Research Article | Open Access
Bounds for the Commutators of Oscillatory Singular Integrals with Rough Kernels
We establish the boundedness for some commutators of oscillatory singular integrals with the kernel condition which was introduced by Grafakos and Stefanov. Our theorems contain various conditions on the phase function.
The homogeneous singular integral operator is defined by where satisfies the following conditions.(a) is homogeneous function of degree zero on ; that is, for any and .(b) has mean zero on , the unit sphere in ; that is,
The oscillatory singular integral we will consider here is defined by If , the operator becomes the singular integral operator .
When is a real polynomial, the boundedness of was first studied by Ricci and Stein  with , and Hu and Pan  obtained the weighted boundedness of . When , , Lu and Zhang proved the boundedness  and this was extended to the case of by Ojanen  and the case of by Fan and Pan .
Theorem A (see ). Let , , and is a convex increasing function for , for some ; then, is bounded on for .
Later, Cheng and Pan  improved the result for by removing the condition .
Theorem B (see ). Let , , and is a convex increasing function for , for some ; then, is bounded on for .
It has been proved that the boundedness of on can be obtained from the boundedness of (see ).
For a function , let be a linear operator on some measurable function space; the commutator between and is defined by .
It has been proved by Hu  that is a sufficient condition for the commutator to be bounded on , which is defined by Recently, Chen and Ding  established the boundedness of the commutator of singular integrals with the kernel condition .
It is natural to ask whether the similar result holds for the commutators of oscillatory singular integrals, which is defined by In this paper, we will give a positive answer to the above question by imposing some conditions on .
We first prove the boundedness of the commutator of singular integral along surfaces, which is defined by
Remark 3. However, for , we can not prove the boundedness of by our method using Lemma 11, since the conditions imposed on in Theorem 1 conflict with Lemma 11. Only when by removing the condition in Theorem 1 can we eliminate the conflict, and is a feasible function. Also, by another method, it is hard to give the boundedness of the maximal operator defined by
Then we give the boundedness of the commutators of oscillatory singular integral .
Let , , , and we have the following result.
Theorem 4. If is bounded on with bound , then is bounded on with bound .
In above theorems, the phase functions are radial. But when Ricci and Stein first studied the oscillatory singular integral , they take , apparently nonradial. In Theorem 7, we will take , and this condition was mentioned in .
Lemma 8. Let be a family of multipliers such that , , and for some constants , and Let be the multiplier operator defined by , . For , denote by the commutator of . Then for any , there exists a positive constant such that
Proof. We assume that . Let and let be a radial function such that , and
for . Set and for positive integer . Let the inverse Fourier transform of . Split as
Let be the convolution operator whose kernel is ; that is, . Recall that . Trivial computation shows that . This via the Young inequality says that Note that . Thus On the other hand, by the Yong inequality, we have Then, using the same argument of the proof of Lemma 2 in  we can prove Lemma 8.
Let the measure on be defined by for all . Define the maximal operator in by .
Lemma 9 (see ). Suppose is bounded on for all . Then, for arbitrary functions , the following vector valued inequality: holds with any .
The maximal function in is defined by We know that the boundedness of is deduced from the boundedness of by method of rotations, and if is as in Theorem 1 or Theorem 2, is a bounded operator on for all (see [23, 24]).
Let be a radial function satisfying with its support in the unit ball and for . The function satisfies for . For , denote by and the convolution operators whose symbols are and , respectively.
Lemma 10 (see ). For the multiplier , , and any fixed , we have where is independent of and .
Let and let be a radial function such that , , and , . Define the multiplier operator by .
Lemma 11. For any , define the operator by , and is monotonic and satisfies condition or :(1);(2), for , and , if , , if . Let , and denote by the commutator of . Suppose satisfying (2). Then for any fixed , ,
Proof. We prove it by using arguments which are essentially the same as those in the proof of Lemma 3.7 in . Two things must be modified:(i)instead of Lemma 3.6 in , we use Lemma 9;(ii)In , , and is the paraproduct of Bony  between two functions and . In the estimate of , we will use the following formulas: by Lemma 10, If satisfies condition (1), we have Thus If satisfies condition (2), we have Thus if , and if ,
Proof of Theorem 1. Let and let be a radial function such that , , and Define the multiplier operator by Let the measure on be defined by for all . Since we get Define the operator , where and the multiplier From the above notation, it is easy to see that where Then by the Minkowski inequality, we get For , we recall By Lemma 2.3 of , we have Denote by the before components truncation of ; that is, Since we get Set , . Recall that by . Straightforward computations lead to Since we get Let be the operator defined by . Denote by and . Similarly, denote by and . Thus via the Plancherel theorem and Lemma 8 it is stated that for any fixed , , Dilation-invariance says that By the proof of Theorem 1 in , we can get So, we have For , by Lemma 2.3 of , if satisfies the hypotheses in Theorem 1, we have When , if , we also have the above estimates (see ). Set , . Recall by . Straightforward computations lead to Let be the operator defined by . Denote by and . Similarly, denote by and . Thus via the Plancherel theorem and Lemma 8 it is stated that for any fixed , , Dilation-invariance says that By the proof of Theorem 1 in , we can get So take , and we have Then, by (50) and (56) we obtain Theorem 1.
Proof of Theorem 2. By (36), we have For , recall ; then, . , and applying Lemma 11, we get for Interpolating between (49) and (58) with , as the proof of Theorem 1 in , we can get For , , and applying Lemma 11, we get for any fixed , , Take ; then, we get For , (55) can be established only when , so interpolating between (55) and (61) with , as the proof of Theorem 1 in , we get Then, by (59) and (62) we obtain Theorem 2.
We begin with a lemma, which plays an important role in proving Theorem 4.
Lemma 12. Let , , and ; then, and .
Proof. We know where and is the square in whose edges are parallel to the axis. So where is the square in whose edges are parallel to the axis. Consider where is the projection on of and is the side length of . Then
Proof of Theorem 4. By Lemma 12, . Using the method in , for and , let . Then by mean value theorem of integrals and Lemma 12, we have Dividing both sides by and letting , we obtain Thus, we obtain Theorem 4.
Proof of Theorem 7. Theorem 7 can be proved by using arguments which are essentially the same as the proof of Theorem 1 in . Only the following two things must be modified.(i)Instead of and , we use (ii)Since is odd and is even with respect to , we get is odd and . So we use the estimates in : Consider
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
The research was supported by NCET of China (Grant: NCET-11-0574), the Fundamental Re-search Funds for the Central Universities (FRF-TP-12-006B), NSF of China (Grant: 10901017), and CSC Funding of China.
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Copyright © 2014 Yanping Chen and Kai Zhu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.