Endpoint Estimates for Oscillatory Singular Integrals with Hölder Class Kernels

Abstract

We prove the uniform and boundedness of oscillatory singular integral operators whose kernels are the products of an oscillatory factor with bilinear phase and a Calderón-Zygmund kernel satisfying a Hölder condition. This Hölder condition appreciably weakens the condition imposed in existing literature.

1. Introduction

Let . We shall consider the following oscillatory singular integral operator:where is a real-valued bilinear form. In past studies of this type of operators, is typically assumed to be a Calderón-Zygmund kernel satisfying a condition away from the diagonal , i.e., there exists an such that(i)for all ,(ii), and for (iii)where

Under conditions (i), (ii), and (iii), Phong and Stein proved the boundedness of for ([1]). The result of Phong and Stein was then extended to operators with polynomial phases by Ricci and Stein ([2]), under the same conditions (i), (ii), and (iii) on , while the weak (1,1) boundedness of such operators was subsequently established by Chanillo and Christ in [3] (for all polynomial phase functions, bilinear or otherwise).

The property of in condition (ii) was instrumental when van der Corput’s lemma, a standard tool in the treatment of oscillatory integrals, was used in past studies, including the seminal papers cited above. There has been widespread interest in finding out what happens when the kernel is replaced by a “rougher” kernel. Many interesting results have been obtained for kernels that are homogeneous and of convolutional type but lack smoothness (i.e., ). See, for example, [4–6].

In this paper we are interested in general kernels for which condition (ii) is replaced by the following weaker condition of Hölder type:

there exists a such thatwhenever andwhenever .

In a recent paper, we were able to prove the following uniform boundedness of for :

Theorem 1 (see [7]). Let and be given as in (1). Suppose that satisfies (i), (ii)′, and (iii). Then, for , there exists a positive which may depend on , , , and but is independent of the bilinear form , such thatfor all .

In this paper we shall investigate the endpoint case and obtain both the weak type (1,1) and Hardy space bounds. We begin with the weak (1,1) result.

Theorem 2. Let and be given as in (1). Suppose that satisfies (i), (ii)′, and (iii). Then, is of weak type (1,1), i.e., there exists a positive such thatfor all and . Moreover, while the constant may depend on , , and , it is otherwise independent of and .

In the statement above, we used to denote the Lebesgue measure of a measurable set .

In order to describe our result on Hardy spaces, let be the Hardy space introduced by Phong and Stein in [1] as a variant of the standard Hardy space suitable for the study of oscillatory singular integrals. It was proved there that under conditions (i), (ii), and (iii) is a bounded operator from to . As an improvement over their result, we have the following.

Theorem 3. Under conditions (i), (ii)′, and (iii), is a bounded operator from to . Moreover, the bound for the operator norm may depend on , , and but is otherwise independent of and .

It is already well-known that analogous results do not hold for when .

The proof of the weak type (1,1) estimate will appear in Sections 2 and 3. It follows a strategy pioneered by C. Fefferman in [8] (see also [3, 4, 9, 10]). The proof of the Hardy space estimate will be given in Section 4.

We now close this section by posing the following natural question which should be of interest to many working in this field: are the and endpoint results for oscillatory singular integrals in Theorems 1–3 still true when the bilinear phase functions are replaced by general polynomials in with real coefficients?

2. Basic Reductions for the Bounds

We shall begin the proof of Theorem 2 with a few reductions. Since the one-dimensional case is relatively easier, we shall focus our attention on . For , can be expressed as

If for all , then . It is well-known that, under the conditions (2), (4), and (6)-(7), is bounded from from to as well as from to for . Thus, from this point on, we may assume that holds for at least one pair . Let

If we let denote the dilation operatorthenSince satisfies (i), , and (iii) with the same constants and as , it suffices to establish (9) under the additional assumption that, for some , (after reindexing the variables if necessary). Clearly, we may also assume that .

For any cube in , let and denote its sidelength and center, respectively. For any , we let denote the cube that has the same center as and sidelength . Also, let .

Let and . Then there is a collection of dyadic cubes with disjoint interiors such that the following are satisfied:

For , let . Let ,and

Then,whereand

It follows from Theorem 1 and a standard argument that (8) remains valid when is replaced by or . Thus, by (14)-(15),andThe set can be treated by a finite overlapping argument. Let . For each , let andClearly,By (2) and , for ,where Thus,The proof of Theorem 2 has thus been reduced to the verification of the following:

3. Estimates for and

For two nonnegative integers and , let

For and , let . LetObserve that

Lemma 4. Let such that . Suppose that satisfies (2) and (6)-(7).(i)For any ,(ii)For any ,(iii)For any satisfying and ,

Proof. We shall omit the arguments for (i) and (ii) because they utilize (2) only and therefore can be found in [3] (see page 152).
Suppose that , and . For any , let where . Thus,whereWhen , we haveIt follows from (2) and (6) thatWhen ,Thus,It follows from the the arguments for thatWhen , we haveandThus,andTherefore,Similarly,The proof of Lemma 4 is now complete.

Let denote the distance between two sets and . For , let

Lemma 5. If and , then

Proof. Let andFor each , letIt follows from (15), (17), and Lemma 4(i) that the cardinality of is bounded by a constant andWhen ,Also, for any and ,Thus, by Lemma 4(ii), (15), and (58),For any ,It follows from Lemma 4(iii), (15), (58), and (61) thatLemma 5 is proved.

For , let .

Lemma 6. If and , then

Proof. The argument is very similar to the proof of the previous lemma. We will point out the differences but omit most of the details.
Let and . LetWhile there is no uniform bound on the cardinality of (unlike ), by using , we still haveBy ,Finally, can be treated the same as , which finishes the proof of Lemma 6.

We now employ a well-known technique to obtain the desired estimates for and (see, for example, [3, 8–11]). By Lemma 5,By Lemma 6,The proof of Theorem 2 is now complete.

4. Boundedness

We shall begin by recalling the definition of the function space .

Definition 7. A measurable function on is called an atom if there exists a cube such that , andA function is in if there exist a sequence in and a sequence of atoms such thatThe norm of is the infimum of over all possible expressions of described in (70).

In order to prove Theorem 3, it suffices to establish that, for every atom ,By using a dilation (as in Section 2) as well as a translation, we can further reduce the task to the verification of (71) under the assumption that, for some and ,Let . ThenwhereBy Theorem 1,By (7),By the vanishing integral property of ,Let denote the partial Fourier transform in the first variable. Then,The proof of Theorem 3 is now complete.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.