Fourier analysis : an introduction

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FOURIER ANALYSIS

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Princeton Lectures in Analysis I Fourier Analysis: An Introduction II Complex Analysis III Real Analysis: Measure Theory, Integration, and Hilbert Spaces

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Princeton Lectures in Analysis

I

FOURIER ANALYSIS an introduction

Elias M. Stein & Rami Shakarchi

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

Copyright © 2003 by Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock, Oxfordshire OX20 1TW All Rights Reserved Library of Congress Control Number 2003103688 ISBN 978-0-691-11384-5 British Library Cataloging-in-Publication Data is available The publisher would like to acknowledge the authors of this volume for providing the camera-ready copy from which this book was printed Printed on acid-free paper. ∞ press.princeton.edu Printed in the United States of America 5  7  9  10  8  6

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To my grandchildren

Carolyn, Alison, Jason E.M.S.

To my parents

Mohamed & Mireille and my brother

Karim R.S.

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Foreword

Beginning in the spring of 2000, a series of four one-semester courses were taught at Princeton University whose purpose was to present, in an integrated manner, the core areas of analysis. The objective was to make plain the organic unity that exists between the various parts of the subject, and to illustrate the wide applicability of ideas of analysis to other fields of mathematics and science. The present series of books is an elaboration of the lectures that were given. While there are a number of excellent texts dealing with individual parts of what we cover, our exposition aims at a different goal: presenting the various sub-areas of analysis not as separate disciplines, but rather as highly interconnected. It is our view that seeing these relations and their resulting synergies will motivate the reader to attain a better understanding of the subject as a whole. With this outcome in mind, we have concentrated on the main ideas and theorems that have shaped the field (sometimes sacrificing a more systematic approach), and we have been sensitive to the historical order in which the logic of the subject developed. We have organized our exposition into four volumes, each reflecting the material covered in a semester. Their contents may be broadly summarized as follows: I. Fourier series and integrals. II. Complex analysis. III. Measure theory, Lebesgue integration, and Hilbert spaces. IV. A selection of further topics, including functional analysis, distributions, and elements of probability theory. However, this listing does not by itself give a complete picture of the many interconnections that are presented, nor of the applications to other branches that are highlighted. To give a few examples: the elements of (finite) Fourier series studied in Book I, which lead to Dirichlet characters, and from there to the infinitude of primes in an arithmetic progression; the X-ray and Radon transforms, which arise in a number of

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FOREWORD

problems in Book I, and reappear in Book III to play an important role in understanding Besicovitch-like sets in two and three dimensions; Fatou’s theorem, which guarantees the existence of boundary values of bounded holomorphic functions in the disc, and whose proof relies on ideas developed in each of the first three books; and the theta function, which first occurs in Book I in the solution of the heat equation, and is then used in Book II to find the number of ways an integer can be represented as the sum of two or four squares, and in the analytic continuation of the zeta function. A few further words about the books and the courses on which they were based. These courses where given at a rather intensive pace, with 48 lecture-hours a semester. The weekly problem sets played an indispensable part, and as a result exercises and problems have a similarly important role in our books. Each chapter has a series of “Exercises” that are tied directly to the text, and while some are easy, others may require more effort. However, the substantial number of hints that are given should enable the reader to attack most exercises. There are also more involved and challenging “Problems”; the ones that are most difficult, or go beyond the scope of the text, are marked with an asterisk. Despite the substantial connections that exist between the different volumes, enough overlapping material has been provided so that each of the first three books requires only minimal prerequisites: acquaintance with elementary topics in analysis such as limits, series, differentiable functions, and Riemann integration, together with some exposure to linear algebra. This makes these books accessible to students interested in such diverse disciplines as mathematics, physics, engineering, and finance, at both the undergraduate and graduate level. It is with great pleasure that we express our appreciation to all who have aided in this enterprise. We are particularly grateful to the students who participated in the four courses. Their continuing interest, enthusiasm, and dedication provided the encouragement that made this project possible. We also wish to thank Adrian Banner and Jose Luis Rodrigo for their special help in running the courses, and their efforts to see that the students got the most from each class. In addition, Adrian Banner also made valuable suggestions that are incorporated in the text.

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FOREWORD

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ix

We wish also to record a note of special thanks for the following individuals: Charles Fefferman, who taught the first week (successfully launching the whole project!); Paul Hagelstein, who in addition to reading part of the manuscript taught several weeks of one of the courses, and has since taken over the teaching of the second round of the series; and Daniel Levine, who gave valuable help in proof-reading. Last but not least, our thanks go to Gerree Pecht, for her consummate skill in typesetting and for the time and energy she spent in the preparation of all aspects of the lectures, such as transparencies, notes, and the manuscript. We are also happy to acknowledge our indebtedness for the support we received from the 250th Anniversary Fund of Princeton University, and the National Science Foundation’s VIGRE program. Elias M. Stein Rami Shakarchi Princeton, New Jersey August 2002

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Preface to Book I Any effort to present an overall view of analysis must at its start deal with the following questions: Where does one begin? What are the initial subjects to be treated, and in what order are the relevant concepts and basic techniques to be developed? Our answers to these questions are guided by our view of the centrality of Fourier analysis, both in the role it has played in the development of the subject, and in the fact that its ideas permeate much of the presentday analysis. For these reasons we have devoted this first volume to an exposition of some basic facts about Fourier series, taken together with a study of elements of Fourier transforms and finite Fourier analysis. Starting this way allows one to see rather easily certain applications to other sciences, together with the link to such topics as partial differential equations and number theory. In later volumes several of these connections will be taken up from a more systematic point of view, and the ties that exist with complex analysis, real analysis, Hilbert space theory, and other areas will be explored further. In the same spirit, we have been mindful not to overburden the beginning student with some of the difficulties that are inherent in the subject: a proper appreciation of the subtleties and technical complications that arise can come only after one has mastered some of the initial ideas involved. This point of view has led us to the following choice of material in the present volume: • Fourier series. At this early stage it is not appropriate to introduce measure theory and Lebesgue integration. For this reason our treatment of Fourier series in the first four chapters is carried out in the context of Riemann integrable functions. Even with this restriction, a substantial part of the theory can be developed, detailing convergence and summability; also, a variety of connections with other problems in mathematics can be illustrated. • Fourier transform. For the same reasons, instead of undertaking the theory in a general setting, we confine ourselves in Chapters 5 and 6 largely to the framework of test functions. Despite these limitations, we can learn a number of basic and interesting facts about Fourier analysis in Rd and its relation to other areas, including the wave equation and the Radon transform.

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PREFACE TO BOOK I

• Finite Fourier analysis. This is an introductory subject par excellence, because limits and integrals are not explicitly present. Nevertheless, the subject has several striking applications, including the proof of the infinitude of primes in arithmetic progression. Taking into account the introductory nature of this first volume, we have kept the prerequisites to a minimum. Although we suppose some acquaintance with the notion of the Riemann integral, we provide an appendix that contains most of the results about integration needed in the text. We hope that this approach will facilitate the goal that we have set for ourselves: to inspire the interested reader to learn more about this fascinating subject, and to discover how Fourier analysis affects decisively other parts of mathematics and science.

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Contents Foreword

vii

Preface

xi

Chapter 1. The Genesis of Fourier Analysis 1

2

3 4

The vibrating string 1.1 Derivation of the wave equation 1.2 Solution to the wave equation 1.3 Example: the plucked string The heat equation 2.1 Derivation of the heat equation 2.2 Steady-state heat equation in the disc Exercises Problem

Chapter 2. Basic Properties of Fourier Series 1 2 3 4 5

6 7

Examples and formulation of the problem 1.1 Main definitions and some examples Uniqueness of Fourier series Convolutions Good kernels Ces`aro and Abel summability: applications to Fourier series 5.1 Ces`aro means and summation 5.2 Fej´er’s theorem 5.3 Abel means and summation 5.4 The Poisson kernel and Dirichlet’s problem in the unit disc Exercises Problems

Chapter 3. Convergence of Fourier Series 1

2

Mean-square convergence of Fourier series 1.1 Vector spaces and inner products 1.2 Proof of mean-square convergence Return to pointwise convergence 2.1 A local result 2.2 A continuous function with diverging Fourier series

1 2 6 8 16 18 18 19 22 27 29 30 34 39 44 48 51 51 52 54 55 58 65 69 70 70 76 81 81 83

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xiv

CONTENTS

3 Exercises 4 Problems Chapter 4. Some Applications of Fourier Series 1 2 3 4 5 6

The isoperimetric inequality Weyl’s equidistribution theorem A continuous but nowhere differentiable function The heat equation on the circle Exercises Problems

Chapter 5. The Fourier Transform on R 1 Elementary theory of the Fourier transform 1.1 Integration of functions on the real line 1.2 Definition of the Fourier transform 1.3 The Schwartz space 1.4 The Fourier transform on S 1.5 The Fourier inversion 1.6 The Plancherel formula 1.7 Extension to functions of moderate decrease 1.8 The Weierstrass approximation theorem 2 Applications to some partial differential equations 2.1 The time-dependent heat equation on the real line 2.2 The steady-state heat equation in the upper halfplane 3 The Poisson summation formula 3.1 Theta and zeta functions 3.2 Heat kernels 3.3 Poisson kernels 4 The Heisenberg uncertainty principle 5 Exercises 6 Problems Chapter 6. The Fourier Transform on Rd 1 Preliminaries 1.1 Symmetries 1.2 Integration on Rd 2 Elementary theory of the Fourier transform 3 The wave equation in Rd × R 3.1 Solution in terms of Fourier transforms 3.2 The wave equation in R3 × R

87 95 100 101 105 113 118 120 125 129 131 131 134 134 136 140 142 144 144 145 145 149 153 155 156 157 158 161 169 175 176 176 178 180 184 184 189

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CONTENTS

4 5

6 7

3.3 The wave equation in R2 × R: descent Radial symmetry and Bessel functions The Radon transform and some of its applications 5.1 The X-ray transform in R2 5.2 The Radon transform in R3 5.3 A note about plane waves Exercises Problems

Chapter 7. Finite Fourier Analysis 1

2

3 4

Fourier analysis on Z(N ) 1.1 The group Z(N ) 1.2 Fourier inversion theorem and Plancherel identity on Z(N ) 1.3 The fast Fourier transform Fourier analysis on finite abelian groups 2.1 Abelian groups 2.2 Characters 2.3 The orthogonality relations 2.4 Characters as a total family 2.5 Fourier inversion and Plancherel formula Exercises Problems

Chapter 8. Dirichlet’s Theorem 1

2

3

4 5

A little elementary number theory 1.1 The fundamental theorem of arithmetic 1.2 The infinitude of primes Dirichlet’s theorem 2.1 Fourier analysis, Dirichlet characters, and reduction of the theorem 2.2 Dirichlet L-functions Proof of the theorem 3.1 Logarithms 3.2 L-functions 3.3 Non-vanishing of the L-function Exercises Problems

Appendix: Integration 1

Definition of the Riemann integral

xv 194 196 198 199 201 207 207 212 218 219 219 221 224 226 226 230 232 233 235 236 239 241 241 241 244 252 254 255 258 258 261 265 275 279 281 281

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CONTENTS

1.1 Basic properties 1.2 Sets of measure zero and discontinuities of integrable functions 2 Multiple integrals 2.1 The Riemann integral in Rd 2.2 Repeated integrals 2.3 The change of variables formula 2.4 Spherical coordinates 3 Improper integrals. Integration over Rd 3.1 Integration of functions of moderate decrease 3.2 Repeated integrals 3.3 Spherical coordinates

282 286 289 289 291 292 293 294 294 295 297

Notes and References

298

Bibliography

300

Symbol Glossary

303

Index

305

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1 The Genesis of Fourier Analysis Regarding the researches of d’Alembert and Euler could one not add that if they knew this expansion, they made but a very imperfect use of it. They were both persuaded that an arbitrary and discontinuous function could never be resolved in series of this kind, and it does not even seem that anyone had developed a constant in cosines of multiple arcs, the first problem which I had to solve in the theory of heat. J. Fourier, 1808-9

In the beginning, it was the problem of the vibrating string, and the later investigation of heat flow, that led to the development of Fourier analysis. The laws governing these distinct physical phenomena were expressed by two different partial differential equations, the wave and heat equations, and these were solved in terms of Fourier series. Here we want to start by describing in some detail the development of these ideas. We will do this initially in the context of the problem of the vibrating string, and we will proceed in three steps. First, we describe several physical (empirical) concepts which motivate corresponding mathematical ideas of importance for our study. These are: the role of the functions cos t, sin t, and eit suggested by simple harmonic motion; the use of separation of variables, derived from the phenomenon of standing waves; and the related concept of linearity, connected to the superposition of tones. Next, we derive the partial differential equation which governs the motion of the vibrating string. Finally, we will use what we learned about the physical nature of the problem (expressed mathematically) to solve the equation. In the last section, we use the same approach to study the problem of heat diffusion. Given the introductory nature of this chapter and the subject matter covered, our presentation cannot be based on purely mathematical reasoning. Rather, it proceeds by plausibility arguments and aims to provide the motivation for the further rigorous analysis in the succeeding chapters. The impatient reader who wishes to begin immediately with the theorems of the subject may prefer to pass directly to the next chapter.

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1 The vibrating string The problem consists of the study of the motion of a string fixed at its end points and allowed to vibrate freely. We have in mind physical systems such as the strings of a musical instrument. As we mentioned above, we begin with a brief description of several observable physical phenomena on which our study is based. These are: • simple harmonic motion, • standing and traveling waves, • harmonics and superposition of tones. Understanding the empirical facts behind these phenomena will motivate our mathematical approach to vibrating strings. Simple harmonic motion Simple harmonic motion describes the behavior of the most basic oscillatory system (called the simple harmonic oscillator), and is therefore a natural place to start the study of vibrations. Consider a mass {m} attached to a horizontal spring, which itself is attached to a fixed wall, and assume that the system lies on a frictionless surface. Choose an axis whose origin coincides with the center of the mass when it is at rest (that is, the spring is neither stretched nor compressed), as shown in Figure 1. When the mass is displaced from its initial equilibrium

m

m

y

0

y

0

y(t)

Figure 1. Simple harmonic oscillator

position and then released, it will undergo simple harmonic motion. This motion can be described mathematically once we have found the differential equation that governs the movement of the mass. Let y(t) denote the displacement of the mass at time t. We assume that the spring is ideal, in the sense that it satisfies Hooke’s law: the restoring force F exerted by the spring on the mass is given by F = −ky(t). Here

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1. The vibrating string

k > 0 is a given physical quantity called the spring constant. Applying Newton’s law (force = mass × acceleration), we obtain −ky(t) = my 00 (t), 00 where we use the notation p y to denote the second derivative of y with respect to t. With c = k/m, this second order ordinary differential equation becomes

(1)

y 00 (t) + c2 y(t) = 0.

The general solution of equation (1) is given by y(t) = a cos ct + b sin ct , where a and b are constants. Clearly, all functions of this form solve equation (1), and Exercise 6 outlines a proof that these are the only (twice differentiable) solutions of that differential equation. In the above expression for y(t), the quantity c is given, but a and b can be any real numbers. In order to determine the particular solution of the equation, we must impose two initial conditions in view of the two unknown constants a and b. For example, if we are given y(0) and y 0 (0), the initial position and velocity of the mass, then the solution of the physical problem is unique and given by y(t) = y(0) cos ct +

y 0 (0) sin ct . c

One can easily verify that there exist constants A > 0 and ϕ ∈ R such that a cos ct + b sin ct = A cos(ct − ϕ). √ Because of the physical interpretation given above, one calls A = a2 + b2 the “amplitude” of the motion, c its “natural frequency,” ϕ its “phase” (uniquely determined up to an integer multiple of 2π), and 2π/c the “period” of the motion. The typical graph of the function A cos(ct − ϕ), illustrated in Figure 2, exhibits a wavelike pattern that is obtained from translating and stretching (or shrinking) the usual graph of cos t. We make two observations regarding our examination of simple harmonic motion. The first is that the mathematical description of the most elementary oscillatory system, namely simple harmonic motion, involves

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Figure 2. The graph of A cos(ct − ϕ)

the most basic trigonometric functions cos t and sin t. It will be important in what follows to recall the connection between these functions and complex numbers, as given in Euler’s identity eit = cos t + i sin t. The second observation is that simple harmonic motion is determined as a function of time by two initial conditions, one determining the position, and the other the velocity (specified, for example, at time t = 0). This property is shared by more general oscillatory systems, as we shall see below. Standing and traveling waves As it turns out, the vibrating string can be viewed in terms of onedimensional wave motions. Here we want to describe two kinds of motions that lend themselves to simple graphic representations. • First, we consider standing waves. These are wavelike motions described by the graphs y = u(x, t) developing in time t as shown in Figure 3. In other words, there is an initial profile y = ϕ(x) representing the wave at time t = 0, and an amplifying factor ψ(t), depending on t, so that y = u(x, t) with u(x, t) = ϕ(x)ψ(t). The nature of standing waves suggests the mathematical idea of “separation of variables,” to which we will return later. • A second type of wave motion that is often observed in nature is that of a traveling wave. Its description is particularly simple:

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1. The vibrating string

u(x, 0) = ϕ(x)

y

u(x, t0 )

x

Figure 3. A standing wave at different moments in time: t = 0 and t = t0

there is an initial profile F (x) so that u(x, t) equals F (x) when t = 0. As t evolves, this profile is displaced to the right by ct units, where c is a positive constant, namely u(x, t) = F (x − ct). Graphically, the situation is depicted in Figure 4.

F (x)

F (x − ct0 )

Figure 4. A traveling wave at two different moments in time: t = 0 and t = t0

Since the movement in t is at the rate c, that constant represents the velocity of the wave. The function F (x − ct) is a one-dimensional traveling wave moving to the right. Similarly, u(x, t) = F (x + ct) is a one-dimensional traveling wave moving to the left.

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Harmonics and superposition of tones The final physical observation we want to mention (without going into any details now) is one that musicians have been aware of since time immemorial. It is the existence of harmonics, or overtones. The pure tones are accompanied by combinations of overtones which are primarily responsible for the timbre (or tone color) of the instrument. The idea of combination or superposition of tones is implemented mathematically by the basic concept of linearity, as we shall see below. We now turn our attention to our main problem, that of describing the motion of a vibrating string. First, we derive the wave equation, that is, the partial differential equation that governs the motion of the string. 1.1 Derivation of the wave equation Imagine a homogeneous string placed in the (x, y)-plane, and stretched along the x-axis between x = 0 and x = L. If it is set to vibrate, its displacement y = u(x, t) is then a function of x and t, and the goal is to derive the differential equation which governs this function. For this purpose, we consider the string as being subdivided into a large number N of masses (which we think of as individual particles) distributed uniformly along the x-axis, so that the nth particle has its x-coordinate at xn = nL/N . We shall therefore conceive of the vibrating string as a complex system of N particles, each oscillating in the vertical direction only; however, unlike the simple harmonic oscillator we considered previously, each particle will have its oscillation linked to its immediate neighbor by the tension of the string.

yn−1

xn−1

yn+1

yn

xn

xn+1 h

Figure 5. A vibrating string as a discrete system of masses

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1. The vibrating string

We then set yn (t) = u(xn , t), and note that xn+1 − xn = h, with h = L/N . If we assume that the string has constant density ρ > 0, it is reasonable to assign mass equal to ρh to each particle. By Newton’s law, ρhyn00 (t) equals the force acting on the nth particle. We now make the simple assumption that this force is due to the effect of the two nearby particles, the ones with x-coordinates at xn−1 and xn+1 (see Figure 5). We further assume that the force (or tension) coming from the right of the nth particle is proportional to (yn+1 − yn )/h, where h is the distance between xn+1 and xn ; hence we can write the tension as ³τ ´ (yn+1 − yn ), h where τ > 0 is a constant equal to the coefficient of tension of the string. There is a similar force coming from the left, and it is ³τ ´ (yn−1 − yn ). h Altogether, adding these forces gives us the desired relation between the oscillators yn (t), namely (2)

ρhyn00 (t) =

τ {yn+1 (t) + yn−1 (t) − 2yn (t)} . h

On the one hand, with the notation chosen above, we see that yn+1 (t) + yn−1 (t) − 2yn (t) = u(xn + h, t) + u(xn − h, t) − 2u(xn , t). On the other hand, for any reasonable function F (x) (that is, one that has continuous second derivatives) we have F (x + h) + F (x − h) − 2F (x) → F 00 (x) h2

as h → 0.

Thus we may conclude, after dividing by h in (2) and letting h tend to zero (that is, N goes to infinity), that ρ

∂2u ∂2u =τ , 2 ∂t ∂x2

or 1 ∂2u ∂2u = , c2 ∂t2 ∂x2

with c =

p

τ /ρ.

This relation is known as the one-dimensional wave equation, or more simply as the wave equation. For reasons that will be apparent later, the coefficient c > 0 is called the velocity of the motion.

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In connection with this partial differential equation, we make an important simplifying mathematical remark. This has to do with scaling, or in the language of physics, a “change of units.” That is, we can think of the coordinate x as x = aX where a is an appropriate positive constant. Now, in terms of the new coordinate X, the interval 0 ≤ x ≤ L becomes 0 ≤ X ≤ L/a. Similarly, we can replace the time coordinate t by t = bT , where b is another positive constant. If we set U (X, T ) = u(x, t), then ∂U ∂u =a , ∂X ∂x

2 ∂2U 2∂ u = a , ∂X 2 ∂x2

and similarly for the derivatives in t. So if we choose a and b appropriately, we can transform the one-dimensional wave equation into ∂2U ∂2U = , 2 ∂T ∂X 2 which has the effect of setting the velocity c equal to 1. Moreover, we have the freedom to transform the interval 0 ≤ x ≤ L to 0 ≤ X ≤ π. (We shall see that the choice of π is convenient in many circumstances.) All this is accomplished by taking a = L/π and b = L/(cπ). Once we solve the new equation, we can of course return to the original equation by making the inverse change of variables. Hence, we do not sacrifice generality by thinking of the wave equation as given on the interval [0, π] with velocity c = 1. 1.2 Solution to the wave equation Having derived the equation for the vibrating string, we now explain two methods to solve it: • using traveling waves, • using the superposition of standing waves. While the first approach is very simple and elegant, it does not directly give full insight into the problem; the second method accomplishes that, and moreover is of wide applicability. It was first believed that the second method applied only in the simple cases where the initial position and velocity of the string were themselves given as a superposition of standing waves. However, as a consequence of Fourier’s ideas, it became clear that the problem could be worked either way for all initial conditions.

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1. The vibrating string

Traveling waves To simplify matters as before, we assume that c = 1 and L = π, so that the equation we wish to solve becomes ∂2u ∂2u = ∂t2 ∂x2

on 0 ≤ x ≤ π.

The crucial observation is the following: if F is any twice differentiable function, then u(x, t) = F (x + t) and u(x, t) = F (x − t) solve the wave equation. The verification of this is a simple exercise in differentiation. Note that the graph of u(x, t) = F (x − t) at time t = 0 is simply the graph of F , and that at time t = 1 it becomes the graph of F translated to the right by 1. Therefore, we recognize that F (x − t) is a traveling wave which travels to the right with speed 1. Similarly, u(x, t) = F (x + t) is a wave traveling to the left with speed 1. These motions are depicted in Figure 6.

F (x + t)

F (x)

F (x − t)

Figure 6. Waves traveling in both directions

Our discussion of tones and their combinations leads us to observe that the wave equation is linear. This means that if u(x, t) and v(x, t) are particular solutions, then so is αu(x, t) + βv(x, t), where α and β are any constants. Therefore, we may superpose two waves traveling in opposite directions to find that whenever F and G are twice differentiable functions, then u(x, t) = F (x + t) + G(x − t) is a solution of the wave equation. In fact, we now show that all solutions take this form. We drop for the moment the assumption that 0 ≤ x ≤ π, and suppose that u is a twice differentiable function which solves the wave equation

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for all real x and t. Consider the following new set of variables ξ = x + t, η = x − t, and define v(ξ, η) = u(x, t). The change of variables formula shows that v satisfies ∂2v = 0. ∂ξ∂η Integrating this relation twice gives v(ξ, η) = F (ξ) + G(η), which then implies u(x, t) = F (x + t) + G(x − t), for some functions F and G. We must now connect this result with our original problem, that is, the physical motion of a string. There, we imposed the restrictions 0 ≤ x ≤ π, the initial shape of the string u(x, 0) = f (x), and also the fact that the string has fixed end points, namely u(0, t) = u(π, t) = 0 for all t. To use the simple observation above, we first extend f to all of R by making it odd1 on [−π, π], and then periodic2 in x of period 2π, and similarly for u(x, t), the solution of our problem. Then the extension u solves the wave equation on all of R, and u(x, 0) = f (x) for all x ∈ R. Therefore, u(x, t) = F (x + t) + G(x − t), and setting t = 0 we find that F (x) + G(x) = f (x). Since many choices of F and G will satisfy this identity, this suggests imposing another initial condition on u (similar to the two initial conditions in the case of simple harmonic motion), namely the initial velocity of the string which we denote by g(x): ∂u (x, 0) = g(x), ∂t where of course g(0) = g(π) = 0. Again, we extend g to R first by making it odd over [−π, π], and then periodic of period 2π. The two initial conditions of position and velocity now translate into the following system: ½ F (x) + G(x) = f (x) , F 0 (x) − G0 (x) = g(x) . 1 A function f defined on a set U is odd if −x ∈ U whenever x ∈ U and f (−x) = −f (x), and even if f (−x) = f (x). 2 A function f on R is periodic of period ω if f (x + ω) = f (x) for all x.

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1. The vibrating string

Differentiating the first equation and adding it to the second, we obtain 2F 0 (x) = f 0 (x) + g(x). Similarly 2G0 (x) = f 0 (x) − g(x), and hence there are constants C1 and C2 so that · ¸ Z x 1 f (x) + g(y) dy + C1 F (x) = 2 0 and

· ¸ Z x 1 G(x) = f (x) − g(y) dy + C2 . 2 0

Since F (x) + G(x) = f (x) we conclude that C1 + C2 = 0, and therefore, our final solution of the wave equation with the given initial conditions takes the form Z 1 1 x+t g(y) dy. u(x, t) = [f (x + t) + f (x − t)] + 2 2 x−t The form of this solution is known as d’Alembert’s formula. Observe that the extensions we chose for f and g guarantee that the string always has fixed ends, that is, u(0, t) = u(π, t) = 0 for all t. A final remark is in order. The passage from t ≥ 0 to t ∈ R, and then back to t ≥ 0, which was made above, exhibits the time reversal property of the wave equation. In other words, a solution u to the wave equation for t ≥ 0, leads to a solution u− defined for negative time t < 0 simply by setting u− (x, t) = u(x, −t), a fact which follows from the invariance of the wave equation under the transformation t 7→ −t. The situation is quite different in the case of the heat equation. Superposition of standing waves We turn to the second method of solving the wave equation, which is based on two fundamental conclusions from our previous physical observations. By our considerations of standing waves, we are led to look for special solutions to the wave equation which are of the form ϕ(x)ψ(t). This procedure, which works equally well in other contexts (in the case of the heat equation, for instance), is called separation of variables and constructs solutions that are called pure tones. Then by the linearity

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of the wave equation, we can expect to combine these pure tones into a more complex combination of sound. Pushing this idea further, we can hope ultimately to express the general solution of the wave equation in terms of sums of these particular solutions. Note that one side of the wave equation involves only differentiation in x, while the other, only differentiation in t. This observation provides another reason to look for solutions of the equation in the form u(x, t) = ϕ(x)ψ(t) (that is, to “separate variables”), the hope being to reduce a difficult partial differential equation into a system of simpler ordinary differential equations. In the case of the wave equation, with u of the above form, we get ϕ(x)ψ 00 (t) = ϕ00 (x)ψ(t), and therefore ψ 00 (t) ϕ00 (x) = . ψ(t) ϕ(x) The key observation here is that the left-hand side depends only on t, and the right-hand side only on x. This can happen only if both sides are equal to a constant, say λ. Therefore, the wave equation reduces to the following ½

(3)

ψ 00 (t) − λψ(t) = 0 ϕ00 (x) − λϕ(x) = 0.

We focus our attention on the first equation in the above system. At this point, the reader will recognize the equation we obtained in the study of simple harmonic motion. Note that we need to consider only the case when λ < 0, since when λ ≥ 0 the solution ψ will not oscillate as time varies. Therefore, we may write λ = −m2 , and the solution of the equation is then given by ψ(t) = A cos mt + B sin mt. Similarly, we find that the solution of the second equation in (3) is ˜ sin mx. ϕ(x) = A˜ cos mx + B Now we take into account that the string is attached at x = 0 and x = π. This translates into ϕ(0) = ϕ(π) = 0, which in turn gives A˜ = 0, and ˜ 6= 0, then m must be an integer. If m = 0, the solution vanishes if B identically, and if m ≤ −1, we may rename the constants and reduce to

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1. The vibrating string

the case m ≥ 1 since the function sin y is odd and cos y is even. Finally, we arrive at the guess that for each m ≥ 1, the function um (x, t) = (Am cos mt + Bm sin mt) sin mx, which we recognize as a standing wave, is a solution to the wave equation. Note that in the above argument we divided by ϕ and ψ, which sometimes vanish, so one must actually check by hand that the standing wave um solves the equation. This straightforward calculation is left as an exercise to the reader. Before proceeding further with the analysis of the wave equation, we pause to discuss standing waves in more detail. The terminology comes from looking at the graph of um (x, t) for each fixed t. Suppose first that m = 1, and take u(x, t) = cos t sin x. Then, Figure 7 (a) gives the graph of u for different values of t.

−2π

(a)

−π

0

π

2π

−π

−2π

0

π

2π

(b)

Figure 7. Fundamental tone (a) and overtones (b) at different moments in time

The case m = 1 corresponds to the fundamental tone or first harmonic of the vibrating string. We now take m = 2 and look at u(x, t) = cos 2t sin 2x. This corresponds to the first overtone or second harmonic, and this motion is described in Figure 7 (b). Note that u(π/2, t) = 0 for all t. Such points, which remain motionless in time, are called nodes, while points whose motion has maximum amplitude are named anti-nodes. For higher values of m we get more overtones or higher harmonics. Note that as m increases, the frequency increases, and the period 2π/m

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Chapter 1. THE GENESIS OF FOURIER ANALYSIS

decreases. Therefore, the fundamental tone has a lower frequency than the overtones. We now return to the original problem. Recall that the wave equation is linear in the sense that if u and v solve the equation, so does αu + βv for any constants α and β. This allows us to construct more solutions by taking linear combinations of the standing waves um . This technique, called superposition, leads to our final guess for a solution of the wave equation (4)

u(x, t) =

∞ X

(Am cos mt + Bm sin mt) sin mx.

m=1

Note that the above sum is infinite, so that questions of convergence arise, but since most of our arguments so far are formal, we will not worry about this point now. Suppose the above expression gave all the solutions to the wave equation. If we then require that the initial position of the string at time t = 0 is given by the shape of the graph of the function f on [0, π], with of course f (0) = f (π) = 0, we would have u(x, 0) = f (x), hence ∞ X

Am sin mx = f (x).

m=1

Since the initial shape of the string can be any reasonable function f , we must ask the following basic question: Given a function f on [0, π] (with f (0) = f (π) = 0), can we find coefficients Am so that

(5)

f (x) =

∞ X

Am sin mx ?

m=1

This question is stated loosely, but a lot of our effort in the next two chapters of this book will be to formulate the question precisely and attempt to answer it. This was the basic problem that initiated the study of Fourier analysis. A simple observation allows us to guess a formula giving Am if the expansion (5) were to hold. Indeed, we multiply both sides by sin nx

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1. The vibrating string

and integrate between [0, π]; working formally, we obtain ! Z π Z π ÃX ∞ f (x) sin nx dx = Am sin mx sin nx dx 0

0

∞ X

=

m=1

Z

Am

m=1

π

0

sin mx sin nx dx = An ·

π , 2

where we have used the fact that ½ Z π 0 if m 6= n, sin mx sin nx dx = π/2 if m = n. 0 Therefore, the guess for An , called the nth Fourier sine coefficient of f , is Z 2 π f (x) sin nx dx. (6) An = π 0 We shall return to this formula, and other similar ones, later. One can transform the question about Fourier sine series on [0, π] to a more general question on the interval [−π, π]. If we could express f on [0, π] in terms of a sine series, then this expansion would also hold on [−π, π] if we extend f to this interval by making it odd. Similarly, one can ask if an even function g(x) on [−π, π] can be expressed as a cosine series, namely ∞ X

g(x) =

A0m cos mx.

m=0

More generally, since an arbitrary function F on [−π, π] can be expressed as f + g, where f is odd and g is even,3 we may ask if F can be written as F (x) =

∞ X

Am sin mx +

m=1

∞ X

A0m cos mx,

m=0

or by applying Euler’s identity eix = cos x + i sin x, we could hope that F takes the form F (x) =

∞ X

am eimx .

m=−∞

3 Take,

for example, f (x) = [F (x) − F (−x)]/2 and g(x) = [F (x) + F (−x)]/2.

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By analogy with (6), we can use the fact that ½ Z π 1 0 if n 6= m eimx e−inx dx = 1 if n = m, 2π −π to see that one expects that 1 an = 2π

Z

π

F (x)e−inx dx.

−π

The quantity an is called the nth Fourier coefficient of F . We can now reformulate the problem raised above: Question: Given any reasonable function F on [−π, π], with Fourier coefficients defined above, is it true that

(7)

F (x) =

∞ X

am eimx ?

m=−∞

This formulation of the problem, in terms of complex exponentials, is the form we shall use the most in what follows. Joseph Fourier (1768-1830) was the first to believe that an “arbitrary” function F could be given as a series (7). In other words, his idea was that any function is the linear combination (possibly infinite) of the most basic trigonometric functions sin mx and cos mx, where m ranges over the integers.4 Although this idea was implicit in earlier work, Fourier had the conviction that his predecessors lacked, and he used it in his study of heat diffusion; this began the subject of “Fourier analysis.” This discipline, which was first developed to solve certain physical problems, has proved to have many applications in mathematics and other fields as well, as we shall see later. We return to the wave equation. To formulate the problem correctly, we must impose two initial conditions, as our experience with simple harmonic motion and traveling waves indicated. The conditions assign the initial position and velocity of the string. That is, we require that u satisfy the differential equation and the two conditions u(x, 0) = f (x)

and

∂u (x, 0) = g(x), ∂t

4 The first proof that a general class of functions can be represented by Fourier series was given later by Dirichlet; see Problem 6, Chapter 4.

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1. The vibrating string

where f and g are pre-assigned functions. Note that this is consistent with (4) in that this requires that f and g be expressible as f (x) =

∞ X

Am sin mx

and

g(x) =

m=1

∞ X

mBm sin mx.

m=1

1.3 Example: the plucked string We now apply our reasoning to the particular problem of the plucked string. For simplicity we choose units so that the string is taken on the interval [0, π], and it satisfies the wave equation with c = 1. The string is assumed to be plucked to height h at the point p with 0 < p < π; this is the initial position. That is, we take as our initial position the triangular shape given by  xh      p

f (x) =

for 0 ≤ x ≤ p

  h(π − x)   

π−p

for p ≤ x ≤ π,

which is depicted in Figure 8.

h

0

p

π

Figure 8. Initial position of a plucked string

We also choose an initial velocity g(x) identically equal to 0. Then, we can compute the Fourier coefficients of f (Exercise 9), and assuming that the answer to the question raised before (5) is positive, we obtain f (x) =

∞ X m=1

Am sin mx

with

Am =

2h sin mp . m2 p(π − p)

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Thus (8)

u(x, t) =

∞ X

Am cos mt sin mx,

m=1

and note that this series converges absolutely. The solution can also be expressed in terms of traveling waves. In fact (9)

u(x, t) =

f (x + t) + f (x − t) . 2

Here f (x) is defined for all x as follows: first, f is extended to [−π, π] by making it odd, and then f is extended to the whole real line by making it periodic of period 2π, that is, f (x + 2πk) = f (x) for all integers k. Observe that (8) implies (9) in view of the trigonometric identity cos v sin u =

1 [sin(u + v) + sin(u − v)]. 2

As a final remark, we should note an unsatisfactory aspect of the solution to this problem, which however is in the nature of things. Since the initial data f (x) for the plucked string is not twice continuously differentiable, neither is the function u (given by (9)). Hence u is not truly a solution of the wave equation: while u(x, t) does represent the position of the plucked string, it does not satisfy the partial differential equation we set out to solve! This state of affairs may be understood properly only if we realize that u does solve the equation, but in an appropriate generalized sense. A better understanding of this phenomenon requires ideas relevant to the study of “weak solutions” and the theory of “distributions.” These topics we consider only later, in Books III and IV.

2 The heat equation We now discuss the problem of heat diffusion by following the same framework as for the wave equation. First, we derive the time-dependent heat equation, and then study the steady-state heat equation in the disc, which leads us back to the basic question (7). 2.1 Derivation of the heat equation Consider an infinite metal plate which we model as the plane R2 , and suppose we are given an initial heat distribution at time t = 0. Let the temperature at the point (x, y) at time t be denoted by u(x, y, t).

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2. The heat equation

Consider a small square centered at (x0 , y0 ) with sides parallel to the axis and of side length h, as shown in Figure 9. The amount of heat energy in S at time t is given by ZZ

H(t) = σ

u(x, y, t) dx dy , S

where σ > 0 is a constant called the specific heat of the material. Therefore, the heat flow into S is ∂H =σ ∂t

ZZ S

∂u dx dy , ∂t

which is approximately equal to σh2

∂u (x0 , y0 , t), ∂t

since the area of S is h2 . Now we apply Newton’s law of cooling, which states that heat flows from the higher to lower temperature at a rate proportional to the difference, that is, the gradient. h

h (x0 , y0 )

(x0 + h/2, y0 )

Figure 9. Heat flow through a small square

The heat flow through the vertical side on the right is therefore −κh

∂u (x0 + h/2, y0 , t) , ∂x

where κ > 0 is the conductivity of the material. A similar argument for the other sides shows that the total heat flow through the square S is

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Chapter 1. THE GENESIS OF FOURIER ANALYSIS

given by "

κh

∂u ∂u (x0 + h/2, y0 , t) − (x0 − h/2, y0 , t) ∂x ∂x # ∂u ∂u + (x0 , y0 + h/2, t) − (x0 , y0 − h/2, t) . ∂y ∂y

Applying the mean value theorem and letting h tend to zero, we find that σ ∂u ∂2u ∂2u = + 2; κ ∂t ∂x2 ∂y this is called the time-dependent heat equation, often abbreviated to the heat equation. 2.2 Steady-state heat equation in the disc After a long period of time, there is no more heat exchange, so that the system reaches thermal equilibrium and ∂u/∂t = 0. In this case, the time-dependent heat equation reduces to the steady-state heat equation (10)

∂2u ∂2u + 2 = 0. ∂x2 ∂y

The operator ∂ 2 /∂x2 + ∂ 2 /∂y 2 is of such importance in mathematics and physics that it is often abbreviated as 4 and given a name: the Laplace operator or Laplacian. So the steady-state heat equation is written as 4u = 0, and solutions to this equation are called harmonic functions. Consider the unit disc in the plane D = {(x, y) ∈ R2 : x2 + y 2 < 1}, whose boundary is the unit circle C. In polar coordinates (r, θ), with 0 ≤ r and 0 ≤ θ < 2π, we have D = {(r, θ) : 0 ≤ r < 1}

and

C = {(r, θ) : r = 1}.

The problem, often called the Dirichlet problem (for the Laplacian on the unit disc), is to solve the steady-state heat equation in the unit

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2. The heat equation

disc subject to the boundary condition u = f on C. This corresponds to fixing a predetermined temperature distribution on the circle, waiting a long time, and then looking at the temperature distribution inside the disc. y u(1, θ) = f (θ)

x

0

4u = 0

Figure 10. The Dirichlet problem for the disc

While the method of separation of variables will turn out to be useful for equation (10), a difficulty comes from the fact that the boundary condition is not easily expressed in terms of rectangular coordinates. Since this boundary condition is best described by the coordinates (r, θ), namely u(1, θ) = f (θ), we rewrite the Laplacian in polar coordinates. An application of the chain rule gives (Exercise 10): 4u =

∂ 2 u 1 ∂u 1 ∂2u + + . ∂r2 r ∂r r2 ∂θ2

We now multiply both sides by r2 , and since 4u = 0, we get r2

∂2u ∂u ∂2u + r = − . ∂r2 ∂r ∂θ2

Separating these variables, and looking for a solution of the form u(r, θ) = F (r)G(θ), we find r2 F 00 (r) + rF 0 (r) G00 (θ) =− . F (r) G(θ)

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Since the two sides depend on different variables, they must both be constant, say equal to λ. We therefore get the following equations: ½ 00 G (θ) + λG(θ) = 0 , r2 F 00 (r) + rF 0 (r) − λF (r) = 0. Since G must be periodic of period 2π, this implies that λ ≥ 0 and (as we have seen before) that λ = m2 where m is an integer; hence ˜ sin mθ. G(θ) = A˜ cos mθ + B An application of Euler’s identity, eix = cos x + i sin x, allows one to rewrite G in terms of complex exponentials, G(θ) = Aeimθ + Be−imθ . With λ = m2 and m 6= 0, two simple solutions of the equation in F are F (r) = rm and F (r) = r−m (Exercise 11 gives further information about these solutions). If m = 0, then F (r) = 1 and F (r) = log r are two solutions. If m > 0, we note that r−m grows unboundedly large as r tends to zero, so F (r)G(θ) is unbounded at the origin; the same occurs when m = 0 and F (r) = log r. We reject these solutions as contrary to our intuition. Therefore, we are left with the following special functions: um (r, θ) = r|m| eimθ ,

m ∈ Z.

We now make the important observation that (10) is linear , and so as in the case of the vibrating string, we may superpose the above special solutions to obtain the presumed general solution: u(r, θ) =

∞ X

am r|m| eimθ .

m=−∞

If this expression gave all the solutions to the steady-state heat equation, then for a reasonable f we should have ∞ X

u(1, θ) =

am eimθ = f (θ).

m=−∞

We therefore ask again in this context: given any reasonable function f on [0, 2π] with f (0) = f (2π), can we find coefficients am so that f (θ) =

∞ X m=−∞

am eimθ ?

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3. Exercises

Historical Note: D’Alembert (in 1747) first solved the equation of the vibrating string using the method of traveling waves. This solution was elaborated by Euler a year later. In 1753, D. Bernoulli proposed the solution which for all intents and purposes is the Fourier series given by (4), but Euler was not entirely convinced of its full generality, since this could hold only if an “arbitrary” function could be expanded in Fourier series. D’Alembert and other mathematicians also had doubts. This viewpoint was changed by Fourier (in 1807) in his study of the heat equation, where his conviction and work eventually led others to a complete proof that a general function could be represented as a Fourier series.

3 Exercises 1. If z = x + iy is a complex number with x, y ∈ R, we define |z| = (x2 + y 2 )1/2 and call this quantity the modulus or absolute value of z. (a) What is the geometric interpretation of |z|? (b) Show that if |z| = 0, then z = 0. (c) Show that if λ ∈ R, then |λz| = |λ||z|, where |λ| denotes the standard absolute value of a real number. (d) If z1 and z2 are two complex numbers, prove that |z1 z2 | = |z1 ||z2 |

and

|z1 + z2 | ≤ |z1 | + |z2 |.

(e) Show that if z 6= 0, then |1/z| = 1/|z|.

2. If z = x + iy is a complex number with x, y ∈ R, we define the complex conjugate of z by z = x − iy. (a) What is the geometric interpretation of z? (b) Show that |z|2 = zz. (c) Prove that if z belongs to the unit circle, then 1/z = z.

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3. A sequence of complex numbers {wn }∞ n=1 is said to converge if there exists w ∈ C such that lim |wn − w| = 0,

n→∞

and we say that w is a limit of the sequence. (a) Show that a converging sequence of complex numbers has a unique limit. The sequence {wn }∞ n=1 is said to be a Cauchy sequence if for every ² > 0 there exists a positive integer N such that |wn − wm | < ²

whenever n, m > N .

(b) Prove that a sequence of complex numbers converges if and only if it is a Cauchy sequence. [Hint: A similar theorem exists for the convergence of a sequence of real numbers. Why does it carry over to sequences of complex numbers?]

P∞

A series n=1 zn of complex numbers is said to converge if the sequence formed by the partial sums SN =

N X

zn

n=1

converges.PLet {an }∞ n=1 be a sequence of non-negative real numbers such that the series n an converges. (c) Show that if {zn }∞ of complex numbers satisfying n=1 is a sequence P |zn | ≤ an for all n, then the series n zn converges. [Hint: Use the Cauchy criterion.] 4. For z ∈ C, we define the complex exponential by ez =

∞ X zn n=0

n!

.

(a) Prove that the above definition makes sense, by showing that the series converges for every complex number z. Moreover, show that the convergence is uniform5 on every bounded subset of C. (b) If z1 , z2 are two complex numbers, prove that ez1 ez2 = ez1 +z2 . [Hint: Use the binomial theorem to expand (z1 + z2 )n , as well as the formula for the binomial coefficients.] 5 A sequence of functions {f (z)}∞ n n=1 is said to be uniformly convergent on a set S if there exists a function f on S so that for every ² > 0 there is an integer N such that |fn (z) − f (z)| < ² whenever n > N and z ∈ S.

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3. Exercises

(c) Show that if z is purely imaginary, that is, z = iy with y ∈ R, then eiy = cos y + i sin y. This is Euler’s identity. [Hint: Use power series.] (d) More generally, ex+iy = ex (cos y + i sin y) whenever x, y ∈ R, and show that |ex+iy | = ex . (e) Prove that ez = 1 if and only if z = 2πki for some integer k. (f) Show that every complex number z = x + iy can be written in the form z = reiθ , where r is unique and in the range 0 ≤ r < ∞, and θ ∈ R is unique up to an integer multiple of 2π. Check that r = |z|

and

θ = arctan(y/x)

whenever these formulas make sense. (g) In particular, i = eiπ/2 . What is the geometric meaning of multiplying a complex number by i? Or by eiθ for any θ ∈ R? (h) Given θ ∈ R, show that cos θ =

eiθ + e−iθ 2

and

sin θ =

eiθ − e−iθ . 2i

These are also called Euler’s identities. (i) Use the complex exponential to derive trigonometric identities such as cos(θ + ϑ) = cos θ cos ϑ − sin θ sin ϑ, and then show that 2 sin θ sin ϕ 2 sin θ cos ϕ

= =

cos(θ − ϕ) − cos(θ + ϕ) , sin(θ + ϕ) + sin(θ − ϕ).

This calculation connects the solution given by d’Alembert in terms of traveling waves and the solution in terms of superposition of standing waves.

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5. Verify that f (x) = einx is periodic with period 2π and that 1 2π

Z

½

π inx

e

dx =

−π

1 0

if n = 0, if n = 6 0.

Use this fact to prove that if n, m ≥ 1 we have 1 π

Z

½

π

cos nx cos mx dx = −π

0 1

if n 6= m, n = m,

0 1

if n 6= m, n = m.

and similarly 1 π

Z

½

π

sin nx sin mx dx = −π

Finally, show that

Z

π

sin nx cos mx dx = 0

for any n, m.

−π

[Hint: Calculate einx e−imx + einx eimx and einx e−imx − einx eimx .] 6. Prove that if f is a twice continuously differentiable function on R which is a solution of the equation f 00 (t) + c2 f (t) = 0, then there exist constants a and b such that f (t) = a cos ct + b sin ct. This can be done by differentiating the two functions g(t) = f (t) cos ct − c−1 f 0 (t) sin ct and h(t) = f (t) sin ct + c−1 f 0 (t) cos ct. 7. Show that if a and b are real, then one can write a cos ct + b sin ct = A cos(ct − ϕ), where A =

√

a2 + b2 , and ϕ is chosen so that cos ϕ = √

a b and sin ϕ = √ . a2 + b2 a2 + b2

8. Suppose F is a function on (a, b) with two continuous derivatives. Show that whenever x and x + h belong to (a, b), one may write F (x + h) = F (x) + hF 0 (x) +

h2 00 F (x) + h2 ϕ(h) , 2

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4. Problem

where ϕ(h) → 0 as h → 0. Deduce that F (x + h) + F (x − h) − 2F (x) → F 00 (x) h2

as h → 0.

[Hint: This is simply a Taylor expansion. It may be obtained by noting that

Z

x+h

F 0 (y) dy,

F (x + h) − F (x) = x

and then writing F 0 (y) = F 0 (x) + (y − x)F 00 (x) + (y − x)ψ(y − x), where ψ(h) → 0 as h → 0.] 9. In the case of the plucked string, use the formula for the Fourier sine coefficients to show that Am =

2h sin mp . m2 p(π − p)

For what position of p are the second, fourth, . . . harmonics missing? For what position of p are the third, sixth, . . . harmonics missing? 10. Show that the expression of the Laplacian 4=

∂2 ∂2 + 2 2 ∂x ∂y

is given in polar coordinates by the formula 4=

∂2 1 ∂ 1 ∂2 + + . ∂r2 r ∂r r2 ∂θ2

Also, prove that

¯ ¯2 ¯ ¯2 ¯ ¯2 ¯ ¯2 ¯ ¯ ¯ ∂u ¯ ¯ ¯ ¯ ¯ ¯ ¯ + ¯ ∂u ¯ = ¯ ∂u ¯ + 1 ¯ ∂u ¯ . ¯ ∂x ¯ ¯ ∂y ¯ ¯ ∂r ¯ 2 r ¯ ∂θ ¯

11. Show that if n ∈ Z the only solutions of the differential equation r2 F 00 (r) + rF 0 (r) − n2 F (r) = 0, which are twice differentiable when r > 0, are given by linear combinations of rn and r−n when n 6= 0, and 1 and log r when n = 0. [Hint: If F solves the equation, write F (r) = g(r)rn , find the equation satisfied by g, and conclude that rg 0 (r) + 2ng(r) = c where c is a constant.]

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u = f1

1

u=0

4u = 0

u=0

u = f0

0

π

Figure 11. Dirichlet problem in a rectangle

4 Problem 1. Consider the Dirichlet problem illustrated in Figure 11. More precisely, we look for a solution of the steady-state heat equation 4u = 0 in the rectangle R = {(x, y) : 0 ≤ x ≤ π, 0 ≤ y ≤ 1} that vanishes on the vertical sides of R, and so that u(x, 0) = f0 (x)

and

u(x, 1) = f1 (x) ,

where f0 and f1 are initial data which fix the temperature distribution on the horizontal sides of the rectangle. Use separation of variables to show that if f0 and f1 have Fourier expansions f0 (x) =

∞ X

Ak sin kx

and

k=1

f1 (x) =

∞ X

Bk sin kx,

k=1

then u(x, y) =

∞ µ X sinh k(1 − y) k=1

sinh k

sinh ky Ak + Bk sinh k

¶ sin kx.

We recall the definitions of the hyperbolic sine and cosine functions: sinh x =

ex − e−x 2

and

cosh x =

ex + e−x . 2

Compare this result with the solution of the Dirichlet problem in the strip obtained in Problem 3, Chapter 5.

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2 Basic Properties of Fourier Series Nearly fifty years had passed without any progress on the question of analytic representation of an arbitrary function, when an assertion of Fourier threw new light on the subject. Thus a new era began for the development of this part of Mathematics and this was heralded in a stunning way by major developments in mathematical Physics. B. Riemann, 1854

In this chapter, we begin our rigorous study of Fourier series. We set the stage by introducing the main objects in the subject, and then formulate some basic problems which we have already touched upon earlier. Our first result disposes of the question of uniqueness: Are two functions with the same Fourier coefficients necessarily equal? Indeed, a simple argument shows that if both functions are continuous, then in fact they must agree. Next, we take a closer look at the partial sums of a Fourier series. Using the formula for the Fourier coefficients (which involves an integration), we make the key observation that these sums can be written conveniently as integrals: Z 1 DN (x − y)f (y) dy, 2π where {DN } is a family of functions called the Dirichlet kernels. The above expression is the convolution of f with the function DN . Convolutions will play a critical role in our analysis. In general, given a family of functions {Kn }, we are led to investigate the limiting properties as n tends to infinity of the convolutions Z 1 Kn (x − y)f (y) dy. 2π We find that if the family {Kn } satisfies the three important properties of “good kernels,” then the convolutions above tend to f (x) as n → ∞ (at least when f is continuous). In this sense, the family {Kn } is an

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“approximation to the identity.” Unfortunately, the Dirichlet kernels DN do not belong to the category of good kernels, which indicates that the question of convergence of Fourier series is subtle. Instead of pursuing at this stage the problem of convergence, we consider various other methods of summing the Fourier series of a function. The first method, which involves averages of partial sums, leads to convolutions with good kernels, and yields an important theorem of Fej´er. From this, we deduce the fact that a continuous function on the circle can be approximated uniformly by trigonometric polynomials. Second, we may also sum the Fourier series in the sense of Abel and again encounter a family of good kernels. In this case, the results about convolutions and good kernels lead to a solution of the Dirichlet problem for the steady-state heat equation in the disc, considered at the end of the previous chapter.

1 Examples and formulation of the problem We commence with a brief description of the types of functions with which we shall be concerned. Since the Fourier coefficients of f are defined by an =

1 L

Z

L

f (x)e−2πinx/L dx,

for n ∈ Z,

0

where f is complex-valued on [0, L], it will be necessary to place some integrability conditions on f . We shall therefore assume for the remainder of this book that all functions are at least Riemann integrable.1 Sometimes it will be illuminating to focus our attention on functions that are more “regular,” that is, functions that possess certain continuity or differentiability properties. Below, we list several classes of functions in increasing order of generality. We emphasize that we will not generally restrict our attention to real-valued functions, contrary to what the following pictures may suggest; we will almost always allow functions that take values in the complex numbers C. Furthermore, we sometimes think of our functions as being defined on the circle rather than an interval. We elaborate upon this below.

1 Limiting ourselves to Riemann integrable functions is natural at this elementary stage of study of the subject. The more advanced notion of Lebesgue integrability will be taken up in Book III.

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1. Examples and formulation of the problem

Everywhere continuous functions These are the complex-valued functions f which are continuous at every point of the segment [0, L]. A typical continuous function is sketched in Figure 1 (a). We shall note later that continuous functions on the circle satisfy the additional condition f (0) = f (L). Piecewise continuous functions These are bounded functions on [0, L] which have only finitely many discontinuities. An example of such a function with simple discontinuities is pictured in Figure 1 (b). y

y

0

L

x

(a)

0

L

x

(b)

Figure 1. Functions on [0, L]: continuous and piecewise continuous

This class of functions is wide enough to illustrate many of the theorems in the next few chapters. However, for logical completeness we consider also the more general class of Riemann integrable functions. This more extended setting is natural since the formula for the Fourier coefficients involves integration. Riemann integrable functions This is the most general class of functions we will be concerned with. Such functions are bounded, but may have infinitely many discontinuities. We recall the definition of integrability. A real-valued function f defined on [0, L] is Riemann integrable (which we abbreviate as integrable2 ) if it is bounded, and if for every ² > 0, there is a subdivision 0 = x0 < x1 < · · · < xN −1 < xN = L of the interval [0, L], so that if U 2 Starting in Book III, the term “integrable” will be used in the broader sense of Lebesgue theory.

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and L are, respectively, the upper and lower sums of f for this subdivision, namely U=

N X

[

sup

j=1 xj−1 ≤x≤xj

f (x)](xj − xj−1 )

and

L=

N X j=1

[

inf

xj−1 ≤x≤xj

f (x)](xj − xj−1 ) ,

then we have U − L < ². Finally, we say that a complex-valued function is integrable if its real and imaginary parts are integrable. It is worthwhile to remember at this point that the sum and product of two integrable functions are integrable. A simple example of an integrable function on [0, 1] with infinitely many discontinuities is given by    1 if 1/(n + 1) < x ≤ 1/n and n is odd, f (x) = 0 if 1/(n + 1) < x ≤ 1/n and n is even,   0 if x = 0. This example is illustrated in Figure 2. Note that f is discontinuous when x = 1/n and at x = 0.

1

0

1 1 5 4

1 3

1 2

1

Figure 2. A Riemann integrable function

More elaborate examples of integrable functions whose discontinuities are dense in the interval [0, 1] are described in Problem 1. In general, while integrable functions may have infinitely many discontinuities, these

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33

functions are actually characterized by the fact that, in a precise sense, their discontinuities are not too numerous: they are “negligible,” that is, the set of points where an integrable function is discontinuous has “measure 0.” The reader will find further details about Riemann integration in the appendix. From now on, we shall always assume that our functions are integrable, even if we do not state this requirement explicitly. Functions on the circle There is a natural connection between 2π-periodic functions on R like the exponentials einθ , functions on an interval of length 2π, and functions on the unit circle. This connection arises as follows. A point on the unit circle takes the form eiθ , where θ is a real number that is unique up to integer multiples of 2π. If F is a function on the circle, then we may define for each real number θ f (θ) = F (eiθ ), and observe that with this definition, the function f is periodic on R of period 2π, that is, f (θ + 2π) = f (θ) for all θ. The integrability, continuity and other smoothness properties of F are determined by those of f . For instance, we say that F is integrable on the circle if f is integrable on every interval of length 2π. Also, F is continuous on the circle if f is continuous on R, which is the same as saying that f is continuous on any interval of length 2π. Moreover, F is continuously differentiable if f has a continuous derivative, and so forth. Since f has period 2π, we may restrict it to any interval of length 2π, say [0, 2π] or [−π, π], and still capture the initial function F on the circle. We note that f must take the same value at the end-points of the interval since they correspond to the same point on the circle. Conversely, any function on [0, 2π] for which f (0) = f (2π) can be extended to a periodic function on R which can then be identified as a function on the circle. In particular, a continuous function f on the interval [0, 2π] gives rise to a continuous function on the circle if and only if f (0) = f (2π). In conclusion, functions on R that 2π-periodic, and functions on an interval of length 2π that take on the same value at its end-points, are two equivalent descriptions of the same mathematical objects, namely, functions on the circle. In this connection, we mention an item of notational usage. When our functions are defined on an interval on the line, we often use x as the independent variable; however, when we consider these as functions

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on the circle, we usually replace the variable x by θ. As the reader will note, we are not strictly bound by this rule since this practice is mostly a matter of convenience. 1.1 Main definitions and some examples We now begin our study of Fourier analysis with the precise definition of the Fourier series of a function. Here, it is important to pin down where our function is originally defined. If f is an integrable function given on an interval [a, b] of length L (that is, b − a = L), then the nth Fourier coefficient of f is defined by 1 fˆ(n) = L

Z

b

f (x)e−2πinx/L dx,

n ∈ Z.

a

The Fourier series of f is given formally3 by ∞ X

fˆ(n)e2πinx/L .

n=−∞

We shall sometimes write an for the Fourier coefficients of f , and use the notation ∞ X

f (x) ∼

an e2πinx/L

n=−∞

to indicate that the series on the right-hand side is the Fourier series of f. For instance, if f is an integrable function on the interval [−π, π], then the nth Fourier coefficient of f is Z π 1 ˆ f (θ)e−inθ dθ, n ∈ Z, f (n) = an = 2π −π and the Fourier series of f is f (θ) ∼

∞ X

an einθ .

n=−∞

Here we use θ as a variable since we think of it as an angle ranging from −π to π. 3 At

this point, we do not say anything about the convergence of the series.

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1. Examples and formulation of the problem

Also, if f is defined on [0, 2π], then the formulas are the same as above, except that we integrate from 0 to 2π in the definition of the Fourier coefficients. We may also consider the Fourier coefficients and Fourier series for a function defined on the circle. By our previous discussion, we may think of a function on the circle as a function f on R which is 2π-periodic. We may restrict the function f to any interval of length 2π, for instance [0, 2π] or [−π, π], and compute its Fourier coefficients. Fortunately, f is periodic and Exercise 1 shows that the resulting integrals are independent of the chosen interval. Thus the Fourier coefficients of a function on the circle are well defined. Finally, we shall sometimes consider a function g given on [0, 1]. Then Z

gˆ(n) = an =

1

g(x)e−2πinx dx

and

g(x) ∼

0

∞ X

an e2πinx .

n=−∞

Here we use x for a variable ranging from 0 to 1. Of course, if f is initially given on [0, 2π], then g(x) = f (2πx) is defined on [0, 1] and a change of variables shows that the nth Fourier coefficient of f equals the nth Fourier coefficient of g. Fourier series are part of a larger family called the trigonometric seP 2πinx/L ries which, by definition, are expressions of the form ∞ n=−∞ cn e where cn ∈ C. If a trigonometric series involves only finitely many nonzero terms, that is, cn = 0 for all large |n|, it is called a trigonometric polynomial; its degree is the largest value of |n| for which cn 6= 0. The N th partial sum of the Fourier series of f , for N a positive integer, is a particular example of a trigonometric polynomial. It is given by SN (f )(x) =

N X

fˆ(n)e2πinx/L .

n=−N

Note that by definition, the above sum is symmetric since n ranges from −N to N , a choice that is natural because of the resulting decomposition of the Fourier series as sine and cosine series. As a consequence, the convergence of Fourier series will be understood (in this book) as the “limit” as N tends to infinity of these symmetric sums. In fact, using the partial sums of the Fourier series, we can reformulate the basic question raised in Chapter 1 as follows: Problem: In what sense does SN (f ) converge to f as N → ∞ ?

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Before proceeding further with this question, we turn to some simple examples of Fourier series. Example 1. Let f (θ) = θ for −π ≤ θ ≤ π. The calculation of the Fourier coefficients requires a simple integration by parts. First, if n 6= 0, then Z π 1 fˆ(n) = θe−inθ dθ 2π −π · ¸π Z π θ 1 1 − e−inθ = + e−inθ dθ 2π in 2πin −π −π =

(−1)n+1 , in

and if n = 0 we clearly have 1 fˆ(0) = 2π

Z

π

θ dθ = 0. −π

Hence, the Fourier series of f is given by f (θ) ∼

X (−1)n+1 n6=0

in

einθ = 2

∞ X

(−1)n+1

n=1

sin nθ . n

The first sum is over all non-zero integers, and the second is obtained by an application of Euler’s identities. It is possible to prove by elementary means that the above series converges for every θ, but it is not obvious that it converges to f (θ). This will be proved later (Exercises 8 and 9 deal with a similar situation). Example 2. Define f (θ) = (π − θ)2 /4 for 0 ≤ θ ≤ 2π. Then successive integration by parts similar to that performed in the previous example yield ∞

f (θ) ∼

π 2 X cos nθ + . 12 n=1 n2

Example 3. The Fourier series of the function π f (θ) = ei(π−θ)α sin πα on [0, 2π] is f (θ) ∼

∞ X

einθ , n+α n=−∞

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1. Examples and formulation of the problem

whenever α is not an integer. Example 4. The trigonometric polynomial defined for x ∈ [−π, π] by N X

DN (x) =

einx

n=−N

is called the N th Dirichlet kernel and is of fundamental importance in the theory (as we shall see later). Notice that its Fourier coefficients an have the property that an = 1 if |n| ≤ N and an = 0 otherwise. A closed form formula for the Dirichlet kernel is DN (x) =

sin((N + 12 )x) . sin(x/2)

This can be seen by summing the geometric progressions N X

ωn

and

n=0

−1 X

ωn

n=−N

with ω = eix . These sums are, respectively, equal to 1 − ω N +1 1−ω

and

ω −N − 1 . 1−ω

Their sum is then sin((N + 12 )x) ω −N − ω N +1 ω −N −1/2 − ω N +1/2 = , = 1−ω sin(x/2) ω −1/2 − ω 1/2 giving the desired result. Example 5. The function Pr (θ), called the Poisson kernel, is defined for θ ∈ [−π, π] and 0 ≤ r < 1 by the absolutely and uniformly convergent series Pr (θ) =

∞ X

r|n| einθ .

n=−∞

This function arose implicitly in the solution of the steady-state heat equation on the unit disc discussed in Chapter 1. Note that in calculating the Fourier coefficients of Pr (θ) we can interchange the order of integration and summation since the sum converges uniformly in θ for

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each fixed r, and obtain that the nth Fourier coefficient equals r|n| . One can also sum the series for Pr (θ) and see that Pr (θ) =

1 − r2 . 1 − 2r cos θ + r2

In fact, Pr (θ) =

∞ X n=0

ωn +

∞ X

ωn

with ω = reiθ ,

n=1

where both series converge absolutely. The first sum (an infinite geometric progression) equals 1/(1 − ω), and likewise, the second is ω/(1 − ω). Together, they combine to give 1 − ω + (1 − ω)ω 1 − |ω|2 1 − r2 = = , 2 (1 − ω)(1 − ω) |1 − ω| 1 − 2r cos θ + r2 as claimed. The Poisson kernel will reappear later in the context of Abel summability of the Fourier series of a function. Let us return to the problem formulated earlier. The definition of the Fourier series of f is purely formal, and it is not obvious whether it converges to f . In fact, the solution of this problem can be very hard, or relatively easy, depending on the sense in which we expect the series to converge, or on what additional restrictions we place on f . Let us be more precise. Suppose, for the sake of this discussion, that the function f (which is always assumed to be Riemann integrable) is defined on [−π, π]. The first question one might ask is whether the partial sums of the Fourier series of f converge to f pointwise. That is, do we have (1)

lim SN (f )(θ) = f (θ)

N →∞

for every θ?

We see quite easily that in general we cannot expect this result to be true at every θ, since we can always change an integrable function at one point without changing its Fourier coefficients. As a result, we might ask the same question assuming that f is continuous and periodic. For a long time it was believed that under these additional assumptions the answer would be “yes.” It was a surprise when Du Bois-Reymond showed that there exists a continuous function whose Fourier series diverges at a point. We will give such an example in the next chapter. Despite this negative result, we might ask what happens if we add more smoothness conditions on f : for example, we might assume that f is continuously

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39

differentiable, or twice continuously differentiable. We will see that then the Fourier series of f converges to f uniformly. We will also interpret the limit (1) by showing that the Fourier series sums, in the sense of Ces`aro or Abel, to the function f at all of its points of continuity. This approach involves appropriate averages of the partial sums of the Fourier series of f . Finally, we can also define the limit (1) in the mean square sense. In the next chapter, we will show that if f is merely integrable, then Z π 1 |SN (f )(θ) − f (θ)|2 dθ → 0 as N → ∞. 2π −π It is of interest to know that the problem of pointwise convergence of Fourier series was settled in 1966 by L. Carleson, who showed, among other things, that if f is integrable in our sense,4 then the Fourier series of f converges to f except possibly on a set of “measure 0.” The proof of this theorem is difficult and beyond the scope of this book.

2 Uniqueness of Fourier series If we were to assume that the Fourier series of functions f converge to f in an appropriate sense, then we could infer that a function is uniquely determined by its Fourier coefficients. This would lead to the following statement: if f and g have the same Fourier coefficients, then f and g are necessarily equal. By taking the difference f − g, this proposition can be reformulated as: if fˆ(n) = 0 for all n ∈ Z, then f = 0. As stated, this assertion cannot be correct without reservation, since calculating Fourier coefficients requires integration, and we see that, for example, any two functions which differ at finitely many points have the same Fourier series. However, we do have the following positive result. Theorem 2.1 Suppose that f is an integrable function on the circle with fˆ(n) = 0 for all n ∈ Z. Then f (θ0 ) = 0 whenever f is continuous at the point θ0 . Thus, in terms of what we know about the set of discontinuities of integrable functions,5 we can conclude that f vanishes for “most” values of θ. Proof. We suppose first that f is real-valued, and argue by contradiction. Assume, without loss of generality, that f is defined on 4 Carleson’s proof actually holds for the wider class of functions which are square integrable in the Lebesgue sense. 5 See the appendix.

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[−π, π], that θ0 = 0, and f (0) > 0. The idea now is to construct a family R of trigonometric polynomials {pk } that “peak” at 0, and so that pk (θ)f (θ) dθ → ∞ as k → ∞. This will be our desired contradiction since these integrals are equal to zero by assumption. Since f is continuous at 0, we can choose 0 < δ ≤ π/2, so that f (θ) > f (0)/2 whenever |θ| < δ. Let p(θ) = ² + cos θ, where ² > 0 is chosen so small that |p(θ)| < 1 − ²/2, whenever δ ≤ |θ| ≤ π. Then, choose a positive η with η < δ, so that p(θ) ≥ 1 + ²/2, for |θ| < η. Finally, let pk (θ) = [p(θ)]k , and select B so that |f (θ)| ≤ B for all θ. This is possible since f is integrable, hence bounded. Figure 3 illustrates the family {pk }. By

p15

p6 p

Figure 3. The functions p, p6 , and p15 when ² = 0.1 construction, each pk is a trigonometric polynomial, and since fˆ(n) = 0 for all n, we must have Z π f (θ)pk (θ) dθ = 0 for all k. −π

However, we have the estimate ¯Z ¯ ¯ ¯ ¯ ¯ f (θ)pk (θ) dθ¯ ≤ 2πB(1 − ²/2)k . ¯ ¯ δ≤|θ| ¯

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2. Uniqueness of Fourier series

Also, our choice of δ guarantees that p(θ) and f (θ) are non-negative whenever |θ| < δ, thus Z f (θ)pk (θ) dθ ≥ 0. η≤|θ| 1/2, that is, sup |f (θ + t) − f (θ)| ≤ A|t|α

for all t.

θ

For more on these matters, see the exercises at the end of Chapter 3.

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At this point it is worthwhile to introduce a common notation: we say that f belongs to the class C k if f is k times continuously differentiable. Belonging to the class C k or satisfying a H¨older condition are two possible ways to describe the smoothness of a function.

3 Convolutions The notion of convolution of two functions plays a fundamental role in Fourier analysis; it appears naturally in the context of Fourier series but also serves more generally in the analysis of functions in other settings. Given two 2π-periodic integrable functions f and g on R, we define their convolution f ∗ g on [−π, π] by (2)

1 (f ∗ g)(x) = 2π

Z

π

f (y)g(x − y) dy. −π

The above integral makes sense for each x, since the product of two integrable functions is again integrable. Also, since the functions are periodic, we can change variables to see that 1 (f ∗ g)(x) = 2π

Z

π

f (x − y)g(y) dy. −π

Loosely speaking, convolutions correspond to “weighted averages.” For Rπ 1 instance, if g = 1 in (2), then f ∗ g is constant and equal to 2π f (y) dy, −π which we may interpret as the average value of f on the circle. Also, the convolution (f ∗ g)(x) plays a role similar to, and in some sense replaces, the pointwise product f (x)g(x) of the two functions f and g. In the context of this chapter, our interest in convolutions originates from the fact that the partial sums of the Fourier series of f can be expressed as follows: SN (f )(x) =

N X

fˆ(n)einx

n=−N

=

µ Z N X 1

2π

n=−N

1 = 2π

Z

−iny

f (y)e

dy einx

−π

Ã

π

¶

π

f (y) −π

= (f ∗ DN )(x),

N X

n=−N

! in(x−y)

e

dy

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3. Convolutions

where DN is the N th Dirichlet kernel (see Example 4) given by N X

DN (x) =

einx .

n=−N

So we observe that the problem of understanding SN (f ) reduces to the understanding of the convolution f ∗ DN . We begin by gathering some of the main properties of convolutions. Proposition 3.1 Suppose that f , g, and h are 2π-periodic integrable functions. Then: (i) f ∗ (g + h) = (f ∗ g) + (f ∗ h). (ii) (cf ) ∗ g = c(f ∗ g) = f ∗ (cg) for any c ∈ C. (iii) f ∗ g = g ∗ f . (iv) (f ∗ g) ∗ h = f ∗ (g ∗ h). (v) f ∗ g is continuous. (vi) f[ ∗ g(n) = fˆ(n)ˆ g (n). The first four points describe the algebraic properties of convolutions: linearity, commutativity, and associativity. Property (v) exhibits an important principle: the convolution of f ∗ g is “more regular” than f or g. Here, f ∗ g is continuous while f and g are merely (Riemann) integrable. Finally, (vi) is key in the study of Fourier series. In general, the Fourier coefficients of the product f g are not the product of the Fourier coefficients of f and g. However, (vi) says that this relation holds if we replace the product of the two functions f and g by their convolution f ∗ g. Proof. Properties (i) and (ii) follow at once from the linearity of the integral. The other properties are easily deduced if we assume also that f and g are continuous. In this case, we may freely interchange the order of

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integration. For instance, to establish (vi) we write Z π 1 [ (f ∗ g)(x)e−inx dx f ∗ g(n) = 2π −π ¶ µZ π Z π 1 1 = f (y)g(x − y) dy e−inx dx 2π −π 2π −π µ ¶ Z π Z π 1 1 −iny −in(x−y) = f (y)e g(x − y)e dx dy 2π −π 2π −π µ ¶ Z π Z π 1 1 = f (y)e−iny g(x)e−inx dx dy 2π −π 2π −π = fˆ(n)ˆ g (n).

To prove (iii), one first notes that if F is continuous and 2π-periodic, then Z π Z π F (y) dy = F (x − y) dy for any x ∈ R. −π

−π

The verification of this identity consists of a change of variables y 7→ −y, followed by a translation y 7→ y − x. Then, one takes F (y) = f (y)g(x − y). Also, (iv) follows by interchanging two integral signs, and an appropriate change of variables. Finally, we show that if f and g are continuous, then f ∗ g is continuous. First, we may write (f ∗ g)(x1 ) − (f ∗ g)(x2 ) =

1 2π

Z

π −π

f (y) [g(x1 − y) − g(x2 − y)] dy.

Since g is continuous it must be uniformly continuous on any closed and bounded interval. But g is also periodic, so it must be uniformly continuous on all of R; given ² > 0 there exists δ > 0 so that |g(s) − g(t)| < ² whenever |s − t| < δ. Then, |x1 − x2 | < δ implies |(x1 − y) − (x2 − y)| < δ for any y, hence |(f ∗ g)(x1 ) − (f ∗ g)(x2 )| ≤ ≤ ≤ ≤

1 2π 1 2π ² 2π ² 2π

¯Z π ¯ ¯ ¯ ¯ f (y) [g(x1 − y) − g(x2 − y)] dy ¯¯ ¯ −π Z π −π π

|f (y)| |g(x1 − y) − g(x2 − y)| dy

Z

|f (y)| dy −π

2π B ,

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3. Convolutions

where B is chosen so that |f (x)| ≤ B for all x. As a result, we conclude that f ∗ g is continuous, and the proposition is proved, at least when f and g are continuous. In general, when f and g are merely integrable, we may use the results established so far (when f and g are continuous), together with the following approximation lemma, whose proof may be found in the appendix. Lemma 3.2 Suppose f is integrable on the circle and bounded by B. Then there exists a sequence {fk }∞ k=1 of continuous functions on the circle so that sup x∈[−π,π]

and

Z

|fk (x)| ≤ B

for all k = 1, 2, . . . ,

π −π

|f (x) − fk (x)| dx → 0

as k → ∞.

Using this result, we may complete the proof of the proposition as follows. Apply Lemma 3.2 to f and g to obtain sequences {fk } and {gk } of approximating continuous functions. Then f ∗ g − fk ∗ gk = (f − fk ) ∗ g + fk ∗ (g − gk ). By the properties of the sequence {fk }, Z π 1 |f (x − y) − fk (x − y)| |g(y)| dy |(f − fk ) ∗ g(x)| ≤ 2π −π Z π 1 ≤ |f (y) − fk (y)| dy sup |g(y)| 2π y −π →0

as k → ∞.

Hence (f − fk ) ∗ g → 0 uniformly in x. Similarly, fk ∗ (g − gk ) → 0 uniformly, and therefore fk ∗ gk tends uniformly to f ∗ g. Since each fk ∗ gk is continuous, it follows that f ∗ g is also continuous, and we have (v). Next, we establish (vi). For each fixed integer n we must have [ f\ k ∗ gk (n) → f ∗ g(n) as k tends to infinity since fk ∗ gk converges uniformly to f ∗ g. However, we found earlier that fck (n)gck (n) = f\ k ∗ gk (n) because both fk and gk are continuous. Hence ¯Z ¯ ¯ 1 ¯¯ π −inx ˆ ˆ |f (n) − fk (n)| = (f (x) − fk (x))e dx¯¯ ¯ 2π −π Z π 1 |f (x) − fk (x)| dx, ≤ 2π −π

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Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

and as a result we find that fck (n) → fˆ(n) as k goes to infinity. Similarly gck (n) → gˆ(n), and the desired property is established once we let k tend to infinity. Finally, properties (iii) and (iv) follow from the same kind of arguments.

4 Good kernels In the proof of Theorem 2.1 we constructed a sequence of trigonometric polynomials {pk } with the property that the functions pk peaked at the origin. As a result, we could isolate the behavior of f at the origin. In this section, we return to such families of functions, but this time in a more general setting. First, we define the notion of good kernel, and discuss the characteristic properties of such functions. Then, by the use of convolutions, we show how these kernels can be used to recover a given function. A family of kernels {Kn (x)}∞ n=1 on the circle is said to be a family of good kernels if it satisfies the following properties: (a) For all n ≥ 1, 1 2π

Z

π

−π

Kn (x) dx = 1.

(b) There exists M > 0 such that for all n ≥ 1, Z π |Kn (x)| dx ≤ M. −π

(c) For every δ > 0, Z δ≤|x|≤π

|Kn (x)| dx → 0,

as n → ∞.

In practice we shall encounter families where Kn (x) ≥ 0, in which case (b) is a consequence of (a). We may interpret the kernels Kn (x) as weight distributions on the circle: property (a) says that Kn assigns unit mass to the whole circle [−π, π], and (c) that this mass concentrates near the origin as n becomes large.6 Figure 4 (a) illustrates the typical character of a family of good kernels. The importance of good kernels is highlighted by their use in connection with convolutions. 6 In the limit, a family of good kernels represents the “Dirac delta function.” This terminology comes from physics.

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4. Good kernels

Kn (y) f (x − y)

f (x)

y=0

(a)

y

(b) Figure 4. Good kernels

Theorem 4.1 Let {Kn }∞ n=1 be a family of good kernels, and f an integrable function on the circle. Then lim (f ∗ Kn )(x) = f (x)

n→∞

whenever f is continuous at x. If f is continuous everywhere, then the above limit is uniform. Because of this result, the family {Kn } is sometimes referred to as an approximation to the identity. We have previously interpreted convolutions as weighted averages. In this context, the convolution Z π 1 f (x − y)Kn (y) dy (f ∗ Kn )(x) = 2π −π is the average of f (x − y), where the weights are given by Kn (y). However, the weight distribution Kn concentrates its mass at y = 0 as n becomes large. Hence in the integral, the value f (x) is assigned the full mass as n → ∞. Figure 4 (b) illustrates this point. Proof of Theorem 4.1. If ² > 0 and f is continuous at x, choose δ so that |y| < δ implies |f (x − y) − f (x)| < ². Then, by the first property of good kernels, we can write Z π 1 Kn (y)f (x − y) dy − f (x) (f ∗ Kn )(x) − f (x) = 2π −π Z π 1 = Kn (y)[f (x − y) − f (x)] dy. 2π −π

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Hence, ¯ ¯ Z π ¯ 1 ¯ ¯ |(f ∗ Kn )(x) − f (x)| = ¯ Kn (y)[f (x − y) − f (x)] dy ¯¯ 2π −π Z

≤

1 2π

|Kn (y)| |f (x − y) − f (x)| dy Z 1 + |Kn (y)| |f (x − y) − f (x)| dy 2π δ≤|y|≤π Z π Z ² 2B ≤ |Kn (y)| dy + |Kn (y)| dy, 2π −π 2π δ≤|y|≤π |y| 0 and all large n we have |(f ∗ Kn )(x) − f (x)| ≤ C², thereby proving the first assertion in the theorem. If f is continuous everywhere, then it is uniformly continuous, and δ can be chosen independent of x. This provides the desired conclusion that f ∗ Kn → f uniformly. Recall from the beginning of Section 3 that SN (f )(x) = (f ∗ DN )(x) , PN

where DN (x) = n=−N einx is the Dirichlet kernel. It is natural now for us to ask whether DN is a good kernel, since if this were true, Theorem 4.1 would imply that the Fourier series of f converges to f (x) whenever f is continuous at x. Unfortunately, this is not the case. Indeed, an estimate shows that DN violates the second property; more precisely, one has (see Problem 2) Z π |DN (x)| dx ≥ c log N, as N → ∞. −π

However, we should note that the formula for DN as a sum of exponentials immediately gives Z π 1 DN (x) dx = 1, 2π −π so the first property of good kernels is actually verified. The fact that the mean value of DN is 1, while the integral of its absolute value is large,

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51

is a result of cancellations. Indeed, Figure 5 shows that the function DN (x) takes on positive and negative values and oscillates very rapidly as N gets large.

Figure 5. The Dirichlet kernel for large N

This observation suggests that the pointwise convergence of Fourier series is intricate, and may even fail at points of continuity. This is indeed the case, as we will see in the next chapter.

5 Ces` aro and Abel summability: applications to Fourier series Since a Fourier series may fail to converge at individual points, we are led to try to overcome this failure by interpreting the limit lim SN (f ) = f

N →∞

in a different sense. 5.1 Ces` aro means and summation We begin by taking ordinary averages of the partial sums, a technique which we now describe in more detail.

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Suppose we are given a series of complex numbers c0 + c1 + c2 + · · · =

∞ X

ck .

k=0

We define the nth partial sum sn by sn =

n X

ck ,

k=0

and say that the series converges to s if limn→∞ sn = s. This is the most natural and most commonly used type of “summability.” Consider, however, the example of the series (3)

1 − 1 + 1 − 1 + ··· =

∞ X

(−1)k .

k=0

Its partial sums form the sequence {1, 0, 1, 0, . . .} which has no limit. Because these partial sums alternate evenly between 1 and 0, one might therefore suggest that 1/2 is the “limit” of the sequence, and hence 1/2 equals the “sum” of that particular series. We give a precise meaning to this by defining the average of the first N partial sums by σN =

s0 + s1 + · · · + sN −1 . N

The quantity σN is called the N th Ces` aro mean7 of the sequence {sk } P∞ th or the N Ces` aro sum of the series k=0 ck . If σN converges to a limit σ as N tends to infinity, we say that the P series cn is Ces` aro summable to σ. In the case of series of functions, we shall understand the limit in the sense of either pointwise or uniform convergence, depending on the situation. The reader will have no difficulty checking that in the above example (3), the series is Ces`aro summable to 1/2. Moreover, one can show that Ces`aro summation is a more inclusive process than convergence. In fact, if a series is convergent to s, then it is also Ces`aro summable to the same limit s (Exercise 12). 5.2 Fej´ er’s theorem An interesting application of Ces`aro summability appears in the context of Fourier series. 7 Note

P∞

that if the series c begins with the term k = 1, then it is common prack=1 k tice to define σN = (s1 + · · · + sN )/N . This change of notation has little effect on what follows.

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We mentioned earlier that the Dirichlet kernels fail to belong to the family of good kernels. Quite surprisingly, their averages are very well behaved functions, in the sense that they do form a family of good kernels. To see this, we form the N th Ces`aro mean of the Fourier series, which by definition is σN (f )(x) =

S0 (f )(x) + · · · + SN −1 (f )(x) . N

Since Sn (f ) = f ∗ Dn , we find that σN (f )(x) = (f ∗ FN )(x), where FN (x) is the N -th Fej´ er kernel given by FN (x) =

D0 (x) + · · · + DN −1 (x) . N

Lemma 5.1 We have FN (x) =

1 sin2 (N x/2) , N sin2 (x/2)

and the Fej´er kernel is a good kernel. The proof of the formula for FN (a simple application of trigonometric identities) is outlined in Exercise R π 15. To prove the rest of the lemma, note 1 that FN is positive and 2π −π FN (x) dx = 1, in view of the fact that a similar identity holds for the Dirichlet kernels Dn . However, sin2 (x/2) ≥ cδ > 0, if δ ≤ |x| ≤ π, hence FN (x) ≤ 1/(N cδ ), from which it follows that Z |FN (x)| dx → 0 as N → ∞. δ≤|x|≤π

Applying Theorem 4.1 to this new family of good kernels yields the following important result. Theorem 5.2 If f is integrable on the circle, then the Fourier series of f is Ces` aro summable to f at every point of continuity of f . Moreover, if f is continuous on the circle, then the Fourier series of f is uniformly Ces` aro summable to f . We may now state two corollaries. The first is a result that we have already established. The second is new, and of fundamental importance.

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Corollary 5.3 If f is integrable on the circle and fˆ(n) = 0 for all n, then f = 0 at all points of continuity of f . The proof is immediate since all the partial sums are 0, hence all the Ces`aro means are 0. Corollary 5.4 Continuous functions on the circle can be uniformly approximated by trigonometric polynomials. This means that if f is continuous on [−π, π] with f (−π) = f (π) and ² > 0, then there exists a trigonometric polynomial P such that |f (x) − P (x)| < ²

for all −π ≤ x ≤ π.

This follows immediately from the theorem since the partial sums, hence the Ces`aro means, are trigonometric polynomials. Corollary 5.4 is the periodic analogue of the Weierstrass approximation theorem for polynomials which can be found in Exercise 16. 5.3 Abel means and summation Another method of summation was first considered by Abel and actually predates the Ces`aro method. P A series of complex numbers ∞ k=0 ck is said to be Abel summable to s if for every 0 ≤ r < 1, the series A(r) =

∞ X

ck r k

k=0

converges, and lim A(r) = s.

r→1

The quantities A(r) are called the Abel means of the series. One can prove that if the series converges to s, then it is Abel summable to s. Moreover, the method of Abel summability is even more powerful than the Ces`aro method: when the series is Ces`aro summable, it is always Abel summable to the same sum. However, if we consider the series 1 − 2 + 3 − 4 + 5 − ··· =

∞ X

(−1)k (k + 1),

k=0

then one can show that it is Abel summable to 1/4 since A(r) =

∞ X k=0

(−1)k (k + 1)rk =

1 , (1 + r)2

but this series is not Ces`aro summable; see Exercise 13.

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5. Ces` a ro and Abel summability: applications to Fourier series

5.4 The Poisson kernel and Dirichlet’s problem in the unit disc To adapt Abel summability to the context of Fourier series, we define P inθ the Abel means of the function f (θ) ∼ ∞ by n=−∞ an e ∞ X

Ar (f )(θ) =

r|n| an einθ .

n=−∞

Since the index n takes positive and negative values, it is natural to write c0 = a0 , and cn = an einθ + a−n e−inθ for n > 0, so that the Abel means of the Fourier series correspond to the definition given in the previous section for numerical series. We note that since f is integrable, |an | is uniformly bounded in n, so that Ar (f ) converges absolutely and uniformly for each 0 ≤ r < 1. Just as in the case of Ces`aro means, the key fact is that these Abel means can be written as convolutions Ar (f )(θ) = (f ∗ Pr )(θ), where Pr (θ) is the Poisson kernel given by (4)

∞ X

Pr (θ) =

r|n| einθ .

n=−∞

In fact, Ar (f )(θ) = =

∞ X

r|n| an einθ

n=−∞ ∞ X

µ

r|n|

n=−∞

1 = 2π

Z

π

1 2π Ã

f (ϕ) −π

Z

π

¶ f (ϕ)e−inϕ dϕ einθ

−π ∞ X

!

r

|n| −in(ϕ−θ)

e

dϕ,

n=−∞

where the interchange of the integral and infinite sum is justified by the uniform convergence of the series. Lemma 5.5 If 0 ≤ r < 1, then

Pr (θ) =

1 − r2 . 1 − 2r cos θ + r2

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The Poisson kernel is a good kernel,8 as r tends to 1 from below. Proof. The identity Pr (θ) = Section 1.1. Note that

1−r 2 1−2r cos θ+r2

has already been derived in

1 − 2r cos θ + r2 = (1 − r)2 + 2r(1 − cos θ). Hence if 1/2 ≤ r ≤ 1 and δ ≤ |θ| ≤ π, then 1 − 2r cos θ + r2 ≥ cδ > 0. Thus Pr (θ) ≤ (1 − r2 )/cδ when δ ≤ |θ| ≤ π, and the third property of good kernels is verified. Clearly Pr (θ) ≥ 0, and integrating the expression (4) term by term (which is justified by the absolute convergence of the series) yields Z π 1 Pr (θ) dθ = 1, 2π −π thereby concluding the proof that Pr is a good kernel. Combining this lemma with Theorem 4.1, we obtain our next result. Theorem 5.6 The Fourier series of an integrable function on the circle is Abel summable to f at every point of continuity. Moreover, if f is continuous on the circle, then the Fourier series of f is uniformly Abel summable to f . We now return to a problem discussed in Chapter 1, where we sketched the solution of the steady-state heat equation 4u = 0 in the unit disc with boundary condition u = f on the circle. We expressed the Laplacian in terms of polar coordinates, separated variables, and expected that a solution was given by (5)

u(r, θ) =

∞ X

am r|m| eimθ ,

m=−∞

where am was the mth Fourier coefficient of f . In other words, we were led to take Z π 1 f (ϕ)Pr (θ − ϕ) dϕ. u(r, θ) = Ar (f )(θ) = 2π −π We are now in a position to show that this is indeed the case. 8 In this case, the family of kernels is indexed by a continuous parameter 0 ≤ r < 1, rather than the discrete n considered previously. In the definition of good kernels, we simply replace n by r and take the limit in property (c) appropriately, for example r → 1 in this case.

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57

Theorem 5.7 Let f be an integrable function defined on the unit circle. Then the function u defined in the unit disc by the Poisson integral (6)

u(r, θ) = (f ∗ Pr )(θ)

has the following properties: (i) u has two continuous derivatives in the unit disc and satisfies 4u = 0. (ii) If θ is any point of continuity of f , then lim u(r, θ) = f (θ).

r→1

If f is continuous everywhere, then this limit is uniform. (iii) If f is continuous, then u(r, θ) is the unique solution to the steadystate heat equation in the disc which satisfies conditions (i) and (ii). Proof. For (i), we recall that the function u is given by the series (5). Fix ρ < 1; inside each disc of radius r < ρ < 1 centered at the origin, the series for u can be differentiated term by term, and the differentiated series is uniformly and absolutely convergent. Thus u can be differentiated twice (in fact infinitely many times), and since this holds for all ρ < 1, we conclude that u is twice differentiable inside the unit disc. Moreover, in polar coordinates, 4u =

∂ 2 u 1 ∂u 1 ∂2u + + , ∂r2 r ∂r r2 ∂θ2

so term by term differentiation shows that 4u = 0. The proof of (ii) is a simple application of the previous theorem. To prove (iii) we argue as follows. Suppose v solves the steady-state heat equation in the disc and converges to f uniformly as r tends to 1 from below. For each fixed r with 0 < r < 1, the function v(r, θ) has a Fourier series ∞ X n=−∞

inθ

an (r)e

where

1 an (r) = 2π

Z

π

v(r, θ)e−inθ dθ.

−π

Taking into account that v(r, θ) solves the equation (7)

∂ 2 v 1 ∂v 1 ∂2v + + = 0, ∂r2 r ∂r r2 ∂θ2

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we find that 1 n2 a00n (r) + a0n (r) − 2 an (r) = 0. r r

(8)

Indeed, we may first multiply (7) by e−inθ and integrate in θ. Then, since v is periodic, two integrations by parts give 1 2π

Z

π

−π

∂2v (r, θ)e−inθ dθ = −n2 an (r). ∂θ2

Finally, we may interchange the order of differentiation and integration, which is permissible since v has two continuous derivatives; this yields (8). Therefore, we must have an (r) = An rn + Bn r−n for some constants An and Bn , when n 6= 0 (see Exercise 11 in Chapter 1). To evaluate the constants, we first observe that each term an (r) is bounded because v is bounded, therefore Bn = 0. To find An we let r → 1. Since v converges uniformly to f as r → 1 we find that Z π 1 f (θ)e−inθ dθ. An = 2π −π By a similar argument, this formula also holds when n = 0. Our conclusion is that for each 0 < r < 1, the Fourier series of v is given by the series of u(r, θ), so by the uniqueness of Fourier series for continuous functions, we must have u = v. Remark. By part (iii) of the theorem, we may conclude that if u solves 4u = 0 in the disc, and converges to 0 uniformly as r → 1, then u must be identically 0. However, if uniform convergence is replaced by pointwise convergence, this conclusion may fail; see Exercise 18.

6 Exercises 1. Suppose f is 2π-periodic and integrable on any finite interval. Prove that if a, b ∈ R, then

Z

Z

b

Z

b+2π

f (x) dx =

b−2π

f (x) dx =

a

a+2π

π

Z

f (x) dx. a−2π

Also prove that

Z

f (x + a) dx = −π

Z

π

π+a

f (x) dx = −π

f (x) dx. −π+a

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6. Exercises

2. In this exercise we show how the symmetries of a function imply certain properties of its Fourier coefficients. Let f be a 2π-periodic Riemann integrable function defined on R. (a) Show that the Fourier series of the function f can be written as f (θ) ∼ fˆ(0) +

X

[fˆ(n) + fˆ(−n)] cos nθ + i[fˆ(n) − fˆ(−n)] sin nθ.

n≥1

(b) Prove that if f is even, then fˆ(n) = fˆ(−n), and we get a cosine series. (c) Prove that if f is odd, then fˆ(n) = −fˆ(−n), and we get a sine series. (d) Suppose that f (θ + π) = f (θ) for all θ ∈ R. Show that fˆ(n) = 0 for all odd n. (e) Show that f is real-valued if and only if fˆ(n) = fˆ(−n) for all n. 3. We return to the problem of the plucked string discussed in Chapter 1. Show that the initial condition f is equal to its Fourier sine series f (x) =

∞ X

Am sin mx

with

Am =

m=1

2h sin mp . m2 p(π − p)

[Hint: Note that |Am | ≤ C/m2 .] 4. Consider the 2π-periodic odd function defined on [0, π] by f (θ) = θ(π − θ). (a) Draw the graph of f . (b) Compute the Fourier coefficients of f , and show that f (θ) =

8 π

X k odd ≥ 1

sin kθ . k3

5. On the interval [−π, π] consider the function

( f (θ) =

0 1 − |θ|/δ

if |θ| > δ, if |θ| ≤ δ.

Thus the graph of f has the shape of a triangular tent. Show that ∞

f (θ) =

X 1 − cos nδ δ +2 cos nθ. 2π n2 πδ n=1

6. Let f be the function defined on [−π, π] by f (θ) = |θ|.

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(a) Draw the graph of f . (b) Calculate the Fourier coefficients of f , and show that

fˆ(n) =

π   2

if n = 0,

 n   −1 + (−1) πn2

if n 6= 0.

(c) What is the Fourier series of f in terms of sines and cosines? (d) Taking θ = 0, prove that

X n odd ≥1

1 π2 = 2 n 8

and

∞ X 1 n=1

n2

=

π2 . 6

See also Example 2 in Section 1.1. 7. Suppose {an }N and {bn }N of complex numbers. n=1 are two finite sequences Pk n=1 P Let Bk = n=1 bn denote the partial sums of the series bn with the convention B0 = 0. (a) Prove the summation by parts formula N X

an bn = aN BN − aM BM −1 −

n=M

N −1 X

(an+1 − an )Bn .

n=M

(b) Deduce from this formula Dirichlet’s test for convergence of a series: if the P partial sums of the series bn are bounded, and {anP } is a sequence of real numbers that decreases monotonically to 0, then an bn converges. 1 P einx is the Fourier series of the 2π-periodic sawtooth n6=0 2i n function illustrated in Figure 6, defined by f (0) = 0, and

8. Verify that

 π x   −2 − 2 f (x) =   π x 2

−

2

if −π < x < 0, if 0 < x < π.

Note that this function is not continuous. Show that nevertheless, the series converges for every x (by which we mean, as usual, that the symmetric partial sums of the series converge). In particular, the value of the series at the origin, namely 0, is the average of the values of f (x) as x approaches the origin from the left and the right.

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6. Exercises

π 2

−π

0

π − π2

Figure 6. The sawtooth function

[Hint: Use Dirichlet’s test for convergence of a series

P

an bn .]

9. Let f (x) = χ[a,b] (x) be the characteristic function of the interval [a, b] ⊂ [−π, π], that is,

½

χ[a,b] (x) =

1 0

if x ∈ [a, b], otherwise.

(a) Show that the Fourier series of f is given by f (x) ∼

b − a X e−ina − e−inb inx + e . 2π 2πin n6=0

The sum extends over all positive and negative integers excluding 0. (b) Show that if a 6= −π or b 6= π and a 6= b, then the Fourier series does not converge absolutely for any x. [Hint: It suffices to prove that for many values of n one has | sin nθ0 | ≥ c > 0 where θ0 = (b − a)/2.] (c) However, prove that the Fourier series converges at every point x. What happens if a = −π and b = π? 10. Suppose f is a periodic function of period 2π which belongs to the class C k . Show that fˆ(n) = O(1/|n|k ) as |n| → ∞. This notation means that there exists a constant C such |fˆ(n)| ≤ C/|n|k . We could also write this as |n|k fˆ(n) = O(1), where O(1) means bounded. [Hint: Integrate by parts.] 11. Suppose that {fk }∞ k=1 is a sequence of Riemann integrable functions on the interval [0, 1] such that

Z

1

|fk (x) − f (x)| dx → 0 0

as k → ∞.

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Show that fˆk (n) → fˆ(n) uniformly in n as k → ∞.

P

12. Prove that if a series of complex numbers is Ces`aro summable to s. [Hint: Assume sn → 0 as n → ∞.]

cn converges to s, then

P

cn

13. The purpose of this exercise is to prove that Abel summability is stronger than the standard or Ces`aro methods of summation.

P∞

(a) Show that if the series n=1 cn of complex numbers converges to a finite limit s, then the series is Abel summable to s. [Hint: Why is it enough to prove the theorem if sN = c1 + PNwhen ns = 0? Assuming PN s =n0, showNthat +1 · · · + cN , then c r = (1 − r) s r + s r . Let N → ∞ n n N n=1 n=1 to show that

X

cn rn = (1 − r)

X

sn rn .

Finally, prove that the right-hand side converges to 0 as r → 1.] (b) However, show that there exist series which are Abel summable, but P that do not converge. [Hint: Try cn = (−1)n . What is the Abel limit of cn ?]

P∞

(c) Argue similarly to prove that if a series n=1 cn is Ces`aro summable to σ, then it is Abel summable to σ. [Hint: Note that ∞ X n=1

cn rn = (1 − r)2

∞ X

nσn rn ,

n=1

and assume σ = 0.] (d) Give an example of a series that is Abel summable but not Ces`aro summable. P [Hint: Try cn = (−1)n−1 n. Note that if cn is Ces`aro summable, then cn /n tends to 0.] The results above can be summarized by the following implications about series: convergent =⇒ Ces` aro summable =⇒ Abel summable, and the fact that none of the arrows can be reversed. 14. This exercise deals with a theorem of Tauber which says that under an additional condition on the coefficients cn , the above arrows can be reversed.

P

(a) If P

cn is Ces`aro summable to σ and cn = o(1/n) (that is, ncn → 0), then cn converges to σ. [Hint: sn − σn = [(n − 1)cn + · · · + c2 ]/n.]

(b) The above statement holds if we replace Ces`aro summable by Abel summable. PN PN n [Hint: Estimate the difference between n=1 cn and n=1 cn r where r = 1 − 1/N .]

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6. Exercises

15. Prove that the Fej´er kernel is given by FN (x) =

1 sin2 (N x/2) . N sin2 (x/2)

[Hint: Remember that N FN (x) = D0 (x) + · · · + DN −1 (x) where Dn (x) is the Dirichlet kernel. Therefore, if ω = eix we have N FN (x) =

N −1 X n=0

ω −n − ω n+1 .] 1−ω

16. The Weierstrass approximation theorem states: Let f be a continuous function on the closed and bounded interval [a, b] ⊂ R. Then, for any ² > 0, there exists a polynomial P such that sup |f (x) − P (x)| < ². x∈[a,b]

Prove this by applying Corollary 5.4 of Fej´er’s theorem and using the fact that the exponential function eix can be approximated by polynomials uniformly on any interval. 17. In Section 5.4 we proved that the Abel means of f converge to f at all points of continuity, that is, lim Ar (f )(θ) = lim (Pr ∗ f )(θ) = f (θ),

r→1

with 0 < r < 1,

r→1

whenever f is continuous at θ. In this exercise, we will study the behavior of Ar (f )(θ) at certain points of discontinuity. An integrable function is said to have a jump discontinuity at θ if the two limits lim f (θ + h) = f (θ+ )

h → 0 h > 0

and

lim f (θ − h) = f (θ− )

h → 0 h > 0

exist. (a) Prove that if f has a jump discontinuity at θ, then lim Ar (f )(θ) =

r→1

R0

f (θ+ ) + f (θ− ) , 2

1 [Hint: Explain why 2π P (θ) dθ = −π r the proof given in the text.]

1 2π

with 0 ≤ r < 1.

Rπ 0

Pr (θ) dθ = 12 , then modify

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(b) Using a similar argument, show that if f has a jump discontinuity at θ, + (θ − ) . the Fourier series of f at θ is Ces`aro summable to f (θ )+f 2 18. If Pr (θ) denotes the Poisson kernel, show that the function u(r, θ) =

∂Pr , ∂θ

defined for 0 ≤ r < 1 and θ ∈ R, satisfies: (i) 4u = 0 in the disc. (ii) limr→1 u(r, θ) = 0 for each θ. However, u is not identically zero. 19. Solve Laplace’s equation 4u = 0 in the semi infinite strip S = {(x, y) : 0 < x < 1, 0 < y}, subject to the following boundary conditions

 

u(0, y) = 0 u(1, y) = 0  u(x, 0) = f (x)

when 0 ≤ y, when 0 ≤ y, when 0 ≤ x ≤ 1

where f is a given function, with of course f (0) = f (1) = 0. Write f (x) =

∞ X

an sin(nπx)

n=1

and expand the general solution in terms of the special solutions given by un (x, y) = e−nπy sin(nπx). Express u as an integral involving f , analogous to the Poisson integral formula (6). 20. Consider the Dirichlet problem in the annulus defined by {(r, θ) : ρ < r < 1}, where 0 < ρ < 1 is the inner radius. The problem is to solve ∂ 2 u 1 ∂u 1 ∂2u + + 2 2 =0 2 ∂r r ∂r r ∂θ subject to the boundary conditions

½

u(1, θ) = f (θ), u(ρ, θ) = g(θ),

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7. Problems

where f and g are given continuous functions. Arguing as we have previously for the Dirichlet problem in the disc, we can hope to write u(r, θ) =

X

cn (r)einθ

with cn (r) = An rn + Bn r−n , n 6= 0. Set f (θ) ∼

X

an einθ

and

g(θ) ∼

X

bn einθ .

We want cn (1) = an and cn (ρ) = bn . This leads to the solution u(r, θ) =

Xµ n6=0

1 ρn − ρ−n

¶

+a0 + (b0 − a0 )

¤

£

((ρ/r)n − (r/ρ)n ) an + (rn − r−n )bn einθ

log r . log ρ

Show that as a result we have u(r, θ) − (Pr ∗ f )(θ) → 0

as r → 1 uniformly in θ,

and u(r, θ) − (Pρ/r ∗ g)(θ) → 0

as r → ρ uniformly in θ.

7 Problems 1. One can construct Riemann integrable functions on [0, 1] that have a dense set of discontinuities as follows. (a) Let f (x) = 0 when x < 0, and f (x) = 1 if x ≥ 0. Choose a countable dense sequence {rn } in [0, 1]. Then, show that the function F (x) =

∞ X 1 n=1

n2

f (x − rn )

is integrable and has discontinuities at all points of the sequence {rn }. [Hint: F is monotonic and bounded.] (b) Consider next F (x) =

∞ X

3−n g(x − rn ),

n=1

where g(x) = sin 1/x when x 6= 0, and g(0) = 0. Then F is integrable, discontinuous at each x = rn , and fails toP be monotonic in any subinterval of [0, 1]. [Hint: Use the fact that 3−k > n>k 3−n .]

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(c) The original example of Riemann is the function F (x) =

∞ X (nx) n=1

n2

,

where (x) = x for x ∈ (−1/2, 1/2] and (x) is continued to R by periodicity, that is, (x + 1) = (x). It can be shown that F is discontinuous whenever x = m/2n, where m, n ∈ Z with m odd and n 6= 0.

2. Let DN denote the Dirichlet kernel N X

DN (θ) =

eikθ =

k=−N

sin((N + 1/2)θ) , sin(θ/2)

and define LN

Z

1 = 2π

π

|DN (θ)| dθ. −π

(a) Prove that LN ≥ c log N +1/2)θ) for some constant c > 0. [Hint: Show that |DN (θ)| ≥ c sin((N|θ| , change variables, and prove that

Z LN ≥ c π

Nπ

| sin θ| dθ + O(1). |θ|

PN −1 R (k+1)π

. To conclude, use the fact that Write the integral as a sum k=1 kπ 1/k ≥ c log n.] A more careful estimate gives k=1

Pn

LN =

4 log N + O(1). π2

(b) Prove the following as a consequence: for each n ≥ 1, there exists a continuous function fn such that |fn | ≤ 1 and |Sn (fn )(0)| ≥ c0 log n. [Hint: The function gn which is equal to 1 when Dn is positive and −1 when Dn is negative has the desired property but is not continuous. Approximate gn in the integral norm (in the sense of Lemma 3.2) by continuous functions hk satisfying |hk | ≤ 1.] 3.∗ Littlewood provided a refinement of Tauber’s theorem:

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7. Problems

(a) If s.

P

cn is Abel summable to s and cn = O(1/n), then

P

(b) As P a consequence, if cn converges to s.

P

cn converges to

cn is Ces`aro summable to s and cn = O(1/n), then

These results may be applied to Fourier series. By Exercise 17, they imply that if f is an integrable function that satisfies fˆ(ν) = O(1/|ν|), then: (i) If f is continuous at θ, then SN (f )(θ) → f (θ)

as N → ∞.

(ii) If f has a jump discontinuity at θ, then SN (f )(θ) →

f (θ+ ) + f (θ− ) 2

as N → ∞.

(iii) If f is continuous on [−π, π], then SN (f ) → f uniformly. For the simpler assertion (b), hence a proof of (i), (ii), and (iii), see Problem 5 in Chapter 4.

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3 Convergence of Fourier Series The sine and cosine series, by which one can represent an arbitrary function in a given interval, enjoy among other remarkable properties that of being convergent. This property did not escape the great geometer (Fourier) who began, through the introduction of the representation of functions just mentioned, a new career for the applications of analysis; it was stated in the Memoir which contains his first research on heat. But no one so far, to my knowledge, gave a general proof of it . . . G. Dirichlet, 1829

In this chapter, we continue our study of the problem of convergence of Fourier series. We approach the problem from two different points of view. The first is “global” and concerns the overall behavior of a function f over the entire interval [0, 2π]. The result we have in mind is “meansquare convergence”: if f is integrable on the circle, then Z 2π 1 |f (θ) − SN (f )(θ)|2 dθ → 0 as N → ∞. 2π 0 At the heart of this result is the fundamental notion of “orthogonality”; this idea is expressed in terms of vector spaces with inner products, and their related infinite dimensional variants, the Hilbert spaces. A connected result is the Parseval identity which equates the mean-square “norm” of the function with a corresponding norm of its Fourier coefficients. Orthogonality is a fundamental mathematical notion which has many applications in analysis. The second viewpoint is “local” and concerns the behavior of f near a given point. The main question we consider is the problem of pointwise convergence: does the Fourier series of f converge to the value f (θ) for a given θ? We first show that this convergence does indeed hold whenever f is differentiable at θ. As a corollary, we obtain the Riemann localization principle, which states that the question of whether or not SN (f )(θ) → f (θ) is completely determined by the behavior of f in an

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arbitrarily small interval about θ. This is a remarkable result since the Fourier coefficients, hence the Fourier series, of f depend on the values of f on the whole interval [0, 2π]. Even though convergence of the Fourier series holds at points where f is differentiable, it may fail if f is merely continuous. The chapter concludes with the presentation of a continuous function whose Fourier series does not converge at a given point, as promised earlier.

1 Mean-square convergence of Fourier series The aim of this section is the proof of the following theorem. Theorem 1.1 Suppose f is integrable on the circle. Then 1 2π

Z

2π 0

|f (θ) − SN (f )(θ)|2 dθ → 0

as N → ∞.

As we remarked earlier, the key concept involved is that of orthogonality. The correct setting for orthogonality is in a vector space equipped with an inner product. 1.1 Vector spaces and inner products We now review the definitions of a vector space over R or C, an inner product, and its associated norm. In addition to the familiar finitedimensional vector spaces Rd and Cd , we also examine two infinitedimensional examples which play a central role in the proof of Theorem 1.1. Preliminaries on vector spaces A vector space V over the real numbers R is a set whose elements may be “added” together, and “multiplied” by scalars. More precisely, we may associate to any pair X, Y ∈ V an element in V called their sum and denoted by X + Y . We require that this addition respects the usual laws of arithmetic, such as commutativity X + Y = Y + X, and associativity X + (Y + Z) = (X + Y ) + Z, etc. Also, given any X ∈ V and real number λ, we assign an element λX ∈ V called the product of X by λ. This scalar multiplication must satisfy the standard properties, for instance λ1 (λ2 X) = (λ1 λ2 )X and λ(X + Y ) = λX + λY . We may instead allow scalar multiplication by numbers in C; we then say that V is a vector space over the complex numbers.

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1. Mean-square convergence of Fourier series

For example, the set Rd of d-tuples of real numbers (x1 , x2 , . . . , xd ) is a vector space over the reals. Addition is defined componentwise by (x1 , . . . , xd ) + (y1 , . . . , yd ) = (x1 + y1 , . . . , xd + yd ), and so is multiplication by a scalar λ ∈ R: λ(x1 , . . . , xd ) = (λx1 , . . . , λxd ). Similarly, the space Cd (the complex version of the previous example) is the set of d-tuples of complex numbers (z1 , z2 , . . . , zd ). It is a vector space over C with addition defined componentwise by (z1 , . . . , zd ) + (w1 , . . . , wd ) = (z1 + w1 , . . . , zd + wd ). Multiplication by scalars λ ∈ C is given by λ(z1 , . . . , zd ) = (λz1 , . . . , λzd ). An inner product on a vector space V over R associates to any pair X, Y of elements in V a real number which we denote by (X, Y ). In particular, the inner product must be symmetric (X, Y ) = (Y, X) and linear in both variables; that is, (αX + βY, Z) = α(X, Z) + β(Y, Z) whenever α, β ∈ R and X, Y, Z ∈ V . Also, we require that the inner product be positive-definite, that is, (X, X) ≥ 0 for all X in V . In particular, given an inner product (·, ·) we may define the norm of X by kXk = (X, X)1/2 . If in addition kXk = 0 implies X = 0, we say that the inner product is strictly positive-definite. For example, the space Rd is equipped with a (strictly positive-definite) inner product defined by (X, Y ) = x1 y1 + · · · + xd yd when X = (x1 , . . . , xd ) and Y = (y1 , . . . , yd ). Then kXk = (X, X)1/2 =

q

x21 + · · · + x2d ,

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which is the usual Euclidean distance. One also uses the notation |X| instead of kXk. For vector spaces over the complex numbers, the inner product of two elements is a complex number. Moreover, these inner products are called Hermitian (instead of symmetric) since they must satisfy (X, Y ) = (Y, X). Hence the inner product is linear in the first variable, but conjugate-linear in the second: (αX + βY, Z) = α(X, Z) + β(Y, Z)

and

(X, αY + βZ) = α(X, Y ) + β(X, Z). Also, we must have (X, X) ≥ 0, and the norm of X is defined by kXk = (X, X)1/2 as before. Again, the inner product is strictly positivedefinite if kXk = 0 implies X = 0. For example, the inner product of two vectors Z = (z1 , . . . , zd ) and W = (w1 , . . . , wd ) in Cd is defined by (Z, W ) = z1 w1 + · · · + zd wd . The norm of the vector Z is then given by q kZk = (Z, Z)1/2 = |z1 |2 + · · · + |zd |2 . The presence of an inner product on a vector space allows one to define the geometric notion of “orthogonality.” Let V be a vector space (over R or C) with inner product (·, ·) and associated norm k · k. Two elements X and Y are orthogonal if (X, Y ) = 0, and we write X ⊥ Y . Three important results can be derived from this notion of orthogonality: (i) The Pythagorean theorem: if X and Y are orthogonal, then kX + Y k2 = kXk2 + kY k2 . (ii) The Cauchy-Schwarz inequality: for any X, Y ∈ V we have |(X, Y )| ≤ kXk kY k. (iii) The triangle inequality: for any X, Y ∈ V we have kX + Y k ≤ kXk + kY k.

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73

The proofs of these facts are simple. For (i) it suffices to expand (X + Y, X + Y ) and use the assumption that (X, Y ) = 0. For (ii), we first dispose of the case when kY k = 0 by showing that this implies (X, Y ) = 0 for all X. Indeed, for all real t we have 0 ≤ kX + tY k2 = kXk2 + 2t Re(X, Y ) and Re(X, Y ) 6= 0 contradicts the inequality if we take t to be large and positive (or negative). Similarly, by considering kX + itY k2 , we find that Im(X, Y ) = 0. If kY k 6= 0, we may set c = (X, Y )/(Y, Y ); then X − cY is orthogonal to Y , and therefore also to cY . If we write X = X − cY + cY and apply the Pythagorean theorem, we get kXk2 = kX − cY k2 + kcY k2 ≥ |c|2 kY k2 . Taking square roots on both sides gives the result. Note that we have equality in the above precisely when X = cY . Finally, for (iii) we first note that kX + Y k2 = (X, X) + (X, Y ) + (Y, X) + (Y, Y ). But (X, X) = kXk2 , (Y, Y ) = kY k2 , and by the Cauchy-Schwarz inequality |(X, Y ) + (Y, X)| ≤ 2 kXk kY k, therefore kX + Y k2 ≤ kXk2 + 2 kXk kY k + kY k2 = (kXk + kY k)2 . Two important examples The vector spaces Rd and Cd are finite dimensional. In the context of Fourier series, we need to work with two infinite-dimensional vector spaces, which we now describe. Example 1. The vector space `2 (Z) over C is the set of all (two-sided) infinite sequences of complex numbers (. . . , a−n , . . . , a−1 , a0 , a1 , . . . , an , . . .) such that X n∈Z

|an |2 < ∞;

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that is, the series converges. Addition is defined componentwise, and so is scalar multiplication. The inner product between the two vectors A = (. . . , a−1 , a0 , a1 , . . .) and B = (. . . , b−1 , b0 , b1 , . . .) is defined by the absolutely convergent series X (A, B) = an bn . n∈Z

The norm of A is then given by à 1/2

kAk = (A, A)

=

X

!1/2

|an |

2

.

n∈Z

We must first check that `2 (Z) is a vector space. This requires that if A and B are two elements in `2 (Z), then so is the vector A + B. To see this, for each integer N > 0 we let AN denote the truncated element AN = (. . . , 0, 0, a−N , . . . , a−1 , a0 , a1 , . . . , aN , 0, 0, . . .), where we have set an = 0 whenever |n| > N . We define the truncated element BN similarly. Then, by the triangle inequality which holds in a finite dimensional Euclidean space, we have kAN + BN k ≤ kAN k + kBN k ≤ kAk + kBk. Thus X

|an + bn |2 ≤ (kAk + kBk)2 ,

|n|≤N

P and letting N tend to infinity gives n∈Z |an + bn |2 < ∞. It also follows that kA + Bk ≤ kAk + kBk, which is the triangle inequality. The P Cauchy-Schwarz inequality, which states that the sum n∈Z an bn converges absolutely and that |(A, B)| ≤ kAk kBk, can be deduced in the same way from its finite analogue.

In the three examples Rd , Cd , and `2 (Z), the vector spaces with their inner products and norms satisfy two important properties: (i) The inner product is strictly positive-definite, that is, kXk = 0 implies X = 0. (ii) The vector space is complete, which by definition means that every Cauchy sequence in the norm converges to a limit in the vector space.

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1. Mean-square convergence of Fourier series

An inner product space with these two properties is called a Hilbert space. We see that Rd and Cd are examples of finite-dimensional Hilbert spaces, while `2 (Z) is an example of an infinite-dimensional Hilbert space (see Exercises 1 and 2). If either of the conditions above fail, the space is called a pre-Hilbert space. We now give an important example of a pre-Hilbert space where both conditions (i) and (ii) fail. Example 2. Let R denote the set of complex-valued Riemann integrable functions on [0, 2π] (or equivalently, integrable functions on the circle). This is a vector space over C. Addition is defined pointwise by (f + g)(θ) = f (θ) + g(θ). Naturally, multiplication by a scalar λ ∈ C is given by (λf )(θ) = λ · f (θ). An inner product is defined on this vector space by Z 2π 1 f (θ)g(θ) dθ. (1) (f, g) = 2π 0 The norm of f is then µ

kf k =

1 2π

Z

2π

¶1/2 2

|f (θ)| dθ

.

0

One needs to check that the analogue of the Cauchy-Schwarz and triangle inequalities hold in this example; that is, |(f, g)| ≤ kf k kgk and kf + gk ≤ kf k + kgk. While these facts can be obtained as consequences of the corresponding inequalities in the previous examples, the argument is a little elaborate and we prefer to proceed differently. We first observe that 2AB ≤ (A2 + B 2 ) for any two real numbers A and B. If we set A = λ1/2 |f (θ)| and B = λ−1/2 |g(θ)| with λ > 0, we get |f (θ)g(θ)| ≤

1 (λ|f (θ)|2 + λ−1 |g(θ)|2 ). 2

We then integrate this in θ to obtain Z 2π 1 1 |(f, g)| ≤ |f (θ)| |g(θ)| dθ ≤ (λkf k2 + λ−1 kgk2 ). 2π 0 2 Then, put λ = kgk/kf k to get the Cauchy-Schwarz inequality. The triangle inequality is then a simple consequence, as we have seen above.

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Of course, in our choice of λ we must assume that kf k 6= 0 and kgk 6= 0, which leads us to the following observation. In R, condition (i) for a Hilbert space fails, since kf k = 0 implies only that f vanishes at its points of continuity. This is not a very serious problem since in the appendix we show that an integrable function is continuous except for a “negligible” set, so that kf k = 0 implies that f vanishes except on a set of “measure zero.” One can get around the difficulty that f is not identically zero by adopting the convention that such functions are actually the zero function, since for the purpose of integration, f behaves precisely like the zero function. A more essential difficulty is that the space R is not complete. One way to see this is to start with the function (

f (θ) =

0 log(1/θ)

for θ = 0, for 0 < θ ≤ 2π.

Since f is not bounded, it does not belong to the space R. Moreover, the sequence of truncations fn defined by (

fn (θ) =

0 for 0 ≤ θ ≤ 1/n, f (θ) for 1/n < θ ≤ 2π

can easily be seen to form a Cauchy sequence in R (see Exercise 5). However, this sequence cannot converge to an element in R, since that limit, if it existed, would have to be f ; for another example, see Exercise 7. This and more complicated examples motivate the search for the completion of R, the class of Riemann integrable functions on [0, 2π]. The construction and identification of this completion, the Lebesgue class L2 ([0, 2π]), represents an important turning point in the development of analysis (somewhat akin to the much earlier completion of the rationals, that is, the passage from Q to R). A further discussion of these fundamental ideas will be postponed until Book III, where we take up the Lebesgue theory of integration. We now turn to the proof of Theorem 1.1. 1.2 Proof of mean-square convergence Consider the space R of integrable functions on the circle with inner product Z 2π 1 f (θ)g(θ) dθ (f, g) = 2π 0

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1. Mean-square convergence of Fourier series

and norm kf k defined by 1 kf k = (f, f ) = 2π

Z

2π

2

|f (θ)|2 dθ.

0

With this notation, we must prove that kf − SN (f )k → 0 as N tends to infinity. For each integer n, let en (θ) = einθ , and observe that the family {en }n∈Z is orthonormal; that is, ( 1 if n = m (en , em ) = 0 if n 6= m. Let f be an integrable function on the circle, and let an denote its Fourier coefficients. An important observation is that these Fourier coefficients are represented by inner products of f with the elements in the orthonormal set {en }n∈Z : (f, en ) =

1 2π

Z

2π

0

f (θ)e−inθ dθ = an .

P

In particular, SN (f ) = |n|≤N an en . Then the orthonormal property of the family {en } and the fact that an = (f, en ) imply that the difference P f − |n|≤N an en is orthogonal to en for all |n| ≤ N . Therefore, we must have X X (2) (f − an en ) ⊥ bn en |n|≤N

|n|≤N

for any complex numbers bn . We draw two conclusions from this fact. First, we can apply the Pythagorean theorem to the decomposition X X f =f− an en + an en , |n|≤N

|n|≤N

where we now choose bn = an , to obtain X X kf k2 = kf − an en k2 + k an en k2 . |n|≤N

|n|≤N

Since the orthonormal property of the family {en }n∈Z implies that X X k an en k2 = |an |2 , |n|≤N

|n|≤N

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we deduce that (3)

kf k2 = kf − SN (f )k2 +

X

|an |2 .

|n|≤N

The second conclusion we may draw from (2) is the following simple lemma. Lemma 1.2 (Best approximation) If f is integrable on the circle with Fourier coefficients an , then X kf − SN (f )k ≤ kf − cn en k |n|≤N

for any complex numbers cn . Moreover, equality holds precisely when cn = an for all |n| ≤ N . Proof. rem to

This follows immediately by applying the Pythagorean theoX

f−

cn en = f − SN (f ) +

|n|≤N

X

bn en ,

|n|≤N

where bn = an − cn . This lemma has a clear geometric interpretation. It says that the trigonometric polynomial of degree at most N which is closest to f in the norm k · k is the partial sum SN (f ). This geometric property of the partial sums is depicted in Figure 1, where the orthogonal projection of f in the plane spanned by {e−N , . . . , e0 , . . . , eN } is simply SN (f ).

f

f − SN (f )

e−N eN

e0

SN (f )

e1

Figure 1. The best approximation lemma

We can now give the proof that kSN (f ) − f k → 0 using the best approximation lemma, as well as the important fact that trigonometric polynomials are dense in the space of continuous functions on the circle.

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1. Mean-square convergence of Fourier series

Suppose that f is continuous on the circle. Then, given ² > 0, there exists (by Corollary 5.4 in Chapter 2) a trigonometric polynomial P , say of degree M , such that |f (θ) − P (θ)| < ²

for all θ.

In particular, taking squares and integrating this inequality yields kf − P k < ², and by the best approximation lemma we conclude that kf − SN (f )k < ²

whenever N ≥ M .

This proves Theorem 1.1 when f is continuous. If f is merely integrable, we can no longer approximate f uniformly by trigonometric polynomials. Instead, we apply the approximation Lemma 3.2 in Chapter 2 and choose a continuous function g on the circle which satisfies sup

|g(θ)| ≤ sup

θ∈[0,2π]

and

Z

|f (θ)| = B,

θ∈[0,2π]

2π

|f (θ) − g(θ)| dθ < ²2 .

0

Then we get Z 2π 1 |f (θ) − g(θ)|2 dθ kf − gk = 2π 0 Z 2π 1 = |f (θ) − g(θ)| |f (θ) − g(θ)| dθ 2π 0 Z 2B 2π ≤ |f (θ) − g(θ)| dθ 2π 0 ≤ C²2 . 2

Now we may approximate g by a trigonometric polynomial P so that kg − P k < ². Then kf − P k < C 0 ², and we may again conclude by applying the best approximation lemma. This completes the proof that the partial sums of the Fourier series of f converge to f in the mean square norm k · k. Note that this result and the relation (3) imply that if an is the nth P 2 Fourier coefficient of an integrable function f , then the series ∞ n=−∞ |an | converges, and in fact we have Parseval’s identity ∞ X n=−∞

|an |2 = kf k2 .

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This identity provides an important connection between the norms in the two vector spaces `2 (Z) and R. We now summarize the results of this section. Theorem 1.3 Let f be an integrable function on the circle with P inθ f∼ ∞ . Then we have: n=−∞ an e (i) Mean-square convergence of the Fourier series Z 2π 1 |f (θ) − SN (f )(θ)|2 dθ → 0 as N → ∞. 2π 0 (ii) Parseval’s identity ∞ X

|an |2 =

n=−∞

1 2π

Z

2π

|f (θ)|2 dθ.

0

Remark 1. If {en } is any orthonormal family of functions on the circle, and an = (f, en ), then we may deduce from the relation (3) that ∞ X

|an |2 ≤ kf k2 .

n=−∞

This is known as Bessel’s inequality. Equality holds (as in Parseval’s identity) precisely when the family {en } is also a “basis,” in the sense P that k |n|≤N an en − f k → 0 as N → ∞. Remark 2. We may associate to every integrable function the sequence {an } formed by its Fourier coefficients. Parseval’s identity guarantees that {an } ∈ `2 (Z). Since `2 (Z) is a Hilbert space, the failure of R to be complete, discussed earlier, may be understood as follows: there P exist sequences {an }n∈Z such that n∈Z |an |2 < ∞, yet no Riemann integrable function F has nth Fourier coefficient equal to an for all n. An example is given in Exercise 6. Since the terms of a converging series tend to 0, we deduce from Parseval’s identity or Bessel’s inequality the following result. Theorem 1.4 (Riemann-Lebesgue lemma) If f is integrable on the circle, then fˆ(n) → 0 as |n| → ∞. An equivalent reformulation of this proposition is that if f is integrable on [0, 2π], then Z 2π f (θ) sin(N θ) dθ → 0 as N → ∞ 0

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2. Return to pointwise convergence

and Z

2π

f (θ) cos(N θ) dθ → 0

as N → ∞.

0

To conclude this section, we give a more general version of the Parseval identity which we will use in the next chapter. Lemma 1.5 Suppose F and G are integrable on the circle with F ∼

X

an einθ

and

G∼

X

bn einθ .

Then 1 2π

Z

2π

F (θ)G(θ) dθ = 0

∞ X

a n bn .

n=−∞

Recall from the discussion in Example 1 that the series converges absolutely.

P∞ n=−∞

a n bn

Proof. The proof follows from Parseval’s identity and the fact that (F, G) =

¡ ¢¤ 1£ kF + Gk2 − kF − Gk2 + i kF + iGk2 − kF − iGk2 4

which holds in every Hermitian inner product space. The verification of this fact is left to the reader.

2 Return to pointwise convergence The mean-square convergence theorem does not provide further insight into the problem of pointwise convergence. Indeed, Theorem 1.1 by itself does not guarantee that the Fourier series converges for any θ. Exercise 3 helps to explain this statement. However, if a function is differentiable at a point θ0 , then its Fourier series converges at θ0 . After proving this result, we give an example of a continuous function with diverging Fourier series at one point. These phenomena are indicative of the intricate nature of the problem of pointwise convergence in the theory of Fourier series. 2.1 A local result Theorem 2.1 Let f be an integrable function on the circle which is differentiable at a point θ0 . Then SN (f )(θ0 ) → f (θ0 ) as N tends to infinity.

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Proof. Define   f (θ0 − t) − f (θ0 )

F (t) =



0

−f (θ0 )

t

if t 6= 0 and |t| < π if t = 0.

First, F is bounded near 0 since f is differentiable there. Second, for all small δ the function F is integrable on [−π, −δ] ∪ [δ, π] because f has this property and |t| > δ there. As a consequence of Proposition 1.4 in the appendix, the function F is integrable on all of [−π, π]. We know that R SN (f )(θ0 ) = (f ∗ DN )(θ0 ), where DN is the Dirichlet kernel. Since 1 DN = 1, we find that 2π Z π 1 SN (f )(θ0 ) − f (θ0 ) = f (θ0 − t)DN (t) dt − f (θ0 ) 2π −π Z π 1 = [f (θ0 − t) − f (θ0 )]DN (t) dt 2π −π Z π 1 F (t)tDN (t) dt. = 2π −π We recall that tDN (t) = where the quotient can write

t sin(t/2)

t sin((N + 1/2)t), sin(t/2)

is continuous in the interval [−π, π]. Since we

sin((N + 1/2)t) = sin(N t) cos(t/2) + cos(N t) sin(t/2), we can apply the Riemann-Lebesgue lemma to the Riemann integrable functions F (t)t cos(t/2)/ sin(t/2) and F (t)t to finish the proof of the theorem. Observe that the conclusion of the theorem still holds if we only assume that f satisfies a Lipschitz condition at θ0 ; that is, |f (θ) − f (θ0 )| ≤ M |θ − θ0 | for some M ≥ 0 and all θ. This is the same as saying that f satisfies a H¨older condition of order α = 1. A striking consequence of this theorem is the localization principle of Riemann. This result states that the convergence of SN (f )(θ0 ) depends only on the behavior of f near θ0 . This is not clear at first, since forming the Fourier series requires integrating f over the whole circle.

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2. Return to pointwise convergence

Theorem 2.2 Suppose f and g are two integrable functions defined on the circle, and for some θ0 there exists an open interval I containing θ0 such that f (θ) = g(θ)

for all θ ∈ I.

Then SN (f )(θ0 ) − SN (g)(θ0 ) → 0 as N tends to infinity. Proof. The function f − g is 0 in I, so it is differentiable at θ0 , and we may apply the previous theorem to conclude the proof. 2.2 A continuous function with diverging Fourier series We now turn our attention to an example of a continuous periodic function whose Fourier series diverges at a point. Thus, Theorem 2.1 fails if the differentiability assumption is replaced by the weaker assumption of continuity. Our counter-example shows that this hypothesis which had appeared plausible, is in fact false; moreover, its construction also illuminates an important principle of the theory. The principle that is involved here will be referred to as “symmetrybreaking.”1 The symmetry that we have in mind is the symmetry between the frequencies einθ and e−inθ which appear in the Fourier expansion of a function. For example, the partial sum operator SN is defined in a way that reflects this symmetry. Also, the Dirichlet, Fej`er, and Poisson kernels are symmetric in this sense. When we break the symmeP inθ try, that is, when we split the Fourier series ∞ into the two n=−∞ an e P P inθ inθ pieces n≥0 an e and n 3Nk , (ii) αk log Nk → ∞ as k → ∞. k

We choose (for example) αk = 1/k 2 and Nk = 32 which are easily seen to satisfy the above criteria. Finally, we can write down our desired function. It is f (θ) =

∞ X

αk PNk (θ).

k=1

Due to the uniform boundedness of the PN (recall that |PN (θ)| = |fN (θ)|), the series above converges uniformly to a continuous periodic function. However, by our lemma we get |S2Nm (f )(0)| ≥ cαm log Nm + O(1) → ∞

Nk−1

3Nk−1

Nk

3Nk

Nk+1

as m → ∞.

3Nk+1

2Nk

Figure 4. Symmetry broken in the middle interval (Nk , 3Nk ) Indeed, the terms that correspond to Nk with k < m or k > m contribute O(1) or 0, respectively (because the PN ’s are uniformly bounded), while the term that corresponds to Nm is in absolute value greater than cαm log Nm because |P˜N (θ)| = |f˜N (θ)| ≥ c log N . So the partial sums of the Fourier series of f at 0 are not bounded, and we are done since this proves the divergence of the Fourier series of f at θ = 0. To produce a function whose series diverges at any other preassigned θ = θ0 , it suffices to consider the function f (θ − θ0 ).

3 Exercises 1. Show that the first two examples of inner product spaces, namely Rd and Cd , are complete.

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[Hint: Every Cauchy sequence in R has a limit.] 2. Prove that the vector space `2 (Z) is complete. [Hint: Suppose Ak = {ak,n }n∈Z with k = 1, 2, . . . is a Cauchy sequence. Show that for each n, {ak,n }∞ k=1 is a Cauchy sequence of complex numbers, therefore it converges to a limit, say bn . By taking partial sums of kAk − Ak0 k and letting k 0 → ∞, show that kAk − Bk → 0 as k → ∞, where B = (. . . , b−1 , b0 , b1 , . . .). Finally, prove that B ∈ `2 (Z).] 3. Construct a sequence of integrable functions {fk } on [0, 2π] such that 1 k→∞ 2π

Z

2π

|fk (θ)|2 dθ = 0

lim

0

but limk→∞ fk (θ) fails to exist for any θ. [Hint: Choose a sequence of intervals Ik ⊂ [0, 2π] whose lengths tend to 0, and so that each point belongs to infinitely many of them; then let fk = χIk .] 4. Recall the vector space R of integrable functions, with its inner product and norm

µ kf k =

1 2π

Z

¶1/2

2π

|f (x)|2 dx

.

0

(a) Show that there exist non-zero integrable functions f for which kf k = 0. (b) However, show that if f ∈ R with kf k = 0, then f (x) = 0 whenever f is continuous at x. (c) Conversely, show that if f ∈ R vanishes at all of its points of continuity, then kf k = 0.

5. Let

( f (θ) =

0 for θ = 0 log(1/θ) for 0 < θ ≤ 2π,

and define a sequence of functions in R by

( fn (θ) =

0 for 0 ≤ θ ≤ 1/n f (θ) for 1/n < θ ≤ 2π.

Prove that {fn }∞ n=1 is a Cauchy sequence in R. However, f does not belong to R.

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Rb

[Hint: Show that a (log θ)2 dθ → 0 if 0 < a < b and b → 0, by using the fact that the derivative of θ(log θ)2 − 2θ log θ + 2θ is equal to (log θ)2 .] 6. Consider the sequence {ak }∞ k=−∞ defined by

½ ak =

1/k 0

if k ≥ 1 if k ≤ 0.

Note that {ak } ∈ `2 (Z), but that no Riemann integrable function has k th Fourier coefficient equal to ak for all k. 7. Show that the trigonometric series

X n≥2

1 sin nx log n

converges for every x, yet it is not the Fourier series of a Riemann integrable function. P sin nx The same is true for nα for 0 < α < 1, but the case 1/2 < α < 1 is more difficult. See Problem 1. 8. Exercise 6 in Chapter 2 dealt with the sums

X n odd ≥1

1 n2

and

∞ X 1 n=1

n2

.

Similar sums can be derived using the methods of this chapter. (a) Let f be the function defined on [−π, π] by f (θ) = |θ|. Use Parseval’s identity to find the sums of the following two series: ∞ X n=0

1 (2n + 1)4

and

∞ X 1 n=1

n4

.

In fact, they are π 4 /96 and π 4 /90, respectively. (b) Consider the 2π-periodic odd function defined on [0, π] by f (θ) = θ(π − θ). Show that ∞ X n=0

1 π6 = (2n + 1)6 960

and

∞ X 1 n=1

n6

P∞

=

π6 . 945

Remark. The general expression when k is even for n=1 1/nk P in terms of π k ∞ 3 is given in Problem finding a formula for the sum n=1 1/n , or P∞ 4. However, k more generally n=1 1/n with k odd, is a famous unresolved question.

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9. Show that for α not an integer, the Fourier series of π ei(π−x)α sin πα on [0, 2π] is given by ∞ X n=−∞

einx . n+α

Apply Parseval’s formula to show that ∞ X n=−∞

π2 1 = . 2 (n + α) (sin πα)2

10. Consider the example of a vibrating string which we analyzed in Chapter 1. The displacement u(x, t) of the string at time t satisfies the wave equation ∂2u 1 ∂2u = , c2 ∂t2 ∂x2

c2 = τ /ρ.

The string is subject to the initial conditions u(x, 0) = f (x)

∂u (x, 0) = g(x), ∂t

and

where we assume that f ∈ C 1 and g is continuous. We define the total energy of the string by 1 E(t) = ρ 2

Z

L

µ

0

∂u ∂t

¶2

1 dx + τ 2

Z

L

µ

0

∂u ∂x

¶2 dx.

The first term corresponds to the “kinetic energy” of the string (in analogy with (1/2)mv 2 , the kinetic energy of a particle of mass m and velocity v), and the second term corresponds to its “potential energy.” Show that the total energy of the string is conserved, in the sense that E(t) is constant. Therefore, 1 E(t) = E(0) = ρ 2

Z

L

1 g(x) dx + τ 2

Z

L

2

0

f 0 (x)2 dx. 0

11. The inequalities of Wirtinger and Poincar´e establish a relationship between the norm of a function and that of its derivative.

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3. Exercises

(a) If f is T -periodic, continuous, and piecewise C 1 with that

Z

T

|f (t)|2 dt ≤ 0

T2 4π 2

Z

RT 0

f (t) dt = 0, show

T

|f 0 (t)|2 dt, 0

with equality if and only if f (t) = A sin(2πt/T ) + B cos(2πt/T ). [Hint: Apply Parseval’s identity.] (b) If f is as above and g is just C 1 and T -periodic, prove that

¯Z ¯ ¯ ¯

0

T

¯2 Z T Z T ¯ T2 2 ¯ f (t)g(t) dt¯ ≤ |f (t)| dt |g 0 (t)|2 dt. 4π 2 0 0

(c) For any compact interval [a, b] and any continuously differentiable function f with f (a) = f (b) = 0, show that

Z

b

|f (t)|2 dt ≤ a

(b − a)2 π2

Z

b

|f 0 (t)|2 dt. a

Discuss the case of equality, and prove that the constant (b − a)2 /π 2 cannot be improved. [Hint: Extend f to be odd with respect to a and periodic of period T = 2(b − a) so that its integral over an interval of length T is 0. Apply part a) to get the inequality, and conclude that equality holds if t−a and only if f (t) = A sin(π b−a )].

Z

∞

sin x π dx = . x 2 0 [Hint: Start with the fact that the integral of DN (θ) equals 2π, and note that the difference (1/ sin(θ/2)) − 2/θ is continuous on [−π, π]. Apply the RiemannLebesgue lemma.] 12. Prove that

13. Suppose that f is periodic and of class C k . Show that fˆ(n) = o(1/|n|k ), that is, |n|k fˆ(n) goes to 0 as |n| → ∞. This is an improvement over Exercise 10 in Chapter 2. [Hint: Use the Riemann-Lebesgue lemma.] 14. Prove that the Fourier series of a continuously differentiable function f on the circle is absolutely convergent. [Hint: Use the Cauchy-Schwarz inequality and Parseval’s identity for f 0 .] 15. Let f be 2π-periodic and Riemann integrable on [−π, π].

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(a) Show that 1 fˆ(n) = − 2π

Z

π

f (x + π/n)e−inx dx −π

hence 1 fˆ(n) = 4π

Z

π

[f (x) − f (x + π/n)]e−inx dx. −π

(b) Now assume that f satisfies a H¨older condition of order α, namely |f (x + h) − f (x)| ≤ C|h|α for some 0 < α ≤ 1, some C > 0, and all x, h. Use part a) to show that fˆ(n) = O(1/|n|α ). (c) Prove that the above result cannot be improved by showing that the function f (x) =

∞ X

k

2−kα ei2 x ,

k=0

where 0 < α < 1, satisfies |f (x + h) − f (x)| ≤ C|h|α , and fˆ(N ) = 1/N α whenever N = 2k .

P

[Hint: For (c), break up the sum as follows f (x + h) − f (x) = 2k ≤1/|h| + iθ 2k >1/|h| . To estimate the first sum use the fact that |1 − e | ≤ |θ| whenever θ is small. To estimate the second sum, use the obvious inequality |eix − eiy | ≤ 2.]

P

16. Let f be a 2π-periodic function which satisfies a Lipschitz condition with constant K; that is, |f (x) − f (y)| ≤ K|x − y|

for all x, y.

This is simply the H¨older condition with α = 1, so by Pthe previous exercise, we see that fˆ(n) = O(1/|n|). Since the harmonic series 1/n diverges, we cannot say anything (yet) about the absolute convergence of the Fourier series of f . The outline below actually proves that the Fourier series of f converges absolutely and uniformly.

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3. Exercises

(a) For every positive h we define gh (x) = f (x + h) − f (x − h). Prove that 1 2π

Z

2π

|gh (x)|2 dx = 0

∞ X

4| sin nh|2 |fˆ(n)|2 ,

n=−∞

and show that ∞ X

| sin nh|2 |fˆ(n)|2 ≤ K 2 h2 .

n=−∞

(b) Let p be a positive integer. By choosing h = π/2p+1 , show that

X

|fˆ(n)|2 ≤

2p−1 1.

Deduce from the result in (c), and the expression for the cotangent function obtained in the previous problem, that x cot x = 1 − 2

∞ X ζ(2m)

π 2m

m=1

x2m .

(e) Conclude that 2ζ(2m) = (−1)m+1

(2π)2m B2m . (2m)!

5. Define the Bernoulli polynomials Bn (x) by the formula ∞

X Bn (x) zexz = zn. z e −1 n! n=0

(a) The functions Bn (x) are polynomials in x and Bn (x) =

n µ ¶ X n k=0

k

Bk xn−k .

Show that B0 (x) = 1, B1 (x) = x − 1/2, B2 (x) = x2 − x + 1/6, and B3 (x) = x3 − 23 x2 + 12 x. (b) If n ≥ 1, then Bn (x + 1) − Bn (x) = nxn−1 , and if n ≥ 2, then Bn (0) = Bn (1) = Bn . (c) Define Sm (n) = 1m + 2m + · · · + (n − 1)m . Show that (m + 1)Sm (n) = Bm+1 (n) − Bm+1 .

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4. Problems

(d) Prove that the Bernoulli polynomials are the only polynomials that satisfy (i) B0 (x) = 1, (ii) Bn0 (x) = nBn−1 (x) for n ≥ 1, (iii)

R1 0

Bn (x) dx = 0 for n ≥ 1, and show that from (b) one obtains

Z

x+1

Bn (t) dt = xn . x

(e) Calculate the Fourier series of B1 (x) to conclude that for 0 < x < 1 we have ∞

B1 (x) = x − 1/2 =

−1 X sin(2πkx) . π k k=1

Integrate and conclude that B2n (x) = (−1)n+1

∞ 2(2n)! X cos(2πkx) , (2π)2n k 2n k=1

∞

B2n+1 (x) = (−1)n+1

2(2n + 1)! X sin(2πkx) . (2π)2n+1 k 2n+1 k=1

Finally, show that for 0 < x < 1, Bn (x) = −

n! X e2πikx . (2πi)n kn k6=0

We observe that the Bernoulli polynomials are, up to normalization, successive integrals of the sawtooth function.

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4 Some Applications of Fourier Series Fourier series and analogous expansions intervene very naturally in the general theory of curves and surfaces. In effect, this theory, conceived from the point of view of analysis, deals obviously with the study of arbitrary functions. I was thus led to use Fourier series in several questions of geometry, and I have obtained in this direction a number of results which will be presented in this work. One notes that my considerations form only a beginning of a principal series of researches, which would without doubt give many new results. A. Hurwitz, 1902

In the previous chapters we introduced some basic facts about Fourier analysis, motivated by problems that arose in physics. The motion of a string and the diffusion of heat were two instances that led naturally to the expansion of a function in terms of a Fourier series. We propose next to give the reader a flavor of the broader impact of Fourier analysis, and illustrate how these ideas reach out to other areas of mathematics. In particular, consider the following three problems: I. Among all simple closed curves of length ` in the plane R2 , which one encloses the largest area? II. Given an irrational number γ, what can be said about the distribution of the fractional parts of the sequence of numbers nγ, for n = 1, 2, 3, . . .? III. Does there exist a continuous function that is nowhere differentiable? The first problem is clearly geometric in nature, and at first sight, would seem to have little to do with Fourier series. The second question lies on the border between number theory and the study of dynamical systems, and gives us the simplest example of the idea of “ergodicity.” The third problem, while analytic in nature, resisted many attempts before the

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1. The isoperimetric inequality

solution was finally discovered. It is remarkable that all three questions can be resolved quite simply and directly by the use of Fourier series. In the last section of this chapter, we return to a problem that provided our initial motivation. We consider the time-dependent heat equation on the circle. Here our investigation will lead us to the important but enigmatic heat kernel for the circle. However, the mysteries surrounding its basic properties will not be fully understood until we can apply the Poisson summation formula, which we will do in the next chapter.

1 The isoperimetric inequality Let Γ denote a closed curve in the plane which does not intersect itself. Also, let ` denote the length of Γ, and A the area of the bounded region in R2 enclosed by Γ. The problem now is to determine for a given ` the curve Γ which maximizes A (if any such curve exists).

small A

large A

Γ

Γ

Figure 1. The isoperimetric problem

A little experimentation and reflection suggests that the solution should be a circle. This conclusion can be reached by the following heuristic considerations. The curve can be thought of as a closed piece of string lying flat on a table. If the region enclosed by the string is not convex (for example), one can deform part of the string and increase the area enclosed by it. Also, playing with some simple examples, one can convince oneself that the “flatter” the curve is in some portion, the less efficient it is in enclosing area. Therefore we want to maximize the “roundness” of the curve at each point. Although the circle is the correct guess, making the above ideas precise is a difficult matter. The key idea in the solution we give to the isoperimetric problem consists of an application of Parseval’s identity for Fourier series. However, before we can attempt a solution to this problem, we must define the

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notion of a simple closed curve, its length, and what we mean by the area of the region enclosed by it. Curves, length and area A parametrized curve γ is a mapping γ : [a, b] → R2 . The image of γ is a set of points in the plane which we call a curve and denote by Γ. The curve Γ is simple if it does not intersect itself, and closed if its two end-points coincide. In terms of the parametrization above, these two conditions translate into γ(s1 ) 6= γ(s2 ) unless s1 = a and s2 = b, in which case γ(a) = γ(b). We may extend γ to a periodic function on R of period b − a, and think of γ as a function on the circle. We also always impose some smoothness on our curves by assuming that γ is of class C 1 , and that its derivative γ 0 satisfies γ 0 (s) 6= 0. Altogether, these conditions guarantee that Γ has a well-defined tangent at each point, which varies continuously as the point on the curve varies. Moreover, the parametrization γ induces an orientation on Γ as the parameter s travels from a to b. Any C 1 bijective mapping s : [c, d] → [a, b] gives rise to another parametrization of Γ by the formula η(t) = γ(s(t)). Clearly, the conditions that Γ be closed and simple are independent of the chosen parametrization. Also, we say that the two parametrizations γ and η are equivalent if s0 (t) > 0 for all t; this means that η and γ induce the same orientation on the curve Γ. If, however, s0 (t) < 0, then η reverses the orientation. If Γ is parametrized by γ(s) = (x(s), y(s)), then the length of the curve Γ is defined by Z b Z b ¡ 0 2 ¢1/2 x (s) + y 0 (s)2 ds. |γ 0 (s)| ds = `= a

a

The length of Γ is a notion intrinsic to the curve, and does not depend on its parametrization. To see that this is indeed the case, suppose that γ(s(t)) = η(t). Then, the change of variables formula and the chain rule imply that Z b Z d Z d 0 0 0 |γ (s)| ds = |γ (s(t))| |s (t)| dt = |η 0 (t)| dt, a

c

c

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1. The isoperimetric inequality

as desired. In the proof of the theorem below, we shall use a special type of parametrization for Γ. We say that γ is a parametrization by arclength if |γ 0 (s)| = 1 for all s. This means that γ(s) travels at a constant speed, and as a consequence, the length of Γ is precisely b − a. Therefore, after a possible additional translation, a parametrization by arc-length will be defined on [0, `]. Any curve admits a parametrization by arclength (Exercise 1). We now turn to the isoperimetric problem. The attempt to give a precise formulation of the area A of the region enclosed by a simple closed curve Γ raises a number of tricky questions. In a variety of simple situations, it is evident that the area is given by the following familiar formula of the calculus: ¯Z ¯ ¯ 1 ¯¯ (1) A = ¯ (x dy − y dx)¯¯ 2 Γ ¯Z ¯ ¯ 1 ¯¯ b ¯ = ¯ x(s)y 0 (s) − y(s)x0 (s) ds¯ ; ¯ ¯ 2 a see, for example, Exercise 3. Thus in formulating our result we shall adopt the easy expedient of taking (1) as our definition of area. This device allows us to give a quick and neat proof of the isoperimetric inequality. A listing of issues this simplification leaves unresolved can be found after the proof of the theorem. Statement and proof of the isoperimetric inequality Theorem 1.1 Suppose that Γ is a simple closed curve in R2 of length `, and let A denote the area of the region enclosed by this curve. Then A≤

`2 , 4π

with equality if and only if Γ is a circle. The first observation is that we can rescale the problem. This means that we can change the units of measurement by a factor of δ > 0 as follows. Consider the mapping of the plane R2 to itself, which sends the point (x, y) to (δx, δy). A look at the formula defining the length of a curve shows that if Γ is of length `, then its image under this mapping has length δ`. So this operation magnifies or contracts lengths by a factor of δ depending on whether δ ≥ 1 or δ ≤ 1. Similarly, we see that

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the mapping magnifies (or contracts) areas by a factor of δ 2 . By taking δ = 2π/`, we see that it suffices to prove that if ` = 2π then A ≤ π, with equality only if Γ is a circle. Let γ : [0, 2π] → R2 with γ(s) = (x(s), y(s)) be a parametrization by arc-length of the curve Γ, that is, x0 (s)2 + y 0 (s)2 = 1 for all s ∈ [0, 2π]. This implies that (2)

1 2π

Z

2π

(x0 (s)2 + y 0 (s)2 ) ds = 1.

0

Since the curve is closed, the functions x(s) and y(s) are 2π-periodic, so we may consider their Fourier series X X x(s) ∼ an eins and y(s) ∼ bn eins . Then, as we remarked in the later part of Section 2 of Chapter 2, we have X X x0 (s) ∼ an ineins and y 0 (s) ∼ bn ineins . Parseval’s identity applied to (2) gives ∞ X

(3)

¡ ¢ |n|2 |an |2 + |bn |2 = 1.

n=−∞

We now apply the bilinear form of Parseval’s identity (Lemma 1.5, Chapter 3) to the integral defining A. Since x(s) and y(s) are real-valued, we have an = a−n and bn = b−n , so we find that ¯ ¯ ¯Z ¯ ∞ ¯ X ´¯ ³ ¯ 1 ¯¯ 2π ¯ ¯ 0 0 ¯ A= ¯ x(s)y (s) − y(s)x (s) ds¯ = π ¯ n an bn − bn an ¯ . ¯ ¯ 2 0 n=−∞

We observe next that (4)

|an bn − bn an | ≤ 2 |an | |bn | ≤ |an |2 + |bn |2 ,

and since |n| ≤ |n|2 , we may use (3) to get ∞ X

A≤π

n=−∞

≤ π,

¡ ¢ |n|2 |an |2 + |bn |2

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2. Weyl’s equidistribution theorem

as desired. When A = π, we see from the above argument that x(s) = a−1 e−is + a0 + a1 eis

and

y(s) = b−1 e−is + b0 + b1 eis

because |n| < |n|2 as soon as |n| ≥ 2. We know that x(s) and y(s) are real-valued, so¢ a−1 = a1 and b−1 = b1 . The identity (3) implies that ¡ 2 |a1 |2 + |b1 |2 = 1, and since we have equality in (4) we must have |a1 | = |b1 | = 1/2. We write a1 =

1 iα e 2

and

b1 =

1 iβ e . 2

The fact that 1 = 2|a1 b1 − a1 b1 | implies that | sin(α − β)| = 1, hence α − β = kπ/2 where k is an odd integer. From this we find that x(s) = a0 + cos(α + s)

and

y(s) = b0 ± sin(α + s),

where the sign in y(s) depends on the parity of (k − 1)/2. In any case, we see that Γ is a circle, for which the case of equality obviously holds, and the proof of the theorem is complete. The solution given above (due to Hurwitz in 1901) is indeed very elegant, but clearly leaves some important issues unanswered. We list these as follows. Suppose Γ is a simple closed curve. (i) How is the “region enclosed by Γ” defined? (ii) What is the geometric definition of the “area” of this region? Does this definition accord with (1)? (iii) Can these results be extended to the most general class of simple closed curves relevant to the problem—those curves which are “rectifiable”—that is, those to which we can ascribe a finite length? It turns out that the clarifications of the problems raised are connected to a number of other significant ideas in analysis. We shall return to these questions in succeeding books of this series.

2 Weyl’s equidistribution theorem We now apply ideas coming from Fourier series to a problem dealing with properties of irrational numbers. We begin with a brief discussion of congruences, a concept needed to understand our main theorem.

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The reals modulo the integers If x is a real number, we let [x] denote the greatest integer less than or equal to x and call the quantity [x] the integer part of x. The fractional part of x is then defined by hxi = x − [x]. In particular, hxi ∈ [0, 1) for every x ∈ R. For example, the integer and fractional parts of 2.7 are 2 and 0.7, respectively, while the integer and fractional parts of −3.4 are −4 and 0.6, respectively. We may define a relation on R by saying that the two numbers x and y are equivalent, or congruent, if x − y ∈ Z. We then write x = y mod Z

or

x = y mod 1.

This means that we identify two real numbers if they differ by an integer. Observe that any real number x is congruent to a unique number in [0, 1) which is precisely hxi, the fractional part of x. In effect, reducing a real number modulo Z means looking only at its fractional part and disregarding its integer part. Now start with a real number γ 6= 0 and look at the sequence γ, 2γ, 3γ, . . . . An intriguing question is to ask what happens to this sequence if we reduce it modulo Z, that is, if we look at the sequence of fractional parts hγi, h2γi, h3γi, . . . . Here are some simple observations: (i) If γ is rational, then only finitely many numbers appearing in hnγi are distinct. (ii) If γ is irrational, then the numbers hnγi are all distinct. Indeed, for part (i), note that if γ = p/q, the first q terms in the sequence are hp/qi, h2p/qi, . . . , h(q − 1)p/qi, hqp/qi = 0. The sequence then begins to repeat itself, since h(q + 1)p/qi = h1 + p/qi = hp/qi, and so on. However, see Exercise 6 for a more refined result. Also, for part (ii) assume that not all numbers are distinct. We therefore have hn1 γi = hn2 γi for some n1 6= n2 ; then n1 γ − n2 γ ∈ Z, hence γ is rational, a contradiction.

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107

In fact, it can be shown that if γ is irrational, then hnγi is dense in the interval [0, 1), a result originally proved by Kronecker. In other words, the sequence hnγi hits every sub-interval of [0, 1) (and hence it does so infinitely many times). We will obtain this fact as a corollary of a deeper theorem dealing with the uniform distribution of the sequence hnγi. A sequence of numbers ξ1 , ξ2 , . . . , ξn , . . . in [0, 1) is said to be equidistributed if for every interval (a, b) ⊂ [0, 1), lim

N →∞

#{1 ≤ n ≤ N : ξn ∈ (a, b)} =b−a N

where #A denotes the cardinality of the finite set A. This means that for large N , the proportion of numbers ξn in (a, b) with n ≤ N is equal to the ratio of the length of the interval (a, b) to the length of the interval [0, 1). In other words, the sequence ξn sweeps out the whole interval evenly, and every sub-interval gets its fair share. Clearly, the ordering of the sequence is very important, as the next two examples illustrate. Example 1. The sequence 0,

1 1 2 1 2 3 1 2 , 0, , , 0, , , , 0, , , · · · 2 3 3 4 4 4 5 5

appears to be equidistributed since it passes over the interval [0, 1) very evenly. Of course this is not a proof, and the reader is invited to give one. For a somewhat related example, see Exercise 8 with σ = 1/2. Example 2. Let {rn }∞ n=1 be any enumeration of the rationals in [0, 1). Then the sequence defined by ½

ξn =

rn/2 if n is even, 0 if n is odd,

is not equidistributed since “half” of the sequence is at 0. Nevertheless, this sequence is obviously dense. We now arrive at the main theorem of this section. Theorem 2.1 If γ is irrational, then the sequence of fractional parts hγi, h2γi, h3γi, . . . is equidistributed in [0, 1). In particular, hnγi is dense in [0, 1), and we get Kronecker’s theorem as a corollary. In Figure 2 we illustrate the set of points hγi, h2γi, √ h3γi, . . . , hN γi for three different values of N when γ = 2.

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N = 10 0

1 N = 30

0

1 N = 80

0

1 √ Figure 2. The sequence hγi, h2γi, h3γi, . . . , hN γi when γ = 2

Fix (a, b) ⊂ [0, 1) and let χ(a,b) (x) denote the characteristic function of the interval (a, b), that is, the function equal to 1 in (a, b) and 0 in [0, 1) − (a, b). We may extend this function to R by periodicity (period 1), and still denote this extension by χ(a,b) (x). Then, as a consequence of the definitions, we find that #{1 ≤ n ≤ N : hnγi ∈ (a, b)} =

N X

χ(a,b) (nγ),

n=1

and the theorem can be reformulated as the statement that Z 1 N 1 X χ(a,b) (x) dx, χ(a,b) (nγ) → N n=1 0

as N → ∞.

This step removes the difficulty of working with fractional parts and reduces the number theory to analysis. The heart of the matter lies in the following result. Lemma 2.2 If f is continuous and periodic of period 1, and γ is irrational, then Z 1 N 1 X f (nγ) → f (x) dx N n=1 0

as N → ∞.

The proof of the lemma is divided into three steps. Step 1. We first check the validity of the limit in the case when f is one of the exponentials 1, e2πix , . . . , e2πikx , . . . . If f = 1, the limit

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surely holds. If f = e2πikx with k 6= 0, then the integral is 0. Since γ is irrational, we have e2πikγ 6= 1, therefore N 1 X e2πikγ 1 − e2πikN γ f (nγ) = , N n=1 N 1 − e2πikγ

which goes to 0 as N → ∞. Step 2. It is clear that if f and g satisfy the lemma, then so does Af + Bg for any A, B ∈ C. Therefore, the first step implies that the lemma is true for all trigonometric polynomials. Step 3. Let ² > 0. If f is any continuous periodic function of period 1, choose a trigonometric polynomial P so that supx∈R |f (x) − P (x)| < ²/3 (this is possible by Corollary 5.4 in Chapter 2). Then, by step 1, for all large N we have ¯ ¯ Z 1 N ¯1 X ¯ ¯ ¯ P (nγ) − P (x) dx¯ < ²/3. ¯ ¯N ¯ 0 n=1

Therefore ¯ ¯ Z 1 N N ¯ ¯1 X 1 X ¯ ¯ f (x) dx f (nγ) − ≤ |f (nγ) − P (nγ)|+ ¯ ¯ ¯ ¯N N n=1 0 n=1 ¯ ¯ Z 1 N ¯ ¯1 X ¯ ¯ P (x) dx¯ + +¯ P (nγ) − ¯ ¯N 0 n=1 Z 1 |P (x) − f (x)| dx + 0

< ², and the lemma is proved. Now we can finish the proof of the theorem. Choose two continuous periodic functions f²+ and f²− of period 1 which approximate χ(a,b) (x) on [0, 1) from above and below; both f²+ and f²− are bounded by 1 and agree with χ(a,b) (x) except in intervals of total length 2² (see Figure 3). In particular, f²− (x) ≤ χ(a,b) (x) ≤ f²+ (x), and Z 1 Z 1 f²+ (x) dx ≤ b − a + 2². b − a − 2² ≤ f²− (x) dx and 0

If SN =

1 N

PN n=1

0

χ(a,b) (nγ), then we get N N 1 X − 1 X + f² (nγ) ≤ SN ≤ f (nγ). N n=1 N n=1 ²

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f²+ f²−

0

a−²

a

a+²

b

b−²

b+²

1

Figure 3. Approximations of χ(a,b) (x)

Therefore b − a − 2² ≤ lim inf SN

and

N →∞

lim sup SN ≤ b − a + 2². N →∞

Since this is true for every ² > 0, the limit limN →∞ SN exists and must equal b − a. This completes the proof of the equidistribution theorem. This theorem has the following consequence. Corollary 2.3 The conclusion of Lemma 2.2 holds for every function f which is Riemann integrable in [0, 1], and periodic of period 1. Proof. Assume f is real-valued, and consider a partition of the interval [0, 1], say 0 = x0 < x1 < · · · < xN = 1. Next, define fU (x) = supxj−1 ≤y≤xj f (y) if x ∈ [xj−1 , xj ) and fL (x) = inf xj−1 ≤y≤xj f (y) for x ∈ (xj−1 , xj ). Then clearly fL ≤ f ≤ fU and Z 0

Z

1

fL (x) dx ≤

Z

1

f (x) dx ≤ 0

0

1

fU (x) dx.

Moreover, by making the partition sufficiently fine we can guarantee that for a given ² > 0, Z

Z

1

0

fU (x) dx −

1 0

fL (x) dx ≤ ².

However, Z 1 N 1 X fL (nγ) → fL (x) dx N n=1 0

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by the theorem, because each fL is a finite linear combination of characteristic functions of intervals; similarly we have Z 1 N 1 X fU (nγ) → fU (x) dx. N n=1 0

From these two assertions we can conclude the proof of the corollary by using the previous approximation argument. There is an interesting interpretation of the lemma and its corollary, in terms of a simple dynamical system. In this example, the underlying space is the circle parametrized by the angle θ. We also consider a mapping of this space to itself: here, we choose a rotation ρ of the circle by the angle 2πγ, that is, the transformation ρ : θ 7→ θ + 2πγ. We want next to consider how this space, with its underlying action ρ, evolves in time. In other words, we wish to consider the iterates of ρ, namely ρ, ρ2 , ρ3 , . . ., ρn where ρn = ρ ◦ ρ ◦ · · · ◦ ρ : θ 7→ θ + 2πnγ, and where we think of the action ρn taking place at the time t = n. To each Riemann integrable function f on the circle, we can also associate the corresponding effects of the rotation ρ, and obtain a sequence of functions f (θ), f (ρ(θ)), f (ρ2 (θ)), . . . , f (ρn (θ)), . . . with f (ρn (θ)) = f (θ + 2πnγ). In this special context, the ergodicity of this system is then the statement that the “time average” N 1 X f (ρn (θ)) N →∞ N n=1

lim

exists for each θ and equals the “space average” 1 2π

Z

2π

f (θ) dθ, 0

whenever γ is irrational. In fact, this assertion is merely a rephrasing of Corollary 2.3, once we make the change of variables θ = 2πx. Returning to the problem of equidistributed sequences, we observe that the proof of Theorem 2.1 gives the following characterization.

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Weyl’s criterion. A sequence of real numbers ξ1 , ξ2 . . . in [0, 1) is equidistributed if and only if for all integers k 6= 0 one has N 1 X 2πikξn e → 0, N n=1

as N → ∞.

One direction of this theorem was in effect proved above, and the converse can be found in Exercise 7. In particular, we find that to understand the equidistributive properties of a sequence ξn , it suffices to estimate P 2πikξn the size of the corresponding “exponential sum” N . For exn=1 e ample, it can be shown using Weyl’s criterion that the sequence hn2 γi is equidistributed whenever γ is irrational. This and other examples can be found in Exercises 8, and 9; also Problems 2, and 3. As a last remark, we mention a nice geometric interpretation of the distribution properties of hnγi. Suppose that the sides of a square are reflecting mirrors and that a ray of light leaves a point inside the square. What kind of path will the light trace out?

P

Figure 4. Reflection of a ray of light in a square

To solve this problem, the main idea is to consider the grid of the plane formed by successively reflecting the initial square across its sides. With an appropriate choice of axis, the path traced by the light in the square corresponds to the straight line P + (t, γt) in the plane. As a result, the reader may observe that the path will be either closed and periodic, or it will be dense in the square. The first of these situations

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will happen if and only if the slope γ of the initial direction of the light (determined with respect to one of the sides of the square) is rational. In the second situation, when γ is irrational, the density follows from Kronecker’s theorem. What stronger conclusion does one get from the equidistribution theorem?

3 A continuous but nowhere differentiable function There are many obvious examples of continuous functions that are not differentiable at one point, say f (x) = |x|. It is almost as easy to construct a continuous function that is not differentiable at any given finite set of points, or even at appropriate sets containing countably many points. A more subtle problem is whether there exists a continuous function that is nowhere differentiable. In 1861, Riemann guessed that the function defined by (5)

R(x) =

∞ X sin(n2 x) n=1

n2

was nowhere differentiable. He was led to consider this function because of its close connection to the theta function which will be introduced in Chapter 5. Riemann never gave a proof, but mentioned this example in one of his lectures. This triggered the interest of Weierstrass who, in an attempt to find a proof, came across the first example of a continuous but nowhere differentiable function. Say 0 < b < 1 and a is an integer > 1. In 1872 he proved that if ab > 1 + 3π/2, then the function W (x) =

∞ X

bn cos(an x)

n=1

is nowhere differentiable. But the story is not complete without a final word about Riemann’s original function. In 1916 Hardy showed that R is not differentiable at all irrational multiples of π, and also at certain rational multiples of π. However, it was not until much later, in 1969, that Gerver completely settled the problem, first by proving that the function R is actually differentiable at all the rational multiples of π of the form πp/q with p and q odd integers, and then by showing that R is not differentiable in all of the remaining cases. In this section, we prove the following theorem.

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Theorem 3.1 If 0 < α < 1, then the function fα (x) = f (x) =

∞ X

n

2−nα ei2

x

n=0

is continuous but nowhere differentiable. The continuity is clear because of the absolute convergence of the series. The crucial property of f which we need is that it has many vanishing Fourier coefficients. A Fourier series that skips many terms, like the one given above, or like W (x), is called a lacunary Fourier series. The proof of the theorem is really the story of three methods of summing a Fourier series. First, there is the ordinary convergence in terms of the partial sums SN (g) = g ∗ DN . Next, there is Ces`aro summability σN (g) = g ∗ FN , with FN the Fej´er kernel. A third method, clearly connected with the second, involves the delayed means defined by 4N (g) = 2σ2N (g) − σN (g). Hence 4N (g) = g ∗ [2F2N − FN ]. These methods can best be visualized as in Figure 5. P Suppose g(x) ∼ an einx . Then: • SN arises by multiplying the term an einx by 1 if |n| ≤ N , and 0 if |n| > N . • σN arises by multiplying an einx by 1 − |n|/N for |n| ≤ N and 0 for |n| > N . • 4N arises by multiplying an einx by 1 if |n| ≤ N , by 2(1 − |n|/(2N )) for N ≤ |n| ≤ 2N , and 0 for |n| > 2N . For example, note that S0 (g)(x) + S1 (g)(x) + · · · + SN −1 (g)(x) N N −1 1 X X = ak eikx N `=0

σN (g)(x) =

|k|≤`

= =

1 X N

(N − |n|)an einx

|n|≤N

¶ |n| an einx . 1− N

X µ |n|≤N

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1

−N

0

N

Partial sums P

SN (g)(x) =

|n|≤N

an einx

1

−N

0

Ces`aro means ³ P σN (g)(x) = |n|≤N 1 −

N

|n| N

´

an einx

1

−2N

−N

0

N

Delayed means ∆N (g)(x) = 2σ2N (g)(x) − σN (g)(x)

Figure 5. Three summation methods

2N

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The proof of the other assertion is similar. The delayed means have two important features. On the one hand, their properties are closely related to the (good) features of the Ces`aro means. On the other hand, for series that have lacunary properties like those of f , the delayed means are essentially equal to the partial sums. In particular, note that for our function f = fα (6)

SN (f ) = 4N 0 (f ),

where N 0 is the largest integer of the form 2k with N 0 ≤ N . This is clear by examining Figure 5 and the definition of f . We turn to the proof of the theorem proper and argue by contradiction; that is, we assume that f 0 (x0 ) exists for some x0 . Lemma 3.2 Let g be any continuous function that is differentiable at x0 . Then, the Ces` aro means satisfy σN (g)0 (x0 ) = O(log N ), therefore 4N (g)0 (x0 ) = O(log N ). Proof. First we have Z Z π 0 0 FN (x0 − t)g(t) dt = σN (g) (x0 ) = −π

π −π

FN0 (t)g(x0 − t) dt,

Rπ where FN is the Fej´er kernel. Since FN is periodic, we have −π FN0 (t)dt = 0 and this implies that Z π 0 FN0 (t)[g(x0 − t) − g(x0 )] dt. σN (g) (x0 ) = −π

From the assumption that g is differentiable at x0 we get Z π |σN (g)0 (x0 )| ≤ C |FN0 (t)| |t| dt. −π

Now observe that FN0 satisfies the two estimates |FN0 (t)| ≤ AN 2

and

|FN0 (t)| ≤

A . |t|2

For the first inequality, recall that FN is a trigonometric polynomial of degree N whose coefficients are bounded by 1. Therefore, FN0 is a trigonometric polynomial of degree N whose coefficients are no bigger than N . Hence |F 0 (t)| ≤ (2N + 1)N ≤ AN 2 .

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For the second inequality, we recall that FN (t) =

1 sin2 (N t/2) . N sin2 (t/2)

Differentiating this expression, we get two terms: sin(N t/2) cos(N t/2) 1 cos(t/2) sin2 (N t/2) − . N sin2 (t/2) sin3 (t/2) If we then use the facts that | sin(N t/2)| ≤ CN |t| and | sin(t/2)| ≥ c|t| (for |t| ≤ π), we get the desired estimates for FN0 (t). Using all of these estimates we find that Z Z |σN (g)0 (x0 )| ≤ C |FN0 (t)| |t| dt + C |FN0 (t)| |t| dt |t|≥1/N |t|≤1/N Z Z dt ≤ CA + CAN dt |t|≥1/N |t| |t|≤1/N = O(log N ) + O(1) = O(log N ). The proof of the lemma is complete once we invoke the definition of 4N (g). Lemma 3.3 If 2N = 2n , then n

42N (f ) − 4N (f ) = 2−nα ei2 x . This follows from our previous observation (6) because 42N (f ) = S2N (f ) and 4N (f ) = SN (f ). Now, by the first lemma we have 42N (f )0 (x0 ) − 4N (f )0 (x0 ) = O(log N ), and the second lemma also implies |42N (f )0 (x0 ) − 4N (f )0 (x0 )| = 2n(1−α) ≥ cN 1−α . This is the desired contradiction since N 1−α grows faster than log N . A few additional remarks about our function fα (x) = are in order.

P∞ n=0

2−nα ei2

n

x

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This function is complex-valued as opposed to the examples R and W above, and so the nowhere differentiability of fα does not imply the same property for its real and imaginary parts. However, a small modification of our proof shows that, in fact, the real part of fα , ∞ X

2−nα cos 2n x,

n=0

as well as its imaginary part, are both nowhere differentiable. To see this, observe first that by the same proof, Lemma 3.2 has the following generalization: if g is a continuous function which is differentiable at x0 , then 4N (g)0 (x0 + h) = O(log N )

whenever |h| ≤ c/N .

P −nα We then proceed with F (x) = ∞ cos 2n x, noting as above that n=0 2 −nα n 42N (F ) − 4N (F ) = 2 cos 2 x; as a result, assuming that F is differentiable at x0 , we get that

|2n(1−α) sin(2n (x0 + h))| = O(log N ) when 2N = 2n , and |h| ≤ c/N . To get a contradiction, we need only choose h so that | sin(2n (x0 + h))| = 1; this is accomplished by setting δ equal to the distance from 2n x0 to the nearest number of the form (k + 1/2)π, k ∈ Z (so δ ≤ π/2), and taking h = ±δ/2n . Clearly, when α > 1 the function fα is continuously differentiable since the series can be differentiated term by term. Finally, the nowhere differentiability we have proved for α < 1 actually extends to α = 1 by a suitable refinement of the argument (see Problem 8 in Chapter 5). In fact, using these more elaborate methods one can also show that the Weierstrass function W is nowhere differentiable if ab ≥ 1.

4 The heat equation on the circle As a final illustration, we return to the original problem of heat diffusion considered by Fourier. Suppose we are given an initial temperature distribution at t = 0 on a ring and that we are asked to describe the temperature at points on the ring at times t > 0. The ring is modeled by the unit circle. A point on this circle is described by its angle θ = 2πx, where the variable x lies between 0 and 1. If u(x, t) denotes the temperature at time t of a point described by the

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angle θ, then considerations similar to the ones given in Chapter 1 show that u satisfies the differential equation ∂u ∂2u = c 2. ∂t ∂x

(7)

The constant c is a positive physical constant which depends on the material of which the ring is made (see Section 2.1 in Chapter 1). After rescaling the time variable, we may assume that c = 1. If f is our initial data, we impose the condition u(x, 0) = f (x). To solve the problem, we separate variables and look for special solutions of the form u(x, t) = A(x)B(t). Then inserting this expression for u into the heat equation we get A00 (x) B 0 (t) = . B(t) A(x) Both sides are therefore constant, say equal to λ. Since A must be periodic of period 1, we see that the only possibility is λ = −4π 2 n2 , where n ∈ Z. Then A is a linear combination of the exponentials e2πinx 2 2 and e−2πinx , and B(t) is a multiple of e−4π n t . By superposing these solutions, we are led to (8)

u(x, t) =

∞ X

an e−4π

2

n2 t 2πinx

e

,

n=−∞

where, setting t = 0, we see that {an } are the Fourier coefficients of f . Note that when f is Riemann integrable, the coefficients an are 2 2 bounded, and since the factor e−4π n t tends to zero extremely fast, the series defining u converges. In fact, in this case, u is twice differentiable and solves equation (7). The natural question with regard to the boundary condition is the following: do we have u(x, t) → f (x) as t tends to 0, and in what sense? A simple application of the Parseval identity shows that this limit holds in the mean square sense (Exercise 11). For a better understanding of the properties of our solution (8), we write it as u(x, t) = (f ∗ Ht )(x),

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where Ht is the heat kernel for the circle, given by (9)

∞ X

Ht (x) =

e−4π

2

n2 t 2πinx

e

,

n=−∞

and where the convolution for functions with period 1 is defined by Z

1

(f ∗ g)(x) =

f (x − y)g(y) dy. 0

An analogy between the heat kernel and the Poisson kernel (of Chapter 2) is given in Exercise 12. However, unlike in the case of the Poisson kernel, there is no elementary formula for the heat kernel. Nevertheless, it turns out that it is a good kernel (in the sense of Chapter 2). The proof is not obvious and requires the use of the celebrated Poisson summation formula, which will be taken up in Chapter 5. As a corollary, we will also find that Ht is everywhere positive, a fact that is also not obvious from its defining expression (9). We can, however, give the following heuristic argument for the positivity of Ht . Suppose that we begin with an initial temperature distribution f which is everywhere ≤ 0. Then it is physically reasonable to expect u(x, t) ≤ 0 for all t since heat travels from hot to cold. Now Z 1 f (x − y)Ht (y) dy. u(x, t) = 0

If Ht is negative for some x0 , then we may choose f ≤ 0 supported near x0 , and this would imply u(x0 , t) > 0, which is a contradiction.

5 Exercises 1. Let γ : [a, b] → R2 be a parametrization for the closed curve Γ. (a) Prove that γ is a parametrization by arc-length if and only if the length of the curve from γ(a) to γ(s) is precisely s − a, that is,

Z

s

|γ 0 (t)| dt = s − a. a

(b) Prove that any curve Γ admits a parametrization by arc-length. [Hint: If Rs η is any parametrization, let h(s) = a |η 0 (t)| dt and consider γ = η ◦ h−1 .] 2. Suppose γ : [a, b] → R2 is a parametrization for a closed curve Γ, with γ(t) = (x(t), y(t)).

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(a) Show that 1 2

Z

Z

b

a

Z

b

(x(s)y 0 (s) − y(s)x0 (s)) ds =

b

x(s)y 0 (s) ds = − a

y(s)x0 (s) ds. a

(b) Define the reverse parametrization of γ by γ − : [a, b] → R2 with γ − (t) = γ(b + a − t). The image of γ − is precisely Γ, except that the points γ − (t) and γ(t) travel in opposite directions. Thus γ − “reverses” the orientation of the curve. Prove that

Z

Z (x dy − y dx) = −

(x dy − y dx). γ−

γ

In particular, we may assume (after a possible change in orientation) that A=

1 2

Z

Z

b

b

(x(s)y 0 (s) − y(s)x0 (s)) ds = a

x(s)y 0 (s) ds. a

3. Suppose Γ is a curve in the plane, and that there exists a set of coordinates x and y so that the x-axis divides the curve into the union of the graph of two continuous functions y = f (x) and y = g(x) for 0 ≤ x ≤ 1, and with f (x) ≥ g(x) (see Figure 6). Let Ω denote the region between the graphs of these two functions: Ω = {(x, y) : 0 ≤ x ≤ 1 and g(x) ≤ y ≤ f (x)}.

y = f (x)

0

1 Ω y = g(x)

Figure 6. Simple version of the area formula

R

With the familiar interpretation that the integral h(x) dx gives the area R1 under the graph of the function h, we see that the area of Ω is 0 f (x) dx −

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R1

g(x) dx. Show that this definition coincides with the area formula A given in 0 the text, that is,

Z

Z

1

1

f (x) dx − 0

0

¯ Z ¯

g(x) dx = ¯¯−

¯ ¯

Γ

y dx¯¯ = A.

Also, note that if the orientation of the curve is chosen so that Ω “lies to the left” of Γ, then the above formula holds without the absolute value signs. This formula generalizes to any set that can be written as a finite union of domains like Ω above. 4. Observe that with the definition of ` and A given in the text, the isoperimetric inequality continues to hold (with the same proof) even when Γ is not simple. Show that this stronger version of the isoperimetric inequality is equivalent to Wirtinger’s inequality, which says that if f is 2π-periodic, of class C 1 , and R 2π satisfies 0 f (t) dt = 0, then

Z

Z

2π

2π

|f 0 (t)|2 dt

2

|f (t)| dt ≤ 0

0

with equality if and only if f (t) = A sin t + B cos t (Exercise 11, Chapter 3). [Hint: In one direction, note that if the length of the curve is 2π and γ is an appropriate arc-length parametrization, then

Z

Z

2π

2

0

2π

(y 0 (s)2 − y(s)2 ) ds.

[x (s) + y(s)] ds +

2(π − A) =

0

0

R 2π

A change of coordinates will guarantee 0 y(s) ds = 0. For the other direction, start with a real-valued f satisfying all the hypotheses of Wirtinger’s inequality, and construct g, 2π-periodic and so that the term in brackets above vanishes.] 5. Prove that the sequence {γn }∞ n=1 , where γn is the fractional part of

µ

is not equidistributed in [0, 1].

³

√

´n

√ ¶n 1+ 5 , 2

³

√

´n

[Hint: Show that Un = 1+2 5 + 1−2 5 is the solution of the difference equation Ur+1 = Ur + Ur−1 with U0 = 2 and U1 = 1. The Un satisfy the same difference equation as the Fibonacci numbers.] 6. Let θ = p/q be a rational number where p and q are relatively prime integers (that is, θ is in lowest form). We assume without loss of generality that q > 0. Define a sequence of numbers in [0, 1) by ξn = hnθi where h·i denotes the

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5. Exercises

fractional part. Show that the sequence {ξ1 , ξ2 , . . .} is equidistributed on the points of the form 0, 1/q, 2/q, . . . , (q − 1)/q. In fact, prove that for any 0 ≤ a < q, one has #{n : 1 ≤ n ≤ N, hnθi = a/q} 1 = +O N q

µ

1 N

¶ .

[Hint: For each integer k ≥ 0, there exists a unique integer n with kq ≤ n < (k + 1)q and so that hnθi = a/q. Why can one assume k = 0? Prove the existence of n by using the fact1 that if p and q are relatively prime, there exist integers x, y such that xp + yq = 1. Next, divide N by q with remainder, that is, write N = `q + r where 0 ≤ ` and 0 ≤ r < q. Establish the inequalities ` ≤ #{n : 1 ≤ n ≤ N, hnθi = a/q} ≤ ` + 1.]

7. Prove the second part of Weyl’s criterion: if a sequence of numbers ξ1 , ξ2 , . . . in [0, 1) is equidistributed, then for all k ∈ Z − {0} N 1 X 2πikξn →0 e N

as N → ∞.

n=1

R1

PN

[Hint: It suffices to show that N1 n=1 f (ξn ) → 0 f (x) dx for all continuous f . Prove this first when f is the characteristic function of an interval.] 8. Show that for any a 6= 0, and σ with 0 < σ < 1, the sequence hanσ i is equidistributed in [0, 1). [Hint: Prove that following

PN

N X n=1

n=1

σ

e2πibn = O(N σ ) + O(N 1−σ ) if b 6= 0.] In fact, note the

Z 2πibnσ

e

−

Ã

N

e 1

2πibxσ

dx = O

N X

! n

−1+σ

.

n=1

9. In contrast with the result in Exercise 8, prove that ha log ni is not equidistributed for any a. PN [Hint: Compare the sum n=1 e2πib log n with the corresponding integral.] 10. Suppose that f is a periodic function on R of period 1, and {ξn } is a sequence which is equidistributed in [0, 1). Prove that: 1 The elementary results in arithmetic used in this exercise can be found at the beginning of Chapter 8.

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(a) If f is continuous and satisfies

R1 0

f (x) dx = 0, then

N 1 X f (x + ξn ) = 0 N →∞ N

lim

uniformly in x.

n=1

[Hint: Establish this result first for trigonometric polynomials.] (b) If f is merely integrable on [0, 1] and satisfies

Z

1

lim

N →∞

0

R1 0

f (x) dx = 0, then

¯2 ¯ N ¯ ¯1 X ¯ ¯ f (x + ξn )¯ dx = 0. ¯ ¯ ¯ N n=1

11. Show that if u(x, t) = (f ∗ Ht )(x) where Ht is the heat kernel, and f is Riemann integrable, then

Z

1

|u(x, t) − f (x)|2 dx → 0

as t → 0.

0

12. A change of variables in (8) leads to the solution u(θ, τ ) =

X

2

an e−n τ einθ = (f ∗ hτ )(θ)

of the equation ∂u ∂2u = ∂τ ∂θ2

with 0 ≤ θ ≤ 2π and τ > 0,

P

boundary condition u(θ, 0) = f (θ) ∼ an einθ . Here hτ (θ) = −n2 τ inθ e e . This version of the heat kernel on [0, 2π] is the analogue n=−∞ P∞ of the Poisson kernel, which can be written as Pr (θ) = n=−∞ e−|n|τ einθ with r = e−τ (and so 0 < r < 1 corresponds to τ > 0).

with

P∞

13. The fact that the kernel Ht (x) is a good kernel, hence u(x, t) → f (x) at each point of continuity of f , is not easy to prove. This will be shown in the next chapter. However, one can prove directly that Ht (x) is “peaked” at x = 0 as t → 0 in the following sense: (a) Show that

R 1/2 −1/2

|Ht (x)|2 dx is of the order of magnitude of t−1/2 as t → 0.

More precisely, prove that t1/2 limit as t → 0. (b) Prove that

R 1/2 −1/2

R 1/2

−1/2

|Ht (x)|2 dx converges to a non-zero

x2 |Ht (x)|2 dx = O(t1/2 ) as t → 0.

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6. Problems

P∞

R∞

2

2

−cn t [Hint: For (a) compare the sum with the integral −∞ e−cx t dx −∞ e 2 2 where c > 0. For (b) use x ≤ C(sin πx) for −1/2 ≤ x ≤ 1/2, and apply the 2 mean value theorem to e−cx t .]

6 Problems 1.∗ This problem explores another relationship between the geometry of a curve and Fourier series. The diameter of a closed curve Γ parametrized by γ(t) = (x(t), y(t)) on [−π, π] is defined by d = sup |P − Q| = P, Q∈Γ

sup

|γ(t1 ) − γ(t2 )|.

t1 , t2 ∈[−π,π]

If an is the nth Fourier coefficient of γ(t) = x(t) + iy(t) and ` denotes the length of Γ, then (a) 2|an | ≤ d for all n 6= 0. (b) ` ≤ πd, whenever Γ is convex. Property (a) follows from the fact that 2an =

1 2π

Rπ −π

[γ(t) − γ(t + π/n)]e−int dt.

The equality ` = πd is satisfied when Γ is a circle, but surprisingly, this is not the only case. In fact, one finds that ` = πd is equivalent to 2|a1 | = d. We re-parametrize γ so that for each t in [−π, π] the tangent to the curve makes an angle t with the y-axis. Then, if a1 = 1 we have γ 0 (t) = ieit (1 + r(t)), where r is a real-valued function which satisfies r(t) + r(t + π) = 0, and |r(t)| ≤ 1. Figure 7 (a) shows the curve obtained by setting r(t) = cos 5t. Also, Figure 7 (b) consists of the curve where r(t) = h(3t), with h(s) = −1 if −π ≤ s ≤ 0 and h(s) = 1 if 0 < s < π. This curve (which is only piecewise of class C 1 ) is known as the Reuleaux triangle and is the classical example of a convex curve of constant width which is not a circle. 2.∗ Here we present an estimate of Weyl which leads to some interesting results. (a) Let SN =

PN n=1

e2πif (n) . Show that for H ≤ N , one has

¯

¯

H ¯N −h ¯ N X ¯ X 2πi(f (n+h)−f (n)) ¯ |SN | ≤ c e ¯ ¯, H ¯ ¯ 2

h=0 n=1

for some constant c > 0 independent of N , H, and f . (b) Use this estimate to show that the sequence hn2 γi is equidistributed in [0, 1) whenever γ is irrational.

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(a)

(b)

Figure 7. Some curves with maximal length for a given diameter

(c) More generally, show that if {ξn } is a sequence of real numbers so that for all positive integers h the difference hξn+h − ξn i is equidistributed in [0, 1), then hξn i is also equidistributed in [0, 1). (d) Suppose that P (x) = cn xn + · · · + c0 is a polynomial with real coefficients, where at least one of c1 , . . . , cn is irrational. Then the sequence hP (n)i is equidistributed in [0, 1). [Hint: For (a), let an = e2πif (n) when 1 ≤ n ≤ N and 0 otherwise. Then write P PH P H n an = k=1 n an+k and apply the Cauchy-Schwarz inequality. For (b), note that (n + h)2 γ − n2 γ = 2nhγ + h2 γ, and use the fact that for each integer h, the sequence h2nhγi is equidistributed. Finally, to prove (d), assume first that P (x) = Q(x) + c1 x + c0 where c1 is irrational, and estimate the exponential sum PN 2πikP (n) . Then, argue by induction on the highest degree term which has n=1 e an irrational coefficient, and use part (c).] 3.∗ If σ > 0 is not an integer and a 6= 0, then hanσ i is equidistributed in [0, 1). See also Exercise 8. 4. An elementary construction of a continuous but nowhere differentiable function is obtained by “piling up singularities,” as follows. On [−1, 1] consider the function ϕ(x) = |x| and extend ϕ to R by requiring it to be periodic of period 2. Clearly, ϕ is continuous on R and |ϕ(x)| ≤ 1 for all x so the function f defined by f (x) =

∞ µ ¶n X 3 n=0

is continuous on R.

4

ϕ(4n x)

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6. Problems

(a) Fix x0 ∈ R. For every positive integer m, let δm = ± 21 4−m where the sign is chosen so that no integer lies in between 4m x0 and 4m (x0 + δm ). Consider the quotient γn =

ϕ(4n (x0 + δm )) − ϕ(4n x0 ) . δm

Prove that if n > m, then γn = 0, and for 0 ≤ n ≤ m one has |γn | ≤ 4n with |γm | = 4m . (b) From the above observations prove the estimate

¯ ¯ ¯ f (x0 + δm ) − f (x0 ) ¯ 1 m ¯ ¯ ≥ (3 + 1), ¯ ¯ 2 δm and conclude that f is not differentiable at x0 .

5. Let f be a Riemann integrable function on the interval [−π, π]. We define the generalized delayed means of the Fourier series of f by σN,K =

SN + · · · + SN +K−1 . K

Note that in particular σ0,N = σN ,

σN,1 = SN

and

σN,N = ∆N ,

where ∆N are the specific delayed means used in Section 3. (a) Show that σN,K =

1 ((N + K)σN +K − N σN ) , K

and

µ

X

σN,K = SN +

1−

N +1≤|ν|≤N +K−1

|ν| − N K

¶ fˆ(ν)eiνθ .

From this last expression for σN,K conclude that |σN,K − SM | ≤

X N +1≤|ν|≤N +K−1

for all N ≤ M < N + K.

|fˆ(ν)|

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(b) Use one of the above formulas and Fej´er’s theorem to show that with N = kn and K = n, then σkn,n (f )(θ) → f (θ)

as n → ∞

whenever f is continuous at θ, and also σkn,n (f )(θ) →

f (θ+ ) + f (θ− ) 2

as n → ∞

at a jump discontinuity (refer to the preceding chapters and their exercises for the appropriate definitions and results). In the case when f is continuous on [−π, π], show that σkn,n (f ) → f uniformly as n → ∞. (c) Using part (a), show that if fˆ(ν) = O(1/|ν|) and kn ≤ m < (k + 1)n, we get |σkn,n − Sm | ≤

C k

for some constant C > 0.

(d) Suppose that fˆ(ν) = O(1/|ν|). Prove that if f is continuous at θ then SN (f )(θ) → f (θ)

as N → ∞,

and if f has a jump discontinuity at θ then SN (f )(θ) →

f (θ+ ) + f (θ− ) 2

as N → ∞.

Also, show that if f is continuous on [−π, π], then SN (f ) → f uniformly.

P

(e) The above arguments show that if cn is Ces`aro summable to s and cn = P O(1/n), then cn converges to s. This is a weak version of Littlewood’s theorem (Problem 3, Chapter 2).

6. Dirichlet’s theorem states that the Fourier series of a real continuous periodic function f which has only a finite number of relative maxima and minima converges everywhere to f (and uniformly). Prove this theorem by showing that such a function satisfies fˆ(n) = O(1/|n|). [Hint: Argue as in Exercise 17, Chapter 3; then use conclusion (d) in Problem 5 above.]

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5 The Fourier Transform on R The theory of Fourier series and integrals has always had major difficulties and necessitated a large mathematical apparatus in dealing with questions of convergence. It engendered the development of methods of summation, although these did not lead to a completely satisfactory solution of the problem.. . . For the Fourier transform, the introduction of distributions (hence the space S) is inevitable either in an explicit or hidden form.. . . As a result one may obtain all that is desired from the point of view of the continuity and inversion of the Fourier transform. L. Schwartz, 1950

The theory of Fourier series applies to functions on the circle, or equivalently, periodic functions on R. In this chapter, we develop an analogous theory for functions on the entire real line which are non-periodic. The functions we consider will be suitably “small” at infinity. There are several ways of defining an appropriate notion of “smallness,” but it will nevertheless be vital to assume some sort of vanishing at infinity. On the one hand, recall that the Fourier series of a periodic function associates a sequence of numbers, namely the Fourier coefficients, to that function; on the other hand, given a suitable function f on R, the analogous object associated to f will in fact be another function fˆ on R which is called the Fourier transform of f . Since the Fourier transform of a function on R is again a function on R, one can observe a symmetry between a function and its Fourier transform, whose analogue is not as apparent in the setting of Fourier series.

Roughly speaking, the Fourier transform is a continuous version of the Fourier coefficients. Recall that the Fourier coefficients an of a function f defined on the circle are given by

Z

(1)

an =

0

1

f (x)e−2πinx dx,

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and then in the appropriate sense we have (2)

∞ X

f (x) =

an e2πinx .

n=−∞

Here we have replaced θ by 2πx, as we have frequently done previously. Now, consider the following analogy where we replace all of the discrete symbols (such as integers and sums) by their continuous counterparts (such as real numbers and integrals). In other words, given a function f on all of R, we define its Fourier transform by changing the domain of integration from the circle to all of R, and by replacing n ∈ Z by ξ ∈ R in (1), that is, by setting Z

(3)

fˆ(ξ) =

∞

f (x)e−2πixξ dx.

−∞

We push our analogy further, and consider the following continuous version of (2): replacing the sum by an integral, and an by fˆ(ξ), leads to the Fourier inversion formula, Z

(4)

∞

f (x) =

fˆ(ξ)e2πixξ dξ.

−∞

Under a suitable hypotheses on f , the identity (4) actually holds, and much of the theory in this chapter aims at proving and exploiting this relation. The validity of the Fourier inversion formula is also suggested by the following simple observation. Suppose f is supported in a finite interval contained in I = [−L/2, L/2], and we expand f in a Fourier series on I. Then, letting L tend to infinity, we are led to (4) (see Exercise 1). The special properties of the Fourier transform make it an important tool in the study of partial differential equations. For instance, we shall see how the Fourier inversion formula allows us to analyze some equations that are modeled on the real line. In particular, following the ideas developed on the circle, we solve the time-dependent heat equation for an infinite rod and the steady-state heat equation in the upper half-plane. In the last part of the chapter we discuss further topics related to the Poisson summation formula, X X f (n) = fˆ(n), n∈Z

n∈Z

which gives another remarkable connection between periodic functions (and their Fourier series) and non-periodic functions on the line (and

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131

their Fourier transforms). This identity allows us to prove an assertion made in the previous chapter, namely, that the heat kernel Ht (x) satisfies the properties of a good kernel. In addition, the Poisson summation formula arises in many other settings, in particular in parts of number theory, as we shall see in Book II. We make a final comment about the approach we have chosen. In our study of Fourier series, we found it useful to consider Riemann integrable functions on the circle. In particular, this generality assured us that even functions that had certain discontinuities could be treated by the theory. In contrast, our exposition of the elementary properties of the Fourier transform is stated in terms of the Schwartz space S of testing functions. These are functions that are indefinitely differentiable and that, together with their derivatives, are rapidly decreasing at infinity. The reliance on this space of functions is a device that allows us to come quickly to the main conclusions, formulated in a direct and transparent fashion. Once this is carried out, we point out some easy extensions to a somewhat wider setting. The more general theory of Fourier transforms (which must necessarily be based on Lebesgue integration) will be treated in Book III.

1 Elementary theory of the Fourier transform We begin by extending the notion of integration to functions that are defined on the whole real line. 1.1 Integration of functions on the real line Given the notion of the integral of a function on a closed and bounded interval, the most natural extension of this definition to continuous functions over R is Z ∞ Z N f (x) dx = lim f (x) dx. −∞

N →∞ −N

Of course, this limit may not exist. For example, it is clear that if f (x) = 1, or even if f (x) = 1/(1 + |x|), then the above limit is infinite. A moment’s reflection suggests that the limit will exist if we impose on f enough decay as |x| tends to infinity. A useful condition is as follows. A function f defined on R is said to be of moderate decrease if f is continuous and there exists a constant A > 0 so that |f (x)| ≤

A 1 + x2

for all x ∈ R.

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This inequality says that f is bounded (by A for instance), and also that it decays at infinity at least as fast as 1/x2 , since A/(1 + x2 ) ≤ A/x2 . For example, the function f (x) = 1/(1 + |x|n ) is of moderate decrease as long as n ≥ 2. Another example is given by the function e−a|x| for a > 0. We shall denote by M(R) the set of functions of moderate decrease on R. As an exercise, the reader can check that under the usual addition of functions and multiplication by scalars, M(R) forms a vector space over C. We next see that whenever f belongs to M(R), then we may define Z

Z

∞

(5)

f (x) dx = lim −∞

N

N →∞ −N

f (x) dx,

where the limit now exists. Indeed, for each N the integral IN = RN f (x) dx is well defined because f is continuous. It now suffices to −N show that {IN } is a Cauchy sequence, and this follows because if M > N , then ¯Z ¯ ¯ ¯ ¯ ¯ |IM − IN | ≤ ¯ f (x) dx¯ ¯ N ≤|x|≤M ¯ Z dx ≤A 2 N ≤|x|≤M x 2A → 0 as N → ∞. ≤ N R Notice we have also proved that |x|≥N f (x) dx → 0 as N → ∞. At this point, we remark that we may replace the exponent 2 in the definition of moderate decrease by 1 + ² where ² > 0; that is, |f (x)| ≤

A 1 + |x|1+²

for all x ∈ R.

This definition would work just as well for the purpose of the theory developed in this chapter. We chose ² = 1 merely as a matter of convenience. We summarize some elementary properties of integration over R in a proposition. Proposition 1.1 The integral of a function of moderate decrease defined by (5) satisfies the following properties:

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(i) Linearity: if f, g ∈ M(R) and a, b ∈ C, then Z ∞ Z Z ∞ f (x) dx + b (af (x) + bg(x)) dx = a −∞

−∞

∞

g(x) dx.

−∞

(ii) Translation invariance: for every h ∈ R we have Z ∞ Z ∞ f (x − h) dx = f (x) dx. −∞

−∞

(iii) Scaling under dilations: if δ > 0, then Z ∞ Z ∞ δ f (δx) dx = f (x) dx. −∞

−∞

(iv) Continuity: if f ∈ M(R), then Z ∞ |f (x − h) − f (x)| dx → 0

as h → 0.

−∞

We say a few words about the proof. Property (i) is immediate. To verify property (ii), it suffices to see that Z

Z

N

N

f (x) dx → 0

f (x − h) dx − Since by

RN −N

f (x − h) dx =

as N → ∞.

−N

−N

R N −h −N −h

f (x) dx, the above difference is majorized

¯Z ¯ ¯Z ¯ ¯ −N ¯ ¯ N ¯ A0 ¯ ¯ ¯ ¯ f (x) dx¯ + ¯ f (x) dx¯ ≤ ¯ ¯ −N −h ¯ ¯ N −h ¯ 1 + N2

for large N , which tends to 0 as N tends to infinity. RN The proof of property (iii) is similar once we observe that δ −N f (δx) dx = R δN −δN f (x) dx. To prove property (iv) it suffices to take |h| ≤ 1. For a preassigned ² > 0, we first choose N so large that Z Z |f (x)| dx ≤ ²/4 and |f (x − h)| dx ≤ ²/4. |x|≥N

|x|≥N

Now with N fixed, we use the fact that since f is continuous, it is uniformly continuous in the interval [−N − 1, N + 1]. Hence

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sup|x|≤N |f (x − h) − f (x)| → 0 as h tends to 0. So we can take h so small that this supremum is less than ²/4N . Altogether, then, Z

Z

∞

N

|f (x − h) − f (x)| dx ≤ −∞

−N

|f (x − h) − f (x)| dx Z + |f (x − h)| dx |x|≥N Z + |f (x)| dx |x|≥N

≤ ²/2 + ²/4 + ²/4 = ², and thus conclusion (iv) follows. 1.2 Definition of the Fourier transform If f ∈ M(R), we define its Fourier transform for ξ ∈ R by Z ∞ f (x)e−2πixξ dx. fˆ(ξ) = −∞

Of course, |e−2πixξ | = 1, so the integrand is of moderate decrease, and the integral makes sense. In fact, this last observation implies that fˆ is bounded, and moreover, a simple argument shows that fˆ is continuous and tends to 0 as |ξ| → ∞ (Exercise 5). However, nothing in the definition above guarantees that fˆ is of moderate decrease, or has a specific decay. In particular, it is not R∞ clear in this context how to make sense of the integral −∞ fˆ(ξ)e2πixξ dξ and the resulting Fourier inversion formula. To remedy this, we introduce a more refined space of functions considered by Schwartz which is very useful in establishing the initial properties of the Fourier transform. The choice of the Schwartz space is motivated by an important principle which ties the decay of fˆ to the continuity and differentiability properties of f (and vice versa): the faster fˆ(ξ) decreases as |ξ| → ∞, the “smoother” f must be. An example that reflects this principle is given in Exercise 3. We also note that this relationship between f and fˆ is reminiscent of a similar one between the smoothness of a function on the circle and the decay of its Fourier coefficients; see the discussion of Corollary 2.4 in Chapter 2. 1.3 The Schwartz space The Schwartz space on R consists of the set of all indefinitely differentiable functions f so that f and all its derivatives f 0 , f 00 , . . . , f (`) , . . .,

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1. Elementary theory of the Fourier transform

are rapidly decreasing, in the sense that sup |x|k |f (`) (x)| < ∞

for every k, ` ≥ 0.

x∈R

We denote this space by S = S(R), and again, the reader should verify that S(R) is a vector space over C. Moreover, if f ∈ S(R), we have f 0 (x) =

df ∈ S(R) dx

and

xf (x) ∈ S(R).

This expresses the important fact that the Schwartz space is closed under differentiation and multiplication by polynomials. A simple example of a function in S(R) is the Gaussian defined by 2

f (x) = e−x , which plays a central role in the theory of the Fourier transform, as well as other fields (for example, probability theory and physics). The reader 2 can check that the derivatives of f are of the form P (x)e−x where P is 2 a polynomial, and this immediately shows that f ∈ S(R). In fact, e−ax belongs to S(R) whenever a > 0. Later, we will normalize the Gaussian by choosing a = π. 1

−1

0

Figure 1. The Gaussian e−πx

1 2

An important class of other examples in S(R) are the “bump functions” which vanish outside bounded intervals (Exercise 4). As a final remark, note that although e−|x| decreases rapidly at infinity, it is not differentiable at 0 and therefore does not belong to S(R).

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1.4 The Fourier transform on S The Fourier transform of a function f ∈ S(R) is defined by Z ∞ ˆ f (ξ) = f (x)e−2πixξ dx. −∞

Some simple properties of the Fourier transform are gathered in the following proposition. We use the notation f (x) −→ fˆ(ξ) to mean that fˆ denotes the Fourier transform of f . Proposition 1.2 If f ∈ S(R) then: (i) f (x + h) −→ fˆ(ξ)e2πihξ whenever h ∈ R. (ii) f (x)e−2πixh −→ fˆ(ξ + h) whenever h ∈ R. (iii) f (δx) −→ δ −1 fˆ(δ −1 ξ) whenever δ > 0. (iv) f 0 (x) −→ 2πiξ fˆ(ξ). (v) −2πixf (x) −→

d ˆ f (ξ). dξ

In particular, except for factors of 2πi, the Fourier transform interchanges differentiation and multiplication by x. This is the key property that makes the Fourier transform a central object in the theory of differential equations. We shall return to this point later. Proof. Property (i) is an immediate consequence of the translation invariance of the integral, and property (ii) follows from the definition. Also, the third property of Proposition 1.1 establishes (iii). Integrating by parts gives Z N Z N h iN f (x)e−2πixξ dx, f 0 (x)e−2πixξ dx = f (x)e−2πixξ + 2πiξ −N

−N

−N

so letting N tend to infinity gives (iv). Finally, to prove property (v), we must show that fˆ is differentiable and find its derivative. Let ² > 0 and consider fˆ(ξ + h) − fˆ(ξ) \ )(ξ)= − (−2πixf h Z ∞ −∞

"

f (x)e−2πixξ

# e−2πixh − 1 + 2πix dx. h

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Since f (x) R and xf (x) are of rapid R decrease, there exists an integer N so that |x|≥N |f (x)| dx ≤ ² and |x|≥N |x| |f (x)| dx ≤ ². Moreover, for |x| ≤ N , there exists h0 so that |h| < h0 implies ¯ ¯ ¯ e−2πixh − 1 ¯ ² ¯ ¯ + 2πix¯ ≤ . ¯ ¯ ¯ h N

Hence for |h| < h0 we have ¯ ¯ ¯ fˆ(ξ + h) − fˆ(ξ) ¯ ¯ ¯ \ − (−2πixf )(ξ)¯ ¯ ¯ ¯ h " #¯ Z N ¯¯ ¯ −2πixh −1 ¯ ¯ −2πixξ e ≤ + 2πix ¯ dx + C² ¯f (x)e ¯ ¯ h −N

≤ C 0 ².

Theorem 1.3 If f ∈ S(R), then fˆ ∈ S(R). The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if f ∈ S(R), its Fourier transform fˆ is bounded; then also, for each pair of non-negative integers ` and k, the expression µ

ξk

d dξ

¶`

fˆ(ξ)

is bounded, since by the last proposition, it is the Fourier transform of 1 (2πi)k

µ

d dx

¶k

[(−2πix)` f (x)].

The proof of the inversion formula Z

∞

f (x) =

fˆ(ξ)e2πixξ dξ

for f ∈ S(R),

−∞

which we give in the next section, is based on a careful study of the 2 function e−ax , which, as we have already observed, is in S(R) if a > 0.

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The Gaussians as good kernels We begin by considering the case a = π because of the normalization: Z

∞

2

e−πx dx = 1.

(6) −∞

To see why (6) is true, we use the multiplicative property of the exponential to reduce the calculation to a two-dimensional integral. More precisely, we can argue as follows: µZ

∞

−πx2

e

¶2

dx

Z

∞

Z

∞

2

e−π(x

=

−∞

−∞ −∞ Z 2π Z ∞

+y 2 )

dx dy

2

e−πr r dr dθ

= Z0∞

= h0

0 2

2πre−πr dr i 2 ∞

= −e−πr

0

= 1, where we have evaluated the two-dimensional integral using polar coordinates. The fundamental property of the Gaussian which is of interest to us, 2 and which actually follows from (6), is that e−πx equals its Fourier transform! We isolate this important result in a theorem. 2 Theorem 1.4 If f (x) = e−πx , then fˆ(ξ) = f (ξ).

Proof. Define Z

F (ξ) = fˆ(ξ) =

∞

2

e−πx e−2πixξ dx,

−∞

and observe that F (0) = 1, by our previous calculation. By property (v) in Proposition 1.2, and the fact that f 0 (x) = −2πxf (x), we obtain Z 0

∞

F (ξ) =

Z −2πixξ

f (x)(−2πix)e

∞

dx = i

−∞

f 0 (x)e−2πixξ dx.

−∞

By (iv) of the same proposition, we find that F 0 (ξ) = i(2πiξ)fˆ(ξ) = −2πξF (ξ).

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If we define G(ξ) = F (ξ)eπξ , then from what we have seen above, it follows that G0 (ξ) = 0, hence G is constant. Since F (0) = 1, we conclude 2 that G is identically equal to 1, therefore F (ξ) = e−πξ , as was to be shown. The scaling properties of the Fourier transform under dilations yield the following important transformation law, which follows from (iii) in Proposition 1.2 (with δ replaced by δ −1/2 ). Corollary 1.5 If δ > 0 and Kδ (x) = δ −1/2 e−πx

2

/δ

cδ (ξ) = e−πδξ2 . , then K

We pause to make an important observation. As δ tends to 0, the cδ gets function Kδ peaks at the origin, while its Fourier transform K c flatter. So in this particular example, we see that Kδ and Kδ cannot both be localized (that is, concentrated) at the origin. This is an example of a general phenomenon called the Heisenberg uncertainty principle, which we will discuss at the end of this chapter. We have now constructed a family of good kernels on the real line, analogous to those on the circle considered in Chapter 2. Indeed, with Kδ (x) = δ −1/2 e−πx we have: R∞ (i) −∞ Kδ (x) dx = 1. R∞ (ii) −∞ |Kδ (x)| dx ≤ M . (iii) For every η > 0, we have

R |x|>η

2

/δ

,

|Kδ (x)| dx → 0 as δ → 0.

To prove (i), we may change variables and use (6), or note that the cδ (0), which is 1 by Corollary 1.5. Since Kδ ≥ 0, it is integral equals K clear that property (ii) is also true. Finally we can again change variables to get Z Z 2 |Kδ (x)| dx = e−πy dy → 0 |x|>η

|y|>η/δ 1/2

as δ tends to 0. We have thus proved the following result. Theorem 1.6 The collection {Kδ }δ>0 is a family of good kernels as δ → 0. We next apply these good kernels via the operation of convolution, which is given as follows. If f, g ∈ S(R), their convolution is defined by Z ∞ (7) (f ∗ g)(x) = f (x − t)g(t) dt. −∞

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For a fixed value of x, the function f (x − t)g(t) is of rapid decrease in t, hence the integral converges. By the argument in Section 4 of Chapter 2 (with a slight modification), we get the following corollary. Corollary 1.7 If f ∈ S(R) , then (f ∗ Kδ )(x) → f (x)

uniformly in x as δ → 0.

Proof. First, we claim that f is uniformly continuous on R. Indeed, given ² > 0 there exists R > 0 so that |f (x)| < ²/4 whenever |x| ≥ R. Moreover, f is continuous, hence uniformly continuous on the compact interval [−R, R], and together with the previous observation, we can find η > 0 so that |f (x) − f (y)| < ² whenever |x − y| < η. Now we argue as usual. Using the first property of good kernels, we can write Z ∞ (f ∗ Kδ )(x) − f (x) = Kδ (t) [f (x − t) − f (x)] dt, −∞

and since Kδ ≥ 0, we find Z

|(f ∗ Kδ )(x) − f (x)| ≤

Z

+ |t|>η

|t|≤η

Kδ (t) |f (x − t) − f (x)| dt.

The first integral is small by the third property of good kernels, and the fact that f is bounded, whileR the second integral is also small since f is uniformly continuous and Kδ = 1. This concludes the proof of the corollary. 1.5 The Fourier inversion The next result is an identity sometimes called the multiplication formula. Proposition 1.8 If f, g ∈ S(R), then Z Z ∞ f (x)ˆ g (x) dx = −∞

∞

fˆ(y)g(y) dy.

−∞

To prove the proposition, we need to digress briefly to discuss the interchange of the order of integration for double integrals. Suppose F (x, y) is a continuous function in the plane (x, y) ∈ R2 . We will assume the following decay condition on F : |F (x, y)| ≤ A/(1 + x2 )(1 + y 2 ).

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Then, we can state that for each x the function F (x, y) is of moderate decrease in y, and similarly for each fixed y the function R ∞F (x, y) is of moderate decrease in x. Moreover, the function F1 (x) = −∞ F (x, y) dy is continuous and of moderate decrease; similarly for the function F2 (y) = R∞ −∞ F (x, y) dx. Finally Z

Z

∞

−∞

F1 (x) dx =

∞

−∞

F2 (y) dy.

The proof of these facts may be found in the appendix. We now apply this to F (x, y) = f (x)g(y)e−2πixy . Then F1 (x) = f (x)ˆ g (x), and F2 (y) = fˆ(y)g(y) so Z ∞ Z ∞ fˆ(y)g(y) dy, f (x)ˆ g (x) dx = −∞

−∞

which is the assertion of the proposition. The multiplication formula and the fact that the Gaussian is its own Fourier transform lead to a proof of the first major theorem. Theorem 1.9 (Fourier inversion) If f ∈ S(R), then Z ∞ fˆ(ξ)e2πixξ dξ. f (x) = −∞

Proof. We first claim that Z

∞

f (0) =

fˆ(ξ) dξ.

−∞ 2 cδ (ξ) = Kδ (ξ). By the multiplication forLet Gδ (x) = e−πδx so that G mula we get Z ∞ Z ∞ f (x)Kδ (x) dx = fˆ(ξ)Gδ (ξ) dξ.

−∞

−∞

Since Kδ is a good kernel, the first integral goes R ∞ to f (0) as δ tends to 0. Since the second integral clearly converges to −∞ fˆ(ξ) dξ as δ tends to 0, our claim is proved. In general, let F (y) = f (y + x) so that Z ∞ Z ∞ ˆ f (x) = F (0) = F (ξ) dξ = fˆ(ξ)e2πixξ dξ. −∞

−∞

As the name of Theorem 1.9 suggests, it provides a formula that inverts the Fourier transform; in fact we see that the Fourier transform is its own

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inverse except for the change of x to −x. More precisely, we may define two mappings F : S(R) → S(R) and F ∗ : S(R) → S(R) by Z

Z

∞

F(f )(ξ) =

−2πixξ

f (x)e

dx

∗

and

∞

F (g)(x) =

−∞

g(ξ)e2πixξ dξ.

−∞

Thus F is the Fourier transform, and Theorem 1.9 guarantees that F ∗ ◦ F = I on S(R), where I is the identity mapping. Moreover, since the definitions of F and F ∗ differ only by a sign in the exponential, we see that F(f )(y) = F ∗ (f )(−y), so we also have F ◦ F ∗ = I. As a consequence, we conclude that F ∗ is the inverse of the Fourier transform on S(R), and we get the following result. Corollary 1.10 The Fourier transform is a bijective mapping on the Schwartz space. 1.6 The Plancherel formula We need a few further results about convolutions of Schwartz functions. The key fact is that the Fourier transform interchanges convolutions with pointwise products, a result analogous to the situation for Fourier series. Proposition 1.11 If f, g ∈ S(R) then: (i) f ∗ g ∈ S(R). (ii) f ∗ g = g ∗ f . \ (iii) (f ∗ g)(ξ) = fˆ(ξ)ˆ g (ξ). Proof. To prove that f ∗ g is rapidly decreasing, observe first that for any ` ≥ 0 we have supx |x|` |g(x − y)| ≤ A` (1 + |y|)` , because g is rapidly decreasing (to check this assertion, consider separately the two cases |x| ≤ 2|y| and |x| ≥ 2|y|). From this, we see that Z `

sup |x (f ∗ g)(x)| ≤ A` x

∞

|f (y)|(1 + |y|)` dy,

−∞

so that x` (f ∗ g)(x) is a bounded function for every ` ≥ 0. These estimates carry over to the derivatives of f ∗ g, thereby proving that f ∗ g ∈ S(R) because µ

d dx

¶k

µ

(f ∗ g)(x) = (f ∗

d dx

¶k

g)(x)

for k = 1, 2, . . ..

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This identity is proved first for k = 1 by differentiating under the integral defining f ∗ g. The interchange of differentiation and integration is justified in this case by the rapid decrease of dg/dx. The identity then follows for every k by iteration. For fixed x, the change of variables x − y = u shows that Z

∞

(f ∗ g)(x) =

f (x − u)g(u) du = (g ∗ f )(x). −∞

This change of variables is a composition of two changes, y 7→ −y and yR 7→ y − h (withR h = x). For the first one we use the observation that ∞ ∞ −∞ F (x) dx = −∞ F (−x) dx for any Schwartz function F , and for the second, we apply (ii) of Proposition 1.1 Finally, consider F (x, y) = f (y)g(x − y)e−2πixξ . Since f and g are rapidly decreasing, considering separately the two cases |x| ≤ 2|y| and |x| ≥ 2|y|, we see that the discussion of the change of order of integration after Proposition 1.8 applies to F . InR this case F1 (x) R= (f ∗ g)(x)e−2πixξ , ∞ ∞ and F2 (y) = f (y)e−2πiyξ gˆ(ξ). Thus −∞ F1 (x) dx = −∞ F2 (y) dy, which implies (iii). The proposition is therefore proved. We now use the properties of convolutions of Schwartz functions to prove the main result of this section. The result we have in mind is the analogue for functions on R of Parseval’s identity for Fourier series. The Schwartz space can be equipped with a Hermitian inner product Z

∞

f (x)g(x) dx

(f, g) = −∞

whose associated norm is µZ

∞

kf k =

¶1/2 |f (x)|2 dx .

−∞

The second major theorem in the theory states that the Fourier transform is a unitary transformation on S(R). Theorem 1.12 (Plancherel) If f ∈ S(R) then kfˆk = kf k. Proof. If f ∈ S(R) define f [ (x) = f (−x). Then fc[ (ξ) = fˆ(ξ). Now let h = f ∗ f [ . Clearly, we have Z

ˆ h(ξ) = |fˆ(ξ)|2

and

∞

h(0) = −∞

|f (x)|2 dx.

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The theorem now follows from the inversion formula applied with x = 0, that is, Z ∞ ˆ dξ = h(0). h(ξ) −∞

1.7 Extension to functions of moderate decrease In the previous sections, we have limited our assertion of the Fourier inversion and Plancherel formulas to the case when the function involved belonged to the Schwartz space. It does not really involve further ideas to extend these results to functions of moderate decrease, once we make the additional assumption that the Fourier transform of the function under consideration is also of moderate decrease. Indeed, the key observation, which is easy to prove, is that the convolution f ∗ g of two functions f and g of moderate decrease is again a function of moderate decrease (Exercise 7); also f[ ∗ g = fˆgˆ. Moreover, the multiplication formula continues to hold, and we deduce the Fourier inversion and Plancherel formulas when f and fˆ are both of moderate decrease. This generalization, although modest in scope, is nevertheless useful in some circumstances. 1.8 The Weierstrass approximation theorem We now digress briefly by further exploiting our good kernels to prove the Weierstrass approximation theorem. This result was already alluded to in Chapter 2. Theorem 1.13 Let f be a continuous function on the closed and bounded interval [a, b] ⊂ R. Then, for any ² > 0, there exists a polynomial P such that sup |f (x) − P (x)| < ². x∈[a,b]

In other words, f can be uniformly approximated by polynomials. Proof. Let [−M, M ] denote any interval that contains [a, b] in its interior, and let g be a continuous function on R that equals 0 outside [−M, M ] and equals f in [a, b]. For example, extend f as follows: from b to M define g by a straight line segment going from f (b) to 0, and from a to −M by a straight line segment from f (a) also to 0. Let B be a

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2. Applications to some partial differential equations

bound for g, that is, |g(x)| ≤ B for all x. Then, since {Kδ } is a family of good kernels, and g is continuous with compact support, we may argue as in the proof of Corollary 1.7 to see that g ∗ Kδ converges uniformly to g as δ tends to 0. In fact, we choose δ0 so that |g(x) − (g ∗ Kδ0 )(x)| < ²/2

for all x ∈ R.

Now, we recall that ex is given by the power series expansion ex = P∞ n n=0 x /n! which converges uniformly in every compact interval of R. Therefore, there exists an integer N so that ² |Kδ0 (x) − R(x)| ≤ for all x ∈ [−2M, 2M ] 4M B −1/2 PN (−πx2 /δ0 )n where R(x) = δ0 . Then, recalling that g vanishes n=0 n! outside the interval [−M, M ], we have that for all x ∈ [−M, M ] ¯Z ¯ ¯ M ¯ ¯ ¯ |(g ∗ Kδ0 )(x) − (g ∗ R)(x)| = ¯ g(t) [Kδ0 (x − t) − R(x − t)] dt¯ ¯ −M ¯ Z M |g(t)| |Kδ0 (x − t) − R(x − t)| dt ≤ −M

≤ 2M B

sup z∈[−2M,2M ]

|Kδ0 (z) − R(z)|

< ²/2. Therefore, the triangle inequality implies that |g(x) − (g ∗ R)(x)| < ² whenever x ∈ [−M, M ], hence |f (x) − (g ∗ R)(x)| < ² when x ∈ [a, b]. Finally, note that g ∗ R is a polynomial in the x variable. Indeed, by RM definition we have (g ∗ R)(x) = −M g(t)R(x − t) dt, and R(x − t) is a polynomial in x since it can be expressed, after several expansions, as P R(x − t) = n an (t)xn where the sum is finite. This concludes the proof of the theorem.

2 Applications to some partial differential equations We mentioned earlier that a crucial property of the Fourier transform is that it interchanges differentiation and multiplication by polynomials. We now use this crucial fact together with the Fourier inversion theorem to solve some specific partial differential equations. 2.1 The time-dependent heat equation on the real line In Chapter 4 we considered the heat equation on the circle. Here we study the analogous problem on the real line.

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Consider an infinite rod, which we model by the real line, and suppose that we are given an initial temperature distribution f (x) on the rod at time t = 0. We wish now to determine the temperature u(x, t) at a point x at time t > 0. Considerations similar to the ones given in Chapter 1 show that when u is appropriately normalized, it solves the following partial differential equation: ∂u ∂2u , = ∂t ∂x2

(8)

called the heat equation. The initial condition we impose is u(x, 0) = f (x). Just as in the case of the circle, the solution is given in terms of a convolution. Indeed, define the heat kernel of the line by Ht (x) = Kδ (x),

with δ = 4πt,

so that Ht (x) =

2 1 e−x /4t (4πt)1/2

and

2 2 Hˆt (ξ) = e−4π tξ .

Taking the Fourier transform of equation (8) in the x variable (formally) leads to ∂u ˆ (ξ, t) = −4π 2 ξ 2 u ˆ(ξ, t). ∂t Fixing ξ, this is an ordinary differential equation in the variable t (with unknown u ˆ(ξ, ·)), so there exists a constant A(ξ) so that u ˆ(ξ, t) = A(ξ)e−4π

2 2

ξ t

.

We may also take the Fourier transform of the initial condition and obtain u ˆ(ξ, 0) = fˆ(ξ), hence A(ξ) = fˆ(ξ). This leads to the following theorem. Theorem 2.1 Given f ∈ S(R), let u(x, t) = (f ∗ Ht )(x)

for t > 0

where Ht is the heat kernel. Then: (i) The function u is C 2 when x ∈ R and t > 0, and u solves the heat equation.

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147

(ii) u(x, t) → f (x) uniformly in x as t → 0. Hence if we set u(x, 0) = f (x), then u is continuous on the closure of the upper half-plane R2+ = {(x, t) : x ∈ R, t ≥ 0}. R∞ (iii) −∞ |u(x, t) − f (x)|2 dx → 0 as t → 0. Proof. Because u = f ∗ Ht , taking the Fourier transform in the x2 2 variable gives u ˆ = fˆHˆt , and so u ˆ(ξ, t) = fˆ(ξ)e−4π ξ t . The Fourier inversion formula gives Z ∞ 2 2 u(x, t) = fˆ(ξ)e−4π tξ e2πiξx dξ. −∞

By differentiating under the integral sign, one verifies (i). In fact, one observes that u is indefinitely differentiable. Note that (ii) is an immediate consequence of Corollary 1.7. Finally, by Plancherel’s formula, we have Z ∞ Z ∞ |ˆ u(ξ, t) − fˆ(ξ)|2 dξ |u(x, t) − f (x)|2 dx = −∞ −∞ Z ∞ 2 2 |fˆ(ξ)|2 |e−4π tξ − 1| dξ. = −∞

To see that this last integral goes to 0 as t → 0, we argue as follows: 2 2 since |e−4π tξ − 1| ≤ 2 and f ∈ S(R), we can find N so that Z 2 2 |fˆ(ξ)|2 |e−4π tξ − 1| dξ < ², |ξ|≥N 2 2 and for all small t we have sup|ξ|≤N |fˆ(ξ)|2 |e−4π tξ − 1| < ²/2N since fˆ is bounded. Thus Z 2 2 |fˆ(ξ)|2 |e−4π tξ − 1| dξ < ² for all small t.

|ξ|≤N

This completes the proof of the theorem. The above theorem guarantees the existence of a solution to the heat equation with initial data f . This solution is also unique, if uniqueness is formulated appropriately. In this regard, we note that u = f ∗ Ht , f ∈ S(R), satisfies the following additional property. Corollary 2.2 u(·, t) belongs to S(R) uniformly in t, in the sense that for any T > 0 ¯ ¯ ¯ ∂` ¯ ¯ ¯ (9) sup |x|k ¯ ` u(x, t)¯ < ∞ for each k, ` ≥ 0. ¯ ¯ ∂x x ∈ R 0 < t < T

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Proof. This result is a consequence of the following estimate: Z Z |f (x − y)|Ht (y) dy + |f (x − y)|Ht (y) dy |u(x, t)| ≤ |y|≤|x|/2

≤

|y|≥|x|/2

2 CN C + √ e−cx /t . (1 + |x|)N t

Indeed, since f is rapidly decreasing, we have |f (x − y)| ≤ CN /(1 + |x|)N 2

when |y| ≤ |x|/2. Also, if |y| ≥ |x|/2 then Ht (y) ≤ Ct−1/2 e−cx /t , and we obtain the above inequality. Consequently, we see that u(x, t) is rapidly decreasing uniformly for 0 < t < T . The same argument can be applied to the derivatives of u in the x variable since we may differentiate under the integral sign and apply the above estimate with f replaced by f 0 , and so on. This leads to the following uniqueness theorem. Theorem 2.3 Suppose u(x, t) satisfies the following conditions: (i) u is continuous on the closure of the upper half-plane. (ii) u satisfies the heat equation for t > 0. (iii) u satisfies the boundary condition u(x, 0) = 0. (iv) u(·, t) ∈ S(R) uniformly in t, as in (9). Then, we conclude that u = 0. Below we use the abbreviations ∂x` u and ∂t u to denote ∂ ` u/∂x` and ∂u/∂t, respectively. Proof. We define the energy at time t of the solution u(x, t) by Z E(t) = |u(x, t)|2 dx. R

Clearly E(t) ≥ 0. Since E(0) = 0 it suffices to show that E is a decreasing function, and this is achieved by proving that dE/dt ≤ 0. The assumptions on u allow us to differentiate E(t) under the integral sign dE = dt

Z R

[∂t u(x, t)u(x, t) + u(x, t)∂t u(x, t)] dx.

But u satisfies the heat equation, therefore ∂t u = ∂x2 u and ∂t u = ∂x2 u, so that after an integration by parts, where we use the fact that u and its

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149

x derivatives decrease rapidly as |x| → ∞, we find dE = dt

Z

£

R

¤ ∂x2 u(x, t)u(x, t) + u(x, t)∂x2 u(x, t) dx

Z

=− R Z

[∂x u(x, t)∂x u(x, t) + ∂x u(x, t)∂x u(x, t)] dx

= −2 R

|∂x u(x, t)|2 dx

≤ 0, as claimed. Thus E(t) = 0 for all t, hence u = 0. Another uniqueness theorem for the heat equation, with a less restrictive assumption than (9), can be found in Problem 6. Examples when uniqueness fails are given in Exercise 12 and Problem 4. 2.2 The steady-state heat equation in the upper half-plane The equation we are now concerned with is (10)

4u =

∂2u ∂2u + 2 =0 ∂x2 ∂y

in the upper half-plane R2+ = {(x, y) : x ∈ R, y > 0}. The boundary condition we require is u(x, 0) = f (x). The operator 4 is the Laplacian and the above partial differential equation describes the steady-state heat distribution in R2+ subject to u = f on the boundary. The kernel that solves this problem is called the Poisson kernel for the upper half-plane, and is given by Py (x) =

1 y 2 π x + y2

where x ∈ R and y > 0.

This is the analogue of the Poisson kernel for the disc discussed in Section 5.4 of Chapter 2. Note that for each fixed y the kernel Py is only of moderate decrease as a function of x, so we will use the theory of the Fourier transform appropriate for these types of functions (see Section 1.7). We proceed as in the case of the time-dependent heat equation, by taking the Fourier transform of equation (10) (formally) in the x variable, thereby obtaining −4π 2 ξ 2 u ˆ(ξ, y) +

∂2u ˆ (ξ, y) = 0 ∂y 2

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with the boundary condition u ˆ(ξ, 0) = fˆ(ξ). The general solution of this ordinary differential equation in y (with ξ fixed) takes the form u ˆ(ξ, y) = A(ξ)e−2π|ξ|y + B(ξ)e2π|ξ|y . If we disregard the second term because of its rapid exponential increase we find, after setting y = 0, that u ˆ(ξ, y) = fˆ(ξ)e−2π|ξ|y . Therefore u is given in terms of the convolution of f with a kernel whose Fourier transform is e−2π|ξ|y . This is precisely the Poisson kernel given above, as we prove next. Lemma 2.4 The following two identities hold: Z

∞

−∞ ∞

e−2π|ξ|y e2πiξx dξ = Py (x),

Z

−∞

Py (x)e−2πixξ dx = e−2π|ξ|y .

Proof. The first formula is fairly straightforward since we can split the integral from −∞ to 0 and 0 to ∞. Then, since y > 0 we have Z

∞

Z −2πξy 2πiξx

e

e

∞

"

2πi(x+iy)ξ

e

dξ = 0

0

−

e2πi(x+iy)ξ dξ = 2πi(x + iy)

#∞

= 0

1 , 2πi(x + iy)

and similarly, Z

0

e2πξy e2πiξx dξ =

−∞

Therefore Z ∞ e−2π|ξ|y e2πiξx dξ = −∞

1 . 2πi(x − iy)

1 1 y − = . 2πi(x − iy) 2πi(x + iy) π(x2 + y 2 )

The second formula is now a consequence of the Fourier inversion theorem applied in the case when f and fˆ are of moderate decrease. Lemma 2.5 The Poisson kernel is a good kernel on R as y → 0.

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151

R ∞Proof.

Setting ξ = 0 in the second formula of the lemma shows that P (x) dx = 1, and clearly Py (x) ≥ 0, so it remains to check the last −∞ y property of good kernels. Given a fixed δ > 0, we may change variables u = x/y so that Z ∞ Z ∞ du y ∞ dx = = [arctan u]δ/y = π/2 − arctan(δ/y), 2 2 2 x +y δ/y 1 + u δ and this quantity goes to 0 as y → 0. Since Py (x) is an even function, the proof is complete. The following theorem establishes the existence of a solution to our problem. Theorem 2.6 Given f ∈ S(R), let u(x, y) = (f ∗ Py )(x). Then: (i) u(x, y) is C 2 in R2+ and 4u = 0. (ii) u(x, y) → f (x) uniformly as y → 0. R∞ (iii) −∞ |u(x, y) − f (x)|2 dx → 0 as y → 0. (iv) If u(x, 0) = f (x), then u is continuous on the closure R2+ of the upper half-plane, and vanishes at infinity in the sense that u(x, y) → 0

as |x| + y → ∞.

Proof. The proofs of parts (i), (ii), and (iii) are similar to the case of the heat equation, and so are left to the reader. Part (iv) is a consequence of two easy estimates whenever f is of moderate decrease. First, we have µ ¶ 1 y |(f ∗ Py )(x)| ≤ C + (1 + x2 ) x2 + y 2 which isR proved (as in the case of the heat equation) by splitting the ∞ integral −∞ f (x − t)Py (t) dt into the part where |t| ≤ |x|/2 and the part where |t| ≥ |x|/2. Also, we have |(f ∗ Py )(x)| ≤ C/y, since supx Py (x) ≤ c/y. Using the first estimate when |x| ≥ |y| and the second when |x| ≤ |y| gives the desired decrease at infinity. We next show that the solution is essentially unique. Theorem 2.7 Suppose u is continuous on the closure of the upper halfplane R2+ , satisfies 4u = 0 for (x, y) ∈ R2+ , u(x, 0) = 0, and u(x, y) vanishes at infinity. Then u = 0.

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A simple example shows that a condition concerning the decay of u at infinity is needed: take u(x, y) = y. Clearly u satisfies the steady-state heat equation and vanishes on the real line, yet u is not identically zero. The proof of the theorem relies on a basic fact about harmonic functions, which are functions satisfying 4u = 0. The fact is that the value of a harmonic function at a point equals its average value around any circle centered at that point. Lemma 2.8 (Mean-value property) Suppose Ω is an open set in R2 and let u be a function of class C 2 with 4u = 0 in Ω. If the closure of the disc centered at (x, y) and of radius R is contained in Ω, then 1 2π

u(x, y) =

Z

2π

u(x + r cos θ, y + r sin θ) dθ 0

for all 0 ≤ r ≤ R. Proof. Let U (r, θ) = u(x + r cos θ, y + r sin θ). Expressing the Laplacian in polar coordinates, the equation 4u = 0 then implies µ ¶ ∂2U ∂ ∂U 0= + r r . ∂θ2 ∂r ∂r If we define F (r) = r

R 2π

1 2π

∂ ∂r

0

µ

r

U (r, θ) dθ, the above gives

∂F ∂r

¶

=

1 2π

Z

2π

− 0

∂2U (r, θ) dθ. ∂θ2

2 The integral of ¡∂ 2 U/∂θ over the circle vanishes since ∂U/∂θ is peri¢ ∂ ∂F odic, hence r ∂r r ∂r = 0, and consequently r∂F/∂r must be constant. Evaluating this expression at r = 0 we find that ∂F/∂r = 0. Thus F is constant, but since F (0) = u(x, y), we finally find that F (r) = u(x, y) for all 0 ≤ r ≤ R, which is the mean-value property. Finally, note that the argument above is implicit in the proof of Theorem 5.7, Chapter 2.

To prove Theorem 2.7 we argue by contradiction. Considering separately the real and imaginary parts of u, we may suppose that u itself is real-valued, and is somewhere strictly positive, say u(x0 , y0 ) > 0 for some x0 ∈ R and y0 > 0. We shall see that this leads to a contradiction. First, since u vanishes at infinity, we can find a large semi-disc of ra+ dius R, DR = {(x, y) : x2 + y 2 ≤ R, y ≥ 0} outside of which u(x, y) ≤ + 1 2 u(x0 , y0 ). Next, since u is continuous in DR , it attains its maximum + M there, so there exists a point (x1 , y1 ) ∈ DR with u(x1 , y1 ) = M , while

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3. The Poisson summation formula

u(x, y) ≤ M in the semi-disc; also, since u(x, y) ≤ 21 u(x0 , y0 ) ≤ M/2 outside of the semi-disc, we have u(x, y) ≤ M throughout the entire upper half-plane. Now the mean-value property for harmonic functions implies Z 2π 1 u(x1 , y1 ) = u(x1 + ρ cos θ, y1 + ρ sin θ) dθ 2π 0 whenever the circle of integration lies in the upper half-plane. In particular, this equation holds if 0 < ρ < y1 . Since u(x1 , y1 ) equals the maximum value M , and u(x1 + ρ cos θ, y1 + ρ sin θ) ≤ M , it follows by continuity that u(x1 + ρ cos θ, y1 + ρ sin θ) = M on the whole circle. For otherwise u(x, y) ≤ M − ², on an arc of length δ > 0 on the circle, and this would give Z 2π 1 ²δ u(x1 + ρ cos θ, y1 + ρ sin θ) dθ ≤ M − < M, 2π 0 2π contradicting the fact that u(x1 , y1 ) = M . Now letting ρ → y1 , and using the continuity of u again, we see that this implies u(x1 , 0) = M > 0, which contradicts the fact that u(x, 0) = 0 for all x.

3 The Poisson summation formula The definition of the Fourier transform was motivated by the desire for a continuous version of Fourier series, applicable to functions defined on the real line. We now show that there exists a further remarkable connection between the analysis of functions on the circle and related functions on R. Given a function f ∈ S(R) on the real line, we can construct a new function on the circle by the recipe F1 (x) =

∞ X

f (x + n).

n=−∞

Since f is rapidly decreasing, the series converges absolutely and uniformly on every compact subset of R, so F1 is continuous. Note that F1 (x + 1) = F1 (x) because passage from n to n + 1 in the above sum merely shifts the terms on the series defining F1 (x). Hence F1 is periodic with period 1. The function F1 is called the periodization of f . There is another way to arrive at a “periodic version” of f , this time by Fourier analysis. Start with the identity Z ∞ f (x) = fˆ(ξ)e2πiξx dξ, −∞

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and consider its discrete analogue, where the integral is replaced by a sum F2 (x) =

∞ X

fˆ(n)e2πinx .

n=−∞

Once again, the sum converges absolutely and uniformly since fˆ belongs to the Schwartz space, hence F2 is continuous. Moreover, F2 is also periodic of period 1 since this is the case for each one of the exponentials e2πinx . The fundamental fact is that these two approaches, which produce F1 and F2 , actually lead to the same function. Theorem 3.1 (Poisson summation formula) If f ∈ S(R), then ∞ X

∞ X

f (x + n) =

n=−∞

fˆ(n)e2πinx .

n=−∞

In particular, setting x = 0 we have ∞ X n=−∞

f (n) =

∞ X

fˆ(n).

n=−∞

In other words, the Fourier coefficients of the periodization of f are given precisely by the values of the Fourier transform of f on the integers. Proof. To check the first formula it suffices, by Theorem 2.1 in Chapter 2, to show that both sides (which are continuous) have the same Fourier coefficients (viewed as functions on the circle). Clearly, the mth Fourier coefficient of the right-hand side is fˆ(m). For the left-hand side we have ! Z 1Ã X ∞ ∞ Z 1 X f (x + n) e−2πimx dx = f (x + n)e−2πimx dx 0

n=−∞

=

n=−∞ 0 ∞ Z n+1 X

f (y)e−2πimy dy

n=−∞ n Z ∞

f (y)e−2πimy dy

= −∞

= fˆ(m), where the interchange of the sum and integral is permissible since f is rapidly decreasing. This completes the proof of the theorem.

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3. The Poisson summation formula

We observe that the theorem extends to the case when we merely assume that both f and fˆ are of moderate decrease; the proof is in fact unchanged. It turns out that the operation of periodization is important in a number of questions, even when the Poisson summation formula does not apply. We give an example by considering the elementary function P f (x) = 1/x, x 6= 0. The result is that ∞ n=−∞ 1/(x + n), when summed symmetrically, gives the partial fraction decomposition of the cotangent function. In fact this sum equals π cot πx, when x is not an integer. P 2 2 2 Similarly with f (x) = 1/x2 , we get ∞ n=−∞ 1/(x + n) = π /(sin πx) , whenever x ∈ / Z (see Exercise 15). 3.1 Theta and zeta functions We define the theta function ϑ(s) for s > 0 by ϑ(s) =

∞ X

2

e−πn s .

n=−∞

The condition on s ensures the absolute convergence of the series. A crucial fact about this special function is that it satisfies the following functional equation. Theorem 3.2 s−1/2 ϑ(1/s) = ϑ(s) whenever s > 0. The proof of this identity consists of a simple application of the Poisson summation formula to the pair f (x) = e−πsx

2

and

2 fˆ(ξ) = s−1/2 e−πξ /s .

The theta function ϑ(s) also extends to complex values of s when Re(s) > 0, and the functional equation is still valid then. The theta function is intimately connected with an important function in number theory, the zeta function ζ(s) defined for Re(s) > 1 by ζ(s) =

∞ X 1 n=1

ns

.

Later we will see that this function carries essential information about the prime numbers (see Chapter 8). It also turns out that ζ, ϑ, and another important function Γ are related by the following identity: Z 1 ∞ s/2−1 −s/2 π Γ(s/2)ζ(s) = t (ϑ(t) − 1) dt, 2 0

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which is valid for s > 1 (Exercises 17 and 18). Returning to the function ϑ, define the generalization Θ(z|τ ) given by ∞ X

Θ(z|τ ) =

2

eiπn τ e2πinz

n=−∞

whenever Im(τ ) > 0 and z ∈ C. Taking z = 0 and τ = is we get Θ(z|τ ) = ϑ(s). 3.2 Heat kernels Another application related to the Poisson summation formula and the theta function is the time-dependent heat equation on the circle. A solution to the equation ∂u ∂2u = ∂t ∂x2 subject to u(x, 0) = f (x), where f is periodic of period 1, was given in the previous chapter by u(x, t) = (f ∗ Ht )(x) where Ht (x) is the heat kernel on the circle, that is, Ht (x) =

∞ X

e−4π

2

n2 t 2πinx

e

.

n=−∞

Note in particular that with our definition of the generalized theta function in the previous section, we have Θ(x|4πit) = Ht (x). Also, recall that the heat equation on R gave rise to the heat kernel Ht (x) =

2 1 e−x /4t 1/2 (4πt)

where Hˆt (ξ) = e−4π ξ t . The fundamental relation between these two objects is an immediate consequence of the Poisson summation formula: 2 2

Theorem 3.3 The heat kernel on the circle is the periodization of the heat kernel on the real line: Ht (x) =

∞ X n=−∞

Ht (x + n).

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3. The Poisson summation formula

Although the proof that Ht is a good kernel on R was fairly straightforward, we left open the harder problem that Ht is a good kernel on the circle. The above results allow us to resolve this matter. Corollary 3.4 The kernel Ht (x) is a good kernel for t → 0. R Proof. We already observed that |x|≤1/2 Ht (x) dx = 1. Now note that Ht ≥ 0, which is immediate from the above formula since Ht ≥ 0. Finally, we claim that when |x| ≤ 1/2, Ht (x) = Ht (x) + Et (x), where the error satisfies |Et (x)| ≤ c1 e−c2 /t with c1 , c2 > 0 and 0 < t ≤ 1. To see this, note again that the formula in the theorem gives X Ht (x) = Ht (x) + Ht (x + n); |n|≥1

therefore, since |x| ≤ 1/2, X 2 1 X −(x+n)2 /4t Et (x) = √ e ≤ Ct−1/2 e−cn /t . 4πt |n|≥1 n≥1

Note that n2 /t ≥ n2 and n2 /t ≥ 1/t whenever 0 < t ≤ 1, so e−cn c 2 c 1 e− 2 n e− 2 t . Hence c 1 X c 2 |Et (x)| ≤ Ct−1/2 e− 2 t e− 2 n ≤ c1 e−c2 /t .

2

/t

≤

n≥1

R The proof of the claim is complete, and as a result |x|≤1/2 |Et (x)| dx → 0 as t → 0. It is now clear that Ht satisfies Z |Ht (x)| dx → 0 as t → 0, η 0 and |A|2 = p 2B/π. In fact, we have ¶ µZ µZ ∞ (x − x0 )2 |ψ(x)|2 dx −∞

∞ −∞

¶ 2 ˆ (ξ − ξ0 )2 |ψ(ξ)| dξ

≥

1 16π 2

for every x0 , ξ0 ∈ R. Proof. The second inequality actually follows from the first by replacing ψ(x) by e−2πixξ0 ψ(x + x0 ) and changing variables. To prove the first inequality, we argue as follows. Beginning with our normalizing asR 2 sumption |ψ| = 1, and recalling that ψ and ψ 0 are rapidly decreasing, an integration by parts gives Z ∞ 1= |ψ(x)|2 dx −∞ Z ∞ d =− x |ψ(x)|2 dx dx −∞ Z ∞ ³ ´ =− xψ 0 (x)ψ(x) + xψ 0 (x)ψ(x) dx. −∞

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4. The Heisenberg uncertainty principle

The last identity follows because |ψ|2 = ψψ. Therefore Z

∞

1≤2

|x| |ψ(x)| |ψ 0 (x)| dx

−∞

µZ

∞

≤2

¶1/2 µZ x2 |ψ(x)|2 dx

−∞

∞

¶1/2 |ψ 0 (x)|2 dx ,

−∞

where we have used the Cauchy-Schwarz inequality. The identity Z ∞ Z ∞ 2 ˆ |ψ 0 (x)|2 dx = 4π 2 ξ 2 |ψ(ξ)| dξ, −∞

−∞

which holds because of the properties of the Fourier transform and the Plancherel formula, concludes the proof of the inequality in the theorem. If equality holds, then we must also have equality where we applied the Cauchy-Schwarz inequality, and as a result we find that ψ 0 (x) = βxψ(x) 2 for some constant β. The solutions to this equation are ψ(x) = Aeβx /2 , where A is constant. Since we want ψ to be a SchwartzR function, we must ∞ take β = −2B < 0, and since we impose the condition −∞ |ψ(x)|2 dx = 1 p we find that |A|2 = 2B/π, as was to be shown. The precise assertion contained in Theorem 4.1 first came to light in the study of quantum mechanics. It arose when one considered the extent to which one could simultaneously locate the position and momentum of a particle. Assuming we are dealing with (say) an electron that travels along the real line, then according to the laws of physics, matters are governed by a “state function” ψ, which we can assume to be in S(R), and which is normalized according to the requirement that Z ∞ |ψ(x)|2 dx = 1. (11) −∞

The position of the particle is then determined not as a definite point x; instead its probable location is given by the rules of quantum mechanics as follows: • The probability that the particle is located in the interval (a, b) is Rb 2 a |ψ(x)| dx. According to this law we can calculate the probable location of the particle with the aid of ψ: in fact, there may be only a small probability 0 0 that the particle is located in a given , b ), but nevertheless it R ∞interval (a 2 is somewhere on the real line since −∞ |ψ(x)| dx = 1.

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In addition to the probability density |ψ(x)|2 dx, there is the expectation of where the particle might be. This expectation is the best guess of the position of the particle, given its probability distribution determined by |ψ(x)|2 dx, and is the quantity defined by Z

(12)

∞

x=

x|ψ(x)|2 dx.

−∞

Why is this our best guess? Consider the simpler (idealized) situation where we are given that the particle can be found at only finitely many different points, x1 , x2 , . . . , xN on the real axis, with pi the probability that the particle is at xi , and p1 + p2 + · · · + pN = 1. Then, if we knew nothing else, and were forced to make one choice as to the position of the P particle, we would naturally take x = N i=1 xi pi , which is the appropriate weighted average of the possible positions. The quantity (12) is clearly the general (integral) version of this. We next come to the notion of variance, which in our terminology is the uncertainty attached to our expectation. Having determined that the expected position of the particle is x (given by (12)), the resulting uncertainty is the quantity Z ∞ (x − x)2 |ψ(x)|2 dx. (13) −∞

Notice that if ψ is highly concentrated near x, it means that there is a high probability that x is near x, and so (13) is small, because most of the contribution to the integral takes place for values of x near x. Here we have a small uncertainty. On the other hand, if ψ(x) is rather flat (that is, the probability distribution |ψ(x)|2 dx is not very concentrated), then the integral (13) is rather big, because large values of (x − x)2 will come into play, and as a result the uncertainty is relatively large. It is also worthwhile toRobserve that the expectation x is that choice ∞ for which the uncertainty −∞ (x − x)2 |ψ(x)|2 dx is the smallest. Indeed, if we try to minimize this quantity by equating to 0 its derivative with R∞ respect to x, we find that 2 −∞ (x − x)|ψ(x)|2 dx = 0, which gives (12). So far, we have discussed the “expectation” and “uncertainty” related to the position of the particle. Of equal relevance are the corresponding notions regarding its momentum. The corresponding rule of quantum mechanics is: • The probability thatR the momentum ξ of the particle belongs to b ˆ 2 the interval (a, b) is a |ψ(ξ)| dξ where ψˆ is the Fourier transform of ψ.

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Combining these two laws with Theorem 4.1 gives 1/16π 2 as the lower bound for the product of the uncertainty of the position and the uncertainty of the momentum of a particle. So the more certain we are about the location of the particle, the less certain we can be about its momentum, and vice versa. However, we have simplified the statement of the two laws by rescaling to change the units of measurement. Actually, there enters a fundamental but small physical number ~ called Planck’s constant. When properly taken into account, the physical conclusion is (uncertainty of position)×(uncertainty of momentum) ≥ ~/16π 2 .

5 Exercises 1. Corollary 2.3 in Chapter 2 leads to the following simplified version of the Fourier inversion formula. Suppose f is a continuous function supported on an interval [−M, M ], whose Fourier transform fˆ is of moderate decrease. (a) Fix L with L/2 > M , and show that f (x) = an (L) =

1 L

Z

P

an (L)e2πinx/L where

L/2

f (x)e−2πinx/L dx = −L/2

Alternatively, we may write f (x) = δ

P∞ n=−∞

1 ˆ f (n/L). L

fˆ(nδ)e2πinδx with δ = 1/L.

(b) Prove that if F is continuous and of moderate decrease, then

Z

∞ X

∞

F (ξ) dξ = −∞

Z

∞

(c) Conclude that f (x) =

lim δ

δ → 0 δ > 0

F (δn).

n=−∞

fˆ(ξ)e2πixξ dξ.

−∞

[Hint: For (a), note that the Fourier series of f on [−L/2, L/2] converges abRN solutely. For (b), first approximate the integral by −N F and the sum by P δ |n|≤N/δ F (nδ). Then approximate the second integral by Riemann sums.] 2. Let f and g be the functions defined by

½ f (x) = χ[−1,1] (x) =

1 0

if |x| ≤ 1, otherwise,

½ and

g(x) =

1 − |x| 0

if |x| ≤ 1, otherwise.

Although f is not continuous, the integral defining its Fourier transform still makes sense. Show that sin 2πξ fˆ(ξ) = πξ

µ and

gˆ(ξ) =

sin πξ πξ

¶2 ,

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with the understanding that fˆ(0) = 2 and gˆ(0) = 1. 3. The following exercise illustrates the principle that the decay of fˆ is related to the continuity properties of f . (a) Suppose that f is a function of moderate decrease on R whose Fourier transform fˆ is continuous and satisfies

µ fˆ(ξ) = O

¶

1 |ξ|1+α

as |ξ| → ∞

for some 0 < α < 1. Prove that f satisfies a H¨older condition of order α, that is, that |f (x + h) − f (x)| ≤ M |h|α

for some M > 0 and all x, h ∈ R.

(b) Let f be a continuous function on R which vanishes for |x| ≥ 1, with f (0) = 0, and which is equal to 1/ log(1/|x|) for all x in a neighborhood of the origin. Prove that fˆ is not of moderate decrease. In fact, there is no ² > 0 so that fˆ(ξ) = O(1/|ξ|1+² ) as |ξ| → ∞. [Hint: For part (a), use the Fourier inversion formula to express f (x + h) − f (x) as an integral involving fˆ, and estimate this integral separately for ξ in the two ranges |ξ| ≤ 1/|h| and |ξ| ≥ 1/|h|.] 4. Bump functions. Examples of compactly supported functions in S(R) are very handy in many applications in analysis. Some examples are: (a) Suppose a < b, and f is the function such that f (x) = 0 if x ≤ a or x ≥ b and f (x) = e−1/(x−a) e−1/(b−x)

if a < x < b.

Show that f is indefinitely differentiable on R. (b) Prove that there exists an indefinitely differentiable function F on R such that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b, and F is strictly increasing on [a, b]. (c) Let δ > 0 be so small that a + δ < b − δ. Show that there exists an indefinitely differentiable function g such that g is 0 if x ≤ a or x ≥ b, g is 1 on [a + δ, b − δ], and g is strictly monotonic on [a, a + δ] and [b − δ, b]. [Hint: For (b) consider F (x) = c

Rx −∞

f (t) dt where c is an appropriate constant.]

5. Suppose f is continuous and of moderate decrease. (a) Prove that fˆ is continuous and fˆ(ξ) → 0 as |ξ| → ∞.

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(b) Show that if fˆ(ξ) = 0 for all ξ, then f is identically 0. [Hint: For part (a), show that fˆ(ξ) =

1 2

R∞

[f (x) − f (x − 1/(2ξ))]e−2πixξ dx. R R For part (b), verify that the multiplication formula f (x)ˆ g (x) dx = fˆ(y)g(y) dy still holds whenever g ∈ S(R).] −∞

2

6. The function e−πx is its own Fourier transform. Generate other functions that (up to a constant multiple) are their own Fourier transforms. What must the constant multiples be? To decide this, prove that F 4 = I. Here F(f ) = fˆ is the Fourier transform, F 4 = F ◦ F ◦ F ◦ F , and I is the identity operator (If )(x) = f (x) (see also Problem 7). 7. Prove that the convolution of two functions of moderate decrease is a function of moderate decrease. [Hint: Write

Z

Z

Z

f (x − y)g(y) dy =

+ |y|≤|x|/2

. |y|≥|x|/2

In the first integral f (x − y) = O(1/(1 + x2 )) while in the second integral g(y) = O(1/(1 + x2 )).] 8. Prove that f is continuous, of moderate decrease, and for all x ∈ R, then f = 0.

R∞ −∞

2

f (y)e−y e2xy dy = 0

2

[Hint: Consider f ∗ e−x .] 9. If f is of moderate decrease, then

Z

R

(14) −R

µ

|ξ| 1− R

¶ fˆ(ξ)e2πixξ dξ = (f ∗ FR )(x),

where the Fej´er kernel on the real line is defined by

  FR (t) =



µ R

sin πtR πtR R

¶2 if t 6= 0, if t = 0.

Show that {FR } is a family of good kernels as R → ∞, and therefore (14) tends uniformly to f (x) as R → ∞. This is the analogue of Fej´er’s theorem for Fourier series in the context of the Fourier transform. 10. Below is an outline of a different proof of the Weierstrass approximation theorem.

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Define the Landau kernels by

  (1 − x2 )n Ln (x) =

0

cn

if −1 ≤ x ≤ 1, if |x| ≥ 1,

R∞

where cn is chosen so that −∞ Ln (x) dx = 1. Prove that {Ln }n≥0 is a family of good kernels as n → ∞. As a result, show that if f is a continuous function supported in [−1/2, 1/2], then (f ∗ Ln )(x) is a sequence of polynomials on [−1/2, 1/2] which converges uniformly to f . [Hint: First show that cn ≥ 2/(n + 1).] 11. Suppose that u is the solution to the heat equation given by u = f ∗ Ht where f ∈ S(R). If we also set u(x, 0) = f (x), prove that u is continuous on the closure of the upper half-plane, and vanishes at infinity, that is, u(x, t) → 0

as |x| + t → ∞.

√ [Hint: To prove that u vanishes at infinity, show that (i) |u(x, t)| ≤ C/ t and (ii) 2 |u(x, t)| ≤ C/(1 + |x|2 ) + Ct−1/2 e−cx /t . Use (i) when |x| ≤ t, and (ii) otherwise.] 12. Show that the function defined by u(x, t) =

x Ht (x) t

satisfies the heat equation for t > 0 and limt→0 u(x, t) = 0 for every x, but u is not continuous at the origin. [Hint: Approach the origin with (x, t) on the parabola x2 /4t = c where c is a constant.] 13. Prove the following uniqueness theorem for harmonic functions in the strip {(x, y) : 0 < y < 1, −∞ < x < ∞}: if u is harmonic in the strip, continuous on its closure with u(x, 0) = u(x, 1) = 0 for all x ∈ R, and u vanishes at infinity, then u = 0. 14. Prove that the periodization of the Fej´er kernel FN on the real line (Exercise 9) is equal to the Fej´er kernel for periodic functions of period 1. In other words, ∞ X

FN (x + n) = FN (x),

n=−∞

when N ≥ 1 is an integer, and where FN (x) =

µ N X n=−N

|n| 1− N

¶ e2πinx =

1 sin2 (N πx) . N sin2 (πx)

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15. This exercise provides another example of periodization. (a) Apply the Poisson summation formula to the function g in Exercise 2 to obtain ∞ X

1 π2 = (n + α)2 (sin πα)2

n=−∞

whenever α is real, but not equal to an integer. (b) Prove as a consequence that ∞ X

(15)

n=−∞

1 π = (n + α) tan πα

whenever α is real but not equal to an integer. [Hint: First prove it when 0 < α < 1. To do so, integrate the formula in (b). What is the precise meaning of the series on the left-hand side of (15)? Evaluate at α = 1/2.] 16. The Dirichlet kernel on the real line is defined by

Z

R

fˆ(ξ)e2πixξ dξ = (f ∗ DR )(x)

so that

DR (x) = χ\ [−R,R] (x) =

−R

sin(2πRx) . πx

Also, the modified Dirichlet kernel for periodic functions of period 1 is defined by

X

∗ DN (x) =

|n|≤N −1

1 e2πinx + (e−2πiN x + e2πiN x ). 2

Show that the result in Exercise 15 gives ∞ X

∗ DN (x + n) = DN (x),

n=−∞

where N ≥ 1 is an integer, and the infinite series must be summed symmetrically. ∗ In other words, the periodization of DN is the modified Dirichlet kernel DN . 17. The gamma function is defined for s > 0 by

Z

∞

e−x xs−1 dx.

Γ(s) = 0

(a) Show that for s > 0 the above integral makes sense, that is, that the following two limits exist:

Z δ → 0 δ > 0

Z

1

e−x xs−1 dx

lim

δ

and

A

e−x xs−1 dx.

lim

A→∞

1

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(b) Prove that Γ(s + 1) = sΓ(s) whenever s > 0, and conclude that for every integer n ≥ 1 we have Γ(n + 1) = n!. (c) Show that

µ ¶ Γ

[Hint: For (c), use

R∞

1 2

=

√

µ ¶ π

and

Γ

3 2

√

π . 2

=

2

−∞

e−πx dx = 1.]

18. The zeta function is defined for s > 1 by ζ(s) = identity π −s/2 Γ(s/2)ζ(s) =

1 2

Z

∞

s

t 2 −1 (ϑ(t) − 1) dt

P∞ n=1

1/ns . Verify the

whenever s > 1

0

where Γ and ϑ are the gamma and theta functions, respectively:

Z

∞ X

∞

e−t ts−1 dt

Γ(s) =

and

ϑ(s) =

0

2

e−πn s .

n=−∞

More about the zeta function and its relation to the prime number theorem can be found in Book II. 19. The following is a variant of the calculation of ζ(2m) = in Problem 4, Chapter 3. (a) Apply the Poisson summation formula to and fˆ(ξ) = e−2πt|ξ| where t > 0 in order to get

P∞ n=1

1/n2m found

f (x) = t/(π(x2 + t2 ))

∞ ∞ X 1 X t = e−2πt|n| . π t2 + n2 n=−∞

n=−∞

(b) Prove the following identity valid for 0 < t < 1: ∞ ∞ 1 X t 1 2 X = + (−1)m+1 ζ(2m)t2m−1 π t 2 + n2 πt π n=−∞

m=1

as well as ∞ X n=−∞

e−2πt|n| =

2 − 1. 1 − e−2πt

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5. Exercises

(c) Use the fact that ∞ z z X B2m 2m = 1 − + z , ez − 1 2 (2m)! m=1

where Bk are the Bernoulli numbers to deduce from the above formula, 2ζ(2m) = (−1)m+1

(2π)2m B2m . (2m)!

20. The following results are relevant in information theory when one tries to recover a signal from its samples. Suppose f is of moderate decrease and that its Fourier transform fˆ is supported in I = [−1/2, 1/2]. Then, f is entirely determined by its restriction to Z. This means that if g is another function of moderate decrease whose Fourier transform is supported in I and f (n) = g(n) for all n ∈ Z, then f = g. More precisely: (a) Prove that the following reconstruction formula holds: f (x) =

∞ X

f (n)K(x − n)

where K(y) =

n=−∞

sin πy . πy

Note that K(y) = O(1/|y|) as |y| → ∞. (b) If λ > 1, then f (x) =

∞ X 1 ³n´ n=−∞

λ

f

λ

³ Kλ x −

n´ λ

where Kλ (y) =

cos πy − cos πλy . π 2 (λ − 1)y 2

Thus, if one samples f “more often,” the series in the reconstruction formula converges faster since Kλ (y) = O(1/|y|2 ) as |y| → ∞. Note that Kλ (y) → K(y) as λ → 1.

Z

∞

|f (x)|2 dx =

(c) Prove that −∞

∞ X

|f (n)|2 .

n=−∞

[Hint: For P part (a) show that if χ is the characteristic function of I, then ∞ fˆ(ξ) = χ(ξ) n=−∞ f (n)e−2πinξ . For (b) use the function in Figure 2 instead of χ(ξ).] 21. Suppose that f is continuous on R. Show that f and fˆ cannot both be compactly supported unless f = 0. This can be viewed in the same spirit as the uncertainty principle.

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1

− λ2

1 2

− 12

λ 2

Figure 2. The function in Exercise 20

[Hint: Assume f is supported in [0, 1/2]. Expand f in a Fourier series in the interval [0, 1], and note that as a result, f is a trigonometric polynomial.] 22. The heuristic assertion stated before Theorem 4.1 can be made precise as follows. If F is a function on R, then we say that the preponderance of its mass is contained in an interval I (centered at the origin) if

Z x2 |F (x)|2 dx ≥

(16) I

1 2

Z x2 |F (x)|2 dx. R

Now suppose f ∈ S, and (16) holds with F = f and I = I1 ; also with F = fˆ and I = I2 . Then if Lj denotes the length of Ij , we have L1 L2 ≥

1 . 2π

A similar conclusion holds if the intervals are not necessarily centered at the origin. 23. The Heisenberg uncertainty principle can be formulated in terms of the d2 2 operator L = − dx 2 + x , which acts on Schwartz functions by the formula L(f ) = −

d2 f + x2 f. dx2

This operator, sometimes called the Hermite operator, is the quantum analogue of the harmonic oscillator. Consider the usual inner product on S given by

Z

∞

(f, g) =

f (x)g(x) dx −∞

whenever f, g ∈ S.

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6. Problems

(a) Prove that the Heisenberg uncertainty principle implies (Lf, f ) ≥ (f, f )

for all f ∈ S.

This is usually denoted by L ≥ I. [Hint: Integrate by parts.] (b) Consider the operators A and A∗ defined on S by A(f ) =

df + xf dx

and

A∗ (f ) = −

df + xf. dx

The operators A and A∗ are sometimes called the annihilation and creation operators, respectively. Prove that for all f, g ∈ S we have (i) (Af, g) = (f, A∗ g), (ii) (Af, Af ) = (A∗ Af, f ) ≥ 0, (iii) A∗ A = L − I. In particular, this again shows that L ≥ I. (c) Now for t ∈ R, let At (f ) =

df + txf dx

and

A∗t (f ) = −

df + txf. dx

Use the fact that (A∗t At f, f ) ≥ 0 to give anotherR proof of the Heisenberg ∞ uncertainty principle which says that whenever −∞ |f (x)|2 dx = 1 then

µZ

¶ ÃZ

∞ 2

∞

2

x |f (x)| dx −∞

−∞

¯ ¯2 ! ¯ df ¯ ¯ ¯ dx ≥ 1/4. ¯ dx ¯

[Hint: Think of (A∗t At f, f ) as a quadratic polynomial in t.]

6 Problems 1. The equation (17)

x2

∂2u ∂u ∂u + ax = 2 ∂x ∂x ∂t

with u(x, 0) = f (x) for 0 < x < ∞ and t > 0 is a variant of the heat equation which occurs in a number of applications. To solve (17), make the change of variables x = e−y so that −∞ < y < ∞. Set U (y, t) = u(e−y , t) and −y F (y) = f (e ). Then the problem reduces to the equation ∂U ∂U ∂2U + (1 − a) = , 2 ∂y ∂y ∂t

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with U (y, 0) = F (y). This can be solved like the usual heat equation (the case a = 1) by taking the Fourier transform in the y variable. One must then compute R∞ 2 2 the integral −∞ e(−4π ξ +(1−a)2πiξ)t e2πiξv dξ. Show that the solution of the original problem is then given by u(x, t) =

1 (4πt)1/2

Z

∞

e−(log(v/x)+(1−a)t)

2

/(4t)

f (v)

0

dv . v

2. The Black-Scholes equation from finance theory is ∂V ∂V σ 2 s2 ∂ 2 V + rs + − rV = 0, ∂t ∂s 2 ∂s2

(18)

0 < t < T,

subject to the “final” boundary condition V (s, T ) = F (s). An appropriate change of variables reduces this to the equation in Problem 1. Alternatively, the substi2 tution V (s, t) = eax+bτ U (x, τ ) where x = log s, τ = σ2 (T − t), a = 12 − σr2 , and

¡

¢2

b = − 12 + σr2 reduces (18) to the one-dimensional heat equation with the initial condition U (x, 0) = e−ax F (ex ). Thus a solution to the Black-Scholes equation is V (s, t) = p

e−r(T −t) 2πσ 2 (T − t)

Z

∞

e

−

(log(s/s∗ )+(r−σ 2 /2)(T −t))2 2σ 2 (T −t)

F (s∗ ) ds∗ .

0

3. ∗ The Dirichlet problem in a strip. Consider the equation 4u = 0 in the horizontal strip {(x, y) : 0 < y < 1, −∞ < x < ∞} with boundary conditions u(x, 0) = f0 (x) and u(x, 1) = f1 (x), where f0 and f1 are both in the Schwartz space. (a) Show (formally) that if u is a solution to this problem, then u ˆ(ξ, y) = A(ξ)e2πξy + B(ξ)e−2πξy . Express A and B in terms of fb0 and fb1 , and show that u ˆ(ξ, y) =

sinh(2π(1 − y)ξ) b sinh(2πyξ) b f0 (ξ) + f0 (ξ). sinh(2πξ) sinh(2πξ)

(b) Prove as a result that

Z

∞

|u(x, y) − f0 (x)|2 dx → 0

as y → 0

|u(x, y) − f1 (x)|2 dx → 0

as y → 1.

−∞

and

Z

∞

−∞

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(c) If Φ(ξ) = (sinh 2πaξ)/(sinh 2πξ), with 0 ≤ a < 1, then Φ is the Fourier transform of ϕ where ϕ(x) =

sin πa 1 · . 2 cosh πx + cos πa

This can be shown, for instance, by using contour integration and the residue formula from complex analysis (see Book II, Chapter 3). (d) Use this result to express u in terms of Poisson-like integrals involving f0 and f1 as follows: u(x, y) =

sin πy 2

µZ

∞ −∞

f0 (x − t) dt + cosh πt − cos πy

Z

∞ −∞

¶

f1 (x − t) dt . cosh πt + cos πy

(e) Finally, one can check that the function u(x, y) defined by the above expression is harmonic in the strip, and converges uniformly to f0 (x) as y → 0, and to f1 (x) as y → 1. Moreover, one sees that u(x, y) vanishes at infinity, that is, lim|x|→∞ u(x, y) = 0, uniformly in y.

In Exercise 12, we gave an example of a function that satisfies the heat equation in the upper half-plane, with boundary value 0, but which was not identically 0. We observed in this case that u was in fact not continuous up to the boundary. In Problem 4 we exhibit examples illustrating non-uniqueness, but this time with continuity up to the boundary t = 0. These examples satisfy a growth condition 2+² at infinity, namely |u(x, t)| ≤ Cecx , for any ² > 0. Problems 5 and 6 show 2 that under the more restrictive growth condition |u(x, t)| ≤ Cecx , uniqueness does hold.

4.∗ If g is a smooth function on R, define the formal power series (19)

u(x, t) =

∞ X

g (n) (t)

n=0

x2n . (2n!)

(a) Check formally that u solves the heat equation. (b) For a > 0, consider the function defined by

½ g(t) =

e−t 0

−a

if t > 0 if t ≤ 0.

One can show that there exists 0 < θ < 1 depending on a so that |g (k) (t)| ≤

k! − 1 t−a e 2 (θt)k

for t > 0.

(c) As a result, for each x and t the series (19) converges; u solves the heat equation; u vanishes for t = 0; and u satisfies the estimate |u(x, t)| ≤ 2a/(a−1) Cec|x| for some constants C, c > 0.

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(d) Conclude that for every ² > 0 there exists a non-zero solution to the heat equation which is continuous for x ∈ R and t ≥ 0, which satisfies u(x, 0) = 2+² 0 and |u(x, t)| ≤ Cec|x| . 5.∗ The following “maximum principle” for solutions of the heat equation will be used in the next problem. Theorem. Suppose that u(x, t) is a real-valued solution of the heat equation in the upper half-plane, which is continuous on its closure. Let R denote the rectangle R = {(x, y) ∈ R2 : a ≤ x ≤ b, 0 ≤ t ≤ c} and ∂ 0 R be the part of the boundary of R which consists of the two vertical sides and its base on the line t = 0 (see Figure 3). Then min

(x,t)∈∂ 0 R

u(x, t) = min u(x, t) (x,t)∈R

and

max u(x, t) = max u(x, t).

(x,t)∈∂ 0 R

(x,t)∈R

t

c

∂0R R

a

b

x

Figure 3. The rectangle R and part of its boundary ∂ 0 R

The steps leading to a proof of this result are outlined below. (a) Show that it suffices to prove that if u ≥ 0 on ∂ 0 R, then u ≥ 0 in R. (b) For ² > 0, let v(x, t) = u(x, t) + ²t. Then, v has a minimum on R, say at (x1 , t1 ). Show that x1 = a or b, or else t1 = 0. To do so, suppose on the contrary that a < x1 < b and 0 < t1 ≤ c, and prove that vxx (x1 , t1 ) − vt (x1 , t1 ) ≤ −². However, show also that the left-hand side must be non-negative. (c) Deduce from (b) that u(x, t) ≥ ²(t1 − t) for any (x, t) ∈ R and let ² → 0. 6.∗ The examples in Problem 4 are optimal in the sense of the following uniqueness theorem due to Tychonoff.

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6. Problems

Theorem. Suppose u(x, t) satisfies the following conditions: (i) u(x, t) solves the heat equation for all x ∈ R and and all t > 0. (ii) u(x, t) is continuous for all x ∈ R and 0 ≤ t ≤ c. (iii) u(x, 0) = 0. 2

(iv) |u(x, t)| ≤ M eax for some M , a, and all x ∈ R, 0 ≤ t < c. Then u is identically equal to 0. 7.∗ The Hermite functions hk (x) are defined by the generating identity ∞ X

hk (x)

k=0

2 2 tk = e−(x /2−2tx+t ) . k!

(a) Show that an alternate definition of the Hermite functions is given by the formula µ ¶k 2 d k x2 /2 hk (x) = (−1) e e−x . dx 2

2

2

2

[Hint: Write e−(x /2−2tx+t ) = ex /2 e−(x−t) and use Taylor’s formula.] Conclude from the above expression that each hk (x) is of the form 2 Pk (x)e−x /2 , where Pk is a polynomial of degree k. In particular, the Her2 mite functions belong to the Schwartz space and h0 (x) = e−x /2 , −x2 /2 h1 (x) = 2xe . (b) Prove that the family {hk }∞ k=0 is complete in the sense that if f is a Schwartz function, and

Z

∞

f (x)hk (x) dx = 0

(f, hk ) =

for all k ≥ 0,

−∞

then f = 0. [Hint: Use Exercise 8.] (c) Define h∗k (x) = hk ((2π)1/2 x). Then

c∗ (ξ) = (−i)k h∗k (ξ). h k Therefore, each h∗k is an eigenfunction for the Fourier transform. (d) Show that hk is an eigenfunction for the operator defined in Exercise 23, and in fact, prove that Lhk = (2k + 1)hk . In particular, we conclude that the functions hk are mutually orthogonal for the L2 inner product on the Schwartz space. (e) Finally, show that ing relation.]

R∞

−∞

[hk (x)]2 dx = π 1/2 2k k!. [Hint: Square the generat-

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8.∗ To refine the results in Chapter 4, and to prove that fα (x) =

∞ X

2−nα e2πi2

n

x

n=0

is nowhere differentiable even in the case α = 1, we need to consider a variant of the delayed means 4N , which in turn will be analyzed by the Poisson summation formula. (a) Fix an indefinitely differentiable function Φ satisfying

½ Φ(ξ) =

1 0

when |ξ| ≤ 1, when |ξ| ≥ 2.

By the Fourier inversion formula, there exists ϕ ∈ S so that ϕ(ξ) ˆ = Φ(ξ). Let ϕN (x) = N ϕ(N x) so that ϕc N (ξ) = Φ(ξ/N ). Finally, set ˜ N (x) = 4

∞ X

ϕN (x + n).

n=−∞

˜ N (x) = by the Poisson summation formula that 4 2πinx ˜ Φ(n/N )e , thus 4N is a trigonometric polynomial of degree ≤ 2N , with terms whose coefficients are 1 when |n| ≤ N . Let

Observe P

∞ n=−∞

˜ N (f ) = f ∗ 4 ˜ N. 4 Note that

˜ N 0 (fα ) SN (fα ) = 4

where N 0 is the largest integer of the form 2k with N 0 ≤ N . ˜ N (x) = ϕN (x) + EN (x) where (b) If we set 4 EN (x) =

X

ϕN (x + n),

|n|≥1

then one sees that: 0 (i) sup|x|≤1/2 |EN (x)| → 0 as N → ∞.

˜ 0 (x)| ≤ cN 2 . (ii) |4 N ˜ 0 (x)| ≤ c/(N |x|3 ), for |x| ≤ 1/2. (iii) |4 N Moreover, ∞.

R

|x|≤1/2

˜ 0 (x) dx = 0, and − 4 N

R |x|≤1/2

˜ 0 (x) dx → 1 as N → x4 N

(c) The above estimates imply that if f 0 (x0 ) exists, then ˜ 0 )(x0 + hN ) → f 0 (x0 ) (f ∗ 4 N

as N → ∞,

whenever |hN | ≤ C/N . Then, conclude that both the real and imaginary parts of f1 are nowhere differentiable, as in the proof given in Section 3, Chapter 4.

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6 The Fourier Transform on R

d

It occurred to me that in order to improve treatment planning one had to know the distribution of the attenuation coefficient of tissues in the body. This information would be useful for diagnostic purposes and would constitute a tomogram or series of tomograms. It was immediately evident that the problem was a mathematical one. If a fine beam of gamma rays of intensity I0 is incident on the body and the emerging density isR I, then the measurable quantity g equals log(I0 /I) = L f ds, where f is the variable absorption coefficient along the line L. Hence if f is a function of two dimensions, and g is known for all lines intersecting the body, the question is, can f be determined if g is known? Fourteen years would elapse before I learned that Radon had solved this problem in 1917. A. M. Cormack, 1979

The previous chapter introduced the theory of the Fourier transform on R and illustrated some of its applications to partial differential equations. Here, our aim is to present an analogous theory for functions of several variables. After a brief review of some relevant notions in Rd , we begin with some general facts about the Fourier transform on the Schwartz space S(Rd ). Fortunately, the main ideas and techniques have already been considered in the one-dimensional case. In fact, with the appropriate notation, the statements (and proofs) of the key theorems, such as the Fourier inversion and Plancherel formulas, remain unchanged. Next, we highlight the connection to some higher dimensional problems in mathematical physics, and in particular we investigate the wave equation in d dimensions, with a detailed analysis in the cases d = 3 and d = 2. At this stage, we discover a rich interplay between the Fourier transform and rotational symmetry, that arises only in Rd when d ≥ 2. Finally, the chapter ends with a discussion of the Radon transform. This topic is of substantial interest in its own right, but in addition has significant relevance in its application to the use of X-ray scans as well

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as to other parts of mathematics.

1 Preliminaries The setting in this chapter will be Rd , the vector space1 of all d-tuples of real numbers (x1 , . . . , xd ) with xi ∈ R. Addition of vectors is componentwise, and so is multiplication by real scalars. Given x = (x1 , . . . , xd ) ∈ Rd we define |x| = (x21 + · · · + x2d )1/2 , so that |x| is simply the length of the vector x in the usual Euclidean norm. In fact, we equip Rd with the standard inner product defined by x · y = x 1 y1 + · · · + x d yd , so that |x|2 = x · x. We use the notation x · y in place of (x, y) of Chapter 3. Given a d-tuple α = (α1 , . . . , αd ) of non-negative integers (sometimes called a multi-index), the monomial xα is defined by αd α2 1 xα = xα 1 x2 · · · xd .

Similarly, we define the differential operator (∂/∂x)α by µ

∂ ∂x

¶α

µ

=

∂ ∂x1

¶α1 µ

∂ ∂x2

¶α2

µ

∂ ··· ∂xd

¶αd

=

1 ∂xα 1

∂ |α| , d · · · ∂xα d

where |α| = α1 + · · · + αd is the order of the multi-index α. 1.1 Symmetries Analysis in Rd , and in particular the theory of the Fourier transform, is shaped by three important groups of symmetries of the underlying space: (i) Translations (ii) Dilations (iii) Rotations 1 See Chapter 3 for a brief review of vector spaces and inner products. Here we find it convenient to use lower case letters such as x (as opposed to X) to designate points in Rd . Also, we use | · | instead of k · k to denote the Euclidean norm.

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177

We have seen that translations x 7→ x + h, with h ∈ Rd fixed, and dilations x 7→ δx, with δ > 0, play an important role in the one-dimensional theory. In R, the only two rotations are the identity and multiplication by −1. However, in Rd with d ≥ 2 there are more rotations, and the understanding of the interaction between the Fourier transform and rotations leads to fruitful insights regarding spherical symmetries. A rotation in Rd is a linear transformation R : Rd → Rd which preserves the inner product. In other words, • R(ax + by) = aR(x) + bR(y) for all x, y ∈ Rd and a, b ∈ R. • R(x) · R(y) = x · y for all x, y ∈ Rd . Equivalently, this last condition can be replaced by |R(x)| = |x| for all x ∈ Rd , or Rt = R−1 where Rt and R−1 denote the transpose and inverse of R, respectively.2 In particular, we have det(R) = ±1, where det(R) is the determinant of R. If det(R) = 1 we say that R is a proper rotation; otherwise, we say that R is an improper rotation. Example 1. On the real line R, there are two rotations: the identity which is proper, and multiplication by −1 which is improper. Example 2. The rotations in the plane R2 can be described in terms of complex numbers. We identify R2 with C by assigning the point (x, y) to the complex number z = x + iy. Under this identification, all proper rotations are of the form z 7→ zeiϕ for some ϕ ∈ R, and all improper rotations are of the form z 7→ zeiϕ for some ϕ ∈ R (here, z = x − iy denotes the complex conjugate of z). See Exercise 1 for the argument leading to this result. Example 3. Euler gave the following very simple geometric description of rotations in R3 . Given a proper rotation R, there exists a unit vector γ so that: (i) R fixes γ, that is, R(γ) = γ. (ii) If P denotes the plane passing through the origin and perpendicular to γ, then R : P → P, and the restriction of R to P is a rotation in R2 . 2 Recall that the transpose of a linear operator A : Rd → Rd is the linear operator B : Rd → Rd which satisfies A(x) · y = x · B(y) for all x, y ∈ Rd . We write B = At . The inverse of A (when it exists) is the linear operator C : Rd → Rd with A ◦ C = C ◦ A = I (where I is the identity), and we write C = A−1 .

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Geometrically, the vector γ gives the direction of the axis of rotation. A proof of this fact is given in Exercise 2. Finally, if R is improper, then −R is proper (since in R3 det(−R) = − det(R)), so R is the composition of a proper rotation and a symmetry with respect to the origin. Example 4. Given two orthonormal bases {e1 , . . . , ed } and {e01 , . . . , e0d } in Rd , we can define a rotation R by letting R(ei ) = e0i for i = 1, . . . , d. Conversely, if R is a rotation and {e1 , . . . , ed } is an orthonormal basis, then {e01 , . . . , e0d }, where e0j = R(ej ), is another orthonormal basis. 1.2 Integration on Rd Since we shall be dealing with functions on Rd , we will need to discuss some aspects of integration of such functions. A more detailed review of integration on Rd is given in the appendix. A continuous complex-valued function f on Rd is said to be rapidly decreasing if for every multi-index α the function |xα f (x)| is bounded. Equivalently, a continuous function is of rapid decrease if sup |x|k |f (x)| < ∞

for every k = 0, 1, 2, . . ..

x∈Rd

Given a function of rapid decrease, we define Z Z f (x) dx = lim f (x) dx, Rd

N →∞ QN

where QN denotes the closed cube centered at the origin, with sides of length N parallel to the coordinate axis, that is, QN = {x ∈ Rd : |xi | ≤ N/2

for i = 1, . . . , d}.

The integral over QN is a multiple integral in the usual sense of Riemann integration. That the limit exists follows from the fact that the integrals R IN = QN f (x) dx form a Cauchy sequence as N tends to infinity. Two observations are in order. First, we may replace the square QN by the ball BN = {x ∈ Rd : |x| ≤ N } without changing the definition. Second, we do not need the full force of rapid decrease to show that the limit exists. In fact it suffices to assume that f is continuous and sup |x|d+² |f (x)| < ∞ x∈Rd

for some ² > 0.

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For example, functions of moderate decrease on R correspond to ² = 1. In keeping with this we define functions of moderate decrease on Rd as those that are continuous and satisfy the above inequality with ² = 1. The interaction of integration with the three important groups of symmetries is as follows: if f is of moderate decrease, then Z Z (i) f (x + h) dx = f (x) dx for all h ∈ Rd , Rd

Z

Z

(ii) δ d Z

(iii)

Rd

f (δx) dx = Rd

f (x) dx for all δ > 0, Z

Rd

f (R(x)) dx = Rd

f (x) dx for every rotation R. Rd

Polar coordinates It will be convenient to introduce polar coordinates in Rd and find the corresponding integration formula. We begin with two examples which correspond to the case d = 2 and d = 3. (A more elaborate discussion applying to all d is contained in the appendix.) Example 1. In R2 , polar coordinates are given by (r, θ) with r ≥ 0 and 0 ≤ θ < 2π. The Jacobian of the change of variables is equal to r, so that Z

Z

2πZ ∞

f (r cos θ, r sin θ) r dr dθ.

f (x) dx = R2

0

0

Now we may write a point on the unit circle S 1 as γ = (cos θ, sin θ), and given a function g on the circle, we define its integral over S 1 by Z

Z

2π

g(cos θ, sin θ) dθ.

g(γ) dσ(γ) = S1

0

With this notation we then have Z Z Z f (x) dx = R2

∞

f (rγ) r dr dσ(γ).

S1 0

Example 2. In R3 one uses spherical coordinates given by   x1

= r sin θ cos ϕ, x2 = r sin θ sin ϕ,  x3 = r cos θ,

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where 0 < r, 0 ≤ θ ≤ π and 0 ≤ ϕ ≤ 2π. The Jacobian of the change of variables is r2 sin θ so that Z f (x) dx = R3 Z 2π Z π Z ∞ f (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ)r2 dr sin θ dθ dϕ. 0

0

0

If g is a function on the unit sphere S 2 = {x ∈ R3 : |x| = 1}, and γ = (sin θ cos ϕ, sin θ sin ϕ, cos θ), we define the surface element dσ(γ) by Z

Z

2πZ π

g(γ) dσ(γ) =

g(γ) sin θ dθ dϕ.

S2

0

0

As a result, Z

Z

Z

∞

f (x) dx = R3

f (rγ) r2 dr dσ(γ).

S2 0

In general, it is possible to write any point in Rd − {0} uniquely as x = rγ where γ lies on the unit sphere S d−1 ⊂ Rd and r > 0. Indeed, take r = |x| and γ = x/|x|. Thus one may proceed as in the cases d = 2 or d = 3 to define spherical coordinates. The formula we shall use is Z Z Z ∞ f (x) dx = f (rγ) rd−1 dr dσ(γ), Rd

S d−1

0

whenever f is of moderate decrease. Here dσ(γ) denotes the surface element on the sphere S d−1 obtained from the spherical coordinates.

2 Elementary theory of the Fourier transform The Schwartz space S(Rd ) (sometimes abbreviated as S) consists of all indefinitely differentiable functions f on Rd such that ¯ µ ¶ ¯ β ¯ ¯ ¯ α ∂ ¯ f (x)¯ < ∞, sup ¯x ¯ ¯ ∂x d x∈R for every multi-index α and β. In other words, f and all its derivatives are required to be rapidly decreasing.

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181

Example 1. An example of a function in S(Rd ) is the d-dimensional 2 Gaussian given by e−π|x| . The theory in Chapter 5 already made clear the central role played by this function in the case d = 1. The Fourier transform of a Schwartz function f is defined by Z fˆ(ξ) = f (x)e−2πix·ξ dx, for ξ ∈ Rd . Rd

Note the resemblance with the formula in one-dimension, except that we are now integrating on Rd , and the product of x and ξ is replaced by the inner product of the two vectors. We now list some simple properties of the Fourier transform. In the next proposition the arrow indicates that we have taken the Fourier transform, so F (x) −→ G(ξ) means that G(ξ) = Fˆ (ξ). Proposition 2.1 Let f ∈ S(Rd ). (i) f (x + h) −→ fˆ(ξ)e2πiξ·h whenever h ∈ Rd . (ii) f (x)e−2πixh −→ fˆ(ξ + h) whenever h ∈ Rd . (iii) f (δx) −→ δ −d fˆ(δ −1 ξ) whenever δ > 0. µ ¶α ∂ (iv) f (x) −→ (2πiξ)α fˆ(ξ). ∂x µ ¶α ∂ (v) (−2πix)α f (x) −→ fˆ(ξ). ∂ξ (vi) f (Rx) −→ fˆ(Rξ) whenever R is a rotation. The first five properties are proved in the same way as in the onedimensional case. To verify the last property, simply change variables y = Rx in the integral. Then, recall that | det(R)| = 1, and R−1 y · ξ = y · Rξ, because R is a rotation. Properties (iv) and (v) in the proposition show that, up to factors of 2πi, the Fourier transform interchanges differentiation and multiplication by monomials. This motivates the definition of the Schwartz space and leads to the next corollary. Corollary 2.2 The Fourier transform maps S(Rd ) to itself. At this point we disgress to observe a simple fact concerning the interplay between the Fourier transform and rotations. We say that a

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function f is radial if it depends only on |x|; in other words, f is radial if there is a function f0 (u), defined for u ≥ 0, such that f (x) = f0 (|x|). We note that f is radial if and only if f (Rx) = f (x) for every rotation R. In one direction, this is obvious since |Rx| = |x|. Conversely, suppose that f (Rx) = f (x), for all rotations R. Now define f0 by ½

f (0) if u = 0, f (x) if |x| = u.

f0 (u) =

Note that f0 is well defined, since if x and x0 are points with |x| = |x0 | there is always a rotation R so that x0 = Rx. Corollary 2.3 The Fourier transform of a radial function is radial. This follows at once from property (vi) in the last proposition. Indeed, the condition f (Rx) = f (x) for all R implies that fˆ(Rξ) = fˆ(ξ) for all R, thus fˆ is radial whenever f is. 2

An example of a radial function in Rd is the Gaussian e−π|x| . Also, we observe that when d = 1, the radial functions are precisely the even functions, that is, those for which f (x) = f (−x). After these preliminaries, we retrace the steps taken in the previous chapter to obtain the Fourier inversion formula and Plancherel theorem for Rd . Theorem 2.4 Suppose f ∈ S(Rd ). Then Z

fˆ(ξ)e2πix·ξ dξ.

f (x) = Rd

Moreover Z

Z

|fˆ(ξ)|2 dξ = Rd

|f (x)|2 dx. Rd

The proof proceeds in the following stages. 2

2

Step 1. The Fourier transform of e−π|x| is e−π|ξ| . To prove this, notice that the properties of the exponential functions imply that 2

2

2

e−π|x| = e−πx1 · · · e−πxd

and

e−2πix·ξ = e−2πix1 ·ξ1 · · · e−2πixd ·ξd ,

so that the integrand in the Fourier transform is a product of d functions, each depending on the variable xj (1 ≤ j ≤ d) only. Thus the assertion

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follows by writing the integral over Rd as a series of repeated integrals, each taken over R. For example, when d = 2, Z

−π|x|2 −2πix·ξ

e

e

Z

dx =

µZ

e

R2

¶

−πx22 −2πix2 ·ξ2

−πx21 −2πix1 ·ξ1

e

ZR

e

e

R 2

dx1

dx2

2

e−πx2 e−2πix2 ·ξ2 e−πξ1 dx2

=

R −πξ12 −πξ22

=e

e

−π|ξ|2

=e

.

As a consequence of Proposition 2.1, applied with δ 1/2 instead of δ, we \ 2 ) = δ −d/2 e−π|ξ|2 /δ . find that (e−πδ|x| 2 Step 2. The family Kδ (x) = δ −d/2 e−π|x| /δ is a family of good kernels in Rd . By this we mean that Z (i) Kδ (x) dx = 1, Rd

Z

(ii) Rd

|Kδ (x)| dx ≤ M (in fact Kδ (x) ≥ 0), Z

(iii) For every η > 0, |x|≥η

|Kδ (x)| dx → 0 as δ → 0.

The proofs of these assertions are almost identical to the case d = 1. As a result Z Kδ (x)F (x) dx → F (0) as δ → 0 Rd

when F is a Schwartz function, or more generally when F is bounded and continuous at the origin. Step 3. The multiplication formula Z

Z

fˆ(y)g(y) dy

f (x)ˆ g (x) dx = Rd

Rd

holds whenever f and g are in S. The proof requires the evaluation of the integral of f (x)g(y)e−2πix·y over (x, y) ∈ R2d = Rd × Rd as a repeated integral, with each separate integration taken over Rd . The justification is similar to that in the proof of Proposition 1.8 in the previous chapter. (See the appendix.) The Fourier inversion is then a simple consequence of the multiplication formula and the family of good kernels Kδ , as in Chapter 5. It also

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follows that the Fourier transform F is a bijective map of S(Rd ) to itself, whose inverse is Z ∗ g(ξ)e2πix·ξ dξ. F (g)(x) = Rd

Step 4. Next we turn to the convolution, defined by Z (f ∗ g)(x) = f (y)g(x − y) dy, f, g ∈ S. Rd

\ We have that f ∗ g ∈ S(Rd ), f ∗ g = g ∗ f , and (f ∗ g)(ξ) = fˆ(ξ)ˆ g (ξ). The argument is similar to that in one-dimension. The calculation of the Fourier transform of f ∗ g involves an integration of f (y)g(x − y)e−2πix·ξ (over R2d = Rd × Rd ) expressed as a repeated integral. Then, following the same argument in the previous chapter, we obtain the d-dimensional Plancherel formula, thereby concluding the proof of Theorem 2.4.

3 The wave equation in Rd × R Our next goal is to apply what we have learned about the Fourier transform to the study of the wave equation. Here, we once again simplify matters by restricting ourselves to functions in the Schwartz class S. We note that in any further analysis of the wave equation it is important to allow functions that have much more general behavior, and in particular that may be discontinuous. However, what we lose in generality by only considering Schwartz functions, we gain in transparency. Our study in this restricted context will allow us to explain certain basic ideas in their simplest form. 3.1 Solution in terms of Fourier transforms The motion of a vibrating string satisfies the equation ∂2u 1 ∂2u = , ∂x2 c2 ∂t2 which we referred to as the one-dimensional wave equation. A natural generalization of this equation to d space variables is (1)

∂2u 1 ∂2u ∂2u + ··· + 2 = 2 2 . 2 ∂x1 ∂xd c ∂t

In fact, it is known that in the case d = 3, this equation determines the behavior of electromagnetic waves in vacuum (with c = speed of light).

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3. The wave equation in Rd × R

Also, this equation describes the propagation of sound waves. Thus (1) is called the d-dimensional wave equation. Our first observation is that we may assume c = 1, since we can rescale the variable t if necessary. Also, if we define the Laplacian in d dimensions by 4=

∂2 ∂2 + · · · + , ∂x21 ∂x2d

then the wave equation can be rewritten as (2)

4u =

∂2u . ∂t2

The goal of this section is to find a solution to this equation, subject to the initial conditions u(x, 0) = f (x)

and

∂u (x, 0) = g(x), ∂t

where f, g ∈ S(Rd ). This is called the Cauchy problem for the wave equation. Before solving this problem, we note that while we think of the variable t as time, we do not restrict ourselves to t > 0. As we will see, the solution we obtain makes sense for all t ∈ R. This is a manifestation of the fact that the wave equation can be reversed in time (unlike the heat equation). A formula for the solution of our problem is given in the next theorem. The heuristic argument which leads to this formula is important since, as we have already seen, it applies to some other boundary value problems as well. Suppose u solves the Cauchy problem for the wave equation. The technique employed consists of taking the Fourier transform of the equation and of the initial conditions, with respect to the space variables x1 , . . . , xd . This reduces the problem to an ordinary differential equation in the time variable. Indeed, recalling that differentiation with respect to xj becomes multiplication by 2πiξj , and the differentiation with respect to t commutes with the Fourier transform in the space variables, we find that (2) becomes −4π 2 |ξ|2 u ˆ(ξ, t) =

∂2u ˆ (ξ, t). 2 ∂t

For each fixed ξ ∈ Rd , this is an ordinary differential equation in t whose solution is given by u ˆ(ξ, t) = A(ξ) cos(2π|ξ|t) + B(ξ) sin(2π|ξ|t),

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where for each ξ, A(ξ) and B(ξ) are unknown constants to be determined by the initial conditions. In fact, taking the Fourier transform (in x) of the initial conditions yields u ˆ(ξ, 0) = fˆ(ξ)

and

∂u ˆ (ξ, 0) = gˆ(ξ). ∂t

We may now solve for A(ξ) and B(ξ) to obtain A(ξ) = fˆ(ξ)

and

2π|ξ|B(ξ) = gˆ(ξ).

Therefore, we find that sin(2π|ξ|t) u ˆ(ξ, t) = fˆ(ξ) cos(2π|ξ|t) + gˆ(ξ) , 2π|ξ| and the solution u is given by taking the inverse Fourier transform in the ξ variables. This formal derivation then leads to a precise existence theorem for our problem. Theorem 3.1 A solution of the Cauchy problem for the wave equation is ¸ Z · sin(2π|ξ|t) 2πix·ξ ˆ e dξ. (3) u(x, t) = f (ξ) cos(2π|ξ|t) + gˆ(ξ) 2π|ξ| Rd Proof. We first verify that u solves the wave equation. This is straightforward once we note that we can differentiate in x and t under the integral sign (because f and g are both Schwartz functions) and therefore u is at least C 2 . On the one hand we differentiate the exponential with respect to the x variables to get ¸ Z · sin(2π|ξ|t) 4u(x, t) = fˆ(ξ) cos(2π|ξ|t) + gˆ(ξ) (−4π 2 |ξ|2 )e2πix·ξ dξ, d 2π|ξ| R while on the other hand we differentiate the terms in brackets with respect to t twice to get ∂2u (x, t) = ∂t2 ¸ Z · sin(2π|ξ|t) 2πix·ξ 2 2 2 2ˆ −4π |ξ| f (ξ) cos(2π|ξ|t) − 4π |ξ| gˆ(ξ) e dξ. 2π|ξ| Rd This shows that u solves equation (2). Setting t = 0 we get Z u(x, 0) = fˆ(ξ)e2πix·ξ dξ = f (x) Rd

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by the Fourier inversion theorem. Finally, differentiating once with respect to t, setting t = 0, and using the Fourier inversion shows that ∂u (x, 0) = g(x). ∂t Thus u also verifies the initial conditions, and the proof of the theorem is complete. As the reader will note, both fˆ(ξ) cos(2π|ξ|t) and gˆ(ξ) sin(2π|ξ|t) are 2π|ξ| functions in S, assuming as we do that f and g are in S. This is because both cos u and (sin u)/u are even functions that are indefinitely differentiable. Having proved the existence of a solution to the Cauchy problem for the wave equation, we raise the question of uniqueness. Are there solutions to the problem 4u =

∂2u ∂t2

subject to

u(x, 0) = f (x)

∂u (x, 0) = g(x), ∂t

and

other than the one given by the formula in the theorem? In fact the answer is, as expected, no. The proof of this fact, which will not be given here (but see Problem 3), can be based on a conservation of energy argument. This is a local counterpart of a global conservation of energy statement which we will now present. We observed in Exercise 10, Chapter 3, that in the one-dimensional case, the total energy of the vibrating string is conserved in time. The analogue of this fact holds in higher dimensions as well. Define the energy of a solution by Z

E(t) = R

¯ ¯2 ¯ ¯ ∂u ¯ ¯ ¯ ¯ + ¯ ∂u ¯ ¯ ¯ ∂x d ∂t

1

¯2 ¯ ¯ ¯ ¯ + · · · + ¯ ∂u ¯ ¯ ∂x

d

¯2 ¯ ¯ dx. ¯

Theorem 3.2 If u is the solution of the wave equation given by formula (3), then E(t) is conserved, that is, E(t) = E(0),

for all t ∈ R.

The proof requires the following lemma. Lemma 3.3 Suppose a and b are complex numbers and α is real. Then |a cos α + b sin α|2 + | − a sin α + b cos α|2 = |a|2 + |b|2 .

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This follows directly because e1 = (cos α, sin α) and e2 = (− sin α, cos α) are a pair of orthonormal vectors, hence with Z = (a, b) ∈ C2 , we have |Z|2 = |Z · e1 |2 + |Z · e2 |2 , where · represents the inner product in C2 . Now by Plancherel’s theorem, Z

¯ ¯2 Z ¯ ∂u ¯ ¯ ¯ dx = ¯ ∂t ¯ d

¯2 ¯ ¯ ¯ ¯−2π|ξ|fˆ(ξ) sin(2π|ξ|t) + gˆ(ξ) cos(2π|ξ|t)¯ dξ.

Rd

R

Similarly, Z

d ¯ X ¯ ∂u ¯ ¯ ∂x d

R j=1

j

¯2 Z ¯ ¯ dx = ¯

¯2 ¯ ¯ ¯ ¯2π|ξ|fˆ(ξ) cos(2π|ξ|t) + gˆ(ξ) sin(2π|ξ|t)¯ dξ.

Rd

We now apply the lemma with a = 2π|ξ|fˆ(ξ),

b = gˆ(ξ)

and

α = 2π|ξ|t.

The result is that Z

E(t) = ZR

¯ ¯2 ¯ ¯ ∂u ¯ ¯ ¯ ¯ + ¯ ∂u ¯ ¯ ¯ ∂x d ∂t

1

¯2 ¯ ¯ ¯ ¯ + · · · + ¯ ∂u ¯ ¯ ∂x

d

¯2 ¯ ¯ dx ¯

(4π 2 |ξ|2 |fˆ(ξ)|2 + |ˆ g (ξ)|2 ) dξ,

= Rd

which is clearly independent of t. Thus Theorem 3.2 is proved. The drawback with formula (3), which does give the solution of the wave equation, is that it is quite indirect, involving the calculation of the Fourier transforms of f and g, and then a further inverse Fourier transform. However, for every dimension d there is a more explicit formula. This formula is very simple when d = 1 and a little less so when d = 3. More generally, the formula is “elementary” whenever d is odd, and more complicated when d is even (see Problems 4 and 5). In what follows we consider the cases d = 1, d = 3, and d = 2, which together give a picture of the general situation. Recall that in Chapter 1, when discussing the wave equation over the interval [0, L], we found that the solution is given by d’Alembert’s formula (4)

f (x + t) + f (x − t) 1 u(x, t) = + 2 2

Z

x+t

g(y) dy. x−t

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with the interpretation that both f and g are extended outside [0, L] by making them odd in [−L, L], and periodic on the real line, with period 2L. The same formula (4) holds for the solution of the wave equation when d = 1 and when the initial data are functions in S(R). In fact, this follows directly from (3) if we note that cos(2π|ξ|t) =

1 2πi|ξ|t (e + e−2πi|ξ|t ) 2

and sin(2π|ξ|t) 1 = (e2πi|ξ|t − e−2πi|ξ|t ). 2π|ξ| 4πi|ξ| Finally, we note that the two terms that appear in d’Alembert’s formula (4) consist of appropriate averages. Indeed, the first term is precisely the average of f over the two points that are the boundary of the interval [x − t, x + t]; the second term is, up to a factor of t, the mean R x+t value of g over this interval, that is, (1/2t) x−t g(y) dy. This suggests a generalization to higher dimensions, where we might expect to write the solution of our problem as averages of the initial data. This is in fact the case, and we now treat in detail the particular situation d = 3. 3.2 The wave equation in R3 × R If S 2 denotes the unit sphere in R3 , we define the spherical mean of the function f over the sphere of radius t centered at x by (5)

Mt (f )(x) =

1 4π

Z

f (x − tγ) dσ(γ), S2

where dσ(γ) is the element of surface area for S 2 . Since 4π is the area of the unit sphere, we can interpret Mt (f ) as the average value of f over the sphere centered at x of radius t. Lemma 3.4 If f ∈ S(R3 ) and t is fixed, then Mt (f ) ∈ S(R3 ). Moreover, Mt (f ) is indefinitely differentiable in t, and each t-derivative also belongs to S(R3 ). Proof. Let F (x) = Mt (f )(x). To show that F is rapidly decreasing, start with the inequality |f (x)| ≤ AN /(1 + |x|N ) which holds for every fixed N ≥ 0. As a simple consequence, whenever t is fixed, we have |f (x − γt)| ≤ A0N /(1 + |x|N )

for all γ ∈ S 2 .

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To see this consider separately the cases when |x| ≤ 2|t|, and |x| > 2|t|. Therefore, by integration |F (x)| ≤ A0N /(1 + |x|N ), and since this holds for every N , the function F is rapidly decreasing. One next observes that F is indefinitely differentiable, and µ

(6)

∂ ∂x

¶α

F (x) = Mt (f (α) )(x)

where f (α) (x) = (∂/∂x)α f . It suffices to prove this when (∂/∂x)α = ∂/∂xk , and then proceed by induction to get the general case. Furthermore, it is enough to take k = 1. Now F (x1 + h, x2 , x3 ) − F (x1 , x2 , x3 ) 1 = h 4π

Z S2

gh (γ) dσ(γ)

where gh (γ) =

f (x + e1 h − γt) − f (x − γt) , h

∂ and e1 = (1, 0, 0). Now, it suffices to observe that gh → ∂x f (x − γt) 1 as h → 0 uniformly in γ. As a result, we find that (6) holds, and by ¡ ∂ ¢α F (x) is also rapidly decreasing, the first argument, it follows that ∂x hence F ∈ S. The same argument applies to each t-derivative of Mt (f ).

The basic fact about integration on spheres that we shall need is the following Fourier transform formula. Z 1 sin(2π|ξ|) Lemma 3.5 e−2πiξ·γ dσ(γ) = . 4π S 2 2π|ξ| This formula, as we shall see in the following section, is connected to the fact that the Fourier transform of a radial function is radial. Proof. Note that the integral on the left is radial in ξ. Indeed, if R is a rotation then Z Z Z −2πiR(ξ)·γ −2πiξ·R−1 (γ) e dσ(γ) = e dσ(γ) = e−2πiξ·γ dσ(γ) S2

S2

S2

because we may change variables γ → R−1 (γ). (For this, see formula (4) in the appendix.) So if |ξ| = ρ, it suffices to prove the lemma with

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ξ = (0, 0, ρ). If ρ = 0, the lemma is obvious. If ρ > 0, we choose spherical coordinates to find that the left-hand side is equal to Z 2πZ π 1 e−2πiρ cos θ sin θ dθ dϕ. 4π 0 0 The change of variables u = − cos θ gives Z 2πZ π Z 1 1 π −2πiρ cos θ e−2πiρ cos θ sin θ dθ dϕ = e sin θ dθ 4π 0 0 2 0 Z 1 1 2πiρu = e du 2 −1 1 £ 2πiρu ¤1 e = −1 4πiρ sin(2πρ) = , 2πρ and the formula is proved. By the defining formula (5) we may interpret Mt (f ) as a convolution of the function f with the element dσ, and since the Fourier transform \ interchanges convolutions with products, we are led to believe that M t (f ) is the product of the corresponding Fourier transforms. Indeed, we have the identity \ ˆ sin(2π|ξ|t) . M t (f )(ξ) = f (ξ) 2π|ξ|t

(7) To see this, write \ M t (f )(ξ) =

µ

Z −2πix·ξ

e R3

1 4π

¶

Z

f (x − γt) dσ(γ)

dx,

S2

and note that we may interchange the order of integration and make a simple change of variables to achieve the desired identity. As a result, we find that the solution of our problem may be expressed by using the spherical means of the initial data. Theorem 3.6 The solution when d = 3 of the Cauchy problem for the wave equation 4u =

∂2u ∂t2

subject to

u(x, 0) = f (x)

and

∂u (x, 0) = g(x) ∂t

is given by u(x, t) =

∂ (tMt (f )(x)) + tMt (g)(x). ∂t

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Proof. Consider first the problem 4u =

∂2u ∂t2

subject to

u(x, 0) = 0

∂u (x, 0) = g(x). ∂t

and

Then by Theorem 3.1, we know that its solution u1 is given by ·

¸ sin(2π|ξ|t) 2πix·ξ gˆ(ξ) e dξ u1 (x, t) = 2π|ξ| R3 · ¸ Z sin(2π|ξ|t) 2πix·ξ gˆ(ξ) =t e dξ 2π|ξ|t R3 = tMt (g)(x), Z

where we have used (7) applied to g, and the Fourier inversion formula. According to Theorem 3.1 again, the solution to the problem 4u =

∂2u ∂t2

subject to

u(x, 0) = f (x)

∂u (x, 0) = 0 ∂t

and

is given by Z

h

i fˆ(ξ) cos(2π|ξ|t) e2πix·ξ dξ R3 µ Z · ¸ ¶ ∂ sin(2π|ξ|t) 2πix·ξ ˆ = t f (ξ) e dξ ∂t 2π|ξ|t R3 ∂ = (tMt (f )(x)). ∂t

u2 (x, t) =

We may now superpose these two solutions to obtain u = u1 + u2 as the solution of our original problem.

Huygens principle The solutions to the wave equation in one and three dimensions are given, respectively, by u(x, t) =

f (x + t) + f (x − t) 1 + 2 2

Z

x+t

g(y) dy x−t

and u(x, t) =

∂ (tMt (f )(x)) + tMt (g)(x). ∂t

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t (x, t)

0

x−t

x

x+t

x

Figure 1. Huygens principle, d = 1

We observe that in the one-dimensional problem, the value of the solution at (x, t) depends only on the values of f and g in the interval centered at x of length 2t, as shown in Figure 1. If in addition g = 0, then the solution depends only on the data at the two boundary points of this interval. In three dimensions, this boundary dependence always holds. More precisely, the solution u(x, t) depends only on the values of f and g in an immediate neighborhood of the sphere centered at x and of radius t. This situation is depicted in Figure 2, where we have drawn the cone originating at (x, t) and with its base the ball centered at x of radius t. This cone is called the backward light cone originating at (x, t). (x, t)

x-space

Figure 2. Backward light cone originating at (x, t)

Alternatively, the data at a point x0 in the plane t = 0 influences the solution only on the boundary of a cone originating at x0 , called the forward light cone and depicted in Figure 3. This phenomenon, known as the Huygens principle, is immediate from the formulas for u given above. Another important aspect of the wave equation connected with these

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x0

Figure 3. The forward light cone originating at x0

considerations is that of the finite speed of propagation. (In the case where c = 1, the speed is 1.) This means that if we have an initial disturbance localized at x = x0 , then after a finite time t, its effects will have propagated only inside the ball centered at x0 of radius |t|. To state this precisely, suppose the initial conditions f and g are supported in the ball of radius δ, centered at x0 (think of δ as small). Then u(x, t) is supported in the ball of radius |t| + δ centered at x0 . This assertion is clear from the above discussion. 3.3 The wave equation in R2 × R: descent It is a remarkable fact that the solution of the wave equation in three dimensions leads to a solution of the wave equation in two dimensions. Define the corresponding means by Z ft (F )(x) = 1 F (x − ty)(1 − |y|2 )−1/2 dy. M 2π |y|≤1 Theorem 3.7 A solution of the Cauchy problem for the wave equation in two dimensions with initial data f, g ∈ S(R2 ) is given by (8)

u(x, t) =

∂ f ft (g)(x). (tMt (f )(x)) + tM ∂t

Notice the difference between this case and the case d = 3. Here, u at (x, t) depends on f and g in the whole disc (of radius |t| centered at x), and not just on the values of the initial data near the boundary of that disc. Formally, the identity in the theorem arises as follows. If we start with an initial pair of functions f and g in S(R2 ), we may consider the corresponding functions f˜ and g˜ on R3 that are merely extensions of f and g that are constant in the x3 variable, that is, f˜(x1 , x2 , x3 ) = f (x1 , x2 )

and

g˜(x1 , x2 , x3 ) = g(x1 , x2 ).

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Now, if u ˜ is the solution (given in the previous section) of the 3-dimensional wave equation with initial data f˜ and g˜, then one can expect that u ˜ is also constant in x3 so that u ˜ satisfies the 2-dimensional wave equation. A difficulty with this argument is that f˜ and g˜ are not rapidly decreasing since they are constant in x3 , so that our previous methods do not apply. However, it is easy to modify the argument so as to obtain a proof of Theorem 3.7. We fix T > 0 and consider a function η(x3 ) that is in S(R), such that η(x3 ) = 1 if |x3 | ≤ 3T . The trick is to truncate f˜ and g˜ in the x3 -variable, and consider instead f˜[ (x1 , x2 , x3 ) = f (x1 , x2 )η(x3 )

and

g˜[ (x1 , x2 , x3 ) = g(x1 , x2 )η(x3 ).

Now both f˜[ and g˜[ are in S(R3 ), so Theorem 3.6 provides a solution u ˜[ of the wave equation with initial data f˜[ and g˜[ . It is easy to see from the formula that u ˜[ (x, t) is independent of x3 , whenever |x3 | ≤ T and |t| ≤ T . In particular, if we define u(x1 , x2 , t) = u ˜[ (x1 , x2 , 0, t), then u satisfies the 2-dimensional wave equation when |t| ≤ T . Since T is arbitrary, u is a solution to our problem, and it remains to see why u has the desired form. By definition of the spherical coordinates, we recall that the integral of a function H over the sphere S 2 is given by Z 1 H(γ) dσ(γ) = 4π S 2 Z 2πZ π 1 H(sin θ cos ϕ, sin θ sin ϕ, cos θ) sin θ dθ dϕ. 4π 0 0 If H does not depend on the last variable, that is, H(x1 , x2 , x3 ) = h(x1 , x2 ) for some function h of two variables, then Mt (H)(x1 , x2 , 0) = Z 2πZ π 1 h(x1 − t sin θ cos ϕ, x2 − t sin θ sin ϕ) sin θ dθ dϕ. 4π 0 0 To calculate this last integral, we split the θ-integral from 0 to π/2 and then π/2 to π. By making the change of variables r = sin θ, we find, after a final change to polar coordinates, that Z 1 h(x − ty)(1 − |y|2 )−1/2 dy Mt (H)(x1 , x2 , 0) = 2π |y|≤1 ft (h)(x1 , x2 ). =M

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Applying this to H = f˜[ , h = f , and H = g˜[ , h = g, we find that u is given by the formula (8), and the proof of Theorem 3.7 is complete. Remark. In the case of general d, the solution of the wave equation shares many of the properties we have discussed in the special cases d = 1, 2, and 3. • At a given time t, the initial data at a point x only affects the solution u in a specific region. When d > 1 is odd, the data influences only the points on the boundary of the forward light cone originating at x, while when d = 1 or d is even, it affects all points of the forward light cone. Alternatively, the solution at a point (x, t) depends only on the data at the base of the backward light cone originating at (x, t). In fact, when d > 1 is odd, only the data in an immediate neighborhood of the boundary of the base will influence u(x, t). • Waves propagate with finite speed: if the initial data is supported in a bounded set, then the support of the solution u spreads with velocity 1 (or more generally c, if the wave equation is not normalized). We can illustrate some of these facts by the following observation about the different behavior of the propagation of waves in three and two dimensions. Since the propagation of light is governed by the three-dimensional wave equation, if at t = 0 a light flashes at the origin, the following happens: any observer will see the flash (after a finite amount of time) only for an instant. In contrast, consider what happens in two dimensions. If we drop a stone in a lake, any point on the surface will begin (after some time) to undulate; although the amplitude of the oscillations will decrease over time, the undulations will continue (in principle) indefinitely. The difference in character of the formulas for the solutions of the wave equation when d = 1 and d = 3 on the one hand, and d = 2 on the other hand, illustrates a general principle in d-dimensional Fourier analysis: a significant number of formulas that arise are simpler in the case of odd dimensions, compared to the corresponding situations in even dimensions. We will see several further examples of this below.

4 Radial symmetry and Bessel functions We observed earlier that the Fourier transform of a radial function in Rd is also radial. In other words, if f (x) = f0 (|x|) for some f0 , then

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fˆ(ξ) = F0 (|ξ|) for some F0 . A natural problem is to determine a relation between f0 and F0 . This problem has a simple answer in dimensions one and three. If d = 1 the relation we seek is Z ∞ (9) F0 (ρ) = 2 cos(2πρr)f0 (r) dr. 0

If we recall that R has only two rotations, the identity and multiplication by −1, we find that a function is radial precisely when it is even. Having made this observation it is easy to see that if f is radial, and |ξ| = ρ, then Z ∞ F0 (ρ) = fˆ(|ξ|) = f (x)e−2πix|ξ| dx −∞ Z ∞ = f0 (r)(e−2πir|ξ| + e2πir|ξ| ) dr 0 Z ∞ cos(2πρr)f0 (r) dr. =2 0

In the case d = 3, the relation between f0 and F0 is also quite simple and given by the formula Z ∞ −1 sin(2πρr)f0 (r)r dr. (10) F0 (ρ) = 2ρ 0

The proof of this identity is based on the formula for the Fourier transform of the surface element dσ given in Lemma 3.5: Z F0 (ρ) = fˆ(ξ) = f (x)e−2πix·ξ dx R3 Z Z ∞ f0 (r) e−2πirγ·ξ dσ(γ)r2 dr = 0 S2 Z ∞ 2 sin(2πρr) 2 f0 (r) = r dr ρr 0 Z ∞ sin(2πρr)f0 (r)r dr. = 2ρ−1 0

More generally, the relation between f0 and F0 has a nice description in terms of a family of special functions that arise naturally in problems that exhibit radial symmetry. The Bessel function of order n ∈ Z, denoted Jn (ρ), is defined as the nth Fourier coefficient of the function eiρ sin θ . So Z 2π 1 eiρ sin θ e−inθ dθ, Jn (ρ) = 2π 0

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therefore eiρ sin θ =

∞ X

Jn (ρ)einθ .

n=−∞

As a result of this definition, we find that when d = 2, the relation between f0 and F0 is Z ∞ (11) F0 (ρ) = 2π J0 (2πrρ)f0 (r)r dr. 0

Indeed, since fˆ(ξ) is radial we take ξ = (0, −ρ) so that Z ˆ f (ξ) = f (x)e2πix·(0,ρ) dx R2 Z 2πZ ∞ = f0 (r)e2πirρ sin θ r dr dθ 0 0 Z ∞ J0 (2πrρ)f0 (r)r dr, = 2π 0

as desired. In general, there are corresponding formulas relating f0 and F0 in Rd in terms of Bessel functions of order d/2 − 1 (see Problem 2). In even dimensions, these are the Bessel functions we have defined above. For odd dimensions, we need a more general definition of Bessel functions to encompass half-integral orders. Note that the formulas for the Fourier transform of radial functions give another illustration of the differences between odd and even dimensions. When d = 1 or d = 3 (as well as d > 3, d odd) the formulas are in terms of elementary functions, but this is not the case when d is even.

5 The Radon transform and some of its applications Invented by Johann Radon in 1917, the integral transform we discuss next has many applications in mathematics and other sciences, including a significant achievement in medicine. To motivate the definitions and the central problem of reconstruction, we first present the close connection between the Radon transform and the development of X-ray scans (or CAT scans) in the theory of medical imaging. The solution of the reconstruction problem, and the introduction of new algorithms and faster computers, all contributed to a rapid development of computerized tomography. In practice, X-ray scans provide a “picture” of an internal organ, one that helps to detect and locate many types of abnormalities.

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After a brief description of X-ray scans in two dimensions, we define the X-ray transform and formulate the basic problem of inverting this mapping. Although this problem has an explicit solution in R2 , it is more complicated than the analogous problem in three dimensions, hence we give a complete solution of the reconstruction problem only in R3 . Here we have another example where results are simpler in the odddimensional case than in the even-dimensional situation. 5.1 The X-ray transform in R2 Consider a two-dimensional object O lying in the plane R2 , which we may think of as a planar cross section of a human organ. First, we assume that O is homogeneous, and suppose that a very narrow beam of X-ray photons traverses this object.

I

O

I0

Figure 4. Attenuation of an X-ray beam

If I0 and I denote the intensity of the beam before and after passing through O, respectively, the following relation holds: I = I0 e−dρ . Here d is the distance traveled by the beam in the object, and ρ denotes the attenuation coefficient (or absorption coefficient), which depends on the density and other physical characteristics of O. If the object is not homogeneous, but consists of two materials with attenuation coefficients ρ1 and ρ2 , then the observed decrease in the intensity of the beam is

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given by I = I0 e−d1 ρ1 −d2 ρ2 where d1 and d2 denote the distances traveled by the beam in each material. In the case of an arbitrary object whose density and physical characteristics vary from point to point, the attenuation factor is a function ρ in R2 , and the above relations become R ρ I = I0 e L . R Here L is the line in R2 traced by the beam, and L ρ denotes the line integral of ρ over L. Since we observe I and I0 , the data we gather after sending the X-ray beam through the object along the line L is the quantity Z ρ. L

Since we may initially send the beam in any given direction, we may calculate the above integral for every line in R2 . We define the X-ray transform (or Radon transform in R2 ) of ρ by Z ρ. X(ρ)(L) = L

Note that this transform assigns to each appropriate function ρ on R2 (for example, ρ ∈ S(R2 )) another function X(ρ) whose domain is the set of lines L in R2 . The unknown is ρ, and since our original interest lies precisely in the composition of the object, the problem now becomes to reconstruct the function ρ from the collected data, that is, its X-ray transform. We therefore pose the following reconstruction problem: Find a formula for ρ in terms of X(ρ). Mathematically, the problem asks for a formula giving the inverse of X. Does such an inverse even exist? As a first step, we pose the following simpler uniqueness question: If X(ρ) = X(ρ0 ), can we conclude that ρ = ρ0 ? There is a reasonable a priori expectation that X(ρ) actually determines ρ, as one can see by counting the dimensionality (or degrees of freedom) involved. A function ρ on R2 depends on two parameters (the x1 and x2 coordinates, for example). Similarly, the function X(ρ), which is a function of lines L, is also determined by two parameters (for example, the slope of L and its x2 -intercept). In this sense, ρ and X(ρ)

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201

convey an equivalent amount of information, so it is not unreasonable to suppose that X(ρ) determines ρ. While there is a satisfactory answer to the reconstruction problem, and a positive answer to the uniqueness question in R2 , we shall forego giving them here. (However, see Exercise 13 and Problem 8.) Instead we shall deal with the analogous but simpler situation in R3 . Let us finally remark that in fact, one can sample the X-ray transform, and determine X(ρ)(L) for only finitely many lines. Therefore, the reconstruction method implemented in practice is based not only on the general theory, but also on sampling procedures, numerical approximations, and computer algorithms. It turns out that a method used in developing effective relevant algorithms is the fast Fourier transform, which incidentally we take up in the next chapter. 5.2 The Radon transform in R3 The experiment described in the previous section applies in three dimensions as well. If O is an object in R3 determined by a function ρ which describes the density and physical characteristics of this object, sending an X-ray beam through O determines the quantity Z ρ, L

for every line in R3 . In R2 this knowledge was enough to uniquely determine ρ, but in R3 we do not need as much data. In fact, by using the heuristic argument above of counting the number of degrees of freedom, we see that for functions ρ in R3 the number is three, while the number of parameters determining a line L in R3 is four (for example, two for the intercept in the (x1 , x2 ) plane, and two more for the direction of the line). Thus in this sense, the problem is over-determined. We turn instead to the natural mathematical generalization of the twodimensional problem. Here we wish to determine the function in R3 by knowing its integral over all planes 3 in R3 . To be precise, when we speak of a plane, we mean a plane not necessarily passing through the origin. If P is such a plane, we define the Radon transform R(f ) by Z R(f )(P) = f. P

To simplify our presentation, we shall follow our practice of assuming that we are dealing with functions in the class S(R3 ). However, many 3 Note that the dimensionality associated with points on R3 , and that for planes in R3 , equals three in both cases.

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of the results obtained below can be shown to be valid for much larger classes of functions. First, we explain what we mean by the integral of f over a plane. The description we use for planes in R3 is the following: given a unit vector γ ∈ S 2 and a number t ∈ R, we define the plane Pt,γ by Pt,γ = {x ∈ R3 : x · γ = t}. So we parametrize a plane by a unit vector γ orthogonal to it, and by its “distance” t to the origin (see Figure 5). Note that Pt,γ = P−t,−γ , and we allow t to take negative values.

Pt,γ

γ 0

Figure 5. Description of a plane in R3

Given a function f ∈ S(Rd ), we need to make sense of its integral over Pt,γ . We proceed as follows. Choose unit vectors e1 , e2 so that e1 , e2 , γ is an orthonormal basis for R3 . Then any x ∈ Pt,γ can be written uniquely as x = tγ + u

where

If f ∈ S(R3 ), we define Z Z (12) f= Pt,γ

R2

u = u1 e1 + u2 e2

with u1 , u2 ∈ R.

f (tγ + u1 e1 + u2 e2 ) du1 du2 .

To be consistent, we must check that this definition is independent of the choice of the vectors e1 , e2 .

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Proposition 5.1 If f ∈ S(R3 ), then for each γ the definition of is independent of the choice of e1 and e2 . Moreover ÃZ ! Z Z

R Pt,γ

f

∞

f −∞

dt =

f (x) dx. R3

Pt,γ

Proof. If e01 , e02 is another choice of basis vectors so that γ, e01 , e02 is orthonormal, consider the rotation R in R2 which takes e1 to e01 and e2 to e02 . Changing variables u0 = R(u) in the integral proves that our definition (12) is independent of the choice of basis. To prove the formula, let R denote the rotation which takes the standard basis of unit vectors4 in R3 to γ, e1 , and e2 . Then Z Z f (x) dx = f (Rx) dx R3 R3 Z = f (x1 γ + x2 e1 + x3 e2 ) dx1 dx2 dx3 R3 ÃZ ! Z ∞

=

f −∞

dt.

Pt,γ

Remark. We digress to point out that the X-ray transform determines the Radon transform, since two-dimensional integrals can be expressed as iterated one-dimensional integrals. In other words, the knowledge of the integral of a function over all lines determines the integral of that function over any plane. Having disposed of these preliminary matters, we turn to the study of the original problem. The Radon transform of a function f ∈ S(R3 ) is defined by Z R(f )(t, γ) = f. Pt,γ

In particular, we see that the Radon transform is a function on the set of planes in R3 . From the parametrization given for a plane, we may equivalently think of R(f ) as a function on the product R × S 2 = {(t, γ) : t ∈ R, γ ∈ S 2 }, where S 2 denotes the unit sphere in R3 . The relevant class of functions on R × S 2 consists of those that satisfy the Schwartz condition in t uniformly in γ. In other words, we define S(R × S 2 ) to be the space of all continuous functions F (t, γ) that are indefinitely 4 Here

we are referring to the vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1).

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differentiable in t, and that satisfy ¯ ¯ ¯ d` F ¯ ¯ ¯ sup |t|k ¯ ` (t, γ)¯ < ∞ ¯ ¯ ∂t t∈R, γ∈S 2

for all integers k, ` ≥ 0.

Our goal is to solve the following problems. Uniqueness problem: If R(f ) = R(g), then f = g. Reconstruction problem: Express f in terms of R(f ). The solutions will be obtained by using the Fourier transform. In fact, the key point is a very elegant and essential relation between the Radon and Fourier transforms. Lemma 5.2 If f ∈ S(R3 ), then R(f )(t, γ) ∈ S(R) for each fixed γ. Moreover, b )(s, γ) = fˆ(sγ). R(f

To be precise, fˆ denotes the (three-dimensional) Fourier transform b )(s, γ) denotes the one-dimensional Fourier transform of of f , while R(f R(f )(t, γ) as a function of t, with γ fixed. Proof. Since f ∈ S(R3 ), for every positive integer N there is a constant AN < ∞ so that (1 + |t|)N (1 + |u|)N |f (tγ + u)| ≤ AN , if we recall that x = tγ + u, where γ is orthogonal to u. Therefore, as soon as N ≥ 3, we find Z du (1 + |t|)N R(f )(t, γ) ≤ AN < ∞. N R2 (1 + |u|) A similar argument for the derivatives shows that R(f )(t, γ) ∈ S(R) for each fixed γ. To establish the identity, we first note that ! Z ∞ ÃZ b )(s, γ) = R(f f e−2πist dt Z

−∞ ∞

Z

Pt,γ

= −∞

R2

f (tγ + u1 e1 + u2 e2 ) du1 du2 e−2πist dt.

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However, since γ · u = 0 and |γ| = 1, we may write e−2πist = e−2πisγ·(tγ+u) . As a result, we find that Z ∞ Z b )(s, γ) = R(f f (tγ + u1 e1 + u2 e2 )e−2πisγ·(tγ+u) du1 du2 dt −∞ R2 Z ∞ Z = f (tγ + u)e−2πisγ·(tγ+u) du dt. −∞

R2

A final rotation from γ, e1 , e2 to the standard basis in R3 proves that b )(s, γ) = fˆ(sγ), as desired. R(f As a consequence of this identity, we can answer the uniqueness question for the Radon transform in R3 in the affirmative. Corollary 5.3 If f, g ∈ S(R3 ) and R(f ) = R(g), then f = g. The proof of the corollary follows from an application of the lemma to the difference f − g and use of the Fourier inversion theorem. Our final task is to give the formula that allows us to recover f from its Radon transform. Since R(f ) is a function on the set of planes in R3 , and f is a function of the space variables x ∈ R3 , to recover f we introduce the dual Radon transform, which passes from functions defined on planes to functions in R3 . Given a function F on R × S 2 , we define its dual Radon transform by Z ∗ (13) R (F )(x) = F (x · γ, γ) dσ(γ). S2

Observe that a point x belongs to Pt,γ if and only if x · γ = t, so the idea here is that given x ∈ R3 , we obtain R∗ (F )(x) by integrating F over the subset of all planes passing through x, that is, Z R∗ (F )(x) = F, {Pt,γ such that x∈Pt,γ }

where the integral on the right is given the precise meaning in (13). We use the terminology “dual” because of the following observation. If V1 = S(R3 ) with the usual Hermitian inner product Z (f, g)1 = f (x)g(x) dx, R3

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and V2 = S(R × S 2 ) with the Hermitian inner product Z Z (F, G)2 = F (t, γ)G(t, γ) dσ(γ) dt, R

S2

then R∗ : V2 → V1 ,

R : V1 → V2 , with

(Rf, F )2 = (f, R∗ F )1 .

(14)

The validity of this identity is not needed in the argument below, and its verification is left as an exercise for the reader. We can now state the reconstruction theorem. Theorem 5.4 If f ∈ S(R3 ), then 4(R∗ R(f )) = −8π 2 f. We recall that 4 =

∂2 ∂x21

∂2 ∂x22

+

∂2 ∂x23

+

is the Laplacian.

Proof. By our previous lemma, we have Z ∞ fˆ(sγ)e2πits ds. R(f )(t, γ) = −∞

Therefore Z

Z

∗

∞

R R(f )(x) = S2

fˆ(sγ)e2πix·γs ds dσ(γ),

−∞

hence Z

Z

∞

fˆ(sγ)(−4π 2 s2 )e2πix·γs ds dσ(γ) Z Z ∞ = − 4π 2 fˆ(sγ)e2πix·γs s2 ds dσ(γ)

4(R∗ R(f ))(x) =

S2

−∞

S2

Z

Z

fˆ(sγ)e2πix·γs s2 ds dσ(γ) −∞ Z Z ∞ − 4π 2 fˆ(sγ)e2πix·γs s2 ds dσ(γ)

= − 4π 2 S2

Z

= − 8π

−∞ 0

Z

∞

2 S2

= − 8π 2 f (x).

0

S2

fˆ(sγ)e

0

2πix·γs 2

s ds dσ(γ)

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In the first line, we have differentiated under the integral sign and used the fact 4(e2πix·γs ) = (−4π 2 s2 )e2πix·γs , since |γ| = 1. The last step follows from the formula for polar coordinates in R3 and the Fourier inversion theorem. 5.3 A note about plane waves We conclude this chapter by briefly mentioning a nice connection between the Radon transform and solutions of the wave equation. This comes about in the following way. Recall that when d = 1, the solution of the wave equation can be expressed as the sum of traveling waves (see Chapter 1), and it is natural to ask if an analogue of such traveling waves exists in higher dimensions. The answer is as follows. Let F be a function of one variable, which we assume is sufficiently smooth (say C 2 ), and consider u(x, t) defined by u(x, t) = F ((x · γ) − t), where x ∈ Rd and γ is a unit vector in Rd . It is easy to verify directly that u is a solution of the wave equation in Rd (with c = 1). Such a solution is called a plane wave; indeed, notice that u is constant on every plane perpendicular to the direction γ, and as time t increases, the wave travels in the γ direction. (It should be remarked that plane waves are never functions in S(Rd ) when d > 1 because they are constant in directions perpendicular to γ).5 The basic fact is that when d > 1, the solution of the wave equation can be written as an integral (as opposed to sum, when d = 1) of plane waves; this can in fact be done via the Radon transform of the initial data f and g. For the relevant formulas when d = 3, see Problem 6.

6 Exercises 1. Suppose that R is a rotation in the plane R2 , and let

µ

R=

a c

b d

¶

denote its matrix with respect to the standard basis vectors e1 = (1, 0) and e2 = (0, 1). (a) Write the conditions Rt = R−1 and det(R) = ±1 in terms of equations in a, b, c, d. 5 Incidentally, this observation is further indication that a fuller treatment of the wave equation requires lifting the restriction that functions belong to S(Rd ).

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(b) Show that there exists ϕ ∈ R such that a + ib = eiϕ . (c) Conclude that if R is proper, then it can be expressed as z 7→ zeiϕ , and if R is improper, then it takes the form z 7→ zeiϕ , where z = x − iy. 2. Suppose that R : R3 → R3 is a proper rotation. (a) Show that p(t) = det(R − tI) is a polynomial of degree 3, and prove that there exists γ ∈ S 2 (where S 2 denotes the unit sphere in R3 ) with R(γ) = γ. [Hint: Use the fact that p(0) > 0 to see that there is λ > 0 with p(λ) = 0. Then R − λI is singular, so its kernel is non-trivial.] (b) If P denotes the plane perpendicular to γ and passing through the origin, show that R : P → P, and that this linear map is a rotation.

3. Recall the formula

Z

Z

Z

∞

F (rγ)rd−1 dr dσ(γ).

F (x) dx = Rd

S d−1

0

Apply this to the special case when F (x) = g(r)f (γ), where x = rγ, to prove that for any rotation R, one has

Z

Z f (R(γ)) dσ(γ) = S d−1

f (γ) dσ(γ), S d−1

whenever f is a continuous function on the sphere S d−1 . 4. Let Ad and Vd denote the area and volume of the unit sphere and unit ball in Rd , respectively. (a) Prove the formula Ad =

2π d/2 Γ(d/2)

R∞

so that A2 = 2π, A3 = 4π, A4 = 2π 2 , . . .. Here Γ(x) = 0 e−t tx−1 dt is the Gamma function. [Hint: Use polar coordinates and the fact that R 2 e−π|x| dx = 1.] Rd

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6. Exercises

(b) Show that d Vd = Ad , hence Vd =

π d/2 . Γ(d/2 + 1)

In particular V2 = π, V3 = 4π/3, . . .. 5. Let A be a d × d positive definite symmetric matrix with real coefficients. Show that

Z

e−π(x,A(x)) dx = (det(A))−1/2 . Rd

R

2

This generalizes the fact that Rd e−π|x| dx = 1, which corresponds to the case where A is the identity. [Hint: Apply the spectral theorem to write A = RDR−1 where R is a rotation and, D is diagonal with entries λ1 , . . . , λd , where {λi } are the eigenvalues of A.] 6. Suppose ψ ∈ S(Rd ) satisfies

µZ

R

|ψ(x)|2 dx = 1. Show that

¶ µZ

¶ 2 ˆ |ξ|2 |ψ(ξ)| dξ

|x|2 |ψ(x)|2 dx Rd

≥

Rd

d2 . 16π 2

This is the statement of the Heisenberg uncertainty principle in d dimensions. 7. Consider the time-dependent heat equation in Rd : ∂u ∂2u ∂2u = , + ··· + 2 ∂t ∂x1 ∂x2d

(15)

where t > 0,

with boundary values u(x, 0) = f (x) ∈ S(Rd ). If (d)

Ht (x) =

2 1 e−|x| /4t = (4πt)d/2

Z e−4π

2

t|ξ|2 2πix·ξ

e

dξ

Rd

is the d-dimensional heat kernel, show that the convolution (d)

u(x, t) = (f ∗ Ht )(x) is indefinitely differentiable when x ∈ Rd and t > 0. Moreover, u solves (15), and is continuous up to the boundary t = 0 with u(x, 0) = f (x). The reader may also wish to formulate the d-dimensional analogues of Theorem 2.1 and 2.3 in Chapter 5. 8. In Chapter 5, we found that a solution to the steady-state heat equation in the upper half-plane with boundary values f is given by the convolution u = f ∗ Py where the Poisson kernel is Py (x) =

1 y 2 π x + y2

where x ∈ R and y > 0.

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More generally, one can calculate the d-dimensional Poisson kernel using the Fourier transform as follows. (a) The subordination principle allows one to write expressions involving the function e−x in terms of corresponding expressions involving the 2 function e−x . One form of this is the identity

Z e

−β

∞

2 e−u √ e−β /4u du πu

= 0

when β ≥ 0. Prove this identity with β = 2π|x| by taking the Fourier transform of both sides. (b) Consider the steady-state heat equation in the upper half-space {(x, y) : x ∈ Rd , y > 0} d X ∂2u

∂x2j j=1

+

∂2u =0 ∂y 2

with the Dirichlet boundary condition u(x, 0) = f (x). A solution to this (d) (d) problem is given by the convolution u(x, y) = (f ∗ Py )(x) where Py (x) is the d-dimensional Poisson kernel

Z Py(d) (x)

e2πix·ξ e−2π|ξ|y dξ.

= Rd

(d)

Compute Py (x) by using the subordination principle and the d-dimensional heat kernel. (See Exercise 7.) Show that Py(d) (x) =

Γ((d + 1)/2) y . π (d+1)/2 (|x|2 + y 2 )(d+1)/2

9. A spherical wave is a solution u(x, t) of the Cauchy problem for the wave equation in Rd , which as a function of x is radial. Prove that u is a spherical wave if and only if the initial data f, g ∈ S are both radial. 10. Let u(x, t) be a solution of the wave equation, and let E(t) denote the energy of this wave

Z E(t) = R

¯ ¯2 X d Z ¯ ∂u ¯ ¯ (x, t)¯ + ¯ ∂t ¯ d j=1

¯ ¯2 ¯ ∂u ¯ ¯ (x, t)¯¯ dx. ¯ d ∂xj

R

We have seen that E(t) is constant using Plancherel’s formula. Give an alternate proof of this fact by differentiating the integral with respect to t and showing that dE = 0. dt

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6. Exercises

[Hint: Integrate by parts.] 11. Show that the solution of the wave equation ∂2u ∂2u ∂2u ∂2u = + + ∂t2 ∂x21 ∂x22 ∂x23 subject to u(x, 0) = f (x) and 1 u(x, t) = |S(x, t)|

∂u ∂t (x, 0)

= g(x), where f, g ∈ S(R3 ), is given by

Z [tg(y) + f (y) + ∇f (y) · (y − x)] dσ(y), S(x,t)

where S(x, t) denotes the sphere of center x and radius t, and |S(x, t)| its area. This is an alternate expression for the solution of the wave equation given in Theorem 3.6. It is sometimes called Kirchhoff ’s formula. 12. Establish the identity (14) about the dual transform given in the text. In other words, prove that

Z Z

Z

(16)

f (x)R∗ (F )(x) dx

R(f )(t, γ)F (t, γ)dσ(γ) dt = R

S2

R3

where f ∈ S(R3 ), F ∈ S(R × S 2 ), and

Z

Z

R(f ) =

f

∗

and

R (F )(x) =

F (x · γ, γ) dσ(γ). S2

Pt,γ

[Hint: Consider the integral

Z Z Z f (tγ + u1 e2 + u2 e2 )F (t, γ) dt dσ(γ) du1 du2 . Integrating first in u gives the left-hand side of (16), while integrating in u and t and setting x = tγ + u1 e2 + u2 e2 gives the right-hand side.] 13. For each (t, θ) with t ∈ R and |θ| ≤ π, let L = Lt,θ denote the line in the (x, y)-plane given by x cos θ + y sin θ = t. This is the line perpendicular to the direction (cos θ, sin θ) at “distance” t from the origin (we allow negative t). For f ∈ S(R2 ) the X-ray transform or twodimensional Radon transform of f is defined by

Z

Z

X(f )(t, θ) =

∞

f= Lt,θ

f (t cos θ + u sin θ, t sin θ − u cos θ) du. −∞

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Calculate the X-ray transform of the function f (x, y) = e−π(x

2

+y 2 )

.

14. Let X be the X-ray transform. Show that if f ∈ S and X(f ) = 0, then f = 0, by taking the Fourier transform in one variable. 15. For F ∈ S(R × S 1 ), define the dual X-ray transform X ∗ (F ) by integrating F over all lines that pass through the point (x, y) (that is, those lines Lt,θ with x cos θ + y sin θ = t):

Z X ∗ (F )(x, y) =

F (x cos θ + y sin θ, θ) dθ.

Check that in this case, if f ∈ S(R2 ) and F ∈ S(R × S 1 ), then

Z Z

Z Z f (x, y)X ∗ (F )(x, y) dx dy.

X(f )(t, θ)F (t, θ) dt dθ =

7 Problems 1. Let Jn denote the nth order Bessel function, for n ∈ Z. Prove that (a) Jn (ρ) is real for all real ρ. (b) J−n (ρ) = (−1)n Jn (ρ). (c) 2Jn0 (ρ) = Jn−1 (ρ) − Jn+1 (ρ).

³

(d)

2n ρ

´

Jn (ρ) = Jn−1 (ρ) + Jn+1 (ρ).

(e) (ρ−n Jn (ρ))0 = −ρ−n Jn+1 (ρ). (f) (ρn Jn (ρ))0 = ρn Jn−1 (ρ). (g) Jn (ρ) satisfies the second order differential equation Jn00 (ρ) + ρ−1 Jn0 (ρ) + (1 − n2 /ρ2 )Jn (ρ) = 0. (h) Show that Jn (ρ) =

∞ ³ ρ ´n X

2

(−1)m

m=0

ρ2m . + m)!

22m m!(n

(i) Show that for all integers n and all real numbers a and b we have Jn (a + b) =

X `∈Z

J` (a)Jn−` (b).

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7. Problems

2. Another formula for Jn (ρ) that allows one to define Bessel functions for non-integral values of n, (n > −1/2) is Jn (ρ) =

Z

(ρ/2)n √ Γ(n + 1/2) π

1

eiρt (1 − t2 )n−(1/2) dt. −1

(a) Check that the above formula agrees with the definition of Jn (ρ) for integral n ≥ 0. [Hint: Verify it for n = 0 and then check that both sides satisfy the recursion formula (e) in Problem 1.]

q (b) Note that J1/2 (ρ) =

2 −1/2 πρ

sin ρ.

(c) Prove that

r lim

n→−1/2

Jn (ρ) =

2 −1/2 ρ cos ρ. π

(d) Observe that the formulas we have proved in the text giving F0 in terms of f0 (when describing the Fourier transform of a radial function) take the form

Z (17)

∞

J(d/2)−1 (2πρr)f0 (r)rd/2 dr,

F0 (ρ) = 2πρ−(d/2)+1 0

for d = 1, 2, and 3, if one uses the formulas above with the understanding that J−1/2 (ρ) = limn→−1/2 Jn (ρ). It turns out that the relation between F0 and f0 given by (17) is valid in all dimensions d. 3. We observed that the solution u(x, t) of the Cauchy problem for the wave equation given by formula (3) depends only on the initial data on the base on the backward light cone. It is natural to ask if this property is shared by any solution of the wave equation. An affirmative answer would imply uniqueness of the solution. Let B(x0 , r0 ) denote the closed ball in the hyperplane t = 0 centered at x0 and of radius r0 . The backward light cone with base B(x0 , r0 ) is defined by LB(x0 ,r0 ) = {(x, t) ∈ Rd × R : |x − x0 | ≤ r0 − t, 0 ≤ t ≤ r0 }. Theorem Suppose that u(x, t) is a C 2 function on the closed upper half-plane {(x, t) : x ∈ Rd , t ≥ 0} that solves the wave equation ∂2u = 4u. ∂t2 If u(x, 0) = LB(x0 ,r0 ) .

∂u ∂t (x, 0)

= 0 for all x ∈ B(x0 , r0 ), then u(x, t) = 0 for all (x, t) ∈

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In words, if the initial data of the Cauchy problem for the wave equation vanishes on a ball B, then any solution u of the problem vanishes in the backward light cone with base B. The following steps outline a proof of the theorem. (a) Assume that u is real. For 0 ≤ t ≤ r0 let Bt (x0 , r0 ) = {x : |x − x0 | ≤ r0 − t}, and also define

µ ∇u(x, t) =

∂u ∂u ∂u ,..., , ∂x1 ∂xd ∂t

¶ .

Now consider the energy integral E(t) = =

Z

1 2

|∇u|2 dx

Z

1 2

Bt (x0 ,r0 )

µ

Bt (x0 ,r0 )

∂u ∂t

¶2 +

¶2 d µ X ∂u ∂xj

j=1

dx.

Observe that E(t) ≥ 0 and E(0) = 0. Prove that

Z

d

0

E (t) = Bt (x0 ,r0 )

∂u ∂ 2 u X ∂u ∂ 2 u 1 + dx − ∂t ∂t2 ∂xj ∂xj ∂t 2 j=1

Z |∇u|2 dσ(γ). ∂Bt (x0 ,r0 )

(b) Show that ∂ ∂xj

·

¸

∂u ∂u ∂u ∂ 2 u ∂ 2 u ∂u = + . ∂xj ∂t ∂xj ∂xj ∂t ∂x2j ∂t

(c) Use the last identity, the divergence theorem, and the fact that u solves the wave equation to prove that

Z 0

E (t) =

d X ∂u ∂u

∂Bt (x0 ,r0 ) j=1

1 νj dσ(γ) − ∂xj ∂t 2

Z |∇u|2 dσ(γ), ∂Bt (x0 ,r0 )

where νj denotes the j th coordinate of the outward normal to Bt (x0 , r0 ). (d) Use the Cauchy-Schwarz inequality to conclude that d X ∂u ∂u j=1

∂xj ∂t

νj ≤

1 |∇u|2 , 2

and as a result, E 0 (t) ≤ 0. Deduce from this that E(t) = 0 and u = 0.

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7. Problems

4.∗ There exist formulas for the solution of the Cauchy problem for the wave equation ∂2u ∂2u ∂2u = + ··· + 2 2 ∂t ∂x1 ∂x2d

with u(x, 0) = f (x) and

∂u (x, 0) = g(x) ∂t

in Rd × R in terms of spherical means which generalize the formula given in the text for d = 3. In fact, the solution for even dimensions is deduced from that for odd dimensions, so we discuss this case first. Suppose that d > 1 is odd and let h ∈ S(Rd ). The spherical mean of h on the ball centered at x of radius t is defined by Mr h(x) = M h(x, r) =

1 Ad

Z h(x − rγ) dσ(γ), S d−1

where Ad denotes the area of the unit sphere S d−1 in Rd . (a) Show that

· 4x M h(x, r) = ∂r2 +

¸

d−1 M h(x, r), r

where 4x denotes the Laplacian in the space variables x, and ∂r = ∂/∂r. (b) Show that a twice differentiable function u(x, t) satisfies the wave equation if and only if

·

¸

∂r2 +

d−1 M u(x, r, t) = ∂t2 M u(x, r, t), r

where M u(x, r, t) denote the spherical means of the function u(x, t). (c) If d = 2k + 1, define T ϕ(r) = (r−1 ∂r )k−1 [r2k−1 ϕ(r)], and let u ˜ = T M u. Then this function solves the one-dimensional wave equation for each fixed x: ∂t2 u ˜(x, r, t) = ∂r2 u ˜(x, r, t). One can then use d’Alembert’s formula to find the solution u ˜(x, r, t) of this problem expressed in terms of the initial data. (d) Now show that u(x, t) = M u(x, 0, t) = lim

r→0

where α = 1 · 3 · · · (d − 2).

u ˜(x, r, t) αr

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(e) Conclude that the solution of the Cauchy problem for the d-dimensional wave equation, when d > 1 is odd, is u(x, t) =

£ ¡ ¢ 1 ∂t (t−1 ∂t )(d−3)/2 td−2 Mt f (x) + 1 · 3 · · · (d − 2) ¡

(t−1 ∂t )(d−3)/2 td−2 Mt g(x)

¢¤

.

5.∗ The method of descent can be used to prove that the solution of the Cauchy problem for the wave equation in the case when d is even is given by the formula u(x, t) =

h ³ ´ 1 ft f (x) + ∂t (t−1 ∂t )(d−3)/2 td−2 M 1 · 3 · · · (d − 2) ³

ft g(x) (t−1 ∂t )(d−3)/2 td−2 M

´i ,

ft denotes the modified spherical means defined by where M ft h(x) = M

Z

2 Ad+1

f (x + ty)

p Bd

dy.

1 − |y|2

6.∗ Given initial data f and g of the form f (x) = F (x · γ)

and

g(x) = G(x · γ),

check that the plane wave given by u(x, t) =

F (x · γ + t) + F (x · γ − t) 1 + 2 2

Z

x·γ+t

G(s) ds x·γ−t

is a solution of the Cauchy problem for the d-dimensional wave equation. In general, the solution is given as a superposition of plane waves. For the case d = 3, this can be expressed in terms of the Radon transform as follows. Let ˜ )(t, γ) = − 1 R(f 8π 2

µ

d dt

¶2 R(f )(t, γ).

Then u(x, t) = 1 2

Z

·

Z ˜ )(x · γ − t, γ) + R(f ˜ )(x · γ + t, γ) + R(f

S2

x·γ+t

x·γ−t

¸ ˜ R(g)(s, γ) ds dσ(γ).

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7. Problems

7. For every real number a > 0, define the operator (−4)a by the formula

Z (2π|ξ|)2a fˆ(ξ)e2πiξ·x dξ

a

(−4) f (x) = Rd

whenever f ∈ S(Rd ). (a) Check that (−4)a agrees with the usual definition of the ath power of −4 (that is, a compositions of minus the Laplacian) when a is a positive integer. (b) Verify that (−4)a (f ) is indefinitely differentiable. (c) Prove that if a is not an integer, then in general (−4)a (f ) is not rapidly decreasing. (d) Let u(x, y) be the solution of the steady-state heat equation d

∂2u X ∂2u + = 0, ∂y 2 ∂x2j

with u(x, 0) = f (x),

j=1

given by convolving f with the Poisson kernel (see Exercise 8). Check that (−4)1/2 f (x) = − lim

y→0

∂u (x, y), ∂y

and more generally that ∂ku (x, y) y→0 ∂y k

(−4)k/2 f (x) = (−1)k lim for any positive integer k.

8.∗ The reconstruction formula for the Radon transform in Rd is as follows: (a) When d = 2, (−4)1/2 ∗ R (R(f )) = f, 4π where (−4)1/2 is defined in Problem 7. (b) If the Radon transform and its dual are defined by analogy to the cases d = 2 and d = 3, then for general d, (2π)1−d (−4)(d−1)/2 R∗ (R(f )) = f. 2

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7 Finite Fourier Analysis This past year has seen the birth, or rather the rebirth, of an exciting revolution in computing Fourier transforms. A class of algorithms known as the fast Fourier transform or FFT, is forcing a complete reassessment of many computational paths, not only in frequency analysis, but in any fields where problems can be reduced to Fourier transforms and/or convolutions... C. Bingham and J. W. Tukey, 1966

In the previous chapters we studied the Fourier series of functions on the circle and the Fourier transform of functions defined on the Euclidean space Rd . The goal here is to introduce another version of Fourier analysis, now for functions defined on finite sets, and more precisely, on finite abelian groups. This theory is particularly elegant and simple since infinite sums and integrals are replaced by finite sums, and thus questions of convergence disappear. In turning our attention to finite Fourier analysis, we begin with the simplest example, Z(N ), where the underlying space is the (multiplicative) group of N th roots of unity on the circle. This group can also be realized in additive form, as Z/N Z, the equivalence classes of integers modulo N . The group Z(N ) arises as the natural approximation to the circle (as N tends to infinity) since in the first picture the points of Z(N ) correspond to N points on the circle which are uniformly distributed. For this reason, in practical applications, the group Z(N ) becomes a natural candidate for the storage of information of a function on the circle, and for the resulting numerical computations involving Fourier series. The situation is particularly nice when N is large and of the form N = 2n . The computations of the Fourier coefficients now lead to the “fast Fourier transform,” which exploits the fact that an induction in n requires only about log N steps to go from N = 1 to N = 2n . This yields a substantial saving in time in practical applications. In the second part of the chapter we undertake the more general theory of Fourier analysis on finite abelian groups. Here the fundamental example is the multiplicative group Z∗ (q). The Fourier inversion formula

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1. Fourier analysis on Z(N )

for Z∗ (q) will be seen to be a key step in the proof of Dirichlet’s theorem on primes in arithmetic progression, which we will take up in the next chapter.

1 Fourier analysis on Z(N ) We turn to the group of N th roots of unity. This group arises naturally as the simplest finite abelian group. It also gives a uniform partition of the circle, and is therefore a good choice if one wishes to sample appropriate functions on the circle. Moreover, this partition gets finer as N tends to infinity, and one might expect that the discrete Fourier theory that we discuss here tends to the continuous theory of Fourier series on the circle. In a broad sense, this is the case, although this aspect of the problem is not one that we develop. 1.1 The group Z(N ) Let N be a positive integer. A complex number z is an N th root of unity if z N = 1. The set of N th roots of unity is precisely n o 1, e2πi/N , e2πi2/N , . . . , e2πi(N −1)/N . Indeed, suppose that z N = 1 with z = reiθ . Then we must have rN eiN θ = 1, and taking absolute values yields r = 1. Therefore eiN θ = 1, and this means that N θ = 2πk where k ∈ Z. So if ζ = e2πi/N we find that ζ k exhausts all the N th roots of unity. However, notice that ζ N = 1 so if n and m differ by an integer multiple of N , then ζ n = ζ m . In fact, it is clear that ζn = ζm

if and only if

n − m is divisible by N .

We denote the set of all N th roots of unity by Z(N ). The fact that this set gives a uniform partition of the circle is clear from its definition. Note that the set Z(N ) satisfies the following properties: (i) If z, w ∈ Z(N ), then zw ∈ Z(N ) and zw = wz. (ii) 1 ∈ Z(N ). (iii) If z ∈ Z(N ), then z −1 = 1/z ∈ Z(N ) and of course zz −1 = 1. As a result we can conclude that Z(N ) is an abelian group under complex multiplication. The appropriate definitions are set out in detail later in Section 2.1.

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Chapter 7. FINITE FOURIER ANALYSIS

ζ3

ζ2 ζ

ζ

4

1 ζ5

ζ8 ζ6

ζ7

Z(9), ζ = e2πi/9

Z(N ), N = 26

Figure 1. The group of N th roots of unity when N = 9 and N = 26 = 64

There is another way to visualize the group Z(N ). This consists of choosing the integer power of ζ that determines each root of unity. We observed above that this integer is not unique since ζ n = ζ m whenever n and m differ by an integer multiple of N . Naturally, we might select the integer which satisfies 0 ≤ n ≤ N − 1. Although this choice is perfectly reasonable in terms of “sets,” we ask what happens when we multiply roots of unity. Clearly, we must add the corresponding integers since ζ n ζ m = ζ n+m but nothing guarantees that 0 ≤ n + m ≤ N − 1. In fact, if ζ n ζ m = ζ k with 0 ≤ k ≤ N − 1, then n + m and k differ by an integer multiple of N . So, to find the integer in [0, N − 1] corresponding to the root of unity ζ n ζ m , we see that after adding the integers n and m we must reduce modulo N , that is, find the unique integer 0 ≤ k ≤ N − 1 so that (n + m) − k is an integer multiple of N . An equivalent approach is to associate to each root of unity ω the class of integers n so that ζ n = ω. Doing so for each root of unity we obtain a partition of the integers in N disjoint infinite classes. To add two of these classes, choose any integer in each one of them, say n and m, respectively, and define the sum of the classes to be the class which contains the integer n + m. We formalize the above notions. Two integers x and y are congruent modulo N if the difference x − y is divisible by N , and we write x ≡ y mod N . In other words, this means that x and y differ by an integer multiple of N . It is an easy exercise to check the following three properties:

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1. Fourier analysis on Z(N )

• x ≡ x mod N for all integers x. • If x ≡ y mod N , then y ≡ x mod N . • If x ≡ y mod N and y ≡ z mod N , then x ≡ z mod N . The above defines an equivalence relation on Z. Let R(x) denote the equivalence class, or residue class, of the integer x. Any integer of the form x + kN with k ∈ Z is an element (or “representative”) of R(x). In fact, there are precisely N equivalence classes, and each class has a unique representative between 0 and N − 1. We may now add equivalence classes by defining R(x) + R(y) = R(x + y). This definition is of course independent of the representatives x and y because if x0 ∈ R(x) and y 0 ∈ R(y), then one checks easily that x0 + y 0 ∈ R(x + y). This turns the set of equivalence classes into an abelian group called the group of integers modulo N , which is sometimes denoted by Z/N Z. The association R(k) ←→ e2πik/N gives a correspondence between the two abelian groups, Z/N Z and Z(N ). Since the operations are respected, in the sense that addition of integers modulo N becomes multiplication of complex numbers, we shall also denote the group of integers modulo N by Z(N ). Observe that 0 ∈ Z/N Z corresponds to 1 on the unit circle. Let V and W denote the vector spaces of complex-valued functions on the group of integers modulo N and the N th roots of unity, respectively. Then, the identification given above carries over to V and W as follows: F (k) ←→ f (e2πik/N ), where F is a function on the integers modulo N and f is a function on the N th roots of unity. From now on, we write Z(N ) but think of either the group of integers modulo N or the group of N th roots of unity. 1.2 Fourier inversion theorem and Plancherel identity on Z(N ) The first and most crucial step in developing Fourier analysis on Z(N ) is to find the functions which correspond to the exponentials en (x) = e2πinx in the case of the circle. Some important properties of these exponentials are:

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Chapter 7. FINITE FOURIER ANALYSIS

(i) {en }n∈Z is an orthonormal set for the inner product (1) (in Chapter 3) on the space of Riemann integrable functions on the circle. (ii) Finite linear combinations of the en ’s (the trigonometric polynomials) are dense in the space of continuous functions on the circle. (iii) en (x + y) = en (x)en (y). On Z(N ), the appropriate analogues are the N functions e0 , . . . , eN −1 defined by e` (k) = ζ `k = e2πi`k/N

for ` = 0, . . . , N − 1 and k = 0, . . . , N − 1,

where ζ = e2πi/N . To understand the parallel with (i) and (ii), we can think of the complex-valued functions on Z(N ) as a vector space V , endowed with the Hermitian inner product (F, G) =

N −1 X

F (k)G(k)

k=0

and associated norm kF k2 =

N −1 X

|F (k)|2 .

k=0

Lemma 1.1 The family {e0 , . . . , eN −1 } is orthogonal. In fact, ½

(em , e` ) =

N 0

if m = `, if m = 6 `.

Proof. We have (em , e` ) =

N −1 X

ζ mk ζ −`k =

k=0

N −1 X

ζ (m−`)k .

k=0

If m = `, each term in the sum is equal to 1, and the sum equals N . If m 6= `, then q = ζ m−` is not equal to 1, and the usual formula 1 + q + q 2 + · · · + q N −1 =

1 − qN 1−q

shows that (em , e` ) = 0, because q N = 1. Since the N functions e0 , . . . , eN −1 are orthogonal, they must be linearly independent, and since the vector space V is N -dimensional, we

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conclude that {e0 , . . . , eN −1 } is an orthogonal basis for V . Clearly, property (iii) also holds, that is, e` (k + m) = e` (k)e` (m) for all `, and all k, m ∈ Z(N ). √ By the lemma each vector e` has norm N , so if we define 1 e∗` = √ e` , N then {e∗0 , . . . , e∗N −1 } is an orthonormal basis for V . Hence for any F ∈ V we have (1)

F =

N −1 X

(F, e∗n )e∗n

as well as

kF k2 =

N −1 X

n=0

|(F, e∗n )|2 .

n=0

If we define the nth Fourier coefficient of F by an =

N −1 1 X F (k)e−2πikn/N , N k=0

the above observations give the following fundamental theorem which is the Z(N ) version of the Fourier inversion and the Parseval-Plancherel formulas. Theorem 1.2 If F is a function on Z(N ), then F (k) =

N −1 X

an e2πink/N .

n=0

Moreover, N −1 X n=0

|an |2 =

N −1 1 X |F (k)|2 . N k=0

The proof follows directly from (1) once we observe that an =

1 1 (F, en ) = √ (F, e∗n ). N N

Remark. It is possible to recover the Fourier inversion on the circle for sufficiently smooth functions (say C 2 ) by letting N → ∞ in the finite model Z(N ) (see Exercise 3).

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1.3 The fast Fourier transform The fast Fourier transform is a method that was developed as a means of calculating efficiently the Fourier coefficients of a function F on Z(N ). The problem, which arises naturally in numerical analysis, is to determine an algorithm that minimizes the amount of time it takes a computer to calculate the Fourier coefficients of a given function on Z(N ). Since this amount of time is roughly proportional to the number of operations the computer must perform, our problem becomes that of minimizing the number of operations necessary to obtain all the Fourier coefficients {an } given the values of F on Z(N ). By operations we mean either an addition or a multiplication of complex numbers. We begin with a naive approach to the problem. Fix N , and suppose that we are given F (0), . . . , F (N − 1) and ωN = e−2πi/N . If we denote th by aN Fourier coefficient of F on Z(N ), then by definition k (F ) the k aN k (F ) =

N −1 1 X kr F (r)ωN , N r=0

and crude estimates show that the number of operations needed to calculate all Fourier coefficients is ≤ 2N 2 + N . Indeed, it takes at most N −1 2 N − 2 multiplications to determine ωN , . . . , ωN , and each coefficient aN requires N + 1 multiplications and N − 1 additions. k We now present the fast Fourier transform, an algorithm that improves the bound O(N 2 ) obtained above. Such an improvement is possible if, for example, we restrict ourselves to the case where the partition of the circle is dyadic, that is, N = 2n . (See also Exercise 9.) Theorem 1.3 Given ωN = e−2πi/N with N = 2n , it is possible to calculate the Fourier coefficients of a function on Z(N ) with at most 4 · 2n n = 4N log2 (N ) = O(N log N ) operations. The proof of the theorem consists of using the calculations for M division points, to obtain the Fourier coefficients for 2M division points. Since we choose N = 2n , we obtain the desired formula as a consequence of a recurrence which involves n = O(log N ) steps. Let #(M ) denote the minimum number of operations needed to calculate all the Fourier coefficients of any function on Z(M ). The key to the proof of the theorem is contained in the following recursion step.

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Lemma 1.4 If we are given ω2M = e−2πi/(2M ) , then #(2M ) ≤ 2#(M ) + 8M. 2M Proof. The calculation of ω2M , . . . , ω2M requires no more than 2M 2 operations. Note that in particular we get ωM = e−2πi/M = ω2M . The main idea is that for any given function F on Z(2M ), we consider two functions F0 and F1 on Z(M ) defined by

F0 (r) = F (2r)

and

F1 (r) = F (2r + 1).

We assume that it is possible to calculate the Fourier coefficients of F0 and F1 in no more than #(M ) operations each. If we denote the Fourier coefficients corresponding to the groups Z(2M ) and Z(M ) by a2M and k M ak , respectively, then we have a2M k (F ) =

´ 1³ M k ak (F0 ) + aM k (F1 )ω2M . 2

To prove this, we sum over odd and even integers in the definition of the Fourier coefficient a2M k (F ), and find 2M −1 1 X kr F (r)ω2M 2M r=0 Ã ! M −1 M −1 1 1 X 1 X k(2`) k(2m+1) = F (2`)ω2M + F (2m + 1)ω2M 2 M `=0 M m=0 ! Ã M −1 M −1 1 X 1 1 X k` km k F0 (`)ωM + F1 (m)ωM ω2M , = 2 M `=0 M m=0

a2M k (F ) =

which establishes our assertion. M k As a result, knowing aM k (F0 ), ak (F1 ), and ω2M , we see that each a2M k (F ) can be computed using no more than three operations (one addition and two multiplications). So #(2M ) ≤ 2M + 2#(M ) + 3 × 2M = 2#(M ) + 8M, and the proof of the lemma is complete. An induction on n, where N = 2n , will conclude the proof of the theorem. The initial step n = 1 is easy, since N = 2 and the two Fourier coefficients are aN 0 (F ) =

1 (F (1) + F (−1)) 2

and

aN 1 (F ) =

1 (F (1) + (−1)F (−1)) . 2

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Calculating these Fourier coefficients requires no more than five operations, which is less than 4 × 2 = 8. Suppose the theorem is true up to N = 2n−1 so that #(N ) ≤ 4 · 2n−1 (n − 1). By the lemma we must have #(2N ) ≤ 2 · 4 · 2n−1 (n − 1) + 8 · 2n−1 = 4 · 2n n, which concludes the inductive step and the proof of the theorem.

2 Fourier analysis on finite abelian groups The main goal in the rest of this chapter is to generalize the results about Fourier series expansions obtained in the special case of Z(N ). After a brief introduction to some notions related to finite abelian groups, we turn to the important concept of a character. In our setting, we find that characters play the same role as the exponentials e0 , . . . , eN −1 on the group Z(N ), and thus provide the key ingredient in the development of the theory on arbitrary finite abelian groups. In fact, it suffices to prove that a finite abelian group has “enough” characters, and this leads automatically to the desired Fourier theory. 2.1 Abelian groups An abelian group (or commutative group) is a set G together with a binary operation on pairs of elements of G, (a, b) 7→ a · b, that satisfies the following conditions: (i) Associativity: a · (b · c) = (a · b) · c for all a, b, c ∈ G. (ii) Identity: There exists an element u ∈ G (often written as either 1 or 0) such that a · u = u · a = a for all a ∈ G. (iii) Inverses: For every a ∈ G, there exists an element a−1 ∈ G such that a · a−1 = a−1 · a = u. (iv) Commutativity: For a, b ∈ G, we have a · b = b · a. We leave as simple verifications the facts that the identity element and inverses are unique. Warning. In the definition of an abelian group, we used the “multiplicative” notation for the operation in G. Sometimes, one uses the “additive” notation a + b and −a, instead of a · b and a−1 . There are times when one notation may be more appropriate than the other, and the examples below illustrate this point. The same group may have different interpretations, one where the multiplicative notation is more suggestive, and another where it is natural to view the group with addition, as the operation.

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Examples of abelian groups • The set of real numbers R with the usual addition. The identity is 0 and the inverse of x is −x. Also, R − {0} and R+ = {x ∈ R : x > 0} equipped, with the standard multiplication, are abelian groups. In both cases the unit is 1 and the inverse of x is 1/x. • With the usual addition, the set of integers Z is an abelian group. However, Z − {0} is not an abelian group with the standard multiplication, since, for example, 2 does not have a multiplicative inverse in Z. In contrast, Q − {0} is an abelian group with the standard multiplication. • The unit circle S 1 in the complex plane. If we view the circle as the set of points {eiθ : θ ∈ R}, the group operation is the standard multiplication of complex numbers. However, if we identify points on S 1 with their angle θ, then S 1 becomes R modulo 2π, where the operation is addition modulo 2π. • Z(N ) is an abelian group. Viewed as the N th roots of unity on the circle, Z(N ) is a group under multiplication of complex numbers. However, if Z(N ) is interpreted as Z/N Z, the integers modulo N , then it is an abelian group where the operation is addition modulo N. • The last example consists of Z∗ (q). This group is defined as the set of all integers modulo q that have multiplicative inverses, with the group operation being multiplication modulo q. This important example is discussed in more detail below. A homomorphism between two abelian groups G and H is a map f : G → H which satisfies the property f (a · b) = f (a) · f (b), where the dot on the left-hand side is the operation in G, and the dot on the right-hand side the operation in H. We say that two groups G and H are isomorphic, and write G ≈ H, if there is a bijective homomorphism from G to H. Equivalently, G and H are isomorphic if there exists another homomorphism f˜ : H → G, so that for all a ∈ G and b ∈ H (f˜ ◦ f )(a) = a

and

(f ◦ f˜)(b) = b.

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Roughly speaking, isomorphic groups describe the “same” object because they have the same underlying group structure (which is really all that matters); however, their particular notational representations might be different. Example 1. A pair of isomorphic abelian groups arose already when we considered the group Z(N ). In one representation it was given as the multiplicative group of N th roots of unity in C. In a second representation it was the additive group Z/N Z of residue classes of integers modulo N . The mapping n 7→ R(n), which associates to a root of unity z = e2πin/N = ζ n the residue class in Z/N Z determined by n, provides an isomorphism between the two different representations. Example 2. In parallel with the previous example, we see that the circle (with multiplication) is isomorphic to the real numbers modulo 2π (with addition). Example 3. The properties of the exponential and logarithm guarantee that exp : R → R+

and

log : R+ → R

are two homomorphisms that are inverses of each other. Thus R (with addition) and R+ (with multiplication) are isomorphic. In what follows, we are primarily interested in abelian groups that are finite. In this case, we denote by |G| the number of elements in G, and call |G| the order of the group. For example, the order of Z(N ) is N . A few additional remarks are in order: • If G1 and G2 are two finite abelian groups, their direct product G1 × G2 is the group whose elements are pairs (g1 , g2 ) with g1 ∈ G1 and g2 ∈ G2 . The operation in G1 × G2 is then defined by (g1 , g2 ) · (g10 , g20 ) = (g1 · g10 , g2 · g20 ). Clearly, if G1 and G2 are finite abelian groups, then so is G1 × G2 . The definition of direct product generalizes immediately to the case of finitely many factors G1 × G2 × · · · × Gn . • The structure theorem for finite abelian groups states that such a group is isomorphic to a direct product of groups of the type Z(N ); see Problem 2. This is a nice result which gives us an overview of the class of all finite abelian groups. However, since we shall not use this theorem below, we omit its proof.

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We now discuss briefly the examples of abelian groups that play a central role in the proof of Dirichlet’s theorem in the next chapter. The group Z∗ (q) Let q be a positive integer. We see that multiplication in Z(q) can be unambiguously defined, because if n is congruent to n0 and m is congruent to m0 (both modulo q), then nm is congruent to n0 m0 modulo q. An integer n ∈ Z(q) is a unit if there exists an integer m ∈ Z(q) so that nm ≡ 1

mod q.

The set of all units in Z(q) is denoted by Z∗ (q), and it is clear from our definition that Z∗ (q) is an abelian group under multiplication modulo q. Thus within the additive group Z(q) lies a set Z∗ (q) that is a group under multiplication. An alternative characterization of Z∗ (q) will be given in the next chapter, as those elements in Z(q) that are relatively prime to q. Example 4. The group of units in Z(4) = {0, 1, 2, 3} is Z∗ (4) = {1, 3}. This reflects the fact that odd integers are divided into two classes depending on whether they are of the form 4k + 1 or 4k + 3. In fact, Z∗ (4) is isomorphic to Z(2). Indeed, we can make the following association: Z∗ (4) Z(2) 1 ←→ 0 3 ←→ 1 and then notice that multiplication in Z∗ (4) corresponds to addition in Z(2). Example 5. The units in Z(5) are Z∗ (5) = {1, 2, 3, 4}. Moreover, Z∗ (5) is isomorphic to Z(4) with the following identification: Z∗ (5) 1 2 3 4

←→ ←→ ←→ ←→

Z(4) 0 1 3 2

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Example 6. The units in Z(8) = {0, 1, 2, 3, 4, 5, 6, 7} are given by Z∗ (8) = {1, 3, 5, 7}. In fact, Z∗ (8) is isomorphic to the direct product Z(2) × Z(2). In this case, an isomorphism between the groups is given by the identification Z∗ (8) 1 3 5 7

←→ ←→ ←→ ←→

Z(2) × Z(2) (0, 0) (1, 0) (0, 1) (1, 1)

2.2 Characters Let G be a finite abelian group (with the multiplicative notation) and S 1 the unit circle in the complex plane. A character on G is a complexvalued function e : G → S 1 which satisfies the following condition: (2)

e(a · b) = e(a)e(b)

for all a, b ∈ G.

In other words, a character is a homomorphism from G to the circle group. The trivial or unit character is defined by e(a) = 1 for all a ∈ G. Characters play an important role in the context of finite Fourier series, primarily because the multiplicative property (2) generalizes the analogous identity for the exponential functions on the circle and the law e` (k + m) = e` (k)e` (m), which held for the exponentials e0 , . . . , eN −1 used in the Fourier theory on Z(N ). There we had e` (k) = ζ `k = e2πi`k/N , with 0 ≤ ` ≤ N − 1 and k ∈ Z(N ), and in fact, the functions e0 , . . . , eN −1 are precisely all the characters of the group Z(N ). ˆ the set of all characters If G is a finite abelian group, we denote by G of G, and observe next that this set inherits the structure of an abelian group.

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ˆ is an abelian group under multiplication defined Lemma 2.1 The set G by (e1 · e2 )(a) = e1 (a)e2 (a)

for all a ∈ G.

The proof of this assertion is straightforward if one observes that the ˆ the dual group trivial character plays the role of the unit. We call G of G. In light of the above analogy between characters for a general abelian group and the exponentials on Z(N ), we gather several more examples of groups and their duals. This provides further evidence of the central role played by characters. (See Exercises 4, 5, and 6.) Example 1. If G = Z(N ), all characters of G take the form e` (k) = ζ `k = e2πi`k/N for some 0 ≤ ` ≤ N − 1, and it is easy to check that e` 7→ ` gives \) to Z(N ). an isomorphism from Z(N Example 2. The dual group of the circle1 is precisely {en }n∈Z (where c1 en (x) = e2πinx ). Moreover, en 7→ n gives an isomorphism between S and the integers Z. Example 3. Characters on R are described by eξ (x) = e2πiξx

where ξ ∈ R.

b to R. Thus eξ 7→ ξ is an isomorphism from R

Example 4. Since exp : R → R+ is an isomorphism, we deduce from the previous example that the characters on R+ are given by eξ (x) = x2πiξ = e2πiξ log x

where ξ ∈ R,

d + is isomorphic to R (or R+ ). and R

The following lemma says that a nowhere vanishing multiplicative function is a character, a result that will be useful later. Lemma 2.2 Let G be a finite abelian group, and e : G → C − {0} a multiplicative function, namely e(a · b) = e(a)e(b) for all a, b ∈ G. Then e is a character. 1 In addition to (2), the definition of a character on an infinite abelian group requires continuity. When G is the circle, R, or R+ , the meaning of “continuous” refers to the standard notion of limit.

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Proof. The group G being finite, the absolute value of e(a) is bounded above and below as a ranges over G. Since |e(bn )| = |e(b)|n , we conclude that |e(b)| = 1 for all b ∈ G. The next step is to verify that the characters form an orthonormal basis of the vector space V of functions over the group G. This fact was obtained directly in the special case G = Z(N ) from the explicit description of the characters e0 , . . . , eN −1 . In the general case, we begin with the orthogonality relations; then we prove that there are “enough” characters by showing that there are as many as the order of the group. 2.3 The orthogonality relations Let V denote the vector space of complex-valued functions defined on the finite abelian group G. Note that the dimension of V is |G|, the order of G. We define a Hermitian inner product on V by (3)

(f, g) =

1 X f (a)g(a), |G| a∈G

whenever f, g ∈ V .

Here the sum is taken over the group and is therefore finite. Theorem 2.3 The characters of G form an orthonormal family with respect to the inner product defined above. Since |e(a)| = 1 for any character, we find that (e, e) =

1 X 1 X e(a)e(a) = |e(a)|2 = 1. |G| a∈G |G| a∈G

If e 6= e0 and both are characters, we must prove that (e, e0 ) = 0; we isolate the key step in a lemma. Lemma 2.4 If e is a non-trivial character of the group G, then P a∈G e(a) = 0. Proof. Choose b ∈ G such that e(b) 6= 1. Then we have e(b)

X a∈G

e(a) =

X a∈G

e(b)e(a) =

X a∈G

e(ab) =

X

e(a).

a∈G

The last equality follows because as a ranges over the group, ab ranges P over G as well. Therefore a∈G e(a) = 0.

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233

We can now conclude the proof of the theorem. Suppose e0 is a character distinct from e. Because e(e0 )−1 is non-trivial, the lemma implies that X e(a)(e0 (a))−1 = 0. a∈G

Since (e0 (a))−1 = e0 (a), the theorem is proved. As a consequence of the theorem, we see that distinct characters are linearly independent. Since the dimension of V over C is |G|, we conclude ˆ is finite and ≤ |G|. The main result to which we now that the order of G ˆ = |G|. turn is that, in fact, |G| 2.4 Characters as a total family The following completes the analogy between characters and the complex exponentials. Theorem 2.5 The characters of a finite abelian group G form a basis for the vector space of functions on G. There are several proofs of this theorem. One consists of using the structure theorem for finite abelian groups we have mentioned earlier, which states that any such group is the direct product of cyclic groups, that is, groups of the type Z(N ). Since cyclic groups are self-dual, using ˆ = |G|, and therefore the characters this fact we would conclude that |G| form a basis for G. (See Problem 3.) Here we shall prove the theorem directly without these considerations. Suppose V is a vector space of dimension d with inner product (·, ·). A linear transformation T : V → V is unitary if it preserves the inner product, (T v, T w) = (v, w) for all v, w ∈ V . The spectral theorem from linear algebra asserts that any unitary transformation is diagonalizable. In other words, there exists a basis {v1 , . . . , vd } (eigenvectors) of V such that T (vi ) = λi vi , where λi ∈ C is the eigenvalue attached to vi . The proof of Theorem 2.5 is based on the following extension of the spectral theorem. Lemma 2.6 Suppose {T1 , . . . , Tk } is a commuting family of unitary transformations on the finite-dimensional inner product space V ; that is, Ti Tj = Tj Ti

for all i, j.

Then T1 , . . . , Tk are simultaneously diagonalizable. In other words, there exists a basis for V which consists of eigenvectors for every Ti , i = 1, . . . , k.

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Proof. We use induction on k. The case k = 1 is simply the spectral theorem. Suppose that the lemma is true for any family of k − 1 commuting unitary transformations. The spectral theorem applied to Tk says that V is the direct sum of its eigenspaces V = Vλ1 ⊕ · · · ⊕ Vλs , where Vλi denotes the subspace of all eigenvectors with eigenvalue λi . We claim that each one of the T1 , . . . , Tk−1 maps each eigenspace Vλi to itself. Indeed, if v ∈ Vλi and 1 ≤ j ≤ k − 1, then Tk Tj (v) = Tj Tk (v) = Tj (λi v) = λi Tj (v) so Tj (v) ∈ Vλi , and the claim is proved. Since the restrictions to Vλi of T1 , . . . , Tk−1 form a family of commuting unitary linear transformations, the induction hypothesis guarantees that these are simultaneously diagonalizable on each subspace Vλi . This diagonalization provides us with the desired basis for each Vλi , and thus for V . We can now prove Theorem 2.5. Recall that the vector space V of complex-valued functions defined on G has dimension |G|. For each a ∈ G we define a linear transformation Ta : V → V by (Ta f )(x) = f (a · x)

for x ∈ G.

Since G is abelian it is clear that Ta Tb = Tb Ta for all a, b ∈ G, and one checks easily that Ta is unitary for the Hermitian inner product (3) defined on V . By Lemma 2.6 the family {Ta }a∈G is simultaneously diagonalizable. This means there is a basis {vb (x)}b∈G for V such that each vb (x) is an eigenfunction for Ta , for every a. Let v be one of these basis elements and 1 the unit element in G. We must have v(1) 6= 0 for otherwise v(a) = v(a · 1) = (Ta v)(1) = λa v(1) = 0, where λa is the eigenvalue of v for Ta . Hence v = 0, and this is a contradiction. We claim that the function defined by w(x) = λx = v(x)/v(1) is a character of G. Arguing as above we find that w(x) 6= 0 for every x, and w(a · b) =

v(a · b) λa v(b) v(1) = = λa λb = λa λb = w(a)w(b). v(1) v(1) v(1)

We now invoke Lemma 2.2 to conclude the proof.

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2.5 Fourier inversion and Plancherel formula We now put together the results obtained in the previous sections to discuss the Fourier expansion of a function on a finite abelian group G. Given a function f on G and character e of G, we define the Fourier coefficient of f with respect to e, by 1 X fˆ(e) = (f, e) = f (a)e(a), |G| a∈G and the Fourier series of f as f∼

X

fˆ(e)e.

ˆ e∈G

Since the characters form a basis, we know that X f= ce e ˆ e∈G

for some set of constants ce . By the orthogonality relations satisfied by the characters, we find that (f, e) = ce , so f is indeed equal to its Fourier series, namely, X f= fˆ(e)e. ˆ e∈G

We summarize our results. Theorem 2.7 Let G be a finite abelian group. The characters of G form an orthonormal basis for the vector space V of functions on G equipped with the inner product (f, g) =

1 X f (a)g(a). |G| a∈G

In particular, any function f on G is equal to its Fourier series X fˆ(e)e. f= ˆ e∈G

Finally, we have the Parseval-Plancherel formula for finite abelian groups.

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Theorem 2.8 If f is a function on G, then kf k2 =

X

|fˆ(e)|2 .

ˆ e∈G

Proof. Since the characters of G form an orthonormal basis for the vector space V , and (f, e) = fˆ(e), we have that X

kf k2 = (f, f ) =

X

(f, e)fˆ(e) =

ˆ e∈G

|fˆ(e)|2 .

ˆ e∈G

The apparent difference of this statement with that of Theorem 1.2 is due to the different normalizations of the Fourier coefficients that are used.

3 Exercises 1. Let f be a function on the circle. For each N ≥ 1 the discrete Fourier coefficients of f are defined by aN (n) =

N 1 X f (e2πik/N )e−2πikn/N , N

for n ∈ Z.

k=1

We also let

Z

1

f (e2πix )e−2πinx dx

a(n) = 0

denote the ordinary Fourier coefficients of f . (a) Show that aN (n) = aN (n + N ). (b) Prove that if f is continuous, then aN (n) → a(n) as N → ∞. 2. If f is a C 1 function on the circle, prove that |aN (n)| ≤ c/|n| whenever 0 < |n| ≤ N/2. [Hint: Write aN (n)[1 − e2πi`n/N ] =

N 1 X [f (e2πik/N ) − f (e2πi(k+`)/N )]e−2πikn/N , N k=1

and choose ` so that `n/N is nearly 1/2.] 3. By a similar method, show that if f is a C 2 function on the circle, then |aN (n)| ≤ c/|n|2 ,

whenever 0 < |n| ≤ N/2.

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As a result, prove the inversion formula for f ∈ C 2 , f (e2πix ) =

∞ X

a(n)e2πinx

n=−∞

from its finite version. [Hint: For the first part, use the second symmetric difference f (e2πi(k+`)/N ) + f (e2πi(k−`)/N ) − 2f (e2πik/N ). For the second part, if N is odd (say), write the inversion formula as f (e2πik/N ) =

X

aN (n)e2πikn/N .]

|n| 1. 10. A group G is cyclic if there exists g ∈ G that generates all of G, that is, if any element in G can be written as g n for some n ∈ Z. Prove that a finite abelian group is cyclic if and only if it is isomorphic to Z(N ) for some N . 11. Write down the multiplicative tables for the groups Z∗ (3), Z∗ (4), Z∗ (5), Z∗ (6), Z∗ (8), and Z∗ (9). Which of these groups are cyclic? 12. Suppose that G is a finite abelian group and e : G → C is a function that satisfies e(x · y) = e(x)e(y) for all x, y ∈ G. Prove that either e is identically 0, or e never vanishes. In the second case, show that for each x, e(x) = e2πir for some r ∈ Q of the form r = p/q, where q = |G|.

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4. Problems

13. In analogy with ordinary Fourier series, one may interpret finite Fourier expansions using convolutions as follows. Suppose G is a finite abelian group, 1G its unit, and V the vector space of complex-valued functions on G. (a) The convolution of two functions f and g in V is defined for each a ∈ G by (f ∗ g)(a) =

1 X f (b)g(a · b−1 ). |G| b∈G

\ ˆ one has (f Show that for all e ∈ G ∗ g)(e) = fˆ(e)ˆ g (e). (b) Use Theorem 2.5 to show that if e is a character on G, then

X

e(c) = 0

whenever c ∈ G and c 6= 1G .

ˆ e∈G

(c) As a result of (b), show that the Fourier series Sf (a) = a function f ∈ V takes the form

P ˆ e∈G

fˆ(e)e(a) of

Sf = f ∗ D, where D is defined by (4)

D(c) =

X

½ e(c) =

ˆ e∈G

|G| 0

if c = 1G , otherwise.

Since f ∗ D = f , we recover the fact that Sf = f . Loosely speaking, D corresponds to a “Dirac delta function”; it has unit mass 1 X D(c) = 1, |G| c∈G

and (4) says that this mass is concentrated at the unit element in G. Thus D has the same interpretation as the “limit” of a family of good kernels. (See Section 4, Chapter 2.) Note. The function D reappears in the next chapter as δ1 (n).

4 Problems 1. Prove that if n and m are two positive integers that are relatively prime, then Z(nm) ≈ Z(n) × Z(m).

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Chapter 7. FINITE FOURIER ANALYSIS

[Hint: Consider the map Z(nm) → Z(n) × Z(m) given by k 7→ (k mod n, k mod m), and use the fact that there exist integers x and y such that xn + ym = 1.] 2.∗ Every finite abelian group G is isomorphic to a direct product of cyclic groups. Here are two more precise formulations of this theorem. • If p1 , . . . , ps are the distinct primes appearing in the factorization of the order of G, then G ≈ G(p1 ) × · · · × G(ps ), where each G(p) is of the form G(p) = Z(pr1 ) × · · · × Z(pr` ), with 0 ≤ r1 ≤ · · · ≤ r` (this sequence of integers depends on p of course). This decomposition is unique. • There exist unique integers d1 , . . . , dk such that d1 |d2 , d2 |d3 , · · · , dk−1 |dk and G ≈ Z(d1 ) × · · · × Z(dk ). Deduce the second formulation from the first. ˆ denote the collection of distinct characters of the finite abelian group 3. Let G G. ˆ is isomorphic to G. (a) Note that if G = Z(N ), then G ˆ ˆ (b) Prove that G\ 1 × G2 = G1 × G2 . ˆ is iso(c) Prove using Problem 2 that if G is a finite abelian group, then G morphic to G. 4.∗ When p is prime the group Z∗ (p) is cyclic, and Z∗ (p) ≈ Z(p − 1).

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8 Dirichlet’s Theorem Dirichlet, Gustav Lejeune (D¨ uren 1805-G¨ ottingen 1859), German mathematician. He was a number theorist at heart. But, while studying in Paris, being a very likeable person, he was befriended by Fourier and other like-minded mathematicians, and he learned analysis from them. Thus equipped, he was able to lay the foundation for the application of Fourier analysis to (analytic) theory of numbers. S. Bochner, 1966

As a striking application of the theory of finite Fourier series, we now prove Dirichlet’s theorem on primes in arithmetic progression. This theorem states that if q and ` are positive integers with no common factor, then the progression `, ` + q, ` + 2q, ` + 3q, . . . , ` + kq, . . . contains infinitely many prime numbers. This change of subject matter that we undertake illustrates the wide applicability of ideas from Fourier analysis to various areas outside its seemingly narrower confines. In this particular case, it is the theory of Fourier series on the finite abelian group Z∗ (q) that plays a key role in the solution of the problem.

1 A little elementary number theory We begin by introducing the requisite background. This involves elementary ideas of divisibility of integers, and in particular properties regarding prime numbers. Here the basic fact, called the fundamental theorem of arithmetic, is that every integer is the product of primes in an essentially unique way. 1.1 The fundamental theorem of arithmetic The following theorem is a mathematical formulation of long division.

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Theorem 1.1 (Euclid’s algorithm) For any integers a and b with b > 0, there exist unique integers q and r with 0 ≤ r < b such that a = qb + r. Here q denotes the quotient of a by b, and r is the remainder, which is smaller than b. Proof. First we prove the existence of q and r. Let S denote the set of all non-negative integers of the form a − qb with q ∈ Z. This set is non-empty and in fact S contains arbitrarily large positive integers since b 6= 0. Let r denote the smallest element in S, so that r = a − qb for some integer q. By construction 0 ≤ r, and we claim that r < b. If not, we may write r = b + s with 0 ≤ s < r, so b + s = a − qb, which then implies s = a − (q + 1)b. Hence s ∈ S with s < r, and this contradicts the minimality of r. So r < b, hence q and r satisfy the conditions of the theorem. To prove uniqueness, suppose we also had a = q1 b + r1 where 0 ≤ r1 < b. By subtraction we find (q − q1 )b = r1 − r. The left-hand side has absolute value 0 or ≥ b, while the right-hand side has absolute value < b. Hence both sides of the equation must be 0, which gives q = q1 and r = r1 . An integer a divides b if there exists another integer c such that ac = b; we then write a|b and say that a is a divisor of b. Note that in particular 1 divides every integer, and a|a for all integers a. A prime number is a positive integer greater than 1 that has no positive divisors besides 1 and itself. The main theorem in this section says that any positive integer can be written uniquely as the product of prime numbers. The greatest common divisor of two positive integers a and b is the largest integer that divides both a and b. We usually denote the greatest common divisor by gcd(a, b). Two positive integers are relatively prime if their greatest common divisor is 1. In other words, 1 is the only positive divisor common to both a and b.

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243

Theorem 1.2 If gcd(a, b) = d, then there exist integers x and y such that ax + by = d. Proof. Consider the set S of all positive integers of the form ax + by where x, y ∈ Z, and let s be the smallest element in S. We claim that s = d. By construction, there exist integers x and y such that ax + by = s. Clearly, any divisor of a and b divides s, so we must have d ≤ s. The proof will be complete if we can show that s|a and s|b. By Euclid’s algorithm, we can write a = qs + r with 0 ≤ r < s. Multiplying the above by q we find qax + qby = qs, and therefore qax + qby = a − r. Hence r = a(1 − qx) + b(−qy). Since s was minimal in S and 0 ≤ r < s, we conclude that r = 0, therefore s divides a. A similar argument shows that s divides b, hence s = d as desired. In particular we record the following three consequences of the theorem. Corollary 1.3 Two positive integers a and b are relatively prime if and only if there exist integers x and y such that ax + by = 1. Proof. If a and b are relatively prime, two integers x and y with the desired property exist by Theorem 1.2. Conversely, if ax + by = 1 holds and d is positive and divides both a and b, then d divides 1, hence d = 1.

Corollary 1.4 If a and c are relatively prime and c divides ab, then c divides b. In particular, if p is a prime that does not divide a and p divides ab, then p divides b. Proof. We can write 1 = ax + cy, so multiplying by b we find b = abx + cby. Hence c|b. Corollary 1.5 If p is prime and p divides the product a1 · · · ar , then p divides ai for some i. Proof. By the previous corollary, if p does not divide a1 , then p divides a2 · · · ar , so eventually p|ai . We can now prove the main result of this section.

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Chapter 8. DIRICHLET’S THEOREM

Theorem 1.6 Every positive integer greater than 1 can be factored uniquely into a product of primes. Proof. First, we show that such a factorization is possible. We do so by proving that the set S of positive integers > 1 which do not have a factorization into primes is empty. Arguing by contradiction, we assume that S 6= ∅. Let n be the smallest element of S. Since n cannot be a prime, there exist integers a > 1 and b > 1 such that ab = n. But then a < n and b < n, so a ∈ / S as well as b ∈ / S. Hence both a and b have prime factorizations and so does their product n. This implies n∈ / S, therefore S is empty, as desired. We now turn our attention to the uniqueness of the factorization. Suppose that n has two factorizations into primes n = p1 p2 · · · pr = q1 q2 · · · qs . So p1 divides q1 q2 · · · qs , and we can apply Corollary 1.5 to conclude that p1 |qi for some i. Since qi is prime, we must have p1 = qi . Continuing with this argument we find that the two factorizations of n are equal up to a permutation of the factors. We briefly digress to give an alternate definition of the group Z∗ (q) which appeared in the previous chapter. According to our initial definition, Z∗ (q) is the multiplicative group of units in Z(q): those n ∈ Z(q) for which there exists an integer m so that (1)

nm ≡ 1 mod q.

Equivalently, Z∗ (q) is the group under multiplication of all integers in Z(q) that are relatively prime to q. Indeed, notice that if (1) is satisfied, then automatically n and q are relatively prime. Conversely, suppose we assume that n and q are relatively prime. Then, if we put a = n and b = q in Corollary 1.3, we find nx + qy = 1. Hence nx ≡ 1 mod q, and we can take m = x to establish the equivalence. 1.2 The infinitude of primes The study of prime numbers has always been a central topic in arithmetic, and the first fundamental problem that arose was to determine whether

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245

there are infinitely many primes or not. This problem was solved in Euclid’s Elements with a simple and very elegant argument. Theorem 1.7 There are infinitely many primes. Proof. Suppose not, and denote by p1 , . . . , pn the complete set of primes. Define N = p1 p2 · · · pn + 1. Since N is larger than any pi , the integer N cannot be prime. Therefore, N is divisible by a prime that belongs to our list. But this is also an absurdity since every prime divides the product, yet no prime divides 1. This contradiction concludes the proof. Euclid’s argument actually can be modified to deduce finer results about the infinitude of primes. To see this, consider the following problem. Prime numbers (except for 2) can be divided into two classes depending on whether they are of the form 4k + 1 or 4k + 3, and the above theorem says that at least one of these classes has to be infinite. A natural question is to ask whether both classes are infinite, and if not, which one is? In the case of primes of the form 4k + 3, the fact that the class is infinite has a proof that is similar to Euclid’s, but with a twist. If there are only finitely many such primes, enumerate them in increasing order omitting 3, p1 = 7, p2 = 11, . . . , pn , and let N = 4p1 p2 · · · pn + 3. Clearly, N is of the form 4k + 3 and cannot be prime since N > pn . Since the product of two numbers of the form 4m + 1 is again of the form 4m + 1, one of the prime divisors of N , say p, must be of the form 4k + 3. We must have p 6= 3, since 3 does not divide the product in the definition of N . Also, p cannot be one of the other primes of the form 4k + 3, that is, p 6= pi for i = 1, . . . n, because then p divides the product p1 · · · pn but does not divide 3. It remains to determine if the class of primes of the form 4k + 1 is infinite. A simple-minded modification of the above argument does not work since the product of two numbers of the form 4m + 3 is never of the form 4m + 3. More generally, in an attempt to prove the law of quadratic reciprocity, Legendre formulated the following statement:

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Chapter 8. DIRICHLET’S THEOREM

If q and ` are relatively prime, then the sequence ` + kq,

k∈Z

contains infinitely many primes (hence at least one prime!). Of course, the condition that q and ` be relatively prime is necessary, for otherwise ` + kq is never prime. In other words, this hypothesis says that any arithmetic progression that could contain primes necessarily contains infinitely many of them. Legendre’s assertion was proved by Dirichlet. The key idea in his proof is Euler’s analytical approach to prime numbers involving his product formula, which gives a strengthened version of Theorem 1.7. This insight of Euler led to a deep connection between the theory of primes and analysis. The zeta function and its Euler product We begin with a rapid review of infinite products. If {An }∞ n=1 is a sequence of real numbers, we define ∞ Y

An = lim

n=1

N →∞

N Y

An

n=1

if the limit exists, in which case we say that the product converges. The natural approach is to take logarithms and transform products into sums. We gather in a lemma the properties we shall need of the function log x, defined for positive real numbers. Lemma 1.8 The exponential and logarithm functions satisfy the following properties: (i) elog x = x. (ii) log(1 + x) = x + E(x) where |E(x)| ≤ x2 if |x| < 1/2. (iii) If log(1 + x) = y and |x| < 1/2, then |y| ≤ 2|x|. In terms of the O notation, property (ii) will be recorded as log(1 + x) = x + O(x2 ). Proof. Property (i) is standard. To prove property (ii) we use the power series expansion of log(1 + x) for |x| < 1, that is, (2)

log(1 + x) =

∞ X (−1)n+1 n=1

n

xn .

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1. A little elementary number theory

Then we have E(x) = log(1 + x) − x = −

x2 x3 x4 + − + ··· , 2 3 4

and the triangle inequality implies |E(x)| ≤

¢ x2 ¡ 1 + |x| + |x|2 + · · · . 2

Therefore, if |x| ≤ 1/2 we can sum the geometric series on the right-hand side to find that µ ¶ x2 1 1 |E(x)| ≤ 1 + + 2 + ··· 2 2 2 µ ¶ 2 x 1 ≤ 2 1 − 1/2 ≤ x2 . The proof of property (iii) is now immediate; if x 6= 0 and |x| ≤ 1/2, then ¯ ¯ ¯ ¯ ¯ log(1 + x) ¯ ¯ ¯ ¯ ¯ ≤ 1 + ¯ E(x) ¯ ¯ ¯ ¯ x x ¯ ≤ 1 + |x| ≤ 2, and if x = 0, (iii) is clearly also true. We can now prove the main result on infinite products of real numbers. P Proposition 1.9 If An = 1 + an and |an | converges, then the prodQ uct n An converges, and this product vanishes if and only if one of Q its factors An vanishes. Also, if an 6= 1 for all n, then n 1/(1 − an ) converges. P Proof. If |an | converges, then for all large n we must have |an | < 1/2. Disregarding finitely many terms if necessary, we may assume that this inequality holds for all n. Then we may write the partial products as follows: N Y n=1

An =

N Y

elog(1+an ) = eBN ,

n=1

P where BN = N n=1 bn with bn = log(1 + an ). By the lemma, we know that |bn | ≤ 2|an |, so that BN converges to a real number, say B. Since

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Chapter 8. DIRICHLET’S THEOREM

the exponential function is continuous, we conclude that eBN converges to eB as N goes to infinity, proving the first assertion of the proposition. Observe also that if 1 + an 6= 0 for all n, the product converges to a non-zero limit since it is expressed as eB . Q Finally observe that the partial products of n 1/(1 − an ) are QN 1/ n=1 (1 − an ), so the same argument as above proves that the product in the denominator converges to a non-zero limit. With these preliminaries behind us, we can now return to the heart of the matter. For s a real number (strictly) greater than 1, we define the zeta function by ζ(s) =

∞ X 1 n=1

ns

.

To see that the series defining ζ converges, we use the principle that P f (n) with Rwhenever f is a decreasing function one can compare f (x) dx, as is suggested by Figure 1. Note also that a similar technique was used in Chapter 3, that time bounding a sum from below by an integral.

y = f (x)

f (n) n−1

n

Figure 1. Comparing sums with integrals

Here we take f (x) = 1/xs to see that ∞ X 1 n=1

ns

≤1+

∞ Z X

n

n=2 n−1

dx =1+ xs

Z

∞ 1

dx , xs

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1. A little elementary number theory

and therefore, (3)

ζ(s) ≤ 1 +

1 . s−1

Clearly, the series defining ζ converges uniformly on each half-line s > s0 > 1, hence ζ is continuous when s > 1. The zeta function was already mentioned earlier in the discussion of the Poisson summation formula and the theta function. The key result is Euler’s product formula. Theorem 1.10 For every s > 1, we have ζ(s) =

Y p

1 , 1 − 1/ps

where the product is taken over all primes. It is important to remark that this identity is an analytic expression of the fundamental theorem of arithmetic. In fact, each factor of the product 1/(1 − p−s ) can be written as a convergent geometric series 1+

1 1 1 + 2s + · · · + M s + · · · . s p p p

So we consider Y pj

Ã

1 1 1 1 + s + 2s + · · · + M s + · · · pj pj pj

!

,

where the product is taken over all primes, which we order in increasing order p1 < p2 < · · · . Proceeding formally (these manipulations will be justified below), we calculate the product as a sum of terms, each term originating by picking out a term 1/pks j (in the sum corresponding to pj ) with a k, which of course will depend on j, and with k = 0 for j sufficiently large. The product obtained this way is (pk11 pk22

1 1 = s, km s n · · · pm )

where the integer n is written as a product of primes n = pk11 pk22 · · · pkmm . By the fundamental theorem of arithmetic, each integer ≥ 1 occurs in this way uniquely, hence the product equals ∞ X 1 n=1

ns

.

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Chapter 8. DIRICHLET’S THEOREM

We now justify this heuristic argument. Proof. Suppose M and N are positive integers with M > N . Observe now that any positive integer n ≤ N can be written uniquely as a product of primes, and that each prime must be less than or equal to N and repeated less than M times. Therefore N X 1

Y µ

1 1 1 + 2s + · · · + M s s s n p p p n=1 p≤N µ ¶ Y 1 ≤ 1 − p−s p≤N µ ¶ Y 1 ≤ . 1 − p−s p ≤

¶

1+

Letting N tend to infinity now yields ∞ X 1 n=1

ns

≤

Yµ p

1 1 − p−s

¶

.

For the reverse inequality, we argue as follows. Again, by the fundamental theorem of arithmetic, we find that Y µ p≤N

1 1 1 1 + s + 2s + · · · + M s p p p

¶

≤

∞ X 1 n=1

ns

.

Letting M tend to infinity gives Y µ p≤N

1 1 − p−s

¶

≤

∞ X 1 n=1

ns

.

Hence Yµ p

1 1 − p−s

¶

≤

∞ X 1 n=1

ns

,

and the proof of the product formula is complete. We now come to Euler’s version of Theorem 1.7, which inspired Dirichlet’s approach to the general problem of primes in arithmetic progression. The point is the following proposition.

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1. A little elementary number theory

Proposition 1.11 The series X

1/p

p

diverges, when the sum is taken over all primes p. Of course, if there were only finitely many primes the series would converge automatically. Proof. We take logarithms of both sides of the Euler formula. Since log x is continuous, we may write the logarithm of the infinite product as the sum of the logarithms. Therefore, we obtain for s > 1 −

X

log(1 − 1/ps ) = log ζ(s).

p

Since log(1 + x) = x + O(|x|2 ) whenever |x| ≤ 1/2, we get −

X£

¤ −1/ps + O(1/p2s ) = log ζ(s),

p

which gives X

1/ps + O(1) = log ζ(s).

p

P P 2 The term O(1) appears because p 1/p2s ≤ ∞ n=1 1/n . Now we let s + tend to 1 from above, namely s → 1 , and note that ζ(s) → ∞ since P∞ PM s s n=1 1/n ≥ n=1 1/n , and therefore

lim inf + s→1

∞ X

1/ns ≥

n=1

M X

1/n

for every M .

n=1

P We conclude that p 1/ps → ∞ as s → 1+ , and since 1/p > 1/ps for all s > 1, we finally have that X

1/p = ∞.

p

In the rest of this chapter we see how Dirichlet adapted Euler’s insight.

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Chapter 8. DIRICHLET’S THEOREM

2 Dirichlet’s theorem We remind the reader of our goal: Theorem 2.1 If q and ` are relatively prime positive integers, then there are infinitely many primes of the form ` + kq with k ∈ Z. Following Euler’s argument, Dirichlet proved this theorem by showing that the series X 1 p p≡` mod q diverges, where the sum is over all primes congruent to ` modulo q. Once q is fixed and no confusion is possible, we write p ≡ ` to denote a prime congruent to ` modulo q. The proof consists of several steps, one of which requires Fourier analysis on the group Z∗ (q). Before proceeding with the theorem in its complete generality, we outline the solution to the particular problem raised earlier: are there infinitely many primes of the form 4k + 1? This example, which consists of the special case q = 4 and ` = 1, illustrates all the important steps in the proof of Dirichlet’s theorem. We begin with the character on Z∗ (4) defined by χ(1) = 1 and χ(3) = −1. We extend this character to all of Z as follows:  

0 if n is even, 1 if n = 4k + 1, χ(n) =  −1 if n = 4k + 3. Note that this function is multiplicative, that is, χ(nm) = χ(n)χ(m) on P s all of Z. Let L(s, χ) = ∞ n=1 χ(n)/n , so that L(s, χ) = 1 −

1 1 1 + s − s + ··· . s 3 5 7

Then L(1, χ) is the convergent series given by 1−

1 1 1 + − + ··· . 3 5 7

Since the terms in the series are alternating and their absolute values decrease to zero we have L(1, χ) 6= 0. Because χ is multiplicative, the Euler product generalizes (as we will prove later) to give ∞ X χ(n) n=1

ns

=

Y p

1 . 1 − χ(p)/ps

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2. Dirichlet’s theorem

Taking the logarithm of both sides, we find that log L(s, χ) =

X χ(p) p

ps

+ O(1).

Letting s → 1+ , the observation that L(1, χ) 6= 0 shows that remains bounded. Hence X 1 X 1 − s p ps p≡3 p≡1

P p

χ(p)/ps

is bounded as s → 1+ . However, we know from Proposition 1.11 that X 1 ps p is unbounded as s → 1+ , so putting these two facts together, we find that X 1 2 ps p≡1 P is unbounded as s → 1+ . Hence p≡1 1/p diverges, and as a consequence there are infinitely many primes of the form 4k + 1.

We digress briefly to show that in fact L(1, χ) = π/4. To see this, we integrate the identity 1 = 1 − x2 + x4 − x6 + · · · , 1 + x2 and get

Z

y 0

dx y3 y5 = y − + − ··· , 1 + x2 3 5

0 < y < 1.

We then let y tend to 1. The integral can be calculated as Z 1 dx π 1 = arctan u|0 = , 2 1 + x 4 0 so this proves that the series 1 − 1/3 + 1/5 − · · · is Abel summable to π/4. Since we know the series converges, its limit is the same as its Abel limit, hence 1 − 1/3 + 1/5 − · · · = π/4. The rest of this chapter gives the full proof of Dirichlet’s theorem. We begin with the Fourier analysis (which is actually the last step in the example given above), and reduce the theorem to the non-vanishing of L-functions.

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Chapter 8. DIRICHLET’S THEOREM

2.1 Fourier analysis, Dirichlet characters, and reduction of the theorem In what follows we take the abelian group G to be Z∗ (q). Our formulas below involve the order of G, which is the number of integers 0 ≤ n < q that are relatively prime to q; this number defines the Euler phifunction ϕ(q), and |G| = ϕ(q). Consider the function δ` on G, which we think of as the characteristic function of `; if n ∈ Z∗ (q), then ½ 1 if n ≡ ` mod q, δ` (n) = 0 otherwise. We can expand this function in a Fourier series as follows: X δ` (n) = δb` (e)e(n), ˆ e∈G

where

1 1 X e(`). δb` (e) = δ` (m)e(m) = |G| m∈G |G|

Hence δ` (n) =

1 X e(`)e(n). |G| ˆ e∈G

We can extend the function δ` to all of Z by setting δ` (m) = 0 whenever m and q are not relatively prime. Similarly, the extensions of the characters ˆ to all of Z which are given by the recipe e∈G ½ e(m) if m and q are relatively prime χ(m) = 0 otherwise, are called the Dirichlet characters modulo q. We shall denote the extension to Z of the trivial character of G by χ0 , so that χ0 (m) = 1 if m and q are relatively prime, and 0 otherwise. Note that the Dirichlet characters modulo q are multiplicative on all of Z, in the sense that χ(nm) = χ(n)χ(m)

for all n, m ∈ Z.

Since the integer q is fixed, we may without fear of confusion, speak of “Dirichlet characters” omitting reference to q.1 With |G| = ϕ(q), we may restate the above results as follows: 1 We use the notation χ instead of e to distinguish the Dirichlet characters (defined on Z) from the characters e (defined on Z∗ (q)).

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2. Dirichlet’s theorem

Lemma 2.2 The Dirichlet characters are multiplicative. Moreover, 1 X δ` (m) = χ(`)χ(m), ϕ(q) χ where the sum is over all Dirichlet characters. With the above lemma we have taken our first step towards a proof of the theorem, since this lemma shows that X 1 p≡`

ps

=

X δ` (p)

ps

p

=

X χ(p) 1 X χ(`) . ϕ(q) χ ps p

P Thus it suffices to understand the behavior of p χ(p)p−s as s → 1+ . In fact, we divide the above sum in two parts depending on whether or not χ is trivial. So we have X 1 p≡`

(4)

ps

=

X χ(p) 1 X χ0 (p) 1 X χ(`) + ϕ(q) p ps ϕ(q) χ6=χ ps p 0

X X χ(p) 1 1 1 X = + χ(`) . s ϕ(q) p not dividing q p ϕ(q) χ6=χ ps p 0

Since there are only finitely many primes dividing q, Euler’s theorem (Proposition 1.11) implies that the first sum on the right-hand side diverges when s tends to 1. These observations show that Dirichlet’s theorem is a consequence of the following assertion. Theorem 2.3 If χ is a nontrivial Dirichlet character, then the sum X χ(p)

ps

p

remains bounded as s → 1+ . The proof of Theorem 2.3 requires the introduction of the L-functions, to which we now turn. 2.2 Dirichlet L-functions We proved earlier that the zeta function ζ(s) = pressed as a product, namely ∞ X 1

ns n=1

=

Y p

1 . (1 − p−s )

P n

1/ns could be ex-

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Chapter 8. DIRICHLET’S THEOREM

Dirichlet observed an analogue of this formula for the so-called L-functions defined for s > 1 by L(s, χ) =

∞ X χ(n) n=1

ns

,

where χ is a Dirichlet character. Theorem 2.4 If s > 1, then ∞ X χ(n) n=1

ns

=

Y

1 , (1 − χ(p)p−s )

p

where the product is over all primes. Assuming this theorem for now, we can follow Euler’s argument formally: taking the logarithm of the product and using the fact that log(1 + x) = x + O(x2 ) whenever x is small, we would get log L(s, χ) = −

X

log(1 − χ(p)/ps )

p

=−

X · χ(p)

−

ps

p

=

X χ(p)

ps

p

µ

+O

1 p2s

¶¸

+ O(1).

If L(1, χ) is finite and non-zero, then log L(s, χ) is bounded as s → 1+ , and we can conclude that the sum X χ(p) p

ps

is bounded as s → 1+ . We now make several observations about the above formal argument. First, we must prove the product formula in Theorem 2.4. Since the Dirichlet characters χ can be complex-valued we will extend the logarithm to complex numbers w of the form w = 1/(1 − z) with |z| < 1. (This will be done in terms of a power series.) Then we show that with this definition of the logarithm, the proof of Euler’s product formula given earlier carries over to L-functions. Second, we must make sense of taking the logarithm of both sides of the product formula. If the Dirichlet characters are real, this argument works

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2. Dirichlet’s theorem

and is precisely the one given in the example corresponding to primes of the form 4k + 1. In general, the difficulty lies in the fact that χ(p) is a complex number, and the complex logarithm is not single valued; in particular, the logarithm of a product is not the sum of the logarithms. Third, it remains to prove that whenever χ 6= χ0 , then log L(s, χ) is bounded as s → 1+ . If (as we shall see) L(s, χ) is continuous at s = 1, then it suffices to show that L(1, χ) 6= 0. This is the non-vanishing we mentioned earlier, which corresponds to the alternating series being non-zero in the previous example. The fact that L(1, χ) 6= 0 is the most difficult part of the argument. So we will focus on three points: 1. Complex logarithms and infinite products. 2. Study of L(s, χ). 3. Proof that L(1, χ) 6= 0 if χ is non-trivial. However, before we enter further into the details, we pause briefly to discuss some historical facts surrounding Dirichlet’s theorem. Historical digression In the following list, we have gathered the names of those mathematicians whose work dealt most closely with the series of achievements related to Dirichlet’s theorem. To give a better perspective, we attach the years in which they reached the age of 35: Euler 1742 Legendre 1787 Gauss 1812 Dirichlet 1840 Riemann 1861 As we mentioned earlier, Euler’s discovery of the product formula for the zeta function is the starting point in Dirichlet’s argument. Legendre in effect conjectured the theorem because he needed it in his proof of the law of quadratic reciprocity. However, this goal was first accomplished by Gauss who, while not knowing how to establish the theorem about primes in arithmetic progression, nevertheless found a number of different proofs of quadratic reciprocity. Later, Riemann extended the study of the zeta function to the complex plane and indicated how properties

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Chapter 8. DIRICHLET’S THEOREM

related to the non-vanishing of that function were central in the further understanding of the distribution of prime numbers. Dirichlet proved his theorem in 1837. It should be noted that Fourier, who had befriended Dirichlet when the latter was a young mathematician visiting Paris, had died several years before. Besides the great activity in mathematics, that period was also a very fertile time in the arts, and in particular music. The era of Beethoven had ended only ten years earlier, and Schumann was now reaching the heights of his creativity. But the musician whose career was closest to Dirichlet was Felix Mendelssohn (four years his junior). It so happens that the latter began composing his famous violin concerto the year after Dirichlet succeeded in proving his theorem.

3 Proof of the theorem We return to the proof of Dirichlet’s theorem and to the three difficulties mentioned above. 3.1 Logarithms The device to deal with the first point is to define two logarithms, one for complex numbers of the form 1/(1 − z) with |z| < 1 which we denote by log1 , and one for the function L(s, χ) which we will denote by log2 . For the first logarithm, we define µ ¶ X ∞ 1 zk log1 = 1−z k k=1

for |z| < 1.

Note that log1 w is then defined if Re(w) > 1/2, and because of equation (2), log1 w gives an extension of the usual log x when x is a real number > 1/2. Proposition 3.1 The logarithm function log1 satisfies the following properties: (i) If |z| < 1, then elog1 ( 1−z ) = 1

(ii) If |z| < 1, then

µ

log1

1 1−z

1 . 1−z

¶

= z + E1 (z),

where the error E1 satisfies |E1 (z)| ≤ |z|2 if |z| < 1/2.

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3. Proof of the theorem

(iii) If |z| < 1/2, then ¯ ¶¯ µ ¯ ¯ 1 ¯log ¯ ¯ 1 1 − z ¯ ≤ 2|z|.

Proof. To establish the first property, let z = reiθ with 0 ≤ r < 1, and observe that it suffices to show that P∞ iθ k (5) (1 − reiθ ) e k=1 (re ) /k = 1. To do so, we differentiate the left-hand side with respect to r, and this gives " Ã∞ !0 # P∞ X iθ k iθ iθ iθ k −e + (1 − re ) (re ) /k e k=1 (re ) /k . k=1

The term in brackets equals à iθ

iθ

iθ

−e + (1 − re )e

∞ X

! iθ k−1

(re )

= −eiθ + (1 − reiθ )eiθ

k=1

1 = 0. 1 − reiθ

Having found that the left-hand side of the equation (5) is constant, we set r = 0 and get the desired result. The proofs of the second and third properties are the same as their real counterparts given in Lemma 1.8. Using these results we can state a sufficient condition guaranteeing the convergence of infinite products of complex numbers. Its proof is the same as in the real case, except that we now use the logarithm log1 . P Proposition 3.2 If |an | converges, and an 6= 1 for all n, then ∞ µ Y n=1

1 1 − an

¶

converges. Moreover, this product is non-zero. Proof. For n large enough, |an | < 1/2, so we may assume without loss of generality that this inequality holds for all n ≥ 1. Then N µ Y n=1

1 1 − an

¶

=

N Y n=1

PN 1 1 elog1 ( 1−an ) = e n=1 log1 ( 1−an ) .

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But we know from the previous proposition that ¯ µ ¶¯ ¯ ¯ 1 ¯log ¯ ¯ 1 1 − z ¯ ≤ 2|z|, P so the fact that the series |an | converges, immediately implies that the limit ¶ µ N X 1 =A lim log1 N →∞ 1 − an n=1 exists. Since the exponential function is continuous, we conclude that the product converges to eA , which is clearly non-zero. We may now prove the promised Dirichlet product formula X χ(n) n

=

ns

Y p

1 . (1 − χ(p)p−s )

For simplicity of notation, let L denote the left-hand side of the above equation. Define ¶ X Y µ 1 −s SN = χ(n)n and ΠN = . 1 − χ(p)p−s n≤N p≤N ´ Q ³ 1 The infinite product Π = limN →∞ ΠN = p 1−χ(p)p converges by −s −s the previous proposition. Indeed, if we set an = χ(pn )pn , where pn is P the nth prime, we note that if s > 1, then |an | < ∞. Also, define

ΠN,M =

Y µ p≤N

¶ χ(pM ) χ(p) 1 + s + · · · + Ms . p p

Now fix ² > 0 and choose N so large that |SN − L| < ²

and

|ΠN − Π| < ².

We can next select M large enough so that |SN − ΠN,M | < ²

and

|ΠN,M − ΠN | < ².

To see the first inequality, one uses the fundamental theorem of arithmetic and the fact that the Dirichlet characters are multiplicative. The

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3. Proof of the theorem

second inequality follows merely because each series verges. Therefore

P∞ n=1

χ(pn ) pns

con-

|L − Π| ≤ |L − SN | + |SN − ΠN,M | + |ΠN,M − ΠN | + |ΠN − Π| < 4², as was to be shown. 3.2 L-functions The next step is a better understanding of the L-functions. Their behavior as functions of s (especially near s = 1) depends on whether or not χ is trivial. In the first case, L(s, χ0 ) is up to some simple factors just the zeta function. Proposition 3.3 Suppose χ0 is the trivial Dirichlet character, ½ 1 if n and q are relatively prime, χ0 (n) = 0 otherwise, and q = pa1 1 · · · paNN is the prime factorization of q. Then −s −s L(s, χ0 ) = (1 − p−s 1 )(1 − p2 ) · · · (1 − pN )ζ(s).

Therefore L(s, χ0 ) → ∞ as s → 1+ . Proof. The identity follows at once on comparing the Dirichlet and Euler product formulas. The final statement holds because ζ(s) → ∞ as s → 1+ . The behavior of the remaining L-functions, those for which χ 6= χ0 , is more subtle. A remarkable property is that these functions are now defined and continuous for s > 0. In fact, more is true. Proposition 3.4 If χ is a non-trivial Dirichlet character, then the series ∞ X

χ(n)/ns

n=1

converges for s > 0, and we denote its sum by L(s, χ). Moreover: (i) The function L(s, χ) is continuously differentiable for 0 < s < ∞. (ii) There exists constants c, c0 > 0 so that L(s, χ) = 1 + O(e−cs ) as s → ∞, and 0

L0 (s, χ) = O(e−c s )

as s → ∞.

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Chapter 8. DIRICHLET’S THEOREM

We first isolate the key cancellation property that non-trivial Dirichlet characters possess, which accounts for the behavior of the L-function described in the proposition. Lemma 3.5 If χ is a non-trivial Dirichlet character, then ¯ ¯ k ¯ ¯X ¯ ¯ χ(n) ¯ ≤ q, ¯ ¯ ¯

for any k.

n=1

Proof. First, we recall that q X

χ(n) = 0.

n=1

In fact, if S denotes the sum and a ∈ Z∗ (q), then the multiplicative property of the Dirichlet character χ gives χ(a)S =

X

χ(a)χ(n) =

X

χ(an) =

X

χ(n) = S.

Since χ is non-trivial, χ(a) 6= 1 for some a, hence S = 0. We now write k = aq + b with 0 ≤ b < q, and note that k X n=1

χ(n) =

aq X n=1

χ(n) +

X aq δ > 1. Moreover, the differentiated series also converges absolutely and uniformly for s > δ > 1, which shows that L(s, χ) is continuously differentiable for s > 1. We

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3. Proof of the theorem

sum by parts2 to extend this result to s > 0. Indeed, we have N X χ(k) k=1

ks

=

N X sk − sk−1

ks · ¸ N −1 X 1 1 sN = sk s − + s s k (k + 1) N k=1 k=1

=

N −1 X k=1

fk (s) +

sN , Ns

where fk (s) = sk [k −s − (k + 1)−s ]. If g(x) = x−s , then g 0 (x) = −sx−s−1 , so applying the mean-value theorem between x = k and x = k + 1, and the fact that |sk | ≤ q, we find that |fk (s)| ≤ qsk −s−1 . P Therefore, the series fk (s) converges absolutely and uniformly for s > δ > 0, and this proves that L(s, χ) is continuous for s > 0. To prove that it is also continuously differentiable, we differentiate the series term by term, obtaining X χ(n) (log n) s . n

Again, we rewrite this series using summation by parts as X £ ¤ sk −k −s log k + (k + 1)−s log(k + 1) , and an application of the mean-value theorem to the function g(x) = x−s log x shows that the terms are O(k −δ/2−1 ), thus proving that the differentiated series converges uniformly for s > δ > 0. Hence L(s, χ) is continuously differentiable for s > 0. Now, observe that for all s large, |L(s, χ) − 1| ≤ 2q

∞ X

n−s

n=2

≤ 2−s O(1), and we can take c = log 2, to see that L(s, χ) = 1 + O(e−cs ) as s → ∞. 0 A similar argument also shows that L0 (s, χ) = O(e−c s ) as s → ∞ with in fact c0 = c, and the proof of the proposition is complete. 2 For

the formula of summation by parts, see Exercise 7 in Chapter 2.

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Chapter 8. DIRICHLET’S THEOREM

With the facts gathered so far about L(s, χ) we are in a position to define the logarithm of the L-functions. This is done by integrating its logarithmic derivative. In other words, if χ is a non-trivial Dirichlet character and s > 1 we define3 Z ∞ 0 L (t, χ) log2 L(s, χ) = − dt. L(t, χ) s We know that L(t, χ) 6= 0 for every t > 1 since it is given by a product (Proposition 3.2), and the integral is convergent because L0 (t, χ) = O(e−ct ), L(t, χ) which follows from the behavior at infinity of L(t, χ) and L0 (t, χ) recorded earlier. The following links the two logarithms. Proposition 3.6 If s > 1, then elog2 L(s,χ) = L(s, χ). Moreover log2 L(s, χ) =

X p

µ

log1

1 1 − χ(p)/ps

¶

.

Proof. Differentiating e− log2 L(s,χ) L(s, χ) with respect to s gives −

L0 (s, χ) − log2 L(s,χ) e L(s, χ) + e− log2 L(s,χ) L0 (s, χ) = 0. L(s, χ)

So e− log2 L(s,χ) L(s, χ) is constant, and this constant can be seen to be 1 by letting s tend to infinity. This proves the first conclusion. To prove the equality between the logarithms, we fix s and take the exponential of both sides. The left-hand side becomes elog2 L(s,χ) = L(s, χ), and the right-hand side becomes ¶ ¡ ¢ Y ¡ ¢ Yµ P 1 1 1 log1 1−χ(p)/p log1 1−χ(p)/p s s p = L(s, χ), e e = = 1 − χ(p)/ps p p 3 The notation log used in this context should not be confused with the logarithm to 2 the base 2.

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3. Proof of the theorem

by (i) in Proposition 3.1 and the Dirichlet product formula. Therefore, for each s there exists an integer M (s) so that log2 L(s, χ) −

X

µ

log1

p

1 1 − χ(p)/ps

¶

= 2πiM (s).

As the reader may verify, the left-hand side is continuous in s, and this implies the continuity of the function M (s). But M (s) is integer-valued so we conclude that M (s) is constant, and this constant can be seen to be 0 by letting s go to infinity. Putting together the work we have done so far gives rigorous meaning to the formal argument presented earlier. Indeed, the properties of log1 show that X p

µ

log1

1 1 − χ(p)/ps

¶

=

X χ(p) p

=

ps

X χ(p) p

ps

Ã

+O

X 1 p

!

p2s

+ O(1).

Now if L(1, χ) 6= 0 for a non-trivial Dirichlet character, then by its integral representation log2 L(s, χ) remains bounded as s → 1+ . Thus P the identity between the logarithms implies that p χ(p)p−s remains bounded as s → 1+ , which is the desired result. Therefore, to finish the proof of Dirichlet’s theorem, we need to see that L(1, χ) 6= 0 when χ is non-trivial. 3.3 Non-vanishing of the L-function We now turn to a proof of the following deep result: Theorem 3.7 If χ 6= χ0 , then L(1, χ) 6= 0. There are several proofs of this fact, some involving algebraic number theory (among them Dirichlet’s original argument), and others involving complex analysis. Here we opt for a more elementary argument that requires no special knowledge of either of these areas. The proof splits in two cases, depending on whether χ is complex or real. A Dirichlet character is said to be real if it takes on only real values (that is, +1, −1, or 0) and complex otherwise. In other words, χ is real if and only if χ(n) = χ(n) for all integers n.

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Chapter 8. DIRICHLET’S THEOREM

Case I: complex Dirichlet characters This is the easier of the two cases. The proof is by contradiction, and we use two lemmas. Lemma 3.8 If s > 1, then Y

L(s, χ) ≥ 1,

χ

where the product is taken over all Dirichlet characters. In particular the product is real-valued. Proof. We have shown earlier that for s > 1 Ã µ ¶! X 1 log1 . L(s, χ) = exp 1 − χ(p)p−s p Hence, Y χ

à XX

µ

1 L(s, χ) = exp log1 −s 1 − χ(p)p χ p ! Ã ∞ XXX 1 χ(pk ) = exp k pks χ p k=1 Ã ! ∞ X XX 1 χ(pk ) = exp . k pks p k=1 χ

¶!

P Because of Lemma 2.2 (with ` = 1) we have χ χ(pk ) = ϕ(q)δ1 (pk ), and hence à ! ∞ Y XX 1 δ1 (pk ) L(s, χ) = exp ϕ(q) ≥ 1, k pks χ p k=1

since the term in the exponential is non-negative. Lemma 3.9 The following three properties hold: (i) If L(1, χ) = 0, then L(1, χ) = 0. (ii) If χ is non-trivial and L(1, χ) = 0, then |L(s, χ)| ≤ C|s − 1|

when 1 ≤ s ≤ 2.

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3. Proof of the theorem

(iii) For the trivial Dirichlet character χ0 , we have |L(s, χ0 )| ≤

C |s − 1|

when 1 < s ≤ 2.

Proof. The first statement is immediate because L(1, χ) = L(1, χ). The second statement follows from the mean-value theorem since L(s, χ) is continuously differentiable for s > 0 when χ is non-trivial. Finally, the last statement follows because by Proposition 3.3 −s −s L(s, χ0 ) = (1 − p−s 1 )(1 − p2 ) · · · (1 − pN )ζ(s),

and ζ satisfies the similar estimate (3). We can now conclude the proof that L(1, χ) 6= 0 for χ a non-trivial complex Dirichlet character. If not, say L(1, χ) = 0, then we also have L(1, χ) = 0. Since χ 6= χ, there are at least two terms in the product Y L(s, χ), χ

that vanish like |s − 1| as s → 1+ . Since only the trivial character contributes a term that grows, and this growth is no worse than O(1/|s − 1|), we find that the product goes to 0 as s → 1+ , contradicting the fact that it is ≥ 1 by Lemma 3.8. Case II: real Dirichlet characters The proof that L(1, χ) 6= 0 when χ is a non-trivial real Dirichlet character is very different from the earlier complex case. The method we shall exploit involves summation along hyperbolas. It is a curious fact that this method was introduced by Dirichlet himself, twelve years after the proof of his theorem on arithmetic progressions, to establish another famous result of his: the average order of the divisor function. However, he made no connection between the proofs of these two theorems. We will instead proceed by proving first Dirichlet’s divisor theorem, as a simple example of the method of summation along hyperbolas. Then, we shall adapt these ideas to prove the fact that L(1, χ) 6= 0. As a preliminary matter, we need to deal with some simple sums, and their corresponding integral analogues. Sums vs. Integrals Here we use the idea of comparing a sum with its corresponding integral, which already occurred in the estimate (3) for the zeta function.

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Chapter 8. DIRICHLET’S THEOREM

Proposition 3.10 If N is a positive integer, then: Z N X 1 dx (i) = + O(1) = log N + O(1). n x 1 1≤n≤N (ii) More precisely, there exists a real number γ, called Euler’s constant, so that X 1 = log N + γ + O(1/N ). n 1≤n≤N Proof. It suffices to establish the more refined estimate given in part (ii). Let Z n+1 1 dx γn = − . n x n Since 1/x is decreasing, we clearly have 0 ≤ γn ≤

1 1 1 − ≤ 2, n n+1 n

P so the series ∞ n convergesR to a limit which we denote by γ. Moren=1 γP over, if we estimate f (n) by f (x) dx, where f (x) = 1/x2 , we find ∞ X

∞ X

1 γn ≤ ≤ n2 n=N +1 n=N +1

Z

∞

N

dx = O(1/N ). x2

Therefore N X 1 n=1

n

Z

− 1

N

Z N +1 ∞ X dx dx =γ− , γn + x x N n=N +1

and this last integral is O(1/N ) as N → ∞. Proposition 3.11 If N is a positive integer, then Z N X 1 dx = + c0 + O(1/N 1/2 ) 1/2 1/2 n x 1 1≤n≤N = 2N 1/2 + c + O(1/N 1/2 ). The proof is essentially a repetition of the proof of the previous proposition, this time using the fact that ¯ ¯ ¯ 1 ¯ 1 C ¯ ¯ ¯ n1/2 − (n + 1)1/2 ¯ ≤ n3/2 .

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3. Proof of the theorem

This last inequality follows from the mean-value theorem applied to f (x) = x−1/2 , between x = n and x = n + 1. Hyperbolic sums If F is a function defined on pairs of positive integers, there are three ways to calculate XX SN = F (m, n), where the sum is taken over all pairs of positive integers (m, n) which satisfy mn ≤ N . We may carry out the summation in any one of the following three ways. (See Figure 2.) (a) Along hyperbolas: Ã

X

SN =

1≤k≤N

X

!

F (m, n)

nm=k

(b) Vertically:  X

SN =

 X



1≤m≤N

F (m, n)

1≤n≤N/m

(c) Horizontally:  X

SN =

 X



1≤n≤N

F (m, n)

1≤m≤N/n

It is a remarkable fact that one can obtain interesting conclusions from the obvious fact that these three methods of summation give the same sum. We apply this idea first in the study of the divisor problem. Intermezzo: the divisor problem For a positive integer k, let d(k) denote the number of positive divisors of k. For example, k d(k)

1 1

2 2

3 2

4 3

5 2

6 4

7 2

8 4

9 3

10 4

11 2

12 6

13 2

14 4

15 4

16 5

17 2

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Chapter 8. DIRICHLET’S THEOREM

(a)

(b)

(c)

Figure 2. The three methods of summation

One observes that the behavior of d(k) as k tends to infinity is rather irregular, and in fact, it does not seem possible to approximate d(k) by a simple analytic expression in k. However, it is natural to inquire about the average size of d(k). In other words, one might ask, what is the behavior of N 1 X d(k) N k=1

as N → ∞?

The answer was provided by Dirichlet, who made use of hyperbolic sums. Indeed, we observe that X d(k) = 1. nm=k, 1≤n,m

Theorem 3.12 If k is a positive integer, then N 1 X d(k) = log N + O(1). N k=1

More precisely, N 1 X d(k) = log N + (2γ − 1) + O(1/N 1/2 ), N k=1

where γ is Euler’s constant. P Proof. Let SN = N k=1 d(k). We observed that summing F = 1 along hyperbolas gives SN . Summing vertically, we find X X SN = 1. 1≤m≤N 1≤n≤N/m

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3. Proof of the theorem

P But 1≤n≤N/m 1 = [N/m] = N/m + O(1), where [x] denote the greatest integer ≤ x. Therefore   X X SN = (N/m + O(1)) = N  1/m + O(N ). 1≤m≤N

1≤m≤N

Hence, by part (i) of Proposition 3.10, SN = log N + O(1) N which gives the first conclusion. For the more refined estimate we proceed as follows. Consider the three regions I, II, and III shown in Figure 3. These are defined by I = {1 ≤ m < N 1/2 , N 1/2 < n ≤ N/m}, II = {1 ≤ m ≤ N 1/2 , 1 ≤ n ≤ N 1/2 }, III = {N 1/2 < m ≤ N/n, 1 ≤ n < N 1/2 }.

n N

I

nm = N

N 1/2

II

III N

1/2

N

m

Figure 3. The three regions I, II, and III

If SI , SII , and SIII denote the sums taken over the regions I, II, and III, respectively, then SN = SI + SII + SIII = 2(SI + SII ) − SII ,

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Chapter 8. DIRICHLET’S THEOREM

since by symmetry SI = SIII . Now we sum vertically, and use (ii) of Proposition 3.10 to obtain   X X  SI + SII = 1 1≤n≤N/m

1≤m≤N 1/2

=

X

[N/m]

1≤m≤N 1/2

=

X

(N/m + O(1))

1≤m≤N 1/2





X

=N

1/m + O(N 1/2 )

1≤m≤N 1/2

= N log N 1/2 + N γ + O(N 1/2 ). Finally, SII corresponds to a square so X X SII = 1 = [N 1/2 ]2 = N + O(N 1/2 ). 1≤m≤N 1/2 1≤n≤N 1/2

Putting these estimates together and dividing by N yields the more refined statement in the theorem. Non-vanishing of the L-function Our essential application of summation along hyperbolas is to the main point of this section, namely that L(1, χ) 6= 0 for a non-trivial real Dirichlet character χ. Given such a character, let F (m, n) =

χ(n) , (nm)1/2

and define SN =

XX

F (m, n),

where the sum is over all integers m, n ≥ 1 that satisfy mn ≤ N . Proposition 3.13 The following statements are true: (i) SN ≥ c log N for some constant c > 0. (ii) SN = 2N 1/2 L(1, χ) + O(1).

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3. Proof of the theorem

It suffices to prove the proposition, since the assumption L(1, χ) = 0 would give an immediate contradiction. We first sum along hyperbolas. Observe that X

χ(n) 1 X = χ(n). (nm)1/2 k 1/2 n|k nm=k For conclusion (i) it will be enough to show the following lemma. ½ X 0 for all k Lemma 3.14 χ(n) ≥ 1 if k = `2 for some ` ∈ Z. n|k

From the lemma, we then get X

SN ≥

k=`2 , `≤N 1/2

1 k 1/2

≥ c log N,

where the last inequality follows from (i) in Proposition 3.10. The proof of the lemma is simple. If k is a power of a prime, say k = pa , then the divisors of k are 1, p, p2 , . . . , pa and X χ(n) = χ(1) + χ(p) + χ(p2 ) + · · · + χ(pa ) n|k

= 1 + χ(p) + χ(p)2 + · · · + χ(p)a . So this sum is equal to  a + 1 if χ(p) = 1,    1 if χ(p) = −1 and a is even,  0 if χ(p) = −1 and a is odd,   1 if χ(p) = 0, that is p|q. In general, if k = pa1 1 · · · paNN , then any divisor of k is of the form pb11 · · · pbNN where 0 ≤ bj ≤ aj for all j. Therefore, the multiplicative property of χ gives X n|k

χ(n) =

N ³ Y

´ a χ(1) + χ(pj ) + χ(p2j ) + · · · + χ(pj j ) ,

j=1

and the proof is complete. To prove the second statement in the proposition, we write SN = SI + (SII + SIII ),

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Chapter 8. DIRICHLET’S THEOREM

where the sums SI , SII , and SIII were defined earlier (see also Figure 3). We evaluate SI by summing vertically, and SII + SIII by summing horizontally. In order to carry this out we need the following simple results.

Lemma 3.15 For all integers 0 < a < b we have

(i)

b X χ(n) n=a

(ii)

n1/2

b X χ(n) n=a

n

= O(a−1/2 ),

= O(a−1 ).

Proof. This argument is similar to the proof of Proposition 3.4; we P use summation by parts. Let sn = 1≤k≤n χ(k), and remember that |sn | ≤ q for all n. Then b X χ(n) n=a

n1/2

=

b−1 X

h i sn n−1/2 − (n + 1)−1/2 + O(a−1/2 )

n=a

Ã

=O

∞ X

! −3/2

n

+ O(a−1/2 ).

n=a

P −3/2 By comparing the sum ∞ with the integral of f (x) = x−3/2 , n=a n −1/2 we find that the former is also O(a ). A similar argument establishes (ii).

We may now finish the proof of the proposition. Summing vertically we find 



SI =

X m 1/2 and w = ρeiϕ with ρ > 0, |ϕ| < π, then log1 w = log ρ + iϕ. [Hint: If eζ = w, then the real part of ζ is uniquely determined and its imaginary part is determined modulo 2π.]

8. Let ζ denote the zeta function defined for s > 1. (a) Compare ζ(s) with

R∞ 1

x−s dx to show that

ζ(s) =

1 + O(1) s−1

as s → 1+ .

(b) Prove as a consequence that

X 1 p

ps

µ = log

1 s−1

¶ + O(1)

as s → 1+ .

9. Let χ0 denote the trivial Dirichlet character mod q, and p1 , . . . , pk the distinct −s prime divisors of q. Recall that L(s, χ0 ) = (1 − p−s 1 ) · · · (1 − pk )ζ(s), and show as a consequence L(s, χ0 ) =

ϕ(q) 1 + O(1) q s−1

as s → 1+ .

[Hint: Use the asymptotics for ζ in Exercise 8.] 10. Show that if ` is relatively prime to q, then

X 1 p≡`

1 = log ps ϕ(q)

µ

1 s−1

¶ + O(1)

as s → 1+ .

This is a quantitative version of Dirichlet’s theorem. [Hint: Recall (4).] 11. Use the characters for Z∗ (3), Z∗ (4), Z∗ (5), and Z∗ (6) to verify directly that L(1, χ) 6= 0 for all non-trivial Dirichlet characters modulo q when q = 3, 4, 5, and 6. [Hint: Consider in each case the appropriate alternating series.]

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Chapter 8. DIRICHLET’S THEOREM

12. Suppose χ is real and non-trivial; assuming the theorem that L(1, χ) 6= 0, show directly that L(1, χ) > 0. [Hint: Use the product formula for L(s, χ).] 13. Let {an }∞ n=−∞ be a sequence of complex numbers such that an = am if n = m mod q. Show that the series ∞ X an n=1

converges if and only if

Pq n=1

n

an = 0.

[Hint: Sum by parts.] 14. The series F (θ) =

X einθ |n|6=0

n

,

for |θ| < π,

converges for every θ and is the Fourier series of the function defined on [−π, π] by F (0) = 0 and

½ F (θ) =

i(−π − θ) i(π − θ)

if −π ≤ θ < 0 if 0 < θ ≤ π,

and extended by periodicity (period 2π) to all of R (see Exercise 8 in Chapter 2). Show also that if θ 6= 0 mod 2π, then the series E(θ) =

∞ X einθ n=1

n

converges, and that 1 E(θ) = log 2

15. To sum the series 0, proceed as follows.

P∞ n=1

µ

1 2 − 2 cos θ

¶

i + F (θ). 2

an /n, with an = am if n = m mod q and

(a) Define A(m) =

q X n=1

an ζ −mn

where ζ = e2πi/q .

Pq n=1

an =

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5. Problems

Note that A(q) = 0. With the notation of the previous exercise, prove that ∞ X an n=1

=

n

q−1 1X A(m)E(2πm/q). q m=1

[Hint: Use Fourier inversion on Z(q).] (b) If {am } is odd, (a−m = −am ) for m ∈ Z, observe that a0 = aq = 0 and show that

X

A(m) =

an (ζ −mn − ζ mn ).

1≤n 0 so that d(k) ≥ c(log k)N for infinitely many k. [Hint: Let p1 , . . . , pN be N distinct primes, and consider k of the form (p1 p2 · · · pN )m for m = 1, 2, . . ..]

3. Show that if p is relatively prime to q, then

Yµ χ

χ(p) 1− s p

¶

µ =

1 1 − pf s

¶g ,

where g = ϕ(q)/f , and f is the order of p in Z∗ (q) (that is, the smallest n for which pn ≡ 1 mod q). Here the product is taken over all Dirichlet characters modulo q. 4. Prove as a consequence of the previous problem that

Y χ

L(s, χ) =

X an n≥1

ns

,

where an ≥ 0, and the product is over all Dirichlet characters modulo q.

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Appendix : Integration This appendix is meant as a quick review of the definition and main properties of the Riemann integral on R, and integration of appropriate continuous functions on Rd . Our exposition is brief since we assume that the reader already has some familiarity with this material. We begin with the theory of Riemann integration on a closed and bounded interval on the real line. Besides the standard results about the integral, we also discuss the notion of sets of measure 0, and give a necessary and sufficient condition on the set of discontinuities of a function that guarantee its integrability. We also discuss multiple and repeated integrals. In particular, we extend the notion of integration to the entire space Rd by restricting ourselves to functions that decay fast enough at infinity.

1 Definition of the Riemann integral Let f be a bounded real-valued function defined on the closed interval [a, b] ⊂ R. By a partition P of [a, b] we mean a finite sequence of numbers x0 , x1 , . . . , xN with a = x0 < x1 < · · · < xN −1 < xN = b. Given such a partition, we let Ij denote the interval [xj−1 , xj ] and write |Ij | for its length, namely |Ij | = xj − xj−1 . We define the upper and lower sums of f with respect to P by U(P, f ) =

N X

[ sup f (x)] |Ij |

and

L(P, f ) =

j=1 x∈Ij

N X j=1

[ inf f (x)] |Ij |. x∈Ij

Note that the infimum and supremum exist because by assumption, f is bounded. Clearly U(P, f ) ≥ L(P, f ), and the function f is said to be Riemann integrable, or simply integrable, if for every ² > 0 there exists a partition P such that U(P, f ) − L(P, f ) < ². To define the value of the integral of f , we need to make a simple yet important observation. A partition P 0 is said to be a refinement of the partition P if P 0 is obtained from P by adding points. Then, adding one

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Appendix: INTEGRATION

point at a time, it is easy to check that U(P 0 , f ) ≤ U(P, f )

and

L(P 0 , f ) ≥ L(P, f ).

From this, we see that if P1 and P2 are two partitions of [a, b], then U(P1 , f ) ≥ L(P2 , f ), since it is possible to take P 0 as a common refinement of both P1 and P2 to obtain U(P1 , f ) ≥ U(P 0 , f ) ≥ L(P 0 , f ) ≥ L(P2 , f ). Since f is bounded we see that both U = inf U(P, f )

and

P

L = sup L(P, f ) P

exist (where the infimum and supremum are taken over all partitions of [a, b]), and also that U ≥ L . Moreover, if f is integrable we must have Rb U = L, and we define a f (x) dx to be this common value. Finally, a bounded complex-valued function f = u + iv is said to be integrable if its real and imaginary parts u and v are integrable, and we define Z b Z b Z b v(x) dx. u(x) dx + i f (x) dx = a

a

a

For example,Rthe constants are integrable functions and it is clear that b if c ∈ C, then a c dx = c(b − a). Also, continuous functions are integrable. This is because a continuous function on a closed and bounded interval [a, b] is uniformly continuous, that is, given ² > 0 there exists δ such that if |x − y| < δ then |f (x) − f (y)| < ². So if we choose n with (b − a)/n < δ, then the partition P given by a, a +

b−a b−a b−a ,..., a + k , . . . , a + (n − 1) ,b n n n

satisfies U(P, f ) − L(P, f ) ≤ ²(b − a). 1.1 Basic properties Proposition 1.1 If f and g are integrable on [a, b], then: Rb Rb Rb (i) f + g is integrable, and a f (x) + g(x) dx = a f (x) dx + a g(x) dx.

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1. Definition of the Riemann integral

(ii) If c ∈ C, then

Rb a

cf (x) dx = c

Rb a

f (x) dx.

(iii) If f and g are real-valued and f (x) ≤ g(x), then Rb ≤ a g(x) dx. Rb Rc Rb (iv) If c ∈ [a, b], then a f (x) dx = a f (x) dx + c f (x) dx.

Rb a

f (x) dx

Proof. For property (i) we may assume that f and g are real-valued. If P is a partition of [a, b], then U(P, f + g) ≤ U(P, f ) + U(P, g)

and

L(P, f + g) ≥ L(P, f ) + L(P, g).

Given ² > 0, there exist partitions P1 and P2 such that U(P1 , f ) − L(P1 , f ) < ² and U(P2 , g) − L(P2 , g) < ², so that if P0 is a common refinement of P1 and P2 , we get U(P0 , f + g) − L(P0 , f + g) < 2². So f + g is integrable, and if we let I = inf P U(P, f + g) = supP L(P, f + g), then we see that I ≤ U(P0 , f + g) + 2² ≤ U(P0 , f ) + U(P0 , g) + 2² Z b Z b g(x) dx + 4². f (x) dx + ≤ a

a

Rb Rb Rb Similarly I ≥ a f (x) dx + a g(x) dx − 4², which proves that a f (x) + Rb Rb g(x) dx = a f (x) dx + a g(x) dx. The second and third parts of the proposition are just as easy to prove. For the last property, simply refine partitions of [a, b] by adding the point c.

Another important property we need to prove is that f g is integrable whenever f and g are integrable. Lemma 1.2 If f is real-valued integrable on [a, b] and ϕ is a real-valued continuous function on R, then ϕ ◦ f is also integrable on [a, b]. Proof. Let ² > 0 and remember that f is bounded, say |f | ≤ M . Since ϕ is uniformly continuous on [−M, M ] we may choose δ > 0 so that if s, t ∈ [−M, M ] and |s − t| < δ, then |ϕ(s) − ϕ(t)| < ². Now choose a partition P = {x0 , . . . , xN } of [a, b] with U(P, f ) − L(P, f ) < δ 2 . Let Ij = [xj−1 , xj ] and distinguish two classes: we write j ∈ Λ if supx∈Ij f (x) − inf x∈Ij f (x) < δ so that by construction sup ϕ ◦ f (x) − inf ϕ ◦ f (x) < ². x∈Ij

x∈Ij

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Appendix: INTEGRATION

Otherwise, we write j ∈ Λ0 and note that X X δ |Ij | ≤ [ sup f (x) − inf f (x)] |Ij | ≤ δ 2 j∈Λ0

j∈Λ0 x∈Ij

x∈Ij

P so j∈Λ0 |Ij | < δ. Therefore, separating the cases j ∈ Λ and j ∈ Λ0 we find that

U(P, ϕ ◦ f ) − L(P, ϕ ◦ f ) ≤ ²(b − a) + 2Bδ, where B is a bound for ϕ on [−M, M ]. Since we can also choose δ < ², we see that the proposition is proved. ¿From the lemma we get the following facts: • If f and g are integrable on [a, b], then the product f g is integrable on [a, b]. 2 This from the ¡ follows ¢ lemma with ϕ(t) = t , and the fact that f g = 1 2 2 4 [f + g] − [f − g] .

• ¯If f is integrable ¯ R on [a, b], then the function |f | is integrable, and ¯R b ¯ b ¯ a f (x) dx¯ ≤ a |f (x)| dx. We can take ϕ(t) = |t| to see that |f | is integrable. Moreover, the inequality follows from (iii) in Proposition 1.1. We record two results that imply integrability. Proposition 1.3 A bounded monotonic function f on an interval [a, b] is integrable. Proof. We may assume without loss of generality that a = 0, b = 1, and f is monotonically increasing. Then, for each N , we choose the uniform partition PN given by xj = j/N for all j = 0, . . . , N . If αj = f (xj ), then we have N 1 X U(PN , f ) = αj N j=1

and

N 1 X L(PN , f ) = αj−1 . N j=1

Therefore, if |f (x)| ≤ B for all x we have U(PN , f ) − L(PN , f ) = and the proposition is proved.

αN − α0 2B ≤ , N N

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1. Definition of the Riemann integral

Proposition 1.4 Let f be a bounded function on the compact interval [a, b]. If c ∈ (a, b), and if for all small δ > 0 the function f is integrable on the intervals [a, c − δ] and [c + δ, b], then f is integrable on [a, b]. Proof. Suppose |f | ≤ M and let ² > 0. Choose δ > 0 (small) so that 4δM ≤ ²/3. Now let P1 and P2 be partitions of [a, c − δ] and [c + δ, b] so that for each i = 1, 2 we have U(Pi , f ) − L(Pi , f ) < ²/3. This is possible since f is integrable on each one of the intervals. Then by taking as a partition P = P1 ∪ {c − δ} ∪ {c + δ} ∪ P2 we immediately see that U(P, f ) − L(P, f ) < ². We end this section with a useful approximation lemma. Recall that a function on the circle is the same as a 2π-periodic function on R. Lemma 1.5 Suppose f is integrable on the circle and f is bounded by B. Then there exists a sequence {fk }∞ k=1 of continuous functions on the circle so that sup x∈[−π,π]

|fk (x)| ≤ B

for all k = 1, 2, . . .,

and Z

π −π

|f (x) − fk (x)| dx → 0

as k → ∞.

Proof. Assume f is real-valued (in general apply the following argument to the real and imaginary parts separately). Given ² > 0, we may choose a partition −π = x0 < x1 < · · · < xN = π of the interval [−π, π] so that the upper and lower sums of f differ by at most ². Denote by f ∗ the step function defined by f ∗ (x) =

sup xj−1 ≤y≤xj

f (y)

if x ∈ [xj−1 , xj ) for 1 ≤ j ≤ N .

By construction we have |f ∗ | ≤ B, and moreover Z

π

(1) −π

Z

|f ∗ (x) − f (x)| dx =

π

(f ∗ (x) − f (x)) dx < ².

−π

Now we can modify f ∗ to make it continuous and periodic yet still approximate f in the sense of the lemma. For small δ > 0, let f˜(x) = f ∗ (x) when the distance of x from any of the division points x0 , . . . , xN is ≥ δ. In the δ-neighborhood of xj for j = 1, . . . , N − 1, define f˜(x) to be the linear function for which f˜(xj ± δ) = f ∗ (xj ± δ). Near x0 = −π, f˜

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Appendix: INTEGRATION

f∗

f˜ x0

x1

x2

x3

x0

x1

x2

x2 − δ

x3 x2 + δ

Figure 1. Portions of the functions f ∗ and f˜

is linear with f˜(−π) = 0 and f˜(−π + δ) = f ∗ (−π + δ). Similarly, near xN = π the function f˜ is linear with f˜(π) = 0 and f˜(π − δ) = f ∗ (π − δ). In Figure 1 we illustrate the situation near x0 = −π. In the second picture the graph of f˜ is shifted slightly below to clarify the situation. Then, since f˜(−π) = f˜(π), we may extend f˜ to a continuous and 2πperiodic function on R. The absolute value of this extension is also bounded by B. Moreover, f˜ differs from f ∗ only in the N intervals of length 2δ surrounding the division points. Thus Z π |f ∗ (x) − f˜(x)| dx ≤ 2BN (2δ). −π

If we choose δ sufficiently small, we get Z π |f ∗ (x) − f˜(x)| dx < ². (2) −π

As a result, equations (1), (2), and the triangle inequality yield Z π |f (x) − f˜(x)| dx < 2². −π

Denoting by fk the f˜ so constructed, when 2² = 1/k, we see that the sequence {fk } has the properties required by the lemma. 1.2 Sets of measure zero and discontinuities of integrable functions We observed that all continuous functions are integrable. By modifying the argument slightly, one can show that all piecewise continuous functions are also integrable. In fact, this is a consequence of Proposition 1.4

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1. Definition of the Riemann integral

applied finitely many times. We now turn to a more careful study of the discontinuities of integrable functions. We start with a definition1 : a subset E of R is said to have measure 0 if for every ² > 0 there exists a countable family of open intervals {Ik }∞ k=1 such that S (i) E ⊂ ∞ k=1 Ik , P∞ (ii) k=1 |Ik | < ², where |Ik | denotes the length of the interval Ik . The first condition says that the union of the intervals covers E, and the second that this union is small. The reader will have no difficulty proving that any finite set of points has measure 0. A more subtle argument is needed to prove that a countable set of points has measure 0. In fact, this result is contained in the following lemma. Lemma 1.6 The union of countably many sets of measure 0 has measure 0. Proof. Say E1 , E2 , . . . are sets of measure 0, and let E = ∪∞ i=1 Ei . Let ² > 0, and for each i choose open interval Ii,1 , Ii,2 , . . . so that Ei ⊂

∞ [

Ii,k

and

k=1

Now clearly we have E ⊂

∞ X

|Ii,k | < ²/2i .

k=1

S∞

i,k=1 Ii,k ,

∞ X ∞ X i=1 k=1

|Ii,k | ≤

and ∞ X ² i=1

2i

≤ ²,

as was to be shown. An important observation is that if E has measure 0 and is compact, then it is possible to find a finite number of open intervals Ik , k = 1, . . . , N , that satisfy the two conditions (i) and (ii) above. We can prove the characterization of Riemann integrable functions in terms of their discontinuities. Theorem 1.7 A bounded function f on [a, b] is integrable if and only if its set of discontinuities has measure 0. 1 A systematic study of the measure of sets arises in the theory of Lebesgue integration, which is taken up in Book III.

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Appendix: INTEGRATION

We write J = [a, b] and I(c, r) = (c − r, c + r) for the open interval centered at c of radius r > 0. Define the oscillation of f on I(c, r) by osc(f, c, r) = sup |f (x) − f (y)| where the supremum is taken over all x, y ∈ J ∩ I(c, r). This quantity exists since f is bounded. Define the oscillation of f at c by osc(f, c) = lim osc(f, c, r). r→0

This limit exists because osc(f, c, r) is ≥ 0 and a decreasing function of r. The point is that f is continuous at c if and only if osc(f, c) = 0. This is clear from the definitions. For each ² > 0 we define a set A² by A² = {c ∈ J : osc(f, c) ≥ ²}. Having done that, we see that the set of points in J where f is disconS tinuous is simply ²>0 A² . This is an important step in the proof of our theorem. Lemma 1.8 If ² > 0, then the set A² is closed and therefore compact. Proof. The argument is simple. Suppose cn ∈ A² converges to c and assume that c ∈ / A² . Write osc(f, c) = ² − δ where δ > 0. Select r so that osc(f, c, r) < ² − δ/2, and choose n with |cn − c| < r/2. Then osc(f, cn , r/2) < ² which implies osc(f, cn ) < ², a contradiction. We are now ready to prove the first part of the theorem. Suppose that the set D of discontinuities of f has measure 0, and let ² > 0. Since A² ⊂ D, we can cover A² by a finite number of open intervals, say I1 , . . . , IN , whose total length is < ². The complement of this union I of intervals is compact, and around each point z in this complement we can find an interval Fz with supx,y∈Fz |f (x) − f (y)| ≤ ², simply because z∈ / A² . We may now choose a finite subcovering of ∪z∈I c Iz , which we denote by IN +1 , . . . , IN 0 . Now, taking all the end points of the intervals I1 , I2 , . . . , IN 0 we obtain a partition P of [a, b] with U(P, f ) − L(P, f ) ≤ 2M

N X

|Ij | + ²(b − a) ≤ C².

j=1

Hence f is integrable on [a, b], as was to be shown. Conversely, suppose that f is integrable on [a, b], and let D be its set of discontinuities. Since D equals ∪∞ n=1 A1/n , it suffices to prove

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2. Multiple integrals

that each A1/n has measure 0. Let ² > 0 and choose a partition P = {x0 , x1 , . . . , xN } so that U(P, f ) − L(P, f ) < ²/n. Then, if A1/n intersects Ij = (xj−1 , xj ) we must have supx∈Ij f (x) − inf x∈Ij f (x) ≥ 1/n, and this shows that X 1 |Ij | ≤ U (P, f ) − L(P, f ) < ²/n. n {j:Ij ∩A1/n 6=∅}

So by taking intervals intersecting A1/n and making them slightly larger, we can cover A1/n with open intervals of total length ≤ 2². Therefore A1/n has measure 0, and we are done. Note that incidentally, this gives another proof that f g is integrable whenever f and g are.

2 Multiple integrals We assume that the reader is familiar with the standard theory of multiple integrals of functions defined on bounded sets. Here, we give a quick review of the main definitions and results of this theory. Then, we describe the notion of “improper” multiple integration where the range of integration is extended to all of Rd . This is relevant to our study of the Fourier transform. In the spirit of Chapters 5 and 6, we shall define the integral of functions that are continuous and satisfy an adequate decay condition at infinity. Recall that the vector space Rd consists of all d-tuples of real numbers x = (x1 , x2 , . . . , xd ) with xj ∈ R, where addition and multiplication by scalars are defined componentwise. 2.1 The Riemann integral in Rd Definitions The notion of Riemann integration on a rectangle R ⊂ Rd is an immediate generalization of the notion of Riemann integration on an interval [a, b] ⊂ R. We restrict our attention to continuous functions; these are always integrable. By a closed rectangle in Rd , we mean a set of the form R = {aj ≤ xj ≤ bj : 1 ≤ j ≤ d} where aj , bj ∈ R for 1 ≤ j ≤ n. In other words, R is the product of the one-dimensional intervals [aj , bj ]: R = [a1 , b1 ] × · · · × [ad , bd ].

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Appendix: INTEGRATION

If Pj is a partition of the closed interval [aj , bj ], then we call P = (P1 , . . . , Pd ) a partition of R; and if Sj is a subinterval of the partition Pj , then S = S1 × · · · × Sd is a subrectangle of the partition P . The volume |S| of a subrectangle S is naturally given by the product of the length of its sides |S| = |S1 | × · · · × |Sd |, where |Sj | denotes the length of the interval Sj . We are now ready to define the notion of integral over R. Given a bounded real-valued function f defined on R and a partition P , we define the upper and lower sums of f with respect to P by X X U(P, f ) = [sup f (x)] |S| and L(P, f ) = [ inf f (x)] |S|, x∈S

x∈S

where the sums are taken over all subrectangles of the partition P . These definitions are direct generalizations of the analogous notions in one dimension. A partition P 0 = (P10 , . . . , Pd0 ) is a refinement of P = (P1 , . . . , Pd ) if each Pj0 is a refinement of Pj . Arguing with these refinements as we did in the one-dimensional case, we see that if we define U = inf U(P, f )

and

P

L = sup L(P, f ), P

then both U and L exist, are finite, and U ≥ L. We say that f is Riemann integrable on R if for every ² > 0 there exists a partition P so that U(P, f ) − L(P, f ) < ². This implies that U = L, and this common value, which we shall denote by either Z Z Z f, f (x) dx, or f (x1 , . . . , xd ) dx1 · · · dxd , R

R

R

is by definition the integral of f over R. If f is complex-valued, say f (x) = u(x) + iv(x), where u and v are real-valued, we naturally define Z Z Z f (x) dx = u(x) dx + i v(x) dx. R

R

R

In the results that follow, we are primarily interested in continuous functions. Clearly, if f is continuous on a closed rectangle R then f is integrable since it is uniformly continuous on R. Also, we note that if f is continuous on, say, a closed ball B, then we may define its integral

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2. Multiple integrals

over B in the following way: if g is the extension of f defined by g(x) = 0 if x ∈ / B, then g is integrable on any rectangle R that contains B, and we may set Z

Z

g(x) dx.

f (x) dx = R

B

2.2 Repeated integrals The fundamental theorem of calculus allows us to compute many one dimensional integrals, since it is possible in many instances to find an antiderivative for the integrand. In Rd , this permits the calculation of multiple integrals, since a d-dimensional integral actually reduces to d one-dimensional integrals. A precise statement describing this fact is given by the following. Theorem 2.1 Let f be a continuous function defined on a closed rectangle R ⊂ Rd . Suppose R = R1 × R2 where R1 ⊂ Rd1 and R2 ⊂ Rd2 with d = d1 + d2 . If we write x = (x1 , x2 ) with xi ∈ Rdi , then F (x1 ) = R R2 f (x1 , x2 ) dx2 is continuous on R1 , and we have µZ

Z

Z

f (x) dx = R

R1

R2

¶

f (x1 , x2 ) dx2

dx1 .

Proof. The continuity of F follows from the uniform continuity of f on R and the fact that Z

|F (x1 ) − F (x01 )| ≤

R2

|f (x1 , x2 ) − f (x01 , x2 )| dx2 .

To prove the identity, let P1 and P2 be partitions of R1 and R2 , respectively. If S and T are subrectangles in P1 and P2 , respectively, then the key observation is that !

Ã

sup f (x1 , x2 ) ≥ sup

S×T

sup f (x1 , x2 )

x1 ∈S

x2 ∈T

and µ

inf f (x1 , x2 ) ≤ inf

S×T

x1 ∈S

¶

inf f (x1 , x2 ) .

x2 ∈T

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Appendix: INTEGRATION

Then, U(P, f ) =

X

[ sup f (x1 , x2 )] |S × T |

S,T S×T

≥

XX S

≥

X

sup [ sup f (x1 , x2 )] |T | × |S|

T x1 ∈S x2 ∈T

µZ

sup

S x1 ∈S

µ

R2

Z

≥ U P1 ,

R2

¶ f (x1 , x2 ) dx2 |S|

¶ f (x1 , x2 ) dx2 .

Arguing similarly for the lower sums, we find that Z Z L(P, f ) ≤ L(P1 , f (x1 , x2 ) dx2 ) ≤ U(P1 , f (x1 , x2 ) dx2 ) ≤ U(P, f ), R2

R2

and the theorem follows from these inequalities. Repeating this argument, we find as a corollary that if f is continuous on the rectangle R ⊂ Rd given by R = [a1 , b1 ] × · · · [ad , bd ], then ÃZ ! ! Z ÃZ Z b1

b2

f (x) dx = R

bd

··· a1

a2

ad

f (x1 , . . . , xd ) dxd . . . dx2 dx1 ,

where the right-hand side denotes d-iterates of one-dimensional integrals. It is also clear from the theorem that we can interchange the order of integration in the repeated integral as desired. 2.3 The change of variables formula A diffeomorphism of class C 1 , g : A → B, is a mapping that is continuously differentiable, invertible, and whose inverse g −1 : B → A is also continuously differentiable. We denote by Dg the Jacobian or derivative of g. Then, the change of variables formula says the following. Theorem 2.2 Suppose A and B are compact subsets of Rd and g : A → B is a diffeomorphism of class C 1 . If f is continuous on B, then Z Z f (x) dx = f (g(y)) | det(Dg)(y)| dy. g(A)

A

The proof of this theorem consists first of an analysis of the special situation when g is a linear transformation L. In this case, if R is a rectangle, then |g(R)| = | det(L)| |R|,

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2. Multiple integrals

which explains the term | det(Dg)|. Indeed, this term corresponds to the new infinitesimal element of volume after the change of variables. 2.4 Spherical coordinates An important application of the change of variables formula is to the case of polar coordinates in R2 , spherical coordinates in R3 , and their generalization in Rd . These are particularly important when the function, or set we are integrating over, exhibit some rotational (or spherical) symmetries. The cases d = 2 and d = 3 were given in Chapter 6. More generally, the spherical coordinates system in Rd is given by x = g(r, θ1 , . . . , θd−1 ) where   x1 = r sin θ1 sin θ2 · · · sin θd−2 cos θd−1 ,     x = r sin θ1 sin θ2 · · · sin θd−2 sin θd−1 ,  2  .. .      xd−1 = r sin θ1 sin θ2 ,   x = r cos θ1 , d with 0 ≤ θi ≤ π for 1 ≤ i ≤ d − 2 and 0 ≤ θd−1 ≤ 2π. The determinant of the Jacobian of this transformation is given by rd−1 sind−2 θ1 sind−3 θ2 · · · sin θd−2 . Any point in x ∈ Rd − {0} can be written uniquely as rγ with γ ∈ S d−1 the unit sphere in Rd . If we define Z f (γ) dσ(γ) = S d−1

Z

π

Z

Z

π

··· 0

0

0

2π

f (g(r, θ)) sind−2 θ1 sind−3 θ2 · · · sin θd−2 dθd−1 · · · dθ1 ,

then we see that if B(0, N ) denotes the ball of radius N centered at the origin, then Z Z N Z f (rγ) rd−1 dr dσ(γ). f (x) dx = (3) B(0,N )

S d−1

0

In fact, we define the area of the unit sphere S d−1 ⊂ Rd as Z ωd = dσ(γ). S d−1

An important application of spherical coordinates is to the calculation R of the integral A(R1 ,R2 ) |x|λ dx, where A(R1 , R2 ) denotes the annulus

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Appendix: INTEGRATION

A(R1 , R2 ) = {R1 ≤ |x| ≤ R2 } and λ ∈ R. Applying polar coordinates, we find Z Z Z R2 |x|λ dx = rλ+d−1 drdσ(γ). S d−1

A(R1 ,R2 )

Therefore Z

( λ

|x| dx = A(R1 ,R2 )

R1

λ+d ωd λ+d [R2

− R1λ+d ] if λ 6= −d, ωd [log(R2 ) − log(R1 )] if λ = −d.

3 Improper integrals. Integration over Rd Most of the theorems we just discussed extend to functions integrated over all of Rd once we impose some decay at infinity on the functions we integrate. 3.1 Integration of functions of moderate decrease For each fixed N > 0 consider the closed cube in Rd centered at the origin with sides parallel to the axis, and of side length N : QN = {|xj | ≤ N/2 : 1 ≤ j ≤ d}. Let f be a continuous function on Rd . If the limit Z lim f (x) dx N →∞ QN

exists, we denote it by

Z

f (x) dx. Rd

We deal with a special class of functions whose integrals over Rd exist. A continuous function f on Rd is said to be of moderate decrease if there exists A > 0 such that |f (x)| ≤

A . 1 + |x|d+1

Note that if d = 1 we recover the definition given in Chapter 5. An important example of a function of moderate decrease in R is the Poisson y kernel given by Py (x) = π1 x2 +y 2. We claim R that if f is of moderate decrease, then the above limit exists. Let IN = QN f (x) dx. Each IN exists because f is continuous hence integrable. For M > N , we have Z |IM − IN | ≤ |f (x)| dx. QM −QN

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3. Improper integrals. Integration over Rd

295

Now observe that the set QM − QN is contained in the annulus A(aN, bM ) = {aN ≤ |x| ≤ bM }, where a and b are constants that depend only on the dimension d. √ This is because the cube QN is contained in the √ annulus N/2 ≤ |x| ≤ N d/2, so that we can take a = 1/2 and b = d/2. Therefore, using the fact that f is of moderate decrease yields Z |IM − IN | ≤ A |x|−d−1 dx. aN ≤|x|≤bM

Now putting λ = −d − 1 in the calculation of the integral of the previous section, we find that µ ¶ 1 1 |IM − IN | ≤ C − . aN bM So if f is of moderate decrease, we conclude that {IN }∞ N =1 is a Cauchy R sequence, and therefore Rd f (x) dx exists. Instead of the rectangles QN , we could have chosen the balls BN centered at the origin and of radius N . Then, if f is of moderate R decrease, the reader should have no difficulties proving that lim N →∞ BN f (x) dx R exists, and that this limit equals limN →∞ QN f (x) dx. Some elementary properties of the integrals of functions of moderate decrease are summarized in Chapter 6. 3.2 Repeated integrals In Chapters 5 and 6 we claimed that the multiplication formula held for functions of moderate decrease. This required an appropriate interchange of integration. Similarly for operators defined in terms of convolutions (with the Poisson kernel for example). We now justify the necessary formula for iterated integrals. We only consider the case d = 2, although the reader will have no difficulty extending this result to arbitrary dimensions. Theorem 3.1 Suppose f is continuous on R2 and of moderate decrease. Then Z F (x1 ) = f (x1 , x2 ) dx2 R

is of moderate decrease on R, and ¶ Z Z µZ f (x) dx = f (x1 , x2 ) dx2 dx1 . R2

R

R

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296

Appendix: INTEGRATION

Proof. To see why F is of moderate decrease, note first that Z Z Z A dx2 |F (x1 )| ≤ + . ≤ 2 2 3/2 R 1 + (x1 + x2 ) |x2 |≤|x1 | |x2 |≥|x1 | In the first integral, we observe that the integrand is ≤ A/(1 + |x1 |3 ), so Z Z A A0 A dx2 ≤ dx ≤ . 2 2 2 3/2 1 + |x1 |3 |x2 |≤|x1 | 1 + |x1 |2 |x2 |≤|x1 | 1 + (x1 + x2 ) For the second integral, we have Z Z A dx2 dx2 A000 00 ≤ A ≤ , 2 3/2 2 3 |x1 |2 |x2 |≥|x1 | 1 + (x1 + x2 ) |x2 |≥|x1 | 1 + |x2 | thus F is of moderate decrease. In fact, this argument together with Theorem 2.1 shows that F is the uniform limit of continuous functions, thus is also continuous. To establish the identity we simply use an approximation and Theorem 2.1 over finite rectangles. Write S c to denote the complement of a set S. Given ² > 0 choose N so large that ¯Z ¯ Z ¯ ¯ ¯ f (x1 , x2 ) dx1 dx2 − f (x1 , x2 ) dx1 dx2 ¯¯ < ², ¯ R2

IN ×IN

where IN = [−N, N ]. Now we know that Z

µZ

Z

IN ×IN

f (x1 , x2 ) dx1 dx2 =

IN

¶

IN

f (x1 , x2 ) dx2 dx1 .

But this last iterated integral can be written as Z µZ

= R

R

¶

f (x1 , x2 ) dx2

µZ

Z

dx1 −

c IN

¶ R

f (x1 , x2 ) dx2

ÃZ

Z

−

! c IN

IN

f (x1 , x2 ) dx2

We can now estimate ¯Z Ã Z ¯ ! µ ¶ ¯ ¯ 1 ¯ ¯ f (x1 , x2 ) dx2 dx1 ¯ ≤ O ¯ ¯ IN ¯ c N2 IN ÃZ Z +C 1≤|x1 |≤N

µ

≤O

1 N

dx1

¶

.

|x2 |≥N

dx1 .

dx2 (|x1 | + |y1 |)3

!

dx1

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3. Improper integrals. Integration over Rd

A similar argument shows that ¯ ¯Z µZ ¶ ¯ ¯ C ¯ ¯ f (x1 , x2 ) dx2 dx1 ¯ ≤ . ¯ ¯ ¯ Ic N R N Therefore, we can find N so large that ¯ ¯Z ¶ Z µZ ¯ ¯ ¯ f (x1 , x2 ) dx1 dx2 − f (x1 , x2 ) dx2 dx1 ¯¯ < ², ¯ IN ×IN

R

R

and we are done. 3.3 Spherical coordinates In Rd , spherical coordinates are given by x = rγ, where r ≥ 0 and γ belongs to the unit sphere S d−1 . If f is of moderate decrease, then for each fixed γ ∈ S d−1 , the function of f given by f (rγ)rd−1 is also of moderate decrease on R. Indeed, we have ¯ ¯ ¯ ¯ ¯f (rγ)rd−1 ¯ ≤ A

B rd−1 ≤ . 1 + |rγ|d+1 1 + r2

As a result, by letting R → ∞ in (3) we obtain the formula Z

Z

Z

f (x) dx = Rd

S d−1

∞

f (rγ) rd−1 dr dσ(γ).

0

As a consequence, if we combine the fact that Z Z f (R(x)) dx = f (x) dx, Rd

Rd

whenever R is a rotation, with the identity (3), then we obtain that Z Z (4) f (R(γ)) dσ(γ) = f (γ) dσ(γ) . S d−1

S d−1

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Notes and References Seeley [29] gives an elegant and brief introduction to Fourier series and the Fourier transform. The authoritative text on Fourier series is Zygmund [36]. For further applications of Fourier analysis to a variety of other topics, see Dym and McKean [8] and K¨orner [21]. The reader should also consult the book by Kahane and Lemari´e-Rieusset [20], which contains many historical facts and other results related to Fourier series. Chapter 1 The citation is taken from a letter of Fourier to an unknown correspondent (probably Lagrange), see Herivel [15]. More facts about the early history of Fourier series can be found in Sections I-III of Riemann’s memoir [27]. Chapter 2 The quote is a translation of an excerpt in Riemann’s paper [27]. For a proof of Littlewood’s theorem (Problem 3), as well as other related “Tauberian theorems,” see Chapter 7 in Titchmarsh [32]. Chapter 3 The citation is a translation of a passage in Dirichlet’s memoir [6]. Chapter 4 The quote is translated from Hurwitz [17]. The problem of a ray of light reflecting inside a square is discussed in Chapter 23 of Hardy and Wright [13]. The relationship between the diameter of a curve and Fourier coefficients (Problem 1) is explored in Pfluger [26]. Many topics concerning equidistribution of sequences, including the results in Problems 2 and 3, are taken up in Kuipers and Niederreiter [22]. Chapter 5 The citation is a free translation of a passage in Schwartz [28]. For topics in finance, see Duffie [7], and in particular Chapter 5 for the BlackScholes theory (Problems 1 and 2). The results in Problems 4, 5, and 6 are worked out in John [19] and Widder [34]. For Problem 7, see Chapter 2 in Wiener [35]. The original proof of the nowhere differentiability of f1 (Problem 8) is in Hardy [12]. Chapter 6 The quote is an excerpt from Cormack’s Nobel Prize lecture [5]. More about the wave equation, as well as the results in Problems 3, 4, and 5 can be found in Chapter 5 of Folland [9].

298

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NOTES AND REFERENCES

October 20, 2007

299

A discussion of the relationship between rotational symmetry, the Fourier transform, and Bessel functions is in Chapter 4 of Stein and Weiss [31]. For more on the Radon transform, see Chapter 1 in John [18], Helgason [14], and Ludwig [25]. Chapter 7 The citation is taken from Bingham and Tukey [2]. Proofs of the structure theorem for finite abelian groups (Problem 2) can be found in Chapter 2 of Herstein [16], Chapter 2 in Lang [23], or Chapter 104 in K¨ orner [21]. For Problem 4, see Andrews [1], which contains a short proof. Chapter 8 The citation is from Bochner [3]. For more on the divisor function, see Chapter 18 in Hardy and Wright [13]. Another “elementary” proof that L(1, χ) 6= 0 can be found in Chapter 3 of Gelfond and Linnik [11]. An alternate proof that L(1, χ) 6= 0 based on algebraic number theory is in Weyl [33]. Also, two other analytic variants of the proof that L(1, χ) 6= 0 can be found in Chapter 109 in K¨orner [21] and Chapter 6 in Serre [30]. See also the latter reference for Problems 3 and 4. Appendix Further details about the results on integration reviewed in the appendix can be found in Folland [10] (Chapter 4), Buck [4] (Chapter 4), or Lang [24] (Chapter 20).

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Bibliography [1] G. E. Andrews. Number theory. Dover Publications, New York, 1994. Corrected reprint of the 1971 originally published by W.B. Saunders Company. [2] C. Bingham and J.W. Tukey. Fourier methods in the frequency analysis of data. The Collected Works of John W Tukey, Volume II Time Series: 1965-1984(Wadsworth Advanced Books & Software), 1984. [3] S. Bochner. The role of Mathematics in the Rise of Science. Princeton University Press, Princeton, NJ, 1966. [4] R. C. Buck. Advanced Calculus. McGraw-Hill, New York, third edition, 1978. [5] A. M. Cormack. Nobel Prize in Physiology and Medicine Lecture, volume Volume 209. Science, 1980. [6] G. L. Dirichlet. Sur la convergence des s´eries trigonometriques qui servent `a representer une fonction arbitraire entre des limites donn´ees. Crelle, Journal f¨ ur die reine angewandte Mathematik, 4:157–169, 1829. [7] D. Duffie. Dynamic Asset Pricing Theory. Princeton University Press, Princeton, NJ, 2001. [8] H. Dym and H. P. McKean. Fourier Series and Integrals. Academic Press, New York, 1972. [9] G. B. Folland. Introduction to Partial Differential Equations. Princeton University Press, Princeton, NJ, 1995. [10] G. B. Folland. Advanced Calculus. Prentice Hall, Englewood Cliffs, NJ, 2002. [11] A. O. Gelfond and Yu. V. Linnik. Elementary Methods in Analytic Number Theory. Rand McNally & Compagny, Chicago, 1965. [12] G. H. Hardy. Weierstrass’s non-differentiable function. Transactions, American Mathematical Society, 17:301–325, 1916. [13] G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers. Oxford University Press, London, fifth edition, 1979. 300

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BIBLIOGRAPHY

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301

[14] S. Helgason. The Radon transform on Euclidean spaces, compact two-point homogeneous spaces and Grassman manifolds. Acta. Math., 113:153–180, 1965. [15] J. Herivel. Joseph Fourier The Man and the Physicist. Clarendon Press, Oxford, 1975. [16] I. N. Herstein. Abstract Algebra. Macmillan, New York, second edition, 1990. [17] A. Hurwitz. Sur quelques applications g´eometriques des s´eries de Fourier. Annales de l’Ecole Normale Sup´erieure, 19(3):357–408, 1902. [18] F. John. Plane Waves and Spherical Mean Applied to Partial Differential Equations. Interscience Publishers, New York, 1955. [19] F. John. Partial Differential Equations. Springer-Verlag, New York, fourth edition, 1982. [20] J.P. Kahane and P. G. Lemari´e-Rieusset. S´eries de Fourier et ondelettes. Cassini, Paris, 1998. English version: Gordon & Breach, 1995. [21] T. W. K¨orner. Fourier Analysis. Cambridge University Press, Cambridge, UK, 1988. [22] L. Kuipers and H. Niederreiter. Uniform Distribution of Sequences. Wiley, New York, 1974. [23] S. Lang. Undergraduate Algebra. Springer-Verlag, New York, second edition, 1990. [24] S. Lang. Undergraduate Analysis. Springer-Verlag, New York, second edition, 1997. [25] D. Ludwig. The Radon transform on Euclidean space. Comm. Pure Appl. Math., 19:49–81, 1966. [26] A. Pfluger. On the diameter of planar curves and Fourier coefficients. Colloquia Mathematica Societatis J´ anos Bolyai, Functions, series, operators, 35:957–965, 1983. [27] B. Riemann. Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe. Habilitation an der Universit¨ at zu G¨ ottingen, 1854. Collected Works, Springer Verlag, New York, 1990.

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BIBLIOGRAPHY

[28] L. Schwartz. Th´eorie des distributions, volume Volume I. Hermann, Paris, 1950. [29] R. T. Seeley. An Introduction to Fourier Series and Integrals. W. A. Benjamin, New York, 1966. [30] J.P. Serre. A course in Arithmetic. GTM 7. Springer Verlag, New York, 1973. [31] E. M. Stein and G. Weiss. Introduction to Fourier Analysis on Euclidean Spaces. Princeton University Press, Princeton, NJ, 1971. [32] E. C. Titchmarsh. The Theory of Functions. Oxford University Press, London, second edition, 1939. [33] H. Weyl. Algebraic Theory of Numbers, volume Volume 1 of Annals of Mathematics Studies. Princeton University Press, Princeton, NJ, 1940. [34] D. V. Widder. The Heat Equation. Academic Press, New York, 1975. [35] N. Wiener. The Fourier Integral and Certain of its Applications. Cambridge University Press, Cambridge, UK, 1933. [36] A. Zygmund. Trigonometric Series, volume Volumes I and II. Cambridge University Press, Cambridge, UK, second edition, 1959. Reprinted 1993.

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Symbol Glossary The page numbers on the right indicate the first time the symbol or notation is defined or used. As usual, Z, Q, R and C denote the integers, the rationals, the reals, and the complex numbers respectively. 4 |z|, z ez sinh x, cosh x fˆ(n), an P f (θ) ∼ an einθ SN (f ) ˜ N , D∗ DN , D R , D N (d)

Pr , Py , Py O, o Ck

f ∗g σN , σN (f ) FN , FR A(r), Ar (f ) χ[a,b] f (θ+ ), f (θ− ) Rd , Cd X⊥Y `2 (Z) R ζ(s) [x], hxi ˜N 4N , σN,K , 4 (d) Ht , Ht , Ht M(R)

Laplacian Absolute value and complex conjugate Complex exponential Hyperbolic sine and hyperbolic cosine Fourier coefficient Fourier series Partial sum of a Fourier series Dirichlet kernel, conjugate, and modified Poisson kernels Big O and little o notation Functions that are k times differentiable Convolution Ces`aro mean Fej´er kernel Abel mean Characteristic function One-sided limits at jump discontinuities Euclidean spaces Orthogonal vectors Square summable sequences Riemann integrable functions Zeta function Integer and fractional parts Delayed means Heat kernels Space of functions of moderate decrease on R

303

20, 185 22 24 28 34 34 35 37, 95, 165 37, 149, 210 42, 62 44 44, 139, 184, 239 52, 53 53, 163 54, 55 61 63 71 72 73 75 98 106 114, 127, 174 120, 146, 209 131

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304

SYMBOL GLOSSARY

fˆ(ξ) S, S(R), S(Rd ) R2+ , R2+ ϑ(s), Θ(z|τ ) Γ(s) kxk, |x|; (x, y), x · y xα , |α|,

October 20, 2007

¡

¢ ∂ α ∂x

S 1 , S 2 , S d−1 ft Mt , M Jn P, Pt,γ R, R∗

Ad , Vd Z(N ) Z/N Z G, |G| G≈H G1 × G2 Z∗ (q) ˆ G a|b gcd(a, b) ϕ(q) χ, χ0 L(s, ³χ) ´ 1 log1 1−z , log2 L(s, χ) d(k)

Fourier transform Schwartz space Upper half-plane and its closure Theta functions Gamma function Norm and inner product in Rd Monomial, its order, and differential operator Unit circle in R2 , and unit spheres in R3 , Rd Spherical mean Bessel function Plane Radon and dual Radon transforms Area and volume of the unit sphere in Rd Group of N th roots of unity Group of integers modulo N Abelian group and its order Isomorphic groups Direct product of groups Group of units modulo q Dual group of G a divides b Greatest common divisor of a and b Number of integers relatively prime to q Dirichlet character, and trivial Dirichlet character Dirichlet L-function Logarithms Number of positive divisors of k

134, 181 134, 180 149 155, 156 165 71, 176 176 179, 180 194, 216 197, 213 202 203, 205 208 219 221 226, 228 227 228 227, 229, 244 231 242 242 254 254 256 258, 264 269

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Index Relevant items that also arise in Book I are listed in this index, preceeded by the numeral I.

Abel means, 54 summable, 54 abelian group, 226 absolute value, 23 absorption coefficient, 199 amplitude, 3 annihilation operator, 169 approximation to the identity, 49 attenuation coefficient, 199

closed rectangle, 289 complete vector space, 74 complex conjugate, 23 exponential, 24 congruent integers, 220 conjugate Dirichlet kernel, 95 convolution, 44, 139, 239 coordinates spherical in Rd , 293 creation operator, 169 curve, 102 area enclosed, 103 closed, 102 diameter, 125 length, 102 simple, 102

Bernoulli numbers, 97, 167 polynomials, 98 Bernstein’s theorem, 93 Bessel function, 197 Bessel’s inequality, 80 best approximation lemma, 78 Black-Scholes equation, 170 bump functions, 162

d’Alembert’s formula, 11 delayed means, 114 generalized, 127 descent (method of), 194 dilations, 133, 177 direct product of groups, 228 Dirichlet characters, 254 complex, 265 real, 265 Dirichlet kernel conjugate (on the circle), 95 modified (on the real line), 165 on the circle, 37 Dirichlet problem

Cauchy problem (wave equation), 185 Cauchy sequence, 24 Cauchy-Schwarz inequality, 72 Ces`aro means, 52 sum, 52 summable, 52 character, 230 trivial (unit), 230 class C k , 44 305

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306 annulus, 64 rectangle, 28 strip, 170 unit disc, 20 Dirichlet product formula, 256 Dirichlet’s test, 60 Dirichlet’s theorem, 128 discontinuity jump, 63 of a Riemann integrable function, 286 divisibility of integers, 242 divisor, 242 greatest common, 242 divisor function, 269, 280 dual X-ray transform, 212 group, 231 Radon transform, 205 eigenvalues and eigenvectors, 233 energy, 148, 187 of a string, 90 equidistributed sequence, 107 ergodicity, 111 Euclid’s algorithm, 241 Euler constant γ, 268 identities, 25 phi-function, 254, 276 product formula, 249 even function, 10 expectation, 160 exponential function, 24 exponential sum, 112 fast Fourier transform, 224 Fej´er kernel on the circle, 53 on the real line, 163 Fibonacci numbers, 122

INDEX

Fourier coefficient (discrete), 236 coefficient, 16, 34 on Z(N ), 223 on a finite abelian group, 235 series, 34 sine coefficient, 15 Fourier inversion finite abelian group, 235 on Z(N ), 223 on R, 141 on Rd , 182 Fourier series, 34, 235 Abel means, 55 Ces`aro mean, 53 delayed means, 114 generalized delayed means, 127 lacunary, 114 partial sum, 35 uniqueness, 39 Fourier series convergence mean square, 70 pointwise, 81, 128 Fourier series divergence, 83 Fourier transform, 134, 136, 181 fractional part, 106 function Bessel, 197 exponential, 24 gamma, 165 moderate decrease, 131, 179, 294 radial, 182 rapidly decreasing, 135, 178 sawtooth, 60, 83 Schwartz, 135, 180 theta, 155 zeta, 98 gamma function, 165

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307

INDEX

Gaussian, 135, 181 Gibbs’s phenomenon, 94 good kernels, 48 greatest common divisor, 242 group (abelian), 226 cyclic, 238 dual, 231 homomorphism, 227 isomorphic, 227 of integers modulo N , 221 of units, 227, 229, 244 order, 228 H¨older condition, 43 harmonic function, 20 mean value property, 152 harmonics, 13 heat equation, 20 d-dimensions, 209 on the real line, 146 steady-state, 20 time-dependent, 20 heat kernel d-dimensions, 209 of the real line, 146, 156 of the circle, 120, 124, 156 Heisenberg uncertainty principle, 158, 168, 209 Hermite functions, 173 operator, 168 Hermitian inner product, 72 Hilbert space, 75 homomorphism, 227 Hooke’s law, 2 Huygens’ principle, 193 hyperbolic cosine and sine functions, 28 hyperbolic sums, 269 inner product, 71 Hermitian, 72

strictly positive definite, 71 integer part, 106 integrable function (Riemann), 31, 281 inverse of a linear operator, 177 isoperimetric inequality, 103, 122 jump discontinuity, 63 Kirchhoff’s formula, 211 L-function, 256 Landau kernels, 164 Laplace operator, 20 Laplacian, 20, 149, 185 polar coordinates, 27 Legendre expansion, 96 polynomial, 95 light cone backward, 193, 213 forward, 193 linearity, 6, 22 Lipschitz condition, 82 logarithm log1 , 258 log2 , 264 mean value propery, 152 measure zero, 287 moderate decrease (function), 131 modulus, 23 monomial, 176 multi-index, 176 multiplication formula, 140, 183 natural frequency, 3 Newton’s law, 3 of cooling, 19 nowhere differentiable function, 113, 126

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308 odd function, 10 order of a group, 228 orthogonal elements, 72 orthogonality relations, 232 orthonormal family, 77 oscillation (of a function), 288 overtones, 6, 13 parametrization arc-length, 103 reverse, 121 Parseval’s identity, 79 finite abelian group, 236 part fractional, 106 integer, 106 partition of a rectangle, 290 of an interval, 281 period, 3 periodic function, 10 periodization, 153 phase, 3 Plancherel formula on Z(N ), 223 on R, 143 on Rd , 182 Planck’s constant, 161 plucked string, 17 Poincar´e’s inequality, 90 Poisson integral formula, 57 Poisson kernel d-dimensions, 210 on the unit disc, 37, 55 on the upper half-plane, 149 Poisson kernels,comparison, 157 Poisson summation formula, 154–156, 165, 174 polar coordinates, 179 polynomials Bernoulli, 98 Legendre, 95

October 20, 2007

INDEX

pre-Hilbert space, 75 prime number, 242 primes in arithmetic progression, 245, 252, 275 probability density, 160 profile, 5 pure tones, 6 Pythagorean theorem, 72 Radon transform, 200, 203 rapid decrease, 135 refinement (partition), 281, 290 relatively prime integers, 242 repeated integrals, 295 Reuleaux triangle, 125 Riemann integrable function, 31, 281, 290 Riemann localization principle, 82 Riemann-Lebesgue lemma, 80 root of unity, 219 rotation, 177 improper, 177 proper, 177 sawtooth function, 60, 83, 84, 94, 99, 278 scaling, 8 Schwartz space, 134, 180 separation of variables, 4, 11 simple harmonic motion, 2 space variables, 185 spectral theorem, 233 commuting family, 233 speed of propagation (finite), 194 spherical coordinates in Rd , 293 mean, 189 wave, 210 spring constant, 3 standing wave, 4

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309

INDEX

subordination principle, 210 subrectangle, 290 summable Abel, 54 Ces`aro, 52 summation by parts, 60 superposition, 6, 14 symmetry-breaking, 83 theta function, 155 functional equation, 155 tone fundamental, 13 pure, 11 translations, 133, 177 transpose of a linear operator, 177 traveling wave, 4 trigonometric polynomial, 35 degree, 35 series, 35 Tychonoff’s uniqueness theorem, 172 uncertainty, 160 unit, 229 unitary transformation, 143, 233

variance, 160 vector space, 70 velocity of a wave, 5 velocity of the motion, 7 vibrating string, 90 wave standing, 4, 13 traveling, 4 velocity, 5 wave equation, 184 d-dimensional, 185 d’Alembert’s formula, 11 is linear, 9 one-dimensional, 7 time reversal, 11 Weierstrass approximation theorem, 54, 63, 144, 163 Weyl criterion, 112, 123 estimate, 125 theorem, 107 Wirtinger’s inequality, 90, 122 X-ray transform, 200 zeta function, 98, 155, 166, 248

COMPLEX ANALYSIS

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Princeton Lectures in Analysis I Fourier Analysis: An Introduction II Complex Analysis III Real Analysis: Measure Theory, Integration, and Hilbert Spaces

Princeton Lectures in Analysis

II

COMPLEX ANALYSIS

Elias M. Stein & Rami Shakarchi

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

Copyright © 2003 by Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock, Oxfordshire OX20 1TW All Rights Reserved Library of Congress Control Number 2005274996 ISBN 978-0-691-11385-2 British Library Cataloging-in-Publication Data is available The publisher would like to acknowledge the authors of this volume for providing the camera-ready copy from which this book was printed Printed on acid-free paper. ∞ press.princeton.edu Printed in the United States of America 5  7  9  10  8  6

To my grandchildren

Carolyn, Alison, Jason E.M.S.

To my parents

Mohamed & Mireille and my brother

Karim R.S.

Foreword

Beginning in the spring of 2000, a series of four one-semester courses were taught at Princeton University whose purpose was to present, in an integrated manner, the core areas of analysis. The objective was to make plain the organic unity that exists between the various parts of the subject, and to illustrate the wide applicability of ideas of analysis to other fields of mathematics and science. The present series of books is an elaboration of the lectures that were given. While there are a number of excellent texts dealing with individual parts of what we cover, our exposition aims at a different goal: presenting the various sub-areas of analysis not as separate disciplines, but rather as highly interconnected. It is our view that seeing these relations and their resulting synergies will motivate the reader to attain a better understanding of the subject as a whole. With this outcome in mind, we have concentrated on the main ideas and theorems that have shaped the field (sometimes sacrificing a more systematic approach), and we have been sensitive to the historical order in which the logic of the subject developed. We have organized our exposition into four volumes, each reflecting the material covered in a semester. Their contents may be broadly summarized as follows: I. Fourier series and integrals. II. Complex analysis. III. Measure theory, Lebesgue integration, and Hilbert spaces. IV. A selection of further topics, including functional analysis, distributions, and elements of probability theory. However, this listing does not by itself give a complete picture of the many interconnections that are presented, nor of the applications to other branches that are highlighted. To give a few examples: the elements of (finite) Fourier series studied in Book I, which lead to Dirichlet characters, and from there to the infinitude of primes in an arithmetic progression; the X-ray and Radon transforms, which arise in a number of

vii

viii

FOREWORD

problems in Book I, and reappear in Book III to play an important role in understanding Besicovitch-like sets in two and three dimensions; Fatou’s theorem, which guarantees the existence of boundary values of bounded holomorphic functions in the disc, and whose proof relies on ideas developed in each of the first three books; and the theta function, which first occurs in Book I in the solution of the heat equation, and is then used in Book II to find the number of ways an integer can be represented as the sum of two or four squares, and in the analytic continuation of the zeta function. A few further words about the books and the courses on which they were based. These courses where given at a rather intensive pace, with 48 lecture-hours a semester. The weekly problem sets played an indispensable part, and as a result exercises and problems have a similarly important role in our books. Each chapter has a series of “Exercises” that are tied directly to the text, and while some are easy, others may require more effort. However, the substantial number of hints that are given should enable the reader to attack most exercises. There are also more involved and challenging “Problems”; the ones that are most difficult, or go beyond the scope of the text, are marked with an asterisk. Despite the substantial connections that exist between the different volumes, enough overlapping material has been provided so that each of the first three books requires only minimal prerequisites: acquaintance with elementary topics in analysis such as limits, series, differentiable functions, and Riemann integration, together with some exposure to linear algebra. This makes these books accessible to students interested in such diverse disciplines as mathematics, physics, engineering, and finance, at both the undergraduate and graduate level. It is with great pleasure that we express our appreciation to all who have aided in this enterprise. We are particularly grateful to the students who participated in the four courses. Their continuing interest, enthusiasm, and dedication provided the encouragement that made this project possible. We also wish to thank Adrian Banner and Jose Luis Rodrigo for their special help in running the courses, and their efforts to see that the students got the most from each class. In addition, Adrian Banner also made valuable suggestions that are incorporated in the text.

FOREWORD

ix

We wish also to record a note of special thanks for the following individuals: Charles Fefferman, who taught the first week (successfully launching the whole project!); Paul Hagelstein, who in addition to reading part of the manuscript taught several weeks of one of the courses, and has since taken over the teaching of the second round of the series; and Daniel Levine, who gave valuable help in proof-reading. Last but not least, our thanks go to Gerree Pecht, for her consummate skill in typesetting and for the time and energy she spent in the preparation of all aspects of the lectures, such as transparencies, notes, and the manuscript. We are also happy to acknowledge our indebtedness for the support we received from the 250th Anniversary Fund of Princeton University, and the National Science Foundation’s VIGRE program.

Elias M. Stein Rami Shakarchi Princeton, New Jersey August 2002

Contents

Foreword

vii

Introduction

xv

Chapter 1. Preliminaries to Complex Analysis 1 Complex numbers and the complex plane 1.1 Basic properties 1.2 Convergence 1.3 Sets in the complex plane 2 Functions on the complex plane 2.1 Continuous functions 2.2 Holomorphic functions 2.3 Power series 3 Integration along curves 4 Exercises Chapter 2. Cauchy’s Theorem and Its Applications

1 1 1 5 5 8 8 8 14 18 24 32

1 Goursat’s theorem 2 Local existence of primitives and Cauchy’s theorem in a disc 3 Evaluation of some integrals 4 Cauchy’s integral formulas 5 Further applications 5.1 Morera’s theorem 5.2 Sequences of holomorphic functions 5.3 Holomorphic functions defined in terms of integrals 5.4 Schwarz reflection principle 5.5 Runge’s approximation theorem 6 Exercises 7 Problems

37 41 45 53 53 53 55 57 60 64 67

Chapter 3. Meromorphic Functions and the Logarithm

71

1 Zeros and poles 2 The residue formula 2.1 Examples 3 Singularities and meromorphic functions 4 The argument principle and applications xi

34

72 76 77 83 89

xii 5 6 7 8 9

CONTENTS

Homotopies and simply connected domains The complex logarithm Fourier series and harmonic functions Exercises Problems

Chapter 4. The Fourier Transform 1 2 3 4 5

The class F Action of the Fourier transform on F Paley-Wiener theorem Exercises Problems

Chapter 5. Entire Functions 1 2 3

4 5 6 7

Jensen’s formula Functions of finite order Infinite products 3.1 Generalities 3.2 Example: the product formula for the sine function Weierstrass infinite products Hadamard’s factorization theorem Exercises Problems

Chapter 6. The Gamma and Zeta Functions 1

93 97 101 103 108 111 113 114 121 127 131 134 135 138 140 140 142 145 147 153 156 159

The gamma function 1.1 Analytic continuation 1.2 Further properties of Γ The zeta function 2.1 Functional equation and analytic continuation Exercises Problems

160 161 163 168 168 174 179

Chapter 7. The Zeta Function and Prime Number Theorem

181

2 3 4

1

Zeros of the zeta function 1.1 Estimates for 1/ζ(s) 2 Reduction to the functions ψ and ψ1 2.1 Proof of the asymptotics for ψ1 Note on interchanging double sums 3 Exercises

182 187 188 194 197 199

CONTENTS

4 Problems Chapter 8. Conformal Mappings 1 Conformal equivalence and examples 1.1 The disc and upper half-plane 1.2 Further examples 1.3 The Dirichlet problem in a strip 2 The Schwarz lemma; automorphisms of the disc and upper half-plane 2.1 Automorphisms of the disc 2.2 Automorphisms of the upper half-plane 3 The Riemann mapping theorem 3.1 Necessary conditions and statement of the theorem 3.2 Montel’s theorem 3.3 Proof of the Riemann mapping theorem 4 Conformal mappings onto polygons 4.1 Some examples 4.2 The Schwarz-Christoffel integral 4.3 Boundary behavior 4.4 The mapping formula 4.5 Return to elliptic integrals 5 Exercises 6 Problems Chapter 9. An Introduction to Elliptic Functions 1 Elliptic functions 1.1 Liouville’s theorems 1.2 The Weierstrass ℘ function 2 The modular character of elliptic functions and Eisenstein series 2.1 Eisenstein series 2.2 Eisenstein series and divisor functions 3 Exercises 4 Problems

xiii 203 205 206 208 209 212 218 219 221 224 224 225 228 231 231 235 238 241 245 248 254 261 262 264 266 273 273 276 278 281

Chapter 10. Applications of Theta Functions

283

1 Product formula for the Jacobi theta function 1.1 Further transformation laws 2 Generating functions 3 The theorems about sums of squares 3.1 The two-squares theorem

284 289 293 296 297

xiv

4 5

CONTENTS

3.2 The four-squares theorem Exercises Problems

Appendix A: Asymptotics 1 2 3 4 5

Bessel functions Laplace’s method; Stirling’s formula The Airy function The partition function Problems

Appendix B: Simple Connectivity and Jordan Curve Theorem 1 2

Equivalent formulations of simple connectivity The Jordan curve theorem 2.1 Proof of a general form of Cauchy’s theorem

304 309 314 318 319 323 328 334 341

344 345 351 361

Notes and References

365

Bibliography

369

Symbol Glossary

373

Index

375

Introduction ... In effect, if one extends these functions by allowing complex values for the arguments, then there arises a harmony and regularity which without it would remain hidden. B. Riemann, 1851

When we begin the study of complex analysis we enter a marvelous world, full of wonderful insights. We are tempted to use the adjectives magical, or even miraculous when describing the first theorems we learn; and in pursuing the subject, we continue to be astonished by the elegance and sweep of the results. The starting point of our study is the idea of extending a function initially given for real values of the argument to one that is defined when the argument is complex. Thus, here the central objects are functions from the complex plane to itself f : C → C, or more generally, complex-valued functions defined on open subsets of C. At first, one might object that nothing new is gained from this extension, since any complex number z can be written as z = x + iy where x, y ∈ R and z is identified with the point (x, y) in R2 . However, everything changes drastically if we make a natural, but misleadingly simple-looking assumption on f : that it is differentiable in the complex sense. This condition is called holomorphicity, and it shapes most of the theory discussed in this book. A function f : C → C is holomorphic at the point z ∈ C if the limit f (z + h) − f (z) h→0 h lim

(h ∈ C)

exists. This is similar to the definition of differentiability in the case of a real argument, except that we allow h to take complex values. The reason this assumption is so far-reaching is that, in fact, it encompasses a multiplicity of conditions: so to speak, one for each angle that h can approach zero. xv

xvi

INTRODUCTION

Although one might now be tempted to prove theorems about holomorphic functions in terms of real variables, the reader will soon discover that complex analysis is a new subject, one which supplies proofs to the theorems that are proper to its own nature. In fact, the proofs of the main properties of holomorphic functions which we discuss in the next chapters are generally very short and quite illuminating. The study of complex analysis proceeds along two paths that often intersect. In following the first way, we seek to understand the universal characteristics of holomorphic functions, without special regard for specific examples. The second approach is the analysis of some particular functions that have proved to be of great interest in other areas of mathematics. Of course, we cannot go too far along either path without having traveled some way along the other. We shall start our study with some general characteristic properties of holomorphic functions, which are subsumed by three rather miraculous facts: 1. Contour integration: If f is holomorphic in Ω, then for appropriate closed paths in Ω
f (z)dz = 0. γ

2. Regularity: If f is holomorphic, then f is indefinitely differentiable. 3. Analytic continuation: If f and g are holomorphic functions in Ω which are equal in an arbitrarily small disc in Ω, then f = g everywhere in Ω. These three phenomena and other general properties of holomorphic functions are treated in the beginning chapters of this book. Instead of trying to summarize the contents of the rest of this volume, we mention briefly several other highlights of the subject. • The zeta function, which is expressed as an infinite series ζ(s) =

∞  1 , ns n=1

and is initially defined and holomorphic in the half-plane Re(s) > 1, where the convergence of the sum is guaranteed. This function and its variants (the L-series) are central in the theory of prime numbers, and have already appeared in Chapter 8 of Book I, where

xvii

INTRODUCTION

we proved Dirichlet’s theorem. Here we will prove that ζ extends to a meromorphic function with a pole at s = 1. We shall see that the behavior of ζ(s) for Re(s) = 1 (and in particular that ζ does not vanish on that line) leads to a proof of the prime number theorem. • The theta function Θ(z|τ ) =

∞ 

2

eπin τ e2πinz ,

n=−∞

which in fact is a function of the two complex variables z and τ , holomorphic for all z, but only for τ in the half-plane Im(τ ) > 0. On the one hand, when we fix τ , and think of Θ as a function of z, it is closely related to the theory of elliptic (doubly-periodic) functions. On the other hand, when z is fixed, Θ displays features of a modular function in the upper half-plane. The function Θ(z|τ ) arose in Book I as a fundamental solution of the heat equation on the circle. It will be used again in the study of the zeta function, as well as in the proof of certain results in combinatorics and number theory given in Chapters 6 and 10. Two additional noteworthy topics that we treat are: the Fourier transform with its elegant connection to complex analysis via contour integration, and the resulting applications of the Poisson summation formula; also conformal mappings, with the mappings of polygons whose inverses are realized by the Schwarz-Christoffel formula, and the particular case of the rectangle, which leads to elliptic integrals and elliptic functions.

1 Preliminaries to Complex Analysis The sweeping development of mathematics during the last two centuries is due in large part to the introduction of complex numbers; paradoxically, this is based on the seemingly absurd notion that there are numbers whose squares are negative. E. Borel, 1952

This chapter is devoted to the exposition of basic preliminary material which we use extensively throughout of this book. We begin with a quick review of the algebraic and analytic properties of complex numbers followed by some topological notions of sets in the complex plane. (See also the exercises at the end of Chapter 1 in Book I.) Then, we define precisely the key notion of holomorphicity, which is the complex analytic version of differentiability. This allows us to discuss the Cauchy-Riemann equations, and power series. Finally, we define the notion of a curve and the integral of a function along it. In particular, we shall prove an important result, which we state loosely as follows: if a function f has a primitive, in the sense that there exists a function F that is holomorphic and whose derivative is precisely f , then for any closed curve γ
f (z) dz = 0. γ

This is the first step towards Cauchy’s theorem, which plays a central role in complex function theory.

1 Complex numbers and the complex plane Many of the facts covered in this section were already used in Book I. 1.1 Basic properties A complex number takes the form z = x + iy where x and y are real, and i is an imaginary number that satisfies i2 = −1. We call x and y the

2

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

real part and the imaginary part of z, respectively, and we write x = Re(z)

and

y = Im(z).

Imaginary axis

The real numbers are precisely those complex numbers with zero imaginary parts. A complex number with zero real part is said to be purely imaginary. Throughout our presentation, the set of all complex numbers is denoted by C. The complex numbers can be visualized as the usual Euclidean plane by the following simple identification: the complex number z = x + iy ∈ C is identified with the point (x, y) ∈ R2 . For example, 0 corresponds to the origin and i corresponds to (0, 1). Naturally, the x and y axis of R2 are called the real axis and imaginary axis, because they correspond to the real and purely imaginary numbers, respectively. (See Figure 1.)

z = x + iy = (x, y) iy i 0

1

x

Real axis

Figure 1. The complex plane

The natural rules for adding and multiplying complex numbers can be obtained simply by treating all numbers as if they were real, and keeping in mind that i2 = −1. If z1 = x1 + iy1 and z2 = x2 + iy2 , then z1 + z2 = (x1 + x2 ) + i(y1 + y2 ), and also z1 z2 = (x1 + iy1 )(x2 + iy2 ) = x1 x2 + ix1 y2 + iy1 x2 + i2 y1 y2 = (x1 x2 − y1 y2 ) + i(x1 y2 + y1 x2 ).

3

1. Complex numbers and the complex plane

If we take the two expressions above as the definitions of addition and multiplication, it is a simple matter to verify the following desirable properties: • Commutativity: z1 + z2 = z2 + z1 and z1 z2 = z2 z1 for all z1 , z2 ∈ C. • Associativity: (z1 + z2 ) + z3 = z1 + (z2 + z3 ); z1 (z2 z3 ) for z1 , z2 , z3 ∈ C.

and (z1 z2 )z3 =

• Distributivity: z1 (z2 + z3 ) = z1 z2 + z1 z3 for all z1 , z2 , z3 ∈ C. Of course, addition of complex numbers corresponds to addition of the corresponding vectors in the plane R2 . Multiplication, however, consists of a rotation composed with a dilation, a fact that will become transparent once we have introduced the polar form of a complex number. At present we observe that multiplication by i corresponds to a rotation by an angle of π/2. The notion of length, or absolute value of a complex number is identical to the notion of Euclidean length in R2 . More precisely, we define the absolute value of a complex number z = x + iy by |z| = (x2 + y 2 )1/2 , so that |z| is precisely the distance from the origin to the point (x, y). In particular, the triangle inequality holds: |z + w| ≤ |z| + |w|

for all z, w ∈ C.

We record here other useful inequalities. For all z ∈ C we have both |Re(z)| ≤ |z| and |Im(z)| ≤ |z|, and for all z, w ∈ C ||z| − |w|| ≤ |z − w|. This follows from the triangle inequality since |z| ≤ |z − w| + |w|

and

|w| ≤ |z − w| + |z|.

The complex conjugate of z = x + iy is defined by z = x − iy, and it is obtained by a reflection across the real axis in the plane. In fact a complex number z is real if and only if z = z, and it is purely imaginary if and only if z = −z.

4

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

The reader should have no difficulty checking that Re(z) =

z+z 2

and

Im(z) =

z−z . 2i

Also, one has |z|2 = zz

and as a consequence

z 1 = 2 whenever z = 0. z |z|

Any non-zero complex number z can be written in polar form z = reiθ , where r > 0; also θ ∈ R is called the argument of z (defined uniquely up to a multiple of 2π) and is often denoted by arg z, and eiθ = cos θ + i sin θ. Since |eiθ | = 1 we observe that r = |z|, and θ is simply the angle (with positive counterclockwise orientation) between the positive real axis and the half-line starting at the origin and passing through z. (See Figure 2.)

z = reiθ

r θ 0

Figure 2. The polar form of a complex number

Finally, note that if z = reiθ and w = seiϕ , then zw = rsei(θ+ϕ) , so multiplication by a complex number corresponds to a homothety in R2 (that is, a rotation composed with a dilation).

5

1. Complex numbers and the complex plane

1.2 Convergence We make a transition from the arithmetic and geometric properties of complex numbers described above to the key notions of convergence and limits. A sequence {z1 , z2 , . . .} of complex numbers is said to converge to w ∈ C if lim |zn − w| = 0,

n→∞

and we write

w = lim zn . n→∞

This notion of convergence is not new. Indeed, since absolute values in C and Euclidean distances in R2 coincide, we see that zn converges to w if and only if the corresponding sequence of points in the complex plane converges to the point that corresponds to w. As an exercise, the reader can check that the sequence {zn } converges to w if and only if the sequence of real and imaginary parts of zn converge to the real and imaginary parts of w, respectively. Since it is sometimes not possible identify the limit of a N to readily 3 sequence (for example, limN →∞ n=1 1/n ), it is convenient to have a condition on the sequence itself which is equivalent to its convergence. A sequence {zn } is said to be a Cauchy sequence (or simply Cauchy) if |zn − zm | → 0

as n, m → ∞.

In other words, given  > 0 there exists an integer N > 0 so that |zn − zm | <  whenever n, m > N . An important fact of real analysis is that R is complete: every Cauchy sequence of real numbers converges to a real number.1 Since the sequence {zn } is Cauchy if and only if the sequences of real and imaginary parts of zn are, we conclude that every Cauchy sequence in C has a limit in C. We have thus the following result. Theorem 1.1 C, the complex numbers, is complete. We now turn our attention to some simple topological considerations that are necessary in our study of functions. Here again, the reader will note that no new notions are introduced, but rather previous notions are now presented in terms of a new vocabulary. 1.3 Sets in the complex plane If z0 ∈ C and r > 0, we define the open disc Dr (z0 ) of radius r centered at z0 to be the set of all complex numbers that are at absolute 1 This

is sometimes called the Bolzano-Weierstrass theorem.

6

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

value strictly less than r from z0 . In other words, Dr (z0 ) = {z ∈ C : |z − z0 | < r}, and this is precisely the usual disc in the plane of radius r centered at z0 . The closed disc D r (z0 ) of radius r centered at z0 is defined by D r (z0 ) = {z ∈ C : |z − z0 | ≤ r}, and the boundary of either the open or closed disc is the circle Cr (z0 ) = {z ∈ C : |z − z0 | = r}. Since the unit disc (that is, the open disc centered at the origin and of radius 1) plays an important role in later chapters, we will often denote it by D, D = {z ∈ C : |z| < 1}. Given a set Ω ⊂ C, a point z0 is an interior point of Ω if there exists r > 0 such that Dr (z0 ) ⊂ Ω. The interior of Ω consists of all its interior points. Finally, a set Ω is open if every point in that set is an interior point of Ω. This definition coincides precisely with the definition of an open set in R2 . A set Ω is closed if its complement Ωc = C − Ω is open. This property can be reformulated in terms of limit points. A point z ∈ C is said to be a limit point of the set Ω if there exists a sequence of points zn ∈ Ω such that zn = z and limn→∞ zn = z. The reader can now check that a set is closed if and only if it contains all its limit points. The closure of any set Ω is the union of Ω and its limit points, and is often denoted by Ω. Finally, the boundary of a set Ω is equal to its closure minus its interior, and is often denoted by ∂Ω. A set Ω is bounded if there exists M > 0 such that |z| < M whenever z ∈ Ω. In other words, the set Ω is contained in some large disc. If Ω is bounded, we define its diameter by diam(Ω) = sup |z − w|. z,w∈Ω

A set Ω is said to be compact if it is closed and bounded. Arguing as in the case of real variables, one can prove the following.

7

1. Complex numbers and the complex plane

Theorem 1.2 The set Ω ⊂ C is compact if and only if every sequence {zn } ⊂ Ω has a subsequence that converges to a point in Ω. An open covering of Ω is a family of open sets {Uα } (not necessarily countable) such that  Uα . Ω⊂ α

In analogy with the situation in R, we have the following equivalent formulation of compactness. Theorem 1.3 A set Ω is compact if and only if every open covering of Ω has a finite subcovering. Another interesting property of compactness is that of nested sets. We shall in fact use this result at the very beginning of our study of complex function theory, more precisely in the proof of Goursat’s theorem in Chapter 2. Proposition 1.4 If Ω1 ⊃ Ω2 ⊃ · · · ⊃ Ωn ⊃ · · · is a sequence of non-empty compact sets in C with the property that diam(Ωn ) → 0

as n → ∞,

then there exists a unique point w ∈ C such that w ∈ Ωn for all n. Proof. Choose a point zn in each Ωn . The condition diam(Ωn ) → 0 says precisely that {zn } is a Cauchy sequence, therefore this sequence converges to a limit that we call w. Since each set Ωn is compact we must have w ∈ Ωn for all n. Finally, w is the unique point satisfying this property, for otherwise, if w satisfied the same property with w = w we would have |w − w | > 0 and the condition diam(Ωn ) → 0 would be violated. The last notion we need is that of connectedness. An open set Ω ⊂ C is said to be connected if it is not possible to find two disjoint non-empty open sets Ω1 and Ω2 such that Ω = Ω1 ∪ Ω2 . A connected open set in C will be called a region. Similarly, a closed set F is connected if one cannot write F = F1 ∪ F2 where F1 and F2 are disjoint non-empty closed sets. There is an equivalent definition of connectedness for open sets in terms of curves, which is often useful in practice: an open set Ω is connected if and only if any two points in Ω can be joined by a curve γ entirely contained in Ω. See Exercise 5 for more details.

8

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

2 Functions on the complex plane 2.1 Continuous functions Let f be a function defined on a set Ω of complex numbers. We say that f is continuous at the point z0 ∈ Ω if for every  > 0 there exists δ > 0 such that whenever z ∈ Ω and |z − z0 | < δ then |f (z) − f (z0 )| < . An equivalent definition is that for every sequence {z1 , z2 , . . .} ⊂ Ω such that lim zn = z0 , then lim f (zn ) = f (z0 ). The function f is said to be continuous on Ω if it is continuous at every point of Ω. Sums and products of continuous functions are also continuous. Since the notions of convergence for complex numbers and points in R2 are the same, the function f of the complex argument z = x + iy is continuous if and only if it is continuous viewed as a function of the two real variables x and y. By the triangle inequality, it is immediate that if f is continuous, then the real-valued function defined by z → |f (z)| is continuous. We say that f attains a maximum at the point z0 ∈ Ω if |f (z)| ≤ |f (z0 )|

for all z ∈ Ω,

with the inequality reversed for the definition of a minimum. Theorem 2.1 A continuous function on a compact set Ω is bounded and attains a maximum and minimum on Ω. This is of course analogous to the situation of functions of a real variable, and we shall not repeat the simple proof here. 2.2 Holomorphic functions We now present a notion that is central to complex analysis, and in distinction to our previous discussion we introduce a definition that is genuinely complex in nature. Let Ω be an open set in C and f a complex-valued function on Ω. The function f is holomorphic at the point z0 ∈ Ω if the quotient (1)

f (z0 + h) − f (z0 ) h

converges to a limit when h → 0. Here h ∈ C and h = 0 with z0 + h ∈ Ω, so that the quotient is well defined. The limit of the quotient, when it exists, is denoted by f  (z0 ), and is called the derivative of f at z0 : f  (z0 ) = lim

h→0

f (z0 + h) − f (z0 ) . h

2. Functions on the complex plane

9

It should be emphasized that in the above limit, h is a complex number that may approach 0 from any direction. The function f is said to be holomorphic on Ω if f is holomorphic at every point of Ω. If C is a closed subset of C, we say that f is holomorphic on C if f is holomorphic in some open set containing C. Finally, if f is holomorphic in all of C we say that f is entire. Sometimes the terms regular or complex differentiable are used instead of holomorphic. The latter is natural in view of (1) which mimics the usual definition of the derivative of a function of one real variable. But despite this resemblance, a holomorphic function of one complex variable will satisfy much stronger properties than a differentiable function of one real variable. For example, a holomorphic function will actually be infinitely many times complex differentiable, that is, the existence of the first derivative will guarantee the existence of derivatives of any order. This is in contrast with functions of one real variable, since there are differentiable functions that do not have two derivatives. In fact more is true: every holomorphic function is analytic, in the sense that it has a power series expansion near every point (power series will be discussed in the next section), and for this reason we also use the term analytic as a synonym for holomorphic. Again, this is in contrast with the fact that there are indefinitely differentiable functions of one real variable that cannot be expanded in a power series. (See Exercise 23.) Example 1. The function f (z) = z is holomorphic on any open set in C, and f  (z) = 1. In fact, any polynomial p(z) = a0 + a1 z + · · · + an z n is holomorphic in the entire complex plane and p (z) = a1 + · · · + nan z n−1 . This follows from Proposition 2.2 below. Example 2. The function 1/z is holomorphic on any open set in C that does not contain the origin, and f  (z) = −1/z 2 . Example 3. The function f (z) = z is not holomorphic. Indeed, we have h f (z0 + h) − f (z0 ) = h h which has no limit as h → 0, as one can see by first taking h real and then h purely imaginary.

10

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

An important family of examples of holomorphic functions, which we discuss later, are the power series. They contain functions such as ez , sin z, or cos z, and in fact power series play a crucial role in the theory of holomorphic functions, as we already mentioned in the last paragraph. Some other examples of holomorphic functions that will make their appearance in later chapters were given in the introduction to this book. It is clear from (1) above that a function f is holomorphic at z0 ∈ Ω if and only if there exists a complex number a such that (2)

f (z0 + h) − f (z0 ) − ah = hψ(h),

where ψ is a function defined for all small h and limh→0 ψ(h) = 0. Of course, we have a = f  (z0 ). From this formulation, it is clear that f is continuous wherever it is holomorphic. Arguing as in the case of one real variable, using formulation (2) in the case of the chain rule (for example), one proves easily the following desirable properties of holomorphic functions. Proposition 2.2 If f and g are holomorphic in Ω, then: (i) f + g is holomorphic in Ω and (f + g) = f  + g  . (ii) f g is holomorphic in Ω and (f g) = f  g + f g  . (iii) If g(z0 ) = 0, then f /g is holomorphic at z0 and (f /g) =

f g − f g . g2

Moreover, if f : Ω → U and g : U → C are holomorphic, the chain rule holds (g ◦ f ) (z) = g  (f (z))f  (z)

for all z ∈ Ω.

Complex-valued functions as mappings We now clarify the relationship between the complex and real derivatives. In fact, the third example above should convince the reader that the notion of complex differentiability differs significantly from the usual notion of real differentiability of a function of two real variables. Indeed, in terms of real variables, the function f (z) = z corresponds to the map F : (x, y) → (x, −y), which is differentiable in the real sense. Its derivative at a point is the linear map given by its Jacobian, the 2 × 2 matrix of partial derivatives of the coordinate functions. In fact, F is linear and

11

2. Functions on the complex plane

is therefore equal to its derivative at every point. This implies that F is actually indefinitely differentiable. In particular the existence of the real derivative need not guarantee that f is holomorphic. This example leads us to associate more generally to each complexvalued function f = u + iv the mapping F (x, y) = (u(x, y), v(x, y)) from R2 to R2 . Recall that a function F (x, y) = (u(x, y), v(x, y)) is said to be differentiable at a point P0 = (x0 , y0 ) if there exists a linear transformation J : R2 → R2 such that (3)

|F (P0 + H) − F (P0 ) − J(H)| →0 |H|

as |H| → 0, H ∈ R2 .

Equivalently, we can write F (P0 + H) − F (P0 ) = J(H) + |H|Ψ(H) , with |Ψ(H)| → 0 as |H| → 0. The linear transformation J is unique and is called the derivative of F at P0 . If F is differentiable, the partial derivatives of u and v exist, and the linear transformation J is described in the standard basis of R2 by the Jacobian matrix of F

 J = JF (x, y) =

∂u/∂x ∂u/∂y ∂v/∂x ∂v/∂y

 .

In the case of complex differentiation the derivative is a complex number f  (z0 ), while in the case of real derivatives, it is a matrix. There is, however, a connection between these two notions, which is given in terms of special relations that are satisfied by the entries of the Jacobian matrix, that is, the partials of u and v. To find these relations, consider the limit in (1) when h is first real, say h = h1 + ih2 with h2 = 0. Then, if we write z = x + iy, z0 = x0 + iy0 , and f (z) = f (x, y), we find that f (x0 + h1 , y0 ) − f (x0 , y0 ) h1 →0 h1 ∂f = (z0 ), ∂x

f  (z0 ) = lim

where ∂/∂x denotes the usual partial derivative in the x variable. (We fix y0 and think of f as a complex-valued function of the single real variable x.) Now taking h purely imaginary, say h = ih2 , a similar argument yields

12

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

f (x0 , y0 + h2 ) − f (x0 , y0 ) h2 →0 ih2 1 ∂f (z0 ), = i ∂y

f  (z0 ) = lim

where ∂/∂y is partial differentiation in the y variable. Therefore, if f is holomorphic we have shown that ∂f 1 ∂f = . ∂x i ∂y Writing f = u + iv, we find after separating real and imaginary parts and using 1/i = −i, that the partials of u and v exist, and they satisfy the following non-trivial relations ∂u ∂v = ∂x ∂y

and

∂u ∂v =− . ∂y ∂x

These are the Cauchy-Riemann equations, which link real and complex analysis. We can clarify the situation further by defining two differential operators     ∂ ∂ 1 ∂ 1 ∂ 1 ∂ 1 ∂ and . = + = − ∂z 2 ∂x i ∂y ∂z 2 ∂x i ∂y Proposition 2.3 If f is holomorphic at z0 , then ∂f (z0 ) = 0 ∂z

and

f  (z0 ) =

∂f ∂u (z0 ) = 2 (z0 ). ∂z ∂z

Also, if we write F (x, y) = f (z), then F is differentiable in the sense of real variables, and det JF (x0 , y0 ) = |f  (z0 )|2 . Proof. Taking real and imaginary parts, it is easy to see that the Cauchy-Riemann equations are equivalent to ∂f /∂z = 0. Moreover, by our earlier observation   1 ∂f 1 ∂f  (z0 ) + (z0 ) f (z0 ) = 2 ∂x i ∂y ∂f = (z0 ), ∂z

13

2. Functions on the complex plane

and the Cauchy-Riemann equations give ∂f /∂z = 2∂u/∂z. To prove that F is differentiable it suffices to observe that if H = (h1 , h2 ) and h = h1 + ih2 , then the Cauchy-Riemann equations imply   ∂u ∂u JF (x0 , y0 )(H) = −i (h1 + ih2 ) = f  (z0 )h , ∂x ∂y where we have identified a complex number with the pair of real and imaginary parts. After a final application of the Cauchy-Riemann equations, the above results imply that (4)   2  2   ∂u 2 ∂v ∂u ∂u ∂u ∂u ∂v det JF (x0 , y0 ) = − = + = 2  = |f  (z0 )|2 . ∂x ∂y ∂x ∂y ∂x ∂y ∂z

So far, we have assumed that f is holomorphic and deduced relations satisfied by its real and imaginary parts. The next theorem contains an important converse, which completes the circle of ideas presented here. Theorem 2.4 Suppose f = u + iv is a complex-valued function defined on an open set Ω. If u and v are continuously differentiable and satisfy the Cauchy-Riemann equations on Ω, then f is holomorphic on Ω and f  (z) = ∂f /∂z. Proof. Write u(x + h1 , y + h2 ) − u(x, y) =

∂u ∂u h1 + h2 + |h|ψ1 (h) ∂x ∂y

v(x + h1 , y + h2 ) − v(x, y) =

∂v ∂v h1 + h2 + |h|ψ2 (h), ∂x ∂y

and

where ψj (h) → 0 (for j = 1, 2) as |h| tends to 0, and h = h1 + ih2 . Using the Cauchy-Riemann equations we find that   ∂u ∂u f (z + h) − f (z) = (h1 + ih2 ) + |h|ψ(h), −i ∂x ∂y where ψ(h) = ψ1 (h) + ψ2 (h) → 0, as |h| → 0. Therefore f is holomorphic and ∂f ∂u f  (z) = 2 = . ∂z ∂z

14

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

2.3 Power series The prime example of a power series is the complex exponential function, which is defined for z ∈ C by ez =

∞  zn n=0

n!

.

When z is real, this definition coincides with the usual exponential function, and in fact, the series above converges absolutely for every z ∈ C. To see this, note that  n  z  |z|n  = ,  n!  n!

 n |z| /n! = e|z| < ∞. In fact, this so |ez | can be compared to the series estimate shows that the series defining ez is uniformly convergent in every disc in C. In this section we will prove that ez is holomorphic in all of C (it is entire), and that its derivative can be found by differentiating the series term by term. Hence (ez ) =

∞ ∞   zm z n−1 = = ez , n n! m! n=0

m=0

and therefore ez is its own derivative. In contrast, the geometric series ∞ 

zn

n=0

converges absolutely only in the disc |z| < 1, and its sum there is the function 1/(1 − z), which is holomorphic in the open set C − {1}. This identity is proved exactly as when z is real: we first observe N 

zn =

n=0

1 − z N +1 , 1−z

and then note that if |z| < 1 we must have limN →∞ z N +1 = 0. In general, a power series is an expansion of the form (5)

∞  n=0

an z n ,

15

2. Functions on the complex plane

where an ∈ C. To test for absolute convergence of this series, we must investigate ∞ 

|an | |z|n ,

n=0

and we observe that if the series (5) converges absolutely for some z0 , then it will also converge for all z in the disc |z| ≤ |z0 |. We now prove that there always exists an open disc (possibly empty) on which the power series converges absolutely. ∞ Theorem 2.5 Given a power series n=0 an z n , there exists 0 ≤ R ≤ ∞ such that: (i) If |z| < R the series converges absolutely. (ii) If |z| > R the series diverges. Moreover, if we use the convention that 1/0 = ∞ and 1/∞ = 0, then R is given by Hadamard’s formula 1/R = lim sup |an |1/n . The number R is called the radius of convergence of the power series, and the region |z| < R the disc of convergence. In particular, we have R = ∞ in the case of the exponential function, and R = 1 for the geometric series. Proof. Let L = 1/R where R is defined by the formula in the statement of the theorem, and suppose that L = 0, ∞. (These two easy cases are left as an exercise.) If |z| < R, choose  > 0 so small that (L + )|z| = r < 1. By the definition L, we have |an |1/n ≤ L +  for all large n, therefore n

|an | |z|n ≤ {(L + )|z|} = rn .  n  r shows that an z n conComparison with the geometric series verges. If |z| > R, then a similar argument proves that there exists a sequence of terms in the series whose absolute value goes to infinity, hence the series diverges. Remark. On the boundary of the disc of convergence, |z| = R, the situation is more delicate as one can have either convergence or divergence. (See Exercise 19.)

16

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

Further examples of power series that converge in the whole complex plane are given by the standard trigonometric functions; these are defined by cos z =

∞  n=0

z 2n , (−1) (2n)! n

∞  z 2n+1 sin z = , (−1)n (2n + 1)! n=0

and

and they agree with the usual cosine and sine of a real argument whenever z ∈ R. A simple calculation exhibits a connection between these two functions and the complex exponential, namely, cos z =

eiz + e−iz 2

and

sin z =

eiz − e−iz . 2i

These are called the Euler formulas for the cosine and sine functions. Power series provide a very important class of analytic functions that are particularly simple to manipulate. ∞ Theorem 2.6 The power series f (z) = n=0 an z n defines a holomorphic function in its disc of convergence. The derivative of f is also a power series obtained by differentiating term by term the series for f , that is, f  (z) =

∞ 

nan z n−1 .

n=0

Moreover, f  has the same radius of convergence as f . Proof. The assertion about the radius of convergence of f  follows from Hadamard’s formula. Indeed, limn→∞ n1/n = 1, and therefore lim sup |an |1/n = lim sup |nan |1/n ,   n an z n and na so that n z have the same radius of convergence, and n hence so do an z and nan z n−1 . To prove the first assertion, we must show that the series g(z) =

∞ 

nan z n−1

n=0

gives the derivative of f . For that, let R denote the radius of convergence of f , and suppose |z0 | < r < R. Write f (z) = SN (z) + EN (z) ,

17

2. Functions on the complex plane

where SN (z) =

N 

an z n

and

EN (z) =

n=0

∞ 

an z n .

n=N +1

Then, if h is chosen so that |z0 + h| < r we have   f (z0 + h) − f (z0 ) SN (z0 + h) − SN (z0 )  − g(z0 ) = − SN (z0 ) h h   EN (z0 + h) − EN (z0 )  + (SN (z0 ) − g(z0 )) + . h Since an − bn = (a − b)(an−1 + an−2 b + · · · + abn−2 + bn−1 ), we see that

    ∞ ∞    EN (z0 + h) − EN (z0 )   (z0 + h)n − z0n   ≤   |a | ≤ |an |nrn−1 , n     h h n=N +1

n=N +1

where we have used the fact that |z0 | < r and |z0 + h| < r. The expression on the right is the tail end of a convergent series, since g converges absolutely on |z| < R. Therefore, given  > 0 we can find N1 so that N > N1 implies    EN (z0 + h) − EN (z0 )    < .   h  (z0 ) = g(z0 ), we can find N2 so that N > N2 Also, since limN →∞ SN implies  (z0 ) − g(z0 )| < . |SN

If we fix N so that both N > N1 and N > N2 hold, then we can find δ > 0 so that |h| < δ implies    SN (z0 + h) − SN (z0 )    − SN (z0 ) <  ,  h simply because the derivative of a polynomial is obtained by differentiating it term by term. Therefore,    f (z0 + h) − f (z0 )   − g(z0 ) < 3  h whenever |h| < δ, thereby concluding the proof of the theorem. Successive applications of this theorem yield the following.

18

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

Corollary 2.7 A power series is infinitely complex differentiable in its disc of convergence, and the higher derivatives are also power series obtained by termwise differentiation. We have so far dealt only with power series centered at the origin. More generally, a power series centered at z0 ∈ C is an expression of the form f (z) =

∞ 

an (z − z0 )n .

n=0

The disc of convergence of f is now centered at z0 and its radius is still given by Hadamard’s formula. In fact, if g(z) =

∞ 

an z n ,

n=0

then f is simply obtained by translating g, namely f (z) = g(w) where w = z − z0 . As a consequence everything about g also holds for f after we make the appropriate translation. In particular, by the chain rule, 



f (z) = g (w) =

∞ 

nan (z − z0 )n−1 .

n=0

A function f defined on an open set Ω is said to be analytic (or have a power at a point z0 ∈ Ω if there exists a power  series expansion) series an (z − z0 )n centered at z0 , with positive radius of convergence, such that f (z) =

∞ 

an (z − z0 )n

for all z in a neighborhood of z0 .

n=0

If f has a power series expansion at every point in Ω, we say that f is analytic on Ω. By Theorem 2.6, an analytic function on Ω is also holomorphic there. A deep theorem which we prove in the next chapter says that the converse is true: every holomorphic function is analytic. For that reason, we use the terms holomorphic and analytic interchangeably.

3 Integration along curves In the definition of a curve, we distinguish between the one-dimensional geometric object in the plane (endowed with an orientation), and its

19

3. Integration along curves

parametrization, which is a mapping from a closed interval to C, that is not uniquely determined. A parametrized curve is a function z(t) which maps a closed interval [a, b] ⊂ R to the complex plane. We shall impose regularity conditions on the parametrization which are always verified in the situations that occur in this book. We say that the parametrized curve is smooth if z  (t) exists and is continuous on [a, b], and z  (t) = 0 for t ∈ [a, b]. At the points t = a and t = b, the quantities z  (a) and z  (b) are interpreted as the one-sided limits z  (a) = lim

h → 0 h > 0

z(a + h) − z(a) h

and

z  (b) = lim

h → 0 h < 0

z(b + h) − z(b) . h

In general, these quantities are called the right-hand derivative of z(t) at a, and the left-hand derivative of z(t) at b, respectively. Similarly we say that the parametrized curve is piecewise-smooth if z is continuous on [a, b] and if there exist points a = a0 < a1 < · · · < an = b , where z(t) is smooth in the intervals [ak , ak+1 ]. In particular, the righthand derivative at ak may differ from the left-hand derivative at ak for k = 1, . . . , n − 1. Two parametrizations, z : [a, b] → C

and

z˜ : [c, d] → C,

are equivalent if there exists a continuously differentiable bijection s → t(s) from [c, d] to [a, b] so that t (s) > 0 and z˜(s) = z(t(s)). The condition t (s) > 0 says precisely that the orientation is preserved: as s travels from c to d, then t(s) travels from a to b. The family of all parametrizations that are equivalent to z(t) determines a smooth curve γ ⊂ C, namely the image of [a, b] under z with the orientation given by z as t travels from a to b. We can define a curve γ − obtained from the curve γ by reversing the orientation (so that γ and γ − consist of the same points in the plane). As a particular parametrization for γ − we can take z − : [a, b] → R2 defined by z − (t) = z(b + a − t).

20

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

It is also clear how to define a piecewise-smooth curve. The points z(a) and z(b) are called the end-points of the curve and are independent on the parametrization. Since γ carries an orientation, it is natural to say that γ begins at z(a) and ends at z(b). A smooth or piecewise-smooth curve is closed if z(a) = z(b) for any of its parametrizations. Finally, a smooth or piecewise-smooth curve is simple if it is not self-intersecting, that is, z(t) = z(s) unless s = t. Of course, if the curve is closed to begin with, then we say that it is simple whenever z(t) = z(s) unless s = t, or s = a and t = b.

Figure 3. A closed piecewise-smooth curve

For brevity, we shall call any piecewise-smooth curve a curve, since these will be the objects we shall be primarily concerned with. A basic example consists of a circle. Consider the circle Cr (z0 ) centered at z0 and of radius r, which by definition is the set Cr (z0 ) = {z ∈ C : |z − z0 | = r}. The positive orientation (counterclockwise) is the one that is given by the standard parametrization z(t) = z0 + reit ,

where t ∈ [0, 2π],

while the negative orientation (clockwise) is given by z(t) = z0 + re−it ,

where t ∈ [0, 2π].

In the following chapters, we shall denote by C a general positively oriented circle. An important tool in the study of holomorphic functions is integration of functions along curves. Loosely speaking, a key theorem in complex

21

3. Integration along curves

analysis says that if a function is holomorphic in the interior of a closed curve γ, then
f (z) dz = 0, γ

and we shall turn our attention to a version of this theorem (called Cauchy’s theorem) in the next chapter. Here we content ourselves with the necessary definitions and properties of the integral. Given a smooth curve γ in C parametrized by z : [a, b] → C, and f a continuous function on γ, we define the integral of f along γ by







b

f (z) dz = γ

f (z(t))z  (t) dt.

a

In order for this definition to be meaningful, we must show that the right-hand integral is independent of the parametrization chosen for γ. Say that z˜ is an equivalent parametrization as above. Then the change of variables formula and the chain rule imply that




b

f (z(t))z  (t) dt =

a




d

f (z(t(s)))z  (t(s))t (s) ds =




c

d

f (˜ z (s))˜ z  (s) ds.

c

This proves that the integral of f over γ is well defined. If γ is piecewise smooth, then the integral of f over γ is simply the sum of the integrals of f over the smooth parts of γ, so if z(t) is a piecewise-smooth parametrization as before, then


f (z) dz = γ

n−1 
ak+1 k=0

f (z(t))z  (t) dt.

ak

By definition, the length of the smooth curve γ is


length(γ) =

b

|z  (t)| dt.

a

Arguing as we just did, it is clear that this definition is also independent of the parametrization. Also, if γ is only piecewise-smooth, then its length is the sum of the lengths of its smooth parts. Proposition 3.1 Integration of continuous functions over curves satisfies the following properties:

22

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

(i) It is linear, that is, if α, β ∈ C, then










(αf (z) + βg(z)) dz = α γ

f (z) dz + β γ

g(z) dz. γ

(ii) If γ − is γ with the reverse orientation, then





f (z) dz = −

γ

f (z) dz. γ−

(iii) One has the inequality


    f (z) dz  ≤ sup |f (z)| · length(γ).   z∈γ γ

Proof. The first property follows from the definition and the linearity of the Riemann integral. The second property is left as an exercise. For the third, note that



b    f (z) dz  ≤ sup |f (z(t))| |z  (t)| dt ≤ sup |f (z)| · length(γ)   t∈[a,b] z∈γ γ

a

as was to be shown. As we have said, Cauchy’s theorem states that for appropriate closed curves γ in an open set Ω on which f is holomorphic, then


f (z) dz = 0. γ

The existence of primitives gives a first manifestation of this phenomenon. Suppose f is a function on the open set Ω. A primitive for f on Ω is a function F that is holomorphic on Ω and such that F  (z) = f (z) for all z ∈ Ω. Theorem 3.2 If a continuous function f has a primitive F in Ω, and γ is a curve in Ω that begins at w1 and ends at w2 , then


f (z) dz = F (w2 ) − F (w1 ). γ

23

3. Integration along curves

Proof. If γ is smooth, the proof is a simple application of the chain rule and the fundamental theorem of calculus. Indeed, if z(t) : [a, b] → C is a parametrization for γ, then z(a) = w1 and z(b) = w2 , and we have







b

f (z) dz = γ




a




a

b

= b

= a

f (z(t))z  (t) dt F  (z(t))z  (t) dt d F (z(t)) dt dt

= F (z(b)) − F (z(a)). If γ is only piecewise-smooth, then arguing as we just did, we obtain a telescopic sum, and we have


f (z) dz = γ

n−1 

F (z(ak+1 )) − F (z(ak ))

k=0

= F (z(an )) − F (z(a0 )) = F (z(b)) − F (z(a)).

Corollary 3.3 If γ is a closed curve in an open set Ω, and f is continuous and has a primitive in Ω, then
f (z) dz = 0. γ

This is immediate since the end-points of a closed curve coincide. For example, the function f (z) = 1/z does not have a primitive in the open set C − {0}, since if C is the unit circle parametrized by z(t) = eit , 0 ≤ t ≤ 2π, we have

2π it ie f (z) dz = dt = 2πi = 0. eit C 0 In subsequent chapters, we shall see that this innocent calculation,  which provides an example of a function f and closed curve γ for which γ f (z) dz = 0, lies at the heart of the theory. Corollary 3.4 If f is holomorphic in a region Ω and f  = 0, then f is constant.

24

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

Proof. Fix a point w0 ∈ Ω. It suffices to show that f (w) = f (w0 ) for all w ∈ Ω. Since Ω is connected, for any w ∈ Ω, there exists a curve γ which joins w0 to w. Since f is clearly a primitive for f  , we have




f  (z) dz = f (w) − f (w0 ). γ

By assumption, f  = 0 so the integral on the left is 0, and we conclude that f (w) = f (w0 ) as desired. Remark on notation. When convenient, we follow the practice of using the notation f (z) = O(g(z)) to mean that there is a constant C > 0 such that |f (z)| ≤ C|g(z)| for z in a neighborhood of the point in question. In addition, we say f (z) = o(g(z)) when |f (z)/g(z)| → 0. We also write f (z) ∼ g(z) to mean that f (z)/g(z) → 1.

4 Exercises 1. Describe geometrically the sets of points z in the complex plane defined by the following relations: (a) |z − z1 | = |z − z2 | where z1 , z2 ∈ C. (b) 1/z = z. (c) Re(z) = 3. (d) Re(z) > c, (resp., ≥ c) where c ∈ R. (e) Re(az + b) > 0 where a, b ∈ C. (f) |z| = Re(z) + 1. (g) Im(z) = c with c ∈ R.

2. Let ·, · denote the usual inner product in R2 . In other words, if Z = (x1 , y1 ) and W = (x2 , y2 ), then Z, W  = x1 x2 + y1 y2 . Similarly, we may define a Hermitian inner product (·, ·) in C by (z, w) = zw.

25

4. Exercises

The term Hermitian is used to describe the fact that (·, ·) is not symmetric, but rather satisfies the relation (z, w) = (w, z)

for all z, w ∈ C.

Show that z, w =

1 [(z, w) + (w, z)] = Re(z, w), 2

where we use the usual identification z = x + iy ∈ C with (x, y) ∈ R2 . 3. With ω = seiϕ , where s ≥ 0 and ϕ ∈ R, solve the equation z n = ω in C where n is a natural number. How many solutions are there? 4. Show that it is impossible to define a total ordering on C. In other words, one cannot find a relation  between complex numbers so that: (i) For any two complex numbers z, w, one and only one of the following is true: z  w, w  z or z = w. (ii) For all z1 , z2 , z3 ∈ C the relation z1  z2 implies z1 + z3  z2 + z3 . (iii) Moreover, for all z1 , z2 , z3 ∈ C with z3  0, then z1  z2 implies z1 z3  z2 z3 . [Hint: First check if i  0 is possible.] 5. A set Ω is said to be pathwise connected if any two points in Ω can be joined by a (piecewise-smooth) curve entirely contained in Ω. The purpose of this exercise is to prove that an open set Ω is pathwise connected if and only if Ω is connected. (a) Suppose first that Ω is open and pathwise connected, and that it can be written as Ω = Ω1 ∪ Ω2 where Ω1 and Ω2 are disjoint non-empty open sets. Choose two points w1 ∈ Ω1 and w2 ∈ Ω2 and let γ denote a curve in Ω joining w1 to w2 . Consider a parametrization z : [0, 1] → Ω of this curve with z(0) = w1 and z(1) = w2 , and let t∗ = sup {t : z(s) ∈ Ω1 for all 0 ≤ s < t}. 0≤t≤1

Arrive at a contradiction by considering the point z(t∗ ). (b) Conversely, suppose that Ω is open and connected. Fix a point w ∈ Ω and let Ω1 ⊂ Ω denote the set of all points that can be joined to w by a curve contained in Ω. Also, let Ω2 ⊂ Ω denote the set of all points that cannot be joined to w by a curve in Ω. Prove that both Ω1 and Ω2 are open, disjoint and their union is Ω. Finally, since Ω1 is non-empty (why?) conclude that Ω = Ω1 as desired.

26

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

The proof actually shows that the regularity and type of curves we used to define pathwise connectedness can be relaxed without changing the equivalence between the two definitions when Ω is open. For instance, we may take all curves to be continuous, or simply polygonal lines.2 6. Let Ω be an open set in C and z ∈ Ω. The connected component (or simply the component) of Ω containing z is the set Cz of all points w in Ω that can be joined to z by a curve entirely contained in Ω. (a) Check first that Cz is open and connected. Then, show that w ∈ Cz defines an equivalence relation, that is: (i) z ∈ Cz , (ii) w ∈ Cz implies z ∈ Cw , and (iii) if w ∈ Cz and z ∈ Cζ , then w ∈ Cζ . Thus Ω is the union of all its connected components, and two components are either disjoint or coincide. (b) Show that Ω can have only countably many distinct connected components. (c) Prove that if Ω is the complement of a compact set, then Ω has only one unbounded component. [Hint: For (b), one would otherwise obtain an uncountable number of disjoint open balls. Now, each ball contains a point with rational coordinates. For (c), note that the complement of a large disc containing the compact set is connected.] 7. The family of mappings introduced here plays an important role in complex analysis. These mappings, sometimes called Blaschke factors, will reappear in various applications in later chapters. (a) Let z, w be two complex numbers such that zw = 1. Prove that    w−z    if |z| < 1 and |w| < 1,  1 − wz  < 1 and also that

   w−z     1 − wz  = 1

if |z| = 1 or |w| = 1.

[Hint: Why can one assume that z is real? It then suffices to prove that (r − w)(r − w) ≤ (1 − rw)(1 − rw) with equality for appropriate r and |w|.] (b) Prove that for a fixed w in the unit disc D, the mapping F : z →

w−z 1 − wz

satisfies the following conditions: 2 A polygonal line is a piecewise-smooth curve which consists of finitely many straight line segments.

27

4. Exercises

(i) F maps the unit disc to itself (that is, F : D → D), and is holomorphic. (ii) F interchanges 0 and w, namely F (0) = w and F (w) = 0. (iii) |F (z)| = 1 if |z| = 1. (iv) F : D → D is bijective. [Hint: Calculate F ◦ F .]

8. Suppose U and V are open sets in the complex plane. Prove that if f : U → V and g : V → C are two functions that are differentiable (in the real sense, that is, as functions of the two real variables x and y), and h = g ◦ f , then ∂g ∂f ∂g ∂f ∂h = + ∂z ∂z ∂z ∂z ∂z and ∂g ∂f ∂g ∂f ∂h = + . ∂z ∂z ∂z ∂z ∂z This is the complex version of the chain rule. 9. Show that in polar coordinates, the Cauchy-Riemann equations take the form 1 ∂v ∂u = ∂r r ∂θ

and

1 ∂u ∂v =− . r ∂θ ∂r

Use these equations to show that the logarithm function defined by where z = reiθ with −π < θ < π

log z = log r + iθ

is holomorphic in the region r > 0 and −π < θ < π. 10. Show that 4

∂ ∂ ∂ ∂ =4 = , ∂z ∂z ∂z ∂z

where is the Laplacian

=

∂2 ∂2 + . ∂x2 ∂y 2

11. Use Exercise 10 to prove that if f is holomorphic in the open set Ω, then the real and imaginary parts of f are harmonic; that is, their Laplacian is zero. 12. Consider the function defined by f (x + iy) =



|x||y|,

whenever x, y ∈ R.

28

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

Show that f satisfies the Cauchy-Riemann equations at the origin, yet f is not holomorphic at 0. 13. Suppose that f is holomorphic in an open set Ω. Prove that in any one of the following cases: (a) Re(f ) is constant; (b) Im(f ) is constant; (c) |f | is constant; one can conclude that f is constant. N N 14. Suppose k {an }n=1 and {bn }n=1 are two finite sequences  of complex numbers. bn with the convention Let Bk = n=1 bn denote the partial sums of the series B0 = 0. Prove the summation by parts formula N 

an bn = aN BN − aM BM −1 −

n=M

N−1 

(an+1 − an )Bn .

n=M

15. Abel’s theorem. Suppose

∞

lim

n=1

r→1, r 0..] (e) Find the radius of convergence of the hypergeometric series F (α, β, γ; z) = 1 +

∞  α(α + 1) · · · (α + n − 1)β(β + 1) · · · (β + n − 1) n z . n!γ(γ + 1) · · · (γ + n − 1) n=1

Here α, β ∈ C and γ = 0, −1, −2, . . ..

29

4. Exercises

(f) Find the radius of convergence of the Bessel function of order r: Jr (z) =

∞

z r 

2

n=0

(−1)n z 2n , n!(n + r)! 2

where r is a positive integer. 17. Show that if {an }∞ n=0 is a sequence of non-zero complex numbers such that lim

n→∞

|an+1 | = L, |an |

then lim |an |1/n = L.

n→∞

In particular, this exercise shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series. 18. Let f be a power series centered at the origin. Prove that f has a power series expansion around any point in its disc of convergence. [Hint: Write z = z0 + (z − z0 ) and use the binomial expansion for z n .] 19. Prove the following:  n (a) The power series nz does not converge on any point of the unit circle.  n 2 (b) The power series z /n converges at every point of the unit circle.  n (c) The power series z /n converges at every point of the unit circle except z = 1. [Hint: Sum by parts.] 20. Expand (1 − z)−m in powers of z. Here m is a fixed positive integer. Also, show that if (1 − z)−m =

∞ 

an z n ,

n=0

then one obtains the following asymptotic relation for the coefficients: an ∼

1 nm−1 (m − 1)!

as n → ∞.

21. Show that for |z| < 1, one has n

z2 z2 z z , + + ··· + + ··· = 2 4 1−z 1−z 1−z 1 − z 2n+1

30

Chapter 1. PRELIMINARIES TO COMPLEX ANALYSIS

and k

2z 2 z 2k z 2 z + . + · · · + + ··· = k 2 1+z 1 + z2 1 − z 1+z Justify any change in the order of summation. [Hint: Use the dyadic expansion of an integer and the fact that 2k+1 − 1 = 1 + 2 + 22 + · · · + 2k .] 22. Let N = {1, 2, 3, . . .} denote the set of positive integers. A subset S ⊂ N is said to be in arithmetic progression if S = {a, a + d, a + 2d, a + 3d, . . .} where a, d ∈ N. Here d is called the step of S. Show that N cannot be partitioned into a finite number of subsets that are in arithmetic progression with distinct steps (except for the trivial case a = d = 1).  za [Hint: Write n∈N z n as a sum of terms of the type 1−z d .] 23. Consider the function f defined on R by f (x) =

if x ≤ 0 , if x > 0.

0 2 e−1/x

n ≥ 1. Prove that f is indefinitely differentiable on R, and that f (n) (0) = 0 for all ∞ n Conclude that f does not have a converging power series expansion n=0 an x for x near the origin. 24. Let γ be a smooth curve in C parametrized by z(t) : [a, b] → C. Let γ − denote the curve with the same image as γ but with the reverse orientation. Prove that for any continuous function f on γ



f (z) dz = − γ

f (z) dz. γ−

25. The next three calculations provide some insight into Cauchy’s theorem, which we treat in the next chapter. (a) Evaluate the integrals
z n dz γ

for all integers n. Here γ is any circle centered at the origin with the positive (counterclockwise) orientation. (b) Same question as before, but with γ any circle not containing the origin.

31

4. Exercises

(c) Show that if |a| < r < |b|, then
γ

2πi 1 dz = , (z − a)(z − b) a−b

where γ denotes the circle centered at the origin, of radius r, with the positive orientation.

26. Suppose f is continuous in a region Ω. Prove that any two primitives of f (if they exist) differ by a constant.

2 Cauchy’s Theorem and Its Applications The solution of a large number of problems can be reduced, in the last analysis, to the evaluation of definite integrals; thus mathematicians have been much occupied with this task... However, among many results obtained, a number were initially discovered by the aid of a type of induction based on the passage from real to imaginary. Often passage of this kind led directly to remarkable results. Nevertheless this part of the theory, as has been observed by Laplace, is subject to various difficulties... After having reflected on this subject and brought together various results mentioned above, I hope to establish the passage from the real to the imaginary based on a direct and rigorous analysis; my researches have thus led me to the method which is the object of this memoir... A. L. Cauchy, 1827

In the previous chapter, we discussed several preliminary ideas in complex analysis: open sets in C, holomorphic functions, and integration along curves. The first remarkable result of the theory exhibits a deep connection between these notions. Loosely stated, Cauchy’s theorem says that if f is holomorphic in an open set Ω and γ ⊂ Ω is a closed curve whose interior is also contained in Ω then
(1) f (z) dz = 0. γ

Many results that follow, and in particular the calculus of residues, are related in one way or another to this fact. A precise and general formulation of Cauchy’s theorem requires defining unambiguously the “interior” of a curve, and this is not always an easy task. At this early stage of our study, we shall make use of the device of limiting ourselves to regions whose boundaries are curves that are “toy contours.” As the name suggests, these are closed curves whose visualization is so simple that the notion of their interior will be unam-

33 biguous, and the proof of Cauchy’s theorem in this setting will be quite direct. For many applications, it will suffice to restrict ourselves to these types of curves. At a later stage, we take up the questions related to more general curves, their interiors, and the extended form of Cauchy’s theorem. Our initial version of Cauchy’s theorem begins with the observation that it suffices that f have a primitive in Ω, by Corollary 3.3 in Chapter 1. The existence of such a primitive for toy contours will follow from a theorem of Goursat (which is itself a simple special case)1 that asserts that if f is holomorphic in an open set that contains a triangle T and its interior, then
f (z) dz = 0. T

It is noteworthy that this simple case of Cauchy’s theorem suffices to prove some of its more complicated versions. From there, we can prove the existence of primitives in the interior of some simple regions, and therefore prove Cauchy’s theorem in that setting. As a first application of this viewpoint, we evaluate several real integrals by using appropriate toy contours. The above ideas also lead us to a central result of this chapter, the Cauchy integral formula; this states that if f is holomorphic in an open set containing a circle C and its interior, then for all z inside C,
1 f (ζ) f (z) = dζ. 2πi C ζ − z Differentiation of this identity yields other integral formulas, and in particular we obtain the regularity of holomorphic functions. This is remarkable, since holomorphicity assumed only the existence of the first derivative, and yet we obtain as a consequence the existence of derivatives of all orders. (An analogous statement is decisively false in the case of real variables!) The theory developed up to that point already has a number of noteworthy consequences: • The property at the base of “analytic continuation,” namely that a holomorphic function is determined by its restriction to any open subset of its domain of definition. This is a consequence of the fact that holomorphic functions have power series expansions. 1 Goursat’s result came after Cauchy’s theorem, and its interest is the technical fact that its proof requires only the existence of the complex derivative at each point, and not its continuity. For the earlier proof, see Exercise 5.

34

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

• Liouville’s theorem, which yields a quick proof of the fundamental theorem of algebra. • Morera’s theorem, which gives a simple integral characterization of holomorphic functions, and shows that these functions are preserved under uniform limits.

1 Goursat’s theorem Corollary 3.3 in the previous chapter says that if f has a primitive in an open set Ω, then
f (z) dz = 0 γ

for any closed curve γ in Ω. Conversely, if we can show that the above relation holds for some types of curves γ, then a primitive will exist. Our starting point is Goursat’s theorem, from which in effect we shall deduce most of the other results in this chapter. Theorem 1.1 If Ω is an open set in C, and T ⊂ Ω a triangle whose interior is also contained in Ω, then
f (z) dz = 0 T

whenever f is holomorphic in Ω. Proof. We call T (0) our original triangle (with a fixed orientation which we choose to be positive), and let d(0) and p(0) denote the diameter and perimeter of T (0) , respectively. The first step in our construction consists of bisecting each side of the triangle and connecting the mid(1) (1) (1) points. This creates four new smaller triangles, denoted T1 , T2 , T3 , (1) and T4 that are similar to the original triangle. The construction and orientation of each triangle are illustrated in Figure 1. The orientation is chosen to be consistent with that of the original triangle, and so after cancellations arising from integrating over the same side in two opposite directions, we have (2)




f (z) dz = T (0)

(1)

f (z) dz +

T1

(1)

f (z) dz +

T2

For some j we must have 

      ≤ 4  f (z) dz  (0)   T

(1)

f (z) dz +

T3

   f (z) dz  , (1)  Tj

(1)

T4

f (z) dz.

35

1. Goursat’s theorem

(1)

T3

T (0)

(1)

(1) T1

T2

(1)

T4

Figure 1. Bisection of T (0)

for otherwise (2) would be contradicted. We choose a triangle that satisfies this inequality, and rename it T (1) . Observe that if d(1) and p(1) denote the diameter and perimeter of T (1) , respectively, then d(1) = (1/2)d(0) and p(1) = (1/2)p(0) . We now repeat this process for the triangle T (1) , bisecting it into four smaller triangles. Continuing this process, we obtain a sequence of triangles T (0) , T (1) , . . . , T (n) , . . . with the properties that 
 
   n     (0) f (z) dz  ≤ 4  T

T

  f (z) dz  (n)

and d(n) = 2−n d(0) ,

p(n) = 2−n p(0)

where d(n) and p(n) denote the diameter and perimeter of T (n) , respectively. We also denote by T (n) the solid closed triangle with boundary T (n) , and observe that our construction yields a sequence of nested compact sets T (0) ⊃ T (1) ⊃ · · · ⊃ T (n) ⊃ · · · whose diameter goes to 0. By Proposition 1.4 in Chapter 1, there exists a unique point z0 that belongs to all the solid triangles T (n) . Since f is holomorphic at z0 we can write f (z) = f (z0 ) + f  (z0 )(z − z0 ) + ψ(z)(z − z0 ) , where ψ(z) → 0 as z → z0 . Since the constant f (z0 ) and the linear function f  (z0 )(z − z0 ) have primitives, we can integrate the above equality using Corollary 3.3 in the previous chapter, and obtain

(3) f (z) dz = ψ(z)(z − z0 ) dz. T (n)

T (n)

36

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

Now z0 belongs to the closure of the solid triangle T (n) and z to its boundary, so we must have |z − z0 | ≤ d(n) , and using (3) we get, by (iii) in Proposition 3.1 of the previous chapter, the estimate 
     ≤ n d(n) p(n) , f (z) dz   T (n)

where n = supz∈T (n) |ψ(z)| → 0 as n → ∞. Therefore


  

T

  f (z) dz  ≤ n 4−n d(0) p(0) , (n)

which yields our final estimate 
 
     ≤ 4n  f (z) dz  (0)   T

T (n)

  f (z) dz  ≤ n d(0) p(0) .

Letting n → ∞ concludes the proof since n → 0. Corollary 1.2 If f is holomorphic in an open set Ω that contains a rectangle R and its interior, then
f (z) dz = 0. R

This is immediate since we first choose an orientation as in Figure 2 and note that


f (z) dz = f (z) dz + f (z) dz. R

T1

T2

T2

R T1

Figure 2. A rectangle as the union of two triangles

2. Local existence of primitives and Cauchy’s theorem in a disc

37

2 Local existence of primitives and Cauchy’s theorem in a disc We first prove the existence of primitives in a disc as a consequence of Goursat’s theorem. Theorem 2.1 A holomorphic function in an open disc has a primitive in that disc. Proof. After a translation, we may assume without loss of generality that the disc, say D, is centered at the origin. Given a point z ∈ D, consider the piecewise-smooth curve that joins 0 to z first by moving in the horizontal direction from 0 to z˜ where z˜ = Re(z), and then in the vertical direction from z˜ to z. We choose the orientation from 0 to z, and denote this polygonal line (which consists of at most two segments) by γz , as shown on Figure 3.

z γz

0

z˜

Figure 3. The polygonal line γz

Define


f (w) dw.

F (z) = γz

The choice of γz gives an unambiguous definition of the function F (z). We contend that F is holomorphic in D and F  (z) = f (z). To prove this, fix z ∈ D and let h ∈ C be so small that z + h also belongs to the disc. Now consider the difference





f (w) dw −

F (z + h) − F (z) = γz+h

f (w) dw. γz

38

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

The function f is first integrated along γz+h with the original orientation, and then along γz with the reverse orientation (because of the minus sign in front of the second integral). This corresponds to (a) in Figure 4. Since we integrate f over the line segment starting at the origin in two opposite directions, it cancels, leaving us with the contour in (b). Then, we complete the square and triangle as shown in (c), so that after an application of Goursat’s theorem for triangles and rectangles we are left with the line segment from z to z + h as given in (d). z+h

z+h

z z

0 (a)

(b)

(c)

(d)

Figure 4. Relation between the polygonal lines γz and γz+h

Hence the above cancellations yield


F (z + h) − F (z) =

f (w) dw η

where η is the straight line segment from z to z + h. Since f is continuous at z we can write f (w) = f (z) + ψ(w) where ψ(w) → 0 as w → z. Therefore (4)



F (z + h) − F (z) = f (z) dw + ψ(w) dw = f (z) dw + ψ(w) dw. η

η

η

η

On the one hand, the constant 1 has w as a primitive, so the first integral is simply h by an application of Theorem 3.2 in Chapter 1. On the other

2. Local existence of primitives and Cauchy’s theorem in a disc

39

hand, we have the following estimate: 
    ψ(w) dw ≤ sup |ψ(w)| |h|.   w∈η

η

Since the supremum above goes to 0 as h tends to 0, we conclude from equation (4) that F (z + h) − F (z) = f (z) , h→0 h lim

thereby proving that F is a primitive for f on the disc. This theorem says that locally, every holomorphic function has a primitive. It is crucial to realize, however, that the theorem is true not only for arbitrary discs, but also for other sets as well. We shall return to this point shortly in our discussion of “toy contours.” Theorem 2.2 (Cauchy’s theorem for a disc) If f is holomorphic in a disc, then
f (z) dz = 0 γ

for any closed curve γ in that disc. Proof. Since f has a primitive, we can apply Corollary 3.3 of Chapter 1. Corollary 2.3 Suppose f is holomorphic in an open set containing the circle C and its interior. Then
f (z) dz = 0. C

Proof. Let D be the disc with boundary circle C. Then there exists a slightly larger disc D which contains D and so that f is holomorphic   on  D . We may now apply Cauchy’s theorem in D to conclude that f (z) dz = 0. C In fact, the proofs of the theorem and its corollary apply whenever we can define without ambiguity the “interior” of a contour, and construct appropriate polygonal paths in an open neighborhood of that contour and its interior. In the case of the circle, whose interior is the disc, there was no problem since the geometry of the disc made it simple to travel horizontally and vertically inside it.

40

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

The following definition is loosely stated, although its applications will be clear and unambiguous. We call a toy contour any closed curve where the notion of interior is obvious, and a construction similar to that in Theorem 2.1 is possible in a neighborhood of the curve and its interior. Its positive orientation is that for which the interior is to the left as we travel along the toy contour. This is consistent with the definition of the positive orientation of a circle. For example, circles, triangles, and rectangles are toy contours, since in each case we can modify (and actually copy) the argument given previously. Another important example of a toy contour is the “keyhole” Γ (illustrated in Figure 5), which we shall put to use in the proof of the Cauchy integral formula. It consists of two almost complete circles, one large

Γ

Γint

Figure 5. The keyhole contour

and one small, connected by a narrow corridor. The interior of Γ, which we denote by Γint , is clearly that region enclosed by the curve, and can be given precise meaning with enough work. We fix a point z0 in that interior. If f is holomorphic in a neighborhood of Γ and its interior, then it is holomorphic in the inside of a slightly larger keyhole, say Λ, whose interior Λint contains Γ ∪ Γint . If z ∈ Λint , let γz denote any curve contained inside Λint connecting z0 to z, and which consists of finitely many horizontal or vertical segments (as in Figure 6). If ηz is any other such curve, the rectangle version of Goursat’s theorem (Corollary 1.2) implies that

f (w) dw = f (w) dw , γz

ηz

and we may therefore define F unambiguously in Λint .

41

3. Evaluation of some integrals

Λ z0 z

γz Λint

Figure 6. A curve γz

Arguing as above allows us to show that F is a primitive of f in Λint  and therefore Γ f (z) dz = 0. The important point is that for a toy contour γ we easily have that


f (z) dz = 0 , γ

whenever f is holomorphic in an open set that contains the contour γ and its interior. Other examples of toy contours which we shall encounter in applications and for which Cauchy’s theorem and its corollary also hold are given in Figure 7. While Cauchy’s theorem for toy contours is sufficient for most applications we deal with, the question still remains as to what happens for more general curves. We take up this matter in Appendix B, where we prove Jordan’s theorem for piecewise-smooth curves. This theorem states that a simple closed piecewise-smooth curve has a well defined interior that is “simply connected.” As a consequence, we find that even in this more general situation, Cauchy’s theorem holds.

3 Evaluation of some integrals Here we take up the idea that originally motivated Cauchy. We shall show by several examples how some integrals may be evaluated by the use of his theorem. A more systematic approach, in terms of the calculus of residues, may be found in the next chapter.

42

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

Rectangular keyhole The multiple keyhole

Semicircle

Indented semicircle

Parallelogram

Sector

Figure 7. Examples of toy contours

Example 1. We show that if ξ ∈ R, then
∞ 2 2 (5) e−πξ = e−πx e−2πixξ dx. −∞ 2

This gives a new proof of the fact that e−πx is its own Fourier transform, a fact we proved in Theorem 1.4 of Chapter 5 in Book I. If ξ = 0, the formula is precisely the known integral2
∞ 2 1= e−πx dx. −∞ 2

Now suppose that ξ > 0, and consider the function f (z) = e−πz , which is entire, and in particular holomorphic in the interior of the toy contour γR depicted in Figure 8. 2 An

alternate derivation follows from the fact that Γ(1/2) = function in Chapter 6.

√

π, where Γ is the gamma

43

3. Evaluation of some integrals

−R + iξ

R + iξ

0

−R

R

Figure 8. The contour γR in Example 1

The contour γR consists of a rectangle with vertices R, R + iξ, −R + iξ, −R and the positive counterclockwise orientation. By Cauchy’s theorem,
(6) f (z) dz = 0. γR

The integral over the real segment is simply
R 2 e−πx dx , −R

which converges to 1 as R → ∞. The integral on the vertical side on the right is
ξ
ξ 2 2 f (R + iy)i dy = e−π(R +2iRy−y ) i dy. I(R) = 0

0

This integral goes to 0 as R → ∞ since ξ is fixed and we may estimate it by 2

|I(R)| ≤ Ce−πR . Similarly, the integral over the vertical segment on the left also goes to 0 as R → ∞ for the same reasons. Finally, the integral over the horizontal segment on top is
−R
R 2 2 2 e−π(x+iξ) dx = −eπξ e−πx e−2πixξ dx. −R

R

Therefore, we find in the limit as R → ∞ that (6) gives
∞ 2 πξ 2 e−πx e−2πixξ dx, 0=1−e −∞

44

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

and our desired formula is established. In the case ξ < 0, we then consider the symmetric rectangle, in the lower half-plane. The technique of shifting the contour of integration, which was used in the previous example, has many other applications. Note that the original integral (5) is taken over the real line, which by an application of Cauchy’s theorem is then shifted upwards or downwards (depending on the sign of ξ) in the complex plane. Example 2. Another classical example is
∞ 1 − cos x π dx = . 2 x 2 0 Here we consider the function f (z) = (1 − eiz )/z 2 , and we integrate over the indented semicircle in the upper half-plane positioned on the x-axis, as shown in Figure 9. + γR

γ + −R

−



R

Figure 9. The indented semicircle of Example 2

+ If we denote by γ + and γR the semicircles of radii  and R with negative and positive orientations respectively, Cauchy’s theorem gives
−

R
1 − eix 1 − eiz 1 − eix 1 − eiz dx + dz + dx + dz = 0. x2 z2 x2 z2 γ
+ γ+ −R R

First we let R → ∞ and observe that    1 − eiz  2    z 2  ≤ |z|2 , + so the integral over γR goes to zero. Therefore

1 − eix 1 − eiz dx = − dz. x2 z2 |x|≥ γ
+

45

4. Cauchy’s integral formulas

Next, note that f (z) =

−iz + E(z) z2

where E(z) is bounded as z → 0, while on γ + we have z = eiθ and dz = ieiθ dθ. Thus

0 1 − eiz dz → (−ii) dθ = −π as  → 0. z2 γ
+ π Taking real parts then yields
∞ −∞

1 − cos x dx = π. x2

Since the integrand is even, the desired formula is proved.

4 Cauchy’s integral formulas Representation formulas, and in particular integral representation formulas, play an important role in mathematics, since they allow us to recover a function on a large set from its behavior on a smaller set. For example, we saw in Book I that a solution of the steady-state heat equation in the disc was completely determined by its boundary values on the circle via a convolution with the Poisson kernel
2π 1 (7) u(r, θ) = Pr (θ − ϕ)u(1, ϕ) dϕ. 2π 0 In the case of holomorphic functions, the situation is analogous, which is not surprising since the real and imaginary parts of a holomorphic function are harmonic.3 Here, we will prove an integral representation formula in a manner that is independent of the theory of harmonic functions. In fact, it is also possible to recover the Poisson integral formula (7) as a consequence of the next theorem (see Exercises 11 and 12). Theorem 4.1 Suppose f is holomorphic in an open set that contains the closure of a disc D. If C denotes the boundary circle of this disc with the positive orientation, then 1 f (z) = 2πi


C

f (ζ) dζ ζ −z

for any point z ∈ D.

3 This fact is an immediate consequence of the Cauchy-Riemann equations. We refer the reader to Exercise 11 in Chapter 1.

46

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

Proof. Fix z ∈ D and consider the “keyhole” Γδ, which omits the point z as shown in Figure 10.

Γδ,

z

C

Figure 10. The keyhole Γδ,

Here δ is the width of the corridor, and  the radius of the small circle centered at z. Since the function F (ζ) = f (ζ)/(ζ − z) is holomorphic away from the point ζ = z, we have


F (ζ) dζ = 0 Γδ,


by Cauchy’s theorem for the chosen toy contour. Now we make the corridor narrower by letting δ tend to 0, and use the continuity of F to see that in the limit, the integrals over the two sides of the corridor cancel out. The remaining part consists of two curves, the large boundary circle C with the positive orientation, and a small circle C centered at z of radius  and oriented negatively, that is, clockwise. To see what happens to the integral over the small circle we write

(8)

F (ζ) =

f (z) f (ζ) − f (z) + ζ −z ζ−z

and note that since f is holomorphic the first term on the right-hand side of (8) is bounded so that its integral over C goes to 0 as  → 0. To

47

4. Cauchy’s integral formulas

conclude the proof, it suffices to observe that

f (z) dζ dζ = f (z) C
ζ − z C
ζ − z
2π ie−it dt = −f (z) e−it 0 = −f (z)2πi , so that in the limit we find
0= C

f (ζ) dζ − 2πif (z) , ζ −z

as was to be shown. Remarks. Our earlier discussion of toy contours provides simple extensions of the Cauchy integral formula; for instance, if f is holomorphic in an open set that contains a (positively oriented) rectangle R and its interior, then
1 f (ζ) f (z) = dζ , 2πi R ζ − z whenever z belongs to the interior of R. To establish this result, it suffices to repeat the proof of Theorem 4.1 replacing the “circular” keyhole by a “rectangular” keyhole. It should also be noted that the above integral vanishes when z is outside R, since in this case F (ζ) = f (ζ)/(ζ − z) is holomorphic inside R. Of course, a similar result also holds for the circle or any other toy contour. As a corollary to the Cauchy integral formula, we arrive at a second remarkable fact about holomorphic functions, namely their regularity. We also obtain further integral formulas expressing the derivatives of f inside the disc in terms of the values of f on the boundary. Corollary 4.2 If f is holomorphic in an open set Ω, then f has infinitely many complex derivatives in Ω. Moreover, if C ⊂ Ω is a circle whose interior is also contained in Ω, then
f (ζ) n! (n) f (z) = dζ 2πi C (ζ − z)n+1 for all z in the interior of C.

48

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

We recall that, as in the above theorem, we take the circle C to have positive orientation. Proof. The proof is by induction on n, the case n = 0 being simply the Cauchy integral formula. Suppose that f has up to n − 1 complex derivatives and that
f (ζ) (n − 1)! (n−1) (z) = dζ. f n 2πi C (ζ − z) Now for h small, the difference quotient for f (n−1) takes the form (9)

f (n−1) (z + h) − f (n−1) (z) = h (n − 1)! 2πi


f (ζ) C

 1 1 1 − dζ. h (ζ − z − h)n (ζ − z)n

We now recall that An − B n = (A − B)[An−1 + An−2 B + · · · + AB n−2 + B n−1 ]. With A = 1/(ζ − z − h) and B = 1/(ζ − z), we see that the term in brackets in equation (9) is equal to h [An−1 + An−2 B + · · · + AB n−2 + B n−1 ]. (ζ − z − h)(ζ − z) But observe that if h is small, then z + h and z stay at a finite distance from the boundary circle C, so in the limit as h tends to 0, we find that the quotient converges to

 

1 n f (ζ) n! (n − 1)! f (ζ) dζ = dζ , 2 n−1 2πi (ζ − z) (ζ − z) 2πi C (ζ − z)n+1 C which completes the induction argument and proves the theorem. From now on, we call the formulas of Theorem 4.1 and Corollary 4.2 the Cauchy integral formulas. Corollary 4.3 (Cauchy inequalities) If f is holomorphic in an open set that contains the closure of a disc D centered at z0 and of radius R, then n! f C , |f (n) (z0 )| ≤ Rn where f C = supz∈C |f (z)| denotes the supremum of |f | on the boundary circle C.

49

4. Cauchy’s integral formulas

Proof. Applying the Cauchy integral formula for f (n) (z0 ), we obtain |f

(n)

 
 n!  f (ζ) (z0 )| =  dζ  n+1 2πi C (ζ − z0 ) 
  n!  2π f (z0 + Reiθ ) iθ = Rie dθ  iθ n+1 2π 0 (Re ) n! f C ≤ 2π. 2π Rn

Another striking consequence of the Cauchy integral formula is its connection with power series. In Chapter 1, we proved that a power series is holomorphic in the interior of its disc of convergence, and promised a proof of a converse, which is the content of the next theorem. Theorem 4.4 Suppose f is holomorphic in an open set Ω. If D is a disc centered at z0 and whose closure is contained in Ω, then f has a power series expansion at z0 f (z) =

∞ 

an (z − z0 )n

n=0

for all z ∈ D, and the coefficients are given by an =

f (n) (z0 ) n!

for all n ≥ 0.

Proof. Fix z ∈ D. By the Cauchy integral formula, we have
f (ζ) 1 dζ , (10) f (z) = 2πi C ζ − z where C denotes the boundary of the disc and z ∈ D. The idea is to write (11)

1 1 1 1

, = = ζ −z ζ − z0 − (z − z0 ) ζ − z0 1 − z−z0 ζ−z0

and use the geometric series expansion. Since ζ ∈ C and z ∈ D is fixed, there exists 0 < r < 1 such that    z − z0     ζ − z0  < r,

50

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

therefore (12)

1−

1

z−z0 ζ−z0

=

n ∞   z − z0 n=0

ζ − z0

,

where the series converges uniformly for ζ ∈ C. This allows us to interchange the infinite sum with the integral when we combine (10), (11), and (12), thereby obtaining 
∞   1 f (ζ) f (z) = dζ · (z − z0 )n . 2πi C (ζ − z0 )n+1 n=0

This proves the power series expansion; further the use of the Cauchy integral formulas for the derivatives (or simply differentiation of the series) proves the formula for an . Observe that since power series define indefinitely (complex) differentiable functions, the theorem gives another proof that a holomorphic function is automatically indefinitely differentiable. Another important observation is that the power series expansion of f centered at z0 converges in any disc, no matter how large, as long as its closure is contained in Ω. In particular, if f is entire (that is, holomorphic on all of C), the theorem ∞ implies that f has a power series expansion around 0, say f (z) = n=0 an z n , that converges in all of C. Corollary 4.5 (Liouville’s theorem) If f is entire and bounded, then f is constant. Proof. It suffices to prove that f  = 0, since C is connected, and we may then apply Corollary 3.4 in Chapter 1. For each z0 ∈ C and all R > 0, the Cauchy inequalities yield |f  (z0 )| ≤

B R

where B is a bound for f . Letting R → ∞ gives the desired result. As an application of our work so far, we can give an elegant proof of the fundamental theorem of algebra. Corollary 4.6 Every non-constant polynomial P (z) = an z n + · · · + a0 with complex coefficients has a root in C. Proof. If P has no roots, then 1/P (z) is a bounded holomorphic function. To see this, we can of course assume that an = 0, and write

a P (z) a0  n−1 + · · · + = a + n zn z zn

51

4. Cauchy’s integral formulas

whenever z = 0. Since each term in the parentheses goes to 0 as |z| → ∞ we conclude that there exists R > 0 so that if c = |an |/2, then |P (z)| ≥ c|z|n

whenever |z| > R.

In particular, P is bounded from below when |z| > R. Since P is continuous and has no roots in the disc |z| ≤ R, it is bounded from below in that disc as well, thereby proving our claim. By Liouville’s theorem we then conclude that 1/P is constant. This contradicts our assumption that P is non-constant and proves the corollary. Corollary 4.7 Every polynomial P (z) = an z n + · · · + a0 of degree n ≥ 1 has precisely n roots in C. If these roots are denoted by w1 , . . . , wn , then P can be factored as P (z) = an (z − w1 )(z − w2 ) · · · (z − wn ). Proof. By the previous result P has a root, say w1 . Then, writing z = (z − w1 ) + w1 , inserting this expression for z in P , and using the binomial formula we get P (z) = bn (z − w1 )n + · · · + b1 (z − w1 ) + b0 , where b0 , . . . , bn−1 are new coefficients, and bn = an . Since P (w1 ) = 0, we find that b0 = 0, therefore

  P (z) = (z − w1 ) bn (z − w1 )n−1 + · · · + b1 = (z − w1 )Q(z), where Q is a polynomial of degree n − 1. By induction on the degree of the polynomial, we conclude that P (z) has n roots and can be expressed as P (z) = c(z − w1 )(z − w2 ) · · · (z − wn ) for some c ∈ C. Expanding the right-hand side, we realize that the coefficient of z n is c and therefore c = an as claimed. Finally, we end this section with a discussion of analytic continuation (the third of the “miracles” we mentioned in the introduction). It states that the “genetic code” of a holomorphic function is determined (that is, the function is fixed) if we know its values on appropriate arbitrarily small subsets. Note that in the theorem below, Ω is assumed connected.

52

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

Theorem 4.8 Suppose f is a holomorphic function in a region Ω that vanishes on a sequence of distinct points with a limit point in Ω. Then f is identically 0. In other words, if the zeros of a holomorphic function f in the connected open set Ω accumulate in Ω, then f = 0. Proof. Suppose that z0 ∈ Ω is a limit point for the sequence {wk }∞ k=1 and that f (wk ) = 0. First, we show that f is identically zero in a small disc containing z0 . For that, we choose a disc D centered at z0 and contained in Ω, and consider the power series expansion of f in that disc f (z) =

∞ 

an (z − z0 )n .

n=0

If f is not identically zero, there exists a smallest integer m such that am = 0. But then we can write f (z) = am (z − z0 )m (1 + g(z − z0 )), where g(z − z0 ) converges to 0 as z → z0 . Taking z = wk = z0 for a sequence of points converging to z0 , we get a contradiction since am (wk − z0 )m = 0 and 1 + g(wk − z0 ) = 0, but f (wk ) = 0. We conclude the proof using the fact that Ω is connected. Let U denote the interior of the set of points where f (z) = 0. Then U is open by definition and non-empty by the argument just given. The set U is also closed since if zn ∈ U and zn → z, then f (z) = 0 by continuity, and f vanishes in a neighborhood of z by the argument above. Hence z ∈ U . Now if we let V denote the complement of U in Ω, we conclude that U and V are both open, disjoint, and Ω = U ∪ V. Since Ω is connected we conclude that either U or V is empty. (Here we use one of the two equivalent definitions of connectedness discussed in Chapter 1.) Since z0 ∈ U , we find that U = Ω and the proof is complete. An immediate consequence of the theorem is the following. Corollary 4.9 Suppose f and g are holomorphic in a region Ω and f (z) = g(z) for all z in some non-empty open subset of Ω (or more generally for z in some sequence of distinct points with limit point in Ω). Then f (z) = g(z) throughout Ω.

53

5. Further applications

Suppose we are given a pair of functions f and F analytic in regions Ω and Ω , respectively, with Ω ⊂ Ω . If the two functions agree on the smaller set Ω, we say that F is an analytic continuation of f into the region Ω . The corollary then guarantees that there can be only one such analytic continuation, since F is uniquely determined by f .

5 Further applications We gather in this section various consequences of the results proved so far. 5.1 Morera’s theorem A direct application of what was proved here is a converse of Cauchy’s theorem. Theorem 5.1 Suppose f is a continuous function in the open disc D such that for any triangle T contained in D
f (z) dz = 0, T

then f is holomorphic. Proof. By the proof of Theorem 2.1 the function f has a primitive F in D that satisfies F  = f . By the regularity theorem, we know that F is indefinitely (and hence twice) complex differentiable, and therefore f is holomorphic. 5.2 Sequences of holomorphic functions Theorem 5.2 If {fn }∞ n=1 is a sequence of holomorphic functions that converges uniformly to a function f in every compact subset of Ω, then f is holomorphic in Ω. Proof. Let D be any disc whose closure is contained in Ω and T any triangle in that disc. Then, since each fn is holomorphic, Goursat’s theorem implies
fn (z) dz = 0 for all n. T

By assumption fn → f uniformly in the closure of D, so f is continuous and

fn (z) dz → f (z) dz. T

T

54

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

 As a result, we find T f (z) dz = 0, and by Morera’s theorem, we conclude that f is holomorphic in D. Since this conclusion is true for every D whose closure is contained in Ω, we find that f is holomorphic in all of Ω. This is a striking result that is obviously not true in the case of real variables: the uniform limit of continuously differentiable functions need not be differentiable. For example, we know that every continuous function on [0, 1] can be approximated uniformly by polynomials, by Weierstrass’s theorem (see Chapter 5, Book I), yet not every continuous function is differentiable. We can go one step further and deduce convergence theorems for the sequence of derivatives. Recall that if f is a power series with radius of convergence R, then f  can be obtained by differentiating term by term the series for f , and moreover f  has radius of convergence R. (See Theorem 2.6 in Chapter 1.) This implies in particular that if Sn are the partial sums of f , then Sn converges uniformly to f  on every compact subset of the disc of convergence of f . The next theorem generalizes this fact. Theorem 5.3 Under the hypotheses of the previous theorem, the se quence of derivatives {fn }∞ n=1 converges uniformly to f on every compact subset of Ω. Proof. We may assume without loss of generality that the sequence of functions in the theorem converges uniformly on all of Ω. Given δ > 0, let Ωδ denote the subset of Ω defined by Ωδ = {z ∈ Ω : Dδ (z) ⊂ Ω}. In other words, Ωδ consists of all points in Ω which are at distance > δ from its boundary. To prove the theorem, it suffices to show that {fn } converges uniformly to f  on Ωδ for each δ. This is achieved by proving the following inequality: (13)

sup |F  (z)| ≤ z∈Ωδ

1 sup |F (ζ)| δ ζ∈Ω

whenever F is holomorphic in Ω, since it can then be applied to F = fn − f to prove the desired fact. The inequality (13) follows at once from the Cauchy integral formula and the definition of Ωδ , since for every z ∈ Ωδ the closure of Dδ (z) is contained in Ω and
F (ζ) 1  dζ. F (z) = 2πi Cδ (z) (ζ − z)2

55

5. Further applications

Hence,


|F (ζ)| 1 |dζ| |F (z)| ≤ 2π Cδ (z) |ζ − z|2 1 1 ≤ sup |F (ζ)| 2 2πδ 2π ζ∈Ω δ 1 = sup |F (ζ)| , δ ζ∈Ω 

as was to be shown. Of course, there is nothing special about the first derivative, and in fact under the hypotheses of the last theorem, we may conclude (arguing (k) as above) that for every k ≥ 0 the sequence of kth derivatives {fn }∞ n=1 converges uniformly to f (k) on every compact subset of Ω. In practice, one often uses Theorem 5.2 to construct holomorphic functions (say, with a prescribed property) as a series (14)

F (z) =

∞ 

fn (z).

n=1

Indeed, if each fn is holomorphic in a given region Ω of the complex plane, and the series converges uniformly in compact subsets of Ω, then Theorem 5.2 guarantees that F is also holomorphic in Ω. For instance, various special functions are often expressed in terms of a converging series like (14). A specific example is the Riemann zeta function discussed in Chapter 6. We now turn to a variant of this idea, which consists of functions defined in terms of integrals. 5.3 Holomorphic functions defined in terms of integrals As we shall see later in this book, a number of other special functions are defined in terms of integrals of the type


f (z) =

b

F (z, s) ds, a

or as limits of such integrals. Here, the function F is holomorphic in the first argument, and continuous in the second. The integral is taken in the sense of Riemann integration over the bounded interval [a, b]. The problem then is to establish that f is holomorphic.

56

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

In the next theorem, we impose a sufficient condition on F , often satisfied in practice, that easily implies that f is holomorphic. After a simple linear change of variables, we may assume that a = 0 and b = 1. Theorem 5.4 Let F (z, s) be defined for (z, s) ∈ Ω × [0, 1] where Ω is an open set in C. Suppose F satisfies the following properties: (i) F (z, s) is holomorphic in z for each s. (ii) F is continuous on Ω × [0, 1]. Then the function f defined on Ω by
1 f (z) = F (z, s) ds 0

is holomorphic. The second condition says that F is jointly continuous in both arguments. To prove this result, it suffices to prove that f is holomorphic in any disc D contained in Ω, and by Morera’s theorem this could be achieved by showing that for any triangle T contained in D we have

1 F (z, s) ds dz = 0. T

0

Interchanging the order of integration, and using property (i) would then yield the desired result. We can, however, get around the issue of justifying the change in the order of integration by arguing differently. The idea is to interpret the integral as a “uniform” limit of Riemann sums, and then apply the results of the previous section. Proof. For each n ≥ 1, we consider the Riemann sum fn (z) = (1/n)

n 

F (z, k/n).

k=1

Then fn is holomorphic in all of Ω by property (i), and we claim that on any disc D whose closure is contained in Ω, the sequence {fn }∞ n=1 converges uniformly to f . To see this, we recall that a continuous function on a compact set is uniformly continuous, so if  > 0 there exists δ > 0 such that sup |F (z, s1 ) − F (z, s2 )| <  z∈D

whenever |s1 − s2 | < δ.

57

5. Further applications

Then, if n > 1/δ, and z ∈ D we have

  n
k/n     F (z, k/n) − F (z, s) ds |fn (z) − f (z)| =    k=1 (k−1)/n n
k/n  ≤ |F (z, k/n) − F (z, s)| ds
0, there exists δ > 0 such that sup |F (z, t1 ) − F (z, t2 )| < 

whenever |t1 − t2 | < δ.

z∈K

Arguing as in the  1 proof of Theorem 5.4, we see that the Riemann sums of the integral 0 F (z, t) dt approximate it uniformly on K. Since each of these Riemann sums is a rational function with singularities on γ, the lemma is proved. Finally, the process of pushing the poles to infinity is accomplished by using the fact that K c is connected. Since any rational function whose only singularity is at the point z0 is a polynomial in 1/(z − z0 ), it suffices to establish the next lemma to complete the proof of Theorem 5.7. / K, then the function Lemma 5.10 If K c is connected and z0 ∈ 1/(z − z0 ) can be approximated uniformly on K by polynomials. Proof. First, we choose a point z1 that is outside a large open disc D centered at the origin and which contains K. Then

 1 1 1 zn =− = − n+1 , z − z1 z1 1 − z/z1 z1 ∞

n=1

where the series converges uniformly for z ∈ K. The partial sums of this series are polynomials that provide a uniform approximation to 1/(z − z1 ) on K. In particular, this implies that any power 1/(z − z1 )k can also be approximated uniformly on K by polynomials. It now suffices to prove that 1/(z − z0 ) can be approximated uniformly on K by polynomials in 1/(z − z1 ). To do so, we use the fact that K c is connected to travel from z0 to the point z1 . Let γ be a curve in K c that is parametrized by γ(t) on [0, 1], and such that γ(0) = z0 and γ(1) = z1 . If we let ρ = 12 d(K, γ), then ρ > 0 since γ and K are compact. We then choose a sequence of points {w1 , . . . , w } on γ such that w0 = z0 , w = z1 , and |wj − wj+1 | < ρ for all 0 ≤ j < . We claim that if w is a point on γ, and w any other point with |w − w | < ρ, then 1/(z − w) can be approximated uniformly on K by polynomials in 1/(z − w ). To see this, note that 1 1 1 =   z−w z − w 1 − w−w z−w ∞  (w − w )n = , (z − w )n+1 n=0

64

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

and since the sum converges uniformly for z ∈ K, the approximation by partial sums proves our claim. This result allows us to travel from z0 to z1 through the finite sequence {wj } to find that 1/(z − z0 ) can be approximated uniformly on K by polynomials in 1/(z − z1 ). This concludes the proof of the lemma, and also that of the theorem.

6 Exercises 1. Prove that


∞




∞

sin(x2 ) dx =

0

√ cos(x2 ) dx =

0

These are the Fresnel integrals. Here, [Hint: Integrate the function e  ∞ −x2 √ e dx = π.] −∞

−z 2

∞ 0

2π . 4

is interpreted as limR→∞

over the path in Figure 14.

R 0

.

Recall that

π

Rei 4

0

R

Figure 14. The contour in Exercise 1




∞

π sin x dx = . x 2 0 ∞ 1 [Hint: The integral equals 2i −∞ 2. Show that

eix −1 x

dx. Use the indented semicircle.]

3. Evaluate the integrals


∞

e−ax cos bx dx

0

by integrating e−Az , A = cos ω = a/A.




∞

and

e−ax sin bx dx ,

a>0

0

√

a2 + b2 , over an appropriate sector with angle ω, with

65

6. Exercises 2

4. Prove that for all ξ ∈ C we have e−πξ =




∞

2

e−πx e2πixξ dx.

−∞

5. Suppose f is continuously complex differentiable on Ω, and T ⊂ Ω is a triangle whose interior is also contained in Ω. Apply Green’s theorem to show that
f (z) dz = 0. T

This provides a proof of Goursat’s theorem under the additional assumption that f  is continuous. [Hint: Green’s theorem says that if (F, G) is a continuously differentiable vector field, then  

∂F ∂G − dxdy. F dx + G dy = ∂y T Interior of T ∂x For appropriate F and G, one can then use the Cauchy-Riemann equations.] 6. Let Ω be an open subset of C and let T ⊂ Ω be a triangle whose interior is also contained in Ω. Suppose that f is a function holomorphic in Ω except possibly at a point w inside T . Prove that if f is bounded near w, then
f (z) dz = 0. T

7. Suppose f : D → C is holomorphic. Show that the diameter d = supz, w∈D |f (z) − f (w)| of the image of f satisfies 2|f  (0)| ≤ d. Moreover, it can be shown that equality holds precisely when f is linear, f (z) = a0 + a1 z. Note. In connection with this result, see the relationship between the diameter of a curve and Fourier series described in Problem 1, Chapter 4, Book I.  f (ζ)−f (−ζ) 1 dζ whenever 0 < r < 1.] [Hint: 2f  (0) = 2πi |ζ|=r ζ2 8. If f is a holomorphic function on the strip −1 < y < 1, x ∈ R with |f (z)| ≤ A(1 + |z|)η ,

η a fixed real number

for all z in that strip, show that for each integer n ≥ 0 there exists An ≥ 0 so that |f (n) (x)| ≤ An (1 + |x|)η , [Hint: Use the Cauchy inequalities.]

for all x ∈ R.

66

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

9. Let Ω be a bounded open subset of C, and ϕ : Ω → Ω a holomorphic function. Prove that if there exists a point z0 ∈ Ω such that ϕ(z0 ) = z0

ϕ (z0 ) = 1

and

then ϕ is linear. [Hint: Why can one assume that z0 = 0? Write ϕ(z) = z + an z n + O(z n+1 ) near 0, and prove that if ϕk = ϕ ◦ · · · ◦ ϕ (where ϕ appears k times), then ϕk (z) = z + kan z n + O(z n+1 ). Apply the Cauchy inequalities and let k → ∞ to conclude the proof. Here we use the standard O notation, where f (z) = O(g(z)) as z → 0 means that |f (z)| ≤ C|g(z)| for some constant C as |z| → 0.] 10. Weierstrass’s theorem states that a continuous function on [0, 1] can be uniformly approximated by polynomials. Can every continuous function on the closed unit disc be approximated uniformly by polynomials in the variable z? 11. Let f be a holomorphic function on the disc DR0 centered at the origin and of radius R0 . (a) Prove that whenever 0 < R < R0 and |z| < R, then 1 f (z) = 2π






2π iϕ

f (Re )Re 0

Reiϕ + z Reiϕ − z

 dϕ.

(b) Show that  Re

Reiγ + r Reiγ − r

 =

R2

R2 − r 2 . − 2Rr cos γ + r 2

[Hint: For the first part, note that if w = R2 /z, then the integral of f (ζ)/(ζ − w) around the circle of radius R centered at the origin is zero. Use this, together with the usual Cauchy integral formula, to deduce the desired identity.] 12. Let u be a real-valued function defined on the unit disc D. Suppose that u is twice continuously differentiable and harmonic, that is,

u(x, y) = 0 for all (x, y) ∈ D. (a) Prove that there exists a holomorphic function f on the unit disc such that Re(f ) = u. Also show that the imaginary part of f is uniquely defined up to an additive (real) constant. [Hint: From the previous chapter we would have f  (z) = 2∂u/∂z. Therefore, let g(z) = 2∂u/∂z and prove that g is holomorphic. Why can one find F with F  = g? Prove that Re(F ) differs from u by a real constant.]

67

7. Problems

(b) Deduce from this result, and from Exercise 11, the Poisson integral representation formula from the Cauchy integral formula: If u is harmonic in the unit disc and continuous on its closure, then if z = reiθ one has u(z) =

1 2π




2π

Pr (θ − ϕ)u(ϕ) dϕ

0

where Pr (γ) is the Poisson kernel for the unit disc given by Pr (γ) =

1 − r2 . 1 − 2r cos γ + r 2

13. Suppose f is an analytic function defined everywhere in C and such that for each z0 ∈ C at least one coefficient in the expansion ∞ 

f (z) =

cn (z − z0 )n

n=0

is equal to 0. Prove that f is a polynomial. [Hint: Use the fact that cn n! = f (n) (z0 ) and use a countability argument.] 14. Suppose that f is holomorphic in an open set containing the closed unit disc, except for a pole at z0 on the unit circle. Show that if ∞ 

an z n

n=0

denotes the power series expansion of f in the open unit disc, then lim

n→∞

an = z0 . an+1

15. Suppose f is a non-vanishing continuous function on D that is holomorphic in D. Prove that if |f (z)| = 1

whenever |z| = 1,

then f is constant. [Hint: Extend f to all of C by f (z) = 1/f (1/z) whenever |z| > 1, and argue as in the Schwarz reflection principle.]

7 Problems 1. Here are some examples of analytic functions on the unit disc that cannot be extended analytically past the unit circle. The following definition is needed. Let

68

Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

f be a function defined in the unit disc D, with boundary circle C. A point w on C is said to be regular for f if there is an open neighborhood U of w and an analytic function g on U , so that f = g on D ∩ U . A function f defined on D cannot be continued analytically past the unit circle if no point of C is regular for f . (a) Let f (z) =

∞ 

n

for |z| < 1.

z2

n=0

Notice that the radius of convergence of the above series is 1. Show that f cannot be continued analytically past the unit disc. [Hint: Suppose θ = 2πp/2k , where p and k are positive integers. Let z = reiθ ; then |f (reiθ )| → ∞ as r → 1.] (b) ∗ Fix 0 < α < ∞. Show that the analytic function f defined by f (z) =

∞ 

n

2−nα z 2

for |z| < 1

n=0

extends continuously to the unit circle, but cannot be analytically continued past the unit circle. [Hint: There is a nowhere differentiable function lurking in the background. See Chapter 4 in Book I.] 2.∗ Let F (z) =

∞ 

d(n)z n

for |z| < 1

n=1

where d(n) denotes the number of divisors of n. Observe that the radius of convergence of this series is 1. Verify the identity ∞ 

d(n)z n =

n=1

∞  n=1

zn . 1 − zn

Using this identity, show that if z = r with 0 < r < 1, then |F (r)| ≥ c

1 log(1/(1 − r)) 1−r

as r → 1. Similarly, if θ = 2πp/q where p and q are positive integers and z = reiθ , then 1 |F (reiθ )| ≥ cp/q log(1/(1 − r)) 1−r as r → 1. Conclude that F cannot be continued analytically past the unit disc.  3. Morera’s theorem states that if f is continuous in C, and T f (z) dz = 0 for all triangles T , then f is holomorphic in C. Naturally, we may ask if the conclusion still holds if we replace triangles by other sets.

69

7. Problems

(a) Suppose that f is continuous on C, and
(16)

f (z) dz = 0 C

for every circle C. Prove that f is holomorphic. (b) More generally, let Γ be any toy contour, and F the collection of all translates and dilates of Γ. Show that if f is continuous on C, and
f (z) dz = 0

for all γ ∈ F

γ

then f is holomorphic. In particular, Morera’s theorem holds under the weaker assumption that T f (z) dz = 0 for all equilateral triangles. [Hint: As a first step, assume that f is twice real differentiable, and write f (z) = f (z0 ) + a(z − z0 ) + b(z − z0 ) + O(|z − z0 |2 ) for z near z0 . Integrating this expansion over small circles around z0 yields ∂f /∂z = b = 0 at z0 . Alternatively, suppose only that f is differentiable and apply Green’s theorem to conclude that the real and imaginary parts of f satisfy the Cauchy-Riemann equations. In general, let ϕ(w)  = ϕ(x, y) (when w = x + iy) denote a smoothfunction with 0 ≤ ϕ(w) ≤ 1, and R2 ϕ(w) dV (w) = 1, where dV (w) = dxdy, and denotes the usual integral of a function of two variables in R2 . For each > 0, let ϕ (z) = −2 ϕ( −1 z), as well as
f (z) =

R2

f (z − w)ϕ (w) dV (w),

where the integral denotes the usual integral of functions of two variables, with dV (w) the area element of R2 . Then f is smooth, satisfies condition (16), and f → f uniformly on any compact subset of C.] 4. Prove the converse to Runge’s theorem: if K is a compact set whose complement if not connected, then there exists a function f holomorphic in a neighborhood of K which cannot be approximated uniformly by polynomial on K. [Hint: Pick a point z0 in a bounded component of K c , and let f (z) = 1/(z − z0 ). If f can be approximated uniformly by polynomials on K, show that there exists a polynomial p such that |(z − z0 )p(z) − 1| < 1. Use the maximum modulus principle (Chapter 3) to show that this inequality continues to hold for all z in the component of K c that contains z0 .] 5.∗ There exists an entire function F with the following “universal” property: given any entire function h, there is an increasing sequence {Nk }∞ k=1 of positive integers, so that lim F (z + Nk ) = h(z)

n→∞

uniformly on every compact subset of C.

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Chapter 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

(a) Let p1 , p2 , . . . denote an enumeration of the collection of polynomials whose coefficients have rational real and imaginary parts. Show that it suffices to find an entire function F and an increasing sequence {Mn } of positive integers, such that (17)

|F (z) − pn (z − Mn )|
0 and Mn > 0 are chosen appropriately with cn → 0 and Mn → ∞. [Hint: The function 2 e−z vanishes rapidly as |z| → ∞ in the sectors {| arg z| < π/4 − δ} and {|π − arg z| < π/4 − δ}.] In the same spirit, there exists an alternate “universal” entire function G with the following property: given any entire function h, there is an increasing sequence {Nk }∞ k=1 of positive integers, so that lim DNk G(z) = h(z)

k→∞

uniformly on every compact subset of C. Here Dj G denotes the j th (complex) derivative of G.

3 Meromorphic Functions and the Logarithm One knows that the differential calculus, which has contributed so much to the progress of analysis, is founded on the consideration of differential coefficients, that is derivatives of functions. When one attributes an infinitesimal increase to the variable x, the function f (x) of this variable undergoes in general an infinitesimal increase of which the first term is proportional to , and the finite coefficient of of this increase is what is called its differential coefficient... If considering the values of x where f (x) becomes infinite, we add to one of these values designated by x1 , the infinitesimal , and then develop f (x1 + ) in increasing power of the same quantity, the first terms of this development contain negative powers of ; one of these will be the product of 1/ with a finite coefficient, which we will call the residue of the function f (x), relative to the particular value x1 of the variable x. Residues of this kind present themselves naturally in several branches of algebraic and infinitesimal analysis. Their consideration furnish methods that can be simply used, that apply to a large number of diverse questions, and that give new formulae that would seem to be of interest to mathematicians... A. L. Cauchy, 1826

There is a general principle in the theory, already implicit in Riemann’s work, which states that analytic functions are in an essential way characterized by their singularities. That is to say, globally analytic functions are “effectively” determined by their zeros, and meromorphic functions by their zeros and poles. While these assertions cannot be formulated as precise general theorems, there are nevertheless significant instances where this principle applies. We begin this chapter by considering singularities, in particular the different kind of point singularities (“isolated” singularities) that a holomorphic function can have. In order of increasing severity, these are: • removable singularities

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• poles • essential singularities. The first type is harmless since a function can actually be extended to be holomorphic at its removable singularities (hence the name). Near the third type, the function oscillates and may grow faster than any power, and a complete understanding of its behavior is not easy. For the second type the analysis is more straight forward and is connected with the calculus of residues, which arises as follows. Recall that by Cauchy’s theorem a holomorphic function f in an open set which contains a closed curve γ and its interior satisfies
f (z) dz = 0. γ

The question that occurs is: what happens if f has a pole in the interior of the curve? To try to answer this question consider the example f (z) = 1/z, and recall that if C is a (positively oriented) circle centered at 0, then
dz = 2πi. C z This turns out to be the key ingredient in the calculus of residues. A new aspect appears when we consider indefinite integrals of holomorphic functions that have singularities. As the basic example f (z) = 1/z shows, the resulting “function” (in this case the logarithm) may not be single-valued, and understanding this phenomenon is of importance for a number of subjects. Exploiting this multi-valuedness leads in effect to the “argument principle.” We can use this principle to count the number of zeros of a holomorphic function inside a suitable curve. As a simple consequence of this result, we obtain a significant geometric property of holomorphic functions: they are open mappings. From this, the maximum principle, another important feature of holomorphic functions, is an easy step. In order to turn to the logarithm itself, and come to grips with the precise nature of its multi-valuedness, we introduce the notions of homotopy of curves and simply connected domains. It is on the latter type of open sets that single-valued branches of the logarithm can be defined.

1 Zeros and poles By definition, a point singularity of a function f is a complex number z0 such that f is defined in a neighborhood of z0 but not at the point

73

1. Zeros and poles

z0 itself. We shall also call such points isolated singularities. For example, if the function f is defined only on the punctured plane by f (z) = z, then the origin is a point singularity. Of course, in that case, the function f can actually be defined at 0 by setting f (0) = 0, so that the resulting extension is continuous and in fact entire. (Such points are then called removable singularities.) More interesting is the case of the function g(z) = 1/z defined in the punctured plane. It is clear now that g cannot be defined as a continuous function, much less as a holomorphic function, at the point 0. In fact, g(z) grows to infinity as z approaches 0, and we shall say that the origin is a pole singularity. Finally, the case of the function h(z) = e1/z on the punctured plane shows that removable singularities and poles do not tell the whole story. Indeed, the function h(z) grows indefinitely as z approaches 0 on the positive real line, while h approaches 0 as z goes to 0 on the negative real axis. Finally h oscillates rapidly, yet remains bounded, as z approaches the origin on the imaginary axis. Since singularities often appear because the denominator of a fraction vanishes, we begin with a local study of the zeros of a holomorphic function. A complex number z0 is a zero for the holomorphic function f if f (z0 ) = 0. In particular, analytic continuation shows that the zeros of a non-trivial holomorphic function are isolated. In other words, if f is holomorphic in Ω and f (z0 ) = 0 for some z0 ∈ Ω, then there exists an open neighborhood U of z0 such that f (z) = 0 for all z ∈ U − {z0 }, unless f is identically zero. We start with a local description of a holomorphic function near a zero. Theorem 1.1 Suppose that f is holomorphic in a connected open set Ω, has a zero at a point z0 ∈ Ω, and does not vanish identically in Ω. Then there exists a neighborhood U ⊂ Ω of z0 , a non-vanishing holomorphic function g on U , and a unique positive integer n such that f (z) = (z − z0 )n g(z)

for all z ∈ U .

Proof. Since Ω is connected and f is not identically zero, we conclude that f is not identically zero in a neighborhood of z0 . In a small disc centered at z0 the function f has a power series expansion f (z) =

∞ 

ak (z − z0 )k .

k=0

Since f is not identically zero near z0 , there exists a smallest integer n

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such that an = 0. Then, we can write f (z) = (z − z0 )n [an + an+1 (z − z0 ) + · · · ] = (z − z0 )n g(z) , where g is defined by the series in brackets, and hence is holomorphic, and is nowhere vanishing for all z close to z0 (since an = 0). To prove the uniqueness of the integer n, suppose that we can also write f (z) = (z − z0 )n g(z) = (z − z0 )m h(z) where h(z0 ) = 0. If m > n, then we may divide by (z − z0 )n to see that g(z) = (z − z0 )m−n h(z) and letting z → z0 yields g(z0 ) = 0, a contradiction. If m < n a similar argument gives h(z0 ) = 0, which is also a contradiction. We conclude that m = n, thus h = g, and the theorem is proved. In the case of the above theorem, we say that f has a zero of order n (or multiplicity n) at z0 . If a zero is of order 1, we say that it is simple. We observe that, quantitatively, the order describes the rate at which the function vanishes. The importance of the previous theorem comes from the fact that we can now describe precisely the type of singularity possessed by the function 1/f at z0 . For this purpose, it is now convenient to define a deleted neighborhood of z0 to be an open disc centered at z0 , minus the point z0 , that is, the set {z : 0 < |z − z0 | < r} for some r > 0. Then, we say that a function f defined in a deleted neighborhood of z0 has a pole at z0 , if the function 1/f , defined to be zero at z0 , is holomorphic in a full neighborhood of z0 . Theorem 1.2 If f has a pole at z0 ∈ Ω, then in a neighborhood of that point there exist a non-vanishing holomorphic function h and a unique positive integer n such that f (z) = (z − z0 )−n h(z). Proof. By the previous theorem we have 1/f (z) = (z − z0 )n g(z), where g is holomorphic and non-vanishing in a neighborhood of z0 , so the result follows with h(z) = 1/g(z).

75

1. Zeros and poles

The integer n is called the order (or multiplicity) of the pole, and describes the rate at which the function grows near z0 . If the pole is of order 1, we say that it is simple. The next theorem should be reminiscent of power series expansion, except that now we allow terms of negative order, to account for the presence of a pole. Theorem 1.3 If f has a pole of order n at z0 , then (1)

f (z) =

a−n+1 a−1 a−n + G(z) , + +···+ (z − z0 )n (z − z0 )n−1 (z − z0 )

where G is a holomorphic function in a neighborhood of z0 . Proof. The proof follows from the multiplicative statement in the previous theorem. Indeed, the function h has a power series expansion h(z) = A0 + A1 (z − z0 ) + · · · so that f (z) = (z − z0 )−n (A0 + A1 (z − z0 ) + · · · ) a−n+1 a−1 a−n + G(z). + +···+ = n n−1 (z − z0 ) (z − z0 ) (z − z0 ) The sum a−n a−n+1 a−1 + + ···+ (z − z0 )n (z − z0 )n−1 (z − z0 ) is called the principal part of f at the pole z0 , and the coefficient a−1 is the residue of f at that pole. We write resz0 f = a−1 . The importance of the residue comes from the fact that all the other terms in the principal part, that is, those of order strictly greater than 1, have primitives in a deleted neighborhood of z0 . Therefore, if P (z) denotes the principal part above and C is any circle centered at z0 , we get
1 P (z) dz = a−1 . 2πi C We shall return to this important point in the section on the residue formula. As we shall see, in many cases, the evaluation of integrals reduces to the calculation of residues. In the case when f has a simple pole at z0 , it is clear that resz0 f = lim (z − z0 )f (z). z→z0

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

If the pole is of higher order, a similar formula holds, one that involves differentiation as well as taking a limit. Theorem 1.4 If f has a pole of order n at z0 , then 1 resz0 f = lim z→z0 (n − 1)!



d dz

n−1 (z − z0 )n f (z).

The theorem is an immediate consequence of formula (1), which implies (z − z0 )n f (z) = a−n + a−n+1 (z − z0 ) + · · · + a−1 (z − z0 )n−1 + +G(z)(z − z0 )n .

2 The residue formula We now discuss the celebrated residue formula. Our approach follows the discussion of Cauchy’s theorem in the last chapter: we first consider the case of the circle and its interior the disc, and then explain generalizations to toy contours and their interiors. Theorem 2.1 Suppose that f is holomorphic in an open set containing a circle C and its interior, except for a pole at z0 inside C. Then
f (z) dz = 2πi resz0 f. C

Proof. Once again, we may choose a keyhole contour that avoids the pole, and let the width of the corridor go to zero to see that

f (z) dz = f (z) dz C

C


where C is the small circle centered at the pole z0 and of radius . Now we observe that
1 a−1 dz = a−1 2πi C
z − z0 is an immediate consequence of the Cauchy integral formula (Theorem 4.1 of the previous chapter), applied to the constant function f = a−1 . Similarly,
1 a−k dz = 0 2πi C
(z − z0 )k

77

2. The residue formula

when k > 1, by using the corresponding formulae for the derivatives (Corollary 4.2 also in the previous chapter). But we know that in a neighborhood of z0 we can write f (z) =

a−n+1 a−1 a−n + +···+ + G(z), n n−1 (z − z0 ) (z − z0 ) z − z0

where G is holomorphic.   By Cauchy’s theorem, we also know that G(z) dz = 0, hence f (z) dz = a−1 . This implies the desired reC
C
sult. This theorem can be generalized to the case of finitely many poles in the circle, as well as to the case of toy contours. Corollary 2.2 Suppose that f is holomorphic in an open set containing a circle C and its interior, except for poles at the points z1 , . . . , zN inside C. Then
N  f (z) dz = 2πi reszk f. C

k=1

For the proof, consider a multiple keyhole which has a loop avoiding each one of the poles. Let the width of the corridors go to zero. In the limit, the integral over the large circle equals a sum of integrals over small circles to which Theorem 2.1 applies. Corollary 2.3 Suppose that f is holomorphic in an open set containing a toy contour γ and its interior, except for poles at the points z1 , . . . , zN inside γ. Then
N  f (z) dz = 2πi reszk f. γ

k=1

In the above, we take γ to have positive orientation. The proof consists of choosing a keyhole appropriate for the given toy contour, so that, as we have seen previously, we can reduce the situation to integrating over small circles around the poles where Theorem 2.1 applies.  N The identity γ f (z) dz = 2πi k=1 reszk f is referred to as the residue formula. 2.1 Examples The calculus of residues provides a powerful technique to compute a wide range of integrals. In the examples we give next, we evaluate three

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

improper Riemann integrals of the form




∞

f (x) dx. −∞

The main idea is to extend f to the complex plane, and then choose a family γR of toy contours so that







lim

∞

f (z) dz =

R→∞

f (x) dx. −∞

γR

 By computing the residues of f at its poles, we easily obtain γR f (z) dz. The challenging part is to choose the contours γR , so that the above limit holds. Often, this choice is motivated by the decay behavior of f . Example 1. First, we prove that




∞

(2) −∞

dx =π 1 + x2

by using contour integration. Note that if we make the change of variables x → x/y, this yields 1 π




∞ −∞

y dx = 2 y + x2




∞

−∞

Py (x) dx.

In other words, formula (2) says that the integral of the Poisson kernel Py (x) is equal to 1 for each y > 0. This was proved quite easily in Lemma 2.5 of Chapter 5 in Book I, since 1/(1 + x2 ) is the derivative of arctan x. Here we provide a residue calculation that leads to another proof of (2). Consider the function f (z) =

1 , 1 + z2

which is holomorphic in the complex plane except for simple poles at the points i and −i. Also, we choose the contour γR shown in Figure 1. The contour consists of the segment [−R, R] on the real axis and of a large half-circle centered at the origin in the upper half-plane. Since we may write f (z) =

1 (z − i)(z + i)

79

2. The residue formula

γR

i 0

−R

R

Figure 1. The contour γR in Example 1

we see that the residue of f at i is simply 1/2i. Therefore, if R is large enough, we have
2πi f (z) dz = = π. 2i γR + If we denote by CR the large half-circle of radius R, we see that 
   B M   f (z) dz ,   ≤ πR 2 ≤ R R  C+  R

+ and R is where we have used the fact that |f (z)| ≤ B/|z|2 when z ∈ CR large. So this integral goes to 0 as R → ∞. Therefore, in the limit we find that
∞ f (x) dx = π, −∞

as desired. We remark that in this example, there is nothing special about our choice of the semicircle in the upper half-plane. One gets the same conclusion if one uses the semicircle in the lower half-plane, with the other pole and the appropriate residue. Example 2. An integral that will play an important role in Chapter 6 is
∞ eax π , 0 < a < 1. dx = x sin πa −∞ 1 + e To prove this formula, let f (z) = eaz /(1 + ez ), and consider the contour consisting of a rectangle in the upper half-plane with a side lying

80

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

2πi

γR

πi

0

−R

R

Figure 2. The contour γR in Example 2

on the real axis, and a parallel side on the line Im(z) = 2π, as shown in Figure 2. The only point in the rectangle γR where the denominator of f vanishes is z = πi. To compute the residue of f at that point, we argue as follows: First, note (z − πi)f (z) = eaz

z − πi z − πi = eaz z . z 1+e e − eπi

We recognize on the right the inverse of a difference quotient, and in fact ez − eπi = eπi = −1 z→πi z − πi lim

since ez is its own derivative. Therefore, the function f has a simple pole at πi with residue resπi f = −eaπi . As a consequence, the residue formula says that
(3) f = −2πieaπi . γR

We now investigate the integrals of f over each side of the rectangle. Let IR denote
R f (x) dx −R

and I the integral we wish to compute, so that IR → I as R → ∞. Then, it is clear that the integral of f over the top side of the rectangle (with

81

2. The residue formula

the orientation from right to left) is −e2πia IR . Finally, if AR = {R + it : 0 ≤ t ≤ 2π} denotes the vertical side on the right, then 

2π  a(R+it)     e  (a−1)R    f  ≤ ,   1 + eR+it  dt ≤ Ce AR 0 and since a < 1, this integral tends to 0 as R → ∞. Similarly, the integral over the vertical segment on the left goes to 0, since it can be bounded by Ce−aR and a > 0. Therefore, in the limit as R tends to infinity, the identity (3) yields I − e2πia I = −2πieaπi , from which we deduce eaπi 1 − e2πia 2πi = πia e − e−πia π , = sin πa

I = −2πi

and the computation is complete. Example 3. Now we calculate another Fourier transform, namely
∞ −2πixξ 1 e dx = cosh πx cosh πξ −∞ where cosh z =

ez + e−z . 2

In other words, the function 1/ cosh πx is its own Fourier transform, a 2 property also shared by e−πx (see Example 1, Chapter 2). To see this, we use a rectangle γR as shown on Figure 3 whose width goes to infinity, but whose height is fixed. For a fixed ξ ∈ R, let f (z) =

e−2πizξ cosh πz

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

γR

2i β

α −R

0

R

Figure 3. The contour γR in Example 3

and note that the denominator of f vanishes precisely when eπz = −e−πz , that is, when e2πz = −1. In other words, the only poles of f inside the rectangle are at the points α = i/2 and β = 3i/2. To find the residue of f at α, we note that 2(z − α) + e−πz (z − α) = 2e−2πizξ eπz 2πz . e − e2πα

(z − α)f (z) = e−2πizξ

eπz

We recognize on the right the reciprocal of the difference quotient for the function e2πz at z = α. Therefore

lim (z − α)f (z) = 2e−2πiαξ eπα

z→α

1 eπξ , = 2πe2πα πi

which shows that f has a simple pole at α with residue eπξ /(πi). Similarly, we find that f has a simple pole at β with residue −e3πξ /(πi). We dispense with the integrals of f on the vertical sides by showing that they go to zero as R tends to infinity. Indeed, if z = R + iy with 0 ≤ y ≤ 2, then |e−2πizξ | ≤ e4π|ξ| ,

3. Singularities and meromorphic functions

and

83

 πz   e + e−πz   | cosh πz| =   2  1 ≥  |eπz | − |e−πz |  2 1 ≥ (eπR − e−πR ) 2 → ∞ as R → ∞,

which shows that the integral over the vertical segment on the right goes to 0 as R → ∞. A similar argument shows that the integral of f over the vertical segment on the left also goes to 0 as R → ∞. Finally, we see that if I denotes the integral we wish to calculate, then the integral of f over the top side of the rectangle (with the orientation from right to left) is simply −e4πξ I where we have used the fact that cosh πζ is periodic with period 2i. In the limit as R tends to infinity, the residue formula gives  πξ  e3πξ e 4πξ I − e I = 2πi − πi πi = −2e2πξ (eπξ − e−πξ ), and since 1 − e4πξ = −e2πξ (e2πξ − e−2πξ ), we find that I =2

eπξ − e−πξ 1 eπξ − e−πξ 2 = 2 = = πξ 2πξ −2πξ πξ −πξ πξ −πξ −πξ e −e (e − e )(e + e ) e +e cosh πξ

as claimed. A similar argument actually establishes the following formula:
∞ 2 sinh 2πaξ sin πa dx = e−2πixξ cosh πx + cos πa sinh 2πξ −∞ whenever 0 < a < 1, and where sinh z = (ez − e−z )/2. We have proved above the particular case a = 1/2. This identity can be used to determine an explicit formula for the Poisson kernel for the strip (see Problem 3 in Chapter 5 of Book I), or to prove the sum of two squares theorem, as we shall see in Chapter 10.

3 Singularities and meromorphic functions Returning to Section 1, we see that we have described the analytical character of a function near a pole. We now turn our attention to the other types of isolated singularities.

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

Let f be a function holomorphic in an open set Ω except possibly at one point z0 in Ω. If we can define f at z0 in such a way that f becomes holomorphic in all of Ω, we say that z0 is a removable singularity for f . Theorem 3.1 (Riemann’s theorem on removable singularities) Suppose that f is holomorphic in an open set Ω except possibly at a point z0 in Ω. If f is bounded on Ω − {z0 }, then z0 is a removable singularity. Proof. Since the problem is local we may consider a small disc D centered at z0 and whose closure is contained in Ω. Let C denote the boundary circle of that disc with the usual positive orientation. We shall prove that if z ∈ D and z = z0 , then under the assumptions of the theorem we have
1 f (ζ) (4) f (z) = dζ. 2πi C ζ − z Since an application of Theorem 5.4 in the previous chapter proves that the right-hand side of equation (4) defines a holomorphic function on all of D that agrees with f (z) when z = z0 , this give us the desired extension. To prove formula (4) we fix z ∈ D with z = z0 and use the familiar toy contour illustrated in Figure 4.

z

z0

Figure 4. The multiple keyhole contour in the proof of Riemann’s theorem

The multiple keyhole avoids the two points z and z0 . Letting the sides of the corridors get closer to each other, and finally overlap, in the limit

3. Singularities and meromorphic functions

85

we get a cancellation:


f (ζ) f (ζ) f (ζ) dζ + dζ + dζ = 0 , C ζ −z γ
ζ − z γ
 ζ − z where γ and γ  are small circles of radius  with negative orientation and centered at z and z0 respectively. Copying the argument used in the proof of the Cauchy integral formula in Section 4 of Chapter 2, we find that
f (ζ) dζ = −2πif (z). ζ −z γ
For the second integral, we use the assumption that f is bounded and that since  is small, ζ stays away from z, and therefore 
   f (ζ)   dζ  ≤ C.   γ
 ζ − z  Letting  tend to 0 proves our contention and concludes the proof of the extension formula (4). Surprisingly, we may deduce from Riemann’s theorem a characterization of poles in terms of the behavior of the function in a neighborhood of a singularity. Corollary 3.2 Suppose that f has an isolated singularity at the point z0 . Then z0 is a pole of f if and only if |f (z)| → ∞ as z → z0 . Proof. If z0 is a pole, then we know that 1/f has a zero at z0 , and therefore |f (z)| → ∞ as z → z0 . Conversely, suppose that this condition holds. Then 1/f is bounded near z0 , and in fact 1/|f (z)| → 0 as z → z0 . Therefore, 1/f has a removable singularity at z0 and must vanish there. This proves the converse, namely that z0 is a pole. Isolated singularities belong to one of three categories: • Removable singularities (f bounded near z0 ) • Pole singularities (|f (z)| → ∞ as z → z0 ) • Essential singularities. By default, any singularity that is not removable or a pole is defined to be an essential singularity. For example, the function e1/z discussed at the very beginning of Section 1 has an essential singularity at

86

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

0. We already observed the wild behavior of this function near the origin. Contrary to the controlled behavior of a holomorphic function near a removable singularity or a pole, it is typical for a holomorphic function to behave erratically near an essential singularity. The next theorem clarifies this. Theorem 3.3 (Casorati-Weierstrass) Suppose f is holomorphic in the punctured disc Dr (z0 ) − {z0 } and has an essential singularity at z0 . Then, the image of Dr (z0 ) − {z0 } under f is dense in the complex plane. Proof. We argue by contradiction. Assume that the range of f is not dense, so that there exists w ∈ C and δ > 0 such that |f (z) − w| > δ

for all z ∈ Dr (z0 ) − {z0 }.

We may therefore define a new function on Dr (z0 ) − {z0 } by g(z) =

1 , f (z) − w

which is holomorphic on the punctured disc and bounded by 1/δ. Hence g has a removable singularity at z0 by Theorem 3.1. If g(z0 ) = 0, then f (z) − w is holomorphic at z0 , which contradicts the assumption that z0 is an essential singularity. In the case that g(z0 ) = 0, then f (z) − w has a pole at z0 also contradicting the nature of the singularity at z0 . The proof is complete. In fact, Picard proved a much stronger result. He showed that under the hypothesis of the above theorem, the function f takes on every complex value infinitely many times with at most one exception. Although we shall not give a proof of this remarkable result, a simpler version of it will follow from our study of entire functions in a later chapter. See Exercise 11 in Chapter 5. We now turn to functions with only isolated singularities that are poles. A function f on an open set Ω is meromorphic if there exists a sequence of points {z0 , z1 , z2 , . . .} that has no limit points in Ω, and such that (i) the function f is holomorphic in Ω − {z0 , z1 , z2 , . . .}, and (ii) f has poles at the points {z0 , z1 , z2 , . . .}. It is also useful to discuss functions that are meromorphic in the extended complex plane. If a function is holomorphic for all large values of

3. Singularities and meromorphic functions

87

z, we can describe its behavior at infinity using the tripartite distinction we have used to classify singularities at finite values of z. Thus, if f is holomorphic for all large values of z, we consider F (z) = f (1/z), which is now holomorphic in a deleted neighborhood of the origin. We say that f has a pole at infinity if F has a pole at the origin. Similarly, we can speak of f having an essential singularity at infinity, or a removable singularity (hence holomorphic) at infinity in terms of the corresponding behavior of F at 0. A meromorphic function in the complex plane that is either holomorphic at infinity or has a pole at infinity is said to be meromorphic in the extended complex plane. At this stage we return to the principle mentioned at the beginning of the chapter. Here we can see it in its simplest form. Theorem 3.4 The meromorphic functions in the extended complex plane are the rational functions. Proof. Suppose that f is meromorphic in the extended plane. Then f (1/z) has either a pole or a removable singularity at 0, and in either case it must be holomorphic in a deleted neighborhood of the origin, so that the function f can have only finitely many poles in the plane, say at z1 , . . . , zn . The idea is to subtract from f its principal parts at all its poles including the one at infinity. Near each pole zk ∈ C we can write f (z) = fk (z) + gk (z) , where fk (z) is the principal part of f at zk and gk is holomorphic in a (full) neighborhood of zk . In particular, fk is a polynomial in 1/(z − zk ). Similarly, we can write f (1/z) = f˜∞ (z) + g˜∞ (z) , where g˜∞ is holomorphic in a neighborhood of the origin and f˜∞ is the principal part of f (1/z) at 0, that is, a polynomial in 1/z. Finally, let f∞ (z) = f˜∞ (1/z). n We contend that the function H = f − f∞ − k=1 fk is entire and bounded. Indeed, near the pole zk we subtracted the principal part of f so that the function H has a removable singularity there. Also, H(1/z) is bounded for z near 0 since we subtracted the principal part of the pole at ∞. This proves our contention, and by Liouville’s theorem we conclude that H is constant. From the definition of H, we find that f is a rational function, as was to be shown. Note that as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles.

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

The Riemann sphere The extended complex plane, which consists of C and the point at infinity, has a convenient geometric interpretation, which we briefly discuss here. Consider the Euclidean space R3 with coordinates (X, Y, Z) where the XY -plane is identified with C. We denote by S the sphere centered at (0, 0, 1/2) and of radius 1/2; this sphere is of unit diameter and lies on top of the origin of the complex plane as pictured in Figure 5. Also, we let N = (0, 0, 1) denote the north pole of the sphere. N S W

0 w

C

Figure 5. The Riemann sphere S and stereographic projection

Given any point W = (X, Y, Z) on S different from the north pole, the line joining N and W intersects the XY -plane in a single point which we denote by w = x + iy; w is called the stereographic projection of W (see Figure 5). Conversely, given any point w in C, the line joining N and w = (x, y, 0) intersects the sphere at N and another point, which we call W . This geometric construction gives a bijective correspondence between points on the punctured sphere S − {N } and the complex plane; it is described analytically by the formulas x=

X 1−Z

and

y=

Y , 1−Z

giving w in terms of W , and X=

x , 2 x + y2 + 1

Y =

y , 2 x + y2 + 1

and

Z=

x2 + y 2 x2 + y 2 + 1

giving W in terms of w. Intuitively, we have wrapped the complex plane onto the punctured sphere S − {N }.

4. The argument principle and applications

89

As the point w goes to infinity in C (in the sense that |w| → ∞) the corresponding point W on S comes arbitrarily close to N . This simple observation makes N a natural candidate for the so-called “point at infinity.” Identifying infinity with the point N on S, we see that the extended complex plane can be visualized as the full two-dimensional sphere S; this is the Riemann sphere. Since this construction takes the unbounded set C into the compact set S by adding one point, the Riemann sphere is sometimes called the one-point compactification of C. An important consequence of this interpretation is the following: although the point at infinity required special attention when considered separately from C, it now finds itself on equal footing with all other points on S. In particular, a meromorphic function on the extended complex plane can be thought of as a map from S to itself, where the image of a pole is now a tractable point on S, namely N . For these reasons (and others) the Riemann sphere provides good geometrical insight into the structure of C as well as the theory of meromorphic functions.

4 The argument principle and applications We anticipate our discussion of the logarithm (in Section 6) with a few comments. In general, the function log f (z) is “multiple-valued” because it cannot be defined unambiguously on the set where f (z) = 0. However it is to be defined, it must equal log |f (z)| + i arg f (z), where log |f (z)| is the usual real-variable logarithm of the positive quantity |f (z)| (and hence is defined unambiguously), while arg f (z) is some determination of the argument (up to an additive integral multiple of 2π). Note that in any case, the derivative of log f (z) is f  (z)/f (z) which is single-valued, and the integral
 f (z) dz γ f (z) can be interpreted as the change in the argument of f as z traverses the curve γ. Moreover, assuming the curve is closed, this change of argument is determined entirely by the zeros and poles of f inside γ. We now formulate this fact as a precise theorem. We begin with the observation that while the additivity formula log(f1 f2 ) = log f1 + log f2 fails in general (as we shall see below), the additivity can be restored to the corresponding derivatives. This is confirmed by the following

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

observation: (f1 f2 ) f  f2 + f1 f2 f f = 1 = 1 + 2, f1 f2 f1 f2 f1 f2 which generalizes to



N k=1 fk

N



k=1 fk

=

N  f

k

k=1

fk

.

We apply this formula as follows. If f is holomorphic and has a zero of order n at z0 , we can write f (z) = (z − z0 )n g(z) , where g is holomorphic and nowhere vanishing in a neighborhood of z0 , and therefore f  (z) n + G(z) = f (z) z − z0 where G(z) = g  (z)/g(z). The conclusion is that if f has a zero of order n at z0 , then f  /f has a simple pole with residue n at z0 . Observe that a similar fact also holds if f has a pole of order n at z0 , that is, if f (z) = (z − z0 )−n h(z). Then −n f  (z) = + H(z). f (z) z − z0 Therefore, if f is meromorphic, the function f  /f will have simple poles at the zeros and poles of f , and the residue is simply the order of the zero of f or the negative of the order of the pole of f . As a result, an application of the residue formula gives the following theorem. Theorem 4.1 (Argument principle) Suppose f is meromorphic in an open set containing a circle C and its interior. If f has no poles and never vanishes on C, then
 1 f (z) dz = (number of zeros of f inside C) minus 2πi C f (z) (number of poles of f inside C), where the zeros and poles are counted with their multiplicities. Corollary 4.2 The above theorem holds for toy contours.

4. The argument principle and applications

91

As an application of the argument principle, we shall prove three theorems of interest in the general theory. The first, Rouch´e’s theorem, is in some sense a continuity statement. It says that a holomorphic function can be perturbed slightly without changing the number of its zeros. Then, we prove the open mapping theorem, which states that holomorphic functions map open sets to open sets, an important property that again shows the special nature of holomorphic functions. Finally, the maximum modulus principle is reminiscent of (and in fact implies) the same property for harmonic functions: a non-constant holomorphic function on an open set Ω cannot attain its maximum in the interior of Ω. Theorem 4.3 (Rouch´ e’s theorem) Suppose that f and g are holomorphic in an open set containing a circle C and its interior. If |f (z)| > |g(z)|

for all z ∈ C,

then f and f + g have the same number of zeros inside the circle C. Proof. For t ∈ [0, 1] define ft (z) = f (z) + tg(z) so that f0 = f and f1 = f + g. Let nt denote the number of zeros of ft inside the circle counted with multiplicities, so that in particular, nt is an integer. The condition |f (z)| > |g(z)| for z ∈ C clearly implies that ft has no zeros on the circle, and the argument principle implies
 ft (z) 1 nt = dz. 2πi C ft (z) To prove that nt is constant, it suffices to show that it is a continuous function of t. Then we could argue that if nt were not constant, the intermediate value theorem would guarantee the existence of some t0 ∈ [0, 1] with nt0 not integral, contradicting the fact that nt ∈ Z for all t. To prove the continuity of nt , we observe that ft (z)/ft(z) is jointly continuous for t ∈ [0, 1] and z ∈ C. This joint continuity follows from the fact that it holds for both the numerator and denominator, and our assumptions guarantee that ft (z) does not vanish on C. Hence nt is integer-valued and continuous, and it must be constant. We conclude that n0 = n1 , which is Rouch´e’s theorem. We now come to an important geometric property of holomorphic functions that arises when we consider them as mappings (that is, mapping regions in the complex plane to the complex plane). A mapping is said to be open if it maps open sets to open sets.

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

Theorem 4.4 (Open mapping theorem) If f is holomorphic and nonconstant in a region Ω, then f is open. Proof. Let w0 belong to the image of f , say w0 = f (z0 ). We must prove that all points w near w0 also belong to the image of f . Define g(z) = f (z) − w and write g(z) = (f (z) − w0 ) + (w0 − w) = F (z) + G(z). Now choose δ > 0 such that the disc |z − z0 | ≤ δ is contained in Ω and f (z) = w0 on the circle |z − z0 | = δ. We then select  > 0 so that we have |f (z) − w0 | ≥  on the circle |z − z0 | = δ. Now if |w − w0 | <  we have |F (z)| > |G(z)| on the circle |z − z0 | = δ, and by Rouch´e’s theorem we conclude that g = F + G has a zero inside the circle since F has one. The next result pertains to the size of a holomorphic function. We shall refer to the maximum of a holomorphic function f in an open set Ω as the maximum of its absolute value |f | in Ω. Theorem 4.5 (Maximum modulus principle) If f is a non-constant holomorphic function in a region Ω, then f cannot attain a maximum in Ω. Proof. Suppose that f did attain a maximum at z0 . Since f is holomorphic it is an open mapping, and therefore, if D ⊂ Ω is a small disc centered at z0 , its image f (D) is open and contains f (z0 ). This proves that there are points in z ∈ D such that |f (z)| > |f (z0 )|, a contradiction.

Corollary 4.6 Suppose that Ω is a region with compact closure Ω. If f is holomorphic on Ω and continuous on Ω then sup |f (z)| ≤ sup |f (z)|. z∈Ω

z∈Ω−Ω

In fact, since f (z) is continuous on the compact set Ω, then |f (z)| attains its maximum in Ω; but this cannot be in Ω if f is non-constant. If f is constant, the conclusion is trivial. Remark. The hypothesis that Ω is compact (that is, bounded) is essential for the conclusion. We give an example related to considerations that we will take up in Chapter 4. Let Ω be the open first quadrant, bounded by the positive half-line x ≥ 0 and the corresponding imagi2 nary line y ≥ 0. Consider F (z) = e−iz . Then F is entire and clearly

93

5. Homotopies and simply connected domains

continuous on Ω. Moreover |F (z)| = 1 on the two boundary lines z = x and z = iy. However, F√(z) is unbounded in Ω, since for example, we 2 have F (z) = er if z = r i = reiπ/4 .

5 Homotopies and simply connected domains The key to the general form of Cauchy’s theorem, as well as the analysis of multiple-valued functions, is to understand in what regions we can define the primitive of a given holomorphic function. Note the relevance to the study of the logarithm, which arises as a primitive of 1/z. The question is not just a local one, but is also global in nature. Its elucidation requires the notion of homotopy, and the resulting idea of simple-connectivity. Let γ0 and γ1 be two curves in an open set Ω with common end-points. So if γ0 (t) and γ1 (t) are two parametrizations defined on [a, b], we have γ0 (a) = γ1 (a) = α

and

γ0 (b) = γ1 (b) = β.

These two curves are said to be homotopic in Ω if for each 0 ≤ s ≤ 1 there exists a curve γs ⊂ Ω, parametrized by γs (t) defined on [a, b], such that for every s γs (a) = α

and

γs (b) = β,

γs (t)|s=0 = γ0 (t)

and

γs (t)|s=1 = γ1 (t).

and for all t ∈ [a, b]

Moreover, γs (t) should be jointly continuous in s ∈ [0, 1] and t ∈ [a, b]. Loosely speaking, two curves are homotopic if one curve can be deformed into the other by a continuous transformation without ever leaving Ω (Figure 6). Theorem 5.1 If f is holomorphic in Ω, then

f (z) dz = f (z) dz γ0

γ1

whenever the two curves γ0 and γ1 are homotopic in Ω. Proof. The key to the proof lies in showing that if two curves are close to each other and have the same end-points, then the integrals over these curves are equal. Recall that by definition, the function F (s, t) = γs (t) is continuous on [0, 1] × [a, b]. In particular, since the image of F , which we

94

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

β

γ0

γs Ω α

γ1

Figure 6. Homotopy of curves

denote by K, is compact, there exists  > 0 such that every disc of radius 3 centered at a point in the image of F is completely contained in Ω. If not, for every  ≥ 0, there exist points z ∈ K and w in the complement of Ω such that |z − w | < 1/. By compactness of K, there exists a subsequence of {z }, say {z k }, that converges to a point z ∈ K ⊂ Ω. By construction, we must have w k → z as well, and since {w } lies in the complement of Ω which is closed, we must have z ∈ Ωc as well. This is a contradiction. Having found an  with the desired property, we may, by the uniform continuity of F , select δ so that sup |γs1 (t) − γs2 (t)| < 

whenever |s1 − s2 | < δ.

t∈[a,b]

Fix s1 and s2 with |s1 − s2 | < δ. We then choose discs {D0 , . . . , Dn } of radius 2, and consecutive points {z0 , . . . , zn+1 } on γs1 and {w0 , . . . , wn+1 } on γs2 such that the union of these discs covers both curves, and zi , zi+1, wi , wi+1 ∈ Di . The situation is illustrated in Figure 7. Also, we choose z0 = w0 as the beginning end-point of the curves and zn+1 = wn+1 as the common end-point. On each disc Di , let Fi denote a primitive of f (Theorem 2.1, Chapter 2). On the intersection of Di and Di+1 , Fi and Fi+1 are two primitives of the same function, so they must differ by a constant, say ci . Therefore Fi+1 (zi+1) − Fi (zi+1) = Fi+1 (wi+1 ) − Fi (wi+1 ),

95

5. Homotopies and simply connected domains

z5 = w 5

z4

D4

w4 z3 w3

D3

z2

D2

w2 z1 w1

D1

z0 = w 0

D0

Figure 7. Covering two nearby curves with discs

hence Fi+1 (zi+1) − Fi+1 (wi+1 ) = Fi (zi+1 ) − Fi (wi+1 ).

(5)

This implies





f− γs1

f= γs2

n n   [Fi (zi+1 ) − Fi (zi )] − [Fi (wi+1 ) − Fi (wi )] i=0

=

n 

i=0

Fi (zi+1) − Fi (wi+1 ) − (Fi (zi ) − Fi (wi ))

i=0

= Fn (zn+1 ) − Fn (wn+1 ) − (F0 (z0 ) − F0 (w0 )) , because of the cancellations due to (5). Since γs1 and γs2 have the same beginning and end point, we have proved that

f= f. γs1

γs2

By subdividing [0, 1] into subintervals [si , si+1] of length less than δ, we may go from γ0 to γ1 by finitely many applications of the above argument, and the theorem is proved.

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

A region Ω in the complex plane is simply connected if any two pair of curves in Ω with the same end-points are homotopic. Example 1. A disc D is simply connected. In fact, if γ0 (t) and γ1 (t) are two curves lying in D, we can define γs (t) by γs (t) = (1 − s)γ0 (t) + sγ1 (t). Note that if 0 ≤ s ≤ 1, then for each t, the point γs (t) is on the segment joining γ0 (t) and γ1 (t), and so is in D. The same argument works if D is replaced by a rectangle, or more generally by any open convex set. (See Exercise 21.) Example 2. The slit plane Ω = C − {(−∞, 0]} is simply connected. For a pair of curves γ0 and γ1 in Ω, we write γj (t) = rj (t)eiθj (t) (j = 0, 1) with rj (t) continuous and strictly positive, and θj (t) continuous with |θj (t)| < π. Then, we can define γs (t) as rs (t)eiθs (t) where rs (t) = (1 − s)r0 (t) + sr1 (t)

and

θs (t) = (1 − s)θ0 (t) + sθ1 (t).

We then have γs (t) ∈ Ω whenever 0 ≤ s ≤ 1. Example 3. With some effort one can show that the interior of a toy contour is simply connected. This requires that we divide the interior into several subregions. A general form of the argument is given in Exercise 4.

Example 4. In contrast with the previous examples, the punctured plane C − {0} is not simply connected. Intuitively, consider two curves with the origin between them. It is impossible to continuously pass from one curve to the other without going over 0. A rigorous proof of this fact requires one further result, and will be given shortly. Theorem 5.2 Any holomorphic function in a simply connected domain has a primitive. Proof. Fix a point z0 in Ω and define
f (w) dw F (z) = γ

where the integral is taken over any curve in Ω joining z0 to z. This definition is independent of the curve chosen, since Ω is simply connected,

97

6. The complex logarithm

and if γ˜ is another curve in Ω joining z0 and z, we would have

f (w) dw = f (w) dw γ

γ ˜

by Theorem 5.1. Now we can write


F (z + h) − F (z) =

f (w) dw η

where η is the line segment joining z and z + h. Arguing as in the proof of Theorem 2.1 in Chapter 2, we find that F (z + h) − F (z) = f (z). h→0 h lim

As a result, we obtain the following version of Cauchy’s theorem. Corollary 5.3 If f is holomorphic in the simply connected region Ω, then
f (z) dz = 0 γ

for any closed curve γ in Ω. This is immediate from the existence of a primitive. The fact that the punctured plane is not simply connected now follows rigorously from the observation that the integral of 1/z over the unit circle is 2πi, and not 0.

6 The complex logarithm Suppose we wish to define the logarithm of a non-zero complex number. If z = reiθ , and we want the logarithm to be the inverse to the exponential, then it is natural to set log z = log r + iθ. Here and below, we use the convention that log r denotes the standard 1 logarithm of the positive number r. The trouble with the above definition is that θ is unique only up to an integer multiple of 2π. However, 1 By the standard logarithm, we mean the natural logarithm of positive numbers that appears in elementary calculus.

98

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

for given z we can fix a choice of θ, and if z varies only a little, this determines the corresponding choice of θ uniquely (assuming we require that θ varies continuously with z). Thus “locally” we can give an unambiguous definition of the logarithm, but this will not work “globally.” For example, if z starts at 1, and then winds around the origin and returns to 1, the logarithm does not return to its original value, but rather differs by an integer multiple of 2πi, and therefore is not “single-valued.” To make sense of the logarithm as a single-valued function, we must restrict the set on which we define it. This is the so-called choice of a branch or sheet of the logarithm. Our discussion of simply connected domains given above leads to a natural global definition of a branch of the logarithm function. Theorem 6.1 Suppose that Ω is simply connected with 1 ∈ Ω, and 0 ∈ / Ω. Then in Ω there is a branch of the logarithm F (z) = logΩ (z) so that (i) F is holomorphic in Ω, (ii) eF (z) = z for all z ∈ Ω, (iii) F (r) = log r whenever r is a real number and near 1. In other words, each branch logΩ (z) is an extension of the standard logarithm defined for positive numbers. Proof. We shall construct F as a primitive of the function 1/z. Since 0∈ / Ω, the function f (z) = 1/z is holomorphic in Ω. We define
logΩ (z) = F (z) = f (w) dw , γ

where γ is any curve in Ω connecting 1 to z. Since Ω is simply connected, this definition does not depend on the path chosen. Arguing as in the proof of Theorem 5.2, we find that F is holomorphic and F  (z) = 1/z for all z ∈ Ω. This proves (i). To prove (ii), it suffices to show that ze−F (z) = 1. For that, we differentiate the left-hand side, obtaining d  −F (z)  ze = e−F (z) − zF  (z)e−F (z) = (1 − zF  (z))e−F (z) = 0. dz Since Ω is connected we conclude, by Corollary 3.4 in Chapter 1, that ze−F (z) is constant. Evaluating this expression at z = 1, and noting that F (1) = 0, we find that this constant must be 1. Finally, if r is real and close to 1 we can choose as a path from 1 to r a line segment on the real axis so that
r dx F (r) = = log r, x 1

99

6. The complex logarithm

by the usual formula for the standard logarithm. This completes the proof of the theorem. For example, in the slit plane Ω = C − {(−∞, 0]} we have the principal branch of the logarithm log z = log r + iθ where z = reiθ with |θ| < π. (Here we drop the subscript Ω, and write simply log z.) To prove this, we use the path of integration γ shown in Figure 8.

z = reiθ

0

1

r

Figure 8. Path of integration for the principal branch of the logarithm

If z = reiθ with |θ| < π, the path consists of the line segment from 1 to r and the arc η from r to z. Then
r
dx dw + log z = x η w 1
θ ireit = log r + dt it 0 re = log r + iθ. An important observation is that in general log(z1 z2 ) = log z1 + log z2 . For example, if z1 = e2πi/3 = z2 , then for the principal branch of the logarithm, we have log z1 = log z2 =

2πi , 3

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

and since z1 z2 = e−2πi/3 we have −

2πi = log(z1 z2 ) = log z1 + log z2 . 3

Finally, for the principal branch of the logarithm the following Taylor expansion holds:

 zn z2 z3 + − ··· = − (−1)n 2 3 n ∞

(6)

log(1 + z) = z −

for |z| < 1.

n=1

Indeed, the derivative of both sides equals 1/(1 + z), so that they differ by a constant. Since both sides are equal to 0 at z = 0 this constant must be 0, and we have proved the desired Taylor expansion. Having defined a logarithm on a simply connected domain, we can now define the powers z α for any α ∈ C. If Ω is simply connected with 1 ∈ Ω and 0 ∈ / Ω, we choose the branch of the logarithm with log 1 = 0 as above, and define z α = eα log z . Note that 1α = 1, and that if α = 1/n, then (z 1/n )n =

n 

1

e n log z = e

n

1 k=1 n

log z

n

= e n log z = elog z = z.

k=1

We know that every non-zero complex number w can be written as w = ez . A generalization of this fact is given in the next theorem, which discusses the existence of log f (z) whenever f does not vanish. Theorem 6.2 If f is a nowhere vanishing holomorphic function in a simply connected region Ω, then there exists a holomorphic function g on Ω such that f (z) = eg(z) . The function g(z) in the theorem can be denoted by log f (z), and determines a “branch” of that logarithm. Proof. Fix a point z0 in Ω, and define a function


g(z) = γ

f  (w) dw + c0 , f (w)

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7. Fourier series and harmonic functions

where γ is any path in Ω connecting z0 to z, and c0 is a complex number so that ec0 = f (z0 ). This definition is independent of the path γ since Ω is simply connected. Arguing as in the proof of Theorem 2.1, Chapter 2, we find that g is holomorphic with g  (z) =

f  (z) , f (z)

and a simple calculation gives

 d  f (z)e−g(z) = 0 , dz so that f (z)e−g(z) is constant. Evaluating this expression at z0 we find f (z0 )e−c0 = 1, so that f (z) = eg(z) for all z ∈ Ω, and the proof is complete.

7 Fourier series and harmonic functions In Chapter 4 we shall describe some interesting connections between complex function theory and Fourier analysis on the real line. The motivation for this study comes in part from the simple and direct relation between Fourier series on the circle and power series expansions of holomorphic functions in the disc, which we now investigate. Suppose that f is holomorphic in a disc DR (z0 ), so that f has a power series expansion f (z) =

∞ 

an (z − z0 )n

n=0

that converges in that disc. Theorem 7.1 The coefficients of the power series expansion of f are given by
2π 1 f (z0 + reiθ )e−inθ dθ an = 2πrn 0 for all n ≥ 0 and 0 < r < R. Moreover, 1 0= 2πrn whenever n < 0.


0

2π

f (z0 + reiθ )e−inθ dθ

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Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

Proof. Since f (n) (z0 ) = an n!, the Cauchy integral formula gives
f (ζ) 1 an = dζ , 2πi γ (ζ − z0 )n+1 where γ is a circle of radius 0 < r < R centered at z0 and with the positive orientation. Choosing ζ = z0 + reiθ for the parametrization of this circle, we find that for n ≥ 0
2π f (z0 + reiθ ) 1 rieiθ dθ an = 2πi 0 (z0 + reiθ − z0 )n+1
2π 1 f (z0 + reiθ )e−i(n+1)θ eiθ dθ = 2πrn 0
2π 1 = f (z0 + reiθ )e−inθ dθ. 2πrn 0 Finally, even when n < 0, our calculation shows that we still have the identity 1 2πrn




2π iθ

f (z0 + re )e 0

−inθ

1 dθ = 2πi


γ

f (ζ) dζ. (ζ − z0 )n+1

Since −n > 0, the function f (ζ)(ζ − z0 )−n−1 is holomorphic in the disc, and by Cauchy’s theorem the last integral vanishes. The interpretation of this theorem is as follows. Consider f (z0 + reiθ ) as the restriction to the circle of a holomorphic function on the closure of a disc centered at z0 with radius r. Then its Fourier coefficients vanish if n < 0, while those for n ≥ 0 are equal (up to a factor of rn ) to coefficients of the power series of the holomorphic function f . The property of the vanishing of the Fourier coefficients for n < 0 reveals another special characteristic of holomorphic functions (and in particular their restrictions to any circle). Next, since a0 = f (z0 ), we obtain the following corollary. Corollary 7.2 (Mean-value property) If f is holomorphic in a disc DR (z0 ), then 1 f (z0 ) = 2π




2π

f (z0 + reiθ ) dθ,

for any 0 < r < R.

0

Taking the real parts of both sides, we obtain the following consequence.

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8. Exercises

Corollary 7.3 If f is holomorphic in a disc DR (z0 ), and u = Re(f ), then
2π 1 u(z0 ) = u(z0 + reiθ ) dθ, for any 0 < r < R. 2π 0 Recall that u is harmonic whenever f is holomorphic, and in fact, the above corollary is a property enjoyed by every harmonic function in the disc DR (z0 ). This follows from Exercise 12 in Chapter 2, which shows that every harmonic function in a disc is the real part of a holomorphic function in that disc.

8 Exercises 1. Using Euler’s formula sin πz =

eiπz − e−iπz , 2i

show that the complex zeros of sin πz are exactly at the integers, and that they are each of order 1. Calculate the residue of 1/ sin πz at z = n ∈ Z. 2. Evaluate the integral


∞ −∞

dx . 1 + x4

Where are the poles of 1/(1 + z 4 )? 3. Show that




∞ −∞

4. Show that




∞ −∞

e−a cos x dx = π , 2 2 x +a a

x sin x dx = πe−a , x2 + a2

for a > 0.

for all a > 0.

5. Use contour integration to show that


∞ −∞

for all ξ real.

π e−2πixξ dx = (1 + 2π|ξ|)e−2π|ξ| (1 + x2 )2 2

104

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

6. Show that




∞ −∞

1 · 3 · 5 · · · (2n − 1) dx · π. = (1 + x2 )n+1 2 · 4 · 6 · · · (2n)

7. Prove that


2π 0

2πa dθ = 2 , (a + cos θ)2 (a − 1)3/2

8. Prove that




2π 0

whenever a > 1.

2π dθ = √ a + b cos θ a 2 − b2

if a > |b| and a, b ∈ R. 9. Show that




1

log(sin πx) dx = − log 2. 0

[Hint: Use the contour shown in Figure 9.]

0

1

Figure 9. Contour in Exercise 9

10. Show that if a > 0, then
∞ 0

π log x dx = log a. x2 + a2 2a

[Hint: Use the contour in Figure 10.] 11. Show that if |a| < 1, then


2π 0

log |1 − aeiθ | dθ = 0.

105

8. Exercises

ia

−R

−



R

Figure 10. Contour in Exercise 10

Then, prove that the above result remains true if we assume only that |a| ≤ 1. 12. Suppose u is not an integer. Prove that ∞  n=−∞

π2 1 = 2 (u + n) (sin πu)2

by integrating f (z) =

π cot πz (u + z)2

over the circle |z| = RN = N + 1/2 (N integral, N ≥ |u|), adding the residues of f inside the circle, and letting N tend to infinity. Note. Two other derivations of this identity, using Fourier series, were given in Book I. 13. Suppose f (z) is holomorphic in a punctured disc Dr (z0 ) − {z0 }. Suppose also that |f (z)| ≤ A|z − z0 |−1+ for some > 0, and all z near z0 . Show that the singularity of f at z0 is removable. 14. Prove that all entire functions that are also injective take the form f (z) = az + b with a, b ∈ C, and a = 0. [Hint: Apply the Casorati-Weierstrass theorem to f (1/z).] 15. Use the Cauchy inequalities or the maximum modulus principle to solve the following problems: (a) Prove that if f is an entire function that satisfies sup |f (z)| ≤ ARk + B

|z|=R

106

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

for all R > 0, and for some integer k ≥ 0 and some constants A, B > 0, then f is a polynomial of degree ≤ k. (b) Show that if f is holomorphic in the unit disc, is bounded, and converges uniformly to zero in the sector θ < arg z < ϕ as |z| → 1, then f = 0. (c) Let w1 , . . . , wn be points on the unit circle in the complex plane. Prove that there exists a point z on the unit circle such that the product of the distances from z to the points wj , 1 ≤ j ≤ n, is at least 1. Conclude that there exists a point w on the unit circle such that the product of the distances from w to the points wj , 1 ≤ j ≤ n, is exactly equal to 1. (d) Show that if the real part of an entire function f is bounded, then f is constant. 16. Suppose f and g are holomorphic in a region containing the disc |z| ≤ 1. Suppose that f has a simple zero at z = 0 and vanishes nowhere else in |z| ≤ 1. Let f (z) = f (z) + g(z). Show that if is sufficiently small, then (a) f (z) has a unique zero in |z| ≤ 1, and (b) if z is this zero, the mapping → z is continuous.

17. Let f be non-constant and holomorphic in an open set containing the closed unit disc. (a) Show that if |f (z)| = 1 whenever |z| = 1, then the image of f contains the unit disc. [Hint: One must show that f (z) = w0 has a root for every w0 ∈ D. To do this, it suffices to show that f (z) = 0 has a root (why?). Use the maximum modulus principle to conclude.] (b) If |f (z)| ≥ 1 whenever |z| = 1 and there exists a point z0 ∈ D such that |f (z0 )| < 1, then the image of f contains the unit disc.

18. Give another proof of the Cauchy integral formula f (z) =

1 2πi


C

f (ζ) dζ ζ−z

using homotopy of curves. [Hint: Deform the circle C to a small circle centered at z, and note that the quotient (f (ζ) − f (z))/(ζ − z) is bounded.] 19. Prove the maximum principle for harmonic functions, that is:

107

8. Exercises

(a) If u is a non-constant real-valued harmonic function in a region Ω, then u cannot attain a maximum (or a minimum) in Ω. (b) Suppose that Ω is a region with compact closure Ω. If u is harmonic in Ω and continuous in Ω, then sup |u(z)| ≤ sup |u(z)|. z∈Ω

z∈Ω−Ω

[Hint: To prove the first part, assume that u attains a local maximum at z0 . Let f be holomorphic near z0 with u = Re(f ), and show that f is not open. The second part follows directly from the first.] 20. This exercise shows how the mean square convergence dominates the uniform convergence of analytic functions. If U is an open subset of C we use the notation 


1/2 |f (z)|2 dxdy

f L2 (U ) = U

for the mean square norm, and f L∞ (U ) = sup |f (z)| z∈U

for the sup norm. (a) If f is holomorphic in a neighborhood of the disc Dr (z0 ), show that for any 0 < s < r there exists a constant C > 0 (which depends on s and r) such that f L∞ (Ds (z0 )) ≤ Cf L2 (Dr (z0 )) . (b) Prove that if {fn } is a Cauchy sequence of holomorphic functions in the mean square norm  · L2 (U ) , then the sequence {fn } converges uniformly on every compact subset of U to a holomorphic function. [Hint: Use the mean-value property.] 21. Certain sets have geometric properties that guarantee they are simply connected. (a) An open set Ω ⊂ C is convex if for any two points in Ω, the straight line segment between them is contained in Ω. Prove that a convex open set is simply connected. (b) More generally, an open set Ω ⊂ C is star-shaped if there exists a point z0 ∈ Ω such that for any z ∈ Ω, the straight line segment between z and z0 is contained in Ω. Prove that a star-shaped open set is simply connected. Conclude that the slit plane C − {(−∞, 0]} (and more generally any sector, convex or not) is simply connected.

108

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

(c) What are other examples of open sets that are simply connected? 22. Show that there is no holomorphic function f in the unit disc D that extends continuously to ∂D such that f (z) = 1/z for z ∈ ∂D.

9 Problems 1.∗ Consider a holomorphic map on the unit disc f : D → C which satisfies f (0) = 0. By the open mapping theorem, the image f (D) contains a small disc centered at the origin. We then ask: does there exist r > 0 such that for all f : D → C with f (0) = 0, we have Dr (0) ⊂ f (D)? (a) Show that with no further restrictions on f , no such r exists. It suffices to / f (D). find a sequence of functions {fn } holomorphic in D such that 1/n ∈ Compute fn (0), and discuss. (b) Assume in addition that f also satisfies f  (0) = 1. Show that despite this new assumption, there exists no r > 0 satisfying the desired condition. [Hint: Try f (z) = (ez/ − 1).] The Koebe-Bieberbach theorem states that if in addition to f (0) = 0 and f  (0) = 1 we also assume that f is injective, then such an r exists and the best possible value is r = 1/4. 1 (c) As a first step, show that if h(z) + c0 + c1 z + c2 z 2 + · · · is analytic and z = ∞ injective for 0 < |z| < 1, then n=1 n|cn |2 ≤ 1.

[Hint: Calculate the area of the complement of h(Dρ (0) − {0}) where 0 < ρ < 1, and let ρ → 1.] (d) If f (z) = z + a2 z 2 + · · · satisfies the hypotheses of the theorem, show that there exists another function g satisfying the hypotheses of the theorem such that g 2 (z) = f (z 2 ). [Hint: f (z)/z is nowhere vanishing so there exists ψ such that ψ 2 (z) = f (z)/z and ψ(0) = 1. Check that g(z) = zψ(z 2 ) is injective.] (e) With the notation of the previous part, show that |a2 | ≤ 2, and that equality holds if and only if f (z) =

z (1 − eiθ z)2

for some θ ∈ R.

[Hint: What is the power series expansion of 1/g(z)? Use part (c).] (f) If h(z) = 1z + c0 + c1 z + c2 z 2 + · · · is injective on D and avoids the values z1 and z2 , show that |z1 − z2 | ≤ 4. [Hint: Look at the second coefficient in the power series expansion of 1/(h(z) − zj ).]

109

9. Problems

(g) Complete the proof of the theorem. [Hint: If f avoids w, then 1/f avoids 0 and 1/w.]

2. Let u be a harmonic function in the unit disc that is continuous on its closure. Deduce Poisson’s integral formula u(z0 ) =

1 2π




2π 0

1 − |z0 |2 u(eiθ ) dθ |eiθ − z0 |2

for |z0 | < 1

from the special case z0 = 0 (the mean value theorem). Show that if z0 = reiϕ , then 1 − r2 1 − |z0 |2 = = Pr (θ − ϕ), iθ 2 |e − z0 | 1 − 2r cos(θ − ϕ) + r 2 and we recover the expression for the Poisson kernel derived in the exercises of the previous chapter. [Hint: Set u0 (z) = u(T (z)) where T (z) =

z0 − z . 1 − z0 z

Prove that u0 is harmonic. Then apply the mean value theorem to u0 , and make a change of variables in the integral.] 3. If f (z) is holomorphic in the deleted neighborhood {0 < |z − z0 | < r} and has a pole of order k at z0 , then we can write f (z) =

a−1 a−k + g(z) + ··· + (z − z0 )k (z − z0 )

where g is holomorphic in the disc {|z − z0 | < r}. There is a generalization of this expansion that holds even if z0 is an essential singularity. This is a special case of the Laurent series expansion, which is valid in an even more general setting. Let f be holomorphic in a region containing the annulus {z : r1 ≤ |z − z0 | ≤ r2 } where 0 < r1 < r2 . Then, f (z) =

∞ 

an (z − z0 )n

n=−∞

where the series converges absolutely in the interior of the annulus. To prove this, it suffices to write

1 f (ζ) f (ζ) 1 dζ − dζ f (z) = 2πi Cr2 ζ − z 2πi Cr1 ζ − z when r1 < |z − z0 | < r2 , and argue as in the proof of Theorem 4.4, Chapter 2. Here Cr1 and Cr2 are the circles bounding the annulus.

110

Chapter 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

4.∗ Suppose Ω is a bounded region. Let L be a (two-way infinite) line that intersects Ω. Assume that Ω ∩ L is an interval I. Choosing an orientation for L, we can define Ωl and Ωr to be the subregions of Ω lying strictly to the left or right of L, with Ω = Ωl ∪ I ∪ Ωr a disjoint union. If Ωl and Ωr are simply connected, then Ω is simply connected. 5.∗ Let g(z) =

1 2πi




M −M

h(x) dx x−z

where h is continuous and supported in [−M, M ]. (a) Prove that the function g is holomorphic in C − [−M, M ], and vanishes at infinity, that is, lim|z|→∞ |g(z)| = 0. Moreover, the “jump” of g across [−M, M ] is h, that is, h(x) =

lim

→0,>0

g(x + i ) − g(x − i ).

[Hint: Express the difference g(x + i ) − g(x − i ) in terms of a convolution of h with the Poisson kernel.] (b) If h satisfies a mild smoothness condition, for instance a H¨ older condition with exponent α, that is, |h(x) − h(y)| ≤ C|x − y|α for some C > 0 and all x, y ∈ [−M, M ], then g(x + i ) and g(x − i ) converge uniformly to functions g+ (x) and g− (x) as → 0. Then, g can be characterized as the unique holomorphic function that satisfies: (i) g is holomorphic outside [−M, M ], (ii) g vanishes at infinity, (iii) g(x + i ) and g(x − i ) converge uniformly as → 0 to functions g+ (x) and g− (x) with g+ (x) − g− (x) = h(x). [Hint: If G is another function satisfying these conditions, g − G is entire.]

4 The Fourier Transform Raymond Edward Alan Christopher Paley, Fellow of Trinity College, Cambridge, and International Research Fellow at the Massachusetts Institute of Technology and at Harvard University, was killed by an avalanche on April 7, 1933, while skiing in the vicinity of Banff, Alberta. Although only twenty-six years of age, he was already recognized as the ablest of the group of young English mathematicians who have been inspired by the genius of G. H. Hardy and J. E. Littlewood. In a group notable for its brilliant technique, no one had developed this technique to a higher degree than Paley. Nevertheless he should not be thought of primarily as a technician, for with this ability he combined creative power of the first order. As he himself was wont to say, technique without “rugger tactics” will not get one far, and these rugger tactics he practiced to a degree that was characteristic of his forthright and vigorous nature. Possessed of an extraordinary capacity for making friends and for scientific collaboration, Paley believed that the inspiration of continual interchange of ideas stimulates each collaborator to accomplish more than he would alone. Only the exceptional man works well with a partner, but Paley had collaborated successfully with many, including Littlewood, P´ olya, Zygmund, and Wiener. N. Wiener, 1933

If f is a function on R that satisfies appropriate regularity and decay conditions, then its Fourier transform is defined by
∞ fˆ(ξ) = f (x)e−2πixξ dx, ξ∈R −∞

and its counterpart, the Fourier inversion formula, holds
∞ fˆ(ξ)e2πixξ dξ, f (x) = x ∈ R. −∞

The Fourier transform (including its d-dimensional variants), plays a ba-

112

Chapter 4. THE FOURIER TRANSFORM

sic role in analysis, as the reader of Book I is aware. Here we want to illustrate the intimate and fruitful connection between the one-dimensional theory of the Fourier transform and complex analysis. The main theme (stated somewhat imprecisely) is as follows: for a function f initially defined on the real line, the possibility of extending it to a holomorphic function is closely related to the very rapid (for example, exponential) decay at infinity of its Fourier transform fˆ. We elaborate on this theme in two stages. First, we assume that f can be analytically continued in a horizontal strip containing the real axis, and has “moderate decrease” at infinity,1 so that the integral defining the Fourier transform fˆ converges. As a result, we conclude that fˆ decreases exponentially at infinity; it also follows directly that the Fourier inversion formula holds. Moreover one can easily obtain from considerations the Poisson summation formula   these ˆ(n). Incidentally, all these theorems are elegant f f (n) = n∈Z n∈Z consequences of contour integration. At a second stage, we take as our starting point the validity of the Fourier inversion formula, which holds if we assume that both f and fˆ are of moderate decrease, without making any assumptions on the analyticity of f . We then ask a simple but natural question: What are the conditions on f so that its Fourier transform is supported in a bounded interval, say [−M, M ]? This is a basic problem that, as one notices, can be stated without any reference to notions of complex analysis. However, it can be resolved only in terms of the holomorphic properties of the function f . The condition, given by the Paley-Wiener theorem, is that there be a holomorphic extension of f to C that satisfies the growth condition |f (z)| ≤ Ae2πM |z|

for some constant A > 0.

Functions satisfying this condition are said to be of exponential type. Observe that the condition that fˆ vanishes outside a compact set can be viewed as an extreme version of a decay property at infinity, and so the above theorem clearly falls within the context of the theme indicated above. In all these matters a decisive technique will consist in shifting the contour of integration, that is the real line, within the boundaries of a horizontal strip. This will take advantage of the special behavior of e−2πizξ when z has a non-zero imaginary part. Indeed, when z is real this exponential remains bounded and oscillates, while if Im(z) = 0, it will 1 We say that a function f is of moderate decrease if f is continuous and there exists A > 0 so that |f (x)| ≤ A/(1 + x2 ) for all x ∈ R. A more restrictive condition is that f ∈ S, the Schwartz space of testing functions, which also implies that fˆ belongs to S. See Book I for more details.

1. The class F

113

have exponential decay or exponential increase, depending on whether the product ξIm(z) is negative or positive.

1 The class F The weakest decay condition imposed on functions in our study of the Fourier transform in Book I was that of moderate decrease. There, we proved the Fourier inversion and Poisson summation formulas under the hypothesis that f and fˆ satisfy |f (x)| ≤

A 1 + x2

and

|fˆ(ξ)| ≤

A 1 + ξ2

for some positive constants A, A and all x, ξ ∈ R. We were led to consider this class of functions because of various examples such as the Poisson kernel y 1 Py (x) = π y 2 + x2 for y > 0, which played a fundamental role in the solution of the Dirichlet problem for the steady-state heat equation in the upper half-plane. There y (ξ) = e−2πy|ξ| . we had P In the present context, we introduce a class of functions particularly suited to the goal we have set: proving theorems about the Fourier transform using complex analysis. Moreover, this class will be large enough to contain many of the important applications we have in mind. For each a > 0 we denote by Fa the class of all functions f that satisfy the following two conditions: (i) The function f is holomorphic in the horizontal strip Sa = {z ∈ C : |Im(z)| < a}. (ii) There exists a constant A > 0 such that |f (x + iy)| ≤

A 1 + x2

for all x ∈ R and |y| < a.

In other words, Fa consists of those holomorphic functions on Sa that are of moderate decay on each horizontal line Im(z) = y, uniformly in 2 −a < y < a. For example, f (z) = e−πz belongs to Fa for all a. Also, the function 1 c f (z) = , π c2 + z 2

114

Chapter 4. THE FOURIER TRANSFORM

which has simple poles at z = ±ci, belongs to Fa for all 0 < a < c. Another example is provided by f (z) = 1/ cosh πz, which belongs to Fa whenever |a| < 1/2. This function, as well as one of its fundamental properties, was already discussed in Example 3, Section 2.1 of Chapter 3. Note also that a simple application of the Cauchy integral formulas shows that if f ∈ Fa , then for every n, the nth derivative of f belongs to Fb for all b with 0 < b < a (Exercise 2). Finally, we denote by F the class of all functions that belong to Fa for some a. Remark. The condition of moderate decrease can be weakened somewhat by replacing the order of decrease of A/(1 + x2 ) by A/(1 + |x|1+ ) for any  > 0. As the reader will observe, many of the results below remain unchanged with this less restrictive condition.

2 Action of the Fourier transform on F Here we prove three theorems, including the Fourier inversion and Poisson summation formulas, for functions in F. The idea behind all three proofs is the same: contour integration. Thus the approach used will be different from that of the corresponding results in Book I. Theorem 2.1 If f belongs to the class Fa for some a > 0, then |fˆ(ξ)| ≤ Be−2πb|ξ| for any 0 ≤ b < a. ∞ Proof. Recall that fˆ(ξ) = −∞ f (x)e−2πixξ dx. The case b = 0 simply says that fˆ is bounded, which follows at once from the integral defining fˆ, the assumption that f is of moderate decrease, and the fact that the exponential is bounded by 1. Now suppose 0 < b < a and assume first that ξ > 0. The main step consists of shifting the contour of integration, that is the real line, down by b. More precisely, consider the contour in Figure 1 as well as the function g(z) = f (z)e−2πizξ . We claim that as R tends to infinity, the integrals of g over the two vertical sides converge to zero. For example, the integral over the vertical segment on the left can be estimated by 

 −R  b   g(z) dz  ≤ |f (−R − it)e−2πi(−R−it)ξ | dt   −R−ib  0
b A −2πtξ e dt ≤ 2 R 0 = O(1/R2 ).

2. Action of the Fourier transform on F

115

−R

R 0

−R − ib

R − ib

Figure 1. The contour in the proof of Theorem 2.1 when ξ > 0

A similar estimate for the other side proves our claim. Therefore, by Cauchy’s theorem applied to the large rectangle, we find in the limit as R tends to infinity that
∞ ˆ (1) f (ξ) = f (x − ib)e−2πi(x−ib)ξ dx, −∞

which leads to the estimate
∞ |fˆ(ξ)| ≤

−∞

A e−2πbξ dx ≤ Be−2πbξ , 1 + x2

where B is a suitable constant. A similar argument for ξ < 0, but this time shifting the real line up by b, allows us to finish the proof of the theorem. This result says that whenever f ∈ F, then fˆ has rapid decay at infinity. We remark that the further we can extend f (that is, the larger a), then the larger we can choose b, hence the better the decay. We shall return to this circle of ideas in Section 3, where we describe those f for which fˆ has the ultimate decay condition: compact support. Since fˆ decreases rapidly on R, the integral in the Fourier inversion formula makes sense, and we now turn to the complex analytic proof of this identity. Theorem 2.2 If f ∈ F, then the Fourier inversion holds, namely
∞ f (x) = fˆ(ξ)e2πixξ dξ for all x ∈ R. −∞

116

Chapter 4. THE FOURIER TRANSFORM

Besides contour integration, the proof of the theorem requires a simple identity, which we isolate. ∞ Lemma 2.3 If A is positive and B is real, then 0 e−(A+iB)ξ dξ = 1 A+iB . Proof. Since A > 0 and B ∈ R, we have |e−(A+iB)ξ | = e−Aξ , and the integral converges. By definition
∞
R e−(A+iB)ξ dξ = lim e−(A+iB)ξ dξ. R→∞

0

However,




R

e

−(A+iB)ξ

0

0

−(A+iB)ξ R e dξ = − , A + iB 0

which tends to 1/(A + iB) as R tends to infinity. We can now prove the inversion theorem. Once again, the sign of ξ matters, so we begin by writing
∞
0
∞ fˆ(ξ)e2πixξ dξ = fˆ(ξ)e2πixξ dξ + fˆ(ξ)e2πixξ dξ. −∞

−∞

0

For the second integral we argue as follows. Say f ∈ Fa and choose 0 < b < a. Arguing as the proof of Theorem 2.1, or simply using equation (1), we get
∞ ˆ f (ξ) = f (u − ib)e−2πi(u−ib)ξ du, −∞

so that with an application of the lemma and the convergence of the integration in ξ, we find
∞
∞
∞ fˆ(ξ)e2πixξ dξ = f (u − ib)e−2πi(u−ib)ξ e2πixξ du dξ 0
0 ∞ −∞
∞ f (u − ib) e−2πi(u−ib−x)ξ dξ du = −∞ 0
∞ 1 = du f (u − ib) 2πb + 2πi(u − x) −∞
∞ f (u − ib) 1 = du 2πi −∞ u − ib − x
f (ζ) 1 dζ, = 2πi L1 ζ − x

2. Action of the Fourier transform on F

117

where L1 denotes the line {u − ib : u ∈ R} traversed from left to right. (In other words, L1 is the real line shifted down by b.) For the integral when ξ < 0, a similar calculation gives




0

1 fˆ(ξ)e2πixξ dξ = − 2πi −∞


L2

f (ζ) dζ, ζ −x

where L2 is the real line shifted up by b, with orientation from left to right. Now given x ∈ R, consider the contour γR in Figure 2.

γR −R + ib

R + ib

0

−R − ib

x

R − ib

Figure 2. The contour γR in the proof of Theorem 2.2

The function f (ζ)/(ζ − x) has a simple pole at x with residue f (x), so the residue formula gives
f (ζ) 1 dζ. f (x) = 2πi γR ζ − x Letting R tend to infinity, one checks easily that the integral over the vertical sides goes to 0 and therefore, combining with the previous results, we get

1 f (ζ) f (ζ) 1 dζ − dζ f (x) = 2πi L1 ζ − x 2πi L2 ζ − x
∞
0 2πixξ ˆ f (ξ)e fˆ(ξ)e2πixξ dξ = dξ + 0 −∞
∞ fˆ(ξ)e2πixξ dξ, = −∞

118

Chapter 4. THE FOURIER TRANSFORM

and the theorem is proved. The last of our three theorems is the Poisson summation formula. Theorem 2.4 If f ∈ F, then 

f (n) =

n∈Z



fˆ(n).

n∈Z

Proof. Say f ∈ Fa and choose some b satisfying 0 < b < a. The function 1/(e2πiz − 1) has simple poles with residue 1/(2πi) at the integers. Thus f (z)/(e2πiz − 1) has simple poles at the integers n, with residues f (n)/2πi. We may therefore apply the residue formula to the contour γN in Figure 3 where N is an integer.

γN −N −

1 2

+ ib

N+

−N − 1 −N −N −

1 2

−1 0

1

− ib

1 2

N

N+

+ ib

N +1

1 2

− ib

Figure 3. The contour γN in the proof of Theorem 2.4

This yields

 |n|≤N


f (n) = γN

f (z) dz. e2πiz − 1

Letting N tend to infinity, and recalling that f has moderate decrease,  we see that the sum converges to n∈Z f (n), and also that the integral over the vertical segments goes to 0. Therefore, in the limit we get

 f (z) f (z) dz − dz (2) f (n) = 2πiz 2πiz −1 −1 L1 e L2 e n∈Z

2. Action of the Fourier transform on F

119

where L1 and L2 are the real line shifted down and up by b, respectively. Now we use the fact that if |w| > 1, then

 1 = w−1 w−n w−1 ∞

n=0

to see that on L1 (where |e2πiz | > 1) we have 1 e2πiz − 1

= e−2πiz

∞ 

e−2πinz .

n=0

Also if |w| < 1, then

 1 =− wn w−1 ∞

n=0

so that on L2 1 e2πiz − 1

=−

∞ 

e2πinz .

n=0

Substituting these observations in (2) we find that   ∞ 

∞    f (n) = f (z) e−2πiz e−2πinz dz + f (z) e2πinz dz L1

n∈Z

= = =

∞


n=0

f (z)e−2πi(n+1)z dz +

n=0 L1 ∞
∞  n=0 ∞ 

=

L2

f (x)e−2πi(n+1)x dx +

fˆ(n + 1) +

n=0 ∞ 

n=0

f (z)e2πinz dz

n=0 L2 ∞
∞ 

−∞

n=0



∞


f (x)e2πinx dz

−∞

fˆ(−n)

n=0

fˆ(n),

n∈Z

where we have shifted L1 and L2 back to the real line according to equation (1) and its analogue for the shift down. The Poisson summation formula has many far-reaching consequences, and we close this section by deriving several interesting identities that are of importance for later applications.

120

Chapter 4. THE FOURIER TRANSFORM

First, we recall the calculation in Example 1, Chapter 2, which showed 2 that the function e−πx was its own Fourier transform:
∞ 2 2 e−πx e−2πixξ dx = e−πξ . −∞

For fixed values of t > 0 and a ∈ R, the change of variables x → t1/2 (x + a) in the above integral shows that the Fourier transform 2 2 of the function f (x) = e−πt(x+a) is fˆ(ξ) = t−1/2 e−πξ /t e2πiaξ . Applying the Poisson summation formula to the pair f and fˆ (which belong to F) provides the following relation: (3)

∞ 

e

n=−∞

−πt(n+a)2

=

∞ 

2

t−1/2 e−πn

/t 2πina

e

.

n=−∞

This identity has noteworthy consequences. For instance, the special case a = 0 is the transformation law for a version of the “theta function”: 2 ∞ if we define ϑ for t > 0 by the series ϑ(t) = n=−∞ e−πn t , then the relation (3) says precisely that ϑ(t) = t−1/2 ϑ(1/t)

(4)

for t > 0.

This equation will be used in Chapter 6 to derive the key functional equation of the Riemann zeta function, and this leads to its analytic continuation. Also, the general case a ∈ R will be used in Chapter 10 to establish a corresponding law for the more general Jacobi theta function Θ. For another application of the Poisson summation formula we recall that we proved in Example 3, Chapter 3, that the function 1/ cosh πx was also its own Fourier transform:
∞ −2πixξ 1 e dx = . cosh πξ −∞ cosh πx This implies that if t > 0 and a ∈ R, then the Fourier transform of the function f (x) = e−2πiax / cosh(πx/t) is fˆ(ξ) = t/ cosh(π(ξ + a)t), and the Poisson summation formula yields (5)

∞  n=−∞

∞  e−2πian t = . cosh(πn/t) cosh(π(n + a)t) n=−∞

This formula will be used in Chapter 10 in the context of the two-squares theorem.

121

3. Paley-Wiener theorem

3 Paley-Wiener theorem In this section we change our point of view somewhat: we do not suppose any analyticity of f , but we do assume the validity of the Fourier inversion formula
∞
∞ fˆ(ξ)e2πixξ dξ if fˆ(ξ) = f (x)e−2πixξ dx, f (x) = −∞

−∞

under the conditions |f (x)| ≤ A/(1 + x2 ) and |fˆ(ξ)| ≤ A /(1 + ξ 2 ). For a proof of the inversion formula under these conditions, we refer the reader to Chapter 5 in Book I. We start by pointing out a partial converse to Theorem 2.1. Theorem 3.1 Suppose fˆ satisfies the decay condition |fˆ(ξ)| ≤ Ae−2πa|ξ| for some constants a, A > 0. Then f (x) is the restriction to R of a function f (z) holomorphic in the strip Sb = {z ∈ C : |Im(z)| < b}, for any 0 < b < a. Proof. Define


fn (z) =

n

fˆ(ξ)e2πiξz dξ, −n

and note that fn is entire by Theorem 5.4 in Chapter 2. Observe also that f (z) may be defined for all z in the strip Sb by
∞ fˆ(ξ)e2πiξz dξ , f (z) = −∞

because the integral converges absolutely by our assumption on fˆ: it is majorized by
∞ A e−2πa|ξ| e2πb|ξ| dξ , −∞

which is finite if b < a. Moreover, for z ∈ Sb
|f (z) − fn (z)| ≤ A e−2πa|ξ| e2πb|ξ| dξ |ξ|≥n

→0

as n → ∞,

and thus the sequence {fn } converges to f uniformly in Sb , which, by Theorem 5.2 in Chapter 2, proves the theorem. We digress briefly to make the following observation.

122

Chapter 4. THE FOURIER TRANSFORM

Corollary 3.2 If fˆ(ξ) = O(e−2πa|ξ| ) for some a > 0, and f vanishes in a non-empty open interval, then f = 0. Since by the theorem f is analytic in a region containing the real line, the corollary is a consequence of Theorem 4.8 in Chapter 2. In particular, we recover the fact proved in Exercise 21, Chapter 5 in Book I, namely that f and fˆ cannot both have compact support unless f = 0. The Paley-Wiener theorem goes a step further than the previous theorem, and describes the nature of those functions whose Fourier transforms are supported in a given interval [−M, M ]. Theorem 3.3 Suppose f is continuous and of moderate decrease on R. Then, f has an extension to the complex plane that is entire with |f (z)| ≤ Ae2πM |z| for some A > 0, if and only if fˆ is supported in the interval [−M, M ]. One direction is simple. Suppose fˆ is supported in [−M, M ]. Then both f and fˆ have moderate decrease, and the Fourier inversion formula applies
M fˆ(ξ)e2πiξx dξ. f (x) = −M

Since the range of integration is finite, we may replace x by the complex variable z in the integral, thereby defining a complex-valued function on C by
M fˆ(ξ)e2πiξz dξ. g(z) = −M

By construction g(z) = f (z) if z is real, and g is holomorphic by Theorem 5.4 in Chapter 2. Finally, if z = x + iy, we have


|g(z)| ≤

M

|fˆ(ξ)|e−2πξy dξ

−M 2πM |z|

≤ Ae

.

The converse result requires a little more work. It starts with the observation that if fˆ were supported in [−M, M ], then the argument above would give the stronger bound |f (z)| ≤ Ae2π|y| instead of what we assume, that is |f (z)| ≤ Ae2π|z| . The idea is then to try to reduce to the better situation, where this stronger bound holds. However, this is not quite enough because we need in addition a (moderate) decay as x → ∞

123

3. Paley-Wiener theorem

(when y = 0) to deal with the convergence of certain integrals at infinity. Thus we begin by also assuming this further property of f , and then we remove the additional assumptions, one step at a time. Step 1. We first assume that f is holomorphic in the complex plane, and satisfies the following condition regarding decay in x and growth in y: (6)

|f (x + iy)| ≤ A

e2πM |y| . 1 + x2

We then prove under this stronger assumption that fˆ(ξ) = 0 if |ξ| > M . To see this, we first suppose that ξ > M and write
∞ fˆ(ξ) = f (x)e−2πiξx dx
−∞ ∞ = f (x − iy)e−2πiξ(x−iy) dx. −∞

Here we have shifted the real line down by an amount y > 0 using the standard argument (equation (1)). Putting absolute values gives the bound
∞ 2πM y−2πξy e dx |fˆ(ξ)| ≤ A 1 + x2 −∞ ≤ Ce−2πy(ξ−M ) . Letting y tend to infinity, and recalling that ξ − M > 0, proves that fˆ(ξ) = 0. A similar argument, shifting the contour up by y > 0, proves that fˆ(ξ) = 0 whenever ξ < −M . Step 2. We relax condition (6) by assuming only that f satisfies (7)

|f (x + iy)| ≤ Ae2πM |y| .

This is still a stronger condition than in the theorem, but it is weaker than (6). Suppose first that ξ > M , and for  > 0 consider the following auxiliary function f (z) =

f (z) . (1 + iz)2

We observe that the quantity 1/(1 + iz)2 has absolute value less than or equal to 1 in the closed lower half-plane (including the real line) and

124

Chapter 4. THE FOURIER TRANSFORM

converges to 1 as  tends to 0. In particular, this shows that f (ξ) → fˆ(ξ) as  → 0 since we may write


∞ 1 |f (ξ) − fˆ(ξ)| ≤ |f (x)| − 1 dx, (1 + ix)2 −∞ and recall that f has moderate decrease on R. But for each fixed , we have |f (x + iy)| ≤ A

e2πM |y| , 1 + x2

so by Step 1 we must have f (ξ) = 0, and hence fˆ(ξ) = 0 after passing to the limit as  → 0. A similar argument applies if ξ < −M , although we must now argue in the upper half-plane, and use the factor 1/(1 − iz)2 instead. Step 3. To conclude the proof, it suffices to show that the conditions in the theorem imply condition (7) in Step 2. In fact, after dividing by an appropriate constant, it suffices to show that if |f (x)| ≤ 1 for all real x, and |f (z)| ≤ e2πM |z| for all complex z, then |f (x + iy)| ≤ e2πM |y| . This will follow from an ingenious and very useful idea of Phragm´en and Lindel¨ of that allows one to adapt the maximum modulus principle to various unbounded regions. The particular result we need is as follows. Theorem 3.4 Suppose F is a holomorphic function in the sector S = {z : −π/4 < arg z < π/4} that is continuous on the closure of S. Assume |F (z)| ≤ 1 on the boundary of the sector, and that there are constants C, c > 0 such that |F (z)| ≤ Cec|z| for all z in the sector. Then |F (z)| ≤ 1

for all z ∈ S.

In other words, if F is bounded by 1 on the boundary of S and has no more than a reasonable amount of growth, then F is actually bounded everywhere by 1. That some restriction on the growth of F is necessary 2 follows from a simple observation. Consider the function F (z) = ez . Then F is bounded by 1 on the boundary of S, but if x is real, F (x) is unbounded as x → ∞. We now give the proof of Theorem 3.4.

125

3. Paley-Wiener theorem 2

Proof. The idea is to subdue the “enemy” function ez and turn it 2 α to our advantage: in brief, one modifies ez by replacing it by ez with α < 2. For simplicity we take the case α = 3/2. If  > 0, let F (z) = F (z)e− z

3/2

.

Here we have chosen the principal branch of the logarithm to define z 3/2 so that if z = reiθ (with −π < θ < π), then z 3/2 = r3/2 e3iθ/2. Hence F is holomorphic in S and continuous up to its boundary. Also |e− z

3/2

| = e− r

3/2

cos(3θ/2)

;

and since −π/4 < θ < π/4 in the sector, we get the inequalities −

3π 3θ 3π π π 0, y > 0}, and the result remains the same. Then, we consider F (z) = f (z)e2πiM z ,

126

Chapter 4. THE FOURIER TRANSFORM

and note that F is bounded by 1 on the positive real and positive imaginary axes. Since we also have |F (z)| ≤ Cec|z| in the quadrant, we conclude by the Phragm´en-Lindel¨of theorem that |F (z)| ≤ 1 for all z in Q, which implies |f (z)| ≤ e2πM y . A similar argument for the other quadrants concludes Step 3 as well as the proof of the Paley-Wiener theorem. We conclude with another version of the idea behind the Paley-Wiener theorem, this time characterizing the functions whose Fourier transform vanishes for all negative ξ. Theorem 3.5 Suppose f and fˆ have moderate decrease. Then fˆ(ξ) = 0 for all ξ < 0 if and only if f can be extended to a continuous and bounded function in the closed upper half-plane {z = x + iy : y ≥ 0} with f holomorphic in the interior. Proof. formula

First assume fˆ(ξ) = 0 for ξ < 0. By the Fourier inversion




∞

f (x) =

fˆ(ξ)e2πixξ dξ ,

0

and we can extend f when z = x + iy with y ≥ 0 by
∞ fˆ(ξ)e2πizξ dξ . f (z) = 0

Notice that the integral converges and that
∞ dξ < ∞, |f (z)| ≤ A 1 + ξ2 0 which proves the boundedness of f . The uniform convergence of
n fˆ(ξ)e2πixξ dξ fn (z) = 0

to f (z) in the closed half-plane establishes the continuity of f there, and its holomorphicity in the interior. For the converse, we argue in the spirit of the proof of Theorem 3.3. For  and δ positive, we set f ,δ (z) =

f (z + iδ) . (1 − iz)2

Then f ,δ is holomorphic in a region containing the closed upper halfplane. One also shows as before, using Cauchy’s theorem, that f ,δ (ξ) = 0

127

4. Exercises

f ,0 (ξ) = 0 for all ξ < 0. Then, passing to the limit successively, one has  (ξ) = 0 for all ξ < 0. for ξ < 0, and finally fˆ(ξ) = f 0,0 Remark. The reader should note a certain analogy between the above theorem and Theorem 7.1 in Chapter 3. Here we are dealing with a function holomorphic in the upper half-plane, and there with a function holomorphic in a disc. In the present case the Fourier transform vanishes when ξ < 0, and in the earlier case, the Fourier coefficients vanish when n < 0.

4 Exercises 1. Suppose f is continuous and of moderate decrease, and fˆ(ξ) = 0 for all ξ ∈ R. Show that f = 0 by completing the following outline: (a) For each fixed real number t consider the two functions


t

A(z) =




f (x)e−2πiz(x−t) dx

and

−∞

∞

B(z) = −

f (x)e−2πiz(x−t) dx.

t

Show that A(ξ) = B(ξ) for all ξ ∈ R. (b) Prove that the function F equal to A in the closed upper half-plane, and B in the lower half-plane, is entire and bounded, thus constant. In fact, show that F = 0. (c) Deduce that


t

f (x) dx = 0, −∞

for all t, and conclude that f = 0.

2. If f ∈ Fa with a > 0, then for any positive integer n one has f (n) ∈ Fb whenever 0 ≤ b < a. [Hint: Modify the solution to Exercise 8 in Chapter 2.] 3. Show, by contour integration, that if a > 0 and ξ ∈ R then 1 π and check that







∞ −∞

∞ −∞

a e−2πixξ dx = e−2πa|ξ| , a2 + x2

e−2πa|ξ| e2πiξx dξ =

a 1 . π a2 + x2

128

Chapter 4. THE FOURIER TRANSFORM

4. Suppose Q is a polynomial of degree ≥ 2 with distinct roots, none lying on the real axis. Calculate
∞ −2πixξ e dx, ξ∈R −∞ Q(x) in terms of the roots of Q. What happens when several roots coincide? [Hint: Consider separately the cases ξ < 0, ξ = 0, and ξ > 0. Use residues.] 5. More generally, let R(x) = P (x)/Q(x) be a rational function with (degree Q) ≥ (degreeP )+2 and Q(x) = 0 on the real axis. (a) Prove that if α1 , . . . , αk are the roots of R in the upper half-plane, then there exists polynomials Pj (ξ) of degree less than the multiplicity of αj so that


∞

R(x)e−2πixξ dx =

−∞

k 

Pj (ξ)e−2πiαj ξ ,

when ξ < 0.

j=1

(b) In if Q(z) has no zeros in the upper half-plane, then  ∞particular, R(x)e−2πixξ dx = 0 for ξ < 0. −∞ (c) Show that similar results hold in the case ξ > 0. (d) Show that


∞

R(x)e−2πixξ dx = O(e−a|ξ| ),

−∞

ξ∈R

as |ξ| → ∞ for some a > 0. Determine the best possible a’s in terms of the roots of R. [Hint: For part (a), use residues. The powers of ξ appear when one differentiates the function f (z) = R(z)e−2πizξ (as in the formula of Theorem 1.4 in the previous chapter). For part (c) argue in the lower half-plane.] 6. Prove that ∞ ∞  1  a = e−2πa|n| 2 2 π n=−∞ a + n n=−∞

whenever a > 0. Hence show that the sum equals coth πa. 7. The Poisson summation formula applied to specific examples often provides interesting identities. (a) Let τ be fixed with Im(τ ) > 0. Apply the Poisson summation formula to f (z) = (τ + z)−k ,

129

4. Exercises

where k is an integer ≥ 2, to obtain ∞  n=−∞

∞ (−2πi)k  k−1 2πimτ 1 = m e . (τ + n)k (k − 1)! m=1

(b) Set k = 2 in the above formula to show that if Im(τ ) > 0, then ∞ 

π2 1 = . 2 2 (τ + n) sin (πτ ) n=−∞ (c) Can one conclude that the above formula holds true whenever τ is any complex number that is not an integer? [Hint: For (a), use residues to prove that fˆ(ξ) = 0, if ξ < 0, and (−2πi)k k−1 2πiξτ fˆ(ξ) = e , ξ (k − 1)!

when ξ > 0.]

8. Suppose fˆ has compact support contained in [−M, M ] and let f (z) = Show that
(2πi)n M ˆ an = f (ξ)ξ n dξ, n! −M

∞ n=0

an z n .

and as a result lim sup(n!|an |)1/n ≤ 2πM. n→∞

 n with In the converse direction, let f be any power series f (z) = ∞ n=0 an z 1/n lim supn→∞ (n!|an |) ≤ 2πM . Then, f is holomorphic in the complex plane, and for every > 0 there exists A > 0 such that |f (z)| ≤ A e2π(M +)|z| .

9. Here are further results similar to the Phragm´en-Lindel¨ of theorem. (a) Let F be a holomorphic function in the right half-plane that extends continuously to the boundary, that is, the imaginary axis. Suppose that |F (iy)| ≤ 1 for all y ∈ R, and |F (z)| ≤ Cec|z|

γ

for some c, C > 0 and γ < 1. Prove that |F (z)| ≤ 1 for all z in the right half-plane.

130

Chapter 4. THE FOURIER TRANSFORM

(b) More generally, let S be a sector whose vertex is the origin, and forming an angle of π/β. Let F be a holomorphic function in S that is continuous on the closure of S, so that |F (z)| ≤ 1 on the boundary of S and α

|F (z)| ≤ Cec|z| for all z ∈ S for some c, C > 0 and 0 < α < β. Prove that |F (z)| ≤ 1 for all z ∈ S. 2

10. This exercise generalizes some of the properties of e−πx related to the fact that it is its own Fourier transform. Suppose f (z) is an entire function that satisfies |f (x + iy)| ≤ ce−ax

2

+by 2

for some a, b, c > 0. Let
fˆ(ζ) =

∞

f (x)e−2πixζ dx.

−∞

Then, fˆ is an entire function of ζ that satisfies  2

|fˆ(ξ + iη)| ≤ c e−a ξ

+b η 2

for some a , b , c > 0.  2 [Hint: To prove fˆ(ξ) = O(e−a ξ ), assume ξ > 0 and change the contour of integration to x − iy for some y > 0 fixed, and −∞ < x < ∞. Then 2 fˆ(ξ) = O(e−2πyξ eby ).

Finally, choose y = dξ where d is a small constant.] 11. One can give a neater formulation of the result in Exercise 10 by proving the following fact. Suppose f (z) is an entire function of strict order 2, that is, 2

f (z) = O(ec1 |z| ) for some c1 > 0. Suppose also that for x real, 2

f (x) = O(e−c2 |x| ) for some c2 > 0. Then |f (x + iy)| = O(e−ax

2

+by 2

for some a, b > 0. The converse is obviously true.

)

131

5. Problems

12. The principle that a function and its Fourier transform cannot both be too small at infinity is illustrated by the following theorem of Hardy. If f is a function on R that satisfies 2

f (x) = O(e−πx )

and

2 fˆ(ξ) = O(e−πξ ),

2

2

then f is a constant multiple of e−πx . As a result, if f (x) = O(e−πAx ), and 2 fˆ(ξ) = O(e−πBξ ), with AB > 1 and A, B > 0, then f is identically zero. (a) If f is even, show that fˆ extends to an even entire function. Moreover, if g(z) = fˆ(z 1/2 ), then g satisfies |g(x)| ≤ ce−πx

and

|g(z)| ≤ ceπR sin

2

(θ/2)

≤ ceπ|z|

when x ∈ R and z = Reiθ with R ≥ 0 and θ ∈ R. (b) Apply the Phragm´en-Lindel¨ of principle to the function F (z) = g(z)eγz

where γ = iπ

e−iπ/(2β) sin π/(2β)

and the sector 0 ≤ θ ≤ π/β < π, and let β → π to deduce that eπz g(z) is bounded in the closed upper half-plane. The same result holds in the lower half-plane, so by Liouville’s theorem eπz g(z) is constant, as desired. (c) If f is odd, then fˆ(0) = 0, and apply the above argument to fˆ(z)/z to deduce that f = fˆ = 0. Finally, write an arbitrary f as an appropriate sum of an even function and an odd function.

5 Problems p

1. Suppose fˆ(ξ) = O(e−a|ξ| ) as |ξ| → ∞, for some p > 1. Then f is holomorphic for all z and satisfies the growth condition |f (z)| ≤ Aea|z|

q

where 1/p + 1/q = 1. Note that on the one hand, when p → ∞ then q → 1, and this limiting case can be interpreted as part of Theorem 3.3. On the other hand, when p → 1 then q → ∞, and this limiting case in a sense brings us back to Theorem 2.1. [Hint: To prove the result, use the inequality −ξ p + ξu ≤ uq , which is valid when ξ and u are non-negative. To establish this inequality, examine separately the cases ξ p ≥ ξu and ξ p < ξu; note also that the functions ξ = uq−1 and u = ξ p−1 are inverses of each other because (p − 1)(q − 1) = 1.]

132

Chapter 4. THE FOURIER TRANSFORM

2. The problem is to solve the differential equation an

dn dn−1 u(t) + a u(t) + · · · + a0 u(t) = f (t) , n−1 dtn dtn−1

where a0 , a1 , . . . , an are complex constants, and f is a given function. Here we suppose that f has bounded support and is smooth (say of class C 2 ). (a) Let


∞

fˆ(z) =

f (t)e−2πizt dt.

−∞

Observe that fˆ is an entire function, and using integration by parts show that |fˆ(x + iy)| ≤

A 1 + x2

if |y| ≤ a for any fixed a ≥ 0. (b) Write P (z) = an (2πiz)n + an−1 (2πiz)n−1 + · · · + a0 . Find a real number c so that P (z) does not vanish on the line L = {z : z = x + ic, x ∈ R}. (c) Set
u(t) = L

e2πizt ˆ f (z) dz. P (z)

Check that n  j=0

 aj

d dt

j


e2πizt fˆ(z) dz

u(t) = L

and



e2πizt fˆ(z) dz =

∞

e2πixt fˆ(x) dx.

−∞

L

Conclude by the Fourier inversion theorem that n  j=0

 aj

d dt

j u(t) = f (t).

133

5. Problems

Note that the solution u depends on the choice c. 3.∗ In this problem, we investigate the behavior of certain bounded holomorphic functions in an infinite strip. The particular result described here is sometimes called the three-lines lemma. (a) Suppose F (z) is holomorphic and bounded in the strip 0 < Im(z) < 1 and continuous on its closure. If |F (z)| ≤ 1 on the boundary lines, then |F (z)| ≤ 1 throughout the strip. (b) For the more general F , let supx∈R |F (x)| = M0 and supx∈R |F (x + i)| = M1 . Then, sup |F (x + iy)| ≤ M01−y M1y ,

if 0 ≤ y ≤ 1.

x∈R

(c) As a consequence, prove that log supx∈R |F (x + iy)| is a convex function of y when 0 ≤ y ≤ 1. 2

[Hint: For part (a), apply the maximum modulus principle to F (z) = F (z)e−z . For part (b), consider M0z−1 M1−z F (z).] 4.∗ There is a relation between the Paley-Wiener theorem and an earlier representation due to E. Borel. (a) A function f (z), holomorphic for all z, satisfies |f (z)| ≤ A e2π(M +)|z| for all if and only if it is representable in the form
e2πizw g(w) dw f (z) = C

where g is holomorphic outside the circle of radius M centered at the origin, and g vanishes at infinity. Here C is any circle centered  at the origin of radius −n−1 with larger than M . In fact, if f (z) = an z n , then g(w) = ∞ n=0 An w n+1 /n!. an = An (2πi) (b) The connection with Theorem 3.3 is as follows. For these functions f (for which in addition f and fˆ are of moderate decrease on the real axis), one can assert that the g above is holomorphic in the larger region, which consists of the slit plane C − [−M, M ]. Moreover, the relation between g and the Fourier transform fˆ is g(z) =

1 2πi




M −M

fˆ(ξ) dξ ξ−z

so that fˆ represents the jump of g across the segment [−M, M ]; that is, fˆ(x) =

lim

→0,>0

See Problem 5 in Chapter 3.

g(x + i ) − g(x − i ).

5 Entire Functions ...but after the 15th of October I felt myself a free man, with such longing for mathematical work, that the last two months flew by quickly, and that only today I found the letter of the 19th of October that I had not answered. The result of my work, with which I am not entirely satisfied, I want to share with you. Firstly, in looking back at my lectures, a gap in function theory needed to be filled. As you know, up to now the following question had been unresolved. Given an arbitrary sequence of complex numbers, a1 , a2 , . . . , can one construct an entire (transcendental) function that vanishes at these values, with prescribed multiplicities, and nowhere else?... K. Weierstrass, 1874

In this chapter, we will study functions that are holomorphic in the whole complex plane; these are called entire functions. Our presentation will be organized around the following three questions: 1. Where can such functions vanish? We shall see that the obvious necessary condition is also sufficient: if {zn } is any sequence of complex numbers having no limit point in C, then there exists an entire function vanishing exactly at the points of this sequence. The construction of the desired function is inspired by Euler’s product formula for sin πz (the prototypical case when {zn } is Z), but requires an additional refinement: the Weierstrass canonical factors. 2. How do these functions grow at infinity? Here, matters are controlled by an important principle: the larger a function is, the more zeros it can have. This principle already manifests itself in the simple case of polynomials. By the fundamental theorem of algebra, the number of zeros of a polynomial P of degree d is precisely d, which is also the exponent in the order of (polynomial) growth of P , namely sup |P (z)| ≈ Rd

|z|=R

as R → ∞.

135

1. Jensen’s formula

A precise version of this general principle is contained in Jensen’s formula, which we prove in the first section. This formula, central to much of the theory developed in this chapter, exhibits a deep connection between the number of zeros of a function in a disc and the (logarithmic) average of the function over the circle. In fact, Jensen’s formula not only constitutes a natural starting point for us, but also leads to the fruitful theory of value distributions, also called Nevanlinna theory (which, however, we do not take up here). 3. To what extent are these functions determined by their zeros? It turns out that if an entire function has a finite (exponential) order of growth, then it can be specified by its zeros up to multiplication by a simple factor. The precise version of this assertion is the Hadamard factorization theorem. It may be viewed as another instance of the general rule that was formulated in Chapter 3, that is, that under appropriate conditions, a holomorphic function is essentially determined by its zeros.

1 Jensen’s formula In this section, we denote by DR and CR the open disc and circle of radius R centered at the origin. We shall also, in the rest of this chapter, exclude the trivial case of the function that vanishes identically. Theorem 1.1 Let Ω be an open set that contains the closure of a disc DR and suppose that f is holomorphic in Ω, f (0) = 0, and f vanishes nowhere on the circle CR . If z1 , . . . , zN denote the zeros of f inside the disc (counted with multiplicities),1 then

(1)

log |f (0)| =

N  k=1

 log

|zk | R



1 + 2π




2π

log |f (Reiθ )| dθ. 0

The proof of the theorem consists of several steps. Step 1. First, we observe that if f1 and f2 are two functions satisfying the hypotheses and the conclusion of the theorem, then the product f1 f2 also satisfies the hypothesis of the theorem and formula (1). This observation is a simple consequence of the fact that log xy = log x + log y whenever x and y are positive numbers, and that the set of zeros of f1 f2 is the union of the sets of zeros of f1 and f2 . 1 That

is, each zero appears in the sequence as many times as its order.

136

Chapter 5. ENTIRE FUNCTIONS

Step 2. The function g(z) =

f (z) (z − z1 ) · · · (z − zN )

initially defined on Ω − {z1 , . . . , zN }, is bounded near each zj . Therefore each zj is a removable singularity, and hence we can write f (z) = (z − z1 ) · · · (z − zN )g(z) where g is holomorphic in Ω and nowhere vanishing in the closure of DR . By Step 1, it suffices to prove Jensen’s formula for functions like g that vanish nowhere, and for functions of the form z − zj . Step 3. We first prove (1) for a function g that vanishes nowhere in the closure of DR . More precisely, we must establish the following identity: log |g(0)| =

1 2π




2π

log |g(Reiθ )| dθ. 0

In a slightly larger disc, we can write g(z) = eh(z) where h is holomorphic in that disc. This is possible since discs are simply connected, and we can define h = log g (see Theorem 6.2 in Chapter 3). Now we observe that |g(z)| = |eh(z) | = |eRe(h(z))+i Im(h(z)) | = eRe(h(z)) , so that log |g(z)| = Re(h(z)). The mean value property (Corollary 7.3 in Chapter 3) then immediately implies the desired formula for g. Step 4. The last step is to prove the formula for functions of the form f (z) = z − w, where w ∈ DR . That is, we must show that

 log |w| = log

|w| R



1 + 2π




2π

log |Reiθ − w| dθ. 0

Since log(|w|/R) = log |w|−log R and log |Reiθ − w| = log R+log |eiθ − w/R|, it suffices to prove that




2π

log |eiθ − a| dθ = 0,

whenever |a| < 1.

0

This in turn is equivalent (after the change of variables θ → −θ) to




2π

log |1 − aeiθ | dθ = 0, 0

whenever |a| < 1.

137

1. Jensen’s formula

To prove this, we use the function F (z) = 1 − az, which vanishes nowhere in the closure of the unit disc. As a consequence, there exists a holomorphic function G in a disc of radius greater than 1 such that F (z) = eG(z) . Then |F | = eRe(G) , and therefore log |F | = Re(G). Since F (0) = 1 we have log |F (0)| = 0, and an application of the mean value property (Corollary 7.3 in Chapter 3) to the harmonic function log |F (z)| concludes the proof of the theorem. From Jensen’s formula we can derive an identity linking the growth of a holomorphic function with its number of zeros inside a disc. If f is a holomorphic function on the closure of a disc DR , we denote by n(r) (or nf (r) when it is necessary to keep track of the function in question) the number of zeros of f (counted with their multiplicities) inside the disc Dr , with 0 < r < R. A simple but useful observation is that n(r) is a non-decreasing function of r. We claim that if f (0) = 0, and f does not vanish on the circle CR , then
R
2π 1 dr = (2) n(r) log |f (Reiθ )| dθ − log |f (0)|. r 2π 0 0 This formula is immediate from Jensen’s equality and the next lemma. Lemma 1.2 If z1 , . . . , zN are the zeros of f inside the disc DR , then




R

0

  N R dr  = n(r) log   . r zk k=1

Proof. First we have N  k=1

   N
R   log   = zk k=1

R

|zk |

dr . r

If we define the characteristic function 1 if r > |zk |, ηk (r) = 0 if r ≤ |zk |, then

N

N
 k=1

= n(r), and the lemma is proved using 
R 
R N
N dr  R dr dr dr = = = ηk (r) ηk (r) n(r) . r r r r 0 0 0

k=1 ηk (r)

R

|zk |

k=1

k=1

138

Chapter 5. ENTIRE FUNCTIONS

2 Functions of finite order Let f be an entire function. If there exist a positive number ρ and constants A, B > 0 such that |f (z)| ≤ AeB|z|

ρ

for all z ∈ C,

then we say that f has an order of growth ≤ ρ. We define the order of growth of f as ρf = inf ρ , where the infimum is over all ρ > 0 such that f has an order of growth ≤ ρ. 2 For example, the order of growth of the function ez is 2. Theorem 2.1 If f is an entire function that has an order of growth ≤ ρ, then: (i) n(r) ≤ Crρ for some C > 0 and all sufficiently large r. (ii) If z1 , z2 , . . . denote the zeros of f , with zk = 0, then for all s > ρ we have ∞  1 < ∞. |zk |s k=1

Proof. It suffices to prove the estimate for n(r) when f (0) = 0. Indeed, consider the function F (z) = f (z)/z where  is the order of the zero of f at the origin. Then nf (r) and nF (r) differ only by a constant, and F also has an of order of growth ≤ ρ. If f (0) = 0 we may use formula (2), namely




R

n(x) 0

1 dx = x 2π




2π

log |f (Reiθ )| dθ − log |f (0)|. 0

Choosing R = 2r, this formula implies




2r

n(x) r

1 dx ≤ x 2π




2π

log |f (Reiθ )| dθ − log |f (0)|. 0

On the one hand, since n(r) is increasing, we have




2r r

dx ≥ n(r) n(x) x




2r r

dx = n(r)[log 2r − log r] = n(r) log 2, x

139

2. Functions of finite order

and on the other hand, the growth condition on f gives
2π
2π ρ iθ log |f (Re )| dθ ≤ log |AeBR | dθ ≤ C  rρ 0

0

for all large r. Consequently, n(r) ≤ Crρ for an appropriate C > 0 and all sufficiently large r. The following estimates prove the second part of the theorem:   ∞     |zk |−s = |zk |−s  |zk |≥1

2j ≤|zk | ρ. Part (ii) of the theorem is a noteworthy fact, which we shall use in a later part of this chapter. We give two simple examples of the theorem; each of these shows that the condition s > ρ cannot be improved. Example 1. Consider f (z) = sin πz. Recall Euler’s identity, namely f (z) =

eiπz − e−iπz , 2i

which implies that |f (z)| ≤ eπ|z| , and f has an order of growth ≤ 1. By taking z = ix, where x ∈ R, it is clear that the order of growth of f is actually equal f vanishes to order 1 at z = n for each  to 1. However, s n ∈ Z, and n=0 1/|n| < ∞ precisely when s > 1. Example 2. Consider f (z) = cos z 1/2 , which we define by cos z 1/2 =

∞  zn . (−1)n (2n)! n=0

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Chapter 5. ENTIRE FUNCTIONS

Then f is entire, and it is easy to see that |f (z)| ≤ e|z|

1/2

,

and the order of growth  of f is 1/2. Moreover, f (z) vanishes when zn = ((n + 1/2)π)2, while n 1/|zn |s < ∞ exactly when s > 1/2. A natural question is whether or not, given any sequence of complex numbers z1 , z2 , . . ., there exists an entire function f with zeros precisely at the points of this sequence. A necessary condition is that z1 , z2 , . . . do not accumulate, in other words we must have lim |zk | = ∞ ,

k→∞

otherwise f would vanish identically by Theorem 4.8 in Chapter 2. Weierstrass proved that this condition is also sufficient by explicitly constructing a function with these prescribed zeros. A first guess is of course the product (z − z1 )(z − z2 ) · · · , which provides a solution in the special case when the sequence of zeros is finite. In general, Weierstrass showed how to insert factors in this product so that the convergence is guaranteed, yet no new zeros are introduced. Before coming to the general construction, we review infinite products and study a basic example.

3 Infinite products 3.1 Generalities Given a sequence {an }∞ n=1 of complex numbers, we say that the product ∞ 

(1 + an )

n=1

converges if the limit lim

N →∞

N 

(1 + an )

n=1

of the partial products exists. A useful necessary condition that guarantees the existence of a product is contained in the following proposition.

141

3. Infinite products

 ∞ |an | < ∞, then the product n=1 (1 + an ) conProposition 3.1 If verges. Moreover, the product converges to 0 if and only if one of its factors is 0. This is simply Proposition 1.9 of Chapter 8 in Book I. We repeat the proof here.  Proof. If |an | converges, then for all large n we must have |an | < 1/2. Disregarding if necessary finitely many terms, we may assume that this inequality holds for all n. In particular, we can define log(1 + an ) by the usual power series (see (6) in Chapter 3), and this logarithm satisfies the property that 1 + z = elog(1+z) whenever |z| < 1. Hence we may write the partial products as follows: N 

(1 + an ) =

n=1

N 

elog(1+an ) = eBN ,

n=1

N

where BN = n=1 bn with bn = log(1 + an ). By the power series expansion we see that | log(1 + z)| ≤ 2|z|, if |z| < 1/2. Hence |bn | ≤ 2|an |, so BN converges as N → ∞ to a complex number, say B. Since the exponential function is continuous, we conclude that eBN converges to eB as N → ∞, proving the first assertion of the proposition. Observe also that if 1 + an = 0 for all n, then the product converges to a non-zero limit since it is expressed as eB . More generally, we can consider products of holomorphic functions. Proposition 3.2 Suppose {Fn } is a sequence of holomorphic functions on the open set Ω. If there exist constants cn > 0 such that  cn < ∞ and |Fn (z) − 1| ≤ cn for all z ∈ Ω, then:

∞ (i) The product n=1 Fn (z) converges uniformly in Ω to a holomorphic function F (z).

(ii) If Fn (z) does not vanish for any n, then F  (z)  Fn (z) = . F (z) Fn (z) ∞

n=1

Proof. To prove the first statement, note that for each z we may argue as in the previous proposition if we write Fn (z) = 1 + an (z), with

142

Chapter 5. ENTIRE FUNCTIONS

|an (z)| ≤ cn . Then, we observe that the estimates are actually uniform in z because the cn ’s are constants. It follows that the product converges uniformly to a holomorphic function, which we denote by F (z). To establish the second part of the theorem, suppose that K is a compact subset of Ω, and let GN (z) =

N 

Fn (z).

n=1

We have just proved that GN → F uniformly in Ω, so by Theorem 5.3 in Chapter 2, the sequence {GN } converges uniformly to F  in K. Since GN is uniformly bounded from below on K, we conclude that GN /GN → F  /F uniformly on K, and because K is an arbitrary compact subset of Ω, the limit holds for every point of Ω. Moreover, as we saw in Section 4 of Chapter 3

 F GN n = , GN Fn N

n=1

so part (ii) of the proposition is also proved. 3.2 Example: the product formula for the sine function Before proceeding with the general theory of Weierstrass products, we consider the key example of the product formula for the sine function:

 ∞   z2 sin πz =z 1− 2 . π n

(3)

n=1

This identity will in turn be derived from the sum formula for the cotangent function (cot πz = cos πz/ sin πz): (4)

π cot πz =

∞  n=−∞

∞  1  2z 1 1 = lim = + . z + n N →∞ z+n z z 2 − n2 |n|≤N

n=1

The first formula holds for all complex ∞ numbers z, and the second whenever z is not an integer. The sum n=−∞ 1/(z + n) needs to be properly understood, because the separate halves corresponding to positive and negative values of n do not converge. Only when interpreted symmetrically, as limN →∞ |n|≤N 1/(z + n), does the cancellation of terms lead to a convergent series as in (4) above.

143

3. Infinite products

We prove (4) by showing that both π cot πz and the series have the same structural properties. In fact, observe that if F (z) = π cot πz, then F has the following three properties: (i) F (z + 1) = F (z) whenever z is not an integer. (ii) F (z) =

1 + F0 (z), where F0 is analytic near 0. z

(iii) F (z) has simple poles at the integers, and no other singularities. Then, we note that the function ∞  n=−∞

 1 1 = lim z + n N →∞ z+n |n|≤N

also satisfies these same three properties. In fact, property (i) is simply the observation that the passage from z to z + 1 merely shifts the terms in the infinite sum. To be precise,

 |n|≤N

 1 1 1 1 = − + . z+1+n z+1+N z−N z+n |n|≤N

Letting N tend to infinity proves the assertion. Properties (ii) and (iii) ∞ are evident from the representation z1 + n=1 z 22z −n2 of the sum. Therefore, the function defined by ∆(z) = F (z) −

∞  n=−∞

1 z+n

is periodic in the sense that ∆(z + 1) = ∆(z), and by (ii) the singularity of ∆ at the origin is removable, and hence by periodicity the singularities at all the integers are also removable; this implies that ∆ is entire. To prove our formula, it will suffice to show that the function ∆ is bounded in the complex plane. By the periodicity above, it is enough to do so in the strip |Re(z)| ≤ 1/2. This is because every z  ∈ C is of the form z  = z + k, where z is in the strip and k is an integer. Since ∆ is holomorphic, it is bounded in the rectangle |Im(z)| ≤ 1, and we need only control the behavior of that function for |Im(z)| > 1. If Im(z) > 1 and z = x + iy, then cot πz = i

eiπz + e−iπz e−2πy + e−2πix = i −2πy , iπz −iπz e −e e − e−2πix

144

Chapter 5. ENTIRE FUNCTIONS

and in absolute value this quantity is bounded. Also

 2(x + iy) 1 1  2z + + ; = 2 2 2 z z −n x + iy x − y 2 − n2 + 2ixy ∞

∞

n=1

n=1

therefore if y > 1, we have   ∞ ∞ 1    2z y   ≤ C + C .  +  2 2 2 z −n  y + n2 z n=1

n=1

Now the sum on the right-hand side is majorized by
∞ y dx , 2 + x2 y 0 because the function y/(y 2 + x2 ) is decreasing in x; moreover, as the change of variables x → yx shows, the integral is independent of y, and hence bounded. By a similar argument ∆ is bounded in the strip where Im(z) < −1, hence is bounded throughout the whole strip |Re(z)| ≤ 1/2. Therefore ∆ is bounded in C, and by Liouville’s theorem, ∆(z) is constant. The observation that ∆ is odd shows that this constant must be 0, and concludes the proof of formula (4). To prove (3), we now let  ∞   sin πz z2 G(z) = and P (z) = z 1− 2 . π n n=1



1/n2 < ∞ guarantee that the prodProposition 3.2 and the fact that uct P (z) converges, and that away from the integers we have P  (z) 1  2z = + . P (z) z z 2 − n2 ∞

n=1

Since G (z)/G(z) = π cot πz, the cotangent formula (4) gives  

 P (z) P (z) P  (z) G (z) − = = 0, G(z) G(z) P (z) G(z) and so P (z) = cG(z) for some constant c. Dividing this identity by z, and taking the limit as z → 0, we find c = 1. Remark. Other proofs of (4) and (3) can be given by integrating analogous identities for π 2 /(sin πz)2 derived in Exercise 12, Chapter 3, and Exercise 7, Chapter 4. Still other proofs using Fourier series can be found in the exercises of Chapters 3 and 5 of Book I.

145

4. Weierstrass infinite products

4 Weierstrass infinite products We now turn to Weierstrass’s construction of an entire function with prescribed zeros. Theorem 4.1 Given any sequence {an } of complex numbers with |an | → ∞ as n → ∞, there exists an entire function f that vanishes at all z = an and nowhere else. Any other such entire function is of the form f (z)eg(z) , where g is entire. Recall that if a holomorphic function f vanishes at z = a, then the multiplicity of the zero a is the integer m so that f (z) = (z − a)m g(z), where g is holomorphic and nowhere vanishing in a neighborhood of a. Alternatively, m is the first non-zero power of z − a in the power series expansion of f at a. Since, as before, we allow for repetitions in the sequence {an }, the theorem actually guarantees the existence of entire functions with prescribed zeros and with desired multiplicities. To begin the proof, note first that if f1 and f2 are two entire functions that vanish at all z = an and nowhere else, then f1 /f2 has removable singularities at all the points an . Hence f1 /f2 is entire and vanishes nowhere, so that there exists an entire function g with f1 (z)/f2(z) = eg(z) , as we showed in Section 6 of Chapter 3. Therefore f1 (z) = f2 (z)eg(z) and the last statement of the theorem is verified. Hence we are left with the task of constructing a function that vanishes at all the points of the sequence {an } and nowhere else.  A naive guess, suggested by the product formula for sin πz, is the product n (1 − z/an ). The problem is that this product converges only for suitable sequences {an }, so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros. For each integer k ≥ 0 we define canonical factors by E0 (z) = 1 − z

and

Ek (z) = (1 − z)ez+z

2

/2+···+z k /k

,

for k ≥ 1.

The integer k is called the degree of the canonical factor. Lemma 4.2 If |z| ≤ 1/2, then |1 − Ek (z)| ≤ c|z|k+1 for some c > 0. Proof. If |z| ≤ 1/2, then with the logarithm defined in terms of the power series, we have 1 − z = elog(1−z) , and therefore Ek (z) = elog(1−z)+z+z

2

/2+···+z k /k

= ew ,

146

Chapter 5. ENTIRE FUNCTIONS

where w = −

∞

n=k+1 z

|w| ≤ |z|

k+1

n

/n. Observe that since |z| ≤ 1/2 we have

∞ 

|z|

n−k−1

/n ≤ |z|

k+1

∞ 

2−j ≤ 2|z|k+1 .

j=0

n=k+1

In particular, we have |w| ≤ 1 and this implies that |1 − Ek (z)| = |1 − ew | ≤ c |w| ≤ c|z|k+1 .

Remark. An important technical point is that the constant c in the statement of the lemma can be chosen to be independent of k. In fact, an examination of the proof shows that we may take c = e and then c = 2e. Suppose that we are given a zero of order m at the origin, and that a1 , a2 . . . are all non-zero. Then we define the Weierstrass product by ∞ 

f (z) = z m

En (z/an ).

n=1

We claim that this function has the required properties; that is, f is entire with a zero of order m at the origin, zeros at each point of the sequence {an }, and f vanishes nowhere else. Fix R > 0, and suppose that z belongs to the disc |z| < R. We shall prove that f has all the desired properties in this disc, and since R is arbitrary, this will prove the theorem. We can consider two types of factors in the formula defining f , with the choice depending on whether |an | ≤ 2R or |an | > 2R. There are only finitely many terms of the first kind (since |an | → ∞), and we see that the finite product vanishes at all z = an with |an | < R. If |an | ≥ 2R, we have |z/an | ≤ 1/2, hence the previous lemma implies  n+1 z  c ≤ n+1 . |1 − En (z/an )| ≤ c   an 2 Note that by the above remark, c does not depend on n. Therefore, the product  En (z/an ) |an |≥2R

defines a holomorphic function when |z| < R, and does not vanish in that disc by the propositions in Section 3. This shows that the function

147

5. Hadamard’s factorization theorem

f has the desired properties, and the proof of Weierstrass’s theorem is complete.

5 Hadamard’s factorization theorem The theorem of this section combines the results relating the growth of a function to the number of zeros it possesses, and the above product theorem. Weierstrass’s theorem states that a function that vanishes at the points a1 , a2 , . . . takes the form ∞ 

eg(z) z m

En (z/an ).

n=1

Hadamard refined this result by showing that in the case of functions of finite order, the degree of the canonical factors can be taken to be constant, and g is then a polynomial. Recall that an entire function has an order of growth ≤ ρ if ρ

|f (z)| ≤ AeB|z| , and that the order of growth ρ0 of f is the infimum of all such ρ’s. A basic result we proved earlier was that if f has order of growth ≤ ρ, then n(r) ≤ Crρ ,

for all large r,

and if a1 , a2 , . . . are the non-zero zeros of f , and s > ρ, then



|an |−s < ∞.

Theorem 5.1 Suppose f is entire and has growth order ρ0 . Let k be the integer so that k ≤ ρ0 < k + 1. If a1 , a2 , . . . denote the (non-zero) zeros of f , then f (z) = eP (z) z m

∞ 

Ek (z/an ),

n=1

where P is a polynomial of degree ≤ k, and m is the order of the zero of f at z = 0. Main lemmas Here we gather a few lemmas needed in the proof of Hadamard’s theorem.

148

Chapter 5. ENTIRE FUNCTIONS

Lemma 5.2 The canonical products satisfy |Ek (z)| ≥ e−c|z|

k+1

if |z| ≤ 1/2

and 

|Ek (z)| ≥ |1 − z| e−c |z|

k

if |z| ≥ 1/2.

Proof. If |z| ≤ 1/2 we can use the power series to define the logarithm of 1 − z, so that Ek (z) = elog(1−z)+

k n=1

z n /n

= e−

∞ n=k+1

z n /n

= ew .

Since |ew | ≥ e−|w| and |w| ≤ c|z|k+1 , the first part of the lemma follows. For the second part, simply observe that if |z| ≥ 1/2, then |Ek (z)| = |1 − z||ez+z

2

/2+···+z k /k

|,

and that there exists c > 0 such that |ez+z

2

/2+···+z k /k

| ≥ e−|z+z

2

/2+···+z k /k|



≥ e−c |z| . k

The inequality in the lemma when |z| ≥ 1/2 then follows from these observations. The key to the proof of Hadamard’s theorem consists of finding a lower bound for the product of the canonical factors when z stays away from the zeros {an }. Therefore, we shall first estimate the product from below, in the complement of small discs centered at these points. Lemma 5.3 For any s with ρ0 < s < k + 1, we have

∞    s   Ek (z/an ) ≥ e−c|z| ,  n=1  except possibly when z belongs to the union of the discs centered at an of radius |an |−k−1 , for n = 1, 2, 3, . . .. Proof. The proof this lemma is a little subtle. First, we write ∞  n=1

Ek (z/an ) =

 |an |≤2|z|

Ek (z/an )

 |an |>2|z|

Ek (z/an ).

149

5. Hadamard’s factorization theorem

For the second product the estimate asserted above holds with no restriction on z. Indeed, by the previous lemma

          E (z/a ) |Ek (z/an )| k n =  |an |>2|z|  |an |>2|z|  k+1 ≥ e−c|z/an | |an |>2|z|

≥ e−c|z|

k+1

 |an |>2|z|

|an |−k−1

.

But |an | > 2|z| and s < k + 1, so we must have |an |−k−1 = |an |−s |an |s−k−1 ≤ C|an |−s |z|s−k−1 . Therefore, the fact that



|an |−s converges implies that

       s  Ek (z/an ) ≥ e−c|z|  |an |>2|z|  for some c > 0. To estimate the first product, we use the second part of Lemma 5.2, and write             k z      Ek (z/an ) ≥ e−c |z/an | . (5) 1 − an   |an |≤2|z|  |an |≤2|z| |an |≤2|z| We now note that







e−c |z/an| = e−c |z| k

k

 |an |≤2|z|

|an |−k

,

|an |≤2|z|

and again, we have |an |−k = |an |−s |an |s−k ≤ C|an |−s |z|s−k , thereby proving that   k s e−c |z/an | ≥ e−c|z| . |an |≤2|z|

It is the estimate on the first product on the right-hand side of (5) which requires the restriction on z imposed in the statement of the

150

Chapter 5. ENTIRE FUNCTIONS

lemma. Indeed, whenever z does not belong to a disc of radius |an |−k−1 centered at an , we must have |an − z| ≥ |an |−k−1 . Therefore      an − z   z 1 −  =     an  an  |an |≤2|z| |an |≤2|z|  ≥ |an |−k−1 |an |−1 |an |≤2|z|

=



|an |−k−2 .

|an |≤2|z|

Finally, the estimate for the first product follows from the fact that  log |an | ≤ (k + 2)n(2|z|) log 2|z| (k + 2) |an |≤2|z|

≤ c|z|s log 2|z| 

≤ c |z|s , for any s > s, and the second inequality follows because n(2|z|) ≤ c|z|s by Theorem 2.1. Since we restricted s to satisfy s > ρ0 , we can take an initial s sufficiently close to ρ0 , so that the assertion of the lemma is established (with s being replaced by s ). Corollary 5.4 There exists a sequence of radii, r1 , r2 , . . ., with rm → ∞, such that   ∞   s   Ek (z/an ) ≥ e−c|z| for |z| = rm .    n=1

Proof. Since



|an |−k−1 < ∞, there exists an integer N so that ∞ 

|an |−k−1 < 1/10.

n=N

Therefore, given any two consecutive large integers L and L + 1, we can find a positive number r with L ≤ r ≤ L + 1, such that the circle of radius r centered at the origin does not intersect the forbidden discs of Lemma 5.3. For otherwise, the union of the intervals

 1 1 In = |an | − , |an | + |an |k+1 |an |k+1

151

5. Hadamard’s factorization theorem

the interval [L, L + 1]. (which are of length 2|an |−k−1 ) would ∞cover all −k−1 (See Figure 1.) This would imply 2 n=N |an | ≥ 1, which is a contradiction. We can then apply the previous lemma with |z| = r to conclude the proof of the corollary.

a3 a1 an

I2 I3

I1

In

a2

Figure 1. The intervals In

Proof of Hadamard’s theorem Let E(z) = z

m

∞ 

Ek (z/an ).

n=1

To prove that E is entire, we repeat the argument in the proof of Theorem 4.1; we take into account that by Lemma 4.2

 k+1 z  |1 − Ek (z/an )| ≤ c   an

for all large n,

 and that the series |an |−k−1 converges. (Recall ρ0 < s < k + 1.) Moreover, E has the zeros of f , therefore f /E is holomorphic and nowhere vanishing. Hence f (z) = eg(z) E(z)

152

Chapter 5. ENTIRE FUNCTIONS

for some entire function g. By the fact that f has growth order ρ0 , and because of the estimate from below for E obtained in Corollary 5.4, we have    f (z)  Re(g(z))   ≤ c ec|z|s = e E(z)  for |z| = rm . This proves that Re(g(z)) ≤ C|z|s ,

for |z| = rm .

The proof of Hadamard’s theorem is therefore complete if we can establish the following final lemma. Lemma 5.5 Suppose g is entire and u = Re(g) satisfies u(z) ≤ Crs

whenever |z| = r

for a sequence of positive real numbers r that tends to infinity. Then g is a polynomial of degree ≤ s. Proof. We can expand g in a power series centered at the origin g(z) =

∞ 

an z n .

n=0

We have already proved in the last section of Chapter 3 (as a simple application of Cauchy’s integral formulas) that
2π 1 an rn if n ≥ 0 iθ −inθ g(re )e dθ = (6) 0 if n < 0. 2π 0 By taking complex conjugates we find that
2π 1 g(reiθ )e−inθ dθ = 0 (7) 2π 0 whenever n > 0, and since 2u = g + g we add equations (6) and (7) to obtain
1 2π u(reiθ )e−inθ dθ, whenever n > 0. an rn = π 0 For n = 0 we can simply take real parts of both sides of (6) to find that
1 2π u(reiθ ) dθ. 2Re(a0 ) = π 0

153

6. Exercises

Now we recall the simple fact that whenever n = 0, the integral of e−inθ over any circle centered at the origin vanishes. Therefore
2π 1 [u(reiθ ) − Crs ]e−inθ dθ when n > 0, an = n πr 0 hence 1 |an | ≤ n πr




2π

[Crs − u(reiθ )] dθ ≤ 2Crs−n − 2Re(a0 )r−n .

0

Letting r tend to infinity along the sequence given in the hypothesis of the lemma proves that an = 0 for n > s. This completes the proof of the lemma and of Hadamard’s theorem.

6 Exercises 1. Give another proof of Jensen’s formula in the unit disc using the functions (called Blaschke factors) ψα (z) =

α−z . 1 − αz

[Hint: The function f /(ψz1 · · · ψzN ) is nowhere vanishing.] 2. Find the order of growth of the following entire functions: (a) p(z) where p is a polynomial. n

(b) ebz for b = 0. z

(c) ee .

3. Show that if τ is fixed with Im(τ ) > 0, then the Jacobi theta function ∞ 

Θ(z|τ ) =

2

eπin τ e2πinz

n=−∞

is of order 2 as a function of z. Further properties of Θ will be studied in Chapter 10. [Hint: −n2 t + 2n|z| ≤ −n2 t/2 when t > 0 and n ≥ 4|z|/t.] 4. Let t > 0 be given and fixed, and define F (z) by F (z) =

∞ 

(1 − e−2πnt e2πiz ).

n=1

Note that the product defines an entire function of z.

154

Chapter 5. ENTIRE FUNCTIONS 2

(a) Show that |F (z)| ≤ Aea|z| , hence F is of order 2. (b) F vanishes exactly when z = −int + m for n ≥ 1 and n, m integers. Thus, if zn is an enumeration of these zeros we have 

1 =∞ |zn |2



but

1 < ∞. |zn |2+

[Hint: To prove (a), write F (z) = F1 (z)F2 (z) where F1 (z) =

N 

(1 − e−2πnt e2πiz )

and

∞ 

F2 (z) =

n=1

(1 − e−2πnt e2πiz ).

n=N+1

Choose N ≈ c|z| with c appropriately large. Then, since 

∞ 

 e

−2πnt

e2π|z| ≤ 1 ,

N+1

one has |F2 (z)| ≤ A. However, |1 − e−2πnt e2πiz | ≤ 1 + e2π|z| ≤ 2e2π|z| . 

2

Thus |F1 (z)| ≤ 2N e2πN|z| ≤ ec |z| . Note that a simple variant of the function F arises as a factor in the triple product formula for the Jacobi theta function Θ, taken up in Chapter 10.] 5. Show that if α > 1, then
Fα (z) =

∞

α

e−|t| e2πizt dt

−∞

is an entire function of growth order α/(α − 1). [Hint: Show that −

|t|α + 2π|z||t| ≤ c|z|α/(α−1) 2

by considering the two cases |t|α−1 ≤ A|z| and |t|α−1 ≥ A|z|, for an appropriate constant A.] 6. Prove Wallis’s product formula π 2·2 4·4 2m · 2m = · ··· ··· . 2 1·3 3·5 (2m − 1) · (2m + 1) [Hint: Use the product formula for sin z at z = π/2.] 7. Establish the following properties of infinite products.

155

6. Exercises

  (a) Show that if |an |2 converges,  then the product (1 + an ) converges to a non-zero limit if and only if an converges.  an (b) Find an example  of a sequence of complex numbers {an } such that converges but (1 + an ) diverges.   (c) Also find an example such that (1 + an ) converges and an diverges.

8. Prove that for every z the product below converges, and cos(z/2) cos(z/4) cos(z/8) · · · =

∞ 

cos(z/2k ) =

k=1

sin z . z

[Hint: Use the fact that sin 2z = 2 sin z cos z.] 9. Prove that if |z| < 1, then (1 + z)(1 + z 2 )(1 + z 4 )(1 + z 8 ) · · · =

∞ 

k

(1 + z 2 ) =

k=0

1 . 1−z

10. Find the Hadamard products for: (a) ez − 1; (b) cos πz. [Hint: The answers are ez/2 z respectively.]

∞

n=1 (1

+ z 2 /4n2 π 2 ) and

∞

n=0 (1

− 4z 2 /(2n + 1)2 ),

11. Show that if f is an entire function of finite order that omits two values, then f is constant. This result remains true for any entire function and is known as Picard’s little theorem. [Hint: If f misses a, then f (z) − a is of the form ep(z) where p is a polynomial.] 12. Suppose f is entire and never vanishes, and that none of the higher derivatives of f ever vanish. Prove that if f is also of finite order, then f (z) = eaz+b for some constants a and b. 13. Show that the equation ez − z = 0 has infinitely many solutions in C. [Hint: Apply Hadamard’s theorem.] 14. Deduce from Hadamard’s theorem that if F is entire and of growth order ρ that is non-integral, then F has infinitely many zeros. 15. Prove that every meromorphic function in C is the quotient of two entire functions. Also, if {an } and {bn } are two disjoint sequences having no finite limit

156

Chapter 5. ENTIRE FUNCTIONS

points, then there exists a meromorphic function in the whole complex plane that vanishes exactly at {an } and has poles exactly at {bn }. 16. Suppose that Qn (z) =

Nn 

k cn k z

k=1

are given polynomials for n = 1, 2, . . .. Suppose also that we are given a sequence of complex numbers {an } without limit points. Prove that there exists a meromorphic function f (z) whose only poles are at {an }, and so that for each n, the difference   1 f (z) − Qn z − an is holomorphic near an . In other words, f has a prescribed poles and principal parts at each of these poles. This result is due to Mittag-Leffler. 17. Given two countably infinite sequences of complex numbers {ak }∞ k=0 and , with lim |a | = ∞, it is always possible to find an entire function F {bk }∞ k→∞ k k=0 that satisfies F (ak ) = bk for all k. (a) Given n distinct complex numbers a1 , . . . , an , and another n complex numbers b1 , . . . , bn , construct a polynomial P of degree ≤ n − 1 with P (ai ) = bi

for i = 1, . . . , n.

(b) Let {ak }∞ k=0 be a sequence of distinct complex numbers such that a0 = 0 and limk→∞ |ak | = ∞, and let E(z) denote a Weierstrass product associated with {ak }. Given complex numbers {bk }∞ k=0 , show that there exist integers mk ≥ 1 such that the series E(z) b0 E(z)  bk +  (a ) z − a E  (z) z E k k k=1 ∞

F (z) =



z ak

mk

defines an entire function that satisfies F (ak ) = bk

for all k ≥ 0.

This is known as the Pringsheim interpolation formula.

7 Problems 1. Prove that if f is holomorphic in the unit disc, bounded and not identically zero, and z1 , z2 , . . . , zn , . . . are its zeros (|zk | < 1), then  n

(1 − |zn |) < ∞.

157

7. Problems

[Hint: Use Jensen’s formula.] 2.∗ In this problem, we discuss Blaschke products, which are bounded analogues in the disc of the Weierstrass products for entire functions. (a) Show that for 0 < |α| < 1 and |z| ≤ r < 1 the inequality    α + |α|z  1+r    (1 − αz)α  ≤ 1 − r holds. (b) Let {αn } be a sequence in the unit disc such that αn = 0 for all n and ∞ 

(1 − |αn |) < ∞.

n=1

Note that this will be the case if {αn } are the zeros of a bounded holomorphic function on the unit disc (see Problem 1). Show that the product f (z) =

∞  αn − z |αn | 1 − αn z αn n=1

converges uniformly for |z| ≤ r < 1, and defines a holomorphic function on the unit disc having precisely the zeros αn and no other zeros. Show that |f (z)| ≤ 1.

3.∗ Show that

 zn is an entire function of order 1/α. (n!)α

 n 4.∗ Let F (z) = ∞ n=0 an z be an entire function of finite order. Then the growth order of F is intimately linked with the growth of the coefficients an as n → ∞. In fact: ρ

(a) Suppose |F (z)| ≤ Aea|z| . Then (8)

lim sup |an |1/n n1/ρ < ∞. n→∞

(b) Conversely, if (8) holds, then |F (z)| ≤ A ea
|z|

ρ+


, for every > 0.

[Hint: To prove (a), use Cauchy’s inequality |an | ≤ ρ

A arρ e , rn

and the fact that the function u−n eu , 0 < u < ρ, attains its minimum value en/ρ (ρ/n)n/ρ at u = n1/ρ /ρ1/ρ . Then, choose r in terms of n to achieve this minimum.

158

Chapter 5. ENTIRE FUNCTIONS

To establish (b), note that for |z| = r, |F (z)| ≤

 cn r n  cn r n ≤ n/ρ n (n!)1/ρ

for some constant c, since nn ≥ n!. This yields a reduction to Problem 3.]

6 The Gamma and Zeta Functions It is no exaggeration to say that the gamma and zeta functions are among the most important nonelementary functions in mathematics. The gamma function Γ is ubiquitous in nature. It arises in a host of calculations and is featured in a large number of identities that occur in analysis. Part of the explanation for this probably lies in the basic structural property of the gamma function, which essentially characterizes it: 1/Γ(s) is the (simplest) entire function1 which has zeros at exactly s = 0, −1, −2, . . .. The zeta function ζ (whose study, like that of the gamma function, was initiated by Euler) plays a fundamental role in the analytic theory of numbers. Its intimate connection with prime numbers comes about via the identity for ζ(s):

 p

 1 1 = , 1 − p−s ns ∞

n=1

where the product is over all primes. The behavior of ζ(s)  for real s > 1, with s tending to 1, was used by Euler to prove that p 1/p diverges, and a similar reasoning for L-functions is at the starting point of the proof of Dirichlet’s theorem on primes in arithmetic progression, as we saw in Book I. While there is no difficulty in seeing that ζ(s) is well-defined (and analytic) when Re(s) > 1, it was Riemann who realized that the further study of primes was bound up with the analytic (in fact, meromorphic) continuation of ζ into the rest of the complex plane. Beyond this, we also consider its remarkable functional equation, which reveals a symmetry about the line Re(s) = 1/2, and whose proof is based on a corresponding identity for the theta function. We also make a more detailed study of the growth of ζ(s) near the line Re(s) = 1, which will be required in the proof of the prime number theorem given in the next chapter.

1 In keeping with the standard notation of the subject, we denote by s (instead of z) the argument of the functions Γ and ζ.

160

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

1 The gamma function For s > 0, the gamma function is defined by
∞ (1) Γ(s) = e−t ts−1 dt. 0

The integral converges for each positive s because near t = 0 the function ts−1 is integrable, and for t large the convergence is guaranteed by the exponential decay of the integrand. These observations allow us to extend the domain of definition of Γ as follows. Proposition 1.1 The gamma function extends to an analytic function in the half-plane Re(s) > 0, and is still given there by the integral formula (1). Proof. It suffices to show that the integral defines a holomorphic function in every strip Sδ,M = {δ < Re(s) < M } , where 0 < δ < M < ∞. Note that if σ denotes the real part of s, then |e−t ts−1 | = e−t tσ−1 , so that the integral
∞ e−t ts−1 dt , Γ(s) = 0

which is defined by the limit lim →0 s ∈ Sδ,M . For  > 0, let




1/

F (s) =

 1/

e−t ts−1 dt, converges for each

e−t ts−1 dt.



By Theorem 5.4 in Chapter 2, the function F is holomorphic in the strip Sδ,M . By Theorem 5.2, also of Chapter 2, it suffices to show that F converges uniformly to Γ on the strip Sδ,M . To see this, we first observe that

∞ −t σ−1 |Γ(s) − F (s)| ≤ e t dt + e−t tσ−1 dt. 0

1/

The first integral converges uniformly to 0, as  tends to 0 since it can be easily estimated by δ /δ whenever 0 <  < 1. The second integral converges uniformly to 0 as well, since 
∞ 
∞
∞   −t σ−1 −t M −1   e t dt ≤ e t dt ≤ C e−t/2 dt → 0,  1/

1/

1/

161

1. The gamma function

and the proof is complete. 1.1 Analytic continuation Despite the fact that the integral defining Γ is not absolutely convergent for other values of s, we can go further and prove that there exists a meromorphic function defined on all of C that equals Γ in the half-plane Re(s) > 0. In the same sense as in Chapter 2, we say that this function is the analytic continuation2 of Γ, and we therefore continue to denote it by Γ. To prove the asserted analytic extension to a meromorphic function, we need a lemma, which incidentally exhibits an important property of Γ. Lemma 1.2 If Re(s) > 0, then (2)

Γ(s + 1) = sΓ(s).

As a consequence Γ(n + 1) = n! for n = 0, 1, 2, . . .. Proof. Integrating by parts in the finite integrals gives




1/

d −t s (e t ) dt = − dt




1/

−t s




e t dt + s

1/

e−t ts−1 dt,



and the desired formula (2) follows by letting  tend to 0, and noting that the left-hand side vanishes because e−t ts → 0 as t tends to 0 or ∞. Now it suffices to check that
∞  ∞ e−t dt = −e−t 0 = 1 , Γ(1) = 0

and to apply (2) successively to find that Γ(n + 1) = n!. Formula (2) in the lemma is all we need to give a proof of the following theorem. Theorem 1.3 The function Γ(s) initially defined for Re(s) > 0 has an analytic continuation to a meromorphic function on C whose only singularities are simple poles at the negative integers s = 0, −1, . . . . The residue of Γ at s = −n is (−1)n /n!. 2 Uniqueness of the analytic continuation is guaranteed since the complement of the poles of a meromorphic function forms a connected set.

162

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

Proof. It suffices to extend Γ to each half-plane Re(s) > −m, where m ≥ 1 is an integer. For Re(s) > −1, we define F1 (s) =

Γ(s + 1) . s

Since Γ(s + 1) is holomorphic in Re(s) > −1, we see that F1 is meromorphic in that half-plane, with the only possible singularity a simple pole at s = 0. The fact that Γ(1) = 1 shows that F1 does in fact have a simple pole at s = 0 with residue 1. Moreover, if Re(s) > 0, then F1 (s) =

Γ(s + 1) = Γ(s) s

by the previous lemma. So F1 extends Γ to a meromorphic function on the half-plane Re(s) > −1. We can now continue in this fashion by defining a meromorphic Fm for Re(s) > −m that agrees with Γ on Re(s) > 0. For Re(s) > −m, where m is an integer ≥ 1, define Fm (s) =

Γ(s + m) . (s + m − 1)(s + m − 2) · · · s

The function Fm is meromorphic in Re(s) > −m and has simple poles at s = 0, −1, −2, . . . , −m + 1 with residues Γ(−n + m) (m − 1 − n)!(−1)(−2) · · · (−n) (m − n − 1)! = (m − 1 − n)!(−1)(−2) · · · (−n) (−1)n . = n!

ress=−n Fm (s) =

Successive applications of the lemma show that Fm (s) = Γ(s) for Re(s) > 0. By uniqueness, this also means that Fm = Fk for 1 ≤ k ≤ m on the domain of definition of Fk . Therefore, we have obtained the desired continuation of Γ. Remark. We have already proved that Γ(s + 1) = sΓ(s) whenever Re(s) > 0. In fact, by analytic continuation, this formula remains true whenever s = 0, −1, −2, . . ., that is, whenever s is not a pole of Γ. This is because both sides of the formula are holomorphic in the complement of the poles of Γ and are equal when Re(s) > 0. Actually, one can go further, and note that if s is a negative integer s = −n with n ≥ 1, then both sides of the formula are infinite and moreover ress=−n Γ(s + 1) = −n ress=−n Γ(s).

163

1. The gamma function

Finally, note that when s = 0 we have Γ(1) = lims→0 sΓ(s). An alternate proof of Theorem 1.3, which is interesting in its own right and whose ideas recur later, is obtained by splitting the integral for Γ(s) defined on Re(s) > 0 as follows:




1

Γ(s) =

e−t ts−1 dt +




0

∞

e−t ts−1 dt.

1

The integral on the far right defines an entire function; also expanding e−t in a power series and integrating term by term gives




1

e−t ts−1 dt =

0

∞  (−1)n . n!(n + s) n=0

Therefore (3)


∞ ∞  (−1)n + e−t ts−1 dt Γ(s) = n!(n + s) 1 n=0

for Re(s) > 0.

Finally, the series defines a meromorphic function on C with poles at the negative integers and residue (−1)n /n! at s = −n. To prove this, we argue as follows. For a fixed R > 0 we may split the sum in two parts ∞ N ∞    (−1)n (−1)n (−1)n = + , n!(n + s) n!(n + s) n!(n + s) n=0

n=0

n=N +1

where N is an integer chosen so that N > 2R. The first sum, which is finite, defines a meromorphic function in the disc |s| < R with poles at the desired points and the correct residues. The second sum converges uniformly in that disc, hence defines a holomorphic function there, since n > N > 2R and |n + s| ≥ R imply    (−1)n  1    n!(n + s)  ≤ n!R . Since R was arbitrary, we conclude that the series in (3) has the desired properties. In particular, the relation (3) now holds on all of C. 1.2 Further properties of Γ The following identity reveals the symmetry of Γ about the line Re(s) = 1/2.

164

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

Theorem 1.4 For all s ∈ C, Γ(s)Γ(1 − s) =

(4)

π . sin πs

Observe that Γ(1 − s) has simple poles at the positive integers s = 1, 2, 3, . . ., so that Γ(s)Γ(1 − s) is a meromorphic function on C with simple poles at all the integers, a property also shared by π/ sin πs. To prove the identity, it suffices to do so for 0 < s < 1 since it then holds on all of C by analytic continuation.
∞ a−1 π v dv = . Lemma 1.5 For 0 < a < 1, 1 + v sin πa 0 Proof. We observe first that
∞ a−1
∞ v eax dv = dx , x 1+v 0 −∞ 1 + e which follows by making the change of variables v = ex . However, using contour integration, we saw in Example 2 of Section 2.1 in Chapter 3, that the second integral equals π/ sin πa, as desired. To establish the theorem, we first note that for 0 < s < 1 we may write
∞
∞ −u −s Γ(1 − s) = e u du = t e−vt (vt)−s dv , 0

0

where for t > 0 we made the change of variables vt = u. This trick then gives
∞ Γ(1 − s)Γ(s) = e−t ts−1 Γ(1 − s) dt 0 
∞ 
∞ −t s−1 −vt −s = e t t e (vt) dv dt 0 0
∞
∞ = e−t[1+v] v −s dvdt
0 ∞ 0 −s v dv = 1+v 0 π = sin π(1 − s) π , = sin πs and the theorem is proved.

165

1. The gamma function

In particular, by putting s = 1/2, and noting that Γ(s) > 0 whenever s > 0, we find that √ Γ(1/2) = π. We continue our study of the gamma function by considering its reciprocal, which turns out to be an entire function with remarkably simple properties. Theorem 1.6 The function Γ has the following properties: (i) 1/Γ(s) is an entire function of s with simple zeros at s = 0,−1,−2,. . . and it vanishes nowhere else. (ii) 1/Γ(s) has growth

   1  c2 |s| log |s|   .  Γ(s)  ≤ c1 e

Therefore, 1/Γ is of order 1 in the sense that for every  > 0, there exists a bound c() so that    1  c2 |s|1+
  .  Γ(s)  ≤ c()e Proof. By the theorem we may write 1 sin πs = Γ(1 − s) , Γ(s) π

(5)

so the simple poles of Γ(1 − s), which are at s = 1, 2, 3, . . . are cancelled by the simple zeros of sin πs, and therefore 1/Γ is entire with simple zeros at s = 0, −1, −2, −3, . . .. To prove the estimate, we begin by showing that
∞ e−t tσ dt ≤ e(σ+1) log(σ+1) 1

whenever σ = Re(s) is positive. Choose n so that σ ≤ n ≤ σ + 1. Then
∞
∞ −t σ e t dt ≤ e−t tn dt 1

0

= n! ≤ nn = en log n ≤ e(σ+1) log(σ+1) .

166

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

Since the relation (3) holds on all of C, we see from (5) that ∞  
∞   (−1)n sin πs sin πs 1 −t −s = + . e t dt Γ(s) n!(n + 1 − s) π π 1 n=0

However, from our previous observation, 
∞ 
∞   −t −s  e t dt ≤ e−t t|σ| dt ≤ e(|σ|+1) log(|σ|+1) ,  1

1

and because | sin πs| ≤ eπ|s| (by Euler’s formula for the sine function) we find that the second term in the formula for 1/Γ(s) is dominated by ce(|s|+1) log(|s|+1) eπ|s| , which is itself majorized by c1 ec2 |s| log |s| . Next, we consider the term ∞  n=0

sin πs (−1)n . n!(n + 1 − s) π

There are two cases: |Im(s)| > 1 and |Im(s)| ≤ 1. In the first case, this expression is dominated in absolute value by ceπ|s| . If |Im(s)| ≤ 1, we choose k to be the integer so that k − 1/2 ≤ Re(s) < k + 1/2. Then if k ≥ 1, ∞  n=0

sin πs (−1)n sin πs = (−1)k−1 + n!(n + 1 − s) π (k − 1)!(k − s)π



+

n=k−1

(−1)n

sin πs . n!(n + 1 − s)π

Both terms on the right are bounded; the first because sin  πs vanishes at s = k, and the second because the sum is majorized by c 1/n!. ∞ (−1)n When k ≤ 0, then Re(s) < 1/2 by our supposition, and n=0 n!(n+1−s)  is again bounded by c 1/n!. This concludes the proof of the theorem. The fact that 1/Γ satisfies the type of growth conditions discussed in Chapter 5 leads naturally to the product formula for the function 1/Γ, which we treat next. Theorem 1.7 For all s ∈ C, ∞

 1 s  −s/n γs =e s 1+ e . Γ(s) n n=1

167

1. The gamma function

The real number γ, which is known as Euler’s constant, is defined by N  1 − log N. N →∞ n n=1

γ = lim

The existence of the limit was already proved in Proposition 3.10, Chapter 8 of Book I, but we shall repeat the argument here for completeness. Observe that


N N N N −1
n+1    1 1 1 1 1 1 dx + , − log N = − dx = − n n x n x N 1 n n=1

n=1

n=1

and by the mean value theorem applied to f (x) = 1/x we have

  1 1  − ≤ 1  n x  n2

for all n ≤ x ≤ n + 1.

Hence ∞ N −1   1 1 − log N = an + n N n=1

n=1

 where |an | ≤ 1/n2 . Therefore an converges, which proves that the limit defining γ exists. We may now proceed with the proof of the factorization of 1/Γ. Proof. By the Hadamard factorization theorem and the fact that 1/Γ is entire, of growth order 1, and has simple zeros at s = 0, −1, −2, . . ., we can expand 1/Γ in a Weierstrass product of the form ∞

 1 s  −s/n As+B =e s 1+ e . Γ(s) n n=1

Here A and B are two constants that are to be determined. Remembering that sΓ(s) → 1 as s → 0, we find that B = 0 (or some integer multiple of 2πi, which of course gives the same result). Putting s = 1, and using

168

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

the fact that Γ(1) = 1 yields e

−A

=

∞   n=1

= lim

N →∞

1 1+ n N   n=1 N

= lim e N →∞

 e−1/n 1 1+ n

 e−1/n

n=1 [log(1+1/n)−1/n]

= lim e−(

N n=1

1/n)+log N +log(1+1/N )

N →∞ −γ

=e

.

Therefore A = γ + 2πik for some integer k. Since Γ(s) is real whenever s is real, we must have k = 0, and the argument is complete. Note that the proof shows that the function 1/Γ is essentially characterized (up to two normalizing constants) as the entire function that has: (i) simple zeros at s = 0, −1, −2, . . . and vanishes nowhere else, and (ii) order of growth ≤ 1. Observe that sin πs has a similar characterization (except the zeros are now at all the integers). However,   while sin πs has a stricter growth estimate of the form sin πs = O ec|s| , this estimate (without the logarithm in the exponent) does not hold for 1/Γ(s) as Exercise 12 demonstrates.

2 The zeta function The Riemann zeta function is initially defined for real s > 1 by the convergent series ∞  1 . ζ(s) = ns n=1

As in the case of the gamma function, ζ can be continued into the complex plane. There are several proofs of this fact, and we present in the next section the one that relies on the functional equation of ζ. 2.1 Functional equation and analytic continuation In parallel to the gamma function, we first provide a simple extension of ζ to a half-plane in C.

169

2. The zeta function

Proposition 2.1 The series defining ζ(s) converges for Re(s) > 1, and the function ζ is holomorphic in this half-plane. Proof. If s = σ + it where σ and t are real, then |n−s | = |e−s log n | = e−σ log n = n−σ . As a consequence, if σ > 1 + δ > 1 the series defining ζ is uniformly ∞  bounded by n=1 1/n1+δ , which converges. Therefore, the series 1/ns converges uniformly on every half-plane Re(s) > 1 + δ > 1, and therefore defines a holomorphic function in Re(s) > 1. The analytic continuation of ζ to a meromorphic function in C is more subtle than in the case of the gamma function. The proof we present here relates ζ to Γ and another important function. Consider the theta function, already introduced in Chapter 4, which is defined for real t > 0 by ϑ(t) =

∞ 

2

e−πn t .

n=−∞

An application of the Poisson summation formula (Theorem 2.4 in Chapter 4) gave the functional equation satisfied by ϑ, namely ϑ(t) = t−1/2 ϑ(1/t). The growth and decay of ϑ we shall need are ϑ(t) ≤ Ct−1/2

as t → 0,

and |ϑ(t) − 1| ≤ Ce−πt

for some C > 0, and all t ≥ 1.

The inequality for t tending to zero follows from the functional equation, while the behavior as t tends to infinity follows from the fact that

 n≥1

2

e−πn

t

≤



e−πnt ≤ Ce−πt

n≥1

for t ≥ 1. We are now in a position to prove an important relation among ζ, Γ and ϑ.

170

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

Theorem 2.2 If Re(s) > 1, then π

−s/2

1 Γ(s/2)ζ(s) = 2




∞

u(s/2)−1 [ϑ(u) − 1] du.

0

Proof. This and further arguments are based on the observation that
∞ 2 e−πn u u(s/2)−1 du = π −s/2 Γ(s/2)n−s , if n ≥ 1. (6) 0

Indeed, if we make the change of variables u = t/πn2 in the integral, the left-hand side becomes 
∞  −t (s/2)−1 e t dt (πn2 )−s/2 , 0

which is precisely π −s/2 Γ(s/2)n−s . Next, note that ϑ(u) − 1  −πn2 u = e . 2 ∞

n=1

The estimates for ϑ given before the statement of the theorem justify an interchange of the infinite sum with the integral, and thus 1 2




∞

u(s/2)−1 [ϑ(u) − 1] du =

0

∞
 n=1

∞

2

u(s/2)−1 e−πn

u

du

0

= π −s/2 Γ(s/2)

∞ 

n−s

n=1

=π

−s/2

Γ(s/2)ζ(s),

as was to be shown. In view of this, we now consider the modification of the ζ function called the xi function, which makes the former appear more symmetric. It is defined for Re(s) > 1 by (7)

ξ(s) = π −s/2 Γ(s/2)ζ(s).

Theorem 2.3 The function ξ is holomorphic for Re(s) > 1 and has an analytic continuation to all of C as a meromorphic function with simple poles at s = 0 and s = 1. Moreover, ξ(s) = ξ(1 − s)

for all s ∈ C.

171

2. The zeta function

Proof. namely

The idea of the proof is to use the functional equation for ϑ, ∞ 

2

e−πn

u

∞ 

= u−1/2

n=−∞

2

e−πn

/u

,

u > 0.

n=−∞

We then could multiply both sides by u(s/2)−1 and try to integrate in u. Disregarding the terms corresponding to n = 0 (which produce infinities in both sums), we would get the desired equality once we invoked formula (6), and the parallel formula obtained by making the change of variables u → 1/u. The actual proof requires a little more work and goes as follows. Let ψ(u) = [ϑ(u) − 1]/2. The functional equation for the theta function, namely ϑ(u) = u−1/2 ϑ(1/u), implies ψ(u) = u−1/2 ψ(1/u) +

1 1 − . 1/2 2 2u

Now, by Theorem 2.2 for Re(s) > 1, we have


π

−s/2

∞

u(s/2)−1 ψ(u) du

Γ(s/2)ζ(s) = 0







1

u

=

(s/2)−1

0




∞

u(s/2)−1 ψ(u) du

ψ(u) du + 1

1

=

u

(s/2)−1

u−1/2 ψ(1/u) +

0




1 1 − 1/2 2 2u

 du +

∞

u(s/2)−1 ψ(u) du

+ 1

=

1 1 − + s−1 s




∞

 (−s/2)−1/2  u + u(s/2)−1 ψ(u) du

1

whenever Re(s) > 1. Therefore ξ(s) =

1 1 − + s−1 s




∞



 u(−s/2)−1/2 + u(s/2)−1 ψ(u) du.

1

Since the function ψ has exponential decay at infinity, the integral above defines an entire function, and we conclude that ξ has an analytic continuation to all of C with simple poles at s = 0 and s = 1. Moreover, it is

172

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

immediate that the integral remains unchanged if we replace s by 1 − s, and the same is true for the sum of the two terms 1/(s − 1) − 1/s. We conclude that ξ(s) = ξ(1 − s) as was to be shown. From the identity we have proved for ξ we obtain the desired result for the zeta function: its analytic continuation and its functional equation. Theorem 2.4 The zeta function has a meromorphic continuation into the entire complex plane, whose only singularity is a simple pole at s = 1. Proof. namely

A look at (7) provides the meromorphic continuation of ζ,

ζ(s) = π s/2

ξ(s) . Γ(s/2)

Recall that 1/Γ(s/2) is entire with simple zeros at 0, −2, −4, . . ., so the simple pole of ξ(s) at the origin is cancelled by the corresponding zero of 1/Γ(s/2). As a consequence, the only singularity of ζ is a simple pole at s = 1. We shall now present a more elementary approach to the analytic continuation of the zeta function, which easily leads to its extension in the half-plane Re(s) > 0. This method will be useful in studying the growth properties of ζ near the line Re(s) = 1 (which will be  needed in ∞ the next chapter). The idea behind it is to compare the sum n=1 n−s ∞ −s with the integral 1 x dx. Proposition 2.5 There is a sequence of entire functions {δn (s)}∞ n=1 that satisfy the estimate |δn (s)| ≤ |s|/nσ+1 , where s = σ + it, and such that
N  1  dx − = δn (s), (8) s s n x 1 1≤n 0 we have ζ(s) − where H(s) =

∞

n=1 δn (s)

1 = H(s), s−1

is holomorphic in the half-plane Re(s) > 0.

173

2. The zeta function

To 

prove the proposition  n+1 −s x dx, and set 1≤n 1,  −s the series n converges to ζ(s), and this proves theassertion when Re(s) > 1. The uniform convergence also shows that δn (s) is holomorphic when Re(s) > 0, and thus shows that ζ(s) is extendable to that half-plane, and that the identity continues to hold there. Remark. The idea described above can be developed step by step to yield the continuation of ζ into the entire complex plane, as shown in Problems 2 and 3. Another argument giving the full analytic continuation of ζ is outlined in Exercises 15 and 16. As an application of the proposition we can show that the growth of ζ(s) near the line = 1 is “mild.” Recall that when Re(s) > 1, ∞Re(s) −σ we have |ζ(s)| ≤ n=1 n , and so ζ(s) is bounded in any half-plane Re(s) ≥ 1 + δ, with δ > 0. We shall see that on the line Re(s) = 1, |ζ(s)| is majorized by |t| , for every  > 0, and that the growth near the line is not much worse. The estimates below are not optimal. In fact, they are rather crude but suffice for what is needed later on. Proposition 2.7 Suppose s = σ + it with σ, t ∈ R. Then for each σ0 , 0 ≤ σ0 ≤ 1, and every  > 0, there exists a constant c so that (i) |ζ(s)| ≤ c |t|1−σ0 + , if σ0 ≤ σ and |t| ≥ 1.

174

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

(ii) |ζ  (s)| ≤ c |t| , if 1 ≤ σ and |t| ≥ 1. In particular, the proposition implies that ζ(1 + it) = O(|t| ) as |t| tends to infinity,3 and the same estimate also holds for ζ  . For the proof, we use Corollary 2.6. Recall the estimate |δn (s)| ≤ |s|/nσ+1 . We also have the estimate |δn (s)| ≤ 2/nσ , which follows from the expression for δn (s) given by (9) and the fact that |n−s | = n−σ and |x−s | ≤ n−σ if x ≥ n. We then combine these two estimates for |δn (s)| via the observation that A = Aδ A1−δ , to obtain the bound  δ  1−δ |s| 2 2|s|δ ≤ , |δn (s)| ≤ nσ0 +1 nσ0 nσ0 +δ as long as δ ≥ 0. Now choose δ = 1 − σ0 +  and apply the identity in Corollary 2.6. Then, with σ = Re(s) ≥ σ0 , we find   ∞   1  1 1−σ0 +   + 2|s| , |ζ(s)| ≤  s − 1 n1+ n=1

and conclusion (i) is proved. The second conclusion is actually a consequence of the first by a slight modification of Exercise 8 in Chapter 2. For completeness we sketch the argument. By the Cauchy integral formula,
2π 1 ζ  (s) = ζ(s + reiθ )eiθ dθ, 2πr 0 where the integration is taken over a circle of radius r centered at the point s. Now choose r =  and observe that this circle lies in the halfplane Re(s) ≥ 1 − , and so (ii) follows as a consequence of (i) on replacing 2 by .

3 Exercises 1. Prove that Γ(s) = lim

n→∞

ns n! s(s + 1) · · · (s + n)

whenever s = 0, −1, −2, . . .. [Hint: Use the product formula for 1/Γ, and the definition of the Euler constant γ.] 2. Prove that ∞  Γ(a + 1)Γ(b + 1) n(n + a + b) = (n + a)(n + b) Γ(a + b + 1) n=1

3 The

reader should recall the O notation which was introduced at the end of Chapter 1.

175

3. Exercises

whenever a and b are positive. Using the product formula for sin πs, give another proof that Γ(s)Γ(1 − s) = π/ sin πs. 3. Show that Wallis’s product formula can be written as !

π 22n (n!)2 = lim (2n + 1)1/2 . n→∞ (2n + 1)! 2

As a result, prove the following identity: Γ(s)Γ(s + 1/2) =

√

π21−2s Γ(2s).

4. Prove that if we take f (z) =

1 , (1 − z)α

for |z| < 1

(defined in terms of the principal branch of the logarithm), where α is a fixed complex number, then f (z) =

∞ 

an (α)z n

n=0

with an (α) ∼

1 nα−1 Γ(α)

as n → ∞.

5. Use the fact that Γ(s)Γ(1 − s) = π/ sin πs to prove that ! |Γ(1/2 + it)| =

2π , eπt + e−πt

whenever t ∈ R.

6. Show that 1+

1 1 1 1 γ + + ··· + − log n → + log 2, 3 5 2n − 1 2 2

where γ is Euler’s constant. 7. The Beta function is defined for Re(α) > 0 and Re(β) > 0 by


1

B(α, β) =

(1 − t)α−1 tβ−1 dt.

0

(a) Prove that B(α, β) =

Γ(α)Γ(β) . Γ(α + β)

176

Chapter 6. THE GAMMA AND ZETA FUNCTIONS




∞

(b) Show that B(α, β) = 0

uα−1 du. (1 + u)α+β

[Hint: For part (a), note that ∞
∞


Γ(α)Γ(β) = 0

tα−1 sβ−1 e−t−s dtds,

0

and make the change of variables s = ur, t = u(1 − r).] 8. The Bessel functions arise in the study of spherical symmetries and the Fourier transform. See Chapter 6 in Book I. Prove that the following power series identity holds for Bessel functions of real order ν > −1/2: ν

Jν (x) =

(x/2) √ Γ(ν + 1/2) π




1 −1

eixt (1 − t2 )ν−(1/2) dt =

∞

x ν  (−1)m

2

m=0



x2 4

m

m!Γ(ν + m + 1)

whenever x > 0. In particular, the Bessel function Jν satisfies the ordinary differential equation   ν2 1 dJν d2 Jν + 1 − Jν = 0. + dx2 x dx x2 [Hint: Expand the exponential eixt in a power series, and express the remaining integrals in terms of the gamma function, using Exercise 7.] 9. The hypergeometric series F (α, β, γ; z) was defined in Exercise 16 of Chapter 1. Show that
1 Γ(γ) tβ−1 (1 − t)γ−β−1 (1 − zt)−α dt. F (α, β, γ; z) = Γ(β)Γ(γ − β) 0 Here α > 0, β > 0, γ > β, and |z| < 1. Show as a result that the hypergeometric function, initially defined by a power series convergent in the unit disc, can be continued analytically to the complex plane slit along the half-line [1, ∞). Note that log(1 − z)

=

−zF (1, 1, 2; z),

ez

=

limβ→∞ F (1, β, 1; z/β),

(1 − z)−α

=

F (α, 1, 1; z).

[Hint: To prove the integral identity, expand (1 − zt)−α as a power series.] 10. An integral of the form


∞

F (z) = 0

f (t)tz−1 dt

177

3. Exercises

is called a Mellin transform, and we shall write M(f )(z) = F (z). For example, the gamma function is the Mellin transform of the function e−t . (a) Prove that


∞

M(cos)(z) = 0

z cos(t)tz−1 dt = Γ(z) cos π 2

for 0 < Re(z) < 1,

z sin(t)tz−1 dt = Γ(z) sin π 2

for 0 < Re(z) < 1.

and


∞

M(sin)(z) = 0

(b) Show that the second of the above identities is valid in the larger strip −1 < Re(z) < 1, and that as a consequence, one has


∞ 0

sin x π dx = x 2




∞

and 0

√ sin x dx = 2π. 3/2 x

This generalizes the calculation in Exercise 2 of Chapter 2. [Hint: For the first part, consider the integral of the function f (w) = e−w wz−1 around the contour illustrated in Figure 1. Use analytic continuation to prove the second part.]

0



R

Figure 1. The contour in Exercise 10

z

11. Let f (z) = eaz e−e where a > 0. Observe that in the strip {x + iy : |y| < π} the function f (x + iy) is exponentially decreasing as |x| tends to infinity. Prove that fˆ(ξ) = Γ(a + iξ),

for all ξ ∈ R.

12. This exercise gives two simple observations about 1/Γ.

178

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

(a) Show that 1/|Γ(s)| is not O(ec|s| ) for any c > 0. [Hint: If s = −k − 1/2, where k is a positive integer, then |1/Γ(s)| ≥ k!/π.] (b) Show that there is no entire function F (s) with F (s) = O(ec|s| ) that has simple zeros at s = 0, −1, −2, . . . , −n, . . ., and that vanishes nowhere else.

13. Prove that ∞  d2 log Γ(s) 1 = 2 ds (s + n)2 n=0

whenever s is a positive number. Show that if the left-hand side is interpreted as (Γ /Γ) , then the above formula also holds for all complex numbers s with s = 0, −1, −2, . . .. 14. This exercise gives an asymptotic formula for log n!. A more refined asymptotic formula for Γ(s) as s → ∞ (Stirling’s formula) is given in Appendix A. (a) Show that d dx




x+1

log Γ(t) dt = log x,

for x > 0,

x

and as a result


x+1

log Γ(t) dt = x log x − x + c.

x

(b) Show as a consequence that log Γ(n) ∼ n log n as n → ∞. In fact, prove that log Γ(n) ∼ n log n + O(n) as n → ∞. [Hint: Use the fact that Γ(x) is monotonically increasing for all large x.]

15. Prove that for Re(s) > 1, ζ(s) =

[Hint: Write 1/(ex − 1) =

∞ n=1

1 Γ(s)




∞ 0

xs−1 dx. ex − 1

e−nx .]

16. Use the previous exercise to give another proof that ζ(s) is continuable in the complex plane with only singularity a simple pole at s = 1. [Hint: Write ζ(s) =

1 Γ(s)




1 0

1 xs−1 dx + ex − 1 Γ(s)




∞ 1

xs−1 dx. ex − 1

179

4. Problems

The second integral defines an entire function, while


1 0

∞  xs−1 Bm dx = , x e −1 m!(s + m − 1) m=0

where Bm denotes the mth Bernoulli number defined by ∞  x Bm m = x . ex − 1 m! m=0

Then B0 = 1, and since z/(ez − 1) is holomorphic for |z| < 2π, we must have lim supm→∞ |Bm /m!|1/m = 1/2π.] 17. Let f be an indefinitely differentiable function on R that has compact support, or more generally, let f belong to the Schwartz space.4 Consider
∞ 1 I(s) = f (x)x−1+s dx. Γ(s) 0 (a) Observe that I(s) is holomorphic for Re(s) > 0. Prove that I has an analytic continuation as an entire function in the complex plane. (b) Prove that I(0) = 0, and more generally I(−n) = (−1)n f (n+1) (0)

for all n ≥ 0.

[Hint: To prove the analytic continuation, as well as the formulas in the second (−1)k  ∞ (k) part, integrate by parts to show that I(s) = Γ(s+k) f (x)xs+k−1 dx.] 0

4 Problems 1. This problem provides further estimates for ζ and ζ  near Re(s) = 1. (a) Use Proposition 2.5 and its corollary to prove ζ(s) =

 1≤n 0. (b) Show that |ζ(1 + it)| = O(log |t|), as |t| → ∞ by using the previous result with N =greatest integer in |t|. Schwartz space on R is denoted by S and consists of all indefinitely differentiable functions f , so that f and all its derivatives decay faster than any polynomials. In other words, supx∈R |x|m |f () (x)| < ∞ for all integers m,  ≥ 0. This space appeared in the study of the Fourier transform in Book I. 4 The

180

Chapter 6. THE GAMMA AND ZETA FUNCTIONS

(c) The second conclusion of Proposition 2.7 can be similarly refined. (d)  Show that if t = 0 and t is fixed, then the partial sums of the series ∞ 1+it are bounded, but the series does not converge. n=1 1/n 2.∗ Prove that for Re(s) > 0 ζ(s) =

s −s s−1




∞ 1

{x} dx xs+1

where {x} is the fractional part of x. 3.∗ If Q(x) = {x} − 1/2, then we can write the expression in the previous problem as
∞ s 1 Q(x) dx. ζ(s) = − −s s−1 2 xs+1 1 Let us construct Qk (x) recursively so that


1

Qk (x) dx = 0, 0

dQk+1 = Qk (x), dx

Q0 (x) = Q(x)

and

Qk (x + 1) = Qk (x).

Then we can write 1 s − −s ζ(s) = s−1 2




∞ 1



 dk Qk (x) x−s−1 dx , dxk

and a k-fold integration by parts gives the analytic continuation for ζ(s) when Re(s) > −k. 4.∗ The functions Qk in the previous problem are related to the Bernoulli polynomials Bk (x) by the formula Qk (x) =

Bk+1 (x) (k + 1)!

for 0 ≤ x ≤ 1.

Also, if k is a positive integer, then 2ζ(2k) = (−1)k+1

(2π)2k B2k , (2k)!

where Bk = Bk (0) are the Bernoulli numbers. For the definition of Bk (x) and Bk see Chapter 3 in Book I.

7 The Zeta Function and Prime Number Theorem Bernhard Riemann, whose extraordinary intuitive powers we have already mentioned, has especially renovated our knowledge of the distribution of prime numbers, also one of the most mysterious questions in mathematics. He has taught us to deduce results in that line from considerations borrowed from the integral calculus: more precisely, from the study of a certain quantity, a function of a variable s which may assume not only real, but also imaginary values. He proved some important properties of that function, but enunciated two or three as important ones without giving the proof. At the death of Riemann, a note was found among his papers, saying “These properties of ζ(s) (the function in question) are deduced from an expression of it which, however, I did not succeed in simplifying enough to publish it.” We still have not the slightest idea of what the expression could be. As to the properties he simply enunciated, some thirty years elapsed before I was able to prove all of them but one. The question concerning that last one remains unsolved as yet, though, by an immense labor pursued throughout this last half century, some highly interesting discoveries in that direction have been achieved. It seems more and more probable, but still not at all certain, that the “Riemann hypothesis” is true. J. Hadamard, 1945

Euler found, through his product formula for the zeta function, a deep connection between analytical methods and arithmetic properties of numbers, in particular primes. An easy consequence  of Euler’s formula is that the sum of the reciprocals of all primes, p 1/p, diverges, a result that quantifies the fact that there are infinitely many prime numbers. The natural problem then becomes that of understanding how these primes are distributed. With this in mind, we consider the

182

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

following function: π(x) = number of primes less than or equal to x. The erratic growth of the function π(x) gives little hope of finding a simple formula for it. Instead, one is led to study the asymptotic behavior of π(x) as x becomes large. About 60 years after Euler’s discovery, Legendre and Gauss observed after numerous calculations that it was likely that (1)

π(x) ∼

x log x

as x → ∞.

(The asymptotic relation f (x) ∼ g(x) as x → ∞ means that f (x)/g(x) → 1 as x → ∞.) Another 60 years later, shortly before Riemann’s work, Tchebychev proved by elementary methods (and in particular, without the zeta function) the weaker result that (2)

π(x) ≈

x log x

as x → ∞.

Here, by definition, the symbol ≈ means that there are positive constants A < B such that x x A ≤ π(x) ≤ B log x log x for all sufficiently large x. In 1896, about 40 years after Tchebychev’s result, Hadamard and de la Vall´ee Poussin gave a proof of the validity of the relation (1). Their result is known as the prime number theorem. The original proofs of this theorem, as well as the one we give below, use complex analysis. We should remark that since then other proofs have been found, some depending on complex analysis, and others more elementary in nature. At the heart of the proof of the prime number theorem that we give below lies the fact that ζ(s) does not vanish on the line Re(s) = 1. In fact, it can be shown that these two propositions are equivalent.

1 Zeros of the zeta function We have seen in Theorem 1.10, Chapter 8 in Book I, Euler’s identity, which states that for Re(s) > 1 the zeta function can be expressed as an infinite product  1 ζ(s) = . 1 − p−s p

183

1. Zeros of the zeta function

For the sake of completeness we provide a proof of the above identity. The key observation is that 1/(1 − p−s ) can be written as a convergent (geometric) power series 1+

1 1 1 + 2s + · · · + M s + · · · , s p p p

and taking formally the product of these series over all primes p, yields the desired result. A precise argument goes as follows. Suppose M and N are positive integers with M > N . Observe now that, by the fundamental theorem of arithmetic,1 any positive integer n ≤ N can be written uniquely as a product of primes, and that each prime that occurs in the product must be less than or equal to N and repeated less than M times. Therefore

 N   1 1 1 1 ≤ 1 + s + 2s + · · · + M s ns p p p n=1 p≤N    1 ≤ 1 − p−s p≤N  1  ≤ . 1 − p−s p Letting N tend to infinity in the series now yields ∞   1  1 ≤ . ns 1 − p−s n=1

p

For the reverse inequality, we argue as follows. Again, by the fundamental theorem of arithmetic, we find that

 p≤N

1 1 1 1 + s + 2s + · · · + M s p p p

 ≤

∞  1 . ns n=1

Letting M tend to infinity gives

 p≤N

1 1 − p−s

 ≤

∞  1 . ns n=1

1 A proof of this elementary (but essential) fact is given in the first section of Chapter 8 in Book I.

184

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

Hence

 p

1 1 − p−s

 ≤

∞  1 , ns n=1

and the proof of the product formula for ζ is complete. From the product formula we see, by Proposition 3.1 in Chapter 5, that ζ(s) does not vanish when Re(s) > 1. To obtain further information about the location of the zeros of ζ, we use the functional equation that provided the analytic continuation of ζ. We may write the fundamental relation ξ(s) = ξ(1 − s) in the form π −s/2 Γ(s/2)ζ(s) = π −(1−s)/2 Γ((1 − s)/2)ζ(1 − s), and therefore ζ(s) = π s−1/2

Γ((1 − s)/2) ζ(1 − s). Γ(s/2)

Now observe that for Re(s) < 0 the following are true: (i) ζ(1 − s) has no zeros because Re(1 − s) > 1. (ii) Γ((1 − s)/2) is zero free. (iii) 1/Γ(s/2) has zeros at s = −2, −4, −6, . . . . Therefore, the only zeros of ζ in Re(s) < 0 are located at the negative even integers −2, −4, −6, . . .. This proves the following theorem. Theorem 1.1 The only zeros of ζ outside the strip 0 ≤ Re(s) ≤ 1 are at the negative even integers, −2, −4, −6, . . .. The region that remains to be studied is called the critical strip, 0 ≤ Re(s) ≤ 1. A key fact in the proof of the prime number theorem is that ζ has no zeros on the line Re(s) = 1. As a simple consequence of this fact and the functional equation, it follows that ζ has no zeros on the line Re(s) = 0. In the seminal paper where Riemann introduced the analytic continuation of the ζ function and proved its functional equation, he applied these insights to the theory of prime numbers, and wrote down “explicit” formulas for determining the distribution of primes. While he did not succeed in fully proving and exploiting his assertions, he did initiate many important new ideas. His analysis led him to believe the truth of what has since been called the Riemann hypothesis:

185

1. Zeros of the zeta function

The zeros of ζ(s) in the critical strip lie on the line Re(s) = 1/2. He said about this: “It would certainly be desirable to have a rigorous demonstration of this proposition; nevertheless I have for the moment set this aside, after several quick but unsuccessful attempts, because it seemed unneeded for the immediate goal of my study.” Although much of the theory and numerical results point to the validity of this hypothesis, a proof or a counter-example remains to be discovered. The Riemann hypothesis is today one of mathematics’ most famous unresolved problems. In particular, it is for this reason that the zeros of ζ located outside the critical strip are sometimes called the trivial zeros of the zeta function. See also Exercise 5 for an argument proving that ζ has no zeros on the real segment, 0 ≤ σ ≤ 1, where s = σ + it. In the rest of this section we shall restrict ourselves to proving the following theorem, together with related estimates on ζ, which we shall use in the proof of the prime number theorem. Theorem 1.2 The zeta function has no zeros on the line Re(s) = 1. Of course, since we know that ζ has a pole at s = 1, there are no zeros in a neighborhood of this point, but what we need is the deeper property that ζ(1 + it) = 0

for all t ∈ R.

The next sequence of lemmas gathers the necessary ingredients for the proof of Theorem 1.2. Lemma 1.3 If Re(s) > 1, then log ζ(s) =

 p−ms p,m

m

=

∞ 

cn n−s

n=1

for some cn ≥ 0. Proof. Suppose first that s > 1. Taking the logarithm of the Euler product formula, and using the power series expansion for the logarithm

 log

1 1−x

 =

∞  xm , m

m=1

186

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

which holds for 0 ≤ x < 1, we find that log ζ(s) = log

 p

 1 = log −s 1−p



p

1 1 − p−s

 =

 p−ms p,m

m

.

Since the double sum converges absolutely, we need not specify the order of summation. See the Note at the end of this chapter. The formula then holds for all Re(s) > 1 by analytic continuation. Note that, by Theorem 6.2 in Chapter 3, log ζ(s) is well defined in the simply connected half-plane Re(s) > 1, since ζ has no zeros there. Finally, it is clear that we have

 p−ms p,m

m

=

∞ 

cn n−s ,

n=1

where cn = 1/m if n = pm and cn = 0 otherwise. The proof of the theorem we shall give depends on a simple trick that is based on the following inequality. Lemma 1.4 If θ ∈ R, then 3 + 4 cos θ + cos 2θ ≥ 0. This follows at once from the simple observation 3 + 4 cos θ + cos 2θ = 2(1 + cos θ)2 . Corollary 1.5 If σ > 1 and t is real, then log |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| ≥ 0. Proof. Let s = σ + it and note that Re(n−s ) = Re(e−(σ+it) log n ) = e−σ log n cos(t log n) = n−σ cos(t log n). Therefore, log |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| = 3 log |ζ(σ)| + 4 log |ζ(σ + it)| + log |ζ(σ + 2it)| = 3Re[log ζ(σ)] + 4Re[log ζ(σ + it)] + Re[log ζ(σ + 2it)]  = cn n−σ (3 + 4 cos θn + cos 2θn ) , where θn = t log n. The positivity now follows from Lemma 1.4, and the fact that cn ≥ 0.

187

1. Zeros of the zeta function

We can now finish the proof of our theorem. Proof of Theorem 1.2. Suppose on the contrary that ζ(1 + it0 ) = 0 for some t0 = 0. Since ζ is holomorphic at 1 + it0 , it must vanish at least to order 1 at this point, hence |ζ(σ + it0 )|4 ≤ C(σ − 1)4

as σ → 1,

for some constant C > 0. Also, we know that s = 1 is a simple pole for ζ(s), so that |ζ(σ)|3 ≤ C  (σ − 1)−3

as σ → 1,

for some constant C  > 0. Finally, since ζ is holomorphic at the points σ + 2it0 , the quantity |ζ(σ + 2it0 )| remains bounded as σ → 1. Putting these facts together yields |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| → 0

as σ → 1 ,

which contradicts Corollary 1.5, since the logarithm of real numbers between 0 and 1 is negative. This concludes the proof that ζ is zero free on the real line Re(s) = 1. 1.1 Estimates for 1/ζ(s) The proof of the prime number theorem relies on detailed manipulations of the zeta function near the line Re(s) = 1; the basic object involved is the logarithmic derivative ζ  (s)/ζ(s). For this reason, besides the nonvanishing of ζ on the line, we need to know about the growth of ζ  and 1/ζ. The former was dealt with in Proposition 2.7 of Chapter 6; we now treat the latter. The proposition that follows is actually a quantitative version of Theorem 1.2. Proposition 1.6 For every  > 0, we have 1/|ζ(s)| ≤ c |t| when s = σ + it, σ ≥ 1, and |t| ≥ 1. Proof. From our previous observations, we clearly have that |ζ 3 (σ)ζ 4 (σ + it)ζ(σ + 2it)| ≥ 1,

whenever σ ≥ 1.

Using the estimate for ζ in Proposition 2.7 of Chapter 6, we find that |ζ 4 (σ + it)| ≥ c|ζ −3 (σ)| |t|− ≥ c (σ − 1)3 |t|− ,

188

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

for all σ ≥ 1 and |t| ≥ 1. Thus (3)

|ζ(σ + it)| ≥ c (σ − 1)3/4|t|− /4 ,

whenever σ ≥ 1 and |t| ≥ 1.

We now consider two separate cases, depending on whether the inequality σ − 1 ≥ A|t|−5 holds, for some appropriate constant A (whose value we choose later). If this inequality does hold, then (3) immediately provides |ζ(σ + it)| ≥ A |t|−4 , and it suffices to replace 4 by  to conclude the proof of the desired estimate, in this case. If, however, σ − 1 < A|t|−5 , then we first select σ  > σ with σ  − 1 = A|t|−5 . The triangle inequality then implies |ζ(σ + it)| ≥ |ζ(σ  + it)| − |ζ(σ  + it) − ζ(σ + it)|, and an application of the mean value theorem, together with the estimates for the derivative of ζ obtained in the previous chapter, give |ζ(σ  + it) − ζ(σ + it)| ≤ c |σ  − σ| |t| ≤ c |σ  − 1| |t| . These observations, together with an application of (3) where we set σ = σ  , show that |ζ(σ + it)| ≥ c (σ  − 1)3/4 |t|− /4 − c (σ  − 1)|t| . Now choose A = (c /(2c ))4 , and recall that σ  − 1 = A|t|−5 . This gives precisely c (σ  − 1)3/4 |t|− /4 = 2c (σ  − 1)|t| , and therefore |ζ(σ + it)| ≥ A |t|−4 . On replacing 4 by , the desired inequality is established, and the proof of the proposition is complete.

2 Reduction to the functions ψ and ψ1 In his study of primes, Tchebychev introduced an auxiliary function whose behavior is to a large extent equivalent to the asymptotic distribution of primes, but which is easier to manipulate than π(x). Tchebychev’s ψ-function is defined by  ψ(x) = log p. pm ≤x

189

2. Reduction to the functions ψ and ψ1

The sum is taken over those integers of the form pm that are less than or equal to x. Here p is a prime number and m is a positive integer. There are two other formulations of ψ that we shall need. First, if we define

Λ(n) =

log p if n = pm for some prime p and some m ≥ 1, 0 otherwise,

then it is clear that



ψ(x) =

Λ(n).

1≤n≤x

Also, it is immediate that ψ(x) =

 log x  log p

p≤x

log p

where [u] denotes the greatest integer ≤ u, and the sum is taken over the primes less than x. This formula follows from the fact that if pm ≤ x, then m ≤ log x/ log p. The fact that ψ(x) contains enough information about π(x) to prove our theorem is given a precise meaning in the statement of the next proposition. In particular, this reduces the prime number theorem to a corresponding asymptotic statement about ψ. Proposition 2.1 If ψ(x) ∼ x as x → ∞, then π(x) ∼ x/ log x as x → ∞. Proof. The argument here is elementary. By definition, it suffices to prove the following two inequalities: (4)

1 ≤ lim inf π(x) x→∞

log x x

and

lim sup π(x) x→∞

log x ≤ 1. x

To do so, first note that crude estimates give ψ(x) =

 log x  p≤x

log p

log p ≤

 log x p≤x

log p

log p = π(x) log x,

and dividing through by x yields π(x) log x ψ(x) ≤ . x x

190

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

The asymptotic condition ψ(x) ∼ x implies the first inequality in (4). The proof of the second inequality is a little trickier. Fix 0 < α < 1, and note that   ψ(x) ≥ log p ≥ log p ≥ (π(x) − π(xα )) log xα , xα 1, we have proved that lim supx→∞ ψ(x)/x ≤ 1. A similar argument with α < 1, then shows that lim inf x→∞ ψ(x)/x ≥ 1, and the proof of the proposition is complete. It is now time to relate ψ1 (and therefore also ψ) and ζ. We proved in Lemma 1.3 that for Re(s) > 1 log ζ(s) =

 p−ms m,p

m

.

Differentiating this expression gives

  Λ(n) ζ  (s) =− (log p)p−ms = − . ζ(s) ns ∞

m,p

n=1

We record this formula for Re(s) > 1 as ζ  (s)  Λ(n) = . ζ(s) ns ∞

(5)

−

n=1

The asymptotic behavior ψ1 (x) ∼ x2 /2 will be a consequence via (5) of the relationship between ψ1 and ζ, which is expressed by the following noteworthy integral formula. Proposition 2.3 For all c > 1 (6)

1 ψ1 (x) = 2πi




c+i∞

c−i∞

xs+1 s(s + 1)



ζ  (s) − ζ(s)

 ds.

To make the proof of this formula clear, we isolate the necessary contour integrals in a lemma.

192

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

Lemma 2.4 If c > 0, then
c+i∞ 1 as 0 if 0 < a ≤ 1, ds = 1 − 1/a if 1 ≤ a. 2πi c−i∞ s(s + 1) Here, the integral is over the vertical line Re(s) = c. Proof. First note that since |as | = ac , the integral converges. We suppose first that 1 ≤ a, and write a = eβ with β = log a ≥ 0. Let f (s) =

esβ as = . s(s + 1) s(s + 1)

Then ress=0 f = 1 and ress=−1 f = −1/a. For T > 0, consider the path Γ(T ) shown on Figure 1.

S(T )

C(T )

0

c

Γ(T )

Figure 1. The contour in the proof of Lemma 2.4 when a ≥ 1

The path Γ(T ) consists of the vertical segment S(T ) from c − iT to c + iT , and of the half-circle C(T ) centered at c of radius T , lying to the left of the vertical segment. We equip Γ(T ) with the positive (counterclockwise) orientation, and note that we are dealing with a toy contour. If we choose T so large that 0 and −1 are contained in the interior of Γ(T ), then by the residue formula
1 f (s) ds = 1 − 1/a. 2πi Γ(T )

193

2. Reduction to the functions ψ and ψ1

Since










f (s) ds = Γ(T )

f (s) ds + S(T )

f (s) ds, C(T )

it suffices to prove that the integral over the half-circle goes to 0 as T tends to infinity. Note that if s = σ + it ∈ C(T ), then for all large T we have |s(s + 1)| ≥ (1/2)T 2 , and since σ ≤ c we also have the estimate |eβs | ≤ eβc . Therefore


  

  C f (s) ds ≤ 2 2πT → 0 T C(T )

as T → ∞,

and the case when a ≥ 1 is proved. If 0 < a ≤ 1, consider an analogous contour but with the half-circle lying to the right of the line Re(s) = c. Noting that there are no poles in the interior of that contour, we can give an argument similar to the one given above to show that the integral over the half-circle also goes to 0 as T tends to infinity. We are now ready to prove Proposition 2.3. First, observe that ψ(u) =

∞ 

Λ(n)fn (u),

n=1

where fn (u) = 1 if n ≤ u and fn (u) = 0 otherwise. Therefore,




x

ψ1 (x) = =

ψ(u) du 0 ∞
x  n=1

=



Λ(n)fn (u) du

0




x

Λ(n)

n≤x

du, n

and hence ψ1 (x) =

 n≤x

Λ(n)(x − n).

194

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

This fact, together with equation (5) and an application of Lemma 2.4 (with a = x/n), gives 1 2πi




c+i∞

c−i∞

xs+1 s(s + 1)



ζ  (s) − ζ(s)




c+i∞ (x/n)s 1 ds ds = x Λ(n) 2πi c−i∞ s(s + 1) n=1

 n Λ(n) 1 − =x x ∞ 

n≤x

= ψ1 (x), as was to be shown. 2.1 Proof of the asymptotics for ψ1 In this section, we will show that ψ1 (x) ∼ x2 /2

as x → ∞,

and as a consequence, we will have proved the prime number theorem. The key ingredients in the argument are: • the formula in Proposition 2.3 connecting ψ1 to ζ, namely 1 ψ1 (x) = 2πi




c+i∞

c−i∞

xs+1 s(s + 1)



ζ  (s) − ζ(s)

 ds

for c > 1. • the non-vanishing of the zeta function on Re(s) = 1, ζ(1 + it) = 0

for all t ∈ R,

and the estimates for ζ near that line given in Proposition 2.7 of Chapter 6 together with Proposition 1.6 of this chapter. Let us now discuss our strategy in more detail. In the integral above for ψ1 (x) we want to change the line of integration Re(s) = c with c > 1, to Re(s) = 1. If we could achieve that, the size of the factor xs+1 in the integrand would then be of order x2 (which is close to what we want) instead of xc+1 , c > 1, which is much too large. However, there would still be two issues that must be dealt with. The first is the pole of ζ(s) at s = 1; it turns out that when it is taken into account, its contribution is exactly the main term x2 /2 of the asymptotic of ψ1 (x). Second, what remains must be shown to be essentially smaller than this term, and so

195

2. Reduction to the functions ψ and ψ1

we must further refine the crude estimate of order x2 when integrating on the line Re(s) = 1. We carry out our plan as follows. Fix c > 1, say c = 2, and assume x is also fixed for the moment with x ≥ 2. Let F (s) denote the integrand xs+1 F (s) = s(s + 1)



ζ  (s) − ζ(s)

 .

First we deform the vertical line from c − i∞ to c + i∞ to the path γ(T ) shown in Figure 2. (The segments of γ(T ) on the line Re(s) = 1 consist of T ≤ t < ∞, and −∞ < t ≤ −T .) Here T ≥ 3, and T will be chosen appropriately large later. c + i∞

1 + i∞

1 + i∞

γ5

γ4

s=1

s=1

γ3

s=1 γ2

γ1 c − i∞ Re(s) = c

1 − i∞ γ(T )

1 − i∞ γ(T, δ)

Figure 2. Three stages: the line Re(s) = c, the contours γ(T ) and γ(T, δ)

The usual and familiar arguments using Cauchy’s theorem allow us to see that
c+i∞
1 1 F (s) ds = F (s) ds. (7) 2πi c−i∞ 2πi γ(T )

196

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

Indeed, we know on the basis of Proposition 2.7 in Chapter 6 and Proposition 1.6 that |ζ  (s)/ζ(s)| ≤ A|t|η for any fixed η > 0, whenever s = σ + it, σ ≥ 1, and |t| ≥ 1. Thus |F (s)| ≤ A |t|−2+η in the two (infinite) rectangles bounded by the line (c − i∞, c + i∞) and γ(T ). Since F is regular in that region, and its decrease at infinity is rapid enough, the assertion (7) is established. Next, we pass from the contour γ(T ) to the contour γ(T, δ). (Again, see Figure 2.) For fixed T , we choose δ > 0 small enough so that ζ has no zeros in the box {s = σ + it, 1 − δ ≤ σ ≤ 1, |t| ≤ T }. Such a choice can be made since ζ does not vanish on the line σ = 1. Now F (s) has a simple pole at s = 1. In fact, by Corollary 2.6 in Chapter 6, we know that ζ(s) = 1/(s − 1) + H(s), where H(s) is regular near s = 1. Hence −ζ  (s)/ζ(s) = 1/(s − 1) + h(s), where h(s) is holomorphic near s = 1, and so the residue of F (s) at s = 1 equals x2 /2. As a result 1 2πi


F (s) ds = γ(T )

1 x2 + 2 2πi


γ(T,δ)

xs+1 F (s) ds. s(s + 1)

We now decompose the contour  γ(T, δ) as γ1 + γ2 + γ3 + γ4 + γ5 and estimate each of the integrals γj F (s) ds, j = 1, 2, 3, 4, 5, with the γj as in Figure 2. First we contend that there exists T so large that


     F (s) ds ≤ x2  2 γ1

and


     F (s) ds ≤ x2 .  2 γ5

To see this, we first note that for s ∈ γ1 one has |x1+s | = x1+σ = x2 . Then, by Proposition 1.6 we have, for example, that |ζ  (s)/ζ(s)| ≤ A|t|1/2 , so 

∞ 1/2   |t|   ≤ Cx2 F (s) ds dt.   t2 γ1 T Since the integral converges, we can make the right-hand side ≤ x2 /2 upon taking T sufficiently large. The argument for the integral over γ5 is the same.

197

2. Reduction to the functions ψ and ψ1

Having now fixed T , we choose δ appropriately small. On γ3 , note that |x1+s | = x1+1−δ = x2−δ , from which we conclude that there exists a constant CT (dependent on T ) such that 
    F (s) ds ≤ CT x2−δ .  γ3

Finally, on the small horizontal segment γ2 (and similarly on γ4 ), we can estimate the integral as follows: 

1 2    1+σ  x   . F (s) ds ≤ C x dσ ≤ C T T   log x γ2 1−δ We conclude that there exist constants CT and CT (possibly different from the ones above) such that   2 2  x ψ1 (x) −  ≤ x2 + CT x2−δ + CT x .   2 log x Dividing through by x2 /2, we see that    2ψ1 (x)  1 −δ     x2 − 1 ≤ 2 + 2CT x + 2CT log x , and therefore, for all large x we have    2ψ1 (x)    ≤ 4. − 1  x2  This concludes the proof that ψ1 (x) ∼ x2 /2

as x → ∞ ,

and thus, we have also completed the proof of the prime number theorem.

Note on interchanging double sums We prove the following facts about the interchange of infinite sums: if {ak }1≤k, 0, there isa positive integer N so that for all K, L > N we have K L   A − k=1 =1 ak  < . (iii) If m → (k(m), (m)) is a bijection from N to N × N, and if we write cm = ak(m)(m) , then A = ∞ k=1 ck . Statement (iii) says that any rearrangement of the sequence {ak } can be summed without changing the limit. This is analogous to the case of absolutely convergent series, which can be summed in any desired order.  The condition (8) says that each sum  ak converges absolutely, and moreover this convergence is “uniform” in k. An analogous situation arises for sequences of functions, where an important question is whether or not the interchange of limits ?

lim lim fn (x) = lim lim fn (x)

x→x0 n→∞

n→∞ x→x0

holds. It is a well-known fact that if the fn ’s are continuous, and their convergence is uniform, then the above identity is true since the limit function is itself continuous. To take advantage of this fact, define bk = ∞ =1 |ak | and let S = {x0 , x1 , . . .} be a countable set of points with limn→∞ xn = x0 . Also, define functions on S as follows: ∞ for k = 1, 2, . . . fk (x0 ) = =1 ak fk (xn ) = g(x) =

n =1

∞ k=1

ak

for k = 1, 2, . . . and n = 1, 2, . . .

fk (x)

for x ∈ S.

By  assumption (8), each fk is continuous at x0 . Moreover |fk (x)| ≤ bk and bk < ∞, so the series defining the function g is uniformly convergent on S, and therefore g is also continuous at x0 . As a consequence we find (i), since ∞ ∞  

ak = g(x0 ) = lim g(xn ) = lim n→∞

k=1 =1

= lim

n→∞

n  ∞ 

ak =

=1 k=1

n ∞  

n→∞

∞ ∞  

ak

k=1 =1

ak .

=1 k=1

For the second statement, first observe that   L ∞ K          ak  ≤ |ak | + |ak |. A −   k=1 =1

k≤K >L

k>K =1

199

3. Exercises

 To estimate bk converges, which implies ∞ the second term, we use the fact that  |a | < /2 whenever K > K , for some K 0 0 . For the first term above, k>K =1 ∞   k  |a | ≤ |a |. But the  argument above guarnote that k k k=1 k≤K >L >L ∞ ∞ antees that we can interchangethese last two sums; also =1 k=1 |ak | < ∞, so that for all L > L0 we have >L ∞ k=1 |ak | < /2. Taking N > max(L0 , K0 ) completes the proof of (ii). The proof of (iii) is a direct consequence of (ii). Indeed, given any rectangle R(K, L) = {(k, ) ∈ N × N : 1 ≤ k ≤ K and 1 ≤  ≤ L}, there exists M such that the image of [1, M ] under the map m → (k(m), (m)) contains R(K, L). When U denotes any open set in R2 that contains the origin, we define for R > 0 its dilate U (R) = {y ∈ R2 : y = Rx for some x ∈ U }, and we can apply (ii) to see that  ak . A = lim R→∞

(k,)∈U (R)

In other words, under condition (8) the double sum summing over discs, squares, rectangles, ellipses, etc.

 k

ak can be evaluated by

Finally, we leave the reader with the instructive task of finding a sequence of complex numbers {ak } such that  k



ak =

 

ak .

k

[Hint: Consider {ak } as the entries of an infinite matrix with 0 above the diagonal, −1 on the diagonal, and ak = 2−k if k > .]

3 Exercises 1. Suppose that {an }∞ n=1 is a sequence of real numbers such that the partial sums An = a1 + · · · + an are bounded. Prove that the Dirichlet series ∞  an ns n=1

converges for Re(s) > 0 and defines a holomorphic function in this half-plane. [Hint: Use summation by parts to compare the (non-absolutely convergent) original series to the (absolutely convergent) series An (n−s − (n + 1)−s ). An estimate for the term in parentheses is provided by the mean value theorem. To prove that the series is analytic, show that the partial sums converge uniformly on every compact subset of the half-plane Re(s) > 0.]

200

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

2. The following links the multiplication of Dirichlet series with the divisibility properties of their coefficients. (a) Show that if {am } and {bk } are two bounded sequences of complex numbers, then  ∞   ∞ ∞  am  bk   cn where cn = mk=n am bk . = s s s m k n m=1 n=1 k=1 The above series converge absolutely when Re(s) > 1. (b) Prove as a consequence that one has (ζ(s))2 =

∞  d(n) ns n=1

ζ(s)ζ(s − a) =

and

∞  σa (n) ns n=1

for Re(s) > 1 and Re(s − a) > 1, respectively. Here d(n) equals the number of divisors of n, and σa (n) is the sum of the ath powers of divisors of n. In particular, one has σ0 (n) = d(n).

3. In line with the previous exercise, we consider the Dirichlet series for 1/ζ. (a) Prove that for Re(s) > 1 ∞  µ(n) 1 , = ζ(s) ns n=1

where µ(n) is the M¨ obius function defined by   1 µ(n) = (−1)k  0

if n = 1 , if n = p1 · · · pk , and the pj are distinct primes , otherwise .

Note that µ(nm) = µ(n)µ(m) whenever n and m are relatively prime. [Hint: Use the Euler product formula for ζ(s).] (b) Show that 

µ(k) =

k|n

1 0

if n = 1, otherwise.

4. Suppose {an }∞ n=1 is a sequence of complex numbers such that an = am if n ≡ m mod q for some positive integer q. Define the Dirichlet L-series associated to {an } by L(s) =

∞  an ns n=1

for Re(s) > 1.

201

3. Exercises

Also, with a0 = aq , let Q(x) =

q−1 

aq−m emx .

m=0

Show, as in Exercises 15 and 16 of the previous chapter, that L(s) =




1 Γ(s)

∞ 0

Q(x)xs−1 dx, eqx − 1

for Re(s) > 1.

Prove as a result that L(s) is continuable into the complex plane, with the only possible singularity a pole at s = 1. In fact, L(s) is regular at s = 1 if and only if  q−1 m=0 am = 0. Note the connection with the Dirichlet L(s, χ) series, taken up in Book I, Chapter 8, and that as a consequence, L(s, χ) is regular at s = 1 if and only if χ is a non-trivial character. 5. Consider the following function ∞  (−1)n+1 ˜ = 1 − 1 + 1 − ··· = ζ(s) . s s 2 3 ns n=1

˜ (a) Prove that the series defining ζ(s) converges for Re(s) > 0 and defines a holomorphic function in that half-plane. ˜ = (1 − 21−s )ζ(s). (b) Show that for s > 1 one has ζ(s) (c) Conclude, since ζ˜ is given as an alternating series, that ζ has no zeros on the segment 0 < σ < 1. Extend this last assertion to σ = 0 by using the functional equation.

6. Show that for every c > 0 1 lim N→∞ 2πi




c+iN c−iN

  1 ds 1/2 as =  s 0

if a > 1, if a = 1, if 0 ≤ a < 1.

The integral is taken over the vertical segment from c − iN to c + iN . 7. Show that the function ξ(s) = π −s/2 Γ(s/2)ζ(s) is real when s is real, or when Re(s) = 1/2. 8. The function ζ has infinitely many zeros in the critical strip. This can be seen as follows.

202

Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

(a) Let F (s) = ξ(1/2 + s),

ξ(s) = π −s/2 Γ(s/2)ζ(s).

where

Show that F (s) is an even function of s, and as a result, there exists G so that G(s2 ) = F (s). (b) Show that the function (s − 1)ζ(s) is an entire function of growth order 1, that is |(s − 1)ζ(s)| ≤ A ea
|s|

1+


.

As a consequence G(s) is of growth order 1/2. (c) Deduce from the above that ζ has infinitely many zeros in the critical strip. [Hint: To prove (a) and (b) use the functional equation for ζ(s). For (c), use a result of Hadamard, which states that an entire function with fractional order has infinitely many zeros (Exercise 14 in Chapter 5).] 9. Refine the estimates in Proposition 2.7 in Chapter 6 and Proposition 1.6 to show that (a) |ζ(1 + it)| ≤ A log |t|, (b) |ζ  (1 + it)| ≤ A(log |t|)2 , (c) 1/|ζ(1 + it)| ≤ A(log |t|)a , when |t| ≥ 2 (with a = 7). 10. In the theory of primes, a better approximation to π(x) (instead of x/ log x) turns out to be Li(x) defined by


x

Li(x) = 2

dt . log t

(a) Prove that x +O Li(x) = log x



x (log x)2

 as x → ∞,

and that as a consequence π(x) ∼ Li(x)

as x → ∞.

[Hint: Integrate by parts in the definition of Li(x) and observe that it suffices to prove  
x x dt = O . 2 (log x)2 2 (log t) To see this, split the integral from 2 to

√

x and from

√

x to x.]

203

4. Problems

(b) Refine the previous analysis by showing that for every integer N > 0 one has the following asymptotic expansion Li(x) =

  x x x x x + 2 · · · + (N − 1)! + O + log x (log x)2 (log x)3 (log x)N (log x)N+1

as x → ∞.

11. Let ϕ(x) =



log p

p≤x

where the sum is taken over all primes ≤ x. Prove that the following are equivalent as x → ∞: (i) ϕ(x) ∼ x, (ii) π(x) ∼ x/ log x, (iii) ψ(x) ∼ x, (iv) ψ1 (x) ∼ x2 /2.

12. If pn denotes the nth prime, the prime number theorem implies that pn ∼ n log n as n → ∞. (a) Show that π(x) ∼ x/ log x implies that log π(x) + log log x ∼ log x. (b) As a consequence, prove that log π(x) ∼ log x, and take x = pn to conclude the proof.

4 Problems 1. Let F (s) =

∞ n=1

an /ns , where |an | ≤ M for all n.

(a) Then lim

T →∞

1 2T




T −T

|F (σ + it)|2 dt =

∞  |an |2 n2σ n=1

if σ > 1.

How is this reminiscent of the Parseval-Plancherel theorem? See e.g. Chapter 3 in Book I.

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Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM

 −s (b) Show as a consequence the uniqueness of Dirichlet series: If F (s) = ∞ , n=1 an n k where the coefficients are assumed to satisfy |an | ≤ cn for some k, and F (s) ≡ 0, then an = 0 for all n. Hint: For part (a) use the fact that 1 2T




T −T

(nm)−σ n−it mit dt →



n−2σ 0

if n = m, if n =

m.

2.∗ One of the “explicit formulas” in the theory of primes is as follows: if ψ1 is the integrated Tchebychev function considered in Section 2, then ψ1 (x) =

xρ x2  − − E(x) 2 ρ(ρ + 1) ρ

where the sum is taken over all zeros ρ of the zeta in the critical strip.  function The error term is given by E(x) = c1 x + c0 + ∞ x1−2k /(2k(2k − 1)), where k=1 1+ < ∞ for every c1 = ζ  (0)/ζ(0) and c0 = ζ  (−1)/ζ(−1). Note that ρ 1/|ρ| > 0, because (1 − s)ζ(s) has order of growth 1. (See Exercise 8.) Also, obviously E(x) = O(x) as x → ∞. 3.∗ Using the previous problem one can show that π(x) − Li(x) = O(xα+ )

as x → ∞

for every > 0, where α is fixed and 1/2 ≤ α < 1 if and only if ζ(s) has no zeros in the strip α < Re(s) < 1. The case α = 1/2 corresponds to the Riemann hypothesis. 4.∗ One can combine ideas from the prime number theorem with the proof of Dirichlet’s theorem about primes in arithmetic progression (given in Book I) to prove the following. Let q and  be relatively prime integers. We consider the primes belonging to the arithmetic progression {qk + }∞ k=1 , and let πq, (x) denote the number of such primes ≤ x. Then one has πq, (x) ∼

x ϕ(q) log x

as x → ∞ ,

where ϕ(q) denotes the number of positive integers less than q and relatively prime to q.

8 Conformal Mappings The results I found for polygons can be extended under very general assumptions. I have undertaken this research because it is a step towards a deeper understanding of the mapping problem, for which not much has happened since Riemann’s inaugural dissertation; this, even though the theory of mappings, with its close connection with the fundamental theorems of Riemann’s function theory, deserves in the highest degree to be developed further. E. B. Christoffel, 1870

The problems and ideas we present in this chapter are more geometric in nature than the ones we have seen so far. In fact, here we will be primarily interested in mapping properties of holomorphic functions. In particular, most of our results will be “global,” as opposed to the more “local” analytical results proved in the first three chapters. The motivation behind much of our presentation lies in the following simple question: Given two open sets U and V in C, does there exist a holomorphic bijection between them? By a holomorphic bijection we simply mean a function that is both holomorphic and bijective. (It will turn out that the inverse map is then automatically holomorphic.) A solution to this problem would permit a transfer of questions about analytic functions from one open set with little geometric structure to another with possibly more useful properties. The prime example consists in taking V = D the unit disc, where many ideas have been developed to study analytic functions.1 In fact, since the disc seems to be the most fruitful choice for V we are led to a variant of the above question: Given an open subset Ω of C, what conditions on Ω guarantee that there exists a holomorphic bijection from Ω to D? 1 For the corresponding problem when V = C, the solution is trivial: only U = C is possible. See Exercise 14 in Chapter 3.

206

Chapter 8. CONFORMAL MAPPINGS

In some instances when a bijection exists it can be given by explicit formulas, and we turn to this aspect of the theory first. For example, the upper half-plane can be mapped by a holomorphic bijection to the disc, and this is given by a fractional linear transformation. From there, one can construct many other examples, by composing simple maps already encountered earlier, such as rational functions, trigonometric functions, logarithms, etc. As an application, we discuss the consequence of these constructions to the solution of the Dirichlet problem for the Laplacian in some particular domains. Next, we pass from the specific examples to prove the first general result of the chapter, namely the Schwarz lemma, with an immediate application to the determination of all holomorphic bijections (“automorphisms” of the disc to itself). These are again given by fractional linear transformations. Then comes the heart of the matter: the Riemann mapping theorem, which states that Ω can be mapped to the unit disc whenever it is simply connected and not all of C. This is a remarkable theorem, since little is assumed about Ω, not even regularity of its boundary ∂Ω. (After all, the boundary of the disc is smooth.) In particular, the interiors of triangles, squares, and in fact any polygon can be mapped via a bijective holomorphic function to the disc. A precise description of the mapping in the case of polygons, called the Schwarz-Christoffel formula, will be taken up in the last section of the chapter. It is interesting to note that the mapping functions for rectangles are given by “elliptic integrals,” and these lead to doubly-periodic functions. The latter are the subject of the next chapter.

1 Conformal equivalence and examples We fix some terminology that we shall use in the rest of this chapter. A bijective holomorphic function f : U → V is called a conformal map or biholomorphism. Given such a mapping f , we say that U and V are conformally equivalent or simply biholomorphic. An important fact is that the inverse of f is then automatically holomorphic.

Proposition 1.1 If f : U → V is holomorphic and injective, then f  (z) = 0 for all z ∈ U . In particular, the inverse of f defined on its range is holomorphic, and thus the inverse of a conformal map is also holomorphic.

207

1. Conformal equivalence and examples

Proof. We argue by contradiction, and suppose that f  (z0 ) = 0 for some z0 ∈ U . Then f (z) − f (z0 ) = a(z − z0 )k + G(z)

for all z near z0 ,

with a = 0, k ≥ 2 and G vanishing to order k + 1 at z0 . For sufficiently small w, we write f (z) − f (z0 ) − w = F (z) + G(z),

where F (z) = a(z − z0 )k − w.

Since |G(z)| < |F (z)| on a small circle centered at z0 , and F has at least two zeros inside that circle, Rouch´e’s theorem implies that f (z) − f (z0 ) − w has at least two zeros there. Since f  (z) = 0 for all z = z0 but sufficiently close to z0 it follows that the roots of f (z) − f (z0 ) − w are distinct, hence f is not injective, a contradiction. Now let g = f −1 denote the inverse of f on its range, which we can assume is V . Suppose w0 ∈ V and w is close to w0 . Write w = f (z) and w0 = f (z0 ). If w = w0 , we have g(w) − g(w0 ) = w − w0

1 w−w0 g(w)−g(w0 )

=

1 f (z)−f (z0 ) z−z0

.

Since f  (z0 ) = 0, we may let z → z0 and conclude that g is holomorphic at w0 with g  (w0 ) = 1/f  (g(w0 )). From this proposition we conclude that two open sets U and V are conformally equivalent if and only if there exist holomorphic functions f : U → V and g : V → U such that g(f (z)) = z and f (g(w)) = w for all z ∈ U and w ∈ V . We point out that the terminology adopted here is not universal. Some authors call a holomorphic map f : U → V conformal if f  (z) = 0 for all z ∈ U . This definition is clearly less restrictive than ours; for example, f (z) = z 2 on the punctured disc C − {0} satisfies f  (z) = 0, but is not injective. However, the condition f  (z) = 0 is tantamount to f being a local bijection (Exercise 1). There is a geometric consequence of the condition f  (z) = 0 and it is at the root of this discrepency of terminology in the definitions. A holomorphic map that satisfies this condition preserves angles. Loosely speaking, if two curves γ and η intersect at z0 , and α is the oriented angle between the tangent vectors to these curves, then the image curves f ◦ γ and f ◦ η intersect at f (z0 ), and their tangent vectors form the same angle α. Problem 2 develops this idea. We begin our study of conformal mappings by looking at a number of specific examples. The first gives the conformal equivalence between

208

Chapter 8. CONFORMAL MAPPINGS

the unit disc and the upper half-plane, which plays an important role in many problems. 1.1 The disc and upper half-plane The upper half-plane, which we denote by H, consists of those complex numbers with positive imaginary part; that is, H = {z ∈ C : Im(z) > 0}. A remarkable fact, which at first seems surprising, is that the unbounded set H is conformally equivalent to the unit disc. Moreover, an explicit formula giving this equivalence exists. Indeed, let F (z) =

i−z i+z

and

G(w) = i

1−w . 1+w

Theorem 1.2 The map F : H → D is a conformal map with inverse G : D → H. Proof. First we observe that both maps are holomorphic in their respective domains. Then we note that any point in the upper halfplane is closer to i than to −i, so |F (z)| < 1 and F maps H into D. To prove that G maps into the upper half-plane, we must compute Im(G(w)) for w ∈ D. To this end we let w = u + iv, and note that   1 − u − iv Im(G(w)) = Re 1 + u + iv   (1 − u − iv)(1 + u − iv) = Re (1 + u)2 + v 2 =

1 − u2 − v 2 >0 (1 + u)2 + v 2

since |w| < 1. Therefore G maps the unit disc to the upper half-plane. Finally, F (G(w)) =

i − i 1−w 1+w i+

i 1−w 1+w

=

1+w−1+w = w, 1+w+1−w

and similarly G(F (z)) = z. This proves the theorem. An interesting aspect of these functions is their behavior on the boundaries of our open sets.2 Observe that F is holomorphic everywhere on C 2 The boundary behavior of conformal maps is a recurrent theme that plays an important role in this chapter.

209

1. Conformal equivalence and examples

except at z = −i, and in particular it is continuous everywhere on the boundary of H, namely the real line. If we take z = x real, then the distance from x to i is the same as the distance from x to −i, therefore |F (x)| = 1. Thus F maps R onto the boundary of D. We get more information by writing F (x) =

i−x 1 − x2 2x = +i , 2 i+x 1+x 1 + x2

and parametrizing the real line by x = tan t with t ∈ (−π/2, π/2). Since sin 2a =

2 tan a 1 + tan2 a

and

cos 2a =

1 − tan2 a , 1 + tan2 a

we have F (x) = cos 2t + i sin 2t = ei2t . Hence the image of the real line is the arc consisting of the circle omitting the point −1. Moreover, as x travels from −∞ to ∞, F (x) travels along that arc starting from −1 and first going through that part of the circle that lies in the lower half-plane. The point −1 on the circle corresponds to the “point at infinity” of the upper half-plane. Remark. Mappings of the form z →

az + b , cz + d

where a, b, c, and d are complex numbers, and where the denominator is assumed not to be a multiple of the numerator, are usually referred to as fractional linear transformations. Other instances occur as the automorphisms of the disc and of the upper half-plane in Theorems 2.1 and 2.4. 1.2 Further examples We gather here several illustrations of conformal mappings. In certain cases we discuss the behavior of the map on the boundary of the relevant domain. Some of the mappings are pictured in Figure 1. Example 1. Translations and dilations provide the first simple examples. Indeed, if h ∈ C, the translation z → z + h is a conformal map from C to itself whose inverse is w → w − h. If h is real, then this translation is also a conformal map from the upper half-plane to itself. For any non-zero complex number c, the map f : z → cz is a conformal map from the complex plane to itself, whose inverse is simply g : w → c−1 w. If c has modulus 1, so that c = eiϕ for some real ϕ, then f is

210

Chapter 8. CONFORMAL MAPPINGS

a rotation by ϕ. If c > 0 then f corresponds to a dilation. Finally, if c < 0 the map f consists of a dilation by |c| followed by a rotation of π. Example 2. If n is a positive integer, then the map z → z n is conformal from the sector S = {z ∈ C : 0 < arg(z) < π/n} to the upper half-plane. The inverse of this map is simply w → w1/n , defined in terms of the principal branch of the logarithm. More generally, if 0 < α < 2 the map f (z) = z α takes the upper halfplane to the sector S = {w ∈ C : 0 < arg(w) < απ}. Indeed, if we choose the branch of the logarithm obtained by deleting the positive real axis, and z = reiθ with r > 0 and 0 < θ < π, then f (z) = z α = |z|α eiαθ . Therefore f maps H into S. Moreover, a simple verification shows that the inverse of f is given by g(w) = w1/α , where the branch of the logarithm is chosen so that 0 < arg w < απ. By composing the map just discussed with the translations and rotations in the previous example, we may map the upper half-plane conformally to any (infinite) sector in C. Let us note the boundary behavior of f . If x travels from −∞ to 0 on the real line, then f (x) travels from ∞eiαπ to 0 on the half-line determined by arg z = απ. As x goes from 0 to ∞ on the real line, the image f (x) goes from 0 to ∞ on the real line as well. Example 3. The map f (z) = (1 + z)/(1 − z) takes the upper halfdisc {z = x + iy : |z| < 1 and y > 0} conformally to the first quadrant {w = u + iv : u > 0 and v > 0}. Indeed, if z = x + iy we have f (z) =

1 − (x2 + y 2 ) 2y +i , (1 − x)2 + y 2 (1 − x)2 + y 2

so f maps the half-disc in the upper half-plane into the first quadrant. The inverse map, given by g(w) = (w − 1)/(w + 1), is clearly holomorphic in the first quadrant. Moreover, |w + 1| > |w − 1| for all w in the first quadrant because the distance from w to −1 is greater than the distance from w to 1; thus g maps into the unit disc. Finally, an easy calculation shows that the imaginary part of g(w) is positive whenever w is in the first quadrant. So g transforms the first quadrant into the desired half-disc and we conclude that f is conformal because g is the inverse of f . To examine the action of f on the boundary, note that if z = eiθ be-

211

1. Conformal equivalence and examples

longs to the upper half-circle, then f (z) =

e−iθ/2 + eiθ/2 i 1 + eiθ . = −iθ/2 = iθ iθ/2 1−e tan(θ/2) e −e

As θ travels from 0 to π we see that f (eiθ ) travels along the imaginary axis from infinity to 0. Moreover, if z = x is real, then f (z) =

1+x 1−x

is also real; and one sees from this, that f is actually a bijection from (−1, 1) to the positive real axis, with f (x) increasing from 0 to infinity as x travels from −1 to 1. Note also that f (0) = 1. Example 4. The map z → log z, defined as the branch of the logarithm obtained by deleting the negative imaginary axis, takes the upper halfplane to the strip {w = u + iv : u ∈ R, 0 < v < π}. This is immediate from the fact that if z = reiθ with −π/2 < θ < 3π/2, then by definition, log z = log r + iθ. The inverse map is then w → ew . As x travels from −∞ to 0, the point f (x) travels from ∞ + iπ to −∞ + iπ on the line {x + iπ : −∞ < x < ∞}. When x travels from 0 to ∞ on the real line, its image f (x) then goes from −∞ to ∞ along the reals. Example 5. With the previous example in mind, we see that z → log z also defines a conformal map from the half-disc {z = x + iy : |z| < 1, y > 0} to the half-strip {w = u + iv : u < 0, 0 < v < π}. As x travels from 0 to 1 on the real line, then log x goes from −∞ to 0. When x goes from 1 to −1 on the half-circle in the upper half-plane, then the point log x travels from 0 to πi on the vertical segment of the strip. Finally, as x goes from −1 to 0, the point log x goes from πi to −∞ + iπ on the top half-line of the strip. Example 6. The map f (z) = eiz takes the half-strip {z = x + iy : −π/2 < x < π/2, y > 0} conformally to the half-disc {w = u + iv : |w| < 1, u > 0}. This is immediate from the fact that if z = x + iy, then eiz = e−y eix .

212

Chapter 8. CONFORMAL MAPPINGS

If x goes from π/2 + i∞ to π/2, then f (x) goes from 0 to i, and as x goes from π/2 to −π/2, then f (x) travels from i to −i on the half-circle. Finally, as x goes from −π/2 to −π/2 + i∞, we see that f (x) travels from −i back to 0. The mapping f is closely related to the inverse of the map in Example 5. Example 7. The function f (z) = − 12 (z + 1/z) is a conformal map from the half-disc {z = x + iy : |z| < 1, y > 0} to the upper half-plane (Exercise 5). The boundary behavior of f is as follows. If x travels from 0 to 1, then f (x) goes from ∞ to 1 on the real axis. If z = eiθ , then f (z) = cos θ and as x travels from 1 to −1 along the unit half-circle in the upper halfplane, the f (x) goes from 1 to −1 on the real segment. Finally, when x goes from −1 to 0, f (x) goes from −1 to −∞ along the real axis. Example 8. The map f (z) = sin z takes the upper half-plane conformally onto the half-strip {w = x + iy : −π/2 < x < π/2 y > 0}. To see this, note that if ζ = eiz , then   1 −1 eiz − e−iz iζ + , = sin z = 2i 2 iζ and therefore f is obtained first by applying the map in Example 6, then multiplying by i (that is, rotating by π/2), and finally applying the map in Example 7. As x travels from −π/2 + i∞ to −π/2, the point f (x) goes from −∞ to −1. When x is real, between −π/2 and π/2, then f (x) is also real between −1 and 1. Finally, if x goes from π/2 to π/2 + i∞, then f (x) travels from 1 to ∞ on the real axis. 1.3 The Dirichlet problem in a strip The Dirichlet problem in the open set Ω consists of solving u = 0 in Ω, (1) u = f on ∂Ω, where  denotes the Laplacian ∂ 2 /∂x2 + ∂ 2 /∂y 2 , and f is a given function on the boundary of Ω. In other words, we wish to find a harmonic function in Ω with prescribed boundary values f . This problem was already considered in Book I in the cases where Ω is the unit disc or the

213

1. Conformal equivalence and examples

f (z) = z α

0

0

f (z) =

1+z 1−z

1

0

0

iπ

f (z) = log z

0

0

f (z) = log z

iπ

1

0

f1 (z) = eiz

0

f2 (z) = iz

f3 (z) =

−1 2



z+

1 z



i 0 −π 2

π 2

0

1

f (z) = sin z

Figure 1. Explicit conformal maps

upper half-plane, where it arose in the solution of the steady-state heat equation. In these specific examples, explicit solutions were obtained in terms of convolutions with the Poisson kernels. Our goal here is to connect the Dirichlet problem with the conformal maps discussed so far. We begin by providing a formula for a solution to the problem (1) in the special case where Ω is a strip. In fact, this exam-

214

Chapter 8. CONFORMAL MAPPINGS

ple was studied in Problem 3 of Chapter 5, Book I, where the problem was solved using the Fourier transform. Here, we recover this solution using only conformal mappings and the known solution in the disc. The first important fact that we use is that the composition of a harmonic function with a holomorphic function is still harmonic. Lemma 1.3 Let V and U be open sets in C and F : V → U a holomorphic function. If u : U → C is a harmonic function, then u ◦ F is harmonic on V . Proof. The thrust of the lemma is purely local, so we may assume that U is an open disc. We let G be a holomorphic function in U whose real part is u (such a G exists by Exercise 12 in Chapter 2, and is determined up to an additive constant). Let H = G ◦ F and note that u ◦ F is the real part of H. Hence u ◦ F is harmonic because H is holomorphic. For an alternate (computational) proof of this lemma, see Exercise 6. With this result in hand, we may now consider the problem (1) when Ω consists of the horizontal strip Ω = {x + iy : x ∈ R, 0 < y < 1}, whose boundary is the union of the two horizontal lines R and i + R. We express the boundary data as two functions f0 and f1 defined on R, and ask for a solution u(x, y) in Ω of u = 0 that satisfies u(x, 0) = f0 (x)

and

u(x, 1) = f1 (x).

We shall assume that f0 and f1 are continuous and vanish at infinity, that is, that lim|x|→∞ fj (x) = 0 for j = 0, 1. The method we shall follow consists of relocating the problem from the strip to the unit disc via a conformal map. In the disc the solution u ˜ is then expressed in terms of a convolution with the Poisson kernel. Finally, u ˜ is moved back to the strip using the inverse of the previous conformal map, thereby giving our final answer to the problem. To achieve our goal, we introduce the mappings F : D → Ω and G : Ω → D, that are defined by   1 1−w i − eπz F (w) = log i and G(z) = . π 1+w i + eπz These two functions, which are obtained from composing mappings from examples in the previous sections, are conformal and inverses to one

215

1. Conformal equivalence and examples

u = f1 z = iy

Ω u = f0

Figure 2. The Dirichlet problem in a strip

another. Tracing through the boundary behavior of F , we find that it maps the lower half-circle to the line i + R, and the upper half-circle to R. More precisely, as ϕ travels from −π to 0, then F (eiϕ ) goes from i + ∞ to i − ∞, and as ϕ travels from 0 to π, then F (eiϕ ) goes from −∞ to ∞ on the real line. With the behavior of F on the circle in mind, we define f˜1 (ϕ) = f1 (F (eiϕ ) − i)

whenever −π < ϕ < 0,

and f˜0 (ϕ) = f0 (F (eiϕ ))

whenever 0 < ϕ < π.

Then, since f0 and f1 vanish at infinity, the function f˜ that is equal to f˜1 on the lower semi-circle, f˜0 on the upper semi-circle, and 0 at the points ϕ = ±π, 0, is continuous on the whole circle. The solution to the Dirichlet problem in the unit disc with boundary data f˜ is given by the Poisson integral3
π 1 u ˜(w) = Pr (θ − ϕ)f˜(ϕ) dϕ 2π −π
0
π 1 1 ˜ = Pr (θ − ϕ)f1 (ϕ) dϕ + Pr (θ − ϕ)f˜0 (ϕ) dϕ, 2π −π 2π 0 where w = reiθ , and Pr (θ) =

1 − r2 1 − 2r cos θ + r2

3 We refer the reader to Chapter 2 in Book I for a detailed discussion of the Dirichlet problem in the disc and the Poisson integral formula. Also, the Poisson integral formula is deduced in Exercise 12 of Chapter 2 and Problem 2 in Chapter 3 of this book.

216

Chapter 8. CONFORMAL MAPPINGS

is the Poisson kernel. Lemma 1.3 guarantees that the function u, defined by u(z) = u ˜(G(z)), is harmonic in the strip. Moreover, our construction also insures that u has the correct boundary values. A formula for u in terms of f0 and f1 is first obtained at the points z = iy with 0 < y < 1. The appropriate change of variables (see Exercise 7) shows that if reiθ = G(iy), then
π
f0 (t) sin πy ∞ 1 ˜ Pr (θ − ϕ)f0 (ϕ) dϕ = dt. 2π 0 2 cosh πt − cos πy −∞ A similar calculation also establishes
π
1 f1 (t) sin πy ∞ ˜ dt. Pr (θ − ϕ)f1 (ϕ) dϕ = 2π 0 2 −∞ cosh πt + cos πy Adding these last two integrals provides a formula for u(0, y). In general, we recall from Exercise 13 in Chapter 5 of Book I, that a solution to the Dirichlet problem in the strip vanishing at infinity is unique. Consequently, a translation of the boundary condition by x results in a translation of the solution by x as well. We may therefore apply the same argument to f0 (x + t) and f1 (x + t) (with x fixed), and a final change of variables shows that 
∞ 
∞ sin πy f0 (x − t) f1 (x − t) u(x, y) = dt + dt , 2 −∞ cosh πt − cos πy −∞ cosh πt + cos πy which gives a solution to the Dirichlet problem in the strip. In particular, we find that the solution is given in terms of convolutions with the functions f0 and f1 . Also, note that at the mid-point of the strip (y = 1/2), the solution is given by integration with respect to the function 1/ cosh πt; this function happens to be its own Fourier transform, as we saw in Example 3, Chapter 3. Remarks about the Dirichlet problem The example above leads us to envisage the solution of the more general Dirichlet problem for Ω (a suitable region), if we know a conformal map F from the disc D to Ω. That is, suppose we wish to solve (1), where f is an assigned continuous function and ∂Ω is the boundary of Ω. Assuming we have a conformal map F from D to Ω (that extends to a continuous

1. Conformal equivalence and examples

217

bijection of the boundary of the disc to the boundary of Ω), then f˜ = f ◦ F is defined on the circle, and we can solve the Dirichlet problem for the disc with boundary data f˜. The solution is given by the Poisson integral formula
2π 1 iθ Pr (θ − ϕ)f˜(eiϕ ) dϕ, u ˜(re ) = 2π 0 where Pr is the Poisson kernel. Then, one can expect that the solution of the original problem is given by u = u ˜ ◦ F −1 . Success with this approach requires that we are able to resolve affirmatively two questions: • Does there exist a conformal map Φ = F −1 from Ω to D? • If so, does this map extend to a continuous bijection from the boundary of Ω to the boundary of D? The first question, that of existence, is settled by the Riemann mapping theorem, which we prove in the next section. It is completely general (assuming only that Ω is a proper subset of C that is simply connected), and necessitates no regularity of the boundary of Ω. A positive answer to the second question requires some regularity of ∂Ω. A particular case, when Ω is the interior of a polygon, is treated below in Section 4.3. (See Exercise 18 and Problem 6 for more general assertions.) It is interesting to note that in Riemann’s original approach to the mapping problem, the chain of implications was reversed: his idea was that the existence of the conformal map Φ from Ω to D is a consequence of the solvability of the Dirichlet problem in Ω. He argued as follows. Suppose we wish to find such a Φ, with the property that a given point z0 ∈ Ω is mapped to 0. Then Φ must be of the form Φ(z) = (z − z0 )G(z), where G is holomorphic and non-vanishing in Ω. Hence we can take Φ(z) = (z − z0 )eH(z) , for suitable H. Now if u(z) is the harmonic function given by u = Re(H), then the fact that |Φ(z)| = 1 on ∂Ω means that u must satisfy the boundary condition u(z) = log(1/|z − z0 |) for z ∈ ∂Ω. So if we can find such a solution u of the Dirichlet problem,4 we can construct H, and from this the mapping function Φ. 4 The harmonic function u(z) is also known as the Green’s function with source z for 0 the region Ω.

218

Chapter 8. CONFORMAL MAPPINGS

However, there are several shortcomings to this method. First, one has to verify that Φ is a bijection. In addition, to succeed, this method requires some regularity of the boundary of Ω. Moreover, one is still faced with the question of solving the Dirichlet problem for Ω. At this stage Riemann proposed using the “Dirichlet principle.” But applying this idea involves difficulties that must be overcome.5 Nevertheless, using different methods, one can prove the existence of the mapping in the general case. This approach is carried out below in Section 3.

2 The Schwarz lemma; automorphisms of the disc and upper half-plane The statement and proof of the Schwarz lemma are both simple, but the applications of this result are far-reaching. We recall that a rotation is a map of the form z → cz with |c| = 1, namely c = eiθ , where θ ∈ R is called the angle of rotation and is well-defined up to an integer multiple of 2π. Lemma 2.1 Let f : D → D be holomorphic with f (0) = 0. Then (i) |f (z)| ≤ |z| for all z ∈ D. (ii) If for some z0 = 0 we have |f (z0 )| = |z0 |, then f is a rotation. (iii) |f  (0)| ≤ 1, and if equality holds, then f is a rotation. Proof. We first expand f in a power series centered at 0 and convergent in all of D f (z) = a0 + a1 z + a2 z 2 + · · · . Since f (0) = 0 we have a0 = 0, and therefore f (z)/z is holomorphic in D (since it has a removable singularity at 0). If |z| = r < 1, then since |f (z)| ≤ 1 we have    f (z)  1    z  ≤ r, and by the maximum modulus principle, we can conclude that this is true whenever |z| ≤ r. Letting r → 1 gives the first result. For (ii), we see that f (z)/z attains its maximum in the interior of D and must therefore be constant, say f (z) = cz. Evaluating this expression 5 An implementation of Dirichlet’s principle in the present two-dimensional situation is taken up in Book III.

2. The Schwarz lemma; automorphisms of the disc and upper half-plane

219

at z0 and taking absolute values, we find that |c| = 1. Therefore, there exists θ ∈ R such that c = eiθ , and that explains why f is a rotation. Finally, observe that if g(z) = f (z)/z, then |g(z)| ≤ 1 throughout D, and moreover f (z) − f (0) g(0) = lim = f  (0). z→0 z Hence, if |f  (0)| = 1, then |g(0)| = 1, and by the maximum principle g is constant, which implies f (z) = cz with |c| = 1. Our first application of this lemma is to the determination of the automorphisms of the disc. 2.1 Automorphisms of the disc A conformal map from an open set Ω to itself is called an automorphism of Ω. The set of all automorphisms of Ω is denoted by Aut(Ω), and carries the structure of a group. The group operation is composition of maps, the identity element is the map z → z, and the inverses are simply the inverse functions. It is clear that if f and g are automorphisms of Ω, then f ◦ g is also an automorphism, and in fact, its inverse is given by (f ◦ g)−1 = g −1 ◦ f −1 . As mentioned above, the identity map is always an automorphism. We can give other more interesting automorphisms of the unit disc. Obviously, any rotation by an angle θ ∈ R, that is, rθ : z → eiθ z, is an automorphism of the unit disc whose inverse is the rotation by the angle −θ, that is, r−θ : z → e−iθ z. More interesting, are the automorphisms of the form α−z ψα (z) = , where α ∈ C with |α| < 1. 1 − αz These mappings, which where introduced in Exercise 7 of Chapter 1, appear in a number of problems in complex analysis because of their many useful properties. The proof that they are automorphisms of D is quite simple. First, observe that since |α| < 1, the map ψα is holomorphic in the unit disc. If |z| = 1 then z = eiθ and ψα (eiθ ) =

w α − eiθ = e−iθ , iθ −iθ e (e − α) w

where w = α − eiθ , therefore |ψα (z)| = 1. By the maximum modulus principle, we conclude that |ψα (z)| < 1 for all z ∈ D. Finally we make

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the following very simple observation: (ψα ◦ ψα ) (z) =

α− 1−

α−z 1−αz α−z α 1−αz 2

α − |α| z − α + z 1 − αz − |α|2 + αz (1 − |α|2 )z = 1 − |α|2 = z, =

from which we conclude that ψα is its own inverse! Another important property of ψα is that it vanishes at z = α; moreover it interchanges 0 and α, namely ψα (0) = α

and

ψα (α) = 0.

The next theorem says that the rotations combined with the maps ψα exhaust all the automorphisms of the disc. Theorem 2.2 If f is an automorphism of the disc, then there exist θ ∈ R and α ∈ D such that f (z) = eiθ

α−z . 1 − αz

Proof. Since f is an automorphism of the disc, there exists a unique complex number α ∈ D such that f (α) = 0. Now we consider the automorphism g defined by g = f ◦ ψα . Then g(0) = 0, and the Schwarz lemma gives (2)

|g(z)| ≤ |z|

for all z ∈ D.

Moreover, g −1 (0) = 0, so applying the Schwarz lemma to g −1 , we find that |g −1 (w)| ≤ |w|

for all w ∈ D.

Using this last inequality for w = g(z) for each z ∈ D gives (3)

|z| ≤ |g(z)|

for all z ∈ D.

Combining (2) and (3) we find that |g(z)| = |z| for all z ∈ D, and by the Schwarz lemma we conclude that g(z) = eiθ z for some θ ∈ R. Replacing z by ψα (z) and using the fact that (ψα ◦ ψα )(z) = z, we deduce that f (z) = eiθ ψα (z), as claimed. Setting α = 0 in the theorem yields the following result.

2. The Schwarz lemma; automorphisms of the disc and upper half-plane

221

Corollary 2.3 The only automorphisms of the unit disc that fix the origin are the rotations. Note that by the use of the mappings ψα , we can see that the group of automorphisms of the disc acts transitively, in the sense that given any pair of points α and β in the disc, there is an automorphism ψ mapping α to β. One such ψ is given by ψ = ψβ ◦ ψα . The explicit formulas for the automorphisms of D give a good description of the group Aut(D). In fact, this group of automorphisms is “almost” isomorphic to a group of 2 × 2 matrices with complex entries often denoted by SU(1, 1). This group consists of all 2 × 2 matrices that preserve the hermitian form on C2 × C2 defined by Z, W  = z1 w1 − z2 w2 , where Z = (z1 , z2 ) and W = (w1 , w2 ). For more information about this subject, we refer the reader to Problem 4. 2.2 Automorphisms of the upper half-plane Our knowledge of the automorphisms of D together with the conformal map F : H → D found in Section 1.1 allow us to determine the group of automorphisms of H which we denote by Aut(H). Consider the map Γ : Aut(D) → Aut(H) given by “conjugation by F ”: Γ(ϕ) = F −1 ◦ ϕ ◦ F. It is clear that Γ(ϕ) is an automorphism of H whenever ϕ is an automorphism of D, and Γ is a bijection whose inverse is given by Γ−1 (ψ) = F ◦ ψ ◦ F −1 . In fact, we prove more, namely that Γ preserves the operations on the corresponding groups of automorphisms. Indeed, suppose that ϕ1 , ϕ2 ∈ Aut(D). Since F ◦ F −1 is the identity on D we find that Γ(ϕ1 ◦ ϕ2 ) = F −1 ◦ ϕ1 ◦ ϕ2 ◦ F = F −1 ◦ ϕ1 ◦ F ◦ F −1 ◦ ϕ2 ◦ F = Γ(ϕ1 ) ◦ Γ(ϕ2 ). The conclusion is that the two groups Aut(D) and Aut(H) are the same, since Γ defines an isomorphism between them. We are still left with the

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task of giving a description of elements of Aut(H). A series of calculations, which consist of pulling back the automorphisms of the disc to the upper half-plane via F , can be used to verify that Aut(H) consists of all maps z →

az + b , cz + d

where a, b, c, and d are real numbers with ad − bc = 1. Again, a matrix group is lurking in the background. Let SL2 (R) denote the group of all 2 × 2 matrices with real entries and determinant 1, namely

SL2 (R) =

 M=

a b c d



" : a, b, c, d ∈ R and det(M ) = ad − bc=1 .

This group is called the special linear group. Given a matrix M ∈ SL2 (R) we define the mapping fM by fM (z) =

az + b . cz + d

Theorem 2.4 Every automorphism of H takes the form fM for some M ∈ SL2 (R). Conversely, every map of this form is an automorphism of H. The proof consists of a sequence of steps. For brevity, we denote the group SL2 (R) by G. Step 1. If M ∈ G, then fM maps H to itself. This is clear from the observation that (4) Im(fM (z)) =

(ad − bc)Im(z) Im(z) = >0 2 |cz + d| |cz + d|2

whenever z ∈ H.

Step 2. If M and M  are two matrices in G, then fM ◦ fM  = fM M  . This follows from a straightforward calculation, which we omit. As a consequence, we can prove the first half of the theorem. Each fM is an automorphism because it has a holomorphic inverse (fM )−1 , which is simply fM −1 . Indeed, if I is the identity matrix, then (fM ◦ fM −1 )(z) = fM M −1 (z) = fI (z) = z. Step 3. Given any two points z and w in H, there exists M ∈ G such that fM (z) = w, and therefore G acts transitively on H. To prove this,

2. The Schwarz lemma; automorphisms of the disc and upper half-plane

223

it suffices to show that we can map any z ∈ H to i. Setting d = 0 in equation (4) above gives Im(fM (z)) =

Im(z) |cz|2

and we may choose a real number c so that Im(fM (z)) = 1. Next we choose the matrix   0 −c−1 M1 = c 0 so that fM1 (z) has imaginary part equal to 1. Then we translate by a matrix of the form   1 b with b ∈ R, M2 = 0 1 to bring fM1 (z) to i. Finally, the map fM with M = M2 M1 takes z to i. Step 4. If θ is real, then the matrix   cos θ − sin θ Mθ = sin θ cos θ belongs to G, and if F : H → D denotes the standard conformal map, then F ◦ fMθ ◦ F −1 corresponds to the rotation of angle −2θ in the disc. This follows from the fact that F ◦ fMθ = e−2iθ F (z), which is easily verified. Step 5. We can now complete the proof of the theorem. We suppose f is an automorphism of H with f (β) = i, and consider a matrix N ∈ G such that fN (i) = β. Then g = f ◦ fN satisfies g(i) = i, and therefore F ◦ g ◦ F −1 is an automorphism of the disc that fixes the origin. So F ◦ g ◦ F −1 is a rotation, and by Step 4 there exists θ ∈ R such that F ◦ g ◦ F −1 = F ◦ fMθ ◦ F −1 . Hence g = fMθ , and we conclude that f = fMθ N −1 which is of the desired form. A final observation is that the group Aut(H) is not quite isomorphic with SL2 (R). The reason for this is because the two matrices M and −M give rise to the same function fM = f−M . Therefore, if we identify the two matrices M and −M , then we obtain a new group PSL2 (R) called the projective special linear group; this group is isomorphic with Aut(H).

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Chapter 8. CONFORMAL MAPPINGS

3 The Riemann mapping theorem 3.1 Necessary conditions and statement of the theorem We now come to the promised cornerstone of this chapter. The basic problem is to determine conditions on an open set Ω that guarantee the existence of a conformal map F : Ω → D. A series of simple observations allow us to find necessary conditions on Ω. First, if Ω = C there can be no conformal map F : Ω → D, since by Liouville’s theorem F would have to be a constant. Therefore, a necessary condition is to assume that Ω = C. Since D is connected, we must also impose the requirement that Ω be connected. There is still one more condition that is forced upon us: since D is simply connected, the same must be true of Ω (see Exercise 3). It is remarkable that these conditions on Ω are also sufficient to guarantee the existence of a biholomorpism from Ω to D. For brevity, we shall call a subset Ω of C proper if it is non-empty and not the whole of C. Theorem 3.1 (Riemann mapping theorem) Suppose Ω is proper and simply connected. If z0 ∈ Ω, then there exists a unique conformal map F : Ω → D such that F (z0 ) = 0

and

F  (z0 ) > 0.

Corollary 3.2 Any two proper simply connected open subsets in C are conformally equivalent. Clearly, the corollary follows from the theorem, since we can use as an intermediate step the unit disc. Also, the uniqueness statement in the theorem is straightforward, since if F and G are conformal maps from Ω to D that satisfy these two conditions, then H = F ◦ G−1 is an automorphism of the disc that fixes the origin. Therefore H(z) = eiθ z, and since H  (0) > 0, we must have eiθ = 1, from which we conclude that F = G. The rest of this section is devoted to the proof of the existence of the conformal map F . The idea of the proof is as follows. We consider all injective holomorphic maps f : Ω → D with f (z0 ) = 0. From these we wish to choose an f so that its image fills out all of D, and this can be achieved by making f  (z0 ) as large as possible. In doing this, we shall need to be able to extract f as a limit from a given sequence of functions. We turn to this point first.

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3. The Riemann mapping theorem

3.2 Montel’s theorem Let Ω be an open subset of C. A family F of holomorphic functions on Ω is said to be normal if every sequence in F has a subsequence that converges uniformly on every compact subset of Ω (the limit need not be in F ). The proof that a family of functions is normal is, in practice, the consequence of two related properties, uniform boundedness and equicontinuity. These we shall now define. The family F is said to be uniformly bounded on compact subsets of Ω if for each compact set K ⊂ Ω there exists B > 0, such that |f (z)| ≤ B

for all z ∈ K and f ∈ F .

Also, the family F is equicontinuous on a compact set K if for every  > 0 there exists δ > 0 such that whenever z, w ∈ K and |z − w| < δ, then |f (z) − f (w)| < 

for all f ∈ F .

Equicontinuity is a strong condition, which requires uniform continuity, uniformly in the family. For instance, any family of differentiable functions on [0, 1] whose derivatives are uniformly bounded is equicontinuous. This follows directly from the mean value theorem. On the other hand, note that the family {fn } on [0, 1] given by fn (x) = xn is not equicontinuous since for any fixed 0 < x0 < 1 we have |fn (1) − fn (x0 )| → 1 as n tends to infinity. The theorem that follows puts together these new concepts and is an important ingredient in the proof of the Riemann mapping theorem. Theorem 3.3 Suppose F is a family of holomorphic functions on Ω that is uniformly bounded on compact subsets of Ω. Then: (i) F is equicontinuous on every compact subset of Ω. (ii) F is a normal family. The theorem really consists of two separate parts. The first part says that F is equicontinuous under the assumption that F is a family of holomorphic functions that is uniformly bounded on compact subsets of Ω. The proof follows from an application of the Cauchy integral formula and hence relies on the fact that F consists of holomorphic functions. This conclusion is in sharp contrast with the real situation as illustrated by the family of functions given by fn (x) = sin(nx) on (0, 1), which is

226

Chapter 8. CONFORMAL MAPPINGS

uniformly bounded. However, this family is not equicontinuous and has no convergent subsequence on any compact subinterval of (0, 1). The second part of the theorem is not complex-analytic in nature. Indeed, the fact that F is a normal family follows from assuming only that F is uniformly bounded and equicontinuous on compact subsets of Ω. This result is sometimes known as the Arzela-Ascoli theorem and its proof consists primarily of a diagonalization argument. We are required to prove convergence on arbitrary compact subsets of Ω, therefore it is useful to introduce the following notion. A sequence {K }∞

=1 of compact subsets of Ω is called an exhaustion if (a) K is contained in the interior of K +1 for all  = 1, 2, . . .. (b) Any compact set K ⊂ Ω is contained in K for some . In particular ∞  Ω= K .

=1

Lemma 3.4 Any open set Ω in the complex plane has an exhaustion. Proof. If Ω is bounded, we let K denote the set of all points in Ω at distance ≥ 1/ from the boundary of Ω. If Ω is not bounded, let K denote the same set as above except that we also require |z| ≤  for all z ∈ K . We may now begin the proof of Montel’s theorem. Let K be a compact subset of Ω and choose r > 0 so small that D3r (z) is contained in Ω for all z ∈ K. It suffices to choose r so that 3r is less than the distance from K to the boundary of Ω. Let z, w ∈ K with |z − w| < r, and let γ denote the boundary circle of the disc D2r (w). Then, by Cauchy’s integral formula, we have f (z) − f (w) =

1 2πi


f (ζ) γ

1 1 − ζ −z ζ −w

 dζ.

Observe that

   1 1  |z − w| |z − w|   ζ − z − ζ − w  = |ζ − z| |ζ − w| ≤ r2 since ζ ∈ γ and |z − w| < r. Therefore |f (z) − f (w)| ≤

1 2πr B|z − w|, 2π r2

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3. The Riemann mapping theorem

where B denotes the uniform bound for the family F in the compact set consisting of all points in Ω at a distance ≤ 2r from K. Therefore |f (z) − f (w)| < C|z − w|, and this estimate is true for all z, w ∈ K with |z − w| < r and f ∈ F ; thus this family is equicontinuous, as was to be shown. To prove the second part of the theorem, we argue as follows. Let {fn }∞ n=1 be a sequence in F and K a compact subset of Ω. Choose a sequence of points {wj }∞ j=1 that is dense in Ω. Since {fn } is uniformly bounded, there exists a subsequence {fn,1 } = {f1,1 , f2,1 , f3,1, . . .} of {fn } such that fn,1 (w1 ) converges. From {fn,1 } we can extract a subsequence {fn,2 } = {f1,2 , f2,2, f3,2 , . . .} so that fn,2 (w2 ) converges. We may continue this process, and extract a subsequence {fn,j } of {fn,j−1 } such that fn,j (wj ) converges. Finally, let gn = fn,n and consider the diagonal subsequence {gn }. By construction, gn (wj ) converges for each j, and we claim that equicontinuity implies that gn converges uniformly on K. Given  > 0, choose δ as in the definition of equicontinuity, and note that for some J, the set K is contained in the union of the discs Dδ (w1 ), . . . , Dδ (wJ ). Pick N so large that if n, m > N , then |gm (wj ) − gn (wj )| < 

for all j = 1, . . . , J.

So if z ∈ K, then z ∈ Dδ (wj ) for some 1 ≤ j ≤ J. Therefore, |gn (z) − gm (z)| ≤ |gn (z) − gn (wj )| + |gn (wj ) − gm (wj )| + + |gm (wj ) − gm (z)| < 3 whenever n, m > N . Hence {gn } converges uniformly on K. Finally, we need one more diagonalization argument to obtain a subsequence that converges uniformly on every compact subset of Ω. Let K1 ⊂ K2 ⊂ · · · ⊂ K ⊂ · · · be an exhaustion of Ω, and suppose {gn,1 } is a subsequence of the original sequence {fn } that converges uniformly on K1 . Extract from {gn,1 } a subsequence {gn,2 } that converges uniformly on K2 , and so on. Then, {gn,n } is a subsequence of {fn } that converges uniformly on every K and since the K exhaust Ω, the sequence {gn,n } converges uniformly on any compact subset of Ω, as was to be shown. We need one further result before we can give the proof of the Riemann mapping theorem. Proposition 3.5 If Ω is a connected open subset of C and {fn } a sequence of injective holomorphic functions on Ω that converges uniformly

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Chapter 8. CONFORMAL MAPPINGS

on every compact subset of Ω to a holomorphic function f , then f is either injective or constant. Proof. We argue by contradiction and suppose that f is not injective, so there exist distinct complex numbers z1 and z2 in Ω such that f (z1 ) = f (z2 ). Define a new sequence by gn (z) = fn (z) − fn (z1 ), so that gn has no other zero besides z1 , and the sequence {gn } converges uniformly on compact subsets of Ω to g(z) = f (z) − f (z1 ). If g is not identically zero, then z2 is an isolated zero for g (because Ω is connected); therefore
 g (ζ) 1 1= dζ, 2πi γ g(ζ) where γ is a small circle centered at z2 chosen so that g does not vanish on γ or at any point of its interior besides z2 . Therefore, 1/gn converges uniformly to 1/g on γ, and since gn → g  uniformly on γ we have

 1 1 gn (ζ) g (ζ) dζ → dζ. 2πi γ gn (ζ) 2πi γ g(ζ) But this is a contradiction since gn has no zeros inside γ, and hence
 gn (ζ) 1 dζ = 0 for all n . 2πi γ gn (ζ)

3.3 Proof of the Riemann mapping theorem Once we have established the technical results above, the rest of the proof of the Riemann mapping theorem is very elegant. It consists of three steps, which we isolate. Step 1. Suppose that Ω is a simply connected proper open subset of C. We claim that Ω is conformally equivalent to an open subset of the unit disc that contains the origin. Indeed, choose a complex number α that does not belong to Ω, (recall that Ω is proper), and observe that z − α never vanishes on the simply connected set Ω. Therefore, we can define a holomorphic function f (z) = log(z − α) with the desired properties of the logarithm. As a consequence one has, ef (z) = z − α, which proves in particular that f is injective. Pick a point w ∈ Ω, and observe that f (z) = f (w) + 2πi

for all z ∈ Ω

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3. The Riemann mapping theorem

for otherwise, we exponentiate this relation to find that z = w, hence f (z) = f (w), a contradiction. In fact, we claim that f (z) stays strictly away from f (w) + 2πi, in the sense that there exists a disc centered at f (w) + 2πi that contains no points of the image f (Ω). Otherwise, there exists a sequence {zn } in Ω such that f (zn ) → f (w) + 2πi. We exponentiate this relation, and, since the exponential function is continuous, we must have zn → w. But this implies f (zn ) → f (w), which is a contradiction. Finally, consider the map F (z) =

1 . f (z) − (f (w) + 2πi)

Since f is injective, so is F , hence F : Ω → F (Ω) is a conformal map. Moreover, by our analysis, F (Ω) is bounded. We may therefore translate and rescale the function F in order to obtain a conformal map from Ω to an open subset of D that contains the origin. Step 2. By the first step, we may assume that Ω is an open subset of D with 0 ∈ Ω. Consider the family F of all injective holomorphic functions on Ω that map into the unit disc and fix the origin: F = {f : Ω → D holomorphic, injective and f (0) = 0}. First, note that F is non-empty since it contains the identity. Also, this family is uniformly bounded by construction, since all functions are required to map into the unit disc. Now, we turn to the question of finding a function f ∈ F that maximizes |f  (0)|. First, observe that the quantities |f  (0)| are uniformly bounded as f ranges in F . This follows from the Cauchy inequality (Corollary 4.3 in Chapter 2) for f  applied to a small disc centered at the origin. Next, we let s = sup |f  (0)|, f ∈F

and we choose a sequence {fn } ⊂ F such that |fn (0)| → s as n → ∞. By Montel’s theorem (Theorem 3.3), this sequence has a subsequence that converges uniformly on compact sets to a holomorphic function f on Ω. Since s ≥ 1 (because z → z belongs to F ), f is non-constant, hence injective, by Proposition 3.5. Also, by continuity we have |f (z)| ≤ 1 for all z ∈ Ω and from the maximum modulus principle we see that |f (z)| < 1. Since we clearly have f (0) = 0, we conclude that f ∈ F with |f  (0)| = s.

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Step 3. In this last step, we demonstrate that f is a conformal map from Ω to D. Since f is already injective, it suffices to prove that f is also surjective. If this were not true, we could construct a function in F with derivative at 0 greater than s. Indeed, suppose there exists α ∈ D such that f (z) = α, and consider the automorphism ψα of the disc that interchanges 0 and α, namely ψα (z) =

α−z . 1 − αz

Since Ω is simply connected, so is U = (ψα ◦ f )(Ω), and moreover, U does not contain the origin. It is therefore possible to define a square root function on U by 1

g(w) = e 2 log w . Next, consider the function F = ψg(α) ◦ g ◦ ψα ◦ f. We claim that F ∈ F . Clearly F is holomorphic and it maps 0 to 0. Also F maps into the unit disc since this is true of each of the functions in the composition. Finally, F is injective. This is clearly true for the automorphisms ψα and ψg(α) ; it is also true for the square root g and the function f , since the latter is injective by assumption. If h denotes the square function h(w) = w2 , then we must have −1 ◦ F = Φ ◦ F. f = ψα−1 ◦ h ◦ ψg(α)

But Φ maps D into D with Φ(0) = 0, and is not injective because F is and h is not. By the last part of the Schwarz lemma, we conclude that |Φ (0)| < 1. The proof is complete once we observe that f  (0) = Φ (0)F  (0), and thus |f  (0)| < |F  (0)|, contradicting the maximality of |f  (0)| in F . Finally, we multiply f by a complex number of absolute value 1 so that f  (0) > 0, which ends the proof. For a variant of this proof, see Problem 7.

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4. Conformal mappings onto polygons

Remark. It is worthwhile to point out that the only places where the hypothesis of simple-connectivity entered in the proof were in the uses of the logarithm and the square root. Thus it would have sufficed to have assumed (in addition to the hypothesis that Ω is proper) that Ω is holomorphically simply connected in the sense that for any  holomorphic function f in Ω and any closed curve γ in Ω, we have f (z) dz = 0. Further discussion of this point, and various equivalent γ properties of simple-connectivity, are given in Appendix B.

4 Conformal mappings onto polygons The Riemann mapping theorem guarantees the existence of a conformal map from any proper, simply connected open set to the disc, or equivalently to the upper half-plane, but this theorem gives little insight as to the exact form of this map. In Section 1 we gave various explicit formulas in the case of regions that have symmetries, but it is of course unreasonable to ask for an explicit formula in the general case. There is, however, another class of open sets for which there are nice formulas, namely the polygons. Our aim in this last section is to give a proof of the Schwarz-Christoffel formula, which describes the nature of conformal maps from the disc (or upper half-plane) to polygons. 4.1 Some examples We begin by studying some motivating examples. The first two correspond to easy (but infinite and degenerate) cases. Example 1. First, we investigate the conformal map from the upper half-plane to the sector {z : 0 < arg z < απ}, with 0 < α < 2, given in Section 1 by f (z) = z α . Anticipating the Schwarz-Christoffel formula below, we write


α

z = f (z) =

z






z

f (ζ) dζ = α 0

ζ −β dζ

0

with α + β = 1, and where the integral is taken along any path in the upper half-plane. In fact, by continuity and Cauchy’s theorem, we may take the path of integration to lie in the closure of the upper half-plane. Although the behavior of f follows immediately from the original definition, we study it in terms of the integral expression above, since this provides insight for the general case treated later. Note first that ζ −β is integrable near 0 since β < 1, therefore f (0) = 0. Observe that when z is real and positive (z = x), then f  (x) = αxα−1 is

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Chapter 8. CONFORMAL MAPPINGS

positive; also it is not finitely integrable at ∞. Therefore, as x travels from 0 to ∞, we see that f (x) increases from 0 to ∞, thus f maps [0, ∞) to [0, ∞). On the other hand, when z = x is negative, then f  (z) = α|x|α−1 eiπ(α−1) = −α|x|α−1 eiπα , so f maps the segment (−∞, 0] to (eiπα ∞, 0]. The situation is illustrated in Figure 3 where the infinite segment A is mapped to A and the segment B is mapped to B  , with the direction of travel indicated in Figure 3. A

A

0

0

B

B

Figure 3. The conformal map z α

Example 2. Next, we consider for z ∈ H,
z dζ f (z) = , (1 − ζ 2 )1/2 0 where the integral is taken from 0 to z along any path in the closed upper half-plane. We choose the branch for (1 − ζ 2 )1/2 that makes it holomorphic in the upper half-plane and positive when −1 < ζ < 1. As a result (1 − ζ 2 )−1/2 = i(ζ 2 − 1)−1/2

when ζ > 1.

We observe that f maps the real line to the boundary of the half-strip pictured in Figure 4. In fact, since f (±1) = ±π/2, and f  (x) > 0 if −1 < x < 1, we see that f maps the segment B to B  . Moreover
x
∞ dx π  f (x) dx when x > 1, and = ∞. f (x) = + 2 2 (x − 1)1/2 1 1 Thus, as x travels along the segment C, the image traverses the infinite segment C  . Similarly segment A is mapped to A . Note the connection of this example with Example 8 in Section 1.2. In fact, one can show that the function f (z) is the inverse to the function

233

4. Conformal mappings onto polygons

A A

B

C

C

−1

1 − π2

B

π 2

Figure 4. Mapping of the boundary in Example 2

sin z, and hence f takes H conformally to the interior of the half-strip bounded by the segments A , B  , and C  . Example 3. Here we take
z dζ , f (z) = 2 )(1 − k 2 ζ 2 )]1/2 [(1 − ζ 0

z ∈ H,

where k is a fixed real number with 0 < k < 1 (the branch of [(1 − ζ 2 ) (1 − k2 ζ 2 )]1/2 in the upper half-plane is chosen to be the one that is positive when ζ is real and −1 < ζ < 1). Integrals of this kind are called elliptic integrals, because variants of these arise in the calculation of the arc-length of an ellipse. We shall observe that f maps the real axis onto the rectangle shown in Figure 5(b), where K and K  are determined by




1

K= 0

dx , K = [(1 − x2 )(1 − k2 x2 )]1/2




1/k

1

dx . [(x2 − 1)(1 − k2 x2 )]1/2

We divide the real axis into four “segments,” with division points −1/k, −1, 1, and 1/k (see Figure 5(a)). The segments are [−1/k, −1], [−1, 1], [1, 1/k], and [1/k, −1/k], the last consisting of the union of the two half-segments [1/k, ∞) and (−∞, −1/k]. It is clear from the definitions that f (±1) = ±K, and since f  (x) > 0, when −1 < x < 1, it follows that f maps the segment [−1, 1] to [−K, K]. Moreover, since


f (z) = K + 1

x

[(1 −

dζ − k2 ζ 2 )]1/2

ζ 2 )(1

if 1 < x < 1/k,

234

Chapter 8. CONFORMAL MAPPINGS −K + iK 

A

B

−1/k

C

−1

D

E

K + iK 

B

E

1

A

−K

1/k

D C

K

(b)

(a)

Figure 5. Mapping of the boundary in Example 3

we see that f maps the segment [1, 1/k] to [K, K + iK  ], where K  was defined above. Similarly, f maps [−1/k, −1] to [−K + iK  , −K]. Next, when x > 1/k we have f  (x) = −

1 , [(x2 − 1)(k2 x2 − 1)]1/2

and therefore, 




x

f (x) = K + iK − 1/k

[(x2

dx . − 1)(k2 x2 − 1)]1/2

However,




∞ 1/k

dx = [(x2 − 1)(k2 x2 − 1)]1/2




1 0

dx , [(1 − x2 )(1 − k2 x2 )]1/2

as can be seen by making the change of variables x = 1/ku in the integral on the left. Thus f maps the segment [1/k, ∞) to the segment [K + iK  , iK  ). Similarly f maps (−∞, −1/k] to [−K + iK  , iK  ). Altogether, then, f maps the real axis to the above rectangle, with the point at infinity corresponding to the mid-point of the upper side of the rectangle. The results obtained so far lead naturally to two problems. The first, which we pursue next, consists of a generalization of the above examples. More precisely we define the Schwarz-Christoffel integral and prove that it maps the real line to a polygonal line. Second, we note that in the examples above little was inferred about the behavior of f in H itself. In particular, we have not shown that f maps H conformally to the interior of the corresponding polygon. After a careful study of the boundary behavior of conformal maps, we

4. Conformal mappings onto polygons

235

prove a theorem that guarantees that the conformal map from the upper half-plane to a simply connected region bounded by a polygonal line is essentially given by a Schwarz-Christoffel integral. 4.2 The Schwarz-Christoffel integral With the examples of the previous section in mind, we define the general Schwarz-Christoffel integral by
z dζ . (5) S(z) = β 1 · · · (ζ − A )βn n 0 (ζ − A1 ) Here A1 < A2 < · · · < An are n distinct points on the real axis arranged in increasing order. The exponents β k will be assumed to satisfy the n conditions βk < 1 for each k and 1 < k=1 βk .6 The integrand in (5) is defined as follows: (z − Ak )βk is that branch (defined in the complex plane slit along the infinite ray {Ak + iy : y ≤ 0}) which is positive when z = x is real and x > Ak . As a result # (x − Ak )βk if x is real and x > Ak , βk (z − Ak ) = βk iπβk if x is real and x < Ak . |x − Ak | e The complex plane slit along the union of the rays ∪nk=1 {Ak + iy : y ≤ 0} is simply connected (see Exercise 19), so the integral that defines S(z) is holomorphic in this open set. Since the requirement βk < 1 implies that the singularities (ζ − Ak )−βk are integrable near Ak , the function S is continuous up to the real line, including the points Ak , with k = 1, . . . , n. Finally, this continuity condition implies that the integral can be taken along any path in the complex plane that avoids the union of the open slits ∪nk=1 {Ak + iy : y < 0}. Now   n       (ζ − Ak )−βk  ≤ c|ζ|− βk   k=1

 for |ζ| large, so the assumption βk > 1 guarantees the convergence of the integral (5) at infinity. This fact and Cauchy’s theorem imply that limr→∞ S(reiθ ) exists and is independent of the angle θ, 0 ≤ θ ≤ π. We call this limit a∞ , and we let ak = S(Ak ) for k = 1, . . . , n.  that the case βk ≤ 1, which occurs in Examples 1 and 2 above is excluded. However, a modification of the proposition that follows can be made to take these cases into account; but then S(z) is no longer bounded in the upper half-plane. 6 Note

236

Chapter 8. CONFORMAL MAPPINGS

Proposition 4.1 Suppose S(z) is given by (5). n (i) If k=1 βk = 2, and p denotes the polygon whose vertices are given (in order) by a1 , . . . , an , then S maps the real axis onto p − {a∞ }. The point a∞ lies on the segment [an , a1 ] and is the image of the point at infinity. Moreover, the (interior) angle at the vertex ak is αk π where αk = 1 − βk . n (ii) There is a similar conclusion when 1 < k=1 βk < 2, except now the image of the extended line is the polygon of n + 1 sides with vertices a1 , a2 , . . . , an , a∞ . The angle nat the vertex a∞ is α∞ π with α∞ = 1 − β∞ , where β∞ = 2 − k=1 βk . Figure 6 illustrates the proposition. The idea of the proof is already captured in Example 1 above. a1 

a∞

an

βk = 2 p

ak A1

An

Ak

R

a∞ a1

1
k

237

4. Conformal mappings onto polygons

Hence

 

arg S (x) = arg



 (x − Aj )

−βj

= arg

j>k

 j>k

e−iπβj = −π



βj ,

j>k

which of course is constant when x traverses the interval (Ak , Ak+1 ). Since
x S  (y) dy, S(x) = S(Ak ) + Ak

we see that as x varies from Ak to Ak+1 , S(x) varies from S(Ak ) = 7 ak to S(Ak+1 ) = ak+1 along  the straight line segment [ak , ak+1 ], and this makes an angle of −π j>k βj with the real axis. Similarly, when An <  x then S  (x) is positive, while if x < A1 , the argument of S  (x) n is −π k=1 βk = −2π, and so S  (x) is again positive. Thus as x varies on [An , +∞), S(x) varies along a straight line (parallel to the x-axis) between an and a∞ ; similarly S(x) varies along a straight line (parallel to that axis) between a∞ and a1 as x varies in (−∞, A1 ]. Moreover, the union of [an , a∞ ) and (a∞ , a1 ] is the segment [an , a1 ] with the point a∞ removed. Now the increase of the angle of [ak+1 , ak ] over that of [ak−1 , ak ] is πβk , whichmeans that the angle at the vertex ak is παk . The proof n when 1 < k=1 βk < 2 is similar, and is left to the reader. As elegant as this proposition is, it does not settle the problem of finding a conformal map from the half-plane to a given region P that is bounded by a polygon. There are two reasons for this. 1. It is not true for general n and generic choices of A1 , . . . , An , that the polygon (which is the image of the real axis under S) is simple, that is, it does not cross itself. Nor is it true in general that the mapping S is conformal on the upper half-plane. 2. Neither does the proposition show that starting with a simply connected region P (whose boundary is a polygonal line p) the mapping S is, for certain choices of A1 , . . . , An and simple modifications, a conformal map from H to P . That however is the case, and is the result whose proof we now turn to. 7 We

denote the closed straight line segment between two complex numbers z and w by [z, w], that is, [z, w] = {(1 − t)z + tw : t ∈ [0, 1]}. If we restrict 0 < t < 1, then (z, w) denotes the open line segment between z and w. Similarly for the half-open segments [z, w) and (z, w] obtained by restricting 0 ≤ t < 1 and 0 < t ≤ 1, respectively.

238

Chapter 8. CONFORMAL MAPPINGS

4.3 Boundary behavior In what follows we shall consider a polygonal region P , namely a bounded, simply connected open set whose boundary is a polygonal line p. In this context, we always assume that the polygonal line is closed, and we sometimes refer to p as a polygon. To study conformal maps from the half-plane H to P , we consider first the conformal maps from the disc D to P , and their boundary behavior. Theorem 4.2 If F : D → P is a conformal map, then F extends to a continuous bijection from the closure D of the disc to the closure P of the polygonal region. In particular, F gives rise to a bijection from the boundary of the disc to the boundary polygon p. The main point consists in showing that if z0 belongs to the unit circle, then limz→z0 F (z) exists. To prove this, we need a preliminary result, which uses the fact that if f : U → f (U ) is conformal, then





|f  (z)|2 dx dy.

Area(f (U )) = U

 This assertion follows from the definition, Area(f (U )) = f (U ) dx dy, and the fact that the determinant of the Jacobian in the change of variables w = f (z) is simply |f  (z)|2 , an observation we made in equation (4), Section 2.2, Chapter 1. Lemma 4.3 For each 0 < r < 1/2, let Cr denote the circle centered at z0 of radius r. Suppose that for all sufficiently small r we are given two points zr and zr in the unit disc that also lie on Cr . If we let ρ(r) = |f (zr ) − f (zr )|, then there exists a sequence {rn } of radii that tends to zero, and such that limn→∞ ρ(rn ) = 0. Proof. If not, there exist 0 < c and 0 < R < 1/2 such that c ≤ ρ(r) for all 0 < r ≤ R. Observe that
 f  (ζ) dζ, f (zr ) − f (zr ) = α

where the integral is taken over the arc α on Cr that joins zr and zr in D. If we parametrize this arc by z0 + reiθ with θ1 (r) ≤ θ ≤ θ2 (r), then


ρ(r) ≤

θ2 (r)

θ1 (r)

|f  (z)|r dθ.

239

4. Conformal mappings onto polygons

We now apply the Cauchy-Schwarz inequality to see that




θ2 (r)

ρ(r) ≤

1/2 




|f (z)| r dθ 2

1/2

θ2 (r)

r dθ

θ1 (r)

.

θ1 (r)

Squaring both sides and dividing by r yields ρ(r)2 ≤ 2π r




θ2 (r)

|f  (z)|2 r dθ.

θ1 (r)

We may now integrate both sides from 0 to R, and since c ≤ ρ(r) on that region we obtain




R

c2 0

dr ≤ 2π r


R
0

θ2 (r) θ1 (r)

|f  (z)|2 r dθdr ≤ 2π



D

|f  (z)|2 dxdy.

Now the left-hand side is infinite because 1/r is not integrable near the origin, and the right-hand side is bounded because the area of the polygonal region is bounded, so this yields the desired contradiction and concludes the proof of the lemma. Lemma 4.4 Let z0 be a point on the unit circle. Then F (z) tends to a limit as z approaches z0 within the unit disc. Proof. If not, there are two sequences {z1 , z2 , . . .} and {z1 , z2 , . . .} in the unit disc that converge to z0 and are so that F (zn ) and F (zn ) converge to two distinct points ζ and ζ  in the closure of P . Since F is conformal, the points ζ and ζ  must lie on the boundary p of P . We may therefore choose two disjoint discs D and D centered at ζ and ζ  , respectively, that are at a distance d > 0 from each other. For all large n, F (zn ) ∈ D and F (zn ) ∈ D . Therefore, there exist two continuous curves8 Λ and Λ in D ∩ P and D ∩ P , respectively, with F (zn ) ∈ Λ and F (zn ) ∈ Λ for all large n, and with the end-points of Λ and Λ equal to ζ and ζ  , respectively. Define λ = F −1 (Λ) and λ = F −1 (Λ ). Then λ and λ are two continuous curves in D. Moreover, both λ and λ contain infinitely many points in each sequence {zn } and {zn }. Recall that these sequences converge to z0 . By continuity, the circle Cr centered at z0 and of radius r will intersect λ and λ for all small r, say at some points zr ∈ λ and zr ∈ λ . This 8 By a continuous curve, we mean the image of a continuous (not necessarily piecewisesmooth) function from a closed interval [a, b] to C.

240

Chapter 8. CONFORMAL MAPPINGS Cr

D

zr ζ

z0

λ

D Λ

λ

zr

ζ

D

P

Λ

Figure 7. Illustration for the proof of Lemma 4.4

contradicts the previous lemma, because |F (zr ) − F (zr )| > d. Therefore F (z) converges to a limit on p as z approaches z0 from within the unit disc, and the proof is complete. Lemma 4.5 The conformal map F extends to a continuous function from the closure of the disc to the closure of the polygon. Proof. By the previous lemma, the limit lim F (z)

z→z0

exists, and we define F (z0 ) to be the value of this limit. There remains to prove that F is continuous on the closure of the unit disc. Given , there exists δ such that whenever z ∈ D and |z − z0 | < δ, then |F (z) − F (z0 )| < . Now if z belongs to the boundary of D and |z − z0 | < δ, then we may choose w such that |F (z) − F (w)| <  and |w − z0 | < δ. Therefore |F (z) − F (z0 )| ≤ |F (z) − F (w)| + |F (w) − F (z0 )| < 2, and the lemma is established. We may now complete the proof of the theorem. We have shown that F extends to a continuous function from D to P . The previous argument can be applied to the inverse G of F . Indeed, the key geometric property of the unit disc that we used was that if z0 belongs to the boundary of D, and C is any small circle centered at z0 , then C ∩ D consists of an arc. Clearly, this property also holds at every boundary point of the

4. Conformal mappings onto polygons

241

polygonal region P . Therefore, G also extends to a continuous function from P to D. It suffices to now prove that the extensions of F and G are inverses of each other. If z ∈ ∂D and {zk } is a sequence in the disc that converges to z, then G(F (zk )) = zk , so after taking the limit and using the fact that F is continuous, we conclude that G(F (z)) = z for all z ∈ D. Similarly, F (G(w)) = w for all w ∈ P , and the theorem is proved. The circle of ideas used in this proof can be used to prove more general theorems on the boundary continuity of conformal maps. See Exercise 18 and Problem 6 below. 4.4 The mapping formula Suppose P is a polygonal region bounded by a polygon p whose vertices are ordered consecutively a1 , a2 , . . . , an , and with n ≥ 3. We denote by παk the interior angle at ak , and define the exterior n angle πβk by αk + βk = 1. A simple geometric argument provides k=1 βk = 2. We shall consider conformal mappings of the half-plane H to P , and make use of the results of the previous section regarding conformal maps from the disc D to P . The standard correspondences w = (i − z)/(i + z), z = i(1 − w)/(1 + w) allows us to go back and forth between z ∈ H and w ∈ D. Notice that the boundary point w = −1 of the circle corresponds to the point at infinity on the line, and so the conformal map of H to D extends to a continuous bijection of the boundary of H, which for the purpose of this discussion includes the point at infinity. Let F be a conformal map from H to P . (Its existence is guaranteed by the Riemann mapping theorem and the previous discussion.) We assume first that none of the vertices of p correspond to the point at infinity. Therefore, there are real numbers A1 , A2 , . . . , An so that F (Ak ) = ak for all k. Since F is continuous and injective, and the vertices are numbered consecutively, we may conclude that the Ak ’s are in either increasing or decreasing order. After relabeling the vertices ak and the points Ak , we may assume that A1 < A2 < · · · < An . These points divide the real line into n − 1 segments [Ak , Ak+1 ], 1 ≤ k ≤ n − 1, and the segment that consists of the join of the two half-segments (−∞, A1 ] ∪ [An , ∞). These are mapped bijectively onto the corresponding sides of the polygon, that is, the segments [ak , ak+1 ], 1 ≤ k ≤ n − 1, and [an , a1 ] (see Figure 8). Theorem 4.6 There exist complex numbers c1 and c2 so that the conformal map F of H to P is given by F (z) = c1 S(z) + c2 where S is the Schwarz-Christoffel integral introduced in Section 4.2.

242

Chapter 8. CONFORMAL MAPPINGS

H

P ak+1

A1

Ak−1Ak

Ak+1

An

ak−1 παk πβk

ak

Figure 8. The mapping F

Proof. We first consider z in the upper half-plane lying above the two adjacent segments [Ak−1 , Ak ] and [Ak , Ak+1 ], where 1 < k < n. We note that F maps these two segments to two segments that intersect at ak = F (Ak ) at an angle παk . By choosing a branch of the logarithm we can in turn define hk (z) = (F (z) − ak )1/αk for all z in the half-strip in the upper half-plane bounded by the lines Re(z) = Ak−1 and Re(z) = Ak+1 . Since F continues to the boundary of H, the map hk is actually continuous up to the segment (Ak−1 , Ak+1 ) on the real line. By construction hk will map the segment [Ak−1 , Ak+1 ] to a (straight) segment Lk in the complex plane, with Ak mapped to 0. We may therefore apply the Schwarz reflection principle to see that hk is analytically continuable to a holomorphic function in the two-way infinite strip Ak−1 < Re(z) < Ak+1 (see Figure 9). We claim that hk never vanishes in that strip. First, if z belongs to the open upper halfstrip, then h (z) F  (z) = αk k , F (z) − F (Ak ) hk (z) and since F is conformal, we have F  (z) = 0 so hk (z) = 0 (Proposition 1.1). By reflection, this also holds in the lower half-strip, and it remains to investigate points on the segment (Ak−1 , Ak+1 ). If Ak−1 < x < Ak+1 , we note that the image under hk of a small half-disc centered at x and contained in H lies on one side of the straight line segment

243

4. Conformal mappings onto polygons

Ak−1

Ak

Ak+1

Figure 9. Schwarz reflection

Lk . Since hk is injective up to Lk (because F is) the symmetry in the Schwarz reflection principle guarantees that hk is injective in the whole disc centered at x, whence hk (x) = 0, whence hk (z) = 0 for all z in the strip Ak−1 < Re(z) < Ak+1 . k  Now because F  = αk h−β hk and F  = −βk αk hk−βk −1 (hk )2 + k −βk  αk hk hk , the fact that hk (z) = 0 implies that

−βk F  (z) = + Ek (z),  F (z) z − Ak where Ek is holomorphic in the strip Ak−1 < Re(z) < Ak+1 . A similar result holds for k = 1 and k = n, namely β1 F  (z) =− + E1 ,  F (z) z − A1 where E1 is holomorphic in the strip −∞ < Re(z) < A2 , and F  (z) βn + En , =− F  (z) z − An where En is holomorphic in the strip An−1 < Re(z) < ∞. Finally, another application of the reflection principle shows that F is continuable in the exterior of a disc |z| ≤ R, for large R (say R > max1≤k≤n |Ak |). Indeed, we may continue F across the union of the segments

244

Chapter 8. CONFORMAL MAPPINGS

(−∞, A1 ) ∪ (An , ∞) since their image under F is a straight line segment and Schwarz reflection applies. The fact that F maps the upper half-plane to a bounded region shows that the analytic continuation of F outside a large disc is also bounded, and hence holomorphic at infinity. Thus F  /F  is holomorphic at infinity and we claim that it goes to 0 as |z| → ∞. Indeed, we may expand F at z = ∞ as F (z) = c0 +

c2 c1 + 2 +··· . z z

This after differentiation shows that F  /F  decays like 1/z as |z| becomes large, and proves our claim. Altogether then, because the various strips overlap and cover the entire complex plane, F  (z)  βk + F  (z) z − Ak n

k=1

is holomorphic in the entire plane and vanishes at infinity; thus, by Liouville’s theorem it is zero. Hence

 βk F  (z) = − . F  (z) z − Ak n

k=1

From this we contend that F  (z) = c(z − A1 )−β1 · · · (z − An )−βn . Indeed, denoting this product by Q(z), we have

 βk Q (z) =− . Q(z) z − Ak n

k=1

Therefore d dz



F  (z) Q(z)

 = 0,

which proves the contention. A final integration yields the theorem. We may now withdraw the hypothesis we made at the beginning that F did not map the point at infinity to a vertex of P , and obtain a formula for that case as well. Theorem 4.7 If F is a conformal map from the upper half-plane to the polygonal region P and maps the points A1 , . . . , An−1 , ∞ to the vertices of p, then there exist constants C1 and C2 such that
z dζ + C2 . F (z) = C1 β βn−1 1 0 (ζ − A1 ) · · · (ζ − An−1 )

245

4. Conformal mappings onto polygons

In other words, the formula is obtained by deleting the last term in the Schwarz-Christoffel integral (5). Proof. After a preliminary translation, we may assume that Aj = 0 for j = 1, . . . , n − 1. Choose a point A∗n > 0 on the real line, and consider the fractional linear map defined by 1 Φ(z) = A∗n − . z Then Φ is an automorphism of the upper half-plane. Let A∗k = Φ(Ak ) for k = 1, . . . , n − 1, and note that A∗n = Φ(∞). Then (F ◦ Φ−1 )(A∗k ) = ak

for all k = 1, 2, . . . , n.

We can now apply the Schwarz-Christoffel formula just proved to find that (F ◦ Φ

−1






z

)(z ) = C1

(ζ −

0

A∗1 )β1

dζ + C2 . · · · (ζ − A∗n )βn

The change of variables ζ = Φ(w) satisfies dζ = dw/w2 , and since we can write 2 = β1 + · · · + βn , we obtain (F ◦ Φ

−1






Φ−1 (z  ) dw βn−1 ∗ β1 ···(w(A∗ −A∗ (w(A∗ n −A1 )−1) n n−1 )−1)

)(z ) = C1 =

C1




0

+ C2

Φ−1 (z  ) dw βn−1 ∗ β1 ···(w−1/(A∗ −A∗ (w−1/(A∗ n −A1 )) n n−1 ))

0

+ C2 .

Finally, we note that 1/(A∗n − A∗k ) = Ak and set Φ−1 (z  ) = z in the above equation to conclude that F (z) =

C1


0

z

(w − A1

)β 1

dw + C2 , · · · (w − An−1 )βn−1

as was to be shown. 4.5 Return to elliptic integrals We consider again the elliptic integral




z

I(z) = 0

[(1 −

dζ − k2 ζ 2 )]1/2

ζ 2 )(1

with 0 < k < 1,

246

Chapter 8. CONFORMAL MAPPINGS

which arose in Example 3 of Section 4.1. We saw that it mapped the real axis to the rectangle R with vertices −K, K, K + iK  , and −K + iK  . We will now see that this mapping is a conformal mapping of H to the interior of R. According to Theorem 4.6 there is a conformal map F to the rectangle, that maps four points on the real axis to the vertices of R. By preceding this map with a suitable automorphism of H we may assume that F maps −1, 0, 1 to −K, 0, K, respectively. Indeed, by using a preliminary automorphism, we may assume that −K, 0, K are the images of points A1 , 0, A2 with A1 < 0 < A2 ; then we can further take A1 = −1 and A2 = 1. See Exercise 15. Next, let  be chosen with 0 <  < 1, so that 1/ is the point on the real line mapped by F to the vertex K + iK  , which is the vertex next in order after −K and K. We claim that F (−1/) is the vertex −K + iK  . Indeed, if F ∗ (z) = −F (−z), then by the symmetry of R, F ∗ is also a conformal map of H to R; moreover F ∗ (0) = 0, and F ∗ (±1) = ±K. Thus F −1 ◦ F ∗ is an automorphism of H that fixes the points −1, 0, and 1. Hence F −1 ◦ F ∗ is the identity (see Exercise 15), and F = F ∗ , from which it follows that F (−1/) = −F (1/) = −K + iK  . Therefore, by Theorem 4.6
F (z) = c1

z

[(1 −

0

dζ + c2 . − 2 ζ 2 )]1/2

ζ 2 )(1

Setting z = 0 gives c2 = 0, and letting z = 1, z = 1/, yields K(k) = c1 K() where




1

K(k) = 0

K  (k) =


1

and

[(1 −

1/k

K  (k) = c1 K  (),

dx , − k2 x2 )]1/2

x2 )(1

dx . [(x2 − 1)(1 − k2 x2 )]1/2

Now K(k) is clearly strictly increasing as k varies in (0, 1). Moreover, a change of variables (Exercise 24) establishes the identity ˜ K  (k) = K(k)

where k˜2 = 1 − k2 and k˜ > 0,

247

4. Conformal mappings onto polygons

and this shows that K  (k) is strictly decreasing. Hence K(k)/K  (k) is strictly increasing. Since K(k)/K  (k) = K()/K  (), we must have k = , and finally c1 = 1. This shows that I(z) = F (z), and hence I is conformal, as was to be proved. A final observation is of significance. A basic insight into elliptic integrals is obtained by passing to their inverse functions. We therefore consider z → sn(z), the inverse map of z → I(z).9 It transforms the closed rectangle into the closed upper half-plane. Now consider the series of rectangles R = R0 , R1 , R2 , . . . gotten by reflecting successively along the lower sides (Figure 10). K + iK  R0 K R1 K − iK  R2 K − 2iK 

Figure 10. Reflections of R = R0

With sn(z) defined in R0 , we can by the reflection principle extend it to R1 by setting sn(z) = sn(z) whenever z ∈ R1 (note that then z ∈ R0 ). Next we can extend sn(z) to R2 by setting sn(z) = sn(−iK  + z) if z ∈ R2 and noting that if z ∈ R2 , then −iK  + z ∈ R1 . Combining these reflections and continuing this way we see that we can extend sn(z) in the entire strip −K < Re(z) < K, so that sn(z) = sn(z + 2iK  ). Similarly, by reflecting in a series of horizontal rectangles, and combining these with the previous reflections, we see that sn(z) can be continued to the complex plane and also satisfies sn(z) = sn(z + 4K). Thus sn(z) is doubly periodic (with periods 4K and 2iK  ). A further examination 9 The notation sn(z) in somewhat different form is due to Jacobi, and was adopted because of the analogy with sin z.

248

Chapter 8. CONFORMAL MAPPINGS

shows that the only singularities sn(z) are poles. Functions of this type, called “elliptic functions,” are the subject of the next chapter.

5 Exercises 1. A holomorphic mapping f : U → V is a local bijection on U if for every z ∈ U there exists an open disc D ⊂ U centered at z, so that f : D → f (D) is a bijection. Prove that a holomorphic map f : U → V is a local bijection on U if and only if f  (z) = 0 for all z ∈ U . [Hint: Use Rouch´e’s theorem as in the proof of Proposition 1.1.] 2. Supppose F (z) is holomorphic near z = z0 and F (z0 ) = F  (z0 ) = 0, while F  (z0 ) = 0. Show that there are two curves Γ1 and Γ2 that pass through z0 , are orthogonal at z0 , and so that F restricted to Γ1 is real and has a minimum at z0 , while F restricted to Γ2 is also real but has a maximum at z0 . [Hint: Write F (z) = (g(z))2 for z near z0 , and consider the mapping z → g(z) and its inverse.] 3. Suppose U and V are conformally equivalent. Prove that if U is simply connected, then so is V . Note that this conclusion remains valid if we merely assume that there exists a continuous bijection between U and V . 4. Does there exist a holomorphic surjection from the unit disc to C? [Hint: Move the upper half-plane “down” and then square it to get C.] 5. Prove that f (z) = − 12 (z + 1/z) is a conformal map from the half-disc {z = x + iy : |z| < 1, y > 0} to the upper half-plane. [Hint: The equation f (z) = w reduces to the quadratic equation z 2 + 2wz + 1 = 0, which has two distinct roots in C whenever w = ±1. This is certainly the case if w ∈ H.] 6. Give another proof of Lemma 1.3 by showing directly that the Laplacian of u ◦ F is zero. [Hint: The real and imaginary parts of F satisfy the Cauchy-Riemann equations.] 7. Provide all the details in the proof of the formula for the solution of the Dirichlet problem in a strip discussed in Section 1.3. Recall that it suffices to compute the solution at the points z = iy with 0 < y < 1. (a) Show that if reiθ = G(iy), then reiθ = i

cos πy . 1 + sin πy

This leads to two separate cases: either 0 < y ≤ 1/2 and θ = π/2, or 1/2 ≤

249

5. Exercises

y < 1 and θ = −π/2. In either case, show that r2 =

1 − sin πy 1 + sin πy

Pr (θ − ϕ) =

and

sin πy . 1 − cos πy sin ϕ

π 1 Pr (θ − ϕ)f˜0 (ϕ) dϕ make the change of variables t = (b) In the integral 2π 0 iϕ F (e ). Observe that i − eπt , i + eπt

eiϕ =

and then take the imaginary part and differentiate both sides to establish the two identities sin ϕ =

1 cosh πt

dϕ π = . dt cosh πt

and

Hence deduce that 1 2π




π


π 1 sin πy f˜0 (ϕ) dϕ 2π 0 1 − cos πy sin ϕ
f0 (t) sin πy ∞ dt. = 2 cosh πt − cos πy −∞

Pr (θ − ϕ)f˜0 (ϕ) dϕ =

0

(c) Use a 0 1 2π

similar argument P (θ − ϕ)f˜1 (ϕ) dϕ. r −π

to prove

the

formula

for

the

integral

8. Find a harmonic function u in the open first quadrant that extends continuously up to the boundary except at the points 0 and 1, and that takes on the following boundary values: u(x, y) = 1 on the half-lines {y = 0, x > 1} and {x = 0, y > 0}, and u(x, y) = 0 on the segment {0 < x < 1, y = 0}. [Hint: Find conformal maps F1 , F2 , . . . , F5 indicated in Figure 11. Note that 1 arg(z) is harmonic on the upper half-plane, equals 0 on the positive real axis, π and 1 on the negative real axis.] 9. Prove that the function u defined by  u(x, y) = Re

i+z i−z

 and

u(0, 1) = 0

is harmonic in the unit disc and vanishes on its boundary. Note that u is not bounded in D. 10. Let F : H → C be a holomorphic function that satisfies |F (z)| ≤ 1

and

F (i) = 0.

250

Chapter 8. CONFORMAL MAPPINGS u=1

F1 u=1

0 u=01 u=1

−1

u=0

u=1 1 F2

u=0 u=0

u=1

iπ

F3 u=1

− π2

u=1

u=1

π 2

0

F4

F5

u=1

0

u=1

1

u=0

u=1

0

u=0

Figure 11. Successive conformal maps in Exercise 8

Prove that

  z − i  |F (z)| ≤  z + i

for all z ∈ H.

11. Show that if f : D(0, R) → C is holomorphic, with |f (z)| ≤ M for some M > 0, then    f (z) − f (0)  |z|   .  ≤  M 2 − f (0)f (z)  MR [Hint: Use the Schwarz lemma.] 12. A complex number w ∈ D is a fixed point for the map f : D → D if f (w) = w. (a) Prove that if f : D → D is analytic and has two distinct fixed points, then f is the identity, that is, f (z) = z for all z ∈ D.

251

5. Exercises

(b) Must every holomorphic function f : D → D have a fixed point? [Hint: Consider the upper half-plane.]

13. The pseudo-hyperbolic distance between two points z, w ∈ D is defined by    z−w   . ρ(z, w) =  1 − wz  (a) Prove that if f : D → D is holomorphic, then ρ(f (z), f (w)) ≤ ρ(z, w)

for all z, w ∈ D.

Moreover, prove that if f is an automorphism of D then f preserves the pseudo-hyperbolic distance ρ(f (z), f (w)) = ρ(z, w)

for all z, w ∈ D.

[Hint: Consider the automorphism ψα (z) = (z − α)/(1 − αz) and apply the −1 .] Schwarz lemma to ψf (w) ◦ f ◦ ψw (b) Prove that 1 |f  (z)| ≤ 1 − |f (z)|2 1 − |z|2

for all z ∈ D.

This result is called the Schwarz-Pick lemma. See Problem 3 for an important application of this lemma.

14. Prove that all conformal mappings from the upper half-plane H to the unit disc D take the form eiθ

z−β , z−β

θ ∈ R and β ∈ H.

15. Here are two properties enjoyed by automorphisms of the upper half-plane. (a) Suppose Φ is an automorphism of H that fixes three distinct points on the real axis. Then Φ is the identity. (b) Suppose (x1 , x2 , x3 ) and (y1 , y2 , y3 ) are two pairs of three distinct points on the real axis with x1 < x2 < x3

and

y1 < y2 < y3 .

Prove that there exists (a unique) automorphism Φ of H so that Φ(xj ) = yj , j = 1, 2, 3. The same conclusion holds if y3 < y1 < y2 or y2 < y3 < y1 .

252

Chapter 8. CONFORMAL MAPPINGS

16. Let f (z) =

i−z i+z

f −1 (w) = i

and

1−w . 1+w

(a) Given θ ∈ R, find real numbers a, b, c, d such that ad − bc = 1, and so that for any z ∈ H

 az + b = f −1 eiθ f (z) . cz + d (b) Given α ∈ D find real numbers a, b, c, d so that ad − bc = 1, and so that for any z ∈ H az + b = f −1 (ψα (f (z))) , cz + d with ψα defined in Section 2.1. (c) Prove that if g is an automorphism of the unit disc, then there exist real numbers a, b, c, d such that ad − bc = 1 and so that for any z ∈ H az + b = f −1 ◦ g ◦ f (z). cz + d [Hint: Use parts (a) and (b).]

17. If ψα (z) = (α − z)/(1 − αz) for |α| < 1, prove that 1 π



D

|ψα |2 dxdy = 1

and

1 π



D

|ψα | dxdy =

1 − |α|2 1 log , |α|2 1 − |α|2

where in the case α = 0 the expression on the right is understood as the limit as |α| → 0. [Hint: The first integral can be evaluated without a calculation. For the second, use polar coordinates, and for each fixed r use contour integration to evaluate the integral in θ.] 18. Suppose that Ω is a simply connected domain that is bounded by a piecewisesmooth closed curve γ (in the terminology of Chapter 1). Then any conformal map F of D to Ω extends to a continuous bijection of D to Ω. The proof is simply a generalization of the argument used in Theorem 4.2. 19. Prove that the complex plane slit along the union of the rays ∪n k=1 {Ak + iy : y ≤ 0} is simply connected. [Hint: Given a curve, first “raise” it so that it is completely contained in the upper half-plane.] 20. Other examples of elliptic integrals providing conformal maps from the upper half-plane to rectangles are given below.

253

5. Exercises

(a) The function


z 0



dζ , ζ(ζ − 1)(ζ − λ)

with λ ∈ R and λ = 1

maps the upper half-plane conformally to a rectangle, one of whose vertices is the image of the point at infinity. (b) In the case λ = −1, the image of


z



0

is a square whose side lengths are

dζ ζ(ζ 2 − 1)

Γ2 (1/4) √ . 2 2π

21. We consider conformal mappings to triangles. (a) Show that


z

z −β1 (1 − z)−β2 dz,

0

with 0 < β1 < 1, 0 < β2 < 1, and 1 < β1 + β2 < 2, maps H to a triangle whose vertices are the images of 0, 1, and ∞, and with angles α1 π, α2 π, and α3 π, where αj + βj = 1 and β1 + β2 + β3 = 2. (b) What happens when β1 + β2 = 1? (c) What happens when 0 < β1 + β2 < 1? (d) In (a), the length of the side of the triangle opposite angle αj π is sin(αj π) Γ(α1 )Γ(α2 )Γ(α3 ). π 22. If P is a simply connected region bounded by a polygon with vertices a1 , . . . , an and angles α1 π, . . . , αn π, and F is a conformal map of the disc D to P , then there exist complex numbers B1 , . . . , Bn on the unit circle, and constants c1 and c2 so that
z dζ + c2 . F (z) = c1 β1 · · · (ζ − B )βn (ζ − B ) 1 n 1 [Hint: This follows from the standard correspondence between H and D and an argument similar to that used in the proof of Theorem 4.7.] 23. If




z

F (z) = 1

dζ , (1 − ζ n )2/n

254

Chapter 8. CONFORMAL MAPPINGS

then F maps the unit disc conformally onto the interior of a regular polygon with n sides and perimeter
π n−2 (sin θ)−2/n dθ. 2 n 0

24. The elliptic integrals K and K  defined for 0 < k < 1 by


1

K(k) = 0

dx ((1 − x2 )(1 − k2 x2 ))1/2

and






1/k

K (k) = 1

dx ((x2 − 1)(1 − k2 x2 ))1/2

satisfy various interesting identities. For instance: ˜ 2 = 1 − k2 and 0 < k ˜ < 1, then (a) Show that if k ˜ K  (k) = K(k). ˜2 y 2 )−1/2 in the integral defining K  (k).] [Hint: Change variables x = (1 − k ˜ 2 = 1 − k2 , and 0 < k ˜ < 1, then (b) Prove that if k 2 K K(k) = ˜ 1+k



˜ 1−k ˜ 1+k

 .

˜ + (1 − k)t ˜ 2 ).] [Hint: Change variables x = 2t/(1 + k (c) Show that for 0 < k < 1 one has K(k) =

π F (1/2, 1/2, 1; k2 ), 2

where F the hypergeometric series. [Hint: This follows from the integral representation for F given in Exercise 9, Chapter 6.]

6 Problems 1. Let f be a complex-valued C 1 function defined in the neighborhood of a point z0 . There are several notions closely related to conformality at z0 . We say that f is isogonal at z0 if whenever γ(t) and η(t) are two smooth curves with γ(0) = η(0) = z0 , that make an angle θ there (|θ| < π), then f (γ(t)) and f (η(t)) make an angle of θ at t = 0 with |θ | = |θ| for all θ. Also, f is said to be isotropic if it magnifies lengths by some factor for all directions emanating from z0 , that is, if the limit lim

r→0

|f (z0 + reiθ ) − f (z0 )| r

exists, is non-zero, and independent of θ.

255

6. Problems

Then f is isogonal at z0 if and only if it is isotropic at z0 ; moreover, f is isogonal at z0 if and only if either f  (z0 ) exists and is non-zero, or the same holds for f replaced by f . 2. The angle between two non-zero complex numbers z and w (taken in that order) is simply the oriented angle, in (−π, π], that is formed between the two vectors in R2 corresponding to the points z and w. This oriented angle, say α, is uniquely determined by the two quantities (z, w) |z| |w|

and

(z, −iw) |z| |w|

which are simply the cosine and sine of α, respectively. Here, the notation (·, ·) corresponds to the usual Euclidian inner product in R2 , which in terms of complex numbers takes the form (z, w) = Re(zw). In particular, we may now consider two smooth curves γ : [a, b] → C and η : [a, b] → C, that intersect at z0 , say γ(t0 ) = η(t0 ) = z0 , for some t0 ∈ (a, b). If the quantities γ  (t0 ) and η  (t0 ) are non-zero, then they represent the tangents to the curves γ and η at the point z0 , and we say that the two curves intersect at z0 at the angle formed by the two vectors γ  (t0 ) and η  (t0 ). A holomorphic function f defined near z0 is said to preserve angles at z0 if for any two smooth curves γ and η intersecting at z0 , the angle formed between the curves γ and η at z0 equals the angle formed between the curves f ◦ γ and f ◦ η at f (z0 ). (See Figure 12 for an illustration.) In particular, we assume that the tangents to the curves γ, η, f ◦ γ, and f ◦ η at the point z0 and f (z0 ) are all non-zero.

η z0 γ

f (z0 ) f ◦η

f ◦γ

Figure 12. Preservation of angles at z0

(a) Prove that if f : Ω → C is holomorphic, and f  (z0 ) = 0, then f preserves angles at z0 . [Hint: Observe that (f  (z0 )γ  (t0 ), f  (z0 )η  (t0 )) = |f  (z0 )|2 (γ  (t0 ), η  (t0 )).] (b) Conversely, prove the following: suppose f : Ω → C is a complex-valued function, that is real-differentiable at z0 ∈ Ω, and Jf (z0 ) = 0. If f preserves angles at z0 , then f is holomorphic at z0 with f  (z0 ) = 0.

256

Chapter 8. CONFORMAL MAPPINGS

3.∗ The Schwarz-Pick lemma (see Exercise 13) is the infinitesimal version of an important observation in complex analysis and geometry. For complex numbers w ∈ C and z ∈ D we define the hyperbolic length of w at z by wz =

|w| , 1 − |z|2

where |w| and |z| denote the usual absolute values. This length is sometimes referred to as the Poincar´ e metric, and as a Riemann metric it is written as ds2 =

|dz|2 . (1 − |z|2 )2

The idea is to think of w as a vector lying in the tangent space at z. Observe that for a fixed w, its hyperbolic length grows to infinity as z approaches the boundary of the disc. We pass from the infinitesimal hyperbolic length of tangent vectors to the global hyperbolic distance between two points by integration. (a) Given two complex numbers z1 and z2 in the disc, we define the hyperbolic distance between them by


1

d(z1 , z2 ) = inf γ

γ  (t)γ(t) dt,

0

where the infimum is taken over all smooth curves γ : [0, 1] → D joining z1 and z2 . Use the Schwarz-Pick lemma to prove that if f : D → D is holomorphic, then d(f (z1 ), f (z2 )) ≤ d(z1 , z2 )

for any z1 , z2 ∈ D.

In other words, holomorphic functions are distance-decreasing in the hyperbolic metric. (b) Prove that automorphisms of the unit disc preserve the hyperbolic distance, namely d(ϕ(z1 ), ϕ(z2 )) = d(z1 , z2 ),

for any z1 , z2 ∈ D

and any automorphism ϕ. Conversely, if ϕ : D → D preserves the hyperbolic distance, then either ϕ or ϕ is an automorphism of D. (c) Given two points z1 , z2 ∈ D, show that there exists an automorphism ϕ such that ϕ(z1 ) = 0 and ϕ(z2 ) = s for some s on the segment [0, 1) on the real line. (d) Prove that the hyperbolic distance between 0 and s ∈ [0, 1) is d(0, s) =

1+s 1 log . 2 1−s

257

6. Problems

(e) Find a formula for the hyperbolic distance between any two points in the unit disc.

4.∗ Consider the group of matrices of the form 

a c

M=



b d

,

that satisfy the following conditions: (i) a, b, c, and d ∈ C, (ii) the determinant of M is equal to 1, (iii) the matrix M preserves the following hermitian form on C2 × C2 : Z, W  = z1 w1 − z2 w2 , where Z = (z1 , z2 ) and W = (w1 , w2 ). In other words, for all Z, W ∈ C2 M Z, M W  = Z, W . This group of matrices is denoted by SU(1, 1). (a) Prove that all matrices in SU(1, 1) are of the form 

a b

b a

 ,

where |a|2 − |b|2 = 1. To do so, consider the matrix  J=

1 0

0 −1

 ,

and observe that Z, W  = tW JZ, where tW denotes the conjugate transpose of W . (b) To every matrix in SU(1, 1) we can associate a fractional linear transformation az + b . cz + d Prove that the group SU(1, 1)/{±1} is isomorphic to the group of automorphisms of the disc. [Hint: Use the following association.]

e

2iθ

 iθ √e 2 z−α 1−|α| −→  −iθ 1 − αz − √αe 2 1−|α|

iθ

− √αe

1−|α|2

−iθ

√e

1−|α|2

 .

258

Chapter 8. CONFORMAL MAPPINGS

5. The following result is relevant to Problem 4 in Chapter 10 which treats modular functions. (a) Suppose that F : H → C is holomorphic and bounded. Also, suppose that ., where {rn } is a bounded seF (z) vanishes when z = irn , n = 1, 2, 3, . . quence of positive numbers. Prove that if ∞ n=1 rn = ∞, then F = 0.  (b) If rn < ∞, it is possible to construct a bounded function on the upper half-plane with zeros precisely at the points irn . For related results in the unit disc, see Problems 1 and 2 in Chapter 5.] 6.∗ The results of Exercise 18 extend to the case when γ is assumed merely to be closed, simple, and continuous. The proof, however, requires further ideas. 7.∗ Applying ideas of Carath´eodory, Koebe gave a proof of the Riemann mapping theorem by constructing (more explicitly) a sequence of functions that converges to the desired conformal map. Starting with a Koebe domain, that is, a simply connected domain K0 ⊂ D that is not all of D, and which contains the origin, the strategy is to find an injective function f0 such that f0 (K0 ) = K1 is a Koebe domain “larger” than K0 . Then, one iterates this process, finally obtaining functions Fn = fn ◦ · · · ◦ f0 : K0 → D such that Fn (K0 ) = Kn+1 and lim Fn = F is a conformal map from K0 to D. The inner radius of a region K ⊂ D that contains the origin is defined by rK = sup{ρ ≥ 0 : D(0, ρ) ⊂ K}. Also, a holomorphic injection f : K → D is said to be an expansion if f (0) = 0 and |f (z)| > |z| for all z ∈ K − {0}. (a) Prove that if f is an expansion, then rf (K) ≥ rK and |f  (0)| > 1. [Hint: Write f (z) = zg(z) and use the maximum principle to prove that |f  (0)| = |g(0)| > 1.] Suppose we begin with a Koebe domain K0 and a sequence of expansions {f0 , f1 , . . . , fn , . . .}, so that Kn+1 = fn (Kn ) are also Koebe domains. We then define holomorphic maps Fn : K0 → D by Fn = fn ◦ · · · ◦ f0 . (b) Prove that  for  each n, the function Fn is an expansion. Moreover, Fn (0) = n k=0 fk (0), and conclude that limn→∞ |fn (0)| = 1. [Hint: Prove that the sequence {|Fn (0)|} has a limit by showing that it is bounded above and monotone increasing. Use the Schwarz lemma.] (c) Show that if the sequence is osculating, that is, rKn → 1 as n → ∞, then {Fn } converges uniformly on compact subsets of K0 to a conformal map F : K0 → D. [Hint: If rF (K0 ) ≥ 1 then F is surjective.] To construct the desired osculating sequence we shall use the automorphisms ψα = (α − z)/(1 − αz). (d) Given a Koebe domain K, choose a point α ∈ D on the boundary of K such that |α| = rK , and also choose β ∈ D such that β 2 = α. Let S denote the square root of ψα on K such that S(0) = 0. Why is such a function well defined? Prove that the function f : K → D defined by f (z) = ψβ ◦ S ◦ ψα

259

6. Problems

√ is an expansion. Moreover, show that |f  (0)| = (1 + rK )/2 rK . [Hint: To prove that |f (z)| > |z| on K − {0} apply the Schwarz lemma to the inverse function, namely ψα ◦ g ◦ ψβ where g(z) = z 2 .] (e) Use part (d) to construct the desired sequence.

8.∗ Let f be an injective holomorphic function in the unit disc, with f (0) = 0 and f  (0) = 1. If we write f (z) = z + a2 z 2 + a3 z 3 · · · , then Problem 1 in Chapter 3 shows that |a2 | ≤ 2. Bieberbach conjectured that in fact |an | ≤ n for all n ≥ 2; this was proved by deBranges. This problem outlines an argument to prove the conjecture under the additional assumption that the coefficients an are real. (a) Let z = reiθ with 0 < r < 1, and show that if v(r, θ) denotes the imaginary part of f (reiθ ), then an r n =

2 π




π

v(r, θ) sin nθ dθ. 0

(b) Show that for 0 ≤ θ ≤ π and n = 1, 2, . . . we have | sin nθ| ≤ n sin θ. (c) Use the fact that an ∈ R to show that f (D) is symmetric with respect to the real axis, and use this fact to show that f maps the upper half-disc into either the upper or lower part of f (D). (d) Show that for r small, v(r, θ) = r sin θ[1 + O(r)], and use the previous part to conclude that v(r, θ) sin θ ≥ 0 for all 0 < r < 1 and 0 ≤ θ ≤ π. (e) Prove that |an r n | ≤ nr, and let r → 1 to conclude that |an | ≤ n. (f) Check that the function f (z) = z/(1 − z)2 satisfies all the hypotheses and that |an | = n for all n. 9.∗ Gauss found a connection between elliptic integrals and the familiar operations of forming arithmetic and geometric means. We start with any pair (a, b) of numbers that satisfy a ≥ b > 0, and form the arithmetic and geometric means of a and b, that is, a1 =

a+b 2

and

b1 = (ab)1/2 .

We then repeat these operations with a and b replaced by a1 and b1 . Iterating this process provides two sequences {an } and {bn } where an+1 and bn+1 are the arithmetic and geometric means of an and bn , respectively.

260

Chapter 8. CONFORMAL MAPPINGS

(a) Prove that the two sequences {an } and {bn } have a common limit. This limit, which we denote by M (a, b), is called the arithmetic-geometric mean of a and b. [Hint: Show that a ≥ a1 ≥ a2 ≥ · · · ≥ an ≥ bn ≥ · · · ≥ b1 ≥ b and an − bn ≤ (a − b)/2n .] (b) Gauss’s identity states that 1 2 = M (a, b) π




π/2 0

dθ . (a2 cos2 θ + b2 sin2 θ)1/2

To prove this relation, show that if I(a, b) denotes the integral on the righthand side, then it suffices to establish the invariance of I, namely  (6)

I(a, b) = I

 a+b 1/2 . , (ab) 2

Then, observe that the connection with elliptic integrals takes the form I(a, b) =

1 1 K(k) = a a




1 0



dx (1 − x2 )(1 − k2 x2 )

where k2 = 1 − b2 /a2 ,

and that the relation (6) is a consequence of the identity in (b) of Exercise 24.

9 An Introduction to Elliptic Functions The form that Jacobi had given to the theory of elliptic functions was far from perfection; its flaws are obvious. At the base we find three fundamental functions sn, cn and dn. These functions do not have the same periods... In Weierstrass’ system, instead of three fundamental functions, there is only one, ℘(u), and it is the simplest of all having the same periods. It has only one double infinity; and finally its definition is so that it does not change when one replaces one system of periods by another equivalent system. H. Poincar´e, 1899

The th