=1
1-11
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Cambridge Studies in Advanced Mathematics Editorial Board D.J.H. Garling, D. Gorenstein, T. tom Dieck, P. Walters
THE LOGARITHMIC INTEGRAL II
21
Already published 1
2 3
4 5
6 7 8
9 10 11
12 13 14 15 16 17 18 19
20
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 I A. Pietsch Eigenvalues and s-numbers S.J. Patterson An introduction to the theory of the Riemann 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
22 23 24
Paul Koosis The logarithmic integral: II M.J. Collins Representations and characters of finite groups H. Kunita Stochastic flows and stochastic differential equations P. Wojtaszczyk Banach spaces for analysts
25
J.E. Gilbert & M.A.M. Murray Clifford algebras and Dirac operators in
21
harmonic analysis A. Frohlich & M.J. Taylor Algebraic number theory K. Goebel & W.A. Kirk Topics in metric fixed point theory 28 J.F. Humphreys Reflection groups and Coxeter groups 29 D.J. Benson Representations and cohomology I 30 D.J. Benson Representations and cohomology II
26 27
log M(t)
I+t2
dt
The logarithmic integral
II
PAUL KOOSIS McGill University in Montreal
AMBRIDGE
UNIVERSITY PRESS
CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi
Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521102544
© Cambridge University Press 1992
This publication 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 1992 This digitally printed version 2009
A catalogue record for this publication is available from the British Library
Library of Congress Cataloguing in Publication data (Revised for volume 2) Koosis, Paul The logarithmic integral. (Cambridge studies in advanced mathematics; 21) Includes bibliographies and indexes. 1. Analytic functions. 2. Harmonic analysis. 3. Integrals, Logarithmic. I. Title. II. Series. QA331.K7393 1988 515.4 85-28018 ISBN 978-0-521-30907-3 hardback ISBN 978-0-521-10254-4 paperback
Remember
Genevieve Bergeron, age 21,
Helen Colgan, age 23, Nathalie Croteau, age 23, Barbara Daigneault, age 22, Anne-Marie Edward, age 21,
Maud Haviernick, age 29, Barbara Maria Klucznik-Widajewicz, age 31,
Maryse Laganiere, age 25, Maryse Leclair, age 23, Anne-Marie Lemay, age 22,
Sonia Pelletier, age 28, Michele Richard, age 21, Annie St-Arneault, age 23,
Annie Turcotte, age 20; murdered in the Montreal Ecole Polytechnique on December 6, 1989.
Contents
Foreword to volume II, with an example for the end of volume I Errata for volume I IX Jensen's Formula Again
A Polya's gap theorem B
Scholium. A converse to Pblya's gap theorem 1 Special case. E measurable and of density D > 0 Problem 29
xi
xxv
I I
8 8
9
2 General case; Y. not measurable. Beginning of Fuchs' construction
13
3 Bringing in the gamma function Problem 30 4 Formation of the group products R;(z)
20 22 24 29
5 Behaviour of (1/x) log I (x - 2)/(x + 2)1
6 Behaviour of (1/x)logIR;(x)I outside the interval [Xi,YY] 7 Behaviour of (1/x)logIRj(x)I inside [Xi, YY] 8 Formation of Fuchs' function F(z). Discussion 9 Converse of Pblya's gap theorem in general case C A Jensen formula involving confocal ellipses instead of circles
31
34
43 52 57
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 DA E Extension of the results in §D to the zero distribution of entire functions f (z) of exponential type with f°°. (log` (f(x)I/(1 +x2))dx convergent
62 64 65 70
87
1 Introduction to extremal length and to its use in estimating harmonic measure Problem 32 Problem 33 Problem 34
88 101
108
109
viii
Contents 2 Real zeros of functions f (z) of exponential type with (log+ I f(x)1/(1 + x2))dx < oo F
110
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
126 127 131
132
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
B
140 146 157 158 158
Problem 38 1 The weight is even and increasing on the positive real axis
159
2 Statement of the Beurling-Malliavin multiplier theorem 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
164 165
sequences A Problem 39
189
159
169 173
195
C The multiplier theorem for weights with uniformly continuous logarithms 1 The multiplier theorem 2 A theorem of Beurling Problem 40
195 195
202 208
D Poisson integrals of certain functions having given weighted E
quadratic norms 209 Hilbert transforms of certain functions having given weighted quadratic norms 225 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 F Relation of material in preceding § to the geometry of unit sphere
in L./HO Problem 44 Problem 45 Problem 46 Problem 47
226 234 248 249 260 272 279 282 292 293 295 296
Contents XI
Multiplier theorems
A Some rudimentary potential theory 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 1
B
ix 298 298 298 311
327 328
329 334 339
Relation of the existence of multipliers to the finitness of a
superharmonic majorant 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 Use of the domains from §C of Chapter VIII 1
1
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 used in Chapter VIII, §C.5 Problem 61 Problem 62 5 Even weights W with II log W(x)/x IIE < ao Problem 63 Problem 64
D Search for the presumed essential condition Example. Uniform Lip I condition on log log W(x) not
341
341 361
362 363 369 370 371
374 383 389 391
395 405 407 412 413
418 443 444 446 451
452 452
1
sufficient
2 Discussion Problem 65 3 Comparison of energies Problem 66 Problem 67
454 467 469 472 483 484
x
Contents 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 Bibliography for volume II Index
487 487 502 511
512 524 558 561
565 566 572
Foreword to volume II, with an example for the end of volume I
Art is long and life is short. More than four years elapsed between completion of the MS for volume I and its publication; a good deal of that time was taken up with the many tasks, often tedious, called for by the production of any decently printed book on mathematics. An attempt has been made to speed up the process for volume II. Three quarters of it has been set directly from handwritten MS, with omission of the intermediate preparation of typed copy, so useful for bringing to light mistakes of all kinds. I have tried to detect such deficiencies on the galleys and corrected all the ones I could find there; I hope the result is satisfactory. Some mistakes did remain in volume I in spite of my efforts to remove
them; others crept in during the successive proof revisions. Those that have come to my attention are reported in the errata immediately following this foreword.
In volume I the theorem on simultaneous polynomial approximation was incorrectly ascribed to Volberg; it is almost certainly due to T. Kriete, who published it some three years earlier. L. de Branges' name should have
been mentioned in connection with the theorem on p. 215, for he gave
(with different proof) an essentially equivalent result in 1959. The developments in §§A and C of Chapter VIII have been influenced by earlier
work of Akhiezer and Levin. A beautiful paper of theirs made a strong
impression on me many years ago. For exact references, see the bibliography at the end of this volume. I thank Jal Choksi, my friend and colleague, for having frequently helped me to extricate myself from entanglements with the English language while I was writing and revising both volumes.
Suzanne Gervais, maker of animated films, became my friend at a bad time in my life and has constantly encouraged me in my work on this book,
Foreword to volume II
xii
from the time I first decided I would write it early in 1983. Although she had visual work enough of her own to think about, she was always willing to examine my drawings of the figures and give me practical advice on how to do them. For that help and for her friendship which I am fortunate to enjoy, I thank her affectionately.
One point raised at the very end of volume I had there to be left unsettled. This concerned the likelihood that Brennan's improvement of Volberg's theorem, presented in article 1 of the addendum, was essentially best possible. An argument to support that claim was made on pp. 578-83; it depended, however, on an example which had been reported, but not described, by Borichev and Volberg. No description was available before Volume I went to press, so the claim about Brennan's improvement could not be fully substantiated. Now we are able to complete verification of the claim by providing the missing example. Its description is found at the end of a paper by Borichev
and Volberg appearing in the very first issue of the new Leningrad periodical Algebra i analiz. We continue using the notation of the addendum to volume I. Two functions have to be constructed. The first, should be decreas< oo and satisfy h(g) >, 1, together with the relation
ing for 0 < 1
=
logh(g)dl;
oo.
f0 The second, F(z), is to be continuous on the closed unit disk and ' its interior, with 8F(z)
IF(e19)l
exp(-h(log(1/Izl))), >
0
in
IzI < 1,
a.e.,
and n
loglF(ei9)Id9
=
-oo.
-n
The function F(z) we obtain will in fact be analytic in most of the unit disk A, ceasing to be so only in the neighborhood of some very small
segments on the positive radius, accumulating at 1. The function h(log(1/x)) will be very much larger than 1/log(1/x) for most of the
Example for addendum to volume I
xiii
x e (0, 1) contiguous to those segments. Three simple ideas form the basis for the entire construction:
(1) In a domain 6V with piecewise analytic boundary having a 90° corner (internal measure) at , say, we have z
for arcs I on 09 containing (see volume I, pp 260-1); (2) The use of a Blaschke product involving factors affected with
fractional exponents to `correct', in an infinitely connected subdomain of A, a function analytic there and multiple-valued, but with single valued modulus; (3) The use of a smoothing operation inside A, scaled according to that disk's hyperbolic geometry.
We start by looking at harmonic measure in domains & = A - [a, 1], where 0 < a < 1. According to (1), if rl > 0 is small (and < 1 - a ), we have
w8(En, 0)
0 small enough to make wg(E,,, 0)/it less than some preassigned amount. Then, if we put 9 = A - [a, 1 - q], we will have, by simple comparison of ww(I., z) and wg(E., z) in ', w1(Iry, 0)/n
2/(2 +,/3) and < 1 (we shall see presently why the first condition is needed), we take a b1, a1 < b1 < 1, so close to I as to make (O's , (I1, 0) II11
If(ei9)I
e19 0 1.
0,
Because the product b(z)f(z) is analytic in -9, we have there 8z
b(z)f(z)
=
0;
the expression on the left may therefore be looked on as a distribution in
A, supported on the slits J of A - .9. In order to obtain a W., function defined in A, we smooth out b(z) f (z); that is our third idea. The smoothing
is scaled according to the square of the gauge for the hyperbolic metric in A, i.e., like 1/(1 - IzI)2. Taking a (C., function iji(p) >, 0 supported on the interval [1/4, 1/2] of the real axis, with /2 1
a(p)p dp
2n'
0 f,
we put, for z e A, G(z)
=
ffo(
Iz-
I
\
(C)
0 - IzI)2 0 -1zD'
(writing, as usual, t' =
d do
+ in). The first thing to observe here is that the expression on the right makes sense. Although b(t') f (l) is defined merely in -9, the slits J,, making up A - -9 are of planar Lebesgue measure zero, so we only need the values of the product in -9 in order to do the integral. The second observation is that G(z) is W,,,, in -9. As a function of t', (I z - I/(1 - I z I )2) vanishes outside the disk I - z I < (1 - I z I)' which, however, lies well within A z
xviii
Foreword to volume II
for z e A, since then IzI + z (1 - I z 1)2 < 1. We may therefore differentiate
under the integral sign with respect to z or z as often as we wish, t/'(p) being W., (its identical vanishing for p near 0 helps here), and I b(t') f (g) I
being
2 -,/3. Indeed, a, as one of the an, is > al which we initially took > 2/(2 + ,/3), while Oz lies in the disk { I C I < IzI + (1 - IzI )2 } whose radius increases with
i works out to 2/(2+,/3), so IzI. For IzI = 2-,/3, that radius
if
< 2-.,/3, [a, b] would lie outside the disk containing A ; < 1, is thus > 2-V3 for z e Bn. Now when 2-.,/3 < IzI - (1 - IzI)2 > 0, so Az is in fact contained in the ring IzI
IzI
IzJ
z
IzI - '2(1 - IzI)2
1/2 but have in fact a > 2/(2+,,/3). Using differentiation, one readily verifies that a' < a and b < b' < 1. We see that Bn (the set of z e 0 for which Az intersects with [a, b] ) is
an oval-shaped region including [a, b] and contained in the ring a' < I z I < b'; its boundary crosses the x-axis at the points a' and b'. When a is close to 1, B. is quite thin in the vertical direction because, if OZ touches the x-axis at all, we must have 13z I < (1 - IzI )2. z
One can specify the an and bn so as to ensure disjointness of the oval regions Bn. The preceding description shows that this will be the case if IzI < b; are disjoint, where (restoring the subscript n) the rings a'
a' b;,
_ ,/(2an - 1), = 2 - ,J(3 - 2bn);
xx i.e.,
Foreword to volume II
if b;, < an+ 1 for n = 1, 2, 3, ....
It
is easy to arrange this in
making the successive choices of the an and bn; all we need is to have
an+1 = an+1 ' + z 1-an+l )2 > b;, + '(1 -bn )2. Here it is certainly true that bn < bn < 1 when 0 < bn < I ; then, however, the extreme right-hand member of the relation is still < 1, and
numbers an+1 < 1 satisfying it are available. There is obviously no obstacle to our making the an increase as rapidly as we like towards 1; we can, in particular, have Y (1 - an) n=1
1, we see that the constants c and K can be adjusted so as to have OF(z) Of
exp
1
log(1/Izl)
within the B at least. But then this holds outside them as well (in A, including in the neighborhood of 0), because aF(z)/az = 0 there. F(z) has now been shown to enjoy all the properties enumerated at the beginning of this exposition except the one involving the function not yet constructed. That construction comes almost as an afterthought. Since the sets B. lie inside the disjoint rings a' < I z I < b;,, we start by
putting h(log(1/Izl)) = 1/log(1/IzI) on each of the latter; in view of the preceding relation, this already implies that 3F(z) Of
exp(-h(log(1/lzl)))
throughout 0, no matter how h(log (1/ I z l )) is defined for the remaining z e A, because the left side is zero outside the B. To complete the definition
of h(l;) for 0 < < oo, we continue to use h(log(1/lzl)) = 1/log(1/IzI) on the range 0 < I z I < a i and then take h(log (1/I z l )) to be linear in
Example for addendum to volume I
xxiii
Jzi on each of the complementary rings bn
n = 1, 2, 3,
an+1>
Izi
.
..
.
The function h(l;) we obtain in this fashion is certainly decreasing ); h(log(1/lzl)) is also > 1 for Izi > b1, because b1 > a1 > 2/(2+,/3) > 1/e. h(log(1/JzJ)) is moreover > l/log(1/Izl) on the complementary rings, for 1/log(1/x) is a convex function of x for 1/e2 < x < 1, and b1 > l/e2. In terms of the variable = log(l/Izl)
(in
we therefore have 1,
The trick in arranging to have
d=
log fo,
oo
is to use linearity of h(log(l/x)) in x on each interval bn < x < an+1 to get lower bounds on the integrals log (1, /6, )
log JIog(1/a +i)
We have indeed
log(l/bn) < log(1/b;) and
for
1
h(log(l/an+1)) = 1/log(1/an+1), so the linearity just mentioned makes h(%) > 1/2log(l/an+1) for (bn+an+1)/2 < e-4 , 1 by taking an+ 1 > b' close enough to 1, and that can in turn be achieved by choosing an+
1
1 = an+ 1 + z (1 - an+ 1)2 sufficiently near 1. We therefore select the successive an in accordance with this requirement in carrying out the inductive procedure followed at the beginning of our construction. That will certainly guarantee that b' < an+ (which we needed), and may obviously be done so as to have Y_,'=, (1 - an) < oo (by making the an tend more rapidly towards 1 we can only improve matters). 1
Once the an have been specified in this way, we will have log(
log J'Iog(l/a+
1)
d >,
1
xxiv
Foreword to volume II
for each n, and therefore log
d=
oo.
f"
Our construction of the functions F(z) and with the desired properties is thus complete, and the gap in the second half of article 2 in the addendum to volume I filled in. This means, in particular, that in the hypothesis of Brennan's result (top of p. 574, volume I), the condition that M(v)/vi"2 be increasing cannot be replaced by the weaker one that M(v)/v1"2 > 2.
January 26, 1990 Outremont, Quebec.
Errata for volume I
Location
Correction
page 66
At end of the theorem's statement, words in roman should be in italic, and words in italic in roman.
pages 85, 87
In running title, delete bar under second M. but
page 102
page 126, line 8
keep it under first one. In heading to §E, delete bar under M" in first and third 6R({M"}) but keep it in second one. In statement of theorem, change determinant to determinate.
page 135, line 11 page 136, line 4 from bottom page 177, line 11
from bottom page 190 page 212 and following even numbered pages up to page 232 inclusive page 230
page 241, line 3
In displayed formula, change w* to w*. In displayed formula I P(xo) 1 2v({xo }) should stand
on the right.
The sentence beginning `Since, as we already' should start on a new line, separated by a horizontal space from the preceding one In last displayed formula, change x" to x" Add to running title: Comparison of Ww(O) to 'W(0 +)
In the last two displayed formulas replace (1 - a2) throughout by I 1 - a2l. Change bb in denominator of right-hand expression to b,2,.
page 270, line 10 page 287
Change F(z) to F(Z). In figure 69, B1 and B2 should designate the lower and upper sides of -9o, not -9.
xxvi
Errata for volume I
page 379, line 8 from bottom page 394, line 3 page 466, last line page 563, line 9 page 574, line 9 from bottom page 604 page 605
Change comma after `theorem' to a full stop, and capitalize `if'.
Change yl to y, Delete full stop. Change `potential' to `potentials'. Delete full stop after `following'. In running title, `volume' should not be capitalized.
In titles of §§C.1 and C.4 change `Chapter 8' to `Chapter VIII'.
IX
Jensen's formula again
The derivations of the two main results in this chapter - P61ya's gap theorem and a lower bound for the completeness radius of a set of imaginary exponentials - are both based on the same simple idea: application of Jensen's formula with a circle of varying radius and moving
centre. I learned about this device from a letter that J.-P. Kahane sent me in 1958 or 1959, where it was used to prove the first of the results just mentioned. Let us begin our discussion with an exposition of that proof. A.
P61ya's gap theorem Consider a Taylor series expansion f (w) = Y- anwn 0
with radius of convergence equal to 1. The function f (w) must have at least one singularity on the circle I w I = 1. It was observed by Hadamard that if many of the coefficients a are zero, i.e., if, as we say, the Taylor series has many gaps, f (w) must have lots of singularities on the series' circle
of convergence. In a certain sense, the more gaps the power series has, the more numerous must be the singularities associated thereto on its circle of convergence.
This phenomenon was studied by Hadamard and by Fabry; the best result was given by P61ya. In order to formulate it, P61ya invented the
maximum density bearing his name which has already appeared in Chapter VI. In this §, it will be convenient to denote by i the set of integers > 0 (and not just the ones > 1 as is usually done, and as we will do in §B!). If E c N, we denote by n,(t) the number of elements of E in [0, t], t ? 0. The Pdlya 1
2
IX A Polya's gap theorem
maximum density of E, studied in §E.3 of Chapter VI, is the quantity
DE = lim
-'i -
Clim sup
njr) - nE(Ar)
(1 -,)r
r- oo
We have shown in the article referred to that the outer limit really does exist for any E, and that DE is the minimum of the densities of the measurable
sequences containing E. In this §, we use a property of DE furnished by the following Lemma. Given e > 0, we have, for p > Er, n£(r + p) - nE(r) P
when r is large enough (depending on E).
Proof. According to the above formula, if N is large enough and =
(I+E)-1/N,
we will have
n..(r) - nE(2r) (1 -A)r
*
R, say. Fix such an N. When r >, R, we certainly have
nE( k 1r) _ nE( -kr)
k-1-2-k)r
fork =
E
DE+2
, Er = (A
N
- 1)r.
Then, if k is the least integer such that (A - k - 1)r > p, we have k >, N, so, nE(t) being increasing, nE(.1-kr)
nE(r + p) - nE(r) p
0. In the formula an
=
1
If(w)w-n-idw
2ir1 IWI
,
(we are, of course, taking an as zero for noE, n > 0) one may, thanks to the analyticity of f (w), deform the path of integration w I = e - '} to the contour I,a shown here:
The quantity c > 0 is fixed once D is given, and independent of 6.
4
IX A P6lya's gap theorem
In the integral around I'b, make the change of variable w = e-S, where s = u + i r with r ranging from - it to it. Our expression then goes over into
f(e-S)e"sds
I
2ni
=
a"
Yn
with this path y,:
(-c,7r) Ln(1-D)
Yd
0
6
a
J7.c(1 -D)
Figure 158
Write F(z)
2ni,f,
f(e-')eZSds
so that F(n) = a" for n e N (and is hence zero for n e % - E); F(z) is of course entire and of exponential type. We break up the integral along ya into three pieces, I, II and III, coming from the front vertical, horizontal and rear vertical parts of ya respectively. On the front vertical part of y, I f(e-s)I < Mb and IeSZI S eax+n(u -D)IyI (writing as usual z = x + iy); hence III
Maeax+n(1 -D)lyl f(e-s)I
On the horizontal parts of yb, S C (a number independent of 6, by the way), and IeSZI \ eax+n(i -D)lyl for x > 0, whence I
1111 5 Ceax+n(l -D)lyj
x > 0.
A Pdlya's gap theorem
5
Finally, on the rear vertical parts of eszI \ e-`X+nlvl for x>0, making 11111 < Ce-`X+nlvl
f(e-s)I
Y&
I
5 C and
x>0.
Adding these three estimates, we get
< (Mb+C)ex +n(1-n)Ivl
IF(z)I
+ Ce-`"+nlvl
for x > 0. Since c > 0, the second term on the right will be < the first in the sector
S=
{z:
13z 1
irD
J)
with opening independent of S. We thus have IF(z)I
0
9
ly used by V. Bernstein in his work on Dirichlet series, and later on by L. Schwartz in his thesis on sums of exponentials. 0010,
Restricting our attention to sequences E of strictly positive integers clearly involves no loss in generality; we do so throughout the present § because that makes certain formulas somewhat simpler. Denote by N the set of integers
> 0 (N.B. this is different from the notation of §A, where N also included 0 ), and by A the sequence of positive integers complementary to E, i.e.,
A=
N - E.
For t > 0, we simply write n(t) for the number of elements of A (N.B.!) in
[0, t]. Put* Z2
C(z) _ H 1 - s n
"CA
;
in the present situation
n(t)-1-D t
for
t -+oo
and on account of this, C(z) turns out to be an entire function of exponential type with quite regular behaviour.
Problem 29 (a) By writing I log C(z) I as a Stieltjes integral and integrating by parts, show
that log
I CGY) I
-+ 7r(1 - D)
IYI
±oo for y (b) Show that for x > 0,
n(xT)Tn(x)) dT
logIC(x)I = 2J p
T
1 - TZ
T
(Hint: First write the left side as a Stieltjes integral, then integrate by parts. Make appropriate changes of variable in the resulting expression.) (c) Hence show that for x > 0,
(n(xT)-Tn (x))
logjC(x)I 5 2n(x)log- + 2 J
T
T
dT I - Tz
with y any number between 0 and 1. * When D = 1, the complementary sequence A has density zero and may even be empty. In the last circumstance we take C(z) w 1; the function f (w) figuring in the construction given below then reduces simply to w/a(1 + w).
IX B Converse to Polya's gap theorem; special case
10
(d) By making an appropriate choice of the number y in (c), show that log I C(x) I < ex for large enough x, e > 0 being arbitrary. (e) Use an appropriate Phragmbn-Lindelof argument to deduce from (a) and (d) that lira sup
log I C(re' 9)
n(1 -D)Isin9I.
(f) Show that in fact log I C(n) n
for
oo in E,
n
and that we have equality in the result of (e). (Hint: Form the function z2
K(z) = rl 1 - Z ; n
net
then, as in (e), lim sup
log I K(re's)
nD l sin 91.
Show that the same result holds if K(re''9) is replaced by K'(re's). Observe
that nzK(z)C(z) = sin nz. Look at the derivative of the left-hand side at points neZ.)
We are going to use the function C(z) to construct a power series Y_ anw" nex
having radius of convergence 1, and representing a function which can be analytically continued into the whole sector I arg w I < mD.
Start by putting
"+» C
1
f (w) -
27Ei
f,-
w; dl; i OD
sin irC
for I arg w I < nD. Given any E > 0, we see, by part (e) of the above problem,
that C(Z + ill)
sin m(z + W
I
cont. cosh nq
(x('D) +E)Inl
1 1 measurable, of density D > 0
11
for real rl, where the constant on the right depends on E.
lµ#+iryl =
IN
At the same time,
I1/2e-7argw
so the above integral converges absolutely and uniformly for w ranging
over any bounded part of the sector
IargwI < aD-2e. The function f (w) is hence analytic in the interior of that sector, and thus finally for I arg w I
< rrD,
since e > 0 was arbitrary. We proceed now to obtain a series expansion in powers of w for f (w), valid for w of small modulus with I arg'wI < nD. For this purpose the method of residues is used. Taking a large integer R, let us consider the integral 1
2ni
fr. sin zr
wC dr
around the following contour FR :
n E
A FR
0
+R
2
A
Figure 160
IX B Converse to P6lya's gap theorem; special case
12
On the top horizontal side of FR, e(n(' -D)+e)R
C(()
0 is arbitrary, and the constant depends on it. The same estimate holds on the lower horizontal side of I'R. Also, if I w I < 1 and I arg w (
< nD - 2e,
we have < e(nD - 2e)R I wC I
for C on the horizontal sides of FR; in this circumstance the contribution of the horizontal sides to the contour integral is thus const. Ree-eR
in absolute value, and that tends to zero as R -> oo. Along they right vertical side of I'R, by part (e) of the problem, en(1-D)IPI I+eR
CQ
const.
sin nC
cosh ntl
with e > 0 arbitrary as before (we write as usual I arg w I
_ + in). For
< nD - 2e and ( on that side, IwSI
0. The function b(z) is constructed so as to vanish at the points of AO belonging to a sparse sequence of the intervals [Xi, YY], and so as to make
ID(x)I
0 outside of those intervals, while J (D(m) l
>, const.ek"
for most of the integers m inside them that don't belong to Ao. As we have just observed, there will be plenty of the latter. 3.
Bringing in the gamma function
For obtaining the function (D(z) mentioned at the end of the preceding article, procedures yielding entire functions will not work.* Fuchs' idea is to construct '(z) by using products of the form 1 - z/n
fl( 1 + z/n
ezz/"
taken over certain sets of positive integers n; these are analytic in the right half plane, but have poles in the left half plane. The exponential factors ensure convergence. * at least, so it seems
3 The gamma function
21
The prototype of such a product is °°
n1
1 - z/n 1 +z/n
this can be expressed in terms of the gamma function. F(z) is the reciprocal of an entire function of order 1 defined by means of a certain infinite product designed to make
F(x + 1) = J txe-`dt 0
for real x > - 1. Starting from this formula, successive integrations by parts yield
f-t`e`dt
1
F(x + 1) =
=
x+1 0
(x+1)(x+2) o
tx+2e `dt
tx+me-[ dt
Jf0,0
The last expression can be rewritten
expi m
(x/k)) Jtedt
k1
ml
x\e -x/k (1k + l
One has, of course, m! = f o tme-` dt, and, for real x tending to oo, Stirling's formula, x
txe `dt
V(27rx) e X
f0,0
is valid. (The latter may be proved by applying Laplace's method to the integral on the left.) Using these relations to simplify the expression just written, we find that mx exp
F(x + 1)
=
lim
(-
k=1
(x/k)
/
k=m1
+ x) e k
We have
(1/k))m x
exp(x k=1
=
exp{ x(
l
1 + 11 m
k=1 k
-
logm) }. JJJ
IX B Converse to Polya's gap theorem; general case
22
By drawing a picture, one sees that as m - oo, m-11 E -
-
log m
k=1 k
increases steadily to a certain finite limit C (called Euler's constant). Therefore, by the preceding formula, I'(x + 1)
=
I
exp (Cx)00f M=1
\
1+x
le-xik.
k
For general complex z, one just defines
=1F(z+1) =
1/exp(Cz)
fl C1+Zn
n=1
e"
By a slight adaptation of the work in Chapter III, §§A, B one easily shows using this formula that
I1/F(z+1)l < KEexp(IzI1+E) for each e > 0. ( 1/f(z + 1) is NOT, by the way, of exponential type, on account of Lindelof's theorem if for no other reason!) Since
sinnz
z21
H n=1
7cz
n2
we have -
1
_z/ n
H 1 + z/n )
e2z1n
_
-
ezcZ sin nz (F(z
+ 1))2.
1[z
Use of this relation together with Stirling's formula for complex z enables us to get a good grip on the behaviour of the right-hand product. Problem 30
To extend Stirling's formula to complex values of z in the right half plane. Write
g(z) = /(2nz) (Z )Z'I'(z + 1). e
(a) Show that g(z) is of order 1 - i.e., that Ig(z)i 5 MEexp(IzI1+E)
3 The gamma function
23
for each e > 0 - in any open sector of the form I arg z I 0, and
is continuous up to the boundary of such a sector. (b) Show that for real y, Ig(iy)I = `/2.e-1"JyJni/(sinhrrIyI). (c) Hence show that g(z) is bounded for 93z > 0. (Hint: Use Stirling's formula to estimate g(x) for x > 0. Then use Phragmen-Lindelof in the first and fourth quadrants.)
(d) Hence show that g(z) 1 uniformly for z tending to oo in any sector I arg z l < n/2 - S, 6>0. (Hint: g(x) --.1 for x oo by Stirling's formula. In view of (c), a theorem of Lindelof may be applied.) (e) Show that g(re t (2"'13)) _,1 as r ---+ oo. (Hint: First show that
r(1 + #r(1- z) = sin ztz/nz. Use this in conjunction with the result from (d), noting that e±(2"i/3) = _ eT("i/3)) (f) Hence show that g(z) -> 1 uniformly for z tending to oo in any sector of the form
largzl
0.
(Hint: Use Phragmen-Lindelof and the theorem of Lindelof referred to in the hint to part (d) again.)
From part (f) of this problem we have in particular r(z + 1)
-
V(2nz)- (Z)Z e
for I arg z I < n/2 and I z I large. This means that
z-2ze2(1 -c)z
jj
1 - z/n\ e2z/n =1(1 +z/n
,.
2sin 7tz
for 91z > 0 when I z I is large. The expression on the left is thus certainly
of exponential type it in the right half plane.
Fuchs takes* the intervals [X;, Y;] constructed in the previous article, * His construction is, of course, needed by us only for the case of non-measurable E, when the sequence AO is certainly available, and indeed infinite.
24
IX B Converse to P6lya's gap theorem; general case
corresponding to a small value of e > 0. He then fixes a large integer L with, however, L < 1/e, and forms the function F(z/L)
Z
=
2z/L
e2(1 -c)z/L
n
1 - z/nL
e 2z/nL
1 + zInL n= 1 CL) According to the above boxed formula, this has very regular behaviour in the right half plane, and is of exponential type ir/L there. Fuchs' idea is now to modify the product on the right side of this last relation by throwing away the factors
1 - zInL
e2zlnL
1 + z/nL)
corresponding to the n for which nL belongs to certain of the intervals [X3, Y3]. Those factors are replaced by others of the form 1 - z/), e2z/z 1 + z/A,)
corresponding to the .1 e AO belonging to the same intervals [X3, Y3 ]. This
alteration of F(z/L) produces a new function, vanishing at the points of AO lying in certain of the [Xi, YJ We have to see how much the behaviour of the latter differs from that of the former. 4.
Formation of the group products R; (z)
We want, then, to remove from the product °° T'-(
1 - z/nL
n=1
1 + zInL
e2z/nL
the group of factors
H (1 - z/nL
e2z/nL
nLE[X;,Y;] \ 1 + z/nL
and to insert
f
).EAon[X.i,Y;]
1 - z/A) e2z/z
( 1 + Z/a,
in their place, doing this for infinitely many of the intervals [X;, YY] constructed in article 2, corresponding to some fixed small e > 0. This amounts to multiplying our original product by expressions of the form
(nL+z\e_2z/nL T7 A-zezz/x L z fl nLE[X;,Y;] zenojn[jX';,r;] (A+Z)
4 The group products Ri(z)
25
As we said at the end of the preceding article, Fuchs takes the integer L < 1/E. Therefore, since
n(YY)-n(Xj) < E(Yj - Xj) by the theorem of article 2, there are fewer 2 e A0 than integral multiples of L in [X;, Y;], and the exponential factors in the above expression do not multiply out to 1. Their presence would cause difficulties later on, and we would like to get rid of them. For this reason, Fuchs brings in a small multiple a; of X;, chosen so as to make 1
qi
A
a3
1
nLe[X ,YJ] nL
AeAo
J,YJ1
with a positive integer q;, and then adjoins to the previous expression an additional dummy factor of the form
(aJ_z)J2z/aJ a-2zInL
e2ei=faJ
Once this is done, the exponential factors
and e2=iz
figuring in the resulting product cancel each other out.
The details in this step involve some easy estimates. In this and the succeeding articles, when concentrating on any particular interval [X;, Y;], we will simplify the notation by dropping the subscript j, writing just X for X3, Y for Y3, a for a;, and so forth.
By the theorem of article 2, n(t) - n(X) < E(t - X) for X < t S Y, and X and Y are half-odd, while the 2 e A0 are integers. Hence, Yn(X)
Y
S
+
n(Y)
J x d t(t)
AeA r X,Y] 2
E(Y - X) + Y
t X dt
c
fX
J x n(t)
=
e to g
2
t 2n(X) dt
(-X
).
In the theorem just mentioned, Y - X >, (a/4)X with a constant a > 0 depending on e. Therefore, if X is a very large X; (which we always assume
henceforth), there will be numbers nL e [X, Y], n e N, and then, as a simple picture shows,
1 nLe[[x,Y] nL
>
1 log L
Y
X+ L
>
1
L
log
Y- 1 X
X
IX B Converse to Polya's gap theorem; general case
26
Thus, when X is large,
I
-
1
1
CL
- e IlogX
leAon[X,Y] A
nLe[X,Y] nL
-
X.
Here, 1/L > e and Y >, (1 + a/4)X, so the right side is bounded below by a constant > 0 depending on L, c and a for X large enough (again depending on those parameters). Take now a small parameter n > 0 which is to remain fixed throughout all the following constructions - later on we will see how rl is to be chosen. Here, we observe that if X is large, there is a number a between qX/2 and q X whose product with the left side of the preceding inequality is an integer. Picking such an a, we call the corresponding integer q, and we have
q
_
a
ZeAoX,Y] A
,Y] nL
nLE
Because riX/2 < a
1X, the inequality
1
X
1
1
5
sYnL
1
X
(YLX
+1)
gives us the useful upper estimate
q
Definition. We write
R(z)
\a+Zj
LEH,Y] \nL±z/
xeA
X,Y](A+z).
When using the subscript j with X, Y, a and q, we also write R j(z) instead of R(z).
Let us establish some simple properties of the group product R(z). In
4 The group products R3(z)
27
the first place, we can put the aforementioned exponential factors (needed for convergence) back into R(z) if we want to: R(z)
1 - z/a)9e2gzla T7 1 + z/a nLE]x,Y]
=
1 + z/nLl e-2zlnL
1 - z/nL
(1 - Z/2'\ e2zIa
X
11
2eAo n [X, Y] I` 1
+ Z/A
This is so by the above boxed relation involving q/a. In the second place, we have the Lemma. If q > 0 is chosen small enough (depending only on L and e) and X is large,
IR(x)I < 1 for0<x1X < X we can, for 0 < x < a, expand the logarithms in powers of x. Collecting terms, we find, thanks to the above boxed formula for q/a, that the coefficient of x vanishes, and we get loglR(x)I
I
= 3 LnLE
N odd
,Y] (nL)N
xEAO
X,Y] AN
q
2xN
aN
N
By a previous inequality for the right side of the boxed formula for q/a,
(LI-E)logX
q
a
-
X.
Hence, since a
L-E X - (N-1) 1
q aN
forN>1.
N-1
log
Y
X
1
N-1XN
28
IX B Converse to Polya's gap theorem; general case
At the same time, we already have
1 (Y-X
1
L
XN
nLe[X,Y] (nL)N
+1).
Y -< (1 + a)X,
Thus, since (1 + a/4)X 1
(06L
1
X
nLe,Y1(nL)N
1 XN-1
while, by the preceding calculation,
[(!)log(l +4 I - XJ (fX)N-I*
aN
Since 1/L - s > 0, for sufficiently small values of g, the second of these quantities exceeds the first for every N >, 2, as long as X is large enough. The required smallness of g is determined here by a, L and e, and therefore really by the last two of these quantities, for a itself depends on E. Under the circumstances just described, all the coefficients in the above
power series expansion of logIR(x)I will be negative. This makes log I R(x) I < 0 for 0 < x < a, Q.E.D.
Another result goes in the opposite direction. Lemma. If g > 0 is taken small enough (depending only on L and c) and X is large,
IR(X/2)I > 1. Proof. logIR(X/2)I
=
glog iX + aI + lenox,Y]
log
YnLE[X,Y]
log
nL+zX nL - ZX
ZX
Since a 5 iX, the first term on the right is g
Y-X+1 )log 1-2g L l+2g
by the previous boxed estimate on q. Recall that in Fuchs' construction, L is an integer, AO consists of integers, and (by the theorem of article 2) X and Y are half-odd integers. The sum
29
5 Behaviour of (1/x) log I (x - ))/(x + 2)1
of the second and third terms on the right in the previous relation can therefore be written as Y
j x
log I t + X d([t/L] - n(t)) t{[Y/L] -ZX
l - [X/L] - (n(Y) - n(X)) } log (2Y+ X ) 2Y - X
_
+ fx 4t
4XX2([t/L]
- [X/L] - (n(t) - n(X)))dt
(we are using the symbol [p] to denote the biggest integer < p). We have
[t/L] - [X/L] >, [(t - X)/L]. Also, n(t) - n(X) < e(t - X) for X < t s, so surely [(t - X)/L]
log x
t+ZX
t-zX
d([t/L] - n(t))
e1(Y-X)-2 1log 3l +2a +2a
Combining this estimate with the one previously obtained, we see that
logIR(X/2)I
>,
(Y
-X){(L-e Ilogl+2a J
- 2 log
3+2o 1+2a
-
log n
Llogl±2g} g
1+21
l-2r1
Because Y - X >, (a/4)X and 1/L - e > 0, the right side will be positive for all large X provided that n > 0 is sufficiently small (depending on L, e and
a, hence on L and e). This does it. 5.
Behaviour of (1/x) log I (x - ))/(x + .)I.
We are going to have to study (1/x)logIRj(x)I for the products R j(z) constructed in the preceding article. For this purpose, frequent use will be made of the
IX B Converse to Polya's gap theorem; general case
30
Lemma. If 2 > 0,
x-2
8
is < 0 for0<x 0 forx>A. OX(' (Xlog x+2 ) Also,
x-2
02
log ) > 0 for x > 0 different from A. OlOx x x+A Proof.
ax x
x-1
x+2
x+2
x-2
22
+ x(X2 - 12).
The right side is > 0 for x > 2. When 0 < x < 2 we rewrite the right side as
110 (1+'1 X2
with
g
l-)
-
21 21
1- j
= x/2, and then expand the quantity in curly brackets in powers of
. This yields
which is < 0 since 0 < < 1. Finally, 02
1
x-2'\
_
4x
(22 - x2)2
alai (x log x + 2 /
>0
forx>0, x962. We are done. Corollary. If 0 < A < A', 1
x
log
x-2
x - A' x +A'
x+A
is a decreasing function of x for 0 < x < 2 and for x > A'.
Proof. By the second derivative inequality from the lemma. In like manner, we have the
Corollary. If 0 < x < x', 110 g
x'
x'-2 x'+2
X
logg
X
X+2
is an increasing function of A for 2 > x' and for 0 < 2 < x.
6 Behaviour of (1/x)logIRj(x)I outside [Xj, YJ
31
Behaviour of (1/x) log I R1(x) I outside the interval [Xi, YY]
6.
Turning now to the group products R j(z) constructed in article 4, we have the Lemma. If the parameter q > 0 is taken sufficiently small (depending only on L and e),
(1/x) log I R;(x) I is decreasing for x 3 YY provided that X; is large
enough.
Proof. Dropping the subscript j, we have, for x >, Y, 1togIR(x)l x
=
x-a)
q
(x+A)
xlog (x+a 1
x - nL
nt.e[x.Y] x
x + nL
- log
Y-
We are going to make essential use of the property
n(Y) - n(t) < e(Y - t), X S t < Y (see theorem of article 2). Since 1/L > e, we have a picture like the following: A,(') n(Y)
I
1t4
t5
Figure 163
3
L
2
t4
tj
tz t
?
LI Y1 L
Number the members of AO in [X, Y] downwards, calling the largest of those A' , the next largest A' , and so forth. We also denote by ti = L[Y/L] the largest integral multiple of L in [X, Y], by ' the next largest one, and
IX B Converse to Polya's gap theorem; general case
32
so on. Since L and the members of Ao are integers while Y is half-odd (theorem, article 2), we in fact have .1i < Y and ti < Y. By the above property of n(t),
'< Y - 1/s, At < Y - 2/e, A,
2
etc. Since, however, L < 1/s, it is also true that t'l
> Y - 1/e,
t2> Y-2/e, and so on. We can pair off each of the ak, X < ) < Y, with tk, and still have some tks left over after all the 2 are taken care of in this way. Indeed, there are
at most n(Y) - n(X) < e(Y - X) of the 2 e A 0 in [X, Y], X being half-odd - in fact there are at most [s(Y - X)] of them, since n(Y) - n(X) is integer-valued. At the same time, there are at least [(Y - X)/L] integral multiples of L in that interval. Since 1/L > e, there are more of the latter than the former, and, after pairing off each 2 with tk, there will still be at least
[(Y-X)/L]-[e(Y-X)] ?
[(_c)(Y_ X)]
integral multiples of L left over in [X, Y]. Now from article 4, q
(Y-X)+
Since Y - X >, (a/4)X, we see that if n/L < 1 /L - e, [(1 /L - e)(Y - X )] is larger than q, provided only that X is big enough. Under these circumstances, there will be more than enough of the points tk left over to pair off with the q -fold point a, after each A' has been paired with the tk corresponding to it. Write n(Y) - n(X) = N. Then, after the pairings just described, the above formula for (1/x)logIR(x)I can be rewritten thus: N
1
X
log I R(x) I
1
=
log
kY1
N+q
+
k=N
1(xlog1
1
,
f/x-a
1
x-tk log - ( x + tk ) ) + Ak x-t
x-'1k
x+a
1
xlog(x+t'
1
x+t
+ k>N+gXlOgx-t'J
6 Behaviour of (1/x) log I R j(x) I outside [Xi, Yj]
33
As we have seen, for 1 < k < N,
Y-
Ak
k
, Y. The same holds good for the terms of the second right-hand sum, because a < X < tk. The terms of the third sum are decreasing for x > Y by the lemma of article 5, since tk < Y. All in all, then, (1/x) logIR(x)I decreases for x > Y. That is what we had to prove.
Lemma. Under the hypothesis of the preceding lemma, (1/x)logIRj(x)I is increasing for aj < x < X j.
Proof. We drop the subscript j and argue as in the last proof. Here, we number the elements of AO and the whole multiples of L belonging to [X, Y] in increasing order, calling the smallest of the former .11, the next smallest A, and so on, and similarly denoting by t1 the smallest
integral multiple of L in [X, Y], by t2 the next smallest one, etc. In the present situation, the property
n(t) - n(X) < E(t - X), X < t E, that tk
<X+
k E
,
Ak.
(The reader may wish to make a diagram like the one accompanying the proof of the preceding lemma.) There are of course fewer Aks than tks, just as in the proof referred to. Denoting by N the number of the former, we can write 1_ -61,mil^11
-
-5 x-a + Y - log
a.
N
1
x+a k=1 x 1log(tk+x 1 + Y_ \tk - x/ k>NX
x
A kk
-
Ak+x
- x log 1
t
x
tk+x
I
for a < x < X. Using the lemma from the preceding article and its first corollary, we readily see that the right-hand expression is an increasing function of x for a < x < X. Done.
34
IX B Converse to Polya's gap theorem; general case
Combining the above two lemmas with those from article 4 we now get the Theorem. If n > 0 is taken small enough (depending only on the values of L and c), and if X j is large, the maximum value of (1/x) log I R,(x) I for x > 0 outside of (X,, YY) is attained for x = X3 or x = YY, and it is positive.
Proof. Under the hypothesis, we have, by the results in article 4, 1logIR(x)I x
5 0, 0,<x 0 (as usual, we suppress the subscript j). The maximum in question is hence > 0, and it is attained for a < x < X or for x > Y. Now use the preceding two lemmas. The theorem follows. 7.
Behaviour of (1/x)log IRR(x)I inside [XI, YI]
The essential step in Fuchs' construction consists in showing that (1/x) log I R, (x) I really gets larger inside [Xi, Y;] than outside that interval.
Lemma. If K > 0 is sufficiently small (depending only on L and e) and (1 + (K/5))Xj < < (1 + (4K/5))X j, is an integer not in A0, we have
,
X loglR.(X.)I +
log IR;OI
2lL-c
J
Ialog-, I
where
Xi,
1 + 6
provided that Xi is large enough (depending on K, L and E).
Proof. As usual, we suppress the subscript j. For X < < Y, we have
X+a) +
1
Y AeAon[X,Y] 1
+ nLe[X,Y]
1
-log Y log
K
X
K+ A
log
X-.1 X+ 2
+nL - -log X + nL X - nL - nL X 1
1
l
7 Behaviour of (1/x) log I Rj(x) I inside [Xj, Y;]
35
Let us start by looking at the third term on the right - the summation over 2.
Taking a K > 0 which is fairly small in relation to a (the number > 0 depending on c such that (1 + (a/4))X < Y < (1 + a)X ), we break up the sum in question as
I
+
>
(1+K)X..
X- 21, 23>22 and so on. As in the proof of the second lemma from the preceding article, we have
.lk>' X
+k. E
It is, in the first place, possible
that some of the
Al
with
, (1 + K)X. Denoting by S the set of indices k for which this occurs (if there are any), we can write 1
(1+K)X-< A-< Y
Zk,k i EKX + I
Zk,kES
)EAO
The first sum on the right is Y
+ 2,,
S
For X
X + k/e, the summand involving Ak is X + k/E 1
- log
e
X+(k - l l/E
t - C(
t-X
1
t+
Xlog t + X
dt,
because it is >, the integrand here for each t c- [X + (k - 1)/e, X + k/e] according to the second corollary of article 5. In our present sum, the index k ranges from the smallest integer >, eKX + 1 up to n(Y) - n(X) (which we
assume is not less than the former quantity; otherwise the sum is just zero). That sum is thus
/_
1X+(n(Y)-n(X))/E
)
-Xlogl t+X)}dt
llog(t-
+
e X + KX
where, by the lemma of article 5, the integrand is negative
(for X
,
(1
log(-
1
X
Y\
(I+K)X
-l
Ac AO
1/sK,
log
-X))
(A +X
- sa log 1- 90(U)-8O(K) = -9 a log ! + O(a)) where the O(a) term depends on a (and thus on s). We lay the sum X,z
L
-
o(logx)
for integers = (1 + 6)X between (1 + (K/5))X and (1 + (4K/5))X. When X /log X is large in relation to L/K, the right side of this inequality reduces to
-
Ka
L
lo
I
gK
-
O(K) L
O(a) L
for small K. Referring to the estimate for
(1+K)X 0 is small. Hence, given K > 0 there are at least (1 1 integers in [(1 +(K/5))Xj, (1 +(4K/5))Xj] to which the lemma applies when X j is large.
It is important that the ratio (1/x)logIRj(x)I has behaviour similar to that described by the preceding lemma when x is near the right endpoint Yj of [X j, Yj]. Arguing very much as in the long proof just given, but using the inequality
n(Y) -n(t)
e(Yj-t), XjtYj,
instead of n(t) - n(X j)
e(t - X), x j
t
Yj,
one establishes the
Lemma. If K > 0 is sufficiently small (depending on L and e) and (1 - (4K/5)) Yj 0 is chosen small enough (depending on L and s) for the results of articles 4 and 6 to apply. We next take an exceedingly sparse sequence of the intervals [X j, Yj] - by this we mean that the ratios X j+ 1 1X j are to increase very rapidly, in a way to be determined presently. The rest of the construction uses only the [X j, Yj] from this sparse sequence. In terms of these, we write
n=
U [X j, Yj]' j
(0, 00)
and finally put (D(z)
2z/L
=
ec2
2c)z1L 11
L
nLcQ
1 - z/nL
e2z/nL
1 + z/nL
nE N
X
f
1 - Z/a j
9i
e2gjz/aj
1+ z/aj
j
H
.
AEAO n Xj,Yj
I - Z/A 1+ Z A
e2z/P
Here, C is Euler's constant (see article 3).
The function O(z) is analytic in 91z > 0 and vanishes at the positive integral multiples of L outside our sparse sequence of intervals [X j, Yj], as well as at the points of AO lying within the latter. It also has a qj-fold zero at each aj. Using the group products defined in article 4 and studied in the previous two articles, we can write O(z) = F(z/L) fl Rj(z),
j
where
F(z) = z-zze(2
1 -z/n
zc)z
n=i
e2zin
l+z/n
= z-2ze2z sin 7rz (f'(z+ 1))2, 7Cz
a function already looked at in article 3. (In this last formula for 1(z), it is of course the product of the Rj(z) corresponding to our sparse sequence of intervals [X j, Yj] that is understood.)
Lemma. If Yj < (D(z)I
Z
0, is given, we have, for µk >, 21z1, 71 - Z/12k e2z/µk
exp
+ Z/1.13
C
2z5
2z3 3123
5123
exp3123 \1
+ °\123///
Clearly,
OI
1X12
I.
Hence
eogz1)µk>-
1 - Z/11k) e2z/µ1l
l
H C 1 +Z/123 21Z1
For 0 < µk < 21z I, since x = ttz > 0, 1 - Z/123
e2x/Pk (1),
e2z/µk
(1 +Z/12k)
whence 1 - z/1 0 depending on L and e. The quantity lim sup n
log I fi(n) I
oo
ne£
is finite and > 0.
n
IX B Converse to Polya's gap theorem; general case
48
Proof. Using the group products R3(z) constructed in article 4, we have (D(z) = F(z/L) f l R3(z) i
with the function F studied in article 3. For each fixed j, the modulus of = ai - z4i nL+z 2-z Ri(z)
ai + z
.,.[X;,,;] nL - Z
H
AEAon[X;,Y; ]
+z)
tends obviously to 1 when z -> oc. R3(z) also tends to 1 when z 0, and in a manner dependent only on the ratio I z I/X' 13, while otherwise independent o f j. To see this, recall that ai > i rjX, so that, for z < 4t1Xi, say, we can expand log Ri(z) in powers of z, as in the proof of the first lemma in article 4. As we saw there, the first degree term in z is absent from this expansion, and we can readily deduce from the latter that
IRi(z)-11
0 depended on L and s (refer to the first lemma in article 7). The right side of the last relation is therefore a certain strictly positive quantity S(L, e) dependent on L and E. The integers e E now under consideration are not divisible by L, so .
sin
i
sin L,
L
and it thence follows from the above inequality that log I
I
>
const. + log I R,(X,) I
+ S(L, E)
X,
for them when X, is large. The l; satisfying this relation are in E and also in the interval
C(1 + 5)Xi, (1 +
)X,].
5
An argument just like the one used to get them, but based on the second lemma of article 7 instead of the first, will similarly give us other E E, this time in
R
1-5)Y,, (i)],
such that
const.
log I
+
log I RI(Y,) I
+ b(L, e),
Y,
provided that X, is sufficiently large. From this and the preceding inequality we see in the first place that lim sup SET
log I _(_)
8 Fuchs' function D(z). Discussion
51
is certainly > 0 by the theorem of article 6 - it is, on the other hand, finite by the preceding lemma. The second statement of our theorem is thus verified.
For the first statement, we confront the two inequalities just obtained with the previous estimate on (log I (D(m) J )/m for in e A0 close to [X,, Y,]. In
view of the behaviour of the product f,IR;(m)I described earlier, we see in that way that lim sup log I (W) I eE
>,
lim sup m-'ao meAo
log I gy(m) I
+ b(L, a).
in
The theorem is now completely proved. We are done.
Discussion. Let us look back and try to grasp the idea behind this and the preceding 6 articles, taken as a whole. On the sequence A0, II(m)I is smaller by a factor of roughly a-8m than on a certain sequence E in the complement N - A0. It seems at first glance as though we had succeeded in `controlling' the magnitude of 1(m) on A0 by causing it to have zeros at the points of the latter contained in an extremely sparse sequence of intervals [Xi, Y3], that is, by using only an insignificantly small part of A0.
This is hard to believe. What is going on? The truth is that we are not so much controlling t(m) on A0 as making it large at the points of N - A0 in the intervals [Xj, YJ ID(m)I is of about the same order of magnitude outside those intervals whether in e Ao or not, as long as L does not divide in. It(m)I is made large inside the [Xi, YA
by what amounts to the removal of some of the zeros that F(z/L) has in them. The latter function vanishes at the points nL, n e N, and behaves like 2 sin ((ir/L)z) on the real axis; 4)(z) is obtained from it by essentially replacing its zeros in each [Xj, YY], which are about (1/L)(YY - Xj) in
number, by the elements of Ao therein, of which there are at most c(Y, - Xj). Since 1/L > a, we are in effect just throwing away some of the zeros that F(z/L) has in each interval [X,, YY] in order to arrive at t(z), and the result of this is to make 1 gy(m) I considerably larger than F(m/L) J at the integers in 0 Ao therein. Outside the [Xj, Yj] (where the modification has taken place), this effect is less pronounced. Its evaluation in the two cases (m inside one of the intervals or outside all of them) depends ultimately on the behaviour of factorials - that is the real origin (somewhat disguised by the use of integrals) of the (crucial) terms in a log 1/Q appearing
in the lemmas of article 7. It is the simple monotoneity properties of (1 /x) log I (x - A)/(x + A) l given
52
IX B Converse to Polya's gap theorem; general case
in article 5 that make the computations work out the way they do; those properties form the basis for Fuchs' construction with the factors 1 - z/a, e2z/
(1+z/A and the resulting appearance of the gamma function. The use of such factors leads of course to functions analytic in the right half plane rather than to entire functions. Analogous constructions with entire functions of exponential type would involve the somewhat more complicated monotoneity properties of (1/x) log I1 - x/A I or of (1/x) log I 1 - x2/221; for such work one should consult Rubel's 1955 paper and especially the one of Malliavin and Rubel published in 1961. Malliavin's very difficult 1957 paper is also based on use of the factors 1 - z/A
ezz/x
1 + z/A
(he works mainly with the logarithms of their absolute values), and is thus
in part a generalization of Fuchs' work. Keeping this in mind should help anyone who wishes to understand Malliavin's article. 9.
Converse of P6Iya's gap theorem in general case
Based on Fuchs' construction, we can now establish the
Theorem. Let 19 N have P6lya maximum density DE > 0. Given any D < DE, there is an analytic function
f(w) = Y, a w" nez
whose expansion in powers of w has radius of convergence 1, and which can be analytically continued across the arc
{e'9: -nD
for large n, so the power series 1)"e-1"4)(n)C(n)w" 7C
1
certainly has a positive radius of convergence. For I wI > 0 sufficiently small and
largwl < in
L
2e)
its sum must then be equal to
-
1 f"+1 2ni t _;,
sin i
wd
g(w)
by the preceding two relations. Let us look more carefully at the power series just written. The function
C(z) _ f 1 -z2 ( Az AEA,
vanishes at the points of A,. Therefore, since N - A, = E u AO our series can be written as 1(Y "EE
+ nEAo
)(_
I)ne-"(D(n)C(n)w".
56
IX B Converse to Polya's gap theorem; general case By our choice of y, limsupie-IND(n)I1/"
=
" cc
1,
ne£
and I C(n) 111" - + 1 as n --* oo with n 0 A 1, according to problem 29, part (f) (article 1). Hence 1
_ Y (- l)"e-Y"(i (n)C(n)w", it ne£ l
the f rst of the two power series into which our original one was split, has radius of convergence 1 and is equal, in { w I < I, to a certain function f (w), analytic there. It is at this point that we apply the main part of the theorem from the preceding article. According to that theorem, if the sequence of intervals [Xi, Y3] used in constructing the Fuchs function 1(z) is sparse enough, lim sup n- X
neAo
log 14)(n) I
lim sup
'
1. There is thus a function h(w), analytic for lwi < e ' and equal there to the sum of this second series.
For Iargwl < rc(DE - 1/L - 2e) and Iwi > 0 small enough, f (w) + h(w)
_ -Y (- 1)"e 7E
Yn(D(n)C(n)wn
1
is, as we have just seen, equal to - g(w), a function analytic in the whole sector
largwl < it D£* - 1
L
-
2c).
A Jensen formula for confocal ellipses
57
The formula g(w) - h(w)
f(w)
thus furnishes an analytic continuation of f (w) from the unit disk into the intersection of our sector with the disk {IwI < ea}:
Figure 165
The function given, for I w ( < 1, by the power series I it nez
( - 1)n e - n(D(n)C(n)wn,
having convergence radius 1 can, in other words, be continued analytically across the are {e19:
191 < ir(DE
- -
i
2e/}
of the unit circle. Here, e > 0 can be as small as we like and L is any large
integer < 1/e. Hence D = D* - 1/L - 2e can be made as close as we like to D. Our theorem is proved.
A Jensen formula involving confocal ellipses instead of circles Suppose that we are only interested in the real zeros of a function f(z) analytic in some disk { z < R} with f (0) 0. If we denote by n(r) the number of zeros of f on the segment [ - r, r], Jensen's formula implies that C.
f'n(P) dp
oP
1
('
2n o
-logIf(0)I
for 0 < r < R. This relation can be used to estimate n(r) for certain values
58
IX C A Jensen formula for confocal ellipses
of r, and that application has been frequently made in the present book: in Chapter III, for instance, and in §A of this one. Such use of it does, however, involve a drawback - it furnishes a kind of average of n(p) for p < r rather than n(r) itself. In order to alleviate this shortcoming, we proceed to derive a similar formula by working with confocal ellipses instead of concentric circles. The standard Joukowski mapping 2(
+ w1
1
w -+ z = -
w
takes { I w I > 11 conformally onto the complement (in C) of the real segment
[- 1, 1], and each of the circles Iwl = R > 1 onto an ellipse 4x2 (R+R-1)2
+
4y2(R-R-1)z
-
1
with foci at 1 and - 1:
Figure 166
For such an ellipse we have the parametric representation -i9
z
=
Rei9+eR
,
2
and as R increases, the ellipse gets bigger. Theorem. Let f(z) be analytic inside and on the ellipse ;s
z
=
Rei9+eR)'
Jensen formula for confocal ellipses
59
R > 1, and, for 1 < r < R, denote by N(r) the number of zeros of f (counting multiplicities) inside or on the ellipse i9
z
=
IIrei9 +
e
Then fIR
N(r)
dr = r
I f,2,,
d9 -
log
27r
1 rr
1
log 1f(x)1 dx.
,/(1-x2)
Proof. Like that of a theorem of Littlewood given at the end of Chapter III in Titchmarsh's Theory of Functions. (Our result can in fact be derived from that theorem.)
Suppose that 1 < ro < r1 5 R, and that f(z) has no zeros inside or on the boundary of the ring-shaped open region bounded by the two ellipses z
//
2(roe'9 1
"sl + ero
/,
/
z
21
-.s erl
Then log f(z) can be defined so as to be analytic and single-valued in the simply connected domain obtained by removing the segment.
(-
i(r1 + r1 1),
-
i(r0 + r0 1))
from that region:
Figure 167
Along the upper and lower sides of the removed segment, arg f will generally
have two different determinations. Indeed, by the principle of argument,
arg f (x + i0) - arg f (x - i0)
=
2rtN(ro)
for
-z(r1+rl 1)
x 0 and ask whether the finite linear combinations of these exponentials are uniformly dense in '(- L, L). If they are, the exponentials are said to be complete on [ - L, L]; otherwise, they are incomplete on that interval. If the are complete on [ - L, L], they are obviously complete on
[ - L, L'] for any L with 0 < L < L. There is thus a certain number A associated with those exponentials, 0 < A < 00, such that the former are complete on [ - L, L] if 0 < L < A and incomplete on [ - L, L] if L > A. We are, of course, not limited here to consideration of intervals centred at the origin; when 0 < A < oo it is immediate (by translation!) that the e';L^t will in fact be complete on any real interval of length < 2A and incomplete on any one of length > 2A. In the extreme case where A = 0, the given exponentials are incomplete on any real interval of length > 0, and, when A = co, they are complete on all finite intervals. Simple examples
show that both of these extreme cases are possible.
Regarding completeness of the exponentials on intervals of length exactly equal to 2A, nothing can be said a priori. There are examples in
which the e"^' are complete on [ -A, A] and others where they are incomplete thereon. Without going into the matter at all, it seems clear that the outcome in this borderline situation must depend in very delicate and subtle fashion on the sequence of frequencies
We will not consider
that particular question in this book; various fragmentary results concerning it may be found in Levinson's monograph and in Redheffer's expository article. What interests us is the more basic problem of finding out how the number A - L. Schwartz called it the completeness radius associated with the A. - actually depends on those frequencies. We would like, if possible, to get a formula relating A to the distribution of the A..
Completeness of sets of imaginary exponentials
63
This important question was investigated by Paley and Wiener, Levinson, L. Schwartz and others. A complete solution was obtained around 1960 by Beurling and Malliavin, whose work involved two main steps:
(i) The determination of a certain lower bound for A, (ii) Proof that the lower bound found in (i) is also an upper bound for A.
The first of these can be presented quite simply using the formula from the preceding §; that is what we will do presently. The second step is much more difficult; its completion required a deep existence theorem established
expressly for that purpose by Beurling and Malliavin. That part of the solution will be given in Chapter X, with proof of the existence theorem itself deferred until Chapter XI. The first step amounts to a proof of completeness of the ei -' on intervals
[ - L, L] with L small enough (depending on the A,,). The idea for this goes back to Szasz and to Paley and Wiener. Reasoning by contradiction, we take an L > 0 and assume incompleteness of the ei "' on [ - L, L]. Duality (Hahn-Banach theorem) then gives us a non-zero complex measure µ on [ - L, L] with L
j -L
e1 A
dµ(t) = 0 0 for the Fourier-Stieltjes transform
for each 2,,, i.e.,
µ(z) _
eizt d t(t) LL
The function µ(z) is entire, of exponential type < L, and bounded on the real axis. Using a familiar result from Chapter III, §G.2, together with the one from the preceding §, one now shows that for small enough L > 0, the zeros A,, of µ(z) cannot (in some suitable sense) be too dense without forcing ,u(z) = 0, contrary to our choice of p. The details of this argument are given in the following article. Before proceeding to it, we should observe how duality can be used to demonstrate
one very important fact: the completeness radius A associated with a sequence is not really specific to the topology of uniform convergence and the spaces W( - L, L). If, in place of 16( - L, L), we take any of the spaces LP( - L, L), I < p < oo, the value of A corresponding to a given sequence of frequencies A,, turns out to be the same.
64
IX D Completeness of sets of imaginary exponentials Suppose indeed that 0 < L < A. The space '( - L, L) is contained in each
of the LP( - L, L) and dense in the latter. Therefore, since uniform convergence on [ - L, L] implies LP convergence thereon, the linear combinations of the e"-, being uniformly dense in '( - L, L), will bell P dense in LP( - L, L). Let, on the other hand, L > A. Then the finite linear combinations of the are not II II P dense in any of the spaces LP(- L, L). To see this, we can II
take an L, A < L < L, and apply duality as above to get a non-zero complex measure µ on [ - L, L'] (sic!) with µ(),") = 0 for each n. If h > 0 is sufficiently small, the function r
w(t) =
1
2h
dµ(T) _,,
is supported on [ - L, L]; (p is clearly bounded, hence in each of the duals Lq(- L, L), 1 < q < oo, to our LP spaces. We have
pz=
sin hz hz
A(Z),
so in particular O(z) # 0, hence qp(t) cannot vanish a.e. on [ - L, L]. By the same token, however, J
L L
e'A"`an(t) dt = O(A) = 0
for each n, so finite linear combinations of the
cannot be dense in
LP( - L, L).
A considerable refinement of the preceding observation is due to L. Schwartz - Levinson also certainly knew of it: Theorem. If, for L > 0, the finite linear combinations of the e" are not uniformly dense in '( - L, L), removal of any one of the exponentials e"-' leaves us with a collection whose finite linear combinations are not dense in L, (- L, L) (hence not dense in any of the LP(- L, L), 1 < p < oo).
Taking the exponentials e'", n e Z, on [ - it, it], we see that this result is sharp. Finite linear combinations of the former are dense in each of the L( - n, n), 1 5 p < co, but can uniformly approximate only those
f e '(- it, it) for which fl-n)=f(n). In order to have (uniform) completeness on [- n, n] we must use one additional imaginary exponential e", A 0 Z (any such one will do!). Problem 31 Prove the above theorem. (Hint: Assuming a non-zero measure µ on
1 Use of formula from §C
65
[ - L, L] with j(A,,) = 0, take, for instance, A,, and look at the function (p supported on [ - L, L] given by qq(t)
=
e"e
eizl=dµ(i) -L
for - L < t _< L. Compute 0(z). ) 1.
Application of the formula from §C
Let us, without further, ado, proceed to this chapter's basic result about completeness.
Theorem. Given the sequence of distinct frequencies A > 0, suppose that for some D > 0 there are disjoint half-open intervals (ak, bk] in (0, oo) such that, for each k, number of A,,, in (ak, bk] bk - ak
and that Y_(bk - ak
2
= oo. k
ak
Then, if 0 < L < iWD, the exponentials ei ' are complete on [ - L, L]. Remark. The second condition on the intervals (ak, bk] has already figured in Beurling's gap theorem (§A.2, Chapter VII).
Proof of Theorem. Assume that the e"^` are not complete on [ - L, L], where 0 < L < 7rD. Then, as in the discussion immediately preceding the present article, there is a non-zero complex measure µ on [ - L, L] with A(A) = 0. The function A(z) is entire, of exponential type < L, and bounded on the real axis, indeed, wlog, (*)
IA(z)I
0.
As we saw at the beginning of this discpssion, ak
co. If µ(z) # 0, we
must also have
ask ->oo.
1
Indeed, such a function µ(z) satisfies the hypothesis of Levinson's theorem (Chapter III, §H.3), after being multiplied by a suitable exponential e'"Z (see the observation at the beginning of §H.2, Chapter III). According to that result, if we denote the number of zeros of µ(z) with modulus < r in the right half plane by n+(r), we have n+(r) r
L' , - for r -> oo,
it
where (here) 0 < L' < L. Therefore, if e > 0, we will certainly have n+(ak)
i
and
CL + E. 2bk
n+(bk)
ri
for all sufficiently large k, since ak and bk tend to oo with k. If now ak < (1 - e)bk for some large enough k, the previous inequalities yield n+(bk) - n+(ak)
0 is small enough. Once we know that a, -+ oo and bk/ak -+ 1 for k ---+ oo, we can be sure that the quantity sup bk2 + 1
M=
k
ak
is finite. Then, since logI2(x)I < 0,
bk log I f (x) k=1
nk
1 +x2
(,*k) yields
- irD(y - r1)2M- L sinh y
dx
tanh (i/2) cosh 211 k=1 C
bk - ak")2 ak
and the right side equals - oo by hypothesis. That is, C°
_00
log
I A(x) I
1 +x2
dx
the relation sought. The proof is complete. Remark 1. Beurling and Malliavin originally proved this result using the ordinary Jensen formula (for circles). For such a proof, a covering lemma for intervals on O is required.
Remark 2. In case the A. are not distinct but µ(z), having the other properties assumed in the proof has, at each of the former, a zero of order
equal to that point's number of occurences in the sequence, we still conclude that µ(z) - 0 by reasoning the same as above. Remark 3. If we assume the apparently stronger condition
-
bk - ak
k=1
bk
2
= 00
on the intervals (ak, bk], the appeal to Levinson's theorem in the above argument can be avoided. In that way one arrives at what looks like a weaker criterion for completeness. That criterion is in fact not weaker, for the condition just written is implied by the divergence of Y-k((bk - ak)/ak)2. Convergence of either series
is actually equivalent to that of the other; see the top of p. 81. Be that as it may, we are in possession of Levinson's theorem. It was therefore just as well to use it.
70
IX D Completeness of sets of imaginary exponentials
2.
Beurling and Malliavin's effective density D,
A certain notion of density for positive real sequences, different from the one used in the first two §§ of the present chapter, is suggested by the result proved in the preceding article. Starting with a sequence A of numbers A. > 0 tending to oo, we denote by n (t) the number* of A. in [0, t] when t >, 0 (as in §A), and take n,(t) as zero for t < 0. Fixing a D > 0, we then consider the set 0D of t > 0 such that n"(r) - nA(t)
T-t
>D
for at least one 'r > t. Since nA(t) = nA(t+), OD is open, and hence the union of a sequence of disjoint open intervals (ak, bk) c (0, co), perhaps only finite in number. It is convenient in the present article to have the index k start from the value zero. The (ak, bk) are yielded by a geometric construction reminiscent of that
of the Bernstein intervals made at the beginning (first stage) of §B.2, Chapter VIII, but different from the latter. Imagine light shining downwards
and from the right, in a direction of slope D, onto the graph of nA(t) vs. t for t >, 0. The intervals (ak, bk) will then lie under those portions of that graph which are left in shadow: nA(t)
a0
bo
a,
bl
a2
b2
t
Figure 170 * We allow repetitions in the sequence A; nA(t) thus counts the points in A with their appropriate multiplicities.
2 The effective density Dn
71
It is clear that each of the points A.,, (where nn(t) has a jump) must lie in oo we may one of the half-open intervals (ak, bk]. Therefore since A. enumerate those intervals so as to have
O-
= oo.
ak
If there are only finitely many (ak, bk), bk must be infinite for the last one of those by (*), so, if that one is contained in (a',, b',), say, surely b'', = oo, making (9D' substantial. Otherwise, (*) consists of infinitely many terms, each of which is finite. Then, in case bo = oo, we are done. When b'0 < oo, however, there must be intervals (a,, b,) with l >, 1 since an + oo and each A. is contained in some interval of (9D.. Taking, then, any 1 >, 1 and
IX D Completeness of sets of imaginary exponentials
72
denoting by
N,, N, + 1,
.
.
,
M,
the indices k - there are some - for which (ak, bk) C (a,,
b,),
we get M,
b; - a,
>,
k=N,
(bk-ak),
so that M,
(b', - a,)2 % Y (bk - ak)Z k=N,
(the possibility that M, = oo is not excluded here). Because 0 < ai 1< ak
for k > N,, we see that
akAdding
kY-,(b_a)2
(baa/2
both sides for the values of l>, 1, we obtain on the right a sum which differs from the one in (*) by at most a finite number of terms (those, if any, for which 1 < k < N 1). The former sum must thus diverge, making
L(
)
=
co,
and (9D- is substantial.
We are done. Definition. If a sequence A of (perhaps repeated) strictly positive numbers has no finite limit point, its effective density Dn is the supremum of the
D > 0 for which the sets (9D corresponding to A in the way described above are substantial. If none of the (9D with D > 0 are substantial, we put
DA = 0. Finally, if A has a finite limit point, we put Dn =
00-
The density Dn was brought into the investigation of completeness for sets of exponentials eiz^` by Beurling and Malliavin; its role there turns out to be analogous to the one played by the Polya maximum density D* in studying singularities of Taylor series on their circles of convergence.
We will see at the end of this article that DA is a kind of upper density, being the infimum of the (ordinary) densities of those measurable sequences containing A that enjoy a certain definite property, to be described presently.
2 The effective density D,,
73
It is convenient to extend our definition of Dn to arbitrary real sequences A.
Definition. If the real sequence A includes 0 infinitely often, DA = oo. Otherwise, DA is the greater of DA, and DA- for the positive sequences
A+ = An(0, oo), A_ _ (-A)n(0, oo)
(sic!).
( - A denotes the sequence { - An} when A = IAn}.) The result from the preceding article can then be reformulated as follows: Theorem. Let A be a sequence of distinct real numbers A. with D,, > 0. Then,
if 0 < L < 7CDn, the exponentials
are complete on [- L, L].
Proof. Consists mainly of reductions to the result referred to. Suppose in the first place that the An have a finite limit point, making DA = oo. We see then as at the beginning of the proof of the theorem from the preceding article that the eiz^` are complete on any finite interval. Having disposed of this trivial case, we look at A+ = A r)(0, oo) and
A. = (-A)n(0, oo). Assume, wlog, that Dn = Dn.; in that case we re-enumerate A so as to make A+ consist of the An with n > 1, and then with n > 1 are already complete on [ - L, L] for claim that the
0 ako for which nA(Tj) - nA(ako +)
>
Ti - ako
D.
There is, indeed, a r1 > ako such that nA(r 1)
- nA(ako +)
T1 - ako
i
D
(see the above diagram, and keep in mind that n,(t) = nA(t +) ). Then, however, r1 E (aka, oo) c (9D, so there is a T2 > T1 with nA(T2) - nA(rl)
>
T2-T1
D,
and similarly a r3 > r2 with HA(TS) - nA(T2)
>
D,
T3 - r2
and so forth. Since nA(t) increases by at least unity at each of its we must have nA(TJ) - oo. But then rj oo. are in the case where A, discontinuities
oo since we
Putting together the inequalities from the chain just obtained, we see that number of A,, in (aka, T;] Tj - akp
>
D
with T; - ) oo. If now the eiz^` were incomplete on [-L, L] for 0 < L < nD, we would as in the previous article get a non-zeror complex 0 for n > 1, and the zeros of ,u(z) measure µ on [-L, L] with in the right half plane would have density < L/ir < D by Levinson's theorem (Chapter III, §H.3). This, however, is incompatible with the with n >, 1 must hence be complete on [-L, L] previous relation. The in the event that one of the bk is infinite. There remains the case where (9D consists of infinitely many finite J.
intervals (ak, bk) with C bk - ak k>1
2
-
oo.
ak
Here, however, number of A. in (ak, bk] bk - ak
=D
2 The effective density Dn
75
of the ei2" on [ - L, L] for 0 < L < rxD is an immediate consequence of the preceding article's for k > 1, and the completeness result.
We are done. Corollary. The completeness radius associated with A is > RDA-
Remark. Work in the next chapter will show that in the corollary we actually have equality. That is the real reason for DA's having been defined as it was. This extension, due also to Beurling and Malliavin, lies much deeper than the results of the present §. We proceed to look at how DA can be regarded as an upper density. The following lemma and corollary will be used in Chapter X.
Lemma. If DA < oo for a sequence of (perhaps repeated) strictly positive numbers A there is, corresponding to any D > DA, a sequence E 2 A of strictly positive numbers for which
f
°° In.(t)-Dtldt
DA, we form the set
(9D = U (ak, bk) k>_0
corresponding to A in the manner described above, with the intervals (ak, bk) enumerated from left to right. By choice of D, (9D cannot be substantial, hence bo < oo and bk - ak
2
< 00. k
l
ak
Let us first find a continuous increasing function µ(t) such that °°
(t)
J0
I nn(t) + /1(t) - Dt I
1+t2
dt
0 and E is a strictly positive sequence such that
° In(t)-AtI Jo
l + t2
dt
A and form a set
(9D = U (ak, bk) k>_0
corresponding to the sequence E, following the procedure used up to now with positive sequences A. According to the definition of DE, it is enough * with appropriate multiplicity
78
IX D Completeness of sets of imaginary exponentials
to show that (9D is not substantial. In the following discussion, we assume that A > 0. When A = 0, the treatment is similar (and easier). Order the interval components (ak, bk) of (9D in the now familiar fashion:
O < ao < bo < a l < b, < a2 < .
.
.
.
It is claimed first of all that bo < oo. Suppose indeed that bo = oo. Then, co such as in the proof of the previous lemma, we obtain a sequence Tj that nr(ao)
D,
Tj -ao n. (T;) % D(T;-ao),
since of course nE(ao) = 0 (see figure near the beginning of this article). This means that n,(t) D(T; - ao) for t > T;, nE(t) being increasing. Therefore, if T; is large enough to make D
AT3
D
Aao
>
T;
(we are taking D > A !), we have
nE(t)-At
DTj-At-Dao > 0
for Tj 5 t < (D/A)T; - (D/A)ao : nE(t)
t
Figure 172
2 The effective density DA
79
Looking at a zj larger than (2D/(D - A))ao, we see from the figure that (D/A)tj-(D/A)ao nE(t) - At
ft'
1 + t2
dt
{ (D - A)zj - Dao }{ 1
Dzj-Dao -z; }
D \2 + Az - Aao JI
CD
(D - A)2 8A
z2 2
1+
A2
zi
and this is >, A(D - A)2/16D2 (say) for large enough zj. Since the zj tend to oo, selection of a suitable subsequence of them shows that
1°° In.(t)-At I o
1+t2
dt
a contradiction. Having thus proved that bo < oo, we are assured of the existence of intervals (ak, bk) with k 3 1, and need to show that 2
bk - ak
< oo.
ak
Considering any one of the intervals (ak, bk) in question,* we denote by Yk the straight line of slope D through (ak, nE(ak)) and (bk, n,(bk)), and look
at the abscissa ck of the point where Yk and the line of slope A through the origin intersect:
Figure 173
* bk < oo by the argument just made for bo
IX D Completeness of sets of imaginary exponentials
80
If Ck lies to the right of the midpoint, (ak + bk)/2, of (ak, bk), we say that the
index k > 1 belongs to the set R. Otherwise, when Ck < (ak + bk)/2 (as in the last picture), we say that k 3 1 belongs to the set S.
Let us first show that ak
bk
2
< 00.
ak
keR
When k e R, the situation is as follows:
ak
Ck
bk
I
Figure 174
It may, of course, happen that ck > bk. In order to allow for that possibility, we work with c'k
=
min(ck, bk).
The preceding figure shows that here f`k At - nE(t) dt ak
>
i(D - A)(ck - ak)(c - ak) > D - A (bk - ak)2
1 + t2
1
since c' - ak i
+
(ck)2
8
1 + bk
(bk - ak) for k e R. However, (ak, Ck) c (ak, bk) with the latter intervals2 disjoint. On adding the previous inequalities for k e R it thus follows by the hypothesis that (bk - ak)2
YkeR
1 + bk
al > 0. The last relation certainly implies that there cannot be infinitely many keR with bk
ak
> 4,
say,
so we must also have
(bk_ak)2 Y-
keR
< 00.
ak
We now show that 2
Y
keS
bk-ak
< 00;
ak
this involves a covering argument. Given k e S, we put
/
bk = min(bk +
DA-A (bk - ck),
b
D-A
k+A
(bk - ak)
so that
DA(bk 2A - ak)
< b-bk k
5
DA(bk-a A k),
and observe that nE(t) > At for bk < t < bk :
ak
Ck
bk
bk
't
IX D Completeness of sets of imaginary exponentials
82
We see that in the present case bkk n,(t) 6
- At
2(D - A)(bk - Ck)(bk - bk)
dt
I +t2
1 + (bk)2
(D - A)2 (bk - ak)2 8A
2
1 + D2 bk
What prevents us now from reasoning as we did when examining the sum Y-keR (bk - ak)2/bk is that the intervals (bk, bk), k e S, may overlap, although of course the (ak, bk) do not. To deal with this complication, we fix for the moment any finite subset S' of S, and set out to obtain a bound independent of S' on the sum (bk - ak)2
kE' 1 + (Dbk/A)2
For this purpose, we select certain of the intervals (ak, bk), k e S', in the following manner. First of all, we take the leftmost of the (ak, bk), k e S', and denote it by (ai, /3k). If (ai, /3j) is (ak,, bk,), say, we denote bk, by /3i; thus,
D2AA01-00 < IIi-I3
'
A was arbitrary. Q.E.D.
Putting together this and the preceding lemma, we immediately obtain the Theorem. Let A be a strictly positive sequence.* Then D, is the infimum of the positive numbers A such that there exist positive sequences E 2 A with
('°° In.(t)-At Idt 0
1 + t2
1 for all the curves y of the family G means that we require our p to make each of those have (gauged) length >, 1. We then look to see how small the (gauged) areas f f _9(p(z))2 dx dy can come out using the different conversion factors p fulfilling that requirement. The infimum of those gauged areas is our quantity A(-9, G).
WARNING Most authors work with the actual extremal length A(-9, G) equal to 1/A(Y, G), although at least one uses ),(-9, G) to denote our A(-9, G) and calls it extremal length. Some write A(.9, G) where we have 1/A(.9, G). Care must therefore be taken when consulting the formulas in other publications not to confound what we call A(-9, G) with its reciprocal.
Here are some practically obvious properties of reciprocal extremal length:
1. If G' c G, A(-9, G') 5 A(Y, G). Indeed, there are certainly at least as many weights p >, 0 admissible for G' as there are for G. 2. A(-9, G) is a conformal invariant; in other words, if p is a conformal say, and d consists of the images under p of the mapping of -9 onto
IX E Real zeros o f f (z) when f °° (log' I f (x)I/(1 + x2)) dx < oo
90
curves belonging to G, we have
G) = A(.9, G). To verify this, observe that if y e G and 0 defined on -9 for which f yp(z)Idzl >, I when y is any curve like
the one in the diagram. That relation must in particular hold when y is a line parallel to the x-axis, so we must have i
Jp(x+i)dx > 1 0
for 0 < y < h.
I Extremal length and harmonic measure
91
From this, by Schwarz' inequality, i
1(p(x+iy))2dx
l
p(x+iy)dx)
>,
( fo
2
>,
1,
so h
ft
h
(p(x + iy)2 dx dy
1'
0
and A(-q, G) >, h/l.
However, the function p(z) - 1/1 is admissible for G, because if y is any curve like the one shown, 1
lvi
dz =
leng thy
=
>, I
1
(!)
l
This p gives us exactly the value h/l for f f 9p2 dx dy. Therefore
A(-9, G)
=
h
l
for this particular situation. It is important to note that the computation just made goes through in the same way, and yields the same result, when we except a finite number of values of y from the requirement (on p(z) ) that p(x + iy) dx Jo0
for 0 < y < h. This means that we obtain the same value, h/l, for A(9, G) when -9 is a rectangle of height h and length I with a finite number of horizontal slits in it, and G consists of the curves in -9 joining Ys vertical sides (and avoiding those horizontal slits):
h
1
Figure 180
92
IX E Real zeros off(z)when$°.(log' I f(x)I/(1 +x2))dx < co
We can now show how extremal length can be used to express the harmonic measure of a single arc on the boundary of a simply connected domain -9. Given such a domain -9 with a Jordan curve boundary 8-9 and an arc a on 8-9, we take any fixed zo c- -9 and consider the family G of curves in .9 which start out from a, loop around z0, and then (eventually) go back to a:
(The curves belonging to G are not required to intersect themselves, although they are allowed to do so.) There is a precise relation between A(9, G) and wg(6, z0), the harmonic measure of a in -9, as seen from z0.
This is due to the conformal invariance enjoyed by both A(-9, G) and wl,(6, z0). Let us first map -9 conformally onto the unit disk A in such a way that zo goes to 0 and a to an arc j of the unit circle with midpoint at - 1:
Figure 182
Then
(OA6, ZO) = wA(6, 0) = I6I/2n,
1 Extremal length and harmonic measure
93
and, if G denotes the family of curves in A which leave &, loop around 0, and then come back to d (see the figure), A(-9, G) = A(A, G)
according to property 2. From this, we already see that A(-9, G) is a function of IQI = 27cwg(o, zo), because the whole configuration used to define A(A, G) is completely determined by the size of the arc d. Calling that
function 0, we have
A(-9, G)
=
zo)),
the relation referred to above.
If the boxed formula is to be of any use, we need some information about 0. With that in mind, we look first at the way A(A, G), equal to A(', G), is obtained. The reflection, y*, of any curve y e G in the real axis also belongs to G, because the arc d is symmetric with respect to the real axis, due to our having chosen it that way:
Figure 183
Therefore, if p > 0 (defined on A) is admissible for G, we not only have $,p(z) I dz >, 1 for any y e G, but also
f- p(z)IdzI
>,
I
for such curves y, i.e.,
j'P(ff)IdzI
>
1.
v
This means that p(z) is also admissible for G, from which it follows that z(p(z) + p(±))
IX E Real zeros off(z)when f °°.(log' I f(x)I/(1 +x2))dx < 00
94
is admissible for G, whenever p is. However,
I
P(z)
P(z)
)2dxdY
0 and mapping the rectangle {z: 0 < ¶3iz < 1, 0 < jz < h/2} conformally onto the quarter circle {z: z I < 1, 9iz > 0, jz > 0} in such a way as to take 1 to 1, hi/2 to i, and 0 to 0: Figure 186
0
1
..
0
1
Under this mapping, the upper right-hand corner of the rectangle goes to a certain point e'Q, 0 < /3 < ir/2, where #3 evidently depends on h. Successive Schwarz reflections in the x and y axes will now yield a conformal mapping
of the enlarged rectangle {z: --1 < biz < 1, - h/2 < 3z < h/2 } * The curves in H' join the left vertical side of Q' to its right one with endpoints omitted. That does not affect the computation; see the observation following it.
1 Extremal length and harmonic measure
97
onto the unit disk A, which takes 0 to 0, the right vertical side of the new rectan flee to the arc (C 'fl, ''), and its left vertical side to the opposite are
(-e'Q, -e-'Q):
Figure 187
From this we see that 2/3/n is equal to the harmonic measure of the rectangle's two vertical sides relative to that rectangle, as seen from 0. It is, however, obvious by the principle of extension of domain that this harmonic measure increases when h does, in fact, grows steadily from 0 towards 1 as h increases from 0 to oo: TY
0
Figure 188
1
'x
98
IX E Real zeros off(z) when$°°.(log' I f(x)I/(1 +x2))dx < ao
Given a, 0 < a < 2it, we may therefore adjust h so as to make /3 = a/4, and there will be only one value of h for which this happens. Taking that value of h (which depends on a), we denote by F,, the inverse of the last of the above conformal mappings (the one from the enlarged rectangle to A). If a is the length of our arc 6 on the boundary of the slit disk S2, the transformation
z -p maps f) conformally onto the rectangle
{z: 0 0, kh(a)
_
n
2log (1/Qa2)'
with a quantity Q tending to a limit # 0 as a -+ 0 Proof. The mapping cp used above to express ' is obtained by putting together a chain of simpler conformal mappings. The whole construction
100
IX E Real zeros off (z) when f °°. (log+ I f (x) I /(I + x2)) dx < oo
can be most easily presented using the following diagram. In it, Cot denotes a quantity asymptotic to some non-zero constant multiple of a for a ---).0.
That constant need not be the same in the different steps.
1@
Figure 191
1
27t
4
log(1/C.a2)
_
it
2log(1/C.a2)
,
Q.E.D.
Theorem. In the preceding theorem, wq(a, zo) lies between two constant multiplies of e-n/an(.Q,c)
(with absolute constants).
Proof. For small values of a = 2nw.9(a, zo), the statement follows immediately from the lemma in view of the relation A(2, G) =
Ii(a).
Because is strictly increasing, when either of the quantities w_q(a, zo), A(Y, G), is not small, the other is not small either. Hence, since
0 < w91(a, zo) S 1, the statement holds generally. Remark. Usually what is used is the inequality
w9(a, zo)
, 0 on -9 admissible
for the family G. To this feature is due the great practical value of the inequality.
Suppose we have a simply connected unbounded domain -9, with reasonably nice boundary. We fix some zo e -9 and take any R > dist(zo, 8-9). The intersection of -9 with the disk I z - zo ( < R will then have a connected component containing zo, which we denote by (OR:
Figure 192
The boundary of (9R consists of part of 8-9 and a certain number of arcs on the circle z - zo I = R. Some of those separate (9R from unbounded
102
IX E Real zeros off(z) when f °°. (log' i f (x) I /(1 + x2)) dx < oo
components of -9 - 9R (which must be present, -9 being assumed unbounded); we call the former distinguished arcs. Let us denote by 2 R the set of points in -9 which can be joined to zo by paths lying entirely in -9 and not crossing any distinguished arc. If 0-9 has sufficient regularity, which we are assuming, -9R will be a bounded domain. It may, however, contain points z with I z - zo I > R, and hence include (9R properly:
Either of these two arcs could have instead been taken as a (r).
Figure 193
We now single out at pleasure one of the distinguished arcs on I z - zo I = R
and call it S(R). S(R) is part of 0-9R. We are interested in estimating the harmonic measure w lR(S(R), zo) from above.
For each r, dist(zo, 0-9) < r < R, the circle of radius r about zo intersects -9R (sic!) in a number of open arcs. One or more of these must separate zo from S(R) in -9R; in other words, any path in -9R from zo to S(R) must pass through it (or them). We choose such an arc and call it r(r) (see the preceding figure). When several choices are possible, this may be done in fairly arbitrary fashion; we do require, however, that the selection be done in such a way as to make the union of the r(r) at least a Borel set in C. As long as a-9 is decent (which we are assuming), this is certainly possible.
Put 9(r) = l o(r)I /r for dist(zo, 82) < r < R; 9(r) is simply the angle subtended by the arc a(r) at zo. For 0 < r < dist(zo, a-9) we take 9(r) = oo (sic!). In our present set-up, we then have the
I Extremal length and harmonic measure
103
Theorem. (due essentially to Ahlfors and Carleman) w.9R(S(R), zo) < Cexp
(_7,f
dr
r 9(r) '
C being an absolute constant.
Proof. We use the preceding theorem, specifying a suitable weight p on -9R admissible for the family G of curves in 2R that loop around zo and have both ends on S(R). In this we are guided by the special computation for a rectangle made earlier.
Denote by S the union of the a(r), dist(zo, 8-9) < r < R. Then S c -9R; note that S need not be connected! We are assuming that it is a Borel set. For z e S, we put k p(z)
= Iz - zol9(Iz - zol)
with a constant k to be presently determined. When z e -9R lies outside the set S, we put p(z) = 0; this is consistent with our having taken 9(r) = 00 for 0 < r < dist(zo, 8Y). Our weight p will be admissible for G provided that
Si
p(z)Idzl
>
1
for each curve y in -9R starting from S(R), looping around zo, and then returning to S(R). In terms of the polar coordinates
re`* = z - zo with origin at zo, we have Idzl
=
V((dr)2+r2(d(p)2)
%
Idrl
(with possible equality) along the curves y; we thus require that
5 p(z)Idri
>,
I
Y
for y e G. Any of these y must, however, pass at least twice through each
of the arcs o(r), dist(zo, 8') < r < R; going and coming back:
104
IX E Real zeros off(z) when f '. (log' I P x) I /(1 + x2)) dx < o0
Figure 194
Our condition on p(z) = k/r9(r) will therefore be met if k is adjusted so as to make 2
k dr
fdR
=
isuzo,a2) r 9(r)
1,
i.e., if k fR
dr
1
=
o r 9(r)
2
Choosing the value of k which satisfies this last relation, we then have k2
ilL (p(z))2
r d(p dr
s r2(9(r))2
R
k
a ist(zo,a2)
a(r)
2
rdcpdr
r2(9(r))2 foR
f
k2
R
disgzo,a2) r9(r)
dr
=
dr
k
r,9(r)
=
k 2*
Here, the very first of the above integrals is by definition >, A(-9, G), so A(1, G)
k 1. The case where n = 1 needs also to be considered when f (z) has order < its treatment is like that of the one for n> 1, and somewhat easier 2;
* It is equal to cos(n/2p)F(1/p)/(p - 1), as is readily seen on putting tP = x, integrating
by parts and then taking the integral of z11P-' e" around a contour consisting
of the positive real and imaginary axes. Here, t(1/p) is clearly > 0 - look at the integral representation for F(x + 1) in §B.3!
I Extremal length and harmonic measure
107
than the latter. Taking, then, n > 1, we have certain curves Y1, Y2
, Yn
going out to oo with f (t;) tending to some limit as l; -> oo along any one of them, these limits being all different. We may obviously take each of the yk to be polygonal, and without self-intersections (just cut off any closed loops that yk may have:
Figure 195
Since f (t') tends to different limits as C -> oo along the different curves yk, two of those cannot intersect at points arbitrarily far out from 0. There is
thus no loss of generality in taking the yk disjoint, and in assuming that the origin does not lie on any one or them. The yk then bound n separate channels, or tracts, starting from a common central neighbourhood of 0 and going out to oo:
-92
Figure 196
0
z4
108
IX E Real zeros off(z)when $°°.(log + I f(x)I/(1+x2))dx < oo
We can index the Yk in such fashion that for k = 1, 2, ... , n - 1, yk and yk+, together bound one of these tracts, denoted by -9k, and that yn and y, bound one, called -9n. The preceding figure shows how things could look when n = 4; in it, the tracts -9, and -93 are shaded. The function f(z) cannot be bounded in any of the tracts 9k. Suppose, for instance, that f(z) is bounded in -9,. Closing up the `base' of -9, in any convenient fashion then gives us a simply connected region, part of whose boundary consists of the curves y, and y2, both going out to oo:
o
Yl Figure 197
f (z), bounded in that region and continuous up to y, and y2, then tends to two limits, say a, and a2, according as z ---+ oo along y, or along y2. In this circumstance, a well known theorem of Lindelof implies that a, = a2. Since, however, a, and a2 are two different asymptotic values of f, we have a contradiction. Problem 33 Prove Lindelof's theorem. (Hint: By means of a conformal mapping, one
may convert the region in question to the upper half plane and the function f to a new one, F(z), analytic and bounded for 3z > 0, continuous
up to R, and having the property that F(x) -> a2 for x
- co while F(x) -+ a, for x --+ oo. Apply the Poisson representation to G(z) _ (F(z) - a1)(F(z) - a2) (sic!), thus showing that G(z) -+0 uniformly for z oo in {r3z >- 0}.)
Having established that P z) cannot be bounded in any of the 22k, it suffices, in order to prove the Denjoy conjecture, to assume that n > [2p] (with f (z) of order p) and deduce that then f (z) must be bounded in some -9k. For this purpose, we take large values of r and look at the intersections
1 Extremal length and harmonic measure
109
Ek(r) of each of the tracts -9k with the circle I z I = r. Each Ek(r) is the union of one or more arcs; we single out one of them, called o(r), in such a way as to ensure that any path from 0 to ak(r) which touches neither yk nor Yk+1 (and hence stays in -9k after once entering that channel) must necessarily cut every Qk(r') with r' < r (the latter being defined for all sufficiently large r').
-
Figure 198
--
This we do for each k, taking care to select the ok(r) for different values of r in such fashion as to make their union a Borel set, which is clearly possible since the yk are polygonal curves. Calling 9k(r) = 1Qk(r)I/r, it is then evident that
+
91(r) + 92(r) +
2n.
The above picture shows that the sum on the left may actually be < 27r. Problem 34
(a) Show that if ro is fixed and large enough and R > ro, we have k=1 J ro
dr r&(r)
n2
R
27G
rQ g-
n
(Hint:
n
k=1
(b) Hence show that for some (fixed) k there must be arbitrarily large values of R for which R
dr
,p r9k(r)
n 27C
R
log r0.
110
IX E Real zeros off (z) when f ' , (log' I f(x) I /(I + xz)) dx < co
(c) Wlog, let the index k in (b) be unity. Take, then, the tract -9, and attach to it a bounded region containing 0 so as to obtain a simply connected unbounded domain -9:
For large R > 0, denote by -9R the set of points in -q which can be reached
by paths in -9 starting at 0 and not crossing a,(R). Show that for each z e -9 there is a number C. such that, for large enough R, R
z)
, [2p] + 1, show that f (z) is bounded in -9, and thus bounded in ',, yielding a contradiction that proves the Denjoy conjecture.
(Hint: f (z) is bounded on 8-9 since the part of that boundary lying outside some large circle consists of points either on y, or on yz. Fix any z e -9, take large values of R for which the conclusion of (b) holds
(with, as we are assuming, the index k = 1), and use the theorem on harmonic estimation (Chapter VII, §B.1) to estimate log I f (z)I in the domains 2'R. Note that on B-9Rn-9 = a,(R), loglf(C)I < O(1)+R°+E with e > 0 arbitrary. Apply the conclusion of (c).)
2.
Real zeros of functions f(z) of exponential type with
I'.(log + If(x)I/(1 + x2)) dx < oo Now that the Ahlfors-Carleman estimate for harmonic measure is at our disposal, we are ready to carry out the extension of the results from the preceding § described at the beginning of the present one. With
(log+ I f (x) 1/(l + x2))dx < oo
2 f (z) of exponential type and
111
that in mind, we turn again to the proof of the theorem in §D. 1, considering,
instead of the Fourier-Stieltjes transform µ(z), an entire function f (z) of
exponential type < L with (*) -'0.0
log+ If x) dx < 1+x2
oo.
Taking f(z) to vanish* at each point of a certain positive sequence we assume as in §D.1 that for some number D > L/rr there is a sequence of disjoint half-open intervals (ak, bk],
ak > 0, such that
number of A,, in (ak, bk] bk - ak
and 2
- ak Y( bk ak k
Our object is to prove that then °°
log
_00
I P x) I dx
1+x2
=
o0
from which it will follow by §G.2 of Chapter III that f (z) - 0. The argument starts out exactly as in §D. 1, and proceeds as it did there until we arrive at the examination off (z) in the ellipses
z = c+rcosh(y+i9) about the midpoint c = (ak + bk)l2
of one of the intervals (ak, bk], where bk - ak
2cosh y
0 much smaller than y, and denoting the number of An
in (c - R, c + R] by N (a quantity >, 2RD ), we find as before that 2rz
N(y-rl)
R/cosh rl. Here, however, the simple inequality log j f (c + r cosh (y +i9)) j< L r sinh y I sin 9
is no longer available for the estimation of the first integral on the right, because f is no longer assumed to be bounded on the real axis. Instead of boundedness, (*) is all we have to work with. Our adaptation of the earlier reasoning to the present circumstances is
nevertheless not altogether thwarted. In the passage from the previous relation to what corresponds to (t) of §D.1, there is a certain amount of leeway. Provided that the constant S > 0 is small enough, it is sufficient to have
logif(c+rcosh(y+i8))I < LrsinhyIsin9I
+
6R
for0
log R ro
- O(1) +
1 fR W(r) + `fi(r) dr.
7 ,o
r
(log' I f(x)I/(1 + x2))dx < oo
2 f(z) of exponential type and
121
In terms of the polar coordinates about i
reie = z - i, the integral on the right has a simple interpretation. It is none other than
I ('( rd0dr JJ f2n(ro ro}, of the union of the arcs a(r) for r > ro. K2 certainly includes the complement
of -9 in the region just mentioned, and may, indeed, include the latter properly:
Figure 206
Writing S2R = S2 n {z: ro < Iz - iI < R},
we see, going over to rectangular coordinates, that 1 f R (p(r) + O(r) dr Ir J ro
=
1 ( ('
dx dy
JJ
r
oRx2+(y-1)2,
whence finally
('R dr ro
r9(r)
log
R ro
- O(1) +
dx dy
JLR
1 + x2 + y2
122
IX E Real zeros off (z) when f °°. (log' I f(x) I /(I + x2)) dx < o0
Substitution of this into the above estimate for harmonic measure yields w.9R(6(R), i)
const. oo. By the last relation, divergence of
fr,
dx dy I + XI + y2
will then force V(i) = 0. As we have seen, however,
Y(x) i Rk for Ck - Rk 1< x < Ck + 3 Rk 3
when k e S', so, when k e S' is large, the rectangle of height Rk with base on 1Ck - Rk, Ck + 3 Rk] must lie in 0 (recall that the intervals 3 (Ck - Rk, Ck + Rk) S (0, 09) are disjoint and that Ck k 00 !):
Rk
Ck - 3 Figure 207
Therefore, fixing a suitable ko, 1
dx dy
J n1+x2+y2
R2 k,S
k>ko
3(1 + c. + Rk)
Ck
Rk
Ck + 3
2 f(z) of exponential type and f °. (log+ I f(x)I/(1 +x2))dx < o0
123
Again, since (ck - Rk, ck + Rk) C (0, oo), ck i Rk, and, since ck k oo,
Ck > some number A > 0 for k >, ko. From the preceding relation we thus get Rk
A2
dx dy
n1+x2+y2
6A2+3
kes'
k->ko
2
CCk
Divergence of the sum on the right therefore makes the integral on the left infinite, which, as we have just shown, implies that V(i) = 0. But V(i) > 0. The right-hand sum must therefore converge, so that I (Rk/ck)2
keS'
L/n, there is a sequence of disjoint intervals (ak, bk], ak > 0, with bk k
ak
-
2
= 00
ak
and
number of An in (ak, bk]
D
bk - ak
for each k, then f (z) - 0. Remark. The number L enters into this theorem solely on account of ($). We may therefore replace the condition, figuring in the hypothesis, that
f be of exponential type < L by the simpler requirement that f be of
* with appropriate multiplicity at any repeated value A.
2 f(z) of exponential type and
$oo.
(log+ I f(x)1/(1 + x2))dx < co
125
(some) exponential type, with lira sup
log I f(iy) I
and
urn sup
y_-1
y
Y_oo
log I f 0y) I
lyl
both < L (see Chapter III, §E). Of course, the latter in fact implies that f (z) is of exponential type < L (see the last theorem in §E.2, Chapter VI and especially the discussion in §B.2 there), so that, logically, we have gained nothing. It is nevertheless often easier in practice to estimate the two limsups than to obtain a good upper bound on f's exponential type by direct examination of that function.
Once the result just stated is available, it is useful to bring in the effective
density D,, discussed in §D.2. Arguing as at the very beginning of the present §, we then obtain the following propositions: Theorem. Let A be a sequence of real numbers on which a non-zero entire function f of exponential type vanishes,* with lim sup log I f 0y) I
and
lim sup log I f 0y) II y
Y_oo
y-. - oo
Iyl
both < L and 1
log + J01 dx
1 tending to oo as x -k + oo and a number A > 0, suppose that °
log WA(x)
_OD
1 +x2
dx
A/1L, (ffA is not II -dense in Ww(R). 11
To compare this improved result with the original one, it suffices to note that there exist positive sequences having ordinary density equal to zero for which the effective density is infinite. Again, we can now see that the entire function (D(z) of exponential type A with 1 °°
log+ I4(x)I
_00
1+x2
dx
oo if w e -9 is fixed and finite. At the same time, Gq(z,w) is supposed by definition to stay > 0. In complex variable theory, there is a standard procedure for adapting the various notions of local behaviour originally defined for the points of
C to the point at co on the Riemann sphere. One first uses a linear fractional transformation to map oo to a finite point a, and then says that a function defined near and at oo has such and such behaviour there if the one related to it through the transformation behaves thus (in the usual accepted sense) at a. The Green's function for a domain -9 on the Riemann sphere having reasonable boundary, with oo c-.9, is defined in accordance with that convention: taking a linear fractional transformation q0 which
maps -9 onto some domain 9 c C, we put G_q(z, w) = Ge((p(z), *p(w)),
z, w e 1.
In this extension of the definition of G q(z, w) to domains -9 containing oo, all of the usual general properties of that function holding for domains
-9 g C are preserved. That is, in particular, true of the important symmetry relation G_q(z, w) = G_q(w, z),
128
IX F Pfluger's theorem and Tsuji's inequality
established at the end of §A.2, Chapter VIII. We see also that when w e -9 is not equal to oo, G q(z, w) is described by just the ordinary specification:
As a function of z, G_q(z, w) is continuous and > O on 9 - {w}, harmonic in -9 - {w} (including at oc), and zero on 89. Near w, it equals log (1/ I z - w l) plus a harmonic function of z.
For w = oo, however, our definition leads to the following description:
G_9(z, oo) is continuous and > 0 on -q - {oo}, harmonic in -9 - { oo }, and zero on 0-9. Near oo, it equals log I z I (sic!)
plus a function harmonic in z (including at oo).
Keeping these characterizations in mind, we easily obtain an integral representation for Gq(z, oo) akin to the one given above for G,?,(z, w) and bounded domains .9. Fix, for the moment, a finite w e -9, and consider the function
h,v(z) = log Iz - wl + G_9(z, w) - G.9(z, oo). According to the descriptions just made hW(z) is bounded and harmonic in -9 (including at z = w and at z = co ), and continuous up to 8'. Therefore, as long as 89 is decent (which we are assuming throughout this § !),
h,(z)
=
f
z)
ag
=
a log I
- w l dco (C, z).
oo, loglz - wI - G_9(z, oo) = 0(1) + log IzI - G_9(z, oo) tends As z to a certain finite limit, which we denote by - yg, so then
hw(z) - G-9(oo, w) - 7-9 Making z
oo in the previous relation, we thus obtain
-y.9 + G.9(oo,w)
=
J
a,
loglC-wIdoo.9(C, oc)
After using the above mentioned symmetry property and then replacing w by z, this becomes
Gg(z,oo)
=
y.9 + f
ag
the integral representation sought.
oo),
I Logarithmic capacity and the conductor potential
129
From the last formula, we see in particular that 1
log
dco.,(C, oo)
=
y_,
for z E a_9.
fa_q
oo) of total mass 1 supported on 0-9 has The positive measure coy( constant logarithmic potential equal to y.9 thereon. y.9 is called the Robin constant for -9.
If we imagine a very long metallic cylinder perpendicular to the x - y plane, cutting the latter, near its own middle, precisely in 0-9, and having its
different pieces joined to each other by thin perfectly conducting wires,
an electric charge placed on it will distribute itself so as to make the
electrostatic potential constant thereon. Near the x - y plane, that equilibrium distribution will depend mainly on z = x + iy and hardly at all on distance measured perpendicularly to the cylinder's cross section; the same is true of the corresponding electrostatic potential. To within a constant factor, the latter*, near the x - y plane, is practically equal to a logarithmic potential in z = x + iy corresponding to a measure giving the amount of electric charge per unit of cylinder generator length. The second of the aboved boxed formulas shows that if the whole cylinder carries one unit of electric charge per unit of length (measured along a generator), the electrostatic potential (see footnote) at equilibrium is 1
log a.9
z
I
oo)
(near the x - y plane); this is called the logarithmic conductor potential (or equilibrium potential) for a-9. The constant value that this potential assumes on 0-9 is equal to the Robin constant y_q. In physics, the capacity of a conductor is the quantity of electricity which must be placed on it in order to raise its (equilibrium) electrostatic potential to unity. That potential is there taken as log L + yg instead of y9 when dealing with a long cylinder of length L bearing one unit of electric charge
per unit of length (see footnote); in this way one arrives at a value L/(log L + yg) for the capacity of the cylinder. Even after division by L * measured from a certain reference value depending on the cylinder's length and net electric charge, but not on its cross-sectional form
130
IX F Pfluger's theorem and Tsuji's inequality
(in order to obtain a capacity per unit length), this quantity shows practically no dependence on y.9, i.e., on the cylinder's cross section, because the
length L is assumed to be very large. That is why mathematicians have agreed on a different specification of logarithmic capacity. Definition. The logarithmic capacity, Cap D Q, of the compact boundary 0-9 is equal to e-vQ
where y is the Robin constant for -9.
The logarithmic conductor potential and measure wg( , oo) corresponding to it are characterized by an important extremal property. From physics, we expect that the equilibrium charge distribution
oo)
on 0-9 should be the positive measure u of total mass 1 carried thereon for which the energy (cf. Chapter VIII, §B.5)
log
I
z1
dlt(z) I
is as small as possible. That is true.
Lemma (goes back to Gauss). If it is a positive measure on 0-9 with
µ(a-9) = 1, Jam
I
1
>,
dµ(g)dµ(z)
y..,
log -z
and equality is realized for u = coq(
,
oo).
Remark. It is not too hard to show that equality holds only for y = co( _q, co). We will not require that fact. Proof of lemma. Since
I
log
1
dcwg(1;, co)
=
zea!2,
e9
we have
Jai I
loglz
1
_Idwq(C,co){dµ(z)-dw.9(z,co)}
=
0
1 Logarithmic capacity and the conductor potential
131
for any measure p on 09 with p(39) = cog(8-9, oc) = 1. Now*
J a J a log z-1 (
-y
dµ(C) dµ(z) CI
_
1
f
z-
Jam
_
CI{dp(i)du(z)-dwq(C,oo)dwp(z,oo)}
1
. log
J
I
(dµ(() + dw.(C, oo))(dµ(z) - dwg(z, cc))
z-
_
I
1
log z- CI
(dµ(C) - dw.(C, oo))(dp(z) - dcog(z, oo) )
I
+ 2J
.
log I
1 CI
l
z
dwq(C, oo) { dp(z) - dco-9(z, oo) 1.
In the last expression, the second double integral is zero, as we have just seen. However, f a.9 (dp(C) - dwq(C, oo)) = 0, so the first double integral in the last expression is positive, according to the scholium and warning just past the middle of §B.5, Chapter VIII. (The argument alluded to there works at least for sufficiently smooth signed measures of total mass zero supported on compact sets. The positivity thus established can be extended to our present signed measures p - wg( , oo), with p not necessarily smooth, by an appropriate limiting argument (regularization).) We see in this way that the first member in the above chain of equalities is positive. That's what we had to prove. The following exercise gives us an alternative procedure for verifying that
JJ.JJ.log Iz - CI
(du(() - dw.g(C, oo))(dp(z) - dcog(z, oo))
>
0
1
when p(09) = 1. Problem 35. (Ahlfors) (a) Show that for large R, dx dy y
!:!<xIz-CIIz-wl
2n log
1
IC-wl
+ 2n log R + C + 8(C, w, R),
where C is a certain numerical constant (its value will not be needed) and 8(C, w, R) uniformly for C and w ranging over any compact set
in C, as R
oo.
(Hint: Wlog, w - C = a > 0. Use the polar
coordinates re's = z - C and, in the double integral thus obtained, integrate r first.) * note that
10sfaslog(1/iz-CJ)dw9(C,oo)du(z)
=
yq, afinitequantity!
IX F Pfluger's theorem and Tsuji's inequality
132
(b) Hence show that log
aj a
1
(dµ(1') - dw,(1, oo))(dp(w) - dco,(w, co))
3
0
K - wl
for any positive measure y on 0-9 having total mass 1(the double integral
may be infinite). Pay attention to the problems of convergence and of possibly getting oo - oo. (Note that under our assumptions on a-9, yg is certainly finite.)
2
A conformal mapping. Pfluger's theorem
Let us fix any finite union E of closed arcs on { j z = 1 } and a
simple closed curve F about 0 lying in the unit disk A. The unit circumference and r bound a certain ring domain A. lying in A, and we denote by G the family of curves 2 lying in Ar and going from the set E to F:
Figure 208
In this article we will be mainly concerned with the reciprocal extremal length A(Ar, G)
(see §E.1 for the definition and elementary properties of reciprocal extremal length). Pfluger found a relation between A(Ar, G) and the logarithmic capacity of E defined in the preceding article. That relation also involves the curve
I of course, but in fairly straightforward fashion. To arrive at Pfluger's result, we construct a conformal mapping of A onto a certain disk with radial slits.
2 A conformal mapping. Pfluger's theorem
133
Let.9 = (Cu { oo }) - E, so that 8-9 = E. With the harmonic measure , z) for this domain, we form the conductor potential for E,
wq(
=
UE(Z)
Iz
dwg(t", oo),
1
J'EE
CI
described in the last article. Our conformal mapping will be constructed from the function VE(Z)
=
UE(Z) + UE C 1) Z
We have, first of all, VE(Z) = VE(l/z). Explicitly, since E s{ I C I = 1 }, VE(Z)
log
= JE
z
(z - l;)(1 - (Z)
+
log Izl
dw.q(l;, oo)
2 JE 1 log
dwq(t;, oo),
1
Iz
CI
from which we see in particular that
VE(z) -+ - oo for z --* 0 and hence also, that
VE(Z) -- - 00, Z -* 00. Since VE(z) is harmonic in E ' E - {0} and VE(z) = 2y_9 for z e E, the previous relations and the maximum principle imply that VE(z) < 2y.9
for z 0 E.
The function VE(z) has a harmonic conjugate VE(z) in {0 < I z I < 1 }. VE(z)
is of course multiple valued there; we proceed to investigate its behaviour. For 0 < r < 1, we have VE(re s)
=
E
log
r 1 + r2 - 2r cos (9 -
r)) dwq(e,t' oo),
from which 8VE(rei9)
8r
_
1
('
1 - r2
r El+r2-2rcos(9-i) dw,(e;`, cc),
134
IX F Pfluger's theorem and Tsuji's inequality
whence, by the Cauchy-Riemann equations, 1 - r2
3VE(rei9)
dcog(e'i, co).
89
From this we see, firstly, that VE(rei) is a strictly increasing function of 9 for each fixed r, 0 < r < 1, and, secondly, that VE(rei9) increases by f2,,
o
1 - r2
El
+r2 - 2r cos(9 - i) oo)
21L
dcoy(e", co) d9
=
2n
E
fE
when 9 goes from 0 to 2n. The determinations of VE(z) are, however, well defined and single valued in the simply connected region {z:
0<argz 1. Proof. Uses the minimum property of the conductor potential. Denoting the unit circumference by K, let us write
I = {IzI> 1}v{oo}, so that 841 = K. If E 9 K, any positive measure y of total mass 1 supported on E is certainly supported on K, so, by the extremal property in question (lemma at the end of the preceding article), Ye
5 JE JE
Choosing y = wq( YS
-, 1,
0
we see that ye = 0. Therefore y.9 > 0, as required.
138
IX F Pfluger's theorem and Tsuji's inequality
Lemma. For I z I < 1, (1 +
(1
IfE(Z)I
I
IzD2.
IZI)2
Proof. By the boxed formula from the theorem, log
fE(Z)
2JE log
z
1
I1- Z I
dwq(C, cc).
The integral on the right evidently lies between 21og(1/(1 + IzI)) and 2log(1/(1 - IzI)). The lemma follows.
Consider now any simple closed curve F about 0 lying in A, and write
Mr
su p (1 ZEr
mr = inf
IzI
-
IZI)2,
IzI
zer (1 + IzI)2
For a finite union E of arcs on the unit circumference, CapE = e-,", and the quantity A(Ar, G) are then related through Pfluger's Theorem. If I z I < (3 - V5) for z e F (so as to make Mr < 1), z the logarithmic capacity of E satisfies the double inequality M1/2e-n/n(er,c)
< CapE
0 is any weight on of admissible for the family H (in the parlance of §E.1), the weight p*(z) on S2 equal to p(z) on S2;- n of and to zero on 52;. n - of is certainly admissible for d, so, by definition (§E.1), A(KIr,G)
0. Descriptions of such constructions are found in many books, including the ones by Nevanlinna, Kahane & Salem, and Tsuji. Problem 36 (a) Let E be compact and composed of a finite number of smooth arcs, and put 2 = (C - E) u { oo }. If f (z), analytic in 2, has modulus < 1 there and is zero at co, show that lim I zf(z) I _< CapE
(Hint: Look at log I f(z) I + G_9(z, oo). ) (b) Let E be a finite union of closed arcs on the unit circumference. Show that
CapE _>
sin (I E I/4).
(Hint: With
= 2n- Ee"e"+z dt, -z 1
(p(z)
ze2,
take
exhp(z)/2
f(z)
=
-
erziW(O)/2
z(erzi(p(.)/2 +e - xi'0'2)
and use the result of part (a).)
Remarks. This proof of the inequality in (b) is from Lebedev's book on the area principle. The inequality becomes an equality for single arcs E. The
supremum of the left-hand limits in (a) for the kind of functions f considered there is called the analytic capacity of E, and the f given in (b) actually realizes that supremum. For more about analytic capacity
and its role in approximation theory the reader should consult the monographs by Garnett and by Zalcman. Combination of the lemma just proved with Pfluger's theorem shows that if E is a finite union of arcs on the unit circumference and I a simple closed curve about 0 on which IzI < 1(3 - ,/5), we have (EI 2n
< 2mr ii2e-n/n(er,c)
where mr = infcEr(IzV(1 + IzI)2), and A(Ar, G) is related to E and r in
3 Application to estimation of harmonic measure
147
the way described at the beginning of article 2. The use of this relation to
estimate harmonic measure for simply connected domains comes immediately to mind on account of the conformal invariance of both harmonic measure and extremal length. Some control on the quantity mr is of course needed if the results obtained are to have any practical value. We use a couple of the elementary properties of univalent functions for that; those are covered in many texts, for instance, the ones by Nehari and by Markushevich. Lemma. Let (9 be a simply connected domain with zo e (9, and put
Ro = dist(zo, 0(9). If u is the circle I z - zo I = Ro/16 and (p is a conformal mapping of (9 onto the unit disk A with cp(zo) = 0, p takes or onto a closed curve F about 0,
such that
IwI
1/4, i.e., R01 p'(zo)I %
I
.
According to the distortion theorem, If(O I
1
dist(zo, 80), we denote by 0, the component of
0 n {Iz-zol < r} 0, is bounded by all or part of 80, and, in the second case, by certain arcs on the circle I z - zo I = r as well. We call containing the point zo.
the union of these E(r) (understanding that E(r) may be empty), and write
0(r) = IE(r)I/r.
For 0 < r < dist(zo, 80), E(r) consists of the entire circumference Iz-zol = r, and then we put 0(r) = oo (sic!) as in §E.1.
152
IX F Pfluger's theorem and Tsuji's inequality
Fixing any R > dist(zo, a(9), we look at harmonic measure coon( for the domain (9R. Concerning the latter, we have the
, z)
Theorem. (Tsuji's inequality)
dr rz
we,,(E(R), z0)
, 0 on (9, admissible for the family S. Then, by definition (§E.1), we'll have
A(OQ,S)
\
Q(p(z))2dxdy.
JJ a
3 Application to estimation of harmonic measure
153
Write
=
E
U E(r)
,
dist(zo,a0)- a as t with y > 0, the inequality just written becomes an equality, showing that
U(iy)/y -> ira for y -p o o. These facts imply that
U(z) = ira jz +
! _. 3z U(t) dt IZ_tI2
for 3z > 0
by an argument exactly like the one used to prove the theorem of §G.1, Chapter III. Plugging the previous relation into the integral on the right
162
X A Meaning of term `multiplier theorem' in this book
then gives
Cm(It1)dt
1
U(x+i) < 0(1) -
n jtj_A(x-t)Z+1'
whence, since m(I t 1) increases with I t 1,
U(x + i)
0(1) -
,A.
Taking a larger 0(1) term of course ensures this estimate's validity for all real x.
The idea now is to observe that the integral
1- z2 2 t
dv(t)
would represent the logarithm of the modulus of an entire function of exponential type, if the increasing function v(t) were integer-valued. Our v(t), of course, is not (it is absolutely continuous!), but one expects that log 0 f`0
z2
1- 2
d[v(t)]
(with [v(t)] designating the greatest integer < v(t) ) should be close to
U(z) = J
1- z2 2
log
t
0
dv(t).
This is indeed true as soon as z gets away from the real axis, and we have the following simple
Lemma. If v(t) is increasing and 0(t) on the positive real axis,
1- z2 2 t
(d[v(t)] - dv(t)) log f max(IxI, lyi) + 1Y1 1 2max(Ixi, 1 Yi)J 1 2iyi
fort=y
0.
Proof. Assuming that 3z :A 0, integrate the left-hand member by parts. Because v(t) = 0(t), the integrated term vanishes, and we obtain
MO - N01) 0 f0,0
at
logl 1 - rz dt.
I Weight even and increasing on positive real axis
163
Fixing z, let us introduce the new variable C = z2/t2. As t runs through C moves in along a certain ray 2' coming out from the origin. When'Rz2 0, the distance I i - t'I decreases as t increases, so, since v(t) - [v(t)] 0, the expression just written is 0, 11 -1; I, for increasing t, first decreases to a (0, oo ),
minimum value 13z21 / 1 z 12 and then increases, tending to 1 as t -% oo :
Figure 222
Hence, since 0 < v(t) - [v(t)] < 1, the expression under consideration is log(Iz12/13221).
We see that f o log I 1 - (z2/t2) I (d[v(t)] - dv(t)) is
log(21XI 1YI
+ 21x1)'
Ixl > IYI
The right-hand side can be represented by the single expression log
max(Ix1,JY1) 21Y1
IYI
+ 2max(Ixl,IYl)
The lemma is proved Using the lemma with our function U(z), we get I
0
'0logll-(x t2 i)2 d[v(t)] < log+lxl + U(x+i),
164
X A Meaning of term 'multiplier theorem' in this book
so, by the relation established above, (x + i)2 2 d[v(t)] t
J0log
1, denote by Ak the positive value of t for which v(t) = k. Then, noting that v(O) = 0, we have log
1- z2 Z t
J 0`0
Z2
d[v(t)]
= log H k=1
1
-Az k
so, putting I
qO(z)
=H
z2
1
k=2
A2
(sic!),
the preceding inequality yields
5 0(1) - 2m(Ix1),
loglq(x+i)1
xel R.
Arguing as we did above for U(z), we find without trouble that loglcp(z)I , 1, increasing when x > 0 and decreasing for x < 0, such that log M(x)
1+x2
dx
ITD,, are given.
Beurling and Malliavin obtained a complete solution of this problem around 1961. Considerable effort had previously been expended on it by
others who had succeeded in finding various constructions of entire functions G, subject always, however, to restrictive assumptions on the sequence A. This was done by Paley and Wiener and then by Levinson; later on, Redheffer obtained a number of results. I have worked on the question myself. Many of the methods devised for these investigations are still of interest even though they were not powerful enough to yield the final definitive conclusion; some of them indeed find service in the present book. The reader who wants to find out more about these matters should consult Redheffer's survey article (in Advances in Math.), which gives a very clear exposition of most of what has been done. There, the delicate question of completeness of the on intervals of length exactly equal to 2nD,, is also discussed. Before going on to article 1, let us indicate how the work will proceed. We are given a sequence A c 118 with DA < oo.* Picking any q > 0, we
wish to construct a non-zero entire function G(z) of exponential type ir(D,, + 3r1), say, such that
G(2) = 0 for AeAt and
IG(x)Idx < oo. Because the distribution of the A e A may be very irregular, it is not * It is best to allow A to have repeated points; that makes no difference for the constructions to follow. t with, of course, appropriate multiplicity at the repeated points of A
168
X B Completeness of sets of exponentials on intervals
advisable to start with the Hadamard product
fl
.ieA
(i_f)e.
z#o
Instead, we first turn to the second lemma of §D.2; Chapter IX, and to its A corollary. Given D > DA, these provide us with a real sequence E for which °°
I nE(t) - Dt I
J_.
1+t2
, 0, and minus
the number of such points in [t, 0) if t < 0. For our purposes, we take D = DA + U. The points of E are already quite regularly distributed. Assuming, wlog,
that 0 0 E, we form the function F(z)
=
fl C 1 -
z e=/A,
ACE
which turns out to be of exponential type. Its behaviour is worked out in article 1. The next step (in article 2) is to prove what is called the little multiplier
theorem. This result (which, strictly speaking, is not a multiplier theorem in the sense adopted for that term at the beginning of the present chapter) gives us a non-zero entire function cp(z) of exponential type rcrl such that °°
log' I F(x)(p(x)I
j_00
1+x2
dx
0 and put D = Dn + rl, we take a real sequence A having, perhaps, repetitions, such that °°
f-.
In.(t)-DtIdt
0, we can write
1-x2
log I F(x) I
=
log dEE+
22
+
(,z _
x
x
A
2
)(log 1+eE
+
Since limt_.(nE+(t)/t) exists, the first sum on the right is ( o(x) for x -> oo by problem 29(d). (For the solution of parts (a) - (e) of that
1 Hadamard product over E
171
problem, it is not necessary that the zeros of the function C(z) considered there be integers - they need only be real and positive.) What is left on the right side of the previous relation can be rewritten as
\logl 1 + t ) - 0(dn,
Jo
dnT
This is integrated by parts, upon which all the integrated terms vanish ( nE + (t) and nE (t) are zero for 0 < t < 1 ! ), and we end with °
nE- ((
x2
fo" x + t
dt.
t2
Since
f
nE. (t) - nE (t) t2
J
dt
nz(t) - Dt
=
t2
JItI1
0,0
dt
with the right-hand integral absolutely convergent and equal to c, the left-hand integral is also absolutely convergent and equal to c, so the
previous expression is - cx for x - oo. We see that
logIF(x)I < cx+o(Ix1)
for x
co.
In like manner, the same is seen to hold for x - - oo. The function e-`ZF(z) is thus in modulus < e°(IXI) on the real axis when x is large, and
has the same growth as F(z) on the imaginary axis; it is, moreover, of exponential type. Our desired result now follows by application of a Phragmen-Lindelof theorem, as in part (e) of problem 29.
We shall have to look more closely at the behaviour of I F(x) I on the real axis. Of course, logIF(x)I
f.(
=
log
x
x
t
t
1-- + - dnE(t),
x c- R.
Regarding integrals like the one on the right, one has the following generalization of the formula derived in problem 29(b):
Lemma. Let v(t), zero on a neighborhood of 0 (N.B.!), be increasing on (- oo, ac)) and O(t) there. Then log
x
x
t
t
1-- + -
dv(t)
=
x2 v(t)
fo
oox-t
t2
dt
172
X B Completeness of sets of exponentials on intervals
at the x c- l where v'(x) exists and is finite, and also at those where v(t) has a jump discontinuity.
Remark. The expression on the right is a Cauchy principal value, viz., v(t)
x2
hm E-o
dt.
t2
1'-.'J"x-t
See the end of §C.1 in Chapter VIII.
Proof. Taking an s > 0, integrate x
(iog 1-JIB
xI,E
+ i )dv(t)
t
by parts. Under the given conditions, the integrated terms corresponding to t = ± oo vanish, and, if v'(x) exists and is finite, the sum of the ones corresponding to t = x ± e tends to zero as a --> 0, leaving us with the right side of the identity in question. When v has a jump discontinuity at x, that identity is valid because each of its sides is then equal to - oo.
Application of the lemma to our function F (formed from nE(t) which vanishes for I t I < 1) yields
logIF(x)I
x2
I
r1>
1 x-t
nEt)dt. t2
In using this relation, we will want to take advantage of the condition I n5(t) - Dt I
1 +t2
Ir1%1
0, and tends to 2D as x --+ ± oo; for it, we certainly have
f
I
-WX
1+x2
Dxloglx+ I dx
x-1
X_k -k - 00, .
and a function b(t), zero on (-1, 1), with the following properties: (i)
(xk+1k
xk12
(xk_xk_1)2
+
k Y00
Yk
1
, 1 and the function S(t) when t >, 1, the constructions on (- oo, -1] being exactly the same. We start by putting x1 = 1. Then, assuming that xk has already been determined (and 8(t) specified on [1, xk) if k > 1 ), let us see how to find xk+l, and how to define b(t) for Xk 1< t < xk+l. As x > Xk increases, the integral
Jxtx xk
=
dt
t2
tog
X
X
-
Xk
x
xk
tends to co, while
remains bounded, by hypothesis. Hence, unless the ratio
JxIIth/Jxt xk
t
Zxkdt t
remains always < it for x > Xk, there is a value of x for which it is equal to our given number q. If equality last obtains for a value x > xk + 1 we call that value xk+ 1; in any other case we put xk+ 1 = xk + 1. We thus have xk+ 1 % xk + 1 and also
I(t)1 Jxk*l xk
2
dt
t
nJ
2
dt,
t
xk
t
-xk
with equality holding when xk+1 > xk+ 1. We have xk+l xk+1
J xk
2
- t dt
t
-
x k+l
t - xk dt
JXk
>
0,
t2
for the difference on the left can be rewritten as
i`(t+c)2di -2 2J
((c lr)
(c+i)2)idT
with c = (xk + xk+ 1)/2, 1 = ('xk+ 1 - xk)/2, and the new variable
176
X B Completeness of sets of exponentials on intervals
r = t - c. Therefore, as x' increases from Xk to Xk + 11 X
k+1 x'- t
I
2
dt
t
1 Xk
increases from -1 fXk+'((t-xk)/t2)dt to rifzk+'((xk+1-t)/t2)dt >
xk+' ((t - xk)/t2) dt, and, by the previous relation, there must be an x' E [Xk, xk + 1 ] for which ( Xk+I A(t) Sk+l x - t J t 2 d t+ t, dt = 0. rl f
f
t2
.Xk
Xk
We denote that value of x' by xk, and put
S(t) = ri(xk - t) for xk < t < xk+ 1' In this way, the function 6(t) is defined piece by piece on the successive intervals [xk, xk+,) and thus on all of [1, oo), since our requirement that xk+ 1 i xk + 1 ensures that Xk k 00. 6(t)
0
x3
t
Figure 223
We must verify properties (i) - (vi) for this b(t) and the sequence {xk}. Property (ii) is obvious, and (v) guaranteed by our choice of the xk. To check (i), observe that when xk+ 1 > xk + 1, (xk+ 12 2
xk + 1
xk)2
xk+ t t
G
Zxk
it Xk
t
dt
xk+, lo(t)I JXk
t2
A
2 The little multiplier theorem
177
whence ("
Xk+1-Xk
Xk+1>Xk+l
Xk+l
2
1, the right side of the previous relation to be (2+,/27c)(D+ri)(Ilk-11 + I41 + IIk+II)IIk1
x2k
We therefore certainly have
JlVk(x)ld oo
, 1. From this we see that I Vk(x)1 k=1
x2
2
dx
7(D+rl) E k=1
xk+2-xk
1
xk
We have
(xk+2-xk-1)2
_
0, there is a sequence S of real numbers lying outside (-1, such that ns(t)
t
-
n
1)
for t --+ ± oo
and, for a suitable real number y, the function cp(z)
=
e-Iz
1-
z
e=/z
Acs
satisfies °°
log+ I F(x)(p(x) I
1+x2
_OD
dx
± oo both exist, and are equal to n(D + rl), a quantity as close as we like to irD. Because the product has a convergent logarithmic integral, one can show by the method of §B.2, Chapter VI, that in fact IF(z)(p(z)I < CEexp(ir(D + n)I3zl + slzl) The one on bounded analytic functions - the recent publication by Garcia-Cuerva and Rubio de Francia is also (and especially!) called to the reader's attention.
184
X B Completeness of sets of exponentials on intervals
for each s > 0. At the same time, F(z)cp(z) vanishes (with appropriate multiplicity) at each point of the given sequence E.
Proof of theorem. Fixing the number ri > 0, we take the function 5(t) corresponding to it furnished by the lemma and v(x), related to 8(t) as in the statement of the preceding theorem. The function U(x) figuring in that result is, as we know, related to our F by the formula
=
U(x)
log I F(x) I
+ Dxlog
x+1 x-1
Let us obtain a similar representation for v(x). Write v(t)
=
rit+6(t), ItI >, 1,
to,
I t I < 1.
By property (ii) of the lemma, v(t) is increasing on (-co, oo), and, by property (iii), v(t) t
-4n
as t--*+oo.
v(t) is in fact piecewise constant, with jump discontinuities at (and only at)
the points Xk, k = ± 1, + 2,... mentioned in the lemma's statement:
X-3
Figure 226
x-2
x2
X3
2 The little multiplier theorem
185
According to the lemma and discussion at the end of the last article, we have r °° J}
=
b(t)
x2
00x-t t2
dt
(logI - -
J
x+1 f .lx-t x2 v(t) dt + rlx log t2 x-1
=
+ x ) dv(t) + rix log t
t
In terms of H(z)
=
JTT
z (log 1-- + 91z )dv(t). t
we thus have our desired representation: v(x)
=
H(x) + rlx log
x+1 x-1
Using this formula with the previous one for U(x), we can reformulate the conclusion of the last theorem to get °°
IlogIF(x)I + H(x) I
f-00
1+x2
dx
0, [p] denotes, as usual, the greatest integer < p, but
when p < 0, we take [p] as the least integer > p, so as to have
[-p] = -
Here, one must be somewhat careful. The expression log
z
+ %z)dv(t)
t
is very sensitive to small changes in v because of the term 9Rz/t in the integrand; replacement of v(t) by [v(t)] usually produces a new term linear in 91z = x which spoils the convergence of the integral involving log IF 1
186
X B Completeness of sets of exponentials on intervals
and H. What we do have is a relation °°
f
(log 1 -
x+it + x)(d[v(t)] - dv(t)) t
t
yx + 2log+Ixl + 0(1),
xeR,
valid with a certain real constant y. To show this, a device from the proof of the theorem in article 1 is used.* Assuming, wlog, that x > 0, we observe that the left-hand member of the
relation in question can be rewritten as
(d[v(t)] - dv(t))
1t
+xt 1
+
( log
f0'0
tI
- )(d[-v(-t)]+dv(-t)-d[v(t)]+dv(t)).
To estimate the first integral we fall back on the lemma from §A.1, according to which it is log+,Ixl
+ 0(1).
The second one we integrate by parts, remembering that v(t) = 0 for It l < 1. When that is done, the integrated terms (involving the differences
(- v(- t)) - [ - v(- t)] and v(t)-[v(t)] ) all vanish, leaving x
r,°° [v(t)]-v(t)+(-v(-t))-[-v(-t)]dt t2
J
+
I
a log 1 +
x+i {(-v(-t) - [-v(-t)]) - (v(t)- [v(t)])} dt. t /I
The first term here is just yx, where
=
f °°
N01 - v(t) + (- v(- t)) - [ - v(- t)]
y
t2
t
dt
(a quantity between -1 and 1). In the second term, the expression in { lies between -1 and 1, while
atlogll+xt tl
, 0, proving the desired inequality when that is the case. A similar argument may be used when x is negative.
Having established our relation, we proceed to construct the entire function cp(z). Take S as the sequence of points where [v(t)] jumps, each of those being repeated a number of times equal to the magnitude of the
jump corresponding to it; S simply consists of some of the points Xk, k = ± 1, ± 2, ... , with certain repetitions. We then put
e-yZf zEs
--)eZ/a (1
A
taking care to repeat each of the factors on the right as many times as the 2 corresponding to it is repeated in S. This function 9(z) is entire, and clearly ('
log(z)I = -yJiz + J
(oi -
1
t
+)d[v(t)] t
The inequality proved above now yields loglcp(x+i)I
< H(x+i) + 2log+lxl + O(1),
xell.
Obviously, ns(t) = [v(t)], so ns(t)
t
n
as t -> ± oo.
Also, °° co
Ins(t)
- ntI l+t 2 dt
± oo.
-
t
We can thus certainly take two different points sl, s2 eS (!), and cp(z)/(z - sl)(z - s2) will still be entire. Let, finally, G(z)
-
Fo(z)gq(z)V1(z)
(z - s1)(z - s2)
This function is entire and of exponential type 1< n(DA + 3n), and
f
IG(x)Idx
27CDn..
Proof. If the eizt are incomplete on (say) the interval [- L, L], there is (as at the beginning of §D, Chapter IX) a non-zero measure ,u on [ - L, L] with L
j -L
eizt dµ(t)
=
0,
2 e A.
The non-zero entire function fL G(z) = eizidu(t) J L
of exponential type < L thus vanishes at each point of A, so, since G is bounded on the real axis, we have
x
z#o
00 1212
by §G.3 of Chapter III.
192
X B Completeness of sets of exponentials on intervals
Let E denote the complete sequence of zeros of G (with repetitions according to multiplicities, as usual). The Hadamard representation for G is then =
G(z)
zlU
1-
Azpe`z fl
eZiµ.
µEE
µ#0
Denote by S the difference set
E - {).EA: 91h,
O};
then
=
G(z)
(ii
Azpe'Z
z eZiµ.
_
zes
µ#0
µ
f7
(ii
- ? eZix.
.tGA
A
91l#0
Take now the function G0(z)
=
Azpe'z 11 C 1 'LEs
p#0
1 ez
/J ?,
u. ji (ii _ ? 1 eZix .t'EA'
ez/z.
(with exponentials ez and not in the second product!). By work done in §H.3 of Chapter III, we see that Go(z) is of exponential type, and that
IG0(x)I 5 IG(x)I,
xetJ ,
so that G0(x) is bounded on the real axis (like G(x) ). Write
B=
lim sup Y-w
log I G°(ly) I
y
and
B'
=
limsu p
logIG0(iy)I Iyl
Observe also that logIG(z)I
0 with the following property: there exist distinct integers nz corresponding to the different non-zero 2 in A such that CO.
Y_ .lEA
A*O
This criterion is due to Redheffer. (Hint: Look again at the constructions in §D.2 of Chapter IX.)
C.
The multiplier theorem for weights with uniformly continuous logarithms The result of Beurling and Malliavin enunciated in §A.2 is broader
in scope than may appear at first sight. One can, for instance, deduce from it another multiplier theorem for weights fulfilling a simple descriptive
regularity condition. This is done in article 1 below; the work depends
on some elementary material from Chapter VI and the first part of Chapter VII. In article 2, the theorem of article 1 is used to extend a result obtained in problem 11 (Chapter VII, §A.2) to certain unbounded measures on R. 1
The multiplier theorem
Theorem (Beurling and Malliavin, 1961). Let W(x) > 1, and let log W(x) be uniformly continuous* on R. Then W admits multipliers iff * Beurling and Malliavin require only that m(s) = ess supxE R flog W(x + s) - log W(x) I befinite for a set of s c- R having positive Lebesgue measure. To reduce the treatment under this less stringent assumption to that of the uniform Lip 1 case handled below, they observe that there must be some M < oo with cuts) _< M on a Lebesgue measurable set E with
EI > 0. But then E - E includes a whole interval (- h, h), h > 0, so _ I is Lebesgue measurable, for then /P
W(s)
=
lim
Ilog W(x+s) - log W(x)lpe-2 "dx
J
Lebesgue measurable (the integrals are by Tonelli's theorem, log W(x + s) - log W(x) being Lebesgue measurable on U82). is also
X C Weights with uniformly continuous logarithms
196
log W(x)
°°
-.0 1 +xZ
0. As we have just seen, for each given x e R, the maximum value of I S(x) I for sums S of the kind just specified is equal
200
X C Weights with uniformly continuous logarithms
to ,I(KN(x)). It is now claimed that when N -> co, the KN(z) converge u.c.c to a certain entire function K(z) of exponential type W1(x) 3
on 68, where W(x) > 1, so 4K(x) >, (' °°
log S21(x)
J
1 + x2
dx
, 2cxo. Then, by the Lip 1 property of a-,
a-(x) >, cxo for (1 - c)xo 5 x 5 (1 + c)xo, so (1 +c)Xo
f
t1 -OXO
a-(x) dx
2c2
(1 +c)2'
for
2 A theorem of Beurling
205
If the boxed relation holds, this cannot happen for arbitrarily large values of x0, and a - (x) must be o(I x I) for x ---+ co. Similarly for x -p - oo. Lemma x+1
e-0-(t) I du(t) I
5
e2.
x-I
Proof. By definition, e-°(x) > rx+1e-Ix-thIdµ(t)I Jx-1
>
e-1 ('x+1Idµ(t)I x-1
This holds a fortiori if o(x) is replaced by -a - (x) 5 a(x). The Lip 1 property of a- now makes a-(t)
'>
X-1 s t 0. Also,
I.f(x)I < 1, so Idv(t)I < Idµ(t)I and
, a(x). Combining this with the previous inequality, we get
T(x) i Q+(x), whence aD
T(x)
f_00 1 + x2
dx
00
by hypothesis. It is now claimed that eizOt dv(t) = 0 for
I Ao I
< A - a.
By the Paley-Wiener theorem, we have
f(t) =
JeitAq)d2, a
where cp is (under the present circumstances) a continuous function on
[- a, a]. Therefore,
f
=
e"Otdv(t)
lim
s-o
-moo ao
a
lim
eUaoteitA(p(2)e-altl dAdµ(t)
b-'0
aD
a
=
lim .5-0 o
e-6111eiAO`f(t)dµ(t) f-.
-a -o
+zne-6111e(zodp(t)4V(1)dA;
.
208
X C Weights with uniformly continuous logarithms
here, for each 6 > 0, absolute convergence holds throughout. The last limit is just gv(A)µa(A + ))dx
lim
which is, however, zero when I AO I
A - a since then #,,(A + )) -+ 0
s a as S -> 0.
uniformly for I A
The claim just established and the integral condition on 2(x) now make
v - 0 by problem 11. That is,
f(x)dp(x) - 0. If the function f vanishes at all on Q8, it does so only at certain points x isolated from each other, for f is entire and not identically zero. What we have just proved is that u, if not identically zero, has all its mass distributed on the points x,. Then there must be one of those points, say xo, for which
µ({xo}) * 0. That, however, cannot happen. If, for instance, xo is a k-fold zero of f (z), we may repeat the above argument using the entire function fo(z)
_
f (z)
(z - xo)"
instead of f; doing so, we then find that
fo(t)du(t) - 0. 0, we thus have
Since fo(xo)
µ({xo}) = 0, a contradiction. The measure t must hence vanish identically. We are done. Problem 40
Show that the result from problem 11 is best possible in the following sense:
If p is a finite complex measure and e - S(x)
=
Je__hhIdP(t)I,
Control of Poisson integrals by weighted norms
209
then, in the case that a(x)
f7. 1+x2
dx
!i(t)/(t - zo)
is bounded on R - so,
then, is
OWN - a) The ratio I0(t)I/I t - /3I is thus both in L1(R) and bounded on R. Therefore it is in L2(R). So, however, is 0 (/3)/(t - J3). The difference
-
fi(t)
0(t)
t-f3
-
a(ft)
t-fl
must hence also be square integrable. Because ffl is entire and of exponential type < a, we now have A
l.i.m. A-+oo
eizt fp(t) dt
=
0
I_A
for almost all A 0 [ -a, a] by the L2 form of the Paley-Wiener theorem (Chapter III, §D). At the same time, when A -. oo, the integrals
fA eiztff(t)dt A
tend, for 2 0, to a certain function of 2 continuous on R - {0}. This, indeed, is certainly true if, in those integrals, we replace fi(t) by fi(t)l(t - l3) e L1(R).
Direct verification shows that the same holds good when fi(t) is replaced by 0(/3)/(t - /3). The statement therefore holds for the difference fi(t) of these functions.
218
X D Control of Poisson integrals by weighted norms
The (continuous) pointwise limit of the expressions A
eizt fp(t) dt,
0,
A
-A
for A -> oo must, however, coincide a.e. with their limit in mean, known to be zero a.e. for I A I >, a, as we have just seen. Hence A
lim A-oo
=
eizt fi(t) dt
f-A
0,
JAI
% a.
That is, lim A-f.o for I A I >,
=
0(R) eat dt
A
0(t)eW dt
J-At - I'
f-"O. (t - N)
a, and finally, 0(fl)S(t)
c if(t)S(t)
=
dt
lim f A A-ooJ-A t-$
oo t-Y
dt
for each of our sums S(t). for the norm appearing in the proof of Let us continue to write the preceding theorem. In terms of this notation, we have for the modulus of the right-hand member of the last relation the upper bound II
I
I0(t)I2
Af-'w(t)It-QI2 dtl/
IISII.
.
The condition that I (t) 12
w(t)(t2 + 1)dt
0. In like manner, A
lim A-+oo
S(t)
-At - I'
dt
s
CP
IC$)I
IISII1
Control of Poisson integrals by weighted norms
219
so finally, since 1
2i3 f1
1t_fl1z,
t- P
KflllsII
2S(t) dt
It-Q1 If E
_
1
t-f
{.3z > 0} is compact,
K
=
+
Ca
Ce
21'(fl)I
214($)I
may be taken to be bounded above on E, since, as explained at first, we can choose ,li with 1 /'(fi)I and 10(f3)I bounded away from 0 on E. Sufficiency is proved.
Necessity. Suppose that for every /3, 3/3 > 0, the last inequality (at the end of the preceding discussion) holds, with some finite K,. Then, by the
previous theorem, we certainly have a non-zero entire function cp of exponential type < a, with I w(t)12
fo
w(t)(t2 + 1)2
dt
0. Then we are done. We Suppose that 3p(f3) can, indeed, argue as at the very start of the previous theorem's proof to obtain, thanks to the last relation, a k(t) with ilk II < oo (and hence w(t)k(t) e L1(R) by Schwarz) such that lim A- ao for I A I
A
I w(t)k(t) -
J A\
1Q eizt dt
t -N
=
0
% a.
Here, the integrable function w(t)k(t) must also be in L2(IIB). Indeed, the
Control of Poisson integrals by weighted norms
221
(bounded!) Fourier transform eizt w(t)k(t) dt
coincides with the L2 Fourier transform of 1/(t - /3) for large 11, and is thus itself in L2. Then, however, w(t)k(t) e L2(R) by Plancherel's theorem.
We may now apply the L2 Paley-Wiener theorem (Chapter III, §D) to the function
w(t)k(t) -
1
t-/3
and conclude from the preceding relation that it coincides a.e. on R with an entire function f (t) of exponential type < a. The function fi(t)
=
(t-j3)f(t) + 1
is also entire and of exponential type a, and i/i(t)
=
(t - /)w(t)k(t)
a.e., t e R.
The above integral relation clearly implies that w(t)k(t) cannot vanish a.e., so 0 # 0. Finally, I 0(t) I 2
dt
=
11k112
1)dt
0. We have 3tT"(t)
_
sin n(t + iry)
_
sinh 711 cos nt
sin n(t + ij)I2
sin' nt + sinh2 ng
Clearly 91T"(t + 2) = fR T"(t) and RT" is W. on the real axis, so RT"(t)
_
X
Y
t e R,
the series being absolutely convergent. Since 91T"(t - and 91T"(t + Z) Z) are odd functions of t, we have ao
=
(RRT")(t)dt
=
0,
and RRT"(t) is a (uniform!) limit of sums
1- 0, by dominated convergence.
At the same time, since each of the functions T (z) "
i
=
sin 76(z + iq)
is analytic and bounded in 3z > 0,
-oo ( 1+ " as
(7r
n91 T"(i)
t Zt) dt
sinh
---* 1 + rl)
Ir
sinh
> 0
-* 0. This does it. It is not hard to see that here, for the sums
S(t) _ Y Aze'a` I2Ii,
the condition
5
E00
1
gives us control on the integrals It.3# 00
when 'R/J values of /3.
S(t) dt
Q12
2,
±
i, ± 2, ... ,
3 sin irl vanishing precisely for such
One may pose a problem similar to the one discussed in this §, but with the sums
S(t) _ Y Aaeizt Ill %a
Control of Hilbert transforms by weighted norms
225
replaced by others of the form Az e'A` Ill a
(i.e., by entire functions of exponential type S a bounded on R !). That seems harder. Some of the material in the first part of de Branges' book is relevant to it. E.
Hilbert transforms of certain functions having given weighted quadratic norms.
We continue along the lines of the preceding §'s discussion. Taking, as we did there, some fixed w >, 0 belonging to L1(IIB), let us suppose that
we are given a certain class of functions U(t), bounded on the real axis, whose harmonic extensions U(z)
=
_' f 0"
n
-3z
t12
Iz
U(t)dt
to the upper half plane are controlled by the weighted norm U(t)12w(t)dt).
J(J _-
I
A suitably defined harmonic conjugate U(z) of each of our functions U(z) will then also be controlled by that norm. As we have seen in Chapter III, §F.2 and in the scholium to §H.1 of that chapter, the U(z) have well defined non-tangential boundary values a.e. on R and thereby give rise to Lebesgue
measurable functions U(t) of the real variable t. Each of the latter is a Hilbert transform of the corresponding original bounded function U(t);
we say a Hilbert transform because that object, like the harmonic conjugate, is really only defined to within an additive constant. The reader
can arrive at a fairly clear idea of these transforms by referring first to the §§ mentioned above and then to the middle of §C.1, Chapter VIII, and the scholium at the end of it. Whatever specification is adopted for the Hilbert transforms U(t) of our functions U, one may ask whether their size is governed by the weighted norm in question when that is the case for the harmonic extensions U(z). To be more definite, let us ask whether there is some integrable function
w(t) > 0, not a.e. zero on F, such that
JTt12t1t
a
worked with in §D. Although the problem, as formulated, no longer refers directly to the harmonic extensions U(z), it will turn out to have a positive solution (for given w) precisely when the latter are controlled by I U(t)12 w(t) dt in {,,)z > 0} (and only then). For this reason, multipliers will again be involved in our discussion.
The work will require some material from the theory of HP spaces. In order to save the reader the trouble of digging up that material elsewhere, we give it (and no more) in the next article, starting from scratch. This is not a book about HP spaces, and anyone wishing to really learn about them should refer to such a book. Several are now available, including (and why not!) my own.* 1.
Hp spaces for people who don't want to really learn about them
We will need to know some things about H1, H,, and H2, and proceed to take up those spaces in that order. Most of the real work involved here has actually been done already in various parts of the present book. For our purposes, it is most convenient to use the Definition. H1(l8), or, as we usually write, H1, is the set off in L1(D) for which the Fourier transform .f (2)
=
f
e;at f(t) dt
J
vanishes for all 2 > 0.
* As much as I want that book to sell, I should warn the reader that there are a fair number of misprints and also some actual mistakes in it. The statement of the lemma on p. 104 is inaccurate; boundedness only holds for r away from 0 when F(0) = 0. Statement of the lemma on p. 339 is wrong; v may also contain a point
mass at 0. That, however, makes no difference for the subsequent application of the lemma. The argument at the bottom of p.116 is nonsense. Instead, one should
say that if BI B. and du' -< daa for each a, then every fa is in Q H21 where 0 is given by the formula displayed there. Hence wH2 = E is 9 O H21 so B I B and da' -< da by reasoning like that at the top of p. 116. There are confusing misprints in the proof of the first theorem on p 13; near the end of that proof, F should be replaced by G.
1 About the spaces H1, H., and H2
227
Lemma. 1ff EH1, eiztf(t) E H1 for each A >, 0. Proof. Clear.
Lemma. If f e H1 and 3z > 0, f (t)/(t - 2) E H1.
Proof. For 3z > 0 (i.e., Ii(- if) < 0 ), we have e-izAeizt d).,
t e R.
0 foo
Therefore, if f e H 1,
f
i
f (t) dt
00
t-z
=
ezze"f(t) d,dt.
J -. J o
The double integral on the right is absolutely convergent, and hence can be rewritten as Joo f-'000 e-ifAe,xt f(t) dt d.1
f(A) d l
=
=
0.
If a ,>0 and f e H 1, e'"t f (t) is also in H 1 by the preceding lemma, so, using it in place of f (t) in the computation just made, we get
f J
ei"t f (t) dt
t-z
00
=
0.
f(t)/(t - z) is thus in H 1 by definition. Theorem. If, for f e H1, we write
f(z) =
1 it
J-
-3z
Iz-t12
f(t) dt
for 3z > 0, the function f (z) is analytic in the upper half plane.
Proof. We have f (z)
=
I f- ( 276
t
l
-t
f(t)
z
By the last lemma, the right side equals
t 27ri
f (t) dt J -mot - z
for 3z > 0, and this expression is clearly analytic in the upper half plane. We are done.
X E Control of Hilbert transforms by weighted norms
228
Theorem. The function f (z) defined in the statement of the preceding result has the following properties: (i) f (z) is continuous and bounded in each half plane {,3z > h}, h > 0,
and tends to 0 as z -> oo in any one of those; (ii)
(iii)
JTao I f (x + iy) I dx
0;
II f II
f(t + iy) -f(t) I dt
.
as y>0;
0
(iv) f (t + iy) -+ f (t) a.e. as y --> 0. Remark. Properties (iii) and (iv) justify our denoting
J
J1°°
3z
_
Iz-t12
f(t)dt
by f (z).
Proof of theorem. Property (i) is verified by inspection; (ii) and (iii) hold because the Poisson kernel is a (positive) approximate identity. Property
(iv) comes out of the discussion beginning in Chapter II, §B and then continuing in §F.2 of Chapter III and in the scholium to §H.1 of that chapter. These ideas have already appeared frequently in the present book.
Theorem. If f (t) e Hl is not zero a.e. on R, we have °°
log- I f(t) I
_. 1+t2 and, for each z,
loglf(z)I
t
0, 5
Iz3
-
tl2loglf(t)Idt,
the integral on the right being absolutely convergent. Here, f(z) has the same meaning as in the preceding two results.
Proof. For each h > 0 we can apply the results from Chapter III, §G.2 to f (z + ih) in the half plane 3z > 0, thanks to property (i), guaranteed by the last theorem. In this way we get log I fl z + ih) I
0. From property (iv) in the preceding theorem and Fatou's lemma, we have, however,
' Iz3 it
t121og
If(t)I dt
lim i0nf
W
-f
W Iz
3z
t
12
log - I f (t + ih) I dt.
Using this and the preceding relation we see, by making h --+0 in our initial one, that - Co
0. It is of course > log I f (z) I in case f (z) = 0. We are done. Corollary. If f(t) e H1 is not a.e. zero, I f(t) I is necessarily > 0 a.e.. Proof. The theorem's boxed inequality makes log - I f (t) I
> - oo a.e..
Definition. Hc(R), or, as we frequently write, H., is the collection of g in L,,(68) satisfying
JT g(t)f(t) dt
=
0
for all feH1. H. is thus the subspace of L., dual of L1, consisting of functions orthogonal to the closed subspace H1 of L1. As such, it is closed, and even w* closed, in LcD.
By definition of H1 we have the Lemma. Each of the functions eizt, A > 0, belongs to H..
Corollary. A function f eL1(O) belongs to H1 iff
g(t)f(t)dt
=
0
-.o
for all geH,. Lemma. If f e H1 and g e Ham,
g(t)f(t) e H1.
Proof. First of all, g f e L 1. Also, when A,> 0, eizt f (t) e H previous lemma, so by definition of H.., g(t)eut f(t) dt
=
0,
JT e'Atg(t) f (t) dt
=
0
JT
for each 2>, 0. Therefore g f e H 1.
Lemma. If g and h belong to H., g(t)h(t) does also.
1
by a
1 About the spaces H1, H. and H2
231
Proof. If f is any member of H1, gf is also in H1 by the previous lemma. Therefore 0. J
This, holding for all f EH1, makes hg E Hc,, by definition.
Theorem. Let g e H,. Then the function g(z)
=
tl2g(t)dt
- _ Iz
is analytic for 3z > 0. Proof. Fix z,
f(t)
3z > 0, and, for the moment, a large A > 0. The function iA
1
=
t-z t+iA
belongs to H1. This is easily verified directly by showing that
e'-`f(t)dt
=
0
-00
for A > 0 using contour integration. One takes large semi-circular contours
in the upper half plane with base on the real axis; the details are left to the reader. By definition of H., we thus have °°
g(t)
iA
1
t-z iA+t
_.0
dt
=
0.
Subtracting the left side from °°
iA
1
g(t)
t - z iA + t
dt
and then dividing by 2i, we see that
3z
iA
L'z_t'2
iA+t
g(t) dt
=
1
1 7r
3z
iA
Iz-t12 iA+t
is analytic for ..3z > 0 (by inspection).
iA
2iJ_t-z iA+t g(t) dt.
For each A > 0, then, z 9a() =
1
9(t) dt
232
X E Control of Hilbert transforms by weighted norms
As A - co, the functions gA(z) tend u.c.c. in {,Zz > 0} to
'f' I
z
Zz
t I2
=
g(t) dt
g(z).
The latter is therefore also analytic there. Remark. For the function g(z) figuring in the above theorem we have, for each z, 3z > 0,
Ig(z)I s IIg1l., where the L.,, norm on the right is taken for g(t) on R. This is evident by inspection. The same reasoning which shows that
f (t + iy) -. f (t) a.e. as y ---) 0 for functions fin H, also applies here, yielding the result that g(t + iy)
-
a.e. as y - 0
+ g(t)
when g e H. Unless g(t) is uniformly continuous, however, we do not have II g(t + iy) - g(t) II
.
0
for y - +0. Instead, we are only able to affirm that g(t + iy) tends w* to g(t)
(in L. (R) ) as y --* 0.
The theorem just proved has an important converse:
Theorem. Let G(z) be analytic and bounded for 3z > 0. Then there is a g E H., such that G(z)
=
!-' f
_
-1zt Iz g(t) dt Iz
for .3z > 0, and Il g ll
c = sup
I G(z) I
.
3z> o
Proof. It is claimed first of all that each of the functions G(t + ih), h > 0, belongs to H. (as a function of t). Take any f c- H,, and put
- f'
.f(z)
=
Iz-3il2 f(t)dt,
Jz > 0.
Our definition of H requires us to verify that
f
G(t + ih) f (t) dt
=
0.
1 About the spaces H1, H. and H2
233
Since
IIf(t+ib) - f(t)II1
0
as b -o 0, it is enough to show that =
G(t + ih) f (t + ib) dt
0
-00
for each b > 0. Fix any such b. According to a previous result, f (z + ib) is then analytic and bounded for 3z > 0, and continuous up to the real axis. The same is true for G(z + ih). These properties make it easy for us to see by contour integration that iA
`°
f (A+t
12 G(t + ih) f (t + ib) dt
=
0
for A > 0; one just integrates 2
iA
iA+z
G(z + ih)f(z + ib)
around large semi-circles in .3z 3 0 having their diameters on the real axis. Since f (t + ib) e L1(R), we may now make A --* oo in the relation just found to get =
G(t + ih) f (t + ib) dt
0
-.
and thus ensure that G(t + ih) e H.(ll ). For each It > 0 the first lemma of §H.1, Chapter III, makes G(z + ih)
=
when 3z > 0. Here, I G(t + ih) I
0 tending to zero and a g in L. with
-s g(t)
G(t +
w*
as n -. oo. From this we see, referring to the preceding formula, that G(z)
=
lim G(z +
1
-3z
n-.0
n
-.Iz-tI2
g(t) dt
234
X E Control of Hilbert transforms by weighted norms
for )z > 0. By the w* convergence we also have
o
n-+co
However, the representation just found for G(z) implies the reverse inquality, so II g ll
co = sup
I G(z) I.
3z> o
As we have seen, each of the functions G(t + ihn) is in H,,. Their w* limit g(t) must then also be in H.. The theorem is proved. Remark. An analogous theorem is true about H1. Namely, if F(z), analytic
for .3z > 0, is such that the integrals
IF(x+iy)Idx J
are bounded for y > 0, there is an f e H1 for which
--
F(z)
Iz3z
f(t)dt,
tI2
3z > 0.
This result will not be needed in the present §; it is deeper than the one just found because L1(R) is not the dual of any Banach space. The F. and M. Riesz theorem is required for its proof; see §B.4 of Chapter VII. Problem 41 Let g e H., and write 1 1 °° g(z)
=
3z
g(t) dt
irJ-.Iz-tl2
for ..3z > 0. (a) If 3c > 0, both functions g(t) - g(c)
and
t-c
g(t) - g(c)
t-c
belong to H.. (Hint: In considering the first function, begin by noting
that 1/(t-c) a H. according to the second lemma about H1. To investigate the second function, look at (g(z) - g(c))/(z - c) in the upper half plane.) (b) Hence show that if f eH1 and
f(z)
=
mlztl2f(t)dt
n
1 About the spaces H1i H,,, and H2
235
for .3z > 0, one has
j
_
.f(c)g(c)
3 t12.f(t)g(t)dt
IC
for each c with 3c > 0. (c) If, for the f (z) of part (b) one has f (c) = 0 for some c, 3c > 0, show that f(t)1(t - c) belongs to H1. (Hint: Follow the argument of (b) using the function g(t) = e", where A _> 0 is arbitrary.)
Theorem. If g(t) E H., is not a.e. zero on R, we have log
I g(t) I
1+t2
_x
0,
3z
log I g(z) I
It1
tl
2
log I g(t) I dt,
the integral on the right being absolutely convergent. Here, g(z) has its usual meaning: g(z)
=
1
f°°
7t
-
1z
g(t) dt.
i z - t 12
Proof. By the first of the preceding two theorems, g(z) is analytic (and of
course bounded) for 3z > 0. Therefore, by the results of §G.2 in Chapter III, for each h > 0, log l g(z + ih) I _
Iz
3z
log l g(t + ih) I dt
when 3z > 0. We may, wlog, take II g II to be < 1, so that lg(z)l < 1 and log I g(z) 15 0 for 3z > 0. As h -* 0, g(t + ih) -* g(t) a.e. accord-
.
ing to a previous remark, so, by Fatou's lemma, lim soup
-' f ' n
I
z
-3z
t
12
log I g(t + ih) l dt
-'
lz3tl2loglg(t)Idt.
The right-hand quantity must thus be >, log l g(z) l by the previous relation, proving the second inequality of our theorem.
X E Control of Hilbert transforms by weighted norms
236
In case g(t) is not a.e. zero, there must be some z, 3z > 0, with g(z) 0 0, again because g(t + ih) - g(t) a.e. for h - + 0. Using this z in the inequality just proved, we see that
'
1
Zz J112loI(t)dt
- oo,
>
whence °°
log Ig(t)I
_Go
1+t2
dt
0, whether g(z) 0 0 or not. We are done.
Come we now to the space H2. Definition. A function f e L2(R) belongs to H2(II8), usually designated as H21 if f (t) dt _00 t - z
=
0
for all z with 3z > 0. H2 is clearly a closed subspace of L2(R). Theorem. If f e H21 the function
=n
f(z)
_.Iz3tl2f(t)dt
is analytic for 3z > 0. Proof. Is like that of the corresponding result for H1.
Theorem. If f e H2, the function f (z) in the preceding theorem has the following properties: (i)
If(z)I
0;
11f II2IV (irZz),
for y > 0;
11 f 11 Z
1 About the spaces H1, H., and H2 0
I f(t+iY)-f(t)12dt
(iii)
JT (iv) Pt + iy) -p f (t) a.e. as y -
237
as
+ 0.
Proof. Property (i) follows by applying Schwarz' inequality to the formula for f (z). The remaining properties are verified by arguments like those used in proving the corresponding theorem about H1, given above.
As is the case for H., (and for H1), these results have a converse:
Theorem. Let F(z) be analytic for 3z > 0, and suppose that
IF(x+iy)I2dx -CO
is bounded for y > 0. Then there is an f eH2 with 3
f '0 F(z)
-
1
t 12
Iz
-3z > 0,
.f (t) dt,
and 00
=
11f II
sup
I F(x + iy) 12 dx.
I
z
Proof. For each h > 0, put 1
Fh(z)
h
2h -n
F(z + s) ds,
3z > 0.
By Schwarz' inequality, IFh(z)I
(2h)-1/2J\J
0,
238
X E Control of Hilbert transforms by weighted norms
We have, for each h and y .> 0,
jjIF(x+s+iy)I2dsdx h
ao
-
IFh(x+iy)I2dx
Zh
-h
00
I F(x + iy) 12 dx
by Schwarz' inequality and Fubini. Since the right side is bounded by a quantity M < oo independent of y (and h), the limit relation just written guarantees that IIfh112 < M
for h > 0, according to Fatou's lemma. Once it is known that the norms II fh II 2 are bounded we can, as in
the proof of the corresponding theorem about H., get a sequence of h > 0 tending to zero for which the fh. converge weakly, this time in L2, to some f e L2(ll ). Then, for each z,
3z > 0,
-BIZ-tl2fhn(t)dt
it
-p
'f'
-aolZ3tl2f(t)dt
it
as n -> oo. At the same time,
Fh,(z) ,) F(z), so we have our desired representation of F(z) if we can show that f cH2. For this purpose, it is enough to verify that when 3z > 0, fID fh(t) dt
_00 t-z
=
0,
since the fh tend to f weakly in L2. However, the fh belong to Ham, and,
when 3z > 0 and A > 0, the function 1
iA
t-z iA+t belongs to H1, as we have noted during the proof of a previous result. Hence iA
fh(t)
iA+t t - z
dt
=
0.
Here, fh(t)/(t - z) belongs to L1, so we may make A --i oo in this relation, which yields the desired one.
1 About the spaces H1, H., and H2
239
We still need to show that II f II 2 = sup,,, o f °°. I F(x + iy) 12 dx. Here, we now know that the function F(z) is nothing but the f (z) figuring in the preceding theorem. The statement in question thus follows from properties (ii) and (iii) of that result. We are done.
Remark. Using the theorems just proved, one readily verifies that H2 consists precisely of the functions u(t) + iu(t), with u an arbitrary real-valued
member of L2(R) and u its L2 Hilbert transform - the one studied in the
scholium to §C.1 of Chapter VIII. The reader should carry out this verification.
Our use of the space H2 in the following articles of this § is based on a relation between H2 and H1, established by the following two results.
Theorem. If f and g belong to H2, f g is in H1. Proof. Certainly fg c- L1, so the quantity
eu`f(t)g(t)dt
I
J
varies continuously with A. It is therefore enough to show that it vanishes
for 2 > 0 (sic) in order to prove that fg c- H1. Let, as usual, 1
f(z)
=
z
°°
It
for 3z > 0, and g(z)
=
n
_
I
z
t
12 g (t)
dt
there. Using the facts that II f (t + ih) - f (t) 112 ---p 0 and II g(t + ih) - g(t) 112 -+ 0
for h -* 0 (property (iii) in the first of the preceding two theorems) and applying Schwarz' inequality to the identity f (t + ih) g(t + ih) - f (t) g(t)
=
[ f (t + ih) - f (t)] g(t) + f (t + ih) [g(t + ih) - g(t)],
one readily sees that 11 fit + ih) g(t + ih) - f (t) g(t) II
1
0
240
as h
X E Control of Hilbert transforms by weighted norms
It is therefore sufficient to check that
=
e'-t f (t + ih) g(t + ih) dt
0
-00
for each h > 0 when A > 0. Fix any such h. By property (i) from the result just referred to, const.
I f (z + ih) I
for
h
1
0, the relation we needed. The theorem is proved.
The last result has an important converse: Theorem. Given cp e H, there are functions f and g in H2 with rp = f g and IIf112 = 119112 = 1101(P110-
Proof. There is no loss of generality in assuming that (p(t) is not a.e. zero on R, for otherwise our theorem is trivial. Putting, then, (P(z)
=
f' nI
-3z
_ oo I z
t2
(p(t) dt
for 3z > 0, we know by previous results that cp(z) is analytic in the upper * Here, the condition A > 0 plays the role that the relation A > A did in the discussion referred to.
I About the spaces H,, H., and H2
241
half plane and that log I w(z)
3z
-
I
log I p(t) I dt
t 12
Iz
there, the integral on the right being absolutely convergent.
Thanks to the absolute convergence, we can define a function F(z) analytic for 3z > 0 by writing l
l
=
F(z)
exp 2nif_
- t2 + ) log I (P(t)I dt j; 1
Ct
zZ
the idea here is that F(z) 0 0 for 3z > 0, with log I F(z) I
=
27r
t 12log I (P(t) I dt,
- ,,I z
one half the right side of the preceding inequality. The ratio G(z)
_
(P(z)
F(z)
is then analytic for 3z > 0, and we have log I G(z) I
5
.Iz3z
_J
--tl2
log I ip(t) I dt
=
log I F(z) I,
i.e., I G(z) I
3z > 0.
< I F(z)I,
By the inequality between arithmetic and geometric means,
3
IF(z)12
It
-
t121 w(t)I dt,
Iz
so, for each y > 0, ('
J -0000 IF(x+iy)12dx
JT IG(x+iy)I2dx 1
0. * One may also appeal directly to that theorem after noting that f 2 e H1.
1 About the spaces H1i He and H2
245
Theorem. Let f c- H2, not a.e. zero on I18, be outer. Then the finite linear combinations of the eizt f (t) with A >, 0 are II
112 dense in H2.
Remark. This result is due to Beurling, who also established its converse. The latter will not be needed in our work; it is set at the end of this article as problem 42.
Proof of theorem. In order to show that the eizt f (t) with A >, 0 generate H2, it suffices to verify that if rp is any element of L2 such that
-
=
eiztf(t)cp(t)dt
0
for all A 3 0, then
=
g(t)(p(t) dt
0
-00
for each g E H2. This will follow if we can show that such a cp belongs to H21 for then the products gcp with g e H2 will be in H1.
Since f and cp e L2, f q c- L1, and our assumed relation makes f (P in H1. The function F(z)
=
-' f
n-
lz
3t
12
f (t)(p(t) dt
is thus analytic for 3z > 0. If cp(t) - 0 a.e. there is nothing to prove, so we may assume that this is not the case. By the preceding corollary, I f(t)l > 0 a.e.; therefore f (t)cp(t) is not a.e. zero on R. Hence, by an earlier result,
-
log I F(z) I
1
z3
t
2
log I f (t)N(t) I dt
when ,,3z > 0, with the right-hand integral absolutely convergent. At the same time, for 3z
c'
1
f(z)
=
f(t)dt
7T
we have
logIf(z)I
=
-
Iz3t2loglf(t)dt
by hypothesis whenever 3z > 0. The integral on the right is certainly > - oo, being absolutely convergent, so F(z)/f(z) is analytic in 3z > 0.
246
X E Control of Hilbert transforms by weighted norms
For that ratio, the previous two relations give F(z)
log
3z
°°
1
7t , Iz-tI2
f (Z)
log I cp(t) I dt,
.3z > 0.
Thence, by the inequality between arithmetic and geometric means, F(z) f(Z)
2
3z > 0,
1
from which, by Fubini's theorem, F(x + iy) -00
2
dx
II w 112.
P x + iy)
According to a previous theorem, there is hence a function 0 E H2 with F(z)
1
f (Z)
it
'3Z tl
Z
2'J(t) dt
for 3z > 0, and F(t + iy)
Pt + iy)
fi(t)
a.e.
as
We have, however, by the formula for F(z),
F(t + iy) --> f (t)cp(t) a.e. as y -* 0, and, for f (z),
f(t + iy) -* f(t) a.e. as y -*0. Therefore, since I f (t) I > 0 a.e., cp(t) = 0(t) a.e., i.e., cp E H2, as we needed to show.
The theorem is proved.
Remark. The function f in H2 appearing near the end of the above factorization theorem for HI is outer. In general, given any function M(t) >, 0 such that °° log- M(t) dt _, 1 +t2
0,
so that, in the first place,
log I F(t + iy) I -> log M(t) a.e.
for y - 0. In the second place, since geometric means do not exceed arithmetic means,
JIF(x+iY)I2dx
0, by an argument like one in the above proof. There is thus an f cH2 with F(z)
=
'
1 7r
Zz
- .Iz - ti
2 f (t) dt,
3z > 0,
and
F(t + iy) -* f (t)
a.e.
as y - 0. Comparing the above two limit relations we see, first of all, that If(t)I
= M(t)
a.e.,
teR.
Therefore logIF(z)I
=
-
IZ3
tl2loglf(t)Idt
248
X E Control of Hilbert transforms by weighted norms
for 3z > 0. Here, our function F(z) is in fact the f (z) figuring in the proof of the last theorem. Hence f is outer. This construction works in particular whenever M(t) = Ig(t)l with g(t) in H2 not a.e. zero on R. Therefore, any such g in H2 coincides a.e. in modulus with an outer function in H2. Problem 42 Prove the converse of the preceding result. Show, in other words, that if f eH2 is not outer, the eiztf(t) with A >- 0 do not generate H2 (in norm II 2) (Hint: One may as well assume that f (t) is not a.e. zero on R. Take then the outer function geH2 with Ig(t)I = If(t)I a.e., furnished by the preceding remark. Show first that the ratio co(t) = f(t)/g(t) - it is of modulus I a.e. - belongs to H.. For this purpose, one may look at II
f(z)/g(z) in 3z > 0. Next observe that =
eizt f (t)w(t)g(t) dt
J
0
for all A -> 0, so that it suffices to show that p(t)w(t)g(t) dt J
7.
cannot be zero for all co e H2. Assume that were the case. Then e"",1i(t)w(t)g(t) dt
=
0,
w(t)0(t)eiztg(t)dt
=
0
f-'W
for all A -> 0 and every 0 e H2.
Use now the preceding theorem (!) and another result to argue that J
w(t)h(t) dt
=
0
for all heH,, making w(t) also in H., together with w(t). This means that w(z)
=
1 it
'3z
-.Iz-tI2
w(t) dt
and w(z) are both analytic in 3z > 0. Since f is not outer, however, Iw(z)I = I f(z)/g(z)I < 1 for some such z. A contradiction is now easily obtained.)
2 The problem; simple reductions of it
249
Remark. The co e H. figuring in the argument just indicated is called an inner function.
2.
Statement of the problem, and simple reductions of it
Given a function w > 0 belonging to L1(Ll ), we want to know whether there is an co >, 0 defined on Ifs, not a.e. zero, such that
f-'l
I U(t) 12 w(t) dt
5 JT
1 U(t) 12
dt
for the Hilbert transforms U(t) (specified in some definite manner) of the functions U(t) belonging to a certain class. Depending on that class, the answer is different for different specifications of U(t). Two particular specifications are in common use in analysis. The first is preferred when dealing with functions U for which only the convergence of U(t) I
dt
2
is assured; in that case one takes U(x)
I f-
=
n -0 Cx
t
+ t2
+1
U(t)dt.
1 really an integral - is a Cauchy principal The expression on the right - not value, defined for almost all real x. (At this point the reader should look again at §H.1, Chapter III and the second part of §C.1, Chapter VIII.) A second definition of 0 is adopted when, for 6 > 0, the integrals U(t) at J't-xI>-bx-t
are already absolutely convergent. In that case, one drops the term t/(t2 + 1) figuring in the previous expression and simply takes U(x)
=
1 it
U(t) dt,
-00 xt
in other words, 1/rc times the limit of the preceding integral for S -+0. This specification of 0 was employed in §C.1 of Chapter VIII (see especially
the scholium to that article). It is useful even in cases where the above integrals are not absolutely convergent for 6 > 0 but merely exist as limits,
250
X E Control of Hilbert transforms by weighted norms
[A
lm A
l oo \J xAa+ -
`ttdt.
Jx+6/ x
This happens, for instance, with certain kinds of functions U(t) bounded
on F and not dying away to zero as t -f ± co. The Hilbert transforms thus obtained are the ones listed in various tables, such as those issued in the Bateman Project series.
If now our question is posed for the first kind of Hilbert transform, it turns out to have substance when the given class of functions U is so large as to include all bounded ones. In those circumstances, it is most readily treated by first making the substitution t = tan (9/2) and then working with functions U(t) equal to trigonometric polynomials in 9 and with certain auxiliary functions analytic in the unit disk. One finds in that way that the question has a positive answer (i.e., that a non-zero co > 0 exists) if and only if 1
(t2 + 1)2 w(t)
dt
0;
2 > 0.
From this we see already that our question (about the existence of non-zero w > 0 ) is without substance for the present specification of the Hilbert transform, when posed for all trigonometric sums U. There can never be an w >, 0, not a.e. zero, such that I U(t)I2w(t)dt
f-00. I U(t)12 w(t) dt f"000
for all such U, when w is integrable. This follows immediately on taking
U(t) = sin At,
U(t) = - cos At
in such a presumed relation and then making 2 --) 0; in that way one concludes by Fatou's lemma that w(t) dt
=
0.
The same state of affairs prevails whenever our given class of functions U includes pure oscillations of arbitrary phase with frequencies tending to zero. For this reason, we should require the class of trigonometric sums U(t) under consideration to only contain terms involving frequencies bounded away from zero, as we did in §D. The simplest non-trivial version of our problem thus has the following formulation: Let a > 0. Under what conditions on the given w >, 0 belonging to L,(P) does there exist an w >, 0, not a.e. zero, such that
JT
I U(t)12w(t) dt
I U(t)I2w(t)dt -00
for all finite trigonometric sums U(t)
_
Y, CA eiz` ? Izl %a
Here, we are dealing with the second kind of Hilbert transform, so, for
252
X E Control of Hilbert transforms by weighted norms
the sum U(t) just written, U(t)
_ Y (- iCz sgn2)eiz`. JAIia
Such functions U(t) can, of course, also be expressed thus: U(t)
= E (Ax cos At + B. sin At). lea
Then U(t)
_
(A1 sin At - BA cos At). 1>a
This manner of writing our trigonometric sums will be preferred in the following discussion; it has the advantage of making the real-valued sums U(t) be precisely the ones involving only real coefficients A., and BA.
We see in particular that if U(t) is a complex-valued sum of the above kind, 91 U(t) and 3 U(t) are also sums of the same form. This means that our relation
JT
I U(t)IZw(t)dt
I U(t)12w(t)dt -00
holds for all complex-valued U of the above form iff it holds for the real valued ones.
Given any trigonometric sum U(t) (real-valued or not) of the form in question, we have
U(t) + i0(t) _
Cze'2' lea
with certain coefficients C2. Conversely, if F(t) is any finite sum like the one on the right,
9IF(t) = U(t) is a sum of the form under consideration, and then
U(t) = 3F(t). These statements are immediately verified by simple calculation. Lemma. Given w > 0 in L1(a8), let a > 0. The relation f_'000 IU(t)12w(t)dt
< JT IU(t)I2w(t)dt
2 The problem; simple reductions of it
253
holds for all trigonometric sums
_
U(t)
(A2 cos At + B. sin At) a
with the function co(t) >, 0 iff
_ (w(t) + (w(t))(F(t))2 dt
JT
(w(t) - (o(t))I F(t)I2 dt
for all finite sums F(t)
Ce'z`
_ z,a
Proof. As remarked above, our relation holds for trigonometric sums U of the given form if JT (U(t))2uw(t)dt
5
J
(U(t))2 w(t) dt
for all such real-valued U. Multiply this relation by 2 and then add to both sides of the result the quantity {(U(t))2w(t) - (U(t))2w(t) - (U(t))2co(t) - (U(t))2o)(t)}dt. -z
We obtain the relation
(w(t) + w(t)) {(U(t))2 - (U(t))2} dt
J
1
J'0 (w(t) -
0)(t)){(U(t))2 + (U(t))2} dt
which must thus be equivalent to our original one (see the remark immediately following this proof).
In terms of F(t) = U(t) + iU(t), the last inequality becomes - 91
(w(t) + co(t)(F(t))2 dt
(w(t) - co(t)) I F(t) I2 dt, -ao
J-
so, according to the statements preceding the lemma, our original relation holds with the trigonometric sums U(t) if the present one is valid for the finite sums F(t)
Czeszc
_ a
X E Control of Hilbert transforms by weighted norms
254
If, however, F(t) is of this form, so is &"F(t) for each real constant y. The preceding condition is thus equivalent to the requirement that - Re2iy
J(w(t) + co(t))(F(t) )2 dt
0 in L1(R) and a > 0, any w > 0 such that I U(t)12 w(t) dt
for all sums U of the form U(t)
= Y (Ax cos At + Bx sin At) x>- a
must satisfy
w(t) < w(t)
a.e. on R.
Proof. Such an (o must in the first place belong to L1(P). For, putting first
2 The problem; simple reductions of it
255
U(t) = sin at, U(t) cos at in our relation, and then U(t) = sin 2at, U(t) = - cos 2at, we get
0
for each continuous function (p >, 0 of compact support. Fix any such (p, and pick an e > 0. Choose first an L so large that (p(t) vanishes identically outside (- L, L) and that II (P II .'
fiti,L (w(t) + w(t)) dt
-a
so Jw
(w(t) - (4t))(SN(t))2 dt
= JT (w(t) - w(t))I FN(t) 12 dt
is >, 0 by the lemma. Using this in the last member of the previous chain
of inequalities, we see that L
(w(t) - (O(t))(SN(t))2 dt -L
for each N, so, by the above limit relation, (w(t) - w(t))Qp(t) dt
>-
- E.
-00
Squeezing e, we see that the integral on the left is > 0, which is what we needed to show to prove the theorem. Done. Lemma. Given w > 0 in L1(l ), let a > 0. A necessary and sufficient condition
that there be an co > 0, not a.e. zero on R, such that I v(t)12w(t)dt J
a
257
2 The problem; simple reductions of it
is that there exist a function p(t) not a.e. zero, 0 5 p(t) < w(t), with w(t)(F(t)
J
)2 dt
f T (w(t) - p(t)) I F(t)I2 dt
I
for all functions F of the form F(t)
=
Y Cx eiz` a
When an w fulfilling the above condition exists, p may be taken equal to it.
When, on the other hand, a function p is known, the co equal to 2 p will work.
Proof. If a function co with the stated properties exists, we know by the previous theorem that 0 S w(t) < w(t) a.e.. Therefore, if U(t) is any sum of the above form,
'f2
1
I U(t)12w(t)dt
°°
I U(t)I2w(t)dt
2
f O I U(t)I2(w(t) - 2 w(t))dt. as
The first condition of the previous lemma is thus fulfilled with
col(t) = 2 w(t) in place of w(t) and
wl(t) = w(t) - 1200 in place of w(t). Hence, by that lemma,
f
(wl(t) + col(t))(F(t))2 dt -OD
(wl(t) - (ol(t))I F(t)12 dt
for the functions F of the form described. This relation goes over into the asserted one on taking p(t) = co(t).
If, conversely, the relation involving functions F holds for some p, 0 5 p(t) < w(t), we certainly have
f
(w(t) + 2 p(t))(F(t) )2 dt
-0000 (w(t) - p(t)) I F(t) 12 dt
-OD
+
p(t)IF(t)I2dt
2
f -00 (w(t) -
2
p(t)) I F(t) I2 dt
258
X E Control of Hilbert transforms by weighted norms
for such F, so, by the previous lemma,
If-
2
I U(t)IZp(t)dt
I U(t)I2w(t)dt --
for the sums U. Our relation for the latter thus holds with w(t) = p(t)/2, and this is not a.e. zero if p(t) is not. Done. Theorem. If, for given w > 0 in L1(R) and some a > 0 there is any co >, 0, not a.e. zero, such that
5 JT
I U(t)I2co(t)dt
f
00
I
U(t)12w(t)dt
for the finite sums U(t)
_
(A,1 cos At + B, sin At), a
we have °°
log- w(t)
-00
1+t2
dt
0. We have reached our promised contradiction. This shows that the
since
integral f °° .(log - w(t)/(1 + t2)) dt must indeed be finite, as claimed. The theorem is proved.
3.
Application of Hp space theory; use of duality
The last theorem of the preceding article shows that our problem can have a positive solution only when °° _00
log - w(t) 1
+t
dt
we may thus limit our further considerations to functions w , 0 in L1(l ) fulfilling this condition. According to a remark at the end of article 1, there is, corresponding to any such w, an outer function cp in H2 with I w(t)
I= ./(w(t))
a.e., t e R.
Theorem. Let w , 0, belonging to L1(18), satisfy the above condition on its logarithm, and let a > 0. In order that there exist an co , 0, not a.e. zero, such that
f
I U(t)IZw(t)dt
f
iU(t)IZw(t)dt
3 Use of Hp spaces and duality
261
for the functions U(t)
_
(A., cos At + B, sin At), a
it is necessary and sufficient that there be a function a(t), not a.e. zero, with
0 0 on a set of positive
measure if o(t) is, the first and main conclusion of our theorem now follows directly from the second lemma of the preceding article. Again, since p(t) = a(t)w(t), the second conclusion also follows by that lemma. We are done.
In order to proceed further, we use the duality between L,(IR) and L.(I8). When one says that the latter space is the dual of the former, one means that each (bounded) linear functional `P on L, corresponds to a unique 0 e L,,, such that `P(F)
=
J
F(t)/i(t)dt
for F E L, . Here, we need the linear functionals on the closed subspace H,
of L,. These can be described according to a well known recipe from functional analysis, in the following way. Take the (w*) closed subspace E of L. consisting of the 1i therein for which f (t)ili(t) dt
=
0
whenever f EH,; the quotient space L./E can then be identified with the dual of H,. This is how the identification goes: to each bounded linear functional A on H, corresponds precisely one subset of L,,, of the form wo + E (called a cosec of E) such that A(.f)
=
J
f (t)t/i(t) dt
e o + E, and only for those i/i. From article 1, we know that E is H. The dual of H, can thus be
whenever f e H, for any
identified with the quotient space L,.,/H,,. We want to use this fact to investigate the criterion furnished by the last result. For this purpose, we resort to a trick, consisting of the introduction of new norms, equivalent to the usual ones, for L, and L. If the inequality in the conclusion of the last theorem holds with any function o, 0 5 Q(t) 1, it certainly does so when a(t)12 stands in place of a(t). According to that theorem, however, it is the existence of such functions a d erent from zero on a set of positive
3 Use of HP spaces and duality
265
measure which is of interest to us here. We may therefore limit our search for one for which the inequality is valid to those satisfying
0 < a(t) < 1/2
a.e..
This restriction on our functions a we henceforth assume.
Given such a a, we then put
If IIi
f
=
(1 - a(t))If(t)Idt 00
for f c -L1; 1 1 f1 1i is a norm equivalent to the usual one on L1, because
UP,
IIfIII
II.fII1.
On L., we use the dual norm 11
110'
=
esssup te68
1 - a(t)
here, the 1 - a(t) goes in the denominator although we multiply by it when defining II
11 i
.
We have
II0II6 < 211011.
11011. 0, write, as in article 1, pp(z)
_
_
Iz3 tlZ(p(t)dt;
by a theorem from that article, T(z) is analytic in the upper half-plane and
cp(t+iy) -> cp(t) a.e. as y ---+ 0. Saying that cp is outer means, as we recall, that loglP(z)I
=
-'f' n
IIZ3z
tl2loglgv(t)I dt,
3z > 0;
p(z) has, in particular, no zeros in the upper half plane. The ratio
R(z) = e2iaz _ (Z) cp(z)
2
276
X E Control of Hilbert transforms by weighted norms
is thus analytic for 3z > 0. Since f (z) is entire and I (p(t) I
>0
a.e.,
R(t + iy) approaches for almost every t e R a definite limit,
'
R(t) = eras f(t)
z
as y - 0. Because our function f is real on 68, and
R(t)
e2ia1(P (t)
cp(t)
have the same argument there, and we see, referring to our requirement on h, that if R(t) were in Hco, we could take for h a suitable constant multiple of R. Usually, however, R(t) is not bounded, so this will not be the case, and we have to do a supplementary construction. We have IR(t)I
-
(f(t))2,
teR,
w(t)
so by hypothesis, IR(t)I _"0
1+t2
dt
0; the previous relation guarantees absolute convergence of the integral on the right, and Q(z) is analytic in the upper half plane, with 93Q(z) > 0 there. The quotient R(z)/Q(z) is thus analytic for 3z > 0. It is now claimed that R(z)
z > 0.
1,
Q(z)
The function f is of exponential type < a and fulfills the above condition involving log+ If (t) 1. Hence, by §G.2, Chapter III, logIf(z)I
< adz + n
-
3z 2loglf(t)I dt,
Iz - tI
which, with the previous formula for loglgp(z)I, yields logIR(z)I
0, an entire f # 0 of exponential type < a satisfying the relation in the statement exists. Then the point
P=
e2 iat
00 + H (PM
will have the property in question as long as II P II = 1. It will indeed frequently happen that II P II , = 1. Should that fail to come about, in which case II(e2iatw(t)l(p(t)) + H0 II
0. As we know, lim l/i(t + iy) = fi(t) y-.o
exists a.e. on R, with fi(t)
e (t) + iv(t) a.e. there, i (t) being the first kind of Hilbert transform described at the beginning of §E.2. Show that, when p < 1, e1
t)
(t + i)2
294
X F Geometry of unit sphere in L./H. belongs to H1. (Hint: By mapping the upper half plane conformally onto the unit disk and using the result from (b), show first of all that when p < 1,
_
e"v'(=)
1
.3z
it
z-tI2
enyu° dt
for 3z > 0, the integral on the right being absolutely convergent. This representation gives us fairly good control on the size of exp(pii(z)) in 3z > 0 - cf. Chapter VI, §A.2 - A.3. Knowing this, show by integrating around suitable contours that if A and S > 0, eiAX
(x+i)z+aexp(po(x+ih)dx
=
0
for any h > 0 - cf proof of theorem that the product of two H2 functions is in H1, §E.1. Now one may make
and use domi-
nated convergence (guaranteed by our representation for exp (ptli(z)) to get
elxx
(x + 1)2
exp(pgi(x+ih))dx
=
0.
Plug the representation for exp(po(z)) into this result and use Fubini's theorem, noting that e"`/(t + i)2 is in H.. Finally, make h-* 0.)
Let us now take a measurable sequence A of integers > 0 having (ordinary) density zero, whose Beurling-Malliavin effective density DA is equal to 1 (see §D.2 of Chapter IX). It is easy to construct such sequences. We need merely pick intervals [ak, bk] with integral endpoints,
0 < a1 < bt < a2 < b2 < a3 < such that bk - 1/bk and (bk - ak)lbk both tend to zero as k --> oc, while bk - ak
2 oo,
bk
and then have A consist of the integers in the [ak, bk]. Using such a sequence A, let us form the functions C(z), B(z) and u(t) considered in problem 44. Then,
u + H., is an extreme point of E even though it is not a support point thereof.
This will follow from problem 44, the first lemma of the present §, and
relation of material in preceding § thereto
295
Problem 46 To show that there can be no non-zero h e H , with I u(t) - h(t) I V(t) _ -v(t) + iv(t) as in problem 45(c). Show that in the present circumstances, 1
I h(t) exp 0 (t) I
l+t2
dt
U(zo + Re19)
uniformly in 9 for r < R tending to R; the same is of course true if we replace V by U on the left. On the circles I z - zoI = r with radii r < R sufficiently close to R we therefore have
U(z) - V(z) 5 e.
Here, both U(z) and the harmonic function V(z) enjoy the local mean value
property in the open disk { I z -zoI < R }. Hence, by what has just been
shown, we have the strong maximum principle for the difference U(z) - V(z) on the smaller disks { I z -zoI < r}. The preceding inequality thus implies that U(z) - V(z) < e on each of those disks, and finally that U(z) - V(z) < e for Iz - zo 1 < R. Squeezing e, we see that
U(z) - V(z) < 0 for Iz-zoI < R.
I Superharmonic functions
301
By working with the difference V(z) - U(z) we can, however, prove the reverse inequality in the same fashion. This means that one must have
U(z) = V(z) for I z - zo I < R, and our proof is finished. It is this argument that the reader will find helpful to keep in mind during the following development.
Next in importance to the harmonic functions as objects of interest in potential theory come those that are subharmonic or superharmonic. One
can actually work exclusively with harmonic functions and the ones belonging to either of the last two categories; which of the latter is singled
out makes very little difference. Logarithms of the moduli of analytic functions are subharmonic, but most writers on potential theory prefer (probably on account of the customary formulation of Riesz' theorem, to be given in article 2) to deal with superharmonic functions, and we follow their example here. The difference between the two kinds of functions is
purely one of sign: a given F(z) is subharmonic if and only if -F(z) is superharmonic.
Definition. A function U(z) defined in a domain -q with - oo < U(z) < 00 there is said to be superharmonic in provided that
(i) lim inf U(z) > U(zo) for zo a -9; Z-Zo
(ii) to each z e that 2"
271
corresponds an rZ,
U(z + pe''9) d 9 < U (z)
0 < rZ < dist(z, ate), such
for 0 < p < rZ.
o
Superharmonic functions are thus permitted to assume the value + 00 at
certain points. Although authors on potential theory do not generally agree to call the function identically equal to + oo superharmonic, we will sometimes find it convenient to do so. Assumption of the value - 00, on the other hand, is not allowed. This restriction plays a serious role in the subject. By it, functions like U(z)
=
3z, 3z > 0, 1- 00, .3z < 0,
are excluded from consideration. It may seem at first sight that an extensive theory could hardly be based on the definition just given. On thinking back, however, to the proof of
302
XI A Some rudimentary potential theory
Gauss' result, one begins to suspect that the simple conditions figuring in the definition involve more structure than is immediately apparent. One
notices, to begin with, that (i) and (ii) signify opposite kinds of local behaviour. The first guarantees that U(z) stays almost as large as U(zo) on small neighborhoods of zo, and the second gives us lots of points z in
such neighborhoods at which U(z) < U(zo). Considerable use of the interplay between these two contrary effects will be made presently; for the moment, let us simply remark that together, they entail equality of liminfz -zp U(z) and U(zo) at the zo e -9.
It is probably best to start our work with superharmonic functions by seeing what can be deduced from the requirement that U(z) > - oo and condition (i), taken by themselves. The latter is nothing other than a prescription for lower semicontinuity in -9; as is well known, and easily verified by the reader, it implies that U(z) has an assumed minimum on each compact subset of -9. Together with the requirement, that means that U(z) has a finite lower bound on every compact subset of .9. This property will be used repeatedly. (I can never remember which of the two kinds of semicontinuity is upper, and which is lower, and suspect that some readers of this book may have the same trouble. That is why I systematically avoid using the terms here, and prefer instead to specify explicitly each time which behaviour is meant.) A monotonically increasing sequence of functions continuous on a
domain -9 tends to a limit U(z) > - oo satisfying (i) there. This is immediate; what is less apparent is a kind of converse:
Lemma. If U(z) > - oo has property (i) in -9 there is, for any compact subset K of -9, a monotonically increasing sequence of functions continuous on K and tending to U(z) there.
Proof. For each n >, 1 put, for z e K,
inf(U(Q) + nIz-t I). CcK
Since U(t') is bounded below on K by the above observation, the functions *p (z) are all > - oo. It is evident that (p.+ ,(z) < U(z) for z e K and each n. To show continuity of q p,, at zo a K, we remark that the function of equal to U(l;) + nIC - zoI enjoys, like U(t;), property (i) and thus assumes its minimum on K. There is hence a L e K such that
nICo-zoI + U(o),
I Superharmonic functions
303
so, if z e K, U(C0)
s
njz-zol + 9 (zo).
In the same way, we see that (p.(zo)
0 such has, as just recalled, a finite lower that U(() > V for I - zo I < rl. bound, say -M, on K. Then, for n > (V + M)/rl, we have n l K - zo l + U(1;) > V f o r e K with I t; - zo l >, rl. But when I C - zo l < n we also have n I ( - zo 1 + U(t4) > V. Therefore for n > (M + V)/7
V
Since, on the other hand, T,,(zo) < U(zo), we see that the convergence in question holds, V < U(zo) being arbitrary. The lemma is proved.
Remark. This result figures in some introductory treatments of the Lebesque integral. Let us give some examples of superharmonic functions. The class of these
includes, to begin with, all the harmonic functions. Gauss' result implies indeed that a function U(z) defined on a domain -9 is harmonic there if and only if both U(z) and - U(z) are superharmonic in -9. The simplest kind of functions U(z) superharmonic, but not harmonic, in -9 are those of the form U(z)
=
log
with zo e .9.
I
I z - zo)
Positive linear combinations of these are also superharmonic, and so, finally, are the expressions U(z)
=
I log Jx
1
Iz
-
dy(t;) CI
formed from positive measures y supported on compact sets K. The reader should not proceed further without verifying the last statement. This involves
304
XI A Some rudimentary potential theory
the use of Fatou's lemma for property (i), and of the handy relation zn
2n
1
log
1
z+pe's-CI
o
d9 =
min log
1
,
log 1 P
(essentially the same as one appearing in the derivation of Jensen's formula, Chapter 1!) for property (ii). Integrals like the above one actually turn out to be practically capable
of representing all superharmonic functions. In a sense made precise by Riesz' theorem, to be proved in article 2, the most general superharmonic function is equal to such an integral plus a harmonic function. By such examples, one sees that superharmonic functions are far from being `well behaved'. Consider, for instance U (z)
= Y an log n
1
IZ - ZnI
formed with the zn of modulus < 1/2 tending to 0 and numbers an > 0 chosen so as to make a,,
loglZ < II
oo.
n
Here, U(0) < oo although U is infinite at each of the zn. In more sophisticated versions of this construction, the zn are dense in { I z I < 1/2} and various sequences of an > 0 with Y,nan < oo are used.
We now allow both properties from our definition to play their parts, (ii)
as well as (i). In that way, we obtain the first general results pertaining specifically to superharmonic functions, among which the following strong minimum principle is probably the most important:
Lemma. Let U(z) be superharmonic in a domain -9. Then, if KI is a (connected) domain with compact closure contained in -9, U(z)
>
inf U(')
for z E f2
scan
unless U(z) is constant on S2.
Proof. As we know, U(z) attains its (finite) minimum, M, on n, and it is enough to show that if U(zo) = Mat some zo E f2, we have U(z) - M on S2. The reasoning here is like that followed in establishing the strong maximum principle for harmonic functions.
I Superharmonic functions
305
Assuming that there is such a z0, we have, by property (ii), 1
M
U(zo)
>
2n
2rz
U(zo + pei9)d9 0
whenever p > 0 is sufficiently small. Here, U(zo + pei9) > M and if, at any 90, we had U(zo + pei90) > M, U(zo + pe'9) would be > M for all 9 belonging to some open interval including 90, by property (i). In that
event, the above right-hand integral would also be > M, yielding a contradiction. We must therefore have U(z0 + pei9) - M for small enough values of p > 0. The rest of the proof is like that of the result for harmonic functions, with E = {zec : U(z) = M} closed in Q's relative topology thanks to property (i). We are done. Corollary. Let U(z) be superharmonic in a domain -9, and let (9 be an open set with compact closure lying in -9. Then, for z e (9, U(z)
>,
inf U(t;). COO
Proof. Apply the lemma in each component of (9. Corollary. Let U(z) be superharmonic in -9, a domain with compact closure.
If lim infz_, U(z) > M at each 4 e 82', one has U(z) >, M in -9.
Proof. Fix any s > 0. Then, corresponding to each C e 8.9 there is an rc,
0 < rs < E, such that
U(z) >, M-e for ze2 and Iz-l;
, M-E. U(z)ishence ? M-e in (9 by the previous corollary. (9, however, certainly includes all points of -9 distant by more than E from 8.9. Our result thus follows on making
E -+ 0. From these results we can deduce a useful characterization of superharmonic functions.
XI A Some rudimentary potential theory
306
Theorem. If U(z) is > - oo and enjoys property (i) in a domain -9, it is superharmonic there provided that for each zo e -q and every disk A of sufficiently small radius with centre at zo, one has U(zo) > h(zo)
for every function h(z) harmonic in A and continuous up to 8A, satisfying
h(() < U(C) on
A.
Conversely, if U(z) is superharmonic in -9 and SZ is any domain having compact closure - -9, every function h(z) harmonic in 52 and continuous up to 852 is < U(z) in 52 provided that h(C) < U(C) on 852.
Figure 232
Proof. For the first part, we take any zo c- -q and verify property (ii) for U there, assuming the hypothesis concerning disks A about zo. Let then 0 < r < dist(zo, 8-9). By the first lemma of this article, there is an increasing sequence of functions un(9), continuous and of period 2n, such that
u,,(9) - U(zo + re'9), Put
re'9) = pei9)
0 < 9 < 2it.
and, for 0 < p < r, take
_
1
2n
r2 - p2
27[Jo r2+p2-2rpcos(9-T)
un(T) dT.
1 Superharmonic functions
307
Then each function hn(z) is harmonic in the disk A of radius r about zo and continuous up to 8A, where we of course have U(C)
If r > 0 is small enough, our assumption thus tells us that
hn(zo) < U(zo) for every n. Now Lebesgue's monotone convergence theorem ensures that 1
hn(zo) n > 2TC
2n
U(zo + re") dr o
as n -- oo. Hence f02 a
1
U(zo + re't) dT
2n
U(zo)
for all sufficiently small r > 0, and property (ii) holds. The other part of the theorem is practically a restatement of the second of the above corollaries. Indeed, if h(z), harmonic in f) and continuous
up to 0 c .9 satisfies h(C) < U(C) on 852, we certainly have lim inf (U(z) - h(z)) > 0
ZzEn
at each t; E 7f) on account of property (i). At the same time, U(z) - h(z) is superharmonic in 52; it must therefore be > 0 there by the corollary in question. This does it.
By combining the two arguments followed in the last proof, we immediately obtain the following inequality: For U(z) superharmonic in -9, 1
U(zo + pe'9)
2
2n
zoe-i, and 0 < r < dist(zo 8!2), r2
-P
2
n o r2+p2-2rpcos(9-T)
U(zo + re") dT, 0 < p < r.
This in turn gives us a result needed in article 2: Lemma. If U(z) is superharmonic in a domain -9 and zo e -9, 1
f02 n
U(zo + re"') d9 27r
is a decreasing function of r for 0 < r < dist(zo, 8-9).
308
XI A Some rudimentary potential theory
Proof. Integrate both sides of the boxed inequality with respect to 9 and then use Fubini's theorem on the right. Along these same lines, we have, finally, the Theorem. Let U(z) be superharmonic in a domain 9, and suppose that zo e 9
and that 0 < R < dist(zo, 8-9). Denoting by A the disk { I z - zo I < R},
put V(z) = U(z) for z e -9 - A. In A, take V(zo + re' 9)
=
1
2xc
2rz
o
R2 - r2 U(zo + Re")dT R2 + r2 - 2rR cos(9 - T)
(for 0 < r < R ). Then V(z) < U(z) and V(z) is superharmonic in -9.
Proof. For z e .9 ' A, the relation V (z) < U (z) is manifest, and for z e A it is a consequence of the above boxed inequality. To verify property (ii) for V, suppose first of all that z e for sufficiently small p > 0, 1
V(z)
=
U(z)
'>
2n
- A. Then,
2"
U(z+pei9)d9. 0
By the relation just considered, the right-hand integral is in turn
i 2n
2a
V(z+pei9)d9; o
V thus enjoys property (ii) at z. We must also look at the points z e A. On OA, the function U(C) is
bounded below, according to an early observation in this article. The Poisson integral used above to define V(z) in A is therefore either infinite for every, r, 0 < r < R, or else convergent for each such r. In the former case, V(z) oo for z e A, and V (trivially) possesses property (ii) at those
z. In the latter case, V(z) is actually harmonic in A and hence, for any given z therein, equal to the previous mean value when p < dist(z, 8A). Here also, V has property (ii) at z. Verifications of the relation V(z) > - oo and of property (i) remain. The first of these is clear; it is certainly true in -9 - A where V coincides with
U, and also true in A where, as a Poisson integral, V(z)
>,
inf U(C) K{-zol=R
with the right side > - oo, as we know. We have, then, to check property (i). The only points at which this can present any difficulty must lie on 8A, for, inside A, V is either harmonic
I Superharmonic functions
309
and thus continuous or else everywhere infinite, and outside A, V coincides
(in -9) with U, a function having the semicontinuity in question. Let therefore I z - zo I = R. Then we surely have lim inf V(C)
=
>
lim inf U(C)
U(z)
=
V(z),
so we need only examine the behaviour of V(C) for C tending to z from
within A. The relation just written holds in particular, however, for C = zo + Re" tending to z on 80. Since U(zo + Re") is also bounded below on 8A, we see by the elementary properties of the Poisson kernel that lim inf V(t') {
>,
U(z)
>,
V(z)
=
V(z).
z
CEO
We thus have lim inf V(z)
for the points z on OA, as well as at the other z e -9, and V has property (i). The theorem is proved.
Our work will involve the consideration of certain families of superharmonic functions. Concerning these, one has two main results. Theorem. Let the Un(z) be superharmonic in a domain -9, with U1(z)/ 1
M, so, since Un enjoys property (i), Un(z) > M in a neighborhood of zo. A fortiori, U(z) > M in that same neighborhood, and lim infz-zp U(z) >, U(zo) on account of the arbitrariness of M. Property (ii) is a consequence of Lebesgue's monotone convergence theorem. Let zo c- -9 and fix any p < dist(zo, 821). Then, by the above boxed inequality, 1
Un(zo)
?
2n
2n
Un(zo + pe''9) d8 0
310
XI A Some rudimentary potential theory
for each n. Here U,(zo + pei9) is bounded below for 0 < 9 < 2ir, so the right-hand integral tends to 1
2a
2n o
U(zo + pe19) d9
n -* oo by the monotone convergence. At the same time, U (zo) --n - U(zo), so property (ii) holds. We are done.
as
A statement of opposite character is valid for finite collections of superharmonic functions. If, namely, U1(z), U2(z), ... , UN(z) are superharmonic in a domain -9, so is min1,k,NUk(z). This observation, especially useful when the functions Uk(z) involved are harmonic, is easily verified directly.
WARNING. The corresponding statement about max1_ M in that punctured neighborhood, so, since M < V(zo) was arbitrary, we have lim inf=-=,,V(z)
>,
V(zo).
To complete verification of V(z)'s superharmonicity in -9 when that function is > - oo there, one may resort to the criterion provided by the first of the preceding theorems. According to the latter, it is enough to show that if zo e -9 and A is any disk centred at zo with radius < dist(zo, 8-9), we have V(zo) >, h(zo) for each function h(z) continuous on A, harmonic in A, and satisfying h(C) < on A. But for any such function h we certainly have h(C) 5 U(,) on 8A for every U e , so, by the second part of the theorem referred to, h(z) < U(z) in A for those U. Hence h(z) S inf U(z) = W(z) UeF
in A, and finally, h being continuous at zo (the centre of A ! ), h(zo)
=
lim h(z)
0 (and in fact for all such p < dist(z, 8.9) by the boxed inequality near the end of the preceding article); on the other hand, lim inf 0 UP(z) >, U(z) by property (i). Thus, for each z e -9, UP(z) --- U(z) as p -> 0.
A lemma from the last article shows that this convergence is actually monotone; the UP(z) increase as p diminishes towards 0.
Concerning the UP, we have the useful
Lemma. If U(z) is superharmonic in a (connected) domain .9 and not identically infinite there, the UP(z) are finite for z e -9 and
0 < p < dist(z, 8.9). Proof. Suppose that 1
Ur(zo)
=
2n
2n
U(zo + re") dt
=
00
0
for some zo e 9 and an r with 0 < r < dist(zo, 8-9). It is claimed that then U(z) - oo in -9. By one of our first observations about superharmonic functions in the
preceding article, U(zo + re") is bounded below for 0 < t < 27r. The above relation therefore makes the Poisson integrals occurring in the boxed inequality near the end of that article infinite, and we must have U(z) _- oo for Iz - zoI < r.
2 Riesz representation of superharmonic functions
313
Let now z' be any point in -9 about which one can draw a circle - of radius r', say - lying entirely* in -9 and also intersecting the open disk of radius r centred at zo. Since U(z) is bounded below on that circle, we have Ur.(z') = oo so, by the argument just made, U(z) - co for I z - z' I < r'. The process may evidently be continued indefinitely so as to gradually fill out the connected open region -9. In that way, one sees that U(z) - 00 therein, and the lemma is proved.
Corollary. If U(z) is superharmonic and not identically infinite in a (connected) domain -9, it is locally L1 there (with respect to Lebesgue measure for R2).
Proof. It is enough to verify that if zo e -9 and 0 < r
0, denote by -9r the set of z e -9 with dist(z, 8-9) > 2r. Let U be superharmonic in -9, then:
((DrU)(z) < U(z) for ze2r;
(Or U) (z) -- U(z) as r ---* 0 for each z e -9; (02rU)(Z) < ((DrU)(z)
for ze-92r.
If also U(z) * oo in the (connected) domain -9, each (DrU)(z) is infinitely differentiable in the corresponding -9r, and superharmonic in each connected component thereof.
Proof. The first two properties of the Or U follow as direct consequences
2 Riesz representation of superharmonic functions
315
of the behaviour, noted above, of the U,(z) together with Bp's normalization. The third is then assured by cp(p)'s being supported on the interval (1, 2).
Passing to the superharmonicity of (D, U, we first check property (ii)
for that function in -9r. This does not depend on the condition that
U(z) # co. Fix any z e fir. For 0 < a < dist(z, 8-9) - 2r we then have zn
1
(4),U)(z+ae"')di/i 2a
z 1
47r2 r
f2n
f2,,
zz
o
0
0
Since pp(p/r) vanishes for p >, 2r,
z + pe''9 lies in -9 and has distance
> a from 8.9 for all the values of p actually involved in the second expression. The argument z + pe''9 + ae''' of U thus ranges over a compact
subset of 9 in that triple integral, and on such a subset U is bounded below, as we know. This makes it permissible for us to perform first the integration with respect to 0. Doing that, and using the boxed inequality from the preceding article, we obtain a value zn
U(z+peis)(p(p/r)pd9dp
1
2rtrz
=
((DrU)(z),
o
o
showing that 0,U has property (ii) at z. Superharmonicity of 'D, U in the
components of -9, thus follows if it meets our definition's other two requirements there. Satisfaction of the latter is, however, obviously guaranteed by the infinite differentiability of s, U in -9 which we now proceed to verify for functions
U(z) # oo in -9. The left-hand member of the last relation can be rewritten as
2nrzJ1
-.
where, as usual, l; = °°
+ irl. Putting z + C
iil', this becomes
°°
'.f .
U(C') rp(I (' - zl /r) di;' drl'.
27r2
Here, cp(I C' - z I /r) vanishes for I C' - z 15 r and I K' - z I > 2r. Looking,
then, at values of z near some fixed zo e -9, - to be definite, at those, say, with I z - zo l
0 for each ('e K. (Here we have been helped by q (p)'s I zo vanishing for 0 < p < 1. ) br U is thus W. in .9r. The theorem is proved.
-
The approximations F,U to a given superharmonic function U are used in establishing the Riesz representation for the latter. That says essentially
2 Riesz representation of superharmonic functions
317
that a function U(z) superharmonic and * oo in and on a bounded domain -9 (i.e., in a domain including
_ I log
U(z)
1
Iz
-
i
dp(C)
with p a (finite) positive measure on
is given there by a formula
+
H(z),
and H(z) harmonic in Y. (Conversely,
expressions like the one on the right are always superharmonic in -9, according to the remarks following the first lemma of the preceding article.)
The representation is really of local character, for the restriction of the measure p figuring in it to any open disk A S -9 is completely determined by the behaviour of U in A (see problem 48 below), and at the same time, the function of z equal to
f .9-" log Iz-TI 1
dp(C)
is certainly harmonic in A. The general form of the result can thus be obtained from a special version of it for disks by simply pasting some of those together so as to cover the given domain -9 ! In fact, only the version for disks will be required in the present chapter, so that is what we prove here. Passage from it to the more general form is left as an exercise to the reader (problem 49).
We proceed, then, to the derivation of the Riesz representation formula
for disks. The idea is to first get it for le,, superharmonic functions by simple application of Green's theorem and then pass from those to the general ones with the help of the (D. U. In this, an essential role is played by the classical Lemma. A function V(z) infinitely differentiable in a domain -9 is superharmonic there if and only if 02V(z)
8x2
+ 82V(z) 8y2
0, there is thus by Egorov's theorem a compact E c K such that V,,(z) -* V(z) uniformly for z E E
as p -> 0, and p(K - E)
0 small enough, we can
ensure that the last quantity on the right, M + E log diam K, denoted henceforth by M', is as close as we like to M. At the same time, when z 0 K, log JK
1 y
Iz - SI
E
dp(i)
lies between s log
1
dist (z, K)
+ diam K
and
E log
1
dist (z, K
For any such fixed z, then,
I log JE
Iz
- (I du(C)
will be arbitrarily close to V(z) when s > 0 is sufficiently small (depending on z ). We see, e > 0 being arbitrary, that we will have V(z) < M at each
3 Pure potentials: maximum principle and continuity
331
z 0 K (thus proving the theorem) if we can deduce that fE log
1
y
Z
dp(i)
< M'
outside E knowing that this holds everywhere on E.
The last implication looks just like the one affirmed by the theorem, so it may seem as though nothing has been gained. We nevertheless have
more of a toehold here on account of the first condition on our set E, according to which log + Iz
dp(i)
p0
=
Jx log Iz
1
1 dµQ
logi
min (log JK
Iz
dp(i)
tends to zero uniformly for z e E as p - 0. Thence, a fortiori (!), P
log+ E
Iz
-
0
dp(1;)
uniformly for z e E
CI
as p -> 0. This uniformity plays an essential role in the following argument.
It will be convenient to write U(z)
=
JE I log
1
dp(C).
Iz
The proof of our theorem has boiled down to showing that if
U(z) < M' for z e E, then U(z) is also < M' at each z 0 E. This is where we use the maximum principle for harmonic functions. In C - E, U is harmonic; also,
U(z) -+ - oo as z - oo unless p(E) = 0, in which case the desired conclusion is obviously true.
The principle of maximum will therefore make U(z) 0; we wish to show that at each z,) E E,
U(z)<M'+76 for the points z in a neighborhood of zo. Thanks to the uniformity arrived at in the preceding construction, we can fix a p > 0 such that
332
XI A Some rudimentary potential theory
P dµ(C) < 6 SI JE whenever z e E. With such a p, which we can also take to be < 1, we have
I log+
IZ
U(z)
m i(log
=
log
iE
d) +
p
flog E
Z
The first integral on the right is < U(z) and hence < M' for z e E; it is, moreover, continuous in z. That integral is therefore < M' + 6 whenever z is sufficiently close to any zo e E; our task thus reduces to verifying that JElog+ Iz P
dp(()
, 1 fulfill the requirement with parameters L, C, and a, and suppose that
log W(t)
°
I+ t,
-
dt
, W(x) also meeting the requirement such that, corresponding to any entire function tp(z) * 0 of
exponential type < A making W(x) I cp(x)I < 1 on R, one has an entire O(z) # 0 of exponential type < mA with W1(x)IO(x)I < const., xeR. Here, for m we can take any integer > 4/a. Remark. As we know, the integral condition on log W follows from the existence of just one entire function co having the properties in question. Proof of theorem. Any entire function cp satisfying the conditions of the hypothesis must in particular have modulus < 1 on the real axis, so, by the second theorem of §G.2, Chapter III, log I tv(z) I
0. Adding to both sides the finite quantity 1
°°
In
-co
rjz log W(t)
Iz-tI
dt
346
XI B Multipliers and the smallest superharmonic majorant
we see, remembering the given relation
+ log W(t)
log I cp(t) I
0.
Put now z = x + iL, and use the fact that log W(t)
>,
a log W(x)
+
log C
for t belonging to an interval of length L containing the point x. Since log W(t) >, 0, the integral on the left comes out 4 (a log W(x) + log C), and we find that 4 log l cp(x + iL) I + log Wi (x) a
,
f' L log W(t) dt (x-t)2+L2 na
= C _ 1 exp f 4
W(x). This function is, on the other hand, infinitely
differentiable, and it satisfies the regularity requirement by the last lemma.
At the same time, x e R. 5 const., is bounded on the real axis, we know by the third
Wj(x) I qp(x + iL)141"
Because 'p
Phragmen-Lindelof theorem of §C, Chapter III that cp(x + iL) is also bounded for x e R. Hence, taking any integer m >, 4/a, we have W1(x)IiIi(x)I
5
const.,
xef8,
with the entire function O(z) = ((p(z + 1L))'",
obviously of exponential type < mA. Done.
The elementary result just proved permits us to restrict our attention to infinitely differentiable weights when searching for the form of the `essential'
1 Local regularity requirement for weights
347
second condition that those meeting the regularity requirement must satisfy in order to admit multipliers. This observation will play a role in the last two §§ of the present chapter. But the main service rendered by the requirement is to make the property of admitting multipliers reduce to a more general one, easier to work with, for weights fulfilling it. In order to explain what is meant by this, let us first consider the situation
where an entire function (p(z) # 0 of exponential type < A with W(x)Ip(x)I < const. on F is known to exist. If the weight W(x) is even, some details of the following discussion may be skipped, making it shorter (although not really easier). One can in fact stick to just even weights (and even functions (p(z) ) and still get by - see the remark following the last theorem in this article - and the reader is invited to make this simplification if he or she wants to. We treat the general case here in order
to show that such investigations do not become that much harder when evenness is abandoned.
Assume that W(x) > I is either continous, or fulfills the regularity requirement (of course, one property does not imply the other). Then, since cp(z) 0 0, W(x) cannot be identically infinite on any interval of length
> 0. By the first of the above lemmas, this means that W(x) is bounded on finite intervals under the second assumption. The same is of course true in the event of the first assumption. The function W(x) is, in particular, bounded near the origin, so if cp(z) has a zero there - of order k, say - the product W(x)cp(x)/xk will still be bounded on R. We can, in other words, assume wlog that (p(0) 0 0, and
hence that 9 has a Hadamard factorization of the form =
cp(z)
Ce1'
C 1 - zlez12.
Following a procedure already familiar to us, we construct from the product on the right a new entire function O(z) having only real zeros (cf.
§H.3 of Chapter III and the first half of the proof of the second Beurling-Malliavin theorem, §B.3, Chapter X).
Denote by A the set of zeros A figuring in the above product with 91A
0. For each A e A we put
_
91C
this gives us real numbers A' with I l' I % I A I . (It is understood here that
each A' is to be taken with a multiplicity equal to the number of times that the corresponding 2 e A figures as a zero of cp.) The number N(r) of
348
XI B Multipliers and the smallest superharmonic majorant
points A' having modulus 0 close to zero). Plugging the above inequality for N(r) into the last integral, we see immediately that lim sup log I 0y) I
, 0. That it does not
exceed A is guaranteed by the estimate on U(z) just found. That estimate
and the fourth theorem of §C, Chapter III, now show that in fact U(z)
< K + A'3z
for 3z > 0;
the function U(z) - K - A',3z is thus harmonic and < 0 in the upper half plane. * any such z must be real - U(z) is finite for 3z # 0 t and, in particular, finite (see preceding footnote) - the integral in the following Poisson representation is thus surely convergent.
I Local regularity requirement for weights
353
By §F.1 of Chapter III we therefore have
U(z) - K - A'z = -biz -
J 1
C
)z dc(t)
Iz-tl2
for 3z > 0, with a constant b > 0 and a certain positive measure v on R. It is readily verified that b must equal zero. Our desired Poisson representation for U(z) will now follow from an argument like the one in §G.1 of Chapter III if we verify absolute continuity of a. For this purpose, it is enough to show that when y --* 0, M
f
I U(x + iy) - U(x) I dx
0
M
for each finite M. Given such an M, we can write (J.2M U(Z)
=
1-
+
y'.RZ
_2M
+ J t1>2M)(lOg
z
+
1Az
) dp(t).
t
The second of the two integrals involved here clearly tends uniformly to (log
flo > 2M
x
1-t
as z = x + iy tends to x, when - M < x < M. Hence, since p(t) is zero on a neighborhood of the origin, the matter at hand boils down to checking that m
2M
-M
-2M
(loglx+iy-tl - loglx-tI)dp(t)
dx
--*
0
as y -> 0. The inner integrand is already positive here, so the left-hand expression is just 2M
M
S-2MJ-M
ti - logix-ti)dxdp(t).
In this last, however, the inner integral is easily seen - by direct calculation,
if need be - to tend to zero uniformly for - 2M < t < 2M as y -* 0. (Incidentally, I MM log I w - x I dx is the negative of a logarithmic potential
generated by a bounded linear density on a finite segment, and therefore continuous everywhere in w.) The preceding relation therefore holds, so a is absolutely continuous, giving us the desired Poisson representation for U(z).
X1 B Multipliers and the smallest superharmonic majorant
354
Once that representation is available, we have, for 3z > 0, U(z)
'3z log W(t)
1
+
i- Iz-t12 1
°° 3z (U(t) + log W(t))dt.
it
-D
+
A',3z
dt
Iz-t12
Since, however, U(t) + log W(t)
K on R, the right side of this relation
must be < K + A'Sz, so we have y'9iz
+
( log 1-
J
z t
Zz log W(t)
1
irJ-
Iz-t12
dt
3z>0.
K+A'Sz,
(Putting z = i, we see by the way that $(log W(t)/(1 + t2)) dt < oo.) By hypothesis, W meets our regularity requirement with parameters L, C, and a; this means that a log W(x) + log C
log W(t)
for t e JX, an interval of length L containing x. Therefore, if
z=x+iL, second integral on the left in the preceding relation is (a/4) log W(x) + (1/4) log C. After multiplying the latter through by 4/a we thus find, recalling that A' S A,
the
ay x
iL + (log 1 - x
+
J
+
log W(x)
K',
d(4p(t)la) x e tFl,
where
a
It is at this point that we apply the last lemma, with v(t)
=
4 a
p(t)
and z = x + iL. If p(t), and hence v(t), vanishes on the neighborhood (- a, a) of the origin, we see on combining that lemma with the preceding
I Local regularity requirement for weights
355
relation that fax
+
(log1_ x + iL
('
J
+
t
K' + log+
+
d [v(t)]
log W(x)
t
x
+ log
L
1 + JxJ+iL a
on !!B, with a certain constant /3. From this we have, afortiori, log
)d[v(t)]
1-x + t
K" + 2 log+ I x l
+
log W(x)
for x e 1,
K" being a new constant. The first two terms on the left add up, however, to logIcp(x)I, where qq(z)
=
(1
- zz
ez1"
is the Hadamard product formed from the discontinuities A of [v(t)], each one taken with multiplicity equal to the height of the jump in that function corresponding to it. Since v(t)
=
5 - + o(1)
4p(t)
t
at
an
for t -+ ± oo (hypothesis!), that product is certainly convergent in the complex plane, and q is an entire function. In terms of it, the previous relation can be rewritten as W(x) k cp(x) J
± oc, besides being bounded on R. That, however, makes
I
y
y2+t2
P(t) dt t
for y - ± oo,
0
as one readily sees on breaking up the integral into two appropriate pieces. We thus have
,0
U(iy) IYI
for y -* ± cc,
and the quantity A' figuring in the above examination of U(z) is equal to zero. By the estimate obtained there, we must then have
U(z) < K for 3z > 0, and exactly the same reasoning (or the evident equality of U(z) and U(z) ) shows this to also hold for .3z < 0. The subharmonic function U(z) is, in other words, bounded above in the complex plane if I p(t)I is bounded.
Such a subharmonic function is, however, necessarily constant. That is a general proposition, set below as problem 52. In the present circumstances, we can arrive at the same conclusion by a simple ad hoc argument. Since
p(t)/t > 0, the previous formula for U(iy) yields, for y > 0, v
U(iy)
'>
YZ
P(t)
Y
dt
>,
P(t) dt.
1
2 _Y t f_Yy2+t2 t If ever p(t) is different from zero, there must be some k and yo, both > 0,
with either p(t) > k for y >, yo or p(t) < - k for y , yo. This, however, would make U(iy) -> oo for y --> oo, contra-
dicting the boundedness of U(z), so we must have p(t) - 0. But then yx
=
U(x)
S K - log W(x),
x c- R,
which contradicts our assumption that W(x) >, I (either for x ---+ 00 or for x --> - oo) unless y = 0. Finally, then, the boundedness of p(t) forces U(x) to reduce to zero, whence log W(x)
=
U(x) + log W(x)
i.e., W(x) is bounded, as we claimed.
1 on R. Suppose there is an increasing function p(t), zero on a neighborhood of the origin, such that p(t) t
'
/i(z) is not very precise. Problem 52 Show that a function V(z) superharmonic in the whole complex plane and bounded below there is constant. (Hint: Referring to the first theorem of §A.2, take the means (1, V)(z) considered there. Assuming wlog that V(z) oo, each of those means is also superharmonic and bounded below
362
XI B Multipliers and the smallest superharmonic majorant in C, and it is enough to establish the result for them. The F,V are also W., so we may as well assume to begin with that V(z) is 16.. That reduction made, observe that if V(z) is actually harmonic in C, the desired result boils down to Liouville's theorem, so it suffices to establish this harmonicity. For that purpose, fix any zo and look at the means 1
1
2,
=
V,(zo)
V(zo + res) d9. I
0
Consult the proof of the second lemma in §A.2, and then show that aV,(zo)
alogr is a decreasing function of r, so that V,(zo) either remains constant for all r > 0 - and hence equal to V(zo) - or else tends to - oo as r -> oo. In the second case, V could not be bounded below in C. Apply Gauss' theorem from §A.1.)
Scholium. The regularity requirement for weights given in the 1962 paper of Beurling and Malliavin is much less stringent than the one we have been using. A relaxed version of the former can be stated thus:
There are four constants C > 0, a > 0, #
W(t)
for
Ce-1x1'(W(x))"
teI,.
The point we wish to make here is that the exponentials in J x V and I x Ifl are in a sense red herrings; a close analogue of the first of the above two theorems, with practically the same proof, is valid for weights meeting the more general condition. The only new ingredient needed is the elementary Paley-Wiener multiplier theorem. Problem 53 Suppose that W(x) >, 1 fulfills the condition just formulated, and that there is an increasing function p(t), zero on a neighborhood of the origin, with p(t) t
5
A
+ o(1)
for t -* ± co
IT
and ex
+
fi log
x
x
1-- + t dp(t) + t
log W(x)
const.
2 The smallest superharmonic majorant M F
363
on 08, where c is a certain real constant. Show that for any rt > 0 there
is an entire function /i(z) # 0 of exponential type < 4A/a + rl making W(x) I i/i(x) I
-
, r where V(z) = (WIF)(z). We see as in the proof of the last lemma that V(z) is a superharmonic majorant of F(z). Hence V(z) >, (9J1F)(z).
But the reverse inequality was already noted above. Therefore,
V(z) = (9RF)(z) Since V(z) is harmonic for IzI < r, we are done.
Let us now look at the set E on which (9JIF)(z) = F(z) for some given continuous function F. E may, of course, be empty; it is, in any event, closed. Suppose, indeed, that we have a sequence of points zk e E and that zk zo. Then, since 9JIF enjoys property (i) (§A.1),
XI B Multipliers and the smallest superharmonic majorant
368
we have (9JIF)(zo)
F(zo), and thus finally (9RF)(zo) = F(zo), making zo c- E. This means that the set of z for which (9JlF)(z) > F(z) is open. Regarding it, we have the important Lemma. (TIF)(z), if finite, is harmonic in the open set where it is > F(z). Note. I became aware of this result while walking in Berkeley and thinking
about a conversation I had just had with L. Dubins on the material of the present article, especially on the notions developed in problems 55 and 56 below. Dubins thus gave me considerable help with this work. Proof of lemma. Is much like those of the two previous ones. Let us show that if (9JRF)(zo) > F(zo) with OF finite, then (93lF)(z) is harmonic in some small disk about zo.
We can, wlog, take zo = 0; suppose, then, that
(9RF)(0) > F(0) + 2j,
say,
where q > 0. Property (i) then gives us an r > 0 such that
(9QF)(z) > F(0) + n < r, and the continuity of F makes it possible for us to choose this r small enough so that we also have for I z I
F(z) < F(0) + ri for
zi
< r.
Form now the superharmonic function V(z) used in the proofs of the last two lemmas. As in the second of those, we certainly have V(z) < (U F)(z),
according to our theorem from §A.1. In the present circumstances, for
JzJ < r, V(z)
=
1 fn. rz - & z t z 2n - lz-re'Lla
(9J1F) (re'ti) di
is > F(0) + rl, whereas F(z) < F(0) + ,j there; V(z) is thus > F(z) for I z I < r. When z I > r, V(z) = (MF)(z) is also > F(z), so V is again a superharmonic majorant of F. Hence V(z) > (9J1F)(z),
2 The smallest superharmonic majorant 9XF
369
and we see finally that V(z) = (9JtF)(z), with the left side harmonic for I z i < r, just as in the proof of the preceding
lemma. Done. Lemma. If 9JIF is finite, it is everywhere continuous.
Proof. Depends on the Riesz representation for superharmonic functions. Take the sets
E=
{z: (93IF)(z) = F(z)}
and (9
=
C - E;
as we have already observed, E is closed and (9 is open. By the preceding lemma, (9RF)(z) is harmonic in (9 and thus surely continuous therein. We therefore need only check continuity of 9JIF at the points of E. Let, then, zo E E and consider any open disk A centered at zo, say the one of radius = 1. In the open set f = An (9 the function (9JlF)(z) is harmonic, as just remarked and on A - Q = A r )E, (9RF)(z) = F(z) depends continuously on z. The restriction of 9JRF .to E is, in particular, continuous at the centre, zo, of A. The corollary to the Evans-Vasilesco theorem (at the end of §A.3) can
now be invoked, thanks to the superharmonicity of (9JtF)(z). After translating zo to the origin (and A to a disk about 0), we see by that result that (MF)(z) is continuous at zo. This does it. Remark. These last two lemmas will enable us to use harmonic estimation to examine the function (9J2F)(z) in §C.
It is a good idea at this point to exhibit two processes which generate (MF)(z) when applied to a given continuous function F, although we will not make direct use of either in this book. These are described in problems 55 and 56. The first of those depends on Problem 54 Let U(z), defined and > - oo in a domain -9, satisfy lim inf UO
>,
U(z)
for z e I. Show that U(z) is then superharmonic in -9 if, at each z therein,
370
XI B Multipliers and the smallest superharmonic majorant one has U(z)
irrZ JJI_zl 0 sufficiently small. (As usual, = + in.) (Hint: For the if part, the first theorem of §A.1 may be used.)
Problem 55 For Lebesgue measurable functions F(z) defined on C and bounded below on each compact set, put (MF)(z)
=
sup
1z J J
r>0 1r
IS-z1 0. Show that there is a v e 2 (depending, in general, on z, R and N) such that 1
2R
VN(z+Rei9)d9 2n
F(z+l')dv().
= c
o
(Hint: First show how to get a Borel function k(9) taking the values 1, 2; 3, ... , N such that VN(z + Re' 9)
=
fc F (z + Re' 9 + C)
372
XI B Multipliers and the smallest superharmonic majorant Then define v by the formula 2n
l
G(1;) dv(1;)
=
G(C + Re's) dµk(s)Q d9 rz
Sc
Jc
o
and verify that it belongs to Q.)
(d) Hence show that
=
V(z)
sup J F(z + l;) dj
(sic!)
)
C
is superharmonic. (Hint: Since F is continuous, we also have V(z) = sup k f c F(z + C)dukG) with the µk from (b). That is, V(z) = limN-. VN(z), where the VN are the functions from (c). But by (c), l
2'
2n o
0 and form the
function
F(z)
=
13z I log W(t)
1
Iz-t12
-co
?T
dt-A13z1,
the expression on the right being interpreted
as log W(x) when
z = x e R. This function F is then continuous and the material of the preceding article applies to it; the smallest superharmonic majorant, 931F, of F is thus at our disposal.
Our object in the present article is to establish a converse to the observation made near the beginning of the last one. This amounts to showing that if 9A F is finite, one actually has an increasing function p, zero on a neighborhood of the origin, such that P(t)
t
1 the function 9JtF corresponding to F(z)
=
1
n -a,
13z l log W(t)
Iz-tla
dt - A 13z l
(where A > 0 ) is finite. If (9J1F)(z) is also harmonic in a neighborhood of the origin, we have
=
(9JtF)(z)
(9JlF)(0) - y9iz - J
( log 1-
z
+
9iz
)dp(t),
t
with a constant y and a certain increasing function p(t), zero on a neighborhood of the origin, such that p(t)
A
t
n
for t -- + co.
Remark. The subsidiary requirement that (9RF)(z) be harmonic in a neighborhood of the origin serves merely to ensure p(t)'s vanishing in such a neighborhood; it can be lifted, but then the corresponding representation for 9RF looks more complicated (see problem 57 below). Later on in this article, we will see that the harmonicity requirement does not really limit applicability of the boxed formula.
Proof of theorem. Is based on the Riesz representation from §A.2; to the superharmonic function (9JtF)(z) we apply that representation as it is formulated in the remark preceding the last theorem of §A.2 (see the boxed
3 How 9RF gives us a multiplier if it is finite
377
formula there). For each R > 0 this gives us a positive measure pR on {I C I < R} and a function HR(z) harmonic in the interior of that disk, such
that ( 9 R F ) (z)
=
log
J
+
- ' I dpR(C) 1
Iz
R
for
HR(z)
1z1
< R.
By problem 48(c), the measures pR and PR. agree in {ICI < R} whenever R' > R; this means that we actually have a single positive (and in general infinite) Borel measure p on C whose restriction to each open disk { I C I < R}
is the corresponding PR (cf, problem 49). This enables us to rewrite the last formula as (9RF)(z)
f
w, 0} and in f 3z < 0}, so, according to the last theorem of §A.2, p cannot have any mass in either of those half planes. By the same token, p has no mass in a certain neighborhood of the origin, 931F being, by hypothesis, harmonic in such a neighborhood. There is thus an increasing function p(t), zero on a neighborhood of the origin, such that p(E)
=
J
dp(t) ErR
for Borel sets E c C, and we have (9RF)(z)
=f
R
1
log
dp(t) tI
IZ
R
+
HR(z)
for Izl < R,
with HR harmonic there. Our desired representation will be obtained by making R --+ oo in this relation. For that purpose, we need to know the asymptotic behaviour of p(t) as t -+ ± oo. It is claimed that the ratio
p(r) - p(- r) r (which is certainly positive) remains bounded when r -+ oo. Fixing any R, let us consider values of r < R. Using the preceding formula and reasoning as in the proof of the last theorem in §A.2, we easily find that
-
2n
('a I
R
(9RF)(rei 9) d9
=
f
R
min log
I
I,
log r d
p(t)
+
HR(0),
378
XI B Multipliers and the smallest superharmonic majorant
from both sides, that
and thence,(' subtracting
-J
fit
R
log'
dp(t)
I
(T?F)(re''9)d9
2a
t1
R
-
(931F)(0)
for 0 < r < R. Here, p(t) vanishes on a neighborhood of the origin, so we can integrate the left side by parts to get
J'p(t)
tP( - t)dt = (lF)(0)
-
21rfnrz(WJ1F)(rei9)d9,
which is, of course, nothing but a version of Jensen's formula. In it, R no longer appears, so it is valid for all r > 0. By the lemma, however, - (9JIF)(z)
= A I Zz i
-
1 fI ..3z I ('JlF)(t)
ir - Iz-t12
dt,
a quantity < AI,3z1, since ('JJ;F)(t) >, log W(t) > 0. Using this in the previous relation, we get
- P(- t) dt S
JrP(t)
(XflF)(0) +
t
2A
r, 77
whence
p(r) - p(- r)
0 is arbitrary. That works out to 2 1tR
oo
-
2R 1
log R
t
r+t r - t
(9RF)(t) dr dt
=
iR
-, `I' (t )(TIF)(t)dt, IRI
XI B Multipliers and the smallest superharmonic majorant
380
where
=
em(u)
log s+l
ds.
Ju
s-1
The last integral can be directly evaluated, but here it is better to use power series and see how it acts when u --* 0 and when u --> co. For 0 < u < 1, expand the integrand in powers of s to get
3u2 + O(u4),
=
`Y(u)
0 < u < 1.
For u > 1, we expand the integrand in powers of 1/s and find that
21og 2 + O I
=
T(u)
2 I,
u > 1.
u/
P(R/I t I)IR thus behaves like 1/R for small values of I t h/R and like R/t2 for
large ones, so, all in all,
2 T(R) Itl
irR
=
F(O)
0
rz
for 0 < r < h.
Proof. We have F(z) = F(z), so 1
2n
f
1
F (re'9)d9
n fo
rz
F(re'9) d9.
It will be convenient to denote the right-hand integral by J(r) and to work with the function G(z)
=
('
1
n
-
3z log W(t) Iz-tI2
dt - A 3z
(sic!)
instead of F(z); we of course also have rz
J(r)
=
1
G(re's) d9.
n
o
In the present circumstances the function G(z) is finite, and hence harmonic, in both the upper and the lower half planes. Moreover, since log W(t)
(- h, h)
0 for Its < h, G(z) (taken as zero on the real interval is actually harmonic* in C - (- oo, - h] - [h, oo) and
hence c,,) in that region. There is thus no obstacle to differentiating under the integral sign so as to get
dJ(r) dr
_
1
rz 8G(re'9)
nr fo
rd9,
Or
0 0 with G(x) = F(x) = log W(x) = 0 for - h < x < h, so, for such x, F(x + i&y)
lim
Gy(x)
Ay
AY-O+
which, by our formula for F, is equal to
1f R
log W(t)
(X-0'
-
-
dt
A.
If, then, this expression is > 0 for I x I < h, we must, by the preceding formula, have
J'(r) > 0 for 0 < r < h. Obviously, J(r) -> F(0) = 0 for r --+ 0. Therefore,
0. This
can be readily seen by putting together some of the above results and then using a simple measure-theoretic lemma. Let us look again at the least superharmonic majorant (9RF)(z) of the function F(z)
=
1
n
1 sz I log I gm(t) I
Iz-t12
dt - A 13z I
2 Weight is an entire function of exponential type
401
formed from the weight W(x) = I g,(x) I used in the preceding proof. Here, I gM(t) I = IM( - t) I , so F(z) = F(- z) and therefore (JlF)(z) = (JRF)(- z) (cf. beginning of proof of first lemma, §B.3). We know that JJiF
is finite, but here, since (WF)(0) = F(0) = log IgM(0)I = 0, we cannot affirm that (931F)(z) is harmonic in a neighborhood of 0 and thus are not able to directly apply the first theorem from §B.3. An analogous result is nevertheless available by problem 57. In the present circumstances, with (JJtF)(z) even, that result takes the form (JJJF)(z)
=C-
log I z2
- t2 I dp(t) - J
of,
log
i
1- z22 t
dp(t),
where p is a certain positive measure on [0, oo) with
p([0,t])
A
t
It
for t ->oo.
Because (D F)(0) < oo, we actually have f('1
- oo,
>
log(t2) dp(t) Jo
so, after changing the value of the constant C, we can just as well write (JJtF)(z)
=C-J
log
1- z2 2 t
0
dp(t).
By the first lemma of §C.5, Chapter VIII (where the function corresponding to our present gm(z) was denoted by G(z) ), we have
=
log I gm(z) I
JI
log
1- z22
dv(t)
for 3z >,
0,
0
v(t) being a certain absolutely continuous (and smooth) increasing function. Taking the function gm(z) to be of exponential type exactly equal to B (so as not to bring in more letters ! ), we also have
=
log I gM(z) I
+
B.3z
f
3z log I gM(t)1
J
Iz-t12
dt
for .3z > 0 by §G.1 of Chapter III.* Referring to the above formula for F(z) = F(2), we see from the last two relations that F(z)
=
log
J 0
1- z2 Z dv(t) t
-
(A + B) 13z 1.
* see also end of proof of lemma at beginning of this article
402
XI C Theorems of Beurling and Malliavin
(9J1F)(z) is, however, a majorant of F(z). Hence, z2
F(z) - (MF)(z) =
", log
J0
-
Z (dv(t) + dp(t)) t
(A+B)l,3zI
-C
is
0,
and thus any two zeros of cpl(z) will be distant by at least it
A+B units, in conformity with Beurling and Malliavin's observation.
2 Weight is an entire function of exponential type
403
Verification of the claim remains, and it is there that we resort to the Lemma. Let y be a finite positive measure on 18 without point masses. Then the derivative µ'(t) exists (finite or infinite) for all t save those belonging to a Borel set Eo with µ(EO) + I Eo I = 0. If E is any compact subset of R, IEI
e
µ'(t) +
1(dµ(t) + dt).
Proof. The initial statement is like that of Lebesgue's differentiation theorem which, however, only asserts the existence of a (finite) derivative µ'(t) almost everywhere (with respect to Lebesgue measure). The present result can nonetheless be deduced from the latter one by making a change
of variable. Lest the reader feel that he or she is being hoodwinked by the juggling of notation, let us proceed somewhat carefully. Put, as usual, µ(t) = f O dy(T), making the standard interpretation of the integral for t < 0. By hypothesis, µ(t) is bounded, increasing and without jumps, so
S(t) = µ(t) + t is a continuous, strictly increasing map of U8 onto itself. S therefore has a continuous (and also strictly increasing) inverse which we denote by T :
T(1(t) + t) = t. If cp(s) is continuous and of compact support we have the elementary substitution formula
=
cp(s) ds
(*)
J
cp(S(t)) (dµ(t) + dt)
-.D
which is easily checked by looking at Riemann sums. The dominated convergence theorem shows that (*) is valid as well for any function cp everywhere equal to the pointwise limit of a bounded sequence of continuous
ones with fixed compact support. That is the case, in particular, for W = XF' the characteristic function of a compact set F, and we thus have IFI
=
XF(S(t))(dµ(t) + dt) = µ(T (F)) +
XF(s) ds = J '0
The quantity
I T (F)
J '0 I T(F) I
is, of course, nothing other than the Lebesgue-Stieltjes
measure fF dT(s) generated by the increasing function T(s) in the usual way. The last relation shows that I T(F)I
IFI
for compact sets F; the measure on the left is thus absolutely continuous
404
XI C Theorems of Beurling and Malliavin
with respect to Lebesgue measure, and indeed T(s + h) - T(s)
0
1
h
for It # 0.
By the theorem of Lebesgue already referred to, we know that the derivative
=
T'(s)
T(s + h) - T(s)
lim
h-o
h
exists for all s outside some Borel set FO with 1F01 = 0. The image Eo = T(F0) is also Borel (T being one-one and continuous both ways) and, for any compact subset C of E0, the previous identity yields
µ(C) + ICI = IS(C)I < IFoI = 0, since C = T(S(C)) and S(C) c S(E0) = FO. Therefore =
µ(E0) + Eo I
0.
Suppose that t E0. Then s = µ(t) + t (for which T(s) = t ) cannot lie in F0, T being one-one, and thus T'(s) exists. For b 0 and any such t (and corresponding s ), write h(s, 6)
=
µ(t + b) - µ(t) + 6.
We have s + h(s, 8) = µ(t + 6) + t + S, so by definition of T, T(s + h(s, b)) = t + 6, and (t)
_
T(s + h(s, 8)) - T(s)
The function µ(t) corresponding t
is
S
µ(t + 6) - µ(t) + 6
h(s, 6)
in any event continuous, so at each s (and
),
h(s, 6) -- 0 as S , 0. b) - µ(t))/8 + 1)-' must, by (t),
Therefore, when t Eo,
exist and equal T'(s). This shows that µ'(t) exists for such t (being infinite in
case T'(s) = 0 ). Take now any continuous function c/i(s) of compact support. Because of the absolute continuity of the measure f FdT(s) already noted, we have J
0 (s) d T(s)
=
J
0 (s) T'(s) ds.
2 Weight is an entire function of exponential type
405
Here, T'(s)
=
limT(s -- + h(s, 8)) - T(s) 3-o
a.e.
h(s, a)
where h(s, S) is the quantity introduced above. The difference quotients i(s) T'(s) ds equals on the right lie, however, between 0 and 1. Hence the limit, for 8 ----+ 0, of
T(s + h(s, 8)) - T(s)
°°
f-.0
O(s)
h(s 6)
ds
by dominated convergence. In this last expression the integrand is continuous and of compact support when 8 0. We may therefore use (*) to make the substitution s = µ(t) + t therein; with the help of (t), that gives us
(µ(t) + t) f-"O.O
811(t)
µ(t + 8)
+b
(du(t) + dt).
The quantity 8/(µ(t + 8) - µ(t) + 8) lies between 0 and 1 and, as we have just seen, tends to 1/(µ'(t) + 1) for every t outside Eo when 6 -+ 0, where µ(E0) + 1 E01 = 0. Another application of the dominated convergence theorem thus shows the integral just written to tend to as 8 - 0. In this way, we see cc i(p(t) + t) (µ'(t) + l )- `(d t(t) + dt) that cc
4'(s) dT(s)
_
O(u(t)+ t) (dy(t) µ'(t) + 1
+ dt)
when 0 is continuous and of compact support. Extension of this formula to functions >/i of the form xF with F compact now proceeds as at the beginning of the proof. Given, then, any compact E, we put F = S(E), making T(F) = E and x,(µ(t) + t) = xF(S(t)) = XE(t); using hi(s) = XF(S) we thus find that jcc dy(t)+dt icc xE(t) IEI = IT(F)I = XF(s)dT(s) = µ'(t) +I The lemma is established.
We proceed to the claim. Problem 58 (a) Show that in our present situation, neither v(t) nor p(t) can have any point masses. (Hint: Concerning p(t), recall that (UlF)(x) is continuous ! )
XI C Theorems of Beurling and Malliavin
406
(b) We take v(0) = p(O) = 0 and then extend the increasing functions v(t) and p(t) from [0, oo) to R by making them odd. Show that for x e E (the set
on which F(x) - (971F)(x) = 0 ), we have 1
-
1+
fog
t)z
(x
-
(dv(t) + dp(t))
(A + B)y
0.
(c) Writing µ(t) = v(t) + p(t), show that for fixed y > 0, y
J log t +
t z ) dµ(t)
x (
_
z
y2 z
µ(x + i) - µ(x - i)
oY+T
(Hint: Since µ({x}) = 0, the left hand integral is the limit, for 8
(l
log
+
y
di.
T
0, of
z
dµ(t).
(t - x)z
2
Here we may integrate by parts to get y2
µ(x + T) - fL(x - T) - (µ(x + b) - µ(x - b)) dT.
h
Y + T
2
T
Now make S -+ 0 and use monotone convergence.) (d) Hence show that for each x e E where µ'(x) exists, we have mp'(x)
A+B.
(Hint: Refer to (b).)
(e) Show then that if F is any compact subset of E,
µ(F) 5 (A + B) F whence
v(F) + p(F)
A+B
IF I.
It
(Hint: Apply the lemma.)
The reader who prefers a more modern treatment yielding the result of part (e) may, in place of (d), establish that
(Dµ)(x)
0 on the real axis and 1 °°
_
log' fi(x) dx I +x2
0, denote by JM the quantity
tI o
logl 1 + x2O(X) Idx.
I
x2
\
M/
Suppose that for some given A > 0, M is large enough to make JM
+ - -.(JM(JM + 7rB))
1 on the real axis and we can use it as a weight thereon. As long as M fulfills the condition in the hypothesis,
j
=
J
lz log P(x) dx
ox
will satisfy the relation J
+
I(J(J + itB))
,In I
5A
for large enough R; we choose and fix such a value of that quantity. Using the weight W(x) = P(x) and the number A, we then form the
function F(z) and the sequence of FN(z) corresponding to it as in the previous two articles, and set out to show that (9RF)(0) = 0. 0 for each N. This is done as before, by verifying that (9RFN)(0) Assuming, on the contrary, that some (9XFN)(0) is strictly larger than some E > 0, we have (9RFN)(0)
1
1 on R, so, by the lemma from the last article, we can get an entire function g(z) of exponential type B/2 (half that of P), having all its zeros in the lower half plane, and such that
g(z)g(z) = P(z). For the entire function G(z) = (g(z))2
of exponential type B with all its zeros in 3z < 0 we then have IG(x)I = P(x) on R, so that f log P(t) dw v(t, 0)
9
=
log I G(t) I
0).
f,9
Here, G(z) satisfies the hypothesis of the second theorem in §C.5 of Chapter VIII, and that result can be used to get an upper bound for the last integral. We can, however, do somewhat better by first improving the theorem, using, at the very end of its proof, the estimate furnished by problem 28(c) in place of the one applied there. The effect of this is to replace the term J(2e J(J + nB/4)) figuring in the theorem's conclusion by
n .,I(J(J + nB)) I
with
J
Jo X
log I G(x) I dx
=
fo xz log P(x) dx,
and in that way one finds that 0)
Yr
(0){J +
I(J(J+irB))
Substituted into the above relation, this yields (97IFN)(0)
1
0, an increasing
410
XI C Theorems of Beurling and Malliavin
,
function p(t), zero on a neighborhood of the origin, such that p(t) t
log P(x)
A
for t -+ + oo
It
+
yx
+
log
X
xdp(t)
t
t
1-- +
f-0,W
h
on 1, y being a certain real constant. In the present circumstances P(x) = P(- x), so, taking the increasing function =
v(t)
21 (P (t) - P(- t))
(also zero on a neighborhood of the origin), we have simply log P(x)
x2 f log 1- tz
+
dv(t)
5
rl
for x e R.
0
Our given function (D is > 0 on the real axis. Therefore P(x) > x2't(x)/M there, and our last relation certainly implies that log(
x2_) M
)
/
+
l og fo"o
1- x2 s dv(t) t
on R. Denote for the moment log z2
D(z)
z
1 - t2 dv(t)
M
by U(z); this function is subharmonic, and, since D is of exponential type B while v(t)
A
t
It
for t -->oo,
we have U(z)
(B+A)IzI +
jzj. Because U(x) q on R, we see by the third Phragmen-Lindelof theorem of §C, Chapter III, that
for large
U(z)
'1 + (A + B)z
for .3z > 0.
To the integral f log 1 - (z2/t2)I dv(t) we now apply the lemma of o
3 Quantitative version of preceding result
411
§A.1, Chapter X, according to which z2
Jo
(d[v(t)] - dv(t))
log5max(IxI, IYI)
0. There is clearly an entire function (p(z) with loglP(z)I
z2
log 1 - Z d[v(t)] ;
J
=
t
0
(p is even and (p(0) = 1. Moreover, in view of the asymptotic behaviour of v(t) for large t, cp(z) is of exponential type A. In terms of gyp, the preceding relation becomes Me(A+B)y+"
{max(IxI,Y)/y + Y/max(IxI,Y)}, 2(X2 + y2)
y > 0.
The fraction on the right is just
with
1
max(, 1) + 1/max(, 1)
2y2
2+1
= IxI/y, and hence is < 1/y2. Thus, putting z
x + ih with
h > 0, we see that
II(x+ih)Qp(x+ih)I
, 0
when working with real signed measures p on [0, oo). It will also be convenient to extend the definition of such functions p to all of P by making them even (sic!) there.
Lemma. Let
IC
. 0
log
x+t
x-t
Idp(t)Ildp(x)I
, 1,
p'(t)dt if IP'(t)I K_ N,
dPN(t)
(N sgn p'(t)) dt otherwise ;
it is claimed that
IIUP -
UPNIIE - 0
forN -->co. By breaking p'(t) up into positive and negative parts, we can reduce the general situation to one in which p'(t) % 0, so we may as well assume this property. Then, for each x > 0, the potentials
=
UPN(x)
Jiog x+t x-t
min(p'(t), N) dt
0
increase and tend to UP(x) as N -* oo. Hence, by monotone convergence, f'O UPN(x)dp(x) 0
-, fo"O
From this, we see that =
II UP - UPN II E
=
E(dp(t) - dPN(t), dp(t) - dPN(t))
J 'O(UP(x) - UPN(x))(p'(x) - min(p'(x), N))dx 0
J(Up(X) - UPN(x)) dp(x)
(!)
0
must tend to zero as N -) oo, verifying our assertion. From what we have just shown, it follows that U, UPN >E
E
N
But we clearly have
JU(x)dPN(x) N 0
JU(x)dP(x) 0
by the given absolute convergence of the integral on the right. The desired first relation will therefore follow if we can prove that
JU(x)dPN(x) 0
=
E
4 Energy and the space S from §C of Chapter VIII
433
for each N; that, however, simply amounts to verifying the relation in question for measures p satisfying the hypothesis and having, in addition, bounded densities p'(x). We have thus brought down by one notch the generality of what is to be proven.
Suppose, then, that p satisfies the hypothesis and that p'(x) is also bounded. For each a, 0 < a < 1, put
p'(t), a 0, put if I U(x)) < M, Msgn U(x) if I U(x)I >, M. U(x)
UM(x) )
Then
I UMW - UM(Y)I
o and we see that it must tend to zero when h -> 0 (by continuity of translation in L2(!P2) ! ), since
and
(ux(z))2 dx dy J J 3z>o
(uy(z))2 dx dy
JJ3z>0
are both finite, according to the observation just made. Thus, 11 u - uhIIE
0
ash
0
4 Energy and the space Sa from §C of Chapter VIII
439
It is now claimed that each function uh(x) is equal (on ll) to a potential Uo(x) of the required kind. Since u(t) is odd, we have
u(x+iy) = u(-x+iy)
for y>0,
so uh(x) is odd. Our condition on u(x) implies a similar one, const
uh/
lx)
xE!!B,
x2+I'
on uh, so that function is (and by far!) in L2(- oo, oo ), and we can apply to it the L2 theory of Hilbert transforms from the scholium at the end of
§C.1, Chapter VIII. In the present circumstances, uh(x) = u(x + ih) is c in x, so the Hilbert transform uh(x)
=
uhlt)
a0
1
n _00 x - t
dt
=
I 7r
m U& -'r) - U& + T) di o
T
is defined and continuous at each real x, the last integral on the right being absolutely convergent. From the Hilbert transform theory referred to (even a watered-down version of it will do here!) we thence get, by the inversion formula, uh(x)
=-
uh(t)
1
x-t
7E
dt
a.e., x e R.
This relation, like the one preceding it, holds in fact at each x e R, for uh(x) is nothing but the value of a harmonic conjugate to u(z) at z = x + ih and is hence ( like uh(x) ) 16,,, in x. Wishing to integrate the right-hand member by parts, we look at the behaviour of uh(t). By the Cauchy-Riemann equations,
uh(x) = ux(x + ih) = - uy,(x + ih). After differentiating the (Poisson) formula for u(z) and then plugging in the given estimate on I u(t) I, we get ( for small h > 0 ) uy(x + ih)
const. h(x2 + 1)'
so
l u (t) h
5
const. t2 + 1
for t c R.
As we have noted, uh(x) is odd. Its Hilbert transform uh(t) is therefore
440
XI C Theorems of Beurling and Malliavin
even, and the preceding formula for u,, can be written
(x it,o
l
t +
x+t)
u,,(t) dt.
Here, we integrate by parts as in proving the lemma of §C.3, Chapter VIII
and the third lemma of the present article. By the last inequality we actually have
o
Instead, physicists are obliged to resort to what they call a double-layer distribution on O (formed from `dipoles'); mathematically, this simply amounts to using the Poisson representation U(z)
=
If 7r
Iz3
- tlz U(t)dt
4 Energy and the space S5 from §C of Chapter VIII
443
in place of the formula U(z)
=
- 1n o
log
z+t U,,(t + i0) dt, z-t
which is not available unless 8U(z)/8y is sufficiently well behaved for 3z ---> 0. ( U(x) may be continuous and the above Dirichlet integral finite, and yet the boundary value U,,(x + i0) exist almost nowhere on R. This is most easily seen by first mapping the upper half plane conformally onto the unit disk and then working with lacunary Fourier series.) Problem 61
Let V(x) be even and > 0, with II V IIE < oo. Given U e Sa, define a function Uv(x) by putting Uv(x)
=
U(x)
if IU(x)I < V(x), if IU(x)I ? V(x);
V(x)sgnU(x)
the formation of Uv is illustrated in the following figure:
x
Figure 247
Show that UV E (Hint: Show that
I Uv(x) - Uv(y)I
-
U(x) and
V(y) s U(x) and U(y)
V(y)
U(y) , V(y).
Problem 62 (Beurling and Malliavin) Let w(x) be even, , 0, and uniformly Lip 1, with. m(x)
, 0, f1.9(t)
=
w,(8-9 n((- 00, -t]v[t, 00)), 0)
as in §C of Chapter VIII, we have q log(1 + t2) dwg(t, 0)
J
=
- q fo"O log(1 + t2)
a2 f2 g(t) dt.
q
By the fundamental result of §C.2, Chapter VIII,
Sli(t)
Y0(0)
S
t
so the last expression is 2q YY(0)
°°
0
dt 1 + t2
=
nq Y.9(0).
Thence,
a
log W"(t) dcog(t, 0)
=
fa
qlog(1 + t2) dw_q(t, 0)
ngYA) +
J
a
+
J
a log W(t) dwq(t, 0)
log W(t) dw0(t, 0),
and our main work is the estimation of the integral in the last member. For that purpose, we may as well make full use of the third theorem in the preceding article, having done the work to get it. The reader wishing to avoid use of that theorem will find a similar alternative procedure sketched in problems 63 and 64 below. According to our hypothesis, log W(x)/x e fj
so, by the theorem referred to, there is, for any q > 0, a potential U'(x)
XI C Theorems of Beurling and Malliavin
448
of the sort considered in the last article, with W(X) II log
- UP(x)
< IIE
and also (by the remark to that theorem) I Uv(x) I
z Knz l+x
for x e R.
Let us now proceed as in proving the theorem of §C.4, Chapter VIII, trying, however, to make use of the difference (log W(x)/x) - UP(x). We have
JloW(t)dw(t
,
0)
=J
t UP(t) dco21(t, 0)
a.9
+
a (log W(t) - tUP(t)) dwg(t, 0).
Because I t UP(t) I < K,7/2 by the last inequality, and w_g(
,
0) is a positive
measure of total mass 1, the first integral on the right is
K,/2. In terms of 0.9(t), the second right-hand integral is
-f
" (log W(t) - t UP(t)) H1.9(t); 0
to this we now apply the trick used in the proof just mentioned, rewriting the last expression as f°° C1ogW(t)
t
0
(log W(t)
- UP(t)
I\
t
- Uo(t)
0
)
The first of these terms can be disposed of immediately. Taking a large number L, we break it up as log w(t) fL o
0_9(t) dt - f"O UP(t))q(t) dt +
t
log fLw
W (t)
t
!Q.9(t) dt.
We use the inequality 0.9(t) S 1 in the first two integrals, plugging the above estimate on UP(t) into the second one. In the third integral, the relation S2q(t) < Yq(0)/t is once again employed. In that way, the sum of these integrals is seen to be
5 L Jo
logtW (t) dt
+ 2 K11 + YA0) fL logtW(t) dt.
5 Even weights W with 11 log W(x)/x 11 E
8
because xp = exp(pl/3), and the latter function has a positive second derivative for p > 8. (Now we see why we use the sequence of points xP beginning with x8 !) The intervals [xp - RP) XP + RP] therefore do not overlap when p > 8, so our desired conclusion, namely, that °°
f
log
I g(x) I
1 +x2
_.O
dx
=
oo
will surely follow if we can establish that
1
P>8 Xp
RP
logIg(xp+s)Ids =
-Rp
with the help of the preceding relation.
-oo
I Uniform Lip 1 condition on log log W(x) insufficient
463
Here, we are guided by a simple idea. Everything turns on the middle term
figuring on the right side of our relation, for the sums of the first and third terms are readily seen to be convergent. To see how the middle term
behaves, we observe that by Levinson's theorem (!), the function g(z), bounded on the real axis and of exponential type A, should, on the average, have about 2
ARP =
A AP
n
7r
zeros on the interval (x p - RP, XP + RP], for all of g's zeros are real. The quantity v(RP, XP + i) is clearly not more than that number of zeros, so the factor in { } from our middle term should, on the average, be
c----AAP R
(approximately). Straightforward computation easily shows, however, that APRPlogRP
-
xP
1
18p
for large values of p. It is thus quite plausible that the series RP log RP
{(
XP
P
should diverge to - oo. This inference is in fact correct, but for its justification we must resort to a technical device. Picking a number y > 1 close to 1 (the exact manner of choosing it will be described presently), we form the sequence
X. =
in = 1, 2, 3. ... .
ym,
We think of {Xm} as a coarse sequence of points, amongst which those of {xP} - regarded as a fine sequence - are interspersed:
m+l
Xm -2
Xm+2
Figure 250
It is convenient to denote by v(t) the number of zeros of g(t) in [0, t] for t >, 0; then, as remarked above, v(RP, XP + i)
Xm, and by h;,,+ 1 the value of that quantity corresponding to the largest xP < X m + i Since, for p > 8, the intervals [xP - 2 OP, XP + AP] don't Z overlap, we have
Xm<xpX,,,+
XP+ 2P)
-
VCxP
V(Xm+1 +h1) m
2P)}
- V(Xm - hm).
According to Levinson's theorem (the simpler version from §H.2 of Chapter III is adequate here), we have At
- o(t)
5
v(t)
0 is arbitrary. We turn to the sum
X,,,<xp-<Xm+i
[AP] RP2log RP xP
for which a lower bound is needed. We have [API RP log RP
N
2 2(3p_2/3x P)2log(bp_213xP)
1
log xP
l
P
18p413
P 1
=
for p --+ oo.
18p
Thence, calling pm the smallest value of p for which xP > X. and p;,,+1 the largest such value with xP < Xm+i, the preceding sum works out to
(18 + o(1)logpm+1 J
Pm
when in is large. In that case, xPm = exp (pm/3) is nearly Xm = ym, and exp((pm+1)1/3) nearly ym+i So then, pm+i/pm is practically equal to ((m + 1)/M)3
'
log p'"+i
1 + 3/m, and
-
3
in
Pm
Thus,
[AP]RPlogRP xP2 for large values of m.
>
(1
6
0(1)1
J
1
in
466
XI D Search for the presumed essential condition
Now combine this result with the one obtained previously. One gets RP log RP
lv(RP, xP + i) - [OP] }
{(
X,<xpSX,+t
2
lxp
1
1
6{(n + 2eIlog 1 - 1 + s}
/
Y
)))
(with s > 0 arbitrary) for large m. We had, however, A/n < 1. It is thus possible to choosey > 1 so as to make
Ay-1
1 - S,
0. Fixing such a y, we can then take an e > 0 small enough to ensure that 6{((A
+ 2s l
/ l0g Y
1111\\\
1
-1+
l e{
< 7
)
The left-hand sum in the previous relation is then 6
7m
for sufficiently large m, so .,/3 n 4
Y {v(R P ,
xP + i) -
[AP] } RP x2 log RP P
does diverge to - oo.
Aside from the general term of this series, our upper bound on log I g(xP + s) I ds P
-Rp
involved two other terms, each of which is
const.Ap/x2 when p is
large. But
eP
,"
9p-4/3
2 P
for p --+ oo,
so the sum of those remaining terms is certainly convergent. The divergence
just established therefore does imply that 1
2 p>8 XP
RP
log I g(xP + s) I ds = - oo, -Rp
467
2 Discussion and hence that °°
log
I g(x) I
1+x2
J_00
dx
00
as claimed, yielding finally our desired contradiction. The weight W(x)
=
eB
I `'(x + i)I
constructed above thus admits no multipliers f of exponential type < 7C, even though it enjoys the regularity property log log W(x) - log log W(x') I
0, we have:
f
w
A
0
isA log
d
dx d
dx
l og A
A 0
log
x+t
dt dx
x-t x+t
t
dt
x-t
t
x+t
dt
x-t
t
2+
x
1 lo g x
3 3
+
x+ A
x- A
- x og x+A x-A 1
l
5 3
+
for x>0, x#A ; for x
>0 , x 0A
Proof. To establish the first relation, make the changes of variable
t
t A
.
3 Comparison of energies
475
and expand the logarithm in powers of then integrate term by term. For the last two relations, we use a different change of variable, putting
s = t/x. Then the left side of the second relation becomes d
dx
l+s 1-s
log Aix
ds S
and this may be worked out for x # A by the fundamental theorem of calculus. The third relation follows in like manner.
Lemma. Let p(t) = f O dp(i) be bounded for 0 < t < oc. Then, for A > 0, the two expressions
x+t
log
it
dt
x-t
log
dp(x),
t
x+t
dp(t)
x-t
dx x
are bounded in absolute value by quantities independent of A.
Proof. Considering the second expression, we have, for large M > A and
any M' > M, JMlog M
=
P(M) log
x+t dp(t) x-t
1 + (x/M')
- p(M) log
1 - (x/M')
1 + (x/M)
1 - (x/M)
M'
+
2x
f
M
t2p(t) _x 2 dt
whenever 0 < x < A. Because I p(t)I is bounded, the right side is equal to x times a quantity uniformly small for 0 < x < A when M and M' are both large. The second expression is therefore equal to the limit, for M - oo, of the double integrals
x+t
x-t
dp(t)
dx x
Any one of these is equal to fA
(f
0
log
x+t dt) dp(x);
x-t
t
here we use partial integration on the outer integral and refer to the third
XI D Search for the presumed essential condition
476
formula provided by the preceding lemma. In that way we get
--
M+t dt
fOA
log
p(M)
M- t
t
A
p(A) folog
fMlogI x + AI p(x)
+
x-A x
A
A +t
dt
A-t
t
dx.
Remembering that
x+t
dt
x-t
t
_
n2
2
for x>0,
we see that the last expression is < 3n2 K/2 in absolute value if p(t) I
< K on [0, cc). Thence, log
x+t
x-t
dp(t)
dx x
31r2
K
2
independently of A > 0. Treatment of the first expression figuring in the lemma's statement is similar (and easier). We are done.
Now we are ready to give our version of Cartan's lemma. So as not to
obscure its main idea with fussy details, we avoid insisting on more generality than is needed for the next article. An alternative formulation is furnished by problem 68 below.
Theorem. Let w(x), even and tending to 0o for x -p + oo, be given by a formula w(x)
=
- J0
log
dv(t),
where v(t), odd and increasing, is ' ' Suppose there is an even function S2(x) f °° log fl(x) 0
x2
dx
L.
log Q(x)
co(x)
Because co(x) -- oo for x --> oo, we can take (and fix) L large enough to also make w(x)
for x >, L.
0
The given properties of v(t) make w(0) = 0 and w(x) infinitely differentiable* on R. Therefore, since w(x) is even, we have w(x) = O(x2) near 0, and, having chosen L, we can find an M such that
-x2M '< w(x) 5 x2M
for 0,<x oo. Thence, by the second theorem of §C.4 and Fatou's lemma,
1.
That generalization is related to some material of independent interest taken up in problems 66 and 67. Let us, as usual, write log
x-t dp(t).
Under our assumption on p, the integral on the right is certainly unambiguously defined because
x-t
min (1, 1/t) dt
3 Comparison of energies
483
is finite for x > 0 and, if K is large enough,
=
da(t)
dp(t) + K min (1, 1/t) dt
is > 0 for t >,O.* The preceding integral is indeed O(xlog(1/x)) for small values of x > 0, so, by applying Fubini's theorem separately to I
I
x+t
I
log
x-t
J0 Jo
di(t)
dx x
and to the similar expression with dv(t) - dp(t) standing in place of da(t), we see that for each finite A > 0, f (UA(x)lx) dx is well defined and equal o to
x+t dx - dp(t).
A
log ii
x-t
o 0
x
By writing dp(t) one more time as the difference of the two positive measures do(t) and K min (1, 1/t) dt, one verifies that the last expression is in turn equal to A
('M
lim J
M- 0
log 0
x+t
x-t
dx x
dp(t).
Problem 66 In this problem, we suppose that the above assumptions on the measure p hold, and that in addition the integrals J A U°(x) dx 0
x
are bounded as A oo. The object is to then obtain a preliminary grip on the magnitude of l p(t)j.
(a) Show that for each M and A. M fA
log 0
0
x+t
x-t
x dx dp(t)
(oA
=
p(M) J log IX ('
+
fo
+-M dx
loglt+Al ptt)dt.
(Hint: cf. proof of second lemma, beginning of this article.) * Only this property of p is used in problems 66 and 67; absolute continuity of that measure plays no role in them (save that p(t) should be replaced by p(t) - p(0 + ) throughout if p has point mass at the origin).
484
XI D Search for the presumed essential condition (b) Hence show that
-> 0
P(M) M
as M -> co. - K(1 + log+t) for t > 0, making
(Hint: p(t) ('M
- J log t + A P(t) dt
e, say. Deduce that for fixed large A and M -+ co,
2P( )A < 0(1) + const.logA. (c) Then show that A U°(x)
=
dx Jo
log
x
fo
t+A
P(t)
t-A
t
dt.
(d) Show that for large t > 1 we not only have p(t) also p(t) < const. log t. (Hint: Wlog, dp(t)
- const. log t but
- dt/t for
t > 1.
Assuming that for some large A we have p(A) >- k log A with a number k > 0, it follows that p(t)
>-
k log A - log A
At the same
time, p(t)
for t > A.
- 0(1) - log+t for 0 < t < A. Use
result of (c) with these relations to get a lower bound on $o(U°(x)/x)dx involving k and log A, thus arriving at an upper bound for k.)
Problem 67 Continuing with the material of the preceding problem, we now assume that lim A-+ao
i
A U°(x) o
dx
x
exists (and is finite). It is proposed to show by means of an elementary Tauberian argument that p(t) then also has a limit (equal to 2/7r2 times
the preceding one) for t --+ oo. Essentially this result was used by Beurling and Malliavin* in their original proof of the Theorem on the * under the milder condition on p pointed out in the preceding footnote - they in fact assumed only that the measure p on [0, oo) satisfies dp(t) >- - const. dt/t there, but then the conclusion of problem 67 holds just as well because the existence
3 Comparison of energies
485
Multiplier.
(a) Show that for a and b > 0,
x+a x + b log dx x-a x-b
log f.'
=
n2 min (a, b).
(Hint: We have log it
x+a x-a
1
°
dt
n _ax- t
Apply the L2 theory of Hilbert transforms sketched at the end of §C.1, Chapter VIII.) (b) Hence derive the formula t+x
Jiog
t - x
=
{ 2log
x+A x+(1 -b)A x+(1 +8)A - log - log x-A x-(1+8)A x-(1-8)A
}dx
7E 2(6A - It-AU+,
valid fort>0, A>0and0
t
for large t.
(log t)2
Thence, putting F1(z)
=
J
log
1 - z2 2 t
0
dv(t),
we see by computations just like those at the beginning of article 1 that F1(z)
0.
x-t
Therefore, because T(t) is increasing, we have BT(x)
F2(x)
fI'O log
1+t dt 1-t t2'
x?0.
The integral on the right is just a certain strictly positive numerical quantity. We can thus pick B large enough (independently of the value of the large number I used in the specification of T) so as to ensure that F2(x)
>
2T(x)
for x , 0.
Fix such a value of B - it will be clear later on why we want the coefficient 2 on the right. Then, taking F(z)
=
F1(z) - Fz(z),
z
- T(f x I) + const.
we will have F(x)
const.
for real values of x.
The function F is given by the formula F(z)
1 - z2 2
=
t
d(v(t) - µ(t)),
in which v(t) - µ(t) is increasing, provided that the parameter 1 entering into the definition of T is chosen properly. Because v(t) and µ(t) are each increasing, with the second function absolutely continuous, this may be
verified by looking at v'(t) - µ'(t). For x,_1 < t < xp with p> 8, we have
v'(t) - µ (t) = RP + BTt(t)
-B
tT(2)
t
dt,
and an analogous relation holds in the interval (0, x8). Choose, therefore, I large eno(u'gh to make
dt
BJ o
t2
= BC f"O log log t dt (log t)2t
0
for t > 0 different from any of the points xi,, and v(t) - µ(t) will be increasing.
It is also clear that v(t) - µ(t)
as t - co.
1
t
Hence
0, to build an increasing al(t) with al(t)/t < n/2 having jumps that will cancel out most of v's, making, indeed, al(t) - v(t) a constant multiple of t for large values of that variable. The property that AP --+ 1 asp--> oo enables us to do this. Given the quantity n > 0, there is a number p(,) such that AP
>
1-2
for p >
We put 1
al(t)
-
0,
t < XP01)
2 -p _1 +
{AP
2 - (l_)}(t_xp_i)
for
xp_1 < t < xP with p > p(rl). This increasing function al(t) is related to v(t) in the way shown by the following diagram:
4 The finite energy condition not necessary
493
g\OQe
t
Figure 253
It is clear that
for t i X AnY
01(t) - v(t) _ Take now G1(z)
0 = f "O
dol(t).
log
We have
1(t)
17
t
2
as t--4oo,
so for large values of Iz1,
The first lemma of §B.4, Chapter VIII, tells us that 2
G1(x) - F1(x)
=
log
J
1 - tXZ I d(a 1(t) - v(t))
0
x+t
x-t
d(v(t) t
al(t)),
xeR.
494
XI D Search for the presumed essential condition
As we have just seen, (v(t)-01(t))/t is constant for t > xp(,); the last expression on the right thus reduces to rv(v)
X
log
ta1(t)
x±tId(v(t)
I.
0
This, however, is clearly bounded (above and below!) for l x say. Therefore
G1(x) - F1(x)
, 2xp(,),
Ixl i 2xP(n).
const.,
This relation does not hold everywhere on l ; G1(x) - F1(x) is indeed infinite at each of the points + xP with 8 < p < p(ry). But at those places
(corresponding to the points where a, (t) - v(t) jumps downwards) the infinities of G1(z) - F1(z) are logarithmic, and hence harmless as far as we are concerned. Besides becoming - oo (logarithmically again) at ± xp(a) , the function G1 (x) - F1 (x) is otherwise well behaved on R, and belongs to L1(- 2xp(,) , 2xp(,)). We can now reason once again as in the proof of the second theorem, §B.1, and deduce from the properties of G1(x) - F1(x) just noted, and from those of G1(z) and F1(z) in the complex plane, pointed out previously, that
G1(z) - Fl(z)
n
=
1
C
+
I..3zl
1
- 2/
°
-.
Iszl(G1(t)-F1(t))dt. Iz-t12
Keeping in mind the behaviour of G1(t) - F1(t) on the real axis, we see by this relation that
G1(x+i) - F1(x+i)
0,
from which it is manifest that F2(x), like T(x), is increasing on [0, co). Again, by Fubini's theorem,
f
o
F2(x) dx x
= B I
log
I
0
o
x+t
x-t
dx T(t) x
t2
dt
=
4 The finite energy condition not necessary 2
B
T(t)
COO
J
2
, 1, the relation just written would make W(x) W(0)
, (1 - /6p)AP
for such x as soon as r exceeds ,/2. Since A p/xp -+ 0 for p - oo we may, for large p, take R = J2 Op in the formula, which, in view of the relation just written, then yields n 1
F(x+i) 5
2r
F(x+i+,/2 APe'9)dS
-
(1 - AP)OplogOP
for xP - 1 '< x 5 xP + 1. From the relations F(x) < const., x E R, and F(z) , 1 meeting our local regularity requirement, with 1 °°
log W(x)
I+ x2
dx
0, the harmonic conjugates V(z)
=
J
-2) dv(t),
arg I
1
arg
1 - Z dµ(t)
o z2
Jco
U(z)
=
t
O
* To show that F(x) is W, for lxi < A, say, take an even 'L function ap(t) equal
5 A and to 0 for for lxi < A, to 1 for I t i
ti
2A. Then, since F(t) is also even, we have,
f2A F(x)
=
((1 - (p(t))F(t)/(x2 - t2)) dt
(2x/n)J A
+
(cp(t)F(t)/(x - t)) dt.
(1/7c) 2A
The first expression on the right is clearly B in x for lxI < A. To the second,
we apply the partial integration technique used often in this book, and get d((p(t)F(t) ). Reason now as in the footnote to the theorem of §D.3. Since d(rp(t)F(t))/dt is' , the integral is also W, (in x) for Ixi < A. (1/7C) J' -'2A log I x - t
518
XI E A necessary and sufficient condition
(where the argument is determined so as to make arg 1 = 0). Here, where v(t) and µ(t) are continuous, we can argue as in the proof of the last lemma to show that V(z) and U(z) are continuous up to the real axis, where they take the boundary values V(x)
U(x)
xv(x),
nµ(x).
Thus, V(x)
- U(x)
nF(x).
Our assumptions on µ(x) are not strong enough to yield as much information about the behaviour of U(z) (or of V(z) ) at the points of R. Consider, however, the difference G(z)
=
V(z) - U(z).
Since F(t) = v(t) - µ(t), we can write ('o°°
G(z)
=
J
log 1 -
z2
2 dF(t) t
for .3z # 0; the integral on the right is, however, absolutely convergent even when z is real. To check this, take any R > I z I and break up that integral into two pieces, the first over [0, 2R] and the second over [2R, oo). Regarding the first portion, note that dF(t) = F'(t) dt with P'(t) continuous and hence bounded on finite intervals (F(t) being '1); for the
second, just use IdF(t)I < dv(t) + dy(t). In this way we also verify without difficulty that G(z) is continuous up to (and on) R, and takes there the boundary value G(x)
=
dP(t).
J oo log 0
By this observation and the one preceding it we see that the function g(z)
=
G(z) + i(V(z) - U(z)),
analytic for .3z > 0, is continuous up to the real axis where it has the boundary value g(x)
=
G(x) - iirP(x).
Bringing in now the function f (z) described earlier, we can conclude that
n f (z) + g(z), analytic in 3z > 0, is continuous up to R and assumes there the boundary value
n f (x) + g(x)
=
xxF(x) + G(x).
1 Five lemmas
519
The right side is obviously real, so we may use Schwarz reflection to continue n f (z) + g(z) analytically across U and thus obtain an entire function. The latter's real part, H(z), is hence everywhere harmonic, with
=
H(x)
nF(x) + G(x)
on R.
For 3z 54 0, we have H(z) = H(z), so* =
H(z)
J
13z I
F(t)
dt + V(z) - U(z).
00
It is now claimed that H(z) is a linear function of ¶Rz and jz; this we verify by estimating the integrals J' , (H(rei9))+d9 for certain large values of r. By the last relation, we have
I-3zIIF(t)Idt
(H(z))+
for 3z
Iz-t1
f'O
+
(V(z))+
+
(U(z))
0. Consider first the second term on the right. Since v(t) is
increasing,
log 11+
0 t one may also just refer to the subharmonicity of U(z)
520
XI E A necessary and sufficient condition
I 1 - (re19/t)e le = (1 - r2/t2)2 + 4(r/t)2 sine 9
11-e W12, so, since µ(t)
increases,
Jiog
=
U(re19)
r2 e2i3 1
t2
0
µ(r)logll -e2isl
+
r2 e2is d u(t).
1 t2
Integration of the two right-hand terms from - it to it now presents no difficulty (Fubini's theorem being applicable to the second one), and both
of the results are zero. Thus, f"U(re13)da > 0 which, substituted with (U(rei9))+ \ const.r into the previous relation, yields
(U(re19))- d9 < const.r. fir
Examination of the first right-hand term in the above inequality for (H(z))' remains. To estimate the circular means of that term - call it P(z) - one argues as in the proof of the first theorem from §B.3, leaning heavily on the convergence of f °° (l F(t) l/(1 + t2)) dt (without which, it is true, P(z) would be infinite!). In that way, one finds that It
P(r e19) d4
5 const. r
- 7[
for a certain sequence of r tending to oo. Combining our three estimates, we get It
(H(r ei9))+ d9
0, have there partial derivatives continuous up to
R (in the present circumstances). Thence, by the Cauchy-Riemann equations, lim
aw(x + iy)
_
- lira
aoO(x + iy) ay
lim y-o+
co(x) - co(x + iv) y
and the last limit is clearly equal to the integral in question. Proof of theorem
1° The necessity. As we saw at the very beginning of §C, there is no loss of generality incurred in taking W(x) ---* oo
for x -+ + oc ;
this property we henceforth assume. If W(x) admits multipliers there is, corresponding to any a > 0, a non-zero
entire function f{z) of exponential type < a with W(x) I f (x) I
for
IxI
>, A,,.
It follows then from the last relation that log W(x)
+
log
J
for I x I> A,,.
0
0
We can evidently choose the A,, successively so as to have
A,,+1 > 2A.; this property will be assumed to hold from now on. For each n, let
0 A,,, use the evident formula
of t (all >
A /,/2
o'
J
2 - i) - log 2
Ilog(x
2x2 2X2
-
l
i)}dµn(t)
dµ#).
The first integral on the right is here clearly >, 0, and the second log W(x) > 0 by the above inequality. This establishes the claim, and
XI E A necessary and sufficient condition
528
shows besides that log W(x)
for I x I >, A,,.
In order to get the function w(x), we first smooth out each of the by multiplicative convolution, relying on the fulfillment by W(x) of the
local regularity requirement*. According to the latter, there are three constants C, L and k > 0, the first two > 0, such that, for any x c- IJ,
log W(x) 5 C log W(t) + k
for the t belonging to a certain interval of length L containing the point x.
Choose, for each n, a small number n > 0 less than both of the quantities An
and 4A,,+t
it is convenient to also have the rl tend monotonically towards 0 as n --* cc. Take then a sequence of infinitely differentiable functions 0 with p supported on the interval [1 - n,,, 1 + such that p
)d
(t t
ry
=
S
I
t+ M0 d
=
1.
1
When L
0sxS
2n the points x with 1 - n,, [x - L/2, x] and the ones with
1
I5
are included in the segment 1< 1 + n in the segment
[x, x + L/2]. One of those segments surely lies in the interval of length L containing x on which the preceding relation involving log W(t) does hold. By that relation and the specifications for p,, we thus have log W(x)
, 0, and forming the open set (9
=
f X > 0:
s(t) - s(x)
t-x
>
for some t > x I ,
we see that (9 can contain no points to the left of a certain ao > 0.
2 Proof of the conjecture from §D.5
535
x
a,
as
b, a2
(9
(9
a3
/(
b3
b2
Figure 260
The set (9 is thus the union of a certain disjoint collection of intervals (ak, bk),
k = 1, 2, 3,
. .
.
,
with
ak>- ao>0 for every k; these, of course, may be disposed in a much more complicated fashion than is shown in the diagram, there being no a priori lower limit to their lengths. Every point x > O for which s'(x) > S/it certainly belongs to (9, so s'(x)
5
S
for x e [0, oo) - (9.
R
For each k > 1, s(bk) - s(ak)
6
bk - ak
it
as is clear from the figure.
It is now claimed that aak2 00
Yk (bk
0 on R, we have U(x)
w(0) - w(x),
>'
x e R,
so, because w has property (i), f1D
-,
1
+() dx > - oo.
Writing $
_
Y N
and recalling that Gn(z) > - fn Iz I, we see, moreover, that U(z) < Q13zI. The convergence of the series Y-k((bk - ak)/ak)Z will be deduced from the
last two inequalities involving U and the fact that $ < 5/e due to our choice of N. (It would in fact be enough if we merely had fi < 6; our having
has required us to been somewhat crude in the estimation of the work with an extra margin of safety expressed by the factor 1/e.) Fixing our attention on any particular interval (ak, bk), let us denote its midpoint by c and its length by 2A, so as to have (ak, bk)
=
(c - A, C + A).
The following discussion, corresponding to the one in §D.1 of Chapter IX,
is actually quite simple; it may, however, at first appear complicated because of the changes of variable involved in it.
We take a certain quantity R > 1 (the same, in fact, for each of the intervals (ak, bk) - its exact size will be specified presently) and then,
2 Proof of the conjecture from §D.5
537
choosing a value for the parameter 1, 2A
> A,
and let us look at the Riesz representation for F(z) (obtained by putting a minus sign in front of the one for the superharmonic function - F(z) !) in the disk {I z I < A). In terms of the variable w = Iz + c, we have this picture:
'c+ 2 (r+ r)
Figure 261
Because
=
flo1
µ,U(w)
- 2 ds(t),
XI E A necessary and sufficient condition
538
we can, after making the change of variable t = it + c, write A
=
F(z)
logIz-rlds(lr+c) +
=
U(lz + c)
h(z),
-A
with h(z) a certain function harmonic for I z I < A. If, then, r > 1 and
Z(r + l/r) < A, the closed region of the z-plane bounded by the ellipse
-
=
z
I
re'
9
2
i9
+ er
(whose image in the w-plane is shown in the above figure) has mass M(r) equal to J(r+ 11r)
+ 1/r) ds(lr+c)
J
=
s1 c +
-
2(r+r I)
s(c
- (r+
(
assigned to it by the measure associated with the Riesz representation for
F(z) just given. By the fifth lemma of the last article we thus have
s(c + 21(r+r-1)) - s(c - Zl(r+r-1))
dr
f
r
J 1R n
U(c + 2(Re' 9+eR9)Id8
/
2n,f
-
J11
R M(r) dr
r
U(i +x ) dx. (I - X')
As in §D.1 of Chapter IX, it is convenient to now write
R = el (thus making y a certain fixed quantity > 0), and to take a number e > 0, considerably smaller than y (corresponding to the quantity denoted by n in the passage referred to). If the parameter 1 is actually A
cosh e ' we will have
s(c+'(r+')) -s(c-2r+r
s(bk) - s(ak)
2
(see once more the preceding figure). By construction of our intervals (ak, bk), the quantity on the right is equal to S (bk - ak) n
=
26
n
A.
2 Proof of the conjecture from §D.5
539
The previous relation thus yields 26'A
R
dr
eey
5
r
7r
M() r r
-dr = 1 1
rc
2n
i
U(c + lcosh(y+i9))d9
_
U ( I x + c)
_1,/(1-x2)
dx
for
coshe
, 0. As we have already noted, ltv1(t) = bt for t >, 0 lying outside all the intervals (ak, bk). When ak < t < bk, we have, since v1(t) increases,
dak co
* By its definition, v2(t) is absolutely continuous with v'2(t) bounded on finite intervals; A(t), on the other hand, has a graph similar to the one shown in fig. 226 (Chapter X, §B.2). These properties make (V(z))+ continuous at the points of R, and the arguments from §§E and G.1 of Chapter III may be used.
548
XI E A necessary and sufficient condition
(by Levinson's theorem). By first dividing out four of the zeros of >(i if
need be (it has infinitely many, being of exponential type 8 > 0 and bounded on 1!1 !) we can finally ensure that in fact IQP(x+i)1I(x)I
,
for xER (xonst)2
with (perhaps another) 0 of the kind described. A relation between V(x + i) and log I cp(x + i) I opposite in sense to the above one is now called for. To get it, observe that
=
V(z)
d( min (v2(t) + 2(t),
f'O log
1) )
0
+
d(v2(t) + 2(t) -
log
1)+.
0f00
Since 2(t) = St/it for 0 < t 5 1 and v'2(t) is certainly bounded there, the first integral on the right is 2 log+ I z l
+ const..
Therefore, when x e E,
V(x + i) 5 2 log+ I x I + const. + fO log
1-
(x + i)2
d(v2(t) +2(t) - 1)+.
t2
0
However, (v2(t) +2(t) - 1)+ 5 [v2(t)+ 2(t)] for t > 0, so, by reasoning identical to that used in proving the lemma of §A.1, Chapter X, we find that the last right-hand integral is
loglcp(x+i)I + log+IxI. Thus, V(x + i)
/i(x) I
0, the second of the two integrals on the right is
5
2
irY(x)
w(x) ;
it is, on the other hand, w(t)
2
°°
7E
- (x - t)2 + (Y(x))2
dt.
For the present purpose this last expression's behaviour is adequately described by the 1967 lemma of Beurling and Malliavin given in §E.2 of Chapter IX. That result shows that for any given n > 0, the integral in question will lie between - n and 0 for a function Y(x) > 0 with 0o
1Y(x)
JJ_O o
dydx
1+x2+y2
0 on which w'(x) exceeds some large K seems to essentially be determined by the behaviour of w(x)/Y(x) and of the integral 1
7C
('Y(x) 2w(x) - w(x + t) - w(x t2
Jo
t)
dt.
Both of these expressions involve local behaviour of co.
I think an investigation along this line is worth trying, but have no time to undertake it now. This book must go to press.
Remark 5 (added in proof). We have been dealing with the notion of multiplier adopted in §B.1, using that term to desiquate a non-zero entire function of exponential type whose product with a given weight is bounded on R. This specification of boundedness is largely responsible for our having had to introduce a local regularity requirement in §B.1. Such requirements become to a certain extent irrelevant if we return to the broader interpretation of the term accepted in Chapter X and permit its use whenever the product in question belongs to some LP(R). This observation, already made by Beurling and Malliavin at the end of their 1962 article, is based on the following analogue of the second theorem in §B.2:
Lemma. Let Q(x) > I be Lebesgue measurable. Suppose, given A > 0, that there is a function p(t), increasing and O(t) on [0, oo), with lim sup (p(t)/t)
- 0), we can rewrite the last integral as fX
x-t
[v(t)]-v(t)dt =
f
t
b
+ L (x_t l
I
x-t
+
t)([v(t)]-v(t))dt t
+ tz + l([v(t)]
v(t)) dt,
where the quantity b
°°
-00
[v(t)] - v(t) dt t(t2 + l)
is finite. Hence, aside from the additive constant log b, I cp(x)I - U(x) is just the Hilbert transform of n([v(x)] - v(x)) which is, however, bounded by it in absolute value. Referring now to problem 45(c) (Chapter X, §F), we see that Icp(x)I'14e-1(x114
0 and that in (0, oo), w(x) > w0(x) on a certain set of disjoint bounded open intervals lying therein.
Continuing with this problem we take just the intervals from (b) that lie in (1, oo), and denote them by (A., with n = 1, 2, 3, .... In
562
XI E A necessary and sufficient condition order to verify the desired property of w(x), it is enough to show that
w2)dx
1 meeting the local regularity requirement, existence of a W,,,, even co with e'(x)
>
on R
W(x)
enjoying the other properties enumerated in the conjecture is necessary. Problem 71 Show that such a proposition would be false. (Hint: Were such an c) to
exist, the preceding constructions would give us an even uniformly Lip 1 w(x) >- w(x) for which
(
w(x)
J
1 +x2
E
0, equal to zero at the origin and O(x2) near there, agreeing with w(x) for I x I > 1, say. Then °°
f
o
w1(x) 2
dx
