Research Article | Open Access
Criterion on -Boundedness for Oscillatory Bilinear Hilbert Transform
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.
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 ). We will give a characterization of nondegenerate polynomials by in Section 2. Now we first give the definition of .
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 Calderón-Zygmund singular integral operators of convolution type, one can see Ricci and Stein . For the multilinear singular operators, one may refer to . In  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 is bounded on .(ii)If satisfies for some polynomials , , and , then the operator extends as a bounded operator from to itself.(iii)The following truncated operator 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 , , and satisfy . Each of the following three statements implies the other two.(i)The bilinear Hilbert transform is bounded from to .(ii)The oscillatory bilinear Hilbert transform is bounded from to for any real-valued polynomial .(iii)The truncated operator 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 Minkowski’s and Hölder’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 Hölder’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.
Corollary 11. Let and . If , satisfy , then there exists a constant such that
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  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  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 , , 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 Hölder’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 Hölder’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-Thörin, 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