#### Abstract

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.

#### 1. Introduction

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 [1] with , and Hu and Pan [2] obtained the weighted boundedness of . When , , Lu and Zhang proved the boundedness [3] and this was extended to the case of by Ojanen [4] and the case of by Fan and Pan [5].

Grafakos and Stefanov [6] introduced a class of kernel functions which contains all satisfying (3) and where is a fixed constant. This kernel condition has been considered by many authors [7–13].

The singular integral along surfaces which is defined by was also studied by many authors [14–18]. Under the condition , Pan et al. [16] established the following Theorem.

Theorem A (see [16]). *Let , , and is a convex increasing function for , for some ; then, is bounded on for .*

Later, Cheng and Pan [14] improved the result for by removing the condition .

Theorem B (see [14]). *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 [5]).

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 [19] that is a sufficient condition for the commutator to be bounded on , which is defined by Recently, Chen and Ding [20] 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

Theorem 1. *Let be a function in satisfying (2) and (3), , radial function with , and is a convex increasing function. If for some , then is bounded on .*

Theorem 2. *Let be a function in satisfying (2) and (3), , radial function . If for some , then is bounded on for .*

*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 **.*

Combining Theorem 4 with Theorems 1 and 2, respectively, we can get the following two theorems immediately.

Theorem 5. *Let be a function in satisfying (2) and (3), , radial function with , and is a convex increasing function. If for some , then is bounded on .*

Theorem 6. *Let be a function in satisfying (2) and (3), , radial function . If for some , then is bounded on for .*

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 [21].

Theorem 7. *Let be a function in satisfying (2) and (3), . If is an odd kernel for some , is an even phase; then, extends to a bounded operator from into itself for .*

#### 2. Lemmas

We give some lemmas which will be used in the proof of Theorems 1 and 2.

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 [22] we can prove Lemma 8.

Let the measure on be defined by for all . Define the maximal operator in by .

Lemma 9 (see [18]). *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 [20]). *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 [20]. Two things must be modified:(i)instead of Lemma 3.6 in [20], we use Lemma 9;(ii)In [20], , and is the paraproduct of Bony [25] 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 ,

#### 3. The Proof of Theorems 1 and 2

* 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 [16], 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 [20], we can get
So, we have
For , by Lemma 2.3 of [16], if satisfies the hypotheses in Theorem 1, we have
When , if , we also have the above estimates (see [14]). 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 [20], 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 [20], 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 [20], we get
Then, by (59) and (62) we obtain Theorem 2.

#### 4. The proof of Theorems 4 and 7

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 [5], 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 [20]. 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 [21]: Consider

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

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.