An Introduction to Harmonic Analysis (second corrected edition)

I

f

11/11100< 1. 7, Show that log

Ie"~.--=-~ I - e i,

iIX

'P

=

( ( i,

Re log e. -

el .. ) )

.

e" - e,/1

is a constant mul-

tiple of the conjugate function of the characteristic function of (a, p) by examining Im(IOg

«' - e~«)). e,l - e'P

8. Let 1 < p < co. Show that there exists a constant cp such that for leLPJ:C), 1 > 0,

Remark: This is an immediate consequence of 1.8; try, however, to prove it using 1.10 rather then 1.8 and then use it in the foIlowing exercise to prove 1.8. Hint: Assume that I is real valued, denote U).. = {/;/(i') > l} and V).. = {t;J(ei') < - l}. Denote by g).. the characteristic function of U)..; deduce from Parseval's formula that (q = p/(P -

AI U)..I ;;::;

1»

I

i

/(e")g;.(e ,) dl

= -

I I(i')g)..

(i') dl :::;; 2n " I

II LP I g;. "L4

"g,dl

and use (1.27) to evaluate L" Repeat for V;.. *9. Prove theorem 1.8 by the method of 1.7.

* 2.

THE MAXIMAL FUNCTION OF HARDY AND LITTLEWOOD

2.1

DEFINITION:

The maximal JUllction ofa functionJeLl(T)is the

function

(2.1)

sup 0<"1>"

I 2111

[t+"

J,-"

I

J(r)dr .

74

An Introduction to Harmonic Analysis

If we allow the value + 00 then Mit) is well defined for all t E T . We shall see presently that M ,( t) is finite for almost all t E T and that AI, is of weak Ll type. This will follow from the following simple

Lemma:

From any family

n = {Ia}

T one can ex-

of intervals on

tract a sequellce {In} of pairwise disjoint intervals, such that (2.2)

III

PROOF: Denote a 1 = sup fell and let I, be any interval of n satisfying II, > la,; let n 2 be the subfamily of all the intervals in o which do not intersect Jl ' Denote a2 = sup fell> and let 12 E O 2 such that > ~ a 2' We continue by induction; having picked I" ... ,lk we consider the family OH 1 of the intervals of n which inter-

I

III

1/21

Ih+11

> iak+! sect none of Ij,j ~ k, and pick IHI EnH ! such that where (/k+' = sUPfell .. We claim that the sequence so obtained satisfies (2.2). In fact, denoting by I n the interval of length

,III·

{In}

41/nl of which In is the center part, we claim that U I n 2 U la' clearly implies (2.2). We notice first that ak ..... 0 and consequently

n Ok n

=

0.

For lEO

let k denote the first index such that I ¢ Ok; then I Ik - 1 -:;t. since /I k - , ~ Is; J k - , • and the lemma is proved.

I ll/l,

2.2

Theorem:

which

0

and,
For fE L'(T) , M, is of weak LI type,

PROOF: Since Mit) ~ M/lI(t), we may assume f~ if AI it) > ,l. let It be an interval centered at t such that

o.

Let A> 0;

(2.3) Thus we cover the set {t;Mlt»,l.} by a family of intervals {It}. Let {In} be a pairwise disjoint subsequence of {It} satisfying (2.2). Then, by (2.2) and (2.3),

(2.4)

l{t;M.r
r

f(r)dr~~ff(r)dr.

).IUln

'T


2.3 Lemma: Let fE L'(T) and let m(}.) =mlfl(Je) be the distributioll fUllction of If Theil

I·

(2.5)

4foo ydnt(y).

/{t;MrC t »2Je}1 ~ -:;

,..

)+

75

III. The Conjugate Function PROOF: Writing Ii = f whenever If hence, by (2.4),

f g + h where g f whenever If I ~ A. and I> }., we have M At) ~ AI get) -I- Mh(t) ~ }. +Mh(t); =

=

Corollary: (a) For 1 < p < 00 there exists a constant cp such that if fE I!'(T), then M fE L!'(T) alld M f ~ cp

I

(b) If

flog+ III E LI(T), thell M f

llu

E

I fill"

LI(T) and

I M f IlL' ~ 2 + 4 JTlJllog+ If Idt. Denote by m(;.) and n(}.) the distribution functions of If I and M f respectively. We can rewrite (2.5) in the form PROOf:

foo). Y dm(y)

2n - n (2),) ~ 4l

(2.6)

hence if f We have

E

I!'(T) , 1 ~ p <

11M f I rp

parts we obtain (l

00,

4 ~ },P

J

Xl

). yP dm(y);

we have }.P(2n - n().» ~ 0 as A. ~

00 .

in J A dno.) = 2n2 1"" ),Pdn(2}'); integrating by

~ 1

~ p

P

P

0 00

0

< CD)

LOCi AP t/n(2}.) = fAP(2n-

n(2).»]~ + foro (2rr- n(2}.)p}.r- 1 d}. ~

J8 P iOCi).p-11OOYdm(y)dA

If p> I

l2n +8 Joor

it

l

tooYdm(Y)dA

.1I1d integrating by parts again we finally obtain:

p

=

I

An Introduction to Harmonic Analysis

76

2.4 Lemma: Let k be a nonnegative even function on (-n,1t), monotone non increasing on (0,1t) and such that f~- k(t) dt = 1. Then for all f E IJ(T)

II

(2.7)

k(t - t)f(r) dt

I; ;

Mit).

The definition (2.1) is equivalent to

PROOF:

M

~t)

= sup 0<"&_

IJ

(Mt - r)f(r)dr

I

where q,,, is the characteristic function of (- h, h) multiplied by 1/211 (so that q,,,dt = 1). A function k satisfying the conditions of the lemma can be uniformly approximated by convex combinations of q,,,, 0 < h ;;; n, and (2.7) is then obvious. ...

f

2.5 LetfeLl(T), letf(ri,) be its Poisson integral and let !(re i,) be the harmonic conjugate, DEFINITION:

The maximal conjugate function of f is the function

l(i')

=

sup

i

1 f{re ,) I·

0
Theorem: Let fE1!(T), 1 < p < 00; then lEI!(T) and

PROOF; Remembering that ](rfi') is the Poisson integral of J(i') and that the Poisson kernel satisfies the condition of lemma 2.4, we obtain I J(ri') 1~ M ,At); hence l(i');;; M J(I) and the theorem fol... lows from 2.3 and 1.8.

2.6 We have defined the conjugate J of a function fE IJ(T) as the boundary value of the harmonic function J(ri') = (Q(r, . ).f)(t) where 2rsint Q(r,t) = 1-2rcost +r2

(2.8)

is the conjugate Poisson kernel. Since the limit (2.9)

Q(l, t)

=

t cossin t 2 t lim Q(r, t) = = - - =cot-2 I-cost. t r~l SID 2

77

HI. The Conjugate Function

is so obvious and so explicit. we are tempted to reverse the order of the operations and write (2.10) The difficulty, however. is that Q(1,t) is not Lebesgue integrable so that the convolution (2.10) is, as yet, undefined. We now aim at showing that, the convolution appearing in (2.10) can be defined as an improper integral and that. with this definition. (2.10) is valid almost everywhere. Lemma: For! E LI(T) and IJ E(r,I)= PROOF:

(2.11)

I

=

1 - r. we have

1 [2,,-3 2nJ3 Q(l.·)f(t-.)dr-/(ri')

I

Write

E(r,t) ~

I +

1 [2,,-3 . 2x Js (Q(l,.) - Q{r,·»f(t - .)d.

I+

I ;n f~3 Q{r••)!(t-.)d·1

=

1

We notice that the function

(2.12)

(1-r)2 sint Q(I, t) - Q(r, I) - (i _ cos 1)(1 _ 2rcos t +r2) =

1- r - l - Q(I,I)p(r,l) +r

is odd and monotone decreasing on (0, x). For IJ< t < x 1- r I - r ( IJ)-l ---Q{1 t) ~ - - sin < x l+r ' -1+r 2

so that (2.13) It follows that

(2.14)

Q(1, t) - Q(r. t) < xp(r, t).

2

E (r,t)+E (r,t).

An f ntroduction to Harmonic Analysis

78

rn order to estimate E 2(r, t) it is sufficient to notice that in (-{},{J) 2 we have 1Q(r, 1)1 < - - - and consequently

1- r

(2.15)

Corollary: sup O<S<"

I 2~rr Js(2n-S Q(I, r)f( t -

r) dr

I ~ f(/I) + 4M1fl (t).

2.7 The estimates (2.14) and (2.15) are clearly very wasteful. They do not take into account the fact that Q(r, t) is odd and we can improve them by writing (2.16)

E](r,l)

1 (It

=

12it Js

(Q(l,r) - Q(r,r»(f(t - r) - f(t

+ r»dr

1

and (2.17)

Elr, I)

1

=

~

;rr loS Q(r, t) (J(t - r) - f(t + r» dr -- 1 _ _

n(l-r)

1

(SIU(t - r) - f(t Jo

+ r»

dr

1

1

where 0 < {}1 < {} (the mean value theorem). At every point t of continuity of f, and more generally, at every point in which the primitive of f is differentiable, we have:

LSI (J(t -

(2.18)

r) - f(t

+ r»dr

=

O(/}l).

By (2.17) it is clear that if (2.18) holds, Eir, t) -+ O.

Theorem: Let fE Ll(T); at every t E T for which (2.18) is valid we have, ({) = 1 - r), £(r, t)

1

~ 12rr

1 2

S

1<-S

Q(l, r)f(t - r) dr -J(ri

l )

1-+ 0

as r --+ 1. PROOF: As in (2.11), £(r,t) ~ £1(r,t) + £2(r,I). We have already remarked that under the assumption (2.18), lim r _ 1 Eir, t) ~ 0 so that

III. The Conjugate Function

79

we can confine our attention to E 1(r, t). For e > 0, let '1 > 0 be such that for 0 < {)1 ~ 11

If:'

(2.19) and write (2.20) 2rr.E 1(r,t)

=

(J(t - r) - J(t

I(f +

+ r»

I

dr < sf}1

L')(Q(l,r)-Q(r,r»(J(t-r)-J(t+r»drl·

The second integral tends to zerO by virtue of the fact that on (1/, n), Q(r, r) ~ Q(I, r) uniformly. Put

[9

1


+ r»

dr,

then, integrating the first integral in (2.20) by parts, we see that it is bounded by [
elf

{)1 d(Q(1,1?1) - Q(r'{)1»

I;

integrating by parts once more and remembering that Q(1,{)1) < rr.liJ!> it follows from (2.13) and (2.19) that E 1(r,t)<10c+o(l) and the theorem is proved.

2.8 Let F be defined on T and assume that for all {) > 0 F is integrable on T,,(-{),{). DEFINITION:

The principal value of fTF(t)dt is PV

1

F(t) dt

T

lim

=

9->0

i

2"-~

F(t) dt.

9

For fE Ll(T) condition (2.18) is satisfied for almost alit eT; since

l(r it) ~ j(e il ) almost everywhere, we obtain, Theorem: LetJEL1(T). The principal value of

21rr.ff(t-r)Cot~dr

exists for almost all t ET, alld, almost everywhere,

- . = PV 2n 1 f(e") 2.9

f

J(t - r)cot r dr.

2

Theorem 2.7 can be used both ways. We can use it to show the

f

existence of the principal value of P V J(t - r) coti dr if we know that

j (e

il

)

exists or to obtain the existence of j(eil ) at points where

80

An Introduction to Harmoruc Analysis PV

f

J(t -7:)cot

~dt"

clearly exists. For instance, if J satisfies a Lipschitz condition at t, that is, if IJ(t + h) - J(t) ~ K h for some K > 0 and IX >0, then

I

I I·

f~IJ(t - t") - J(t + t") Icot idt"< ro and it follows that /(eit ) exists and 7·

1 ('

J(e~ =21t

(2.21)

Jo

U(t-t")-J(t

t"

+ t"»cot idt".

If J satisfies a Lipschitz condition uniformly on a set E c T, that is, if for some K > 0 and IX > 0

IJ(t + h) - J(t) I ~ K Ih I'"

for all Ie E,

then the integrals (2.21) are uniformly bounded and

2~

L'

U(t - -r) - J(t

+ t")}cotidt" -+ I(e i -)

uniformly in teE as B -+ O. It follows, rexamining the proof of 2.7, that /(r t!') -+ I(eit) uniformly for teE as r -+ 1. In particular, if E is an interval, it follows that I(i') is continuous on E. 2.10 Conjugation is not a local operation; that is. it is not true that ifJ(t) = g(t) in some interval I. then /(t) = get) on I. or equivalently. that if J(t) = 0 on I, then I(t) = 0 on I. However,

Theorem: IJ J(t) =0 on an interval I, then I(t) is analytic on I . PROOF: By the previous remarks I is continuous on I. Thus the function F = J + il is analytic in D and is continuous and purely imaginary on I. By Schwarz's reflection principle F admits an analytic extension through I, and since F(it ) = il(e1 on 1, the theorem follows. <411

Remark: Using (2.21) we can estimate the successive derivatives of j at points tel and show that I is analytic on 1 without the use of the "complex" reflection principle. EXERCISES FOR SECfION 2 1. Assume / e LiP.(T), 0 < a ~ 1. Show that Ie Lip...(T) for all a.' < a.. *2. Assume /eLip,Jn, 0 < rx < I. Show that Ie LiP.(T). Hint:

81

III. The Conjugate Function - +h)-J(t) 7. 1(1

=

-

If

2n.

r (f(t+h -r)-f(t+h»cot-dr-

2

-~ f 2n

f

21t

=

O(h a) +

(f(t - r) - f(t»cot 2h

-

(f(t - r) - f(t

~ elr

2

+ h»cot r --1-"- dr 2

• 2h

-

f

21t - 2h

(f(t - r) - f(t»cot ~ elr 2

2h

r

21t

=

O(ha ) 1-

-

(f(r - r) - (I») (r+1z cot ----- - - cot - dr2

• 2h

- (/(1

r)

2h -

+ h)

21t-2h

- f(t»

1

r

2

+17

cot --- dr. 2

2h

.3. Assume fE C"(T), n~ I. Show thatje Cn- 1(T) andjln-1)E LiPiT) for all a < I. 4. Localize exercises 1-3, that is. assume that 1 satisfies the respective conditions on an open interval leT and show that the conclusions hold in I. *3. THE HARDY SPACES

In this section we study some spaces of functions holomorphic in the unit disc D. these spaces are closely related to spaces of functions on T and we obtain, for example, a characterization of I! functions and of measures whose Fourier coefficients vanish for negative values of n. We also prove that jf for some f E O(T) , j(eit ) is sum mabIe then S[lJ == S[J], and, finally, we obtain results concerning the absolute convergence of some classes of Fourier series. We start with some preliminary remarks about products of Moebius functions.

3.1 LetO
=1_~lrr_=-Z)_ defines,asisweJl ( (1 - z()

known, a conformal representation of l> onto itself, taking ( into zero and zero into The important thing for us now is that b(z, () van< r, then ishes only at z = ( and I b(z, 0 I = 1 on I z I = 1. If 0 <

1(I.

b(r

Z

()

-, -

r

=

r

at z =(, and b(z,O) = z.

-

I(-~«( I (r-i - -z)zO- IS. hoIomorphlc. ,In Ib(~, ~)I

= 1 on

1

z

1

I I ~ r, Z

1(I

.

val1ls hes on Iy

=r. For (= 0 we define

82

An Introduction to Harmonic Analysis

II

Let 1 be holomorphic in z ~ r and denote its zeros there by (1) ···,(k (counting each zero as many times as its multiplicity). The function 11(z)

~ l(z) (Q

b

I

U, ~,,)r

1

is holomorphic in

I

for

I

is zero-free and satisfies 11(z) = l1(z) is harmonic in ! z ~ r we have

I

log 111(0) 1

=

1 2n

Iz I ~ r,

Iz! = r. Since log 111(z)!

f 10gI11(re

it )

I dt

and if we assume, for simplicity, that 1(0) ~ 0, the above formula is equivalent to Poisson-lensen's lormula:

We implicitly assumed that 1 has no zeros of modulus r; however, since both sides of the formula depend continuously on r, the above is valid even if 1 vanishes on z = r. The reader should check the form that Poisson-Jensen's formula takes when 1 vanishes at z =0 .

II

k

k

I I_I = L

The term log TI r (n I

log(r I

(til-I) is positive, and remov-

I

ing it from (3.1) we obtain lensen's inequality (3.2)

log 11(0)!

~ 1n flOg 11(r it) 1dt ;

or, if 1 has a zero of order s at z

logl~_~

z-j(z)

=

0,

I + log (r ~ S

)

1 2n

f

i

10gI1(re ')!dt.

Another form of Jensen's inequality is: let 1 be holomorphi:- in 1 z I ~ r and let (~, ... ,,~ be (some) zeros of 1 in ~ r, counted each one at most as many times as its multiplicity. Then

Izi

log 11(0) 1 +

(3.3)

Z

log(r I (~1-1)

~ 2~ flog 11(reit ) I dt.

Inequality (3.3) is obtained from (3.1) by deleting some (positive) terms of the form log (r 1(n 1- I) from the left-hand side. '1.2 1.

Let p > 0 and let 1 be holomorphic in D. We introduce the notation I.

Ill. The Conjugate Function

83

IfO
or, in other words, hi!, r) is a monotone nondecreasing function of n r. The case p = 2 is particularly obvious since for fez) = I~ anz , 2 we have hi!,r) = r2n. We show now that the same is true n for aU p> O.

Ig"la I

Lemma: Let! be holomorphic in D and p > O. 1 hen hp(f,r) is a monotone nondecreasing functi01l of r.

We reduce the case of an arbitrary positive p to the case p = 2. Let r 1 < r < 1. Assume first that! has no zeros on z ~ r and consider the function t g(z) = (f(Z»PI2; then PROOF:

II

;1tJ !J(rl,eit)IPdt

~2~

JI

g(rt,eitWdt

~2~ =

J

~

21t

Ig(re

it

)1 2 dt

J

it !J(re ) \p dt

or hp(f, rl) ~ hp(f, r). If! has zeros inside z < r but not on z = r, we denote the zeros, repeating each according to its multiplicity, by (I> ... , (k, and write

II

II

!1(Z) = fez)

" (Z (n))_1 . (11 b r' r

For Iz I < r we have \!(z) \ < \!I(Z) I, for \ z I = r we !J I(Z) I, and!1 is zero-free in Iz I ~ r. It follows that h/f,r l ) < hp(ft>rJ) ~ hp(fl,r)

=

have If(z)

1=

hp(j,r).

Since hp(f, r) is a continuous function of r, the same is true even if does have zeros on z \ = r, and the lemma is proved.
I

f

Lemma: Let {en} be a sequence of complex numbers satisfying (n 1 alld L (1 - Cn 00. Then the product

3.3

I I<

(3.5)

I I) <

B(z) =

TI b(z,(.)=zm II 1

{n#O

_e:«(._=-~ __

1(.1 (l -

z(.)

converges absoilltely and uniformly in every disc of the form

Izl~r<1.

t

Any branch

of(f(z»p12.

An Introduction to Harmonic Analysis

84

PROOF:

It is sufficient to show that

verges uniformly in

Iz I : : ; r < 1.

But

L

(1 -

and the series converges since

L

/1 -

C~eC

z)_ len (1- zen)

I

I

con-

Ien I) < 00 .

The product (3.5), often called the Blaschke product corresponding to {en}, is clearly holomorphic in D and it vanishes precisely at the points en. Nothing prevents, of course, repeating the same complex number a (finite) number of times in {en}' so that we can prescribe not only the zeros but their multiplicities as well. Since all the terms in (3.5) are bounded by I in modulus, we have B(z) < 1 in D.

I

I

3.4 We now introduce the spaces H P (H for Hardy) and N (N for Nevanlinna). DEFINITION: The space lIP, p> 0, is the (linear) space of all functions f holomorphic in D, such that

(3.6)

If

II~p

=

lim hif, r)

sup hif, r) <

=

r~l

00.

0
T he space N is the space of all functions Jholomorphic in D, such that (3.7)

I f liN

=

sup

O
,f- flog

_7[

+

it

If(r e

)

Idt < 00.

Remarks: (a) For p ~ 1, IIIIHP as defined in (3.6) is a norm and we shall see later that HP endowed with this norm, can be identified with a closed subspace of L!'(T). For p < 1, II lI~p satisfies the triangle inequality and is homogeneous of degree p. It can be used as a metric for HP; II IIHP is homogeneous of degree one but does not satisfy the triangle inequality. II liN is not homogeneous and does not satisfy the triangle inequality. (b) If p' < p we have N:::l HP' :::l HP. The space H2 has a simple characterization:

Lemma: is finite. PROOF:

Letf(z) =I:anzn; thenfEH2 if,andonly if, h if, r)

=

'I.: 1an 2r2n , It follows that II f 1

'I.:la I

2

n

II~' = I~ 1an 12 ....

85

III. The Conjugate Function An immediate consequence is that in this case a., e inl. tegral of j(i') ....,

L:

r is the Poisson in-

3.5 Lemma: Let fEN, and denote its zeros in D by (I> (2'"'' counting each one according to its multiplicity. Then L (1 - I < 00.

q)

Remark: The convergence of the series L (1 -I (n I) is equivalent I (n I hence to the boundedness to the convergence of the product (below) of the series I log I (n I (not counting the zeros at the origin, if any).

n

PROOF: We may assume f(O);& o. By Jensen's inequality (3.3), if M is fixed and r sufficiently close to 1 }.{

10gl!(0) I-II f 11.v ~ Letting r

L logl ~n I - Mlogr. I

-+ 1 we

obtain

and since M is arbitrary the lemma follows.

3.6 If we combine lemma 3.3 with 3.5 we see that iffE N, the Blaschke product corresponding to the sequence of 7..eros of f is a well-defined holomorphic function in D, having the same zeros (with the same mUltiplicities) as f and satisfying I 8(z) I < 1 in D. If we write F(z) = j(z)(B(z)f 1 then F is holomorphic and satisfies I F(z) I > If(z) I in D. We shall refer to f = BF as the callonical factorization of f· Theorem:

Let f

E

H P, P > 0, and let f = BF be its canonica I fac-

torization. Then FE H P and II F IIHP ~ II f IllfP' PROOF:

The Blaschke product B has the form N

B(z)

=

lim N _ ",zmTIb(z, (n) . 1

If we write Fiz) ~f(z)(zmn~ b(z,(,))-t,then F",convergesto Funiformly on every disc of the form I z I ~ r < 1. Since the absolute value of the finite product appearing in the definition of F N tends to one uniformly as I z 1, it follows from lemma 3.2 that F N E /fP and

1-+

"FN IIHP

=

IlfIIHP' Let r hp(F, r)

=

< 1, then

lim h/FN, r) ~ lim N-oo

N-OO

I F IIJp = I f II,fp· N

An Introduction to Harmonic Analysis

86

r)

f II:p

r

FilliP

filliP,

Now h/F, ~ I for all < 1 is equivalent to I ~I and since the reverse inequality is obvious, the theorem is proved.

~

Theorem 3.6 is a key theorem in the theory of H P spaces. It allows us to operate mainly with zero-free functions which, by the fact of being zero-free, can be raised to arbitrary powers and thereby move from one HP to a more convenient one. This idea was already used in the proof of lemma 3.2. Our first corollary to theorem 3.6 deals with Blaschke products. 3.7 Corollary: Let B be a Blaschke product. Then I B(e it ) 1 = I almost everywhere. it PROOF: Since I B(z) 1 < 1 in D it follows from lemma 1.2 that B(e )

exists as a radial (actually: nontangentiaI) .Iimit for almost all t E T. The canonical factorization of f = B is trivial, the function F is identically one, and consequently

Since 1 B(eit)1 ~ 1, the above equality can hold only if I B(il) 1 = 1 almost everywhere. ~ 3.8 Theorem: Assume fE lIP, p> O. Then the radial limit lim, .... d(reil ) exists for almost all tE T and, denoting it by f(e· t ), we have

PROOf: The case p = 2 follows from 3.4. For arbitrary p > 0, let f = BF be the canonical factorization of f, and write G(z) = (F(z)Y/2. Then G belongs to H2 and consequently G(rei,) -> G(ei~ for almost all t E T; ~t every such t, F(rit) converges to some F(e i ,) such that 1FCit) Ip/2 = 1G(e il ) I. Since B has a radial limit of absolute value one almost everywhere we see that f(i') = limf(r eit ) exists and /fCe i ,) IP/2 = I G(eit ) 1 almost everywhere. 1 il IIGII ~2 ~II ) 1P dt Now I "~p = I F _n c(i') 12 dt = -2n and the proof is complete.

f

Ilftp=

Ilf(e

3.9 The convergence assured by theorem 3.8 is pointwise convergence almost everywhere. For p;:;; 2 we know that f is the Poisson

87

III. The Conjugate Function "

,

,

integral of f(e') and consequently f(re") converges to f(e't) in the J!(T) norm. We shall show that the same holds for p = 1 (hence for p ~ 1); first, however, we use the case p~ 2 to prove:

Theorem: fEH P' ,

Let 0 < p < p' and supposefEHP andf(e i') ElJ"(T), Then

PROOF: As before, if we writef = BF, G(z) = (F(z)Y12, then G E H2 and G(ei') E L2P 'IP(T). G is the Poisson integral of G( eit ) and consequently GEHzp'Ip which means FEHP', hencefEHP'. ~

Corollary: Let fELI(T) alld assume JELI(T); then (f + iJ)EHI, PROOF: We know (corollary 1.6) that (f+ij)EW for all cx< 1 and by the assumption (f. + ij)(eit ) E LI(T). ~

3.10 Theorem: Every function f ill HI can be written as a product f = fd2 with fl.JZ E HZ . Let f = BF be the canonical factorization of Fl/z, fz =BFl/z.

PROOF:

fl

=

3.11

f.

Write ~

We can now prove:

Theorem: Let fE H' and let fCe i ,) be its boul/dary value. Then f is tlte Poisson integral of f(e it ). We prove the theorem by showing that f(re i,) converges to fe/') in the L' norm, This implies that if fez) = Lanzn, then an is the Fourier coefficient of f(i') which is clearly equivalent to f being the Poisson integral of f( it). Write f =fdz with f j EH 2, j = 1,2; we know that as r~ 1, PROOF:

lifiri\- fieit) Ilu -+ O.

f(re it ) - f(e it )

Now, =

f,(reit)fz(re i')

-

fl(eit)/z(/');

adding and subtracting 1,(ei')/2(r eit ) and using the Cauchy-Schwarz inequality, we obtain

IIf(reit ) - f(i')

liLt

it ~ IIJzIIL2I1fl(re ) - fl(eit)lIu

+ II II IILllllireit ) - f2(i t) IIL2 and the proof is complete.

88

An Introduction to Harmonic Analysis

Remark: See exercise 2 at the end of the section for an extension of the theorem to the case 0 < p < 1. Corollary: Let feL1(T) and leL1(T). Then S[l]

=

S[J].

From 3.9 and the theorem above follows that l(r eit) is the .... Poisson integral of l( i~ (see remark 1.4). PROOF:

3.12 Theorem: Assllme p ~ I . A function f belongs to H P if, and only if, it is the Poisson illtegral of some f(i t ) E l!'(T) satisfying j(lI)

(3.8)

=

0

for all n < O.

Let f E H P ; by theorem 3. I 1, f is the Poisson integral of f(it) and (3.8) is clearly satisfied. On the other hand, let fE IJ(T) and assume (3.8); the Poisson integral off, PROOF;

00

f(re it )

=

2:o j(lI)r" e

is hoIomorphic in D, and since that fez) E HP.

1/

00

int

=

fer e

2: j(n)z" 0

it

) 1/ LP

~

1/

f( eit ) II LP, it follows ....

3.13 For p > 1, we can prove that every fEll P is the Poisson integral of f(e it) without appeal to theorem 3.11 or any other result obtained in this section. We just repeat the proof of lemma 1.2 (which is the case p =00 of 3.12): if fEH P , IIf(re it )/lLP is bounded as r .....d; we it can pick a sequence rn ~ I such that f.(e ) = fern eit ) converge weakly in I!(T) to some f(e it ). Since weak convergence in IJ implies convergence of Fourier coefficients, it is clear that (3.8) is satisfied and that the function f with which v.e started is the Poisson integral of f(i t ). For p = 1 the proof as given above is insufficient. Ll(T) is a subs pace of M(T), the space of Borel measures on T, which is the dual of C(T), and the argument above can be used to show that every f E H 1 is the Poisson integral of some measure Jl on T. This measure has the property (3.9)

{l(n)

=

0,

for all n < 0 .

All that we have to do in order to complete the (alternative) proof of theorem 3.12 in the case p = I is to prove that the measures satisfying (3.9), often called allalytic measures, are absolutely continuous with respect to the Lebesgue measure on T.

III. The Conjugate Function

Theorem: (F. and M. Riesz): fying (3.9)

lien)

=

89

Let Jl be a Borel measure on T satisfor all n <0.

0

Then Jl is absolutely continuous with respect to Lebesgue measure. We first prove:

Lemma: Let E c: T be a closed set of measure zero. There exists a functioll

(ii)

I
(3.10)

1

=

PROOF: Since E is closed and of measure zero we can construct a function !/t(eil) on T such that !/t(ei ,) > 0 everywhere, !/t(ei ,) is coni tinuously differentiable in each component of T" E, !/t(e ,) ~ co as t i approaches E, and !/t E [}(T). The Poisson integral !/t(z) of !/t(e ,) is positive on D and !/t(z) ~ co as z approaches E. The conjugate function is continuous in [)"E (see the end of section 2) and consequently,

~ is holomorphic in D and continuous in jj" E. At every point where !/t(z) < co we have I~(z) I < 1, if we put q;(z)

and as !/t(z) -+ (3.10).

=

00,

!/t(~(~ ~~~{z~ i'

then

rp(z) ~ 1. If we define cp(z)

=

1 on E then

~ satisfies

....

Assume that Jl satisfies the condition (3.9). We can assume P(O) = 0 as well (otherwise consider JL - A(O) dt) and it then follows from Parseval's formula that PROOF OF THE THEOREM:

(3.11)

ff

(j, Jl >=

dll

=

0

for every IE C(T) which is the boundary value of a holomorphic function in D or, equivalently, sllch that J(n) = 0 for alI negative n. Let E c: T be closed and of (Lebesgue) measure zero. Let ~ be a function satisfying (3.10). Then, by (3.11)

f ~md'L f ~m m-"" =

and by (3.10)

lim

0

for alI In> 0

dJl = Jl(E).

An Introduction to Harmonic Analysis

90

Thus p(E) = 0 for every closed set E of Lebesgue measure zero and, since p is regular, tht: theorem follows. .... 3.14 Theorem 3.13 can be given a more complete form in view of the following important

Theorem: Let fEHP, p>

O,J~

0; then

log IJ(it) IE V(T).

Remarks: The same conclusion holds under the weaker assumption fE N. We state it for H P since we did not prove the existence of f(i t) for fE N (cf. [28], Vol. I, p. 276). Since plog+ If ~ Ifl p , it we already know that log+ J(e ) EL'(T). Thus the content of the it theorem is that J(e ) cannot be too small on a large set.

I

I

I

PROOF: Replacing f by z-mJ, if J has a zerO of order m at z we may assume f(O) r!: O. Let r < 1; then, by Jensen's inequality

=

0,

(where log-x = -logx if x < 1 and zero otherwise). It follows that Jlloglf(reit)lldt is bounded as r-+ I and the theorem follows from .... Fatou's lemma.

Corollary: If fr!: 0 is in HP, f(e i') call vanish only on a set of measure zero. Combining theorem 3.13 with our last corolJary we obtain that analytic measures are equivalent to Lebesgue's measure (i.e., they all have the same null sets).

3.15 Theorem: Let E ,be a closed proper subset of T. Any COIItilluous Junction on E can be approximated uniJormly by Taylor polYllomials. t We denote by C(E) the algebra of all continuous functions on E endowed with the supremum norm. The theorem claims that the restrictions to E of Taylor polynomials are dense in C(E). Consider a measure Ji carried by E orthogonal to all zn, n = 0, 1, .... PROOF:

t We use the term "Taylor polynomial" to designate trigonometric polynomials of the form L~aneint.

m. The Conjugate Function

91

pen) = 0 sO that f1. is analytic; hence I' = I dt with by E. By theorem 3.14 I = 0; hence there is no nontrivial functional on C(E), which is orthogonal to all Taylor polynomials, and the theorem follows from the Hahn-Banach theorem. ~ Now (zn,ji.) =

I E HI ,J carried

3.16

We finish this section with another application of 3.10. Theorem: (Hardy): Letf(z) = I,g'a nz nEH I • Then I,'flanln-1
Remark: (3.8), then

The theorem can also be stated: "Let

IE Ll(T)

satisfy

I, f If(n) In-I 51/f11z.l."

If F(e il ) is a primitive of I(e i ,), then F is continuous on T and consequently its Fourier series is Abel summable to F at every t E T. In particular I, 'f (anlnV tends to a finite limit as r -+ 1. If we assume an;?; 0 for all n then I,a"ln is clearly convergent (compare with the proof of 1.4.2). In the general case write 1= Id2 with 11 E H2 and f2 E H2. Writing liz) = I,Aj,n zn we have an = I,~=oA1.kA2.n-k and consequently lanl ~ a: = I,~=oIAI,kIIA2n-kl. The functions!f(z) = I, IAi.nl zn are clearly in H2 so that, by the Cauchy-Schwarz inequality, I*(z) = Ji(z)/i(z) = I, a:znE HI. Since an*;?; 0 it follows from the first part of the proof that I,'f (a: In) < 00 and the theorem follows PROOF:

I I

from an ~

a: .

~

3.17 LetfeH I and assume 1hat f(i') is of bounded variation on T. [f I,..., I,;:'ane int then I,~inaneint is the Fourier-Stieltjes series of til. Thus the measure df satisfies the condition of theorem 3.13 and dt and /,(z) is in HI. Combining this with 3.16 consequently dl = we obtain:

r

Theorem: Let leHI and assume that f(eit)is of bounded variation on T. Then f(e i') is absolutely continuous and I,lf(n)1 ~ 00.

An equivalent form of the theorem is (see 3.9): Theorem: Let 1,/ E LI(T) and assume that both I and I are 01 boullded variation. Then they are both absolutely continuous alld

L~GOjj(n)l<

00.

EXERCISES FOR SECTION 3 1. Deduce theorem 3.13 (F. and M. Riesz) from theorem 3.1 1. 2. Show that for all p>O, if fEH P, then flf(it)-f(rlWdt-+O as

92

An Introduction to Harmonic Analysis

r-+ 1. Hint: Reduce the general case to the case in which I is zero-free. In the case that I is zero-free, \\Tite 11 =/t; then /1 E H 2p and

hence show that if the statement is valid for 2p is also valid for p. Use the fact that it is valid for p;;;;' 2. 3. Let E be a closed set of measure zero on T. Let rp be a continuous function on E. (a) Show that there exists a function <1>, holomorphic in D and continuous on D such that <1>(eil ) = rp(e il ) on E, (b) Show that <1> can be chosen satisfying the additional condition sUPzeDI(z)1

=

sUPe'feEI rp(eil)l.

Him: Construct by successive approximation using 3.15 and lemma 3.13. 4. Let IEL1(T) be absolutely continuous and assume f'log+If'1 EL1(T). Prove that I IJ(n) I < 00.

Chapter IV

Interpolation of Linear Operators and the Theorem of Hausdorff-Young

Interpolation of norms and of linear operators is really a topic in functional analysis rather than harmonic analysis proper; but, though less so than ten years ago, it still seems esoteric among authors in functional analysis and we include a brief account. The interpolation theorems that are the most useful in Fourier analysis are the RieszThorin theorem and the Marcinkiewicz theorem. We give a general description of the complex interpolation method and prove the RieszThorin theorem in section 1. In the second section we use Riesz-Thorin to prove the Hausdorff-Young theorem. We do not discuss the Marcinkiewicz theorem although it appeared implicitly in the proof of theorem 111.1.7. We refer the reader to Zygmund ([28 J chap. XII) for a complete account of Marcinkiewicz's theorem. 1. INTERPOLATION OF NORMS AND OF LINEAR OPERATORS

1.1 Let B be a normed linear space and let F be defined in some domain n in the complex plane, taking values in B. We say that F is holomorphic in n if, for every continuous linear functional J1 on B, the numerical function h(z} =
I

I I

93

Analysi~

94

An Introduction to Harmonic

(1.1)

IIFII = sup {II F(iy) 110' IIF(I +iy)lll}' y

•

For 0 < a < 1, the set BB,. = {F EBB; F(a) = O} is a linear subspace of BB. We shall say that 110 and are consistent if BB~ is closed in BB for all < a < I. A convenient criterion for consistency is the following:

I III

I

°

Lemma: Assume that for every fE B, r i:- 0, there exists afunctional IL continuous with respect to both 110 and such that (f.f1.) i:- O. Then 110 alld 1/ are consistent.

°

I

I Ill'

I

III

PROOF: Let <:x < 1 and let Fn E fJ8,., Fn -+ F in :fl. Let IL be an arbitrary linear functional continuous with respect to both norms. The functions (F.(z),f1.) are bounded on the strip Q and tend to (F(2),J1) uniformly on the lines z = iy and z = 1 + iy. By the theorem of Phragmen-LindelOf the convergence is uniform throughout nand in particular (F(a),J1) = limn~o:>(Fn(a),J1) = O. Since this is true for every functional J1 it follows that F(a) = 0, that is, FE f!$. and the .: lemma is proved. ~

I lit

Remark: The condition of the lemma is satisfied if!1 110 and both majorize a third norm" 112' This follows from the Hahn-Banach theorem: iff i:- 0, there cxists a functional f1. continuous with respect to 112 such that (f,IL) i:- O. It is clear that if > 112' f1. is continuous with respect to " j ' j = 0, 1 .

I

I II} I

I

1.2 We interpolate consistent norms on B as follows: for 0 < a < 1, the quotient space BljBl,. is algebraically isomorphic to 8 (through the mapping F -+ F(a». Since :!I,. is closed in ff9, Blj/4,. has a canonical quotient norm which we can transfer to 8 through the forementioned isomorphism; we denote this new norm on B by The usefulness of this method of interpolating norms comes from the fact that it permits us to interpolate linear operators in the following sense:

I I •.

Theorem: Let 8 (resp. 8') be a normed linear space with two CO/lsistent norms I!o and (resp. II~ alldll II~)· Denote the interpolating norms by (resp. II~), 0 < a < t. Let S be a linear • transformation from B to B' which i~ bounded as

I

(1.2)

I III I I

(B,

I

I

I II)! (8; I II;),

j

= 0, 1 .

95

IV. Interpolation of Linear Operators

Then S is bounded as (1.3)

(B, II Ila)

alld its norm II S

1111

-!

(B', II II),

satisfies

I S 1111 ~ I S I!~ -II I S II~ .

(1.4)

PROOF: We denote by 89' the space of holomorphic B'-valued functions which is us~d in defining I I!~. The map B ~ B' can be extended to a map 14-+ fig' by writing SF(z) = 8{F(z». To show that SF so defined is holomorphic, we consider an arbitrary functional JI, continuous with respect to II II~ or I and notice that (SF(z),Jl) = (F(Z),S*JI). Since SF(z) is clearly bounded it follows that SF E fJI' • Let /E B, II / lIa = 1; then there exists an FE fJI such that F(rx) = / and such that I F I < 1 + 8. Applying S to F, we obtain

!I:,

I S/II~ ~

II SF II' ~ (1 + e) max (II Silo'" Sill);

hence II S 1111

~ max(1I Silo' II S !It>

which proves the continuity of (1.3). To prove the better estimate (1.4), we consider the function eQ(z-a) F(z), where ell = I S 110 II S II; I. We have

I s/II:

~ "S(e (Z-II)F(z» Q

=

II'

sup{e-Q<xIISF(it)II~, e

Q

(1-

lIl

IlSF(1 + it)

II:}

t

~ (1 + 8)sup{e-

=

QII

Q

II Silo' e (1-II)!! Sill}

(1+e)IISII~-allslI~·

..

Remark: The idea of using the function eD (Z-II) goes back to Hadamard (the "three-circles theorem"); it can be used to show that, for every /E B, (1.5)

1.3 A very important example of interpolation of norms is the following: let (X, dx) be a measure space, let I ~ Po < PI ~ CIJ, and let B be a subspace of J!o (') U'(dx). We claim that the norms II 110 and II III induced on B by 1.!'°(dx) and U'(dx), respectively, are consistent. By lemma 1.1, all we have to show is that, given / E B, / # 0,

96

An Introduction to Harmonic Analysis

there exists a linear functional J1, continuous with respect to both norms, such that # 0; we can take as J1 the functional defined by =jfgd1 where gEVnL"'(dl) has the propertyt that fg> 0 whenever

\II> o.

B LPO (")

Let (X, dl) be a measure space, = I!'(dl) (with t ~ Po < PI ~ (0). Denote by " the norms induced by J!'(dl) , and by II 1111 the interpolating norms. Then II II. coincides with the norlll induced all B by l!'a(dl) where Theorem:

t

_

(1.6)

___ J'OPI_ __

Pil - poa

+ PI(t

(

_ a)

Po

= 1_ a

of

I

PI

Let fEB and IlfIILPa~l. Consider F(z) = i = e ., and

PROOf:

where

f

= 00

\II

a=poai°;'&I~)

(= I-J--a

WehaveF(a)=fand consequently Ilfll.~ = so that

\llp·,PO

II F(iy)

110

=

)

\11

• 0

(:-11)+

lei.,

if PI =(0).

IIFII;nowIF(iy)1 = \III-a.

(f \lIPa dl

t

Po

~ I;

similarly 111'(1 + iy) III ~ 1 (use the same argument if PI < 00 and check directly if PI = (0); hence ~ 1. This proves II II,,;?;; II P • ° In order to prove the reverse inequality, we denote by qo, ql the conjugate exponents of Po and PI and notice that the exponent conjugate

Ilfllll

IIL

to p" is

(1.6') We now set B' = Co (") Lq'(dl) and denote by go' the corresponding space of holomorphic B'-valued functions. LetfE B and assume 111 IIL P • > I; then, since 8' is dense in C·, there exists agE B' such that IILO. ~ 1 and such that j fgdl > 1. As in the first part of this proof, there exists a function G E go' such that G(a) = g and II G II (with respect to qo, q I) is bounded by t. Let FE tB such that F(a) = j. The function h(z) = j F(z)G(z)dl (remember that for each ZEn, F(Z)EB and G(z)EB') is holomorphic and bounded in n (see Appendix). Now Il(a) > 1, hence, by the Phragmen-LindelOf

h

t If we writef= I f I eicp with real-valued

!P. we may takeg = min (l,lfIPo)eiCP.

IV. Interpolation of Linear Operators

I I I I I

97

theorem, h(z) must exceed 1 on the boundary, However, on the boundary lI(z) ~ F" ~ F!! so that 1/ F 1. This proves ~ 1 and it follows that and 1/ are identical. ..

l/fll"

IIG I I I II"

I/u.

II>

1.4 As a corollary to theorems 1.2 and 1.3, we obtain the RieszThorin theorem. Theorem: Let (X, dr) and (~, dl}) be two measure spaces. Let B = !!,o fl lJ"(dl) and 8' = fl I!".(dl}) alld let S be a linear transformation from B to B', cOlltinuous as (B, (B', j = 0, I, where (resp. II~) is the norm induced by I!"(dx) (resp. I!'J(dlJ». Then S is contilluous as

eo

I Ilj

I Ilj);

I

(8,11

II,,); (B', II

I Iii),

II~),

where II 1111 (resp. II II~) is the norm induced by J!·(dx) (resp. I!~(dl}), p" and p~ are defined in (1.6».

A bounded linear transformation S from one normed space 8 to another can be completed in one and only one way, to a transformation having the same norm, from the completion of B into the completion of the range space of S. Thus, under the assumption of 1.4, S can be extended as a transformation from If·(dr) into IT; (dl}) with norm satisfying (1.4). The same remark is clearly valid for theorem 1.2. *1.5 As a first application of the Riesz-Thorin theorem we give Bochner's proof of M. Riesz' theorem III.1.8. We show that !!'(T) admits conjugation in the case that p is an even integer. It then follows by interpolation that the same is true for all p ~ 2. Using duality we can then obtain the result for all p > 1. Let f be a real- valued trigonometric polynomial and assume, for simplicity, 1(0) = O. As usual we denote the conjugate by f and put f" = t(f + if)· P is a Taylor polynomiaIt and its constant term is zero; the same is clearly true for (10) P, P being any positive integer. Consequently 1 2n (f°(tW dt = O.

f

Assume now that p is ev:!n, p = 2k, and consider the real part of the identity above; we obtain: t We use the term "Taylor polynomial" to designate trigonometric polynomials of the fonn r~aneinl.

98

An Introduction to Harmonic Analysis

~

2n

f

(J)2k

dt - (2k) ~ f (J)2k- 2r dt + (2k) ~ f (/)2k-Y4 dt 2 2n 4 2n Z

_ ... =0. By HOlder's inequality

hence

or, denoting we have

which implies that Y is bounded by a constant depending on k (i.e., on p). Thus the mapping i ~ J is bounded in the l!(T) norm for all polynomials i, and, since polynomials are dense in L"(T), the theorem follows. EXERCISES FOR SECfION I I. Prove inequality (1.5).

2. Let {an} be a sequence of numbers. Find min(L conditions L an l 2 = I, L an 14 = a.

I

I

Iani) under the

2. THE THEOREM OF HAUSDORFF-YOUNG

The theorem of Riesz-Thorin enables us to prove now a theorem that we stated without proof at the end of J.4 (theorem 1.4.7); it is known as the Hausdorff-Young theorem:

Theorem: Let 1 ~ p ~ 2 alld let q be the conjugate exponent, that is, q = p/(p-l). If fEI!(T) then L Il(n)lq < 00. More pre-

2.1

cisely

(L

Il(n)lq)l,q~

"i"V'. :F

The mappingf ~ {/(n)} is a transformation of functions ~n the measure space (T,dt) into functions on (Z,dll), Z being the group of integers and £ill the so-called counting measure, that is, the PROOf:

IV. Interpolation of Linear Operators

99

measure that places a unit mass at each integer. We know that the norm of the mapping as LI(T)! L""(Z) = t'" is 1 (1.1.4) and we know that it is an isometry of VeT) onto U(Z) = t2 (1.5.5). It follows from the Riesz-Thorin theorem that :F is a transformation of norm ~ 1 from J!(T) into I!(Z) = t q > which is precisely the statement of our theorem. We can add that sinC£ the exponentials are mapped with ~ no loss in norm, the norm of f! 1 on H(T) into t q is exactly 1. 2.2 Theorem: Let 1 ~ P ~ 2 and let q be the cOlljugate exponent. If {an} E t P thell there exists a fUllction fE JJ(T) such that an = 1(11) • Moreover, Ilflll,. ~ la n lp)l/P.

(2:

PROOF: Theorem 2.2 is the exact analog to 2.1 with the roles of the groups T and Z reversed. The proof is identical: if {an} E t 1 then f(t) = 2: anein , is continuous on T and l(n) = an' The case p = 2 is again given by theorem 1.5.5 and the case 1 < P < 2 is obtained by interpolation. ~

*2.3 We have already made the remark (end of ].4) that theorem 2.1 cannot be extended to the case p > 2 since there exist continuous functions f such that 2: 11(11) 12 -. = 00 for all & > O. An example of "" ein logn . such a function isf(t) = 2: --1/2 -. -)2 ern, (see [2H] , vol. I, p. 199); n=2 11 (ogll another example is get) = I m- 2r m/2fm(t) where fm are the RudinShapiro polynomials (see exercise 1.6.6, part c.) We can try to explain the phenomenon by a less explicit but more elementary construction. The first remark is that this, like many problems in analysis, is a problem of comparison of norms. It is sufficient, we claim, to show that, given p < 2, there exist functions g such that II g II 00 ~ 1 and I I§(II) IP is arbitrarily big. If we assume that, we may assume that our functions g are polynomials (replace g by (Jig) with sufficiently big n) and then, taking a sequence Pr-+2, gj satisfying II gill", ~ 1 and Ln",=_",\§/II)lpJ>2 J , we can writef= Lj=lr2eimj'gj(t) where the integers m j increase fast enough to ensure that eimj' git) and eimktgit) have no frequencies in common if j i:- k. The series defining f converges uniformly and for any p < 2 we have

100

An Introduction to Harmonic Analysis

One way to show the existence of the functions g above is to show that, given e > 0, there exist functions g sati:;fying (2.1)

IIgll",~l,

IldL2~L

I

I

sup g(lI) <e. n

In fact, if (2.1) is valid then

2:

Ig(n)lp~

eP-zL Ig(n)12>i ep - 2 ,

and if e can be chosen arbitrarily small, the corresponding g will have I 1g(n)IP arbitrarily large. Functions satisfying (2.1) are not hard to find; however, it is important to realize that when we need a function satisfying certain conditions, it may be easier to construct an example rather than look for one in our inventory. We therefore include a construction of functions satisfying (2.1). The key remark in the construction is simple yet very useful: if P is a trigonometric polynomial of degree N, f E IJ(T) and }.. > 2N is an integer, then the Fourier coefficients of ~t) = f(}..t)P(t) are either zero or have the form j(m)P(k). This follows from the identity 97(n) = I;'m+k=J(m)P(k) and the fact that there is at most one way to write n = Am + k with integers m, k such that 1 k 1 ~ N < A/2. Consider now any continuous function of modulus 1 on T, which is not an exponential (of the form ein .); for example t{!(t)=eicos t. Since $(n) 12 = 1 and the sum contains more than one term, it follows that suP. 1$(n) I = p < 1. Let M be an integer such that pM < E. Let 'I < 1 be such

II

that '1M> t. Let fji = O"N(t{!) , where the order N is high enough to ensure tI < 1
TIf=1

*2,4 We can use the polynomials satisfying (2.1) to show also that theorem 2.2 does not admit an extension to the case p> 2. fn fact, we can construct a trigonometric series I an e int which is 1Iot a FourierSlieltjes series, and such that I 1 anl P < ro for all p> 2. Let gj be a trigonometric polynomial satisfying (2.1) with e = 2 - i. Since now p> 2 we have

2:

1iifl)jP ~ e P - z

L

1gj(fl) 12 ~

r

j(p-2)

Ij

and consequently, for any choice of the integers mi , eimj'g/t) = Iane int does satisfy 1an Ip < ro for all p > 2. We now choose the integers

L

IV. Interpolation of Linear Operators

101

I1!J increasing very rapidly in order to well separate the blocks corresponding to j eimJ'g, in the series above. If we denote by N j the degree ofthe polynomial gj' we can take mjso that 3Nj > mj-1 + 3N j _ 1 • If anein'is the Fourier-StieItjes series of a measure Jl then

In, -

L

Jl

* eim,lyNJ = j eimjlgj

(YNJ being de la Vallee Poussin's kernel)

and consequently

which is impossible. We have thus proved

Theorem: (a) There exists a continuous fUllction f such that for all p < 2, l1(n)lp = roo (b) There exists a trigonometric series, I an e,n" which is not a p Fourier-Stieltjes series, such that anl < ro for all p > 2.

L

LI

*2.5 We finish this section with another construction: that of a set E of positive measure on T which carries no function with Fourier coefficients in t P for any p < 2. Such a set clearly must be totally disconnected and therefore carries no continuous functions. Its characteristic function, however, is a bounded function the Fourier coefficients of which belong to no t P , P < 2.

Theorem: There exists a compact set Eon T such that E has positive measure alld such that' the only functi01l f carried by E with I IJ(j) IP < ro for some p < 2, is f == 0 . First, we introduce the notation t

and prove:

Lemma: Let e> 0, 1 ~ p < 2. There exists a closed set E.,p c T havi1lg the following properties: (1) The measure of E•. p is > 21< - e. (2) Iff is carried byE, p then

11I1!st

OO

PROOF:

ellflls,p.

Let y> O. Put

epi t) = t

~

Notice that

y - 21< {

-1-y-

iI lis!' is the same

o< t < y y~ t

as II

IIA(T) •

~

mod 2IT

2IT mod 2IT .

102

An rntroduction to Harmonic Analysis

Then, by theorem 2.1 !..-I

I tpy lI,to ~ I tpyllu ~

(2.3)

1 1 where - + - = 1.

2rry

p

q

We notice that ~/O) = 0 so that, if we choose the integers )..1' A2 , ••• , AN increasing fast enough, every Fourier coefficient of L~ tp/AA is essentially a Fourier coefficient of one of the summands. It then follows that 1 N n q.!.-I 9 ' (2.4) tp,(Ai) II ~ 2N tpy II N

2:

I lI,t

Ft9

1

We take a large value for N and put y =

=

$(t)

1 N

e/ Nand

N

2:

tppjt).

1

Then, by (2.3) and (2.4), it follows that .!..- I !..-I .!.-I .!.-.!. q q N = 4rrE" N "

II $IIFto~ 4rrl

so that if N is large enough

I $11:r(0 <

n

E.

We can take

N

E.,p

= {t;(t) =

I}

=

{t; tppjt)

= I}.

1

Since tpy(..1) #- 1 on a set of measure y, it follows that

IE.,pl ~ 2rr - Ny

= 2rr -

Now if f is carried by E.,p' then for arbitrary n,

e.

'

](n) ;rrf e-intf(t)dt = ;nf e-inlf(t)$(t)dt. =

It follows from Parseval's formula that

and the proof of the lemma is complete.

-n,

Take en = 3 Pn = 2 - en and E= 1 ECn,Pn' The measure of E is clearly positive, and if f is carried by E and P < 00 for some P < 2, it follows that PROOf OF THE THEOREM:

n:=

or all

II

II! II§t

large enough; hence 1 = 0 and so f = O.

103

IV. Interpolation of Linear Operators EXERCISES FOR SECTION 2

1. Verify that r(11I+l)/2fm ([m as defined in exercise 1.6.6, part c) satisfy (2.1) when r(m+1)/2 < e. 2. Show that if N> e -1 and that if mn increases fast enough, then g, defined by: g(t) =eiNmntfor 2nn/N ~!~2n(n + I)/N, n=l, ... ,N sat isfies (2.1). 3. Let {all} be an even sequence of positive numbers. A closed set E c T is a set of type U(a,,) if the only distribution Jl carried by E and satisfying {1(n) = o(a,,) as I n I -+ (», is Jl = O. Show that if a" -+ 0 there exist sets E of positive measure which are of type U(a,,). Hint: For 0 < a < 1r we write (see exercise 1.6.3):

t..(t)

__ {01-a- 1I

t

l

III a

~

~

a

It I

;5

n.

We have t.. E A(T), II t.. I AU) = I, and ~a(O) = a/2n. Choose nj so that I n I> nJ implies aj < 1O-~; put E j = {t; t. 2 -)(2n/) = o} and 1 j E = E j • Notice that lEI;;;; 2n 2 - > O. If Jl is carried imr by E we have, for all m and j, < e t. 3-) (2nJ'), Jl > = 0 since t. 3-) (2n/) vanishes in a neighborhood of E. By Parseval's formula

L

ni"'=1

3- i - - -

.

0= <e,mtt.3._;(2ni),Jl)

=

-fi.(m)

21[

.......--..... -----j (k){1(m +2nj k)

+ I t. 3 k*O

if nj > I m I and if I{1(n)l ~ an we have for k :f- 0, I p(m + 2njk) I < 10- J, hence (3- i /2n) I fi.(m) I < lO- i . Letting j-+ (» we obtain p(m) = 0, and, m being arbitrary, Jl = O.

Chapter V

Lacunary Series and Quasi-analytic Classes

The theme of this chapter is that of 1.4, namely, the study of the ways in which properties of functions or of classes of functions are reflected by the Fourier series. We consider important special case~ of the following general problem: let A be a sequence of integers and B a homogeneous Banach space on T; denote by B" the closed subspace of B spanned by {i).I}A e" or, equivalently, the space of all IE B with Fourier series of the form e" a l eill • Describe the properties of functions in B" in terms of their Fourier serics (and A). An obvious example of the above is the case of a finite A in which ~1I the functions in B" are polynomials. If A is the sequence of nonnegative integers and B = J!(T), 1 ~ P < if.), then B" is the space of boundary values of functions in the corresponding HP. Tn the first section \\<e consider lacunary sequences A and show, for instance, that if A is lacunary a la Hadamard then (V(T»" = (V(T»" and every bounded function in (V(T»" has an absolutely convergent Fourier series. In the second section we prove the Denjoy-Carleman theorem on the quasi-analyticity of classes of infinitely differentiable functions and discuss briefly some related problems.

LA

l. LACUNARY SERIES

1.1 A scquence of positive integers Un} is said to be Hadamardlacunary, or simply lacunary, if there exists a constant q > 1 such that }'n+l > qAn for all n. A power series LanzA .. is lacunary if the 104

V. Lacunary Series and Quasi-analytic Classes

105

sequence P.} is, and a trigonometric series is lacunary if all the frequencies appearing in it have the form ± A. where Pn} is lacunary. The reason for mentioning Hadamard's name is his classical theorem stating that the circle of convergence of a lacunary power series is a natural boundary for the function given by the sum of the series within its domain of convergence. The general idea behind Hadamard's theorem and behind most of the results concerning lacunary series is that the sparsity of the exponents appearing in the series forces on it a certain homogeneity of behavior. 1.2 Lacunarity can be used technically in a number of ways. OUf first example is "local"; it illustrates how a Fourier coefficient that stands apart from the others is affected by the behavior of the function in a neighborhood of a point. Lemma: Let fE Ll(T) and assume that 1(j) = 0 for all j satisfying 1 ~ I no - j I ~ 2N. Assume that f(t) = O(t) as t -+ O. Then

(1.1)

IJ(no)1 ~ 2rr4(N-l

It- f(l) I + N-zllfIIL')' I

sup Irl
PROOf: We use the condition J(j) = 0 for j satisfying j ~ 2N as follows: let gN be any polynomial of degree 2N 1~ satisfying gN(O) = 1, then

Ino - I

KN IIi/K;, KN

We take gN = I membering inequality

I KNllzl =

(3.10)

t (1- )~llr

being Fejer's kernel of order N; reof chapter I and noticing that

>

~

we obtain

gN(t) ~ 2rr 4 N- 3 t- 4 •

We now write

The first integral is bounded by

106

An Introduction to Harmonic Analysis

The second integral is bounded by

The third integral is bounded by

Adding up the three estimates we obtain (Ll). Corollary: Let P.n} be a lacunary sequence and I'" I ancos A.nt be in V(T). Assume that I is differentiable alone point. Then an = O(A;I). PROOF: Assume that I is differentiable at t = o. Replacing it, if ne· cessary, by 1-1(0) cos t - 1'(0) sin t we can assume 1(0) = 1'(0) = o. It follows that I(t) = oCt) as t -? o. The lacunarity condition on {An} is equivalent to saying that there exists a positive constant c such that, for all n, none of the numbers j satisfying I ~ An - j ~ cA.n is in the sequence; hence !(j) = 0 for all such j. Applying the lemma with no = A.n and 2N = cA.n we obtain a.,. = 2!().n) = 0(..1.; 1). ~

I

Corollary: The differentiable.

Weierstrass lunction

I

I

2 -ncos 2 nt is nowhere

I I

The condition an = o().; 1) clearly implies that I an < ro. It is not hard to see (see Zygmund [28], chap. 2, §3, 4) thatf(t) = I an cos A•.t is then in LipaCT) for all Cf. < 1 and that it is differentiable on a set having the power of the continuum in every interval. Thus, for a lacunary series, differentiability at one point implies differentiability on an everywhere dense set. This is one example of the "certain homogeneity of behavior" mentioned earlier. We can obtain a more striking result if instead of differentiability we consider Lipschitz conditions. For instance, if 0 < Cf. < 1, a lacunary series that satisfies a Lip.. condition at a point satisfies the same condition everywhere (see exercise 1 at the end of the section).

1.3 Another typical use of the condition of lacunarity is through its arithmetical consequences. A useful remark is that if Aj+ 1 ~ qAJ

V. Lacunary Series and Quasi-analytic Classes

107

with q ~ 3, then every integer n has at most one representation of the '1jAj where '11 = -1,0,1. With this remark in mind we form n = consider products of the form

L

N

(1.2)

PN(t)

= 11 (1 + ajcos(A/ + rp) 1

the a /s being arbitrary complex numbers and rpj E T . The Fourier coefficients of a factor 1 + ajcos(Al + rpj) are: 1 for 1J = 0, tajei., for n = Aj' !aje-i'!'J for n = -Aj' and zero elsewhere. If we assume the lacunarity condition with q ~ 3, it follows that PN(II) = 0 unless n = L'1jAj, with '1j = -1,0,1, in which case PN(n) = IT~j*o!ajeiqj,!,J; in particular PN(O) = 1. If we compare the Fourier series of P N+ 1 to that of PN we see that PN+ 1 contains P N as a partial sum, and contains two more blocks: !aN+ lei'!'N+' ei )." + ,I PN and taN+l e-i'!'N+'e-iJ.N+.IPN. The frequencies appearing in the first block lie within the interval (AN+l- L~Aj' AN+l + I~A) c c (AN+l (q - 2)/(q - 1), AN+lq/(q - 1» and the second block is symmetric to the first with respect to the origin. No matter what coefficients a j we take, the (formal) infinite product P =

'" (1 + ajcos(Ajt + rpj» 11 1

can be expanded as a well-defined trigonometric series, and if the product converges in the weak-star topology of M(T) to a function f or a measure fJ, then the corresponding trigonometric series is the Fourier series of f (resp. fJ). We shall refer to the finite or infinite products described above as Riesz products. Two classes of Riesz products will be of special interest. 1. The coefficients aj are all real and aj ~ 1. In this case 1 + ajcos(Ajt + rp;) ~ 0 hence PN(t) ~ 0 for all N. It follows that PN = PN(O) = 1 and, taking a weak-star limit point, it follows that P is a positive measure of total mass 1 (i.e., that the trigonometric series formally corresponding to P is 'the Fourier-Stieltjes series of a positive measure of mass 1). 2. The coefficients aj are purely imaginary (in which case we shall write P = IT(1 + iajcos(Ajt + rp) with aj real) and satisfy \a j l 2 < 00. In this case 1 ~ + iajcos(Ajt + ~ 1 + a; and 1 ~ PH(t) ~ IT ~ (1 + a;) < 00. Since the PH are uniformly bounded

I I

I IlL'

I

I

I

11

rpj)1

108

An Introduction to Harmonic Analysis

we can pick a sequence N j such that P N , converge weakly to a bounded function P whose Fourier series is the formal expansion of P. 1.4 The usefulness of the Riesz products can be seen in the proof of

Lemma: Let f(t) = r.~N cjei ).}I with A_ j = -Aj, Al > 0 alld, for some q> 1, Aj + 1 > qA j , j = 1,2,,,.,N. Theil (1.3) and (1.4) where Aq alld Bq are cOllstants depending only on q. We remark first that it is sufficient to prove (1.3) and (1.4) in the case thatfis real valued (i.e., c_ j = cj ) since we can then apply them separately to the real and imaginary parts of arbitrary f, thereby at most doubling the constants Aq and Bq. Assume first that q ~ 3. In order to prove (L.3) we consider the Riesz product p(t) = n~ (I + cos (Ajt + Pj» where PJ is defined by We have pilL' = I and consequently the condition Cjei'PJ = 11/2rrJP(t)f(t)dtl~II!II(X). Since P(Aj ) = tei'P} we obtain from Parseval's formula: 1C; ei'PJ = ! Cj ~ and (1.3) follows with Aq = 2 for real-valued f and Aq = 4 in the general case. For the proof of (1.4) we consider a Riesz product of the second type. We remark that Il!lIi2=IlcjI2andifwetakeaj=lcjl'lIfll~! and Pj such that iCjei'Pj = cj then pet) = (I + ia j cos (A} + Pj» is uniformly bounded by + aJ )t~ etID; = et . By Parseval's formula PROOF:

I

Icjl.

I I I II! I

I

I I

n(1

(X)

,

n

which is (1.4) with Bq = 2et . Again if we put Bq = 4et then (1.4) is valid for complex-valued functions as well. If 1 < q < 3 and we try to repeat the proofs above, we face the difficulty that, having set the product P the way we did, we cannot assert that P(A) is 1- e''P' (or fia j ei'P' , in case 2) since Aj may happen to satisfy nontrivial relations of the form Aj = L tlk)'k with 11k = 0, ± 1. We can, however, construct the Riesz products for subsequences of Vj }. Let M = Mq be an integer large enough so that

V. Lacunary Series and Quasi-analytic Classes MIl q > 3, 1 - ---- > -, and qM - 1 q

(1.5)

1+

-M

q

109

t

-1 < q.

-

For k ~ m < M write ).~m) = Am+jM and notice that A~~l > qMA~m). By the remark concerning the frequencies appearing in a Riesz product it follows that all the frequencies n appearing in any product corres-

11111- Ajm) I < q-~): 1

ponding to {Ajm)} satisfy

for some j, hence

by (1.5) if k > 0, k '1= m (mod M), ± }ot does not appear as a frequency in a Riesz product constructed on {Ajm)}. It follows that if

pet) then 1 -2n

J -

= I1 (1 + am+iMcOS()·m+iM t + rpm+j,\f»

P(t)f(t)dt

'"

= ~"

1 -a .· (e'CPm+iMc m+JM . 2 m+JM

. + e-'CP,"+JM .) c -m-JM

and repeating the two constructions used above we obtain

~

(1.3')

(2

(1.4')

Icm+jMI ~ 4 111 I

'fJ

ICm +jMI2)t ~ 4etllfllLI.

Adding (1.3') and (1.4') for m = I, ... ,M we obtain (1.3) and (1.4) with Aq = 4Mq and Bq = 4e t M q. ~

L I I

Theorem: Let {AA be lacunary. (a) If cjei).Jt is the Fourier series of a bounded function, then L cj < ro. (b) If ciei).jt is a Fourier series, then Icj l2 < 00.

L

PROOF:

Writef(t) ~ L

Cj

L

ei).jt and apply (1.3) resp. (1.4) to

u.cr. t).

~

1.5 The role of the Riesz products in the proof of lemma 1.4 may become clearer if we consider the statements obtained from 1.4 by duality. For an arbitrary sequence of ¥1tegers A, we denote by C... the space of all continuous functions f on T such that j(lI) = 0 if n rf.A. CA is clearly a closed subspace of C(T). DEFINITION: A set of integers A is a Sidon set if every an absolutely convergent Fourier series.

fE

C... has

It follows from the closed-graph theorem that A is a Sidon set if, and only if, there exists a constant K such that (1.6) for every polynomial

;' Ij(n) I < K fE C ....

Ilf/l",

110

An Introduction to Harmonic Analysis

Lemma: A set (of integers) A is a Sidon set if, and only if,for every bounded sequence {d"h eA there exists a measure JL E M(T) such that P.(l) = d). for A EA. PROOF: Let A be a Sidon set and {d).} a bounded sequence on A. The mapping f ~ eAI(A) J;. is a well-defined linear functional on CA' By the Hahn-Banach theorem it can be extended to a functional on C(T), that is, a measure p. For this measure JL we have

L).

(1.7)

for all lEA.

Assume, on the other hand, that the interpolation (1.7) is always possible. LetfE CA and write d). = sgnj(A). Then, by Parseval's formula L II(A) I = L lu,) J). is summable to <J, JL) where JL is a measure which satisfies (1.7). Since for series with positive terms summability is equivalent to convergence, L II(A) 1 < 00 and the proof is complete. ~ The statement of part (a) of theorem 1.4 is that lacunary sequences are Sidon sets, and the Riesz product is simply an explicit construction of corresponding interpolating measures. 1.6 The statement of part (b) of theorem 1.4 is that for lacunary A, (V(T»A = (U(T)A' Every sequence {d).} such that eA 1d" 12 < 00 defines, as above, a linear functional on (U(T)h which, by 1.4, is a closed subspace of V(T). Remembering that the dual space of Ll(T) is LOO(T), we obtain, using the Hahn-Banach theorem, that there exists a bounded measurable function g such that

L).

(1.8) Here, again, Riesz products (of type 2) provide explicit construction of such functions g. One can actually prove the somewhat finer result:

L).

Theorem: Let A be lacunary and assume that eA 1d" 12 < 00. Then there exists a continuous function g such that (1.8) is valid. We refer the reader to exercise 6 for the proof.

EXERCISES FOR SECTION I 1. Let Pn} be lacunary and let I,...., L an cos )'nt. Assume that I satisfies a Lip" condition with 0 < (X < 1 at t = to. Show that an = O(A;") as n ~ <Xl. Deduce that IE Lip,,(T).

V. Lacunary Series and Quasi-analytic Classes

III

2. Let UII} be a sequence of integers and assume that for some 0< (X < I the following statement is true: if /(t) = 2: a" e;;."t satisfies a Lip" condition at one point, then jE Lip,,(T). Show that {A,,} is lacunary. Hint: If Un} is not lacunary it contains a subsequence {Jik} such that lim Jilt-II 1l2k = 1 and lim Ji2k+ II Ji2k = <x>. For an appropriate sequence Uk' /(/) = I:a,,(COSIl 2kt - cos Ji2k-lt) satisfies a Lip" condition at I = 0 but / ¢: Lip,,(T). 1 3. Let /eL (T), / '" 2: a"cosA"t with {A,,} lacunary. Assume /(t) = 0 for I t I ~ 11, 11 being a positive number. Show that / is infinitely differentiable. 4. Show directly, without the use of Riesz products, that if An+ 1 > 4A." and /(1) = L~ anCOSA"/, is real-valued, then sup IJ(/) I >

(2: I ani·

Hint: Consider the sets {t; an cos Ant> I an 1/2} . 5. If dn ~ 0 as n ~ <X> we can write d n = on"''' where {on} is bounded 1 and = ~(n) for some IJI EL (T). (See theorem 1.4.1 and exercise 1.4.1.) Deduce that if A is a lacunary sequence and d;. ~ 0 as I AI ~ <x> , there exists a function gEL l(T} such that

"'n

for AEA.

I

6. If 2: dn 12 < <X> there exist sequences {o,,} and {"'n} such that 2 d n = 0""',,, 2:,1 < <x> , and = v;(n) for some", EL1(T). Remembering that the convolution of a summable function with a bounded function is continuous, prove theorem 1.6. 7. Assume Aj+l/)'j ~ q > J. Show that there exists a number M = Mq such that every integer n has at most one representation of the form n = 2: l1/mj + jM where '1j = -1,0,1, and 1 ~ mj ~ M. Use this to show that the product TIi=l (1 + 2:~=1 dm+jMCOS(Am+jMt + 'l'm+jM» has (formally) the Fourier coefficient t dk e i , . at the point Ak' Show that if 0 < d" ~ II Mq for all k, then the product above is the Fourier Stieltjes series of a positive measure which interpolates {t dk ei'k} on{ Ak } • 8. Assume Aj+t!A.j ~ q> 1 and d j l 2 < <x>. Find a product analogous to that of exercise 7), which is the Fourier series of a bounded function, and which interpolates {dj } on pj}. *9. Show that the following condition is sufficieI1t to imply that the sequence A is a Sidon set: to every sequence {d.} such that d. = I there exists a measure Ji E M(T) such that

onl

"'n

LI

I I

AeA.

112

An Introduction to Harmonic Analysis

*10. Show that a finite union of lacunary sequences is a Sidon set. *2. QUASI-ANALYTIC CLASSES

2.1

We consider classes of infinitely differentiable functions on T. Let {Mn} be a sequence of positive numbers; we denote by C*{Mn} the class of all infinitely differentiable functions j on T such that for an appropriate R > 0 (2.1)

n = 1,2, ....

We shall denote by C# {M.} the class of infinitely differentiable functions on T satisfying: (2.2)

n

=

1,2, .. ,

for some R (depending on f). The inclusion C*{Mn} £; C#{Mn} is obvious; on the other hand, since the mean value of derivatives on T is zero, we obtain SUPt eT/J(n)(t) 1 ;;;; IIf(n+l>IIL2 and consequently C# {Mn} £; C*{M"+I'} Thus the two classes are fairly close to each other. Examples: If Mn = 1 for all n, then C#{Mn} is precisely the class of all trigonometric polynomials on T. If Mn = n 1, C# {Mn} is precisely the class of all functions analytic on T. (See exercise 1.4.3.) We recall that a sequence {cn } , Cn > 0, is log convex if the sequence {log cn } is a convex function of n. This amounts to saying that, given k < I < m in the range of n, we have (2.3)

log c1

;;;;

m-I --k log Ck

m-

I-k

+ -m--k

log Cm

or equivalently (2.4) 2.2 The identity IIf(n>IIL2 =(2: 11(j)12/n)t allows an expression of condition (2.2) directly in terms of the Fourier coefficients of f. Also it implies

Lemma: Let f be N times differentiable on T. Then the sequence {IIf(n) IIu} is monotone increasing and log cOl/vex for 1 ;;;; n;;;; N.

IIfn)IILl (L

The fact that = IJ(j)12/n)~ is monotone increasing is obvious. In order to prove (2.4) we write p = (m -k)/(m -I), q = (m-k)j(l-k); then lip + lIq = 1 and by Holder's inequality PROOF:

V. Lacunary Series and Quasi-analytic Classes

L 1a j 12j21 = L (I a

j

12/p/

k P /

)(1 a jI2, q/m, q) ~

113

IIf(k)IIi'! IIf m)W q

which is exactly (2.4). rt follows from lemma 2.2 (cf. exercise 2 at the end of this section) that for every sequence {M~} there exists a sequence {Mil} which is monotone increasing and log convex such that C#{Mn} = C#{M~}. Thus, when studying classes C# {M.} we may assume without loss of generality that {Mn} is monotone increasing and log convex; throughout the rest of this section we always assume that, for k < I < m, (2.5)

2.3 For a (monotone increasing and log convex) sequence {Mn} we define the associated function T(r) by (2.6)

T(r) = inf Mnr-n n~O

We considersequencesM. which increase fasterthatRnfor all R>O; hence the infimum in (2.6) is attained and we can write T(r) = minn~oMnr-n. If we write 111 = M~ I, and 11. = M._ tlMn for /l > 1; then 11ft is a monotone-decreasing sequence since by (2.5), 11.+dl1n = M;/Mn_IMn+ 1 < 1; we have M"r -n = (11 jr) - 1 and consequently

n;

(2.6')

T(r) =

11

(l1jr)-l.

Ili'> 1

The function T(r) was implicitly introduced in 1.4; thus it follows from f.4.4 that if fE C# {Mn} then, for the appropriate R > 0

11u)1 ~ ,(jR-I), and exercise 1.4.6 is essentially an estimate for oCr) in the case Mn = /lIZ•• 2.4 An analytic function on T is completely determined by its Taylor expansion around any point to E T, that is, by the sequence {J(n)(to)}:'=o' In particular if j<')(t o) = 0, n = 0,1,2, ... , it follows that f = 0 identically. A class of infinitely differentiable functions on T is quasi-analytic if the only function in the class, which vanishes with all its derivatives at some 10 E T, is the function which vanishes identically. DEFINITION:

The main result of this section is the so-called Denjoy-Carleman

114

An Introduction to Harmonic Analysis

theorem which gives a necessary and sufficient conditions for the quasi-analyticity of classes C# {Mn } .

Theorem: Let {Mn} be monotone increasing and log cOl/vex. Let r(r) be the associated function (2.6). The following three conditions are equivalent: (i)

C# {M,,} is quasi-analytic

(ii)

J

roo logr(r) dr

= _ 00

1 + r2

I

(iii) The proof will consist in establishing the three implications (ii) => (i) (theorem 2.6 below), (i) => (iii) (theorem 2.8), and (iii) => (ii) (lemma 2.9). We begin with: 2.5 Lemma: Let cp{z) =f 0 be holomorphic alld bounded in the half plane Re(z) > 0 and continuous on Re(z) ~ O. Then

d Jo(" log 1cp(± iy) I f::;..yy2 > The function F(O =

PROOf:

cp(~ ~

-

00.

DiS holomorphicand bounded

,=

in the unit disc D (and is continuous on l> except possibly at I). By IIl.3.14 we have F(ej~1 dt > - 00. The change of variables that we have introduced gives for the boundaries i' = (iy - l)/(iy + 1) or t = 2 arc cot y.

J;log I

Consequently dt =

Io

">

-2 t + y2 dy and

log 1tp(iy) 1

similarly jo"'log!lf( -iy)1

d' _·J·z

I+y

=

-t_ dY_ 2 > +y

IX

.

1 log 1F(e") 1dt > - 00; 20

-

00

and the lemma is proved.

...

Theorem: A sufficient condition for the qllasi-al/alyticity or en logr(r) . the class c# {'I'fn } i~ that -1--- ' dr = - 00 where r{r) IS defined , + r2

2.6

uy

(2.6).

f

v. PROOF:

Lacunary Series and Quasi-analytic Classes

Let fE

ell' {M.}

and assume that fn)(O) =

°

lIS

n = 0,1, ....

Define p(z)

1

fh

= 2n Jo

Integrating by parts we obtain, z p(z) =

-.!.

27t

e-z'l"(t) dt.

*0

[-=-!z e-z'l'(t)] 2" + 2nz 1_ 0

°

fo2l< e-Z'!'(t)dt

Jc

*

and since 1(0) = 1(27t} = the first term vanishes for all z 0 (we have used the same integration by parts in 1.4; there we did not assume 1(0) = but considered only the case z = irn, that is, e-ztl is 2x-periodic). Repeating the integration by parts n times (using/(j)(O) = fj)(27t)

°

=

°

for j ~ n), we obtain

1 2

p(z) =

1 2nz.

0 "

e-zy
For Re(z)~O, le-x'i ~ Ion (0,2n) and consequently for n = 0, I, ... hence /p(z)1 ~ -c(lzl) or logl cp(z) 1 < 10gl-c(lzl)·

It follows that

Y 2 jnog 1cp(iy) 1-1_3+y

= -

00

and

by

lemma

p(z)=O. Since q;(in)=j(lI) it follows that/=O.

2.5 <4

Then f(t) = L~w CP(k)ikl

_sin!!J_k J1. k· j is carried by [-1,1] (mod2n), it is in-

finitely diJferentiable and

11/(0)11 L2 ~ 2

2.7

Lemma:

Assume J1.j> 0, I~ J1.j ~ 1. Write CP(k) =

a)

rI 0

n~J1.j-t.

PROOF: All the factors in the product defining P(k) are bounded by 1 so that the product either converges or diverges to zero (actually it converges for all k) and rp(k) is well defined. q{k) clearly tends to zero faster than any power of k so that the series defining converges

r

116

An Introduction to Harmonic Analysis

uniformly and f is infinitely differentiable. We have p(O)

= 1 so

that

f¢O. The sequence {Sin,JLkjk}<Xl is the sequence of Fourier coefJL) k=-<Xl l

1t 1 < J1.j ; elsewhere

ficients of the function f .(t) = {nJLl ) 0

if we write

N --k-,we sinJLjk h r() ~ (k ikl =fo*fl*···*f (k PN)= aveJNt=i..JPN)e N o ~ -00 and the support of fN is equal to [ - L~ JL j, L~JL j] (mod 2n). Since

n

fN converges uniformly to f, the support of f

is contained

in

[ - L;JLj, L; JLj] (mod 2n). Finally, since 11/(n> IIZ2 = L 1P(k) 12 k2n n l 1 P(k) I ~ n~ (JLjkfl = (n~ JLj)-lk- - , we obtain IIf(n>/IL2

and

~ (n~ JLj) -1(Lk"'O k -2)t and the proof is complete.

2.8

<4

Theorem: A necessary condition for the quasi-analyticity of

of the class C# {M,,} is that L MM"

= 00.

,,+1 PROOF:

Assume that

L M-'!~ <

Without loss of generality we

00.

n+1

may assume

L MM~ < !

(replacing Mn by M: = MnR n does

not

n+t

change the class C# {M,,}

while

L :~ =

R- 1

"+1

nnoJLj

l M j _ l J=. Thcn . > 2 JLO=JL.=4, JLj=-M;'

-I

L AlMn

). Write

n+.

M" = 16)Jli 1

an d t h e

function f defined by lemma 2.7 has a zero of infinite order (actually vanishes outside of[ -1,1]), is not identically zero, andlE C# {Mn } . <4

2.9

Lemma: Under the assumption of theorem 2.4 we have

PROOF:

'" . JLn = M,,_I ' A s be lore we wnte -x,r-' W e d efi ne t h e countmg n

function vll(r) of {JL,,} by:

vll(r)

= the

number of elements JLj such that JLjr ~ e,

and recall that -r(r) = fIIlJ'> 1 (JLjr)-I; hence - logr(r) =

L JIlT> 1

10g(J1.f)~

.L pjr~e

log (JLjr) ~ ..#(r).

V. Lacunary Series and Quasi-analytic Classes Thus for k = 2,3, ... (2.7)

I

eHI

ek

k

:-.I
i

e '"

dr >

~. Jt(l); ek

2 e2

ek

117

on the other hand,

L

(2.8) e 1 - Ic

JLj

~ (J!(l) -

J!(e k - l »e2 -

k

~ e J~~'1 2


and the theorem follows by summing (2.7) and (2.8) with respect to k, k = 2,3, ....

Remark: Theorems 2.6, 2.8, and lemma 2.9 together prove theorem 2.4. We see in particular that if C# {Mn} is not quasi-analytic, it contains functions (which are not identically zero) having arbitrarily small supports. For further reading, generalizations, and related topics we mention

[17]. EXERCISES FOR SECTION 2

c;

I. Show that {C'n} is log convex if, and only if, ;? Cn-l Cn + 1 for all n. 2. (a) Let {c:},;';.J be a log-convex sequence for all IX belonging to some index set I. Assume that Mn -= sUPa cl c~ < ex) for all n. Prove that {Mn} is log convex. (b) Let {M;} be a sequence of positive numbers. Let {c:} be the family of all log-convex sequences satisfying c: ~ M~ for all n. Put Mn =sup~c:. Then C#{Mn} =C#{M~}. 3. Let M j -;;; j! for infinitely many values of j. Show that C*{ Mn} and C#{Mn} are quasi-analytic. Hint: Assuming IE C*{Mn} and j(k'(O) ~ 0 for all k, use Taylor's expansion with remainder to show

1==0.

{II

I

4. We say that alul/ction rp E Coo(T) is quasi-analytic zf C# 'P(") L2} is quasi-analytic. Let IE Coo(T); show that if the sequence Pj} increases fast enough and if we set / 1(1)

=

2:

~

j

A2j
/(k)/kr,

then both II and 12 ~ I-II are quasi-analytic. Thus, every infinitely differentiable function is the sum of two quasi-analytic functions.

An Introduction to Harmonic Analysis

118

5. Show that C# {n! (log n)"n} is quasi-analytic if 0 ;£; IX ;£; 1. and is nonquasi-analytic if IX> 1. 6. Let r(r) be the function associated with a sequence {Mn}. (a) Show that (r(r» -1 is log-convex function of r. (b) Show that Mil = max, r" r(r).

7. Let {wlI } be a log-convex sequence, n=O,I, .... Wn ~ I, and let A {wn }

I I

be the space of all IE C(T) such that f {ron} that with the norm

I 1I{"n}'

=

L li(n) IWlnl < co. Show

A{Wn } is a Banach space. Show that a

necessary and sufficient condition for A {wlI } to contain functions with arbitrarily small support

. "co -log2 - < co. Wn

IS L.,,1

n

8. Let {w n} be log convex, wn ;::::: 1 and assume that con --. than any power of n. The sequences

CO

faster

{~~}OO

tend to zero as wlnl 11=-00 n co. Show that the subspace that Uk' k = 0, 1, '" generate in Co (the space of sequences tending to zero at 0') is uniformly dense Uk =

1 1--'

in Co if, and only if,

t

1.

If {an}

E

t

1

L

10g;an

n

is orthogonal to

Uk'

~ k

oc>. =

Hint: The dual space of Co is

0, ... , the function f(t) =

L an e int, COn

which clearly belongs to A{ con}. has a 7ero of infinite order. 9. Let I be as in exercise 1.3. Show that f

=

°

identically.

Chapter VI

Fourier Transforms on the Line

In the preceding chapters we studied objects (functions, measures, and so on) defined on T. Our aim in this chapter is to extend the study to objects defined on the real line R. Much of the theory, especially the V theory, extends almost verbatim and with only trivial modifications of the proofs; such results, analogous in statement and in proof to theorems that we have proved for T, often are stated without a proof. The difference between the circle and the line becomes more obvious when we try to see what happens for I! with p > 1. The (Lebesgue) measure of R being infinite entails that, unlike VeT) which contains most of the Unatural" function spaces on T, Ll(R) is relatively small; in particular H(R) $V(R) for p> 1. The definition of Fourier transforms in VCR) has now a much more special character and a new definition (i.e., an extension of the definition) is needed for LJ(R) , p > 1 . The situation turns out to be quite different for p ~ 2 and for p > 2. If P ~ 2, Fourier transforms of functions in H(R) can be defined by continuity as functions in U(R), q = pl(p - I); however, jf p > 2, the only reasonable way to define the Fourier transform on H(R) is through duality and Fourier transforms are now defined as distributions. The plan of this chapter is as follows: in section 1 we define the Fourier transform in Ll(R) and discuss its elementary properties. We also mention the connection between Fourier transforms and Fourier coefficients and prove Poisson's formula. In section 2 we define Fourier-Stieltjes transforms and obtain various characterizations of Fourier-Stieltjes transforms of arbitrary and positive measures. In section 3 we prove PlanchereI's 119

120

An Introduction to Harmonic Analysis

theorem and the Hausdorff-Young inequality, thus defining Fourier transforms in H(R) , 1 < p ~ 2. In section 4 we use Parseval's formula, that is, duality, to define the Fourier transforms of tempered distributions and study some of the properties of Fourier transforms of functions in 1!'(R) , p ~ co. Sections 5 and 6 deal with spectral analysis and synthesis in L%(R). In section 5 we consider the problems relative to the norm topology and show that the class of functions for which we have satisfactory theory is precisely that of Bohr's almost periodic functions. In section 6 we study the analogous problems for the weak-star topology. Sections 7 and 8 are devoted to relations between Fourier transforms and analytic functions. Finally, section 9 contains Kronecker's theorem (which we have already used in chapter II) and some variations on the same theme. 1. F9URIER TRANSFORMS FOR L 1(R)

1.1 We denote by OCR) the space of Lebesgue integrable functions on the real line. For fE L1(R) we write

/If/lU(R) = r:lf(X)ldX, and when there is no risk of confusion, we write

/If/lL'

or simply

II f /I instead of /I f /I L'(R) •

The Fourier transform J off is defined by

(Ll)

for all real ~. t

This definition is analogous to L(t .5), and the disappearance of the factor 1/2n is due to none other than our (arbitrary) choice to remove it. It was a natural normalizing factor for the Lebesgue measure on T; but, at this point, it seems arbitrary for R. The factor 1/2n will reappear in the inversion formula and some authors, seeking more symmetry for the inversion formula, write 11.j2n in front of the integral (1.1) so that the same factor appear in the Fourier transform and its inverse. The added symmetry, however, may increase the possibility of confusion between the domains of definition of a function and its transform. In VeT) the functions are defined on T whereas

t Throughout this chapter, integrals with unspecified limits of integration are always to be taken over the entire real line.

VI. Fourier Transforms on the Line

121

the Fourier transforms are defined on the integers; in D(R) the functions are defined on R and the domain of definition of the Fourier transforms is again the real line. It may be helpful to consider two copies of the real line: one is R and the other, which will serve as the domain of definition of Fourier transforms of functions in D(R) , we denote by R. This notation is in accordance with that of chapter VII. Most of the elementary properties of Fourier coefficients are valid for Fourier transforms.

Theorem:

Let J,gED(R). Then

(t+'g)(~)

(a) (b)

=

J(~) + g(~)

For any complex number IX /'.

(IXf)(e) = rxl(e) (c)

If j i5 the complex conjugate off, then

Iw = 1(-~) (d)

Denote J,(x)

(e)

(f)

=

f(x - y), y E R. Theil Jy{~)

= l(~) e -i{y

11ml ~f

If(x)ldx =

Ilfll

For reai-!1alued A, A>O, deflote
=

Jf(AX);

then

PROOF: The theorem follows immediately from (Ll). Parts (a) through (e) are analogous to the corresponding parts of r. 1.4. Part (f) is obtained by a change of variable y = AX:


=I f(Ax)e-i(~IA»).xdAX =I f(y)e-i({1i.1Tdy =1(1).

Theorem:

-co<~
PROOf:

Let fE V(R). Then

1 is


uniformly continuous in

An Introduction to Harmonic Analysis

122 hence

The integral on the right of (1.2) is independent of ~, the integrand is bounded by 21/(x) and tends to zero everywhere as o.
,,-+

I

1.3 The following are immediate adaptations of the corresponding theorems in chapter I. Theorem: Let J,gEL1(R). For almost all x, f(x - y)g(y) is integrable (as a function 01 y) and, il we write hex) = f l(x - y) g(y) dy , then hE VCR) and

II h II ~

11I1111 gil;

moreover,

for all As in chapter I we denote h

= 1* g,

~.

call h the convolution of I and

g, and notice that the convolution operation is commutative, asso-

ciative, and distributive. 1.4

Let l, hE VCR) and

Theorem:

hex) with integrable

H(~).

PROOF The function Fubini's theorem,

I H(~) i~x d~

Then

(h *I)(x) =

(1.3)

I

= 2n:

2~

I HmJWeiX~ d~

H(~)l(y)

is integrable in

1 (h*/)(x) = f h(x-y)f(y)dy =2n: =

~

2n:

. (~,

y), hence, by

II H(~)i~Xe-i~YI(y)d~dy

f H(~)ei~X f e-i~YI(y)dyd~

=

~ fHmJ(~)elt;Xd~. 2n:

VI. Fourier Transforms on the Line 1.5

Theorem:

123

Let fEVeR) and define F(x) = J:oof(y)d y .

Then,

if F E VCR)

we have

1m =

(1.4)

all real ¢ ~ O.

:¢J(¢)

An equivalent statement of the theorem is: if F, F' E IJ (R), then

Fm = i¢/(¢) . 1.6 Theorem: Let fE VCR) and xf(x) ferentiable and

E

VCR). Then J is dif-

d ----------deJW = (-ixf)(l;,).

(1.5) PROOF:

(1.6)

J(l;,

+ h~ -/(l;) =

J

f(x)e-i{X

e-

I

ihX h-1)dX.

I

The integrand in (1.6) is bounded by xf(x) (which is in V(R) by assumption) and tends to - ix/ex) e -i~x pointwise, hence (Lebesgue) it converges to -ixf(x)e-i{x in the VCR) norm. This implies that as

-+

h 0 the right-hand side of (1.6) converges to (-iif)(l;) and the theorem follows.
Theorem (Riemann-Lebesgue lemma):

lim

l(l;)

For fE VCR)

= o.

(~( .... oo

PROOF: If g is continuously differentiable and with compact support we have, by 1.5 and 1.1, Il;,g(e) ~ g' IIL'(R); hence lim({( .... "" gW O. For arbitrary IE V(R), let I: > 0 and g be a continuously differentiable, V(R) < 1:. We have compactly supported function such that IJ(l;)-g(e)1 O, we obtain

I I

ooIJ(l;,)I
lim({( ....

""Jm=O.

I¢I-+

I

1=

II! - gil I I


1.8 We denote by A(R) the space of all functions rp on R, which are the Fourier transforms of functions in V(R). By the results above, A(R.) is an algebra of continuous functions vanishing at infinity, that is, a subalgebra of Co(R) , the algebra of all continuous functions on :it which vanish at infinity. We introduce a norm to A(R) by transferring to it the norm of U(R), that is, we write

An Introduction to Harmonic Analysis

124

IIlIIA(R)

Ilfllt.I(R)' It follows from 1.3 that the norm I IL~(R) is multiplicative, that is, satisfies the inequality: Ii ~I f{J2IL~(R) ~ I f{JI IIA(It) I f{J211 A(R)' The norm I IIA(R) is not equivalent to tAhe supremum norm; consequently, A(R) =

is a proper subalgebra of Co(R).

1.9 A summability kernel on the real line is a family of continuous functions {k).} on R, with either discrete or continuous parameter t A. satisfying the following:

f

k;.(x) dx

=

1

I klIlLI(R) = 0(1)

(1.7)

lim ). .... 00

as A. .....

Ikl(x) Idx =

f

),.<,>6

OC!

0, for all b > O.

A common way to produce summability kernels on R is to take a function fE VCR) such that f(x)dx = 1 and to write k;.(x) = if(AX) for A> O. Condition (1.7) is satisfied since, introducing the change of variable y = AX, we obtain

f

f

f

k;.(x)dx =

Ilk).11

=

Ik).(x)ldx

=

fI

f(y)dy

=

I

k;.Cx) dx =

f

1 If(y)ldy =

Ilfll

and

f

)''<'>0

r

J,Y,>)'J

If(y) Idy ..... 0

as A .....

OC! •

The Fejer kernel on R is defined by

K;.(x) = AK(lx),

A>O,

where (1.8)

K(x) =

J.. (~!n X/~)2 2n

x/2

=

-.!..

fl

21t_l

(1

_I eI)

ei{x

de.

The second equality in (1.8) is obtained directly by integration. By the previous remark it is clear that the only thing that we have to check, in

t The indexing parameter). is often continuous, that is, real valued; however, it should not be considered as an element of R so that no confusion with the notation of 1.1.d should arise.

VI. Fourier Transforms on the Line

125

order to establish that {Kl} is a summability kernel, is that This can be done directly, for example, by contour integration, or using the information that we have about the Fejer kernel of the circle, that is, that for all 0 < ~ < 7T

f K(x) dx = 1.

(1.9)

lim

-.!...

n-+oo

2n

Since f K(x)dx

K;.(x)

+

-0

1)

(~ + I)XI2)2 dx =

_1_ (Sin

n

+1

= f K;.(x)dx,

1 21T(n

JO

1.

smxl2

we may take I.

(Sin (n + 1)XI2)2 --- ---- ---

xl2

'

= /I + 1,

in which case

. that If . (j> 0 IS . small and notice

enough, the ratio of 2nK;.(x) to the' integrand in (1.9) is arbitrarily (j. More precisely, we obtain (.l. = n + L): close to one in x

I 1<

-.! __ (~~(n.+ l)X/2)2 dx ( ~in15 ~)22n~ JO_6n+1 SInxl2

<

JO KAx)dx -6

I"

< -.!... _1_ (Sin (n. + I)X/~)2 dx. 2n _" n + 1 smxl2 Letting n-+oo we see that fK(x)dx=lim'hXlf~6Kix)dx is a number between sin 215115 2 and I; since (j > 0 is arbitrary f K(x) dx = 1 .

1.10 Theorem: on R, then

Let fE Ll(R) and let {k.. } be a slImmability kernel

lim A-+ ':0

PROOF:

IIf- k.. *fllL'\R) = O.

Repeat the proof of theorem 1.2.3 and lemma 1.2.4.

...

1.11 Specifying theorem 1.10 to the Fejer kernel and using theorem lA, we obtain

Theorem:

Let fEV(R), then

(1.10) in the V(R) norm.

Corollary (the uniqueness theorem): all ~ER; then f= O.

1m = 0 for

Let fE Ll(R) and assume

126

An Introduction to Harmonic Analysis

1.12 If it happens that 1 is Lebesgue integrable, the integral on the right-hand side of (1.10) converges, uniformly in x, to 1/2rrflee) ei~x de. We see that f is equivalent to a uniformly continuous function and obtain the so-called "inversion formula": (1.11) An immediate consequence of (1.11) is

K:w = max ( 1 - I; I, 0)

(1.12)

and, by theorem 1.3,

(Ll 3)

Combining this with theorem 1. t 0, we obtain

Theorem: The functiolls with compactly carried Fourier transforms form a dense subspace of D(R). This theorem is analogous to the statement that trigonometric polynomials form a dense subspace of V(T). 1.13 Besides the Fejer kernel we mention the following: De la Vallee Poussin's kernel (1.14)

whose Fourier transform is given by I, ( LlS)

V;.(fJ

2-

=

{ 0,

~J,

lei ~A ). ~ Iel ~ 21 21 ~ IeI·

Poisson's kernel

Pix) = AP(AX) , where (1.16)

127

VI. Fourier Transforms on the Line and (1.17) And finally Gauss' kernel

G;.(X)

=

lG(AX) ,

where (1.18) and (1.19) To the inversion formula (1.11) and the summability in norm (theorems 1.10 and 1.11), one should add results about pointwise summability. Both the statements and the proofs of section 1.3 can be adapted to V(R) almost verbatim and we avoid the repetition. 1.14 As in chapter I, we can replace the VCR) norm, in the statement of theorems 1.10 and 1.11, by the norm of any homogeneous Banach space Be VCR). As in chapter I, a homogeneous Banach space is a space of functions which is invariant under translation and such that for every IE B, /y (defined by fix) = f(x depends continuously on y. The assumption Be LI(R) is more restrictive than was the assumption B c VeT) in chapter I; it excludes such natural spaces as l!'(R), p> I. We can obtain a reasonably general theory by considering homogeneous Banach space of locally summable functions, that is, functions which are Lebesgue integrable on every finite interval. We denote by .rt> the space of all measurable functions f on R such that

y»

Ilfll£" = sup >'

>'+ 1

i

I

Ii(x) dx <

00

Y

and by .!l'c the subspace of !I! consisting of all the functions satisfy as y~O.

f

which

Theorem: If B is a homogeneous Banach space of locally summable functions on R and if convergence in B implies convergence ill measure, then the ,:t> /lorm i~ majorized by the B norm and, in particular, Bc.!l'c·

128

An Introduction to Harmonic Analysis

I

If the 2' norm is not majorized by liB, we can choose n < rn and 3 • Replacing a sequence f" E B such that fn by Yn) (if necessary) we may assume dx > 3". Since In liB --+ 0, f" converges to zero in measure and it follows that if 11 j --+ co fast enough which belongs to B, is not integrable on (0,1).
I

11f"IIB

f,,(x -

11f,,11:e>

nIf,,(x) I

L f"j'

We can now extend theorem 1.10 to homogeneous Banach spaces of locally summable functions (see exercises 11-14 at the end of this section); theorem 1.11 can be generalized only after we extend the definition of the Fourier transfonnation. 1.15 We finish this section with a remark concerning the relation between Fourier coefficients and Fourier transforms. Let f E lJ(R) and define qJ by

qJ(t)

= 2n

i

f(t

+ 2nj).

j=-oo

t is here a real number, but it is clear that qJ(t) depends only on t mod 2n so that we can consider qJ as defined on T. qJ is clearly summable on T and we have

If n is a rational integer, we have

'" qJ(n)

=

1 2n

f2" -int Jo qJ(t)e dt = =

Jor

~,

h

j~OO

f

f(t

+ 2nj)e- i"t dt

I(x) e -inx dx

= j(n),

q;

so that is simply the restriction to the integers of j. Similarly, if we write f;.(x) = Af(.tx) and: (1.20)

qJit)

= 2n j~OO

/;.
+ 2nj),

we obtain, using 1.1, (1.21)

~(n) =

j(;).

The preceding remarks, as simple as they sound, link the theory of Fourier integrals to that of Fourier series, and we can obtain a

VI. Fourier Transforms on the Line

129

great many facts about Fourier integrals from the corresponding facts about Fourier series. (For examples, see exercises 5 and 6 at the end of this section.) An application to the above procedure is the very important formula of Poisson: (1.22)

In order to establish Poisson's formula, to under~tand its meaning and its domain of validity, all that we need to do is simply rewrite it as 0()

(1.23)

rp;.(O)

=

2

fj;,(n).

11= -00

If rpiO), as defined by (1.20), is well defined and if the Fourier series of rp;. converges to rp;.(O) for t = 0, then (1.23) and (1.22) are valid. One enhances the generality of (1.22) considerably by interpreting the sum on the right as lim

i (1 - I~I)J( ~),

N .... oo -N

that is, using C-l summability instead of summation. Using Fejer's theorem, for instance, one obtains that, with this interpretation, (1.22) is valid if t = 0 is a point of continuity of rp;.. We remark that the continuity off and is not sufficient to imply (1.22) even if both sides of (1.22) converge absolutely (see exercise 15).

J

EXERCISES FOR SECTION I 1. Perform the integration in (1.8). 2 1 2. Prove that 2n ) dx = I by contour integration.

J(~i:;;2

3. Prove (1.17). Hint: Use contour integration. 4. Prove (1.19). Hint: Use 1.5 and 1.6 to show that G(~") satisfies the equation d/dl;GW = -(';/2)G(.;). 5. Let jEL 1 (R) and let 'P;.(t) be defined as in (1.20). Show that Ii m).-+ to II 'P).IIL'(T) = II j IIL1(R); hence deduce the uniqueness theorem from (1.21). 6. Prove theorem 1.7 using (1.21), the uniform continuity of Fourier transforms and the Riemann-Lebesgue lemma for Fourier coefficients.

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An Introduction to Harmonic Analysis

7. Show that A(R) contains every twice continuously differentiable function with compact support on R. Deduce that A(R) is uniformly dense in Co(R); however, show that A(R) ¥: Co(R). 8. Let Ie L l(R) be continuous at x = 0 and assume that j(l;) ~ o. eER. Show thatjELl(lt) and/(0)=I/211:fjmde. Hint: Use the analog to Fejer's theorem and the fact that for positive functions C-l summabiIity is equivalent to convergence. 9. Show that CorlL I(R). with the norm II III = sUPx I/(x) I +11 I IILleR)' is a homogeneous Banach space onR and conclude that if Ie Co rlLI(R) then I(x) = lim;.; 00 I/21C f~;. (1- Ie I/ ..t) jm i~x de uniformly. 10. Let I be bounded and continuous on R and let {k;.} be a summability kernel. Show that k;.. I = k;.(x - y)/(y) dy converges to I uniformly on compact sets on R. II. Let Ie .!l'c and let 'P be continuous with compact support; write

f

rp·1

=

J


Interpreting the integral above as an .!l'c-valued integral. show that 'P • le.!l'c and II 'P • I 11.2' ~ II 'P Ib(R) III 11.2" Use this to define g. I for geLl(R) and Ie !l'c. 12. Show that if IE.Pc then III e.Pc (notice. however. that exp (ix log I x D¢ .!l'c) and. using exercise II. prove that if IE .!l'c and geLl(R) then for almost all xeR, g(y)/(X-Y)EL 1 (R) and g*l. as defined in exercise II, is equal to g(y)/(x - y) dy. 13. Let le.!l' and let gELl(R). Prove that for almost all xeR g(y)/(x - y)eLl(R) and that h(x) = g(y)/(x - y)dy satisfies

f

f

I h I~ ~ II g IILI(R)II I

I~ . 14. Let {k;.} be a summability kernel in Ll(R) and let B <;;.Pc be a homogeneous Banach space. Show that for every IE B, II k;. *1-/118--+ 0, and conclude that if leBrlLl(R),J=lim;. ... oof~;.(1-I.;I/A)j\e)ei~X de; in the B norm.

15. Con.,truct a continuous function IELl(R) such that jeLI(R). 1(211:n) = 0 for all integers n, j(O) = 1 and j(n) ~- 0 for all integers II ¥: O. Hints: (a) We denote :I I i!A(R) = 1/21C f'~oo I j WI de. Let g be continuous \\ilh support in [0.2n] and such that geLleR). Write gN(X)

=

-

I

--

L~

N+lj~_N

UI)

.

( 1 - ------ g(x - 211J). NTl

131

VI. Fourier Transforms on the Line

Show that gNW = (N + 1)-1 KN(e)gW where KN is the 211-periodic Fejer kernel, and deduce that II gN IIA(R) -t 0 as N -t co. (b) Let gU) be nonnegative continuous functions such that /'.

'"

L gU)(x) 00

gU> EL l(R) and such that

{]

0

< x < 211

=

.

0

J

Nj-t co fast enough II g'vJIIA(R);£ 2- i , and f = sired properties.

Then

if

otherwise

Lj=tgV]

has the de-

2. FOURIER-STIELTJES TRANSFORMS

2.1 We denote by M(R) the space of all finite Borel measures on R. M(R) is identified with the dual space of Co(R)-the (supremumnormed) space of all continuous functions on R which vanish at infinity-by means of the coupling (2.1)

(f,p.) =

f

fECo(R), p.EM(R).

fdp.

f

The total mass norm on M(R) is defined by II p.IIM(R) = I dp.1 and is identical to the "dual space" norm defined by means of (2.1). The mapping f -t f(x)dx identifies V(R) with a closed subspace of M(R). The convolution of a measure p. E M(R) and a function tp E Co(R) is defined by the integral (2.2)

(p.$tp)(x) =

f

tp(x-y)dll(Y)'

and it is clear that 11$ tp E Co(R) and that II p. $ tp IloD 2! II JlIIM(R) II tp II 00 • The convolution of two measures, 11, v E M(R), can be defined by means of duality and (2.2), analogously to what we have done in 1.7, or directly by defining

Vt *V)(E) =

f

Il(E - y)dv(y)

for every Borel set E. Whichever way we do it, we obtain easily that II p. $V IIM(R)

2! II P.IIM(R) II v IIM(R)'

2.2 The Fourier-Stieltjes transform of a measure 11 E M(R) is defined by: (2.3)

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An Introduction to Harmonic Analysis

It is clear that if J1 is absolutely continuous with respect to Lebesgue measure, say dJ1 = f(x) dx, then = lw. Many of the properties of Ll Fourier transforms are shared by Fourier-Stieltjes transforms: if J1, v E M(R) then !fi(~)! ~ II J111 M(R), Ais uniformly continuous, and A~) = fi(~) \I(~). A departure from the theory of V Fourier transforms is the failing of the Riemann-Lebesgue lemma (the same way it fails for M(T»; the Fourier-Stieltjes transform of a measure J1 need not vanish at infinity.

pm

Theorem: (Parseval's formula):t Let It E M(R) and let f be a continuous function in VCR) such that IE LI(R). Then

J

f(x)dJ1(x) =

(2.4) PROOF:

2~J 1(~)fi(-~)d~.

By (1.11)

hence

The assumption 1(~) E V(R) justifies the change of order of integration (by Fubini's theorem); however, it is not really needed. Formula (2.4) is valid under the weaker assumption 1(~)P( -~) E V(lt), and is valid for all bounded continuous fE V(R) if we replace the

fA_;. (t -' ,~ I) lmp( -~) d~ (cf.

integral on the right by lim 21 • ;._", TC ClSe t.to).

I\.

exer-

'

2.3 The problem of characterizing Fouricr-Stidtjes transforms among bounded and uniformly continuous functions on R is very hard. As far as local behavior is concerned this is equivalent to characterizing A(R): every E A(R) is a Fourier-Stieltjes transform, and on the other hand, if J1 C M(R) and V;. is de la Vallee Poussin's kernel (1.14), then

1

!I

11* V;. E V(R) and = fiW for ~ ~ A.. The following theorem is analogous to 1.7.3:

p.V;.W

Theorem:

Let 'P be continuous <1>;.(x) =

011

R, define <1>;. by:

~ f;'_;. (1 _Ill)rp(~)i~xd~. A-

2n

t Notice that (2.4) is equivalent to

f f(x) d;(x) = 1/2:,; fiwid.;)d;.

133

VI. Fourier Transforms on the Line

Then cp is a Fourier-Stieltjes transform if, and only if, <1>;. E VCR) for all A. > 0, and II <1>;.IILI(R) is bounded as A -+ ~. PROOF: If cp = fJ. with f1. E M(R), then <1>;. = J1* K.I.> K;. denoting Fejer's kernel. It follows that for all A > 0, <1>;. E Ll(R) and

II <1>;.IILI ~ II J1 IIM(R). If we assume that <1>;. E VCR) with uniformly bounded norms, we consider the measures <1>;. dx and denote by J1 a weak-star limit point of <1>;. dx as A -+~. We claim that cp = {l and since both functions are continuous, this will follow if we show that

(2.5)

for every twice continuously differentiable g with compact support. For such g we have g = G with G E /} () Co(R); by Parseval's formula

=

lim 2n: A-+ 00

f

G(x)<1>;.(x) dx = 2n:

f

G(x) dJ1(x)

and the proof is complete. Remark: The application of Parse val's formula above is typical and is the, more or less, standard way to check that weak-star limits in M(R) are what we expect them to be. Nothing like that was needed in the case of M(T) since weak-star convergence in M(T) implies pointwise convergence of the Fourier-Stieltjes coefficients (the exponentials belong to C(T) of which M(T) is the dual). The exponentials on R do not belong to Co(R) and it is false that weak-star convergence in M(R) implies pointwise convergence of the Fourier-Stieltjes transforms (cf. exercise 1 at the end of this section.) However, the argument above gives: Lemma: Let IlnEM(R) alld assume that Iln-+J1 in the weak-star topology. Assume also that fin(~) -+ cp(~) pointwise, cp being continuous on R. Then fl = cp.

134

An Introduction to Harmonic Analysis

2.4 A similar application of Parse val's formula gives the foUowing useful criterion: Theorem: A function cp, defined and continuous on R, is a FourierStieltjes transform if, and only if, there exists a constant C such that

(2.6)

f

I 2In ]mcp(-~)d~ I ~ Csu",p I!(x) I

for every continuous fE Ll(R) such that] has compact support. PROOF: If cp = p, (2.6) follows from (2.4) with C = I Jl.IIM(R). If (2.6) holds, f -+ 1/21tJ]( ~)cp( -~) d~ defines a bounded linear functional on a dense subspace of Co(R), namely on the space of all the functionsfE Co () VCR) such that] has a compact support. This functional has a unique bounded extension to Co(R), whkh, by the Riesz representation theorem, has the form 1-+ I(x) dJl.(x). Moreover, II Jl.IIM(R) ~ C. Using (2.4) again we see that fl- cp is orthogonal to all the continuous, compactly supported functions] with fE VCR), and consequently cp = fl. ...

J

Remark: The family {J} of test functions for which (2.6) should be valid can be taken in many ways. The only properties that have been used are that {J} is dense in Co(R) and {l} is dense in Co(R). Thus we could require the validity of (2.6) only for (a) functions f such that J is infinitely differentiable with compact support; or (b) functions I which are themselves infinitely differentiable with compact support, and so on.

2.5 With measures on R we can associate measures on T simply by integrating 21t-periodic functions. Formally: if E is a Borel set on T (T being identified with ( -n, 1t1) we denote by En the set E + 2nn and write £ = U En; if Jl. E M(R) we define Jl.T(E) = Jl(E).

It is clear that IlT is a measure on T and that, identifying continuous functions on T with 21t-periodic functions on R,

(2.7) The mapping Jl. -+ JiT is an operator of norm one from M(R) onto M(T), and its restriction to VCR) is the mapping that we have discussed in section U5. It follows from (2.7) that fl(n) = flrCn) for all n;

VI. Fourier Transforms on the Line

135

thus the restriction of a Fouricr-Stieltjes transform to the integers gives a sequence of Fourier-Stieltjes coefficients.

Theorem: Afunction cp, defined and continuous on it, is a FourierStieltjes transform if, and only if, there exists a constant C > 0, such that for all A> 0, {cp(An)}:'= -co are the Fourier-Stieltjes coefficients of a measure of norm ~ Con T. PROOF:

If cp = fi with Jl E M(R) we have cp{n) = fi.(n) = fl-r(n) with Writing dJl(x/A) for the measure satisfying

I JlT I ~ I JlII·

f

f(x) dJl (

we have

~)

I Jl(X/A) IIM(R) = I JlIIM(R)

=

f

f(AX) dJl(x)

and

p(xp.)(~)

= (leO) so that

./'-..

cp(An) = Jl(x/Ah(n) and the "only if" part is established . . In order to establish the converse we use 2.4. Let f be continuous and integrable on R and assume that] is infinitely differentiable and compactly supported. We want to estimate the integral 1/27r f]( ~)P< -~) d~ and, since the integrand is continuous and compactly supported, we can approximate the integral by its Riemann sums. Thus, for arbitrary s > 0, if A is small enough:

Now, (A/2n)](An) are the Fourier coefficients of the function tfrit) = L ~ = -oof«t + 2nm)/).) on T, and since the infinite differentiability of] implies a very fast decrease of f(x) as x 00, we see that if A is sufficiently small

I 1-+

(2.9) Assuming that cp(}.n) = (lill), Jl;. E M(T) and from Parseval's formula

I Jl;.IL'J(T) ~ C, we obtain

I:n2 j(An)rp(-An) I by (2.8) and (2.9)

I 2~ f j(~)cp( -~)d~ I ~ Csup If(X) I + (C + 1) s

136

An Introduction to Harmonic Analysis

and since e > 0 is arbitrary, (2.6) is satisfied and the theorem follows from theorem 2.4. ~ 2.6 Parseval's formula also offers an obvious criterion for determining when a function 'P is the Fourier-StieItjes transform of a positive measure. The analog to 2.4 is

Theorem: A function T, bounded and continuolls 011 it, is the F ourier-Stieltjes transforlll of a positive measure on R if, and only if, (2.10)

for every nonnegative function compactly supported.

I

which

i~

infinitely differentiable and

PROOF: ParsevaI's formula clearly implies the "only if" part and also the fact that if we assume 7J = fl with J1- E M(R), then J1- is a positive measure. In order to complete the proof we show that (2.10) implies (2.6), with C = q:{0), for every real-valued, compactly supported infinitely differentiable f (and consequently with C = 2rp(0) for complexvalued f). As usual, we denote by K;.(x) the Fejer kernel (1.8) and notice that

1

}. - K;.(X) =

K

1 1 (Sin h/2 ). 2 . . (AX) = 2rc -- hj2 IS nonnegatIve and tends to

1/2rc, as A-+O, uniformly on compact subsets of R. By (1.12) the Fourier transform of K(AX) is ;.-1 max(1 ~ IIA, 0) and consequently

-I

when A-+ 0 (using the continuity of 'P(~) at ~ = 0)

f

(2.11)

1~

1 K;(~)'P( -~)d~ -+ 'P(O).

If I is real-valued and compactly supported and e > 0, then, for sufficiently small }. and all x,

2rc(e

+ sup III) K(h) -

hence, by (2.10) and (2.11), if

(2.12)

;n

I

j E V (it)

j( ~)'P( -~) d~

rewriting (2.12) for -

f

f(x) ~ 0;

~

rp(0)(2e

and letting

E -+

+ sup II

I),

0 we obtain:

137

VI. Fourier Transforms on the Line

Remark: The assumption that cp is bounded is superfluous (cf. exercise 6 at the end of this section). 2.7

The analog to 2.5 is:

Theorem: . A function cp, defined and continuous on R, is the Fourier-Stieltjes transform of a positive measure, if and only if, for all A. > 0, {rp(A.n)}:=_ao are the Fourier-Stieltjes coefficients of a positive measure on T. PROOF: The "only if" part follows as in 2.5. For the "if" part we notice first that if cp(}.n) = p;.(n) with Ji). ;;:; 0 on T, then I~Ji).11 = CP(O) and consequently, by 2.5, cp is a Fourier-Stieltjes transform. Using the continuity of cp, we can now establish (2.10) by approximating the integral by its Riemann sums as in the proof of 2.5. ~

2.8 DEFI~ITI0N: A function cp defined on R is said to be positive definite if, for every choice of ~ h ... , ~N E it and complex numbers Zl' ... ,7 N , we have N

.L

(2.13)

j,k

~

cp(~J - ~k)Zj~

;;:;

O.

1

Immediate consequences of (2.13) are: (2.14) and (2.15) In order to prove (2.14) and (2.15), we take N ZI = 1, Z2 = z; then (2.13) reads q{O)(l

= 2,

+ 1Z 12) + cp(~)z + cp( -~)z;;:; 0; cp(~) + cp( -~) real and taking

~I = 0, ~2 =~,

taking Z = 1, we get Z = i, we get i( CP(~) - cp( real, hence (2.14). If we take z such that zcp(~) = cp(~) we obtain:

m

-I

1

which establishes (2.15).

Theorem (Bochner): Afunctioll rp defined 011 R is a Fourier-Stieltjes transform of a positive measure if, lIlId only if, it is positive definite and continuous.

138 PROOF: Zt> ... , ZN

An Introduction to Harmonic Analysis Assume first cp = fl with Jl ~ O. Let be complex numbers; then

(2.16) =

I Ii

ej, ... , eN E Rand

zie-i~JXI2dJl(X) ~

°

I

so that Fourier-Stieltjes transforms of positive measures are positive definite. If, on the other hand, we assume that cp is positive definite, it follows that for all A. > 0, {1p(A.n)} is a positive definite sequence (cf. 1.7.6)' By Herglotz' theorem 1.7.6, q:(A.n) = flin) for some positive measure Jl;. on T, and by theorem 2.7, 'P = fi for some positive Jl E M(R). ~ 2.9 For the convenience of future reference we state here the analog to Wiener's theorem 1.7.11. The theorem can be proved either by essentially repeating the proof of 1.7.11 or by reducing it to 1.7.11. We leave the proof as an exercise (exercise 7 at the end of this section) to the reader.

Theorem:

Let JlEM(R). Then

In particular, a necessary and sufficient condition for the continuity of Jl is

EXERCISES FOR SECTION 2 I. Denote by 0" the measure of mass one on R concentrated at x = n. Show that limn _ oo On = 0 in the weak-star topology of M(R) and conclude that weak-star convergence of a sequence of measures does not imply pointwise convergence of the Fourier-StieItjes transforms. 2. Writing ~n = n- 1(01 + 0l +... + on) show that ~n-+ 0 in the weak-star topology and flit;) converges for every t;E R; however, Iimflie) is not identically zero.

139

VI. Fourier Transforms on the Line

I

3. Let /I,.E M(R) and assume lim;.-+oosuPn!lxl>;.1 d/ln = O. Show that if /In -+ /I in the weak-star topology, {In(e) -+ flU;) uniformly on compact subsets of R. 4. Let /In E M(R) such that II /In II ~ 1. Assume that fln converges pointwise to a continuous function qJ. Show that qJ = fl for some /I E M(R) such that I /I I ~ t. 5. Show that there exists a uniformly continuous bounded function qJ which is not a Fourier-Stieltjes transform, and such that {qJOn)} is a sequence of Fourier-Stieltjes coefficients for every A. > O. Hint: Construct a continuous function with compact support which is not a Fourier transform of a summable function. 6. Show that if qJ is continuous on it and (2.10) is valid, then qJ is positive definite. Conclude from (2.t5) that the boundedness assumption of 2.6 is superfluous. 7. Prove theorem 2.9. 8. Express /I{[a,b]} and I/{(a, b)} in terms of fl. ([a,bJ is the closed interval with endpoints a and b, and (a, b) is the open one.) 3. FOURIER TRANSFORMS IN LP(R); 1 < p

~

2

The definition of the Fourier transform (i.e., Fourier coefficients) for functions in various function spaces on T, was largely simplified by the fact that all these spaces were contained in V(T). The fact that the Lebesgue measure of R is infinite changes the situation radically. If p> 1 we no longer have I!' c: V, and, if we want to have Fourier transforms for functions in L"(R) (or other function spaces on R), we have to find a new way to define them. [n this section we consider the case 1 < P ~ 2 and obtain a reasonably satisfactory extension of the Fourier transformation for this case.

3.1 We start with L2 (R).

Lemma: Let f be continuous and with compact support on R; then 21n;

f IJ(~) 12 d~ f =

it(x) 12 dx.

We give two proofs. PROOF I: Assume first that the support off is included in ( - n;, n;). By theorem 1.5.5,

1 211:

f

itex) 12 dx =

n

~ ~oo

1

1

2/(n)

11

140

An Introduction to Harmonic Analysis

and replacing f by e- i'1.Xf we have

f

(3.1)

I!(x) 12 dx

=

1 2n

n

~cc II(n + a) 12;

integrating both sides of(3.1) with respect to a on 0

~ a

< 1, we obtain

If the support of f is not included in (-n, n), we consider g(x) = A. 112f(,h ). If A. is sufficiently big, the support of g is included in (-n,n) and, since g(~)=A.-1/2IWA.), we obtain

f l!(xWdx f Ig(x)1 dx =2~ f Ig(~)12de JIJwI2d~. 2

=

= 2in

~

f

Write g = f * j( -x); we have g(O) = f(x)j(x) dx = 2 • If we know that flJm 12 de < 00 and gee) = (e.g., if we assume thatfis differentiable), it follows from the inversion formula (UI) that PROOF

=

II:

Ilwl

fl!(x) 12 dx

In the general case we may apply Fcjcr's theorem and obtain

and, since the integrand is nonnegative, its C-1 summability is equiv~ alent to its convergence and the proof is complete. DEFINITIO]>;:

For g E L2CR) we write

Theorem (Plancherel):

There exists a unique operator :F from

VCR) onto v(Jl) having the properties: (3.2)

Remark:

:Ff=l

for fEV nV(R).

In view of (3.2) we shall often write

I

instead of :Ff.

141

VI. Fourier Transforms on the Line

We notice first that Ll (\ Ll(R) is dense in Ll(R) and consequently any continuous operator defined on Ll(R) is determined by its values on V n Ll(R). This shows that there exists at most one operator satisfying (3.2) and (3.3). By the lemma, (3.3) is satisfied if I is continuous with compact support, and since continuous functions with compact support are dense in V n Ll(R) (with respect to the PROOF:

I IIL'(R) +:1 IIL2(R»,

norm (3.3) holds for all IE l! n U(R}. The mapping 1- J clearly can be extended by continuity to an isometry from L2(R) into L1(R). Finally, since every twice differentiable compactly supported function on R is the Fourier transform of a bounded integrable function on R (1.5 and the inversion formula), it follows that the range ofI ~ is dense in U(R) and hence coincides with it. ..

1

1

Remarks: (a) Given a function IE VCR) we define as the limit (in Ll(R» of 1n, where J. is any sequence in l! n U(R) which converges to I in L1(R). As such a sequence we can take J.(x)

=

/(X) {0

Ixl < n Ix I ~ n

and obtain the following form of Plancherel's theorem: the sequence

(3.4) converges, in L2(R), to a function which we denote by j, and for which (3.2) and (3.3) are valid. (b) The mapping I ~ J being an isometry of L2(R) onto PCR) , clearly has an inverse. Using theorem 1.11 and the fact that we have an isometry, we obtain the inverse map by I = limJ(n) in U(R) where (3.5) (c) Parseval's formula (3.6) for f, g E VCR), follows immediately from (3.3) (and in fact is equivalent to it).

3.2 We turn now to define Fourier transforms for functions in l!(R), 1 < p < 2. Using the Riesz-Thorin theorem and the fact that 3': I ~

1

142

An Introduction to Harmonic Analysis

has norm 1 as operator from VCR) into L<XJ(R) and from L2(R) onto L2(R), we obtain as in IV.2:

Theorem: (Hausdorff-Young): IE L1 n VCR). Then

Let 1 < p < 2, q

r

f 11(eW de ~ (f I/(x) IPdxt q

(;7t

= p!(p -

1) and

P .

For IE L!'(R), 1 < p < 2, we now define1by continuity; for example, as the limit in Lq(R) of r"'ne-j~XI(x)dx. The mapping :F:/-+1so defined is an operator of norm t from lJ'(R) into Lq(R); however, it is no longer an isometry and the range is not the whole of Lq(R). (see exercise 10 at the end of this section). 3.3 The fact that for p < 2, :F is not an invertible operator from L!'(R) onto VCR) makes the inversion problem more delicate than it is for V. The situation in the case of L!'(R) is similar to that which we encountered for 1!'(T). We have inversion formulas both in terms of summability and in terms of convergence. The summability result can be stated in terms of general summability kernels without reference to the Fourier transform as we did in 1.10 for VCR); and in fact the statement of theorem 1.10 remains valid if we replace in it VCR) by H(R), 1 ~ P < 00. For p ~ 2 we can generalize theorem 1.1 ~. We first check (see exercise 9 at the end of this section) that if g E 1J'(R) and IE VCR) then

1* g is a well-defined element in L(R) ano

j;g = 1 g.

This is particularly simple if we take for Ithe Fejer kernel K. . : we have

JGg

(1_1~I)g

=

and, since KA * g is clearly bounded (KA E [}(R), q = p!(p hence belongs to I! n e'eR) c V(R), it follows that

K}. * g = 27t ~ fA (1 - BJ)\ g(e)ei/;x de 2 -A

and from the general form of theorem I. t 0 we obtain:

Theorem:

Let g E L!'(R), 1 ~ p g =

2; then

lim...!.... f}. (1 _l~J) gwi{.< de A-OO

in the H(R) norm.

~

27t

_)

).

1» and

VI. Fourier Transforms on the Line

143

Corollary: The Junctions with compactly supported Fourier transJorms Jorm a dense subspace oj l!(R).

* 3.4

The analog to the inversion given by 3.1, remark (b) (i.e., convergence rather than summability) is valid for 1 < p < 2 but not as easy to prove as for p = 2; it corresponds to theorem 11.1.5 and can be proved either through the study of conjugate harmonic functions in the half-plane, analogous to that done for the disc in chapter III, or directly from Il.1.5. The idea needed in order to obtain the norm inversion formula for H(R), 1 < p < 2, from 11.1.5 is basically the one we have used in proof I of lemma 3.l.

Lemma:

For 1 < p < co, there exist cOllstants Cp such that

Jor every Junction J with compact support and every N> O. PROOF:

Inequality (3.7) is equivalent to the statement that, for

M-+co (3.8)

(f:)SN(J,X)\PdX rIP ~ CpI\JIILP(R)'

Writing tpM(X) = M'/PJ(Mx) we see that check that

I tpM IILP(R) = I J I

LP(R)

and

(3.9) In view of (3.9), (3.8) is equivalent to (3.10) As M -+ co, tr..e support of tpM shrinks to zero and consequently the lemma will be proved if we show that (3.8) is valid, with an appropriate Cp , for all J with support contained in (-7t,n) for M = 1 (or any other fixed positive number) and for all integers N. We now write 1

(3.11)

S,.,(J,x) =

(,

• 0

2: -2n1 len + '1.)

N-l

-N

ei(n+»xda

144

An Introduction to Harmonic Analysis

I

and notice that ~';/(1/2Tt)!(n + cx)e is a partial sum of the Fourier series of J(x) e -ia:c (it is carried by (- rr, rr) which we now identify with T). As an L"(T)-valued function of IX, the integrand in (3.lt) is clearly continuous t and, by B.l.2 and 11.1.5, it is bounded in L"(T) by a constant mUltiple of IIJi"xIILP(T) = (2rr)-J/PIIJIILP(R). We therefore obtain inx

and the proof is complete. Corollary:

For 1 < p ~ 2, inequality (3.7) is valid Jor all JE L"(R).

PROOF: Write J = limn-> en/" with /" E l!(R), In having compact SUpports and the limit being taken in the U(R) norm. By theorem 3.2, ! = lim!. in Lq(R) and consequently, for fixed N > 0, SN(J, x) = = lim. SN(/", x) uniformly in x. It follows that

II SN(f) II LP(R) ~

liminfiISN(/,,)IIt.P\R) ~ Cplim II In IILr(R) = CpIlJIILP(R). n

Theorem:

~

•

Let JE i!'(R), 1 < p ~ 2. Theil lim IISN(f)-JIIt.P(R) = N->oo

o.

PROOF: {SN} is a uniformly bounded family of operators which converge to the identity, as N --. 00, on all functions with compactly supported Fourier transform, and hence, by corollary 3.3, it converges ~ strongly to the identity.

EXERCISES FOR SECTION 3 1. Let B £; fL'c be a homogeneous Banach space on R and let /E B. Show that: (a) for every q.> EL '(R), q.> * / can be approximated (in the B norm) by linear combinations of translates of I, that is, given e > 0, there exist numbers Y"···'YnER and complex numbers A1,···,A. such that 1Iq.>*/- L;:lAJY"'B<e, where/ix) -j(x-y). (b) For every y ER, J;, can be approximated by functions of the form EL1(R). Deduce that a closed subspace H of B is trullslatioll invariallt (Le., IE H implies Iy E H for all yE R) if, and only if, IEH implies rp*!ElI for every q.>EL1(R).

t Note that f, having a compact a support, is in V(R) and! is therefore continuous.

VI. Fourier Transforms on the Line

145

2. Let F, GEL 2(R) and assume FW = 0 implies G(~) = 0 for almost all ~ E R. Show that, given e > 0, there exists at wice-ditferentiable compactly supported function such that

I ~F -

G

Ilo(R) < e.

3. Let f,gELz(R) and assume that J(~) = 0 implies iW = 0 for almost all ~ E R. Show that g can be approximated on L2(R) by linear combinations of translates of /. Hint: Use exercises 1 and 2 and Plancherel's theorem. 4. A measurable set E c It is a 0 set if it coincides with its set of points of density. Show that finite intersections of 0 sets are again 0 sets. For a subspace H S L 2 (R), define E(H) as the union of all the 0 sets which are supports ofsomejwith/E H. Show that this establishes a one-toone correspondence between the closed translation invariant subspaces of L 2(R) and the 0 sets on

It, and that

E(HI n Hz) £(Hl

+ H 2)

=

E(H1 )n E(H2 ) ,

=

£(Hl) V E(H2 )

HI .1 H2

<;:>

V

a set of measure zero,

E(Ht)n E(H2 ) =

0.

5. Show that every closed translation invariant subspace of L 2(R) is singly generated. 6. Let fELZ(R). Show that the translates of f generate L2(R) if, and only if, J(~) ¥= 0 almost everywhere. 7. The information obtained through exercises 2 through 6 can be obtained very easily by duality arguments (i.e., using the Hahn-Banach theorem). For instance, by Plancherel's theorem, exercise 6 is equivalent to the statement that {f(~ i~X}X€R spans L 2(R) if, and only if, f(~) ¥= 0 almost everywhere. By the Hahn-Banach theorem U(~)ei~X}x€R does not span L 2 CR) if, and only if, there exists a function IjI E L 2eR) which does not vanish identically, such that JJ(~)IjI{~) ei~x de = 0 for all x E R, which. by the uniqueness theorem, is equivalent to: fiji = 0 identically, that is, f vanishes on the support of IjI. Use the same method to prove exercises 3 through 5. 8. Both the "if" and the "only if" parts of exercise 6 are based on Plancherel's theorem and are both false for U(R), p < 2. Assuming the existence of a measure Jl carried by a closed set of measure zero and such that /I.EL q for all q> 2, construct a function fELl n L""(R) such that J(~) ¥= 0 almost everywhere and such that the translates of f do not span e(R) for any p < 2. Hint: Put 11 on R.

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An Introduction to Harmonic Analysis

9. Show that if IE L I (R) and gEL P(R), then j.g = 1g. We denote by:FU the space of all functions 1 such that IE e(R), (thus §'L I ~ A(R»). By definition:

10. If pEM(R) and
",m P

,>0


IIn

{0

e :;;, 0,

I

p :;;, C p
';>0 , :;;, 0,

then '" ",.fJ EffL P • 13. Let IE LP(R), 1 < p :;;, 2. Show that h(x) = rr- I I(x - y) sin y/ y dy is well defined and continuous on R, h ELP(R) and h I/LP(R) ~CpI//IILP(R)' 14. Show that, for 1 s.p < 2, the nor.!,lls II d9"LP and II fPllu(R) are not equivalent. Deduce that :FLP rfL9(R) (q = p/(p - 1». Hint: See

I

J

IV.2.

4. TEMPERED DISTRIBUTIONS AND PSEUDO-MEASURES

In the previous section we defined the Fourier transforms for functions in L!'(R), 1 < p ~ 2, by showing that on dense subspace OIl which 3': 1-> J is already well defined (e.g., on IJ n L2(R», we have the norm inequality

and consequently there exists a unique continuous extension of 3', as an operator from L"(R) into LqCR). If p > 2 this procedure fails. It is not hard to see that not only is it impossible to extend the validity of the Hausdorff-Young theorem for p> 2, but also there is no homogeneous Banach space B on R such that for some p > 2, some constant C and all IE V n L"'(R),

VI. Fourier Transforms on the Line

147

So, a different procedure is needed if we want to extend the notion of Fourier transforms to I!'(R), p > 2. Clearly, we try to extend the notion, keeping as many of its properties as possible; in particular, we would like to keep some form of the inversion formula and the very useful Parseval's formula. We realize immediately that, since the Fourier transforms of measures are bounded functions, if any reasonable form of inversion is to be valid, the Fourier transforms of some bounded functions will have to be measures; and once we accept the idea that Fourier transforms need not be functions but could be other objects, such as measures, the procedure that we look for is given to us by Parseval's formula. So far we have established Parseval's formula for various function spaces as a theorem following the definition of the Fourier transforms of functions in the corresponding spaces. In this section we consider Parseval's formula as a definition of Fourier transform for a much larger class of objects. Having proved Parseval's formula for I!'(R), 1 ~ p ~ 2, we are assured that our new definition is consistent with the previous ones.

4.1 We denote by S(R) the space of all infinitely differentiable functions on R which satisfy: (4.1)

lim xnJU)(x) = 0

for all n ;;:;; 0, j ;;:;; O.

Ixl-+oo

S(R) is a topological vector space, the topology being given t by the family of seminorms (4.2)

This topology on S(R) is clearly metri7able and S(R) is complete, in other words, S(R) is a Frechet space. DrnNITION: A tempered distribution on R is a continuous linear functional on S(R). We denote the space of tempered distributions on R, that is, the dual of S(R), by S*(R).

t A sequ~nce of functionsfmE S(R) converges to f if limm-+uo IIfm - f for all j ~ 0 and 1/ ;;:;; O. The metric in S(R) can be defined by: dist(f, g) =

L -.- -1£:-::: g ~j~

j.n~O

2J+n

1 + II f

- gil j.n

Ilj.n =

0

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An Introduction to Harmonic Analysis

S(R) is a natural space to study within the theory of Fourier transforms. By theorems 1.5 and 1.6 we see that if jE S(R) then j E S(R) (the analogous space on R) and, e"j(j)(e) being the Fourier transform

(_it+jdn(Xlx~X»i'

we see that the mappingj--+jis continuous from

S(R) into S(R). By the inversion formula this mapping is onto SeR) and is bicontinuous. We now define p, for /1 E S*(R), as the tempered distribution on R satisfying


(4.3)

for all jE S(R). The space of tempered distributions on R is quite large. Every function g which is measurable and locally summable, and which is bounded at infinity by a power of x can be identified with a tempered distribution by means of:

<J, g) =

J

j(x)g(x) dx

jES(R)

and so can every gEl!'(R), for any p ~ 1, and every measure P E M(R); thus our definition has a very satisfactory domain. However, the cange of the definition is as large and this is clearly a disadvantage; it gives relatively little information about the Fourier transform. We thus have to supplement this definition with studies of the following general problem: knowing that a distribution pES*(R) bas some special properties, what can we say about jll Much of what we have done in the first three sections of this chapter falls into this category: if Jl is (identified with) a summable function, then p is (identified with) a function in CorR); if P is a measure, ft is a unifonnly continuous bounded function; if P E l!'(R) with 1 < p ~ 2, then fl E L'l(R), q = p/(P - 1). We shall presently obtain some information about Fourier transforms of functions in l!'(R) with 2 < p ~ 00, but we should not leave the general setup without mentioning the notion of the support of a tempered distribution. 4.2 DEFINITION: A distribution l' E S*(R) vanishes on an open set OcR, if
VI. Fourier Transforms on the Line

149

Lemma: Let 0 1,0 2 be open on R alld let K be a compact set, K c 0 1 U O 2 • Theil there exist two compactly supported C~ functions 'PI

and 'P2 satisfying: support of 'Pj

C

OJ and PI

+ 'P2 == 1 on

K.

Let U j C OJ have the following properties: U, is open, OJ is compact and included in 0 i' and K CUI U U 2. Denote the characteristic function of U I by 1/1 I and tha t of U 2 " U I by 1/12. Let 8 > 0 be smaller than the distance of K to the boundary of U I U U z and also smaller than the distance of U j to the complement of OJ, j = 1,2. Let b(x) be an infinitely differentiable function carried by (-8,8) and whose integral is 1. Then we can take 'Pj = I/Ii * b. ~ PROOF:

Corollary:

If v E S*(R) vanishes on 0 1 and on O 2 , it vanishes on

0 1 uO z· PROOF: LetfES(R) have a compact support included in 0 1 U02 • Denote the support of f by K and let 'PI' 'Pz be the functions described in the lemma. Then 'Pj E S(R), f = f( 'PI + cpz) = f'Pl + ('Pz. Now
Our corollary clearly implies that the union of any finite number of open sets on which v vanishes has the same property, and since our test functions all have compact support, the same is valid for arbitrary unions. The union of all the open sets on which v vanishes is clearly the largest such set. 4.3 DEFINITION: The support 1:(v) of VE S*(R) is the complement of the largest open set 0 C R on which v vanishes. Remarks: (a) If v is (identified with) a continuous function g then 1:(v) is the closure of {x;g(x)"* O}. If v is a measurable function g then 1:(v) is the closure of the set of points of density of {x; g(x) O}. The set of poi nts of density of {x; g(x) o} is a finer notion of support which may be useful (cf. exercise 3.4). (b) The definition of 1:( v) implies that if 'P E S(R) and if the support of cp is compact and disjoint from 1:( v) then <'P, v) = O. It may be useful to notice that if 1/1 E S(R) and if [) is infinitely differentiable with compact support and b(O) = 1, then 1/1 = lim.. _ o [)(AX)I/I in S(R), and consequently if the support of 1/1 is disjoint from L( v) (but not necessarily compact) we have <1/1, v) = Iim.. _ o <[)(AX)I/I, v) = O. In particular, if 1:(v) = 0 then v = o.

"*

"*

An Introduction to Harmonic Analysis

150

(c) Let B 2 S(R) be a function space and assume that every IE B with compact support can be approximated in the topology of B by functions ff'n E S(R) such that the supports of gJn tend to that off Let v E S*(R) and assume that v can be extended to a continuous linear functional on B. If IE B has a compact support disjoint from rev), then (J, v) = O. 4.4 S(R) is an algebra under pointwise multiplication. The productlv of a function IE S(R) and a distribution v E S*(R) is defined by (g,jv) = (gj, v),

gES(R),

that is, the multiplication by I in S*(R) is the adjoint of the multiplication by I in S(R). From the definition!. above, it is clear that r(jv) s; r(j) () r( v). 4.5 We denote by:Fil = $/'"L!'(R) the space of distributions on Ii which are Fourier transforms of functions in L!'(R), I < p ~ 00 (we keep the notation A(R) for ffU(R». YJ1 inherits from Lf(R) its Banach space structure; we simply put I J II"LP = II IIILP{R); and we can identify $/'"U with the dual of §Lq if q = p/(p - I) < 00. In particular, ,FLoo is the dual of A(R). This identification may be considered as purely formal: writing (J,g) = (/,g) carries the duality from (I!'(R),LJ(R» to (iFl!, $/'"Lq ); however, we have already made enough formal identifications to allow a somewhat clearer meaning to the one above. Having identified functions with the corresponding distributions, we clearly have S(R) c .'Fl! and, if p < 00, S(R) is dense in §U; consequently, every continuous linear functional on §L P is canonically identified with a tempered distribution. The identification of :FLq as the dual of $/'"l!' now becomes a theorem stating that a distribution v E S*(R) is continuous on S(R) with respect to the norm induced by :Fl! if, and only if, v EffLJ. We leave the proof as an exercise to the reader. We now confine our attention to .'FL"'. If J1 is a measure in MeR) then J1 is the Fourier transform of the bounded function hex) oo oo = ei~xdJ1(~); thus M(R) c§L • The elements of ffL are commonly referred to as pseudo-measures. It is clear that M(R) is a relatively small part of.:TC); for instance, if '{J E Loo is not uniformly continuous on R, (p cannot be a measure.

J

The convolutioll hI * h2 of the pseudo-measures hI and L"'(R», is the Fourier transform of hI h2 •

DEFINITION:

11 2 • (h j

E

151

VI. Fourier Transforms on the Line

Again we reverse the roles; we take something which we have proved for measures, as a definition for the larger class of pseudo-measures. Thus, if hi and hz happen to be measures, hi *h z is their (measure theoretic) convolution. 4.6 Another case in which we can identify the convolution is given by Lemma:

Let h]EL"'(R) and h2ELI nL"°(R); then

(4.4) We remark first that h1h z E Ll n Loo(R) and consequently so that we can talk about its value at ~ E R. If h]ES(R) we have PROOF:

* hl

1,]

=

h;ii; E A(R)

(h]*h2)(~)=

I hl(x)hz(x)e-i~Xdx I hz(~

= 2!

=

- 1/)11I(YI) dYi

2~

II hl(101lz{x)i(q-~)X

=


dxdYl

r;~('1»'

17),

Since S(R) is dense in Loo (R) in the weak-star topology (as dual of LI(R», and since both sides of (4.4) depend on hi continuously with respect to the weak-star topology, (4.4) is valid for arbitrary h] ELoo(R).
IJh]ELoo(R) and hlEUnL"'(R), then

* hz)£; "[. (iiI) + "[.(h z)·

"[.(h i

4.7 This corollary can be improved: Lemma:

Assume hl>hzELOO(R). Then

"[. (hI PROOF:

* hz)

£;

"[.(h])

+ "[.(h z).

Consider a smooth function jE A(R) with compact support

disjoint from "[.( hl )+ "i.( h2 ). We have to show that <J. (/0 = 0 which is

the

same

as

f JhIhz dx

g; - ~ E "[.( hI)} and, by ~oE"[.(fh;)n"[.(hz), then there

set

that

~o = Ylo -

YI I>

that

/'-.

=

O. Now "[.(Ii]) is the

/'-.

=

from "[.(h z), proved.

<J,(/;;) =
hence

=

4.6, "[.<1JII) £; "[.(j) + "[.(h I). If exist YloE"[.(j) and YI] E"[.(h]) such is, Ylo = ~o + '11 which contradicts

"[.(]) n ("[.(h 1) + "[.( hz»

0.

A

<j"h t ,!lz)

It

follows

that 2)

"[.(j iii) is

disjoint

= 0 and the lemma is

An Introduction to Harmonic Analysis

152

4.8 The reader might have noticed that we were using not only the duality between LI(R) and LOO(R) but also the fact that a multiplication by a bounded function is a bounded operator on VCR). Another operation between VCR) and Leo (R) which we have used is the convolution that takes V x La) into LCO(R). Passing to Fourier transforms we see that .'FL"" is a module over A(R) the multiplication of a pseudomeasure by a function in A(R) being the adjoint of the multiplication in A(R). This extends the notion of mUltiplication introduced in 4.4. 4.9 Let k be an infinitely differentiable function on R, carried by [-I,IJ and such that fk(~)d~=1. For JEA(R) we set JA = )'k()'O *J = ). k(hOj( ~ - dl1· h. is infinitely differentiable, r.(h.) £; r.(f) + [ - 1/).,1/)'], and as ). ~ 00, j). ~ in A(R). By 4.3, remark (c) it follows that if v Eff La) andjE A(R) has a compact support disjoint from L(V), we have v) = O. Further, if JE A(R) and r.(j)n r.(v) = 0, it follows that «I ~ fA)j~ v) = 0 for all ). > 0, and letting }. ~ 00, we obtain <J. v) = O. For convenient reference we state this as:

to

f

J

<J,

Let vEff'L"" alld jEA(R).

Lemma:

If

-I 1

r.(j)nr.(v)=0

then


Let vEffLoo and jEA(R); then

r.(jv)

£;

r.(j) n r.( v).

4.II We show now that a pseudo-measure with finite support is a measure. Using the multiplication by elements of A(R) we see that a pseudo-measure with finite support is a linear combination of pseudomeasures carried by one point each; thus it would be sufficient to prove: Theorem: PROOF:

and qJ.W

A pseudo-measure carried by one point is a measure.

Let hE L""(R) and assume r.(h) = {O}. If 'PI' qJ2 E.A(R) = qJi~) in a neighborhood of ~ = 0, then <'1'1' h) = <'Pl, h ).

Put c = <'P, f,) where 'P is any function in A(R) such that qJ(~) = 1 near ~ = 0. As usual we denote by K the Fejer kernel and recall that KW = sup(1 - 1~ I, 0). By lemma 4.6 we have

153

VI. Fourier Transforms on the Line (4.5) "-

II

For ~ ~ I we clearly have II K(O = O. If -1 < ~1 < ~2 < 0 we have K(~2 - 1'/) - K(~l - 1'/) = ~2 - ~t for 1'/ near zero. By (4.5) and the "-

"-

definition of c we conclude that hK(~2) - hK(~l) "-

= C(~2 -

el), and "-

since hK(~) is continuous, upon letting ~l -+ -1 we obtain hK(O = e(l +"-e) for -1~ ~ ~ ~ O. Repeating the argument for 0 ~ ~ ~ 1 we obtain h K(O = eK(~) and by the uniqueness theorem h(x) = e a.e. ft follows that is the measure of mass c concentrated at the origin.
r,

4.12 We add just a few remarks concerning distributions in :FI!, 2 < p < 00. There is clearly no inclusion relation between .~I! and :FL00 but it might be useful to notice that 10cally:Fl! £; :Fl!' if p ~ p' and in particular all distributions in .~I! are locally pseudomeasures. (We recall that a tempered distribution v belongs locally to a set G c S*(R) if for every ~ E R there exists p. E G such that r (p. - v) does not contain 0 If \' EjOU and ~ Ell we may take A. > ~ and consid~r Jl = V"v where V" is de la Vallee Poussin's kernel (V"m = 1 for ~ ~ A, = 2 - ~ A-1 for A < ~ < 2A, and = 0 for ~ > 2A). It is clear that v = p. on (- A.A.), that is, r (Jl - v) () ( - A, A.) = 0 and

II

II

II

II

II

&

if v = J with f E 1!(R), then Jl = and V" *f E IT () L"'(R) since V). E V () VCR), q = p/(p - 1). fn particular, if rev) is compact, say r( v) c ( - A, A). then p. = v; we have thus proved: Theorem:

If v EjOI! and rev) is compact, then v E.'FLoo •

4.13 If v E.'FI! () :FLoo we can consider the repeated convolution of v with itself; writing v = with fE U() e'eR), the convolution of v with itself m times is the Fourier transform of f m, and if m > p, m times fm E VCR) so that ~ E A(R). In particular, assuming v '# 0, r(H'" * v) £; rev) + '" + rev) contains an interval. As an immediate consequence we obtain:

J

Theorem: Let v E:FL~ p < 00, and let J be an Ope/l interval on such that J () rev) '# 0. Then J () rev) is a basis for R.

R

Theorems 4.11 and 4.13 are equivalent to the following approximation theorems:

154

An Introduction to Harmonic Analysis

4.11'

Theorem:

Let ';ER and denote 1(0

{f;f E A(R),fW = O}

10 (';)

{f;fES(R),';¢~(f)}.

Then loW is dense in 1(';) in the A(R) topology. 4.13'

Theorem:

Let E c

R be

closed, and denote

10(E) = {J;fES(R),~(f)

n

0}.

E=

Asmme that E + E + ... + E (m times) has no interior. Let 1 < p and q = p/(p - I). Then 10CE) is (norm) dense in g;/J.

~

m

The proofs of 4.11' and 4.13' are essentially the same and follow immediately from the Hahn-Banach theorem (and 4.11, 4.13, respectively). A linear functional on A(R) which annihilates loW is a pseudohence is constant multiple of the Dirac measure supported by measure at .;, and hence annihilates IC ';). A linear functional on :?i'U which annihilates 10(E) is an element of Yf!' supported by E; hence it must be zero.

{n

EXERCISES FOR SECTION 4 I. 2. 3. 4.

Deduce 4.11 from 4.11'. Deduce 4./3 from 4.13'. What is a function h ELoo such that ~ If IE A(R) and vEg;L oo , then

Illv 1/ §L'"

~

(h)

is finite?

I fllA(ihil vlI:n

oo •

5. Let IELoo(R). Show that ~(Re(f)) £; ~(j) U (-'£(j». 6. (Bernstein) Let hELoo(R) and assume that ~(fi) £;[ -k,k]. Show that h is infinitely differentiable and that II ,,<m) II 'Xl ::;;; k mII h II "". 7. Let Iz EL oo(R), Iz ~ O. Assume that h(x t- y) = Iz(x)h(y) a.e. in (x, y). Show that h(x) = /~o" for some ';0 E :R. 8. Assume Vj EffL 00, j = 0, I, "', and Vj - 1'0 in the weak-star topology. Let U be an open set such that U n ~(v) = 0 for infinitely many j's. Show that U n ~(vo) = 0. 9. Fourier transforms of functions in Co(R) are called pseudoIlinctionf (on R). (a) Show that if IELICR), then I is a pseudo-function. (b) Show that if /j are pseudo-functions on ft, Ii Ij li.F l .oo < c and }'j ro fast enough. then /""{/j converges (weak-star) in §L oo . 10. Let h(x) =sinx 2 • Show that /";li-.o (weak-star) as roo

L

I}.I-

VI. Fourier Transforms on the Line

155

S. ALMOST-PERIODIC FUNCTIONS ON THE LINE

The usefulness of Fourier series of functions on T is largely due to the information they offer about approximation of the functions by trigonometric polynomials. On the line, trigonometric polynomials do not belong to many of the function spaces in which we are interested, for example, to l!'(R) for p < co; and the positive results, which we had for I!'(R), I ~ P ~ 2, were in terms of trigonometric integrals rather than polynomials. Trigonometric polynomials do belong to e'eR), and in this section we characterize the functions that are uniform limits of trigonometric polynomials.

5.1 DEFINITION: Let I be a complex-valued function on R and let e > O. An e-almost-period of I is a number r such that sup I/(x - r) - I(x) I <

1:.

x

Examples: r = 0 is a trivial I:-almost-period for all e > 0; if I is periodic then its period, or any integral multiple thereof, is an e-almostperiod for all e > 0; if I is uniformly continuous, every sufficiently small r is an e-almost-period. 5.2 DEFINITION: A function I is (uniformly) almost-periodic on R ifit is continuous and iffor every e > 0 there exists a number A = A(e,!) such that every interval of length A on R contains an I:-almost-period of 1. We denote by AP(R) the set of all almost-periodic functions on R.

Examples: (a) Continuous periodic functions are almost-periodic. (b) We shall show (see 5.7) that the sum of two almost-periodic functions is almost-periodic; hence 1= cosx + cos 1I:X is almostperiodic (see also exercise 1 at the end of this section); noticing, however, that lex) = 2 only for x = 0, we see that I is not periodic. (c) If J is almost-periodic, so are al for any complex number a, and I(h) for any real A.

III, /,

5.3

Lemma:

Almost-periodic lunetiolls are bOllntled.

PROOI': Let I be almost-periodic. Take e = 1 and let A = A(I,f). For arbitrary x E R let't' be a I-almost-period in the interval ['( - A, x]. We have 0 ~ x - r ~ A and I ((\') - r(x - r) I < I, consequently

I/(x) I ~suPo~Y:<;AI/(Y)1

+ 1.

~

An Introduction to Harmonic Analysis

156 Corollary:

Iff is almost-periodic, so is

f~.

I

PROOF: Without loss of generality we may assume If(x) ~! for allxER. Wehavej2(x - t) - j2(x) = U(x-t) + f(x»U(x - t) - f(x» which implies that, for every e > 0, e-almost-periods of f are also e-almost-periods of f2.
5.4

Lemma:

Almost-periodic functions are uniformly continuous.

PROOF: Let f be almost-periodic, e> 0, A =A(eI3,/). Since f is uniformly continuous on [O,A], there exists '10> 0 such that for all

I'll < '10 sup

If(x

+ ']) -

I

f(x) <

~3 .

O~:c~A

Let Y E R; we can find an eI3-almost-period of f say interval [y - A, y], and writing fey

+ '1) -

fey) = (f(y +U(y - t

+ '1) + '1) -

fey -

t

t,

within the

+ '1» +

fey - t»

+ (f(y

- t) - f(y»,

we see that each of the three summands is bounded by e13; the first and the third since t is an eI3-almost-period, and the second since o ~ y - t ~ A and < '10' Thus, if < '10' I!(x + '1) - f(x):1 < e for all x, and the proof is complete. ...

I'll

lId

5.5 For a function fE L"'(R) we denote by Wo(f) the set of all translates of f; Woe/) = {1,}YER· t Theorem: A function fEL"'(R) is almost-periodic if, and only if, Wo(f) is precompact (in the norm topology of C'(R». PROOF: We recall that a set in a complete metric space is precompact (i.e., has a compact closure) if, and only if, it is totally bounded, that is, if for every e > 0, it can be covered by a union of a finite number of balls of radius f.. Assume first that f is almost-periodic and let us show that Woe/) is totally bounded in e'eR). Let e > 0 be given and let A =A( e/2,/); by the uniform continuity of f we can find numbers '1t. ... , '1M in [0, A] such that if O~Yo ~ A, inf) ~j~M 1/1,0 - f"", '" < e12. For arbitrary y E R let t be an eI2-almost-period of f in [y - A, y] ; writing yo=y-t we obtain O~Yo~A and <eI2; con-

//I,-holl",

t

Remember the notation Iy(x) = I(x - y).

VI. Fourier Transforms on the Line

157

sequently, infl ~J&M IIh - hj 1100 < e and Wo(f) is covered by the union of baIls of radius e, centered at/~J' j = 1, ... ,M. Assume now that Wo(f) is precompact. Let e > 0 and let 0 1 , ••• ,OM be balls of radius e/2 such that Wo(f) c U~ OJ. We may clearly assume that 0/'\ Wo(f) # 0 and hence pickhjEOj, j = 1, ... ,M; the balls of radius e centered at hJ clearly cover Wo(f). We claim that every interval J of length A = 2max J ~j ~M 1Yj contains an e-almost-period of /. In fact, denoting by y the middle of J, there exists a jo such that IIh-h}olloo <e; writing T = Y-Yio it is obvious that TEJ and, on the other hand,

I

Hence, all that we have to do in order to complete the proof is show that, under the assumption that Wo(f) is precompact, / is continuous. t We show directly that it is uniformly continuous, that is, that limq_.o I /~ - /11 00 = ,0. If this were false, there would be an e > 0 and a sequence 1/" -. 0 such that I h. - f II 00 > e; by the precompactnes~ there would exist a bubsequence {'7nj} such that/~nJ converges in Loo{R) 1D some functioD g which would clearly satisfy g ~ e. But, since for arbitrary f E LOO(R) (actually for arbitrary measurable !), '1~O imp1iesf{x-1J)-.f(x)inmeasure,t we see that g(x)=f(x) a.e. and this contradiction completes the proof. ~

IIf - II",

5.6 DEfiNITION: The translation convex hull, W(f), of a function feLoo(R) is the closed convex hull of Ul"l~l Wia!). Equivalently, it is the set of uniform limits of functions of the form

(5.1)

Remark: If J is uniformly continuous we can define W(f) as the closure of the set of aU functions of the form (5.1')

cp *f

with

cp E V(R),

II

cp

lILl(R) ~

I.

Another observation that will be useful later is: (5.2)

t

That is: / is equal a.e. to a continuous function. That is: for every finite intervall and every e > 0, the measure of in {x;l!(x - 1)-/(x) I > e} tends to zero as 1) ~ o.

t

158

An Introduction to Harmonic Analysis

By its very definition WU) is convex and closed in LOO(R). Since W(f)::> Wo(f), it is clear that if W(f) is compact then Wo(f) is precompact; the converse is also true: if Wo(f) is precompact, there exist for every e> 0, a finite number of translates {!,)7:1 such that every translate off lies within less than e from h j for some 1 ~ j ~ M. Thus, every function of the form (5.1) lies within e of a function having the form ~ 1 we can j fyj with L~lbjl ~ 1. In the unit disc pick a finite number of points {ck }:= 1 such that every b in the unit disc lies within eM - J II f /I L~(R) from one of the Ck'S; thus every combination j/yj' j I ~ 1 lies within e of some

L7=lb

Ibl

Lb

Lib

(5.3)

It foIlows that W(n is covered by the union of M N baUs of radius 36 centered at the functions of the form (5.3); hence W(f) is precompact and being closed it is compact. We have proved:

Lemma: W(n is compact if, and ollly if, Wo(f) is precompact, that is, if, and only if: fE AP(R). 5.7

Theorem:

AP(R) is a closed subalgebra of Loo(R).

PROOF: In order to show that AP(R) is a subspace, we have to show that if f, g <= AP(R) so does f + g. We clearly have W(f + g) s W(i) + W(g) and since, by 5.6, WU) and W(g) are both compact, W(f) + W(g) is compact and hence W(j + g) is precompact. Since W(f + g) is closed, it is compact, and by 5.6, f + g E AP(R). It follows from the corollary 5.3 that f2, g2, (f + g)2 E AP(R) and consequently f g = 1«(f + g)2 - j2 - g2) is almost-periodic and we have proved that AP(R) is a subalgebra of LOO(R). In order to show that it is closed, we consider a function f in its closure. Since f is the uniform limit of continuous functions, it is continuous. Given e> 0 we can find agE AP(R) such that f 00 < e/3, and if t is an e/3 almostperiod of g we have

I

fT -

f

= Ut -

gil

g t)

+ (g t - g) + (g - f) ,

hence II ff - f 1100 < e/3 + e/3 + [;/3 = e and t is an c-almost-period of f. Thus every interval of length A(e/3,g) contains an e-almost-period of J, and f is almost-periodic. ~

159

VI. Fourier Transforms on the Line 5.8 DEFINITION: the form

A trigonometric polynomial on R is a function of

f(x) The numbers

~j

= )~

......1

aJi{jX,

are called the frequencies off.

By theorem 5.7, all trigonometric polynomials and alI uniform limits of trigonometric polynomials are almost-periodic. The main theorem in the theory of almost-periodic functions states that every almost-periodic function is the uniform limit of trigonometric polynomials, and actually gives a recipe, analogous to Fejer's theorem for periodic functions, for finding the approximating polynomials (see 5.18). 5.9 set

DEFI~1TI0N:

The norm spectrllm of a function hE LOO(R) is the

q(h) = g;~ER, aei~XEW(h) for sufficiently small a:;l:O}. q(h) may welI be empty even if h :;I: 0; for instance, if hE Co(R) we have W(h) c Co(R) and consequently q(h) = 0. We notice that from (5.2) and our definition above it follows immediately that (5.4)

Lemma:

If hE LOO(R) then q(h) S I( h).

PROOF: Since hy = e'{Y h it is clear that I( hy) = I( h) and consequently I(j) s I(ii) for any fE W(h). If f = a i{x, then J = ab~ (<5~ being the measure of mass one concentrated at ~) and I(j) = g}; thus if ~Eq(h) then ~EI(Ji). ~

5.10 Lemma: Let h be bounded and uniformly continuous. Assume that '1K('1x) * h converges uniformly as '1-t 0 to a limit which is not identically zero. Then 0 E q(h). PROOF:

Writing

I(g~)c[-'1,'1]

g~

=

'1 K('1x) * h we have

gq =

--~~

f«~f'1)h, hence

and it follows that I(lim~->og'l)={O}. By 4.11, lim,,->og'l is a constant, and by the remark following definition 5.6, g'lE W(h) and hence lim'l->og"E W(h); now, Iimg" being a constant different from zero, we obtain 0 E a(h). ~ Corollary: Let J-L be a measure on :it and assume p({O}):;I: O. Let h(x) = fei<xdp(~) (so that J-L= h); then OEq(h).

An Introduction to Harmonic Analysis

160

PROOF: Keeping the notations above, we have i" = KWYJ)p and consequently i" tends to p({O})c5 o in MeR) which implies gil -+ p({O}) uniformly. ~

It is clear that 0 E R plays no specific role in 5.10; if p(gn '" 0 we have ~Ea(h) (h as above). Also, it is not essential to use Fejer's kernel: if FE V(R), and if we assume that YJF(YJx) * h converges uniformly to a nonvanishing limit, it follows that 0 E a(II). This can be seen as follows: given a sequence en -+ 0, we can write F = Gn+ lin such that Gn,H n E V (R), Gn has compact support, say included in (- Cn, cn), and I Hn IIL'(R) < en' Writing Gn,,,(x) = YJGn(YJx), H n,,, = YJH.(YJx) and noticing that I H n,,, * h < en II h II, we obtain

5.11

Remarks:

ik"(R)

~'----,

limn .... 00 Gn,,,* h=lim" .... oF,,* h. Remembering that r(Gn~~ -YJcn,YJcn) we obtain, letting YJ-+O faster than Cn -+ 00, r(limF,,*h) = to} as before. The condition of existence of a uniform limit of F" * h as YJ -+ 0 can clearly be replaced by the less stringent condition of the existence of a nonvanishing limit point, that is, a limit of some sequence F". * h with YJn -+ O. We restate these remarks as:

Lemma: Let fEAP(R) and assume Orj:a(f); then for all FE V(R) lim 1/ YJF(flx) *fll LOO(R) = O. " .... 0

Let FEV(R); with no loss of generality we assume It follows that YJF(YJx) *f E W(f) and, if it did not tend to zero as YJ -+ 0, it would have, W(f) being compact, other limit points. By the preceding remarks this would imply 0 E aU). ~ PROOF:

II F II LI(R) ~ 1.

5.12 Lemma 5.11 has the following converse:

J

Lemma: Let fE AP(R), FE V(R) and F(x) dx '" O. If for some sequence YJn-+O, Iim n-+ oo IIYJnF(YJnx)*fIILoo(R) =0, then Orf:a(f). We notice first that for any translate of J, hence for any linear combination of translates, and hence for any gEW(f), we have PROOF:

limn .... "" II YJnF(YJn x ) *g I L"'(R) = O. If g = const, YJnF(YJnx) * g = F(O)g and consequently the only constant in W(f) is zero, that is, 0 ¢ a(f). ~

5.13 Theorem: To every fEAP(R) there corresponds a unique number M(f\ called the mean value of J, having the property that 0rj:(J(f - M(f».

VI. Fourier Transforms on the Line

161

PROOF: We have seen before that uniform limit points of I'/K(I'/x)*f as 1'/-+0 are necessarily constants. Since I'/K(l'/x).fE W(f) and since W(/) is compact, there exists a number a such that for an appropriate sequence I'/n-+O, '1nK(l'/nx)*f converges uniformly to a. Since K(O) = I, I'/n K(I'/nx) * (f - a) -+ 0 uniformly; hence, by 5.12, 0 ¢ a(f - a). If P is another number such that 0 ¢ a(f - p) we obtain, using 5.11, that

I'/K(I'/x)*[(f - a) - (f -

P)]

= "K(f/x)*(f - a) -I'/K(I'/x)*(f-P)

converges to zero uniformly as 1'/-+0. However,I'/K(l'/x)*[(f-a)-(f- P)] = P- a identically and consequently P= a. Thus the property o¢ (J(f - a) determines a uniquely and we set M(f) = a. ~ Corollary: If fE AP(R) and FE V (R), then I'/F(I'/x) * f converges uniformly as 1'/ -+ 0 to f(O)M(f).

In particular, taking F(x)

IXI < 1

= {:

Ixi ~ 1

writing T= f/-l, and

evaluating the convolution at the origin, we obtain: Corollary:

(5.5)

For fE AP(R), M(f) =

1 lim 2T T--+oo

JT f(x) dx . -T

Using the mean value we can determine the norm spectrum of f completely. By (5.4) it is clear that ~ E a(f) if, and only if, oE (J(f e- ii!,,) and consequently (5.6)

By our definition of M(f) and by corollary 5.10 it is clear that if j is a measure then j({O}) = M(f) and similarly (5.6')

j({~}) = M(fe-i~");

thus we can recover the discrete part of]. We shall soon see that j has no continuous part when fE AP(R). 5.14 The mean value clearly has the basic properties of a translation invariant integral, namely: (5.7) (5.8)

M(f + g)

=

M(/)

+ M(g),

M(af) = aM(f),

162

An Introduction to Harmonic Analysis

(S.9)

M(h)

= M(})

(where hex)

= I(x -

y».

It is also positive:

Lemma: Assume IE AP(R), I(x) zero. Then M(}) > O. PROOF:

~

0 on R, and I not identically

By (S.9) we may assume 1(0) > 0 and consequently, if

a> 0 is small enough, I(x) > a on -a < x < a. Let A = A(a/2,}); every interval of length A contains an aI2-almost-period of f, say 't, and I(x) > al2 in (. - a,. + a). It follows that the integral of lover any intervai of length A is at least a2 ; hence M(j) ~ a2 /A. ...

5.15

We define the inner product of almost-periodic functions by:

(S.10)


= M(fg)

and claim that with the inner product so defined, AP(R) is a preHilbert space, that is, satisfies all the axioms of a Hilbert space except for completeness. The bilinearity of 0 unless 1=0 has been established in S.14. In this pre-Hilbert space, the exponentials {ei~X}~ ER form an orthonormal family, since if ~= 17 if ~ # 17. We now introduce the notation t

(S.II) that is.J( {~}) are the Fourier coefficients of I relative to the orthonormal family {ej~X}~ ER' Bessel's inequality now reads

2: IJ({~})12 ~ (f,/)M

(S.12)

~ E

=

M(1/12)

R

and it follows that J({~}) = 0 except possibly for a countable set of ~'s. Combining this with (S.6) we obtain that for all IE AP(R), (J(f) is countable.

J

t If is a measure on R, (5.11) agrees with (5.6 '). By abuse of language we shall sometimes refer to J({ ¢}) for arbitrary JE AP(R), as the mass of the pseudomeasurc Jat ~.

VI. Fourier Transforms on the Line

163

We have not discussed yet the question whether the orthonormal family {i~Xh e R is complete or not. It is well known that the completeness of an orthonormal family is equivalent to the uniqueness theorem, that is, to the statement that zero is the only element all of whose Fourier coefficients (relative to the orthonormal family under consideration) vanish. Thus in our case the completeness of {ei~Xh eft follows from LeI IE AP(R), 1=1= O. Then a(f) =1= 0.

Theorem (uniqueness theorem):

PROOF: Assume first that 1 is a measure on R. Saying that a(/) =1= 0 is equivalent to saying that 1 has a nonvanishing discrete part. By Wiener's theorem 2.9, if lis continuous then lim (2T) -lf~T I/(x) 12 dx =0 and, by 5.14, 1= O. We now show that if IE AP(R) there is always a (nonzero) function h E W(!) such that Ii is a positive measure on R. Since by the previous argument (J(h) would then be nonempty, and since a(h) £ a(/), the theorem will follow. 'Xl ~ 1 we put We construct h as follows: assuming"

I"

1 hT{x) = 2T

IT

-T

J(y)/(x

+ y)dy

and, noticing that liT E W(/) for all T> 0 and remembering that W(!) is compact, we take h to be a limit point of hT as T --+ 00, say h = lim hT". We have h(O) = lim hr.,(O) = M(1J12) =1= 0, hence h =1= O. h is clearly continuous and we claim that it is positive definite: if Xj E Rand Zj are complex numbers, j = t, ... , N, then

1 = lim 2T n-+cr,

n

JTH I) ," zj/(x j + y) 12dy ~ O. -Tn

--I

Thus, by Bochner's theorem 2.8, h is the Fourier transform of a pdsitive measure or, equivalently, jj is a positive measure and the proof is complete. ... Corollary: (5.13)

For IE AP(R)

An Jntroduction to Harmonic Analysis

164

Theorem: II IE AP(R) and lEM(R), then II1IlM(R) = LI1({~})I, and I(x) = Ll({~}) ei~x.

5.16

1= Ll({~})8~,

By (5.6)" Ll({~})8~ is the discrete part of 1, hence II1({~}) I ~ "lIlM(R). If we denote the continuous part of 1 by fl, then fl is the Fourier transform of the almost-periodic function f - Ll({~})ei~X and, as in the first part of the proof of 5.15, Il = 0; hence/= Ll({~})ei~X and II 1 I Mdt) = LI1({~})I. ~ PROOF:

5.17 For arbitrary IE AP(R), the series Ll({~})ei~", to which we refer as the Fourier series of J, converges to f in the norm induced by the bilinear form )M. Our next goal is to show that, as in the caSe of periodic functions, the Fourier series of any IE AP(R) is summable to I in the uniform norm. Before describing the summablity process we introduce the mean convolution / * g of two almost-periodic functions.

<., .

M

Lemma:

Let J,gEAP(R). Define: T

(5.14)

(f *g)(x)

= My(f(x- y)g(y»

= lim 21rJ T~~

M

f(x- y)g(y)dy.

-T

I tt g is almost-periodic and, if M( Ig I) ~ 1, then f * g E W(f). Furthermore, if g is a trigonometric polynomial, then Then

We may clearly assume from the beginning that It follows that for all T~ufficientJy large PROOF:

2iT

f:/(X -

Mqgf) < 1.

y)g(y)dyE W(f)

and, combining the compactness of W(f) with the fact that pointwise (5.14) is well defined, we obtain 1* g as the uniform limit of M

(2Trlf~rf(x - y)g(y)dy. lfg(x)

= Li({~})e~ (the sum being finite)

then and since / tteiF.x = M,.(f(x - y) ei~ = Mt(.f(t)ei~("-t~ =]({ ~})ei~",

the proof is complete.

VI. Fourier Transforms on the Line

165

Remark: Notice that the function h introduced in the proof of 5,15 is nothing but 1* i( - x). M

5.18 Lemma: Given a finite number 01 points ~I""'~NER and an 6 > 0, there exists a trigonometric polynomial B having the lollowing properties: (a) 8(x)

~

0

(b) M(B) = 1 (c) B(gj}) > 1 - e

for j

= 1,

.. " N .

PROOF: We notice first that if ~ I' ... , ~N happen to be integers and if m is an integer larger than 6- 1 max 1~j I, then the Fejer kernel of

order m,

namely

Km

=

f (1 - _1_k_I_) e +1

ikx

-m

has all the pro-

m

perties mentioned. In the general case let ..1.1"'" Aq be a basis for ~ h .•• , ~N; that is, ..1.1" •. , Aq are linearly independent over the rationals and every j can be written in the form j = l A j.k)'k with integral A j •k • Let 61 > 0 be such that (1 - el)q > 1 - e, and let

e

e L

n1

1 I;

we contend that B = = I Km(Akx) has all m > e; I maXj,k A j.k the required properties. Property (a) is obvious since B is a product of nonnegative functions. In order to check (b) and (c) we rewrite Bas

the summation extending over 1k I 1~ m, .. ,,1 kq 1~ m. Because of the independence of the A/s there is no regrouping of terms having the same frequency and we conclude from (5.15) that 8 ({O}) = the constant term in (5.15) = M(B) = 1, which establishes (b), and

which establishes (c).

Theorem: Let IEAP(R). Then I can be approximated uniformly by trigonometric polynomials Pn E W(f), Since a(J) is countable we can write it as gJ/~ I' For each n let Bn be the polynomial described in the lemma for ~h ... , ~n and F. = l/n. Write Pn = 1* Bn. By 5.17, Pn E W(/) and taking account PROOF:

M

166

An Introduction to Harmonic Analysis

of (c) above, limP.({e j }) =j(gj}) for every ~;E(J(j). If ~¢;(J(j) we have p.({~}) = j({ ~}) = 0 for all n. It follows that jf g is a limit point of p. in We!), then g( {~}) = j({~}) for all ~ and by the uniqueness theorem g = f. Thus,fis the only limit point of the sequence Pn in the compact space W(j) and it follows that p. converge to f (in norm, i.e., uniformly.) ~ Corollary: Every closed translation iI/variant sub5pace of AP(R) is spanned by exponentials.

5.19 We finish this section with two theorems providing sufficient conditions for functions to be almost-periodic. Though apparently different they are essentially equivalent and both are derived from the same principle. We start with some preliminary definitions and lemmas. For Iz E AP(R), we say, by abuse of language, that h is an almostperiodic pseudo-measure. DEFINITION: A pseudo-measure v is almost-periodic at a point ~o E ft, if there exists a function q; E A(R),
It is clear that v is almost-periodic at ~o if, and only if, I/Iv is almostperiodic for every 1/1 E A(R) whose support is sufficiently close to ~o (e.g., within the neighborhood of ~o on which the function q; above is equal to one). In particular, v is almost-periodic at every ~o ¢,L(v).

Lemma: Let vE :FLoc; and assume that rev) is compact and that v is almost-periodic at every point of L(v). Then v is almost-Periodic.

By a standard compactness argument we see that there exists an '1 > 0 such that I/Iv is almost-periodic for evcry t/I E A(R) which is supported by an interval of length '1. Let t/I j E A(R) have their supports contained in intervals of length '1, j = 1,2, ... , N, and such that I~ t/lj= 1 on a neighborhood of rev). By the assumption concerning the supports of 1/1;, I/Ijl' is almost-periodic for all j, and consequently PROOF:

is almost-periodic.

VI. Fourier Transforms on the Line

167

5.20 Theorem: Let hE L""(R) and assume that I.(/,) is compact and that h is almost-periodic at every ~ E except, possibly, at ~ = O. Then h EAP(R).

it

5.21 Theorem (Bohr): Let hE Loo(R) and assume that it is differentiable and that h'EAP(R). Then hEAP(R) These two theorems are very closely related. We shall first show how theorem 5.20 follows from 5.21, and then prove 5.21. 5.20: We begin by showing that if I.( iz) is compact, then /'. , = i~1I (see exercise 4.6). LetfES(R) be such that 1m = 1 in a neighborhood of I.(iz). We have fz = Jfi and consequently h = j * h or hex) = fj(x - y) h(y)dy. Since h is bounded and j E S(R) we can differentiate under the integral sign and obtain that h is (infinitely) differentiable and that h' = f' * h. Remembering PROOF OF

h is differentiable and h'

-

/'.

/'.

/'.

that f'(~) = i~ in a neighborhood of I.(Iz), we obtain h' = j'Ji = i~ fi. By theorem 4.1l' there exists a sequence {!PII} in VCR) such that /'. q;n<~) = 0 in a neighborhood of ~ =0, and such that J' 11..f.(R) .... 0

Ii

/'.

II.'F?""

II Til -

This implies (exercise 4.4) that ~nfi - 11' 0, that is, h' is the uniform limit of f[i" * h. Now, since ~" vanishes in a neighborhood of = 0, it follows from 5.19 that !p" * h EAP(R); by 5.7, h' EAP(R), and by 5.21 II E AP(R). ~

e

5.21: Since h is clearly continuous we only have to show that for every e > 0 there exists a constant A(e, h) such that every interval of length A(e, h) contains an e-almost period of h. In view of 5.7 we may consider the real and the imaginary parts of h separately, so that we may assume that h is real valued. Denote PROOF OF

M = sup hex),

(5.16)

r

Let e> O. Let

Xo

and

Xl

m = inf hex). x

bc real numbers such that

(5.17)

we put e1

= ___ 8__

and claim that if r is an 81-almost period of 4l x t - x ol h' then h(xo - r) < In + 8/2. In order to see this we write

An Introduction to Harmonic Analysis

168

h(x i - 't) - h(xo - 't) =

i

X •

h'(x - 't)dx

%0

=

(5.18)

I

x. h'(x)dx +

IX' (1I'(x -

%0

't) - h'(x»dx

Xo

= hex I) - h(x o) +

i

X•

(h'(x - 't) - h'(x»dx,

Xo

I

and, since the last integral is bounded by x I from (5.17) and (5.18) that hex I

-

Xo

Ie

l

= el4

it follows

e 2

't) - h(xo - 't) > M - m - -

and, remembering that h(xi -'t) ~ M we obtain that h(xo - 't) < m +eI2. We now use the points {xo - r}, where 't is an ed2-almost-period of h' as reference points. Let Al = A(ed2,h') and define e2 by e2 = min(etf2,eI2A I ). We claim that every e2-almost-period of h' is an e-almost-period of h. In order to prove it let x E R and let 't I be an e2 -almost-period of h'; we take 'to to be an eI/2-almost-period of h' such that x ~ Xo - 'to ~ X +A I , and write h(x - 't 1 )

-

+ h(xo -

(5.19)

hex) = hex - 't l ) 'to - 't l

) -

-

h(xo - 'to - 't l )

h(xo - 'to)

h(x o - 'to - 'tt) - h(xo - 'to)

+

+ h(x(j -

'to) - h(x)

LXO- 'O(h'(Y) - h'(y -

T 1»

dy.

'to and 'to + 'tt are both el-almost-periods we have m ~ h(xo - 'to - r.) ~ m + el2 and 111 ~ h(xo - 'to) ~ m + el2 and consequently h(xo - 'to - 't 1) - h(xo - 'to) ~ e12. The integral in e. (5.19) is bounded by e2A I ~ el2 and it follows that h(x-rl)-h(x) Thus, every interval of length A(e2' h') contains an e-almost-period

Since

I

of h and the proof is complete.

I

I

1<

...

5.22 Theorem: Let h EL
This is a corollary of 5.20. The set of points ~ such that Iz is not almost-periodic at ~ is a subset of r.(h) and, by 5.20, has no isolated points. Since a countable set contains no nonempty perfect sets, Ii is almost-periodic at every ~ E R and, by 5.19, h E AP(R). ... PROOF:

VI. Fourier Transforms on the Line

169

EXERCISES FOR SECl'ION 5 I. Show that I = cos 21rx + cos x is almost-periodic by showing directly that given E > 0, there exists an integer M such that at least one of any M consecutive integers lies within E from an integeral multiple of 21r.

2. Let h EL oo(R). Show that u(h) contains every isolated point of

r.(fz). 3. Let /,g E A P(R). Show that

1* g = :E]( {~} )t({ ~}) i~x . M

4. Let Ie AP(R) and assume that W(f) is minimal in the sense that if he W(f) and h =F 0 then ale W(h) for sufficiently small a. Show that I is a constant multiple of an exponential. 5. Let Ie AP(R) and assume that Show that I' e A P(R) .

If is uniformly continuous.

6. Show that the assumption that r.(h) is compact is essential in the statement of theorem 5.22. Hint: Consider discontinuous periodic functions. 7. Show that in the statement of theorem 5.22, the assumption that r.(h) is compact can be replaced by the weaker condition that h be uniformly continuous. 8. Deduce 5.21 from 5.20. 9. Let P be a trigonometric polynomial on R, and let e> o. Show that the;e exists a positive '1 = I'/(P, e) such that jf Q EL oo(R), II Q II ~ 1 and r.(Q) c (-1'/,1'/), then range(P

+ Q) + (-e,e)

:2 range(P)

+ range(Q).

I Q' II 00 ~ '1; see exercise 4.6. oo 10. Let hE :!FL • ~o E R. and {'In} a sequence tending to zero. Show

H i1lt: The conditions on Q imply that

that if K('In- t(~ - ~o»h tends to a limit (weak-star), then the limit has the form ao~o. Introducing the notation a = h({ ';o}, K, {'In}) show that h( {';o}, K, {'In}) 12 < 00 where the summation extends over all ~o Eft such that weak-star-limn .... co K('1n- 1(.; - ';o»h exists.

LI

11. Let hE :!FLco • Show that for all ';0 E ft, except possibly countably many, weak-star-lim~~oK('1_1(~ - ~o»fi exists and is equal to zero. 12. Show that if h EL
AP(R) s;;; B

170

An Introlduction to Harmonic Analysis

6. THE WEAK-STAR SPlECTRUM OF BOUNDED FUNCTIONS

6.1 Given a function hE L "'(R), we denote by [h] the smallest translation invariant subspace: of L"'(R) that contains II; that is, the span of {hy}yeR' We denote by [h] the norm closure of [h] in LCO(R), and by the weak-star closure of [h] in LOO(R). Our definition 5.9 of the norm spectrum of h is clearly equivalent to

Dilw.

«(1(h) =

g; ei~x E [h ]}

and we define the weak-star spectrum by (1w ••(h) = g;ei~XE[h]w'}'

The problem of weak -star spectral analysis is: given hE L"'(R), find (1w.(h). The problem of weak-star spectral synthesis is: for hE LCO(R), is it true thalt h belongs to the weak-star closure of span {el~X} ~.l1w'(h)? The conesponding problems for the unifonn topology were studied in section 5.. We have obtained some information about (1(h) for arbitrary hand ccomplete information in the case that h was almost-periodic (see (5.6;»; we proved that the norm spectral synthesis is valid for h if, anld only if, hE A P(R). The problem of weakstar spectral analysis adnnits the following answer: Theorem:

For liE LOO(R.), (1",.(11) = r(Ji).

The subspace of Ll(R) orthogonal to [h] is composed of all the functions f E L'(R) satisfying PROOF:

J

f(x)h(x - y) dx

=0

for all yER

which is equivalent to (6.1)

f * fz( - x)

=

o.

We denote this subspace of L'(R) by [h].L. By the Hahn-Banach tlheorem, ej~x E

[hI.•

for allfE [h].l. We thus have an equiwalent definition of common zeros of {j; f E [11t].L} .

if, and only if,

(1

w.(h) as the set of all

VI. Fourier Transforms on the Line Assume eo rf:'L(h); if e> Oissmallenough(eo - e,eo + e)n'L(h) and hence iffE VCR) and the support ofl is contained in (eo-e, eo we have

<j,h) = JI(X)h(x)dX

171

=0 + e),

= O.

We claim that I is orthogonal not only to h. but also to all the translates of h. hence to [h]. This follows from (6.2)

J f(x)h(x-y)dx = J I(x+y)h(x)clx,

{1,

and since the Fourier transform of f(x + y) is ei hence supported by (eo - e, eo + e), both sides of (6.2) must vanish. There are many functions 1 in A(R) supported by (eo - e,eo + e) such that l(eo) '# 0; it follows that eo is not a common zero of {!;/E[h].L}, hence eo rf: CTw.(h); this proves CTw.(h) S 'L (h). In the course of the proof of the converse inclusion we shaH need the following lemma, due to Wiener. The proof of the lemma will come in chapter VIIl (see VIII.6.3). Lemma: Assume l,jl EA(R) and assume that the support of II is contained in a bounded interval U on which I is bounded away from zero. Theil II = gl for some gEI!(R).

To prove 'L(h) S CT .....(h), we have to show that if eo if. CTw.(h), then h vanishes in some neighborhood of eo. Now, since eo 1: CTw.(h), there exists a function fE LI(R) ~atisfying (6.1) and such that I(eo) '# 0 and consequently I is bounded away from zero on some neighborhood U of eo. We contend that Ii vanishes in U, a contention that will be proved if we show that if fl E LI(R) and the support of II is contained in U then fl * h( -x) = O. By Wiener's lemma there exists a function g E Ll(R) such that II = ij or equivalently II = g *1. Now fl*h(-x)=(g*/)*h(-x) = g*(f*fi(-:""'x}) =0

and the proof is complete. 6.2 The Hahn-Banach theorem, used as in the foregoing proof, gives a convenient restatement of the problem of spectral synthesis. We introduce first the following notations: if E is a closed set on it we denote

172

An Introduction to Harmonic Analysis

(6.3)

1(E)

= {.f;fE

V(R),j(~

=0

on E}

and

(6.4)

neE)

= {g; gEL""(R)

and

(J. g)

=

0 for allfE f(E)}.

feE) is clearly the orthogonal complement in VCR) to the span of {ei~Xh e E and n(E) is the ortlnogonal complement in L""(R) of feE).

By the Hahn-Banach theorem neE) is precisely the weak-star closure of span {ei:xh eE and the prOlblem of (weak-star) spectral synthesis for hE L""(R) can be formulmted as: is it true that hE n(uw.(h»? Equivalently, is it true that fOir A(R),

IE

(6.5)

I(~

=0

on

cr1 w .(h) ~

(J. h) = O?

or: is it true that, (/ E A(R» (6.6)

I(~ = 0 om r.(h) :;. jiz

= O?

(The equivalence of (6.5) and «6.6) follows from (6.2).)

Theorem: Let jEA(R) and izEffLcc and assume that I(~)=o on r.(h). Then r.(jh) is a perfect subset ofr.(j) () bdry(r.(iz». By 4.10, r.Uh) £; r.(/) () r.(iz) and since I vanishes on r.(h), no interior point of r.Ch) is in r.(/). Let be an isolated point of r.(jiz); with no loss of generality we may assume =0 and that [-q, '1] contains mo other point of r.(jh). Writing K:(e> = K('l-le) = sup(O, I ;:::J!1- 1e we see that r.(~Jh) = {O} and consequently (see 4.11) K,,/h = ab, a being a constant # 0 and [) being the unit mass concen1trated at e= O. By 4.11' there exists a function g E VCR) such that .t vanishes in a neighborhood of = 0, say in (-'1"T/1)' and such that IIg-fllv(R) < (l a I/2) Illtll~J,(R) (remember thatj(O) =0). Since /I' K;I~ 1, we have g)hll"Lco < lal/2 and, mUltiplying everything b'!y KQt7 we obtain, (remembering that ~ Kq,,~ = 0) I a I = II abll"Loo < I ,a 1/2 which is a contradiction. Thus r. (jlz) has no isolated points amd the proof is complete.
eo

eo

l)

e

I K:u -

Corollary: If r.(fi) has countlC1ble boundary then h admits weakstar spectral synthesis; that is,. hE n(u...(h». We recall that if r.(I;) itself, mnd not just its boundary, is countable, and if h is uniformly continUiOus, then hE AP(R) (theorem 5.22), that is, admits norm spectral synthesis.

173

VI. Fourier Transformations on the Line

Weak-star spectral synthesis is closely related to the structure of closed ideals in A(R), and we shall discuss it further in chapter VIII. In particular, we shaH show that weak-star spectral synthesis in ~Loo is not always possible. 7. THE PALEY-WIENER THEOREMS

7.1 Our purpose in this section is to study the relationship between properties of analyticity and growth of a function on R, and the growth of its Fourier transform on R. The situation is similar to, though not as simple as, the case of functions on the circle. We have seen in chapter I (see exercise 1.4.4) that a function J, defined on T, is analytic if, and only if, J(n) tends to zero exponentially as n co. The simplicity of this characterization of analytic functions on T is due to the compactness of T. If we consider the canonical identification of T with the unit circle in the complex plane (i.e. t ~..+eit), then a function f is analytic on T (i.e., is locally the sum of a convergent power series) if, and only if, f is the restriction to T of a function F, holomorphic in some annulus, concentric and containing the unit circle. This function F is automatically bounded in an annulus containing the unit circle, and the Fourier series of f is simply the restriction to T of the Laurent expansion of F. Considering R as the real axis in the complex plane, it is clcar that a fUJIctionfis analytic on R if, and only if, it is the restriction to R of a Cunction_ F. holornorphic in some domain containing R; however, < a}, this domain need not contain a whole strip {z;z = x + nor need F be bounded in strips around R or on R itself (cf. exercises 1 through 3 at the end of this section). If we assume exponential decrease of J at infinity we can deduce more than just the analyticity of f on R; in fact, writing

1 1-'

iy, Iyl

F(z) =

;11:

f J(e)i~% de,

we see that if J(e) = O(e-"I~I) for some a> 0, F is well defined and holomorphic in the strip {z;jyl < a}, is bounded in every strip {z; IYI < al},a t < a, and, by the inversion fonnulat, FI!!- = f. Under the same assumption we obtain also that, since JE I!(R), f E I!(R);

t

FIR denotes the restriction of F to R.

An rntroduction to Harmonic Analysis

174

and since forlYI
two con-

f is the restriction to R of a functioll F holomorphic in the strip {z; Iy I < a} and sati.~fying

(I)

f

(7.1)

I F(x

+ iy) 12 dx ~ eDI~~ E

(2)

(2)

PROOF:

=;.

const

VCR) .

(1); write

(7.2)

F(z) =

f jWej~: d~'

-~

2n

'

then by the inversion formula FIR = f; F is well defined and! holomorphic in {z; I y < a}, and, by PlanchereI's theorem:

I

JI

F(x

+ iy) 12 dx

=

~1t

f

Ij(e) 12 e2~y d~

~ III eal {lllz2(RJ'

(1),*(2); writef"(x)= F(x + ;y)(thusf = fo), and consider the Fourier transforms lr We want to show that j,(~) = l(Oe-~Y since then, by Plancherel's theorem and (7.1), we would have JI1(~) 12e2~Yde < const for Iy I < a, which clearly implies (2). Notice that if we assume (2') then, by the first part of the proof, we do have j,(~) = j( e) e -~y. For A> 0 and z in the strip {z; y I < a} we put:

I

(7.3)

Giz) = K,,*F

L:

F(z - u)K;.(u) du,

K denoting Fejer's kernel. G" is clearly holomorphic in the: strip {z; Iy I < a} and"""we Anotice that gA,,(X) = G;.(x + iy) = K). *.t;, and hence gA.,W = K;.( ~)J,.( e). Now since gA.y( e) has a compact sl:lpport (contained in [-..1.,..1.]) we havc gA.,.{O = gA.O(e)e-~Y and conseqrucntly if I I < A, j/O = l(~) e -~y. Since A> 0 is arbitrary, the above holds
e

e

VI. Fourier Transforms on the Line

175

We may clearly replace the "symmetric" conditions of 7. I by unsymmetric ones. Thus the assumption (7.1) for - l I l < Y < a, with a,a l > 0, is equivalent to: (eal~ + e-a~) JEL2(R) For fE U(R) the following two con-

7.2 Theorem (Paley-Wiener): ditions are equivalent:

(1) There exiHs a function F, holomorphic in the upper half-plane {z;y > O}, and satisfying:

f

(7.4)

1F(x

and (7.5)

lim y!O

f

+ iy) 12 dx < const, IF(x

j(~)

(2)

=

+ iy) -

y

>0

f(x)1 2dx = O.

for ~ < o.

0

PROOF: (2) '* (1); define F(z), for y > 0, by (7.2). F is clearly holomorphic, F(x + iy) is the inverse Fourier transform of e -~,. j and, by Plancherel's theorem,

I F(x + iy) I L'(R) =

IIle-~YilL'(R) ~

I j IIv(R)'

which establishes (7.4), and also

I F(x + iy) - f IIO(R) = II/(e -~y as y

1)

I L2(R) -+ 0

J, O.

(1)~->(2);

write fl(X) = F(x

1111 e-~YIIL2(R) =

IIF(x

+ i).

By 7.1:

+ i + iy)IIL2(R)

for -1 < y <

00

and, in particular, by (7.4):

f Ill(OI2e-2~Yd~ ~

(7.6)

const.

Letting y -+ 00, (7.6) clearly implies jl(~) = 0 for ~ < O. By 7.1, the Fourier transform of F(x + iy) is 11(~)ew-Y); hence, by (7.5), j(~)

= jl(~)e~,

and J(~)

=0

for ~ < O.


*7.3 The foregoing proofs yield more information than that ,stated explicitly. The proof of the implication (2) '* (1) also shows that F is bounded for y:S I: > 0 since 11(~)e-{Yld~ is then bounded .. In the proof (1) '* (2) no mention of f is needed nor is the assumptIOn

J

An Introduction to Harmonic Analysis

176

(7.5); if we simply assume that F is holomorphic in the Upper half-plane and satisfies (7.4), We obtain, keeping the notations of true proof above, that e: E UCR) and, denoting by f the function in I3(R) of which 11 e: is the Fourier transform, we obtain (7.5) as a conse~quence (rather than assumption). The Phragmen-Lindelof theorem a1Bows a further improvement:

11

Lemma: LeI F be holomorphic in a neighborhood of the closed upper half-plane {z; y ~ O} and assume that

f

(7.7)

2

IF(x)1 dx <

00

and

I

I

lim r-1log+ F(ri 8 ) = 0

(7.8)

, .... 00

for all 0 <

e< n.

Then (7.4) is valid.

Let p be continuous with compact support on Rand Write G(z) = p*F = F(z - u)g:(u)dur; then is hoI om orphic in {z; y ~ OJ, satisfies the condition (7.8)" and, on R, t PROOF:

/I p

IILl ~ 1.

G

f':!oo

I

I I FIRIiLl

By the Phragmen-Lindelof theorem we have G(z) ~ throughout the upper half-plane, which means that, if y > 0, IfF(x+iY)'1(-x)dxl;£IIFIRIIL2. This being true for every p(continuous and with compact support) such that ~ 1, it follows that

lip Ilv

7.4 Theorem: LeI F be all el1lire (UIICtiOIl alld a > O. Thefollowing two conditions on Fare equivalelll: (1) E U(R) 0/1(1

FIR

IF(z) I = o(e"lzl)

(7.9)

(2) There exists a function IE L2 CR) , (7.10)

F(z)

=

21 n

lee) = 0 for I~I > a, such that

f" l(~)i~: de. _a

t FIR denotes the restriction of F to R.

177

VI. Fourier Transforms on the Line PROOF:

(2):::. (1); if (7.10) is valid we have

IF(z) I ~ I l(Oe-~)lIIL'
f/ 2

Now

and consequently

FIR

which is clearly stricter than (7.9). The square summabiIity of follows from PIancherel's theorem. (1),->(2); assume first that is bounded. The function G(z) = eiaz F(z) is entire, satisfies (7.9) in the upper half-plane, and G(iy) ~ 0 as y ~ 00. By the Phragmen-Lindelof theorem, G is bounded in the upper half-plane and, writing g = G it follows from lemma 7.3 and theorem 7.2 that g is carried by (0,00). Writing I = we clearly have l(e) = gee + a) which implies 1(0 = 0 for < -a. Similarly, considering G 1 = e-iazF, we obtain ](e) = 0 for a and, writing lI(z) = 1/2n: l(e) ei~z de, we obtain, by the inversion theorem, H = so that H = F and (7.10) is established. fn the general case, that is without assuming that F is bounded on R, we consider F./z) = ffJ * F = F(z - u)ffJ{u)du where ffJ is an arbitrary continuous function with compact support. FIp satisfies the conditions in (1) and is bounded on R. Writing lip = F Ip we have lie) = l(e)p(e) and lie) = 0 if ~ a. Since ffJ is arbitrary this impliesl( = 0 for > a and the proof is completed as before. ~

FIR

IR.

e

FIR

J

e>

IR FIR

J

e)

IeI

I I>

IR'

EXERCISES FOR SECTION 7 1. Show that F(z) = I:;=';12-n[(Z + n)2 + n-1r 1 is analytic on R and FIRELI nLoo(R); however, F is not holomorphic in any strip {z;lylO. 2. Show that for a proper choice of the constants {an} and {b n} the function

is entire, GIREL1(R), but G is unbounded on R.

178

An Introduction to Harmonic Analysis

,-

is entire, H /R ELI nL co(R); however, H is unbounded on any line y = const '# O. 4. Let F be holomorphic in a neighborhood of the strip {z; y ~ a} and assume F(x + iy) 12 dx < const for y ~ a. Show that for z in the interior of the strip: 3. Show that H(z) = e-e

z2

F(z)

II

II

II

=

~-;2n:r

fco (F(U ;: ia) _ F(u .+ ia~) du. -co

5. Let p be a measure on

U -

R,

F(z) =

ta -z

U

+ ra -

supported by [ -

f e-i~z

0,

z

aJ.

Define:

dpW.

Show that F is entire and satisfies F(z) = O(eal)'I). Give an example to show that F need not satisfy (7.9). Hint: F(z) = cos liZ. 6. Let v be a distribution on R, supported by [- a, aJ. Define F(z) = e -i~z dv. Show that F is entire and that there exists an integer N such that

I

as / z

1-+

00.

7. Titchmarsh's cOllvolution theorem: (a) Let F be an entire function of exponential type (Le., F(z) = O(eafzl) for some a > 0) and assume that F(x) ~ I for all real x and that F(iy) is real valued. Assuming that F is unbounded in the upper halfplane, show that the domain D = {z; y > 0, F(z)1 > 2} is symmetric with respect to the imaginary axis, is connected, and its intersection with the imaginary axis is unbounded. Hint: Phragmcn-LindeIOf. (b) Let F t and F2 both have the properties of F in part (a) and denote the corresponding domains by Dl'D Z ' respectively. Show that Dl n D]. '# 0 and deduce that Fl F]. is unbounded in the upper halfplane. (c) Let fj EL 2eit), j = 1,2, and assume that fj are both real valued and carried by [ - a, 0]' Show that if It * 12 vanishes in a neighborhood of ~ = 0, so does at least one of the functions fj. Remark: Titchmarsh's theorem is essentially statement (c) above. The assumption that Ij are real valued is introduced to ensure that the corresponding F j , defined by an integral analogous to (7.10), is real valued on the imaginary axis. This assumption is not essential; in fact, part (c) is an immediate consequence of the Paley-Wiener theorems in the case It = /2 (in which case part (b) is trivial), and the full part (c) can be obtained from it quite simply (see [18J).

I

I

I

179

VI. Fourier Transforms on the Line *8. THE FOURIER-CARLEMAN TRANSFORM

We sketch briefly another way to extend the domain of the Fourier transformation. There is no aim here at maximum generality and we describe the main ideas using L"'(R) as an example, although only minor modifications are needed in order to extend the theory to functions of polynomial growth at infinity or, more generally, to functions whose growth at infinity is slower than exponential. For more details we refer the reader to [3].

8.1

For hE C'(R) We write F 1(h,O =

LO", e-i~"h(x)dx

F 2 (1I,0 =

-

, = ~

+ i'1,

'1 > 0

(8.1)

L'" e-i~XIz(x)dx

F l(h, () and F 2(1z, () are clearly holomorphic in their respective domains of definition, and it is apparent from (8.1) that if '1 > 0, then Fl(h,~+i'1)-F2(h,~- i'1) is the Fourier transform of e-~Ixlh. Hence if h EV(R) we obtain

(8.2)

lim (F l(h, ~-o+

e+ i'1) -

F 2(1z,

e- i'1» =

Ii(~)

uniformly. Since e-~Ixlh tends to h in the weak-star topology for any Ii is allowed to be a pseudo-measure and the limit is in the weak-star topology of :F L'" as dual of A(R). Let us consider the case h coV(R). If I is an interval on R disjoint from the support of ft, and D is the disc of which 1 is a diameter, and if we define the function F in D by

hE L"'(R), (8.2) is valid for every hE Lco(R) provided

(11.3)

'I ~ 0, then it follows from (8.2) that F(IJ,O is well defined and continuous in D and it is holomorphic in D" I. It is well known that this implies (e.g., by Morera's theorem) that F(h,O is holomorphic in D. We see that in the case hELl n L"'(R), F 1(11,0 and F 2(II,O are analytic continuations of each other through the complement of L(li) on :R On the other hand, if F l(h, 0, and F 2 (h, 0 are the analytic continuation

An Introduction to Harmonic Analysis

180

of each other through an open interval I, ,Ii(e) = F1(h, 0 - Fih, () = 0 on I, and I r.I.(Ii) = 0. Denoting by c(h) the set of concordance of (Fl(h,O,Fih,O), that is, the set of p.oints on :it in the neighborhood of which F I(h, () and F 2(h, 0 are amalytic continuations of each other, we can state our result as

Lemma: Assume hE V r. LOOCR); then r(li) is the complement of c(h). 8.2

We now show that the same is true without assuming h EVCR).

Theorem: of c(h).

For every bounded function II, r(Ji) is the complement

PROOF: Let E be a compact subselt of c(h); then, as fl--'O+ Fl(h, (+ ifl) - Fih,( - ifl) --. 0 uniformlly for (EE. If/EACH.) and the support of / is contained in E, then

which proves r(Ji) r. c(h) = 0. The fact that eo ¢ r(h) implies eo E c(h) iis obtained from lemma 8.1 and the following simple lemma about removable singularities:

8.3 Lemma: Let I be an interval on the real line, D the disc in the ( plane of which I is a diameter, F a h())lomorphic function defined in D"-./, satisfying the growth condition (8.5)

Assume that there exists a sequence of fiunctions {<1>J, holomorphic in D, satisfying (8.5) (with a constant independent of j) and such that <1>J{(} ~ F(O in D "-.1. Then F call be extended to afunction holomorphic in D. Let DI be a concentric disc properly included in D and D2 a concentric disc properly included in D 1. Denote by (I' e2 the points of intersection of the boundary of /)11 with I. The functions 2 t<1>iO are uniformly boumded on the boundary of D 1 , hence in D 1 , and consequently <1>j are: uniformly bounded in D 2 • The Cauchy integral formula now shows ahat <1>j converge uniformly in D2 to a holomorphic function which aglrces with F on D2 "-. 1. Since D2 is an arbitrary concentric disc in D, the lemma follows. ~ PROOF:

«( - elt«( - e

vr. Fourier Transforms 8.4

Lemma:

on the Line

181

Let hE L"'(R); thm

IF.(h,ol

~

IF 2(h,OI ~

'1>0 ( = ~ +i'1, '1 < O.

Il h ll",'1-' 11"11",1'11-·

(= ~+i'1,

PROOF:

and similarly for F 2. We can now finish the proof of theorem 8.2. We have to show that if ~o ¢: I.(iz), then ~o E c(h). Assume ~o ¢ I.(h); by lemma 4.7, ~o has an interval I about it which does not intersect I.(hK;) provided }. is large enough (K" = the Fejcr kernel and I.( K,,) = [ - t p., 1/ A]). If D is the disc for which I is a diameter, it follows from lemma 8.1 that the pair (FI(hK",O, F 2 (hK",0) defines holomorphic functions <1>" in D, which clearly converge, as ). -+ 00, to (F,(h,O, Fiiz,m on D"T. By lemma 8.4 we can apply lemma 8.3 and the theorem follows. whose only singularity in the finite ( plane is at the point ( = O. By lemma 8.4 and the Phragmen-Lindelof theorem, tends to zero at infinity and has a simple pole at ( = O. Hence, for some constant c, <1>(0 = c/i(, which is the Fourier-Carleman transform of the constant c.

m

9. KRONECKER'S THEOREM

9.1 Theorem (Kronecker); Let AI ,A2, ... ,}." be real numbers, independent over the rationals. Let Cil' ••• ,Cin be real lIumbers and e> O. Then there exists a real number x such that

(9.1)

j

= 1,2, ... , n.

Kronecker's theorem is equivalent to Theorem: Let AI' ... , An be real numbers independent over the rationals, AO = 0, and let ao,a., ... ,an be any complex numbers. Then

9.2

An Introduction to Harmonic Analysis

182

(9.2) We first establish the equivalence of 9.\ and 9.2 and then obtain 9.2 as a limit theorem. PROOF THAT 9.1 * 9.2: Write aj:::: rjei~j, 1'j ~ O. By 9.1, there exist values of x for which ei)j~ - ei(~o-2j)1 is small, j = 1, ... , II. For these values of x, La jei) JX is close to ei~o L~rj' ....

I

9.27- 9. t: Consider the polynomial \ + L~ e - i~J eO.)", and notice that its absolute value can be close to n + 1 only if all the summands are close to I, that is, only if(9.l) is satisfied. .... PROOF THAT

Remark: If )'1' ... , Am 1r are linearly independent over the rationals, 2K we can add the condition e / X - I < f. which essentially means that We can pick x in (9.1) to be an integer. Theorem 9.2 is a limiting case of theorem 9.3", below. The idea in the proof is that used in the proof of lemma V.1.3, that is, the application of Riesz products and of the inequality

I

I

(9.3) which is clearly valid for f, g E AP(R) (see (5.5». Actually, we use (9.3) for polynomials only, in which case the existence of the limit (5.5) and the fact that it equals the constant term are obvious, and this section is essentiaIly independent of section 5. For the sake of clarity we state 9.3 N fir!>t for N = I, as

9.3 Theorem: properties: (a)

""

Let AI' ... ,An be real numbers having the following

- 0,

Cj

= -\,0, I,

=>-

Cj

= 0 for all j.

Ak ,

8j

=

-1,0,1,

=>-

8f

= 0 for j # k.

L, CjA] 1

(b)

i

['./j

=

I

Then, for any complex numbers ai' ... , am (9.4)

VI. Fourier Transforms on the Line

183

TI• (1 + cos(}.jX + C(j»

g(X) ==

•

g is a nonnegative trigonometric polynomial whose frequencies all

have the form ~>jAj' ej = -1,0,1. By (a), the constant term in g is 1, hence M(g) = g = 1. By (b), the constant term (which is the same as the mean value) of fg is t rj and, by (9.3),

M(I I)

Ii=.

1 "2 j

9.3 N Theorem: ing properties:

Let

L" rj ~ supifl· ~

1

he real numbers having the follow-

}'l' "',}n

n

(a)

)'e).j=O,

°

integers,

lejl ~ N

=> ej =

ej integers,

lejl ~ N

=> ej = 0 for j =P

f.j

"""-

•

for allj.

n

(b)

)' e/j = }'k'

""I Then,for any complex numbers

k.

u., ... ,an ,

Virtually identical to that of 9.3; wc only have to replace g as defined there by PROOF:

n KN(Ajx + a) n

g(x) =

•

reader. It is clear that if )'1, ... , An are linearly independent, the conditions

of 9.3 N are satisfied for all N and consequently we obtain (9.4 N ) for all N, hence (9,2). This completes the proof of theorem 9.2 and hence of Kronecker's theorem. *9.4 The extension of theorem 9.1 to infinite, linearly independent sets presents a certain number of problems, not all of which are solved.

184

An Introduction to Harmonic Analysis

We restrict our attention to compact linearly independent sets E and ask under what conditions is it possible to approximate uniformly on E every function of modulus 1, by an exponential. The obvious answer is that this is possible if, and only if, E is finite; this follows from Kronecker's theorem ("if") and the fact that uniform limits of exponentials must be continuous on E, and if E is infinite (and compact) not all functions of modulus 1 are continuous ("only if"). We therefore modify our questions and ask under what condition is it possible to approximate uniformly on E every contilluous function of modul us 1 by an exponential. We do not ha ve a satisfactory answer to this question; for some sets E the approximation is possible, for others it is not, and we introduce the following: A compact set E c R is a Kronecker set if every continuous function of modulus 1 on E can be approximated on E uniformly be exponentials. The existence of an infinite perfect Kronecker set is not hard to establish by a direct construction. We choose, however, to prove it by a less dircct method which also may be uscd to obtain finer results (see [14]). DEFINITION:

Theorem: Let E be a perfect totally disconnected set on R. Denote by CR(E) the space of continuous, real-valued fUllctions 011 E. Then there exists a set G of the first categoryt in CR(E) such that every If' E CR(E)'-. G maps E homeomorphically onto a Kronecker set. PROOF: A function If' E CR(E) maps E homeomorphicaIly onto a Kronecker set if, and only if, for every continuous function II of modulus 1 on E and for every e > 0, there exists a real number A. such that

(9.5)

sup

I

eUop(x) -

lI(x)

I<

1:.

xeE

We show first that if we fix hand e, the set of functions If' for which (9.5) holds for an appropriate A. is everywhere dense in CR(E). For this, let tjJ E CR(E) and let 11 > 0. We take A. = 1011-1 and write E as a union of disjoint closed subsets E}, j = 1,···,N, tbe E/s being small enough so that the variation of either h or ei).tjI on E} does not exceed

t II If'

CR(E) is a complete metric space, the metric being determined by the norm 1/00 = Supx"E IIf'(x) I •

185

VI. Fourier Transforms on the Line

&/3. Let e i ' J be a value assumed by h on E j and ei/l J a value as~ 1t and ~ 1t sumed by /;'0/1 on Ej ; we may clearly assume for all j. We now define

letjl

(9.6)

lP(x)

Ipjl

= tjJ(x) + r:.j - pj A.

We have IPE CR(E) and IIIP - tjJ 1I:xl ~ 21t/A < 11; also, checking on each Ej , it is clear that (9.5) holds. It follows that the set G(h, e) of all IP E CR(E) for which (9.5) holds for no A. E R, a set which is clearly closed, is nondense. Taking a sequence of continuolls functions of modulus 1, say {hll}' which is dense in the set of all such functions, and taking a sequence of positive numbers {em} such that em -+ 0, it is clear that G = Un.m G(h ll , em) is of the first category. Also, if IP rf: G, then every hn can be approximated uniformly on E by ei;'