A note on singular integrals associated with a variable surface of revolution

variable surface of revolution

著者 Fan Dashan Shan, Sato Shuichi

journal or

publication title

Mathematical Inequalities and Applications

volume 12

number 2

page range 441-454

year 2009-04-01

A NOTE ON SINGULAR INTEGRALS ASSOCIATED

WITH A VARIABLE SURFACE OF REVOLUTION

DASHAN FAN AND SHUICHI SATO

Abstract. We proveL

pboundedness of certain singular integral

operators associated with a variable surface of revolution assum-ing a boundedness of related lower dimensional maximal operators. The singular integrals are dened by rough kernels satisfying cer-tain size and cancellation conditions.

1. Introduction

Let 2L

1(Sn;1), where Sn;1 is the unit sphere in

Rn (n
2) with

the Lebesgue surface measure d . We assume that

Z

Sn;1

(
)d (
) = 0:

(1.1)

De ne a singular integral

Sf(x) = p:v:Z Rn (y0) jyj ;nf(x ;y)dy

for f 2S(Rn) (the Schwartz space), where y 0 =y=

jyjfor y 2Rn nf0g.

Let ^f() = R

Rnf(x)e

;2 ihx
idx be the Fourier transform of f, where hidenotes the inner product inRn. Then it is known that (Sf)^() =

m(0) ^f(), where m(0) = ; Z Sn;1 (
)h i_{2 sgn(}h 0
i) + logjh 0
ij i d (
):

By using this expression of the Fourier transform of Sf, we can show that S extends to a bounded operator on L2(

Rn) if sup 0 2Sn ;1 Z Sn;1 j(
)jlog ; jh 0
ij ;1
d (
)<1: (1.2)

By Young's inequality, (1.2) follows from the condition  2LlogL(Sn ;1),

where LlogL(Sn;1) is the Zygmund class of all those functions  on

2000 Mathematics Subject Classi cation. Primary 42B20.

Sn _{which satisfy}

Z

Sn;1

j(
)jlog(2 +j(
)j)d (
)<1:

Furthermore, if  2 LlogL(Sn

;1), by the method of rotations of

Calder on-Zygmund (see 1]) we can show thatS extends to a bounded operator on Lp₍Rn) for all p2 (11) (see also 18] for the best

possi-bility of the class LlogL(Sn;1) ).

In the article 10], Grafakos-Stefanov considered the following con-dition: sup 2Sn ;1 Z Sn;1 j(
)j ; log; jh
ij ;1

 d (
)<1 for  >1, (1.3)

which is stronger than (1.2), and proved the Lp _{boundedness of S}

de ned by  satisfying (1.3) for a certain range of p depending on . It has been improved by Fan-Guo-Pan 4] as follows:

Theorem A.

Suppose that  satis es (1:3) for  > 1. Then S is bounded on Lp₍

R

n_{) for p}

2(2=(2;1)2).

For s
1, let s denote the collection of measurable functions h(t)

onR + = ft2R :t
0g satisfying khk s = sup u>0  u;1 Z u 0 jh(t)j s_dt 1=s <1: For b 2  1 and  2 L

1(Sn;1) satisfying (1.1), we de ne a singular

integral of R. Feerman type (see 9]):

Sbf(x) = p:v: Z Rn b(jyj)(y 0) jyj ;nf(x ;y)dy:

In this note, we study singular integrals associated with a variable surface of revolution. We shall prove theLp_{boundedness of the singular}

integrals under certain conditions and as an application we shall extend Theorem A to the case of Sb onR

2.

We write f(xz), x 2 Rn, z 2 Rm for functions on Rn +m, n

2,

m
1. De ne a singular integral with a rough kernel by

Tf(xz) = p:v:Z Rn b(jyj)(y 0) jyj ;nf(x ;yz;(jyjz))dy (1.4) initially for f 2S(Rn +m), where b 2 1 and (tz) = (1(tz)2(tz):::m(tz))

is a suitable continuous mapping from R +

Rm to Rm such that the

To study the mapping property of T, we consider two lower dimen-sional maximal functions associated with the function :

(g)(z) = sup_u> 0 u;1 Z u 0 jg(z;(tz))jdt (1.5) M(h)(sz) = sup_u> 0 u;1 Z u 0 jh(s;tz;(tz))jdt: (1.6)

It is known that theLr _{boundedness ofM implies that of (see 13]).}

LetH1(Sn;1) denote the Hardy space onSn;1. The spaceLlogL(Sn;1)

is a proper subspace ofH1(Sn;1). The following result has been shown

by 7]:

Theorem B.

Suppose that 2H

1(Sn;1) satis es (1:1), b is bounded

and M is bounded on Lr(Rm

+1) for all r > 1. Then T is bounded on

Lp₍Rn

+m) for all p

2(12].

See 13], 8], 6], 12], 14] for the case where (tz) is independent of the z variable. We consider the singular integral T in (1.4) de ned by  satisfying a condition similar to (1.3) and prove Lp _boundedness

of it for a certain range of p contained in (11). Let ' be a positive,

continuous function on (01) such that

(1) 'is increasing, (2) '(s)=sis decreasing, (3) '(st)c('(s) +'(t)),

(4) '(2s)c'(s).

A prime example of '(t) is the function (log(a+t))_{, where  > 0,} a
2 and a may depend on. De ne a function by

(t) = sup_s>

0

min(1st)

'(s) = '(11=t) (1.7)

where the last equality follows from (1) and (2) of the properties of '. We consider the following two conditions on :

sup 2Sn ;1 Z Sn;1 j(
)j' ; jh
ij ;1
d (
)<1 (1.8) sup 2Sn ;1 ZZ Sn;1 Sn ;1 j(
)(!)j' ; jh
;!ij ;1
d (
)d (!)<1: (1.9) For 2L

1(Sn;1), the condition (1.3) is equivalent to (1.8) with'(t) =

(log(a+t)) _(a
2). In Section 2, we shall see that the condition (1.8)

implies (1.9) when n = 2.

Theorem 1.

Let 0 <  <1=2, 2=(1;2)< s. Suppose that b 2s, P

j1'(2

j₎; <

1,  satis es (1:9) and M is bounded on Lr(Rm +1)

for all r > 1. Then T is bounded on Lp₍Rn

+m) for s=(s(1

;);1)<

p2.

We shall see that the proof of theL2 boundedness of Theorem 1 only

uses the Lr _{boundedness of . Thus we have the following:}

Corollary 1.

Let the numbers , s and the functions ', b,  be as in Theorem 1. If  is bounded on Lr(Rm) for all r > 1, then T is

bounded on L2( Rn

+m).

For examples of functions(tz) for which the maximal operator

is bounded onLr₍Rm), see 2]. When the function is independent of

z, we have the following result:

Theorem 2.

Let s 2 (12], 0 <  < 1=s

0, where s0 = s=(s

;1). Let

be independent of the z variable. Suppose that P

j1'(2

j₎; < 1,

b 2s,  satis es (1:9) and M is bounded on Lr(Rm

+1) for all r >1.

ThenT is bounded onLp₍Rn

+m) for 2s=((3

;2)s;2)< p <2s=((2;

1)s+ 2).

When (t) =P(t), where P(t) is a polynomial mapping from R + to Rm, the boundedness ofM can be found in 16, pp. 476-478]. Taking

f(xz) =g(x)h(z) and = 0 in Theorem 2, we have the following:

Corollary 2.

Let the numbers , s and the functions ', b,  be as in Theorem 2. Then Sb is bounded onLp(Rn) for the same range of p as

in Theorem 2.

By Lemma 1 in Section 2, the condition (1.9) follows from (1.8) when

n = 2. Therefore, by taking'(t) = (log(a+t)) _{in Theorems 1 and 2,}

we have the following two results.

Theorem 3.

Let n = 2. Let 2 < , 2=(;2) < s. Suppose that

b 2s,  satis es (1:3) and M is bounded on Lr(Rm

+1) for all r >1.

Then T is bounded onLp₍Rn

+m) for s=(s(

;1);)< p2.

Theorem 4.

Let n= 2. Suppose that s 2(12], s

0 < , b

2s. Let

be independent of the z variable. Suppose that  satis es (1:3) and M

is bounded onLr₍Rm

+1) for allr >1. ThenT is bounded onLp( Rn

+m)

for 2s=((3;2)s;2)< p <2s=((2;)s+ 2).

Also, Theorem 4 implies the following:

Corollary 3.

Let n= 2. Let the numbers , s and the functions b,  be as in Theorem 4. ThenSb is bounded onLp(Rn) for the same range

In 10], it was shown that there exists a function  which satis es the condition (1.3) for all >0 but does not belong toH1(Sn;1) and there

exists an f 2H

1(Sn;1) which does not satisfy (1.3) for any  >1.

If we combine the proof of Theorem 1 with that of Theorem B in 7], we have the following:

Theorem 5.

Suppose that  2 H

1(Sn;1) satis es (1:1), b

2 s for

some s > 2 and M is bounded on Lr(Rm

+1) for all r > 1. Then T is

bounded on Lp₍Rn

+m) for s0 < p 2.

In this note, C is used to denote non-negative constants which may be dierent in dierent occurrences. In section 2 we shall give several results which will be used to prove Theorems 1{4. We shall prove Theorems 1 and 2 in Section 3. Finally, in Section 4, we shall give a related result for the Marcinkiewicz integrals.

2. Results for the proofs of Theorems 1{4

The following lemma is for the case n = 2, which is the reason we con ne ourselves to the case n= 2 in Theorems 3, 4 and Corollary 3.

Lemma 1.

Let n= 2. Let ' be as in Section1 and let 2L

1(Sn;1), 
0. Suppose that sup 2Sn ;1 Z Sn;1 (
)'; jh
ij ;1
d (
)<1: Then sup 2Sn ;1 ZZ Sn;1 Sn ;1 (
)(!)'; jh
;!ij ;1
d (
)d (!)<1:

Proof. Put ~(t) = (costsint). Then the condition assumed for  is equivalent to sup u2R Z 2 0 ~(t)';

jcostcosu+ sintsinuj ;1
dt= sup u2R Z 2 0 ~(t)'; jcos(t;u)j ;1
dt (2.1) = sup_u 2R Z 2 0 ~(t)'; jsin(t;u)j ;1
dt <1:

Also, the conclusion of the lemma is equivalent to sup_u 2R Z 2 0 Z 2 0 ~(t)~(s)'; jcos(t;u);cos(s;u)j ;1
dtds = sup u2R Z 2 0 Z 2 0 ~(t)~(s)'; j2sin((t+s;2u)=2)sin((t;s)=2)j ;1
dtds <1:

Thus by (3) of the properties of ', it suces to show that sup_u 2R Z 2 0 ~(t)'; jsin(t=2;u)j ;1
dt <1:

This can be shown by applying (1), (4) of the properties of'and (2.1) as follows: sup u2R Z 2 0 ~(t)'; jsin(t=2;u)j ;1
dt Csup u2R Z 2 0 ~(t)'; j2sin((t;u)=2)cos((t;u)=2)j ;1
dt =Csup u2R Z 2 0 ~(t)'; jsin(t;u)j ;1
dt <1:

This completes the proof. Fork2Z, letIk = 2k2k

+1), where

Zdenotes the set of all integers.

Then T(f)(xz) = 1 X k=;1 Tkf(xz) where Tkf(xz) = Z Rn Ik(jyj)b(jyj)jyj ;n(y0)f(x ;yz;(jyjz))dy:

Let F be the Fourier transform acting on the x variable. Then it is

easy to see that

F(Tkf)(z) = Z Ik(jyj)b(jyj)jyj ;n (y0 )F(f)(z;(jyjz))e ;2 ihy
idy:

By applying the boundedness of the maximal operator as in 7], we

have Lemmas 2 and 3 below for the estimates of F(Tkf).

Lemma 2.

Let b2s for some s >2. Suppose that  is bounded on

Lr₍Rm) for all r >1. Then kF(Tkf)()kL2 (Rm) Cj2 k_ jkF(f)()kL2 (Rm)

where C is a constant independent of k2Z and 2Rn.

Proof. Note that

F(Tkf)(z) = Z Ikb(t) Ff(z;(tz)) Z Sn;1 (
)e;2 ith
id (
)dt=t: (2.2)

We de ne F(z) =Ff(z). Then by (1.1) and Holder's inequality we have jF(Tkf)(z)j Z Ik    b(t)F(z ;(tz)) Z Sn;1 (
); e;2 ith
i ;1
d (
)     dt=t Ckk 1 Z Ik jb(t)F(z;(tz))jjjdt Ckk 1 kbk s  2k    (jFj s0 )(z) 1=s 0 : By the L2=s 0 boundedness of , we have Z Rm jF(Tkf)(z)j 2 dz  1=2 Ckk 1 kbk s  2k        (jFj s0 ) 1=s 0     L2 (Rm) Ckk 1 kbk s  2k   kFkL2 (Rm)

which completes the proof of Lemma 2.

Lemma 3.

Letb 2s for somes >2 and let  satisfy (1:9). Suppose

that  is bounded on Lr(Rm) for all r > 1. Let 1 < q < 2s=(s+ 2).

Then kF(Tkf)()kL2 (Rm) C' ; j2 k_ j
;1=q 0 kF(f)()kL2 (Rm)

where C is a constant independent of k2Z and 2Rn.

Proof. Forqsatisfying 1< q <2s=(s+2), by using Holder's inequality in (2.2), we have jF(Tkf)(z)j Z Ik jb(t)F(z;(tz))j q _dt=t 1=q Lq 0
k Ckbk s ; (jFj sq=(s;q))(z)
(s;q)=(sq) Lq0
k

where F is as in the proof of Lemma 2 and

Lr
k= Z Ik     Z Sn;1 (
)e;2 ith
id (
)     r dt=t  1=r : Note that Lq0
k kk (q 0 ;2)=q 0 1 L 2=q 0 2
k :

A direct computation shows L 2 2
k = ZZ Sn;1 Sn ;1 (
) (!) " Z 2k +1 2k exp (;2ith
;!i) dt=t # d (
)d (!) C ZZ Sn;1 Sn ;1 j(
)(!)jmin ; 1j2 k h
;!ij ;1
d (
)d (!) =:I: By (1.7) we have I C ZZ Sn;1 Sn ;1 j(
)(!)j' ; jh
;! 0 ij ;1
(2;k jj ;1)d (
)d (!) C'(2 k jj) ;1 ZZ Sn;1 Sn ;1 j(
)(!)j' ; jh
;! 0 ij ;1
d (
)d (!):

Since  satis es (1.9), we have

L 2 2
k C'(2 k jj) ;1: (2.3) Therefore we have jF(Tkf)(z)jCkbk s'(2 kjj) ;1=q 0 ; (jFjsq= (s;q))(z)
(s;q)=(sq) 

which proves Lemma 3, since

 ( (jFj sq=(s;q)))(s;q)=(sq)   L2 (Rm) CkFkL2 (Rm):

When the function is independent of the z variable, we have the following two lemmas.

Lemma 4.

Suppose that the functionis independent of thezvariable:

(tz) =(t). Then j(Tkf)^(!)jCkbk 1 kk 1 j2 k_ jjf^(!)j

where C is a constant independent of k2Z, 2Rn and ! 2Rm.

Proof. It is easy to see that

(Tkf)^(!) = ^f(!)Ak(!) where Ak(!) = Z Ik(jyj)b(jyj)(y 0) jyj

;ne;2 i(hy
i+h(jyj)
!i)dy:

By (1.1) we have Ak(!) = Z Ik(jyj)b(jyj)(y 0) jyj ;ne;2 ih(jyj)
!i ; e;2 ihy
i ;1
dy

Lemma 5.

Suppose that the function is independent of the z vari-able. Let b 2s for some s2(12] and let  satisfy (1:9). Then

j(Tkf)^(!)jC' ; j2 k_ j
;1=s 0 jf^(!)j

where C is a constant independent of k2Z, 2Rn and ! 2Rm.

Proof. Let Ak(!) be as in the proof of Lemma 4 and Lr
k as in the

proof of Lemma 3. By Holder's inequality we have

jAk(!)jCkbk s Ls 0
k Ckbk s L 2=s 0 2
k 

which combined with the estimate (2.3) completes the proof of Lemma 5.

3. Proofs of Theorems 1 and 2

Let ! 2C 1 0 (

Rn) be a radial function supported in f : 1=2<jj

2g. We assume that 1 X j=;1 !(2j_)2 = 1 for all 6 = 0. LetTk be as in Section 2. Decompose

Tf =X j X k Sj +k(Tk(Sj+kf)) ! =X j Ujf

where the operator Sj is de ned by

F(Sjf)(z) =F(f)(z)!(2 j_) and Ujf = X k Sj +k(Tk(Sj+kf)):

By the Littlewood-Paley theory we have the following:

Lemma 6.

Let p2(11). Then k X k Skfk kLp (Rn +m ) C       X k jfkj 2 ! 1=2       Lp(Rn +m )  (3.1) where fk2S(Rn +m) andf

k= 0 for all but a nite number of values of k" also       X k jSkfj 2 ! 1=2       Lp(Rn +m ) CkfkLp (Rn +m ) (3.2)

where f 2S(Rn m).

Now, we prove Theorem 1. LetDj =f 2Rn : 2 ;j;1

jj2 ;j+1

g.

Then, by (3.1) with p= 2 and the Plancherel theorem we have

kUjfk 2 L2 (Rn +m ) C X k Z Rn +m jTk(Sj +kf)(yz) j 2 dy dz (3.3) =CX k Z Rm Z Rn  !(2j +k)   2 jF(Tkf)(z)j 2 d  dz C X k Z Dj+k Z Rm jF(Tkf)(z)j 2 dz  d:

Let j
0. Since the Lr boundedness of M implies the Lr

bound-edness of  (see 13]), by Lemma 2 and (3.3) we see that

kUjfk 2 L2 (Rn +m ) C X k Z Rm Z Dj+k jF(f)(z)j 2 j2 k_ j 2d ! dz C2 ;2j Z Rm X k Z Dj+k jF(f)(z)j 2 d ! dz C2 ;2j Z Rn +m jF(f)(z)j 2 d dz: Thus we have kUjfkL2 (Rn +m ) C2 ;j kfkL2 (Rn +m ) for j
0. (3.4)

Next, let j <0. Then, using Lemma 3 and (3.3), we have

kUjfkL2 (Rn +m ) C'(2 jjj);1=q 0 kfkL2 (Rn +m ) for 1< q <2s=(s+ 2): (3.5) Lets0 < r 2. We shall prove kUjfkLr (Rm +n ) CkfkLr (Rm +n ): (3.6)

By interpolating between the estimates (3.4), (3.6) and between (3.5), (3.6), we complete the proof of Theorem 1, sincekT(f)kp 

P

jkUjfkp.

By Lemma 6, to prove (3.6) it suces to show that

      1 X k=;1 jTkfkj 2 ! 1=2       Lr(Rm +n ) C       1 X k=;1 jfkj 2 ! 1=2       Lr(Rm +n ) for r2(s 02]: (3.7)

Lemma 7.

For h2u, u >1, 2L (Sn ), let Qh(f)(xz) = sup_k 2Z Z 2kjyj<2k +1 jh(jyj)(y 0) jjyj ;n jf(x;yz;(jyjz))jdy:

Suppose that M is bounded on Lr(Rm

+1) for all r > 1. Then Q

h is

bounded on Lp₍Rn

+m) for p > u0.

Proof. The proof is similar to that of 8, Lemma 2.4]. By Holder's inequality we have Qh(f)(xz)Ckhk u Z Sn;1 j(
)j  M
(jfj u0 )(xz) 1=u 0 d (
) where M
(F)(xz) = sup_k 2Z 2;k Z 2k +1 2k jF(x;t
z;(tz))jdt:

In 7], it was proved that the boundedness of M implies

kM
(g)kLp (Rn +m ) CkgkLp (Rn +m ) for allp >1 with C independent of
2Sn ;1. Therefore, if p > u0, kQh(f)kLp (Rn +m ) Ckhk u     Z Sn;1 j(
)j  M
(jfj u0 )(xz) 1=u 0 d (
)     Lp(Rn +m ) Ckhk u Z Sn;1 j(
)j   M
( jfj u0 )   1=u 0 Lp=u0 (Rn +m ) d (
) Ckhk u kk 1 kfkLp (Rn +m ):

This completes the proof.

Now, we can prove (3.7) by applying the method of 7]. We have the following estimates:    sup k2Z jTkfkj     Lr(Rn +m ) C    sup k2Z jfkj     Lr(Rn +m ) for all r > s0 (3.8)       X k2Z jTkfkj r ! 1=r       Lr(Rn +m ) C       X k2Z jfkj r ! 1=r       Lr(Rn +m ) for allr > s0 (3.9) since supk2Z jTkfkj  Qb(supk 2Z jfkj), jTkfkj  Qb(jfkj) and Qb is bounded on Lr₍Rn

+m) for r > s0 by Lemma 7, where Q

b is de ned

as in Lemma 7 by using the functions b,  and of Theorem 1. Thus, (3.7) follows by interpolating between the estimates (3.8) and (3.9). This completes the proof of (3.7) and hence the proof of Theorem 1.

Now, we turn to the proof of Theorem 2. Let Vj be de ned in the

same way as Uj in the proof of Theorem 1 by using the functionsb, 

and satisfying the conditions assumed in Theorem 2. Then, arguing as in the proof of Theorem 1 and using Lemmas 4 and 5, we can get estimates similar to (3.4) and (3.5)" that is,

kVjfkL 2 (Rn +m ) C2 ;j kfkL 2 (Rn +m ) for j
0. (3.10) kVjfkL2 (Rn +m ) C'(2 jjj);1=s 0 kfkL2 (Rn +m ) for j <0. (3.11) Also we have kVjfkLr (Rm +n ) CkfkLr (Rm +n ) for j1=r;1=2j<1=s 0. (3.12)

This follows from Lemma 6 and the estimate

      X k jTkfkj 2 ! 1=2       Lr(Rn +m ) C       X k jfkj 2 ! 1=2       Lr(Rn +m ) for j1=r;1=2j<1=s 0, where T

k is as in the proof of Theorem 1, with

everything adapted for the present case. The proof of this estimate is essentially the same as that of (3.14) of 8] (see also 5, Theorem 7.5]). Applying interpolation to the estimates (3.10){(3.12) and noting

kT(f)kp  P

jkVjfkp, we complete the proof of Theorem 2.

4. Marcinkiewicz integral analog

We can also consider the Marcinkiewicz integral

(f)(x) = Z 1 0     Z jyjt (y0 )jyj ;n+1b (jyj)f(x;y)dy     2 t;3dt ! 1=2 :

The following theorem related to the above operator is from Theorem 1 in 11].

Theorem C

(11])

.

Let fK j tg ;1<j0 be a sequence of functions on Rn(01) such that kK j tk 1 C2j, supp(K j t)ft2j ;1 jxjt2j +1 g

for all t >0 and ;1< j 0, and such that the operator

(f)(x) = 0 @ Z 1 0      0 X j=;1 K_tj f(x)      2 t;1dt 1 A 1=2 is bounded on L2( Rn). Suppose that

(a) for each non-positive integerj, the maximal operatorsupt>0    Kj t   f(x)  

(b) there exists some positive constant  > 1=2 such that for any non-positive integer j and t >0

  K^ j t()    C2 j; log(2 +jt2 j_ j)
; :

Then the operator  is bounded on Lp₍Rn) for 4=(4;1)< p < 4.

Now we write (f)(x) = 0 @ Z 1 0      0 X j=;1 Kj
t f(x)      2 t;1dt 1 A 1=2 with Kj 
t(x) =t ;1 jxj ;n+1b( jxj)(x 0) ft2j ;1< jxjt2jg(x):

For simplicity in the discussion, we only study the case b2 2.

Theorem 6.

Let n = 2, b 2 

2. If a function  is homogeneous of

degree 0 on Rn and satis es (1:1) and (1:3) with  > 1, then   is

bounded on Lp₍Rn) for all 2 p < 2. If we further assume that b is

bounded, then  is bounded on L

p₍Rn) for all 2=(2;1)< p <2.

Proof. First, it is easy to see that

kK j 
t k 1 C2 j kbk  1 kk 1 C2 j kbk  2 kk 1:

Second, using the argument in Lemmas 3 and 5, we nd that

  K^ j 
t()    C2 j=2 kbk  2L 2
j
t where (L2
j
t) 2 = ZZ Sn;1 Sn ;1 (
) (!) " t;1 Z t 2j t2j ;1 exp (;2irh
;!i) dr # d (
)d (!):

Thus usingt2j _{to replace 2}k _{in (2.3), we obtain that}

  K^ j 
t()    C2j ; log(2 +jt2jj)
;=2 : The L2 boundedness of

 is obvious from this estimate and (1.1). We

omit the detail. Therefore, by combining the results and checking the proof of Theorem C, we can get the rst part of the conclusion of The-orem 6 (the boundedness of the maximal operator supt>0

   Kj t   f(x)  

is not needed for this case). It is easy to see that the maximal operator supt>0    Kj t   f(x)  is bounded onLp(

Rn) with boundC(np)2j for all

1< p < 1if b is bounded. Thus the second part of the conclusion of

We refer to 17] for related results on the Marcinkiewicz integrals (see also 15]).

References

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Department of Mathematics, University of Wisconsin-Milwaukee, Milwaukee, Wisconsin 53201, U.S.A.

E-mail address: [email protected]

Department of Mathematics, Faculty of Education, Kanazawa Uni-versity, Kanazawa 920-1192, Japan