# Estimates for Unimodular Multipliers on Modulation Hardy Spaces

Authors: Jiecheng Chen; Dashan Fan; Lijing Sun; Chunjie Zhang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

It is known that the unimodular Fourier multiplierseit|Δ|α/2,α>0, are bounded on all modulation spaces Mp,qs for 1≤p,q≤∞. We extend such boundedness to the case of all 0<p,q≤∞ 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 Mp1,qs→Mp2,qs boundedness for the operator eit|Δ|α/2, for the case 0<α≤2 and α≠1. Finally, we investigate the boundedness of the operator eit|Δ|α/2 for α>0 and obtain the local well-posedness for the Cauchy problem of some nonlinear partial differential equations with fundamental semigroup eit|Δ|α/2.


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## Body

## 1. Introduction

A Fourier multiplier is a linear operatorHμ  whose action on a test function f on ℝn is formally defined by
(1)Hμf(x)=1(2π)n∫ℝn‍μ(ξ)f^(ξ)ei2πξ·xdξ.
The function μ  is called the symbol or multiplier of  Hμ.In this paper, we will study the unimodular Fourier multipliers with symboleit|ξ|α for t∈ℝ+. They arise when one solves the Cauchy problem for dispersive equations. For example, for the solution u(t,x) of the Cauchy problem
(2)i∂tu+|Δ|α/2u=0,(t,x)∈ℝ+×ℝn,u(0,x)=u0(x),
we have the formula u(t,x)=(eit|Δ|α/2u0)(x).  Here  Δ=Δx  is the Laplacian and  eit|Δ|α/2  is the multiplier operator with symbol  eit|ξ|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 Fourier multipliers generally do not preserve any Lebesgue spaceLp, except for p=2. The Lp-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  Mp,qs(ℝn)  were first introduced by Feichtinger [2–4] to measure smoothness of a function or distribution in a way different from Lp 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 multiplierseit|Δ|α/2  on the modulation spaces has been investigated in [1, 8–15]. Particularly, one has the following results.Theorem A (see [11]).
Lets∈ℝ,1≤p, q≤∞, α>1/2and α≠1. One has, for  t≥1,
(3)∥eit|Δ|α/2f∥Mp,qs⪯tn|1/2-1/p|∥f∥Mp,qs+γ+∥f∥Mp,qs,
where
(4)γ=n(α-2)|12-1p|.Here (and throughout this paper), we use the notationA⪯B  to mean that there is a positive constant  C  independent of all essential variables such that  A≤CB.Theorem B (see [15]).
Let0<p<1, 0<q≤∞, α>n(1/p-1) and s∈ℝ. Then eit|Δ|α/2 is bounded from Mp,qs to Mp,q if and only if
(5)s≥max{0,α-2}n(1p-12).In this paper, we use a different method from [15] to prove the following theorem, which, in particular, uses the modulation Hardy spaces μp,qs that will be later defined in Section 2.Theorem 1.
Lets∈ℝ, 0<q≤∞, t≥1. For a positive α≠1, denote γ=n(α-2)|1/p-1/2|. Let α0=2/(2+n) if n is even and α0=3/(3+n) if n is odd.(i)
Assumeα>α0. If   p>n/(n+α)  and  n≥2,  one has
(6)∥eit|Δ|α/2f∥Mp,qs(ℝn)⪯tn|1/p-1/2|∥f∥Mp,qs+γ(ℝn)+∥f∥Mp,qs(ℝn).Particularly, the above inequality holds for all  p>0  if  αis a positive even number. (ii)
For any0<p≤∞, one has
(7)∥eit|Δ|α/2f∥Mp,qs(ℝn)⪯tn|1/p-1/2|(∥f∥Mp,qs+γ(ℝn)+∥f∥μp,qs-δ(ℝn)),for any  δ≥0.
Here(8)γ=n(α-2)|1p-12|.(iii)
Assumen=1.  If  α>1/2, then
(9)∥eit|Δ|α/2f∥Mp,qs(ℝ)⪯t|1/p-1/2|∥f∥Mp,qs+γ(ℝ)+∥f∥Mp,qs(ℝ)for all  p>2/3.We want to make a few remarks on Theorem1. First, (iii) in Theorem 1 says that when  n=1, compared to the case  n≥2  in (i), one obtains a larger range of  α  and a smaller range of  p. We do not know if there is a unified formula regarding  α  and  p  for all dimension  n≥1. Second, in the proof we will see that, in the low frequency parts of the definition of  Mp,qs, the fractional Schrödinger semigroup has a growth  tn|1/p-1/2|  when  t  is growing, but it gains an arbitrary regularity. In the high frequency part, the semigroup can be controlled by
(10)max{1,(t|k|α-2)n|1/p-1/2|},
at each piece of its decomposition with frequency  k. This phenomenon was also more precisely observed in [1, 15] (see also [11]). Thirdly, the case α=1 was studied in [8, 16].Since theLp norm is dominated by the  Hp  norm and the Riesz transforms are bounded on  Hp, by the Riesz transform characterization of the Hp (see Section 2), we easily obtain the following corollary.Corollary 2.
Lets∈ℝ, 0<p,q≤∞and α>0,α≠1. One has for  t≥1(11)∥eit|Δ|α/2f∥μp,qs⪯tn|1/p-1/2|∥f∥μp,qs+γ,
where
(12)γ=n(α-2)|1p-12|.Our next result shows that the asymptotic factortn|1/p-1/2|  in Theorem 1 is the best for all  p>0, at least for α=2.Theorem 3.
Letα=2. The asymptotic factor tn|1/p-1/2|  in Theorem 1 is the best. Precisely, for  t>1, if
(13)∥eit|Δ|f∥Mp,qs⪯tδ∥f∥Mp,qs,
then
(14)δ≥n|1p-12|.In the next theorem, we state some mixed norm estimates.Theorem 4.
Let0<α≤2and α≠1. For 0<p1≤1≤p2≤∞, suppose nr(1/2-1/p2)>1.(i)
Ifr≥q,   then
(15)∥∥eit|Δ|α/2f∥Mp2s,q∥Lr(0,∞)⪯∥f∥Mp1,qs+(2-α)/r.(ii)
Ifr≤q,   then
(16)∥∥eit|Δ|α/2f∥Mp2s,q∥Lr(0,∞)⪯∥f∥Mp1,rs+(2-α)/r.We consider the following linear Cauchy problem with negative power:(17)i∂tu+|Δx|-α/2u=0,(t,x)∈ℝ+×ℝn,α>0,u(0,x)=u0(x).We give the grow-up rate of the solution to the above Cauchy problem in the modulation spaces.Theorem 5.
Assumeα>0 and  γ=n(α+2)|1/2-1/p|.(i)
Let1≤p<2.  One has that for any  0<r<np/(n+α([n/2]+1)(2-p))(18)∥u(t)∥Mp,qs(ℝn)⪯(1+t)([n/2]+1)|2/p-1|∥u0∥μr,qs-γ(ℝn)+∥u0∥Mp,qs(ℝn).(ii)
For anyr≤2≤p,  one has
(19)∥u(t)∥Mp,qs(ℝn)⪯∥u0∥μr,qs(ℝn)+(1+t)n|1/p-1/2|∥u0∥Mp,qs-γ(ℝn).Now, we study the following Cauchy problem of the nonlinear dispersive equations (NDE):(20)i∂u∂t-|Δ|-α/2u+F(u)=0,u(0,x)=u0(x),
where   F(u)=|u|2ku   for some positive integer  k. For  T>0, the space  Y=C([0,T],Mp,1s)  is defined by
(21)C([0,T],Mp,1s)={u(t,x):∥u∥C([0,T],Mp,1s)=sup0≤t≤T∥u(t,·)∥Mp,1s<∞}.We obtain the quantitative forms about the solution to the above Cauchy problem of the nonlinear dispersive equations.Theorem 6.
Let2>p>1, γ=n(α+2)|1/2-1/p|, α>0   and   assume
(22)npn+α([n/2]+1)(2-p)>1.
Assumeu0∈Mr,1s-γ∩Mp,1s  for any
(23)1<r<npn+α([n/2]+1)(2-p).
There existsT>0  such that the above Cauchy system (NDE) has a unique solution  u∈C([0,T],Mp,1s),  where  T  depends on the norm   ∥u0∥Mr,1s-γ  and   ∥u0∥Mp,1s.According to the inclusions of modulation space (see Proposition2.5 in [13]), we know the space of initial data   Mr,1s-γ∩Mp,1s=Mr,1s-γ  if  r≤p.Theorem 7.
Letp≥2,γ=n(α+2)|1/2-1/p|. Assume
(24)p2k+1≥1
and  u0∈Mr,1s  for any
(25)p2k+1≤r<2.
There existsT>0  such that the above Cauchy system (NDE) has a unique solution  u∈C([0,T],Mp,1s),  where  T  depends on the norm   ∥u0∥Mr,1s.The rest of the paper is organized as follows. In Section2, 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  Mp1,qs→Mp2,qs  boundedness for the operator  eit|Δ|α/2  in the case  α≤2 and α≠1, including negative  α.


## 2. Preliminaries

### 2.1. The Definitions

The modulation space is originally defined by Feichtinper in 1983 on the locally compact Ablian groupsG. When  G=ℝn, 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].Letσ be a fixed nonnegative-valued function in  𝒮(ℝn)  with support in the cube  [-4/5,4/5]n and satisfy  σ(ξ)=1  for any  ξ  in the cube  [-2/5,2/5]n. By a standard constructive method, we may assume that for all  ξ∈ℝn,
(26)∑k∈ℤn‍σk(ξ)=1,
where σk is the k-shift of σ that is defined by
(27)σk(ξ)=σ(ξ-k),σ0(ξ)=σ(ξ).For eachk∈ℤn, we use σk(ξ) as its symbol of a smooth projection πk on the frequency space. Precisely, for any  f∈𝒮′(ℝn), we have
(28)πkf^=σkf^.LetX  be a Banach space of measurable functions on  ℝn with quasi-norm  ∥·∥X. We define the modulation space
(29)Ms(X,ℓq)(ℝn)={fis  measurable:∥f∥Ms(X,ℓq)<∞},
where
(30)∥f∥Ms(X,ℓq)=∥〈k〉s∥πk(f)∥X∥ℓq,〈k〉=1+|k|.
By definition, we have the inclusion
(31)Ms(X,ℓq)⊂Mr(X,ℓq)ifr≤s.
It is known that the definition of the modulation space Ms(X,ℓq)  is independent of the choice of functions σ. In this paper, we are particularly interested in the cases X=Lp  and  X=Hp, where Lp is the Lebesgue space and Hp is the real Hardy space. For all  0<p,q≤∞, we call Ms(Lp,ℓq) the modulation spaces and Ms(Hp,ℓq) the modulation Hardy space. As a usual notation
(32)Ms(Lp,ℓq)=Mp,qs=Mp,qs(ℝn),
we similarly define
(33)Ms(Hp,ℓq)=μp,qs=μp,qs(ℝn).
By the definition and known properties of Hp, we have that for all  1<p<∞,
(34)μp,qs(ℝn)=Mp,qs(ℝn),
and for all  0<p≤1,
(35)μp,qs(ℝn)⊂Mp,qs(ℝn).
For simplicity in notation, we denote
(36)μp,q(ℝn)=μp,q0(ℝn),Mp,q(ℝn)=Mp,q0(ℝn).
The following imbedding relation can be found in Proposition 5.1.5 of [18]. Let  s1,s2∈ℝ, 0<p1,p2,q1,q2≤∞. If
(37)s2≤s1,p1≤p2,q1≤q2
then
(38)Mp1,q1s1⊂Mp2,q2s2.


### 2.2.Hp Spaces

It is well known that the Hardy spaceHp(ℝn)  coincides with the Lebesgue space Lp(ℝn)  when  1<p<∞. For  0<p≤1, the space Hp(ℝn)  has many characterizations. We will use its Riesz transform characterization in this paper. For an integer  L≥0and multi-index  J=(j1,…,jN)∈{1,2,…,n}L, let  RJ  denote the generalized Riesz transform
(39)RJ(f)=Rj1Rj2RjL(f),
where each  Rj(f)  is the  jth  Riesz transform of  f  if j≠0and R0(f)=f. It is known that for  p>(n-1)/(n-1+L)  and all  f∈Hp∩L2,
(40)∥f∥Hp≃∑J‍∥RJ(f)∥Lp,
where   ∑J‍  is a sum of finite terms.The operatorπkeit|Δ|α/2 is a convolution. We have
(41)∥πkeit|Δ|α/2f∥Hp≃∑J‍∥πkeit|Δ|α/2RJ(f)∥Lp.
Also it is well known that  RJ  is bounded on  Hp  spaces for any  0<p<∞.


### 2.3. Some Lemmas and Known Results

Lemma 8.
Let0<p<∞and t≥1. Suppose that there is an integer  N>0,  such that for all test functions  f(42)∥πkeit|Δ|α/2f∥Lp⪯tb1∥f∥Hp
for  |k|<N  and
(43)∥πkeit|Δ|α/2f∥Lp⪯tb2|k|d∥f∥Hp
for  |k|≥N. Here  b1≥b2≥0 and d  is a real number. Then for  f∈Hp∩L2, one has
(44)∥eit|Δ|α/2f∥Mp,qs⪯tb1∥f∥Mp,qs-ν+tb2∥f∥Mp,qs+d,
where   ν is an arbitrary positive number.Proof.
The casep≥1 is proved in [11]. It suffices to show the lemma for  0<p≤1. By the Riesz transform characterization of  Hp, for  |k|≥N,  we have
(45)∥πkeit|Δ|α/2f∥Hp≃∑J‍∥πkeit|Δ|α/2RJ(f)∥Lp⪯∑J‍tb2|k|d∥RJf∥Lp≃tb2|k|d∥f∥Hp.
By checking the Fourier transform, we have the identity
(46)πk=∑|j|∞≤1‍πk+jπk,
where
(47)j=(j1,…,jn),|j|∞=max{|j1|,…,|jn|}.
So for  |k|≥N-1, one has
(48)∥πkeit|Δ|α/2f∥Hp=∥∑|j|∞≤1‍πk+jπk(Sα(t)f)∥Hp⪯∑|j|∞≤1‍∥πk+jeit|Δ|α/2(πkf)∥Hp⪯tb2|k|d∥πkf∥Hp.
A similar argument shows that for  |k|<N-1,
(49)∥πkeit|Δ|α/2f∥Hp⪯tb1∥πkf∥Hp
for any  f∈Hp∩L2.  The rest of the lemma easily follows from the definition of the modulation spaces.Lemma 9 (see [18, 19]).
LetE⊂ℝn  denote an open set and  ψ∈C0∞(E). If  ϕ∈C∞(E)  and the rank of the matrix
(50)(DxjDxiϕ(x))i,j=1n
is at least  m>0  for all  x∈supp(ψ), then
(51)|∫ℝn‍eiλϕ(x)ψ(x)dx|⪯|λ|-m/2∥ψ∥C2n.Lemma 10.
Let2≥α>0and α≠1. Suppose that  σ~  is a  C∞ function with support in  [-1,1]n. Then
(52)∥ℱ-1(σ~(ξ)eit|ξ|α)∥L∞(ℝn)⪯min{1,t-n/2}.Proof.
The caseα=2  is known [20]. It then suffices to show that for  0<α<2,
(53)∥ℱ-1(σ~(ξ)eit|ξ|α)∥L∞(ℝn)⪯t-n/2
for large  t. Let  Γ  be a standard bump radial function supported in the set
(54){ξ:12<|ξ|≤2}
and satisfying, for all  ξ≠0,
(55)∑j‍Γ(2j|ξ|)=1.
Noting the support condition of  σ~, we write
(56)|ℱ-1(σ~(ξ)eit|ξ|α)(x)|≃|∫ℝn‍σ~(ξ)eit|ξ|αei(x·ξ)dξ|≃|∑j=-∞∞‍∫ℝn‍Γ(2j|ξ|)σ~(ξ)eit|ξ|αei(x·ξ)dξ|⪯|∑j∈A1∪A2∪A3∞‍2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|,
where the sets  Ak,  k=1,2,3  are defined by
(57)A1={j≥0:2-j(α-1)-5≥|x|αt},A2={j≥0:n2-j(α-1)+5≤|x|αt},A3={j≥0:25|x|αt>2j(1-α)>2-5|x|αtn}.
For  j∈A1, we use polar coordinates to write
(58)2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ=2-jn∫Sn-1‍{∫0∞‍rn-1Γ(r)σ~(2-jrξ|ξ|)×eit2-jαrαei2-jr(x·(ξ/|ξ|))dr{(2-jrξ|ξ|)}}dξ′,
where  dξ′  is the induced Lebesgue measure on the unit sphere  Sn-1.  When  n  is even, taking integration by parts for  n/2  times on the inside integral, we obtain
(59)∑j∈A1‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2∑j∈A1‍2-jn(1-α/2)⪯t-n/2.
When  n  is odd,  we use integration by parts for n/2+1/2  times on the inside integral,
(60)|∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯|∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|n/(n+1)⪯t-n/22jαn/2.
Again we obtain that for odd  n,
(61)∑j∈A1‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2.
For  j∈A2, without loss of generality, we assume  |x1|≥(|x|/n).  Perform integration by parts on the  ξ1  variable for suitable amount of times. We similarly obtain
(62)∑j∈A2‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2.
For  j∈A3, invoking Lemma 9, we obtain
(63)|∫ℝn‍Γ(|ξ|)σ~(2-jξ)ei(t2-jα|ξ|α+2-j(x·ξ))dξ|⪯min{1,2jαn/2tn/2}.
Noting that  A3contains no more than  (10/|1-α|)+log2n numbers of  j, it is easy to check
(64)∑j∈A3‍2-jnmin{1,2jαn/2tn/2}⪯t-n/2.
The lemma is proved.Lemma 11 (see [21, pages 163–171]).
Let0<p<2  and
(65)N=[n(1p-12)]+1.
Suppose that  Tm  is a Fourier multiplier with symbol  m. If  m  is a bounded function which is of class  CN  in  ℝn∖{0}  and if
(66)|(∂∂ξ)Jm(ξ)|≤(A|ξ|-1)|J|for|J|≤N
with  A≥1, then  Tm  is a bounded operator on  Hp  and
(67)∥Tm∥Hp→Hp⪯An(1/p-1/2).Lemma 12.
Let|k|≥1 and α≠1. For all 1≤p≤∞, one has
(68)∥πkeit|Δ|α/2f∥Lp⪯∥πkf∥Lpift|k|α-2≤1,∥πkeit|Δ|α/2f∥Lp⪯(t|k|α-2)n|(1/p)-(1/2)|∥πkf∥Lpift|k|α-2≥1.This lemma can be found in Section 4.2 of [11].Lemma 13.
LetF  be a compact subset in  ℝn, and let  0<p≤q≤∞. There exists a constant  C  depending only on the diameter of  F  and  p, such that
(69)∥f∥Lq≤C∥f∥Lp
for all  f∈Lp satisfying  suppf^⊂F.This lemma is the Nikol'skij-Triebel inequality, see Proposition1.3.2 in [20] (also Lemma 2.5 in [22]).Lemma 14.
Let0<p≤1  and  F,F′ be compact subsets of ℝn. Then there exists a constant  C  depending only on the diameters of  F,F′  and  p, such that
(70)∥|f|*|g|∥Lp≤C∥f∥Lp∥g∥Lp
for all  f, g∈Lp  satisfying  suppf^⊂F  and  suppg^⊂F′.This is Lemma 2.6 in [22] (see also Proposition 1.5.3 in [20]).Lemma 15 (Pitt's theorem).
If1<p≤2  and  0≤α<1/p′, then
(71)(∫ℝn‍|f^(x)|p′|x|-αnp′dx)1/p′⪯(∫ℝn‍|f(x)|p|x|αnpdx)1/p.Lemma 16.
Lets∈ℝ  and  p,pj≥1(j=1,2,…,m)  satisfy
(72)1p=1p1+1p2+⋯+1pm.
Then one has
(73)∥∏j=1m‍uj∥Mp,1s⪯∏j=1m‍∥uj∥Mpj,1s.This result is a particular case of Lemma 2.5 in [8].


## 2.1. The Definitions

The modulation space is originally defined by Feichtinper in 1983 on the locally compact Ablian groupsG. When  G=ℝn, 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].Letσ be a fixed nonnegative-valued function in  𝒮(ℝn)  with support in the cube  [-4/5,4/5]n and satisfy  σ(ξ)=1  for any  ξ  in the cube  [-2/5,2/5]n. By a standard constructive method, we may assume that for all  ξ∈ℝn,
(26)∑k∈ℤn‍σk(ξ)=1,
where σk is the k-shift of σ that is defined by
(27)σk(ξ)=σ(ξ-k),σ0(ξ)=σ(ξ).For eachk∈ℤn, we use σk(ξ) as its symbol of a smooth projection πk on the frequency space. Precisely, for any  f∈𝒮′(ℝn), we have
(28)πkf^=σkf^.LetX  be a Banach space of measurable functions on  ℝn with quasi-norm  ∥·∥X. We define the modulation space
(29)Ms(X,ℓq)(ℝn)={fis  measurable:∥f∥Ms(X,ℓq)<∞},
where
(30)∥f∥Ms(X,ℓq)=∥〈k〉s∥πk(f)∥X∥ℓq,〈k〉=1+|k|.
By definition, we have the inclusion
(31)Ms(X,ℓq)⊂Mr(X,ℓq)ifr≤s.
It is known that the definition of the modulation space Ms(X,ℓq)  is independent of the choice of functions σ. In this paper, we are particularly interested in the cases X=Lp  and  X=Hp, where Lp is the Lebesgue space and Hp is the real Hardy space. For all  0<p,q≤∞, we call Ms(Lp,ℓq) the modulation spaces and Ms(Hp,ℓq) the modulation Hardy space. As a usual notation
(32)Ms(Lp,ℓq)=Mp,qs=Mp,qs(ℝn),
we similarly define
(33)Ms(Hp,ℓq)=μp,qs=μp,qs(ℝn).
By the definition and known properties of Hp, we have that for all  1<p<∞,
(34)μp,qs(ℝn)=Mp,qs(ℝn),
and for all  0<p≤1,
(35)μp,qs(ℝn)⊂Mp,qs(ℝn).
For simplicity in notation, we denote
(36)μp,q(ℝn)=μp,q0(ℝn),Mp,q(ℝn)=Mp,q0(ℝn).
The following imbedding relation can be found in Proposition 5.1.5 of [18]. Let  s1,s2∈ℝ, 0<p1,p2,q1,q2≤∞. If
(37)s2≤s1,p1≤p2,q1≤q2
then
(38)Mp1,q1s1⊂Mp2,q2s2.


## 2.2.Hp Spaces

It is well known that the Hardy spaceHp(ℝn)  coincides with the Lebesgue space Lp(ℝn)  when  1<p<∞. For  0<p≤1, the space Hp(ℝn)  has many characterizations. We will use its Riesz transform characterization in this paper. For an integer  L≥0and multi-index  J=(j1,…,jN)∈{1,2,…,n}L, let  RJ  denote the generalized Riesz transform
(39)RJ(f)=Rj1Rj2RjL(f),
where each  Rj(f)  is the  jth  Riesz transform of  f  if j≠0and R0(f)=f. It is known that for  p>(n-1)/(n-1+L)  and all  f∈Hp∩L2,
(40)∥f∥Hp≃∑J‍∥RJ(f)∥Lp,
where   ∑J‍  is a sum of finite terms.The operatorπkeit|Δ|α/2 is a convolution. We have
(41)∥πkeit|Δ|α/2f∥Hp≃∑J‍∥πkeit|Δ|α/2RJ(f)∥Lp.
Also it is well known that  RJ  is bounded on  Hp  spaces for any  0<p<∞.


## 2.3. Some Lemmas and Known Results

Lemma 8.
Let0<p<∞and t≥1. Suppose that there is an integer  N>0,  such that for all test functions  f(42)∥πkeit|Δ|α/2f∥Lp⪯tb1∥f∥Hp
for  |k|<N  and
(43)∥πkeit|Δ|α/2f∥Lp⪯tb2|k|d∥f∥Hp
for  |k|≥N. Here  b1≥b2≥0 and d  is a real number. Then for  f∈Hp∩L2, one has
(44)∥eit|Δ|α/2f∥Mp,qs⪯tb1∥f∥Mp,qs-ν+tb2∥f∥Mp,qs+d,
where   ν is an arbitrary positive number.Proof.
The casep≥1 is proved in [11]. It suffices to show the lemma for  0<p≤1. By the Riesz transform characterization of  Hp, for  |k|≥N,  we have
(45)∥πkeit|Δ|α/2f∥Hp≃∑J‍∥πkeit|Δ|α/2RJ(f)∥Lp⪯∑J‍tb2|k|d∥RJf∥Lp≃tb2|k|d∥f∥Hp.
By checking the Fourier transform, we have the identity
(46)πk=∑|j|∞≤1‍πk+jπk,
where
(47)j=(j1,…,jn),|j|∞=max{|j1|,…,|jn|}.
So for  |k|≥N-1, one has
(48)∥πkeit|Δ|α/2f∥Hp=∥∑|j|∞≤1‍πk+jπk(Sα(t)f)∥Hp⪯∑|j|∞≤1‍∥πk+jeit|Δ|α/2(πkf)∥Hp⪯tb2|k|d∥πkf∥Hp.
A similar argument shows that for  |k|<N-1,
(49)∥πkeit|Δ|α/2f∥Hp⪯tb1∥πkf∥Hp
for any  f∈Hp∩L2.  The rest of the lemma easily follows from the definition of the modulation spaces.Lemma 9 (see [18, 19]).
LetE⊂ℝn  denote an open set and  ψ∈C0∞(E). If  ϕ∈C∞(E)  and the rank of the matrix
(50)(DxjDxiϕ(x))i,j=1n
is at least  m>0  for all  x∈supp(ψ), then
(51)|∫ℝn‍eiλϕ(x)ψ(x)dx|⪯|λ|-m/2∥ψ∥C2n.Lemma 10.
Let2≥α>0and α≠1. Suppose that  σ~  is a  C∞ function with support in  [-1,1]n. Then
(52)∥ℱ-1(σ~(ξ)eit|ξ|α)∥L∞(ℝn)⪯min{1,t-n/2}.Proof.
The caseα=2  is known [20]. It then suffices to show that for  0<α<2,
(53)∥ℱ-1(σ~(ξ)eit|ξ|α)∥L∞(ℝn)⪯t-n/2
for large  t. Let  Γ  be a standard bump radial function supported in the set
(54){ξ:12<|ξ|≤2}
and satisfying, for all  ξ≠0,
(55)∑j‍Γ(2j|ξ|)=1.
Noting the support condition of  σ~, we write
(56)|ℱ-1(σ~(ξ)eit|ξ|α)(x)|≃|∫ℝn‍σ~(ξ)eit|ξ|αei(x·ξ)dξ|≃|∑j=-∞∞‍∫ℝn‍Γ(2j|ξ|)σ~(ξ)eit|ξ|αei(x·ξ)dξ|⪯|∑j∈A1∪A2∪A3∞‍2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|,
where the sets  Ak,  k=1,2,3  are defined by
(57)A1={j≥0:2-j(α-1)-5≥|x|αt},A2={j≥0:n2-j(α-1)+5≤|x|αt},A3={j≥0:25|x|αt>2j(1-α)>2-5|x|αtn}.
For  j∈A1, we use polar coordinates to write
(58)2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ=2-jn∫Sn-1‍{∫0∞‍rn-1Γ(r)σ~(2-jrξ|ξ|)×eit2-jαrαei2-jr(x·(ξ/|ξ|))dr{(2-jrξ|ξ|)}}dξ′,
where  dξ′  is the induced Lebesgue measure on the unit sphere  Sn-1.  When  n  is even, taking integration by parts for  n/2  times on the inside integral, we obtain
(59)∑j∈A1‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2∑j∈A1‍2-jn(1-α/2)⪯t-n/2.
When  n  is odd,  we use integration by parts for n/2+1/2  times on the inside integral,
(60)|∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯|∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|n/(n+1)⪯t-n/22jαn/2.
Again we obtain that for odd  n,
(61)∑j∈A1‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2.
For  j∈A2, without loss of generality, we assume  |x1|≥(|x|/n).  Perform integration by parts on the  ξ1  variable for suitable amount of times. We similarly obtain
(62)∑j∈A2‍|2-jn∫ℝn‍Γ(|ξ|)σ~(2-jξ)eit2-jα|ξ|αei2-j(x·ξ)dξ|⪯t-n/2.
For  j∈A3, invoking Lemma 9, we obtain
(63)|∫ℝn‍Γ(|ξ|)σ~(2-jξ)ei(t2-jα|ξ|α+2-j(x·ξ))dξ|⪯min{1,2jαn/2tn/2}.
Noting that  A3contains no more than  (10/|1-α|)+log2n numbers of  j, it is easy to check
(64)∑j∈A3‍2-jnmin{1,2jαn/2tn/2}⪯t-n/2.
The lemma is proved.Lemma 11 (see [21, pages 163–171]).
Let0<p<2  and
(65)N=[n(1p-12)]+1.
Suppose that  Tm  is a Fourier multiplier with symbol  m. If  m  is a bounded function which is of class  CN  in  ℝn∖{0}  and if
(66)|(∂∂ξ)Jm(ξ)|≤(A|ξ|-1)|J|for|J|≤N
with  A≥1, then  Tm  is a bounded operator on  Hp  and
(67)∥Tm∥Hp→Hp⪯An(1/p-1/2).Lemma 12.
Let|k|≥1 and α≠1. For all 1≤p≤∞, one has
(68)∥πkeit|Δ|α/2f∥Lp⪯∥πkf∥Lpift|k|α-2≤1,∥πkeit|Δ|α/2f∥Lp⪯(t|k|α-2)n|(1/p)-(1/2)|∥πkf∥Lpift|k|α-2≥1.This lemma can be found in Section 4.2 of [11].Lemma 13.
LetF  be a compact subset in  ℝn, and let  0<p≤q≤∞. There exists a constant  C  depending only on the diameter of  F  and  p, such that
(69)∥f∥Lq≤C∥f∥Lp
for all  f∈Lp satisfying  suppf^⊂F.This lemma is the Nikol'skij-Triebel inequality, see Proposition1.3.2 in [20] (also Lemma 2.5 in [22]).Lemma 14.
Let0<p≤1  and  F,F′ be compact subsets of ℝn. Then there exists a constant  C  depending only on the diameters of  F,F′  and  p, such that
(70)∥|f|*|g|∥Lp≤C∥f∥Lp∥g∥Lp
for all  f, g∈Lp  satisfying  suppf^⊂F  and  suppg^⊂F′.This is Lemma 2.6 in [22] (see also Proposition 1.5.3 in [20]).Lemma 15 (Pitt's theorem).
If1<p≤2  and  0≤α<1/p′, then
(71)(∫ℝn‍|f^(x)|p′|x|-αnp′dx)1/p′⪯(∫ℝn‍|f(x)|p|x|αnpdx)1/p.Lemma 16.
Lets∈ℝ  and  p,pj≥1(j=1,2,…,m)  satisfy
(72)1p=1p1+1p2+⋯+1pm.
Then one has
(73)∥∏j=1m‍uj∥Mp,1s⪯∏j=1m‍∥uj∥Mpj,1s.This result is a particular case of Lemma 2.5 in [8].


## 3. Proof of Theorem1

The operatorπkeit|Δ|α/2 is  a convolution operator with the symbol σ(ξ-k)eit|ξ|α. This symbol is a C∞ function on ℝn∖{0}  with compact support. Clearly for any  N>0and ξ≠0, we have that for  |k|≤10,
(74)|(∂∂ξ)Jσ(ξ-k)eit|ξ|α|⪯{(1+t)|ξ|-1}|J|for|J|≤N.So Lemma11 implies the following estimate.Proposition 17.
Let0<p<∞. For any  k  with  |k|≤10,  one has
(75)∥πkeit|Δ|α/2f∥Hp⪯max{1,tn|1/p-1/2|}∥f∥Hp.By the proof of Lemma8 and Proposition 17, we have that for all  |k|≤9,
(76)∥πkeit|Δ|α/2f∥Hp⪯max{1,tn|1/p-1/2|}∥πkf∥Hp.The following proposition extends Lemma12 to all  p∈(0,1].Proposition 18.
Let0<p≤1.  For any  k  with  |k|≥1,  for any  α∈ℝ∖{0,1},  one has
(77)∥πkeit|Δ|α/2f∥Lp⪯max{1,(t|k|α-2)n(1/p-1/2)}∑|j|∞≤1‍∥πk+jf∥Lp.Proof.
The proof uses the same idea used in proving the casep≥1 which was represented in [11]. For the convenience of the reader, we present its proof.
LetΩk  be the kernel of  πkeit|Δ|α/2. Then
(78)Ωk=ℱ-1(σ(ξ-k)eit|ξ|α).
By Lemma 14 and (46), we have
(79)∥πkeit|Δ|α/2f∥Lp⪯∥Ωk∥Lp∑|j|∞≤1‍∥πk+jf∥Lp.
Thus to prove the proposition, it suffices to show
(80)∥ℱ-1(σ(ξ-k)eit|ξ|α)∥Lp⪯max{1,(t|k|α-2)n(1/p-1/2)}.
For simplicity, we prove the casen=2. The proof for  n≥3,  is tedious but shares the same idea as that for  n=2.
First we study the caset|k|α-2>1. For i=1,2,  and  k=(k1,k2), if  α≥2  we denote
(81)Ci(k)=αt(|ki|+1){(|k1|+1)2+(|k2|+1)2}(α/2)-1,Di(k)=αt(|ki|-1){(|k1|-1)2+(|k2|-1)2}(α/2)-1.
If  α<2,  we denote
(82)Ci(k)=|α|t(|ki|+1){(|k1|-1)2+(|k2|-1)2}(α/2)-1,Di(k)=|α|t(|ki|-1){(|k1|+1)2+(|k2|+1)2}(α/2)-1.
Also, fori=1,2and j∈ℕ,  we define sets
(83)Fi={xi∈ℝ:Di(k)-t|k|α-2<|xi|≤Ci(k)+t|k|α-2},Gi,j={xi∈ℝ:Ci(k)+t|k|α-2+j-1xi∈ℝ:<|xi|≤Ci(k)+t|k|α-2+j},Hi,j={xi∈ℝ:Di(k)-t|k|α-2-jxi∈ℝ:<|xi|≤Di(k)-t|k|α-2-j+1}.
It is easy to check
(84)Length(Fi)⪯t|k|α-2,Length(Gi,j)=Length(Hi,j)=1.
Let
(85)Ki,j=Gi,j∪Hi,j.
We have for  i=1,2,
(86)χFi(xi)+∑j=1∞‍χKi,j(xi)≡1.
Write
(87)∫ℝ2‍|ℱ-1(σk(ξ)eit|ξ|α)(x)|pdx=∫ℝ2‍|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx⪯∫ℝ2‍χF1(x1)χF2(x2)|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑j=1∞‍∫ℝ2‍χK1,j(x1)χF2(x2)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑m=1∞‍∫ℝ2‍χK2,m(x2)χF1(x1)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑j=1∞‍∑m=1∞‍∫ℝ2‍χK1,j(x1)χK2,m(x2)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx=II1+II2+II3+II4,
where
(88)ℱ-1(σ(ξ)eit|ξ+k|α)(x)≃∫ℝ2‍ei(t|ξ+k|α+〈x,ξ〉)σ(ξ)dξ.
It is easy to check that ift|k|α-2≥1  and  ξ∈ supp(σ), the phase function
(89)Φ(t,k,x,ξ)=t|ξ+k|α+(x·ξ),
satisfies
(90)|detDξjDξiΦ|≥t|k|α-2.
So by Lemma 9,  we have
(91)II1⪯(t-1|k|2-α)p∫ℝ2‍χF1(x1)χF2(x2)dx⪯(t-1|k|2-α)p(t|k|α-2)2=t2-p|k|(α-2)(2-p).
Observe the easy fact that ifx∈K1,j  and  ξ∈ supp (σ), for any integer  L,
(92)∂L∂ξlL(σ(ξ)Φξi(t,k,x,ξ))=O(1j+t|k|α-2)L,i,l=1,2.
Perform integration by parts on  ξ1 and  ξ2 variables both for  L  times such that  Lp>1. An easy computation shows that
(93)II4⪯∑j=1∞‍∑m=1∞‍∫ℝ2‍χK1,j(x1)χK2,m(x2)∑j=1∞‍∑m=1∞‍×1(j+t|k|α-2)Lp(m+t|k|α-2)Lp⪯∑j≥t|k|α-2∞‍∑m≥t|k|α-2∞‍1jLpmLp×∫ℝ2‍χK1,j(x1)χK2,m(x2)dx⪯1.
The estimates forII2  and  II3  are exactly the same. We only estimate  II2. Take integration by parts on  ξ1  variable for  L  times with  pL>1. Again, a simple computation shows that
(94)II2⪯∑j≥t|k|α-2∞‍1jLp∫ℝ2‍χK1,j(x1)χF2(x2)dx⪯(t|k|α-2)-Lp+2⪯1,
if we chose a suitably large  L. These estimates on  IIj, j=1,2,3,4, indicate
(95)∥ℱ-1(σ(ξ-k)eit|ξ|α)∥Lp⪯t2/p-1|k|(α-2)(2/p-1)=(t|k|α-2)2(1/p-1/2),
provided  t|k|α-2≥1.
We now turn to show the caset|k|α-2<1. For i=1,2,  and  k=(k1,k2), let  Ci(k)Di(k)  be the numbers defined above. For i=1,2and j∈ℕ,  we define sets
(96)𝔉i={xi:Di(k)-1<|xi|≤Ci(k)+1},𝔊i,j={xi:Ci(k)+j<|xi|≤Ci(k)+1+j},ℌi,j={xi:Di(k)-1-j<|xi|≤Di(k)-j}.
It is easy to check
(97)Length(𝔉i)⪯1,Length(𝔊i,j)=Length(ℌi,j)=1.
Let
(98)𝔎i,j=𝔊i,j∪ℌi,j.
Thus,
(99)∫ℝ2‍|ℱ-1(σk(ξ)eit|ξ|α)(x)|pdx⪯∫ℝ2‍χ𝔉1(x1)χ𝔉2(x2)|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑j=1∞‍∫ℝ2‍χ𝔎1,j(x1)χ𝔉2(x2)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑m=1∞‍∫ℝ2‍χ𝔎2,m(x2)χ𝔉1(x1)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx+∑j=1∞‍∑m=1∞‍∫ℝ2‍χ𝔎1,j(x1)χ𝔎2,m(x2)×|ℱ-1(σ(ξ)eit|ξ+k|α)(x)|pdx.
Using the same argument as we used before, we can show
(100)∥ℱ-1(σ(ξ-k)eit|ξ|α)∥Lp⪯1.
We complete the proof of Proposition 18.We are now in a position to prove Theorem1.Proof.
By an argument involving interpolation and duality, it suffices to show the casep≤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 Theorem1, by Proposition 18 and the definition of the modulation spaces, it suffices to show
(101)∥π0Sα(t)f∥Lp⪯max{1,tn(1/p-1/2)}∑|j|∞≤1‍∥πjf∥Lp.
Again, by Lemma 14, the proof of the inequality in (101) can be reduced to show that for  t≥1,
(102)(∫ℝn‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯tn(1/p-1/2).
We show (iii) first. The proof of  n=1  may illustrate the method.  When  n=1(103)(∫ℝ‍|∫ℝ‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯(∫|x|<t‍|∫ℝ‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p+(∫|x|≥t‍|∫ℝ‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p.
By Hölder's inequality and the Plancherel theorem, the first term above
(104)(∫|x|<t‍|∫ℝ‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯(∫ℝ‍|σ(ξ)eit|ξ|α|2dξ)1/2t(1/p-1/2)⪯t(1/p-1/2).
For the second term, performing integration by parts, we obtain
(105)(∫|x|≥t‍|∫ℝ‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p=2(∫|x|≥t‍|∫0∞‍σ(ξ)eitξαcosξxdξ|pdx)1/p=2(∫|x|>t‍1|x|p|{sinξxdξ{∫0+∞‍}|p}∫0+∞‍{iαtσ(ξ)ξα-1+σ′(ξ)}eit|ξ|α×sinξxdξ{∫0+∞‍}|pdx{∫|x|>t‍1|x|p})1/p≤2(∫|x|>t‍1|x|2p/(2-p)dx)(2-p)/2p×({×eit|ξ|αsinξxdξ{∫0+∞‍{iαtσ(ξ)ξα-1+σ′(ξ)}}|2dx}∫|x|>t‍|∫0+∞‍{iαtσ(ξ)ξα-1+σ′(ξ)}×eit|ξ|αsinξxdξ{∫0+∞‍{iαtσ(ξ)ξα-1+σ′(ξ)}}|2dx)1/2⪯t(∫|x|>t‍1|x|2p/(2-p)dx)(2-p)/2p×(∫R1‍(|σ(ξ)||ξ|α-1+|σ′(ξ)|)2dξ)1/2⪯t1/p-1/2,
since
(106)2p(2-p)>1.
Now we return to show (i) of Theorem1. We will prove only the case  α<1. Write
(107)(∫ℝn‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯(∫|x|<t‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p(∫|x|≥t‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p=J1+J2.
Using Hölder's inequality and the Plancherel theorem, we obtain
(108)J1⪯(∫ℝn‍|σ(ξ)eit|ξ|α|2dξ)1/2tn(1/p-1/2)⪯tn(1/p-1/2).
Fori,j∈{1,2,…,n}, we denote sets
(109)Et={x∈ℝn:|x|≥t},Et,i={x∈Et:|xi|≥|xj|∀j≠i}.
We now write
(110)J2⪯∑i=1n‍(∫Et,i‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p.
To show (102), it now suffices to show that for each  i,
(111)I=(∫Et,i‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯tn(1/p-1/2).
Using the Leibniz rule, for any positive integerL, we have
(112)∂i(L)(σ(x)eit|x|α)=∑k=0L‍CLk∂i(L-k)(σ(x))∂i(k)(eit|x|α).
Here, an easy induction argument shows that, for k≥1,
(113)∂i(k)(eit|x|α)=eit|x|α∑j=1k‍tjφjα-k(k,j)(x),
where φjα-k(j.k)∈C∞(Rn-{0})  is a homogeneous function of degree jα-k for each  j. We now write
(114)∂i(L)(σ(x)eit|x|α)=eit|x|α(∂i(L)(σ(x))+∑j=1L‍tj|x|jα-LΦj,L,α(x)),
where
(115)Φj,L,α(x)=∑k=jL‍CLk∂i(L-k)(σ(x))φjα-k(k,j)(x|x|)|x|L-k.
By the definition, it is easy to see that each  Φj,L,α  is an  L∞ and C∞(Rn-{0}) function with support in the cube  [-1/2,1/2]n.
LetL=[(n/2)+α]+1. Performing integration by parts on  ξi  variables for  L times, we have
(116)I=(∫Et,i‍|∫Rn‍σ(ξ)eit|ξ|αei(x·ξ)dξ|pdx)1/p⪯(∫Et,i‍|x|-Lp|∫Rn‍∂i(L)(σ(ξ))eit|ξ|αei(x·ξ)dξ|pdx)1/p+∑j=1L‍tj(∫Et,i‍|x|-Lp|∫Rn‍|ξ|jα-LΦj,L,α+∑j=1L‍tj∫Et,i‍|x|-Lp×(ξ)eit|ξ|αei(x·ξ)dξ{∫Rn‍|ξ|jα-LΦj,L,α}|px{∫Et,i‍|x|-Lp})1/p=∑j=0L‍Ij.
We first estimate eachIj, 0≤j≤L-1. Recall that we assume  p∈(0,2). Let  q=2/p,  so pq′=2p/(2-p). By the choice of  L  and the assumption
(117)p>nn+α,
it is easy to see  Lpq′>n. Therefore, by Hölder's inequality, we obtain
(118)I0≤(∫|x|≥t‍|x|-Lpq′dx)1/(pq′)×(∫Rn‍|∫Rn‍∂i(L)(σ(ξ))eit|ξ|αei(ξ·x)dξ|2dx)1/2⪯t-L+n(1/p-1/2)∥∂i(L)σ∥2⪯t-L+n(1/p-1/2).
For each  j=1,2,…,L-1, by the choice of  L, the assumption on  p, and an easy computation, it is not difficult to see that we may obtain a number αj in the interval [0,n/2)  satisfying
(119)L-αj>n(1p-12),αj+jα-L>-n2,αj+j-L≤0.
By Hölder's inequality and Pitt's theorem, for each  j, we obtain
(120)Ij≤tj(∫Et,i‍|x|-(L-αj)p||x|-αj∫Rn‍|ξ|jα-LΦj,L,α(ξ)eit|ξ|α×ei(x·ξ)dξ{∫Rn‍|ξ|jα-L}|pdx{∫Rn‍|ξ|jα-LΦj,L,α})1/p≤(∫|x|≥t‍|x|-(L-αj)pq′dx)1/(pq′)×(∫Rn‍||x|-αj∫Rn‍|ξ|jα-LΦj,L,α(ξ)eit|ξ|αei(x·ξ)dξ|2dx)1/2⪯t-L+αj+j+n(1/p-1/2)(∫Rn‍||x|αj|x|jα-LΦj,L,α(x)|2dx)1/2⪯t-L+αj+j+n(1/p-1/2).
Combining all the estimates, we have
(121)(∫Et,i‍|∫Rn‍σ(ξ)eit|ξ|αei(x·ξ)dξ|pdx)1/p⪯IL+∑j=0L-1‍t-L+αj+j+n(1/p-1/2)⪯IL+tn(1/p-1/2).
It remains to estimate  IL.
It is easy to see that the choice ofL  and the condition
(122)α>α0={2n+2ifniseven3n+3ifnisodd
in the theorem imply Lα-L>-n/2. So, by Hölder's inequality and Pitt's theorem again, we obtain
(123)IL=tL(∫Et,i‍|x|-Lp|∫Rn‍|ξ|Lα-LΦL,L,α(ξ)eit|ξ|αei(x·ξ)dξ|pdx)1/p⪯tL(∫|x|≥t‍|x|-Lpq/dx)1/(pq′)×(∫Rn‍|∫Rn‍|ξ|Lα-LΦL,L,α(ξ)eit|ξ|αei(x·ξ)dξ|2dx)1/2⪯tn(1/p-1/2)∥|ξ|Lα-LΦL,L,α(ξ)∥2⪯tn(1/p-1/2).
This completes the proof of (102).
Whenα=2,4,…,  we have
(124)(∫Et,i‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯tL(∫Et,i‍1|x|pL|∫|ξ|≤2‍ω(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p
for any integer  L,  where  ω(ξ)  is a C∞  function. Thus it is trivial to see that
(125)(∫Et,i‍|∫ℝn‍σ(ξ)eit|ξ|αei(ξ·x)dξ|pdx)1/p⪯tn(1/p-1/2)
for all  p>0. This proves (i) in Theorem 1.


## 4. Proof of Theorem3

Recall that the functionσ  chosen in the definition of the modulation space is flexible. We may choose
(126)σ=∏j=1n‍θj,
where each  θj  is a nonnegative valued function in  S(ℝ)  with support in  [-4/5,4/5]  and satisfies  θj(ξj)=1  if  ξj∈[-2/5,2/5]  and
(127)∑kj∈ℤ‍θj,kj(ξj)=1,θj,kj(ξj)=θj(ξj+kj).For simplicity, we work on the case0<p<2. Let  ϕ∈C∞(ℝ)  be a nonnegative function with support in the set  [1/16,1/8]. For
(128)x=(x1,x2,…,xn),
define a function  Φ  on  ℝn  by
(129)Φ(x)=∏j=1n‍ϕ(xj).
and an f∈S(ℝn)  by  f=ℱ-1(Φ). Let  k∈ℤn. It is easy to see that
(130)πk(f)(x)=ϵ(k)f(x),
where  ϵ(k)=1  if  k=0  and   ϵ(k)=0  if  k≠0.  Similarly we have
(131)πk(eit|Δ|f)(x)=ϵ(k)(eit|Δ|f)(x).
Suppose that we have some  δ  such that
(132)∥eit|Δ|f∥Mp,qs(ℝn)⪯tδ∥f∥Mp,qs(ℝn).
By the choice of  f,  we have
(133)∥eit|Δ|f∥Mp,qs(ℝn)=∥π0eit|Δ|f∥Lp(ℝn)⪯tδ∥f∥Mp,qs(ℝn)=tδ∥f∥Lp(ℝn).
On the other hand,
(134)∥π0eit|Δ|f∥Lp(ℝn)=(∫ℝn‍|∏j=1n‍∫ℝ‍ϕ(ξj)eiψ(ξj,xj)dξj|pdx)1/p,
where the phase function  ψ(ξj,xj)  is defined by
(135)ψ(ξj,xj)=tξj2+ξjxj.
Since
(136)∂∂ξjψ(ξj,xj)=2tξj+xj,
The critical point of  ψ(ξj,xj)  is at
(137)ξj*=-xj2t.
Thus, by the stationary phase method (see [19, Proposition 3, page 334]), an easy computation gives that, as  t→∞,
(138)(∫ℝn‍|∏j=1n‍∫ℝ‍ϕ(ξj)eiψ(ξj,xj)dξj|pdx)1/p≥(∏j=1n‍∫|xj|≤2t‍|∫ℝ‍ϕ(ξj)eiψ(ξj,xj)dξj|pdxj)1/p≃t-n/2(∏j=1n‍∫|xj|≤2t‍|ϕ(-xj2t)|pdxj)1/p+O(tn(1/p-1/2)-1/2)≃tn(1/p-1/2).
Thus the inequality
(139)∥eit|Δ|f∥Mp,qs(ℝn)⪯tδ∥f∥Mp,qs(ℝn)
implies
(140)tn(1/p-1/2)⪯tδ,ast→∞.
This shows the conclusion.


## 5. Applications

### 5.1. OperatorsSα,β(t)

Letα≠1and β∈ℝ. In [23], to investigate the absolute convergence for multiple Fourier series, Wainger studied the oscillating multipliers  Sα,β(t)  with symbol
(141)mα,β(t,ξ)=eit|ξ|α(1+|ξ|2)β/2.
In [24], Miyachi proved that in the case  α>0 and α≠1, for  0<p<∞,
(142)∥Sα,β(1)f∥Hp(ℝn)⪯∥f∥Hp(ℝn)
if and only if
(143)β≥nα|1p-12|.
By Theorem 1 and its proof, we not only obtain the boundedness of  Sα,β(t)  on  μp,qs(ℝn)  for any β∈ℝ, but also gain a regularity of β if β>0.Theorem 19.
Let0<p<∞ and α≠1. One has for  t≥1(144)∥Sα,β(t)f∥μp,qs⪯tn|1/p-1/2|∥f∥μp,qs+γ-β+∥f∥μp,qs,
where
(145)γ=n(α-2)|1p-12|.Particularly, ifβ=n(α-2)|1/p-1/2|, we have
(146)∥Sα,β(t)f∥μp,qs⪯tn|1/p-1/2|∥f∥μp,qs.Proof.
The proof of Theorem19 is the same as the proof of Theorem 1. We skip it.


### 5.2.Mp1,qs→Mp2,qs Boundedness

We will study theMp1,qs→Mp2,qs  boundedness for the operator  eit|Δ|α/2  in the case  0<α≤2and α≠1. The other cases of αwill be addressed in another paper. First we estimate  πkeit|Δ|α/2for  |k|≤10.Proposition 20.
Letα≠1, 0<α≤2 and |k|≤9. One has for all  0<r≤1≤p≤∞,
(147)∥πkeit|Δ|α/2f∥Lp⪯(1+t)-n(1/2-1/p)∥πk(f)∥Lr.Proof.
By Lemma9, we know
(148)∥πkeit|Δ|α/2∥L1→L∞⪯(1+t)-n/2.
On the other hand, it is known from [11]
(149)∥πkeit|Δ|α/2∥Lp→Lp⪯(1+t)n|1/p-1/2|
for all  1≤p≤∞. Interpolating these two inequalities with the energy inequality
(150)∥πkeit|Δ|α/2∥L2→L2⪯1,
we find that for any  1≤p≤∞,
(151)∥πkeit|Δ|α/2∥Lp→L∞⪯(1+t)-n(1/p-1/2).
Let  1≤p≤∞. For  f∈L1, we have
(152)∥πkeit|Δ|α/2f∥Lp=sup∥g∥Lp′=1|<πkSα(t)f,g>|=sup∥g∥Lp′=1|<f,(πkeit|Δ|α/2)*g>|⪯sup∥g∥Lp′=1∥(πkeit|Δ|α/2)*g∥L∞∥f∥L1⪯(1+t)n(1/p-1/2)∥f∥L1.
This shows that for  |k|≤10,
(153)∥πkeit|Δ|α/2f∥Lp⪯(1+t)-n(1/2-1/p)∥f∥L1.
Now if  0<r≤1, we obtain that for all  1≤p≤∞,
(154)∥πkeit|Δ|α/2f∥Lp⪯∑|ℓ|≤1‍∥πk+ℓSα(t)πk(f)∥Lp⪯(1+t)-n(1/2-1/p)∥πk(f)∥L1⪯(1+t)-n(1/2-1/p)∥πk(f)∥Lr.
The last inequality is from Lemma 13.As these discussions, we obtain the following corollary.Corollary 21.
Let0<α≤2,  α≠1, |k|≤10  and  0<p1≤1≤p2≤∞, one has
(155)∥πkeit|Δ|α/2f∥Hp2⪯∥πkf∥Hp1ift≤1,∥πkeit|Δ|α/2f∥Hp2⪯tn(1/p2-1/2)∥πkf∥Hp1ift>1,
where for convenience, one denotes  L∞=H∞.Proposition 22.
Let0<α≤2 and α≠1. One has for  |k|≥9  and all  0<r≤1≤p≤∞(156)∥πkeit|Δ|α/2f∥Lp⪯(1+t|k|α-2)-n(1/2-1/p)∥πk(f)∥Lr.Proof.
By Lemma9, it is trivial to see that
(157)∥πkeit|Δ|α/2∥L1→L∞⪯min{1,t-n/2|k|-n(α-2)/2}.
Interpolating this inequality with the energy inequality, we find that for any  p≤2,
(158)∥πkeit|Δ|α/2∥Lp→L∞⪯(1+t|k|α-2)-n(1/p-1/2).
So by duality, for  |k|≥9,
(159)∥πkeit|Δ|α/2f∥Lp⪯(1+t|k|α-2)n(1/p-1/2)∥f∥L1.
Now for any  0<r≤1, we obtain that for all  1≤p≤∞,
(160)∥πkeit|Δ|α/2f∥Lp⪯∑|ℓ|≤1‍∥πk+ℓSα(t)πk(f)∥Lp⪯(1+t|k|α-2)-n(1/p-1/2)∥πk(f)∥L1⪯(1+t|k|α-2)n(1/p-1/2)∥πk(f)∥Lr.As a consequence the proposition, we have the following.Corollary 23.
Let0<α≤2, α≠1, |k|≥9  and  0<p1≤1≤p2≤∞, one has
(161)∥πkeit|Δ|α/2f∥Hp2⪯∥πkf∥Hp1ift|k|α-2≤1,∥πkeit|Δ|α/2f∥Hp2⪯|k|n(α-2)(1/p2-1/2)tn(1/p2-1/2)∥πkf∥Hp1ift|k|α-2≥1.By Corollaries21 and 23 and the definition of the modulation spaces, we now obtain the μp1,qs→Mp2,qs  boundedness of  eit|Δ|α/2  on the modulation spaces.Theorem 24.
Let0<α≤2, α≠1, and 0<p1≤1≤p2≤∞, one has that
(162)∥eit|Δ|α/2f∥Mp2,qs⪯(1+t)n(1/p2-1/2)∥f∥μp1,qs+γ,
where
(163)γ=n(α-2)(1p2-12).Next, we give the proof of Theorem4.Proof.
By the definition of the modulation space, we know(164)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)={∫0∞‍[∑k‍(1+|k|)sq∥πkeit|Δ|α/2f∥Lp2q]r/qdt}1/r.
If   r/q≥1, write
(165)(1+|k|)sq∥πkSα(t)f∥Lp2q=gk(t).
By the Minkowski inequality, we have
(166)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(∫0∞‍|gk(t)|r/qdt)q/r}1/q={∑k‍(∫0∞‍(1+|k|)sr∥πkeit|Δ|α/2f∥Lp2rdt)q/r}1/q={∑k‍(1+|k|)sq(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r}1/q.
When  |k|≤10,
(167)(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∥πkf∥Hp1r∫0∞‍(1+t)-rn(1/2-1/p2)dt)q/r⪯∥πkf∥Hp1q.
When  |k|≥9,
(168)(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∫0|k|2-α‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r+(∫|k|2-α∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∥πkf∥Lp1r∫0|k|2-α‍dt)q/r+∥πkf∥Lp1q(∫|k|2-α∞‍(t|k|α-2)rn(1/2-1/p2)dt)q/r⪯∥πkf∥Lp1q|k|(2-α)q/r+∥πkf∥Lp1q(|k|(α-2)rn(1/p2-1/2)∫|k|2-α∞‍t-rn(1/2-1/p2)dt)q/r⪯∥πkf∥Lp1q|k|(2-α)q/r.
This shows that if  r≥q, then
(169)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(1+|k|)sq〈k〉(2-α)q/r∥πkf∥Lp1q}1/q=∥f∥μp1,qs+(2-α)/r(ℝn).
Ifr/q≤1,  then
(170)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∫0∞‍∑k‍(1+|k|)sr∥πkeit|Δ|α/2f∥Lp2rdt}1/r.
As in the previous case,
(171)∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt⪯∥πkf∥Hp1
for  |k|≤10,  and
(172)(1+|k|)sr∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt⪯(1+|k|)sr∥πkf∥Lp1r|k|(α-2)rn(1/2-1/p2)×∫|k|2-α∞‍t-rn(1/2-1/p2)dt+(1+|k|)sr∥πkf∥Lp1r∫|k|2-α∞‍t-rn(1/2-1/p2)dt⪯(1+|k|)sr∥πkf∥Lp1r〈k〉(α-2).
This shows that
(173)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(1+|k|)sr∥πkf∥Lp1r〈k〉(α-2)}1/r=∥f∥μp1,rs+(2-α)/r(ℝn).
The theorem is proved.


### 5.3. Schrödinger Equation

Consider the Cauchy problem of the linear free Schrödinger equation(174)i∂tu=Δxu,(t,x)∈ℝ+×ℝnu(0,x)=u0(x).The formal solution to this equation is given by(175)u(t,x)=1(2π)n∫ℝn‍eit|ξ|2u0^(ξ)e2πiξ·xdξ=(e-itΔu0)(x).
By Theorem 1, we obtain the growth rate as t→∞  for the solution to the linear free Schrödinger equation.Theorem 25.
Letu(t,x)  be the solution of the above Cauchy problem of the Schrödinger equation. For  0<p≤∞, one has
(176)∥u(t,·)∥Mp,qs(ℝn)⪯(1+tn(1/p-1/2))∥u0∥Mp,qs(ℝn),
where the asymptotic factor  tn(1/p-1/2)  is sharp as  t→∞.


### 5.4. Linear Cauchy Problem with Negative Power

We start with the following linear Cauchy problem with negative power:(177)i∂tu+|Δx|-α/2u=0,(t,x)∈ℝ+×ℝn,α>0,u(0,x)=u0(x).The formal solution to this equation is given by(178)u(t,x)=1(2π)n∫ℝn‍eit|ξ|αu0^(ξ)e2πiξ·xdξ=(e-it|Δ|-α/2u0)(x).Proposition 26.
Letα>0  and  1≤p≤2. One has
(179)∥π0e-it|Δ|-α/2f∥Lp(ℝn)⪯(1+t)([n/2]+1)|2/p-1|∑|j|∞≤1‍∥πjf∥Hr(ℝn),
where
(180)1r>n+α([n/2]+1)(2-p)np.Proof.
We only prove the case of oddn,  since the proof for even  n  is similar.  For any fixed (181)β>α([n2]+1),
we write
(182)π0e-it|Δ|-α/2f(x)=1(2π)n∫ℝn‍σ(ξ)|ξ|βeit|ξ|-α|ξ|-βf^(ξ)e2πiξ·xdξ≃π0|Δ|β/2e-it|Δ|-α/2Iβf,
where Iβ  is the Riesz potential of order  β. The kernel of    π0|Δ|β/2e-it|Δ|-α/2  is
(183)Ωα,β(x)=∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ.
We first show
(184)∥Ωα,β*f∥L1⪯(1+t)[n2]+1∥f∥L1.
To this end, by Young's inequality, we need to show
(185)∥Ωα,β∥L1(ℝn)⪯(1+t)[n/2]+1.
It suffices to show the case   t≥1. As the same argument in the proof of Theorem 1, with the Schwarz inequality we have
(186)∥Ωα,β∥L1(ℝn)⪯tn/2+∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx.
Performing integration by parts  (n+1)/2  times on the second term, without loss of generality, we may write
(187)∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx⪯t(n+1)/2∫|x|>t‍1|x|(n+ε)/2⪯tn+12×|1|x|(1-ε)/2∫ℝn‍ψ(ξ)eit|ξ|-ασ~(ξ)e2πiξ·xdξ|dx,
where  σ~(ξ)  is a C∞ function supported in  [-1,1]n  and  ψ(ξ)  is a function satisfying
(188)|ψ(ξ)|⪯|ξ|β-((n+1)α/2)-((n+1)/2).
Choose a small  ε>0  such
(189)2β-α(n+1)-ε>0.
Thus by Schwarz's inequality and the Pitt's theorem, we obtain
(190)∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx⪯t(n+1)/2({e2πiξ·xdξ{1|x|(1-ε)/2∫ℝn‍}|2}∫ℝn‍|1|x|(1-ε)/2∫ℝn‍ψ(ξ)eit|ξ|-ασ~(ξ)⪯tn+12×e2πiξ·xdξ{1|x|(1-ε)/2}|2dx)1/2,t(n+1)/2(∫ℝn‍|σ~(ξ)|2|ξ|-n-ε|ξ|2β-α(n+1)dξ)1/2⪯t(n+1)/2.
Combining these estimates, we have
(191)∥π0e-it|Δ|-α/2f∥L1⪯t(n+1)/2∥Iβf∥L1⪯t(n+1)/2∥f∥L-β1,
where   L-β1  denotes the  L1  Sobolev space of order  -β.  On the other hand, we have the easy energy estimate
(192)∥π0e-it|Δ|-α/2f∥L2⪯∥f∥L2.
An interpolation yields that for all  1≤p≤2,
(193)∥π0e-it|Δ|-α/2f∥Lp⪯t(n+1)(1/p-1/2)∥f∥L-γp,
where
(194)γ=β(2p-1).
We now use the Sobolev imbedding theorem and the almost orthogonality of  {πk}  to obtain
(195)∥π0e-it|Δ|-α/2f∥Lp(ℝn)⪯t(n+1)(1/p-1/2)∑|j|∞≤1‍∥πjf∥Hr(ℝn)
with
(196)1r=1p+β(2/p-1)n=n+β(2-p)np.
Since   β   is an arbitrary number larger than  α([n/2]+1),  the proposition now follows from Lemma 13.Lemma 27.
Letα>0  and  2≤p≤∞. One has
(197)∥π0e-it|Δ|-α/2f∥L2(ℝn)⪯∑|j|∞≤1‍∥πjf∥Hr(ℝn),
for any  r≤2.Proof.
The almost orthogonality (Identity (46)) and the energy estimate give
(198)∥π0e-it|Δ|-α/2f∥L2(ℝn)⪯∑|j|∞≤1‍∥πjf∥L2(ℝn).
Thus the lemma follows from Lemma 13.Now we are ready to give the proof of Theorem5.Proof.
The proof of (i) can be obtained from Propositions18 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.


### 5.5. Nonlinear Cauchy Problem with Negative Power

Now, we study the following Cauchy problem of the nonlinear dispersive equations (NDE):(199)i∂u∂t-|Δ|-α/2u+F(u)=0,u(0,x)=u0(x),
where   F(u)=|u|2ku   for some positive integer  k. For  T>0, the space  Y=C([0,T],Mp,1s)  is defined by
(200)C([0,T],Mp,1s)={u(t,x):∥u∥C([0,T],Mp,1s)=sup0≤t≤T∥u(t,·)∥Mp,1s<∞}.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 Theorem6.Proof.
In the proof, the lettersCj, j=1,2,3 denote some positive constants that are independent of all essential variables. We write the Cauchy problem in the equivalent form
(201)u(t,·)=eit|Δ|-α/2u0-i∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ,
and consider the mapping
(202)ℑu=eit|Δ|-α/2u0-i∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ.
We want to show that  ℑ  is a contraction.
By Theorem5 and Lemma 12,
(203)∥eit|Δ|-α/2u0∥Y≤C1(1+T)([n/2]+1)|2/p-1|∥u0∥Mr,1s-γ(ℝn)+C1∥u0∥Mp,1s(ℝn).
By Lemma 16, there is a constant  A2k+1>0  for which we have
(204)∥|u(t,·)|2ku(t,·)∥Mr,1s≤A2k+1∥u(t,·)∥M(2k+1)r,1s2k+1.
Thus, by Theorem 5 and Lemma 12,
(205)∥∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ∥C([0,T],Mp,1s)⪯sup0<t≤T∫0t‍∥ei(t-τ)|Δ|-α/2F(u(τ,·))∥Mp,1sdτ≤C1sup0<t≤T∫0t‍(1+(t-τ))([n/2]+1)|2/p-1|≤C1sup0<t≤T∫0t‍×∥|u(τ,·)|2ku(τ,·)∥Mr,1sdτ≤C2MkT(1+T)([n/2]+1)|2/p-1|sup0≤t≤T∥u(t,·)∥Mr(2k+1),1s2k+1⪯C2MkT(1+T)([n/2]+1)|2/p-1|sup0≤t≤T∥u(t,·)∥Mp,1s2k+1.
The last inequality is because
(206)r(2k+1)≥p
and the imbedding relation   Mp,1s⊂Mr(2k+1),1s.
As a consequence,(207)∥ℑu∥C([0,T],Mp,1s)=sup0≤t≤T∥ℑu(t,·)∥Mp,1s⪯C3(1+T)([n/2]+1)|2/p-1|×(∥u0∥Mr,1s-γ(ℝn)+Tsup0≤t≤T∥u(t,·)∥Mp,1s2k+1)+C1∥u0∥Mp,1s(ℝn).
Fix an  L>0  such that
(208)C3∥u0∥Mr,1s-γ(ℝn)+C1∥u0∥Mp,1s(ℝn)<L4.
Let  BL  be the closed ball of radial  L  centered at the origin in the space  C([0,T],Mp,1s).  Now we choose  T0  such that for any   u∈BLand T∈(0,T0],
(209)∥ℑu∥C([0,T],Mp,1s)≤C3(1+T)([n/2]+1)|2/p-1|(∥u0∥Mr,1s-γ(ℝn)+TL2k+1)+C1∥u0∥Mp,1s(ℝn)≤L.
By this choice,  ℑ  is a mapping form BLinto BL.
Furthermore, an easy computation gives that(210)∥ℑu-ℑv∥C([0,T],Mp,1s)≤C3(1+T)([n/2]+1)|2/p-1|×Tsup0≤t≤T∥u(t,·)-v(t,·)∥Mp,1s(2k+1)L2k≤12∥u-v∥C([0,T],Mp,1s),
for all T∈(0,T0] if T0 is suitably chosen. Finally, using the Banach contraction mapping theorem, we obtain a fixed point in  BL. This is the solution of the Cauchy problem. We prove the theorem.Using a similar argument with the help of (ii) in Theorem5, we can prove Theorem 7.


## 5.1. OperatorsSα,β(t)

Letα≠1and β∈ℝ. In [23], to investigate the absolute convergence for multiple Fourier series, Wainger studied the oscillating multipliers  Sα,β(t)  with symbol
(141)mα,β(t,ξ)=eit|ξ|α(1+|ξ|2)β/2.
In [24], Miyachi proved that in the case  α>0 and α≠1, for  0<p<∞,
(142)∥Sα,β(1)f∥Hp(ℝn)⪯∥f∥Hp(ℝn)
if and only if
(143)β≥nα|1p-12|.
By Theorem 1 and its proof, we not only obtain the boundedness of  Sα,β(t)  on  μp,qs(ℝn)  for any β∈ℝ, but also gain a regularity of β if β>0.Theorem 19.
Let0<p<∞ and α≠1. One has for  t≥1(144)∥Sα,β(t)f∥μp,qs⪯tn|1/p-1/2|∥f∥μp,qs+γ-β+∥f∥μp,qs,
where
(145)γ=n(α-2)|1p-12|.Particularly, ifβ=n(α-2)|1/p-1/2|, we have
(146)∥Sα,β(t)f∥μp,qs⪯tn|1/p-1/2|∥f∥μp,qs.Proof.
The proof of Theorem19 is the same as the proof of Theorem 1. We skip it.


## 5.2.Mp1,qs→Mp2,qs Boundedness

We will study theMp1,qs→Mp2,qs  boundedness for the operator  eit|Δ|α/2  in the case  0<α≤2and α≠1. The other cases of αwill be addressed in another paper. First we estimate  πkeit|Δ|α/2for  |k|≤10.Proposition 20.
Letα≠1, 0<α≤2 and |k|≤9. One has for all  0<r≤1≤p≤∞,
(147)∥πkeit|Δ|α/2f∥Lp⪯(1+t)-n(1/2-1/p)∥πk(f)∥Lr.Proof.
By Lemma9, we know
(148)∥πkeit|Δ|α/2∥L1→L∞⪯(1+t)-n/2.
On the other hand, it is known from [11]
(149)∥πkeit|Δ|α/2∥Lp→Lp⪯(1+t)n|1/p-1/2|
for all  1≤p≤∞. Interpolating these two inequalities with the energy inequality
(150)∥πkeit|Δ|α/2∥L2→L2⪯1,
we find that for any  1≤p≤∞,
(151)∥πkeit|Δ|α/2∥Lp→L∞⪯(1+t)-n(1/p-1/2).
Let  1≤p≤∞. For  f∈L1, we have
(152)∥πkeit|Δ|α/2f∥Lp=sup∥g∥Lp′=1|<πkSα(t)f,g>|=sup∥g∥Lp′=1|<f,(πkeit|Δ|α/2)*g>|⪯sup∥g∥Lp′=1∥(πkeit|Δ|α/2)*g∥L∞∥f∥L1⪯(1+t)n(1/p-1/2)∥f∥L1.
This shows that for  |k|≤10,
(153)∥πkeit|Δ|α/2f∥Lp⪯(1+t)-n(1/2-1/p)∥f∥L1.
Now if  0<r≤1, we obtain that for all  1≤p≤∞,
(154)∥πkeit|Δ|α/2f∥Lp⪯∑|ℓ|≤1‍∥πk+ℓSα(t)πk(f)∥Lp⪯(1+t)-n(1/2-1/p)∥πk(f)∥L1⪯(1+t)-n(1/2-1/p)∥πk(f)∥Lr.
The last inequality is from Lemma 13.As these discussions, we obtain the following corollary.Corollary 21.
Let0<α≤2,  α≠1, |k|≤10  and  0<p1≤1≤p2≤∞, one has
(155)∥πkeit|Δ|α/2f∥Hp2⪯∥πkf∥Hp1ift≤1,∥πkeit|Δ|α/2f∥Hp2⪯tn(1/p2-1/2)∥πkf∥Hp1ift>1,
where for convenience, one denotes  L∞=H∞.Proposition 22.
Let0<α≤2 and α≠1. One has for  |k|≥9  and all  0<r≤1≤p≤∞(156)∥πkeit|Δ|α/2f∥Lp⪯(1+t|k|α-2)-n(1/2-1/p)∥πk(f)∥Lr.Proof.
By Lemma9, it is trivial to see that
(157)∥πkeit|Δ|α/2∥L1→L∞⪯min{1,t-n/2|k|-n(α-2)/2}.
Interpolating this inequality with the energy inequality, we find that for any  p≤2,
(158)∥πkeit|Δ|α/2∥Lp→L∞⪯(1+t|k|α-2)-n(1/p-1/2).
So by duality, for  |k|≥9,
(159)∥πkeit|Δ|α/2f∥Lp⪯(1+t|k|α-2)n(1/p-1/2)∥f∥L1.
Now for any  0<r≤1, we obtain that for all  1≤p≤∞,
(160)∥πkeit|Δ|α/2f∥Lp⪯∑|ℓ|≤1‍∥πk+ℓSα(t)πk(f)∥Lp⪯(1+t|k|α-2)-n(1/p-1/2)∥πk(f)∥L1⪯(1+t|k|α-2)n(1/p-1/2)∥πk(f)∥Lr.As a consequence the proposition, we have the following.Corollary 23.
Let0<α≤2, α≠1, |k|≥9  and  0<p1≤1≤p2≤∞, one has
(161)∥πkeit|Δ|α/2f∥Hp2⪯∥πkf∥Hp1ift|k|α-2≤1,∥πkeit|Δ|α/2f∥Hp2⪯|k|n(α-2)(1/p2-1/2)tn(1/p2-1/2)∥πkf∥Hp1ift|k|α-2≥1.By Corollaries21 and 23 and the definition of the modulation spaces, we now obtain the μp1,qs→Mp2,qs  boundedness of  eit|Δ|α/2  on the modulation spaces.Theorem 24.
Let0<α≤2, α≠1, and 0<p1≤1≤p2≤∞, one has that
(162)∥eit|Δ|α/2f∥Mp2,qs⪯(1+t)n(1/p2-1/2)∥f∥μp1,qs+γ,
where
(163)γ=n(α-2)(1p2-12).Next, we give the proof of Theorem4.Proof.
By the definition of the modulation space, we know(164)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)={∫0∞‍[∑k‍(1+|k|)sq∥πkeit|Δ|α/2f∥Lp2q]r/qdt}1/r.
If   r/q≥1, write
(165)(1+|k|)sq∥πkSα(t)f∥Lp2q=gk(t).
By the Minkowski inequality, we have
(166)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(∫0∞‍|gk(t)|r/qdt)q/r}1/q={∑k‍(∫0∞‍(1+|k|)sr∥πkeit|Δ|α/2f∥Lp2rdt)q/r}1/q={∑k‍(1+|k|)sq(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r}1/q.
When  |k|≤10,
(167)(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∥πkf∥Hp1r∫0∞‍(1+t)-rn(1/2-1/p2)dt)q/r⪯∥πkf∥Hp1q.
When  |k|≥9,
(168)(∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∫0|k|2-α‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r+(∫|k|2-α∞‍∥πkeit|Δ|α/2f∥Lp2rdt)q/r⪯(∥πkf∥Lp1r∫0|k|2-α‍dt)q/r+∥πkf∥Lp1q(∫|k|2-α∞‍(t|k|α-2)rn(1/2-1/p2)dt)q/r⪯∥πkf∥Lp1q|k|(2-α)q/r+∥πkf∥Lp1q(|k|(α-2)rn(1/p2-1/2)∫|k|2-α∞‍t-rn(1/2-1/p2)dt)q/r⪯∥πkf∥Lp1q|k|(2-α)q/r.
This shows that if  r≥q, then
(169)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(1+|k|)sq〈k〉(2-α)q/r∥πkf∥Lp1q}1/q=∥f∥μp1,qs+(2-α)/r(ℝn).
Ifr/q≤1,  then
(170)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∫0∞‍∑k‍(1+|k|)sr∥πkeit|Δ|α/2f∥Lp2rdt}1/r.
As in the previous case,
(171)∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt⪯∥πkf∥Hp1
for  |k|≤10,  and
(172)(1+|k|)sr∫0∞‍∥πkeit|Δ|α/2f∥Lp2rdt⪯(1+|k|)sr∥πkf∥Lp1r|k|(α-2)rn(1/2-1/p2)×∫|k|2-α∞‍t-rn(1/2-1/p2)dt+(1+|k|)sr∥πkf∥Lp1r∫|k|2-α∞‍t-rn(1/2-1/p2)dt⪯(1+|k|)sr∥πkf∥Lp1r〈k〉(α-2).
This shows that
(173)∥∥eit|Δ|α/2f∥Mp2s,q(ℝn)∥Lr(0,∞)⪯{∑k‍(1+|k|)sr∥πkf∥Lp1r〈k〉(α-2)}1/r=∥f∥μp1,rs+(2-α)/r(ℝn).
The theorem is proved.


## 5.3. Schrödinger Equation

Consider the Cauchy problem of the linear free Schrödinger equation(174)i∂tu=Δxu,(t,x)∈ℝ+×ℝnu(0,x)=u0(x).The formal solution to this equation is given by(175)u(t,x)=1(2π)n∫ℝn‍eit|ξ|2u0^(ξ)e2πiξ·xdξ=(e-itΔu0)(x).
By Theorem 1, we obtain the growth rate as t→∞  for the solution to the linear free Schrödinger equation.Theorem 25.
Letu(t,x)  be the solution of the above Cauchy problem of the Schrödinger equation. For  0<p≤∞, one has
(176)∥u(t,·)∥Mp,qs(ℝn)⪯(1+tn(1/p-1/2))∥u0∥Mp,qs(ℝn),
where the asymptotic factor  tn(1/p-1/2)  is sharp as  t→∞.


## 5.4. Linear Cauchy Problem with Negative Power

We start with the following linear Cauchy problem with negative power:(177)i∂tu+|Δx|-α/2u=0,(t,x)∈ℝ+×ℝn,α>0,u(0,x)=u0(x).The formal solution to this equation is given by(178)u(t,x)=1(2π)n∫ℝn‍eit|ξ|αu0^(ξ)e2πiξ·xdξ=(e-it|Δ|-α/2u0)(x).Proposition 26.
Letα>0  and  1≤p≤2. One has
(179)∥π0e-it|Δ|-α/2f∥Lp(ℝn)⪯(1+t)([n/2]+1)|2/p-1|∑|j|∞≤1‍∥πjf∥Hr(ℝn),
where
(180)1r>n+α([n/2]+1)(2-p)np.Proof.
We only prove the case of oddn,  since the proof for even  n  is similar.  For any fixed (181)β>α([n2]+1),
we write
(182)π0e-it|Δ|-α/2f(x)=1(2π)n∫ℝn‍σ(ξ)|ξ|βeit|ξ|-α|ξ|-βf^(ξ)e2πiξ·xdξ≃π0|Δ|β/2e-it|Δ|-α/2Iβf,
where Iβ  is the Riesz potential of order  β. The kernel of    π0|Δ|β/2e-it|Δ|-α/2  is
(183)Ωα,β(x)=∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ.
We first show
(184)∥Ωα,β*f∥L1⪯(1+t)[n2]+1∥f∥L1.
To this end, by Young's inequality, we need to show
(185)∥Ωα,β∥L1(ℝn)⪯(1+t)[n/2]+1.
It suffices to show the case   t≥1. As the same argument in the proof of Theorem 1, with the Schwarz inequality we have
(186)∥Ωα,β∥L1(ℝn)⪯tn/2+∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx.
Performing integration by parts  (n+1)/2  times on the second term, without loss of generality, we may write
(187)∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx⪯t(n+1)/2∫|x|>t‍1|x|(n+ε)/2⪯tn+12×|1|x|(1-ε)/2∫ℝn‍ψ(ξ)eit|ξ|-ασ~(ξ)e2πiξ·xdξ|dx,
where  σ~(ξ)  is a C∞ function supported in  [-1,1]n  and  ψ(ξ)  is a function satisfying
(188)|ψ(ξ)|⪯|ξ|β-((n+1)α/2)-((n+1)/2).
Choose a small  ε>0  such
(189)2β-α(n+1)-ε>0.
Thus by Schwarz's inequality and the Pitt's theorem, we obtain
(190)∫|x|>t‍|∫ℝn‍|ξ|βeit|ξ|-ασ(ξ)e2πiξ·xdξ|dx⪯t(n+1)/2({e2πiξ·xdξ{1|x|(1-ε)/2∫ℝn‍}|2}∫ℝn‍|1|x|(1-ε)/2∫ℝn‍ψ(ξ)eit|ξ|-ασ~(ξ)⪯tn+12×e2πiξ·xdξ{1|x|(1-ε)/2}|2dx)1/2,t(n+1)/2(∫ℝn‍|σ~(ξ)|2|ξ|-n-ε|ξ|2β-α(n+1)dξ)1/2⪯t(n+1)/2.
Combining these estimates, we have
(191)∥π0e-it|Δ|-α/2f∥L1⪯t(n+1)/2∥Iβf∥L1⪯t(n+1)/2∥f∥L-β1,
where   L-β1  denotes the  L1  Sobolev space of order  -β.  On the other hand, we have the easy energy estimate
(192)∥π0e-it|Δ|-α/2f∥L2⪯∥f∥L2.
An interpolation yields that for all  1≤p≤2,
(193)∥π0e-it|Δ|-α/2f∥Lp⪯t(n+1)(1/p-1/2)∥f∥L-γp,
where
(194)γ=β(2p-1).
We now use the Sobolev imbedding theorem and the almost orthogonality of  {πk}  to obtain
(195)∥π0e-it|Δ|-α/2f∥Lp(ℝn)⪯t(n+1)(1/p-1/2)∑|j|∞≤1‍∥πjf∥Hr(ℝn)
with
(196)1r=1p+β(2/p-1)n=n+β(2-p)np.
Since   β   is an arbitrary number larger than  α([n/2]+1),  the proposition now follows from Lemma 13.Lemma 27.
Letα>0  and  2≤p≤∞. One has
(197)∥π0e-it|Δ|-α/2f∥L2(ℝn)⪯∑|j|∞≤1‍∥πjf∥Hr(ℝn),
for any  r≤2.Proof.
The almost orthogonality (Identity (46)) and the energy estimate give
(198)∥π0e-it|Δ|-α/2f∥L2(ℝn)⪯∑|j|∞≤1‍∥πjf∥L2(ℝn).
Thus the lemma follows from Lemma 13.Now we are ready to give the proof of Theorem5.Proof.
The proof of (i) can be obtained from Propositions18 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.


## 5.5. Nonlinear Cauchy Problem with Negative Power

Now, we study the following Cauchy problem of the nonlinear dispersive equations (NDE):(199)i∂u∂t-|Δ|-α/2u+F(u)=0,u(0,x)=u0(x),
where   F(u)=|u|2ku   for some positive integer  k. For  T>0, the space  Y=C([0,T],Mp,1s)  is defined by
(200)C([0,T],Mp,1s)={u(t,x):∥u∥C([0,T],Mp,1s)=sup0≤t≤T∥u(t,·)∥Mp,1s<∞}.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 Theorem6.Proof.
In the proof, the lettersCj, j=1,2,3 denote some positive constants that are independent of all essential variables. We write the Cauchy problem in the equivalent form
(201)u(t,·)=eit|Δ|-α/2u0-i∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ,
and consider the mapping
(202)ℑu=eit|Δ|-α/2u0-i∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ.
We want to show that  ℑ  is a contraction.
By Theorem5 and Lemma 12,
(203)∥eit|Δ|-α/2u0∥Y≤C1(1+T)([n/2]+1)|2/p-1|∥u0∥Mr,1s-γ(ℝn)+C1∥u0∥Mp,1s(ℝn).
By Lemma 16, there is a constant  A2k+1>0  for which we have
(204)∥|u(t,·)|2ku(t,·)∥Mr,1s≤A2k+1∥u(t,·)∥M(2k+1)r,1s2k+1.
Thus, by Theorem 5 and Lemma 12,
(205)∥∫0t‍ei(t-τ)|Δ|-α/2F(u(τ,·))dτ∥C([0,T],Mp,1s)⪯sup0<t≤T∫0t‍∥ei(t-τ)|Δ|-α/2F(u(τ,·))∥Mp,1sdτ≤C1sup0<t≤T∫0t‍(1+(t-τ))([n/2]+1)|2/p-1|≤C1sup0<t≤T∫0t‍×∥|u(τ,·)|2ku(τ,·)∥Mr,1sdτ≤C2MkT(1+T)([n/2]+1)|2/p-1|sup0≤t≤T∥u(t,·)∥Mr(2k+1),1s2k+1⪯C2MkT(1+T)([n/2]+1)|2/p-1|sup0≤t≤T∥u(t,·)∥Mp,1s2k+1.
The last inequality is because
(206)r(2k+1)≥p
and the imbedding relation   Mp,1s⊂Mr(2k+1),1s.
As a consequence,(207)∥ℑu∥C([0,T],Mp,1s)=sup0≤t≤T∥ℑu(t,·)∥Mp,1s⪯C3(1+T)([n/2]+1)|2/p-1|×(∥u0∥Mr,1s-γ(ℝn)+Tsup0≤t≤T∥u(t,·)∥Mp,1s2k+1)+C1∥u0∥Mp,1s(ℝn).
Fix an  L>0  such that
(208)C3∥u0∥Mr,1s-γ(ℝn)+C1∥u0∥Mp,1s(ℝn)<L4.
Let  BL  be the closed ball of radial  L  centered at the origin in the space  C([0,T],Mp,1s).  Now we choose  T0  such that for any   u∈BLand T∈(0,T0],
(209)∥ℑu∥C([0,T],Mp,1s)≤C3(1+T)([n/2]+1)|2/p-1|(∥u0∥Mr,1s-γ(ℝn)+TL2k+1)+C1∥u0∥Mp,1s(ℝn)≤L.
By this choice,  ℑ  is a mapping form BLinto BL.
Furthermore, an easy computation gives that(210)∥ℑu-ℑv∥C([0,T],Mp,1s)≤C3(1+T)([n/2]+1)|2/p-1|×Tsup0≤t≤T∥u(t,·)-v(t,·)∥Mp,1s(2k+1)L2k≤12∥u-v∥C([0,T],Mp,1s),
for all T∈(0,T0] if T0 is suitably chosen. Finally, using the Banach contraction mapping theorem, we obtain a fixed point in  BL. This is the solution of the Cauchy problem. We prove the theorem.Using a similar argument with the help of (ii) in Theorem5, we can prove Theorem 7.



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*Source: 982753-2013-03-18.xml*