Estimates for Unimodular Multipliers on Modulation Hardy Spaces

Δ x is the Laplacian and e α/2 is the multiplier operator with symbol e α (see [1] for its definition). The cases α = 1, 2, 3 are of particular interest because they correspond to the (half-) wave equation, the Schrödinger equation, and (essentially) the Airy equation, respectively. Unimodular Fouriermultipliers generally do not preserve any Lebesgue space L, except for p = 2. The L-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.Themodulation spacesM p,q (R)were first introduced by Feichtinger [2–4] to measure smoothness of a function or distribution in a way different from L spaces, and they are now recognized as a useful tool for studying pseudodifferential operators [5–7]. We will recall the precise definition of modulation spaces in Section 2 below. Recently, the boundedness of unimodular Fourier multipliers e α/2 on the modulation spaces has been investigated in [1, 8–15]. Particularly, one has the following results.


Introduction
A Fourier multiplier is a linear operator whose action on a test function on R 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 ∈ R + . They arise when one solves the Cauchy problem for dispersive equations. For example, for the solution ( , ) of the Cauchy problem + |Δ| /2 = 0, ( , ) ∈ R + × R , (0, ) = 0 ( ) , (2) we have the formula ( , ) = ( |Δ| /2 0 )( ). Here Δ = Δ is the Laplacian and |Δ| /2 is the multiplier operator with symbol | | (see [1] for its definition). The cases = 1, 2, 3 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 = 2. 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 , (R ) were first introduced by Feichtinger [2][3][4] 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 [5][6][7]. We will recall the precise definition of modulation spaces in Section 2 below.
Theorem A (see [11]). Let ∈ R,1 ≤ , ≤ ∞, > 1/2 and ̸ = 1. One has, for ≥ 1, 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 0 < < 1, 0 < ≤ ∞, > (1/ − 1) and ∈ R. Then |Δ| /2 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.
Particularly, the above inequality holds for all > 0 if is a positive even number.
We want to make a few remarks on Theorem 1. First, (iii) in Theorem 1 says that when = 1, compared to the case ≥ 2 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 ≥ 1. Second, in the proof we will see that, in the low frequency parts of the definition of , , the fractional Schrödinger semigroup has a growth |1/ −1/2| 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 = 1 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.
In the next theorem, we state some mixed norm estimates.
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.

Preliminaries
2.1. The Definitions. The modulation space is originally defined by Feichtinper in 1983 on the locally compact Ablian groups . When = R , the modulation space can be equivalently defined by using the unit-cube decomposition to the frequency space (see Appendix A 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 S(R ) with support in the cube [−4/5, 4/5] and satisfy ( ) = 1 for any in the cube [−2/5, 2/5] . By a standard constructive method, we may assume that for all ∈ R , where is the -shift of that is defined by For each ∈ Z , we use ( ) as its symbol of a smooth projection on the frequency space. Precisely, for any ∈ S (R ), we havê=̂.
Let be a Banach space of measurable functions on R with quasi-norm ‖ ⋅ ‖ . We define the modulation space where ( ,ℓ ) = ⟨ ⟩ ( ) ℓ , By definition, we have the inclusion It is known that the definition of the modulation space ( , ℓ ) 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 0 < , ≤ ∞, we call ( , ℓ )

The operator
|Δ| /2 is a convolution. We have Also it is well known that is bounded on spaces for any 0 < < ∞.

Lemma 8. Let 0 < < ∞ and ≥ 1. Suppose that there is an integer > 0, such that for all test functions
for | | < and for | | ≥ . Here 1 ≥ 2 ≥ 0 and is a real number. Then for ∈ ∩ 2 , one has where is an arbitrary positive number.
Proof. The case ≥ 1 is proved in [11]. It suffices to show the lemma for 0 < ≤ 1. By the Riesz transform characterization of , for | | ≥ , we have By checking the Fourier transform, we have the identity So for | | ≥ − 1, one has A similar argument shows that for | | < − 1, for any ∈ ∩ 2 . The rest of the lemma easily follows from the definition of the modulation spaces.
Lemma 9 (see [18,19]). Let ⊂ R denote an open set and ∈ ∞ 0 ( ). If ∈ ∞ ( ) and the rank of the matrix is at least > 0 for all ∈ supp( ), then (52) Proof. The case = 2 is known [20]. It then suffices to show that for 0 < < 2, for large . Let Γ be a standard bump radial function supported in the set and satisfying, for all ̸ = 0, Noting the support condition of̃, we write where the sets , = 1, 2, 3 are defined by For ∈ 1 , we use polar coordinates to write where is the induced Lebesgue measure on the unit sphere −1 . When is even, taking integration by parts for /2 times on the inside integral, we obtain When is odd, we use integration by parts for /2+1/2 times on the inside integral, Again we obtain that for odd , For ∈ 2 , without loss of generality, we assume | 1 | ≥ (| |/ ). Perform integration by parts on the 1 variable for suitable amount of times. We similarly obtain For ∈ 3 , invoking Lemma 9, we obtain Noting that 3 contains no more than (10/|1 − |) + log 2 numbers of , it is easy to check The lemma is proved.
This lemma can be found in Section 4.2 of [11].

Proof of Theorem 1
The operator is a convolution operator with the symbol ( − ) | | . This symbol is a ∞ function on R \{0} with compact support. Clearly for any > 0 and ̸ = 0, we have that for | | ≤ 10, So Lemma 11 implies the following estimate.
Proof. The proof uses the same idea used in proving the case ≥ 1 which was represented in [11]. For the convenience of the reader, we present its proof.
Let Ω 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 = 2. The proof for ≥ 3, is tedious but shares the same idea as that for = 2.
First we study the case | | −2 > 1. For = 1, 2, and = ( 1 , 2 ), if ≥ 2 we denote If < 2, we denote Journal of Function Spaces and Applications 7 Also, for = 1, 2 and ∈ N, we define sets It is easy to check We have for = 1, 2, Write where It is easy to check that if | | −2 ≥ 1 and ∈ supp( ), the phase function So by Lemma 9, we have Observe the easy fact that if ∈ 1, and ∈ supp ( ), for any integer , Perform integration by parts on 1 and 2 variables both for times such that > 1. An easy computation shows that The estimates for 2 and 3 are exactly the same. We only estimate 2 . Take integration by parts on 1 variable for times with > 1. Again, a simple computation shows that if we chose a suitably large . These estimates on , = 1, 2, 3, 4, indicate provided | | −2 ≥ 1.

Journal of Function Spaces and Applications
We now turn to show the case | | −2 < 1. For = 1, 2, and = ( 1 , 2 ), let ( ) ( ) be the numbers defined above. For = 1, 2 and ∈ N, 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 ≤ 1. 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 ≥ 1, We show (iii) first. The proof of = 1 may illustrate the method. When = 1 By Hölder's inequality and the Plancherel theorem, the first term above For the second term, performing integration by parts, we obtain Journal of Function Spaces and Applications 9 Now we return to show (i) of Theorem 1. We will prove only the case < 1. Write Using Hölder's inequality and the Plancherel theorem, we obtain For , ∈ {1, 2, . . . , }, 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 ≥ 1, is a homogeneous function of degree − for each . We now write We first estimate each , 0 ≤ ≤ − 1. Recall that we assume ∈ (0, 2). Let = 2/ , so = 2 /(2 − ). By the choice of and the assumption it is easy to see > . Therefore, by Hölder's inequality, we obtain For each = 1, 2, . . . , −1, 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 [0, /2) satisfying − > ( 1 − 1 2 ) , + − > − 2 , 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 − > − /2. So, by Hölder's inequality and Pitt's theorem again, we obtain This completes the proof of (102). When = 2, 4, . . ., we have for any integer , where ( ) is a ∞ function. Thus it is trivial to see that for all > 0. This proves (i) in Theorem 1.

Proof of Theorem 3
Recall that the function chosen in the definition of the modulation space is flexible. We may choose define a function Φ on R by and an ∈ (R ) by = F −1 (Φ). Let ∈ Z . It is easy to see that where ( ) = 1 if = 0 and ( ) = 0 if ̸ = 0. Similarly we have Suppose that we have some such that By the choice of , we have Journal of Function Spaces and Applications

11
On the other hand, where the phase function ( , ) is defined by Since The critical point of ( , ) is at * = − 2 . (137) Thus, by the stationary phase method (see [19,Proposition 3, page 334]), an easy computation gives that, as → ∞, Thus the inequality implies This shows the conclusion.

Operators , ( ).
Let ̸ = 1 and ∈ R. In [23], to investigate the absolute convergence for multiple Fourier series, Wainger studied the oscillating multipliers , (t) with symbol In [24], Miyachi proved that in the case > 0 and ̸ = 1, for 0 < < ∞, if and only if By Theorem 1 and its proof, we not only obtain the boundedness of , ( ) on , (R ) for any ∈ R, but also gain a regularity of if > 0.
Theorem 19. Let 0 < < ∞ and ̸ = 1. One has for ≥ 1 Proof. The proof of Theorem 19 is the same as the proof of Theorem 1. We skip it.

Journal of Function Spaces and Applications
Let 1 ≤ ≤ ∞. For ∈ 1 , we have This shows that for | | ≤ 10, Now if 0 < ≤ 1, we obtain that for all 1 ≤ ≤ ∞, The last inequality is from Lemma 13.
As these discussions, we obtain the following corollary.
Proof. By the definition of the modulation space, we know If / ≥ 1, write By the Minkowski inequality, we have When | | ≤ 10, When | | ≥ 9, This shows that if ≥ , then If / ≤ 1, then As in the previous case, for | | ≤ 10, and This shows that The theorem is proved.

Schrödinger Equation.
Consider the Cauchy problem of the linear free Schrödinger equation The formal solution to this equation is given by By Theorem 1, we obtain the growth rate as → ∞ for the solution to the linear free Schrödinger equation.

Linear Cauchy Problem with Negative Power.
We start with the following linear Cauchy problem with negative power: The formal solution to this equation is given by Proposition 26. Let > 0 and 1 ≤ ≤ 2. One has where Proof. We only prove the case of odd , since the proof for even is similar. For any fixed we write where is the Riesz potential of order . The kernel of We first show To this end, by Young's inequality, we need to show It suffices to show the case ≥ 1. As the same argument in the proof of Theorem 1, with the Schwarz inequality we have Performing integration by parts ( + 1)/2 times on the second term, without loss of generality, we may write wherẽ( ) is a ∞ function supported in [−1, 1] and ( ) is a function satisfying Choose a small > 0 such Thus by Schwarz's inequality and the Pitt's theorem, we obtain Combining these estimates, we have 0 − |Δ| − /2 where 1 − denotes the 1 Sobolev space of order − . On the other hand, we have the easy energy estimate An interpolation yields that for all 1 ≤ ≤ 2, Since is an arbitrary number larger than ([ /2] + 1), the proposition now follows from Lemma 13.
Proof. The almost orthogonality (Identity (46)) and the energy estimate give Thus the lemma follows from Lemma 13.
Now we are ready to give the proof of Theorem 5.
Proof. The proof of (i) can be obtained from Propositions 18 and 26, and the definition and the modulation spaces. Similarly, the proof of (ii) follows by Proposition 18, Lemma 27 and the definition of the modulation spaces.
Our proof will follow the same method used in [8], or, more precisely, the idea introduced in an earlier paper [13]. Now we give the proof of Theorem 6.
Proof. In the proof, the letters , = 1, 2, 3 denote some positive constants that are independent of all essential variables. We write the Cauchy problem in the equivalent form The last inequality is because Fix an > 0 such that