Journal of Function Spaces and Applications
Volume 2013 (2013), Article ID 982753, 16 pages
http://dx.doi.org/10.1155/2013/982753
Research Article

## Estimates for Unimodular Multipliers on Modulation Hardy Spaces

1Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China
2Department of Mathematics, University of Wisconsin-Milwaukee, Milwaukee, WI 53201, USA
3School of Science, Hangzhou Dianzi University, Hangzhou 310016, China

Received 23 November 2012; Accepted 23 January 2013

Copyright © 2013 Jiecheng Chen et al. 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.

#### Abstract

It is known that the unimodular Fourier multipliers are bounded on all modulation spaces for . We extend such boundedness to the case of all and obtain its asymptotic estimate as t goes to infinity. As applications, we give the grow-up rate of the solution for the Cauchy problems for the free Schrödinger equation with the initial data in a modulation space, as well as some mixed norm estimates. We also study the boundedness for the operator , for the case and Finally, we investigate the boundedness of the operator for and obtain the local well-posedness for the Cauchy problem of some nonlinear partial differential equations with fundamental semigroup .

#### 1. Introduction

A Fourier multiplier is a linear operator whose action on a test function on is formally defined by The function is called the symbol or multiplier of .

In this paper, we will study the unimodular Fourier multipliers with symbol for . They arise when one solves the Cauchy problem for dispersive equations. For example, for the solution of the Cauchy problem we have the formula . Here is the Laplacian and is the multiplier operator with symbol (see [1] for its definition). The cases are of particular interest because they correspond to the (half-) wave equation, the Schrödinger equation, and (essentially) the Airy equation, respectively.

Unimodular Fourier multipliers generally do not preserve any Lebesgue space , except for . The -spaces are not the appropriate function spaces for the study of these operators and the so-called modulation spaces are good alternative classes for the study of unimodular Fourier multipliers. The modulation spaces were first introduced by Feichtinger [24] to measure smoothness of a function or distribution in a way different from spaces, and they are now recognized as a useful tool for studying pseudodifferential operators [57]. We will recall the precise definition of modulation spaces in Section 2 below.

Recently, the boundedness of unimodular Fourier multipliers on the modulation spaces has been investigated in [1, 815]. Particularly, one has the following results.

Theorem A (see [11]). Let ,, , and . One has, for , where

Here (and throughout this paper), we use the notation to mean that there is a positive constant independent of all essential variables such that .

Theorem B (see [15]). Let , , and . Then is bounded from to if and only if

In this paper, we use a different method from [15] to prove the following theorem, which, in particular, uses the modulation Hardy spaces that will be later defined in Section 2.

Theorem 1. Let , , . For a positive , denote . Let if n is even and if is odd.(i)Assume . If and , one has Particularly, the above inequality holds for all if is a positive even number. (ii) For any , one has for any .
Here (iii) Assume . If , then for all .

We want to make a few remarks on Theorem 1. First, (iii) in Theorem 1 says that when , compared to the case in (i), one obtains a larger range of and a smaller range of . We do not know if there is a unified formula regarding and for all dimension . Second, in the proof we will see that, in the low frequency parts of the definition of , the fractional Schrödinger semigroup has a growth when is growing, but it gains an arbitrary regularity. In the high frequency part, the semigroup can be controlled by at each piece of its decomposition with frequency . This phenomenon was also more precisely observed in [1, 15] (see also [11]). Thirdly, the case was studied in [8, 16].

Since the norm is dominated by the norm and the Riesz transforms are bounded on , by the Riesz transform characterization of the (see Section 2), we easily obtain the following corollary.

Corollary 2. Let , and . One has for where

Our next result shows that the asymptotic factor in Theorem 1 is the best for all , at least for .

Theorem 3. Let . The asymptotic factor in Theorem 1 is the best. Precisely, for , if then

In the next theorem, we state some mixed norm estimates.

Theorem 4. Let and . For , suppose .(i)If , then (ii)If , then

We consider the following linear Cauchy problem with negative power:

We give the grow-up rate of the solution to the above Cauchy problem in the modulation spaces.

Theorem 5. Assume and .(i)Let . One has that for any (ii)For any , one has

Now, we study the following Cauchy problem of the nonlinear dispersive equations (NDE): where for some positive integer . For , the space is defined by

We obtain the quantitative forms about the solution to the above Cauchy problem of the nonlinear dispersive equations.

Theorem 6. Let , , and assume
Assume for any
There exists such that the above Cauchy system (NDE) has a unique solution , where depends on the norm and .

According to the inclusions of modulation space (see Proposition in [13]), we know the space of initial data if .

Theorem 7. Let . Assume and for any
There exists such that the above Cauchy system (NDE) has a unique solution , where depends on the norm .

The rest of the paper is organized as follows. In Section 2, we recall or establish some necessary lemmas and known results. Sections 3 and 4 are devoted to the proofs of Theorems 1 and 3, respectively. Finally, in Section 5, we give some applications including the boundedness for the operator in the case and , including negative .

#### 2. Preliminaries

##### 2.1. The Definitions

The modulation space is originally defined by Feichtinper in 1983 on the locally compact Ablian groups . When , the modulation space can be equivalently defined by using the unit-cube decomposition to the frequency space (see Appendix in [13], also [14, 17]). The following definition is based on the unit-cube decomposition introduced in [13].

Let be a fixed nonnegative-valued function in with support in the cube and satisfy for any in the cube . By a standard constructive method, we may assume that for all , where is the -shift of that is defined by

For each , we use as its symbol of a smooth projection on the frequency space. Precisely, for any , we have

Let be a Banach space of measurable functions on with quasi-norm . We define the modulation space where By definition, we have the inclusion It is known that the definition of the modulation space is independent of the choice of functions . In this paper, we are particularly interested in the cases and , where is the Lebesgue space and is the real Hardy space. For all , we call the modulation spaces and the modulation Hardy space. As a usual notation we similarly define By the definition and known properties of , we have that for all , and for all , For simplicity in notation, we denote The following imbedding relation can be found in Proposition of [18]. Let , . If then

##### 2.2. Spaces

It is well known that the Hardy space coincides with the Lebesgue space when . For , the space has many characterizations. We will use its Riesz transform characterization in this paper. For an integer and multi-index , let denote the generalized Riesz transform where each is the Riesz transform of if and . It is known that for and all , where is a sum of finite terms.

The operator is a convolution. We have Also it is well known that is bounded on spaces for any .

##### 2.3. Some Lemmas and Known Results

Lemma 8. Let and . Suppose that there is an integer , such that for all test functions for and for . Here and is a real number. Then for , one has where is an arbitrary positive number.

Proof. The case is proved in [11]. It suffices to show the lemma for . By the Riesz transform characterization of , for , we have By checking the Fourier transform, we have the identity where So for , one has A similar argument shows that for , for any . The rest of the lemma easily follows from the definition of the modulation spaces.

Lemma 9 (see [18, 19]). Let denote an open set and . If and the rank of the matrix is at least for all (), then

Lemma 10. Let and . Suppose that is a function with support in . Then

Proof. The case is known [20]. It then suffices to show that for , for large . Let be a standard bump radial function supported in the set and satisfying, for all , Noting the support condition of , we write where the sets ,   are defined by For , we use polar coordinates to write where is the induced Lebesgue measure on the unit sphere . When is even, taking integration by parts for times on the inside integral, we obtain When is odd, we use integration by parts for times on the inside integral, Again we obtain that for odd , For , without loss of generality, we assume . Perform integration by parts on the variable for suitable amount of times. We similarly obtain For , invoking Lemma 9, we obtain Noting that contains no more than numbers of , it is easy to check
The lemma is proved.

Lemma 11 (see [21, pages 163–171]). Let and Suppose that is a Fourier multiplier with symbol . If is a bounded function which is of class in and if with , then is a bounded operator on and

Lemma 12. Let and . For all , one has

This lemma can be found in Section 4.2 of [11].

Lemma 13. Let be a compact subset in , and let . There exists a constant depending only on the diameter of and , such that for all satisfying .

This lemma is the Nikol'skij-Triebel inequality, see Proposition in [20] (also Lemma 2.5 in [22]).

Lemma 14. Let and be compact subsets of . Then there exists a constant depending only on the diameters of and , such that for all , satisfying and .

Lemma 15 (Pitt's theorem). If and , then

Lemma 16. Let and satisfy Then one has

This result is a particular case of Lemma 2.5 in [8].

#### 3. Proof of Theorem 1

The operator is a convolution operator with the symbol . This symbol is a function on with compact support. Clearly for any and , we have that for ,

So Lemma 11 implies the following estimate.

Proposition 17. Let . For any with , one has

By the proof of Lemma 8 and Proposition 17, we have that for all ,

The following proposition extends Lemma 12 to all .

Proposition 18. Let . For any with , for any , one has

Proof. The proof uses the same idea used in proving the case which was represented in [11]. For the convenience of the reader, we present its proof.
Let be the kernel of . Then By Lemma 14 and (46), we have Thus to prove the proposition, it suffices to show
For simplicity, we prove the case . The proof for , is tedious but shares the same idea as that for .
First we study the case . For , and , if we denote If , we denote
Also, for and , we define sets It is easy to check Let We have for , Write where
It is easy to check that if and supp, the phase function satisfies So by Lemma 9, we have
Observe the easy fact that if and supp , for any integer , Perform integration by parts on and variables both for times such that . An easy computation shows that
The estimates for and are exactly the same. We only estimate . Take integration by parts on variable for times with . Again, a simple computation shows that if we chose a suitably large . These estimates on , , indicate provided .
We now turn to show the case . For , and , let be the numbers defined above. For and , we define sets It is easy to check Let Thus, Using the same argument as we used before, we can show We complete the proof of Proposition 18.

We are now in a position to prove Theorem 1.

Proof. By an argument involving interpolation and duality, it suffices to show the case . Using Proposition 18, the inequality in (76) and the definition of the modulation spaces, we easily obtain (ii) in Theorem 1.
To show (i) and (iii) in Theorem 1, by Proposition 18 and the definition of the modulation spaces, it suffices to show Again, by Lemma 14, the proof of the inequality in (101) can be reduced to show that for , We show (iii) first. The proof of may illustrate the method. When By Hölder's inequality and the Plancherel theorem, the first term above For the second term, performing integration by parts, we obtain since
Now we return to show (i) of Theorem 1. We will prove only the case . Write Using Hölder's inequality and the Plancherel theorem, we obtain
For , we denote sets We now write To show (102), it now suffices to show that for each ,
Using the Leibniz rule, for any positive integer , we have Here, an easy induction argument shows that, for , where is a homogeneous function of degree for each . We now write where By the definition, it is easy to see that each is an and function with support in the cube .
Let . Performing integration by parts on variables for times, we have
We first estimate each , . Recall that we assume . Let , so . By the choice of and the assumption it is easy to see . Therefore, by Hölder's inequality, we obtain For each , by the choice of , the assumption on , and an easy computation, it is not difficult to see that we may obtain a number in the interval satisfying By Hölder's inequality and Pitt's theorem, for each , we obtain Combining all the estimates, we have It remains to estimate .
It is easy to see that the choice of and the condition in the theorem imply . So, by Hölder's inequality and Pitt's theorem again, we obtain This completes the proof of (102).
When , we have for any integer , where is a function. Thus it is trivial to see that for all . This proves (i) in Theorem 1.

#### 4. Proof of Theorem 3

Recall that the function chosen in the definition of the modulation space is flexible. We may choose where each is a nonnegative valued function in with support in and satisfies if and

For simplicity, we work on the case . Let be a nonnegative function with support in the set . For define a function on by and an by . Let . It is easy to see that where if and if . Similarly we have Suppose that we have some such that By the choice of , we have