i-z (t - z0) - (z - z0) <-* 1-¾ ^(*-^)n+1'
and the series converges uniformly for t with a <t < b. Hence
ff, _ rbjL- f ff (2~2°)n
-11di

411
holds for z S {z : \z — zq\ < 6}, which implies /(z) is analytic in {z : \z — zq\ <
5}. Since zq is an arbitrary point in D, we obtain that /(z) is analytic in D.
For z G D, /(z) can also be represented explicitly by
f{z)z=ft^-z=l!di°g{t~z)
log
where the single-valued branch is defined by arg ( fz£ ) \z=xo>b = 0. Let Ti
and T2 be two continuous curves connecting z = xq > b and z = c in the upper
half plane and the lower half plane respectively. Then the limit of /(z) as z
tends to c from the upper half plane is
log
+ iAFlarg-
-b
log
c-b
+ iri,
while the limit of /(z) as z tends to c from the lower half plane is
log
+ iAr2arg:
-b
log
c-b
in.
5330
For each z G U — {z : Imz > 0} define
«)=&L
sin t
dt.
Determine which points a £ St have the following property: there exist e > 0
and an analytic function / on D(a, e) such that /(z) = g(z) for all z G U D
D(a,e).
(/nrfiana)
Solution.
Let r be the half unit circle in the lower half plane whose direction is
defined from point z = — 1 to point z = 1, and define a function
/w = Si/r
sin2 2
dt.
It follows from the Cauchy integral theorem that when z G U, f(z) = g(z).
With a similar reason as in problem 5328, /(z) is analytic in the complement
of T. Hence we obtain that for any a G M, a -£■ ±1, there exists e > 0

412
(e < min{|a—1|, |a+l|}) such that /(z) is analytic in D(a,e) = {z : \z — a\ < e}
and f(z) - g(z) for all z G U D D(a,e).
When a = ±1, such a f(z) does not exist. The reason is as follows: As
sin21 _ sin2 2 — sin2 z sin2 z
t — z t — z t — z '
. 2 + ■ 2
where sin ^1^1° z is an analytic function of two variables for (t, z) £<E xW, we
know that
V ' 2iriJ_1 t-z
is analytic for z 6 (F. But
1 Z"1 sin2z „ sin2z /1 ,, ., . sin2 z, z —1
^-: / & = „ . / ologli - z) = „ . log -,
which has branch points z = ±1, hence g(z) can not be analytically continued
toD(±l,e).

413
SECTION 4
THE MAXIMUM MODULUS AND
ARGUMENT PRINCIPLES
5401
Let oGff, \a\ < 1, and consider the polynomial
P{z)= | + (1- \af)z-ar\
Show that \P(z)\ < 1 whenever \z\ < 1.
Solution.
{Indiana)
When \z\ — 1,
P(z)
= | + (i-M»)*-p
= 4(i-l«l2) + ^-5z)]
ra a.
Re(- - az) = Re[ (az)} = Re[ ] = 0,
Im( az)
z
<2\a\
Hence when \z\ — 1,
4, a
\P(z)\2 = (l_|a|2)2 + (ImP(--az)])2
z z
< (1 - 2|a|2 + |a|4) + |a|2 = 1 - \a\2 + |a|4 < 1.
By the maximum modulus principle, |P(z)| < 1 whenever \z\ < 1.
5402
Let / be holomorphic in the unit disk {|z| < 1}, continuous in {|z| < 1}
and |/(z)| = 1 whenever \z\ = 1. Prove that / is a rational function.
(SUNY, Stony Brook)

414
Solution.
If f(z) has infinite many zeros, by the isolatedness of the zeros of
holomorphic functions, the zeros must have limit points on the boundary of the unit
disk. But it will violate the fact that / is continuous in {\z\ < 1} and |/(z)| = 1
whenever \z\ = 1. Hence / has only finite zeros in the unit disk. Denote all
these zeros by zi, Z2, • • •, zn, multiple zeros being repeated, and define
F(z) = f(z)/H
Jt=i
Zk
1 - ZjfcZ
Then F(z) is holomorphic in {|z| < 1}. continuous in {\z\ < 1} and |.F(z)| = 1
when \z\ — 1. By the maximum modulus principle, |.F(z)| < 1 in {\z\ < 1}.
Since F(z) has no zero in {\z\ < 1}, j^t is also holomorphic in {\z\ < 1},
continuous in {\z\ < 1} and
— 1 when \z\ = 1. Application of the
maximum modulus principle to jriLy yields |F(z)| > 1 in {\z\ < 1}. Hence
\F(z)\ = 1 holds in {\z\ < 1}, which implies F(z) = e'a with a a real number.
So we obtain
>w = «\n(rf£)
Jfc=l
5403
Let / be a continuous function on U — {z : \z\ < 1} such that / is analytic
in U. If / = 1 on the half-circle 7 = {eie : 0 < 6 < ir}, prove that /=1
everywhere in U.
(Indiana)
Solution.
Define F(z) = (/(z) —1)(/(-z) — 1), then F(z) is also continuous on U and
analytic in U. When z S dU, we have either /(z) — 1 = 0 or /(—z) — 1 = 0.
Hence F(z) = 0 holds for all z S £/", which implies either /(z) — 1 = 0 or
/(-z) — 1 = 0. Since /(z) — 1 = 0 is equivalent to /(—z) — 1 = 0, we obtain
/(z) = 1 for all zeU.
Remark. The condition that "/ = 1 on the half-circle 7" can be weakened
to that "/ = 1 on an arc 7 = {e'e : 0 < Q < ^}, where n is a natural number".
In this case, the proof is the same except that F(z) is defined by
F(z) = (/(z) - l)(/(ze^') - l)(/(ze^) -1)--- (/(ze^«) - 1).

415
5404
Let S denote the sector in the complex plane given by S = {z : — j <
argz < |}. Let S denote the closure of 5. Let / be a continuous complex
function on S which is holomorphic in S. Suppose further
(1) |/(z)| < 1 for all z in the boundary of S;
(2) \f(x + iy)\< e^ for all x + iy G S.
Prove that |/(z)| < 1 for all z£S.
(SUNY, Stony Brook)
Solution.
Let F(z) = e~ezf(z), where £ > 0 is an arbitrary fixed number. Then
F(z) is also continuous on S and analytic in S. When z is on the boundary
of S, \F(z)\ = e~"\f(x)\ < 1. When \z\ - +oo (-f < argz < f), \F(z)\ <
e-ex . e^/x _^ q gv (.^ jnaximum modulus principle, we have |.F(z)| < 1 for
all z S S, which implies |/(z)| < \eez\ = eex for all z S S. Because e > 0 can
be arbitrarily chosen, letting £ —+ 0, we obtain |/(z)| < 1 for all z S S.
5405
Let if be a compact, connected subset of (F containing more than one point
and let / be a one-to-one conformal map oi(U\K onto A = {z : |z| < 1} with
/(oo) = 0. If p is a polynomial of degree n for which |p(z)| < 1 for z 6 K,
prove that
IpOOI < l/(*)l"n for zeW\K.
(Indiana)
Solution.
Because / is a one-to-one conformal map oi(U\K onto A with /(oo) = 0,
it has a simple zero at z = oo. Since p is a polynomial of degree n, it has a
pole of order n at z = oo. Hence the function F(z) = p(z)/"(z) is analytic in
W\K which contains point z = oo. As /(z) maps(E\K onto A = {z : \z\ < 1},
we have lim |/(z)| = 1. Together with |p(z)| < 1 for z 6 K, we know that the
z—>iv
limit of \F{z)\ when z tends to K can not be larger than 1. Apply the maximum
modulus principle to F(z) on(D\K, we obtain |.F(z)| < 1 for z S @\K, which
implies \p(z)\ < \f(z)\~n for all z £W\K.

416
5406
Suppose / and g (non-constant functions) are analytic in a region G and
continuous on the closure G of the region. Assume that G is compact. Prove
that |/| + \g\ achieves its maximum value on the boundary of G.
(Iowa)
Solution.
Assume that |/| + \g\ achieves its maximum value c (c > 0) at zq G G, we
prove that if zq G G, then / and g must be constants.
Let
l/(2o)| = /(2o)e^, |ff(z0)| = (/(*„)*•'*'.
Then for fixed <j> i and <^2,
F(z) = /(z)e^1 + ff(z)e1'^
is analytic in G and continuous on G. It follows from
I^COI < \m\ + \g(z)\<c,
F(z0) = /(20^+g(zo)e"h = \f(z0)\ + \g(zo)\ = c
and zq G G that
F(z) = /(z)e* + g(z)e^
must be the constant c.
Without loss of generality, we assume that / is not a constant, and try to
lead to a contradiction. Since the image of an open set {z : \z — zq\ < 6} C G
under / is an open set which contains point /(z0), /(z) assumes all the values
/(z) = /(z0) + eei4> for small e > 0 and 0 < <j> < 2ir in {z : \z - z0\ < 6}. Then
when 4> + 4>\ ^- 0, x, we have
\f(z)\ + \g(z)\ = |/(z)| + |c-/(z)e^|
= |/(z0) + ee'*| + |c - /(zo)e^1 - ee'(*+*>) |
= |ee''(^-*i) + /(zo)e^11 + lee^*1* - ff(z0)e^2|
> f(zo)e^ + ff(z0)e^ = c,
which contradicts that Zo is a maximum value point of |/| + \g\. Hence / must
be a constant, which also implies g is a constant too.

417
5407
Suppose /(z) is an entire function with
1
l/WI <
|Rez|
all
Show that /(z) is identically 0.
Solution.
For any R > 0, consider function
(Iowa)
g(z) = (z - Ri)(z + Ri)f(z).
When \z\ = i?, and Imz > 0, denote by 6 the angle between the line
perpendicular to the imaginary axis and the line passing through z and Ri.
Then 0 < 6 < f, and
z — Ri n n-
secfl < V2.
Rez
When \z\ = R, and Imz < 0, denote by 8 the angle between the line
perpendicular to the imaginary axis and the line passing through z and — Ri. Then
0 < 0 < f, and
z + Ri
Rez
sec 6 < V2.
It follows from the above discussion that when \z\ = R,
(z - Ri)(z + Ri)
\g(z)\ = \(z-Ri)(z + Ri)f(z)\<
Rez
< 2V2R.
By the maximum modulus principle, when \z\ < i?,
1/(2)1 =
ff(*)
(z - Ri)(z + Ri)
<
2V2R
R2-\z\-
Now fixing z, and letting R —> +00, we obtain /(z) = 0. Since R can be
arbitrarily large, we have /(z) = 0 for all z S <F.

418
5408
Suppose / is analytic on {2; 0 < \z\ < 1} and
|/(2)| < log X
\Z\
Show that / = 0.
(Indiana)
Solution.
Denote the Laurent expansion of / on {2; 0 < \z\ < 1} by
00
/(2)= 52 ««2"'
n = —00
where
^-2^,^2-+^-
It follows that
27r./H=r|2+ | r" r
When n < 0, letting r —+ 0, we have
an = 0 (n =-1,-2,---),
which implies 2 = 0 is a removable singularity of /. In other words, / can be
extended to an analytic function of the unit disk.
Since log nr = 0 when \z\ = 1. By the maximum modulus principle, we
obtain
/ = 0.
5409
Let / be an analytic function on D = {2 : \z\ < 1}, /(D) C D and /(0) = 0.
(a) Prove that 1/(2) + f(—z)\ < 2\z\2 for all 2 in D and if equality occurs
for some non-zero 2 in D, then /(2) = e'°22.

419
(b) Prove that
I f1
/ f(x)dx
\J-i
<
(Indiana)
Solution.
(a) Let F(z) = f(z) + /(-z), then F(0) = 0,
nO)=lim^) = limfM_/L£)1=0.
£-►0 Z z->0 V z —z
£M
Hence -jp is analytic in D, and when z tends to dD, the limit of
not be larger than 2. By the maximum modulus principle, |/(z) + /(— z)\ <
2|z|2 holds for all z e D.
If equality occurs for some non-zero z in D, we have
/(z) + /(-z) = 2e'<V,
where a is a real constant.
Let
it follows from
that
/(z) = £>nz",
n = l
/(z)+/(-z) = 2eiaz2
a2 = e , 04 = a6 =
Because |/(z)| < 1 for z S D, we have
1 /* --
lim-/ |/(rei(,)|2d0 = ][>n|2<l.
Since a2 = e'a, the other coefficients must be zero, which implies /(z) = eiaz2.
(b)
/ /(*)&
1./-1
\ f° f1 I
/ /(x)da;+ / /(x)da;
U-i ./o I
1/-1 I /1
/ (/(^) + /(-^))^ < / 2x2dx
\Jq \ JO

420
5410
If/ is analytic and |/(z)| < 1 on {z : \z\ < 1}, prove that /(z) has a fixed
point.
(Rutgers)
Solution.
Let F(z) = f(z) - z and G(z) = -z.
When \z\ = 1,
\F(z)-G(z)\ = \f(z)\<l=\G(z)\.
By Rouche's theorem, F(z) and G(z) have the same number of zeros in
{z : \z\ < 1}. Since G(z) has only one simple zero in {z : \z\ < 1}, we conclude
that /(z) — z has one zero in {z : \z\ < 1}, which implies that /(z) has a fixed
point in {z : \z\ < 1}.
5411
Let /(z) = z+e~z, A > 1. Prove or disprove: /(z) takes the value A exactly
once in the right half-plane. If the answer is yes, is the point necessarily real?
Justify.
(Iowa)
Solution.
Let R be a sufficiently large real number such that R > 2A. Take a closed
curve T on the right half-plane, where
r = {z = x + iy : x = 0, -R < y < R} U {z : \z\ = R, -- < argz < -}.
Define
F(z) = X-z-e~z
and
G(z) = A - z.
When z G T,
\F(z)-G(z)\ = \e-'\<K\G(z)\.
Since G(z) has exactly one zero inside T, it follows from Rouche's theorem that
F(z) has exactly one zero inside T. Because R can be arbitrarily large, F(z)
has exactly one zero in the right half-plane. Hence /(z) takes value A exactly
once in the right half plane.

421
Take 2 = x > 0. We have
F(x) = X-x-e~x,
which is a real-valued function of real variable x. Since F{x) is continuous and
F(0) > 0,
lim F(x) = —00,
X—* + oo
there must exist xoj 0 < %o < +00, such that F(xq) = 0. In other words, the
point 2 in the right half-plane such that /(2) = A is necessarily real.
5412
Suppose / is analytic in a region which contains the closed unit disc {2 :
\z\ < 1}. Assume / is non-zero on the unit circle {2 : \z\ — 1}. Let C denote
the unit circle traversed in the counterclockwise sense. Suppose that
17\ dz = 2'
(1) hjc
<2> ssX'^H0-
and
(3) hi
z"\;i-dz = -.
c m 2
Find the location of the zeros of / in the open unit disc {2 : \z\ < 1}.
(Iowa)
Solution.
Assume z\, 22, •■■,zn are the zeros of/(2) in {2 : \z\ < 1}, multiple zeros
being repeated. Then
n
/(*) = «/(*) ![(*-*;)>
where 3(2) is analytic and has no zero in {2 : \z\ < 1}. We have
2^,/c /(2) 2ir?yc

422
It follows from (1) that n = 2, i.e., /(z) has two zeros in the unit disk. Then
for f(z) = (z - zi)(z - z2)g(z),
±.[£&* = ' /(_!_+_!_+a£iu
2ir« Jc /(z) 2ir? Jc \z - zx z - z2 ff(z) /
= zi + z2 = 0
and
1 /" 2/'(2)J 1 [ ( z2 z2 z2g'(z)\ ,
ziHdz = i^ \ — + — + ~tV)dz
Jc f{z) 2*1 Jc \z- zi z-z2 g(z) J
2iri Jc f(z)
r2 A- z2 - -
which show that z1?2 = ±2• Hence z = ±^ are the only zeros of /(z) in the
unit disc.
5413
(a) How many roots does this equation
z4 + z + 5 = 0
have in the first quadrant?
(b) How many of them have argument between ~ and £■?
[Indiana -Purdue)
Solution.
(a) Let R be sufficiently large such that when \z\ = i?,
|z4 + 5| > \z\.
Set
/(z) = z4 + z + 5
and
g(z) = z4 + 5.
Choose a closed curve
r = {z = x + iy;Q < x < R,y = 0} U {z : \z\ = R,0 < argz < ^}
U{z = x + iy : x = 0,0 < y < R}.

423
It is obvious that
\m-g(z)\<\g(z)\
holds when z G T. By Rouche's theorem, the numbers of the zeros of / and
g inside T are equal. Since g has only one zero inside T, f has also one zero
inside T. Noting that R can be arbitrarily large, we know that
z4 + z + 5 = 0
has one root in the first quadrant.
(b) Let R be sufficiently large such that when \z\ — R, *jkA is approximately
zero. Set
/(z) = z4 + z + 5
and
and
rx = {z = x + iy: x = 0,0 < y< R},
T2 = {z = re*1 :0< r < R},
T3 = {z : \z\ = R, - < argz < -}.
It is easy to see that Im/(z) > 0 when
2G(r1ur2)\{z = o},
/(0) = 5, and
f(Ri) G {w : 0 < argio < e}, / {Re1*') G {w : -it — e < argio < ir}
where e > 0 is very small. We also know that
z + 5'
Ar5arg/(z) = Araargz + Araarg I 1 +
where Araarg/(z) denotes the change of arg/(z) when z goes continuously
from Re*' to Ri along T3. It is obvious that
Ar.argz4 = ir,
while
Araarg 1 + —j-

424
is very small. Let r = Ti U r2 U T3 is taken once counterclockwise, it follows
from the above discussion that
Ararg/(z) = 2ir.
By the argument principle, the number of the roots of /(z) = 0 inside T is
equal to
h i mdz=siir los/w = sArais/W = '■
Hence
f(z) = z4 + z + 5 = 0
has exactly one root in the domain
{z : - < argz < -}.
5414
Prove that the equation sin z = z has infinitely many solutions in W.
(Indiana)
Solution.
Let
/(z) = sinz— z
and z — x + iy, then /(z) can be written as
/(2) = ^ ~2i~" ~Z=l ^y~" ~ ^^ ~{X + iy)-
For any fixed natural number n, choose a positive number t ^> log n and a
closed contour r = rx U T2 U T3 U T4 in the counterclockwise sense, where
Ti — {z = x + iy : 2nir < x < 2(n + l)ir,y = 0},
T2 = {z = x + iy : a; = 2(n+l)ir,0 < y <t},
T3 = {z = x+ iy : 2nir < x < 2(n+l)ir,y = t},
and
T4 = {z = x + iy : x = 2mr, 0 < y < t}.

425
Then we consider the image of T under w = f(z):
f(Ti) = {w = u + iv: -2(n + l)ir < u < -2wr, v = 0}
with the direction from the right to the left;
/(r2) = {w - u + iv : u = -2(n + l)ir, 0 < v < - (e* - e~l) -1}
with the direction upwards; /(T3) lies in the annulus
| w : -e' - ( -e-* + t + 2(n + l)x j < \w\ < -e* + l-e'1 + t + 2(n + l)ir J i
starting from
w = -2(n + l)x + i I ~el - -~e~l - t
1,* X
2e-2'
and ending at
w - -2nir + il -e* - -e~l - t J
in the counterclockwise sense;
/(r4) = {w = u + iv : u = -2mr, 0 < v < - (e* - e-') - i}
with the direction downwards.
Hence the winding number of f(T) around w — 0 is 1. By the argument
principle, /(2) — sin 2 — 2 has one zero inside the contour T. Since n is
arbitrarily chosen, we conclude that sin 2 = 2 has infinitely many solutions in
<F.
Remark. This problem can also be proved by Hadamard's theorem.
Assume that
/(2) = sin 2 — 2
has only finite zeros in W, and denote all the zeros by 21; z2, • • •, zn, multiple
zeros being repeated. By Hadamard's theorem, /(2) can be written as
/(2) = e««p(z),
where
n
k = l
and g(z) is a polynomial.

426
It is obvious that /(z) is an entire function of order A = 1, where
loglog<max|/(z)|l
x= BE te L
r-» oo log r
which implies that g(z) must be a polynomial of degree 1. Hence we have
sinz-z = ea*+6p(z).
Let z = x+iy and x be fixed. By letting y —* +oo and y —> —oo respectively,
and comparing the increasing order on both sides, we obtain that Ima < 0 in
the former case and that Ima > 0 in the latter case. This contradiction implies
that sin z = z has infinite many solutions in (F.
5415
(a) Let / be a non-constant analytic function in the annulus {1 < \z\ < 2}
and suppose that |/| = 5 on the boundary. Show that / has at least two zeros.
(b) If / is meromorphic in the annulus, is the statement in part (a) still
true?
(Stanford)
Solution.
(a) Let D = {z : 1 < \z\ < 2} and 3D = rx U T2, where rx = {z : \z\ = 2}
is in the counterclockwise sense, and 1^ = {z : \z\ = 1} is in the clockwise
sense. Because / is non-constant analytic in D and |/| = 5 when z 6 dD, we
know that both /(1^) and /(r2) must be {w : \w\ = 5} in the counterclockwise
sense. Hence ^:Ar1arg/(z) > 1 and ^rAr:!arg/(z) > 1. In other words,
—Aaflarg/(z) > 2,
which shows by the argument principle that / has at least two zeros in D.
(b) If / is meromorphic in D, the statement in (a) is not true. It might
occur that /(Ti) and /(T2) are two subarcs of {w : |u>| = 5}, or both /(Ti)
and /(T2) are {w : \w\ — 5} in the clockwise sense. In the latter case, / has no
zero in D. The following is a counterexample. Let ^(¢) be a conformal map of

427
onto {w : \w\ > 5} with the normalization g(0) = oo, <?'(0) > 0. Then
is a non-constant meromorphic function in D with |/| = 5 when z S dD. But
/ has no zero in D.
5416
Let n be a positive integer, and let P be a polynomial of exact degree 2n:
P(z) = a0 + aiz + a2z2 + \- a2„z2n,
where each ay £ <F, and ct2n 7^ 0. Suppose that there is no real number x such
that P(x) — 0, and suppose that
lim f ^\dx = 0.
Prove that P has exactly n roots (counted with multiplicity) in the open upper
half plane {z EW : Imz > 0}.
(Indiana)
Solution.
Let r > 0 be sufficiently large such that when \z\ — r,
\a2nz2n\ > \a0 + aiz + ■ ■ ■ + a2„-iz2n-1\.
Take a closed contour r = Ti U T2 in the counterclockwise sense, where
Ti = {z = reie : 0 < 9 < ir}
and
T2 = {z = a; + iy : — r < x < r, y = 0}.
Then the number of zeros of P(z) inside T is equal to
2iri Jr P(z) 2tti JTl P(z) 2vi JT2 P(z)
It is already known that
lim f EMdz = lim f ^-dx = 0.

428
We also have
EAlLd7. = — f d\n<*P(r\ =
2ir
h I W)dz = h idlogP(2) - i^gpw
i i /
= ^Ariarg (a2„z2n) + ^Ariarg f 1 +
ao + aizH ha2n-i22" n
a2nz2n
Note that
—AFlarg (a2„z2n) =n
and
1 a A . ao + aizH ha2n-i22"-1
r—*oo
lirn _Ariarg ^ + -" 'a2nz2n^ ) = o,
we obtain that P has exactly n roots (counted with multiplicity) in the open
upper half plane.
5417
Consider the function
/(z =l + - + --j+ •••+- —.
z 2! zJ n! z"
(a) What does the integral
±f
W*
count?
(b) What is the value of the integral for large n and fixed r?
(c) What does this tell you about the zeros of /(z) for large n?
(Courant Inst.)
Solution.
(a) Let
.1, . . 1 >2 1 .., 1
no =/(^) = ^¢ + 21^ + 3^+ ••• + ;/"•
From
i_/ ZMd2 = -1-( f'^] dc
l*iJ\*\=r f(Z) 2™J\(\=i f(l
"2^7K|=i F(C)
2« ./w=r /(z) 27ri7K|=i /(i) -C

429
we know that the negative of
J_ f L
^ij\z\=T /I
w*
\z\ = r /00
represents the number of zeros of F(Q in {|C| < £}, which is just the number
of zeros of /(2) in {\z\ > r}.
(b) When n —> 00, ■F'(C) converges to e^ uniformly in any compact subset
offf. Let
min \e^\ = m,
\n\=y
then m > 0.
When n is sufficiently large,
\F(0 -eC\ <m< \ec\
for |C| = £, which implies the numbers of zeros for F(Q and e^ in {|C| < ^}
are equal. Since e^ has no zero in (F, we obtain
2xi JM=r f(z)
{z)dz = 0
| = r /(*)
for fixed r and large n.
(c) From the above discussion, we conclude that for any fixed r > 0, when
n is sufficiently large, there is no zero of /(2) in {\z\ > r}. In other words, all
the n zeros of /(2) are in {\z\ < r}.
5418
(a) Suppose that /(2) is analytic in the closed disk \z\ < R, and that there
is a unique, simple solution z\ of the equation /(2) = w in {\z\ < R}. Show
that this solution is given by the formula
2™ ./M=.R
l\z\=R /(*) ~ W
(b) Show that, if the integer n is sufficiently large, the equation
— (f)"
has exactly one solution with \z\ < 2.

430
(c) If z\ is the solution in (b), show that
lim(z1-l)» = -.
Solution.
(a) Let
f(z) -w = (z- Zx)Q(z),
where Q(z) is analytic and has no zero in {\z\ < R}. Then
f(z) - w
Hence
(Courant Inst.)
= [log(/(z) - w)}' = [log(z - zx) + log Q(z)}' =
i +Q'(z)
z - zx Q(z) '
-Lf
2™J\z\ = R
zf'(z)
m-
dz
-Lf
-Lf
2™ J\z\=
\z\=R Z~ Zl
Z
(b) Let
>\z\=R Z - ZX
fn(z) = z- l-(f)", </(*) = *-l,
1 /" zQ'fVK
-^2 + ^-: / !;,\ dz
2« y|2|=iJ <5(z)
dz = zi.
and
re = {|z| = 2-e}.
For fixed large n, we choose e > 0 sufficiently small such that when z £ re,
\fn(z)-9(z)\
= (l--)n<l-e<|ff(z)|.
Hence /„(z) and g(z) have the same number of zeros in {|z| < 2 — e}, and the
number is 1. Since £ can be arbitrarily small, the equation z = 1 + (|) has
exactly one solution (denoted by z[n') in {\z\ < 2}.
(c) f„(x) is a continuous real-valued function for 1 < x < §. When n is
sufficiently large, we have /„(1) < 0 and /„ (|) > 0.
Hence we have z<n) G (1, §). It follows from
*<»> - 1
,(»)
<5
that
lim z<n) = 1,

431
which implies
lim (z[n) - l) i = lim V
Let
5419
O = JD(0,l)\{i,-|>.
Find all analytic functions / : fi —+ fi with the following property: if 7 is
any cycle in fi which is not homologous to zero (mod fi), then / * 7 is not
homologous to zero (mod fi).
(Indiana)
Solution.
Since / is analytic in fi and bounded by \f(z)\ < 1, the points 2 — ±|
must be the removable singularities of /. Let
7i = {l2- 2l = £}' 72 = {k+ 2I = e>'
where e > 0 is small, and the directions of jt and 72 are both in the
counterclockwise sense. Since 71, 72 are not homologous to zero (mod fi), / + 71 and
/ *72 are also not homologous to zero (mod fi). As e tends to zero, /(71) and
/(72) will tend to either w = \ or w — — i, because otherwise, / * ^x or / * 72
will be homologous to zero (mod fi). Hence we obtain
«4»=4
Now we claim that the case that /(|) — /(— -|) will not happen. If, for
example,
f(\) = f(-\) = \,
we assume that z = \ is a zero of /(2) — ^ of order n and 2 = — | is a zero of
/(2) — I of order m, then
/ * (77171 - 7172)
is homologous to zero (mod fi), while mji — nj2 is not homologous to zero
(mod fi), which is a contradiction. Thus we obtain either

432
or
/(2) = _2' /(_2)=2-
In the case of
/(3) =2' ^~2^ = ~2'
we consider the function
i/(z) ■ 1 - iz
^)=1-^-^-^-1
which is analytic in £)(0,1) and satisfies |.F(z)| < 1. It follows from F(— |) = 1
that F(z) = 1, which implies that /(z) = z.
In the case of
,1, 1 ., 1, 1
/(
we consider the function
/(V_ 2' K 2* ~ 2'
1//-^ "i 1.
G(z) - ' v_/ ■ 2 • Z 2
l+i/(z) 1-iz
which is also analytic in £)(0,1) and satisfies |G(z)| < 1. It follows from
G(—|) = —1 that G(z) = —1, which implies that /(z) = —z. Thus we
conclude that the functions which satisfy the requirements of the problem are
/(z) = z and /(z) = -z.

433
SECTION 5
SERIES AND NORMAL FAMILIES
5501
Let
CO
n = 0
have a radius of convergence r and let the function /(z) to which it converges
have exactly one singular point zq, on \z\ — r, which is a simple pole. Prove
that
lim an/an+i = z0.
n—*oo
(Indiana)
Solution.
Assume that the residue of /(z) at Zq is A, and define
A
F(z) = f(z)
z- z0
Then F(z) is analytic on {z : \z\ < r}. In other words, the Taylor expansion
of F(z) at z = 0 has a radius of convergence larger than r. Hence the power
series
00 A
F(z) = Y>nz"
ti> 2"2°
n=0 n=0 0
= Ef^ + ^r)2"
n=0 V ^0/
is convergent at z = Zq, which implies
lim ( an + —rj-) zj = 0.
It follows that
lim a„Zg = ^ 0
n-»oo Zq

434
and
and we obtain
lim an+iZo+1 = ^0,
n-»oo Zq
lim
n-oo an + 1
20-
5502
(1) Show that the series
n>l
is convergent for 1 ^ a G<F with \a\ = 1.
(2) Show that this series converges to log(l — a) for such a.
Solution.
(1) Let a = e", t G (0,2ir), then
Ea" ^--v cos ni + i sin ni
n>l
For t G (0,2ir) we have
n
^ cos fci
and
(Minnesota)
Jt=i
n>l
I sin | - sin ^±i< I
I 2^h4 |
<
sin;
2_] sin fci
Jfc=l
t 2n+1t
2 sin |
<
sin ■
Because - tends to zero monotonically, by Dirichlet's criterion we know that
both J2 ^^r1 and £ 5»L«i converge, which shows that - £ *£■ is convergent
n>l " n>l n>l
for l^a G<F with |a| = 1.
(2) Let
/w = -Et (N<1)-
n>l
Differentiating term by term, we have
n>l

435
Integrating both sides on the above identity, we obtain /(z) = log(l — z), for
\z\ < 1.
Let a = e'\ z = re** where 0 < r < 1, 0 < 2 < 2ir. It follows from Abel's
limit theorem that
eint ^ (reH)n
_y?L = -ye—= lim-y
*-^ n t—' n r-»i- ^-^
. .. _ n
n>l n>l n>l
lim log(l - re'*) = log(l - e") = log(l - a).
Consider a power series
5503
oo 1
y -zn!.
I—* n.
1 n
T=l
Show that the series converges to a holomorphic function on the open unit
disk centered at origin. Prove that the boundary of the disk is the natural
boundary of the function.
( Columbia)
Solution.
First of all, we prove the following proposition: If the radius of convergence
of
/(z) = ][>nz"
n=0
is equal to 1 and a„ > 0 for all n, then z = 1 is a singular point of /(z).
Assume the proposition is false, i.e., z = 1 is a regular point of /, then for
fixed x G (0,1) there exists a small real number 8 > 0 such that the power
series expansion of / at point x is convergent at z = 1 + 5. Suppose the series
is
oo
Y,h(z-x)k,
k=0
where
6t = Z^)=^n(„_i)...(„_fc+i)ana!»-fc.
n=k

436
Thus
E°° i t \k s^s^. n(n-l)---(n-k+l) . ,
k\
k=0 k=On=k
is convergent at z = 1 + 6. Noting that when z = 1 + 6 the right side in the
above identity is a convergent double series with positive terms, and hence the
order of summation can be changed, we assert that when z = 1 + 5,
oo oo
El t \k V* V n(n- 1)- •• (n- k +1) k ,
bk(z~x)k = 2^2^^ '—^ >-an{z-xfxn
k\
k=0 k=On=k
oo n
E ^--^ n(n-1)-•• (n-k+1) k ,
a"E^ V L{z-x)kxn
n=0 k=0
- ~Y^anzn,
n=0
which contradicts the statement that the radius of convergence of
oo
n=0
is equal to 1.
Now we return to the power series
oo .
F(z) = £-z">.
n=l
It follows from
F" -A" 1
lim \ — = 1
n—*oo v ^
that the radius of convergence of
oo 1
V-z"!
^ n
n=l
is equal to 1. By the above proposition, z = 1 is a singular point of F(z). For
any natural numbers p and q,
p-l . oo .
n-") = £^T! + £^"!-
n—l n=p

437
Since z = 1 is a singular point of
n=p
it is also a singular point of F(ze r> K'). In other words, z = e~ r> " is a singular
point of ^(z). Since the set {e r> *' : p,q = 1,2,-• •} is dense on {|z| = 1}, we
conclude that the unit circle {\z\ = 1} is the natural boundary of F(z).
Remark. By the above discussion, the boundary of the unit disk is also the
oo
natural boundary of the function £) ^z"! although the series is absolutely
n = l
and uniformly convergent on the closure of the unit disk.
5504
Suppose / is analytic in U = {\z\ < 1} with /(0) = 0 and |/(z)| < 1 for all
z £U. If the sequence {/„} is defined by composition
/»(*) = /(/(■■■/(*))■■•)
and
for all z £U, prove that either g(z) = 0 or g(z) = z.
(Indiana-Purdue)
Solution.
By Schwarz's lemma, it follows from /(0) = 0 and |/(z)| < 1 that |/(z)| <
\z\ for all z G U, and if |/(z)| = \z\ for some z ^ 0, then /(z) = e'°z where a
is a real number.
In the case when /(z) = eiaz. /„(z) = einaz. Since /„(z) is convergent,
we obtain a — 0, which implies that /(z) = z and #(z) = z.
In other cases, we have
< 1
for all zeU. Let 0 < r < 1. Then
max
l*l<r
/(*)
A< 1.

438
For all z G {|z| < r}, we have
1/(2)1 < A|z|,
I/2WI = l/(/(*))| < A|/(z)| < A2|z|,
l/»(*)| = |/(/„-i(z))|<A|/n_1(z)|<A"H,
Hence /„(z) converges to zero uniformly in-{|z| < r}. Since 0 < r < 1 is
arbitrarily chosen, we obtain g(z) = 0 for all z £ U.
5505
Let {fn}T=i be a sequence of analytic functions in a domain D which
converges uniformaly on compact subsets of D to a function / on D.
(a) Prove that if /„(z) 7^ 0 for all n > 1 and z £ D, then either / is
identically zero in D or /(z) ^ 0 for all z £ D.
(b) If each /„ is one-to-one on D, show that / is either constant or one-to-
one on D.
(UC, Irvine)
Solution.
(a) First of all, we know from Weierstrass' theorem that / is analytic on
D. Suppose / is not identically zero, but has a zero point zq S D. Since the
zeros of a non-zero analytic function are isolated, there exists r > 0, such that
/(z) ^ 0 when
z G {z : 0 < \z - z0| < r} C D.
Let m be the minimum value of |/(z)| on
{z : \z - z0\ - r}.
Then m > 0. As {/„} converges to /(z) uniformly on compact subsets of D,
we know that for sufficiently large n,
|/n(z)-/(z)|<m<|/(z)|
holds on {z : |z — Zo| = r}. It follows from Rouche's theorem that /„ and
/ have the same number of zeros in{z:|z — zq\ < r}. Since z0 is a zero of
/, f„ must have a zero in {z : |z — Zo| < r}. which is a contradiction to the
assumption that /„(z) ^. 0 for all z G X>.

439
(b) Suppose / is not a constant, and is not one-to-one on D. Then there
exist 2i,22 G D (21 ^ 22), such that /(21) = /(22) (denote it by a). Choose
r > 0 sufficiently small, such that
{2 : |2 - 2i| <r}C\ {z:\z- z2\ < r} = 0,
{2 : \z - 2i| < r} U {2 : |2 - 22| < r} C D,
and /(2) — a 7^ 0 in {2 : 0 < |2 — 2i| < r} U {2 : 0 < |2 — 22| < r}. Let m be
the minimum value of \f(z) — a\ on {2 : |2 — 2i| = r ov \z — z2| = ?"}• Then
m > 0. With the same reason as in (a), when n is sufficiently large,
|(/n(2) - a) - (/(2) - a)| = \fn(z) - /(2)| < m < \f(z) - a\
holds on {2 : |2 — 2i| = r or |2 — 22| = r}. It follows from Rouche's theorem
that /„ (2) — a and /(2) — a have the same number of zeros in {2 : \z — 2X | < r}
and {2 : \z — z2\ < r} respectively. In other words, there exists
zi S {2 : |2 — 2X| < r}
and
22 G {2 : |2 - 22| < r},
such that /„(2'i) - a = 0 and /n(22) - a = 0, which implies /n(2'i) = /n(22)
(zi 7^ z2)- This is a contradiction to the assumption that /„ is one-to-one on
D.
5506
Let flcffbea bounded domain, and let {/„} be a sequence of analytic
automorphisms of D such that
lim f„(a) -bedD
n—*oo
for some point a G D. Prove that
lim fn(z) = b
n—»00
for every 2 £D.
(Indiana)

440
Solution.
Take ao G D, ao ■£ a. If {/„(ao)} does not converge to b, there exists a
subsequence of {fn(ao)} converging to 60 i=- b. Without loss of generality, we
assume
lim fn(a) = b£dD,
n-»oo
lim fn(a0) = b0^b.
n—»00
Since {fn(z)} is a normal family, there is a subsequence {fnk(z)} converging
uniformly on compact subsets of D to f(z). Because /(a) ^ /(ao), f(z) is a
non-constant analytic function of D.
Let r be sufficiently small such that f(z) — b has no zero in {2 : 0 < \z — a\ <
r} C D, then m = min{|/(z) — b\ : \z - a\ — r} > 0. Since {f„k} converges
uniformly to / on {2 : \z — a\ = r}, when k is sufficiently large,
!/»»(*) " /OOI = !(/«»(«) - b) - (f(z) -b)\<m< \f(z) - b\
on {2 : \z — a\ = r}. By Rouche's theorem, fHk(z) — b has zero(s) in {2 :
\z — a\ < r}, which is a contradiction to the fact that /„fc does not assume the
value b £ dD in £) because /„fc is an automorphism of D.
5507
Which of the following families are normal, and which is compact? Justify
your answers.
(a) T = {/ : / is analytic in £>, /(0) = 0, diam /(£>) < 2}
(b) Q = {g : g is analytic in D,ff(0) = l,Re{s} > 0, diam ff(D) > 1}.
Here the diameter of a set S is diam S = sup{|2 — C| : 2, C G S}.
(Indiana)
Solution.
(a) For any / G T, it follows from /(0) = 0 and diam /(D) < 2 that
|/(2)| < 2, which shows that T is normal.
Let {/n} be a sequence of functions in T. Then there exists a subsequence
{fnk} converging uniformly in compact subsets of D to /(2), which obviously
satisfies the conditions that /(2) is analytic in D and /(0) = 0. For any two
fixed points 2,(6D,we have
!/»»(*)-/»»(01 < 2

441
because diam fnk(D) < 2. We choose a compact subset K C D such that
z,C £ K. It follows from the uniform convergence of {fnk} on K that
|/(z)-/(C)|<2.
Since z, C S D can be arbitrarily chosen, we obtain diam f(D) < 2, hence
/(z) £ T, which shows that T is also compact.
(b) Let {gn} be any sequence of functions in Q. Then for G„(z) = e~gn^z\
we have |G„(z)| < 1. Hence there exists a subsequence {G„fc} converging
uniformly in compact subsets of D to a function G(z) which is either a constant
or a non-constant analytic function in D. If G(z) is a constant, then the
constant is e_1 because
G(0) = lim e-s"(°) = e-1;
n—>co
if G{z) is non-constant analytic, since Gn(z) ^ 0 for all z S D, by Hurwitz's
theorem, we have G(z) 7^ 0 for all z £ D. Hence we can define an analytic
function g(z) = — log G(z), where the single-valued branch is chosen by g(0) =
— logG(O) = 1, and we conclude that
gn„(2) = - log Gn„(z)
converges uniformly in compact subsets of D to g(z), which shows that family
Q is normal. But family Q is not compact. First we can choose a sequence of
functions gn(z) in Q as follows: gn(z) is a conformal mapping of D onto
1 1
fi„ = {w : \w - 1| < -} U {w : \w - 3| < 1} U {w : \Imw\ < -, 1 < Reio < 3}
4 n
satisfying <?n(0) = 1, g'n(0) > 0. By the Riemann mapping theorem, such a
mapping gn exists and is unique, and it is obvious that gn satisfies all the
conditions required by the family Q. Because the domain sequence {fi„ } converges
to fi = {w : \w — 1| < ^} which is called the kernel of {fi„} with respect to
w = 1, by Caratheodory's theorem, {gn(z)} converges uniformly in compact
subsets of D to g(z) which is a conformal mapping of D onto fi. Since diam
g(D) = |, g(z) does not belong to the family Q, which shows that Q is not
compact.
5508
Suppose that 1 < p < 00 and c > 0 is a real number. Let T be the set of
all analytic functions / on {\z\ < 1} such that

442
sup f \f(rei0)\Pde<c.
0<r<l7o
Show that T is a normal family.
(Illinois)
Solution.
It suffices to prove that the functions in T are uniformly bounded on every
compact set of {z\ < 1}. We prove the assertion by contradiction. If it is
not the case, then there exist zn £ £), /„ £ T such that zn —> zq £ D and
/n(Zn) ~* OO-
Let 1 — |z0| = 3r. Then when n is sufficiently large, \zn — z0\ < r. By
Cauchy integral formula,
^) = ^/ T~^ (2r<p<3r).
Hence
\fn(Zn)\ < £- [K\f„(zo + peie)\d6
2t Jo
= -^-l(r\fn(zo + peie)\pM
(2ir)p \Jo
where
Then
P 9
i/n(zn)|p r pdP < r r \uz0+pj*y?p&pM
J2r J2r JO
< jf ^K\f(peiS)\"pdpde<C-.
As n —+ oo, the left side of the above inequality tends to infinity, while the
right side of the inequality is a constant. The contradiction implies that T is
a normal family.

443
5509
(a) Let / be holomorphic for \z\ < R and satisfy /(0) = 0, /'(0) ^ 0,
/(z) ^. 0 for 0 < \z\ < r < R. Let C be the circle \z\ = p where p < r. Show
that
9{w) = hjc
define a holomorphic function of w for
tf'(t)dt
c /(*) - «>
|io| < m = min |/(pe'e)|,
and that z = g(w) is the unique solution of
/(z) = 10
that tends to zero with w.
(b) Find the Taylor's expansion of g(w), and apply this to find the explicit
series expansion of the root of the equation
z3 + 3z - w = 0
that tends to zero with w.
(Harvard)
Solution.
(a) It follows from
w\ < m — min \f(pe,0)\
0
that when t S C,
1 1 ^y> w»
/(*) " - /(*) (l - 3¾) „^o
/(*)"
Hence
9
If tf'(t)dt ^ (I f tf'jt) \
which implies that g(w) is holomorphic in {w : \w\ < m}.

444
Let T be the image of C under / where C is the circle {z : \z\ = p} taken
once counterclockwise. Because
\w\ < m = min|/(peie)|,
0
the winding number
n(r,0) = n(T,w),
which shows that /(z) and /(z) — w have the same number of zeros in {z :
\z\ < p). Since z = 0 is the only simple zero of / in {z : \z\ < p}, we know that
/(z) = w has a unique solution in {z : \z\ < p}. Denote the unique solution
by z\, then
f(t) -w=(t- Zl)Q(t)
where Q(t) is analytic and has no zero in {t : \t\ < p}, and
jm- = [iog(/W - „)r. [iog(* - zl}+log Qit)Y = T±-+m.
Hence
ff(U,) ~ 2iri 7c /(<) -w~ 2iriJct-z1 + 2m Jc Q(t)
- Zi,
-dt
which shows that g(w) is just the unique solution of/(z) = w. As the constant
term in the Taylor expansion of g(w) is
± [ *!&*
2iriJc f(t)
which is obviously zero, we assert that the unique solution g(w) tends to zero
together with w.
(b) Let
/(z) = z3 + 3z,
then
1 tf'(t)dt ^
n=l
where
^= 2Wo7^*= 2^1^+37 *'

445
After some computation, we obtain 02¾ = 0 and
«2fc-l
J_/ I
2iri Jc 32k~H2k-1
42r<k-2
t <--2k \ 3
2\ fc-2
+ c.
Jfc-1
-2k
2\ fc-l
dt
33¾
-(3(¾+(7.^-1
2k J
5510
Find an explicit formula for a meromorphic function / whose only
singularities are simple poles at —1, —2, —3, • • • with residue n at z = — n. Prove in
detail that your function has all the required properties.
(Illinois)
Solution.
By Mittag-Leffler's theorem, we construct
/(z) = V (-5--1+^) = v.
For any natural number N, when |z| < N, n> 2N,
i(n + z)
Hence
n(n + z)
00
E
<
2N2
n(n + z)
n = 2N
converges uniformly in {\z\ < N} to a function which is analytic in {\z\ < N}.
In addition,
2JV-1 2
V —?
„tt "(n + z)
is a meromorphic function whose only singularities in {|z| < N} are simple
poles at z = —1, —2, • • •, — N + 1 with residue n at z = —n. So /(z) is analytic
in {\z\ < N}\{-1, -2, • • •, -N + 1}, and z = -1, -2,---, -N + l are its simple
poles with residue n at z = —n.
Because JV can be chosen arbitrarily large, it is obvious that /(z) has all
the required properties of the problem.

446
5511
(a) Does there exist a sequence of polynomials {P„} such that Pn(z) —+ -V
uniformly on the annulus 1 < |z| < 2? If Yes, give an explicit formula for the
Pn; if No, explain why not.
(b) Does there exist an entire function g whose zero-set is {^fn(l + i) : n =
0,1, 2,3, • • •}? If Yes, give an explicit formula for g; if No, explain why not.
(Illinois)
Solution.
(a) No. If there exists a sequence of polynomials {Pn} such that Pn(z) —* ~r
uniformly on {1 < jzj < 2}, then for any e G (0, \), there exists N > 0 such
that when n > N, \Pn(z) - jjr| < e holds for all z G {1 < \z\ < 2}. Multiply
both sides by |z|2, we have
|z2P„(z) - 1| < e|z|2 < 4e < 1 for z G {1 < \z\ < 2}.
Because z2Pn(z) — 1 is an analytic function in {|z| < 2}, it follows from the
maximum modulus principle that
|z2P„(z) - 1| < 1
holds for all z G \\z\ < 2}. The contradiction follows by taking z = 0 in the
inequality.
(b) Yes. The function g can be chosen as
oo , .
9(z) = ,J| l_i li+Ki)1,
where an = ^/n(l + i).
For any R > 0, let \z\ < R and choose N > R2. Then when n> N,
log[(l-— )e*+*<*)2] = log(i_-1)+^. +1(^.)2
an an an z an
_ _Z(JL\3_... ( z \m
3 an m an
It is easy to see that
log (1 )ea„T2U„^
R3
n3/2'

447
^^ 3
Because 52 -575- converges, we know that
n=N
V^ log (l )e^r+2(<.*„)
n=N
is analytic in {z : \z\ < R}, which implies
n=N Qn
£ an ' 2V an >
is analytic in {z : \z\ < i?}. Hence g(z) is analytic in {z : \z\ < i?}, and its
zeros in {z : \z\ < i?} are 0,ai,a2,-•• ,at (\ 1 < fc < ^-). Since R can
be arbitrarily large, we see that g(z) is an entire function with the required
zero-set.
5512
State whether the following statement is True or False, and prove your
assertion.
For each positive integer n there exists an entire function /„ such that
max |Re/n(z) — log |zII < —.
i<|*|<2 n
(Indiana)
Solution.
False.
We prove the assertion by contradiction.
If for each positive number n there exists an entire function /„ such that
^iRe/^-loglzH^,
then for 1 < \z\ < 2, we have
-l<Re/„(z)< l + log2.
Define Fn(z) = etn^. Then Fn(z) are entire functions with no zeros, and
- < \Fn(z)\ = eRe'»(*) < 2e

448
for 1 < \z\ < 2. By the maximum modulus principle, |.F„(z)| < 2e for \z\ < 2.
Hence {Fn(z)} is a normal family in {z : \z\ < 2}, and there exists a
subsequence {F„k(z)} converging locally uniformly to an analytic function F(z) in
{z : |z| < 2}. Since |.Fn(z)| > \ for 1 < \z\ < 2, F(z) cannot be identically
zero, and by Hurwitz's theorem F(z) has no zero in {z : \z\ < 2}. But we have
for 1 < |z| < 2,
\F(z)\ = lim |F„fc(z)| = lim eRe'"*W = elosW = Izl,
which implies that F(z) = az with \a\ = 1 in {z : \z\ < 2}. This is a
contradiction to the fact that F(z) has no zero in {z : \z\ < 2}.
5513
Let G = D\(-1,0], where D = {z : \z\ < 1}.
(a) Give a single-valued definition for z' in G.
(b) Why should there exist a sequence of polynomials Pn such that
lim Pn(z) = zi
n—»oo
for all z in G?
(c) Can the polynomials be chosen so that there exists a constant M with
l-fn(z)| < M for all z S G and all n? Justify your answer.
(Indiana)
Solution.
(a) z' is defined by e'log'. In domain G, single-valued branch of log z can
be chosen. For example, a single-valued branch of z' in G can be defined by
arg2|o<;<i = 0.
(b) Choose
111 1
Kn = {z : — < \z\ < 1 , —w H— < argz < ir }
n n n n
where n > 2. Then Kn C Kn+i, and
lim Kn = G.
n—»oo
Because the complement of Kn is connected and contains z = oo, we know by
Runge's theorem that there exists a sequence of polynomials which converges

449
uniformly on Kn to z'. In other words, we can find a polynomial Pn(z) such
that
|P„(z)-z'|< i
n
for all z G Kn. Hence {Pn(z);n = 1,2,---} converges to z! uniformly on
compact subsets of G.
(c) No. If there exists M with |P„(z)| < M for all z G G and all n,
then because P„(z) are continuous on D, \Pn(z)\ < M for all z G D and
all n. It follows that {Pn(z)} is a normal family in D, and there exists a
subsequence Pnit (z) which converges uniformly on compact subsets of D to an
analytic function /(z) in D. Since P„(z) converges to z' in G, hence z! = /(z)
for z £ G, which implies that z' can be extended to a single-valued analytic
function in D. It is obvious impossible, so the contradiction is obtained.
5514
(a)
(b)
Prove that
7T2
sin 7rz
Use this to show that
7T COt 7TZ =
oo
^ (z
n=—oo v
1 °°
Z -^ 2
n=l
1
-n)2'
2z
:2 — n2
Justify your steps.
(c) Develop Trcot-Trz in a Laurent series about the origin directly and by
oo oo
use of (b), with enough terms to find the values of ^ ^r an<l X) ^-
n=l n=l
{Harvard)
Solution.
(a) Let
/OO
sin2 7tz
The singular part of / at z = n (n = 0, ±1, ±2, • • •) is (zlny Now we consider
the series

450
For any natural number N and \z\ < N,
1
(z - nf
^
holds for n > 2N and n < —2N. It follows from the convergence of JZ -%
n=2N "
n = -2N oo
and J2 ^- that ^ (z-n)* 1S analytic in
— oo n — — oo
{|«| < JV}\{« = 0, ±1, • • •, ±(JV - 1)}.
Because N can be arbitrarily large, we obtain the result that
oo
V -
^ (z-n)2
n= — oo v '
is a meromorphic function which has the same singularities as /(z).
Let
1
,(z) = /(z)- J] "
(z - n)2'
Then g(z) is an entire function.
As /(z) and
oo 1
n= —oo v '
are both periodic functions with period equal to 1, we restrict z in the strip
{z : 0 < Rez < 1}.
It is obvious that
As the convergence of
is uniform for
lim /(z) = 0.
Iroj->±oo
oo ^
2-» 77^
(z - n)2
|Imz| > 1,
the limit of the series for Imy —► ±oo can be obtained by taking the limit in
each term and the limit is also zero. Hence g(z) is a bounded entire function,

451
which implies that g(z) is a constant. It is obvious that the constant must be
zero. Thus we obtain the identity
sin 7TZ
00 1
= y —-
(z - n)2
(1)
(b) Let
F(z) — irctgwz.
The singular part of F at z = n (n — 0, ±1, ±2, • • •) is -^-. Now we consider
the series
z ■<-—'Vz —n z + nj z ■<-—' z2 — n2
n—l n—1
With similar discussion to that in (a), we know that
1 ^ 2z
z *—' z-* — n2
n=l
is a meromorphic function which has the same singularities as F(z).
Let
1 2z
*"(*)= 7 + E ^r=j + GW-
z *—' z" — nz
n=l
Then G(z) is an entire function. Differentiating both sides of the above identity,
we obtain
Sin 7TZ z
1 °°
n=l
1 1
+
(z-n)2 (z + n)2
- G'(z)
Comparing this identity with (1), we have G'(z) = 0 which implies that G = c
(c is a constant). For
1 °°
*(*)= 7 + E
2z
z '—' z2 — n2
n = l
+ C,
it follows from the fact that F(z) and
1 f, 2z
Z' — W
are both odd functions that c = 0. Hence we obtain
2z
-K COt 7TZ
(2)
n=l

452
(c) The Laurent expansion of 7r cot irz about the origin is
1 *2 t4 3
7T COt 7T2 = 2 2 — ■ • ■ . (3)
z 3 45 y '
It follows from (2) and (3) that around the origin,
00 1 2 4
El _ ^ "" 2
22-n2 """IT- 90Z ' (4)
n — l
Take 2 = 0, we obtain
00 1 2
El _ *"
n=l
After differentiating (4) on both sides, we can also obtain
00 4
El _ it
,^4 ~ 90'
5515
Let 21,-- -,2n be distinct complex numbers. Let f and g be polynomials,
/ of degree < n — 2 and
fif(2) = (2- 2l)---(2 - Zn).
(a) Show that
(b) Show that there exists a polynomial of degree < n — 2 with f(zj) — a,j
if and only if
U 9'{Zj)
(c) Given a sequence of complex numbers 21,22,--- such that \zn\ —► 00,
does there exist an entire function f with f(zj) — ap. Can you write this
function down?
(Harvard)

Solution.
(a) Take R sufficiently large such that
2i,z2,---,z„ G {\z\ < R}.
Because
fif(z) = (z - zi)(z - z2) • • • (z - zn)
is of degree n, while /(z) is of degree < n — 2,
/ Mdz=lim/ Mdz = 0.
Since
we obtain
(b) If /(z) is a polynomial of degree < n — 2 with /(¾) = a/
1,2,---, n), then by (a), we have
y^ Jvzi) y^ aJ _ n
If a\, a,2, ■ ■ •, an are n complex numbers such that
n
a,-
we construct the function /(z) by
'« = £;£
9(z)
~{9'(zj) (z~zi)
For each j,
■2-Q- = (z - zx) • • • (z - z>_0(z - zy+i) • • • (z - zn)
Z Zj
is a polynomial of degree n — 1, and the coefficient of zn_1 is 1. Since

454
the coefficient of zn_1 of /(z) is zero. In other words, /(z) is a polynomial of
degree < n — 2.
Because
lim —-
»(*)
= ay,
while for k -£ j,
ajfc g(z)
0,
g'(zk) z- zjt
/(z) satisfies the condition f(zj) — a.j (j = 1, 2, • • •, n).
(c) For the given sequence z\, Z2, • • •, such that \zn | —► oo, by the Weierstrass
theorem about the canonical product of entire functions, we can construct an
entire function g(z) with simple zeros zi, Z2, • • •. Then we define
OO OO / .
/(z) = £<*) = £^-^/^---^,
n=l n = l Z Zn g (Zn)
where yn is chosen such that when \z\ < ^jp,
\un(z)
2-2n ff'(Zn)
1
Because |zn| —► oo, for any R > 0, there exists JV > 0 such that \zn\ > 2i?
when n> N. Hence
K(z)\ < -2
OO
holds for all \z\ < R when n > N. In other words, £) Wn(^) converges uni-
n=l
formly for all \z\ < R, so that /(z) is analytic in {|z| < R}. Since i? can be
arbitrarily large, /(z) is an entire function.
It is easy to see that
lim un(z) = lim eT.(*-«.) jl£L . _^L_ =
z^zn z^zn z-zn g'(zn)
while for k -£ n,
Uk(zn) = 0,
which implies that /(z) is an entire function satisfying the required condition.

Part VI
Partial Differential Equations

457
SECTION 1
GENERAL THEORY
6101
a) Let A = (a»j)", = 1 be a real matrix. Show that
I (x, Ax)dx = —.—2__ trace A.
J\x\<i n(n + 2)
'|x|<l
Here (•, •) is the dot product of vectors in Mn and un is the area of the unit
sphere in Mn.
b) Show that u G Cl(Mn) implies
/ (Aufdx - V] / |D,ju|2<£c-
Solution.
a) It is not difficult to verify that
and
Therefore,
/ dijXjXjdx = 0, Vi ■£ j
J\*\<i
I x\dx = • • • = I x2n dx.
J\x\<l J\x\<l
I (x,Ax)dx = I y^^agx^dx
J\x\<l ^|x|<l>._1
/ I*
J\x\<i
— trace A / |x|2da;
n J\x\<
— trace A.
n(n + 2)
(Iowa)

458
sjjudx
dx.
b) First let u G Co°(2Rn). By applying Green's formula, we get immediately
I (Aufdx - YL f Duu-Dj
JlRn ij=.1JlRn
n .
= E / \D*rt
As C^(Mn) is dense in H§(Mn), the conclusion is true for u G C$(]Rn).
6102
Let T be a distribution on M and suppose that T' = 0 on JR. Show that
T = const; i.e., show that there is a number a such that
T{<j>) = f a<f>dx for all <f> G Cg°(2R).
(Indiana)
Solution.
T' = 0 if and only if
T(^') = 0, V^G(70M(1R).
It is easy to verify that a function <f> in C£° (2R) is the derivative of a function
in C^(JR) if and only if
/ <£da; = 0.
J m.
im.
Take a function p G C^(M) such that
/ /sda; = 1.
J m.
Then it is easy to verify that for any <j> G C£° (M)
tj> = (f> — I <f>dxp
Jm.
satisfies the condition
J ipdx - 0.

459
Hence T(ij>) = 0 and
TO) = J a<f>dx, \f<t> G C0°°(iR)
Jm.
where a = T(p).
6103
Let u G H"(Mn) with s > n/2. Show that lim u(a;) = 0.
|x|—»00
(Cincinnati)
Solution.
We show first that u £ Ll(]Rn) if u G H3(Mn) with s > n/2. In fact,
u G H*(lRn) if and only if (1 + |£|2)'/2G G I2(2Rn). And it is clear that
(1 + \£\2)-s/2 G L2(Mn) if s > n/2. Therefore, we have u G L1(Mn).
Then we get immediately
as \x\ —* oo.
6104
Let u be defined for f G 2)((0,1)) by
<«,« = E/»>(i).
Determine whether w is a distribution on (0,1), and support your answer.
(Indiana)
Solution.
It is clear that u, defined above, is a linear functional in 2)((0,1)). Let {/¾}
be a sequence in 2)((0,1)) such that
fk -> 0 in 2)((0,1)), as k -► oo.
Then we have a compact interval [a, b] C (0,1) such that
SUPP/* C [a,b], Vfc

460
and
fk ~* ® uniformly on [a, b] as k —+ oo,
for any nonnegative integer n.
Let N be a positive integer such that
1
N<a-
Then we have
<«./») = E ^B) UJ-^ °' as k^°°-
Therefore, it is continuous in 25((0,1)), and is a distribution on (0,1).
6105
Let
B = {(x,y) \x2 + y2 < 1}.
For which p > 1 does the function
u(x,y) =
y/x2 + y2
belong to Wl'P(B)?
Solution.
Let
1
uy(x,y) = -
It is clear that ux,uy £ Ll(B).
We show that
^T¥ (*2 + </2)3/2'
xy
(x2 + y2)3!2'
9 A d
— u = ux and — u = uy
ox ay
in the sense of distributions, that is
I u<f>xdxdy — — I ux<j>dxdy
Jb Jb
(Iowa)

461
and
I u(f>ydxdy = — I Uy<j>dxdy
Jb Jb
for all <f> G Co°(B). By Be we denote the ball centred at the origin with the
radius e < 1. Let fie = B\B£ and Se = dBc. Then we have
I u<j>xdxdy = — / ux<t>dxdy + / u<f> cos(~rt, x)ds
Jn. Jnt Js.
for any <f> G C^(B). Letting e —* 0 in the above inequality, we get J^m = ux.
Similarly, -^-u = Uy.
It is not difficult to verify that
u,ux,uy G 1^(5)
for 1 < p < 2. Therefore, u G W^B) for 1 < p < 2. And we can verify that
uy £ L2(B),
which means that u ¢ W1,P{B) for p = 2.
6106
Let £ be a smooth hypersurface dividing Mn into two disjoint open regions
Q,\ and ^2- Denote by v = v(x) the unit normal to £ pointing into 0,2-
Suppose v : Mn —* Mn solves the equation
divufz) = b(x) for x G Mn - £ (*)
in the classical (C1) sense where b is a continuous function. Assume
furthermore that v lies in both C°(fii) and C°(02) (but not necessarily C°(2Rn)!!)
with
v+(xo) = lim v(x), v~(xo) = lim -u(:c)
x€fl! x€fl2
for each x0 G ^. Derive a necessary and sufficient condition in terms of v+,
v~ and v for v to be a distribution solution of (*) on all of Mn.
(Indiana)
Solution.
v is a distribution solution of (*) in Mn if and only if

462
-/ v-V<j>dx= J b<j>dx, V<£€C£°(.ZRn). (1)
J]Rn J]Rn
By Green's formula, we have
— / v • V<f>dx = — I v • V<j>dx — I v ■ V<t>dx
Jm.n Jnx J(i2
— — I 4>(v+ - v~) • vds + I divv • <f>dx.
's JR
Hence the equation (1) is equivalent to
I 4>(v+ - v~) • vds = 0, V<£ G C^(Mn).
Therefore v is a distribution solution to (*) on all of Mn if and only if
(v+ — v~) • v — 0 on E.
6107
Recall the definition of the curl operator in two dimensions acting on a
vector field U(x,y) = (u(x,y),v(x,y)):
V x U = uy — vx.
(a) Give a definition of what it means for a (not necessarily continuous)
vector field U to have curl zero in the sense of distributions.
(b) Suppose U takes the form
U(x v)=l Uu (I,!/)efil'
u(x,y)^<[ ^ (a!)1/)Gn2)
where fti and fi2 are open sets such that fix U O2 = M2, T = dtli f~l 5¾ is a
smooth curve, and U\ and U2 are (different) constant vectors. If U has curl
zero in 1R2 in the sense of distributions, describe as completely as possible the
curve T.
(Indiana)
Solution.
(a) Let u,v e Iloc(2R2). Then
curl «7 = 0

463
in the sense of distributions is defined as
(-u4>y + v<j>x)dxdy = 0, M<j> G C0°°(iR2).
/(
(b) Let Ui = (ui,vi), U2 — (u2,V2), where ui, «i, «2 and -u2 are constants.
Then curl U = 0 if and only if
/ {-ui<j)y + vi<f>x)dxdy + / (~U2<f>y + v24>x)dxdy = 0, V<£ G C™{M2).
Integrating by parts, we have
/(ui - «2)^da! + («1 - t>2)M/ = 0, V<£ G C^(M2).
Therefore, T is a straight line defined by
(«1 — U2)dx + (vi — v2)dy = 0.
6108
Let p be a polynomial in n variables, and assume that the set {x : p(ix) = 0}
is bounded. Prove that there exists a tempered distribution E on Mn such that
p(D)E — 6 G S. Here S is the Schwartz class of rapidly decreasing functions,
and 6 is the Dirac distribution defined by 6 • <f> — <f>(0).
{Indiana)
Solution.
As the set {£ : p(i£) = 0} is bounded, there exists a positive constant R
such that
tf:K*)=0}Ctf:K|<.R}.
Construct a function XR G C%° (2Rn) such that
X*(0=L V£G{£:|£|<i?}.
Set
^"t (i-x*(0)afe, Kl>*
It is clear that £ G S'(2Rn) D C°°(2Rn), where S'(2Rn) denotes the space of
tempered distributions on Mn, and that
-SmJ 0, |£| <iJ,
**)m ~ { 1 - x*(0, Kl > a

464
It is easy to see that
p(*£)%) - 1 G CST(Mn).
Therefore, we get
p(D)E -6eS,
where E = F'^E) G S'(Mn).
6109
Let P(D) be an mth order linear constant coefficient partial differential
operator on Mn; i.e.,
P(D) = J2 C°D°
\a\<m
where a denotes a multi-index a = (a\, c*2, • • •, an) and the Ca are constants.
Suppose
P(%) = J2 C««)a + ° for alU € Mn - {0}.
\a\<m
If w G 5' satisfies P(D)u = 0, prove that it must be a polynomial. (Here S'
denotes the space of tempered distributions on Mn.)
(Indiana)
Solution.
Let w(£) be the Fourier transform of u. Transforming the equation, we get
p«)S(0 = o.
From the transformed equation, we can see that
suppw = {0}.
Therefore, w(£) must be finite linear combinations of the derivatives of the
Dirac delta function 8((,) at {0}, i.e.,
u($) = J2 ««*(o)(0.
\a\<N
where a = (ai, a2, • • •, an) is a multi-index and aa are constants. Hence
u(x)= Yl ^-1^)¾ E ««(-«)°
|a|<JV ^ ' \a\<N
is a polynomial.

465
6110
(a) Show that
1
lim
t_0+ J\-Kt
-'lit _
= 6(x)
in 2)'.
(b) Show how to define - as a distribution (i.e., define the Cauchy Principle
Value and show that it is in V).
(c) Calculate rigorously the derivative of log |a;| and express the result in
terms of your answer to (b).
(d) Calculate the Fourier transform of the distribution £.
(Indiana)
Solution.
(a) It is well-known that
/oo i
-^e~x^4tdx = 1 Mt > 0.
-oo V47T<
For any <f> S V, we have
/OO I 1 i»00
-==e-'9l*t^{x)dx = — / e-* <f>(2Viy)dy.
-oo V47rf V71" ./-oo
Then it is not difficult to verify that
If00 1
/ -^=6-^/4^)^ _ ^,(0)
7-00 V47Ti
1 I f°° 2
= lim — / e"» (^2^/¾) - W))dy = 0.
t^o+ Vtt \J_0O
Hence
lim
1
,-»/« -
= 6(x).
<->o+ V47r<
(b) We define the distribution £ by its Cauchy Principle Value Vp^ as
follows:
<ypI, *) = lim f f£ ^-dx + r ^Idx) w e v.
The linearity of the functional Vp^ is obvious. Further, suppose that <j>n —► 0
in P. Let supp^n C (—a, a) Vn. Then we have
(Vp 1, <f,n) = lim f ^(»=) M x)dx ^ 0>
1
a;'

466
Therefore, Vp^ is a distribution in V.
(c) For any <f> £V, we have
,00
(log \x\,<f>'(x)) = I log \x\<j)'(x)dx
J—00
= lim ( / +/ J log |x|<^'(x)tfa3
= £lim ((¢(-6) - ¢(€)) loge - (/_ ' +1") l-4>(x)dx\
Hence £log|x| = ypI.
(d) By F(f) we denote the Fourier transform of a tempered distribution f
For any <f> £ S, we have
('(v4)'*) = (^:^(^
i»0O -I *0O
= —2i lim / -/ sin(:c£)<^(:c)d:cd£
«-> o+Js £,/-00
/00 *M -I
<f>(x) I - sin(a;£)d£d:E
00 ./0 £
= —«ir / (signx)^(a;)dx.
«/—00
Here, we have applied the integral equality
1 f°° 1
- / - 8in(a:£)d£ = 1 for x > 0.
Therefore, we obtain
-wrsigna;.
6111
a) Let
Lu(x) = ^2 aaDau(x) xEMn
\a\<m

467
where the a'as are constants, and a is a multi-index. What is a fundamental
solution of LI When is it unique?
b) If /+ denotes the Heaviside step function
r m _ J 1. « > 0.
/+({) -1 o, «< o.
Show that
F(a!,t) = ^/+(<)/+(<-|a!|)
is a fundamental solution of the wave operator W(u) = m(( — uxx on M2.
c) Express the solution of the Cauchy problem
utt — uxx = 4>(x,t), — oo < x < oo, t > 0
u(a:,0) =/(a:), u*(a:,0) = fif(a:), -oo < x < oo
(fi9i<f> sufficiently nice) in terms of the above fundamental solution. Simplify
your result to obtain D'Alembert's solution of this problem.
(Indiana)
Solution.
a) A distribution E £ V'(Mn) is called a fundamental solution of the
differential operator L if
LE(x) = 6(x),
where 6(x) is the Dirac function.
Let L* be the formal adjoint of L defined by
L*v(x) = J2 (-l)WDa(aav(x)).
\a\<m
If L*S is dense in S, where S is the space of rapidly decreasing functions, then
the operator L has at most one fundamental solution in S'.
b) We need only to show that
J F(x,t) {j^ - |^) dxdt = ¢(0,0) V^ £ C^(M2).
In fact
2/ F(x,t)(<j)tt - 4>xx)dxdt
Jm?
= / (<i>tt — <t>xx)dxdb
J\x\<t

468
= -/ dt <f>xx(x,t)dx + / dx I <j)tt{x,t)dt
Jo J-t J-co J-x
i»0O *0O
+ dx I 4>tt{x,t)dt
Jo Jx
i»0O *0O
= / (4>x(~t,t) - <j>x(t,t))dt - I (<f>t(-x,x)+<f>t(x,x))dx
Jo Jo
i»0O *0O
= -/ 0,(M) + 4>x(t,t))dt - (fr(-t,t) - fa(-t,t))dt
Jo Jo
= 2^(0,0).
c) The solution of the Cauchy problem
utt — Uxx — <t>(x,i), x S M,t > 0
u(x,0) = w((a;,0) = 0
is given by
u(x,t) = f dr f 4(x-Z,t-T)F{Z,T)dZ
Jo J -oo
= I I dr f 4>(x - i,t - T)dZ
1 Jo J\i\<r
1 ft rx+(t-r)
= ; ^ 4>{Z,r)&Z.
L Jo Jx-(t-r)
The solution of the Cauchy problem
utt — uxx = 0, x G M,t > 0
u(a;, 0) = 0, ut(x,Q) = g(x)
is given by
/oo
g(x-(,)F{i,t)di
-oo
1 t*
1 fx+t
Therefore, the solution of the original Cauchy problem can be obtained as
rx+t 1 fx+t
I ft f*+* I f*+t
«MO

469
1 ft rx + (t-r)
+ - dr 4>($,r)dt.
* JO Jx-{t-T)
This is just the D'Alembert's formula when <j> = 0.
6112
Let u be the solution of the initial problem
utt = — uXjXjXkXk, x £ M ,t > 0,
u(x, 0) = 0, ut(x, 0) = 4>(x),
where summation is understood in the pde, and <f> is a C°° function on M3
with compact support. Assume that there is a constant C such that
||M(-,i)||L = (JR»)<Ci1/4/logi,
and prove that
J 4>(x)dx - 0.
(Indiana)
Solution.
By applying Fourier transform to the problem, we have
G» + |£|42 = 0,
£(¢,0) = 0, fc«,o) = £(o,
where u and <j> denote the Fourier transforms of u and <j> with respect to x,
respectively.
By solving the above transformed problem, it yields that
a«,*) = ^8in(Ki2t).
Therefore, we have
j HU)\2dt= J ^fsm\\i\H)di.
Jm* Jr' m
Set 77 = t = £ in the integral in the right side of the above equality. It is reduced
to
I m,t)M=i*[ w-h)\2S^f±dv.

470
This implies that
£(0) = I <j>{x)dx = 0.
Jm.>
In fact, if |<^(0)| = 2a ■£ 0, then there exists a positive constant r such that
tol>«> V^ G M3, |£| < r.
And when 2 > ^, we have
2 . [ sin2(|77|2) ,
> a2*= / —rnr1^^-
This contradicts the condition satisfied by the solution u, given in the problem.
6113
Given <f> S S (rapidly decreasing functions) over 2R1, consider the solution
u of the Schrodinger Equation
iut + uxx - 0 (x G 2R1, t > 0),
u(a;,0) = 4>(x).
Show that
/|x|<* ^|f|<
Here <j> denotes the Fourier Transform of <f>
lim f \u(x,t)\2dx = f |fe)|2^-
<^°°./|x|<* 7|f|<i
(Indiana)
Solution.
It can be verified that the Poisson's formula for the Cauchy problem of the
heat equation applies to the Schrodinger equation. So we have
u(x,t)= —-7=e~«' / ^y)e{ "'** dy.
zVTi Jm.1
And then
12
/ 4>(yy"~" dy dx.

471
Set x = 2£t, we get
/ \u(x,t)\2dx = -z- I \ 4>{y)eiW^V) dy
J\x\<t Zlr J\i\<h WRX
= -T [ I / <t>(y)e~iiy+i" dy
^ hi I/ ^y)e~iiVdv
= i l?(OI2d£. asi^o
Here the Fourier transform of <f> is defined by
¢(0 = -^=/ 4>(y)e-^dy.
d*
d*
d£

472
SECTION 2
ELLIPTIC EQUATIONS
6201
Let u £ 0 satisfy w G C2(Mn), Aw = 0 on Mn. Show that
i2dx
Jm.n
does not exist
{Indiana)
Solution.
There exists a point a;0 G Mn such that u(x°) ^ 0. Applying the mean-
value property for the harmonic function w, we have
Kx°) = Z~~T I u(x)dSx
UnPn~X J\x-x°\=p '
l\x-x°\-p
where un is the surface area of the unit sphere in Mn. Schwarz's inequality
gives
2
MV) < (—4rr) / w2(x)dSx / dSx
\Unp ) ./|x-x°| = p ./|x-x°|=p
= —[ / w2(x)dSx,
that is
'|x—X°\=p
f u2(x)dSx>u)npn-1u2(x°).
J\x—x°\=p
Then we have
/ w2(x)d:c > / w2(x)dx
./JR" ./|x-x0|<r
= [If u2(x)dSx)dp
J0 \J\x-x<>\ = p J
w„uV) /V-1^
Jo
Dl=p
>
Un-u\x»)rn,
n

473
for any r > 0. The conclusion follows as r —► oo.
6202
Let u G C0(O) be weakly harmonic in an open set 0, i.e., the relation
/ u&4>dx - 0
holds for all <j> G Cg(fi). Show that u is then harmonic in 0.
Solution.
Set
^, |as| < 1
0,
where C is a constant such that
1«.
it
I p(x)dz
, v , is = 1.
/JR"
For a; G 0, set
^0/) = ^(^)
Then it is easy to see that
Pe(y) € c0°°(fi),
provided e < dist(a;,dfi). Therefore, we have
i.e.
where
/ u(y)&yPe(y) = 0,
Jsi
0 = e-» / w(2/)Ax/9 (^0 d» = Axw£(x),
(Iowa)
This implies that for any compact set K C 0, ue is harmonic in K if e <
distaff, dfi).
It is well-known that
u€ —* u uniformly in if, as e —» 0.
Therefore it is harmonic in fi.

474
6203
Let f(x,y) be a locally bounded function, harmonic in a; 6 M2 and
continuous in y G M2. Show that f is continuous in (x,y), i.e., as a function on
2R4.
(Indiana)
Solution.
Let (x°,y°) G M4 and R > 0 be given. Then there exists a constant M > 0
such that
\f(x,y)\<M, V(x,2/)G2R4, |aj - x°\2 + \y - y°\2 < 2R2.
Applying the mean-value theorem to the harmonic function -^-, we have
df
dx
i t-R2 J\x-i\<§ dxi
V(£,2,)G2R4,K-*0|<f,l2/-2/°l<#-
By Green's formula, we get
dx~^y)
4
^R2
•/|.-f| = $
2/)cos(n, x1)dSj
< A-f, ^,y)em\\C-x0\<^,\y-y°\<R.
For g^-, we have the same estimation.
These estimations imply that f(x,y) is continuous at x — x° uniformly
with respect to y near y°. Then the conclusion that f(x, y) is continuous at
(x°,y°) G M4 follows immediately.
6204
Suppose u is a function defined on Mn such that
(i) u is bounded and continuous on the half-space xn > 0 of Mn,
(ii) u = 0 on xn = 0,
(iii) u is harmonic in xn > 0.
Prove that u = 0 on x„ > 0.
(Indiana)

475
Solution.
Redefine the function u in the half-space xn < 0 as an odd function of xn:
u(x',xn) = -u(x',-xn), \fxn < 0
where x' = (xi, ■ • ■ ,xn-i). By using the uniqueness theorem of the Dirichlet
problem and the Poisson's formula of the Dirichlet problem in a ball, we can
verify that the redefined function u is harmonic in any ball centred at the
origin in Mn. Therefore, u is bounded and harmonic in iR". Thanks to the
Liouville's theorem, we get u = 0 immediately.
6205
Let u(x) be C2 on the half-space 2R" = {x : xn > 0} and continuous on
the boundary <92R" = {x : xn = 0}, and let g(x) be a compactly supported,
C1 function on d2R™ — {x : xn — 0}. If u, bounded, satisfies
| A« = 0, xn> 0
I u = g, xn = 0, v ;
show that its "tangential" derivatives -^, j — 1,---,71- 1, are bounded in
magnitude by max|Vgf| on all of 2R" = {x : xn > 0}. (Warning: you must
justify any assumptions of regularity you make on the solution u.)
(Indiana)
Solution.
By the uniqueness of the bounded solution to the Dirichlet problem (3), we
can show that the unique bounded solution to the problem (3) is given by
u(x*x)-^[ ^^
where x' = (xi, ••• ,xn_i), £' = (£l5 • • • ,£n_i) and un is the surface measure
of the unit sphere in Mn.
It is not difficult to verify from the above formula that
,n — 1
and
2xn t
Wn 7jR»-l (¾2
d?
+ |a;'-^|2)"/2

476
Therefore, we have
0
dxj
u(x',xn)
< max
£-9(0
, j = 1, ---,71- 1,
which are just we want to show.
6206
Show that the problem
Am = -1 for \x\ < 1, \y\ < 1,
u = 0 for \x\~ 1,
du du
dx dy
- 0 for \y\ = 1
has at most one solution. (Here u = u(x,y) is a function of two variables x
and y, and is a classical solution).
(Cincinnati)
Solution.
We need only prove that the problem
Am = 0 for \x\ < 1, \y\ < 1,
m = 0 for \x\ = 1,
----- = 0 for \y = 1
ox dy
has the unique solution m = 0. It is easy to see that
I uAudxdy = 0.
./|x|<l,|y|<l
Integrating the above integral by parts, we have
0=/ uAudxdy
J\x\<l,\y\<l
+ LXu^yix,1)~u^yix,~1})dx

477
J\x\<i,\y\<i \\dx) \dy) )
= LXud£{x,l)~u^{x,~l})dx
J\x\<i,\y\<i \\dx) \dy) )
= |(u2(l, 1) - u2(-l, 1) - u2(l, -1) + u2(-l, -1))
ilxKi.lyKi \\dxj \dy) j
i|x|<l,|y|<l \\
Then we get u = 0 immediately.
£) + (¾) I ^
6207
Let fi be a bounded normal domain in 2R3, and suppose a is such that a
nontrivial solution exists to
f -V2m = a2u in fi,
\ u = 0 onSft. W
(a) Show that a is a real number, and that u can be chosen to be real-
valued.
(b) Suppose ati -£ a.2 are such that nontrivial real solutions u\ and U2
satisfy (*). Show that
/ ui(T*)u2(T*)(fo = 0.
Jq,
(Indiana-Purdue)
Solution.
(a) Multiplying the equation in the problem (*) by the complex conjugate
of the solution u, and integrating the resulting equation in fi, we obtain
I IVu^d-u = a2 I \u\2dv,
Jq Jq
which implies that a is a real number. Then it is clear that u can be chosen
to be real-valued.

478
(b) Suppose that mi and u2 are chosen to be real-valued. Multiplying the
equation satisfied by u%
—V «i = a\ui in fi
and the equation satisfied by U2
—V2«2 = a\u2 in 0
by m2 and ui, respectively, and integrating the resulting equations in fi, we
can obtain
/ Vui • Vu2<iv — a\ J uiv,2dv
Ja Ja
and
/ Vui • Vu2<iv = af I uiU2<iv.
Ja Ja
From the above integral equalities, we get immediately
(a\ — a\) I uiU2<iv ~ 0,
Jn
which implies the conclusion of the problem.
6208
Let fi C IRn be an open set. Let u be a continuous function on fi. Prove
the equivalance of these two notions of "subharmonic".
(i) For every x G fi and r > 0 such that the ball B(x, r) centered at x with
radius r satisfies B(x,r) CC fi one has
u(x) < r / u(y)dSy.
unrn-1 7as(x,r)
(Here un denotes the surface area of the unit sphere in Mn and B(x, r) C C fi
means that the closure of B is contained within 0,.)
(ii) For every ball B CC fi and every harmonic function /i S C°(B) such
that m < /i on dB, one has u < h in B.
(/ndiana)
Solution.
Let u be subhamonic in the sense of (i). By a standard argument, we can
show that u can not take its maximum in fi unless u is a constant.

479
Let h be the function given in (ii). Then v = u — h is also subharmonic in
the sense of (i). Hence that »<0on dB implies that v < 0 in B.
Conversely, let u be subhamonic in the sense of (ii). We define the function
h as follows:
f A/i- 0 in B(x,r),
\ h = u on dB(x, r).
From the definition of subhamonicity in the sense of (ii), and the mean-value
theorem, we have
u(x) < h(x) = ——[ / u{y)dSy.
wn" JdB(x,r)
This completes the proof.
6209
Let fi C Mn be an open set, and let u be harmonic in Q, and continuous in
n.
a) Show that ^- is harmonic in fi for each i.
b) Let Bcfi, where B is the open ball centered at x of radius r. Show
that
Ti
|Vu(x)| < -sup{|u(2/)| : \x - y\ = r}.
c) Show that, if f2 is bounded and x G fi,
|Vu(»)| < ^-sup{|w(2/)| : y G ft}
where d(x) = dist(x, 5ft).
(/ndiana)
Solution.
a) The conclusion is clear.
b) Applying the mean-value theorem for the harmonic function J^ in the
ball B, we have
k{x)=^Luxiiy)dy>
where oj„ is the surface area of the unit sphere in Mn, hence wn/n is the volume
of the unit ball.
By using the Green's formula, the above equality can be rewritten as
37» = Z~H I u(2/)cos(n>,*,-)dS.
Oxi u>nr JqB

480
And therefore, we have
du
OXi
< ~/ |u(2/)|dS
wn~ JdB
< -sup|u(j/)|.
r as
c) The conclusion can be obtained from the result of the part b)
immediately.
6210
Let fi C M3 be bounded and open with smooth boundary. Let u G C1(0)fl
C2(ft) solve
Au + k2u=0 in ft (Jb >0).
Derive an appropriate mean-value property for the solution.
(Indiana)
Solution.
First we look for the fundamental solutions of the equation. Let r = \x\.
Then the spherically symmetric solutions depending only on r satisfy
d2 2
-r-ziru) + k ru = 0.
dr2
For this ordinary differential equation, there are two linearly independent
solutions:
u = - cos(fcr), - sin(fcr).
r r
Let u be a solution to the equation Am + k2u = 0. By the fundamental
solution -r^- cos kr, it is not difficult to verify that
u(x0) = -hLSu{x)^(^cos{krxox))
cos(fcrxox) — u(x)I dSx, Vx° G fi,
rxox an J
where rxox = \x — x°\.
Let R < dist(x°, d£l) be a fixed positive constsant. By B(x°, R) we denote
the ball centered at x° with radius R. Applying the above integral formula in

the domain B(x°, R), we get
u(x°) = -■:- \u(x)x I cos(krxox)\
4* JdB(x°,R)\ orxox \rxox ')
cos(krxox)- u(a;) I dSx.
drxox
< x"X
Now we look for a function g(x, x°), which satisfies
(Ax + k2)g(x,x°) = 0, xeB(x°,R),
g(x, x°) = cos(krxox), x G 8B(x°, R).
It is easy to see that the function
g(x,x°) = cot(kR)s{n{krx°x)
7*x°x
satisfies the above conditions. For this function, it is easy to verify that
°=~i I Ux)-^-g(x,x0)-g(x,x°)-^u(x))dSx.
47r JdB(x°,R) V orxox orxox )
Subtracting the equation (2) from the equation (1), we obtain
u(x°) - --7- u(x)- ( cos(krxoa
47r JdB(x°,R) orxox \rxox
— cot(kR) s'm(krxox) I dSx
7*x°x /
k f
- a p ■ n m I u(x)dSx.
This is a mean-value theorem for equation Am + k2u = 0.
6211
Let u € C2(M2) satisfy
Aw= u+ 1.
Prove that its spherical mean, defined by
)
v(x) = im / u(y)dSy, x^o
<"""R JdB(0.\x\)

482
and -y(O) = w(0), also satisfies (1), for x ^. 0. Here, B(0, r) denotes the ball of
radius r, dSy the element of arc length.
(Indiana)
Solution.
Set r — \x\. Then we have
v(.r) - it I u(rw)dSu
ilr JdB(0,l)
and
8_
dr
1 /
-v(r) = — / Y]ux,(rw)w,-d5a,
l* JdB(o,i)jr[
2*r J9B(o,r) JT[
Applying Green's formula, we get
—v(r) --— / Au(y)dy
dr 2irr JB(0>r)
= ^/(/ Hy)+l)dSy)dp.
tor J0 \JaB(o,p) J
Then it follows that
= r(v(r) + 1).
Therefore
Av(r) =-^. [rfrv(r)) = <r) + L
The proof is completed.
6212
Let B = {(x,y) \ x2 + y2 < 1}. Prove that the problem
Am = = in B, u = 0 on dB
y/x2 + y2

483
has exactly one weak solution u in Hq(B).
(Iowa)
Solution.
Let
x y , , 1
v = =, w = , = and /
\Jx2 + y2' \A2 + 2/2 \A2 + V2
It is easy to see that v,w £ L2(B) and
dv dw
ox ay
in the sense of distributions (see the solution to the problem 6105). This
implies that f G H~1(B). Hence the problem admits a unique weak solution
u G H^B).
6213
(a) Let fi be a bounded domain in Mn with smooth boundary dfi, and
f G £2(fi), and g G L2(d£l) be given. Consider the problem
/ (gradit) • (grad-y) + Xuvdx = / fvdx + / gvda Vw G W1,2(Q) (1)
Ja Ja Jen
where A > 0. Show that the problem (1) has a unique solution u G W1,2(Q,).
(b) Will the conclusion of part (a) be valid if A = 0? Explain.
(c) Soppose that /, g and d£l are sufficiently smooth, write problem (1) in
the classical form.
(Cincinnati)
Solution.
(a) Define the inner product in W1,2(fi) by
(u,v)i - / [(gradit) • (grad-u) + Xuv]dx, Vw,v G W1'2^).
Jn
And denote the corresponding norm in W1,2(fi) by || • ||!. Then by applying
Cauchy inequality and the trace theorem in Sobolev's spaces, we get
I fvdx + I
gvda
an
< ||/IU*(n)HU*(n) + IMlLH^IMU^an)
< C\\v\\u

484
where C is a constant. Therefore,
F(v) = / fvdx + / gvda
Jn, Jen
is a bounded linear functional in W 1,2(fi). Thanks to the Riesz theorem, there
exists a function w G W1'2^) such that
F(v) = (ri,v)i, Vt»€W1,2(fi).
Then w is a solution to the problem (1). The uniqueness is clear.
(b) By the classical form of the problem, it is easy to see that the conclusion
of the part (a) is false if A = 0 even for smooth f and g.
(c) Suppose that /, g and dCl are sufficiently smooth. Integrating by parts,
we have
I (gradw) • (grad-y)da; = / — vda — I (Au)vdx.
Jn Jen on Jfi
Therefore, the classical form of the problem (1) is
—Aw + Aw = / in fi,
du or,
— = g on Oil.
an
6214
Let flbea unit ball in Mn centered at zero. Consider the following problem
Find w G -Ho(fi) sucn tnat
J Vw • V$dx = I <j>dx\ V<£ G fl"o(fi),
Jn Jnn{xn=o}
where Mn 3 x = (xi, • • •, x„_i, xn) — (x', xn). Prove the existence of a unique
solution of that problem.
Denote u+ = u |nn{x„>o} and w~ = w |nn{x„<o}> an<l assume that w* are
regular enough. Show that Aw* = 0 and that
du~ du+ f n1
(Cincinnati)

485
Solution.
Define the scalar product in Hq(CI) by
(m, v)i= / Vu-Vvdx.
Jn
By the Cauchy inequality and the trace theory in Sobolev spaces, we can
get
4>dx'
Jsir\{xn--
:0}
< (f dx) (f \4>\2dx')
\/nn{x„=o} J \Jnn{xn=o} J
< cu\\u
where 11 -1|x is the norm in Hq(Q,). According to the Riesz theorem, there exists
u G -ffo(fi) such that
(u,^)i= f 4>dx', W<f>€H^(Q).
Jsin{xn=o]
This proves the existence of the solution to the problem.
The uniqueness of the solution to the problem is clear.
Assume that u* are regular enough. Write the equation in the following
form
J Vu+-V<£da;+( Vu" • V<t>dx = I 4>dx',
Jsi+ Jn_ Jnn{x„=o}
where fi+ = fi D {xn > 0} and fi_ = fi fl {xn < 0}. Integrating the above
equation by parts, we have
— I Am+ • <f>dx — / Am~ • <f>dx
+ I (if- - IT-)6*' = I ^x', W G H^Sl).
From the above equation, we get immediately that
.n±,
-1 onfin{a;„ = 0}.
Aw* = 0 in fi±,
du~ du+
dxn dxn
The proof is completed.

486
6215
a) Let u G W0' (ft) satisfy
I Vw • V<j>dx > 0 V<£ G Wo'2(SI),<f,>Q.
Show that u > 0 a.e. in ft.
b) Let it G W1,2(Sl) satisfy (I) above, show that
inf u > inf u (essinf)
n - an '
Solution.
a) This is a corollary of the conclusion in b) of this problem.
b) If inf u = — oo, the proposition is true.
Assume that inf u = I > — oo. Let
an
4>(x) — max{! — u, 0}.
Then it can be verified that 4> G W0' (ft) and
V*-\0, u>l.
Now we have
I Vw • V<j>dx = - J \V$\2dx < 0.
J(i Jn
From the above inequality and (I), it holds that
j \V<t>\2dx = 0.
By the Poincare inequality with the above estimate, we obtain
j \<f>\2dx = 0.
Hence
¢ = 0 a.e. in ft
This implies that u > I a.e. in ft.

487
6216
Consider functions un,vn, w G W2,2(£l) D Wq,2(£1), such that
-Awn + un = w,
-Aun - nvn = 0,
in fi. Prove that un —► w strongly in L2(Cl), asn-» +oo.
(Cincinnati)
Solution.
Multiplying the first equation and the second equation by un and vn
respectively, and then integrating the resulting equations in fi, we get
- / Awn -undx + \\un||£2 = I wundx,
Jo, Jii
- / Aun ■ vndx - n\\vn\\l2 = 0.
Jsi
By using Green's formula and noting that un,vn G W0' (0,), both equations
above can be written as
/ V-y„ -Vundx + \\un \\2L2 = / wundx,
Jo, Jo,
/ Vwn -Vvndx - n\\vn\\2L2 = 0.
Jsi
Then we have
lUnllis + nIWIl,2 = I WUndx
= / wuni
Jo,
1 .. „2 1,, „2
< 2^1^2+2"Wn"L2'
Therefore, we obtain
{un} is bounded in L (fi) (1)
and
vn -» 0 strongly in I2(fi). (2)
Multiplying the first equation by -u„ and integrating it on fi, we have
llVwn||£» + / unVndx = / WVndx.
Jo, Jo,

488
By noting (1) and (2), it follows from above equality that
vn -> 0 strongly in Wq'2(SI). (3)
Multiplying the first equation and the second equation by Aun and Avn
respectively, and then integrating the resulting equations, we can get
- / AvnAundx - ||Vw„|||,2 = / wAundx,
Jn Jn
- / AunAvndx + n\\Vun\\2L2 = 0.
Jn
Hence
||Vu„||jr,a + n||V«„||jr,a = / Vu> -Vundx.
Jil
This implies that
{un} is bounded in Wq'2(CI). (4)
Multiply the first equation by Avn. By similar consideration, we can get
finally
-||AwJ|?.a - / Vwn -Vvndx= / Vw • Vvndx.
Jil Jil
"n||Ls
By noting (3) and (4), it follows from the above equality that
Awn —» 0 strongly in I2(fi).
This gives us the conclusion of the problem.
6217
Assume fi C Mn is a bounded open set with smooth boundary and let
f G C£°(fi). Suppose Uk G C°°(fi) satisfies Auk + Uk = f and has the property
that for some positive number M,
J
Jil
Uk(x)2dx < M
for k = 1,2,---. Prove that there exist a function u G C°°(fi) satisfying
Am + u = f and a subsequence {ukj} such that Ukj converges uniformly to u
on each compact subset of fi.
(Indiana)

489
Solution.
As {wjt} is bounded in L2(Jl), there exists a subsequence {wjtj} such that
Ukj converges to u weakly in L2(Cl), and therefore in Z>'(fi). Then Am + u= f
in the sense of the distribution. By the regularity theorem of the elliptic
equations, we have u G C°°(fi).
Now we show the second conclusion. Set vk = uk — mi. It is clear that vk
satisfies
A-yjt + vk — 0 in fi
and
I \vk(x)\2dx < C, Vfc =1,2,---.
Hereafter, by C we denote various constants.
Let fii be a subdomain of fi such that fix CC fi and R — dist(5fi, fii)
fixed. For any a; G fii, by B(x,r) we denote the ball with the radius r (< i?)
centered at x. By the mean-value theorem for the equation Am + m = 0 (for
the case n = 3, see the solution to the problem 6210), we have
h(r)vk(x)- / vk(y)dSy,
JdB(x,r)
where h(r) is a continuous function of r. Integrating the above equality from
0 to R with respect to r, we find that
vk(x)=\ h(r)dr) / vk(y)dy.
\Jo ) JB{x,R)
Here R can be chosen so small that JQ h(r)dr -£ 0. Therefore, it holds that
hk(a!)|<C, Vs €«!,*= 1,2,.-.. (1)
Set vk = wk + zk, where wk and zk are defined by
Awk = —vk in iii,
wk = 0 on 5fii
and
Azjt = 0 in Hi,
zk = vk on 5fii,

490
respectively. By applying the solution formula of the Dirichlet's problem for
the Poisson's equation, we can get from the estimation (1)
|Vi»fc(a:)| < C, Vz €^1,4= 1,2,---.
It is clear that
max|zfc(a;)| < C, Vfc =1,2,---.
Hi
And by applying the mean-value property for the harmonic solutions, we can
obtain from the above estimation
|Vzjfc(a;)| < C in any compact subset of fii.
Then by using the Ascoli-Arzela theorem, we prove the second conclusion
of the problem immediately.
6218
Let fi be a bounded Lipschits domain in Mn, and
I(v) = l[ \Vv\2dx,
z Jn
A = {v e flo(^) ■■h1<v <h2 a.e. in fi},
where hi, h2 : fi —► fit are given smooth functions.
(a) Show that if A ^ 0, there exists u El A satisfying
I(u) = minl(v).
(b) Show that
J VuV(v - u)dx > 0 \/v eA-
Ja
Solution.
(a) Let {un} be the minimizing sequence of I(v) in A, i.e.,
lim I(un) - inf I(v).
n—>oo v€A
Define the norm of v in Hq (fi) as
1/2
(Iowa)
;Hi=(/jv*i
da;

491
Then we get the boundedness of {11 un \ | i}. Therefore, there exists a subsequence
of {un}, for simplicity we still denote it by {u„}, such that
un —- u in Hq(£1)
and
un —► u in L2(ti).
And there exists a subsequence of {un}, denoted still by {u„}, such that
un —► u a.e. in fi.
Therefore, we get u £ A and that
||w||i < liminf ||w„||i = lim 2I(un)
n—>co n—>oo
by the weak lower semicontinuity of ||i»||i. This implies that
I(u) = min/(-y).
The conclusion of (a) is proved,
(b) It is easy to verify that
u + t(v-u)eA, \/v eA,0 <t<i.
Let
F(t) = I V(u + t(v - u)) ■ V(u + t(v - u))dx.
Jn
Then F(t) has the minimum at t = 0. Therefore we have
F'(0) > 0.
This implies
/ Vw • V(w - u)dx > 0.
The proof is completed.
6219
Let fi C Mn, n > 1, be a bounded domain with smooth boundary. Let
u G Cx(fl) be harmonic in fi.

492
a. Prove that
maxlVul = maxlVwI.
(Note: In both parts a and b of this question, you may cite, without proof,
any standard maximum principles you are using.)
b. Suppose there exist functions $x, $2 € C2(fi) fl C1(0) such that
A$! < 0, A$2 > 0, $i = u = $2 on dft.
Prove that
maxlVitl < max ( max|V$i|,max|V$2l I •
n V dn en ' J
(Indiana)
Solution.
a. The conclusion is an immediate corollary of the following inequality:
A(|V«|2) = 2 £ u2XiXj > 0.
b. For the function $2 ~ u, we have
A($2 - u) > 0 in ft
and
$2 - u = 0 on dVl.
From the maximum principle, it holds that
sup($2 — u) = max($2 — u) = 0,
n an
i.e.,
$2(») < u(x), Vx G ft.
Similarly, we have
u(x) < $i(a:), Vx G ft.
Therefore, it holds that
¢2(¾) - $2Qc°) «(») - «(»")
|x — x°\ ~ \x — x°|
|x — xu\
This inequality implies the conclusion of the part b.

493
6220
Let ft bean open subset of Mn. Suppose u S C2(fi) is a solution of the
equation Am = it3 with the property that |Vu(a;)| < 1 for each x S 30. Prove
that |Vw(a;)| < 1 for all x 6 0.
(Indiana)
Solution.
Set
V|2
U\ .
It is easy to see that w G C2(0) (~1 C^O) and
n
Aw = 2 ^2 uliXj + 6w2|Vw|2 > 0.
Therefore, w takes its maximum on ft at some point on d£l. Hence w < 1 on
dQ, implies that w < I'm 0,.
6221
Consider (a nonlinear) equation
—Am = Am2(1 — u) in ft,
u = 0 on dO, (1)
where A > 0 is a parameter, and 80 sufficiently smooth.
(a) Prove that for any solution u of (1), 0 < u < 1.
(b) Show that if A > 0 is sufficiently small, the only solution of (1) is u = 0.
(Cincinnati)
Solution.
(a) If u < 1 is false, then there exists x° S ft such that u takes maximum
at x° and
u(x°) > 1.
For the maximum point x°, it is clear that
Au(x°) < 0,
i.e.,
-Au(x°) > 0.

494
This contradicts the assumption that
Au2(l - u)(x°) < 0.
Hence u < 1.
As Au2(l — u) > 0, u can not take minimum in fi unless u = 0. Hence
w> 0.
(b) Multiplying the equation by u and integrating the resulting equality on
fi, we can get
/ |Vw|2da; = A / w3(l - u)dx.
Ja Ja
By using the conclusion of (a), it holds that
J |Vw|2da; < A / \u\2dx.
Applying Poincare inequality, we obtain from the above inequality
/ |Vw|2da; < AC / |Vw|2da;,
where C is a constant. If A < C~1, we have u = 0 immediately.
6222
Let fibea bounded normal domain in 2R" with smooth boundary dCl. Let
It be the exterior unit normal on dCl. Show that the only solution of the
boundary value problem for the biharmonic equation:
A(Aw) = 0 in ft
u = 0 on 5ft,
— = 0 on 5ft,
on
is the trivial one u = 0. Assume u G C4(0).
(Indiana-Purdue)
Solution.
Integrating by parts and taking the boundary conditions into account, we
have
0=/ A(Au)udx

495
= I -r-(AuWS - [ V(Au) • Vudx
Jan on Jn
= I (-H-(&u)u-Au~)dS+ f \Au\2dx
Jan \dn on J Ja
= J \Au\2dx.
Jn
Therefore, u is harmonic in 0. Taking into account u = 0 on 30, we get u = 0
immediately.

496
SECTION 3
PARABOLIC EQUATIONS
6301
Consider the Cauchy problem for the Heat Operator in JR}\
ut = uxx (—oo < x < oo, t > 0)
u(x, 0) = f(x) (—oo < x < oo),
where / is bounded, continuous, and satisfies
/oo
|/(x)|2dx < oo.
■oo
Show that there exists a constant C such that
Q
l«0M)l < £1/4
for all —oo < x < oo, t > 0.
(Indiana)
Solution.
The solution to the problem is given by the Poisson's formula
1 f°° (,-y)a
u(x,t) =-—= f(y)e ■» dy.
Zy-Kl J -oo
By the Cauchy inequality, we have
W'"1 £ STB (£ lml'dy)' (£«"^^)*•
Set y = x + \f2tr] in the second integral of the above inequality. Then we get
immediately
I / M C
KM) I < ^/4-

497
6302
Let fi C -ffi" be an open set with smooth boundary and suppose u S
C°°(fi x [0, oo)) is a solution of the equation
Ut — Am = /
with u = 0 on d£l x [0,oo). Assume that
lim f f(x,t)2dx-0.
Let
rj>(t) = I u(x,t)2dx.
Prove that r/>(t) —> 0 as t —► +oo.
{Indiana)
Solution.
Multiplying the equation by u and then integrating the resulting equation
on fi, we can get
Jt Qll«(-,*)l|£i(n)) + IIVu(-,<)ll£'(n) = jjudx-
The Poincare inequality gives
ll«(-,*)lli»(„)<q|Vu(.,t)||i =(n)»
where C > 0 is a constant. For any e > 0, it holds that
Taking £ = 1/C, we obtain
^(i) + i^)<q|/(-,i)||£2(n).
Solving this differential inequality, we have
W) <e-^(Q) + C /'e-*('-r)||/(-,r)||£2(n)dr.
Jo

498
It is not difficult to verify that
J e-£(t-r)||/(.5 T)||2a(n)dT - 0, as t -+ +0O
JO
if ||/(-)2)||f,2(m —* 0 as t —► +00. This completes the proof.
6303
Consider the Cauchy problem
ut — uxx (—00 < x < 00, t > 0),
u(*,0) = /(*),
where / is bounded and continuous.
a) Using the Fourier Transform, construct explicitly a fundamental solution
for this problem, and write down the solution u.
b) State a maximum principle for this problem. Why is the Tychanov
non-uniqueness example possible?
c) Suppose in addition that / £ £1(2R). Show that
lim u(x, t) = 0.
d) Show how to solve the problem
ut = uXx (x > d,t > 0)
u(x,0) = f(x) (z>0)
u(0,<) = 0 (<>0)
by using a) and an appropriate extension of /.
(Indiana)
Solution.
a) Let «(£,£) be the Fourier transform of u(x,t) with respect to x.
Transforming the following Cauchy problem with the initial data 6(x):
ut = uxx (—00 < x < 00, t > 0)
u(x,0) = 6(x),
we get
u* + £2u = 0,
6(^,0) = 1.

499
The solution to the transformed problem is given by
u(t,t)
e
-et
Then the inverse Fourier transform of u gives the fundamental solution to the
problem
u(x,t) = ± [ efa«-«a*de = -^=e- "2
2t Jjri 2vxi
From this fundamental solution, we can get the explicit representation for
the solution u(x,t) to the general Cauchy problem with the initial data f(x)
1 f (»-y)a
b) The maximum principle can be stated as follows: Suppose that u(x,t)
is bounded and continuous on M x [0,T], and satisfies the heat equation in
M x (0,T] with the initial data f(x). Then it holds that
inf f(x) < u(x,t) < sup f(x) for (x,t) G M x (0,T].
The Techanov non-uniqueness example is possible for some unbounded
solutions.
c) The conclusion is clear.
d) Defined an odd function <f>(x) as follows:
Mx) _ J /(*), x > 0.
^ ' ~ I -/(-*), * < o.
Then by solving the following Cauchy problem:
Uf = uxx ( — oo < x < oo,t > 0),
m(x,0) = <^(a:),
we get the solution to the initial-boundary problem
u(x,t) = —= / f(y) I e « - e «1 d».
6304
Consider the problem

500
ut — uxx (x € M, t > 0)
u(aj, 0) = «o(as) € L2(M).
Find a bound on J_ \ux(x,t)\2dx in t > 0.
{Indiana)
Solution.
From the Poisson's formula for the Cauchy problem of the heat equation,
we can find that
1 /*°° c»-
ux(x,t) = -, / uo(y)(x-y)e- *■»' dy.
— I ?<nl VHZ — «ir
Then by using the Cauchy inequality, we have
K(M)|2<I^y \uo(y)\2e-i^L-dy. J (x-y)2e-L^L-dy.
And therefore, it holds that
/OO -^ yOO 2 »0O 2
|wx(a;,i)|2da; < ^t^IKI^jr) / e~ *< dy ■ y2e-^dy
1.. ,2
2i
Iuo||l»(k)-
We can also get the result by the method of energy integrals.
Multiplying the heat equation by u and tut respectively, and integrating
the resulting equations in M x (0, T), we can obtain
1 y»oo y»j y»oo -I i»oo
-/ |w(x,T)|2da;+ / / |wx(a;,i)|2da;di = - / \u0(x)\2dx
^ J-oo JO J-oo *• J -OO
and
l-T /.oo ,T i>oo
I I t\ut(x,t)\2dxdt = I I tuxx(x,t)ut(x,t)dxdt.
J0 J-oo Jo J-oo
Adding the double of the second equality to the first one, we have
1 f°° fT f°°
- / \u(x,T)\2dx+ / / \ux(x,t)\2dxdt
^ J-oo Jo J-oo
F t°°
+2 / / t\ut(x,t)\2dxdt
Jo J-oo
■t yoo fT yoo
= o I \uo(x)\2dx + 2 / / tuxx(x,t)ut(x,t)dxdt.
^ J-oo Jo J-oo

501
It can be calculated that
fT ,00
2 / I tuxx(x,t)ut(x,t)dxdt
Jo J-00
,T ,00
= —2 / / tux(x,t)uxt(x,t)dxdt
Jo J-co
/•T j / i>oo \ »T y>oo
= -/ — I / |wx(a;,i)|2da;i I di + / / |ux(a;,i)|2da;di
/00 i»T i»oo
|wx(a;,T)|2da;+ / / \ux(x,t)\2dxdt.
■00 «/o J —00
Here we have applied the assumption
/00
\ux(x,e)\2dx = 0,
-00
otherwise an approach to the function uo(x) by a sequence of functions in
C5°(2R") can be used.
From the above two equalities, we have
1 f°° f°°
- \u(x,T)\2dx + T \ux(x,T)\2dx
* J-co J-00
,T ,00
+2 / I t\ut(x,t)\2dxdt
Jo J-co
1 f°°
= 0 / \u0(x)\2dx.
6 J-00
Therefore, we get an estimation
,00 i ,00
/ \ux(x,t)\2dx < — / |w0(a;)|2da;, V< > 0.
J-00 ^* J-00
6305
For the indicated domain fi (interior of the parabola) let Q,? denote the
points (x,t) with x £ fi and t < T. Let Si denote the open line-segment
forming the top of dfir, and let B2 — 8CIt\Bi. Let u(x,t) S C°(Ot) with
Mxx;M* € C°(£It U Si) be a solution of
uxx — ut + a(x,t)u — f(x,t)

502
with a, / G C°(0). Show that if a < 0, / < 0 in Ot, then every non-constant
solution u assumes its negative minimum (if one exists) on B2. What can you
say if a < 0, / > 0 in fiT?
(Iowa)
1
\ aT J
Fig.6.1
Solution.
Suppose that u assumes its negative minimum at (x°,t°) G 0TUBi. Then
we have
u(x°,t°)<0, ut(x°,t°)<0
and
uxx(x°,t0) > 0.
Hence
Mxx(^0,i0)-«t(Ai0) + «(^0,i°WAi0) > 0.
This contradicts f(x°,t°) < 0. (Here a < 0 should be assumed in fiT U Si).
If a < 0, / > 0 in fi^, we can sav that every non-constant solution assumes
its positive maximum (if there exists one) on Z?2-
6306
Consider the boundary-value problem
uxx + it( = 0 (0 < x < -k and t > 0),
B.C. u(0,t) = 0,
Ux(ir,t) = 0,
I.C. u(x, 0) = f(x).
Use the sequence {fn(x) = ^-j sin (^^x)} and an appropriate space of
continuous functions to decide whether this problem is well-posed.
(Indiana -Purdue)

503
Solution.
When n is even integer, fn(x) is the eigenfunction of the corresponding
two-point boundary value problem. By setting u = fn{x)T„(t), we can find
the solution u„ to the problem with the initial data f(x) — fn(x)
. JN 2 fn+1 \ (a±±\2t
Let
= max |/(a;)|
for all / G C[0, x]. It is clear that
||/„||-»0 as n -»oo.
But for any fixed t > 0
||w„(-, t)|| —>■ cxd asn-KX).
This implies that the problem is not well-posed in the space of continuous
functions with the norm defined above.
6307
In Probability Theory one encounters the Ornstein-Uhlenbeck process, in
which the particles of Brownian motion are subjected to an elastic force. One
is required to solve
ut = uxx - xux (x £ M,t > 0)
m(k,0) = uo(x).
Assume u0 G S and find u. (You should be able to find u explicitly; if not,
leave your result in the form of a one-dimensional integral.)
{Indiana)
Solution.
By «(£,£) we denote the Fourier transform of u with respect to x. It is easy
to see that
„. . d _. , d ^ ,du
F(xux) = t-grF(ux) = -^(£«) = -« - £"j£ •

504
Then the transformed equation and initial condition are in the following form
*?_*|| = (i-Os teM,t>o,
2tf,0) =
absolving the above Cauchy problem for the first order linear partial
differential equation, we can obtain
8tf,t) = e«-i(«a<-iKaGo(e**)-
It is not difficult to verify that
F-\uo(e^)) = e-*uo(e-'a:)
and
V V2x(e» - 1)
Therefore, we get the solution to the problem
1 f (*-y)2
y2x(e2t - 1) Jiri
Remark. From the above solution formula, we can find a convenient
transformation of variables to the problem. In fact, set
r=i(l-e-2'), £ = e-'*.
Then the problem is transformed into the following
r = 0 : u = uo(£)-
The above solution formula can be obtained easily from the Poisson's formula
for the Cauchy problem of the heat equation.
6308
Let L be the usual heat operator
d_
let <t> G D(2R"), and let h be a fixed point of Mn. Consider the problem
^-1^ + -+^),

505
(Lu)(x,t) = u(x-h,t), xeMn,t>Q,
u(x,o) = 4>(x), xemn
for an unknown function u(x,t).
a. Derive an explicit representation for a solution u(x,t) of the above
problem in terms of the initial data. (You should prove that your u is well-
defined, but you need not prove that it represents a smooth solution.)
b. Assume that <j> is the Fourier transform of a function which vanishes in
the ball of radius R > 0 centered at the origin. Show then that the solution u
satisfies
(Indiana)
Solution.
a. Let m(£, t) be the Fourier transform of u with respect to x. Transforming
the above problem we obtain
r «( + (K|2-e-*'h«)« = o, temn,t>o,
\ 2(^,0) = ^), xeMn
which admids the following solution
2(^i) = ^)e-(l«l2-ex p(--fc.O)t.
The inverse Fourier transform of u gives us the explicit representation for the
solution u(x,t) to the problem
(A) Jm.*
It is clear that eix<~^-^^^-^1 G Ll(Mn) and £(£) G S. Therefore, the
above representation is well defined.
b. By the Plancherel theorem, we have
n«(-.*)iii'(H») = (2*rn\\u(;t)\\h(iR«)
= (2x)-" / |fe)|2e2(cos(h-«H«P)<^
= (2x)-« / |^)j2e2(cos(h-«)-l«l2)'^
< (2^)-^1^ / 1^)12¾
„2(l-iJ2)*M /ii2
This is just the conclusion of the problem.

506
6309
Let fi C Mn be a bounded domain and assume u(x,t) > 0 is a function
with u S C2(0 x [0,oo)), which solves the equation
ut — Am = — u
(heat conduction with heat loss due to radiation) with the boundary condition
u \dn= 0. Prove that we can find a constant C such that
E(l)= f u2(x,l)dx<C
Jn
regardless of the initial value u(x, 0).
(Iowa)
Solution.
Multiplying both sides of the equation by u and integrating on fi, we have
/ utudx — / Aw • udx — — J u5dx.
Jn Jn Jn
By using the Green's formula and noting the boundary condition, we get from
above equality
-— E(t) + I \Vu\2dx -- I u5dx,
2 dt J ^ J ^
where E(t) = Jnu2dx.
The Holder inequality gives
(1)
i.e.,
E(t) < ( J u5dx\ * ( f dx\
= |fi|* ( I u5dxj ,
f u5dx> \Q\-*Et(t).
Jn
Using above estimation, we get from (1)
—*?(*)<-|nr«j?4(*)-

507
The above inequality can be written as
^jr*(t)>3|nr*.
Integrating the above inequality from 0 to t, we have
E~i(t) > E-*(0) + 3\Q\-h.
It follows that
E(i)<(±y\n\.
This completes the proof.
6310
Let
u € C2(B(0, r) x (0, T]) n C(B(0, r) x [0, T])
satisfy
ut + v? + sin(u) = Axm, B(0,r) x (0,T],
u = /i, 3B(0,r) x [0, T],
u = g, B(0,r)x{0}
with g, h continuous and g, h < 1. Prove that maxu < 1 on B(0,r) x [0,T].
(/nrfiana)
Solution.
If the conclusion of the problem is false, then there exists (x°, t°) S B(0, r) x
(0,T] such that u takes the maximum at (x°,t°) and
w(x0,i°)> 1.
It is clear that
ut(x°,t°)>0 and Ax(x°,t°) < 0.
Therefore, we must have
(M(-AxM)(x°,io)>0. (1)
But from the equation we get
(uj-A.uXa!0,*0) = -(«3 + 8in(u))(a!0,t0)
f-l-sinl<0, if u(x°,t°) = l,
~ { -u3(x°,t0)-sin u(x°,t°) < 0, if u(x°,t°) > 1.
This contradicts (1).

508
6311
Let u G C2(fi x (0,T)) D C°(0 x [0,T]) be a solution to the problem
ut — Am - m3 iG!l,i>0,
w = 0 zGdfi,i>0,
u(a;,0) = /(¾) iGfi,
where fi is a smooth bounded domain in 2R".
(a) Prove that
IMIl'(0)(*) < ||/||l'(0) for all* € (0,T].
(b) Prove that
IMIl-(o)(*) < ||/||l-(0) for all t G (0, T}.
(c) Prove that the solution to this boundary value problem is unique.
(d) Show that for / G I4(ft) D H1^), the following bound also holds:
\H\tHn)(t) + ll«lllr^n)C*) < \\f\\Un) + WfWhmy
(Indiana)
Solution.
(a) Multiplying the equation by u, and integrating the resulting equation
on fi, we get
I utudx = / Am ■ udx — I u4dx.
Jii Ja Jsi
By applying Green's formula, we obtain from above integral equality
_d 1
dl~2
I u2dx = - J |Vw|2da; - I u4dx < 0.
Jii Ja Ja
Then we get the conclusion immediately.
(b) First we show that u can not take the positive maximum in fi x (0,T].
In fact, if u take the positive maximum at (x°,t°) G fi x (0,T], then we must
have
ut(x°,t°) > 0 and Au(x°,t°) < 0.
Hence
(u» - Au)(x°,t°) > 0.

509
This contradicts the inequality
-u3(x°,t°)<0.
In a similar way, we can show that u can not take the negative minimum
infix (0,T].
Therefore, we have
IMU-(n)(*) < ll/IU-(n), ViG(0,T].
(c) Let mi and M2 be both solutions to the boundary value problem, and
w = mi — M2- The w solves the following problem:
{wt = Aw — (u\ + M1M2 + u\)w, x £ ft, t > 0,
w = 0, i6 3fi,t>0,
io(a:,0) = 0, i£fi.
By a similar deduction as done in (a), we can get
HHlL'(n)(*) < 0> Vt €(0,31
for above problem.
This implies that Mi = M2, and the uniqueness of the solution is proved.
(d) Multiplying the equation by ut and m3 respectively, and integrating the
resulting equations, we get
and
[u>dx=[ An-utdx-[u>utdx
I utu — J Am • u3dx — I u dx.
Jn Jq, Ja
By applying Green's formula to the first terms in the right side of the both
equalities above, we can obtain
and
— (- f u4dx] =- f 3M2|VM|2da; - / u6dx < 0.
dt V4 Jn J Jn Jn
Adding the both inequalities above, we have
^-( f u4dx + I \Vu\2dx ) < 0.
dt \Jn Jn J ~

510
This differential inequality, combining with (a) if necessary, implies the
conclusion of (d).
6312
For each p S 2R", let A(p) be a real positive definite n x n matrix with
entries a,-j-(p). Assume A 6 C1(2R"). Let fi C Mn be a bounded open set.
Prove (from first principles) the following comparison principle:
Let u and « beC2 solutions to the inequalities
ut < aij{Vu)uXiXj for (x,t) 6 fi x (0,oo)
and
vt > aij(Vv)vx.Xj for (x,t) 6 fi x (0, oo),
respectively. Suppose, furthermore, that u and v are continuous on fi x [0, oo)
and satisfy the condition «<»on the parabolic boundary
(fix {0})u(dft x [0,oo)).
Then u < v on fi X [0, oo).
(Indiana)
Solution.
Set w = u - v and
h(x,t) = W TT^CVf + 0(V« - Vt>))dfc>x<Xi.
• ,• Jo ouXk
Then from the given differential inequalities, we can obtain a differential
inequality for w as follow:
n n
^ atj(Vu)u>XjX_,. + ^6jt(a;,i)u;Xfc - iot > 0.
By applying the maximum principle to the above differential inequality, we
find _
w < 0 on fi x [0, oo).
This is the result we need.

511
6313
For ft C Mn bounded and open, let u G C2(ft x [0, 00)). Assume atj(x,t,
z,p) and bi(x,t, z) are continuous functions of their arguments and assume
that
n
J2 ay(M>*>P)6£> > \(x,t,z,p)\t\2 > 0 for every £ G Mn - {0},
t,i=i
where A is a positive function. Suppose u satisfies the inequality
n n
wt - 2J OtjOM, u, Vu)ux.Xj. - 2_,&,(:c,2,m)mX; > 0 for a; G ft, 2 > 0.
«,y=i «'=1
Prove that for any T > 0
_min u(a;,i) = mmu(x,t)
Slx[0,T] Q
where Q - (ft x {0}) U (dft x [0,T]).
(Indiana)
Solution.
First we prove that if w G C2(ft x [0,oo)) satisfies
n n
u)( - ^2 aij(x,t,u,'Vu)wx.Xj - '^2bi(t,x,u)wx; > 0, at (x,t) G ft x (0,T],
i,j = l «' = 1
(1)
then (x,t) can not be the minimum point of w in ft x (0,T]. In fact, at the
minimum point (x,t), it must hold that
wt < 0, wXi — 0
and
n
y~l a,ij(x,t,u, Vu)wXiXj > 0,
which contradicts (1).
Now suppose that u takes the minimum at (x°,t°) G ft x (0, T] and
u(x°,t°) < minw.

512
Take a > 0 large enough so that
aan(x,t,u, Vw) + bi(x,t, u, Vw) > 0, for (x,t) G fi x [0,T].
Then
w = u- eeaXl
takes its minimum at some point in fi x (0, T], if £ > 0 is sufficiently small.
However, for the function w we have
n n
w* - ^2 a<j(:M,u, Vu)iuXi.Xj. - 7^6,(3:,^, n)wx,-
> ea(an(a;,i,M, Vw)a + bi(x,t, u))eaXl
> 0, forall(a;,i)Gfi x (0,T].
This is in contradiction with w having minimum point in fi x (0, T].

513
SECTION 4
HYPERBOLIC EQUATIONS
6401
Assume that g is smooth and in £1(2R"), n arbitrary. Let u(£,t) be the
Fourier transform (with respect to x) of the solution to the n-dimensional wave
equation:
utt — c2Au = 0,
u(a:,0) = 0,
ut(x,0) = g(x).
Show that the Lp norm of u(-,t) at time t for p > n is bounded by a constant
times tl~nlP.
{Indiana)
Solution.
Transforming the equation and the initial conditions, we have
utt + c2\t\2u=0,
S(£,0) = 0,
and
u(U)=~9($)sMc\m-
Then
g G £1(2R") implies that g(£) is bounded in Mn. Therefore, it holds with a
positive constant M that
Set j] — c£t in the integrand. Then we get

514
p > n ensures the convergence of the integral. The above estimation gives the
conclusion of the problem.
6402
Consider the Cauchy problem for the wave equation
Utt-Au = 0, u = u(x,t), xeMn,t>0
u(x,0) - u0(x),
M((a;,0) = ui(x).
Assume uq,ui S S.
(a) Use the Fourier-transform to show that the total energy at time t
E{t) =\ I u2(x,t) + \Vxu(x,t)\2dx
is constant in time.
(b) Let Eq denote the constant energy in (a), i.e., E(t) = Eq- Show that
lim I u2(x,t)dx — lim / \Vxu(x,t)\2dx — E0.
t^'00J]Rn t^'0OJjrtn
(Iowa)
Solution.
(a) Taking the Fourier-transforms of the wave equation and of the initial
conditions with respect to the variable x ~ (xi, ■ ■ ■, xn), we have
utt + \t\2u(t,t) = (},
«(£,o) = «o(£),
2,(^0) = ^(0,
where £ = (£1,---,61)- Solving the above initial problem of the ordinary
differential equation, we get
G(£, t) = MO cos(K|<) + ^ sin(K|i)
and
ut(t,t) = -So(OKhin(K|i) + «i(0cos(K|i). (1)

515
Therefore
/ (R£,*)I2 + KI2K^)I2K = I (MOI2 + KI2MOI2K-
Then we can complete the proof of (a) by the Plancherel theorem,
(b) From (1), we have
\ut(t,t)\2 = |Go(£)|2K|2sin2(|£|i) +1^(01^082(1^1^)
-2Re(M0wi) sin(|£|i) cos(|£|i)
= ^|2o(^)|2|^|2(l - cos(2^|t)) + i|Si(^)|2(l + cos(2|^|<))
—Re(woMi) sin(2|£|2)
= ^(I2i(0l2 + KI2|2o(OI2) ~ ^o(OI2cos(2|^|i)
+ ^|«i(0|2cos(2|£|i) - Re(So5i)8in(2K|t).
Integrating the above equation on Mn and noting that
lim / |G0(0|2K|2cos(2pK = Km / |Gi(£)|2 cos(2p)d£
= lim / Re(G0Mi)sin(2|^|i)c^
t^°° JjR*
= o,
we obtain
lim/ |^,i)|2^=i/ (|fii(0l2 + KI2M0l2K-
Then the Plancherel theorem gives us
lim / Ut(x,t)dx - E0.
Combining this with (a), we finish the proof of (b).
6403
Consider the initial problem for the wave equation in three space variables:
uXlx2 + mX2x2 + Miju — utt = 0; x G M ,2 > 0,
u(x,G) = <j>{x); xeM3,
ut(x,Q) = rl>(x); x£M3.

516
(a) Write down the formula for the solution of the problem.
(b) If <j> and r/> vanish outside a ball of radius 3 centered at the origin, find
the set of points in M3 where you are sure that u vanishes when t ~ 10.
(c) If 4> vanishes everywhere in 2R3, and
{0, for \x\ < 1,
Jb, for 1 < \x\ < 2,
0, for 2 < |x|,
where k is a constant, find w(0, t) for alH > 0. (Your answer should be explicit,
no integrals).
(Indiana-Purdue)
Solution.
(»)
(b) By applying the domain of dependence for the solutions to the wave
equation, we can verify that u vanishes when t — 10 for \x\ < 7 or \x\ > 13.
In fact, when \x\ < 7, it holds that
\y\ > 3 for all y 6 {y : \y - x\ - 10}.
Therefore
f 4(y)ds = f i>(y) = 0.
J\y-x\ = 10 J\y-x\=10
Similarly, we can show that the above equality holds when \x\ > 13.
(c) It is easy to verify that
u
TO, 0 < t < 1,
,(0, <) = < kt, 1 <t < 2,
I 0, 2 < t.
6404
Let fi be the upper-half space in 2R"
fi = {(xi,x2,x3) £ 1R ;x3 > 0}

517
2 + ^-2 + ^7^ - 1SI7 =0; a=Gfi,i>0,
Consider the initial-boundary value problem
d2u d2u d2u d2u
dx\ + dx\ + dx2 m2
u(x,0) = </>(a;), ut(x,0) = r/>(x); x£$l,
u(x,t) = 0, x£dtt,t>0.
(a) Write down a formula for the value u(x,t) of the solution at (x,t),
x £ fi, when
t < x3.
(b) Find a formula for the value u(x,t) of the solution at (x,t), x S fi for
alii > 0.
(Indiana-Purdue)
Solution.
(a) When x 6 fi and 0 < t < x3, the domain of dependence for the value
u(x, t) at (x, t) is in fi. The value u(x, t) is given by the formula for the Cauchy
problem of the equation:
(b) Define $(¾) and ¢(¾) as both odd functions with respect to x$ in M3
as follows:
¢(3,)-/^1^2,3:3), x3 > 0,
\ -$(xi,x2,-i
X3), x3 < 0,
x3>0,
2:3 < 0,
[ -1/-(2:1,2:2,-2:3),
respectively. Then the value u(x, t) is given by
"(M)=^/,.,,./(^+1 (jk /,_„./H •
If both <f> and V" satisfy certain compatibility conditions, it is easy to see
that u(x, t) given by the above formula satisfies the wave equation and the
initial conditions. When x3 = 0,
f <Sl(y)dS = f rKvuV2,V3)dS
J\y-x\=t Jyi = y/t2-(yi-xi)2-(y2-X2)2
~ ^(2/1,2/2,-2/3)^ = 0.
Jyi = -y/t2-(yi-xi)2-(y2-x2)2

518
In a similar way, we have
J $(y)ds — 0, when x3 = 0.
J\y-x\=t
Therefore, u(x, t) satisfies the boundary condition on x3 = 0.
6405
Let u(x,t) be a solution of class C2 to the Cauchy problem
utt = Au (x£lRn,t>Q),
u(x,Q) = 4>(x), ut(x,0) = i>(x); 4>,i>eC°°,
with compact support. Define the local energy by
ER(u(t)) = I f (u2t+ \Vu\2)dx.
L J\x\<R
(i) Use the energy indentity to show that
£jj-t(«(0)) < ER(u(t)) < ER+T(u(0)).
Conclude that the total energy £?oo(w(i)) is conserved,
(ii) When n = 3, evaluate lim ER(u(t)).
t—»oo
(iii) Do (ii) above when n — 1.
(iv) Let n — 2, <f> = 0 and suppose that if) = 0 for \x\ > k, where x =
(^1,^2)- Show that u satisfies
u(x,t)\ <c(t + r + 2k)-1/2(t-r + 2k)-V2 [ \r/>(x)\dx
(Indiana)
/|x|
for some constant c, on the set r < t — 4k. (Here r — \x
Solution.
(i) ForiG [0,T], we have
ER-T+t(u(t)) =lf f (u2 + \Vu\2)dSdp.
* JO J\x\ = p

519
By the above expression, we can calculate that
^-T+t(u(*)) = \f (u2t + \Vu\2)dS
al z J\x\ = R-T+t
+ / (ututt + Vw • Vut)dx
J\x\<R-T+t
= \ I («? + |V«|2)d5
+ / (utAu + Vu-Vut)dx
= I f (u2 + \Vu\2 + 2utp)dS
1 J\x\=R-T+t \ on J
= -/ y2(ut cos(n, xt) + uXi)2dS > 0.
1 J\X\ = R-T + t i=1
Hence
^«-t(«(0)) < ER„T+t(u(t)) < ER(u(t)).
Similarly, we can show that for t G [0, T]
-%!-((«(*)) < 0.
Hence
ER+T_t(u(t)) < ER+T(u(0)), Vt G [0,T].
Then the estimation
ER(u(t)) < ER+T{u(0))
is obtained from the clear inequality
ER(u(t)) < ER+T-t(u(t)), V* G [0, T].
It is clear that the total energy E^uft)) equals to .Z?oo(w(0)).
(ii) From the formula of the solution to the problem
it is easy to verify
£^(2)) = 0 ifi>i? + a,

520
where a is the radius of a ball in which supp<^> U supp^ is contained. Hence
lim ER(u(t)) = 0.
t—*QO
(iii) By the formula
1 1 fx+t
u(x,t) = -0(x -t) + <j>(x+t))+-J i>(y)dy,
it is easy to evalute
lim ER(u(t)) = 0.
t—>CO
(iv) When n — 2, the solution is given by
** J\y-x\<t \/t2- \y-A
If \x\ < t — 4k, it is easy to see that
{y: \y\<k}c {y : \y-x\ <t}.
Hence the formula of the solution can be rewritten as
u(x,t)= — j
'\y\<k \A2- \y~x\2
When \x\ < t — 4k and \y\ < k, we have
\y — x\ < \x\ + 2k,
and hence
dy.
<
yjt2 - \y-x\2 " y/(t+ \x\ + 2k)(t- \x\-2k)
It is easy to verify that
V t-\x\-2k
Therefore, we obtain
V3
t - |a;| + 2fc /-,,
u(x,t)\ < y^(t+\x\ + 2k)-Ht-\x\ + M)-* [ \i>(y)\dy
2* J\y\<k
I»l5
W- \x\ >4k.

521
6406
Let fi C Mn be a bounded domain and assume u(x,t) is a function with
u £ C2(0 x [0, oo)), which solves the equation
utt — Am = u
with the boundary condition u\en = 0. With
E(t) = - I u2(x,t) + \Vxu(x,t)\2dx
1 J a
prove that there is a C < oo such that
E(t) < exp(Ct)E(0)
for all t > 0.
(Iowa)
Solution.
Multiplying the equation by m and integrating the resulting equation with
respect to x in fi, we have
—E(t) = / uutdx.
at Ja
Here we have applied the Green's formula:
/ ut&vdx = / VXW( • Vxudx.
Ja Ja
By the Poincare inequality, we can conclude that
/ uutdx < - I (u2(x,t)+ u2(x,t))dx
\J(l 2 Jn
< CE(t).
Thus we obtain a differential inequality for E(t)
jtE(t) < CE(t).
Then E(t) < exp(Ct)E(Q) is an immediate consequence of the above
differential inequality.

522
6407
(a) Find the explicit form of the solution of the initial value problem
V2u = utt, xeM3,t>(i,
u(a:,0) = —, xeM3,
r
f'(r)
ut(x,o) = -^-^-, xem3,
where r2 = x^+x^ + x^, and /(77) is a C3 function on the real line with compact
support such that
/(77) = 0, 77 < 0,
f{(/')2 + ^/'-/)2} ^=1,
where /'(77) is the derivative of / with respect to 77.
(b) What is its energy at time t — 10.
(Indiana-Purdue)
Solution.
Find the spherically symmetric solution to the problem. In the symmetric
case, the equation can be written as
d2u 2 du
or2 r or
Set w = ru. Then w solves the following problem:
wrr = wtt, r > 0,i > 0,
w = 0, r = 0,
w = f(r), wt = -f'(r), t = 0.
Solving this problem, and then we can get
u(xt)-l lrf(T-t), r>t,
(b) The energy of the problem is given by
E(t) =\ I (u2 + \Vu\2)dx.
It is easy to verify that
E(t) - E{0), Vt > 0.

And therefore
523
i»00
■/ (f'(r)2 + r-2(f-rf')2)dr
Jo
2\
I r2dr
= 2x
= 2x.
6408
Consider the wave equation in M3:
utt - Am = 0 for x 6 2R3,i > 0,
u(x,0) =0,
w*(z,0) = g(x),
where g S Co°(2R3). Prove that there exists a constant C depending only on
the given data such that
Q
sup |u(a;,2)| < —, t > 0.
(Indiana)
Solution.
Let i? and M be positive constants such that
suppgf C BR = {x G 2R3, \x\ < R}
and
|ff(a;)|<Af, VxeM3.
Prom the Poisson's formula, we have
4*t Ji«-x|=t
'|y-x| =
It is easy to verify that
The area of the intersection {y S M3, \y — x\ = t} D Bj?
< The area of dBR.

524
Therefore, we obtain
\u(x,t)\<-—-M-^R2 = ^^-, t>0.
4x1 t
6409
Let m S C2(Mn+1(x,t)) be asolution of the equation utt — Au = 0. Suppose
also that u — Ut = 0 on the ball B = {(x, 0) : |x| < to} in the plane 2 = 0.
Prove that u vanishes in the conical region
D - {(x,t) : 0 < t < t0 and \x\ <t0- t}.
(Indiana)
Solution.
Let
fir = {(x,t) € iR"+1, |x| <t0-t,t = r}
for 0 < r < to- Multiplying the wave equation by ut, and integrating the
resulting equation with respect to x in fir, we can obtain
2
Set
1 t f\ t n
- / -^-(ut + \VU\2)(X> r)dx -I ut V] wx>. cos(n, Xi)dS = 0.
2 ./nr or J9iir ^
^W= 5 I (u2 + \Vu\2)(x,T)dx.
1 Jilr
It is not difficult to verify that
dE(r) It 3 . 2 .„ l2w XJ
-\l (u2+\Vu\2)(x,r)dS.
Therefore, we have
—j—t — -» / (u2 + |Vn|2 - 2ut y^ttxj cos(n,Xj))dS
l y "
= -- / V](w(Cos(n, a;,-) - uXi)2dS < 0, V0 < r < t0.
2 ^anr ,=1

525
It is clear that E(0) — 0. Hence
E(t) = 0, V0 < r < t0.
This implies that
u - 0 in D.
6410
Let u(x,t) be a solution of the Cauchy problem.
utt = Au (¾ G M3,t > 0),
«(s,0) = /0=), ut(x,0) -=g(x).
Assume that /,(/6 C^°(iR3).
Prove that if
lim I u2dx = 0,
then
/ grdx = 0.
Solution.
Let i? > 0 such that
supp/, suppgf C BR = {x G JR3, |x| < R}-
It is not difficult to verify that
supptt(-,i) C {x G M3,t- R < \x\ < t + R}
for any t > R.
By using Green's formula, we get the following equality from the equation
I Uttdx = / Auda; = 0.
Jm.* Jm*
Therefore, we have
d f
— / utdx = 0,
at JR*
I Utdx = I gdx
JiRi Jut*
(Indiana)

526
1/2
ana
I udx = / fdx+t / gfda; (1)
•/jR* ^jR* ^jR4
by the initial conditions.
In the other side, by Cauchy inequality it follows that
I udx < I I u2dx)(/ dx
JR.* KJlR* J \Jt-R<\x\<t+R t
= Qxi?y (3i2 + i?2)^||«||L2(JRS), (2)
for any t > R. This contradicts (1) provided that
I gdx ^ 0.
Jm.*
6411
Consider the solution of the wave equation
% = An (xeM3,t>Q)
with data u(x, 0) = 0, u*(a;, 0) = ij>(x) G C£°.
a) Use Fourier transform to show that there is a positive constant c such
that
Km / u dx — c.
b) Show that there is a positive constant Ci such that
t ■ max|w| > Ci
X
for all sufficiently large t.
(Indiana)
Solution.
a) Let u denote the Fourier transform of u with respect to the variable
x — (xi, ■■ ■, xn). By taking the Fourier transforms of the wave equation and
the initial conditions, we can get

527
Therefore
Noting
we have
Then we get
where
f u2(x,t)dx = (2x)-3 f \u(i,t)\2d4
JlR* JlR*
Jm* Kr
lim f Kr2|^(O|2cos(2^|iK = 0.
lim I u2(x,t) = c,
if v 7*0.
b) Let fi(i) be the support of u(x,t) with respect to the variable x =
(2:1,2:1,2:3) for t > 0. If the conclusion is false, then there exists a sequence
{£„} such that
t„ —► 00, as n —> 00
and
This gives
tn -max|u(:M„)| < |fi(*n)| '•
X
Then we have
I U2(x,tn)dx < I \n(t„)\-H-2dx
= t~2 —► 0, asn^oo.
This contradicts (a). The proof of (b) is completed.

528
6412
Consider the Cauchy problem for the wave equation
u«-Au = 0 (xeMn,t>0),
u(x,0) - 4>(x), ut(x,0) = ip(x).
a) Let n = 3; $,i/> G Cg°(2R3). Let 0 < 1. Determine the large time
behavior of the integral
I u (x,t)dx.
J\x\<9t
b) Let n = 2, <j> = 0, V G C§°(2R2). Show that for any 0 < 1, we have
sup \u(x,t)\ = 0{t~l) as 2 —► oo.
|x|<0*
(Indiana)
Solution.
a) Assume that R > 0 is so large that
supp<£ U suppV" C {x G JR3, |x| < R}.
From the Huygens's principle for the 3-dimensional wave eauation, we conclude
that if t > R and
\x\<t — R,
then
u(x,t) = 0.
Therefore, when t > y^, it holds that
u(x,t) = 0, V\x\<6t,
which implies that
t R
/ u4(x,t)dx — 0, when t > -.
J\x\<0t 1 ~ V
b) From the Poisson's formula, the solution to the 2-dimensional problem
is given by
, ., 1 f ^(2/1,2/2) ,
u(Xl, x2, t) = — , 2-dy.
to J\y-x\<t v*2 -\y- x\2

529
Let R > 0 be so large that
suppV" C {x G M2, \x\ < R}.
When t > -^g, it is clear that if \x\ < Ot and \y\ < R then
}y-x\<6t + R
and
Vt2~\y~x\2> yft2-{et + Ry.
Therefore, we have
i / -m . 1 f 1^(2/1,2/2)1 ,
2x ^|y|<i* V* - (e< + #)
< C =, Vi> —^r.lajl <9t,
where C is a constant. The proof is completed.
6413
The linear transport equation is following PDE for a scalar function / of
seven variables x, v, t (x G M3,v G 2R3,i G 1R\)'-
ft + v-Vxf = 0.
Consider the initial-value problem for this; i.e., let
f(x, v, 0) =f (x, v); f given, /€ C1.
0
a) Show that, if /> 0 for all x, v, then the same is true for / at all later
times.
b) Define the local density p by
p(x,t) = I f(x,v,t)dv.
/jr»
o
Suppose that 0 <f (x,v) < h(x), where h G £1(2R3). Show that
sup p =0(2-3) as t —> oo.
X
{Indiana)

530
Solution.
a) The characteristic curves of the equation are defined by
dx
—- = v
dt
and along which it holds that
Therefore, the forward characteristic curve departing from t = 0, x = a is
x = a + tv. (2)
From (1) and (2), we get the solution to the initial-value problem
f(x,v,t)=f(x-tv,v). (3)
It is clear that / > 0 if /> 0.
b) From (3), we have
P = f(x,v,t)dv
JlR*
= / f (x — tv,v)dv
< I h(x — tv)dv
Jm.*
= — I h(u)du.
This completes the proof.
6414
Solve the Cauchy problem
xux + yuuy = -xy (x > 0)
u = 5 on xy = 1,
if a solution exists.
(Indiana)
Solution.
The system of the characteristic equations for the problem is given by

531
dx __ dy du
ds
with the initial conditions
= x, — = yu, — = -xy
ds ds ds
s = 0 : x = a, 2/=-, u = 5.
a
From the above Cauchy problem, we have immediately
x = ae*
and
du .du d . , .
u— = -aeg— = —j-(,ae y + u).
ds ds ds
This implies that
1 i
~uz + u + xy=C,
where C is a constant. The initial data gives C = ^-. Then we obtain
immediately
u= -l + ^/38-2a;j/.
6415
Compute an explicit solution u(x, t) to the initial-value problem
ut + (¾)2 = 0,
u(x,0) = x2.
(Indiana)
Solution.
Set v = ux. Differentiating the equation and the initial condition with
respect to x, we get
vt + 2vvx = 0,
v(x,0)=2x. (1)
The characteristic curve of the quasilinear equation (1) is given by
dx
-dt=2^

532
along which it holds that
£ = 0.
dt
From the above characteristic equations, we can obtain immediately
2x
4<+l
Hence
x2
U :
4< + l
6416
Solve the following initial value problem.
Ut + uux = 0, x £ M, t > 0,
( 1, x<Q,
u(x,Q) = I 1-x, 0<x<l,
[ 0, x>l.
Show that the continuous solution exists only for a finite time, and find the
discontinuous entropy solution, giving explicitly the discontinuity curve and
the Rankine-Hugoniot jump condition along it.
(Indiana-Purdue)
Solution.
The characteristic curve departing from t = 0, x = /3 is given by
dx
— = «, t = 0 : x = /3, (1)
along which u is a constant, i.e.,
^ = 0, t = 0:u = u°C9), (2)
where u°(x) = u(x, 0). From (1) and (2), we have
x = u°(fi)t+P, u = u0((3),
and
r t + p, /3<o,
x = { (l-p)t + p, 0</3<l,
1/3, P > 1,
1, /3 < 0,
1-/3, 0</3<l,
0, /3 > 1.

533
For 0 < t < 1, 0 < /3 < 1, we have
/3 =
x — t
and hence
u= 1
1-t
X — t 1 — X
1-t 1-t
Therefore we get the continuous solution to the problem
1, x <t,
'X
-t 1
0, x > 1,
u- { ^Ef, t < X < 1,
for 0 < t < 1 (see Fig.6.2).
t=i
Fig.6.2
When t = 1, the continuous solution to the problem blows up.
For t > 1, the Rankine-Hugoniot condition along the discontinuous curve
is given by
dx (u\ ul\
i.e.,
dx 1
- = -(M++M_).
Noting m+ = 0, w_ = 1, we have
dx 1
"S" ~ 2'
Therefore, the discontinuous curve is
x = -(* + 1), t>l

534
and the entropy solution to the problem is given by
f 1, x<Ut + l),
u-\ 0, *> J(t + 1).
6417
a) Let u(x,t) be a smooth solution of
-£ + a(x,u)-^--bu=0 forO<i<T,
at ox
where b is a constant and a is smooth. Show that if
a < u(x, 0) < /3 for all x,
then
aebt <u(x,t) </3eM
for all x and all t G (0, T).
b) Let u(x,t) be smooth solution of
— + a(u)— = 0 forO<i<T,
at ox
u(x, 0) = u0(x),
where wo and a are smooth. Show that, if x(t,y) is the characteristic through
(y, 0), then
Ux(Mt,y),t) = a , u*^]—rm-
(-^a(u0(x))\x=y) t + 1
i) Use this to compute the largest T for which a smooth solution can exist
up to time T.
ii) Show that, if 0 < ii < t2 < T, then
/OO i»0O
\ux(x,t2)\dx — I \ux(x, ti)\dx.
-oo ^ — oo
(Indiana)
Solution.
a) Let u = ebtv. Then v satisfies the following equation
ov / bt \ov
— + a(x, ebtv — = 0.
at ox

535
The solution v maintains a constant along a characteristic of the above
equation, defined by the equation
dx i bt \
Therefore, it holds that
inf v(x, 0) < v(x,t) < sup-y(a;, 0).
x X
This gives the estimation in the problem immediately.
b) u(x,t) is a constant along the characteristic x = x(t,y). Hence
u(x(t,y),t) = My) Vt€(0,T).
Differertiating above equality with respect to y, we have
ux(x(t,y),t)-^x(t,y) = u'0(y) Vt G (0,T).
The characteristic x(t,y) through (y, 0) is given by the following Cauchy
problem:
/ mx(^y) = a(uo(y)),
\ x{0,y) =y.
Differentiating this problem with respect to y, we get
/ ^(^(*,»))=a'M»)K(»),
\ ^(0,2/) = 1.
Then it follows from above problem that
Therefore, we have
■^-x(t,y) = a'(u0(y))u'0(y)t + 1.
oy
ux(x(t,y),t) = —— ° ■
a'(uo(y))u'0(y)t + 1
This is just that we want to show.
i) Let m — inf a'(uo(y))u'0(y). From the above expression, we can get the
y
following conclusions easily.
If m > 0, the Cauchy problem admits a global smooth solution in (0, oo);

536
If —oo < m < 0, the largest T for which the problem admits a smooth
solution in (0, T) is given by
mT+1 = 0.
ii) For any fixed t G (0,T), by the change of variables x = x(t,y) in the
integral, we have
/OO i»00 Q
\ux(x,t)\dx = / \ux(x(t,y),t)\—x(t,y)dy.
-oo ./-oo oy
Taking the expressions of ux(x(t,y),t) and -§rx(t,y) into account, we obtain
/OO i»00
\ux(x,t)\dx = / \u'o(y)\dy.
■OO J — OO
This completes the proof of the problem.
6418
(Uniqueness of weak solutions) Consider the Cauchy problem
f ut + ux = 0, onJRx [0,T],
(2)
u(x,0) = uo(x), on M x {0}.
A function w G £P(2R x [0, T]) is defined to be a weak solution of (2) iff
tT y + oo
/ / u(y,8)(-<f>t(y,s)-$x(y,8))dyd8
JO J -oo
+ I u0(y)4>(y, o)dy = o (25)
J — oo
for all test functions <f>(x,t) G C°°(M x [0,T]) which are compactly supported
and vanish on {t = T}.
If «o(k) = 0, prove that the only LP solution of (2') is u = 0.
(Indiana)
Solution.
Consider the following Cauchy problem:
f jt + fa = f(t,x), onJRx [0,T],
\ 4>(x,T) = 0, onJRx {T},

537
where / G C°°{1R x [0, T]) with a compact support.
It is not difficult to show that the above Cauchy problem admits a solution
<t> G C°°(IR x [0, T}) which is compactly supported. Therefore, we have
,T ,+ 00
/ / u(y, s)f(y, s)dyds = 0 (*)
JO J -oo
for all function / G C°°(]R x [0,T]) with a compact support, provided uq{x) =
0. As the space C^°(2R x [0,T]), consisting of all C°°-functions with compact
support in M x [0,T], is dense in Lq(M x [0,T]), where q = p/(p - 1), the
equality (*) is valid for all functions / G Lq(M x [0, T]). Hence we have u = 0.
6419
Consider the Cauchy problem for a function u = u(x,t), where x G M and
t > 0.
f ut + a(x,t)ux = b(x,t),
\ u(x,0)-uo(x) eC(M).
a) Show that if b = 0, then TV(u(-,i)) < TV(m0(-)) for each t > 0 (where
TV = "total variation").
b) Show that when 6^0, one still has the bound
TV(u(;t))<TV(u0)+ ( TV(b(-,s))ds.
Jo
(Indiana)
Solution.
The characterstic curves of the equation are given by the equation
dx
^ = aOM),
along which the solutions to the equation satisfy
du ,,
-=&(M).
a) Suppose that 6 = 0. Then u is a constant along a characteristic curve.
We divide the line t = r into subintervals by the points • • ■, x_„, ■ ■ •, x_i, xo>
xi, ■ • •, xn, ■ ■ -. Suppose that the backward characteristic curve through (a;,-, r)
meets the line t = 0 at (£j, 0) (i = • • •, —n, • • •, —1,0,1, ■ • •, n, ■ • •). Then we
have
u(xi,r) = u0(£i),

538
and
+00 +00
Y^ \u(xi+i,T)-u(Xi,T)\ = ^2 |Mo(6-+l)-Mo(6)|-
i= — 00 i= — 00
Therefore, we get
TV(u(-,t))<TV(u0(-)), Vr>0.
b) Let {xi} and {£,•} be given as in a). When b ^ 0, we have
u(xi,r) — uo(£«) + / 6(aT,(s), s)ds,
Jo
where by x = aT,(i) we denote the characteristic curve through (x,-,r). From
the above inequalities, we get
+00
X |u(a;i+i,T)-u(a;,-,T)|
i = —00
+ 00 «r +00
< X K(£>-+i)- Mo(6)l+ / X l&(£«"+i(*)>*)-&(£••(*)>*)lds
< TV(u0)+ [T TV(b(-,s))ds.
Jo
Then we can complete the proof easily.

539
Abbreviations of Universities in This Book
Cincinnati University of Cincinnati
Columbia Columbia University
Courant Inst. Courant Institute of Mathematical Sciences,
New York University
Harvard Harvard University
Illinois University of Illinois (Urbana)
Indiana Indiana University (Bloomington)
Indiana-Purdue Indiana University-Purdue University (Indianapolis)
Iowa University of Iowa (Iowa City)
Minnesota University of Minnesota (Minneapolis)
Rutgers Rutgers University
Stanford Stanford University
SUNY, Stony Brook State University of New York (Stony Brook)
Toronto University of Toronto
UC, Irvine University of California (Irvine)

Major American Universities Ph.D.
Qualifying Questions and Solutions
PROBLEMS and
SOLUTIONS in
MATHEMATICS
This book contains a selection of more than 500
mathematical problems and their solutions from
the Ph.D. qualifying examination papers of more
than ten famous American universities. The
problems cover six aspects of graduate school
mathematics: Algebra, Topology, Differential
Geometry, Real Analysis, Complex Analysis and
Partial Differential Equations. The depth of
knowledge involved is not beyond the contents of
the textbooks for graduate students, while solution
of the problems requires deep understanding of
the mathematical principles and skilled techniques.
For students this book is a valuable complement
to textbooks; for lecturers teaching graduate school
mathematics, a helpful reference.
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