Graduate Texts in Mathematics 12

Managing Editors: P. R. Halmos C. C. Moore

Richard Beals

Advanced Mathematical Analysis Periodic Functions and Distributions, Complex Analysis, Laplace Transform and Applications

Springer Science+Business Media, LLC

Richard Beals Professor of Mathematics University of Chicago Department of Mathematics 5734 University Avenue Chicago, Illinois 60637

Managing Editors

P. R. Halmos Indiana University Department of Mathematics Swain Hali East Bloomington, Indiana 47401

AMS Subject Classification

C. C. Moore University of California at Berkeley Department of Mathematics Berkeley, California 94720

46-01, 46S05, 46C05, 30-01,43-01 34-01,3501

Library of Congress Cataloging in Publication Data

Beals, Richard, 1938-Advanced mathematical ana!ysis.

(Graduate texts in mathematics, v. 12) 1. Mathematica! analysis. 1. TitIe. II. Series.

QA300.B4 515 73-6884

AII rights reserved.

No part of this book may be trans!ated or reproduced in any form without written permission from Springer-Verlag.

© 1973 by Springer Science+Business Media New York Originally published by Springer-Verlag New York Inc in 1973

ISBN 978-0-387-90066-7 ISBN 978-1-4684-9886-8 (eBook) DOI 10.1007/978-1-4684-9886-8

to Nancy

PREFACE

Once upon a time students of mathematics and students of science or engineering took the same courses in mathematical analysis beyond calculus. Now it is common to separate" advanced mathematics for science and engi­neering" from what might be called "advanced mathematical analysis for mathematicians." It seems to me both useful and timely to attempt a reconciliation.

The separation between kinds of courses has unhealthy effects. Mathe­matics students reverse the historical development of analysis, learning the unifying abstractions first and the examples later (if ever). Science students learn the examples as taught generations ago, missing modern insights. A choice between encountering Fourier series as a minor instance of the repre­sentation theory of Banach algebras, and encountering Fourier series in isolation and developed in an ad hoc manner, is no choice at all.

It is easy to recognize these problems, but less easy to counter the legiti­mate pressures which have led to a separation. Modern mathematics has broadened our perspectives by abstraction and bold generalization, while developing techniques which can treat classical theories in a definitive way. On the other hand, the applier of mathematics has continued to need a variety of definite tools and has not had the time to acquire the broadest and most definitive grasp-to learn necessary and sufficient conditions when simple sufficient conditions will serve, or to learn the general framework encompass­ing different examples.

This book is based on two premises. First, the ideas and methods of the theory of distributions lead to formulations of classical theories which are satisfying and complete mathematically, and which at the same time provide the most useful viewpoint for applications. Second, mathematics and science students alike can profit from an approach which treats the particular in a careful, complete, and modern way, and which treats the general as obtained by abstraction for the purpose of illuminating the basic structure exemplified in the particular. As an example, the basic L2 theory of Fourier series can be established quickly and with no mention of measure theory once L 2(O, 21T) is known to be complete. Here L2(O, 21T) is viewed as a subspace of the space of periodic distributions and is shown to be a Hilbert space. This leads to a dis­cussion of abstract Hilbert space and orthogonal expansions. It is easy to derive necessary and sufficient conditions that a formal trigonometric series be the Fourier series of a distribution, an L2 distribution, or a smooth function. This in turn facilitates a discussion of smooth solutions and distri­bution solutions of the wave and heat equations.

The book is organized as follows. The first two chapters provide back­ground material which many readers may profitably skim or skip. Chapters 3, 4, and 5 treat periodic functions and distributions, Fourier series, and applications. Included are convolution and approximation (including the

vii

viii Preface

Weierstrass theorems), characterization of periodic distributions, elements of Hilbert space theory, and the classical problems of mathematical physics. The basic theory of functions of a complex variable is taken up in Chapter 6. Chapter 7 treats the Laplace transform from a distribution-theoretic point of view and includes applications to ordinary differential equations. Chapters 6 and 7 are virtually independent of the preceding three chapters; a quick reading of sections 2, 3, and 5 of Chapter 3 may help motivate the procedure of Chapter 7.

I am indebted to Max 10deit and Paul Sally for lively discussions of what and how analysts should learn, to Nancy for her support throughout, and particularly to Fred Flowers for his excellent handling of the manuscript.

Richard Beals

TABLE OF CONTENTS

Chapter One Basis concepts

§1. Sets and functions . §2. Real and complex numbers §3. Sequences of real and complex numbers §4. Series §5. Metric spaces §6. Compact sets §7. Vector spaces

Chapter Two Continuous functions

§1. Continuity, uniform continuity, and compactness §2. Integration of complex-valued functions §3. Differentiation of complex-valued functions . §4. Sequences and series of functions . §5. Differential equations and the exponential function §6. Trigonometric functions and the logarithm §7. Functions of two variables §8. Some infinitely differentiable functions

Chapter Three Periodic functions and periodic distributions

§1. Continuous periodic functions §2. Smooth periodic functions . §3. Translation, convolution, and approximation §4. The Weierstrass approximation theorems §5. Periodic distributions §6. Determining the periodic distributions §7. Convolution of distributions §8. Summary of operations on periodic distributions

Chapter Four Hilbert spaces and Fourier series

§1. An inner product in <'C, and the space 22 §2. Hilbert space §3. Hilbert spaces of sequences §4. Orthonormal bases . §5. Orthogonal expansions . §6. Fourier series

ix

5 10 14 19 23 27

34 38 42 47 51 57 62 67

69 72 77 81 84 89 94 99

103 109 113 116 121 125

x Table of Contents

Chapter Five Applications of Fourier series

§1. Fourier series of smooth periodic functions and periodic dis-tributions 131

§2. Fourier series, convolutions, and approximation 134 §3. The heat equation: distribution solutions 137 §4. The heat equation: classical solutions; derivation 142 §5. The wave equation . 145 §6. Laplace's equation and the Dirichlet problem 150

Chapter Six Complex analysis

§1. Complex differentiation §2. Complex integration §3. The Cauchy integral formula §4. The local behavior of a holomorphic function §5. Isolated singularities §6. Rational functions; Laurent expansions; residues §7. Holomorphic functions in the unit disc .

Chapter Seven The Laplace transform

§1. Introduction §2. The space 2 . §3. The space 2' §4. Characterization of distributions of type 2' §5. Laplace transforms of functions . §6. Laplace transforms of distributions §7. Differential equations .

Notes and bibliography

Notation index

Subject index

155 159 166 171 175 179 184

190 193 197 201 205 210 213

223

225

227

Advanced Mathematical Analysis

Chapter 1

Basic Concepts

§1. Sets and functions

One feature of modern mathematics is the use of abstract concepts to provide a language and a unifying framework for theories encompassing numerous special cases and examples. Two important examples of such concepts, that of "metric space" and that of "vector space," will be taken up later in this chapter. In this section we discuss briefly the concepts, even more basic, of "set" and of "function."

We assume that the intuitive notion of a "set" and of an "element" of a set are familiar. A set is determined when its elements are specified in some manner. The exact manner of specification is irrelevant, provided the elements are the same. Thus

A = {3, 5, 7}

means that A is the set with three elements, the integers 3, 5, and 7. This is the same as

A = {7, 3, 5}, or

A = {n I n is an odd positive integer between 2 and 8}

or A = {2n + 1 I n = 1,2, 3}.

In expressions such as the last two, the phrase after the vertical line is sup­posed to prescribe exactly what precedes the vertical line, thus prescribing the set. It is convenient to allow repetitions; thus A above is also

{5, 3, 7, 3, 3},

still a set with three elements. If x is an element of A we write

xEA or A3X.

If x is not an element of A we write

x¢ A or A ~ x.

The sets of all integers and of all positive integers are denoted by 71. and 71. + respectively:

71. = {O, 1, -1,2, -2,3, -3, ... }, 71.+ = {I, 2,3,4, ... }.

As usual the three dots ... indicate a presumed understanding about what is omitted.

1

2 Basic concepts

Other matters of notation:

o denotes the empty set (no elements). Au B denotes the union, {x I x E A or x E B (or both)}.

A ('\ B denotes the intersection, {x I x E A and x E B}.

The union of AI> A2, ... , Am is denoted by m

Al U A2 U Aa U ... U Am or U A j ,

1=1

and the intersection by m

Al ('\ A2 ('\ Aa ('\ ... ('\ Am or n A j •

j=1

The union and the intersection of an infinite family of sets AI, A2 ... indexed by 7L. + are denoted by

co co

UA, and nAI. j=1 1=1

More generally, suppose J is a set, and suppose that for each j E J we are given a set AI. The union and intersection of all the Aj are denoted by

UAj and nAI. jel leI

A set A is a subset of a set B if every element of A is an element of B; we write

A c B or B::::> A.

In particular, for any A we have 0 c A. If A c B, the complement 0/ A in B is the set of elements of B not in A:

B\A = {x I x E B, x ¢ A}.

Thus C = B\A is equivalent to the two conditions

Au C = B, A('\C=0.

The product of two sets A and B is the set of ordered pairs (x, y) where x E A and y E B; this is written A x B. More generally, if Al> A 2 , ••• , An are the sets then

Al X A2 X ••• x An

is the set whose elements are all the ordered n-tuples (Xl> X2, ••• , xn), where each XI E A j • The product

AxAx···xA

of n copies of A is also written An. A/unction from a set A to a set B is an assignment, to each element of A,

of some unique element of B. We write

/:A-+B

Sets and functions 3

for a function f from A to B. If x E A, then f(x) denotes the element of B assigned by f to the element x. The elements assigned by f are often called values. Thus a real-valued function on A is a function f: A -r ~, ~ the set of real numbers. A complex-valued function on A is a functionf: A -r C, C the set of complex numbers.

A function f: A -r B is said to be 1-1 ("one-to-one") or injective if it assigns distinct elements of B to distinct elements of A: If x, YEA and x =1= y, thenf(x) =1= f(y). A functionf: A -r B is said to be onto or surjective if for each element y E B, there is some x E A such thatf(x) = y. A function f: A -r B which is both 1-1 and onto is said to be bijective.

If f: A -r Band g: B -r C, the composition of f and g is the function denoted by g 0 f:

g 0 f: A -r C, g 0 f(x) = g(f(x)), for all x E A.

Iff: A -r B is bijective, there is a unique inverse functionf- 1 : B -r A with the properties: f- 1 0 f(x) = x, for all x E A; f 0 f-1(y) = y, for all y E B.

Examples

Consider the functions f: Z -r Z +, g: Z -r Z, h: 7L. -r Z, defined by

f(n) = n2 + 1, nEZ, g(n) = 2n, nEZ, h(n) = 1 - n, n E Z.

Thenfis neither 1-1 nor onto, g is 1-1 but not onto, h is bijective, h- 1(n) =

1 - n, andfoh(n) = n2 - 2n + 2.

A set A is said to be finite if either A = 0 or there is an n E 7L. +, and a bijective function f from A to the set {I, 2, ... , n}. The set A is said to be countable if there is a bijective f: A -r Z +. This is equivalent to requiring that there be a bijective g: 7L. + -r A (since if such anf exists, we can take g = f- 1 ;

if such a g exists, take f = g-I). The following elementary criterion is convenient.

Proposition 1.1. If there is a surjective (onto) function f: Z+ -r A, then A is either finite or countable.

Proof Suppose A is not finite. Define g: Z+ -r Z+ as follows. Let g(1) = 1. Since A is not finite, A =1= {f(1)}. Let g(2) be the first integer m such thatf(m) =1= f(1). Having defined g(1), g(2), ... , g(n), let g(n + 1) be the first integer m such thatf(n + 1) ¢ {f(1),f(2), ... ,f(n)}. The function g defined inductively on all of Z + in this way has the property that fog: Z + -r A is bijective. In fact, it is 1-1 by the construction. It is onto becausefis onto and by the construction, for each n the set {f(1),f(2), ... ,f(n)} is a subset of {f 0 g(l),f 0 g(2), .. . ,f 0 g(n)}. 0

Corollary 1.2. If B is countable and A c B, then A is finite or countable.

4 Basic concepts

Proof If A = 0, we are done. Otherwise, choose a functionf: 71.+ ~ A which is onto. Choose an element Xo EA. Define g: 71.+ ~ A by: g(n) = fen) if/en) E A, g(n) = Xo iff(n) r# A. Then g is onto, so A is finite or countable. 0

Proposition 1.3. If AI, A2, A3, ... are finite or countable, then the sets

n ..

UAi and UAi i=1 i=1

are finite or countable.

Proof We shall prove only the second statement. If any of the Ai are empty, we may exclude them and renumber. Consider only the second case. For each Aj we can choose a surjective functionjj: 71.+ ~ Aj. Definef: 71.+ ~ Ui=1 Ai by f(l) = fl(1), f(3) = fl(2), f(5) = fl(3), . . ·,f(2) = f2(1), f(6) = f2(2), f(10) = f2(3), ... , and in general f(2 j(2k - 1» = fAk), j, k = 1,2, 3, .... Any x E Ui=l Aj is in some Ai> and therefore there is k E 71.+ such that jj(k) = x. Thenf(2 i(2k - 1» = x, sofis onto. By Proposition 1.1, Uj.,1 Ai is finite or countable. 0

Example

Let Q be the set of rational numbers: Q = {min I mE 71., n E 71. +}. This is countable. In fact, let An = {j/n I j E 71., _n2 ~ j ~ n2}. Then each An is finite, and Q = U:'=1 An.

Proposition 1.4. If AI> A2, .. . , An are countable sets, then the product set Al x A2 X ••• x An is countable.

Proof Choose bijective functions jj: A j ~ 71. +, j = 1, 2, ... , n. For each mE 71.+, let Bm be the subset of the product set consisting of all n-tuples (Xl' X2, ... , xn) such that each jj(Xj) ~ m. Then Bm is finite (it has mn ele­ments) and the product set is the union of the sets Bm. Proposition 1.3 gives the desired conclusion. 0

A sequence in a set A is a collection of elements of A, not necessarily distinct, indexed by some countable set J. Usually J is taken to be 71.+ or 71.+ u {O}, and we use the notations

(an):'=1 = (al> a2, a3,·· .), (an):' = 0 = (ao, al> a2,· .).

Proposition 1.5. The set S of all sequences in the set {O, I} is neither finite nor countable.

Proof Suppose f: 71. + ~ A. We shall show that f is not surjective. For each mE 71.+, f(m) is a sequence (an.m):'=l = (al.m, a2.m,' .. ), where each an.m is 0 or 1. Define a sequence (an):'=1 by setting an = 0 if an.n = 1, an = 1 if an•n = O. Then for each m E 71.+, (an):'=1 '" (an.m):'=l = f(m). Thusfis not surjective. 0

Real and complex numbers s

We introduce some more items of notation. The symbol ~ means "implies"; the symbol <= means "is implied by"; the symbol-¢> means "is equivalent to."

Anticipating §2 somewhat, we introduce the notation for intervals in the set IR of real numbers. If a, b E IR and a < b, then

Also,

(a, b) = {x I x E IR, a < x < b}, (a, b] = {x I x E IR, a < x ~ b}, [a, b) = {x I x E IR, a ~ x < b}, [a, b] = {x I x E IR, a ~ x ~ b}.

(a, (0) = {x I x E IR, a < x}, (-00, a] = {x I x E IR, x ~ a}, etc.

§2. Real and complex numbers

We denote by IR the set of all real numbers. The operations of addition and multiplication can be thought of as functions from the product set IR x IR to IR. Addition assigns to the ordered pair (x, y) an element of IR denoted by x + y; multiplication assigns an element of IR denoted by xy. The algebraic properties of these functions are familiar.

Axioms of addition

AI. (x + y) + z = x + (y + z), for any x, y, Z E IR. A2. x + y = y + x, for any x, y E IR. A3. There is an element 0 in IR such that x + 0 = x for every x E IR. A4. For each x E IR there is an element -x E IR such that x + (-x) = O.

Note that the element 0 is unique. In fact, if 0' is an element such that x + 0' = x for every x, then

0' = 0' + 0 = 0 + 0' = O.

Also, given x the element - x is unique. In fact, if x + y = 0, then

y = y + 0 = y + (x + (-x)) = (y + x) + (-x) = (x + y) + (-x) = 0 + (-x) = (-x) + 0 = -x.

This uniqueness implies -( -x) = x, since (-x) + x = x + (-x) = o.

Axioms of multiplication

Ml. (xy)z = x(yz), for any x, y, z E IR. M2. xy = yx, for any x, y E IR. M3. There is an element I =F 0 in IR such that xl = x for any x E IR. M4. For each x E IR, x =F 0, there is an element x- 1 in IR such that

xx- 1 = 1.

6 Basic concepts

Note that 1 and X-I are unique. We leave the proofs as an exercise.

Distributive law

DL. x(y + z) = xy + XZ, for any x, y, Z E IR.

Note that DL and A2 imply (x + y)z = xz + yz. We can now readily deduce some other well-known facts. For example,

O·x = (0 + o)·x = O·x + O·x,

so O·x = O. Then

x + (-I)·x = l·x + (-I)·x = (1 + (-I»·x = O·x = 0,

so (-I)·x = -x. Also,

(-x)·y = « -1)·x)·y = (-I)·(xy) = -xy.

The axioms AI-A4, MI-M4, and DL do not determine IR. In fact there is a set consisting of two elements, together with operations of addition and multiplication, such that the axioms above are all satisfied: if we denote the elements of the set by 0, 1, we can define addition and multiplication by

0+0 = 1 + 1 = 0, 0·0 = 1·0 = 0·1 = 0,

0+1 = 1 + 0,= 1, 1·1 = 1.

There is an additional familiar notion in IR, that of positivity, from which one can derive the notion of an ordering of IR. We axiomatize this by intro­ducing a subset P c IR, the set of "positive" elements.

Axioms of order

01. If x E IR, then exactly one of the following holds: x EP, x = 0, or -XEP.

02. If x, YEP, then x + YEP. 03. If x, YEP, then xy E P.

It follows from these that if x # 0, then x2 E P. In fact if x E P then this follows from 03, while if -x EP, then (_X)2 EP, and (_X)2 = -(x( -x» _(_x2) = x2• In particular, 1 = PEP.

We define x < y if y - x E P, X > y if y < x. It follows that x E P <0>

X > O. Also, if x < y and y < z, then

Z - x = (z - y) + (y - x) EP,

so X < z. In terms ofthis order, we introduce the Archimedean axiom.

04. If x, y > 0, then there is a positive integer n such that nx = x + x + ... + x is > y.

(One can think of this as saying that, given enough time, one can empty a large bathtub with a small spoon.)

Real and complex numbers 7

The axioms given so far still do not determine IR; they are all satisfied by the subset 0 of rational numbers. The following notions will make a distinc­tion between these two sets.

A nonempty subset A c IR is said to be bounded above if there is an x E IR such that every YEA satisfies y ::; x (as usual, y ::; x means y < x or y = x). Such a number x is called an upper bound for A. Similarly, if there is an x E IR such that every YEA satisfies x ::; y, then A is said to be bounded below and x is called a lower bound for A.

A number x E IR is said to be a least upper bound for a nonempty set A c IR if x is an upper bound, and if every other upper bound x' satisfies x' ~ x. If such an x exists it is clearly unique, and we write

x = lubA.

Similarly, x is a greatest lower bound for A if it is a lower bound and if every other lower bound x' satisfies x' ::; x. Such an x is unique, and we write

x = glbA.

The final axiom for IR is called the completeness axiom.

05. If A is a nonempty subset of IR which is bounded above, then A has a least upper bound.

Note that if A c IRis bounded below, thenthesetB = {x I x E IR, -x E A} is bounded above. If x = lub B, then - x = glb A. Therefore 05 is equiva­lent to: a nonempty subset of IR which is bounded below has a greatest lower bound.

Theorem 2.1. 0 does not satisfy the completeness axiom.

Proof Recall that there is no rational p/q, p, p E 7l., such that (p/q)2 = 2: in fact if there were, we could reduce to lowest terms and assume either p or q is odd. Butp2 = 2q2 is even, so p is even, so p = 2m, m E 7L.. Then 4m2 = 2q2, so q2 = 2m2 is even and q is also even, a contradiction.

Let A = {x I x EO, x 2 < 2}. This is nonempty, since 0, 1 EA. It is bounded above, since x ~ 2 implies x2 ~ 4, so 2 is an upper bound. We shall show that no x E 0 is a least upper bound for A.

If x ::; 0, then x < I E A, so x is not an upper bound. Suppose x > 0 and x 2 < 2. Suppose h E 0 and 0 < h < I. Then x + h E 0 and x + h > x. Also, (x + h)2 = x 2 + 2xh + h2 < x2 + 2xh + h = x 2 + (2x + I)h. If we choose h > 0 so small that h < I and h < (2 - x2)/(2x + 1), then (x + h)2 < 2. Then x + h E A, and x + h > x, so x is not an upper bound of A.

Finally, suppose x E 0, X > 0, and x 2 > 2. Suppose hE 0 and 0 < h < x. Then x - hE 0 and x - h > O. Also, (x - h)2 = x 2 - 2xh + h2 > x 2 -2xh. If we choose h > 0 so small that h < I and h < (x2 - 2)/2x, then (x - h)2 > 2. It follows that if yEA, then y < x-h. Thus x - h is an upper bound for A less than x, and x is not the least upper bound. 0

We used the non-existence ofa square root of2 in 0 to show that 05 does not hold. We may turn the argument around to show, using 05, that there is

8 Basic concepts

a real number x > 0 such that x2 = 2. In fact, let A = {y lYE IR, y2 < 2}. The argument proving Theorem 2.1 proves the following: A is bounded above; its least upper bound x is positive; if x2 < 2 then x would not be an upper bound, while if x2 > 2 then x would not be the least upper bound. Thus x2 = 2.

Two important questions arise concerning the above axioms. Are the axioms consistent, and satisfied by some set IR? Is the set of real numbers the only set satisfying these axioms?

The consistency of the axioms and the existence of IR can be demonstrated (to the satisfaction of most mathematicians) by constructjng IR, starting with the rationals.

In one sense the axioms do not determine IR uniquely. For example, let 1R0 be the set of all symbols xO, where x is (the symbol for) a real number. Define addition and multiplication of elements of 1R0 by

XO + yO = (x + y)O, xOyO = (xy)o.

Define po by XO E po -¢> X E P. Then 1R0 satisfies the axioms above. This is clearly fraudulent: 1R0 is just a copy of IR. It can 'be shown that any set with addition, multiplication, and a subset of positive elements, which satisfies all the axioms above, is just a copy of IR.

Starting from IR we can construct the set C of complex numbers, without simply postulating the existence of a "quantity" j such that j2 = - I. Let Co be the product set 1R2 = IR x IR, whose elements are ordered pairs (x, y) of real numbers. Define addition and multiplication by

(x, y) + (x', y') = (x + x', Y + y'), (x, y)(x', y') = (xx' - yy', xy' + x'y).

It can be shown by straightforward calculations that Co together with these operations satisfies AI, A2, Ml, M2, and DL. To verify the remaining algebraic axioms, note that

(x, y) + (0,0) = (x, y). (x,y) + (-x, -y) = (0,0),

(x, y)(1, 0) = (x, y), (x, y)(x/(x2 + y2), - y/(x2 + y2» = (1, 0) if (x, y) i= (0, 0).

If x E IR, let XO denote the element (x, 0) E Co. Let jO denote the element (0, 1). Then we have

(x, y) = (x,O) + (0, y) = (x,O) + (0, 1)(y, 0) = XO + ;Oyo.

Also, (i0)2 = (0, 1)(0, 1) = (-1,0) = _1°. Thus we can write any element of Co uniquely as XO + jOyO, x, y E IR, where (i0)2 = -1°. We now drop the superscripts and write x + jy for XO + jOyO and C for Co: this is legitimate, since for elements of IR the new operations coincide with the old: XO + yO = (x + y)O, xOyO = (xy)o. Often we shall denote elements of C by z or w. When

Real and complex numbers 9

we write z = x + iy, we shall understand that x, yare real. They are called the real part and the imaginary part of z, respectively:

z = x + iy, x = Re (z), y = 1m (z).

There is a very useful operation in C, called complex conjugation, defined by:

z* = (x + iy)* = x - iy.

Then z* is called the complex conjugate of z. It is readily checked that

(z + w)* = z* + w*, (zw)* = z*w*, (z*)* = z, z*z = x2 + y2.

Thus z*z =P 0 if z =P O. Define the modulus of z, Izl, by

Izl = (Z*Z)1/2 = (x2 + y2)1I2,

Then if z =P 0,

z = x + iy.

1 = z*zlzl-2 = z(z*lzl-2),

or

Adding and subtracting gives

z + z* = 2x, z - z* = 2iy if z = x + iy.

Thus

Re (z) = !(z + z*), 1m (z) = !i-1(z - z*).

The usual geometric representation of C is by a coordinatized plane: z = x + iy is represented by the point with coordinates (x, y). Then by the Pythagorean theorem, Izl is the distance from (the point representing) z to the origin. More generally, Iz - wi is the distance from z to w.

Exercises

1. There is a unique real number x > 0 such that x3 = 2. 2. Show that Re(z + w) = Re(z) + Re (w), Im(z + w) = Im(z) + Im(w). 3. Suppose z = x + iy, x, y E R Then

Izl ~ Ixl + Iyl·

4. For any z, WEe,

Izw*1 = Izllwl·

5. For any z, WEe,

Iz + wi ~ Izl + Iwl·

10 Basic concepts

(Hint: Iz + WI2 = (z + w)*(z + w) = IzI2 + 2 Re (zw*) + Iw12; apply Ex­ercises 3 and 4 to estimate IRe (zw*)I.)

6. The Archimedean axiom 04 can be deduced from the other axioms for the real numbers. (Hint: use 05).

7. If a > 0 and n is a positive integer, there is a unique b > 0 such that bn = a.

§3. Sequences of real and complex numbers

A sequence (Zn):'=l of complex numbers is said to converge to Z E C if for each 8 > 0, there is an integer N such that IZn - zi < 8 whenever n ;::: N. Geometrically, this says that for any circle with center z, the numbers Zn all lie inside the circle, except for possibly finitely many values of n. If this is the case we write

Zn -+ Z, or lim Zn = z, or lim Zn = Z. n .... ao

The number Z is called the limit of the sequence (Zn):'= l' Note that the limit is unique: suppose Zn -+ Z and also Zn -+ w. Given any 8 > 0, we can take n so large that IZn - zi < 8 and also IZn - wi < 8. Then

Iz - wi :s; Iz - znl + IZn - wi < 8 + 8 = 28.

Since this is true for all 8 > 0, necessarily Z = w. The following proposition collects some convenient facts about con-

vergence.

Proposition 3.1. Suppose (Zn):'=l and (Wn):'=l are sequences in C.

(a) Zn -+ Z if and only if Zn - Z -+ O. (b) Let Zn = Xn + iyno Xn, Yn real. Then Zn -+ Z = x + iy if and only if

Xn -+ x and Yn -+ y. (c) Ifzn-+zandwn-+w, then Zn + Wn-+Z + w. (d) If Zn -+ Z and Wn -+ w, then ZnWn -+ ZW. (e) If Zn -+ Z ¥- 0, then there is an integer M such that Zn ¥- 0 if n ;::: M.

Moreover (zn -l):'=M converges to Z-l.

Proof (a) This follows directly from the definition of convergence.

(b) By Exercise 3 of §3,

IXn - xl + IYn - yl :s; IZn - zi :s; 21 xn - xl + 21Yn - YI·

lt follows easily that Zn - Z -+ 0 if and only if Xn - x -+ 0 and Yn - Y -+ O. (c) This follows easily from the inequality

I(z + wn) - (z + w)1 = I(zn - z) + (wn - w)1 :s; IZn - zi + IWn - wi.

(d) Choose M so large that if n ;::: M, then IZn - zi < 1. Then for n~M,

Sequences of real and complex numbers

Let K = 1 + Iwi + Izi. Then for all n ~ M,

IZnwn - zwl = IZn(wn - w) + (zn - z)wl ::; IZnllwn - wi + IZn - zllwl ::; K(iwn - wi + IZn - zi).

Since Wn - W ~ 0 and Zn - Z ~ 0, it follows that ZnWn - ZW ~ o.

11

(e) Take M so large that IZn - zi ::; tizi when n ~ M. Then for n ~ M,

IZnl = IZnl + tizi - -lizi ~ IZnl + Iz - znl - -lizi ~ IZn + (z - zn)1 - tizi = tlzl·

Therefore, Zn 1= O. Also for n ~ M.

IZn-1 - z- 11 = I(z - Zn)Z-IZn -11 ::; IZ - Znl·IZI-l·(tIZi)-1 = Klz - znl,

where K = 2Izl-2. Since Z - Zn ~ 0 we have zn -1 - Z-1 ~ o. 0

A sequence (Zn):' = 1 in C is said to be bounded if there is an M ~ 0 such that IZnl ::; M for all n; in other words, there is a fixed circle around the origin which encloses all the zn's.

A sequence (xn):' = 1 in Iffi is said to be increasing if for each n, Xn ::; Xn + 1;

it is said to be decreasing if for each n, Xn ~ Xn+1.

Proposition 3.2. A bounded, increasing sequence in Iffi converges. A bounded, decreasing sequence in Iffi converges.

Proof. Suppose (Xn)~=l is a bounded, increasing sequence. Then the set {xn I n = 1,2, ... } is bounded above. Let x be its least upper bound. Given I!! > 0, X - I!! is not an upper bound, so there is an N such that XN ~ X-I!!. lfn ~ N, then

so IXn - xl ::; I!!. Thus Xn ~ x. The proof for a decreasing sequence is similar. 0

If A c Iffi is bounded above, the least upper bound of A is often called the supremum of A, written sup A. Thus

sup A = lubA.

Similarly, the greatest lower bound of a set B c Iffi which is bounded below is also called the infimum of A, written inf A :

inf A = glb A.

Suppose (Xn)~= 1 is a bounded sequence of reals. We shall associate with this given sequence two other sequences, one increasing and the other decreasing. For each n, let An = {xn' X n +1, X n +2, • •• }, and set

X~ = inf Am

12 Basic concepts

Now An ::::> An+ l , so any lower or upper bound for An is a lower or upper bound for An + l • Thus

Choose M so that IXnl :::; M, all n. Then - M is a lower bound and M an upper bound for each An. Thus

(3.1) - M :::; x;. :::; x~ :::; M, all n.

We may apply Proposition 3.2 to the bounded increasing sequence (X;'):'=l and the bounded decreasing sequence (x~):' = I and conclude that both converge. We define

lim inf Xn = lim x;', lim sup Xn = lim x~.

These numbers are called the lower limit and the upper limit of the sequence (xn);' = 1> respectively. It follows from (3.1) that

(3.2) -M:::; lim inf Xn :::; lim sup Xn :::; M.

A sequence (Zn);'= I in C is said to be a Cauchy sequence if for each 8 > ° there is an integer N such that IZn - zml < 8 whenever n ~ Nand m ~ N. The following theorem is of fundamental importance.

Theorem 3.3. A sequence in C (or IR) converges if and only ifit is a Cauchy sequence.

Proof Suppose first that Zn -+ z. Given 8 > 0, we can choose N so that IZn - zi < t if n ~ N. Then if n, m ~ N we have

IZn - zml :::; IZn - zi + Iz - zml < t8 + 1-8 = 8.

Conversely, suppose (Zn):'= I is a Cauchy sequence. We consider first the case of a real sequence (Xn):'=l which is a Cauchy sequence. The sequence (Xn):'=l is bounded: in fact, choose M so that IXn - xml < 1 if n, m ~ M. Then ifn ~ M,

IXnl :::; IXn - xMI + IXMI < 1 + IXMI·

Let K = max {lXII, Ix2 1, ... , IXM-II, IXMI + I}. Then for any n, IXnl :::; K. Now since the sequence is bounded, we can associate the sequences (X;'):'=l and (X~):'=l as above. Given 8 > 0, choose N so that IXn - xml < 8 if n, m ~ N. Now suppose n ~ m ~ N. It follows that

Xm - 8 :::; Xn :::; Xm + 8, n~m~N.

By definition of x;. we also have, therefore,

n~m~N.

Letting x = lim inf Xn = lim x;., we have

Xm - 8 :::; X :::; Xm + 8, m ~ N,

Sequences of real and complex numbers 13

or IXm - xl :::;; e, m ~ N. Thus Xn ~ x. Now consider the case of a complex Cauchy sequence (Zn):'=l. Let

Zn = Xn + iYn, Xn, Yn e Iht Since IXn - xml :::;; IZn - zml, (Xn):'=l is a Cauchy sequence. Therefore Xn ~ X E Iht Similarly, Yn ~ Y E IR. By Proposition 3.1(b), Zn ~ x + iy. 0

The importance of this theorem lies partly in the fact that it gives a cri­terion for the existence of a limit in terms of the sequence itself. An im­mediately recognizable example is the sequence

3, 3.1, 3.14, 3.142, 3.1416, 3.14159, ... ,

where successive terms are to be computed (in principle) in some specified way. This sequence can be shown to be a Cauchy sequence, so we know it has a limit. Knowing this, we are free to give the limit a name, such as "17".

We conclude this section with a useful characterization of the upper and lower limits of a bounded sequence.

Proposition 3.4. Suppose (xn):' = 1 is a bounded sequence in Iht Then lim inf Xn is the unique number x' such that

(i)' for any e > 0, there is an N such that Xn > x' - e whenever n ~ N, (ii)' for any e > 0 and any N, there is an n ~ N such that Xn < x' + e.

Similarly, lim sup Xn is the unique number x" with the properties

(i)" for any e > 0, there is an N such that Xn < x" + e whenever n ~ N, (ii)" for any e > 0 and any N, there is an n ~ N such that Xn > X-e.

Proof We shall prove only the assertion about lim inf Xn. First, let x~ = inf {xn' Xn+l> . .. } = inf An as above, and let x' = lim x~ = lim inf Xn. Suppose e > o. Choose N so that x~ > x' - e. Then n ~ N implies Xn ~ x~ > x' - e, so (i)' holds. Given e > 0 and N, we have x~ :::;; x' < x' + -le.

Therefore x' + '1e is not a lower bound for AN, so there is an n ~ N such that Xn :::;; x' + -le < x' + e. Thus (ii)' holds.

Now suppose x' is a number satisfying (i)' and (ii)'. From (i)' it follows that ipf An > x' - e whenever n ~ N. Thus lim inf Xn ~ x' - e, all e, so lim inf Xn ~ x'. From (ii)' it follows that for any N and any e, inf AN < x' + e. Thus for any N, inf AN :::;; x', so lim inf Xn :::;; x'. We have lim inf Xn = x'. 0

Exercises

1. The sequence (1/n):'=l has limit O. (Use the Archimedean axiom, §2.) 2. If Xn > 0 and Xn ~ 0, then xnl/2 ~ o. 3. If a > 0, then al/n ~ 1 as n ~ 00. (Hint: if a ~ 1, let al/n = 1 + Xn.

By the binomial expansion, or by induction, a = (1 + xn)n :::;; 1 + nxn. Thus Xn < n-la ~ o. If a < 1, then al/n = (bl/n)-l where b = a-I> 1.)

14 Basic concepts

4. lim n1/n = 1. (Hint: let n1/n = 1 + Yn. For n ~ 2, n = (1 + Yn)n ~ 1 + nYn + !n(n - I)Yn2 > !n(n - I)Yn2 , so Yn2 ::; 2(n -'1)-1-+ O. Thus Yn -+ 0.)

5. If Z E C and /z/ < I, then zn -+ 0 as n -+ 00.

6. Suppose (Xn):=1 is a bounded real sequence. Show that Xn -+ x if and only if lim inf Xn = x = lim sup X n•

7. Prove the second part of Proposition 3.4. 8. Suppose (Xn):", 1 and (an):= 1 are two bounded real sequences such that

an-+a > O. Then

lim inf anXn = a ·lim inf Xn> lim sup anXn = a ·lim sup X n•

§4. Series

Suppose (Zn):=1 is a sequence in C. We associate to it a second sequence (Sn):=l, where

n

Sn = L Zn = Z1 + Z2 + ... + Zn' m=l

If (Sn):=1 converges to s, it is reasonable to consider s as the infinite sum 2::= 1 zn· Whether (Sn):= 1 converges or not, the formal symbol 2::= 1 Zn or 2: Zn

is called an infinite series, or simply a series. The number Zn is called the nth term of the series, Sn is called the nth partial sum. If Sn -+ S we say that the series 2: Zn converges and that its sum is s. This is written

(4.1) 00

s = L Zn· n=l

(Of course if the sequence is indexed differently, e.g., (zn):=O, we make the corresponding changes in defining Sn and in (4.1).) If the sequence (Sn):= 1 does not converge, the series 2: Zn is said to diverge.

In particular, suppose (Xn):= 1 is a real sequence, and suppose each Xn ~ O. Then the sequence (sn):= 1 of partial sums is clearly an increasing sequence. Either it is bounded, so (by Proposition 3.2) convergent, or for each M > 0 there is an N such that

i Xm > M whenever n ~ N. m=l

In the first case we write

(4.2)

and in the second case we write

(4.3)

00

L Xn < 00 n=l

00

L Xn = 00. n=1

Series 15

Thus (4.2) -¢> 2: Xn converges, (4.3) -¢> 2: Xn diverges.

Examples

1. Consider the series 2::=1 n -1. We claim 2::=1 n -1 = 00. In fact (symbolically),

L>-l = 1 + 1 + t + -!- + t + i + ~ + i + ... 2': 1 + 1 + (-!- + -!-) + (i + i + i + i) + ... = 1 + 1 + 2(-!-) + 4(i) + 8(+6) + ... = 1 + 1 + 1 + ... = 00.

2. 2::=1 n- 2 < 00. In fact (symbolically),

L>-2 = 1 + (1)2 + (t)2 + ... + (~)2 + ... ~ 1 + (1)2 + (1)2 + (t)2 + (t)2 + (t)2 + (-!-)2 + ... = I + 2(1)2 + 4(-!-)2 + 8(i)2 + ... = I + 1 + -!- + i + ... = 2.

(We leave it to the reader to make the above rigorous by considering the respective partial sums.)

How does one tell whether a series converges? The question is whether the sequence (sn): = 1 of partial sums converges. Theorem 2.3 gives a necessary and sufficient condition for convergence of this sequence: that it be a Cauchy sequence. However this only refines our original question to: how does one tell whether a series has a sequence of partial sums which is a Cauchy sequence? The five propositions below give some answers.

Proposition 4.1. If 2::= 1 Zn converges, then Zn --+ O.

Proof If 2: Zn converges, then the sequence (sn): = 1 of partial sums is a Cauchy sequence, so Sn - Sn -1 --+ O. But Sn - Sn -1 = Zn. D

Note that the converse is false: lin --+ 0 but 2: lin diverges.

Proposition 4.2. If 1 Z 1 < I, then 2::=0 zn converges; the sum is (I - z) -1 . If Izi 2': 1, then 2::=0 zn diverges.

Proof The nth partial sum is

Sn = 1 + Z + Z2 + ... + zn-1.

Then sn(l - z) = 1 - zn, so Sn = (l - zn)/(l - z). If Izl < 1, then as n --+ 00, zn --+ 0 (Exercise 5 of §3). Therefore Sn --+ (l - Z)-l. If Izl 2': 1, then Iznl 2': 1, and Proposition 4.1 shows divergence. D

The series 2::=0 zn is called a geometric series.

Proposition 4.3. (Comparison test). Suppose (zn):= 1 is a sequence in C and (an):=l a sequence in IR with each an 2': O. If there are constants M, N such that

IZnl ~ Man whenever n 2': N,

16 Basic concepts

and ifL a" converges, then L ZIt converges.

Proof. Let SIt = L:' ~ 1 Zm, bIt = L:' = 1 a". If n, m ~ N then

Is" - sml = I i Zn I ~ i Iz,,1 J=m+1 J=m+1

" ~ M L: a" = M(b" - bm). J=m+1

But (b,,):=l is a Cauchy sequence, so this inequality implies that (S,,):=l is also a Cauchy sequence. 0

Proposition 4.4. (Ratio test). Suppose (Z,,):=1 is a sequence in C and suppose ZIt =1= 0, all n.

(a) If

then L ZIt converges. (b) If

then L ZIt diverges.

Proof. (a) In this case, take r so that lim sup IZ"+l/Z,,1 < r < 1. By Proposition 3.4, there is an N so that IZ"+l/Z,,1 ~ r whenever n ~ N. Thus if n > N,

Iz,,1 ~ rlzn-11 ~ r'rlz"_21 ~ ... ~ r"-NlzNI = Mr",

where M = ,-NlzNI. Propositions 4.2 and 4.3 imply convergence. (b) In this case, Proposition 3.4 implies that for some N, IZn+1/znl ~ 1 if

n ~ N. Thus for n > N.

IZnl ~ IZ"-ll ~ ... ~ IZNI > O.

We cannot have ZIt -+ 0, so Proposition 4.1 implies divergence. 0

Corollary 4.5. If ZIt =1= 0 for n = 1,2, ... and if lim IZn+1/z,,1 exists, then the series L ZIt converges if the limit is < 1 and diverges if the limit is > 1.

Note that for both the series L lin and L I/n2 , the limit in Corollary 4.5 equals 1. Thus either convergence or divergence is possible in this case.

Proposition 4.6. (Root test). Suppose (Z,,):=1 is a sequence in C.

(a) If

then L ZIt converges.

(b) If lim sup Iz,,111n > 1,

then L ZIt diverges.

Series 17

Proof. (a) In this case, take r so that lim sup IZnl1fn < r < 1. By Propo­sition 3.4, there is an N so that I Zn Ilfn ~ r whenever n ;::: N. Thus if n ;::: N, then IZnl ~ rn. Propositions 4.2 and 4.3 imply convergence.

(b) In this case, Proposition 3.4 implies that IZnl1fn ;::: 1 for infinitely many values of n. Thus Proposition 4.1 implies divergence. D

Note the tacit assumption in the statement and proof that (lznI1fn);,= 1 is a bounded sequence, so that the upper and lower limits exist. However, if this sequence is not bounded, then in particular IZnl ;::: 1 for infinitely many values of n, and Proposition 4.1 implies divergence.

Corollary 4.7. If lim IZnl1fn exists, then the series L Zn converges if the limit is < I and diverges if the limit is > 1.

Note that for both the series L lin and L Iln2 , the limit in Corollary 4.7 equals 1 (see Exercise 4 of §3). Thus either convergence or divergence is possible in this case.

A particularly important class of series are the power series. If (an);'=o is a sequence in C and Zo a fixed element of C, then the series

(4.2)

is the power series around Zo with coefficients (an);'= o. Here we use the conven­tion that WO = 1 for all w E C, including w = O. Thus (4.2) is defined, as a series, for each Z E C. For Z = Zo it converges (with sum ao), but for other values of Z it mayor may not converge.

Theorem 4.8. Consider the power series (4.2). Define R by

R = 0 if(lanI 1fn);'=1 is not a bounded sequence, R = (lim sup lanI1fn)-1 if lim sup lanl 1fn > 0, R = 00 if lim sup lanl 1fn = O.

Then the power series (4.2) converges if Iz - zol < R, and diverges if Iz - zol > R.

Proof. We have

(4.3)

Suppose Z =I Zo0 If (lanI1fn);'=1 is not a bounded sequence, then neither is (4.3), and we have divergence. Otherwise the conclusions follow from (4.3) and the root test, Proposition 4.6. D

The number R defined in the statement of Theorem 4.8 is called the radius of convergence of the power series (4.2). It is the radius of the largest circle in the complex plane inside which (4.2) converges.

Theorem 4.8 is quite satisfying from a theoretical point of view: the radius of convergence is shown to exist and is (in principle) determined in all

18 Basic concepts

cases. However, recognizing lim sup lanl1/n may be very difficult in practice. The following is often helpful.

Theorem 4.9. Suppose an =J Of or n 2': N, and suppose lim lan+1/anl exists. Then the radius of convergence R of the power series (4.2) is given by

R = (lim lan+1/anJ)-l iflim lan+1lanl > O. R = 00 iflim lan+1Janl = O.

Proof. Apply Corollary 4.5 to the series (4.2), noting that if z =1= Zo then

lan+1(z - zo)n+lJan(z - zo)nl = lan +1Janl·lz - zol. 0

Exercises

I. If 2:'= 1 Zn converges with sum sand 2:'= 1 Wn converges with sum t, then 2:'= 1 (zn + wn) converges with sum s + t.

2. Suppose 2 an and 2: bn each have all non-negative terms. If there are constants M > 0 and N such that bn 2': Man whenever n 2': N, and if 2: an =

00, then 2 bn = 00.

3. Show that 2::'=1 (n + 1)/(2n2 + 1) diverges and 2::'=1 (n + 1)/(2n3 + 1) converges. (Hint: use Proposition 4.3 and Exercise 2, and compare these to 2 lin, 2: Iln2.)

4. (2k-Test). Suppose a1 2': a2 2': ... 2': an 2': 0, all n. Then 2:'=1 an < 00 -¢> 2k= 1 2k a2k < 00. (Hint: use the methods used to show divergence of 2 lin and convergence of 2 IJn2.)

5. (Integral Test). Suppose a1 2': a2 2': ... 2': an 2': 0, all n. Suppose f: [1,00) -J-iR is a continuous function such thatf(n) = an, all n, andf(y) ::; f(x) if y 2': x. Then 2:'=1 an < 00 -¢> II'" f(x) dx < 00.

6. Suppose p > O. The series 2:'=1 n- P converges if p > I and diverges if p :s; 1. (Use Exercise 4 or Exercise 5.)

7. The series 2::'=2n-1(logn)-2 converges; the series 2::'=2n-1(logn)-1 diverges.

8. The series 2::'=0 znJn! converges for any z E C. (Here O! = 1, n! = n(n - I)(n - 2)· .... 1.)

9. Determine the radius of convergence of

00

2 n!zn, n=O

'" L n! znJ(2n)! n=O

10. (Alternating series). Suppose IXll 2': IX21 2': •.. 2': IXnl, all n, Xn 2': 0 if n odd, Xn ::; 0 if n even, and Xn -J- O. Then 2: Xn converges. (Hint: the partial sums satisfy S2 ::; S4 ::; S6 ::; •.• ::; S5 ::; S3 ::; SI')

11. 2::'= 1 (_l)njn converges.

Metric spaces 19

§5. Metric spaces

A metric on a set S is a function d from the product set S x S to ~, with the properties

Dl. d(x, x) = 0, d(x, y) > 0, if x, YES, x =1= y. D2. d(x, y) = d(y, x), all x, YES. D3. d(x, z) ~ d(x, y) + dey, z), all x, y, z E S.

We shall refer to d(x, y) as the distance from x to y. A metric space is a set S together with a given metric d. The inequality D3 is called the triangle inequality. The elements of S are often called points.

As an example, take S = ~2 = ~ X ~,with

(5.1) d«x, y), (x', y'» = [(x - X')2 + (y - y')2]l/2.

If we coordinatize the Euclidean plane in the usual way, and if (x, y), (x', y') are the coordinates of points P and P' respectively, then (5.1) gives the length of the line segment PP' (Pythagorean theorem). In this example, D3 is the analytic expression of the fact that the length of one side of a triangle is at most the sum of the lengths of the other two sides. The same example in different guise is obtained by letting S = C and taking

(5.2) d(z, w) = Iz - wi as the metric. Then D3 is a consequence of Exercise 5 in §2.

Some other possible metrics on ~2 are:

dl«x, y), (x', y'» = Ix - x'i + Iy - y'l, d2«x, y), (x', y'» = max {Ix - x'l, Iy - y'I}, da«x, y), (x', y'» = 0 if (x, y) = (x', y'), and 1 otherwise.

Verification that the functions dl , d:1" and da satisfy the conditions Dl, D2, D3 is left as an exercise. Note that da works for any set S: if x, YES we set d(x, y) = 1 if x =1= y and 0 if x = y.

A still simpler example of a metric space is IR, with distance function d given by

(5.3) d(x, y) = Ix - yl.

Again this coincides with the usual notion of the distance between two points on the (coordinatized) line.

Another important example is ~n, the space of ordered n-tuples x = (Xl> X2, ... , xn) of elements of R There are various possible metrics on ~n like the metrics dl , d2 , da defined above for ~n, but we shall consider here only the generalization of the Euclidean distance in ~2 and ~a. If x = (Xl' X2, ... , xn) and y = (Yl> Y2, ... , Yn) we set

(5.4) d(x, y) = [(Xl - Yl)2 + (X2 - Y2)2 + ... + (xn - Yn)2]1/2.

When n = 1 we obtain ~ with the metric (5.3); when n = 2 we obtain ~2 with the metric (5.1), in somewhat different notation. It is easy to verify that

20 Basic concepts

d given by (5.4) satisfies 01 and 02, but condition 03 is not so easy to verify. For now we shall simply assert that d satisfies 03; a proof will be given in a more general setting in Chapter 4.

Often when the metric d is understood, one refers to a set S alone as a metric space. For example, when we refer to IR, C, or IRn as a metric space with no indication what metric is taken, we mean the metric to be given by (5.3), (5.2), or (5.4) respectively.

Suppose (S, d) is a metric space and T is a subset of S. We can consider T as a metric space by taking the distance function on TxT to be the restriction of d to TxT.

The concept of metric space has been introduced to provide a uniform treatment of such notions as distance, convergence, and limit which occur in many contexts in analysis. Later we shall encounter metric spaces much more exotic than IRn and C.

Suppose (S, d) is a metric space, x is a point of S, and r is a positive real number. The ball of radius r about x is defined to be the subset of S consisting of all points in S whose distance from x is less than r:

B,(x) = {y lYE S, d(x, y) < r}.

Clearly x E B,(x). If 0 < r < s, then B,(x) c Bs(x).

Examples

When S = IR (metric understood), B,(x) is the open interval (x - r, x + r). When S = 1R2 or C, B,(z) is the open disc of radius r centered at z. Here we take the adjective "open" as understood; we shall see that the interval and the disc in question are also open in the sense defined below.

A subset A c S is said to be a neighborhood of the point XES if A con­tains B,(x) for some r > 0. Roughly speaking, this says that A contains all points sufficiently close to x. In particular, if A is a neighborhood of x it contains x itself.

A subset A c S is said to be open if it is a neighborhood of each of its points. Note that the empty set is an open subset of S: since it has no points (elements), it is a neighborhood of each one it has.

Example

Consider the interval A = (0, I] c IR. This is a neighborhood of each of its points except x = 1. In fact, if ° < x < 1, let r = min {x, 1 - x}. Then A => B,(x) = (x - r, x + r). However, for any r > 0, B,(I) contains 1 + -!-r, which is not in A.

We collect some useful facts about open sets in the following proposition.

Proposition 5.1. Suppose (S, d) is a metric space.

(a) For any XES and any r > 0, B,(x) is open. (b) If Ab A2, ••• , Am are open subsets of S, then n~=l Am is also open.

Metric spaces 21

(c) If (An)nEB is any collection of open subsets of S, then UnEB An is also open.

Proof (a) Suppose y E Br(x). We want to show that for some s > 0, Bs(Y) c Br(x). The triangle inequality makes this easy, for we can choose s = r - dey, x). (Since y E Br(x), s~ positive.) If z E Bs(Y), then

d(z, x) ::; d(z, y) + d(y, x) < s + dey, x) = r.

Thus z E Br(x). (b) Suppose x E n~ = 1 Am. Since each Am is open, there is rem) > 0 so that

Br(m)(x) C Am. Let r = min {r(l), r(2), ... , r(n)}. Then r > 0 and Br(x) C

Br(m)(x) C Am, so Br(x) C n~=l Am. (Why is it necessary here to assume that A10 A 2 , ••• is afinite collection of sets?)

(c) Suppose x E A = UnEB An. Then for some particular fl, x E An. Since An is open, there is an r > 0 so that Br(x) C An C A. Thus A is open. 0

Again suppose (S, d) is a metric space and suppose A C S. A point XES is said to be a limit point of A if for every r > 0 there is a point of A with distance from x less than r:

Br(x) n A ¥= 0 if r > O.

In particular, if x E A then x is a limit point of A. The set A is said to be closed if it contains each of its limit points. Note that the empty set is closed, since it has no limit points.

Example

The interval (0, 1] C IR has as its set of limit points the closed interval [0,1]. In fact if 0 < x ::; 1, then x is certainly a limit point. If x = 0 and r > 0, then Br(O) n (0, I] = (-r, r) n (0, I] ¥= 0. If x < ° and r = lxi, then BrCx) n (0, I] = 0, while if x> I and r = x - I, then Br(x) n (0, I] = 0.

Thus the interval (0, I] is neither open nor closed. The exact relationship between open sets and closed sets is given in Proposition 5.3 below.

The following is the analogue for closed sets of Proposition 5.1.

Proposition 5.2. Suppose (S, d) is a metric space.

(a) For any x E Sandanyr > 0, the closed ball C = {y lYE S, d(x, y) ::; r} is a closed set.

(b) If A 10 A 2 , ••• , An are closed subsets of S, then U~ = 1 Am is closed. (c) If(Ap)pEB is any collection of closed subsets of S, then nnEB An is closed.

Proof (a) Suppose z is a limit point of the set C. Given e > 0, there is a point y E B.(z) n C. Then

d(z, x) ::; d(z, y) + dey, x) < e + r.

Since this is true for every e > 0, we must have d(z, x) ::; r. Thus z E C.

22 Basic concepts

(b) Suppose x ¢ A = U::'=l Am. For each m, x is not a limit point of Am, so there is r(m) > 0 such that Br(mlx) n Am = 0. Let

r = min {r(1), r(2), ... , r(n)}.

Then Br(x) n Am = 0, all m, so B.(x) n A = 0. Thus x is not a limit point of A.

(c) Suppose x is a limit point of A = npeB An. For any r > 0, Br(x) n A :I 0. But A cAp, so Br(x) n Ap :I 0. Thus x is a limit point of All' so it is in All' This is true for each p, so x E A. 0

Proposition 5.3. Suppose (S, d) is a metric space. A subset A c S is open if and only if its complement is closed.

Proof Let B be the complement of A. Suppose B is closed, and suppose x E A. Then x is not a limit point of B, so for some r > 0 we have Br(x) n B = 0. Thus Br(x) c A, and A is a neighborhood of x.

Conversely, suppose A is open and suppose x ¢ B. Then x E A, so for some r > 0 we have Br(x) c A. Then Br(x) n B = 0, and x is not a limit point of B. It follows that every limit point of B is in B. 0

The set of limit points of a subset A c S is called the closure of A; we shall denote it by A - . We have A c A - and A is closed if and only if A = A - . In the example above, we saw that the closure of (0, I] c ~ is [0,1].

Suppose A, B are subsets of S and A c B. We say that A is dense in B if Be A -. In particular, A is dense in S if A - = S. As an example, Q (the rationals) is dense in IR. In fact, suppose x E IR and r > O. Choose a positive integer n so large that lin < r. There is a unique integer m so that min < x < (m + 1)ln. Then d(x, min) = x - min < (m + 1)ln - min = lin < r, so min E Br(x). Thus x E Q - .

A sequence (x,,);'= 1 in S is said to converge to XES if for each e > 0 there is an N so that d(x", x) < e if n ~ N. The point x is called the limit of the sequence, and we write

lim x" = x or x" -+x. " .... 00

When S = ~ or C (with the usual metric), this coincides with the definition in §3. Again the limit, if any, is unique.

A sequence (x,,);, = 1 in S is said to be a Cauchy sequence if for each e > 0 there is an N so that d(x", xm) < e if n, m ~ N. Again when S = IR or C, this coincides with the definition in §3.

The metric space (S, d) is said to be complete if every Cauchy sequence in S converges to a point of S. As an example, Theorem 3.3 says precisely that ~ and C are complete metric spaces with respect to the usual metrics.

Many processes in analysis produce sequences of numbers, functions, etc., in various metric spaces. It is important to know when such sequences converge. Knowing that the metric space in question is complete is a powerful tool, since the condition that the sequence be a Cauchy sequence is then a

Compact sets 23

necessary and sufficient condition for convergence. We have already seen this in our discussion of series, for example.

Note that IRn is complete. To see this note that in IRn,

max {Ixj - Yjl,j = I, ... ,n}:::; d(x,y):::; n·max{lxj-Yjl,j = I, ... ,n}.

It follows that a sequence of points in IRn converges if and only if each of the n corresponding sequences of coordinates converges in IR. Similarly, a se­quence of points in IRn is a Cauchy sequence if and only if each of the n corresponding sequences of coordinates is a Cauchy sequence in IR. Thus completeness of IRn follows from completeness of IR. (This is simply a general­ization of the argument showing C is complete.)

Exercises

I. If (S, d) is a metric space, XES, and r :2: 0, then

{y lYE S, d(y, x) :> r}

is an open subset of S. 2. The point x is a limit point of a set A c S if and only if there is a

sequence (xn);:'= 1 in A such that Xn --+ x. 3. If a sequence (xn);:'= 1 in a metric space converges to XES and also

converges to YES, then x = y. 4. If a sequence converges, then it is a Cauchy sequence. 5. If (S, d) is a complete metric space and A c S is closed, then (A, d) is

complete. Conversely, if B c Sand (B, d) is complete, then B is a closed subset of S.

6. The interval (0, I) is open as a subset of IR, but not as a subset of C. 7. Let S = Q (the rational numbers) and let d(x, y) = Ix - yl, x, y E Q.

Show that (S, d) is not complete. 8. The set of all elements x = (Xl> X2, ... , xn) in IRn such that each Xj is

rational is a dense subset of IRn. 9. Verify that IRn is complete.

§6. Compact Sets

Suppose that (S, d) is a metric space, and suppose A is a subset of S. The subset A is said to be compact if it has the following property: suppose that for each x E A there is given a neighborhood of x, denoted N(x); then there are finitely many points Xl, X2, ... , Xn in A such that A is contained in the union of N(Xl), N(X2), ... , N(xn). (Note that we are saying that this is true for any choice of neighborhoods of points of A, though the selection of points Xl, X2, . .. may depend on the selection of neighborhoods.) It is obvious that any finite subset A is compact.

24 Basic concepts

Examples

1. The infinite interval (0, 00) c III is not compact. For example, let N(x) = (x - I, x + I), x E (0, 00). Clearly no finite collection of these intervals of finite length can cover all of (0, 00).

2. Even the finite interval (0, I] c III is not compact. To see this, let N(x) = (-!-x,2), x E (0, I]. For any Xl, X2, ... , XlI E (0, 1], the union of the intervals N(xj) will not contain y if y :::; -!- min {Xl> X2, ... , x lI}.

3. The set A = {o} U {I, -!-, 1, t, ... } c IR is compact. In fact, suppose for each x E A we are given a neighborhood N(x). In particular, the neigh­borhood N(O) of 0 contains an interval (-e, e). Let M be a positive integer larger than lie. Then lin E N(O) for n ~ M, and it follows that A c N(O) U

N(l) U NH) U ... U N(IIM).

The first two examples illustrate general requirements which compact sets must satisfy. A subset A of S, when (S, d) is a metric space, is said to be bounded if there is a ball Br(x) containing A.

Proposition 6.1. Suppose (S, d) is a metric space, S =1= 0, and suppose A c S is compact. Then A is closed and bounded.

Proof Suppose y 1= A. We want to show that y is not a limit point of A. For any x E A, let N(x) be the ball of radius -!-d(x, y) around x. By assump­tion, there are X1 ,X2, ... ,Xll EA such that A c U~=IN(xm)' Let r be the minimum of the numbers -!-d(XI, y), ... , -td(xn , y). If x E A, then for some m, d(x, xm) < -td(xm' y). But then

d(xm,y):::; d(xm' x) + d(x,y) :::; !d(xm, y) + d(x, y).

so d(x, y) > !d(xm, y) ~ r. Thus Br(y) (') A = 0, and y is not a limit point of A.

Next, we want to show that A is bounded. For each x E A, let N(x) be the ball of radius 1 around x. Again, by assumption there are Xl, X2, ... , Xn E A such that A c U~= I N(xm). Let

r = 1 + max {d(xl> X2), d(Xb X3), ... , d(XI' x lI)}.

If YEA then for some m, dey, Xm) < 1. Therefore dey, Xl) :::; dey, Xm) + d(xm' Xl) < 1 + d(xm' Xl) :::; r, and we have A c Br(XI)' 0

The converse of Proposition 6.1, that a closed, bounded subset of a metric space is compact, is not true in general. It is a subtle but extremely important fact that it is true in IIln, however.

Theorem 6.4. (Heine-Borel Theorem). A subset ofllln or ofe is compact if and only if it is closed and bounded.

Proof. We have seen that in any metric space, if A is compact it is necessarily closed and bounded. Conversely, suppose A c IRn is closed and bounded. Let us assume at first that n = 1. Since A is bounded, it is contained

Compact sets 25

in some closed interval [a, b]. Suppose for each x E A, we are given a neigh­borhood N(x) of x. We shall say that a closed subinterval of [a, b] is nice if there are points XI. X2, ... , Xm E A such that UT~ 1 N(xj) contains the inter­section of the subinterval with A; we are trying to show that [a, b] itself is nice. Suppose it is not. Consider the two subintervals [a, c] and [c, b], where c = -tea + b) is the midpoint of [a, b]. If both of these were nice, it would follow that [a, b] itself is nice. Therefore we must have one of them not nice; denote its endpoints by ab bb and let C1 = -t(a1 + b1). Again, one of the intervals [ab cd and [Cb b1] must not be nice; denote it by [a2, b2]. Continu­ing in this way we get a sequence of intervals [am, bm], m = 0, 1,2, ... such that [ao, bo] = [a, b], each [am, bm] is the left or right half of the interval [am-I. bm- 1], and each interval [am, bm] is not nice. It follows that ao ~ a1 ~ ... ~ am ~ bm ~ ... ~ b1 ~ bo and bm - am = 2-m(bo - ao)-+O. There­fore there is a point x such that am -+ x and bm -+ x. Moreover, am ~ X ~ bm, for all m. We claim that x E A; it is here that we use the assumption that A is closed. Since [am, bm] is not nice, it must contain points of A: otherwise A n [am, bm] = 0 would be contained in any Uj~l N(xj). Let

Clearly Xm -+ x, since am -+ x and bm -+ x. Since A is closed, we get x E A. Now consider the neighborhood N(x). This contains an interval (x - e,

X + e). If we choose m so large that bm - am < e, then since am ~ x ~ bm this implies [am, bm] c N(x). But this means that [am, bm] is nice. This contra­diction proves the theorem for the case n = 1.

The same method of proof works in IRn, where instead of intervals we use squares, cubes, or their higher dimensional analogues. For example, when n = 2 we choose M so large that A is contained in the square with corners ( ± M, ± M). If this square were not nice, the same would be true of one of the four equal squares into which it can be divided, and so on. Continuing we get a sequences of squares So J Sl J S2 J ... , each of side -t the length of the preceding, each intersecting A, and each not nice. The intersection n~~o Sm contains a single point x, and x is in A. Then N(x) contains Sm for large m, a contradiction. Since as metric space C = 1R2, this also proves the result for C. 0

Suppose (Xn):~ 1 is a sequence in a set S. A subsequence of this sequence is a sequence of the form (Yk):' ~ I. where for each k there is a positive integer nk so that

n1 < n2 < . . . < nk < nk + 1 < ... , Yk = xnk·

Thus, (Yk):'~ 1 is just a selection of some (possibly all) of the xn's, taken in order. As an example, if (xn):~ 1 C IR has Xn = (-l)nfn, and if we take nk = 2k,then(xn):~1 = (-l,-t, -t,t, -t,···)and(Yk):'~l = (-t,t,!,···)· As a second example, let (xn):~ 1 be an enumeration of the rationals. Then for any real number x, there is a subsequence of (Xn):~l which converges to x.

26 Basic concepts

Suppose (8, d) is a metric space. A set A c 8 is said to be sequentially compact if, given any sequence (xn):'= 1 C A, some subsequence converges to a point of A.

Examples

1. Any finite set is sequentially compact. (Prove this.) 2. The interval (0, 00) c IR is not sequentially compact; in fact let

Xn = n. No subsequence of (Xn):'=l converges. 3. The bounded interval (0, 1] c IR is not sequentially compact; in fact

let Xn = lin. Any subsequence of (Xn):'=l converges to 0, which is not in (0, 1].

Proposition 6.3. 8uppose (8, d) is a metric space, 8 -# 0, and suppose A c 8 is sequentially compact. Then A is closed and bounded.

Proof. Suppose x is a limit point of A. Choose Xn E B1/n(X) n A, n = 1,2,3, .... Any subsequence of (Xn):'=l converges to x, since Xn ~ x. It follows (since by assumption some subsequence converges to a point of A) that x E A. Thus A is closed.

Suppose A were not bounded. Take x E 8 and choose Xl E A such that x rf B1(x). Let r1 = d(x, Xl) + 1. By the triangle inequality, B1(X1) c BT'(X), Since A is not bounded, there is X2 E A such that X2 rf BTl (X). Thus also d(x1> X2) ~ 1. Let r2 = max {d(x, Xl), d(x, X2)} + 1 and choose X3 E A such that X3 rt= BTl (X). Then d(x1' x3) ~ 1 and d(x2' x3) ~ 1. Continuing in this way we can find a sequence (xn):'= 1 C A such that d(xm' xn) ~ 1 if m #- n. Then no subsequence of this sequence can converge, and A is not sequentially compact. 0

Theorem 6.4. (Bolzano-Weierstrass Theorem). A subset A oflRn or ofe is sequentially compact if and only if it is closed and bounded.

Proof. We have shown that A sequentially compact implies A closed and bounded. Suppose A is closed and bounded, and suppose first that n = 1. Take an interval [a, b] containing A. Let c = -t(a + b). One (or both) of the subintervals [a, c] and [c, b] must contain Xn for infinitely many integers n; denote such a subinterval by [a1> bd, and consider [a1' cd, [C1> bd where C1 = -t(a1 + b1). Proceeding in this way we can find intervals [am, bm] with the properties [ao, bo] = [a, b], [am, bm] c [am-1> bm- 1], bm - am = 2 - m(bo - ao), and [am, bm] contains Xn for infinitely many values of n. Then there is a point x such that am ~ x, bm ~ x. We choose integers n1> n2, ... so that xnl E [a1' bd, n2 > n1 and x n2 E [a2, b2], n3 > n2 and xna E [a3, b3], etc. Then this subsequence converges to x. Since A is closed, x E A.

The generalization of this proof to higher dimensions now follows as in the proof of Theorem 6.2. 0

Both the terminology and the facts proved suggest a close relationship between compactness and sequential compactness. This relationship is made precise in the exercises below.

Vector spaces 27

Exercises

1. Suppose (xn):= 1 is a sequence in a metric space (S, d) which converges to XES. Let A = {x} U {Xn}:= l' Then A is compact and sequentially compact.

2. Let 0, the rationals, have the usual metric. Let A = {x I x E 0, X2 < 2}. Then A is bounded, and is closed as a subset of 0, but is not compact.

3. Suppose A is a compact subset of a metric space (S, d). Then A is sequentially compact. (Hint: otherwise there is a sequence (Xn):= 1 in A with no subsequence converging to a point of A. It follows that for each x E A. there is an rex) > 0 such that the ball N(x) = Br<x)(x) contains Xn for only finitely many values of n. Since A is compact, this would imply that {I, 2, 3, ... } is finite, a contradiction.)

4. A metric space is said to be separable if there is a dense subset which is countable. If (S, d) is separable and A c S is sequentially compact, then A is compact. (Hint: suppose for each x E A we are given a neighborhood N(x). Let {Xl> x 2 , X3, ... } be a dense subset of S. For each x E A we can choose an integer m and a rational r m such that x E Brm(xm) c N(x). The collection of balls Brm(xm) so obtained is (finite or) countable; enumerate them as C1,

C2 , .••• Since each Cj is contained in some N(x), it is sufficient to show that for some n, Ui= 1, C j ::::> A. If this were not the case, we could take Yn E A, Yn ¢; Ui=l C j , n = 1,2, .... Applying the assumption of sequential com­pactness to this sequence and noting how the Cj were obtained, we get a contradiction.)

§7. Vector spaces

A vector space over IR is a set X in which there are an operation of addi­tion and an operation of multiplication by real numbers which satisfy certain conditions. These abstract from the well-known operations with directed line segments in Euclidean 3-space.

Specifically, we assume that there is a function from X x X to X, called addition, which assigns to the ordered pair (x, y) E X X X an element of X denoted x + y. We assume

VI. (x + y) + z = x + (y + z), all x, y, Z E X. V2. x + y = y + x, all x, y E X. V3. There exists 0 E X such that x + 0 = x, all X. V4. For all x E X, there exists -x E X such that x + (-x) = O.

We assume also that there is a function from ~ x X to X, called scalar multiplication, assigning to the ordered pair (a, x) E ~ X X an element of X denoted ax. We assume

VS. (ab)x = a(bx), all a, b E ~, X E X. V6. a(x + y) = ax + ay, all a E ~, x, Y E X. V7. (a + b)x = ax + bx, all a, b E ~,x E X. V8. Ix = x, all x E X.

28 Basic concepts

Summarizing: a vector space over IR, or a real vector space, is a set X with addition satisfying VI-V4 and scalar multiplication satisfying VS-VS. The elements of X are called vectors and the elements of IR, in this context, are often called scalars.

Similarly, a vector space over C, or a complex vector space, is a set X together with addition sstisfying VI-V4 and scalar multiplication defined from C x X to X and satisfying VS-VS. Here the scalars are, of course, complex numbers.

Examples

1. IR is a vector space over IR, with addition as usual and the usual multiplication as scalar multiplication.

2. The set with one element 0 is a vector space over IR or C with 0 + 0 = 0, aO = 0, all a.

3.lRn is a vector space of IR if we take addition and scalar multiplication as

(Xl> X2, ... , xn) + (Yl> Y2, ... , Yn) = (Xl + Yl> X2 + Y2, ... , Xn + Yn), a(xI' X2, ... , xn) = (axl> aX2, ... , aXn).

4. C is a vector space over IR or C with the usual addition and scalar multiplication.

5. Let S be any set, and let F(S; IR) be the set whose elements s are the functions from S to IR. Define addition and scalar multiplication in F(S; IR) by

(f + g)(s) = f(s) + g(s), (af)(s) = af(s),

Then F(S; IR) is a vector space over IR.

SES, S E S, a E IR.

6. The set F(S; C) of functions from S to C can be made a complex vector space by defining addition and scalar multiplication as in 5.

7. Let X be the set of all functionsf: IR ~ IR which are polynomials, i.e., for some aD, al> ... , an E IR,

all x E IR.

With addition and scalar multiplication defined as in 5, this is a real vector space.

S. The set of polynomials with complex coefficients can be considered as a complex vector space.

Let us note two elementary facts valid in every vector space: the element o of assumption V3 is unique, and for any x E X, Ox = O. First, suppose 0' E X has the property that x + 0' = x for each x E X. Then in particular 0' = 0' + 0 = 0 + 0' = 0 (using V2 and V3). Next, if x E X, then

Ox = Ox + 0 = Ox + [Ox + (-Ox)] = [Ox + Ox] + (-Ox) = (0 + O)x + (-Ox) = Ox + (-Ox) = O.

Vector spaces

Note also that the element -x in V4 is unique. In fact if x + Y = 0, then

Y = Y + 0 = Y + [x + (-x)] = [y + x] + (-x) = [x + y] + (-x) = 0 + (-x) = (-x) + 0 = -x.

This implies that (-I)x = -x, since

x + (-I)x = [I + (-I)]x = Ox = o.

29

A non-empty subset Y of a (complex) vector space X is called a subspace of X if it is closed with respect to the operations of addition and scalar multi­plication. This means that if x, Y E Y and a E 18, then x + y and ay are in Y. If so, then Y itself is a vector space over 18, with the operations inherited from X.

Examples

l. Any vector space is a subspace of itself. 2. The set {o} is a subspace. 3. {x I Xn = O} is a subspace of IRn. 4. In the previous set of examples, the space X in example 7 is a subspace

of F(IR; IR). 5. Let X again be the space of polynomials with real coefficients. For each

n = 0, I, 2, ... , let Xn be the subset of X consisting of polynomials of degree ::; n. Then each Xn is a subspace of X. For m ::; n, Xm is a subspace of Xn.

Suppose Xl> X2, ... , Xn are elements of the vector space X. A linear combination of these vectors is any vector x of the form

where al> a2' ... , an are scalars.

Proposition 7.1. Let S be a nonempty subset of the vector space X, and let Y be the set of all linear combinations of elements of S. Then Y is a subspace of X andY:::> S.

IfZ is any other subspace of X which contains the set S, then Z :::> Y.

Proof If x, y E Y, then by definition they can be expressed as finite sums x = L ajxj, y = L bjYj, where each Xj E S and each Yj E S. Then ax = L (aaj)x j is a linear combination of the x/s, and x + Y is a linear combination of the x/s and the y/s. If XES, then x = Ix E Y. Thus Y is a subspace containing S.

Suppose Z is another subspace of X containing S. Suppose x E Y. Then for some Xl> x2 , ••• , Xn E S, X = L ajXj. Since Z is a subspace and Xl> X2, ... , Xn E Z, we have alxl> a2x2, ... , anXn E Z. Moreover, alxl + a2x2 E Z, so (alxl + a2x2) + aaXa E Z. Continuing we eventually find that x E Z. 0

We can paraphrase Proposition 7.1 by saying that any subset S of a vector space X is contained in a unique smallest subspace Y. This subspace is

30 Basic concepts

called the span of S. We write Y = span (S). The set S is said to span Y. Note that if S is empty, the span is the subspace {O}.

Examples

Let X be the space of all polynomials with real coefficients. Let fm be the polynomial defined by fm(x) = xm. Then span {fo.!l' ... .!n} is the subspace Xn of polynomials of degree ~ n.

A linear combination alXl + a2X2 + ... + anxn of the vectors Xl> X2, ... , Xn is said to be nontrivial if at least one of the coefficients al> a2, ... , an is not zero. The vectors Xl> X2, ... , Xn are said to be linearly dependent if some non­trivial linear combination of them is the zero vector. Otherwise they are said to be linearly independent. More generally, an arbitrary (possibly infinite) subset S is said to be linearly dependent if some nontrivial linear combination of finitely many distinct elements of S is the zero vector; otherwise S is said to be linearly independent. (Note that with this definition, the empty set is linearly independent.)

Lemma 7.2. Vectors Xl> X2, ... , Xn in X, n ~ 2, are linearly dependent if and only if some Xj is a linear combination of the others.

Proof If Xl, X2, ... , Xn are linearly dependent, there are scalars al> a2, ... , an, not all 0, such that L a,xj = 0. Renumbering, we may suppose al # 0. Then Xl = L1=2 (-al-laj)x,.

Conversely, suppose Xl, say, is a linear combination L1=2 bjxj. Letting al = 1, and aj = -bj for j ~ 2, we have L ajxj = 0. 0

The vector space X is said to be finite dimensional if there is a finite subset which spans X. Otherwise, X is said to be infinite dimensional. A basis of a (finite-dimensional) space X is an ordered finite subset (Xl> X2, ... , xn) which is linearly independent and spans X.

Examples

1. IRn has basis vectors (el> e2, ... , en), where el = (1, 0, 0, ... , 0), e2 = (0, 1,0, ... , 0), ... , en = (0,0, ... ,0, 1). This is called the standard basis in IRn.

2. The set consisting of the single vector 1 is a basis for C as a complex vector space, but not as a real vector space. The set (1, i) is a basis for C as a real vector space, but is linearly dependent if C is considered as a complex vector space.

Theorem 7.3. A finite-dimensional vector space X has a basis. Any two bases of X have the same number of elements.

Proof Let {Xl> x2, ••• , xn} span X. If these vectors are linearly inde­pendent then we may order this set in any way and have a basis. Otherwise

Vector spaces 31

ifn ;:::; 2 we may use Lemma 7.2 and renumber, so that Xn is a linear combina­tion "17= f ajxj. Since span {Xl> X2, ... , xn} = X, any X E X is a linear com­bination

X = i bjxI = ~1 bjxj + bn(ni,l alx/) j=l /=1 /=1

n-1 = 2: (bj + bnaj)xj.

/=1

Thus span {Xl' X2, ... , Xn-1} = X. If these vectors are not linearly inde­pendent, we may renumber and argue as before to show that

span{Xl>X2,""Xn- 2} = X.

Eventually we reach a linearly independent subset which spans X, and thus get a basis, or else we reach a linearly dependent set {Xl} spanning X and con­sisting of one element. This implies Xl = 0, so X = {O}, and the empty set is the basis.

Now suppose (Xl> X2, ... , xn) and (Yl> Y2, ... , Ym) are bases of X. and suppose m :::; n. If n = 0, then m = O. Otherwise Xl "# O. The y/s span X, so Xl = "1 ajYj. Renumbering, we may assume a1 "# O. Then

Thus Y1 is a linear combination of Xl> Y2, ... , Ym' It follows easily that span {XH Y2,' .. , Ym} ::::> span {Yl> Y2,· .. ,ym} = X. If m = 1 this shows that span {Xl} = X, and the linear independence of the x/s then implies n = 1. Otherwise X2 = bXI + "11'= 2 bjYj. The independence of Xl and X2 implies some bi "# O. Renumbering, we assume b2 "# O. Then

Y2 = b2- 1(X2 - bX1 - i biYi ). 1=3

This implies that

span {Xl> X2, Y3, ... , Ym} ::::> span {Xl> Y2, ... , Ym} = X.

Continuing in this way, we see that after the y/s are suitably renumbered, each set {Xl> X2,"" Xb Yk+l>"" Ym} spans X, k :::; m. In particular, takitlg k = m we have that {Xl> X2, ... , xm} spans X. Since the x/s were assumed linearly independent, we must have n :::; m. Thus n = m. 0

If X has a basis with n elements, n = 0, 1, 2, ... , then any basis has n elements. The number n is called the dimension of n. We write n = dim X.

The argument used to prove Theorem 7.3 proves somewhat more.

Theorem 7.4. Suppose X is a finite-dimensional vector space with dimen­sion n. Any subset X which spans X has at least n elements. Any subset of X which is linearly independent has at most n elements. An ordered subset of n elements which either spans X or is linearly independent is a basis.

32 Basic concepts

Suppose (Xl' X2, ... , xn) is a basis of X. Then any X E X can be written as a linear combination X = 2: ajxj • The scalars aI, a2, ... , an are unique; in fact if X = 2: bjxj, then

o = x - x = L (aj - bj)Xi'

Since the x/s are linearly independent, each aj - bj = 0, i.e., ai = bj. Thus the equation x = 2: ajxj associates to each x E X a unique ordered n-tuple (aI' a2, ... , an) of scalars, called the coordinates ofx with respect to the basis (Xl' ... , xn). Note that if x and y correspond respectively to (ab a2, ... , an) and (bb b2, ... , bn), then ax corresponds to (aal, aa2, ... , aan) and x + y corresponds to (al + bl , a2 + b2, ... , an + bn). In other words, the basis (Xl> X2, ... , xn) gives rise to a function from X onto IRn or en which preserves the vector operations.

Suppose X and Yare vector spaces, either both real or both complex. A function T: X -+ Y is said to be linear if for all vectors x, x' E X and all scalars a,

T(ax) = aT(x), T(x + y) = T(x) + T(y).

A linear function is often called a linear operator or a linear transformation. A linear function T: X -+ IR (for X a real vector space) or T: X -+ e (for X a complex vector space) is called a linear functional.

Examples

1. Suppose X is a real vector space and (Xl> x2, ... , xn) a basis. Let T(2: ajxj) = (aI' a2, ... , an). Then T: X -+ IRn is a linear transformation.

2. Let T(z) = z*, Z E C. Then T is a linear transformation of e into itself if e is considered as a real vector space, but is not linear if e is considered as a complex vector space.

3. Letjf: IRn -+ IR be defined by jf(Xl' X2, ... , xn) = Xj' Thenjf is a linear functional.

4. Let X be the space of polynomials with real coefficients. The two func­tions S, T defined below are linear transformations from X to itself. If f(x) = I7=0 aixi, then

" S(f)(x) = L U + 1)-lajxi+1, 1=0 " T(f)(x) = L jal x i - 1.

i=l

Note that T(S(f» = J, while S(T(f» = fif and only if ao = O.

Exercises

1. If the linearly independent finite set {Xb X2, ... , x,,} does not span X, then there is a vector X,,+1 E X such that {Xb X2, ... , x,,, X"+1} is linearly independent.

Vector spaces 33

2. If X is finite-dimensional and Xl, X2, ••• , Xn are linearly independent, then there is a basis of X containing the vectors Xl> X2, ... , Xn•

3. If X is a finite-dimensional vector space and Y c X is a subspace, then Y is finite-dimensional. Moreover, dim Y ::::; dim X, and dim Y < dim X unless Y = X.

4. If Y and Z are subspaces of the finite-dimensional vector space X and Y n Z = {O}, then dim Y + dim Z ::::; dim X.

5. Suppose Y and Z are subspaces of the finite-dimensional vector space X, and suppose span (Y u Z) = X. Then dim Y + dim Z ~ dim X.

6. If Y is a subspace of the finite dimensional vector space X, then there is a subspace Z with the properties Y n Z = fO}; dim Y + dim Z = dim X; any vector X E X can be expressed uniquely in the form X = Y + Z, where Y E Y, Z E Z. (Such subspaces are said to be complementary.)

7. Prove Theorem 7.4. 8. The polynomials fm(x) = xm, 0 ::::; m ::::; n, are a basis for the vector

space of polynomials of degree ::::; n. 9. The vector space of all polynomials is infinite dimensional. 10. If S is a non-empty set, the vector space F(S; JR) of functions from

S to ~ is finite dimensional if and only if S is finite. 11. If X and Yare vector spaces and T: X -+ Y is a linear transformation,

then the sets

NCT) = {x I x E X, T(x) = O} R(T) = {T(x) I x E X}

are subspaces of X and Y, respectively. (They are called the null space or kernel ofT, and the range ofT, respectively.) Tis 1-1 ifand only ifN(T) = {O}.

12. If X is finite dimensional, the subspaces N(T) and R(T) in problem 11 satisfy dim N(T) + dim R(T) = dim X. In particular, if dim Y = dim X, then Tis 1-1 if and only if it is onto. (Hint: choose a basis for N(T) and use problem 2 to extend to a basis for X. Then the images under T of the basis elements not in N(T) are a basis for R(T).)

Chapter 2

Continuous Functions

§1. Continuity, uniform continuity, and compactness

Suppose (S, d) and (S', d') are metric spaces. A function f: S -)- S' is said to be continuous at the point XES if for each e > ° there is a 0 > Osuch that

d'(f(x),J(y)) < e if d(x, y) < O.

In particular, if Sand S' are subsets of ~ or of C (with the usual metrics) then the condition is

If(y) - f(x)1 < e if Iy - xl < O.

(This definition is equivalent to the following one, given in terms of con­vergence of sequences: f is continuous at x if f(xn) ~ f(x) whenever (xn);:'= 1

is a sequence in S which converges to x. The equivalence is left as an exercise.)

Recall that we can add, multiply, and take scalar multiples of functions with values in C (or ~): iff, g: S ~ C and a E C, XES, then

(f + g)(x) = f(x) + g(x), (af)(x) = af(x) , (fg)(x) = f(x)g(x),

(flg)(x) = f(x)/g(x) if g(x) -# 0.

Proposition 1.1. Suppose (S, d) is any metric space, and suppose f, g: S ~ C are functions which are continuous at x. Then f + g, af, and fg are continuous at x. Ifg(x) -# 0, then fig is defined in a ball BrCx) and is continuous at x.

Proof Continuity of f + g and af at x follow from the definition of continuity and the inequalities

l(f + g)(y) - (f + g)(x) I = If(y) - f(x) + g(y) - g(x) I :$; !f(y) - f(x)! + !g(y) - g(x)!,

I(af)(y) - (~f)(x)1 = lallf(y) - f(x)l·

To show continuity offg at x we choose 01 > ° so small that if dey, x) < 01

then !f(y) - f(x)! < 1. Let M be the larger of !g(x)1 and !f(x)! + 1. Given e > 0, choose 0 > ° so small that

If(y) - f(x) I < e12M, Ig(y) - g(x) I < el2M

34

Continuity, uniform continuity, and compactness

if dey, x) < O. Then dey, x) < 0 implies

l(fg)(y) - (fg)(x) I = If(y)g(y) - f(x)g(x) I

But also

= If(y)g(y) - f(y)g(x) + f(y)g(x) - f(x)g(x) I ~ If(y)llg(y) - g(x) I + If(y) - f(x) I I g(x) I ~ If(y)I'e/2M + M·e/2M = If(y)I'e/2M + e/2.

If(y) I = If(y) - f(x) + f(x) I ~ If(y) - f(x) I + If(x) I < 1 + If(x) I ~ M,

so l(fg)(y) - (fg)(x) I < e.

35

Finally, suppose g(x) =F O. Choose r > 0 so that Ig(y) - g(x) I < tlg(x)1 if dey, x) < r. Then if dey, x) < r we have

I g(x) I = Ig(y) + g(x) - g(y)1 ~ Ig(Y)1 + tlg(x)l,

so Ig(Y)1 ~ tlg(x)1 > O. Thus l/g is defined on Br(x). Since the product of functions continuous at x is continuous at x, we only need show that l/g is continuous at x. But if y E Br(x), then

II/g(y) - l/g(x) I = Ig(y) - g(x)I/lg(y)llg(x)1 ~ Klg(y) - g(x)l,

where K = 2/1 g(x)i2. Since g is continuous at x, it follows that l/g is. 0

A function f: S -+ S' is said to be continuous if it is continuous at each point of S.

The following is an immediate consequence of Proposition 1.1.

Corollary 1.2. Suppose f, g: S -+ C are continuous. Then f + g, af, and fg are also continuous. Ifg(x) =F 0, all x, thenflg is continuous.

A functionf: S -+ S' is said to be uniformly continuous if for each e > 0 there is a 0 > 0 such that

d'(f(x),J(y» < e if d(x, y) < 8.

In particular, if S, S' c C, then this condition reads

If(y) - f(x) I < e if Iy - xl < 8.

The distinction betwe encontinuity and uniform continuity is important. If f is continuous, then for each x and each e > 0 there is a 8 > 0 such that the above condition holds; however, 8 may depend on x. As an example, let S = S' = ~,f(x) = x 2• Then If(y) - f(x)1 = ly 2 - x2 1 = Iy + xlly - xl­If Ixl is very large, then Iy - xl must be very small for If(y) - f(x) I to be less than 1. Thus this function is not uniformly continuous. (However it is clear that any uniformly continuous function is continuous.)

In one important case, continuity does imply uniform continuity.

36 Continuous functions

Theorem 1.3. Suppose (S, d) and (S', d') are metric spaces and suppose I: S ~ S' is continuous. If S is compact, then I is uniformly continuous.

Proof Given e > 0, we know that for each XES there is a number 8(x) > 0 such that

d'(f(x),f(y» < te if d(x, y) < 28(x).

Let N(x) = Bo(x)(x). By the definition of compactness, there are points Xl, X2, ... , Xn E S such that S C U N(xj). Let 8 = min {8(Xl)' 8(X2), ... , 8(xn)}, and suppose d(x, y) < 8. There is some Xt such that X E N(xt). Then

d(Xh x) < S(Xj) < 28(xt), d(Xh y) < d(Xh x) + d(x, y) < 8(xt) + 8 ::::;; 28(xj),

so

d'(f(x),f(y» ::::;; d'(f(x),f(Xt» + d'(f(xt),f(y» ::::;;~+~=~ 0

There are other pleasant properties of continuous functions on compact sets. A function I: S ~ C is said to be bounded if/(S) is a bounded set in C, i.e., there is an M ~ 0 such that

I/(x) I ::::;; M, all XES.

Theorem 1.4. Suppose (S, d) is a metric space and suppose I: S ~ C is continuous. Then I is bounded and there is a point Xo E S such that

I/(xo) I = sup {1/(x) I Ix E S}.

IfI(S) c Ill, then there are X+, x_ E S such that

I(x+) = sup {/(x) 1 XES}, I(x-) = inf{f(x) 1 XES}.

Proof For each XES, there is a number 8(x) > 0 such that I/(y) - I(x) I < 1 if Y E Bo(x)(x) = N(x). Choose X1o" ., Xn such that S c U N(xj). If XES then x E N (Xt) for some i, and

I/(x) I ::::;; I/(Xj) I + I/(x) - I(xt) I < I/(Xi) I + 1.

Thus we can take M = 1 + max{l/(xl)I, ... , I/(xn)l} and we have shown that I is bounded.

Let a = sup {l/(x) I I XES}, and suppose I/(x) I < a, all XES. Then for each XES there are numbers a(x) < a and e(x) > 0 such that 1/(y) I ::::;; a(x) if y E B.(x)(x) = M(x). Choose Y1o' .. , Ym such that S c U M(Yt), and let al = max {aCYl), ... , a(Ym)} < a. If XES then x E M(Yt) for some i, so

I/(x) I < a(Yt) ::::;; al < a.

This contradicts the assumption that a = sup {1/(x) I I XES}. Thus there must be a point Xo with I/(x) I = a.

The proof of the existence of x+ and x_ when/is real-valued is similar, and we omit it. 0

Continuity, uniform continuity, and compactness 37

Both theorems above apply in particular to continuous functions defined on a closed bounded interval [a, b] c IR. We need one further fact about such functions when real-valued: they skip no values.

Theorem 1.5. (Intermediate Value Theorem). Suppose f: [a, b] ~ IR ;s continuous. Suppose either

f(a) ~ c ~ feb) or feb) ~ c ~ f(a).

Then there is a point Xo E [a, b] such that f(xo) = c.

Proof We consider only the case f(a) ~ c ~ feb). Consider the two intervals [a, !(a + b)], [-t(a + b), b]. For at least one of these, c lies between the values of f at the endpoints; denote this subinterval by [aI, bd. Thus f(al) ~ c ~ f(b l ). Continuing in this way we get a sequence of intervals [an, bn] with [an + 1> bn+l ) c [an, bn], bn+l - an+l = t(bn - an), andf(an) ~ c ~ f(bn). Then there is Xo E [a, b] such that an ~ Xo, bn ~ Xo. Thus

f(xo) = limf(an) ~ c. f(xo) = limf(bn) ~ c.

Exercises

o

1. Prove the equivalence of the two definitions of continuity at a point. 2. Use Theorem 1.6 to give another proof of the existence of V2. Prove

that any positive real number has a positive nth root, n = 1,2, .... 3. Supposef: S~ S', where (S, d) and (S', d') are metric spaces. Prove

that the following are equivalent: (a) fis continuous; (b) for each open set A' c S',j-l(A') is open; (c) for each closed set A' c S',j-l(A') is closed. 4. Find continuous functionsJj: (0, 1) ~ lR,j = 1,2,3, such that

fl is not bounded, f2 is bounded but not uniformly continuous, f3 is bounded but there are no points x+, x_ E (0,1) such thatf3(x+) =

sup {!3(X) I XE(O, 1)},j3(x_) = inf{f3(x) I XE(O, I)}. 5. Suppose f: S ~ S' is continuous and S is compact. Prove thatf(S) is

compact. 6. Use Exercise 5 and Theorem 6.2 of Chapter 1 to give another proof of

Theorem 1.4. 7. Use Exercise 3 of Chapter 1, §6 to give a third proof of Theorem 1.4.

(Hint: take (Xn);'=l c S such that lim If(xn)I = sup {f(x) I XES}, etc.) 8. Suppose (S, d) is a metric space, XES, and r > O. Show that there is

a continuous functionf: S ~ IR with the properties: 0 ~ fey) ~ 1, all YES, fey) = 0 if y rt Br(x),j(x) = 1. (Hint: takef(y) = max {I - r-ld(y, x), O}.

9. Suppose (S, d) is a metric space and 'suppose S is not compact. Show that there is a continuousf: S ~ IR which is not bounded. (Hint: use Exercise 8.)

38 Continuous functions

§2. Integration of complex-valued functions

A partition of a closed interval [a, b] c IR is a finite ordered set of points P = (xo, Xl> ••• , xn) with

a = Xo < Xl < ... < Xn = b.

The mesh of the partition P is the maximum length of the subintervals [XI-I, XI]:

IPI = max {XI - XI-l I i = 1,2, ... , n}.

Iff: [a, b] -+ C is a bounded function and P = (xo, Xl> ••• , xn) is a parti­tion of [a, b], then the Riemann sum off associated with the partition P is the number

n

S(f; P) = 2: f(xl)(x; - XI-I)' 1=1

The function f is said to be integrable (in the sense of Riemann) if there is number Z E C such that

lim S(f; P) = z. IPI .... O

More precisely, we mean that for any e > 0 there is a 8 > 0 such that

(2.1) IS(f;P) - zl < e if IPI < 8.

If this is the case, the number z is called the integral off on [a, b] and denoted by

ff or ff(X)dx.

If f: [a, b] -+ C is bounded, suppose If(x) I :::;; M, all X E [a, b]. Then for any partition P of [a, b],

IS(f; P)I :::;; 2: If(Xi) I (XI - XI-I) :::;; M 2: (XI - XI-l) = M(b - a).

Therefore, iff is integrable,

(2.2) If f I :::;; M(b - a), M = sup {If(x) I I X E [a, b]}.

Recall that f: [a, b] -+ C is a sum f = g + ih where g and hare real­valued functions. The functions g and h are called the real and imaginary parts off and are defined by

g(X) = Re (f(x», h(x) = 1m (f(x», X E [a, b].

We denote g by Refand h by 1m!

Proposition 2.1. A boundedfunctionf: [a, b] -+ C is integrable if and only if the real and imaginary parts off are integrable. If so,

ff= r Ref + if 1m! a a a

Integration of complex-valued functions 39

Proof Recall that if Z = x + iy, x, y E lR, then

(2.3) 11xl + 11yl ::; Izi ::; Ixl + Iyl·

If P is any partition of [a, b], then

S(f; P) = S(Ref; P) + is(Imf; P),

and S(Ref; P), S(Imf; P) are real. Let z = x + iy, x, y real, and apply (2.3) to S(f; P) - z. We get

1IS(Ref; P) - xl + 1IS(Imf; P) - yl ::; IS(f; P) - zl ::; IS(Ref; P) - xl + IS(Imf; P) - YI.

Thus S(f; P) -+ z as IP 1-+ 0 ifand only if S(Ref; P) -+ x and S(Imf; P)-+ yas I P 1--; o. 0

Proposition 2.2. Suppose f: [a, b] -+ C and g: [a, b] -+ C are bounded integrable functions, and suppose c E C. Then f + g and cf are integrable, and

f (f + g) = f f + r g, r cf = c r f Proof For any partition P of [a, b],

S(f + g; P) = S(f; P) + S(g; P), S(cf; P) = cS(f; P).

The conclusions follow easily from these identities and the definition. 0

Neither of these propositions identifies any integrable functions. We shall see shortly that continuous functions are integrable. The following criterion is useful in that connection.

Proposition 2.3. A boundedfunctionf: [a, b] -+ C is integrable if and only if for each e > 0 there is as> 0 such that

(2.4) IS(f;P) - S(f; Q)I < e if IPI, IQI < S.

Proof Suppose f is integrable, and let z = J: f For any e > 0 there is a S > 0 such that S(f; P) is in the disc of radius 1e around z if I P I < S. Then IPI < S, IQI < S implies S(f;P), S(f; Q) are at distance < e.

Conversely, suppose for each e > 0 there is as> 0 such that (2.4) holds. Take partitions Pn with IPnl < lin, n = 1,2,3, ... , and let Zn = S(f;Pn). It follows from our assumption that (Zn):=l is a Cauchy sequence. Let z be its limit. If n is large, IPnl is small and S(f; Pn) is close to z, and if I QI is small, S(f; Q) is close to S(f; Pn). Thus S(f; Q) -+ z as I QI-+ o. 0

Theorem 2.4. Iff: [a, b] -+ C is continuous, it is integrable.

Proof We know by §1 that f is bounded and uniformly continuous. Given e > 0, choose S > 0 so that If(x) - f(y) I < e if Ix - yl < S. Suppose P, Q are partitions of [a, b] with I P I < S, I Q I < S. Suppose P = (xo, Xl> ... ,

xn). Let P' be a partition which includes all points of P and of Q, P' = (Yo, Y1, ... , Ym). We examine one summand of S(f; P). Suppose

Xj-1 = Y1-1 < Y1 < ... < Yk = Xi·

40 Continuous functions

Then

k(X,)(Xi - Xj-l) - ~/(YI)(YI - YI-l) I = 1~1 (f(Xj) - f(YI»(YI - YI-l) I

k

~ e L (YI - YI-l) = e(XI - XI-l), 1=1

since each IXi - YII < 8. Adding, we get

IS(f; P) - S(/; P')I < e(b - a).

Similarly,

IS(f; Q) - S(/; P')I < e(b - a),

so IS(f; P) - S(f; Q)I < 2e(b - a).

By Proposition 2.3, I is integrable. 0

We now want to consider the effect of integrating over subintervals.

Proposition 2.5. Suppose a < b < c and I: [a, c] -+ C is bounded. Then lis integrable if and only ifit is integrable as alunction on [a, b] and on [b, c]. lfso, then

f/= f/+ ff. Proof. Suppose I is integrable on [a, b] and on [b, c]. Given e > 0,

choose 8 > 0 so that

(2.5) IS(f;p) - {II < ~e, IS(f; Q) = f/l < ~e if P, Q are partitions of [a, b], [b, c] respectively, IPI < 8, I QI < 8. Suppose P' is a partition of [a, c], IP'I < 8. If b is a point of P', then P' determines partitions P of [a, b] and Q of [b, c], IP I < 8, 101 < 8. It follows from (2.5) that

(2.6) IS(f; P') - .C I - r I I < e.

If b is not a point of P', let P" be the partition obtained by adjoining b. Then (2.6) holds with P" in place of P'. Suppose I/(x) I ~ M, all x E [a, b]. The sums S(f; P') and S(f; PH) differ only in terms corresponding to the sub­interval determined by P' which contains b. It is easy to check, then, that

IS(f; P') - S(f; P")I < 28M. Thus

IS(f;p') - ('/- f/l < e + 28M

Integration of complex-valued functions 41

if IP'I < S. It follows thatfis integrable with integral I:f + I:f Conversely, suppose f is not integrable on [a, b] or [b, cl; say it is not

integrable on [a, b]. Then there is an e > 0 such that for any S > 0 there are partitions Pl' P2 of [a, b] with IPll < S, IP2 1 < S, but

IS(I; Pl) - S(I; P2)1 > e.

Let Q be a partition of [b, c] with I QI < S, and let P~, P~ be the partitions of [a, c] containing all points of Q and of Plo P2 respectively. Then IP~I < S, IP~I < S, and

IS(I; PD - S(I; P~)I = IS(I; Pl) - S(f; P2)1 > e.

By Proposition 1.3,fis not integrable. 0

Supposef: [ao, bol -?- C is integrable, and suppose a, b E [ao, bolo If a < b, then f is integrable on [a, b]. (In fact f is integrable on [ao, b], therefore on [a, b], by two applications of Proposition 2.5.) If b < a, thenfis integrable on [b, a] and we define

We also define

Then one can easily check, case by case, that for any a, b, c E [ao, bo],

(2.7) LCf= ff+ ff It is convenient to extend the notion of the integral to certain unbounded

functions and to certain functions on unbounded intervals; such integrals are called improper integrals. We give two examples, and leave the remaining cases to the reader.

Suppose f: (ao, b] -?- C is bounded and integrable on each subinterval [a, b], ao < a < b. We set

(2.8)

if the limit exists. Suppose f: [a, (0) -?- C is bounded and integrable on each subinterval

[a, b], a < b < 00. We set

(2.9) fcc f= lim fbf a b-+oo a

if the limit exists; this means that there is a Z E C such that for each E > 0

if b ~ bee).

42 Continuous functions

Exercises

1. Let f: [0, 1] ~ ~ with f(x) = ° if x is irrational, f(x) = 1 if x is rational. Show thatfis not integrable.

2. Letf: [0, 1] ~ ~ be defined by: f(O) = 0, f(x) = sin (l/x) if x i= 0. Sketch the graph. Show that f is integrable.

3. Supposej, g: [a, b] ~ IC are bounded,fis integrable, and g(x) = f(x) except on a finite set of points in [a, b]. Then g is integrable and f: g = f: f

4. Suppose f: [a, b] ~ IC is bounded and is continuous except at some finite set of points in [a, b]. Show that f is integrable.

5. Suppose f: [a, b] ~ IC is continuous and f(x) ~ 0, all x E [a, b]. Show that f:f = ° impliesf(x) = 0, all x E [a, b].

6. Supposef: [a, b] ~ IC is bounded, integrable, and real-valued. Suppose

If f 1= M(b - a), where M = sup {If(x) I I x E [a, b]}.

Show that f is constant. 7. Do Exercise 6 without the assumption thatfis real-valued. 8. Let f: [0, 1] ~ IC be defined by: f(x) = ° if x = ° or x is irrational,

f(x) = l/q if x = p/q, p, q relatively prime positive integers. Show that f is continuous at x if and only if x is zero or irrational. Show thatfis integrable and f:f= 0.

§3. Differentiation of complex-valued functions

Suppose (a, b) is an open interval in ~ and thatf: (a, b) ~ IC. As in the case of a real-valued function, we say that the functionfis differentiable at the point x E (a, b) if the limit

(3.1) lim fey) - f(x) y .... x y - x

exists. More precisely, this means that there is a number Z E C such that for any e > 0, there is a 8 > ° with

(3.2) l(f(y) - f(x))(y - X)-l - zl < e if ° < Iy - xl < 8.

If so, the (unique) number z is called the derivative off at x and denoted variously by

f'(x) , Df(x), or df (x). dx

Proposition 3.1. Iff: (a, b) ~ C is differentiable at x E (a, b) then f is continuous at x.

Differentiation of complex-valued functions 43

Proof Choose 8 > ° so that (3.2) holds with z = f'(x) and e = 1. Then when Iy - xl < 8 we have

If(y) - f(x) I :,; If(y) - f(x) - z(y - x)1 + Iz(y - x)1 < (1 + Izj)ly - xl·

As y -* x,j(y) -* f(x). 0

Proposition 3.2. Thefunctionf: (a, b) -* C is differentiable at x E (a, b) if and only if the real and imaginary parts g = Re f and h = 1m fare differentiable at x. If so, then

f'(x) = g'(x) + ih'(x).

Proof As in the proof of Proposition 2.l, the limit (3.l) exists if and only if the limits of the real and imaginary parts of this expression exist. If so, these are respectively g'(x) and h'(x). 0

Proposition 3.3. Suppose f: (a, b) -* C and g: (a, b) -* C are differenti­able at x E (a, b), and suppose c E C. Then the functions f + g, cf, and fg are differentiable at x, and

(f + g)'{x) = f'(x) + g'(x), (cf)'(x) = cf'(x), (fg)'{x) = f'(x)g(x) + f(x)g'(x).

If g(x) =F ° then fig is differentiable at x and

(fjg)'(x) = [f'(x)g(x) - f(x)g'(x)]g(X)-2.

Proof This can be proved by reducing it to the (presumed known) theorem for real-valued functions, using Proposition 3.2. An alternative is simply to repeat the proofs, which are no different in the complex case. We shall do this for the product, as an example. We have

(fg)(y) = (fg)(x) = f(y)g(y) - f(x)g(x) = [fey) - f(x)]g(y) + f(x) [g(y) - g(x)].

Divide by (y - x) and let y -* x. Since g(y) -* g(x), the first term converges to f'(x)g(x). The second converges to f(x)g'(x). 0

We recall the following theorem, which is only valid for real-valued functions.

Theorem 3.4. (Mean Value Theorem). Suppose f: [a, b] -* IR is contin­uous, and is differentiable at each point of (a, b). Then there is aCE (a, b) such that

f'(c) = [feb) - f(a)](b - a)-I.

Proof. Suppose first thatf(b) = f(a). By Theorem 1.5 there are points c+ and c_ in [a, b] such thatf(c+) ~ f(x), all x E [a, b] andf(c_) :,; f(x), all x E [a, b]. If c+ and c_ are both either a or b, thenfis constant andf'(c) = 0,

44 Continuous functions

all c E (a, b). Otherwise, suppose c+ E (a, b). It follows that [fey) - f(c+)] x (y - c +) - 1 is :::; ° if y < c + and ~ ° if y > c +. Therefore thelimit as y -+ c + is zero. Similarly, if c_ "# a and c_ "# b, thenf'(c_) = 0. Thus in this case f'(c) = ° for some c E (a, b).

In the general case, let

g(x) = f(x) - (x - a)[f(b) - f(a)](b - a)-l.

Then g(a) = f(a) = g(b). By what we have just proved, there is aCE (a, b) such that

0= g'(c) =f'(c) - [feb) - f(a)](b - a)-l. o Corollary 3.5. Supposef: [a, b] -+ C is continuous, and is differentiable at

each point of (a, b). Iff'(x) = 0for each x E (a, b), thenfis constant. Proof Let g, h be the real and imaginary parts off Then g'(x) = h'(x) =

0, X E (a, b). We want to show g, h constant. If [x, y] c [a, b], Theorem 3.1 applied to g, h on [x, y] implies g(x) = g(y), hex) = hey). 0

Theorem 3.6. (Fundamental Theorem of Calculus). Suppose f: [a, b] -+

C is continuous and suppose c E [a, b]. The function F: [a, b] -+ C defined by

F(x) = LX f

is differentiable at each point x of (a, b) and

F'(x) = f(x).

Proof Let g be the constant function g(y) = f(x), y E (a, b). Given e > 0, choose 0 so small that If(y) - g(y)1 = If(y) - f(x) I < eif Iy - xl < O. Then

(3.3) F(y) - F(x) = i Y f - LX f = f f = f g + f (1- g)

= f(x)(y - x) + f (I - g).

If Iy - xl < 0, then

If (f - g) I:::; Iy - xl sup {If(r) - f(x)lllr - xl < Iy - xl}

< elY - xl· Thus dividing (3.3) by (y - x) we get

I[F(y) - F(x)](y - X)-l - f(x) I < e. o Theorem 3.7. Suppose f: [a, b] -+ ~ is continuous and differentiable at each

point of (a, b) and supposef'(x) > 0, all i E (a, b). Thenfis strictly increasing

Differentiation of complex-valued functions 4S

on (a, b). For each y E [f(a)f(b)] there is a unique point x = g(y) E [a, b] such that f(x) = y. The function g = f- 1 is differentiable at each point of (f(a),f(b» and

g'(y) = [f'(g(y))]-I.

Proof If x, y E [a, b] and x < y, application of Theorem 3.4 to [x, y] shows that f(x) < f(y). In particular, f(a) < f(b). By Theorem 1.6, if f(a) ::; y ::; f(b) there is x E [a, b] withf(x) = y. Sincefis strictly increasing, x is unique. Letting g = f- 1 we note that g is continuous. In fact, suppose y E (f(a),f(b» and e > O. Take y', y" such that

f(a) =:;; y' < y < y" =:;;f(b)

and y" - y ::; e, y - y' =:;; e. Let x' = g(y'), x = g(y), x" = g(y"). Then x' < x < x". Let S = min {x" - x, x - x'}. If Ix - wi < S then w E (x', x"), sof(w) E (y', y"), so If(w) - f(x) I = If(w) - yl < e. Continuity atf(a) and f(b) is proved similarly.

Finally, let x = g(y), x' = g(y'). Then

(3.3) g(y') - g(y) = x' - x .

y' - y f(x') - f(x)

As y' -+ y, we have shown that x' -+ x. Thus (3.3) converges to f'(X)-l = [f'(g(y»]-I. 0

Proposition 3.8. (Chain rule). Suppose g: (a, b) -+ IR is differentiable at x, and suppose f: g«a, b) -+ IC is differentiable at g(x). Then the composite function fog is differentiable at x and

(3.4) (f 0 g)'(x) = f'(g(x»g'(x).

Proof We have

(3.5) f 0 g(y) - f 0 g(x) = f(g(y» - f(g(x»

= f(g(y» - f(g(x» .g(y) - g(x) .(y _ x) g(y) - g(x) y - x

if g(y) =F g(x). If g'(x) =F 0 then g(y) =F g(x) if y is close to x and y =F x. Taking the limit as y -+ x we get (3.4). Suppose g'(x) = O. For each y near x either g(y) = g(x), so f 0 g(y) - f 0 g(x) = 0, or (3.5) holds. In either case, [fog(y) - fog(x)](y - X)-1 is close to zero for y near x. 0

Proposition 3.9. (Change of variables in integration). Suppose g: [a, b] -+ IR is continuous, and is differentiable at each point of (a, b). Supposef: g([a, b]) -+ IC is continuous. Then

fg(b) fb f = (fo g)g'.

g(a) a

Proof Define F: g([a, b]) -+ IC and G: [a, b] -+ IC by

F(y) = fY J, g(a)

G(x) = LX (fog)g'.

46 Continuous functions

We want to prove that F(g(b)) = G(b); we shall in fact show that Fog = G on [a, b]. Since F 0 g(a) = G(a) = 0, it suffices to prove that the derivatives are the same. But

(F 0 g)' = (F' 0 g)g' = (f 0 g)g' = G'. o A functionJ: (a, b) -+ C is said to be differentiable if it is differentiable at

each point of (a, b). If J is differentiable, then the derivative f' is itself a function from (a, b) to C which may (or may not) have a derivative f"(x) = (f')'(x) at x E (a, b). This is called the second derivative ofJat x and denoted also by

Higher derivatives are defined similarly, by induction:

J<°)(x) = J(x), J<l)(X) = f'(x), J<k+l)(X) = (f(k»)'(X) , k = 0, 1,2, ....

The functionJ: (a, b) -+ C is said to be of class C k, or k-times continuously differentiable, if each of the derivatives f,f', ... ,J(k) is a continuous function on (a, b). The function is said to be a class Coo, or infinitely differentiable, if

J<k) is continuous on (a, b) for every integer k ~ 0.

Exercises

1. Show that any polynomial is infinitely differentiable. 2. Show that the Mean Value Theorem is not true for complex-valued

functions, in general, by finding a differentiable function J such that J(O) = ° = J(l) butf'(x) "" ° for ° < x < 1.

3. State and prove a theorem analogous to Theorem 3.7 whenf'(x) < 0, all x E (a, b).

4. Suppose f, g are of class C k and c E Co Show that J + g, cf, and Jg are of class Ck.

5. Suppose p is a polynomial with real coefficients. Show that between any two distinct real roots of p there is a real root of p'.

6. Show that for any k = 0, 1,2, ... there is a function/: IR -+ IR which is of class C k, such that J(x) = ° if x :s; 0, J(x) > ° if x > 0. Is there a function of class Coo having this property?

7. Prove the following extension of the mean value theorem: ifJand g are continuous real-valued functions on [a, b], and if the derivatives exist at each point of (a, b), then there is c E (a, b) such that

[feb) - J(a)]g'(c) = [g(b) - g(a)]f'(c).

Sequences and series of functions 47

8. Prove L'Hopital's rule: iff and g are as in Exercise 7 and if

limf'(x)[g'(x)] -1 x-a

exists andf(a) = g(a) = 0, then

limf(x)[g(x)] -1 x-a

exists and the two limits are equal.

§4. Sequences and series of functions

Suppose that S is any set and thatf: S ----+ C is a bounded function. Let

If I = sup {If(x) I I XES}.

A sequence of bounded functions (fn)';= 1 from S to C is said to converge uniformly to f if

lim Ifn - fl = 0, n- 00

This sequence (fn)'; = 1 is said to be a uniform Cauchy sequence if for each e > ° there is an integer N so that

(4.1) Ifn - fml < e if n, m 2': N.

It is not difficult to show that if the sequence converges uniformly to a functionf, then it is a uniform Cauchy sequence. The converse is also true.

Theorem 4.1. Suppose (fn):=l is a sequence of bounded functions from a set S to C which is a uniform Cauchy sequence. Then there is a unique bounded functionf: S ----+ C such that (fn):=l converges uniformly to f If S is a metric space and each fn is continuous, then f is continuous.

Proof For each XES, we have

Ifn(x) - fm(x)1 ::; Ifn - fml·

Therefore (fn(x)):= 1 is a Cauchy sequence-in C. Denote the limit by f(x). We want to show that (fn):= 1 converges uniformly to the function f defined in this way. Given e > 0, take N so large that (4.1) holds. Then for a fixed m2':N,

Ifm(x) - f(x) I = Ifm(x) - lim fn(x) I n- 00

= lim Ifm(x) - fn(x) I ::; e. n- 00

Thus Ifm - fl ::; e if m 2': N, and (fn):=l converges uniformly to f If the sequence also converged uniformly to g, then for each XES,

Ifn(x) - g(x) I ::; Ifn - gl

48 Continuous functions

so In(x) -+ g(x) as n -+ 00, and g = f The function lis bounded, since

I/(x) I = I/(x) - Im(x) + Im(x) I ::; e + I/m(x) I ::; e + I/ml, so III::; e + I/ml, ifm ~ N.

Finally, suppose each In is continuous on the metric space S. Suppose XES and e > O. Choose N as above. Choose 0 > 0 so small that

if dey, x) < o. Then

I/(y) - I(x) I ::; I/(y) - IN(Y) I + I/N(Y) - IN(X) I + I/N(X) - I(x) I ::; II - INI + I/N(Y) - IN (X) I + II - INI < 3e if dey, x) < o.

Thus I is continuous. 0

The usefulness of the notion of uniform convergence is indicated by the next theorem and the example following.

Theorem 4.2. Suppose (fn)': = 1 is a sequence 01 continuous complex-valued lunctions on the interval [a, b], and suppose it converges uniformly to f Then

fb I = lim fb In. a n-+oo a

Proof By (2.2),

If In - f I I = If (fn - I) I ::; lin - 11·lb - al·

As n -+ 00, this -+ O. 0

Example

For each positive integer n, letln : [0, I] -+ lR be the function whose graph consists of the line segments joining the pairs of points (0,0), «2n)-I, 2n); «2n)-I, 2n), (n-I, 0); (n-I, 0), (1, 0). Then In is continuous, In(x) -+ 0 as n -+ 00 for each x E [0, I], but f: In = I, all n.

Here we are interested particularly in sequences of functions which are partial sums of power series. Associated with the sequence (an):'=o in C and the point Zo E C is the series

00

(4.2) 2: an(z - zo)n, ZEC. n=O

Recall from §3 of Chapter I that there is a number R, 0 ::; R ::; 00, such that (4.2) converges when Iz - zol < R and diverges when Iz - zol > R; R is called the radius 01 convergence of (3.2). The partial sums

n

(4.3) In(z) = 2: am(z - zo)m m=O

Sequences and series of functions 49

are continuous functions on C which converge at each point z with Iz - zol < R to the function

00

(4.4) f(z) = L: am(z - zo)m, Iz - zol < R. m=O

Theorem 4.3. Let R be the radius of convergence of the power series (4.2)­Then the function f defined by (4.4) is a continuous function. Moreover, the functionsfn defined by (4.3) converge tofuniformly on each disc

Dr = {zllz - zol < r}, 0 < r < R.

Proof We prove uniform convergence first. Given ° < r < R, choose s with r < s < R. Take w with Iw - zol = s. By assumption, L an(w - zo)n converges. Therefore the terms of this series -+ O. It follows that there is a constant M so that

lan(w - zo)nl :::;; M, n = 0, 1, ....

Since Iw - zol = s, this means

(4.5) lanl :::;; Ms-n, n = 0, 1, ... ,

Now suppose z E Dr and m < n. Then

Ifn(z) - fm(z) I = Ii alz - zo)f I m+1 n

:::;; L: la,llz - zol' m=l

_ ~ -1 _ sm+1 - sn+1 Msm+1 -ML.,.S -M I-S :::;; I-S'

n+1

where S = rls < 1. As m -+ 00 the final expression on the right -+ 0, so (fn)': = 0 is a uniform Cauchy sequence on Dr. It follows that it converges to f uniformly on Dr and that f is continuous on Dr. Since this is true for each r < R, f is continuous. 0

In particular, suppose Xo E IR. The power series 00

(4.6) L: an(x - xo)n n=O

defines a continuous function in the open interval (xo - R, Xo + R). Is this function differentiable?

Theorem 4.4. Suppose the power series (4.6) has radius of convergence R' Then the function f defined by this series is differentiable, and

00

(4.7) f'(x) = L: nan(x - xo)n-l, Ix - xol < R. n=l

50 Continuous functions

Proof To simplify notation we shall assume Xo = O. We claim first that the two series

<Xl

(4.8) L n2 lan lr n - 2

n=2

converge uniformly for Ixl :s; r < R. Take r < s < R. Then (4.5) holds. It follows that

i Imamxm-11 :s; M i ms-mrm-l m=l m=l

8 = rls < 1. Take e > 0 so small that (I + e)8 < l. By Exercise 4 of Chapter I, §3, m :s; (1 + e)m for all large m. Therefore there is a constant M' so that

m = 1,2, ....

Then

i Imamxm-11 :s; rMM' i (1 + e)m8m. n=l m=l

This last series converges, so the first series in (4.8) converges uniformly for Ixl :s; r. Similarly, m2 :s; (I + e)m for large m, and the second series in (4.8) converges uniformly for Ixl :s; r.

Let g be the function defined by the first series in (4.8). Recall that we are taking Xo to be O. We want to show that

(4.9) [fey) - f(x)](y - X)-l - g(x) -+ 0 as y -+ x.

Assume Ixl < r, Iyl < r. Then the expression in (4.9) is

<Xl

(4.10) L an[yn - xn - nxn-1(y - x)](y - X)-l. n=2

Now

where

Thus

if Ixl :s; r, Iyl:s; r. Then

yn _ xn - nxn-1(y - x) = (y _ x)[yn-l + xyn-2 + ... + xn- 1 _ nxn-1] = (y _ x)[(yn-l _ xn- 1) + (yn-2 _ xn-2)x + ... +(y _ x)xn-2

+ xn - 1 _ xn - 1 ]

= (y - x)2[gn_l(X, y) + gn-2(X, y)x + ... + gl(X, y)xn- 2],

Differential equations and the exponential function 51

so Iyn - xn - nxn- 1(y - x)1 :::; Iy - xI 2·n2rn- 2.

It follows that for Ixl :::; r, Iyl :::; r we have co

I [f(y) - f(x)](y - X)-l - g(x) I :::; 2: lanllY - xl 2n2rn- 2 ly - xl- 1 n=2

co

= Iy - xl 2: n2lanlrn- 2 = Kly - xl, n=2

K constant. Thus f'(x) = g(x). 0

Corollary 4.5. The function f in Theorem 4.5 is infinitely differentiable, and

co

(4.11) J<k)(X) = 2: n(n - I)(n - 2)··· (n - k + I)an(x - xo)n-k, n=k

Ix - xol < R.

Proof. This follows from Theorem 4.4 by induction on k. 0

In particular, if we take x = Xo in (4.11) then all terms ofthe series except the first are zero and (4.11) becomes

(4.12)

This means that the coefficients of the power series (4.6) are determined uniquely by the function f (provided the radius of convergence is positive).

Exercises

1. Find the function defined for Ixl < 1 by f(x) = 2::'=1 xnjn. (Hint: f(x) = f: 1'.)

2. Show that iff is defined by (4.6), then

Ix co

Xo f = n~o (n + l)-lan(X - xo)n+1.

3. Find the function defined for Ixl < 1 by f(x) = 2::'=1 nxn- 1. 4. Supposethereisasequence(xn):'=lsuchthatlxn+l - xol < IXn - xol,

xn ~ xo, and f(xn) = 0 for each n, where f is defined by (4.6). Show that f(x) = 0 for all x. (Hint: show that ao, al> a2, ... are each zero.)

§5. Differential equations and the exponential function

Rather than define the general notion of a "differential equation" here, we shall consider some particular examples. We begin with the problem of finding a continuously differentiable function E: IR ~ C such that

(5.1) E(O) = I, E'(x) = E(x), X E IR.

52 Continuous functions

Suppose there were such a function E, and suppose it could be defined by a power series

00

E(x) = L a"x". ,,=0

Then (5.1) and Theorem 4.4 imply

co 00

L na"x"-l = L a"x", ,,=1 ,,=0

or 00 00

L (n + l)a,,+lx" = L a"x". ,,=0 ,,=0

Since the coefficients are uniquely determined, this implies

a,,+l = a,,/(n + 1), n = 0, 1,2, ....

But

ao = E(O) = 1,

so inductively

a" = (n!)-l = [n(n - 1)(n - 2)·· .1]-1.

We have shown that if there is a solution of (5.l) defined by a power series, then it is given by

00

(5.2) E(x) = L (n!)-lx". ,,=0

The ratio test shows that (5.2) converges for all real or complex x, and application of Theorem 4.4 shows that E is indeed a solution of (5.1). We shall see that it is the only solution.

Theorem 5.1. For each a, c E C there is a unique continuously differenti­able function f: IR ~ C such that

(5.3) f(O) = c, f'(x) = af(x), XE!R.

This function is

co

(5.4) f(x) = cE(ax) = c L (n!)-la"x". ,,=0

Proof The function given by (5.4) can be found by the argument used to find E, and Theorem 4.4 shows that it is a solution of (5.3). To show uniqueness, supposefis any solution of (5.3), and let g(x) = E( -ax), so that

g(O) = 1,

Thenfg is differentiable and

g'(x) = - ag(x).

(fg)' = f'g + fg' = afg - afg = O.

Differential equations and the exponential function 53

Thereforefg is constant and this constant value isf(O)g(O) = c. If c "# 0, this impliesfg is never zero, so g is never zero. Then for any c,

f(x) = c/g(x), all x.

Thus f is unique. 0

This can be extended to more complicated problems.

Theorem 5.2. For each a, c E IC and each continuous function h: IR -?- IC there is a unique continuously differentiable function f: IR -?- IC such that

(5.5) f(O) = c, I'(x) = af(x) + hex), XE IR.

Proof Let fo(x) = E(ax), g(x) = E( -ax). As in the preceding proof, fog is constant, == 1. Therefore neither function vanishes. Any solution f of (5.5) can be written as

f = fdo, where f1 = gf.

Then

I' = fVo + fd~ = fifo + af,

so (5.5) holds if and only if

f1(0) = c, fHx)fo(x) = hex).

These conditions are equivalent to

f1(X) = c + {' gh.

Thus the unique solution of (5.5) is given by

(5.6) f(x) = cfo(X) + fo(x) f' gh

= cE(ax) + E(ax) f' E( -at)h(t) dt.

Now we consider equations involving the second derivative as well.

o

Theorem 5.3. For any b, c, do, d1 E IC and any continuous function h: IR -?- IC there is a unique f: IR -?- IC of class C2 which satisfies

(5.7)

(5.8)

f(O) = do, 1'(0) = dt>

f"(x) + bl'(x) + cf(x) = hex), X E IR.

Proof To motivate the proof, we introduce two operations on functions of class C1 from IR to IC: given such a function g, let

Dg =g', Ig = g.

If g is of class C2, let D2g = g". Then (5.8) can be written

D2f + bDf + clf = h.

54 Continuous functions

This suggests the polynomial Z2 + 2hz + c. We know there are roots aI, a2 of this polynomial such that

all z E C.

Thus

From the properties of differentiation it follows that

Let

(D - al1)[(D - a21)f] = D2f - (al + a2)Df + a2ad =/" + hf + cf

We have shown that f is a solution of (S.8) if and only if

(D - al1)g = g' - alg = h.

If also (S.7) holds, then g(O) = 1'(0) - a2f(0) = d1 - a2do. Thus f is a solution of (S.7), (5.8) if and only if

(S.9) f(O) = do, I' - ad= g,

where

(S.10)

But (S.lO) has a unique solution g, and once g has been found then (S.9) has a unique solution. It follows that (S.7), (5.8) has a unique solution. 0

Now we return to the function E,

co

E(z) = L (n!)-lzn, z E C. n=O

Define the real number e by

co

(S.11) e = E(I) = L (n!)-l. n=O

Theorem 5.4. The function E is a function from !R to !R of class C co •

Moreover,

(a) E(x) > 0, x E !R, (b) for each y > 0, there is a unique x E !R such that E(x) = y, (c) E(x + y) = E(x)E(y), all x, y E !R, (d) for any rational r, E(r) = eT•

Proof Since E is defined by a power series, it is of class C co. It is clearly real for x real, and positive when x ~ O. As above, E(x)E( -x) = 1, all x, so also E(x) > 0 when x < O.

Differential equations and the exponential function 55

To prove (b), we wish to apply Theorem 1. Taking the first two terms in the series shows (since y > 0) that E(y) > 1 + Y > y. Also, E(y-l) > y-l, so

E( _y-l) = E(y-l)-l < (y-l)-l = y.

Thus there is x E (- y-l, y) such that E(x) = y. Since E' = E> 0, E is strictly increasing and x is unique.

We have proved (c) when x = - y. Multiplying by E( -x)E( - y), we want to show

E(x + y)E(-x)E(-y) = 1, all x, y E 1ft

Fix y. This equation holds when x = 0, and differentiation with respect to x shows that the left side is constant.

Finally, repeated use of (c) shows that

Thus

E(nx) = E(x)n, n = 0, ± I, ± 2, ....

e = E(l) = E(n/n) = E(l/n)n, e1/n = E(l/n), n = 1,2,3, ....

em/n = (e1/n)m = E{l/n)m = E(m/n). o Because of (d) above and the continuity of E, it is customary to define

arbitrary complex powers of e by

co

(5.12) e2 = E(z) = L (nl)-lzn. n=O

The notation

(5.13) e2 = expz

is also common. We extend part of Theorem 5.4 to the complex exponential function.

(Recall that z* denotes the complex conjugate of z E C.)

Theorem 5.5. For any complex numbers z and w,

(5.14) E(z + w) = E(z)E(w) , E(z*) = E(z)*.

Proof. The second assertion can be proved by examining the partial sums of the series. To prove the first assertion, recall that we showed in the proof of Theorem 5.1 that E(zx)E( - zx) is constant, x E IR. Therefore E(z)E( -z) = E(0)2 = 1. We want to show

E(z + w)E( -z)E( -w) = I, all z, WE C.

Let

g(x) = E«z + w)x)E(-zx)E(-wx),

Differentiation shows g is constant. But g(O) = I. 0

X E 1ft

56 Continuous functions

The notation (S.12) and the identity (S.14) can be used to consolidate expressions for the solutions of the differential equations above. The unique solution of

1(0) = c, I' = al+ h

is

(S.lS) I(x) = ceax + f: ea(X-t)h(t) dt.

The unique solution of

1(0) = do, 1'(0) = d1 , f" + df' + cl2 = h

is given by

(S.16)

where

and aI, a2 are the roots of Z2 + bz + c.

Exercises

1. Find the solution of

1(0) = 1, 1'(0) = 0, f" - 21' + 1 = 0

by the procedure in Theorem S.3, and also by determining the coefficients in the power series expansion off.

2. Let J, g be the functions such that

1(0) = 1, g(O) = 0,

1'(0) = 0, g'(O) = 1,

f" + bl' + cl = 0, f" + bg' + cg = O.

Show that for any constants dh d2 the function h = dd + d2 g is a solution of

h" + bh' + ch = O.

Show that conversely if h is a solution of this equation then there are unique constants dh d2 E C such that h = dd + d2g. (This shows that the set of solutions of (*) is a two-dimensional complex vector space, and (J, g) is a basis.)

3. Suppose h(x) = L::'=o dnxn, the series converging for all x. Show that the solution of

1(0) = 0 = 1'(0), f" + bl' + cl = h

Trigonometric functions and the logarithm 57

is of the form 2:;=0 anxn, where this series converges for all x. (Hint: deter­mine the coefficients ao, al, ... inductively, and prove convergence.)

4. Suppose ZS + bz2 + cz + d = (z - al)(z - a2)(z - as), all z E C. Discuss the problem of finding a functionfsuch that

f(O) = eo, 1"(0) = e2, f'" + dl" + c!' + d = O.

§6. Trigonometric functions and the logarithm

In §5 the exponential function arose naturally from study of the differen­tial equation!, = f In this section we discuss solutions of one of the simplest equations involving the second derivative: I" + f = O.

Theorem 6.1. There are unique functions S, C: IR -+ C of class C2 such that

(6.1)

(6.2)

S(O) = 0, S'(O) = 1,

C(O) = 1, C'(O) = 0,

S" + S = 0,

C" + C = O.

Proof Existence and uniqueness of such functions is a consequence of Theorem 5.3. 0

Let us obtain expressions for Sand C using the method of Theorem 5.3. The roots of Z2 + 1 are z = ± i. Therefore

where

Thus

(6.3)

g(x) = eix.

Sex) = LX e-Hx-t)eit dt = e- ix I: e21t dt

= e-IX(2i)-le2it I~ = (2i)-le-IX(e2IX - 1),

1 Sex) = Ii (elX - e- 1X).

A similar calculation gives

(6.4)

Theorem 6.2. The functions S, C defined by (6.3) and (6.4) are rea/­valuedfunctions of class coo on R Moreover,

(a) S' = C, C' = - S, (b) S(X)2 + C(X)2 = 1, all x E IR,

58 Continuous functions

(c) there is a smallest positive number p such that C(p) = 0, (d) ifp is the number in (c), then

Sex + 4p) = Sex), C(x + 4p) = C(x), all x E R

Proof Since the exponential function exp (ax) is of class Coo as a func­tion of x for each a E C, Sand C are of class Coo. Since (exp (ix))* = exp ( - ix), Sand C are real-valued. In fact,

C(x) = Re (eiX), Sex) = 1m (eIX),

so

e1x = C(x) + is(x).

Differentiation of (6.3) and (6.4) shows S' = C, C' = - S. Differentiation of S2 + C 2 shows that S(X)2 + C(X)2 is constant; the value for x = ° is 1.

To prove (c), we suppose that C(x) =1= ° for all x > 0. Since C(I) = 1 and C is continuous, the Intermediate Value Theorem implies C(x) > 0, all x > 0. Since S' = C, S is then strictly increasing for x :?: 0. In particular, Sex) :?: S(l) > 0, all x :?: 1. But then

° < C(x) = C(l) + IX C'(t) dt = C(l) - LX Set) dt

::;; C(1) - r S(1) dt = C(1) - (x - I)S(l), x:?: 1.

But for large x the last expression is negative, a contradiction. Thus C(x) = ° for some x > 0. Let p = inf {x I x > 0, C(x) = a}. Then p :?: 0. There is a sequence (xn)l' such that C(xn) = 0, P ::;; Xn ::;; p + lin. Thus C(p) = 0, and p is the smallest positive number at which C vanishes.

To prove (d) we note that

1 = S(p)2 + C(p)2 = S(p)2,

so S(p) = ± 1. But Sea) = ° and S' = C is positive on [O,p), so S(p) > 0, Thus S(p) = 1. Consider Sex + p) as a function of x. It satisfies (6.2), and so by uniqueness we must have

Sex + p) = C(x), x E lR.

Similarly, - C(x + p) considered as a function of x satisfies (6.1), so

C(x + p) = -Sex), xER

Then

sex + 4p) = C(x + 3p) = -sex + 2p) = -C(x + p) = Sex), C(x + 4p) = -sex + 3p) = - C(x + 2p) = sex + p) = C(x). 0

We define the positive number 7r by

7r = 2p,

Trigonometric functions and the logarithm 59

p the number in (c), (d) of Theorem 6.3. We define the functions sine and cosine for all Z E C by

(6.5) 1 co

sinz = _.(eI2 - e- 12) = 2: [(2n + I)!]-1(-I)"z2"+1, 2l ,,=0

1 co cos z = - (e12 + e- 12) = 2: [2n!]-1( -I)"z2".

2 ,,=0 (6.6)

Note that because of the way we have defined 7r and the sine and cosine functions, it is necessary to prove that they have the usual geometric sig­nificance.

Theorem 6.3. Let y: [0, 27r] -+ 1R2 be defined by

yet) = (cos t, sin t).

Then y is a 1-1 mapping of [0, 27r] onto the unit circle about the origin in 1R2. The length of the arc of this circle from yeO) to yet) is t. In particular, the length of the unit circle is 27r.

Proof. We know from Theorem 6.2 that (cos t)2 + (sin t)2 = 1, so (cos t, sin t) lies on the unit circle. The discussion in the proof of Theorem 6.2 shows that on the interval [0, -!7r], cos t decreases strictly from 1 to 0 while sin t increases strictly from 0 to 1. Therefore, y maps [0,1-7r] into the portion of the circle lying in the quadrant x :?; 0; y :?; 0 in a 1-1 manner. Further­more, suppose 0 :5: x :5: 1, 0 :5: Y :5: 1, and x2 + y2 = 1. By the Intermediate Value Theorem and the continuity of cosine, there is a unique t E [0, !w] such that cos t = x. Then sin t :?; 0, (sin t)2 = 1 - x2 = y2, and y :?; 0, so sin t = y. Thus y maps [0, -tw] onto the portion of the circle in question.

Since cos (t + trr) = -sin t and sin (t + -tw) = cos t, the cosine de­creases from 0 to -1 and the sine decreases from 1 to 0 on [!7r, 7r]. As above, we find that y maps this interval 1-1 and onto the portion of the circle in the quadrant x :5: 0, Y :?; O. Continuing in this way we see that y does indeed map [0,27r) 1-1 onto the unit circle.

The length of the curve y from yeO) to yet) is usually defined to be the limit, if it exists, of the lengths of polygonal approximations. Specifically, suppose

o = to < t1 < t2 < ... < t" = t.

The sum of the lengths of the line segments joining the points y(tl - 1) and y(tl), i = 1,2, ... , n is

" (6.7) 2: [(cos tl - cos tl_1)2 + (sin tl - sin tl_1)2]1/2.

1=1

By the Mean Value Theorem, there are t; and t7 between tt-1 and ti such that

costt - costi_ 1 = -sint;(tl- tt-1), sin tl - sin tl- 1 = cos t7(t1 - tl- 1).

60 Continuous functions

Therefore, the sum (6.7) is

n L [(sin t;)2 + (cos t~)2](t1 - tl _ 1).

1=1

Since sine and cosine are continuous, hence uniformly continuous on [0, tl, and since (sin t)2 + (cos t)2 = 1, it is not hard to show that as the maximum length It I - tl - 1 1-+ 0, (6.7) approaches t. 0

This theorem shows that sine, cosine, and 1T as defined above do indeed have the usual interpretation. Next we consider them as functions from Cto C.

Theorem 6.4. The sine, cosine, and exponential/unctions have the/ollowing properties:

(a) exp (iz) = cos z + i sin z, all z E C, (b) sin (z + 21T) = sin z, cos (z + 21T) = cos z, exp (z + 21Ti) = exp z,

all z E C, (c) if W E C and w =ft 0, there is a z E C such that w = exp (z). If also

w = exp (z'), then there is an integer n such that z' = z + 2n1Ti.

Proof The identity (a) follows from solving (6.5) and (6.6) for exp (iz). By Theorem 2.2 and the definition of 1T,

exp (21Ti) = cos 21T + i sin 21T = 1.

Then since exp (z + w) = exp z exp w we get

exp (z + 21Ti) = exp z.

This identity and (6.5), (6.6) imply the rest of (b). Suppose w E C, w =ft 0. Let r = Iwl. If x, yare real,

lexp (iy)12 = Icos y + i sin Yl2 = (cos y)2 + (sin y)2 = 1.

Therefore

lexp (x + iy)12 = lexp x exp (iy)1 = lexp xl = exp x.

To have exp (x + iy) = w, then, we must have exp x = r. By Theorem 5.4 there is a unique such x E R. We also want exp (iy) = ,-lW = a + bi. Since ir- 1wl = 1, a2 + b2 = 1. By Theorem 6.3 there is a unique y E [0, 21T) such that cos y = a, sin y = b. Then exp (iy) = cos y + i sin y = a + ib. We have shown that there are x, y E R such that if z = x + iy,

exp z = exp x exp (iy) = r·r- 1 w = w.

Suppose z' = x' + iy', x', y' real, and exp z' = w. Then

r = Iwl = lexp z'l = lexp x'i = exp x',

x' = x. There is an integer n such that

2n1T S y' - y < 2n1T + 21T,

Trigonometric functions and the logarithm 61

or

y' = y + 2mr + h, h E [0, 2TT).

Since exp z' = exp z, we have

exp (iy) = exp (iy') = exp (i(y + 2nTT + h» = exp (iy + ih),

so

1 = exp (- iy) exp (iy + ih) = exp (ih).

Since hE [0, 2TT), this implies h = O. Thus y' = y + 2nTT. 0

The trigonometric functions tangent, secant, etc., are defined for complex values by

tan z = sin z/cos z, Z E C, cos z -1= 0,

etc. If w, Z E C and w = exp z, then z is said to be a logarithm of w,

z = log w.

Theorem 6.4 shows that any w -1= ° has a logarithm; in fact it has infinitely many, whose imaginary parts differ by integral multiples of 27T. Thus log w is not a function of w, in the usual sense. It can be considered a function by restricting the range of values of the imaginary part. For example, if w -1= ° the z such that exp z = w, 1m z E [a, a + 2TT) is unique, for any given choice of a E~.

If x > 0, it is customary to take for log x the unique real y such that exp y = x. Thus as a function from (0, (0) to ~, the logarithm is the inverse of the exponential function. Theorem 3.7 shows that it is differentiable, with

Thus

(6.8)

d -(logx) = dx

- eYI = e-10gx = x- 1 ( d )-1 dy y=logx •

log x = f: t- 1 dt, x> 0.

Exercises

1. Prove the identities

sin (z + w) = sin z cos w + cos z sin w, cos (z + w) = cos z cos w - sin z sin w

for all complex z, w. (Hint: use (6.5) and (6.6).)

62 Continuous functions

2. Show that tan x is a strictly increasing function from (-1-17, -trr) onto IR. Show that the inverse function tan -1 x satisfies

~ (tan- 1 x) = (1 + X2)-1 dx

3. Show that J: to (1 + x2) -1 dx = 17.

4. Show that

log (1 + x) = LX (1 + f)-1 dt,

5. Show that

-1 < x < 00.

to ( l)n-1 log (1 + x) = L - xn ,

n=1 n -1<x<l.

(Hint: use Exercise 4.)

§7. Functions of two variables

Suppose A is an open subset of 1R2, i.e., for each (xo, Yo) E A there is an open disc with center (xo, Yo) contained in A:

some r > O.

In particular, A contains (x, Yo) for each x in the open interval Xo - r < x < Xo + r, and A contains (xo, y) for each y in the open interval Yo - r < y < Yo + r.

SupposeJ: A -? C. It makes sense to ask whether J(x, Yo) is differentiable as a function of x at Xo. If so, we denote the derivative by

Other common notations are

Similarly, ifj(xo, y) is differentiable at Yo as a function of y we set

Dd(xo, Yo) = lim [f(xo, y) - J(xo, Yo)](Y - YO)-1. V-Yo

The derivatives Dd, Dd are called the first order partial derivatives of f The second order partial derivatives are the first order derivatives of the first order derivatives:

D1Y = D1(Dd), D1Dd = D1(Dd),

D22J = D2(Dd),

D2Dd= D2(Dd).

Functions of two variables 63

Other notations are

o2j oyox' etc.

Higher order partial derivatives are defined similarly. An (n + I)-order partial derivative of I is DIg or D2g, where g is an n-order partial derivative of f. The function I: A -l> C is said to be n-times continuously differentiable, or 01 class cn if all the partial derivatives of I of order ::5: n exist and are con­tinuous at each point of A. If this is true for every integer n, then I is said to be infinitely differentiable, or 01 class Coo.

Theorem 7.1. (Equality of mixed partial derivatives). If I: A -l> C is 01 class C 2, then DIDd = D2DIf.

Proof. Suppose (a, b) EA. Choose r > 0 so small that A contains the closed square with center (a, b), edges parallel to the coordinate axes, and sides of length 2r. Thus (x, y) E A if

Ix - al ::5: rand Iy - bl ::5: r.

In this square we apply the fundamental theorem of calculus to I as a function of x with y fixed, and conclude

I(x, y) = LX Dd(s, y) ds + I(a, y).

Let g(y) = I(a, y). We claim that

(7.1) Dd(x, y) = LX D2Dd(s, y) ds + g'(y).

If so, then differentiation with respect to x shows DIDd = D2DIf. To prove (7.1) we consider

e- 1(f(x, y + e) - I(x, y)) - IX D2Dd(s, y) ds - g'(y)

= r [e- 1(Dd(s,y + e) - Dd(s,y)) - D2Dd(s,y)]ds a

+ [e- 1(g(y + e) - g(y)) - g'(y)].

The second term in brackets -l> 0 as e -l> O. If I is real-valued, we may apply the Mean Value Theorem to the first term and conclude that for each sand y and for each small e, there is a point y' = y'(s, y, e) between y and y + e such that

(7.2) e- 1(Dd(s, y + e) - Dd(s, y)) - D2Dd(s, y) = D2Dd(s, y') - D2Dd(s, y).

Now Iy' - yl < e, so I(s, y') - (s, y)1 < E. Since D2Dd is uniformly con­tinuous on the square Ix - al ::5: r, Iy - hi ::5: r, it follows that the maximum value of (7.2) converges to zero as E -l> O. This implies convergence to zero of

64 Continuous functions

the integral of (7.1) with respect to s, proving (7.1) when / is real. In the general case, we look at the real and imaginary parts of/separately. 0

Remarks. In the course of proving Theorem 7.1 we have, in effect, proved the following. If/is a complex-valued function of class CI defined on a rectangle Ix - al < r1> Iy - hi < r 2 , then the derivative with respect to yof

is

LX Dd(s, y) ds.

Similarly, the derivative with respect to x of

f /(x, t) dt

is r Dd(x, t) dt.

We need one more result of this sort: if a = h, rl = S1> then

F(y) = f /(s, y) ds

is defined for Iy - al < rl. The derivative is

(7.3) f Dd(s, y) ds + /(y, y).

In fact

fy fY+8 F(y + e) - F(y) = a [f(s, y + e) - /(s, y)] ds + I /(S, y) dy.

Divide by e and let e -+ O. By the argument above, the first integral converges to

f Dd(s, y) ds.

In the second integral, we are integrating a function whose values are very close to /(y, y), over an interval oflength e. Then, dividing bye, we see that the limit is /(y, y).

We need two results on change of order of integration.

Theorem 7.2. Suppose / is a continuous complex valued/unction on the rectangle

A = {(x,y) I a::; x ::; h, c ::; y ::; d}.

Functions of two variables 65

Then the lunctions

g(x) = r I(x, t) dt, hey) = f I(s, y) ds

are continuous, and

f g(x) dx = f hey) dy.

Proof The preceding remarks show that g and h are not only continuous but differentiable. More generally,

r I(s, t) dt, LX I(s, t) ds

are continuous functions of s and of t respectively. Define

Fl(x, y) = r {( I(s, t) dt} ds,

F2(x, y) = r {LX I(~, t) dS} dt.

We want to show that Fl(b, d) = F2(b, d). The remarks preceding this theo­rem show that

Therefore, F2 - Fl is constant along each vertical line segment in the rectangle A. Similarly, DlFl = DlF2, so F2 - Fl is constant along each horizontal line segment. Since Fl(a, c) = F2(a, c) = 0, Fl == F2. D

The next theorem describes the analogous situation for integration over a triangle.

Theorem 7.3. Suppose I is a continuous complex-valued lunction defined on the triangle

A = {(x, y) I 0 :$ x :$ a,O :$ y :$ x}.

Then r {LX I(x, y) dY} dx = {a {( I(x, y) dX} dy.

Proof Consider the two functions of t, 0 :$ t :$ a, defined by

f {LX I(x, y) dy } dx, f {f I(x, y) dX} dy.

By the remarks following Theorem 7.1, the derivatives of these functions with respect to tare

f/(t, y) dy, f I(t, y) dy + f {O} dy.

66 Continuous functions

Thus these functions differ by a constant. Since both are zero when t = 0, they are identical. 0

Finally we need to discuss polar coordinates. If (x, y) E 1R2 and (x,y) =F (0,0), let

Then

so there is a unique 8, 0 ~ 8 < 2'17, such that cos 8 = r- 1x, sin 8 = r- 1y. This means

x = rcos 8, r = (x2 + y2)1/2,

y = r sin 8 8 = tan- 1 (yLx).

Thus any point p of the plane other than the origin is determined uniquely either by its Cartesian coordinates (x, y) or by its polar coordinates r, 8. A function defined on a subset of 1R2 can be expressed either as f(x, y) or g(r, 8). These are related by

(7.4)

(7.5)

f(x,y) = g((x2 + y2)1/2, tan-1 (y/x),

g(r, 8) = fer cos 8, r sin 8).

Theorem 7.4. Suppose f is a continuous complex-valued function defined on the disc

DB = {(x,y) I x2 + y2 < R}.

Suppose g is related to f by (7.5). Then

fB {ICB2-112)1/2 } LB {L2" } f(x, y) dx dy = g(r, 8)r d8 dr. -B _(B2_112)1/2 0 0

Proof Look first at the quadrant x ~ 0, y ~ O. For a fixed y ~ 0, if x ~ 0 then x = (r2 - y2)1/2. Proposition 3.9 on coordinate changes gives

We integrate the integral on the right over 0 ~ y ~ R, and use Theorem 7.3 to get

LB {f f((r 2 - y2)1/2, y)(r2 - y2)-1/2 dy}r dr.

Let y = r sin 8, 8 E [0,1-'17]. Then (r 2 - y2)1/2 = r cos 8. We may apply Proposition 3.9 to the preceding integral of g·r over 0 ~ 8 ~ tn-. Similar arguments apply to the other three quadrants. 0

Some infinitely differentiable functions 67

Exercises

1. Suppose A c 1R2 is open and suppose g: A ~ C and h: A ~ C are of class CI. Show that a necessary condition for the existence off: A ~ C such that

Dd=g, Dd=h

is that D2 g = DIh. 2. In Exercise 1,supposeAisadisc{(x,y) I (x - XO)2 + (y - YO)2 < R2}.

Show that the condition D2 g = DIh is sufficient. (Hint: consider

f(x, y) = fX g(s, y) ds + fY h(xo, t) dt.) Xo Yo

§8. Some infinitely differentiable functions

In §4 it was shown that any power series with a positive radius of con­vergence defines an infinitely differentiable function where it converges:

0)

f(x) = 2: an(x - xo)n. n=O

We know

an = (n!) -IJ(n)(xo)'

In particular, if all derivatives off are zero at Xo, then f is identically zero. There are infinitely differentiable functions which do not have this property.

Proposition 8.1. There is an infinitely differentiable function f: IR ~ IR such that

Proof We definefby

f(x) = 0, f(x) > 0,

x ~ 0, x> 0.

f(x) = 0, x ~ ° f(x) = exp (-Ijx), x > 0.

Near any point x -# 0, f is the composition of two infinitely differentiable functions. Repeated use of the chain rule shows that f is, therefore, infinitely differentiable except possibly at zero.

Let us show thatfis continuous at 0. If y > 0, then

ex>

eY = 2: (n!)-Iyn < (m!)-Iym, m = 0, I, .... n=O

Thus if x> 0,

° < f(x) = exp (-Ijx) = exp (IjX)-l < m!(ljx)-m = m! x m,

68 Continuous functions

m = 0, 1, .... In particular,J(x) --+ 0 as x --+ O. It is easy to show by induction that for x > 0,

J<k)(X) = Pk(x- 1)f(x),

where Pk(x) is a polynomial of degree :::;; k + 1. Of course, this equation also holds for x < O. Suppose we have shown thatJ<k) exists and is continuous at 0; then of courseJ<k)(O) = O. We have

(8.1) 1[f(k)(X) - J<k)(0)]x- 11 = lJ<k)(X)x- 11 = IX- 1Pk(X- 1)f(x)l·

Since Pk is of order :::;; k + 1 and

If(x) I :::;; (k + 3)! X- k- 3 ,

it follows that the right side of (8.1) converges to zero at x --+ O. ThusJ<k + 1)(0) exists and is zero. Similarly, J<k + 1)(X) = PH 1(x- 1)f(x) --+ 0 as x --+ 0, so J<k+1) is continuous. 0

Note that all derivatives of the preceding function vanish at zero, but f is not identically zero. Therefore f does not have a convergent power series expansion around zero.

Corollary 8.2. Suppose a < b. There is an infinitely differentiable function g: IH --+ IH such that

g(x) = 0, g(x) > 0,

xrt(a, b), xE(a,b).

Proof Letfbe the function in Theorem 8.1 and let

g(x) = f(x - a)f(b - x).

This is positive where x - a > 0 and where b - x < 0, and is zero else­where. It is clearly of class C '" . 0

Corollary 8.3. Suppose a < b. There is an infinitely differentiablefunction h: IH --+ IH such that

h(x) = 0, o < h(x) < 1,

hex) = 1,

x:::;; a, 0< x < b, x 2! b.

Proof Let g be the function in Corollary 8.2 and let

h(x) = c LX", g(t) dt,

where c > 0 is chosen so that h(b) = 1. Then h' = cg 2! 0, h is constant outside (a, b), etc. 0

Chapter 3

Periodic Functions and Periodic Distributions

§1. Continuous periodic functions

Suppose u is a complex-valued function defined on the real line IR. The function u is said to be periodic with period a "" ° if

u(x + a) = u(x)

for each x E IR. If this is so, then also

u(x + 2a) = u«x + a) + a) = u(x + a) = u(x) , u(x - a) = u«x - a) + a) = u(x).

Thus u is also periodic with period 2a and with period -a. More generally, u is periodic with period na for each integer n. If u is periodic with period a "" 0, then the function v,

vex) = u(lalx/27T)

is periodic with period 27T. It is convenient to choose a fixed period for our study of periodic functions, and the period 27T is particularly convenient. From now on the statement" u is periodic" will mean "u is periodic with period 27T." In this section we are concerned with continuous periodic functions. We denote the set of all continuous periodic functions from IR to C by re. This set includes, in particular, the functions

sin nx, cos nx, exp (inx) = cos nx + i sin nx

for each integer n. The set re can be considered a vector space in a natural way. We define the

operations of addition and scalar multiplication by

(1)

(2)

(u + v)(x) = u(x) + v(x) ,

(au)(x) = au(x),

u, V E re, x E IR;

u E re, a E C, x E IR.

it is easily checked that the functions u + v and au are periodic. By Proposi­tion 1.1 of Chapter 2, they are also continuous. Thus u + v Ere, au Ere. The axioms VI-V8 for a vector space are easily verified. We note also that there is a natural multiplication of elements of re,

(uv)(x) = u(x)v(x), u, V Ere, X E IR.

The set re may also be considered as a metric space. Since the interval [0, 27T] is a compact set in IR and since u E re is continuous,

sup I u(x) I < 00. xe[O.2,,]

69

70 Periodic functions and periodic distributions

We define the norm of u E re to be the real number lui where

(3) lui = sup I u(x) I = sup lu(x)l. xelll xe[O.2,,]

The norm (3) has the following properties:

(4) lui ~ 0, and lui = 0 only if u(x) = 0, all x;

(5) laul = lallul, a E C, u E re; (6) lu + vi ::;; lui + lvi, u, v Ere. The properties (4) and (5) are easily checked. As for (6), suppose x E R Then

I(u + v)(x) I = lu(x) + v(x) I ::;; I u(x) I + I v(x) I ::;; lui + Ivl· Since this is true for every x E IR, (6) is true.

To make re a metric space, we set

(7) d(u, v) = lu - vi.

Theorem 1.1. The set re of continuous periodic functions is a vector space with the operations defined by (1) and (2). The set re is also a metric space with respect to the metric d defined by (7), and it is complete.

Proof. As we noted above, checking that re satisfies the axioms for a vector space is straightforward. The axioms for a metric space are also easily checked, using (4), (5), and (6). For example,

d(u, w) = lu - wi = I(u - v) + (v - w)1 ::;; lu - vi + Iv - wi = d(u, v) + d(v, w).

Finally, suppose (un):'= 1 is a Cauchy sequence of functions in re. By Theorem 4.1 of Chapter 2, there is a continuous function u: IR --+ C such that IUn - ul--+ O. Clearly u is periodic, so u Ere and re is complete. 0

Sets which are simultaneously vector spaces and metric spaces of this sort are common enough and important enough to have been named and studied in the abstract. Suppose X is a real or complex vector space. A norm on X is a function assigning to each u E X a real number lui, such that

lui ~ 0, and lui = 0 implies u = 0; laul = lallul, a scalar, u E X; lu + vi ::;; lui + lvi, u, VEX.

A normed linear space is a vector space X together with a norm lui. Associated to the norm is the metric

d(u, v) = lu - vi. If the normed linear space is complete with respect to this metric, it is said to be a Banach space.

In this terminology, Theorem 1.1 has a very brief statement: ri is a Banach space.

Continuous periodic functions 71

Suppose X is a complex normed linear space. A linear functional F: X --+ C is said to be bounded if there is a constant c ;;:: 0 such that

IF(u) I ~ clul, all u E X.

Proposition 1.2. A linear functional F on a normed linear space X is continuous if and only if it is bounded.

Proof Suppose F is bounded. Then

IF(u) - F(v) I = IF(u - v)1 ~ clu - vi, so IF(u) - F(v) I < e if lu - vi < c- 1e.

Conversely, suppose F is continuous. There is as> 0 such that lui = lu - 01 ~ S implies

IF(u) I = IF(u) - F(O) I ~ 1.

For any u -# 0, U E X, the vector v = 81ul- 1u has norm 8. Therefore

I F(u) I = IF(8- 1 Iulv)1 = 8- 1 IuIIF(v)1 ~ 8- 1 Iul, and F is bounded. D

It is important both in theory and practice to determine all the con­tinuous linear functionals on a given space of functions. The reason is that many problems, in theory and in practice, can be interpreted as problems about existence or uniqueness of linear functionals satisfying given con­ditions. The examples below show that it is not obvious that there is any way to give a unified description of all the continuous linear functionals on ~. In fact one can give such a description (in terms of Riemann-Stieltjes integrals, or integrals with respect to a bounded Borel measure), but we shall not do this here. Instead we introduce a second useful space of periodic functions and determine the continuous linear functionals on this second space.

Exercises

1. Suppose (an):'= _ 00 is a (two sided) sequence of complex numbers such that

~ la,,1 < 00; n= - co

here we take the infinite sum to be 00 00 00

L la,,1 = L la,,1 + L la_"I· ,,=-00 ,,=0 ,,=1

Show that the function u defined by 00

u(x) = L a" exp (inx) n= -00

72 Periodic functions and periodic distributions

is continuous and periodic 2. Suppose u: IR --+ C is a continuous function and suppose there is a

constant M such that u(x) = ° if Ixl ~ M. Show that for any x the series

00

vex) = 2: u(x + 2mr) n= - 00

converges. Show that the function v is in ~. 3. If u E~, define the real number lui' by

r2" lui' = (27T)-1 I u(x) I dx.

o

Show that lui' is a norm on ~ and that lui' ~ lui. 4. Suppose d' is the metric associated with the norm lui' in Exercise 3.

Show that ~ is not complete with respect to this metric. (Hint: take a sequence of functions (un):= 1 of functions in ~ such that

o ~ un(x) ~ 1, un(x) = 0, un(x) = 1,

X E IR, n = 1,2, ... , X E [0, 7T/2 - I/n] U [37T/2 + I/n, 27T], x E [7T/2, 37T/2].

Then I Un - Urn I' --+ ° as n, m --+ 00. If u E~, there is an open interval (7T/2 - 0,7T/2 + 0) on which either I u(x) I > t or lu(x) - 11 > t. Show that IUn - ul' > 0/67T for large values of n.)

5. Which of the following are bounded linear functionals on ~, with respect to the norm lui?

(a) F(u) = u(7T/2), (b) F(u) = f:" sin nx u(x) dx,

(c) F(u) = f:" (U(X»2 dx,

(d) F(u) = 17u(0) + f:" u(x) dx. (e) F(u) = -3Iu(0)1.

6. Suppose X is a normed linear space. Let X' be the set of all bounded linear functionals on X. Then X' is a vector space. For FE X', let

IFI = sup {IF(u)11 u E X, lui ~ I}.

Show that IFI is a norm on X'. Show that for any u E X and FE X',

IF(u)1 ~ IFllul· Show that X' is a Banach space with respect to this norm.

§2. Smooth periodic functions

Suppose u: IR --+ C is a continuous periodic function, and suppose that the derivative

Du(x) = u'(x)

Smooth periodic functions 73

exists for each x E IR. Then Du is also periodic:

Du(x + 27T) = lim h-1[u(x + 27T + h) - u(x + 27T)] " ... 0

= lim h-1[u(x + h) - u(x)] = Du(x). " ... 0

In particular, if u is infinitely differentiable and periodic, then each deriv­ative Du, D2U, . .. , DkU,. .. is in ~.

We shall denote by gJ the subset of ~ which consists of all functions u E ~ which are smooth, i.e., infinitely differentiable. Such a function will be called a smooth periodic function. If u is in &! then the derivatives Du, D2U, . .. are also in fJ'.

The set gJ is a subspace of ~ in the sense of vector spaces, so it is itself a vector space. The function Isin xl is in C(j but not in &! so gJ i: ~. We could consider gJ as a metric space with respect to the metric on ~ given in the previous section, but we shall see later that gJ is not complete with respect to that metric. To be able to consider gJ as a complete space we shall intro­duce a new notion of convergence for functions in ~

A sequence of functions (Un):'=1 c gJ is said to converge to u E gJ in the sense of gJ if for each k = 0, 1,2, ... ,

I Dkun - DkUI-+ ° as h -+ 00.

(Here DOu = u.) We denote this by

Un -+ U (gJ).

Thus (Un):'=1 converges to u in the sense of gJ if and only if each derivative of Un converges uniformly to the corresponding derivative of u as n -+ 00.

A sequence of functions (Un):'=1 is said to be a Cauchy sequence in the sense of gJ if for each k = 0, 1, ... , (Dkun):, = 1 is a Cauchy sequence in ~. Thus

for each k. When there is no danger of confusion we shall speak simply of "con­

vergence" and of a "Cauchy sequence," without referring to the "the sense of fJ'." The statement of the following theorem is to be understood in this way.

Theorem 2.1. The set gJ of all smooth periodic functions is a vector space. If(un):'=1 c gJ is a Cauchy sequence, then it converges to afunction u E ~

Proof. As noted above, gJ is a subspace of the vector space ~: if u, v E gJ

then u -I- v E &! au E fJ'. Thus gJ is a vector space.

Suppose (Un):'=l is a Cauchy sequence. For each k the sequence of derivatives (Dkun):'=l is a Cauchy sequence in ~. Therefore it converges uniformly to a function Vk E~. For k = 0, 1,2, ... ,

D"un{x) = D"un(O) + LX D"+1un(t) dt.

74 Periodic functions and periodic distributions

By Theorem 4.2 of Chapter 2,

Vk(X) = lim Dkun(X) = lim Dkun(O) + lim fX Dk+lvn(t) dt n-+Qj n-+oo n .... oo 0

= Vk(O) + J: Vk+l(t) dt.

Therefore DVk = Vk+1' all k. This means that if u = Vo, then Vk = Dku and IDkun - DkUI-+ 0 as n -+ 00. Thus Un -+ u (in the sense of fJI). 0

The remainder of this section is not necessary for the subsequent develop­ment. We show that there is no way of choosing a norm on fJI so that con­vergence as defined above is equivalent to convergence in the sense of the metric associated with the norm. However, there is a way of choosing a metric on fJI (not associated with a norm) such that convergence in the sense of fJI is equivalent to convergence in the sense of the metric. Finally, we introduce the abstract concept which is related to fJI in the way that the concept of "Banach space" is related to ~.

Suppose there were a norm lui' on fJI such that a sequence (Un);'=l c fJI converges in the sense of fJI to u E fJI if and only if

IUn - ul' -+ O.

Then there would be a constant M and an integer N such that

(1) lui' ~ M(lul + IDul + ... + IDNUI), all u E!!l!

In fact, suppose (1) is false for every M, N. Then for each integer n there would be a Un E fJI such that

Let

Then

if n ~ k,

so Vn -+ 0 in the sense of!!l! But Ivnl' = 1, all n. This shows that the norm lui' must satisfy (I) for some M, N. Now let

Wn(X) = n- N- 1 sin nx, n = 1,2, ....

Then Wn E fJI and

k ~ N.

Thus by (1),

But IDN+IWnl = 1, all n, so (Wn);'=l does not converge to 0 in the sense of!!l! This contradicts our assumption about the norm lui'.

Smooth periodic functions 75

Although we cannot choose a metric on 9, associated with a norm, which gives the right notion of convergence, we can choose a metric as follows. Let

00

d'(u, v) = L 2-k-1IDku - Dkvl[l + IDku - Dkvll- 1. k=O

The term of this sum indexed by k is non-negative and is smaller than 2-/(-1. Thus

d'(u, v) < 1, u, V E fIJ.

It is clear that

d'(u, v) ::::: 0, d'(u, v) = ° implies u = v, d'(u, v) = d'(v, u).

The triangle inequality is a little more difficult. Let

d(u, v) = lu - vi, d*(u, v) = d(u, v)[l + d(u, V)]-l.

The reader may verify that

d*(u, w) ~ d*(u, v) + d*(v, w).

Then

GO

d'(u, w) = L 2- k- 1d*(DkU, DkW )

k=O

~ ... ~ d'(u, v) + d'(v, w).

Theorem 2.2. A sequence offunctions (Un)~=1 C fIJ is a Cauchy sequence in the sense of fIJ if and only if it is a Cauchy sequence in the sense of the metric d'. Thus (fIJ, d') is a complete metric space.

Proof Suppose (un)~= 1 is a Cauchy sequence in the sense of!!Ji. Suppose e > ° is given. Choose k so large that 2 - k < e. Choose N so large that if m ::::: Nand n ::::: N, then

IDiUn - Diunl < -te, Then if m, n ::::: N,

00

j = 0, 1, ... , k.

d'(um, un) = 2: 2- i - 1d*(Dium, Djun) j=O

k 00

< 2: 2- f - 1(!e) + 2: 2- j - 1

j=O j=k+1 < !e + 2 - k -1 < !e + -le.

Conversely, suppose (Un)~= 1 is a Cauchy sequence in the sense of the metric d'. Given an integer k ::::: ° and an e > 0, choose N so large that if m, n ::::: N then

76 Periodic functions and periodic distribution

For m,n ~ N,

IDkun - Dkuml = 2k+2.2-k-l.2-1IDkun - Dkuml < 2k+2·2-k-l·d*(DkUn> Dkum) < 2k+2d'(un, um) < e.

Thus (Un):'=1 is a Cauchy sequence in the sense of Bl! The same argument shows that d'(un, u) -+ 0 if and only if Un -+ U (91').

Thus (f!J', d') is complete. 0

There is an important generalization of the concept of a Banach space, which includes spaces like f!J'. Let X be a vector space over the real or complex numbers. A seminorm on X is a function u -+ lui from X to IR such that

lui ~ 0, laul = lallul, lu + vi ::::; lui + Ivl· (Thus a seminorm is a norm if and only if lui = 0 implies u = 0.) Suppose there is given a sequence of semi norms on X, lUll> luI2"'" with the property that

(2) all k implies u = O.

Then we may define a metric on X by

00

d'(u, v) = L 2-klu - vlk[1 + lu - vlk]-I. 1<=1

If X is complete with respect to the metric d', it is said to be a Frechet space. Note that

d'(un, v) -+ 0 as n -+ 00

is equivalent to

IUn - vlk-+ O as n-+oo, In particular, if we take X = f!J' and

Iulk = IDk-lul,

for all k.

then d' agrees with d' as defined above. Thus Theorems 3.1 and 3.2 say that f!J'is a Frechet space.

Exercises

1. Which of the following are Cauchy sequences in the sense of f!J'?

un(x) = n- 3 cos nx,

vn(x) = (nl)-l sin nx, n

wn(x) = L (ml)-1 sin mx. m-l

Translation, convolution, and approximation 77

2. Suppose X is a complex vector space with a sequence of seminorms lull> lul2' ... , satisfying (2). Let d' be the associated metric. Show that a linear functional F: X --+ C is continuous if and only if there are a constant M and an integer N such that

alluEX.

§3. Translation, convolution, and approximation

The aim of this section and the next is to show that the space f!J' of smooth periodic functions is dense in the space CC of continuous periodic functions; in other words, any continuous periodic function U is the uniform limit of a sequence (Un):'=l of smooth periodic functions. Even more important than this theorem is the method of proof, because we develop a systematic procedure for approximating functions by smooth functions.

The idea behind this procedure is that an average of translates of a function U is smoother than u itself, while if the translated functions are translated only slightly, the resulting functions are close to u. To illustrate this the reader is invited to graph the following functions from IR to IR:

U1(X) = lxi, U2(X) = tlx - el + tlx + el, U3(X) = tlx - el + tlxl + tlx + el,

where e > O. If U E CC and t E IR, the translation of u by t is the function Ttu,

Ttu(x) = u(x - t), X E IR.

Then Ttu Ere. The graph of Tt is the graph of u shifted t units to the right (i.e., shifted I t I units to the left, if t < 0). In these terms the functions above are

More generally, one could consider weighted averages of the form

(1)

where

ao + a1 + ... + aT = 1,

and most of the t k are near O. If

o ~ to < t1 < ... < tk = 2'IT

and we set

then (1) becomes

(2)

78 Periodic functions and periodic distributions

The natural continuous analog of (2) is the symbolic integral

12" V = 0 b(t)Ttu dt,

defining the function

(3) 12 " vex) = 0 b(t)u(x - t) dt.

We wish to study integrals of the form (3). If u, v E <iff, the convolution of u and v is the function u * v defined by

(4) I 12" (u * v)(x) = 27T 0 u(x - y)v(y) dy.

It follows readily from (4) that

(5) I f2" lu * vi ::s; lul 27T 0 I v(x) I dx ::s; lullvl·

Proposition 3.1. If u, v E <iff, then u * v E <iff. Moreover

(6)

(7)

(8)

(9)

(10)

u * v = v * u,

(au) * v = a(u * v), a E 18,

(u + v) * W = U * W + v * W,

(u * v) * W = U * (v * w),

WE<iff,

Tt(u * v) = (Ttu) * v = u * (Ttv).

Proof We begin with part of (10).

(11) Tt(u * v)(x) = (u * v)(x - t) = L I u(x - t - y)v(y) dy

= 2~ J 1tu(x - y)v(y) dy = (T,u) * v.

Therefore,

(12) lTt(u * v) - u * vi = I(Ttu - v) * vi ::s; ITtu - ullvl,

where we have assumed (7) and (8). Now u is uniformly continuous on [0, 27T] and is periodic; it follows easily that u is uniformly continuous on IR. There­fore ITtu - ul--+ 0 as t --+ O. Then (12) implies continuity of u * v. Also,

so u * v is periodic. The equality of (6) follows from a change of variables in (4): let y' =

x - y, and use the periodicity of u and v. Equalities (7), (8), and (9) are easy computations. The last part of (10) follows from (11) and (6). 0

Translation, convolution, and approximation 79

Note that

rl[U(t + x) - U(X)] = t- 1[L tu - u](x).

Lemma 3.2. If u E &! then

It- 1(L tu - u) - Dul-+ 0 as t -+ O.

Proof By the Mean Value Theorem,

rl[Ltu - u](x) - Du(x) = Du(y) - Du(x)

where y = yet, x) lies between x and x + t. Since Du is uniformly continuous, (-l(T -tU - u) converges uniformly to Du as (-+ O. 0

Corollary 3.3. If u E &! then

r 1(L tu - u) -+ u (.9')

as t -+ O.

Proof It is easy to see that

Dk(T -tu) = Tt(Dku).

Then

Dk[t-l(Ltu - u)] = r 1(L t DkU - Dku),

which converges uniformly to D(Dku) = Dk(Du). 0

Proposition 3.4. If u E .9' and v E ~ then u * v E .9' and

(13) all k.

Proof By Proposition 3.1,

(14) (-l[Lt(u * v) - (u * v)] = [r1(Ltu - u)] * v.

By Lemma 3.2 and (5), the expression on the right in (14) converges uni­formly to (Du) * v as t -+ O. Thus

D(u * v) = (Du) * v,

and u * v has a continuous derivative. But, Du E &! so we also have

D2(U * v) = D«Du) * v) = (D2u) * V.

By induction, (13) holds for all k. 0

Corcllary 3.5. Suppose (vn)1' c~, V E~, and IVn - vl-+ O. Then for each u E&!

u * Vn -+ U * v (.9').

Proof For each k,

Dk(U * Vn - U * v) = Dku * (vn - v).

80 Periodic functions and periodic distributions

It follows from this and (5) that

I Dk(U * Vn - U * v)1 -+ 0,

all k. 0 Having established the general properties of convolution, let us return

to the question of approximation. Suppose U E f{f. If (IPn)~= 1 is a sequence of functions in &! then each function

Un = IPn * U

is smooth. Thinking of Un as a weighted average of translates of u, we can expect Un to be close to U if IPn has average value 1 and is concentrated near o and 27T (as a function on [0, 27T]).

A sequence (IPn)~=l c f{f is said to be an approximate identity if

(i) IPn(x) ~ 0, all n, x; (ii) 1/27T f:" IPn(X) dx = 1, all n;

(iii) for each 0 < 8 < 7T,

[2"-6 IPn(X) dx -+ 0 as n -+ 00 • .16

Theorem 3.6. Suppose (IPn)i' c f{f is an approximate identity. Then for each U E f{f,

IlPn * U - ul-+O as n-+oo.

Moreover, if U E f!J1 then

IPn * U -+ U (f!J1) as n -+ 00.

Proof Since (27T)-1 J:" IPn(Y) dy = 1 and IPn * U = U * lPn, we have

27T1(lPn * u)(x) - U(x) I = If" u(x - Y)lPn(Y) dy - u(x) f"lPn(Y) dy \

If:" [u(x - y) - U(X)]IP,,(Y) dy \

~ II: \ + If"-6 \ + \{:-6 \ ~ sup IT.u - ul (f 6 IPn + f2" IPn) + 21uI [2"-6 IPn

Isl,,26 0 2,,-6 J 6

~ sup ITsu - ul + 21ul r2"-6 IPn.

Isl,,26 .16

Given 8 > 0, we first choose D > 0 so small that 8 < 7T and

ITsu - ul ~ 7T8

Then choose N so large that

r2"-6

21u l .16

IPII < 27T8,

if lsi ~ D.

n ~ N.

The Weierstrass approximation theorems

If n :2: N, then

l'Pn * u - ul < e.

This proves the first assertion. Now suppose u E fYJ. For each k,

Dk('Pn * u) = Dk(U * 'Pn) = (Dku) * 'Pn = 'Pn * (Dku),

which converges uniformly to Dku. Thus

'Pn * u ~ u (&).

81

o In the next section we shall construct a sequence in & which is an approxi­

mate identity. It will follow, using Proposition 3.4 and Theorem 3.6, that & is dense in ~.

Exercises

1. Let e,cCx) = exp (ikx), k = 0, ± 1, ± 2, .... These functions are in & (see §6 of Chapter 2). Suppose u E~. Show that

where

I J2" ak = 211' 0 e-lkYu(y) dy.

2. Show that ek * Lie = I, and ej * e-k = ° ifj:F k.

§4. The Weierstrass approximation theorems

A trigonometric polynomial is a function of the form

n

(1) 'P(x) = L ak exp (ikx), k=-n

where the coefficients ak are in C. The reason for the terminology is that for k > 0,

exp (± ikx) = [exp (± iX)]k = (cos x ± i sin X)k.

Therefore any function of the form (I) can be written as a polynomial in the trigonometric functions cos x and sin x. Conversely, recall that

cos x = -t[exp (ix) + exp ( - ix)],

sin x = t [exp (ix) - exp (-ix)].

Therefore any polynomial in cos x and sin x can be written in the form (I).

82 Periodic functions and periodic distributions

Lemma 4.1. There is a sequence (<Pn)r' of trigonometric polynomials which is an approximate identity.

Proof. We want to choose a non-negative trigonometric polynomial 11' such that

11'(0) = <p(2n) = 1, <p(x) < 1 for 0 < x < 2n.

Then successive powers of 11' will take values at points near 0 and 2n which are relatively much greater than those taken at points between 0 and 2n. We may take

<p(x) = -t(1 + cos x)

and set

where Cn is chosen so that

f" <Pn(x) dx = 2n.

We need to show that for each 0 < 8 < n,

(2) f2"-6

6 <Pn(X) dx ~ 0 as n ~ 00

There is a number r, 0 < r < 1, such that

(3) 1 + cos x < r (1 + cos y)

if

(4) X E [8, 2n - 8], Y E [0, -!8].

Then (3) and (4) imply

<Pn(x) = cn(1 + cos x)n :::;; r n<pn(y),

so

or

X E [8, 2n - 8]

Thus <Pn ~ 0 uniformly on [8, 2n - 8]. 0

Lemma 4.2. If 11' is a trigonometric polynomial and u E 'i&', then 11' * u is a trigonometric polynomial.

Proof. This follows from Exercise 1 of the preceding section. 0

Theorem 4.3. The trigonometric polynomials are dense in the space ct of continuous periodic functions, and in the space (!} of smooth periodic functions.

The Weierstrass approximation theorems 83

That is, ifu E ~ and v E~, there are sequences (un)f and (vn)f of trigonometric polynomials such that

and

vn ~ v (.9).

Proof Let (Pn)!' be a sequence of trigonometric polynomials which is an approximate identity, as in Lemma 4.1. Let

un = Pn * u, vn = Pn * v.

By Lemma 4.2, the functions Un and Vn are trigonometric polynomials. By Theorem 3.6, Un ~ u uniformly and Vn ~ v in the sense of ~ 0

Note that if u, v are real-valued, then so are the sequences (un)!', (vn)f constructed here.

Corollary 4.4. .9 is dense in ~.

Theorem 4.3 is due to Weierstrass. There is a better-known approxima­tion theorem, also due to Weierstrass, which can be deduced from Theorem 4.3.

Theorem 4.5. (Weierstrass polynomial approximation theorem). Let u be a complex-valued continuous function defined on a closed interval [a, b] c IR. Then there is a sequence (Pn)f of polynomials which converges uniformly to u on the interval [a, b].

Proof Suppose first that [a, b] = [0, 7T]. We can extend u so that it is a function in ~; for example, let u( -x) = u(x), X E [0, 7T] and take the unique periodic extension ofthis function. Then there is a sequence (un)!' oftrigono­metric polynomials converging uniformly to u. Now the partial sums of the power series

L: (m!)-l(ikx)m = exp (ikx)

converge to exp (ikx) uniformly on [0,7T]. Therefore for each n, we may replace the functions exp (ikx) in the expression of the form (1) for Un by partial sums, so as to obtain a polynomial Pn with

X E [0, 27T].

Then Pn ~ u uniformly on [0,27T]. In the case of an arbitrary interval [a, b], let

vex) = u(a + (b - a)x/7T) , X E [0, 7T].

Then v is continuous on [0,7T], so there is a sequence (qn)!' of polynomials with qn ~ v uniformly on [0, 7T]. Let

Y E [a, b].

84 Periodic functions and periodic distributions

Then Pn is also a polynomial and Pn -+ U uniformly on [a, b]. 0

Exercises

1. Suppose u E <'C and suppose that for each integer k,

J, 2,.

o u(x) exp (ikx) dx = O.

Show that u = O. 2. Suppose u: [a, b] -+ <'C is continuous, and for each integer n ~ 0,

f u(x)xn dx = O.

Show that u = O.

§5. Periodic distributions

In general, a "distribution" is a continuous linear functional on some space of functions. A periodic distribution is a continuous linear functional on the space ~ Thus a periodic distribution is a mapping F: f!IJ -+ C such that

F(au) = aF(u), F(u + v) = F(u) + F(v),

F(un) -+ F(u)

aEC, UEf!IJ; U, V E f!IJ; if Un -+ U (f!IJ).

If v is a continuous periodic function defined on [0, 27T], then we define a linear functional F = Fv by

(1) 1 f2" Fv(u) = 27T 0 v(x)u(x) dx, U E <'C.

Then Fv: <'C -+ C is linear, and

Therefore Fv is continuous on <'C. Its restriction to the subspace f!IJ is a periodic distribution. We say that a periodic distribution F is a function if there is a v E <'C such that F = Fv' If so, we may abuse notation and write F = v.

Note that different functions v, WE <'C define different distributions. In fact, suppose Fv = Fw. Choose (un)i c f!IJ such that Un -+ W* - v* uni­formly, where w*(x) = W(x) * , the complex conjugate. Then

ih ih o = 27T(Fw(Un) - Fv(un)) = 0 (w(x) - v(x))un(x) dx -+ 0 Iw(x) - v(x)12 dx,

so W = v.

Periodic distributions 85

Not every periodic distribution is a function. For example, let 8: re ~ C be defined by

(2) 8(u) = u(O), u Ere.

Then the restriction of 8 to fYJ is a periodic distribution. It is called t~ 8-distribution, or Dirac 8-distribution. To see that it is not a function, let

un(X) = H + 1- cos x)n.

Then 8(un) = I, all n. But un(x) ~ 0 uniformly for x E fe, 27T - e], any e > O. Also 0 ~ un(x) ~ I, all x, n. It follows from this that for any v Ere, Fiun) ~ o. Thus 8 f= Fv.

The set of all periodic distributions is denoted by fYJ'. We consider fYJ' as a vector space in the usual way: if F, G E fYJ', u E.9, a E C, then

(F + G)(u) = F(u) + G(u) , (aF)(u) = aF(u).

Note that if v, ware continuous periodic functions, then

A sequence (Fn)f £ fYJ' is said to converge to FE fYJ' in the sense of fYJ' if

Fn(u) ~ F(u), all u E [lJ.

We denote convergence in the sense of fYJ' by

Fn ~ F (fYJ'),

or simply by

when it is understood in what sense convergence is understood. We want to define operations of complex conjugation, reversal, transla­

tion, and differentiation for periodic distributions. For any such operation there is a standard procedure for extending the operation from functions to distributions. For example, if VEe, the complex conjugate function v* is defined by

v*(X) = V(X) * . Then

I f2" (I f2" )* Fv.(u) = 27T 0 v(x)*u(x) dx = 27T 0 v(x)u*(x) dx = (Fv(u»*.

Then we define F* for an arbitrary FE fYJ' by

(3) F*(u) = F(u*)*,

Similarly, if v E re we define the reversed function v by

v(x) = v( -x).

86 Periodic functions and periodic distributions

Then

I f211 I f211 F;,(u) = 27T 0 v( -X)U(X) dx = 27T 0 V(X)U( -x) dx = Flu).

We define F-, FE &', by

(4) F-(u) = F(u), UE~

If v E f{;' and t E IR, recall that the translate Ttv is defined by

Ttv(x) = vex - t).

Then

I f211 I f211 27T 0 Ttv(x)u(x) dx = 27T 0 vex - t)u(x) dx

I f211 = 27T 0 v(x)u(x + t) dx = Fv(T -tu).

We define TtF, FE &', by

(5) UE~

If v E & and u E &', then integration by parts gives

1 f211 I f211 27T 0 Dv(x)u(x) dx = - 27T 0 v(x)Du(x) dx = - Fv(Du).

We define DF, FE &', by

(6) (DF)(u) = -F(Du), u E ~

Then inductively,

(7) UE~

Each of the linear functionals so defined is a periodic distribution. For example, if Un ~ U (&) then DUn ~ Du (&). It follows that

(DF)(un) = -F(Dun) ~ -F(Du) = DF(u),

so the derivative DFis continuous. Similarly, F*, F, TtF, and DkFare in &'. In particular, let us take F = 8. Then

(8)

(9)

(10)

0= 0* = a Tto(u) = u(t),

Dk8(u) = (_I)kDku(O), UE~

Proposition 5.1. The operations in &' defined by equations (3), (4), (5), and (6) are continuous, in the sense that if Fn ~ F (&') then

Fn * ~ F* (&'), Fn - ~ F- (&'),

TtFn ~ TtF (&'), DFn ~ DF (&').

Periodic distributions 87

Proof Each of these assertions follows trivially from the definitions. For example, if u E fll then

(DFn)(u) = -Fn(Du) --? -F(Du) = DF(u).

Thus DFn --? DF (fll'), etc. 0

Recall that if u E fll then Du is the limit of the "difference quotient" t-l(Ltu - u).

Proposition 5.2. If FE fll', then

(11) t-l(LtF - F) --? F (fll')

as t --? O.

Proof Suppose u E f?lJ. By definition,

(12) t-l(LtF - F)(u) = t-1F(Ttu) - t-1F(u) = -F(t-l[u - Ttu]).

Now

(13)

An argument like that proving Lemma 3.2 and Corollary 3.3 shows that the expression in (13) converges to Du in the sense of fll as t --? O. From this fact and (12) we get (11). 0

As an example,

t-l(Lto - o)(u) = t-l[u(-t) - u(O)]--? -Du(O) = (Do)(u).

The real and imaginary parts of a function v E C(? can be defined by

Re v = !(v + v*),

1m v = ~ (v - v*).

Similarly, we define the real and imaginary parts of a periodic distribution F by

ReF = -HF + F*),

ImF = ~(F - F*).

F is said to be real if F = F*. A function v E C(? is said to be even if vex) = v( -x), all x; it is said to be odd if vex) = -v( -x), all x. These conditions may be written

v = D, v = -D.

Similarly, we say a periodic distribution F is even if

F= F-;

88 Periodic functions and periodic distributions

we say F is odd if

Exercises

1. Which of the following define periodic distributions?

(a) F(u) = Du(1) - 3u(27T). (b) F(u) = J:" (u(xW dx.

(c) F(u) = J:" u(x) dx.

(d) F(u) = J: u(x)(1 + xf dx.

(e) F(u) = - J:" D3U(x)lcos 2xl dx. (f) F(u) = L:7=o ajDiu(tj). (g) F(u) = L:i=o (j!)-l Dju(O).

2. Verify (8), (9), (10) 3. Express the distributions in parts (a) and (f) of Exercise 1 in terms of the

8-distribution and its translates and derivatives. 4. Compute DFwhen Fis the distribution in part (c) or (d) of Exercise l. 5. Show that Re F and 1m F are real. Show that F = Re F + i 1m F. 6. Show that F real and ureal, u E.9, imply F(u) is real. 7. Show that F even and u odd, u E.9, imply F(u) = o. Show that F odd

and u even, u E.9, imply F(u) = O. 8. Show that any FE f!jJ' can be written uniquely as F = G + H, where

G, HE&" and G is even, H is odd. 9. Suppose that v E '6' is differentiable at each point of IR and Dv = w is

in '6'. Show that

in other words, if F = v, then DF = Dv. 10. Suppose v is a continuous complex-valued function defined on the

interval [0, 27T] and that Dv = w is continuous on (0, 27T) and bounded. Define Fv E &" by

1 f2" Fv(u) = 27T 0 v(x)u(x) dx, u E &'.

Show that

1 f2" DFv(u) = (v(O) - V(27T»U(0) + 27T 0 w(x)u(x) dx.

In other words,

DFv = Fw + [v(O) - V(27T)]8.

11. Let vex) = I sin !xl. Compute

(D2Fv)(U), UE&'.

Determining the periodic distributions 89

§6. Determining the periodic distributions

We know that any continuous periodic function v may be considered as a periodic distribution Fv. The derivatives DkFv are also periodic distributions, though in general they are not (defined by) functions. It is natural to ask whether all periodic distributions are of the form Dk Fv, v Ere. The answer is nearly yes.

Theorem 6.1. Suppose F is a periodic distribution. Then there is an integer k ~ 0, a continuous periodic function v, and a constant function J, such that

(I)

The proof of this theorem will be given later in this section, after several other lemmas and theorems. First we need the notion of the order of a periodic distribution. A periodic distribution F is said to be of order k (k an integer ~ 0) if there is a constant c such that

JF(u)J ~ c{JuJ + JDuJ + ... + JDkUJ}, all u E &1J.

For example, 8 is of order o. If v E re then DkFv is of order k. It is true, but not obvious, that any FE f!JJ' is of order k for some integer k ~ O.

Theorem 6.2. If F E f!JJ', then there is an integer k ~ 0 such that F is of order k.

Proof If F is not of order k, there is a function Uk E f!JJ such that

JF(Uk)J ~ (k + l){JukJ + JDUkJ + ... + JDkUkJ}.

Let

Vk = (k + 1)-l{JUkJ + JDUkJ + ... + JDkukl}-lUk.

Then we have

(2)

while

(3)

Suppose now that F were not of order k for any k ~ O. Then we could find a sequence (Vk)k'=l c f!JJ satisfying (2) and (3) for each k. But (3) implies

Vk -+ 0 (f!JJ).

Then (2) contradicts the continuity of F. Thus F must be of order k, some k. 0

Lemma 6.3. Suppose FE f!JJ' is of order O. Then there is a unique con­tinuous linear functional F1 : re -+ re such that

all u E &1J.

Proof By assumption there is a constant c such that

JF(u)J ~ cJuJ, u E &1J.

90 Periodic functions and periodic distributions

If U E'i&', there is a sequence (un):'= 1 C & such that Un --? U uniformly. Then

so (F(un»:'= 1 is a Cauchy sequence. Let

(4)

We want to show that Fl(U) is independent of the particular sequence used to approximate u. If (vnH" c & and Vn --? U uniformly, then

IUn - vnl--? 0

so

Thus

The functional F1 : 'i&' --? 'i&' defined by (4) is easily seen to be linear. It is continuous (= bounded), because

I F(u) I = lim IF(un) I ~ c lim lunl = clul. Conversely, suppose F2: 'i&' --? 'i&' is continuous and suppose F2(U) = F(u),

all u E f!JJ. For any u E'i&', let (un);." c & be such that Un --? U uniformly. Then

D

(The remainder of this section is not needed subsequently.)

Lemma 6.4. Suppose FE &' is of order 0, and suppose F(w) = 0 if w is a constant function. Then there is a function v E 'i&' such that

D2Fv = F.

Proof Let us suppose first that F = Ft , where f E 'i&'. We shall try to find a periodic function v such that D2v = f Then we must have

Dv(x) = Dv(O) + IX f(t) dt = a + C f(/) dl, o .0

where a is to be chosen so that v is periodic. We may require v(O) = O. Then

vex) = r Dv(t)dt = r [a + ff(S)dS] dt

= ax + fox ff(S)dSdl.

We use Theorem 7.3 of Chapter 2 to reverse the order of integration and get

vex) = ax + LX f(s)(x - s) ds.

Determining the periodic distributions 91

Let

(x - s)+ = 0 if x < s, (x - s)+ = X - S if x ;?: s.

Then

(5) f2"

vex) = ax + 0 f(s)(x - s)+ ds.

By assumption onf,

f2"

o hf(s) ds = 0,

Now we want to choose a in (5) so that v is periodic. This will be true if v(271-) = 0, i.e.,

fb fb o = 27Ta + 0 (27T - s)f(s) ds = 27Ta - 0 sf(s) ds.

Thus

I f2" a = 27T 0 sf(s) ds

and

(6) I f2" vex) = 27T 0 f(s)[xs + 27T(X - s)+] ds.

Now suppose only that FE &' is of order 0 and that F(w) = 0 if w is constant. Let Fl be the extension of F to a continuous linear functional on <'C. Let

Then ux(O) = 27TX = UxC27T). We can extend Ux so that it is a continuous periodic function of s. Then (6) suggests that we define a function v by

(7)

We want to show that v E <'C and D2Fv = F. It is easy to check that

Therefore

Iv(x) - v(y)1 s IF1(uX ) - Fl(Uy ) I s clux - uyl s C27Tlx - yl, vex + 27T) = F1(uX +2") = F1(uX ) = vex).

Thus v E <'C. Let us compute DFv. If w E .9, then

92 Periodic functions and periodic distributions

Approximate the integral by Riemann sums. These give expressions of the form

(8) rl L: [v(Xj + t) - v(Xj)]w(Xj)(Xj - Xj-l)

= FI(L: W(Xj)(Xj - Xj_l)t-I.[Ux/+t - UX /]),

since FI is linear. As partitions (xo, Xl, ... , xn) of (0,21T) are taken with smaller mesh, the functions on which FI acts in (8) converge uniformly to the function gt. Here

gtCs) = r t-l[ux_t(s) - uxCs )]w(x) dx.

Now It-l(ux_t - ux)1 :5 21T. For fixed S E (0, 21T), and ° < X < s,

rl(ux+t - ux) -+ S as t -+ 0.

This convergence is uniform for X in any closed subinterval of (0, s). Similarly,

t-l(ux+t - ux) -+ S + 21T as t -+ 0,

uniformly for X in any closed subinterval of (s, 21T). It follows that

(9) 21TDFv(w) = lim rl[Ltv - v](w) = FI(g), t-+o

where

r211 r2" g(s) = s 10 W(X) dx + 21T 10 W(X) dx.

Then

where

r211 f211 h(s) = s 10 Dw(x) dx + 21T s Dw(x) dx = 4rrW(21T) - 21TW(S).

Since FI applied to a constant function gives zero, we have

21T(D2FV)(W) = -FI(h) = 21TFI(W) = 21TF(w).

Thus D2Fv = F. 0

Lemma 6.5. Suppose FE &' and suppose F(w) = ° if w is a constant function. Then there is a unique G E &' such that DG = F and G(w) = ° if w is a constant function. If F is of order k ~ 1, then G is of order k - 1.

Proof If u E .9, it is not necessarily the derivative of a periodic function. We can get a periodic function by setting

I x X I211 IX Su(x) = 0 u(t) dt - 21T 0 u(t) dt = 0 u(t) dt - xFe(U),

Determining the periodic distributions 93

where

e(x) = 1, all x,

Then

D(Su) = u - Fe(u)e.

It follows that if DG = F and G(e) = 0, then

(10) G(u) = G(u - Fe(u)e) = G(D(Su» = - DG(Su) = - F(Su).

Thus G is unique. To prove existence, we use (10) to define G. Since S: fJ' ~ fJ' is linear, G is linear. Also

ISul :=:; 41Tlul, I D(Su) I :=:; 21ul,

I Dk(SU) I = IDk-1UI, k;::; 2.

Then if Un ~ U (fJ') we have

G(un) = -F(Sun) ~ -F(Su) = G(u).

Thus G E fJ". Also

DG(u) = - G(Du) = F(S(Du» = F(u).

Finally, suppose F is of order k ;::; 1. Then

IG(u)1 = IF(Su) I :=:; c{ISul + I DSul + ... + IDkSul} :=:; 51TC{lul + IDul + ... + IDk-lul},

and G is of order k - 1. 0

Corollary 6.6. If G E fJ" and DG = 0, then G = Ft , where! is constant.

Proof Again let e(x) = 1, all x. Let! = (21T)-lG(e)e, and

B = G - Ft.

Then DB = ° and B(e) = 0. By Lemma 6.5 (uniqueness), B = 0. Thus G = Ft. 0

FinaIIy, we can prove Theorem 6.1. Suppose FE fJ". Take an integer k so large that F is of order k - 2 ;::; 0. Again, let e(x) = 1, all x, ! = (21T)-lF(e)e, and

Fo = F - Ft.

Then Fo is of order k - 2 and Fo(e) = 0. By repeated applications of Lemma 6.5 we can find Flo F2 , ••• , Fk - 2 E fJ" so that

Fie) = 0,

and Fj is of order k - 2 - j. Then Fk - 2 is of order 0. By Lemma 6.4, there is a v E <'C such that D2 Fv = Fk _ 2' Then

DkFv = Dk- 2Fk_2 = Dk-1Fk_1 = ... = Fo

= F - Ft. o

94 Periodic functions and periodic distributions

Exercises

1. To what extent are the functions v and f in Theorem 6.1 uniquely determined?

i.e.,

2. Find v E 'fj such that D2 Fv = F, where

F = 8 - T,,8,

F(u) = u(O) - u(7r).

3. Find v E'fj and a constant function f such that

8 = D2Fv + Ft.

§7. Convolution of distributions

Suppose v E'fj and u E fJ'. The convolution v * u can be written as

1 f2" (v * u)(x) = (u * v)(x) = 27r 0 u(x - y)v(y) dy

I f2" = 27r 0 v(y)u(y - x) dy = Fv(Txu);

here again u(x) = u( -x). Because of this it is natural to define the convolu­tion of a periodic distribution F and a smooth periodic function u by the formula

(I) (F * u)(x) = F(Txu).

Proposition 7.1. IfF is a periodic distribution and u is a smooth periodic function, then the function F * u defined by (1) is a smooth periodic function. Moreover,

(2) (aF) * u = a(F* u) = F* (au), FEfJ",uE&!aEC;

(3) (F + G) * u = F * u + G * u, FEfJ", uEfJ';

(4) F* (u + v) = F* u + F* v, FE fJ", u, V E fJ';

(5) Tt(F* u) = (TtF) * u = F* (Ttu), FEfJ", UEfJ';

(6) D(F* u) = (DF) * u = F* (Du), FE fJ", u E&.

Proof The identities (2)-(5) follow from the definition (1) by elementary manipUlations. For example,

Tt(F * u)(x) = (F * u)(x - t) = F(Tx_tu) = F(T -tTxu) = (TtF)(Txu) = «TtF) * u)(x).

Also,

Convolution of distributions

This proves (5). It follows that

(F * u)(x + 27T) = (F * (T2"U))(X) = (F * u)(x).

We know that

Therefore

t- 1 [(F* u)(x + t) - (F* u)(x)] = t- 1[L t(F* u)(x) - (F* u)(x)] = t- 1[L tF - TF](Txu)~ DF(Txu) = «DF) * u)(x).

95

This shows that F * u is differentiable at each point x E ~, with derivative (DF) * u(x). By induction, Dk(F * u) = (DkF) * u. Thus F * u E fJ. Finally, using (5) again,

t- 1[(F*u)(x + t) - (F*u)(x)] = F* [t-1(L t y - u)](x) = F(TAt- 1(L tu - u)n ~ F(TxCDur) = (F * (Du))(x).

By induction, Dk(F * u) = F * (DkU), all k. We leave the proofs of (2), (3), (4) as an exercise. 0

As an example:

(7) s * u = u,

In using (1) to define F * u, we departed from the procedure in §4, where operations on distributions were defined in terms of their actions on functions. Suppose u E '(j', V E ~ W E.9. Then

(8) Fv*u(w) = Fu*v(w) = (27T)-2 {2" f" u(x - y)v(y)w(x) dy dx

= (27T)-2 f" f(y){f" a(y - x)w(x) dX} dy

= Fv(a * w).

This suggests that we could have defined F * u as a distribution by letting it assign to WE fjJ the number F(u * w). We shall see that this distribution corresponds to the function defined by (1).

Lemma 7.2. If u E '(j' and v E '(j', then w = u * v is the uniform limit of the functions wn, where

n

(9) Wn = (27T)-2 L n- 1v(27Tm/n)T2"mmU. m=1

Proof Let Xmn = 2m7T/n. Then it is easy to see that

27T(Wn(X) - w(x)) = i fxmn [v(xmn)u(x - x mn) - v(y)u(x - y)] dy, n=l Xm-l,n

96 Periodic functions and periodic distributions

Now u, v are uniformly continuous and over the range of integration of the m-th summand,

Iy - xmnl < 27T/n.

Therefore IWn - wl-+ 0 as n -+ 00. 0

Corollary 7.3. If U E gJ and v E~, then the functions Wn given by (9) converge to W = U * v in the sense of gJ as n -+ 00.

Proof Since Dk(Txu) = TiDkU), Dkwn is the corresponding sequence of functions for (Dku) * W = Dk(U * w). Therefore Dkwn -+ Dkw uniformly as n -+ 00, for each k. 0

Proposition 7.4. IfF E gJ' and u, v E &, then

F,(v) = F(a * v),

where

f= F* u.

Proof Let w = a * v and let Wn be the corresponding function defined by (9), with a replacing u. Then Wn -+ a * v (gJ), so

F(a * v) = lim F(wn)

But n

F(wn) = (27T)-Z L n-lv(27Tm/n)F(Tz1Imtnu) m=l

1 f211 -+ 27T 0 v(x)f(x) dx = F,(v). o

We shall now define the convolution of two periodic distributions F, G by

(10) (F * G)(u) = F(G- * u),

If G = Fv'/ = F * v, then Proposition 3.4 shows that F, = F * G. In general, we must verify that (10) defines a periodic distribution. Clearly F * G: gJ' -+ C is linear. If Un -+ U (gJ) and G is of order k, then

I(G- * un)(x) - G- * u(x) I = I(G- * (un - u»(x)I = IG-(Tiun - u)-) I :$; c{ITiun - u)1 + IDTiun - u)1 + ... + IDkTiun - u)1} = c{lun - ul + ID(un - u)1 + ... + IDk(Un - u)I}·

Thus G- * Un -+ G- * u uniformly. Similarly, for each j, Df(G- * un) = G- * Djun -+ Df(G- * u) uniformly. Thus

G - * Un -+ G- * u (gJ),

Convolution of distributions 97

so

(F * G)(un) -+ (F * G)(u).

This shows that F * G E f!JJ'. As an example,

(11)

In the course of showing that F * G is continuous, we have given an argument which proves the following.

Lemma 7.5. If FE f!JJ', (un)f c &, and Un -+ U (f!JJ), then F * Un -+

F * u (f!JJ).

Corollary 7.6. Suppose (fn)f C &, (gn)'f C .9, and set

Suppose

Fn -+ F (f!JJ') and Gn -+ G (f!JJ').

Then

Fn * G -+ F* G (f!JJ')

and

F * Gn -+ F * G (f!JJ').

Proof Suppose u E.9. Then

(Fn * G)(u) = Fn(G~ * u) -+ F(G~ * u) = (F * G)(u)

Also, Gn ~ * u -+ G~ * u (f!JJ) so

(F * Gn)(u) = F(Gn ~ * u) -+ F(G~ * u) = (F * G)(u). 0

We can now prove approximation theorems for periodic distributions analogous to those for functions.

Theorem 7.7. Suppose (CPnH" C f!JJ is an approximate identity, and suppose FE f!JJ'. Let Fn = FIn' where fn = F * CPn' Then Fn -+ F (f!JJ').

In particular, there is a sequence (fn)f of trigonometric polynomials such that FIn -+ F (f!JJ').

Proof We have, by Proposition 7.4,

Fn(u) = F(CPn * u), u E.9.

But (CPn):=l is also an approximate identity, so

CPn * U -+ u (f!JJ).

Therefore

Fn(u) -+ F(u).

98 Periodic functions and periodic distributions

If (<Pn); = 1 is an approximate identity consisting of trigonometric poly­nomials, then the functions In = F * <Pn are also trigonometric polynomials. In fact, let

Then

But

so

Thus

is a trigonometric polynomial. 0

Finally, we prove the analog of Proposition 7.1 and Proposition 3.1.

Proposition 7.8. Suppose F, G, HE fYJ', a E C. Then

(12)

(13)

(14)

(15)

(16)

(17)

F*G = G*F,

(aF) * G = a(F* G) = F* (aG),

(F + G) * H = F * H + G * H,

(F*G)*H= F*(G*H),

Tt(F * G) = (TtF) * G = F * (TtG),

D"(F* G) = (D"F) * G = F* (D"G).

Proof All of these identities except (12) and (15) follow from the definitions by a sequence of elementary manipulations. As an example, we shall prove part of (16):

[Tt(F * G)](u) = F * G(Ltu) = F(G- * Ltu) = F((T -tG-) * u) = F((TtG)- * u) = (F * TtG)(u).

Here we used the identity

(18)

To prove (12) and (15) we use Theorem 7.7 and Corollary 7.6. First, suppose G = Fg , g E f!JJ. Take (fn)i c fYJ such that

Fn = F'n -+ F (fYJ').

Summary of operations on periodic distributions 99

Let

h,. =/,. *g,

It follows from (8), (10), and Corollary 7.6 that

H,. = F,. * G __ F * G (,gil').

But

h,. = g */,.,

so also

H,. = G * F,. __ G * F (,gil').

Thus (12) is true when G = Fg , g E ~ In the general case, take (g,.)i c ,gil so that

Then, in the sense of ,gil',

F * G = lim F * G,. = lim G,. * F = G * F.

The proof of (15) is similar. In the first place, (15) is true when

since

F * (G * H) = F * Fgoh = FIO(goh) = F(fog)Oh = (F * G) * H.

We then approximate an arbitrary F by Fin and get (15) when G = Fg, H = Fh • Then approximate G, H successively to get (15) for all F, G, HE ,gil'. The rest of the proof is left as an exercise.

Exercises

1. Prove the identities (2), (3), (4). 2. Prove the identities (7), (11). 3. Prove the identities (13), (14), (16), (17), (18) directly from the de­

finitions. 4. Prove the identities in Exercise 3 by approximating the distributions

F, G, H by smooth periodic functions.

§8. Summary of operations on periodic distributions

In this section we simply collect for reference the definitions and results concerning ,gil'. The space ,gil is the set of infinitely differentiable periodic functions u: IR __ C. We say

u,. __ u (,gil) if I DkU,. - DkUI -- 0, all k.

100 Periodic functions and periodic distributions

A periodic distribution is a mapping F: &' -+ C with

F(u + v) = F(u) + F(v), F(au) = aF(u), F(un) -+ F(u) if Un -+ U (&').

If v E ~, the space of continuous periodic functions, then Fv E &" is defined by

1 i2n Fv(u) = 2'17 0 v(x)u(x) dx.

The 8-distribution is defined by

8(u) = u(O).

The sum, scalar multiple, complex conjugate, reversal, and translation of distributions are defined by

(F + G)(u) = F(u) + G(u), (aF)(u) = aF(u),

F*(u) = (F(u*»* (u*(x) = u(x)*), F-(u) = F(u) (u(x) = u( -x»,

(TtF)(u) = F(T -tu) (Ttu(x) = u(x - t».

Derivatives ate defined by

We say

if

In particular,

Then

Ifv E~ then

IfvEfI!

(DkF)(U) = ( -1)kF(Dku).

all u E &'.

8 = 8* = 8-(Tt8)(u) = u(t),

(Dk8)(u) = (_l)k Dku(O).

(Fv)* = F"., (Fv)- = F1J ,

Tt(Fv) = Fw, where w = Tty.

Dk(Fv) = Fw where w = Dkv.

The convolution F * u is the function

(F * u)(x) = F(T:;.fl), u E &'.

Summary of operations on periodic distributions

Then F * u E f/J. If v E~, then

Fv * u = v * u.

If (IPn)l' c f!Jl is an approximate identity, then

F * IPn --* F (f!Jl').

More precisely,

FIn --* F (f!Jl') , where In = F * IPn.

In particular,

I) * u = U, U Ef!Jl.

The convolution F * G is the distribution

(F* G)(u) = F(G~ * u), u Ef/J.

In particular if v E.9! then

F * Fv = FI, where 1= F * v.

Clearly

If Fn --* F (f!Jl')

then

Fn * G --* F * G (f!Jl').

The convolution of distributions satisfies

F*G=G*F, (aF) * G = a(F* G) = F* (aG),

(F + G) * H = F * H + G * H, (F * G) * H = F * (G * H),

Tt(F * G) = (TtF) * G = F * (TtG), Dk(F* G) = (DkF) * G = F* (DkG).

101

A periodic distribution F is real if F = F*. Any FE f!Jl' can be written uniquely as

In fact

F= G + iH, G, H real.

G = Re F = !(F + F*),

1 H = 1m F = 2i (F - F*).

A periodic distribution F is even if F = F~ and odd if F = - F~. Any FE f!Jl' can be written uniquely as

F= G + H, G even, H odd.

102 Periodic functions and periodic distributions

In fact G = t(F + F-), H = t(F - F-).

The 8-distribution is real and even. A periodic distribution is of order k if there is a constant c such that

IF(u)1 :s; c(lul + IDul + ... + IDkUI),

Any FE f!)J' is of order k for some k. If v E <'C and f is constant,

DkFv + Ff E f!)J'.

all u Ef!JJ.

Conversely, if F E f!)J' is of order k + 2, k ~ 0, then there are v E <'C and con­stant function f such that

Chapter 4

Hilbert Spaces and Fourier Series

§1. An inner product in CC, and the space L2

Suppose u and v are in '1&', the space of continuous complex-valued periodic functions. The inner product of u and v is the number (u, v) defined by

(1) 1 f.211

(u, v) = 27T 0 u(x)v(x)* dx.

It is easy to verify the following properties of the inner product:

(2) (au, v) = a(u, v) = (u, a*v),

(3) (Ul + U2, v) = (Ub v) + (U2' v),

(4) (u, VI + V2) = (u, VI) + (u, V2),

(5) (v, u) = (u, v)*,

(6) (u, v) ~ 0, (u, u) = 0 only if u = o. We define Ilull for u E 'I&' by

(7) ( 1 f.211 )1/2 IIuli = (u, U)1/2 = 27T 0 lu(x)12 dx .

Lemma 1.1. If u, v E '1&', then

(8) I(u, v)1 ~ IIuli IIvll·

Proof. If v = 0 then

(u, v) = (u,Ov) = O(u, v) = 0,

and (8) is true. Suppose v #- O. Note that for any complex number a,

(9) 0 ~ (u - av, u - av) = (u, u) - (av, u) - (u, av) + (av, av) = IIull 2 - a(u, v)* - a*(u, v) + laI 211vll 2.

Let

Then (9) becomes

o ~ IIull 2 - 21(u, v)1211vll -2 + I(u, v)J211vll -2,

and this implies (8). 0

The inequality (8) is known as the Schwarz inequality. Note that only the properties (2)-(6) were used in the proof, and no other features of the inner product (1).

103

104 Hilbert spaces and Fourier series

Corollary 1.2. The junction u -+- II u II is a norm on 7/.

Proof Recall that this means that II u II satisfies

(10)

(11)

(12)

Ilull 2: 0, Ilull = 0 only if u = 0,

Ilaull = lal Ilull, a E C,

Ilu + vii :::; Ilull + IlvII-Property (10) follows from (6) and property (11) follows from (2). To prove (12), we take the square and use the Schwarz inequality:

Ilu + vl1 2 = (u + v, u + v) = IIul12 + (u, v) + (v, u) + IIvl12 :::; IIul12 + 211ullllvil + II vl1 2 = (Ilull + Ilv11)2. 0

The new norm on 7/ is dominated by the preceding norm:

(13) Ilull :::; lui = sup {lu(x)I}·

It is important to note that 7/ is not complete with respect to the metric associated with this new norm. For example, let Un: IR -+-IR be the periodic function whose graph contains the line segments joining the pairs of points

(0,0),

(~7T - ~,J) , G 7T, 0); G 7T, 0), (27T, 0).

Then

so (un)'f is a Cauchy sequence in the new metric. However, there is no U E 7/ such that

Ilun - ull-+- o. In order to get a complete space which contains 7/ with this inner product,

we turn to the space of periodic distributions. Suppose (un)! C 7/ is a Cauchy sequence with respect to the metric induced by the norm Ilull, I.e.

Ilun - urnll-+- 0 as n, m -+- 00.

An inner product in 'C, and the space L2 lOS

Let (Fn)i be the corresponding sequence of distributions:

Thus if v EEl!

1 f2" Fn(v) = 217 0 un(x)v(x) dx = (v, u:),

where again u: denotes the complex conjugate function. By the Schwarz inequality,

JFn(v) - Fm(v)J = J(v, u: - u:)J ~ JJvJJ JJun - umJJ ~ JvJJJun - umJJ.

Therefore (Fn(v»:f has a limit. We define

(14) F(v) = lim Fn(v).

The functional F: fYJ -+ t:C defined by (14) is clearly linear, since each Fn is linear. In fact, F is a periodic distribution. To see this, we take N so large that

ifn, m ;:: N.

Let

Then for any n ~ N,

while if n > N,

Therefore

so

JJunJJ = JJun - UN + UNJJ ~ JJun - UNJJ + JJUNJJ < 1 + JJUNJJ ~ M.

JFn(v)J = J(u, u:)J ~ JJvJJ JJunJJ ~ MJvJ,

JF(v)J = lim JFn(v)J ~ MJvJ.

We have proved the following lemma.

Lemma 1.2. If (un)i c t:C is a Cauchy sequence with respect to the norm JJuJJ, then the corresponding sequence of distributions

Fn = Fun

converges in the sense of fYJ' to a distribution F, which is of order O.

It is important to know when two Cauchy sequences in t:C give rise to the same distribution.

Lemma 1.3. Suppose (un)i c t:C and (vn)i c t:C are Cauchy sequences with respect to the norm JJuJJ. Let Fn = Fun and Gn = Fun be the corresponding distributions, and let F, G be the limits:

Fn -+ F (fYJ') and Gn -+ G (fYJ').

106 Hilbert spaces and Fourier series

Then F = G if and only if

Proof Let Wn = Un - Vn and let Hn = Fn - Gn = Fwn. We want to show

Hn -+ 0 (£31>') if and only if llwnll-+ O.

Suppose

Then for any U E 9,

Conversely, suppose

(15) Hn -+ 0 (£31>').

Given e > 0, take N so large that n, m ~ N implies

(16)

Fix, m ~ N. Then if n ~ N we use (16) to get

IIwml1 2 = (wm' wm) = (wm' Wm - wn) + (wm' wn) = (wm' Wm - wn) + Hn(wm) ~ ellwmll + I Hn(wm)l·

Letting n -+ 00, from (15) we get

or m ~ N. o

We define V to be the set consisting of all periodic distributions F with the property that there is a sequence (un)f such that

Ilun - umll-+ 0 as n, m -+ 00,

Fun -+ F (£31>').

If (un)f C 'C is such a sequence, we say that it converges to F in the sense of V and write

Un -+F (V).

Lemma 1.2 can be rephrased: if

Un -+F (V), Vn -+ G (V),

then

F=G if and only if Ilun - vnll -+ O.

Clearly L2 is a subspace of £31>' in the sense of vector spaces. In fact, if

Un -+ F (V) and Vn -+ G (V),

An inner product in ~, and the space V 107

then

Un + Vn -+ F + G (V).

We may extend the inner product on 'C to V as follows. If

un-+F (V), Vn -+ G (V),

let

(17)

The existence of this limit is left as an exercise. Lemma 1.2 shows that the limit is independent of the particular sequences (un)f and (vn)f. That is, if also

Un -+ F (V), v' -+ G (V)

then

(18)

Theorem 1.3. The inner product in V defined by (17) satisfies the identities (2), (3), (4), (5), (6). If we define

(19) IIFII = (F, F)1!2,

then this is a norm on L2. The space V is complete with respect to this norm.

Proof The fact that (2)-(6) hold is a consequence of (17) and (2)-(6) for functions. We also have the Schwarz inequality in V:

I(F, G)I ::;; IIFIIIIGII· It follows that IIF II is a norm.

Finally, suppose (Fn)f C L2 is a Cauchy sequence with respect to this norm. First, note that if

un-+F (V)

and v E 'C, and if we take Vn = v, all n, then

It follows that

i.e., F can be approximated in L2 by functions. Therefore, for each n = 1, 2, . . . we can find a function Vn E 'C such that

Then

Ilvn - vmll = IlFvn - Fvmll ::;; IlFvn - Fnll + IlFn - Fmll + IlFm - Fvmll < IlFn - Fmll + n- 1 + m- 1 -+ 0 as n, m -+ 00.

108 Hilbert spaces and Fourier series

Thus there is an FE L2 such that

But then

v" -+ F (L2).

IIF" - FII ::; IIF" - Fvnll + IlFvn - FII < n- l + IlFvn - FII-+O.

Exercises

o

1. Carry out the proof that C(;' is not complete with respect to the norm

lIull· 2. Show that the limit in (17) exists. 3. Show that (18) is true. 4. Suppose I: [0, 217] -+ ~ is such that

Define

I(x) = 1, I(x) = 0,

X e [a, b), x ¢ [a .. b).

F(v) = 2~ f" I(x)v(x) dx = 2~ f vex) dx, b E ~ Show that FEV.

5. Suppose I: [0,217] -+ IC is constant on each subinterval [Xl-lo XI),

where

o = Xo < Xl < ... < X" = 217.

Define

1 (2" F(v) = -2 I(x)v(x) dx. 17 .0

Show that FEV. 6. Show that 8, the 8-distribution, is not in V. 7. Show that if FE V there is a sequence (u,,)i of smooth periodic

functions such that

u" -+F (L2).

8. Let T t denote translation. Show that if u E C(;' then

IITtu - T.ull-+ 0 as t -+ s.

If FE L2, show that

IITtF - T.FII-+ 0 as t -+ s. 9. For any FeV, show that

IIFII = sup {IF(u)11 u E.9! lIuli ::; I}.

Hilbert space 109

§2. Hilbert space

In this section we consider an abstract version of the space L2 of §l. This clarifies the nature of certain theorems. In addition, the abstract version describes other spaces which are obtained in very different ways.

Suppose H is a vector space over the real or complex numbers. An inner product in H is a function assigning to each ordered pair of elements u, v E H a real or complex number denoted by (u, v), such that

(au, v) = a(u, v), a a scalar, (U1 + U2, v) = (Ub v) + (U2' v),

(v, u) = (u, v)* ((v, u) = (u, v) in the real case), (u, u) > 0 if u =F O.

The argument of Lemma 1.1 shows that

(I) I(u, v)1 :::; lIullllvll, where

(2)

Then lIuli is a norm on H. If H is complete with respect to the metric associ­ated with this norm, then H is said to be a Hilbert space. In particular, V is a Hilbert space. Clearly any Hilbert space is a Banach space.

A more mundane example than L2 is the finite-dimensional vector space eN of N-tuples of complex numbers, with

N

(a, b) = L anb: n=l

when

a = (ab a2, ... , aN),

In this case the Schwarz inequality is

1~1 anb: \2 :::; ~1 lanl2 n~l Ibnl2.

Notice in particular that if we let

then

A still more mundane example is the plane 1R2, with

a·b = a1b1 + a2b2

when a = (ab a2), b = (bb b2); here we use the dot to avoid confusing the inner product with ordered pairs. It is worth noting that the law of cosines of trigonometry can be written

(2) a·b = lallbl cos 0,

110 Hilbert spaces and Fourier series

where e is the angle between the line segments from 0 to a and the line segment from 0 to b. Therefore la·bl = lallbl if and only if the segments lie on the same line. Similarly, a· b = 0 if and only if the segments form a right angle.

Elements u and v of a Hilbert space H are said to be orthogonal if the inner product (u, v) is zero. In 1R2 this means that the corresponding line segments are perpendicular. We write

u.lv

when u and v are orthogonal. More generally, U E H is said to be orthogonal to the subset S c H if

u.l v, all v E S.

If so we write

u.l S.

Ifu.l v then

(3) Ilu + vl1 2 = (u + v, u + v) = (u, u) + (u, v) + (v, u) + (v, v)

= IIul1 2 + Ilv11 2 •

In 1R2, this is essentially the Pythagorean theorem, and we shall give the identity (3) that name in any case. Another simple identity with a classical geometric interpretation is the parallelogram law:

(4) Ilu - vl1 2 + Ilu + vl1 2 = 211ul1 2 + 211v11 2 •

This follows immediately from the properties of the inner product. In 1R2

it says that the sum of the squares of the lengths of the diagonals of the parallelogram with vertices 0, u, v, u + v is equal to the sum of the lengths of the squares of the (four) sides.

When speaking of convergence in a Hilbert space, we shall always mean convergence with respect to the metric associated with the norm. Thus

Un -+ U means Ilun - ull-+ O.

The Schwarz inequality shows that the inner product is a continuous function.

Lemma 2.1. [I(un)l, (Vn)l cHand

then

(Un> vn) -+ (u, v).

Proof Since the sequences converge, they are bounded. In particular, there is a constant M such that Ilvnll ::;; M, all n. Then

I(un, vn) - (u, v)1 = I(un - u, vn) + (u, Vn - v)1 ::;; Ilun - ull Ilvnll + Ilull Ilvn - vii ::;; Mllun - ull + Ilullllvn - vll-+ o. 0

Hilbert space 111

Corollary 2.2. If u E H, S c H, and u 1.. S, then u is orthogonal to the closure of S.

The general theory of Hilbert space essentially rests on the following two geometric lemmas.

Lemma 2.3. Suppose Hl is a closed subspace of the Hilbert space H, and suppose u E H. Then there is a unique v E Hl which is closest to u, in the sense that

Ilu - vii :::; Ilu - wll, Proof The set

{Ilu - will wEHl} is bounded below by O. Let dbe the greatest lower bound of this set. For each integer n > 0 there is an element Vn E Hl such that

If we can show that (vn)1" is a Cauchy sequence, then it has a limit v E H. Since Hl is closed, we would have v E Hl and Ilu - vii = d as desired.

Geometrically the argument that (vn}1° is a Cauchy sequence is as follows. The midpoint i(vn + vm) of the line segment joining Vn and Vm has distance ~ d from u, by the definition of d. Therefore the square of the length of one diagonal of the parallelogram with vertices u, Vn, Vm, Vn + Vm is nearly equal to the sum of squares of the lengths of the sides. It follows that the length Ilvn - vmll of the other diagonal is small. Algebraically, we use (4) to get

o :::; IIVn - vm l1 2 = 211Vn - ul1 2 + 211Vm - ul1 2 - II(vn + Vm) - 2ul1 2

< 2(d + n- l )2 + 2(d + m- l)2 - 411!(Vn + Vm) - ul1 2 :::; 2(d + n- l )2 + 2(d + m- l)2 - 4d2 ~ O.

To show uniqueness, suppose that v and w both are closest to u in the above sense. Then another application of the parallelogram law gives

Ilu - i(v + w)112 = -!-Ilu - vI12 + -!-Ilu - wl1 2 - illv - wl1 2 = d 2 - Ilv - w11 2.

Since the left side is ~ d 2 , we must have v = w. D

As an example, take H = [R2, Hl a line through the origin. The unique point on this line closest to a given point u is obtained as the intersection of Hl and the line through u perpendicular to Hl. This connection between perpendicularity (orthogonality) and the closest point is also true in the general case.

Lemma 2.4. Under the hypotheses of Lemma 2.3, the element v E Hl is closest to u if and only if

112 Hilbert spaces and Fourier series

Proof. First, suppose v E HI is closest to u, and suppose WE HI' We want to show (u - v, w) = 0, and we may assume W "# O. Let Ul = U - v. For any a E C, v + aw E HI' Therefore

Ilul - awII2 = lIu - (v + aw)1I2 ~ lIu - vll2 = lIu1112, or

Let

(5)

Then (5) becomes

Thus (u1, w) = O. Conversely, suppose u - v .1 HI> and suppose W E HI' Then v - WE HI>

so (u - v) .1 (v - w).

The Pythagorean theorem gives

lIu - wI12 = lI(u - v) + (v - w)112 = lIu - vl12 + Ilv - wI12 ~ lIu - v1l 2• 0

Corollary 2.5. Suppose HI is a closed subspace of a Hilbert space H. Then either HI = H, or there is a nonzero element u E H such that u .1 HI'

Proof. If HI "# H, take Uo E H, UO ¢ HI' Take Vo E HI such that Vo is closest to Uo. Then u = Uo - Vo is nonzero and orthogonal to H. 0

As a first application of these results, we determine all the bounded linear functionals on H. The following theorem is one of several results known as the Riesz Representation Theorem.

Theorem 2.6. Suppose H is a Hilbert space and suppose v E H. The mapping Lv: H -+ C (or IR) defined by

4(u) = (u, v), uEH,

is a bounded linear functional on H. Moreover, if L is any bounded linear functional on H, then there is a unique v E H such that L = Lv'

Proof. Clearly 4 is linear. By the Schwarz inequality

ILv(u)I ~ IIvllllull· Thus 4 is bounded.

Suppose L is a bounded linear functional on H. If L = 0 we may take v = O. Otherwise, let

Hl = {u E H I L(u) = O}.

Hilbert spaces of sequences 113

Then H1 is a subspace of H, since L is linear; H1 is closed, since L is con­tinuous. Since L -:f 0, H1 is not H. Take a nonzero U E H which is orthog­onal to Hlo and let

v = Ilull-2L(u)u.

Then also v 1- H 1 , so

4(w) = L{w),

Moreover,

4(u) = Ilull- 2L{u){u, u) = L(u).

If w is any element of H,

w - L{U)-lL(w)u E H 1•

Thus any element of H is of the form

au + W1

for some a E C (or IR) and W1 E H 1• It follows that 4 = L. To show uniqueness, suppose 4 = Lw. Then

o = 4(v - w) - Lw(v - w) = Ilv - w11 2.

Exercises

1. Prove (2).

o

2. Suppose FE &J'. Show that FE V if and only if there is a constant e such that

I F(u) I ::::; ellull, all u E fiJ.

3. Let H be any Hilbert space and let H1 be a closed subspace of H. Let

H2 = {u E H I u 1- H1}.

Show that H2 is a closed subspace of H. Show that for any u E H there are unique vectors U1 E H1 and U2 E H2 such that

u = U1 + U2.

§3. Hilbert spaces of sequences

In this section we consider two infinite dimensional analogs of the finite dimensional complex Hilbert space CN. Recall that if

x = (al> a2, ... , aN) E CN

then N

IIxl1 2 = L: lanl2• n=l

114 Hilbert spaces and Fourier series

Let 1+2 denote the set of all sequences

x = (an)! C C

such that

(1)

If

we set

co

(2) (x, y) = L anb~, n=l

provided this series converges.

Theorem 3.1. The space 1+2 of complex sequences satisfying (1) is a Hilbert space with respect to the inner product (2).

Proof Suppose

x = (an)! El+2 and y = (bn)! E/+2.

If a E C, then clearly

As for x + y, we have

2: Ian + bnl2 ::; L (lanl2 + 21anbnl + Ibnl2)

::; 2 L lanl2 + 2 L Ibnl2 < 00.

Thus 1+ 2 is a vector space. To show that the inner product (2) is defined for all x, y E 1+ 2, we use the inequality (1) from §2. For each N,

Therefore

J-l lanb~1 ::; (Jl lanl2t2 C~l Ibnl2f2

::; C~l lanl2 f2 C~ Ibnl2f2

and (2) converges. It is easy to check that (x, y) has the properties of an inner product. The only remaining question is whether 1+2 is complete.

Suppose

m = 1,2, ....

Hilbert spaces of sequences 115

Suppose (xm)i is a Cauchy sequence in the metric corresponding to the norm

Ilxll = (x, X)1/2. For each fixed n,

lam." - a1'.,,12 :$; Ilxm - x1'l12 -? 0

as m, p -? 00. Thus (am,,,):= 1 is a Cauchy sequence in C, and

am." -? a" as m -? 00.

Let x = (a,,)i. We want to show that xEl+2 and Ilxm - xll-?O. Since (xmHO is a Cauchy sequence, it is bounded:

all m.

Therefore for any N, N N

L la,,12 = lim L: lam,,,12 :$; K2. n=1 m .... oo n=1

Finally, given e > 0 choose M so large that m, p ~ M implies

Ilxm - x1'll < e.

Then for any N and any m ~ M,

Thus

ifm ~ M. o It is often convenient to work with sequences indexed by the integers,

rather than by the positive integers; such sequences are called two-sided sequences. We use the notation

Let 12 denote the space of two-sided sequences

x = (anY~oo c C

such that

(3)

Here a two-sided infinite sum is defined to be the limit

if this limit exists. If

00 N

L Cn = lim L: en n=-oo M,N~+oo n=-M

116 Hilbert spaces and Fourier series

we let 00

(4) (x, y) = 2: anb:, n= - 00

provided the series converges.

Theorem 3.2. The space J2 of two-sided complex sequences satisfying (3) is a Hilbert space with respect to the inner product (4).

The proof of this theorem is very similar to the proof of Theorem 3.1.

Exercises

1. Prove Theorem 3.2 2. Let em E 1+2 be the sequence

with am,n = 0 if m i= n, an,n = 1. Show that

(a) JJemJJ = 1; (b) em.l ep ifm i= p; (c) (x, em) --+ 0 as m --+ 00, for each x E 1+ 2;

(d) the set of linear combinations of the elements em is dense in 1+ 2; (e) if x E 1+ 2 there is a unique sequence (bn}r' c (; such that

3. Show that the unit ball in 1+ 2, the set

B = {x E 1+ 2JJJXJJ ~ I},

is closed and bounded, but not compact. 4. Show that the set

C = {x E/+ 2J x = (an)i, each JanJ ~ n- 1}

is compact; C is called the Hilbert cube.

§4. Orthonormal bases

The Hilbert space (;N is a finite dimensional vector space. Therefore any element of (;N can be written uniquely as a linear combination of a given set of basis vectors. It follows that the inner product of two elements of (;N

can be computed if we know the expression of each element as such a linear combination. Conversely, the inner product makes possible a very convenient way of expressing a given vector as a linear combination of basis vectors.

Orthonormal bases 117

Specifically, let en E CN be the N-tuple

en = (0,0, ... ,0,1,0, ... ,0),

where the 1 is in the n-th place. Then {el , e2, ... , eN} is a basis for CN. More-over it is clear that

(1)

If x = (ab a2, ... , aN) E eN then the expression for x as a linear combination of the basis vectors en is

(2)

Because of (1),

Thus we may rewrite (2) as

N

(3) X = 2: (x, en)en. n=l

N

(x, y) = 2: anb~. n=l

Using (3) and the corresponding expression for y, we have

N N

(4) (x, y) = 2: (x, en)(y, en)* = 2: (x, en)(en, y). n=l n=l

In particular, N

(5) JJXJJ2 = 2: J(x, enW· n=l

The aim of this section and the next is to carry this development over to a class of Hilbert spaces which are not finite dimensional. We look for infinite subsets (en)l' with the properties (1), and try to write elements as convergent infinite sums analogous to (3).

A subset S of a Hilbert space H is said to be orthonormal if each U E S has norm 1, while

(U, v) = ° if u, v E S, U f= v.

The following procedure for producing orthonormal sets is called the Gram­Schmidt method.

Lemma 4.1. Suppose {Ub U2, ... } is a finite or infinite set of elements of a Hilbert space H. Then there is a finite or infinite set S = {eb e2, ... } of elements of H such that S is orthonormal and such that each Un is in the sub­space spanned by {eb e2, ... , en}.

118 Hilbert spaces and Fourier series

(If S has m elements, m < 00, we interpret the statement as saying that Un E span {elo' .. , em} when n ;::: m.)

Proof The proof is by induction. If each Ui = 0, we may take S to be the empty set. Otherwise let VI be the first nonzero Uh and let

el = II V 111- IV 1. Then {e1} is orthonormal and Ul E span {e1}. Suppose we have chosen elo ... , em such that {elo ... , em} is orthonormal and Ulo "" Um E

span {elo ... , em}. If each Ui E span {elo ... , em} we may stop. Otherwise choose the first j such that Uj is not in this subspace. Let

N

Vm+ 1 = Uj - L (u), en). n=1

Since {elo ... , em} is orthonormal, it follows that

::; n::; m.

Since uj 1= span {elo' .. , em}, Vm+l =I- 0. Let

em+l = Ilvm+1 11- 1vm+l'

Then {e1, ... , em+ 1} is orthonormal and Um+l is in the span. Continuing, we get the desired set S. 0

Note that completeness of H was not used. Thus Lemma 4.1 is valid in any space with an inner product.

An orthonormal basis for a Hilbert space H is an orthonormal set S c H such that span (S) is dense in H. This means that for any U E H and any e > 0, there is a v, which is a linear combination of elements of S, such that Ilu - vii < e.

A Hilbert space H is said to be separable if there is a sequence (un)f C H which is dense in H. This means that for any U E H and any e > 0, there is an n such that Ilu - unll < e.

Theorem 4.2. Suppose H is a separable Hilbert space. Then H has an orthonormal basis S, which is finite or countable.

Conversely, if H is a Hilbert space which has a finite or countable ortho­normal basis, then H is separable.

Proof Suppose (un)f is dense in H. By Lemma 4.1, there is a finite or countable orthonormal set S = {elo e2, ... } such that each Un is a linear combination of elements of S. Thus S is an orthonormal basis.

Conversely, suppose S is a finite or countable orthonormal basis for H. Suppose H is a complex vector space. Let T be the set of all elements of H of the form

Orthonormal bases 119

where N is arbitrary, the en are in S, and the an are complex numbers whose real and imaginary parts are rational. It is not difficult to show that T is countable, so the elements of T may be arranged in a sequence (un)!? Any complex number is the limit of a sequence of complex numbers with rational real and imaginary parts. It follows that any linear combination of elements of S is a limit of a sequence of elements in T. Since S is assumed to be an orthonormal basis, this implies that (vn)1' is dense in H. D

To complete Theorem 4.2, we want to know whether two orthonormal bases in a separable Hilbert space have the same number of elements.

Theorem 4.3. Suppose H is a separable Hilbert space. If dim H = N < 00, then any orthonormal basis for H is a basis for H as a vector space, and therefore has N elements.

IfH is not finite dimensional, then any orthonormal basisfor H is countable.

Proof Suppose dim H = N < 00, and suppose S cHis an ortho­normal basis. If el> ... , eM are distinct elements of Sand

then

m = 1, ... ,M.

Thus the elements of S are linearly independent, so S has :0; N elements. Let S = {e1, e2, ... , eM}' We want to show that S is a basis. Let H1 be the subspace spanned by S. Given U E H, let

M

U1 = 2: (u, en)en. n=l

Then m = 1, ... ,M.

It follows that U - U1 is orthogonal to the subspace H1. The argument used to prove Lemma 2.4 shows that U1 is the element of H1 closest to u. But by assumption on S, there are elements of H1 arbitrarily close to u. Therefore U = U1 E H1, and S is a basis.

The argument just given shows that if H has a finite orthonormal basis S, then S is a basis in the vector space sense. Therefore if H is not finite di­mensional, any orthonormal basis is infinite. We want to show, therefore, that if H is separable and not finite dimensional then any orthonormal basis is at most countable. Let (un)!? be dense in H, and let S be an orthonormal basis. For each element e E S, there is an integer n = nee) such that

lie - unll < 2- 1/2.

Suppose e, f E Sand e #- f. Let n = nee), p = n(f). Then

lie - fl12 = IIel12 + (e, f) + (f, e) + IIfl12 = 1 + 0 + 0 + 1 = 2,

120 Hilbert spaces and Fourier series

so Ilun - Up II = Ilun - e + e - (+ (- upll

~ lie - (II - Ilun - e + ( - upll ~ lie - (II - Ilun - ell - II( - upll > 21/2 - 2-1/2 - 2-1/2 = 0

Thus Un =F Up. We have shown that n(e) =F n«() if e =F (, so the mapping e -+ n(e) is a 1-1 function from S to a subset of the integers. It follows that S is finite or countable. 0

Exercises

1. Let Xl be the vector space of continuous complex-valued functions defined on the interval [-I, 1]. If u, v E Xl> let an inner product (u, V)l be defined by

(u, V)l = f1 u(x)v(x)* dx.

Let un(x) = xn-l, n = 1,2, .... Carry out the Gram-Schmidt process of Lemma 4.1 to find polynomials Pn, n = 1, 2, 3, 4 such that Pn is of degree n - 1, Pn has real coefficients, the leading coefficient is positive, and

(Pn, Pm)l = 1 if n = m, (Pn, Pm)l = 0 if n =F m.

These are the first four Legendre polynomials. 2. Let X2 be the set of all continuous functions u: IR -+ C such that

L: lu(x)j2e- x dx < 00.

Show that X2 is a vector space. If u, v E X2 , show that the integral

(u, V)2 = L: u(x)v(x)e- X dx

exists as an improper integral, and that this defines an inner product on X2•

Show that there are polynomials Pn, n = I, 2, 3, . . . such that Pn is of degree n - 1 and

(Pn, Pm)2 = 1 if n = m, (Pn, Pm)2 = 0 if n =F m.

Determine the first few polynomials of such a sequence. Except for constant factors these are the Laguerre polynomials.

3. Show that there is a sequence (Pn)i' of polynomials such that Pn is of degree n - 1 and

L: Ipn(x)j2e-<1/2)X2 dx = 1,

L: Pn(x)Pm(X)e-<1/2)X2 dx = 0 if n =F m.

Except for constant factors these are the Hermite polynomials.

Orthogonal expansions 121

4. Suppose H is a finite dimensional complex Hilbert space, of dimension N. Show that there is a linear transformation U from H onto eN such that

(Uu, Uv) = (u, v), all u, v E H.

(Hint: choose an orthonormal basis for H.) 5. Suppose H and H' are two complex Hilbert spaces of dimension

N < 00. Show that there is a linear transformation U from H onto H' such that

(Uu, Uv) = (u, v), all u, v E H.

6. In the space [2 of two-sided complex sequences, let en be the sequence with entry 1 in the nth place and all other entries O. Show that (en)'~ co is an orthonormal basis for [2.

7. Show that there is a linear transformation U from [2 onto 1+ 2 such that

(Ux, Uy) = (x, y), all x, y E [2

§5. Orthogonal expansions

Suppose H is a Hilbert space of dimension N < 00. We know that H has an orthonormal basis {el , e2 , ••. , eN}' Any element u E H is a linear combination

(1)

and as in §4 we see that

(2)

It follows that if u, v E H then N

(3) (u, v) = L (u, en)(e", v). n=l

In particular,

N

(4) IIul1 2 = L I(u, en}!? n=l

The expression (1) for u E H with coefficients given by (2) is called the orthogonal expansion of u with respect to the orthonormal basis {eb ... , eN}'

We are now in a position to carry (1)-(4) over to an infinite-dimensional separable Hilbert space.

Theorem 5.1. Suppose H is a Hilbert space with an orthonormal basis (en)l'. Ifu E H, there is a unique sequence (an)l' of scalars such that

(5)

122 Hilbert spaces and Fourier series

in the sense that

(6)

The coefficients are given by

(7)

and they satisfy

00

(8) 2: lan l2 = Ilu11 2• n=l

More generally, if 00

(9) and v = 2: bnen n=l

then

00 00

(10) (U, v) = 2: anb! = 2: (u, en)(en. v). n=l n=l

Conversely, suppose (an)r' is a sequence of scalars with the property

(11)

Then there is a unique element U E H such that (5) is true.

Proof First let us prove uniqueness. Suppose (an)f is a sequence of scalars such that (6) is true. Let

(12)

Since the sequence (en)f is orthonormal,

(UN' en) = an

U sing Lemma 2.1 we get

if N ~ n.

an = lim (UN' en) = (u, en). N-oo

Thus (an)f is unique. To prove existence, set an = (u, en). Define UN by (12). Then

l~n~N.

This implies that U - UN is orthogonal to the subspace HN spanned by {eb ... , eN}' Now given e > 0, there is a linear combination v of the en such that

Ilu - vii < e.

Orthogonal expansions 123

Then there is an No such that v E HN when N::::: No. As in the proof of Lemma 2.3, the facts that UN E HN and that U - UN ..l HN imply

Thus UN --+ u. Since the en are orthonormal,

N

IIuNI12 = (UN' UN) = L lan l2. n=l

Thus 00

IIuI12 = lim IIuNI12 = L lan l2. n=l

More generally, suppose U and v are given by (9). Let UN be defined by (12), and let VN be defined in a similar way. Then by Lemma 2.1,

N 00

(u, v) = lim (UN' VN) = lim L anb: = L anb:. n=l n=l

Finally, suppose (anH" is any sequence of scalars satisfying (II). Define UN by (12). All we need do is show that (uN)r' is a Cauchy sequence, since we can then let U be its limit. But if N > M,

N

(13) IIUN - UMI12 = (UN - UM, UN - UM) = L lan l2. n=M+l

Since (12) is true, the right side of (13) converges to zero as M, N --+ 00. 0

It is convenient to have the corresponding statement for a Hilbert space with an orthonormal basis indexed by all integers. The proof is essentially unchanged.

Theorem 5.2. Suppose H is a Hilbert space with an orthonormal basis (enY~ 00. If U E H, there is a unique two-sided sequence (an)~ 00 of scalars such that

(14)

in the sense that

(15)

The coefficients are given by

(16)

and they satisfy

00

(17) L lan l2 = IIuI12 n= - 00

124 Hilbert spaces and Fourier series

More generally, if

then 00 00

(18) (u, v) = 2: anb: = 2: (u, en)(en, v). -00 -00

Conversely, suppose (anY~~ 00 is a sequence of scalars with the property

Then there is a unique u E H such that (IS) is true.

The equations (5), (6), or (14), (IS) give the orthogonal expansion of u with respect to the respective orthonormal bases. The identity (8) or (17) is called Bessel's equality. It implies Bessel's inequality:

N

(19) 2: I(u, en)12 ~ I/ul/ 2

n=l

or N

(20) 2: I(u, en)12 ~ I/ul/ 2. n= -N

The identity (10) or (18) is called Parseval's identity. The coefficients an given by (6) or (14) are often called the Fourier coeffiCients ofu with respect to the respective orthonormal basis.

Exercises

1. Suppose Hand H1 are two infinite-dimensional separable complex Hilbert spaces. Use Theorems 4.2,4.3, and 5.1 to show that there is a linear transformation U from H onto Hl such that

(Uu, Uv) = (u, v),

Show that U is invertible and that

(U-IUb U-IV1) = (Ul' vl),

all u, v E H.

Such a transformation U is called a unitary transformation, or a unitary equivalence.

2. Let U: 1+ 2 -+ 1+ 2 be defined by

U«a1, a2, a3, ... )) = (0, a1, a2, a3, ... ).

Show that U is a I-I linear transformation such that

(Ux, Uy) = (x, y),

Show that U is not onto.

Fourier series 125

§6. Fourier series

Let L2 be the Hilbert space introduced in §1. Thus V consists of each periodic distribution F which is the limit, in the sense of L2, of a sequence (unH" c <C, i.e.,

Ilun - umll-+ 0, Fun -+ F (.9").

We can identify the space C(j of continuous periodic functions with a subspace of L2 by identifying the function u with the distribution Fu. Then

1 f2" IJFul1 2 = IIul1 2 = 217 0 lu(x)12 dx.

In particular, we may consider the two-sided sequence of functions en,

en(x) = exp (inx),

as elements of L2.

n = 0, ± 1, ±2, ...

Lemma 6.1. The sequence of functions (en)~ <x" considered as elements of V, is an orthonormal basis for V.

Proof Clearly Ilenll = 1. If m =1= n, then

f" en(x)em(x)* dx = f" exp (inx - imx) dx

= [i(n - m)]-l exp (i(n - m)x)I~" = 0.

Thus (en)~ 00 is an orthonormal set. Now suppose FE L2. Given B > 0, there is a function u E C(j such that

IIFu - FII < e.

There is a linear combination v = L anen such that

Iv - ul < e.

Then

IJFv - FII ~ IJFv - Full + IJFu - FII = Ilu - vii + IJFu - FII < Iv - ul + e < 2e. o

Since (en)~ 00 is an orthonormal basis, we may apply Theorem 5.2 and obtain orthogonal expansions for distributions in V.

Theorem 6.2. Suppose FE L2. There is a unique two-sided sequence (an)~ 00 c C such that

00

(1) F = 2: an exp (inx) n= -00

126 Hilbert spaces and Fourier series

in the sense that the functions

N

(2) 2: anen -+ F (.,<l'2) as N -+ 00. n= - 00

The coefficients are given by

(3)

and they satisfy

al

(4) 2: lanl2 = IIFI12. -al

More generally, if al al

F = 2: an exp (inx) and G = 2: bn exp (inx) -00 -00

in the sense of (2), then

al al

(5) (F, G) = 2: anb: = 2: F(e_n)G*(en). -co -co

Conversely, suppose (anY~~ al c C and

-al

Then there is a unique FE V such that (1) is true, in the sense of(2).

Proof. In view of Lemma 6.1 and Theorem 5.2, we only need to verify that

The second identity follows from the first and the definition of G*. To prove the first, take (um)i c C,

um-+F (V).

Then

(F,Fe.) = lim (um, en) = limF"m(e_ n) = F(e_ n). 0

The an In (3) are called the Fourier coefficients of the distribution F. The formal series on the right in (1) is called the Fourier series of F.

Suppose F = F" where u E r"C. Then (2) is equivalent to

(6) 12" N

o Iu(x) - n=~N an exp (inx)12 dx -+ 0,

where

(7) 1 f2" an = 21T 0 u(x) exp ( - inx) dx.

Fourier series 127

In fact, (6) and (7) remain valid when u is simply assumed to be an integrable function on [0, 21T]. In this case the an are called the Fourier coefficients of the function u, and the formal series

00

2: an exp (inx) -00

is called the Fourier series of the function u. The fact that (6) and (7) remain valid in this case is easily established, as follows. If u: [0, 21T] --* C is integ­rable, then again it defines a distribution F,. by

1 J2" F,.(v) = 21T 0 u(x)v(x) dx.

Then an extension of the Schwarz inequality gives

( 1 f2" )1/2 IF,.(v) I :;; 21T 0 lu(x)12 dx IIvll = IIull IIvll·

By Exercise 2 of §2, F,. E V. When u E~, it is tempting to interpret (1) for F,. as

00

u(x) = 2: an exp (inx). -00

In general, however, the series on the right may diverge for some values of x, and it will certainly not converge uniformly without further restrictions on u. It is sufficient to assume that u has a continuous derivative.

Lemma 6.3. If (an) ~ 00 c C has the property

00

(8) 2: lanl < 00, -00

then the functions

converge uniformly to a function u E~, and (an)~ 00 is the sequence of Fourier coefficients of u.

Proof Since I en(x) I = lexp (inx)1 = 1 for all x E C, it follows from (8) that the sequence offunctions (UN)! is a uniform Cauchy sequence. Therefore it converges uniformly to a function u E~. For N ~ m we have

(UN' em) = am·

Thus

am = lim (UN' em) = (u, em)

is the mth Fourier coefficient of u. 0

128 Hilbert spaces and Fourier series

Theorem 6.4. If u E ~ and u is continuously differentiable, then the partial sums of the Fourier series of u converge uniformly to u. Thus

00

u(x) = L (u, en) exp (inx), all x E IR. -00

Moreover, ifu E ~ then the partial sums converge to u in the sense off!IJ.

Proof Let v = Du. There is a relation between the Fourier coefficients of v and those of u. In fact integration by parts gives

(9) 1 f2" bn = (v, en) = 21T 0 Du(x) exp (-inx) dx

in f2" = 21T 0 u(x) exp ( - inx) dx

= in(u, en) = inan. We can apply the Schwartz inequality for sequences to show L lanl < 00.

In fact

( 00 )1/2

= laol + 2 n~1 n- 2 II Dull < 00.

By Lemma 6.3, the partial sums of the Fourier series of u converge uniformly to u.

Now suppose u E ~ and let

Then (9) shows that

N

UN = L anen· n= -N

N N

DUN = L inanen = L bnen n= -N n=-N

is the Nth partial sum of the Fourier series of Du. Similarly, DkuN is the Nth partial sum of Dku. Each Dku is continuously differentiable, so each DkuN -+ Dku uniformly. 0

Fourier series expansions are very commonly written in terms of sine and cosine functions, rather than the exponential function. This is partic­ularly natural when the function u or distribution F is real. Suppose FE L2. Let

(10) bn = 2F(cos nx), cn = 2F(sin nx).

Since e -n(x) = cos nx - i sin nx

Fourier series

we have

a" = t(b" - ie,,).

Also

e_" = -e".

Thus for n > 0,

a"e,,(x) + a_fIe_flex) = a,,(cos nx + i sin nx) + a_,,(cos nx - i sin nx) = btl cos nx + e" sin nx.

Then

N N

L: a" exp (inx) = tbo + L: btl cos nx + e" sin nx. n.= -N n=1

The formal series

.. (11) tbo + L: btl cos nx + e" sin nx

,,=1

129

is also called the Fourier series of F, and the coefficients btl, e" given by (10) are also called the Fourier coefficients of F. If F is real, then btl, e" are real, and (11) is a series of real-valued functions of x. Theorems 6.2 and 6.4 may be restated using the series (11).

Exercises

1. Find the Fourier coefficients of the following integrable functions on [0,2'IT]:

(a) u(x) = 0, X E [0, 'IT], u(x) = 1, x E ('IT, 2'IT]. (b) u(x) = 0, X E [0, 'IT], u(x) = X - 'IT, X E ('IT, 2'IT]. (c) u(x) = Ix - 'lT1. (d) u(x) = (x - 'IT)2. (e) u(x) = x (f) u(x) = Icos xl. 2. Suppose u E re and suppose btl, e" are as in (10). Show that if u is even

then e" = 0, all n. Show that if u is odd, then btl = 0, all n. (It is convenient to integrate over [-'IT, 'IT] instead of [0, 2'IT].) Show that if u is real then

1 b 2 1 ~ (b 2 2) _ 1 f.2" ( )2 d 4 0 + 2: L.. ,,+ e" - 2" u x x. ,,=1 'IT 0

3. Suppose u E re, N

UN = L: (u, e _")e,,. ,,=-N

130 Hilbert spaces and Fourier series

Show that

where

DN(X) = sin (N + ·!-)x/sin -tx.

The function DN is called the Dirichlet kernel. Thus

4. Extend the result of Exercise 3 to the partial sums of the Fourier series of a distribution FE L2.

5. For FE fJ", define

an = F(e_ n),

Show that FE L2 if and only if

-00

n = 0, ± 1, ±2, ....

6. Suppose FE fJ" and DF E L2. Show that F = Fu for some continuous function u. (Hint: find the Fourier coefficients of u.)

Chapter 5

Applications of Fourier Series

§1. Fourier series of smooth periodic functions and of periodic distributions

If u is a smooth periodic function with Fourier coefficients (anY~ ex" then we know that the sequence (anr~ 00 uniquely determines the function u; in fact, the partial sums

(1) N

us(x) = L an exp (inx) -N

of the Fourier series converge to u in the sense of.9. Therefore it makes sense to ask: what are necessary and sufficient conditions on a two sided sequence (an)'~~ 00 c C that it be the sequence of Fourier coefficients of a function u E!Y? The question is not hard to answer.

A sequence (an)~ 00 c C is said to be of rapid decrease if for every r > ° there is a constant c = c(r) such that

(2) all n =f. 0.

Theorem 1.1. A sequence (an)~ 00 c C is the sequence of Fourier coeffi­cients of a function u E!Y if and only if it is of rapid decrease.

Proof Suppose first that u E!Y has (an)~ en as its sequence of Fourier coefficients. Given r > 0, take an integer k ::::: r. In proving Theorem 6.4 of Chapter 4 we noted that (inan)~ 00 is the sequence of Fourier coefficients of Du. It follows that

«(in)ka,,)~ 00

is the sequence of Fourier coefficients of Dku. But then

Inlklanl :::; IDkul,

which gives (2) with c = 1 Dku I. Conversely, suppose (an)~ 00 c C is a sequence which is of rapid decrease.

Define functions UN by (1). From (2) with r = 2 we deduce

-00

By Lemma 6.3 of Chapter 4, the functions (1) converge uniformly to u E ~,

and (an)'~~ 00 are the Fourier coefficients of u. Also

N

DkuN = L (in)kanen. -N

131

132 Applications of Fourier series

and (2) with r = k + 2 implies

-00

Thus each derivative DkuN also converges uniformly as N ~ 00. Therefore UE~ 0

If F is a periodic distribution which is in V, then we have defined its Fourier coefficients by

(3) e -n(x) = exp ( - inx).

Since e _ n E fJ! the expression in (3) makes sense for any periodic distribution F, whether or not it is in L2. Thus given any FE f!jJ', we define its Fourier coefficients to be the sequence (an)'~ 00 defined by (3). We know that if all the an are zero, then F = 0 (see Chapter 3, Theorem 4.3). Therefore, FE f!jJ'

is uniquely determined by its Fourier coefficients, and we may ask: what are necessary and sufficient conditions on a sequence (anY~~ 00 c IC that it be the sequence of Fourier coefficients of a periodic distribution F? Again, the answer is not difficult.

A sequence (a,,)':: 00 c IC is said to be of slow growth if there are some positive constants c and r such that

(4) all n "# O.

Theorem 1.2. A sequence (an)':: 00 c IC is the sequence of Fourier co­efficients of a distribution FE f!jJ' if and only if it is of slow growth.

Proof Suppose first that FE f!jJ' has (an)':: 00 as its sequence of Fourier coefficients. Recall that for some integer k, F is of order k. Thus for u E fJ!

I F(u) I ::; c(lul + IDul + ... + IDkUI)·

With u = e -n this means

lanl = I F(u) I ::; cll + Inl + ... + Inlk) < 2clnl lc•

Thus (an)':: 00 is of slow growth. Conversely, suppose (an)':: 00 c IC is a sequence which is of slow growth.

Then there is an integer k > 0 such that

(5) n"# O. Let bo = 0 and

n"# O. Let

From (5) we get 00

2: Ibnl < 00. -00

Fourier series of smooth periodic functions and distributions 133

Therefore VN converges uniformly to v E ce. Let

This is a distribution; we claim that its Fourier coefficients are the (anr~~ 00'

In fact, for n i= 0,

Also

F(e_ n) = DkFv(e_n) = FvC( _D)ke_n) = (in)kFv(e_n) = (in)kbn = an.

o In the course of the preceding proof we gave a second proof of the

characterization theorem for periodic distributions, Theorem 6.1 of Chapter 3. In fact, the whole theory of f!J1 and f!J1' in Chapter 3 can be derived from the point of view of Fourier series. We shall do much of such a derivation in this section and the next. An important feature of such a program is to express the action of F E f!J1' on u E f!J1 in terms of the respective sequences of Fourier coefficients.

Theorem 1.3. Suppose FE f!J1' has (an)~ 00 as its sequence of coefficients, and suppose u E f!J1 has (bn)~ 00 as its sequence of Fourier coefficients. Then

(6)

Proof Let

We know

Therefore

But

00 00

F(u) = 2: anb_ n = 2: a_nbn. -co -CD

N

UN (X) = 2 bn exp (inx). -N

F(u) = lim F(UN)'

N N N F(UN) = 2: bnF(en) = 2: a_nbn = 2: anb- n·

-N -N -N o

Implicit in the proof of Theorem 1.3 is the proof that the series in (6) converges. A more direct proof uses the criteria in Theorem 1.1 and 1.2. In fact,

lanl ::; clnlr, n i= 0, Ibnl ::; c'lnl-r- 2, n =1= °

and convergence follows.

134 Applications of Fourier series

Corollary 1.4. If FE &' has Fourier coefficients (an)~ 00, and if FN is the distribution defined by the function

then

N

L: an exp (inx), -N

Proof With U, UN as in Theorem 1.3,

Exercises

1. Compute the Fourier coefficients of Sand DkS.

o

2. If FE &' has Fourier coefficients (an)~ co, compute the Fourier coeffi­cients of

F* ,

3. Give necessary and sufficient conditions on the Fourier coefficient (an)~ 00 of a distribution F that F be real; or even; or odd.

4. Suppose (!Pm)i c ~ is an approximate identity. Let

(am,n);'= - 00

be the sequence of Fourier coefficients of !Pm. Show that

m ~ 1, lim am,n = 1, m __ oo

n = 0, ± 1, ±2, ... ; all n.

§2. Fourier series, convolutions, and approximation

Recall that if F, G E &' and u E.q;, the convolutions F * u and F * G are defined by

F * u(x) = F(Txu), (F * G)(u) = F(G- * u).

We want to compute the Fourier coefficients of the convolutions in terms of the Fourier coefficients of F, G, and u.

Theorem 2.1. Suppose FE &' has Fourier coefficients (an)~ 00" suppose G E &' has Fourier coefficients (bn)~ co; and suppose u E & has Fourier coeffi­cients (cn)~ co' Then F * u has Fourier coefficients (ancn)~ co and F * G has Fourier coefficients (anbn)~ co'

Proof Note that

(Txen)(Y) = e_n(y - x) = en(x)e_n(y).

Fourier series, convolutions, and approximation 135

Therefore

and

Taking the limit as N -+ 00, we find that F * u has Fourier coefficients (ancn)~ co'

Now

Therefore

(G- * en)(x) = G-(Txen) = en(x)G-(en) = en(x)G(en) = b_nen(x).

o Using Theorem 2.l and Theorems 1.1 and 1.2, we may easily give a

second proof that F * u E f7J. Similarly, if u E f7J and G = Fu, then

in fact F * G and F * u have the same Fourier coefficients. The approximation theorems of Chapter 3 may be proved using Theorem

2.1 and the following two general approximation theorems.

Theorem 2.2. Suppose (urn)! C &. Suppose the Fourier coefficients of Urn are

(am,n):= - co'

Suppose that for each r > 0 there is a constant c = c(r) such that

all m, all n 1= O.

Suppose, finally, that for each n,

am,n -+ an as m -+ 00.

Then (an)~ 00 is the sequence of Fourier coefficients of a function u E.9. More­over,

um -+ u (f7J).

Proof The conditions imply that also

n 1= O.

Thus (an)~ co is the sequence of Fourier coefficients of a function u E.9. Given e > 0, choose N so large that

co

c(2) L n- 2 < e. n=N

136 Applications of Fourier series

Choose M so large that m ~ M implies

N

L: lam•1I - alii < 8. -N

It follows that if m ~ M then

co

L: lam•1I - alii < 58. -co

Since

this implies

ifm ~ M.

Thus Um -+ U uniformly. A similar argument shows each Dkum -+ Dku. 0

Theorem 2.3. Suppose (Fm)f C f?I". Suppose the Fourier coefficients of Fm are

(am.II):'~ - co.

Suppose that for some r > 0 there is a constant c such that

lam.1I1 ~ clnlr, Suppose, finally, that for each n,

all m, all n # O.

am•1I -+ all and m -+ 00.

Then (allY~ co is the sequence of Fourier coefficients of a distribution FE f?I". Moreover,

Fm -+ F (f?I").

Proof. The conditions imply that also

lalll ~ clnlr, all n # O.

Thus (allr~co is the sequence of Fourier coefficients of a distribution FEf?I". Take an integer k ~ r + 2, and let

bm•o = bo = 0, bm•1I = (in)-kam.lI ,

bll = (in)-kall, n#O n # O.

As in the proof of Theorem 1.2, (bm•II):'= _ co is the sequence of Fourier coefficients of a function Vm E C(/, with

Similarly, (bll)'~ co is the sequence of Fourier coefficients of v E C(/ with

The heat equation: distribution solutions 137

The hypotheses of the theorem imply

jbm.nj ~ cjnj-2, n #- 0, bm•n --+ bn as m --+ 00.

As in the proof of Theorem 2.1, these conditions imply that

Vm --+ v uniformly as m --+ 00.

Also,

It follows that F m --+ F (.9"). 0

Exercises

1. Suppose (9lm)l' c '?? is an approximate identity. Use the theorems of this section and Exercise 4 of §l to prove that:

9lm * u --+ u (.9') F * 9lm --+ F (.9")

ifuE.9'; if FE .9".

2. State and prove a theorem for V which is analogous to Theorem 1.2 for .9' and Theorem 1.3 for .9".

3. Use the result of Exercise 2 to show that if (9lm)l' c '?? is an approxi­mate identity and FE V, then

F * 9lrn --+ F (V).

4. Prove the converse of Theorem 2.2: if Urn --+ U (.9') then the Fourier coefficients satisfy the hypotheses of Theorem 2.2.

§3. The heat equation: distribution solutions

Many physical processes are approximately described by a function U"

depending on time and on position in space, which satisfies a type of partial differential equation called a heat equation or diffusion equation. The simplest case is the following. Find u(x, t), a continuous function defined for x E [0, 7T] and for t ~ 0, satisfying the equation

(1) o ( 0)2 ot U(X, t) = K ax U(x, t), X E (0, 7T), t > 0,

the initial condition

(2) u(x, 0) = g(X), X E [0, 7T],

138 Applications of Fourier series

and one of the following two sets of boundary conditions:

(3) u(O, t) = U(7T, t) = 0, t ;;:: 0;

(3)' o 0

ox U(O, t) = oX U(7T, t) = 0, t> 0.

The function U describes the temperature distribution in a thin homo­geneous metal rod of length 7T. The number x represents the distance of the point P on the rod from one end of the rod, the number t represents the time, and the number u(x, t) the temperature at the point P at time t. Equation (1) expresses the assumption that the rod is in an insulating medium, with no heat gained or lost except possibly at the ends. The constant K > ° is pro­portional to the thermal conductivity of the metal, and we may assume units are chosen so that K = 1. Equation (2) expresses the assumption that the temperature distribution is known at time t = 0. Equation (3) expresses the assumption that the ends of the rod are kept at the constant temperature 0, while the alternative equation (3)' applies if the ends are assumed insulated. Later we shall sketch the derivation of Equation (1) and indicate some other physical processes it describes.

Let us convert the two problems (1), (2), (3) and (1), (2), (3)' into a single problem for a function periodic in x. Note that if (2) and (3) are both to hold, we should have

(4) g(O) = g(7T) = o.

Let g( - x) = - g(x), X E ( -7T, 0). Then g has a unique extension to all of ~ which is odd and periodic (period 27T). Because of (4) the resulting function is still continuous. Suppose u were a function defined for all x E ~ and t ;;:: 0, periodic in x, and satisfying (1) for all x E ~, t > 0. Then u(x,O) = g(x) is odd. If g is smooth, then also

is odd, and we might expect that u is odd as a function of x for each t ;;:: 0. If this is so, then necessarily (3) is true.

Similarly, if (2) and (3)' are both to hold, we would expect that if g is smooth then

Dg(O) = Dg(7T) = 0.

In this case, let g( -x) = g(x), X E (-7T, 0). Then g has a unique extension to all of ~ which is even and periodic. If g is of class C 1 on [0, 27T], the ex­tension is of class Cl on R Again, if u were a solution of (1), (2) which is periodic in x for all t, we might expect u to be even for all t. Then (3)' is necessarily true.

The above considerations suggest that we replace the two problems

The heat equation: distribution solutions l39

above by the single problem: find u defined for x E IR, t 2:: 0, such that u is periodic in x,

(5) o ( 0)2 ot u(x, t) = ox u(x, t), X E JR, t > 0,

(6) u(x, 0) = g(x), X E JR,

where g E C{} is given. It is convenient and useful to ask for solutions of an analogous problem

for periodic distributions. We formulate this more general problem as foIIows. Suppose that to each t in some interval (a, b) E JR we have assigned a distri­bution Ft E gJ'. If s E (a, b), G E gJ', and

as t ~ s, it is natural to consider G as the derivative of Ft with respect to t at t = s. We do so, and write

G = dd Ftl . t t=s

Our formulation of the problem for distribution is as foIIows: given G E gJ', find distributions Ft E gJ' for each t > 0, such that

(7) dd Ftl = D2F., aIls> 0, t t=s

(8) Ft~G (gJ') as t~O.

Theorem 3.1. For each G E gJ' there is a unique family (Ft)t>o c gJ'

such that (7) and (8) hold. For each t > 0, Ft is a function ut(x) = u(x, t) which is infinitely differentiable in both variables and satisfies (5).

Proof Let us prove uniqueness first. Suppose (Ft)t>o is a solution of (7), (8). Each F t has Fourier coefficients (an(t»~ '("

t > 0, n = 0, ± 1, ....

Then

as t ~ s. In other words,

(9) t > 0.

As t~Owehave

(10)

The unique function a(t), t > ° which satisfies (9) and (10) is

(11)

140 Applications of Fourier series

This shows that an(t) are uniquely determined, so the distributions Ft are uniquely determined.

To show existence, we want to show that the an(t) defined by (11) are the Fourier coefficients of a smooth function for t > 0. Recall that if y > ° then

so

Therefore

m = 0, 1~ 2, ....

But

for some c and k. Therefore

lan(t)1 ::;; cm! Inl k- 2m ·t-m, m = 0, 1,2, ....

It follows that for each t > 0, (an(t))~ 00 is the sequence of Fourier coeffi­cients of a function Ut E &. Then u, is the uniform limit of the functions

N

un(x, t) = 2: an(t) exp (inx)

Then

As above,

-N

N

= 2: bn exp (inx - n2t). -N

N

= 2: an,:,,(t) exp (inx). -N

la (t)1 <_ cm'·lnlk+21+r-2m·t-m, n,l,r m = 0, 1,2, .. '.

It follows that each partial derivative of UN converges as N --+ 00, uniformly for x E ~ and t ;;::: S > 0. Thus

U(x, t) = u,(x)

is smooth for x E ~, t > 0. For each N, UN satisfies (5). Therefore U satisfies (5).

Let F t be the distribution determined by Ut. t > 0. Thus the Fourier coefficients of F t are the same as those of Ut.

It follows from (11) that

The heat equation: classical solutions; derivation 141

By Theorem 2.3, therefore, (8) is satisfied. Finally, each partial derivative of u is bounded on the region x E IR, t 2 8 > 0. It follows from this and the mean value theorem that

(t - S)-l(Ut - us) -+ :t Us

uniformly as t -+ s > 0. Therefore (7) also holds. 0

Exercises

1. In Theorem 3.1, suppose G is even or odd. Show that each Ft is re­spectively even or odd. Show that if G is real, then each Ft is real.

2. Discuss the behavior of Ft as t -+ 00.

3. Formulate the correct conditions on the function u if it represents the temperature in a rod with one end insulated and the other held at constant temperature.

4. In Theorem 3.1, suppose G = Fw, the distribution defined by a function w E fJ1. Show that the functions Ut -+ W (.9') as t -+ 0.

5. Let 00

gt(x) = L exp (-n2t + inx) , n= - co

Show that gt E fJ1. In Theorem 3.1, show that

Ut = G * gt.

6. With g, as in Exercise 5, show that

gt * g. = gt+8'

t > 0, X E IR.

7. The backwards heat equation is the equation (1) considered for t :s; 0, with initial (or" final") condition (2). Is it reasonable to expect that solutions for this problem will exist? Specifically, given G E .9", will there always be a family of distributions (Ft)t < 0 C .9" such that

dd Ft I = D2F., t t=s

all s < 0,

Ft -+ G (.9") as t -+ O?

8. The Schrodinger equation (simplest form) is the equation

OU .02U ot = I ox2'

Consider the corresponding problem for periodic distributions: given G E .9", find a family (Ft)t>o of periodic distributions such that

dd Ft I = iD2F., t t=s

all s > 0,

Ft -+ G (.9") as t -+ 0.

Discuss the existence and uniqueness of solutions to this problem.

142 Applications of Fourier series

§4. The heat equation: classical solutions; derivation

Let us return to the problem given at the beginning of the last section: find u(x, t), continuous for x E [0, 27T] and t ~ 0, and satisfying

(1) :t u(x, t) = (:xrU(X, t), x E (0, 7T), t > 0;

(2) u(x,O) = g(x), where g E C[O, 7T) is given;

(3) u(O, t) = U(7T, t) = 0, t ~ ° or

(3)' o 0

ox u(O, t) = oX U(7T, t) = 0, t > 0.

Such a function u is called a classical solution of the problem (1), (2), (3) or (1), (2), (3)', in contrast to the distribution solution for the periodic problem given by Theorem 3.1. In this section we complete the discussion by showing that a classical solution exists and is unique, and that it is given by the distribution solution. We consider the problem (1), (2), (3), and leave the problem (1), (2), (3)' as an exercise.

Given g E C[O, 7T] with g(O) = g(7T) = ° (so that (3) is reasonable), extend g to be an odd periodic function in 't', and let G = Fg• By Theorem 3.1, there is a function u(x, t) = ut(x) which is smooth in x, t for x E ~, t > 0, satisfies (1) for all such x, t, and which converges to G in the sense of f!jJ' as t --+ 0. By Exercise 1 of §3, u is odd as a function of x for each t > 0. Since u is also periodic as a function of x, this implies that (2) holds when t > 0. Ifwe knew that Ut --+ g uniformly as t --+ 0, it would follow that the restriction of u to ° ~ x ~ 7T is a classical solution of (I), (2), (3). Note that this is true when (the extension of) g is smooth: see Exercise 4 of §3. Everything else we need to know follows from the maximum principle stated in the following theorem.

Theorem 4.1. Suppose u is a real-valued classical solution of (1), (2). Thenfor each T > 0, the maximum value ofu(x, t) in the rectangle

° ~ x ~ 7T, ° ~ t ~ T

is attained on one of the three edges t = 0, x = 0, or x = 7T.

Proof Given e > 0, let

vex, t) = u(x, t) - et.

It is easy to see that the maximum value of v is attained on one of the edges in question. Otherwise, it would be attained at (xo, to), where

Xo E (0, 7T),

For v to be maximal here, we must have

to > 0.

The heat equation: classical solutions; derivation 143

But then

:2': e > 0,

so Xo cannot be a maximum for vex, to) on [0, 217]. Now u and v differ by at most eT on the rectangle. Therefore for any

(x, t) in the rectangle,

u(x, t) ~ M + 2eT,

where M is the maximum of u for t = 0, x = 0, or x = 17. Since this is true for every e > 0, the conclusion follows. 0

Theorem 4.2. For each continuous g with g(O) = g(17) = 0, there is a unique classical solution of problem (1), (2), (3).

Proof Note that u is a classical solution with initial values given by g if and only if the real and imaginary parts of u are classical solutions with initial values given by the real and imaginary parts of g, respectively. There­fore we may assume that g and u are real-valued. Applying Theorem 4.1 we see that

U(x, t) ~ Igl, all x E [0,17], t :2': 0.

Applying Theorem 4.1 to - u, which is a solution with initial values given by -g, we get

-U(x, t) ~ Igl, all x, t. Thus

Iu(x, t)1 ~ 1 g I, all x, t.

This proves uniqueness. To prove existence, let g be extended so as to be odd and periodic. Let

(<Pm)i c fJ' be an approximate identity. Let

gm = <Pm *gE~

We can choose <Pm to be even, so that gm is odd. Let Um be the distribution solution given by Theorem 3.1 for gm as initial value, and let U be the distri­bution solution with g as initial value. Then we know um(x, t) -+ gm(x) uniformly with respect to x at t -+ 0, so we may consider Um as continuous for t :2': 0, X E R Moreover, since gm -+ g uniformly as m -+ 00, it follows that um(x, t) -+ u(x, t), at least in the sense of fJ", for each t > O. On the other hand, since 1 gm - g 1 -+ 0 we find that for x E [0, 17]

Ium(x, t) - up(x, t)1 -+ 0

as m, p -+ 00, uniformly in x and t, t > O. Thus we must have Um -+ U

uniformly. It follows that U has a continuous extension to t = 0, and there­fore that U (restricted to x E [0, 7T]) is a classical solution. 0

A second proof of the uniform convergence of U to g as t -+ 0 is sketched in the exercises below.

144 Applications of Fourier series

The heat equation, (1), may be derived as follows. Again we consider a homogeneous thin metal rod in an insulating medium. Imagine the rod divided into sections of length e, and suppose x is the coordinate of the midpoint of one section. We consider only this and the two adjacent sections, and approximate the temperature distribution at time t by assuming u to be constant in each section. The rate of flow of heat from the section centered at x + e to that centered at x is proportional to the temperature difference u(x + e, t) - u(x, t), and inversely proportional to the distance e. The temperature in each section is the amount of heat divided by the volume, and the volume is proportional to e. Considering also the heat flow from the section centered at x - e to that centered at x, we get an approximate ex­pression for the rate of change of temperature at (x, t):

a e at u(x, t) ~ Ke- 1[u(x + e, t) - u(x) + u(x - e, t) - u(x)]

or

a at u(x, t) ~ Ke- 2[u(x + e, t) - u(x - e, t) - 2u(x)].

Consider the expression on the right as a function of e and let e -+ O. Two applications of L'H6pital's rule give

(4) a ( 0)2 at u(x, t) = K ox u(x, t).

Note that essentially the same reasoning applies to the following general situation. A (relatively) narrow cylinder contains a large number of individual objects which move rather randomly about. The random motion of each object is assumed symmetric in direction (left or right is equally likely) and essentially independent of position in the cylinder, past motion, or the presence of the other objects. As examples one can picture diffusion of molecules or dye in a tube of water kicked about by thermal motion of the water molecules, or the late stages of a large cocktail party in a very long narrow room. If u(x, t) represents the density of the objects near the point x at time t, then equation (4) arises again. Boundary conditions like (3)' correspond to the ends of the cylinder being closed, while those like (3) correspond to having one way doors at the ends, to allow egress but not ingress.

Exercises

I. Suppose G = Fw, W E ~ and suppose (Ut)t>o is the family of functions in Theorem 3.1. Show that if W ~ 0, then each Ut is ~ O.

2. As in Exercise 5 of §3, let

co

gt(x) = L exp (-n2t + inx), t> o. -co

The wave equation 145

Show that

1 f2n 211" 0 gt(x) dx = 1.

Show that if w E fJJ and w ~ 0, then gt * w ~ O. Show, by using anyapproxi­mate identity in .9, that gt ~ O.

3. Show that (gt)t> 0 is an approximate identity as t ~ 0, i.e., (in addition to the conclusions of Exercise 2) for each 0 < 8 < 11",

12,,-6

lim gt(x) dx = O. t .... o 6

(Hint: choose WE fJJ such that W ~ 0, w(O) = 0, and w(x) = 1 for 8 ~ x ~ 211" - 8. Then consider gt * w(O).)

4. Use Exercise 3 to show that ifw E~, G = Fw and (Ut)t>o is the family of functions given in Theorem 3.1, then Ut ~ w uniformly as t ~ O.

5. In a situation in which heat is being supplied to or drawn from a rod with ends at a fixed temperature, one is led to the problem

:tu(x,t) = (:xYU(X,t) +/(X,/), X E(O, 11"), t > 0,

u(x,O) = g(x), u(O, t) = u(11", I) = O.

Formulate and solve the corresponding problem for periodic distributions, and discuss existence and uniqueness of classical solutions.

§5. The wave equation

Another type of equation which is satisfied by the functions describing many physical processes is the wave equation. The simplest example occurs in connection with a vibrating string. Consider a taut string of length 11" with endpoints fixed at the same height, and let u(x, t) denote the vertical displacement of the string at the point with coordinate x, at time t. If there are no external forces, the function u is (approximately) a solution of the equation

(1) X E (0, 11"), t> O.

Here c is a constant depending on the tension and properties of the string. The condition that the endpoints be fixed is

(2) u(O, t) = u(11", I) = 0, t ~ O.

To complete the determination of u it is enough to know the position and velocity of each point of the string at time t = 0:

(3) u(x, 0) = g(x), 8 81 u(x, 0) = hex), X E [0, 11"].

We shall discuss the derivation of (1) later in this section.

146 Applications of Fourier Series

As in the case of the heat equation we begin by formulating a corre­sponding problem for periodic distributions and solving it. The conditions (2) suggest that we extend g, h to be odd and periodic and look for a solution periodic in x. The procedure is essentially the same as in §3.

Theorem 5.1. For each G, HE f!/J' there is a unique family (F,)t>o C f!/J' with the following properties:

(4)

(5)

(6)

exists for each s > 0,

all s > 0,

Ft --+ G (f!/J') as t --+ 0,

; Ftl --+ H (f!/J') as s --+ 0. s

Proof Let (bn)'~ co be the sequence of Fourier coefficients of G, and let (en)::' co be the sequence of Fourier coefficients of "H. If (Ft)t>o is such a family of distributions, let the Fourier coefficients of Ft be

(an(t))~ co'

As in the proof of Theorem 3.1, conditions (4), (5), (6) imply that an is twice continuously differentiable for t > 0, an and Dan have limits at t = 0, and

D2an(t) = -n2an(t), an(O) = bn, Dan(O) = cn·

The unique function satisfying these conditions (see §6 of Chapter 2) is

(7)

(8)

n i= 0,

Thus we have proved uniqueness. On the other hand, the functions (7), (8) satisfy

(9)

(10)

(11)

(12) Dan(t) --+ Cn as t --+ 0, all n.

It follows from (9) and Theorem 1.2 that (an(t))~ co is the sequence of Fourier coefficients of a distribution Ft. It follows from (10) and Theorem 2.3 that (5) is true. It follows from (11), the mean value theorem, and Theorem 1.2 that

The wave equation 147

exists for s > 0 and has Fourier coefficients (Dan(s»~ <Xl. Then (12) gives (6). Finally,

It follows that

(~rFtl t= s

exists, s > O. The choice of an(t) implies that (4) holds. 0

Let us look more closely at the distribution Ft in the case when G and H are the distributions defined by functions g and h in .9. First, suppose h = O. The Fourier series for g converges:

<Xl

g(x) = L an exp (inx). -<Xl

Inequality (9) implies that the Fourier series for Ft also converges; then Ft

is the distribution defined by the function Ut E ~ where <Xl

u(x, t) = u,(x) = L an cos nt·exp (inx) -<Xl

= ~ ~ an[exp (int) + exp ( - int)] exp (inx) -<Xl

1 <Xl 1 <Xl

= :2 L an exp (in(x + t» + :2 L an exp (in(x - t», -00 -00

or

(13) u(x, t) = !g(x + t) + tg(x - t).

It is easily checked that u is a solution of

82u 82u 8t 2 = 8x2 '

u(x, 0) = g(x), 8u 8t (x,O) = O.

Next, consider the case g = o. Again (an(t»~ <Xl is the sequence of Fourier coefficients of a function Ut E P,

u(x, t) = utCx) = L n- 1cn sin nt·exp (inx) + cot. n .. O

To get rid of the n -1 factor, we differentiate:

: u(x, t) = ~ iCn sin nt exp (inx) x -<Xl

= ~ ~ cn[exp (int) - exp ( - int)] exp (inx) -00

1 <Xl 1 <Xl

= :2 L Cn exp (in(x + t» - :2 L Cn exp (in(x - t» -00 -00

= th(x + t) - -!-h(x - t).

148 Applications of Fourier series

Integrating with respect to x gives

(13)' 1 fx+t

u(x, t) =:2 x-t hey) dy + aCt)

where aCt) is a constant to be determined so that u is periodic as a function of x. But the periodicity of h implies that aCt) should be taken to be zero. Then u so defined in a solution of

u(x,O) = 0, au at (x, 0) = hex).

Equation (l) for the vibrating string may be derived as follows. Approxi­mate the curve Ut representing the displacement at time t by a polygonal line joining the points ... (x - e, u(x - e, t», (x, u(x, t», (x + e, u(x + e, t», .... The force on the string at the point x is due to the tension of the string. In this approximation the force due to tension is directed along the line seg­ments from (x, u(x, t» to (x ± e, u(x ± e, t». The net vertical component of force is then proportional to

(:r r u(x, t) ~ c2e- 2[u(x + e) + u(x - e) - 2u(x)]

where c is constant. We take the limit as e -+ O. Two applications of L'Hopital's rule to the right side considered as a function of e give (1).

The constant c2 can be seen to be equal to

where r is the diameter of the string or wire, T is the tension, and K is pro­portional to the density of the material. Let us suppose also that the length of the string is wi instead of w. The solution of the problem corresponding to (1), (2), (3) when g is real and h = 0 can be shown to be of the form

co

(14) u(x, t) = L bn cos (cntfl) sin (nx/I). n=l

A single term

bn cos (cnt/I) sin (nx/I)

represents a "standing wave", a sine curve with n maxima and minima in the interval [0, wi] and with height varying with time according to the term cos (cnt/I). Thus the maximum height is bn, and the original wave is repeated after a time interval of length

2wl/cn.

Thus the frequency for this term is

cn/2wl = n(KT)1/2/2wlr,

an integral multiple of the lowest frequency

(15)

The wave equation 149

In hearing the response of a plucked string the ear performs a Fourier analysis on the air vibrations corresponding to u(x, t). Only finitely many terms of (14) represent frequencies low enough to be heard, so the series (14) is heard as though it were a finite sum

N

(16) .L bn cos (cnt/I) sin (nx/l). n=l

In general, if the basic frequency (15) is not too low (or high) it is heard as the pitch of the string, and the coefficients b1 , b2 , ••• , bN determine the purity: a pure tone corresponds to alI but one bn being zero. Formula (15) shows that the pitch varies inversely with length and radius, and directly with the square root of the tension.

Exercises

1. In Theorem 5.1, suppose DG and H are in V. Show that

are in L2 for each t, and

IIDFtl12 + II~FtIl2 = IIDGI12 + IIHI12.

2. In the problem (1), (2), (3) with c = 1 suppose g, h, and u are smooth functions.. Show, by computing the derivative with respect to t, that

f I:x u(x, t) r dx + f I:t u(x, t) 12 dx

is constant. (This expresses conservation of energy: the first term represents potential energy, from the tension, while the second term represents kinetic energy.)

3. Use Exercise 2 and the results of this section to prove a theorem about existence and uniqueness of classical solutions of the problem (1), (2), (3).

4. Show that if H(f) = 0 when f is constant, then the solution (Ft)t>o in Theorem 5.1 can be written in the form

Ft = !(TtG + LtG) + !(LtSH - TtSH).

Here again Tt denotes translation, while S is the operator from &' to &' defined by

SH(u) = H(v),

where

f2" X f2" vex) = x vet) dt + 27T 0 vet) dt.

150 Applications of Fourier series

This is the analog for distributions of formulas (13) and (13)'.

§6. Laplace's equation and the Dirichlet problem

A third equation of mathematical and physical importance is Laplace's equation. In two variables this is the equation

(1) 02U 02U ox2 + oy2 = o.

A function u satisfying this equation is said to be harmonic. A typical problem connected with this equation is the Dirichlet problem for the disc: find a function continuous on the closed unit disc

{(x,y) I x2 + y2 ::; I}

and equal on the boundary to a given function g:

u(x, y) = g(x, y),

A physical situation leading to this problem is the following. Let u(x, y, t) denote the temperature at time t at the point with coordinates (x, y) on a metal disc of radius 1. Suppose the temperature at the edge of the disc is fixed, though varying from point to point, while the interior of the disc is insulated. Eventually thermal equilibrium will be reached: u will be inde­pendent of time. The resulting temperature distribution u is approximately a solution of (1), (2).

To solve this equation we express u and g by polar coordinates:

where

x = rcos 0,

Then

u = u(r, 0), g = g(O)

y = r sin 0; 0= tan- 1 (yx- 1).

au = au or + au 00 = cos 0 au _ sin 0 au, ax or ax 00 ax or r 00

and similarly for oujoy. An elementary but tedious calculation gives

02U 02U 02U I au 1 02U - + - = - + - - + --. ox2 oy2 or2 r or r2 002

Thus we want to solve

(3) 02U aU 02U

r2 or2 + r or + 002 = 0, o ::; r < 1, o E IR,

(4) u(I,O) = g(O), o E IR,

Laplace's equation and the Dirichlet problem 151

where u and g are periodic in e. We proceed formally. Suppose g has Fourier coefficients (bn)~ "" and u,(e) = u(r, e) has Fourier coefficients

O<r<1.

Then (3) leads to the equation

(5)

From (4) we get

(6)

O<r<1.

and since uo(e) is constant we want

(7) n # O.

We look for a solution an(r) of the form bnrc, where e = e(n) is a constant, for each n. This will be a solution if and only if

e(e - 1) + e - n2 = 0,

or c2 = n2 • Then (7) gives e = Inl. We are led to the formal solution

00

u(r, e) = L: bnr1nl exp (ine). -00

Formally, this should be the convolution of g with the distribution whose Fourier coefficients are

(r Inl)~ 00'

For r < 1 these are the Fourier coefficients of a function Pr E f!JJ. In fact,

00

(8) Pr(e) = 2: rlnlelno -00

= 1 + i (reiO)n + i (re-iO)n 1 1

= (1 - r2)11 - re1o l- 2 = (1 - r2)(1 - 2r cos e + r2)-1.

Note that Pr(e) is an approximate identity as r ~ 1: the first expression on the right in (8) shows

2~ {" Pr(e) de = 1

and the last expression shows that pre e) ;;:: 0 and

12"--lim Pr(e) de = 0 r-+1 _

for each 0 < 8 < 7T. The function

per, e) = Pr(e)

is called the Poisson kernel for the Dirichlet problem in the unit disc.

152 Applications of Fourier series

Theorem 6.1. Suppose F is a periodic distribution. There is a unique function u(r, 8) defined and smooth in the open unit disc, satisfying (3), and such that the distributions defined by the functions ur(8) = u(r, 8) converge to F in the sense of f!IJ' as r -+ I. Moreover, ifF = Fg where g E '?l, then Ur -+ g uniformly as r -+ I. The functions Ur are given by convolution with the Poisson kernel:

ur = F*Pr •

Proof Uniqueness was proved in the derivation above. Let Ur = F * P,. Since P, is an approximate identity, we do have Ur -+ F (f!IJ') as r -+ I, and u, -+ g uniformly if F = Fg , g E '?l. We must show that u is smooth and satisfies (3). Note that when F = Fg , g E Iif, then explicitly

I f2" u(r, 8) = 27T 0 per, 8 - cp)g(cp) dcp

and we may differentiate under the integral sign to prove that u is smooth. Moreover, since per, 8) satisfies (3), so does u.

Finally, suppose u has merely a distribution F as its value on the boundary. NotethatifO:s; r,s < 1 then (by computing Fourier coefficients, for example)

In particular, choose any R > 0, R < 1. It suffices to show that u is smooth in the disc r < R and satisfies (3) there. But when r < R,

s = rR- 1 < I

and

u, = F * P, = F * (Pr * p.) = (F * PR) * p. = UR * p •.

Since PR E &, UR is a smooth function of 8. Then

I f2" u(r, 8) = 27T 0 P(rR-l, 8 - CP)UR(CP) dcp,

° :s; r < R. Again, differentiation under the integral sign shows that u is smooth and satisfies (3). 0

The preceding theorem leads to the remarkable result that a real-valued harmonic function is (locally) the real part of a function defined by a con­vergent power series in z = x + iy.

Theorem 6.2. Suppose u is a harmonic real-valued function of class C2 defined in an open subset of 1R2 containing the point (xo, Yo). Then there is a function f defined by a convergent power series:

00

fez) = L an(z - zo)n, Iz - zol < e, Zo = Xo + iyo, o

such that

u(x, y) = Ref(x + iy)

Laplace's equation and the Dirichlet problem 153

when

Ix + iy - zol < e. Proof Suppose first that (xo, Yo) = (0, 0) and that the set in which u is

defined contains the closed disc r + y2 :s; 1. Let

g( 0) = u( cos 0, sin 0), 0 E IR.

Then u is the unique solution of the Dirichlet problem in the unit disc with g as value on the boundary. If (bn)'~ 00 are the Fourier coefficients of g, then we know that in polar coordinates u is given by

00

L r In1bn exp (inO). -00

Since u is real, g is real. Therefore bn = b! n and the series is

Let I be defined by

where

Then

00

I(z) = L anzn o

an = 2bn, n > O.

u(x,y) = Re (~an(reI8)n) = Re (J(reIB))

= Re(J(x + iy)), r + y2 < 1.

In the general case, assume that u is defined on a set containing the closed disc of radius e centered at (xo, Yo), and let

Ul(X, y) = u(xo + ex, Yo + ey).

Then Ul is harmonic in a set containing the unit disc, so

11 defined by a power series in the unit disc. Then

u = Ref,

where

is defined by a power series in the disc of radius e around Zo = Xo + iyo. 0

154 Applications of Fourier series

Exercises

1. Prove the converse of Theorem 6.2: if

00

fez) = 2: an(z - zo)n, Iz - zol < R, o

and we let

u(x,y) = Ref(x + iy), Ix + iy - zol < R,

then u is harmonic. 2. There is a maximum principle for harmonic functions analogous to

the maximum principle for solutions of the heat equation discussed in §4. (a) Show that if u is of class C2 on an open set A in 1R2 and

02U 02U

ox2 + oy2 > 0

at each point of A, then u does not have a local maximum at any point of A. (b) Suppose u is of class C2 and harmonic in an open disc in 1R2 and

continuous on the closure of this disc. Show, by considering the functions

uB(x, y) = u(x, y) + ex2 + ey2

that u attains its maximum on the boundary of the disc. 3. Use the result of Exercise 2 to give a second proof of the uniqueness

of the solution of the Dirichlet problem for a continuous boundary function g. 4. Suppose u is continuous on the closed disc x 2 + y2 ~ R and harmonic

in the open disc x 2 + y2 < R. Give a formula for u(x, y) (or u(r, 8)) for x2 + y2 < R in terms of the values of u for x2 + y2 = R. Give formulas for the derivatives ou/or and ou/o8 also.

S. Suppose u is defined on all of 1R2 and is harmonic. Use the result of Exercise 4 to show that if u is bounded, then it is constant.

Chapter 6

Complex Analysis

§1. Complex differentiation

Suppose n is an open subset of the complex plane C. Recall that this means that for each Zo E n there is a 8 > ° so that n contains the disc of radius 8 around Zo:

ZEn if Iz - zol < 8.

A function J: n ~ C is said to be differentiable at ZEn if the limit

lim Pe....:C,--w )c.....-----"--J('--"z ) w-+z w - Z

exists. If so, the limit is called the derivative of J at Z and denoted j'(z). These definitions are formally the same as those given for functions

defined on open subsets of IR, and the proofs of the three propositions below are also identical to the proofs for functions of a real variable.

Proposition 1.1. If J: n ~ C is differentiable at ZEn, then it is con­tinuous at z.

Proposition 1.2. Suppose J: n ~ C, g: n ~ C and a E IC. IJ J and g are differentiable at z = n, then so are the Junctions af, J + g, and Jg:

(aJ),(z) = aj'(z) (f + g)'(z) = j'(z) + g'(z)

(fg)'(z) = j'(z)g(z) + J(z)g'(z).

IJ also g(z) =I- 0, then fig is differentiable at z and

(f/g)'(z) = [j'(z)g(z) - J(z)g'(z)]g(Z)-2.

Proposition 1.3 (Chain rule). Suppose J is differentiable at z E C and g is differentiable at J(z). Then the compositive Junction go J is differentiable at z and

(g 0 J),(z) = g'(f(z))j'(z).

The proof of the following theorem is also identical to the proof of the corresponding Theorem 4.4 of Chapter 2.

Theorem 1.4. Suppose J is defined by a convergent power series:

00

J(z) = 2: an(z - zo)n, Iz - zol < R. n=O

155

156 Complex Analysis

Then f is differentiable at each point z with I z - Zo I < R, and

co

f'(z) = L: nan(z - zo)n-1. n=1

In particular, the exponential and the sine and cosine functions are differentiable as functions defined on C, and (exp z)' = exp z, (cos z)' = - sin z, (sin z)' = cos z.

A remarkable fact about complex differentiation is that a converse of Theorem 2.4 is true : iff is defined in the disc I z - Zo I < R and differentiable at each point of this disc, then f can be expressed as the sum of a power series which converges in the disc. We shall sketch one proof of this fact in the Exercises at the end of this section, and give a second proof in §3 and a third in §7 (under the additional hypothesis that the derivative is continuous). Here we want to give some indication why the hypothesis of differentiability is so much more powerful in the complex case than in the real case. Consider the function

fez) = z*, or f(x + iy) = x - iy.

Take t E IR, t =F O. Then

t- 1 [f(z + t) - fez)] = 1, (it)-l[f(Z + it) - fez)] = -1,

so the ratio

w- 1 [f(w) - fez)]

depends on the direction of the line through wand z, even in the limit as w -+ z. Therefore this functionfis not differentiable at any point.

Givenf: n -+ C, define functions u, v by

u(x, y) = Ref(x + iy) = ff(x + iy) + ·!(f(x + iy»*,

vex, y) = Imf(x + iy) = ~ (f(x + iy»* - ~f(x + iy).

Thus f(x + iy) = u(x, y) + iv(x, y).

We shall speak ofu and v as the real and imaginary parts off and write

f= u + iv.

(This is slightly incorrect, since f is being considered as a function of z E

nee, while u and v are considered as functions of two real variables x, y.)

Theorem 1.5. Suppose nee is open, f: n -+ C, f = u + iv. Then iff is differentiable at z = x + iy E n, the partial derivatives

(1) au ox'

au oy'

ov ox'

Complex differentiation 157

all exist at (x, y) and satisfy

(2) o 0

ox u(x, y) = oy vex, y)

(3) o 0 oy u(x, y) = - ox vex, y).

Conversely, suppose that the partial derivatives (1) all exist and are con­tinuous in an open set containing (x, y) and satisfy (2), (3) at (x, y). Then f is differentiable at z = x + iy.

The equations (2) and (3) are called the Cauchy-Riemann equations. They provide a precise analytical version of the requirement that the limit definingf'(z) be independent of the direction of approach. Note that in the examplef(z) = z* we have

ou = 1 ox '

ov =-1 oy ,

ou=ov=o oy ox .

Proof Suppose f is differentiable at z = x + iy. Then

(4) lim t-1[f(z + t) - fez)] = f'(z) = lim (it)-l[f(z + it) - fez)]. t-+o t-+o

The left side of (4) is clearly

:x u(x, y) + i :x vex, y),

while the right side is

-i :yu(x,y) + :yv(x,y).

Equating the real and imaginary parts of these two expressions, we get (2) and (3).

Conversely, suppose the first partial derivatives of u and v exist and are continuous near (x, y), and suppose (2) and (3) are true. Let

h=a+ib

where a and b are real and near zero, h t= O. We apply the Mean Value Theorem to u and v to get

fez + h) - fez) = fez + a + ib) - fez + a) + fez + a) - fez)

= :y u(x + a, y + t1b)b + i :y vex + a, y + t2b)b

158

where 0 < tl < 1, j = 1, 2, 3, 4. Because of (2) and (3),

[:x u(x, y) + i ;y vex, Y)]h

Complex Analysis

= ;y u(x, y)b + i :y vex, y)b + :x u(x, y)a + i ! vex, y)a.

Therefore

h-l[f(z + h) - fez)] - [:x u(x, y) + i ;y vex, y)] is a sum of four terms similar in size to

a a oy u(x + a, y + tlb) - oy u(x, y).

Since the partial derivatives were assumed continuous, these terms ~ 0 as h~O. 0

A function f: n ~ C, n open in C, is said to be holomorphic in n if it is differentiable at each point of n and the derivative/' is a continuous function. (Actually, the derivative is necessarily continuous if it exists at each point; later we shall indicate how this may be proved.) Theorem 1.5 has the following immediate consequence.

Corollary 1.6. f: n ~ C is holomorphic in n if and only if its real and imaginary parts, u and v, are of class Cl and satisfy the Cauchy-Riemann equations (2) and (3) at each point (x, y) such that x + iy En.

Locally, at least, a holomorphic function can be integrated.

Corollary 1.7. Suppose g is holomorphic in a disc Iz - zol < R. Then there is afunction!, holomorphicfor Iz - zol < R, such that/' = g.

Proof Let u, v be the real and imaginary parts of g. We want to deter­mine real functions q, r such that

f= q + ir

has derivative g. Because of Theorem 1.5 we can see that this will be true if and only if

oq or -=-=U oX oy ,

oq or -=--= -v oy ox .

The condition that u, - v be the partial derivatives of a function q is (by Exercises 1 and 2 of §7, Chapter 2)

au ov oy = -ox·

The condition that v and u be the partial derivatives of a function r is

ov ou oy = ox'

Complex integration 159

Thus there are functions q, r with the desired properties. 0

Exercises

1. Let/(x + iy) = x2 + y2. Show that lis differentiable only at z = O. 2. Suppose I: C ~ ~. Show that I is differentiable at every point if and

only if I is constant. 3. Let 1(0) = 0 and

x + iy =I O.

Show that the first partial derivatives of I exist at each point and are both zero at x = y = O. Show that/is not differentiable (in fact not continuous) at z = O. Why does this not contradict Theorem 1.5?

4. Suppose I is holomorphic in n and suppose the real and imaginary parts u, v are of class C2 in n. Show that u and v are harmonic.

5. Suppose I is as in Exercise 4, and suppose the disc Iz - zol :$; R is contained in n. Use Exercise 4 together with Theorem 6.2 of Chapter 5 to show that there is a power series 2: an(z - zo)n converging to I(z) for Iz - zol < R.

6. Suppose g is holomorphic in n, and suppose n contains the disc Iz - zol :$; R. Let I be such that f'(z) = g(z) for Iz - zol :$; R (using Corollary 1.7). Show that the real and imaginary parts of I are of class C2.

7. Use the results of Exercises 5 and 6, together with Theorem 1.4, to prove the following theorem.

If g is ho10morphic in nand Zo E n, then there is a power series such that 00

g(z) = L: an(z - zo)n, Iz - zol < R, n=O

for any R such that n contains the disc of radius R with center zo0

§2. Complex integration

Suppose nee is open. A curve in n is, by definition, a continuous function" from a closed interval [a, b] c ~ into n. The curve" is said to be smooth if it is a function of class C1 on the open interval (a, b) and if the one-sided derivatives exist at the endpoints:

(t - a)-1[,,(t) - ,,(a)] converges as t ~ a, t > a; (t - b)-1[,,(t) - ,,(b)] converges as t~b, t < b.

The curve " is said to be piecewise smooth if there are points ao, a1> ... , ar

with a = ao < a1 < ... < ar = b

160 Complex Analysis

such that the restriction of y to [aj-I. aj] is a smooth curve, 1 :::;; j :::;; r. An example is

yet) = Zo + e exp (it), t E [0, 27T];

then the image

{yet) I t E [0, 27T]}

is the circle of radius e around zoo This is a smooth curve. A second example is

yet) = t, yet) = 1 + i(t - 1), yet) = 1 + i - (t - 2), yet) = i - (t - 3)i,

t E [0, 1], tE(1,2], t E (2,3], t E (3, 4].

Here y is piecewise smooth and the image is a unit square. Suppose y: [a, b] ~ Q is a curve and f: Q ~ C. The integral off over y,

is defined to be the limit, as the mesh of the partition P = (to, tI. ... , tn) of [a, b] goes to zero, of

n

2 f(y(tl))[y(tl) - y(tl - 1)]. j=l

Proposition 2.1. If y is a piecewise smooth curve in Q and f: Q ~ C is continuous, then the integral off over y exists and

(1) {f = f f(y(t))y'(t) dt.

Proof The integral on the right exists, since the integrand is bounded and is continuous except possibly at finitely many points of [a, b]. To prove that (1) holds we assume first that y is smooth. Let P = (to, t1 , •• ", tn) be a partition of [a, b]. Then

(2) 2f(y(tt))[y(tt) - y(tt-1)] = 2f(y(tl))y'(tt)[t1 - ti-d + R,

where

Applying the Mean Value Theorem to the real and imaginary parts of yon [tl - 1 , ttl and using the continuity of y', we see that

R~O as the mesh IPI ~O. On the other hand, the sum on the right side of (2) is a Riemann sum for the integral on the right side of (1). Thus we have shown that the limit exists and (1) is true.

Complex integration 161

If )I is only piecewise smooth, then the argument above breaks down on intervals containing points of discontinuity of )I'. However, the total contri­bution to the sums in (2) from such intervals is easily seen to be bounded in modulus by a constant times the mesh of P. Thus again the limit exists as IPI-? ° and (1) is true. 0

Note that the integral I71 depends not only on the set of points

{)I(t) I t E [a, b]}

but also on the "sense," or ordering, of them. For example, if

)ll(t) = exp (it), )l2(t) = exp ( - it), t E [0, 217],

then the point sets are the same but

Furthermore, it matters how many times the point set is traced out by)l: if

)la(t) = exp (int), then

f l=nJ. f. 73 71

A curve )I: [a, b] -? n is said to be constant if )I is a constant function. If so, then

LI= 0, allf.

A curve)l: [a, b] -? n is said to be closed if )I(a) = y(b). (All the examples given so far have been examples of closed curves.) Two closed curves )10, )11: [a, b] -? n are said to be homotopic in n if there is a continuous function

such that

r:[a,b] x [O,I]-?n

r(t,O) = )lo(t), r(a, s) = r(b, s),

r(t, 1) = )ll(t), all s E [0, 1].

all t E [a, b],

The function r is called a homotopy from )10 to )II' If r is such a homotopy, let

)I,(t) = r(t, s), S E (0,1).

Then each )I. is a closed curve, and we think of these as being a family of curves varying continuously from )10 to )110 within n.

Theorem 2.2 (Cauchy's Theorem). Suppose nee is open and suppose 1 is holomorphic in n. Suppose )10 and )II are two piecewise smooth closed curves in n which are homotopic in n. Then

162 Complex Analysis

The importance of this theorem can scarcely be overestimated. We shall first cite a special case of the theorem as a Corollary, and prove the special case.

Corollary 2.3. Suppose n is either a disc

{z liz - zol < R}

or a rectangle

Suppose f is holomorphic in nand Y is any piecewise smooth closed curve in n. Then

Lf=O. Proof It is easy to see that in this case y is homotopic to a constant

curve Yo, so that the conclusion follows from Cauchy's Theorem. However, let us give a ,different proof. By Corollary 1.7, or by the analogous result for a rectangle in place of a disc, there is a function h, holomorphic in n, such that h' = f But then

f f = Jb !(y(t))y'(t) dt = r h'(y(t))y'(t) dt y a a

= C [h 0 y]'(t) dt = h(y(b)) - h(y(a)) = 0, 'a

since y(a) = y(b). 0

Proof of Cauchy's Theorem. Let r be a homotopy from Yo to Yl and let

YB(t) = ret, s), t E [a, b], S E (0,1).

Assume for the moment that each curve YB is piecewise smooth. We would like to show that the integral off over YB is independent of s, ° ;5; s ;5; 1.

Assume first that r is of class C2 on the square [a, b] x (0,1), that the first partial derivatives are uniformly bounded, and that

Y; ~ y~ as s ~ 0, ,,~-+y~ as s-+ 1

uniformly on each interval (c, d) c [a, b] on which y~ or y~, respectively, is continuous. These assumptions clearly imply that

Complex integration 163

is a continuous function of s, s E [0, 1]. Furthermore under these assumptions we may apply results of §7 of Chapter 2 and differentiate under the integral sign when s E (0, 1) to get

F'(s) = f :s (J(r(t, s» :t ret, S») dt

Jb a a = j'(r(t, S»"8 ret, S)"8 ret, s) dt

a S I

+ f J(r(l, s» a:~s ret, s) dt

= f :t [J(r(t, s» :s r(l, s) J dt

a a = J(r(b, s» as reb, s) - J(r(a, s» as rea, s)

=0

(since reb, s) = rea, s». Thus F(O) = F(I). Finally; do not assume r is differentiable. We may extend r in a unique

way so as to be periodic in the first variable with period b - a, and even and periodic in the second variable with period 2; then r: IR x IR ---+ Q. Let If!: IR ---+ IR be a smooth function such that

<p(x) ~ 0, all x,

I <p(x) dx = 1

<p(x) = ° if Ixl ~ 1.

Let <Pn(x) = n<p(nx), n = 1, 2, 3, .. " Then

I <Pn(X) dx = 1,

Let

(t, s) E IR x R Then (see the arguments in §3 of Chapter 3) r n is also periodic with the same periods as r, and r n ---+ r uniformly as n ---+ 00. It follows from this (and the fact that r(1R x IR) is a compact subset of Q) that r n(1R x IR) c Q if n is sufficiently large. Furthermore, the argument above shows that

f J=1 J, "O.n Yl,n

where

Ys.n(t) = r n(t, s), IE [a,b], S E [0, 1].

164 Complex Analysis

All we need do to complete the proof is show that

(3) f f -+ f f as n -+ 00, )I,"n fa

s = 0 or 1.

Note that we may choose the homotopy r so that Ys = Yo for 0 ~ s ~ ! and Ys = YI for j- ~ s ~ I; to see this, let robe any homotopy from Yo to YI in n, and let

ret, s) = Yo(t), ret, s) = r o(t,3s - 1), ret, s) = YI(t),

o ~ s ~ t, t < s <j-, t~s~1.

If r has these properties, then when n ~ 3 we have

s = 0 or 1.

It follows that Ys,n -+ Ys uniformly, s = 0 or 1. It also follows, by differen­tiating with respect to t, that y~,n is uniformly bounded and

on each interval of [0, 1] where y~ is continuous, s = 0 or 1. Therefore (3) is true. 0

Exercises

1. Suppose YI: [a, b] -+ nand Y2: [c, d] -+ n are two piecewise smooth curves with the same image:

C = {YI(t) I t E [a, b]} = {Y2(t) I t E [c, d]}.

Suppose these curves trace out the image in the same direction, i.e., if

s, t E [a, b], s', t' E [c, d]

and

and

s < t

then

s' < t'.

Show that for any continuousf: n -+ C (not necessarily holomorphic),

Complex integration 165

This justifies writing the integral as an integral over the point set C:

J J= i J(z)dz, Yl C

where we tacitly assume a direction chosen on C. 2. Suppose C in Exercise 1 is a circle of radius R and suppose IJ(z)1 :$; M,

all z E C. Show that

IL J(z) dz I :$; M·2rrR.

3. Let y(t) = Zo + eelt, t E [0, 2rr], where e > 0. Show that

L (z - zo)-l = 2rri.

4. Let Yn(t) = exp (int), t E [0, 2rr], n = 0, ± 1, ± 2, .... Show that

f Z-l = 2nrri. Yn

5. Let n = {z Eel z ¥- O}. Use the result of Exercise 4 to show that the curves Yn and Ym are not homotopic in n if n ¥- m. Show that each Yn is homotopic to Yo in C, however.

6. Use Exercise 4 to show that there is no function/, defined for all z ¥- 0, such thatf'(z) = z-l, all z ¥- 0. Compare this to Corollary 1.7.

7. Let n be a disc with a point removed:

n = {z I Iz - zol < R, z ¥- Zl},

where Iz - zll < R. Let Yo and Yl be two circles in n enclosing Zh say

Yo(t) = Zl + eelt,

Yl(t) = z + relt,

t E [0, 2rr], t E [0, 2rr].

Here Iz - zll < r < Rand e > ° is chosen so that IYl(t) - zll > e, all t. Construct a homotopy from Yo to Yl'

8. Suppose n contains the square

{x + iy 10:$; x, y :$; I}.

SupposeJis differentiable at each point of n; here we do not assume thatf' is continuous. Let C be the boundary of the square, with the counterclockwise direction. Show that

LJ(z)dz = 0.

This extension of Corollary 2.3 is due to Goursat. (Hint: for each integer k > 0, divide the square into 4k smaller squares with edges of length 2- k •

166 Complex Analysis

Let Ck,l,"" Ck ,2k be the boundaries of these smaller squares, with the counterclockwise direction, Then show that

J J(z) dz = 2: J J(z) dz. C f C".I

It follows that if

II J(Z)dZI = M > 0,

then for each k there is aj = j(k) such that

where C k = Ck,f' Let Zk be the center of the square with boundary C k •

There is a subsequence Zkn of the sequence Zk which converges to a point Z of the unit square. Now derive a contradiction as follows. Since J is differentiable at z,

J(w) = J(z) + J'(z)(w -. z) + r(w)

where

Jr(w)(w - Z)-lJ ~ 0 as w ~ z

Therefore for each e > 0 there is a 8 > 0 such that if Co is the boundary of a square with sides of length h lying in the disc Jw - zJ < 8, then

§3. The Cauchy integral formula

There are many approaches to the principal results of the theory of holomorphic functions. The most elegant approach is through Cauchy's Theorem and its chief consequence, the Cauchy integral formula. We begin with a special case, which itself is adequate for most purposes.

Theorem 3.1. Suppose J is holomorphic in an open set n. Suppose C is a circleor rectangle contained in n and such that all points enclosed by Care in n. Then if w is enclosed by C,

(1) J(w) = -21 • f J(z)(z - W)-l dz. 7T1 C

(Here the integral is taken in the counterclockwise direction on C.)

Proof Let Yo: [a, b] ~ n be a piecewise smooth closed curve whose image is C, traced once in the counterclockwise direction. Given any positive e which is so small that C encloses the closed disk of radius and center w, let C. be the circle of radius e centered at w.

The Cauchy integral formula 167

We can find a piecewise smooth curve YI: [a, b] -+ .0 which traces out Cs once in the counterclockwise direction and is homotopic to Yo in the region .0 with the point w removed. Granting this for the moment, let us derive (1). By Exercise 1 of §2, and Cauchy's Theorem applied to g(z) = J(z)(z - w)-I, we have

L J(z)(z - W)-I dz = f g = f g = L J(z)(z - W)-l dz, o fu h ~

so

(2) r J(z)(z - W)-I dz = f J(w)(z - W)-I dz + f [few) - J(z)](z - W)-I dz. Jo 0 8 0 8

By Exercise 2 of §2, the first integral on the right in (2) equals 27Tif(w). Since Jis differentiable, the integrand in the second integral on the right is bounded as e -+ O. But the integration takes place over a curve of length 27Te, so this integral converges to zero as e -+ O. Therefore (1) is true.

Finally, let us construct the curve YI and the homotopy. For t E [a, b], let YI(t) be the point at which Ce and the line segment joining w to Yo(t) intersect. Then for 0 ~ s ~ 1, let

It is easily checked that Yo and r have the desired properties. 0

The preceding proof applies to any situation in which a given curve YI is homotopic to all small circles around WEn. Let us make this more precise. A closed curve Y in an open set n is said to enclose the point WEn within .0 if the following is true: there is a I) > 0 such that if 0 < e ~ I) then Y is homotopic, in .0 with w removed, to a piecewise smooth curve which traces out once, in a counterclockwise direction, the circle with radius e and center w. We have the following generalization of Theorem 3.1.

Theorem 3.2. Suppose J is holomorphic in an open set .0. Suppose Y is a piecewise smooth closed curve in .0 which encloses a point w within .0. Then

(3) J(w) = -21 . f J(z)(z - W)-I dz. 7Tl y

Equation (3) is the Cauchy integral Jormula, and equation (1) is essentially a special case of (3). (In section 7 we shall give another proof of (1) when C is a circle.)

The Cauchy integral formula makes possible a second proof of the result of Exercise 5, §1: any holomorphic function can be represented locally as a power series.

Corollary 3.3. Suppose J is holomorphic in .0, and suppose .0 contains the disc

{z liz - zol < R}.

168 Complex Analysis

Then there is a unique power series such that

co

(4) J(w) = 2: anew - zo)n, all w with Jw - zoJ < R. n=O

Proof Suppose 0 < r < R, and let C be the circle of radius r centered at Zo, with the counterclockwise direction. If Jw - zoJ < rand z E C then

J(w - zo)(z - ZO)-lJ = S < 1.

We expand (z - W)-l in a power series:

z - w = z - Zo - (w - zo) = (z - zo)[1 - (w - zo)(z - ZO)-l],

so

The sequence of functions

N

gN(Z) = J(z) 2: (w - zo)n(z - zo)-n-l n=O

converges uniformly for z E C to the function

J(z)(z - W)-l.

Therefore we may substitute in equation (l) to get (4) with

(5) an = -21 . f J(z)(z - zo)-n-l dz. wI C

This argument shows that the series exists and converges for Jw - zoJ < r. The series is unique, since repeated differentiation shows that

(6)

Since r < R was arbitrary, and since the series is unique, it follows that it converges for all Jw - zoJ < R. 0

Note that our two expressions for an can be combined to give

(7) n' f j<n)(zo) = -2 '. J(z)(z - zo)-n-l dz. wI c

This is a special case of the following generalization of the Cauchy integral formula.

Corollary 3.4. Suppose J is holomorphic in an open set containing a circle or rectangle C and all the points enclosed by C. If w is enclosed by C then the nth derivative oj J at w is given by

(8) n' f j<n>(w) = -'. J(z)(z - W)-l dz. 2m c

The Cauchy integral formula 169

Proof Let C1 be a circle centered at wand enclosed by C. Then by the Cauchy integral theorem, and the argument given in the proof of Theorem 3.1, we may replace C by C1 in (8). But in this case the formula reduces to the case given in (7). D

A functionf defined and holomorphic on the whole plane C is said to be entire. The following result is known as Liouville's Theorem.

Corollary 3.5. Iff is an entire function which is bounded, then f is constant.

Proof We are assuming that there is a constant M such that If(z) 1 ::;; M, all z E C. It is sufficient to show that f' == 0. Given w E C and R > 0, let C be the circle with radius R centered at w. Then

f'(w) = 2~f f(z)(z - W)-2 dz, 1T1 c

so

If'(w) 1 ::;; 2~·M.R-2.21TR = MR-l.

Letting R ---'J- Owe getf'(w) = 0. D

A surprising consequence of Liouville's Theorem is the "Fundamental Theorem of Algebra."

Corollary 3.6. Any nonconstant polynomial with complex coefficients has a complex root.

Proof Suppose p is such a polynomial. We may assume the leading coefficient is 1:

p(z) = zn + an_1zn- 1 + ... + al(z) + ao.

It is easy to show that there is an R > ° such that

(8) if Izl ~ R.

Now suppose p has no roots: p(z) i= 0, all z E Co Thenf(z) = p(Z)-l would be an entire function. Then f would be bounded on the disc Izl ::;; R, and (8) shows that it would be bounded by 2R-n for Izl ~ R. But thenfwould be constant, a contradiction. D

Exercises

1. Verify the Cauchy Integral Formula in the form (1) by direct com­putation whenf(z) = eZ and C is a circle.

2. Compute the power series expansion (4) in the following cases. (Hint: (6) is not always the simplest way to obtain the an.)

(a) fez) = sin z, Zo = 1-1T. (b) eZ , Zo arbitrary.

170

(c) J(z) = Z3 - 2Z2 + z + i, Zo = -i. (d) J(z) = (z - 1)(z2 + 1)-\ Zo = 2.

Complex Analysis

3. Derive equation (8) directly from (1) by differentiating. 4. Suppose J is an entire function, and suppose there are constants M

and n such that

all z E C.

Show that J is a polynomial of degree :s; n. Show that, conversely, any polynomial of degree :s; n satisfies such an inequality.

5. The Cauchy integral formula can be extended to more general situa­tions, such as the case of a region bounded by more than one curve. For example, suppose n contains the annulus

A = {z I r < Iz - zol < R},

and also the two circles

C1 = {z liz - zol = r}, C2 = {z I Iz - zol = R}

which bound A. Give C1 and C2 the counterclockwise direction. Then if w E A, and J is holomorphic in n, show that

(*) J(w) = -21 . f J(z)(z - W)-1 dz - -21 • f J(z)(z - W)-1 dz. 7Tl C2 7Tl Cl

(Hint: choose a E C, lal = 1 so that w does not lie on the line segment L = {zo + ta I r :s; t :s; R} joining C1 to C2 • There is a curve y tracing out L, then C2 in the counterclockwise direction, then L in the reverse direction, then C1 in the clockwise direction. This curve is homotopic in n to any small circle about w. Moreover, the integral of J(z)(z - W)-1 over yequals the right side of (*), since the two integrations over L are in the opposite directions and cancel each other.)

6. Extend the result of Exercise 5 to the following situation: n contains a circle or rectangle C, together with all points enclosed by C except Zl> Z2, ... , Zm' Let Cl> C2, ... , Cm be circles around these points which do not intersect each other or C. If J is holomorphic in nand WEn is enclosed by C, then

1 f m 1 f J(w) = 2' J(z)(z - W)-1 dz - L -2' J(z)(z - W)-1 dz; 7Tl C j=1 7Tl CJ

again, all integrals are taken in the counterclockwise direction. 7. Suppose J: n -,)- C is merely assumed to be differentiable at each

point of the open set n, and suppose n contains a rectangle C and all points enclosed by C. Suppose w is enclosed by C. Modify the argument of Exercise 8, §2, to show that the integral in (1) remains unchanged if we replace C by any rectangle enclosed by C and enclosing w. Therefore show that (1) holds in this case also.

8. Use the result of Exercise 7 to show that if J is differentiable at each point of an open set n, the derivative is necessarily a continuous function.

The local behavior of a holomorphic function 171

§4. The local behavior of a holomorphic function

In this section we investigate the qualitative behavior near a point Zo E IC of a function which is holomorphic in a disc around Zoo Iff is not constant, then its qualitative behavior near Zo is the same as that of a function of the form

where ao and am are constants, am =1= 0, and m ;:::: 1.

Lemma 4.1. Suppose f is holomorphic in an open set Q and Zo E Q. If f is not constant near zo, then

(1)

where ao and am are constants, m ;:::: 1, and his holomorphic in Q with h(zo) = 1.

Proof Near zo, f is given by a power series expansion

(2) fez) = ao + a1(z - zo) + ... + an(z - zo)n + .. '. Let m be the first integer;:::: 1 such that am =1= 0. Then (2) gives (1) with

00

h(z) = L: anam -1(Z - zo)n-m. n=m

This function is holomorphic near zo, and h(zo) = 1. On the other hand, (1) defines a function h in Q except at zo, and the function so defined is holomorphic. Thus there is a single such function holomorphic through­out Q. 0

Our first theorem here is the Inverse Function Theorem for holomorphic functions.

Theorem 4.2. Suppose f is holomorphic in an open set Q, and suppose Zo E Q,f'(zo) =1= 0. Let Wo = f(zo). Then there is an e1 > ° and a holomorphic function g defined on the disc Iw - wol < e1 such that

g({w Ilw - wol < e1D is open, f(g(w» = w if Iw - wol < e1'

In other words,ftakes an open set containing Zo in a 1-1 way onto a disc about wo, and the inverse function g is holomorphic.

Proof We begin by asking: Suppose the theorem were true. Can we derive a formula for g in terms off? The idea is to use the Cauchy integral formula for g, using a curve y around Wo which is the image by f of a curve around Zo, because then we may take advantage of the fact that g(f(z» = z. To carry this out, let 0 > ° be small enough that Q contains the closed disc Iz - zol :s; 0; later we shall further restrict O. Let

yo(t) = Zo + oett , t E [0, 21T], yet) = f(Yo(t»,

172 Complex Analysis

and let C be the circle of radius 0 around zoo Assuming the truth of the theorem and assuming that y enclosed WI> we should have

1 I2" = 2-: g(y(t»y'(t)[y(t) - wd- 1 dt 1T1 0

1 f2" = 2-: yo(t)J'(Yo(t»y~(t)[f(Yo(t» - wd- 1 dt

1T1 0

or

(3) g(W1) = -21 . f zJ'(z)[f(z) - wd- 1 dz. 1T1 c

Our aim now is to use (3) to define g and show that it has the desired prop­erties. First, note that (1) holds with m = l. We may restrict 0 still further, so that Ih(z)1 ~ 1- for z E C. This implies that fez) =f. Wo if z E C. Then we may choose e > 0 so that

fez) =f. WI if IWI - wol < e and z E C.

With this choice of 0 and e, (3) defines a function g on the disc I WI - Wo I < e. This function is holomorphic; in fact it may be differentiated under the integral sign.

Suppose

IZI - zol < 0, f(zl) = WI> and IWI - wol < e.

We can, and shall, assume that 0 is chosen so small that J'(z) =f. 0 when Iz - zol < O. Then

fez) - WI = fez) - f(zl) = (z - zl)k(z)

where k is holomorphic in Q and k(Zl) = J'(Zl) =f. O. Therefore k is nonzero in Q. We have

But the right side is the Cauchy integral formula for

Zd'(Zl)k(Zl)-l = Zl'

Thus g(f(Zl» = Zl for Zl near zo, and we have shown that f is 1-1 near zoo Also

1 = (g 0 f)'(zo) = g'(wo)f'(zo),

so g'(wo) =f. O. Therefore g is 1-1 near woo We may take eo > 0 so small that eo :::;; e, and so that g is I-Ion the disc Iw - wol < eo. We may also assume el :::;; eo so small that

Iw - wol < el implies Ig(w) - zol < 0

The local behavior of a holomorphic function

and If(g(w)) - wol < eo·

But then for Iw - wol < e1 we have

g(f(g(w))) = g(w),

and since g is I-Ion Iw - wol < eo this implies

f(g(w)) = w.

173

o As an example, consider the logarithm. Suppose Wo E C, WO "# O. We

know by results of §6 of Chapter 2 that there is a Zo E C such that eZo = Wo. The derivative of e Z at Zo is eZo = Wo "# O. Therefore there is a unique way of choosing the logarithm

Z = log w,

Z near zo, in such a way that it is a holomorphic function of w near Wo. (In fact, we know that any two determinations of log w differ by an integral multiple of 27Ti; therefore the choice of log w will be holomorphic in an open set n if and only if it is continuous there.) By definition a branch of the logarithm function in n is a choice Z = log w, WEn, such that Z is a holomorphic function of w in n.

As a second example, consider the nth root, n a positive integer. If Wo "# 0, choose a branch of log w holomorphic in a disc about Wo. Then if we set

w1 /n = exp (~log w), this is a holomorphic function of w near Wo and

(wl/n)n = exp (n.~ log w) = exp (log w) = w.

We refer to w1/n as a branch of the nth root. There are exactly n branches of the nth root holomorphic near Wo. In

fact, suppose

Then for any choice of log Zo and log Zl>

exp (n log zo) = exp (n log Zl),

so n log Zo = n log Zl + 2m7Ti,

some integer m. This implies

(4) Zo = exp (log zo) = ... = Zl exp (2m7Tin-l).

Since exp (2m7Tin- 1) = exp (2m'7Tin- 1)

if and only if (m - m')n- 1 is an integer, we get all n distinct nth roots of Wo by letting Z 1 be a fixed root and taking m = 0, 1, ... , n - I in (4).

174 Complex Analysis

We can now describe the behavior of a nonconstant holomorphic function near a point where the derivative vanishes.

Theorem 4.3. Suppose f is holomorphic in an open set 0, and suppose Zo E O. Suppose f is not constant near zo, and let m be the first integer ~ 1 such that j<m)(zo) '# O. Let Wo = f(zo). There are e > 0 and 8 > 0 such that if

0< Iw - wol < e

there are exactly m distinct points z such that

Iz - zol < 8 and fez) = w.

Proof. By Lemma 4.1,

fez) = Wo + (z - zo)mh(z),

where h is holomorphic in 0 and h(zo) '# O. Choose a branch g of the mth root function which is hoi om orphic near h(zo). Then near zo,

fez) = Wo + [(z - zo)(g(h(z))]m = Wo + k(z)m,

where k is holomorphic near zoo Then k(zo) = 0, k'(zo) = g'(h(zo)) '# O. For z near zo, z '# zo, we have

fez) = w if and only if k(z) = (w - wo)l/m

for some determination of the mth root of w - woo We can apply Theorem 4.2 to k: there are e, 8 so that

k(z) = t

has a unique solution z in the disc Iz - zol < 8 for each t in the disc It I < e1/m• But if

0< Iw - wol < e

then w - Wo has exactly m mth roots t, all with It I < e1/m• 0

The following corollary is called the open mapping property of holo­morphic functions.

Corollary 4.4. If 0 is open and f: 0 ~ C is holomorphic and not constant near any point, thenf(O) is open.

Proof. Suppose Wo Ef(O). Then Wo = f(zo), some Zo E O. We want to show that there is an e > 0 such that the disc Iw - wol < e is contained in f(O). But this follows from Theorem 4.3. 0

Exercises

1. Use Corollary 4.4 to prove the Maximum Modulus Theorem: if f is holomorphic and not constant in a disc Iz - zol < R, then g(z) = I f(z) I does not have a local maximum at zoo

Isolated singularities 175

2. Withfas in Exercise 1, show that g(z) = I f(z) I does not have a local minimum at Zo unless f(zo) = O.

3. Use Exercise 2 to give another proof of the Fundamental Theorem of Algebra.

4. Suppose z = log w. Show that Re z = log Iwl. 5. Supposefis holomorphic near Zo andf(zo) #- O. Show that log If(z) I

is harmonic near zoo 6. Use Exercise 5 and the maximum principle for harmonic functions to

give another proof of the Maximum Modulus Theorem. 7. Use the Cauchy integral formula (for a circle with center zo) to give

still another proof of the Maximum Modulus Theorem. 8. Use the Maximum Modulus Theorem to prove Corollary 4.4. (Hint:

let Wo = f(zo) and let C be a small circle around Zo such that fez) #- Wo if z E C. Choose e > 0 so that If(z) - wol ;::: 2e if z E C. If Iw - wol < e, can (f(z) - W)-l be holomorphic inside C?

9. A set 0 c C is connected if for any points zo, Zl E 0 there is a (con­tinuous) curve y: [a, b] -'>- 0 with yea) = zo, y(b) = Zl' Suppose 0 is open and connected, and supposefis holomorphic in O. Show that iffis identically zero in any nonempty open subset 0 1 c 0, thenf == 0 in O.

10. Let 0 be the union of two disjoint open discs. Show that 0 is not connected.

§5. Isolated singularities

Suppose f is a function holomorphic in an open set O. A point Zo is said to be an isolated singularity off if Zo 1: 0 but if every point sufficiently close to Zo is in O. Precisely, there is a a > 0 such that

z EO if 0 < Iz - zol < O.

For example, 0 is an isolated singularity for fez) = z-n, n a positive integer, and for g(z) = exp (l/z). On the other hand, according to the definition, 0 is also an isolated singularity for the functionfwhich is defined by fez) = 1, z #- 0 and is not defined at O. This example shows that a singularity may occur through oversight: not assigning values to enough points. An isolated singularity Zo for f is said to be a removable singularity iff can be defined at Zo in such a way as to remain holomorphic.

Theorem 5.1. Suppose Zo is an isolated singularity for the holomorphic function f It is a removable singularity if and only iff is bounded near zo, i.e., there are constants M, a > 0 such that

(1) If(z) I :$ M if 0 < Iz - zol < o.

Proof Suppose Zo is a removable singularity. Then f has a limit at Zo, and it follows easily that (1) is true.

176 Complex Analysis

Conversely, suppose (1) is true. Choose r with 0 < r < 8 and let 8 > 0 be such that 0 < 8 < r. Let C be the circle with center Zo and radius r. Given w with 0 < Jw - zoJ < r, choose 8 so small that 0 < 8 < Jw - zoJ, and let C~ be the circle with center Zo and radius 8. By Exercise 5 of §3,

f(w) = -21 .f f(z)(z - W)-l dz - -21 .f f(z)(z - W)-l dz. TTl c TTl c.

Since f is bounded on C independent of 8, the second integral goes to zero as 8-+0. Thus

(2) f(w) = -21 • f f(z)(z - W)-l dz, TTl C

o < Jw - zoJ < r.

We may define f(zo) by (2) with w = zo, and then (2) will hold for all w, Jw - zoJ < r. The resulting function is then holomorphic. 0

An isolated singularity Zo for a function f is said to be a pole of order n for J, where n is an integer ~ 1, iff is of the form

(3) f(z) = (z - ZO)-lIg(Z)

where g is defined at Zo and holomorphic near Zo, while g(zo) =1= O. A pole of order 1 is often called a simple pole.

Theorem 5.2. Suppose Zo is an isolated singularity for the holomorphic function f. It is a pole of order n if and only if the function

(z - zoYi(z)

is bounded near zo, while the function

(z - ZO)"-Y(Z)

is not.

Proof. It follows easily from the definition that if Zo is a pole of order n the asserted consequences are true.

Conversely, suppose (z - zoYi(z) = g(z) is bounded near zo0 Then Zo is an isolated singularity, so we may extend g to be defined at Zo and holo~ morphic. We want to show that g(zo) =1= 0 if (z - ZO)"-lJ(Z) is not bounded near Z00 But if g(zo) = 0 then by Lemma 4.1,

g(z) = (z - zo)mh(z)

for some m ~ I and some h holomorphic near Z00 But then (z - ZO)II-Y(Z) = (z - zo)m-1h(z) is bounded near Z00 0

An isolated singularity which is neither removable nor a pole (of any order) is called an essential singularity. Note that if Zo is a pole or a removable singularity, then for some a E C or a = 00

f(z) -+ a as z -+ Zo

This is most emphatically not true near an esseutial singularity.

Isolated singularities 177

Theorem 5.3 (Casorati-Weierstrass). Suppose Zo is an isolated singularity for the holomorphic function f If Zo is an essential singularity for J, then for any e > 0 and any a E C there is a z such that

Iz - zol < e, If(z) - al < e.

Proof Suppose the conclusion is not true. Then for some e > 0 and some a E C we have

If(z) - al ~ e where 0 < Iz - zol < e.

Therefore h(z) = (f(z) - a)-l is bounded near zoo It follows that h can be extended so as to be defined at Zo and holomorphic near zoo Then for some m ~ 0,

h(z) = (z - zo)mk(z)

where k is holomorphic near Zo and k(zo) # O. We have

o < Iz - zol < e.

Therefore Zo is either a removable singularity or a pole for f 0

Actually, much more is true. Picard proved that if Zo is an isolated essen­tial singularity for J, then for any e > 0 anti any a E C, with at most one exception, there is a z such that 0 < Iz - zol < e andf(z) = a. An example is fez) = exp (liz), z # 0, which takes any value except zero in any disc around zero.

Isolated singularities occur naturally in operations with holomorphic functions. Suppose, for example, that f is holomorphic in nand Zo E n. Iff(zo) # 0, then we know thatf(z)-l is holomorphic near zoo The function fis said to have a zero of order n (or mUltiplicity n) at zo, n an integer ~ 0, if

J<k)(ZO) = 0, J<fI)(ZO) # O.

o :$; k < n.

(In particular, f has a zero of order zero at Zo if f(zo) # 0.) A zero of order one is called a simple zero.

Lemma 5.4. If f is holomorphic near Zo and has a zero of order n at Zo, then fez) -1 has a pole of order n at zo0

Proof. By Lemma 4.1,

(4) fez) = (z - zo)flh(z),

where h is holomorphic near Zo and h(zo) # O. The desired conclusion follows. 0

For example, the function

sec z = (cos Z)-l

is holomorphic except at the zeros of cos z, where it has poles (of order 1). The same is then true of

tanz = sinz(cosz)-l = sin z sec Z.

178 Complex Analysis

It is convenient to assign the" value" 00 at Zo to a holomorphic function with a pole at Zoo Similarly, if Zo is a removable singularity for f we shall consider f as being extended to take the appropriate value at zoo With these conventions we may work with the following extension of the notion of holomorphic function.

Suppose Q c C is an open set. A function f: Q -?- C U {co} is said to be meromorphic in Q if for each point Zo E Q, either Zo is a pole off, or f is holomorphic in the disc Iz - zol < 8 for some 8 > 0.

Theorem 5.5. Suppose Q is open and f, g are meromorphic in Q. Suppose a E C. Then the functions

af, f+g, fg

are meromorphic in Q. If Q is connected and g -;;J. ° in Q, then fig is mero­morphic in Q.

Proof We leave all but the last statement as an exercise. Suppose Q is connected, i.e. for each zo, Z1 E Q there is a continuous curve y: [0, 1] -?- Q

with yeO) = zo, y(l) = Z1' Given any Zo E Q, either g(Z)-1 is meromorphic near Zo or g vanishes in a disc around Zoo We wnnt to show that the second alternative implies g == ° in all of Q.

Suppose g vanishes identically near Zo E Q and suppose Z1 is any other point of Q. Let y be a curve joining Zo to Z 1: yeO) = zo, y(l) = Z l' Let A be the subset of the interval [0, I] consisting of all those t such that g vanishes identically near yet). Let c = lub A. There is a sequence (tn)!) C A such that tn ->- C. If g did not vanish identically near y(c) then either g(y(c)) =f. 0, or y(c) is a zero of order n for some n, or y(c) is a pole of order n for some n. But then (see (3) and (4)) we could not have g identically zero near y(tn) for those n so large that y(tn) is very close to y(c). Thus g vanishes identically near c. This means that c = 1, since otherwise g vanishes identically near y(c + e) for small e > 0. Therefore g(Z1) = g(y(l)) = g(y(c)) = 0.

We know now that given Zo E Q, either g(zo) =f. ° or Zo is either a zero or pole of order n for g. It follows that either g(Z)-l is holomorphic near zo, or, by (4), that Zo is a pole of order n, or, by (3), that Zo is a zero of order n. Thus g(Z)-1 is meromorphic in Q, and sofjg is also. 0

Exercises

1. Prove the rest of the assertions of Theorem 5.5. 2. Show that tan z is merom orphic on all of C, with only simple poles.

What about (tan Z)2? 3. Determine all the functions f, meromorphic in all of C, such that

If(z) - tan zl < 2

at each point z which is not a pole either off or of tan z.

Rational functions; Laurent expansions; residues 179

4. Suppose f has a pole of order n at zoo Show that there are an R > 0 and a 0 > 0 such that if Iwl > R then there are exactly n points z such that

fez) = w, Iz - zol < o. (Hint: recall Theorem 4.3.)

5. A functionfis said to be defined near 00 if there is an R > 0 such that

{z Ilzl > R}

is in the domain of definition off The function f is said to be holomorphic at 00 if 0 is a removable singularity for the function g defined by

g(z) = f(1/z),

I/z in the domain off Similarly, 00 is said to be a zero or pole of order n for f if 0 is a zero or pole of order n for g. Discuss the status of 00 for the following functions:

(a) fez) = zn, n an integer. (b) fez) = eZ •

(c) fez) = sin z. (d) fez) = tan z.

6. Suppose Zo is an essential singularity for f, while g is meromorphic near Zo and not identically zero near zoo Is Zo an essential singularity for fg? What if, in addition, Zo is not an essential singularity of g?

§6. Rational functions; Laurent expansions; residues

A rational function is the quotient of two polynomials:

fez) = p(z)/q(z)

where p and q are polynomials, q "¥= O. By Theorem 5.4, a rational function is meromorphic in the whole plane C. (In fact it is also meromorphic at 00;

see Exercise 5 of §5.) It is easy to see that sums, scalar multiples, and products of rational

functions are rational functions. Iff and g are rational functions and g "¥= 0, thenf/g is a rational function. In particular, any function of the form

(1) fez) = a1(z - zo)-l + aiz - ZO)-2 + ... + an(z - zo)-n

is a rational function with a pole at zoo We can write

(2) fez) = p((z - zo)-l)

where p is a polynomial with p(O) = O. It turns out that any rational function is the sum of a polynomial and rational functions of the form (I).

Theorem 6.1. Suppose f is a rational function with poles at the distinct points Z1> Z2, ... , Zm and no other poles. There are unique polynomials Po, P1, ... , Pm such that plO) = 0 if j =f. 0 and

180 Complex Analysis

Proof We induce on m. If m = 0, thenfhas no poles. Thusfis an entire function. We havef = plq where P and q are polynomials and q ;t; O. Suppose P is of degree rand q is of degree s. It is not hard to show that there is a constant a such that

t,S-'!(z) --? a as z --? 00.

This and Exercise 4 of §3 imply that f is a polynomial. Now suppose the assertion of the theorem is true for rational functions

with m - I distinct poles, and suppose f has m poles. Let Zo be a pole of J, of order r. Then

where h is holomorphic near zoo Near zo,

co

h(z) = L: an(z - zo)n. n=l

Therefore

f(z) = ao(z - zo)-r + al(z - zo)l-r + ... + ar-I(z - zo)-l + k(z),

where k is holomorphic near zoo Now k is the difference of two rational functions, hence is rational. The function g = f - k has no poles except at zo, so the poles of k are the poles off which differ from zoo By the induction assumption, k has a unique expression of the desired form. Therefore f has an expression of the desired form.

Finally, we want to show that the expression (3) is unique. Suppose Po is of degree k. The coefficient bk of Zk in Po can be computed by taking limits on both sides in (3):

lim z-kf(z) = lim Z-kpO(Z) = bk. Z-+1Xl 2-+00

Therefore the coefficient of Zk-l can be computed:

lim Zl-k[f(z) - bkzk] = bk-b etc. " .... 00

Continuing in this way, we determine all coefficients of Po. Similarly, if PI is of order r then the coefficient of the highest power of (z - ZI)-l in (3) is

lim (z - ZI)'!(Z) = Cn 2-+21

the coefficient of the next power is

lim (z - ZIY-l[f(z) - Cr(Z - ZI)-r]. 2-+21

All coefficients may be computed successively in this way. 0

The expression (3) is called the partial fractions decomposition of the rational function f

Rational functions; Laurent expansions; residues 181

Let us note explicitly a point implicit in the preceding proof. If f has a pole of order r at Zo, then

'" fez) = L: bn(z - zo)-n, o < Iz - zol < 8, 11.=-"

some 8 > O. This generalization of the power series expansion of a holo­morphic function is called a Laurent expansion. It is one case of a general result valid, in particular, near any isolated singularity.

Theorem 6.2. Suppose f is holomorphic in the annulus

A = {z I r < Iz - zol < R}.

Then there is a unique two sided sequence (an)~ '" c C such that

(4) '" fez) = L: an(z - zo)n, r < Iz - zol < R. -'"

Proof. Suppose r < Iz - zol < R. Choose rl> R1 such that

r < r1 < Iz - zol < R1 < R.

Let C1 be the circle of radius r1 and C2 the circle of radius R1 centered at zo0 By Exercise 5 of §3,

fez) = -21 .f f(w)(w - zo)-1 dw - -21 .f f(w)(w - zo)-l dw TTl C2 TTl Cl

= f2(Z) + f1(Z).

Here f2 and f1 are defined by the respective integrals. We consider f2 as being defined for Iz - zol < R1 and f1 as being defined for Iz - zol > r1. Thenf2 is holomorphic and has the power series expansion

'" (5) f2(Z) = L: an(z - zo)n, n=O

Moreover, by the Cauchy integral theorem we may increase R1 without changing the values of f2 on Iz - zol < R1; thus the series (5) converges for Iz - zol < R.

The functionf1 is holomorphic for Iz - zol > r1. Again, by moving the circle CI> we may extend f1 to be holomorphic for Iz - zol > r. To get an appropriate series expansion we proceed as in the proof of Corollary 3.3. When Iz - zol > r1 and Iw - zol = r1,

(w - Z)-l = «w - zo) - (z - ZO))-l = -(z - zo)-l[1 - (w - zo)(z - ZO)-1]-1

'" = - L: (w - zo)n-1(z - zo)-n. n=1

The series converges uniformly for w E CI> so

'" (6) f1(Z) = L: a_n(z - zo)-n

n=1

182 Complex Analysis

where

a_ n = 2~.J f(w)(w - Zo)n-l dw. TTl 01

Equations (5) and (6) together give (4). Finally, we want to prove uniqueness. Suppose (4) is valid. Then the

power series

n= - 00

converges for r < Iz - zol < R, so it converges uniformly on any smaller annulus. It follows that if C is any circle with center zo, contained in A, then (3) may be integrated term by term over C. Since

Ie (z - zo)n dz

is zero for ni:l and 2TTi for n = -1, this gives

(7) a-I = 2~' J fez) dz. TTl 0

More generally we may multiply fby (z - zo)-m-l and integrate to get

(8) am = -21 .J f(z)(z - zo)-m-l dz, TTl 0

all m. Thus the coefficients are uniquely determined. D

In particular, Theorem 6.2 applies when f has an isolated singularity at zoo In this case the Laurent expansion (3) is valid for

o < Iz - zol < R,

some R > O. The coefficient a -1 is called the residue off at zoo Equation (7) determines the residue by evaluating an integral; reversing the viewpoint we may evaluate the integral if we can determine the residue. These observa­tions are the basis for the" calculus of residues." The following theorem is sufficient for many applications.

Theorem 6.3. Suppose C is a circle or rectangle. Suppose f is holomorphic in an open set Q containing C and all points enclosed by C, except for isolated singularities at the points ZI, Z2, ... , Zm enclosed by C. Suppose the residue of fat Zj is bj • Then

(9) fez) dz = 27Ti(b1 + bJ + ... + brn). o

Proof Let Cl> ... , Cm be nonoverlapping circles centered at Zl> ... , Zm and enclosed by C. Then

Rational functions; Laurent expansions; residues 183

Applying Exercise 6 of §3, we get (9). 0

If fhas a pole at Zo, the residue may be computed as follows. If n is the order of the pole,

'" fez) = (z - zo)-nh(z) = (z - zo)-n L: bm(z - zo)m,

m=O

so the residue at Zo is

In particular, at a simple pole m = 1 and the residue is h(zo).

Let us illustrate the use of the calculus of residues by an example. Suppose we want to compute

1= i'" (1 + t 2)-1 dt

and have forgotten that (1 + t 2)-1 is the derivative oftan- 1 t. Now

Let CR, R > 0, be the square with vertices ± Rand ± R + Ri. Let fez) = (1 + Z2)-1. The integral off over the three sides of CR which do not lie on the real axis is easily seen to approach zero as R -+ 00. Therefore

f:", (1 + t 2)-1 dt = 1!~",rR (1 + t 2)-1 dt

= lim i fez) dz. R-+oo OR

For R > l,fis holomorphic inside CR except at z = i, where it has a simple pole. Since

fez) = (z - i)-1(Z + i)-1,

the residue at i is (2i)-1. Therefore when R > 1,

f fez) dz = 21Ti(2i)-1 = 1T. OB

We get I = !1T (which is tan- 1 (+00) - tan- 1 0, as it should be). We conclude with some further remarks on evaluating integrals by this

method. Theorem 6.3 is easily shown to be valid for other curves C, such as a semicircular arc together with the line segment joining its endpoints, or a rectangle with a portion or portions replaced by semicircular arcs. The method is of great utility, depending on the experience and ingenuity of the user.

184 Complex Analysis

Exercises

1. Compute the partial fractions decomposition of

2. Find the Laurent expansion of exp (lIz) and sin (liz) at 0. 3. Compute the definite integrals

{'", (t + 1)(t 3 - 2it2 + t - 2i)-1 dt

fa'" t 2(t 4 + 1)-1 dt.

4. Show that S: t- 1 sin t dt = Pr. (Hint: this is an even function, and it is the imaginary part of J(z) = z- 1etz• Integrate J over rectangles lying in the half-plane 1m z ? 0, but with the segment - e < t < e replaced by a semicircle of radius e in the same half-plane, and let the rectangles grow long in proportion to their height.)

5. Show that a rational function is holomorphic at 00 or has a pole at 00.

6. Show that any flfnction which is merom orphic in the whole plane and is holomorphic at 00, or has a pole there, is a rational function.

7. Show that if Re z > 0, the integral

r(z) = fa'" t Z - 1t:-t dt

exists and is a hoi om orphic function of z for Re z > 0. This is called the Gamma Junction.

8. Integrate by parts to show that

r(z + 1) = zr(z) , Rez> 0.

9. Define r(z) for -1 < Re z ::s;; 0, z ¥- 0, by

r(z) = z- 1r(z + 1).

Show that r is meromorphic for Re z > -1, with a simple pole at zero. 10. Use the procedure of Exercise 9 to extend r so as to be merom orphic

in the whole plane, with simple poles at 0, -1, - 2, .... 11. Show inductively that the residue of r at - n is ( -1)n(n!) -1.

12. Is r a rational function?

§7. Holomorphic functions in the unit disc

In this section we discuss functions hoi om orphic in the unit disc D = {z I Izl < I} from the point of view of periodic functions and distributions. This point of view gives another way of deriving the basic facts about the local theory of holomorphic functions. It also serves to introduce certain spaces of holomorphic functions and of periodic distributions.

Holomorphic functions in the unit disc 185

Note that iff is defined and holomorphic for Iz - zol < R, then setting

fleW) = f(zo + Rw)

we get a function fl holomorphic for I wi < 1. Similarly, iff has an isolated singularity at zo, we may transform it to a function with an isolated singu­larity at zero and holomorphic elsewhere in the unit disc. Since f can be recovered fromfl by

fez) = h.(R-l(Z - zo)),

all the information about local behavior can be deduced from study of fl instead.

Supposefis holomorphic in D. Then the function

g(r, 8) = f(re iO)

is periodic as a function of 8 for 0 ::::; r < I and constant for r = O. It is also differentiable, and the assumption thatfis holomorphic imposes a condition on the derivatives of g. In fact

h-l[g(r + h, 8) - g(r, 8)] = h-l[f(reiO + heiO) - f(re iO)] = eiO(heiO)-l[f(reiO + hef8 ) - f(re i8)].

Letting h -+ 0 we get

Similarly,

so

h- 1 [g(r18 + h) - g(r, 8)] = h-l[f(rei(o+It») - f(re iO)] '" l' (re IO) • h -1 [rei(8 + It) - re f8 ]

Combining these equations we get

(1) . 8g 8g zr- =-.

8r 88

Now let grC8) = g(r, 8), 0 ::::; r < 1. Since gr is continuous, periodic, and continuously differentiable as a function of 8, it is the sum of its Fourier series:

00

(2) gr(8) = L an(r)efRO• -00

The coefficients an(r) are given by

I f2" air) = 27T 0 g(r, 8)e- fR8 d8.

186 Complex Analysis

It follows that air) is continuous for 0 :::; r < 1 and differentiable for o < r < 1, with derivative

, ( ) _ 1 J 2" og ( e) - inD de an r - 21T 0 or r, e .

Using (1) and integrating by parts we get

(3) n =F 0,

and a~(r) = O. Thus ao is constant. The equation (3) may be solved for an as follows, n =F O. The real and imaginary parts of an are each real solutions of

u'(r) = ,-lnu(r).

On any interval where u(r) =F 0 this is equivalent to

so on such an interval u(r) = ern, e constant. Since an is continuous on [0, 1) and vanishes at 0 if n =F 0 (because go is constant), we must have air) = anrn, with an constant and an = 0 if n < O.

We have proved the following: ifJis holomorphic in D, then

00

(4) J(reiD) = L: anrneinD, O:::;r<1. n=O

Thus 00

J(z) = L: anzn, Izl < 1. n=O

Suppose J is holomorphic in D and defined and continuous on the closure: {z Ilzl :::; I}. Then the functions an(r) = anrn are also continuous at r = 1, and an = an(1) is the nth Fourier coefficient of gl. It follows that gT is a convolution:

(5)

where QT is the periodic distribution with Fourier coefficients bn = rn, n > 0 and bn = 0, n < O. Then

co 00

QT(e) = L: rneinD = L: (reiD)n n=O n=O

or

(6)

Equation (5) can be written

Holomorphic functions in the unit disc 187

Setting w = rei9 and z = eit, we recover the Cauchy integral formula

(7) 1 J' few) = -2' f(z)(z - W)-l dz, TTL c

where C is the unit circle. Thus (5) may be regarded as a version of the Cauchy integral formula.

Note that Qr is a smooth periodic function when ° ::;; r < 1. Therefore (5) defines a function in the disc if gl is only assumed to be a periodic distri­bution. In terms of the Fourier coefficients, if gl has Fourier coefficients (anY~~ 00 then those of gr are (an(r »~ 00 where an(r) = 0, n < 0, an(r) = anrn, n ;::: 0. These observations and the results of §§I and 2 of Chapter 5 leads to the following theorem.

Theorem 7.1. Suppose F is a periodic distribution with Fourier coefficients (an)~ 00, where an = 0for n < 0. Letf be the function defined in the unit disc by

(8) f(rei9) = (F * Q,)(8),

with Qr given by (6). Then f is holomorphic in the unit disc and

00

(9) fez) = 2: anzn, Izl < 1. n=O

Moreover, F is the boundary value of J, in the sense that the distributions Fr defined by the functions J,( 8) = f(rei9) converge to F in the sense of fIJ' as r -+ 1.

Conversely, suppose f is holomorphic in the unit disc. Then f is given by a convergent power series (9). If the sequence (an)O' is of slow growth, i.e., if there are constants c, r such that

(10) n > 0,

then there is a distribution F such that (8) holds. If we require that the Fourier coefficients ofF with negative indices vanish, then F is unique and is the boundary value off in the sense above.

Condition (10) is not necessarily satisfied by the coefficients of a power series (9) converging in the disc. An example is

n > 0.

Thus the condition (10) specifies a subset of the set of all holomorphic functions in the disc. This set of holomorphic functions is a vector space. Theorem 7.1 shows that this space corresponds naturally to the subspace of fIJ' consisting of distributions whose negative Fourier coefficients all vanish.

Recall that FE fIJ' is in the Hilbert space 22 if and only if its Fourier coefficients (an)~ co satisfy

(11)

For such distributions there is a result exactly like the preceding theorem.

188 Complex Analysis

Theorem 7.2. Suppose F is a periodic distribution with Fourier coefficients an = 0 for n < o. If FE V, then F is the boundary value of a function J,

co

(12) f(z) = L: anzn, Izl < 1 n=O

with

(13)

The functions f,( 0) = f(re'9) converge to F in the sense of V as r - 1, and

1 J 2" 11F112 = sup -2 If(re'9)12 dO. 0:$1<1 7T 0

Conversely, suppose f is defined in the unit disc by (12). Suppose either that (13) is true or that

(14) (,2"

sup If(reI9)12 dO < 00. 0:$1:$1,0

Then both (13) and (14) hold, and the boundary value off is a distribution FEV.

Proof. The first part of the theorem follows from Theorem 7.1 and the fact that (11) is a necessary condition for F to be in L2. The second part of the theorem is based on the identity

(15) 1 f2" co

- If(reI9)1 2 dO = L: lan l2r2n, 27T 0 n=O

o :s; r < 1,

which is true because the Fourier coefficients off, are anrn for n ~ 0 and zero for n < O. If (13) is true then (14) follows. Conversely, if (13) is false, then (15) shows that the integrals in (14) will increase to 00 as r _ 1. Thus (13) and (14) are equivalent. By Theorem 7.1, if (13) holds then f has a distribution F as boundary value. The an are the Fourier coefficients of F, so (13) implies FE V. 0

The set of hoi om orphic functions in the disc which satisfy (14) is a vector space which can be identified with the closed subspace of V consisting of distributions whose negative Fourier coefficients are all zero. Looked at either way this is a Hilbert space, usually denoted by

H2 or H2(D).

Exercises

1. Verify that co

f(z) = L: n-r..zn n=l

converges for Izl < I but that the coefficients do not satisfy (10).

Holomorphic functions in the unit disc 189

2. Suppose J is holomorphic in the punctured disc 0 < Izl < 1. Carry out the analysis of this section for

g(r, () = J(re!8)

to deduce:

(a) J(z) = L:'~ _ ao anzn, 0 < Izl < 1, (b) If IJ(z)1 ~ Mlzl-m for some M, m, then

ao

J(z) = 2: anzn. n .. -m

Chapter 7

The Laplace Transform

§1. Introduction

It is useful to be able to express a given function as a sum of functions of some specified type, for example as a sum of exponential functions. We have done this for smooth periodic functions: if u E &! then

"" (1) u(x) = L a"el """,

-"" where

(2) 1 f2" a" = 217 0 u(x)e- I""" dx.

Of course the particular exponential functions which occur here are precisely those which are periodic (period 217). If u: ~ -+ C is a function which is not periodic, then there is no such natural way to single out a sequence of exponential functions for a representation like (1). One might suspect that (1) would be replaced by a continuous sum, i.e., an integral. This suspicion is correct. To derive an appropriate formula we start with the analogue of (2). Let

(3) g(z) = i: u(t)e-Ilt dt,

when the integral exists. (Of course it may not exist for any z E C unless restrictions are placed on u.)

If we are interested in functions u defined only on the half-line [0, (0), we may extend such a function to be zero on (-00,0]. Then (3) for the extended function is equivalent to

(4) g(z) = L"" u(t)e- zt dt.

If u is bounded and continuous, then the integral (4) will exist for each z E C which has positive real part. More generally, if a E ~ and e-atu(t) is continuous and bounded for t > 0, then the integral (4) exists for Re z > a. Moreover, the function

(5) g =Lu

is holomorphic in this half plane:

g'(z) = - L"" tu(t)e- zt dt,

190

Rez > a.

Introduction

In particular, let

Then for Re z > Re w,

t> 0; t ::;; 0, WECo

Luw(z) = 100 e(w-z)t dt = (z - W)-l.

191

The operator L defined by (3) or (4) and (5) assigns, to certain functions on IR, functions holomorphic in half-planes in C. This operator is clearly linear. We would like to invert it: given g = Lu, find u. Let us proceed formally, with no attention to convergence. Since Lu is holomorphic in some half-plane Re z > a, it is natural to invoke the Cauchy integral formula. Given z with Re z > a, choose b such that

a < b < Rez.

Let C be the vertical line Re w = b, traced in the upward direction. We consider C as "enclosing" the half-plane Re w > b, though traced in the wrong direction. A purely formal application of the Cauchy integral formula then gives

g(z) = Lu(z) = --21 . f J(w)(w - Z)-l dw 7Tl c

= -21 . f J(w)LuwCz) dw. 1Tl C

If L has an inverse, then L -1 is also linear. Then we might expect to be able to interchange L -1 and integration in the preceding expression, to get

(6) u(t) = -21 . f g(w)ewt dw, 1Tl C

or

1 foo u(t) = - g(a + is)e(a+is)t ds. 21T _ 00

Thus (3) or (4) and (6) are our analogues of (2) and (1) for periodic functions. It is convenient for applications to interpret (3) and (6) for an appropriate

class of distributions F. Thus if Fis a continuous linear functional on a suitable space !l' of functions, we interpret (3) as

The space !l' will be chosen in such a way that each such continuous linear functional F can be extended to act on all the functions ez for z in some half-plane Re z > a. Then the function LF will be holomorphic in this half­plane. We shall characterize those functions g such that g = LF for some F, and give an appropriate version of the inversion formula (6).

192 The Laplace Transform

The operator L is called the Laplace transform. It is particularly useful in connection with ordinary differential equations. To see why this might be so, let u' = Du be the derivative of a function u. Substitution of u' for u in (3) and integration by parts (formally) yield

(7) [Lu'](z) = zLu(z).

More generally, suppose p is a polynomial

p(z) = amzm + am_IZm- 1 + ... + aIz + ao.

Let p(D) denote the corresponding operator

p(D) = amDm + am_IDm-1 + ... + aID + ao,

i.e.,

Then formally

(8) [Lp(D)u](z) = p(z)Lu(z).

Thus to solve the differential equation

p(D)u = v

we want

p(z)Lu(z) = Lw(z).

From (6), this becomes

(9)

As we shall see, all these purely formal manipulations can be justified.

Exercises

1. Show that the inversion formula (6) is valid for the functions uw , i.e., if a > Re wand C = {z I Re z = a} then

_1_. f (z - w)-Iezt dz = ewt, 2m c

= 0,

t> 0;

t < 0.

2. Suppose u: [0, (0) -+ C is bounded and continuous and suppose the derivative u' exists and is bounded and continuous on (0, (0). Show that

f" e-ztu'(t) dt = z f" e-ztu(t) dt - u(O),

Re z > 0. Does this conflict with (7)?

The space:e 193

§2. The space !l'

Recall that a function u: IR -+ C is said to be smooth if each derivative D"u exists and is continuous at each point of IR, k = 0, 1, 2, . . .. In this section we shall be concerned with smooth functions u which have the property that each derivative of u approaches 0 very rapidly to the right. To be precise, let 2 be the set of all smooth functions u: IR -+ C such that for every integer k ~ 0 and every a E IR,

(1) lim eatD"u(t) = O. t-+ + 00

This is equivalent to the requirement that for each integer k > 0, each a E IR, and each ME IR the function

(2)

is bounded on the interval [M, 00). In fact (1) implies that (2) is bounded on every such interval. Conversely if (2) is bounded on [M,oo) then (1) holds when a is replaced by any smaller number a' < a.

It follows that if u E 2; then for each k, a, M we have that

(3) lulk,a,M = sup {eatID"u(t)11 t E [M, oo)}

is finite. Conversely, if (3) is finite for every integer k ~ ° and every a E IR, M E IR, then u E Y.

The set of functions 2 is a vector space: it is easily checked that if u, v E 2 and bEe then bu and u + v are in Y. Moreover,

(4)

(5)

Ibulk,a,M = Ibllulk,a,M'

lu + VI",a,M :::;; lul",a,M + Ivl",a,M'

A sequence of functions (un)l' c: 2 is said to converge to u E 2 in the sense of 2 if for each k, a, M,

IUn - Uka,M -+ ° as n -+ 00.

If so, we write

Un -+ U (2).

The sequence (un)'f c: 2 is said to be a Cauchy sequence in the sense of 2 if for each k, a, M,

As usual, a convergent sequence in this sense is a Cauchy sequence in this sense. The converse is also true.

Theorem 2.1. 2 is a vector space. It is complete with respect to conver­gence as defined by the expressions (3); i.e., if (un) l' c: 2 is a Cauchy sequence in the sense of 2; then there is a unique u E 2 such that Un -+ U (2).

194 The Laplace Transform

Proof Let (un)f be a Cauchy sequence in the sense of !l'. Taking (3) with a = 0, we see that each sequence of derivatives (Dlcun)f is a uniform Cauchy sequence on each interval [M, 00). It follows, by Theorem 4.1 of Chapter 2, that there is a unique smooth function u such that Dlcun -+ DIcU uniformly on each [M, 00). Now let a be arbitrary. Since (eatDlcun)f is also a uniform Cauchy sequence on [M, 00) it follows that this sequence converges uniformly to eatDlcu. Thus u E.fR and Un -+ U (.fR). 0

It follows immediately from the definition of .fR that certain operations on functions in .fR give functions in !l'. In particular, this is true of differen­tiation:

ifuE.fR;

translation: T.u E.fR ifuE.fR

where

(T.u)(t) = u(t - s);

and complex conjugation:

u* E.fR if u E.fR

where

u*(t) = u(t)*.

It follows that if u E 2, so are the real and imaginary parts:

Re u = !(u + u*),

i 1m u = 2 (u* - u).

Moreover, if u E 2, so is the integral of u taken from the right:

S+u(t) = - fO u(s) ds.

In fact, DS+u = u so

(6) k ~ 1.

For k = 0, note that for t ~ M, a > 0,

/S+u(t)/ ~ {Xl /u(s)/ ds ~ /U/O.a.M {' e- as ds

= a-1/u/O.a.Me- at. Thus

(7) a> O.

The finiteness of

/S+U/O.a.M

for a ~ 0 follows from finiteness for any a > O. Thus S+u E!l'.

The space .I£ 195

Lemma 2.2. The operations of differentiation, translation, complex con­jugation, and integration are continuous from !l' to !l' with respect to conver­gence in the sense of !l'. Moreover, if u E !l' then the difference quotient

as s.....,..O.

Proof These statements chiefly involve routine verifications. We shall prove the final statement. Given an integer k ~ 0, let v = Dku. Suppose t ~ M and 0 < Is I :;:; 1. The Mean Value Theorem implies

s-l[L sv(t) - vet)] = Dv(r)

where It - rl < lsi. Then

Dk{S-l[L su(t) - u(t)]} - Dk Du(t) = Dv(r) - Dv(t) = Dk+lu(r) - Dk+lU(t).

But

(8) t ~ -M.

The left side of (8) converges to zero as s .....,.. 0, uniformly on bounded inter­vals. It follows that

as s .....,.. 0, for any a' < a. 0

The functions ez,

(9) t E IR,

are not in!l' for any Z E C. However, they may be approximated by functions from !l' in a suitable sense.

Lemma 2.3. Suppose Re z > a. There is a sequence (un)r c !l'such that

for each integer k ~ 0 and each ME IR.

Proof Choose a smooth function rp such that rp(t) = 1 if t :;:; 1 and rp(t) = 0 if t ~ 2. (The existence of such a function is proved in §8 of Chapter 2.) Let

rpn(t) = rp(tfn) ,

Then Un is smooth and vanishes for t ~ 2n, so Un E!l'. We shall consider in detail only the (typical) case k = 1.

eat(Dun(t) - Dez(t)) = eat(rpn(t)DezCt) + ez(t)Drpn(t) - DezCt)) = (1 - rpn(t))ze<a-Z)t + Drpn(t)e<a-Z)t.

196 The Laplace Transform

Now both 1 - CPn(t) and DCPn(t) are bounded independent of t and n, and vanish except on the interval [n, 2n]. Therefore

leat(Dun(t) - Deit)) I ~ c exp (na - n Re z),

c independent of nand t. Thus

all M. The argument for other values of k is similar. 0

The following lemma relates the lui k,a,M for different values of the indices k, a, M.

Lemma 2.4. Suppose k, k' are integers, and

o ~ k ~ k', a ~ a', M;::: M'.

Suppose also either that k = k' or that a' > O. Then there is a constant c such that

(10) luka,M ~ clulk',a',M" all u E Ii'.

Proof It is sufficient to prove (10) in all cases when two of the three indices are the same. The case k = k', a = a' is trivial. The case k = k', M = M' is straightforward. Thus, suppose a = a' > 0 and M = M'. Let k' = k + j and set v = Dk'U. We may obtain Dku from v by repeated inte­grations:

We use (7) repeatedly to get

If u E .P, we set

lulk,a,M = IDkulo,a,M ~ a-Jlvlo,a,M -JIDk' I = a u O,a,M

= a-Jlulk',a,M'

lulk = lulk,k,-k = sup {lektDku(t)1 I t ;::: -k}.

Then the following is an easy consequence of Lemma 2.4.

Corollary 2.5. Suppose (un)!X) c Ii'. Then

Un -+ U (2)

if and only if for each integer k ;::: 0,

lun-ulk-+O as n-+oo.

Exercises

1. Show that u(t) = exp ( - t 2) is in Ii'. 2. Show that if u, v E 2, then the product uv is in Ii'.

D

The space 2' 197

3. Show that u E Y, Z E C implies ezu E 2-4. Complete the proof of Lemma 2.2.

§3. The space 2"

A linear functional on the vector space !£' is a function F: !£' --+ C such that

F(au) = aF(u), F(u + v) = F(u) + F(v).

A linear functional F on !£' is said to be continuous if

un --+ U (!£')

implies

The set of all continuous linear functionals on !£' will be denoted !£". An element FE!£" will be called a distribution of type !£", or simply a distribution. An example is the a-distribution defined by

(1) a(u) = u(O).

A second class of examples is given by

(2) F(u) = (' eztu(t) dt,

Z E C.

Suppose f: IR --+ C is a continuous function such that for some a E IR, MEIR,

(3)

(4)

We may define

by

(5)

f(t) = 0, t ~ M,

e-a!j(t) is bounded.

Flu) = t: f(t)u(t) dt.

In fact the integrand is continuous and vanishes for t ~ M. If a' > a then on [M, 00) we have

If(t)u(t) I = le-a!j(t)lle(a-a')t Ilea'tu(t)1 < clul ' e(a-a')t - a,a ,M ,

where c is a bound for le-a!j(t)l. Therefore the integral (5) exists and

(6) 1F,(u) I ~ clulo.a'.M LX) e(a-a')t dt.

198 The Laplace Transform

It follows from (6) that F, is continuous on Z, i.e., F, E fl". We say that FE!l" is a function, or is defined by a function, if there is a

continuous functionfsatisfying (3) and (4), such that F = F,. Suppose FE!l" is defined by f The translates Taf and the complex

conjugate function f* also define distributions. It is easy to check that if g = Tat. i.e., g(t) = f(t - s), t E~, then

Flu) = F,(Lsu), u E!l'.

Similarly,

F,*(u) = F,(u*)*, u E!l'.

We shall define the translates and the complex conjugate of any arbitrary FE!l" by

(7)

(8)

(T.F)(u) = F(T -au),

F*(u) = F(u*)*,

U E!l'.

u E!l'.

Similarly, the real and imaginary parts of F E!l" are defined by

(9)

(10)

ReF = t(F + F*),

i 1m F = 2 (F* - F).

We say that F is real if F = F*. Suppose FE!l" is defined by f and suppose the derivative Df exists, is

continuous, and satisfies (3) and (4). Integration by parts gives

FD'(U) = -F(Du), u E!l'.

Therefore we shall define the derivative DF of an arbitrary FE !l" by

(11) DF(u) = -F(Du), U E!l:

Generally, for any integer k ;;:: 0 we define

(12)

Proposition 3.1. The set !l" is a vector space. ifF E !l", then the translates TaF, the complex conjugate F*, and the derivatives Dk F are in !l".

Proof All these statements follow easily from the definitions and the continuity of the operations in!l'; see Lemma 2.2. For example, DkF:!l' -+ C is certainly linear. If u" -+ u (.!l'), then

DkU" -+ Dku (!l'), so

Thus DkF is continuous. 0

The space 2' 199

A sequence (Fn)'~ c 2' is said to converge to FE 2' in the sense of 2' if for each u E 2

Fn(u) -+ F(u) as n -+ 00.

We denote this by

Fn -+ F (2').

The operations defined above are continuous with respect to this notion of convergence.

Proposition 3.2. Suppose

Fn -+ F (2'), Gn -+ G (2'),

and suppose a E C, s E IR. Then

aFn -+ aF (2'), Fn + Gn -+ F + G (2'),

ToFn -+ ToF (2'), D"Fn -+ D"F (2').

Moreover, the difference quotient

s-I[T -oF - F] -+ F (2')

as s-+O.

Proof. All except the last statement follow immediately from the defi­nitions. To prove the last statement we use Lemma 2.2:

s-l[LoF - F](u) = F(S-l[Tou - uD -+ F( - Du) = (DF)(u). o

The following theorem gives a very useful necessary and sufficient con­dition for a linear functional on 2 to be continuous.

Theorem 3.3. Suppose F: 2 -+ C is linear. Then F is continuous if and only if there are an integer k ;;:: 0 and constants a, M, K c IR such that

(13) !F(u)! :::; K!U!".a.M, all u E!l'.

Proof. Suppose (13) is true. If Un -+ U (2) then

!F(un) - F(u)! :::; K!un - U!k.a.M -+ O.

Thus F is continuous. To prove the converse, suppose that (13) is not true for any k, a, M, K.

In particular, for each positive integer k we may find a v" E 2 such that

!F(v,,)! ;;:: k!Vk!k = k!v!"."._k ¥- O.

Let Uk E 2 be

200

Then

lu"l" = k-1,

But Corollary 2.5 implies that

The Laplace Transform

Since F(u,,) does not converge to 0, F is not continuous. 0

The support of a function u: IR -+ IC is defined as the smallest closed subset A of IR such that u(t) = 0 for every t ¢= A. Another way of phrasing this is that t is not in the support of u if and only if there is an e > 0 such that u is zero on the interval (t - e, t + e). The support of u is denoted

supp (u).

Condition (3) on a function f can be written

supp (I) c [M, (0).

The support of a distribution FE 2' can be defined similarly. A point t E R is not in the support of F if and only if there is an e > 0 such that

F(u) = 0

whenever u E 2 and supp (u) c (t - e, t + e). We denote the support of F also by

supp (F).

Theorem 3.3 implies that any FE 2' has support in a half line.

Corollary 3.4. If FE 2', there is M E IR such that

supp (F) c [M, (0).

Proof Choose k, a, M such that (13) is true for some K. If u E 2 and

supp (u) c (-00, M),

then luka.M = 0

so F(u) = O. Therefore each t < M is not in the support of F. 0

Exercises

1. Compute the following in the case F = S:

T.F(u), F*(u), ReF,

2. Show that if F is given by (2), then

DF= S + zF.

ImF,

3. Suppose F = DiS. For what constants k, a, M, K is (l3) true?

Characterization of distributions of type .:£' 201

4. Find supp (DkS). 5. Prove that if F is defined by a function/, then

supp (F) = supp (f).

§4. Characterization of distributions of type IF'

Iff: IR ---+ C is a function satisfying the two conditions (3) and (4) of §3 and if k is an integer ~ 0, then

(1)

is a distribution of type 'p'. In this section we shall prove that, conversely, any FE 'p' is of the form (1) for some k and some function f The proof depends on two notions: the order of a distribution and the integral (from the left) of a distribution.

A distribution FE 'p' is said to be of order k if (13) of §4 is true, i.e., if for some real constants a, M, and K,

(2) I F(u) I ::;; Klu ",-a,M' all u E 2.

By Theorem 3.3, each FE 'p' is of order k for some k ~ 0. Suppose FEL' is defined by a functionf Let g be the integral of ffrom

the left:

get) = {oo f(s) ds = {f(S) ds,

where supp (f) c [M, (0). If u E.P and v = S+u, then Dv = u. Integration by parts gives

Flu) = Loooo g(t)v'(t) dt = - L""oo g'(t)V(t) dt = - F;(S +u).

For an arbitrary FE 'p' we define the integral of F (from the left), denoted S_F, by

Proposition 4.1. If FE.P' then the integral S _ F is in 'p' and

(3)

If F is of order k ~ 1, then S _F is of order k - 1.

Proof Clearly S _F is linear. The continuity follows from the definition and the fact that S + is continuous in 2. The identity (3) is a matter of manipulation:

D(S_F)(u) = -S_F(Du) = F(S+Du) = F(u),

and similarly for the other part. 0

U E!l',

202 The Laplace Transform

To prove that every FE 2' is of the form (1), we want to integrate F enough times to get a distribution defined by a function. To motivate this, we consider first a function J: IR -? C such that the first and derivatives are continuous and such that

supp (I) c [M, (0),

some M. Then by integrating twice and changing the order of integration (see §7 of Chapter 2) we get

Let

(4)

J(t) = f", DJ(s) ds = f", f", D2f(r) dr ds

= f", f D2f(r) ds dr

= f", (t-r)D2J(r)dr.

h(t) = Itl, = 0,

t :::; 0, t> O.

Then our equation is

J(t) = L"'", her - t)D2J(r) dr

or

(5)

We would like to interpret (5) as the action of the distribution defined by D2f on the function Tth; howev~r Tth is not in .P. Nevertheless, h can be approximated by elements of .P.

Lemma 4.2. Let h be defined by (4). There is a sequence (h n)'{! c 2 such that

(6) Ihn - hl o•a•M -? 0 as n -? 00

Jor each a E IR, ME IR.

Proof There is a smooth function rp: IR -? IR such that 0 :::; rp(t) :::; 1 for all t and rp(t) = 1, t :::; -2, rp(t) = 0, t 2': -1; see §8 of Chapter 2. Let

hn(t) = rp(tln)h(t) = rpn(t)h(t).

Then hn is smooth, since rpn is zero in an interval around 0 and h is smooth except at O. Also hn(t) = h(t) except in the interval (-2In, 0), and

IhnCt) - h(t)1 :::; 21n, t E (-2In, 0).

Thus hn E 2 and (6) is true. 0

Characterization of distributions of type 2' 203

Theorem 4.3. Suppose FE.P' is of order k - 2, where k is an integer ~ 2. Then there is a unique function f such that

F = Dk(F,).

Proof. Suppose first that k = 2, F is of order O. Let h be the function defined by (4) and let (h,.)i C .P be as in Lemma 4.2. Choose a, M, K E IR such that

(7) IF(u) I ~ Klulo.a.M' alluEL.

We may suppose a ~ O. For each s E IR the translates T.h,. also converge to h:

(8) IT.h,. - T .hlo.a.M -+ 0 as n -+ 00.

It follows from (7) and (8) that for each s E IR

F(T .h,.) converges as n -+ 00.

Let f(s) be the limit of this sequence. Then

(9) If(s) I ~ lim KIT .h,.lo.a.M = KIT .hlo.a.M'

But

(10)

(11)

Thus

,. .... IX)

ITshlo•a•M = 0

IT.hlo.a.M ~ (s - M)eaS

ifs ~ M,

ifs> M.

supp (f) C [M, (0)

and for any a' > a there is a constant c such that

If(s) I ~ cea's, all s E IR.

If f is continuous, it follows that f defines a distribution FJ E !R'. Suppose s < t. Taking limits we get

Thus f is continuous.

If(t) - f(s)1 ~ KITth - T.hl o•a.M ~ Keat(t - s).

Letfis) = F(T .h,.). Then/b) = 0 if s ~ M. For s > M,

If,.(s) - f(s) I ~ KITsh,. - T.hl o•a•M ~ 2KeaS(s - M)/n.

Therefore if u E!l;

where G,. is the distribution defined by f,.. Then

204 The Laplace Transform

Let Vn be the function defined by

Vn(t) = L: Ti1n(t)D2u(s) ds

= {"", hn(t - s)D2u(s) ds.

Since D2u E.P, and hn(t - s) = 0 if s :0; t, the integral converges. Moreover, it is not difficult to see that the integral is the limit of its Riemann sums

1 N2 Vn.N(t) = N L hn(t - m/N)D2u(m/N),

m= -N2

in the sense that

IVn.N - vnlo.a.M -+ 0 as N -+ 00.

In fact, for t ~ M,

:0; ~ i'" It - sle- as ds = ~ e- at,

where c and c' are independent of t and N. Therefore

F(vn) = lim F(vn N) N •

Now let

Then

In fact, for t ~ M

Therefore

v(t) = L"'", h(t - s)D2u(s)ds.

IVn - Vlo.a.M -+ 0 as n -+ 00.

D2Ff (u) = FlD2u) = lim Gn(D2u) = lim F(vn) = F(v).

Laplace transforms of functions

But

Thus D2F, = F.

vet) = {' (s - t)D2u(s) ds

= -LX) Du(s) ds = u(t).

205

Now suppose F is of order k - 2 > O. Let G = (S_)k-2F. Then G is of order 0, and by what we have just shown, there is an I such that

D2F, = G.

But then

DkF, = Dk- 2G = F.

Finally, we must prove uniqueness. This is equivalent to showing that DF = 0 implies F = O. But F = S_(DF), so this is the case. 0

Exercises

1. Show that Dko is of order k but not of order k - 1. 2. Find the functionl of Theorem 4.3 when F = o. Compute DF, = S _ o. 3. Show that if supp (F) c [M, (0), then supp (S_F) c [M, (0), and

conversely. 4. Suppose FE 2' and the support of F consists of the single point O.

Show that F is of the form

where the ak's are constants.

§5. Laplace transforms of functions

Suppose thatlis a function which defines a distribution of type 2', i.e., I: ~ -+ C is continuous, and

(1)

(2)

supp (f) c [M, (0),

I/(t)1 :::;; Keat, all t.

If Z E C, let e" be as before:

e,,(t) = e-"t, t E ~.

If Re Z = b > a then

206 The Laplace Transform

Sincef(t) = 0 for t < M, the integral

(3) L: f(t)e,,(t) dt = f: f(t)e,,(t) dt

exists when Re z > a. The Lap/ace transform of the function f is the function Lf defined by (3):

(4) Lf(z) = L"oo e-2tf(t) dt, Re z > a.

Theorem 5.1. Suppose f: IR - C is continuous and satisfies (I) and (2). Then the Laplace transform Lf is holomorphic in the half plane

{z 1 Rez > a}.

The derivative is

(5) (Lf)'(z) = - Loooo e- 2ttf(t) dt.

The Laplace transform satisfies the estimate

(6) ILf(Z) 1 s; K(Re z - a)-1 exp (M(a - Re z», Rez> a.

Proof

(w - z)-1[Lf(w) - Lf(z)] = Loo g(w, z, t)f(t) dt

where

g(w, z, t) = (w - z)-l[e- wt - e- 2t].

Suppose Re z and Re w are ~ b > a. Let

h(s) = exp [-(1 - s)z - sw]t, Os;ss;1.

Then

g(w, z, t) = (w - z)-1[h(l) - h(O)].

An application of the Mean Value Theorem to the real and imaginary parts of h shows that

Ih(l) - h(O) 1 S; clw - zle-b't, t ~ 0

where b' = b if b > 0, b' = max {Re w, Re z} otherwise. Thus as w _ z,

Ig(w, z, t)f(t)1 S; c1e- Bt, where e > O.

Moreover,

g(w, z, t)f(t) _ - te-alJ(t)

as w _ z, uniformly on each interval [M, N]. It follows that Lfis differen­tiable and that (5) is true.

Laplace transforms of functions 207

The estimate (6) follows easily from (1) and (2):

I Lf(z)I ::;; f: If(t)e- 2t l dt

::;; K C exp tea - Re z) dt "M

= K(Re z - a)-l exp (M(a - Re z». o We want next to invert the process: determine J, given Lf

Theorem 5.2. Suppose f satisfies the conditions of Theorem 5.1, and let g = Lf. Given b > max {a, O}, let C be the line

{z IRe z = b}.

Then f is the second derivative of the function F defined by

(7) F(t) = ~. f e2tz- 2g(z) dz. 2m c

Proof By (6), g is bounded on the line C. Therefore the integral (7) exists. Moreover, if

then the gN are bounded uniformly on the line C and converge uniformly to g. Thus F(t) is the limit as N ~ ex) of FN , where

= IN {~.f e2(t-s)z-2 dZ}f(S) ds. -N 2m c

Let us consider the integral in braces. When s > t the integrand is holo­morphic to the right of C and has modulus::;; klzl- 2 for some constant k. Let Cn be the curve consisting of the segment {Re z = b I Iz - bl ::;; N} and the semicircle {Re z > b liz - bl = N}. Then the integral of

over Cn in the counterclockwise direction is zero, and the limit as R ~ ex) is

208 The Laplace Transform

Thus the integral in braces vanishes for s > t. When s < t, let C~ be the reflection of CB about the line C. Then for R > b,

= t - s.

Taking the limit as R ~ 00 we get

-21 .f ez<t-')z-2 dz = t - s, s < t. 7T1 C

Thus ,

F(t) = f"" (t - s)J(s) ds.

It follows that D2F = f. 0

We can get a partial converse of Theorem 5.1.

Theorem 5.3. Suppose g is holomorphic in the half plane

{z IRe z > a}

and satisfies the inequality

(8) I g(z) I :s; K(I + Izl)-2 exp (- M Re z).

Then there is a unique continuous Junction J: R ~ C with the properties

(9)

(10)

(11)

suPpJ c: [M',oo), some M',

IJ(t)1 :s; Kebt, all t,Jor some b,

LJ(z) = g(z) Jor Re z > b.

Moreover, we may take M' = Min (9) and any b > a in (10) and (11).

Proof. Choose b > a and let C(b) be the line {z IRe z = b}. Let

(12) J(t) = -21 • f eztg(z) dz. 'lT1 C(b)

It follows from (8) that the integral exists and defines a continuous function. It follows from (8) and an elementary contour integration argument that (12) is independent of b, provided b > a. Moreover, (8) gives the estimate

(13) IJ(t)1 :s; Ceb<t-M), b > a,

where C is independent of band t. This implies (10). If t < M we may take b ~ +00 in (13) and get

J(t) = 0, t < M.

Thus the Laplace transform of J can be defined for Re z > a. If Re w > a, choose

a < b < Rew.

Laplace transforms of functions 209

Then

Lf(w) = f"" e- wt{-21·f eztg(z) dZ} dt. M 7r1 C(b)

Since

le-wtHtg(z)1 ~ e(1 + IZi)-2 exp [-Mb + t(b - Re w)]

for t E IR and z E C, we may interchange the order of integration. This gives

Lf(w) = _1 . f g(Z){f"" e(a-w)t dt} dz 2m C\b) M

= - ~ f g(z)e(a-W)M(z - W)-l dz. 2m C\b)

In the half plane Re z > b we have

Ig(z)e(a-W)MI ~ e(1 + Izi)-2.

Therefore a contour integration argument and the Cauchy integral formula give

Lf(w) = g(z)e(a-w)Mla_w = g(w).

Finally, we must show thatfis unique. This is equivalent to showing that Lf == 0 implies f == O. But this follows from Theorem 5.2. 0

Exercises

1. Letf(t) = 0, t < O;f(t) = teat, t > O. Compute Lfand verify that

f(t) = -21 . f eatLf(z) dz, 7r1 C

where C is an appropriate line. 2. Show that the Laplace transform of the translate of a function f

satisfies L(Tsf)(z) = e"sLf(z).

3. Suppose bothfand Dfare functions satisfying (1) and (2). Show that

L(Df)(z) = zLf(z).

4. Compute Lf when

f(t) = 0,

where n is a positive integer.

t < 0; f(t) = tn, t > 0,

5. Supposefsatisfies (1) and (2), and let g = ewf. Show that

Lg(z) = Lf(z - w).

210 The Laplace Transform

6. Compute Lf when

f(t) = 0, t < 0; f(t) = ewtt",

7. Compute Lfwhenf(t) = 0, t < 0;

f(t) = f~ sin s ds, t > O.

t > O.

§6. Laplace transforms of distributions

Suppose FE 2'. Theorem 3.3 states that there are constants k, a, M, K such that

(1) alluE2.

If Re z > a, then Lemma 2.3 states that there is a sequence (u,,)i C 2 such that

(2)

where e2(t) = e- 2 • Now (1) and (2) imply that (F(U,,»:=l is a Cauchy sequence in C. We shall define the Laplace transform LF by

(3) LF(z) = lim F(u,,). " .... '"

In view of (2) we shall write, symbolically,

(4) LF(z) = F(e2 )

even though e" 1= 2. Note that if (vn)i c 2 and

Jv" - e2 JIc,a,M -+ 0 as n -+ 00

then

JF(v,,) - F(u,,)J ;:S; KJv" - U"JIc,a,M -+ 0

as n -+ 00. Thus LF(z) is independent of the particular sequence used to approximate e2 •

Proposition 6.1. Suppose F, G E 2' and b E C. Then on the common domain of definition

(5)

(6)

(7)

(8)

(9)

Moreover,

(10)

L(bF) = bLF;

L(F + G) = LF + LG;

L(T.F)(z) = e- 28LF(z);

L(DlcF)(z) = zlcLF(z);

L(S_F)(z) = z-lLF(z), z =F O.

L(ewF)(z) = LF(z + w),

Laplace transforms of distributions 211

where ewF is the distribution defined by

(11) U E!l'.

If F is defined by a function J, then

(12) LF = Lf.

Proof. The identities (5) and (6) follow immediately from the definitions. If (un)f c !f' satisfies (2) then the sequence (Lsun)f approximates Lse;, in the same sense. But

so

L(TsF)(z) = lim TsF(un) = lim F(T -sun) = e-;'S lim F(un) = e- 2SLF(z).

This proves (7), and the proofs of (8), (9) and (10) are similar. Note that U E !f' implies ewu E !f' and

implies

ewun ~ ewu (!f').

Therefore (11) does define a distribution. Finally, (12) follows from the definitions. D

We can now generalize Theorems 5.1 and 5.2 to distributions.

Theorem 6.2. Suppose FE!f" and suppose F satisfies (1). Then the Lap/ace transform LF is holomorphic in the half plane

{z IRe z > a}.

Moreover,

(13)

where f is the function defined by

(14) f(t) = ....!..... f e2tz- k - 2LF(z) dz. 2m 0

Here C is the line {z I Re z = b}, where b > max {a, O}.

Proof. We know by Theorem 4.3 that there is a functionfsuch that

F = Dk+2F,.

It was shown in the proof of Theorem 4.3 that for any b > max {a, O} there is a constant c such that

If(t)1 ~ cebt, all t.

212 The Laplace Transform

Therefore Lfis holomorphic for Re z > max {a, O}. By Proposition 6.1,

(IS) LF(z) = zk+2Lf(z), Re z > max {a, O}.

Therefore LF is hoI om orphic in this half plane. This completes the proof of the first statement in the case a ;::: O. When a < 0, let G = eaF. Then (1) implies

Thus by the argument just given, LG is holomorphic for Re z > O. Since LF(z) = LG(z - a), LF is holomorphic for Re z > a.

Now let C be the line {z I Re z = b}, where b > max {a, O}. Letfbe the function such that (13) is true. Then by Theorem 5.2 and equation (15),f is the second derivative of the function

(16) get) = ~f e2tz- k- 4LF(z) dz. 2m c

From the definition of LF it follows that on C

(17)

Using (17) we may justify differentiating (16) twice under the integral sign to get (14). 0

Theorem 6.2 implies, in particular, that if LF == 0 then F = o. Given a holomorphic function g, how can one tell whether it is the

Laplace transform of a distribution?

Theorem 6.3. Suppose g is holomorphic in a half plane

{z IRe z > a}.

Then g is the Laplace transform of a distribution FE 2' if and only if there are constants k, a, M, Kl such that

(18) I g(z) I :$; K1(l + Izi)k exp (-M Re z), Rez> a.

Proof. Suppose g = LF, where FE 2'. Then there are k, a, M, K such that (I) is true. Then Re z > a implies

where Kl = KeaM•

Conversely, suppose (18) is true. Take b > max {a, O}. We may apply Theorem 5.3 to

h(z) = Z-k-2g(Z)

to conclude that

h = LJ,

Differential equations

where I is continuous,

Let F = Dk +2f. Then

suppl c [M, (0), I/(t)1 ~ cebt•

LF(z) = Zk+2h(z) = g(z),

213

Rez > b.

Since this is true whenever b > max {a, O}, the proof is complete in the case a ~ O.

When a < 0, let

gl(Z) = g(z + a).

Then gl is holomorphic for Re z > 0 and satisfies

It follows that gl = LFI for Re z > 0, some FI e fl". Then

g =LF, o

Exercises

1. Compute the Laplace transforms of Dk8, k = 0, 1,2, ... and of T.8, seR.

2. Compute the Laplace transform of F when

F(u) = (" elDtu(t) dt.

§7. Differential equations

In §§5, 6 of Chapter 2 we discussed differential equations of the form

U'(x) + au(x) = I(x),

and of the form

U"(x) + bu'(x) + cu(x) = I(x).

In this section we turn to the theory and practice of solving general nth order linear differential equations with constant coefficients:

(1)

214 The Laplace Transform

where the aj are complex constants. Using D to denote differentiation, and understanding DO to be the identity operator, DOu = u, we may write (I) in the form

(1)'

Let p be the polynomial

n

p(z) = L akzk. k=O

Then it is natural to denote by p(D) the operator

n

(2) p(D) = L akD". "=0

Equation (1) becomes

(1)" p(D)u =f.

We shall assume that the polynomial is actually of degree n, that is

an =F 0.

Before discussing (1)" for junctions, let us look at the corresponding problem for distributions: given HE 2', find FE 2' such that

p(D)F = H.

Theorem 7.1. Suppose p is a polynomial oj degree n > 0, and suppose HE 2'. Then there is a unique distribution FE 2' such that

(3) p(D)F = H.

Proof. Distributions in 2' are uniquely determined by their Laplace transforms. Therefore (3) is equivalent to

(4) L(P(D)F)(z) = LH(z), Rez> a

for some a E IR. But

L(p(D)F)(z) = p(z)LF(z).

We may choose a so large that p(z) =F ° if Re z ::c: a, and so that LH is holo­morphic for Re z > a and satisfies the estimate given in Theorem 6.3. Then we may define

g(z) = p(z)-lLH(z), Rez > a.

Then g is holomorphic, and it too satisfies estimates

I g(z) I ::;; K(I + Izl)" exp (-M Re z), Rez> a.

Theorem 6.3 assures us that there is a unique FE 2' such that LF = g, and then (4) holds. 0

Differential equations 215

The proof just given provides us, in principle, with a way to calculate F, given H. Let us carry out the calculation formally, treating F and H as though they were functions:

where

(5)

F(t) = -21 . f et2LF(z) dz 7f1 C

= ~. f etzp(z)-lLH(z) dz 2m C

= _I . f f et2p(z)-le- S2H(s) ds dz 2711 C R

= f [_1 . f e<t-S)Zp(z)-l dZ]H(S) ds R 2m C

= L G(t - s)H(s) ds.

G(t) = J.... f et2p(z)-1 dz 2m C

and C is a line Re z = b > a.

We emphasize that the calculation was purely formal. Nevertheless the integral (5) makes sense if p has degree ~ 2, and defines a function G. Equivalently,

(5)' G(t) = lim -21 . f etzp(z) -1 dz, R-.. o 7f1 CB

where CR is the directed line segment from b - iR to b + iR, R > O. We shall show that the limit (5)' also exists when p has degree 1, except when t = O. The function defined by (5)' is called the Green's function for the operator p(D) defined by the polynomial p. Our formal calculation suggests that G plays a central role in solving differential equations. The following two theorems provide some information about it.

Theorem 7.2. Suppose p is a polynomial of degree n ~ 1; suppose Zl, Z2, ... , Zr are the distinct roots of p, and suppose that Zj has mUltiplicity mi. Then (5)' defines afunction G for all t # o. Thisfunction is a linear combination of the functions g;lc, 1 ~ j ~ r, 0 ~ k < mi' where

gilc(t) = 0, t < 0; gjlc(t) = tic exp (zjt), t > o.

Proof Suppose t < O. Let DR denote the rectangle with vertices b ± iR, (b + Rl/2) ± iR. When Re Z ~ b,

(6) le2tp(z)-11 ~ c(t)(1 + Izj)-n exp t(Re Z - b),

216 The Laplace Transform

where c(t) is independent of z. Now the line segment CB is one side of the rectangle DB, and the estimates (6) show that the integral of e2tp(z)-1 over the other sides converges to 0 as R --+ 00. On the other hand, the integral over all of DB vanishes, because the integrand is holomorphic inside DB. Thus the limit in (5)' exists and is 0 when t < 0 (and also when t = 0, if n> 1).

Suppose t > O. Let DB now be the rectangle with vertices b ± iR, (b - Rl/2) ± iR, with the counterclockwise direction, and suppose R is so large that DB contains all roots of p(z). Then the integral of e 2tp(z)-l over DB is independent of R, and the integral over the sides other than CB tends to 0 as R --+ 00. Thus again the limit in (5)' exists, and

(5)" G(t) = -21 .J. e2tp(z)-ldz, 17' CB

t > O.

Now we may apply Theorem 6.3 of Chapter 6: G(t) is the, sum of the residues of the meromorphic function e2tp(z)-l. The point z, is a pole of order m" so near z, we have a Laurent expansion

p(Z)-l = L: bm(z - zJ)m. mO!:-m,

Combining this with

e2t = e"lt L: (ml)-l(z - z,)mtm, m .. O

we see that the residue (the coefficient of(z - Z,)-l in the Laurent expansion) at ZI is a linear combination of

tIc exp (Z,t),

moreover, it is the same linear combination whatever the value of t. 0

Suppose f is a complex-valued function defined on an interval (a, b). We write

f(a+) = limf(a) t ... a

when the limit on the right exists as t approaches from the right. We take the Green's function for p(D) to be 0 at t = 0; when n > 1 this

agrees with (5)'.

Theorem 7.3. Let p be a polynomial of degree n > 0, with leading co­efficient all :F O. Let G be the Green's function for p(D). Then G is the unique function from IR to C having the following properties:

(7) all derivatives D"G exist and are continuous when t :F 0;

(8) the derivatives D"G exist and are continuous at 0 when k ~ n - 2;

(9) G(t) = 0, t ~ 0;

(10) p(D)G(t) = 0, t > 0;

(11) aID"-1G(0+) = 1.

Differential equations 217

Proof We know that G is a linear combination of functions satisfying (7) and (9), so G does also. When t > 0 we may differentiate (5)" and get

(12) DkG(t) = -1.f zke2tp(z)-1 dz. 2m DR

Thus

DkG(O+) = -21 .f Zkp(Z)-ldz. 7rl DR

We may replace DR by a very large circle centered at the origin and conclude that

k ~ n - 2.

Therefore (8) is true. Let us apply the same argument when k = n - 1. Over the large circle the integrand is close to

zn-1(a Z )-1 = a -lZ-l n n n ,

so

Finally, (12) gives

p(D)G(t) = -21 . f ezt dz = 0, 7rl DR

t> O.

Now we must show that G is uniquely determined by the properties (7)-(11). Suppose G1 also satisfies (7)-(11), and let 1= G - G1• Then I satisfies (7)-(10); moreover Dn-y(o) = O. We may factor

p(z) = an(z - Zl)(Z - Z2)' .. (z - zn),

where we do not assume that the Zj are distinct. Let 10 = J, and let

k > O.

Then eachlk is a linear combination of DiJ, 0 ~ j ::;; k, so

Moreover,

Thus

so

Then

k~n-l.

In = (D - zn)(D - Zn-1) ... (D - zl)1 = an -lp(D)1 = 0.

In-1(0) = 0, Dln-1 - Znln-1 = In = 0,

In-1 = 0.

In-2(0) = 0, Dln-2 - zn-1/n-2 = 0,

218 The Laplace Transform

so fn-2 = O. (We are using Theorem 5.1 of Chapter 2). Inductively, each fk = 0, k ::;; n, so f = 0 and G = G1• 0

Now let us return to differential equations for functions.

Theorem 7.4. Suppose p is a polynomial of degree n > 0, and suppose f: [0, (0) ~ C is a continuous function. Then there is a unique solution u: [0, (0) ~ C to the problem

(13)

(14)

p(D)u(t) = f(t),

DiU(O+) = 0,

t > 0;

O::;;j::;;n-l.

This solution u is given by

(15) u(t) = 1: G(t - s)f(s) ds,

where G is the Green'sfunctionfor the operator p(D).

Proof We use properties (7)-(11) of G. Let u be given by (15) for t ~ O. Then successive differentiations yield

(16) Du(t) = G(O+ )f(t) + f DG(t - s)f(s) ds

= I: DG(t - s)f(s) ds, ... ,

(17) Dku(t) = 1: DkG(t - s)f(s) ds, k ::;; n - 1,

(18) Dnu(t) = an -If(t) + s: DnG(t - s)f(s) tis.

Thus

p(D)u(t) = f(t) + f p(D)G(t - s)f(s) ds

= f(t).

Moreover, (17) implies (14). Thus u is a solution. The uniqueness of u is proved in the same way as uniqueness of G. 0

We conclude with a number of remarks.

1. The problem (13)-(14) as a problem for distributions: Let us define f(t) = 0 for t < O. Iff does not grow too fast, i.e., if for some a E IR

e-alJ(t) is bounded,

then we may define a distribution HE .fR' by

H(v) = f f(t)v(t) dt, V E.fR.

Differential equations 219

Suppose u is the solution of (13)-(14). Then it can be shown that u defines a distribution F, and

(19) p(D)F = H.

Thus we have returned to the case of Theorem 7.1.

2. If the problem (13)-(14) is reduced to (19), then the proof of Theorem 7.1 shows that the solution may be found by determining its Laplace trans­form. Since there are extensive tables of Laplace transforms, this is of practical as well as theoretical interest. It should be noted that Laplace transform tables list functions which are considered to be defined only for t ;:::: 0; then the Laplace transform of such a function f is taken to be

Lf(z) = (" e- 2!J(t) dt.

In the context of this chapter, this amounts to settingf(t) = 0 for t < 0 and considering the distribution determined by J, exactly as in Remark 1.

3. Let us consider an example of the situation described in Remark 2. A table of Laplace transforms may read, in part,

f Lf

sin t (Z2 + 1)-1 sinh t (Z2 - 1)-1

(As noted in Remark 2, the function sin t in the table is considered only for t ;:::: 0, or is extended to vanish for t < 0.)

Now suppose we wish to solve:

(20)

(21)

u"(t) - u(t) - sin t = 0, t > 0;

u(O) = u'(O) = O.

Let p(z) = Z2 - 1. Our problem is

p(D)u = sin t, t > 0; u(O) = Du(O) = O.

The solution u is the function whose Laplace transform is

p(z)-lL(sin t)(z) = (Z2 - 1)-1(Z2 + 1)-1.

But

(Z2 - 1)-1(Z2 + 1)-1 = !(Z2 - 1)-1 - !(Z2 + 1)-1.

Therefore the solution to (20)-(21) is

u(t) = t sinh t - t sin t, t ;:::: o.

220 The Laplace Transform

4. In cases where the above method fails, either because the given function! grows too fast to have a Laplace transform or because the function Lu cannot be located in a table, one may wish to compute the Green's function G and use (IS). The Green's function may be computed explicitly if the roots of the polynomial p are known (of course (5)" gives us G in principle). In fact, suppose the roots are Zlo Z2, ... , Zr with multiplicities ml, m2, ... , mr• We know that G for t > 0, is a linear combination of the n functions

Thus

t> 0,

where we must determine the constants Cjk' The conditions (8) and (11) give n independent linear equations for these n constants. In fact,

G(O+) = 2: CjO,

DG(O+) = .2 ZjCjO + 2: Cjlo

etc. 5. The more general problem

(22)

(23)

p(D)u(t) = Jet),

Dku(O +) = bk,

t> 0;

O:S.k<n

may be reduced to (13)-(14). Two ways of doing this are given in the exercises. 6. The formal calculation after Theorem 7.1 led to a formula

F(t) = f G(t - s)H(s) ds

which it is natural to interpret as a convolution (see Chapter 3). A brief sketch of such a development is given in the exercises.

Exercises

1. Compute the Green's function for the operator p(D) in each of the following cases:

p(Z) = Z2 - 4z - 5 p(z) = Z2 - 4z + 4 p(z) = Z3 + 2Z2 - Z - 2 p(z) = Z3 - 3z + 2.

Differential equations 221

2. Solve for u:

u"(t) - 4u'(t) + 4u(t) = et, t > 0, u(O+) = u'(O+) = O.

3. Solve for u:

u"'(t) - 3u'(t) + 2u(t) = t 2 - cos t, t > 0, u(O+) = u'(O+) = u"(O+) = 0

4. Let Uo: (0, 00) ~ C be given by

n-l uo(t) = L (k!)-1bk t k •

k=O

Show that

O:S;k:s;n-1.

5. Suppose uo: (0, 00) ~ IR is such that

O:s;k:S;n-1.

Show that u is a solution of (22)-(23) if and only if u = Uo + Uh where U1 is the solution of

P(D)Ul(t) = f(t) - p(D)uo(t), t > 0, DkU1(0+) = 0, 0 :s; k :s; n - 1.

6. Show that problem (22)-(23) has a unique solution. 7. Show that the solution of

p{D)u{t) = 0, Dku{O+) = 0, DJu(O+) = 1

t > 0, o :s; k < j and j < k :s; n - 1,

is a linear combination of functions t k exp zt. 8. Show that any solution of

p(D)u(t) = 0, t > 0

is a linear combination of the functions t k exp zt, where z is a root of p(D) with multiplicity greater than k, and conversely.

9. Suppose u: (0, 00) ~ C is smooth and suppose Dku(O+) exists for each k. Suppose also that each Dku defines a distribution Fk by

Show that

DFo = F1 + u(O+ )8,

222 The Laplace Transform

and in general

k=l

DkFO = Fk + 2 Dfu(O+ )Dk-l-fS. j=O

10. In Exercise 9 let u(t) = G(t), t > 0, where G is the Green's function for p(D). Show that

p(D)Fo = S.

11. Use Exercise 9 to interpret the problem (22)-(23) as a problem of finding a distribution (when the function f defines a distribution in 'p'). Discuss the solution of the problem.

12. Use Exercise 11 to give another derivation of the result of Exercise 5. 13. Again let

aCt) = u(-t), Tsu(t) = u(t - s).

If F E 'p' and u E.P, set

(a) Suppose F = Fv , where v: IR -+ IC is continuous, vet) = ° for t :::; - M, and e-atv(t) is bounded. Show that for each u E.P' the convolution integral

iJ * u(t) = J iJ(t - s)u(s) ds

exists and equals

F* u(t).

(b) Show that for each FE 'p' and u E 2, the function F * u is in !l'. 14. If F, HE 'p', set

(F * H)(u) = F(D * u), u E!l'.

Show that F * HE .P'. 15. Compute (DkSr * u, u E!l'. Compute (DkS) * F, FE 'p'. 16. Show that

L(F * H) = L(F)L(H).

17. Let G be the distribution determined by the Green's function for p(D). Show that

is

LG = p(Z)-l ..

18. Show that the solution of

p(D)F = H

where G is as in Exercise 17.

NOTES AND BffiLIOGRAPHY

Chapters 1 and 2. The book by Kaplansky [9] is a very readable source of further material on set theory and metric spaces. The classical book by Whittaker and Watson [25] and the more modern one by Rudin [18] treat the real and complex number systems, compactness and continuity, and the topics of Chapter 2. Vector spaces, linear functionals, and linear transformations are the subject of any linear algebra text, such as Halmos [7]. Infinite sequences and series may be pursued further in the books of Knopp [10], [11]. More problems (and theorems) in analysis are to be found in the classic by Polya and Szego [15].

Chapters 3, 4, and 5. The Weierstrass theorems (and the technique of approx­imation by convolution with an approximate identity) are classical. A direct proof of the polynomial approximation theorem and a statement and proof of Stone's generalization may be found in Rudin [18].

The general theory of distributions (or" generalized functions") is due to Laurent Schwartz, and is expounded in his book [20]. The little book by Lighthill [12] discusses periodic distributions and Fourier series. Other references for dis­tribution theory imd applications are the books of Bremermann [2], Liverman [13], Schwartz [21], and Zemanian [27].

Banach spaces, Frechet spaces, and generalizations are treated in books on functional analysis: that by Yosida [26] is comprehensive; the treatise by Dunford and Schwartz [4] is exhaustive; the sprightly text by Reed and Simon [16] is oriented toward mathematical physics. Good sources for Hilbert space theory in particular are the books by Halmos [6], [8] and by Riesz and Sz.-Nagy [17].

The classical L 2-theory of Fourier series treats L 2 (0, 21T) as a space of functions rather than as a space of distributions, and requires Lebesgue integration. Chap­ters 11 through 13 of Titchmarsh [24] contain a concise development of Lebesgue integration and the P-theory. A more leisurely account is in Sz.-Nagy [14]. The treatise by Zygmund [28] is comprehensive.

Chapter 6. The material in §1--§6 is standard. The classic text by Titchmarsh [24] and that by Ahlfors [1] are good general sources. The book by Rudin [19] also treats the boundary behavior of functions in the disc, related to the material in §7.

Chapter 7. The Laplace transform is the principal subject of most books on "operational mathematics" and" transform methods." Doetsch [3] is a compre­hensive classical treatise. Distribution-theoretic points of view are presented in the books of Bremermann [2], Erdelyi [5], Liverman [13], and Schwartz [21].

Bibliography

1. AHLFORS, L. V.: Complex Analysis, 2nd ed. New York: McGraw-Hill 1966. 2. BREMER MANN, H. J.: Distributions, Complex Variables, and Fourier Trans­

forms. Reading, Mass.: Addison-Wesley 1966.

223

224 Notes and Bibliography

3. DOETSCH, G.: Handbuch der Laplace-Transformation, 3 vols. Basel: Birk­hauser 1950-1956.

5. ERDELYI, A.: Operational calculus and generalized functions. New York: Holt, Rinehart & Winston 1962.

4. DUNFORD, N., and SCHWARTZ, J. T.:Linear Operators, 3 Vols. New York: Wiley-Interscience 1958-1971.

6. HALMos, P. R.: Introduction to Hilbert Space, 2nd. ed. New York: Chelsea 1957.

7. HALMOS, P. R.: Finite Dimensional Vector Spaces. Princeton: Van Nostrand 1958.

8. HALMOS, P. R.: A Hilbert Space Problem Book. Princeton: Van Nostrand 1967.

9. KAPLANSKY, I.: Set Theory and Metric Spaces. Boston: Allyn & Bacon 1972. 10. KNOPP, K. : Theory and Applications of Infinite Series. London and Glasgow:

Blackie & Son 1928. 11. KNOPP, K.: Infinite Sequences and Series. New York: Dover 1956. 12. LIGHTHILL, M. J. : Introduction to Fourier Analysis and Generalized Functions.

New York: Cambridge University Press 1958. 13. LIVERMAN, T. P. G.: Generalized Functions and Direct Operational Methods.

Englewood Cliffs, N.J.: Prentice-Hall 1964. 14. Sz.-NAGY, B.: Introduction to Real Functions and Orthogonal Expansions.

New York: Oxford University Press 1965. 15. POLYA, G., and SZEGO, G.: Problems and Theorems in Analysis. Berlin­

Heidelberg-New York: Springer 1972. 16. REED. M., and SIMON, B.: Methods of Modern Mathematical Physics, vol. 1.

New York: Academic Press 1972. 17. RIESZ, F., and Sz.-NAGY, B.: Functional Analysis. New York: Ungar 1955. 18. RUDIN, W.: Principles of Mathematical Analysis, 2nd ed. New York:

McGraw-Hill 1964. 19. RUDIN, W.: Real and Complex Analysis. New York: McGraw-Hill 1966. 20. SCHWARTZ, L.: Theorie des Distributions, 2nd ed. Paris: Hermann 1966. 21. SCHWARTZ, L.: Mathematics for the Physical Sciences. Paris: Hermann;

Reading, Mass.: Addison-Wesley 1966. 22. SOBOLEV, S. L.: Applications of Functional Analysis in Mathematical Physics.

Providence: Amer. Math. Soc. 1963. 23. SOBOLEV, S. L.: Partial Differential Equations of Mathematical Physics.

Oxford and New York: Pergamon 1964. 24. TITCHMARSH, E. c.: The Theory of Functions, 2nd ed. London: Oxford

University Press 1939. 25. WHITTAKER, E. T., and WATSON, G. N.: A Course of Modern Analysis, 4th

ed. London: Cambridge University Press 1969. 26. YOSIDA, K.: Functional Analysis, 2nd ed. Berlin-Heidelberg-New York:

Springer 1968. 27. ZEMANIAN, A. H.: Distribution Theory and Transform Analysis. New York:

McGraw-Hill 1965. 28. ZYGMUND, A.: Trigonometric Series, 2nd ed. London: Cambridge University

Press 1968.

C Q IR Z z+ L2 rc ,!l7 ,!l7' ~

~'

D L U * v, F * u, F * G Un ->- U (,!l7) Un ->- U (~)

Un ->- F(L2) F7} ->- F (,!l7') Fn ->- F(~')

NOTATION INDEX

complex numbers, 8 rational numbers, 4 real numbers, 5 integers, 1 positive integers, 1 Hilbert space of periodic distributions, 106 continuous periodic functions, 69 smooth functions of fast decrease at + 00, 193 distributions acting on L, 197 smooth periodic functions, 73 periodic distributions, 84 differentiation operator, 72, 86, 198 Laplace transform operator, 192, 206, 210 convolution, 78, 94, 96, 100, 101, 222 193 73 106 199 85, 100

225

SUBJECT INDEX

approximate identity, 80

ball, in metric space, 20 Banach space, 70 basis, 30 Bessel's equality, inequality, 124 Bolzano-Weierstrass theorem, 26 bounded function, 36 -linear functional, 71 - sequence, 11 - set, 7, 24 branch of logarithm, 173

Casorati-Weierstrass theorem, 177 Cauchy integral formula, 166, 187 Cauchy-Riemann equations, 157 Cauchy sequence, in a metric space, 22 - in 2,193 - in 9,73 - of numbers, 12 - uniform, 47 Cauchy's theorem, 161 chain rule, 45, 155 change of variables in integration, 45 characterization of distributions in 2',

203 - of periodic distributions, 89, 102 class C\ Coo, 46 closed set, 21 closure, 22 compact set, 23 - in \Rn , IC, 24 comparison test, 15 complement, of set, 2 complementary subspace, 33 complete metric space, 22 completeness axiom for real numbers,

7 completeness of'ti', 70 - of V, 107 - of \R, IC, 12 - of \Rn , 23 complex conjugate of complex num­

ber, 9

227

- of distribution in 2', 198 - of function, 38 - of periodic distribution, 85, 100 composition of functions, 3 connected set, 175 continuous function, 35 continuity, at a point, 34 - uniform, 35 convergence, in a metric space, 22 - in Hilbert space, 110 - in 2',199 - in V, 106 - in 9, 73 - in 9', 85, 100 - of numerical sequences, 10 - of series, 14 convolution, in 2', 222 - in 9', 96, 101 - of functions, 78 - of functions with periodic distri-

butions, 94, 100 coordinates of vector, 32 countable set, 3 curve, 159 - smooth, piecewise smooth, 159

~- distribution, 85, 197 dense set, 22 derivative, of distribution in 2', 198 - of function, 42, 155 - of periodic distribution, 86, 100 differentiable function, 42, 155 differential equations, first order and

second order, 51-56 - higher order, 213-222 diffusion equation, 137 - derivation, 144 dimension, of vector space, 31 Dirac 8- distribution, 85, 197 Dirichlet kernel, 130 Dirichlet problem, 150 distribution, of type 2', 197 -, periodic, 84, 100 divergence of series, 14

228

essential singularity, 176 even function, 87 even periodic distribution, 87, 101

finite dimensional vector space, 30 Fourier coefficients, 124 - in V, 126, 129 - of a convolution, 134 - of periodic distributions, 132 Fourier series, 126, 129 Frechet space, 76 function, 2 - bounded, 36 - class CIe, Coo, 46 - complex-valued, 3 - continuous, 35 - differentiable, 42, 155 - holomorphic, 158 - injective, 3 - infinitely differentiable, 46 - integrable, 38 - meromorphic, 178 -1-1,3 - onto, 3 - periodic, 69 - rational, 179 - real-valued, 3 - smooth, 73 - surjective, 3 - uniformly continuous, 35 fundamental theorem of algebra, 169 fundamental theorem of calculus, 44

gamma function, 184 geometric series, 15 glb,7 Goursat's theorem, 165 Gram-Schmidt method, 117 greatest lower bound, 7 Green's function, 215

IP,188 harmonic function, 150 heat equation, 137 - derivation, 144 Heine-Borel theorem, 24 Hermite polynomials, 120 Hilbert cube, 116

Subject Index

Hilbert space, 109 holomorphic function, 158 homotopy, 161

imaginary part, of complex number, 9 - of distribution in ,[£', 198 - of function, 38 - of periodic distribution, 87, 101 improper integral, 41 independence, linear, 30 inf, infimum, 11 infinite dimensional vector space, 30 infinitely differentiable function, 46 inner product, 103, 109 integrable function, 38 integral, 38 - improper, 41 - of distribution in ,[£', 201 intermediate value theorem, 37 intersection, 2 interval, 5 inverse function, 3 inverse function theorem, for holo­

morphic functions, 171 isolated singularity, 175

kernel, of linear transformation, 33

Laguerre polynomials, 120 Laplace transform, of distribution,

210 - of function, 192,206 Laplace's equation, 150 Laurent expansion, 181 least upper bound, 7 Legendre polynomials, 120 L'H6pital's rule, 47 lim inf, lim sup, 12 limit of sequence, 10,22 limit point, 21 linear combination, 29 - nontrivial, 30 linear functional, 32 - bounded, 71 linear independence, 30 linear operator, linear transformation,

32

Subject Index

Liouville's theorem, 169 logarithm, 61, 173 lower bound, 7 lower limit, 12 lub,7

maximum modulus theorem, 174 maximum principle, for harmonic

functions, 154 - for heat equation, 142 mean value theorem, 43 meromorphic function, 178 mesh, of partition, 38 metric, metric space, 19 modulus, 9

neighborhood, 20 norm, normed linear space, 70 null space, 33

odd function, 87 odd periodic distribution, 88, 101 open mapping property, 174 open set, 20 order, of distribution in 2', 201 - of periodic distribution, 89, 102 - of pole, 177 - of zero, 177 . orthogonal expansion, 121, 124 orthogonal vectors, 110 orthonormal set, orthonormal basis,

117

parallelogram law, 110 Parseval's identity, 124 partial fractions decomposition, 180 partial sum, of series, 14 partition, 38 period,69 periodic distribution, 84, 100 periodic function, 69 Poisson kernel, 151 polar coordinates, 66 pole, simple pole, 176 power series, 17 product, of sets, 2 Pythagorean theorem, in Hilbert

space, 110

radius of convergence, 17 rapid decrease, 131 ratio test, 16 rational function, 179 rational number, 4 real part, of complex number, 9 - of distribution in 2', 198 - of function, 38

229

- of periodic distribution, 87, 101 real distribution in 2', 198 real periodic distribution, 87, 101 removable singularity, 175 residue, 182 Riemann sum, 38 Riesz representation theorem, 112 root test, 16

scalar, 28 scalar multiplication, 27 Schrodinger equation, 141 Schwarz inequality, 103, 109 semi norm, 76 separable, 27 sequence, 4 sequentially compact set, 26 series, 14 simple pole, simple zero, 177 singularity, essential, 176 - isolated, 175 - removable, 175 slow growth, 132 smooth function, 73 span, 30 standard basis, 30 subset, 2 subsequence, 24 subspace, 29 sup, 11 support, 200 supremum, 11

translate, of distribution in 2', 198 - of periodic distribution, 86, 100 - of function, 77 triangle inequality, 19 trigonometric polynomial, 81

230

uniform Cauchy sequence of func-tions,47

uniform continuity, 35 uniform convergence, 47 union, 2 unitary transformation, 124 upper bound, 7 upper limit, 12

vector, vector space, 28

wave equation, 145 -- derivation, 148

Subject Index

Weierstrass approximation theorem, 82

Weierstrass polynomial approxima­tion theorem, 83

zero, of holomorphic function, 177

Graduate Texts in Mathematics

Soft and hard cover editions are available for each volume.

For information

A student approaching mathematical research is often discouraged by the sheer volume of the literature and the long history of the subject, even when the actual problems are readily understandable. Graduate Texts in Mathematics, is intended to bridge the gap between passive study and creative understanding; it offers intro­ductions on a suitably advanced level to areas of current research. These introduc­tions are neither complete surveys, nor brief accounts of the latest results only. They are textbooks carefully designed as teaching aids; the purpose of the authors is, in every case, to highlight the characteristic features of the theory.

Graduate Texts in Mathematics can serve as the basis for advanced courses. They can be either the main or subsidiary sources for seminars, and they can be used for private study. Their guiding principle is to convince the student that mathematics is a living science.

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Vol. 4

Vol. 5

Vol. 6

Vol. 7

Vol. 8

Vol. 9

Vol. 10

Vol. 11

TAKEUTI/ZARING: Introduction to Axiomatic Set Theory. vii, 250 pages. 1971.

OXTOBY: Measure and Category. viii, 95 pages. 1971.

SCHAEFER: Topological Vector Spaces. xi, 294 pages. 1971.

HILTON/STAMMBACH: A Course in Homological Algebra. ix, 338 pages. 1971.

MAC LANE: Categories for the Working Mathematician. ix, 262 pages. 1972.

HUGHES/PIPER: Projective Planes. xii, 291 pages. 1973.

SERRE: A Course in Arithmetic. x, 115 pages. 1973.

TAKEUTI/ZARING: Axiomatic Set Theory. viii, 238 pages. 1973.

HUMPHREYS: Introduction to Lie Algebras and Representation Theory. xiv, 169 pages. 1972.

COHEN: A Course in Simple-Homotopy Theory. xii, 114 pages. 1973.

CONWAY: Functions of One Complex Variable. xiv, 314 pages. 1973.

In preparation

Vol. 12 BEALS: Advanced Mathematical Analysis. xii, 248 pages. Tentative publication date: November, 1973.

Vol. 13 ANDERSON/FuLLER: Rings and Categories of Modules. xiv, 370 pages approximately. Tentative publication date: October, 1973.

Vol. 14 GOLUBITSKy/GUILLEMIN: Stable Mappings and Their Regularities. xii, 224 pages approximately. Tentative publication date: October, 1973.

Vol. 15 BERBERIAN: Lectures In Functional Analysis and Operator Theory. xii, 368 pages approximately. Tentative publication date: January, 1973.

Vol. 16 WINTER: The Structure of Fields. xii, 320 pages approximately. Tenta­tive publication date: January, 1973.