1 Introduction

As the development of singular integral operators (see [1, 2]), their commutators have been well studied. In [3, 4], the authors prove that the commutators generated by the singular integral operators and BMO functions are bounded on L p ( R n ) for 1<p<. Chanillo (see [5]) proves a similar result when singular integral operators are replaced by the fractional integral operators. In [6, 7], some Toeplitz-type operators related to the singular integral operators and strongly singular integral operators are introduced, and the boundedness for the operators generated by BMO and Lipschitz functions is obtained. In [6], some singular integral operators satisfying a variant of Hörmander’s condition are introduced, and the boundedness for the operators is obtained (see [6], [20]). In this paper, we prove the sharp maximal function inequalities for the Toeplitz-type operator related to some singular integral operators satisfying a variant of Hörmander’s condition. As an application, we obtain the weighted boundedness of the Toeplitz-type operator on Lebesgue and Morrey spaces.

2 Preliminaries

First, let us introduce some notations. Throughout this paper, Q will denote a cube of R n with sides parallel to the axes. For any locally integrable function f, the sharp maximal function of f is defined by

f # (x)= sup Q x 1 | Q | Q | f ( y ) f Q | dy,

where, and in what follows, f Q = | Q | 1 Q f(x)dx. We say that f belongs to BMO( R n ) if f # belongs to L ( R n ) and define f BMO = f # L . It has been known that (see [2])

f f 2 k Q BMO Ck f BMO .

Let M be the Hardy-Littlewood maximal operator defined by

M(f)(x)= sup Q x 1 | Q | Q | f ( y ) | dy.

For η>0, let M η (f)=M ( | f | η ) 1 / η . For kN, we denote by M k the operator M iterated k times, i.e., M 1 (f)=M(f) and

M k (f)=M ( M k 1 ( f ) ) when k2.

Let Φ be a Young function and Φ ˜ be the complementary associated to Φ. We denote the Φ-average by, for a function f,

f Φ , Q =inf { λ > 0 : 1 | Q | Q Φ ( | f ( y ) | λ ) d y 1 }

and the maximal function associated to Φ by

M Φ (f)(x)= sup x Q f Φ , Q .

The Young functions to be used in this paper are Φ(t)=t(1+logt) and Φ ˜ (t)=exp(t), the corresponding average and maximal functions denoted by L ( log L ) , Q , M L ( log L ) and exp L , Q , M exp L . Following [2], we know that the generalized Hölder inequality and the following inequalities hold:

1 | Q | Q | f ( y ) g ( y ) | d y f Φ , Q g Φ ˜ , Q , f L ( log L ) , Q M L ( log L ) ( f ) C M 2 ( f ) , f f Q exp L , Q C f BMO

and

f f Q exp L , 2 k Q Ck f BMO .

The A p weight is defined by (see [1])

A p = { w L loc 1 ( R n ) : sup Q ( 1 | Q | Q w ( x ) d x ) ( 1 | Q | Q w ( x ) 1 / ( p 1 ) d x ) p 1 < } , 1 < p < ,

and

A 1 = { w L loc p ( R n ) : M ( w ) ( x ) C w ( x ) , a.e. } .

Given a weight function w, for 1p<, the weighted Lebesgue space L p (w) is the space of functions f such that

f L p ( w ) = ( R n | f ( x ) | p w ( x ) d x ) 1 / p <.

Definition 1 Let Φ={ ϕ 1 ,, ϕ l } be a finite family of bounded functions in R n . For any locally integrable function f, the Φ sharp maximal function of f is defined by

M Φ # (f)(x)= sup Q x inf { c 1 , , c l } 1 | Q | Q | f ( y ) i = 1 l c i ϕ i ( x Q y ) | dy,

where the infimum is taken over all m-tuples { c 1 ,, c l } of complex numbers and x Q is the center of Q. For η>0, let

M Φ , η # (f)(x)= sup Q x inf { c 1 , , c l } ( 1 | Q | Q | f ( y ) i = 1 l c j ϕ i ( x Q y ) | η d y ) 1 / η .

Remark We note that M Φ # f # if l=1 and ϕ 1 =1.

Definition 2 Given a positive and locally integrable function f in R n , we say that f satisfies the reverse Hölder condition (write this as fR H ( R n )) if for any cube Q centered at the origin, we have

0< sup x Q f(x)C 1 | Q | Q f(y)dy.

Definition 3 Let φ be a positive, increasing function on R + , and there exists a constant D>0 such that

φ(2t)Dφ(t)for t0.

Let w be a weight function and f be a locally integrable function on R n . Set, for 1p<,

f L p , φ ( w ) = sup x R n , d > 0 ( 1 φ ( d ) Q ( x , d ) | f ( y ) | p w ( y ) d y ) 1 / p ,

where Q(x,d)={y R n :|xy|<d}. The generalized Morrey space is defined by

L p , φ ( R n , w ) = { f L loc 1 ( R n ) : f L p , φ ( w ) < } .

If φ(d)= d η , η>0, then L p , φ ( R n ,w)= L p , η ( R n ,w), which is the classical weighted Morrey spaces (see [8, 9]). If φ(d)=1, then L p , φ ( R n ,w)= L p ( R n ,w), which is the weighted Lebesgue spaces (see [6]).

As the Morrey space may be considered as an extension of the Lebesgue space, it is natural and important to study the boundedness of the operator on the Morrey spaces (see [5, 811]).

In this paper, we study some singular integral operators as follows (see [12]).

Definition 4 Let K L 2 ( R n ) and satisfy

K L C , | K ( x ) | C | x | n ,

there exist functions B 1 ,, B l L loc 1 ( R n {0}) and Φ={ ϕ 1 ,, ϕ l } L ( R n ) such that | det [ ϕ j ( y i ) ] | 2 R H ( R n l ), and for a fixed δ>0 and any |x|>2|y|>0,

| K ( x y ) i = 1 l B i ( x ) ϕ i ( y ) | C | y | δ | x y | n + δ .

For f C 0 , we define the singular integral operator related to the kernel K by

T(f)(x)= R n K(xy)f(y)dy.

Moreover, let b be a locally integrable function on R n . The Toeplitz-type operator related to T is defined by

T b = j = 1 m T j , 1 M b T j , 2 ,

where T j , 1 are T or ±I (the identity operator), T j , 2 are the bounded linear operators on L p (w) for 1<p< and w A 1 , j=1,,m, M b (f)=bf.

Remark Note that the classical Calderón-Zygmund singular integral operator satisfies Definition 4 (see [13], [19]). Also note that the commutator [b,T](f)=bT(f)T(bf) is a particular operator of the Toeplitz-type operators T b . The Toeplitz-type operators T b are the non-trivial generalizations of the commutator. It is well known that commutators are of great interest in harmonic analysis and have been widely studied by many authors (see [12, 14]). The main purpose of this paper is to prove the sharp maximal inequalities for the Toeplitz-type operator T b . As the application, we obtain the weighted L p -norm inequality and Morrey space boundedness for the Toeplitz-type operators T b .

3 Theorems and lemmas

We shall prove the following theorems.

Theorem 1 Let T be the singular integral operator as Definition 4, 0<r<1 and bBMO( R n ). If T 1 (g)=0 for any g L u ( R n ) (1<u<), then there exists a constant C>0 such that, for any f C 0 ( R n ) and x ˜ R n ,

M Φ , r # ( T b ( f ) ) ( x ˜ )C b BMO j = 1 m M 2 ( T j , 2 ( f ) ) ( x ˜ ).

Theorem 2 Let T be the singular integral operator as Definition 4, 1<p<, w A 1 and bBMO( R n ). If T 1 (g)=0 for any g L u ( R n ) (1<u<), then T b is bounded on L p (w).

Theorem 3 Let T be the singular integral operator as Definition 4, 0<D< 2 n , 1<p<, w A 1 and bBMO( R n ). If T 1 (g)=0 for any g L u ( R n ) (1<u<), then T b is bounded on L p , φ ( R n ,w).

To prove the theorems, we need the following lemmas.

Lemma 1 ([[1], p.485])

Let 0<p<q< and for any function f0. We define that for 1/r=1/p1/q,

f W L q = sup λ > 0 λ | { x R n : f ( x ) > λ } | 1 / q , N p , q ( f ) = sup E f χ E L p / χ E L r ,

where the sup is taken for all measurable sets E with 0<|E|<. Then

f W L q N p , q (f) ( q / ( q p ) ) 1 / p f W L q .

Lemma 2 (see [2])

We have

1 | Q | Q | f ( x ) g ( x ) | dx f exp L , Q g L ( log L ) , Q .

Lemma 3 (see [15])

Let T be the singular integral operator as Definition 4. Then T is bounded on L p (w) for 1<p<, w A 1 and weak ( L 1 , L 1 ) bounded.

Lemma 4 (see [12])

Let 1<p<, 0<η<, w A and Φ={ ϕ 1 ,, ϕ l } L ( R n ) such that | det [ ϕ j ( y i ) ] | 2 R H ( R n l ). Then, for any smooth function f for which the left-hand side is finite,

R n M η (f) ( x ) p w(x)dxC R n M Φ , η # (f) ( x ) p w(x)dx.

Lemma 5 (see [5, 11])

Let 1<p<, w A 1 and 0<D< 2 n . Then, for any smooth function f for which the left-hand side is finite,

M ( f ) L p , φ ( w ) C f L p , φ ( w ) .

Lemma 6 Let 1<p<, 0<η<, w A 1 , 0<D< 2 n and Φ={ ϕ 1 ,, ϕ l } L ( R n ) such that | det [ ϕ j ( y i ) ] | 2 R H ( R n l ). Then, for any smooth function f for which the left-hand side is finite,

M η ( f ) L p , φ ( w ) C M Φ , η # ( f ) L p , φ ( w ) .

Proof For any cube Q=Q( x 0 ,d) in R n , we know M(w χ Q ) A 1 for any cube Q=Q(x,d) by [3]. By Lemma 4, we have, for f L p , φ ( R n ,w),

Q | M η ( f ) ( y ) | p w ( y ) d y = R n | M η ( f ) ( y ) | p w ( y ) χ Q ( y ) d y R n | M η ( f ) ( y ) | p M ( w χ Q ) ( y ) d y C R n | M Φ , η # ( f ) ( y ) | p M ( w χ Q ) ( y ) d y = C ( Q | M Φ , η # ( f ) ( y ) | p M ( w χ Q ) ( y ) d y + k = 0 2 k + 1 Q 2 k Q | M Φ , η # ( f ) ( y ) | p M ( w χ Q ) ( y ) d y ) C ( Q | M Φ , η # ( f ) ( y ) | p w ( y ) d y + k = 0 2 k + 1 Q 2 k Q | M Φ , η # ( f ) ( y ) | p w ( Q ) | 2 k + 1 Q | d y ) C ( Q | M Φ , η # ( f ) ( y ) | p w ( y ) d y + k = 0 2 k + 1 Q | M Φ , η # ( f ) ( y ) | p M ( w ) ( y ) 2 n ( k + 1 ) d y ) C ( Q | M Φ , η # ( f ) ( y ) | p w ( y ) d y + k = 0 2 k + 1 Q | M Φ , η # ( f ) ( y ) | p w ( y ) 2 n k d y ) C M Φ , η # ( f ) L p , φ ( w ) p k = 0 2 n k φ ( 2 k + 1 d ) C M Φ , η # ( f ) L p , φ ( w ) p k = 0 ( 2 n D ) k φ ( d ) C M Φ , η # ( f ) L p , φ ( w ) p φ ( d ) ,

thus

( 1 φ ( d ) Q M η ( f ) ( x ) p w ( x ) d x ) 1 / p C ( 1 φ ( d ) Q M Φ , η # ( f ) ( x ) p w ( x ) d x ) 1 / p

and

M η ( f ) L p , φ ( w ) C M Φ , η # ( f ) L p , φ ( w ) .

This finishes the proof. □

Lemma 7 Let T be the singular integral operator as Definition 3 or the bounded linear operator on L r (w) for any 1<r< and w A 1 , 1<p<, w A 1 and 0<D< 2 n . Then

T ( f ) L p , φ ( w ) C f L p , φ ( w ) .

The proof of the lemma is similar to that of Lemma 6 by Lemma 3, we omit the details.

4 Proofs of theorems

Proof of Theorem 1 It suffices to prove that for f C 0 ( R n ) and some constant C 0 , the following inequality holds:

( 1 | Q | Q | T b ( f ) ( x ) C 0 | r d x ) 1 / r C b BMO j = 1 m M 2 ( T j , 2 ( f ) ) ( x ˜ ),

where Q is any cube centered at x 0 , C 0 = j = 1 m i = 1 l g j i ϕ i ( x 0 x) and g j i = R n B i ( x 0 y) M ( b b Q ) χ ( 2 Q ) c T j , 2 (f)(y)dy. Without loss of generality, we may assume T j , 1 are T (j=1,,m). Let x ˜ Q. Fix a cube Q=Q( x 0 ,d) and x ˜ Q. Write

T b (f)(x)= T b b 2 Q (f)(x)= T ( b b 2 Q ) χ 2 Q (f)(x)+ T ( b b 2 Q ) χ ( 2 Q ) c (f)(x)= f 1 (x)+ f 2 (x).

Then

( 1 | Q | Q | T b ( f ) ( x ) C 0 | r d x ) 1 / r C ( 1 | Q | Q | f 1 ( x ) | r d x ) 1 / r + C ( 1 | Q | Q | f 2 ( x ) C 0 | r d x ) 1 / r = I + II .

For I, by Lemmas 1, 2 and 3, we obtain

( 1 | Q | Q | T j , 1 M ( b b 2 Q ) χ 2 Q T j , 2 ( f ) ( x ) | r d x ) 1 / r | Q | 1 T j , 1 M ( b b 2 Q ) χ 2 Q T j , 2 ( f ) χ Q L r | Q | 1 / r 1 C | Q | 1 T j , 1 M ( b b 2 Q ) χ 2 Q T j , 2 ( f ) W L 1 C | Q | 1 M ( b b 2 Q ) χ 2 Q T j , 2 ( f ) L 1 C | Q | 1 2 Q | b ( x ) b 2 Q | | T j , 2 ( f ) ( x ) | d x C b b 2 Q exp L , 2 Q T j , 2 ( f ) L ( log L ) , 2 Q C b BMO M 2 ( T j , 2 ( f ) ) ( x ˜ ) ,

thus

IC j = 1 m ( 1 | Q | Q | T j , 1 M ( b b 2 Q ) χ 2 Q T j , 2 ( f ) ( x ) | r d x ) 1 / r C b BMO j = 1 m M 2 ( T j , 2 ( f ) ) ( x ˜ ).

For II, we get, for xQ,

| T j , 1 M ( b b Q ) χ ( 2 Q ) c T j , 2 ( f ) ( x ) i = 1 l g j i ϕ i ( x 0 x ) | | R n ( K ( x y ) i = 1 l B i ( x 0 y ) ϕ i ( x 0 x ) ) ( b ( y ) b 2 Q ) χ ( 2 Q ) c ( y ) T j , 2 ( f ) ( y ) d y | k = 1 2 k d | y x 0 | < 2 k + 1 d | K ( x y ) i = 1 l B i ( x 0 y ) ϕ i ( x 0 x ) | | b ( y ) b 2 Q | | T j , 2 ( f ) ( y ) | d y C k = 1 2 k d | y x 0 | < 2 k + 1 d | x x 0 | δ | y x 0 | n + δ | b ( y ) b 2 Q | | T j , 2 ( f ) ( y ) | d y C k = 1 d δ ( 2 k d ) n + δ ( 2 k d ) n b b 2 Q exp L , 2 k + 1 Q T j , 2 ( f ) L ( log L ) , 2 k + 1 Q C b BMO M 2 ( T j , 2 ( f ) ) ( x ˜ ) k = 1 k 2 k δ C b BMO M 2 ( T j , 2 ( f ) ) ( x ˜ ) ,

thus

II 1 | Q | Q j = 1 m | T j , 1 M ( b b Q ) χ ( 2 Q ) c T j , 2 ( f ) ( x ) C 0 | dxC b BMO j = 1 m M 2 ( T j , 2 ( f ) ) ( x ˜ ).

This completes the proof of Theorem 1. □

Proof of Theorem 2 By Theorem 1 and Lemmas 3-4, we have

T b ( f ) L p ( w ) M r ( ( T b ( f ) ) L p ( w ) C M Φ , r # ( T b ( f ) ) L p ( w ) C b BMO j = 1 m M 2 ( T j , 2 ( f ) ) L p ( w ) C b BMO j = 1 m T j , 2 ( f ) L p ( w ) C b BMO f L p ( w ) .

This completes the proof. □

Proof of Theorem 3 By Theorem 1 and Lemmas 5-7, we have

T b ( f ) L p , φ ( w ) M r ( T b ( f ) ) L p , φ ( w ) C M Φ , r # ( T b ( f ) ) L p , φ ( w ) C b BMO j = 1 m M 2 ( T j , 2 ( f ) ) L p , φ ( w ) C b BMO j = 1 m T j , 2 ( f ) L p , φ ( w ) C b BMO f L p , φ ( w ) .

This completes the proof. □