# Inclusion Relations betweenα-Modulation Spaces and Triebel–Lizorkin Spaces

Authors: Mohammad Ali; Hasan Baddour; Boushra Darrag
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we obtain conditions of the inclusion relations betweenα-modulation spaces and Triebel–Lizorkin spaces.


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## Body

## 1. Introduction

The modulation spaceMp,qs was first introduced by Feichtinger [1] in 1983 by the short-time Fourier transform. modulation space has aclose relationship with the topics of time-frequency analysis (see [2]), and it has been regarded as a appropriate space for the study of partial differential equations (see [3–5]).Theα-modulation space is introduced by Gröbner [6] to link Besov and modulation spaces by the parameter 0≤α≤1. One can find some basic properties about α-modulation spaces in [7, 8]. Among many features of the α-modulation spaces, an interesting subject is the inclusion between α-modulation and function spaces, have been concerned by many authors to this topic, see [8–11]. As applications, α-modulation spaces are quite recently applied to the field of partial differential equations. In [12], Misiolek and Yoneda proved locally ill-posedness of the Euler equations in the frame of α-modulation spaces. Furthermore, Han and Wang [13] proved a global well-posedness for the nonlinear Schrödinger equations on α-modulation spaces, and also in [14] studied the Cauchy problem for the derivative nonlinear Schrödinger equation on α-modulation spaces.Remark 1.
Modulation spaces are specialα-modulation spaces in the case α=0, so our theorems also works well in the special case α=0.In this research, we are interested in studying the inclusion relations betweenα-modulation spaces Mp,qs,α and Triebel–Lizorkin spaces Fp,r for p≤1, which greatly improve and extend the results for the inclusion relations between local Hardy spaces and α-modulation spaces obtained by Kato in [10].


## 2. Preliminaries

The notationX≲Y denotes the statement that X≤CY with a positive constant C that may depend on n,α,p,q,s,r. The notation X∼Y means the statement X≲Y≲X, and the notation X≃Y stands for X=CY. For a multi-index k=k1,k2,...,kn∈Zn, we denote |k|∞:=maxi=1,2,...,nki, k=k1+⋯+kn and <k>:=1+|k|21/2.LetS:=SRn be the Schwartz space and S′:=S′Rn be the space of tempered distributions. We define the Fourier transform Ff and the inverse Fourier transform F−1f of f∈SRn by(1)Ffξ=f̂ξ=∫Rnfxe−2πix.ξdxandF−1fx=f̂−x=∫Rnfξe2πix.ξdξ.We give some definitions and properties of sequences.Definition 2.
Let0<p,q≤∞,s∈R,α∈0,1. Let {ak}k∈Zn be a sequence, we denote its ℓps,α (quasi-) norm(2)
a
k
ℓ
p
s
,
α
=
∑
k
∈
Z
n
a
k
p
k
s
p
/
1
−
α
1
/
p
;
0
<
p
<
∞
,
s
u
p
k
∈
Z
n
a
k
k
s
/
1
−
α
;
p
=
∞
,and letℓps,α be the (quasi-) Banach space of sequences whoseℓps,α (quasi-) norm is finite.Letbjj∈N be a sequence, we denote its ℓps,1 (quasi-) norm(3)bjℓps,1=∑j∈N0bjp2jsp1/p;0<p<∞,supj∈N0bj2js;p=∞,and letℓps,1 be the (quasi-) Banach space of sequences whose ℓps,1 (quasi-) norm is finite.Letckk∈Zn be a sequence, we denote its ℓps,0 (quasi-) norm(4)ckℓps,0=∑k∈Znckpksp1/p;0<p<∞,supk∈Znckks;p=∞,and letℓps,0 be the (quasi-) Banach space of sequences whose ℓps,0 (quasi-) norm is finite.We recall some embedding lemmas about sequences defined above.Lemma 3 (sharpness of embedding, for uniform decomposition).
Suppose
0
<
p
,
q
≤
∞
,
s
1
,
s
2
∈
R. Then(5)
ℓ
p
s
1
,
0
⊂
ℓ
q
s
2
,
0
,holds if and only if
(6)
s
2
≤
s
1
,
1
q
+
s
2
n
<
1
p
+
s
1
n
,
o
r
s
1
=
s
2
,
p
=
q
.Lemma 4 (sharpness of embedding, for dyadic decomposition).
Suppose
0
<
p
,
q
≤
∞
,
s
1
,
s
2
∈
R
Then(7)
ℓ
p
s
1
,
1
⊂
ℓ
q
s
2
,
1
,holds if and only if
(8)
s
2
<
s
1
o
r
s
1
=
s
2
,
1
q
≤
1
p
.Lemma 5 (sharpness of embedding, forα-decomposition).
Suppose0<p,q≤∞,s1,s2∈R, α∈0,1. Then(9)
ℓ
p
s
1
,
α
⊂
ℓ
q
s
2
,
α
,holds if and only if
(10)
1
−
α
q
+
s
2
n
<
1
−
α
p
+
s
1
n
o
r
s
1
=
s
2
,
1
q
≤
1
p
.We recall some definitions of the function spaces treated in this paper.Suppose thatc>0 and C>0 are two appropriate constants, which relate to the space dimension n, and a Schwartz functions sequence ηkαk∈Zn satisfies(11)ηkαξ≥1,ifξ−kα/1−αk<ckα/1−α;suppηkα⊂ξ:ξ−kα/1−αk<Ckα/1−α;∑k∈Znηkαξ≡1,∀ξ∈Rn;Dδηkαξ≲k−αδ/1−α,∀ξ∈Rn,δ∈Z+⋃0n.Thenηkαk∈Zn constitutes a smooth decomposition of Rn. The frequency decomposition operators associated with theabove function sequence are defined by(12)□kα≔F−1ηkαF,fork∈Zn. Let 0<p,q≤∞,s∈R,andα∈0,1. Then the α-modulation space associated with the above decomposition is defined by(13)Mp,qs,αRn=f∈S′Rn:fMp,qs,αRn=∑k∈Znksq/1−α□kαfLpq1/q<∞,with the usual modifications whenq=∞. For simplicity, we denote Mp,qs=Mp,qs,0,Mp,q=Mp,q0,0, and ηkξ=ηk0ξ.Remark 6.
We recall that the above definition is independent of the choice of exactηkα (see [8], proposition 2.3). Also, for sufficiently small δ>0, one can construct a function sequence ηkαk∈Zn such that ηkαξ=1 and ηkαξηℓαξ=0 if k≠ℓ, when ξ lies in the ball Bkα/1−αk,kα/1−αδ (see [15, 9, Appendix A]).To define the Bosov spaces and Triebel–Lizorkin spaces, we introduce the dyadicdecomposition ofRn. Let φξ be a smooth bump function supported in the ball {ξ:|ξ|<3/2} and be equal to 1 on the ball {ξ:|ξ|≤4/3}. For integers j∈N, we define the Littewood–Paley operators(14)Δjf^=φ2−jξ−φ2−j+1ξf^ξ,(15)Δ0f^=φξfξ.Let0<p,q≤∞ and s∈R. Then the Besov spaces is defined by(16)Bp,qsRn=f∈S′Rn:fBp,qs=∑j=0∞2jsqΔjfLpq1/q<∞.Let0<p<∞,0<r≤∞, and s∈R. Then the Triebel–Lizorkin spaces is defined by(17)Fp,rsRn=f∈S′Rn:fFp,rs=∑j=0∞2jsrΔjfr1/rLp<∞.LetQn be the collection of all cubes Qvk in Rn with sides parallel to the axes, centered at 2−vk, and with side length 2−v, where k∈Zn and v∈N0.LetQ be a cube in Rn and m>0, then mQ is the cube in Rn concentric with Q and with side length m times the side length of Q. We write v,k<v′,k′ if v≥v′ and(18)Qv,k⊂2Qv′,k′withQv,k,2Qv′,k′∈Qn.Leta∈R, then a+=maxa,0 and a stands for the largest integer less than or equal to a.Definition 7 (see [16]).
Lets∈R, 0<p≤1<r≤∞. Let K and L be integers with(19)
K
≥
s
+
1
+
a
n
d
L
≥
m
a
x
n
1
p
−
1
−
s
,
−
1
.(1) The (complex-valued) functionax is called a s-atom if
(20)
s
u
p
p
a
⊂
5
Q
,
for someQ=Q0,k∈Qn and
(21)
D
α
a
x
≤
1
f
o
r
α
≤
K
.
(2) LetQ=Qvk∈Qn. The (complex-valued) function ax is called a Q,s,p,r-atom if (20) is satisfied,
(22)
D
α
a
x
≤
Q
−
1
/
r
+
s
/
n
−
α
/
n
f
o
r
α
≤
K
,
and
(23)
∫
R
n
x
β
a
x
d
x
=
0
for
β
≤
L
.
(3) The distributiong∈S′ is called an s,p,r-atom if
(24)
g
=
∑
μ
,
l
<
v
,
k
d
μ
l
a
μ
l
x
c
o
n
v
e
r
g
e
n
c
e
i
n
F
p
,
r
s
,
for somev∈N0 and k∈Zn, where aμlx is a Qμl,s,p,r-atom and dμl are complex numbers with
(25)
∑
μ
,
l
<
v
,
k
d
μ
l
r
1
/
r
≤
Q
v
k
1
/
r
−
1
/
p
,
with usual modification ifq=∞.Lemma 8 (see [16]).
Lets∈R, 0<p≤1<r≤∞. Let K and L be fixed integers satisfying (19). Then f∈S′ is an element of Fp,rs if and only if it can be represented as
(26)
f
=
∑
j
=
1
∞
μ
j
a
j
+
λ
j
g
j
c
o
n
v
e
r
g
e
n
c
e
i
n
S
′
,
whereaj are s-atoms, gj are s,p,r-atoms, μj and λj are complex numbers with(27)∑j=1∞μjp+λjp1/p≲fFp,rs.We also give the following lemma for inclusion relations between Besov andα-modulation spaces [8].Lemma 9.
Let0<p,q≤∞, and s∈R. Then the following tow statement are true:
(1)Mp,qs,α⊂Bp,q⇔s≥0∨nα−11p−1q∨nα−11−1p−1q.
(2)Bp,q⊂Mp,qs,α⇔s≤0∧nα−11p−1q∧nα−11−1p−1q.Lemma 10 (Young’s inequality).
(1)
Let0<p≤1,R>0,suppf^,suppg^⊆Bx,R⊆Rn. We have(28)f∗gLp≤CRn1/p−1fLpgLp.
for allf,g∈SRn and R>0, where C independent of x∈Rn.
(2)
Let1≤p,q,r≤∞ satisfy 1+1/q=1/p+1/r. Then we have(29)f∗gLq≲fLpgLr.The following Bernstein multiplier theorem will be used in our proof.Lemma 11 (Bernstein multiplier theorem).
Let
0
<
p
≤
1, ∂δf∈L2for|δ|≤n1−α1/p−1/2+1. Then,(30)F−1fLp≲∑δ≤n1−α1/p−1/2+1∂δfL2.


## 3. Main Results

Now, we state our main results as follows.Theorem 12.
Let0<p≤1, 0<q,r≤∞, s∈R, and 0≤α<1. Then, Mp,qs,αRn⊂Fp,rRn holds if and only if either of the following conditions is satisfied.(1)
p
≥
q
,
s
≥
0
,
1
r
≤
1
q;
(2)
p
<
q
,
s
>
n
1
−
α
1
p
−
1
q.Theorem 13.
Let0<p≤1, 0<q,r≤∞, s∈R, and 0≤α<1. Then, Fp,rRn⊂Mp,qs,αRn holds if and only if either of the following conditions is satisfied.(1)
p
>
q
,
s
<
−
n
1
−
α
1
p
+
1
q
−
1;
(2)
p
≤
q
,
s
≤
−
n
1
−
α
1
p
+
1
q
−
1.We prove the following two propositions used for the proof of the Theorem12.Proposition 14.
Let0<p<∞, 0<q,r≤∞, s∈R, and 0≤α<1. Then we have(1)
M
p
,
q
s
,
α
⊂
F
p
,
r
⇒
ℓ
q
s
/
1
−
α
,
0
⊂
ℓ
p
0
,
0
;
(2)
F
p
,
r
⊂
M
p
,
q
s
,
α
⇒
ℓ
p
0
,
0
⊂
ℓ
q
s
/
1
−
α
,
0.Proof.
Takef to be a smooth function whose Fourier transform has compact small support, denote fk^=f^.−k. We define
(31)
H
N
=
∑
k
∈
Z
n
c
k
f
k
N
;
f
k
N
x
=
f
k
x
−
k
N
.For a truncated (only finite nonzero items) nonnegative sequenceckk∈Zn, where N is some large integer.By the definition ofα-modulation space Mp,qs,α, we have (32)HNMp,qs,α=∑k∈Zn□kαHNLpqksq/1−α1/q∼∑k∈ZnckfkNLpqksq/1−α1/q∼∑k∈ZnckqfLpqksq/1−α1/q∼∑k∈Znckqksq/1−α1/q∼ckℓqs/1−α,0.On the other hand, we use the orthogonality offkNk∈Zn as N→∞, we obtain(33)HNFp,r∼HNLp=∫Rn∑k∈ZnckfkNpdx1/p⟶N→∞∫Rn∑k∈Znckfkpdx1/p∼∑k∈Znmgdckpdx1/p∼ckℓp0,0.Hence,(34)HNMp,qs,α∼ckℓqs/1−α,0,limN→∞HNFp,r∼ckℓp0,0,Thus, we obtainℓqs/1−α,0⊂ℓp0,0, if Mp,qs,α⊂Fp,r hold.On the other hand, we obtainℓp0,0⊂ℓqs/1−α,0 if Fp,r⊂Mp,qs,α hold.Proposition 15.
Let0<p<∞, 0<q,r≤∞, s∈R and 0≤α<1. Then we have(35)
M
p
,
q
s
,
α
⊂
F
p
,
r
⇒
ℓ
q
0
,
α
⊂
ℓ
r
0
,
α
.Proof.
Takef to be a nonzero smooth function whose Fourier transform has small support, such that □kαfk=fk and □ℓαfk=0 if k≠ℓ, where we denote fk^=f^ξ−kα/1−αk/kα/1−α. Denote(36)
E
=
∑
k
∈
Z
n
a
k
f
k
.For a truncated (only finite nonzero items) nonnegative sequenceakk∈Zn. We have(37)EFp,r=∑k∈ZnΔjEr1/rLp∼∑k∈Znakfkr1/rLp∼∑k∈Znakrfkr1/rLp∼akℓr0,α.On the other hand,(38)EMp,q0,α=∑k∈Zn□kαELpq1/q=∑k∈ZnakfkLpq1/q=∑k∈ZnakqfLpq1/q∼akℓq0,α.Hence,(39)EMp,q0,α∼akℓq0,α,EFp,r∼akℓr0,α,Thus, we obtainℓq0,α⊂ℓr0,α, if Mp,q0,α⊂Fp,r.Proof of Theorem12.
We divide this proof into two parts.Necessary. For p≥q, using Proposition 14 and Lemma 3 to deduce ℓqs/1−α0⊂ℓp0,0,whichimpliess≥0.On the other hand, we use Proposition15 and Lemma 5 to deduce ℓq0,α⊂ℓr0,α, which implies 1/r≤1/q.Forq≥p, using Proposition 14 and Lemma 3 to deduce ℓqs/1−α,0⊂ℓp0,0, which implies s>n1−α1/p−1/q.Sufficiency. For p≥q. We have 1/r≤1/q, and then Fp,q⊂Fp,r. using Lemma 9, we obtain Mp,qs,α⊂Bp,q. Thus we deduce that(40)Mp,qs,α⊂Bp,q⊂Fp,q⊂Fp,r.Which is the desired conclusion.Forp<q. Use Mp,p0,α⊂Fp,p obtained above to deduce that(41)Mp,qn1−α1/p−1/q+2ε,α⊂Mp,pε,α⊂Fp,pε⊂Fp,r,for anyε>0,r∈0,∞.We prove the following two propositions used for the proof of the Theorem13.Proposition 16.
Let0<p<∞, 0<q,r≤∞, s∈R, and 0≤α<1. Then we have(1)
M
p
,
q
s
,
α
⊂
F
p
,
r
⇒
ℓ
q
n
1
−
α
/
q
+
n
α
1
−
1
/
p
+
s
,
1
⊂
ℓ
p
n
1
−
1
/
p
,
1
;
(2)
F
p
,
r
⊂
M
p
,
q
s
,
α
⇒
ℓ
p
n
1
−
1
/
p
,
1
⊂
ℓ
q
n
1
−
α
/
q
+
n
α
1
−
1
/
p
+
s
,
1
.Proof.
Letg be a nonzero Schwartz function whose Fourier transform has compact support in {ξ:3/4≤|ξ|≤4/3}, satisfying gξ=1 on {ξ:7/8≤|ξ|≤8/7}. Set gj^ξ:=g^ξ/2j. By the definition of Δj, we have Δjgj=gj for j≥0, and Δlgj=0 if l≠j. Denote(42)
A
j
˜
=
k
∈
Z
n
:
□
k
α
g
j
=
F
−
1
η
k
α
,
A
j
=
k
∈
Z
n
:
□
k
α
g
j
≠
0
,we have|Aj|∼Aj˜∼2jn1−α for j≥N, where N is a sufficiently large number. We define(43)GN=∑j≥NbjgjN;gjNx=gjx−jN,for a truncated (only finite nonzero items) nonnegative sequencebjj=0∞.We first prove that the inclusionMp,qs,α⊂Fp,r implies ℓqn1−α/q+nα1−1/p+s,1⊂ℓpn1−1/p,1. By the definition of α-modulation space, we obtain that(44)GNMp,qs,α=∑k∈Znksq/1−α□kαGNLpq1/q≲∑j≥N∑k∈Ajksq/1−αbjq□kαgjNLpq1/q≲∑j≥N∑k∈Ajbjqksq/1−αF−1ηkαLpq1/q∼∑j≥N∑k∈Ajbjqksq/1−αknαq/1−α1−1/p1/q∼∑j≥NAjbjqkq/1−αs+nα1−1/p1/q∼∑j≥Nbjq2jn1−α.2jqs+nα1−1/p1/q∼∑j≥Nbjq2jqs+nα1−1/p+n1−α/q1/q∼bjℓqs+nα1−1/p+n1−α/q,1.On the other hand, we turn to the estimate ofGNFp,r, using the orthogonality of gjN as N→∞, we obtain(45)GNFp,r=∑j∈N0ΔjGNr1/rLp=∑j=N∞bjgjNr1/rLp⟶N→∞∫Rn∑j=N∞bjgjpdx1/p≃∑j=N∞bjp2jn1−1/pp1/p=bjj≥Nℓpn1−1/p,1.Hence,(46)GNMp,qs,α≲bjℓqs+nα1−1/p+n1−α/q,1,limN→∞GNFp,r∼bjℓpn1−1/p,1.Thus, ifMp,qs,α⊂Fp,r, we obtain the desired inclusion(47)ℓqn1−α/q+nα1−1/p+s,1⊂ℓpn1−1/p,1.Next we prove that the inclusionFp,r⊂Mp,qs,α implies ℓpn1−1/p,1⊂ℓqn1−α/q+nα1−1/p+s,1. By the definition of α-modulation space, we obtain that(48)GNMp,qs,α=∑k∈Znksq/1−α□kαgNLpq1/q≥∑j≥N∑k∈A˜jksq/1−αbjq□kαgjNLpq1/q=∑j≥N∑k∈A˜jbjqksq/1−αF−1ηkαLpq1/q∼∑j≥N∑k∈A˜jbjqksq/1−αknαq/1−α1−1/p1/q∼∑j≥NA˜jbjqkq/1−αs+nα1−1/p1/q∼∑j≥Nbjq2jn1−α.2jqs+nα1−1/p1/q∼∑j≥Nbjq2jqs+nα1−1/p+n1−α/q1/q∼bjℓqs+nα1−1/p+n1−α/q,1.Hence,(49)GNMp,qs,α≳bjℓqs+nα1−1/p+n1−α/q,1.On the other hand, by the same argument of the previous proof, we deduce that(50)limN→∞GNFp,r∼bjℓpn1−1/p,1.Thus, ifFp,r⊂Mp,qs,α holds, we obtain the desired inclusion(51)ℓpn1−1/p,1⊂ℓqn1−α/q+nα1−1/p+s,1.Proposition 17.
Let0<p≤1. We have the following inclusion relation:(52)
F
p
,
∞
n
1
−
α
2
/
p
−
1
⊂
M
p
,
q
0
,
α
.Proof.
We first verify(53)
a
M
p
,
q
0
,
α
≲
1
,for anyn1−α2/p−1-atom a. Tack a to be an n1−α2/p−1-atom as in Definition 7 (with s=n1−α2/p−1). Observing that K≥n1−α2/p−1+1≥n1−α1/p−1/2+1, we have(54)∂δa≤1,forδ≤n1−α1/p−1/2+1. By the Bernstein multiplier theorem, we have the following estimate of a:(55)aMp,q0,α∼F−1aLp≲∑δ≤n1−α1/p−1/2+1∂δaL2≲1.Next, we turn to the estimate of ans,p,∞-atom for Fp,∞n1−α2/p−1. By Definition 7, an s,p,∞-atom g can be represented by(56)g=∑μ,l<v,kdμlaμlxconvergenceinFp,∞n1−α2/p−1,for somek∈Zn and v∈N0, where aμlx are Qμl,s,p,∞-atoms and dμl are complex numbers with(57)supμ,l<v,kdμl≤Qvk−1/p,for a fixedτ≤ν, we denote(58)gτ=∑τ,l<v,kdτlaτlx.Then,g can be represented by(59)g=∑τ≤νgτconvergenceinFp,∞n1−α2/p−1.We now concentrate on the estimate ofgτ. By Definition 7, we have.(60)∂δaτl≤Qτl1−α2/p−1−δ/n=2−τn1−α2/p−1−δ/n,forδ≤n1−α1/p−1/2+1≤K. Recalling suppaτl⊂5Qτl, we use (60) and the almost orthogonality of aτl to deduce that(61)∂δgτ=∑τ,l<v,kdτl∂δaτlx≲supτ,l<v,kdμl2−τn1−α2/p−1−δ/n≲Qvk−1/p2−τn1−α2/p−1−δ/n,for allδ≤n1−α1/p−1/2+1. By the Bernstein multiplier theorem, we deduce that(62)gτMp,q0,α∼F−1gτLp≲∑δ≤n1−α1/p−1/2+1∂δgτL2≲∑δ≤n1−α1/p−1/2+1Qvk1/2−1/p2−τn1−α2/p−1−δ/n.By a dilation argument, we have(63)gτMp,q0,α∼F−1gτLp≲Qvk1/2−1/p2−τn1−α1/p−1/2.Thus,(64)gMp,q0,αp=∑τ≤νgτMp,q0,αp≤∑τ≤νgτMp,q0,α≤Qvkp/2−1∑τ≤ν2−τn1−α1−p/2≲Qvkp/2−1.2−vn1−α1−p/2∼1.By Lemma8, we have(65)fMp,q0,α=∑j=1∞μjaj+λjgjMp,q0,α≤∑j=1∞μjpajMp,q0,α+λjpgjMp,q0,α1/p≲∑j=1∞μjp+λjp1/p≲fFp,∞n1−α2/p−1,which is the desired conclusion.Proof of Theorem13.
We divide this proof into two parts.Sufficiency.For p≥q, by Lemma 9, we obtain Bp,∞⊂Mp,∞n1−α1−1/p,α. Using Fp,∞⊂Bp,∞, we deduce that(66)Fp,∞⊂Mp,∞n1−α1−1/p,α.In addition, we haveFp,∞n1−α2/p−1⊂Mp,p0,α by Proposition 17. By potential lifting, we obtain(67)Fp,∞⊂Mp,pn1−α1−2/p,α.Thus, the desired conclusion can be deduced by a standard interpolation argument between (66) and (67).Forp<q, recalling Fp,∞⊂Mp,pn1−α1−2/p,α obtained in Proposition 17, we deduce that(68)Fp,r⊂Fp,∞⊂Mp,pn1−α1−2/p,α⊂Mp,qn1−α1−1/p−1/q−ε,α,for anyε>0,r∈0,∞.Necessity. We use Proposition 16 to deduce inclusion relation ℓpn1−1/p,1⊂ℓqn1−α/q+nα1−1/p+s,1. Then, Lemma 4 yields that s≤−n1−α1/p+1/q−1 for p≤q, while the inequality is strict for p>q.



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*Source: 9640742-2020-01-30.xml*