#### Abstract

We investigate the bilinear Hilbert transform with oscillatory factors and the truncated bilinear Hilbert transform. The main result is that the -boundedness of the two operators is equivalent with , and . In addition, we also discuss the boundedness of a variant operator of bilinear Hilbert transform with a nontrivial polynomial phase.

#### 1. Introduction

In this paper, we mainly discuss the following operator defined by where is a real-valued polynomial defined on and are smooth functions with compact support.

Clearly, when , the operator becomes the normal bilinear Hilbert transform defined as

If for some one-variable polynomials , , and , then clearly the boundedness of and is equivalent. For this case, we introduce the following definitions.

*Definition 1. *For a polynomial defined on , one calls degenerate, if
holds with three one-variable polynomials , , and .

In fact, a simple verification shows that if the degree of is less than 3, then is degenerate. However, when , there exist nondegenerate polynomials with the degree . We will prove this in Section 2. It should be pointed out that Definition 1 may be extended to such polynomials defined on (see [1]). We will give a characterization of nondegenerate polynomials by in Section 2. Now we first give the definition of .

*Definition 2. *Let be a polynomial with the degree . For , denote by

For the operator defined by (1), the following result has been obtained in [1].

Theorem A. *Let be a real-valued polynomial. Then the operator extends as a bounded operator from to with , , and such that ; furthermore, the operator norm depends only on the degree of .*

In the study of oscillatory singular integrals, the boundedness of the original operator without oscillatory factors may imply the boundedness of the corresponding truncated operator. For the CalderÃ³n-Zygmund singular integral operators of convolution type, one can see Ricci and Stein [2]. For the multilinear singular operators, one may refer to [1]. In [3] Lu and Zhang established a criterion for the boundedness of the following operator: where is a real-valued polynomial defined on , , and satisfies(1) is homogeneous of the degree 0 on ;(2) has mean value zero on ;(3) for some .

For the sake of clarity, we also present the theorem obtained by Lu and Zhang as follows.

Theorem B. *Suppose that is defined by (5) and . Each of the following statements implies the other two.*(i)*If is not of the form for some polynomials and in , then the operator
*â€‰*is bounded on .*(ii)*If satisfies for some polynomials , , and , then the operator
*â€‰*extends as a bounded operator from to itself.*(iii)*The following truncated operator
*â€‰*is bounded on .*

For the bilinear Hilbert transform and the corresponding truncated operator, we can formulate some analogous results as Theorem B.

Theorem 3. *Suppose that satisfy . If the truncated operator
**
is bounded from to , then the bilinear Hilbert transform
**
is bounded from to . Moreover, when , , the converse also holds; that is, if extends to a bounded operator from to , then so does .*

Theorem 4. *Let be a real-valued polynomial and let , , and satisfy . If is degenerate with respect to , for all in the sense of Definition 1 and the following operator
**
is bounded from to , then the truncated operator
**
is bounded from to .*

We summarize the above two theorems as follows.

Theorem 5. *Suppose that , â€‰, and satisfy . Each of the following three statements implies the other two.*(i)*The bilinear Hilbert transform
*â€‰*is bounded from to .*(ii)*The oscillatory bilinear Hilbert transform
*â€‰*is bounded from to for any real-valued polynomial .*(iii)*The truncated operator
*â€‰*is bounded from to .*

Next, we will consider a variant operator of bilinear Hilbert transform defined by for . Clearly, the operator has no singular property at origin, so it is the key at .

We will use the power decay property of a bilinear functional to obtain the following theorem.

Theorem 6. *Suppose that is a real-valued nondegenerate polynomial with the degree and for some . If , , and satisfy , then the operator defined by (16) is bounded from to ; that is,
**
where is a positive constant.*

#### 2. Some Lemmas

Throughout the paper, we always assume that , , and , unless the contrary is explicitly stated. We first introduce some lemmas which are useful in the proof of the main theorems.

Lemma 7. *Suppose that defined on is a homogeneous polynomial of the degree with . Then is degenerate if and only if for all .*

* Proof. *By Definition 1, if is degenerate, then
where , , and are homogeneous single-variable polynomials of the degree . Observe that
This implies that
for all .

Now we assume that for all . Make changes of variables
By the chain rule, we have that

Since is a homogeneous polynomial with the degree , is a homogeneous polynomial of the degree . There, therefore, exists a homogeneous polynomial of the degree such that
It follows from (23) that
The assumption , for all , implies that
for some constants and . Integrating both sides of the equality (25) with respect to gives that
for some constant . Hence is degenerate.

*Remark 8. *For and , and are nondegenerate polynomials. Furthermore, when is a general polynomial of the degree , we can write as sums of homogeneous polynomials,
where is a homogeneous polynomial with the degree . If is degenerate, then each is degenerate for . In fact, if is degenerate, then . It follows that for each . Consequently, each is degenerate.

Lemma 9. *Let be a polynomial with the degree . For any , set
**Then, one has
**
for each .*

*Proof. *By the same change of variables as in the proof of Lemma 7, we obtain
We can use the chain rule as in (22) to differentiate both sides of (28) by the differential operator and easily yield the desired result.

Lemma 10. *Suppose that and are two positive integers and that , , are linear mappings from to . Assume that for satisfy
**
If the linear mapping : defined by
**
is onto for , , then one has
**
where is a bounded measurable function with compact support and for .*

*Proof. *For the case , applying generalized Minkowskiâ€™s and HÃ¶lderâ€™s inequality gives that
Now we deal with the case . We first show that the following estimate
holds uniformly for , where will be decided by the and the norm of , .

Since , by HÃ¶lderâ€™s inequality, it is clear that we have that
Since the linear mapping : is onto, we can make change of variables , , and then . We also have
where is the volume of the unit ball in and is a fixed number such that for all , .

Note the following fact. For any and , it follows that
where is the characteristic function of the ball centered at with radius .

In view of the equality (38), taking integration on both sides of (35) with respect to over , we conclude that
The proof is therefore concluded.

As a special case of Lemma 10, we immediately obtain Corollary 11.

Corollary 11. *Let and . If , satisfy , then there exists a constant such that
*

Actually, the estimate of Corollary 11 may also be found in [4, 5].

Generally speaking, the power decay estimates of oscillatory integrals are necessary to investigate the boundedness of oscillatory integral operators. The following lemma to appear in [5] is just an oscillatory integral and will be used to prove Lemma 14.

Lemma 12. *Suppose that is a real-valued polynomial with the degree and that there exists a positive integer number such that
**
Define the functional
**
where is a smooth function with compact support, , and . Then there exists some constant such that the following power decay estimate
**
holds, where the constant is independent of , and .*

*Remark 13. *It should be pointed out how the constant depends on the cut-off function , provided that other conditions are not changed. Indeed, if, for the collection , every function is uniformly supported in a bounded set and has an upper bound independent of for each positive integer , then the constant can be chosen such that the estimate (43) holds uniformly for . The decay estimate (43) and its other variants have been systematically investigated by many scholars. For certain polynomials with some other conditions, the power exponent can be explicitly given. One may refer to [6] and other references appearing in this paper.

Many oscillatory integral operators have power decay estimates which are indispensable to study mapping properties. The following lemma is significant and will be useful in the next section.

Lemma 14. *Suppose that is a real-valued nondegenerate polynomial with the degree and that there is some , , such that . Let
**
where has support contained in . If , â€‰, and satisfy
**
then there exists some such that
**
for , , and .*

*Proof. *Choose a nonnegative function such that
for every .

To obtain the decay estimate (46), it suffices to show that the following estimate
holds uniformly in , where denotes an interval with the center at and the radius 5.

If inequality (48) has been established, we may use HÃ¶lderâ€™s inequality to obtain (46).

In fact, since , at most five terms are not zero in the summation for any fixed . Thus, if , then we have
and if , then it follows that

Taking the summations of both sides of the inequality (48) with respect to over and using the two inequalities (49) and (50), we notice that the summation of the left side of (48) is greater than .

For the summation of the right side of (48), using HÃ¶lderâ€™s inequality, we obtain that

Next let us turn to the proof of the inequality (48). By Lemma 10 or Corollary 11, we know that is bounded from to and its norm is independent of . By the multilinear interpolation of Riesz-ThÃ¶rin, it is enough to show that the estimate (46) holds for and . Thus we merely need to show that the inequality (48) holds with the case and . Furthermore, since has a compact support, it suffices to prove that (48) holds for .

This is equivalent to proving the following inequality:

By the change of variables, we can easily deduce that
where

Consequently, for some , we can rewrite the above integral as

Observe that integrals and are taken in the domain
Recall and , respectively.

We clearly conclude that

For , by Lemma 9, it is easily verified that
holds for each .

Thus we have that
for some .

Invoking Lemma 12, we obtain, for , that
since the cut-off function is uniformly supported in and its norms have bounds independent of for each positive , and the constant is independent of . Therefore, the Cauchy-Schwartz inequality implies that
where we use the Fubini theorem in the proof of the above inequality.

Consequently, choosing for and combining the above estimates for and , we immediately obtain that
where . This is just the inequality (52).

Lemma 15. *Suppose that is a real-valued polynomial and , , and satisfy
**
If the operator defined by (1) satisfies
**
for all , then the truncated operator
**
is also bounded from to .*

* Proof. *To prove Lemma 15, we will show that
holds uniformly with respect to .

Fix and set
It is clear that .

Now for fixed , set

If , then . We hence obtain that

For , when , we have . It immediately follows that

Finally, both and imply . We clearly have

Similarly, we also have that
provided that and .

Therefore, if , then the equalities (69), (71), and (72) together with (70) imply that
Clearly the inequality (66) is easily obtained by combing (73) together with Corollary 11. We conclude the proof by integrating both sides of (66) with respect to .

Lemma 16. *Suppose that , , and satisfy
**
Define the truncated operator as
**
If the operator defined by (9) satisfies
**
for all , then the truncated operator is also bounded from to for every ; moreover, the operator norm of is independent of .*

* Proof. *Clearly it implies from the definitions of and that . By a simple dilation argument, we will prove that , the norm of the operator defined by (75), is independent of and equals .

Indeed, a simple computation implies that
It follows that
where the dilation operator is defined by
for all and .

Denote the norm of the operator by as follows:

Taking supremum over all for (78), we have
and naturally .

Lemma 17. *Suppose that is a real-valued polynomial and , , and satisfy
**
If the operator defined by (1) satisfies
**
for all , then the following operator
**
is also bounded from to for every ; moreover, the operator norm of is independent of .*

*Proof. *By a simple dilation argument, we will prove that , the norm of the operator , is independent of and equals and naturally equals .

Indeed, a simple computation implies that
Taking supremum over all for (86), we immediately have
and hence .

*Remark 18. *Lemma 17 shows that, for the oscillatory integral operator, the operator norm only depends on the degree of the polynomial but is independent of the coefficient of the polynomial.

#### 3. Proof of the Main Results

*Proof of Theorem 3. *Associated with each positive , denote the operator by
where ; hence, we have by the definition of in Theorem 3. By the dilation argument as in the proof of Lemma 16, we can easily conclude that if can be extended to a bounded operator from to , then so is and the operator norm is independent of . Observe that is dense in all for . For , we regard as a complete metric space. Since we can approximate by finite linear combinations of characteristic function of intervals, we may assume that for some finite open interval . By the Lusin theorem, for any closed interval there exists a function such that , , and for . Hence we obviously have

Fix temporarily and choose a large such that both and are contained in . If , we see that either or must hold for every . Hence we conclude that
where and are operator norms of and from to , respectively.

Obviously it follows that

The validity of the converse follows from Lemma 15.

This finishes the proof of Theorem 3.

*Proof of Theorem 4. *We assume that
holds for some polynomial with being degenerate for all . We wish to obtain
where the constant is independent of and .

Since is degenerate with respect to , we can rewrite as
for some polynomials , , and .

For , it follows that
where

For the polynomial , when and , we immediately have