Real-Variable Methods in Harmonic Analysis (Pure and Applied Mathematics)

00

We start out in an upbeat note because of the positive result, i.e., statement (1.6), for the case p = 2. On the other hand, the case p = 00 was ruled out,

even with L"( T) replaced by C ( T), by the results in Section 2 of Chapter I. Moreover, since for 1 =zp < 00 the Lp(T) spaces are Hilbert only when p = 2, a different approach is needed here. Our first result, although elementary, is of interest since it reduces the question of norm convergence to one of establishing the continuity of a convolution operator in Lp(T); this task seems more accessible. Proposition 2.1. Let 1 d p < 00. Then the following are equivalent: (i) For every f E Lp(T ) , lirnn+,llsn(f) -flip = 0. (ii) IIsn(flllps cpllfllp,for some constant cp independent of n and f:

III. Norm Convergence of Fourier Series

52

Pruuf. (i) implies (ii). If (i) holds, then in particular Isn(f, x)l s cf < a.e. in T, for every f in L P ( T ) .Putting X = Y = L P ( T )in the uniform boundedness principle, Theorem 2.3 of Chapter I, this estimate gives

(ii) implies (i). Let f E Lp(T). Then given E > 0, let g denote a trigonometric polynomial such that Ilf - g ( l ps E. Clearly, sn(g,x) + g(x) as n + co; in fact, both expressions coincide if n 2 degree g. Therefore by Minkowski’s inequality gllp + Ilg - sn(g)llp + Ilsn(g) - s n ( f ) l l p =I+J+K, say. As we pointed out, J = 0, provided n is sufficiently large. Also K = I t s n ( f - g)ll, s c p l l f - gllp d cP&.Therefore Ilf- s n ( j 9 l l p & ( I +c p ) , E arbitrary, provided n is sufficiently large. Proposition 2.1 suffices to rule out norm convergence for p = 1. Indeed, since ((2KN11 = 1, all N , we have that

Ilf-

Ilf-

sn(flIlp

Now we take a closer look at sn(2KN, x); it equals ~ K *N2Dn(x) = aN(2Dn,x), x E T. Moreover, since by Lebesgue’s Theorem 4.2 of Chapter 11, aN(2Dn,x ) + 2Dn(x)a.e. in T, from (2.1) and Proposition 2.1 in Chapter I it follows that I a Ln (4/.rr2)In n, n large. Therefore (ii) of Proposition (2.1) cannot hold and convergence in norm cannot occur. The next reduction in making the problem of norm convergence tractable is to consider an analytic expression which roughly corresponds to “onesided Fourier series.” This is motivated by noting that

-

03

S n U XI

=

1 Xl-n,n](k)Ckeikr k=-a

and is achieved by studying a new object, the conjugate operator.

3. THE CONJUGATE MAPPING To each f E L( T ) ,f

- CcjeiJr,we associate the trigonometric series ‘jJ-i)(sgn j )c,eijt

where 1 0 -1

if j > 0, if j = 0, if j < 0.

(3.1)

3. The Conjugate Mapping

53

If there is an integrable function with Foucer series (3.1), we call this function the conjugate to f and denote it by f: For instance, iff E Lz( T) then also f E L2(T) and IIj(ll, s Ilfl12. This is quite simple, for in this case - 1~01'< CI(-i)(sgnj)cj12 = MI: and by Riesz-Fischer there is a unique L2 function g such that c j ( g ) = (-i)(sgnj)cj; this function is In general we say that Lp(T) admits conjugation if to every f E Lp(T ) , f C cjeiJr7there corresponds a function g E Lp(T) with Fourier coefficients c j ( g ) = (-i)(sgn j)cj and- llgllp s c p ~ ~ f ~ ~ p 7 where cp is a constant independent o f f ; in this case g = f: Closely related to the concept of conjugation is that of projection. Given f E L( T ) , f C cjeiJr7we define its projection pf to be the trigonometric m series Cje0cjeii'. We then say that Lp(T) admits projections if, for every f E Lp(T), w i s the Fourier series of an Lp(T ) function g and llgllp s c p ~ ~ f ~ ~ p , where cp is a constant independent off: The relation between these concepts is given by

-

-

Proposition 3.1. The following statements are equivalent

(i) Lp(T) admits conjugation. (ii) Lp(T) admits projections.

Proof. (i) implies (ii). Let f~ L P ( T ) , f- C cjei'', and put g = c,/2 + (f+ Since Lp admits conjugation, g E L P ( T ) ,and llgllp s cpllfllp. By inspection, we observe that c j ( g ) = c j ( f ) ,j k 0, and cj(g) = 0 if j < 0; thus g = pf and we are done. (ii) implies (i). Conversely, now let

8)/2.

Since Lp admits projections, g E Lp(T) and llgllp c cpllfll,. Again by inspection we notice that c j ( g ) = (-i)(sgn j)cj(f), and therefore g = f The main reason for the study of the conjugate operator is given by Theorem 3.2. The following statements are equivalent (i) For every f E Lp(T ) , lim.+mlls.(f) -flip = 0. (ii) Lp(T) admits conjugation.

Proof. By combining Propositions 2.1 and 3.1, it suffices to prove that the following pair of statements are equivalent: (i') Ilsn(f)llpC c p ~ ~ f cp \ ~independent p, of n and f: (ii') Lp(T) admits projections.

54

I l l . Norm Convergence of Fourier Series

-

(i') implies (ii'). let f E Lp(T ) , f C cJe"'. Since by identity (3.2) of Chapter I cJ(e-'"'f) = c,,, a simple computation gives 2n

etnrsn(e-lnxt ) =

C cJe'J'.

(3.2)

J=o

We denote the polynomial on the right-hand side of (3.2) above by Pnf,the n th partial projection of j Therefore IIPnfllp = IISn(e-'"'f)llp

~~Ilfll~,

all

n.

(3.3)

Next we show that the sequence Pnf is actually Cauchy in Lp(T). Indeed, given E > 0, let g be a trigonometric polynomial of degree N such that IIf- gIIp &+ Then by (3.3) llpn(f) - pn(g)lIp = IIPn(f- g)llp cP&. Now, if n, m > N / 2 , P n ( g ) = P , ( g ) and consequently

IIpn(f) - Pm(f)Ilp s Ilpn(f) - P n k I I l p + llPrn(n - pm(g)llp s 2~p&, and the sequence is Cauchy. Since L P ( T )is complete, there is a function h E L P ( T )such that limn+wllPn(S)- hllp = 0. Moreover, by (3.3), llhllp S c p ~ ~Itf is~ clear ~ p that limn+wcJ(Pnf) = cJ,j 2 0. Also by Theorem 1.1 of Chapter I, c,(h) = limn+wcJ(Pnf),j 2 0. Consequently, h = PJ; and we are done. (ii') implies (i'). We rewrite (3.2) as e'"'sn(e-'"'f, t ) = P f ( t ) - e ' ( 2 n + l ) ' p ( e - ' ( 2 n + ' ) * f ,

t).

(3.4)

Consequently, multiplying through by e-'"' and replacing f by eln'f (3.4) becomes sn(fTt ) = e-'"'P(e'"% t ) - e'(n+l)'P(e-'(n+l)'f;t ) .

Whence it follows that IIs,(f)llp s IIP(e'"'f)ll, with cp independent of n and$ W

+ IIP(e-'(n+l)tf)llp

(3.5) 2cpllfllp,

Corollary 3.3. L( T ) does not admit conjugation.

Proof. Since, by the example following Proposition 2.1, (i) of Theorem 3.2 does not hold, (ii) cannot hold either.

4. MORE ON INTEGRABLE FUNCTIONS

We already know that L( T) does not admit conjugation; in this section we show that, in fact, the situation is as hopeless as it could be.

55

4. More on Integrable Functions

- 1 cje"', such that the

Example 4.1. There exists an integrable function f; f trigonometric series C(-i)(sgnj)cjeiJr (a) converges everywhere in T and (b) is not a Fourier series.

The construction of this example is achieved through a sequence of results of independent interest, which are, therefore, stated separately. As a byproduct of the construction we are able to show that although the Fourier coefficients of an integrable function tend to 0, they do so arbitrarily slowly. We begin introducing a particular kind of sequence, one which is relatively simple to handle. Definition 4.2. We say that a real sequence {a,}, n

+ a,,,

- 2a, 5 0,

L0

n

all

3

is convex if

1.

(4.1)

Examples of convex sequences abound; indeed, it suffices to put a, = + ( n ) , where is a convex, real-valued function. A basic property of these sequences is

+

Proposition 4.3. If {a,}:=, is positive, convex, and bounded, then it is n ( a , - , - a,) = 0, and decreasing, m

Cj(aj-l

+ aj+l - 2aj) = a, - lim a,. n+m

j=1

Proof. Convexity is clearly equivalent to ajPl- aj 3 aj - a j + l ,all j.

(4.2)

We begin by showing that a,-1 - a, L 0, for all n.

(4.3)

Suppose that (4.3) does not hold. Then there exists a value of n such that a,-l - a, < 0, and by (4.2) we also have that lajpl- ail 2 la,-l - a , [ , all j 3 n. Therefore, since for m > n m-1

a, - a, =

C lajpl- ail z (rn - n ) ) a , _ , - a,l,

j=n

it follows that limm+ma, = co,which is not true. Consequently, (4.3) holds and the sequence is decreasing. Let a = a,. Then

a, - a

=

(a, - a,)+ . .

+ (anpl- a,) +

* *

,

III. Norm Convergence of Fourier Series

56

where the series on the right has monotone decreasing terms and converges. It is therefore well known and readily seen that limn+mn(a,-, - a,) = 0. Finally, since n

s, =

1j ( a j - ]+ aj+]- 2aj) = a, - a, - n(a, - a , - ] ) , j=1

limn+ms,

=

a, - limn+ma,.

We would like to show that given a positive, convex sequence a, = 0, there exists a unique f E L( T) such that

f

- 1 aljleiJ‘.

(4.4)

One of the problems we are faced with now is that if even we had = in (4.4) it would be nontrivial to show that f E L( T). On the other hand, this difficulty is rather easily overcome if we can also write the relation (4.4) as

where A, 2 0 and the KnPl’sare the FejCr kernels of order n - 1, because then the Fourier coefficients off are essentially known (once the An’s are fixed) and the convergence and integrability of the series can be readily established by the fact that we are only dealing with positive quantities. More precisely, f will be integrable provided that 1 A, < 00 since by the identity (2.7) in Chapter I1

Moreover, by the formula (2.2) in Chapter 11, it also follows from (4.5) at once that

The An’s are still unknown, whereas the cj’s are precisely the aljl’s; therefore, in order to complete our proof we must be able to show that the

57

4. More on Integrable Functions

linear system with infinitely many unknowns given by (4.6) above, i.e., W

(4.7) n=j+l

can be solved when the aj's are convex and tend to 0. Written in matrix form, (4.7) becomes

[I=[;

1 1 1 -

2

(4.8)

0

Thus by "inverting" the matrix in (4.8) by means of elementary operations, we readily see that under our assumptions this system of equations is equivalent to

8;

1 0 0

=[

:;:I[I!]

..-

. .

.

In other words, the solution is given by

A, = n(a,-,

+ a,,,

-2

4,

n

3

1.

(4.9)

Also by Proposition (4.3), 1 A, < co,which is one of the required conditions. Summing up, we have Proposition 4.4. Given a convex, positive sequence of real numbers { c ~ } ~ ~ which tends to zero, there is a positive, integrable function f such that f 1 cljle"'.

-

Proof. Put f(t) = C :=,

n(c,-,

+ c , + ~ - 2cn)K,-,(t).

H

We collect one more observation, this one dealing with Fourier series with odd coefficients. Proposition 4.5. Let f E L( T ) ,f all n 3 0.Then I:='= c, , / n < co.

- C c,e'"', and assume that 0 =s

c, = --c-,,

Proof. Since co = 0, by Proposition 3.1 of Chapter I, F ( t ) = f(-r, t , f ( s )ds is a periodic, continuous function and c n ( F )= (l/in)c,, In1 3 1. In particular, since F is continuous at 0, by FejCr's theorem in Chapter 11, limn+m a,(F, 0) = F ( 0 ) .

III. Norm Convergence of Fourier Series

58

Now, by formula (2.2) in Chapter I1

= c,(F)

+ f f (1 -*):,

(4.10)

,=I

and this expression converges to F ( 0 ) as N

+ 00.

Therefore

say, also converges as N + 00. But since f E L( T ) , c, + 0 as n + a,and by Proposition 1.2 of Chapter I1 also c, + O( C, l), i.e., BN + 0 as N + 00. Thus limN-,mAN exists. Construction of Example 4.1. Let cj = (1/(2 In1j l ) ) , Ijl 3 2. By Proposition 4.4 there is a positive, integrable function f such that f C cjeijt.Since the sequence cj is even we also have f C,z2 cos(jt)/ln(j). The formal, or distributional, conjugate off is given by the trigonometric series Cljla2(- i ) x (sgn j)(1/2 in1jl))eiJ'orZ,z2 (sin jt)/(ln j ) . This is an everywhere convergent trigonometric series (cf. 7.23 below), which by Proposition 4.5 is not a l / j In j diverges. Fourier series since Finally, observe that if { c , } ~ is = ~any complex sequence which converges to 0 as n + 00, then there is a positive, real sequence {a,}:=o which is convex, ~ ,n. The integrable functionf of Proposidecreases to zero, and a, L ~ c , ,all tion 4.4 corresponding to this sequence of an's,then, has Fourier coefficients which tend to 0 at the rate of the c,'s. There is yet a different way to state this property.

-

-

z2:,

Proposition 4.6. Let f e Lp(T), 1 s p < 00. Then we can find a positive, even sequence { A j } , Aljl increasing to a,and a function g E Lp(T ) such that cj(g) = Aljlcj(f), all i-

Prmf. For each n, let m, be an integer such that Ilf - amn(f)llp 6 2-". This choice is possible on account of Theorem 2.7 of Chapter 11. We may assume that m, increases to 00. Then put

5. Integral Representation of the Conjugate Operator

59

where the limit is taken in the Lp sense. By Proposition 1.1 in Chapter I

c 00

ci(g) = (1

+

min( 1,"-))m,

+1

ci(f).

n=l

The above proof clearly holds for C ( T) as well.

5. INTEGRAL REPRESENTATION OF THE CONJUGATE OPERATOR A key ingredient in the consideration of the partial sums or CesPro means of an integrable function is the integral (actually a convolution) representation of the linear operator determined by these mappings. We now attempt a similar approach for the conjugate mapping given for f 1 cjeii' by

-

f + C(-i)(sgnj)cieijr.

(5.1)

i

The first step in this direction is to study the behavior of the partial sums and CesPro means of the distribution F with Fourier series

-

>

F C(-i)(sgn j)eii'. (5.2) These are denoted by 2Gn(t) and 2 k n (t) as they correspond to the conjugate Dirichlet and FejCr kernels respectively. It is not hard to obtain the explicit expression for these kernels. Indeed,

2Gn(t) =

C (-i)(sgn

n

j)egr = ( - i ) l ( e g r - e-"').

lilsn

j=1

To evaluate this, observe that the geometric sum ireint/2( e-int/2 einr/2) eijr = ,it (1 - ein') eir/2(e-ir/2 - eir/2) (1 - eir) j=1

-

+1) 1/2 sin( n tl2) sin( t/2) * By combining this with the expression obtained by replacing t by --t above, we see that i(n

sin( nt/2) (-i)(ei(n+l)r/Z - e - i ( n + l ) r / 2 = 2 sin(t/2) = 2 sin(n/2)

+

sin((n l)t/2) 2 sin( t/2)

- cos(t/2) - cos((n + 1/2)t) 2 sin( t/2)

1

2 sin( t/2)

'

n = 0, l , . . . .

(5.3)

III. Norm Convergence of Fourier Series

60 Thus

fin(t) is an odd function bounded by

Also as in estimate (1.17) of Chapter I it follows that Ifin(t)l

o <~t< l .rr, all n.

s ,r/ltl,

As the for conjugate Fejir kernel i n ( t )=

fio(t)

k,(t)we have

+ - + E"(t) *

*

n+l

- cos( t/2)

2 sin( t / 2 )

1

--

(n

+ 1)

j=o

cos((j + 1/2)t) 2 sin( t/2) *

(5.6)

The last sum in (5.6) equals

- 2 cos((n - sin((n -

+ l)t/2) sin((n + l)t/2) 2 sin( t/2)

+ 1)t)

2 sin( t / 2 )

'

Consequently,

Thus

in( t ) is also an odd function bounded

by

Moreover, by estimate 1.16 of Chapter I it also readily follows that

This is all we need to know about these kernels. Now, one way to insure that the mapping (5.1) be bounded in Lp(T) is to show that the distribution F with Fourier series (5.2) is actually the

5. Integral Representation of the Conjugate Operator

61

Fourier series of a finite measure p on T. By Theorem 3.5 of Chapter 11, this will be the case if (and only if)

11 kn 11

1

s

all

C,

n.

(5.10)

Rewrite

Since as it is readily seen

I

Isin((. + 1)t) T s (n 1) sint 2t(n 1)

+

+

for O < t < ~ 1 2 ,

we have

I1

-

sin((n + 1)t) ( n + 1) sint

1 2

provided

2-

T

T

n+l

2

-< t < - .

Consequently,

= 3n(JZ/2

- In(sin(T/(n

+ I)))) +

00

as n + 00. This makes it impossible for (5.10) to hold. Let us return, then, to our original program of obtaining an integral representation for (5.1). Assume first that f = ClilrN cjeii*is a trigonometric polynomial of degree siV and let

;,,(A

t)=

1(-i)(sgnj)cjeijr.

(5.11)

lilsn

By the convolution formula (3.3) of Chapter I we readily see that

'I

2fin * f ( X ) = -

T

f ( x - t ) f i n ( t )dt.

;,,(A x ) = (5.12)

T

Since &(t) is odd (5.12) can be rewritten as either

7 -1

f ( x + t)s,,(t) dt or -

)fin(?)dt.

(5.13)

Moreover, since the integrand in the last integral of (5.13) is even we also have

(5.14)

III. Norm Convergence of Fourier Series

62

Since f is a trigonometric polynomial, fn(f,x) actually coincides with f(x) for n 2 N. In particular, limn+.a,;,(A x) = f(x) everywhere. On the other hand, since f(x

+ t) -f(.

1 cj(eij(x+t)) = (-2i) 1 cjeijxsin(jt)

-t)=

eij(x-t)

lilQ N

1sljlsN

it is clear that (f(x + t) -f(x - t))/(sin(t/2)) is bounded, and hence integrable, as a function of t in [0,TI. Therefore, by the Riemann-Lebesgue theorem,

.l'p,,mf

)

x + t) - f ( x - t) cos( ( n + i ) t ) dt + 0 sin(t/2)

as n + a,since this integral can be expressed as the nth Fourier sine and cosine coefficients of integrable functions. Whence by letting n + a in (5.14) we see that f(x) is given by the absolutely convergent integral (5.15) It now becomes apparent that the problem we set to solve may be a difficult one for it is not clear that the integral in (5.15) converges for arbitrary integrable, or even p-integrable, functions f. Indeed, as we shall see below, there is a continuous, periodic function f such that

Thus the existence of the integral in (5.15) is due not to the smallness of thef(x + t) -f(x - t) for small t, but rather to the cancellation of positive and negative values. Proposition 5.1 (Lusin). (5.16) holds.

Proof. Following Kaczmarz, we construct first a continuous function g with period 1 satisfying the following three properties, to wit:

5. Integral Represenfation of the Conjugate Operator

63

Clearly condition (iii) above is the only one which is nontrivial. Nevertheless, let g be chosen so that g ( x + t ) - g ( x - t ) is not identically equal to 0 as a function of t, for each x in [0,1], while at the same time it verifies conditions (i) and (ii) above. For example, the function g with g(0) = g ( 1) = 0 and g, = 1 which is interpolated linearly in (0, and ($, 1 ) will do. It is clear that for the periodic extension of this function with period 1 , which we also call g, we have

t)

(t)

since the integral, as a continuous periodic function of x now, does not assume the value 0 for any x and has therefore a positive lower bound. Also by (i)

I

Ig(x + t ) - g ( x - t)l dt s 2.

(5.18)

C0,lI

Let us replace x and t by nx and nt respectively and consider

I " = [ [1/n,11

1

g(nx

+ nt) - g(nx - nt) t

I

dt

Y =

I

[O,ll

Ig(nx+ t ) - g(nx - t)l

t+n-1

} dt.

(5.19)

Since the term { .} in (5.19) is of order In n, we see that property (iii) holds for g on account of (5.17) and (5.18). By (ii) it also follows that

Consequently, combining (iii) and (5.20), we see at once that

We now turn to construct J: The idea is to set f ( x ) = C;='=,&kg(nkx), where &k > 0, &k < 00, and nk are to be chosen in such a way that there

III. Norm Convergence of Fourier Series

64

is an appropriate interplay between the (5.16) holds for J; note that

&k'S

and the Ink's. To show that

It only remains now to choose the &k'S and nk's so that the right-hand side of (5.22) goes to infinity with k; the reader can verify that nk = 2(k!)2 &k = 1/k!, will do. Therefore J k + CO as k + 00 and (5.16) holds for a function f of period 1. The desired result follows without difficulty from this. We need one more notation before we go on. By the convolution formula (3.3) of Chapter I we readily see that if

(5.23) then also

1I,

G,,(f; X ) = 21Zn * f ( x ) = Moreover, since

f(X

- t)k,,(t) dt.

(5.24)

k,,(t)is odd (5.24) can be rewritten as either (5.25)

Furthermore, since the integrand in the last integral of (5.25) is even, we also have

(5.26)

65

6. The Truncated Hilbert Transform 6. THE TRUNCATED HILBERT TRANSFORM

Proposition 5.1 motivates the consideration of a new integral operator, namely, the truncated Hilbert transform H E f which ; is given by an absolutely convergent integral and which we expect will converge, in some sense, to f: Definition 6.1. For f E L( T) and 0 < E < n-, let

As usual, we do the L2 case first. Proposition 6.2. Suppose f

E

-fll~

LZ(T).Then lirn,,,+IlH,f

= 0.

Proof. We begin by showing that

IIKf 112 6 cllf 112 with c independent of E and f: In first place observe that 1 2tan(t/2)

HEf(X) = -

+'I n-

E
dt f(x - t ) - = I t

-'> t

dt

+J

say. It will therefore suffice to show that (6.2) holds with I and J in place of H,f there. Now, I and J represent convolution operators with integrable functions, and as such they are bounded in L2(T). It only remains to check that the norms are independent of E. For this purpose put

and

42(f)= x[-7r,-E)u(€,7rJ(t)(l/t). If we can show that the sequences {cj(&)} and {cj(&)} of the Fourier and & are uniformly bounded, by A say, then by relation coefficients of (1.7) of Chapter 11, 1/2

111112

=

/If* 241112 = 2 ( ~ ~ ~ j ( ~ ~ ~ ~~ A~ Ij I( ~~I ~I ~) ,~ 2 )

III. Norm Convergence of Fourier Series

66

and similarly for J. The bound for Icj(+,)l is readily obtained since

As for the c~(+~)’s,to fix ideas suppose that j > 0 and consider

Now we invoke the well-known and readily seen fact that

A independent of 4, b. Therefore, also in this case I C ~ ( + ~4) ~A, and (6.2) follows. Next, we show that the conclusion holds for polynomials p. Indeed, by (5.14) and (6.1) we readily see that lim H , p ( x ) = p’(x) &+O

a.e. in

T.

(6.3)

Since clearly by (5.14) and (6.1) again, IH,p(x)l, Ip’(x)l S K for x in T, by the Lebesgue dominated convergence theorem we get that llH,p - p’l12+ 0, Finally, suppose that f Then

E

as

E

+. O+.

(6.4)

L2(T) and let p be a trigonometric polynomial.

llHsf-5112 4 IIHEf-

HEP112+ IIHEP -p’Il*+

= IIHe(f-P)112+ = Il

I I m -p’ll2+

115 -5112 II(P -f)”ll2

+ I2 + 13,

say. By (6.2) we have that I , S cllf - p1I2, and as observed in the comments following (3.1), I3 S IIp - f l 1 2 as well. These observations suffice to show that 11 H E f - fl12 can be made small provided E is sufficiently small. Indeed, first choose p so that Ilf - pl12 < 7, 7 arbitrary; this is possible by Remark 2.5 in Chapter 11. Once p has been chosen, let E be small enough so that I, S 7 ; this is possible by (6.4). Thus adding the estimates for the 4 ’ s we see that for E sufficiently small IIHEf- f l 1 2 s c7 + 7 + 7 = ( c + 2)7. The study of the behavior of HEffor f merely integrable is more subtle, as is the study of the existence of norm and pointwise limits. In this direction

6. The Truncated Hilbert Transform

67

we begin by proving a qualitative result relating H,f(x) to Gn(J;x); the quantitative relation will be discussed later. Theorem 6.3 (Lebesgue). Let f E L( T). Then

lim Gn(A x) - HlInf(x) = 0

a.e.

n-m

Moreover if f ( x ) exists, then Gn(J x) = un@x).

Proof. It is quite similar to the proof of Theorem 2 of Chapter 11. ,.I first place from the definitions it follows that

say. Now, on account of (5.7) 111 s cn

j

If(x + t ) - f ( x - t)l dt,

(6.6)

CO, 1/ n 1

and this expression is o( 1) as n + co at each Lebesgue point x of A that is a.e. in T. Also from (5.8) it readily follows that

and also this expression is o(1) as n += 00 a.e. in T, as we have already shown in the proof of the Lebesgue theorem 4.2 of Chapter 11. The remark concerning Gn(J;x) follows at once from the relation Gn(A x) = * 2Kn(x), which is readily seen to hold by a direct examination of the Fourier series

1

of3

I

Corollary 6.4. Let f E L ( T ) . Then lime,,+ H,f(x) exists, a.e. in T, if and only if limn,m&n(f,x) exists, a.e. in T. Moreover, these limits coincide whenever both exist. Proof. Only the sufficiency has to be proven. Given integer such that l / ( n + 1) =z E < l/n. Then

E

> 0 let n be the

ZZI. Norm Convergence of Fourier Series

68 and

s cn

I

+

~ f ( x t ) -f(x

- t)l

dt.

(6.8)

[0,1/ n 1

Since the last integral tends to 0 at each Lebesgue point x o f f , that is a.e. in T, it follows that, as E -* 0 and as n + “0, H , f ( x ) and H , , J ( x ) are equiconvergent. By Theorem 6.3 also 6,,(f, x) and H , , J ( x ) are equiconvergent as n + co.

Corollary 6.5. Let f 2 ( T). Then

H , f ( x ) = f ( x ) a.e.

Proof. Since f E L2(T), f is also in L2(T).Therefore by Theorem 6.3 and Theorem 4.2 of Chapter 11, &,,(A x) = andx) + f ( x ) , as n -* co, a.e. in T. Moreover, by Corollary 6.4, H , f ( x ) is equiconvergent with 6,,(f, x). H

To deal with the truncated Hilbert transform in the general case we need a new idea, namely, the Calderbn-Zygmund decomposition, which is developed in the next chapter. 7. NOTES; FURTHER RESULTS AND PROBLEMS The study of some important integral equations, similar to the ones we consider in Chapter XVII, can be reduced to that of a system of linear equations in an infinite number of unknowns by the introduction of a set S verifying the assumption (ii) of Theorem 1.8. Thus the simplest space which we introduce consists of those complex sequences { c , } such that CIcnl*< 00, this is called the elementary “Hilbert coordinate space,” or I*(Z). Parseval’s identity, or relation (iv) of Theorem 1.8, leads to the recognition of the fact that actually L2(T) and 12(Z)have the same linear and metric structure, and that they may be therefore considered as realizations of an “abstract” space which is precisely defined in terms of these common properties, namely, linearity, existence of the notions of inner product and norm, and, very importantly, the Riesz-Fischer property of completeness (Proposition 1.7). These are the Hilbert spaces H of interest to us. By dimension of H we meant the smallest cardinal number equivalent to that of a set S which verifies (ii) of Theorem 1.8, or, more precisely, that of a set of elements in H whose finite linear combinations are dense in H. The dimension of a separable Hilbert space is at most countable and it is quite elementary to construct ONS there by means of the Gram-Schmidt

69

7. Notes; Further Results and Problems

process. Both geometric and analytic methods are available in dealing with Hilbert spaces, and this makes for a rich and powerful theory. Some purely analytic methods, such as the one used to deal with the concept of "almost orthogonality" given in 7.14 below, are quite important in harmonic analysis.

Further Results and Problems

Suppose {xj}j"=, c H satisfy (i) of Theorem 1.8, i.e. it is a maximal ONS, and suppose that { y j } z lc H is such that Cjllxj - fill2 < 1. Then { y j } is complete. In other words, if y in H is such that (y, y j ) = 0 for all j , then y = 0. j 2 1. 7.2 Let {xj}j"=,c H satisfy (i) of Theorem 1.8, and let yj = xi Then { y j } is complete. 7.3 (Paley-Wiener) Let { x j } F l c H satisfy (iv) of Theorem 1.8, and suppose that {yi} is a sequence of elements in H such that 7.1

for all scalars A j , j 2 1. Then each x E H can be written as a linear combination of the yj's, i.e., x = C z l cjyj, cj scalar, where the sum converges in H. cj(xj - yi) converges in H whenever cpj (Hint: Since the sum converges, the mapping T ( c j cjxj) = cj(xj is well defined, has norm 11 TI1 s A < 1, and consequently I - T is invertible.) When H = L 2 ( T )the Paley-Wiener result seems to indicate that only under small perturbations of the integers, i.e., only when IA,, - nl is small, m span L2(T). This is because the condition does the system { eiA I},,=-m

CJzl

cj

x)

cJzl

n

whenever s 1 suffices to insure that f E L2(T) can be written as f(t ) = C,, c,eiAnr,in L2.Surprisingly, the following is true. Kadec's itheorem: If {A,,} is a real sequence and IA,, - nl =sL < all n, then (7.1) holds. The

a,

proof is elementary in the sense that it is entirely computational and sharp. The reader may consult Young's book [ 19801 for this and other interesting results in nonharmonic Fourier series. 7.4 Let A denote the Banach algebra of continuous functions f with absolutely convergent Fourier series normed by l l f l l ~ = Cjlcj(f)l< 00. Show that A is a proper subset of C ( T) and that A = L2 * L2, i.e., each f in A can be written a s f = g * h, g, h E L 2 ( T ) .

III. Norm Convergence of Fourier Series

70

Suppose that f is absolutely continuous and that D ~ L2(T). E Show that f E A and llfllA c Ico(f)l + cllDf 112, c an absolute constant. (Hint: c j ( f ) = ( i j ) c j ( f ) / ( i j ) , # j 0.) 7.6 Suppose f E L( T) and C:=l wl(J w / n ) 2 C 00. Then f E L2(T ) . c j ( f ) c j ( g ) is not necessarily convergent for all f E L( T) and g E 7.7 L2(T ) or even g E C ( T ) ; is the same true for C(lcj(f)l Icj(g)l)2-",E > O? 7.8 Let f I: b, = sin nx,, f E L2(T). Then

7.5

-

c 00

7.9

let

- x ) f ( x ) dx.

+

E La( T) and square-summable sequences x = {xn}, y = {y,,}, N = Cn(+) and put Ajv(x, V ) = An+rnXnYrn* Then

For An

( b n / n ) = (1/2m)T: [

n=l

N

(Hint: If p ( t ) = CnEOx,,e-int, then A N ( x ,x ) = (1/2w) j , p ( t ) 2 + ( t ) dt and IAN(x, x)l S 114 [IrnCn=olxn12. The result now follows with not much difficulty from the identity AN(x, y ) = (AN(x + y, x + y ) - A,(x - y , x - y ) ) / 4 . ) 7.10

(Hilbert)

(Hint: For + ( t ) = i ( w - t)e-", c n ( + ) = l/(n + 1) and ((+llm = w ; forfurther results in this direction see Duren [1970].) 7.11 (Wiener) Let p be a measure in T with discrete part C ujSx,,where the xi's are distinct points in T and the uj's are complex scalars. Then ( pCjluj12. ) ~ 2 In particular, a necessary and 1imNJl/(2N + 1)) ~ l n l s N ~ c j = sufficient condition for the continuity of p is that the above limit be 0. (Hint: Let k = C iijiS-.,; by the convolution formula, p * k ( 0 ) = Clujl2. On the other hand, Icj(p)12 = Inl-zN

x

cj(p

InlsN

D N ( f )d ( p * k ) ( ? ) ; finally show that

go to 0 as N + 00.)

Next we discuss a couple of results of the so-called Littlewood-Paley type. We shall return to them in Chapter XII.

7. Notes; Further Results and Problems

71

Given f~ Lz( T), Put F ( x ) = (Cr=llsfl(J x) - u,(Jx)I2/n)'". Then in particular, F ( x ) < 00 a.e. (Hint: Find an appropriate expression for s,(J x ) - u,(Jx ) and use relation (1.7).) 7.13 (Kaczmarz-Zygmund) Given f E L2(T) put

7.12

llFll2

=s I l f l 1 2 ;

Then llK(f)112s cllfl12; in particular K ( J x ) < co a.e. 7.14 (Almost Orthogonality) Let {fk}?=] be a sequence of 27r periodic, real-valued functions verifying the following properties uniformly in k: (i) Ifk(t)l d 1; (ii) j T f k ( t )dt = 0; (iii) there exist a number 77, 0 < 7 d 1, and a constant A such that Ifk(x) - f k ( t ) l d Alx - tl". Furthermore let {hk} be a Hadamard sequence, i.e., hk+]/hk3 8 > 1, all k Under these conditions, there is a constant c = c(v,8 ) such that for all sequences {rk}?=l,

(Hint: Assume first that the sequence { r k } is finite; the infinite sum is ~ kitt )suffices to show understood in the L~sense. If Ij,k = j C 0 , 2 a l $ ( ~ j t ) f k (dt, ' , some constant c and 0 < E d 1, since then the that lIj,kl S c ~ - " J - ~for integral in question is bounded by Cj,klIj&l lqykl and we are done. It is also enough to do the case j < k, for the case k > j is symmetric with this and the case j = k is trivial. For each k fixed, then, and 1 d j < k, write Ij,k as a sum of integrals Jq over the intervals [27rq/hk,27r(q + I)/&], q 3 0; there will also be an error term, possibly, corresponding to [27rq]/hk,27r] where q1 is the largest integer such that q ] / h k < 1. The contribution of the error term is readily seen to be <2?7/hk. Moreover, since

the desired conclusion follows by summation. This proof, which is the prototype of many making use of properties of Hadamard, or lacunary, sequences, is from Meyer [1979], and we shall return to it later on.) 7.15 Assume that C cj is (C, 1) summable to a finite limit and that its partial sums s, verify s, = o ( p , ) , where {l/p,} is convex and tends to 0. Then cj/pj converges. Consequently, if f E L"( T),f C cfleinr, then C(c,/(l Inl)")e'"' converges a.e., whenever 77 > 0. (Hint: Sum by parts twice and invoke Corollary 2.2 of Chapter I.) 7.16 Lp(T ) = L( T)* Lp(T ) , 1 =s p < 00. (Hint: the proof of Proposition 4.6 requires the following adjustment: If a, = l/(ln( n + l))', then C a, = 00, but a , / n < 00; now let { m,}be as in the proposition and put g ( x ) = f ( x ) + k limk-racCn=l a , ( f ( x ) - wmn(Jx)), where the limit is taken in the Lp sense.

CJzo +

-

ZZZ. Norm Convergence of Fourier Series

72

Next note that the sequence {l/c, a, min(1, Ij ( ( m ,+ 1))) is even, convex, and tends to 0, and invoke Proposition 4.4. This idea, in the case p = 1, is Rudin's [19581 and for general p , Edwards's [ 19671.) 7.17 Does the above result hold for C k (T), k 3 O? How about D'? 7.18 If 6, is the conjugate Dirichlet kernel, then lim,+m[l/(ln n)I Sro,

& ( t ) dt = 1.

7.19 Let f E L( T ) have a jump discontinuity f(x + 0) -f(x - 0) = d, x E T. Then lim,+m(;,(A x))/(ln n) = - d / m Obtain as a corollary that at any point x where f has a jump, ;,,(A x) diverges and that if &(A x) = O(ln n), thenf cannot have a jump discontinuity at x. (Hint: Use (5.12).) 7.20 Let f E L(T ) , then f + $ cannot have a simple discontinuity at any point x in T. (Hint: If x is a simple discontinuity o f f + zf then by Fejir's Theorem 2.1 of Chapter I1 a,(f+zf x) remains bounded. On the other hand, if f(x + 0) -f(x - 0) = d > 0, say, then by 7.19 ;,,(A x) + -00, and again by FejCr's theorem a,(X x) = &,,(A x) + -00.) 7.21 (Localization) Iff vanishes in an interval Z c T, then ;,,(A x) converges uniformly in any subinterval of the interior of I. (Hint: It is a straightforward variant of 5.8 and 5.9 of Chapter 11.) 7.22 The distribution p.v. (;tan(t/2)) has Fourier coefficients c k = ( - i ) sgn k (Hint: Apply (5.14) to f ( x ) = e 7.23 If a real sequence bk decreases to 0, then the trigonometric series bk sin kt converges everywhere. Moreover, for any 6 > 0 the series converges uniformly in 6 < 1x1 < T. (Hint: As in 5.8 of Chapter I use Abel's summation formula and estimate (5.5) this time. Since all sine terms vanish when t = 0 the series converges now everywhere.) The question of the uniform convergence of the above series must be decided in a different manner. 7.24 If a real sequence bk decreases to 0,then for the uniform convergence of 1bk sin kt in T it is necessary and sufficient that kbk + 0 as k +. 0O. (Hint: The necessity follows at once upon observing that l ~ ~ ~ mbk+ sin l kxl can be made uniformly small and letting x = rr/4m there. As for the sufficiency, on account of 7.23 it is enough to show that, if r , ( t ) = bk sin kt and E , = maxkankbk, then Ir,( t)l S CE, for tin [ O , ~ / 4 ] Since . it is possible to find an integer N such that 1/ N < t 1 / ( N - l), then write I".)

c';p=,

N-1

r n ( t )=

C+C

(k=n

bk sin(kt) = rn,l(t)+ rn,z(t)>

k:N)

say, where r,.l(t) = 0 if N s n. To estimate r,.l(t), use that [sin ktl to estimate r,J t ) , distinguish two cases.)

S

kltl;

7. Notes; Further Results and Problems

73

There exists a trigonometric series that converges uniformly but not absolutely in T. (Hint: Cz=’=,(sin(kt))/(kIn k).) 7.26 If bk is a real sequence decreasing to 0, and kbk is bounded, then the partial sums of the trigonometric series ITzl bk sin(kt) are bounded. In particular IC;=,(sin(kt)/k)l 6 C, all t. 7.27 If the real sequence bk decreases to 0, the condition kbk + 0 is necessary and sufficient for C bk sin( kt) to be the Fourier series of a continuous function. (Hint: It is enough to prove the necessity of the condit) of the above series converge uniformly, then tion; if the (C, 1) means a,,( un(7r/2n) + 0 and 7.25

The result follows without much difficulty from this.)

CHAPTER

IV The Basic Principles

1. THE CALDERON-ZYGMUND INTERVAL DECOMPOSITION

In this section we discuss one of the most important topics of harmonic analysis, namely, the Calderbn-Zygmund decomposition of an integrable function. This is a principle which roughly states that an integrable function f can be written as a sum of two functions, f = g b say, where the “good part” g is essentially bounded and the “bad part” b has a cancellation property and bounded averages over a particular collection of open, disjoint subintervals of T. Let us be more precise. In first place we assume that f is a nonnegative function, for otherwise we can apply the result to to include arbitrary real- or complex-valued functions as well. Let then A > 0 be a number exceeding f T , the average off over T, i.e.,

+

In

We will work with the family of open, dyadic subintervals of T, i.e., those intervals obtained by successively subdividing T into equal, open subintervals. The first subdivision of T thus consists of TI = (-‘IT, 0) and T2 = (0, ‘IT). Observe that since

then at least one of the averages f T , or f T ; does not exceed A. Also,

74

1. The Caldero'n-Zygmund Interval Decomposition

75

so that even if an average exceeds A, it remains bounded by 2A. We can now get the selection process for intervals going. In case the average off over a subinterval does not exceed A, then we continue subdividing that interval. If, on the other hand, the average off over an interval does exceed A, then we separate that interval and rename it I , . Clearly,

Suppose, then, that after k steps Z,,12,.. . ,Z, have been separated. The intervals {I}which have not been renamed and separated at that step have the property that h < A. We therefore let each Z in this collection play the role of T above and subdivide it into two equal, open subintervals. The average off over either of these intervals does not exceed 2A, and over at least one of them it does not exceed A. Those subintervals of {I}for which . . ,I,,,, the average off exceeds A and is bounded by 2A are renamed Z,,+,,. say, and separated. These are the intervals obtained in the (k + 1)-step. This selection procedure thus produces a collection of open, disjoint, dyadic subintervals (4) of T with the following properties: 1

f ( y ) d y <2A,

allj

and

Moreover, if we put R = U Z , then the following assertion holds: for almost all x in T\Q there is a sequence dp = { I } consisting of dyadic subintervals of T which converges to x and (1.3)

Clearly, (1.3) holds only a.e. in T\Q as it does not necessarily hold for those points, such as 0, which are the endpoints of the dyadic subintervals of T. We call the collection {I,} the Calderbn-Zygmund interval decomposition at level A. Referring to the functions g and b alluded to above we set

IV. The Basic Principles

76

and b ( x ) = f ( x ) - g ( x ) . It is clear that b ( x ) = 0 off R, that r and that

As for g, we readily see that

g(x)G2A,

x in R.

By (1.3) above it is clear that to estimate g in T\O, we need to consider the behavior of the averages off: This we do with the study of our next topic, the Hardy-Littlewood maximal function.

2. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION Suppose that f is an integrable, periodic function, defined in T, and put

where the supremum is taken over the collection 9 = {I}of those open intervals I centered at x, of length ~ 2 7 ~ . It is readily seen that M F ( x ) is lower semicontinuous, i.e., OA = { x E T : M f ( x ) > A } = {Mf>A}, is open for every A > 0. Indeed, let { x , } be a sequence of points in T\OA which converges to x ; we show that x E T\O, as well. It clearly suffices to show that for every interval I = (x - 7, x + T ) , 0 < T S T,we have

Let In = ( x , - T , x n + T ) and put fn(Y) = ~ ( Y ) X ~ , ~ I ( where Y), I,, AZ = ( Z n \ I ) u (Z\I,,)is the symmetric difference of I,,and I. Since Ifn(y)l s If(y)l and f,,(y) + 0 as n + m, by the Lebesgue dominated convergence theorem it follows that

Moreover, since x , E T\OA

77

2. The Hardy-Littlewood Maximal Function Now, by (2.2),

Thus letting n + 00 above, and on account of (2.1), we obtain that (1/(11) j I ( f ( y ) ldy s A, which is precisely what we wanted to show. So Mf(x) is a positive, measurable function, but is it integrable? A simple example shows that this is not necessarily the case. Indeed, let f(y) = x(o,,,z)(y)(d/dy)(l/ln(l/y)). Then f(y) 3 0 and f E U T ) . Now for x in (-4,O) we have Mf(x) 3

2.11

I,

f(v) dy,

all

77

> 0-

x- rl,x+rl)

In particular setting 77

= 21x1,

we get

and this function is not integrable in a neighborhood of 0. To deal with this inconvenience we introduce the weak-L class of Marcinkiewicz. We say that a measurable function f ( x ) is in wk-L( T) provided there is a constant c such that AIb

E

T : If(x)l> A l l = Al{lfl>

A l l S c,

all A > 0.

(2.3)

The infimum of the constants c appearing in (2.3) is called the "wk-L norm" off; although this quantity does not satisfy the usual requirement of a norm. By Chebychev's inequality we see that integrable functionsf are in wk- L( T), with norm not exceeding llflll. On the other hand the function f ( x ) = (l/x)(,q,,l)(x)) is in wk-L( T), although it is not even locally integrable in a neighborhood of 0. We can now prove the Hardy-Littlewood maximal theorem, namely, Theorem 2.1. Suppose f E L( T). Then Mf E wk-L( T) and the wk-L norm of Mf does not exceed 72rrllfll,. More explicitly A ({Mf > A}[

G

36

[

If(y)I dy,

all

A > 0.

(2.4)

T

Proof. Since Mf(x) = M(lfl)(x), we may assume that f is a real-valued, nonnegative function. Moreover, since hI{Mf> A}l 6 A ~ T all , A, we may also assume that ( 1 / 2 ~ j)T f ( y )dy < h/18, for otherwise there is nothing to prove.

ZV. The Basic Principles

78

Let then fT < A/18, and invoke the Calder6n-Zygmund interval decomposition at level A/18. We then have the collection (1,) of open, disjoint, dyadic subintervals of T as well as the functions g and b,f = g + b, verifying properties (1.1)-(1.7). Now, it is clear that M is a sublinear mapping, i.e., M ( f , + f z ) ( x ) s M f , ( x )+ Mf2(x),and, consequently,

M f ( x ) s M g ( x )+ M b ( x ) ,

all x E T.

(2.5)

For each of the intervais 4, let 21, denote the concentric interval with 4 with measure (or length) twice that of 1,. Finally, put = 21,. With the aid of (2.5) we will be able to show that

a* uj

Mf(x)s A,

XE

T\a*.

(2.6)

If this is the case, then {Mf> A} c a* and by (1.2), Al(Mf> A } \ S h C 1 2 1 , 1 = 2 A C l l i l i

i

and we are done. It thus remains to prove (2.6). Let Z be an interval centered at x E T\a*; we begin by showing that

This is not hard. Since b = 0 off a, we have that

where the sum extends only over those j’s with Z n 1, # 0.We divide these 4 ’ s into two mutually exclusive families, namely, (i) IZ n 1,l 3 11,1/2, we call these intervals Z(’)’s, and (ii) IZ n 41 < 11,\/2, we call these intervals Z(’)’s. For the Z‘”’s we immediately see that

2. The Hardy-Littlewood Maximal Function

79

For the Z(2)'swe invoke the fact that x E T\n*. Since in particular x iZ 24, allj, we have [ I n 41 2 141/2. Thus we see that

=s 16(h/18)1I n $1.

Whence collecting estimates (2.8) and (2.9) we obtain

which is precisely (2.7). Since I is arbitrary in (2.7) this means that M b ( x ) s 16h/18,

x

E

T\n*.

(2.10)

I, g ( y ) dy.

Again for an interval I centered at x, we estimate now (1/111) By the definition (1.4) of g we readily see that

I,

g ( Y ) dv

=

c

I I n 41

jr,m + j p dY

dY = A1 + A2,

j.1 n 1,# 0

say. Also from (1.1) it readily follows that (2.11)

To estimate A2 we again resort to the geometry of the situation. In first place it will suffice to consider the integral extended to the open set I\uj 4 c I. This set can be written as a countable, disjoint union of open intervals J k , say. Each of the Jk's such that Jk does not contain an endpoint of I (and there may be only two of these at that) is, in turn an a.e. union of disjoint, dyadic subintervals Jkn,say, of T which were subdivided in the Calder6n-Zygmund decomposition process. Thus by (1.l), (2.12) On the other hand, if .&,, contains an endpoint of I, let n be the integer ~ , the such that (27~/2"+')=s lJkl< (27~/2").Then J k c J ~ u, J ~where are disjoint, dyadic subintervals of T, each of measure s27r/(2n + l), which

ZV. The Basic Principles

80

were not separated in the Calderbn-Zygmund decomposition process. Thus we have

=z

WWIJk,l+

IJk21)

< (A/18)21JkI.

(2.13)

Combining the estimates (2.12) and (2.13), we readily see that (2.14)

Thus adding (2.11) and (2.14) we get

s 2(A/l8)lZl.

Since Z is arbitrary from this estimate it follows that M g ( x ) s 2(A/18),

all x

E

T.

(2.15)

Finally, by estimating the right-hand side of (2.5) by (2.10) and (2.15), we obtain that

M f ( x ) s 16A/18

+ 2A/18 = A,

x E T\R*,

which is precisely (2.6). An interesting corollary to the maximal theorem is a version of the Lebesgue differentiation theorem, which we hope will be useful in estimating the right-hand side of (1.3). Observe, however, that what we really n5ed is a “noncentered” version of our results. Let us do this then. Let M f ( x ) denote the noncentered maximal function defined by

r x. It is trivial to where the Z’s are open intervals of length ~ 2 7 containing see that l@f is lower-semicontinuous and therefore meas_urable. Moreover, the Hardy-Littlewood maximal theorem also gives that Mf E wk-L( T) with norm s2167r11fl11. This is easy to see since for each open interval Z containing x the interval I, = (x - IlZl, x $111)contains Z. Therefore

+

2. The Hardy-Littlewood Maximal Function

81

and, consequently, A&x) s 3 M f ( x ) . Thus {A@> A} = { M f > A / 3 } and the desired estimate follows at once from (2.4). We can now state, and prove

Theorem 2.2 (Lebesgue Differentiation Theorem). Suppose that f E L( T ) . For each x E T let $x = {I,}be a family of open intervals containing x, converging to x as (Z,I + 0. Then

'I

lim I L l + O I,[

I

f(y) dy = f ( x )

a.e. in

T.

I,

Proof. With no loss of generality we may assume that f is real valued. It is also obvious that the desired conclusion holds true for a trigonometric polynomial, or even a continuous function J Such functions are dense in L ( T ) , and we may assume that they are also real-valued when f is (cf. Remark 2.5 of Chapter 11). Let

Clearly, + ( A x ) = +(f - p , x ) for every real-valued trigonometric polynomial p. Moreover, since +(g, x ) s 2&(x), we also have that 4(J;x) s 2 @ ( f - p ) ( x ) , all x in T. Thus for each A > 0, {+(f)> A } = { M ( f - p ) > A/2}, and by Theorem 2.1

I{d(f) > A } I

(4A)ll.f-Plll.

(2.16)

But since the right-hand side of (2.16) can be made arbitrarily small it follows that > A}[ = 0, each A > 0. This in turn implies that for a.e. x in T

I{+(f)

1 f(y) dy = lim inf and, consequently, lim~Ix~+o(l/~Ix~) j I x f ( y )dy exists for almost all x E T. It is now immediate to see that this limit is actually f ( x ) a.e. Indeed, for A > 0 Put

Now, again for an arbitrary trigomometric polynomial p, we have

IV. The Basic Principles

82 Therefore

QfO)= { G ( f - P )> = %,-I -k

u {IP

-A

>

%A27

say. By choosing p appropriately, as was done above, we see at once that 1%,-1 can be made arbitrarily small. But a similar conclusion holds for %x2 since by Chebychev’s inequality

1421; Whence I%,(A)l

j T

I f ( x ) - P ( X ) l dx = 2 4 f - P l l l .

= 0 for each A

> 0 and

‘I

lim - f ( y ) dy = f ( x ) , a.e. in T. W I I,] 1, Theorem 2.2, in one of its simplest formulations can be restated as folows. Let X ( t ) denote the characteristic function of the interval (-4,f) and let X E ( f ) = (2T/E)X(t/E). Then IL-0

In particular, iff is nonnegative (and in the general case also by replacing s u p o < E < 2 *~xE)I ~ f S 6 f f ( x ) .Thus the convolution with the kernal ~ ~ ( t )which , roughly represents a “bump” at the origin of width E, height ~ T / Eand , total mass 1, is controlled by the Hardy-Littlewood maximal operator, and the differentiation of the integral, namely, statement (2.17) holds a.e. in T. The natural questions then is, under what general conditions on an arbitrary kernel will (2.17) remain valid. The particular examples we have in mind are those kernels which look like the Fejer K, kernels; also note that the story is totally different for the Dirichlet kernels 0,since they have variable sign and more importantly IID,II, + 00 as n + 00. More precisely, then, for an integrable function on the line R, under that conditions does

f by If\),

+

(2.18) for x a.e. in T? A closer look at the proof of Theorem 2.2 indicates that a positive answer to (2.18) depends upon the following two properties of (i) sup,lj,f(x - t ) n + ( n t ) dtl E wk-L( T) whenever f~ L( T); $ ( t ) dt)f(x), for a dense class of (ii) limn+mJ , f ( x - t ) n + ( n t ) dt = smooth functions f E L( T ) .

+:

(I,

2. The Hardy-Littlewood Maximal Function

83

+

if we restrict ourselves to (ii) requires no further assumption on trigonometric polynomials. In fact, by linearity, we may assume that f ( t ) = e"', and in this case

I

e"x-')n+(nt) dt = e"

(-m,TrI

1.

X ~ - n + r , n ~ , ( t ) e - i i f / dt. n+(t)

If we denote by +,( t ) the integrand of the above integral it readily follows that I+,(t)l s 1+(t)l and limn+m& ( t ) = + ( t ) a.e.; consequently, by the Lebesgue dominated convergence theorem (ii) holds. To prove (i) it actually suffices to show that the sup in question is majorized by a wk-L function; a constant multiple of M f ( x ) will do. This requires the following argument.

+

Proposition 2.3. Let be an integrable function in R which admits a nonincreasing, even, integrable majorant T ; more precisely, I+(t)l =sd t ) ,

where 7 is even,

I,

t E R,

7 ( t )dt < 00 and ~ ( s z) v ( t ) , 0 < s < t. Then

Proof. Clearly,

m

r

1n 7 ( n 7 ~ / 2 ~ + ' ) k=O

m

I f ( x - t)1 dt

lrl=?r/2*

=

Ak,

(2.19)

k=O

say. It is also readily seen that the integral in Ak is bounded by

and, consequently, Ak s ( 2 7 ~ / 2 ~ ) n ~ ( n r r / 2 " " ) M f ( x ) all ,

k, n, x.

Thus summing over k we get (2.20)

ZV. The Basic Principles

84

Moreover, since

I

tq(t)s 2

q ( s ) ds,

t 20,

(t/2*11

the sum on the right-hand side of (2.20) is bounded by

8 f I &=O

q ( s ) ds s 4

(fl?r/zk+',fl?r/

I,

q ( s ) ds,

(2.21)

2'1

independently of n. Our conclusion follows combining (2.19)-(2.21). Corollary 2.4. Under the assumptions of Proposition 2.3, for f

lim

n-tm

I

f(x

- t ) n + ( n t ) dt =

(IR

+ ( t ) dt)f(x),

E

L( T ) ,

x a.e. in

T.

Corollary 2.4 gives, in particular, a new proof of the result covered by Fejir's theorem, Theorem 4.2 in Chapter 11, but it does not enable us to identify the set of x's where convergence does occur. To see that this is the case, it suffices to note that an account of the estimates (2.8) and (2.9) of Chapter 11, K , ( t ) s c n 4 ( n t ) , where 4(t) = ,yFo,l)(-t) + t-2Xll,m)(t), 4 even. Because of the importance of this particular application, we state it separately as Proposition 2.5. Suppose f E L( T) and let a*(f)E wk-L( T) and

A}I

~ J { a * (> f)

(.*(A x ) = sup,la,,(f,

x ) l . Then

all A > 0.

s cllf((,,

Proof. Follows at once from Proposition 2.3. 3. THE CALDERON-ZYGMUND DECOMPOSITION

We may now return to the Calder6n-Zygmund decomposition of an integrable function f and express it in its most useful form. Theorem 3.1 (The Calderhn-Zygmund Decomposition at Level A). Let f E L ( G ) and let A > (1/27r) j T l f ( t ) l dt. Then there exists a sequence (4) of open, disjoint, dyadic subintervals of T such that If(x)l

s A, 1

AG -

141

I

I,

x

a.e. in

If(y)l dy

T\

s 2A,

uI j all j

(3.1) (3.2)

3. The Caldero’n-Zygmund Decomposition

and if 0,= R =

85

u 4, then

Moreover, if we set

(3.4) and

then f ( x ) = g ( x ) + b ( x ) ; the “good” function g and the “bad” function b are called the Calder6n-Zygmund decomposition off at level A and have the following properties lg(x)l s 2A, llgllp s

x a.e. in

(2~)P-’Ilflll,

(3.6)

T,

P<

1

0 0 9

(3.7)

and

Proof. In view of the results discussed in the first section of this chapter it only remains to prove (3.6) and (3.7), which follows at once from (3.6). By the definition of g, Ig(x)l s 2h in R so we must show that this estimate holds a.e. in T\R as well. But this is an immediate consequence of the Lebesgue differentiation Theorem 2.2 since (a.e.) to x in T\n there corresponds a sequence {I,} of dyadic, open intervals containing x, which converges to x as (I,( + 0 and such that

Thus If(x)l s A a.e. in T\R, and sincef(x) follows.

=

g ( x >there, our conclusion

IV. The Basic Principles

86

4. THE MARCINKIEWICZ INTERPOLATION THEOREM

The Hardy-Littlewood maximal operator Mf maps L( T) into wk-L( T) and it also maps La( T ) into itself, with norm 1, since

What can we say about the behavior of Mf for Lp functions J; 1 < p < oo? Since these functions are in particular integrable, Mf is a well defined, wk-L function, but can we say something more precise about this function? It is convenient to cast this question in a general setting; first we present some definitions. An operation f + Tf is said to be sublinear if T ( f , +fz)and T(cfJ are well defined whenever Tfl and Tfzare defined, c is a scalar, and in this case

Such an operation T is said to be of (strong-) type ( p , p ) , 1 s p s co, if the mapping is defined in Lp(T) and for some constant c

II TfllP

c c Ilf

(4.3)

IlP

with c independent o f f ; the smallest constant in (4.3) above is the norm of T as a bounded mapping in Lp. Similarly, an operation T is said to be of weak-type ( p , p), 1 s p < 00, if the mapping is defined in Lp and for a constant c API{ITfl> h}l

S

cPIlfll;,

all h > 0

(4.4)

with c independent off; the smallest constant in (4.4) is called the weak-type norm of T as an operator of weak-type ( p , p ) . Since we will consider operations T naturally defined in a couple of spaces, it is convenient to introduce the sum and intersection of the Lebesgue spaces. More specifically, let LPo(T ) + Lpl(T) = {measurable, complexvalued functionsf: f = fo + fl,fo E Lpo(T), f,E Lpl(T)}, and introduce there the norm IlfllLPO+LPL

= inf{Ilfollp, + l l f l l I P l ~ ,

(4.5)

where the infimum is taken over all pairs fo E Lpo(T ) ,fl E LP1(T) such that f = fo + A ; then Lpoi LPl also becomes a Banach space. Similarly, in Lpo(T ) n LPl(T) we introduce the norm

IIfll LPOnLP'

= max(llfllm, IlfllPl)

(4.6)

4. The Marcinkiewicz Interpolation Theorem

87

and also Lpo n Lpl becomes a Banach space. Because only the linear structure of these spaces will be used at this time, we postpone discussing further properties until 7.3. We begin by proving a mixed weak-type-strong-type version of the Marcinkiewicz interpolation theorem; this is precisely what we need to handle the maximal function.

Theorem 4.1. Assume that a sublinear operator T is defined in LPo+ LP1 and is simultaneously of weak-type ( p o , p o ) with norm
Proof. To compute the Lp norm of a function g we use the expression 2dgll;

=P

I

AP-’l{lgl > A l l dA,

(4.7)

[O,m)

which is readily obtained from the identity

by Tonelli’s theorem. In the first place, T is well defined in Lp, since, as is readily seen by considering large and small values of functions in Lp separately, Lp E Lpo + LP1.This consideration is not strictly necessary in our particular case as the underlying space is of finite measure and, consequently, Lp G Lpo; nevertheless, we prefer to give an argument that extends to the infinite measure case, such as the line R, with no modifications. Thus iff E Lp, we also have f = fo + f,, E Lp; i = 0, 1, and, by the sublinearity of T,

ITfWl =s l T f O W l + lTfdt)l ax. (4.8) first as this is the easiest case. From (4.8) it follows at

We assume p1 = 03 once that for A > 0,

mil’

{ITA > A ) {ITXOI’ A / 2 ) “ A m . (4.9) Moreover, since 11 T’II, S clJJfllloo, the second set in the right-hand side of (4.9) is empty provided that llfi llm

=s h/2c1.

(4.10)

This suggests that we actually consider a family of decompositions f = +f1,* parametrized by A. On account of (4.10) it is natural to set

f0,A

fl,A(t)

=

(s(‘)

( A / 2 c , )sgnf

if If(t)l S A/2c1, otherwise,

(4.11)

IV; The Basic Principles

88

and fo,*(t) = f ( r ) - f i , * ( t ) . By (4.9), (4.11) and the weak-type estimate, it follows that 1{1Tfl >

l{lTh,*>lh/2)1

(2/A)po@~~f0,A~~%

(4.13)

Whence, upon dividing by 2~ and taking pth roots, (4.13) reads

1I Tf (Ip

G

cPo/Pc;-PdP 4P1IP ( P -Po)l/p O

Il f

IIP.

(4.14)

But since p o / p = 1 - 7,1 - p o / p = 7 and p l / ps e"", 1 < p < co, we can rewrite (4.14) as

IITfll,

1

s c ( p -pO)(I-S)/po~~-~~:llfllp.

(4.15)

c S 4e1Ie independent off and T, and we are done in this case.

Next, suppose that p 1 < ax As usual we putf using (4.11) above we set

=

+ fi,*, but rather than (4.16)

4. The Marcinkiewicz Interpolation Theorem where E is a parameter yet to be chosen, at once that 2~11TfII;

C

P

I

AP-'I{l

89 =f-f1,*.

%,I,

From (4.8) we see

> A/2H dA

COP)

+P = I0

I

~P-'l~l~fl,hl

> A/2)1 dh

[O,oO)

+ 11,

(4.17)

say. The estimate for I. is carried out as before and it now reads (4.18)

To estimate I, we introduce the function (4.19)

Clearly, for t > 0

With this notation we have

whence, by integrating by parts, we readily see that

lo 00

Il = pAP-P14(A,A )

-

+(A, A ) dAp-p~ =J

+ K,

(4.20)

say. We estimate each term separately and do J first. For this purpose let J ( A ) = p A p - p l @ ( A , A). Since J ( 0 ) 3 0 the lower limit in J can be disregarded. As for J(co), we begin by observing that

IV. The Basic Principles

90

and consequently

If(t)Ipld t / A p l - p .

J s c lim sup A+m

(4.23)

I{,flSeA)

We now show that the right-hand side of (4.23) is 0. There are two cases, depending upon whether f is in Lpl or not. In the former case the numerator of (4.23) is a bounded function of A, whereas the denominator goes to 00 with A, and the limit is 0. On the other hand, i f f is not in Lpl, then the expression in (4.23) is indeterminate of the form co/co as A + co; this calls for L'Hopital's rule. The numerator of (4.23) can be written as

and the limit in question actually equals

But this expression is readily seen to be 0 since by Chebychev's inequality (EA)Pl{lf

>4 1

(4.24)

If(t)l"dt (lfl>&*)

and the right-hand side of (4.24) goes to 0 as A + 00 whenever f It remains to estimate K. On account of (4.21)

E

Lp.

(4.25)

Thus combining (4.17), (4.18), (4.20), and (4.25) we finally obtain

The time has come to select E, a prudent choice being that which makes both summands in (4.26) equal. This gives & =~ ~ ~ ~ ~ , - P ~ ~ q - (po))l/(P1-Po) c ~ p ~ / c ~ ( p

5. Extrapolation and the Zygmund L In L Class

91

and consequently the constant in the right-hand side above is 1 -

Po P L - P

1

CPP

( P- P o )

PI P - P o

P , - Po

1 pl-p -~

1 P-Po -~

PPI-PO

PPI-PO P1

P PI - P o C1

It is now a simple matter to verify that the constant c is as it should be. Because of its importance we emphasize the following particular case of the interpolation theorem. Proposition 4.2. Assume that 4 is an integrable function which satisfies the conditions of Proposition 2.3, and let F ( x ) = sup,lj,f(x - t ) n 4 ( n t ) dtl. Then C IlFllP

( p - l)l/P

Ilf IlP

1
Proof. Apply Theorem 4.1 with po = 1, p1 = 00.

W

Next we consider the Marcinkiewicz interpolation theorem in its original version, namely,

+

Theorem 4.3. Assume that a sublinear operator T is defined Lpo LP1,and is simultaneously of weak-types ( p o , po) and ( p l , pl) with norm d co, cl, respectively. If po < p < p1 and l/p = (1 - 7 ) / p 0+ v/pl, 0 < 7 < 1, then T is af type (p, p) with norm

where c is an absolute constant.

Proof. The proof is similar to, yet simpler than, that of Theorem 4.3, since to estimate the term corresponding to Tf1.A we simply use the fact that

5. EXTRAPOLATION AND THE ZYGMUND L l n L CLASS

It is an easy pursuit to characterize those sublinear operations T which are simultaneously of weak-type ( 1 , l ) and of type (00,00). Indeed, we have

+

Proposition 5.1. A sublinear mapping T defined in L L" is simultaneously of weak-type ( 1 , l ) and of type (00, 00) if and only if there are constants

IV. The Basic Principles

92 c,,c2such thatforallfE L + L ” a n d h > O

J

I{lTfl > All c J A

KIA > ~ Idt.I

(5.1)

C.h/cz,m)

In this case c1 and c2 bound the norm of T as a weak-type (1,l) and type mapping, respectively.

(00,oo)

Proof. The necessity of the condition follows at once from (4.12). Conversely, if (5.1) holds and f E L( T), then the weak-type assertion obtains by replacing A/c2 by 0 in the integral above. Also, if f~ Lm(T), then [{If1 > t}l = 0 as soon as t 2 Ilfllrn, and the integral vanishes if A 2 ~ ~ l l f 1 1 ~ consequently, the same is true for [{ITfl> A}I and 11 Tfllm G c2llfllrn.

Although the estimate (5.1) is equivalent to the weak-type (1,l) and type statements, it contains additional information concerning the integrability of T j This information, which is actually “extrapolated” from (5.1), allows us to decide, for instance, under what conditions Tf is integrable. Lp(T), but is this the optimal result? Clearly Tf E L( T) wheneverf E The answer is no, and to make a precise statement we introduce the Zygmund class Lln L. (00,oo)

up,l

Definition 5.2. We say that a measurable function f is in L In L( T) provided that

where In+ t =

otherwise.

We can now prove Theorem 5.3. Assume that T is a sublinear operation defined in L + Lm which is simultaneously of weak-type (1,l) and of type (00, a).Then T maps Lln L into L and

1I Tflll 6 c + c

I,

If(t)l ln+lf(t)l dt

(5.2)

where c is an absolute constant independent off:

Proof. Since /{IT‘ constant c I

=

G

I

1}1 d 27r it actually suffices to show that for some

ITf(t)ldt

{\Tf\>1)

c

I

T

If(t)l ln+lf(t)ldt.

(5.3)

3;

5. Extrapolation and the Zygmund L In L Class

93

That Theorem 5.3 is sharp is readily seen by observing that for the HardyLittlewood maximal operator the reverse inequality to (5.2) holds, namely, Theorem 5.4

(Stein). Suppose f E L( T) is such that Mf E L( T). Then

f~ Lln L(T). Proof. We begin by showing that for A > 1

InT

j{tA>A!f(t)l dt s I{Mf> A H .

(5.4)

Indeed, let { 4) be the Calder6n-Zygmund interval decomposition at level A ; recall that in particular

By the first inequality in (5.5), we see that second one, that

uj4G {Mf>A},

and, by the

r

u

Moreover, since by (5.6) {Id> A } E 4 the estimate (5.4) follows at once from (5.7). Now integrating (5.4) we obtain

S

2

1

1{Mf> h}l dA

W,m)

and the desired conclusion follows from this with no difficulty.

IV. The Basic Principles

94

6. THE BANACH CONTINUITY PRINCIPLE AND 8.e. CONVERGENCE

In this section we discuss some of the basic ingredients in the proof of an a.e. convergence result. Let B denote either a Lebesgue L p ( T )space, 1 s p s co or C ( T ) . We consider a sequence of operators, much like the partial sums of a Fourier series, defined in B and continuous in measure. More precisely, each of the operators T in the sequence verifies the following properties

1 Tf(x)l < co

a.e. for each f E B

(6.1)

and if f n , f are in B and i.e., for each

E

llfn - f l l B

+ 0,

then '''7

+

Tf in measure,

>0

For a sequence { T'} of such operators we set

and

In case

then also

We are interested in showing that property (6.6) by itself gives the continuity in measure for T*fand that this in turn may be used to obtain (6.5). We begin by considering the continuity result (Banach Principle). Assume {T,} is a sequence of linear mappings in B which verifies property (6.6) above. Then there is a positive, decreasing function C(A), A > 0, which tends to 0 as A + 03 and such that for all f in B

Theorem 6.1

I{T*f>

~ I l f l l ~ H c(A),

all A > 0.

(6.7)

6. The Banach Continuity Principle and a.e. Convergence

95

Proof. It suffices to establish the conclusion forf with l l f l l B = 1. Fix E > 0. Then by (6.6) to each f in B there corresponds an integer n = n ( f ) such that I{T*f> n } l < E. In other words, we have a,

B

=

a,

U{ f :I { T * ~ > n=l

n}l

=S E } =

U B,, n=l

(6.8)

say. Observe that each B, is closed in B. Indeed, since m

B, =

fi {f:I { T % ~ n}l > s

EJ

N=l

and since the operators T& are also continuous in measure, each of the sets in the above intersection is closed, and so is B,. Statement (6.8) expresses then the fact that B is a countable union of closed sets and by the Baire category theorem (cf. 7.32 below) one of the Bn's contains a ball. In other words, there are an integer n, f a E B and 7 > 0 such that

I{T*f> n}l s

E

whenever

llfo

-

fll

ZG

(6.9)

3.

By translating this ball so that it is centered at the origin, (6.9) can be restated as Iu-*(fo

+ r l f ) > n}l

E

when

l l f l l B ZG

(6.10)

7.

Moreover, since T* is sublinear, it follows that T*f(x) s T*(fo+ r l f ) ( x ) / ~+ T*fo(x)/v, and, consequently, by (6.10) we see that for l l f l l B 1

I{ T*f > 2 n / ~ } n}l + I{ T*fo> n}l s 2

~ .

(6.1 1)

But E > 0 above is arbitrary, whence (6.11) implies that, if C(A) = supllfllrrGll{ T*f> A}[, then limA+mC ( h ) = 0, which is precisely what we wanted to show. W In most applications, the a.e. convergence of { TJ(x)} can be established forf in a dense class of functions in B. In this case the Banach principle gives Proposition 6.2. Suppose { T,} is a sequence of operators in B which verifies (6.6). Then the set Bo = { f B: ~ TJ(x) is a.e. convergent} is closed in B. In particular if Bo is dense, then { TJ(x)} is a.e. convergent for everyf E B. Proof. Let f be in the norm closure of Bo. We must show that f E Bo as well. Let T ~ ( x = ) lim sup,,,,,I TJ(x) - T'(x)l. We will be done if we can ) 0 a.e. Since T ~ ( xs ) 2T*f(x), (6.7) gives show that T ~ ( x =

I{T.> A lIfll~)I

C(A/2).

(6.12)

IV. The Basic Principles

96

Now for f E B and g E Bo, T ~ ( x= ) ~ (f g)(x) for almost all x, and by (6.12) [{~f > A Il f - gllB}I s C(A/2). Upon choosing A = 1 / and ~ g so that gll, S E 2,we get &}Ic C ( 1 / 2 ~ )Since . C ( 1 / 2 ~+ ) 0 with E, we see at once that T ~ ( x= ) 0 a.e.

\If-

1{~f>

Before we are in a position to apply the Banach principle to the particular case of Fourier series, we need to have some control on the growth of the partial sums s , ( f ;x) of L2 functions f: We will assume we know that s,(J x) = O(A,)

a.e.,

all f E L2(T),

where A, is a fixed, nondecreasing sequence tending to then show that for functions f 1 cjevx satisfying

-

00

(6.13)

with n and will (6.14)

the Fourier series off converges a.e. in T. Clearly, (6.14) is more effective the slower is the growth of the An’s. The best result we give here is the Kolmogorov-Seliverstov Theorem in 7.27; the order of the An’s there is (In n)”2 as n + 00.

Proposition 6.3. Suppose that (6.13) holds; i f f E L2(T) verifies the additional assumption (6.14), then limn+ms,(J x) exists a.e. Proof. Forh E L2(T)put T’h(x) = s,(h,x)/A,, T*h(x) = sup,IT,h(x)(.On account of Theorem 6.1, there is a function C(A) + 0 as A + 00 such that (6.15)

I U * h > AllhlI*>l c CO).

We proceed now along the lines of the proof of Proposition 6.2 and introduce T ~ ( x= ) lim supl~,h(x)l.

(6.16)

n-m

Since T ~ ( xs) T*h(x), (6.15) holds with T in place of T*, i.e., for all h E L2(T) and A > 0, 1 { ~ h> Allh112}1s C(A). But T ~ ( xis ) not affected if we replace h by h - s , ( h ) , so we also have 1 { ~ h> Allh - sn(h)ll2}[e C(A) and consequently > A l l c C(A//lh - sn(h)ll2). (6.17)

1w

Invoking (1.6) of Chapter I11 it follows that the right-hand side of (6.17) goes to 0 as n + 00 and, therefore, 1 { ~ h> A}l = 0 for each A > 0; thus lim (s,(h, x)/A,)

n+m

=

o

a.e. in

T, h E L ~T( ) .

(6.18)

It is now an easy matter to complete the proof. In first place observe that cjAljlevx. by the Riesz-Fischer theorem there is an L2 function g

-c

6. The Banach Continuity Principle and a.e. Convergence

97

Moreover, as is readily seen

= I,

+ Jn,

say. By (6.18) J, + 0 as n + co, a.e. in T. Furthermore since 11 sj(g ) 11 s Ilsj(g)1I2s cllgllz, it also follows that 1 -1lg112 A0

< a,

and the series I, converges (absolutely) a.e. in T. This gives the a.e. convergence of s,(J x ) . It is also quite simple to show that limn+ms,(f; x ) = f ( x ) a.e. The proof we just gave is interesting in that the convergence result follows from a growth estimate; it is not in the spirit of the proof of the a.e. convergence for f * # J , , ( x )which , follows from a maximal estimate instead. The Banach principle itself, on the other hand, is a result in this direction but a rather incomplete one because of the unknown nature of the function C ( A ) .To address this situation we begin by proving an interesting result of independent interest.

Lemma 6.4 (Calderh). Suppose that { E k } is a sequence of subsets in T such that 1 I&/ = a.Then there is a sequence {xk} c T such that almost every x in T is in infinitely many of the sets (xk + E k ) .

Proof. The assumption on the IEkl’sis equivalent to the fact that the product nrz1(l - (1&1/27r)) diverges to 0. With the xk’s yet to be chosen, the subset of those x’s in T which belong to infinitely many of the (xk + &)’S is (6.19) We now show that for an appropriate choice of the xk’s the complement of the set in (6.19) has measure 0. With the notation Fk = T\(xk + &) this complement is U : = p = l ( n k a n Fk). So we will be done once we prove that we can choose the xk’s so that (6.20) We consider the case n = 1 first and take a look at F = Fk. Let xj denote the characteristic function of 4, 1 =zj s m. Then the characteristic function x of F is given by ~ ( t=)Xl(t - X I )

* * *

Xm(t - xm)

(6.21)

98

IV. The Basic Principles

and the measure IF[ is

I T x ( t )d t

=

ITxl(t- X I )

* *

* X m ( t - X m )dt.

(6.22)

+

Interpreting y, as a function of the m 1 variables t, x , , . . . ,x , now, and integrating (1/(27r)")x( t ) over all variables, each one ranging over T, we note that

(6.23)

The expression on the right-hand side of (6.23) can be made arbitrarily small, <$say, provided m is sufficiently large on account of the divergence of the infinite product. Now that m has been fixed it is easy to choose x l , . ..,x , so that IF1 < $. Indeed, suppose that for every x l , . . . ,x,, each one ranging over T, the integral in (6.22) exceeds f.Then dividing by (27r)" and integrating the expression in (6.22) we readily see that the integral in (6.23) exceeds $, and this is not the case. Thus there are points xi'), . . . ,x:! in T so that n ( x f ) iEk)l < f; more generally, it is clear that we can choose points x i , . . . ,x2: in T such that

. . ,m, = 0, mi increasing to soon as mj-,> n,

j = 1,2,.

CO.

Consequently, for each n, and as

(6.20) holds, and we are done.

We can now prove Theorem 6.5 (Calderbn). The Fourier series of each f E Lz( T) converges a.e. in T if and only if the mappingf+ s*(s) = supnIsn(f)lis of weak-type (292).

Proof. That the condition is sufficient follows by an argument analogous to that of Proposition 6.2, and we say no more. Conversely, if the Fourier

6. The Banach Continuity Principle and a.e. Convergence

99

series of each f E L2( T) converges a.e. in T, then s*(f, x ) < 00 and by the Banach continuity principle, there is a function C ( A )+ 0 as A + 00 and I{s*(fl > Allfl12}1 < C ( A ) ; it only remains to show that C ( A )= O(AP2). Suppose this is not the case, i.e., corresponding to each integer N 2 0 there is an L2 function f N with IlfN112 = 1 and a scalar AN > 0 such that IENI = I{s*(f~) > ANII 3 2N/AZ,.

(6.24)

By the density of the trigonometric polynomials in L2 we may take f N to be a polynomial p N of norm 1. Also note that since by (6.24) 2"/A&

=S 2?r,

all

N,

(6.25)

AN + 00 with N. Out of the AN'S we construct a new sequence { p j } as follows: if ko = 1 and for N 2 1, kN denotes the smallest integer such that 1 =Z 2NkN/h%6 2?r, (6.26)

set pj

= ANX[~,N=,k,_,,~:,N_,k",(~), j , N

'1-

(6.27)

This is a complicated way of saying that the pj's consist of successive blocks of AN'S, the first k, terms being equal to A , , the next k2 terms equal to A2, and so on. We also introduce { u j } , replacing AN by 2N in (6.27). By (6.25) and (6.26), we readily verify the following two properties of the sequence {pj}:

(6.28)

and (6.29)

Note that (6.24) can also be written in an obvious fashion as lEjl= I{s*(pj)

(6.30) where the pj's are the polynomials of norm 1 corresponding to the pj's (which are, after all, the AN'S renamed). By (6.29), 1 41= 00 and, consequently by Lemma 6.4, there are points { x , ,. . . ,xi, . ..} in T so that almost every x in T belongs to infinitely many of the sets (xi + Ej). Choose now a sequence mi increasing to 00 so rapidly that the polynomials e""f&(x) have no common terms, and, finally, consider PjIl2 v j / p ; ,

feimppj(x- xi)

(6.31)

CLi Next, we show that the trigonometric series (6.31) converges in L2 to a function f, of which it necessarily is the Fourier series, and that s,(f, x) i=l

IV. The Basic Principles

100

does not converge for almost every x in T; this will clearly produce a contradiction. In first place

as N, M + 00, and (6.31) is the Fourier series of an L2 function$ Moreover, on account of the definition of Ej, s*( p j , x - x j ) > pi any time x E (xj + E j ) ; this means in particular that some complete block arbitrarily advanced in the tail of the Fourier series off is in modulus 31. Since almost every x in T belongs to infinitely many of the ( x j + Ej)'s,the Fourier series o f f cannot converge on a subset of T of full measure.

7. NOTES; FURTHER RESULTS AND PROBLEMS Regardless of the form it assumes, the Calder6n-Zygmund decomposition is an indespensable tool in the study of the continuity of operators which arise in the course of normal events in harmonic analysis. The idea of breaking a function into pieces and then handling each resulting term by an appropriate technique lies at the heart of every important result in this book. The development of the different topics in this chapter is closely intertwined. The continuity of the conjugate operator in Lp(T), 1 < p < W, was established by M. Riesz in 1927; his proof relied exclusively on "complex variable methods." We will touch briefly on this subject in Section 3 of Chapter V and in Chapter VII; Chapter VII in Zygmund's treatise [1968] contains the full scope of this technique. Very shortly after this, Kolmogorov and Zygmund produced substitute results for the L( T) case, Kolmogorov [1925] gave a noncomputational proof of the fact that the conjugate of an integrable function is in the class wk-L( T) (this is the starting point for the material discussed in Section 6), and Zygmund [ 19291 introduced the class L In L( T) in order to discuss a sharp condition that insures that the conjugate of a function in that class is actually integrable. At about the same time Hardy and Littlewood were interested in maximal inequalities to solve a question in the theory of functions; this problem stated in a simple form is the following: suppose A > 0, f(z) is regular in IzI < 1 and F ( x ) = supo
I,

7. Notes; Further Results and Problems

101

out to be yes and the applications, including the boundedness of the Hardy-Littlewood maximal operator in Lp(T), 1 < p < 00, are extremely important. Concerning the method of proof of this result, which relies on averaging of sequences, this is what the authors had to say: “The problem is most easily grasped when stated in the language of cricket, or any other game in which a player compiles a series of scores of which an average is recorded,” [ 19301. From this point on things get even better as the authors discuss, among other things, the maximal theorem for 1 < p < co, the estimate a * ( f x) , s c M f ( x ) , and Theorem 5.3, again for the maximal operator. The weak-type ( 1 , l ) estimate, which seems to be the most natural setting for these results was proved by F. Riesz [1932] a couple of years later; his proof makes use of the rising sun lemma given in 7.1 below. This lemma is the one-dimensional version of the Calder6n-Zygmund decomposition, which first appeared in an article dedicated to M. Riesz on the occasion of his 65th birthday [1952]. The Marcinkiewicz interpolation Theorem 4.3 seems to be the last paper announced by the author. This was in 1939 at the age of 29; at about this time Marcinkiewicz was mobilized as a reserve officer in the Polish army, was taken prisoner on the Eastern front, and vanished without trace. The proof of the theorem was communicated by letter to Zygmund and the complete proof, as well as important extensions, first appeared in Zygmund’s article [1956]. The results of Kolmogorov and Calder6n, expressing the relation between the finiteness a.e. and the type of maximal operators were cast in definite form by Stein in the early 1960’s. Stein’s result roughly states that for a sequence of linear operators { T k } defined in L p ( X ) ,1 S p S 2, where X is a finite measure space, the existence of a family of mixing, measurepreserving mappings on X which commute with the Tk’s implies that the following conditions are equivalent: (i) for each f E L p ( X ) , T * f ( x ) < co a.e., and (ii) f + T*f is of weak-type ( p , p). The technique of proof is quite interesting and uses the Rademacher functions. We present in 7.30 and 7.31 a similar result, due to Sawyer [ 19661, for positive operators; the conclusion holds now, however, for 1 < p < CO.

Further Results and Problems (F. Riesz, Rising Sun Lemma) Givenf E C ( T), let H = {x E T: there is a point y in T, y < x, such that f ( y ) < f ( x ) } . Then H can be written as a union of countably many disjoint intervals I k with endpoints ak and bk such that f ( a k )< f ( b k ) .All the I k ’ s are open except possibly that with 7.1

I K The Basic Principles

102

bk = T. ( H i n t : That H is open, except possibly for T, is immediate; thus H is of the desired form. Now assume that f ( a k ) > f ( b k ) and reach a contradiction; we can actually say more, namely f(a k ) = f(b k ) unless bk = T.) 7.2 ForfE L P ( T ) , l < p < c o , a n d h > l l f l l l l e t

denote the "bad" function in the Calder6n-Zygmund decomposition of f at level A. Show that limN-rmllC,E1bi - blip = 0. (Hint: It sufficesto show :I, bi(x)l, lb(x)I S c$(x).The function there is an Lp function c$ such that,C =

1(If(x)I+ 2A)Xrj(x)+ If(x)I + Ig(x>L i

where g is the "good" part, does the job.) 7.3 Let 1 < p < q < 00 and show that Lp(T) n L4(T ) and Lp(T ) + L"( T ) are Banach spaces. (Hint: All the norm properties for the intersection are obvious except for the completeness. If {f,}is Cauchy in Lp n L", then it is also Cauchy in Lp and L" and limf,(LP) = limf,(L") = limf,(LP n L"). For the sum there are two nonobvious properties; (i) that the norm off = 0 implies f = 0 a.e. and (ii) completeness. To show (i) observe that there are sequences {g,} E Lp and {h,} E L" such that g, + h, = f; all n, and mp~ ~ h = , ~0.~Theng, q h, = g, + h, and g, - g, = h, - h, conl i m ~ ~ g=n l~i ~ verges to -g, (in Lp) and to h, (in L", and thus also in L p ) ;this means that f = g, + h, = 0. To show.(ii) just prove that if C l l f n l l t ~ + L< q 00, then N lim,,,~,=, f, exists in Lp L".) 7.4 Let l / p + l/p' = 1, l / q + l / q ' = 1 , l < p , q < 00. Consider the relation ,between the space of bounded linear functionals on Lp n L" or the space dual to Lp n L4 and Lp' L"'; do the same for the functionals on Lp + L" and Lp' n L4'. (Hint: It may be helpful to think about the relationship between the norm in Lp n L4 and jIo,m)l{ljl > A}l d max(hP, h 4 ) and the > A}[ d min(Ap, A").) norm in Lp + L" and jco.m~l{ljl 7.5 For 0 < a < 00, 0 < p < 1, the following statements are equiv .lent: (i) For every measurable set E E T, jE1f(x)l"dx s c@IP. (ii) f E wk-L"/"-B)( T). How is the constant c in (i) related to the weak norm o f f ? (Hint: (i) implies (ii) Put E = {lfl > A}. (ii) implies (i). Use

+

+

+

I,If(x)l"dx

6

I

lo,m)

min{lEl, I{x E E: If(x)l

'A}l} dA*.) up,,

Show by means of an example that Lln L strictly contains Lp. Show that condition (5.1) is actually equivalent to the single condition l{ITfl > 111 s c1 J[l/c2,m)l{lfl > t ) dt.

7.6 7.7

7. Notes; Further Results and Problems

103

a ( t )d t / t = 7.8 Let a ( t ) be a nondecreasing function such that O ( b ( s ) ) ,and let A ( t ) = jro,,) a ( s ) ds, B ( t ) = jro,t),b ( s ) ds. If T is a sublinear ) mapping simultaneously of weak-type ( 1 , l ) and type ( c o , ~ ~then I,A(ITf(x)l)dx s c + c I, B(If ( x ) l ) dx. (Hint: Theorem 5.3 corresponds to the case a ( t) = 1, b( t ) = In+ t.) 7.9 Let {g,,} be a sequence of wk-L functions with norm (uniformly) ~1 and furthermore let {c,,} be a sequence of positive numbers, C c,, = 1, and C c,,Iln c,l = K < 00. Then AI{C c,,g,, > A}[ S 2 ( K + 2 ) , all A > 0. (Hint: For each n > 1, put u , ( x ) = g,,(x) if g,,(x) < h / 2 and 0 otherwise, u,(x) = g,,(x) if g,,(x) > A/2cn and 0 otherwise, and w,,(x)= g , ( x ) ( u , ( x ) +-u.(x)). Put u ( x ) = C c,,u,,(x)and similarly for u ( x ) and w ( x ) .The following properties are readily verified: u ( x ) < A / 2 , I{u f O}l S C,l{gn > A/2cfl}l< 2 / A and

Finally estimate

This result is from Stein-N. Weiss [1969].) 7.10 Let { T , } be a sequence of sublinear operators defined in L ( T ) and assume that the weak-type ( 1 , l ) norm of these operators is (uniformly) S l . If { c is a sequence of positive real numbers so that c,, Iln c,, I < 00, then T = c,T, is of weak-type (1,l). 7.11 A sublinear operator T defined in Lp + L" is simultaneously of weak-type ( p , p ) and of type (00, co) if and only if there are constants cl, c2 such that for all f E Lp + L" and A > 0

.t:

1{1 Tfl > A l l s

I

6

[{lfl > s}IsP-l ds.

lc2A.m)

Extrapolate and compare with 7.7. 7.12 A sublinear operator T defined in Lp + Lmis simultaneously of types ( p , p ) and (00, co) if and only if there are constants cl, c, such that for all f E L p + L" and h > O

[

1{1Tfl > s}IsP-l ds s c1

[AS)

7.13 Show that for t large,

j

[a&m)

[{lfl > s}IsP-' ds.

IV. The Basic Principles

104

Extend the Marcinkiewicz theorem to the case where the operator T in question is simultaneously of types ( p o , p o ) and ( p l , pl). What can you say about the norm of T, a s a bounded mapping in Lp, po < p < pl? What if T is of type ( p o , p o ) ,weak-type (pl,pl) 1 < po < p1 < a? 7.15 State and prove the “abstract” version of the Marcinkiewicz interpolation theorem, i.e., T acts now on Lpo(X)+ L p ~ ( X where ), ( X , d p ) is an arbitrary measure space. Show that the requirement p o , p l , p > 1 is not necessary, i.e., we may actually assume p o , p l , p > 0. 7.16 There are ways to sharpen the conclusion of the Marcinkiewicz interpolation theorem by considering, for instance, a finer scale of intermediate spaces between the Lebesgue Lp classes. We may consider the Lorentz spaces Lps4(T ) consisting of those measurable f s such that

7.14

llfllp,q

=

(P

I

tos)

I{lA > A}14/PAq-1

dA)Il4 < co-

State and prove such a result. The work of O’Neil [1968] is relevant here. Can we use L In L as an endpoint in an interpolation theorem rather than as an intermediate space? Compare with 7.7 above. The results of Torchinsky [1976] are also of interest in this question. 7.18 A linear operator T defined in L + LP1, 1 < p1 < co, is said to be of pseudo-type ( 1 , l ) if to each f E L + LP1 and A > 0 there correspond a measurable set G c T and a function g E L( T ) such that [GIS cllflll/A, Ig(x>lc CAY llgll1 C l l f l l l and I T \ G I T ( f - g)(x)l dx S cllfIl1; all constants c above are independent off and A. Prove that if T is of pseudo-type ( 1 , l ) and of weak-type (pI,pl),then T is of weak-type ( 1 , l ) and consequently of type ( p , p ) for 1 < p < pl. (Hint: Estimate 1{1Tfl > 2A}l S 1{1T(f- g)l > A}/ + [{ITgl> A}l = 111 + IJI, say. IJI can be easily handled on account of the weak-type ( pl, pl) assumption. Also 111 G I GI + ((T \ G ) n { I T ( f - g)1 > A}/. This result is of interest in that it brings the CalderbnZygmund decomposition into play. We will have more to say about this later on; the proof is Cotlar’s (see Cotlar and Cignoli [1974]).) 7.19 A sublinear operator T is of weak-type (p, p) if and only if for each measurable subset E c T and 0 < r < p Kolmogorov’s inequality holds, namely, for all f E Lp(T ) 7.17

(Hint: Compare with 7.5.) Assume that T and S are sublinear operators and that T is majorized by S in the following sense: if C(x, r) = {y E T : r s Ix - yI =s2 r } , to each x in T and f~ L ( T ) there corresponds 0 < i < r such that ITf(x)lS inf,,,ccq,r)lSf(y)l.Then, if S is of weak-type (p, p) for some p > 0, so is T. 7.20

105

7. Notes; Further Results and Problems

(Hint: Let 0 < q < p. Then ITf(x)I" s infy,c(x,r)lSf(y)lq and averaging over C ( x , f )

where I ( x , T) denotes the interval centered at x of length 4T. The above estimate then gives ITf(x)I" s cM(ISFlq)(x),with c independent of x and f now; therefore, it will suffice to show that iVf((Sf(')(~)"~ is of weak-type ( p , p ) . But this is quite easy since by Proposition 5.1 ((M(lS''

> A'}I s cA-"

({S'

> t l / q } (dt

lcA4,W)

< CA-"

I

IlfllpPt-P/' dt

= CA - p

IIf 1;

;

[cAq,m)

the result is also Cotlar's.) 7.21 Assume that T and S are sublinear operators and that T is majorized by S in the following sense: to each x in T and f E L( T) there corresponds 0 < F < 7r such that if I ( x , f ) denotes the interval centered at x of length 2f, then

Then, if S is of weak-type ( p , p ) for some p > 0, so is T. (Hint: Use (7.1) now, this also is Cotlar's observation.) 7.22 With the notation of Section 6, we say that E G T is a set of divergence for B if there exists f E B whose Fourier series diverges at every point of E. Prove that E is a set of divergence for B if and only if there is f E B such that s*(f; x ) = 00 for x E E. (Hint: The condition is clearly sufficient; to show that it is necessary let g E B have a divergent Fourier series at each point of E and letf E B and { A j } be the function and sequence corresponding to g constructed in Proposition 4.6. Then, as in the proof of Proposition 6.3, we see that Is,(g, x ) - s,(g, x ) / s 2s*(f; x)A;>l and the Fourier series of g converges whenever s*(J x ) < 03. This observation is Katznelson's [ 19681.) 7.23 E is a set of divergence for B if and only if there exists a sequence of trigonometric polynomials { p n } such that IIpnllE1< 03

and

sup s*(pn,x ) = n

for x

E

E.

IV. The Basic Principles

106

(Hint: The proof is very much like that of Theorem 6.5. To show the necessity assume that the polynomials p,, of degree N,, exist and with integers k. such that k,, > knPl NnP1N,,, put f(x) = C,,eiknxp,,(x);then

+

smn+m(.L

+

X) - s m n - m - l ( . L

X) = eimnxsm(pn,

X)

whenever rn < N,, and the partial sums o f f diverge on E. Conversely, if E is a set of divergence for B there are a monotonic sequence p,, +. 00 and f E B such that Is,,(f,x)l > p,, infinitely often, for every x in E. Let now {A,,} be a sequence of integers such that - aAn(f)IIB < 2-" and choose integers 7, such that p,,. > 2 sup, s*(aAn,x). If we set now p,, = (2K2,,,+2* (f- aAn), then 1 IIpnIIB< co, and if x E E and n is an integer such that Is,,(f,x)l > p,,, then for some j,

[If

7j

< n < vj+l and Isn(Pj, x)l=

Isn(.L

x) - s n ( a * , ( f ) ,

x ) l > pn/2-

This proof is also from Katznelson's book [1968].) 7.24 Show that if the Ej's are sets of divergence for B, then so is E = uFlEj. 7.25 (Kolmogorov) Let {vk} be an Hadamard sequence, i.e., nk+l/n!, 3 8 > 1, and assume that f E L( T) has a lacunary series, i.e., f 1 c,,ke'nkx. Then show that s*(S, x) 6 ca*(f,x), where c depends only on 8, and limk+ms,,,(f, x) = f(x) a.e. (Hint: Observe in the first place that, if 1,cnk = s( C, l), then also C c,,, = s. This is not hard to see since we may suppose that s = 0, and in that case the partial sums s,,, of the series in question verify the following relation: if j > k, then (nj - nk)Snk = njan,-l - n p n k - 1 . Thus ( n j - nk)snk= o(nj) + o(&) = o(nj - nk), and Snk = o(1). Also Isnk[6 ( ( n j + nk)/(nj - nk))(T*d ((8 + i)/(8 - l))a*.) 7.26 (Kolmogorov) Let { nk} be a fixed Hadanard sequence and f E L ~T() . Then limk+ms,,,(.L x) = f ( x ) a.e., and, if n*(Jx) = supkIsn,(f, x>l, then Iln*(f)II,s cllfll, where c depends only on 8. (Hint: Both statements follow from the estimate

-

c

where c depends only on 8; but this is easy to obtain from an appropriate expression for s , ( f , x) - a,,(f,x) (cf. 5.4 of Chapter 11). The second assertion also requires the observation that suPIsnk(x)12 k

1Isnk(x) - ank(x)12+ u*(.L x)*) k

7.27 (Kolmogorov-Seliverstov) Iff E L2(T), then s , ( f , x) = o((1n n)'12) a.e. Furthermore, the function T*f(x) = ~up,,==~(1s,,(f, x)l/(ln n)'12) is in L2(T) and 11 T*fI12s cllfl12, with c independent of J: (Hint: It suffices to

7. Notes; Further Results and Problems

107

, ( ~ , n)’/’), with a prove the norm estimate for T % f ( x )= S U P , , ~ ~ ( I Sx)l/(ln constant independent of N. Let n ( x ) be any step function taking integer values 2 S n ( x ) d N, and put A ( x ) = l/ln n ( x ) ;it is enough to show that

I

=

I

A ( X ) S 2 , ( X ) ( f ;x

) dx S cllfll:

T

since T L f ( x ) = A(~)”’s,,(~,(f,x). By the estimates (1.15) and (1.17) in Chapter I, jTIDnCx)(x)I dx S c In N. Now I equals

&

jTA’”(x)sn(x)(f, x ) W )dx,

with

114112 = 1.

Put $ ( x ) = A’/’(x)+(x).Then

say. Clearly

and the estimate obtains using the above bound for jTIDnCx,(x)I dx. This shows that s,(f, x ) = O((1n n)”’) a.e. To refine 0 to o we pass to lim sup instead.) 7.28 (Plessner) I f f E L( T) and

is in L2(T), then the Fourier series off converges a.e. (Hint: Show that the assumption is actually equivalent to ~,n,,l~cn(f)~’ In n < co.) 7.29 With the same notation and assumptions of 7.14 of Chapter 111, show that if

IV. The Basic Principles

108

and 8 such that

then there is a constant c, depending only on

Furthermore IIs*I12 6 cllfl12 and sN(x)+ f ( x ) a.e. (Hint:The proof is an interesting application of an “averaging” method. Let I = [x, x + 27r/ tN]. If y E I, then N

lsN(x) - sN(Y)I

Irk1 If,(@)

-fk(fky)l

k=O

and consequently lsN(x)I S IsN(y)l+ c(CIrk12)1’2.By averaging now over I, it follows that

say, where RN+,(y)in B is the “tail” of the series. Clearly, C is of the right order and A S c M f ( x ) .To do B change variables y = x + s/ tN, 0 S s S 21r, and apply 7.14 of Chapter 111 to RN+I(x + S/tN) = C:=”=,+, rkfk(x+ S/tN). ~ follows at once. Thus the first assertion is proved; that I I s * ~S~ c(Clrk12)1’2 To prove the a.e. convergence apply the norm result to R:(X) = SUpj3ksN\Sj(X)- sk(x)I; since we have ro = * * = rN = 0, 11~511~ + o as N + 00. Finally, note that since Rf 3 3 R53 2 0 and R & ( x ) dx + 0 we also have limN+mRK(x) = 0 a.e. This result is also from Meyer’s work [1979].) 7.30 Suppose {S,}ueIis a family of transformations from T into T verifying the following properties: (i) each S, is measure preserving, i.e., if E is a measurable subset of T, then [S;’(E)l = IEl, and (ii) the family is mixing, i.e., if El and E2 are measurable subsets of T and r > 1, then there is an S, in the family such that IE, n S;’(E2)1 S rlE,(IE21.(Observe that if the S,’s were such that IE, n S,’(E2)l = IElllE2),then E l and S,’(E2) would be probabilistically independent; our assumption is not so restrictive.) Prove that if { E k }is a sequence of measurable subsets of T such that IlEkl = 00, then there exists a sequence {S,} of S,’s such that almost every x in T is in infinitely many of the sets S;’(Ek). 7.31 (Sawyer) Suppose that { Tk}is a sequence of linear operators defined in Lp(T) for some 1 s p =S 00 which verifies the following assumptions: (i) each T k is continuous in measure, (ii) each Tkis positive, i.e., iff 3 0, then Tkf 3 0, (iii) there is a family {S,},EI verifying the assumptions of 7.30 so that the Tk’sand Su’s commute in the following sense: i f f € L p ( T )and

I,

-

-

-

7. Notes; Further Results and Problems

109

( S , f ) ( x ) = f ( S , x ) , then for each Tk, TkS, = S,Tk; more precisely, for each x E T, T k ( S , f ) ( x ) = ( S , ( T k f ) ( x ) . Prove that for such a sequence of operators the following conditions are equivalent: (a) supkl Tkf(x)I = T * f ( x ) is of weak-type ( p , p) and (b) for each f E Lp(T ) , T * f ( x ) < 00 a.e. (Hint: As in the proof of Calderh’s theorem, assume that T* is not of weak-type ( p , p) and reach a contradiction; 7.30 is an important step in the proof which follows de Guzmh’s presentation [ 19811. Sawyer’s principle has wider applications than it may seen on the surface; in fact, although the T k ’ s themselves may not be positive, they may be of the form Tk = PkK where the Pk’s are positive and T is some fixed, bounded, linear operator in Lp. Then the estimate for T* will automatically follow from the one for P*.) 7.32 (Baire Category Theorem) A metric space is said to be of first category if it can be written as a countable union of sets that are nowhere dense. If a metric space is not of first category, then we say it is of second category. The rational numbers on the line with the usual metric are of first category; on the other hand, the real line is of second category. This last assertion is a special case of the Baire category theorem: A complete metric space X is of second category. (Hint: Suppose X = UGl Xi,where the Xjare nowhere dense, that is, the sets have no interior points. Fix a ball B ( x o , 1); since XI does not contain it, there is a point x1 E B(x,,, l ) \ x l ; from this point on the proof is identical to that of the uniform boundedness principle, Theorem 2.3 of Chapter I.)

xj

CHAPTER

V The Hilbert Transform and Multipliers

1. EXISTENCE OF THE HILBERT TRANSFORM

OF INTEGRABLE FUNCIIONS This section covers one of the basic results in harmonic analysis, namely, the (a.e.) existence of the principal value integral defining the Hilbert transform of a summable function. This limit will not, in general, be a locally integrable function, though. The proof we present here is purely real variable and it serves as an inspiration for the extension of these results to the Euclidean n-dimensional case.

Theorem 1.1. Suppose that f E L ( T ) . Then

'I

lim HEf(x) = lim -S+O

E+O

7r

f(x - 2 ) 2 tan(t/2) dt

the Hilbert transform exists for almost every x in T. We call this limit Hf(x), off, and the relation (1.1) is denoted by

Hf(x)

= p.v. - -

dt.

The principal value notation (P.v.) above emphasizes that a symmetric neighborhood about the origin is deleted before the limit is taken; the origin corresponds to the singularity of the convolution kernel here. Proof. The idea of the proof is to show that the p.v. integral converges a.e. in the complement of a sequence of sets with measure approaching 0; once this is achieved, clearly the p.v. integral exists also in the union of these sets, namely, in a subset of T of full measure. 110

111

1. Hilbert Transform of Integrable Functions

In first place we may, and do, assume that f is nonnegative. Let A be a the average of I over T, will do. We then invoke large constant; A > the Calderbn-Zygmund decomposition o f f at level A, Theorem 3.1 of Chapter IVYand obtain a sequence of open, disjoint subintervals ( 4 ) of T and a decomposition f = g b, induced by these intervals. Clearly, for each E > 0 H , f ( x ) = H,g(x) H,b(x), x in T. Moreover, since g E L 2 ( T ) ,by Corollary 6.5 in Chapter 111, lim,,o H,g(x) = g ( x ) exists a.e. in T. Let, as usual, R* = U2 4 , IR*ls(2/A) j T l f ( y ) l dy. Suppose we can show that

If

If IT,

+

+

lim H,b(x) E+O

exists a.e. in

( 1-21

T\R*.

Then by choosing a sequence Ak + a,and if {RE} denotes the corresponding sequence of open sets, the p,v. integral in (1.2) exists, for the corresponding b’s of course, a.e. in the complement of R z and the same is true for the p.v. integral corresponding to J: Thus the Hilbert transform off exists T\R?), which is a subset of T of full measure. a.e. in Uk( Returning to the proof of (1.2), then, it suffices to show that { H , b ( x ) } is Cauchy a.e. in T\R* as E + 0. In other words, we must show that lim \H,b(x) - H,b(x)J= 0,

a.e. for x in

T\O*.

(1.3)

&,V+O

We break the proof of (1.3) into two steps. In the first place, we show that lim suplH,b(x) - H,,b(x)l < a,

a.e. for x in

T\R*,

(1.4)

E,11+0

and once we know that this lim sup is finite we show, in the second step, that it is 0. For E > 7 > 0 fixed, note that H,b(x) - H,b(x) equals

Moreover, since b ( t ) = C j b(t),yIj(t)and x f? 4 for any j , we can divide the 4’s into 2 families, namely, those intersecting [ x - E, x - 7 ) and those intersecting ( x + 7, x + E ] ; all other intervals contribute nothing to the integral in (1.5) and may be disregarded. As the argument for each family is identical we only consider those intervals 4 intersecting ( x + 7, x E ] . These intervals can in turn be separated into three subfamilies, to wit

+

(i) Possibly an 4 containing x + 7 , call it 1;. (ii) Possibly an 4 containing x + E, call it 1;. (iii) Those I’s totally contained in ( x + 7 , x + E ] , call them still 4.

112

V . The Hilbert Transform and Multipliers

The main contribution to (1.5) comes from the integrals extended over the 4’s; the other terms represent basically “error” terms. We make this statement precise. Let L; = length of 17, x; = center of 1; and put

The following two geometric observations are clear: (a) (b)

Since x E 21;, then r] > L;/2. ( x 7,x + E ] n 1; c ( x + 7 , x

+

+ 7 + LT) G ( x + 7,x + 377).

Thus

IA7l zs

I

Ib(‘)l

(x+1),x+31))

-

dt.

(1.6)

2ltan((x - t)/2)1

-

Moreover since tan u 77 when u 7 and 7 is small, the denominator of the integrand in (1.6) is of order c / 7 and it follows that Ib(x + t)l di

Ib(t)l dt = 77 C

Ib(x+t)ldt.

s-j

77

(1),3r))

(1.7)

(031))

Now b ( x ) = 0 since x E T\a* and (1.7) can finally be rewritten as

The expression appearing in the right-hand side of (1.8) is a familiar one, as we have previously encountered it in Proposition 4.1 of Chapter 11; it actually is o ( 1 ) as 7 + 0 at every Lebesgue point x of b that is a.e. in T. Similarly, C

c3E

J

Ib(x + t ) - b(x)l dt

= o(1)

(0.3~)

also a.e. in T. So these are indeed error terms.

as

E

+O

(1.9)

1. Hilbert Transform of Integrable Functions

113

To estimate the main contribution we invoke the cancellation property = center of 4

jI, b(t) dt = 0 and rewrite with xi

(1.10) say. Furthermore, a simple computation making use of elementary trigonometric identities gives k(x, t, x j ) =

sin( ( t - xj)/2) sin((x - t)/2) sin((x - xj)/2)

(1.11)

and to estimate k(x, t, xi) we refer once again to the geometry of the situation. In the first place, notice that if t, xj E 4 and x E 24, then (1.12) Ij I

I

,

t

\

I

x.

J

21j 1

,

X

Indeed, since with Lj = length of 4 we have Ix - xjl > Lj/2, then we see at once that Ix - tl Ix - xi[ + Ixj - tl G Ix - xjl + Lj s 31x - xi[. Exchanging the role of xi and t we also note that Ix - xjl s 31x - tl, and combining these two estimates (1.12) follows. Moreover, since we are dealing with small values in the above expressions, we may replace the sines there by their arguments and on account of (1.12) (and estimate (1.16) of Chapter I) we observe that (1.13) This is all we need to complete the estimate of the Aj's. Indeed, by (1.10) and (1.13) it follows that

V. The Hilbert Transform and Multipliers

114

with c independent of x and j . Combining these bounds with the ones obtained for the interval [ x - E, x - 7 )we finally see that for almost every x in T\R*,

I H A x ) - H,b(x)l

=

L&,,(X),

(1.14)

say. Thus

say. In view of (1.15) estimate (1.4) will hold once we show that L ( x ) < 00 a.e. For this purpose we intoduce a majorant of L ( x ) , the Marcinkiewicz function A(J x) associated to R = I j and given by

U

(1.16)

We claim that

A(J x ) < 00

a.e. in

T\R*.

(1.17)

The usual procedure for this is to note that the stronger statement r (1.18) I = J A(Jx)dx
actually holds. First, observe that

where c is independent of j. Whence by Tonelli’s theorem we get that

thus (1.18) obtains and the first step of our proof is complete. By the way, Chebychev’s inequality applied to (1.19) gives at once that

r

( 1.20)

2. The Hilbert Transform in Lp(T), 1 S p < 00

115

To see that (1.3) also holds is now straightforward. Note that we actually have L(x) c c

f 4IX

j=N+1

- xjl

1

rj

If(t)l dt,

each

N

5

(1.21)

0,

since by the fact that x E T\UICl lj for each N > 0, we may always miss the first N intervals in the family (4) as soon as E , 77 > 0 are small enough in (1.14). Furthermore, since the right-hand side of (1.21) is the tail of a convergent series for almost every x in T\R* we obtain that L(x) = 0 for those x's by letting N + 00 in (1.21); this means that (1.3) holds. Remark 1.2. We isolate two facts from the proof above. First, by putting E = m in (1.14) we see that for almost every x in T\R*

IH,b(x)l

'I

W f x) ; +77

Ib(t)l dt +

(X+,,X+3,)

C ( A U x) + Mb(x)),

77

I

Ib(t)l dt

(x-3,T-V)

( 1.22)

71 > 0

with c independent of 7. Moreover, once again by (1.14) but taking now limits as 77 + 0, we note that for almost every x in T\R* (1.23)

IWX)l zs c A(f, x). 2. THE HILBERT TRANSFORM IN Lp(T),1

00

Although the Hilbert transform Hf is well defined for any f E L(T), it is not in general integrable. The following simple example illustrates this point: let f E L( T) be a positive function, vanishing off [O,m/2). Then for x E [-w/2, O),

Since x < 0 and t > 0 above, tan((x - t)/2) = -tan((lxl+ t)/2) and consequently

116

V. The Hilbert Transform and Multipliers

But as we have already seen in Section 2 of Chapter IV, to make this integral large it suffices to put f(t) = d(l/ln(l/t))/dt 2 0; then f E L( T) but Hf is not integrable. Nevertheless, as in the case of the Hardy-Littlewood maximal function, we have the following substitute weak-type result: Theorem 2.1. Let f E L( T). Then for each 0 < E < T there is a constant c, independent of E andf, such that for all A > 0

hl{lH&fl> A l l c C l l f l l l .

(2.2) Proof. We may assume that A > llflll; we then invoke the Calder6nZygmund decomposition at level A and write, in the notation of Theorem 1.1, f = g b. Also for x E T\R*,

+

H&f(X)

= H,g(x)

+ H&b(X).

(2.3)

Moreover, since {IHJ(x)l>

A}

c Q* u {x E T\Q* :IHef(x)I > A},

(2.4)

it suffices to estimate the measure of each set on the right-hand side of (2.4). In first place

IQ*l

2 C l l i l c 4Tllflll/A and this bound is of the right order. Also by (2.3) {x

E

(2.5)

T\R*: IH&f(x)l > A }

c {x E T\R*: IH,g(x)l> h/2} u {x E T\R*:

IH,b(x)l> h/2)

=IuJ,

say. To estimate Z we make use of the L2 result for H,g. From Chebychev’s inequality, estimate (6.2) in Chapter I11 and estimate (3.7) in Chapter IV,

which in turn implies that 1 1 1s C l l f l l l / A ,

and this term is also of the right order. As for the set J, by estimate (1.22) of Remark 1.2, we note that it is contained in {x

E

T\R*: A(J x) > A/4} u {x

E

T\Q*: M b ( x ) > A/4}

= J1 u Jz,

say. By estimate (1.20), lJll

s Cllflll/A.

(2.7)

2. The Hilbert Transform in Lp(T), 1 ==p < 00

117

Also by the maximal theorem and estimate (3.9) of Chapter IV we obtain cIIbIIi/A C I I f I I i / A (2.8) Thus, by combining (2.5)-(2.8), and since all constants are independent of E and f; the desired conclusion follows. IJ2I

Observe that actually H , f ( x ) , being the convolution of integrable funcH,flll = 00 in tions, is also integrable for each E > 0; nevertheless, limE+.o~~ general. A reasoning essentially analogous to that of Theorem 2.1 obtains Theorem 2.2 (Kolmogorov). Let f~ L ( T ) . Then there is a constant c, independent off; such that for all A > 0

> All c Cllflll.

(2.9) Proof. We use estimate (1.23) of Remark 1.2 in this case; the proof is simpler than that of Theorem 2.1 since the term J2 does not appear. AI{IHfI

It is now an easy task to complete the consideration of the Lp result as well. We do the case 1 < p < 2 first, which in view of the L2and weak-type (1,l) estimates we expect to be rather straightforward. Theorem 2.3. Suppose f E Lp(T ) , 1 < p < 2. Then there is a constant cp = O(1/( p - l)), independent of E and f; such that llH€fllP cpllfll,.

(2.10)

Theorem 2.4 (M. Riesz). Suppose f E Lp(T), 1 < p < 2. Then there is a constant cp = O(l/(p - l)), independent off; such that

IIHfllP

(2.11)

CPllfllP.

Moreover, H f ( x ) = f ( x ) a.e. and consequently Lp(T) admits conjugation. Proof of Theorem 2.4. Let l / p = (1 - 7)+ 7/2,0 < 7 < 1; then by Theorem 4.1 in Chapter IV (with po = 1, p, = 2 there) H is an Lp bounded mapping with norm O( l / ( p - 1)('-')). Furthermore, since ( p - 1)' d 1 and actually tends to 1 as p + 1, cp = O(l/(p - 1)) as anticipated. In order to show that H f ( x ) coincides with f ( x ) a.e., we consider H ( f - a , ( f ) ) ( x ) , which equals H f ( x ) - c?,,(f; x ) a.e. since o n ( fE>L 2 ( T ) . Now by (2.11)

IIW-

Gn(f>IIp

cpllf-

an(f)Ilp,

(2.12)

with cp independent of n. But the right-hand side of (2.12) goes to 0 as n + 00 by FejCr's Theorem 2.7 of Chapter 11, and so does the left-hand side. Thus, on account of Proposition 1.1 of Chapter I,

c j ( H f )= limn+.m cj(c?"(f)), allj.

K The Hilbert Transform and Multipliers

118

Moreover, since c j ( G n ( f ) >= (1 - Ijl/(n + l))(-i)(sgnj)cj(f) when Ijl d n and 0 otherwise, we see at once that the above limit is (-i)(sgn j)cj(f); this implies that f = Hf E Lp(T). H The case 2 < p S 00 remains to be discussed. When p = 00 we may, and will, give more than one answer; this is because f E L“(T) does not, in general, imply that Hf E La( T), as the simple example in Remark 2.7 below shows. Before discussing this negative result, we consider the positive results for 2 < p < 00. They are readily obtained by a “duality” argument.

Theorem 2.5. Suppose f E Lp(T), 2 < p < 00. Then there is a constant cp = . O(p), independent of E and f; such that IIHEfllP 6 CPllfIlP.

(2.13)

Theorem 2.6. Suppose f E Lp(T), 2 < p < 00. Then there is a constant cp = O ( p ) ,independent off; such that llHf

IlP

s

CPIlfIlP.

(2.14)

Moreover, Hf(x) = f ( x ) a.e. and consequently Lp(T) admits conjugation.

Proof of Theorem 2.6. Since Lp c L2, we already have Hf = f E L2(T) by the results in Section 6 of Chapter 111; it only remains to prove that Hf E Lp(T) as well. By Theorem 3.3 of Chapter I1 it suffices to show that IIantf)IIp

s

CpIIfIIp

(2.15)

with cp = O(p) independent of n and f ; this is not hard to do. Indeed, let g be a trigonometric polynomial with llgllp. d 1, l/p + l/p’ = 1. Then by Parseval’s identity (iv) of Theorem 1.8 in Chapter 111 it follows that

Whence by Holder’s inequality, estimate (3.4) of Chapter 11, and Theorem 2.4 we obtain

III

cp*llfllp* (2.15) follows now by the converse to Holder’s inequality but with cp,instead of cpthere. This is only a minor inconveniencebecause since p = pf/(p’ - l), pf near 1 and c,. = O( 1/( p’ - l)), then also cp,= cP = O(p). l l ~ n ~ ~ ~ l l p ~ l l f l l p

Remark 2.7. The order of the constant cp in Theorem 2.7 is sharp. Indeed,

2. The Hilbert Transform in Lp(T), 1 < p < w let f ( t ) = x[o,ll(t). Then clearly c(ln(sin(x/2)1- lnlsin((1 - x)/2)l), and

= c(

I[,.,

l f 11,

sPe-’ ds)

119

= 1,

all

-=. r ( p + c

p

3

i)lIP.

1,

Hf(x)

=

(2.16)

Here r ( p + 1) denotes the gamma function of order p and by known estimates from (2.16) we see at once that lim infp+mllHfllp/p3 c > 0. Note in passing that, although Hf is not bounded, Hf E Lp,Hfis exponentially integrable, and IHf(x)l has logarithmic growth. These properties remain true for the Hilbert transform of every L“ function, as we will prove shortly (however, the last property will have to wait until Chapter VIII). Since Lp(T) admits conjugation for 1 < p < co, IlS,,(fl -flip + 0 as n + 00. We also have in this case

np<m

Proposition 2.8. Assume that f E L P ( T ) ,1 < p < co.Then as n + co. Proof. Since ;,(A n+m.

-f

=

(sn(n-n-,

IIfn(f)

To complete the discussion of the case p

--711p =

IIs”(f) -flip + 0

CpIISn(f)

-fIIp+=

0 as

1 we show

Proposition 2.9. Assume that f E L( T ) and 0 < p < 1. Then IIHfll, c p ~ ~ cp f ~=~0(1/1 l , - p ) independent off: A similar estimate holds for HE with cp also independent of E.

Proof. Since Hf E wk-L( T) we are in a position to use 7.5 of Chapter IV; we put (Y = p and /3 = 1 - p there and note that the constant is of the right order. W One of the first results we discovered was that 11 H E f - HfII, + 0 as E + 0. It is reasonable to expect that a similar result will hold for Lp norms, 1 < p < 00, as well as substitute results for p = 1; we prove this next. Proposition 2.10. Assume that f E Lp(T ) , 1 < p < a.Then limll HJ - Hfll, E+O

= 0.

(2.17)

Proof. For an arbitrary f~ L P ( T )let g be a trigonometric polynomial so that Ilf- gllp < 7 is arbitrarily small. Then by Minkowski’s inequality

V . The Hilbert Transform and Multipliers

120

we have

IIH€f-H f l l P

s

I I H E f - H&llP + IIKg - HgII,

+

llm?-

s c p l l f - gllp + IIKg - Hgllp + cpllf-

WllP

SllP

s 2crl+ IIH&g- HgIIp. - HgII, = 0. But since g Consequently, it suffices to show that limE+ol[H&g is a trigonometric polynomial, the proof for estimate (6.4) of Proposition 6.2 in Chapter I11 also works in this case.

Proposition 2.11. Assume that f E L( T ) . Then

lim H E f = Hf;

in measure,

E-0

(2.18)

and - Hfll, = 0, limllHEf €’O

0 < p < 1.

Proof. The convergence in the Lp metric follows as above and (2.18) is an immediate consequence of this. W

The truncated Hilbert transforms, as well as the Hilbert transform itself, are uniformly controlled by the “maximal Hilbert transform H*” given by H * f ( x ) = SUP IH€f(X)l.

(2.19)

O<€ca

The behavior of H* is similar to that of H itself. The next two results make this precise. Theorem 2.12. Assume that f E Lp(T ) , 1 < p < 00. Then there is a constant c independent off such that

H*f(x)

c ( M f ( x )+ M ( ~ f ) ( x ) ) .

Thus IIH*fllp s cpllfllp with cp = O ( l / ( p - 1)’) as p as p + co. Proof. H E f can be readily estimated when

s c

E

3

+

(2.20)

1+ and cp = O ( p )

1 since

If(t)l dt s c M f ( x ) . T

Now if 0 < E < 1 and n is the integer such that l / ( n + 1 ) s

E

H & f ( x )= ( H & f ( x ) H l / n f ( x ) ) + ( H l / n f ( ~-) G n ( f ; + Gn(f;x ) = I + J + K ,

< l / n , write

3. Limiting Results

121

say. Thus IHEf(x)ls 111 + IJI + IKI, and it suffices to estimate each term separately; we do 111 first. By taking sup over n in estimate (6.8) in Corollary 6.4 of Chapter 111, it follows that 111 S c M f ( x ) . The same is true for IJI if we consider estimates (6.6) and (6.7) in Theorem 6.3 of Chapter I11 instead; the former gives c M f ( x ) directly whereas the latter must be combined with Proposition 2.3 of Chapter IV first (with +( t ) = t-2,yCl,m)( t ) there). To bound JKI we invoke Proposition 2.3 of Chapter IV. This gives IKI s c M ( H f ) ( x ) and (2.20) holds; the norm estimate with the growth order of the cp’s obtains from (2.20) at once. W We also have the weak-type estimate when f is merely integrable Theorem 2.13. Assume f~ L( T), then H * ~ wk E - L( T ) and there is a constant c, independent o f f , such that A l { H * f > A } [ S cllfll,,

all A > 0.

(2.21)

Proof. As in the proof of the weak type for Hf itself we invoke the Calder6n-Zygmund decomposition at level A sufficiently large, and write f=g+b,R*=U21,..Then H * f ( x )< H * g ( x )+ H * b ( x )

and it suffices to show that Al{H*g

’

S cllfll1

(2.22)

(2.23)

and Al{x E T\R*: H * b ( x ) > A/2}1 S cllflll.

(2.24)

This is straightforward. Indeed, by Theorem 2.13 11 H*gll, S cllgll, S CA 11f [I1 and (2.23) follows from Chebychev’s inequality. Moreover, by estimate (1.22) in Remark 1.2 we note that H * b ( x ) S c ( A f ( x )+ M b ( x ) ) ,x E T\R* and (2.24) follows at once on account of estimate (1.2), the fact that the maximal operator is of weak-type (1, l), and )Ib 11 S c 1) f 11 Remark 2.14. We may now combine Theorem 2.13 with the Marcinkiewicz interpolation theorem to find that, in Theorem 2.12, cp = O ( l / ( p - 1)); thus the analogy with the Hilbert transform itself is complete.

3. LIMITING RESULTS Assume f E L( T ) ,f

- 1 cje? Still the question remains as to when 1(-i )(sgn j )cje (3.1) gx

V. The Hilbert Transform and Multipliers

122

is a Fourier series. Suppose Hf E L ( T ) . Then since Hf = f * p.v.(l/(tan(t/2))) (in D') by 7.22 in Chapter I11 we get that c j ( H f ) = (-i)(sgnj)cj and (3.1) is a Fourier series in this case. To decide when Hf is integrable we proceed as we did for the maximal operator and search for an extrapolation result first. Proposition 3.1. A sublinear mapping T defined in L + Lp, 1 < p < 00, is simultaneously of weak-types (1,l) and ( p , p ) if and only if there are constants c, , c,, c3 such that for all f E L Lp and A > 0

+

14 Tf I > A l l

I,,

CI(+

s"-'l{lf

I > s)l

ds

.c2A )

+

A

[

I{lfI > 41 d s ) .

(3.2)

[s*,00)

Proof. The necessity follows from the decomposition (4.14) of Chapter IV with E = 1 there. As for the sufficiency, if (3.2) holds and f E Lp say, then in the first integral we just replace c2A by 00 in the limit of integration, and this term is of the right order. It is also readily seen that the second integral does not exceed

which is also of the right order. The weak-type (1,l) case is handled similarly. Finally note that (3.2) is equivalent to the single condition obtained by setting A = 1 there. We can now extrapolate. Proposition 3.2. Assume T is a sublinear operation defined in L + Lp, 1 < p < co, which is simultaneously of weak-types (1,l) and ( p , p ) . Then T maps L In L into L and there are constants A, B independent off such that

11 Tflll

A

+B

IT

If(t)lln+lf(t)l dt.

(3.3)

Proof. It sufficesto show that ~~1~,1,1~~ Tf (t)l dt is bounded by the right-hand side of (3.3). But this is immediate since by (3.2) it suffices to show that the same is true for

(3.4)

3. Limiting Results

123

and (3.5) That the expression in (3.5) is of the right order was already proved in Theorem 5.3 of Chapter IV; the expression in (3.4) is quite easy to handle. Indeed, suppose as we may, that c2 Q 1 and observe that (3.4) is dominated by 1 sP-l ds dA

I,,,,, P I,, ,

I,, +LC2*)

sp-'({lfl > s}l ds dA

+

A+ sA

J

~P-'IWI > S)I

1 dA ds

[S/C2.-3)

IC2.W)

+B J

[

){If1

AP

> s}) ds.

lc2,m)

Proposition 3.2 clearly applies to H E , H, and H*. There also is some evidence, furnished by estimate (2.1), that, as in the case of the maximal operator, some kind of converse result may be true. Indeed, for positive, integrable functions f vanishing off [0, m / 2 ) with Hf E L( T), we have that

*>

IHf(x)l dx

3 c

[

(O,m/21

lro,xlf(r) dt dx

f(t)

= C! [O.r/21

=c

x

I

1 -dxdt sr/21 X

f ( t ) ln(m/2t) dt,

CO,r/21

which is readily seen to imply the integrability of f ( t ) ln+f(t) in [ 0 , r / 2 ) . The positivity o f f is important here. We shall return to this question in Chapter VII and remove the restriction on the support off then. There is yet another extrapolation result and it corresponds to the spaces near L", since, unlike the maximal operator, the Hilbert transform is not bounded in La( T). We may search for this result by a duality argument as follows: for trigonometric polynomials f and g observe that

V. The Hilbert Transform and Multipliers

124

On account of Young's inequality (see 5.2 of Chapter I), we expect now some kind of exponential integrability for f when f~ L". The precise statement for general operators T that behave like the Hilbert transform is

Theorem 3.3. Assume T is a sublinear operator bounded in Lp,1 < po s p < 00, with norm O ( p ) as p + 00. Then there exist constants c and A independent o f x such that (3.6)

Proof. We estimate the integral in (3.6) by

c

+c

" 1 1 G A k ( k+ l)kl(fll&, kapo

which is readily seen to be dominated by an absolute constant provided A is sufficiently small, since ( k / e ) k k ! . H This general theorem indeed applies to H E , H, and H*, but is there a more precise result for the particular case of the Hilbert transform? In other words, can we be more explicit about the value of A in the conclusion of Theorem 3.3? In order to do this and to illustrate the simplicity and power of the so-called complex method, we prove the following result Proposition 3.4 ~ / 2 Then .

(Zygmund). Assume f is a real-valued function, If(x)l s evCx)lc o s ( f ( x ) ) dx < 2 cos (;PIT

f(t) dt

)

s 2.

Proof. Suppose first that f is a trigonometric polynomial, f ( x ) = and put F ( x ) = i ( f ( x )- i f ( x ) ) .F ( x ) has the following properties

c,:

(3.7)

qe""

cje""). (i) It equals i(co+ 2 (ii) If F ( z ) = i(co 2 C,:, cjzJ),then F ( z ) is a harmonic function in Iz( s 1 (actually it is analytic there). (iii) F ( 0 ) = ico = i((1/27r) j T f ( f ) d t ) , (co(s 7 ~ 1 2 . ) also harmonic and analytic in IzI s 1. (iv) e F ( =is By the mean value property of harmonic functions, to be proved in Chapter VII, we have

+

3. Limiting Results

125

Whence replacing F ( e k ) by its explicit form and taking real parts in (3.8) we see that

e R x )cos(f(x)) dx

= Re =

((1/27r)

ef(x)e*(x) dx JT

-

Re( eF(’))= cos( c,).

(3.9)

Also, replacing f by -f in (3.9), we see that

& j, ,-m

cos(f(x)) dx

(3.10)

= cos(c,)

and adding (3.9) and (3.10) we get

kI,

(3.11)

eliCx)lcos(f(x)) dx 6 2 cos(c,).

Next we would like to show that (3.11) still holds for an arbitrary, real-valued function f with If(x)l s 7r/2. To see this, consider the FejCr polynomials a,,(f;x) = f * 2K,,(x); a,,(f,x) is real valued, Ila,,(f )llm ( ~ / ~ ) I I W , I=I I m/2, and c,(a,,{f)) = c,, all n. Furthermore, since Ila,,(f)-fl12 + 0, then also IlG,,(f) -fllz + 0 and there is a subsequence nk -+ co such that a,,,(f;x) +f(x) and G,,,(f; x) + f(x) a.e. in T. Since (3.11) holds with f replaced by u,,,(f)there and the right-hand side is independent of nk, by Fatou’s lemma we see that (3.11) holds for f as well. Corollary 3.5 (Zygmund). Suppose f is real valued and dx < co. Then

I,

Proof. Let

llfllm

=

~

~


L < r / 2 . By (3.7)

j, eIj(x)l

45-

dx < -( 0 0 . cos L

That this result is sharp follows at once by looking at f ( x ) = h(2x(-T/,v)(x)- 1). Corollary 3.6. Let

llfllm

< 7r/2, then (3.12)

and we have finished.

m

V. The Hilbert Transform and Multipliers

126

4. MULTIPLIERS

The fact that H is a bounded mapping in L2 constitutes an important step in establishing the continuity of the Hilbert transform in Lp, 1 < p < co, as well. The proof of the Lz result depends on the following 2 facts: for f C cjeqx, Hf C (-i)(sgnj)cjeqx and { ( - i ) sgnj} E l“(2) = I”, the (Banach) space of bounded sequences. In fact there is nothing special about { ( - i ) sgnj} as any I“ sequence will do. More precisely, suppose that { A j } E I” and let T be the (linear) mapping defined by

-

-

These T’s are called “multiplier operators,” since they are obtained by pointwise, or coordinatewise, multiplication of the Fourier coefficients of f by the “multiplier sequence” or, plainly, “multiplier” { A j } . To show that T is bounded in L2is quite easy, since by the Riesz-Fischer theorem Tf E L2and

Furthermore, if limn+mlAjnI = supjlAjI, by putting f ( x ) = evnx in (4.2) we readily see that the norm 11 TI1 of T as a mapping in L2is

It is also natural to consider the possibility of obtaining Lp continuity results for multipliers for values of p other than 2. It is convenient to introduce the notation Mp for the collection of bounded Lp multipliers, 1 S p G 00, i.e., a sequence { A j } E Mp if and only if for trigonometric polynomials f; (4.4)

with c independent o f f ; the infimum over the c’s in (4.4) is called the “norm” of the multiplier. Short of characterizing M p , which still remains an open question for 1 < p < 00, p # 2, we are often satisfied to decide whether a given individual sequence { A j } belongs to Mp or not. To illustrate this we give this typical, although not sharp, result.

4. Multipliers

127

Proposition 4.1 (Hirschman). Let { A j } be a bounded sequence so that [Ail = O(ljIpa),0 < a < 1. Then { A j } E Mp, ( 1 - a ) / 2 < l/p < ( 1 + a ) / 2 .

Proof. Our assumption implies that sup

k 3 0.

lAjl = 0(2-"k),

(4.5)

~'
For trigonometric polynomials f Tkf(X)

- 1 cje""and k 0 put 1 Ajcjevx. 3

=

2'
Clearly, Tk is of type (p, p) for all p 5 1, but we need a good estimate on its norm; this can be easily done for p = 1,2. In first place, observe that IITkfll1

s

c

IAjI lcjl

C2k2-"kIlfll~

(4.7)

zks1jl<2'+'

and IITkfll:

=

1

C2-2akllfll:-

(4.8)

~ ~ j ~ 2 ~ c j ~ 2 ~ ~ < l j l < ~ ~ + '

(4.7) and (4.8) give that the norm of Tk in L is of order 2(1-u)k and that in L2 is of order 2-"k. We are now in a position to invoke the Marcinkiewicz interpolation theorem 7.14 of Chapter IV: Let 0 < 7 < 1 and put l/p = (1 - 7)+ 7/2. Then the norm 11 T k l l of Tk as a mapping in Lp is of order m

11 ~ ~ 1 C2(1-~)(1-~)k2--u~k 1 = C2(1-a-~)k

and

2 IITkII <

00

k=O

provided 7 > 1 - a. Thus restricting our consideration to those ~ ' s and , consequently looking at those p's so that 1/2 6 l/p < (1 + a)/2, we obtain that Tf = Aoco I: Tkf = is also' bounded =, in Lp.A similar argument applies with (4.7) replaced by

+

11 Tkfll"

s C2(1-")k Ilf

llm

(4.9)

and by (4.8) and (4.9) we may now interpolate between L2 and L" to obtain those p's so that (1 - a ) / 2 < l/p 1/2. Notice that in this case, as well as in the Hilbert transform, the set of l/p's for which { A j } E Mp is a symmetric interval about i; we shall return to this observation shortly. There is yet another way to arrive at the multiplier problem, and it is motivated by the integral, or convolution representation of H. First a definition: for a function f and a E T let ~ , denote f the translation o f f given by T , ~ ( x )= f(x + a ) , x E T. More generally, if F is a distribution we define the distribution T,F by means of T,F( U ) = F( T - , u ) ,

u E C"( T ) .

(4.10)

128

V . The Hilbert Transform and Multipliers

We say that a linear operator T is “translation invariant” if it commutes with translations; more precisely, T,(

Tfl = T ( T , ~ ) , all

aE T

(4.11)

for all trigonometric polynomials f; say. We are interested in obtaining an intrinsic characterization of these operators and in this direction we have Proposition 4.2. Assume T is a linear operator which verifies (4.11) and which, in addition, admits a continuous extension to some Lp, 1 s p s co. Then there exists a sequence { A j } so that for all trigonometric polynomials f C cjeVx,

-

Tf (x) =

c

AjcjeVx.

(4.12)

Moreover, sup(AjlS 11 TII,

the norm of

T in

Lp.

(4.13)

i

Clearly, if A is the distribution 1 Ajevx and f is a trigonometric polynomial, then Tf = A * $ Proof. For each j let $ = T ( e V x )then , {$} is a sequence of Lp functions each with norm s 11 TII. By the translation invariance and linearity of T we readily see that T,$

=

7,(T(eVx))= T(T,eVx)= T(eV(x+a)) = e@‘J.

(4.14)

We compare now the Fourier coefficients of the functions in (4.14). Since by (3.1) of Chapter I we have that Ck(7,fl = e i h c k ( f ) ,(4.14) obtains eihck($) = ei’”ck($),

all j , k.

(4.15)

Clearly, the solution to (4.15) is ck($) = 0 , j # k, and the functions$ = AjeVx, all j . Thus [Ail = 11 T(eVx)ll,S 11 TI/, all j . By the linearity of T we see that for trigonometric polynomials J; T ( 1 cjeVx)= 1 cjT(eVx) = 1 AjcjeVx, and also Tf = A * $ W Corollary 4.3. Assume T is as in Proposition 4.2, then for trigonometric polynomials f; g, Tf * g = T (f * g ) = f * Tg.

Thus translation operators T correspond to multipliers, and vice versa; this suggests our next result. Proposition 4.4. Suppose { A j } E I“. Then { A j } E Mp if and only if for all trigonometric polynomials f, g and with l / p + l/p’ = 1, (4.16)

with c independent off and g.

4. Multipliers

129

Proof. Just use the convolution representation

C Ajcj(f)evx = A * $

By (4.16) we see at once that for p 2 1 { A j } E Mp

if and only if

{Aj} E Mpr,

l/p

+ l/p' = 1,

(4.17)

and from this is clear that

M p = Mp,,

l/p

+ l/p' = 1

(4.18)

Moreover, we also have Proposition 4.5. Assume { A j } E Mp. Then { A j } E M, for min(p,p') max( p, p'); thus

M P sMq,

lspsqS2.

G

r =s

(4.19)

Proof. By (4.17) we may assume that 1 S p S 2 and by the Marcinkiewicz interpolation theorem 7.14 of Chapter IV we see at once that { A j } E M,, p s r S p', and we are done. Consequently, the interval of l/p's for which { A j } E Mp is symmetric about f. Proposition 4.6. M2 = I".

Proof. By (4.2) it suffices to show that L2 multipliers are bounded; this follows at once from (4.13). Also combining (4.2) and (4.13) we see that if T denotes the multiplier operator associated with { A j } , then IITI( = supjlAjl. Proposition 4.7. MI = { { A j } : there is a finite measure p so that Aj SM.

= cj(p)}=

Proof. Since convolution with a finite measure p is continuous in L, it suffices to show that if { A j } E M I , then Aj = c j ( p ) for some finite measure p. But this is easy since Ila,( A) 11 = (\A* 2K, (1 G c ( ( 2 K (,1 = c, all n, and by Theorem 3.5 of Chapter I1 the Aj's are the Fourier coefficients of a finite measure p so that ( 1 / 2 ~ I,) ldpI G c. Combining (4.19) with Propositions 4.6 and 4.7 we obtain the important relation

SMSM,LM,C_I",

1SpSqS2

(4.20)

and the inclusions are proper for 1 < p < q < 2 (this last statement may require some thought; 5.40 and 5.41 below are relevant here). M p carries the structure of a Banach algebra, but we will not pursue this here. We will, however, prove that it is a dual (Banach) space.

V . The Hilbert Transform and Multipliers

130

- 1 hie"", IIA * f l i p

Proposition 4.8. Assume { A j } E Mp, 1 s p < a,A cllfll,. Then, there is a sequence {&} E C ( T ) so that

IIbn * A l p

~llfllp,

all

6

n7f

and limII&*f-A*f~l,=O,

f€LP(T).

n-m

(4.21)

+,,

Proof. Let = a , ( A ) . Then by Corollary 4.3 Ilan(A) *flip = Ilan(A* f ) l l p =s ~ ( ( f l ) ~ ,all n , f : Furthermore, by Theorem 2.7 of Chapter 11, Ila,,(A) * f - A *flip = Ila,(A - A *flip + 0 as n + co.

*n

We still need one more definition. Iff E Lp, g E Lp',then f * g E C ( T ) and Ilf* gll.. s Ilfllpllg[lp,.The same is true for finite sums of convolutions of this kind and also for infinite sums provided that, in addition, (4.22) k

c:='=,fk

for then it is readily seen that * & ( x ) converges absolutely and uniformly to a continuous function. For 1 s p s 2 let A,( T ) = Ap = {h E C ( T): h = C;=, fk * gk, and (4.22) holds}. Ap is a Banach space normed by the infimum of the expression in (4.22) for all decompositions c f k * gk of h. The space A: of bounded linear functionals L on Ap is also a Banach space with norm

and we have Theorem 4.9

(Figa-Talamanca). Mp is isometric and isomorphic to A;,

1
Prmf. In order to simplify the notations we identify the multiplier { A j } in Mp with the operator Tf = A *J; A Ajeqx;we write in this case T E Mp and we also have 11 TI1 = infimum over the c's in (4.16). We construct a mapping @ :Mp + A; with the following properties

-c

(i) @ is well defined, (ii) CP is an isometry, and (iii) @ is onto. This is how to do it: for T

E

M p and h

=

1fk * gk

in Ap put

m

(4.23)

4. Multipliers

131

Notice that since Tfk

*gk(0) =

&I

(4.24)

T f k ( x ) g k ( - x ) dx

T

the series in (4.23) converges absolutely and

11 Tll

I@T(h)l

cIlfkllpllgkllp’*

(4.25)

We want to show that the expression in (4.23) is independent of the * g k = 0, then the sum in representation of h; in other words, if h = (4.23) is also 0. Let (Pn = u,(A), then by Proposition 4.8, C k I I + n *fkllpllgkllp’ < 00, and & ( d n *fk) * g k ( X ) converges absolutely and uniformly as n + 00. Thus

cfk

c

Tfk * g k ( x ) 7

(4.26)

k

where the last equality is justified by (4.21). Moreover, since (P,, is a trigonometric polynomial for each fixed n, we may invoke (4.16) and the boundedness of the convolution with 4” and note that

c

(dn

*.h)* g k ( x )

4 n * (6* g k ) ( X )

= k

k

which combined with (4.26) gives that 1, Tfk * g k ( 0 ) = 0, and @( T ) h = 0, thus proving (i). From the proof above it readily follows that I@(T)hl=S IITII IlhllAp and consequently II@( T)II C 11 TI[.To prove the opposite inequality observe that by (4.16) given E > 0 there are functions f E Lp and g E Lp’ of norm 1 SO that IITII JTf* g(0)I + E. Then h = f * g E Ap, llhllAps 1, and IIT(( J@(T)hJ-f E s II@(T)JJ llhllAp+ E s II@(T))I+ E and (ii) also holds. TO show ( 5 ) let h* E A , * , ~ Lp E and for trigonometric polynomials g and g ’ ( x ) = g ( - x ) consider the map g + h * ( f * g’). It is clear that Ih*(f* g ) 1 Ilh*l( Ilf* 5llA,,s Ilh*ll,llfllpllgllp,so that e a c h f e Lp determines a bounded linear functional on Lp of norm S Ilh*ll Ilfll,. By Proposition 3.2 in Chapter I1 there is a unique Lp function Tf,say, of norm sIIh*(l Ilfll, so that

h * (f*8 )

=-

‘I

2.rr

T

T f ( X ) g ( - x ) dx

=

Tf * g ( 0 )

= @ ( T ) ( f *g

).

V. The Hilbert Transform and Multipliers

132

Since Tf is clearly linear, by Proposition 4.2 it only remains to show that T is translation invariant; this is easy since for each trigonometric polynomial g, T ( T , ~* )g(O) = h * ( ~ , f *g) = h * ( f * ~ ~=gTf* ) T,g(O) = T,( T f ) * g(O), and we have finished. The case p = 1, not covered by the above result, can be done in several ways. The quickest is to observe that by 7.17 of Chapter 111, C ( T ) = L ( T )* C ( T ) , and consequently A , = C ( T ) ; the proof that AT = .9M follows without much difficulty from this.

5. NOTES; FURTHER RESULTS AND PROBLEMS Lusin considered the question of the a.e. existence of the p.v. integral defining the Hilbert transfohn (Theorem l . l ) , as well as the integral in (5.16) of Chapter 111, and concluded that the cancellation due to the positive and negative values in these expressions is an important fact which plays a fundamental role in the study of the convergence of Fourier series. He was not satisfied with his own proof of the a.e. convergence of Hf for f E L2(T ) , since he felt that the use of the Riesz-Fischer theorem made it a complex, rather than a real, variables proof. It was Besicovitch who in 1926 first established the existence of Hf by purely real methods, and f E L( T ) .The interplay between? and the a.e. convergence of Fourier series is given by the following result, also due to Lusin: In order that the Fourier series of an L2(T ) function f fao+ 1 aj cos j x + bj sin j x converge a.e. in T it is necessary and sufficient that

-

lim lim *-w

E+o+

I

J(x

[E,Tr]

+ t ) -?(x t

-t)

cos nt dt = 0 .

(5.1)

As all the ingredients needed to prove this fact were discussed in this chapter, the reader is invited to furnish a proof. Inspired by (5.1) and the a.e. existence of the integral P.V.

I

CO.rl

g ( t + x ) - g ( t - x ) dt t

for g E L2(T ) , Lusin conjectured that the Fourier series of any f E L2(T) converges a.e. He based this hypothesis on the uniform distribution of the positive and negative values of cos nx and sin nx in T (which suffices to show that the Fourier coefficients of any summable function tend to 0). As for the multiplier question, it seems to have originated with some results of Fekete in 1923. The closest to a characterization of Mp is the following

5. Notes; Further Results and Problems

133

result of Stein: a function K ( x ) belongs to V,, 2 == q < 00 ( V is for Verblunsky) if and only if K E Lq( T) and sup IIc K ( bk - * ) - K ( ak - * ) 11 < 00. Here the sum is taken over any finite collection of nonoverlapping intervals, and the sup is taken over all such collections of intervals. Corresponding to a bounded sequence {A,,}, put K ( x ) = ~ , z o ( h , / i n ) e i " xand , let 2 < q < co. Then if the multiplier corresponding to {A,} belongs to Mp for all q' s p 6 q, K ( x ) E V,. Conversely, if K E V,, then the multiplier belongs to M,,, q' < p < q.

,

Further Results and Problems 5.1

Suppose that f

E

L( T) and that F is a primitive off: Then

'I

F ( x ) = p.v. - T

f(x

+ t)llnlsin(t/2)11 dt,

x

a.e. in

T

T

(Hint: Carry out an integration by parts in the integral

-i

j

T

F(x+ t)- F ( x - t) dt; 2 tan( t/2)

I.,771

this result is Lusin's.) 5.2 ForfE L(T),I{lfl > A}I = o(l/A)_ash + 0O.(Hint: Writef=fl + f 2 , where llfllll < E and f 2 E L2; now ({lf21 > A/2}1 s ~ / ( h / 2=) o(l/A) ~ and for fi use the weak-type ( 1 , l ) estimate.) 5.3 Assume that f E L ( T ) and 0 < p < 1. Then Is,(f, x)lp dx 5 cp\lf\\i. What is the order of cp as p + 1-? (Hint: Take a look at identity (3.5) in Chapter 111.) 5.4 Assume that f~ L ( T ) and O < p < 1. Then Ils,(f) -flip + 0 as n + m. 5.5 Assume that f,f E L( T) and that IIsn(f)lll s A, all n. Then there is a constant B, independent of n, such that )IC,,(f)11 s B. (Hint: Observe that and by ernX ) = (-si(.t x ) ) / ( n + I), G n U X ) = unt..f ; n ( L X) stein's inequality in Chapter 11, IIs'n(f)lll== cn.) 5.6 Assume thatf; f E L( T) and that in addition 11 s, (f3 - f l + 0 as n + 00. Show that ,;I (f)+ 0 as n + 00. (Hint: First choose N so large that IIs,(f) - s N ( f ) ) I< 1 E for n > N, by Bernstein's inequality

s,

fal

Ib'ncn - sIN(f)ll1 6 ten;

next from the estimate Ign(J; x) - &"(f;x)I s Is',(f;x) x ) l / ( n + 1) + JsIN(f;x)l/(_n+ 1) obtain that l i m n + m ~ ~ ~ n (&,(f)lll f) 0. Finally, recall that since f E L, &,,(A x) = u,(J x) a.e. and Ilun(f)-fill + 0 as n + 00.)

V. The Hilbert Transform and Multipliers

134

5,

Assume thatf E L( T ) ,0 < p < 1. Then Is’(f,x)Ipdx is the order of cp as p + 1-? 5.8 Assume that f E L( T ) and 0 < p < 1. Show that

5.7

cpllfll!. what

1im,,+J~~,,(fl- H.11, = 0. (Hint: For trigonometric polynomials g and large enough n

I&(f,

I&(f-g, x)lP+ I&(x)

X ) - Hf(x)IP

- Hf(x)l”)*

Suppose f~ L ( T ) . Then there is a sequence nk + 00 so that simultaneously s,,(f, x ) + f ( x ) a.e. and &(f, x ) + H f ( x ) a.e. (Hint: Use 5 . 4 and 5.8.) 5.10 (Parseval’s relation) Assume f E Lp(T ) , g E Lp’(T ) , 1 < p , p’ < 00, l/p l/p’ = 1. Then

5.9

+

(Hint: Since Ils,,(f)- f l ,

+0

as n + 00,

Compare with 5.18 in Chapter 11.) Does the conclusion in 5.10 above hold iff E L( T ) and g E La(T ) ? Does it hold iff E L( T ) and g is of bounded variation? 5.12 The trigonometric series 1 c,,einxis the Fourier series of an Lp(T ) function, 1 < p < 00, if and only if for every g E Lp’(T ) , 1 c,c-,(g) converges. (Hint: We must only show that the condition is sufficient. If a , ( x ) denote the ( C , 1) means of 1 c,,eiR*and T , ( g ) = T,, the (C, 1) means of C c,c-,,(g), then T, = ( 1 / 2 7 ~ ) a , , ( x ) g ( x )dx. But T,, is a linear functional in Lp’of norm Ila, lip; moreover, since the series converges, it is all the more (C, 1) summable and the values T, are bounded for each g E Lp’, i.e., IT,^ cg < a.Therefore by the Banach-Steinhaus theorem 2.3 in Chapter I, the norms of the functionals are bounded, i.e., Ila,,llp5 c < 0O.) 5.13 Suppose we know that for some p, 1 < p < 2, f + f is of type (p, p) when restricted to characteristic functions of measurable sets (operators with this property are said to be of restricted type (p, p)). Then show that f + f is of weak-type (pryp’), l/p + l / p ’ = 1, and consequently also of type (r, r ) , p < r < p’. What conclusion follows from a restricted weak-type ( p , p ) assumption? (Hint: For a trigonometric polynomial f let A,+= {f > A}, BA = { f < -A}. Then 5.11

I,

AIA,I < j A A j ( x dx ) = - j , f ( . ) ~ , ( x ) dx =Z CII flip'

repeat the argument for BA.)

11x4

Ilp

=

c l l f l l p ‘ lAAll’p;

135

5. Notes; Further Results and Problems

-

There is a bounded function f(x) (sin nx)/n = ( T - x)/2 whose conjugate f(x) (cos nx)/n = -lnlsin(x/2)1 is unbounded. 5.15 There is a continuous function f(x) = C:+(sin nx)/(n In n) whose conjugate f(x) c:=,(cos nx)/(n In n) has a Fourier series which converges everywhere except at 0. (Hint: To show the continuity invoke 7.24 in Chapter 111: actually, the partial sums off diverge to 00 at 0, which means that the series also diverges (C, 1) to co there.) 5.16 Assume J; f E L“( T) and that in addition ~ ~ s n ( f ) ~S~ m A, all n. Then s B. there is a constant B such that also Ils’,,(f)llDO 5.17 Assume now that J; f~ C ( T ) and limn-rwllsn(f)- f l l w = 0; then limn+wll.Sn(A -f1lw = 0. 5.18 Assume f E C ( T ) , 1lfllrn s 1. Then there are constants cl, c, independent of n so that I{Isn(f)l > A}l s qe-4”. 5.19 Suppose f E L In L and show that IIs, , 5.14

- c:=’=,

-

and

5.20 Parseval’s relation holds for f _E L In L( T) and g E Lm(T). 5.21 Assume f E C(T). Then eAlf(x)l dx < 00 for each A > 0. Show that as p + m. (Hint: Write f = f - g g, where g is a IIfII, = o ( p )

I,

+

trigonometric polynomial with Il f - gllm sufficiently small.) 5.22 There is an absolute constant A. > 0 such that if l l f l l w G 1, then for 0 < A S Ao, I, eAISn(f.X)I &9 T eA15m(f.x)I h < c, where c depends only on A. Iff E C ( T), the estimate holds for each A > 0. (Hint: Consider, as usual, the modified partial sums s’f(f; x) off given by

and observe that rewriting sin n t = sin n( t + x) cos nx - cos n( t + x) sin nx it follows that s’f(J; x) = &(x) sin nx - &,(x) cos nx, where gn(x)and h,(x) aref(x) cos nx andf(x) sin nx, respectively, Ig,,(x)l, Ih.(x)l 4 1. Therefore,

5.23

Assume f is as in 5.22. Then there is an absolute constant A. > 0, so

V . The Hilbert Transform and Multipliers

136

l / p + l / p ' = 1 . If A is sufficiently small, and p < 00 sufficiently large so that Ap' is still small, the integral above is bounded and the term IIsn(f) - f l i p = o(l).) 5.24 If elf(x)'"dx < 00, then eAli(x)Isdx < 00 provided /3 = a / ( a + 1) and A is sufficiently small (Hint: A way to do this is by dualizing the result If(x)l(ln+lf ( x ) l ) a dx < 00 implies If(x)l(ln(2 + I ] ( X ) [ ) ) ~ - ' dx < 00, A > 0. Of course, there is always a direct proof.) 5.25 Let A ( u ) be a continuous, nondecreasing function, A ( 0 ) = 0, such that lim sup,,m A ( 2 u ) / A ( u )= co. Then there exists a function f so that A(lf(x)l) dx < 00, yet A ( l j ( x ) l )dx = 00. (Hint: Construct by induction an increasing sequence of positive numbers uk so that A ( 2 u k )> 2 2 k A ( ~> k )2, k 3 1, and uk 2 2 4 - 1 , k 3 2. Next define positive integers 1 s n, < n2 < * * so that 2-k-' < 2-"kA(uk)s 2-k, k 3 1 , and define the function x E [-2-"k, -2-nk+l ) k = 1,2, ..., x E T\[2-"1,0). =

IT

I,

IT

IT

I,

I,

-

{ :,"'

It is readily seen that

IT A(f ( x ) ) dx <

00.

f may also be written

cT='=, (uk - Uk-I)x[-2-"k,O)(x), and consequently (since for 0 < 7 < 5-/2,

5. Notes; Further Results and Problems

137

which gives 1, A ( l f ( x ) l )dx = +a. A similar, but more involved, argument shows that there is a function h, ” ” lim A(ls,(h, x ) - h(x)l) dx = a; A(lh(x)l)dx < co yet n+m

J

T

J

these constructions are from Oswald [ 19821.) 5.26 Corresponding to a collection ( 4 ) of disjoint subintervals of T we associate the function

this is closely related to the Marcinkiewicz function introduced in (1.16). Show that there is a constant c, independent of A > 0, such that I(6 > h } l < c(C Ihl)e-*. (Hint: Let E = E,, = (6 > A}; if IEl = 0 there is nothing to prove. Otherwise set + ( x ) = x E ( x ) / ( l E lln(4~lIEI))3 0, 11+11, = l/ln(4~/lEI),ll+llm = 1/(IEl W4r/lEl))- Thus

M + ( t ) dt

=

A,

say (by Proposition 2.3 in Chapter IV). One way to estimate A is to observe that it does not exceed c + c 1, + ( x ) In+ +(x) dx which obtains the estimate in the conclusion without the factor 1141. A sharper estimate for A is

which gives A s c(1 + ln((C 14l)/IEl)), and we are done. The first estimate above is, for instance, in Hunt’s work [ 19721. The full result is in Muckenhoupt [1983], where background on the problem is given. Notice that a similar conclusion holds for

138

V . The Hilbert Transform and Multipliers

Assume f~ L"(T), ~ ~ sf 7,~ and ~ m put E = { l f l > 9); then [El s Ilfllm)e-S/cllfll-,where c is an absolute constant independent of f and 7. (Hint: Let F = {If1 > Ilfllm}. Then since for appropriate constants cl, c2 > 0, e" S cl(eu - 1 - u) for u 2 c2, we see that

5.27 c(

Ilflll/

Now if 17 > Ilfllm, E c F, and we are done; this observation is Muckenhoupt's [1983].) 5.28 Assume f E L( T) and for 7 > 0, A > 0, let E = { x E T :lf(x)l > AT, M f ( x ) s 7 ) . Then IEl S ~ ~ ~ f ~ ~ where ~ e - c~is' independent ~ / q , off, A and 7. (Hint: We may assume that 7 > llflll, for otherwise E is empty, and that A 2 2, for otherwise the result follows from the weak-type estimate for .f We may also assume that Mf( P ) s 7 and write the open set { Mf > 417) = u4,where the Ts are disjoint open intervals, and put

and b=f-g. Note that M f ( x ) > 7 for x E U 3 4 , and, consequently, E s E , u E 2 , where El = {lg"l > A7/2} and E2 = { x E T\U34: 16(x)1> h17/2}. To show that lEll 6 cllfll le-A/c/7we use 5.27, and to estimate lE21 we use 5.26; this proof is also Muckenhoupt's and extends a result of Hunt [1972].) 5.29 The most general multiplier is a sequence { A j } with the property that for each f E C"( T ) , f cjeVx, Ajcje""E D', we denote this class (C", D'). Show that the characterization of (C", D') is also as general as possible, namely, { A j } E (C", D') if and only if the sequence { A j } is tempered. 5.30 Another natural question is how to characterize (C, M ) , i.e., to identify those sequences { A j } which transform continuous functions into finite measures. This was done by Gaudry [1966] as follows: let B = { h e C ( T ) : h=C,:,lf;*gj, CIIlf;IImllgjll,
-

5. Notes; Further Results and Problems

139

disjoint supports and the norm of the sum is the sum of the norms; this observation is Brown's [1977].) 5.32 Suppose that 1 s p < 00 and that {A;} is a sequence of Mp multipliers which verify (i) limk+mA; = Aj exists for all j , and (ii) ICj A ; C j ( f l C j ( g ) l s ckIIfIIpIIgIIp,, for all trigonometric polynomials g, l / p + l/p' = 1, and limk+.mc k c L. Show that under these conditions { A j } E Mp and its multiplier norm S L . 5.33 Suppose that the sequence { A j } is such that ( ~ 2 ' L ~ , j , ~ 2 k + ~ I A j i / 2 ) i ' 2 ~ 2 - =c~independent , of k and 0 < a < 4. Then { A j } E Mp for 2/( 1 + 2 a ) < p < 2/(1 - 2 a ) . This result is Hirschman's. 5.34 Suppose F ( z ) is an analytic function in IzI < r and that { A j } E Mp, 1 c p c 00, with multiplier norm A < r. Show that { F ( A j ) }E Mp also. (Hint: By 4.13 [ A j [G A < r; for a trigonometric polynomialf(x) = 1 cje"", we consider

and

Assume that { A j } E Mp, 1 C p c 00, and consider the sequence { A j + N } , where N is a fixed integer. Show that also { A j + N } E Mp, with multiplier norm equal to that of { A j } , independently of N. (Hint: If T denotes the multiplier corresponding to { A j } , andf(x) = 1 cjevxis a trigonometric polynomial, then 1 Aj+fijevx= eciNXT(eiNrf)(x).) 5.36 Let { A j } be the sequence of 1's for N , C j C N 2 , and 0's otherwise, where N,,N2 are arbitrary integers. Show that { A j } E Mp, 1 < p < 00, with multiplier norm independent of N l and N 2 ,and that { A j } g MI.(Hint: Use 5.3J and appropriate results on projections.) 5.37 We say that a function 4 defined on R is in the class V, if C,"_-, I + ( j ) - + ( j + 111 = A < a.Show that if 4 E V,, then { 4 ( j ) }E Mp, 1 < p < 00, with norm sl+(O)l + A. (Hint: For trigonometric polynomials f, g put $ ( j ) = c,(flc.(g), where N = max(degreef, degree g ) and $ ( - N - 1) = 0. Then 5.35

-N

-N

V. The Hilbert Transform and Multipliers

140 N-1

=

'2( + ( j ) - + ( j

- l ) ) + ( j )+

+(w+I(N

-N

and consequently N-1

14 s

c

- + ( j - 111

Iml+ I + ( N - +(O)l I+(".

-N

The conclusion follows readily from this, since by 5.36 l+(j)l G c(If Ilpllgllp,, c independent of j . This result is due to Steckin.) 5.38 (Rudin-Shapiro polynomials) We define the trigonometric polynomials p,, Qm inductively as follows: Po = Qo = 1 and

Prn+l(x)= pm(x) + e i Z m X ~ r n ( x ) , Qm+l(x) = Prn(x) - ei2mxQm(x)*

+

Thus, P l ( x ) = 1 + ek, P 2 ( x )= 1 + e k ei2x- ei3x,and so on. These polynomials have the following properties: (i) Ipm+1(x)12+ IQm+l(x)12 = 2(Ipm(x)12+ I Q m ( X ) 1 2 ) , quently, I P , ( X ) ~ ~ + I Q , , , ( x ) ~ ~ = Y+';I I P , I ~ ~ s 2' m+1)/2. ,

and,

(ii) for In1 < 2", cfi(Pm+l) = c,(P,); hence there is a sequence consisting of *l such that P,(x) = C',"r,' Eneinx.

conse{E,}:=~

For 1 s p d 2 the norm 11 P, 11 M p of the Rudin-Shapiro polynomial P, as a (convolution) mapping in Lp(T) is of order 2m(1/p--1/2); more precisely, there are constants cl, c2 independent of m so that

5.39

m(l/p-1/2)

C12m(l/~--1/2)IIpmIIMp s ~ 2 2

(Hint: IIPmIIM, G llprnllco s 2' m+1)/2 , 11 P, 1I M 2 G 1; by the Marcinkiewicz interpolation theorem 7.14 in Chapter IV, for l/p = 1 - 7 ) + 7 / 2 , 0 < 7) < 1, 11 P, IJMp s 2"'(1-7)'2.The converse inequality follows by combining the estimates \\P,\\~,3 l l ~ ~ 1 c2'"/', 1~ IIpmIIp'

=

IIpm

* 2D2m-lIIp's

IIpmIIMp,I12D2m-,IIp,

cIIPmIIMp2m/p'.)

5.40 For 1 < p < q < 2, M p 5 Mq. (Hint: The following proof uses some functional analysis, an argument which does not is given in 5.41 below. Suppose to the contrary that Mq = Mp. Then the injection i : Mq + M p has a closed graph; indeed, if a sequence { T,}of multiplier operators converges to 0 in Mq and to T in M p , then, for all trigonometric polynomials f we have Tnf + 0 in measure and Tf.+ Tf also in measure; thus Tf = 0 for all such S, and T = 0. This proves that i has a closed graph and is continuous. 3 ~ 2 " ' ( ' / ~ -+ ~ /00~ )with m.) On the other hand, IIPmllMp/llPmIIMq

5. Notes; Further Results and Problems

141

5.41 It is often more convenient to work with Rudin-Shapiro measures, for it is then simpler to patch multipliers without making use of more sophisticated techniques such as the Littlewood-Paley theory we will discuss later on. The construction, carried out by Kahane [1970] goes as follows. We proceed again by induction; first put po = vo = Dirac delta centered at the origin, and having chosen measures p,, v, supported in a finite set S, choose x, E T so that x, + S, are disjoint, and let p m + ,= p, + vm( * + xmIy v,+, = p, - v,( * x,). It is easy to verify inductively that there are sequences { y k }c T and fixed *l valued sequences { r k } and {sk} so that p, = C',"=,rk8(yk),v, = skS(yk).An important consideration is the norm of these measures as convolution mappings in Lp(T). A first step is to estimate I I P m I I M , and IIpu,IIM2; clearly, I I p m I I M , c C Z ~ = ~ I I ~ ~ ~ (cY 2". ~ ) IoI n Mthe , other hand, by the inductive definitions and the parallelogram law, the following relation holds: IIp, * fll< + 11 v, * fll< = 2,+' llfll ;, and IIp, 11 Mz G 2(m+1)/2. By interpolation we see that IIp,IIMp G c2'"lP', and since the opposite inequality also holds, by 5.39 above, then lip,11 M p 2"'lp'. It is now possible to construct an explicit example of a convolution operator in Mq\Mp, 1 < p < q < 2. Choose disjointly supported translations pm of p,. It is readily seen that the multiplier norm properties of the pm's are identical to those of the p,'s. Finally, consider T = C:=, 2-"lPp,, this series converges in Mq, and if TN = 2-"lpp, denotes its sequence of partial sums, then and for 2 > 4 > PY IITvIIMq2 ( C , =NO 2 - - m q / p 2n ) 1 / q II TllMq2 2m('--q")''q. Now, if T were to belong to Mp, then by Theorem 7.14 of Chapter IVYthe norm 1) TIIMqmust remain bounded as 4 + p + , and clearly the above estimates show that this is not so. Actually, estimates in the same spirit give an interesting result of Zafran [ 19751: for each p, 1 s p < 2 there is a multiplier which is of weak-type (p, p) but not of type (p, p); for p = 1 the multiplier is the Hilbert transform, but for the other values of p things are a bit more complicated. This construction is due to Cowling-Fournier [1976], and the reader is encouraged to consult their work to grasp the complete scope of their methods, which extend these and other results to locally compact groups.) 5.42 Show that if the multiplier associated with { A j } is of weak-type (p, p), 1 s p < 00, then !Ails c < 00. How does c depend on the weak-norm of the multiplier? (Hint: (IT(einx)l > r ] } is empty for r ] sufficiently large.)

+ I',"=,

-

I:=,

(I:=,

CHAPTER

Paley's Theorem and Fractional Integration

1. PALEY'S THEOREM The time has come to consider some basic questions concerning Fourier coefficients of Lp functions; for instance, we are interested in identifying among the sequences { c j } with limljl+mcj = 0 those which correspond to Fourier coefficients of Lp functions and to make a more precise statement about their summability properties. The two results we have encountered so far are Icj(fil

s

IIfIIl

(1.1)

and

(1.1) states that the operationf + { c j ( f i }is of type (1, a), i.e., it maps L ( T ) into I " ( 2 ) ; (1.2) states that it is of type (2,2). Since the interpolation results in Chapter IV do not cover this case, we first attempt to restate the problem at hand as one to which they do apply. Following Zygmund we introduce the weighted sequence spaces l ; ( Z ) = 1; and wk-lr(Z) = wk-1;. 1; consists of those sequences { c j } so that the quantity IIcjll,: = (xjlcjlpp(j ) ) ' l P < 00,0< p < co (with the obvious interpretation when p = a),and wk-1; consists of those sequences { c j } for which there is a constant c > 0, so that, if OA = { jE 2:lcjl > A}, then AP p ( j ) s c, all A. An important example is p p (j) = (1 + 1j l ) - p ;note that I" G l:, whenever p > 1. With this notation (1.1) and (1.2) can be restated as follows: if to each f E L + L2,f 1 cjeiix, we assign the sequence {cjp2(j)-''2}, then this mapping is bounded from L2 to lt2and from L into wk-l,,.

xjEoA

-

142

1. Paley’s Theorem

143

Only the weak-type statement requires proof; this is easy since 0,= { j E 2 : lcjlpz(j)-l’z> A} G { j E 2: ~ ~ ( j <) llflll/A} ” ~ and consequently

This observation readily leads to Theorem 1.1 (Hardy-Littlewood). Assume f~ Lp(T), f p < 2. Then there is a constant c independent o f f such that IIcjII 1

~

=~ IICjpAj)-’’’II ~

14

cjevx, 1 <

cIIfIIp,

more precisely, (CIcjIp(l + IjI)p-z)l’p c cIIfIIp

with c

=

(1.3)

O ( l / ( p - 1)) as p + 1+.

Proof. Just invoke the abstract version of the Marcinkiewicz interpolation Theorem 4.1 in Chapter IV. There is a statement symmetric to Theorem 1.1, namely, Theorem 1.2. Assume { c j } E lz2-q, 2 < q < 00. Then there is a function = and llfllqC c ~ ~ c j ~ with ~ ~ ~ c2 = - qO ( q ) as q + CO, c independent off: In other words

f~ L q ( T ) such that cj

IIfII,

cj(n

c(CIcjIq(l + ljI)q-z)l/q.

Proof. We dualize Theorem 1.1. Put

and note that for trigonometric polynomials g,

=A,

say.

(1.4)

Vl. Paley’s Theorem and Fractional Integration

144

By Holder’s inequality with indices q and its conjugate p, 1 < p < 2, we get that IAl does not exceed

s cllgllp llcjlll~2-q~

where c is the constant of Theorem 1.1. The converse to Holder’s inequality now gives that

II g n 114 s c II cj II 1z2--q and the conclusion follows by FejCr’s Theorem 3.3 in Chapter 111. To obtain the sharp version of the Hardy-Littlewood theorem we need the notion of rearranged sequence. Suppose { c j } is such that limljl+mcj = 0; then choose among the cj’s with j 3 0 one with largest lcjl and rename it c:. At the same time rename the corresponding character evx, C$~(X). If there is more than one index with largest Icjl, just choose any; observe however that since limj+, cj 5 0 there can only be finitely many. Remove cg and repeat now the process for the remaining cj’s, and so on. This procedure obtains a new sequence with Icgl 3 l c T l 3 * + 0 as well as an ONS {4j}zo. Similarly, we obtain {cT},T& so that lcT,l 2 lcT2l 2 . * * + 0, and also a corresponding ONS {C$j},Ti-m. Note that {4j}7=-m is an ONS consisting of bounded functions, and since these were the only properties we actually used in proving Theorem 1.1 we also have

--

Theorem 1.3 (Paley). Assume f

E

Lp(T ) ,f

-1

cjevx,1

(CICTI”(1 + lA)p-z)l’p fs c l l f ,

< p s 2. Then (1.5)

with c independent off and O ( l / ( p - 1)) as p + 1’

Proof. Fix f E Lp(T), note that 1 cjevx = C c ? ~ ~ ( and x ) apply the version of the Hardy-Littlewood theorem corresponding to {C$j}. Note that (1.5) is sharper than (1.3) in the sense that the expression on the left-hand side is largest among all possible rearrangements of the given cj’s. Next, to pass to unweighted sequences, we have the following.

+

Proposition 1.4. For l / p l / q = 1, 1 < p s 2, lL2-p G P. More precisely, there is a constant c > 0 independent of the sequences involved, so that

I I c j I I ~ ‘C~I I C j I I 1 ~ Z - - P *

1. Paley's Theorem

145

+

+

Proof. Since llcjll 1 = ~,:~lcTlq ~ ~ ~ - , l=cI ~ IJ,~say, we may assume is nonnegative and nonincreasing and { c ~ } , is ~ ~nonnegative - ~ that { cj}po and nondecreasing. To estimate I observe that m

m

m

cllcj II ?Eppllcj

=c

II cj II k-;

Clearly, an identical estimate holds for J and we have finished. This observation gives at once Theorem 1.5 (Hausdorif-Young). C cjeux,and l / p + l / q = 1. Then

Suppose f E Lp(T ) , 1 < p < 2, f

(CIC~I~)~/~ s CII~II~

-

(1.7)

with c independent of J: Moreover, if {cj} E lp, then there is a function f E L q ( T )so that cj = cj(f) and

llfllq 9 cllcjll~p

(1.8)

with c independent of {cj}. Proof. Suppose f E Lp,then by (1.1) lim(jl+mcj = 0 and by Paley's theorem 1.3, 11 c? 11 ,4-ps c Ilfll,. Moreover, by Proposition 1.4, 11 cj 11 p S c 11 11 [ L ~and ~. (1.7) obtains. (1.8) follows now by a duality argument in the spint of Theorem 1.2.

CT

The Hausdorff-Young theorem states that in the case of Fourier sequences, estimates (1.1) and (1.2) suffice to obtain an L P ( T ) - P ( Z )result as well, l/p + l / q = 1, 1 < p < 2. A similar conclusion holds for arbitrary sublinear operations T simultaneously of types (1, CO) and ( 2 , 2 ) ; we prove a rather general result in this direction since it serves to illustrate the role of Young's functions in interpolation. First we present some definitions.

146

VI. Paley's Theorem and Fractional Integration

For a function A( s) defined in s 3 0 we say that A( s)/s increases provided that (i) (ii) (iii) (iv)

0 s A(s) s 00, A(0) = 0, A(s)/s increases (in the wide sense), A is left-continuous, and A is nontrivial, i.e., 0 A(s) co for s > 0.

+

+

If A( s)/ s increases and we set

then A. is convex, increasing, 0 at 0, and nontrivial (in fact A, is positive and finite in the same set as A); A. was introduced by Jodeit as the regularization of A. Moreover, A,(s) s A(s) s A0(2s). Functions which have the properties of A. are called Young's functions; cf. 5.2 in Chapter I. It is often convenient to use A rather than A, because, for example, min(A(s), B ( s ) ) satisfies (i)-(iv) but is not in general convex when A and B are convex. Let A(s)/s increase. The Young's complement of A is given by

a

A(u)= SUP(US- A(s))

(1.10)

sao

and the inequality su

s A(s) + A(u)

(1.11)

is called Young's inequality. The inverse of A is defined on [0,0O] by A-'(u) = inf{s: A(s) > u } ( i n f 0 = +a). (1.12) A-' is positive and finite for u > 0, A-'(oo) = 00, A-' is nondecreasing and right-continuous, and A-'( u ) / u decreases. Moreover, A(A-'(u)) s u s A-'(A(u)), u 3 o (1.13) and A(s) = sup{u: A-'(u) < s}, sup 0 = 0. If p and q are conjugate exponents, then the relation prototype of s s A-'(s)A-'(s) d 2s.

S ' / ~ S ' / ~=

s is the

(1.14)

For our purpose there is a special operation to mention, namely,

mR(s) = sup(l/A(l/u)),

s = 0, s > 0,

(1.15)

s<:{

where 1/0 = 00 and 1/00 = 0. Then RAR is left-continuous and nontrivial, and R A R ( s ) / s increases. The effect of this operation is to reverse the

1. Paley 's Theorem

147

behavior of A at 0 with that at

a),

and vice versa. Since

(RAR)-'(u ) = l/A-'( l / u ) ,

(1.16)

-

(1.17)

it is readily seen that

RAR(s) RAR(s)

in the sense that there are constants cl, c2 independent of s so that R A R ( c , s ) s RA-R(~)s R A R ( c 2 s ) .Let now (1.18)

where (i) a is nonnegative, continuous, and strictly increasing from 0 to a) with u and (ii) a ( u ) / u is nonincreasing. If we denote by n ( u ) the inverse function to d ( u ) and put (1.19)

then the equivalence RA-R(sI4)

d

B ( s ) 4 RA-R(s)

( 1.20)

holds. p ) and (Y, 0 ) be arbitrary We are now ready to interpolate. Let (X, measure spaces and assume that T is a sublinear mapping defined in L ( X ) + L 2 ( X )with values in the v-measurable functions on Y.The question we address is the following: if T is a sublinear operation simultaneously ) (2,2) with norm d 1, i.e., if of types ( 1 , ~ and (1.21)

II TfllLt d IlflL:

( 1.22)

and the function A(s) given by (1.18) above is such that A(s)/s increases and A( s)/s2 decreases, what integrability condition obtains for Tf provided that fx A(lf(x)l) dp(x) d 1, say? In this direction we have

Theorem 1.6. Under the circumstances described above and if A(s)/s with s and A(s)/s

+ a) also

with s, then

-*

0

VI. Paley’s Theorem and Fractional Integration

148

provided that B is defined by (1.19) and the right-hand side of (1.23) does not exceed 1. Proof. The proof is clearest when we treat the function B as an unknown and discover its exact form by making use of (1.18), (1.21) and (1.22) to insure that (1.23) actually holds. Let then B’(u ) = b( u ) 5 0, suppose that ~ ( u=) b ( u ) / u is nonincreasing, and put 7 = ~ ( 0 + ) By . Fubini’s theorem and the sublinearity of T we get

I

> 2s))b(s) ds =

.({ITfl

I

s.({ITfl > 2s))

C0,W)

I

M U )

LOP)

+T I

sv({lTfl> 2s))

[O,m)

sv({lTfl> 2s)) dsdT(u) + K = IL0.m)

l,w)

=J+K,

(1.24)

say. We bound J first. Observe that with the notation fi(x) = min(z, If(x)l) sgnf(x), f’(x) = f(x) -fi(x), it readily follows that J does not exceed

I,.,

,,,,,I

+

sv({lTf’l

I I

> sl) ds d d u )

s.({ITfil>

s)) dsdrl(u) = J1

+ J2,

COP)

say, where z is to be regarded as a monotone function of u, yet to be chosen. To treat J1, note that by (1.21), IITSIImsIlflll, so that v((lTf1 > s)) = 0 if s 2 z and llflll = J[z,w) p ( { l n > s}) ds < z. Since

I,

AClfCX)l> d p ( x ) =

it readily follows that relation

I

a(s)p({lfl > s)) ds

llf’lll

s l/a(z). Thus if we define now l/a(z(u)) = u

it follows that J , = 0. Moreover, since

1,

COP)

z ( u ) by the

(1.25)

149

1. Paley’s Theorem

we also have 5 2 sz

j j

In case

r)

= 0,

dr)(u)P ( { l A > sl) ds.

s

C0.m)

[o,z-’(s))

there is n o term K , and d r ) ( u ) = s , , ~ ( z - ’ ( s ) )s a ( s )

if r ) ( s )s l / s d ( l/s), that is, if

b ( s )S l/d(l/s).

(1.26)

Hence in this case (1.23) holds since

I,

B(lh(Y)l)M Y ) zz I[o,oo) v({lhl

’s})/d(l/s) ds.

Now suppose we define b ( s ) as in (1.19), i.e., with equality in place of inequality in (1.26). Then r ) = lims+ob ( s ) / s = lim,,m ,,~(s)/s = inf, r ) ( s ) / s . In case r ) > 0, we have A ( s ) 3 $ys2 so that f E L 2 ( X )and by (1.22)

On the other hand, returning to the estimate for the J’s, when r) > 0 we dr)(v) = s ( r ) ( u - ’ ( s ) )- 7); also the negative term and note that s the finiteness of K allow some cancellation. That is,

I[o,u-~(s)) J

+K

sJ

+ 4K = J

a ( s ) p ( { l f l > s}) ds, [OS)

(1.23) holds, and the proof is complete.

H

In addition to the Hausdorff -Young theorem, which follows with the choice A ( s ) = sp, 1 < p < 2, Theorem 1.6 gives the following two results of Hardy and Littlewood. Example 1.7. Let A ( s ) = s ln(1 + s) so that E L In L( T). Then A(s) s( es - 1) and

-

f

RA-R (s)

- (el’s

S

- 1)

Theorem 1.6 then gives that, iff

e-’/”

- (s2

I, A ( l f ( x ) l )dx <

00

for small values, for large values.

- 1 cjegx and

r)

> 0, then

means that

VI. Paley's Theorem and Fractional Integration

150 Let now A ( s ) = s/ln(l RA-R

+ s);

since RAR is A in the preceeding example,

- s ( e s - 1) -

es

for small values, for large values,

and in this case we get that, if {cj} is a sequence such that

then there is a function f E L so that cj = c j ( f ) and each k > 0.

1,

ek'f(X)I dx < 00 for

2. FRACTIONAL INTEGRATION

-

We have seen that iff E L, f C qei'", co = 0, then the Fourier series of its indefinite integral is Co l j + o ( l / (ij))cjevx.It is therefore natural to consider also the "fractional" analogue of the above formula, namely,

+

1 1-cje'ix,

o < < 1. (r

Ijl"

j+o

We see at once that (2.1) is the Fourier series of a function with the same integrability properties as f, since by Proposition 4.4 in Chapter I11 there is a summable function g with cj(g) = ljl-", j # 0. But we expect a better behavior from the series in (2.1), and, to get an idea of what this might be, we investigate whether we actually know something else about g. Consider the integrable function h ( x ) = IXI-~, 0 < 77 < 1. We note that its Fourier coefficients are given by the even sequence

'I

Cj(h)= -

?r

[O,r)

-cosjxdx xl)

and since the integral in (2.2) has a limit L # 0 as Ijl+ 00, we see that essentially cj(h) L./lj\l-q for ljl large. Thus g ( x ) lxlU-' near the origin, and in this case g E L'( T ) for r < 1/(1 - a).Therefore by Young's convolution theorem (cf. 4.30 below), (2.1) is the Fourier series of an Lqfunction, with l / q > l/p - a,whenever f E Lp(T) and 1 d p < l/a; this still is not sharp. The precise statement is the following: for 0 < a < 1 consider the Riesz fractional integral I, given by

-

-

I J - ( X ) = I,-dt,Itl1-a

x E T.

(2.3)

151

2. Fractional Integration We then have

Theorem 2.1 (Hardy-Littlewood, Sobolev). Assume that f E Lp(T), 1 < p < 1/a. Then Iuf E Lq(T ) , l/q = l/p - a and there is a constant c = c( a,p , q ) independent off such that

IIIafllq

(2.4)

CllfIlP.

Proof. Since IZuf(x)l d Z u ( l ~ ) ( xwe ) , may suppose t h a t f a 0. For 0 < 77 < T write

say. To estimate J note that + ( t ) = ~ f ~ a - l , y - l , lis~ (integrable, f) and, if + , ( t ) = ( l / T ) + ( t / T ) , then J = 77a j , f ( x - t ) + , ( f ) dt. Thus, by (a simple variant of) Proposition 2.3 in Chapter IV, J

C177uMf(X),

(2.6)

where c1 = 2/a is independent off: To bound K , note that since (1 - a ) p ' > 1, Holder's inequality implies that

where c2 = (2q/p')'/". Therefore, adding (2.6) and (2.7), we see that

Iuf(x)

C(rl"Mf(X)+ 77'-1~pllfllp~.

(2.8)

Now we seek to choose 7 to minimize the right-hand side of (2.8), a good choice being that 77 which makes both summands equal, i.e., ' 77 = ( I l f l l p / ~ f ( x ) ) " .

(2.9)

Note that this choice of 77 is all right as long as it does not exceed T ; however, if it does, it is because M f ( x ) c ~ - ' / ~ J ( f l l ~and for these x's, by setting q = T in (2.8), we see that I.uf(x) c cllfll,.

(2.10)

On the other hand, (2.9) gives that Iaf( x ) c c I1fll

X)

(2.11)

VI. Paley's Theorem and Fractional Integration

152

Since we prefer not to have to determine which estimate applies, we just add (2.10) and (2.11) and obtain that for all x in T I A X )

s ~ c l l f I l ~ " M f ~ x > '+ - "llfll,).

(2.12)

This is all we need to complete the proof. Indeed, since (1 - a p ) q = p , by (2.2) it readily follows that

IIIafIIq

~(IIfII~pIl"Mfll~-" + Ilfllp),

and since by the maximal theorem IIMfll, complete.

S

(2.13)

cllfll, for 1 < p , the proof is

Theorem 2.1 is best possible in the sense that an estimate of the form IIZ,Jllr s cllfllp implies r d q. This is readily seen as follows: for f(t) = ~ ( - , , ~ ) and ( t ) 1x1 d q/2 we note that

whence

and this can only hold as q + O+ provided that ( l / r ) + (Y B l/p. We must still consider p = 1 and p = l/a. The estimate (2.11) suffices to handle the case p = 1 quite easily, namely, Theorem 2.2. Supposef E L( T) and l/q = 1 - a ;then Imf E wk-L4(T) and there is a constant c, independent off; so that for A > 0

I{Lf>All Furthermore, iff B so that

E

(2.14)

C(llflll/A)q.

L In L( T), then IJ E Lq(T) and there are constants A,

IIZfll,

A+

j

lf(t)lln+lf(t)l dt.

(2.15)

T

Prmf. To prove (2.14) we may assume that A > 2~llf11~, where c is the constant in (2.12) above, for otherwise the conclusion is obvious. Then {IJ > A} E { ~ l l f l l : M f ' -> ~ A} and by the maximal theorem

I{Lf> A l l =S cl{M > ( ~ / ~ k f ~ ~ ? ) ~ ' ( ~ - ~ ) } ~ d

cllfll

llfll

Y/('-a)A

= c( llfll I l A ) q *

2. Fractional Integration

153

Also by (2.12), Theorem 5.3 in Chapter IV, and Young's inequality uOul-O's au (1 - a)u (cf. 5.2 in Chapter I), we get IIZOfllq s C ~ ~ f ~ +~ CllfIll ~ ~s cllfll~ ~ ~ +f cllMflll ~ ~ and ~ -we mare done.

+

Whereas the result concerning the weak-type is sharp, namely, there is an integrable function f so that I O f &Lq, l / q = 1 - a ( f ( t )= l / t ln(l/t)l+'/q for 0 < t < f = 0 otherwise will do) the L In L + Lq result is not, cf. 4.28. We consider next whether there is a maximal operator which in some sense controls the Riesz potentials as the Hardy-Littlewood maximal operator controls the Hilbert transform. First note that for f 3 0 and ?T>6>0

therefore a natural candidate for the task at hand is (2.16)

with 71. = a.The continuity properties of M , are readily established. Indeed, assume that 1 d p s 1/77 and j,lf( t ) p dt = 1. By Holder's inequality we see that

s

IZIn-lip(

[ If( I

t)lPdt)'/q

for any 0 < E < 1 since the integral s l . By choosing l / p - 7, we get at once from (2.17) that

E

=

1 - vp, or

(2.17) E/P

=

M,f(x) s M(Jfp)(x)'/"-"

(2.18)

I{M,,f > h}l

(2.19)

and consequently d

~/h'/('/~-"),

i.e., M, is a sublinear operation of weak-type ( p , q ) with l / q = l / p - 77, 1 d p d 1/77. This result can be improved to give the type ( p , q ) estimate for 1 < p s 1/ v ( p = 1/77 we already have) and this requires, of course, an interpolation result. Theorem 2.3. Let 0 < po s qo s co,0 < p1 s q1 s m, po < p l , qo # ql. If the sublinear operator T defined in Lpo(X)+ L p l ( X )is simultaneously of weaktype (PO, 40) with norm s c o and of type ( p l , q l ) with norm s cl, then T is

154

VI. Paley's Theorem and Fractional Integration

Of type (p7 9)s where l / p = ( l - V ) / p O + 7/p17 l / q = - 7 ) / 9 0 + 7/91 with norm S C C ~ - ~0C <:7, < 1; here c is a constant independent offwhich remains bounded as 7 + 1- but tends to infinity as 7 + O+.

P m f . The proof follows along what are well-known lines by now. With the notation of Theorem 1.6 we putf = fi + f and choose z as the monotone function given by z-l(s) = s P ( l / P ~ - l / P , ) / q ( l / q ~ - l / q , ) .

The same ideas as in Theorem 1.6 work here.

As for the relation between I, and M, it is given by Theorem 2.4 (Welland). Suppose that f E L( T ) , 0 < E < a < a Then there is a constant c independent off so that

II d ( x )I s c( Ma -€Ax)K +E f ( X ) ) ] I 2 .

+

E

< 1.

(2.20)

Proof. Suppose a function 9 defined in R verifies I+(u)l s Jl(u), where Jl is an even, nonincreasing function in [0, a),~,(Jl(u)/1u1') du < 00, 0 < y < 1. Then an argument identical to that of Proposition 2.3 in Chapter IV gives that

This observation implies, in the notation of Theorem 2.1, that J s cqEMaJ(x). Also, a simple computation shows that K s c7-"M,+,f(x) and, consequently, II,f(x)l

C(rlEMII--Ef(X)

+ s-EM,+€f(x)).

(2.21)

Minimizing (2.21) gives that the optimal choice of 7 = (M,+J(x)/ M,-,f( x))"~';furthermore, note that unlike Theorem 2.1 this choice always ) . is now clear that (2.20) holds and works since MGf(x) G ( 2 ~ ) ' M ~ - , , f ( xIt the proof is complete. The estimate (2.20) gives Theorem 2.1, but the proof is conceptuallly different since it relies on a "nondiagonal" interpolation result. We conclude this section by discussing briefly a result that arises naturally in connection with the restriction of Riesz and other potentials defined in Euclidean n-space to lower-dimensional manifolds. More precisely, we consider integral operators of the form (2.22)

2. Fractional Integration

155

where k is a positive kernel, such as ( x - y(”-’ when n = 1,and are interested , Y)) continuity properties of T. We impose the following in the ( L p ( x ) L4( restrictions on k, namely,

lkll = sup sup A r p ( { xE X: k(x,y ) > A}) < co

(2.23)

y c Y A>O

and

lk12 = sup sup h S v ( { yE Y: k(x, y ) > A}) < 00.

(2.24)

X S X A>O

We then have

Theorem 2.5 (Adams). Assume Y is a regular Bore1 measure on Y so that u ( K ) < 00 for each compact K . If k satisfies (2.23) and (2.24) above, 1 s, r < co (not both s and r simultaneously l), p < q and s / p ’ + r / q = 1, then there is a constant c independent off such that A q v ( { F > A}) s

~ f ( ~ ) ~ ~ d p ( ~ )(2.25) ) ~ ’ ~ . X

Consequently, interpolating (2.25) gives that the strong inequality also holds in the same range of p’s and 4’s.

Proof. We may assume f 2 0. Let EA = {y E Y : F ( y ) > A} and let K be an arbitrary compact subset of EA. Then by familiar arguments it follows that

”

c

156

VI. Paley's Theorem and Fractional Integration

This is all we need. Indeed, the above estimates give that the innermost integral in (2.26) does not exceed

s min( v ( K ) ,A-slk12)1/P(A-rl kl, ~ ( K ) ) " ~ ' l l f l l ~ , ~ ,

and, consequently, Av(K)s

llfllL,P,

cv(~)l-(l/S)(~-r/P')~k~~/P'~k~~/S)(~-r/P')

which on account of the relation between p and q gives (2.25) but with K in place of { F > A} there. The desired statement follows now by the regularity of v. The restriction p < q is essential in the above proof, but not for the validity of the result; the reader should consult the work of Kerman and Sawyer [ 19851 in this direction. Nevertheless, returning to the trace results, we introduce the following definitions. A positive Borel measure Y on T is said to be P-dimensional, 0 s /3 < 1, if for each subinterval I G T, v(1) s clIls, with c independent of I. Also, denote by LP, the class of functions f so that there exists 4, I1411p < 00, and f ( x ) = + ( t ) / l x - tl'-" dt. We then have

I,

Proposition 2.6. Assume v is a P-dimensional Borel measure. Then L: G L 4 ( v ) ,provided that P / q = l / p - a. Proof. Note that IxI1-@E wk-L1/(l-a)and apply Theorem 2.5. 3. MULTIPLIERS

Now we turn our attention briefly to the spaces M: consisting of bounded Lp-L4 multipliers; more precisely, we say that a sequence { A j } E M : if and only if for every trigonometric polynomial X

111 Ajcj(neGXIIq cIIfIIp

(3.1) with c independent off; the infimum over the c's in (3.1) is called the norm of the multiplier in M:. As indicated in the case p = q, M : can be identified with the (Banach) space of bounded linear transformations T from Lp into L4 which commute with translations. There are some simple identifications and inclusions, for instance Proposition 3.1. Suppose 1 s p < CO, 1 < q =S 00,l/p + l / p ' = l / q + l / q ' 1. Then there is an isometric linear isomorphism of M: onto M$.

=

3. Multipliers

157

Proof. Assume 1 < p, q < 00 first, and corresponding to T the adjoint T’: L4‘+ Lp’ as the operator verifying

E

M:, we define

for all trigonometric polynomials f, g. It is clear that T’ is bounded from Lq‘ to Lp‘ with the same norm as T, and it is also simple to verify that it commutes with translations. By the reflexivity of the Lebesgue spaces involved we see that T -* T’ is onto and we are finished in this case. The remaining cases require an ad hoc argument which is left for the reader to carry out (cf. 4.34). H As in (4.13) in Chapter V it is readily seen that if { A j } E MP, and T f -

C Ajcj(f)evx,then, SuPIAjl s II TII.

(3.2)

j

In fact since the underlying space is of finite measure the following also holds: if 2 s p < 00 and 1 < q S 2, then the following is true: (i) T is a translation invariant operator from Lp into Lq; (ii) There exists a unique sequence { A j } verifying (3.2) so that Tf 1 Ajcj(f)evxfor each trigonometric polynomial f:

-

This result may be restated as follows: if 2 d pl, p2 < 00 and 1 < q l , q2 d 2, then there is an isometric linear isomorphism of M$ onto M 3 . As for the symmetry we expect to hold in this case as well, we have the following result. Let 1 d p < 00, 1 < q d 00. If 1 d r, s d 00 is such that the point (l/r, l / s ) in the square 0 S a, b s 1 is symmetric with respect to the line a + b = 1 to the point (l/p, l/q), then there exists an isometric linear isomoprhism of M z onto M:. Some inclusions, such as M: E Mi, p C r, q > s are trivial, and also some noninclusion results hold, as follows. Let 1 s p, q, r, s s 00, and suppose that p S q and that min( r, s’) < min( p , 4’). Then M : is not included in M:. For further results in this direction, as well as extensions to more general settings the reader is referred to Larsen’s monograph [1971] and to Cowling and Fournier’s article [ 19761. Examples of multipliers in the (most interesting) case p S q are easy to come by; indeed by Young’s inequality L‘ c M:, l / q 1 = l/p l/r (the inclusion is proper), and a simple argument combining Theorems 1.3 and 1.5 shows that, for p d 2 C q and l / q = l/p - a, any sequence { A j } such that / A j [ = O ( 1 f m )as Ijl+ 00 is in M;. In some particular instances, such as M:, it is possible to characterize these spaces (cf. 4.35).

+

+

158

VI. Paley’s Theorem and Fractional Integration

Finally, we point out that M: is also a dual space; in fact, let A: = { h E L‘(T): h = CT=’=,h * gk, Ilfkllpllgkllq-< 00); A: is endowed with the natural norm. We then have

Theorem 3.2 (Figa-Talamanca-Gaudry). For 1 d p, q < co,the space M: is isometrically isomorphic to (A:)*, the dual of A:. 4. NOTES; FURTHER RESULTS AND PROBLEMS

In 1912, W. H. Young extended Parseval’s theorem to the relation

between a function f and its sequence of Fourier coefficients {cj}; the exponents p and q had to verify the relation l/p + l/q = 1 with q an even integer. His proof went roughly as follows: Given a trigonometric polynomial f ( x ) = C cjeVx, consider f * f ( x ) = C Icj(2eVxand observe that (Clcjr) = ( 1 / 2 ~ ) f(x)12dx. By Young’s convolution theorem we have * 711: d c llfll :/3, and, consequently, we obtain the above estimate for q = 4, except for the constant c > 1 appearing on our right-hand side. To obtain the whole range of Young’s original proof we keep convolving with f: Note in turn that, for appropriate indices p, q, r, Young’s convolution theorem can be obtained from the Fourier transform inequality. Young’s original statement is sharp in T, and in fact Hardy and Littlewood characterized the extremal functions: they are multiples of the characters. However, on the real line a much sharper inequality was obtained by Babenko in 1961 for the special values of q = even integer. Recently, Beckner obtained sharp Lp inequalities for both the n-dimensional Fourier transform (the constant is (A,)“, where Ap = (p1’p/p‘1’p‘)1’2)and for Young’s convolution theorem (the constant for Lp * Lq E L‘ is (A&&,,)“). In 1913, Young published the dual to his original result in which sum and integral switch sides; it was ten years later when F. Hausdorff showed that the only restriction was q 2 2. At about the same time F. Riesz observed that the trigonometric systems could be replaced by an arbitrary, uniformly bounded ONS. The “complex” interpolation theorem of M. Riesz had all these results as a first application. The generalization of the result to the case when the powers p and q are replaced by an appropriately related pair of positive, increasing functions was first considered by Mullholland; the complex method of interpolation did not apply in this context. The Marcinkiewin interpolation theorem with powers replaced by convex functions was first considered by Zygmund.

[If

I&*

4. Notes; Further Results and Problems

159

For f~ L ( T ) we set F,(x) = j(-,,xlf(t) dt, and in general F,(x) = F,-;(t) dt, a = 2,. . . . It is readily seen that actually

J(-m,xl

and consequently we may extend this notion, with this definition, to all values of a > 0; this concept is due to Riemann and Liouville. The proof given here of the Hardy-Littlewood, Sobolev inequality is due to Hedberg and is interesting because it reduces the result to a "main diagonal" interpolation result, which is the easiest case. Suppose that J, denotes the Bessel potential of order a in n-dimensional Euclidean space. Stein showed that iff 2 0, then

where CY > (n - k ) / p and p = p k is the restriction of Lebesgue measure to Rk,considered as a subspace of R". In this case, the inequality can be restated as

and the result therefore refers to the restriction, or trace of J,f to a lower dimensional space.

Further Results and Problems

Assume 1 < q < 2 and { A , } ~ = , is a sequence of positive numbers. Show that, if C n"-'A, < 00, there is a continuous function g so that the An's are the Fourier sine coefficients of g and x-"g(x) E L[O, T ] . (Hint: Note that since C A, < a,C A, sin nx converges uniformly to a continuous function g. Use Fatou's lemma to bound

4.1

x-"lg(x)l dx d

1A,

[0,*1

=

1nl-"A,

x-"lsin nxl dx .*I

I -

x "Isinxl dx.

[Omrl

The proof gives more, namely, Ix - al-"(g(x) - g(x)) E L[O,T I for all a E [0, TI.A similar result obtains for cosine series for 1 < 7 < 3 now, and the corresponding functions f verify x-"(f(x) -f(O)) E L[O, T ] and also

VI. Paley’s Theorem and Fractional Integration

160

Ix - a l - ” ( f ( x ) - f ( a ) ) E L[O, P] when 1 < q < 2. This result, as well as the next eight, is from Boas’s book [1967].) 4.2 Assume 0 S v s 1 and A, decreases to 0. Then 1 A, sin nx converges to g ( x ) and x-”g E L[O, P] if and only if 1 nq-’An converges. (Hint: To show the necessity, note that the partial sums of 1 n“-’ sin nx are uniformly bounded and of orderb(x-”) for each q ;separate the cases 0 < q < 1 and 7 = 0 , l . Thus if x - ” g ( x ) is integrable, by the dominated convergence theorem we get that

To prove the necessity, we sum by parts and invoke 4.1 above. The corresponding result for cosines is also true.) 4.3 Is the converse to 4.1 true? 4.4 Hardy’s inequality 4.6 in Chapter I also holds in this general form: if p , r > 1 and f is integrable and positive, then

the best constant c is ( p / ( r - l))p.State and prove a similar result for sequences. 4.5 If q < 2, then F2 f[o,tl s-” sin s ds decreases in (0, T ) and the same s-”(l - cos s) ds if q < 3. (Hint: Elementary calis true for t”-3 I[,,,, culus.) 4.6 If A, 3 0, p > 1, s > 0, and c < sp - 1, then (4.1)

implies n=l

Moreover, if c > -1, then (4.2) implies (4.1). (Hint: Sum by parts and invoke Hardy’s inequality for sequences at the appropriate places.) 4.7 If the An’s are the Fourier sine coefficients of the continuous function g, 1 < p < c o , l / p < < ( l / p ) + l and A , > O , then I x - a I - ” x ( g ( x ) - g ( a ) ) E LP[O,r) for every a, 0 s a < P, if and only if nP“- p -2 (C“,,k h k ) ’ < 00 (or equivalently 1 nPT-2(CsP_nh k ) P < 00). (Hinr: We do the necessity first; assume x - ” g ( x ) E Lp[O,P]. By 4.1 and 4.3 above, C n”-lhn converges. Now if E > 0,

E

s-”

sin ns ds

I 1I =

n”-’

(sn,xn)

s-“ sin s ds

I

=

O(n”-’),

161

4. Notes; Further Results and Problems

I(,,,,

I(,,,, s-“

uniformly in E and x. Hence s - “ g ( s ) ds = 1 A,, verges uniformly in E, and by letting E + O+ we obtain

sin ns ds con-

”

and each term on the right-hand side of (4.3) is positive. Applying 4.4 to the left-hand side of (4.3) we get that

1 (1 I A,x-’

C03971

s-”

sin ns d s ) pdx < a,

[O,X)

which implies that also

j[,,,(~

nAnx-’(nx)A-2

I

s-” sin u du

[O,nx)

Y

< a.

(4.4)

Now, we only decrease the integral in (4.4) by replacing the sum by Cfl=”, and then replacing ( nx)s-2 s-” sin u du by its minimum (here use 4.5); ~ ~ ~ , d x < co. Conversely, for any N > 0, thus, ~ [ o , f f ) ( nA,x’-”)P

I[o,nx)

and both terms are easily estimated. There is a corresponding statement for cosines.) 4.8 Assume that A,, decreases to 0 and denote by f either the Fourier cosine or sine series with coefficients A,, (cosine and sine results involving Lp integrability are usually equivalent because the series are conjugate to each other). Then for 1 < p < 00 and - l / p ’ < r] < l / p , x - ” g ( x ) E Lp[O,P ] if and only if 1 np”+p.-2AI:converges. (Hint: We deduce the cosine version from the sine version of 4.7 above; the sine version is obtained similarly. To show the necessity, assume for simplicity that A. = 0, then F(x)=

I

f ( t ) dt =

1C I A , sin nx

C0,X)

and by4.4, x - ” - ’ F ( x ) E Lp[O,P ) . Thus by4.7 1 nPT-2(1F=,hk)’ < 00, which gives the desired conclusion since the hk’S decrease. To prove the sufficiency notethat 2 f ( x ) sin x = C(A. - A,+z) sin((n 1 ) x ) ;therefore, wemay invoke 4.7 again (with 17 + 1 in place of 71 there) and obtain x - ” - ’ ( x f ( x ) ) E Lp[O,P ) , i.e., x - “ f ( x ) E Lp[O,P ) , provided that

+

VI. Paley’s Theorem and Fractional Integration

162

converges. But since C;=,lAk - Ak+21 = A, + A,+l we are done.) Assume f(x) is a positive, decreasing function in [0, T ) , 1 < p < 00 and -l/p’ < r] < l/p. Furthermore, assume that the An’s are either the Fourier cosine or sine coefficients off: Show that n-qplAnlPconverges if and only if xpq+p-2f(x)pis integrable over [0, T ) . (Hint: Dualize 4.8.) 4.10 Assume w(x) > 0 for x E T is such that for all A > 0, J{W(X)>*) w(x)-’ dx c A/A2, with A independent of A. If 1 < p 6 2 and { c j } E lP, then there is f e L(T) such that c j ( f ) = cj and 4.9

(j I f ( ~ ) I ~ w ( xdx)) ~ - ~ c C(II c ~ ~ ) ’ / ~ . 1/2

T

Also if 2 c q < 00 and f

- C cjeVx,then

(XI cj14)

d

(jTlf(x) pw (x)”-’ dx) l/q.

Can you think of “rearranged” statements such as Paley’s theorem? The Hausdorff-Young theorem and Paley’s theorem are not equivalent; indeed, let co = c1 = cP1= 0, cj = (ljl 111Ijl)-~/~, ljl S 2. Then C c;” = 00, but C c;”l j14/3-2< 00. for each 4.12 There is a continuous function f so that C lcj(f)12--s= E > 0. (Hint: Putf(x) =, :,I jp22-jI2(4(x)- &,(x)), where the 4’s are the Rudin-Shapiro polynomials introduced in 5.38 of Chapter V. The result, which, in particular, shows that there is no Hausdorff-Young theorem for p > 2, is Carleman’s, the example is from Katznelson’s book.) 4.13 The Hausdoe-Young inequality is best possible in the sense that if ‘ 00, then r p’. (Hint: Take a look for f E Lp(T ) , 1 < p < 2, ~ l c , , ( f ) l< at the function ]XI-”, 0 < r) < 1, which is in L P ( T )provided that r]p < 1, and its Fourier coefficients cj l j [ - ( l - q ) ,which are in I‘ only when r(1 77) 1.) 4.14 Assume that f E L In L( T), f C cjeiix,and show that there are constants A, B such that 4.11

-

’

-

c&1/1 1+

S

A+B

I,

(f(x)(ln+(f(x)(dx.

Is the conclusion still true if we replace lcjl by c . in the above sum?

+

4.15 I f f - 1 cjeVxand lcjl < l/ljl 1, then e A t f l lis l integrable, provided 0 < A is sufficiently small. (Hint: Use Hausdoe-Young to estimate llfll,, p = 2,3, . . . .) 4.16 Suppose { A j } E MPq, 1 s p c 2, 1 < q d 2. Show that I(1+

~j~)-q‘~p‘-E~Aj~q’ < 00 for each E > 0. (Hint: There is an Lp function with c j ( f ) ljl-’/p‘-’, n # 0. Since we assume that 1 Ajcj(f)eVxE Lq(T ) we may invoke the Hausdorff-Young theorem.)

-

163

4. Notes; Further Results and Problems

Assume that T is a sublinear operator simultaneously of weak-type (2,2) and of type (1, a), in both cases with norm c 1. Let B( t ) = b(s) ds, where b is positive and nondecreasing and A(t) = ( c o , t ) a ( s ) ds, where a is continuous and strictly increasing from 0 to 00 with s. Show that if t f[o,t) b(s)(ds/s2) d 1/2d(2/t), for t > 0, then B(ITf(y)l/2) dv(y) 1 whenever A(If(x)l) dp(x) c 1. (Hint: d is defined in Theorem 1.6; the result is from Jodeit-Torchinsky [ 19711.) 4.18 Assume that A(s) is as in Theorem 1.6, that {nk}is a fixed Hadamard sequence, i.e., nk+Jnk 3 A > 1, and that I,A(lf(x)l) dx d 1. Show that limk+OD snk(Ax) = A x ) a.e. (Hint: Cj MR(lCj(snkW - gnk(jI)l/2)< 00 (cf. 7.25 in Chapter IV).) 4.19 The real Laplace transform F of f given for x > 0 by F(x) = e-mf(u) du is a linear mapping simultaneously of types (1, 00) and (2.2). 4.20 The "local" control that M,f exhibits over Z,f in (2.20) may also be expressed in terms of the so-called "good-A inequalities," about which we will have more to say later on; these results are quite important in the understanding of the weighted norm inequality problem. In this setting, the good-A inequality assumes the following form: Let Z c T, A > 0, and, for positive constants E, S, let E = {x E I: Z,f(x) > EA, M,f(x) C SA}. Then there are constants E,, and k, depending only on a,so that iff is a nonnegative function and Z contains a point where Z,f(x) C A, then IEId Put g = fX21, h = f - g. We may k(S/E)l/('-a)lZl for all E > E ~ (Hint: . assume there is a point t E Z so that Maf( t ) d SA, for otherwise E is empty and there is nothing to prove. Also observe that by Theorem 2.2 there is a constant c (depending only on a) so that

4.17

Ico,t)

I,

I,

I[o,m)

Next let s E Z be the point where Iaf(x) < A. Since for some constant L > 1 and for x E Z and y iZ 21 we have that 1s - yl C Llx - yl, it is immediate to see that Z,h(x) s L'-"Z,f(s) C LA. If we pick now so = 2L, then for E a so the set {x E Z: Z,h(x) > ~ h / 2 is } empty. This result is from MuckenhouptWheeden [ 19741.) 4.21 Assume that f 2 0. Then for 0 < a, S < 1, IIZasfllr d c ~ ~ f ~ ~ ~ - s ~ ~ Z a f ~ where l / r = (1 - S ) / p S / q , 1 < p < q < 00. (Hint: As in (2.5) bound the integral defining Z,sf(x) by J + K ;estimate K by v)7e(6-1) Jlt-xpTf( t)lx - tl"-' dt d va(s-l)I a f(x) and minimize. This result and the next two are from Hedberg's paper [ 19721.)

+

:,

VI. Paley's Theorem and Fractional Integration

164

Suppose that f s 0, then IIIaS(ff)llrs c ~ ~ f [ ~ ~ - ' ~ ~where Z a f ~0~ < ~ ,a, 6 < 1 , O 1as t = 1 is Problem 4.21 and 0 < t < 1 uses similar ideas. As before, bound the integral defining Z a S ( f ) by J + K; J s c ~ " ~ M ( f ' ) (To x ) estimate . K choose s < p so that t < 6 + ( 1 - 6 ) s and use Holder's inequality to show that K < C Q - ~ ~ / ( ~ - ' ) M ( ~ ) now ( X ) ;minimize.) 4.23 Suppose that f E Lp and let p = l / a > 1. Then there is a constant c (which may be computed explicitly; this is left for the reader) so that for each interval I E T and E > 0 4.22

J, exp(

c

-1

-E

1")

dt < ce1zl.

(Hint: Assume Ilfll, = 1 , estimate II,f(x)l s J + c(In(~I~/2q)'-'~", and minimize.) 4.24 The condition qo f q1 is essential in the interpolation theorem 2.3. (Hint: For f:[ 0 , 1 ] + C consider the mapping

f

=

T f ( x )- F(x) = x-l"

J

f ( t ) dt. [O,Xl

Then T is of weak-type (p, 2) when 1 s p s 2, but is not of strong-type ( p , 2) for any p in that range.) 4.25 For f ' s as above consider the continuity properties of the mappings f ( x ) + T f ( x ) = x" jlo,x) z"f(t) dt, where a, q are real numbers and not necessarily positive. Construct examples of operators of weak-type ( p, q ) but not of type (p, q) for any 1 s p s q < co, and consider whether the interpolation theorem holds for q < p. 4.26 Suppose T is an operator with the following property: for each A > 0, T may be written as TI + T,, where TI maps L p ( x )into L"( Y ) with norm s A / 2 and T2 is of weak-type (p, p) with norm 0. Show that T is of weak-type (p, q). (Hint: For I l f l l , = 1 we have v({lTfl> A}) s v({lTlfl > A/2}) + v ( { \ T f \ > A/2}) C 0 + C A " - ~ A - ~= c K q . This gives a different proof of Theorem 2.1, 2.2.) 4.27 Assume T is a sublinear operator of weak-type ( p o , qo),0 < po qo < co. A sufficient condition that T also be of weak-type (p, q), where l / p = l / q = l/p, - l / q o , p 3 p > 1, is that for every sequence (4) of pairwise disjoint intervals and every function h in Lp(T ) supported in 4 and such that h ( x ) dx = 0, all j , the following estimate holds: I{x E T\U2<: I T h ( x ) l > A}\ C ~ ( l l h l l ~ / A (Hint: ) ~ . We may restrict ourselves to nonnegative functions f with llf11, = 1 and A large, >1, say. By the Calderbn-Zygmund decomposition of f" at level A we obtain a sequence (4)

I,

u

165

4. Notes; Further Results and Problems

of pairwise disjoint intervals such that

and Let now A x ) = Cjfi,xI:(x)+ f ( x ) x T \ u I f ( x ) ,b ( x ) = f ( x ) - A x ) . The estimate for the term involving b is easy; also I{[ Tgl > A}I S c(llgll,/A)qo,which together with g ( x ) =S 2A4'P a.e., llgllp =S 1, and the relations between p and q, gives the desired conclusion. This result is from Baishanski and Coifman's paper [1978].) 4.28 Suppose T is a sublinear operator simultaneously of weak-type (1, q ) and (p, q l ) , with 1 < q, p < q l , q < q l . Then there are constants A, B independent o f f such that 9 T

I Tf(x)I4dx

A + B(

If(x)l(ln'lf(x)()'/'

dx) .

(4.5)

T

(Hint: By (4.16) in Chapter IV, it readily follows that there are constants c, , c,, cj so that for A > 0,

Once we show that

is dominated by the right-hand side of (4.5) we are done, and this is not hard to do. The result is O'Neil's [1966]; however, in the particular case of the fractional integral, it had already been noted by Zygmund.) 4.29 Let 0 s l / q i = pi =S l/pi = ai < 00, i = 0, 1, a. f a l , Po # P I , and let y = E X + y be the equation of the line passing through the points (ai, pi). Assume that the generalized Young's functions A, B are given by A ( u ) = b ( s ) ds, with a and b monotone and further a(s) ds, B ( u ) = assume that, if M = max(qo, 4 , ) and rn = min(qo, q l ) , then B ( u ) / u M decreases and B ( u ) / u " increases and

I

CO?)

( B ( s ) / s " + ' )ds = O ( B ( u ) / u " )

VI. Paley's Theorem and Fractional Integration

166

and

( B ( s ) / s M + 'ds ) = 0(B(u)/uM). Then if B-'( u) = A-'( u E ) u Ythere , is a constant c, independent off, so that J, B(ITf(y)l/c)d v ( y ) s 1 whenever 1 , A ( f[ ( x ) l )d p ( x ) s 1. (Hint: Set the monotone function z - l ( s ) = B-'(A(s)'/').) 4.30 (Young's Convolution Theorem. Lp * L' E Lq, l/q + 1 = l/p + l/r. (Hint: Fix f E Lp(T ) , 1 < p < 00, then convolution with f gives a bounded mapping from L into Lp and from Lp' into L"; now interpolate. Other proofs give a better constant.) 4.31 More generally, wk-LP * L' E Lq, same range of p, q, r's as in 4.30. 4.32 Assume that l/p + l/q > 1, 1 < p, q S 2; then Lp * Lq is not included in U,,,L", l / r + 1 = l / p + l / q . (Hint: For k > 0 put a,, = l/(k + 1)2/p2(k+')/p' for 2k 6 n < 2k+', and extend a, as an even sequence for all n's and similarly for {b,,} with q in place of p. Since 1uPnlnlp-2, 1bzln1q-2 < a there are functions f C u,eim E Lp and g 1b,,eimE L4. However, since ~ ( a n b , , ) S l n ~=S00- Zfor each s > r, f * g cannot be in L". This example is from Quek and Yap's paper [1983].) 4.33 If a > O , O < E < 1, 1 < p < &/a,1 s q < wand f E L" and M E I p E f Lq,then IJ E L' and 11 I, f 11 g cII M E I p11f"I" Ilfll;-"/". (This result is from Adams' work [ 19751.) 4.34 Show that M i = 9Lp, 1 < p G a, i.e. { A j } E Mi if and only if there is a function C#J E Lp(7') so that cj(C#J)= Aj?.allj . (Hint: The sufficiencyis obvious; as for the necessity, if A C Ajevx,then IIA * K,,l[ps C ~ ~ for K , , ~ ~ ~ all n. What does the statement imply concerning A:?) 4.35 The space of multiplier sequences from LP(T) into C(T) is g ~ " , l / p l/p' = 1, 1 s p < 00, and so is M%.What does the statement say about A%?

-

-

+

-

CHAPTER

VII Harmonic and Subharmonic Functions

1. ABEL SUMMABILITY, NONTANGENTIAL CONVERGENCE We refer here to yet another classic, and very important, summability method. This method requires the identification of T with eD, the boundary of the unit disk D = { z E C :Iz( < 1) in the complex plane. Given 0 s a < 7 ~ / 2we define the set Cn,(O) with vertex at 1, that is, e”, and opening a as the convex hull of the disk of radius sin a and { 1}\{ l},

u W O )

This set corresponds to the notion of “nontangential” approach to 1. It is readily seen that points z in Cn,(O) satisfy the condition 1 (1- zJ/(1- 121) S 2 max(l/(l - sin a),l/cos a).Thus it is equivalent, and often simpler, to consider instead the “cone”

r,(o) = { z E D : 11

- zl/(i

-

l z l ) c a},

a a 1.

(1.1)

We are now ready for

Definition 1.1. Given a

2

1 and a function f(z) defined in D, we say that 167

VII. Harmonic and Subharmonic Functions

168 \

f converges (Abel) nontangentially of order lim

Z+

i,zer,(o)

(Y

to L as z + 1 provided that

f(z) = L,

and we denote this by lim,,,f(z) = L ( A , ) . When a = 1 we call the approach “radial,” for then z takes values on the radius joning 0 to 1. Similarly, in the case of series we have Definition 1.2. Let { c i } be a sequence of complex numbers. We say that the m series cj is (Abel) nontangentially convergent of order a ( 2 1) to s, and denote this by CJys0cj = s(A,), provided that for f(z) = cjzJ, limze1f(z) = s(A,). The usual algebraic properties hold in this case as well. For instance, if so and s1 are finite numbers a n d x cj = so(A,),C dj = sl(A,), then also C(cj + d j ) = so + sl(A,).

CJzo

It would be reassuring to know that convergent series also converge nontangentially to the same limit; in fact more is true. Proposition 1.3. Suppose

CJco cj = s(C, 1).

Then also

,:C,

cj = s(A,),

lS(Y
Proof. Let s, and a, denote the partial sums and Ceslro means of order n of { c j } ;our assumption implies that s, = o ( n ) . In the first place, observe that n-I

n

1cjr‘ = ( 1 - 1sir‘ + s,zn. 2)

j=O

j=O

Therefore, if the limit of either side in (1.2) exists as n + CO, then the limit of the other side also exists and they are equal. Moreover, since snzn = o( l), then also m

m

j=O

j=O

1c j 2 = ( 1 - z) 1s j 2

(1.3)

provided either series converges. Summing by parts once again, and since ncr,,_lzn-l = o(1) for z E D as n + co, a similar argument gives m

W

j=O

j=O

1cjzJ = (1 - Z ) ’ C ( j + 1)a;zJ =f(z),

(1.4)

say, again provided either sum converges. Now, it is readily seen that the right-hand side of (1.4) converges absolutely for IzI < 1 with sum s c / ( l - IzI)’, and so the left-hand side is also finite. Next we show that f(z) tends to s nontangentially. In the process we make use of the identity m

1 = (1 - z ) ’ C < j j=O

+ l)zj,

l z ~
(1.5)

169

1. Abel Summability, Nontangential Convergence

More precisely, given E > 0 we show there exists S > 0 so that If(z) - SI S E provided z E r,(O) and 11 - zI s 8. Indeed, combining (1.4) and (1.5), it readily follows that

c m

f(z) - s

+

= (1 - z)' (jro

(j

+ l ) ( q - S ) z i = I + J,

j=N+l)

say. Now, since m

m

j=N+1

j=O

1( j + I)+ s C ( j +I)$ =-( I -1 r)"

if we choose N so that laj - SI s & / 2 a 2for j that for z in ra(0),

3

N

+ 1, then it readily follows

Now that N has been fixed, we note that

j=O

provided S is small enough. We are interested in applying Proposition 1.3 to Fourier series, more specifically to Lebesgue's theorems 4.2 in Chapter I1 and 6.3 in Chapter 111. Since the expression appearing in the Definition 1.2 is one sided, given f E L, f C cjeij*,we introduce the notations

-

C o ( t )= co,

G(t)= c-je-l* + cje'",

j s l

(1.6)

j 2 1.

(1.7)

and eo(t) = 0,

c(t) =

(-i)(-cjeKij'

+ cjeij*),

Proposition 1.3 asserts that actually

j=O

and m

ej(t) = Hf(t)(A,)

almost every t in

T.

( 1 -9)

j=O

We may, and do, assume that (1.8) and (1.9) hold simultaneously. More precisely, if z = rek E ra(0),then for almost every t in T (1.10)

170

VII. Harmonic and Subharmonic Functions

and (1.11)

We would like to unravel (1.10) and (1.11) and express them in terms of the original cj's. First, since r,(O) is symmeric, that is, z E r,(O) if and only if Z E ro(O), and as is readily seen

then for almost every t in T the following is true: (1.12)

Similarly, by (1.11) now, and for the same t's,

"+" statement

Whence, by subtracting the "-"statement in (1.13) from the in (1.12) and rearranging the expressions involved, we obtain

= f(t )

- iHf( t ) ,

a.e. in

T.

(1.14)

Similarly, by adding the "-"statement in (1.12) and the "+" statement in (1.13) and rearranging the expressions involved, we obtain

=f(t)

+ iHf(t),

a.e. in

T.

(1.15)

We have thus arrived at one of the most interesting and important results in this chapter, namely, Theorem 1.4. Suppose f E L( T ) , f

- C cjeGr.Then for almost every t in T (1.16)

and (1.17)

2. The Poisson and Conjugate Poisson Kernels

171

Prmf. (1.16) follows by adding (1.14), (1.15), and (1.17), by subtracting (1.14) from (1.15), and invoking the symmetry of r,(O) to change -x intox. H

2. THE POISSON AND CONJUGATE POISSON KERNELS Expressions (1.16) and (1.17) in Theorem 1.4 correspond to the convolution off with the functions P ( z) and Q( z) with (absolutely convergent) Fourier series given by m

z = re",

O < r < 1 (2.1)

and W

(-i)(sgn j)cjrlj'evL,

P(z) = Q(Z) =

z

=

reiL,

o
(2.2)

j=-00

respectively; these functions are called the Poisson and conjugate Poisson kernel. When convenient, we view these functions as defined in T and denote them by P r ( t ) and Q r ( t ) ,'i.e., as a family of kernels indexed by r, 0 r < 1. The first task at hand is to obtain an explicit expression for these kernels. By summing the geometric series in (2.1) we get

which easily reduces to

Similarly, by summing (2.2) we see that

which also readily reduces to

172

VII. Harmonic and Subharmonic Functions

The interaction between P and Q is given by the expression P which on account of (2.3) and (2.5) equals

+ iQ,

This is an analytic function of z in D ; we will have more to say about this later on. By (2.4) we note that Pr(t) is a periodic, positive, even function of t E T, monotone in [0, T),which is dominated either by

1-2 l+r 2 <(l-r)*-l-r 1-r or, on account of estimate (1.16) in Chapter I, by

Also, by (2.1), (2.10) The reader may have noted the similarity of FejCr's kernel K , ( t ) with the Poisson kernel P,( t) for r = 1 - 1/( n + 1); this hints at a similarity of results as well. For instance by (the continuous version of) Proposition 2.3 in Chapter IV there is a constant c, independent off, so that

(2.1 1) But on account of statement (1.16) of Theorem 1.4 we expect a stronger result, of a nontangential nature, to hold. Before considering this point we list some properties of the conjugate Poisson kernel. By (2.6), Q r ( t ) is an odd function and I Qr(t)l is dominated by either

crltl/(l - r)'

or

c/ltl.

(2.12)

Also by (2.5) (2.13) As for 11 Qrlll, it does not remain bounded as r + 1- (cf. 6.4); this is all we need to know about the kernels. 1 6 a < 00. In Now some definitions. Let I',(x) = { z E D : eVkz E ra(0)}, other words T,(xi is the cone r,(O) rotated so that its vertex lies at ek rather than at 1; a similar definition is given for n,(x).

173

2. The Poisson and Conjugate Poisson Kernels

Definition 2.1. Assume f E L and 1 s a < 00. We introduce the nontangential maximal function off * Pr of order a at x by Na(f*p r , x ) = :UP I f * Pr(t)l* (2.14) z = r e .zeT,(x) N , is called the radial maximal function.

We then have Theorem 2.2. Assume that f E L and 1 s a < m. Then there is a constant c,, independent off, such that (2.15)

N ( f *pr, x) s c,Mf(x).

Proof. The radial maximal function was already considered in (2.11), and the nontangential case is a straightforward computation. Indeed, since each z = reir in T , ( x ) is of the form reicx+'),reis E r,(O), it readily follows that

'I

f* P r ( t ) = I = -

2T

T

'I

f(u)P,(x+s-~)du=2T

~(x+u)~~(s-u)~u,

T

and consequently )I1 d

$-(I I +

C-TGO)

.(fI)

+ u)lP,(s - u ) du = J + K ,

(2.16)

[O,.rr)

say. As the estimate for both summands in (2.16) is obtained in a similar way, we only do K . Let F,(u) = F ( u ) = j,o,,,lf(x u)l du; then F is absolutely continuous and F ' ( u ) = If(x u)l a.e. Thus, by integrating by

+

+

parts the integral in K, we note that it equals

F ( u ) P r ( s- u ) ] : -

I

f ( u ) P : ( s - u ) du

=

K1 + K 2 ,

[O,r)

say. K1 offers no difficulty: since F ( 0 ) = 0, F ( T ) s c M f ( x ) and P,(s - T )s c,, we see at once that K1 s c,Mf(x), which is of the right order. Moreover, since for 0 < u < T F( u)/sin( u/2) s c M f ( x ) ,

also K 2 s c(

I[,.,

)

sin( ;)P:(s - u ) du M f ( x ) .

(2.17)

To bound the integral in (2.17) we distinguish two cases, namely, 0 < r =S f, and 4 < r < 1. The latter is pretty easy since then the integral does not exceed

174

VII. Harmonic and Subharmonic Functions

As for the former we observe that the integral there is dominated by

say. To estimate K4 we note that the integrand uP:(u) is even and, consequently, K4 = -2

J

uP:(u)du

= -~uP,(u)],"

+

[%a)

= -272Pr(

72)

+ 272 s 272.

Finally, to bound K 3 , observe c ( r - 11 cll - re"[; and consequently

+

that

Jro,

a) ~

r ( u du )

flsl s rlsl

(2.18)

clr - reis[s

1 11 - reisl d c + c,. K3
Corollary 2.3. Assume that f E Lp(T ) , 1 C p s 00. Then there is a constant c = c,,~ independent off so that I I N a ( f * Pr)IIp ~ l l f l l p , 1 < P s 00 (2.20)

and

A({Na(f* pr)'

AII s cIIfII1.

Corollary 2.4. Suppose f E Lp(T ) , 1 < p s c = c, independent off so that

X ( f *Qr, and consequently for 1 < p < 00,

(2.21) 00.

Then there is a constant

x) s c W J ) ( x )

(2.22)

I I X ( f * Q J l l P s CllfIlP. (2.23) Proof. Since am(f) * Q r ( x )= &"(f)* P , ( x ) and 1 < p , it follows that also f * Q,(x) = f * P,(x) and the conclusion follows from Theorem 2.2 and Corollary 2.3. As for convergence in Lp we have Proposition 2.5. Assume f E Lp(T ) , 1 s p < a.Then

(i) llf * Prllp s Ilfllp r < 1, (ii) lim,,,Ilf* Pr - f l i p = 0, and (iii) liml=,ix,,,,,r,(0)llf* Pr(x + * ) -flip = 0.

175

2. The Poisson and Conjugate Poisson Kernels

A similar result holds for p

provided that f E C( T).

= 00

Proof. (i) and (ii) have a counterpart, with identical proof, in the case of FejCr's kernel. As for (iii), in case p < 00, the proof is a simple application of Fatou's lemma since, by Theorem 2.2, Pr(x + t ) -f(t>l S c M f ( t ) and, by Theorem 1.4, limz=re~x-tl,zc~,(0)f* Pr(x + t) =f(t) a.e. The case p = 00 also follows easily.

If*

As for the conjugate Poisson Kernel, we have Theorem 2.6. Assume f E L( T), then

(f*Q,(x + I)- H , . J ( t ) )

lim

== re'x-.i,zd-,(o)

=0

a.e.

(2.24)

and there is a constant c = c, independent o f f such that

If*

sup z = relx,zc

Consquently, iff

r, ( 0 ) E

Qr(x + t ) - H , - f ( t ) l s cMf(t) a.e.

(2.25)

Lp(T), 1 < p < 00, then also Qr(x + .) - H l - r f ~ = ~ p0.

/If*

lim

(2.26)

z = retx+ i , z d - , ( 0 )

Proof. (2.24) and (2.25) obtain by arguments similar to those invoked in the case of FejCr's kernel, combined now with the ideas in Theorem 2.2 to incorporate the non-tangential convergence; we sketch the proof of the radial case. First, observe that, since Q1(x) = (2 sin x)/[2(1 - cos x)] = l/tan(x/2), x # 0, we have

f* Qr(t)

- H,-rf(t) =-

2~

1

f(t-x)Qr(x)dx

1xlsl-r

I

+1 2~

= Zl(t)

say. By (2.12) we see that

IIl(t)l s

-1

I-r
f(t - x)(Qr(x) - ~ l ( x )dx )

+ Z*(t),

(2.27)

C

1- r

If(t Ixlrl-r

- x)l dx

and consequently I Z , ( t ) l s cMf(t)and IZl(t)l = o(l), as r + 1-, at each Lebesgue point t of f: Moreover, since Qr(x) - Ql(x) = ((1 - r)/(l + r))Q,(x)P,(x) is an odd and monotone decreasing function in (0, T),and by (2.12), 1-r 1-r 1 ---Q,(X) s --< T, for 0 < x < T, l+r 1 r sin(x/2)

+

VZZ. Harmonic and Subharmonic Functions

176 then we may estimate IZ2(t)l

c j y t - x)lP,(x) dx

cW(t).

Also an argument similar to Theorem 6.3 in Chapter I11 obtains that 12(t)= o(l), as r + 1-. Corollary 2.7. Suppose f dent off so that

E

L( T ) . Then there is a constant c = c, indepen-

Na(f*Or, t )

c ( J f * f ( t )+ M f ( t ) ) .

(2.28)

3. HARMONIC FUNCTIONS The Laplacian A is the second order partial differential operator given by

A=-1 - + a'- + - -a2 r2 at2 ar2

1 a r ar'

(3.1)

It is readily seen that A( r'j'evt)= 0, all j , and consequently by superposition A ( P r ( t ) )= A ( f * P r ( t ) )= 0, f~ L, re" E D. This operator is also familiar in Cartesian coordinates, i.e., in terms of x = r cos t, y = r sin t, and it is given by A = - a2 + - a' ax2 ay2' In what follows we use (3.1) and (3.2) interchangeably. Functions u which verify Au = 0 (in D) are called harmonic (in D). We begin by showing that in some sense the only harmonic functions are those which arise as Poisson integrals. More precisely, we have

Proposition 3.1. Assume u E C 2 ( D )is harmonic. Then there is a sequence of complex numbers { c j } such that u(re") =

1 cjrljleu*,

reit

E

(3.3)

D.

j=-m

Proof. By assumption the partial derivatives au/dx, au/ay exist and are continuous in D. We construct a function u(x,y ) such that for ( x ,y ) E D (or more precisely x + iy E D ) ,

a

a

-u(x, y ) = --u(x, aY ax

y).

(3.4)

177

3. Harmonic Functions

A pair of functions (u, u ) verifyinng (3.4) is called a Cauchy-Riemann pair and the function o is called a harmonic conjugate, or conjugate, to u ; our claim is that to each harmonic function u in D we may assign a conjugate o. The construction in 0, or any simply connected domain for that matter, is quite easy. Indeed, fix (xo,y o ) in D and note that u is well defined by means of the path integral 4 x 9

Y ) - 4 x 0 , Yo) =

j

(X,Y)

du,

(3.5)

(X0,YO)

where the path in (3.5) is totally contained in 0,provided that du is an exact differential for then the integral in (3.5) is path independent. But if we put

then this occurs since

a "(aY -Gu) a

= - ( - u )a, ax ax

as u is harmonic. Thus (3.5) determines a conjugate u to u and by (3.4) the function f ( z ) = u(z) + iu(z) is analytic in D. Therefore, there is a sequence of complex constants { d j } such that m

f(z)=

Iz( < 1.

djz', j=O

Moreover, since u ( z ) = Ref(z)

=

;(c 1 "

djz'

m

+ @),

it readily follows that with

,

c. = ' J - j ,

j < 0,

co = $ ( d o + do),

and

cj = f d j , j

> 0.

By the way, the proof also shows that u(re") = 1 cirlJ'egr,with c; (-i)(sgnj)cj, j # 0, 4, = ( - i / 2 ) ( d 0 - do).

=

Remark 3.2. If f ( z ) = u ( z ) + i u ( z ) is analytic in a region containing 0, by the results discussed in this chapter it follows that

178

VII. Harmonic and Subharmonic Functions

where z = reir,0 d r < 1. This formula exhibits the close connection between an analytic function f and its real part u. Next we discuss the analog to Fejir's theorem

Theorem 3.3. Suppose u is harmonic in D and that, in addition, for some 1
Then there is a function f E Lp(T ) , Ilfll, lim

C

A, such that u(re") = f

* Pr(t),

a.e.

u(rei(x+r) ) =f(t)

re"+l,re'xEr,(0)

and lim re'x+l,re'xEr,(o)

11 u(

) - f l p = 0,

1 < p < 00.

(3.6)

Proof. By Theorem 1.4 and Proposition 2.5 it suffices to show that u(reif) = f * P r ( t ) ,f~ L P ( T ) ,Ilfll, S A. By Proposition 3.1 it follows that u(re") = 1 cjrl.ileVr, and we are reduced to showing that the cj's are the Fourier coefficients of an Lp function$ Let a,,(t)denote the Ceskro means of order n of the cj's. By Fejir's theorem 3.3 in Chapter I1 it is enough to prove that ~ ~ as, A, , ~all ~ n. p We write a,,(?) = (a,,(t) - an* P r ( t ) )+ an* P r ( t ) = Z ( t ) + J ( t ) ,

say, and take a closer look at each summand. Observe that

say. Since I , ( t ) is a trigonometric polynomial of degree n, we have that lllllps (1 - r)IIIlllp = o(l),

as r + 1 - .

As for J ( t ) , note that it actually equals u(r-) * 2K,(

IlJllP

S

H~(r*)llpll2K"IIl s A,

I)

(3.7)

and, consequently,

all r, n.

(3.8)

3. Harmonic Functions

179

Thus, by combining (3.7) and (3.8), we get that ~ ~ c T6, A, ~and ~ ~ we have finished. As is usually the case, (3.6) holds forp = 00 providedf E C ( T). As for p = 1 we have Theorem 3.4. Suppose u is harmonic in D and that, in addition,

Then there is a finite Bore1 measure p with total variation 11pl1 6 A, such that u(re") = p * Pr(t).Furthermore, if +(t) =

[

ddx),

t E

T,

then also lim

u(rei(x+r) ) = +'(t)

a.e.

reix+l,re'xEr,(0)

If u( re") 2 0, then there is a positive measure p such that u( reir)= p * Pr(t). Finally if { u ( r e k ) } r < is l Cauchy in L(T) as r -* 1, then there is an integrable function f such that u(re") = f * Pr(t);f satisfies (3.6) with p = 1. P m f . The existence of p and the statement concerning the nontangential convergence follow as in Theorem 3.3. Note that we cannot assert in general that u(re") = +'* Pr(t);the simplest example of this is the Poisson kernel Pr(t)itself which has nontangential limits 0 at every point save 0 and yet corresponds to p = Dirac delta at the origin. Now, if u is a nonnegative harmonic function, u(re") = C cjr'j'egr,then

[;u(reiI)l dt and u(re") = p

=

I,

u(rei1) dt

= 2m,,

< co

* Pr(t). Therefore

u ( r - )* 2K,,(t) = u,,(p)* P r ( t )3 0,

all n, r

and a,,(p, t) =

lim u,,(p)* P J t ) 2 0

r-I

a.e.

(3.9)

By Theorem 3.5 in Chapter 11, (3.9) implies that p is a nonnegative measure. Finally, in case { u( re")} is Cauchy in L(T) as r + 1-, we infer two facts, (r~) = to wit: since L(T) is complete there i s f e Lsuch that l i m r + l - ~ ~ u -fill 0 and by the first part of the theorem there is a measure p so that u( rei') = p * Pr(t). Since for each fixed s, 0 < s < 1, Ps(x - t ) E L"( T), it follows at

VII. Harmonic and Subharmonic Functions

180

once that limr+l u(r.) * P,(x) = f * Ps(x). On the other hand, since (as is readily seen by direct examination of the Fourier series) Pr * P, = Prs,0 < r, s < 1, we also have that u ( r * )* Ps(x) = p * Prs(x), which tends to p * Ps(x) = u(se") as r + 1-. Thus u(se*) = f * P,(x) and we are done. It is a natural question to consider what integrability properties, if any, are satisfied by the conjugates to harmonic functions verifying the assumptions of Theorem 3.3 or 3.4. The answer is straightforward. Theorem 3.5. Suppose that u is harmonic in D and that, in addition, for somelSp
&

lu( reir)lpdt)

I"

=A

< 00.

T

Let u denote the conjugate to u so that u ( 0 ) = 0. Then there is a constant c= independent of u such that

IT

N,(u, x)" dx s CAP,

1 < p < 00,

(3.10)

and, for all A > 0, Al{N,(u) > A}l

C

cA,

p

=

1.

(3.11)

Proof. Suppose first p > 1; then u(re") = f * Pr(t ) , f E Lp(T), and by inspection of the Fourier series, u(re") = f * Q , ( t ) = Hf* Pr(t). Whence N,(u, x) s N,(Hf*Pryx) 6 cM(Hf)(x) and 1 < P < 00. CIIM(Hf)llp S cllfllp, In case p = 1, we note that f* Qr(t) = (f*Qr(t) - H,-J(t)) + H,-J(t), and consequently by Theorem 2.6 N,(u, x) S c(Mf(x) + H*f(x)). Thus (3.11) also holds and the proof is complete.

I I N Y ( ~ ) l l pS

Two remarks about the above result: the estimate (3.10) points to a satisfactory state of affairs when 1 < p < 00 (a proof of a similar statement which does not rely on the Hilbert transform is sketched in below) and the estimate (3.11) does not. More precisely, if we denote by H P ( D )the Hardy HP space consisting of those harmonic functions u in D such that

I$(:;

IlP

lu(re")Ipdt)

c A < 00,

0