The Logarithmic Integral: Volume 1 (Cambridge Studies in Advanced Mathematics) (v. 1)

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CAMBRIDGE STUDIES IN ADVANCED MATHEMATICS 12 EDITORIAL BOARD D.J.H. GARLING, W. FULTON, T. TOM DIECK, P. WALTERS

THE LOGARITHMIC INTEGRAL I

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W.M.L. Holcombe Algebraic automata theory K. Petersen Ergodic theory P.T. Johnstone Stone spaces W.H. Schikhof Ultrametric calculus J: P. Kahane Some random series of functions, 2nd edition H. Cohn Introduction to the construction of class fields J. Lambek & P.J. Scott Introduction to higher-order categorical logic H. Matsumura Commutative ring theory C.B. Thomas Characteristic classes and the cohomology of finite groups M. Aschbacher Finite group theory J.L. Alperin Local representation theory P. Koosis The logarithmic integral 1 A. Pietsch Eigenvalues and s-numbers S.J. Patterson An introduction to the theory of theRiemann zeta-function H.J. Baues Algebraic homotopy V.S. Varadarajan Introduction to harmonic analysis on semisimple Lie groups W. Dicks & M. Dunwoody Groups acting on graphs L.J. Corwin & F.P. Greenleaf Representations of nilpotent Lie groups and their applications R. Fritsch & R. Piccinini Cellular structures in topology H. Klingen Introductory lectures on Siegel modular forms P. Koosis The logarithmic integral 11 M.J. Collins Representations and characters of finite groups H. Kunita Stochastic flows and stochastic differential equations P. Wojtaszczyk Banach spaces for analysts J.E. Gilbert& M.A.M. Murray Clifford algebras and Dirac operators in harmonic analysis A. Frohlich & M.J. Taylor Algebraic number theory K. Goebel & W.A. Kirk Topics in met ric fixed point theory J.F. Humphreys Reflection groups and Coxeter groups D.J. Benson Representations and cohomology 1 D.J. Benson Representations and cohomology 11 C. Aliday & V. Puppe Cohomological methods in transformation groups C. Soule et al Lectures onArakelov geometry A. Ambrosetti & G. Prodi A primer of nonlinear analysis J. Palis & F. Takens Hyperbolicity, stability and chaos at homoclinic bifurcations M. Auslander, 1. Reiten & S. G. Smale Representation theory ofArtin algebras Y. Meyer Wavelets and operators 1 C. Weibel An introduction to homological algebra W. Bruns & J. Herzog Cohen-Macaulay rings V. Snaith Explicit Brauer induction G. Laumon Cohomology ofDrinfield modular varieties I E.B. Davies Spectral theory and differential operators J. Diestel, H. Jarchow & A. Tonge Absolutely summing operators P. Mattila Geometry of sets and measures in Euclidean spaces R. Pinsky Positive harmonic functions and diffusion G. Tenenbaum Introduction to analytic and probabilistic number theory C. Peskine An algebraic introduction to complex projective geometry I Y. Meyer & R. Coifman Wavelets and operators R. Stanley Enumerative combinatories 1. Porteous Clifford algebras and the classical groups M. Audin Spinning tops V. Jurdjevic Geometric control theory H. Voelklein Groups as Galois groups J. Le Potier Lectures on vector bundles D. Bump Automorphic forms G. Laumon Cohomology ofDrinfeld modular varieties II M. Brodmann & R. Sharp Local cohomology

"0 log M(t) dt -00 1 + t

f

THE LOGARITHMIC INTEGRAL I

PAUL KOOSIS McGill University in Montreal formerly at the University of California, LosAngeles

CAMBRIDGE UNIVERSITY PRESS

PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE

The Pitt Building, Trumpington Street, Cambridge CB21RP, United Kingdom CAMBRIDGE UNIVERSITY PRESS

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www.cambridge.org Information on this title: www.cambridge.org/9780521309066 © Cambridge University Press 1988 This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

First published 1988 First paperback edition (with corrections) 1998 A catalogue record for this book is available from the British Library

Library of Congress Cataloguing in Publication data Koosis, Paul. The logarithmic integral. (Cambridge studies in advanced mathematics; 12) 1. Analytic functions. 2 Harmonic analysis. 3. Integrals, logarithmic. I. Title. II. Series, QA331.K7393 1988 515.4 85-28018 ISBN-13 978-0-521-30906-6 hardback

ISBN-10 0-521-30906-9 hardback ISBN-13 978-0-521-59672-5 paperback ISBN-10 0-521-59672-6 paperback

Transferred to digital printing 2005

Pour le Canada

Notice

In this paperback edition of volume I a number of small errors - and some actual mathematical mistakes - present in the original hard-cover version have been corrected. Many were pointed out to me by Henrik Pedersen, my former student; it was he who observed in particular that the hint given for Problem 28 (b) was ineffective. I wish to express here my gratitude for the considerable service he has thus rendered.

Let me also call the reader's attention to two annoying oversights in volume II. In the statement of the important theorem on p. 65, the condition that the quantities ak all be > 0 was inadvertently omitted. On p. 406 it would be

better, in the last displayed formula, to replace the difference quotient now standing on the right by

µ(x + AX) -µ(x - AX) 2lx

March 22,1997 Outremont, Quebec

Contents

Preface Introduction I

xv xvii

Jensen's formula Problem 1

1

5

II Szego's theorem

7

A The theorem

7

B The pointwise approximate identity property of the Poisson kernel Problem 2

III

10

13

Entire functions of exponential type

A Hadamard factorization Characterization of the set of zeros of an entire function of exponential type. Lindelof's theorems Problem 3 C Phragmbn-Lindelof theorems

15 15

B

D The Paley-Wiener theorem E Introduction to the condition log, If (x)I D

F

1+x2

19

22 23 30 37

dx < o0

Representation of positive harmonic functions as Poisson integrals 1. The representation 2. Digression on the a.e. existence of boundary values

G Return to the subject of §E 1 Functions without zeros in 3z > 0 2 Convergence of f°°.(log -If(x)I/(1+x2))dx 3 Taking the zeros in 3z > 0 into account. Use of Blaschke products Problem 4

39

39 43 47 47 49 52 58

x

Contents

H Levinson's theorem on the density of the zeros 1 Kolmogorov's theorem on the harmonic conjugate 2 Functions with only real zeros 3 The zeros not necessarily real Problem 5 Quasianalyticity Quasianalyticity. Sufficiency of Carleman's criterion 1 Definition of the classes 2 The function T(r). Carleman's criterion B Convex logarithmic regularization of and the necessity of Carleman's criterion 1 Definition of the sequence Its relation to and T(r) 2 Necessity of Carleman's criterion and the characterization of quasianalytic classes C Scholium. Direct establishment of the equivalence between the three conditions

IV A

°° log T(r)

E u

1+r2

dr < oo,

< 00

and F,M _ 1 /M < o0

83 83

89

92

"

Problem 6 D The Paley-Wiener construction of entire functions of small exponential type decreasing fairly rapidly along the real axis

E Theorem of Cartan and Gorny on equality of ''({M}) and an algebra W Problem 7

V The moment problem on the real line A Characterization of moment sequences. Method based on extension of positive linear functionals to be a moment B Scholium. Determinantal criterion for sequence

96 97 102 103

109 110 116

C Determinacy. Two conditions, one sufficient and the other necessary 1 Carleman's sufficient condition 2 A necessary condition Problem 8

126 126 128 131

D M. Riesz' general criterion for indeterminacy 1 The criterion with Riesz' function R(z) 2 Derivation of the results in §C from the above one

132 132 142

VI Weighted approximation on the real line A Mergelian's treatment of weighted polynomial approximation 1 Criterion in terms of finiteness of ((z) 2 A computation

145

3 Criterion in terms of J°°. (log Sl;(t)/(1 + t2) )dt

147 147 150 153

Contents

xi

B Akhiezer's method, based on use of W,(z) 1 Criterion in terms of f "'. (log W (x)/(1 + x2))dx 2 Description of II log W (t)

1+t2

dt <

158 158

II w limits of polynomials when 00

3 Strengthened version of Akhiezer's criterion. Pollard's theorem C Mergelian's criterion really more general in scope than Akhiezer's. Example D Some partial results involving the weight W explicitly. E Weighted approximation by sums of imaginary exponentials with exponents from a finite interval 1 Equivalence with weighted approximation by certain entire function of exponential type. The collection 8. 2 The functions c14(z) and WA(z). Analogues of Mergelian's and Akhiezer's theorems 3 Scholium. P61ya's maximum density 4 The analogue of Pollard's theorem F L. de Branges' description of extremal unit measures orthogonal to the ei2"/ W(x), - A < 2 < A, when 'A is not dense in %w(18) 1 Three lemmas 2 De Branges' theorem 3 Discussion of the theorem 4 Scholium. Krein's functions Problem 9 Problem 10 G Weighted approximation with LP norms H Comparison of weighted approximation by polynomials and by functions in 8A 1 Characterization of the functions in cw(A +) 2 Sufficient conditions for equality of 'w(0) and 'w(0 +) 3 Example of a weight W with 1w(0) ''w(0 +) # W w(1d)

160

163 165

169 171 171

173

175 180

184 187 190 198 203 209 209

210 211

212 219 229

VII How small can the Fourier transform of a rapidly decreasing non-zero function be?

A The Fourier transform vanishes on an interval. Levinson's result 1 Some shop math 2 Beurling's gap theorem Problem 11 3 Weights which increase along the positive real axis 4 Example on the comparison of weighted approximation by polynomials and that by exponential sums 5 Levinson's theorem

233 234 234 236 238 239 243

247

xii

Contents B The Fourier transform vanishes on a set of positive measure Beurling's theorems 1 What is harmonic measure? 2 Beurling's improvement of Levinson's theorem 3 Beurling's study of quasianalyticity 4 The spaces 5p(-90), especially .91(-90) 5 Beurling's quasianalyticity theorem for LP approximation by functions in 9'p(90)

C Kargaev's example 1 Two lemmas 2 The example

D Volberg's work Problem 12 1 The planar Cauchy transform Problem 13 2 The function M(v) and its Legendre transform Problem 14(a) Problem 14(b) Problem 14(c) 3 Dynkin's extension of F(e') to { Iz 15 1) with control on

250 251 265 275 280 292 305 306 312 316 318 319 322 323 327 336 336

I FZ{z)1

338

4 Material about weighted planar approximation by polynomials 5 Volberg's theorem on harmonic measures 6 Volberg's theorem on the logarithmic integral 7 Scholium. Levinson's log log theorem

343 348 356 374

VIII Persistence of the form dx/(1 + x2) A The set E has positive lower uniform density 1 Harmonic measure for -9 Problem 15 2 Green's function and a Phragmen-Lindelof function for -9 Problem 16 Problem 17(a) Problem 17(b) 3 Weighted approximation on the sets E Problem 18 4 What happens when the set E is sparse Problem 19

B The set E reduces to the integers Problem 20 1 Using certain sums as upper bounds for integrals corresponding to them 2 Construction of certain intervals containing the zeros of p(x)

384 386 387 400 400 404 411 413 424 432 434 443 445 446 447 454

Contents 3 Replacement of the distribution n(t) by a continuous one 4 Some formulas Problem 21 5 The energy integral Problem 22 6 A lower estimate for 1.11 log I 1- (x2/t2)Idp(t)(dx/x2) o 7 Effect of taking x to be constant on each of the intervals Jk 8 An auxiliary harmonic function Problem 23 9 Lower estimate for f n f o log 11- (x2/t2)Idp(t)(dx/x2) 10 Return to polynomials Problem 24 11 Weighted polynomial approximation on 7L C Harmonic estimation in slit regions 1 Some relations between Green's function and harmonic measure for our domains .9 Problem 25 2 An estimate for harmonic measure Problem 26 3 The energy integral again 4 Harmonic estimation in 9 5 When majorant is the logarithm of an entire function of exponential type Problem 27 Problem 28

xiii 468 473 478 478 484 487 492 495 497 506 516 518 522 525 526 540 540 545 548 553 555 561 568

Addendum

Improvement of Volberg's theorem on the logarithmic integral. Work of Brennan, Borichev, Joricke and Volberg 1 Brennan's improvement, for M(v)/v1/2 monotone increasing 2 Discussion

3 Extension to functions F(ei,) in L1(-rr,n) 4 Lemma about harmonic functions Bibliography for volume 1 Index Contents of volume II

570 570 574

582 590 596 600 603

Preface

The two volumes that follow make up what is meant primarily as a book for reading. One reason for writing them was to give a connected account of some of the ideas that have dominated my mathematical activity for many years. Another, which was to try to help beginning mathematicians interested in analysis learn how to work by showing how I work, seems now less important because my way is far from being the only one. I do hope, at any rate, to encourage younger analysts by the present book in their efforts to become and remain active. I have loved .(log M(t)/(1 + t2)) dt - the logarithmic integral-ever since I first read Szego's discussion about the geometric mean of a function and the theorem named after him in his book on orthogonal polynomials, over 30 years ago. Far from being an isolated artifact, this object plays an important role in many diverse and seemingly unrelated investigations about functions of one real or complex variable, and a serious account of its appearances would involve a good deal of the analysis done since 1900. That will be plain to the reader of this book, where some of that subject's developments in which the integral figures are taken up. No attempt is made here to treat anything like the full range of topics to which the logarithmic integral is relevant. The most serious omission is that of parts of probability theory, especially of what is called prediction theory. For these, an additional volume would have been needed, and we already have the book of Dym and McKean. Considerations involving HP spaces have also been avoided as much as possible, and the related material from operator theory left untouched. Quite a few books about those matters are now in circulation. Of this book, begun in 1983, all but Chapter X and part of Chapter IX was written while I was at McGill University; the remainder was done at UCLA. The first 6 chapters are based on a course (and seminar) given

Preface

xvi

at McGill during the academic year 1982-83, and I am grateful to the mathematics department there for the support provided to me since then out of its rather modest resources. Chapters I-VI and most of the seventh were typed at that department's office. Chapter VII and parts of Chapter VIII are developed from lectures I gave at the Mittag-Leffler Institute (Sweden) during part of the spring semesters of 1977 and 1983. I am fortunate in having been able to spend almost two years all told working there. Partial support from the U.S. National Science Foundation was also given me during the first year or two of writing. I thank first of all John Garnett for having over a long period of time encouraged me to write this book. Lennart Carleson encouraged and helped me with research that led eventually to some of the expositions set out below. I thank him for that and also for my two invitations to the Mittag-Leffler Institute. For the second of those I must also thank Peter Jones who, besides, helped me with at least one item in Chapter VII. The book's very title is from a letter to me by V.P. Havin, and I hope he does

not mind my using it. I was unable to think of anything except the mathematical expression it represents! It was mainly John Taylor who arranged for me to come to McGill in the fall of 1982 and give the course mentioned above. Since then, a good part of my salary at McGill has been paid out of research grants held by him, Jal Choksi, Sam Drury, or Carl Herz. Taylor also came to some of the lectures of my course as did Georg Schmidt. Robert Vermes attended all

of them and frequently talked about their material with me.

Dr Raymond Couture came part of the time. The students were Janet Henderson, Christian Houdre and Tuan Vu. These people all contributed to the course and helped me to feel that I was doing something of value by giving it. Vermes' constant presence and evident interest in the subject were especially heartening. Most of the typing for volume I was done by Patricia Ferguson who typed Chapters I through VI and the major part of Chapter VII, and by Babette Dalton who did a very fine job with Chapter VIII. I am beholden to S. Gardiner and P. Jackson of the Press' staff and finally to Dr Tranah,

the mathematics editor, for their patience and attention to my desires regarding graphic presentation. The beautiful typesetting was done in India. August 13, 1987 Laurel, Comte Argenteuil, Quebec.

Introduction

The present book has been written so as to necessitate as little consultation by the reader as reasonably possible of other published material. I have

hoped to thereby make it accessible to people far from large research centres or any `good library', and to those who have only their summer vacations to work on mathematics. It is for the same reason that references,

where unavoidable, have been made to books rather than periodicals whenever that could be done. In general, I consider the developments leading up to the various results in the book to be more important than the latter taken by themselves; that is why those developments are set out in more detail than is now customary. My aim has been to enable one to follow them by mostly just reading the text, without having to work on the side to fill in gaps. The reader's active participation is nevertheless solicited, and problems have been given. These are usually accompanied by hints (sometimes copious), so that one may be encouraged to work them out fully rather than feeling stymied by them. It is assumed that the reader's background includes, beyond ordinary undergraduate mathematics, the material which, in North America, is called graduate real and complex variable theory (with a bit of functional analysis). Practically everything needed of this is contained in Rudin's well-known manual. My own preference runs towards a more leisurely approach based on Titchmarsh's Theory of Functions and the beautiful Lecons d'analysefonctionnelle of Riesz and Nagy (now available

in English). Alongside these books, the use of some supplementary descriptive material on conformal mapping (from Nehari, for instance) is advisable, as is indeed the case with Rudin as well. The Krein-Milman theorem referred to in Chapters VI and X is now included in many books; in Naimark's, for example (on normed algebras or rings), and in Yosida's.

In the very few places where more specialized material is called for,

xviii

Introduction

additional references will be given. (Exact descriptions of the works just

mentioned together with those cited later on can be found in the bibliographies placed at the end of each volume.) Although the different parts of this book are closely interrelated, they may to a large extent be read independently. Material from Chapter III is, however, called for repeatedly in the succeeding chapters. For finding one's way, the descriptions in the table of contents and the page headings should be helpful; indices to each volume are also provided. Throughout

volume I, various arguments commonly looked on as elementary or well-known, but which I nonetheless thought it better to include, have been set in smaller type, and certain readers will miss nothing by passing over them.

The book's units of subdivision are, successively, the chapter, the § (plural §§) and the article. These are indicated respectively by roman numerals, capital letters and arabic numerals. A typical reference would be to `§B.2 of Chapter VI', or to `Chapter VI, §B.2'. When referring to another article within the same §, that article's number alone is given (e.g., `see article 3'), and, when it's to another § in the same chapter, just that §'s designation (e.g., `the discussion in § B') or again, if a particular article in that § is meant,

an indication like `§ B.T. Theorems, definitions and so forth are not numbered, nor are formulas. But certain displayed formulas in a connected development may be labeled by signs like (*), (f), &c, which are then used to refer to them within that development. The same signs are used over again in different arguments (to designate different formulas), and their order is not fixed. A pause in a discussion is signified by a horizontal space in the text. About mistakes. There must inevitably be some, although I have tried as hard as I could to eliminate errors in the mathematics as well as misprints. Certain symbols (bars over letters, especially) have an unpleasant tendency to fall off between the typesetters' shop and the camera. I think (and hope) that all the mathematical arguments are clear and correct, at least in their

grand lines, and have done my best to make sure of that by rereading everything several times. The reader who, in following a given development,

should come upon a misprint or incorrect relation, will thus probably see

what should stand in its place and be able to continue unhindered. If something really seems peculiar or devoid of sense, one should try suspending judgement and read ahead for a page or so - what at first appears bizarre may in fact be quite sound and become clear in a moment. Unexpected turnings are encountered as one becomes acquainted with this book's material. It is beautiful material. May the reader learn to love it as I do.

I

Jensen's formula

On making the substitution t = tan (9/2) and then putting M(t) = P(9), the expression 1

°° log M(t)

7r

_ ,D

1 + t2

dt

goes over into 1

21r

f

log P(9) d9.

We begin this book with a discussion of the second integral.

Suppose that R > 1 and we are given a function F(z), analytic in { z I < R}. If F(z) has no zeros for I z 151 we can define a single valued function log F(z), analytic for I z I S R', say, where 1 < R' < R. By Cauchy's formula we will then have l og F(0) =

s:'

log F(e's) d9,

so, taking the real parts of both sides, we get loglF(0)I

= 2.

logIF(e'9)Id9.

What if F(z) has zeros in I z I <, 1? Assume to begin with that there are none on I z I = 1, and denote those that F(z) does have inside the unit disk by al, a2, .... a zero is repeated according to its multiplicity in such an enumeration. Put (D(z) =

F(z)

(z-a1)(z-a2)...(z-a)'

2

1 Jensen's formula

Then (D(z) has no zeros in { I z I < 1 }, so, by the special case already treated,

log(0) I =± f_. logl(D(ei9)ld9 a

1

1

n

R

logJF(e's)Id9-k12 f Iogje"-akld9.

=2

-n

Here we make a side calculation. For I ak I < 1 we have "

1

2 J1ogIe_akId9=_JlogI1-dkei9Id9, andthis = log 1= 0 by the case already discussed (F(z) without zeros in Izl < 1)! Combined with the previous relation this yields

logIfi(0)I=2n

log I F(0)

k=1

log I

2n -n

I F(e19) I d9.

The sum on the left can be written differently. Call n(r) the number of zeros of F(z) in I z 15 r (counting multiplicities). Then, if F(0) # 0,

- k=1 log l ak l = f 1 n(r)r dr. o

Indeed, since n(r) = 0 for r > 0 close to 0,

f 1 nr)

r

o

dr = n(1) log 1-

log r dn(r)

Y log I ak I

k=1 n

f o1

We therefore have

log F ()I+ 0 l

n(r)

if

r dr=2rz

f ol

loglF(e's)Id9.

In case F(z) is regular in a disk including { I z I

F(0) # 0 we can (provided that F(z) # 0 for variable in the preceding relation and get log F()I+ 0

n( r)

l

o fR

This is Jensen's formula.

r

1

dr=2

f

Iz

R} in its interior and R) make a change. of

Jensen's formula

3

The validity of Jensen's formula subsists even when F(z) has zeros on the circle I z I = R. To see this, observe that then F(z) will not have any zeros on the circles I z I= R' with R' < R and sufficiently close to R, for F(z) is analytic in a disk { I z I < R + n}, n > 0, and not identically zero (F(0) :A 0). So, for such R',

log F ()I+ 0 I

If

f R' n(r) dr=2 _rzl°gIF(R'e"s)Id9. o

r

As R'-+R, the left side clearly tends to log I F(0) I + f (n(r)/r) dr - the o because n(r) integral on the left is a continuous function of its upper limit is bounded. We need therefore merely verify that

f

log I F(R'e''') I d9 -+ J

log I F(Re''') I d9

as R'-+ R. The idea here is the same whether F(z) has several zeros on I z I = R or only one, and in order to simplify the writing we just treat the latter case. Suppose then that F(a) = 0 where I a I = R, and there are no other zeros in a ring of the form {R - n I z I < R + n}, n > 0. On this ring we then have I F(z) I > cont. I z - a I', if m is the multiplicity of the zero at a, so, since I F(z) I is also bounded above there, IlogIF(R'ei9)II

const.+mlog +

1Re'9 -al

for R - n < R' S R. (Here, for p > 0, log+ p denotes log p if p 3 1 and 0 if p < 1.) The expression on the right is, however, < cont. + m log+(1 /I Rei9 - a 1), independently of R', when the latter quantity is close

to R:

Figure 1

4

1 Jensen's formula

(The constants of course will be different; the relation between them need not concern us here.) In other words, for R'-+ R the expressions I log I F(R'ei9) I

I

are bounded above by the fixed function const. +

m log+ (1/I Re9 - a I) of 9, which however, has a finite integral over [ - 7t, 7r], as we easily check directly. Since also log I F(R'ei9) I - log I F(Rei9) I

pointwise as R'-+ R, we have

logiF(R'ei9)Id9-

logIF(Rei9)Id9 j:,

fX

by Lebesgue's dominated convergence theorem. This is what we needed to

complete our derivation of Jensen's formula. (We see that the same computation which shows that

f logIF(Rei9)Id9 > -oo J

R

also establishes the convergence of fR , logjF(R'ei9)jd9 to that quantity as R'-+ RD Here is a first application of Jensen's formula. Theorem. Suppose that F(z) is analytic and # 0 for I z I < 1, and that the integrals

Jlo+ I F(rei9)I d9 ar e bounded for 0 < r < 1. Then for any ro, 0 < ro < 1, the integrals

f.

log- I F(rei9) I d9

are bounded for ro < r < 1. Notation. For p >, 0, we write (as remarked above) log+ p = max (log p, 0).

We also take log - p = - min (log p, 0), so that log - p > 0 and log p = log p - log - p. (Everybody means the same thing by log+ p, but, regarding log p, usage is not uniform.)

Proof of theorem. Without loss of generality (henceforth abbreviated `wlog'), let F(0) 96 0. (Otherwise work with F(z)/z' for a suitable k instead of F(z).) By Jensen's formula, R

- oo
log+IF(rei9)Id9 -R

--

1

27[

log-IF(rei9)Id9, 0
Jensen's formula

5

By hypothesis, the right-hand side is

const. -

1

2n

f

logF(rei9) I d9. ,,

The desired result follows by transposition. Corollary. Under the hypothesis of the theorem, suppose that F(ei9) = lim F(rei9) F-1

exists a.e. Then

Jlo_IF(e')Idi9 < co. Pr oof. Fatou's lemma. Remark 1. Actually, the hypothesis of the theorem forces a.e. existence of lim F(rei9).

r-1

This is a fairly deep result, and depends on Lebesgue's theorem on a.e. existence of derivatives of functions of bounded variation. In the situations we will mostly consider, the existence of this limit can be directly verified ('by inspection'), so the deeper result will not be needed. Therefore we do not prove it now. The interested reader can work up a proof by using the subharmonicity of log+ I F(z) I together with an argument from Chapter III, §F.1, so as to produce a positive measure v on [ - n, n] for which I F(z) I

exp

1

ei9+z

l

{2Jei9_z dv() }

IzI < 1.

))

After this, one applies results from §F.2 of Chapter III to the analytic function 1(z) within I I on the right, and then to the ratio F(z)/O(z). Remark 2. The idea of the corollary is that if I F(z) I is not too big in { I z I < 1 }

(especially if IF(z)I is bounded there), then the boundary values IF(ei9)I cannot be too small unless F = 0.

Problem I (a) Let F(z) be entire, F(O) = 1, and I F(z)I

KeA1Z1 for all z, where A and K are

constants. If n(R) denotes the number of zeros of F having modulus 5 R,

6

I Jensen's formula show that, for all R, n(R) 5 eAR + const. (Here, the constant depends on K.)

*(b) Show that in the relation established in (a) the coefficient eA of R cannot

in general be diminished. (Hint. Fix R = m/e with m a large integer. Compute the maximum value of (x/R)`Re - Z for x > 0. Then look at a function which has m equally spaced zeros on the circle I z = R and no others.)

II

Szego's theorem

The theorem Szego's theorem is a beautiful result in approximation theory, obtained with the help of Jensen's formula. Its proof also uses a limit A.

property of integrals involving the Poisson kernel (for the unit disk) which

is now taught in many courses on real variable theory. The reader who does not remember that result will find it in §B, together with its proof. Theorem (Szego). Let w(9) >, 0 belong to L1(- n, n). Then the infimum of

1 - Y-

anBins

w(9) d9,

n>O

taken with respect to all possible finite sums Eoa"e'"s, is equal to log w(9) d9

exp 2n

.

Note: In R log' w(9) d9 is finite if wEL1(- n, n). So I' R log w(9) d9 either converges, or else diverges to - oo.

Proof of theorem. By the inequality between arithmetic and geometric means, 1 - E a"gins w(9) d9 n>0

exp Zn

Nlog

1-E

a"eons

Jensen's formula applied to F(z) =1expression is always R

exp(2n

logw(9)d9

+logw(9) Id9}.

n>O

;

oanz" shows that this last

II A The theorem

8

the desired infimum is thus > the latter quantity. We must establish the reverse inequality. Write wN(9) = max (w(9), a-N). By Lebesgue's monotone convergence theorem and the finiteness of f " log' w(9) d9 we have R

1

2n

R

1

log wN(9) d9

log w(9) d9.

N 21r J _

-R

It will therefore be enough to show that for any N and any S > 0 there exists some finite sum 1- Ek > 0 Ake'k9 such that R

J

1

R 1

R

kE Ake" w(9)d9 < exp(2_ J

log wN(9)d9)+S. -R

To this end, put first of all f,,eit

FN(z) =exp

zlog(WN(t)dt

eu - z

2n

for I z I < 1. We have

/

(*)

FN(0) = exp I

l l

rR

log

1

\\2a J

I dt I. wN(t)/ 1

/

Since wN(t) ,>eN, I FN(z)I ,<eN for Izl < 1. Indeed, taking real parts of the logarithms of both sides of (*) gives us 1 og I FN(reis) I =

1

't

2n _1-r

2

1- r2 log 2rcos(9- t)

1

WN(t)

) dt.

On the right side we recognize the Poisson kernel (that's the real reason for using (e't + z)/(e't - z) in (*), aside from the fact that we want FN(z) to be analytic in { I z I < 11). As one knows, 1

2n

R

_R

1-r2 1 + r2 - 2ncos (9 -

t) dt = 1;

the integrand is obviously positive. We see that logIFN(rei) I 5 N by the previous formula.

Now we use another, much finer property of the Poisson kernel, established in §B below. According to the latter,

_ r2 log _ R 1 + r2 12r cos (9 - t)

fn 27c

dt ---+ log wN(t)

W N(9)

for almost all 9 as r --.1. So I FN(re'9) I -1 /wN(9) a.e. for r -+ 1. However, I FN(z)I is bounded above and w(9)EL1(- n, n). Therefore, by dominated

Szego's theorem

9

convergence, R

R

IFN(rei9)lw(9)d9

2n

W((1')d9

2n

as r-> 1. The right-hand side is clearly < 1. Givens > 0 we can therefore get an r < 1 with (t)

2n

f

IFN(rei9)Iw(9)d9
Fix such an r.

By the very form of the right side of (*), FN(z) is analytic in { ( z I < 11; it therefore has a Taylor expansion there. And, by (*k), FN(O) 0. Letting S(z) be any partial sum of the Taylor series for FN(z), we see that for our fixed r, S(rei9)

is of the form 1- E Ake'k9

FN(0)

k>O

the sum on the right being finite. Since FN(z) is regular in { I z I < 11 and r < 1, we see by (t) that we can choose the partial sum S(z) so that 1

2n

I S(rei' I w(9) d9 < 1 + 2e.

-R

Hence 1

2n

R

Y_ Akeik9 w(9) d9 f- I I - k>O R

i9

fR

w(9) d9 <

(1 + 2e) exp

12 \

109 wN(t) dt

by (* ).

-R

This is enough, and we are done. Remark. This most elegant result was extended by Kolmogorov, and then by Krein, who evaluated the infimum of 1

2n

RR

1 - Z anein9 dµ(9) n>O

for all finite sums Y_n>0a,, ein9 when µ is any finite positive measure. It turns

10

11 B Poisson kernel a pointwise approximate identity

out that the singular part of y (with respect to Lebesgue measure) has no influence here, that the infimum is simply equal to

JRftlog(d9)d9}. exp{in'

_

I do not give the proof of this result. It depends on the construction of Fatou-Riesz functions which, while not very difficult, is not really part of the material being treated here. The interested reader may find a proof in many books; some of the older ones which have it are Hoffman's and Akhiezer's (on approximation theory). The newer books by Garnett (on bounded analytic functions), and by me (on Hp spaces) both contain proofs.

B.

The pointwise approximate identity property of the Poisson kernel Theorem. Let P(9)eLt(- n, n), and, for r < 1, write

U(re') -

1

2n

''

_R

1-r2 1 + r2 - 2r cos (9 - t)

P(t) dt.

For almost every 9, U(z) tends to P(9) uniformly as z tends to e'9 within any sector of the form

Iarg(1 -e-"sz)I < a <

2

Figure 2

Remark. We write `U(z) - P(9) a.e. for z --e'5.' Some people say that U(z) - P(9) a.e. for z tending non-tangentially to ei9, others say that * It is clear that for z of modulus > sin a in such a sector we have I arg z - 91 5 K(1- I z 1) with a constant K depending on a.

Poisson kernel a pointwise approximate identity

11

U(z)-+P(9) uniformly within any Stoltz domain as z--+ei9 (for almost all 9). Of course, the theorem includes the result that U(re'9) P(9) a.e. for r -+ 1, used in proving Szego's theorem. P(9) for r-+ 1 if 19r - 91 K(1- r), whenever (d/d9) J P(t) dt exists and equals P(9), hence for almost every 9, by Lebesgue's differentiation theorem. The rapidity of the convergence will be seen to depend only on the value of K measuring the opening of the sector with vertex at ei9, and not on the particular choice of Sr satisfying the above relation for each value of r. Without loss of generality, take 9 = 0, and assume that

Proof of theorem. We will show that U(rei9)

f('s

P(t) dt = 9P(0) + o(I 91) Jo

for 9-+0 (from above or below!). Pick any small S > 0 and write 1 - r2

1

2n _A1+r2-2rcos(9r-t)

P(t) dt

as 1

a

1

a4iti«+2n ,f _a)1+ As r-+1, 19r1 becomes and remains' <6/2, so (1-r2)/(1+r2-2rcos(9,-t))-+0 r2-2rcos(9,-t)P(t)dt.

(27r

uniformly for S < It I < n, and the first integral tends to zero. The second is treated by partial integration. Writing J(9) = J°P(t) dt, that second integral becomes

1-r2

1

21t

J(S)1 + r2 - 2rcos (9, - S) a

1

2n

1-r2 S)1 + r2 - 2rcos (9, + 6)]

1 - r2

8

J(t)at

J(-

1+r2-2rcos(9,-t)

dt.

The two integrated terms in square brackets tend to zero as r -+ 1. Since J(t) _ P(0)t+o(1t1), the integral equals f6p(o)t

r

2

8t 1+r2-2rcos(9,-t)

2n

dt

r2 1 T,Ooto dt. +2n at 1+r2-2rcos(9,-t))

Here, the first term is readily seen (by reverse integration by parts!) to equal 0(1) +

(O) P2n

1 - r2

f,,I+r

which tends to P(0) as r --> 1.

2-2rcos(9,-t)dt=o(1)+P(0)(1-o(1)),

11 B Poisson kernel a pointwise approximate identity

12

To estimate the second term, we have to use the fact that

1-r2 1+r2-2rcos(9,-t) is a monotone function of t on each of the intervals - S < t < 9, and 9, < t < S. (We are supposing that r is so close to 1 that - S < 9, < S.) Given any e > 0, we can choose b > 0 so small to begin with that the second term is in absolute value 8 1 - r2 dt. Ot( 1+r2-2rcos(9,-t))

Writing t = 9, - t, this becomes

f. f'+'I'-q'I 27t

s,

a

8

1 - r2

at(1+r2-2rcos2)

dt.

We break this up as

1-r2

8

l*)

2no+a)li-9.1

2nJo_a +

,l

8t 1+r2-2rcosr

dT;

in the second integral,

a(

r2

1

1 +r 2 - 2r cos t

OT

)
so that second integral is 5 s,+a

e

2n 819,1 27t

u

J o

t

0

1- r2

Or

1 + r2 - 2r cost

8t 1 + r2 - 2r cos t

dt dt.

Here, the first term is e(2 + 0(1)) (see above treatment of expression involving P(0) !), and the second is

5

819,I 1+r

2n 1-r

This last, however, is (K/7r)e in view of our condition on 9,. We see that the second integral in (*) is 5 (K/it + i + o(1))e for r close enough to 1. The first integral in (*) is similarly treated, and seen to also be _< (K/it + z + o(1))8 for r close to 1. In this way, we have found that the expression

1f

r2

2ivao(Itl)at(1+r2-2rcos(9,-t) dt is in absolute value 5 (1 + 2K/it + o(1))e if b > 0 is small enough to begin with, and r close enough to 1. However, according to the calculation at the beginning

Poisson kernel a pointwise approximate identity

13

of this proof, the sum of the last expression and P(0) differs by o(1) from U(rei9) when r -. 1. So, since e > 0 is arbitrary, we have established the desired result.

Remark. Suppose that U(re .s) =

1

1 - r2

2n

1 + r2 - 2r cos (9 - t)

dµ(t)

with a finite (complex valued) measure µ. Form the primitive

49) = f 9 d t(t). o

Then it is still true that, wherever the derivative µ'(9) exists and is finite, we

have U(z) -y'(9) for z -e'9. (Hence lim,.1 U(re''9) exists and is finite a.e. by Lebesgue's differentiation theorem.) The proof of this slightly more general result is exactly the same as that of the above one. Problem 2 purpose of this problem is to derive, from Szego's theorem, the following result. Let w(x) 3 0 be in L t (- co, oo) and let a > 0. There are finite sums S(x) of the form S(x) _ E,13aA, eizz with The

J1 - S(x)I w(x) dx arbitrarily small ii f $°° .(log w(x)/(1 + x2)) dx = - oo. In case f °°,(log w(x)/(1 + xz))dx = - oo, we can, given any bounded continuous function cp(x), find finite sums S(x) of the above mentioned form with f °°

tp(x) - S(x) I w(x) dx arbitrarily small. Establish-

ment of this result is in a series of steps. (a) Let a > 0 and let p be a positive integer. There are numbers A. with x

(1-iax)

p=-A t-x " oo

i+x

the series on the right being uniformly convergent for - oo < x < oo. (Hint:

Put w = (i - z)/(i + z) and look at where f (w) = z/(1 - iaz) is regular in the w-plane. Little or no computation is used in doing (a).)

(b) Let A > 0. There are finite sums Sk(x), each of the form Y_,,oC"((i - x)/(i + x))", such that jSk(x)I ,<2 on R and Sk(x) k eizs For each a > 0 the series for u.c.c.* on R. (Hint` eiz" = limQ_,0+e'z" /(' exp (i lx/(1- iax)) is uniformly convergent for - oc < x < oo. Little or no computation here.) (c) Given any integer n > 0 there are finite sums Tk(x) of the form Y-x, 0 A,, eiz" with I Tk(x) l ,
* u.c.c means uniform convergence on compacta.

14

II B Poisson kernel a pointwise approximate identity on R. (Hint: Start from the integral formula ri

= i

(1 + x)2

e-ze"&

= -f

ile - xei2" dA,

0

&c.)

(d) Given w > 0 in L,(R), denote by d the class of bounded continuous cp defined on R such that 1% 1
small with suitable finite sums S(x) of the form E.,,oA"((i-x)/(i+x))". Call d the set of bounded continuous (p for which finite sums T(x) = exist making f °°,p I cp(x) - T(x)lw(x) dx arbitrarily small. Prove that sat= S. (e) Let a > 0, and denote by 9 the set of bounded continuous cp such that EA30Ade'dx

there are finite sums a(x) = Z,,Axeizx with f I ip(x) - u(x)I w(x) dx arbitrarily small. Prove that if 1 e.F then 9 contains all bounded continuous p, and this happens if f (log w(x)/(1 + x2)) dx = - oo. (Hint: If 1 e.F, then d (of part (d)) includes all et'x with .l > - a, hence, by iteration, all e'Zx with 2 > - 2a, with A > - 3a, &c. Sod includes all integral powers ((i - x)/(i + x))" with positive and negative n. These are enough to approximate e-1°x(p(x) for any bounded continuous (p.)

III

Entire functions of exponential type

An entire function f (z) is said to be of exponential type if there is a constant

A such that I f (z) I S const.e'1Z1

everywhere. The infimum of the set of A for which such an inequality holds (with the constant in front on the right depending on A) is called the type of f (z).

Entire functions of exponential type come up in various branches of analysis, partly on account of the evident fact that integrals of the form

f.

eUZ

dµ(A)

are equal to such functions whenever K is a compact subset of C. In this chapter we establish some of the most important results concerning them,

which find application throughout the rest of the book. We are not of course attempting to give a complete treatment of the subject. Fuller accounts are contained in the books by Boas and by Levin. A.

Hadamard factorization As in Chapter I, we denote the number of zeros of f (z) having

modulus <, r by n(r) (each zero being counted according to its multiplicity). We sometimes write n1(r) instead of n(r) when several functions are being dealt with.

Theorem. If f (z) is entire and of exponential type, n(r) < Cr + 0(1).

Proof. See Problem 1(a), Chapter I. If I f(z) I S const.e"1Z1 we can take C=eA.

111 A Hadamard factorization

16

Theorem (Hadamard factorization). Let f (z) be entire, of exponential type, and denote by {zn} the sequence of its zeros # 0 (multiplicities counted by repetition), so arranged that O< IZI I'< IZ2I'
z\

1- - I em",

f (z) = Cz'`ebz

ZnJJJ

n

the product being uniformly convergent on compact subsets of C.

Terminology. Henceforth we abbreviate the last phrase as `u.c.c. converg-

ent on C'. Proof of theorem. By working with f(z)/z' instead of f (z) (if necessary), we first reduce the situation to one where f (O) 0. Then n(r) < Kr for some K. If, with a zero zn of f, we have I zn I > 2R, then, for I z I S R,

log

1-

z

= - -z

ezizA

zn

zn

I (Z

- 2 zn

2(zn)2

2

+

z

Zn

3(zn)3 J

(1+0(1)) =-12 (L), zn

(We are using the branch of the logarithm which is zero at Therefore 1

log I( I Zn

)ez/z }

z

2 Zn

20+00)),

whence (assuming always that I z I < R),

E

l

)

1 + 0(1)

Zn JJJ

))

2

(IlogI\1--z 1 + 0(1) R2

2

1+0(1) 2

R Iz,I>-2R

°° dn(t)

f2

R t2

J-

R2 1

2R) 2

+ 2 S2Rt(t3 ) dt J}

Zn

2

1).

Hadamard factorization

17

1+O(1)R2 ('°2Kdt = 1+0(1) 2

KR.

2

J2R t2

This inequality establishes absolute and uniform convergence of

1-

log

z

ez/z"

Zn

Izn132R

for I z I < R, and hence the uniform convergence of

H

(i_-f_)ezIzn1

Zn

> 2R

for such values of z. Write P(z) = fln(1- z/zn)ez/z"; according to what has just been shown, P(z) is an entire function of z. Since f (O) 0, f (z)/P(z) is entire and has no zeros in C. There is thus an entire function (p(z) with 1'(z) = ev(z)

P(z)

and it is claimed that qp(z) = a + bz with constants a and b. To show that cp(z) has the asserted form, we use the fact that f (z) is of exponential type in conjunction with the inequality n(r) < Kr in order to get some control on 1 RRp(z) I for large I z I. For I z I < R, e9tw(z)

f (z)

_

1

z

1 F11

1-- ez/z"

1--

II

Zn

Iz,I<2R

z

Izn132R

= I.1I, say. ez/z"

Zn

The computation made above shows that I log III < CR with some constant C (we estimated I log {I - z/zn} ez/z" I!), and it suffices to estimate I. For I we use a trick. The ratio /i(z) =f (z)/1 11z"1 <2R(1- z/zn)ez/Z" is entire,

so, by the principle of maximum,
H

Iez/z"I

>, exp{ -4R Y

lznl<2R

1I zn152R

111

1/Izn1} O())

rr

= exp - 4R J o R d

I

l

rrf02

t

t)

= exp { - 2n(2R) - 4R

R t(2)

dt I.

Since n(t) = 0 for 0 < t < I zl I, a quantity >0, and n(t)/t < K, the last expression is e-4R(%IogR+o(l))

III A Hadamard factorization

18

At the same time, for I z I = 4R and I zn 15 2R,

so

e-4R(KIogR+O(1)) Zn

<2R

when I z I = 4R. Because f (z) is of exponential type we therefore have IO(z) I \ e4KR(1ogR+O(1))

IzI=4R,

whence e4KR(1ogR+O(1))

1

and finally, since a""(=) = I.11,

%(p(z) 5 4KR log R + O(R) for

IzI 5 R

in view of the fact that log II 5 CR. At this point, we use a device already applied to the study of log I F(rei9) I

near the end of Chapter I. By analyticity of (p(z), ft

9t(p(Re''') d9

91(p(0) 2rz

=

1

-R

rft

2n -R

[91 (p(Rei9)] +

d9 -

If 2n

[9t(p(Re's)]

_

d9

R

(with self-evident notation). Therefore,

J R I9t(p(Re's)I d9 = ft

{

['3t(p(Rei9)]+ + [9w(Rei9)] -} d9

fX

By the one-sided inequality just found for 91(p(z), IzI < R, this last expression

is S 16nKR log R + O(R) + 0(1). Now we can conclude the proof. Since T(z) is entire, Oz) = E,ynZn, 0

so CO

YnRne-'ns+ 29ty0 + EynRne'ns

291(p(Rei9) _ 1

1

Set of zeros of a function of exponential type

Using the relations f ft eik$e-i19d9 =

0, k o 1

X

g

we et, for n >, 2,

"

-1

Y" =

2n,k=l

19

nR" -ft

99(p(Re'9)e-1n9d9.

Therefore ft

IY"I5-l

-ftl91(p(Rei9)Id9

which, by the above work, is < R-"(16KR log R + O(R) + O(1)). Making R - oo, we see that y" = 0 for n > 2. Our power series for cp(z) thus reduces to the linear expression yo + ylz, and finally f (z) = e'(z)p(z) = eyoeY,z j

(1- z

Iez/z",

the required representation. We are done. B.

Characterization of the set of zeros of an entire function of exponential type. Lindelof s theorems

While establishing the Hadamard factorization in the preceding § we found that II < ecR

which was to be expected (having started with a function of exponential growth), but we could only show that

I

eocR log R>

This, however, forced 9(z) to be a first degree polynomial, whence, in fact,

I < e0

,

because the method used to estimate II showed at the same time that II > e- O(R).

The refinement on our estimate of I from a°(RlosR) to e°(R) is due to the fact that I f (z) I < e°(Izl) for large I z I. Otherwise, the R log R growth is best possible, and if we only know that n(r) < Kr, we can only conclude that

H 1 1- z) ez/z"

"\\

e0azu°sizu

Z"

for IzI large, most of the contribution coming from the factors with Iz"I <21zI.

20

111 B Lindelof 's characterization of the set of zeros

The fact that f (z) is of exponential type imposes not only the growth condition n(r) 5 O(r) (for large r), but also a certain symmetry in the distribution of the zeros zn. This symmetry is a deeper property of that set than the growth condition. Theorem (Lindelof). Let f (z) be entire, of exponential type, with f (0):A 0, and denote by {zn} the sequence of zeros of f (z), with, as usual, each zero repeated therein according to its multiplicity. Put

S(r) _ Y

-. 1

iznl <_ r Zn

Then I S(r) I

is bounded as r- oo.

Proof. By double integration. Since n(r) < Kr, f being of exponential type, we clearly have

for R ,
I S(r) - S(R) I < 2K

whence 2R

f

S(r)r dr = 3R2S(R) + O(R2).

R

We proceed to calculate the integral on the left. Provided that f (z) has no zeros for I z = r, we have, by the calculus of residues, 1

1

2ni

1Z.1 <, z

r2i f'(rei9) irei9d9 0 f(rei9) re's

f'(0) f (O)

By the Cauchy-Riemann equations, f '(z)

f(z)

(-a--i-Ologlfwi' ax awhence,

putting z = rei9,

1

S(r)= 2n

a

fo2z

(ax

-ay ) i

This holds for all save a finite number of values of r on the interval [R, 2R].

Multiply by rdr and integrate from R to 2R. We find JR2S(R) =

12R R

2n

S(r)rdr + O(R2)

Jf R<
('Ox-i')Ioglf(z)ldxdy+O(R2).

Set of zeros of a function of exponential type

21

Since S(r) = S(r + ), there is no loss of generality in assuming that f (z) R or on I z I = 2R. We may then apply Green's theorem to the double integral on the right (this is justified by first excising a small disk of radius p, say, about each of the zn in the annulus R < I z I < 2R, and then making p -+ 0), obtaining for it the value has no zeros on I z

1

2n

2'

(2Rloglf(2Rei9)I-Rloglf(Re''9)I)e-'ed9,

o

whence, by the previous relation, R f2a 2 R2 I S(R)

I<,2n

(21 log I f (2Reis) I I + I log I f (Rei9) I I) d9 0

+ O(R2).

Here, by rJensen's inequality (see Chapter I),

f02 nlog+I.f(re'8)Id9-2irloglf(0)I, J2,,

o

which is < 4nAr + 0(1) if I f (z) I < const.eAIZ1. Combined with the preceding,

this yields

RI2 S(R) I
IS(R)I5O(1) for R ->oo.

Q.E.D.

The result just proven has an easy converse. Theorem (also due to Lindelof!). Let

a
I_.1 SrZn

remain bounded in absolute value as r -+ oo. Then the product

C(z) _ f n

z I--

e

Zi=n

Zn

is equal to an entire function of exponential type.

Terminology. C(z) is frequently called a canonical product.

22

111 B Lindelof 's characterization of the set of zeros

Proof of theorem. Uniform convergence of the product on compact subsets of C has already been shown during the establishment of the Hadamard factorization (§A). Let R be given, and, for I z I = R, write I C(z)I = I F1

l 1- Z )e'/'-I-I, Zn

Iz.1<2R \

11- i Ie=12^I = I II, ZnJJJ

2 R \\

say. It has already been shown that II < e°(R) while we were deriving the Hadamard factorization, so we need only consider I. Clearly,

5f

1

1+ R )-expjRI IZn1

Izn142R

} Ize142RZn

By hypothesis, the exponential factor on the right is < eO(R), and we need only estimate the product.

The logarithm of that product is R R

log 1+ 1

1z.142R

\

IZn1

\

I=

f211

R

log 1+ R

0

= n(R) log +

I dn(t)

t

2R R

n(t)

o R+t t

dt,

since n(t) is zero for t near 0. Plugging in n(t) < Kt, we see that the last expression is 5 KR log + 2KR so that, finally, z

logI < KRlog2+2KR+O(R)=O(R). Since II has a similar estimate, we see that I C(z) I < e°(' for I z I = R. We're done. Here is an important consequence of the above results.

Theorem. Let f (z) and g(z) be entire and of exponential type. If the ratio f(z)/g(z) is also entire, it is of exponential type.

Proof. Combine the Hadamard factorization theorem with the two Lindelof theorems.

Problem 3 Let p be an integer > 1; suppose that f (z) is entire with f (0):0 0 and that Ce"'-l'.

If(z)I 5 Prove that n1(r) < Kr° and that the sums

TO

I=^. z

are bounded. (Hint: In studying T(r), express f'(reie)/f(rei9) in terms of (a/ar) log I f (rei') I and (8/09) logl f (re''') I, assuming, of course, that f has no

zeros on IzI = r.)

Phragmen-Lindelof theorems

C.

23

Phragmkn-Lindelof theorems The entire functions of exponential type one meets with in the

following chapters (and, for that matter, in many parts of analysis where they find application) have their size on the real axis subject to some restriction. During the remainder of this chapter we will be concerned with such functions, and we start here by seeing what it means to impose boundedness on R. Some of the following material is contained in textbooks on elementary complex variable theory; we include it for completeness.

Theorem (extended maximum principle). Let.9 be a domain in C not equal to all of C, and suppose that f (z) is analytic and bounded in -9. Assume that, for each e8-9, limsupl f(z)I < m. Then I f(z)I 5 m in -9. Z-.{

ZE1J

Remark. If -9 is a bounded domain, this is the ordinary maximum principle, and then the assumption that f (z) is bounded in -9 is superfluous. When -9 is unbounded, however, this assumption is really necessary, as the simplest examples show. Pick any n > 0 and fix it. According to the hypothesis, we can find a p > 0 such that I f (z) I < m + n for ze2? and I z I < p; we fix such a p and write

Proof of theorem. Wlog, say that

.9P=.9 n{Izl>p}. The open set -9p may not be connected, but that doesn't matter; its boundary consists of part of 8-9 and the arcs of {I z I = p} lying in -9.

Figure 3

24

III C Phragmen-Lindelof theorems

By choice of p, limsup I f(z)I < m + q for [0.9P. Z-C ZE1p

Take now a small e > 0 and consider, in -9p, the subharmonic function

vE(z)=logIf(z)I -elogIz1. is < log I f (z)1 + e log (1/p) for z e-9p, and this is in turn log I f (z) I + q if a is chosen sufficiently small, which we assume henceforth. Referring to the previous relation, we see that

The right side

(s)

limsup v,(z) < log (m + q) + q =-c

for each Cet2P. Let zo asp. Since f (z) is bounded in .9, say I f (z) I < M there, we can find an R > I zo I (depending of course on e) so large that vE(z) < log M - e log I z I is < log m for ze-9P and I z I = R. (This is the crucial step in the proof.) Denoting by -9,,, , the bounded open set -9pn{IzI
Figure 4

Since 2p,R is a bounded open set, we therefore have, by the (ordinary) maximum principle, vE(z) < log (m + q) + q, ze2p,R. This holds in particular for z = zo, so

logl f(zo)I 0 could be chosen as small as we pleased. Therefore,

loglf(zo)I 0 was arbitrary, If(zo)I < M. Q.E.D.

Phragmen-Lindelof theorems

25

Remark. The peculiar reasoning followed in the above proof is called a Phragmen-Lindelof argument. Most Phragmen-Lindelof theorems are proved in the same way. Note the special role played by the harmonic function s log I z I; a function used in this way is called a Phragmen-Lindelof function.

Theorem (Phragmen-Lindelof). Let f (z) be analytic in a sector S of opening 2y, and suppose that I f(z) I <

Ce'IzI'

in S, where a < n/27. If, for every t; e OS, limsup I f (z) I < m, then I f (z) I < m zeS

in S.

Figure 5

Proof. By making a change of variable, we may reduce our situation to the case where S is the sector

{z:-y<argz 0 fixed but arbitrary, consider, in S, the subharmonic function V(Z) = log I.f(z) I - e9R(z')

(Note: z1 is certainly analytic and single valued in S.) For z = re" in S, we have 91(zs) = rP cos #a > rfl cos fly,

26

111 C Phragmen-Lindelof theorems

and cos fly > 0 since O< fly < n/2. Therefore, in the first place, ve(z) < log l f (z)I in S, so, for 1Ce8S, limsup ve(z) < log M. Z-C ZEs

In the second place, since

loglf(z)I

<0(1)+AIzIa

in S and l3 > a, we have vE(z) < log m

for zeS whenever Izi is large enough (how large depends on e!). Suppose now that zoCS. With our fixed s>0, choose an R > IzoI so large that v,(z)
is < log m

Z-'C ZES

for any C on the boundary of the bounded region S n { I z I < R},

so, by the principle of maximum, vZ(z) < log m throughout that region. In particular, v,(zo) < log m, so log I f (zo) I < log m + r. R(zo). Now, keeping zo fixed, squeeze s. Get

If(zo)I <m, as required.

Important remark. The preceding two theorems remain valid if we merely suppose that log I f (z) I is subharmonic instead of taking f (z) to be analytic. The proofs are exactly the same.

Phragmen-Lindelof theorems

27

In the hypothesis of the second of the above two results we required a < n/2y (with strict inequality); this in fact cannot be relaxed to the condition a < n/2y. What happens when a = n/2y is seen from the following result, which, for simplicity, is stated for the case where 2y = n (the only

one which will arise on our work). We give its version for subharmonic functions. Theorem. Let u(z) be subharmonic f o r 3z > 0 with u(z) 5 A l z I + o( I z j) there when I z I

is large. Suppose that, for each real x, limsup u(z)'< 0. z-x 3z>0

Then u(z) < A3z for 3z > 0. Proof. Take any e > 0. The function ve(z) = u(z) - (A + s).3z is subharmonic in the first quadrant and v6(z) 5 O(; z I) there when I z I is large. If t; lies on the boundary of the first quadrant, we clearly have limsupvE(z)'< M z-C

3z >0

for some M since u(iy) 5 Ay + o(y) for y > 0 and large.

Figure 7

The first quadrant has opening < iv, so by the preceding theorem (or rather by its version for subharmonic functions), ve(z) < M throughout that region. We see in like manner that vE(z) is bounded above in the second quadrant, so, finally, vE(z) is bounded above for 3z > 0. However, for x real, limsupvE(z) < 0. z-,x

3z>0

III C Phragmen-Lindelof theorems

28

Therefore, by the version for subharmonic functions of the first theorem in this §, ve(z) < 0

for 3z > 0.

That is, u(z) < (A + e),Zz

for

3z > 0.

Squeeze e. Get u(z) < A 3z, ,3z > 0.

Q.E.D.

Corollary. Let f (z) be analytic inz > 0, continuous up to the real axis, and satisfy If(Z)I < CeAlzI

for 3z > 0. If If(x) I < M for real x, then I.f (z) I <

McA3Z

when 3z > 0. Proof. Apply the theorem to u(z) = log I f (z)/M I.

Remark. The example f (z) = e-'AZ shows that the inequality furnished by

the corollary cannot be improved. (Note also the relation between this particular function - or rather log I f (z) I - and the Phragmen-Lindelof function (A + e).3z used in proving the theorem. That's no accident!) The preceding theorem has an extension with a more elaborate statement, but the same proof. We give the version for analytic functions. Theorem. Let f (z) be analytic in ,3z > 0 and continuous in 3z > 0. Suppose that

(i) logIf(z)I 0, (ii) If (x) I < M, - ao < x < 00, (iii) limsup(logl f(iy)I )/y = A.

Then, for 3z >, 0, I f (z) I < McA3Z.

Remark. The growth off on the imaginary axis is thus enough to control the exponential furnished by the conclusion, as long as If(z) I has at most some finite exponential growth in 3z > 0.

The proof of this result is exactly like that of the preceding one. It is enough to put u(z) = log I f(z)IM I and then copy the preceding argument word for word.

Phragmen-Lindelof theorems

29

Any sector of the form 0 < arg z < a or a < arg z < x has opening < n. Looking at the reasoning used to establish the above two theorems, we see that we can even replace (iii) in the hypothesis of the preceding one by (iii)' limsup(log I f (Re"') I )/R sin a = A for some a, 0 < a < ?C, R-oo

and the same conclusion holds good.

Theorem. Let f (z) be analytic for 3z > 0 and continuous for Zz > 0. Suppose that If (z) I S CeAlz1 for 3z > 0, that If (x) I is bounded on the real axis,

and that f(x)--+0 as x -+oo. Then f (x + iy) -> 0 uniformly in each strip 0 < y

Las x --I' oo.

Proof. If, say, I f (x) I < M on R, we have I f (z) I McA3z for 3z > 0 by the corollary preceding the above theorem. Take any B > A and some large K,

and look at the function 9x(z) = z + iK

e'BZ f(z)

in 3z > 0. Since B > A and K > 0, we have 19x(z)I 1< M, 3z 1> 0. We can, however, do better than this. Givens > 0, we can find a Y so large that e-(B-A)Y < e/M; take such a Y and fix it. Then,

19x(z)I 5

z

z + iK

e-(B-A)3zM

<e

for 3z >, Y as long as K > 0, Choose now X > 0 so large that I f (x) I < e for x > X; this we can do because f (x) -+ 0 as x - oo.

X+iY

III D The Paley- Wiener theorem

30

Having fixed X and Y, we now take K so large that J (X + iy)/(X + iy + iK)I M < e for 0 < y <, Y; fixing this K we will then have 19K(X+iY)I

1<<

X +'Y

e. (B-A)YM

X+iy+iK

<s

for 0 < y < Y By choice of Y the same inequality also holds if y > Y. Finally, 19K(x)I5If(x)I <s for x > X. We see that 19K(z)I <s on the boundary of the quadrant {biz > X, 3z > 0}. However, I gK(z) 15 M in that quadrant, so, by the first theorem of this §,19x(2) I < s throughout it. Let then ¶Rz > X and rjz > 0. We have

If(z)I =

z + iK I eB3z19K(z) I < z

z+iKl eB z

s.

Suppose that 0 <, 3z < L. Then, if 1z > max (X, K) we have, by the previous relation, I f (z) I < 2eBL s.

Here, s > 0 is arbitrary. Therefore f (x + iy) -,. 0 uniformly for 0 < y S L as x ->oo. We are done.

D.

The Paley-Wiener theorem Theorem. Let f (z) be entire and of exponential type A. Suppose that

If(x)I2dx < oo. -00

Then there is a function cp(,)eL2(- A, A) with

f(z)=-f 2 A-A e

d2.

Remark. If f (z) is given by such an integral, it is obviously of exponential type A and belongs to L2(- oo, oo) on account of Plancherel's theorem. So the converse of the theorem is evident. The two results (the theorem and its converse) taken together constitute the celebrated and much used PaleyWiener theorem.

Proof of theorem. Is essentially based on Plancherel's theorem, combined with contour integration and the third and fifth results of the previous §. An easy but rather fussy preliminary reduction is necessary.

Paley- Wiener theorem

31

Plancherel's theorem says that M

P(2) = l.i.m. M-+ao

eiz" f (x) dx

-M

exists and belongs to L2(- oo, oo), and that, for xeR, 1

P X) =

l.i.m.

27E M-+ao

-M

e -;"x(p(,%) d, .

(Here, `l.i.m.' stands for `limit in mean (square)', and denotes a limit in L2(- oo, oo).) Our main task is to show that T(Z) - 0 a.e. for A> A and

A<-A. To this end, let us introduce the function

fh fh(z)

2h -h

f (z + t) dt;

fh(z) is clearly entire. Because f (z) is of exponential type A, we have, for any A' > A, If (z) I <, const.e"IZI,

and from this it is clear that also I fh(z) I <

const.eA'1Z1

(with a different constant). By Schwarz' inequality we also have Ifh(x)I

J(2hfx

If(t)IZdt)I h

so, since f(x)eL2(- oo, oo), fh(x) is bounded on R and in fact for x - ± oo. (This is the main reason for doing (1/2h) fh h on f !) If we call I fh(x) I = Mh, we see by the previous inequality for I fh(z) I and the third theorem (p. 27) of the previous § (applied in each half plane 3z > 0 and 3z < 0), that I fh(z) 15

MheA'111

Here, A' can be any number > A, so in fact (*)

Ifh(z)ISMhe"i1i.

The fifth theorem of § C (p. 29) shows moreover that (*)

uniformly for - L < y < L when x-). ± oo.

32

111 D The Paley- Wiener theorem

In order to prove that M

cp(A) = l.i.m. J M-00

eu" f(x) dx

M

vanishes a.e. for 2 > A and for 2 < - A it is more than sufficient to show the same thing with f (x) replaced by fh(x) in the right-hand integral, h > 0 being arbitrary. That's because "' e'zx.fh(x) sin 2h dx = h
which we can check using Fubini's theorem and the fact that

I fh

sin 2,h

- h e dx - A

2h

Taking a large M, look at f MM e"x fh(x) dx, assuming that 2 > A. Let y consist of the three upper pieces of the rectangular contour shown. Mi Y

-M+Li

M+Li

-M

0

M

Figure 9

By Cauchy's theorem, m

f eiax.fh(x)dx = JM

f eiz=.fh(z)dz. y

The contribution to 1. from the top horizontal portion of y has absolute value M I

e'x(x+'M) fh(x + iM) dx

-M

and this, by (*), is 2M.ljlhe-(o-A)M

a quantity tending to zero as M -+ oo, since 2 > A. Fix any large number L. If M > L, we write the contribution to f y from * To do this, one should start from the second formula at the top of p. 31 and conclude by applying Plancherel's theorem.

Paley- Wiener theorem

33

the right-hand vertical portion of y as

- \Jo+ JL /

fh(M + iy)-idy.

The second integral is in modulus e-(x-A)y, =

Mh

e-(A-A)L

Mh,

A-A

fL"O

again by (*), and we can make the quantity on the right as small as we like by taking L large. The first integral, however, has modulus

f I 'e-zylfh(M+iy)Idy Jo

and this, for any fixed L, tends to 0 as M --, oo according to (*). We see that the contribution from the right vertical portion of y tends to zero as M -+ oo; that of the left vertical portion does the same, as a similar argument shows. In fine, f1eizzfh(z)dz-+0 as i.e., fM

-M

e"fh(x) dx - 0 as M - o0

when A > A. For A < - A we establish the same result using a similar argument and this contour: -M

0

M x

-Mi

Figure 10 Thus, SMMeiZ" f h ( x ) dx - 0 pointwise in A f o r J A I >A as M -* co. However,

for some sequence of Ms tending to co, the integrals in question must tend a.e. to Mly . I.m 00

f

'"

-M

sin Ah

eiAxfh(x) dx =

h

90.).

(L2 convergence of a sequence implies the a.e. pointwise convergence of some subsequence to the same limit.) This means that (sin 2h/lh)gp(1) = 0

34

111 D The Paley- Wiener theorem

a.e. for I A I > A, whence tp(A) = 0 a.e. for I A I > A. (sin th/Ah vanishes only

on a countable set of points!) The Fourier-Plancherel inversion formula now gives us, for xeR,

1

M

e - iXzgp(A) d1. f(x) = l.i.m. M-'aD 2n -M

21 fA

e-

i-p(2) d). a.e.

In fact, we have A

P z) =

e-,:z(p(A) d l

2n

-A

for all complex z. That's because each of the two sides is an entire function

of z. Since these two entire functions coincide a.e. on 68, they must be everywhere equal by the identity theorem for analytic functions. Our theorem is completely proved. If we refer to the fourth theorem of §C (p. 28), we see that we can give the result just proved a more general formulation. The statement thus obtained,

which we give as a corollary, also goes under the names of Paley and Wiener.

Corollary. Let f(z) be entire and of (some) exponential type, with f(x)eL2(R), and let limsuploglf(iY)I = b, Y

limsup

log l f (iY) I

Y- - OD

= - a.

IYI

Then

f(z) =

27r

f ba e

-'z'9(A)

where cpeL2(a,b).

Proof. If f is of exponential type A, say, we certainly have fA

f(z)= I

e-i2z(p(,.)d2

-A

by the theorem, so, if xeR, If(x)I

<2n

J(2A

f AA IP(i)I2d2),

Paley- Wiener theorem

35

a finite quantity, i.e., f is bounded on P if we only assume f eL2(P), provided that it is entire and of exponential type. Applying the fourth theorem of §C in

each of the half planes _(z > 0, 3z < 0, we now see that I f (z) 15 const.eb, 3z > 0; If (z)I < const.e-°13Z1, .3Z<0. Symmetrize by taking g(z) = e"Z f (z) with y = (b + a)/2. Then g(z) is also

entire, g(x)eL2(P), and, by the previous relations, Ig(z)I < const.e(b-')13'112

in both upper and lower half planes, i.e., g(z) is of exponential type (b - a)/2.

(We see at this point that (b - a)/2 cannot be < 0 unless f (z) - 0 - the reader is urged to think out why this is so.) Use the above theorem once more, this time for g(z). We find 1

g(z)=2n

(1/2)(b-a)

e-; 0()d2

-(1)2)(6-a)

with a certain (ieL2. Going back to f (z) = e-'yZg(z), we have

f(z)=I establishing the result with gyp(A) = O(% - y).

Scholium. The Paley-Wiener theorem has more content than meets the eye. Suppose that cpeL2(a, b); then

f(z)= I

be-iZxgp(.)d2

Ja

is entire, of exponential type, and belongs to L. on P. We can also easily verify directly that limsup Y- ID

log I P X + iy) I

y

and

limsup r_-00


log I f (X + iY) I

\-a

IYI

for each real x. These inequalities remain true as long as q vanishes a.e. outside [a, b]. If, however, we take for [a, b] the smallest closed interval containing (P's support - the so-called supporting interval for 9 - the inequalities become equalities!

36

111 D The Paley- Wiener theorem

Without loss of generality, take x = 0, and suppose, for instance, that limsup log I f0y) = b' < b. Y_ OD

Y

The above corollary then shows that P(,l) =1.i.m.

f M eux f(x) dx

Mao J M

in fact vanishes a.e.for 2 > b'. The support of cp would thus be contained in

[a, b'], so [a, b] would not be Bp's supporting interval, and we have a contradiction. If [a, b] is the supporting interval of cp, we must therefore have limsup

log I f (iy) = b.

Y _'O

Y

It is clear that this b can only come from the portions

_ fb 1

27r

e-iz'gp(2)d2

b-

of the integral giving f (z), s > 0. We know that (p(2) cannot vanish identically a.e. on any interval of the form [b - e, b], but it is still quite conceivable that fbb-e

eyzrp(1)

dl

could come out much smaller than eby for large y on account of cancellation.

The Paley-Wiener theorem teaches us that such cancellation cannot take

place. This is a remarkable and deep property of (square) integrable functions q (2).

There are versions of the Paley-Wiener theorem for other spaces besides L2(R). The following is frequently used. Theorem. Let gp(2)('eLl(R) have compact support, put

e"w(2)d2,

f(z)=2 J and suppose that limsup Y-00

log I.f (iy) Y

= b,

limsup log l f 0y) I IYI y_ -'0

Then p (A) vanishes a.e. outside [a, b].

_

- a.

Condition f'-,, (log' I f(x)I/(1 + x2))dx < oo

37

Proof. This would be part of the corollary to the Paley-Wiener theorem, save that f (x) is not necessarily in L2(R). For h > 0, put 'Ph(A) =

2h

then II (Ph - (P II 1

f

T (A + 'C) d-r; h

0 as h -> 0 and we need only show that (Ph(2) = 0 for

)4[a-h,b+h]. Write fh(z) = 2n

f

00

d2;

then

fh(z) =

sink hz

f (z),

z

so, since f (x) is clearly bounded on R (c, being in L1(R)), fh(x)eL2(R). By the

hypothesis we now have limsup

log I fh(OY) I

y-°°

Y

limsuplog1fh(iY)I

y-'-°°

= b + h,

= -a+h.

IYI

Therefore Kph(),) = 0 outside [a - h, b + h], by Paley-Wiener. We're done. Remark. The same result holds (with almost the same proof) if we replace 9(A)d. (with coeL1 and of compact support) by dµ(.1), p being any finite signed measure of compact support. E.

Introduction to the condition

1',,, (iog + I f(x) I /(1 + x 2)) dx < oo The entire functions of exponential type considered in the previous § certainly satisfy this condition, as do those arising in the study of many questions in analysis. We will meet repeatedly with such functions in the following chapters of this book, and the rest of the present chapter is mainly concerned with them. It turns out that the boundedness condition

_00 ,0

log, If W1 1 +x2

dx
implies many results for entire functions f of exponential type. The following simple result is very useful, and all that one needs for many investigations.

38

HI E The condition f '-,,(log' I f(x)I/(1 +x2))dx < oo

Theorem. Let f(z) be regular in 3z > 0 and continuous up to the

real

axis. Suppose that log I f (z) I < O(I z I) for I z I large when 3z > 0, that

limsup

log If (iy) I = A, y

and that -00 00

log, If W1 dx

1+x2

< oo.

Then, for .rjz > 0,

logl f(z)I S A,3 z+ fO.3zlog+If(t)I dt. Iz-t 12 7r

Proof...3z/I z - tI2 = 91(i/(z - t)) is, for each tell, a positive and harmonic function in 3z > 0. For fixed z with positive real part we have, by calculus,

f_'

1 7r

3z Z

IZ

dt =1,

and, if z --+ xo a I8, z

sup

Ii-xol_a

-

1

IZ - t1 2

t2 + 1

0

for each S > 0. Therefore, if P(t) is any positive continuous function with

-

+(ti2 dt < oo, 1

we have by the usual elementary approximate identity argument (no need to refer to Chapter II, §B, here!),

1 °°z n

2

-.Iz-t1

P(t)dt

P(xo)

for In our present situation P(t) = log+ If (t) I is continuous on R, so if we put UO z = I

00

zlog+If(t)I

Iz-t1 2

dt

for ,rjz > 0, U(z) is positive and harmonic in the upper half plane and

Positive harmonic functions-representation as Poisson integrals

39

U(z) log' If (xo)I for z xoaR. We see that in 3z>0, log If (z)l- U(z) is subharmonic, is < 0(I z I) for large I z I, and has boundary values 5 0 everywhere on R. Moreover,

log If (iy) I - U(iy) < Ay + o(y)

for y -+ oo. The fourth theorem of §C (p. 28) (or rather its version for subharmonic functions) now yields without further ado log I f (z) I - U(z) < A,,3z,

3z > 0,

that is,

log f(z)I,0. We are done. Later on we will give some refined versions of this result. Their derivation requires more effort.

F.

Representation of positive harmonic functions as Poisson integrals

In order to proceed further with the discussion begun in §E, it is simplest to apply the Riesz-Evans-Herglotz representation for positive harmonic functions, although its use can in fact be avoided. We explain that representation here, together with some of its function-theoretic consequences.

1.

The representation

Theorem. Let V(w) be positive and harmonic for IwI < 1. There is a finite positive measure v on [ - it, n] with V(w) = 2n

I

J

R

w

-Ie I 2 dv(t),

I w I < 1.

Sketch of Proof. By the ordinary Poisson formula, if R < 1, we have, for Iwl < R, 1

V (w) = 27t

fR R2 - Iwlz -RIw - Re" l2

V(Re" )dt,

that is, for Iwl < 1,

(:)

,, I

V(Rw) = 2 J

I -Ie1I2 V(Re")dt.

40

111 F Positive harmonic functions

In particular, rz

V(Re")dt = V(0) < oo,

2n

rz

no matter how close R < 1 is to 1. We must now use some version of Tychonoff's theorem in order to obtain the measure v.

Take any sequence {R^} tending monotonically to 1, for example R. = 1-1/n. The functions cp^(t) = V(R^e") are all >- 0 and have bounded integrals over [ - it, n] by (,*, ); we can therefore (by using Cantor's diagonal process) extract a subsequence of

these functions, which we also denote by {cp^} (so as not to write subscripts of subscripts!), having the property that

L(G) = lim f

._.

G(t)rp^(t)dt rz

exists and is finite for G ranging over a countable dense subset of '( - it, n). If, however, G and G'e%(- it, n) and 11 G - G'11 < e, we have, for every n, rz

G(t)cp^(t)dt -

G'(t)(p^(t)dt

f.rz

fS

ecp^(t)dt = 2neV(0),

J rz

R

for every Ge'(- n, n), and IL(G)I<2nV(0)IIGII L is thus a bounded linear functional on W( - it, n); it is moreover positive because,

so in fact L(G) = lim^~ f ,,G(t)cp^(t)dt exists

if Gc-W (- n, n) and G >,O, L(G),>0 since cp^(t) 3 0 for each n. By the Riesz representation theorem there is thus a positive finite measure v on [ - it, n] with rz

L(G) = J

G(t)dv(t),

Ge'(- n, n).

rz

Taking in particular G(t) _ (1/2iX 1- I w 12)/ I w - e" 12 with a fixed w, I w I < 1, we

obtain 1

2n

*

1-IwI2

-rzlw-eU12 1

~°°2n 1

= ,21r lim -

dv(t) = L(G)

1* 1- Iw12 -rzlw *

-e I

1-Iw12 -rzlw-eit12

cp^(t)dt

V(Redt.

Referring to (*), we see that the last expression equals lim^V(R^w) = V(w), V being certainly continuous for Iwl < 1. This completes the proof.

Scholium. Once we know that the measure v giving rise to the desired representation exists, we see that the passage to a subsequence of the p^(t) in

1 Their representation as Poisson integrals

41

the above proof was not really necessary (although we are only able to see this once the proof has been carried out!). Suppose G(t) is any continuous function on [ - n, n] with G( - n) = G(n);

then, by the elementary approximate identity property of the Poisson kernel 1

1 - IwI2

1

('"

1- RZ

G( 9) d 9, G(tO

21J_ le"-Rei9I2 rz

uniformly(' for - n S t < n as R -+ 1, so, by Fubini's theorem,

J

J

j n

as R --> 1. This simple fact can frequently be used to get information about the measure v. The reader should think through what happens with the argument just given when Ge'(- n, n) but G(- n) 96 G(n). Here is a hint: we at least have

1 - R2 2n -n le ` -Rei9 1

('"

J

G(t)' 2

G(9)d9

G(n) +2 G(- n)

t # - n, n,

t= _n , n,

as R -1, although the convergence is no longer uniform. The integrals on the left are, however, bounded.

Terminology. The situation of the scholium is frequently described by saying that V(Rei9)d9 -> dv(9)

for R -1, or by writing `V(Rei9) - dv(9) as R -> 1' (with a half arrow).

Theorem. Let v(z) be positive and harmonic in 3z > 0. There is a positive number a and a positive measure µ on l with °°

dµ(1)2

_1+t

< o0

42

III F Positive harmonic functions

such that

3z

v(z) = a3z + 1 f 00 it

2

dµ(t) for

,3z > 0.

-.Iz-t1

Proof. From the previous theorem by making the change of variable

i-z z--'w=i+z

which takes 3z > 0 conformally onto the open unit disk. Everybody should do this calculation at least once in his or her life, so let us give the good example. The conformal mapping just described takes v(z) to a positive harmonic function V(w) = v(z) defined for I w I < 1, so we have

fa

V(w) = 2n

R

Ieit 1 w12 dv(T)

with a positive measure v according to the result just proved. We write T here because t will denote a variable running along the real axis. We have w = (i - z)/(i + z), and the real t corresponds in a similar way to

elt - i

t

i + t

Therefore 2

1-1w12

Ie;t-w12

i-z i+z

i-t i+t

2

43z(t2+ 1) (1i+z12-1i-z12)Ii+tI2 = I (i - txi + z) - (i + t)(i - z)12 = 12i(z t)12

-

_

_z

Iz-tl2

(1 + t2).

Since a t'* = - 1 corresponds to t = oo, we see that 1

2

J

1 - 1w12

°°

1

dv T= 7r

f

3z

dp(t),

-DIz-t12

where dµ(t) = 2(1 + t2)dv(T). We are finally left with the (possible) point masses coming from v at - n

and iv; their contribution gives us the term adz with a >, 0. Recalling that v(z) = V(w), we see that the proof is complete.

2 Digression on a.e. existence of boundary value

43

Remark. IfF(t) is a continuouss function of compact support,

b( t)dµ(t) = ylim J

JT

CD(t)v(t + iy)dt.

To see this, just use the approximate identity property of (1/7t)(,z/Iz - t12) (§E) - compare with the above scholium. 2.

Digression on the a.e. existence of boundary values

The representation derived in the preceding article can be combined with the result in Chapter II, §B, to obtain some theorems about

the a.e. existence of (non-tangential) finite boundary values for certain classes of harmonic and analytic functions defined in { I w I < 1) or in {,3z > 01. Although this is not a book about boundary behaviour or Hp spaces, it is perhaps a good idea to show here how such results are deduced, especially since that can be done with so little additional effort. Theorem. Let V(w) be positive and harmonic in {IwI < 11. Then, for almost every 9, the non-tangential boundary value

lim V(w) W

ei9

exists and is finite.

Proof. By the previous article, V(w)

_1 2n

1-Iw12 -Rlw-a"IZdv(r) c

where v is a finite positive measure. A theorem of Lebesgue says that v'(9)

d9 (J:dv(t))

exists and is finite a.e. And by the remark at the end of Chapter 2, §B, V(w)-+v'(9) as w-L-ei9 wherever v'(9) exists and is finite. We're done. Corollary (Fatou). Let F(w) be analytic and bounded for I w I < 1. Then

lira F(w) exists for almost all S.

Proof. If IF(w)I < M in {I wl < I), M + 91F(w) and M + 3F(w) are both positive and harmonic there.

III F Positive harmonic functions

44

Notation. Let F(w) be analytic and bounded for { I W I < 11. The nontangential limit

lim F(w) w -L-ei9

(which, by the corollary, exists a.e.) is denoted by F(ei9). The function F(ei9),

thus defined a.e., is Lebesgue measurable (and, of course, bounded). Theorem. Let F(w) be analytic and bounded for IWI < 1. Then F(w)

_

C' F(e''g)ei9d 9

1

2n

I

ei9-w

'

IwI<1.

Remark. Thus, the boundary values F(ei) (which are defined a.e.) serve to recover F(w).

Proof. For each R < 1, we have, by Cauchy's theorem, 1

F(Re'9)e19d9

2n

e"-- w

F(Rw)=- l

,

Iwl<1.

Fix w, and take R = R" with R. n' 1. We have I F(R"ei9)I <, M, say, and F(R"ei) n F(e19) a.e. by the corollary. The result follows by Lebesgue's dominated convergence theorem. Lemma. Let F(w) be analytic for Iwl < 1 and suppose that IF(w)I < 1 there. Let tai = 1, and take

E={9:F(ei9)=a,-n
Remark. The result is also true when loci < 1. But then the proof is more difficult.

Proof of lemma. Take, wlog, a = 1. We must then prove that F(w) _- 1 if

IEI>0. The function ((F(w) + 1)/2)" is analytic in {I w I < 11 and in modulus 5 1

there, so, by the above theorem applied to it,

(F(O)+1)"=2n f ""(F(e'2+1)"d9 2 Here, (F(ei9) + 1)/2 = 1 if 9eE, and, if 9, - it < 9 5 iv, is not in E, I(F(ei) + 1)/21 < 1, so ((F(e19) + 1)/2)" n + 0. We see by bounded conver* We follow the customary practice of denoting the Lebesgue measure of E S R by JEl.

2 Digression on a.e. existence of boundary values

45

gence that

f".(F(e"')+ I

I)nd,4

2

I > 0. Then the last relation combines with the

I

rz

previous to yield 1

=

n IEI

+o(1),

n-> oo,

after extracting an nth root. Since the right side tends to 1 for n -+ 00 we have finally F(0) = 1.

However, IF(w)I < 1, IwI < 1. Therefore F(w) -= 1 there by the strong maximum principle, Q.E.D. Theorem. Let f(t)eL1(-n,it), and put, for I w I < 1, I

G(w) = 2n J

-n n e"

+ w.f (t)dt.

lim G(w) exists and is finite a.e.

Then

ei9

w

Proof. Wlog, f (t) >, 0. Notice that G(w) is certainly analytic in {IwI < 1 },

and that JiG(w)=21

n

_I

_n

Iw-e

z 1I2f(t)dt

is >, 0 there. (Compare Chapter II, §A!) The function F(w) =

G(w) - 1 G(w) + 1

is therefore analytic and in modulus < 1 for IwI < 1. So, by a previous corollary,

F(e'9) = lim F(w) w

exists a.e. It follows that, whenever this limit exists,

lim G(w) wtei9

111 F Positive harmonic functions

46

must also exist and equal the finite quantity 1 + F(ei9)

1- F(ei9) unless F(ei') = 1. However, F(ei) can equal 1 only on a set of measure zero by the lemmaotherwise G(w) would equal oo everywhere in {I wI < 1), which is absurd. So limwt e;9G(w) exists and is finite a.e., as required. Scholium. Write w = re19 and suppose that feL1(- n, n) is real-valued. Then we have

1 fn e" + w 2n

f(t)dt

1 fa

1 - r2

2n _ 1 + r2 - 2rcos (S- t) f(t)dt

+

2r sin (9 - t) f(t)dt. 2n f-,, 1 + r2 - 2r cos (9 - t) i

R

We see that both

1f

.s U(re) =2n-

R

1 - r2 f (t) dt 1 + r2 - 2r cos (9 - t)

and

U(rei9) =

1

R

2n J_,,

2r sin (9 - t)

1+r2-2rcos(9-t)

f(t)dt

are harmonic in { 1 w I < 11, U(w) being equal to 91G(w) there, and U(w) equal

to ,3G(w), with G(w) the analytic function considered in the above theorem. U(w) is frequently called a harmonic conjugate to U(w); it has the property

that U(w)+iO(w) is analytic in {Iwl < 1}. It is an easy exercise to see that

any two harmonic conjugates to U(w) must differ by a constant; the particular one we are considering has the property that U(0) = 0.

By Chapter II, §B, we already know that

lim U(w) w -ZL-4,9

exists and is finite a.e.; it is in fact equal to f (9) almost everywhere. The above theorem now tells us that limwte;9U(w) also exists and is finite a.e., indeed,

Return to subject of §E -functions without zeros in 3z > 0

47

under the present circumstances, U(w) = 3G(w). This conclusion is so important that it should be stated as a separate Theorem. Let f eLl (- n, n). Then, for almost every cp, the limit of

1 fx 2r sin (9 - t) f(t) dt 2n _,,1 + r2 - 2r cos (9 - t) exists and is finite for reis

e'm

Notation. The non-tangential limit in question is frequently denoted by J(cp); for obvious reasons we often call 7 the harmonic conjugate of f. It is

also called the Hilbert transform of f. We will come back to the consideration of 7 later on in this chapter.

G.

Return to the subject of §E

1.

Functions without zeros in 3z > 0 Theorem. Let f (z) be analytic in ,3z > 0 and at the points of the real axis. Suppose that

log If(z)I50(Iz1) for 3z > 0 and I z I large, and that ('

log,

j

f (x) I

1 +I+x2

dx < cc.

Then, if f (z) has no zeros in 3z > 0,

logIf(z)I=A3z+IJ 7r

3zlog1 IZt)1dt 1Z

A = limsup log I f 0y) I y-ao

y

Remark. f (z) is allowed to have zeros on R.

Proof of theorem. With 1

U(z) =

n

3z log+ If(t)Idt Iz- t12

we have by §E log I f (z) I - A3z - U(z)

0

48

111 G Return to the subject of §E

for )z > 0. Since f (z) has no zeros in 3z > 0, v(z) = log l f (z) l - A 3z - U(z) is harmonic there, and we have, by §F.1,

a3z - I

v(z)

-

Iz

" t I2 dµ(t)

for 3z > 0, where a >, 0 and u is a positive measure on R. We use the remark at the end of §F.1 to obtain the description of U. According to that remark, if 'V(t) is continuous and of compact support,

fT D(t) du(t) = Y-O+ -00

1(t)v(t + iy) dt. w

In view of the formula for U(z), we also have

JT t(t)log + I f (t) I dt = lim Y0+ J 00

'(t)U(t + iy) dt. a,

Therefore

JT

(D(t)(log+ I f (t) I dt - dµ(t))

= lim Y-0+

4 (t)(U(t + iy) + v(t + iy)) dt f-,000

= lim

y-+0+ J00-.

(D(t)(log I f(t + iy) I - Ay) dt.

Under the hypothesis of the present theorem (analyticity up to and on R),

the last limit is just

f

'1(t) log I f (t) I dt.

Indeed, we easily verify directly (using dominated convergence) that

fi Iloglf(t+iy)I as y - 0 for any finite interval J on R. (The argument is essentially the same

as that used in the proof of Jensen's formula, Chapter I.) We thus have

f

F(t)(log+If(t)Idt-dµ(t))= J1(t) log I f(t)I dt

for each continuous function V of compact support, and hence log+ I f (t) I dt - dµ(t) = log l f (t) I dt.

2 Convergence of I'.(log - I f(x)I/(1 +x2))dx

49

Therefore, for 3z > 0, logI f(z) I = A,3z + U(z) + v(z)

=(A-a),3z+ a1

-

3zlog1f(t)I dt,

Iz -tl

by the formulas for U(z) and v(z). In order to complete the proof, we must show that a = 0. To see this, recall that by §F.1 the positive measure p introduced above satisfies

f

dµ(t)

o

1+t2< oo.

(We are already tacitly using this property - without it the formulas for v(z) and especially for log I f (z) I make no sense!) Therefore, by the evaluation of d1(t) just made, (*)

F

Iloglf(t)II dt 00

1+t2

<

00.

The formula for log I f (z) I just obtained now yields

logIf(iy)I Y

=A+ -aln (°°

log' f(t)Idt. t2 +Y2

Making y -> oo, we see from (*) that log I f (iy) I/y -+ A - a. Since we called

A =limsuP logIf(iY)I Y vwe have a = 0, and the theorem is proved.

Remark. Under the conditions of the present theorem, we see that limsup(logI f(iy)I/y) is actually a limit. Y- '0

2.

Convergence off °°

(log- If(x) I /(1 + x2)) dx

We are going to extend the work of the previous article to functions

f (z) having zeros in 3z > 0. For this purpose, we need some preparatory material. Lemma. Let S(w) be positive and superharmonic in { 1 w I < 1 } and suppose that lim,..l S(re") = S(e") exists a.e. Then, if I w I < 1,

in ,I

1w

_leiit I2 S(e'`) dT < S(w).

Remark. The assumption on the a.e. existence of the radial limit S(e") is

111 G Return to the subject of §E

50

superfluous. This is a consequence of a difficult theorem of Littlewood, which can be found in the books of Tsuji and Garnett. In our applications, this existence will, however, be manifest, so we may as well require it in the hypothesis of the lemma. Proof of lemma. Let I wo I < 1. By superharmonicity (mean value property)

If r increases towards 1 we can therefore

S(wo).

find

a

with

sequence

IwI<1

and w - wo

such

that

S(wo).

By superharmonicity of S(w) for I w I < 1 we have, for each r < 1, ('n

1

dr 5

27C J -n I wn - eiT12

Also, ((1 - I W. 12)/I W. - e'T is > 0 and tends to ((1- 1w012)/1w0 - e'T 12)S(e'T) for almost all T as n -,, oo (separate convergence

of the two factors!). Therefore, by Fatou's lemma, 1-IWO12

1

2n 1

5 liminf -

-,,I w - eir12

2n

di

and we are done. Lemma. If v(z) is subharmonic and 5 0 in,3z > 0 and v(t) = lime ,,1v(z) exists

a.e. on R, then, for 3z > 0,

v(z) < I f _-. z" _ 7E

I

v(t) dt. I2

c

Proof. Apply the previous lemma to S(w), given by the formula S

i-z _

i+z

-v(z),

and use the calculation made to establish the second theorem of §F.1. From this lemma we have first of all the very important and much used Theorem. Let f (z) be analytic for 3z > 0 and continuous up to R. Suppose also that logI f(z) I <, O(1z1) for Iz1 large, 3z>0, and that °°

log + 1 f (x) I

-00

1 +x2

dx < oo.

2 Convergence of f °°.(log - I f(x)I/(1 +x2))dx

51

Then, unless f (z) = 0,

(' °°

J

log I f (x) I dx 1 +x2

< oo.

Proof. Without loss of generality, f(i)

0; otherwise work with

(z+i)kf(z) instead of f (z) if f should have a zero of multiplicity k at i. By the theorem of §E, if we write A = limsup, (log I f(iy) I/y), the function v(z) = log l f (z) I - A3z is

ij

o

.3z l og+

I

f(t) I dt

0 for z > 0. It is not, however, harmonic there as in the previous

subsection, but merely subharmonic.

For z-+teR, the right-hand integral in the previous formula tends to log+ I f (t) I, since log+ I f (x) I is continuous on R. Therefore, when z -> t,

v(z)-+logIf(t)I-log+If(t)I= - logIf(t)I We may now apply the preceding lemma with v(t) = -log -I f(t) 1. Since f(i) 0, we find that

- oo < v(i) < -

1

°°

log I f(t) I dt. 1 + t2

We are done. Theorem. Under the hypothesis of the preceding result,

loglj(z)I< A+1

IZOgItI(t)Idt

for Zz > 0, where A = limsup log I f (ly) I , Y_00

y

the integral on the right being absolutely convergent.

52

111 G Return to the subject of §E

Remark. This is an improvement of the result in §E, where we have log + I f (t) I >, 0 instead of the (signed) quantity log I f (t) I in the right-hand integral.

Proof. Taking v(z) as in the proof of the preceding theorem we have, by the discussion there,

- 1r

v(z)

-

5

'3z log

1 f(t) 1

dt,

Iz-t1

,3z > 0, according to the above lemma. Adding 1

Adz+

°°

3zlog+lf(t)I

oo

Iz-t12

dt

to both sides of this inequality gives the desired result.

3.

Taking the zeros in .3z > 0 into account. Use of Blaschke products

Theorem. Let f (z) be analytic in J 3z > 0} and continuous up to R, and suppose that logIf(z)I 0, and that °°

_00

log, I P x) I

1+x2

dx < oo.

Assume also that f(0)

0.

the sequence of zeros of f (z) in 3z > 0 (with repetitions Denote by according to multiplicities). Then

I3" fr< oo.

Remark. The requirement that f (O) 0 0 is essential.

Proof of theorem. Since f (O) 0 0 and f (z) is continuous up to R, the z cannot accumulate at 0, i.e., c for some c > 0. We may, wlog, take c = 3, for, if c is smaller than 3, we can work with f ((3/c)z) instead of f (z).

3 Functions with zeros in ,Zz > 0. Blaschke products

53

The integral i r00 t+t2+1)log+lf(t)ldt

t

(z

J

1 is absolutely convergent for 3z > 0 because of our condition log, If(t)I dt -OD f'O

1 + t2

< oo;

it therefore represents a function analytic in that half plane whose real part

is none other than

_

' f 00

Iz3 tlZlog+lf(t)Idt.

From this observation and the result of §E we see that eiAZf(z)

g(z) =

exp1- J - (J_ t+tz+l )log+jf(t)jdt} l

W

is analytic and in modulus < 1 for 3z > 0, where the constant A is defined in the usual fashion. We have g(i) 0, since (here) all the I zn I are > 3. For each N, apply the principle of maximum to

n

in 3z > 0. Since I g(z) I < 1 there, we find

i-z

0 < Ig(i)I

i - Zn

2z

For each n, we have Ii - Zn

1-Z

1 - Win IznIZ

2z

Here, however, I z,, I > 3, so the last expression is certainly

< 1- 1 + 1/Iznl Iznlz' as is evident if we look at the image of the circle I co I = 1/I z I (< 3) under the linear fractional transformation 2

ZZn w1+COIZnl2.

111 G Return to the subject of §E

54

Figure 11

Substituting into the previous inequality and taking logarithms, we get

-00 <

log I g(i) I

< ylog(1-Iz II+I1'I

IZZ)

and this is in turn N

3 N .3zn

JZn

2IznI

Iznl + 1 Iz12

2

IZ7I2

since I zn I > 3.

We thus have N 3z" < 2 10g IznZI2

<3

1

< 00

lg(i)I

for all N, and our theorem follows on making N -> oo. Theorem. Let f(z) be an entire function of exponential type with

f°° log, I f (x) I dx < oo. _00 1+x2 There is an entire function g(z) of exponential type with no zeros in 3z > 0 and Ig(x)I = If(x)I for - oo < x < oo.

Proof. Let (),,} denote the set of zeros of f (z) in 3z > 0, and {µn} all the other zeros of f (z) (repetitions according to multiplicities, as usual). Wlog, f (0)

0, otherwise work with f (z)/z' instead of f (z). The Hadamard

factorization of f (z) can then be written (

f(z)=Ce"Rim{ fl 11"1
z An

fl Iµ"I5R

(1

z ne=iµ"

µJ

3 Functions with zeros in 3z > 0. Blaschke products

By the previous theorem, the sums theorem (§B), the sums

55

and by Lindelof's

+ Iµ,l
IAnl
are bounded in absolute value. The sums

+Y

1

IA.ISR"n

1

IPn14 RPn

are therefore also bounded in absolute value, so, by the (easy) converse of Lindelof's theorem, the products z z I e /zn H

Cebz H 11-

'Tn

IznI
z

(1 - - ez/A. µn

converge u.c.c. in the complex plane to an entire function g(z) of exponential type as R --' oo.

For real x, Ox) __

f (x) a

1..14 R

x/A.) exp 2i (1 -

JAn 111212 x

,

the product being u.c.c. convergent on R in view of the condition

Y 1i2 < co . The right side of the above expression is clearly of modulus 1 on R, so If(x)I = Ig(x)I there. And g(z) has no zeros in 3z > 0.

Theorem (Riesz-Fejer). Let F(z) be entire and of exponential type, let F(x) >, 0 on R, and suppose that

log' F(x)

1+x2

dx < oo.

Then there is an entire function f(z) of exponential type without zeros in ,Zz > 0 such that F(z) =f(z) f (zl. In particular, F(x) = I f (x) I2, x e R.

Proof. Since F(x) is real on R, F(z) = F(z) by the Schwarz reflection principle, so, if 32 > 0, A is a zero of F(z) iffIis also a zero thereof, with the same multiplicity. Because F(x) > 0 on R, any real zero of F must have even multiplicity. Denote by {An} the set of zeros of F(z) in 3z > 0 (repetitions according to

56

III G Return to the subject of §E

multiplicities); then, by the observations just made, the Hadamard factoriz-

ation of F must take the form

/

r

z

11-

F(z) = Ce"Z li m {

z )e'/`--' 'ln

H

z

i-z

e=/z

znISR

z (1__)2e2}.*nHere,

the a are certain real numbers corresponding to the possible real Using the fact that zeros of F(z) - of course, there may not really be any F(x) >, 0 we easily check that C > 0 and that b is real. oo, and As in the proof of the preceding theorem, we this, together with the Lindelof theorems of §B, implies that the products

-

JCebz/2 f 1- A.z

eZ/r^ f

z

i -- ez/an an

converge u.c.c. in C to a certain entire function f (z) of exponential type as R -+ oo. We clearly have

F(z) =f (z)f (f),

and f (z) has no zeros in 3z > 0. As required. Theorem. Let f (z) be entire and of exponential type and suppose that (' °° log+ I f (x)1

j - ao 1+x2

dx < ao .

the set of zeros off (z) in 3z > 0 (repetitions according to Denote by multiplicities), and put A = limsup Y_ 00

log If (iY)1

y

Then, for ,3z > 0,

logI f (z)1= AZz +

log

1 - z/2

1 - z/X

+Ir -

Iz-t121og1f(t)Idt.

Proof. With the entire function g(z) used in proving the theorem before the

last one, put

j

-2iy z}. * As long as F(O) # 0. Otherwise, an )additional factor zzk appears on the right, and the description of f(z) following in text must be modified accordingly.

3 Functions with zeros in jz > 0. Blaschke products

57

We have, of course, (*)

34Z < oo,

so G(z) is an entire function of exponential type because g(z) is, and, for x e R, I f (x) I. G(z) is without zeros for,3z > 0, and, in view of the description of g(z) given where it was introduced,

I G(x) I = I g(x) I

i

1-

G(z) =f (z)' lm 1znl< R

1

- z/.1 )

Using (*) and the fact that A. -+ oo, we readily verify directly that the infinite product

1 - z/.1 1 - z/R is u.c.c. convergent for 3z >, 0; in the upper half plane, G(z) evidently equals

f (z) divided by this infinite product.

For ,3z > 0,1(1-

I > 1, therefore I G(z) I > I f (z) I. Hence, if

we call

A' = limsup log I G(iy) I y

o0

Y

we have

y-y

A' 3 limsup log I f (iY) I = A. Apply now the theorem of article 1 to G(z), which has no zeros in 3z > 0. Since I G(t) I = If (t) I for real t, we find that

logIG(z)I =A'Zz+ 1 7r

-z -.IZ -t12

log1f(t)Idt

for ,3z > 0. In view of the relation between G(z) and f (z), this yields log If (z) I= A'.3z + Y log

+n

11- Z/.1 I - Z/

logIf(t)Idt, Zz>0.

I3z

We will be done when we show that A' = A. Indeed, we have already seen that A',> A so it is only necessary to prove that A',< A. To this end, consider the functions

'

"'7

I - Z/2.

GN(z) =f(Z) Ill (1

- zA.

58

111 H Levinson's theorem on density of the zeros

For each fixed N, I GN(x)I = W4 (x) I on IR, and GN(i y)/ f (iy) -1 for y -+ co, since the product on the right has only a finite number of factors. Because of this,

limsup

log I GN(iY) (

v- 00

y

log I f (iy) I = limsup = A, y-. 00

y

whence, by the second theorem of article 2,

logiGN(z)ISA.3z+it

IZ-tlzloglf(t)ldt

for ,3z > 0. Make now N- oo ; then GN(z) N G(z) u.c.c. in 3z > 0, so finally

log IG(z)ISA,,3z+n

Iz-tlzloglf(t)ldt

-

there. However, the left side equals

A z+

Iz-tl2loglf(t)Idt.

Therefore A'< A, whence finally A' = A, and the theorem is proved.

(.

Remark. The expression

I

1 - z/2 1 - z/.I.

is called a Blaschke product.

Problem 4 Let 9 and f eLl(l) be functions of compact support. Let [a,b] be the supporting interval for cp (that's the smallest closed interval outside of which (p = 0 a.e.), and denote by [a', b'] the supporting interval for '. Prove that the supporting interval for 9 is precisely [a + a', b + b']. (Note: By pp

we mean the convolution qq(A - T)1I/(T)dT.)

((P * i)(A) = J

H.

Levinson's theorem on the density of the zeros We are going to close this chapter by proving a version of

Levinson's theorem on the distribution of the zeros of entire functions of exponential type for such functions f which also satisfy the condition

('°° log+lf(x)Idx J

1 +x2

< co.

This version is in fact due to Miss Cartwright and, although sufficient for

1 Kolmogorov's theorem on the harmonic conjugate

59

most applications in analysis, is not the most general form of Levinson's theorem. For the latter one should consult the books by Boas or Levin, or, for that matter, the one by Levinson himself. The proof to be given here depends on Kolmogorov's theorem on the

harmonic conjugates of integrable functions, so we turn first to the establishment of that result. 1.

Kolmogorov's theorem on the harmonic conjugate

Let f (z) be analytic and bounded for 3z > 0. An obvious application of a result of Fatou (the corollary in §F.2) shows that the boundary value

f(t)=slim f(t+iy) exists for almost every real t. In the application of the following lemma to be given below, this fact can also be verified directly; the reader interested in economy of thought may therefore include it in the hypothesis if he or she

wants to. Lemma. Let f (z) be bounded and analytic in 3z > 0. Then

.f(z) =

-'f' 71

Z

Zz

there.

Proof. If z lies inside the contour r shown below, we have, by Cauchy's theorem, f (z)

=1 C .f (c) dC 2ni Jr C - z

III H Levinson's theorem on density of the zeros

60

We fix R and make h - 0 through some sequence of values; since If (t + ih) I S C say and f (t + ih) --*f (t) a.e. for h -+ 0, we have, by dominated convergence,

RRf(tZdt.

t RRf(t+ihzdt-

Similarly,

('"f(ih±ReiRe'sd9- f f( o

ih+Re -z

Jo Te -z

iRe'sd9

as h -4 0. We thus see that (*)

f(z)=2TCi

j_Rt `tZdt+2nJ'f(Re19) a

R e' sd9.

Taking the relation 0

__

f fG)

1

2ni

JrC-zd 0, we see in the same way that

and making h

1 ('R f (t)dt O=-

1

2ni J-R t-i +2n

('n f (Reps) o

s Re's - zRed9.

Subtract this equation from (*). We get 1

J-Rf(z)=1tIt-ZIZf(t)dt+n fo(Re'

-z)(Re's-zd9.

Since f is bounded, the second term on the right is O(1/R) for large R, so, making R --> oo, we end with I foo Q.E.D. - I z 3z f (t) dt. f (z) =

Scholium. The reader is invited to obtain the lemma from the second theorem of §F.1 and the remark thereto (on representation of positive harmonic functions in 3z > 0). Suppose now that u(t) is real valued and that

-1+tt2dt < oo. I

Then the integral

-Oz

-c

fO (

1

t

+

t2

+

1

u(t) dt

1 Kolmogorov's theorem on the harmonic conjugate

61

converges absolutely for ,3z > 0 and equals an analytic function of z - call it F(z) - there. (RRF)(z) is simply the by now familiar harmonic function 1

°°

n

-3z

Iz-

tl2 u(t)dt;

(3F)(z) is equal to

I f- (9?z-t n - 00

Iz

t

tI2 + t2 + 1

u(t)dt.

Let us call the former expression U(z) and the latter U(z). Both are real valued and harmonic in {.3z > 0}, and the latter is a harmonic conjugate of the former for that region since U(z) + iO(z) is equal to the function F(z), analytic therein. In order to examine the boundary behaviour of U(z) and U(z) we may first map {,3z > 0} onto { I wl < 1 } by taking w = (i - z)/(i + z) and then appeal to the results in Chapter II §B and in §F.2 of this chapter. From the first of those §§ we see that one simply has U(t + iy) - u(t) at almost every t e IIB when y - 0. The behaviour of U(t + iy) for y -> 0 is less

transparent. According to the last theorem and scholium following it in §F.2, U(t + iy) must, however, tend to a definite finite limit for almost every t e R as y -* 0. It is not very easy to see how that limit is related to our original function u; we get around this difficulty by denoting the limit by u(t) and calling u the Hilbert transform (or `harmonic conjugate') of u. Under certain circumstances one can in fact write a formula for u(t) and

verify almost by inspection that U(t + iy) tends to u(t) (as given by the formula) when y When this happens, we do not need to use the general result of §F.2 to establish existence of lim,,-o U(t + iy). That will indeed be the case in the application we make here; the reader who is merely interested in

arriving at Levinson's result may therefore include existence of the appropriate Hilbert transforms in the hypothesis of Kolmogorov's theorem,

to be given below. It is, however, true that the Hilbert transforms in question do always exist a.e. Here is a situation in which the existence of lim,,..o U(t + iy) is elementary.

Suppose that the integral (o, u(xo - r) - u(xo + T)

Jo

dT

T

is absolutely convergent; this will certainly be the case, for instance, if u(t) is

III H Levinson's theorem on density of the zeros

62

Lip I (or even Lip a, a > 0) at x0. Then, if we write

u(x) = o

1 u(xo - r) - u(xo

1

t

o

7 1

+9 di

tu(t) (' + Jrx° +It2+Idt+7 J 1

tx01> 1

X0_1 1

t

1

x

t+t2+1

0

1 Ju(t)dt,

we easily verify that

iJ(xo + iy) = JT((X0

- t)2+

1

u

Y2+t2+lJ(t)dt

tends to u(xo) as y-+0. Just break up the integral on the right into two pieces, one with x0 - I < t < x0 + 1 and the other with I t - x0I > 1. The first piece is readily seen to tend to the sum of the first two right-hand terms in the formula for u(xo) when y -> 0, and the second piece tends to the third term. In proving Levinson's theorem, the functions u(t) which concern us are of the form u(t) = log+ if (t)I or u(t) = log- if (t)I with f (z) entire and such

that IlogIf(t)II cD

1 +t2

dt < oo.

The function log+ I f (t) I is certainly Lip 1 at every point of R - f (z) is analytic! And log- If (t) I is Lip 1 at all the points of R save those isolated

ones where f has a zero. In either case, then, u(xo) is defined by the elementary procedure just described for all x0elB except those belonging to some countable set of isolated points. Our purpose in dwelling on the above matter at such length has been to

explain that the proof of Levinson's theorem to be given below does not

really depend on deep theorems about the existence of the Hilbert transform. The question of that existence is, however, close enough to the subject at hand to require our giving it some attention. The reader who

wants to learn more about this question should consult the books of Zygmund or Garnett (Bounded Analytic Functions) or my own (on HP spaces). There is also a beautiful real-variable treatment in Garsia's book on almost everywhere convergence.

Without further ado, let us now give Kolmogorov's theorem. Let u(t) be real valued, let J°°

ut

oo,

1 Kolmogorov's theorem on the harmonic conjugate

63

and put x

il(x) = lira

1

2

2+ Z t

t

(x-t) +y

y-0+7[

t +1

u(t)dt,

the limit on the right existing a.e. Then, if A > 0,

tZ+ dt

4 1

{ia(01>A)

°°

_

Zu(t)I

t+1

dt.

Proof. Consider first the special case where u(t) >, 0; this is actually where

most of the work has to be done. The following argument was first published by Katznelson, and is due to Carleson. Wlog, u(t)

1

1+t2

it

_

dt=1.

Put

F(z)=i

f- (z 1 t+t2+1)u(t)dt;

this function is analytic in 3z > 0 and has positive real part there. Also, F(i) = 1. For almost all tel', 9IF(t + iy) -* u(t) and

ZF(t + iy) --) u(t) as

Fix now any 2 > 0, and take f(z)=1+F(z)-2

F(z) + 2

for 3z > 0; f (z) is analytic there and has modulus at most 2. For almost every teR, f (t) = limy.o+ f (t + iy) exists, and can be expressed in terms of u(t) + iu(t).

By the lemma, for 3z > 0, f(z)=7E

_'Iz tl2f (t)dt,

64

111 H Levinson's theorem on density of the zeros

therefore *

()

1

91f(t)

7rJ_'0 1+t2

dt = 91f(i) = 91

1+

1-A) 1+.1

=

2

1+.1

The transformation F --+f =1 + (F - 2)/(F + 2) makes the half-plane 91F > 0 correspond to the circle If - 1 I < 1 and takes the two portions of the imaginary axis where IFI > 2 onto the right half of the circle

If-1I=1.

I +i

0

1 -i

Figure 13

From the picture we therefore see that 9? f (t) >,I whenever

I F(t)I =

I u(t) + iu(t)I > A, hence, surely whenever I u(t)I >, A. Since we always have 91f(t) > 0, we get from (*), 1

dt

n ft IQ(,H3d}1+t2

\

2

2

2 ('u(t)dt

1+2 < 21

?r2J_w1+t2'

since we assumed that the integral alone is equal to it. By homogeneity, we therefore get

dt f(MOI->4 -f -+t 2

2 ('° u(t)dt J

2

for the case where u(t) > 0. In the general case of real u, write u(t)=u+(t)-u_(t) and observe that u(t) = u + (t) - u _ (t), whence ( t: Iii(t)I > A) c {t: I u+(t)I >, 2/2} u {t:Iu-(t)I > 2/2).

2 Functions with only real zeros

65

The inequality just established may be applied to each of the functions u+, u_ and we obtain the desired result by adding, since °°

u+(t)

u(t)Idt

°

-.+t2dt+

f

u_(t) 1+t2dt1+t2 fo

Kolmogorov's theorem is proved.

When A -+ oo, the right-hand side of the inequality furnished by Kolmogorov's theorem may be replaced by o(1/2). Corollary. Let u(t) be real-valued and $00 ,,(Iu(t)I/(1 + t2))dt < oo. Then

dt

('

2 = oI

1

«\

J{Iu(t)I>x} 1 +t

for A-+oo.

Proof. Take any e > 0. We can find a continuously differentiable function q of compact support with

J

1 u(t) - 001 dt < e. 1 + t2

Referring to the discussion preceding the statement of the above theorem, we see that ip can be readily computed; in fact O(x)

1

'cp(x-T)-(p(x+T)dr

f

n

o 1

+7C

T

('"+1 trp(t)dt

1

('

f

J x-1 t2+1 +n t-xl>1

t

Vt+iT+

1(p(t)dt.

Because cp'(t) is bounded and of compact support, the expression on the right is bounded; it is even 0(1/ I x 1) for large x. So I O(x) I < M, say (M, of course, depends on (p, hence on e!), and, if A > 2M, the set {t: l u(t)I > A} is

included in {t: I u(t) - ip(t)I > ,/2}. Applying the theorem to the function u - gyp, we therefore find that dt

8e

2< J{iu(t)i>x} 1 + t for 2 > 2M. e, however, was arbitrary. We're done. 2.

Functions with only real zeros If we want to study the distribution of the zeros of an entire function

f (z) of exponential type with C°° log+If(x)I dx < oo, J_0D 1 +x 2

66

III H Levinson's theorem on density of the zeros

and we put limsup limsup 00

loglf (iy)I = A, y

loglf(ly)I=A', Iyl

Y~there is no loss of generality in assuming that A = A'. The latter situation may

always be arrived at by working with euA -

(z)

instead off (z); here, the new function has the same zeros as f (z) and equals f (z) in modulus on the real axis. We begin by looking at such functions f which have only real zeros. Theorem. Let f (z) be entire and of exponential type, have only real zeros, and satisfy the condition C

J

log, If (W)I 1 + x2

dx < oo.

Suppose that

limsup log I f (iy) I = limsup log I f (iy) I

y-''O

Y--00

y

= A.

I

y

For t % 0, let v(t) be the number of zeros off on [0, t], and, if t < 0, take v(t) as minus the number of zeros off in [t, 0). (In both cases, multiplicities are counted.) Then v(t)/t - A/ic for t -* oo and for t --> - oo.

Proof. By the theorem of §G. 1, (*)

logIf(z)I=A.3z+n1

f

ID.

-

Iz3tl2logIf(t)Idt

for 3z > 0, and by the same token, for ,3z < 0,

logIf(z)I =A13I +I

_. Iz13 tl2loglf(t}Idt.

71

From these two relations, we see that the function f (z)/ f (z-), analytic in

{,3z>01, has constant modulus equal to

1

there. Therefore in fact

f (z)/ f (i) - fi, a constant of modulus 1, for 3z > 0. Making z -- any point x of the real axis where f (x) 96 0, we see that f(x)/f(x) = P. This means that any continuous determination of arg f (x) on a zero free interval (for f) is constant on that interval.

2 Functions with only real zeros

67

Since f (z) 0 in {,3z > 01, we can define a (single valued) analytic branch of log f (z) in that half plane, and then take arg f (z) as the harmonic function Slog f (z) there. For x e R such that f (x)00, define arg f (x) as limy 0 + arg f (x + iy); as we have just seen, this function arg f (x) is constant

on each interval of R where f(x) # 0. If x increases and passes through a zero xo of f , arg f (x) clearly jumps down by it x the multiplicity of the zero x0.

Therefore

arg f (x) = - irv(x) + const. for real x with f (x) 0. From (*) we see that, in {,Zz > 01, the harmonic function log I If (z) I is the

real part of the analytic one

- iAz + -if"'.( z

t+ t2 + 1 )loIf(t)Idt.

1

It is, at the same time, the real part of log f (z) there. The imaginary part of the latter analytic function must therefore differ by a constant from that of the former one in {3z > 0}, and we have arg f (z)

A91z +

I f 00 ( 91z

tI 2

+ t2 +

I

log I If (t) I dt + const. 1

there. Taking z = x + iy with x not a zero off and making y - 0, we obtain

v(x) = n x - i2 yly0

_ oo \(x t)2

+ y2

+ t2 +

1

)loglf(t)Idt

+ const., in view of the relation between arg f (x) and v(x). Write

0(x) = -lim li 0+ it Y

loglf (t)I dt - \(x t)2 + y2 + t2 + 1

so that v(x) = (A/n)x - A(x) + const., save perhaps when f (x) = 0. (The limit

in question certainly exists if f (x) 0; see the discussion just preceding Kolmogorov's theorem in the previous article.)

In the course of proving the theorem of §G.1 we showed that the condition ('°° log+If (W)I

dx < cc

f"0 1 +x2 (which is part of our hypothesis) actually implies that Iloglf(t)II

f

1 + t2

dt < co;

68

III H Levinson's theorem on density of the zeros

we have of course been tacitly using the latter relation all 'along, since without it, (*) and the formulas following therefrom would not make much sense. We can therefore apply Kolmogorov's theorem, and especially its corollary, to u(t) =1ogl f (t) I. We have 0(x) = (1/n)u(x) with this u, and therefore, by the corollary, (*) I dx (1) J flax>i>A} 1 + x2

=o J

for large A. In order to prove that

for x -> ± oc,

it is enough to show that A(x)/x - 0, x - ± oo, and we restrict ourselves to

the situation where x -+ oo, since the other one is treated in the same manner. Pick any y > 1, as close to 1 as we please, and any s > 0. For large n, we have r"+1

dx

Y-1

1 + x2

yn+1 ,

so, taking.A = cy" in (*) and making n -> oo, we see that there must be some

x"e[y",yn1] with IA(x")I,<sy" if n is large enough. Since v(x)= (A/n)x - 0(x) + const. is increasing (by its definition!), we have, for y" < x < yn` 1 with n large,

-A(y"+1 _y"-1) < -0(x)

0(x"+1)

71

+ A (y"+2 _ y") It

Figure 14

69

3 The zeros not necessarily real

Thus, if n is large enough and y" < x <, y"+ 1, i + A (y2 - 1)y", I A(x) I < By""

1

and

A(X)

A

15 (y2

x

y

+ By.

Since y > 1 and e > 0 are arbitrary, we see that A(x)/x -+ 0 when x -+ oo, as required. Our proof is now complete.

3.

The zeros not necessarily real

Given an entire function f(z), let us denote by n+(r) the number

of its zeros with real part >,0 having modulus <,r, and by n_(r) the number of its zeros with real part < 0 having modulus < r. As usual, n(r) = n+ (r) + n_(r)

is the total number of zeros off with modulus < r, and multiple zeros of f are counted according to their multiplicities in reckoning the quantities n + (r), n_(r) and n(r).

Theorem (Levinson). Let the entire function f (z) of exponential type be such

that

log+If(x)I dx < oo, 1 +x2 and suppose that limsup

logIf(iy)I

Y- 00

Y

= limsup log If(iy)I = A.

v--

I

I

Y

Then n+(r)

A

r

n

n_(r)

A

r

it

and

as r- oo. Given any 6 > 0, the number of zeros off with modulus < r lying outside both of the two sectors I arg z I < 6, 1 arg z - it I < 6 is o(r) for large r.

III H Levinson's theorem on density of the zeros

70

Proof. Without loss of generality, f (O) # 0, for if f (O) = 0 we can work with a suitable quotient f (z)/z' instead of f (z). We may thus just as well take f (0) = 1 in what follows. Denote by {A,,} the sequence of zeros of f(z), each zero being repeated

in that sequence according to its multiplicity. By the first theorem of §G.3,

<

"

00,

3A^>Oz Ian I

and similarly (referring to the lower half plane),

E

> - oo.

2

Hence, I -3A. I

< 00.

IAn12

Take any S > 0. From the previous relation, we have

< cc.

Y 6
This certainly implies that the number of A. with S 5 I arg A. I S n - S and

I An I < r is o(r) for r - oo; the last affirmation of the theorem is thus established.

To get the rest (and main part) of the theorem, we follow an idea due to Levinson himself and compare f (z) with another entire function having only real zeros to which the conclusion of the theorem in the previous subsection can be applied. The Hadamard factorization of our function f has the form z

f (z) = e`Z f 1 - - ez1A n

an

Corresponding to each A. we now compute a real number a according to the formula

n

if perchance A. is pure imaginary we put A = co. Let us now write z

T(z) = e`Z fl 1- Al n

n

ezJz^.

3 The zeros not necessarily real

71

We must, first of all, show that the product just written converges u.c.c. in C. But

C1-? lez/z^ = I 1-? eZ/z n

Here,

e

n

1/I A. I') < oo. Also

I A,, I

A,, I

with n(r), the number of A.

having modulus <, r, at most O(r) (see §A), so f (1 -

does converge u.c.c. in C, by §A. The product defining (p(z) therefore converges u.c.c. in C, and 9(z) is an entire function whose zeros are the real numbers ,% .

We want to show that (p(z) is of exponential type. This can be done most easily by appealing to the Lindelof theorems of §B, and the reader is invited to see how that goes. One can also make a direct verification without resorting to the Lindelof theorems by proceeding as follows. In the first place, we clearly have

IOx)I S If(x)I for x e I8, so, since f (z) is of exponential type, I cp(x) I is at most eofl D on the

real axis. Consider now z = iy; here, gp(iy) I =

e-Y3r

1+

Y

2 )1/2 e'34/I41Z.

n

The right side is easily seen to be 5 e°j'''" ; the only place where calculation is

required is in the evaluation of +(y/2,)Z). To compute this sum, write it as a Stieltjes integral and integrate by parts, using the fact that the number of A, with absolute value < r is O(r); we find without trouble that the sum is O( I y I ). (A very similar calculation was made in proving the second (easy) Lindelof theorem of §B.)

Having seen that I cp(x) I < e°fl" D and I gp(iy) I <, e°(1''D, we must examine qp(z) I for general complex z. According to the discussion at the beginning of §B, from the fact that the number of A' with modulus 5 r is O(r) we can only

deduce an inequality of the form loglw(Z)I 5 O(IzllogIZU,

valid for large IzI. At this point, however, we can apply the second Phragmbn-Lindelof theorem of §C. Look at p(z) in each of the quadrants I, II, III and IV. Take, say, the first one. For proper choice of the complex (!) constant y, el,(P(z)

will be bounded on both the positive real and positive imaginary axes. For

72

111 H Levinson's theorem on density of the zeros

z lying in the first quadrant and I z I large, we will certainly have I eyzcp(z) I <

coazueaizuosizu < eizi3n

say.

0

Figure 15

Therefore, because the opening of the quadrant, 90°, is less than inn radians, the function bounded on the sides of that quadrant, is bounded in its interior, and we see that I cp(z) I < const.e1v0z1

in I. A similar argument works in each of the remaining three quadrants, and (p(z) is therefore of exponential type. Knowing (p to be of exponential type, we refer once more to the property I (p(x) 15 I f(x) 1, x e R, in order to obtain the condition (' °°

log+ I p(x) I

J -00

1 + x2

dx < oo,

from the similar one assumed for f in the hypothesis. We are in a position to apply the theorem of the previous article with our function gyp. Write B = limsup log I (P(iy) I ,

y-

y

B' = limsup

log

i iy) I ,

I

i

Y

and call v(t) the number of points A' (counting multiplicities) in [0, t] if t > 0. For t < 0, let v(t) be minus that number of points An in [t, 0). The theorem of the previous article is directly applicable to the function ei(B-B')z/2(P(z)

3 The zeros not necessarily real

73

and tells us that

v(t)

B + B'

t

2n

for

t -4 + oo.

It is now claimed that (B + B')/2 = A, the common value of limsup,.,logI f(iy)I/y and limsup,,- logIf(iy)I/IYI. First of all, for real y,

1-

=

2

T.

=

1-2y3li+l l2 YZ

('+)(1

2y3 I2n12+y2

Noting that

_

1

+C 3An 12

1

1An12)

1An12

1An12

we see that the last expression is in turn 1

A,

2(1

+CIA"12)2)C1 _ IA2I2 +y2

).

n

Since (1- s)es < 1 for real s we also have

2y3.. s exp 1 -14I2+y2

2yJAn

IAn12+y2

.

Comparing the product representations for f (iy) and (p(iy), we see from the

y nR1

inequalities just written that

2

Il(iy)I < Because

IPrY)III(1+(IAn12))1/2

.exp

Y3(11An12+y2).

En(I ,3An I / I An 12) < oo, i2+ny2l=o(IYI)

for

y- ±oo.

The same is true for the logarithm of the product on the right side of the previous relation. To verify that, denote by N(t) the number of the quantities IAn12/I.32nI lying between 0 and t (counting repetitions in the usual way),

and rewrite

+(y.3)2)1/2 log(1 n

1A.12

74

III H Levinson's theorem on density of the zeros

as

2 fo"O

log

+ t2

dN(t).

Since E 13.% I / 12 12 < 00, we have N(t) = o(t) for t - oo, from which the

integral just written is easily seen to be o(I y I) for y - ± oo after an integration by parts. In view of these facts, the above relation between I f (i y) I and I rp(iy) I shows that loglf(iy)I 5logl(q(iy)I +o(IyI)

for y - ± oo. Therefore A = limsup y-0o

log I f(1y) I

y

\ limsup log I p(ly) I

y

= B,

y

and, in like manner, A < B'. We have thus proved that A < (B + B')/2.

We wish now to prove the reverse inequality. Take any 6 > 0. We showed at the very beginning of this demonstration that, for large r, all but o(r) of the original zeros A,, of f (z) with modulus < r lie in one of the two sectors

Iargzl
Figure 16

3 The zeros not necessarily real

75

For A,, in either of those two sectors,

1A.1 < 1).;,1 = I%,,I2/I91U.I < I1.Isec(,

so, for r > 0 and large,

v(r) < n+(r) 5 v(r sec 6) + o(r),

- v(- r) 5 n _ (r)

v( - r sec S) + o(r).

(Recall that v(t) is by its definition negative for t < 0!) Since 6>0 is arbitrary, these relations and the known asymptotic behaviour of v(t) for t -> ± oo imply that n+(r) r

B+B'

n_(r)

B + B'

r

2n

2Tr

and

for r -+oo. According to the theorem of §E (the first and simplest one of its kind!),

logIf(z)I
-", Iz-t1 2 log+lf(t)ldt -3z

n

for 3z > 0, and similarly, referring to the lower half plane, loglf(z)I < A13zI + 1 7r

1"z

J-.Izt1

2

log + I f(t) I dt

for 3z < 0. Taking any 6 > 0, we see from these two formulas that logl f(Re''')l <, AR I sin 91 + o(R)

holds uniformly in each of the two sectors

S <9 <7r-S, 7r+3 <9 <2n-S when R-> oo on account of the finiteness off

(log +I f(t)I/(1 + t2))dt.* In the remaining sectors 191 < S, 19 - it 1 < 6 we surely have

logI f(Re'9)I < KR.

Therefore, for large R, f Zn

Zn

logl f(Re'9)Id9 <

2 R+o(R)+2 KR,

0

* In the integrals figuring in the two preceding relations the denominator of the integrand, Iz - t12' is, for z = Re''9, equal to R2 + t2 - 2Rtcos 9. When

6<9(1-cosb)(t'+R2).

76

III H Levinson's theorem on density of the zeros

or, since 6 > 0 is arbitrary, rzn 1

2n

I d9 <

lf

f

o 0

2

R + o(R)

71

for R -oo.

Now apply Jensen's formula, recalling that n(r) = n+(r) + n_(r). By what has already been proved, we know that n(r)

B + B'

r

it

for r - oo. Therefore

Rn(r)dr=B+B R+o(R) J0

r

it

for large R. The left-hand side is, however, equal to 1

1zn

2n

o

logl f(Re''')Id9

which, as we have just seen, is < (2A/n)R + o(R) for large R. Therefore (B+B')/2<, A. The reverse inequality has already been shown. Therefore, (B + B')/2 = A, which means that n+(r)

r

-A --it

and

n_(r.)

r

-

A

it

as r -> oo. The first (principal) affirmation of our theorem is thus established,

and we are done. Remark. The above proof of the Cartwright version of Levinson's theorem depends on the elementary material of §§A and C, Kolmogorov's result, the formulas in §§E and G.1, and the first (easy) theorem in §G.3. The more delicate results in §§G.2 and G.3 (the one involving Blaschke products, in particular) are not used, nor are Lindelof's theorems.

Problem 5 Let f (z) be entire and of exponential type, with

°° log+If(x)Idx
loglf(iY)I = A. Y

3 The zeros not necessarily real

77

(a) Show that log I f (Re's) I 0

R

A sin 9

d9 -+0

as R - oo. (b) Let be the sequence of zeros of f (z) in {3z > 0} (repetitions according to multiplicities), and put

B(z)-fl(1-z/A 1 - z/I. ) for 3z > 0. Show that E J R logy B(Rei9) I d9 0

for R -+ oo.

IV

Quasianalyticity

One of the first applications of the relation

logIf(x)Idx J°D

1 +x2

< oo

established in §G.2 of the previous chapter was to the study of quasianalyticity. This subject may now be treated by purely real-variable methods (thanks, in particular, to the work of Bang); the older functiontheoretic approach of Carleman and Ostrowski is still, however, an excellent illustration of the power of complex-variable technique when applied in the investigation of real-variable phenomena, and it will be outlined here.

The material in the present chapter is due to Denjoy, Carleman, Ostrowski, Mandelbrojt and H. Cartan. We are only presenting an introduction to the subject of quasianalyticity; there are, for instance, other notions of that concept besides the one adopted here. One such, due to

Beurling, will be taken up in Chapter VII, but for others, the reader should consult the books of Mandelbrojt and, regarding more recent work, Kahane's thesis published in the Annales de 1'Institut Fourier in the early 1950s. Mandelbrojt's books on quasianalyticity also contain, of course, more elaborate treatments of the material given below. His 1952 book is the most complete, but his two earlier ones are easier to read. A.

Quasianalyticity. Sufficiency of Carleman's criterion

1.

Definition of the classes o'( {

Suppose that f (x) is infinitely differentiable on R. The familiar example

1 The classes ,1({M"}) f(x)-Jexp(-1/x2),

0,

79

x 0, x = 0,

shows that, if, at some x0, f and all of its derivatives vanish, we cannot conclude that f (x) = 0. Under certain restrictions on f and its derivatives, however, such a conclusion may become legitimate. Consider, for example,

functions f subject to the inequalities J f I"I(x) (,
x e U8,

on their successive derivatives. By looking, for instance, at Taylor's formula

with Lagrange's form of the remainder, we see that f (x) =f (x0) +

n_1

n.

(x - xe)"

for Ix - xoI < 1/K, xo being any point of R, and this means that f is in fact the restriction to a8 of a function analytic in 1.3zI < K. Such a function

cannot vanish together with all its derivatives at any point of R without being identically zero. Are there perhaps some other systems of inequalities which, imposed on the successive derivatives of f, will imply the uniqueness property in question without, however, forcing f to be actually analytic? This question (which, like so many others in analysis, comes from mathematical physics) was raised at the beginning of the present century. The answer turns out

to be yes, and the classes of functions thus obtained which, without necessarily being themselves analytic, share with the latter ones the property of being uniquely determined by their values and those of their successive

derivatives at any point, are called quasianalytic. (Note: There are also pseudoanalytic functions in analysis. Those have nothing to do with the present discussion.)

Definition. Given any interval I s R and a sequence of numbers M. > 0, we say that a function f, infinitely differentiable on I, belongs to the class WI({M"})

if there are two numbers c and p, depending on f, such that l f I"I(x)I ,
for

xeI

and n = 0, 1, 2,3,.... Remarks. The number c is introduced mainly for convenience, because we want '1({M"}) to be a vector space. The number p is introduced because, in the case I = R, we want f(px) to belong to WR({M"}) when f belongs to that class.

80

IV A Quasianalyticity. Sufficiency of Carleman's criterion

Scholium. Suppose I has a finite endpoint, say a, but that we are not taking

a to be in I. Then, by requiring I f"W I '< cP"M"

for xel, we obtain the existence of lim f (")(X)

x-a xeI

for n = 0, 1, 2,..., so that f and all its derivatives may be defined by continuity

at a. We will then still have I f`"'(a)I < cp"Mn

for n = 0, 1, 2. .... This means that for the classes 'I({Mn}) as we have defined them, we may always assume that the intervals I are closed.

Definition. A class %I({Mn}) is called quasianalytic if, given any xae1, the only fec1({Mn}) such that n = 0, 1, 2.... ,

.f (")(xo) = 0,

has f(x)-0, xel. Now we have the problem: which classes 'I({Mn}) are quasianalytic and which are not? 2.

The function T(r). Carleman's criterion

The quasianalyticity of the class 'o I({Mn}) turns out to be governed by the function r"

T(r) = supMn n3o

defined for r > 0, whose use is due to Ostrowski.

Theorem. If Jo (logT(r)/(1 +r2))dr= oo,

the

class

WIQ M.))

is

quasianalytic.

Proof. Suppose that xoel and that feci({Mn}) and f ("j(xo) = 0 for n = 0, 1, 2,.... To prove that f (x) - 0 on I it is enough to show that f (x) - 0 on any interval J e I having xo as an endpoint. Without loss of generality, take J = [0, 1] and suppose that xo is 1, i.e., 1 only that f(")(1) = 0, n = 0, 1, 2,.... (Having an interval J of length

2. The function T(r) and Carleman's criterion

81

means that the parameter p in the bounds on sup.,-,, I f (")(x) I gets changed, while the M. remain unchanged.) We have

n = 0, 1, 2.... and 0,<x<, 1.

f (")(x) < cp"Mn,

For 91a > 0, put q,(cr) = J ,

t°f (t) dt.

0

pp(a) is clearly analytic for 9ti > 0 and continuous for 9ta > 0, and 916 > 0.

I (P(o) I <' cMO,

We are going to show that (p(a) - 0. By the theorem of §G.2 in the preceding

chapter (applied, of course to the right half plane instead of the upper half plane), this will certainly follow from the relation °°

log I cp(iT) I

-00

1+T

f

2

dT=-oo,

which we now set out to establish. Since f(1) = 0, we have, when 9ta > 0, (P(o)

f (t)

Q+1 1

a + 1 fo,

0 '

Q+1 0

ta+ f'(t) dt

to+ f'(t) dt.

Again, f'(1) = 0, so a similar integration by parts gives us 1 0a) = (a + 1)(Q + 2)

1

to+z

0

Repeating this process yields

6-

()

(- 1r (a+1)(6+2)...(a+n) fo

1

to+n (n) t

f Odt.

Therefore, for 93a > 0,

1001 <' la+ 1II0r+12I...IQ+nI cP"Mn, n= 1,2,3,.... We have already seen that an analogous inequality holds for n = 0. Putting Or = it with T real, we get IW(iT)I

<

cp.Mn, ITI"

n=0,1,2,...,

IV A Quasianalyticity. Sufficiency of Carleman's criterion

82

that is, Mn

1

1 "1)" I

(P(ii) 15 c,

n = 0,1, 2, ... .

Since by definition

T(')

= su

P

1

n ->O

I

ICI

Mn \ P /

we see that c,

w(ii)IT(I pl) so that

log l gO(ii) 15 log c - log T((PI

.

Since T(r) >, 1/Mo is bounded below (wlog MO > 0, for otherwise surely

f - 0), the relation °° log T(r)

f

1 + r2

0

dr= co

implies that 0°°

log T(r) 1 + pzrs dr = oo.

Therefore log+ iz) I

di

it log c - f

= n log c - 2p

°° log T(r) ,1 dr o 1 + p2r2

f

1ogT+li2/P) dT

= - oo,

as claimed above.

For this reason, q (a) - 0 in 91o , 0 and in particular q (0) = p(1) _ = 0. In other words,

rp(2) =

i

I0

tkf(t)dt=0, k=0,1,2,....

By Weierstrass' theorem on polynomial approximation, this makes the (continuous!) function f(t) vanish identically on [0, 1]. That's what we had to prove. We're done.

Convex logarithmic regularization - definition of B.

Convex logarithmic regularization of necessity of Carleman's criterion

83

and the

We are going to see that the converse of the theorem at the end of the previous § is true. This requires us to make a preliminary study of the geometrical relationship between the sequence {Mn} and the function T(r). 1.

Definition of the sequence

Its relation to {Mn} and T(r)

By the definition of T(r) given at the beginning of §A.2, we have log T(r) = sup(n log r - log Mn); no

moreover (unless c,({Mn}) consists only of the function identically zero on I, which situation we henceforth exclude from consideration),

log T(r) ? - log MO > - oo. The function log T(r) clearly increases with log r. This description of log T(r)

is conveniently shown by the following diagram:

Figure 17

We see that - log T(r) is the y-intercept of the highest straight line of slope log r that lies under all the points (n, log Mn). It is convenient at this time to introduce the highest convex curve, II, lying under all the points (n, log II is the so-called Newton polygon of that collection of points

(first applied by Isaac Newton in the computation of power series expansions of algebraic functions!). The straight line of ordinate

84

1 V B Use of the convex logarithmic regularization

n log r - log T(r) on the above diagram is, for each fixed r > 0, the supporting line to n having slope log r.

At any abscissa n, the ordinate of n is simply the supremum of the ordinates of all of its supporting lines, since II is convex. Therefore, the ordinate of II at n is sup(n log r - log T(r)). r>0

This quantity is henceforth denoted by log M"; and {M"} is called the convex logarithmic regularization of the sequence {M"}; log M. is clearly a convex function of n. The following diagram shows the relation between log M. and log M":

n

points (n, log M") marked o

points (n, log M") marked .

Figure 18

It is evident that M. < M. for each n. The Newton polygon II is also the highest convex curve lying under all the points (n, log M"), as the figures show. Therefore, by the above geometric characterization of log T(r) in terms of supporting lines, we also have log T (r) = sup [n log r - log M"], n;o r"

T(r) = sup. CO Mn

We see that the function T(r) cannot distinguish between the sequence {M"} and its convex logarithmic regularization {M"}. It will be convenient to consider the ordinates of the Newton polygon

1 Definition of {M"} - its relation to {M"} and T(r)

85

II at non-integer abscissae v. We denote this ordinate simply by II(v); II(v) is a convex, piecewise linear function. II(v)

V

9-

Figure 19

Lemma. If T(r) is finite for all finite r, then M,',/" is an increasing function of n when n > some no.

Proof. Under the hypothesis, the slope H'(v) of n must tend to oo as v -+ oo. Otherwise, for some ro < oo, H'(v) would remain < log ro for all v, and certainly no straight line of slope log r, with r > ro, could lie below II.

n(v)

I slope = log r

-1-1

P

Figure 20

86

I V B Use of the convex logarithmic regularization

This means that T(r) = oo for r > ro, contrary to hypothesis. Because fI'(v) -> oo as v --' oo we certainly have II(v) - oo. Pick any vo with II(vo)/vo > 0, and draw a line 2' through the origin with slope bigger than n(vo)/vo. Then . certainly passes above the point (vo, ll(vo)), lying on the convex curve Il.

Figure 21

However, since 11'(v) -+ oo as v -* oo, . cannot lie above H forever. Let v, be the last abscissa to the right where H cuts . - there is such a last abscissa because H is convex. The figure shows that, at v,, II cuts Y from below. This

means that

n(vt) V,

< n'(vt).*

Figure 22 * If necessary, turn 2' slightly about 0 to ensure that (v 11(v,)) is not a vertex of II.

I Definition of {M"} - its relation to {M"} and T(r)

87

It is claimed that 11(v)/v increases for v > v1.

To see this, observe that d

II(v)

dv

v

vII'(v) - rim.

)

v2

so it is enough to verify that vfl'(v) - fl(v) > 0 for v > v1. However, since II(v) is convex, d(vfl'(v) -1(v)) = vdfl'(v) is certainly > 0, so vfl'(v) - II(v) increases. And v1II'(v1) -11(v1) is > 0 by our construction. So vII'(v) - fl(v) remains > 0 for v > v1, and therefore 1I(v)/v increases for such v.

Let no be any integer > v1. Then, if n > no,

log Mn -1I(n) n

1I(n + 1)

n

- log M" +

n+1

1

n+1

This proves the lemma. Lemma. Suppose that M,1,y" - oo for n - oo. Then, for sufficiently large n, M,1,is increasing as a function of n, and M1/n \

Mn+l M"

Proof. Since (log Mn)/n =1I(n)/n tends to co with n, the slope of the convex curve 1I cannot remain bounded as v - oo, and 1I'(v) -+ oo, v -+ oo. We are

therefore back in the situation of the previous lemma, and the argument

used there shows that 1I(v)/v increases for large v, and in particular (log M")/n increases for large n. The reasoning used in the above proof also showed that 1I'(v) > 1(v)/v if v is large. Therefore, for large n, 1I'(n) > H(n) n

Since, for v > n, 1I'(v) > 1I'(n), we thus have, by the mean value theorem,*

fl(n + 1) - fl(n) >

fl(n), n

log Mn+ 1 - log Mn >

log M" n

for large n. We're done. * Note that all the vertices of n have integer abscissae.

88

I V B Use of the convex logarithmic regularization

1M < oo, then

Corollary. If

_n-1 Y Mn < oo. n=1

Proof. Clear. Theorem. If f o (log T(r)/(1 + r2))dr < oo, then 00

< oo. n=1

Proof. (Rudin). Since T(r) is increasing, convergence of f o (log T(r)/(1 + r2))dr certainly implies that T(r) < oo for all finite r, so, by the first of the above lemmas, is increasing for n > no, say.

As we saw during the discussion preceding the above two lemmas, T(r) = sup k>o (r'190, and this is >, e" when r >, If n >, no so that Mnl" 5 MR+i+1), we therefore get 1e

log T(r)

r2

J eM,

n

dr

1

1

a

Similarly, l og T(r) r2

fe"O

n

dr >, e -

M-

Using these inequalities and taking an arbitrary m > no, we find that log T(r) 1:0.1/fto

r

2

el*i+"log T(r)

dr

"=no

f-

+

r log T(r) Jer2

no M-1/no

e

m n(M-1./n-M-1/(n+1))+M-1/m n+1

n

e n=no

no

dr

dr

1m-1

>-

2

e

m

m

+

n=no+leMn1/n

We see that m

Mn 1/" n=no+1

e

°°

e

1/no f.o

log T(r)

r

2

dr.

Since log T(r) is bounded below, the hypothesis makes the integral on the

2 Necessity of Carleman's criterion

89

right finite. Therefore, making m -+ oo, we get 00

y IV IJn<00,

no+1

Q.E.D.

Corollary. If $o (log T(r)/(1 + r2))dr < oo, n=1 Mn

< 00.

Proof. By the theorem and the corollary just before it. 2.

Necessity of Carleman's criterion and the characterization of quasianalytic classes Using the work of the preceding article, we can now establish the

Theorem. If $o (log T(r)/(1 + r2))dr < oo, any interval I of positive length.

is not quasianalytic for

Proof. Take the convex logarithmic regularization the corollary at the end of the last article, we have n-1

of

then, by

< 00

Mn

1

The following construction works with the ratios µ = Mn_ 1/Mn. Picking any a > 0, we fix an no > 1 such that 00

no

For each n, sin is entire, of exponential type and bounded by I in absolute value on the real axis. Hence, by Phragmen-Lindelof (§C of Chapter III), sin

e"'l3z'.

µnZ

will therefore equal a non-zero entire

The product

function Ip(z) with I p(z) I S e`11, if only it is convergent. However, if I small,

Since

oo, we also have

I is

00; the product in question is

therefore u.c.c. convergent in the complex plane.

90

IVB Use of the convex logarithmic regularization

Put

f (Z) = M n-1 o

sin (E/no)z

sin µnz

2n0

H

no

(E/no)z

µnz

f (z) is a non-zero entire function with

If(z)I < M.,,-,e3El3:l and

I.f(x)Idx < oo.

This last relation and the boundedness of f on the real axis certainly make fl. I f (x)12 dx < oo, and f (z) is obviously of exponential type < 3E. We therefore conclude by the Paley-Wiener theorem (Chapter II, §D) that (*)

eaf(x) dx

F(2) = J 00

vanishes identically for A < - 3e and for A > 3e. (Vanishes identically there

and not just a.e., because here, f(x) being in L1(R), F(2) is continuous.) Because f (z) * 0, F(1) cannot be everywhere zero. As we just remarked, F(2) is continuous on R. It is even infinitely differentiable there. Indeed, if k < no,

Ixk.f(x)I < Mno-1(n )k

sin (E/no)x

2no - k

(E/no)x

E

certainly belongs to L1(R), so we can differentiate (*) k times with respect to A under the integral sign. In this way we see that F(k)(1) =

f T (ix)keux f(x) dx J

00

is in absolute value (no )kM". E

-1 J

si n(E/no)xIn0+1

dx,

(E/no)x

a finite constant, for k < no. (Remember, we took no > 1.) When k > no, we can start to use products of the factors (sin µx)/µx, n > no, to absorb powers of x: no no-1 sin(E/no)x ^0+1

Ixk f(x)I < (E)

(E/no)

1

M^o-1 µno

.

.

1 .µk

2 Necessity of Carleman's criterion nono - 1 sin (e/no)x

no

= M

no k

e

Therefore I

no + l

(e/no)x

E

91

sin (e/no)x (e/no)x

Mno

MMk

Mno

M

no-1

k-i

no+ 1

I < Mk (no/e)"° -1 f '. I sin (e/no)x/(e/no)x I"°

dx fork >, no.

We see that we can choose c in such a way that F(k"(A,) 11
on

R

for all k (including the finite number of values from 0 to no - 1). We know, however, that Mk 1< Mk for each k. Therefore our function F belongs to 'R({Mn} ), does not vanish identically, but is identically zero outside the interval [- 3e, 3e]. This means that F and all of its derivatives must vanish at both points 3e and - 3e. Since F(2) # 0 on [ - 3e, 3e], the class c' ({Mn}) cannot be quasianalytic when I = [- 3e, 3e]. By translating F, we see that the same is true when I is any interval of length 6e. Here, s > 0 is arbitrary, so we're done. The work just done can now be combined with that in §A.2 to give a complete characterization of the quasianalytic clases 'I({Mn} ).

Theorem. Given any interval I of positive length, the class 'o I({M"}) is quasianalytic zany one of the following equivalent relations holds:

(a) fa (log T(r)/(1 + r2))dr = oo /

1/n =

oo (/b) EnMn (C) En(Mn- 1/Mn) = 00.

Proof. By the preceding material and logic-chopping. We know that (a) implies quasianalyticity of 'I({Mn}) for any I by the theorem of §A.2. Also, not-(a) implies that F_nMn 'In < oo by the theorem of

the preceding subsection, and this last relation by itself implies that

(I)

Y_ (Mn-1/Mn)<00 n

according to the corollary immediately preceding that theorem.

Now, the proof of the preceding theorem is entirely based on the condition (f). Therefore, (t) by itself implies that WI({Mn}) is not quasianalytic (for any I). So quasianalyticity of WI({Mn}) implies (c) (the negation of (j')) which implies (b) which implies (a), which, however, itself implies the

quasianalyticity of WI({Mn}) as we have already -seen. So complete equivalence of the latter property with any of the conditions (a), (b) and (c) is fully established. Q.E.D.

92

IV C Scholium

C.

Scholium, Direct establishment of the equivalence between conditions the three f (log T(r)l(1 + r2)) dr < oo, o Y"Mn ""<0o and "Mn_1/11I"
These three conditions are of course equivalent according to the theorem at the end of the preceding §. In §B.1 we gave direct arguments, based upon the geometric properties of the Newton polygon II of the set of points (n, log M"), to show that the first condition implied the second, and the second the third. The above establishment of the reverse impli-

cations depended, however, on the theorem of §A.2 as well as on the construction used to prove the first result of §B.2; in other words, on lots of complex variable theory. Since the relationship between T(r) and the convex logarithmic regularization {M"} is strictly graphical, i.e., geometric, we should also give a direct proof of the reverse implications. Let's do that. The reasoning used involves a peculiar change of variable and a summation by parts.

Theorem. Y_ 1(M log T(r) J0

1

+ r2

1/M") < oo implies that

dr
Proof. Look at the Newton polygon 11, denoting, as in §B.1, its ordinate corresponding to the (real) abscissa v by II(v):

H(')

0

Figure 23

H consists of certain straight segments, and each vertex of II has a certain integer abscissa. Therefore, on each segment [n - 1, n], lI is a straight line

Equivalence of conditions on T(r) and on

93

with the constant slope II'(v) = II(n) - II(n - 1) = log

M"

Mtl-1

.

If, then,

Y Mn M -1 < oo, 1- oo, and so the slope n,(v) of II must tend to 00 as v -+ oo. Either, then, n has an infinite number of vertices, or else, if it has we must have

only finitely many, its last side must be vertical and have infinite slope. We examine only the first of these situations; treatment of the second is similar (and easier). We are dealing, then, with a Newton polygon II having an infinite number of vertices and thus an infinite number of straight sides whose slopes increase without limit. Denote by v1 the abscissa of the first vertex of 1I where two sides of positive slope meet, and by v2, v3, etc. those of the successive vertices lying to the right of (v1, 1(v1)).

Figure 24

We call Pk the slope of the side of n meeting the vertex (vk, II(vk)) from the left; thus, II'(v) = pk for Vk_ 1 < v < vk, and therefore 1/M = e-tk for (Keep in mind that the vertex abscissae vk are integers.) Since

vk _ 1 < n <.

Pk k oo and Pk+1 > Pk (convexity of n), we can break up °°

log T(r) 2

J

expp ,

r

dr

94

IV C Scholium

into a sum of integrals of the form fexPPk+, log T(r)

dr.

rz

exppk

Make the change of variable log r = p and put

log T(r) = T(p).

Integrating by parts between rk = ePk and rk + 1 = ePk+1, we have ('k+ i log T(r) dr = log T(rk)

f

J rk

r2

_ log T(rk+ 1)

rk

rk + 1

'k+1 1

+ frk

= e-PkT(pk) - e-Pk+1T(Pk+1) +

r

Pk+ I

e-PdT(P) Pk

However, for Pk 1< P 1< Pk + 1, T(P) = VkP - H(Vk):

-log TO = -r(p)

Figure 25

Therefore

f

Pk+1 Pk

e-Pdr(P)

a-Pdp

= Pk

=

vk(e-Pk

-

dT(r)

e-Pk+1).

Equivalence of conditions on T(r) and on

95

Adding, we thus get r2T(r) dr = kYm (e-Pkr(pk)

J

.,

-

e-Pk+1 T(Ilk+ 1))

M

+ Y- vk(e-Pk-e-Pk+1). k=1

The first sum on the right telescopes, and to the second we apply summation by parts. In this way, we see that the right side of the previous relation equals e-P'T(pl) - e-Pm+1T(pm+) + ,Q

m

+

k=2

(vk-vk-11`

vle-P1

)e-vme-Pm+1

Recall that log T(r) is increasing, so, if T(pm+1) = log T(rm+1) remains < 0

for m - co, (' °° log T (r)

Jo

1+r2

dr

will certainly be < co. There is thus no loss of generality in assuming T(Pm+1)%0 for large m. If that is the case, we may drop the two terms prefixed by - signs from the previous expression and make m -> co, getting finally log T(r)

r

2

dr <,

log T(r1)

rl

°°

+v1e'iP'+ Y-

(Vk-Vk-1)e-Pk'

k=2

since, as we know, rm+ 1 = ePm+1-, co for m - co. As we saw at the beginning of this discussion, we have 1/M = e - Pk for vk -1 < n 5 vk; there are clearly Vk - vk - 1 such values of n. Therefore the

preceding inequality becomes °° log T(r)

f

1

dr <

log T(r1)

rz

r1

+ v1 e-P' +

9"-1

>,, M

converges, and thence

Hence f (log T(r)/r2) dr < oo if

$o (log T(r)/(1 + r2))dr < co, T(r) being increasing. Q.E.D. This theorem, combined with that of §B.1, establishes equivalence of the two conditions

JloT(r) dr < oo 0

1+r2

and

M" -1 n

M.

< 00

96

IV C Scholium

since, as we saw in §B.1, the latter condition is implied by the inequality M 1jn < oo. In order to obtain full equivalence of our three conditions, it is still necessary to show that En(Mn_ 1/Mn) < oo also implies the relation

1/" < oo. It was in order to establish such an implication that Carleman proved his celebrated inequality which says that 00

00

1 /n

n=1

R=1

with an absolute constant c for any sequence of numbers ak > 0. Given this fact, we need only observe that

Mo M1

M"-1

Mo Mn

M. Since Ma/" -n 1, the desired implication follows directly from the inequality.

Problem 6 Prove Carleman's inequality using Lagrange's method of undetermined multipliers. (Hint: Take x 1,x2, .... xN > 0; the problem is to find the maximum of E,=1(x1x2 . . .

subject to the condition that F-k=1 xk = I.

Show first that the maximum is attained at a place where all the xk are strictly positive for 1 <, k < M, say, M < N, and the xk with M < k <, N (if there are any) all vanish. The effect of this is to merely lower N, so we may always take the maximum to be an internal one, obtained for

xk>0,1
and show that at the presumed maximum, Ei (x1x2

x,,)1Jn = A by

adding equations. The whole problem reduces to getting a bound on A. To this end, write each of the N equations involving A, and in them, make the substitutions

xr=r, Sr>0. r Pick out the equation obtained by doing a/axk in the Lagrange procedure, with k so chosen that Sk (at the sought maximum) is > all the other This will give you the estimate k n(

\t/n'

which yields a bound on A independent of k.)

Construction, entire functions of exponential type

D.

97

The Paley-Wiener construction of entire functions of small exponential type decreasing fairly rapidly along the real axis

Suppose we are given an increasing function S(r), defined for r > 0 and (say) > 1 there. Do there exist any non-zero entire functions (p(z) of exponential type such that 1

S(Ixl)' xeR?

Iw(x)I

According to the first theorem of Chapter III, §G.2, there are no such (p if ( °° log S(r)

Jo f 1+r2

dr = oo.

If, however, the integral on the left is convergent, we can use the construction in §B.2 (applied in proving the first theorem of that article) to obtain such (p with, indeed, arbitrarily small exponential type. This application requires us to go a little further with the graphical work of §§B.1 and C. As far as the problem taken up in this § is concerned, there is no loss of generality in assuming that S(r) __ 1 for 0 < r < 1, say.

Lemma. Let S(r) be increasing on [0, oo), with S(r) = 1 for 0 < r < 1, and suppose that

(''* log S(r)

J

0

dr <

r2

oo.

Then there is an increasing function T(r) 3 S(r) with log T(r) a convex function of log r and also log T(r) 2

r Proof. Just put

dr < oo.

o f00

log T(r) =

(".log S(P) dp Jo

P

Then, since S(p) is increasing and >, 1,

log TO

log S(r) er LP = log S(r). J P

Again,

d log TO d log r

=

rd log T(r)

dr

= log S(er),

98

IV D Paley- Wiener construction of certain entire functions

an increasing function of r, so log T(r) is convex in log r. Finally,

f

T (r)

°° log

o

f °°

dr _

Jo

r

r" log Jo

fp'/° log S(p) d2

P

Jo

S(P) p)

d p dr

pr

dp _

(' °° a log S(P) 0

dp < oo.

P

Suppose now that we are given a function S(r) satisfying the hypothesis of the lemma. We may, if we like, first obtain the function T(r) and then search for entire functions (p satisfying the inequality (*)

I (Ax) I

1

<

7'(1x1)

Any such tp will also satisfy 1

I (Ax) I <

S(Ixl)

and hence solve our original problem. Our task thus reduces to the construc-

tion of entire functions qp of exponential type satisfying (*), given that T(r) > 1, that log T(r) is a convex function of log r, and that 0°°

log T(r) r2

dr < oo.

J

Starting with such a function T(r), put

M"=sup

r"

r>0 T(r)

for

n=0,1,2,....

Since log M. = supr> o {n log r - log T(r) }, the sequence {log M"} is already convex in n. Now take lr

'(r) = sup r . n;o

n

Lemma. For r > 1, log T(r) < log T (r) < log T(r) + log r.

Proof. Uses graphs dual to the ones employed up to now to study the sequence {M"}. Because log T(r) is a convex function of log r, we have the following picture:

Construction, entire functions of exponential type

99

log T(r)

slope = n -

-log M,

slope =n+1

-log Mn+1

Figure 26

The supporting line to the graph of log T(r) vs log r with integral slope n has ordinate n(log r) - log M. at the abscissa log r. It is therefore clear that log T(r), the largest ordinate of those supporting lines with integral slope, must lie below log T(r). This proves one of our desired inequalities. To show the other one, take any r > 1, and look at any supporting line through the point (log r, log T(r)) of our graph. Since log T(r) is increasing, the slope, v, of that supporting line must be > 0.

Figure 27

100

IV D Paley- Wiener construction of certain entire functions

If [v] denotes the largest integer < v, it is clear from the figure that v log r - log M1,, ? log T(r).

Therefore, log T(r) 5 ([v] + 1)log r - log M1,,1

= [v]log r - log Mt,1 + log r <, log T (r) + log r

by definition of the function T(r). We are done.

Theorem. If T(r) >, 1 is increasing, with and log T(r) a convex function of log r, and if °°

log T(r)

dr < oo

T(r) - 1 for

0
,

JO

then, given any rl > 0 there is a non-zero entire function (p(z) of exponential

type < 2q with

IOx)I < T(Ixl)' xER. Proof. Form the sequence {M"} and then the function '(r) in the manner described above. According to the preceding lemma, it is enough to find an entire function (p # 0 of exponential type < 2i with I T(x) I < 1 on k and Iw(x)I

Ixl7'(Ixl)

for

Ixl>1,xc-R.

The lemma and the hypothesis taken together tell us that log T(r)

(t) flo"

r

2

dr < co.

Also, since log Mn is here a convex function of n, the sequence {Mn} is identical with its convex logarithmic regularization {Mn}. Therefore (t) implies that M"-1 < ao

n=1 Mn

by the corollary at the end of §B.1. Write µ" = Mn _ 1 /M" and take N so large that 00

n=N

µn<11

Construction, entire functions of exponential type

101

Then put f (z) = MN -1

(sin (, /N)z N (n/N)z

sin µ z n=N

We see as in the proof of the first theorem of §B.2 that f(z) is entire, not identically zero, and of exponential type < n + n = 2r/. Arguing as in §B.2, we see also that I xk f (x) I is bounded on P for each positive integer k and moreover, when k , N, that Ixk+lf(x)I

N

Ixkf(x)I =

I

Ixl

N

MN-1 IxIµNpN+i... L

=

N NMk 'l

Ixl'

whence

i (n )N'

IM-if(x)1

xeR.

I

Since similar inequalities hold also for k = 0, 1, ... , N - 1, we have k

lMk

If(x)I

lx1

for x e P and k = 0,1, 2, ... , C being a certain constant. Given x e R, take the supremum of the left-hand side for k=0,1,2 ..... Referring to the definition of fi(r), we get fi(I x I) I f (x) 15 C/I x 1, i.e.

If(x)I 5

C

Ixlfi(Ixl)'

xeR.

It is now evident that we can take cp as a suitable constant multiple of f, and P will satisfy the required conditions. The theorem is proved. Now we may refer to the lemma at the beginning of this §, and to the discussion given there. In that manner, we deduce from the result just established the following. Corollary. Let S(r) , 1 be increasing. A necessary and sufficient condition for there to exist entire functions q, 0 of exponential type with 4)( x)I

1

S(Ixl)

on

is that a`°

f

lo g S(r)

1+r2

dr < oo.

U8

IV E The Cartan-Gorny theorem

102

If that condition is met, there are entire cp # 0 of arbitrarily small exponential type satisfying the inequality in question.

This result, which is due to Paley and Wiener, has found extensive use. Generalizations of it will be taken up in Chapters X and XI.

E.

Theorem of Cartan and Gorny on equality of c6R({Mn }) and ccR({M"}). WR({M"}) an algebra The criteria for quasianalyticity of the class 1e,({Mn}) given in

§B.2 all depend on the convex logarithmic regularization {Mn} of {Mn} rather

than on the latter sequence itself. This makes it seem plausible that our initial consideration of classes W,({Mn}) with completely general sequences {Mn} was in fact superfluous, and that any such class is in reality equal to

one of the form

with a sequence having fairly regular behaviour. We are going to verify this hunch for the case where I = R by proving that c'R({Mn}) always equals 'R({Mn} D. A similar result holds when the interval I is not the whole real line, but then the regularized sequence {M;} such that '91({Mn}) =',({M;,}) is no longer necessarily {Mn}. In that circumstance, which we do not treat here, the regularization process used to pass from {Mn} to is more complicated than the one yielding {Mn}. Interested

readers may find a discussion of this and other related matters in Mandelbrojt's 1952 book. Lemma (S. Bernstein). Let P(z) be a polynomial of degree n, and suppose that

IP(x)l 5 M for - R < x < R. If a > 1, we have IP(z)I 5 Ma"for all z of the form R(w + w-1) with 1 < I wi < a.

i

Remark. For fixed a, the set of z in question fills out a certain ellipse with

foci at ± R. Proof of lemma Under the conformal mapping w -. z = iR(w+w-1), the region I w I > 1 goes onto the complement of the segment [ - R, R], and the unit circumference is taken onto that segment. With z related to w in the manner described, put, for I w I > 1, P(z)

f (w) = w f (w) is then certainly analytic outside the unit circle, and continuous up to it.

Equality of ',({M"}) and 'R({M"})

w-plane

103

z-plane

Figure 28

For large I w 1, z - Rw/2. Since P(z) is of degree n, we see that f (w) is bounded in { I wI > 1). Therefore, by the extended principle of maximum (Chapter III, §C!), we have, in that region,

If Ml 5 sup if(w)I. 1W1=1

However, for Iw1=1, z=(R/2)(w+1/w) is a real number x on the segment [-R,R], so If (w)I=IP(x)I is <, M. Hence If(w)I<M for IwI>, 1, i.e. IP(z)I = If (w)I Iwl" 5 MIwI". For 1 < Iwl < a, the right side is 5 Ma". We are done.

Problem 7 (a) Let P be a polynomial of degree n -1 with IP(x)I 5 M for - R 5 x < R. Show that IP'(0)I 1

en R

M.

(Hint: Apply Cauchy's inequality, using a circle of suitably chosen radius with its center at 0.) (b) Let f (x) be infinitely differentiable on R, and bounded thereon. Suppose that each off's derivatives is also bounded on 18, and write Bk = sup If (II(x) I x

Show that there is a constant C independent of n such that Bl5CBo-')IB,',/"

for n =1, 2,3,.... (Hint: To show that If'(o)I s CB(n- 1>'"B,',l",

take

x"-i P(x)=f(o)+xf'(o)+...

+(n-1)tf(n-1)(0),

104

IV E The Cartan-Gorny theorem and apply (a) to P(x) with a suitably chosen R, using Lagrange's formula

for the remainder to estimate sup...... IP(x)I.) (c) By iterating the result found in (b), show that Bk < CkB(p"-k)I"Bk1"

for 1 < k < n-1.

(Hint. f"(x) is df'(x)/dx, and so forth.)

Remarks on problem 7. In the result of (a), the factor e is not necessary. The inequality without e requires a more sophisticated proof. The result of (c) was first established (independently) by Gorny and by Cartan. In it, the factor Ck may be

replaced by 2. This improvement, due to Kolmogorov, is quite a bit deeper. A discussion of it is found in Mandelbrojt's 1952 book. Another treatment is in the complements near the end of Akhiezer's book on the theory of approximation.

The final result of the last problem is used to establish the following

Theorem (Cartan, Gorny). Let M > 0, and let logarithmic regularization of

be the convex

Then

WRl

Proof. Since M. < M,,, it is manifest that 'R({Mn}) s task is to prove the opposite inclusion.

so our real

For each N = 2,3,4,..., put Mn(N) =

Mn 0 N,

and form the convex logarithmic regularization usual fashion:

0

of

nN

n

points (n, log M. (N)) marked 0

Figure 29

points (n, log M (N)) marked

in the

Equality of 'R({Mn}) and 4R({Mn})

105

For 0 < n < N, log Mn(N) is the ordinate at n of IIN, the highest convex polygon lying under the first N + 1 points (n, log Mn), n = 0, 1, ... , N. As N -> co, the polygons TIN go down towards 11, the Newton polygon of the set of all the points (n, log Mn), n = 0, 1, 2,.... This means that M, (N) -+ M_ n for

each n as N -+ cc. Let f e%R({ Mn } ). In order to show that f E'R({Mn }) (which will complete

the proof of our theorem) it is enough, according to the observation just made, to show that there are constants a and a independent of N with

If("'(x)I
for every x and n = 0,1, ... , N.

Pick any N, which we fix for the moment, and denote by nk, 0= n0 < n1 < n2 < < np = N, the abscissae of the vertices of IIN. Since fe4R({Mn}), we have I flk>(x)I
If("'(x)I
xeR,

If("j'(x)I

and If("j+i)(x)I <,bpnJ+1Mnj+1'

(*s)

xeR.

Now apply result (c) of problem 7 to the function g(x) =f ("j'(x).

With (*) and (;), that result yields If("'(x)I =

Ig(n-njx)I \ C"-nj{bpnjMnj}("j+l-n)l(nj+l-nj)

l

x

U

nj+l

}(n-nj)/(nj+l-nj)

for xeR. Here, C is an absolute constant which we can wlog take ? 1. However, M(nj+i-n)l(nj+i-nj)M(n-nj)l(nj+1-") = Mn(N). nj

nj+i

106

IV E The Cartan-Gorny theorem 0 0

nN

0 0

0 (n,log M" (N)) AV

4

n

n1

n'+t

Figure 30

So the preceding relation becomes

xeR,

If(")(x)I

or, since C >, 1, 18°1(x)1
xeR.

We have thus established the desired inequality for n = 0, 1, 2,..., N with a = b and a = Cp. As remarked above, this is enough to prove the theorem, and we are done. The result just established has an important theoretical consequence. Theorem.
fg Proof. Let f and g belong to 'R({Mn} ); since M. < Mn it is enough to prove that f gE'R({Mn} ).

By the above theorem, we certainly have f and g in 4R({Mn}), so, for

n>0, If("'(x)I
Ig(")(x)I
(4_)' (f(x)g(x))

n

= kO to

k

)f(k)(x)g(n-k)(x),

Equality of cR({Mn}) and 'R({Mn})

107

and, by the inequalities just given, the sum on the right is in modulus

(n) k=o k However, since log Mn is a convex function of m, log M" - log M" -k >, log Mk - log Mo, SO MkMn-k 1< MoMn.

Figure 31

The preceding sum is therefore n

a2 k(`

(n) pkan-kMoMn k

= a2Mo(p + )"Mn,

in other words,

1d 1" (j-) (f(x)g(x)) 5 a2Mo(p + Q)"Mn, xeR, when n =1, 2,3,.... We also clearly have If (x)g(x) I 5 a2Mo on R. Therefore f ge(R({Mn}), and the theorem is proved. Here is a good place to end our elementary discussion of quasianalyticity. Several ideas introduced in this chapter find applications in other parts of

analysis, and will be met with again in this book. The Paley-Wiener construction in §D has various uses, and is the starting point of some

108

IV E The Cartan-Gorny theorem

important further investigations, which we will take up in Chapters X and XI. The whole notion of convex regularization, which played such a big role

in this chapter, turns out to be especially important. A similar kind of regularization is used in some work of Beurling, and in the proof of Volberg's theorem. Those matters will be studied in Chapter VII.

V

The moment problem on the real line

The moment problem on IIB, known also as the Hamburger moment problem, consists of two questions: 1. Given a numerical sequence So, S, , S2,..., when is there a positive

Radon measure µ on R with all the integrals J°°,,lxI"du(x), k,>0, convergent, such that X dµ(x),

Sk =

k=0,1,2,3,...?

J

2. If the answer to 1 is yes, is there only one positive measure µ on P with Sk= J

xkdu(x),

k=0,1,2,3,...?

00

When the answer to 1 is yes, {Sk} is called a moment sequence and the numbers Sk are called the moments of the measure y. If, for a moment sequence {Sk}, the answer to 2 is yes, we say that {Sk} is determinate. If the answer to 2 is no, the moment sequence {Sk} is said to be indeterminate. The study of various kinds of moment problems goes back to the second half of the last century, when Tchebyshev and Stieltjes investigated the moment problem on the half-line [0, oo). Stieltjes' research thereon led him to invent the integral bearing his name. A lot of familiar ideas and notions in analysis did in fact originate in work on the moment problem,

and the subject as we now know it has many of its roots in such work. The following discussion will perhaps give the reader some perceptions of this relationship. It is really only question 2 (the one involving uniqueness of µ) that has to do with the subject of this book, mainly through its connection with the material in the previous and next chapters. It would not, however, make much sense to discuss 2 without at first dealing with 1 (on the

110

V A Characterization of moment sequences

existence of u). We give what is essentially M. Riesz' treatment of 1 in §A. Most of the rest of the material in this chapter is also based on M. Riesz' work. The reader should not believe that our discussion of the moment problem

reflects its real scope. We do not even touch on some very important approaches to it. There is, for instance, a vast formal structure involved with the recurrence relations of orthogonal polynomials and the algebra of continued fractions which is relevant to other parts of analysis such as

Sturm-Liouville theory and the problem of interpolation by bounded analytic functions, as well as to the moment problem. Our subject is also connected in various ways with the theory of operators in Hilbert space, Krein's work being especially important in this regard. It would require a whole book to deal with all of these matters. There is a very good book, namely, the one by Akhiezer (The Classical Moment Problem). That book has been translated into English; unfortunately, both the Russian and the English versions are now out of print and very hard to find - at present there is one copy that I know of in the city of Montreal! The older work of Tamarkin and Shohat is somewhat

more accessible. The reader may also be interested in looking at the original papers by M. Riesz; they are in the Arkiv for Matematik, Astronomi

och Fysik, and appeared around 1922-3. I am indebted to Professor R. Vermes for showing me those papers.

A.

Characterization of moment sequences. Method based on extension of positive linear functionals

Theorem. There is a positive measure p on P with Sk=

xkdu(x),

k=0,1,2,3,...,

- o0

if and only if N (*)

N

L E j=0 i=0

KK

i 0

for any N and any choice of the real numbers 41, 11...IbN Proof. (M. Riesz) The condition (*) is certainly necessary, for if N

P(x) _ Y_ Kixi i=OS

Use of extension of a positive linear functional

111

is any real polynomial of degree N, we certainly have (P(x))2 dµ(x) > 0; -00

the integral is, however, clearly equal to the left side of (*). The real work is to prove (*) sufficient. Denote by Y the set of real polynomials, and for P(x)EY put N

L(P) _ Y_ Skak, k=0

where 5k=oakxk = P(x). L is then a real linear form on the vector space 9; it is claimed that L is positive on 9, i.e., that if PE9 and P(x) >, 0 on R, we have L(P) > 0. Take a real polynomial P(x) which is non-negative on R. By Schwarz' reflection principle, P(z-) = P(z), so if aOR is a root of P, so is a, and d has the same multiplicity as a. Again, every real root of P must have even multiplicity. Factoring P(x) completely, we see that P(x) must be of the form Ig(x)l2 (for real x), where g(x) is a certain polynomial with complex coefficients. We can write g(x) = R(x) + iS(x)

where R and S are polynomials with real coefcients, and then we will have P(x) = (R(x))2 + (S(x))2, so that L(P) = L(R2) + L(S2).

However, if, for example, R(x) _

bixi,

whence

the coefficient of xk in (R(x))2 is L(R2) _ ESk( k

i+j=k

iSj) _ zz`Si+jSibj I

j

which is > 0 by (*). In the same way, we see that L(S2) >, 0, so finally L(P) > 0 as asserted. In order to prove the sufficiency of (*), we have to obtain a positive measure p on ll with Sk = f °° xk dp(x) for k > 0; for this purpose, M. Riesz brought into play the rather peculiar space 8 = 9 + eo consisting of all sums P(x) + 9(x), where Peg and q eW1, the set of real continuous functions on R tending to zero as x -i ± oo. Our linear form L is defined and positive on the vector subspace 9D of 4', and the idea is to extend L to all of 8, in such fashion that it remain positive on this larger space. The extension is carried out inductively.

112

V A Characterization of moment sequences

Let us take any fixed countable subset {gyp,,: n=1,2,3,...} of 'o, dense ,, and, for n = 1, 2,..., call

therein with respect to the usual sup-norm II

II

9 the vector subspace of a generated by Y and (p,,..., p,,. In order to have a uniform notation, we write go = Y; then go c g, C B2 c , so the union &cc, = U og is also a vector subspace of 9. We first extend L from go = P to 4. so as to keep it positive on Ba. Suppose, for n > 0, that L has already been extended tog,, and is positive

thereon. We show how to extend it totnn+, so that it stays positive. In case we have 4',,, and then nothing need be done. We Because L is already defined

must examine the situation where cp,, on the two quantities

A=sup{Lf: fe4' and f and

B = inf {Lg: ge' and g >1 are available. (If peradventure there were no feln with f < (pn+, we would put A = - oo. And if there were no ge4' with g >, we'd take B = oo. Neither of these possibilities can, however, occur in the present situation as we shall see immediately.) It is important to verify that

-oo
- Ik P,,+1II. 1 < q,,+,(x)

Il vn+ ILD 1,

L(1) and B < II q,+ ,L(1). Moreover, if f and ge8 are such that f < g (note that we have just seen that there are such functions f and g!), we have g - f e4' and g -f > 0, so L(g -f) >, 0 by

so A >, - II

II

positivity of L on

II

i.e.,

L(f) < L(g).

This shows that A < B. Take any number c with

A
which, taking L as linear, gives an extension of L from 9 to claimed that L as thus extended is still positive on 8.+,.

It is

Use of extension of a positive linear functional

113

1 can be expressed as h + a(pi+ 1 with hE8 and

Any element of

a = 0, then we already know L(h + 1) = L(h) ae R. If h + is >, 0. If h + acp + 1 > 0 and a > 0, (1 /a)h + t p + 1 > 0, i.e., - (1 /a)h < ( p , + 1.

Since the left side belongs to 8,,, we have by the definition of A that

L(-lhl5 aJ

A
= L(wn+1),

whence 0 < L((1/a)h) + 1) ,>0. There 1) and finally L(h + remains the case where h + a 1 0 Therefore L(-(1/a)h) > B 3 c = and (1/a)h e L(- (1/a)h) - L((pi+ 1) > 0, and finally L(h + acpn+ 1) ,>0, since - a > 0. Taking 1) = c thus produces a positive extension of L from d to as asserted. Indefinite continuation of this process yields a positive extension of L from 1, = 9 to dom. Once this has been done, however, we may extend L from B., to all of d by continuity.

Take any fed'; we have f = P + p where Peg and peWo. Since the functions p used in forming the d are dense in To, there is some subsequence of them,

with

ao>0 for k -oo.

.

Then I'm,,we have all the

L is positive on 8., which includes

I L((pnk) - L((pn)I < II ink - (p.,11.-L(1),

because

- II wnk - P, II ao < wnk(x) - p ,(x) <

II wnk - (P,,, II ,.

II k 0, so the limit in question does exist. If {1f k} is any other sequence of functions in 9°,,, with II p - Ok II k 0, we have -,0, so, by the argument just given, Y'k II Here,

II (pnk - (p

1

II

L(0k)-u(P,,k) k 0, and limk..L('k) exists, equalling We see in this way that the latter limit is independent of the choice of the particular sequence, of cp used to approximate tp in norm II

II , so it makes sense to

define L((p) = lim

k-ao

We can then put L(f) = L(P) + L(ip), and L is in this way extended so as to be a linear form on 9.

114

V A Characterization of moment sequences

L as thus extended is positive on 9. Suppose that f e9 is non-negative on R. Writing as above f = P + cp with Pe91 and cpeWO, we can take a sequence {q,,, j of the p with I I q p - q, II g 0, and we'll have L(f) = L(P) + lim L(cpnk). k- oo

Since P + T

0, we see that

P+ II(P -(Pnklloo+(pnki0;

this function, however, belongs to &., (the sum of the first two terms is a polynomial!). Therefore, L being positive on cg L(P)+ 11 (p-(Pnkll.-L(1)+L((Pnk)%0,

and, making k -> oc, we get L(f) 3 0, as claimed. The linear form L is in particular positive on WO. Therefore, by F. Riesz' representation theorem, there is a positive measure p on l with L(p) =

J

(p(x) dµ(x) -0000

L(f) = f-0000 f ( x) dµ(x)

for all f in 9. This seems at first unlikely, because there is so little connection

between the two vector spaces 9 and WO used to make up & - there doesn't seem to be much hope of relating L's behaviour on 9 to that on WO. The formula in question turns out nevertheless to be correct. In order to accomplish the passage from 1o to 9, M. Riesz used a trick (which was later codified by Choquet into the so-called `method of adapted cones'). Let us start with an even power x2k of x, and show that x2k dp(x)

is finite and equal to L(x2k). For each large N, take the function (PNEW defined thus:


x2k,

Ixl,
0,

Ixj >, 2N,

a linear function on [N, 2N]

and on [- 2N, - N].

Use of extension of a positive linear functional

115

/There is clearly a/ quantity EN, tending to zero as N -+ oo, such that

(t)

x2k

ENX2k+2,

xE!R.

1< (PN(X) +

(That's the /main idea here!) We also have xER,

coN(x)<x2k,

with equality for - N < x 5 N, and of course c°N >1 0. Since the measure µ is positive, fN

x2k

dµ(x) =

TN(x) dµ(x)

N

J

N

N 00

(PN(x) dµ(x) = L((PN)

L(x2k),

J

the last inequality holding because L is positive on 8. At the same time, by (t) and the positivity of L, r (x2k)

<1 L((QN) + ENL(X2k+2),

whilst 2N

,/

'-'1(pN) =

2N

x2k dµ(x)

gpN(x) dµ(x)

J- 2N

- 2N

Combining these two relations with the preceding one, we get /'2N

N x2kd4u(x) < L(x2k) < J

N

J

2N

'1

l

1

Making N -p oo, we see that L(x2k) = l

x2k

J

dp(x),

since EN -+ 0. Because I x Ik < 1 + x2k and L is finite (!), this reasoning also

shows that all the integrals I x Ik dµ(x) J

are convergent.

We must still treat the odd powers of x. This can be done by going through an argument like the one just made, working with x2k + x2k+ 1 + x2k+2 (a non-negative function of x!) instead of with x2k. In that way, we can conclude that

L(x2k+ x 2k+1 + x 2k+2 ) =

(x 2k + x 2k+1 + x 2k+2) dµ(x),

116

V B Scholium

whence, using what we already know for the even powers of x, L(x2k+ 1) =

fx2k+ 1 du(x). l

.

The relation L(x") = f o x" dµ(x) is now established for n = 0, 1, 2, 3.... . However, L(x") = S" according to our original definition of the linear form L! So S" =

x" dµ(x),

and {S"} is a moment sequence. We are done. Remark. The argument at the beginning of the above proof can be followed so as to establish a general theorem about the extension of positive linear

functionals on real linear spaces e with positive cones. (A positive cone in 9 is a cone A' with vertex at 0 such that &=.7l''-.f, and a functional on JI is called positive if it takes non-negative values on Y.) The reader is invited to formulate and prove such a theorem. Remark. The reader's attention is directed to the similarity of the inductive extension procedure used in the above proof and the inductive step in the

proof of the Hahn-Banach theorem. What is the relation of the general theorem alluded to in the previous remark and Hahn-Banach? Can either one be obtained from the other?

B.

Scholium. Determinantal criterion for {S"} to be a moment sequence

The necessary and sufficient condition for {Sk) to be a moment sequence furnished by the theorem of the preceding §, namely, that the forms N

N YY

YY

Y_ E Si+jSicj

i=0 j=0

be positive, is equivalent to another one involving the principal determinants of the infinite matrix [So,

S1,

S21

S3

S21 2S1,S3,

S4,

S4,

S51

S21

S3,

S31

S41

Characterization by means of determinants

117

This determinantal condition played an important role in the older investigations on the moment problem, and we give it here for the sake of completeness. Matrices like the one just written are called Hankel matrices and were extensively studied towards the end of the last century, in particular, by Frobenius. We need two lemmas from linear algebra. Lemma. Given the symmetric matrix

S=

so,o

So,i

SO, N + 1

S1,0

Sill

S1,N+1

LSN+1,0

SN+1,1

...

SN+1,N+iJ

where N - 1 (sic!), the form N+1 N+1

z

Y- j=0 Z SijSibj i=0 is strictly (sic!) positive definite if and only if all the principal determinants det I

50,0

50,1

x1 0

51

,

,

...

1

SM,1

...

50,M si

'

m

SM.M

are strictly positive for M = 0, 1, ... , N + 1.

Remark and warning. If we replace `strictly positive definite' by `positive

definite' and merely require the principal determinants to be >, 0, the corresponding statement is false. Example: 1

0

S= 0

0

0 0

0

0

-1

.

The danger of this pitfall (in which I myself landed during one of my lectures!) was pointed out to me by Professor G. Schmidt.

Proof of lemma. If the quadratic form in question is strictly positive definite, then so is each of the forms MM

EE

KK

i=0 j=0

for M = 0,1, ... , N + 1. This means that the characteristic values of the matrix of any such form are all strictly positive. But the product of the characteristic values of the form just written is equal to the determinant figuring in the lemma's statement. So, in one direction, the lemma is clear.

118

V B Scholium

To go in the opposite direction, we argue by induction on N. For N = - 1

we have the quadratic form

whose determinant is just s0,0. In

this case, the desired result is manifest. Let us therefore assume that the lemma is true with N standing in place of N + 1, and then prove that it is also true as stated, with N + 1. We are given that the determinants in question are all > 0 for M = 0, 1, ... , N + 1.

In particular, then, we have so,o > 0 (this is the place in the proof where strict positivity of the principal determinants is used!) so we may

wlog take so,o = 1, since multiplication of the quadratic form by a constant > 0 does not affect its strict positivity. With this normalization, our (N + 2) x (N + 2) matrix S take the form 1

...

62

61

6N+1

61

S=

62

S' LaN + 1

J

I

where S' is a certain (N + 1) x (N + 1) symmetric matrix. To show that S is strictly positive definite, it is enough to show that the matrix T congruent to it equal to 1

- 61 - 62

-6N+1

Q)

0

0

..

1

1' -61 -62 "'

1

0

...

0

1

0...

0 0

1

0

0 0

0

1

0

0

...

LO

0

XSx

11

... ...

-6N+1 0 0 1

is strictly positive definite. Observe first of all that

0 r--1

I T=

0

0

0

...

01

T'

L0I

I

with a certain (N + 1) x (N + 1) symmetric matrix T'. It is therefore clear that T will be strictly positive definite if T' is. On account of the particular

triangular forms of the matrices standing on each side of S in (*), we have, however, for any principal minor 1

0

0

0

t1,1

t1,2

0

t2,1

t2,2

0

tR,l

tR,2

...

0

...

t1,R

...

t2,R

Characterization by means of determinants

t10,R

119

0

1

1-

- U1 SO,R

0

1

La

0

X ...

SR,O

SR ,R

The determinant of the matrix on the left is therefore equal to det

CSO,O

50,R 1

L SR, 0

SR,R

which by hypothesis is > 0 for 15 R < N + 1. The determinant of the lefthand matrix is, however, just ...

t1,R

tl,l

det ...

tR,R

tR,1

i.e., the determinant of the Rth principal minor of the (N + 1) x (N + 1) symmetric matrix T'. Those determinants are therefore all > 0, so, since T' has one row and one column less than T, it is strictly positive definite by our induction hypothesis. So, therefore, is T, and hence S, as we wished to show. The lemma is proved. Kronecker's lemma. Let a sequence so, s1, S21 n > 0, the matrix SO

S1

S1

S2

Sn

5n+1

... ...

... be given, and denote, for

Sn

Sn+ 1

Stn

by On. Suppose there

is

a number m > 1 such that det A. 0 0 for

n = 0, 1, ... , m - 1 while det A. = 0 for all n > m. Then there are numbers ao, a1, ... , am - 1 such that, for all p >, 0, Sm+p+am-15m-1+p+ ... +IXOSp=O.

Proof. Since det A. = 0, there is a.non-trivial relation of linear dependence so

ao

S1

5m

Sm

s1

+ IX1

S2

Sm+ 1

+ ... + am

SM 1

52rn

=0

120

V B Scholium

0, we cannot have am = 0,

between the columns of Om. Since det Am -1

and may as well take a.= 1. Then the desired relation clearly holds for p = 0, 1, 2,..., m, and we want to show that it holds for p > m. This we do by induction. Write, for p >, 0,

Em+p=Sm+p+am-1Sm-l+p+ "' +t2 Sp,

and assume that Em+p = 0 for p = 0,1, ... , r - 1, with r - 1 >,m, r

i.e.

m + 1. Let us then prove that Em +, = 0. We have det A, = 0. Since r > m + 1, we can write Sm

Am-1

Sr

Sm+1

Sm+1

Sm+r-1

S2m -1

A, =

Sm

Sm+r

S2m

LSr

...

Sm+r

I

S2r

J

Denote by Qk the kth column of this matrix, whose initial column is called the zeroth one. The determinant of the matrix is then unchanged if, for each k 3 m we add to ok the linear combination am-1Qk-1 +am-2ok-2+ " +aOQk-m

of the m columns preceding it. These column operations convert A, to the matrix Em

...

E2m-1

"*

Em+r-I

E2m

"'

Em+r

Em+r

"'

E, 1

Am-1 Sm

LS,

S2m - 1

"'

Sm4*- 1

I

E2r J

which, by our induction hypothesis, equals 0

0

0

0

... ...

0

0

...

0

S2m- 1

0

...

0

Em+r

0

Em+r

Sm+r-1

Em+r

"'

0 0

Am-1 Sm

[S'

"'

Y- E 21

Characterization by means of determinants

121

The determinant of this latter matrix is, however, just 0

0

Em+r

Em+r+l

...

Em

+1

Em+r

= ± det

"'

E2r

Am-1(Em+r)r-m+ 1 This quantity, then, is equal to det Ar which we

know must be zero since r >, m + 1. But, according to the hypothesis of the lemma, det Am _ i 0. Hence Em+r = 0, which is what we wanted to prove. The lemma is established. Now we are able to prove the main result of this §. Theorem. Given matrices

sequence

a

rSO

S1

Si

S2

Sn

Sn+1

An =

of numbers

so, s1, S2, ... ,

form

the

A necessary and sufficient condition for the Sk to be the moments of a non-zero

positive measure i is that either

(i) all the quantities det An are > 0 (sic!) for n = 0, 1, 2,..., or else

(ii) for some m >, 1, det An > 0 for n = 0,1, ... , m - 1, while det An = 0

for all nom. Remarks. The condition that det An > 0 for n >, 0 is necessary, but not sufficient for {sk} to be a moment sequence. Case (ii) of the theorem is degenerate and, as we shall see, happens if the Sk are the moments of a positive measure supported on a finite set of points. Proof of theorem. Suppose, in the first place, that we have a positive nonzero measure i on R with Sk= J

k=0,1,2.....

xkdu(x),

Then, as we observed at the beginning of the proof of the theorem in §A, n

n zz

f

r

E Si+jSitj = J _

i=0 j=0

t Skxk)du(x) n

2

k=0

If p is not supported on a finite set of points, the integral on the right can

122

V B Scholium

only vanish when

are strictly positive definite. so in this case all the forms Here, det A,, > 0 for all n > 0 by the first of the above two lemmas.

Suppose now that our positive measure µ is supported on m points, call them xi, x2, ... , X. If n < m, the polynomial z E bkxk

k=O

vanishes at each of those points only when o

0, so, if

µ({xP}) > 0 for 1 < p < m, the form n

n

E E si+jbibj zz

YY

i=0 j=0

is strictly positive definite when 0 5 n < m. By the first lemma, then, det A. > 0, 0 < n < m. Consider now a value of n which is >, m. We can then take the polynomial xn-m(x - xi)(x - x2)...(x - xm)

which vanishes on the support of µ. Rewriting that polynomial as n

E Ykxk k=OS

we must therefore have n

n zz

xx

E E si+jSiSj = 0,

i=0 j=0

although here n = 1 0. For such n, our quadratic form, although positive definite, is not strictly so, and hence at least one characteristic value of the matrix A. must be zero. This makes det A,, = 0 whenever n > m. Our theorem is proved in one direction. Going the other way, suppose, first of all, that we are in case (i). Then, by

the first lemma, all the quadratic forms n

n

x K

Y, E si+jSiSj

i=0 j=0

are (strictly) positive definite, so {sk} is a moment sequence by the theorem

of §A. The argument just given shows that, here, no positive measure of

Characterization by means of determinants

123

which the Sk are the moments can be supported on a finite set of points. It remains for us to treat case (ii). By the theorem of §A, we will be through

when we show that all the forms

I n

si+j'ibj zz

i=0 j=0

xx

are positive definite. For 0 < n < m we do have det A,, > 0, so we can by the first lemma conclude that those forms are positive definite for such n. To handle the forms with n > m we must apply Kronecker's lemma. According to that result, we have some quantities ao, al, ... , am-1 such that cOSp+a1Sp+1+...+am-1Sm-1+p+Sm+p=O

for p > 0. For n > in, our matrix A,, takes the form sm

Sn

Am-1

...

Sm

Lsn

S2m-1

Sn+m-1

...

S2m

Sn+m

"'

Sn+m

S2n

The (n + 1) x (n + 1) matrix

1

0

0

0

0

0

0

0

...

0

0

1

0

0

ao

a1

am-1

1

0

0

aO

am-2

am-1

1

a0

...

0 0

... am-1 1

is non-singular. Therefore, positive definiteness of An is implied by that of the

product

V B Scholium

124

1

0

0

1

... ...

0

0

01

...

Sn

...

Sm+n

0 Sm

...

S2m

L Sn

...

Sn+m

0

...

CK0

Sm

am-1

1

... 1

0

0

1

...

CXm-1

0

1

...

0(0

Stn J

0

CCO

X

O

...

1

0

...

0

1

...

0

...

LO

I «m-1

1

J

Using the relation furnished by Kronecker's lemma, we see that the product is just 0

..

0

...

0

...

0

0

...

LO

..

0

0

..

01

OJ

This matrix is certainly positive definite (although, of course, not strictly so!), because Am-1 is, as we already know. So On is positive definite (not strictly) also for n >1 m, and the proof is finished for case (ii). We are all done.

Remark. Since large determinants are hard to compute, the theorem just proved may not seem to be of much use. It does, in any event, furnish the complete answer to a rather interesting question. Suppose we lift the requirement that the measures considered be positive

in our statement of the moment problem. Let us, in other words, ask which real sequences {Ak} can be represented in the form A,

k=0,1,2,...,

with real signed measures r such that $°°. I x Ik l dr(x) I < oo for every k >, 0. The rather surprising answer turns out to be that every real sequence {Ak} can be so represented.

In order to establish this fact, it is enough to show that, given any real

Characterization by means of determinants

125

sequence {Ak}, two moment sequences {Sk} and {Sk} can be found with

k =0,1,2,....

Ak=Sk - Sk,

We use an inductive procedure to do this.

Take first So > 0, and sufficiently large so that So = AO + So

is

also > 0. Put S1 = 0 and S1 = A1. It is clear that if S2 > 0 is large enough, and S' = A2 + S2, both the determinants det S0

ISI

S1

l

S

S,

det

J,

S2

,

S1

S2

will be strictly positive. Now just keep going. We can take S3 = 0 and S'3 = A3. Because the above two determinants are > 0, we can find S4 > 0 large enough so that det

rSS

0

S1

S2

1

S 2

S3

2

S3

S4

and

So

S1

det S1

S2

S3

S2

S3

S4

S2

are both > 0, where S'4 = A4 + S4. There is clearly nothing to stop the continuation of this process. For each odd k we take S. = 0 and Sk = Ak. If k = 2m + 2, we can adjust Sk > 0 so as to make the corresponding (m + 2) x (m + 2) determinants involving the S, and Si, 0 < 1 < k, both > 0 (with, of course, Sk = Ak + SO by merely taking account of the S, and S;

already gotten for 0 < I S k - 1. This is because the preceding step has already ensured that so det

S1

...

Sm

Si

> 0. S2mJ

Sm

and So

det

S1

S1

LS,m

...

S;

2The > 0.

...

S1.1

sequences {Sk} and {Sk} arrived at by following this procedure indefinitely are moment sequences according to the above theorem, and their construction is such that Ak = S'k - Sk for k = 0, 1, 2,.... That is what we needed. The result just found should have some applications. I do not know of any.

126

V C Determinacy. Two conditions

C.

Determinacy. Two conditions, one sufficient and the other necessary Having discussed the circumstances under which {Sk} is a moment

sequence, we come to the second question: if it is, when is the positive measure with moments Sk unique? In this §, we derive some simple partial answers to this question from earlier results.

1.

Carleman's sufficient condition

Theorem (Carleman). A moment sequence {Sk} is determinate provided that cc. S2k2k

k=O

Proof. Suppose we have two positive measures, p and v, with Sk =

-

r"0

xk du(x) =

J-

xk dv(x),

k = 0,1, 2, ... .

We have to show that It = v, and, as is well known, this will be the case if the Fourier-Stiieltjes transform .f W= 2 J

eux(dp(x) - dv(x))

vanishes identically on R. It is now claimed that f (A) is infinitely differentiable on III and in fact belongs to a quasianalytic class thereon (see previous chapter). Observe that

x2k(dp(x) + dv(x)) = S2k < 00; therefore all the integrals 21 JI

-x

(ix)k"x e (dp(x) - dv(x))

are absolutely convergent (at least, first of all, for even k 0 and hence for all k > 0), since the measures p and v are positive (here is where we use their positivity!). This means that f (A) is infinitely differentiable on R, and

that fck>(A) =

(ix)keiz"(dp(x) - dv(x)).

1 Carleman's sufficient condition

127

For 2 e 08 we have

If(k)(2)I s 2

Ixlk(dp(x)+dv(x)),

a finite quantity independent of A. Denote sup.ER I f (")(A) I by Mn. Then,

Bringing in, as in §B.1 of the previous chapter, the convex logarithmic regularization {Mn} of the sequence {M"}, we see that M2k < S2k, so 00

n=0

Mn

1/n >

00

00

2k1/2k > k=0

S2k1/2k'

k=0

The last sum on the right is, however, infinite by hypothesis. Therefore,

by the second theorem of §B.2, Chapter IV, the class ',({Mn}) is quasianalytic.

However, f(2)E'( {M"}) and f (k)(0)

=

2

(ix)k(du(x) - dv(x)) = lk(Sk - Sk) = 0 - ao

for k = 0, 1, 2.... according to our initial supposition. Therefore f (A) = 0 on R, as required, and we are done. Scholium. If {Sk} is a moment sequence, log S2k is a convex function of k. This is an elementary consequence of Holder's inequality. Taking, namely,

a positive measure y with Sk =

Xkdu(x),

we have, for r, s ,>0 and 0<2 < 1, ao

Ixlzr+(l-z)sdu(x) <

x

IxIrdu(x)

w

1

Ixlsdu(x)

x

f-00 T'0 so, if r = 2k, s = 21, and 2r + (1 - 2)s is an even integer, 2m say, with (say) 2k < 2m < 21, we find that -2m)/(21-2k).S(2m-2k)/(21-2k) S2m < S(2 2k 21

the asserted convexity of 109 S2k.

128

V C Determinacy. Two conditions

An obvious adaptation of the work in Chapter IV, §§B.1 and C now shows that this convexity has the following consequence: Theorem. Let S(r) = supk,o(r2k/S2k) for r > 0. Then Y_kS2k112k = oo iff

dr = oo. fO'O 1 + r2

Corollary. If f (log S(r)/(1 + r2)) dr = oo with S(r) as defined in the theorem, o sequence {Sk} is determinate. then the moment

Proof. Combine the preceding theorem with that of Carleman. A necessary condition

2.

Theorem. Let w(x) >, 0, suppose that I'. I x Ikw(x) dx < oo for k = 0, 1, 2, ... , and put

Sk = J

xkw(x) dx,

k=0,1,2.....

If

fcc

1 _+X

2

log I w(x)) dx

the moment sequence

Remark. Since

{Sk\\}

< oo,

is indeterminate.

(w(x)/(1 + x2)) dx 5

w(x) dx < oo, we have

flog+ w(x) dx
aao.o

in any case, by the inequality between arithmetic and geometric means.

Proof of theorem. According to Problem 2 (at the end of §B, Chapter II!),

if

.

f °° (log w(x)/(1 + x2)) dx > - oo, the infimum of Axeizxlw(x)dx taken over all finite sums ,Aae'Zx is f°°.I strictly positive. By the Hahn-Banach theorem and the known form of linear functionals on L1(µ) for a-finite measures y, we get a Borel function

a-'x-

pp(x), defined on

{x: w(x) > 0}

and essentially bounded on that set, with f _'000 rp(x) e -'xw(x) dx

0

2 A necessary condition

129

(hence cpw is not almost everywhere zero!), whilst

dx = 0,

(*)

2 ? 0.

f-0000

Under the conditions of this theorem, w(x) > 0 a.e., so (p(x) is in fact defined

a.e. on R and essentially bounded there, i.e., (peLc0(R). Without loss of generality, I w(x)I 5

i

a.e.,

xeR.

Differentiating (*) successively with respect to A (which we can do, since the integrals f °° I x I"w(x) dx are all finite for k > 0) and looking at the resulting derivatives at 2 = 0, we find that

.

k = 0,1, 2, ... .

JTxkcp(x)w(x) dx = 0,

The functions 9Zbp(x) and .39(x) can't both be zero a.e.; say, wlog, that 9t9(x) isn't zero a.e. Then, from the preceding relation, we have

JTx(x)w(x)dxO,

k = 0,1, 2, ... ,

so that ('

xk(1- 91(p(x)) w(x) dx,

Sk = J

k = 0,1, 2, ... ,

-00

as well as Sk = J

xkw(x) dx,

k = 0,1, 2, ... .

Here, I *p(x) I < i a.e. but 91 p(x) is not a.e. equal to zero; therefore (1- 9i(p(x))w(x) dx is the differential of a certain positive measure on R, different from the positive measure with differential w(x) dx, but having the same moments, Sk, as the latter. The moment sequence {Sk} is thus indeterminate. Q.E.D.

Corollary. Let T(r) be > 1 for r > 0, and bounded near 0. Suppose that log T(r) iss' a convex function of log r, and that Jo

,0

T(r

)dr < oo for

k,0.

The moment sequence ao

Sk

-

xk

T(Ixl)

dx,

k=0,1,2,...,

130

V C Determinacy. Two conditions

is determinate iff 00

k=0

5-112k = 00. 2k

Remark. Here, of course, the S. with odd k are all zero.

Proof of corollary. The if part follows by Carleman's theorem (preceding article). To do only if, suppose that S- 1/2k

< 00.

k

By the formula for the Sk, we have first of all /__2k

00

1*)

S2k = 0(1) + 2

(/ 1

dx x T(x)/x2) X2'

Put

M.

x" s up

=

(T(x)/x2

log M. is then a convex function of n, and we proceed to apply to it and to T(x) some of the work on convex logarithmic regularization from the preceding chapter. From (*), we see that S2k < o(1) + 2M2k,

whence surely

Y-kM2k112k

< oo. This certainly implies that MZk2k

k 00,

so, since log M" is a convex function of n, the proof of the second lemma

in §B.1 of Chapter IV shows that the expression M,1,"" is eventually increasing. Therefore the convergence of EkM2k1121 implies that

YM 11" < oo. n

Taking, for x > 0, x"

P(x) = supMn-, n,>0

we will then get log P(x) (1')

ii

2

x

dx < oo

2 A necessary condition

131

by the second lemma of §B.1, Chapter IV and the theorem of §C in that chapter. Since log T(x) is a convex function of log x, so is log (T(x)/x2). The second

lemma of §D, Chapter IV therefore shows that, for all sufficiently large x. log (T(x)/x2) < log P(x) + log x,

in view of the relations between T(x)/x2, M. and P(x). (The convex function

log T(x) of log x must eventually be increasing, since all the integrals fo (x'/T(x))dx, k > 0, converge!) Referring to (t), we see that log T(x) 1

x

fo- 1+x2

dx <

co,

logT(Ixl)dx <

cc.

Indeterminateness of {Sk} now follows by the above theorem.

Example. The sequence of moments xke-lxla

Sk(a) =

dx

is determinate for a > 1 and indeterminate if 0 < a < 1. (Note: In applying exa the above results it is better to work directly with T(x) = for a >, 1 as well as for 0 < a < 1. Otherwise one should express Sk(a) in terms of the Ffunction and use Stirling's formula.)

Problem 8 The moment sequence

Sk=

xke-x1102xdx

is determinate, but the Taylor series Y,0'o (Sn/n!)(iAr of f eizxe-x/'°axdx does not converge for any A # 0. (Hint: To see that the Taylor series can't converge for A # 0, estimate S. from below for large n. To do this, write

i

- Je_dx S1 with (pn(x) = x/log x - n log x, and use Laplace's method to estimate the integral. To a first approximation, the zero x0 of qp;,(x) has x0 - n log n, and this yields a good enough approximation to gp;;(xo). To get a lower bound for e-1^(x°), compute (pn(x) for x = n log n + n log log n.)

132

V D Determinacy. M. Riesz' general criterion

D.

M. Riesz' general criterion for indeterminacy Let {Sk} be a moment sequence. If we put r2k

S(r) = sup - for

r > 0,

k30 S2k

then, according to the corollary at determinate when (' °°

lo g S(r) 1 + r2

the end of §C.1,

is

{Sk}

dr = oo.

If, on the other hand, there is a density w(x) >, 0 with Sk = fo"00 xkw(x) dx,

k=0,1,2,...,

{Sk} is indeterminate provided that

dx log C w(x)/ f-00. 1 + x2

< oo,

as we have seen in §C.2. Both conditions involve integrals of the same form, containing, however, different functions. This leads one to think that they might both be reflections of some general necessary and sufficient condition expressed in terms of the integral which is the subject of this book. As we shall now see, that

turns out to be the case. 1.

The criterion with Riesz' function R(z) Given a moment sequence {Sk}, we take a positive measure y on R

having the moments Sk, and, for zeC, put ('

R(z) = sup { I P(z)12: P a polynomial with J

l I P(x)12 dp(x) < 1 }.

It is only the sequence of Sk which is needed to get R(z) and not the measure

µ itself of which they are the moments; indeed, if N

P(Z) _ Y_ CkZk k=0

with the ckEC,

=

J--

I P(x)12 dµ(x)

N

N

Y_ Y_ Si+ jc,c;.

i=o J=O

I Use of the function R(z)

133

Thus, R(z) (which may be infinite at some points) depends just on the sequence Al; it turns out to govern that moment sequence's determinacy. Marcel Riesz worked with the reciprocal p(z) = 1/R(z) instead of with R(z),

and the reader should note that, in literature on the moment problem, results are usually stated in terms of p(z).

Theorem (M. Riesz). Given a moment sequence {Sk} and its associated function R(z), Al is indeterminate if R(x) < oo on a non-denumerable subset of R. Conversely, if {Sk} is indeterminate, R(x) < oo everywhere on R and 1 J

log' R(x) dx < oo. 1 +x2

Proof. For the first (and longest) part of the proof, let us suppose that R(x) < oo for all x belonging to some non-denumerable subset E of R. We must establish indeterminacy of {Sk}. Take any positive measure u with Sk = f °° xk du(x), k = 0, 1, 2,..., and let us first show that y cannot be supported on a finite set of points. Suppose, on the contrary, that y were supported on {x 1, x2, ... , xN}, say. Put PM(x) _ M(x - x1)(x - x2)...(x - xN); then,

f

_"0IPM(x)12du(x)=0,

but, if x :A x 1, x2, ... or XN, PM(x) -> oo as M - oo, so R(x) = oo. In that

case, R(x) could not be finite on the non-denumerable set E. Having established that µ is not supported on a finite set, let us apply Schmidt's orthogonalization procedure to the sequence 1, x, X2,. .. and the n >, 0, measure p, obtaining, one after the other, the real polynomials with of degree n such that

JTkpn=0 for

k=0,...,n-1,

when n > 1. Of course, the construction of the really only depends on the Sk, and not on the particular positive measure p of which they are the moments. Since no such y can be supported on a finite set of points, the orthogonalization process never stops, and we obtain a non-zero p for each n. These orthogonal polynomials will be used presently. Pick any x0 with R(xo) < oo. We are going to construct a positive measure v on 11 having the moments Sk, but such that v({xo}) , 1/R(xo) (sic!).

134

V D Determinacy. M. Riesz' general criterion

In order to obtain v, let us take any large N, and try to find M points xl, X2i .... xM different from xo, with M = N - 1 or N (it turns out that either possibility may occur) such that the Gauss quadrature formula (*)

-OD

P(x) du(x) = Y µkP(Xk) k=0

holds for all (complex) polynomials P of degree < M + N; here, the µk are supposed to be certain coefficients independent of P. Assume for the time being that we can obtain a quadrature formula (*) for every large N, and consider the situation for any given fixed N. In the first place, the coefficients Uk are all > 0. To see this, pick any k, 0 5 k <, M,

and write

Qk(x) _ fl (x - xi). i#k

05i-< M

The polynomial Qk is of degree M, so P(x) = [Qk(x)]2 is of degree 2M 5 M + N, and we can apply (*) to it, getting J11D

(Qk(xk))2µk =

-00

[Qk(x)]2 dµ(x)

The right side is surely > 0, for y is not supported on any finite set of points. Therefore µk > 0. Using the polynomial

q(x) _

Q0(x) y µoQo(xo)

of degree M, we have, by (*) applied to (q(x))2, J

(q(x))2 du(x) = 1,

whilst (q(xo))2 = 1/µo. Therefore, since q(x) is a real polynomial, surely R(x0)1>

1

µo

by definition of R(z), i.e., µo % 1/R(xo)

Let VN be the discrete positive measure supported on the set xo, x1,..., xM defined by the llrelations VN(lxkl) = µk,

k = 0, 1, ... , M;

1 Use of the function R(z)

135

according to (*) we will then have xk dµ(x) =

Sk = J

xk dVN(x) f-'*00

00

for 0 < k < M + N, hence certainly for k = 0, 1, 2,..., 2N - 1. And, as we have just seen, VN({xo}) 1> 1/R(x°).

Given any fixed k, the integrals (x2 + 1)k dvN(x) -CO

can, according to what has just been shown, be expressed in obvious fashion in terms of the S,, as soon as N > k. They are hence bounded, and

this means we can find some sequence of N's tending to oo, and finite positive measures v(k) on R, k = 0, 1, 2,..., with, for each k, (x2 + 1)k dVN(x) - dv(k)(x)

w*

as N -+ co through that sequence. (See Chapter III, §F. 1). Let I > 0; then, since (x2 + 1) `is bounded and continuous on R, the w* convergence just mentioned certainly implies that (x2 + 1)k-`dVN(x) -(x2 + 1) -'dv(k)(x),

so, ifl=0,1,...,k, dv(k-`)(x) = (x2 + 1)-`dv(k)(x)

and thus dv(k)(x) = (x2 +

Put v(0) = v. By the preceding relation, (x2 + 1)kdv(x) = dv(k)(x) for k = 0, 1, 2,..., so, since the measures v(k) are all finite, we have (x2 + 1)k dv(x) < oo -.

for k >, 0. It is now claimed that the S,, are the moments of the measure v. Fixing any n, take a k > n. Then x" dv(x)

=

x J_cG (x2 + 1)k dv(k)(W)

By the above mentioned w* convergence, the integral on the right is just the

limit of

fo.

x" (x2 + 1)k (x 2 + 1)k d VN (x)

136

V D Determinacy. M. Riesz' general criterion

as N goes to co through its special sequence of values. Each of the latter integrals, however, = J°°.x"dvN(x) which, by (*), is just S" as soon as 2N - 1 > n. Therefore f_"OOD x"dv(x)=S"

for any n > 0, as claimed. We have, moreover, vN({xo}) > 1/R(xo) by our construction. Therefore, since the VN are positive measures, of which a subsequence tends w* to v, we certainly have (t)

v({xo}) > 1/R(xo).

In this way, we have obtained a positive measure v having the moments Sk and satisfying (t), where xo is any one of the points in the non-denumerable set E on which R(x) < oo. From this fact it follows, however, that Al cannot be determinate. We have, indeed, a positive measure v with moments Sk satisfying (t) for each xoeE, and, in the case of determinacy, those measures v would have to be

all the same. In other words, there would be a single measure v with v({xo}) > 0 for a non-denumerable set of points xo. But that is nonsense. So, if R(x) is finite on a non-denumerable set, we can establish indeterminacy using the quadrature formula (*). Everything turns, then, on the establishment of that formula, to which we will immediately direct our attention. There is, however, one remark which should be made at this point, even though it has no bearing on the proof, namely, that in (t) we in fact have equality, v({xo}) = 1/R(xo).

To see this, suppose that v({xo}) > 1 /R(xo). We can get a polynomial P with I P(x)IZ du(x) = JT I P(x)I2 dv(x) =1

J

but I P(xo) I2 as close as we like to R(xo). Then, however, IP(x)I2dv(x) >, IP(xo)I2v({xo}) -00

would be > IP(xo)I2/R(xo), and hence > 1, a contradiction. We see that the function R(x) gives the solution to a certain extremal problem: 1/R(xo)=max{µ({xo}):,u a positive measure with the moments Sk}.

I Use of the function R(z)

137

We have now to prove the quadrature formula (*). For this purpose we use the orthogonal polynomials described at the beginning of the present demonstration; the idea goes back to Gauss. Take any xo e F and any positive integer N. We can surely find two real numbers a and /3, not both zero, such that Q(x) = aPN(x) + I3PN+ 1(x)

vanishes at x0. The polynomial Q is certainly not identically zero, and in fact it is of degree N or N + 1, depending on whether pN(xo) = 0 or not. It is this uncertainty in the degree of Q which forces us to bring in the number M; we take M = (degree of Q) - 1; thus, M = N - 1 or N. Q(x), being of degree M + 1, vanishes at x0 and at M other points; it is claimed that these points are real and distinct. This statement will be seen to rest entirely on the relation J

(§)

P(x)Q(x) d u(x) = 0,

valid for any polynomial P of degree <, N - 1, which is an obvious consequence of the formula for Q and the orthogonality property of the polynomials po(x).

Suppose, to begin with, that Q(x) has the real zeros xo,... , x, (with repetitions according to multiplicities, as in Chapter III), and no others, and that r < M - 1. Then, if

P(x) = (x - xo)(x - x1)...(x - xr),

P(x)Q(x) will not change sign on R, so, for a suitable constant c

0,

cP(x)Q(x) > 0 on R. Therefore $ °° cP(x)Q(x)dy(x) > 0, for y is not supported

on a finite set of points. However P(x) has degree r + 1 < M < N - 1, so $ -. cP(x)Q(x) du(x) = 0 by (§). We have reached a contradiction, showing that Q(x) must have at least M real zeros (counting multiplicities), including x0. However, Q(x) is of degree M + 1. Therefore Q can have at most one non-real zero. The coefficients of Q are real, however, like those of PN and PN+ 1. Hence, non-real zeros of Q must occur in pairs, and Q cannot have just one such zero. This shows that all the zeros of Q are real.

The real zeros of Q are distinct. Suppose, for instance, that Q has at least a double zero at ao; denote the remaining (real) zeros of Q by a1,. .. , am -1; it is, of course, not excluded that some of them coincide with

ao. Put P(x) = (x - al)(x - a2)... (x - am -1); since Q(x) has the factor (x - ao)2, P(x)Q(x) does not change sign on R. Thus,

138

V D Determinacy. M. Riesz' general criterion

for a suitable constant c, cP(x)Q(x) dp(x) > 0 -CO

as before, and this contradicts (§) since P is of degree M - 1 < N. Denote now the real and distinct zeros of Q by x0, xl, ... , xM, and let us complete the proof of the quadrature formula (*). Take any polynomial P(x) of degree < M + N. Then, long division of P(x) by Q(x) yields P(x) = D(x)Q(x) + R(x)

where, since degree of Q = M + 1, the degree of R is < M and the degree of D is < N - 1. This last fact implies, by (§), that

f

D(x)Q(x) dµ(x) = 0,

J

P(x)du(x) =

so

R(x)dµ(x).

J

Now, since degree of R < M, Lagrange's interpolation formula gives us M

R(x) =

/ R(Xk)

k=0 Q (xk)(x - Xk)

Q(),

P(xk) Qlx), k=0 Q (xk)(X - Xk)

R(x) = m

since R(xk) = P(xk) at each zero Xk of Q. Therefore J

P(x)du(x)= J

R(x) du(x)

M

P(xk)

Y-

\k=0Q'(XkXX-Xk)

00

Y, P(xk)Yk,

k=0

where µk =

Q(x)

fo. Q'(xk)(X - Xk)

dp(x),

k = 0, 1, ... , M.

Our quadrature formula (*) is thus established, and therewith, the first part of the theorem. Proof of the second part of the theorem is quite a bit shorter. Here, we

suppose that {Sk} is an indeterminate moment sequence, and use that property to obtain information about R(z).

I Use of the function R(z)

139

We have, then, two different positive measures u and v with xk dp(x) =

Sk =

xk dv(x),

k = 0, 1, 2, ... .

J

J

Denote by a the positive measure Z(µ + v) and by T the real signed measure Z(µ - v). Then also Sk

k=0,1,2,...,

f-0000 xkda(x),

=

so that, if p(x) is any polynomial,

(tt)

I p(x) I Z d u(x) =

J

1 p(x)12 da(x).

J

For the signed measure T,

JTxkdt(x)=o,

k = 0, 1, 2. ... .

There is a trick based on this identity which, according to M. Riesz, goes

back to Markov who used it in studying the moment problem around 1890. The same idea was used by Riesz himself and then, around 1950, by Pollard in the study of weighted polynomial approximation (see next chapter). Take any polynomial P and any zoo R. Then P(x) - P(zo) x - zo is also a polynomial in x, so, by the identity just written, °°

J

P(x)

- P(zo) dT(x) = 0.

x - zo

_OC-

From this we have (§§)

dT(t)

P(z) I

J- t - z

_ r

00

J

P(t)&(t) 00

t-z

whenever zOR. The function

F(z) _

°° dT (t)

f-00 t - z

is clearly analytic for .3z > 0; moreover, it cannot be identically zero there.

140

VD Determinacy. M. Riesz' general criterion

Indeed, .3F(z) = J

. z-t1 I3z

2

d T(t),

T being real, so by the remark at the end of §F.1, Chapter III, w*

ndT(x)

.Z5F(x + iy) dx

for y -+0 + . Therefore F(z) - 0 for .3z > 0 would make t = 0, which is, however, contrary to the initial assumption that p 0 v. Since F(z) * 0 in {3z > 0}, we can use (§§) to get a formula for P(z) in that half plane:

_

1

. P(t)dr(t) t -Z

('°°

P(z)

F(z) J

_

In particular, if z = + i with

real,

1 co

I

I

P(t)I IdT(t)I

I

I

d6(t),

do(t). Let now P = p2, where p is any polynomial. By the preceding relation and (if), I p(i; + i) 12 < (1 / I F( + i) I) f °° I p(t) 12 d 4t), so, since IdT(t)I

by definition of R(z), R(l; + i)

1

The analytic function F(z) is clearly bounded in {,3z

1 } and continuous

up to the line 3z = 1. Since, as we have seen, F(z) * 0 there, we have, applying the first theorem of §G.2, Chapter III to the half plane {,Jz > 1}, °°

log I F( + i) I

_-

2+1

di < oo.

Combined with the previous inequality, this yields ($)

°° _OD

log+ R(t; + i)

i+l

d < oo.

Using this result, we can now estimate R(x) on the real axis. Let p(z) be any polynomial with I p(t)I2dµ(t) < 1;

1 Use of the function R(z)

141

then, by definition (!),

Ip( + i)I2 < R( + i). On the other hand, by the theorem of §E, Chapter III, applied to the half plane 3z < 1, 1 °° log+ Ip( + i)I di; logIp(x)I < n

-,,

for xER. (Note that p(z) is an entire function of exponential type zero! The reader who does not wish to resort to the result from Chapter III may of course easily verify the inequality for polynomials p(z) directly.) These two relations yield

2log Ip(x)I <

l(g+

f it

R)2 + 1)d

,

xeR,

whence, taking the supremum of 2 log I p(x) I for such polynomials p,

logR(x) <

log+ R(g + i)

1

_

(x

- )2 + 1

xeR,

(by definition again!). We can, of course, replace log R(x) by log+ R(x) in this

inequality, since the right-hand side is 3 0. We see from ($) that the integral on the right in the relation just obtained is < oo for each xeR. That is, R(x) < oo for every real x if our moment sequence is indeterminate, this is part of what we wanted to prove. Again, log+ R(x)

1+x2

1

dx < -

log+ R([; + i)

7C

1

x2+1

d dx

_J it

-1

°° 21og+R( + i)

J_-

S2+4

and the last integral is finite by ($). This shows that log+R(x) 2 +1

dx < oo,

and the second part of our theorem is completely proved. Corollary. The moment sequence {Sk} is determinate iff for the function R(z) associated with it,

°° log+R(x) _. 1+x2 dx = oo.

142

V D Determinacy. M. Riesz' general criterion

Remark. The corollary does not give the full story. What the theorem really says is that there is an alternative for the function R(x): either R(x) = oo everywhere on R save, perhaps, on a countable set of points, or else R(x) < oc everywhere on R and

log' R(x)

2+1

dx < oo.

Scholium. Take the normalized orthogonal polynomials Pn(x) corresponding to a positive measure y with the moments Sk. Like the pn(x) used in the first

part of the proof of the above theorem, the P,, are gotten by applying Schmidt's orthogonalization procedure (with the measure µ) to the successive powers 1, x, x2, x3,...; here, however, one also imposes the supplementary conditions

f

-,0[Pn(x)]2 du(x) = 1,

making each P,(x) a constant multiple of pn(x). One of course needs only the Sk to compute the successive P, .

It is easy to express R(x) in terms of these P,,; we have, in fact, 00

R(x) _ Y (Pn(x))2. n=o

Proof of this relation may be left to the reader - first work out RN(x) = max { I p(x)12: p a polynomial of degree

N with JT I p(t)12dp(t)= 1} by writing p(t) = n=0o,P,(t) and using Lagrange's method; then make N -> oc. The boxed formula seems at first sight to break down if any u with the moments Sk is supported on a finite number of points, say M. In that, case, the formula can, however, be saved by taking P,(x) = oo for n > M and x lying outside the support of µ. This makes sense, because the only polynomial of degree n >, M orthogonal to the powers 1, x,. .. , xM -1 with respect to a measure supported on M points is zero, hence can't be normalized. The vain attempt to normalize it gives us the form 0/0, which we are of course

at liberty to take as o0 outside the support of that measure. 2.

Derivation of the results in §C from the above one

Let us first deduce Carleman's theorem in §C.1 from that of M. Riesz. According to the second theorem of §C.1 (in the scholium of that

2 Results in §C deduced from the one of last article

143

article), Carleman's theorem is equivalent to the following statement: the moment sequence {Sk} is determinate provided that °° log S(r)

(*)

l+r

Jo

dr = oo,

where r2k

S(r) = sup - for r > 0. k30 S2k

To verify this, observe that, if the Sk are the moments of a positive measure u, the polynomials qk(x) = xk/1/ S2k satisfy J(qk(x))2d1z(x)

= 1,

so surely R(x) 3 (gk(x))2 for each k, by definition of R(z). Therefore R(x) > S(Ixl). Also, S(I x I) > I ISO > 0, so log S( I x l) is bounded below. It is thus clear that (*) implies

log' R(x)

1 -00

1 + x2

dx = oo.

The moment sequence {Sk} is therefore determinate by the corollary to Riesz' theorem. Consider now the theorem of §C.2. We are given a positive integrable function w(x) with

`° logw(x)dx _OD

l+x

>

and want to prove that the moment sequence Sk = f%xkw(x)dx is indeterminate using Riesz' theorem. Observe that the integrability of w(x) makes f °°. (w(x)/(1 + x2))dx < oo, so, surely, f °°. (log+ w(x)/(1 + x2))dx < oo. Our other assumption on w therefore implies that Jclow(x)d*

*

- 00

1 +x2

x < oo.

144

V D Determinacy. M. Riesz' general criterion

Take any polynomial p with f',, I p(x)I2w(x)dx 5 1. Then, surely, foo 1

Ip(x)12w(x)

dx <

rc

1 TC

for any real , whence, by the inequality between arithmetic and geometric means,

°° 2loglp(x)I+logw(x) dx (x - )2 + 1 R f-0, 1

1 fO 2log I p(x) I

1J°°

dx <

n -.(x-S)2+1

log

1

0,

it

log w(x) 7-.(x-S)2+ldx.

The left-hand integral is, however, > 21og l p( + i) I by the second theorem of §G.2, Chapter III. (p(z) is entire, of exponential type zero. For polynomials, the fact in question may also be easily verified directly.) We therefore have

log lp( + i)I2 ' lfO.

- (xog )2(+)l dx,

7t

and, taking the supremum over such polynomials p, °

log R( + i)

_

it

g

( )

x dx.

(xo )2+

1

Here, one. may, of course, replace log R( + i) by log+R( + i) on the left. For xoaFl,

log R(xo) <

If' log + R( + i) d zr

_

,(xo-S)2+1

,

just as in the proof of the second part of Riesz' theorem. Substituting in the previous inequality on the right and changing the order of integration, we get finally

log R(xo) <

1

`

21og-

w(x) dx,

n J_cj(xo-x)2 +4

xoeR.

But the integral on the right is finite by (*). Therefore R(xo) < 00 for each xoe 08, so the moment sequence {Sk} is indeterminate by Riesz' theorem. We are done.

VI

Weighted approximation on the real line

In the study of weighted approximation on 08, we start with a function W(x) > 1, henceforth called a weight, defined for - oo < x < oo. We usually

suppose that W(x) - oo for x -- ± oo, but do not always assume W continuous, and frequently allow it to be infinite on some large sets. Given a weight W, we take the space Ww(R) consisting of continuous functions (p(x) defined on t with qp(x)/W(x) --> 0 for x -> ± oo, and write cp(x)/W(x)I for cpeWw(68). Being presented with a certain II W II w = subset 8 of 4,(R), we then ask whether of is dense in'w(IJ) in the norm II w - this is the so-called weighted approximation problem. The following preliminary observation will be used continually. II

Lemma. S is

II

11 w-dense in (Ww(R) iff, for some coR, all the functions

and

1

(x - c) " W (x)

n = 1, 2,3,...,

1

(x - On W (x)'

can be approximated uniformly on R by functions of the form f (x)/W(x) with

fE6. Proof. Only if is manifest. For if, take any function cpE(ew(l) and first

construct a continuous function

i/i

of compact support such that

l(cp(x) - i/i(x))/W(x)I <e/3 on R. We can, for instance, put

Ox), /i(x) =

2A

Ixl < A,

- lxl qp(x), A < Ixl 5 2A,

A

0,

(xI > 2A;

the desired relation will then hold if A is taken large enough, since pp(x)/W(x) --* 0 for x -> ± oo.

146

VI Weighted approximation on ll

By the appropriate version of Weierstrass' theorem, linear combinations

of the functions (x - c) -", (x - c)-", n = 1,2,3,..., can now be used to approximation fi(x) uniformly on R, so we can get such a linear combination a(x) with I O(x) - a(x)l < e/3 on R, whence (since W(x) > 1!), I ( fi(x) - o(x))/W(x) I < e/3 there.

If, now, we can find an f e9 with I a(x)/ W(x) - f (x)/W(x) I < s/3 for we'll have 11 a - f II w < e/3. Then, since II (P - 0 Il w <, s13 and

x e (R,

11 0 - a II w 5 s/3, we obtain II (p -f II w < e. This establishes the if part of the lemma. In the first weighted approximation problem we consider, the so-called Bernstein approximation problem, 9 consists of polynomials. Then, of course, we must impose on the weight W the supplementary requirement that

x"

W(x)

)0 as x -+ ± oo

for all n > 0. In another problem, whose treatment is similar to that of Bernstein's, 6' consists of all finite linear combinations of the exponentials e'A" with - a < A 5 a for some given positive a. If a weight W satisfies the supplementary condition, it is natural to compare the solution of Bernstein's problem with those of the latter one for different values of a. One

may also study approximation using weighted L. norms instead of the weighted uniform norm II II w These questions are taken up in the present chapter. Some of the methods applied in studying them resemble closely the one used to prove the second part of Riesz' theorem (previous chapter, §D. 1). There is indeed a relation

between the material of this chapter and the determinacy problem discussed in the preceding one, and results obtained in the study of either subject may sometimes be applied to the other. I know of no book entirely devoted to the matters mentioned above; the

one by Nachbin has very little concerning them and is really about something else. One who wishes to go into the subject should first read Mergelian's Uspekhi paper and then Akhiezer's; both have been translated into English. There is material in the complements at the end of the second edition of Akhiezer's book on approximation theory, and also in de Branges' book. It is worthwhile to study S. Bernstein's original investigations on weighted polynomial approximation; most of his papers on this are in volume two of his collected works. Some of the results given near the end of the present chapter are from a paper of mine published around 1964.

Mergelian's treatment - criterion involving finiteness of O(z)

147

Mergelian's treatment of weighted polynomial approximation Let W be a weight such that

A.

x"

W(x)

0

as x - ± oo for n = 1, 2, 3, ... ;

the solution of Bernstein's approximation problem for W turns out to be governed by the quantity 52(z) = sup { I P(z) I: P a polynomial and

P(t)

(t - i)W(t)

introduced by Mergelian. Note the similarity between the definition of this quantity and that of the Riesz function R(z) given in §D.1 of the preceding chapter. Note especially that the condition I P(t)/(t - i) W(t) I 5 1 and not the seemingly more natural one I P(t)/W(t)I < 1 is used in defining O(z). About this, more later. Criterion in terms of finiteness of Q(z)

1.

Theorem (Mergelian). The polynomials are II II w-dense in 4w(O) ii f S2(zo) = co for one non-real zo, and, if this happens, then O(z) = oo for all nonreal z.

Proof (Mergelian). Only if Suppose the polynomials are dense in Ww(l ). Then, given any z0 , we can find polynomials such that the quantities O

-Q"(t)

W(it)

S "

IFFY

tend to zero as n -> oo. Put

1 - (t -

b oo, and 1/S is for each n a polynomial, zo)W(t)I 5 1 for to R. There is obviously a number K(zo) > 0 depending only on zo such that I

1

K(zo)

\ It-zoI t-i

S K(zo) for teU.

148

VIA Weighted polynomial approximation. Mergelian's treatment

Therefore I i)W(t)I <, K(zo) on R, so, since n oc, we see that S2(zo) = co, establishing the only if part. If: When zoOIJ, it is convenient to work with the quantity M(zo) = sup { I P(zo) 1: P a polynomial

and

P(t) (t - zo) W(t)


One sees by using the number K(zo) brought in during the above argument that S2(zo) = oo if M(zo) = oo, as long as z0OQB. One advantage of introducing M(z) lies in its continuity property: (*)

1

1

K-zl

M(z)

MG)

1321 13C1

To verify this, take any polynomial P with I P(t)/(t - t;)W(t)I 5 I on R and I P(l;)I close to M(C). For the polynomial in t P(t) _ P(0 (t - z) + PG)

Q(t) =

we have Q(z) = P(C), whilst, for tel', I Q(t)/(t - z)W(t)I I P(t)/(t - OW(t) I + I P(')I I (z - ')/(t - z)(t - t;)W(t)I and this is 51 + I P(0 I I z - ' I/ 13z 1 13Y 1, since, as we are always assuming, W(x) > 1.

Put now

1 +IN)I Iz-(1

R(t) =

13z1 13Y1

R is a polynomial in t, and IR(t)/(t - z)W(t)I

I

on R. Therefore

I R(z) I < M(z); however, I R(z) I

I P(D I

=

1+1P(0113z1 I3YI

Thence, 1

1

M(z)

I R(z) I

_

Iz-Cl

1

I

I + 13z 1130 I'

so, since we can have I PQ I as close as we like to M(C), we get 1

M(z)

1<

1

MG)

+ 13z 1 13 1

This relation and the similar one obtained by reversing the roles of z and 4' in

the argument just made give us (*).

1 Criterion involving the finiteness of S2(z)

149

Armed with (*), we proceed with the if part of our proof. Suppose then that zoOR and that fl(zo) = oo; this means that M(zo) = oo, so we can find polynomials Pn(t) with I Pn(t)/(t - zo)W(t) 151 on F whilst I n' oo. For the polynomials

Qn(t) -

Pn(Z0) - Pn(t)' (t - Z0)Pn(ZO)

we will have, for teR, sup teR

1

Qn(t) I

(t-zo)W(t)

W(t)

1

IPn(zo)I'

a quantity tending to 0 as n -+ oo. Therefore the function (t - zo)-' can be approximated as closely as we like, in the norm II II w, by polynomials. It is now claimed that in

1(zo + pe''')d19 = oo

f

0

for each p > 0. Since fl(zo) = co, there are polynomials qn(t) with I qn(t)/(t - i)W(t)I 5 1 on F (N.B. Here it is t - i in the denominator and not t - zo !), and I gn(zo)I k oo. For each n, I gn(z)I 5 U (z) by definition, therefore 2n

n(zo+pe')d9

f0

2n

f

> 2nlq,.(z0)I.

0o

Since the quantity on the right -+ oo with n, we have Q). If 0 < p < I Zzo I, there is a zP, I zp - zo I = p, with f (zp) = oo. Indeed, (;) implies the existence of a sequence of points tk with I (k - zo I = p and Q(l k) k oc. Suppose, wlog, that Ck k zP. Comparison of the definitions of

and M(t;) shows immediately that

S2) < sup teR

t-i t - (

MG).

Here, the supremum is clearly bounded above for I l; - zo I = p since 13zo I - p > 0; therefore our choice of the sequence {gk} makes M(t'k) k co. Because (,k k ZP, we then get M(z,,) = oo by (*), since IX I % I3zo I - P > 0 on the circle I t; - zo I = p. Thus, Q (z.) = oo. (A mistake I made here while

lecturing was pointed out to me by Dr Raymond Couture.) We thus have points z , for which O (z,) = oo with I zP - zo 1= p, when p > 0 is sufficiently small.

For each such z,, 1/(t - Z.) can be approximated in

II

11 w-norm by

VIA Weighted polynomial approximation. Mergelian's treatment

150

polynomials in t; this is shown by the argument used above for IN - zo). We can therefore obtain a sequence of points z zo tending to zo, such that each of the functions 1

(t - Z.)W(t) is the uniform limit, on f18, of polynomials in t divided by W(t). This fact makes it possible for us to show (by taking limits of difference quotients of successively higher order) that each of the expressions 1

(t - Z0)m W(t)

m1,2,3,...,

can be uniformly approximated on R by polynomials in t divided by W(t) 1).

Now we have

Q(zo) = S2(zo),

for,

if P is

a polynomial with

I P(t)/(t - i)W(t)I z 1 on R, the polynomial P*(t) whose coefficients are the

complex conjugates of the corresponding ones of P(t) also satisfies I P*(t)/(t - i)W(t)I <, 1, teR. Therefore in the present case Q(zo) = oo, so, by the above discussion, each of the functions l /(t - zo)'"W(t), m = 1, 2,3,..., can be uniformly approximated on 18 by polynomials in t

divided by W(t). As we just saw, the same is true for the functions 1/(t - zo)mW(t). According to the general lemma given at the very beginning of this chapter, polynomials must hence be II II w-dense in 9W(R). The if part

of our theorem is thus established. We are done. 2.

A computation

In the next article and later on, as well, we will need a formula for sup

It - il

td

t-z

Lemma. When 3z > 0,

t-i

Iz+il+Iz - il

t - z

23z

sup reR

Proof. I(t - i)/(t - z)I = I I - (z - i)/(t - i)I -1. In order to simplify the writing, put z - i = (; then we have to calculate inf,,R 11 - (/(t - i) I. The linear

2 A computation

151

fractional transformation t -> 1/(t - i) takes the real axis into the circle having the segment [0, i] as diameter:

0

Q

Figure 32

Therefore, as t ranges over the real axis, C/(t - i) ranges over the circle y, with

segment [0, l;i] as diameter:

Figure 33

152

VIA Weighted polynomial approximation. Mergelian's treatment

Since ,3z > 0, 3C > - 1, so SR(gi) < 1. Therefore the point 1 must lie outside the circle y,:

Figure 34

Our quantity inf,ER 11- t;/(t - i) I, which is simply the distance from 1 to yt,

can thus be read off from the diagram:

Figure 35

3 Criterion in terms off'. (log S2(t)/(1 + t2))dt

153

We see that infl1 teR

-

-

t-i

1

-1_1 1

-2I' - radius of y,

21

2

1

iz

2

2

Finally,

sup teR

t-il t-z

=

jz+iI+Iz-iI

1

1z

F

2

I1

izl.

-2+2

2z

proving the lemma.

Corollary. For 3z > 0, t-i sup teR

t-z

This inequality will be sufficient for our purposes. Criterion in terms of f `° (log 11(t)/(1 + t2)) dt

3.

We return to the consideration of the quantity C (t) introduced at the beginning of this §, and to its connection with weighted polynomial approximation. Theorem (Mergelian). Polynomials are dense in leW(R) iff (*)

('`° logf2()dt J

l+t2

= oo.

Remark. Since W (t) , 1 we always have log O (z) , 0, i.e., f2(z) , 1, because 1 is a polynomial (!), and I 1/(t - i) W(t) I 5 I on R.

Proof of theorem Only if: We must show that, if (*) fails, the polynomials can't be dense in Ww(OB).

Assume, then, that log f2(t) J

dt < oo,

1 + t2

and take any polynomial P(t) with I P(t)/(t - i)W(t)I 5 1 on R. Then, by a very simple version of the second theorem of §G.2, Chapter III (which, for

154

VIA Weighted polynomial approximation. Mergelian's treatment

polynomials, can be easily verified directly),

If'

log I P(i) I

log +P(2) I

dt.

Here, by definition (compare with the proof of the second part of M. Riesz' theorem in §D.1, Chapter V), I P(t) I < n(t), so


log I P(i) I

g n(t) dt, lo1

and, taking the supremum of log I P(i) I over all polynomials P subject to the

condition given above, we get

log 0(i) <

J

1i + 2) dt

< oo.

The quantity f (i) isnthus finite. Therefore polynomials cannot be dense in lew(R) by the first Mergelian theorem of article 1. If. Supposing that polynomials are not dense in W (R), we must show that (*) is false. If polynomials are not dense in Ww(R), the Hahn-Banach theorem (whose validity does not, by the way, depend on' ' (R)'s being complete!) furnishes us with a bounded linear functional on 4w(R) which is not identically zero thereon, but is zero at each of the polynomials. It is convenient to denote the value of this linear functional at a member p of W, (R) by the expression L( ap(t) W(t)

the reason for this is that then we will simply have

< C sup 00

L

teR

W(t)

with some constant C, for peWw(l ). (N.B. We are NOT writing L(T(t)/W(t)) as f°°,,(ap(t)/W(t))dp(t) with a Radon measure p. That's because we are not assuming any continuity of W(t)

here, so the existence of such a measure p is problematical.) Let us continue with this part of the proof. We have our linear functional L, such that

(t)

)

L I P(t) = 0 W(t)

\

3 Criterion in terms off'-. (log11(t)/(1 + t2))dt

155

for every polynomial P, whilst (§)

960

w(t)

for some tpoeWH,(l ).

Consider the function

F(z) = L

1

G(t - z)W(t)

defined whenever zOIII. In the first place, F(z) cannot vanish identically for both 3z > 0 and 3z < 0. Suppose it did. A simple modification of the general lemma given at the beginning of this chapter (whose verification is left to the reader) guarantees, for each s > 0, the existence of a finite linear combination pe(t) of the fractions 1/(t - c), c0 IR, such that II po - (PE II w < s for the

function po figuring in (§). If, then, F(c) = L(1/(t - c)W(t)) = 0 for every cOR, we'd have L(cpe(t)/W(t)) = 0, whence I L(9o(t)/W(t))I < CE by (*k). Squeezing c, we get a contradiction with (§). Wlog, F(z) is not identically zero in {,,3z > 0}. It is analytic there. To see this, observe that if zOR, the difference quotient

(t-z-Az)-1 -(t-z)-' tends to (t - z) L and (*),

_

1

(t - z)(t - z - Az)

Az

uniformly for t e Il8 as Az -> 0. Therefore, by the linearity of

F(z + Az) - F(z) Az

-

1

L" (t - z)2 W (t)

1

as Az -> 0, since W(t) > 1 on R. This shows that F'(z) exists at every z

l and

establishes analyticity of F(z) in {,3z > 0}. From (*), we get

IF(z)I,,1. Since F(z) # 0 in the upper half plane, the first theorem of §G.2, Chapter III,

shows that

(tt)

f

log- IF(x + i)I 1 +x2

dx < ao.

We can now bring in the Markov-Riesz-Pollard trick already used in proving the second part of Riesz' theorem in §D.1 of the previous chapter.

156

VIA Weighted polynomial approximation. Mergelian's treatment

Take any polynomial P(t) and any fixed z, 3z > 0. Then P(t) - P(z) t-z

is also a polynomial in t, so, applying (t) to it, we get

L

P(t) - P(z) (t - z)W(t)

- 0,

i.e., in terms of F(z) = L(1/(t - z)W(t)),

F(z)P(z) = L

P(t)

(t - z)W(t)

We can thus write (§§)

1 L P(t) (t - z)W(t) J

P(z) = F(z)

for 3z > 0, provided z is not a zero of the analytic function F(z). The idea now is to use (§§) together with (1") in order to show that (' ° log fl(t)

J

I + t2

dt < oo.

Take any polynomial P(t) such that P(t) (t - i) W(t)

on

1

Then, I P(t)/(t - z)W(t) I

R.

suptER (t - i)/(t - z) I which, by the previous article,

(1 + I z D/Zz for 3z > 0 (see the corollary there). Putting z = x + i, x e R, we thus get, from Q), is

JLI

P(t) (t-x-i)W(t)l )

C(1 +.,/(X2 + 1)).

Referring to (§§), we see that

IP(x+i)I

I F(x+i)I

for any polynomial P with I P(t)/(t - i)W(t)I 5 1 on R. Taking the supremum of I P(x + i) I over such P, we find that S2(x + i)

C(1 + J(x2 + 1)) F(x + i) I

I

3 Criterion in terms off °°. (log fl(t)/(l + t2))dt

157

that is, writing C' = log C, ($)

logQ(x+i) 5 C'+log(1+,/(x2+1))+log- IF(x+i)I.

We use the last relation in conjunction with (tt) in order to get a grip on log O(t) for real. t. The procedure being followed here is like the one used in proving the second part of Riesz' theorem (§D. 1, previous chapter). I call it a hall of mirrors argument because it consists in our first going up to the line 3z = 1 from the real axis and then, going back down to the real axis again. Our reason for engaging in this roundabout manoeuvre is that we do not have any simple way of controlling I F(z) I when z gets near R (unless we bring in Hp spaces, whose use we are avoiding as much as possible!). Let P be a polynomial. By the second theorem of §G.2, Chapter III, °°

log I P(t) I

n

_

"

1(xglP()2 +

)I

dx.

If also I P(t)/(t - i)W(t) I ,<1, we have, of course, I P(x + i) I ,<S2(x + i), so, taking the supremum of logIP(t)I for such P,

°° log f (x + i) dx. log S2(t) 5 If 7 J _ . (t x)2 + 1 Plug in ($) on the right, multiply by 1/(t2 + 1), and integrate t from - co to co. After changing the order of integration and using the identity 1

it

_

dt

°°

_

((t - x)2 + 1)(t2 + 1)

2

x2 + 4'

we obtain

f

zlog(lx +(42+1))dx ll +(Z)dt < nC'+J .+f'° 2log-IF(x+i)I x

_00

dx.

The first integral on the right is obviously finite. The second is also finite

by (tt). The integral log f (t) 1

t2

dt

is therefore finite, contradicting (*). This completes the proof of the if part of the theorem, and we are done.

158

VI B Akhiezer's method based on use of W, (z)

B.

Akhiezer's method, based on use of W*(z)

The function O (z) introduced by Mergelian, which, as we have seen, indicates by its size whether or not the polynomials are dense in Ww(R), is equal to sup { I P(z) I : P a polynomial and I P(t) 15 I t - i I W(t) for t e 118}. The presence of the multiplier I t - i I in front of W(t) is disconcerting,

and it would seem more natural to work with the quantity W,k(z) = sup {I P(z)I: P a polynomial and I P(t) 15 W(t) on 08}.

On the real axis, W,k(x) < W(x), and so W,k(x) is a kind of lower regularization of W(x) by polynomials. (Recall that the idea of using some kind of

lower regularization occurred already in the study of quasianalyticity (Chapter IV); the convex logarithmic regularization which turned out to be useful there is not the same, however, as the regularization by polynomials

dealt with here.)

We are always assuming that W(x) > 1. Therefore, since

1

is a

polynomial (!), we certainly have W,1,(z) >, I.

Criterion in terms of $

1.

(log W;k(x)/(1 + x2))dx

The following theorem, due to Akhiezer, is implicity contained in the work of S. Bernstein, who was in possession of all the elements of the proof. Bernstein, who devoted much effort to the study of the problem

bearing his name, was apparently unable to see that a solution was within his reach, and never formulated this next result. Theorem (Akhiezer). Let W(x) be continuous. Then the polynomials are

dense in'w(R) iff

_

log W,(x)

1+x 2

dx = oo.

Remark. As we shall see from the proof, the continuity requirement on W(x) can be much relaxed. What is really needed here is that W(x) be finite on a set of points which is not too sparse. Proof of Theorem.

If: Comparison of the definitions of W,k(z) and Q (z) shows that W,k(x)

.

S2(x). Therefore, if f "0 (log W,k(x)/(1 + x2))dx = oo, we certainly (logC1(x)/(1 + x2))dx = oo, so polynomials are dense in Ww(R) by

have f Mergelian's second theorem (§A.3). Note that the continuity of W plays no role here. Only if: Assuming that f °°0" (log W. (x)/(1 + x2))dx isfinite, we show that any collection of polynomials P with III' II w < 1 forms a normal family in the complex plane. For this, a hall of mirrors argument like the one at the end of §A.3 is used. If P is any polynomial with 11 P 11 w 5 1, the second theorem of

1 Criterion in terms off °°. (log W*(x)/(1 +x2))dx

159

§G.2, Chapter III and the very definition of W,1, give, for real , log 1 P(t) 1

1

log IP( + i) I '< n

+1

dt 5

1

°°

log

n

W(t) +Idt.

Taking the supremum of log I P( + i) I for such P, we find, as usual,

log W*( + i) ' I f '

- (tog 2*+t)1 dt. Now suppose that 3z < 0. Using again the second theorem of §G.2, Chapter III, but this time in the half plane 3z 51, we see that for any ,

polynomial P,

logIP(z)I 5

I rf (1 + 13z') IOg lp + i)I d J-00

If also II P II w < 1, we have I P( + i)1 < W*(c + i), so, by the inequality just

found for the latter function (which, by the way, is > 1), 1

log I P(z)

15 nz

°°

°°-m

(1+I,3z1)log W*(t)

Changing the order of integration, and using the identity

2+IZzl

(1+I.,3zl)d

('°°

It+2i-z12 valid for ,3z 5 0, we find that log I P(z) I

S

(2 +

1

13z 1) log W(t) dt.

It+2i-X12

n JT

OD

Apply now the corollary from §A.2. In the present situation, where 3z S 0, we get sup tER

t-i

2

t+2i- z

( 1+Iz__2iI

l

2

2+I3z )

,

1

whence, by the preceding, for 3z 5 0, (*)

log1P(z)I S

(3 + 1zl)2

1

2+I.3zl

-n

°°

f-.

log W*(t)

1+t2

dt

whenever P is a polynomial with II P II w < 1

For such polynomials P, however, (*) is also valid for 3z 3 0. This is seen by an argument just like the above one, working first with log I P( - i)I

instead of log I P( + i) I, and then using the second theorem of §G.2, Chapter III in the half plane 3z >, - 1 instead of the half plane 3z 5 1. The polynomials P with II P 11 w < 1 thus satisfy (*) in the whole complex

160

VIB Akhiezer's method based on use of W,k(z)

plane. Since 1-.(log W (t)/(1 + t2))dt is finite, such polynomials form a normal family in C. Once we know that the polynomials P with II P II family in C, it is manifest that polynomials cannot be II

1 do form a normal -dense in W,(R).

II

Suppose, indeed, that tpe'w(R) and that we have polynomials P with II P - (p 11 w -* 0. We may wlog take II W it w < 1, then II P. 11 w <, 1 for all

sufficiently large n, hence, wlog, for all n. These therefore form a normal family in C, so a subsequence of them must tend u.c.c. in C to some entire function c(z). At every xeR, 1'(x) and p(x) must coincide, since W(x), being continuous, must be finite on 68 (!). The function which is 11 II w-approximable by polynomials must thus coincide on 118 with some entire function. Since lots of continuous ppe'w(R) don't do that, we see that polynomials cannot be II II w-dense in

We have finished the only if part of Akhiezer's theorem, which is now completely proved.

Remark. We see already from the argument at the very end of the above proof that we need merely assume W(x) < oo on some closed subset of ER with a finite limit point, instead of the continuity of W on ER, and then the property log W*Z t)

l+t

dt < 00

will surely imply that the polynomials are not II II w-dense in ww(If8). Even this assumption on W(x) can be very much weakened, as we shall see in the next article. Description of II II w limits of polynomials when f°°", (log W,k(t)/(1 + t2))dt < oc

2.

A small refinement of the calculations made in proving the only if part of the previous theorem yields an elegant result. Theorem (Akhiezer). If f '.(log W,(t)/(1 + t2))dt < oo, every function in W (l8) which can be II Il w-approximated by polynomials coincides, on the subset of 18 where W(x) < oo, with some entire function of zero exponential type.

Proof. We start from the estimate of log I P(z) I found in the preceding article for polynomials P with II P II w 5 1. As we saw there, if 3z < 0 and P is a polynomial with II P II w log I P(z) I

<

1f J

1,

(2 + l3z') log

W(t) dt.

It+2i-X12

Take any E > 0. Since f °° (log W(t)l(1 + t2))dt < oo, there is a finite M.

2 Description of 11

II w limits of polynomials

161

such that

(t)

log W'(t)

1

dt < s;

it

we then break up the integral of the preceding relation into the sum of two, one over

{teR:

log W*(t) ,<ME}

and the other over the set where log W*(t) > M,. We obtain in this way log I P(z) I

Mt +

1 71

(2 + 1.3z 1) log W* (t)

(f(log

J

W*(()>M.)

It+21-ZI

2

dt.

Apply now the corollary from §A.2. We find, by virtue of (t), that

logIP(z)I S Mt+(1

+Iz-2i1)28; (2+IZzI)

this holds whenever 3z < 0 if P is a polynomial with II P II w 51 One can, of course, use exactly the same kind of reasoning for the half plane 3z > 0. We see in this way that if P is any polynomial with II P II w < 1,

the relation (fit)

logIP(z)I 5 Me+(3+IZI)2 2+I,3zl

e

holds in the entire complex plane. This inequality we refine still more by use of a Phragmbn-Lindelof argument.

Figure 36

162

VIB Akhiezer's method based on use of W,(z)

Take the two sectors S and S' where I y I < z I x I ; what is important here is that S and S' have opening < 90°. Outside both S and S', I x 15 21 Y I, so I z I < 31y1, and (ft) gives (§)

log I P(z) I '< ME + 9e(1 +1,3z D.

This also holds on the boundaries of S and S', where it can be rewritten thus:

log I P(z) I 5ME+9E+2cIRz1. Let us consider the sector S. Inside S, log I P(z) I - 2ERz is subharmonic,

and < const. I z 12 for large IzI by (tt). On the boundary of S, log IP(z)I -2s9iz 5 ME + 9E as we have just seen. So, since the opening of S is < 90°, this last inequality must in fact hold throughout S, by the second Phragmen-Lindelof theorem of §C, Chapter III. We thus get logIP(z)I'< ME+9E+2E1¶zI in S. The same reasoning applies to S'. Referring to (§), which holds outside S and S', and contenting ourselves with a result slightly worse than what we actually have, we see that

logIP(z)I <ME+9E(1 +IzI) throughout C, whenever II P II w 5 1. If, now, F(z) is any u.c.c. limit of polynomials P with II P II w 51, we must also have logl(D(z)I'< ME + 9E(1 + IzI)

for all complex z. Since s > 0 is arbitrary (with M. depending on s through (t)), we see that the entire function 1(z) must be of exponential type zero. Any gpeWw(R) with 1 1 91 1w < 1 which is the 11 w-limit of polynomials must, on the set of x where W(x) < ac, coincide with such an entire function b(z), as we saw at the end of the preceding subsection. We are done. II

Remark. Let tpeWw(R) be such that there exist polynomials P. with II (0 - P. 11w n' 0. Then, as the above theorem shows, if ('°° log W Zt)

J

l+t

dt < oo,

qp(x) must, on {x: W(x) < cc}, coincide with an entire function F(z) of exponential type zero. Sometimes it is useful to know that F(z) satisfies

3 Strengthened result. Pollard's theorem

163

the more precise condition log l 4(z) I S log ll'D II w+ M, + 9s(1 + I z l ),

z e C.

Here, s > 0 is arbitrary and M, depends only on s (through (t)), and not on (D. This fact follows immediately from the proof just given - we need only note that I I ' I I w = I (D I I w, so that II K -1 q II w < 1 for every K > II (D II w Remark. Given that (log W,(t)/(1 + t2)) dt < oo, is it true that for every entire function `V(z) of exponential type zero whose restriction, `I'(x), to

R belongs to 'w(18), we do have a sequence of polynomials P with 0? As we shall see later on, the answer to this question is no

II`I' - P. Il w

for some weights W(x) with seemingly rather regular behaviour. Theorem. Suppose that W(xk) < eo for a sequence of points xk going to oo, and, if n(t) denotes the number of the points xk in [0, t], suppose that

limsup r-ao

Then, if f

n(t)

t

> 0.

(log W,(t)/(1 + t2))dt < co, the polynomials are not dense in

Ww(!f8).

Proof. Take any function .peWw(18) such that cp(xo) =1 but cp(xk) = 0 for k >, 1. Then cp cannot be II II w-approximated by polynomials.

If, indeed, it could be so approximated, the preceding theorem would furnish an entire function (D(z) of exponential type zero with D(xk) = P(Xk)

for k > 0. Then in particular I (xo) = 1, so (D (z) # 0. At the same time (D(xk) = 0 for k > 1, so, if N(r) denotes the number of zeros of (D(z) with modulus <, r, N(r) > n(r) - 1, and limsup,-. (N(r)/r) would have to be > 0 by hypothesis. This, however, is impossible. For, fi(z) being of exponential type zero and * 0, we must have N(r) = o(r) for r-+ oo by an easy application of Jensen's formula (see problem 1(a) in Chapter 1!). The theorem is proved. Remark. We shall soon see that the condition limsup,... (n(t)/t) > 0 in this theorem cannot be relaxed much. 3.

Strengthened version of Akhiezer's criterion. Pollard's theorem

The Bernstein approximation problem was also studied by the American mathematician Pollard, whose work was largely independent of Akhiezer's and Mergelian's. Pollard published one of the first solutions,

I think in fact before the appearance of the other two mathematicians'

164

VIB Akhiezer's method based on use of W,k(z)

articles. In one direction, the criterion given by him strengthens that furnished by the Akhiezer theorem of article 1. The way this happens is shown by the following

Theorem (due, essentially, to Pollard). If PX2)

sup

If

11 I

I

dx : P a polynomial and II P II w< I j

is finite, then °°. (log W* (x)/(1 +x2))dx < oo.

Proof. As x - ± cc, W(x) -> co faster than any power of x. So, if we take a suitable constant C,

W(x) =

W(x) C

Ix-il

is , 1 on R. W(x) obviously grows faster than any power of x as x -). ± cc, and we may consider weighted polynomial approximation with the weight W. To this situation we apply the Mergelian theorems in §§A.1 and A.3.

Put, for zeC, (z) = sup {IPzI: P a polynomial and (t I

I

S 1 on 1B

P)tff'(t)

note that i (z) is just CW*(z). According to the theorem of §A. 1, (i))< 00 implies that polynomials are not II 11g.-dense in les (I8), and, by §A.3, the latter fact makes

f"0llg(2)dt< Jlog W*(t)

1+t2

dt <

cc,

cc.

In order to show this last relation, it is therefore enough to verify that (i) < cc, or, what comes to the same thing, that W*(i) < oo. Take a sequence of polynomials {Pn(z)} with II P II w 1 and I n W* (i); by §G.2 of Chapter III, I

log l p"(0 I<

1 f '0 log Pi(t) I dt.

Under the hypothesis, however, the integrals on the right are bounded above. Therefore W*(i) < co, which is what we needed. We are done. In the course of the argument just given, we established a subsidiary result, important in its own right. We state it as a

Mergelian's criterion more general than Akhiezer's

165

Corollary. If W,k(i) < oo, then f* w (log W, (t)l(l +t2))dt < co.

Remark. Sometimes it is easier to get an upper bound for _

loglP(t)l dt 1+ t2

when P is a polynomial with II P w II < 1 than to try to directly obtain good

estimates on W,(x). If we can show that the upper bound is finite, the II w-approximable by polynomials given in description of functions II

article 2 is available, and hence the consequences of that description. For this reason, the result proved here is quite useful.

Mergelian's criterion really more general in scope than Akhiezer's. Example

C.

Given a weight W(x) >, 1 on 01 which goes to oo faster than any power of x as x -* ± oo, we can form the two functions

fl(z)=sup I P(z) I : P a polynomial and

{

P(t) < 1 on P } , (t - i) W(t)

and

W*(z) = sup { I P(z) I : P a polynomial and I P(t) I < 1 on P I.

Mergelian's second theorem (§A.3) says that the condition

f

log S1(t)

1 + t2

dt = oo

is necessary and sufficient for polynomials to be Akhiezer's theorem (§B.1) says that the condition °°

log W, (t) 1 + t2

II

II,-dense in 'w(R).

dt = oo

is always sufficient for the

II

II w-density of polynomials in ' (R), and

necessary for that density to hold provided that W(x) has a certain regularity. As we saw in §B.1, continuity of W is enough here; it suffices in fact that W(x) be finite on an infinite closed set with a finite point of accumulation. The work of §B.2 shows that the set on which W(x) is finite need not even have a finite point of accumulation; it is enough

that the set be infinite and not too sparse. As long as number of points in the set and in [ - r, r] limsup r- ao

r

is positive, the necessity of Akhiezer's criterion (involving log good.

holds

166

VI C Mergelian's criterion more general than Akhiezer's

These successive relaxations in the regularity required of W(x) for Akhiezer's criterion to hold make us hope that perhaps all restrictions on

W's regularity may be dispensed with. Maybe the lower polynomial regularization Wk(x) of W(x) is all we need for the study of Bernstein's problem, no matter what the behaviour of the latter function is, and we can forget about fl(x) altogether. Do we really need CI(x) at all in order to have a completely general test for II w-density of polynomials in II

Ww(R)?

We do. Here is an example of a weight W(x) such that flog 1'V*(t)

dt < co,

1 + tz

but nevertheless 12(i) = oo, making the polynomials II II w-dense in'w(R).

Our construction is based on use of the entire function z 2) 00

S(z) _ f 1 - 4n

n=t

of exponential type zero. The weight W(x) will be identically infinite outside

the set of points xk = sgn k- 2 1k1;

k=±1,±2,...;

these are just the zeros of S(/z). On that set we take W(xk) = C1/IxkI-IS'(xk) I

with a constant C chosen so as to make W(xk) > 1 for all k. We start with the asymptotic evaluation of S'(xk) for large IkI; on account of symmetry we need only consider positive values of k. For k > 1, then,

2 ton
S'(2k)

k

Here is a trick which can be used to good effect in many calculations of this kind. Factor each ratio 4k/4n with k > n out of the first product. One finds that S'(2k)

=

_2k7l 4k-n. l1 tin
=

2(- 1)k 2k

2 (k-

For large k, this is

.v (- 1)k2(k-' 2(s(1))2,

k-t

t)k1=1( 1-

H

k

n

C1-4k

n>k

t4n
4`

fl w

mj-jl

1

1- 4m

.

1-4 n

Example

167

and we see that IS'(xk)I behaves like a constant multiple of

for k -). ± oo. This evidently tends to oo faster than any power of xk as k -+ ± oo; Ixkl(Ikj-2)

the same is then true of W(xk). We need also to consider the partial products

SN(Z) = II

(1

-Z)

=1

of S(z). For 1 < k 5 N, we have SN(2k) _

/

-k1

(i_);

k

k1 1-4k
of this with thefrsst of the above formulas for S'(2 k) shows that

for 1 5 k
ISN(xk)I % IS'(xk)I

On account of this fact and of the growth of I S'(xk) I for k - ± oo, we have, for any polynomial P, P(xk)

Ni

P(xk)

0,

//

SN(xk)(Z - xk)

N

--- S'(xk)(z - xk)

as long as z is different from all the xk.

N.B. A prime next to a summation sign means that there is no term corresponding to the value zero of the summation index. (This convention is fairly widespread, by the way.) Let us fix any polynomial P. As soon as 2N > degree of P, we have, by Lagrange's interpolation formula, N

P(z)

=

`SN(Z)

P(xk) -N SNlxkX7-- xk)'

E'

/

provided that z is different from all the xk. Fixing such a z, and making N -*oo, we get, by virtue of the previous relation, (*)

P(z) = S(z) ,

/

P(xk)

oo S (XA;XZ - Xk)

This formula is valid, then, for any polynomial P and any z different from all the xk. We estimate W*(i). Take any polynomial P with II P II w <, 1, i.e., with IP(xk)I

Substituting into (*), we find that I P(i) I < CS(i)

,j l xk l

-00 11-xkl

2 -k/2 = 2CS(i)

S 2CS(i) 1

.`/2-1

168

VI C Mergelian's criterion more general than Akhiezer's

Taking the supremum of I P(i) I over all such P, we see that 2CS(i)

W*(I) < 72---l < oo. According to the corollary in §B.3, this implies that (' °°

J

log W*(t) l + t2

dt < oo.

It is now claimed that f (i) = oo. To see this, consider the polynomials PN(x) =

since PN(i)/JxN

N S(i) > 0, it is clear that PN(i) N oo. It is therefore

enough to show that PN(x)

(x-i)W(x)

1

<

2CS(1)

on 68 in order to conclude that i2(i) = oo. We have, in other words, to verify

that IPN(Xk)I <

IXk -'I W(Xk) 2CS(1)

for k = ± 1, ± 2,.... This is true for 1 < I k I < N because then PN(xk) = 0. Suppose, therefore, that k > N. Then I4k

PN(/x)k I = -Jx N fI

1 n,N4

n

-1 < l

N

l

I4kn

-1

l

/ V XN.XkIS'(Xk)I

/

2fl 1

1

n>k

4

I

4n

)

V/XN.XkIS'(Xk)I

2S(1)

Taking symmetry into account, we see that, for I k I > N, IPN(Xk)I <

-LXkl

1

2S(1)IXk-1IS/IXkIIS(Xk)I

=

IXk -'I W(Xk) 2CS(1)

We thus have IPN(xk)I < IXk - iI W(xk)/2CS(1) for all k and every N, and this, as we have seen, ensures that S2(i) = oo.

Results explicitly involving W

169

Because 0(i) = oc, polynomials are II w-dense in 'Bw(R) by the first Mergelian theorem (§A.1). However, as we already have shown, II

log W*(t)

C

1 + t2

J -01

dt < oo.

Application of Akhiezer's criterion to 'w(R) with the weight W considered here would therefore lead to a false result.

D.

Some partial results involving the weight W explicitly

Let us see how much information about the II II w-density of polynomials in Ww(R) can be obtained by direct examination of the weight W(x) itself. Here, first of all, is an easy negative result. Theorem (T. Hall). If $°°. (log W(x)/(1 + x2))dx < oo, the polynomials are not II

II w-dense in 6w(R).

Proof. Trivially, fl(x) 5 I x - i I W(x) for x e R, so, if the above integral with log W(x) converges, so is 1 °° logo2(x)dx

-- 1+x

<

oo.

The desired result now follows from Mergelian's second theorem, §A.3. One is, naturally, very interested in finding simple conditions on W 11w-density of polynomials in 'w(1 ). In this which will guarantee direction, we begin with a very old result. II

Theorem (S. Bernstein). Let W(x) = Y_ o Anx2n where the A. > 0 and the series is everywhere convergent. If log W(?) -01)

l+x

dx = oo,

polynomials are II 11w-dense in Ww(R).

Proof. The polynomials N

PN(x) _ E Anx2" n=o

N

W(x) for 51, because all the An are > 0. Clearly PN(x) each x, so here W*(x) = W(x). Our result now follows from Akhiezer's satisfy II PN II w

theorem. (§B.1).

Corollary. The polynomials are II II w-dense in Ww( ) for W(x) =

xt

e.

170

VI D Partial results explicitly involving the weight W

Corollary. The polynomials are II Il w-dense in 'w(R) for W(x) = eNNN.

Proof. Use the fact that x2

cosh

Zee"N 5 cosh x <el' and work with the weight x4

Theorem. Let W(x) >, 1 be even, and suppose that, for x > 0, log W(x) is a convex function of log x. Then, if ('°° log W(x)

J

1

+ x2

dx = oo,

polynomials are II IIw-dense in w(R).

Proof. Starts out like that of the corollary in §C.2, Chapter V, with the use of some material from convex logarithmic regularization. Write r>o W(r)

for n = 0, 1, 2,...,

and then put, for r > 0, r2n

T(r) = sup-. n3o S2n

Since log W (r) is a convex function of log r, we have, by the proof of the second lemma in Chapter IV, §D, that (*)

W (r)

< T(r) <, W(r)

r whenever r is sufficiently large. (It's W(r)/r2 on the left and not W(r)/r because we use the even powers of r in forming T(r).) Take any fixed number A between 0 and 1, and form the function 00

S(x) = (1- A2)

n=0

x2n

A2nS2n

We see from the definition of T(r) and (*) that S(x) < W(x) for I x I sufficiently

large, since W is even. Therefore, as in the proof of the above theorem of Bernstein (the numbers Stn are all positive!), we at least have W*(x) >, CS(x)

for some C > 0 chosen so as to make CS(x) < W(x) for all x. Referring again to (*), we see, however, that

S(x) i (1-A2)T(2Ixl) i (1-12)W(2I2I)

Sums of imaginary exponentials - the collection .9A

171

for I x I sufficiently large. Hence, taking a big enough, log C ' log S(x) -+f dx

(' °° log W*(x)

f

x

a

dx

a

log C 1 1 - .12 a + a-log A2

a

x

)-2f °° logx2 x dx + J

('

log W(Ax) dx. x2

a

The last integral on the right equals 1 f "0 (log which is clearly infinite if f 10.(log W(x)/(1 + x2))dx = cc. So f (log W*(x)/(1 +x2))dx = oo, and the result follows by Akhiezer's theorem.

Remark. Is the theorem still true if the even function W(x) is merely required to be increasing for x > 0? An example to be given in Chapter VII shows that the answer to this question is no.

Weighted approximation by sums of imaginary exponentials with exponents from a finite interval

E.

Let W(x) > 1 be a weight which is now merely assumed to tend to oo as x -) ± oo. We fix some A > 0 and ask whether the collection of finite sums of the form

['[r

Cxe i tx

-A
II

Ilw-dense in'H,(l ). It turns out that the theorems of Mergelian and

Akhiezer given in §§A and B above have complete analogues in the present

situation. We will be able to see this in the present § without having to repeat most of the details from the preceding discussion. 1.

Equivalence with weighted approximation by certain entire functions of exponential type. The collection 9A

If u(t) is a finite sum of the form _ A «
Therefore the useful Markov-Riesz-Pollard trick applied, for instance, in the proof of the second Mergelian theorem (§A.3) is not available for such sums. For this reason, the following result is very important. Lemma. If W(x) >, 1 and W(x) -+ co for x- + oo, every entire function of exponential type 5 A, bounded on the real axis, can be II II w-approximated by finite sums of the form C eilx Y_

-A
a

172

VIE Weighted approximation. Sums of imaginary exponentials

Proof. Take any entire function f (z) of exponential type 5 A, bounded on R. Since W (x) - eo for x --> ± oc, sup if (x) -.f (Px)

W(x)

xcR

)0

for p -1 by continuity of f . Given s > 0, fix a p < 1 such that the above supremum is < e. If h > 0 is small enough, we will also have (f(px)_f(px)sinhhx x)

sup xef8

W (X)

< s;

we take such an h so small that 'inhz 9(z) = f (Pz)

z

is of exponential type < A, and fix it. We thus have II f - g II w < 2s. However, g, besides being of exponential type < A, is also in L2 on the real axis. We can therefore apply the PaleyWiener theorem (Chapter III, §D) to g, obtaining

f

g(x) =

A

JA

e)AxG(2)d2

with some GeL2(- A, A). This property of G also makes GeL1(- A, A), by Schwarz' inequality. For large integers N, put fir- 1

9N(x) =

Y- e'(-A+(2klN)A)x k=0

1-A+((21c+2)/N)A

G(A)d2. -A+(2k/N)A

For each N, 19N(x) 11< 11 G II 1 for x e t, and we clearly have gN(x)

N g(x) u.c.c.

on R. Therefore 11 9 - 9N II W N 0, so, taking N large enough, we get

IIf -9NIIW < IIf -911w+119-9NIIW < 3e. Since gN(x) is a finite sum of the form Cze'Az Y-

-A
we are done. Definition. &A denotes the set of entire functions of exponential type < A, bounded on the real axis. Since every finite sum E-ASA,ACAe'AX certainly belongs to G'A, the above

lemma has the obvious

2 Analogues of Mergelian's and Akhiezer's theorems

173

Corollary. Let cpe'w(I8). There are finite sums a(x) of the form Czeiix Y_

-ASA6A

making 11 9 - o II w arbitrarily small if and only if there are

with

11(P-fn11 w-0.

Remark. What is important here is that, if fegA and zoeC, the ratio (f (t) - f (zo))/(t - zo) also belongs to (f A . The functions "A(z) and WA(z). Analogues of Mergelian's and Akhiezer's theorems

2.

In analogy with the definition of Q (z) (beginning of §A), we put S2A(z) = suP { I f(z) I: fedA and

f(t) (t - i)W(t)

Remark. A slight extension of the argument used to prove the lemma in the preceding subsection shows that f'A(z) is already obtained if we use only finite sums f (t) of the form E_A«sACAe'A` in taking the supremum on the right. Verification of this fact is left to the reader.

Observe that, if f E9A, so is the function f * defined by the formula f *(z) =f (z-). This makes c2A(z) = nA(zJ). As we have already noted, when At) e9A, the quotient (f (t) - f (zo))/(t - zo) also belongs to 'A. So, by the way, does

f(t)-f(zo)(t-zi) t - zo

belong to 9A then. These evident facts make it possible for us to virtually copy the proof of the first Mergelian theorem as given in §A. 1, replacing the collection of polynomials by 9A. Keeping the lemma from the previous subsection in mind, we obtain, in this way, the Theorem. If f'A(z) = oo for one non-real z, the collection Off, nite sums of the

form Y_

Cze'Zx

A
II w-dense in'w(R). Conversely, if such sums are II II w-dense in'w(1 ), SZA(z) = oo for all non-real z. The second theorem in §G.2 of Chapter III applies to the functions f egA, is

II

174

VIE Weighted approximation. Sums of imaginary exponentials

and we have, for them, (*)

logIf(z)I < AIZ5zI+-I f

I3IZlog1 f(t) t 12

Using this relation we can copy the proof of Mergelian's second theorem (§A.3), to get Theorem. The finite sums of the form 'w(IIB) if and only if j°° log OA(X)

_

1+x

dx = oo

E_A<.t
are

II

llw-dense in

.

To obtain analogues of Akhiezer's theorems (§§B.1 and B.2), we write

WA(z)=sup{I f(z)I: fe9Aand Ilf IIw51}. As in the formation of fIA(z), we can limit the set of functions f occurring on

the right to the ones expressible as finite sums E,_A,z,ACxeizx. WA(x) is thus a lower regularization of W(x) by such finite sums. Arguing as in §§B.1, B.2, with use of (*) in the appropriate places, we find:

Theorem. For continuous W, finite sums of the form Cze';L"

Y-

-A«
1+x2

dx = co.

Theorem. If f °°. (log WA(x)/(1 + x2)) dx < oc, any function in'w(R) which IIw-approximated by finite sums of the form Y-_A 0, an inequality of the form

can be

II

l(D(z)I 5 II)IIwMEexp(Al3zl+slzl). Here, M. depends only on e, and is independent of the particular function (D arising in this manner.

.

Corollary. Let J0 (log WA(x)/(1 + x2)) dx < oo, and denote by E the set of points on R where W(x) < oo. If either

r-.or

limsup

number of points in E - [0, r] >

A it

3 P6lya's maximum density

175

or

limsup

number of points in E n [- r, 0] > r

r- m

then '9A cannot be II

A it

,

II w-dense in Ww(!!8).

Proof. Is based on a result much deeper than the one needed for the corresponding proposition about weighted polynomial approximation (end of §B.2).

Suppose, wlog, that W(xk) < co where 0 < x0 < xl < x2 < , and that n/x > A/rt. (If the set E has a finite limit point, one can give a much simpler argument.) Take any continuous bounded cp (belonging thus to Ww(!!B)) with rp(xo) = 1 and cp(xk) = 0 for k > 1; it is claimed that such a function (p cannot be II II w-approximated by functions in 49A. If it could, we would, by the theorem, get an entire function C(z) with (D(xk) = Oxk), k >, 0 (hence C(xo) = 1 so that (D # 0) satisfying, for every e > 0, an inequality of the form I C(z)l < CE exp (A 13z l + e l z l ).

This certainly makes CF of exponential type < A. We also have

(t)

flog+lCF(x)l °° _00

1+x

dx < oo

.

Indeed, there is a sequence of functions

with

II(p-fnllw V o (hence, wlog, II f II w < 1), and fe(z) n' d>(z) u.c.c. (That's how one shows there is such a function 4) - see §§B.1 and B.2!) Since II f" II w < 1 we have by definition I WA(z), and thus finally IC(z)I < WA(z). We are, however, assuming that 10-0. (log WA(x)/(1 + x2))dx < oo, and, in the last integral, we may replace log by log +, because WA(z) >, 1. (Note that 1 edA!) Therefore (t) holds.

The hypothesis of Levinson's theorem, from §H.3 of Chapter III is thus satisfied. If n+(r) denotes the number of zeros of C(z) in the right half plane having modulus < r, that theorem says that lim,. n+(r)/r exists, and here

has a value < A/7t. However, C(xk) = cp(xk) = 0 for k >, 1, so certainly n+(xk) 3 k. Our assumption that limsupk.,,k/xk > A/n therefore leads to a contradiction. The corollary is proved. 3.

Scholium. Polya's maximum density

We have not really used the full strength of Levinson's theorem in proving the corollary at the end of the preceding article. One can in fact

176

VIE Weighted approximation. Sums of imaginary exponentials

replace the assumption that limsup

number of points in E o [0, r] > A r

r- ao

1i

by a weaker one, and the corollary's conclusion will still apply. Suppose we have any increasing sequence of points Xk >, 0, some of which

may be repeated. For r > 0, denote the number of those points on [0, r] (counting repetitions) by N(r), and, for each positive A < 1, put N(r) - N(Ar) (1 - A)r r-oo

Dz = limsup

Note that if limsupr- N(r)/r = D is finite, we certainly have D < DA for each A < 1, as simple verification shows. Lemma. lima- 1 Dz exists (it may be infinite).

Proof. Let 0 < A < A'< 1, Writing A/A' = p, we have the identity N(r) - N(Ar)

-

1 - A' N(r) - N(A'r)

1-A (1-A')r

(1-A)r

A' - A N(A'r) - N(,uA'r)

+ 1-A

(1-,u)A'r

whence (*)

Dx

1-

Dz.

+ 1-

D

Since N(r) is increasing we also have, for 0 < A < A'< 1, N(r) - N(Ar)

1 - A' N(r) - N(A'r)

1-A (1-A')r

(1-A)r so

DA

1-

DA,.

Suppose first of all that limsupx-1 DA = oo. Then, if we have Dxo > M, AO, 0 < AO < 1, the previous relation shows that

say, for some

DA > (1/(1 + A0))D,0 > ZM for

Ao < A < Ao.

However, substituting

A= Ao and A'= ../A0 in (*), we get Dxo - DIxo, so also D,/xo >, M. Then,

by the reasoning just given, DA > M/2 for AO < A 5 VAo. This same argument can evidently be repeated indefinitely, getting Dx /4 > M, Dx > M/2 for VAo 5 A < A114 , and so forth. Hence D, > M/2 for Ao < A < 1, so, since M was arbitrary, Dz -+ co as A

Consider now the case where limsupx- Dz = L is finite, and, picking any E > 0, take any AO, 0< Ao < 1, such that DA0 > L - s but Dx < L + E

3 P61ya's maximum density

177

for ,o < A < 1. Putting A = A0 in (*), we find, for A0 5 A'< 1,

L-s 5

1

-

1-A 0

Dz.

1-.10(L+e),

+

0

that is,

1-A'

1-AoL - E -

1-A'

AO

DAI 1-AzE < 1-A 0 -

and

Dx,>L- 1+.1'-2%0 1-A' For 1o < A' S 'l0, the right-hand side is so we see that in fact

L - 3e < Dz. < L + s

for A o 5

L-(1+2 JAo)e > L-3e, 2' S.2o.

As we already saw, D,/,to > Dzo. Therefore DIzp > L - s, and we may repeat the last argument with ,Ao instead of A0 to conclude that

L-3E < DA. < L+e for ",/Ao
L-3e < D. < L+E forA0, 0 was arbitrary, Dz --> L for 2 -> 1. The lemma is proved.

Definition. D* = lima 1 DA = limx.1 _ (limsup,_,, (N(r) - N(Ar))/(1 - A)r) is called the maximum density of the sequence {xk}.

Since, as we have already remarked,

Dz > D = limsup ,moo

N(r)

r

we certainly have D < D*. Simple examples (furnished by sequences with large gaps) show that D* may be much larger than D. Therefore, if, in any theorem whose hypothesis requires D to exceed some value, we can replace D by D*, we obtain a stronger result thereby. This observation applies to the corollary at the end of the preceding article. Theorem. Given a weight W(x) > 1 tending to 0o as x -- ± oo and a number A > 0, suppose that

F

log WA(x)

l+x2

dx < 00

178

VIE Weighted approximation. Sums of imaginary exponentials

and that W(xk) < ao on a strictly increasing sequence of points Xk 1> 0. If the maximum density D* of the sequence {xk} is > A/n, then &A is not II

II w-dense in Ww(R).

Proof. Taking the index k of the sequence {xk} to start from the value k = 0, we begin as in the proof of the corollary by choosing a 1pElew(l8) with gp(xo) = 1 and cp(xk) = 0 for k > 1, and argue that, if cp could be II w-approximated by functions in ' (D1), there would be an entire function '(z) of exponential type < A with II

JlogI(x)I -00

1 + x2

dx <

and t(xk) = cp(xk), k >, 0. Letting n+(r) be the number of zeros of cb(z) with real part >, 0 having modulus 5 r, we have, by Levinson's theorem (§H.3,

Chapter III), that n+(r) r

-- some D ATC

as r-> oo.

The Xk with k > 1 are zeros of t(z); therefore, if N(r) denotes the he number of such xk in [0, r], we have, for each 2 < 1, N(r) - N(Ar) S n+(r) -n+().r). In view of the limit relation just written, the quantity on the right equals (1 -.l)Dr + o(r) for large r, so we get

for each .1 < 1. Therefore D* = lim..1 Dx is also < D < A/n, contradicting our assumption that D* > A/n. We are done.

Remark. Towards the end of Chapter IX, we will see that the theorem remains true when we replace the maximum density D* of the sequence {xk} by a still larger density associated with that sequence. The maximum density D* associated with an increasing sequence of positive numbers {xk} has an elegant geometric interpretation. be an increasing sequence of positive numbers, some Definition. Let of which may be repeated, and let v(r) denote the number of points in the interval [0, r], counting repeated ,, according to their multiplicities is called measurable if lim,. , v(r)/r exists and as usual. The sequence is finite. The value of that limit is called the density of We have then the Theorem (Polya). Let the maximum density D* of the increasing sequence of

positive numbers xk be finite. Then any measurable sequence of positive

3 Pdlya's maximum density

179

numbers containing all the Xk has density >, D*, and there is such a measurable

sequence whose density is exactly D*. If D* = oo, there is no measurable sequence (of finite density) containing all the xk.

Proof. If {xk} is contained in an increasing sequence of numbers 1;n >, 0, and if, with v(r) denoting the number of points cfn in [0, r], v(r)/r - + D for r -* oo, we see, just as in the proof of the preceding theorem, that D* <, D. The first and last statements of the present theorem are therefore true. To

complete the proof we must, when D* < oo, show how to construct a measurable sequence of density D* containing the points Xk. The idea here is

transparent enough, but the details are a bit fussy. Call N(r) the number of points xk (counting repetitions) in [0, r], and write 2 = 2We have An T 1, so, if we put n

En =Sup { I D* - D,1 I:

An < 2 < 11,

en decreases monotonically to zero as n-> oo, since, according to the lemma

at the beginning of this subsection, Dz -. D* for 2 -* 1. Referring to the definition of DA and taking 2 = 2,,, we see that for each n there is an rn with

N(r) - N(2nr) (1)

(1 - An)Y

< D* + 2E for

r >,

It is convenient in what follows to write An = 1/2n = 21/2". For each n, take an integral power Rn of An_, (sic!) which is > rn and large enough so

that (*g)

(An -

1.

We require also that Rn > Rn _ 1 if n > 1 so as to make the sequence {Rn} increasing. An+1 Rn+1

Rn

An Rn

j(n)

An Rn

f(n)

fan)

Rn+1

A.2+1 Rn+1

fl(n+1) fin+1)

Figure 37

Using the numbers Rn and An we construct certain intervals, in the following manner. Given n, we have, from R,, up to Rn+,, the intervals (Rro AnRn], (AnRm An Rn],... , lAn -1Rn, An Rn], say, with AR,,=R,,1. From Rn onwards, each of the intervals splits into two, both of the form R,,+1] After Rn+2, each of those

180

VIE Weighted approximation. Sums of imaginary exponentials

splits further into two, and so forth. We denote the intervals of the form lying between R and by J. Consider any of the intervals Jp J. Since A = we have, by (t) and the choice of the inequality N(APR,,) - N(AP

1

Rn) < D*+2En APR,- AP-'R. N(A -' Rn) of points Xk in JP °. If the ratio for the number N(A

on the left is < D*, let us throw new points into JP(") until we arrive at a total number of such points (including the xk already eJP )) lying between

(A R - A -'

R - An -1 Rn)(D* +

This we can do, thanks to (,*k).

In this manner we adjoin points to the sequence {xk} in each of the intervals JP n) lying between R and Rn+ 11 to the extent necessary. We do that for every n. When finished, we have a new sequence of points containing all the original xk. It is claimed that this new sequence is measurable, and of density D*. For r > 0, call v(r) the number of points of our new sequence in [0, r]. Suppose that R > R lies in one of the intervals JP ) with m n, and, since the e, decrease monotonically,

(R-Rn)D*-(D*+2sn)IJp'"'I 5 v(R)-v(R (R (D* + 2En) I J(m) I, as is evident from our construction. Because A. -> 1 and the last relation shows that JP"''I < (A. - 1)RAm 5 (A. D* - E 1<

v(R) -

R-R

<, D*+3E n

as soon as R is sufficiently large >

D*-2E 5

v(R)

This means that we have

< D*+4E

for R large enough, so, since E can be taken as small as we like, v(R)

D*

for

R - co.

Our new sequence is thus measurable and of density D*, which is what was needed. We are done. 4.

The analogue of Pollard's theorem

Returning from the above digression to the main subject of the * The upper index m of the interval J('°) containing R tends to oo with R.

4 The analogue of Pollard's theorem

181

present §, let us complete our exposition of the parallel between weighted approximation by linear combinations of the eizx, - A < A < A, and that by polynomials. To do this, we need the analogue of the Pollard theorem in §C.3.

In the present situation, we cannot just copy the proof given for W,k(x)

in §C.3. That's because we now suppose merely that W(x) -a oo for x--> ± oo, and no longer assume the growth of W(x) to be more rapid than that of any power of x as x -* ± oo. This means that we no longer necessarily have W(x)/l x - i I co for x -> ± oo, or even W(x)/l x - i I >, const. > 0 on R. The method of the proof in §B.3 can, however, be adapted to the treatment of the present case. Theorem. Let W(x) ? 1 and W(x) -+ oo for x -- ± oo, and suppose A > 0. If WA(i) < oo, then log WA(t)

1+tz dt

Proof. As in §C.3, put W(x) = W(x)/Ix - il. For each fixed z0

IJ

,

the

ratio 1/(t - zo)W(t) is bounded above on R and -* 0 as t -> ± oo. Let us define W,,(D) as the set of functions cp continuous on 11 for which I pp(x)/W (x) I

is bounded and tends to zero as x -> ± oo (just as in the

situation where W(x) >, 1), and put

Il (pll, = sup xcR

for such gyp. As we have just seen, all the functions 1/(x - zo), zoo R, do belong to leN,(R). Denote by '?A the set of functions f (t) in (ffA such that t f (t) also belongs to O'A; '?A is just the set of entire functions f of exponential type < A with

f (t) and t f (t) both bounded on R. There are plenty of such functions; sin A(t - zo)/(t - zo) is one for each complex zo. We have 07A c 4W(I! ). It is claimed that, if WA(i) < eo, ('A is not 11 ,-dense in WK,(R). To see this, it is enough to verify that the function 1/(t - i) (which belongs to 'µ,(l )) is not the 11 II,-limit of functions in 'A. Suppose, for i > 0, that we had an f e''A with II

tt

l -f(t) ii

q.

,

Then

1- (t - i) f (t)

1-(t-i)f(t)

W(t)

(t - i)W(t)

sn

182

VIE Weighted approximation. Sums of imaginary exponentials

for toR, so, putting G(t)_(1-(t-i)f(t))/n, we would have a Ge8A 1 and G(i) = 1/q. This means that 1/?l would have to be < WA(i), son cannot be smaller than 1/WA(i), and our assertion holds.

(because fe'?A) with II G II w

Assuming henceforth that WA(i) < oo, we see by the Hahn-Banach theorem (same application as in §A.3) that there is a linear form L on the functions of the form p/4V, cpeWW(R), with
I1 w11i,,

and L(f/W) = 0 for all fe A, whilst L(1/(t - i)W(t)) * 0. Let now GC-'A (sic!), and take any fixed zOOR. Since the function G(t) - G(zo)

t - zo belongs to '?A (!), we have

_

G(t) - G(zo) (t - zo) W(t))

L

0,

whence (the Markov-Riesz-Pollard trick again!),

I

G(z°)

(§)

\ (t - zo)W(t) G(t)

)IL( (t - zo)W(t) I

provided that the denominator on the right is different from zero. If II G II w

< 1, we have, since

_

G(t)

(t - zo)W(t)

I t - i I G(t)

t - zo W(t)'

that G(t)

t - zo

t-i sup w

tEd

t-zo

The quantity on the right was worked out in §A.2 and seen there (in the corollary) to be <, (1 + I zo I)/I.3zo I. Therefore, if II G II w < 1, LI

) (t - zo)W(t) G(t)

1 +IzoI

G(t)

t - zo

ii,

Calling

(D(z) = L

1

(t - z)W(t)

for zOR

13zo I

4 The analogue of Pollard's theorem

183

we see from this last relation and from (§) that, for Ge 'A, (§§)

IG(z)I <,

1+IzI

if

IIGIlw,<1.

I ,3zi I t(z) I

We know that 1(i) # 0. Also, 4)(z) is analytic for 3z > 0. That's because

t-.I

(t-z-Az)-i-(t-z)-t

-

It - iIAz

1

(tZ)2

Liz

- (t-z)2(t-z-Az)

tends to zero uniformly for - oo < t < oo as provided that z0R. Since It - i I W(t) = W(t) is >, 1 on R, this implies that D(z + Az) - (D(z) AZ

-->

L(

1 (t - z)2W(t) /

when Az -> 0 as long as z OR, and thus establishes analyticity of F(z) in the

upper and lower half planes. The function 4)(z), analytic and not - 0 for 3z > 0, is not quite bounded in {.3z > 11; it is, however, not far from being bounded in the latter region. (Here, by the way, lies the main difference between our present situation and the one discussed in §A.3.) We have, for t-i

1(t - z)W(t)l

I (t - z)W(t)I

I

t-z

since W(t) >, 1, whence, by §A.2, II1/(t-z)IIw<(1+Iz1)/I,,3z1, so

I

I + Izl

The function *(z)/(z + i) is thus analytic and bounded in {3z > 1} and continuous in the closure of that half plane; it is certainly not identically zero there because D(i) # 0. Therefore, by §G.2 of Chapter III, °°

(if)

I

_-

1(x+i)

1

+x21og1

x+i

dx < oo.

By the definition of WA and (§§) we now obtain

WA(x + i) = sup {IG(x+i)I: Ge A and IIGIIw<1}

1+Ix+i1

Ix+iI

ID(x+i)I

2I(D(x+i)I'

and

log WA(x + i) < log 2 + log-

(D(x + i)

x+i

184

VI F De Branges' description of extremal annihilating measures

Now we are in the hall of mirrors again! Take any Ge

A

with II G II w < 1.

Then, on the one hand, log I G(x + i) I < log WA(x + i),

while, on the other, log l G(%) 15 A +

If'

n - 1(

x)z + 11 l dx

for l; c -R, according to the second theorem of §G.2, Chapter III, applied in

the half plane ,3z <, 1. Substituting into this last inequality the two preceding it, we find 1

log I G(1;) I

A + log 2 + 7C

°°

_

00

IJ(x + i)/(x + i)I

log

(l; - x)s + 1

I for such G, we see that

and, since log WA(l;) is the supremum of log I

logWA(l;) 5 A+log2+

1

dx,

'0 logb(x+i)/(x+i)

dx.

Using Fubini's theorem in the usual way together with this relation, we finally get f °° log WA(S)

J

i;2+1

Q S (A+log2)+

1

2

(' `°

-

xz

+4log-00

dx.

The integral on the right is, however, finite by (tt). The theorem is thus proved, and we are done. Remark. Thus, if WA(i) < oo, the fourth theorem of article 2 (Akhiezer's description) and the discussion in article 3 related to it apply.

F.

L. de Branges' description of extremal unit measures orthogonal to the e'2"l W(x), - A < I < A, when gA is

not dense in 'w(R). We have now about finished with the individual treatment of uniform weighted approximation by polynomials or by functions in 8A. There remains one thing, however.

In his study of the situation where linear combinations of the eiz', -A< A,< A, (or polynomials) are not II II w-dense in Ww(U8), Louis de Branges obtained a beautiful description (valid for weights which are not too irregular) of the extremal unit measures orthogonal to the functions

De Branges' description of extremal annihilating measures

185

eizx/W(x), - A 5 A 5 A (or to the polynomials divided by W). We should not end the present discussion without giving it.

For the treatment of de Branges' description, we assume that W(x) is continuous (and finite) on a certain closed unbounded subset E of R, and that W(x) = oo on R - E. As always,

00.

W(x) >' 1.

In this circumstance, the ratios cp(x)/W(x) with cpeWw(R) are continuous

on the locally compact set E, and tend to zero whenever x-> ± oo in E. We can write II w II w = sup I `p(t) teE W(t)

for

cpe'w(R),

and we see that the correspondence cps-+cp/W is an isometric isomorphism between Ww(F1) and '0(E), the usual Banach space of functions continuous

and zero at oc on the locally compact Hausdorff space E. The bounded linear functionals on 'o(E) are given by the Riesz representation theorem. Therefore the II II w-bounded linear functionals on W (R) are all of the form

JE14'(t) dµ(t) =

y((t) dµ(t)

J

with totally finite complex Radon measures y supported on E. We consider in the following discussion the case where linear combinations-of the ei ', - A s A S A, are not II II w-dense in ' (R). We could also treat the situation where polynomials are not II 11w-dense in W,(R) and obtain a result analogous to the one to be found for approximation by exponentials; here, of course, one needs to make the supplementary assumption that x"/W(x) -* 0 for every n 3 0 as x -. ± oo in E.

Granted, then, that linear combinations of the e"", - A < A S A, are not II 11w-dense in Ww(R), there must, by the Hahn-Banach theorem, be some non-zero II II w-bounded linear functional on'w(ll) which is ortho-

gonal to (i.e., annihilates) all the e'Ax, - A 5 A < A, or, what comes to the same thing (lemma of §E.1), to all the functions fe&A. According to the above description of such linear functionals, there is thus a non-zero totally finite Radon measure p on with E'

(*)

fE W(X) dp(x) = J

W(AX)

du(x) = 0 for fe''A.

The idea now is to try to obtain a description of the non-zero measures y on E satisfying (*). In the first place, if a complex measure p satisfies (*). so do its real and

186

VI F De Branges' description of extremal annihilating measures

imaginary parts. That's because u = 2(f + f *) and v = (1/2i)(f - f *) both belong to 40A if f does (and conversely). However, u(x) and v(x) are both real-valued on 18 (recall that f *(z) =f()), so, if p is any complex measure on E satisfying (*), we have, given fe 'A, JEW(x)dp(x)

W(x)dp(x)

0,

= JE = whence (taking real and imaginary parts) J E W(x)

= 0,

d92µ(x) = JE4"(x)

J E W (x) d9tµ(x)

=J E W (x)

dp(x) = 0,

and thus

d 3p(x) = 0.

W((x)

I

E

This means that a description of the real-valued (signed) p on E satisfying (*) provides us, at the same time, with one for all such complex measures p. Our investigation thus reduces to the study of the real signed measures p on E satisfying (*).

Notation. Call E the set of finite real-valued Radon measures p on E satisfying (*) and such that

IIpII=I

Idp(x)I<, 1.

E

The set E is convex and w*-compact (over '0(E)). We can therefore apply to it the celebrated Krein-Millman theorem, which says that E is the w*-closed convex hull of its extreme points. (Recall: an extreme point p of E is a member thereof which cannot be written as Ap, + (1 - 442 with 0 < A < 1 and measures p, and p2 in E different from p.) More

explicitly, we can, given any pEE, find a sequence of finite convex combinations pN = Y_ Ak(N)vk(N) k

of extreme points vk(N) of E (the 2k(N) > 0 and Y-kAk(N) = 1 for each N) with

d4N(x) -k dp(x) w*

as N - oo .

We can, in fact, even do better - Choquet's theorem furnishes a represen-

I Three lemmas

187

tation for y as a kind of integral over the set of extreme points of E. (The book of Phelps is an excellent introduction to these matters. In the present situation, E is metrizable, so a particularly simple and elegant form of Choquet's theorem applies to it.) The point here is that a good description of the extreme points of E will already tell us a great deal about all the members of E and thus, in turn, about the complex-valued y satisfying (*). Knowledge of those extreme points therefore takes us a long way towards a complete description of the measures y which satisfy (*). What Louis de Branges found is an explicit description of the extreme points of E. We now set out to explain his work. 1.

Three lemmas

The main idea behind the following development is contained in

De Branges' lemma. Let y be an extreme point of E and h a bounded real-valued Borel function. Suppose that

J -I

W(ax) h(x)

dy(x) = 0

for all f eoffA. Then h(x) is a.e. (Idyl) equal to a constant. Remark. It is enough to assume that h is essentially (I dy I) bounded, as will

be clear in the proof. Proof. We start out by observing once and for all that an extreme point y of Y. is never the zero measure. That's because

O=ZV+Z(-v) where, for v, we can take any non-zero member of E. In the situation of this §,

F. has non-zero elements. In view of this fact, we must have f °° I dy(t) I = 1 for any extreme point y

of E. Otherwise we could write

y = (1- IIit II)'o+ IIy1I(II11), with 0 < II y II < 1 and the measures 0 and y/ l p II both belonging to E. Now we take a function h as in the hypothesis, and an extreme point y of E. The relation (*) holds, therefore

f f(x) (h(x) + C) dy(x) = 0 E

for every constant C. Since h is bounded, we will have h(x) + C >, 0 for

188

VI F De Branges' description of extremal annihilating measures

suitable C; we may therefore just as well assume that h(x) is positive to begin

with, since otherwise we would only need to replace h by h + C in the following argument. We have, then, a positive bounded h satisfying the hypothesis. Unless h(x) - 0 a.e. (I dp 1) (in which case the lemma is already proved), we have h(t)I dp(t)I > 0. SE

Multiplication of h by a positive constant will then give us a new positive function like the h in the hypothesis, which we henceforth also denote by h, fulfilling the condition

1.

h(t)ldp(t)1 = 1.

Since his bounded, there is a A, 0 < 2 < 1, with 0 5 2h(x) < 1. Picking such a A, we have Ah(t)

1.

1 1

I

1Ah

I dp(t) I= Je

=1

1 A

f I dp(t) I -1 E

since J E I dp(t)1= 1. Also, f (x) 1 - Ah(x)

fEW(x) 1-A

l A A

2(t) I dp(t) I

SEh(t)dt) I= 1,

dp(x) = 0

for all fE'eA by the hypothesis and the property (*). In view of the previous

relation, we see that the measure P2 on E such that 1 1

dµ2(1) =

2 h(t)

dp(t)

belongs to E.

The same is true for the measure pl on E with dpi(t) = h(t)dp(t). However,

dp(t) = 2dpI(t) +(I -A)dp2(t),

and we assumed that p was an extreme point of E. Since 0
1 Three lemmas

189

Lemma. Let µ be an extreme

point

of E,

let FeL1(Idµl), and

suppose that f EF(t)dp(t) = 0. Then there are f e (fA with .fn(t)

_ F(t)

SE W(t)

Idµ(t)I n *

0.

Proof. By duality. The usual application of the Hahn-Banach theorem shows that the infimum, for f ranging over 'A, of F(t) _ .f(t) W(t)

1,

I dµ(t) I

is equal to the supremum of 1.

F(t)h(t) d,u(t)

for Borel ffunction h such that I h(t) I

(t)

1 a.e. (I dI) and

SEh(t)dt )= 0

whenever f eolA.

Since y is a real measure and any f egA can be written as u + iv with u and v in 'A and real on R, we see that, if It satisfies (t) for all f e 'A, so do 91h and 3h. De Branges' lemma now shows that these latter functions are constant a.e. (I dµ I) if h is bounded; in other words, the functions h over which the above mentioned supremum is taken are all constant a.e. (Idol). But then f Eh(t)F(t)du(t) = 0 for such functions h, according to our assumption on F. So the supremum in question is zero, and the infimum is also zero. Done. Lemma. Let p f (t) 1. W(t)

0 belong to E. The functions f (z) in 'A with I dp(t) I <,

1

form a normal family in the complex plane. The limit F(z) of any u.c.c. convergent sequence of such functions f is an entire function of exponential

type 5 A with J

log+IF(t)I l + t2

dt < oo.

Proof. Since p is real, the function (D (z) = f E(dp(t)/(t - z) W(t)), which is

190

VI F De Branges' description of extremal annihilating measures

analytic in both half planes 3z > 0, .3z < 0, cannot vanish identically in either (otherwise it would be 0). 1(z) is bounded for 3z > 1 and for 3z 5 - 1. If now f E9A, the function of t, (f (t) - f (z))/(t - z), also belongs to Of A,

making fE((f(t)-f(z))/(t-z))(dp(t)/W(t))=0. Therefore, if zoR,

Pz) =

1

fE f(t)dµ(t) (t - z) W(t)

(D(z)

(the Markov-Riesz-Pollard trick again!). When f E I f (t)/W(t) I I dp(t) I < 1, this yields, for z = x ± i, I f (x ± i) I < 11/4)(x ± i) I, and, by §E of Chapter III, loglf(C)I

<, AI3(+11+ f

IXT-lllog+;(x±i)ldx

,I

Al .3 + 11 + . fco

I

.,

R

l I log 1((x + i) I

dx.

Since F(z) is # 0 both in :3z > 0 and in 3z < 0, we have

f(1/(1 +x2))log-I I(x+i)Idx < oo and

f

(I /(I + x2))log- I(D(x

- i)Idx < cc.

From here on, the proof is like that of Akhiezer's second theorem (§B.2 see also §E.2, especially the proof of the corollary at the end of that article).

2.

De Branges' theorem

Lemma. Let p be an extreme point of E. Then p is supported on a countably infinite subset of R without finite limit point.

Proof. As we saw in proving de Branges' lemma (previous article), an extreme point p of E cannot be the zero measure. Such a measure p cannot have compact support. Suppose, indeed, that p were supported on the compact set K s E; then we would have ei l'

K W(x) du(x)

f

= 0, - A < A < A.

Here, since K is compact, we can differentiate with respect to A under the integral sign as many times as we wish, obtaining (for . = 0)

dp(x) = 0,

n = 0, 1, 2, ... .

2 De Branges' theorem

191

From this, Weierstrass' theorem would give us g(x)

dp(x) = 0

fK W(x) for all continuous functions g on K. Then, however, p would have to be zero,

since K g E, and W(x) is continuous (and < oc) on E. The fact that p does not have compact support implies the existence of a finite interval J containing two disjoint open intervals, I, and 12, with Idp(t)1 > 0

and

J

Idp(t)I > 0.

f,

This means that we can find a Borel function q, identically zero outside 1, u12 (hence identically zero outside J) with Igp(t)I equal to non-zero constants on each of the intervals 1,, I2, such that cp(t)dy(t) = 0. -.0

On account of this relation we have, by the second lemma of the preceding

article, a sequence of functions ffle A with fn(t) _ w(t) W(t)

Idy(t)I - 0.

The third lemma of the above article now shows that a subsequence of the f (z) converges u.c.c. in C to some entire function F(z) of exponential type A, and we see by Fatou's lemma that F(t) - w(t) I dy(t) I W(t)

= 0,

F(t) = cp(t) a.e. (Idyl). W(t)

By its construction, cp is not a.e. (I dp 1) equal to zero, hence F(z) # 0. The

function p does, however, vanish identically outside the finite interval J. Therefore F(t) = 0 a.e. (Idyl), t0J.

W(t)

Since F(z) # 0 is entire, F can only vanish on a certain countable set without

finite limit point. We see that y, outside J, must be supported on this countable set, consisting of zeros of F.

192

VI F De Branges' description of extremal annihilating measures

Because the support of p is not compact, there is a finite interval J', disjoint from J, and containing two disjoint open intervals Ii and I2 with

f

Idp(t)I>0,

f

1dp(t)I>0.

Repetition of the argument just made, with J' playing the role of J, shows now that p, outside J' (and hence in particular in J!) is also supported on a countable set without finite limit point. Therefore the whole support of y in E

must be such a set, which is what we had to prove. Remark. The support of y must really be infinite. Otherwise it would be compact, and this, as we have seen, is impossible. Now we are ready to establish the Theorem (Louis de Branges). Let W (x) >, 1 be a weight having the properties stated at the beginning of this §, and let E be the associated closed set on which W(x) is finite. Suppose that f'A is not II 11w-dense in 16w(R), and let p bean extreme point

of the set E of real signed measures v on E such that Idv(t)I <, 1 and

f -E

dv(t) = 0 for all fegA.

Then p is supported on an infinite sequence without finite limit point, lying in E. There is an entire function S(z) of exponential type A having a simple zero at each point x,, and no other zeros, with

W(xn) S

Moreover,

f

log+ I SW

l+x dx < o0

and

lim y-a'D

log I S(iY) I Y

=

lim log I S(iY) I y-'-'D IYI

= A.

Proof. Let us begin by first establishing an auxiliary proposition.

2 De Branges' theorem

193

Given an extreme point p of E, suppose that we have a sequence of functions fn E 'A such that I fn(x) -fm(x) I f 00

I du(x) 1

W(x)

0.

n'M

We know from the third lemma of the preceding article that a subsequence of the fn tends u.c.c. in C to some entire function F of exponential type 5 A,

and here it is clear by Fatou's lemma that (*)

0,

whence surely F(x)

dp(x) fo. W(x)

=0

for all n. Our auxiliary proposition says that the relations

since $`°,,,

f°°

F(x) - F(a) du(x)

x- a

- 0,

W(x)

(x

b)

F(x) - F(a) dµ(x)

x- a

f -0000

W(x)

=0

also hold for the limit function F; here, a and b are arbitrary complex numbers. Both of these formulas are proved in the same way, and it is enough to deal here only with the second one. F(z) u.c.c., whence

Wlog, °°

f

I fn(x) -fn(a) - (F(x) - F(a)) I W(x)

- oo

(* ),

I dµ(x)1

0,

n

since W(x) > 1. From this is clear that

x-b (fn(x)fn(a)_1)+1(a)) I J Ix-ai, i

I

x-a

(x)

I Wdp(

10.

Also, the u.c.c. convergence of fn(z) to F(z) makes

fn(z) -fn(a) (z z-a

- F(a) - b) - F(z)z-a

(z

- b)

n

u.c.c., by the elementary theory of analytic functions (Cauchy's formula!). Therefore we also have

f

I dµ(x) I

194

VI F De Branges' description of extremal annihilating measures

and finally Idp(x)I fn(x) -fn(a)(x - b) F(x) - F(a) (x - b)

x-a

-00

x-a

W(x)

n

0.

Since, however, the f e ffA, we have (x - b) (fn(x) - fn(a))/(x - a) e -i A, whence

`° fn(x) -f"(a)(x

x-a

1-00

b)du(x) W(x)

=0

for each n. Referring to the previous relation, we see that F(a)

u(x) = 0,

-O

W( X)

as we set out to show.

Now we turn to the theorem itself. According to the lemma at the beginning of this article, our measure p is supported on a countable set {xn} c E without finite limit point, and, by the remark following that lemma, p({xn})

O for infinitely many of the points xn. There is thus no loss

of generality in supposing that p({xn}) 0 0 for each n. Take any two points from among the x,,, say xo and x,, and put

r(x)0 =

1

p({xo)xxo-x1) 1

p({x, } Xx, - x0)'

and p(x))= 0 for x J

xo or x,. Then

cp(x)dp(x) = 0,

so, as in the proof of the preceding lemma, there is a sequence of f e gA with

-OD

MX) - ip(x) I dp(x) I -n -. 0 W(x)

and, wlog, fn(z)

n

F(z) u.c.c., F being some entire function of exponential

type <, A. We see that F(x) (*)

W(x)

= cp(x) a.e. (I dp I ),

2 De Branges' theorem

195

whence °°

(t)

00

I f"(x) - F(x) I W(x)

I dµ(x)1- 0.

From (*) and the definition of cp, we have F(xo) * 0, F(x1) 0 0, so F(z) # 0.

For the same reasons, however, F(x) vanishes at all the other points x", n 0,1. Put S(z) = F(z)(z - xo)(z - x1).

Then S, like F, is an entire function of exponential type < A. S(z) vanishes at each of the points x,, in the support of y. Finally,

('°° log+z(2)Idx I

-OD

< co,

since, by the third lemma of the preceding article, the function F has this property. Let us compute the quantities S'(x"). We already know that W(x0)

S'(xo) = F(xo)(xo - x1) = W(xo)(p(xo)(xo - x1) = µ({xo} )' and similarly S'(x1) = W(x1)/µ({x1} ). Take any other point x", n

0, 1, and

form the function

(x - x0)(x - x")

F(x) (x - xl). x

Since F(x") = 0, (t) implies, by our auxiliary proposition, that

J

F(xX)n

x

(x - x1) W(x) = 0.

The function S(x)/(x - xo)(x - x") vanishes at all the Xk, save xo and x". The

previous relation therefore reduces to S'(x0)µ({x0})

S'(xn),u({xn})

_

(x0 - xn) W(x0) + (xn - x0) W(xn) i.e.,

S'(x ")

W(x")µ({x"}) = 1, and finally S'(x")= W(x")/p({x"}). The function S(z) can have no zeros apart from the x". Suppose, indeed,

196

VI F De Branges' description of extremal annihilating measures

then we would also have

that S(a) = 0 with a different from all the F(a) = 0, so, in the identity

_

du(x)

S(x)

°°

°°

F(x)

.(x-xo)(x-a) W(x)

J-cx-a(x-x,)

du(x) W(x),

the right-hand integral would have to vanish by our auxiliary proposition. The quantity (F(x)/(x - a))(x - x,) is, however, different from 0 at only one of the points x in µ's support, namely, at xo, where F(xo)

xo - a (xo - xi) =

S'(xo)

xo-a

We would thus get S'(xo) .µ({xo})

xo - a W(xo)

- 0,

i.e., in view of the computation made in the previous paragraph, 1

which is absurd. The function S(z) thus vanishes once at each As we have already seen, A, and ('°°

log +z2x)I

and only at those points. S(z) is of exponential type

dx < oo.

To complete our proof, we have to show that log I S(iy) I/I y I --' A for

y->±o0. In order to do this, let us first derive the partial fraction decomposition _

1

S(Z) -

1

(z - x,,)S'(xn)

is surely convergent because is a finite measure, and W(x) >, 1. Take the function G(t) = F(t)(t - x,) = S(t)/(t - xo), and, for fixed z, observe that Note that Y_n I

I

G(t) - G(z) _ F(t)(t - x,) - F(z)(z - x,)

t-z

t-z

_ F(t) - F(z)

t-z

(t - x,) + F(z).

2 De Branges' theorem

197

By our auxiliary proposition, F(z)

F(tt

J

-z

(t - x1) W(t) = 0,

and of course

f

_°°00

=0

F(z) dµ(t) W(t)

since 1 ESA. Therefore G(t) - G(z) du(t) t - z W(t)

Sco

_

0,

or, since G(t) vanishes at all the x save x0, G(x0) µ({xo})

x0 - z W(x0)

_

1

G(z)

(x - z)S'(xn)

0.

This is the same as S'(x0)

_

1

S(z)

1

z - x0

x0 - z S'(x0)

(x - Z)S'(xn) '

or 1

=

(z -

S(z)

1,

the desired relation. From the result just found we derive a more general interpolation formula. A. Then (ei t - e"')/(t - z) belongs, as a function of t, to dA,

Let - A 5 so

7e;zr _ era=

t-z

co

dy(t) = 0. W(t)

In other words, eixx

1

eiAz

(xn - Z)S'(xn)

(xn - Z)S'(xn)

According to our previous result, the right-hand side is just - e"/S(z). Therefore

eizz

S(z)

e;ax

n (z - xn)S'(xn)

for

-A<,.I<,A.

198

VI F De Branges' description of extremal annihilating measures

(An analogous formula with eizt replaced by any f (t)ES'A also holds, by the

way - the proof is the same.) In the boxed relation, put 2

A and take z = iy, y > 0. We get

e- Mx

eAy

(iy -

S(iy) =

I < oo and the x are real, the right side tends to 0 for

Since Y_ I

y-> oo. Thus, liminf log I S(iy) I y 00

A.

y

But limsupy-log IS(iy)I/y 5 A since S is of exponential type <,A. Therefore log I S(iy) I/y -+ A for y -+ oo.

On taking 2 = A in the above boxed formula and making y - - co, we see in like manner that log I S(iy) I

A

for y -. - oo .

lyl

De Branges' theorem is now completely proved. We are done. 3.

Discussion of the theorem

De Branges' description of the extreme points of E is a most beautiful result; I still do not understand the full meaning of it. (log' IS(x)l/(1 + x2)) dx < oo and S(z) is of exponential type, Since of S, on which the extremal measure p corresponding to the set of zeros S is supported, has a distribution governed by Levinson's theorem (Chapter III; here the version in §H.2 suffices). Because log I S(py) I

-* A for y - ± oo,

Iy1

we see by that theorem that

number of x in [0, t] t

-

A

as

it

t-co

and

number of x in [ - t, 0]

A

t

it

as

t ->oo.

The zeros of S(z) are distributed roughly (very roughly!) like the points n

An,

n=0,±1,±2,±3,....

3 Discussion of the theorem

199

(We shall see towards the end of Chapter IX that a certain refinement of this description is possible; we cannot, however obtain much more information

about the actual position of the points De Branges' result is an existence theorem. It says that, if W is a weight of the kind considered in this § such that the eizx, - A < A ` A, are not

II w-dense in 'w(!!l), then there exists an entire function 1(z) of

II

exponential type A with log I'D(iy) I

--> A

log + I t(x) I

y -> ± 00,

for

l+x a

1

IYI

dx < oo,

and I on a set of points x with x,, - (n/A)n for n - ± oo. I >, It suffices to take t(x) = S'(x) with one of the functions S(z) furnished by the theorem. (There will be such a function S because here E is not reduced to {0}, and will have extreme points by the Krein-Millman theorem!) If is the set of zeros of S, we have W(x,,) I S'(xn) I

so I

°°

J

-

I du(x) I

= 1,

00

Let us verify that

I> (' °°

=

log + x2 x) I

dx < co.

Our function S(z) is of exponential type; therefore, so is S'(z). The desired relation will hence follow in now familiar fashion via Fubini's theorem and §E of Chapter III from the inequality

log+IS'(x+i)I

f'0-00

1+ xZ

which we proceed to establish (cf. the hall of mirrors argument at the end of §E.4).

Since S(z) is free of zeros in 3z > 0, we have there, by §G.1 of Chapter III, log I S(z) I

= A.3z +

1

('

7t J

3z log I SZt)

_ -t Iz

(I t

I

ro

= A3z + rz

-x

3

t

1

z

log I S(t) I dt.

For the same reason one can define an analytic function log S(z) in 3z > 0. Using the previous relation together with the Cauchy-Riemann equations

200

VI F De Branges' description of extremal annihilating measures

we thus find that

-

S'(z)

-

d log S(z)

a

ax -' ay

dz

S(z)

-iA-

f

log I S(z) I

l(gIS(t))Idt,

3z>0,

whence, taking z = x + i, J S'(x + i)

A+

IS(x+i)

°°

1

7r

I log I S(t) I

(

_m(xt)2+1

dt.

Here, since log+ ISW I

t2+1

dt < oo,

we of course have log- IS(t)I

t2+1

dt <

co,

(Chapter III, §G.2) so

nIf' IlogIS(t)II _. l+t dt = say C, 2

a finite quantity. By §B.2 we also have

ti

2

t-i-xl < (Ix1+2)2,

teR,

so °°

1

IIog IS(t)I I

(t - x) 2 +

n

dt < C(IxI+2)2 1

and thence, by the previous relation,

Is(x+1 1 I S(x +T) I

< A+C(IxI+2)2.

This means, however, that log I S'(x + i) I < log (A + Q x I + 2)2) + log I S(x + i)I,

from which

°° log+IS'(x+i)I

-.

x2+

1

dt

f

°° log+IS(x+i)Idx 1+x2

_OD

+ f°° log+ (A+ C(I X1 + 2)2) _OD

x2+1

dx.

3 Discussion of the theorem

201

Both integrals on the right are finite, however, the first because ('°° log + ISWI

1+x2

J -00

dx < oo,

and the second by inspection. Therefore log+ I S'(x + i) I

('

1 + x2

,J

dx < oo,

which is what we needed to show. We still have to check that log I S'(iy) I

y --> ± co.

for

A

IyI

There are several ways of doing this; one goes as follows. Since the limit relation in question is true for S, we have, for each e > 0,

IS(z)I 5 Mexp(AI zI+sIzl) (see discussion at end of §B.2). Using Cauchy's formula (for the derivative) with circles of radius 1 centered on the imaginary axis, we see from this

relation that IS'(iy)I 5 const.e(A+Enyi,

so, since e > 0 is arbitrary, limsup

log I S'(iy) I

Y-±00

lyl

5 A.

However, S(iy) = S(0) + i fo S'(irl) dri. Therefore the above limit superior

along either direction of the imaginary axis must be A, otherwise log I S(iy) I/I Y I could not tend to A as y --> ± co. By a remark at the end of §G.1, Chapter III, it will follow from this fact that the ratio log I S'(iy) I/I Y I actually tends to A as y - ± co, if we can verify that S'(z) has only real zeros.

To see this, write the Hadamard factorization (Chapter III, §A) for S: z

( I - -) e

S(z) = Ae" n

xn

(We are assuming that none of the zeros x,, of S is equal to 0; if one of them is, a slight modification in this formula is necessary.) Here, as we know, all the

xn are real, therefore S(iy)

S(- iy)

= e-Zy

202

VI F De Branges' description of extremal annihilating measures

Since log I S(iy) I/y and log I S(- iy) I/y both tend to the same limit, A, as y -> oc, we must have 3c = 0, i.e., c is real. Logarithmic differentiation of the

above Hadamard product now yields S'(z)

11

1

Yn CZ - x

S(z)

x,

whence S'(z)

_

S(z)

)z - l z - xn I2 .

The expression on the right is < 0 for 3z > 0 and > 0 for )z < 0; S'(z) can hence not vanish in either of those half planes. This argument (which goes back to Gauss, by the way), shows that all the zeros of S'(z) must be real, as

required. We have now finished showing that the function t(z) = S'(z) has all the properties claimed for it. As an observation of general interest, let us just mention one more fact: the zeros of S'(z) are simple and lie between the zeros x of S(z). To see that, differentiate the above formula for S'(z)/S(z) one more time, getting d

S'(z)

dz (S(z)

(z - Xn) _

From this it is clear that S'(x)/S(x) decreases strictly from 00 to - 00 on each open interval with endpoints at two successive points x,,, and hence vanishes precisely once therein. S'(z) therefore has exactly one zero in each such interval, and, since all its zeros are real, no others.

This property implies that the (real) zeros of S'(z) have the same From that it is easy to obtain another

asymptotic distributions as the proof of the limit relation log I S'(iy) I

- A,

y --) ± oo .

lyl Just use the Hadamard factorization of S'(z) to write log I S'(iy) I as a Stieltjes

integral, then perform an integration by parts in the latter. The desired result follows without difficulty (see a similar computation in §H.3, Chapter III). Let us summarize. If, for a weight W(x), the e'-x, - A < 2 < A, are not Ilw-dense in lew(R), Louis de Branges' theorem furnishes entire functions of precise exponential type A having convergent log' integrals which are at the same time large (>, W in absolute value) fairly often, namely on a set of points x with x ' (n/A)n for n -+ ± oo. These points II

4 Krein's functions

203

xn are of course located in the set E where W(x) < oo; the theorem, unfortunately, does not provide much more information about their position, even though some refinement in the description of their asymptotic distribution is possible (Chapter IX). One would like to know more about the location of the xn. Scholium. Krein's functions

4.

Entire functions whose reciprocals have partial fraction decompositions like the one for 1/S(z) figuring in the proof of de Branges' theorem

arise in the study of various questions. They were investigated by M.G. Krein, in connection, I believe, with the inverse Sturm-Liouville problem. We give some results about such functions here, limiting the discussion

to those with real zeros. More material on Krein's work (he allowed complex zeros) can be found, together with references, in Levin's book. Theorem. Let S(z) be entire, of exponential type, and have only the real simple zeros {xn}. Suppose that S(z) -> oc as z -+ oo along each of four rays

arg z = ak, with

0
2

and that also Y_ I 1 /S'(xn) I < 00. n

Then 1

_Y

S(z) -

1

n (z - xn)S'(xn)

Proof. The function

L(z) _ Y,

S(z)

n (z - xn)S'(xn)

is entire, since S(xn) = 0 for each n and Y_n I 1/S'(xn) I < oo.

I claim that L(z) is of exponential type. Clearly, L(z) 15 const.

I S(z)1

I3zl

so the growth of L(z) is dominated by that of S(z) outside the strip 1.3z I <, 1.

For I Z5z I < 1, one may use the following trick. The function JI L(z) I is

204

VI F De Branges' description of extremal annihilating measures

subharmonic, therefore _11I L(z) I

1

2n

,II L(z + 2e'°) I dO.

-n

Substituting the preceding inequality into the integral on the right, we

obtain for it a bound of the form const.eKIZin f" dO/,Jl jz + 2 sin O I, and this is clearly 5 const.eKJzJ/2 for I,3z I< 1. We see in this way that I L(z) I <, CeKIZI for all z.

We have L(xk) = 1 at each xk. Therefore

=Y

L(z) - 1

-

1

n (z - xn)S'(xn)

S(z)

1

S(z)

is entire; as the ratio of two entire functions of exponential type it is also

of exponential type by Lindelof's theorem. (See third theorem of §B, Chapter III.) Since Y_n I I/S'(xn) I < oo, 0

1

"

(z - xn)S'(xn)

as z --> oo along each of the rays arg z = ak, k =1, 2, 3 and 4, and by hypothesis

1/S(z) -* 0 for z -> oo along each of those rays. Therefore (L(z) -1)/S(z) is certainly bounded on each of those rays, so, since it is entire and of exponential type, it is bounded in each of the four sectors separated by them (and having opening < 180°) according to the second Phragmen-Lindelof

theorem of §C, Chapter III. The entire function (L(z) - 1)/S(z) is thus bounded in C, hence equal to a constant, by Liouville's theorem. Since, as we have seen, it tends to zero for z tending to oo along certain rays, the constant must be zero. Hence 1

(Z - Xn)S'(xn)

-

1

S(z)

= 0,

Q.E.D.

Remark. The hypothesis of the theorem just proved is very ungainly, and one would like to be able to affirm the following more general result: Let S(z), of exponential type, have only the real simple zeros x,,, let EnI 1/S'(xn)I < oo, and suppose that S(iy)-+ oo for y-+ ± oo. Then 1 _ 1

S(z) -

(z - xn)S'(xn)

One can waste much time attempting to prove this statement, all in vain, because it is false! In order to lay this ghost for good, here is a counter example.

4 Krein's functions

205

Take Z

S(z) _ H 1 - 2" 00

e

since Y_'2-" < oo, S(z) is of exponential type. One readily computes I S'(2") I by the method used in §C, and finds that 2("(n-3)/2)e2"(S(1))2

IS'(2")1 _ e

for n -+ oo; we thus certainly have 00

Y_ I1/S'(2")I < oo.

It is also true that I S(iy) 12 = rj (1 + Y")

00

for y - ± co. However, fl- (1 + I z 1/2") < e° for z - oo, again by convergence of Ei 2-" (see calculations in §A, Chapter III!). So, for x real and negative,

S(x) = e""I 1 + 1XI ) 2"

<_

e-1x1+0(1x1),

and, for x -+ - 00, 1/S(x) tends to oo like an exponential (!). Therefore 1/S(x)

certainly cannot equal

i °°

1

(x - 2")S'(2")

which tends to zero as x - - co.

Theorem (Krein). Let an entire function S(z) have only the real simple zeros x"; suppose that "I1/S'(x")I < oo and that

_ S(z) 1

1

(z - x")S'(x")

Then S(z) is of exponential type, and

°° log+

1 S(x) I _. 1+x2 dx
Remark. In particular, for functions S(z) satisfying the hypothesis of the

206

VI F De Branges' description of extremal annihilating measures

previous theorem, we have (' °° log, I S(x) I

J

1 + x2

dx <

oo.

The reader who wants only this result may skip all but the last paragraph of the following demonstration.

Proof of theorem. Without loss of generality, Y_ I the assumed representation for 1/S(z),

I = 1, whence, by

S(z)

Given any h > 0, the reciprocal 1/S(z) is thus bounded and non-zero in each of the half-planes {Zz >, h}, {Sz< -h}, as well as being analytic in slightly

larger open half planes containing them. The representation of §G.1, Chapter III, therefore applies in each of those half planes, and we find that in fact

Jlog1/S(t+ih)I 1 + t2

dt < oo,

and that log

1

S(z)

'f'

_ - An( Sz - h) + n

(jz - h)logI 1/S(t + ih)I I z - t - ih 12

dt

for :3z > h, while 17 (log- I 1/S(t - ih) I/(1 + t2) ) dt < oo and logSS(z)I

= -Bh(I3zI-h)+ Jf

-(1 3z1-h)logI1/S(t-ih)1

Iz-t+ih12

dt

when Sz < - h. Here, A,, and B,, are constants which, a priori, depend on h. In fact, they do not, because, by the remark at the end of §G.1, Chapter III, lim,,-Ally) 1og11/S(iy)l exists and equals -A,,, with a similar relation involving B,, for y --> - oo. All the numbers - A,, for h > 0 are thus equal to the limit just mentioned, say to - A, and all the B,, are similarly equal to some number B. For each h > 0 we thus have, for .3z > h, log I S(z) I

=

h) +

A(3z - h) +

h) log I S(t + ih) I 1z-t-1 .h 12

1

it 1

(' °° (3z - h) log+ I S(t + ih) I

J

_

1 z-t-t -

i

tit

Similar relations involving B, which we do not bother to write down, hold

4 Krein's functions

207

for 3z < -h. Let us fix some value of h, say h = 1. We have

f°° log 'IS(t±i)Idt 2

-

slog- I1/S(t±i)Idt, 1 + t2

both of which are finite. Knowing this we can, by using the two inequalities for log I S(z) I involving integrals with log+, in {,rjz > 1 } and in {3z < - 11, verify immediately that S(z) is of exponential growth at most in each of the two sectors S < arg z < n - S, n + S < arg z < 2n - S, S > 0 being arbitrary. This verification proceeds in the same way as the corresponding one made while proving Akhiezer's second theorem, §B.2.

It remains to show that S(z) is of at most exponential growth in each of the two sectors I arg z I < S, I arg z - n I < S. This can be done by choosing

S < n/4 and then following the Phragmen-Lindelof procedure used at the end of the proof of Akhiezer's second theorem, provided that we know that I S(z) I < exp (O(1 z 12)) for large I z I in each of those two sectors. This property we now proceed to establish. The method followed here is like that used to discuss L(z) in the proof of the previous theorem. For h > 0, we have I S(t + ih) I >, h, so 1

('°° log- IS(t+ih)I

nJ

1 + t2

- CO

dt

log+

1

h

At the same time,

10 logIS(t+ih)Idt

A+n

= logIS(i+ih)I,

and, since i + ih lies on the positive imaginary axis, we already know that log I S(i + ih) I < C(h + 1) for h > 0, for the positive imaginary axis lies in the sector S < arg z < it - S where (at most) exponential growth of S(z) is clear. Because log+ I S(t + ih) I = log I S(t + ih) I + log I S(t + ih) 1, the above

relations yield, for h > 0. 1

°°

log+IS(t+ih)1 t2 (t

R

dt < - A + C(h + 1) + log+1

h

In like manner, 1 fo log+t2S+ 1 ih) I

dt <, - B + C'(h + 1) + log+ h

when h > 0. From these two inequalities we now find, for large (!) R, that 1 1R f log+IS(t+iy)I

nJRJ

00

-oo

1+ t2

dt dy 5 const.R 2

208

VI F De Branges' description of extremal annihilating measures

(note that f R R log+ (1/I y I) dy < const!). This inequality yields, in turn R

R

J-R -R

log + I S(z) I dx dy < const.R4

for large R.

Let zo be given. Since log+ I S(z) I is subharmonic,

log+ I S(zo) 15

1

,tIzol

[1 log+ I S(z) I dx dy

2

1z-20161201

4fR

("R

log+ I S(z)1 dx dy,

k. 2

R

R

By what we have just seen, the expression on

where R = 21 zo 1.

is <, (11R2)O(R4) = 0(1 zo 12) for large values of I zo I, i.e., I S(zo) I < exp (0(1 zo 12)) when I zo I is large. This is what we wanted to show; as

the right

explained above, it implies that S(z) is actually of exponential growth in the two sectors I arg z I < S, 1 arg z - n I < S, and hence, finally, that the entire function S(z) is of exponential type.

We still have to show that °° log+ -D

I S(x) I

1+x2

dx
That is, however, immediate. In the course of the argument just completed, we had (taking, for instance, h = 1) the relation

log+IS(t+i)I J

1 + t2

dt < 00.

Because S is of exponential type, the desired inequality follows from this one by §E of Chapter III (applied in the half plane 3z < 1) and Fubini's theorem, in the usual fashion (hall of mirrors). We are done.

Remark. We remind the reader that, since the functions S(z) considered here have no zeros either in 3z > 0 or in 3z < 0, the representation of §G.1, Chapter III holds for them in each of those half planes. That is,

-

log I S(z) I

= A,3z + - J ' 3z log I S(t) I dt for 3z > 0,

log I S(z) I

= B 13z l +

and

f "0 13z I logtlls(t) 1 n

-00

1Z

dt for 3z < 0.

4 Krein's functions

209

Problem 9

Let x_

xn, let Ei 1/x.1 < oo, and suppose that E'

I1/S'(xn)I < oo,

where Z2

S(z) _ f 1- xj The x are assumed to be real. Show that

. W

1

S(z)

1

(z - xn)S'(xn)'

and hence that S(z) is of exponential type, and that log, I S(x) I

1+x2

dx < oo.

(Hint: First put SR(z) =

l

z2/x.1), and show that one can make

R - oo in the Lagrange formula

I=

SR(z) //

1x R (Z - Xfl

Rlxn)

so as to obtain S(z)

°°

(z - xn)S'(xn)

At this point, one may either invoke Krein's theorem, or else look at the Poisson representation of the (negative) harmonic function logI 1/S(z)I in a suitable half-plane {Zz > H}, noting that here I S(z) I < S(i I z 1).)

Problem 10 Let S(z) be entire, of exponential type, and satisfy the rest of the hypothesis

of the first theorem of this article. That is, S has only the real simple zeros xn, EnI1/S'(x,JI < oo, and S(z) --. Qo for z tending to 00 along four rays, one in the interior of each of the four quadrants. Suppose also that the two limits (which exist by the above discussion) of loglS(iy)I/IYI, for

y -+ oo and for y -. - oo, are equal, say to A > 0. The purpose of this problem is to prove that ei.ix 51

=0

for

-A
n S'(xn)

(a) If F(z)

S(z)eizx

(z - xn)S'(xn)'

210

VI G Weighted approximation in LP norm show that F(z) is entire and of exponential type A, and that log+ IF(x)I 1 + x2

J

dx < oo.

(Hint: Refer to the trick with L(z), pp.203-4.) (b) Show that Q(z) = (F(z) - eizz)/S(z) is entire and of exponential type, and that log+ I Q(x) I

1 +x2

dx < co.

(c) If - A < A < A and Q(z) is the function constructed in (b), show that 0 for y -. ± oo when - A < A < A. Use this fact and the result proved in (b) to show that Q(z) _- 0. (Hint: First show that Q(iy)

limsup

\0

log I Q(Re;w) I

R

R-+

0 or it. Then use boundedness of Q on the imaginary axis and apply the Poisson representation for logiQ(z)I (or else Phragmen-Lindelof) in the right and left half-planes.) 0 for - A 5 A < A. (Hint: Show this for - A < A < A (d) Y_ % and argue by continuity. Here, one may observe that if cp

eid"n

SIX.)

eiAx

ao

lim yy M

/ Y - (iY - xn)S'(xn)

i

and use the result of (c).)

G.

Weighted approximation with L. norms The results established in §§A-E apply to uniform weighted

approximation, i.e., to approximation using the norm IIw11w = supl `V(t) tcR W(t)

One may ask what happens if, instead of this norm, we use a weighted LP one, viz. cp(x)

11(p11w,P= f

W(x)

P

dx;

here, p is some number >, 1. The answer is that all the results except for de Branges' theorem (§F) carry over with hardly any change, not even in the proofs. Here, we of course have to assume that, for x - ± oo, W(x) -+ oo rapidly enough to make

(W(x) )dx < oo. J

Comparison of Ww(0) to 'w(0 +)

211

(some weakening of this restriction is possible; compare with the discussion in §E.4.) It is enough to merely peruse the proofs of Mergelian's and Akhiezer's theorems, whether for approximation by polynomials or by functions in G9A, to see that they are applicable as is with the norms II II w,P Here the functions cl(z), S2A(z), W,k(z) and WA(z) have evidently to be defined using the appropriate norm 11w. And it is no longer II w,P instead of necessarily true that W*(x) s W(x). II

II

Verification of all this is left to the reader. In general, in the kind of approximation problem considered here (that of the density of a certain simple class of functions in the whole space), it makes very little difference which LP norm is chosen. If the proofs vary in difficulty, they are hardest for the L 1 norm or for the uniform one. Here, the continuous functions (with the uniform norm) play the role of `limp-,,LP', and not L., which is not even separable.

H.

Comparison of weighted approximation by polynomials and by functions in gA

We now turn to the examination of the relations between the II II wclosed subspaces of ' (O) generated by the polynomials and by the linear combinations of the eizx, - A <, . < A, for A > 0. In order to consider the former subspace, it is of course necessary to assume that

x"/W(x) -* 0 for x - ± oo when n > 0. This we do throughout the present §.

We also use systematically the following Notation. 'Ww(O) is the II II w-closure of the set of polynomials in lew(U8). For

A > 0, 9w(A) is the II

II w-closure of the set of finite linear combinations of

the eizx; - A < . < A. (Equivalently, w(A) is the II

11w-closure of Chi A; see

§E.1.)

It also turns out to be useful to introduce some intersections: Definition. For A > 0 (sic!),

cw(A +) = n %' (A'). A'>A

212

VI H Spaces `6W(A) and `1,,,(A+). Comparison of `6W(0) to `6w(0+)

In this §, we shall be especially interested in 'w(0 + ), the set of functions in

'w(°) which can be II II w-approximated by entire functions of arbitrarily small exponential type. We clearly have (Cw(A) c lew(A +) for A > 0. But also:

Lemma. 'w(0) c Ww(O +).

Proof. We have to show that 'w(0) c 'w(A) for every A > 0. Fix any such A.

We have x/ W(x) -* 0 for x - + oo. Therefore, for the functions ei(A+h)x - ei lx

fh(x) =

h

h > 0,

we have II fh II w <, const., h > 0, and fh(x)/W(x) --* 0 uniformly for h > 0

as x -> ± oo. Since fh(x) -> xeizx u.c.c. in x for h -* 0, we thus have II fh(x) - xeizx II w , 0 as h -> 0, and xeizx e'w(A) if - A < A < A. By iterating this procedure, we find that x"eizx e lew(A) for n = 0, 1, 2, 3,...

if - A < A < A. In particular, then, all the powers x", n = 0, 1,2_., belong to W (A), so Ww(O) c'w(A), as required.

Remark. This justifies the notation 'w(0) for the nomials in

II

Ilw-closure of poly-

' (R).

Once we know that 'w(0) c Ww(0 + ), it is natural to ask whether Ww(0) = Ww(0+) for the weights considered in this §, and, if the equality does not hold for all such weights, for which ones it is true. In other words, if a given function can be Il ll w-approximated by entire functions of arbitrarily small exponential type, can it be II Ilw-approximated by polynomials? This

question, which interested some probabilists around 1960, was studied by Levinson and McKean who used the quadratic norm II II w, 2 (§G) instead of II II w' and, simultaneously and independently, by me, in terms of the uniform norm II Ilw I learned later, around 1967, that I.O. Khachatrian had done some of the same work that I had a couple of years before me, in a

somewhat different way. He has a paper in the Kharkov University Mathematics and Mechanics Faculty's Uchonye Zapiski for 1964, and a short note in the (more accessible) 1962 Doklady (vol. 145). The remainder of this § is concerned with the question of equality of the subspaces 'w(0) and 'w(0 + ). It turns out that in general they are not equal, but that they are equal when the weight W(x) enjoys a certain regularity. 1.

Characterization of the functions in T w(A +)

Akhiezer's second theorem (§§B.2 and E.2) generally furnishes only a partial description of the functions in 'w(A) when that subspace does not

1 Characterization of the functions in ',i,(A +)

213

coincide with WW(I8). One important reason for introducing the intersections W,,,(A +) is that we can give a complete description of the functions

belonging to any one of them which is properly contained in 4(O). Lemma. Suppose that f (z) is an entire function of exponential type with

If(z)I < CEexp(AI 3zl+elzl) for each e > 0. Then, if 6 > 0, the Fourier transform FS(A) =

e - alXle'axf (x) dx

belongs to L1(11), and, if A'> A, (*)

J

IF,,(A)IdA->0 for 6->0. e-bixif(x)

Proof. For each 6 > 0, is in L, (R) (choose e < 6 in the given condition on f (x)), so Fb(2) is continuous and therefore integrable on [ - A', A']. The whole lemma will thus follow as soon as we prove (*). Fix A' > A, and suppose for the moment that 6 > 0 is also fixed. Take an

e > 0 less than both 6/2 and (A' - A)/2. If A > A', we then have, for

y=3z%0, le-szeiAZ.f(z)I < C e(A-A')y-bx+EIz1

,

and, for x = 9iz ,>0, this is in turn < Let us now apply Cauchy's theorem using the following contour FR: CEe-Ex-'Y.

Figure 38

214

VI H Spaces WK,(A) and `t (A+). Comparison of 16,(0) to 16,(0+)

We have frRe-aZe'ZZf(z)dx=0. For large R, Je-"ei"f(z)I is, by the CEe-eR/,/2

preceding inequality, <

on the circular part of FR. Therefore the

portion of our integral taken along this circular part tends to zero as R -> oo, and we see that

e-axe"xf(x)dx 0 foo

=i

e-gave-xyf(iy)dy

0 J 00

This formula is valid whenever A > A'> A and S > 0. By integrating around the following contour

y

iR

-R

0

} x

Figure 39

we see in like manner, on making R -+ oo, that o

eaxe;xx f(x)dx

-00

= -i

e'aye-zyf(iy)dy o

whenever 6 > 0 and ).. >, A' > A. Combining this with the previous formula we get e-61xle'zx f(x)dx

Fa().) =

=2

J -'000

e - zy sin Syf (iy)dy; o

this holds whenever 2 >, A' > A and S > 0.

Take now any q, 0 < ri < (A' - A)/2, and fix it for the following computation. Since, for y > 0, 1 f (iy) l < C e" "')Y, the formula just derived

I

Characterization of the functions in (ew(A +)

yields, for A

215

A',

IFs(A)I

Cne(A+n-A)yIsinSyldy.

<, 2 f 'O 0

By Schwarz' inequality, the right-hand side is in turn

2Cfo'O e-(x-A-n)vdy fo e-(a-A-n)vsin2 or the second integral under the radical we have e-(z-A-n)ysin2bydy

byd

e-cz-A-n)y(1-e2;ay)dy

= Z'l J o

262

(A-A-n)IA-A-n-2ibI2* Therefore, for A > A', 2 .,12C,6 IFa(A)I

2 1/2C,16

(2-A-n)IA-A-n-2ibl

(A-A-n)2.

And

IA'

IFa(A)Id2 S

2.,,/2Cnb

<1

A'- A - n

2 112C,16

n

We see now that f AIF,5(A)IdA---+O for

6-+0.

Working with contours in the lower half plane, we see in the same way that A

IF3(A)IdA-->0

as

We have proved (*), and are done. Theorem (de Branges). Let f (z) be entire,, and suppose that If(z)I S CEeAII+Ei=i

for each s > 0. Then,

if fElew(R), fEcw(A +). Proof. We have to show that, if f Elew(l), then in fact f E'w(A) for each A' > A; this we do by duality. Fix any A' > A. According to the Hahn-Banach theorem it is enough to

216

VI H Spaces W(A) and `1' (A+). Comparison of WW(0) to (6,(0+)

show that if L is any bounded linear functional on functions of the form cp(t)/W(t) with (pe'w(l ), and if eat L = 0 for - A' < 2 A' ,

,

W(t)

then

f(t)

L W(t)

= 0.

To see this, observe in the first place that Il f (t) - e-aiti f (t) II w

-4 0 for 6 - 0,

so surely f (t) W(t)

L

I=

syo L

e

(t)

W(t)

Our task thus reduces to showing that the limit on the right is zero; this we do with the help of the above lemma. Writing, as in the lemma, e-ai:icixx e"xf(x)

Fa(2) = 00

we have F6eL1(R) as we have seen. Hence, by the Fourier inversion formula, I f 00 e-bit if(t)

=

e-,atFa(2)d2. 2n

In order to bring the functional L into play, we approximate the integral on the right by finite sums.

Put SN(t) = l

(k+1)/N

N2-1

e-i(k/N)t

1

2

=-N2

Fa(2)d2; . k/N

l

since FaeL1(R),

SN(t)

2

e-;ztFa(2)dA

= e-aI`If(t)

u.c.c. in t as N --* x,-and, at the same time, SN(t) 15 11F,'11, on l for all N. Therefore, since W(t) --+ oo for t -+ ± co, 11 e 1111f(t)-SN(t)Ilw

N 0,

1 Characterization of the functions in 'w(A +)

217

so, by the boundedness of L,

L

e "`If (t)

= Ni

W(t)

L

W(t) ) m (SN(t)

However, IIe-ixt_e-;x'tIIw-->0 when IA-A'I-->0, so L(e-izt/W(t)) is a continuous function of A on !!B as well as being bounded there (note that Ieiztl =1!). Hence, since F6(A)eL1(R), we have

CSN(t)

-

1

N2-1

(k+l)/N Ce-i(k/N)t)Jk/N

27rNZL k_

L

ao

1

2n

F(A)dA

W(t)

/e tlt\

LI W(t)

f

F6(A) dA

,,,,

for N - oo. In view of the previous relation, we thus get L

(e-11"f(t))

1

W(t)

2n

(e

00

_

,

iM

\W(t)) Fb(A) dA.

We are assuming that e ilt

L

=0

W(t)

for

- A',< 2

A'.

The integral on the right thus reduces to 2n fAj,A

Le

-'At

W(t)

F6(A)dA.

Here, as already noted, IL(e-izt/W(t))I < const.,

Ac-R,

so the last integral is bounded in absolute value by const.

I Fb(A) I dA.

J

This, however, tends to 0 by the lemma as b-+ 0. We see that

L(e-t1f(t))

0

W(t)

for S -> 0, which is what was needed. The theorem is proved. Remark. Since we are not supposing anything about continuity of W(t), we are not in general permitted to write

L f(t)

as JTdt.

W(t)

218

VI H Spaces `PW(A) and `f (A+). Comparison of W(0) to `6 (0+)

with a finite (complex-valued) Radon measure on R. This makes the above proof appear a little more involved than in the case where use of such a measure is allowed. The difference is only in the appearance, however. The argument with a measure is the same, and only looks simpler. Thanks to the above result, we can strengthen Akhiezer's second theorem so as to arrive at the following characterization of the subspaces c9w(A +). Recall the definition (§E.2): WA(z) = suP { I f(Z)1: fe-'A and 11f II w <, 11.

Then we have the Theorem. Let A > 0. Either

F

log WA-(x)

1 +x2

00

dx = o0

for every A' > A, in which case Ww(A +) is equal to 1w(R), or else 'w(A+) consists precisely of all the entire functions f such that f(x)/W(x) --> 0 for x -+ ± oo and If(z)I 5 CEeAi3i+Eiz1

for each e > 0. Remark. In the second case, Ww(A +) may still coincide with 'w(ll ). (If, for example, the set of points x where W(x) < oo is sufficiently sparse. See §C and end of §E.2)

Proof. For the Mergelian function S2A(z) defined in §E.2, we have '2A(z) > WA(z), so, if the first alternative holds, C

J

log S2A.(x)

1 + xZ

dx = 00

for every A' > A. Then, by Mergelian's second theorem, ' ' (A') = Ww(R) for each A'> A, so Ww(A +) = %w(R). The supremum WA,(z) is an increasing function of A' for each fixed z by virtue of the obvious inclusion of 6A' in 'A" when A' S A". Therefore, if the second alternative holds, we have

(t)

log WA.(x)

1 +xZ

dx < oo

for each A' A0, some number larger than A. Let c > 0 be given, wlog e < AO - A, and put S = s/2. Then, if f e' ,(A + ), surely fe ' (A'), where A' = A + S. For this A', (t) holds, so, by Akhiezer's

2 Sufficient conditions for Wµ,(0) to equal 'w(0 +)

219

second theorem (§E.2), we have If(z)I < KseA'i3Zi+sizi.

Therefore If(z)I < KE12eA13zi+Eizi.

Saying that f(x)/W(x) -> 0 for x -' ± oo is simply another way of expressing the fact that fe'w(l ). Thus, in the event of the second alternative, all the functions f in 'w(A +) have the two asserted properties. However, any entire function f with those two properties does belong to cw(A +). For such a function will be in Ww(IIB), and then must belong to'w(A +) by the preceding theorem. The subspace W (A +) thus consists precisely of the functions having the two properties in question (and no others) when the second alternative holds. We are done. Corollary. For the intersections Ww(A +) the following alternative holds: Either Ww(A+) = Ww(118), or, if 16w(A+)

c6w(If8), the former space consists

precisely of the entire functions f (z) belonging to W,(R) with If(z)I < CEeAl-11+Eizi

for each s > 0. Remark. Even when c'W(A+)='Ww(IEB), all the functions in cPw(A+)

may have the form described in the second clause of this statement. That happens when 'w(R) consists entirely of the restrictions of such functions to the set of real x where W(x) < oo. See remark following the statement of the preceding theorem.

2.

Sufficient conditions for equality of Wµ,(0) and W w(O + )

Lemma. Let w(z) = cfl(z - ak), where the ak are distinct, with 3ak < 0. Let g(z) be an entire function of exponential type <,A with (x + i)g(x) I bounded for real x. Then, for x e R, e-iAx9(x) w(x)

-.

9(ak)/e-,Aak

k =1 w (ak) lx - ak) e

°°

g(t + (i/A)) sin A(x - t) dt. w(t + (i/A)) x - t - (i/A)

Proof. (z + i)g(z) is of exponential type < A and is bounded on R, hence has modulus < const.eA13'1 by the third Phragmen-Lindelof theorem of §C,

220

VI H Spaces SD w(A) and '(A+). Comparison of'w(0) to Ww(0+)

Chapter III. Hence eAl3ZI

(*)

Ig(z)I < const.lz+il'

We are going to use (*) together with some contour integration. Fix b > 0, take any large R, and let FR be the following contour:

Figure 40

If R is large enough for FR to encircle all the ak and the real point x, the calculus of residues gives g(()eiA(x_')

I

rR W(S)(x - b)

27x1

dr

N

ll

g(adeiA(x-ak)

k=1 W (ak)(x - ak)

S

g(x) - w(x)

By (*), Ig(C)e-iA;I is 0(1/(1(I - 1)) = O(1/R) on the semi-circular part of I"R,

so, as R -+oo, the portion of the integral taken along that part of the contour tends to zero. Therefore g(t+ib)eiA(x-t-ib)

1

00

27ri J

-. w(t + ib)(x - t - ib)

dt =

g(x)

g(adeMlx-ak)

w(x) -

w''(ak))(x - ak)

We rewrite this is relation as follows: eAb

(*) *

27ci

oc) f

J

g(t + dt w(t + ib)(x - t - ib) ib)eiA(x-t)

=

g(x) w(X)

_

g(ak)eiA(x-ak)

k W (ak)(X - ak)

2 Sufficient conditions for WW(0) to equal 'W(0 +)

221

Let now r' be the contour obtained by reflecting rR in the line 3z = b:

Figure 41

We have g(S )e -

iA(x - {)

rR w(0(x - t;)

dt = 0.

Here, Ig(t;)e'"41 = O(1/R) on the semi-circular part of FR, so, making

R -oo, we get °°

g(t +

oo

ib)e-;A(x-t-'n)

w(t+ib)(x-t-ib) dt = Q

that is,

g(t+ib)e-i"(x-t) f-O'. w(t + ib) (x - t - ib)

dt = 0.

Multiplying the last relation by eAn/2ni and subtracting the result from the left side of (*), we find

e" it

f

°

g(t + ib) sin A(x - t)

x w(t + ib)(x - t - ib)

dt

g(x)

- w(x) -

g(ak)e,A(x-ak)

k w'(ak)(x - ak)

Now put b = 1/A and multiply what has just been written bye -iAx. After

222

VI H Spaces W ,(A) and `t? w(A+). Comparison of 'Ww(0) to 'w(0+)

taking absolute values, we see that g(x)e - iAs

g(ak)e - iAak

-

w(x)

w'(ak)(x - ak)

g(t + (i/A)) sin A(x - t) dt, w(t + (i/A)) x - t - (i/A)

5

Q.E.D.

Corollary. Let w(z) be as in the lemma, and suppose that f(z) is entire, of exponential type 5 A/2, and bounded on P. Then there is a polynomial P(z) of degree less than that of w(z), such that sin (Ax/2 P(x) - e - iAs (Ax/2) :-,f(x)

Ke f (t + (i/A)) -sup it tER w(t + (i/A))

w(x)

forxeP.

Here K is an absolute numerical constant, whose value we do not bother to calculate. Proof. Put g(z) = (sin (Az/2)/(Az/2)) f (z); then g(z) satisfies the hypothesis of the previous lemma, so, with the polynomial N w(x)g(ak)e - iAak P(x) =

kl w'(ak)(x - ak) '

we get, forxeP, P(x) - e iAxg(x) (fi)

e

5

- sup it

w(x)

(ER

f(t + (i/A)) w(t + (i/A))

sin A(x - t)

x

x - t - (i/A)

dt.

In the integral on the right, make the substitutions At/2 = i, Ax/2 =. That integral then becomes 2 - aD

sin (i + (i/2)) z + (i/2)

sin 2(g - t)

2( -r)-1

dt.

By Schwarz, this last is 2

°°

(iJ -

sin (t + (i/2)) 2

i + (i/2)

I

sin 2 2(T - ) dr l

di J - ao

4(i -)2

/

2 Sufficient conditions for 'w(0) to equal 'w(0 +)

223

a finite quantity - call it K - independent of , hence (clearly) independent of A and x.

The right-hand side of (t) is thus bounded above by f (t + (i/A)) K e sup w(t + (i/A)) Ic feR

and the corollary is established. Theorem. Let W(x) = Y_o a2kx2k where the a2k are all >, 0, with ao >, 1 and a2k > 0 for infinitely many values of k. Then 'Kw(0) = cw(0+).

Remark. We require ao >, 1 because our weights W(x) are supposed to be 1. We require a2k > 0 for infinitely many k because W(x) is supposed to go

to oo faster than any polynomial as x --> ± oo.

Proof of theorem. Let (pe'w(0+). Then there are finite sums an(,)elxx

E

fn(x) =

-1/2n4A41/2n

with IIfn - wllw

0.

n

We put gn(x) = (sin (x/2n)/(x/2n)) fn(x), and set out to apply the above corollary with f = fn and suitable polynomials w. Note that fn is entire, of exponential type 1/2n, and bounded on the real axis. Since II fn - (p II wn - 0, we also have lie -,x/ngn(x) - (P(x) li w - n' 0,

in view of the fact that W(x) - oo for x -> ± co.* Choose any s>0. The norms II fn II w must be bounded; wlog I fn(x)/W(x)l < 1, say, for xe68 and every n. For the function cpew, (R) we of course have p(x)/W(x) n 0 for x-+ ± oo, so, since ll fn - (p II w n 0, there must be an L (depending on e) such that W""( I

<s for

lxl>L

whenever n is sufficiently large. Take such an L and fix it. Pick any n large enough for the previous relation to be true, and fix it for the moment. Our individual function fn(x) is bounded on 11 (true, with perhaps an enormous bound!), so, for some No, we will surely have I fn(x)' E a2kx2k < r. l

for

l x l > A, say,

* Note that Ile-"I"(sin (x/2n)/(x/2n))cp(x) - cp(x)Ilw -, 0 for any cpeWw(R).

224

VI H Spaces `6W(A) and ,(A+). Comparison of WW(0) to `Cw(0+)

where, wlog, A > L, the number chosen above. Also, N

Y0

1

a2kx2k

W(x),

the sums on the left being monotone increasing with N (a2k I> 0!). Therefore,

LWI

a2kx2k

l

N I fn(x)/ W(x) I

uniformly for - A < x < A, and, if N >, No is large enough, we have, in view of the previous inequality, 2

Ifn(x)'N ,a2kx2k

for

XEQB

0

(since II fn II w < 1), and also

(tt)

IL()

< 2E for Ixl>,L.

Fix such an N for the moment (it depends of course on n which we have already fixed!), and call N

V(x) _ "' a2kx2k /

0

Because V(x) >, 1 on R, we can find another polynomial w(x), with all its zeros inz < 0, such that I w(x) I = V(x), x c- R. There is no loss of generality in supposing that the zeros of w are distinct.

There are, in any case, a finite number (2N) of them, lying in the open lower half plane. Separating each multiple zero (if there are any*) into a cluster of simple ones, very close together, will change w(x) to a polynomial

w(x) having the new zeros, and such that

(1-6)Iw(x)I S IOW I 5 (1 +6)Iw(x)I

on E, with 6 > 0 as small as we like. One may then run through the following argument with w in place of w; the effect of this will merely be to render the final inequality worse by a harmless factor of (1+S)/(1-S). Let us proceed, then, assuming that the zeros of w are simple. Desiring, as we do, to use the above corollary, we need an estimate for I f"(t + in) rcR w(t + in)

sup

The function e'Zl2nf"(z)/w(z) is analytic and bounded for Zz > 0, and continuous up to R. Therefore we can use Poisson's formula (lemma of * and there are! - all zeros of w(z) are of even order!

2 Sufficient conditions for IW(0) to equal Bµ,(0 +)

225

§H.1, Chapter III), getting e'(I +"")l2"f(t+in) "

_

1

w(t+in)

1[

eix/2"f(W

n

°°

"

_""(t-x)2+n2

w(x)

)

dx.

Since I w(x) I = V(x), we see, by (§) and (tt), that the integral on the right is in

absolute value

2-

n

c(t - x)2 +n 2

n

dx +

n

2E

Jxl3c(t - x)2 +n 2

7r

dx.

This is in turn

arctant+L - arctant-L 5 2( ir\ n n

+ 2E <, 4L+2E, nn

)

so we have

4L tel'. --+2E, nn

e-1/2f"(t + in) w(t + in)

Apply now the corollary with f = f" and A = 1/n. According to it and to the inequality just proved, there is a polynomial P"(x) (depending, of course, partly on our w(x) whose choice also depended on the n we have taken!) such that, for xelB, P"(x) - e -;x/"g"(x) w(x)

Ke

1/2

(4 L itn

it

+28

1

where (as we recall),

9"(x) =

sin (x/2n) (x/2n) f"(x)

Therefore, since Iw(x)I = V(x) 5 W(x) (!), P"(x) - e-ix/"g"(x) l

W(x)

Ke3/2 4 L it

C7r n

+2s

xEUB.

Our number L depended only on s, and the intermediate partial sum V(X) V(x)

N`

= L0

a2kx2k

(which depended on n) is now gone. The only restriction on n (which was

kept fixed during the above argument) was that it be sufficiently large (how large depended on L). For fixed L, then, there is, for each sufficiently

large n, a polynomial P"(x) satisfying the above relation. If such an n is

226

VI H Spaces `Pw(A) and S

(A+). Comparison of 16w(O) to `6w(0+)

also > 2L/7te, we will thus certainly have P,,(x) - e-is/ngn(X)

4e

W(x)

Ke3/2

n

for xeR. Let us return to our function cpe'w(0+). For each sufficiently large n, we have a polynomial Pn(x) with II Pn -

ll w '< II 9(X) - e

-'xJ"gn(x)

II w +

II e

-,"'"gn(x)

- Pn(x)11 w,

which, according to the inequality we have finally established, is 4e

Ke3i2 n

+

I l e-ix/ngn(X) - w(X) I l w.

However, Il e-ix1ngn(x) - lp(x) II

w -' n

0 by choice of our functions fn.

Therefore 11 Pn - rP 11 w < 8e(Ke312/rt) for all sufficiently large n. Since e > 0

was arbitrary, we have, then, 11 P. - 9 II w

0, and tp el'w(0). This proves that 16w(O+) c 'Ww(O). Since the reverse inclusion is always true, we are done.

Remark. An analogous result holds for approximation in the norms II w,,, 1 < p < oo. There, a much easier proof can be given, based on duality and the fact that the Hilbert transform is a bounded operator on L,,(IIB) for 1

proof.

We can apply the technique of convex logarithmic regularisation developed in Chapter IV together with the theorem just proved so as to obtain another result in which a regularity condition on W(x) replaces the explicit representation for it figuring above. Theorem. Let W(x) > 1 be even, with log W(x) a convex function of log x for x > 0. Suppose that for each A > 1 there is a constant CA such that x2 W(x) 1< C W(AX),

x e L.

Then Ww(O) = Ww(0+).

Remark. Speaking, as we are, of 'Pw(O), we of course require that x"/W(x) -0 for x-> ± oc and all n > 0, so W(x) must tend to oo fairly rapidly as x -* ± oo. But one cannot derive the condition involving numbers

A> 1 from this fact and the convexity of log W(x) in logIxI. Nor have I been able to dispense with that ungainly condition.

Proof of theorem. Let us first show that, if cpe'w(IIB) and we write 9x(x) = tp(.%2x) for A < 1, then 11 WA !-I,--O as A -+ 1.

2 Sufficient conditions for 'w(0) to equal 'w(0 +)

227

We know that log W(x) tends to co as x -> ± oo. Hence, since that function is convex in log x for x > 0, it must be increasing in x for all sufficiently large x. Take any q e4w(ll); since cp is continuous on O we certainly have I cp(x) - (px(x) I --) 0 uniformly on any interval [ - M, M] as 2 1. Also, I 9(x)/W(x) I < e for I x I sufficiently large. Choose M big enough so that this inequality holds for I x I >, M/4 and also W(x) increases for

x? M/4. Then, if i< A< 1 and I x i>,M, p(22x) W(x)

TAX) W(x)

p(22x)

W(22x)

< E,

as well as I q(x)/W(x) I < e, so Ox) - (px(x) W(x)

< 2e

for I x I > M and i < 2 < 1. Making 2 close enough to 1, we get the quantity on the left < 2s for - M < x <, M also, so II 4P - Px II w < 2e.

Take now any cpe'Pw(0+). We have to show that (p also belongs to 'w(0), and, by what we have just proved, this will follow if we establish that gpe9w(0) for each 2 < 1. We proceed to verify that fact. We may, wlog, assume that W(x) = 1 for I x I < I and increases for x > 1.

For n = 0, 1, 2,..., put

S. = sup

r"

.>o W(r)

and, then, for r > 0, write r2"

T(r) = sup S2" n;lto

.

Since log W(r) increases for r > 1, the proof of the second lemma from §D of

Chapter IV shows that W2

r

r)

5 T(r) < W(r)

for

r >1

(cf. proof of second theorem in §D, this chapter). Take now m x2"+2

(§§)

S(x) = 1 +"F

Sz"

Then, by the preceding inequalities, for I x I > 1,

S(x) '> x2 T(I x l) % W(x)

228

VI H Spaces (6 w(A) and Sw(A+). Comparison of ( (0) to `6w(0+)

whilst, for any ).., 0 < A < 1, 22n(x/A)2n

S(x) = 1 + x2


1x2

+

2 TI I XI

Stn

0

x

2

+1-,12W

.

X

The first of these relations* clearly also holds for I x I < 1, because W(x) 1 there. So does the second. For, the inequality between its last two members is true for I x I >, A, while T(I x I/A) is, by its definition, increasing when 0 < I x I < A, and W(x/A) constant for such x. We thus have W(x)

1+ xz

S(x)

1

22

W

(X)

for all x. According to the hypothesis, there is a constant Kx for each A < 1 with 1

X2A2

W(x)

<-

We may of course take Kz >, 1, and thus get finally ($)

W(x) 5 S(x) < 2K2W( 2),

xc-R.

Given our function cpc-W,(0 +), we have a sequence of functions f,,

fn E offwith II(v-fn1IW

0.

n

Thence, by (1), afortiori, II ki -fn lls

n

0,

so q e's(0+) as well. Now, however, S(x) has the form (§§), so we may apply

the previous theorem, getting cpEWs(0). There is thus a sequence of polynomials P,,(x) with

IIw-PnIIs n o. From this we see, by (1) again, that - Pn(x) sup gp(x) W(x/22)

- 0,

* i.e., that between S(x) and W(x)

3 Example of a weight W with Ww(O)

rew(O +) 0 lew(1)

229

i.e., sup

pp(22x) - Pn(22x)

W(x)

XER

-> 0 n

for each 2,0
Remark. Some regularity in W(x) is necessary for the equality of Ww(O) and 'w(0+); exactly what kind is not yet known. In the next article we give an example showing that the behaviour of W(x) cannot depart too much from that required in the above two theorems if we are to have 'w(0) = 'w(0+). In the first part of the next chapter we will give another example, of an even weight W(x), increasing for x > 0, such that `ew(0) 0 ('(O +)=`Pw(R)

Example of a weight W with 6w(O)

3.

16w(O+)

16w(R)

The idea for this example comes from a letter J.-P. Kahane sent me in 1963. Take 00

S(z) = fl ( 1 - zz 4n

1

'

pick any fixed number A,, 1 < Al < 2, and write

C(z) =

(i_)ni_). Al

z

-4

This function S(z) is the same as the one used in §C, and C(z) differs from it only in that the two zeros, - 2 and 2, of S(z) closest to the origin have been

moved slightly, the first towards - 1 and the second towards 1. Let us write 2 _ 1 = - A l , and, for I n I > 2, An = (sgn n)2'n'. Then

C(z) = 1

1(

1-

zz An

and

-' 2nS(2n) =

2_1S(A1) C'(21)

For large n, we clearly have C'(1,,) ^

4 2 1

S'(2n),

< 0.

230

VI H Spaces 6,(A) and `' (A+). Comparison of 16,(0) to 16,(0+)

where S'(A,,) was studied in §C. There we found that

IS'(An)I = IS'(2")I -

const.2("-1)z

for large n, so surely (in view of the evenness of C(z)), Y-,

< co for p=0,1,2,3,...

1A.1°

- ao I C/(An)1

Use of the Lagrange interpolation formula now shows, as in §C (where an analogous result was proved with S'(A,,) in place of C'(A")), that

P(z) = C(z) >'

P(An)

- - (z - An)C'(A")

for any polynomial P. Taking P(z) = zP+ 1 and then putting z = 0 gives us 00

2Pn

i

= 0,

=0,1,2.....

We are ready to construct our weight W. Taking a large constant K (chosen so as to make W(x) come out > 1), put W(x) = KS(1) for I x 15 1. For Ix1 > 1, make W(x) = Kx2IS(x)I when x lies outside all the intervals [2"(1 -2 -4n ),

2"(1 +2 - 4n) ]

Finally, if 2"(1 - 2 4n) as

1x1 5 2"(1 + 2-4") for some n >, 1, define W(x)

We see first of all that xS(x)/W(x) --*0 for x--+ ± oo, so xS(x)eWW(O ). Hence, since S(z), and therefore zS(z), is of exponential type zero we have

xS(x)E'w(0+) by the first theorem of article 1. We need some information about the asymptotic behaviour of S(x) for x - co. This may be obtained by the method followed in §C. Suppose that x = 2"a with 1/..j2 < a <, ,/2. Then we have n-1

4na2

= rI 1- 4k

I S(x) I

n-1

= I1-a21

fi

k-1

00

4na2

x11-a21x k=n+1 4na2

4k

n-1

fi -

1=1

1

1

4`a2

1 -

rj i=1

4k

(I_ a2

-, 4i

and this last is I1

-- LX )

-a21(42a")nu1S(I S(a) = Ii

1S(I

0)

S(a).

3 Example of a weight W with Ww(O)

Ww(O+)

ew(R)

231

Thence, for large n,

\2-3ni

-

const.2n2-5n,

so, for 2n(1 -2-4n) <, IxI < 2"(1+2 - In) W(x)

const.2n2 - 3n

and, in particular, const.2n2 - 3n

W(An)

Comparing this with the relation I C'(2n) I - const.V-1)2, valid for large n,

which we already know, we see that

-

W(2n)

Y'

< oo.

I CIA.) I

This permits us to define a finite signed Radon measure p on the set of points

An, n = ± 1, ± 2,..., by putting W//(2n)

C '(A,,)

Then, forp=0, 1,2,...,

fo- W (x)

AP

ao

XP

dp(x)

which is zero, as we have seen, whilst

f

W(x) dµ(x) _

Y-1

0.

So xS(x) e'w(0 +) can't be in W (O), and Ww(O) 0 W w(O +). In the present example, 'w(0+) is a proper subspace of Ww(R). Indeed, this is almost immediate. By the above asymptotic computation of S(x) we clearly have

W(x) =

const.Ixj`921x1+o(=)

with a quantity 9(x) varying between two constants. Therefore, log W(x)

J

I+xzdx
so we surely have '(A)

Ww(R) for each A by an obvious extension of

232

VI H Spaces `D,y(A) and '6,(A+). Comparison of `i7 w(0) to I%I9w(0+)

T. Hall's theorem (p. 169). Hence Ww(O+) our example is finished.*

'W(R). The construction of

Remark. Let fl(x) = H '(I + x2/4"). The asymptotic evaluation of 12(x) for x -+ oo can be made in fashion similar to that for S(x), and is, in fact, easier than the latter. As is clear after a moment's thought, here one also obtains 12(x) - const. I x Ib0521xi+Q(x)

for

x i -> co

with a certain cp(x) varying between two constants. Thus, O(x) = cont. I x I*(x), W(x)

xE ll,

where A 5 iji(x) 5 B, say. However, len(O) = Wn(0 + ). This follows from the first theorem of the previous article, in view of the evident fact that i2(x) = 1 + a2x2 + a4x4 + with a2k > 0. The difference in behaviour of 12(x) and W(x) is small in comparison to their size, and yet W (O) = Wn(O +) although '(O) 'e,(0 +).

The question of how a weight W's local behaviour is related to the equality of l ' (O) and lew(0 +) merits further study.

* W(x) has jump discontinuities among the points ±(2"±2-a"), n-> 1, but a continuous weight with the same properties as W is furnished by an evident elaboration of the procedure in the text.

VII

How small can the Fourier transform of a rapidly decreasing non-zero function be?

Let us consider functions F(x) e L1(Q8) whose modulus goes to zero rapidly as x -> oo, in such/fashion that JTco 1 + x2

IF(t)Idtldx

log I J X"O

The general theme of this chapter is that, for such a function F, the Fourier transform

(2) = f

e'2xF(x) dx

cannot be too small anywhere unless F vanishes identically. The first result of this kind (obtained by Levinson) said that if (for such an F) P vanishes throughout an interval of positive length, then F = 0. This was refined by Beurling, who proved that F(A) cannot even vanish on a set

of positive measure unless F = 0. Analogues of these theorems hold for measures as well as functions F; they, and the methods used to establish them, have various important consequences, some of which apply to material already taken up in the present book. These things have been known for more than 20 years. Until recently, the only developments since the sixties in the subject matter of this chapter had to do mainly with aspects of its presentation. That state of affairs was changed in 1982 by the appearance of a remarkable result, due to A.L. Volberg, which says that if 00

f(9) = Y I(n)i"s -00

has I.T(n)I < e-M(")

for n>0

234

VII A The Fourier transform vanishes on an interval

with M(n) sufficiently regular and increasing, and if M(n) i

n

-

then n

f

logl f(9)Id9 > - c

unless f (9) = 0. The proof of this uses new ideas (coming from the study of weighted planar approximation by polynomials) and is very long; its inclusion has necessitated a considerable extension of the present chapter. I still do not completely understand the result's meaning; it applies to the unit circle and seems to not have a natural analogue for the real line which would generalize Levinson's and Beurling's theorems. There are not too many easily accessible references for this chapter. The earliest results are in Levinson's book; material relating to them can also be found in the book by de Branges (some of it being set as problems). The main source for the first two §§ of this chapter consists, however, of the famous mimeographed notes for Beurling's Standford lectures prepared by

P. Duren; those notes came out around 1961. Volberg published his theorem in a 6-page (!) Doklady note at the beginning of 1982. That paper is

quite difficult to get through on account of its being so condensed. A.

The Fourier transform vanishes on an interval. Levinson's result

Levinson originally proved his theorem by means of a complicated argument, involving contour integration, which figured later on as one of the main ingredients in Beurling's proof of his deeper result. Beurling

observed that Levinson's theorem (and others related to it) could be obtained more easily by the use of test functions, and then de Branges simplified that treatment by bringing Akhiezer's first theorem from §E.2 of Chapter VI into it. I follow this procedure in the present §. The particularly convenient and elegant test function used here (which has several other applications, by the way) was suggested to me by my reading of a paper of H. Widom.

I Some shop math

235

Some shop math

1.

The circle of radius R about 0 lies under the two straight lines of slopes ± tan y passing through the point iR sec y. Therefore, if A > 0,

A J(R2-x2) < ARsecy-(Atany)IxI,

-R<,x<,R.

Consider any function co(x) >, 0 such that

Iw(x)-w(x')I < (Atany)Ix-x'I. If we adjust R so as to make AR sec y = w(0), we have, for - R < x < R, co(x) > w(0) - (A tan y) I x I,

which, by the above, is >, A,./(R2 - x2). The function cos (A V(x2 - R2)) is, however, in modulus 5 I for I x I > R, and for - R 5 x 5 R it equals cosh (A V(R2 - x2)) < exp (A /(R2 - x2)). Therefore, for xeR, co(x) >, log I cos (A

'/(x2 -R 2))I

when

R=

w(0)

Asec y

Let us apply these considerations to a function W(x) >, 1 defined on 11; and satisfying

log W(x)-log W(x')I < CIx-x'I there. Taking any fixed A > 0, we determine an acute angle y such that A tan y = C. Suppose xOeO is given. Then we translate xo to the origin, using the above calculation with

w(x - x0) = log W(x). We see that

Icos(A \/((x - xo)2 - R2))I < W(x) for xeR,

236

VII A The Fourier transform vanishes on an interval

where

R-

log W(x0)

-

A sec y

log W(x0) 1/(A2 + C2)

Here, cos (A ../((z - x0)2 - R2)) is an entire function of z because the Taylor development of cos w about the origin contains only even powers of w. It is clearly of exponential type A, and, for z = x0, has the value

=

ieAR

cosh AR

2(W(x0))AisJ(AZ+c2)

Recall now the definition of the Akhiezer function WA(x) given in Chapter VI, §E.2, namely WA(x) = sup { I f (x) I : f entire of exponential type 5 A, bounded on R and I f(t)/W(t)I < 1 on R}.

In terms of WA, we have, by the computation just made, the Theorem. Let W(x) >, 1 on IR, with

log W(x)-log W(x')I 5 Clx-x'I for x and x' E III. Then, if A > 0, WA(x) '>

Corollary. Let Then, if

W(x),>1,

log W(z) -00

z(W(x))A1.11A2+C2)'

l+x

with

x -R. log W(x)

uniformly

Lip l

on

R.

dx = oo,

we have log WA(x)

1+x2

dx = 00

for each A > 0. According to Akhiezer's first theorem (Chapter VI, §E.2), this in turn implies the Theorem. Let W(x) >, 1, with log W(x) uniformly Lip 1 on R, and W(x)

tending to 0o as x - ± oo. If °°

_

log W(x) dx 1 +x2

= oo,

linear combinations of ei ', - A < A < A, are, for each A > 0,

II

II w-dense

in Ww(R). 2.

Beurling's gap theorem As a first application of the above fairly easy result, let us prove the

following beautiful proposition of Beurling:

2 Beurling's gap theorem

237

Theorem. Let p be a totally finite complex Radon measure on R with I dp(t) I = 0 on each of the disjoint intervals (an, bn), 0 < a1 < b1 < a2 < b2 <

,

and suppose that

(*)

f

(bn nan)2 = 00. L1\ a

If µ(A) = f °° , e'* xdp(x) vanishes identically on some real interval of positive

length, then p - 0. Remark. This is not the only time we shall encounter the condition (*) in the present book. Proof of theorem (de Branges). We start by taking an even function T(x) >, I

whose logarithm is uniformly Lip 1 on P, and which increases to oc so slowly as I x I - oo, that T(x) I dp(x) I

< oo.

(Construction of such a function T is in terms of the given measure p, and is left to the reader as an easy exercise.) For each n, let b be the lesser of bn and 2an. Then, given that (*) holds, we

also have

,,(bna an)2

=

0c).

n

Indeed, this sum certainly diverges if the one in (*) does, when (b - an)/an differs from (bn - an)/an for only finitely many n. But the sum in question also diverges when infinitely many of its terms differ from the corresponding ones in (*), since (b - an)/an = 1 when b' = 2a,,. and on each one of those Let co(x) be zero outside the intervals (an, intervals let the graph of co(x) vs x be a 45° triangle with base on (an, w(x)

a,

Figure 43

b,

a2

b2

b2

a3

b3

x

238

VII A The Fourier transform vanishes on an interval

The function co(x) is clearly uniformly Lip I on R. Put W(x) = eu'(x)T(x). Then W(x) >, 1 and log W(x) is uniformly Lip I on 11; also, W(x) -+ oo for x -. ± oo. Since I du(x) I = 0 throughout each interval and co(x) is zero outside those intervals,

T(x)Idµ(x)I < oo.

W(x) I dµ(x)I = J

J

The complex Radon measure v with dv(x) = W(x) dp(x)

is therefore totally finite. Suppose now that µ(2) vanishes on some interval; say, wlog, that

eizxdp(x) = 0 for

- A <, A 5 A.

_OD

This can be rewritten as **)

fOD

e

1(**)

W(x)

dv(x) = 0, - A

2

A.

However,(' log W(x) = w(x) + log T(x) > w(x), and J a,

co

x2

1

b

2

J

41

2a

which is infinite, as we saw above. Therefore log W(x)

1+x2

dx = oo,

and linear combinations of the e'', - A < A < A, are II II w-dense in Ww () by the second theorem of the preceding article. Referring to (*k), we thence see that v - 0, i.e., dv(x) = W(x)du(x) =- 0 and p = 0. Q.E.D.

Problem 11 Let p be a finite complex measure on R, and put e - e(x) =

JT

e-1x "Idp(t)I

Suppose that f °°.(a(x)/(1 + x2))dx = oo. Then, if 2(A) vanishes identically

on any interval, µ - 0 (Beurling). (Hint. Wlog, I'.Idp(t)I < I so that

3 Weights increasing along positive real axis a(x) > 0. Assuming that µ(A)

239

0 for - A S A S A, write the relation

1,

and use the picture

Figure 44

to estimate the supremum of If (x) I for entire functions f of exponential

type 5 A, bounded on R, and such that

f

I f(t)I Idµ(t)I < 1.)

Remark. Beurling generalized the result of problem 11 to complex Radon measures p which are not necessarily totally finite. This extension will be taken up in Chapter X. 3.

Weights which increase along the positive real axis

Lemma. Let T(x) >, 1 be defined and increasing for x >, 0, and

denote by T(x) the largest minorant of T(x) with the property that I log T(x) - log T(x')I < I x - x' j for x and x',>- 0. If S' (log T(x)/x2)dx = co, then also f i (log T(x)/x2)dx = oo.

Proof. The graph of log T(x) vs x is obtained from that of log T(x) by means of the following construction:

240

VII A The Fourier transform vanishes on an interval

Figure 45

One imagines rays of light of slope 1 shining upwards underneath the graph of log T(x) vs x. The graph of log T(x) is made up of the portions of the former one which are illuminated by those rays of light and some straight segments of slope 1. Those segments lie over certain intervals [a,,, on the x-axis, of which there are generally countably many, that for all n. The cannot necessarily be indexed in such fashion that b < open intervals

are disjoint, and on any one of them we have

log T(x) = log

(x - a.).

On [0, oo) - U (a,,, b,,), T(x) and T(x) are equal. In order to prove the lemma, let us assume that I' (log-T(x)/x2)dx < oo is and then show that I' (log T(x)/x2)dx < oo. If, in the first place, we have, since any of the aforementioned intervals with 1 < a. < log 0, f'2 (' T(x) J log X21 dx > J x x2a, dx >

d

= log2-1. > 0. We can therefore only have finitely many intervals with b. > 2a and a > I if f (log T(x)/x2)dx is finite. This being granted, consider any other of the intervals with a >, 1.

3 Weights increasing along positive real axis

241

Y

a

bn

Figure 46

By shop math, Jb"

fb

1

X log T(x)dx Z

_

62n a"

1

log T(an) + log T(ba)

b

2

_

I

,

(b,, - an) log T(b,,) 2bn

At the same time, since T(x) increases, 1b" 1

-log T(x)dx

(bn - an)log T(bn)

a

Ja"

(ba - an)log T(ba)

8

2b2

when bn < 2an. Therefore, for all the intervals (aa, bn) with a,,,> I and b,, < 2a,,, hence, certainly, for all save a finite number of the (a,,, bn) contained in [1, oo), we have b"

i I

Jan x

b" fa.

log T(x)dx < 8

z log T(x)dx.

The sum of the integrals f A" (1/x2 )log T(x)dx for the remaining finite number of (an, ba) in [ 1, oc) is surel y finite - note that none of those intervals can have

infinite length, for such a one would be of the form (a,, oo), and in that case we would have

-

a, x

dx = oo,

2log T(x)dx faO

,

x 2

242

VII A The Fourier transform vanishes on an interval

contrary to our assumption on T(x). We see that

Y f'"IogT(x)dx X22

<

oo,

since '-log T(x)

Y 8

a

an,

x

2

dx

as

is finite.

On the complement

E=[1,oc)n

U(a,,,ba),

T(x) = T(x) by our construction. Hence log T(x) SE

< 00.

x2

The whole half line [ 1, oo) can differ from the union of E and the (a,,, with as > 1 by at most an interval of the form [1, which happens when there is an m such that a< I < If there is such an m, however, b, must be finite

(see above), and then b log T(x)

x

1

2

dx < oo.

Putting everything together, we see that °° log T(x)

f

1

x

2

dx < oo,

which is what we had to show. We are done. Corollary. Let W(x) > I be defined on P and increasing for x >, 0. If log W(x) X2

dx = oo,

1

we have 109 WA(x) 2

1

x

dx = cc

for each of the Akhiezer functions WA, A > 0 (Chapter VI, §E.2).

Proof. Let, for x > 0, T(x) be the largest minorant of W(x) on [0, oo) with

I log T(x) - log T(x')I < I x - x' I

4 Example to §H of Chapter VI.

243

there, and put T(x) = T(O) for x < 0. By the lemma, $ O (log W(x)/x2 )dx = oo

implies that $i (log T(x)/x2)dx = co. Here, log T(x) is certainly uniformly Lip 1 (and > 0) on 68, so, by the corollary of article 1, we see that (' °°

109 TA(x)

dx = 00

x2

1

for each A > 0. We have T(x) < W(x) + T(0) (the term T(0) on the right being perhaps needed for negative x). Therefore TA(x)

1<

(1+T(0))WA(x),

and (' °° log WA(x)

J

fi

xz

dx = co

for each A > 0 by the previous relation. Q.E.D. From this, Akhiezer's first theorem (Chapter VI, §E.2) gives, without further ado, the following Theorem. Let W(x) > 1 on F, with W(x) -> co for x --> ± oo. Suppose that W(x) is monotone on one of the two half lines ( - oc, 0], [0, oo), and that the integral of log W(x)/(1 + x2), taken over whichever of those half lines on which monotoneity holds, diverges. Then 'K,r(A) = 'w(F) for every

A>0, so (w(0+)=Ww(R). Remark. The notation is that of §E.2, Chapter VI. This result is due to Levinson. It is remarkable because only the monotoneity of W(x) on a half line figures in it. 4.

Example on the comparison of weighted approximation by polynomials and that by exponential sums

If W(x) > 1 tends to oo as x -* ± oo, we know that W w(A) is properly contained in 'w(F) for each A > 0 in the case that (' °°

J

log W(x)

1+x2

dx < co.

(See Chapter VI, §E.2 and also the beginning of §D.) The theorem of the previous article shows that mere monotoneity of W(x) on [0, oo) without any

additional regularity, when accompanied by the condition ° log W(x)

dx = oo,

Jo 1 +x2 already guarantees the equality of Ww(A) and 'w(R) for each A > 0.

VII A The Fourier transform vanishes on an interval

244

The question arises as to whether this also works for W,(O), the II

II w-closure of the polynomials in lew(QB). (Here, of course we must assume

that x"/ W(x) --* 0 as x --). ± oo for all n > 0.) The following example will show that the answer to this question is NO. We start with a very rapidly increasing sequence of numbers An. It will be sufficient to take Al = 2,

Az=e and, in general, 2 = Let us check that An > for n > 1. We have e2 > 22 = 4, and (d/dx)(e" - x2) = e" - 2x is > 0 for x = 2. Also (d2/dx2)(e" - x2) =e'-2>0 for x,>2, so ex - x2 continues to increase 1

strictly on [2, oo). Therefore e" > x2 for x , 2, so An = ez"-1 > A'-,. We note that 1 is turn , since the numbers are , 2. We proceed to the construction of the weight W. For 0 5 x <, A1, take log W(x) = 0, and for 1 < x < An with n > 1 put log W(x) = 1/2 (by the computation just made we do have 22,,_ 1 < We then specify log W(x) on the segments [A _ 1, 22n _ 1] by making it linear on each of them, and finally define W(x) for negative x by putting W(- x) = W(x). Here is the picture:

J

log W (X)

A2

Al

2

T,

2X,

2A2

A2

X3

Figure 47

W(x) is > 1 and increasing for x, 0, and, for large n, + 1 log W(x) 2z

x

2

dx =

(n + 1)2 2

G1

A.

1

An+1

X

4 Example to §H of Chapter VI

245

(n + 1). Therefore $o (log W(x)/(1 + x2))dx = oo, so,

is > (n + 1)An/8In =

by the theorem of thes previous article, 1 (A) ='w(O) for each A > 0 and 'w(0 + ) = 16w(R). For 2An 1 5 I x I<, 2A, W(x) > eni '12 = )n/2

x/2In/2.

n

Hence xP

W(x)

_, 0

x ---+ ± oo

as

for every p > 0, and it makes sense to talk about the space 'w(0). It is claimed that ',(O) 96 Ww(R).

To see this, take the entire function Qo

22

1 - An?

C(z) = 1

Because the An go to oo so rapidly, C(z) is of zero exponential type. For n > 1,

IC'(2n)I =

2An (,.)2(In)2... aAnl)2 A, A2

1

k=1

Since the ratios A

+

f 1-j"

2

n-1 X

ao

9k

I2

1=n+1

nn

1/Ai are always > 2 and --> oc as j --> oo, the two products

written with the sign fl on the right are both bounded below by strictly positive constants for n > 1 and indeed tend to 1 as n -> oo. The product standing before them,

2A2.A2...A2n An-2' 1

2

n

1

far exceeds 22"-2 because 1, > Aj-

.

.

Therefore we surely have

),n-2

for large n.

At the same time, W(An) = en'--1/2 = An12, whence, for large n,

< IC'(An)I \ W(An)

A:12

-

1

,,n-2 = Ann/2)-2

Since the sequence {An} tends to oo, we thus have W(An)

IC'R')I

< Oo.

246

VII A The Fourier transform vanishes on an interval

For n = 1, 2,3.... it is convenient to put , _,, = - A,,. Let us then define a discrete measure u supported on the points An, n = ± 1, ± 2,..., by putting

AIM)

C (An)

The functions W(x) and C(x) are even, hence

< oo

du(x) I -00

by the calculation just made. We can now verify, just as in §H.3 of Chapter VI, that (t)

XP

f

W(x)

dp(x) = 0

for

p = 0, 1, 2, ... .

The integral on the right is just the (absolutely convergent) sum a,

AP

- ao Cu.), and we have to show that this is zero for p 3 0. Taking N

CN(Z) = fl I

n=1\

1 - A?2J n

(cf. §C, Chapter VI), we have the Lagrange interpolation formula 71

=

2 ,CN(z)

-N (Z -

An)

C N( .)

valid for 0 < l < 2N. Fix 1. Clearly, Therefore, since F_".. I

I C'N(Af)I % I C'(A,,) I

for - N < n 5 N.

I < oo, we can make N --> oo in the preceding

relation and use dominated convergence to obtain Z'

W

AnC(z)

(z - An)C'(.ln)'

Putting ! = p + 1 and specializing to z = 0, the desired result follows, and we

have (t). Our measure y is not zero. The strict inclusion of cw(0) in'Vw(R) is thus a

consequence of (t), and the construction of our example is completed. Let us summarize what we have. We have found an even weight W (x) 3 1,

increasing on [0, oo) at a rate faster than that of any power of x, such that Ww(0) #'w(R) but 19w(0 +) = `'w(I8). This was promised at the end of §H.2, Chapter VI. In §H.3 of that chapter we constructed an even weight W with Ww(0) 0 'w(0 +) and Ww(0 +):A 1Bw(118).

5 Levinson's theorem

247

Scholium

As the work of Chapter VI shows, the condition

° log W(x)

f

1+x2

dx < 00

is sufficient to guarantee proper inclusion in'w(R) of each of the spaces 'w(0) and '(A), A > 0 (for Ww(O) see §D of that chapter). The question is, how much regularity do we have to impose on W(x) in order that the contrary property

(tt)

('

J

_

log W(x)

1 +x 2

dx = 00

should imply that Ww(0) = Ww(R) or that Ww(A) = w(R) for A > 0? As we saw in the previous article, monotoneity of W(x) on [0, oo) is enough for (tt) to make 'w(A) = Ww(68) when A > 0, in the case of even weights W. In §D, Chapter VI, it was also shown that (tt) implies 'w(O) = 'w(R) for even weights W with log W(x) convex in loglxl. The example just given shows that logarithmic convexity cannot be replaced by monotoneity when weighted polynomial approximation is involved, even though the later is good enough when we deal with weighted approximation by exponential sums.

We have here a qualitative difference between weighted polynomial approximation and that by linear combinations of the e'-I", - A S 2 A, and in fact the first real distinction we have seen between these two kinds of approximation. In Chapter VI, the study of the latter paralleled that of the former in almost every detail.

The reason for this difference is that (for weights W which are finite reasonably often) the II II v-density of polynomials in ' ,(118) is governed by the lower polynomial regularization W*(x) of W, whereas that of c'A is determined by the lower regularization WA(x) of W based on the use of

entire functions of exponential type 5 A. The latter are better than polynomials for getting at W(x) from underneath. As the example shows, they are qualitatively better. 5.

Levinson's theorem

There is one other easy application of the material in article 1 which should be mentioned. Although the result obtained in that way has been superseded by a deeper (and more difficult) one of Beurling, to be given in the next §, it is still worthwhile, and serves as a basis for Volberg's very refined work presented in the last § of this chapter.

248

VII A The Fourier transform vanishes on an interval

Theorem (Levinson). Let y be a finite Radon measure on R, and suppose that 0 1 + x2 foo

log

(f

dx = co. I

X

dp(t) I

Then the Fourier-Stieltjes transform fOD

eiAx dp(x) 00

cannot vanish identically over any interval of positive length unless p - 0.

Remark 1. Of course, the same result holds if 0 00

1 + x2

log

I

x

dµ(t)

I) dx = ao.

Remark 2. Beurling's theorem, to be proved in the next §, says that under the stated condition on log (f I dp(t) I ), P(,) cannot even vanish on a set of positive measure unless p - 0.

Proof of theorem. It is enough, in the first place, to establish the result for absolutely continuous measures p. Suppose, indeed, that p is any measure

satisfying the hypothesis; from it let us form the absolutely continuous measures p,,, h > 0, having the densities dph(x)

dx

1 ('x+h

- h f.x

dp(t).

Then 1 - e - ixn

an(d) =

iAh

Q(A),

so a,,(2) vanishes wherever a(A) does. Also,

f.'* Idp,,(t)I < f-00 Idp(t)I

for x>0, so

/

,l0 1 + x2

log l $ Idp,,(t)I

)dx = 00

for each h > 0 by the hypothesis. Truth of our theorem for absolutely continuous measures would thus make they,, all zero if µ(A) vanishes on an interval of length > 0. But then p - 0. We may therefore take y to be absolutely continuous. Assume, without

5 Levinson's theorem

249

loss of generality, that

Idµ(t)I < 1 -.

and that µ(.l) = 0 for - A <, .1 <, A, A > 0. For x >, 0, write W(x) = (f '0 I d y(t)1) -112, and, for x < 0, put -112

W(x)

d11(t) I I

The function W(x) (perhaps discontinuous at 0) is >, 1 and tends to co as x -> ± oo. It is monotone on (- oo, 0) and on [0, oo), and continuous on each of those intervals (in the extended sense, as it may take the value 00). By integral calculus (!), we now find that

o

f

W(x) l dµ(x) I=

J

fOOO

d

fI d

°°d j

2

(t) j

fo-

I dµ(t)

and, in like manner, 0

W(x) I dy(x) I

< oo.

-CO

The measure v with dv(x) = W(x)du(x) is therefore totally finite on R. (If W(x) is infinite on any semi-infinite interval J, we of course must have d,u(x) = 0 on J, so dv(x) is also zero there.) For - A < A 5 A, fOD

(§)

ei;x

- W(x) dv(x) = 11(A) = 0. However, by hypothesis,

/ Jo

11+x2)dx = 2f

1+x2logl

XIdy(t)I)dx

oo,

o

so, since W(x) is increasing on [0, oo), 16w(A) is 11 w-dense in lew(O1) according to the theorem of article 3. Therefore, by (§), II

7

(p(x) d u(x) = J

l

W(Ox)

x)

dv(x) = 0

for every continuous cp of compact support. This means that µ = 0. We are done. The proof of Volberg's theorem uses the following

Corollary. Let f(9)-Y_°°.f(n)e'"9 belong to L1(-n,n), and suppose that f(9) = 0 a.e. on an interval J of positive length. If I f(n)I

<e-M(") for

250

VII B Fourier transform zero on a set of positive measure

n > 0 with M(n) increasing, and such that M(n) n2

1

oo,

then f(9)=-0, - Tr< 9 hn. Proof. Take any small h > 0 and form the convolution n

fn(9) = hI

_h(l-Ih1)f(9-t)dt.

If h < 1 (length of J), f,,(9) also vanishes identically on an interval of positive length

From the rudiments of Fourier series, we have

fn(9) _ Y

(sin (nh/2)12 f(n)elns J

nh/2

00

The sum on the right can be rewritten in evident fashion as f °. e'''" du(x) with a (discrete) totally finite measure µ. Let x > 0 be given. If n is the next integer > x we have, since M(n) increases, Csi Ih/22)21f(0I

Idµ(t)I = J

)

13n

x

<

e

4

-M(n)

<

7212

coast. -M(n) h2

e

I>- n

Because Y_ i M(n)/n2 = oo, we see that fo'D

1 +x2

log

Idµ(t)I)dx I

and conclude by the theorem that f,, - 0. Making h - 0, we see that f - 0, Q.E.D.

B.

The Fourier transform vanishes on a set of positive measure. Beurling's theorems

Beurling was able to extend considerably the theorem of Levinson given at the end of the preceding §. The main improvement in technique which made this extension possible involved the use of harmonic measure.

Harmonic measure will play an increasingly important role in the remaining chapters of this book. We therefore begin this § with a brief general discussion of what it is and what it does.

I What is harmonic measure? 1.

251

What is harmonic measure?

Suppose we have a finitely connected bounded domain .9 whose boundary, 0-9, consists of several piecewise smooth Jordan curves. The Dirichlet problem for -9 requires us to find, for any given (p continuous on 0.9, a function U4,(z) harmonic in -9 and continuous up to a-9 with U,(t') = cp(() for t; Ea-9. It is well known that the Dirichlet problem can always be solved for domains like those considered here. Many books on complex variable theory or potential theory contain proofs of this fact, which we henceforth take for granted. Let us, however, tarry long enough to remind the reader of one particularly easy proof, available for simply connected domains -9. There, the Riemann mapping theorem provides us with a conformal mapping F of -9 onto the unit disk { I w I < 11. Such a function F extends continuously up to a-9 and maps the latter in one-one fashion onto { I co I = 11; this is true by a famous theorem of Caratheodory and can also be directly verified in many cases where 0-9 has a simple explicit description (including all the ones to be met with in this book).

F F-1

Figure 48

a-9

Denote by F-' the inverse mapping to F. The function O(w) = cp(F-'((o)) is then continuous on { I w I = 1 }, and, if U, is the harmonic function sought

which is to agree with cp on 0-9, V(w) = U4,(F-'(w)) must be harmonic in { I w I < 1 } and continuous up to { I co I = 11, where V(w) must equal 0(w). A function V with these properties (there is only one such) can, however, be obtained from cli by Poisson's formula: z

V(w) = 2n f-1 _1

ICII2

IW

O(w)Idwl.

Going back to -9, and writing z = F-'(w),

= F-'(w), we get

z

U'(z)

2n a I F(z)

c'(C) I II

IZ

dF()

252

VII B Fourier transform zero on a set of positive measure

for ze.9. This is a formula for solving the Dirichlet problem for -9, based on the conformal mapping function F. Knowledge of this formula will help us later on to get general qualitative information about the behaviour near 8-9 of certain functions harmonic in -9 but not continuous up to 8-9, even when -9 is not simply connected. Let us return to the multiply connected domains -9 of the kind considered

here. If cp is real and continuous on 8-9 and U,,, harmonic in -9 and continuous on 9, agrees with T on 8-9, we have, by the principle of maximum,

- II'II.

U,(z) 5 IIwII

for each ze-9; here we are writing

IIwII. = supIq )I CCOO

This shows in the first place that there can only be one function Um corres-

ponding to a given function cp. We see, secondly, that there must be a (signed) measure µZ on 8-9 (depending, of course, on z) with

(*)

U4(z) = I

The latter statement is simply a consequence of the Riesz representation theorem applied to the space Since U1. can be found for every (pe16(8-9) (i.e., the Dirichlet problem for -9 can be solved!) and since, corresponding to each given gyp, there is only one U(,' there can, for any

ze-9, be only one measure µZ on 8-9 such that (*) is true with every cpe'6(r ). The measure µZ is thus a function of ze-9, and we proceed to

make a gross examination of its dependence on z. If cp(l;) >,0 we must have U4,(z) ,>0 throughout -9 by the principle of maximum. Referring to (*), we see that the measures uz must be positive. Also, 1 is a harmonic function (!), so, if cp(C) - 1, U4,(z) - 1. Therefore J

a

dµ.(C) = 1

for every ze-9. Let t;oE0_9 and consider any small fixed neighborhood 'V of C o. Take any continuous function cp on 0-9 such that 0, pp(C) - 1 for C 0 Y' n 8-9, and 0< p(() 5 1 on Y' n 8-9.

1 What is harmonic measure?

253

Since UV is a solution of the Dirichlet problem, we certainly have

for z-0.

U4,(z) -* gq(Co) = 0

The positivity of the u therefore makes 0

for z ---+ Co. Because all the µZ have total mass 1, we must also have o

1

for

z-o

When z is near l;oE0-9, 1u has almost all of its total mass (1) near t;o (on a-9). This is the so-called approximate identity property of the µZ.

There is also a continuity property for the p applying to variations of z in the interior of -9.

254

VII B Fourier transform zero on a set of positive measure

Take any zoE-9, write p = dist(zo, 8-9) and suppose that I z -zoI < p. Then, if q is continuous and positive on 8-9, J

pQdpz(C)

lies between

p - 1z -zoI p + Iz - zol and

L

p + 1z - zol

p - Iz - zoI

Ja-9

w(C)dµ=O(C)

(d z 0()

This is nothing but Harnack's inequality applied to the circle { I z - zoI < p U4(z) being harmonic and positive in that circle. (The reader who does not

recall Harnack's inequality may derive it very easily from the Poisson representation of positive harmonic functions for the unit disk given in Chapter III, §F.1.) These inequalities hold for any positive (pe'(8 '), so the signed measures P - IZ - zoI

µz - p+Iz-zolµzO p+Iz - zoI

P-Iz-zo l µ zp - iuz are in fact positive. This fact is usually expressed by the double inequality

P - Iz - zoI

p+I

z - zo I

P+Iz - zoI dy.O(C) < du.(C) < P- Iz - zoI dµZO(C) What is important here is that we have a number K(z, zo), 0 < K(z, zo) < oo,

depending only on z and zo (and s!), such that 1

K( ZU)

dµ2O(y

b) <

dµz(r

y

S) < K(z, zo)d4u 0(()

Such an inequality in fact holds for any two points z,zo in -9; one needs

only to join z to zo by a path lying in -9 and then take a chain of overlapping disks -9 having their centres on that path, applying the previous special version of the inequality in each disk. In order to indicate the dependence of the measures µZ in (*) on the domain -9 as well as on ze , we use a special notation for them which is now becoming standard. We write doo1(t', z)

for

dp.(C),

1 What is harmonic measure?

255

so that (*) has this appearance:

U'(z) =

Ia (P(C)dw_,(C, z).

J

We call co,( , z) harmonic measure for -9 (or relative to -9) as seen from z. co,( z) is a positive Radon measure on 8-9, of total mass 1, which serves

to recover functions harmonic in -9 and continuous on 1 from their boundary values on 8-9 by means of the boxed formula. That formula is just the analogue of Poisson's for our domains -9. If E is a Borel set on 8-9,

(o,(E, z) = f

z)

E

is called the harmonic measure of E relative to -9 (or in -9), seen from z. We have, of course,

0 < w.,(E, z) < 1. Also, for fixed E c 8-9, coa(E, z) is a harmonic function of z. This almost obvious property may be verified as follows. Given E s 8-q, take a sequence of functions cpn e le(7-9) with 0 5 cpn 1< 1 such that I XE(b) - gpn(C) I

zo)

0

as

for the characteristic function XE of E. Here, zo is any fixed point of which may be chosen at pleasure. Since dco1(l;, z) < K(z, zo)d(o.(C, zo) as we have seen above, the previous relation makes

UwJz) = I (p.(C)dw.(C, z)

w.,(E, z)

51 for every ze-9; the convergence is even u.c.c. in -9 because 0 < there for each n. Therefore w1,(E, z) is harmonic in ze-9 since the U,.(z) are. Harmonic measure is also available for many unbounded. domains -q. Suppose we have such a domain (perhaps of infinite connectivity) with a decent boundary 8-9. The latter may consist of infinitely many pieces, but each individual piece should be nice, and they should not accumulate near any finite point in such a way as to cause trouble for the solution of the

Dirichlet problem. In such case, 8-9 is at least locally compact and, if cpeleo(8f) (the space of functions continuous on 8-9 which tend to zero

256

VII B Fourier transform zero on a set of positive measure

as one goes out towards oo thereon), there is one and only one function UV harmonic and bounded in -9, and continuous up to 8_q, with U4,() = (p(s),

e8_9. (Here it is absolutely necessary to assume boundedness of U. in -9 in order to get uniqueness; look at the function y in 3z > 0 which takes

the value 0 on R. Uniqueness of the bounded harmonic function with prescribed boundary values is a direct consequence of the first Phragmen-

Lindelof theorem in §C, Chapter III.) Riesz' representation theorem still holds in the present situation, and we will have (*) for cpel'0(8-q). The examination of the pz carried out above goes through almost without change, and we write dj (C) = dco9(t;, z) as before, calling co ( , z) the harmonic measure for -9, as seen from z. It serves to recover bounded functions harmonic in -9 and continuous up to 8!2 from their boundary values, at least when the latter come from functions in '0(8_q). Let us return for a moment to bounded, finitely connected domains 3. Suppose we are given a function f (z), analytic and bounded in .9, and continuous up to 8-9. An important problem in the theory of functions is to obtain an upper bound for If (z)I when ze.9, in terms of the boundary values f(t;), Ce8-9. A verb useful estimate is furnished by the

Theorem (on harmonic estimation). For ze-9, (t)

log l f (z) 15 JI log I f

a

Proof. The result

is

I

z).

really a generalization of Jensen's inequality.

Take any M > 0. The function

VM(z) = max(loglf(z)I, -M) is continuous in

and subharmonic in -9. Therefore the difference

VM(z) - f

VM(S)d(ojS, z)

is subharmonic in -9 and continuous up to 8-9 where it takes the boundary value Vm(C) - Vm(C) = 0 everywhere. Hence that difference is s 0 throughout -9 by the principle of maximum, and log I f (z) I < VM(z) < J f VM(C)dwjC, z)

for ze-9. On making M -> oo, the right side tends to J

a logIf(()ldw.(C,z)

by Lebesgue's monotone convergence theorem, since log If (C) I, and hence

I What is harmonic measure?

257

the VM(C), are bounded above, If (z) I being continuous and thus bounded on the compact set !2. The proof is finished.

The result just established is true for bounded analytic functions in unbounded domains subject to the restrictions on such domains mentioned above. Here the boundedness of f (z) in -9 becomes crucial (look at the functions a-'"z in 3z > 0 with n- oo!). Verification of this proceeds very

much as above, using the functions VM(C). These are continuous and bounded (above and below) on 891, so the functions HM(z) = I VM(C)dw.(C, z) J aci

are harmonic and bounded in.9, and for each t;oe9 we can check directly, by using the approximate identity property of cog( , z) established in the above discussion, that HM(z) ---> VM((o)

for

z -Co.

(It is not necessary that VM(C) belong to '0(82) in order to draw this conclusion; only that it be continuous and bounded on 82.) The difference VM(z) - HM(z)

is thus subharmonic and bounded above in .9, and tends to 0 as z tends to any point of 8-9. We can therefore conclude by the first Phragmen-Lindelof

theorem of §C, Chapter III (or, rather, by its analogue for subharmonic functions), that VM(z) - HM(z) S 0 in .9. The rest of the argument is as above.

The inequality (t) has one very important consequence, called the theorem on two constants. Let f(z) be analytic and bounded in a domain -9

of the kind considered above, and continuous up to 8.9. Suppose that If (C) l < M on 8.9, and that there is a Borel set E c 8-9 with If (C) l -< some

number m (< M) on E. Then, for ze-9,

Deduction of this inequality from (t) is immediate. Much of the importance of harmonic measure in analysis is due to this formula and to (t). For this reason, analysts have devoted (and continue to devote) considerable attention to the estimation of harmonic measure. We shall see some of this work later on in the present book. The systematic use of harmonic measure in analysis is mainly due to Nevanlinna, who also gave us the name for it. There are beautiful examples of its application

258

VII B Fourier transform zero on a set of positive measure

in his book, Eindeutige analytische Funktionen (now translated into English), of which every analyst should own a copy.

Before ending our discussion of harmonic measure, let us describe a few more of its qualitative properties. The first observation to be made is that the measures ow9( , z) are absolutely continuous with respect to arc length on 8-9 for the kind of domains considered here. This will follow if we can show that uo.q(E,,, zo)

0 for

zoeI

when the E lie on any particular component F of 8-9 and Jr.

xE.(OIdCI

n' 0.

(Here, XE. denotes the characteristic function of

We do this by

comparing c( , z) with harmonic measure for a simply connected domain; the method is of independent interest and is frequently used. Let d be the simply connected domain on the Riemann sphere (including perhaps oo), bounded by the component F of 8f and including all the points of -9.

Figure 51

If cpe'(0!) is positive, and zero on all the components of 821 save IF, we have

fr

z) % J

a q,(C)dw.,(C, z)

for ze21. Indeed, both integrals give us functions harmonic in -9 (c 9 !),

I What is harmonic measure?

259

and continuous up to 0.9. The right-hand function, Uq(z), equals cp(l;) on

I- and zero on the other components of a9. The left-hand one - call it V(z) for the moment - also equals cp(z) on r but is surely >, 0 on the other components of a9, because they lie in f and (p > 0. Therefore V(z) > U,,(z) throughout -9 by the principle of maximum, as claimed. This inequality holds for every function (p of the kind described above, whence, on I', dwa(l;, z)

z)

for

zE9.

This relation is an example of what Nevanlinna called the principle of extension of domain.

Let us return to our sets E. c IF

I dC I ) 0;

with fr

in order to verify that wa(En, z)

n

i0,

zE9,

it is enough, in virtue of the inequality just established, to check that w,(E,,, zo) -p0 for each zoe9. Because 9 is simply connected, we may, however, use the formula derived near the beginning of the present article.

Fixing zoe1, take a conformal mapping F of 9 onto {IwI < 1} which sends zo to 0. From the formula just mentioned, it is clear that wa(En,zo)

2nJ XE.,(O)IdF(C)I.

r

The component F of a9 is, however, rectifiable; a theorem of the brothers

Riesz therefore guarantees that the mapping F from F onto the unit circumference is absolutely continuous with respect to arc length. For domains .9 whose boundary components are given explicitly and in fairly simple form (the sort we will be dealing with), that property can also be verified directly. We can hence write w,(En, zo)

H(g) = 2n JXE(O dl' r

with dF(t;) dC

in

L1(I', IdCl),

260

VII B Fourier transform zero on a set of positive measure

zo) - 0 when

and from this we see that

Sr

0.

XE (OIdcl

The absolute continuity of ow,( , z) with respect to arc length on 8-9 is thus verified.

The property just established makes it possible for us to write

w,(E,z) = L XE(S)d

z)

1dt;I

for

E g 8-9 and ze . It is important for us to be able to majorize the integral on the right by one of the form KZ

f

XEG) I dC I

(with K. depending on z and, of course, on -9) when dealing with certain kinds of simply connected domains .9. In order to see for which kind, let us, for fixed zo e-9, take a conformal mapping F of -9 onto { I w I < 1 } which sends zo to 0 and apply the formula used in the preceding argument, which here takes the form w_q(E,zo) =

27r f

zEG)IdF()IldCl.

If the boundary 8-9 is an analytic curve, or merely has a differentiably turning tangent, the derivative F'(z) of the conformal mapping function will be continuous up to 8-9; in such circumstances IdF(r;)/d(I is bounded on 8-9 (the bound depends evidently on zo), and we have a majorization

of the desired kind. This is even true when 8-9 has a finite number of corners and is sufficiently smooth away from them, provided that all those corners stick out.

F(fo)

F

a-9 Figure 52

I What is harmonic measure?

261

In this situation, where 8-9 has a corner with internal angle a at (o, F(z) = F(CO) + (C + o(1))(z - Co)"'" for z in 5 (sic!) near CO; we see that F'(Co) = 0 if a < it, and that F'(t;) is near 0 if l; e 8-9 is near t;o (sufficient smoothness of 8.9 away from its corners is being assumed). In the present case, then, I F'Q I is bounded on 89, and an estimate

w,(E,zo) 5 K..J

a xEG)IdC I

does hold good. It is really necessary that the corners stick out. If, for instance, a > it, then I F'(t;o)I = oo, and IF'(C)I tends to oo for 4 on 8-9 tending to Co:

Figure 53

a-9 Here, we do not have w9(E, zo) < const.

XEG) I d( I

J

for sets E 9 0-9 located near t;o. Let us conclude with a general examination of the boundary behaviour of w,(E, z) for E c 8-9. Consider first of all the case where E is an arc, o, on one of the components of 8-9. Then the simple approximate identity property of co( .,, z) established above immediately shows that if

z--0Cea

and t; is not an endpoint of a, while w'(Q, z) --> 0

if

z - e 8-9 ^ a

and is not an endpoint of a. If zed tends to an endpoint of a, we cannot say much (in general) about w,(Q, z), save that it remains between 0 and

262

VII B Fourier transform zero on a set of positive measure

1. These properties, however, already suffice to determine the harmonic function co,(a, z) (defined in -9) completely. This may be easily verified by using the principle of maximum together with an evident modification of

the first Phragmen-Lindelof theorem from §C, Chapter III; such verification is left to the reader. One sometimes uses this characterization in order to compute or estimate harmonic measure. Of course, once w,(a, z) is known for arcs a c 8-9, we can get co(E, z) for Borel sets E by the standard

construction applying to all positive Radon measures. What about the boundary behaviour of w,(E, z) for a more general set E? We only consider closed sets E lying on a single component I of 8-9; knowledge about this situation is all that is needed in practice. Take, then, a closed subset E of the component IF of 8-9. In the first place, w,(E, z) 5 w_,(I', z). When z tends to any point of a component r' of 8.9 different from r, w,(I , z) tends to zero by the previous discussion

(F is an arc without endpoints!) Hence w,(E, z) - 0 for z - if t; E8.9 belongs to a component of the latter other than r. Examination of the boundary of w,(E, z) for z near I' is more delicate.

Figure 54

Take any point p on F lying outside the closed set E (if E were all of F, we could conclude by the case for arcs handled previously), and draw a curve y lying in -9 like the one shown, with its two endpoints at p. Together,

the curves y and IF bound a certain simply connected domain 9 c -9. We are going to derive the formula

w,(E, z) = J w,(E, t;)

z) + w,,(E, z),

Y

valid for ze9. Take any finite union Gll of arcs on F containing the closed

set E but avoiding a whole neighborhood of the point p, and let >li be

I What is harmonic measure?

263

any function continuous on F with 0 < 4i(t;) < 1, /i(t;) - 0 outside V, and O(C) - 1 on E. Since 0 is zero on a neighborhood of p, the function

UO(z) = f r

z)

tends to zero as z -+p. Write U,,(t;) = fi(C) for ceI; the function U,d(C) then

becomes continuous on r u y =as, so V(z) = fee U, ,(l;)dw,(t;,z) is harmonic in d and continuous up to 89, where it takes the boundary value U,,(z). For this reason, the function UO(z) - V(z),

harmonic in d, is identically zero therein, and we have Ud,(C) dwe(C, z) +

Jr O(C) dwe(C, z) = V(z) = Ur,(z) ('

=J

r

O(C) dco9(C, z)

for zEd. Making the covering Gll shrink down to E, we end with

w.,(E, z) =

a (E, 2;) dwe(C, z) + wd(E, z), JY

our desired relation. The function w9(E, l;) is continuous on y and zero at p, because w9(E, z) 5 each of the functions Ud,(z) considered above. The function harmonic in d with boundary values equal to w9(E, t;) on y and to zero on r is therefore continuous on y u t = 8d, so

ii

d o,(l;, z)

tends to zero when zed tends to any point off . Referring to the previous relation, we see that w9(E, z) - we(E, z)

)0

whenever zed tends to any point of F. The behaviour of the first term on the

left is thus the same as that of the second, for z -i oeF. Because if is simply connected, we may use conformal mapping to study we(E, z)'s boundary behaviour.

264

VII B Fourier transform zero on a set of positive measure

F

Figure 55

Let F map S conformally onto A={ I w I< 1); F takes E g r onto a certain closed subset E' of the unit circumference, and we have we(E, z) = w°(E', F(z))

for zeS (see the formula near the beginning of this article). Assume that r is smooth, or at least that E lies on a smooth part of F. Then it is a fact (easily verifiable directly in the cases which will interest us - the general result for curves with a tangent at every point being due to Lindelof) that F preserves angles right up to r, as long as we stay away from p:

F

Figure 56

This means that if ze(f tends to any point t'o of E from within an acute angle with vertex at CO, lying strictly in S (we henceforth write this as

'P°

-2L-+ Co' ), the image w = F(z) will tend to F(C0) e Efrom within such an angle lying in A. However,

`z

w (E',w) =

f2. 1

2n Jo

1-Iw12

1w-e'pI2XE(e)drp.

A study of the boundary behaviour of the integral on the right was made

2 Beurling's improvement of Levinson's theorem

265

in §B of Chapter II. According to the result proved there, we(E, w) -' XE'(wo)

as w - coo, for almost every coo on the unit circumference. In the present situation (E closed) we even have coo(E, w)

oO

whenever w -> a point of the unit circumference not in E. Under the conformal mapping F, sets of (arc length) measure zero on F correspond precisely to sets of measure zero on { I w I = 11. (As before, one can verify this statement directly for the simple situations we will be dealing with. The general result is due to F. and M. Riesz.) In view of the angle preservation just described, we see, going back to S, that, for almost every t'oeE,

as z -

co,(E, z) -> 1

o,

and that )0 as z --i CO

o),(E, z)

for 1; 0 e r not belonging to E. Now we bring in (*). According to what has just been shown, that relation

tells us that

co,(E,z)- +1

as

z -

Co

for almost every t;0EE, whilst

o.,(E,z)i0

as z -->lo

for 0EI'' E, except possibly when CO = p. By moving p slightly and taking a new curve y (and new domain 49) we can, however, remove any doubt about that case. Referring to the already known boundary behaviour of co.,(E, z) at the other components of 8-9, we have, finally,

co.g(E, z) -

0

as z>CO EB-9 -E,

1

as

z

t;o for almost every (0c-E.

This completes our elementary discussion of harmonic measure.

2.

Beurling's improvement of Levinson's theorem

We need two auxiliary results. Lemma. Let u be a totally finite (complex) measure on Il, and put

µ(t) =

e'x' dµ(t) J

266

VII B Fourier transform zero on a set of positive measure

(as usual). Suppose, for some real .10, that

e-rxe1xaj2(2)d.

-0

to and xo

d.1 - 0

I

J -00

for all X e R and all Y> 0. Then p - 0. Proof. If we write dpA0(t) = e'A0` dp(t), we have µ.(t + ).0) = µxo(T), and the

identical vanishing of pxo clearly implies that of p. In terms of µxo, the two relations from the hypothesis reduce to e-rxe'xTp"xo(r)dr =- 0, J0co 0

J - 00

ertesxpt 2xo(t)dr =- 0,

valid for X ell and Y> 0. Therefore, if we prove the lemma for the case where .10 = 0, we will have p - 0. We thus proceed under the assumption that A0 = 0. By direct calculation (!), for X e R and Y > 0, Y

1 foo

(X+t)2+Y2 = 2.-w The integral on the right is absolutely convergent, so, multiplying it by dp(t), integrating with respect to t, and changing the order of integration, we find

f -'*. (X + t)2 + Y2 dp(t) = 2J

. e-ru'e'xrµ(2)d a.

Under our assumption, the integral on the right vanishes identically for X e IR and Y > 0. Calling the one on the left Jr(X), we have, however,

Jr(-X)dX ---> ndp(X) w* for Y -+ 0. Therefore du(X) - 0, and we are done. Lemma. Let M(r) > 0 be increasing on [0, co), and put M*(r) = min (r, M(r))

for r>, 0. Then, if J

o(r2 dr = 00 T+00

267

2 Beurling's improvement of Levinson's theorem we also have

= ao.

M+(r2dr

Jo

Proof. Is like that of the lemma in §A.3. The following diagram shows that M(r) = M*(r) outside of a certain open set (9, the union of disjoint intervals (an, bn), on which M*(r) = r.

y=r

0

a3

b,

a,

b3

b2

a2

r

Figure 57

It is enough to show that

Ar = (-n If M*(r)

f

r

on[1,oo)

, dr = oo,

we are already finished; let us therefore assume that this last integral is finite.

We then surely have

fb

at a

M*(r) dr r2

=L a

< oo,

log 1

an

r

an

so bn/an -+ I which, fed back into the last relation, gives us b" - an

< co. an

l

an

268

VII B Fourier transform zero on a set of positive measure

Since, however, M(r) is increasing, we see from the picture that M(r)

f

Jb,,

dr

.b."

<

M\bn)

b,,

d- % bn

a,, r

aabn - an

-

as - ar

Therefore

Ib"M(r) an dr Y- J a.31 r

< 00

by the previous relation, so, since we are assuming 0°°

f

M(r)

1+r2

dr = (Do

which implies °° M(r)

5i7

dr = m

(M being increasing), we must have r 2

dr = co,

SE

where

E _ [1, 00) ^' U (an, ba) The set E is either equal to the complement of (9 in [1, cc) or else differs therefrom by an interval of the form [ 1, bk) where (ak, bk) is a component of (9 straddling the point 1 (in case there is one). Since M*(r) = M(r) outside (9, we thus have

M.0 dr = 00 2 IE

r

(including(' in the possible situation where bk = co), and therefore J

M *(r) i

dr = co

r

as required. Theorem (Beurling). Let It be a finite complex measure on F such that 10

1

1

dx = co.

2 Beurling's improvement of Levinson's theorem

If A(A) =

f

I

J

e;at

269

dp(t)

vanishes on a set E c l8 of positive measure, then µ = 0. Proof. In the complex A-plane, let _q be the strip

{0<32<1}; we work with harmonic measure co_,( , A) for _q (see article 1).

Figure 58

Because I E I > 0, E contains a compact set of positive Lebesgue measure;

there is thus no loss of generality in assuming E closed and bounded. According to the discussion at the end of the previous article, we then have (0,(E, A)

)1

)o for almost every A0 in the set E (of positive measure). There is

as A

thus certainly a A0eE with

w,(E,20+iT)-01 for t-+0+; we take such a 20 and fix it throughout the rest of the proof. We are going to show that AO

erxe"xµ(A) d.1

=0

- 00

e-YAeixx,2(A)dA __ 0 lAo

270

VII B Fourier transform zero on a set of positive measure

for Y > 0 and X c- R; by the first of the above lemmas we will then have ,u - 0 which is what we want to establish. The argument here is the same for any value of.1o. In order not to burden the exposition with a proliferation of symbols, we give it for the case where A0 = 0, which we henceforth assume.

We have, then, P(2) = 0 on the closed set E, OeE, and w,(E, ii) ---+ I for

t->0+.

Consider the second of the above two integrals. Under the present circumstances, it is equal to eizzPW) dal

= F(Z),

fo'o

say, where Z = X + iY. The function F(Z) defined in this fashion is analytic

for )Z > 0 and bounded in each half plane of the form 3Z > h > 0. By §G.2 of Chapter III, we will therefore have F(Z) _- 0 for 3Z > 0 provided that

r°°logIF(X+i) I

f

1 + X2

J0

dX = - oo.

This relation we now proceed to establish. Take a number A > 0 (later on, A will be made to depend on X), and write µ(A) = µA(2) + PA(2),

with e t dµ(t)

PAW = 1_A

and A

PA(A)

=-

eiztdµ(t)

00

The function PAW) is actually defined for 3I >, 0 and analytic when 3A > 0. PA(2) is not, in general, defined for 3A > 0; when A is large, it is, however,

very small on the real axis in view of our assumption on I dp(t) I f -X 00

in the hypothesis. We think of PA(A) as a correction to PAW) on R. Wlog,

f

Idµ(t)I <,

1.

2 Beurling's improvement of Levinson's theorem

271

Then, writing

f" dµ(t)I = e-M("), we have IOAO-)I < e-m(A),

Ae68,

with M(A) > 0 and increasing for A > 0. Going back to

F(X + i) =

f0`0 e-Aeixzµ(A)d2,

we see that the latter differs from ei(x+i)APA())d . 0 f,O

by a quantity in modulus

f

e-xIPA(.Z)Idl

0

As we have already remarked, this is very small when A is large. Showing that I F(X + i) I gets small enough for (*) to hold thus turns out to reduce to the careful estimation of e1

+i)APA(A) dA

We use Cauchy's theorem for that. Taking t as shown,

r

r

9tX

Figure 59

VII B Fourier transform zero on a set of positive measure

272

we have

f

eia(x+i),

(1) d2 = f eiA(X+i)pA(A)G

r

OOO

because IA(2) is analytic for 3A > 0 and bounded in the strip

0<32'< 1, with Ieiz(x+i)I going to zero like a-"x there as 91A->oo. The integral along F breaks up as

if

1

e-ire-'xpA(ii)dT +

0ei(ax-i)e-Qe-x/A(a +i)da

Jo

say. Since

< 1, J

we have

jco IeL4 AA(t)I

eiA(t+A) dµ(t)I

=

51

-A

for 3, ,>0. In particular, for o E IR, 11A(a + i) 15 eA, and

Je_

IIII

e'da = eA-x

0

To estimate I, we use the theorem on two constants given in the previous article. As we have just seen, eiAZIA(A) is in modulus 5 1 on the closed

strip !2; it is also continuous there and analytic in -9. However, on E

R, 1i(A) = 0 (by hypothesis!), so PA(S) = N(A) - PA(A)

Thus, for AeE, Ie'A'AA(2)I

= I PA(2)I 5

e-M(A)

According to the theorem on two constants we thus have

e- M(A)wo(E,A). I I - wa(E,A)

I eiA-'f*A(2)1

for AE-9, i.e., eA

I PA(S) I

e-M(A)w9(E,A) e

AE-Q.

Substituting this estimate into I, we find III

eAr-Xr-M(A)wo(E,it)dT.

<\ JO

PA(2) there.

2 Beurling's improvement of Levinson's theorem

273

Recall, however, that cw,,(E, ii) -* 1 for T -* 0. For this reason T + ow,(E, iT) has a strictly positive minimum - call it 0 -for 0 < T 5 1; 0 does not depend on A or X. Suppose X > 0. Then take A = X/2. With this value of A, the previous relation becomes III

S

e (X/2)T - M(X/2)m_g(E,iT) dT 5 e-OM.(X12)

fo,

where M*(r) = min (r, M(r)).

At the same time,

IIII \ e-x72 for A = X/2, according to the estimate made above. Therefore, for X > 0, ei(x+i)xµx12(A)dl = fO'O

Jr fr

is in modulus

III+IIII 5 e`* However, the first of the last two integrals differs from F(X + i) by a quantity in modulus < e-M(X/2) as we have seen. So, for X > 0, I F(X + i)I <, e-OM.(xi2) + e-x12 + e-M(x12)

There is no loss of generality in assuming 0 < 1. Then we get F(X + i) 15 3e X > 0.

Returning to (*), which we are trying to prove, we see that C °° log I F(X + i) I 0

1 + X2

log 3 - aM*(X/2)

dX fOOO

1 + X2

dX,

and the integral on the left will diverge to - oo if

'C' (*)

J

M (XIZ)dX = co.

0

Here,

M(A) = log

f-

I

so f 00 (M(A)/(1 + A2)) dA = co by the hypothesis. Therefore o +(A2 dA = 00 J0

VII B Fourier transform zero on a set of positive measure

274

for M*(A) = min (A, M(A)) by the second lemma, i.e., (' °° 2M*(X/2)

4 + XZ

0

dX = ao,

implying (,*k), since M*(A) > 0.

We conclude in this fashion that (*) holds, whence F(Z) - 0 for 3Z > 0, i.e.,

e-YAe;x.µ(2)di

-0

fooo

for Y > 0 and X e IJ

.

One shows in like manner that f 0 '0 eYZeiX zµ(A) d2 - 0 for Y > 0 and X e R;

here* one follows the above procedure to estimate ezeixzj(%)dA

F. (again for X > 0!) using this contour: 3X

i

Figure 60

9iA

Aside from this change, the argument is like the one given. The two integrals in question thus vanish identically for Y > 0 and X C- R.

This, as we remarked at the beginning of our proof, implies that p - 0. We are done.

Remark 1. The use of the contour integral in the above argument goes back to Levinson, who assumed, however, that µ(2) = 0 on an interval J instead of just on a set E with I E I > 0. In this way Levinson obtained his theorem, given in § A.5, which we now know how to prove much more easily using test functions. By bringing in harmonic measure, Beurling was able to

replace the interval J by any measurable set E with I E l > 0, getting a qualitative improvement in Levinson's result. * In which case the integral just written is an analytic function of X - iY

3 Beurling quasianalyticity

275

Remark Z. What about Beurling's gap theorem from §A.2, which says that if the measure p has no mass on any of the intervals with 0 < al < bl < a2 < b2 < . and co, then µ(2) can't vanish identically on an interval J, IJI > 0, unless u - 0? Can one improve this result so as to make it apply for sets E of positive Lebesgue measure instead of just intervals J of positive length? Contrary to what happens with Levinson's theorem, the answer here turns out to be no. This is shown by an example of P. Kargaev, to be given in § C. 3.

Beurling's study of quasianalyticity

The argument used to establish the theorem of the preceding article can be applied in the investigation of a kind of quasianalyticity.

Let y be a nice Jordan arc, and look at functions 9(t;) bounded and continuous on y. A natural way of describing the regularity of such 9 is to measure how well they can be approximated on y by certain analytic functions. The regularity which we are able to specify in such fashion is not necessarily the same as differentiability; it is, however, relevant to the study of a quasianalyticity property considered by Beurling, namely, that of not being able to vanish on a subset of y having positive (arc-length) measure without being identically zero. A clue to the kind of regularity involved here comes from the observation

that a function q having a continuous analytic extension to a region bordering on one side of y possesses the quasianalyticity property just described. We may thus think of such a cp as being fully regular. In order to make this notion of regularity quantitative, let us assume that the arc y is

part of the boundary 8-9 of a simply connected region -9.

Figure 61

So as to avoid considerations foreign to the matter at hand, we take 8.9 as `nice' - piecewise analytic and rectifiable, for instance. Given 9 bounded and continuous on y, define the approximation index M(A) for 0 by functions analytic in -9 as follows:

VII B Fourier transform zero on a set of positive measure

276

e-MA) is the infimum of

f(t;)I for f analytic in -9 and

continuous on -9 such that I f(z) I <e'',

ze_q.

We should write M,(A, (p) instead of M(A) in order to show the dependence of the approximation index on cp and -9; we prefer, however, to use a simpler

notation. When A is made larger, we have more competing functions f with which to try to approximate cp on y, so e-M(A) gets smaller. In other words, M(A) increases with A and we take the rapidity of this increase as a measure of the regularity of cp. Note that if cp actually has a bounded continuous extension to 9 which is analytic in -9, we have M(A) = oo beginning with a certain value of A. Such a function cp cannot vanish on a set of positive (arc-length) measure on y without being identically zero, as we have already remarked (this comes, by the way, from two well-known results of F. and M. Riesz). We see that if M(A) grows rapidly enough, q will surely have the quasi-

analyticity property in question. The approximation index M(A) is a conformal invariant in the following sense. Let F map -9 conformally onto taking the arc y of 0-9 onto the arc y s e9, and let ip be the function defined on y by the relation O(F(t')) = p((), t' e y. Then has the same approximation index M(A) for functions analytic in as (p has for functions analytic in -9. This is an evident consequence of the use of the sup-norm in defining M(A). Our quasianalyticity property is also a conformal invariant. This follows from the famous theorem of F. and M. Riesz which says that as long as 0-9 and 8_, are both rectifiable, a conformal mapping F of -9 onto takes sets of arc-length measure zero on 0-9 to such sets on 89, and conversely. If 8-9

and 0 are really nice, that fact can also be verified directly. Without further ado, we can now state the Theorem (Beurling). Suppose that, for a given bounded continuous (P on y 9 8-9, the approximation index M(A) for co by functions analytic in satisfies

" J i

AA) 2

dA = oo.

Then, if E 9 y has positive (arc-length) measure, and cp(t;) - 0 on E, cp - 0 on y.

Proof. By the above statements on conformal invariance, it is enough to establish the result for the special case where -9 is the strip 0 < 3I < 1 in the 2-plane and y is the real axis. The fact that 8-9 is not rectifiable here makes no difference. We need only show that a set of measure > 0 on the rectifiable

3 Beurling quasianalyticity

277

boundary of our original nice domain corresponds under conformal mapping to a set of measure > 0 on the boundary of the strip. This may be checked by first mapping the original domain onto the unit disk A (whose boundary is rectifiable) and appealing to the theorem of F. and M. Riesz

mentioned above. One then maps A conformally onto the strip; that mapping is, however, easily obtained explicitly and thus seen by inspection to take sets of measure > 0 on 7A to sets of measure > 0 on the boundary of the strip. We have, then, a function cp bounded and continuous on the real axis; wlog I q 15 1 there.

Figure 62

There is a set E s R (which we may wlog take to be closed) with I E I > 0*

and (p - 0 on E. According to the definition of M(A) there is for each A > 0 a function fA(1) analytic in -9 and continuous on there and

(t)

I q(A) - fA(A) I

< 2e - M(A),

with I fA(2) I < eA

) E III.

By the discussion at the end of article 1, we certainly have c0.9(E, A) -+ 1

as

A

Ao

for some Aoe E. There is no loss of generality in takingAo = 0 (we may arrive

at this situation by sliding -9 along the real axis!), and this we henceforth assume. We have, then, co,(E, ir)

i1

as

T --+ 0 + ,

just as in the proof of the theorem in article 2. * We are denoting the Lebesgue measure of E c 08 by IEl.

278

VII B Fourier transform zero on a set of positive measure

In order to show that cp(2) - 0 on R it is enough to prove that e- YiAJ

d

.

=0

-co

for some Y > 0 and all real X, for then the function e- YIAI(p(,) (which belongs to L1(R)) must vanish a.e. on R by the uniqueness theorem for Fourier transforms. We do this by verifying separately that eu(x+;Y)q (2) dA

=0

for

Y>0

and X e R,

fow

and that 0

e;x(x+;r) p(,)d2

=0

for

Y<0 and XeR.

-co

Considering the first relation, write, for Y > 0, o

F(X + iY) = f

0

the function F(Z) is analytic for 3Z > 0 and bounded in each half plane 3Z > h > 0. We want to conclude that F(Z) - 0 for 3Z > 0. Beginning here, we can practically copy the proof of the theorem in the previous article. In that proof, we replace µ(2) NA(2)

by

qv(A),

by fA(A)

and jA(2) by cp(A) - fA(1). Everything will then be the same, almost word for word. True, instead of the inequality IPA(2)I <- e-M(A) used above, we here have (t), but the extra factor of 2 makes very little difference. We also have to

find an inequality for IfA(A)I in the strip

which will play the role of the

relation IPA(W)I < eA3A used previously. Our function fA satisfies I fA(A)I < eA

on

and I fA(2)I 5 Iw(2)1 + 2e-MCA> < 1 + 2e-M(O)

for

2ER

by (t), M(A) being increasing. Therefore e>AZfA(2)I , 1+2e- MCO>,

Ac-0-9,

and we conclude that this inequality holds throughout 9 by the extended principle of maximum (first theorem of § C, Chapter III). In other words, I fA(2)I < (1 +

2e-M(o))eA3a

for Ac-!P, and this plays the same role as the abovementioned inequality on

3 Beurling quasianalyticity

279

PA(.l), the constant factor I +2e-"fj') being without real importance. Repeating in this way the argument from the previous article, we see that the hypothesis

f

c M(A) A

dA = 00

of our present theorem implies that F(Z) - 0 for 3Z > 0. The fact that 0

eiA(X+ir)q,(2)d2

- 0.

-00

for Y < 0 and X e IF also follows by a simple modification of that argument, as indicated at the end of the proof we have been referring to. We are done.

Corollary. Let p be a finite measure on F such that 10 1

1+x

Z

dx = co,

log I

and suppose that O(2) is analytic in a rectangle

{a<9U 0, and µ(A) = f '. eU` dp(t) coincides with O(2) on E, then µ(A) - O(A) on the whole segment [a, b].

Remark. For E an intervals [a, b], this result was proved by Levinson.

Proof of corollary. Without loss of generality, assume that h = 1, that

Ii(A) I S i on the rectangle in question, and that f c ,,I dp(t) 15 ?.

a+i

0

Figure 63

a

b+i

7

b

917

280

VII B Fourier transform zero on a set of positive measure

Take our rectangle as the domain -9 of the theorem, with (a, b) as the arc y, and put (p(al)=µ(.l)- i/i(A),

a
For A > 0, write

fA(A) = JTA e t dµ(t)

-

the function is analytic in .9 and continuous on 9, and for A > 0,

IfA(.)I < ZeA+Z < eA,

.E !',

while for a<,
<

A I dµ(t)I = e-M(A),

J - r4

where M(A) = log (1/f _ A I dµ(t) I ).

According to our hypothesis, cp(A) - 0 on E 9 (a, b) = y with I E I > 0, and also

J1dA

= oo.

i

Therefore (p().) - 0 on (a, b) by the theorem, and, by continuity, µ(A) - O(A) Q.E.D. for a < A < b. 4.

The spaces -W,(-90), especially 9oi(20) Suppose that F(9) - Y_ '-. a"e'"'9 belongs to L2( - it, n) and that -1

2 to g\\E°-.Iak12 I 1

I= oo.

We would like, in analogy with the theorem of article 2, to be able to affirm that F(9) - 0 a.e. if F(9) vanishes on a set of positive measure. The trouble is that F is not necessarily bounded on [ - it, n], so we cannot work directly

with the uniform norm used up to now in the present §. At least two ideas for getting around this difficulty come to mind; one of them is to establish LP variants of the results in article 2 and 3. Such versions are

no longer conformally invariant. Beurling gave one for rectangular domains; one could of course use his method to obtain similar results for other regions. In this and the next subsection we stick to rectangles. Given a rectangle -9o with sides parallel to the axes, Beurling considers approximation in L. norm by certain functions analytic in 190, belonging to a space .'P(9) to be defined presently. We need some information about

4 The spaces Yp(-90), especially 91(-90)

281

those functions which, strictly speaking, comes from the theory of Hp spaces. Although this is not a book about Hp spaces, we proceed to sketch that material. In various special situations (including the one mentioned at the beginning of this article), easier results would suffice, and the reader is encouraged to investigate the possibilities of such simplification. If -90 is the rectangle {(x, y): xeIO, yEJO},

C0

U

Figure 64

10 we denote by Y,(-9O) the set of functions f (z) analytic in -90 with sup yeJo

I f(x + iy) Ip dx

f

lo

finite, and write sup

-jp(J) =

Yejo J Io

I f(x + iy)11 dx

for such f. We are only interested in values of p > 1, and, for such p, ) is a norm. Note that the compactness of To makes 9p(9O) s Y1(s0) for p > 1.

gyp(

Lemma (Fejer and F. Riesz). Let f (w) be regular and bounded for 3w > 0, continuous up to the real axis, and zero at oo. Then °° 0

1

If(iv)Idv 5 -2 _. If(u)Idu.

Proof. Under our assumptions on f, we have, for v > 0, f(iv)

_

-

0-

I

f (u) du

21ri

u - iv

1

21ri

('

f (u) du

J_ u+iv'

282

VII B Fourier transform zero on a set of positive measure

.

as long as I' I f (u) I du < oo, which is the only situation we need consider (see proof of lemma in § H.1, Chapter III). Adding, we get R iv)

_

1

7ri

uf(u)du _ cc u2+v2

whence

f

I f(iv) I dv

l u l l f(u)1

1

f°°JoI'

u2+v2

o

dvdu =

If"

2 -°°

If(u)Idu. Q.E.D.

Lemma. Let F(z) be analytic in a rectangle -9 and continuous up to 9. If A is a straight line joining the midpoints of two opposite sides of -9, we have

IA IF(z)I Idzl `

If 2

a.9

Proof.

Ip

Figure 65

Let cp map -9 conformally onto 3w > 0 in such a way that A goes onto the positive imaginary axis, and, for ze-q and w = (p(z), put (AZY

When w = 9(z) --f oo, p'(z) must tend to oo (otherwise the upper half plane would be bounded!), so f (w) must tend to zero, F(z) being continuous

on 9. We may therefore apply the previous lemma to f. This yields IA

IF(z)I ldzl =

IA

FOI Iw'(z)dzl = f-Jf(iv)Jdv

lf-cc

5 2 - If(u)Idu

4,,g)

I (d(C)d(l = 2 f a.I F (C)I l dal, Q.E.D.

Lemma (Beurling). Let go be the rectangle { - a < ¶2z < a, 0 < cz < h}, and

let f e.1(-90). Then, if - a < x < a,

f lf(x+iY)IdY (l+h)l(f). a-Ixl

4 The spaces Y,(-90), especially 5" (-90)

283

Proof. Wlog, let x > 0. Taking any small S > 0 we let -9,, for 0 < l < a - x, be the rectangle shown in the figure: y

hi

-o

-a

0

x

x

a

Figure 66

Applying the previous lemma to 9, we find that

f

h-hlf(x+iy)Idy

-'f, 2

S

If(C)Ild(j. !2,

Multiply both sides by dl and integrate I from (a - x) to a - x ! We get a-x a_ 2x h S

If(x+iy)Idy <-

if 2

S

a-x)/2

f

If(O)IldCldl.

3 2,

The lower horizontal sides of the -91 contribute at most

-,a -x f'-x 4

a 4

-(a-x)

to the expression on the right, and the top horizontal sides of the -9, contribute a similar amount. The right vertical sides give h

a-x

S

1

2

f(x+iy+l)Idldy (a-x)/2

S

and the left vertical sides make a similar contribution. The sum of these last two amounts is 1

2

(a-x)

h-S S

J-(a-x)

If(x+iy+l)Idldy

2(h-28)j1(f).

All told, we thus have

fa - x 1 22S(f2

a-x)/2

If(C)Ildgidl faLOj

_ _ a2x+h

284

VII B Fourier transform zero on a set of positive measure

so by the previous relation we see that fah - 8

I f(x + iy) I dy 5 11+

h-2a1 a -x

Making 8 -+ 0, we obtain the lemma for the case x ? 0. Done.

h

Figure 67 Io

Let feYl(-9o), and let be a rectangle lying in -90, in the manner shown - the vertical sides of being at positive distance, say a > 0, from those of moo. We proceed to investigate the boundary behaviour of f in -9. In order to do this, it is convenient to take 0 as the point of intersection of the diagonals of -9. This setup makes it easy for us to imitate the discussion

at the beginning of § F.1, Chapter III.

Figure 68

4 The spaces °,(9o), especially 91(90)

285

For 0 < 2 < 1 denote by -9x the rectangle {2z: zE9} (see diagram). -9, c -9 which, in turn, has the above described disposition inside 90. Since f e9l(-90), we have, by the preceding lemma,

IfQI Idyl < 2-i1(f) + 2(1+ h/Io1(f), calling h the height of -90. In other words,

x

JIf()4)dCI

(*)

for 0 < 2 < 1, where K depends on -9 and on f. Fix any z0 e9 and let 2 < 1. The function f(2z) is certainly analytic (hence

harmonic!) in -9 and continuous up to 89 (when z ranges over 9, the argument of f(2z) actually ranges over 9.). Therefore, by the discussion of article 1,

f(2zo) =

I

f(2C) dwgG, zo),

denoting, as usual, harmonic measure for -9 by co,( , z). Since the corners of .9 makes angles (of 90°) less than 180°from inside, we know by article 1 that dwq(t;, z0)/I dl; I is bounded (and indeed continuous) on 8-9, and the preceding formula can be rewritten thus:

f(2zo) =

d"'J(C,

Ja

Idyl

(In order to compute elliptic functions!)

I explicitly, we would have to resort to

We can now argue by (*) that there is a certain complex valued measure u

on 8.9 such that f(.C)IdCI -s dy(e)

w*

when 2-41 through a certain sequence of values, and thereby deduce from the previous relation that

(t)

f(zo) =

(See proof of first theorem in § F.1, Chapter III.) This, of course, holds for

any z0e9. Let qp be a conformal mapping of -9 onto { I w I < 1 } and let the function F,

analytic in the unit disk, be defined by the formula F((p(z)) =f (z), ze9. If v varying

is the complex measure on { I w I = 1 } such that dv(9(C)) = dp(i) for

286

VII B Fourier transform zero on a set of positive measure

on 8-9, (t) becomes

(tt)

F(w) =

1 - I wI 22 dv(w),

1

2n

w-wI

I wI < 1. The integral on the right therefore represents an analyticfunction of w for I w I < 1. From this it follows by the celebrated theorem of the brothers Riesz that v must be absolutely continuous, i.e., (§)

dv(w) = i((o) I dw l

with some L 1-function ,li on the unit circumference. By Chapter II, § B, and

(tt) we now have F(w) - O(w) as w - co for almost every co on the unit circumference. Write g(C) = '((p(1;)) for Ce89. Then, going back to.9, we see by the discussion in article 1 that

f (z) - g(C) as

z

for almost every t'e0.9. Plugging (§) into (tt) and then returning to (t), we find that

f (zo) =

Jdw9(Zo)gg)IdI a9

IdCl

We have already practically obtained the

Theorem. Let fe °1(9o). Then lim f (z) which we call f

exists for almost every l; on the horizontal sides of .9o. If -9 is a rectangle in -9o, disposed in the manner indicated above,

f

If(C)lld(l < oo,

a9

and, for ze-9, f (z) = J f (C) dwg(C, z). a9

If B1 and B2 denote the horizontal sides of -9Q, we have

fe,

<, -j1(f),

f

< -j1(f)

BI

B2

4 The spaces Yp(-90), especially <9'1('0)

287

Proof.

B2

B1

Figure 69

The first statement holds because limz --/- f (z) exists for almost all C on the boundary of any rectangle -9 lying in .90 in the manner shown; this we have just seen. Of course, if C lies on the vertical sides of such a rectangle

-9, we know anyway that lim,f(z) (without the angle mark!) exists and equals f (C), since those vertical sides lie in 90, where f is given as analytic. The second statement therefore follows from (*) and the first one, by Fatou's

lemma. (In using (*), one must take 0 as the point of intersection of the diagonals of -9.) In view of what has just been said, the third statement is merely another way of expressing the formula immediately preceding this theorem. There remains the fourth statement. Considering, for instance, the upper horizontal

side B2 of 90, we have f (z - i/n)

)f(z) for almost all zeB2 (first

statement!). Therefore, by Fatou's lemma,

f

I f (z) I dz < liminf J

B2

n-OD

dx. e2

The integrals on the right are all < 01(f) (by definition), at least as soon as 1/n < the height of -90. We are done. Theorem. Let I be any interval properly included within the base of -90,

in the manner shown:

VII B Fourier transform zero on a set of positive measure

288

Y

x

Figure 70

Then, iffeY,(9o),

f, If(z+i8)-f(z)Idx --> 0 as 6-0. Proof. To simplify the writing, we take the base of -9o to lie on the x-axis as shown in the figure. In view of the preceding theorem, we may assume that, at the endpoints a and b of I, lime .a f (z) and limZ f b f (z) exist and are finite. (Otherwise,

just make I a little bigger.) Then, if we construct the rectangle d g -9o

with base on I, in the way shown in the figure, f (z) will be continuous on the top and two vertical sides of d, right up to where the latter meet I. And by exactly the same argument as the one used to establish the third statement of the preceding theorem, we can see that

f(z) = fa (C)dws((,z) for zed. No w let e > 0 be given, and take a continuous function g(t) defined on 89 which coincides with f (l;) on the top and vertical sides of d and is specified on I in such a way that

f

c,

For zed, put g(z) =

z);

J ad

4 The spaces 9' (-90), especially .1(_90)

289

g(z) is at least harmonic in 9 (N.B. not necessarily analytic there!), and, by the discussion in article 1, continuous up to 89, where it takes the boundary values

For xel and small S > 0, P x + ib) -f(x) = P x + ib) - g(x + ib) + g(x + ib) - g(x) + g(x) - f (x).

We are interested in showing that 1, If (x + ib) -f (x) j dx is small if S > 0 is small enough. We already know that f I lg(x) - f (x) I dx < c, and, by continuity of g on I, f, I g(x + ib) - g(x) I dx < s if S > 0 is small. We will therefore be done if we verify that

lg(x+ib)-f(x+ib)Idx < s. SI

Since f (t;) = g(C) on M - I, ('

f (x + ib) - g(x + ib) = J (f (i) -

x + ib).

I

However, 9 lies in the upper half-plane and I on the real axis, so, by the principle of extension of domain used in article 1, for x + ib e off,

x + ib) S -

(x

b

+ b2 z

on I, the right-hand expression being the differential of harmonic measure for {,3z > 0} as seen from x + ib. Thus, for xel,

I f (x + ib) - g(x + ib) 15 n

fI l f O -

Idl I

(x

)z + bz

And I f (x + ib) - g(x + ib) I dx f,I

5

IT

I

-oo

If( )-9( )I(x-1;)z+b2dxd s.

fI

This does it.

Corollary. Let fell(_90) and let G(z) be any function analytic in a region including the closure of a rectangle 9 like the one used above lying in _90's

290

VII B Fourier transform zero on a set of positive measure

interior. Then

Ia6 G(C)fQd( = 0.

J

Proof. Use Cauchy's theorem for the rectangles with the dotted base together with the above result: Of

Figure 71

I Note that the integrals along the vertical sides of .9 are absolutely convergent by the third lemma of this article.

We need one more result - a Jensen inequality for rectangles S like the one used above. Theorem. Let fe<So1(.90), and let S be a rectangle like the one shown:

Figure 72

Then, for zE9, log I f(z) I < fee log

(C, z).

as

Proof. This would just be a restatement of the theorem on harmonic

4 The spaces .p(_90), especially 51(-q0)

291

estimation from article 1, except that f (z) is not necessarily continuous up to the base of S. There are several ways of getting around the difficulty caused by this lack of continuity; in one such we first map 4° conformally onto the unit disk and then use properties of the space H1. Functions in H1 can be expressed as products of inner and outer factors, so Jensen's inequality holds for them. In order to keep the exposition as nearly self-contained as possible, we give a different argument, based on Szego's theorem (§A, Chapter II!), whose idea goes back to Helson and Lowdenslager. Given zo a d, take a conformal mapping ip onto { 1 w I < 1) that sends zo to 0, and define a function F(w) analytic in the unit disk by means of the formula

zed.

F((p(z)) = f (z),

The relation ZEd,

f (z) = f f(C) dw,(C, z), a9

used in proving the above theorem, goes over into ,

F(w) = 2i f

Iw -Ie IIIZ F(e'L)ds,

with F(e") = f (q -1(ei')) defined almost everywhere on the unit circumference and in L1 (see discussion preceding the first theorem of this article).

From this last relation, we have ('2"

IF(pe;s)-F(e's)Id9

-> 0

J0

as p -+ 1. Also, for each p < 1, by Cauchy's theorem. Hence

102"e'"9F(pe'9)d9

= 0 when n =1, 2,3,...

fo2w

e'"9F(e'9)d9 = 0

for n = 1, 2,3,..., and, finally, 1 2x (l 2n

1 + Y A"e'"s IF(e's)d9 = F(0) ">o

o

///

for any finite sum E">oAe'"s. Thus,

f 2n 1

IF(0)I <

2,

0

1 + Y_ A"eins IF(e'9)Id9 ">o

292

VII B Fourier transform zero on a set of positive measure

for all such finite sums. By Szego's theorem, the infimum of the expressions on

the right is 1

exp

\

Zn

logIF(eis)Id9).

2n

0

Therefore, 1

log I F(0) I

<

27t

2,

0

log l

F(eie) I d 9,

or, in terms of f and zo = cp-1(0): log I f(zo) I < fee log I f(C) I dco,(C, zo).

T hat's what we wanted to prove. 5.

Beurling's quasianalyticity theorem for LP approximation by functions in Yp(9o). Being now in possession of the previous article's somewhat ad hoc

material, we are able to look at approximation by functions in 9pPo) (p > 1) and to prove a result about such approximation analogous to the one of article 3. Y

Figure 73

Throughout the following discussion, we work with a certain rectangular

domain -90 whose base is an interval on the real axis which we take, wlog, as [ -a, a]. If p >, 1, .9' (-90) c 991(-90), so we know by the first theorem of the previous article that, for functions f in .9'p(-9o), the nontangential boundary values f (x) exist for almost every x on [ - a, a]. As in the proof of that theorem we see by Fatou's lemma (there applied in

5 An Lp version of Beurling quasianalyticity

293

the case p = 1) that fa

IJ (x)Ipdx < (,,(J ))p,

fe.p(_9o)

a

The `restrictions' of functions f E.9'p(!20) to [- a, a] thus belong to Lp(- a, a), and we may use them to try to approximate arbitrary members of Lp(- a, a) in the norm of that space. In analogy with article 3, we define the Lp approximation index Mp(A) for any given

e-'P(^) is the infimum of If-aa I cp(x) - f (x)I'dx for f e6Pp(2o) with

i(f)

e".

Mp(A) is obviously an increasing function of A, and we have the following

Theorem (Beurling). Let cpeLp(- a, a), and let its Lp approximation index Mp(A) (for -9o) satisfy

JMP(A)dA = oo. 1

AZ

If cp(x) vanishes on a set of positive measure in [ - a, a], then cp(x) - 0 a.e.

on [ -a, a]. Proof. We first carry out some preliminary reductions. We have Yp(_90) 9 9'1(_90), Lp(-a, a) s L1(-a, a), and, by Holder's inequality, 41(f) 5 a(p- l)"pip(f) and I I c p -f I I 1 < a(p- l i p I I c p -f l i p for fe.'p(_Q0) and cpELp(-a,a). (We write 11 IIp for the Lp norm on [-a,a]).

From these facts it is clear that, if peLp(-a,a) has Lp approximation a"I(p-1)p (sic!) is index Mp(A), the L1 approximation index M1(A) Of Mp(A). It is therefore enough to prove the theorem for p = 1, for it will then follow for all values of p > 1. Suppose then that f °(M1(A)/A2)dA = oo with M1(A) the L1 approxi-

mation index for (peL1(-a, a), and that 9 vanishes on a set of positive measure in [ - a, a]. In order to prove that q = 0 a.e. on [ - a, a], it is enough to show that it vanishes a.e. on some interval J c [ - a, a] with positive length. To see this, take any very small fixed i > 0 and write cpn(x) = Zg

rp(x + t) dt

-n

294

VII B Fourier transform zero on a set of positive measure

for - a + q < x <, a - q.

cpn(x) is then continuous on the interval [- a + q, a - q], and vanishes identically on an interval of positive length therein as long as 21 < I JI. Corresponding to any fe5"1(-9o) we also form the function

-

n

+ t) dt;

- of (z let us check that fn(z) is analytic in the rectangle Qn with base f n(z)

[- a + 2q, a - 2q] having the same height as -9o, and is continuous on -9n.

a-2n a

-a -a+217

x

Figure 74

The analyticity of fn(z) in -9n is clear; so is continuity up to the vertical sides of .9n. The boundary-value function f (x) belongs to L1(- a,a), so fn(x)

is continuous on [ - a + 2q, a - 2q]. Let, then, - a + 21l <, xo 5 a - 2q, and suppose that x, also on that closed interval, is near xo and that y > 0 is small. We have If(x0) -fn(x+iy)I <,Ifn(xo) -fn(x)I + Ifn(x)-fn(x+iy)I. The first term on the right is small if x is close enough to xo. The second is 1

rx+n

1

f(a-n

2q x-n

which, by the second theorem of the preceding article, tends to zero (independently of x!) as y -* 0. Thus f ,,(x + iy) - fn(xo) as x + iy --p xo from

within 9n, and continuity of fn up to the lower horizontal side of .9n is established. Continuity of fn up to the upper horizontal side of -9n follows in

like manner, so fn(z) is continuous on 9n.

The functions fn are thus of the kind used in article 3 to uniformly approximate continuous functions given on [ - a + 2q, a + 2q]. By

5 An LP version of Beurling quasianalyticity

295

definition of M1(A), we can find an f in .?1(-90) with o1(f)<eA and S'

I T(x) -f(x)Idx 5 2eM'(A). With this f, I f,,(z)I 5 (1/2rl)eA for ze and Iwn(x) -.f,,(x)I

e-Mi(A)

a + 2r1, a - 2rl]. The uniform approximation index M(A) for rpp, (and on the domain -9,,) is thus > M1(A). Therefore, under the hypothesis of the present theorem, f °° M(A) 1

A

dA = oo,

so, since cp,,(x) vanishes identically on an interval of positive length in [- a + 2r1, a - 21] (when q > 0 is small enough) we have cp,(x)=_0,

-a+2rl<x
by the theorem of article 3. However, as ry -+ 0, cp,,(x) - p (x) a.e. on (- a, a). From what has just been

shown we conclude, then, that V(x) - 0 a.e. on (- a, a) if it vanishes a.e. on an interval J of positive length lying therein, provided that

JM1(A)dA = oo. t

Our task has thus finally boiled down to the following one. Given (pEL1(- a, a) with L1 approximation index M1(A) (for .9o) such that

JM1(A)dA =

00,

t

prove that (p vanishes a.e. on an interval of positive length in (- a, a) if it vanishes on a set of positive measure therein. Let us proceed. It is easy to see that the increasing function M1(A) is continuous (in the extended sense) - that's because, if A < 1 is close to 1, Af approximates (p almost as well as f does in L1(- a, a). Since f i (M1(A)/A2) dA = oo we may therefore, starting with a suitable Al > 0, get an increasing sequence of numbers A. tending to 0o such that

M1(A.+1)= 2Mt(A.).* Assume henceforth that cp(x) = 0 on the closed set Eo

[ - a, a] with

* We are allowing for the possibility that M,(A) = oo for large values of A; this happens when (p(x) actually coincides with a function in .11,(20) on (- a, a), and then it is necessary to take A, with M,(A1) = oo. We will, in any event, need to have A, large -see the following page.

296

VII B Fourier transform zero on a set of positive measure

I Eo I >0*. For each A >0 there is an f e .9'1(90) with o 1(f) < eA and a

Iw(x)-f(x)Idx 5

2e-MI(A)

-a

In particular, f(A) If(x)Idx < 2e-"s', E.

so, if AA = {xeEo: If(x)I > e-M,(A)/2l 2e-M1(A)i2. Taking the sequence of numbers An just we have IAAI 5 described, we thus get ao

U

AA.

Ln 1

n

We can choose A 1 large enough so that this sum is

< IE-I; 2

then the set

E = Eo - U in) has measure > I E0I/2, and, by its construction, for each n there is an fn e 6"1(-90) with ol(fn) 4 eA,,,

fa -a

Iw(x)-fn(x)Idx <

and

Ifn(x)I

for xeE. Take now a number b, 0 < b < a, sufficiently close to a so that

IEn[-b, b]I > 0, and construct the rectangle manner shown:

with base on [ - b, b], lying within -90 in the

* where I El denotes the Lebesgue measure of E S R

5 An LP version of Beurling quasianalyticity

297

Figure 75

Take a closed subset F of E n (- b, b) having positive measure; this set F will remain fixed during the following discussion.

As we saw at the end of article 1,

co,(F,x+iy)-+1 as y--,. 0 + for almost every xeF. Let cl and c2, c, < c2, be two such x's for which this is true. We are going to show that T (x) = 0 a.e. for cl 5 x <- c2;

according to what has been said above, this is all we need to do to finish the proof of our theorem. The desired vanishing of qp will follow if 4)(i,) =

CZ

eiAX (x) dx

is identically zero. F is, however, an entire function of exponential type bounded on the real axis. Hence, by §G.2 of Chapter III, = 0 provided that

f

2

log

I

d), = oo.

'00

We proceed to verify this relation. The reasoning here resembles that of article 2, but is more complicated. Take one of the functions f (later on, n will be made to depend on 2), and write CZ

O(A) =

CZ

eizx((p(x) - fn(x)) dx + J

dx = I + II, say.

298

VII B Fourier transform zero on a set of positive measure

Here, for 2 > 0, III5

J C2

I (v(x) -fi(x) I dx 5 2 e _ 1 '

,

and the real work is to estimate II.

-a

-b c,

b

c2

a

x

Figure 76

Let IF be a fixed contour in .9 consisting of three sides of a rectangle we have with base on [c1, c2]. Because C2

,

dx =

eizz f (z) dz

Jr.

by the corollary to the second theorem of the previous article. In order to estimate the integral on the right, we use the inequality log I fe(z) 15 fa.9 log I

I dw, (C, z),

z

furnished by the third theorem in the preceding article. This we further break up so as to obtaii'nn the following for zed:

log I ff(z) 15 J

log

z) + J log I

f Here, II denotes 8-9 of -9:

I

fn(s) I dw,,(C, z).

-b,b)- F

(- b, b), i.e., the vertical and top horizontal sides

5 An LP version of Beurling quasianalyticity

299

n

-a

-b

c,

b

C2

a

x

Figure 77

Consider the first integral on the right in (*). It equals a certain function u(z) harmonic in .9. Take any harmonic conjugate v(z) of u(z) for the region -9

and put 9n(z) =

the function

eu(z)+io(z)

ze21;

is analytic in 1.9 and we have

logI9n(z)I = J I

(C, z),

ze21.

and

analytic in -9 with

n

In the same way we get functions

zE-9,

f

F

and

f

ze21.

-66)-F

In terms of these functions, (*) becomes

(t)

Ifn(z)I <

ze21.

Our idea now is to estimate sup=Er

in order to get a bound on f r e"Z

I and J r I dz I I, supzEr I dz for A > 0. The third of these I

I

quantities will give us the most trouble.

We first look at

zef. For C on HI, the Poisson kernel

dco,,(t;, z)/I dl; I goes to zero when ze2l tends to any point of (- b, b), and does so uniformly for l; E- TI and z tending to any point of [c1, c2]. From this we see, by reflecting the harmonic function dw (C, z)/I dC I of z across (- b, b), that there is a certain constant C, depending only on the geometric configuration of F and .9, such that dco..(t;, z)

Id(I

5 C.3z,

zef , t;eH.

VII B Fourier transform zero on a set of positive measure

300

n

Figure 78

Substituting this into the above formula for log I gn(z) I, we get log 19n(z) 15 C3z fn log I fn(S) I I d I n

for z e I', whence, by the inequality between arithmetic and geometric

(zeF.

means,* 1

I9nZ)I CIril

Write now III I C = B. Then we have Biz

zeI,

Ign(z)I < const. n

where the constant is independent of n. Here, fnebt(-90) and e'^ Thence, by the third lemma of the preceding article, if h denotes the height of .9o,

JIfn()III

< Li(fn)+{1+a

+1'+a-

,Ictl}'t(fn)

czl},t(fn) 5 KeA^

with a constant K independent of n. Plugging this into the previous relation, we find that

zer'

I9n(z)I < const.eBA^3z,

the constant in front on the right being independent of n. To estimate I hn(z) I on r we simply use the fact that I

e-M=IA^>!2

for

eFcE

* in the following relation, In I is used to designate the linear measure (length) of n

5 An LP version of Beurling quasianalyticity

301

and get

exp(J

ZE-9. F

Substituting the estimates for I gn(z) I and I hn(z) I which we have already found into (t), we obtain le"zfn(z)I < const.e(BA

(*)

Z)3ze-w9(F.z)M1 (A,)/2Ikn(z)I

for zEI'. Thus, in order to get a good upper bound for eizzf

Jr

(z) dz

,

it suffices to find one for f r Ikn(z) I I dz I which is independent of n.

We have fa

Iw(x)-fn(x)Idx <

2e-M1(An)

a

Wlog,

f

a

1

Ikv(x)Idx

2,

a

therefore, for all sufficiently large n, fa

(tt)

I.fa(x)Idx 5 1. -a

We henceforth limit our attention to the large values of n for which this relation is true. The formula for log I kn(z) I can be rewritten

=

log I ka(z) I

f

log P(C) dco,,(t;, z),

aQ'

where

P(O = f

CE(-b,b)^,F,

I.

11 elsewhere on 8-9.

From this, by the inequality between arithmetic and geometric means, we get

I ka(z) 15 f

a

z) , 1 + f

-bn

l .fn(s) I

z),

ze

.

However, for - b < i; < b, we can apply the principle of extension of

302

VII B Fourier transform zero on a set of positive measure

domain to compare z) with harmonic measure for {3z > 0} as we did in proving the second theorem of the preceding article. This gives us

door(, z) 5

1

zd

2

ir Iz -

_ b < < b,

i

so the previous inequality becomes

Ik,,(z)I < 1 +

If

6

-blz

3z BIZ

Ifs( )Id

,

zed.

Denoting by h' the height of -9, and using this last relation together with Fubini's theorem, we see that, for 0 < y < h',

f

2b+

I

6

6

(in view of (tt)).

In other words, (sic!), and the d 1-norm of k for -9 is < 2b + 1 independently of n. Use now the third lemma of the previous article for -9.

x

Figure 79

On account of what has just been said, we get

Jr

l

b-

h

I

b-Ic21)))

Idz) < const., Jr,

independently of n. Let us return to (*). It is at this point that we choose n according to the value of A > 0. We are actually only interested in large values of A. For any such one, we refer to the sequence described above, and take n as the integer for which 2BA < A < 2BAi+1. For this n, (*) becomes

Iei'f(z)I <

zeF.

5 An LP version of Beurling quasianalyticity

303

Recall that the two feet c, and c2 of r were chosen so as to have

lim w9(F, c, + iy) = lim w9(F, c2 + iy) = 1.

y-o+ Therefore

B,3z + 1 c)9(F, z)

has a strictly positive minimum, say II, on r. fi depends only on the geometric configuration of -9 and I'. From the preceding relation, we have, then, when

2BAn < 2 < 2BAn+n being large,

zef.

I ei ' fn(z)I

Now use (Jr §). We get

dfor

e

2BA <, 2 < 2BAn+1; this, then, is our desired estimate for I III.

Now C2

e'

I(D(A)I =

III + IIII

p(x) dx

IC'

where I I I < 2e-M,(A"), as we saw near the start of the present discussion. We

may just as well take fi < 1 (which is in fact true any way); then, by the estimate for IIII just found, we have, for large n, 2BAn < A < 2BA +, .

I J(A)I <

)

Our aim here is to show that log

J 1'0

I

I

dA = oo,

or, what comes to the same thing, that f0.0 A2 log

dA = oo

for some large 2o. In view of the above inequality for I b(2)I, this holds if 2BAn+ I n

2BA

M I (A.)) A2

dl=oo,

i.e., if (§§)

E min n

M> (An))

I '4n

- An+i I

= 00.

304

VII B Fourier transform zero on a set of positive measure

We proceed to establish this relation. Our hypothesis says that 'o JM1(A)dA = co. 1

The function M1(A) is increasing, so, by the second lemma of article 2, we also have

()

Jmin(AM1(A))dA = 00. A2

Divide N, the set of positive integers, into three disjoint subsets:

R={ni1: An+1<, 2An}, S = In > 1: An+1 > 2An and An < M1(An)I, T = In i 1: An+1 > 2An and M1(An) < An}.

By ($), one of the three sums JA.+1 min(A,M1(A)) dA, neR

A2

A A

L

+1 min(A,M1(A)) A2

neS. An

r[

A,

dA,

+1 min (A, M1(A))

dA

z

must be infinite. Suppose the first of those sums is infinite. Recall that the An were chosen so as to have M, (A., 1) = 2M1(An). Therefore, if ncR and An <, A < An+ 1,

min(A,M1(A))

min(A,,+1,M1(An+1)) 5 2min(An,M1(An)),

" + 1 min (A, M1(A)) An

AZ

dA

2min(A.,M1(An))

1-

{ A.

1

nER.

An+1

In the present case, then, we certainly have (§§). If the second of the sums in question (the one over S) is infinite, the set S cannot be finite. However, for neS,

min(An,M1(An)) 1 An

so (§§) holds when S is infinite.

1

An+1

An+1 -An An+1

> 1, 2'

305

Kargaev's example

There remains the case where the third sum (over T) is infinite. Here, for n c- T and A. < A < A,, 1 we have

min(An,M1(An)) = M1(An) = iM1(An+1)

iM,(A),

so, for such n, 1

A

min (A., MI (As)) An - An1

>1

+1

A.

+ MI(A)

AA

A2

I ('A_+, min (A, M1(A)) dA. A2

2

A.

Hence, if the sum of the right-hand integrals for ne T is infinite, so is that of the left-hand expressions, and (§§) holds. The relation (§§) is thus proved. This, however, implies that 1

1

J

2

log

1

d), = oo

1"0

as we have seen, which is what we needed to show. The theorem is completely proved, and we are done.

.

Corollary. Let f(9) - E'_ ane'"'9 belong to L2( - it, it), and suppose that -1

Y

-oo

1 n2

log

1

'"_.Iakl2

If f(9) vanishes on a set of positive measure, then f =_ 0 a.e. Let the reader deduce the corollary from the theorem. He or she is also encouraged to examine how some of the results from the previous article can be weakened (making their proofs simpler), leaving, however, enough to establish an L2 version of the theorem which will yield the corollary.

Kargaev's example In remark 2 following the proof of the Beurling gap theorem (§B.2), it was said that that result cannot be improved so as to apply C.

to measure µ with µ(A) vanishing on a set of positive measure, instead of on a whole interval. This is shown by an example due to P. Kargaev which we give in the present §. Kargaev's construction furnishes a measure u with gaps (an, bn) in its

support, 0 < al < bl < a2 < b2 < , such that a,

00

306

VII C Kargaev's example

while P (A) = 0 on a set E with I E I > 0. His method shows that in fact the relative size, (b" - a")/an, of the gaps in µ's support has no bearing on µ(2.)'s capability of vanishing on a set of positive measure without being identically

zero. It is possible to obtain such measures with (bn - an)/an n oo as rapidly as we please. In view of Beurling's gap theorem, there is thus a qualitative difference between requiring that µ(.) vanish on an interval and merely having it vanish on a set of positive measure. The measures obtained are supported on the integers, and their construction uses absolutely convergent Fourier series. The reasoning is elementary and somewhat reminiscent of the work of Smith, Pigno and McGehee on Littlewood's conjecture. 1.

Two lemmas

Let us first introduce some notation.. functions

denotes the collection of

Go

-00

with the series on the right absolutely convergent. For such a function f(9) we put 00

Ilf 11

= Y_00lanl

and frequently write f (n) instead of a" (both of these notations are customary). d, 11

11 is a Banach space; in fact, a Banach algebra because, if

f and g ed, then f (9)g(9) e .4, and IIfg11 < Ilf 11 IIglI.

On account of this relation, 0(f) e d for any entire function F if fed. We will be using some simple linear operators on 4.

,

Definition. If f(9) = E'_ J(n)ein9 belongs to sad, 00

(P+fX9) = Y f(n)e'"9 n=0

and P_ f = f - P+ f. We frequently write f+ for P+ f and f_ for P_.f. Observe that, for fed, II P+f II < Ilf 11 and II P-f II < If II.

Definition. For N an integer >, 1 and fed, (HNf)(9) =f (N9). (The H stands for `homothety'.)

1 Two lemmas

307

The following relations are obvious: HN(f 9) = (HNf) (HN9),

f' 9 e sad,

IIHNfII = IIfII, P+(HNf) = HN(P+f), and HA(f) = (D(HN f) for fell and t an entire function. Lemma. For each integer N > 1 and each S > 0 there is a linear operator TN,,, on sad together with a set EN,,'

[0, 2n) such that:

(i) For each fed, g = TN,, ,f has g(n) = 0 for - N < n < N (sic!); (ii) For each f ed, (TN,a f)(9) =f (9) for 9EEN,a; (iii) II TN,af II < C(b) II f II with C(b) depending only on S and not on N;

(iv) I EN,a I = 2it(1 - b).

Proof. The idea is as follows: starting with an fed, we try to cook functions g+(9) and g - (9) in sad, the first having only positive frequencies and the second only negative ones, in such a way as to get 9+(9)eiNs +

9-(9)e-

WS

`almost' equal to f (9). We take a certain p1ied (to be described in a moment) and write (*)

q = e+(O+-*-)

According to the observations preceding the lemma, gesl. Our construction of TN,s and EN,B is based on the following identity valid for fes4:

f = ((fq)+e-Z'"+)e'd' + ((fq)-e2'0-)e-''.

To check this, just observe that the right-hand side is (fq)+e-+0+-0-) +

(fq)-e'(0--0+)

= ((fq)+ + (fq)-)q-' = fq-q-' = f. Here is the way we choose 0. Take any 21r-periodic W00-function 4pa(9)

with a graph like this on the range 0 < 9 < 2it:

308

VII C Kargaev's example

0

2w-w6 2x

7r&

Figure 80

Then put 0 = HNcOB; 0 thus depends on N and 6. Note that p e.4 because qa is infinitely differentiable (I spa(n) I < 0(1 n j - k) for every k > 0 !). Therefore

belongs to 4. With qed related by (*) to the 0 just specified, put, for fed, TN,a.f = ((fq)+ e-24,

+)e1NS+((fq)-e2"I-)e-iNs

TN,,, obviously takes d into sad; let us show that there is a set EN,a s [0, 27[)

independent off such that (ii) holds. The set

AN,s = {9, 0<9<2ir: 0
9 e [0, 2it)'

ON,a;

i.e., eiO(q) = e,NS,

9 e [0, 2n) - AN, a.

Put, therefore, EN,,, = [0, 2n) - AN,,'; then, by comparing the formula for TN, af with the boxed identity following (*), we see that (TN, a f) (9) =f(9) for 9eEN,a, proving (ii). We also have (iv), since

IEN,aI = 2n-IAN,6I = 2n-2nb.

1 Two lemmas

309

We come to (i). The function (fq)+ only has non-negative frequencies in its Fourier series. The same is true for e-2'0+. Indeed, the latter function equals

(2i/i ) 2 -(2ifi)3

+

with the series convergent in the norm 11 II, and each power ('+)" has a Fourier series involving only frequencies 0. The Fourier series of the thus only involves frequencies >0, and finally, product (fq)+e-2i0+

that for ((fq)+e-2f0+)e'N9

only has /frequencies >, N. One verifies in the same way that ((fq)-e2''-)e-iN8

has a Fourier series involving only the frequencies < - N, and (i) now follows from our definition of TN,,,. There remains (iii). We have, for example, II(fq)+e-"+II

< 11(fq)+IIIIe-'11 11

lfgll Ile

2'y+II

5 Ilf ll

Ilgll Ile 2iy+ll

Here, e-2NO+ = e-2iP+HNwa = e-2iHNP+wa = H N

e-2hP+rpb

,

according to the elementary relations preceding the lemma, so Ile-2i'+II

IIHNe-2,P+,p6ll

=

=

Ile-2,P.wall,

a finite quantity, depending on S but not on N. In like manner, IIqII

=

=

e'(O+-O-) II

HNe'(P+`b-P-"6)II II

= II ei(P+(P,-P-`°"' II,

a finite quantity depending on S but independent of N. We thus have II(fq)+e-2`y+e'N911

=

11(fq)+e-2`11+II

< Asllf11,

where A8 depends only on S.

The norm

II (fq) - e2i-' - e -'N'9 11

is handled in exactly the same way,

and found to be < Bb 11 f 11 with Bb depending only on 6. Referring to the definition of TN,b, we see that (iii) holds. The lemma is thus proved.

We now take two positive integers L and N; N will usually be much larger than 2L.

310

VII C Kargaev's example

Definition. 2L+1

U'

.,12(N, L) =

k=-2L-1

[Nk - L, Nk + L].

Here, the prime next to the union sign means that the term corresponding to the value k = 0 is omitted. For N > 2L, . &(N, L) is the union of 4L + 2 separate intervals, each of length 2L: 2L

2L

2L

-(2L+1)N -2N

-N

0

2L

2L

N

2N

2L

(2L + ON

Figure 81

In proving the following lemma we use another linear operator on a.

Definition. For f ea, put L

(SLf)(9) _

f(n)ein9

n=-L

Observe that II SLf II < II f II and II f - SLf II

0 as L -+ co whenever

f ea. We also have

P+SLf = SLP+f. Lemma. For each S > 0 and pair N, L of positive integers there is a linear operator TN a on sat such that (1) For any f ea, the Fourier coefficients g(n) of g = TN a f are all zero when n0. i(N, L); (2) For f ed, II TN of II < C(b) II f II with C(b) independent of N and L; (3) If TN,s is the operator furnished by the previous lemma, we have II TN,0f - TN,,af II

0

uniformly in N as L--* oo, for each fed and b > 0. Remarks. Actually, the spectrum of T N a f is contained in a smaller set than

.,k(N, L) when fed. It is the uniformity with respect to N in property 3 which will turn out to be especially important in Kargaev's construction. Proof of lemma. Fix b > 0 and take the function q,, used in proving the preceding lemma - here we just denote it by 'p. In terms of

q0 =

ei(w+-(P-),

1 Two lemmas

311

we observe that the definition of TN,a f given in the proof of the previous lemma can be rewritten thus: (fHNg0)-(HNe21`),e-.Ns

TN,af =

Put Tvaf = (SLf-HNSLgo)+(HNSLe-""+) -HNSLgo)-(HNSLe2i,_),e-.Ns

+ (SLf

Since 119 - SL9II -->0 as L-* oo for every ged, T a f is clearly a kind of approximation to TN,af We proceed to verify property (1). The Fourier coefficients of SLf are all zero save for those with index in the set

{-L, -L+1, ..., 0,

1, ..., L}.

The non-zero Fourier coefficients of HNSLgo have their indices in the set

{ - NL, - N(L - 1), ..., - N, 0, N, ..., NL}. Therefore the Fourier coefficients of (SLf 'HNSLgo)+ with index outside the set

{0, 1,

..., L} u {N-L, N-L+1, ..., N, N+1, ..., N+L}

u {2N-L, 2N-L+1, ..., 2N+L} u u {NL-L, NL-L+1, ..., NL+L} are surely zero.

Again, the Fourier coefficients of HNSLe-2i'+ are all zero save for those

with index in the set {0, N, 2N, ..., LN}. So, finally, the Fourier coefficients of -HNSLgo)+(HNSLe-2i(p+)eM

(SLf

(the first of the two terms making up TN a f) are all zero, save for those with index in the union of intervals 2L+ I

[N, N + L] u U [Nk - L, Nk + L]. k=2

Treating the second term of TN a f in the same way, we see that property 1 holds (and that indeed more is true regarding the spectrum of TN af). To check property (2), we have, for the first term of TN af, II (SLf 1<

-HNSLgo)+(HNSLe-2,w+).etN9

II

II SLf 11 11 HNSLg0 11 11 HNSLe-2iI'+ II II f 1111 SLg0 1111 SLe-2ip+ II

5

II f 1111 q0 1111

e-"'P+

II;

312

VII C Kargaev's example

we have used the fact that II HN9 II = 119 11 for gesi. In the extreme righthand member of the chain of inequalities just written, the factors II qO II and II a-2iw+ 11 are finite and only involve p = cpa; therefore they depend only on b.

The second term of TN of is handled in exactly the same fashion, and,

putting together the estimates obtained for both terms, we arrive at property 2. Verification of property (3) remains. This is somewhat long-winded. It is really nothing but an elaborate version of the argument presented in good elementary calculus courses to show that the limit of a product equals the product of the limits. In order not to lose sight of the main idea, let's just compare the first terms of TN,a f and TN a f . The difference of these first terms

has norm equal to /

-

11(SLf.HNSLgO)+(HNSLe-2io+).e"N9

< II (SLf' HNSLgo)+ - (f HNgo)+ 11

+ < 11

a-2i'

II

({ J

.HNgo)+(HNe-2iW+)e"

II

HNSLe-2iv+ 11

(fHNgo)+ 11

11

11 HNe-2i'+ - HNSLe-2i"+ II

1111(SLf -f )HNgo + (SLf) (HNSLgo - HNgO)11

+

SLe-2i(p+

11 f 11 11 q0 11

11 e-2(P+ -

II

< 11e-2'W+II{IISLf-f II IIgo11 + 11f 1111SLgO-gO11}

+

e-2iv+ 1 1 1 1 1 1 1 qO I I

Il

-

SLe-2'W+ II

This last expression does not involve N at all, and, for fixed fed, tends to ei(O+-w-).) zero as L- oo. (It depends on S through the functions cp and q0 = The difference of the second terms of TN of and TN,af is treated in the same way, and we see that property (3) holds. The lemma is proved, and we are done. 2.

The example

Theorem (Kargaev). Let A c Z. Suppose that for each positive integer L there is some positive integer NL with

A 2 .,#(NL, L) n 7L, where the sets Y(N, L) are those defined in the previous article. Then, given

e > 0 and g e d we can find a gE e d such that (i) II 9E II < Kt 11g II, where KE depends only on e;

(ii) 4,.(n) = 0 for n0A; (iii) gt(9) = g(9) for 9 e [0, 2n) - A, where I Al < 21CE.

Proof. Taking s > 0, we put S = E/2" and E =

1) with C(S) from

2 The example

313

property (2) of the second lemma in the previous article. There is no harm in supposing that C(8) > 1; this we do in the following construction. The function gE is obtained from a given ged by a process of successive approximations, using the operators TN,,, and TN a from the two lemmas of the preceding article. According to the second of those lemmas, we can choose an L1 such that (*)

II TN,'1g-TN,a,9II

1<

8111911

for all values of N simultaneously. If we take any positive integer N, the Fourier coefficients h(n) of h = TN a,g all vanish for noJ((N, L1) by that second lemma. The hypothesis now furnishes a value of N such that

.i(N,L1)n7Z c A. Fix such a value of N, calling it N1. Then, if we put h1 = TN,',b,g, we have 1i1(n) = 0 for n4A. Let us also write r1= TN,,,,,g - hl. Then (*) says that el 11 g 11, and, by the first lemma of the preceding article, II r1 II

g(19)-h1(19)-r1(19) = g(9) - (TN,,a,9)(9) =

0

for 9 eEN,,,,,, a certain subset of [0,27r) with I EN,,a, I = 21r(1 - S1).

We proceed, treating r1 the way our given function g was just handled. First use the second lemma to get an L2 such that II TN a2r1 - TN,s2r1 II

1<

8211911

for all positive N simultaneously, then choose (and fix) a value N2 of N for A, such choice being possible according to the which .I1(N2, L2) n Z hypothesis. Writing hz

=

T(L2)

N2,a2

r1

and

r2 = TN2,52r1 - h2,

we will have h2(n)=0 for n0.1(N2i L2) by the second lemma, hence, a fortiori, 1i2(n) = 0 for n0A. Our choice of L2 makes 11r211

8211911,

and, by the first lemma, we have r1(9) _ (TN2,a2r1)(9),

i.e., r1(9) _ h2(9) + r2(9) for 9eEN2,62, a subset of [0, 2x) with IEN2,b2I = 27T(1 - b2) According to the preceding step, we then have

g(9) = h1(9)+h2(9)+r2(9)

for 9 e EN,,,', n

EN2,62.

And'i1(n) + £2(n) = 0 for nOA. Suppose that functions h1, h2i ... , hk _ 1 and rk _ 1 (in d) and positive

VII C Kargaev's example

314

integers N1, N2, ... , Nk - 1 have been determined with

I I rk- I I I I< Ek -1 II 9 II,

h;(n)=0 for n0A, j=1,2,...,k-1, and g=h1+h2+...+hk-1+rk_1 on the intersection nj-11 ENj,,j. Then choose Lk in such a way that II TN,akrk-1 - TN,6krk-1II

1<

Ek1I9II

simultaneously for all N (second lemma), and afterwards pick an Nk with .1&(Nk, Lk) r' Z c A (hypothesis). Putting I.

k

= T (Lk) Nk,Skr k-1

and

rk = TNk,bkrk- 1 - hk, see that lik(n) = 0 for n0A, that II rk II <1Ek II 9 II, and g = h1 + h2 + + hk-1 + hk + rk on n;= 1 ENj,Bj (first lemma). Observe now that, by the second lemma, we also have

we

Ilhk11 = II TN, akrk-1II

1<

C(Sk)IIrk-111

s

1<

that

C(6k)Ek-1II9II

II9II/2k-I

for k > 2 on account of the way the numbers Ek were rigged at the beginning of this proof. The series h 1 + h2 + h3 + therefore converges in the space .4 (hence uniformly on [0, 2n] ). Putting oo

9E(9) = Y hk(9), k=1

we have (1+C(E/2))IIgII,

IIgII

and 0E(n) = 0 for n A since, for such n, we have /ik(n) = 0 for every k. Finally, since Irk(19)I

1<

1< 80911 k' 0,

IIrkII

we have k

Y hj(9) + rk(9) k 9E(9) j=1 uniformly for 0 < 9 < 2n, so g,,(9) = g(9) on the intersection

E = j=1 n OD

ENj,aj.

Here, since I ENj,aj I = 2ir(1 - 6 j) and the sets ENj,,jj all lie in [0, 2ir), we have

IEI >

21z(1-61-62-63-...)

= 27[(1 - E).

The theorem is proved.

=

21r(1-2-4-...)

2 The example

315

Our example is now furnished by the following

Corollary. There exists a non-zero measure it having gaps (a,,,

in its

support, with

0
Y(b,aa,)2

= ao

(and the ratios (b, - a,)/a, even tending to oo as rapidly as we want!), while µ(.l) = 0 on a set of positive measure.

Proof. For 1= 1, 2, 3,..., take the sets 21+1

11,

=

U'

[N,k-1, N,k+1]

k=-21-1

(term with k = 0 omitted), with the positive integers N, so chosen that N, > 21 and that N,+ 1 is much larger than (21 + 1)(N, + 1). There is no obstacle to our taking N,+ 1 as large as we wish in relation to (21 + 1)(N, + 1) for each 1.

Put

A = U co(,ff , r) Z); !=1

it is clear that A satisfies the hypothesis of the theorem. Choose any ge.4 such that g(9) > 0 for ir/2 < 9 < 3n/2 and g(9) = 0 for 0 < 9 < ir/2 and for 37r/2 < 9 < 2n.

3a

1

Figure 82

316

VII D Volberg's work

There are plenty of such functions g; we fix one of them. Apply the theorem with e = 4, getting a function ge in of with de(n) = 0 for noA and ge(9) = g(9) for all 9 e[0, 2n) outside a set of measure < n/2. Then certainly ge(9) must be > 0 on a set in [0, 2n) of measure > n/2 (hence, in particular, ge # 0), while at the same time ge(9) = 0 on a set of measure >, n/2 lying in [0, 27r).

We have ge(9) _

9e(n)ein9

nEA

with Y 19e(n) I

< oo,

neA

so, if we define a measure y supported on A g 7L by putting p(E) = Y-nEEd,(n),

we have u # 0, but µ(9) = ge(9) vanishes on a set of positive measure. The support (s A) of y has the gaps ((21 + 1)N, + 1, N,+1 -1-1) in it. By choosing N,+ 1 sufficiently large in relation to (21 + 1) (N, + 1) for each 1, we can make the ratios

(N1+1 -1- 1)-((21+ 1)N,+l) (21+ 1)N,+1

go to oo as rapidly as we please for l -+ oo. We are done.

D.

Volberg's work Let f (S) e L, (- 7r, 7r); say

P9) ^

anetn8.

Y_ 00

Suppose that the Fourier coefficients an with negative indices n are small enough to satisfy the relation -1

(*)

1

Y -log log`E"-lak)

= oo.

According to a corollary to Levinson's theorem (§ A.5), f(9) then cannot vanish on an interval of positive length unless f =_ 0. If we also assume (for instance) that Y-k I ak I < oo, Beurling's improvement of Levinson's theorem

(§B.2) shows that f(9) cannot even vanish on a set of positive measure without being identically zero when (*) holds. It is therefore natural to ask how small I f(9)I can actually be for a non-

Volberg's work

317

zero f whose Fourier coefficients a satisfy (*), or something like it. Suppose

for instance, that I an l < e - Ma"u

n < O,

with a regularly increasing M(m) for which M(m)

o0

m

1

Volberg's surprising result is that if the behaviour of M(m) is regular enough, then we must have

JlogIf()Id9 > -oo unless f - 0. Very loosely speaking, this amounts to saying that if f # 0 and

00

n 2 log I f (n)

then

f

n

loglf(9)Id9 > -oo, n

at least when the decrease of 1 f (n) I for n -- - oo is sufficiently regular. If one logarithmic integral (the sum) diverges, the other must converge !

One could improve this result only by finding a way to relax the regularity conditions imposed on M(m). Indeed, if p(9) >,0 is any function in L1( - n, n) with

fn l og p(9) d9 > - co, we can get a function

P9) -.

anein8

Y0

such that I f(9)1 = p(9) a.e. by putting

f(9)= lim exp ,is

z
^f1

it e"+zlogp(t)dt

2n -,ne- z

(see Chapter II, § A). Here, the Fourier coefficients of negative index are all zero, i.e., for n < 0,

lanl <

e-m(Inb

318

VII D Volberg's work

with M(I n I) - oo. This means that from the condition

n<0,

Ian1 . e-mCnu, with

one can never hope to deduce a more stringent restriction on the smallness of

If(9)I than

log If(9)Id9 > - oc. J

Also, if M(m) is increasing, from a less stringent condition than M(m)

n

- 00

one can never hope to deduce any limitation on the smallness of I f(9)1 for functions f # 0 with I an I < e-"OnD, n < 0. That is the content of

Problem 12 Let M(m) > 0 be increasing for m > 0, and such that M(m) Y-

< oo.

m

Given h, 0 < h < n, show that there is a function f (S), continuous and of period 2n, with f (9) = 0 for h < 191-< n but f * 0, such that l and < e-M""",

n

0 (sic!),

for the Fourier coefficients an of f(9). (Hint: Use the theorems of Chapter IV, § D and Chapter III, § D. Take a suitable convolution.)

It is important to note that Volberg's theorem relates specifically to the unit circle; its analogue for the real line is false. Take, namely, F(x) = e

that

log F(x) dx = - oo. Here,

E(2) = J

eixxe-x2dx = - 00

()e214 2

-x2,

so

1 The planar Cauchy transform

319

so

/

0

-

l

+ 2 log

G =

1

oo.

and even 10 _

1+2

log

I dA = oo.

$

This example shows that a function /and its Fourier transform can both get very small on l (in terms of the logarithmic integral). 1.

The planar Cauchy transform Notation. If G(z) is differentiable as a function of x and y we write aOz

= G=(z) = as Z) _ i as Z) Y

and aG(z)

of

= G-(z) GA Z) =

aG(z)

+i

aG(z)

ay

Nota bene. Nowadays, most people take aG/az and aG/az as one-half of the respective right-hand quantities.

Remark. If G = U + iV with real functions U and V, the equation Gz = 0 reduces to $ Ux = Vy,

Uy= -Vx, i.e., the Cauchy-Riemann equations for U and V. The condition that Gi = 0 in a domain -9 is thus equivalent to analyticity of G(z) in -9. Theorem. Let F(z) be bounded and W1 in a bounded domain .9, and put

G(z) = i

f F(Z)d drl

if

where, as usual, l; =1; + iii. Then G(z) is W1 in -9 and aG(z) = Of

F(z),

ze-9.

Remark. The integral in question converges absolutely for each z, as is seen by

320

VII D Volberg's work

going over to the polar coordinates (p, qi) with C - z = pe'g'.

G(z) is called the planar Cauchy transform of F(z).

Proof of theorem. We first establish the differentiability of G(z) in -9. Let zoe2 with dist (zo, 8.9) = 3p, say. Take any infinitely differentiable function cp(C) of l; with 0 <
1' K-zoI 'p'

00

0, IC-zol>,2p.

i

P

2P

IFz0I

Figure 83

We can write

((

G(z) 27r

(?(()F(0 d do

z-

iC-zo152p

+

1

(1- _(_))F(P d do

r ('

2J[ J

JIC-zol>P

z-

CE.9

The second integral on the right is obviously a 16., function of z for I z - zo I < p; it remains to consider the first one. After a change of variable, the latter can be rewritten as

F1(z - w)

1

2n

ffc

w

du dv

1 The planar Cauchy transform

321

(where w = u + iv, as usual) with Fl(t) = p(()F(t'). Here, F1(C) is of compact support, and has as much differentiability as F(C). Hence, since du d v

< 00

JJIwI
let in those variables. We have shown that G(z) is W'

1

in the neighborhood of any zo e -9; there

remains the evaluation of Gz(zo) in terms of F. This turns out to be surprisingly difficult if we try to do it directly, and we resort to the following dodge.

Let r > 0 be small, and zoeQ. By the differentiability of G(z) at zo, G(zo + rei9) = G(zo) + Gx(zo)rcos 9 + G,,(zo)rsin 9 + o(r)

= G(zo) + G(z0)re i9 + i

GZ(zo)Ye-'9 + o(r).

i

Multiplying the last expression by e'9 d9 and integrating 9 from 0 to 2n, we find the value nrGZ(zo) + o(r); therefore zR

1

GZ(zo) = lim

G(z0 + rei9)ei9 d9.

nr

o

Plugging in the expression for G in terms of F and changing the order of integration, this becomes

GZ(zo) = h m

zR o

1

2-JfJf Jf

F(C)e'9

z0 + reT-

d9 d5 drl.

However, 2R

f

0

irei9d9

_

rei9-(C-zo)

2ni,

)0,

IK-zol r.

Therefore, by the previous relation, we have GZ(zo) = lim '2 f f

,o irr

;-zol
F(C) d

F(zo),

F having been assumed to be 1O 1 in -9. We are done.

Corollary. Let .9 be a bounded domain. Suppose that F(z) is '2 in -9, that I F(z) I > 0 there, and that there is a constant C such that 8F(z)

8i

< C I F(z) I I

'

7

322

VII D Volberg's work

Then

(D(z) = F(z) exp t

I

d dtl )

( ('

is analytic in 9, and lq)(z)I lies between two constant multiples of IF(z)I therein.

Proof. F2(z)/F(z) is W, in 9 and bounded there by hypothesis, so we can apply the theorem, which tells us first of all that t(z) is differentiable in -9, and secondly that (1 F-(z)1 8(D(z) Fe(z) - F(z) F(z) exp 27r 8z

-

z)

0

there. The Cauchy-Riemann equations for M)(z) and 3(D(z) are thus satisfied (see remark at the beginning of this article), so t(z) is analytic in .9. If R is the diameter of -9, we easily check that

e-c'IF(z)I < J(D(z)l

ec'

I F(z) I

for zE-9. This does it. The corollary has been extensively used by Lipman Bers and by Vekua in the study of partial differential equations. Volberg also uses it so as to bring analytic functions into his treatment.

Problem 13 Show that the condition that I F(z) I > 0 in -9 can be dropped from the hypothesis of the corollary, provided that we maintain the assumption that jF,(z) I _
on the set where F(z) = 0. Hence show that a function F satisfying the inequality IF;(z)I _< CI F(z) I can have only isolated zeros in.9, unless F =_ 0

there. (Hint. On E = {ze_9: F(z) = 0}, assign any constant value to the ratio Fe(z)/F(z). The function D(z) defined in the statement of the corollary is surely analytic in -9 - E; it is also analytic in E° (if that set is non-empty)

because it vanishes identically there. To check existence of

'F'(zo) = lim z-.z°

D(z) - (D(zo)

Z - Zo

at a point zo e 8E n.9, note that both F(zo) and FZ(zo) must vanish, so, near zo,

F(z) = iF.(zo)(z-z0)+o(Iz-z01). If Fz(zo) = 0, D'(zo) exists and equals zero. If F=(zo) 0 0, IF(z)1>0 in some punctured neighborhood 0< l z - zo l < rl of zo, so such a punctured neighborhood is included in -9 - E.)

2 The function M(v) - its Legendre transform h(g) 2.

323

The function M(v) and its Legendre transform h(4)

As explained at the beginning of this chapter, Volberg's work deals with functions

f(9) -

°° anein9 00

for which the an with negative index are very small; more precisely,

Ja-.J

<, e-ti1(n),

n>O,

where M(n) is increasing and such that M(n) n2-

It will be convenient to assume throughout this § that M(v) is defined for all real values of v > 0 and not just the integral ones, and is increasing on [0, cc). We do not, to begin with, exclude the possibility that M(0) = - cc. Whether this happens or not will turn out to make no difference as far as our final result is concerned. Volberg's treatment makes essential use of a weight w(r) > 0 defined for 0 < r < 1 by means of the formula log

1

w(r)

= sup M(v) - v log 1 ). v>o

r

It is therefore necessary to make a study of the relation between M(v) and the function

h(i;) = sup (M(v) - vl;), v>0

defined for > 0, and to find out how various properties of M(v) are We take up these matters in the present article. connected to others of (sometimes called the Legendre The formula for the function transform of M(v)) is reminiscent of material discussed extensively in Chapter IV, beginning with § A.2 therein. It is perhaps a good idea to start by showing how the situation now under consideration is related to that of Chapter IV, and especially how it differs from the latter. Our present function M(v) can be interpreted as log T(v), where T(r) is the Ostrowski function used in Chapter IV. (M(n) is not, as the similarity in letters might lead one to believe, a version of the {M,,} - or of log Mn -

324

VII D Volberg's work

from Chapter IV!) Suppose indeed that

f(9) - Y_ - o0 !1

is infinitely

anein9

differentiable and in the class 1({Mn}) considered in

Chapter IV - in order to simplify matters, let us say that

I f'n'(q) 15 Mn,

n , 0.

We have an = 27r

n

e-in9f(9)d9

and the right side, after k integrations by parts, becomes 1

2

-n

(in) -kf(k)(9) d9

when n 0 0. Using the above inequality on the derivatives of f (9) in this integral, we see that

lanl 5 inf Mk

T(Inl)

where, as in Chapter IV, T(r)

=Sup- kr

for

k3o Mk

r>0.

On putting T(v) = e"', we get lanI

<

e-MUnD

This connection makes it possible to apply the final result of the present § to certain classes W'({Mn}) of periodic functions, of period 2ir. But that application does not show its real scope. The inequality for the an obtained by assuming that f E'({Mn}) is a two-sided one; it shows that the an go to

zero rapidly as n - ± oc. The hypothesis for the theorem on the logarithmic integral is, however, one-sided; it is only necessary to assume that

Ia -nI < e - M(n) for n > 0 in order to reach the desired conclusion. There is another essential difference between our present situation and that of Chapter IV. Here we look at the function sup (M(v) >o

2 The function M(v) - its Legendre transform

325

i.e., in terms of T(v), sup (log T(v) v>o

There we used the convex logarithmic regularisation {Mn} given by log M = sup ((log v)n - log T(v)). v>o

There is, first of all, a change in sign. Besides this, the former expression involves terms vL , linear in the parameter v, where the latter has terms linear in log v. On account of these differences it usually turns out that the function considered here tends to oc for 0, whereas log Mn usually tended to oc for n--* co.

Let us begin our examination of h(i;) by verifying the statement just made about its behaviour for Lemma. If M(v) -+ oo for v -4 oc,

oo for -+ 0.

Proof. Take any vo. Then, if 0 < < 1M(vo)/vo,

M(vo) - vo > i M(vo).

Q.E.D.

The function

sup(M(v) v>o

as the supremum of decreasing functions of l;, is decreasing. As the supremum

of linear functions of g, it is convex. The upper supporting line of slope to the graph of M(v) vs v has ordinate intercept equal to h(g):

h(t)

v

Figure 84

326

VII D Volberg's work

From this picture, we see immediately that

M*(v) = inf (h(g) + v) C>0

is the smallest concave increasing function which is > M(v). Therefore, if M(v) is also concave, M*(v) = M(v). We will come back to this relation later on. Here is a graph dual to the one just drawn:

0

Figure 85

We see that M*(v) is the ordinate intercept of the (lower) supporting line to the convex graph of h(g) having slope - v. Volberg's construction depends in an essential way on a theorem of Dynkin, to be proved in the next article, which requires concavity of the function M(v). Insofar as inequalities of the form la-.J <, e-M(n)

are concerned, this concavity is pretty much equivalent to the cruder property that M(v)/v be decreasing. It is, first of all, fairly evident that the concavity of M(v) makes M(v)/v decreasing (and even strictly decreasing, save in the trivial case where M(v)/v const.) for all sufficiently large v. We have, in the other direction, the following Theorem. Let M(v) be > 0 and increasing for v > 0, and denote by M*(v) the smallest concave majorant of M(v). If M(v)/v is decreasing.

M*(v) < 2M(v).

2 The function M(v) - its Legendre transform

327

Problem 14(a) Prove this result. (Hint: The graph of M*(v) vs v coincides with that of M(v), save on certain open intervals (a,,, on each of which M*(v) is linear, with and

Figure 86

may, of course, be disposed like the contiguous intervals to the The Cantor set, for instance. Consider any one of them, say, wlog, (a,, b1):

M(v)

0

Figure 87

a,

b,

v

328

VII D Volberg's work For a, 5 v < b,, (v, M(v)) must lie above the broken line path APB, and (v, M*(v)) lies on the segment AB. Work with the broken line path AQR, where OR is a line through the origin parallel to AB.)

Because of this fact, the Fourier coefficients a" of a given function which satisfy an inequality of the form

la-"I <

e-M("),

n > 1,

with an increasing M(v) > 0 such that M(v)/v decreases also satisfy la-nl <, e-M*(n)/2

n> 1

with the concave majorant M*(v) of M(v). Clearly, Ei M*(n)/n2 = oc if Y_;M(n)/n2 = co. This circumstance makes it possible to simplify much of the computational work by supposing to begin with that M(v) is concave as well as increasing.

A further (really, mainly formal) simplification results if we consider only functions M(v) for which M(v)/v -+0 as v -+ oo (see the next lemma). As far as Volberg's work is concerned, this entails no restriction. Since we

will be assuming (at least) that M(v)/v is decreasing, lim,,-,(M(v)/v) certainly exists. In case that limit is strictly positive, the inequalities

Ia-"I

<,

e-M(n),

n> 1,

imply that 00

F(z) _ Y, anz" - 00

is analytic in some annulus {p < I z I < 1}, p < 1. This makes it possible for us to apply the theorem on harmonic estimation (§B.1), at least when F(z) is continuous up to { I z I = 1 } (which will be the case in our version of Volberg's result). We find in this way that JIogF(e19)1d9

> - oo

unless F(z) - 0, using a simple estimate for harmonic measure in an annulus. (If the reader has any trouble working out that estimate, he or she may find it near the very end of the proof of Volberg's theorem in article 6 below.) The conclusion of Volberg's theorem is thus verified in the special case that limv-,,(M(v)/v) > 0. For this reason, we will mostly only consider functions M(v) 0 in the present §. for which

2 The function M(v) - its Legendre transform

329

Once we decide to work with concave functions M(v), it costs but little to further restrict our attention to strictly concave infinitely differentiable M(v)'s. Given any concave increasing M(v), we may, first of all, add to it a bounded strictly concave increasing function (with second derivative < 0

E

V

Figure 88

on (0, oo)) whose graph has a horizontal asymptote of height E, and thus obtain a new strictly concave increasing function M1(v), with M;(v) < 0, differing by at most c from M(v). We may then take an infinitely differentiable positive function cp supported on [0, 1] and having f otp(t)dt = 1, p 0-

0

1

t

Figure 89

and form the function 1

M2(v) =

"

h 0

M1(v + t)cp(T/h) dT,

using a small value of h > 0. M2(v) will also be strictly concave with

330

VII D Volberg's work

MZ(v) < 0 on (0, oc), and increasing, and infinitely differentiable besides for 0 < v < oc. It will differ by less than e from M1(v) for v >, a when a is any given number > 0, if It > 0 is small enough (depending on a). That's because

0<M'1(v)<M'1(a), a. Our function M2(v), infinitely differentiable, increasing, and strictly concave, thus differs by less than 2e from M(v) when v is large. This, however, means that h2() = supv>o(M2(v) - vi;) differs by less than 2e from sup (M(v) - vl;) v>0

for small values of > 0, the suprema in question being attained for large values of v if

is small:

I)

Figure 90

Hence, in studying the order of magnitude of for near zero (which is what we will be mainly concerned with in this §), we may as well assume to begin with that M(v) is strictly concave and infinitely differentiable.

When this restriction holds, one can obtain some useful relations in connection with the duality between M(v) and h(c).* Lemma. If M(v) is strictly concave and increasing with M(v)/v -+0 for v -+ oo, there is for each i; > 0 a unique v = such that M(v) has a derivative for > 0, and

Proof. Since M(v)/v -* 0 as v - oo, the supporting line of slope c to the graph of M(v) vs v does touch that graph somewhere (see preceding diagram), say at (v1, M(v1)). Thus, M(vt) - v1

.

* In the following 3 lemmas, it is tacitly assumed that > 0 ranges over some small interval with left endpoint at the origin, for they will be used only for such values of . This eliminates our having to worry about the behaviour of M(v) for small v.

2 The function M(v) - its Legendre transform

331

Suppose that v2 0 v, and also M(v2) v2

> v, . Then

M(v2) = M(v1) + (v2 - v1) Therefore, for v, < v < v2, by strict concavity of M(v),

M(v) > M(v1) + (v - v1), i.e.,

M(v) - vl; > M(v1) - v1 = This, however, contradicts the definition of with

so there can be no v2

v1

M(v2) - v2b

Since M(v) is already concave, it is equal to its smallest concave majorant, M*(v), i.e.,

M(v) = inf (h(1;) + l;v).

>o is convex, so if it does not have a derivative at a point The function o > 0, it has a corner there, with two different supporting lines, of slopes - v1 and - v2, touching the graph of

h

()

M*(v2)

M*(vl)

to

Figure 91

t

332

VII D Volberg's work

Those two supporting lines have ordinate intercepts equal to M*(v,) and to M(v,) and M(v2). But then h(i;o)=M(v,)-v,l;o= M(v2) - v2o, which we have already seen to be impossible. h'(co) must M*(v2), i.e.,

-

therefore exist, and it is now clear that derivative must have the value the slope of the unique supporting line to the graph of vs at the point (co,

Lemma. If M(v) is differentiable and strictly concave and M(v)/v --* 0 for

V -oo, dM(v) dv

=

for v = v()

.

Proof. is the abscissa at which the supporting line of slope graph of M(v) vs v touches that graph.

to the

Recall that, for the strictly concave functions M(v) we are dealing with here, we actually have M"(v) < 0 on (0, oc) - refer to the above construction of M,(v) and M2(v) from M(v). Lemma. If M(v) is twice continuously differentiable and M"(v) < 0 on (0, oo), exists for l; > 0. and if M(v)/v --* 0 for v -* oo,

Proof. M(v) is certainly strictly concave, so, by the preceding two lemmas, exists and we have the implicit relation

M'( Since M"(v) exists, is continuous, and is < 0, we can apply the implicit function theorem to conclude that exists and equals - 1/M"(-h'(1;)). Volberg's construction, besides depending (through Dynkin's theorem) on the concavity of M(v), makes essential use of one additional special property, namely, that

-K < for some K > 1 as --* 0. Let us express this in terms of M(v). Lemma. For concave M(v), the preceding boxed relation holds with some

K>lfor -*Oiff M(v) >, const.vKI("') for large v. Proof. Since M(v) is concave, it is equal to inff> o vl;). If the boxed relation holds and v is large, this expression is >, inff> o (const.l; -K + v)

2 The function M(v) - its Legendre transform

333

whose value is readily seen to be of the form const.v'/(K+1) To go the other way, compute

supv>0(const.vKi(K+1)_v0)

Remark. One might think that the concavity of M(v) and the fact that 00

M(n)/n2 = co together imply that M(v) v° with some positive p (say p = Z) for large v. That, however, is not so. A counter example may easily be constructed

by building the graph of M(v) vs v out of exceedingly long straight segments chosen one after the other so as to alternately cut the graph of v° vs. v from below and from above.

Here is one more rather trivial fact which we will have occasion to use. Lemma. For increasing M(v), M(O) for

and hence lim,_

> 0

is finite if M(0) > - oo.

Proof. h(g) is decreasing, so lime,,h(g) exists, but is perhaps equal to - oo. The rest is clear. was The principal result on the connection between M(v) and published independently by Beurling and by Dynkin in 1972. It says that, if a > 0 is sufficiently small (so that log h(I;) > 0 for 0 < < a), the converd is equivalent to that of f 1 (M(v)/v2) dv (compare with gence of r o log the material in §C of Chapter IV). More precisely:

Theorem. If M(v) is increasing and concave, and

h(c) = sup (M(v) v>o

there is an a > 0 such that log

d < oo

J0a

iff °° M(V)

J

dv < co.

1

Proof. In the first place, if limy ,,,M(v)/v = c > 0, the function h(i;) = supv, 0(M(v) - v[;) is infinite for 0 < < c. In this case, the integrals involved in the theorem both diverge. For the remainder of the proof we may thus suppose that M(v)/v -> 0 as v -+ oo. Again, by the first lemma of this article,

oo for -> 0 unless M(v) is

334

VII D Volberg's work

bounded for v -+ oo, and in that case both of the integrals in question are

obviously finite. There is thus no loss of generality in supposing that oo for c -+ 0, and we may take an a > 0 with h(a) > 2, say. These things being granted, let us, as in the previous discussion, approximate M(v) to within e on [A, oo), A > 0, by an infinitely differentiable strictly concave function ME(v), with ME (v) < 0. If e > 0 and A > 0 are small enough, the corresponding function sup (ME(v) v>o

approximates h(l:) to within 1 unit (say) on (0, a]. But then f0a

dg

fo" log

and

log hg(%) d

converge simultaneously, and the same is true for the integrals (' °°

M(V) v V

dv

and

M ZV) dv.

fl'o V It is therefore enough to establish the theorem for ME(v) and in other words, we may, wlog, assume to begin with that M(v) is infinitely differentiable and strictly concave, with M"(v) < 0, and that M(v)/v ---+0 for v--* oo. J

1

In these circumstances, we can use the relations furnished by the instead of preceding lemmas. It is convenient to work with log I I

log h(g), so for this purpose let us first show that f0a

d and

log

f0a

log I h'(c) I d

converge simultaneously. First of all,

h(a) + (a - )Ih'(i;)I < h(a) + ajh'(g)I by the convexity of h

()

0

Figure 92

for 0<
2 The function M(v) - its Legendre transform h(g)

335

Therefore convergence of the second integral implies that of the first. Again,

for 0 < < a,1

>,2, so

f>

h1

2

for such : h ()

t

t/2

Figure 93

So, since Jo I log I dl < oo, convergence of the first integral implies that of

the second. We have

h(c) = and

with

d log I

I

Therefore

=

v()

Taking a number b, 0 < b < a, and integrating by parts, we find that = a log h'(a) I- b log I h'(b) I-

log I h'(db

Jb

=

M(v(a))

M(v(b)) + rV(a) M(v)

v(a)

Here,

v(b)

J

V(b)

dv.

V2

is decreasing, so v(b) > v(a). Turning things around, we thus have

f log I h'(@) I d + b log l h'(b) I - a log I h'(a) I a

M(v(b)) v(b)

M(v(a)) + C°(b)M(v) dv. v(a) J v2 v(a)

336

VII D Volberg's work

M(v)/v is decreasing (concavity of M(v) !) and, as b see, then, that

0, v(b) -> oo. We

I d < o0

log l J0a

if

dv < oo

M2 v(a)

v

Also, I h'(l;) I decreases, so b log I h'(b) I 1 v(b)

M(V)

f o log I

I d. Therefore

dv

V

is bounded above for b -+ 0 if f o log I f a)(M(v)/v2)dv < oo. We are done.

I d < oo, i.e.,

Problem 14(b) Let H(l:) be decreasing for l; > 0 with H() - oo for

0, and denote by

h(c) the largest convex minorant of H(l;). Show that, if, for some small a > 0,

r log h(l;) do < oc, then f o log H(1;) dl; < oo. Hint: Use the following o

picture:

H(s)

0

4

Figure 94

Problem 14(c) If M(v) is increasing, it is in general false that Ji (M(v)/v2) dv < oo makes I' (M*(v)/v2) dv < oo for the smallest concave majorant M*(v) of M(v). (Hint: In one counter example, M*(v) has a broken line graph with vertices on the one of v/log v (v large).)

2 The function M(v) - its Legendre transform h(g)

337

be decreasing for 1; > 0 and tend to co as t; - 0. For

Theorem. Let

v>0, put

M(v) = inf

l;v).

C>0

Then

d < co

log J0a

for some (and hence for all) arbitrarily small values of a > 0 iff M2 1

dv < oo.

v

Proof. As the infimum of linear functions of v, M(v) is concave; it is obviously increasing. The function

h(l;) = sup (M(v) - vi;) v>0

is the largest convex minorant of H(i;) because its height at any abscissa is the supremum of the heights of all the (lower) supporting lines with slopes - v < 0 to the graph of H: H(Q) M(v)

e

Figure 95

Therefore f o log co makes f o log h(t;) dg = oo by problem 14(b), so in that case f i (M(v)/v2)dv = oo by the preceding theorem. If, on the other hand, f i (M(v)/v2)dv does diverge, fo log h(l;) d = oo for each small enough a > 0 by that same theorem, so certainly ro log dg = oo for such a. This does it.

338

VII D Volberg's work

3.

Dynkin's extension of F(e'5) to {I z I < 1) with control on I F,{z) I

As stated near the beginning of the previous article, a very important role in Volberg's construction is played by a weight w(r) > 0 defined for 0 < r < 1 by the formula

w(r) =

exp(-h(log')), r

where, for 1; > 0,

h(l;) = sup (M(v) - vl;). v>o

Here M(v) is an increasing (usually concave) function such that I' (M(v)lv2) dv = oo; this makes h(g) increase to oo rather rapidly as decreases towards 0, so that w(r) decreases very rapidly towards zero as r -+ 1.

A typical example of the kind of functions M(v) figuring in Volberg's theorem is obtained by putting

M(v) --

V

log v

for v > e2, say, and defining M(v) in any convenient fashion for 0 < v < e2 so as to keep it increasing and concave on that range. Here we find without difficulty that 2

e

eli4

l; - +0,

for

and w(r) decreases towards zero like exp

(1

\-

- r)2 elul -r) e

as r -> 1; this is really fast. It is good to keep this example in mind during the following development.

Lemma. Let M(v) be increasing and strictly concave for v > 0 with M(v)/v --+ 0 for v -> oo, put

sup(M(v)-vl;), v>o

and write w(r) = exp (- h(log (1/r))) for 0 < r < 1. Then 1

f

o

forn>, 1.

r"+2w(r)dr > const. e-M(") n

3 Dynkin's extension of F(e'B) to the unit disk

339

Proof. In terms of = log (1/r), r"w(r) = exp (- h(g) - en). Since M(v) is strictly concave, we have, by the previous article, inf

>o

i;n) = M(n),

the infimum being attained at the value c =1;n = M'(n). Put r" = e ". Then, r.nw(rn) = e-M("). Because w(r) decreases, we now see that r

rn+2w(r)dr > w(rn)

rn+2dr 0

rn3

n+3

n+3

(r"w(rn))

a-M(n)

Here, rn3 = e-3M'(n)

and this is > e-3M'(1) since M'(v) decreases, when n >, 1. From the previous relation, we thus find that e - 3M'(1)

01

f rn+2w(r)dr

4n

CM(n)

Q.E.D.

for n >, 1,

Theorem (Dynkin (the younger), 1972). Let M(v) be increasing on (0, 00),

as well as strictly concave and infinitely differentiable on (0, 00), with M"(v) < 0 there and M(v)/v -+ 0 for v -+ oo. Let M(0) > - oo. For 0 < r < 1, put

w(r) = exp (- h(log 111,

\

r/JJ/

is related to M(v) in the usual fashion. Suppose that

where

00

F(eis)

E

ane"n8

00

is continuous on the unit circumference, and that 00

YIn2a-nleM(") < 00. 1

Then F has a continuous extension F(z) onto {IzI < 1) with F(z) continuously differentiable for I z I < 1 and 18F(z)/ez 15 const.w(I z 1),

I z I < 1.

Remark. The sense of Dynkin's theorem is that rapid growth of M(n)

340

VII D Volberg's work

to co for n - oo (which corresponds to rapid growth of

to co for c

tending to 0) makes it possible to extend F continuously to { I z (< I } in such a way as to have I OF(z)/8z I dropping off to zero very quickly for I z I -> 1.

Proof of theorem. We start by taking a continuously differentiable function f2(e"), to be determined presently, and putting* (*)

°°

G(z) = Y anz" +

1

27c

2i 0

1 S2(e")r'w(r)rdr dt Jo

re" - z

for I z I z 1. The reason for using the factor r 2 with w(r) will soon be apparent. The idea now is to specify S2(e") in such fashion as to make G(e's) have the

same Fourier series as F(e's). If we can do that, the function G(z) will be a continuous extension of F(e's) to { I z I < 11.

To see this, observe that our hypothesis certainly makes the trigonometric series ane'ns 00

absolutely convergent, so, since F(e's) is continuous, 00

Y_ anBins 0

must also be the Fourier series of some continuous function, and hence the power series on the right in (*) a continuous function of z for I z I < 1. According to a lemma from the previous article, the property M(0) > - 00 makes

bounded below for > 0 and hence w(r) bounded above in (0, 1). The right-hand integral in (*) is thus of the form

if(' 2n

b(C) d drl

1c1
C-z

with a bounded function b(C). (Here, we are writing C _ + it/ which conflicts with our frequent use of to denote log (1/r). No confusion should thereby

result.) It is well known that such an integral gives a continuous function of z; that's because it's a convolution on 1182, with

d dt1 ICI
KI

finite for each finite R. We see in this way that the function G(z) given by (*) will be continuous for I z I 5 1. If also the Fourier series of G(e's) and F(e's) coincide, those two functions must obviously be equal.

We wish to apply the theorem of article 1 to the right-hand integral in (*). In order to stay honest, we should therefore check continuous * The power series on the right in (*) may not actually be convergent for IzI = 1, but does represent a continuous function for Izl _< 1, as will be clear in a moment.

3 Dynkin's extension of F(e'B) to the unit disk

341

differentiability (for ICI < 1) of the function b(C) figuring in that double integral, viz., b(re") = r2w(r)S2(e"),

because continuous differentiability (at least) is required in the hypothesis of the theorem. It is for this purpose that the factor r2 has been included; at 0. Thanks to it, the desired that factor ensures differentiability of property of b(C) follows from the continuous differentiability of S2(e") together with the continuity of rw'(r) on [0, 1) which we now verify. We have rw'(r) = h'(log (1/r))w(r) for 0 < r < 1. By a lemma in the previous article, h"(i;) exists for each > 0 since M"(v) < 0. This certainly makes h'(log (1/r)) continuous for 0 < r < 1, so rw'(r) is continuous for such r. When r -> 0, w(r) increases and tends to a finite limit (since M'(0) > - co), and h'(log(1/r)) increases (convexity of remaining, however, always 0. Hence rw'(r) tends to a finite limit as r -> 0, and (with obvious definition of rw'(r) at the origin) is thus continuous at 0. Having justified the application of the theorem from article 1 by this rather fussy argument, we see through its use that aG(z)

is

2

_ - Izl w(Izl)z(e )

Of for

Izl < 1

z = Izle'9

after taking account of the fact that

(anzu') Y- an Zn

)

0

0,

IZI < 1.

=

This relation certainly makes aG(z)

const.w(I z l )

of

for I z I < 1, so, if G(z) does coincide with F(z) for I z I = 1, we will have the theorem on putting F(z) = G(z) for I z I < 1. Everything thus depends on our being able to determine a continuously differentiable S2(e") which will make

()

1

2i

2lt J o

1 S2(e")r2w(r)rdrdt

f

o

have the Fourier series

-00

re " _ e 1s

342

VII D Volberg's work

Expanding the integrand from (*K) in powers of re", we obtain for that expression the value f2x e '' 00 1

-

n

Y1(f 1 rn+2w(r)drl( o

in9

o

We see that we need to have W

f2(e") , Y bne'n` - Co

where, for n > 1,

(t)

bl-n - - rorn+a-n

We may choose the bn with positive index in any manner compatible with the continuous differentiability of Q(e"); let us simply put them all equal to zero. By the lemma, the right side of (t) is in modulus 5 const.Ina_nIeM(n)

for n >, 1. The b,n given by (t) therefore satisfy 0

Y I mb, I

< oo

-Co

according to the hypothesis of our theorem. This means that there is a function f2(e") satisfying our requirements whose differentiated Fourier series is absolutely convergent. Such a function is surely continuously differentiable; that is what was needed. The theorem is proved. Remark 1. We are going to use the extension of F to { I z I < 1 } furnished by Dynkin's theorem in conjunction with the corollary at the end of article 1. That corollary involves the integral 1

2n

FCQ dl; drl

(l; - z)

where, on the open set -9, I

I > 0 and I F{C) I <, const.I F(l;)1. The theorem

of article 1 is used to take a/8f of this integral, and the hypothesis of the corollary requires that F(C) be '2 in -9 in order to guarantee the legitimacy of that theorem's application. Our extension F(z) furnished by Dynkin's theorem is, however, only ensured to be' ' for IzI < 1. Are we not in trouble?

Not to worry. All that the corollary really uses is continuous differentiability of the quotient Fc)/F(t) in -9. Our F is, however, W, in -9, where

4 Weighted planar approximation by polynomials

343

it is also 0 0. And, from the proof of Dynkin's theorem, FJC) = - I K I2w(I K I)1(eit) for C = I t; I e" with I C I < 1. The expression on the

right is, however, ', for I C I < 1; we indeed checked that it had that property during the proof (as we had to do in order to justify using the theorem of article 1 to show that FC(C) was equal to it!). We are all right.

Remark 2. Under the conditions of Volberg's theorem, there is no essential distinction between the functions M(v) and 2M(v), and em(v) goes to infinity much faster than any power of v as v -> oo. (We will need to require that M(v) > const. v" for some a > i as has already been remarked in article 2.) In the application of Dynkin's theorem to be made below, we will therefore be able to replace the condition 00

oo 1

figuring in its hypothesis by const.e-21(n) ,

n >, 1,

or even (after a suitable unessential modification in the description of w(r)) by

Ia-.J < const.e-M("),

4.

n > 1.

Material about weighted planar approximation by polynomials Lemma. Let w(r) > 0 for 0 < r < 1, with

w(r)r dr < oo. fo,

If F(z) is any function analytic in { I z I < 1} such that I F(z)12w(Iz I) dx dy < oo, JJx1 < 1

there are polynomials Q(z) making

if

I F(z) - Q(z)IZw(Iz1)dxdy

Izi <1

arbitrarily small.

Proof. The basic idea is that w( I z I) depends only on the modulus of z.

344

VII D Volberg's work

Givens > 0, take p < 1 so close to 1 that I F(z)12w(I z l) dx dy < E. JJp
Note that if 0 < .1 < 1 and 0 < r < 1, 2n

2n

I F(rei')12 d9.

I F(.1re"s)12 d9 0

o

Therefore, JrPJ1

IF(.1reis)12w(r)rd9dr

I

IF(2z)12w(Iz1)dxdy = JJp
o 1

2" v

I F(rei9)12w(r)r d9 dr < C.

o

Once p < 1 has been fixed, F(z) is uniformly continuous for I z I < p, so f f I=I

Jjzl <1

I F(z) - F(.1z) I2w(I z I) dx dy < 5s.

The Taylor series for F(.1z) converges uniformly for IzI51. We may therefore take a suitable partial sum Q(z) of that Taylor series so as to make I F(.1z) - Q(z)12w(I z l) dx dy < B. J'.czI <1

Then F(z) - Q(z) I2w(I z l) dx dy < 16s. Izl < 1

That does it.

At this point, we begin to make systematic use of a corollary to the theorem of Levinson given in §A.5. Oddly enough, Beurling's stronger results from §B are never called for in Volberg's work. Theorem on simultaneous polynominal approximation (Kriete, Volberg). Let,for 0 < r< 1,

w(r) = exp(- H(log 1 r

\\ I

I

4 Weighted planar approximation by polynomials

345

where H(1;) is decreasing and bounded below on (0, oo), and suppose that

d = oo

log J0a

for all sufficiently small a > 0.

Let E be any proper closed subset of the unit circumference, let p(e'o) e L2(E), and suppose that f (z) is analytic in { I z I < 11, and such that .

f(z)12w(Izl)dxdy < oo.

Then there is a sequence of polynomials

Ip(e;s)-Pn(eis)I2d9+

J

with

ifIf(z)-P,,(z)I2w(Iz1)dxdy -+ 0.

E

zI < 1

Proof. We use the fact that the collection of functions F(z) analytic in {IzI < 1} with $$121<11F(z)I2w(IzI)dxdy < oo forms a Hilbert space if we

bring in the inner product

. = ffzl
F(z)I2w(Iz1)dxdy < L form a normal family in the open unit disk. However, the weight w(r) we are using is strictly positive and decreasing on (0, 1). Hence, for any r < 1, the previous relation makes

if

F(z) I' dx dy < zI
r

)).

It is well known that such functions F(z) form a normal family in {I z I < r}.

Here, r < 1 is arbitrary. Let us turn to the proof of the theorem, reasoning by duality in the Hilbert space L2(E) Q Yt° where 0 is the Hilbert space just described.* Suppose, then, that there is a p(e''9) e L2(E) and an f (z) e Y for which the conclusion of the theorem fails to hold. There must then be a non-zero element (q, F) of L2(E) $" orthogonal in that space to all the elements of the form (P(e''9), P(z)) with polynomials P. We are going to obtain a contradiction by showing that in fact q = 0 and F = 0. * We are dealing here with the direct sum of L2(E) and .'.

346

VII D Volberg's work

The orthogonality in question is equivalent to the relations

d9 + J

f

6E

f

F(z) z"w(I z l) dx dy = 0,

n = 0, 1, 2,3 .....

z1 < 1

Define q(ei9) for all of { I z I = 1} by making it zero for ei9 0 E. Then q(e19)eL2( - it, it), and if we write (*)

anein9

q(e`9) ,., 00

we find from the previous relation that 2n

a"

f fz1<1 F(z) z"w(I z l) dx dy,

n > 0.

Since

ft

IF(z)I2w(Iz1)dxdy < oo, zI<1

the integral on the right is in modulus const.

fo, r 2nw(r)r dr

by Schwarz' inequality. However, in terms of = log (1/r) and the function

r2nw(r) = e-(H()t2n) l; v) by M(v) as in the last theorem of article 2, we Denoting info, o see that the right side is < e-1(2n). The preceding expression is therefore 5 const.e - Mc2">i2, i.e.

(*)

IanI <, const.e-M(2")/2,

n, 1.

Since E is a proper closed subset of { I z I = 1 }, its complement on the unit

circumference contains an arc J of positive length. The function q(ei9) vanishes outside E, hence on J, and certainly belongs to L1(- it, it). Also, by the last theorem in article 2, f °° M(V) 1

dv = ao

v

on account of our hypothesis on

Therefore

4 Weighted planar approximation by polynomials

347

M(v) being increasing, so, by virtue of the corollary at the end of §A.5, (*) and

(*) imply that q(ei9) = 0 a.e. We see that a" = 0 for all n, which means that

f fizi<1 F(z) z"w(I z l) dx dy = 0 for n = 0, 1, 2,.... Since H(l;) is bounded below on (0, oo), w(r) is bounded above for 0 < r < 1, and we can invoke the lemma, concluding that polynomials are dense in the Hilbert space. The previous relation therefore implies that F(z) _- 0. We have thus reached a contradiction by showing that q = 0 and F = 0. The theorem is proved.

Remark. Some applications involve a weight

w(r) = where, for

exp(-h(log ))

> 0,

h(1;) = sup(M(v) - vl;), V>o

the function M(v) being merely supposed increasing, and such that M(0) > - 00. In this situation, we can, from the condition M(n) z n

conclude that the rest of the above theorem's statement is valid.

This can be seen without appealing to the last theorems of article 2. We have here, with =log(1/r), rz"w(r) = e-(h( )+z"C)

and, since, for any l; > 0, M(v) - vl;

for each v > 0, h() + 2n1 > M(2n). We now arrive at (*) in the same way as above, so, since M(v) is increasing, M(n) 1

n

= oo makes

00 M( 2n) i 2n

= 00

and we can conclude by direct application of the corollary from §A.5. (Here,

boundedness of w(r) is ensured by the condition M(0) > - oo.)

348

VII D Volberg's work

Remark on a certain change of cariable

If the weight w(r)=exp(-H(log(1/r))) satisfies the hypothesis of the theorem on simultaneous polynomial approximation, so does the weight

w(r')=exp(-H(Llog(1/r))) for any positive constant L. That's simply because ra o

d=

1

raL

Lo

d

That theorem therefore remains valid if we replace the weight w(r) figuring in its statement by w(rL), L being any positive constant.

We will use this fact several times in what follows. 5.

Volberg's theorem on harmonic measures

The result to be proved here plays an important role in the establishment of the main theorem of this §. It is also of interest in its own right.

Definition. Let (9 be an open subset of { I z < 11, and J any open arc of { I z I = 11. We say that (9 abuts on J if, for each eJ, there is a neighborhood V" of C with

V;n{IzI < l} c (9.

Figure 96

5 Volberg's theorem on harmonic measures

349

Now we come to the Theorem on harmonic measures (Volberg). Let, for 0 < r < 1, w(r) = exp (- H(log (1/r))), where is decreasing and bounded below on (0, oo), and tends to 0o sufficiently rapidly as i; -* 0 to make w(r) = O((1- r)2) for r-* 1. (In the situation of Volberg's theorem, we have H(1;) > const. -` with c > 0, so this will certainly be the case.) Assume furthermore that

log H(l;) dl; = oo J0a

for all sufficiently small a > 0. Let 0 be any connected open set in { I z I < 1) whose boundary is regular enough to permit the solution of Dirichlet's problem for (9. Suppose that there are two open arcs I and J of positive length on { I z I = 1 } such that:

(i) 80 n J is empty; (ii) C abuts on I. Then, if ooo( ,z) denotes harmonic measure for (9 (as seen from ze(9), we have

log

dco (t;, z0) = oo

w(ICI)

nd[r

{

1

for each z0 e (9.

Remark 1. The integral is taken over the part of 8(9 lying inside { I z I < 11.

Remark 2. The assumption that (9 abuts on an arc I can be relaxed. But the proof uses the full strength of the assumption that 8(9 avoids J. Proof of theorem. We work with the weight wl(r) = w(r). By the theorem on

simultaneous polynomial approximation and remark on a change of variable (previous article), there are polynomials Pa(z) with IP,,(e's)I2di ---* 0 {ICI=1}

and at the same time

JJ

P(z)- 112wi(Izl)dx dy

0.

zI < 1

The second relation certainly implies that

JizHl for some C < oo, and all n.

C

350

VII D Volberg's work

Take any zo, I zo I < 1; we use the last inequality to get a uniform upper estimate for the values IP,,(zo)I. Put p = 20 - IzoI).

Figure 97

We have lPn(ZO)l2

'<

np2JJlz-z

I
wl(r) decreases, so the right side is 1

S

_P) ffz-zol
IPn(z)IZwi(Izl)dxdy

which, in turn, is C

np2wi(1 - p)

by the above inequality.

Here, w1(1-p)=w(1-3p+3p2-p3) is > w(1-2p)=w(Izo1) when 3p2 - p3 < p, i.e., for p < (3 - J5)/2. Also, w(r) is bounded above (H(1;) being bounded below), so, for (3 - .x/5)/2 < p < z, w1(1 - p) ,>w(,/5 - 2) ,>const.w(1 zo l ). The result just found therefore reduces to

I Pn (z0 )I2 <

const. (1 - IZOI)2w(IZO1)'

5 Volberg's theorem on harmonic measures

351

with the right-hand side in turn const.

\ (w(zo))2

according to the hypothesis. Thus, since zo was arbitrary, logI w(Izl) I,

(*)

A similar (and simpler) argument, applied to (*)

P"(z) -' 1,

IzI<1. 1, shows that

IzI < 1.

Let us now fix our attention on 06) n { 12; I = 1), which we henceforth denote by S, in order to simplify the notation. The open unit disk A includes 0, therefore, by the principle of extension of domain (see §B.1), for zoe(9, dco (C, zo) < dcon(C, zo)

for 1' varying on S. In other words,

(t)

dwo(C, zo) < K(zo) I dC j

for l; on S with a number K(zo) depending only on zo. Since (9 abuts on the arc I with III > 0, we surely have

wo(S, zo) > 0

for each zoe(9 by Harnack's theorem. Since S c {II;I = 1}''J ((i) of the hypothesis), we have, by (t) and the relation between arithmetic and geometric means,

Is

log I

I door,(eis,

zo)

iwo(S,zo)log1 wo(S

iwo(S,zo)logGO'M S zo)J{I{I= t} _

for zoe(9. This last expression, however, tends to - oo as n -oo because

I{ICI=i}

IP,,(ec9)12d9 n

)0

At the same time, for any zoe(9,

n b 0 by (,*k). Therefore, by the theorem on harmonic estimation in §B.1 (whose extension to possibly

352

VII D Volberg's work

infinitely connected domains 0 of the kind considered here presents no difficulty, at least for polynomials we see that Pn(e;s) I

log I

log I

dwo(e's, zo) + J

JS

I dcoo(t;, zo)

0 ne

f

-n 0.

log e

As we have just shown, the first of the two integrals in the left-hand member tends to - oo as n -+ oo. Hence the second must tend to oo as n -+ oo (!). However, by (s), log

1

w(IM

n0. So we musthave for Ce8(9LnA log(WOM) dcoc(1',zo) = oo for

zoE(9.

The theo rem is proved.

Remark. The result just established holds in particular for weights w(r) = exp (- h(log (1/r))) with h(l;) = sups, o(M(v) - vl;) for l; >O, M(v) being increasing, provided that M(0) > - oo, that Z'(M(n)/n2) = oo, and that M(v) -+ oo as v -a oo fast enough to make w(r) = O((1 - r)2) for r -.1. See remark following the theorem on simultaneous polynomial approximation (previous article).

Corollary. Let the connected open set 0 and the weight w(r) be as in the theorem (or the last remark). Let G(z) be analytic in (9 and continuous up to 80, and suppose that, for some p, 0 < p < 1, we have I

I < w(II) for t' e 80 with 1 -p < ICI < 1

(sic!).

Then G(z) - 0 in (9. Proof. Take any zoe(9. Since G(z) is continuous on (9, it is bounded there, so, since w(r) decreases, we surely have G(z) < const.w(I z I) for ze(9 and I z I < 1 - p. Therefore our hypothesis in fact implies that I G(C) l

< Cw(I

I)

for

1' e 80 n { I I; 1<11

with a certain constant C. Write 0(9 n { I C I < 1 } = y, and denote the intersection 00 - y of 00 with the unit circumference by S as in the proof of the theorem. By the theorem on harmonic estimation (§B.1), we have(' logIG(zo)I <, J logIG(C)Idwo(C,zo)+ J S

y

5 Volberg's theorem on harmonic measures

353

If M is a bound for G(z) on a, the first integral on the right is < log M. The second is

log C + f logw(I lI) dcoo(C, zo) y

by the above inequality. The integral just written is, however, equal to - oo by the theorem on harmonic measures. Hence G(zo) = 0, as required. Scholium. L. Carleson observed that the result furnished by the theorem on harmonic measures cannot be essentially improved. By this he meant the following:

If w(r) = exp ( - h(log (1/r))) with bounded below on (0, oc), and if

fa log

d

strictly decreasing, convex, and

< o0

0

for all sufficiently small a > 0, there is a simply connected open set t9 in t = { I z I < 1) fulfilling the conditions of the above theorem for which log(w(IM)dw°G,zo) < oo,

ZOe 9.

To see this, observe that the convergence of f o log

dl; for all

sufficiently small a > 0 implies that

loglh'(l;)ldc < oo

(§) J0a

for such a. (See the proof of the second theorem in article 2.) We use (§) in

order to construct a domain d like this

Figure 98

354

VII D Volberg's work

for which

L

,1 ae' nA

log( WO'C 1)

)dcoo(C, z0) < co,

zoe(9,

with w(r) = exp (- h(log (1/r))). It is convenient to map our (as yet

undetermined) region (9 conformally onto another one, .9, by taking z = re's to cp = i log (l/z) = 9 + i log (1/r) = 9 + il;. Here has its usual significance.

Figure 99

If, in thisL mapping, the point z0-9 goes over to pee, we have, clearly, log( ne

1

\w(ICI)

)dcozo)

=f

p)

We see that it is enough to determine the equation l; = l;(9) of the upper bounding curve of .9 (see picture) in such fashion as to have ('a

il;(9), p) < oo 0

when pet. The easiest way to proceed is to construct a function (9) = l;(oc - 9), making the upper bounding curve symmetric about the vertical line through its midpoint. Then we need only determine an increasing

function (9) on the range 0 <, 9 < a/2 in such a way that ('Bo

h(l;(9))dw.9(9 + il;(9), p) < oo 0

for some 00 > 0 and some pE2i. From this, the same inequality will follow for every pet by Harnack's theorem, and we can arrive at the full result by adding two such integrals.

355

5 Volberg's theorem on harmonic measures

Figure 100

For 0 < 0 < 00, we have, integrating by parts, ao

p) = h(c(60))w(O0) -

J e

eo

I

w(9) denotes the harmonic measure (at p) of the segment of the upper bounding curve having abscissae between 0 and 9, viz.,

co(9) =

dog.q(i + ii(i), p). J0

Making 6 --* 0 and remembering that want is

decreases, we see that what we

00

(§§)

oo.

I

0

For co(9) we may use the Carleman-Ahlfors estimate for harmonic measure in curvilinear strips, to be derived in Chapter IX.* According to that, if p is fixed and to the right of 00,

w(9) < const.exp

n

feo di l

s(i)I

for 0 < 9 < 00. The most simple-minded way of ensuring (§§) is then to cook * See Remark 1 following the third theorem of §E.1 in that chapter. The upper bound arrived at by the method explained there applies in fact to a harmonic measure larger than w(9).

356

VII D Volberg's work

the positive increasing function l(r) so as to have

log I h'(g(9)) I- f9 0

(T)

=

const.

It is the relation (§) which makes it possible for us to do this. In order to avoid being fussy, let us at this point make the additional (and not really restrictive) assumption that h'(c) is continuously differentiable.

Then we can differentiate the previous equation with respect to 9, getting

d

d

d9

for our unknown increasing function

(9). Calling

the inverse to

the function (9), we have

d

n

dg

In view of (§), this has the solution

90 = 1 71

(log I h'(t) I - log I

i )dt

o

with 9(0) = 0. Since h'(t) is < 0 and increasing (i.e., log I h'(t) I decreases), we

see that the function 9(g) given by this formula is strictly increasing, and therefore has an increasing inverse j(9) for which (§§) holds. This completes our construction, and Carleson's observation is verified. Let us remark that one can, by the same method, establish a version of the Levinson log log theorem which we will give at the end of this § (accompanied, however, by a proof based on a different idea). V.P. Gurarii showed me this simple argument (Levinson's original proof of the log log

.theorem, found in his book, is quite hard) at the 1966 International Congress in Moscow. 6.

Volberg's theorem on the logarithmic integral

We are finally in a position to undertake the proof of the main result of this §. This is what we will establish:* Theorem. Let M(v) be increasing for v >, 0. Suppose that M(v)/v is decreasing, that

M(v) > const.v" * A refinement of the following result due to Brennan is given in the Addendum at the end of the present volume.

6 Volberg's theorem on the logarithmic integral with some a >

z

357

for all large v, and that

M(n) 2 n

Let 00

F(ei9) - Y_ a

ein9

00

be continuous and not identically zero. Then, if

Ia-nI < cm(") for n,>1, we have

J1oIF(e)Id 9> - oo. Remark. Volberg states this theorem for functions F(ei9 )eL, (- n, n).* He replaces our second displayed condition on M(v) by a weaker one, requiring only that v-*M(v) --) 00 for v -> oo, but includes an additional restrictive one, to the effect that v112M(v112) < const.M(v)

for large v. This extra requirement serves to ensure that the function h(g) = sup (M(v) - vl;) v>o

satisfies the relation (h(l;))1-` with some K > 1 and c > 0 for small i; > 0; here we have entirely dispensed with it.

Proof of theorem (essentially Volberg's). This will be quite long. We start by making some simple reductions. First of all, we assume that M(v)/v -> 0 for v -- oo, since, in the contrary situation, the theorem is easily verified directly (see article 2). According to the first theorem of article 2, our condition that M(v)/v

decrease implies that the smallest concave majorant M*(v) of M(v) is S 2M(v); this means that the hypothesis of the theorem is satisfied if, in it, we replace M(v) by the concave increasing function M*(v)/2. There is thus no loss of generality in supposing to begin with that M(v) is also concave. We may also assume that M(0) >, 3. To see this, suppose

that M(0) < 3; in that case we may draw a straight line 2' from (0, 3) tangent to the graph of M(v) vs. v: * See the addendum for such an extension of Brennan's result.

358

VII D Volberg's work

M(v)

3

0

Figure 101

If the point of tangency is at (vo, M(vo)), we may then take the new increasing concave function M0(v) equal to M(v) for v > vo and to the height of £ at the abscissa v for 0 < v < vo. Our Fourier coefficients a will satisfy

Ia_nI S

const.e_M0(n)

for n >, 1, and the rest of the hypothesis will hold with M0(v) in place of M(v).

We may now use the simple constructions of M1(v) and M2(v) given in

article 2 to obtain an infinitely differentiable, increasing and strictly concave function M2(v) (with M2" (v) < 0) which is uniformly close (within

unit, say) to M0(v) on [0, oo ). (Here uniformly close on all of [0, 00) because our present function M0(v) has a bounded first derivative on (0, oo).) We will then still have

Ia_RI <

const.e-M2(n)

for n > 1, and the rest of the hypothesis will hold with M2(v) in place of M(v).

Since M(v) >, const.v°` for large v, where a > 2, we certainly (and by far!) have n4 exp (- M2(n)/2)

0,

Therefore 00

Y- In2a_

n=1

e"2(n)12

< 00.

n - oo.

6 Volberg's theorem on the logarithmic integral

359

So, putting M(v) = M2(v)/2, we have 00

< 00

In2a-nIea

with a function M(v) which is increasing, strictly concave, and infinitely differentiable on (0, 00), having M"(v) < 0 there. The hypothesis of the

theorem holds with M(v) standing in place of M(v). Besides, M(0) (= limv-0M(v)) is > 1 since Mo(0) > 3. Later on, this property will be jo.

helpful technically. Let us henceforth simply write M(v) instead of M(v). Our new function M(v) thus satisfies the hypothesis of Dynkin's theorem (article 3). We put, as usual, sup (M(v) - vc) for v>0

> 0,

and then form the weight

0
w(r) = exp(-h(log 1

I, rJJJJ

Because M(0) >,1, we have h(i;) >,1 for i >O, so w(r),<1/e. Applying Dynkin's theorem, we obtain a continuous extension F(z) of our given 1} with F(z) continuously differentiable in the open

function F(ei9) to { IzI

unit disk and IF(z)

const.w(Izl),

IzI < 1.

of We note here that the properties of M(v) assumed in the hypothesis certainly make w(r) -+ 0 (indeed rather rapidly) as r --> 1; see article 2. Let

Bo = {z: IzI < 1 and IF(z)I < w(Izl)}.

We cover each of the closed sets

Bon{1-n

'<

IzI S 1-n 1

1,

n=1,2,3,...,

by a finite number of open disks lying in {IzI < 11 on which

F(z)I < 2w(Izl) This gives us altogether a countable collection of open disks lying in the unit

circle and covering Bo; the closure of the union of those disks is denoted by B. We then have IF(z)I <, 2w(Izl),

zeB,*

" including for zeB of modulus 1, as long as we take w(1) = 0! See argument for Step 1, p. 361

360

VII D Volberg's work

and

IF(z)I > w(Izl) for z0B and IzI < 1. Put

(9 = {IzI<1}n(-B); (9 is an open subset of the unit disk and I F(z) I > w(I z I) therein, as we have

just seen. We will see presently that (9 fills much of the unit disk. Let us at this point simply observe that (9 is certainly not empty. If, indeed, it were empty, B would fill the unit disk and we would have IF(z)I <2w(Iz1) for I z I < 1. The fact that w(r) -.0 for r -> 1 would then make F(ei9) __ 0, contrary to our hypothesis, by virtue of the continuity of F(z) on { I z I <, 1 } . Although the open set (9 may have an exceedingly complicated structure, the Dirichlet problem for it can be solved. This will follow from a well-known result in elementary potential theory (for a proof of which see, for instance,

pp. 35-6 of Gamelin's book, the latter part of the one by Kellog, or any other work on potential theory - I cannot, after all, prove everything!), if we show that the Poincare cone condition for (9 is satisfied at every point C of 0(9, i.e., if, for each such 1;, there is a small triangle with vertex at lying outside (9. To check this, observe that the small disks used to build up B can accumulate only at points of the unit circumference. Hence, if I R I < 1 and t( a 8(9, l; is on the boundary of one of those small disks, inside of

which a triangle with vertex at may be drawn. If, however, ICJ= 1, we may take a triangle lying outside the unit disk with vertex at . Since IFZ(z)I < Cw(Iz1) in {IzI < 11 while IF(z)I > w(Iz1) in (9, we have

8F(z) F(z) 8z 1

5 C,

ze(9.

Volberg's idea is to take advantage of this relation and use the function (D (z) = F(z) exp {

1

(' ('

27 o

d drl l F(C) ( - z)

. (N.B. Again we are writing t; = + irl, in conflict with the notation = log(1/r) used in discussing h(c).) According to Remark 1 to Dynkin's theorem (end of article 3), F(z) has enough differentiability in {IzI < 1} for us to be able to use the corollary from the end of article 1. By that corollary, (z) is analytic in (9, and IF(z)I lies between two constant on { I z I

1}

multiples of I F(z) I there (and, actually, on { I z I < 11 as the last part of that corollary's proof shows). We thus certainly have I 'D(z) I > 0 in (9 since IF(z)I > w(Iz1) there.

Volberg now applies the theorem on harmonic estimation (§B.1) to

6 Volberg's theorem on the logarithmic integral

361

(D (z) and the open set (9 in order to eventually get at

f"

log I F(ei9) I d9.

His procedure is to show that the whole unit circumference is contained in 0&, and that the part of 80 lying inside the unit disk is so unimportant as to make harmonic measure for (9 act like ordinary Lebesgue measure on the unit circumference. We will carry out this program in 5 steps. Before going to step 1, we should, however, acknowledge that here the theorem on harmonic estimation will be used under conditions somewhat more general than those allowed for in §B.1. The open set (9 may not even be connected, and its components may be infinitely connected! Nevertheless, extension of the theorem in question to the present situation involves

no real difficulty - 1(z) is continuous on { I z I < 1) and the Dirichlet problem for 0 is solvable. We proceed. Step 1. B n { I tC I = 1 } contains no arc of positive length.

Assume the contrary. By construction of B, each of its points on the unit circumference is a limit of a sequence of z,, having modulus < 1 for which I F(zn) I < 2w(I zn l ). Since w(r) - O for r-+ 1 we thus have F(e'9) = 0

for every point of the form ei9 in B.

If now this happens for each point belonging to an are J of positive length on the unit circumference, we will have F(ei9) - 0 on J. At the same time, the Fourier coefficients an of F(ei9) satisfy Ia_nI


for n > 1 by hypothesis. Therefore F(ei9) vanishes identically by the corollary to Levinson's theorem at the end of §A.5. This, however, is contrary to our hypothesis.

Before going on, we note that the properties of M(v) came into play in the preceding argument only when we looked at the Fourier coefficients

of F(ei9) and not when we brought in w(r) = exp(- h(log(1/r)), even is related to M(v) in the usual way. Any other weight w(r) > 0 though tending to zero as r -.1 would have worked just as well. Thanks to what we found in step 1, { It'I = 1} n( - B) is non-empty (and even dense on the unit circumference). It is open, hence equal to a countable union of disjoint open arcs Ik on {I1;1 = 1}. (B, remember, is closed.) 0 = { Izl < 11 n(- B) clearly abuts on each of the I k- (See definition, beginning of article 5.) Take any po, 0 < po < 1, and denote by S2k(po) (or just by f1k, if it is not necessary to keep the value of po in mind) the connected component of (9n{po < Izl < 1} abutting on Ik.

362

VII D Volberg's work

Step 2. All the 52k(po) are the same. In other words, Uk52k(po) is connected.

Assume the contrary. Then we must have two different arcs Ik - call them I1 and I2 - for which the corresponding components 01 and n2 are disjoint.

Figure 102

The function (D (z) is analytic in 521 and continuous on its closure. As stated above, it's continuous on { I z I < 11 - that's because F(z) is, and because the ratio F appearing in the formula for t(z) is bounded on the region (9 over which the double integral figuring therein is taken (see beginning of the proof of the theorem in article 1). The points C on 8521 with po < ICI < 1 (sic!) must belong to B, therefore IF(C)I < 2w(I(I) for them. So, since AI F(z) I < I(D(z)I < A'IF(z)I for I z I < 1, we have I(D(C)I < const.w(It;I)

forCe8s21andpo
We have w(r)=exp(-h(log(1/r))) with h(C) decreasing and bounded below for i; > 0. In the present case, where h(g) = sup,> o(M(v) and f i (M(v)/v2)dv = oo, f o log h(l;)dc = oo for all sufficiently small a > 0 (next to last theorem of article 2 - in the present circumstances we could even bypass that theorem as in the remark following the one of article 4). Finally, the condition (given!) that M(v) > const.v" for large v, with for small > 0, where A> 1, by a lemma of article a > 2, makes 2. Therefore (and by far!) w(r) = O((1 - r)2) as r--+ 1. In our present situation, 521 abuts on I1 and 8521 avoids 12. Here, all the conditions of Volberg's theorem on harmonic measures (previous article) are fulfilled. Therefore, by the corollary to that theorem, 4)(z) - 0 in f2 This, however, is impossible since 521 s 0 on which ID(z)I > 0.

6 Volberg's theorem on the logarithmic integral

363

As we have just seen, the union UkK2k(Po) is connected. We denote that

union by )(po), or sometimes just by Q. Q(po) is an open subset of (9 lying in the ring po < I z I < 1 and abutting on each arc of { I1; I =1 } contiguous

to {ICI =1} nB.

Step 3. If ICI = 1, there are values of r < 1 arbitrarily close to 1 with r e c2(po), and hence, in particular, with I F(rC) I > w(r). Take, wlog, C = 1, and assume that for some a, po < a < 1, the whole segment [a, 1] fails to intersect Q. The function ../(z - a)/(1 - az) can then be defined so as to be analytic and single valued in 52, and, if we introduce the new variable s =

z - a

J 1 - az'

the mapping z -+ s takes S2 conformally onto a new domain - call it R/ - lying in {IsI < 11:

s-plane

z-plane

Figure 103

In terms of the variable s, write F(z) = W(s), 'Y(s)

zen;

is obviously analytic in ill/ and continuous on its closure. If

s e ill/ has I s I > ,Ia, we have, since z =

s2 + a 1 + as2'

that 1 - I z I S ((1 + a)/(1 - a))(1 - I s I2); proof of this inequality is an elementary exercise in the geometry of linear fractional transformations

which the reader should do. Hence, for s e 0,/ with IsI > Va,

IzI , 1-i+a(1-IsI2),

364

VII D Volberg's work

and, if I s I is close to 1, this last expression is > IsIL, where we can take for L a number > 2(1 + a)/(1 - a) (depending on the closeness of Is I to 1). The same relation between I s I and I z I holds for s e a fl., with I s I close to 1.

Suppose IsI < 1 is close to 1 and s e aS2ll. The corresponding z then lies on a& with IzI < 1, so I I (z) I ,
the arc it < arg s < 2it on { I s I = 11 - that's why we did the conformal mapping z s ! Here, the weight w(rL) is just as good (or just as bad) as w(r) - see the remark on a certain change of variable at the end of article 4. We can therefore apply the corollary of the theorem on harmonic measures (end of article 5) to `F(s) and the domain n,/, and conclude that `Y(s) - 0 in This, however, would make '(z) 0 in f2 which is impossible, since fl g (9 where I (D(z) I > 0. Step 3's assertion must therefore hold. The result just proved certainly implies that 852(po) includes the unit circumference. Since the Dirichlet problem can be solved for 0, it can be solved for S2. It therefore makes sense to speak of the harmonic measure wn(E, z) of an arbitrary closed subset E of { 11' I = 1 } (s 8S2) relative

to Q, as seen from zef2. As was said above, our aim is to show that wn(E, zo) > k(zo) I E I for such sets E; the analyticity of (D(z) in 12 together with

the fact that I V(z) I is > 0 and lies between two constant multiples of I F(z) I

there will then make JlogIF(ei9)Id9

> - co

rz

by the theorem on harmonic estimation (§B.1), IF(z)I being in any case bounded above in the closed unit disk. According to Harnack's theorem, the desired inequality for wn(E, zo) will follow from a local version of it, which is thus all that we need establish. At this point the condition, assumed in the hypothesis, that M(v)

const.v" with some a > i for large v, begins to play a more

important role in our construction. We have already made some use of that property; it has not yet, however, been used essentially. According to a lemma in article 2, the condition is equivalent to the property that h(g) >, small > 0. There is, in other words, an rl, 0 < rl < Z, with

5 const.(h(1))"I

6 Volberg's theorem on the logarithmic integral

365

> 0. We fix such an ry and put

for small

for >0. Since h(1;) is decreasing, so is Also, the property 1 (due to the condition M(0) > 1) makes H([;) < whence, a fortiori, h(l;)

for

> 0.

We have 2a

0a

log H() d

=

log h(x) dx = o0

11

2

o

for all sufficiently small a > 0 by a theorem in article 2, since, as we are

assuming, Ei M(n)/n2 = oo. Using

let us form the new weight

w1(r) =

exp(-Hi logl)), 0,
\

Then

wl(r) > w(r), 0<,r<1; we still have, however,

wl(r) = O((1 - r)2) for r is close to 1, wl(r) is much larger than w(r). Starting with w1(r), we proceed to construct a new open set (91 9:0 on which I F(z) I > w1(I z I) in much the same fashion as (9 was formed by use of w(r).

Take first the set Bo = {z: IzI < 1 and IF(z)I < w1(Izl)}. Since w1(r) > w(r), Bo contains the set B0 used above in the construction of B. Note that on each of the little open disks used to cover B0 and

form the set B - call those disks Aj- we have I F(z) I < 2w 1(I z I) since I F(z) I < 2w(I z I) on them. If the L also cover Bo, we take B,=B. Otherwise, we form the difference

Bo = Bo n U Aj and cover each closed set

B0"n{1-n < Izl 5

1-n+1}'

n=1,2,3,...,

366

VII D Volberg's work

by a finite number of open disks Ak(n) lying in { I z I < 11, on which I F(z) I < 2w1(I z 1). We then take B1as the closure of the union

(u) u (U&n)). For zeB1, IF(z)I

2w1(IzI). B1 contains B and has clearly the same

general structure as B; B1 includes all the points z of {IzI < 1 } for which IF(z)I <, w1(Iz1).

We now put

(91 = {IzI<1}n-.B1. The set

(91

is open and contained in

(9

since B1 2 B. For ze(91i

IF(z)I > w1(Izl), and on a&1 n { IzI < 1} we have IF(z)I < 2w1(Izl) since the

points of the later set must belong to B1. The function D(z) introduced above is thus analytic in (91 (and continuous on its closure) and, since A I F(z) I

< I ID(z) I S A' j F(z) I for I z I < 1, satisfies

Ic(z)I > const.w1(Izl),

ze(91,

as well as

I b(z)I < const.w1(IzI) for ze8(91 and IzI < 1 (sic!).

Our new weight w1(r) and the function H(g) to which it is associated fulfill the conditions for the theorem on harmonic measures (article 5). Hence,

in view of the above two inequalities satisfied by '(z), there is nothing to prevent our going through steps 1, 2, and 3 again, with (91 in place of 6) and w1(r) in place of w(r). We henceforth consider this done.

Once step 3 for (91 and w1(r) is carried out, we know that for each t;, ICI = 1, there are r < 1 arbitrarily close to 1 for which IF(rC)I > wl(r). The open set (9, was brought into our discussion in order to obtain this result, which will be used to play off w1(r) against w(r). Having now served its purpose, (91 will not appear again. Given t; o, I Co I = 1, consider any p < 1 for which (*)

IF(PC0)I > w1(P)

and form the domain S2(p2); this is the connected (step 2) component of 9 { p 2
Step 4. If, for given t;o of modulus 1, (*) holds with p close enough to 1, we

have pro E fj(P2) Assuming the contrary, we shall obtain a contradiction. Wlog, Co = 1.

6 Volberg's theorem on the logarithmic integral

367

When p -> 1, the ratio

wt(p)/w(p) = exp(h(logP/

(h(2 log P

tends to oc since h(log(1/p)) tends to oo then, his decreasing, and 0 < q < 1. Hence, if I F(p) I > w,(p) and p > 1 is close enough to 1, we surely have IF(p)I > 2w(p), i.e., p0B, so pe(9. The point p must then belong to some component of 0 n {p2 < IzI < 11, so, if it is not in the component 11(p2), abutting on the Ik, of that intersection, it must be in some other one, which we may call -9. -9, being disjoint from fl(p2), can thus abut on none of the arcs Ik of {II;I =1} contiguous to {ICI = 1}nB.

It is now claimed that 89n{p2
(sic!)

is contained in B. Let C be in that intersection; if I C I < 1, l; e 8(9 n {IzI < 1) g B,

so suppose that I C I = 1. If I; 0B, then 1; must lie in some contiguous arc Ik' say C e1, . Then, for some open disk VV with centre at 2;, V n {IzI < 1 } c (9.

The intersection on the left is, however, connected, and, since t;e8-9, it

contains some points from the (connected!) open set 2. Therefore, Vin {IzI < 1} must lie entirely in -9, and 9 intersects with the component 11(p2) of (9 n {p2 < IzI < 1 } abutting on I,. But it doesn't! This contradiction

shows that we must have CeB, as claimed. Because 0!2 n {p2 < I Z I , 11 is contained in B, we have

I1(O)I < const.IF(C)I < const.w(ICI) for ICEB_9 and p2 < IzI 51 (sic!).* The function O(z) is of course analytic in

9

0, and continuous up to 8-9. In order to help the reader follow the argument, let us try to draw a picture of -9:

* We are taking w(1)=0. See footnote, p. 359.

368

VII D Volberg's work

(N.B. -9 can't really look like this as we can see by applying an argument like the one of step 3 to S2(p2). One of the reasons why the present material is so hard is the difficulty in drawing correct pictures of what can happen.) Call F = a-9r, {p2 < I z 15 l }, and denote harmonic measure for by w.,( , z). Since I (D(z) I is bounded above on the unit disk and I O(ff) I <, const.w(I t;1) on F, we have, by the theorem on harmonic estimation (§B.1),

log 1 F(P) 15 const. + w,(T', p) log w(p2),

for ICI > p2 on I and w(r) decreases. An almost trivial application of the

principle of extension of domain shows that w,(f, p) is larger than the harmonic measure of the circle {It;I = 1}for the ring {p2 < IzI < 1}, seen from p. However, that harmonic measure is Z! We thus see by the previous

inequality that log I gy(p) I

<, const. +

i log w(p2),

i.e., since

Icb(P)I ? const.IF(p)I > const.wl(p) by (*), that

/ 2h(

const. + P2

2 (h(lo8)) P 1

in terms of the function h, with a constant independent of p. Now 0 <-I < 1 and h(1;) -+ oo for - 0. This means that the relation just obtained is impossible for values of p sufficiently close to 1. Therefore, if p < 1 is close enough to 1 and I F(p) I > wl(p), we must have p e 1(p2), the conclusion we desired to make. (Part of the idea for the preceding argument is

due to Peter Jones.) Take now any Co, 1(o 1 = 1, and pick a p < 1 very close to 1 such that IF(PCo)I > wi(P)

(which is possible, as we have already observed), and that therefore pCo e S2(p2) (by step 4, just completed). We are going to show that if E is a

closed set on the arc of the unit circumference going from t40e''°gp to t'oe-''OB°, then (0ncv2)(E, PCo) > QC0,p)IEI

with some constant C(t'o, p) depending on Co and on p. Here, of course, z) denotes harmonic measure for the domain fl(p2). wn(v2)(

6 Volberg's theorem on the logarithmic integral

369

In order to do this, we write yo = 8S2(p2)n {p2 < Izi < 1} (sic!)

and carry out Step 5. If p, chosen according to the above specifications, is close enough to 1, dcon(p2)((, PCo) I

I

is as small as we please.

In order to simplify the notation, let us write w(

z)

for

wn(p2)(

,

z)

during the remainder of the present discussion. The proof of our statement uses almost the full strength of the property that )1- 2 ry

coast.

for small

> 0 (with 0 < n < 2), equivalent to our condition that

M(v) > const.v° (with a > 2) for large v.

Take, wlog, (U = 1. Then, if p e fl(p2), we have, by the theorem on harmonic estimation (§ B.1),

logl(D(P)I <

f

log ID(C)Idw(C,p),

ac(p2)

D(z) being analytic in fl(p2) and continuous up to that set's boundary. The subset y,, of f (p2) is of course contained in B, so, for C e y,, I b(C) I< const. I F(C) I< const.w(I K j);

that is,

log l c(C) I S cont. - It I log

I)

The function 1 c(z) I is in any event bounded on { I z

together with the previous one, yields

log l F(P) I< const. -

Jh(logj-)dwP).

1 }, so this relation,

370

VII D Volberg's work

If p is chosen in such a way that we also have I F(p) I > w,(p), this becomes

jh(log)dwp)

(*)

Since 1/ <

const. + try for small

(h(log-))l. > 0, we have (when p < 1 is

close to 1),

i-2n vo 1

P) <, const. f'P h log

1

I

I

I

dco(i;, P)

I)

Rewrite (!) the right-hand integral as -n

g (h(Io))' 1

1

11

const.f

p).

Jv, ((iog))

decreases, the expression just

Since y, lies in the ring {p2 < I z I < 11 and

written is const.

(h(log_))1dw,P),

1

oo for % - 0, we see that

so, since

dw(C, p)

1

J v,1

ICI

<

8°

J hlog (h(log-))n vo

with a quantity 6° going to zero for p side, we obtain finally

Jvo I

11(1

I) doo(C, p)

1

I

1. Plugging (*) into the right-hand

a°(1 + (h(log(1/P2)))"

Here, the right side tends to zero as p

Step 5 is finished.

Now we can make the local estimate of w(E, z) for closed E on the unit circumference, promised just before step 5. Taking any fixed Co, I Co I = 1, we pick a p < 1 very close to 1, in such fashion that the conclusions reached in

steps 4 and 5 apply. Let E be any closed set on the arc of the unit circumference going from Coe1lo8° to Coe-''°B°

In order to keep the notation simple, consider, as before, the case where Co = 1.

6 Volberg's theorem on the logarithmic integral

371

Figure 105

Denote harmonic measure for the ring { p2 < I z I < 1 } by Co( , z). If we compare to(E, z) with the harmonic measure of E for the sectorial box shown in the above diagram, we see immediately that (t)

w(E,P) > C

IEI 1-p

with a numerical constant C independent of I E I and of p. Put, as in step 5, yo = 0Q(p2)n{p2 < Izl < 1}, and continue to denote the harmonic measure for fl(p2) by co(

,

z). Since

S2(p2) c {p2 < Izl < 1} while

asj(p2) 2 {ICI= I}, we can apply a formula established near the end of § B. 1, getting

6t') f(E,dco(t', z)

co(E, z) = to(E, z) Yp

for zeS2(p2). Comparing h(E, z) with harmonic measure of E for the whole unit disk, we

get the estimate w(E,

IEI

n(1-ICI)

372

VII D Volberg's work

Substitute this into the integral in the above formula for co(E, z), specialize to z = p, and use (t) in the first right-hand term of that formula. We find

w(E,P) > IEI(1

-

C P

1

fy

oI

p)

It was, however, seen in step 5 that p < 1 could be chosen in accordance with our requirements so as to make the integral in this expression small. For a

suitable p < 1 close to 1, we will thus have (and by far!) (o (E, p) >

C

2(1 -p) IEI

provided that the closed set E lies on the (shorter) arc from e''o8P to a-''°gp on the unit circle.

This is our local estimate. What it says is that, corresponding to any I C I = 1, we can get a p, < 1 such that, for closed sets E lying on the smaller arc Jg of the unit circle joining leub0 to le - ilogPS,

(tt)

ws4P4,)(E, PAC) > CC I E I,

with C, > 0 depending (a priori) on C. Observe that, if 0 < p < pS , S2(p) (the component of (9n {p < Izl < l} abutting on the arcs Ik) must by definition contain 12(p; ). Therefore (§)

wn(P)(E, PcC) > (ORP2)(E,

by the principle of extension of domain when E s { I z I = 1), a subset of both

boundaries 00(p), ac (p;). A finite number of the arcs J, serve to cover the unit circumference;

denote them by J1, J2,...,J,,, calling the corresponding values of p and the corresponding pc's P l, p2, ... , p; , j = 1, 2,. . ., n, and denote the least of the C5j by k, which is thus > 0. If E is a closed subset of JJ, (tt) and (§) give 1',

wn(P)(E, p Jl; ) > k I E 1.

Fix any zo c- 0(p). Using Harnack's inequality in 0(p) for each of the pairs of points (zo, p j ( J ), j = 1, 2, ... , n, we obtain, from the preceding relation, (§§)

wn(p)(E, zo) > K(zo)IEI

for closed subsets E of any of the arcs J 1, J2, ... ,

Here, K(zo) > 0

depends on zo. Now we see finally that (§§) in fact holds for any closed subset E of the unit circumference, large or small. That is an obvious consequence of the additivity of the set function wn(P)( , zo), the arcs JJ forming a covering

of {It;I=1}. We are at long last able to conclude our proof of Volberg's theorem

6 Volberg's theorem on the logarithmic integral

373

on the logarithmic integral. Our chosen zo in n (p) lies in (9, therefore I'1(zo)I > 0. By the theorem on harmonic estimation applied to the function D(z) analytic in fl(p) and continuous on { I z I < 11,

- co < log I (D(zo) 15 J

log 14)(C) I dwn( )(C, zo) n(o

const. +

log I F(t') I dcoa( ) (l;, zo).

J is=1

According to (§§), this last is in turn

const.-K(zo) f log-It(ei9)Id9, log I (D(ei9) I being in any case bounded above. Thus,

Jl0g_I()1d9

< co.

However, 14)(ei9) I lies, as we know, between two constant multiples of I F(ei9) I.

Therefore

f.

f.

log- IF(ei9)Id9 < oo,

logIF(ei9)Id9 > -oo.

Volberg's theorem is thus completely proved, and we are finally done.

Remark. Of the two regularity conditions required of M(v) for this theorem, viz., that M(v)/v be decreasing and that M(v) >, const.v" for large v with some a>-!, the first served to make possible the use of Dynkin's result (article 3) by means of which the analytic function F(z) was brought into the proof. Decisive use of the second was not made until step 5, where we estimated YP1-

dw(C, p) ICI

in terms of $ h(log(1/I CI))dcw(C, p).

Examination of the argument used there shows that some relaxation of the condition M(v) > const.v" (with a > Z) is possible if one is willing to replace it by another with considerably more complicated statement.

The method of Volberg's proof necessitates, however, that M(v) be

374

VII D Volberg's work

at least > const.v for large v. For, by a lemma of article 3, that relation is equivalent to the property that 1

= O(h(o))

-+0, and we need at least this in order to make the abovementioned estimate for p < 1 near 1. We needed JYO(1/(1 - KI))dw(C, p) in the computation following step 5, where we got a lower bound on w(E, p). The integral came in there on account of the inequality for

CUE 0 <

IEI

ir(1- KI)

for harmonic measure ah(E, () (of sets E c {ICI = 1 }) in the ring {p2 < I C I < I). And, aside from a constant factor, this inequality is best possible. 7.

Scholium. Levinson's log log theorem

Part of the material in articles 2 and 5 is closely related to some older work of Levinson which, because of its usefulness, should certainly be taken up before ending the present chapter. During the proof of the first theorem in §F.4, Chapter VI, we came up

with an entire function L(z) satisfying an inequality of the form IL(z)I < const.e' /I,3zI, and wished to conclude that L(z) was of exponential type. Here there is an obvious difficulty for the points z lying near the real axis. We dealt with it by using the subharmonicity of I L(z) I and convergence of 1

Iyi--Idy

J

in order to integrate out the denominator 13z I from the inequality and

thus strengthen the latter to an estimate IL(z)I < const.e"'Z' for z near R. A more elaborate version of the same procedure was applied in the proof of the second theorem of §F.4, Chapter VI, where subharmonicity of log I S(z) I was used to get rid of a troublesome term tending to 00 for z approaching the real axis.

It is natural to ask how far such tricks can be pushed. Suppose that f (z) is known to be analytic in some rectangle straddling the real axis, and we are assured that I f(z)I < const.L(y)

7 Levinson's log log theorem

375

in that rectangle, where, unfortunately, the majorant L(y) goes to co as y-+0. What conditions on L(y) will permit us to deduce finite bounds on f(z)I, uniform in the interior of the given rectangle, from the preceding relation? One's first guess is that a condition of the form f a log L(y) dy < oo will do, but that nothing much weaker than that can suffice, because log I f(z) I is subharmonic while functions of I f (z) I which increase more slowly than

the logarithm are not, in general. This conservative appraisal turns out to be wrong by a whole (exponential) order of magnitude. Levinson found that it is already enough to have a

J

-a

log log L(y) dy < co,

and that this condition cannot be further weakened.

Levinson's result is extremely useful. One application could be to eliminate the rough and ready but somewhat clumsy hall of mirrors argument from many of the places where it occurs in Chapter VI. Let us, for instance, consider again the proof of Akhiezer's first theorem from §B.1 of that chapter. If, in the circumstances of that theorem, we have II P II w S 1 for

a polynomial P, the relation

-

ogIP(z)I 5 7E

IZI3tllzloglP(t)Idt

-

\

' f 0" n

IZI -t1I2

log W+(t)dt

and the estimate of sup,,, I (t - i)/(t - z) I from §A.2 (Chapter VI) tell us immediately that M(I+IzI)z

logIP(z)I

I3z l

where

M

=

1

n

log

W(t) dt. l+tz

Taking any rectangle

-9R = {z: IsRzl
I,3zl

z

L(y) = exp M R I

15

'

2}

376

VII D Volberg's work

we have IP(z)I
JloglogL(y)dY < oo, so the result of Levinson gives us a control on the size of IP(z)I in the interior of fR (even right on the real axis). In this way we can see that the polynomials P with II P II w < 1 form a normal family in any strip straddling the real axis. The relation log I P(z) l < M(1 + I z l )2/I Zz I already shows that those polynomials form a normal family outside such a strip, and the main part of the proof of Akhiezer's first theorem is complete. One can easily envision the possibility of other applications like the one just shown to situations where the hall of mirrors argument would not be available. There is thus no doubt about the worth of the result in question; let us, then, proceed to its precise statement and proof without further ado. Levinson's log log theorem. Consider any rectangle

-9 = {z: a<x
J-6

log log L(y) dy < co.

Then there is a decreasing function m(b), depending only on L(y) and finite

for 6 > 0, such that, if f(z) is analytic in -9 and if

If(z)I < L(3z) there, we also have I f (z) I< m(dist.(z, a_q)) for z e.9.

Remark. In this version (due to Y. Domar), no regularity properties whatever are required of L(y). The assumption that L(y) >, e is of course made merely to ensure positivity of log log L(y). Proof of theorem (Y. Domar). Denote by u(A) the distribution function for log L(y); i.e.,

µ(A) = I{y: -bA}1.* Let zoe- and R < dist(zo, a-9), and write u(z) = logI f(z) 1. Since u(z) is * We are continuing to denote by AEI the Lebesgue measure of sets E S R.

377

7 Levinson's log log theorem subharmonic in !2, we have u(zo)

u(z) dx dy. l2 J nR J i=-zol
We are going to show that, if u(zo) is large and zo far enough from 8-9, this inequality leads to a contradiction when u(z) < log L(iz) in -9. Call 0 the disk {z: lz - zol < R}.

Figure 106

Use 11 E 11 to denote the two-dimensional Lebesque measure of E c C (since

is being used for one-dimensional Lebesque measure of sets on R). Then, by boxing 0 into a square of side 2R in the manner shown, we see from the inequality u(z) < log L(iz) that 11 {zEA: u(z) > M/2} 11 < 2Rp(M/2)

for M > 0.

Suppose now that u(zo) ? M, but that at the same time we have u(z) < 2M on A. From the previous subharmonicity relation we will then have

M (*)

u(zo)

+

MR Ml ( µ (M/2). 5ZE0: u(Z)i 2 j < M +

20 M 1I

2

Because log log L(y) is integrable on [ - b, b] we certainly have µ(A) - 0 for A - oo. We can therefore take M so large that µ(M/2) is much smaller than dist(zo,821). With such a value of M, put

R = Ro

16

µ(M12).

The right side of (*) will then be < M. This means that if u(zo) >, M, the assumption under which (*) was derived is untenable, i.e., that u(z) > 2M

somewhere in 0, say for z = z1, lzi - zo1 < Ro.

378

VII D Volberg's work

Supposing, then, that u(zo) >, M, we have a z1, I z1 - zo l < R0, with u(z1) > 2M. We can then repeat the argument just given, making z1 play the role of z0, 2M that of M,

R1 = 16µ(M) n that of Ro, and {z: I z - z1 I < R1 } that of A. As long as R1 > dist(z1, 8.9), hence, surely, provided that

Ro + R1 < dist (zo, 8fi),

we will get a z2, 1z2 - z1 l < R 1, with U(Z2)>4M. Then we can take R2 = (16/n)µ(2M), have 4M play the role held by 2M in the previous step, and keep on going. If, for the numbers

Rk =

16

µ(2k - I M),

It

we have 00

Y Rk < dist (zo, 89),

k=0

the process never stops, and we get a sequence of points ZkE-9, IZk-Zk-1I a point zn3 E -9.

The function f (z) is analytic in !2, so u(z) = log I f (z) I is continuous (in the extended sense) at z,,), where u(z,,) < oo. This is certainly incompatible

with the inequalities U(Zk) > 2kM when zk k z,,. Therefore we cannot have u(zo) >, M if we can take M so large that Y_o Rk < dist (zo, 8-9), i.e., that

(t)

16 -/1(2"'M) < dist (zo, 8_q).

k=0 it

In order to complete the proof, it suffices, then, to show that the left-hand

sum in (t) tends to zero as M - oo. By Abel's rearrangement, "

µ(2k- IM) =

fy( M\ 2) - µ(M) + 2{µ(M) - µ(2M)}

k

+ 3{µ(2M) - µ(4M)} + + n{(2 " - 2 M) -,u(2 n- I M) } + (n + 1)µ(2n - I M).

7 Levinson's log log theorem

379

Remembering that µ(A) is a decreasing function of µ, we see that as long as M > 4, the sum on the right is 2"M

n -1

k1

f

2k

log 2 - log (M/4) log 2

'M

C °°

+

(- dµ(2) )

log 2 - log(M/4) log 2

2" iM

I log 2

r-

JM/2

log dµ(2) =

(- dµ(A))

1

log log L(Y) dY log 2 osuy), M/2 -b
b

Since f b log log L(y) dy < oo, the previous expression, and hence the lefthand side of (t), tends to 0 as M -+oo. Therefore, given zoe2 , we can get

an M sufficiently large for (t) to hold, and, with that M, u(zo) < M, i.e., If (zo) I < eM. Let, then, 00

m(8) = inf t eM: Y

16µ(2k-1M)

k = 0 7<

<S )))

As we have just seen, m(5) is finite for S > 0; it is obviously decreasing. And, if f (z) is analytic in -9 with I f (z) I ,
Remark. This beautiful proof is quite recent. The procedure of the scholium at the end of article 5 will yield the same result for sufficiently regular majorants L(y). We now consider the possibility of relaxing the condition 6

-b

log log L(y) dy < o0

required in the above theorem. If, for some majorant L(y), the conclusion of the theorem holds, any set of polynomials P with I P(z) I < L(3z) for ze-q must form a normal family in -9. This observation enables us to give a simple proof of the fact that the requirement. 6

log log L(y) dy < o0 is essential in Levinson's result, at least for majorants L(y) of sufficiently regular behaviour. Theorem. Let L(y) be continuous and > e for 0 < I y I

b, with L(y) - o0

380

VII D Volberg's work

for y -0 and L(y) decreasing on (0, b). Suppose also that ('6

f log log L(y) dy = oo. Jo

Then there is a sequence of polynomials P"(z) with, for Zz 0 0,

IP.(z)l < const.L(3z) on the rectangle

2 = {z: -1<9iz<1 and -b<3z
P"(z) -) "

1,

zed and 3z > 0,

-1, ze-9and3z<0.

Thus the conclusion of Levinson's theorem does not hold for the majorant L(y).

Remark. We only require L(y) to be monotone on one side of the origin, on the side over which the integral of log log L(y) diverges. Levinson already had this result under the assumption of more regularity for L(y).

Proof of theorem (Beurling, 1972 - compare with the proof of the theorem on simultaneous polynomial approximation in article 4). The last

sentence in the statement follows from the existence of a sequence of polynomials P,, having the asserted properties. For, if the conclusion of Levinson's theorem held, the sequence {P"} would form a normal family in -9 and there would hence be a function analytic in -9, equal to + 1 above the

real axis, and to - 1 below it. This is absurd. Put 1,

P(z)

ze.9 and 3z > 0

- 1 , zed and 3z < 0,

and let us argue by duality to obtain a sequence of polynomials P"(z) for which sup ZED'

I tp(z) - P"(z)1

L(3z)

n

0.

Such P" will clearly satisfy the conclusion of our theorem. Note that, since L(y) -> oo for y - 0, the ratio cp(z)/L(,3z) is continuous on .9 if we define L(0) to be oo, which we do, for the rest of this proof. Therefore, if a sequence of polynomials P. fulfilling the above condition

7 Levinson's log log theorem

381

does not exist, we can, by the Hahn-Banach theorem, find a finite complexvalued measure µ on with zn

(§)

II, L(iz)

dµ(z) = 0 for n = 0, 1, 2,...,

whilst Oz)

f fo L(3z)

dµ(z) # 0.

The proof will be completed by showing that in fact (§) implies Oz) JJ

dµ(z) = 0.

L(3z)

Given any measure µ satisfying (§), write

dv(z) = L(iz) dµ(z) The measure v has very little mass near the real axis, and none at all on it. For each complex .t, the power series for eizz converges uniformly for ze9, so from (§) we get

JJedv(z) = 0. Write now

-9+ =

n {3z > O}

(sic!)

-9_ _

n{3z<0}

(sic!).

and

Then put
f

e'zzdv(z),

f9+ euz dv(z);

(A)

JJ

since v has no mass on III, the previous relation becomes (fit)

4) +(A) + 'F-(A) = 0.

We are going to show that in fact each of the left-hand terms in (tt) vanishes identically.

382

VII D Volberg's work

Consider d>+(A). Since .9+ is a bounded domain; I +(2) is entire and of exponential type. It is also bounded on the real axis. Indeed, for A > 0,

i'D+(A)I < ff9+ e-A3ZIdv(z)I <

JJIdv(z)I ,

and for 2 < 0 we can, on account of (tt), use the relation +(2) (D_(A) and make a similar estimate involving -9 _. These properties of(A.) and the theorem of Chapter III, §G.2, imply that (D +(A) = 0 provided that logla)+(2)I

1 °°

J_00

1+2z

00.

dA

We proceed to establish this relation. Write

H(y) _

Jlog L(y), log L(b),

0 < y < b, y > b.

Then H(y) is decreasing for y > 0 by hypothesis. For 2 > 0, eizz

x(3Z)

+(A)I = f f+ L(3)z dµ(z) < fL+ e

I

Idµ(z)

If, as in article 5, we put

M(A) = inf (H(y) + yi), y>o

we see by the previous relation that (§§)

I4+(2)I < const.e-M

,

2 > 0.

Since H(y) is decreasing for y > 0 and > 1 there, and b

b

log H(y) dy = 0

log log L(y) dy = 00 o

by hypothesis, we have

f MZ-) d2 = o0 according to the last theorem of article 2. This, together with (§§), gives us (*), and hence 1+(A) = 0.

Referring again to (tt), we see that also 0 -(A) - 0. Specializing to 2 = 0 (!), we obtain the two relations

f L. LQz) dµ(z) = 0,

IL - L(iz)

dµ(z) = 0,

7 Levinson's log log theorem

383

from which, by subtraction,

f f Oz) L(iz)dµ(z) = 0, what we had set out to show. The proof of our theorem is thus finished, and we are done. And thus ends this long (aye, too long!) seventh chapter of the present book.

VIII

Persistence of the form dx/(1 + x2)

Up to now, integrals like °°

log I F(Z) I

_ l +X

dx

have appeared so frequently in this book mainly on account of the specific form of the Poisson kernel for a half plane. If w(S, z) denotes the harmonic

measure (at z) of S s I for the half plane (3z > 0), we simply have co(S, i)

t2 .

Suppose now that we remove certain finite open intervals - perhaps infinitely many - from R, leaving a certain residual set E, and that E looks something like R when seen from far enough away. E should, in particular,

have infinite extent on both sides of the origin and not be too sparse. Denote by -9 the (multiply connected - perhaps even infinitely connected) domain C - E, and by co.,( , z) the harmonic measure for -9.

Figure 107

It is a remarkable fact that a formula like the above one for co(S, i) subsists, to a certain extent, for coy( , i), provided that the degradation suffered by R

Persistence of the form dx/(1 + x2)

385

in its reduction to E is not too great. We have, for instance, w9(J (_1 L, >) 5 CE(a)

r

dt

J JnE 1 + t2

for intervals J with I J n E I >, a > 0, where CE(a) depends on a as well as on the set E. In other words, dw9(t, i) still acts (crudely) like the restriction of dx/(1 + x2) to E. It is this tendency of the form dx/(1 + x2) to persist when we reduce R to certain smaller sets E (and enlarge the upper half plane to -9 = C - E) that constitutes the theme of the present chapter. The persistence is well illustrated in the situation of weighted approximation (whether by polynomials or by functions of exponential type) on the sets E. If a function W(x) >, 1 is given on E, with W(x) - oo as x - ± o0 in E (for weighted polynomial approximation on E this must of course

take place faster than any power of x), we can look at approximation on E (by polynomials or by functions of exponential type bounded

on R) using the weight W. It turns out that precise formal analogues of many of the results established for weighted approximation on 68 in §§A, B

and E of Chapter VI are valid here; the only change consists in the replacement of the integrals of the form °°

log M(t)

1+t2

dt

occurring in Chapter VI by the corresponding expressions f log M(t) E

1 + t2

dt.

The integrand, involving dt/(l + t2), remains unchanged. This chapter has three sections. The first is mainly devoted to the case where E has positive lower uniform density on R - a typical example is furnished by the set OD

E = U [n-p, n+p] n=-00

where 0
arises when p = 0, i.e., when E = Z. There is of course no longer any harmonic measure for -9 = C - Z. It is therefore remarkable that something

nevertheless remains true of the results established in §A. If P(z) is a polynomial such that 1+n21og+IP(n)I 5 n

VIII A The set E has positive lower density

386

with ry > 0 sufficiently small (this restriction turns out to be crucial!), we still have, for zEC, IP(z)I <, K(z,n),

where K(z,q) depends only on z and ry, and not on P. The proof of this fact is very long, and hard to grasp as a whole. It uses specific properties of polynomials. Since -9 has no harmonic measure, a corresponding statement with log+ IP(z)I replaced by a general continuous subharmonic function of at most logarithmic growth is false. We return in §C to the study of harmonic estimation in -9 when its boundary, E, does not reduce to a discrete set. Here, we assume that E contains all xel1 of sufficiently large absolute value, that situation being general enough for applications. The purpose of §C is to connect up the behavior of a Phragmen-Lindelof function for -9 (i.e., one harmonic in -9 and acting like I ,3z I there, with boundary value zero on E) to that of harmonic measure for -9. There is a quantitative relation between the former and the latter. Harmonic measure still acts (very crudely!) like the restriction of dt/(1 + t2) to E. This § is independent of §B to a large extent,

but does use a fair amount of material from §A. Results obtained in it are needed for Chapter XI.

The set E has positive lower uniform density

A.

During most of this §, we consider sets E of the special form 00

U [an - Sn, an + bn], n= - co

the intervals [an - bn, an + bn] being disjoint. We will assume that there are four constants, A, B, b and A, with

0
0
for all n. The following notation will be used throughout:

En = [an - bn, an + bn], (On = (an + bn, an+ 1 - bn+ 1),

-9 =C - E. Here is a picture of the setup we are studying:

I Harmonic measure for .9 = C - E

E_2

E_1

a-2

0-1

EO °

00

387

E

0,

Ez

02

E3

Figure 108

The above boxed conditions on the a and b clearly imply the existence

of two constants C, and C2 > 0 (each depending on the four numbers A, B, S and A) such that: (i) if k o l and xE(9k, x'E(9,, we have C,1 k -11 ,
- x 15 C21 k - ll ;

(ii) if k 0 1 - 1, 1, or 1 + I and xEEk, x'EE,, we have

C,Ik-115Ix-x'I1
only bounded above. It is the lengths of the Ek that are bounded above and away from zero. Heavy use will be made of properties (i) and (ii) during the following development. Clearly, if E is any set for which the above boxed condition holds (with given A, B, 6 and A), so is each of its translates E + h (with the same constants A, B, b and A). The properties (i) and (ii) are thus valid for each of those translates, with the same constants C, and C2 as for E. For this reason there is no real loss of generality in supposing that OE(9o, and we will frequently do so when that is convenient. 1.

Harmonic measure for -9

The Dirichlet problem can be solved for the kind of domains -9 we are considering and (at least, certainly!) for continuous boundary data on E given by functions tending to 0 as x -> ± oo in E. Let us, without going into too much detail, indicate how this fact can be verified. Take large values of R, and put

-9R = -9n^-{(-oo,-R]v[R,oo)}:

Figure 109

388

VIII A The set E has positive lower density

Each of the regions -9R is finitely connected, and the Dirichlet problem can be solved in it. This is known; it is true because the straight segment boundary components (`slits') of 2 R are practically as nice as Jordan curve boundary components. One can indeed map 2 R conformally onto a region bounded by Jordan curves by using a succession of Joukowski transformations, one for each slit (including the infinite one through oo):

k, 4

rigure iiu The inverse to the conformal mapping thus obtained does take the Jordan curve boundary components back continuously onto the original slits, so the Dirichlet problem can be solved for -9R if it can be solved for regions bounded by a finite number of Jordan curves. (This same idea will be used again at the end of article 2, in proving the symmetry of Green's function.)

Once we are sure that the Dirichlet problem can be solved in each 9R we can, by examining how certain solutions behave for R --* oo, convince ourselves that the Dirichlet problem for -9 is also solvable, at least for boundary data of the abovementioned kind. Details of this examination are left to the reader.

Since -9 is regular for the Dirichlet problem, harmonic measure is available for it. We know from the rudiments of conformal mapping theory that a slit should be considered as having two sides, or edges. Given (say)

an interval J c one of the boundary components E. of -9, we should distinguish between two intervals coinciding with J: J+ (lying on the upper side of En), and J_ (lying on the lower side of En):

Figure 111

1 Harmonic measure for .9 = C - E

389

It makes sense, then, to talk about the two harmonic measures: WWQ(J+, z)

(which tends to zero when z tends to an interior point of J_), and (o,JJ

_, z).

In most of our work, however, separation of J into J+ and J_ will serve no purpose. It will in fact be sufficient to work with the sum (0.9(j +' z) + (oJJ _, z).

This harmonic function tends to 1 when z tends from either side of the real axis to an interior point of J, and it is what we take as the harmonic measure (O-'(J' z)

of J. The harmonic measure co9(S, z) of any S

E is defined in the same way.

Consider now any of the boundary components Ek of E = 0-9, and write cok(Z) = (O(Ek, Z)

for the harmonic measure of Ek, as seen from ze2. We are going to show that there is a constant C, depending on the four numbers A, B, 8 and A associated with E, such that wk(x)

< (l -) + 1

for xe(9,.

((9,, recall, is the part of.9n ll lying between E, and E,,,.) The proof is due to Carleson; one or two of its ideas go back to earlier work. We need two lemmas, the first of which could almost be given as an exercise. Lemma. Denote by S2k(

, z)

harmonic measure for the domain

-9k = {3Z>0}V(9kV{ 3Z
Figure 112

390

VIII A The set E has positive lower density

There is a constant K', depending only on the numbers B and S associated with the set E, such that for xe(9k and tEB-9k lying outside both of the segments Ek

and Ek+l we have K'

dS2k(t, x)

(x

-

dt.

t)2

Proof. By conformal mapping of -9k onto the unit disk. Calling mk the midpoint of Ok, we apply to ze!2k the chain of mappings

z-)S = We write w = cp(z) and, if t is on a-9k, denote cp(t) by w. (In the latter case we must of

course distinguish between points t lying on the upper side of a-9k and those on its lower side - see the preceding remarks. On this distinction depends the choice of the branch of / to be used in computing w = cp(t).)

For t on

a.9k outside both Ek and Ek+, we have

where r = 2(t - mk)/Iakl satisfies the inequality

Irl-1 /

2min(IEki,IEk+,I) > 46 B

I(9kI

in view of the relations I E, I = 26, > 26, I G I < ak+ 1 - ak < B. In terms of r,

- d2 ( 1 ± /(r2 - 1) )

dw =

(T2T2

1)

Idwl =

r l

For t outside both Ek and Ek+i, the expression on the right is

1 + 4S/e #Idrl <

)

46/B

T2

-

(B (4S + 1)

tIdrl TZ

by the inequality for Irl - 1.

Let xEOk. Then, remembering that df2k(t,x)

is

the harmonic measure

of two infinitesimal intervals [t, t + dt] lying on 0-9k-one on the upper edge and one on the lower, we see that d)k(t, x) = with Idwl

1

it

1 - 104, I(o(x)-wI'

Idwl,

being given by the above formula. Since, for t4EkuEk+i,

I Harmonic measure for .9 = C - E

391

IT I -1 > 46/B, the image, w, of t on the unit circumference must lie outside two arcs thereof entered at 1 and at - 1, and having lengths that depend

on the ratio 46/B. We do not need to know the exact form of this dependence.

Figure 113

Hence, since - 1 < cp(x) < 1, it is clear from the picture that I tp(x) - CO I is a positive quantity depending on the lengths of the excluded arcs about 1 and - 1, and thence on the ratio 46/B. The factor (1- I(p(x)I2)/lcp(x) - wI2 in the above formula for dQ2k(t, x) is thus bounded above by a number depending on 45/B when to E, v Ek+ 1. Substituting into that formula the inequality for Idwl already found, we get I d2 I

dS1k(t, x) 5 C ,2

for t O Ek u Ek + 1, with a constant C depending on 46/B. In terms oft = mk + iI Ok I T,

the right side is dt

Cldkl 2

CB

dt

(t - mk)2 \ 2 (t-Mk )2'

Since xe!k and took,

It-xl -< 21t-mkl:

x

mk

t

Figure 114 So finally,

digk (t x1 < 2BC

dt

,

(1 - X)2'

Q.E.D.

392

VIII A The set E has positive lower density

Computational lemma (Carleson, 1982; see also Benedicks' 1980 Arkiv paper). Let Ak,, > 0 for k, le7L. Suppose there are constants K and A, with 0 < A < 1, such that (*)

K A k'I < (1-k)2+1'

and 00

Y Ak,, 5 2

for all

k.

1= - 00

Then there is a number q > 0 depending on K and 2 such that, for any

sequence {y,} with 0 5 y, < I and 0 5 y, < 11(12 + 1), we have. 00

for all

Ak,,y, < 1 sup y, 00

k,

1

and 00

1+2

1

2

k2 + 1

Ak,IYI S

Remark. The first of the asserted inequalities is manifest; it is the second that is non-trivial. If the constant K is small enough, the second inequality is

also clear; it is when K is not small that the latter is difficult to verify. Proof of lemma. As we have just remarked, the first inequality is obvious (by (*)); let us therefore see to the second, endeavoring first of all to prove it for large values of I k 1, say I k I > some k0.

Assume, wiog, that k > 0, and take some small number It, 0 < µ < 2, about whose precise value we will decide later on. Write the sum CO

Ak,,y, 00

as

E+ I4 <µk

+ IIIO pk

II-kl;,sk

Y

= I + II + III,

II - kI
say. Use of (*) together with the inequality 0 5 y, < I (where n is as yet unspecified) gives us first of all K 11< (1 - µ)2k2 +

1.2kµ

We get III out of the way by combining (*) and the inequality

1 Harmonic measure for .9 = C - E

393

0 < y, < 1/(12 + 1), with the result that A

III <

(1-µ)2k2+1*

It is the middle sum, II, that gives us trouble. We break II up further as

+

Y-

I-< -Nk

+

Y, ukSl
Yk/2515(1-N)k

+

Y

.

1?(1+µ)k

The first of these sums is K

1 1

K 1 k2 + 1mpk m2

,,k 12 + 1 (k -1)2 +l

2K µk(k2 + 1).

5

The second is similarly K K 5 + 1 (k - I )2 + 1 (k/2)2 + 1

k/2

Mk <

12

1

k m2

8K µk(k2 + 4)

The third sum is K

8K

,-+µ)k12+1 (l-k)2+1

2K uk(k2+1)'

1

K

1

All told, then, 20K

II <

µk(k2 + I)'

Adding this last estimate to those already obtained for I and III, we get finally

2kugK Ak.IYI < (I _ µ)2(k2 00

20K A I). + 1) + uk(k2 + 1) + (1 -µ)2(k2 +

The idea now is to first put n equal to a very small quantity rlo, and then,

assuming k is large, put µ = 1/rlo/2k; this will also be small for large enough k. For such large k, the previous inequality will reduce to

0

21lo2K+20rlo2K+A Ak11YI <

Y-

1=

(I - 11)2(k2 + 1)

oo

Choosing first ho'/2 small enough and then taking ko so large that 1/rlo 2ko

394

VIII A The set E has positive lower density

is also small, we will make the right-hand side of this inequality

1+)

1

2

k2+1

for I k I >, ko by putting u = 1/go/2 I k 1. When q < qo and 0,
t=-0o

1+A

1

2

k2 + 1

Ak,1Y, 5

for

Ikl,ko.

With such y,, however, the sum on the left is also < Aq. So, taking finally 1

min (ion k2 0+1

makes the left side S ((1 + 2)/2X1/(k2 + 1)) for ski < ko as well, i.e., Ak,1Yi

1<

1

2

k2 + 1

for all k, provided that 0 < y, 5 q and 0 < yi < 1/(12 + 1). The lemma is proved.

Theorem (Carleson, 1982; see also Benedicks' 1980 Arkiv paper). In the domain -9, the harmonic measure wo(z) of the component Eo of 8-9 satisfies wa(x)

k2 + 1

for XEGk,

with a constant C depending only on the four numbers A, B, 6 and a associated

with E=8.9. Proof (Carleson). Call uk the maximum value of coa(x) on (9k; we are to show that uk

C

\ k2 + 1

For k = - 1 and k = 0 this is certainly true if we put C = 2; we may therefore restrict our discussion to the values of k different from - 1 and 0. As in the first of the above lemmas, denote by -9k the domain

{.3z>0}take{3z<0} and by S2 k( , z) the harmonic measure for .9k.

in 9:

-9k is of course contained

I Harmonic measure for 9 = C - E

395

N,

Figure 115

wo(z) is thus harmonic in -9k; since it is clearly bounded there and continuous up to 8-9k, we may recover it from its values on 8-9k by the Poisson formula wo(Z) =

wo(t)dS2k(t, Z),

ZE2k.

Since we are assuming that k - 1, 0, we have (by definition of harmonic measure!) coo(t) =0 for tEEkuEk+! In fact, wo(t) is identically zero on all the E, save Eo, where wo(t) = 1. The above formula hence becomes

dnk(t, z) + Y J wo(t)d)k(t, z),

wo(z) = J

l*k "

Eo

R being the disjoint union of the intervals 0, and E,. Let x, be the point in (91 where wo(x) assumes its maximum u, therein, and write Ak,! =

I

df2k(t,xk)

Then, by the previous relation, Uk

1<

i2k(Eo, xk) + Y Ak,,u1. l*k

Here, the integrals

I

di2k(t, xk) = SZk(Eo, xk)

a nd

f dflk(t, xk) = Ak,,,

10 k,

are taken over sets disjoint from Ek and Ek+l, whereas xkE(k. We may therefore apply the first lemma to estimate dilk(t, xk) in these integrals, getting ' k(EO,

xk)

dt

K' Eo

(t - t

396

VIII A The set E has positive lower density

and

AU < K'

dt

2

of (t - xk)

I o k,

where K' is a constant depending on the numbers B and 6 associated with E. By properties (i) and (ii), given at the beginning of this §, we have

(t-xk)2 >, C;k2 for

teEO

(t-xk)2 > C2(k-I)2

for tE(91.

and

So, since I E0 I = 260 < 2A and 1(9l 1 < B, the preceding relations become

K

!Gk( 1

SEO, xk) < k2 + 1 and

K

(*)

, Ak,l S (k - 1)+1 2

1A k;

here, K is a constant depending on the four numbers A, B, 8 and A. The numbers Ak,l also satisfy the inequality (*)

>Ak,l
l#k

with A depending only on the ratio 8/B. Indeed,

Y Ak,l = EI#kQ01, Xk)

I#k

Ek - Ek + 1, xk ). Since I ak I 26, I Ek+ 1 I >26, a simple change of variable shows that the latter quantity is less than the is S S2k(8-9k

harmonic measure of the set 1 + 48/B <, I t I < oo on the boundary of the

domain C - (- oo, - 1] - [1, oo), seen from some point in (- 1, 1):

Figure 116

I Harmonic measure for -9 = C - E

397

And that harmonic measure is clearly at most equal to some number 2 < I depending on 46/B. Let us return to the inequality 1< f1k(EO, Xk) + Y- Ak,lul I*k

Uk

established above. By plugging into it the estimates just found, we get K Uk -

(I)

Ak,IUI < k z + 11 l*k

where the Ak,l are > 0, and satisfy (*) and Q). This has been proved for k - 1 and 0, but it also holds (a fortiori!) for those values of k, provided that we take K > 2. Then our (unknown) maxima Uk 1> 0 will satisfy (t) for all k; this we henceforth assume.

The idea now is to invert the relations (t) in order to obtain bounds on the uk. It is convenient to define Ak,, for l = k by putting Ak,k = 0. Then, calling

(tt)

Vk = Uk - Y- Ak,IUI, l

we can recover the Uk from the Vk by virtue of Q). Write A(',)= Ak,/; then put 00

Ak,l =

j= - OD

Ak,JA;,l,

and in general 00

Ak,iA."i

Ak,,1 1)_

The numbers A;"i are >, 0 (since the Ak,l are), and from

), we have

00

(§)

Y AI" < 2". l= - ao

This makes it possible for us to invert (tt), getting Uk = Vk +

Akz/ vl + ... +

f1k1/ vI + l

l

Ak"iv, + I

the Neumann series on the right being absolutely convergent. Since the

Al are > 0 and VI

12+1

398

VIII A The set E has positive lower density

by (f) and (if ), the previous relation gives Ak

1

uk

We proceed to examine the right-hand side of this inequality. We have °°

const.

1

1

(k-1)2+1 I2+1 ' k2+1 (look at the reproduction property of the Poisson kernel y/((x - t)2 + y2) on which the hall of mirrors argument used in Chapter 6 is based!). Hence, by (*), there is a constant L with 0

Ak"

L"

(k-l)2+1,

and the summand 1

Akn` 12+1

on the right side of the above estimate for Uk is const. L"

k2+1 We have, however, to add up infinitely many of these summands. It is here that we must resort to the computational lemma. Call °°

vk") =

Ak"i =-.l2

+1'

we certainly have vk") >, 0. According to the computational lemma, there is an q > 0 depending on A and the K in (*) such that 1+).

00

E

1= - 00

Ak,ryr

2

1

k2 + 1

if 0 ,
there is an M depending on m such that vkm)

_=

Ak ,)

M

12+1

k2+l'

I Harmonic measure for -9 = C - E

399

and we may of course suppose that M > 1. Apply now the computational lemma with Yl = vim)/M; we get (after multiplying by M again - this trick works because n/M < rl (m+1) =

1+2 M

(m)

``lk,lvl

vk

2

12+1

We also have, of course, 0 \ vkm+1) < An by (*).

We may now use the computational lemma again with 2

Yl = (I+ 1)M

v(m+1) l

note that here 2qA

(I+A)M <

0

ri.

After multiplying back by (1 +))M/2, we find v(m+ 2) k

(12M

lvl(m+l)

_

'`1k

2

I

1 k2 +

l'

In this fashion, we can continue indefinitely and prove that (M+ p ) Vk

z

1+.1 p M k2+1

2

for p = 1, 2, 3,.... Therefore, since 2 < 1, v(kn)

n=m

5

2M (1 - 2)(k2 + +

1).

This, however, implies that 1

uk \ k2+C

K{k2+l+vkl)+...+vkm-1)+ Y vkn)

n=m

1

!),

400

VIII A The set E has positive lower density

with a certain constant C, since (n)

vk

const. L" k2 + 1

\

for each n. We have proved that wo(x) (which is at most Uk on ()k) is C <1

k2 + 1

for xe(9k.

Q.E.D

Problem 15 In this problem, the set E is as described at the beginning of the present §, with the boxed condition given there. (a) Let UR(z) be the harmonic measure (for -9) of the subset E n [ - R/2, R/2] of 8.9, seen from ze2. Show that there is a number a > 0 depending only on the four quantities A, B, 6 and A associated with E, such that UR(z) < i

for Izl = R and 13zl -< aR. (Hint: First look at UR(z) for Izl = R and I..3zl < 1; then use Harnack.) (b) Let VR(z) be the harmonic measure (for -9) of

E n I( - oo, - R/2] u [R/2, oo) },

seen from ze.9. Show that there is a number Q > 0 depending only on A, B, 6 and A such that VR(z) >, 13 for I z I = R. (Hint: Use (a) and Harnack.) (c) For R > 0, call 2 R = -9n { Izl < R} and let (WR(z) be the harmonic measure

of {Izl = R} for 2R, as seen from zefR. Prove Benedicks' lemma, which says that w (0) R

C

1< R

with a constant C depending only on the four quantities A, B, S and A. (Hint: Compare the VR(z) of (b) with WR(z) in SR.) 2.

Green's function and a Phragmin-Lindelof function for -9

A Green's function is available for domains -9 of the kind considered here. Let us remind the reader who may not remember that, for given wee, the Green's function G(z, w) is a positive function of z, harmonic in -9 save at w where it acts like log

1

Iz - wI

which is bounded in -9 outside of disks of positive radius centered at w, and tends to zero when z tends to any point on the boundary E of -9. Existence

2 Green's function and a Phragmen-Lindelof function

401

of G(z, w) for our domains.9 follows by standard general arguments, found in many books on complex variable theory. For the sake of completeness, we will show that G(z, w) is symmetric in z and w at the end of this article. The last part of the argument given there may easily be adapted so as to furnish an existence proof for G(z, w).

Theorem. Let -9 = C - E, where E s l has the properties given at the beginning of this §. Assume that 0 e (90. Then there is a constant C, depending only on the four numbers A, B, S and A associated with E, such that G(x, 0) ,
Proof. Draw a circle r with diameter running from the left endpoint of E0 to the right endpoint of E1:

Figure 117

Let us show first of all that there is a number a, depending only on S, A and B, such that

G(z, 0) 5 a,

z e I".

To see this, observe first of all that G(z, 0) < g(z, 0), the Green's function for (C - E1) u { oo }. This follows by looking at the difference g(z, 0) - G(z, 0)

on E. The latter is harmonic and bounded in .9, since the logarithmic poles of g(z, 0) and G(z, 0) at 0 cancel each other out. It is thus enough to get an upper bound for g(z, 0) on F, and that bound will also serve for G(z, 0) there. Translation along R to the midpoint of E1 followed by scaling down, using the factor 2/1E11, takes (C - E1) u { oo } conformally onto the standard domain e = (C - [ - 1,1 ]) u { co 1:

402

VIII A The set E has positive lower density

Figure 118

In this reduction, 0 goes to a point p on the real axis, p < - 1, and the circle F goes to another, y, having [q, 1] as its diameter, where q < p. g(z, 0) is of course equal to the Green's function for .9 with pole at p. We have IPI-1
B

,

and

191-IPI ,

21E01

26

IE11

0

Therefore the Green's function for 9 with pole at p is bounded above on y by some number a depending on B/& and 26/A. (The nature of this

dependence could be worked out by mapping 9 conformally onto {1
This being verified, we take the centre m of r and, with R equal to that circle's radius, map the exterior of F conformally onto the domain 9 just considered by taking z to w = {(z - m)/R + R/(z - m) }. That

i and each of the components mapping takes t to the slit Ei = [- 1, 1] E of 8-9, n 0 0, 1, onto segments E' on the real axis. The function qp(z) =

i { (z - m)/R + R/(z - m) } thus takes -9r = !2c {Iz-m1>R}

conformally onto a domain

-9'=C

U'E;:

2 Green's function and a Phragmen-Lindelof function

403

Figure 119

Define a harmonic function Q(w) for wet' by putting f2((p(z)) = G(z, 0) for zE-9r. S2(w) is evidently bounded in.9', and has boundary value zero on cp(E )) of a-', save on E'1. f2(w) is, however, each of the components

continuous up to the latter one, and on it Q(w) '< a, since E i = cp(i') and G(z, 0) < a on F. We therefore have f2(w) 5 aow_,'(E'l, w) in -9', where co,,( , w) denotes harmonic measure for .9'.

The set oo

E' = U' En, -00

has, however, the properties specified for our sets E at the beginning of this §. Indeed, for real x, 1

OX) OX) = 2R x

m 2R +

ON x/ O

when I x I is large, with R lying between the two numbers 26 and B/2 + 20.

Hence, each of the intervals E (n 0 0) is of the form [a -

a' +

where, for certain numbers A', B', y' and A' depending on the original A, B, E and A,

404

VIII A The set E has positive lower density

0 < A'
0<6'<8;,
and find that (o.,'(E1' u)

<

1

C, -+U 2'

uER,

with a constant C depending on A', B', 6' and A' and hence, finally, on A, B, 6 and A. Thence, in view of the previous relation,

<

aC'

for

1 + u2

G(x 0) <

uc-R,

aC' 1 + ((P(x))2

for real x lying outside the circle r. Using the fact that Oe0(whence I mI 5 B + 2A ), the bounds on R given above, and the asymptotic formula for p(x), we see that the right side of the preceding inequality is in turn

with a constant C depending only on A, B, 6 and A. Thus G(x, 0)

C

I+ x2

for real x outside of E0, Uo and E1. But this also holds on Eo and E1 since G(x, 0) = 0 on those sets! So it holds for real x outside of 00, which is what we had to prove. We are done. Problem 16 Let E s R fulfill the conditions set forth at the beginning of this §, and assume that Oe90. Let ow_,( , z) be harmonic measure for 9 = C - E. Prove

Benedicks' theorem, which says that there is a constant C depending only on the four numbers A, B, S and A associated with E, such that, for t in any component E. = [a - b,,, a +

S]

2 Green's function and a Phragmen-Lindelof function

405

of E, andn96 0, 1,

dt

C

dco,,,,(t, 0) <

J(S2 - (t -

1 + t2

an)2)'

(This is a most beautiful result, by the way!) (Hint: Let G be the Green's

function for .9. According to a classical elementary formula, if, for instance, we consider points t+ lying on the upper edge of En, we have dw.Q(t+, 0)

=

dt

1

Gy(t

27t

+, 0) =

G(t + iAy, 0)

lim AY-O+

2nAy

since G(t, 0) = 0. (Green introduced the functions bearing his name for this very reason!) Take the ellipse t given by the equation (x - an)z

y2

2S

S

and compare G(z, 0) with

U(z) = log

a.)'

z Snan

+

((z

-1/

on r. Note that G(x,0) and U(x) both vanish on En, that U(z) is harmonic in the region & between E. and 17, and that G(z, 0) is at least subharmonic there (not necessarily harmonic because some of the Ek with k # n may intrude into 8).)

The work in Chapter VI frequently involved entire functions of exponential type bounded on the real axis. If f (z) is such a function, of exponential type A say, and we know that

f(x)I < 1,

xeOB,

we can deduce that

f(z)I < e' 3 for all z. This follows by the third Phragmen-Lindelof theorem of Chapter

III, §C, whose proof depends on the availability of the function I3zl, harmonic and > 0 in each of the half planes {,3z > 0}, {,rjz < 0}, and zero on the real axis. Suppose now that we are presented with such a function f (z), known to be bounded (with, however, an unknown bound) on O8, such that, for some closed E s O8,

f(x)I < 1, xeE. If there is a function Y(z), harmonic in -9 = C - E, having boundary value zero on E and such that Y(z) >, 13z I, ze , we can argue as in the proof of the

VIII A The set E has positive lower density

406

Phragmen-Lindelof theorem just mentioned, and conclude that If(z)I < eAYiZ)

for ze.9 if f(z) is of exponential type A.* We are therefore interested in the existence of such functions Y(z) for given closed sets E s R. In order to avoid situations involving irregular boundary points for the Dirichlet problem, whose investigation has nothing to do with the material of the present book, we limit the following discussion to closed sets E which can be expressed as disjoint unions of segments on Ft not accumulating

at any finite point. We do not, however, assume in that discussion that the sets E have the form specified at the beginning of this §.

Definition. A Phragmen-Lindelof function Y(z) for .9 = C - E is one harmonic in 9 and continuous up to E, such that (i)

Y(x) = 0,

xeE

ze-9, (ii) Y(z) ? 13Z 1, (iii) Y(iy) = I Y I + o(I Y I) for y -+ ± oo .

It turns out that for given closed E

Ft of the form just described, the

existence of Y(z) is governed by the behaviour of the Green's function G(z, w)

for -9 = C - E. Before going into this matter, let us mention a simple example (not without its own usefulness) which the reader should keep in mind.

Figure 120 * In fact, boundedness off on R is not necessary here. If I f(x)I < 1, xeE, and If(z)I IyI tan S} and -9 n {x < - Iy1 tan S}, even though the latter domains are not full sectors. Therefore v is bounded above in -9, so by the first theorem of §C, Chapter III, v(z)0 in -9. Hence I f (z) I < exp(A sec S Y(z)), zeQ, and, making b -+0, we get I f (z) I < exp(A Y(z)).

2 Green's function and a Phragmen-Lindelof function

407

Here, E = (- oo, - 1]u[1, oo). In .9 = C ' E, we can put Y(z) = :S(NAz2 - 1)),

using the branch of the square root which is positive imaginary for ze( - 1, 1). It is easy to check that this Y(z) is a Phragmen-Lindelof function for -9. The Green's function G(z, w) for one of our domains -9 enjoys a symmetry property:

G(z, w) = G(w, z),

z, we-9.

The reader who does not remember how this is proved may find a proof,

general enough to cover our situation, at the end of this article. It is convenient to define G(z, w) for all z and w in (which here is just C!) by taking G(z, w) = 0 if either z or w belongs to 8-9. Then we have G(z, w) = G(w, z)

for

z, we.

(N.B. G(z, w) as thus defined is not quite continuous from x .9- to [0, oo] (sic!). We can take sequences {z"} and {w"} of points in -9, both tending to limits on E = 8-9, but with I z" - w,, +0 sufficiently rapidly to make G(z", w")

oo.)

The connection between G(z, w) and Y(z) (when the latter exists) can be made to depend on the elementary formula R

log

lim R-+oo

1 - z

-R

t

dt = xI3zl,

which may be derived by contour integration. The reader is invited (nay, urged!) to do the computation. Here is the result.

Theorem. A Phragmen-Lindelof function Y(z) exists for -9, a domain of the kind considered here, iff G(z, t)dt < oo -00

for some ze2, G being the Green's function for that domain. If the integral just written converges for any such z, it converges for all ze.9, and then Y(z) = 13z I +

I f- -

G(z, t) dt.

Remark. In his 1980 Arkiv paper, Benedicks has versions of this result for R"+1

408

VIII A The set E has positive lower density

Proof of theorem. The idea is very simple, and is expressed by the identity f 00 13z 1

+

G(z, t) dt -00

_

J-AI

+

logltl+logJz-tI+G(z,t) dt log!

7r

E'o

-zz

2 dt +

1

G(z, t) dt,

an obvious consequence of the formula just mentioned. Here, A > 0 is arbitrary. Suppose, indeed, that the left-hand integral is convergent. That integral then equals a positive harmonic function in each of the half planes {3z > 0}, {,)z < 0}, and we can use Harnack to show that it is o(Iyl) for z = iy and

y -. ± oo. Denoting the left side of our identity by Y(z), we thus see that Y(z) is harmonic in the upper and lower half planes and has property (iii). It is clear that Y(z) has the properties (i) and (ii). Only the harmonicity of Y(z) at points of -9 n I1 remains to be verified. This, however, can be checked in the neighborhood of any such point by taking A > 0 sufficiently large and looking at the right side of our identity. The first right-hand term will be harmonic in -9 n ( - A, A), because, for each t therein, the

logarithmic pole of G(z,t) at t is cancelled by the term loglz-fl. The second term on the right is clearly harmonic for Izi < A, and the third harmonic in -9 n (-A, A). This explanation will probably satisfy the experienced analyst. The general mathematical reader may, however, well desire more justification, based if possible on general principles, so that he or she may avoid having to search through specialized books on potential theory. We proceed to furnish this justification. Its details make the following development somewhat long. Let us begin with a preliminary remark. Suppose we have any open subset 0 of -9, and a compact F C -!P

from 0. By the symmetry of G, G(z, w) = G(w, z)

is, for each fixed ze0, continuous (as a function of w) on F, so, if µ is any finite positive measure on F, the integral

f, G(z, w)dp(w) is obtainable as a limit of Riemann sums in the usual way. As a function of z, any one of those sums is positive and harmonic in 0. So the integral, being a pointwise limit of such functions (of z), is itself a positive and harmonic function of z in 0. We will make repeated use of this observation.

2 Green's function and a Phragmen-Lindelof function

409

Suppose, now, that I'. G(z, t)dt < oo for some non-real z, say wlog that G(i, t)dt < oo. For each N, the function

HN(z) = J

G(z, t)dt

N N

is positive and harmonic in both {3z > 0} and {.3z < 0} according to the remark just made. Therefore, since G(z, t) > 0, HN+I(z) > HN(z), and

H(z) = lim HN(z) N-oo

is either harmonic (and finite!) in {3z > 0} or else everywhere infinite there. Because H(i) < oo, the first alternative holds, and H(z) is then also finite (and harmonic) in

3z < 0, since obviously G(z, t) = G(ff, t)

for real t, E = a-9 being on R. Consider now some real x0OE. Take A > max(lxo1,1). The integrals A A

G(xo, t)dt and f AAAG(i, t)dt are both finite, so we can show that

G(xo, t)dt

and 1'. G(i, t)dt are either both finite or else both infinite by comparing G(xo, t) dt Li A

and

G(i, t) dt.

In -qA=_qn{IzJ
is

the limit of the

G(z, t)dt, SA

ti_
each of which is positive and harmonic in -9A. So SItJ,AG(z,t)dt

is either harmonic (and finite) in 2A, or else everywhere infinite there. It is thus finite for

if and only if it is finite for z = xo. We see that 1% G(xo, t)dt < oo if f'. G(i, t)dt < oo, and, if this inequality holds, z=i

H(z) = J

G(x, t)dt

is finite for every ze9. If H(z) is finite, let us show that

H(iy)=o(lyl) for y- ±oo.

410

VIII A The set E has positive lower density

Pick any large N, and write (*)

H(iy) = fit

G(iy, t)dt

+J

G(iy, t)dt.

I,>N

ISN

Since H(i) < oo we can, given any s > 0, take N so large that fItI, N G(i, t) dt < s. The function f 1113 N G(z, t)dt is, by the previous discussion, positive and harmonic in {3z > 0). Therefore, by Harnack's theorem,*

G(iy, t)dt < y J

J 1tI3N

G(i, t)dt < cy ItI

N

for y> 1. This takes care of the second term on the right in (*). The first term from the right side of (*) remains; our claim is that it is bounded. This

(and more) follows from a simple estimate which will be used several times in the proof. Take any component Eo of E = 8-9, and put -9p = (C - Eo) u { oo }. If Eo is of infinite length, replace it by any segment of length 2 thereon in this last expression. We

have .9 s -9o, so, if g(z, w) is the Green's function for .90, G(z, w) < g(z, w),

z, we-9

(cf. beginning of the proof of first theorem in this article). We compute g(z, w) by first

mapping - 9 o conformally onto the unit disk {IIC < 1), thinking of t' as a new coordinate variable for 19o:

Figure 121

Say, for instance, that Eo is [ - 1, 1] so that we can use z = ( + 1/c). Then, if t c-.90 is * Actually, by the Poisson representation for (Zz > 0) of functions positive and harmonic there. Using the ordinary form of Harnack's inequality gives us a factor of 2y instead of y on the right. That, of course, makes no difference in this discussion.

2 Green's function and a Phragmen-Lindelof function

411

real, we can put t = Z(T + 1/T), where -1 < T < 1, and, in terms of and T,

1-T g(z, t) = log r

S-T '

the expression on the right being simply the Green's function for the unit disk. If N > 1, we have G(z, t)dt

J

<J

Irish

g(z, t)dt, 1

N

It

since G(z, t) and g(z, t) vanish for tEE0 = [ - 1, 1]. For 1 < I tl < N, the parameter T satisfies CN 1< ITI < 1, CN > 0 being a number depending on N which we need not calculate. Also, for such t, 1 -T 2

1

dt = - 2

dT. TZ

Therefore,

(t)

G(z, t) dt 5 1

J

lo g

2 fCN Shisl

IrIsN

I - Tl;

b-T

dT. TZ

Since CN > 0, the right side is clearly bounded for I C I < 1; we see already that the first

right-hand term of (*) is bounded, verifying our claim. As we have already shown, the second term on the right in (*) will be , By for y , 1 if N is large enough. Combining this result with the preceding, we have, from (*),

O(1)+ey, y> 1,

H(iy)

so, since e > 0 is arbitrary, H(iy) = o(I yi ), y -> cc. Because H(i) = H(z), the same holds good for y - cc.

Having established this fact, let us return for a moment to (t). For each T, CN_< ITI<1, logl-T

-+0 as

ICI

Starting from this relation, one can, by a straightforward argument, check that log

1 - TC

1 -T2 Z

yy

CN<1rI<1

dT--0

T

as ICI-+1. (One may, for instance, break up the integral into two pieces.)

Problem 17 (a) Carry out this verification This means, by (t), that G(z, t) dt -+0 tI

N

412

VIII A The set E has positive lower density

when ZE2 tends to any point of E0. We could, however, have taken Eo to be any of the

components of E with finite length, or any segment of length 2 on one of the unbounded ones (if there are any); that would not have essentially changed the above argument.* Hence G(z, t) dt

tends to zero whenever z tends to any point on E = 8-9 (besides being bounded in 9).

We can now prove that

H(z) = J

G(z, t) dt -. 0

whenever z tends to any point xo of E. Given such an xo, take a circle y about xo so small that precisely one of the components of E (the one containing x0) cuts y, passing into its inside:

Figure 122

If xo is an endpoint of one of the components of E, our picture looks like this:

Figure 123

Call -9, the part of -9 lying inside y, and E, the part of E therein. * As long as Eo s { I t < N}. If this is not so, we can increase N until the argument in the text applies. Since that only makes the integral in question larger, the one corresponding to the original value of N must (a fortiori!) have the asserted

2 Green's function and a Phragmen-Lindelof function

413

E Figure 124

Fix any z, e-9y. Then, there is a constant K, depending on z,, such that, for any function V(z), positive and harmonic in -9,, and continuous on its closure, with V(x) = 0 on EY, we have V(z) < K V(z,) for I z - xo < i radius of y.

Problem 17 (b) Prove the statement just made. This being granted, choose N so large that

G(z,,t)dt < E/K, e being any number ('> 0. For each M > N, the function

G(z,t)dt

VM(z) = J NsIti M

is positive and harmonic in .9y, and certainly continuous up to y n .9. Also, VM(z) <, f ItI, M G(z, t) which, by the previous discussion, tends to zero whenever z tends to any point of E. VM(z) is therefore continuous up to EY, where it equals zero.

By the above statement, we thus have

G(z,, t)dt < E

VM(z) < KVM(z,) -< K J tI>N

for I z - xo I < i radius of y. This holds for all M > N, so making M -+ oo, we

get f Jti , N G(z, t) dt < E for I z - xo I < 1 radius of y. Hence, since

H(z) = J

G(z, t)dt + f ItI
G(z, t)dt,

Id>- - N

and, as we already know, the first integral on the right tends to zero when z -+ x0, we must have H(z) < 2E for z close enough to x0. This shows that H(z) -+0 whenever z

tends to any point of E.

414

VIII A The set E has positive lower density

We now see by the preceding arguments that 1

Y(z) = I zI + -H(z) it

enjoys the properties (i), (ii) and (iii) required of Phragmen-Lindelof functions, and is also harmonic in both the lower and upper half planes, and continuous everywhere.

Therefore, to complete the proof of the fact that Y(z) is a Phragmen-Lindelof function for -9, we need only verify that it is harmonic at the points of -9 n R. For this purpose, we bring in the formula 1 CA

= lim -

1.3z l

A-'

z

log 1 - -

R -A

dt

t

mentioned earlier. From it, and the definition of H(z), we get (*)

Y(z) = 13zI + 1

G(z, t)dt

n

A

A(log-+loglz-tl+G(z,t)

= 7'1 f-

Idt

Itl 1

°°

+ -

z2

log

1- z dt +

n' fA

t

if n

G(z, t) dt. tI>A

The number A > 0 may be chosen at pleasure.

Let x0 egr l

;

pick A larger than Ix0I. The function J logll -z2/t2Idt

is certainly harmonic near x0; we have also seen previously that f10AG(z,t)dt is harmonic in -9 n{IzI
(log lz-tl+G(z,t))dt A

at x0. Take a S > 0 such that (xo - 56, x0 + 56) c.9. According to observations already made, G(z, t) dt

is harmonic for I z - x0I < S; so is (clearly)

logIz - tldt. JI, -xol>a ItI
We therefore need only check the harmonicity of xo+ 8

(log lz-tI+G(z,t))dt. Sxo-4

2 Green's function and a Phragmen-Lindelof function

415

Here, we must use the symmetry of G(z, w). In order not to get bogged down in notation, let us assume that x0 = a > 0 and that the segment [ - 2a, 6a] lies entirely in -9. The general situation can always be reduced to this one by suitable translation. It

will be enough to show that 2.

(log Iz-tI+G(z,t))dt

f

o

is harmonic for I z - a I < a. For each fixed z,

loglz-wI+G(z,w) = logjw-zI+G(w,z) is a certain harmonic function, hz(w), of wee; this is where the symmetry of G comes in. (h=(w) is harmonic in w even at the point z, for addition of the term logj w - zI removes the logarithmic singularity of G(w, z) there.) Hence, if p < dist (w, E), zx

1

hz(w) = 2

hz(w + pe'9)d9.

This relation makes a trick available. In it, put w = t where 0 < t < 2a, and use

p=t+2a. We get 2x

h=(t) =

1

2n

hjt + (t + 2a)e's)d9, 0

whence, ('2a

J

f

1

2a

2x

(logIz-t(+G(z,t))dt = -

0

Figure 125

2ir

h=(t+(t+2a)e"9)d9dt. o

0

416

VIII A The set E has positive lower density

The double integral on the right can be expressed as one over the region

S = {K-2aI<4a}n{KI>2a} shown in the above picture. Indeed, the mapping (t, 9)

n)

given by . + in =1' = t + (t + 2a)e'o takes {0 < t < 2a} x {O < 9 < 2rz} in one-one fashion onto S, and the Jacobian a(, n) 3(t, 9)

works out to (t + 2a)(1 + cos 9)

+ 2a.

Hence, I

- J2j.2" o

h(t + (t + 2)e19) d9dt =

2-Iff h)

o

s

n

+ 2'

so, by the previous relation, ('2a

(log I z - t j + G(z, t)) dt 0

log I z- C I

1

+ 2a

2a ,) ,Is

1

d do +

2n ,1

G(z, l')

s + 2a d dn.

Here, we have

d do

if

f22 f211 o

,I

d9dt
o

so both of the above double integrals must equal harmonic functions of z in the disk ( I z - a I < a}, disjoint from 9. (This follows for the second of those integrals by the remark at the very beginning of this proof.) We see that the left-hand expression is

harmonic in z for z near xo = a. According, then, to (;) and the observations immediately following, the same is true for Y(z). We have finished proving that Y(z) is a Phragmen-Lindelof function for -9 if the

integral I. G(z, t)dt is finite for any z therein. The second half of our theorem thus remains to be established. That is easier. In the second half, we assume that.9 has a Phragmen-Lindelof function Y(z), and

set out to show that G(z, t) dt < oo _w

for each ze.9, G being that domain's Green's function.

2 Green's function and a Phragmen-Lindelof function

417

Given any A > 0, consider the expression

+n

YAW = 13Z I

IA A G(z, t) dt

A

A( logII+loglz-tl+G(z,t) Idt

1 n

z2

1- 2 dt. t

From the preceding arguments, we know that the first integral on the right is harmonic for ze

- proof of this fact did not depend on the convergence of

G(z, t) dt.

-M

What we have already done also tells us that YA(z) tends to zero when z tends to any

point of E (again, whether f °°. G(z, t)dt converges or not) and that, for any fixed A, fA G(z, t) dt A

is bounded in the complex plane. The expression z2

2 dt

l og

t

JA

is evidently subharmonic in the complex plane. The function YA(z) given by the above formula is thus subharmonic, and zero on E,

and moreover, YA(z) = 13ZI + O(1),

Ze-q.

Our Phragmen-Lindelof function Y(z) (presumed to exist!) is, however, harmonic and >, 13z I in.9, and zero on E The difference YA(z) - Y(z) is therefore subharmonic and bounded above in 9, and zero on E. We can conclude by the extended maximum

principle (subharmonic version of first theorem in § C, Chapter III) that . In other words,

YA(z) - Y(z) < 0 for ze 1

13z I +

It

A

G(z, t) dt < Y(z). A

Fixing ze2 and then making A -+ oo, we see that

_. G(z, t) dt s Y(z) -13zI < 00. 71

418

VIII A The set E has positive lower density

This is what we wanted. The second half of the theorem is proved. We are done.

We apply the result just proved to domains .9 of the special form described at the beginning of the present §, using the first theorem of this article. In that way we obtain the important

Theorem (Benedicks). If E is a union of segments on F fulfilling the conditions given at the beginning of this § (involving the four constants A, B, b

and A), there is a Phragmen-Lindelof function for the domain .9 = C - E.

Proof. Assume wlog that 0 e -9, and call (9o the component of lf8 - E containing 0. By the first theorem of the present article,

for te00, and clearly

G(t,O) < log' II +0(1),

ted0.

ThereforeJT (symmetry again!)

G(0,t)dt =

G(t,0)dt < oo. J

-0000

Now refer to the preceding theorem. We are done. This result will be applied to the study of weighted approximation on sets E in the next article. We cannot, however, end this one without keeping our promise about proving symmetry of the Green's function. So, here we go:

Theorem. In .9 = C - E, G(z, w) = G(w, z). Proof. Let us first treat the case where E consists of a finite number of intervals, of finite or infinite length. (If E contains two semi-infinite intervals at opposite ends of R,

we consider them as forming one interval passing through oo.) We first proceed as at the beginning of article 1, and map -9 (or -9v{ cc }, if oo OE) conformally onto a bounded domain, bounded by a finite number of analytic Jordan curves. This useful trick simplifies a lot of work; let us describe (in somewhat more detail than at the beginning of article 1) how it is done.

2 Green's function and a Phragmen-Lindelof function

419

Suppose that E,, E21... , EN are the components of E. First map (Cu { co }) - E, conformally onto the disk (I z I < 11; in this mapping, E, (which gets split down its middle, with its two edges spread apart) goes onto { I z I = 1}, and E2, ... , EN are taken onto analytic Jordan arcs, A2..... AN respectively, lying inside the unit disk. (Actually, in our situation, where the Ek lie on R, we can choose the mapping of (C u { oo }) - E, onto { I zI < 1 } so that U8 - E, is taken onto ( -1,1). Then A2, ..., AN will be segments on (- 1, 1).) In this fashion, .9 is mapped conformally onto

{IzI<1}- A2-A3-...,,,AN' Now map (Cu { oo }) - A 2 conformally onto { 1w I < 1). In this transformation, z I =1 } goes onto a certain analytic Jordan curve W, lying inside the unit disk, A. (after having its two sides spread apart) goes onto { I w I = 1 }, and, if N > 2, the arcs A3,...,AN go onto other analytic arcs A3,...,AN, lying inside {Iwl < 1}. (A3,...,AN

are indeed segments on (- 1, 1) in our present situation, if this second conformal mapping is properly chosen.) So far, composition of our two mappings yields a conformal transformation of -q onto the region lying in { I w I < I), bounded by the unit circumference, the analytic Jordan curve W,, and the analytic Jordan arcs A3,._ , A'N (in the case where N > 2). It is evident how one may continue this process when N > 2. Do the same thing with A3 that was done with A2, and so forth, until all the boundary components are used up. The final result is a conformal mapping of .9 onto a region bounded by the unit circumference and N - 1 analytic Jordan curves situated within it. Under conformal mapping, Green's functions correspond to Green's functions. Therefore, in order to prove that G(z, w) = G(w, z), we may as well assume that G is the Green's function for a bounded domain it like the one arrived at by the process just described, i.e., with ail consisting of a finite number of analytic Jordan curves. For

such domains it we can establish symmetry using methods going back to Green himself. (Green's original proof - the result is due to him, by the way - is a little different from the one we are about to give. Adapted to two dimensions, it amounts to the observation that

G(z, w) = log

1

Iz-wl

+

fan log I C - w I dwn(C, z)

1

= log

I Iz-wl + JIanan J

where wn(

, z) is the harmonic measure for il. This argument can easily be made rigorous for our domains fl. The interested reader may want to consult Green's collected papers, reprinted by Chelsea in 1970.) If C e Oil and the function F is W, in a neighborhood of t', we denote by OF(g) On;

the directional derivative of F in the direction of the unit outward normal nt to ail at C:

420

VIII A The set E has positive lower density

Figure 126

If weft is fixed, G(z, w) is harmonic as a function of zef2 (for z away from w) and continuous up to 8f2, where it equals zero. Analyticity of the components of Oil means that, given any t'o e 8f2, we can find a conformal mapping of a small disk

centered at So which takes the part of ia) lying in that disk to a segment r on the real axis. If we compose G(z, w) with this conformal mapping for zef2 near Co, we see, by Schwarz' reflection principle, that the composed function is actually harmonic in a neighborhood of a, and thence that G(z, w) is harmonic (in z) in a neighborhood of o. G(z, w) is, in particular, a W. function of z in the neighborhood of every point on oft. This regularity, together with the smoothness of the components of Of), makes it possible for us to apply Green's theorem. Given z and wei2 with z 0 w, take two small non-intersecting circles yz and yw lying in f2, about z and w respectively. Call lithe domain obtained from fl by removing therefrom the small disks bounded by y. and yw:

2 Green's function and a Phragmen-Lindelof function

421

ri), where _ + iri, and by

Denote by grad the vector gradient with respect to '-'the dot product in R2. We have aGg, w)

aG(s, z)

JI

an'

Since the vector-valued function G(1;, w) grad G(l;, z) - G(t, z) grad G(l, w)

of C is W. in and on f', (W. on ail by what was said above), we can apply Green's theorem to the second of these integrals, and find that it equals

'sly

div (G(C, w) grad G(l4, z) - G({, z) grad G(C, w)) d dn,

where div denotes divergence with respect to (l;, rl). However, by Green's identity, div (G(1;, w) grad G(C, z) - G(C, z) grad G(1;, w))

= G(t, w)V 2G(C, z) -

z)V 2G(C, w),

where Q2 = a2/a 2 + a2/aq2. Because zoQ' and woQ',

z) and G(t,, w) are

harmonic in 1', 1; efY. Hence

V2G(C,z) = V2G(C,w) = 0,

CeH',

and the above double integral vanishes identically. Therefore the first of the above line integrals around OSY must be zero. Now ai' = ail u y= u y and G(C, w) = G(t', z) = 0 for C e Q. That line integral therefore reduces to w)

aG( C, z)

y= yan,

anS

)Idyl,

which must thus vanish. Near z, G(l;, z) equals log (1 /I l; - z I) plus a harmonic function of t'; with this in mind we see that the integral around y. is very nearly 2nG(z, w) if the radius of yz is small. The integral around yw is seen in the same way to be very nearly equal to - 2nG(w, z) when that circle has small radius, so, making the radii of both y= and yw tend to zero, we find in the limit that 27rG(z, w) - 2nG(w, z) = 0,

i.e., G(z, w) = G(w, z) for z, well. This same symmetry must then hold for the Green's functions belonging to finitely connected domains -9 of the kind we are considering. How much must we admire George Green, self taught, who did such beautiful work isolated in provincial England at the beginning of the nineteenth century. One wonders what he might have done had he lived longer than he did.

VIII A The set E has positive lower density

422

AN ESSAY ON THE

APPLICATION OF

MATHEMATICAL ANALYSIS TO THE THEORIES OF ELECTRICITY AND MAGNETISM.

BY

GEORGE GREEN. *ottingbam; PIX TRD '"R T

AV=Oz, NT r. W>®Zzmovm.

SOLD BY HAMILTON, ADAMS & Co. 33, PATERNOSTER ROW ; LONGMAN & Co.; AND W. JOY, LONDON; J. DEIGHTON, CAMBRIDGE; AND S. BENNETT, H. BARNETT, AND W. DEARDEN, NOTITNGHAM.

1828.

Once the symmetry of Green's function for finitely connected domains -9 is known,

we can establish that property in the general case by a limiting argument. By a slight modification of the following procedure, one can actually prove existence of the Green's function for infinitely connected domains = C - E of the kind being considered here, and the reader is invited to see how such a proof would go. Let us, however, content ourselves with what we set out to do. Put E. = E n [ - R, R] and take 2R = (C u { oo }) - ER. With our sets E, ER consists of a finite number of intervals, so -9R is finitely connected, and, by what we have just shown, GR(z, W) = GR(w, z) for the Green's function GR belonging to QR. (Provided, of course, that R is large enough to make IERI > 0, so that -QR has

a Green's function! This we henceforth assume.) We have -9R ? -9, whence, for

z,we.9, G(z, w) < GR(z, w).

2 Green's function and a Phragmen-Lindelof function

423

If we can show that GR(Z, w) - G(z, w)

for z, w e -9 as R -+ oo, we will obviously have G(z, w) = G(w, z).

To verify this convergence, observe that W) < GR(Z, w)

for z and win 1R (hence certainlyforz,we2 !)when R',> R, because then -9R. g 9R. The limit G(z, w) = lim GR(z, W) Rim

thus certainly exists for z, wE91, and is -> 0. If we can prove that -(;(z, w) = G(z, w), we will be done. Fix any wE2i. Outside any small circle about w lying in -9, l`i(z,w) is the

limit of a decreasing sequence of positive harmonic functions of z, and is therefore itself harmonic in that variable. Let x0EE. Take R > Ixol; then, since 0-< (i(z, w) < GR(z, w) for ze91 and GR(z, w) -.0 as z -+xo, we have

G(z, w) -+0

for z - * xo.

If we fix any large R, we have, for ze21,

0 < C(z, w) < GR(z, w) = log

1

Iz-wI

+ O(1).

Therefore, since

G(z,w) = log

1

Iz-WI

+O(1),

we have C(z, w) < G(z, w) + O(1),

ZE-9.

However, this last inequality can be turned around. Indeed, for zE91 and every sufficiently large R, GR(z, w) % G(z, w),

from which we get

G(z,w) 3 G(z,w),

ze21

on making R -+ oo.

We see finally that 0 < G(z, w) - G(z, w) < 0(1) for ze-9 (at least when z # w) and z # w); the difference in question is, moreover, harmonic in z (for tends, according to what we have shown above, to zero when z tends to any point of E = 891. Hence

C(z, w) - G(z, w) = 0,

zE-9,

424

VIII A The set E has positive lower density

GR(z, w) -. G(z, w)

for zE.9 when R -. oo, which is what we needed to establish the symmetry of G(z, w).

We are done. 3.

Weighted approximation on the sets E

Let E be a closed set on R, having infinite extent in both directions and consisting of (at most) countably many closed intervals not accumulating at any finite point. Suppose that we are given a function W(x) >, 1, defined and continuous on E, such that W(x) -> oo for x -+ ± co in E. Then, in analogy with Chapter VI, we make the

Definition. 'w(E) is the set of functions cp defined and continuous on E, such that -p(x)

W(x) -4

0

for

x - + -oo in E.

And we put IIwIIW,E = sup ox) xeE

W(x)

for cpE'w(E).

For A > 0, we denote by 'w(A, E) the II II w,E-closure in 'w(E) of the collection of finite sums of the form Cxe'xx Y_

-ASa6A

Also, if, for every n > 0, x"

W(x)

i0

as

x - + oo in E,

we denote by 'w(0, E) the II II w,E-closure in 'w(E) of the set of polynomials.

We are interested in obtaining criteria for equality of the Ww(A, E), A > 0, (and of 16w(0, E)) with 'Ww(E). One can, of course, reduce our present

situation to the one considered in Chapter VI by putting W(x) = co on IIB - E and working with the space''(O). The equality in question is then governed by Akhiezer's theorems found in §§B and E of Chapter VI, according to the remark in §B.1 of that chapter (see also the corollary at the end of §E.2 therein). In this way, one arrives at results in which the set E does not figure explicitly. Our aim, however, already mentioned at

3 Weighted approximation on E

425

the beginning of the present chapter, is to show how the form log W*(x)

1 +x2

dx,

occurring in Akhiezer's first theorem, can, in the present situation, be replaced by log W,k(x)

f

E

1 +x2

dx

when dealing with certain kinds of sets E. That is the subject of the following discussion. Our results will depend strongly on those of the preceding two articles.

Lemma. Let A > 0, and suppose that there is a finite M such that (*)

(' logIS(x)I dx f-+ x2 E

<M

for all finite sums S(x) of the form aae' z x Y, -ASZSA with II S 11 w,E <

1. Then there is a finite M' such that

f log, IS(2 )I dx (§) E

\

1 +X2

M'

for such S with IISIIw,E51 Proof. Given a sum S(x) of the specified form with II S II w,E < 1, we wish to show that (§) holds for some M' independent of S. Let us assume, to begin with, that the exponents A figuring in the sum S(x) are in arithmetic progression, more precisely, that N

S(x) = Y,

n=-N

Cne'nhx

where h = A/N, N being some large integer. There is then another sum N

T(x) = I ane'nhx n= -N

(which is thus also of the form

Y-_A11

1 + S(x)S(x) = T(x)T(x)

ACxe'zx ) such that

for xell.

This we can see by an elementary argument, going back to Fejer and

426

VIII A The set E has positive lower density

Riesz. For xel, we have 2N

Y yneinhx

I+ S(x)S(x) =

n=-2N

with certain coefficients y,,. Write, for the moment, eikx = y. ,

then

1 + S(x)S(x) = R(C),

where 2N

R(C) _

Y ynS"

n=-2N

is a certain rational function of C. We have R(C) >, 1 for I l; ( = 1, so, by the Schwarz reflection principle,

R(1/t) = R(C). Therefore, if a, 0 < I a I < 1, is a zero of R(C), so is 1/i, and the latter has

the same multiplicity as a. Also, if -m denotes the least integer n for which yn j4 0, we must have y,, = 0 for n > m (sic!), as follows on comparing

the orders of magnitude of R(t;) for C -0 and for C oo. The polynomial t'mR(C) is thus of degree 2m, and of the form M

const. H (t' - ak) C -

`

k=1

).

1

ak

Thence,

R(C) = C

k=1

( - ak)

-

- ak

and C > 0 since R(C) >, 1 for l; = 1. Going back to the real variable x, we

see that m

1 + S(x)S(x) = C fl (eikx - ak)(e - ihx _ ik) = T(x)T(x), k=1

where m

TO = C#e - iNhx r7 l(eihx _ ak) kj=j1

is of the form

3 Weighted approximation on E

427

N

Y ane n= -N

infix

since m < 2N. Once this is known, it is easy to deduce (§) for sums S(x) of the special form just considered with II S II 1,E 5 1. Take any such S; we have another sum T(x) of the same kind with 1 + I S(x) I2 = I T(x) 12 on R. Since W(x) > 1 on E, the condition 11 S II w,E 1< 1 implies that II T II w,E 1< -,/2, i.e., II T/,/21I w,E 1<

1.

For this reason, T(x)/,/2 satisfiesfE (*), by hypothesis. Hence log+T(x2)I

J

dx < M +

lo g X2 dx.

But log I T(x) I = log /(1 + IS(X)12) > log+ I S(x) 1. Therefore

log+X(2)I

dx 5 M+nlogV2,

E

and we have obtained (§). We must still consider the case where the exponents A in the finite sum aaeixx = S(x) Y-

-A"15A

are not in arithmetic progression, the condition II S II w, E < 1 being, however,

satisfied. Here, we may associate to each A figuring in the expression just written a rational multiple A' of A, with IA' - AI exceedingly small. Since W(x) -* oo for x -+ ± oo in E, the sum

S'(x) =

Y aze"

-A,
II w,E-norm to S(x) (depending will then be as close as we like in on the closeness of the individual A' to their corresponding A). In II

this way, we can get a sequence of sums Sn(x) of the form in question, each one having its exponents in arithmetic progression, such that 11 Sn 11 w,E < 1 and Sn(x) n > S(x) u.c.c. on R. By what we have already shown, f E(1og+ I Sn(x) I/(1 + x2)) dx 5 M + rz log.,/2 for each n. Therefore log

1,

x)I

+X2

dx 5 M + it log.,/2

by Fatou's lemma. We are done.

For the sets E described at the beginning of the present § we can

428

VIII A The set E has positive lower density

establish analogues, involving integrals over E, of the Akhiezer and Pollard

theorems given in Chapter VI, §§E.2 and E.4. These are included in the following theorem which, in one direction, assumes as little as possible and concludes that WW(A, E) c AW(E) properly. In the other, it assumes the proper inclusion and asserts as much as possible. For zeC, denote by WA,E(z) the supremum of IS(z)I for the finite sums

S(z) = Y

aAe'AZ

-A
with II S II W,E 51. (If we agree that W(x) - oo on R - E, WA,E(z) and

WW(A, E) reduce respectively to the function WA(z) and the space 'W(A) already considered in Chapter VI, §§E.2ff.)

Theorem. Let E be one of the sets described at the beginning of this §, the conditions involving the four numbers A, B, S and A being fulfilled, If, for some C > 0, the supremum of logIS(x)I

f

E

1+x2

dx

for all finite sums S(x) _

aze'Ax

Y-

-c
with II S IIW,E 1< 1 is finite, then Ww(C, E) c AW(E) properly.

If, conversely, that proper inclusion holds, then log WC,E(x)

1.

dx < -

Proof. All the work here is in the establishment of the statement. Define W(x) on all of F by putting it equal to 00 on I8 ' E. This makes

it possible for us to apply results about 'w(C) from Chapter VI, §E, in the present situation. According to the Pollard theorem of Chapter VI, §E.4, and the remark thereto, we will have W (C) 'eW(l) as soon as Wc(i) < co; in other words, 1' (C, E) 'w(E) provided that

Wc,E(i) < c0. We proceed to show this inequality, using the results of articles 1 and 2. According to the lemma, our hypothesis for the first part of the theorem implies that (*)

(' log, IS2x)I dx

f

JE

1+x

M' < 00

3 Weighted approximation on E

429

for all sums S(x) of the stipulated form with II S II w.E < 1 We have 00

E = U [a -

a +

n= - cc

where

0
Given any finite sum Y S(z) _

aze''lZ

c
with II S IIW,E S 1, let us put

vs(z) =

1

f/2 log+ I S(z + t) I dt a/2

(using the 6 associated to E by the above inequalities). The function vs(z) is then continuous and subharmonic in the complex plane. Obviously, I S(z) I < const.eC13z1

(the constant may be enormous, but we don't care!), so

vs(Z) < O(1)+CI.3zI, zeC.

(t)

Put now

r n=-ro

2

2

On the component

E = [an_ 2 , a + 2 Jl of E we have vs(x)

JE,log+IS(t)Idt, where (as usual) En = [an -

a + 6n]-

Denoting the right side of the previous relation by vn, we have (§§)

vs(x) <

The set E (like E) is one of the kind specified at the beginning of the

present §; the numbers A, B, 6/2 and A/2 are associated to

it. The

VIII A The set E has positive lower density

430

results of the previous two articles are therefore valid for E' and the domain -9' = C - E. We can, in particular, apply Carleson's theorem from article 1. Assume, wlog, that

0 e Coo = (ao+., a1-21/. Then, if we denote by wn(z) the harmonic measure of E in 2' (as seen from z),

that theorem tells us that K w'(0) <

I +n2

with a constant K depending on A, B, 6/2 and A/2. By Harnack's theorem, there is thus a function K(z), continuous in 9', such that

ze9'

wn(z) 5 IK(n)2'

(see discussion near the beginning of §B.1, Chapter VII). Using properties (i) and (ii) from the beginning of this § we see that the quantities vn introduced above satisfy v"

l+n

log+ I S(t) I

a E"

l + t2

dt

'

a being a certain constant depending only on the set E. Combined with the previous estimate, this yields vnwn(z) < aK(z)

log

E

+S2t)I dt,

zed',

so, for the sum

P(z) _

x Y-

v"wn(z),

n=-00

we have

(fit)

P(z) < o M'K(z),

ze9',

by virtue of (*). Because IS(x)I is bounded on R, the v" (which are >, 0, by the way) are bounded. The series used to define P(z) is therefore u.c.c. convergent, so

that function is continuous up to E' = 89' as well as being positive and harmonic in 9'. On the component En of E', P(x) takes the constant value vn. Hence the function VS(Z) - P(z)

3 Weighted approximation on E

431

is subharmonic in -9' and continuous up to E', where it is <, 0 by (§§). It is, moreover, 5 0(1) + C13z1 by (t). Now according to Benedicks' theorem (article 2), a Phragmen-Lindelof function Y(z) is available for .9'. The function

vs(z) - P(z) - CY(z)

is subharmonic and bounded above in -9' and continuous up to E' = 8-9' where it is < 0. It is thence 5 0 throughout -9' by the extended maximum principle (Chapter III, §C). Referring to (tt), we see that ($)

vs(z) S aM'K(z) + CY(z),

z c- .9'.

Let p=min (Z, 6/2). Since log+ IS(z)I is subharmonic, we have 1

log+IS(i)I 5rzp2

log+ I S(z) I dx dy. i=-1i
0

E'

E'

Figure 128

The integral on the right is l+p

52 1

1-p

rzP

d/2

log+ I S(x + iy) I dx dy = -6/2

2 8

nP

l+p

1-p

vs(iy)dy

Plugging ($) into the last expression, we obtain l+p b (aM'K(iy) + CY(iy)) dy log+ I S(i) 15 P2 l -° This, then, is valid for any finite sum S(z) of the form Y-

aze'zz

-cszsc

with II S II w.E < I

The right side of the inequality just found is a finite quantity, dependent on M' and C, and on the set E (through a, K(iy) and Y(iy) ); it is, however,

432

VIII A The set E has positive lower density

completely independent of the particular sum S(z) under consideration. Therefore Wc,E(i), the supremum of IS(i)I for such sums S with IIS II w,E 1 1,

is finite. This is what we needed to show in order to infer the proper inclusion of 'w(C, E) in Ww(E). The first part of our theorem is thus proved.

There remains the second part. That, however, is not new! Putting, as before, W(x) equal to oo on l - E, which makes 1Bw(C, E) coincide with the subspace lew(C) of lew(If8) considered in Chapter VI, §E, we have, in the notation of that §, log Wc(x)

l+ x2

dx < oo

if 16w(C) 96 ' (Fl), according to Akhiezer's theorem (Chapter VI, §E.2).

Our function Wc,E(x) is simply Wc(x). Hence, since E s L'8 (!), log WC,E(x)

f

E

1 + x2

dx < 00

when W w(C, E) 0 Ww(E).

The theorem is completely proved. We are done.

Remark. If we do not assume anything about the continuity of a weight W(x) >, 1 defined on E, it is still possible to characterize the equality of 'Bw(C, E) with W (E) by an analogue of Mergelian's second theorem involving an integral over E. The establishment of such a result proceeds very much along the lines of the proof just finished, and is left to the reader.

Problem 18 Let E be a closed set on l of the kind specified at the beginning of this

§. Show that there are two constants a, b, depending on E, such that, for any entire function f (z) of exponential type < C, bounded on R, we have

°° log(1+If()IZ)

_. 7(Hint:

l +x2

dx 5 aC + b

log(l+If(x)12) dx. l +x2 e

One may apply the third and fourth theorems from Chapter III,

§G.3, and reason as in the above proof. Another procedure is to use the proof of the lemma in §E.1 of Chapter VI so as to first approximate f (z) by finite sums S(z) of the form considered above, having exponents in arithmetic progression.)

If W(x), continuous and >, 1 on E, is such that xn

W(x)

_..0

for

x -- ± oo

in

E

3 Weighted approximation on E

433

and n = 1, 2,3,..., we denote by W0,E(z) the supremum of IP(z)I for all polynomials P with II P II w,E 5

1.

Theorem. Let E 9 P be a set of the kind specified at the beginning of this

§, and let W(x), continuous and >, 1 on E, tend to oc faster than any power of x as x - ± oo in E. If, for polynomials P(z) with 11 P II w.E 5 1, the integrals (' log I P(x) I JE 1 + x2

dx

are bounded above, then '(O , E) is properly contained in'w(E). If' w(0,E) is properly contained in lew(E), then ( log Wo,E(x)

J

f

E

1 +x2

dx
Proof. The second part reduces (as at the end of the preceding demonstra-

tion) to a known result of Akhiezer (in this case from §B.1, Chapter VI) on putting W(x) = oo on P - E. Hence only the first part requires discussion here.

According to Pollard's theorem (Chapter VI, §B.3), proper inclusion of 'w(0, E) in 'w(E) will certainly follow if the integrals ('°°

log(1 + IP(x)I2)

1+x2

J

dx

are bounded above for P ranging over the polynomials with II P 11 w,E 5 1.

It is therefore enough to show this, under the assumption that

SE1+

M, say,

for any polynomial P with 11 P II w,E 1 1 We may, first of all, argue as in the proof of the above lemma to conclude

that our assumption implies a seemingly stronger property: we have log(1 + IP(x)1dx 2) 1 + x2 E

2M + n log 2

for the polynomials P with 11 P II W,E 5 1. The proof will therefore be complete if we can verify that

log(i f°°

+1PZx)12)dx

5 bJElog(1 +IP(x)IZ)dx x

VIII A The set E has positive lower density

434

for polynomials P, b being a certain constant depending on the set E. This we do, using the result of problem 18. Take any polynomial P, of degree N, say. With an arbitrary n > 0, put

fn(Z) =

(sin nZ Z JN P(Z)i

,,(z) is then entire, of exponential type Nn, and bounded on R. By problem 18, we thus have f °°ao log (l +1 f,,x2 (x) I2)

-

dx < aNn + b J

1

log(' + I fn(x) I2) dx. 1 + x2

Here, I f (x) I < I P(x) I on R and f,,(x) - P(x) as n - 0, so the desired

inequality follows on making n - 0. We are done. What happens when the set E is sparse

4.

The sets E described at the beginning of this § have the property that

IErII/III>c>0 for all intervals I on i8 of length exceeding some L. In other words, their lower uniform density is positive. One suspects that the continual occurrence

of the form dx/(1 + x2) in the integrals over E figuring in the preceding article is somehow connected with this positivity. As a first step towards finding out whether our hunch has any basis in fact, let us try to see what happens to the form dx/(1 + x2) when E becomes sparse. We do this in the special case where

E=U oo

a

n and p > 1. This example was worked out by Benedicks

(see his preprint), and all the material in the present article is due to him.

In order that there may be no doubt, we point out that the sets E now under consideration are no longer of the sort described at the beginning of the present §.

Lemma. Let S be the square

{(x,y): -a<x
and

-a
and denote by H the union of its two horizontal sides, and by V the union of

its two vertical sides. Then, if - a < x < a, ws(H, x) < cos(H, 0)

4 What happens when E is sparse

435

and

ws(V, x) > ws(V, 0),

where, as usual, cos( , z) denotes harmonic measure for S.

Proof (Benedicks). Let, wlog, 0 < xo < a and consider the harmonic function 4(z) = cos(H, z) - cos(H, z + xa)

defined in the rectangle

T = {(x,y): -a<x
a

Figure 129

It is clear by symmetry that, for zeS, ws(H, z) = ws(H, z), and also ws(H, x + iy) = cos(H, - x + iy). Therefore A( - xo + iy) = 0 on the i vertical bisector of T (see figure). Again, on T's right vertical side,

A(a - xo + iy) = cos(H, a - xo + iy) - ws(H, a + iy)

= ws(H, a - xa + iy) > 0 (and similarly, on the opposite side of T,

A(-a+iy) = -ws(H,-a+xo+iy) < 0). It is clear on the other hand that A(z) = 0 on the top and bottom sides of T (1 - 1 = 0). By the principle of maximum we thus have A(z) > 0 in the right half of T; in particular, A(0) = ws(H, 0) - ws(H, xo) > 0, and ws(H, xo) lemma.

ws(H, 0), proving the first inequality asserted by the

VIII A The set E has positive lower density

436

The second inequality follows from the first one because ws(H, z) + ws(V, z) = 1 in S and clearly cos(V, 0) = cos(H, 0). We are done.

Lemma (Benedicks). Let E c R be any `reasonable' closed set (for instance, a finite union of closed intervals), let S be the square of the preceding lemma, and put

12 = Sn-E. If H denotes the union of the two horizontal sides of S and V that of the vertical ones, we have con(V, 0) 5 con(H, 0)

for the harmonic measure wn(

, z) associated with the domain 12.

H V

V

Proof. By a formula derived near the end of §B.1, Chapter VII, for zei1,

wa(H, z) = ws(H, z) - f E

wn(V, z) = ws(V, z) - f

)

z)

z).

E

From the previous lemma, ws(H, c) 5 ws(H, 0) = ws(V, 0) 5 ws(V,

)

for real lying in S; in particular, for SEE. Substituting this relation into the preceding ones and then making z = 0, we get wn(V, 0) 5 wn(H, 0). Q.E.D.

4 What happens when E is sparse

437

Corollary. In the above configuration, wn(8S, 0) < 2con(H, 0).

Proof. Clear. Lemma. Let p > 1 and put 00

E = U [InV'sgnn-8, InIPsgnn+S], 6 > 0 being taken small enough so that the intervals figuring in the union do not intersect. With x0 > 0, let Sxa be the square x

2 <9z<

32x0

, - 2x0 <3z< x02, 1'

and fl, the domain

S,0n-E. For large x0, the harmonic measure conxo(

,

z) associated with S2xp satisfies

log x0 wn: (aSxo, x0) <, const. x0t/p

Proof (Benedicks). By use of a test function and application of the preceding corollary. Y

xo + ixo/2

0

Note: E is not shown to scale Figure 131

438

VIII A The set E has positive lower density

The function z'1P (taken as positive on the positive real axis) is analytic for %z > 0; so, therefore, is sin nz 1/P.

In Rlz > 0, this function vanishes only at the midpoints nP of the intervals making up E, and, at x = nP, d(sin nx'"P) dx

_

n

- (-1) pnP-

This means that, if we take xo > 0 large and put Co we have J sin nx' /PI >,kCoS for x outside E on the interval (x0/2, 3xo/2), k > 0 being a constant depending on p, but independent of xo and S. Recalling the behaviour of the Joukowski transformation

w -->w+

/(w2

1),

we see that for a suitable definition of ,/, the function

v(z) = log

J(sin21rz" - 1

sin nz' l p

kCoS

(kCob)2

+

is positive and harmonic in flxO.

For this reason, when xEI8nS

X°,

v(x) > inf v(l() wnx (H, x), (eH

H denoting the union of the two horizontal sides of 8Sx0. However, v(l;)

>, const.x iP

for cnH

as is easily seen (almost without computation, if one refers to the above diagram). Also,

v(x) S logkC 0

S = (1-1/p)logxo+O(1),

x e R nf2X0

Therefore (O nX (H, x) < const.

log xo x1 1P

,

x c R n i2

0

Since xo lies at the centre of the square Sx0, the corollary to the previous lemma gives wa (BSxo, x0) < 2cons (H, x0).

Combining this and the preceding relations, we obtain the desired result.

439

4 What happens when E is sparse Theorem (Benedicks). Let G be the Green's function for the domain co

-9 = C-E = C_ U [InIPsgnn-8, InIPsgnn+S], where p > 1 and 6 > 0 is small enough so that the intervals in the union do not intersect. Then, for real x of large modulus, G(x,i)
IxI(P+1)/P' with a constant C depending on p and S.

Proof. G(z, i) is certainly bounded above - by M say - in the sector {0 < 13zl < 91z}. Given xo > 0, the square Sx0 considered in the previous lemma lies in that sector, so G(t;, i) < M on BS, G(z, i) is, moreover, harmonic in Ox, c -9 and zero on E, whence G(x0, i) 5 M wnxo(BSxo, xo). By the last lemma we therefore have (*)

G(xo, i) < const.l

I

l oI l/P

for large xo > 0. Benedicks' idea is to now use Poisson's formula for the half plane, so

as to improve (*) by iteration. Take any fixed a with 0 < a < 1/p. Then (*) certainly implies (by symmetry of E) that G(x,i)

const. 1, IxI"+

xeOB,

G(x, i) being at any rate bounded on the real axis. The function G(z, i) is in fact bounded and harmonic in 3z < 0, so G(z, i) =

-

3 I

tII 2 G(t, i) dt,

3z < 0.

zI

Plugging in the previous relation, we get

G(z, i) < const.

L(X_t)2+y2ItI+l'

y<0.

Let, wlog, x > 0. Then, the integral just written can be broken up as Since 1110 (IyI/((x - t)2 + y2) )dt = iv, the second of these f -x/2 + terms is obviously O(x-") for large x. The first, on the other hand, is

const.x2+4 Y2 x'-"

440

VIII A The set E has positive lower density

so, all in all, G(z,i) < const. I x I

for 0> y > -IxI, IxI being large. The inequality just found remains true, however, for 0 < y < IxI, in spite of the logarithmic singularity that G(z, i) has at i. This follows from the fact that 0 < a < 1/p < 1 and the relation G(z, i) - G(i, i)

= log

To verify the latter, just subtract the right side from the left. The difference is harmonic in 3z > 0 and bounded there (the logarithmic poles at i cancel each other out). It is also clearly zero on R, so hence zero for 3z > 0. For large I z I, log i (z + i)/(z - i) I = O(1/ I z I ), and we see that

G(z,i) <

for 0,
const.Ixl_a

since this inequality is true for 0 > y > - IxI . Suppose that xO > 0 is large; we can use the previous lemma again. By what has just been shown, G(l;, i) 5 const. I xO I -",

t; eOSxQ.

Arguing as at the beginning of this proof, we get G(x0, i) < const.IxOI -awaxa(BSxa, x0) <, const.

log I x0I Ix01°`Ix0I11"

Hence, since 0 < a < l/p, we have

G(x,i) <

const.

,

IxI2a + 1

The exponent a in the inequality we started with has been improved to 20C. If now 2a < 1, we may start from the inequality just obtained and repeat

the above argument, ending with the relation

G(x,i) <

const. IxI3a + 1

xeR.

The process may evidently be continued so as to yield successively the estimates

G(x,i) S

const. IxI"a+ 1

xeR,

with n = 3,4,..., as long as (n - 1)a < 1. Choosing a, 0 < a < l/p, to not

4 What happens when E is sparse

441

be of the form 1/m, m = 1, 2,3,..., we arrive at an estimate

(,)

lxlna+l, cont.

G(x,i)

xeR,

where na is the first integral multiple of a strictly > 1. Because the exponent na in (*) is > 1, we have

G(t,i)dt < co. As before, for y < 0, we can write Ixl/2

G(z, i) =

1

IYI

1z-ti IXuz

For ItlSlxl/2,

2 G(t,i)dt + 1 7t

IYI

tIIxI/2 (z-tl2

G(t) i) dt.

IYI/lz-t12'< 1/lxl, so the first term on the right is

const./Ixl in view of the preceding relation. The second is 5 const./lxl"°` = o(1/Ixl) by (*). Thence, for Ixl large,

G(z, i) 5

cost.

y < 0.

,

1XI

Using the relation G(z, i) - G(% i) = log

i+zl

i-z

as above, we find that in fact

cont.

G(z, i)

I

for I z I large.

xi

Take this relation and apply the preceding lemma one more time. For large x0, we have

cont./xo

G(C, i)

on 3Sx0.

Therefore G(xo, 1)

cont. x0

wp:o(aS

const.log xo 0, xU)

This is what we wanted to prove. We are done. Corollary.

A Phragmen-Lindelof function Y(z) exists for the domain

C

U [InI"sgnn-S, Inl"sgnn+S].

442

VIII A The set E has positive lower density

Proof. By the theorem, we certainly have

G(x, i) dx < oo. -.

The result then follows by the second theorem of article 2.

Remark. Although the theorem tells us that, on the real axis, G(x, i) < const.1xlglxlp when Ixl is large, the inequality G(z,i)

c Ixlt.

1<

valid for I z I large, obtained near the end of the theorem's proof, cannot be improved in the sector 0 < IYI < I x 1.

0 we have, for large Ixl,

Indeed, since G(t, i)

i) = -' f ' G(t, i)dt n _ (x - t)21 + x2

G(x - iIx1, 4 131r x

G(t,i)dt I

f

4

xl/z

13n x J - m I

xl/2

G(t,i)dt.

I

A better bound on G(z, i) can be obtained if IYI is much smaller than Ixl. The following result is used in the next exercise.

Lemma. For large xl, G(z,i) < const.

logl xl

Ixl'

+

01< IYI
i/p '

Proof. Taking wlog x > 0, consider first the case where y < 0. By the theorem,

G(z, i) =

'

I

ly'

_ . x - t)2 + y2

71

G(t, i)dt IYI

const. foo - °°

(x -

t)2 + y2

log+ ltl + I I t l'

+ 11p

+1

dt

As usual, we break up the right-hand integral into ('x/2

-x/ 2 +

f

tI3 x/2

The first term is < const.lyl/x2 (because 1 + 1/p> I !), and this is

4 What happens when E is sparse

443

5 const./x' + 1/p for I y I 5 x' -1/p. The second term is clearly

const.logx x1+1/p

'

This handles the case of negative y. For 0 < y < x1 -1/p, use the relation z+i z-i

G(z, i) - G(z, i) = log

already applied in the proof of the theorem. Note that the right hand side is 91logl

I +('/z)

2-

I - (i/z)

Izl

for large Izj. For 0 < 3z <

+ 0\ Izl

3

Ix11-1/p this is

1

5 const. 1x11+l/p The lemma thus follows because it is true for negative y. In the following problem the reader is asked to work out the analogue, for our present sets E, of Benedicks' beautiful result about the ones with positive lower uniform density (Problem 16). Problem 19 If t is on the component [n" - b, np + b] of

-9 = C- U [Iklpsgnk-b, IkIpsgnk+b], show that

dw,(t,i) dt

const.

1

<- t'+ '1p+1 \/(b2-(t-np)2)'

where co( ,2,z) denotes harmonic measure for .9. Here, the constant depends only on p > 1 and b > 0.

Remark. The result is due to Benedicks. We see that the factor log ti in the estimate for G(t, i) furnished by the above theorem disappears when we evaluate harmonic measure. Hint for the problem: One proceeds as in the solution of Problem 16, here comparing G(z, i) with U(z)

log z

np

6

+

((z

bnp I2

-1/

444

VIII A The set E has positive lower density on the ellipse r. with foci at n° ± 6 and semi-minor-axis equal to nP - 'S:

Figure 132

By Problem 19, we have, for the harmonic measure of the component

E. = [InlPsgnn-S, InlPsgnn+S] of E=a-9, const. < InIp+1+1

(o (En) i)

Using this estimate, one can establish a result corresponding to the first part of the first theorem in article 3. Theorem. Let W(x) > 1 be continuous on 00

E = U [InlPsgnn-6, InlPsgnn+S], -00

and suppose that W(x) -> oo for x - ± oo in E. If, for some C > 0, the supremum of log I S(t) I

El+ItIl+1/P dt for S ranging over all finite sums of the form

S(t) = with 11 S 11 W,E 1<

E

_c
a2e'"

1 is finite, then Ww(E, C) 0 Ww(E).

The set E reduces to the integers

445

Proof. We have the above boxed estimate for the harmonic measure (in = C ' E) of the components of E, and a previous corollary gives us a Phragmen-Lindelof function Y(z) for -9. Using these facts, one proceeds exactly as in the proof of the first theorem of article 3.

Remark. The sparsity of the set E involved here has caused the form dt/(1 + t2) occurring in the result of article

3

to be replaced by

Remark. The statement of the above theorem goes in only one direction, unlike that of the corresponding one in article 3. There, since we were dealing with the restriction of the form dt/(1 + t2) to E, we were able to obtain a converse by simply appealing to Akhiezer's theorem from §E.2 of Chapter VI. In the present situation we can't do that, because we are dealing with dt/(1 + I tI' +'IP) instead of dt/(1 + t2), and 1/p < 1. It would be interesting to see whether (as seems likely) the converse is true here. In case W(x) -* oo faster than any power of I x I as x -* ± oo in E, we can formulate a result like the above one for polynomial approximation on E in the weight W. The statement of it is exactly like that of the first part of the second theorem in article 3, save that the integrals logIP(x)I dx SE 1 + x2

figuring there are here replaced by log I P(x) I IxI'+i/c dx. E1+

J

The proof runs much like that of the result in article 3. Details are left to the reader.

B.

The set E reduces to the integers Consider the set 00

EP = U [n - p, n +p], n= -m

where 0, 0, the results of §§A.1-A.3 apply to EP, and there is, in particular, a constant bP such that the inequality f °°

log(1+IPZx)I2) dx < bP fEP log (1 +IPZx)I2) dx

used in proving the second theorem of §A.3, holds for polynomials P.

446

VIII B The set E reduces to the integers

For this reason, given any M, the set of polynomials P such that log JE P

I P(x) I

1+x

dx 5 M

forms a normal family in the complex plane. Suppose now that p = 0. Then E. = Z, and the proof in §A of the above inequality involving bP, available when p > 0, cannot be made to work so as

to yield a relation of the form fOD log (I + IP(x)I2)

1+x2

°° log (I + I P(n)I2)

dx < C Y-

1+n2

That proof depends on the properties of harmonic measure for 2., = C - EP worked out in §A.1 (for p > 0); there is, however, no harmonic measure for -9 = C - Z. This makes it seem very unlikely that the set of polynomials P satisfying log+ I P(n) I

I+n

<M

for arbitrary given M would form a normal family in the complex plane, and

it is in fact easy to construct a counter example to such a claim. Take simply N

2

PN(x) = (1 - x2)[N/IogN] fl

1

k=1

x

for N >, 2, with [p] denoting the greatest integer 5 p as usual. Then it is not hard to verify that log+ Pn2n)I

(*)

Y

\ 20

for N > 8.

At the same time, PN() ->- 2[N/IogN] _i c0. N

Problem 20 Prove (*).

(Hint:-log+IPN(n)I 1n

N

LlogN

log(n2 - 1) n2

n=N+1 m

1

N

+ Y 2 Y- log N+1n

n2

k2-1

k=1

After replacing the sums on the right by suitable integrals and doing

1 Certain sums acting as upper bounds for integrals

447

some calculation, one obtains the upper bound

2+

2

°

2+2

log N +

1

log

(+1\d -1

.

Here, the integral can be worked out by contour integration.)

This example, however, does not invalidate the analogue (with obvious statement) of Akhiezer's theorem for weighted polynomial approximation

on Z. In order to disprove such a conjecture, one would (at least) need similar examples with the number 20 standing on the right side of (*) replaced by arbitrarily small quantities > 0. No matter how one tries to construct such examples, something always seems to go wrong. It seems impossible to diminish the number in (*) to less than a certain strictly positive quantity without forcing boundedness of the IPN(i)I. One comes in such fashion to believe in the existence of a number C > 0 such that the set of polynomials P with log+IP(n)I

Y-4 l+n 2


does form a normal family in the complex plane.

This partial extension of the result from §A.3 to the limiting case EP = Z turns out to be valid. With its help one can establish the complete analogue of Akhiezer's theorem for weighted polynomial approximation on Z; its interest is not, however, limited to that application. The extension is easily reduced to a special version of it for polynomials P of the particular form

P(x) = fl

x2 1 --

k

X k

with real roots Xk > 0, and most of the real work is involved in the treatment,

of this case, taking up all but the last two of the following articles. The investigation is straightforward but very laborious; although I have tried hard to simplify it, I have not succeeded too well. The difficulties are what they are, and there is no point in stewing over them. It is better to just take hold of the traces and forge ahead. 1.

Using certain sums as upper bounds for integrals corresponding to them

Our situation from now up to almost the end of the present § is as follows: we have a polynomial P(z) of the special form

P(z) = fl i k

z2 z

xk

,

448

VIII B The set E reduces to the integers

where the Xk are > 0 (in other words, P(z) is even, with all of its zeros real, and P(O) = 1), and we are given an upper bound for the sum log+IP(m)I

1+m2

00

or, what amounts to the same thing here, for EWlog+ I P(m)I

From this information we desire to obtain a bound on IP(z)I for each complex z. The first idea that comes to mind is to try to use our knowledge about the preceding sum in order to control the integral

-.

log+ I P(x) I

1+x2

dx;

we have indeed seen in Chapter VI, §B.1, how to deduce an upper bound on IP(z)I from one for this integral. This plan, although probably too simple to be carried out as it stands, does suggest a start on the study of our problem. For certain intervals I c (0, oo), (' log I P(x) I

f

x2

J

dx

I

is comparable with

MeInz

log+IP(m)I m2

We have d2 log I P(x) I

dx2

_ -21: x2 + xk < 0, (x2 - x2 )2 k

so log I P(x) I is concave (downward) on any real interval free of the zeros

± xk of P. This means that, if a < b and P has no zeros on [a, b], ('6

loglP(x)ldx 5 (b-a)logIP(m)I a

for the midpoint m of [a, b]:

1 Certain sums acting as upper bounds for integrals

449

Figure 133

Of course f b log I P(x) I dx is not the integral we are dealing with here. If, however, a > 0 is large and b - a not too big, the presence of the factor 1/x2 in front of logIP(x)I does not make much difference. A similar formula still holds, except that m is no longer exactly the midpoint

of [a, b].

Lemma. Let 0 < a < m < b, and suppose that P has no zeros on [a, b]. Then JbloIP(x)Idx 2 (log l P(m) 1)

x2

logba - =ma --

m b

Proof. Let M denote the slope of the graph of log I P(x) I vs. x at x = m. Then, since log I P(x) I is concave on [a, b], we have there log I P(x) I < log I P(m) I + M(x - m)

(see the previous figure). Hence logIP(x)I x2

dx \

(logIP(m)I)

f

-'Z\dx. + MJf 6i1 at x x x a

450

VIII B The set E reduces to the integers

The second term on the right is

Mloga - M(m-b

)

We will be interested in situations where the number m figuring in the above boxed relation is a positive integer, and where one of the two numbers

a, b (a < m 5 b) is to be found, the other being given. Regarding these, we have two estimates. Lemma. If m > 7 and m -1 < a 5 m, the number b > m such that log

b

a

=

m

m

a

b

is < m+2. Proof. Write p = a/b; then 0 < p 51, and the relation to be satisfied becomes log(1/p) = (m/a)(1- p). If a = m, this is obviously satisfied for p = 1, i.e., m = b; otherwise 0 < p < 1, and we have log

m

1

p

1-p

a

Now 1

log- = so the preceding relation implies that m

a

1> 1+2(l-p),

i.e.,

1-p <, 2m-a a and

3a - 2m P

a

Therefore b

_

a

p

5

a

z

3a-2m'

1 Certain sums acting as upper bounds for integrals

451

and

(2m-a)(m-a)

b-m <

3a-2m

m+1 5 m-3*

Here the right-hand side is 5 2 for m > 7. We are done. Lemma. If m >, 2 and m < b < m + 1, the number a < m such that

logba - =ma- -

m

b

is > m-2. Proof. Put p = alb as in proving the preceding lemma; here, it is also convenient to write

Then 0 < p 51 and 0 < y 51. In terms of y and p, our equation becomes 1

Y logy =P -y.

When y < 1, we must also have p < 1, and then

= p log (1/p)

1-p

Y

This yields, for 0 < p < 1, dy dp

-

109010-0 -P) _ 2+3(l-P)+4 ll-P) Z+... (1_p)2

Hence, since the value y = 1 corresponds to p = 1, we have, for 0 < y < 1,

i(1 - p) < 1 - y, i.e.,

p >, 1- 2(1- y). It was given that m , b _- m + 1, so

1_ Y

b-m

1

b

m+1

(the middle term here is a monotone function of b). Therefore, by the

452

VIII B The set E reduces to the integers

previous relation, 2

P '>

1

m+l'

and finally,

a = pb > pm % m-m+

1> m-2.

We are done. Theorem. Let 6 5 a < b. There is a number b*, b < b* < b + 3, such that (' 6log_P(x)Idx x

log+IP(m)I

5

a<m
m

provided that P has no zeros on [a, b*]. The sum on the right is taken over

the integers m with a < m < b*.

NO.

Definition. During the rest of this §, we will say that b* is well disposed with respect to a.

Proof. By repeated application of the first two of the above lemmas. Let the integer m1 be such that m1 - 1 a < m1; then ml > 7, so, by the second lemma, we can find a number a1, m1 < a1 s m1 + 2, with

loga1 =m1 a

a

-

ml a1

We have a1 a, a1
. -X2 logIP(x)dx

< logIP(m1)I

Jaldx

Here, a1 - a < 3 and m1/a 5 6, so C°' dx °

5

x2 \ mi

Therefore, f a, logIP(x)I dx x2

5log+IP(m1)I m2

1 Certain sums acting as upper bounds for integrals

453

If now a1 ,> b, we simply put b* =a I and the theorem is proved. Otherwise, a1 < b and we take the integer m2 such that m2 -15 a1 < m2. Since a1 > m1, m2 > ml, and we can find an a2, m2 < a2 m2 + 2, with

log a2 = a1

-

m2 a1

M2. a2

We have a2 5 a1 + 3 < b + 3, and, by -the first lemma, f a2,logIP(x)I

J

x

('a2 dx

51og+IP(m2)I

a, x

m22

dx < logIP(m2)I J z.

2

just as in the preceding step, provided that P has no zeros on [a1, a2]. If a2 >, b, we put b* = a2. If not, we continue as above, getting numbers a3 > a2, a4 > a3, and so forth, ak+ 1 K, ak + 3, until we first reach an a, with a, > b. We will then have a, < b + 3, and we put b* = a,. There are integers mk, m2 < m3 < < m,, with ak _ 1 < Mk < ak, k = 3, ... , 1, and, as in the previous steps, log I P(x) I

('ak

X k

2

dx

5log

I P(mk) I 2

Mk

1

for k = 3,..., 1, as long as P has no zeros on [ak _ 1, ak].

Write ao = a. Then, if P has no zeros on [a, b*] = [ao, all,

dx = ` fak

f w log I P(x) I a

x

k=1

i

k=1

ak-I

51og+ I P(mk) I Mk2

log I P(x) I

dx

x

E

5 log+ I P(m) I

a<m
m

2

meZ

We are done. In the result just proved, a is kept fixed and we move from b to a point b* well disposed with respect to a, lying between b and b + 3. One can obtain the same effect keeping b fixed and moving downward from a. Theorem. Let 10 < a < b. There is an a*, a - 3 < a* <, a, such that b is well disposed with respect to a*, i.e., (b logIP(x)I x J a*

f

dx < 5 , a'<m< b

log+IP(m)I m

provided that P(x) has no zeros on [a*, b].

The proof uses the first and third of the above lemmas, and is otherwise very much like the one of the previous theorem. Its details are left to the reader.

454

VIII B The set E reduces to the integers

2.

Construction of certain intervals containing the zeros of P(x)

We have seen in the preceding article how certain intervals 1 c (0, oo) can be obtained for which f log I P(x) I

x

dx 5 5

mel

log, I P(m) I m

as long as they are free of zeros of P. Our next step is to split up (0, oo) into two kinds of intervals: zero-free ones of the sort just mentioned and then some residual ones which, together, contain all the positive zeros of P(x).

The latter are closely related to some intervals used earlier by Vladimir Bernstein (not the S. Bernstein after whom the weighted polynomial approximation problem is named) in his study of Dirichlet series, and it is to their construction we now turn. Pop-

As is customary, we denote by n(t) the number of zeros Xk of P(x) in the

interval [0, t] for t > 0 (counting multiplicities as in Chapter III). When t < 0, we take n(t) = 0. The function n(t) is thus integer-valued and increasing. It is zero for all t > 0 sufficiently close to 0 (because the Xk > 0), and constant

for sufficiently large t (P being a polynomial). The graph of n(t) vs. t consists of some horizontal portions separated by jumps. At each jump, n(t) increases by an integral multiple of unity. In this quantization must lie the essential difference between the behaviour of subharmonic functions of the special form log I P(x) I with P a polynomial,

and that of general ones having at most logarithmic growth at oo, for which there holds no valid analogue of the theorem to be established in this

§.

(Just look at the subharmonic functions

rl log I PN(z) I ,

where

rl > 0 is arbitrarily small and the PN are the polynomials considered in Problem 20.) During the present article we will see precisely how the quantization affects matters. For the following work we fill in the vertical portions of the graph of n(t) vs. t. In other words, if n(t) has a jump discontinuity at to, we consider the vertical segment joining (to, n(to - )) to (to, n(to + )) as forming part of that graph. Our constructions are arranged in three stages.

First stage. Construction of the Bernstein intervals We begin by taking an arbitrary small number p > 0 (requiring, say, that p < 1/20). Once chosen, p is kept fixed during most of the discussion

of this and the following articles. Denote by (9 the set of points toeR with the property that a straight line of slope p through (to, n(to)) cuts or touches the graph of n(t) vs. t only

once. 0 is open and its complement in t consists of a finite number of

455

2 Inclusion of zeros of P(x) in special intervals Jk

closed intervals Bo, B1, B2,... called the Bernstein intervals for slope p associated with the polynomial P(x). (Together, the Bk make up what V. Bernstein called a neighborhood set for the positive zeros of P - see page 259 of his book on Dirichlet series. His construction of the Bk is different from the one given here.) It is best to show the formation of the Bk by a diagram:

n (t) oQi

Bo

B,

J

t

Figure 134

We see that all the positive zeros of P (points of discontinuity of n(t)) are contained in the union of the Bk. Also, taking any Bk and denoting it by [a, b]: The part of the graph of n(t) vs. t corresponding to the values of tin Bk lies between the two parallel lines of slope p through the points (a, n(a)) and (b, n(b)).

For a closed interval I = [a, l3], say, let us write n(I) for n(fi +) - n(a - ). The statement just made then implies that n(Bk)/PI BkI 5 1

for each Bernstein interval. An inequality in the opposite sense is less apparent. Lemma (Bernstein). For each of the Bk, n(Bk)/PI BkI

1>

1/2.

Proof. It is geometrically evident that a line of slope p which cuts (or touches) the graph of n(t) vs. t more than once must come into contact with some vertical portion of it - let the reader make a diagram.

456

VIII B The set E reduces to the integers

Take any interval Bk, denote it by [a, b], and denote the portion of the graph of n(t) vs. t corresponding to the values a < t < b by G. We indicate by L and M the lines of slope p through the points (a, n(a)) and (b, n(b)) respectively. According to our definition, any line N of slope p between

L and M must cut (or touch) the graph of n(t) vs. t at least twice, and

hence come into contact with some vertical portion of that graph. Otherwise such a line N, which surely cuts G, would intersect the graph only once, at some point with abscissa toe(a,b); to would then belong to (9 and thus not to Bk. The line N must in fact come into contact with a vertical portion of G, for, as a glance at the preceding figure shows, it can never touch any part of the graph that does not lie over [a, b]. In order to prove the lemma, it is therefore enough to show that if n(Bk)lplBkl < 1/2.

there must be some line N of slope p, lying between L and M, that does not come into contact with any vertical portion of G. Let V be the union of the vertical portions of G, and for X e V, denote by n(x) the downward projection, along a line with slope p, of the point

X onto the horizontal line through (a, n(a)).

1

a

t

k

Bk

13'

Figure 135

In this figure, I Bk I = PS and n(Bk) = QS. The result, II(V), of applying Il

to all the points of V is a certain closed subset of the segment PR, and, if we use I I to denote linear Lebesgue measure, it is clear that

If(V)I < IVI/p. We

have

i

pRS = QS,

n(Bk) = QS < p l Bk l = 2P-PS, Z and therefore PR > z'PS > QS/p = I V I/p. With the so,

if

2 Inclusion of zeros of P(x) in special intervals Jk

457

preceding relation, this yields

III(V)I < PR. There is thus a point Y on PR not belonging to the projection II(V). If, then, N is the line of slope p through Y, N cannot come into contact with V. This line N lies between L and M, so we are done. Second stage. Modification of the Bernstein intervals

The Bernstein intervals Bk just constructed include all the positive zeros of P(x), and I 2

<

n(Bk)

pIBkI

< 1.

We are going to modify them so as to obtain new closed intervals Ik g (0, co )

containing all the positive zeros of P(x), positioned so as to make f log I P(x) I

Ji

x

log, I P(m) I

dx < 5

m

mel

for each of the interval components I of (0, oo) ^ U lk. k

(Note that Bo need not even be contained in [0, oo).) For the calculations which come later on, it is also very useful to have all the ratios n(Ik)/Ilk) the same, and we carry out the construction so as to ensure this.

Specifically, the intervals Ik, which we will write as [ak, Nk] with k = 0, 1,2.... and 0 < ao < Ilo < a, < /i, < , are to have the following properties: (i) All the positive zeros of P(x) are contained in the union of the Ik,

k=0,1,2,...,

(ii) n(lk)/pIlkl =i, (iii) For ao < t < Qo,

n(llo) - n(t) <

1

pap (Qo -

t),

and, for ak < t < /lk with k > 1,

n(t) - n(ak) <

1

n(Nk) - n(t) < 1

pap (t - ak), p

ap 3p

- t),

(recall that we are assuming 0 < p < io),

VIII B The set E reduces to the integers

458

(iv) For k> 1, ak is well disposed with respect to /'k_1 (see the preceding article).

Denote the Bernstein intervals Bk, k = 0, 1, 2,..., by [ak, bk], arranging the indices so as to have bk -I < ak. We begin by constructing 10. Take ao as the smallest positive zero of P(x); a0 is the first point of discontinuity of n(t) and a0 < a0 < b0. We have n(Bo)

1

p(bo - ao) > plBol

n(B0)

2

n([ao,bo]) p(bo - ao)

by the lemma from the preceding (first) stage. For r > b0, let JT be the interval [ao, T]. As we have just seen, n(J,)lpI JLl > 1/2

for T = b0. When T increases from b0 to a1 (assuming that there is a Bernstein interval B1; there need not be!) the numerator of the left-hand ratio remains equal to n(B0), while the denominator increases. The ratio itself therefore decreases when T goes from b0 to a1, and either gets down to i in (b0, a1), or else remains > i there. (In case there is no Bernstein interval B1 we may take a1 = oc, and then the first possibility is realized.) Suppose that we do have n(JL)lpI J,I = i for some T, b0 < T < a1. Then we put /l0 equal to that value of T, and property (ii) certainly holds for to = [ao, /30]. Property (iii) does also. Indeed, by construction of the Bk, the line of slope p through (/30, n(/30)) cuts the graph of n(t) vs. t only once, so the portion of the graph corresponding to values of t < #0 lies entirely to the left of that line (look at the first of the diagrams in this article). That is,

n(llo) - n(t) < p(llo - t),

t < /lo,

whence, a fortiori, n(flo) - n(t)

p3p (fio - t),

t < #0

(since 0 i for bo < T < a1. Then that ratio is still i for T = b1. This is true because n(B1)lpl B1I > i (lemma from the preceding stage), and

n([ao,bl]) = n(a1-)-n(a0-)+n(B1), while

bl-a0 = a1-a0+IB1I.

2 Inclusion of zeros of P(x) in special intervals Jk

459

Thus, in our present case, n(JL)/pI J,I is >, i for t = b, and again decreases

as t moves from b, towards a2 > b,. (If there is no interval B2 we may take a2 = oo) If, for some -re[b,, a2), we have n(JT)/PI JAI = }, we take /30

equal to that value of t, and property (ii) holds for l0 = [ao, /30]. Also, for /0e[b,, a2), the part of the graph of n(t) vs. t corresponding to the values t < /l0 lies on or entirely to the left of the line of slope p through (/30, n(/3o)), as in the situation already discussed. Therefore, n($0) - n(t) < (p/(1 - 3p)) (/30 - t) for t < #0 as before, and property (iii) holds for lo. In case n(JT)/p I JL I

still remains > i for b, < t < a2, we will have

n(JT)/P I Jt I % z for t = b2 by an argument like the one used above, and we look for flo in the interval [b2, a3). The process continues in this way, and we either get a /o lying between two successive intervals Bk, Bk+1 (perhaps coinciding with the right endpoint of Bk), or else pass through

the half open interval separating the last two of the Bk without ever bringing the ratio n(JT)/p I Jt I down to 2. If this second eventuality occurs, suppose that B, = [a,, b,] is the last Bk; then n(JT)/PI JtI > i for t = b, by the reasoning already used. Here, n(JT) remains equal to n([0, b,]) for t > b, while I JL I increases without limit, so a value /o of t > b, will make n(Ji)/P I J, I = -. There is then only one interval Ik, namely, to = [ao, Qo], and our construction is finished, because properties (i) and (ii) obviously hold, while (iii) does by the above reasoning and (iv) is vacuously true. In the event that the process gives us a Qo lying between two successive

Bernstein intervals, we have to construct I, = [a,, /3,]. In these circumstances we must first choose a, so as to have it well disposed with respect to fio, ensuring property (iv) for k = 1.

00.

It is here that we make crucial use of the property that each jump in n(t) has height > 1.

Assume that bk < fio < ak+ 1 We have p(/30 - a0) = 2n(10) >,2 with

0 < p < o; therefore #0>40 and there is by the first theorem of the preceding article a number a,, ak+, < a, < ak+, + 3, which is well disposed with respect to fio.

Now a, may well lie to the right of ak+,. It is nevertheless true that n(a1) = n(ak+, -), and moreover n(t) - n(a1) <

P 1

3P (t

- a,)

for t > a, .

The following diagram shows how these properties follow from two facts:

460

VIII B The set E reduces to the integers

that n(t) increases by at least 1 at each jump, and that 1/p > 3: slope = p/(1 -3p)

o P-

t

3

Io

ak+1

al ak+1 +3 Bk+ I

Figure 136

For this choice of al, properties (i)-(iv) will hold, provided that /3,, a2 and so forth are correctly determined.

We go on to specify /3,. This is very much like the determination of /30. Since

n(bk+1)-n(al) = n(bk+l)-n(ak+l-) = n(Bk+1), we certainly have n([a,,bk+1])

_

n(Bk+1)

p(bk+l -al)

p(bk+l -al)

n(Bk+1)

1

pIBk+1I

2

by the lemma from the preceding stage. For T 1> bk+,, denote by JT the interval [a,, T]; then is >, i for T = bk+, and diminishes as T increases along [bk+1, ak+2). (If there is no Bk+2 we take ak+2 = oo.) We may evidently proceed just as above to get a r > bk+,, lying either in a half open interval separating two successive Bernstein intervals or else beyond all of the latter, such that n(J' )/pIJTI = z. That value of T is taken as f,. The part of the graph of n(t) vs. t corresponding to values of t 5 /3, lies, as before, on or to the left of the line through (/f,, n(/3,)) with slope p. Hence, a fortiori, n(131)-n(t)

1

pap (/31 -t)

for t

/31

We see that properties (ii) and (iii) hold for Io and I, = [a,, /31]. If Io u I, does not already include all of the Bk, /3, must lie between

2 Inclusion of zeros of P(x) in special intervals Jk

461

two of them, and we may proceed to find an a2 in the way that a, was found above. Then we can construct an I2. Since there are only a finite number of Bk, the process will eventually stop, and we will end with a finite number of intervals Ik = [ak, Nk] having properties (ii)-(iv). Property (i) will then also hold, since, when we finish, the union of the Ik includes that of the Bk.

Here is a picture showing the relation of the intervals Ik to the graph of n(t) vs. t: n(t)

P1i, 1/2

p111/2

it

pllo1/2

0

ao

go

'- I,

a,

01

a2

P2

L-12-J

Figure 137

Let us check the statement made before starting the construction of the Ik, to the effect that f logIP(x)I I

dx < 5

x

mCI

log, m

for each of the interval components I of the complement (0, oo)

U Ik. k

Since, for k > 0, ak is well disposed with respect to /3k_,, this is certainly true for the components I of the form ($k_,, ak), k 1 (if there are any!), by the first theorem of the preceding article. This is also true, and trivially so, for I = (0, ao), because

IP(x)I = II

< 1

k

for 0 < x < ao, all the positive zeros Xk of P(x) being , ao. Finally, if I, is the last of the Ik' our relation is true for 1= (/3,, co ). This follows because

462

VIII B The set E reduces to the integers

we can obviously get arbitrarily large numbers A > $, which are well disposed with respect to (3,. We then have (A logIp(X)I _2

dx \ 5

log+IP(m) Q,<m
m2

for each such A by the first theorem of the preceding article, and need only make A tend to oo. Third stage. Replacement of the first few intervals Ik by a single one if

n(t)/t is not always < p/(1 - 3p) Recall that the problem we are studying is as follows: we are presented with an unknown even polynomial P(z) having only real roots and such that P(O) = 1, and told that log+ I P(m) I m2

is small. We are asked to obtain, for zeC, a bound on IP(z)I depending on that sum, but independent of P. As a control on the size of IP(z)I we will use the quantity n(t) sup -. r>0 t

A computation like the one at the end of §B, Chapter III, shows indeed that logIP(z)I '< nIzIsupn(t). r>0

t

We are therefore interested in obtaining an upper bound on sup,>o(n(t)/t) from a suitable (small) one for log+ I P(m) I In2

Our procedure is to work backwards, assuming that sup,> o(n(t)/t) is not small and thence deriving a strictly positive lower bound for the sum. We begin with the following simple

Lemma. If sup,>o(n(t)/t) > p/(1-3p), we have II I/fo > 3 for the interval Io = [ao, fio] arrived at in the previous stage of our construction.

Proof. Let us examine carefully the initial portion of the last diagram given above:

2 Inclusion of zeros of P(x) in special intervals Jk

463

4

,W

n (t)

r-----------------

/ p I/01/2

t

Figure 138

We see that, for t > 0, n(t) t

PIIoIl

max {_p 1 -3p' 2ao J'

whether or not the first term in curly brackets is less than the second. Here, PIIoI

IIo1/so

P

2 1 - IIOI//log

2ao

and this is < p < p/(1- 3p) (making the above maximum equal to p/(l - 3p) ) if I Io I/Qo < 3. Done.

Our construction of the intervals Ik involved the parameter p. We now bring in another quantity, q, which will continue to intervene during most of the articles of this §. For the time being, we require only that 0 < q < 3 and take the value of q as fixed during the work that follows. From time

to time we will obtain various intermediate results whose validity will depend on q's having been chosen sufficiently small to begin with. A final decision about q's size will be made when we put together those results. In accordance with the above indication of our procedure, we assume henceforth that n(t)

su I>o t

>

p

1 -3p.

By the lemma we then certainly have

IIOI/Io>q, since we are taking 0 < q < 3. This being the case, we replace the first few intervals Ik by a single one, according to the following construction.

464

VIII B The set E reduces to the integers

Let w(x) be the continuous and piecewise linear function defined on [0, oo) which has slope 1 on each of the intervals Ik and slope zero elsewhere,

and vanishes at the origin:

Figure 139

The ratio co(t)/t is continuous and tends to zero as t - co since there are only a finite number of Ik. Clearly, w(t)/t < 1, so, if t belongs to the interior of an Ik, w(t) dt t d

_

1

t

w(t)

> 0;

t2

i.e., w(t)/t is strictly increasing on each Ik. We have

w(#0)/#0 = 1101/fo > n, so, in view of what has just been said, there must be a largest value of t (> /i0) for which

w(t)lt = n and that value cannot lie in the interior, or be a left endpoint, of any of the intervals Ik. Denote by d that value of t. Then, since d > flu, there must be a last interval Ik - call it I. - lying entirely to the left of d. If Im is also the last of the intervals Ik we write

do=d, co = (1 - ,1)d, and denote the interval [co, do] by Jo. In this case all the positive zeros of P(x) (discontinuities of n(t)) lie to the left of do. It may be, however, that Im is not the last of the Ik; then there is an interval

Im+1 = [«m+1, I'm+1],

2 Inclusion of zeros of P(x) in special intervals Jk

465

and we must have d < am+, according to the above observation. Since

d > fo > 2/p > 40 (remember that we are taking 0 < p < o), we can apply the second theorem of article 1 to conclude that there is ado,

d-3<do
co = do-r1d and denote by Jo the interval [co, do]. The intervals Im+,, Im+2+ are also relabeled as follows:

Im+l = 4 Im+2 = J2

and we write a.,,=c,, lm+ l = dl, am+2 = c2, Nm+2 = d2, and so forth, so as to have the uniform notation

In the present case, I3m 5 d < am+, (sic!) so, referring to the previous (second) stage of our construction, we see that the part of the graph of n(t) vs. t corresponding to values of t < d lies entirely to the left of, or on, the line of slope p (sic!) through (d, n(d)). By an argument very much like

the one near the end of the second stage, based on the fact that n(t) increases by at least 1 at each of its jumps, this implies that do, although it may lie to the left of d, still lies to the right of all the zeros of P(x) in

Io,...,Im' and that n(do) - n(t)

1 p 3p (do - t)

for t < do.

(The diagram used here is obtained by rotating through 1800 the one from the argument just referred to.) We have, in the first place, co >, (1 -1)d - 3 > 0, because n < 3 and d>CPo>40. In the second place,

1J01/do % IJoI/d = n, by choice of d. Also, Vol

IJoI

do < d-3

since d > 40.

i1d

40x1

d-3 < 37

466

VIII B The set E reduces to the integers

Finally, n(do)

_

PIJ0I

1

2

Indeed, both do and d lie strictly between all the discontinuities of n(t) in Io, I1, ..., Im and those in Im+,, 1m+2, ... (or to the right of the last I. if our construction yields only one interval J0), so m

n(do) =

k=O

/

m

1

n(I4) =

2 kY PIIkI

by property (ii) of the Ik. And

E I41 = w(d) = nd = IJO1

k=O

by the choice of d and the definition of Jo. Thus n(do) = claimed.

z

p I Jo I, as

Denote by J the union of the Jk, and put for the moment t(t) = 1 [0, t] r J I

.

The function Co(t) is similar to w(t), considered above, and differs from the latter only in that it increases (with constant slope 1) on each of the Jk instead of doing so on the 1k. The ratio th(t)/t is therefore increasing on each Jk (see above), so in particular w(t) t

5

w(do)

IJOI

do

do

401

< 37

for teJo = [co, do]. This inequality remains (trivially) true for 0 < t < co, since Cb(t) = 0 there. It also remains true for do 5 t < d, for w(t) is constant on that interval. And finally,

w(t) = w(t)

for t > d,

so w(t)lt = w(t)/t < n for such t by choice of d. Thus, we surely have w(t)

t

< 2n

fort ? 0.

The quantity on the left is, however, equal to I JO I/do > 1 for t = do. The purpose of the constructions in this article has been to arrive at the

intervals Jk, and the remaining work of this § concerned with even polynomials having real zeros deals exclusively with them. The preceding discussions amount to a proof of the following

2 Inclusion of zeros of P(x) in special intervals Jk

467

Theorem. Let p, 0 < p < io, and q, 0 < h < 3, be given, and suppose that p

n(t) su rp t

1 -3p*

Then there is a finite collection of intervals Jk = [ck, dk],

k 3 0, lying in

(0, oo), such that (i) all the discontinuities of n(t) lie in (0, do) u Uk11Jk; (ii)

n(do)

n(Jk)

=

p I Jo I

=

p I Jk I

1

2

for k, l

(if there are intervals Jk with k >, 1);

(iii) for 0 < t < do, n(do) - n(t)

1

p3p (do -

t),

whilst, for ck < t < dk when k >, 1,

n(t) - n(ck) 1

1

p 3p (t - ck)

and

n(dk) - n(t) 5

p3p (dk - t); 1

(iv) for k > 1, ck is well disposed with respect to dk - 1 (if there are Jk with

k > l); (v) for t > 0, 1

t

[0, t] n U Jk < 2q, k30

and the quantity on the left is > q for t = do.

Remark 1. The Jk with k > 1 (if there are any) are just certain of the Ir from the second stage. So, for k > 1, the above property (ii) is just property (ii) for the Ir.

Remark 2. By property (iv) and the theorems of article 1, we have k

f

log I P(x) I

x

log+ I P(M) I

dx 5 5 dk_1<m
rn

for each of the intervals (dk_ 1, CO with k >, 1 (if there are any). And, if J1

468

VIII B The set E reduces to the integers

is the last of the Jk, I Jf °° log I P(x) 2 x d,

dx < 5

d,<m<w

log + I P(m) m2

See the end of the second stage of the preceding construction. Here is a picture of the graph of n(t) vs. t, showing the intervals Jk: slope = pl(1-3p)

4

n(t)

:plJ,l/2

iplJol/2

co

0

Jo

d' p c1\ j

Ji-,

di

p

t

I.

Figure 140

3.

Replacement of the distribution n(t) by a continuous one

Having chosen p, 0 < p < 1/20, and rl, 0 < rl < 3, we continue with our program, assuming that n(t)

su r>o t

>

p

1 - 3p'

our aim being to obtain a lower bound for log+ 1

m2 IP(m)I

Our assumption makes it possible, by the work of the preceding article, to get the intervals

Jk = [ck, dk] c (0,oo), k=0,1,..., related to the (unknown) increasing function n(t) in the manner described by the theorem at the end of that article. Let Ji be the last of those Jk; during this article we will denote the union

(do, cl)u(dl, c2)u...u(di-i, ci)u(di, co) by t - see the preceding diagram. (Note that this is not the same set (9 as

3 Replacing n(t) by a continuous distribution

469

the one used at the beginning of article 2!) Our idea is to estimate log+IP(m)I 2

m from below, this quantity being certainly smaller than t meo

interested in. According to Remark 2 following the theorem about the Jk, we have log+ P(m) I

I fe log 1 P(x) 1

In 2

Meo

X

2

dx.

What we want, then, is a lower bound for the integral on the right. This is the form that our initial simplistic plan of `replacing' sums by integrals finally assumes. In terms of n(t), log I P(x) I

= f log

= Y log

Jo

k

x2 1- z dn(t), t

so the object of our interest is the expression

1- x22 dn(t)dx 2. x t

Here, n(t) is constant on each component of (9, and increases only on that set's complement.

We are now able to render our problem more tractable by replacing n(t) with another increasing function µ(t) of much more simple and regular

behaviour, continuous and piecewise linear on R and constant on each of the intervals complementary to the Jk. The slope p'(t) will take only two values, 0 and p/(1 - 3p), and, on each Jk, µ(t) will increase by p1Jkl/2. What we have to do is find such a µ(t) which makes r

S fo

f

l- t2 du(t)az 2

o0

log

smaller than the expression written above, yet still (we hope) strictly positive. Part of our requirement on µ(t) is that µ(t) = n(t) for te(9, so we will have

Jioi_ t2 dµ(t) - f,0 log 1- tZ dn(t) x2

0

0

_

(do

f

log

x2

1-t2

d(µ(t) - n(t))

0

+

('dk

x2

log 1 -

d(µ(t) - n(t)). t2 k,l ck We are interested in values of x in (9, and for them, each of the above J

470

VIII B The set E reduces to the integers

terms can be integrated by parts. Since µ(t) = n(t) = 0 for t near 0 and µ(do) = n(do), U(ck) = n(ck) and µ(dk) = n(dk) for k >, 1, we obtain in this way the expression do

2x2

fo x2 - t2

µ(t) - n(t) dt t

+

f dk 2x2 ck X 2

_t

k,>l

u(t) - n(t) dt. t

2

Therefore

JJiogi - t2

l

2dx

do

0x

0

d(µ(t) - n(t))

µ(t) - n(t)

t2

fdk Ck

+

t

dt p(t)

2dx

JeX -t 2

k>1

dx

n(t)

t

2

dt

,

and we desire to find a function µ(t) fitting our requirements, for which each of the terms on the right comes out negative. Put F(t) = 2 dx

fe x2 -t2 for to(9. We certainly have F(t) > 0 for 0 < t < do, so the first right-hand term, which equals d0

J

F(t)

µ(t)

- n(t) dt t

is 5 0 if µ(t) <, n(t) on [0, do]. Referring to the diagram at the end of the previous article, we see that this will happen if, for 0 < t < do, µ(t) has the form shown here:

n(t)

pIJoI/2

7 µ(t) 0

Figure 141

do

CO

Jo

do

3 Replacing n(t) by a continuous distribution

471

For k >, 1, we need to define u(t) on [Ck, dk] in a manner compatible with our requirements, so as to make ° k F(t) u(t)

- n(t) dt
fCk

0.

Here, (9 includes intervals of the form (Ck - 8, Ck) and (dk, dk + b)

where S > 0, so, when t e (ck, dk),

F(t) - - oo for t -> ck and F(t) - oo for

t - dk. Moreover, for such t,

F'(t) = 4t f

o (x2 dxt2)2

> 0,

so there is precisely one point tke(ck, dk) where F(t) vanishes, and F(t) < 0 for ck < t < tk, while F(t) > 0 for tk < t < dk. We see that in order to make dk

F(t) µ(t) - n(t)

Ick

t

dt < 0,

it is enough to define µ(t) so as to make

µ(t) i n(t)

for CI, < t < tk

µ(t) < n(t)

for tk < t < dk.

and

The following diagram shows how to do this: slope = pl(1-3p)

nN

Ck

Hv

tk

'rk

Jk

Figure 142

Sk

dk

..

O

VIII B The set E reduces to the integers

472

We carry out this construction on each of the Jk. When we are done we will have a function µ(t), defined for t,>- 0, with the following properties:

(i) µ(t) is piecewise linear and increasing, and constant on each interval component of (0, 00)

U Jk; k->O

(ii) on each of the intervals Jk, µ(t) increases by p 14112;

(iii) on Jo, µ(t) has slope zero for co < t < So and slope p/(l - 3p) for b0 < t < do, where (do - So)/(do - co) = (1 - 3p)/2;

(iv) on each Jk, k >, 1, µ(t) has slope zero for yk < t < Sk and slope p/(1 - 3p) in the intervals (Ck, yk) and (Sk, dk), where

Ck
yk-ck+dk-Sk _ 1-3p. dk - Ck

J o log

(v) J0

'

2

x2 1-i2

dµ(t)

d Z \ Jro

f

0

log

x2 1-t2

dx dn(t) x2.

o

Here is a drawing of the graph of µ(t) vs. t which the reader will do well to look at from time to time while reading the following articles:

-

~-rI"-, 1

IJol%2 ;

o n co

So do

Jo-l

S

Cj 7,

8, dl $Z

l-J2

c2 72

b2d2

t Sl

Figure 143

In what follows, we will in fact be working with integrals not over 0, but

over the set 0 = (0, CO) u d = (0, x) - U k> 0 Jk (see the diagram). Since our function µ(t) is zero for t

log 1Jo`0

co, we certainly have

x2 I

r2

dp(t) 1< 0

4 Formulas

473

for 0 < t < co. Hence, by property (v), s dx x2 I dµ(t)

log 1

fa fo

t

r S JJ o

log I P(x)1 x

dx

for our polynomial P. And, as we have seen at the beginning of this article,

the right-hand integral is in turn

Slog+IP(m)I m2

i

What we have here is a Theorem. Let 0 < p < 1/20 and 0 < ri < 3, and suppose that n(t) >

p

su t>o t

1-3p'

Then there are intervals Jk c (0, oo), k ,>O, fulfilling the conditions enumerated in the theorem of the preceding article, and a piecewise linear increasing function µ(t), related to those Jk in the manner just described, such that log

Jnfo

x2 1-t2

dx x

S

log+ I P(/11)

d,u(t) 2

1

m2

for the polynomial P(x). Here, U Jk.

S2 = (0, 00)

k30

Our problem has thus boiled down to the purely analytical one of finding

a positive lower bound for

fn f0'0 log

i

1- x2 dµ(t)dz t

when µ(t) has the very special form shown in the above diagram. Note that here I J01/do % n according to the theorem of the preceding article. 4.

Some formulas

The problem, formulated at the end of the last article, to which we have succeeded in reducing our original one seems at first glance to be rather easy - one feels that one can just sit down and compute

f

logll - t2ldµ(t)d2.

n Jo

474

VIII B The set E reduces to the integers

This, however, is far from being the case, and quite formidable difficulties

still stand in our way. The trouble is that the intervals Jk to which u is related may be exceedingly numerous, and we have no control over their positions relative to each other, nor on their relative lengths. To handle our task, we are going to need all the formulas we can muster. Lemma. Let v(t) be increasing on [0, oo), with v(0) = 0 and v(t) = O(t) for t-+0 and for t -+oo. Then, for xeR, x2

1- t2 dv(t) = - x J OO log 0

x+t dlvtt)I. x-t

Proof. Both sides are even functions of x and zero for x = 0, so we may as well assume that x > 0. If v(t) has a (jump) discontinuity at x, both sides are clearly equal to - oo, so we may suppose v(t) continuous at x. We have x+t

o"

f log

x-t

d(v')) = J_1og____dv(t) lIx+tlJ

Using the identity

Zlog

ft

x+t dt = - I log x+t x-t x-t

1- x2 2 t

we integrate the second term on the right by parts, obtaining for it the value

-

2v(x)Glog 2

+

i to g x+t + I

J

x-t

0

2

log 1- t2

/

dv(t),

taking into account the given behaviour of v(t) near 0. Hence Cx

log o

x+t d (v(t)) x- t t

2v(x)log2 x

"

1

x fo

In the same way, we get

x+t

x-t =

(2v(x)lo2) x

-

1

°°

x fX log

dv(t).

Adding these last two relations gives us the lemma.

x2

log 1- 2 dv(t). t

4 Formulas

475

Corollary. Let v(t) be increasing and bounded on [0, oo), and zero for all t sufficiently close to 0. Let w(x) be increasing on [0, oo), constant for all sufficiently large x, and continuous at 0. Then log

dv(t)

dx - dw(x) x2

x+t Jo

Jo log

x-t

d (v(t))d(o(x) t x

Proof. By the lemma, the left-hand side equals

x+t

x-t

`0 log

0 fo f"o

d v(t)) dw(x) - dx t x

Our condition on v makes log

x+t

x-t

d

v(t)

t

x dx

absolutely convergent, so we can change the order of integration. For t > 0,

x+t f* log 0

x-t

dx

x

assumes a constant value (equal to 7t2/2 as shown by contour integration see Problem 20), so, since in our present circumstances

J0d()

= 0,

the previous double integral vanishes, and the corollary follows.*

In our application of these results we will take

v(t) = 1 -3p p

µ(t),

µ(t) being the function constructed in the previous article. This function v(t) increases with constant slope 1 on each of the intervals [ak, dk], k >, 0, and [ck, yk], k > 1, and is constant on each of the intervals complementary to those. Therefore, if SL = (0, 00) - U [Ck, yk] ' U [Sk, dk] k31

k30

* The two sides of the relation established may both be infinite, e.g., when v(t) and co(t) have some coinciding jumps. But the meaning of the two iterated integrals in question is always unambiguous; in the second one, for instance, the outer integral of the negative part of the inner one converges.

476

VIII B The set E reduces to the integers

(note that this set

Jt Jo

includes our f'), we have

1- xti

log P

JJlog

-3p

o

0

i

1- x2 t

dv(t)

dx - dv(x) x2

The corollary shows that this expression (which we can think of as a first approximation to

1- tyI d1t(t)dz

log

)

is equal to

x+t d (v(t)) dv(x)

P

1 - 3p Jo fooo log

x-t

x

t

This double integral can be given a symmetric form thanks to the Lemma. Let v(t) be continuous, increasing, and piecewise continuously differentiable on [0, oo ]. Suppose, moreover, that v(0) = 0, that v(t) is constant for t sufficiently large, and, finally*, that (dldt)(v(t)%t) remains bounded when t -+ 0+. Then,

Jo Jo log

x+t d(v(t)) zx)dx x-t 2

= - 4 (v(0))2.

Proof. Our assumptions on v make reversal of the order of integrations in the left-hand expression legitimate, so it is equal to

x+t z2dxd(vtt)) x-t

Sc

Jo

Since

(f

log

o

v(t)

+

-1

+1 d -

n2

1;-1

2

t

(which may be verified by contour integration), we have log fo`0

+I

-1 t

d

7r 2

2

v' (0)

for t - 0, and integration by parts of the outer integral in the previous * This last condition can be relaxed. See problem 28(b), p. 569.

4 Formulas

477

expression yields the value

- 2 (v'(0))2 - f

v tt) dt

(' J

log 0

O'O

+

-1

t

d 1 dt.

Under the conditions of our hypothesis, the differentiation with respect to t can be carried out under the inner integral sign. The last expression thus becomes 7r

log

v(t)

2

2 2 (v'(0))2 o

2

-

t

d dt

+1 d

fOOO

tt)

(v'(0))Jo

0

x+t x d v(x) dx dt. fO'O

log

x-t

t dx

x

x

In other words

log 7E

x+t

x-t

d i v(t) I v(x) dx

t Jx2

t+x

2

- 2 (v'(0))2 - fo"O fO'O log

t-x

dC

v(x)

x

v(t)

/

t2

dt.

The second term on the right obviously equals the left-hand side, so the lemma follows. Corollary. Let v(t) be increasing, continuous, and piecewise linear on [0, oo),

constant for all sufficiently large t and zero for t near 0. Then x2 1-i2

dv(t)

dx - dv(x)

0 log 0 f'O fOO

=j

log 0

0

x+t d(vtt))d(vxx) ). x - t

Proof. By' the previous corollary, the left-hand expression equals

fa fo

loglx+tIdl vtt) Idxx).

In the present circumstances, v'((0) exists and equals zero. Therefore by the lemma log

x+t

x-t

d l vtt)

I

x) dx = 0,

and the previous expression is equal to

flog fo'O

x+t

x-t dl

vtt))dl v(x)

I.

478

VIII B The set E reduces to the integers Problem 21 Prove the last lemma using contour integration. (Hint: For 3z > 0, consider the analytic function log((z

F(z) It J o

t t )d(vt)

and examine the boundary values of 9?F(z) and 3F(z) on the real axis. ± Then look at $r((F(z))2/z) dz for a suitable contour I'.) 5.

The energy integral

The expression, quadratic in d(v(t)/t), arrived at near the end of the previous article, namely, Jo"O Jo'O loglx±tld\vtt)/d(vxx)/, has a simple physical interpretation. Let us assume that a flat metal plate of infinite extent, perpendicular to the z-plane, intersects the latter along

the y-axis. This plate we suppose grounded. Let electric charge be continuously distributed on a very large thin sheet, made of nonconducting material, and intersecting the z-plane perpendicularly along the positive x-axis. Suppose the charge density on that sheet to be constant along lines perpendicular to the z-plane, and that the total charge contained in any rectangle of height 2 thereon, bounded by two such lines intersecting

the x-axis at x and at x + Ax, is equal to the net change of v(t)l t along [x, x + Ax]. This set-up will produce an electric field in the region lying to the right of the grounded metal plate; near the z-plane, the potential function for that field is equal, very nearly, to u(z) = J 'O log 0

z+t z-t

d(vtt)).

The quantity

Ju(x)d()

JJlog x+t x-t d(vtt))d(vxx))

is then proportional to the total energy of the electric field generated by our distribution of electric charge (and inversely proportional to the height of the charged sheet). We therefore expect it to be positive, even though charges of both sign be present at different places on the non-conducting sheet, i.e., when d(v(t)/t)/dt is not of constant sign. Under quite general circumstances, the positivity of the quadratic form

in question turns out to be valid, and plays a crucial role in the computations of the succeeding articles. In the present one, we derive two formulas, either of which makes that property evident.

5 The energy integral

479

The first formula is familiar from physics, and goes back to Gauss. It is convenient to write v(t)

p(t)

t

Lemma. Let p(t) be continuous on [0, oo), piecewise W3 there (say), and differentiable at 0. Suppose furthermore that p(t) is uniformly Lip 1 on [0, oo) and tp(t) constant for sufficiently large t. If we write

u(z) =

flog 0

z+t z-t

dp(t),

we have

x+t

x-t

Sc

dp(t)dp(x) = n

f 0

{(ux(z))2

J o

+ (u,,(z))2} dx dy.

Remark 1. Note that we do not require that p(t) vanish for t near zero, although p(t) = v(t)lt has this property when v(t) is the function introduced in the previous article.

Remark 2. The factor 1/n occurs on the right, and not 1/2n which one might expect from physics, because the right-hand integral is taken over the first quadrant instead of over the whole right half plane (where the `electric field' is present). The right-hand expression is of course the Dirichlet integral of u over the first quadrant. Remark 3. The function u(z) is harmonic in each separate quadrant of the zplane. Since log

z+w

z-w

is the Green's function for the right half plane, u(z) is frequently referred to as the Green potential of the charge distribution dp(t) (for that half plane).

Proof of lemma. For y > 0, we have uy(z)

=

y J0((x±t+y2

_

Y

(x - t)2 + y2 dP(t),

and, when x > 0 is not a point of discontinuity for p'(t), the right side

480

VIII B The set E reduces to the integers

tends to - ap'(x) as y - 0 + by the usual (elementary) approximate identity property of the Poisson kernel. Thus,

u,,(x + i0) = - np'(x), and

=-

log X + t dp(t)dp(x)

Jo Jo

u (x)u,,(x + iO)dx. fo,

n left-hand double integral At the same time, u(iy) = 0 for y > 0, so the from the previous relation is equal to 1

1

u(x)u,,(x + iO)dx -

n Jo

n

u(iy)ux(iy)dy. fo`0

We have here a line integral around the boundary of the first quadrant. Applying Green's theorem to it in cook-book fashion, we get the value Jfo-

/

l

\\Y

JJ

I a (u(z)u,(z)) + ax (u(z)ux(z)) I dx dy,

which reduces immediately to °°

J°°

((uy(z))2 + (ux(z))2)dxdy

n fo

0

(proving the lemma), since u is harmonic in the first quadrant, making uV2u = 0 there. We have, however, to justify our use of Green's theorem. The way to do that here is to adapt to our present situation the common 'non-rigorous' derivation of the theorem (using squares) found in books on engineering mathematics. Letting

-9A denote the square with vertices at 0, A, A + iA and iA, we verify in that way without difficulty (and without any being created by the discontinuities of p'(x) = - u,(x + i0)/ir ), that

(uuxdy - uuydx) = J

J

f

a9A

(ux' + uy) dx dy.*

9A

The line integral on the left equals

r

- J oA u(x)uy(x + iO) dx +

(uuxdy - uuydx),

J rA

where I'A denotes the right side and top of -9A: * The simplest procedure is to take h > 0 and write the corresponding relation involving u(z + ih) in place of u(z), whose truth is certain here. Then one can make h 0. Cf the discussion on pp. 506-7.

5 The energy integral

481

Y

r,,

A

0 Figure 144

We will be done if we show that

IrA

(uuxdy - uuydx) -.0

for A -+ oo.

For this purpose, one may break up u(z) as

Jo

logl-z+'Idp(t) + fm logl-

dp(t),

M being chosen large enough so as to have p(t) = C/t on [M, oo). Calling the first of these integrals ul(z), we easily find, for IzI > M (by expanding the logarithm in powers of t/z), that const.

lu,(z)I 5

IzI

and that the first partial derivatives of ul(z) are O(1/Izl2). Denote by u2(z) the second of the above integrals, which, by choice of M, is actually equal to z+t

dt

z-1

t2

The substitution t = I z I T enables us to see after very little calculation that this expression is in modulus const.

loglzl IZI

for large IzI. To investigate the partial derivatives of u2(z) in the open first quadrant, we take the function

F(z) =

('°°

log

f

M

z+t dt

(z-t)t2

,

482

VIII B The set E reduces to the integers

analytic in that region, and note that by the Cauchy-Riemann equations,

_ -CF'z()

aaxz)-iaay(z) there. Here,

F(z) =

dt

`°

t2(z + t)

nr r

-

dt Mt

2(Z

- t)

The first term on the right is obviously O(1/Izl) in modulus when Rtz and 3z > 0. The second works out to °°

1

1

1

1

1

/z-M

zM+Zlogl

zt2+z2t+z2(z-t))dt

,y

M

using a suitable determination of the logarithm. This is evidently O(1/IzI) for large Iz1, so IF'(z)I = 0(1/Iz1) for z with large modulus in the first quadrant. The same is thus true for the first partial derivatives of u2(z). Combining the estimates just made on ul(z) and u2(z), we find for u = ul + u2 that lz lu(z)I < const.logIZI

I ux(z) I

cont.

1

IzI 1

Iui,(z)I < const.lzI when 91z > 0, .3z > 0, IzI being large. Therefore

r loA )

SrAX

- uudx) = O 1\

J

for large A, and the line integral tends to zero as A -+ oo. This is what was needed to finish the proof of the lemma. We are done.

Corollary. If p(t) is real and satisfies the hypothesis of the lemma, ( 0D

foo,

log

Ix+t

J o"

x-t

dp(t)dp(x) > 0.

Proof. Clear. Notation. We write

E(dp(t), do(t)) = Joo Jo o

log

x+ t Ix-t

dp(t)do(x)

5 The energy integral

483

for real measures p and a on [0, oo) without point mass at the origin making

both of the integrals log

x+t

x-t

log

dp(t) dp(x), 0

0

x+t

x-t da(t) da(x)

absolutely convergent. (Vanishing of p({0}) and a({0}) is required because log I (x + t)/(x - t) I cannot be defined at (0, 0) so as to be continuous there.)

Note that, in the case of functions p(t) and a(t) satisfying the hypothesis of the above lemma, the integrals just written do converge absolutely. In terms of E(dp(t), da(t)), we can state the very important

Corollary. If p(t) and a(t), defined and real valued on [0, oo), both satisfy the hypothesis of the lemma, I E(dp(t), da(t))I 5

I(E(dp(t),

da(t))).

Proof. Use the preceding corollary and proceed as in the usual derivation of Schwarz' inequality.

Remark. The result remains valid as long as p and a, with p({0}) = a({0}) = 0, are such that the abovementioned absolute convergence holds. We will see that at the end of the present article. Scholium and warning. The results just given should not mislead the reader

into believing that the energy integral corresponding to the ordinary logarithmic potential is necessarily positive. Example: 2.

2rz

2,

1

2itlog- dqp = -4nZlog2 !

logl2e's-2e"'I d9dcp = 0

0

0

It is strongly recommended that the reader find out exactly where the argument used in the proof of the lemma goes wrong, when one attempts to

adapt it to the potential 2rz

J1og1211d19. u(z) = fo For 'nice' real measures p of compact support, it is true that log

JJc

1

Iz

wI

dp(z)dp(w) >, 0

provided that fcdp(z)=0. The reader should verify this fact by applying a suitable version of Green's theorem to the potential 1. log (1/I z - w I) dp(w).

The formula for E(dp(t),dp(t)) furnished by the above lemma exhibits that quantity's positivity. The same service is rendered by an analogous

484

VIII B The set E reduces to the integers

relation involving the values of p(t) on [0, oo). Such representations go back to Jesse Douglas; we are going to use one based on a beautiful identity of Beurling. In order to encourage the reader's participation, we set as a problem the derivation of Beurling's result. Problem 22

(a) Let m be a real measure on R. Suppose that h > 0 and that f fis ,

dm(q) converges absolutely. Show that

h

f 7. (m(x + h) - m(x))2 dx = JT J

(h - I -?I )+

dm(q).

(Hint: Trick: x+h

rx+h

(m(x + h) - m(x))2 = Jx

Jx

(b) Let K(x) be even and positive,'62 and convex for x > 0, and such that K(x) for x - oo. Show that, for x # 0,

K(x) = J (h-Ix1)+K"(h)dh. 0

K (x)

x

0 Figure 145

(Hint: First observe that K'(x) must also -+0 for x -+ oo.) (c) If K(x) is as in (b) and m is a real measure on R with dm(rl) absolutely convergent, that integral is f °° f °°,,K(1 - q equal to [m(x + h) - m(x)]2K"(h) dh dx

J J

J

Cn(Y)-n(x)]ZK"(Ix-vl)dydx.

5 The energy integral

485

(Hint: The assumed absolute convergence guarantees that m fulfills, for each h > 0, the condition required in part (a). The order of integration in K"(hxm(x + h) - m(x))2 dh dx

may be reversed, yielding, by part (a), an iterated triple integral. Here, that triple integral is absolutely convergent and we may conclude by the help of part (b).)

Lemma. Let the real measure p on [0, oo), without point mass at the origin, be such that log

x+t

dp(t) dp(x)

x-t

is absolutely convergent. Then

log

x-t dp(t) dp(x) Joco

(p(x) - P(Y) )2 x2 + y2 y)2 dx dx -Y (x + Jo

Proof. The left-hand double integral is of the form

fo J o

k

(t) dP(x) dP(t),

where

k(i) = log I 1 + i

-i

so we can reduce that integral to one figuring in Problem 22(c) by making the substitutions x = e4, t = e", p(x) = p(t) = m(ri), and ki

t

I = K( - q) = log coth(

2

K(h), besides being obviously even and positive, tends to zero for h - oo. Also

K'(h) =

-

tanh 2

2

coth 2

2

486

VIII B The set E reduces to the integers

and

K"(h) = 4 sech2 2+ 4 cosech2

2>

0,

so K(h) is convex for h > 0. The application of Beurling's formula from problem 22(c) is therefore legitimate, and yields

x+t 0 log 0 f'O f'O

f '0

dp(t) dp(x)

x-t

f

00

K(I

-

K"(I

=

J

(note that the first of these integrals, and hence the second, is absolutely convergent by hypothesis). Here, 1

sinh2

+ cosh2

2

2 n

sinh2 (--) cosh2 (\ 2 " e2 2ry

cosh(- n) Sinh2(g

- q)

2e a

+e

(e24 -e 2n)2

/

J

'

so the third of the above expressions reduces to °°

r

e24+e2"fm()-m(,1))2e4e"d

(e + e")2 -j- e"

fM

do

x2 + t2 fp(x) - p(t) )2 dxdt. x-t Jo Jo (x+t)2

We are done. Remark. This certainly implies that the first of the above corollaries is true for any real measure p with p({0}) = 0 rendering absolutely convergent the double integral used to define E(dp(t), dp(t)). The second corollary is then also true for such real measures p and a.

The formula provided by this second lemma is one of the main ingredients in our treatment of the question discussed in the present §. It is the basis for the important calculation carried out in the next article.

6 Lower estimate for fa f log 11- (x2/t2) I dµ(t) dx/x2 o 6.

log 1- t2 dµ(t)

A lower estimate for f

487

d2

I

-Jo

We return to where we left off near the end of article 4, focusing our attention on the quantity

ff a

i x2

log

t

O

dµ(t)

d2 ,

where µ(t) is the function constructed in article 3 and

!a = (0, a) - {x: µ'(x) > 01. Before going any further, the reader should refer to the graph of µ(t) found near the end of article 3. As explained in article 4, we prefer to work not with µ(t), but with

v(t) = 1 -3p P

µ(t);

the graph of v(t) looks just like that of µ(t), save that its slanting portions all have slope 1, and not p/(1- 3p). Those slanting portions lie over

certain intervals [ck, yk], k , 1,

[Sk, dk],

k , 0, contained in the

Jk = [Ck, dk], and

SL = (0, co) ^' U [Sk, dk] ^' U [ck, yk] k,0

k>, l

This set S is obtained from the one f shown on the graph of p(t) by adjoining to the latter the intervals (co, So) JO and (yk, 5) c Jk, k ,1. By the corollary at the end of article 4,

i x2

log

r

fa f0`0

dµ(t)

d

z

°°

[-log 1-3p J. JO P

P

1

t2 x2

dv(t)

dx - dv(x) 2

r°° r°° log

13pJo

0

and this is just 1

E(

Pap E(di

x)

, ) being the bilinear form defined and studied in the previous article. This identification is a key step in our work. It, and the results of article

488

VIII B The set E reduces to the integers

5, enable us to see that x2

1-r2 fa f0'0 log

is at least positive (until now, we were not even sure of this). The second lemma of article 5 actually makes it possible for us to estimate that integral from below in terms of a sum,

f(Yk_ck)2 A k>

+E

(dkdk1k)2,

k >-O

like one which occurred previously in Chapter VII, §A.2. In our estimate, that sum is affected with a certain coefcient. On account of the theorem of article 3, we are really interested in log

1- x2 2 t

rather than the quantity considered here. It will turn out later on that the passage from integration over S) to that over C1 involves a serious loss, in whose evaluation the sum just written again figures. For this reason we have to take care to get a large enough numerical value for the coefficient mentioned above. That circumstance requires us to be somewhat fussy in the computation made to derive the following result. From now on, in order to make the notation more uniform, we will write Yo = co.

Theorem. If v(t) = ((1 - 3p)/p)µ(t) with the function µ(t) from article 3, and the parameter q > 0 used in the construction of the Jk (see the theorem, end of article 2) is sufficiently small, we have

E(d(vit)),

d(v(t)) A- ck

(2-log2-Krl) k,0

Yk

2

+ (dk

J

Sk

2

dk

Here, K is a purely numerical constant, independent of p or the configuration

of the A. Remark. Later on, we will need the numerical value

i - log 2 = 0.80685....

6 Lower estimate for $of o log I 1 - (x2/t2) I dy(t) dx/x2

489

Proof of theorem. By the second lemma of article 5 and brute force. The lemma gives

E\d\vtt)/, d(vtt)) v(x)

v( y)

x

Y

0o

ff ( 0

x2 + 2

x-y

o

2

YZdxdy

(x+y)

r() v(x)

('

v(y)

2

> 2Yk,oJ rk J rk Vx- - yY

dxdy.

T Yk

Ck

7k

x

dk

6k

ak

11

- Jk

Figure 146

On each interval Jk = [ck, dk] we take Yk = Ck + 2(Yk

Ck)

Sk = dk - 2(dk - 8k)

(see figure). Since

Yk-Ck+dk-6k _ 1-3p dk - Ck

1

<2

2

(properties (iii), (iv) of the description near the end of article 3) we have yk < 6'. Therefore, for each k, (

)

2

-

(

)

x JJk JJk

Vx

-Y

dxdy

I

If,,

Ck+ J

dk

6.

J

v(x)

v(y)

xx

y

dk 6"

--

y

2

dx d Y.

490

VIII B The set E reduces to the integers

We estimate the second of the integrals on the right - the other one is handled similarly. We begin by writing dk

fdk

f

ak

v(x)

v(x)

x

y

x-y

'

k

2

) dx dy v(x)

fkk+fk

+f

k

xx

v(Y)

2

-Yy

dxdY

I

Of the three double integrals on the right, the first is easiest to evaluate. Things being bad enough as they are, let us lighten the notation by dropping, for the moment, the subscript k, putting 6'

for

6'k.

S

for

Sk

d

for

dk.

and

Since v'(x) =1 for bk = S <x
x

= l + v(S)-a, x

S<x
Using this, we easily find that V(X)

v(y)

xx-yy

Jda ,la

2

)dxdy =

(()2(j.,5)2

In terms of j = U ((Ck,Yk)U(5k,dk)) k>0

and

J = U Jk, k30

we have clearly

v(t) = I[0, t]nfl <' I[0, t]nJl,

t>0.

The right-hand quantity is, however, < 2qt by construction of the Jk (property

(v)

in the theorem at the end of article

2).

Therefore

6 Lower estimate for Info logy 1 -(x2/t2)Idit(t)dx/x2

491

VOW = v(dk)/dk 5 2q, and the integral just evaluated is (1 -2 q)2 (

6)2

d d

We pass now to the second of the three double integrals in question, continuing to omit the subscript k. To simplify the work, we make the changes of variable

y=6-t,

x=6+s,

and denote d - 6 = 6 - 6' by A. Then v(x)

v(y)

/v(6) + S

2

fe

S Jrad

dx dy X

Jo

Jo

Y

v(6) \2

f= b+ss+t

S-t

ds dt,

o

since v(y) = v(6) for 6' < y < 6 (see the above figure). The expression on the right simplifies to fA 0

e 0

s

v(6)

(6 + s)(t + s)

(6 - t)(6 + s)

2

ds dt

which in turn is

zd J

() t+s) dtds - 2v6 0 J 0X e

e

z

e2 T2

e

6'6J 0 fo

(1 -log 2) -

t+sdsdt

4i1e2 616

We have (we have again used the fact that v(6)/6 5 v(d) >, v(d) - v(6) = d - 6 = e, so, since v(d)/d 5 2r1,

6 = d-e >, (1-2q)d and

6' = d-2e >, (1-4q)d. By the computation just made we thus have v(x) S

J

d

a..a

v(y)

2

y

y

x- y

dxdy )41l

(1

1 -2

(1

-log2-(1-2q)(1-40)(d

-

-40)(d d

6)2.

For the third of our three double integrals we have exactly the same

492

VIII B The set E reduces to the integers

estimate. Hence, restoring now the subscript k, v(x) dk

x

dk

ii

-

v(Y)

2

y

(3-21og2-15x1dk

dxdy >

Sk

k

)

)z f

Cd

as long as n > 0 is sufficiently small. In the same way, one finds that v(x)

Ck

v(y)

2

x-Y

Ck

dx dy

(3 - 21og 2 - Krl) A - ck Yk

l2

/

for small enough n > 0, K being a certain numerical constant. Adding this to the previous relation gives us a lower estimate for v(x)

v(y) 1 2

fJkfJk( xx - Yy

ax ay;

adding these estimates and referring again to the relation at the beginning of

this proof, we obtain the theorem.

Q.E.D.

From the initial discussion of this article, we see that the theorem has the following

Corollary. Let µ(t) be the function constructed in article 3 and !a be the complement, in (0, oo), of the set on which µ(t) is increasing. Then, if the parameter q > 0 used in constructing the Jk is sufficiently small,

1J.

zt x2

log

1

p3p(2-log 2-Krl)

YkYkckZ+(dkdkk)2).

Here K is a numerical constant, independent of p or of the particular configuration of the Jk.

In the following work, our guiding idea will be to show that 1"1o log 11- x2/t2 I dµ(t)(dx/x2) is not too much less than the left-hand

integral in the above relation, in terms of the sum on the right. 7.

Effect of taking x to be constant on each of the intervals Jk

We continue to write

fl = (0, oc) - J,

7 Effect of assuming x constant on each Jk

493

where J = Uk,oJk with Jk = [ck, dk], and = (0, oo) - J, with U ((Ck, A)U(Sk, dk)) k_> 0

being the set on which µ(t) is increasing. The comparison of with interest, of our object f of o log I I - x2/t2Idµ(t)(dx/x2), f n f o log I 1 - x2/t2 I dµ(t)(dx/x2) is simplified by using two approximations

to those quantities. As in the previous article, we work in terms of

v(t) = 1 - 3p

µ(t)

P

instead of p(t). Put

u(z) =

log 0f"o

z+t dl vtt)). z-t

Then, by the corollary to the first lemma in article 4, x2

°°

fn Jo log

I

dx

1 - t2 dµ(t) x2

_

p

u(x)

1 _-3 p

dx X

J

and

JJ°

log

x2 1- a t

dµ(t) d 2

=1

p3

u(x)

.

p fj

Our approximation consists in the replacement of u(x)

Si

dx x

1

by

k,0 d k Jk

u(x)dx

and of u(x)

Si

dx x

1

dk

Ik

+

by k-Odk \

ck

u(x)dx. Sk

To estimate the difference between the left-hand and right-hand quantities we use the positivity of the bilinear form E( , ), proved in article 5.

494

VIII B The set E reduces to the integers

Theorem. If the parameter n > 0 used in the construction of the

Jk is

sufficiently small. Ju(x) J

-Y1

dx x

k30dk

u(x) dx

fj,,

1 / fiR

dx

f u(x) z -

kO

J

dk

+

u(x)dx 6k

kk

are both

Cn+E(dl vtt) ), d(vtt))

I,

where C is a purely numerical constant, independent of p < -2L o or the configuration of the Jk.

Remark. Here, E(d(vtt) ),

J,"(x)ax

d(vtt)))

X

according to the corollary at the end of article 4. Proof. Let us treat the second difference; the first is handled similarly. Take

Ck<x
Ox) =

1

X

0

-

1

,

Sk < x < dk,

k >, 0;

Wk'

elsewhere.

(Recall that yo = co, so (co, yo) is empty.) The second of the expressions in question is then just the absolute value of

x-t d() q (x)dx, x+t

u(x)cp(x) dx 0 f"O

i.e., of E(d(v(t)/t), gp(t)dt), in the notation of article 5. By the second corollary in that article and the remark at the end of it,

J(E(d(v(tt)),d(v(tt))))

E(d(vrt)),,p(t)dt II -<

x V (E(cp(t) dt, (p(t) dt)).

8 An auxiliary harmonic function

495

The function cp(x) is surely zero outside of the Jk, and, on Jk, dk - x IJki

0 5 9W 5

xdk

xdk

with I Jk I/dk < 2n as in the proof of the theorem of article 6. Therefore,

0<J'o log x+t x-t

x+t

gp(t)dt < 2n f'0 log

0

x-t

0

and ID

E(cp(t)dt, (p(t)dt) = J

OD

log

J 0

0

dt

211 7T

t

x+t

x-t gp(t)dtcp(x)dx

dk-xdx

2

k30JJk xdk

jr22nE nYIJkI2 k30Ckdk

We have ck = dk -I Jk I % (1 - 2n)dk (see above), and, according to property (iv) from the list near the end of article 3, 2 IJkI = dk - Ck =

{(Yk - Ck) + (dk - 60)

1 - 3p

Since we are assuming (throughout this §) that p < 20, this makes

0)2 IJkI2 < 2 17/) {(Yk-Ck)2+(dk-(Sk)2},

(

yielding, by the preceding relation, t(Yk-Ck)2+(dk-'k)2 k1

I

\

j. 1

1

2rl

Substitute this inequality into the previous estimate and then apply the theorem from the preceding article. One obtains q (t)dt)

6n2n

(1- 2n)(3 -log 2 - Kn)

E(dl v(t)),d1 v(t))).

\t

\tJ

Using this in the above inequality for I E(d(v(t)/t), co(t) dt) I, we immediately

arrive at the desired bound on the difference in question. We are done. 8.

An auxiliary harmonic function

We desire to use the lower bound furnished by the theorem of article 6 for

f

Jiu(x)dx

E(d(vtt)),d(vtt)))

496

VIII B The set E reduces to the integers

in order to obtain one for f,u(x)(dx/x), the quantity of interest to us. Our plan is to pass from

sj

U(X)

dk

yk

dx x

to

+

k30 dk

u(x)dx dk

ck

and from I

fd,,

Y

dx

to

u(x) dx

k30 d k

ii

k

u(x)

x ;

according to the result of the preceding article (whose notation we maintain here), this will entail only small losses (relative to $-u(x)(dx/x) ), if it > 0

is small. This procedure still requires us, however, to get from the first sum to the second. The simplest idea that comes to mind is to just compare corresponding terms of the two sums. That, however, would not be quite right, for in $dku(x) dx, the integration takes place over a set with larger Lebesgue measure than in (f k + $ak)u(x) dx. In order to correct for this discrepancy, one should take an appropriate multiple of the second integral and then match the result against the first. The factor to be used here is obviously 2

1 -3p' since (article 3),

Yk-Ck+dk-6k _ 1 -3p dk - Ck

2

We are looking, then, at dk

fk

u(x)dx -

(JTh

2

1 - 3p

u(x)dx -

lu(x)dx

bk /

k

bk

Yk

('dkl

+J

1 +3p I

3P

fdk

Yk

+ ck

u(x)dx. bk

From now on, it will be convenient to write

-

1+3p 1 - 3p'

A is > I and very close to 1 if p > 0 is small. It is also useful to split up

8 An auxiliary harmonic function

497

each interval (yk, Sk) into two pieces, associating the left-hand one with (ck, yk) and the other with (Sk, dk), and doing this in such a way that each piece has 2 times the length of the interval to which it is associated. This is of course possible because

_ 1+3p

ak - yk

I - 3p

Yk - Ck + dk - Sk

=;

we thus take gk E (yk,, Sk) with gk = Yk + 2(Yk - Ck) (and hence also gk = ak - ).(dk - Sk) ), and look at each of the two differences k

u (x)dx

- A f Yk u(x) dx,

f:'

ak

J 9k

Ck

u(x)dx - A

dk

u(x) dx

dk

separately; what we want to show is that neither comes out too negative, for we are trying to obtain a positive lower bound on Jju(x)(dx/x). Ck

7k

dk

sk ik

Figure 147

It is a fact that the two differences just written can be estimated in terms of E(d(v(t)/t), d(v(t)/t)).

Problem 23 (a) Show that for our function

u(z) = J

log

z+t d(v(t) ), z-t

one has

E(d\vtt)/'

d\vtt)//

41n 2

f f

(u(x)-

u(y)/ 2

-y

dx dy .

This is Jesse Douglas' formula - I hope the coefficient on the right is

correct. (Hint: Here, u(x)= -(1/x)fe log I1 -x2/t2Idv(t) belongs to L2(- oo, oo) (it is odd on II), so we can use Fourier-Plancherel transforms. In terms of T

12(2) = f

eiztu(t)dt

we have

I f.

u(x + iy) = 2n

A(2)d i

498

VIII B The set E reduces to the integers for y > 0 (the left side being just the Poisson harmonic extension of the function u(x) to 5z > 0), and u(x + h) - u(x)

1

h

2n

U( h e-x e - izh

f

(All the right-hand integrals are to be understood in the l.i.m. sense.) Use Plancherel's theorem to express

ru (x + h) - u(x)2

J

h

I\

('

dx

J

and

[(ux(z))2 + (u(z))2] dx

in terms of integrals involving I u(A)12, then integrate h from - oo to oo and y from 0 to oc, and compare the results. Refer finally to the first lemma of article 5.) (b) Show that f'9k

u(x)dx - A J

Yku(x)dx Ck

k

u(xX_u(.Y))2dydx

/((1+2)4_'_24\ 12.(Yk-Ck)

f 8ku(x)dx

J

JYk 9k(

- 2 J dku(x)dx. 'k

9k

(Hint: Trick: Yk

9k

u(x)dx - .1 J Yk

fYk

1

9k

u(x)dx =

[u(y) - u(x)] dy dx.

J Ck

A

Ck

k

Yk

(c) Use the result of article 6 with those of (a) and (b) to estimate

1 f"k Y_

1

kio dk

Yk

/

rvk

('dk\

\\\

Ck

bk 111

u(x)dx - 2(J + J

Iu(x)dx

in terms of E(d(v(t)/t), d(v(t)/t)).

By working the problem, one finds that the difference considered in part (c) is in absolute value < C f ju(x)(dx/x) for a certain numerical constant C. The trouble is, however, that the value of C obtained in this way comes

out quite a bit larger than 1, so that the result cannot be used to yield a positive lower bound on f ,,u(x)(dx/x), A being near 1. Too much is lost in following the simple reasoning of part (b); we need a more refined argument that will bring the value of C down below 1. Any such refinement that works seems to involve bringing in (by use

8 An auxiliary harmonic function

499

of Green's theorem, for instance) certain double integrals taken over portions of the first quadrant, in which the partial derivatives of u occur. Let us see how this comes about, considering the difference 9k

u(x) dx - A

(Yk

u(x) dx.

J Ck

f":

The latter can be rewritten as u(Ck + X)Sk(X) dx,

where Ak = Yk - ck, and Sk(X)

A,

0 < X < Ak,

1,

Ak<x<(l+A)Ak.

Suppose that we can find a function V(z) = Vk(z), harmonic in the half-strip

Sk = {z: 0 < lz < (l + i)Ak and 3z > O} and having the following boundary behaviour:

V,,(x + io) = - Sk(x),

0 < x < (1 + A)Ak

(V),(x + i0) will be discontinuous at x = Ak),

VV(iy) = 0,

y >0,

Vx(ly+(l+,.)Ak) = 0,

Figure 148

y>0.

500

VIII B The set E reduces to the integers

Then the previous integral becomes (- u(Ck + z)VY(z)dx + u(Ck + z)V,,(z)dy),

8Sk being oriented in the usual counterclockwise sense. Application of Green's theorem, if legitimate (which is easily shown to be the case here, as we shall see in due time), converts the line integral to (u y(ck + z)VY(z) + uX(ck + z)Vx(z))dxdy J fsk

+

ff

u(Ck + Z) IVYY(z) + Vxx(z)] dx dy.

$k

The harmonicity of V in Sk will make the second integral vanish, and finally the difference under consideration will be equal to the first one. Referring to the first lemma of article 5, we see that the successful use of this procedure in

order to get what we want necessitates our actually obtaining such a harmonic function V = Vk and then computing (at least) its Dirichlet integral

fiSk

(VX + V')dxdy.

We will in fact need to know a little more than that. Let us proceed with the necessary calculations.

Our harmonic function Vk(z) (assuming, of course, that there is one) will depend on two parameters, Ak and A = (1 + 3p)/(l - 3p). The dependence on

the first of these is nothing but a kind of homogeneity. Let v(z, A) be the function V(z) corresponding to the special value n/(1 + A) of Ak, using the value of A figuring in Vk(z); v(z, A) is, in other words, to be harmonic in the half-strip

S = {z: 0 < 91z < n and 3z>01 with vx(z, A) = 0 on the vertical sides of S and

A, 0<x<1 vY(x + i0, A) =

-1,

,

1+A<x«.

8 An auxiliary harmonic function

501

On the half-strip Sk of width (1 +.i)Ak shown previously, the function

1 (1 + A)Akv(nz/(1 + A)Ak, A) 7C

is harmonic, and its partial derivatives clearly satisfy the boundary conditions on those of Vk(z) stipulated above. We may therefore take

Vk(z) = I (1 + A)Akv(1CZ/(1 + A)Ak, a,) ;

this permits us to do all our calculations with the standard function v. Note that we will have, by simple change of (variables, axk)2dxdy =

((1 +2)Ak)2JJs(v.x(z,,))2dxdy

fiSk ( and Jsk[CaY)+]2dxdy

I

=

while oVk

11,. 8y

dx dy

C(1 + ;)ok)2

v ifs

Lemma. Given 2 , 1, we can find a function v(z, A) harmonic in S whose partial derivatives satisfy the boundary conditions specified above. Ifs > 0 is

502

VIII B The set E reduces to the integers

given, we have, for all 2 >, 1 sufficiently close to 1, (vx(z, 2 2 dx

n s

i

dy < 4 (1 +

1

33

/

('

1

+ 53 + ... + e), 1

1

J I(vy(z, A))+72 dx dy < 21 1 + 33 + 53 + ... + s

I,

and

f fs

lvi,(z,2)Idxdy < C,

C being a numerical constant, whose value we do not bother to calculate.

Remark. In the next article we will need the numerical approximation

42 1+33+53+"')

< 0.4268.

Proof of lemma. The method followed here (plain old `separation of variables' from engineering mathematics) was suggested to me by Cedric Schubert. We look for a function v represented in the form OD

v(z,2) = YA"(2)e-' cosnx 1

The series on the right, if convergent, will represent a function harmonic in S

(each of its terms is harmonic!), and, for y > 0,

vx(z,2) _ -YnA"(2)e-"ysinnx i

will vanish for x = 0 and x = it, for the exponentially decreasing factors a-"'' will make the series absolutely convergent. For y = 0, by Abel's theorem, v,,(x + i0, 2)

Y nA"(2) cos nx

at each x for which the series on the right is convergent. Let us choose the A"(2) so as to make the right side the Fourier cosine series of the function it

A,

s(x,A) =

-1,

0<x<1+.1' it

1+.1<x
We know from the very rudiments of Fourier series theory that this is

8 An auxiliary harmonic function

503

accomplished by taking

-

n

2

it

s(x, A) cos nx dx,

I

o

and that the resulting cosine

does converge to

series

s(x, A)

for

0 < x < n/(1 + A) and for n/(1 + A) < x < n. We can therefore get in this way a function v(z A) meeting all of our requirements.

Let us continue as long as we can without resorting to explicit computations. For fixed y > 0, Parseval's formula yields n(vY(z, 2))zdx

J0

=

it

00

Y

and, in like manner, I0 [vY(z,A)

- vi,(z,1)]z dx =

it

00

Y

Integrating both sides of this last relation with respect to y, we find that

(vy(z,2.)-vy(z,l))zdxdy =

n00

fon

By Parseval's formula, we have, however,

=

I

2

n

o[s(x,2) -s(x,1)]zdx,

and it is evident that the right-hand integral tends to zero as 2-41. Hence, by the preceding relation,

JJ[vy(z,2)_vy(z,l)]2dxdy --> 0 0

0

for A->1. Now clearly I (vY(z, 2))+ - (vY(z, 1))+ 15 I v,(z, A) - vy(z,1) I;

the result just obtained therefore implies that n

fo",

oo

[(vY(z1))+]zdxdy C(())+]-- J0J0,

as A -> 1. We see in the same fashion that (vx(z, 2))z dx dy =

which -+

7E 00

4

y

as 2-* 1.

504

VIII B The set E reduces to the integers

For our purpose, it thus suffices to make the calculations for the limiting case A = 1. Here, n/

rz

cosnxdx = fox

TCn

rz/2

so

7r(1+33+53+...

whence, if A >, 1 is sufficiently close to 1, n

°°

1

1

(vx(z,A))2dxdy < 411+33+53+...+E .

0

0

Again 1,

0<x<2,

1) =

v,,(x + i0,

-1, 2<x
so by symmetry, for y > 0,

Jfo 1

Hence,

[(vy(z, l))+]2dxdy =

fox

1

> 0,

0
(V (Z' 1))2dxdy

fo'O

00

a J0

"0

J 0n

(v v(z, 1))2 dx dy = g Y n(Ar,(1))2

2(1+

1

1

+ ... J

33 + 53

Therefore, by the above observation, n

J[(v,,(z,A))4- ]2dxdy < 21

for). 1 close enough to 1. We are left with the integral f 0 $o I v, (z,1) jdx dy. This, by Schwarz'

8 An auxiliary harmonic function

505

inequality, is n

4

S fOQ( n fo (vy(z, A))2 dx I dy

=

n2(An(A))2e-2ny

o!

0

dy

e-y/2(-Zn 2(t Anl/,1))2e-c2n-1)y

Jo

co

(f

n"

e-ydY'f

0co

dy /lf

2

n2(An(A))2e(2n-1)Ydy)

2

0

(L22* ' 2n - 1 (A.(A))2

J2 J(n(An(2))2).

We have already seen that the sum inside the radical in the last of these terms tends to a definite (finite) limit as A -+ 1. So

JJvy(zA)IdxdY 0

is certainly bounded for A > I near 1. The lemma is proved. Referring to the remarks made just before the lemma and to the boxed numerical estimate immediately following its statement, we obtain, regarding our original functions Vk, the following Corollary. Given A >, I there is, for each k, a function Vk(z) (depending on A),

harmonic in Sk = {z: 0 < 91z < (1 + A)1, and 3z > 0}, with 8Vk(z)/8x = 0 on the vertical sides of Sk and, on the latter's base, 8Vk/8y taking the boundary values A and - 1 along (0, A k) and (A., (1 + A)Ak) respectively.

If A >, 1 is close enough to 1, we have.

it f

sk

a dx dy < 0.44(1 + A)2Ak, M), nffJ()+]2dxdy

<_ 0.22(1+A)2A2 Y and 3Vk

ll.. ay

dx dy

all + A)200

a being a certain numerical constant.

506

VIII B The set E reduces to the integers

9.

Lower estimate f o r 101 o l o g ! - x2/t2I d1u(t)(dx/x2)

We return to the termwise comparison of

1

Y-

k,odk

dk

1 1 -3p kaodk

u(x) dx and

2

ck

('Yk

J

+

C ck

rdk \ i u(x) dx,

J

ak

which, as we saw in the first half of the preceding article, leads to the task of estimating f9k

u(x) dx - A

u(x) dx Ck

k

and f 6k

u(x) dx - A

dk

u(x) dx

f bk

Jk

from below. The notation of the previous two articles is maintained here. Following the idea of the last article, we use the harmonic function Vk(z) described there to express the first of the above differences as a line integral

J ask , - u(ck

+ z) a k(z) dx + u(ck + z) a XZ) dy

around the vertical half-strip Sk whose base is the segment [0, (1 + A)Ak] = [0, (1 + A)(yk - ck)] of the real axis. By use of Green's theorem, this line integral is converted to a ux(Ck + Z)

k(Z)

+

+ Z) a Vk(Z)ldx uy(Ck

dy,

Y

thanks to the harmonicity of Vk(z) in Sk. The justification of the present application of Green's theorem proceeds as follows. We have /' (1 +A)(Yk-Ck)

u(ck + x)(Vk)Y(x + i0) dx 0

(1 +A)(Yk-Ck)

= lim

u(ck + x + ih)(Vk)y(x + ih) dx,

k-.0 o

because u(z) is continuous up to the real axis, and, as one verifies by referring to the computations with v and vY near the end of the previous article, r(1 +.Z)(Yk-Ck)

[(Vk)Y(x + ih) - (Vk)Y(x + 10) 2 dx -+ 0 0

for h-+0.

9 Lower estimate for fn fo log I 1 - (x2/t2)Idu(t)dx/x2

507

However, for h > 0 and 0 < x < (1 + A)(yk - ck),

J

h

a

- u(ck + z)

aVk(Z) ay

aY

k

dy = - u(ck + x + ih)(Vk)Y(x + ih),

since const.lol Ilzl,

IU(Ck+z)I <

ZESk,

z

by an estimate used in proving the first lemma of article 5, and Vk(z), together with its partial derivatives, tends (exponentially) to zero as z - oo in Sk (see the calculations

at end of the previous article). Again, since aVk(z)/ax = 0 on the vertical sides of Sk, (I+)XYk-Ck) u(Ck

ax

a

+

z)

av Z ax

)

dx = 0,

y > 0.

By integrating y in this formula from h to 0o and x in the previous one from 0 to (1 +A)(yk - CO, and then adding the results, we express 0 l + A)(k - Ck)

U(Ck + x + ih)(Vk),(x + ih) dx Jo

as the sum of two iterated integrals. For h > 0, both of the latter are absolutely convergent, and the order of integration in one of them may be reversed. Doing this and remembering that V' V, = 0 in S, we see that the sum in question boils down to 0+.Z)(Yk-Ck)

('a0

( ux(ck + z) - + U,(ck + z) a Jk

OX

Jo

yz) dx dy. O

)

Making h-+0 in this expression finally gives us the corresponding double integral over Sk (whose absolute convergence readily follows from the first lemma in article 5

and the work at the end of the previous one by Schwarz' inequality). This, together with our initial observation, shows that the double integral over Sk

is equal to r() + A)(Yk -Ck)

U(Ck + x)(Vk),(x + iO) dx, 0

a quantity clearly identical with the above line integral around aSk.* In this way, we see that our use of Green's theorem is legitimate.

The line integral is, as we recall (and as we see by glancing at the preceding expression), the same as f "9k

u(x) dx - A, Yk

Yk

u(x) dx.

J Ck

* and actually coinciding with the original expression on p. 499 (the second one displayed there) from which the line integral was elaborated

508

VIII B The set E reduces to the integers

That difference is therefore equal to ux(ck + z) a

a

k (Z) + uy(ck + z) k(Z))dx dy.

ax

ay

What we want is a lower bound for the difference, and that means we have to find one for this double integral. Our intention is to express such a lower bound as a certain portion of E(d(v(t)/t), d(v(t)/t)), the hope being that when all these portions are added (and also all the ones corresponding to the differences ak

u(x) dx - .1 J 9k

dk

u(x) dx ),

J bk

we will end with a multiple of E(d(v(t)/t), d(v(t)/t)) that is not too large. In view, then, of the first lemma of article 5, we are interested in getting a lower

bound in terms of f f

n

[(ux(ck + Z))2 + (UY(ck + z))2] dx dy.

$

The present situation allows for very little leeway, and we have to be quite careful.

We start by writing ux(ck + z) a k(z) dx dy

ax

$k

- hrff$k Ca axx')ZdxdyI X

V\

ftsk (ux(ck + z))2 dx dy).

According to the corollary at the end of the last article, the right side is in turn - (0.44)(1 + A) (Yk - ck) J

\

71

ff

(Ux (ck + Z))2

dx dy)

$k

provided that A = (1 + 3p)/(1 - 3p) is close enough to 1(recall that the Ok of the previous article equals Yk - ck).

For the estimation of uy(ck + z)

JJSk

a yk(Z)

ay

dx dy,

9 Lower estimate f o r f 0 f p log 11- (x2/t2)Idµ(t)dx/x2

509

we split up Sk into two pieces, aYk(z) OV

Sk = I zesk:

>0 }

and

Sk = S,-S' We have

(,V

aYZ))2 dx dY/

-V \

J J sk

x

JC J J

sk

(uy(ck

+ z))2 dx dy),

which, by the corollary of the preceding article, is - (0.22)(1 + 2)(Yk - ck) JC 1

ff

(uy(ck +

z))2 dx dy)

Sk

for 2 close enough to 1. In this last expression, the integral involving uy may,

if we wish, be replaced by one over Sk, yielding a worse result. We are left with

fk

uy(ck

+ z) OVk(z) dx dy, 8y

in which 8Vk(z)/ay < 0. To handle this integral, we recall that

u(z) =

log

I

0

z+t z-t

which makes u y(z) fO'O

(x + t)2 + y 2

(x - t)2 + y 2

]d(v(t)), t

with the quantity in brackets obviously negative for x, y and t > 0. Since v(t)lt < 2n by our construction of the intervals Jk, we have d

w(t)) t

-

tdv(t)

v(t) dt t2

- 2g

dt t

,

510

VIII B The set E reduces to the integers

and therefore, for x and y > 0, y

°°

U(z) S2 y

(x - t)2 + y2

o

= 2r1 lim .5-0

21l7t

z y (x - t)+ f- . t +S x z

lim a-0 x2

t

dt

- (x + t)2y + y2

z

+(y+6)z =

27ur

z

dt t

dt x

xz +y z

(We have simply used the Poisson representation for the function `.1(1/(z + 0)), harmonic in .3z > 0.) Thus, 27rrl,

ZESk,

Uy(Ck + Z) Ck

whence

Jic

+ z) u''(ck

0Vk(z)

ay

dx dy >,

-

2

OVk(Z)

ffk-

Ck

ay

dx dy.

For 2 close to 1, the right side is

i - 27roul (1 +')2(Yk - Ck)2 Ck

by the corollary from the previous article, a being a numerical constant. Combining the three estimates just obtained, we find with the help of Schwarz' inequality that

JJSk

C

(Ck +

Z)ax+

uy(ck +

z)a ayZ))dx dy (ux(Ck + z))2 dxdy/

1 J - (0.44)-I(l +A) (Yk - ck)J ('$k

('

- (0.22)1(1 + 2)(Yk - Ck) \ \

27Carl

(uy(ck + z))2 dx dy)

JJ $k

(1 + 2)2(Yk - ck)2 Ck

- (0.66)1(1 +2)(Yk - CO

x

UJJSkuuxk + z))2 + (UY(Ck+ z))2)dx dy l

27Gatl

(1 + 2)2(Y, - Ck)2 Ck

9 Lower estimate f o r f01 log I 1 - (x2It2) I dµ(t)dx/x2 o

511

provided that 2 is close enough to 1. The double integral on the left is nothing but a complicated expression for the first of the two differences with

which we are concerned - that was, indeed, our reason for bringing the function Vk(z) into this work. Hence the relation just proved can be rewritten

f

9k

u(x) dx - A f

Yk

u(x) dx

ck

Yk

('

- (0.66)1(1+ 2)(Yk - Ck)

('

V \- J o J

9k ((ux(z))2

- gnarl (1 +

2)2(Yk

+ (u(z))2) dx dy

l

- Ck)2

Ck

The difference f Sk u(x) dx - 2 f dk u(x) dx can also be estimated by the method of this and the preceding articles. One finds in exactly the same way

as above that fAk Jk

rdk

u(x) dx - 2

u(x) dx Jbk

- (0.66)(1 + 2)(dk - ak)

(1 I

\-

J

L ((u(z))2 + (u(z))2)dx

27raq

dy)

(1 + A)2(dk - 6k)2 9k

for A close enough to 1. The following diagram shows the regions over which the double integrals involved in this and the previous inequalities are taken:

Figure 150

512

VIII B The set E reduces to the integers

We now add the two relations just obtained. After dividing by dk and using Schwarz' inequality again together with the fact that Ck

1< Yk 1< gk < ak < dk < (1 + 2n)Ck,

we get, recalling that A = (1 + 3p)/(l - 3p), 1

dk

" J Ck

_ (0.66)1 23p(1+2r1)

_

z

dkak)2)

Ykck) +(dk

C\Yk ak

x

8(1(1

3

(ux+uy)dxdY)

(-

Jck

p)2)nL\

Yk`k)2+(dkdkak)2

for ).. close enough to 1, in other words, for p > 0 close enough to zero.

We have now carried out the program explained in the first half of article 8 and at the beginning of the present one. Summing the preceding relation over k and using Schwarz' inequality once more, we obtain, for small p > 0, dk

ckYk

u( x) dx k>OdkJc,

-3pk>Odk C

3P

X J\ n k Yk - ck

89tCC(1 + 2Yj)

D

Jo

2

(dk _

+

dk

2

Sk) )

k(ux+uy)dxdY

J

)

+ dk - k)2).

q Y_

k->O((

u(x) dx

/ _ 2 ((Yk Yk ck)

- (0.66)1(1 + 2r1)1 2

(1 -3 p)2

+

2

Yk

dk

To the right-hand expression we apply the theorem of article 6 together with its remark and the first lemma of article 5. In this way, we find that the right side is

(1+2r1)E\d(vtt)) - K'r1E1 d(vtt))

d(vtt)))

d(vtt)))

for small enough positive values of r1 and p, K and K' being certain

513 9 Lower estimate for 1. 1 log I 1 - (x2/t2)Idp(t)dx/x2 o numerical constants independent of p and of the configuration of the Jk.

According to the theorem of article 7, the left-hand difference in the above relation is within 1 - 3p Cn+E(d(vtt)),

d(vtt)))

of

dx

2

dx

('r"

u(x)

x

sJ

1- 3P J u(x)

x

for small enough n > 0, where C is a numerical constant independent of p or the configuration of the Jk. So, since Ju(x) dx

-

E(d(vtt)),d(vtt)

/

(see remark to the theorem of article 7), what we have boils down, for small

enough p and n > 0, to ('

f

JJ

u(x)

dx x

2

1-

1- 3p (

0.66

(0.80)

An

B

n)

x E(d(vtt)),d(vtt))) with numerical constants A and B independent of p and the configuration of

the J. Here, 0.66

J( 0.80

0.9083-,

so, the coefficient on the right is 2

1- 3p

(0.0917 - An -

Not much at all, but still enough!

We have finally arrived at the point where a value for the parameter n must be chosen. This quantity, independent of p, was introduced during the third stage of the long construction in article 2, where it was necessary to take 0 < n < 3. Aside from that requirement, we were free to assign any value we liked to it. Let us now choose, once and for all, a numerical value > 0 for n, small enough to ensure that all the estimates of articles 6, 7 and the present one hold good, and that besides

0.0917 - An - BVn > 1/20.

514

VIII B The set E reduces to the integers

That value is henceforth fixed. This matter having been settled, the relation finally obtained above reduces to

Ju(x) dx 1 x

J

I 10(1-I 3p) E\d\v(t)/,d(v(t) t

To get a lower bound on the right-hand member, we use again the inequality

E(d(vtt)), d\vtt)) z

2

(0.80-Kq) E \\yk-ckl/ k,0(( A + \dkdk- k/ (valid for our fixed value of n!), furnished by the theorem of article 6. In article 2, the intervals Jk were constructed so as to make d0 - co = I JO I >, ryd0 (see property (v) in the description near the end of that

article), and in the construction of the function u(t) we had

do-60 _ 1-3p do - c0

2

(property (iii) of the specification near the end of article 3). Therefore do - 60 > 1- 3p 2

do

which, substituted into the previous inequality, yields

E\d\t) vt/,d(vtt))

>,

(0.80-K?)I

\2 1

2 3p

2.

We substitute this into the relation written above, and get

1.

U(X)

dx x

(1- 3p)c

with a certain purely numerical constant c. (We see that it is finally just the ratio IJ01/d0 associated with the first of the intervals Jk that enters into these last calculations. If only we had been able to avoid consideration

of the other A. in the above work!) In terms of the function µ(t) = (p/(1- 3p))v(t) constructed in article 3, we have, as at the beginning

9 Lower estimate for fn fo log I 1 - (x2/t2)I dy(t)dx/x2

515

of article 7, log

1- x2 i2 dy(t)d2 = I

p3 P

fj

u(x)dx .

By the preceding boxed formula and the work of article 3 we therefore have the

Theorem. If p > 0 is small enough and if, for our original polynomial P(x), the zero counting function n(t) satisfies

sup

n(t)

p

t

1 - 3p'

then, for the function u(t) constructed in article 3, we have

i x2

log

t

dp(t)

d

% PC,

c being a numerical constant independent of P(x). Here,

S2 = (0, oo) - U Jk, k>0

where the Jk are the intervals constructed in article 2. In this way the task described at the very end of article 3 has been carried

out, and the main work of the present § completed. Remark. One reason why the present article's estimations have had to be so delicate is the smallness of the lower bound on

E\d\vtt)/,d\vtt)// obtained in article 6. If we could be sure that this quantity was considerably

larger, a much simpler procedure could be used to get from fu(x)(dx/x) to f ru(x)(dx/x); the one of problem 23 (article 8) for instance. It is possible that E(d(v(t)/t), d(v(t)/t)) is quite a bit larger than the lower bound we have found for it. One can write

E\d\vtt)/,d\vtt)//

=

ff4lo

1

g

1- x2/t2

dt dx.

If the intervals Jk are very far apart from each other (so that the cross terms

i 1

Jjkflf

log

1

1 - x2/t2

dt dx,

VA 1,

516

VIII B The set E reduces to the integers

are all very small), the right-hand integral behaves like a constant multiple of (P1)2 2 Y_ k>0

(

dk

)

dk

log

IJkI

When h > 0 is taken to be small, this, on account of the inequality IJkI /dk <, 2r1, is much larger than the bound furnished by the theorem of article 6, which is essentially a fixed constant multiple of

o y

IJkI

k

2

dk

I have not been able to verify that the first of the above sums can be used to give a lower bound for E(d(v(t)/t), d(v(t)/t)) when the Jk are not far apart. That, however, is perhaps still worth trying. 10.

Return to polynomials

Let us now combine the theorem from the end of article 3 with the one finally arrived at above. We obtain, without further ado, the Theorem. If p > 0 is sufficiently small and P(x) is any polynomial of the form

i

1- x2 xk

with the xk > 0, the condition n(t)

su >o t

>

p

1-3p

for n(t) = number of xk (counting multiplicities) in [0, t] implies that log+

mI 2P(m) I > / cp 5

Here, c > 0 is a numerical constant independent of p and of P(x). Corollary. Let Q(z) be any even polynomial (with, in general, complex zeros) such that Q(0) = 1. There is an absolute constant k, independent of Q, such

that, for all z, log I Q(z) I IzI

k 00 log+ I Q(m) I 1

m2

provided that the sum on the right is less than some number y > 0, also independent of Q.

10 Return to polynomials

517

Proof. We can write z

1-yk

Q(Z) k

S

Put Xk = I bk I and then let

7

P(z) =

z2

1-

11

xk

we have I P(x) 15 I Q(x) I on R, so

-

log+

I P(m) I

m

1

log + I Q(m) I

S

m

1

To P(x) we apply the theorem, which clearly implies that sup

10 °° log+ IP(m)I -Y c m2

n(t)

t>o t

1

for n(t), the number of xk in [0, t], whenever the sum on the right is small

enough. For zeC, z

log i Q(Z) 15

log 1 +

I

I

z

I dn(t),

and partial integration converts the last expression to IZI2+t2dt <

0 t f'0nt)

iIzIsuppntt). t>O

In view of our initial relation, we therefore have 10?[ - log+ I Q(m) I

log I Q(z) I

C

Iz1

1

m2

whenever the right-hand sum is small enough. Done. Remark 1. These results hold for objects more general than polynomials. Instead of IQ(z)I, we can consider any finite product of the form

2r z2

0

.1k

bk

k

where the exponents 2k are all > some fixed a > 0. Taking IP(x)I as

fl

2

X2JAk

I

k

518

VIII B The set E reduces to the integers

with xk = IRkl, and writing

n(t) _

Y,

Ak

Xke[O,tj

(so that each `zero' Xk is counted with `multiplicity' 2k), we easily convince

ourselves that the arguments and constructions of articles 1 and 2 go through for these functions IP(x)I and n(t) without essential change. What was important there is the property, valid here, that n(t) increase by at least some fixed amount a > 0 at each of its jumps, crucial use having been made

of this during the second and third stages of the construction in article 2. The work of articles 3-8 can thereafter be taken over as is, and we end with analogues of the above results for our present functions IP(x)I and IQ(z)I.

Thus, in the case of polynomials P(z), it is not so much the singlevaluedness of the analytic function with modulus IP(z)I as the quantization of the point masses associated with the subharmonic function log I P(z) I that is essential in the preceding development.

Remark 2. The specific arithmetic character of Z plays no role in the above work. Analogous results hold if we replace the sums log+IP(m)I

log, IQ(m)I

m2

m2

by others of the form log+IP(2)I 12

log+ IQ(2)I A2

'

'

being any fixed set of points in (0, oo) having at least one element in each interval of length >, h with It > 0 and fixed. This generalization requires some rather self-evident modification of the work in article 1. The reasoning in articles 2-8 then applies with hardly any change. A

Problem 24 Consider entire functions F(z) of very small exponential type a having the special form z2

F(z) = fj 1 - Z k

xk

where the Xk are > 0, and such that log+IF2 J

1 + x2

dx < oo.

10 Return to polynomials

519

Investigate the possibility of adapting the development of this § to such functions F(z) (instead of polynomials P(z)). Here, if the small numbers 21l and p are both several times larger than a, the constructions of article 2 can be made to work (by problem 1(a), Chapter I!), yielding an infinite number of intervals Jk. The statement of the second lemma from article 4 has to be modified. I have not worked through this problem.

We now come to the principal result of this whole §, an extension of the above corollary to general polynomials. To establish it, we need a simple

Lemma. Let a > 0 be given. There is a number M,, depending on a such that, for any real valued function f on 1 satisfying log+lf(n)I

1+n2

a,

we have rig log i

l + n2(f (n)Mf

n))2

5 6a

and 1

2

log I+ (f (n) .fz-n))2) J

6a

Proof. When q >, 0, the function log(1 + q) - log+q assumes its maximum for q = 1. Hence

log (1+q) < log2+log+q,

q>0.

Also,

log+ (qq') < log+ q + log+ q',

q, q'

>, 0.

Therefore, if M >, 1, log ( 1 +

n2(f (n) +f (- n))2I M2

5 log 2 + 21og+ n + 2log+(If (n)I + If (- n)I)

31og2+2logn+2max(log+If(n)I, log+lf(-n)I) forn>, 1. Given a > 0, choose (and then fix) an N sufficiently large to make

31og2+21ogn n>N

n2

< a

520

VIII B The set E reduces to the integers

Then, if f is any real valued function with log+If(n)I l + n2 we will surely have

a,

00

log(l+n2(f(n)+f(-n))21

Y 21 n>Nn

l

m2

< 5a

by the previous relation, as long as M > 1. Similarly,

I+ (f(n)-f-n))2) < 5a m2

121og1

Y-

n>Nn

for such f, if M > 1. Our condition on f certainly implies that

log' If(n)I < all +n2), so

If(n)I +If(-n)I <

2e(1+N2)«

for 1 5 n 5 N. Choosing M« >, 1 sufficiently large so as N

1

n(i+4n2e212) 2 M«

1

/

to have


will thus ensure that N 1

1n2 logl l +n2(f(n)+f(-n))2) M

< a

and

Yri21og1

I+ (f (n) f2-n))2)

< a.

Adding each of these relations to the corresponding one obtained above, we have the lemma.

Theorem. There are numerical constants a° > 0 and k such that, for any polynomial p(z) with Ip(n)I - log+1+n2

-00

= a <, a , °

we have, for all z, I p(z) I

< K«e3k«Izl,

where K« is a constant depending only on a (and not on p).

10 Return to polynomials

521

Proof. Given a polynomial p, we may as well assume to begin with that

p(x) is real for real x - otherwise we just work separately with the polynomials (p(z) + p(2))/2 and (p(z) - p(z))/2i which both have that property. Considering, then, p to be real on 18 and assuming that it satisfies the

condition in the hypothesis, we take the number Ma furnished by the lemma and form each of the polynomials

Q1(z) = 1 + z2(p(z) + p( - z))2 Ma

Qz(z) = 1 + (p(z) - p(- z))2 Ma

The polynomials Q1 and Q2 are both even, and

Q1(0) = Q2(0) = 1. By the lemma, 00 1

n log' IQ1(n)I ,

6a

and

Y n2 log+ I Q2(n)I < 6a, since (here) Q1(x) >, 1 and Q2(x) 3 1 on R. If a > 0 is small enough, these inequalities imply by the above corollary

that (logIQl(z)I)/Izl and (logIQ2(z)I)/Izl are both <6ka, k being a certain numerical constant. Therefore 1+ z2(p(z)

+ p( - z))2

e 6kalzl

Ma

and 1

p(- z))2 + (p(z) -M2

eek.i=l.

From these relations we get I z2(p(z) + p( - z))2I 5 Ma (l +

ebk«i2i)

2Mae6kalzl,

whence

I p(z) + p(- z) 15

,/2M,,eskaizi

for I z I > 1,

VIII B The set E reduces to the integers

522

and similarly I p(z) - p(- z) I <

1/2M,e3kajzI

Hence I p(z) I

j2Mae3kalzi

<, 2.

for I z 1 , 1, and from this, by the principle of maximum, for IzI < 1.

Ip(z)I < 2,J2Mae3ka The theorem therefore holds with

Ka = 2../2Mae3ka

Q.E.D.

Corollary. If a > 0 is small enough, the polynomials p(z) satisfying

-

log+ I p(n)I

1+n2


form a normal family in the complex plane, and the limit of any convergent sequence of such polynomials is an entire function of exponential type < 3ka, k being an absolute constant. It is thus somewhat as if harmonic measure were available for the domain C - Z, even though that is not the case.

it.

Weighted polynomial approximation on 7L

Given a weight W(n) > 1 defined on Z, we consider the Banach space (ew(Z) of functions cp(n) defined on 77 for which ('(n) W(n)

0

as

n

± oo,

and write II w 11 Wz = sup

l w(n) I

nEIC W(n)

for such cp. (This is the notation of §A.3.) Provided that nk

W (n)

-E0

as

n -++oo

for each k = 0, 1,2,3,..., we can form the 11 11 w,, closure, 'w(0, Z), of the set

11 Weighted polynomial approximation on the integers

523

of polynomials in n, in 16w(Z). The Bernstein approximation problem for Z

requires us to find necessary and sufficient conditions on weights W(n) having the property just stated in order that ' (O,ZZ) and Ww(7L) be the same.

The preceding work enables us to give a complete solution in terms of the Akhiezer function W*(n) = sup { I p(n) I: p a polynomial and p w, < 1 }

introduced in §B.1 of Chapter VI. Theorem. Let W(n), defined and ? 1 on 7L, tend to oo faster than any power of n as n -+ ± oo. Then Ww(0, 7L) 16w(7L) if and only if log W*(n) oo.

+n 2

Proof. Let us get the easier if part out of the way first - this is not really new, and depends only on the work of Chapter VI, §B. 1.

As in §A.3, we take W(x) to be specified on all of O by putting W(x) = oo for xO71, and define W,k(z) for all zeC using the formula W*(z) = sup { I p(z) I : p a polynomial and p

,<

1} .

Then Ww(ZL) can be identified in obvious fashion with the space Ww(R) constructed from the (discontinuous) weight W(x), and 'w(0, 7Z) identified with 'w(0), the closure of the set of polynomials in 'w(R). Proper inclusion of Ww(0,7L) in'w(Z) is thus the same as that of 16w(O) in Ww(R), and we

can apply the if part of Akhiezer's theorem from §B.1 of Chapter III (whose validity does not depend on the continuity of W(x) !) to conclude that J

log W,(t)

t+ 2

dt < co

1

when that proper inclusion holds. If p is any polynomial with II p II wa < 1, the hall of mirrors argument at the

beginning of the proof of Akhiezer's theorem's only if part shows that

for xc-R. Taking the supremum over such polynomials p gives us 1

l

ogW(n)

(n2 log)2+4dt,

F

ne71.

524

VIII B The set E reduces to the integers

Therefore, since log W,,(t) > 0 (1 being a polynomial!), we have log W,k(n)

< 1 1'

1+n2

nJ

,,

1

2 log W*(t)

(n-t)2+1

dt.

The inner sum over n may easily be compared with an integral, and we find

in this way that the last expression is const.

°°

log W*(t) 1 + t2

dt.

This, however, is finite when 'H,(0, Z) 0 Vw(Z), as we have just seen. The if

part of our theorem is proved. For the only if part, we assume that - log W*(n) < 00, 1 +n2

_.

and show that the function (po(n)

=

1,

n=0

to, n 00,

cannot belong to ' (0,7L). We do this using the corollary to the first theorem of the preceding article. It is not necessary to resort to the second theorem given there. Suppose, then, that we have a sequence of polynomials p,(z) with

II4Po-PllIw2 - 0. This implies in particular that PI(0)

(Po(0) = 1,

so there is no loss of generality in assuming that p,(0) = 1 for each 1, which we do. The polynomials Q1(z) = z (P,(z) + Pi( - z))

then satisfy the hypothesis of the corollary in question. We evidently have II P, II wa < C for some C, so, by definition of W,1, I p,(n)I < CW,k(n) for neZ and therefore

IQ,(n)I <

4C(W,k(n)+W,k(-n)),

neZ.

Also, p,(n) - cpo(n) = 0 for each non-zero neZ, so, given any N, we will have I Q,(n) I< 1

for 0< I n I< N

when 1 is sufficiently large.

VIII C Harmonic estimation in slit regions

525

Taking any a > 0, we choose and fix an N large enough to make Y nz log+(-1QW,k(n) + W*

n))) < a,

N

this being possible in view of our assumption on W. By the preceding two relations we will then have Y-nz 1og+ I QI(n)I =

n2

log+ I QI(n)i < a

for sufficiently large values of 1. If a > 0 is sufficiently small, the last condition implies that I Q1(z)I <

e"'1Z1

by the corollary to the first theorem of the preceding article, with k an absolute constant. This must therefore hold for all sufficiently large values of 1.

A subsequence of the polynomials Q1(z) therefore converges u.c.c. to a certain entire function F(z) of exponential type < ka. We evidently have F(0) = 1 (so F # 0 !), while F(n) = 0 for each non-zero neZ. However, by problem 1(a) in Chapter I (!), such an entire function F cannot exist, if a > 0 is chosen sufficiently small to begin with. We have thus reached a contradiction, showing that cpo cannot belong to 'W(0,7L). The latter space is thus properly contained in 'W(ZL), and the only if part of our theorem is proved. We are done.

Harmonic estimation in slit regions We return to domains -9 for which the Dirichlet problem is solvable, having boundaries formed by removing certain finite open intervals from R. Our interest in the present § is to see whether, C.

for -9 (the reader from the existence of a Phragmen-Lindelof function should perhaps look at §A.2 again before continuing), one can deduce any estimates or the harmonic measure for 2 . We would like in fact to be able to compare harmonic measure for -9 with YQ,(z). The reason for this desire is the

following. Given A > 0 and M(t) >, 0 on 0-9, suppose that we have a function v(z), subharmonic in -9 and continuous up to 0-9, with

v(z) < const. - A 13z1, and

v(t) 5 M(t),

tea-9.

ze

,

526

VIII C Harmonic estimation in slit regions

Then, by harmonic estimation M(t)dw1,(t, z) - AY1,(z),

v(z) <,

ze2

,

fa s

where (as usual) col( , z) denotes (two-sided) harmonic measure for -9 (see

§A.1). It would be very good if, in this relation, we had some way of comparing the first term on the right with the second. As we shall see below, such comparison is indeed possible. In order to avoid fastidious justification arguments like the one occurring in the proof of the second theorem from §A.2, we will assume throughout that 8-9 consists of R minus a finite number of (bounded) open intervals. The results obtained for this situation can usually be extended by means of a simple limiting procedure to cover various more general cases that may arise in practice. The domains -9 considered here thus look like this:

Figure 151

As in §A, we shall frequently denote 8-9 by E. E is a closed subset of R which, in this §, will contain all real x of sufficiently large absolute value. 1.

Some relations between Green's function and harmonic measure for our domains -9

During the present §, we will usually denote the Green's function for one of the domains .9 by G9(z, w), instead of just writing G(z, w) as in §A.2ff. We similarly write Y1,(z) instead of Y(z) for _9's Phragmen-Lindelof function. Our domains .9 have Phragmen-Lindelof functions. Indeed, for fixed z e- and real t, G9,(z, t) = G,(t, z) vanishes for t outside the bounded set OB - E. (We are using symmetry of the Green's function, established at the

I Relations between Green's function and harmonic measure 527

end of §A.2.) If we take zO R, G,(t, z) is also a continuous function of teR. The integral G,(z, t) dt

is then certainly finite, and the existence of the function Y, hence assured by the second theorem of §A.2. According to that same theorem, f 00

YY(z) = 13z I + i 7E

G,(z, t) dt. -00

This formula suggests that we first establish some relations between G,(z, t) and co1,( , z) before trying to find out whether the latter is in any way governed by Y,,,(z).

We prove three such relations here. The first of them is very well known.

Theorem. For wed,

G,(z, w) = log Iz

I wl

+ fE

log I t - w I dco1,(t, z).

Proof. The right side of the asserted formula is identical with log

I

Iz-w1

+ Jf

log I t- w l dco1,(t, z),

al

and, for bounded domains _q, this expression clearly coincides with G,(z, w) -just fix we-9 and check boundary values for z on 0-9 ! (This argument, and the formula, are due to George Green himself, by the way.) In our situation, however, -9 is not bounded, and the result is not true,

in general, for unbounded domains. (Not even for those with `nice' boundaries; example:

= {IzI> 1}u{00}. ) What is needed then in order for it to hold is the presence of `enough' ag near oo. That is what we must verify in the present case. Fixing we-9, we proceed to find upper and lower bounds on the integral

f, log I t - w l dw,(t, z). In order to get an upper bound, we take a function h(z), positive and harmonic in -9 and continuous up to 0-9, such that

h(z) = log+IzI+O(1).

528

VIII C Harmonic estimation in slit regions

In the case where E includes the interval [ - 1, 1] (at which we can always arrive by translation), one may put

h(z) = log lz + "/(z2 - 1)I using, outside [ - 1, 1], the determination of ,/ that is positive for z = x > 1. For large A > 0, let us write

h,(z) = min (h(z), A).

The function hA(t) is then bounded and continuous on E, so, by the elementary properties of harmonic measure (Chapter VII, §B.1), the function of z equal to

1.

hA(t)duo, (t, z)

is harmonic and bounded above in -9, and takes the boundary value hA(z)

for z on 8-q. The difference f EhA(t)dco,(t, z) - h(z) is thus bounded above in _q and < 0 on 89. Therefore, by the extended principle of maximum (Chapter III, §C), it is 5 0 in -9, and we have h(z),

ze-9.

SE

For A' >,A, hA.(t) >,hA(t). Hence, by the preceding relation and Lebesque's monotone convergence theorem,

1.

h(t) dw1,(t, z) < h(z),

ze-9;

that is, 1.

log+Itldco9(t,z) S log+Izl+O(1)

for ze2i. When wed is fixed, we thus have the upper bound 1.

loglt-wldow,(t,z) < log+Izl+O(1)

for z ranging over -9. We can get some additional information with the help of the function h(z). Indeed, for each A,

h(t) dw (t, z) < h(z)

h A(t) dco2(t, z) 5

JE

fE

when ze-9. As we remarked above, the left-hand expression tends to hA(xo)

1 Relations between Green's function and harmonic measure 529

whenever z --> xo e 8-9; at the same time, the right-hand member evidently tends to h(xo). Taking A > h(xo), we see that

1.

h(t)dw.,(t, z) --> h(xo) e 8-9. On the other hand, for fixed we-9,

for

log l t - w l - h(t) is continuous and bounded on 8-9. Therefore

1.

(log I t - w I - h(t)) dw9(t, z) ---i log I xo - w I - h(xo)

when z --b xo a 8.9, so, on account of the previous relation, we have

1.

log I t- w I dwq(t, z)

log I xo - w I

for z -> xo e 8-9. To get a lower bound on the left-hand integral, let us, wlog, assume that 91z > 0, and take an R > 0 sufficiently large to have (- oo, - R] u [R, oo) c E. Since -9 2 {,Zz > 0}, we have, for I t I > R, dw,(t, z) >' 1

3z

dt

n Iz - tlz

by the principle of extension of domain (Chapter VII, §B.1), the right side being just the differential of harmonic measure for the upper half plane. Hence,

1.

loglt+ildw,(t,z) >

log It+iIdw,,(t,z) {ItI3R}

-If

3zloglt+il dt IZ- tl z

dt - 0(1). If'-. ,3zloglt+il Iz-tI2

7r

The last integral on the right has, however, the value log I z + i I, as an elementary computation shows (contour integration). Thus, 1.

loglt+ildw.,(t,z) > loglz+il-0(1)

530

VIII C Harmonic estimation in slit regions

for 3z > 0, so, for fixed wee,

loglt-wldco,(t,z) > log+lzl-O(1),

ze-9.

JE

Taking any wee, we see by the above that the function of z equal to log

+ J log I t- w l dco f(t, z)

1

Iz - wI

E

is harmonic in -9 save at w, differs in -9 by 0(1) from log (1/I z - w l) + log' I z I, and assumes the boundary value zero on 8.9. It is in particular bounded above and below outside of a neighborhood of w (point where it becomes infinite), and hence >, 0 in -9 by the extended maximum

principle. The expression just written thus has all the properties required of a Green's function for -9, and must coincide with G1,(z, w). We are done. It will be convenient during the remainder of this § to take duw-, (t,z)as defined on all of R, simply putting it equal to zero outside of E. This enables us to simplify our notation by writing oo.,(S, z) for w,(S r) E, z) when S c R.

Lemma. Let OeY, and write w.9(x) =

J(o ([x, c), 0), x>0,

w,((-00,x], 0), x<0

(note that coa(x) need not be continuous at 0). Then, for 3z 0 0, x

0) _

- (x

t) z

wi(t) sgn t dt. +yz y

Proof. By the preceding theorem and symmetry of the Green's function (proved at the end of §A.2), we have

G.,(z, 0) = G.,(O, z) = log ± + I log I t - z I do) ,(t, 0). Thanks to our convention, we can rewrite the right-hand integral as

(J°

)log lt-zldwq(t,0).

Let us accept for the moment the inequality const.

Itl+l' postponing its verification to the end of this proof. Then partial integration

tx

I Relations between Green's function and harmonic measure 531 yields

loglt-zldw,(t,0) = co.,(0+)loglzl + fo

Jo It

and

ZIZw,(t)dt,

floglt-zldco,(t,0) = w,(0-)loglzI - J_'t_z'2tt. 00

w,(0 +) + w,(0 -) = w_,((- oo, oo), 0) = w,(E, 0) = 1, so, adding, we get

G.,(z, 0) = log

+ - log I t - z I dw,(t, 0) II

Z

II

= loglZl +logIZI+f

It- 12w,(t)sgntdt

x-t2w,(t)sgntdt,

fo.Iz-tI as claimed. We still have to check the above inequality for w,(t). To do this, pick an R > 0 large enough to have

(- oo, - R] u [R, oo) g E, and take a domain 9 equal to the complement of

(- oo, - R] u [R, oo) in C. Then -9 c 4ff, so, by the principle of extension of domain (Chapter VII, §B.1), wi(t) + t) < wB((- oo, - t] u [t, oo), 0) for t > R. The

quantity on the right can, however, be worked out explicitly by mapping

6 conformally onto the unit disk so as to take - R to - 1, 0 to 0 and R to 1. In this way, one finds it to be 5 CR/t (with a constant C independent of R), verifying the inequality in question. Details are left to the reader - he or she is referred to the proof of the first lemma from §A.1, where most of the

computation involved here has already been done. The integral figuring in the lemma just proved, viz.,

-

JT"Iz-tI2

t) sgn t dt

532

VIII C Harmonic estimation in slit regions

is like one used in the scholium of §H.1, Chapter III, to express a certain harmonic conjugate. It differs from the latter by its sign, by the absence of the constant 1/it in front, and because its integrand involves the factor (x - t)/I z - t I2 instead of the sum

x-t

t

Iz-tI2 + t2+1 In §H of Chapter III, the main purpose of the term t/(t2 + 1) was really to ensure convergence; here, since wi(t) is O(1/(ItI + 1)), we already have convergence without it, and our omission of the term t/(t2 + 1) amounts merely to the subtraction of a constant from the value of the integral. Since

harmonic conjugates are only determined to within additive constants anyway, we may just as well take

x-tt Iz wi(t) sgn t dt

1 1,00

n

_, Iz-

as the harmonic conjugate of 1

z

°°

,Iz-tl2w, (t)sgntdt

n

in {3z > 0}. This brings the investigation of the former integral's boundary behavior on the real axis very close to the study of the Hilbert transform already touched on in Chapter III, §§F.2 and H.1. In our present situation, we already know that, for real x 0 0, lim G,(x + iy, 0) = G1,(x, 0) Y-0

exists. The identity furnished by the lemma hence shows, independently of the general considerations in the articles just mentioned, that lim

0°

-c

x-t 2 w,(t) sgn t dt

Z -tI

exists (and equals - G,(x, 0) ) for real x 0. According to an observation in the scholium of §H.1, Chapter III, we can express the preceding limit as an integral, namely

w_,(x-T)sgn(x-T)-w,(x+r)sgn(x+T)dT. fo'O

T

That's because this expression converges absolutely for x 0 0, on account of the above inequality for w_,(t) and also of the

Lemma. Let 0e9. Then w_ (t) is Lip i for t > 0 and for t < 0.

I Relations between Green's function and harmonic measure 533 Proof. The statement amounts to the claim that

cw,(1,0) < const.,/III for any small interval I c E. To show this, take any interval JO E and consider small intervals I s JO. Letting be the region (L u { oo }) ' JO, the usual application of the principle of extension of domain gives us w0(I, 0) _< cor(I, 0),

with, in turn, w,(I, 0) 5 const. wr(1, oo) by Harnack's theorem. To simplify the estimate of the right side of the last inequality, we may take JO to

be [- 1, 1]; this just amounts to making a preliminary translation and change of scale - never mind here that 0e-9 ! Then one can map d onto the unit disk by the Joukowski transformation

z --+ which takes oo to 0, -1 to -1, and 1 to 1. In this way one easily finds that wd(I, oo) < const.../III, proving the lemma.

Remark. The square root is only necessary when I is near one of the endpoints of JO. For small intervals I near the middle of JO, co,(1, oo) acts like a multiple of I 11.

By the above two lemmas and related discussion, we have the formula (0,(X - t) sgn (x - t) - co,(x + T) sgn (x + t)

G.9(x, 0) = -

dT,

T fo valid for x :0 if 0 belongs to -9. It is customary to write the right-hand member in a different way. That expression is identical with

- lim

co.,(t) sgn t

dt. 6-0 JI(-XJ>_b x - t If a function f (t), having a possible singularity at ae R, is integrable over each set of the form { I t - a I > S}, 6 > 0, and if lim f (t) dt 8-o J It-al>6

exists, that limit is called a Cauchy principal value, and denoted by f-000 f (t) dt

or by

f (t) dt.

V.P.

J

-000

534

VIIIC Harmonic estimation in slit regions

It is important to realize that f (t) dt is frequently not an integral in the ordinary sense. In terms of this notation, the formula for G1(x, 0) just obtained can be expressed as in the following

Theorem. Let 0e-9. Then, for real x 0 0, (t) sgn t (0_,(t)

°°

x-t

_00

dt,

where co1,(t) is the function defined in the first of the above two lemmas.

This result will be used in article 3 below. Now, however, we wish to use it to solve for col(t) sgn t in terms of G1(x, 0), obtaining the relation Gi (x, 0)

cul(t) sgn t =

dx.

n2

By the inversion theorem for the L2 Hilbert transform, the latter formula is indeed a consequence of the boxed one above. Here, a direct proof is not very difficult, and we give one for the reader who does not know the inversion theorem. Lemma. f °°

. I GQ(x + iy, 0) - G1(x, 0) I dx --- 0 for y -4 0.

Proof. The result follows immediately from the representation 1

G1(x+iy, 0) _

y>0,

J

by elementary properties of the Poisson kernel, in the usual way. The representation itself is practically obvious; here is one derivation. From the first theorem of this article,

G®(t,0) = log ItI

+ JElogIs

-

0)

and 1

G.,(z,0) = loglZl+ ,log Is-zIdcu_(s,0). For .YJz > 0, we have the elementary formula 1

=

°° _w

7r

3zlogls-tllogls-z

dt,

seR.

1 Relations between Green's function and harmonic measure 535 Use this in the right side of the preceding relation (in both right-hand terms !), change the order of integration (which is easily justified here), and then refer to the formula for Gy,(t, 0) just written. One ends with the relation in question.

Lemma. Let 0e-9. Then Gu(x, 0) is Lip for x > 0 and for x < 0.

i

Proof. The open intervals of R - E belong to .9, where G,,,(z, 0) is harmonic (save

at 0), and hence W.. So G,(x,0) is certainly W1 (hence Lip 1) in the interior of each of those open segments (although not uniformly so!) for x outside any neighborhood of 0. Also, G,(x, 0) = 0 on each of the closed segments making up E; it is thus

surely Lip 1 on the interior of each of those. Our claim therefore boils down to the statement that

IG,,(x,0)-G,(a,0)1 < const.,/Ix-al near any of the endpoints a of any of the segments making up E. Since Ga(a, 0) = 0,

we have to show that

G.,,(x,0) < const.,/Ix-al for x e 118 - E near such an endpoint a. Assume, wlog, that a is a right endpoint of a component of E and that x > a. Pick b < a such that

[b, a] c E and denote the domain (C u { co }) - [b, a] by '. We have .9 c e, so GQ(x, 0) < G,(x, 0)

by the principle of extension of domain. Here, one may compute G,(x, 0) by

mapping f onto the unit disk conformally with the help of a Joukowski transformation. In this way one finds without much difficulty that

G,(x, 0) 5 const., j(x - a) for x > a, proving the lemma. (Cf. proof of the lemma immediately preceding the previous theorem.)

Theorem. Let Oe!2. Then, for x 96 0, co9(x) sgn x =

lZ

it

°°

f

00

G.9(t, 0)

x

-t

dt,

where w.,(x) is the function defined in the first lemma of this article.

Proof. By the first of the preceding lemmas, for G.9 (t + ih, 0) =

w tY(t - Y)2 + h2

(

and h > 0, ) sgn

536

VIII C Harmonic estimation in slit regions

Multiply both sides by

x-t (x-t)2+y2 and integrate the variable t. We get 00

x-t

(x-t)2 +y 2Gg(t+ih, O)dt

f

°°

x

tZ

t

_(x-t)z + y2

sgn i d dt.

+h 2

Suppose for the moment that absolute convergence of the double integral has been established. Then we can change the order of integration therein. We have, however, for y > 0,

(x-t) t-I; J_(x_t)2+y2(t_)2+h2dt

y+h

= -1r (x-x)2+(y+h)2'

as follows from the identity `°

1

1

_.x+iy-t

+ih -

tdt = 0,

verifiable by contour integration (h and y are > 0 here), and the semigroup

convolution property of the Poisson kernel. The previous relation thus becomes C0D

x-t

_

(x - t)2 + y2 = IT

G,(t + ih, 0) dt y+h

f-0. (x-Y)2+(y+ h)2

w

sgn di;.

Fixing y > 0 for the moment, make h -> 0. According to the third of the above lemmas, the last formula then becomes x

t

fo. (x-t)2+y2

G (t, 0) dt

it

Now make y --> 0, assuming that x 0 0. Since co

Y

+Y 2

sgn

is continuous at x, the

right side tends to it2w,(x) sgn x.

Also, by the fourth lemma, G_(t, 0) is Lip i at x. The left-hand integral

I Relations between Green's function and harmonic measure 537

therefore tends to the Cauchy principal value G12 (t,0)

f°

dt

x-t

(which exists !), according to an observation in §H.1 of Chapter III and the discussion preceding the last theorem above. We thus have 12

(0a(x) sgn x =

for x

G-9(t'

n

O)

dt

0, as asserted.

The legitimacy of the above reasoning required absolute convergence of

t-

x-t (x-t)2+y2

which we must now establish. Fixing y and h > 0 and x e R, we have

x-t

t-

const.

(Iti+1)(I -tI+1)

(x - t)2 + y2 (t - S)2 + h2 Wlog, let

> 0. Then

L(II

dt

+ 1)(I-tI+1) s 2

dt

°°

(c+1)(I-tI+1),

0

which we break up in turn as 21

2f4/212

+ + In the first of these integrals we use the inequality

1r-tI/2,

and, in the second,

t/2,

taking in the latter a new variable s = t - . Both are thus easily seen to have values const.

log 1

+1 + 1

In the third integral, use the relation

t-

t/3.

This shows that expression to be In fine, then, °°

E.

x-t

const.1/(l + 1).

t-

(x-t)2+y2 (t-z)2+h2

dt

const.

log+II+1 ICI+1

538

VIII C Harmonic estimation in slit regions

for fixed x e Il and y, h > 0. From the proof of the first lemma in this article, we know,

however, that const.

S

sgn f l =

I

I

I+ 1

Absolute convergence of our double integral thus depends on the convergence of °°

(ICI+1)2

which evidently holds. Our proof is complete.

Notation. If -9 is one of our domains with Oe9, we write, for x > 0, f1Q(x) = owe((- oo, - x] u [x, co), 0).

Further work in this § will be based on the function 52... For it, the theorem

just proved has the

Corollary. If Oe9,

Q-9(x) = 2 n

xG.,(t, 0)

_cOx-t

dt

for x > 0.

Proof. When x > 0, SUx) = co.q(x) + ow.q(- x).

Plug the formula furnished by the theorem into the right side.

Scholium. The preceding arguments practically suffice to work up a complete treatment of the L2 theory of Hilbert transforms. The reader who

has never studied that theory thus has an opportunity to learn it now. If f eL2( - oo, oo), let us write

u(z) =

'

-f _ . I Z 3Z- t 12 f(t) dt IT

and

u(z)

=

x-t

1

°°

n

- ' I Z - t 12

f (t) dt

1 Relations between Green's function and harmonic measure 539 for 3z > 0; u(z) is a harmonic conjugate of u(z) in the upper half plane. By

taking Fourier transforms and using Plancherel's theorem, one easily checks that

Iu(x+iy)I2dx 5

1 1f1 1z

-00

for each y > 0. Following a previous discussion in this article and those of §§F.2 and H.1, Chapter III, we also see that

7(x) = lim u"(x + iy) Y-0

exists a.e. Fatou's lemma then yields 117112 s 11f 112

in view of the previous inequality.

It is in fact true that 17(x) - u(x + iy) 12 dx - + 0 -0D

for y - 0. This may be seen by noting that J

Iu(x+iy)-u(x+iy')IZdx

f-0000 Iu(x+iY)

- u(x + iy') 12 dx

for y and y' > 0, which may be verified using Fourier transforms and Plancherel's theorem. According to elementary properties of the Poisson kernel, the right-hand integral is small when y > 0 and y' > 0 are, as long as feL2. Fixing a small y > 0 and then making y'-+ 0 in the left-hand integral, we find that

JT 11(x)-u(x+iy)I'dx is small by applying Fatou's lemma. Once this is known, it is easy to prove that Ax)

f (x) a.e.

by following almost exactly the argument used in proving the last theorem above. (Note that (log+ 1 I + 1)/(I c I + 1) E L2(- oo, oo). ) This must then imply that 11 f 112

1<

11.112,

540

VIII C Harmonic estimation in slit regions

so that finally = (11112.

11f112

To complete this development, we need the result that the Cauchy principal value f (t) dt -"x-t °°

1

7t

exists and equals 7(x) a.e. That is the content of Problem 25 Let f eL,( - co, oo), p >, 1. Show that

x-t

1

(x-t)2+y2 tends to zero as y x+y

1

J .Y

x-y

f(r)dt -

f(t)

If 7

t-xl>yx-tdt

0 if

U(t)-f(x)Idt , 0

for y -.0, and hence for almost every real x. (The set of x for which the last conditionholds is called the Lebesgue set of f .) (Hint. One may wlog take f to be of compact support, making Ii f II i < oo. Choosing a small b > 0, one considers values of y between 0 and b, for which the difference in question

can be written as

yT(f(x-r)-f(x+t))dr

it

fo

T2

+

y2

I (f a it

+

T T2+y2 b

y

' (f(x-T)-f(x+T))dT. T

If the stipulated condition holds at x, the first of these integrals clearly -4 0 as y --> 0. For fixed b > 0, the integral from S to oo is < 2y2 II f II , /63 and as y -4 0. The integral from y to S is in absolute value this 2

Y2

('alf(x-T)-f(x+T)IdT. T3 y

Integrate this by parts.)

2.

An estimate for harmonic measure

Given one of our domains

with Oe-9, the function f2.,(x) = owq(( - oo, - x] u [x, oo), 0) is equal to 2 7t 2

f

xG,(t, 0) dt x2 - t2

-9

2 An estimate for harmonic measure

541

by the corollary near the end of the preceding article. The Green's function

G,(t, 0) of course vanishes on 0-9 = R n (- -9), and our attention is restricted to domains -9 having bounded intersection with R. The above Cauchy principal value thus reduces to an ordinary integral for large x, and we have

09(x) -

z 2

7r x

-cc

0) dt

for x - cc,

i.e., in terms of the Phragmen-Lindelof function YY(z) for.9, defined in §A.2, 2YY(0)

nx

x - co.

It is remarkable that an inequality resembling this asymptotic relation holds for all positive x; this means that the kind of comparison spoken of at the beginning of the present § is available. Theorem. If Oe-9, YY(O) for x > 0.

SZ,(x) 5

x

Proof. By comparison of harmonic measure for .9 with that for another smaller domain that depends on x.

Given x > 0, we let Ex = E v (- oo, - x] u [x, oo) and then put Qx=C - Ex:

Figure 152

We have -9x -9. On comparing wax((- cc, - x] v [x, cc), C) with co-,((- oo, - x] u [x, co), t) on Ex, we see that the former is larger than

542

VIIIC Harmonic estimation in slit regions

the latter for l; e2 . Hence, putting l; = 0, we get n2 (x) '< S2gx(x).

Take any number p > 1. Applying the corollary near the end of the previous article and noting that G,x(t, 0) vanishes for teE. ( - oc, - x] u [x, oo), we have

x(Px)

=

2 1 x pxG,x(t, 0) dt. 2 J -x P 2 x2 - t2

71

Since Qx c -9, G9x(t, 0) < G,(t, 0), so the right-hand integral is n2(P22P

1)x fx

G_(t, 0) dt

-x

2(P22P

7r

1)xfo ao

Gq(t, 0) dt.

By the formula for YY(z) furnished by the second theorem of §A2, we thus get fk'x(Px)

2p '< 7r(p2 - 1)

x

In order to complete the proof, we show that f1,x(px)/S2,x(x) is bounded below by a quantity depending only on p, and then use the inequality just established together with the previous one. To compare S1,x(px) with S2,x(x), take a third domain 00 = C - ((- 00, - x] U [x, 00))'

Figure 153

Note that _9x c & and 8-9x = Ex consists of 09 together with the part of E lying in the segment [ - x, x]. Fort'e2x (and p > 1), a formula from §B.1 of Chapter VII tells us that

wax((- cc, - Px] U [Px, cc), ()

= wA- cc, - Px] U [Px, cc), (0,((- oo, - Px] U [Px, oo), t)dw_,x(t, C), End

2 An estimate for harmonic measure

543

whence, taking t; = 0, f22x(Px) = (OX - cx, - Px] U [Px, cc), 0) we(( - oo, - px] u [px, oo), t)d(o_qx(t, 0). Er

Also,

O's(x) = 1 -

dco,x(t, 0).

fEng

The harmonic measure co,(( - oo, - px] u [px, oo), t) can be computed explicitly by making the Joukowski mapping x J(x2 l;->w=-Z-1

of9onto A={IwI<1}:

Figure 154

This conformal map takes [ - x, x] to the diameter [ - 1, 1], and 0 to 0. The union of the (two-sided!) intervals (- oo, - px] and [px, oo) on 09 is taken onto that of two arcs, a and a', on { I w I =1}, the first symmetric about i and the second symmetric about - i. For beg, u0,(( - oo, - px] v [px, oo), C) is the sum of the harmonic measures of these two arcs in A, seen from the point w therein corresponding to C. When C = t

is real, this sum is just 2coe(a, u), u being the point of (- 1, 1) corresponding to t. However, from the rudiments of complex variable theory, the level lines of (os(a, w) are just the circles through the endpoints of a. From a glance at the following diagram, it is hence obvious that cos(a, u) has its maximum for - 1 < u < 1 when u = 0:

544

VIII C Harmonic estimation in slit regions

Going back to .9, we see that w6(( - oo, - Px] U [Px, 00), t) <' w8((- 00, - Px] U [Px, 00), 0)

when - x < t <, x. Plugging this into the above formula for f

x(px), we

rind

that f.x(Px) '> we(( - cc, - Px] U CPx, cc), 0) X I.

1- f

d(o,,x(t, 0) 1

The quantity in curly brackets is just S2,x(x), so we have

49 ((x))

>' co(( - cc, - Px] U LPx, oo), 0).

Here, the right side clearly depends only on p; this is the relation we set out to obtain. From the inequality just found together with the two others established at the beginning of this proof, we now get

49(x) \ S2-.,x(x) \

n_Qx(PX)

we((- oo, - Px] U LPx, cc), 0) 2p

Y.9 (0)

n(PZ - 1)wg(( - 00, - Px] U CPx, 00), 0)

x

The front factor in the right-hand member depends only on the parameter

p; let us compute its value. The two arcs a and a' both subtend angles 2aresin(I'/p) at 0. Therefore

w6(( - oo, - px] u [px, oo), 0) = 2wo(a, 0) = 2 arcsin

1

and the factor in question equals P

(p2 - 1)aresin

1

P

It is readily ascertained (put 1/p = sin a !) that the expression just written decreases for p > 1. Making p - oo, we get the limit 1, whence

'(x) < Y,(0)/x,

S2,

Q.E.D.

Remark. An inequality almost as good as the one just established can be obtained with considerably less effort. By the first theorem of the preceding

2 An estimate for harmonic measure

545

article, we have, for y > 0,

'/ t Co

G.9(iy, 0) = logy +

f-

log I iy - t I dco9(t, 0)

'co

=J

log

I

a quantity clearly > Q9(y)log 0)

1 + _ I dwq(t, 0),

\\\2.

On the other hand,

yy +tt2 dt

= n ,1

as in the proof of the third lemma from that article. Here, the right side is

5 7ty- i('°° 1

YY(0)

0)dt =

,

Y

so the previous relation yields

(Y)

2

Y9(0)

log 2

y

.

Problem 26 For 0 < p < z, let EP be the union of the segments

2n-1 2

- p,

l

2n-1 +P1 2

ne7L;

these are just the intervals of length 2p centered at the half odd integers.

Denote the component [(2n - 1)/2 - p, (2n - 1)/2 + p] of E, by J. (it would be more logical to write J (p) ). 2, = C - E. is a domain of the kind considered in §A, and, by Carleson's theorem from §A.1, K, w o(J

0) , n2 + I

The purpose of this problem is to obtain quantitative information about the asymptotic behaviour of the best value for K. as p -+ 0. (a) Show that Y., o(0) - (1/n) log(1/p) as p - 0. (Hint. In -Q,, consider the harmonic function log

cos sin n

+

-1

Ircos2 J 1\ sine Rp

(b) By making an appropriate limiting argument, adapt the theorem just proved to the domain a', and hence show that f2 o(x) < Y o(0)/x

for x > 0.

546

VIII C Harmonic estimation in slit regions (c) For n >, 1, show that CO o(J"+1, 0) < w o(Jn, 0).

(Hint:

0

0

0

0

0

1

l

0

0

0

Figure 156

(d) Hence show that, for n >, 3, (09P (J,,, 0) '<

I C log 1P I n2 .)

/

\

with a numerical constant C independent of p. o(Jk+1, 0).) (e) Show that the smallest constant K. such that w for all n satisfies

(Hint: Q, (n) 3

0) 5 K./(n2 + 1)

1

Ko 3 C'logP

with a constant C' independent of p. (Hint. This is harder than parts (a)-(d). Fixing any p > 0, write, for large R, ER = EP u (- oo, - R] u [R, oo), and then put 2R = C - ER. As R -> oo, G,R(t, 0) increases to G,o(t, 0), so Y9R(0) increases to Y-,(0). For

each R, by the first theorem of the previous article,

G$ (z, w) = log R

1

Iz-wI

+J

log I w - s I dwgR(s, z), ER

whence z

GSR(t, 0) + GAR(- t, 0) =

log 1- sZ dw-QR(s, 0). JER

t

2 An estimate for harmonic measure

547

Fix any integer A > 0. Then I A-AG, (t, 0)dt is the limit, as R -+ oo, of j,. f Alog I 1 - (s2/t2)I dt dw,R(s, 0). Taking an arbitrary large M, which for the moment we fix, we break up this double integral as fM foA A +

f, > M JO

-M

To study the two terms of this sum, first evaluate A

s2

Z dt;

log

f

t

0

for Isi > A this can be done by direct computation, and, for Isi < A, by using the identity s2

10A log

1- 2

dt = - l

t

log

1- Z t

A

Regarding I MM f o log i 1 - (s2/t2) I dt da R(s, 0), we may use the fact that w-@R(S, 0) -+ w.9°(S, 0) as R -+ co for bounded S R, and then plug in the

inequality

w0 (J°, 0) 5 Kv/(n2 + 1) together with the result of the computation just indicated. In this way we easily see that limR f MM f o A, M, and p.

CKv with a constant C independent of

In order to estimate A

log JIsI> M Jo

$2

1- Z t

dt dw9R (s, 0),

use the fact that S2

R()s 5

s

5

Y°(0) s

(where Y. (0), as we already know, is finite) together with the value of the 0 inner integral, already computed, and integrate by parts. In this way one finds an estimate independent of R which, for fixed A, is very small if M is

large enough. Combining this result with the previous one and then making M -+ oc, one sees that

fA J-A

G,P(t,0)dt e CKo

with C independent of A and of p.)

Remark. In the circumstances of the preceding problem G., (z, 0) must, when p -> 0, tend to co for each z not equal to a half odd integer, and it is

548

VIII C Harmonic estimation in slit regions

interesting to see how fast that happens. Fix any such z 0 0. Then, given

p > 0 we have, working with the domains .9R used in part (e) of the problem, G,o(z, 0) = lim G9R(z, 0). R-oo

Here,

G'9R(z, 0) = log

I

= O(1)+ J

I+

log J z- t J dw,R(t, 0)

log+Itldco,R(t,0),

where the O(1) term depends on z but is independent of R, and of p, when

the latter is small enough. Taking an M > 1, we rewrite the last integral on the right as f,I<M

ItI>_M

and thus find it to be

log M - J

log t df2'R(t)

= logM+f R(M)logM+ fm t (t)dt. Plug the inequalities f fR(t) < YYR(0)/t and Y,,(0) < Y. o(0) into the expression on the right. Then, referring to the previous relation and making R -> oo, we see that G"(z, 0) '< O(1) + log M + Y9p(0)

By part

(a)

log M + 1 M

of the problem, Y, (0) = 0(1)+(1/n)log(1/p). Hence,

choosing M = (1/n)log(1/p)loglog(1/p) in the last relation, we get G,o(z, 0) < O(1) + log log P + log log log I .

This order of growth seems rather slow. One would have expected G.,o(z,0) to behave like log(1/p) for small p when z is fixed. 3.

The energy integral again

The result of the preceding article already has some applications

to the project described at the beginning of this

§.

Suppose that

3 The energy integral again

549

the majorant M(t) > 0 is defined and even on P. Taking M(t) to be identically zero in a neighborhood of 0 involves no real loss of generality. If M(t) is also increasing on [0, co), the Poisson integral

L v(t)dco,(t, 0) for a function v(z) subharmonic in one of our domains -9 with OE_9 and satisfying

v(t) 5 M(t),

t C- 0-9,

has the simple majorant M(t) YY(0) fo"O

dt.

T he entire dependence of the Poisson integral on the domain -9 is thus expressed by means of the single factor YY(O) occurring in this second expression.

To see this, recall that coy(( - oo, - t] u [t, oo), 0) = 0,(t) for t > 0; the given integral

majoration on v(t) therefore makes f o M(t)dfl,(t), which here is equal to

the

Poisson

c (t)dM(t). fO'O

Since M(t) is increasing on [0, oo), we may substitute the relation S0t) 5 YY(0)/t proved in the preceding article into the last expression, showing it to be Y.9(0) J0

dM(t)

= YIP J0 M(t)dt.

This argument cannot be applied to general even majorants M(t) > 0,

because the relation Q(t) < YY(0)/t cannot be differentiated to yield dcou(t, 0) 5 (YY(O)/t2) dt. Indeed, when x c- 8-9 = E gets near any of the

endpoints a of the intervals making up that set, dco,(x,0)/dx gets large like a multiple of Ix - aI -1/2 (see the second lemma of article 1 and the remark following it). We are not supposing anything about the disposition of these intervals except that they be finite in number; there may otherwise

be arbitrarily many of them. It is therefore not possible to bound f °° M(t)dco,(t, 0) by an expression involving only f (M(t)/t2)dt for

.

o properties of general even majorants M(t) >, 0; some additional regularity M(t) are required and must be taken into account. A very useful instrument for this purpose turns out to be the energy introduced in §B.5 which has

550

VIII C Harmonic estimation in slit regions

already played such an important role in §B. Application of that notion to matters like the one now under discussion goes back to the 1962 paper of

Beurling and Malliavin. The material of that paper will be taken up in Chapter XI, where we will use the results established in the present §. Appearance of the energy here is due to the following Lemma. Let OE-9. For x

0,

G.(x, 0) + G,,(- x, 0) = 1

log

x fo

x+t x

Proof. By the second theorem of article 1, wi(t) sgn t

G.,(x, 0)

dt

x-t

for x 0 0,

where

wi(t) =

coy((- co, t], 0), w_At, cc), 0),

t < 0, t > 0.

Thence, 2t sgn t (t)2 (t)

G.,(x, 0) + G.,(- x, 0) _

t -x 2t

_x2

dt

f2t) ( dt,

since co.,(t) + coq(- t) = Q .(t) for t > 0.

Assuming wlog that x > 0, we take a small e > 0 and apply partial integration to the two integrals in

(

fo'

o

+

J x+E

/ t2 2x2 tf

(t)dt,

getting

tMtog t-x C

x

t+x

)(JO-E

+ 1x'+ E)

+

+

(

x-t

Jo-E+fx+E)xloglxt

d(tn,(t)).

The function fLL(t) is I for t > 0 near 0 and O(1/t) for large t; it is moreover

Lip 2 at each x > 0 by the second lemma of article 1. The sum of the

3 The energy integral again

551

integrated terms therefore tends to 0 as e - 0, and we see that

r

2tx2 t2

i2.9(t)dt = I 1 log

x+t

x-t d(tS2.q(t)).

Since the left side equals G,(x, 0) + G.,(- x, 0), the lemma is proved. In the language of §B.5, x(G9(x, 0) + G.9( - x, 0)) is the Green potential of d(tf2.,(t)). Here, since we are assuming -9 n U8 = I8 - E to be bounded, °°

1

S2.9(x) = n2

f

2x x2 - t2 G,,(t, 0) dt

has, for large x, a convergent expansion of the form

so that d(t(2,(t))

2a3

i3

4a5

+ t5 +

1 I dt

for large t. Using this fact it is easy to verify that

flog xx-+ t oJ

d(tS2.,(t)) d(xf (x))

o

is absolutely convergent; this double integral thus coincides with the energy

E(d(tf2.9(t)), d(ti2.(t))) defined in §B.5.

Theorem. If Oe-q, E(d(tS29(t)), d(tQ29(t))) <

n(Y.9(0))2.

Proof. By the lemma, the left side, equal to the above double integral, can be rewritten as x [G,(x, 0) + G.,,(- x, 0)] d(xf2.,(x)). fo"O

Here, G.,(x, 0) + G.,,(- x, 0) ,>0 and 0,(x) is decreasing, so the last expression is

<J

o

[G,(x,0) +G.,(- x,0)]xf,(x)dx. 0

VIII C Harmonic estimation in slit regions

552

From the theorem of the preceding article we have xS2_,(x) S Y1,(0), so this

is in turn G_,(x, 0) dx,

Y.9(0)

J which, however equals n(Y9(0))2 by the second theorem of §A.2. We are done.

This theorem will be used in establishing the remaining results of the

present §. For that work it will be convenient to have at hand an alternative notation for the energy E(dp(t), dp(t)).

Suppose that we have a real Green potential

x+t

u(x) = J "O log

x-t

0

dp(t).

If the double integral

x+t 0 f'O o log f'O

x-t

dp(t)dp(x)

used to define E(dp(t), dp(t)) is absolutely convergent, we write

0.

II

u

II E

= E(dp(t), dp(t)).

If we have another such Green potential log

v(x)

fo

Ix-t x+t da(t), t

we similarly write

E = E(dp(t), da(t)) <

,

>E is a bilinear form on the collection of real Green potentials of

this kind; according to the remark at the end of §B.5 it is positive definite. The reader may wonder whether our use of the symbol II U II E to denote 1(E(dp(t), dp(t))) is legitimate; could not the same function u(x) be the Green potential of two different measures? That this cannot occur

4 Harmonic estimation

553

is easily seen, and boils down to showing that if p(x) is not constant, the Green potential

x+t

u(x) = J

'O log 0

x-t

dp(t)

cannot be =- 0 on [0, oo) (provided, of course, that the double integral used to define E(dp(t), dp(t)) is absolutely convergent). Here, we have E(dp(t), dp(t)) = f o u(x)dp(x). Hence, if u(x) =_ 0, the left-hand side is also zero. Then, however, p(x) is constant by the second lemma of §B.5. 4.

Harmonic estimation in -9

We are now able to give a fairly general result of the kind envisioned at the beginning of this §. Suppose we have an even majorant M(t)>,O with M(0) = 0. In the case where M(x)/x is a Green potential

J log

x+t

x-t dp(t)

with the double integral defining E(dp(t), dp(t)) = II M(x)/x IIE absolutely convergent, the following is true:

Theorem. Let M(t) be a majorant of the kind just described. Given one of our domains .9 containing 0, suppose we have a function v(z), subharmonic in .9 and continuous up to 0-9, with v(z) S A I .Zz I + O(1)

for some real (sic!) A, and

v(t) 5 M(t)

for t c- a.9.

Then

v(0) < Y2(0) { A +

JM(t)dt

+

M(x) x

Remark. The assumptions on v(z)'s behaviour can be lightened by means of standard Phragmen-Lindelof arguments (see footnote near beginning

of §A.2, after problem 16). Such extensions are left to the reader; what we have here is general enough for the applications in this book.

554

VIII C Harmonic estimation in slit regions

Proof of theorem. The difference v(z) - A YY(z)

is (by the definition of YY(z) in §A.2) subharmonic and bounded above in -9, and continuous up to 8-9, where it coincides with v(z). Hence, by harmonic majoration (Chapter VII, §B.1),

v(z) - AYY(z) <, f

<, J

z)

ae9

M(t)dcw,(t, z)

a

for

ze-9.

Taking z = 0, we see that we have to estimate $a,M(t)dco,(t, 0), which, in view of the definition of SZ,(t), equals - So M(t)df2,(t), M(t) being even. The trick here is to write

-J

M(t) dfZ.,(t) =

' M(t)njt) dt - f

J0

0

(t)

o

Since M(t) > 0, the first integral on the right is o YA(0) fo

MZt) dt

by the theorem of article 2. In view of our assumption on M(t), the second

right-hand integral can be rewritten

-Jo

log fOO

t+x dp(x) d(tQ_,(t)) = - E(dp(t), d(tQ.9(t))). t-x

Using Schwarz' inequality on the positive definite bilinear form E( , (see remark, end of §B.5), we see that the last expression is in modulus d(tf)_,(t))))

J(E(dp(t),

which, by the result of the preceding article, is 5 Putting our two estimates together, we get M(t)dw (t, 0) S Y,(0) { f M(t) dt Jai

)

o

II M(x)/x II EJx YY(0).

+ /I

x El. M(x)

As we have seen v(O) - A Y ,(0) is S the left-hand integral. The theorem is thus proved.

Remark. This result shows that for special majorants M(t) of the kind described, the entire dependence of our bound for v(0) on the domain -9 is

5 Majorant is the logarithm of an entire function

555

expressed through the quantity Y.,(0), Y_, being the Phragmen-Lindelof function for -9.

When majorant is the logarithm of an entire function of exponential type

5.

The result in the preceding article can be extended so as to apply to certain even majorants M(x) of the form x f0`0 log

x+t

x-t

dp(t)

for which the iterated integral log ,10 fo'O

x+t

x-t

dp(t) dp(x)

is not absolutely convergent. This can, in particular, be done in the important special case where M(x) = log I G(x)I

with an entire function G of exponential type,

1

at 0, having even

modulus >, 1 on R, and such that log G(x)

C

J

1

+xz dx < oo.

-,000

Then the right side of the boxed formula at the end of the previous article can be simplified so as to involve only Y,(0), f o (M(t)/t2)dt, and the type of G.

The treatment of any majorant M(x), even or not, of the form log+ IF(x)I

with F entire, of exponential type, and such that C °°

log+ I F(x) I

J -00

1 +x2

dx < a),

can be reduced to that of one of the kind just described. Indeed, to any such M(x) corresponds another, M ,(x) =log I G(x) I with G entire and of exponential type, such that M1(x) >, M(x)

for

I x I >, 1,

M1(x) = M1(- x) % 0,

M1(0)=0, and

°° M1(x)

x Jo

VIIIC Harmonic estimation in slit regions

556

To see this, put first of all

'(z) = 'D(z) is then entire and of exponential type, even, and > 1 on l with 1(0) = 1. Clearly b(x) 1 + x2 f `° log

dx < co

in view of the similar property of F, and for I x 1 > 1. '(x) >, IF(x)l' By the Riesz-Fejer theorem (the third one in §G.3 of Chapter III), there is an entire function G(z) of exponential type, having all its zeros in 3z < 0 (since here (D(x) >, 1), such that

D(z) = G(z)G(z). The majorant M ,(x) = log I G(x) I then has the required properties.

The result to be obtained in this article regarding even majorants of the abovementioned kind can thus be used in studying

log I G(x) I

problems involving the more general ones of the form log' I F(x) I.

For entire functions

G(z)

of exponential type

with G(0) = 1,

G(x) I = I G(- x) I > 1, and °°

_00

log I G(x) I

1+x2

dx < oo,

log I G(x) I has a simple representation as a Stieltjes integral. When dealing only with the modulus of G on III, we may, by the second theorem of §G.3,

Chapter III, assume that G(z) has all its zeros in the lower half plane. Forming, for the moment, the entire function 4'(z) = G(z)G(2), we see that

O(x) = t(- x) on R so that fi(z) = 4)(- z), and every zero of O(z) is also one of D(- z). The zeros of V(z) are just those of G(z) together with their complex conjugates, so, since all the former lie in Y3z < 0, we have G(- .Z) = 0

whenever G(.1) = 0. The zeros of G(z) thus fall into three groups: those on

the negative imaginary axis, those in the open fourth quadrant, and the reflections of these latter ones in the imaginary axis. The Hadamard factorization (Chapter III, §A) of G(z) can therefore be written

G(z)=e"fl(1+1?

)eiZiµk

.

n(1-f)eijz^(1+-)e

='z,

5 Majorant is the logarithm of an entire function

557

where the µk > 0, 931 > 0 and 32n > 0. One (or even both!) of the two products occurring on the right may of course be empty. Since I G(x) I = I G(- x) I, a is pure imaginary. We also know, by the first

theorem of §G.3, Chapter III, that 1

and

Lr

k µk

both converge. The exponential factors figuring in the above product may therefore be grouped together and multiplied out separately, after which the expression takes the form e'bzH

1+

k

-

1- -z

z

1µk

1+-nz ,

Xn

n

with b real. Here, we are only concerned with the modulus I G(x) I, xE IR; No. we may hence take b = 0. This we do throughout the remainder of this article, working exclusively with entire functions of exponential type of the form

I+- 11( I

G(z) =

+Wk)

k

z

(I +

-1;n

n)'

where the µk > 0, 9i2n > 0 and 32n > 0. The products on the right are of course assumed to be convergent. Our Stieltjes integral representation for such functions is provided by the Lemma. Let G(z), of exponential type, be of the form just described. Then, for

z5z>0, 2

log 1 -

=

log I G(z) I

f0'0

t2

dv(t)

with an increasing function v(t) given by dv(t)

1

dt

n

JAn

JAn

µk

k Uk+t2 +

n

(IA.-tl2+IAn+tl2

Proof. Fix z, 3z > 0. Then log 11 + z/A, I is a harmonic function of A in {.32 > 0}, bounded therein for A away from 0, and continuous up to F save at A = 0 where it has a logarithmic singularity. We can therefore apply Poisson's formula, getting

log 1 +

z A

_ 7r

j

log '0

z I1+tl2dt

t

558

VIII C Harmonic estimation in slit regions

for 32 > 0, from which

+ log

log

1- z 1 J-00

=

log

no

Z2 1-2 t

t12

IA+

+IA_

dt. t12

Similarly, for µ > 0, log

1

1+? iµ

°°

nJ o

log

1-2z2t

p2 + t2 dt.

We have log I G(z) I

= Y log 1 +.Z + YI log 1+T + log 1µk

k

n \

A,,

z n

1

When 3z > 0, we can rewrite each of the terms on the right using the formulas just given, obtaining a certain sum of integrals. If I'.Rz I < 3z, the order of summation and integration in that sum can be reversed, for then log

Z2 1-2 t

0,

tER.

This gives

log I G(z) I

= fO'O log

z2 1-2 dv(t), t

at least for 19Rz I < 3z, with v'(t) as in the statement of the lemma.

Both sides of the relation just found are, however, harmonic in z for 3z > 0; the left one by our assumption on G(z) and the right one because f o log 11 + y2/t2 I dv(t), being just equal to log l G(iy) I for y >0, is convergent

for every such y. (To show that this implies u.c.c. convergence, and hence harmonicity, of the integral involving z for 3z > 0, one may argue as at the beginning of the proof of the second theorem in §A, Chapter III.) The two sides of our relation, equal for I'Rz I < 3z, must therefore coincide for 3z > 0 and finally for 3z > 0 by a continuity argument. Remark. Since G(z) has no zeros for 3z 3 0, a branch of log G(z), and hence

of arg G(z), is defined there. By logarithmic differentiation of the above boxed product formula for G(z), it is easy to check that d arg G(t)

= - 7CV (t)

dt

with the v of the lemma. From this it is clear that v'(t) is certainly continuous

(and even WJ on R.

5 Majorant is the logarithm of an entire function

559

In what follows, we will take v(O) = 0, v(t) being the increasing function

in the lemma. Since v'(t) is clearly even, v(t) is then odd. With v(t) thus specified, we have the easy Lemma. If G(z), given by the above boxed formula, is of exponential type, the

function v(t) corresponding to it is 5 const.t for t > 0. Proof. By the preceding lemma,

I - tz dv(t) = log I G(z) I I

for 3z > 0, the right side being < K I z I by hypothesis, since G(0) = 1. Calling

the left-hand integral U(z), we have, however, U(z) = U(IJ, so U(z) < K I z I

for all z. Reasoning as in the proof of Jensen's formula, Chapter I (what we are dealing with here is indeed nothing but a version of that formula for the subharmonic function U(z)), we see, for t 96 0, that

f"Iogll

1

2n

R

-

rei9

t

Id8 =

r,

log

I t l< r,

Itl

Itl,r.

0,

Thence, by Fubini's theorem, 1

r log I t I dv(t).

U(rei9) d9 =

I

-r

-,,

Integrating the right side by parts, we get the value 2 f o(v(t)/t) dt, v(t) being odd and v'(0) finite. In view of the above inequality on U(z), we thus have V(t) r

Jo t

dt < i Kr.

From this relation we easily deduce that v(r) < Chapter I. Done.

i eKr as in problem 1,

Using the two results just proved in conjunction with the first lemma of §B.4, we now obtain, without further ado, the Theorem. Let the entire function G(z) of exponential type be given by the above boxed formula, and let v(t) be the increasing function associated to G in

the way described above. Then, for x > 0, log I G(x) I

= - x J OO log 0

x+t

x-t

d(v(t)

I.

560

VIII C Harmonic estimation in slit regions

For our functions G(z), (log I G(x) I )lx is thus a Green potential on (0, oo).

This makes it possible for us to apply the result of the preceding article to majorants M(t) = log I G(t) 1.

With that in mind, let us give a more quantitative version of the second of the above lemmas. Lemma. If G(z), given by the above boxed formula, is > 1 in modulus on R and of exponential type a, the increasing function v(t) associated to it satisfies v( t)

<

e

-a + e I' logIG(x)Idx. J

t?0.

- OD

Remark. We are not striving for a best possible inequality here. Proof of lemma. The function U(z) used in proving the previous lemma is subharmonic and < K I z I. Assuming that °° _OD

log I G(t) I 2

dt < o0

t

(the only situation we need consider), let us find an explicit estimate for K. Under our assumption, we have, for .3z > 0,

logIG(z)I < azz+I f by §E of Chapter III. When - y

Iz

-3z

t12logIG(t)Idt

x < y, we have, however, for z = x + iy, t2

3

t2 2

t2

2 +2 -2xt+2x2

Iz-t12 = t2-2xt+x2+y2

t c R,

,

whence, log I G(t) I being >, 0,

< ay +

log I G(z) I

2y f'o log I G(t) I _ 00

t

2

dt.

Thus, since U(z) = U(zr) = log I G(z) I for 3z >, 0,

U(z) < I a +

\

2

log I G(t) I

--

dt

13Z I

00

in both of the sectors I `Rz I < 13z I

Because U(z) < const. I z I we can apply the second Phragmen-Lindelof

5 Majorant is the logarithm of an entire function

561

theorem of §C, Chapter III, to the difference

U(z) -

t2

7E

in the 90° sector 13z 15 91z, and find that it is < 0 in that sector. One proceeds similarly in 9Iz U(z) 1<

13z 1, and we have

(a+jl0t)ldt)lzI

for I3zI < 19tzI.

Combining the two estimates for U(z) just found, we get

U(z) , K Izl with

K = a+ 2it

_ co

log 16(t) I dt. t2

This value of K may now be plugged into the proof of the previous lemma. That yields the desired result.

Problem 27 Let 4(z) be entire and of exponential type, with D(0) = 1. Suppose that I(z)

has all its zeros in 3z <0 and that I1(x)I > 1 on R. Show that then loglo(x)I x2

_co

dx < oo.

(Hint: First use Lindelof's theorem from Chapter III, §B, to show that the Hadamard factorization for 1(z) can be cast in the form

1(z) =

fl(I - ^

the 32,, < 0. Taking 'P(z) = d>(z) exp (- iz3c), a losl`P(z)l/ay ,>0 for y>0, and then look at 1/`Y(z). ) where

show that

Suppose now that we have an entire function G(z) given by the above boxed representation, of exponential type a and >, 1 in modulus on P. If the double integral f0°0

Io° °

J

log

Ix+t

)d('

is absolutely convergent, we may, as in the previous two articles, speak of the

562

VIII C Harmonic estimation in slit regions

energy

El

d(vtt)/'

of`

in terms

d( t)

the Green potential logIG(x)I

-J

X

log

x+t

x-t

this is just II (log I G(x) I )/x II I according to the notation introduced at the end of article 3.

To Beurling and Malliavin is due the important observation that II (log I G(x) I )lx II E can be expressed in terms of a and f o (log I G(x) I/xz)dx

under the present circumstances. Since log I G(t) I 30 and v(t) increases, we have indeed

II(log IG(x)I)/xIlI = -

/v(x)

0°° log I G(x) I

d1

X

log I G(x) I

v(x)

xz

x

o

, su v(x))J0 x>o x

lo

I

dx -dv(x)

g I G(x)

xz

/

dx.

Using the preceding lemma and remembering that I G(x) I is even, we find

that

log I G(x) I

x

2

tea

E

2

+

2e (' - log I G(x) I dx 1

n

xz

o

J

1

log I G(x) I

xz

Jo

dx.

Take now an even majorant M(t) >, 0 equal to log I G(t) I, and consider

one of our domains -9 with Oe9. From the result just obtained and the boxed formula near the end of the previous article, we get

f M(t) duoq(t, 0) <

I J+

J

2eJI J+ 4 I

with

J =

C' log I G(t) I

Jo

tz

dt = J o

M(t) t2 dt,

I },

5 Majorant is the logarithm of an entire function

563

at least in the case where log

x+t

dlvtt)Idlvzx)I

x-t

is absolutely convergent. On the right side of this relation, the coefficient Y,(0) is multiplied by a factor involving only a, the type of G, and the integral

f (M(t)/t2) dt (essentially, the one this book is about!). o It is very important that the requirement of absolute convergence on the above double integral can be lifted, and the preceding relation still remains true. This will be shown by bringing in the completion, for the norm II IIE, of the collection of real Green potentials associated with absolutely con-

vergent energy integrals - that completion is a real Hilbert space, since IIE comes from a positive definite bilinear form. The details of the argument take up the remainder of this article. Starting with our entire function G(z) of exponential type and the increasing function v(t) associated to it, put II

Q(x)

= - Ioxloglx+tldrvtt)), ` f

=log G(x)

and, for n = 1, 2, 3,. - -,

I j'

n

log 1 -

Qn(x) = z

0

x2 t2

dv(t).

In terms of vn(t) =

0 5 t < n, t > n,

(v(t), 1 v(n),

we have t

Qn(x) = - f0'0 log x + t l

0

by the first lemma of §B.4; evidently, Q.(x) --> Q(x) u.c.c. in [0, oo) as n -+ oo.

Each of the integrals

+t

fo fo loglxx-t dl ntt)/d\vnxx)/ is absolutely convergent. This is easily verified using the facts that

d\v t)/ -

4v"(0)dt

564

VIIIC Harmonic estimation in slit regions

near 0 (v(t) being W. by a previous remark), and that d(

I = - vt2) dt

,(t)

for t > n.

Lemma. If I G(x) I >, 1 on R, the functions Q,,(x) are >, 0 for x > 0, and II Qn II E < 2 V'(0)

Proof. Fort>0, logI1-x2/t2I,>0 when x>,,/2t, so fon

log

xQi(x) =

is >,0 for x

.

x2 1-t2

dv(t)

J2n. Again, for 0,<x,<,,./2t, log 11- x2/t2 15 0, so, for

0K, x
and finally xQn(x), equal to logIG(x)I minus this integral, is >0 since I G(x) I > 1.

The second lemma of §B.4 is applicable to the functions vn(t). Using it and

the positivity of Qn(x), already established, we get IIQnIIE= f'O f,* log O

o

o

x+t

x-t d ( n(t) )d \ vnxx)/

Qn(x)1 "z2)dx-dvxx)l

Jo

Qn(x) nz2)dx

= 4 (v (0))2 = 4 (V,(0))2.

We are done. Theorem. Let G(z) be an entire function of exponential type a, 1 at 0, with I G(x) I even and > 1 on 08, and such that 1 °° log G(x)

_.

1

+x2I dx < oo.

If -9 is one of our domains containing 0, we have

5 Majorant is the logarithm of an entire function

JloIG(t)Idw(tO) S Y9(0) { J l

\

a.9

2eJ I J +

\

565

4 /I)1

where

f

log I G(t) I

tz

0

Proof. According to the discussion at the beginning of this article we may, without loss of generality,* assume that G(z) has the above boxed product representation. Beginning as in the proof of the theorem from the preceding article, we have log Iz (x) I

J log I G(x) I dco.,(x, 0) =

!0.,(x) dx

fO'O

°°logIG(x)I

Iox

d(xfL(x)).

The first term on the right is of course °° log I G(x) I Y'2(0) ) fo

xz

dx

by the theorem of article 2, log I G(x) I being positive. The second, equal to

- focan be looked at in two different ways. In the first place, for x > 0,

Q(x) = lim Qn(x) n-oo

with the functions Qn(x) introduced above. Also, for each n,

* Dropping the factor exp(ibz) from the second displayed expression on p. 557 can only diminish the overall exponential type, for, if G(z) is given by the boxed formula on that page, the limsups of logIG(iy)I/IyI for y tending to oo and to - oo are equal. To see that, observe that the limsup for y-. oo is actually a limit (see remark, p. 49), and that G(z)/G(z) = B(z) is a Blaschke product like the one figuring in the remark on p. 58. The argument of pp. 57-8 shows, however, that then the limsup of logI B(iy)I/y for y-, oo is zero.

566

VIII C Harmonic estimation in slit regions n

Qn(x) 5 1

xo

log

x2 dv(t) +2 t

< 1J

log 1 +

2

dv(t),

x > 0.

0

Since v(t) < Kt, the right-hand member comes out < xK on integrating by parts. This, together with the preceding lemma, shows that

0<

nK

for x > 0.

However, for large x,

d(xf2,,(x)) _

( const.

+

O( 51)dx x

x

(see just before the theorem of article 3). Therefore

J

Q(x)d(xf2.q(x)) = lim J n-oo

O'O

Qn(x)d(x!nq(x)) ooo

by dominated convergence.

The right-hand limit can also be expressed as an inner product in a certain real Hilbert space. The latter - call it .5 - is the completion with respect to the norm II IIE of the collection of real Green potentials

u(x) = f

log

0o

x+t

x-t

dp(t)

such that fOO fOD

log 0

0

x+t

x-t Idp(t)Ildp(x)l < oo;

the positive definite bilinear form < , >E extends by continuity to Sa for which it serves as inner product. For each n, we have

Qn(x)d(xf.q(x)) = EI dl "tt)

I,

d(Q.9(t))) = E,

fOOO

where

P(x) = x(G.,(x, 0) + G.,(- x, 0)) ;

here only Green potentials associated with absolutely convergent energy integrals are involved. By the lemma, however, II Qn IIE <

2

v'(O),

5 Majorant is the logarithm of an entire function

567

so a subsequence of {Q,,}, which we may as well also denote by {Qn}, converges weakly in Sa to some element q of that space. (Here, we do not need to `identify' q with the function Q(x), although that can easily be done.) In view of the previous limit relation, we see that

Q(x)d(xfL(x)) = lim E = E n-oo

0 J,*

Thence, by Schwarz' inequality and the result of article 3, Q(x)d(xCZ9(x))J <, IIgIIEIIPIIE f0`0

= II q II Ey (E(d(tS2c(t)), d(tc2 (t)))) < J7rY9(O) II q IIE Returning to the beginning of this proof, we see that l og G(x) X2 I

J

a

log I G(x) I dw9(x, 0) < Y9(O) fo

I

dx

+ s/1rY9(0)IIgIIE,

and thus need an estimate for II q IIE The obvious one, II q IIE < liminfn_. II Q. IIE 1< nv (O)/2, is not good enough to give us what we

want here, so we argue as follows. The weak convergence of Q,, to q in Sj implies first of all that

IIgIIE = lim E n _OD

Fix any n; then, by weak convergence again, llm

E = llm E

k- oo

k- oo

x

p

Here, d(vn(x)/x) is just - (v(n)/x2) dx for x > n, so, since 0 < Qk(x) we have, by dominated convergence, kim JU

\ )= Qk(x)d(vn

-JO'Q(x)d\ 0

nK,

\t

C°nx) x)

which, Q(x) being positive, is Q(x) vn(Z) S fOOO

dx.

X

Again, vn(x) < v(x) for x > 0, so finally v(x)

Q(x)

E 1< fO,O

x2

° log I G(x) I v(x)

dx = fO

x2

dx

x

for each fixed n. The right-hand integral was already estimated above,

VIII C Harmonic estimation in slit regions

568

before the preceding lemma, and found to be

( ira ne

4

00 log I G(x)1 dx'

+ fo

z

J°° log IG(x) I c2

dx.

o

This quantity is thus >

E'

II q II

giving us an upper

bound on II q IIE.

Substituting the estimate just obtained into the above inequality for $ a9 log I G(x) I daw9(x, 0), we have the theorem. The proof is complete.

Corollary. Let G(z) and the domain -q be as in the hypothesis of the theorem.

If v(z), subharmonic in 9 and continuous up to 8.9, satisfies

te8.9,

v(t) 5 log I G(t)I, and

v(z) < AI.3zI + o(l) with some real A, we have

v(0) < Y9(0) { A + J +

I

J

/

(2eJ1 J +

ira

4

)

\l

I },

where

J _ J0f

log I G(x) I

xz

dx

and a is the type of G.

This result will be used in proving the Beurling-Malliavin multiplier theorem in Chapter XI. Problem 28 Let G(z), entire and of exponential type, be given by the above boxed product formula and satisfy the hypothesis of the preceding theorem. Suppose also that 1081G(iY)I

--+a for y--+±ao.

IYI

The purpose of this problem is to improve the estimate of II (log I G(x) I )/x II

E

obtained above.

(a) Show that v'(0) = a/x + 2J/n2 and that v(t)/t - a/x as t - co. Here, J has the same meaning as in'the statement of the theorem. (Hint. For the second relation, one may just indicate how to adapt the argument from §H.2 of Chapter III.)

5 Majorant is the logarithm of an entire function

569

(b) Show that °° logI G(x) I v() dx

x

Jo

X2

= ,21 (v'(0))Z 4

lim r

ac

VW

)2).

t

(Hint. Integral on left is the negative of

f.

f."

log

X+t

lx-t

v(t)

v(x)

d(t) X2

dx.

Here, direct application of the method used to prove the second lemma of §B.4 is hampered by (d/dt)(v(t)/t)'s lack of regularity for large r, however, the following procedure works and is quite general. For small > 0 and large L one can get E, 0<e<8,andR>L making L

R

logl

fR e

X+t

dx d ( tt)) v(x) X2

x-t

nearly equal to the above iterated integral. The order of integration can now be reversed and then the second mean value theorem applied to show that f ft, and f L f s are both small in magnitude when & > 0 is e small and L large. Our initial expression is thus closely approximated b(ys

' J

ILlogIx-t

i

z2) dxd( t)).

Apply to this a suitable modification of the reasoning in the proof of the aforementioned lemma, and then make 8 --4 0, L 00.) (c) Hence show that

jloIG(x)) v(x) x2

o

x

dx =

1 n2

J(J + na)

so that

x+t Jo

(,lo log

x-t

t/

d(v(t)))d(X))

\x

5 12J(J+na). It

Addendum

Improvement of Volberg's Theorem on the Logarithmic Integral. Work of Brennan, Borichev, Joricke and Volberg. Writing of §D in Chapter VII was completed early in 1984, and some copies of the MS were circulated that spring. At the beginning of 1987 I learned, first from V.P. Havin and then from N.K. Nikolskii, that

the persons named in the title had extended the theorem of §D.6. Expositions of their work did not come into my hands until April and May of 1987, when I had finished going through the second proof sheets for this volume. In these circumstances, time and space cannot allow for inclusion of a

thorough presentation of the recent work here. It nevertheless seems important to describe some of it because the strengthened version of Volberg's theorem first obtained by Brennan is very likely close to being best possible. I am thankful to Nikolskii, Volberg and Borichev for having made sure that the material got to me in time for me to be able to include the following account. The development given below is based on the methods worked out in §D of Chapter VII, and familiarity with that § on the part of the reader is assumed. In order to save space and avoid repetition, we will refer to §D frequently and use the symbols employed there whenever possible. 1.

Brennan's improvement, for M(v)/v1/2 monotone increasing

Let us return to the proof of the theorem in §D.6 of Chapter

VII, starting from the place on p. 359 where and the weight w(r) = exp(-h(log(1/r))) were brought into play. We take over the notation used in that discussion without explaining it anew. What is shown by the reasoning of pp. 359-73 is that unless F(ei9)

M(v)/v112 is increasing

1

571

vanishes identically,

log I F(ei') I d9 > - oo f.

provided that

h(1;) > const. -(1+a)

with some 6 > 0 as l; -- 0, and that

log h(g) d = oo 0 Ja

for small a> 0. Brennan's result is that the first condition on h can be replaced by the requirement that be decreasing for small > 0. (The second condition then obviously implies that oo as --> 0.) Borichev and Volberg made the important observation that Brennan's result is yielded by Volberg's original argument. To see how this comes about, we begin by noting that in §D.6 of Chapter VII, no real use of the property const. is made until one comes to step 5 on p. 369.

Up to then, it is more than enough to have h(g) > const. -` with some c > 0 together with the integral condition on log Step 5 itself, however, is carried out in rather clumsy fashion (see p. 370). The reader was probably aware of this, and especially of the wasteful manner of using that step's conclusion in the subsequent local estimate of cw(E, z) (pp. 370-2). At the $41/(1- IC 1)) dw(l;, p) was used where its top of p. 372, the smallness of smallness in relation to 1/(1 - p) would have sufficed! Instead of verifying the conclusion of step 5, let us show that the quantity f dw(C, pro)

(1-p)

1-1cl ,P P

can be made as small as we please for p sufficiently close to 1 chosen according to the specifications at the bottom of p. 368, under the assumption divergent. that l;h(l;) decreases, with the integral of log

The original argument for step 5 is unchanged up to the point where the relation (*)

h I log J" \

I

I

I dw(i;, p) < const. + (h(log (1/p2)))" 111

is obtained at the top of p. 370; here it can be chosen at pleasure in the interval (0,1), the construction following step 3 (pp. 365-6) and subsequent

572

Addendum. Improvement of Volberg's theorem

carrying out of step 4 being in no way hindered. Write now

P(c) = under the present circumstances P(1;) is decreasing for small > 0. Since yo, recall, lies in the ring { p2 < I C I < 11, we then have, for p near 1, doi(t;, p) Yp

1-1c1

<2

dco(l;, P)

2 1))

Yolog(1/IC1)

2 P(2log(1/p))

dw(t;, p)

",P(log(1/ICI)) h(log (1/1 C 1))dw(C, p). vo

Referring to (*), we see that the last expression is

5

P(21og(1/p)){const. + (h(2log(1/p)))"}.

Here, the monotoneity of P(i;) makes it tend to oo for -0; otherwise f o log h() di; would be finite for small a > 0 as already remarked. The function h(i;) also tends to oo for --> 0, so, for p close to 1 the preceding quantity is 3 3I h(2 log (1/p))1" = P(2log(1/P)) (log(1/PZ))"

3

(1-p)"

We thus have Jdco((,P) yo

l - ICI

< 3(1-p)-" = 0(1/(1-P))

for values of p tending to I chosen in the way mentioned above, and our substitute for step 5 is established. This, as already noted, is all we need for the reasoning at the top of p. 372. The local estimate for co(E, p) obtained on pp. 370-2 is therefore valid, and proof of the relation

JlogIF(ei)Id9 > -oo ft

is completed as on pp. 372-3. It may well appear that the argument just made did not make full use of the monotoneity of h(g). However that may be, this requirement does not seem capable of further significant relaxation, as we shall see in the next two articles. At present, let us translate our conclusion into a result involving the majorant M(v) figuring in Volberg's theorem (p. 356).

I

M(v)/v112 is increasing

573

In the statement of that theorem, two regularity properties are required of the increasing function M(v) in addition to the divergence of Y_' M(n)/n2,

namely, that M(v)/v be decreasing and that

M(v) > const. v" for large v, where a> 1/2. The first of these properties is (for us) practically equivalent to concavity of M(v) by the theorem on p. 326. The concavity is needed for Dynkin's theorem (p. 339) and is not at issue here. Our interest is in replacing the second property by a weaker one. That being the object, there is no point in trying to gild the lily, and we may as well phrase our result for concave majorants M(v). Indeed, nothing is really lost by sticking to infinitely differentiable ones with M"(v) < 0 and M'(v) -+ 0 for v --> oo,

as long as that simplifies matters. See the theorem, p. 326 and the subsequent discussion on pp. 328-30; see also the beginning of the proof of the theorem in the next article. With this simplification granted, passage from the result just arrived at to one stated in terms of M(v) is provided by the easy Lemma. Let M(v) be infinitely differentiable for v > 0 with M"(v) < 0 and M'(v) -> 0 for v --> oo, and put (as usual) sup (M(v) - vl ). v>o

Then i;h(i;) is decreasing for small > 0 if and only if M(v)/v112 is increasing for large v.

Proof. Under the given conditions, when 1; > 0 is sufficiently small, h(1;) = M(v) - vl; for the unique v with M'(v) = g by the lemmas on pp. 330 and 332. Thus, M'(v)h(M'(v)) = M(v)11 '(v) - v(M,(v))2,

so, since M'(v) tends monotonically to zero as v -* oo, h(1;) is decreasing for small > 0 if and only if the right side of the last relation is increasing for large v. But dv

(M(v)M'(v) - v(M'(v))2) = M"(v)M(v) - 2vM"(v)M'(v)

_ -2v312M"(v)dv(M(2)). Since M"(v) < 0, the lemma is clear.

Referring now to the above result, we get, almost without further ado,

574

Addendum. Improvement of Volberg's theorem

the

Theorem (Brennan). Let M(v) be infinitely differentiable for v > 0, with M"(v) < 0, M(Z)

increasing for large v,

and 00

Y M(n)/n2 = oo. 1

Suppose that an eine

F(e's) ao

is continuous, with Ia"I < const.e-M(l"h)

forn<0.

Then, unless F(ei,) vanishes identically,

logIF(e'1)Id1 > - oo. f.

Indeed, this follows directly by the lemma unless limy.. M'(v) > 0. Then, however, the theorem is true anyway - see p. 328. 2.

Discussion

Brennan's result really is more general than the theorem on p. 356. That's because the hypothesis of the former one is fulfilled for any

function F(ei9) satisfying the hypothesis of the latter, thanks to the following

Theorem. Let M(v), increasing and with M(v)/v decreasing, satisfy the condition Y_i M(n)/n2 = oc and have M(v) > const. v +a for large v, where 6 > 0. Then there is an infinitely differentiable function M0(v), with M"(v) < 0,

M0(v) < M(v) for large v, M0(v)/v112 increasing, and Ei Ma(n)/n2 = oo. Proof. By the theorem on p. 326 we can, wlog, take M(v) to be actually concave. It is then sufficient to obtain any concave minorant M,k(v) of M(v)

2 Discussion

575

with M(v)/v112 increasing and fl (M,k(v)/v2)dv divergent, for from such a

minorant one easily obtains an M0(v) with the additional regularity affirmed by the theorem. The procedure for doing this is like the one of pp. 229-30. Starting with

an M*(v), one first puts M1(v) = M,(v) + v112 and then, using a W. function qp(T) having the graph shown on p. 329, takes

Mo(v) = c

M 1(v -

dT

0f,

for v > 1 with a suitable small constant c. This function M0(v) (defined in any convenient fashion for 0 < v < 1) is readily seen to do the job. Our main task is thus the construction of an M*(v). For that it is helpful to make a further reduction, arranging for M(v) to have a piecewise linear graph starting out from the origin. That poses no problem; we simply replace our given concave function M(v) by another, with graph consisting of a

straight segment going from the origin to a point on the graph of the original function followed by suitably chosen successive chords of that graph. This having been attended to, we let R(v) be the largest increasing minorant of M(v)/v112 and then put M*(v) = v1I2R(v); this of course makes M,k(v)/v1/2 automatically increasing and M*(v)'< M(V)-

Thanks to our initial adjustment to the graph of M(v), we have M(v)/v112 --* 0 for v -* 0. Hence, since M(v) >, const. v++' for large v, R(v)

must tend to oo for v - oo, and coincides with M(v)/v'12 save on certain disjoint intervals (ak, I'k) MOO (Xkl/2

(0, cc) for which

= R(V) = M(#k), #1/2

ak 1< V < F'k.

Concavity of M,k(v) follows from that of M(v). The graph of M,(v) coincides with that of M(v), save over the intervals (ak, fik), where it has concave arcs (along which M,(v) is proportional to v1/2), lying below the corresponding arcs for M(v) and meeting those at their endpoints. The former graph is thus clearly concave if the other one is. Proving that Y_1 M,k(n)/n2 = oo is trickier. There would be no trouble at all here if we could be sure that the ratios f3k/ak were bounded, but we cannot assume that and our argument makes strong use of the fact that

6 > 0 in the condition M(v) > const. v We again appeal to the special structure of M(v)'s graph to argue that the local maxima of M(v)/v112, and hence the intervals (ak,flk), cannot accumulate at any finite point. Those intervals can therefore be indexed

576

Addendum. Improvement of Volberg's theorem

from left to right, and in the event that two adjacent ones should touch at their endpoints, we can consolidate them to form a single larger interval and then relabel. In this fashion, we arrive at a set-up where

0 < al < 91 < a2 < #2 <

,

with M,(v) = M(v) outside the union of the (perhaps new) (ak, flk), and

(/3)1 /2

112

V

=

Q

11'I(tk=

for ak

v

Ca k

It is convenient to fix a fo with 0 < /0 < al . Then, since M(v)/v decreases, M(a1) < (a1//3o)M(/3o), so, by the preceding relation, 1/2

/3

M(N1) =

al

M(al) 5

N1

1/2 al

((Xl

Qo

M(fl0)

In like manner we find first that M(a2) 5 (a2//31)M(f 1) and thence that M(F'2) S (l32/a2)1/2(a2/N1)M(f1) which, substituted into the previous, yields

21/2

(fl)l/2(fl)

Q

M(f32)

(a2)

(0ti

N1

f30M(/30)

Continuing in this fashion, we see that #n)1/2

M($)

/fan

1

Rn-1 an-1

\an

(al

Nn-2

o

Now by hypothesis, M(ln) > C = 1. Use this with the relation just found and then divide the resulting inequality by a +a, noting that an

Nn-1 an-1

/'1 al

n-1 an-1 fl.-2 fl.-1

a1 #0

_ N

One gets

0

a (P±) "2

al

C#n I'n-1 an-1 an

/'n-1 fl n-1

an

I'n-lan-1

M(fi0)

...(!Y '2

#1 al p0

I'0 1/2+a

061

After cancelling (fn/an)1/2 from both sides and rearranging, this becomes I'nl'n-1 anan-1

N1 al

a

<

an an-1 a

Nn-1Pn-2

al #-aMoo) a

a0) /

+a

PO

There is of course no loss of generality here in assuming 6 < 1/2. The last

2 Discussion

577

formula can be rewritten log I

1-26

k Isc+

log Gk- ak

\ak/

k=1

k=1

1

where c = (116) log (M(fio)/fo'z+a) is independent of n, and this estimate

makes it possible for us to compare some integrals of M(v)/vz over complementary sets. Since M(v)/v is decreasing, we have M(v) dv v

ffl,."-I

a dv =

M(an) an

ffln - 1

M(an) log an

v

an

fn-1

and at the same time,

f M(y) z dv <

J

=

M(an) f Q dv

v

J

an

a

v

M(an) an

log

/'n an

From the second inequality,

fInM(V) -

n=1 a

dv \

V2

11'r(an) log,n n=1

an

an

and partial summation converts the right side to

Ni (M(an)

-

M(;+ 1))

an

n=1 {(

an+1

log Ek

I k=1

+

ak

M(aN) aN

log 1k k=1

ak

The ratios M(an)/an are, however, decreasing, so we may apply the estimate

obtained above to see that the last expression is 1

1<

M(an)

Y-

n=1

-

M((Xn+1) l J 1 - 28 an+1

an

+ M(aN aN

J 1 26

1- 26 N 28

y log Qak + C /'k-1

k=1

tog

k=1

Qak N k-1

+c

which, by reverse summation by parts, boils down to

an1

1 - 26 N M(a) n 26

n=1

log

a

P.-

+c

M(a1)

al

This in turn is 1 - 26 N C'" M(v) Y jQn vzdv + c M(a1) 26

n=1

a1 -, by the first of the above inequalities, so, since M(v) = M,(v) on each of

578

Addendum. Improvement of Volberg's theorem

the intervals [fln _ 1,

we have finally

? dv+c M(a1) 2 dv< 1-2S2S NJ ('M (v) al

N

fM(v)

n=1

V

a^

Adding n=1

n=1 #nV

(M*(v)/v2) dv to both sides

J

(M(v)/v2) dv =

J

n=1

Bn

t

of this relation one gets (a fortiori!) MZV)

dv < c M«a l) +

2S feo

J fl o

My (v) dv,

and thence ffl.M(V)

dv

cM(al) +

2SJea

My

(v)dV.

In the present circumstances, however, divergence of Ei M(n)/n2 is equivalent to that of the left-hand integral and divergence of E1 M*(n)/n2 equivalent to that of the integral on the right. Our assumptions on M(v) thus make Y_° M*(n)ln2 = oo, and the proof of the theorem is complete.

The second observation to be made about Brennan's theorem is that its monotoneity requirement on M(v)/v'12 is probably incapable of much further relaxation. That depends on an example mentioned at the end of Borichev and Volberg's preprint. Unfortunately, they do not describe the construction of the example, so I cannot give it here. Let us, in the present addendum, assume that their construction is right and show how to deduce from this supposition that Brennan's result is close to being best possible in a sense to be soon made precise. The example of Borichev and Volberg, if correct, furnishes a decreasing function h(g) with 1 and f o log d = oo together with F(z), bounded and 16 , in { I z I < 11 and having the non-tangential boundary value F(ei9) a.e. on { I z I = 11, such that 8F(z) Of

5 exp\\\l -h\I loglZl

while

JlogIF(ei)Ida = -00 although F(ei9) is not a.e. zero.

I

I

J JJJ

for I z I < 1,

2 Discussion

579

The procedure we are about to follow comes from the paper of Joricke and Volberg, and will be used again to investigate the more complicated situation taken up in the next article. In order that the reader may first see its main idea unencumbered by detail, let us for now make an additional assumption that the function F(z) supplied by the Borichev- Volberg construc-

tion is continuous up to I z I = 1. At the end of the next article we will see that a counter-example to further extension of the L, version of Brennan's result given there can be obtained without this continuity. Assuming it here enables us to just take over the constructions of §D.6, Chapter VII. The present function F(z) is to be subjected to the treatment applied to the one thus denoted in §D.6, beginning on p. 359. We also employ the symbols

w(r) =

exp(-h(log')), r

0, B, 0, fl, &c with the meanings adopted there. Starting with F(z), we construct a continuous function g(ei9) on {IzI = 1} and a concave increasing majorant M(v) having the following properties: g(e,9)

(i) (ii)

(iii) (iv)

J

# 0,

log Ig(e")Id9 = - oo, M(n)/nz = oo,

M(v)/v'/2 % 2, x

(v)

g(e'9) - E an ein9

with I an I

<, const. a mr n for n < 0.

00

It is clear from this how close Brennan's result comes to being best possible

provided that the above assumptions are granted. 1, The weight w(r) we are now using is decreasing, and, since goes to zero rapidly enough for the reasoning followed in steps 2 and 3 of §D.6, Chapter VII to carry over without change. But the argument made for step 1 on p. 361 requires modification. Here, since F(z) is continuous on the closed unit disk and 0 0 on its circumference, there is a non-empty open arc I of that circumference on which I F(ei9) I is bounded away from zero. Then, because w(r) -) 0 for r -+ 1, the open set &' must have a component - call it (,-' - abutting on I.

If, at the same time, B contained a non-void open arc J of the unit circumference, we would have 00'r J = 0. In that event one could reason

580

Addendum. Improvement of Volberg's theorem

with the analytic function 4)(z) as at the bottom of p. 362, because 11(C) I <, const. w(I C I) on 8(9' n { I C I < 11. In that way, one would find that b(z) - 0 in (9', making F(ei9) - 0 on I, a contradiction. Hence no such arc as J can exist. Once steps 1, 2 and 3 are carried out, we fix any p, 0 < p < 1 and take

the connected set 52 = f2(p) c {p < I z I < 1 } described at the top of p. 363. As pointed out on p. 364, 852 includes the whole unit circumference. Fix now a zo e 52. Given an integer n > 0, let us apply Poisson's formula to the function z"I(z), harmonic (since analytic!) in S2 and continuous up to 852. We get ('

zo) an

where, as usual, wn( , ) is harmonic measure for 0. The boundary 852 consists of the unit circumference together with

y = 852n{IzI < l}, so the last relation can be rewritten fn

efns(D(e(s) dwn(e°s zo)

= zo D(zo) - J

n

zo). y

Let us first examine the right side of this formula. With log Iol=

o > 0, the first term on the right has modulus I'D(zo) I e-"4°

Concerning the second term, we recall that by the construction of (9, I I

) I < const. w(I i; I) on y, including on any arcs thereof lying on

{ I C I = p} and in (9, as long as the constant is chosen large enough. Therefore,

writing

M(v) = inf

i;v)

4>o

we have, since w(ICI) = exp(-h(l;)) with I C"F(C) I< const. e- M("),

=log(1/Il I),

C E Y.

Harmonic measure of course has total mass 1. Our second term is hence 5 const. a-M(") in magnitude, and we find that altogether, for n > 0, emns I (eis) down(ei9,

zo)

It will be seen presently that a-'(n) dominates for large n, so that the latter term can be dropped from this last relation. On account of that,

2 Discussion

581

we next turn our attention to M(v). This function is concave by its definition, and, since h(g) >, 1/c, easily seen to be > 2v1/2 and thus enjoy property (iv) of the above list. Because is decreasing and fo log h(1;) dl; = co,

we have J° (M(v)/v2) dv = oo by the theorem on p. 337. That, however, implies that Y_i M(n)ln2 = co, which is property (iii). We look now at the measure t(ei)dwn(ei9, z0) appearing on the left in the preceding relation. In the first place, dwn(ei9, zo) is absolutely continuous

with respect to d9 on { I t' I =1 }, and indeed S C d9 there, the constant C depending on zo. This follows immediately by comparison of dwn(ei9,zo) with harmonic measure for the whole unit disk. We can therefore write D(ei9)dwn(ei9 zo) = g(ei9)d9

with a bounded function g, and have just the moduli of 2ng(ei9)'s Fourier coefficients (of negative index) standing on the left in the above relation.

In fact, dwn(ei9, zo) has more regularity than we have just noted. The derivative dwn(ei9, zo)/d9 is, for instance, strictly positive in the

interior of each arc Ik of the unit circumference contiguous to B's intersection therewith. To see this one may, given Ik' construct a very shallow sectorial box 5 in the unit disk with base on Ik and slightly shorter than the latter. A shallow enough 5 will have none of 8S2 in its interior

since S2 abuts on Ik. One may therefore compare dwn(ei9, z) with harmonic measure for 5 when zed and ei9 is on that box's base, and an application of Harnack then leads to the desired conclusion. From this we can already see that Ig(ei9)I is bounded away from zero inside some of the arcs Ik, for instance, on the arc I used at the beginning of this discussion. But there is more - g(ei9) is continuous on the unit circumference. That follows immediately from four properties: the continuity of D(ei9), its vanishing for ei9eB, the boundedness of dwn(ei9, zo)/d9,

and, finally, the continuity of this derivative in the interior of each arc Ik contiguous to B n { I I = 11. The first three of these we are sure of, so it suffices to verify the fourth. For that purpose, it is easiest to use the formula dwn(ei9, zo)

d9

dwo(ei9 zo) d9

-

dws(ei9, ) dwn(", zo), d9

where we( , zo) is ordinary harmonic measure for the unit disk A (cf. p. 371). For ei9 moving along an arc Ik, dwo(ei9, C)/d9 = (1- I K I2)/2it K K - e1912

varies continuously, and uniformly so, for t; ranging over any subset of A

Addendum. Improvement of Volberg's theorem

582

staying away from ei9. Continuity of dwn(ei9, z0)/d9 can then be read off from the formula since y has no accumulation points inside the I, The function g(ei9) is thus continuous, in addition to enjoying property (i) of our list. Verification of properties (ii) and (v) thereof remains. Because dwn(ei9, zo)/d9 < C and I I (ei9) l lies between two

constant multiples of IF(ei9)j, property (ii) holds on account of the analogous condition satisfied by F and the relation of g(ei9) to F(ei9). Passing to property (v), we note that an earlier relation can be rewritten ein9g(ei9)d9

const. (e-"40 + e-M(")),

n > 0.

M(v)/v eventually decreases and tends to a limit 1 > 0 as v -* oo. Were 1 > 0, the right side of the inequality just written would be const. a-"'° with 10 = min 0. Such a bound on the left-hand integral would, with property (ii), force g(ei9) to vanish identically - see the bottom of p. 328. Our g(ei9), however, does not do for large n. The right that, so we must have l = 0, making M(n) < side of our inequality can therefore be replaced by const. a-M("), and property (v) holds. The construction is now complete. By concavity of M(v),

It is to be noted that the only objects we actually used were the with its specified properties and I(z), analytic in a certain domain & g { z < 11 land continuous up to satisfying function

/

I O(C) I

< const. exp

(- h ( log

l

l

)

on 0(9 r K l < 11 and II

J > 0 on

some arc of { I I = 1 } included in 8(9. I have a persistent nagging feeling

that such functions and bi(z), if there really are any, must be lying around somewhere or at least be closely related to others whose constructions are already available. One thinks of various kinds of functions meromorphic in the unit disk but not of bounded characteristic there; especially do the ones described by Beurling at the eighth Scandinavian mathematicians' congress come back continually to mind. This addendum, however, is already being written at the very last moment. The imminence of press time leaves me no opportunity for pursuing the matter. 3.

Extension to functions F(ei9) in L1(-ir, n).

The theorem of p. 356 holds for L1 functions F(ei9) not a.e. zero, as does Brennan's refinement of it given in article 1 above. A procedure

for handling this more general situation (absence of continuity) is worked out in the beautiful Mat. Sbornik paper by Joricke and Volberg. Here we

3

F(e19) in L1(-ir, ir)

583

adapt their method so as to make it go with the development already familiar from §D.6, Chapter VII, hewing as closely as possible to the latter.

Our aim is to show that

J1ogIF(eId 9 > for any function F(ei9) e L,(-7r,-r) not a.e. zero and satisfying the hypothesis of Brennan's theorem. Let us begin by observing that the treatment of this case can be reduced to that of a bounded function F. Suppose, indeed, that co

F(ei9) - Y an ei"9 -00

belongs to L1, with I an I <, const. e-M(IfI) for n < 0. The series En <0 anem9

is then surely absolutely convergent, so 00

Y a nein9 0

is also the Fourier series of an L1 function, which we denote by F+(ei9) (this belongs in fact to the space H1). For I z I < 1, put 00

F+(z) = Y az"; 0

for this function, analytic in { I z I < 11, we have (Chapter II, §B!), a.e. as z -L-+ei9.

F+(z) -, F+(ei9)

Using the integrable function log+ IF+(ei9)I > 0, we now form n

b(z) =

I

i9

e19±zlog+IF+(ei9)Idy9,

_n

analytic and with positive real part for I z I < 1. According to the third theorem and scholium of §F.2, Chapter III, b(z) tends for almost every 9 to a limit b(ei9) as z -L--+ ei9, with

91b(ei9) = log+ IF+(ei9)I a.e.

A standard extension of Jensen's inequality to H1 also tells us that log IF+(z)I < 91b(z),

I zI < 1

(cf, pp. 291-2 where this was proved and used for z = 0). We next perform the Dynkin extension (described on pp. 339-40) on the continuous function F_(e'9) _

anein9

584

Addendum. Improvement of Volberg's theorem

This gives us F_(z), W. in the unit disk and continuous (hence bounded!) up to its boundary, with

/

/

const. exp l- h I log

l l

l

I

),

IzI < 1,

a ai(z) I

where, in the present circumstances, h(1;) = sup (M(v)/2 - vl;) v>O

(see remark 2, p. 343). As usual, we write

w(r) = exp - h I log r

\

\

I

I;

JJ JJ

then, putting

F(z) = F -(z) + F+(z) for IzI < 1, we have aF(z) I

\ cont. w(I z 1)

aZ

there, and F(z) --* F(ei9)

a.e. for z -/-+ e's.

The bounded function spoken of earlier is simply F0(z) = e-b(z)F(z).

It is bounded in the unit disk by one of the previous relations; another

tells us that F0(z) has a non-tangential boundary value Fo(ei9) = F(ei9) exp (- b(ei9)) equal in modulus to I F(ei9) I/max (I F+(e19)1,1) at almost

every point of the unit circumference. Then, since F(ei9)EL, is not a.e. e_b(z), zero, neither is Fo(ei9). We note finally that by analyticity of aFO(z)/az = e-b(z)3F(z)/az, making 5Fo(z)

az

5 const. w(Izl),

IzI < 1 .

1

Given that M(v) satisfies the hypothesis of Brennan's theorem, our function h(l;) enjoys the two properties used in the first part of article 1, namely, that lh(l;) decreases and that fo log dl; = oo for small a > 0. If, now, we can deduce from these together with the preceding relation that

F(ei9) in L,(-iv,ir)

3

585

the bounded function F0(z), not a.e. zero for I z I = 1, satisfies JlogIFo(&5)1d19 > - oo,

w e will certainly have the same conclusion for

logIF(e'9)I = logIFo(e19)I + log+IF±(e;9)I The rest of our work deals exclusively with F0(z). In order to stay as close as possible to the notation of §D.6, Chapter VII, we denote the bounded function F0(z) by F(z) from now on. Using this new F(z), we f i r s t form the sets B c { I z I ,< 11 and ( 9 c { I z I < 1 } as on pp. 359-60, and then the function (D (z) introduced on p. 360. The latter, analytic in (9, is actually defined on the whole unit disk, and has there at least as much continuity as F(z) besides lying in modulus between two

constant multiplies of IF(z)l. It has, in particular, a non-tangential boundary value'1(ei9) a.e. on the unit circumference, and this does not vanish a.e. The construction of B ensures that

Idi(t')I 5 const.w(ICI)

on 8(9n {ICI < 1}

(indeed, on B), and our task amounts to showing that

JlogI(e)Id>

- co

rz

on account of these properties. What makes the present situation more complicated than the one studied in §D.6 of Chapter VII is that cF(z) need no longer be continuous up to the whole unit circumference. This causes the notion of abutment introduced

on p. 348 to be less useful here for the examination of our set 0 than it was in §D.6, and we have to supplement it with another, that of fatness. The latter, based on the famous sawtooth construction of Lusin and Privalov, helps us to take account of F(z)'s non-tangential boundary behaviour. To describe what is meant by fatness, we need to bring in a special kind of domain together with some notation; both will also be used further on. Corresponding to each point e'°` on the unit circumference, we have an open set S., consisting of the z with 1 /2 < I z I < 1 lying in the open 60° sector having vertex at e'a and symmetric about the radius from 0 out to that point. Given any subset E of {Il I = 1} we then write

SE = U SQ. e1 .E

586

Addendum. Improvement of Volberg's theorem

It is evident that if we take any SE and a p, 1/2 < p < 1, the intersection

SEn{p
SEkn{p
SEn{p
Now we can state the Definition. An open subset ill of the unit disk is called fat if it contains a sawblade biting on a closed E c { I C I = 1 } with I E I > 0. In that circumstance we also say that 0& is fat at E.

Equipped with these tools, we endeavour to investigate the set 0 according to the procedure of §D.6, Chapter VII. In this, some modifications are necessary; we have, in the first place, to skip over step 1 (p. 361). Then, taking p, 1/2 < p < 1, we construct a set S1(p), proceeding differently, however, than as we did on pp. 361-3.

There is, by the properties of b(z), a closed subset E0 of the unit circumference, I E0 I > 0, such that, for the non-tangential boundary values cli(l;), we have, wlog, I (D(() I > 1,

e E0.

Egorov's theorem enables us to in fact pick E0 so as to have 11(z) I > 1 for zESE( with p' < IzI < 1 when p' > p is sufficiently close to 1. But the construction of B and 0 makes I F(z) I < const. w(I z I) on B, hence on { I z I < 11 ' V. Therefore, since w(r) -* 0 for r --)1, we must have

SEon{p' 0; a suitable closed subset E of E' then has

3

F(ei9) in L1(- 7r, ir)

587

I E I > 0, and there is a sawblade of depth 1- p' biting on E and contained in d. We now take n(p) as the component of 0 n {p < I z I < 1 } including that sawblade; fl(p) is fat at E.

For the present set fl(p) there is a substitute for step 2 of p. 362: Step 2'. .S2(p) includes the whole unit circumference.

This we establish by reductio ad absurdum. Let us write KI for S2(p), and put

y = a)n{IzI<1},

r=aS2_y; F is thus the part of as2 lying on the unit circumference. Assume that there

is on the latter a non-empty open arc J with J n r = 0; we will then deduce a contradiction. For that it is quicker to fall back on the device used in the second half

of article 2 than to adapt Volberg's theorem on harmonic measures (p. 349) to the present situation. Fixing zoeS2, we can say that zocp(zo)

=J

"I

dwn(C, zo)

for n > 0,

an

whence eins(D(e;,v)dcon(ei9,z0)

= zoO(zo)

Sr

-J

dwn(C, zo),

n >, 0.

Y

Here we are using Poisson's formula for the bounded function l;"cD(C) harmonic (even analytic) in S2 and continuous up to y, but not necessarily up to F, where it is only known to have non-tangential boundary values a.e. Such use is legitimate; we postpone verification of that, and of a corresponding version of Jensen's inequality, to the next article, so as not to interrupt the argument now under way.

As in article 2, dwn(ei9, zo) is absolutely continuous and <, C d8 on r, and we obtain a bounded measurable function g(ei9) by putting g(e's) =

(D(e")dcon(ei9'z0)

d9

for ei9 F r

and (here!) taking g(ei9) to be zero outside F. From the preceding relation

588

Addendum. Improvement of Volberg's theorem

we then see, as in article 2, that fn

ein9g(e") d0

e-M1(n))

-n

for n > 0, where o > 0 and

Ml(v) = inf

cv).

4>o

This function is increasing and concave, so the right side of the last inequality can be replaced by const. a-MZ0) for large n, with M2(n) equal o ) or else to M,(n). In either (M,(v)/v) either to ion (in case event, M2(n) increases and Y_ i M2(n)/n2 = oo on account of the properties of (See the theorem of p. 337 - M,(n) is actually equal to M(n)/2 in the present set-up.) Now we can apply Levinson's theorem, since g(ei9) vanishes on the arc J. The conclusion is that g(ei9) __ 0 a.e. But g(ei9) does not vanish a.e. Indeed, S2 contains a sawblade.9 biting on a

closed set E, I E I > 0, where 11(ei9) I > 1. Thence,

Ig(ei9)Id9 = f E

zo) >, wn(E,zo)

JE

Harnack's theorem assures us that the quantity on the right is > 0 if, for some z, Ee, wn(E, z,) > 0. However, by the principle of extension of domain, wn(E,z,) > w,(E,z,). At the same time, M is rectifiable, so a conformal mapping of & onto the unit disk must take the subset E of M, having linear measure > 0, to a set of measure > 0 on the unit circumference. (This follows by the celebrated F. and M. Riesz theorem; a proof can be found in Zygmund or in any of the books about HP spaces.) We therefore have w,,(E, z,) > 0, making wn(E, zo) > 0 and hence, as we have seen, JEIg(ei9)Id9 > 0. Our contradiction is thus established. By it we see that the arc J cannot exist, i.e., that F is the whole unit circumference, as was to be shown.

With step 2' accomplished, we are ready for step 3. One starts out as on p. 363, using the square root mapping employed there. That gives us a domain 52,/, certainly fat at a closed subset E", of E, (the image of E under our mapping), with I E" I > 0 (recall the earlier use of Egorov's theorem). Thereafter, one applies to QI/ the argument just made for Q in doing step 2'. The weight w,(r) is next introduced as on p. 365, and the sets B, and 01 constructed (pp. 365-6). After doing steps 2' and 3 again with these objects, we come to step 4.

F(ei9) in L1(-ic,n)

3

589

Joricke and Volberg are in fact able to circumvent this step, thanks to a clever rearrangement of step 5. Here, however, let us continue according

to the plan of §D.6, Chapter VII, for the work done there carries over practically without change to the present situation. What is important for step 4 is that a t', I1; I = 1, not in B must, even here, lie on an arc of the unit circumference abutting on (9. Such a CAB must thus, as on p. 367, have a neighborhood VV with

V,n{IzI
log 14)(C) I dw(C, p) an(v2)

(notation of p. 369) available in the present circumstances, where continuity of 1(z) up to { I( I = 1 } may fail. The legitimacy of this will be established

in the next article; granting it for now, we may proceed exactly as at the beginning of article 1. From here on, one continues as on pp. 370-2, and reaches the desired conclusion that J' n log l '(ei9) I d9 > - oo as on p. 373, after one more application of our extended Jensen inequality. We thus arrive at the Theorem. Let F(ei9) e L1(- ir, 7C) not be zero a.e., and suppose that F(ei9) - Y, an e1n9 - ao

with l an l

<, const. e-M(InI),

n < 0.

Suppose that M(v) is concave, that M(v)/v'12 is increasing for large v, and

that 00

M(n)ln2 = cc. Then

J'logIF(e1)Id 9 > - oo.

590

Addendum. Improvement of Volberg's theorem

Remark. In their preprint, Borichev and Volberg consider formal trigonometric series Y, aneins 00

in which the an with negative n satisfy the requirement of the theorem, but the an with n > 0 are allowed to grow like eM(n' as n , oo. Assuming more regularity for M(v) (M(v) >, const. v" with an a < I close to I is enough), they are able to show that under the remaining conditions of the theorem, all the an must vanish if 0

lim inf f7r log Y_ an eins + -.0

oc' Y, anrn

d9 = -oo.

1

Before ending this article let us, as promised in the last one, see how

the example of Borichev and Volberg shows that the monotoneity requirement on M(v)/v1"2 cannot, in the above theorem at least, be relaxed to M(v)/v112 >, C > 0, even though continuity up to {ICI = l } should fail for the function F(z) supplied by their construction. The reader should refer back to the second part of article 2. Corresponding to the bounded function F(z) used there, no longer assumed continuous up to { I C I = 1) but having at least non-tangential boundary values a.e. on that circumference, one can, as in the preceding discussion, form the sets B, (9 and 92(p) and do step 2'. One may then form the function g(ei9) as in article 2; here it is bounded and measurable at least. The work of step

2' shows that g(ei9) is not a.e. zero, while properties (ii)-(v) of article 2 hold for it (for the last one, see again the end of that article). This is all we need. 4.

Lemma about harmonic functions

Suppose we have a domain 92 regular for Dirichlet's problem,

lying in the (open) unit disk A and having part of 892 on the unit circumference. As in the last article, we write

r=892n8A and y=892r A. For the following discussion, let us agree to call C, I l; I = 1, a radial accumulation point of 92 if, for a sequence {rn} tending to 1, we have rnC e SZ

for each n. We then denote by t' the set of such radial accumulation points, noting that F' IF with the inclusion frequently proper.

4 Lemma on harmonic functions

591

Lemma. (Joricke and Volberg) Let V(z), harmonic and bounded in 52, be continuous up to y, and suppose that lira V(() r-'1

r;en

exists for almost all CeI'. Put v(C) equal to that limit for such C, and to zero for the remaining CeF. On y, take v(C) equal to V(C). Then, for ze52,

V(z) =

v( t;) dcon(C, z).

fen

Proof. It suffices to establish the result for real harmonic functions V(z), and, for those, to show that V(z) S

zef2,

v(() an fan

since the reverse inequality then follows on changing the signs of V and v. By modifying v(C) on a subset of F having zero Lebesgue measure, we

get a bounded Borel function defined on 852. But on F, we have dwn(C, z) < C. I dC l (see articles 2 and 3), so such modification cannot alter

the value of fan v(C) dwn(2;, z). We may hence just as well take v(C) as a bounded Borel function (on a) to begin with. That granted, we desire to show that the integral just written is > V(z).

For this it seems necessary to hark back to the very foundations of integration theory. Call the limit of any increasing sequence of functions continuous on 892 an upper function (on aO). There is then a decreasing sequence of upper functions v(C) such that

J..

wnQ

z)

L

v(C) dwa(l;, z),

ze92.

Indeed, corresponding to any given ze52, such a sequence is furnished by a basic construction of the Lebesgue-Stieltjes integral, wn( , z) being a Radon measure on 8fl. But then that sequence works also for any other ze52, since dcon(C, z') <, C(z, z')dwn(C, z) (Harnack). Our inequality involving v and V will thus be established, provided that we can verify

V(z) < J

dwn(t;, z), an

ze52,

592

Addendum. Improvement of Volberg's theorem

for each n. Fixing, then, any n, we write simply w(C) for %(C) and put

W(z) =

J an

w(C) dwn(C, z)

for zefZ, making W(z) harmonic there. Our task is to prove that

V(z) 5 W(z),

zecL

It is convenient to define W(z) on all of S2 by putting

W(O = w(D,

t'e an.

At each (e& we then have lim inf W(z) 3 W(1') =-c ZEn

by the elementary approximate identity property of harmonic measure, since w(2;), as limit of an increasing sequence of continuous functions, satisfies lim inf w(l;) >, w(ho) c-so

for o easy.

cEan

The function W(z) enjoys a certain reproducing property in S2. Namely, if the domain -9 c f is also regular for Dirichlet's problem, with perhaps (and especially!) part of 0-9 on On, we have

W(z) = J

W(C)dow,(C,z)

for ze-9.

To see this, take an increasing sequence of functions fk(C,) continuous on ail and tending thereon, and let Fk(Z) =

fk(S)dwa(S,z), JL

zef.

Then the Fk(z) tend monotonically to W(z) in f by the monotone convergence theorem. That convergence actually holds on [I if we put Fk(C) = fk(t;) on ail; this, however, makes each function Fk(z) continuous on S2 besides being harmonic in n. In the domain -9, we therefore have

Fk(z) =

J

a Fk(C)

z)

for each k. Another appeal to monotone convergence now establishes the corresponding property for W. Fix any zoefZ; we wish to show that V(zo) < W(zo). For this purpose,

4 Lemma on harmonic functions

593

we use the formula just proved with -9 equal to the component f2r of n n { I z I < r} containing zo, where I zo I < r < 1. Because 0 is regular for

Dirichlet's problem, so is each f2,; that follows immediately from the characterization of such regularity in terms of barriers, and, in the circumstances of the last article, can also be checked directly (cf. p. 360). We write IF, = 8f2, n f2,

making IF, the union of some open arcs on { t; I = r}, and then take ,., Fr;

Yr = 8f2r

y, is a subset (perhaps proper) of y n{ I I S r}. The function V(z), given as harmonic in S2 and continuous up to y, is

certainly continuous up to 852,. Therefore, since V(() = v(() on y 2 y we have, for z e f2 V (z) =

f

v(() dwn. ((, z)

rr

+ J rr V (C) dwn.((, z).

At the same time, (b' by the reproducing property of W, W(z)

= J rr

W() dwnr((, z),

W() dwr,(C, z) +

z e f2r.

J

We henceforth write w,( , ) for con,( , ). Then, since on y, c 852, W(C) = w(C) is > v(1'), the two last relations yield W(z) - V(z) > frr (W(C)

-

for z e 52,. Our idea is to now make r

in this inequality.

For I I = 1, define

_

W(rl;) - V(r() 0 otherwise.

if rl' e IF,,

Since V(z) is given as bounded, the functions A,(t;) are bounded below. Moreover (and this is the clincher),

lim inf Ar(() > 0 a.e., r-'1

I C I =1.

That is indeed clear for the C on the unit circumference outside F' (the set of radial accumulation points of f2); since for such a C, rC cannot even belong to f2 (let alone to Fr) when r is near 1. Consider therefore a (C-17, and take any sequence of r < 1 tending to 1 with, wlog, all the r,,C in 52

594

Addendum. Improvement of Volberg's theorem

and even in their corresponding IFr,,. Then our hypothesis and the specification of v tell us that

V (r.C) - v(C), except when l; belongs to a certain set of measure zero, independent of

{rn}. For such a sequence {rn}, however, lim inf W(rnC) >, W(C) = w(C) n-co

as seen earlier, yielding, with the preceding,

lim inf Or (t;) > w(C) - v(t;) > 0. n- oo

The asserted relation thus holds on F' as well, save perhaps in a set of measure zero.

Returning to our fixed zoef2, we note that for (1 + Izo1)/2 < r < 1 (say), we have, on Fr, dwr(C, zo) < K l dC l

with K independent of r (just compare co,.( , ) with harmonic measure for { I z I
for these values of r, with 0 < p,(l;) < K (and p,(t;) = 0 for rl;OI ,

),

such that Jr

r(W(?) -

dw,(C, zo) = J

I dS 1

ici=i

Here the products A,(t;)rp,(l;) are uniformly bounded below since the 0,(l;) are. And, by what has just been shown, lim inf, ,(t;)rpr(t;)

r-1

0

a.e., I C I = 1.

Thence, by Fatou's lemma (!), lim inf J r- 1

I dt; I

> 0.

icl=1

We have seen, however, that when r> I zo I,

W(zo) - V(zo) is

the left-hand integral in the previous relation. It follows therefore that W(zo) - V(zo) '> 0, as was to be proven.

We are done.

4 Lemma on harmonic functions

595

Remark 1. When V(z) is only assumed to be subharmonic in f but satisfies otherwise the hypothesis of the lemma, the argument just made shows that

V(z) < J f v(C)dwn(l;,z)

for zefl.

an

Remark 2. In the applications made in article 3, the function V(z) actually has a continuous extension to the open unit disk A with modulus bounded, in A - fZ, by a function of z tending to zero f o r z ---+ 1. That extension also has non-tangential boundary values a.e. on 8A. In these circumstances the lemma's ad hoc specification of v(t;) on IF ' I" is superfluous, for the non-tangential limit of V(z) must automatically be zero at any C e IF - t' where it exists.

Remark 3. To arrive at the version of Jensen's inequality used in article

3, apply the relation from remark 1 to the subharmonic functions VM(z) = log+ I Mcb(z)1, referring to remark 2. That gives us max I log I b(z) I, log M

I < fan max (log I F(l;) I, log M)dwn(2C, z) J

for zec2. Then, since 14)(z)1 is bounded above, one may obtain the desired

result by making M - oo. Addendum completed June 8, 1987.

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Benedicks, M. Positive harmonic functions vanishing on the boundary of certain domains in I8". Arkiv for Mat. 18 (1980), 53-72. Benedicks, M. Weighted polynomial approximation on subsets of the real line. Preprint, Uppsala Univ. Math. Dept., 1981, l2pp. Bernstein, S. Sobranie sochineniL Akademia Nauk, USSR. Volume 1, 1952; volume II, 1954. Bernstein, V. Lecons sur les progres recents de la theorie des series de Dirichlet.

Gauthier-Villars, Paris, 1933. Bers, L. An outline of the theory of pseudo-analytic functions. Bull. AMS 62 (1956), 291-331. Bers, L. Theory of Pseudo-Analytic Functions. Mimeographed lecture notes, New

York University, 1953. Beurling, A. Analyse spectrale des pseudomesures. C.R. Acad. Sci. Paris 258 (1964),

406-9. Beurling, A. Analytic continuation across a linear boundary. Acta Math. 128 (1972), 153-82. Beurling, A. On Quasianalyticity and General Distributions. Mimeographed lecture notes, Stanford University, summer of 1961. Beurling, A. Sur les fonctions limites quasi analytiques des fractions rationnelles. Huitieme Congres des Mathematiciens Scandinaves, 1934. Lund, 1935,

pp.199-210. Beurling, A. and Malliavin, P. On Fourier transforms of measures with compact support. Acta Math. 107 (1962), 291-309. Boas, R. Entire Functions. Academic Press, New York, 1954. Borichev, A. and Volberg, A. Uniqueness theorems for almost analytic functions. Preprint, Leningrad branch of Steklov Math. Institute, 1987, 39pp.

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Brennan, J. Functions with rapidly decreasing negative Fourier coefficients. Preprint, University of Kentucky Math. Dept., 1986, l4pp. Carleson L. Estimates of harmonic measures. Annales Acad. Sci. Fennicae, Series A.I. Mathematica 7 (1982), 25-32. Cartan, H. Sur les classes de fonctions definies par des inegalites portant sur leurs derivees successives. Hermann, Paris, 1940. Cartan, H. and Mandelbrojt. S. Solution du probleme d'equivalence des classes de fonctions indefiniment derivables. Acta Math. 72 (1940), 31-49. Cartwright, M. Integral Functions. Cambridge Univ. Press, 1956. Choquet, G. Lectures on Analysis. 3 vols. Benjamin, New York, 1969. De Branges, L. Hilbert Spaces of Entire Functions. Prentice-Hall, Englewood Cliffs, NJ, 1968. Domar, Y. On the existence of a largest subharmonic minorant of a given function. Arkiv far Mat. 3 (1958), 429-40. Duren, P. Theory of HP Spaces. Academic Press, New York, 1970. Dym, H. and McKean, H. Gaussian Processes, Function Theory and the Inverse Spectral Problem. Academic Press, New York, 1976.

Dynkin, E. Funktsii s zadannoi otsenkoi 8 f /8z i teoremy N. Levinsona. Mat. Sbornik 89 (1972), 182-90. Functions with given estimate for 8 f /az and N. Levinson's theorem. Math. USSR Sbornik 18 (1972), 181-9. Gamelin, T. Uniform Algebras. Prentice-Hall, Englewood Cliffs, NJ, 1969. Garnett, J. Bounded Analytic Functions. Academic Press, New York, 1981. Garsia, A. Topics in Almost Everywhere Convergence. Markham, Chicago, 1970 (copies available from author). Gorny, A. Contribution a 1'etude des fonctions derivables d'une variable reelle. Acta Math. 71 (1939), 317-58. Green, George, Mathematical Papers of. Chelsea, New York, 1970. Helson, H. Lectures on Invariant Subspaces. Academic Press, New York, 1964. Helson, H. and Lowdenslager, D. Prediction theory and Fourier Series in several variables. Part 1, Acta Math. 99 (1958),165-202; Part II, Acta Math. 106 (1961), 175-213. Hoffman, K. Banach Spaces of Analytic Functions. Prentice-Hall, Englewood Cliffs, NJ, 1962.

Joricke, B. and Volberg, A. Summiruemost' logarifma pochti analiticheskol funktsii i obobshchenie teoremy Levinsona-Kartrait. Mat. Sbornik 130 (1986), 335-48.

Kahane, J. Sur quelques problemes d'unicite et de prolongement, relatifs aux fonctions approchables par des sommes d'exponentielles. Annales Inst. Fourier 5 (1953-54), 39-130. Kargaev, P. Nelokalnye pochti differentsialnye operatory i interpoliatsii funktsiami s redkim spektrom. Mat. Sbornik 128 (1985), 133-42. Nonlocal almost differential operators and interpolation by functions with sparse spectrum. Math. USSR Sbornik 56 (1987), 131-40. Katznelson, Y. An Introduction to Harmonic Analysis. Wiley, New York, 1968 (Dover reprint available). Kellog, 0. Foundations of Potential Theory. Dover, New York, 1953. Khachatrian, 1.0. 0 vzveshonnom priblizhenii tselykh funktsii nulevol stepeni mnogochlenami na deistvitelnoi osi. Doklady A.N. 145 (1962), 744-7. Weighted approximation of entire functions of degree zero by polynomials on the real axis. Soviet Math (Doklady) 3 (1962), 1106-10.

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Bibliography for volume I Koosis, P. Harmonic estimation in certain slit regions and a theorem of Beurling and Malliavin. Acta Math. 142 (1979), 275-304. Koosis, P. Introduction to Hp Spaces. Cambridge University Press, 1980. Koosis, P. Solution du probleme de Bernstein sur les entiers. C.R. Acad. Sci. Paris 262 (1966), 1100-2. Koosis, P. Sur l'approximation ponderee par des polynomes et par des sommes d'exponentielles imaginaires. Annales Ecole Norm. Sup. 81 (1964), 387-408. Koosis, P. Weighted polynomial approximation on arithmetic progressions of intervals or points. Acta Math. 116 (1966), 223-77. Levin, B. Raspredelenie kornei tselykh funktsii. Gostekhizdat, Moscow, 1956. Distribution of Zeros of Entire Functions (second edition). Amer. Math. Soc., Providence, RI, 1980. Levinson, N. Gap and Density Theorems. Amer. Math. Soc., New York, 1940, reprinted 1968. Levinson, N. and McKean, H. Weighted trigonometrical approximation on the line with application to the germ field of a stationary Gaussian noise. Acta Math. 112 (1964), 99-143. Lindelof, E. Sur la representation conforme d'une aire simplement connexe sur l'aire d'un cercle. Quatrieme Congres des Mathematiciens Scandinaves, 1916.

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Mandelbrojt, S. Series de Fourier et classes quasi-analytiques de fonctions. Gauthier-Villars, Paris, 1935. McGehee, 0., Pigno, L. and Smith, B. Hardy's inequality and the L' norm of exponential sums. Annals of Math. 113 (1981), 613-18. Mergelian, S. Vesovye priblizhenie mnogochlenami. Uspekhi Mat. Nauk 11(1956), 107-52. Weighted approximation by polynomials. AMS Translations 10 Set 2 (1958), 59-106. Nachbin, L. Elements of Approximation Theory. Van Nostrand, Princeton, 1967. Naimark, M. Normirovannye koltsa. First edition: Gostekhizdat, Moscow, 1956. First edition: Normed Rings, Noordhoff, Groningen, 1959. Second edition: Nauka, Moscow, 1968. Second edition: Normed Algebras. Wolters-Noordhoff, Groningen, 1972. Nehari, Z. Conformal Mapping. McGraw-Hill, New York, 1952.

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Phelps, R. Lectures on Choquet's Theorem. Van Nostrand, Princeton, 1966. Pollard, H. Solution of Bernstein's approximation problem. Proc. AMS 4 (1953), 869-75. Riesz, F. and M. Uber die Randwerte einer analytischen Funktion. Quatrieme Congres des Mathematiciens Scandinaves, 1916. Uppsala, 1920, pp. 27-44. [Note:

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Index

Akhiezer's description of entire functions arising in weighted approximation 160, 174 Akhiezer's theorems about weighted polynomial approximation 158ff, 424, 523

Akhiezer's theorems on weighted approximation by sums of imaginary exponentials 174, 424, 432, 445 approximation index M(A), Beurling's 275 approximation index MP(A), Beurling's 293 approximation, weighted 145ff, 385, 424 see also under weighted approximation Benedicks, M. 434ff Benedicks' lemma on harmonic measure for slit regions bounded by a circle 400 Benedicks' theorem on existence of a Phragmen-Lindelof function 418, 431

Benedicks' theorem on harmonic measure for slit regions 404 Bernstein approximation problem 146ff Bernstein intervals associated with a set of points on (0, co) 454ff Bernstein's lemma 102 Bernstein's theorem on weighted polynomial approximation 169 Beurling, A. and Malliavin, P. 550, 568 Beurling quasianalyticity 275ff Beurling quasianalyticity for. LP functions 292ff Beurling-Dynkin theorem on the Legendre transform 333 Beurling's approximation indices see under approximation index Beurling's gap theorem 237, 305

Beurling's identity for certain bilinear forms 484 Beurling's theorem about Fourier-Stieltjes transforms vanishing on a set of positive measure 268 Beurling's theorem about his quasianalyticity 276 Beurling's theorem on his LP quasianalyticity 293 boundary values, non-tangential 10, 43ff, 265, 269, 286ff

canonical product 21 Carleman's criterion for quasianalyticity 80 its necessity 89 Carleman's inequality 96 Carleson's lemma on linear forms 392, 398

Carleson's theorem on harmonic measure for slit regions 394, 404, 430 Cartan-Gorny theorem 104 Cauchy principal value, definition of 533 Cauchy transform, planar 320ff class of infinitely differentiable functions 79 its quasianalyticity 80 convex logarithmic regularization of a sequence 8311, 92ff, 104ff, 130, 226

de Branges' lemma 187 de Branges' theorem 192 discussion about 198ff density, of a measurable sequence 178 Dirichlet integral 479, 500, 510ff Dirichlet problem 251, 360, 387, 388 Dynkin's extension theorem 339, 359, 373 energy of a measure on (0, co) 562, 568

479ff, 549ff,

Index

601

bilinear form associated thereto

482, 487, 494ff, 508, 512ff, 551, 552, 553, 563, 566

formulas for 479, 485, 497, 512 positivity of 482, 493 entire functions of exponential type 15ff arising in weighted approximation 160, 174, 218, 219, 525 as majorants on subsets of It 555ff, 562, 564, 568 coming from certain partial fraction expansions 203ff, 205 see also under Hadamard factorization exponential type, entire functions of see preceding extension of domain, principle of 259, 289 301, 368, 372, 529, 531 extension of positive linear functionals 111ff, 116 extreme point of a convex w* compact set of measures 186ff

Fejer and Riesz, lemma of 281 function of exponential type, entire see also under entire functions function T(r) used in study of quasianalyticity 80ff

15ff

gap theorem, Beurling's 237 Gauss quadrature formula 134, 137ff Green, George, homage to 419-22 Green's function 400ff, 406, 407, 410, 418ff, 439, 479, 526ff, 547ff, 550

estimates for in slit regions 401, 439, 442, 548

symmetry of 401, 415, 418ff, 530 Green potential 479, 551, 552, 553, 560, 562, 563, 566

Hadamard factorization for entire functions of exponential type 16, 19, 22, 54, 56, 70, 201, 556, 561 157, 158, 184, 208, 375, 523

hall of mirrors argument

Hall, T., his theorem on weighted polynomial approximation 169 Hankel matrix 117 harmonic conjugate 46, 59, 61 existence a.e. of 47, 532, 537 see also under Hilbert transform harmonic estimation, statement of theorem on it 256 harmonic functions, positive, representations for in half plane 41 in unit disk 39 harmonic measure 251ff approximate identity property of 253, 261

boundary behaviour of 261ff, 265 definition of 255 in curvilinear strips, use of estimate for 355 in slit regions 385, 389ff, 394, 403, 404, 430, 437, 443, 444, 446, 522, 525ff, 530, 541, 545ff, 554, 562, 565

Volberg's theorem on

349, 353, 362, 364,

366

Harnack's inequality 254, 372, 410, 430 Hilbert transform 47, 61, 62, 63, 65, 532, 534, 538ff

Jensen's formula 2, 4, 7, 21, 76, 163, 291, 559

Kargaev's example on Beurling's gap theorem 305ff, 315 Kolmogorov's theorem on the harmonic conjugate 62ff Krein's theorem on certain entire functions 205 Krein-Milman theorem, its use 186, 199 Kronecker's lemma 119 Legendre transform h(1;) of an increasing function M(v) 323ff Levinson (and Cartwright), theorem on distribution of zeros for functions with real zeros only 66 general form of 69 use of 175, 178 Levinson's log log theorem 374ff, 376, 379ff

Levinson's theorem about Fourier-Stieltjes transforms vanishing on an interval 248, 347, 361 Levinson's theorem on weighted approximation by sums of imaginary exponentials 243 Lindelof's theorems about the zeros of entire functions of exponential type, statements 20, 21 Lindelof's theorem on conformal mapping 264 log log theorem see under Levinson Lower polynomial regularization W*(x) of a weight W(x), its definition 158 Lower regularization WA(x) of a weight W(x) by entire functions of exponential type -< A 175, 428 for Lip 1 weights

236

for weights increasing on [0, oo) 242 Lower regularizations WA,E(x) of a weight W(x) corresponding to closed unbounded sets E s 118 428

Markov-Riesz-Pollard trick 139, 155, 171,182,190

602

Index

maximum principle, extended, its statement 23 measurable sequence 178 Mergelian's theorems about weighted polynomial approximation 147ff Mergelian's theorems on weighted approximation by sums of imaginary exponentials 173, 174, 432 moment problem see under Riesz moment sequences definition of 109 determinacy of 109, 126, 128, 129, 131, 141, 143

indeterminacy of 109, 128, 133, 143 Riesz' characterization of 110 same in terms of determinants 121

representations for positive harmonic functions see under harmonic functions Riesz, F. and M. 259, 276, 286 Riesz' criterion for existence of a solution to moment problem 110, 121 Riesz' criterion for indeterminacy of the moment problem 133 Riesz-Fejer theorem 55, 556 simultaneous polynomial approximation, Volberg's theorem on 344, 349 slit regions (whose boundary consists of slits along real axis) 384, 386ff, 401, 402, 418, 430, 439, 441, 525ff, 540ff, 545, 553, 564, 568

see also under harmonic measure spaces Ww(0) and Ww(O+) 212 conditions on W for their equality 223, 226

Newton polygon 83ff non-tangential limit 11

weights W for which they differ 2296 244ff

Paley and Wiener, their construction of certain entire functions 100 Paley and Wiener, theorem of 31 Ll version of same 36 Phragmen-Lindelof argument 25, 405, 406, 553

Phragmen-Lindelof function

25, 386, 406, 407, 418, 431, 441, 525ff, 541, 555

spaces of functions used in studying weighted approximation, their definitions Ww(R)

145

W ,(O), `',(A), Ww(A+)

211

'w(E), %w(A, E), Ww(0, E) 424 W w(Z),'w(0, Z) 522 spaces 91P(.90) 281ff

Szego's theorem 7, 291, 292 extension of same by Krein 9

Phragmen-Lindelof theorems first 23 second 25

two constants, theorem on

third 27 fourth 28

Volberg's theorems on harmonic measures 349, 353, 362,

fifth

29

364, 366

Poisson kernel for half plane 38, 42, 384, 534, 536, 539 for rectangle 299 for unit disk 7, 8, 10ff pointwise approximate identity property of latter 10 Pollard's theorem 164, 433 for weighted approximation by sums of imaginary exponentials 181, 428 Polya maximum density for a positive increasing sequence

Polya's theorem

257

176ff

178

on simultaneous polynomial approximation 344, 349 on the logarithmic integral 317ff, 357 w* convergence 41 weight 145ff weighted approximation 145ff, 385, 424 weighted approximation by polynomials 147ff, 169, 247, 433, 445 on Z 447ff, 523 see also under Akhiezer, Mergelian weighted approximation by sums of imaginary exponentials 171ff

on closed unbounded subsets of P 428, quasianalytic classes' their characterization 91 quasianalyticity, Beurling's 275ff quasianalyticity of a class 80 Carleman's criterion for it 80 necessity of same 89

444

with a Lip 1 weight 236 with a weight increasing on [0, oc) 247

see also under Akhiezer, Mergelian well disposed, definition of term 452

243,

Contents of volume II

IX

Jensen's Formula Again

A Polya's gap theorem B

Scholium. A converse to P61ya's gap theorem 1 Special case. E measurable and of density D > 0 Problem 29

2 General case; E not necessarily measurable. Beginning of Fuchs' construction 3 Bringing in the gamma function Problem 30 4 Formation of the group products R,(z)

5 Behaviour of 1 log x log IX+AI 1

6 Behaviour of - log I R J(x)I outside the interval [Xi, YY] X 1

7 Behaviour of - log I R3(x)I inside [X;, YY] X

8 Formation of Fuchs' function sb(z). Discussion 9 Converse of P61ya's gap theorem in general case C A Jensen formula involving confocal ellipses instead of circles D A condition for completeness of a collection of imaginary exponentials on a finite interval Problem 31 1 Application of the formula from §C 2 Beurling and Malliavin's effective density D,,. E Extension of the results in §D to the zero distribution of entire functions f (z) of exponential type with J(log+ If(x)I/(1 + x2))dx convergent

604

Contents of volume II 1 Introduction to extremal length and to its use in estimating harmonic measure Problem 32 Problem 33 Problem 34 2 Real zeros of functions f(z) of exponential type with

i

(log+lf(x)I/(1+x2))dx < co

F Scholium. Extension of results in §E.1. Pfluger's theorem and Tsuji's inequality

1 Logarithmic capacity and the conductor potential Problem 35 2 A conformal mapping. Pfluger's theorem 3 Application to the estimation of harmonic measure. Tsuji's inequality Problem 36 Problem 37 X Why we want to have multiplier theorems A Meaning of term `multiplier theorem' in this book Problem 38 1 The weight is even and increasing on the positive real axis 2 Statement of the Beurling-Malliavin multiplier theorem B Completeness of sets of exponentials on finite intervals 1 The Hadamard product over E 2 The little multiplier theorem

3 Determination of the completeness radius for real and complex sequences A Problem 39 C The multiplier theorem for weights with uniformly continuous logarithms 1 The multiplier theorem 2 A theorem of Beurling Problem 40

D Poisson integrals of certain functions having given weighted quadratic norms E Hilbert transforms of certain functions having given weighted quadratic norms 1 Hp spaces for people who don't want to really learn about them Problem 41 Problem 42 2 Statement of the problem, and simple reductions of it 3 Application of Hp space theory; use of duality 4 Solution of our problem in terms of multipliers Problem 43

Contents of volume 11

605

F Relation of material in preceding § to the geometry of unit sphere in

LOJ/H Problem 44 Problem 45 Problem 46 Problem 47 XI Multiplier theorems

A Some rudimentary potential theory 1 Superharmonic functions; their basic properties 2 The Riesz representation of superharmonic functions Problem 48 Problem 49 3 A maximum principle for pure logarithmic potentials. Continuity of such a potential when its restriction to generating measure's support has that property Problem 50 Problem 51 B

Relation of the existence of multipliers to the finiteness of a superharmonic majorant

1 Discussion of a certain regularity condition on weights Problem 52 Problem 53 2 The smallest superharmonic majorant Problem 54 Problem 55 Problem 56 3 How 931F gives us a multiplier if it is finite Problem 57 C Theorems of Beurling and Malliavin 1 Use of the domains from §C of Chapter VIII

2 Weight is the modulus of an entire function of exponential type Problem 58 3 A quantitative version of the preceding result Problem 59 Problem 60 4 Still more about the energy. Description of the Hilbert space Sj used in Chapter VIII, §C.5

Problem 61 Problem 62 5 Even weights W with 11 log W(x)/xIIE < x Problem 63 Problem 64

D Search for the presumed essential condition 1 Example. Uniform Lip I condition on log log W(x) not sufficient

606

Contents of volume II 2 Discussion Problem 65 3 Comparison of energies Problem 66 Problem 67 Problem 68 4 Example. The finite energy condition not necessary 5 Further discussion and a conjecture

E A necessary and sufficient condition for weights meeting the local regularity requirement 1 Five lemmas 2 Proof of the conjecture from §D.5 Problem 69 Problem 70 Problem 71