Estimates for singular integrals and extrapolation

extrapolation

著者 Sato Shuichi journal or publication title Studia mathematica volume 192 number 2009 page range 219-233 year 2009-01-01 URL http://hdl.handle.net/2297/23947 doi: 10.4064/sm192-3-2

ESTIMATES FOR SINGULAR INTEGRALS AND

EXTRAPOLATION

SHUICHI SATO

Abstract. In this note, we study singular integrals with rough kernels, which belong to a class of singular Radon transforms. We prove certain estimates for the singular integrals that are useful in an extrapolation argument. As an application, we prove L

p boundedness of the singular

integrals under a certain sharp size condition on their kernels.

1. Introduction

Let  be a function in L1(Sn;1) satisfying Z

Sn;1

( )d
( ) = 0

(1.1)

where d
denotes the Lebesgue surface measure on the unit sphere Sn;1 in Rn. In this note we assume n 2. For s 1, let s denote the collection

of measurable functions h onR + = ft 2R :t >0g satisfying khk s = sup j2Z Z 2j +1 2j jh(t)j s_dt=t ! 1=s <1

where Zdenotes the set of integers. We note that s t if s > t. In this

note we always assume h 2

1. Let P(y) = (P1(y)P2(y):::Pd(y)) be a

polynomial mapping, where each Pj is a real-valued polynomial onR

n_{. We} consider a singular integral operator of the form:

T(f)(x) = p:v:Z Rn f(x;P(y))K(y)dy= lim
!0 Z jyj>
f(x;P(y))K(y)dy (1.2)

for an appropriate function f on Rd, where K(y) = h(jyj)(y 0)

jyj

;n, y0 = jyj

;1y. Then, T(f) belongs to a class of singular Radon transforms. See

Stein 20], Fan-Pan 8] and Al-Salman-Pan 1] for this singular integral. When h = 1 (a constant function), n = d and P(y) = y, we also write

T(f) = S(f). Let ^f() = R

Rdf(x)e

;2ihxidx be the Fourier transform

2000 Mathematics Subject Classi cation. Primary 42B20, 42B25.

Key Words and Phrases. Singular integrals, singular Radon transforms, maximal functions, extrapolation.

of f, where hi denotes the inner product in Rd. 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
( ):

Using this, we can show that S extends to a bounded operator on L2 if

 2 LlogL(Sn

;1), where LlogL(Sn;1) denotes the Zygmund class of all

those functions  on Sn;1 which satisfy Z

Sn;1

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

Furthermore, if  2LlogL(Sn

;1), by the method of rotations of

Calderon-Zygmund (see 2]) it can be shown that S extends to a bounded operator on Lp _{for allp}2(11).

When n=dand P(y) =y, R. Feerman 10] proved that ifh is bounded and  satises a Lipschitz condition of positive order on Sn;1, then the

singular integral operator T in (1.2) is bounded on Lp _{for 1 < p <} 1.

Namazi 13] improved this result by replacing the Lipschitz condition by the condition that  2 Lq(Sn

;1) for some q > 1. In 7], Duoandikoetxea

and Rubio de Francia developed methods which can be used to study map-ping properties of several kinds of operators in harmonic analysis including the singular integrals considered in 13]. Also, see 6, 25] for weighted Lp boundedness of singular integrals, and 21, 22] for background materials.

For the rest of this note we assume that the polynomial mapping P in (1.2) satises P(;y) = ;P(y) and P 6= 0. We shall prove the following

result:

Theorem 1.

Let 2Lq(Sn ;1), q

2(12] and h2 s, s 2(12]. Suppose

 satises (1:1). Let T be as in (1:2). Then we have

kT(f)kLp (Rd) Cp(q;1) ;1(s ;1) ;1 kkLq (Sn ;1 ) khk s kfkLp (Rd)

for all p 2 (11), where the constant Cp is independent of qs and h.

Also, the constant Cp is independent of polynomials Pj if we x deg(Pj) (j = 12:::d).

See 15, 16, 17] for relevant results. In Al-Salman-Pan 1], theLp bounded-ness ofT was proved under the condition that  is a function inLlogL(Sn;1)

satisfying (1.1) and h 2 s for some s > 1 (1, Theorem 1.3]). Also it is

noted there that estimates like those in Theorem 1 (with s being xed) can be used to prove the same result by applying an extrapolation method, but such estimates are yet to be proved (see 1, p. 156]). In 1], the authors

also considered singular integrals dened by certain polynomial mappings

P which do not satisfy the condition P(;y) =;P(y).

As a consequence of Theorem 1 we can give a dierent proof of 1, The-orem 1.3] via an extrapolation method in fact, we can get an improved result. For a positive number a and a function h onR

+, let La(h) = sup_j 2Z Z 2j +1 2j jh(r)j(log(2 +jh(r)j)) a _dr=r:

We dene a class La to be the space of all those measurable functions h on R + which satisfy La(h)< 1. Also, let Na(h) = X m1 ma₂m_dm(h) wheredm(h) = supk2Z2 ;k jE(km)jwithE(km) =fr2(2k2k +1] : 2m;1 < jh(r)j  2mg for m 2, E(k1) = fr 2 (2k2k +1] : 0 < jh(r)j  2g. We

denote by Na the class of all those measurable functionsh onR

+ such that

Na(h) < 1. Then we readily see that Na(h) < 1 implies La(h) < 1.

Conversely, if La+b(h)<

1 for some b > 1, then Na(h)< 1. To see this,

note that 2m_ma+b 2;k jE(km)jC Z E(km) jh(r)j(log(2 +jh(r)j)) a+b dr=r CLa +b(h) for m 2 thus Na(h) 2d 1(h) +CLa+b(h) P m2m ;b < 1. By Theorem

1 and an extrapolation method we have the following:

Theorem 2.

Suppose  is a function inLlogL(Sn;1) satisfying (1:1) and

h 2N 1. Let T be as in (1:2). Then kT(f)kLp (Rd) CpkfkLp (Rd)

for all p2(11), where Cp is independent of polynomials Pj if the

polyno-mials are of xed degree.

By Theorem 2 and the remark preceding it we see that T is bounded on

Lp _{for all p} 2 (11) if  is as in Theorem 2 and h 2 La for some a > 2.

When n=d, P(y) =y,  is as in Theorem 2 and his a constant function, it is known that T is of weak type (11) see 5, 18, 4, 9, 11, 12, 19, 23, 24]. In Section 2, we shall prove Theorem 1. Applying the methods of 7] involving the Littlewood-Paley theory and using results of 8, 14], we shall prove Lp _{estimates for certain maximal and singular integral operators} re-lated to the operator T in Theorem 1 (Lemmas 1 and 2). Lemma 1 is used to prove Lemma 2. By Lemma 2 we can easily prove Theorem 1. A key

idea of the proof of Theorem 1 is to apply a Littlewood-Paley decompo-sition adapted to a suitable lacunary sequence depending on q and s for which  2 Lq(Sn

;1) and h

2 s. The method of appropriately choosing a

lacunary sequence was inspired by 1], where, in a somewhat dierent way from ours, a similar method was used to study several classes of singular integrals.

We shall prove Theorem 2 in Section 3. Finally, in Section 4, we consider the maximal operator

T (f)(x) = sup_N
> 0     Z
<jyj<N f(x;P(y))K(y)dy     (1.3)

where P and K are as in (1.2). We shall prove analogs of Theorems 1 and 2 for the operator T.

Throughout this note, the letter C will be used to denote non-negative constants which may be dierent in dierent occurrences.

2. Proof of Theorem 1

Let h be as in Theorem 1. We consider the singular integral T(f) dened in (1.2). Let 2 and Ek = fx 2 Rn : k < jxj  k

+1

g. Then

T(f)(x) = P 1 ;1
k

f(x), where f
kg is a sequence of Borel measures on Rd such that
kf(x) = Z Ekf(x ;P(y))K(y)dy: (2.1) We note that (
kf)^() = ^f() Z Eke ;2ihP(y)iK(y)dy: We write P(y) = X` j=1 Qj(y) Qj(y) = X jj=N(j) ay (a 2R d<sub>)</sub> where Qj 6= 0, 1  N(1) < N(2) <  < N(`),  = ( 1:::n) is a multi-index, y _{= y}1 1 :::y n n and jj = 1+ +n. Let m = N (m) and m = (q;1)(s;1)=(2qsN(m)) for 1m`. Put P (m)(y) = Pm j=1Qj(y)

and dene a sequence (m) = f (m) k g of positive measures on Rd by (m) k f(x) = Z Ekf ; x;P (m)(y)  jK(y)jdy

for m = 12:::`. Also, dene (0) = f (0) k g by  (0) k = ( R EkjK(y)jdy),

where  is Dirac's delta function on Rd. For a sequence  =fkg of nite

Borel measures on Rd, we dene the maximal operator 

supkjjkjf(x)j, where jkjdenotes the total variation. We consider maxi-mal operators ; (m)   (0m`). We also write ; (`)   = . Let Lj() = (ha (j1) iha (j2) i:::ha (jrj) i) where f(jk)g rj k=1 is an enumeration of fg jj=N(j) for 1  j  `. Then

Lj is a linear mapping from Rd to Rrj. Let sj = rankLj. There exist

non-singular linear transformations Rj :Rd !Rd and Hj :Rsj !Rsj such

that

jHjds

jRj()jjLj()jCjHjds

jRj()j

where dsj() = (1:::sj) is the projection and C depends only on rj (a

proof can be found in 8]). Let f
(m)

k g(0 m `) be a sequence of Borel

measures on Rd such that

(m) k f(x) = Z Ekf ; x;P (m)(y)  K(y)dy form= 12:::`, while
(0) k = 0. Let'2C 1 0 ( R) be supported infjrj1g

and '(r) = 1 for jrj < 1=2. Dene a sequence  (m) = f (m) k g of Borel measures by ^ (m) k () = ^
(m) k ()km();
^ (m;1) k ()km;1() (2.2) for m = 12:::`, where km() = ` Y j=m+1 ' kjjHjds jRj()j  if 0m `;1 and k` = 1. Then
k=
(`) k = P` m=1 (m) k . We note that km()'; _kmjHmds mRm()j  = km;1() (1 m`): For 1 m`, let T (m) (f) = P k(m) k f. Then T = P` m=1T (m) . For p 2 (11) we put p 0 = p=(p ; 1) and (p) = j1=p;1=p 0 j. Let 2(01). Then we have the followingLp estimates for (

(m)) and T (m)

.

Lemma 1.

For p >1 + and 0j `, we have  ( (j) ) (f)  Lp(Rd) C(log )kkLq (Sn ;1 ) khk s  1; ;=(2q 0s0 )  ;2=p kfkLp (Rd): (2.3)

Lemma 2.

For p2(1 + (1 + )= ) and 1m`, we have kT (m) (f)kLp (Rd) C(log )kkLq (Sn ;1 ) khk s  1; ;=(2q 0s0 )  ;1;(p) kfkLp (Rd):

The constants C in Lemmas 1 and 2 are independent of qs2(12], 2

Lq_(Sn;1),h

2 s, and the coecients of the polynomialsPk(1k d).

We prove Lemma 2 rst, taking Lemma 1 for granted for the moment. Let

A = (log )kkLq (Sn ;1 ) khk s and B = ; 1; ;m m  ;1 = ; 1; ;=(2q 0s0 )  ;1 . Then, we have the following estimates:

k (m) k kc 1A ( k (m) k k=j (m) k j(Rd)) (2.4) j^ (m) k ()jc 2A ; kmjLm()j  ;m  (2.5) j^ (m) k ()jc 3A ; k+1 m jLm()j m  (2.6)  ( (m))(f)   p CpAB 2=p kfkp for p >1 +  (2.7)

for some constantsci(1i3) andCp, where we simply writekfkLp (Rd) = kfkp.

Now we prove the estimates (2.4){(2.7). First we see that

k (m) k kC  k
(m) k k+k
(m;1) k k  (2.8) Ckk 1 Z k+1 k jh(r)jdr=r C(log )kk 1 khk 1:

From this (2.4) follows. To prove (2.5), dene

F(r) = Z Sn;1 ( )exp(;2ihP (m)(r ) i)d
( ):

Then, via Holder's inequality, for s2(12] we see that j
^ (m) k ()j=      Z k +1 k h(r)F(r)dr=r      (2.9)  Z k +1 k jh(r)j s_dr=r ! 1=s Z k +1 k jF(r)j s0 dr=r ! 1=s 0 C(log ) 1=s khk s kk (s 0 ;2)=s 0 1 Z k +1 k jF(r)j 2 dr=r ! 1=s 0 :

We need the following estimates for the last integral:

Lemma 3.

Let 1 < q  2 and  2 Lq(Sn

;1). Then there exists a

con-stant C > 0 independent of q  and the coecients of the polynomial components of P(m) such that

Z k+1 k jF(r)j 2 dr=r C(log ) ; kmjLm()j  ;1=(2q 0N (m)) kk 2 q:

Proof. Take an integer  such that 2 _< 2 . By the proof of Propo-sition 5.1 of 8] we have Z k +1 k jF(r)j 2 dr=r = Z 1  F( kr)   2 dr=r  X j=0 Z 2j +1 2j  F( kr)   2 dr=r  X j=0 C Z 2 1  F(2j kr)  q 0 dr=r 2=q 0  X j=0 C; 2jN(m) kN(m) jLm()j  ;1=(2N(m)q 0 ) kk 2 q C(log ) ; kN(m) jLm()j  ;1=(2N(m)q 0 ) kk 2 q: This completes the proof of Lemma 3.

By (2.9) and Lemma 3 we have j
^ (m) k ()j  CA ; kmjLm()j  ;m . Also, we have k
(m;1)

k kCA by (2.8). We can prove the estimate (2.5) by using

these estimates in the denition of (m)

k in (2.2) and by noting that ' is compactly supported. Next, to prove (2.6), using (1.1) when m = 1, we see that j^ (m) k ()j      ^
(m) k ();
^ (m;1) k ()  km()   +   (km() ;km ;1()) ^
(m;1) k ()    Ckk 1 k+1 m jLm()j Z k +1 k jh(r)jdr=r+Ck
(m;1) k kkmjLm()j C(log )kk 1 khk  1 k+1 m jLm()j

where to get the last inequality we have used (2.8). By this and (2.8), we have j^ (m) k ()jC(log )kk 1 khk  1 ; k+1 m jLm()j c

for allc2(01], which implies (2.6). Finally, the estimate (2.7) follows from

Lemma 1 since  ( (m) ) (f)  p C  ( (m) ) (jfj)   p+C  ( (m;1) ) (jfj)   p CAB 2=p kfkp

for p > 1 + , where the rst inequality can be seen by change of variables and a well-known result on maximal functions (see 8]).

Let fkg 1

;1 be a sequence of non-negative functions in C 1((0 1)) such that supp(k) ;k;1 m ;k+1 m ], X k k(t) 2 = 1 j(d=dt) j_k(t) jcj=t j _{(j = 12:::) for all t >0}

where the constantscj are independent ofm (this is feasible sincem 2). Dene an operator Sk by (Sk(f))^() =k; jHmds mRm()j  ^ f() and let V(m) j (f) = 1 X ;1 Sj+k  (m) k Sj +k(f)  :

Then by Plancherel's theorem and the estimates (2.4){(2.6) we have

  V (m) j (f)    2 2  X k C Z D(j+k) j^ (m) k ()j 2 jf^()j 2d (2.10) CA 2 min; 1;2m(jjj;2) m  X k Z D(j+k) jf^()j 2d CA 2 min; 1;2m(jjj;2) m  kfk 2 2 where D(k) =f ;k;1 m jHmdsmRm()j ;k+1 m g.

Applying the proof of Lemma on 7, p. 544] and using the estimates (2.4) and (2.7), we can prove the following.

Lemma 4.

Let u 2 (1 + 2]. Dene a number v by 1=v;1=2 = 1=(2u).

Then we have the vector valued inequality

     X j (m) k gkj 2  1=2     v (c 1Cu) 1=2AB1=u      X jgkj 2  1=2     v (2.11)

where the constants c1 and Cu are as in (2:4) and (2:7), respectively.

By the Littlewood-Paley theory we have

kV (m) j (f)kp cp       X k j (m) k Sj +k(f) j 2 ! 1=2       p  (2.12)       X k jSk(f)j 2 ! 1=2       p cpkfkp (2.13)

where 1< p <1andcp is independent ofmand the linear transformations

RmHm. Suppose that 1+ < p4=(3; ). Then we can ndu2(1+ 2]

such that 1=p= 1=2 + (1; )=(2u). Let v be dened by u as in Lemma 4.

Then by (2.11){(2.13) we have kV (m) j (f)kv CAB 1=u kfkv: (2.14)

Since 1=p = =2 + (1; )=v, interpolating between (2.10) and (2.14), we

have kV (m) j (f)kp CAB (1;)=umin ; 1;m(jjj;2) m  kfkp:

It follows that kT (m) (f)kp  X j kV (m) j (f)kp CAB (1;)=u(1 ; ;m m );1 kfkp (2.15) CAB 2=p kfkp

where we have used the inequalityP

min 1;m(jjj;2) m  5 ; 1; ;m m  ;1 . We also have kT (m) (f)k 2  P kV (m) j (f)k 2  CABkfk 2 by (2.10), since B (1; ;m m );1

. By duality and interpolation, we can now get the con-clusion of Lemma 2.

Next, we give a proof of Lemma 1. We prove Lemma 1 by induction on

j. Now we assume (2.3) for j = m ;1, 1  m  `, and prove (2.3) for

j = m. Let ' 2 C 1 0 (

R) be as above. Dene a sequence  (m) = f (m) k g of Borel measures on Rd by ^ (m) k () = ' ; kmjHmdsmRm()j  ^ (m;1) k (): Then by (2.3) with j =m;1, we have

 ( (m))(f)   p C  ( (m;1))(f)   p CAB 2=p kfkp (2.16)

for p >1 + . Furthermore, we have the following:

k (m) k k+k (m) k kCk (m;1) k k+k (m) k kCkk 1 Z k+1 k jh(r)jdr=r (2.17) C(log )kk 1 khk 1 CA j^ (m) k ();^ (m) k ()jC(log )kk 1 khk 1 ; k+1 m jLm()j m (2.18) CA ; k+1 m jLm()j m  j^ (m) k ()jCA ; kmjLm()j  ;m  (2.19) j^ (m) k ()jC(log )khk 1 kk 1 ; kmjLm()j  ;m CA ; kmjLm()j  ;m : (2.20)

To see (2.18) we note that

j^ (m) k ();^ (m) k ()j j^ (m) k ();^ (m;1) k ()j+    ; '; _kmjHmdsmRm()j  ;1  ^ (m;1) k ()   :

Thus arguing as in the proof of (2.6), we have the rst inequality of (2.18). The estimate (2.19) follows from the arguments used to prove (2.5). Also, we can see the rst inequality of (2.20) by the denition of (m)

Sincek(m ) (f)k 1

CAkfk

1, by taking into account an interpolation,

it suces to prove (2.3) with j = m for p 2 (1 + 2]. Dene a sequence

(m) = f (m) k g of Borel measures by  (m) k =(m) k ; (m) k . Let gm(f)(x) =  X    (m) k f(x)    2 1=2 : Then ((m))(f) gm(f) + ( (m))( jfj): (2.21)

Thus, by (2.16), to get (2.3) with j = m it suces to prove kgm(f)kp 

CAB2=p

kfkp for p2 (1 + 2] with an appropriate constant C. By a

well-known property of Rademacher's functions, this follows from

 U (m)
(f)   p CAB 2=p kfkp (2.22) for p 2(1 + 2], where U (m)
(f) = P kk(m) k f with  = fkg, k = 1 or ;1, and the constant C is independent of .

The estimate (2.22) is a consequence of the following:

Lemma 5.

We dene a sequence fpjg 1 1 by p

1 = 2 and 1=pj+1 = 1=2 +

(1; )=(2pj) for j 1. (We note that 1=pj = (1;aj)=(1 + ), where

a = (1; )=2, so fpjgis decreasing and converges to 1+ .) Then, for j 1

we have  U (m)
(f)   pj CjAB 2=pj kfkp j: Proof. Let U(m) j (f) = 1 X k=;1 kSj+k  (m) k Sj +k(f)  :

Then by Plancherel's theorem and the estimates (2.17){(2.20), as in (2.10) we have   U (m) j (f)    2 CAmin ; 1;m(jjj;2) m  kfk 2: (2.23) It follows that   U (m)
(f)    2  P jkU (m) j (f)k 2 CABkfk 2. If we denote by

A(s) the assertion of Lemma 5 forj =s, this proves A(1).

Now we derive A(s + 1) from A(s), which will complete the proof of Lemma 5 by induction. Using (2.21), we see that

((m) ) (f)( (m) ) (jfj) + ( (m) ) (jfj)gm(jfj) + 2( (m) ) (jfj):

Note that A(s) implies kgm(f)kps  CAB 2=ps kfkps. By this and (2.16) we have  ( (m))(f)   ps kgm(jfj)kps + 2  ( (m))( jfj)   ps CAB 2=ps kfkps: (2.24)

By (2.17), (2.23) and (2.24), we can now apply the arguments used in the proof of (2.15) to get A(s+ 1). This completes the proof of Lemma 5.

Now we prove (2.22) for p 2 (1 + 2]. Let fpjg 1

1 be as in Lemma 5.

Then we have pN+1 < p

 pN for some N. Thus, interpolating between

the estimates of Lemma 5 for j =N and j =N + 1, we have (2.22). This proves (2.3) for j =m.

Finally, we can easily see that ((0))(f)

 C(log )kk 1 khk  1 jfj (see

(2.17)), which implies the estimate (2.3) for j = 0. Therefore, by induction we have (2.3) for all 0 j `. This completes the proof of Lemma 1.

Now we can prove Theorem 1. Since 2 (01) is arbitrary, by taking

= 2q0s0 in Lemma 2 we have kT (m) 2q 0s0(f) kp Cp(q;1) ;1(s ;1) ;1 kkqkhk s kfkp

for all p 2 (11). This completes the proof of Theorem 1, since T = P`

m=1T (m)

.

3. Proof of Theorem 2

Theorem 2 can be proved by Theorem 1 and an extrapolation argument. Let T(f) be the singular integral in (1.2). We also write T(f) = Th (f).

We x q2(12], 2Lq(Sn ;1), p

2(11) and a function f with kfkp 1

and put S(h) =kTh (f)

kp. Then we have the subadditivity:

S(h+k)S(h) +S(k): (3.1) Set E1 = fr 2 R + : 0< jh(r)j  2g and Em = fr 2 R + : 2 m;1 < jh(r)j 

2mg for m 2. Then, applying Theorem 1, we see that

S(hEm)C(q;1) ;1(s ;1) ;1 kkqkhEmk s (3.2)

for s 2 (12], where E denotes the characteristic function of a set E.

Now we follow the extrapolation argument of Zygmund 26, Chap. XII, pp. 119{120]. First, note that

khEmk  1+1=m 2mdm= (m+1) m (h)

for m 1, where dm(h) is as in Section 1. Using this and (3.2) we see that

X m1 S(hEm)C(q;1) ;1 kkq X m1 mkhEmk  1+1=m C(q;1) ;1 kkq X m1 m2m_dm=(m+1) m (h):

Recalling the denition of Na(h), we have X m1 m2m_dm=(m+1) m (h) = X dm(h)<3 ;m m2m_dm=(m+1) m (h) + X dm(h)3 ;m m2m_dm=(m+1) m (h)  X m1 m2m₃;m 2= (m+1)+ X m1 m2m_dm(h)3m=(m+1) C(1 +N 1(h)):

Therefore, by (3.1) we see that

S(h) X m1 S(hEm)C(q;1) ;1 kkq(1 +N 1(h)): (3.3) Next, x h 2 N 1, p

2 (11) and f with kfkp  1 and let R() = kTh

(f)

kp. Put em =
(Fm) for m 1, where Fm = f 2 Sn

;1 : 2m;1 < j( )j 2mg for m 2 and F 1 = f 2 Sn ;1 : j( )j 2g. We decompose  as  = P 1 m=1m, where m = Fm ;
(Sn ;1);1 R Fmd
. We note that R md
= 0,kmkr C2me 1=r

m for 1< r <1. Now, by (3.3) and the

subadditivity of R() we see that

R() X m1 R(m)C(1 +N 1(h)) X m1 mkmk 1+1=m C(1 +N 1(h)) X m1 m2m_em=(m+1) m =C(1 +N1(h)) 0 @ X em<3 ;m + X em3 ;m 1 A C(1 +N 1(h)) X m1 m2m₃;m 2= (m+1)+ X m1 m2m_em3m=(m+1) ! C(1 +N 1(h)))  1 + Z Sn;1 j( )jlog(2 +j( )j)d
( ) :

This completes the proof of Theorem 2.

4. Estimates for maximal functions

For the maximal operatorT in (1:3) we have a result similar to Theorem

1.

Theorem 3.

Let q 2(12], s 2(12] and  2Lq(Sn ;1), h

2 s. Suppose

 satises (1:1). Then we have

kT (f) kLp (Rd) Cp(q;1) ;1(s ;1) ;1 kkLq (Sn ;1 ) khk s kfkLp (Rd)

As Theorem 1 implies Theorem 2, we have the following theorem as a consequence of Theorem 3.

Theorem 4.

Let  be a function in LlogL(Sn;1) satisfying (1:1) and h 2 N 1. Then kT (f) kLp (Rd) CpkfkLp (Rd) for all p2(11).

As in the cases of Theorems 1 and 2, the constants Cp of Theorems 3 and 4 are also independent of polynomialsPj if we x deg(Pj) (j = 12:::d). When  is as in Theorem 4 and h2 s for somes >1, theLp boundedness

of T was proved in 1]. When n =d,P(y) =y, 

2Lq for someq >1 and

h is bounded, the Lp _{boundedness of T} is due to 3].

We use the following result to prove Theorem 3.

Lemma 6.

Let (m) = f

(m)

k g(1m`), where the measures  (m)

k are as in(2:2). Let 2(01) and let positive numbersA= (log )kkLq

(Sn ;1 ) khk s, B =; 1; ;m m  ;1 be as above. We dene T m(f)(x) = sup_k 2Z      1 X j=k (m) j f(x)      : (4.1) Then, for p2(2(1 + )=( 2 ; + 2)(1 + )= ) =:I we have kT  m(f)kp CA ; B1+(p)+B2=p+1;=2  kfkp where C is independent of qs 2 (12],  2 Lq(Sn ;1), h 2 s, and the

coecients of the polynomials Pj (1j d).

Proof. LetT(m)

(f) =P

k(m)

k fbe as in Lemma 2. Let a function'be as in

the denition of (m)

k in (2.2). Dene 'k by ^'k() =';

kmjHmdsmRm()j 

. Let  be the delta function as above. Following 8], we decompose

1 X j=k (m) j f ='kT (m) (f);'k k;1 X j=;1 (m) j f ! +(;'k) 1 X j=k (m) j f ! : It follows that T m(f)sup k  ' kT (m) (f)  + 1 X j=0 N(m) j (f) (4.2) whereN(m) j (f) = supk   'k   (m) k;j;1 f   +supk   ( ;'k)  (m) j+k f   . By Lemma 2 we have    sup k  ' kT (m) (f)       p CAB 1+(p) kfkp for p2(1 + (1 + )= ). (4.3)

Also, by (2.7) we see that kN (m) j (f)kr CAB 2=r kfkr for r >1 + . (4.4)

On the other hand, we have

N(m) j (f) X k   (;'k)  (m) j+k f    2 ! 1=2 + X k   'k  (m) k;j;1 f    2 ! 1=2 :

Therefore, by the estimates (2.5), (2.6) and Plancherel's theorem, we get

kN (m) j (f)k 2 CA ;mj m ; 1; ;2m m  ;1=2 kfk 2 (4.5)

(see 8, p. 820]). For p 2 I we can nd r 2 (1 + 2(1 + )= ) such that

1=p= (1; )=r+ =2, so an interpolation between (4.4) and (4.5) implies

that kN (m) j (f)kp CAB 2(1;)=r ; 1; ;2m m  ;=2 ;mj m kfkp: (4.6)

Therefore, by (4.2), (4.3) and (4.6), for p2I we have kT  m(f)kp CA  B1+(p)+B2(1;)=r+1 ; 1; ;2m m  ;=2  kfkp:

This implies the conclusion of Lemma 6, since (1; ;2m

m );1

 B and

2(1; )=r+ =2 + 1 = 2=p+ 1; =2.

Proof of Theorem 3. Note that T(f)  2T  0(f) + 2  (jfj), where T  0(f) is

dened by the formula in (4.1) with f (m)

j g replaced by the sequence f
jg

of measures in (2.1) and 

= ((`)) is as in Lemma 1. We note that

T 0(f)  P` m=1T 

m(f). Now, Lemma 6 implies that

kT  m(f)kp C(log )  1; ;=(2q 0s0 )  ;3 kkqkhk s kfkp

for p2I. By using this with = 2

q0s0

, since 2(01) is arbitrary, we can

conclude that kT  2q 0s0 m(f)kp Cp(q;1) ;1(s ;1) ;1 kkqkhk s kfkp

for p 2 (11). Also, by Lemma 1  

satises a similar estimate when

= 2q0s0

. Collecting results, we have Theorem 3. Remark. Let M(f)(x) = sup_t> 0 t;n Z jyj<t jf(x;P(y))jj(y 0) jjh(jyj)jdy:

It is easy to see that M(f)  C 

(f), where C is independent of 2. Therefore, by Lemma 1 we can prove results similar to Theorems 1 and 2 for the maximal operatorM. In 1],Lp_{boundedness ofM was proved under the} condition that  2LlogL(Sn

;1) andh

P(y) = y, it is known that M is of weak type (11) if  2 LlogL(Sn )

and h is bounded (see 5, 4]).

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Department of Mathematics, Faculty of Education, Kanazawa Univer-sity, Kanazawa 920-1192, Japan