Boundedness of Sublinear Operators with Rough Kernels on Weighted Morrey Spaces

When Ω is a smooth kernel and T Ω a standard CalderónZygmund singular integral operator which has been fully studied by many papers, a classical survey work; see, for example, [1]. For simplicity of notation, Ω is always homogeneous of degree zero and satisfies (3) and (4) throughout this paper if there are no special instructions. Here and in what follows, for x 0 ∈ R, r > 0, and λ > 0, B = B(x 0 , r) denotes the ball centered at x 0 with radius r and λB =


Introduction and Main Results
Given a function Ω over the unit sphere  −1 of R  ( ≥ 2) equipped with the normalized Lebesgue measure  and   = /||, a Calderón-Zygmund singular integral operator with rough kernel was given by −        () and a related maximal operator where Ω is homogeneous of degree zero and satisfies When Ω is a smooth kernel and  Ω a standard Calderón-Zygmund singular integral operator which has been fully studied by many papers, a classical survey work; see, for example, [1].For simplicity of notation, Ω is always homogeneous of degree zero and satisfies (3) and (4) throughout this paper if there are no special instructions.Here and in what follows, for  0 ∈ R  ,  > 0, and  > 0,  = ( 0 , ) denotes the ball centered at  0 with radius  and  = ( 0 , ).When Ω satisfies some size conditions, the kernel of the operator  Ω has no regularity, and so the operator  Ω is called rough singular integral operator.In recent years, a variety of operators related to the singular integrals for Calderón-Zygmund, but lacking the smoothness required in the classical theory, have been studied.Duoandikoetxea [2] studied the norm inequalities for  Ω in homogeneous case on weighted   (1 <  < ∞) spaces.For more corresponding works, we refer the reader to [3][4][5][6][7][8] and the references therein.
Let   = { ∈ R  : || ≤ 2  } and let   =   \  −1 for  ∈ .Throughout this paper, we will denote by   the characteristic function of the set . Inspired by the works of [6,13], in this paper, we consider some sublinear operators under some size conditions (the following ( 7) and ( 8)) which are more general than (5): when supp  ⊆   and || ≥ 2 +1 with  ∈ Z and when supp  ⊆   and || ≤ 2 −1 with  ∈ Z, respectively.It is worth pointing out that  Ω satisfies conditions ( 7) and (8).Also, condition (5) implies the size conditions ( 7) and ( 8 The topic of this paper is intended as an attempt to study the boundedness of sublinear operators with rough kernels which satisfy ( 7) and ( 8) on weighted Morrey spaces.We first recall some definitions and notations for weighted spaces.The Muckenhoupt classes   and  (,) [14] contain the functions  which satisfy respectively, where 1/ + 1/  = 1.For  = 1, the  1 and  (1,) (1 <  < ∞) weights are defined by respectively.Here ess sup and the following essinf are the abbreviations of essential supremum and essential infimum, respectively.Clearly,  ∈  1 if and only if there is a constant  > 0 such that In [15], Komori and Shirai introduced a weighted Morrey space, which is a natural generalization of weighted Lebesgue space, and investigated the boundedness of classical operators in harmonic analysis.Let 1 ≤  < ∞, 0 <  < 1 and let  be a weight function.Then the weighted Morrey space  , () is defined by where () = ∫  ().For  ∈   (1 ≤  < ∞), if  = 0, then  ,0 () =   () while  = 1 implies  ,1 () =  ∞ ().Now, we formulate our major results of this paper as follows.
When  = 1, we have the following theorem.Theorem 2. Let 1 <  < ∞ and 1/+ < 1 and let T Ω satisfy (7) and (8).Then if T Ω is bounded from  1 () to  1,∞ () with  ∈  1 , there exists a constant  > 0 such that for all  > 0 and all balls ,  ({ ∈  : In the fractional case, we need to consider a weighted Morrey space with two weights which is also introduced by Komori and Shirai in [15].Let 1 ≤  < ∞, 0 <  < 1.For two weights  1 and  2 , If  1 =  2 = , we write  , ( ) and 1/ +  < 1, then there exists a constant  > 0 such that for all  > 0 and all balls , ({ ∈  : where 1 <  < ∞. We emphasize that ( 15) and ( 16) are weaker conditions than the following condition: for any integral function  with compact support.Condition ( 18) is satisfied by most fractional integral operators with rough kernels, such as the fractional integral operators of Muckenhoupt and Wheeden [16]: For some mapping properties of T ,Ω on various kinds of function spaces, see [17][18][19] and the references therein.
We end this section with the outline of this paper.Section 2 contains the proofs of Theorems 1 and 3; this part is partly motivated by the methods in [20] dealing with the case of the Lebesgue measure.In Section 3, we extend the corresponding results to commutators of certain sublinear operators.

Boundedness of Sublinear Operators
Proofs of Theorems 1 and 3 depend heavily on some properties of   weights, which can be found in any papers or any books dealing with weighted boundedness for operators in harmonic analysis, such as [1].For the convenience of the reader we collect some relevant properties of   weights without proofs, thus making our exposition self-contained.(a) There exists a constant  such that where  satisfies this condition; one says  satisfies the doubling condition.
(b) There exists a constant  > 1 such that where  satisfies this condition; one says  satisfies the reverse doubling condition.
(c) There exist two constants  and  > 1 such that the following reverse Hölder inequality holds for every ball  ⊂ R  : (d) For all  > 1, one has (e) There exist two constants  and  > 0 such that for any measurable set  ⊂ if  satisfies (24); one says  ∈  ∞ .
(f) For all  <  < ∞, one has The following lemma about the rough kernel Ω is essential to our proofs.One can find its proof in [21].  ) . (28) For the term , by (8) we have where We distinguish two cases according to the size of  and  to get the estimates for T Ω, .

Proof of Theorem
Combining these inequalities for  and , we have completed the proof of Theorem 2.
Proof of Theorem 3. We can use the similar arguments as in the proof of Theorem 1 and Theorem 2. For the proof of (), it suffices to show that 1   () For the term , by the similar arguments as that of Theorem 1, we obtain We have completed the proof of ().
We will omit the proof of () since we can prove it by using  (1,) condition and the weak type estimates of T ,Ω similar to the proof of Theorem 2.

Boundedness of Commutators
We say that  is a BMO(R  ) function if the following sharp maximal function is finite: where the supreme is taken over all balls  ⊂ R  and An early work about BMO(R  ) space can be attributed to John and Nirenberg [22].For 1 <  < ∞, there is a close relation between BMO(R  ) and   weights: Given an operator  acting on a generic function  and a function , the commutator   is formally defined as Since  ∞ (R  ) ⊊ BMO(R  ), the boundedness of   is worse than  (e.g., the singularity; see also [23]).Therefore, many authors want to know whether   shares the similar boundedness with .There are a lot of articles that deal with the topic of commutators of different operators with BMO functions on Lebesgue spaces.The first results for this commutator were obtained by Coifman et al. [24] in their study of certain factorization theorems for generalized Hardy spaces.In the present section, we will extend the boundedness of T Ω and T ,Ω to T Ω, and T ,Ω, , respectively.Theorem 6.Let  , , , and  be as in Theorem 1. Suppose that the sublinear operator T Ω satisfies condition (5) for any integral function  with compact support.If T Ω, is bounded on   () with  ∈ BMO(R  ), then T Ω, is bounded on  , ().Theorem 7. Let , , , , , and  be as in Theorem 3() and let the sublinear operator T ,Ω satisfy condition (18) for any integral function  with compact support.If T ,Ω, maps   (  ) into   (  )with  ∈ BMO(R  ), then T ,Ω, is bounded from  , (  ,   ) to  ,/ (  ).
The following lemmas about BMO(R  ) functions will help us to prove Theorems 6 and 7.
Let  ∈  ∞ and 1 <  < ∞.Then the following statements are equivalent: Proof.We will consider two cases.
Case 1 ( >   ).In this case,  ∈  /  .Using Hölder's inequality and Lemma 5 to the left-hand side of (51), we have Combining ( 23) with (49), we have In the same manner we can see that It follows immediately that Therefore  ≤  ( + 1) where  > 1 is a constant that appeared in (21).
Case 2 ( =   ).In this case,  ∈  1 .We can prove (51) by a similar analysis as in the proof of Theorem 1 (in the case  =   ) and Case 1.
Having disposed of the previous preliminary step, we can now return to the proofs of Theorems 6 and 7.
Proof of Theorem 6.The task is now to find a constant  such that for fixed ball  = ( 0 , 1), we can obtain According to (64) and (69), we have completed the proof of Theorem 6.
Proof of Theorem 7. The proof of Theorem 7 is similar to that of Theorem 6, except using  ∈  (,) .We omit its proof here.