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

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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
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 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)d9O R

i9

fR

w(9) d9
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
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' 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

) 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 , 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 , 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

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 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

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) 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
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 A and

A 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

0, 3z < 0, we now see that I f (z) 15 const.eb, 3z > 0; If (z)I < const.e-°13Z1, .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,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 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


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 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 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
0. Blaschke products

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

55

and by Lindelof's

+ Iµ,l 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 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 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,


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 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 , 0 and continuous for 9ta > 0, and 916 > 0.

I (P(o) I 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, no.

We see that we can choose c in such a way that F(k"(A,) 11 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)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
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 , 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 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 , 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(x)I , 1, 18°1(x)1 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, 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 - 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
, 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 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 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
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
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 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 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 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

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 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 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) , 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 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* 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
± 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 , 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} 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 ± 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, 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 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

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. 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 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 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
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 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, 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 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 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,
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.

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

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 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 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 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

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 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

, 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)

, 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 2,00

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 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
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


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 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 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 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 , 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


,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)
, 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."


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 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

00. 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.

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 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, 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 < 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 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 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
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


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
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 !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
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
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, 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


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 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 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 < 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 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 _ 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


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 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
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

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 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 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 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

'
const.w(1 zo l ). The result just found therefore reduces to

I Pn (z0 )I2
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
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

0,1 for i >O, so 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 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 , 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, 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
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 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 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 00}V(9kV{ 3Z 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
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 0}take{3z 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 ,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' ®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
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, 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 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 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 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) 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)
- 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) 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
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
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
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

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

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 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 , 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 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


$, which are well disposed with respect to (3,. We then have (A logIp(X)I _2

dx \ 5

log+IP(m) Q,<m0 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<doo 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) , 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 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

CkO, 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

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, 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 -
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 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 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 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 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.

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
, 1 sufficiently large so as N

1

n(i+4n2e212) 2 M«

1

/

to have

0 and k such that, for any polynomial p(z) with Ip(n)I - log+1+n2

-00

= a 0 is sufficiently small, the last condition implies that I Q1(z)I
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 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
, 3, (09P (J,,, 0) '
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)))
, 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)


, 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
(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)
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, 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
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
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<eL 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

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, 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
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


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 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 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 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') , 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 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.

Bibliography for volume I

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Akhiezer, N.I.0 vzveshonnom priblizhenii nepreryvnykh funktsii na vsei chislovoi

osi. Uspekhi Mat. Nauk 11 (1956), 3-43. On the weighted approximation of continuous functions by polynomials on the entire real axis. AMS Translations 22 Ser. 2 (1962), 95-137. Akhiezer, N.I. Theory of Approximation (first edition). Ungar, New York, 1956. Lektsii po teorii approksimatsii (second edition). Nauka, Moscow, 1965. Vorlesungen fiber Approximationstheorie (second edition). Akademie Verlag, Berlin, 1967.

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.

Bibliography for volume I

597

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.

Khachatrian, I.O. 0 vzveshonnom priblizhenii tselykh funktsii nulevol stepeni mnogochlenami na deistvitelnoi osi. Kharkovskil Universitet, Uchonye Zapiski 29, Ser. 4 (1963), 129-42.

598

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.

Uppsala, 1920, pp. 59-90. [Note: The principal result of this paper is also established in the books by Tsuji and Zygmund (second edition), as well as in my own (on Hp spaces).] Mandelbrojt, S. Analytic Functions and Classes of Infinitely Differentiable Func-

tions. Rice Institute Pamphlet XXIX, Houston, 1942. Mandelbrojt, S. Series adherentes, regularisation des suites, applications. Gauthier-Villars, Paris, 1952.

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.

Nevanlinna, R. Eindeutige analytische Funktionen (second edition). Springer, Berlin, 1953. Analytic Functions. Springer, New York, 1970. Paley, R. and Wiener, N. Fourier Transforms in the Complex Domain. Amer. Math. Soc., New York, 1934.

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:

The material of this paper can be found in the books by Duren, Garnett, Tsuji, Zygmund (second edition) and myself (on Hp spaces).]

Riesz, F. and Sz-Nagy, B. Leyons d'analyse fonctionnelle (second edition). Akademiai Kiado, Budapest, 1953. Functional Analysis. Ungar, New York, 1965.

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Riesz, M. Sur le probleme des moments. First note: Arkiv for Mat., Astr. och Fysik 16 (12) (1921), 23pp. Second note: Arkiv for Mat., Astr. och Fysik 16 (19) (1922), 21pp. Third note: Arkiv Jor Mat., Astr. och Fysik 17 (16) (1923), 52pp. Rudin, W. Real and Complex Analysis (second edition). McGraw Hill, New York, 1974.

Shohat, J. and Tamarkin, J. The Problem of Moments. Math. Surveys No. 1, Amer. Math. Soc., Providence, RI, 1963.

Szego, G. Orthogonal Polynomials. Amer. Math. Soc., Providence, RI, 1939; revised edition published 1958. Titchmarsh, E. Introduction to the Theory of Fourier Integrals (second edition). Oxford Univ. Press, 1948. Titchmarsh, E. The Theory of Functions (second edition). Oxford Univ. Press, 1939; corrected reimpression, 1952. Tsuji, M. Potential Theory in Modern Function Theory. Maruzen, Tokyo, 1959; reprinted by Chelsea, New York, 1975. Vekua, I. Obobshchonnye analiticheskie funktsii. Fizmatgiz, Moscow, 1959. Generalized Analytic Functions. Pergamon, London, 1962. Volberg, A. Logarifm pochti-analiticheskoi funktsii summiruem. Doklady A.N. 265 (1982),1297-302. The logarithm of an almost analytic function is summable. Soviet Math (Doklady) 26 (1982), 238-43. Volberg, A. and Erikke, B., see Joricke, B. and Volberg, A. Widom, H. Norm inequalities for entire functions of exponential type. Orthogonal Expansions and their Continuous Analogues. Southern Illinois Univ. Press, Carbondale, 1968, pp. 143-65. Yosida, K. Functional Analysis. Springer, Berlin, 1965. Zygmund, A. Trigonometric Series (second edition of following item). 2 vols. Cambridge Univ. Press, 1959; now reprinted in a single volume. Trigonometrical Series (first edition of preceding). Monografje matematyczne, Warsaw, 1935; reprinted by Chelsea, New York, in 1952, and by Dover, New York, in 1955.

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