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R E S E A R C H

Open Access

The closedness of shift invariant

subspaces in

L

p,q

(

R

d+1

)

Qingyue Zhang

1*

*Correspondence:

[email protected]

1College of Science, Tianjin

University of Technology, Tianjin, China

Abstract

In this paper, we consider the closedness of shift invariant subspaces inLp,q(Rd+1). We first define the shift invariant subspaces generated by the shifts of finite functions in

Lp,q(Rd+1). Then we give some necessary and sufficient conditions for the shift invariant subspaces inLp,q(Rd+1) to be closed. Our results improve some known results in (Aldroubi et al. in J. Fourier Anal. Appl.7:1–21,2001).

MSC: 42C15; 42C40; 41A58

Keywords: Mixed Lebesgue spacesLp,q(Rd+1); Closedness of shift invariant subspaces; Shift invariant subspaces

1 Introduction and main result

Lp,q(Rd+1) (1 <p,q< +) are called mixed Lebesgue spaces which generalize Lebesgue

spaces [2–6]. They are very important for the study of sampling and equation problems, since we can consider functions to be independent quantities with different properties [5–8]. Recently, Torres, Ward, Li, Liu and Zhang studied the sampling theorem on the shift invariant subspaces inLp,q(Rd+1) [6–8]. In this environment, we study the closedness of shift invariant subspaces inLp,q(Rd+1).

The closedness is an expected property for shift invariant subspaces, which is widely considered in the study of shift invariant subspaces. de Boor, DeVore, Ron, Bownik and Shen studied the closedness of shift invariant subspaces inL2(Rd) [911]. And Jia,

Mic-chelli, Aldroubi, Sun and Tang discussed the closedness of shift invariant subspaces in

Lp(Rd) [1,12,13]. In this paper, we consider the closedness of shift invariant subspaces in Lp,q(Rd+1).

In order to provide our main result which extends the result in [1], we introduce some definitions and notations.

The definition ofLp,q(Rd+1) is as follows.

Definition 1.1 For 1 <p,q< +∞.Lp,q=Lp,q(Rd+1) is made up of all functionsf satisfying

fLp,q=

R

Rd

f(x,y)qdy

p

q

dx

1

p

< +∞.

(2)

We define mixed sequence spacesp,q(Zd+1) as follows:

p,q=p,qZd+1 =

c:cp,q=

n∈Z

l∈Zd

c(n,l)q

p

q1p

< +∞

.

Given a functionf, define

fLp,q:=

k1∈Z

[0,1]d

k2∈Zd

f(·+k1,x2+k2)

q dx2

1/q

Lp[0,1]

.

For 1≤p,q≤ ∞, letLp,q =Lp,q(Rd+1) be the linear space of all functionsf for which

fLp,q <∞. The norms are defined above and with usual modification in the case of

porq=∞.Lp,q is a generalization ofLp (the definition ofLp see [14, Sect. 1]). Clearly,

for 1≤p,q≤ ∞, one hasL∞,∞LandL∞,∞Lp,qL1,1.

Letfˆ(ω) denote the Fourier transform offL1(Rd+1):

ˆ

f(ω) =

Rd+1f(x)e –iωxdx.

For a given sequencecand a functionφ,c∗sdφ=k∈Zd+1c(k)φ(·–k) is called

semi-convolution ofcandφ.

Assume thatBis a Banach space. (B)(r)denotesrcopiesB×B× · · · ×BofB. IfC=

(c1,c2, . . . ,cr)T∈(B)(r), then one defines the norm ofCbyC(B)(r)=

r

j=1cjB. WCp,q

(1≤p,q≤ ∞) consists of all distributions whose Fourier coefficients belong to

p,q. Whenp=q= 1,WC1,1becomes the Wiener classWC.

Suppose that = (θ1,θ2, . . . ,θr)T and = (ψ1,ψ2, . . . ,ψs)T are two vector functions

which satisfyθj(ω)ψj(ω) (1≤jr, 1≤js) are integrable. One defines

[,](ω) =

k∈Zd+1

θj(ω+ 2)ψj(ω+ 2)

1≤jr,1js

.

Remark 1.2 By [14, Theorem 3.1 and Theorem 3.2], [,](ω)∈WC for any ,

L∞,∞ LL2. Therefore, for anyL∞,∞, using the continuity of [,](ω) and

rank[,](ω) =rank((ω+ 2))k∈Zd+1, one obtains, for any n≥0, the set n={ω: rank((ω+ 2))kZd+1>n}is open.

The following proposition shows that the shift invariant subspaces inLp,q(1 <p,q<)

are well defined.

Proposition 1.3([8, Lemma 2.2]) LetθLp,q,where1 <p,q<∞.Then,for any cp,q,

c∗sdθLp,qcp,qθLp,q.

Definition 1.4 For= (θ1,θ2, . . . ,θr)T∈(L∞,∞)(r), the multiply generated shift invariant

subspace in the mixed Lebesgue spacesLp,qis defined by

Vp,q() =

r

j=1

k∈Zd+1

cj(k)θj(·–k) :cj=

cj(k) :k∈Zd+1p,q, 1≤jr

(3)

The following is our main result.

Theorem 1.5 Assume= (θ1,θ2, . . . ,θr)T∈(L∞,∞)(r)and1 <p,q<∞.Then the following four conditions are equivalent.

(i) Vp,q()is closed inLp,q.

(ii) There exist some positive constantsC1andC2satisfying

C1[,](ω)≤[,](ω)[,](ω)TC2[,](ω),ω[–π,π]d+1.

(iii) There exist constantsB1,B2> 0satisfying

B1fLp,q≤ inf

f=rj=1cj∗sdφj

r

j=1

cjp,qB2fLp,q, ∀fVp,q().

(iv) There is= (ψ1,ψ2, . . . ,ψr)T∈(L∞,∞)(r)satisfying

f =

r

j=1

k∈Zd+1

f,ψj(·–k)

θj(·–k)

=

r

j=1

k∈Zd+1

f,θj(·–k)

ψj(·–k), ∀fVp,q().

The paper is organized as follows. In the next section, we give some three useful lemmas and two propositions. In Sect.3, we give the proof of Theorem1.5. Finally, concluding remarks are presented in Sect.4.

2 Some useful lemmas and propositions

In this section, we give three useful lemmas and two propositions which are needed in the proof of Theorem1.5.

Proposition 2.1([1, Lemma 1]) Let∈(L2)(r).Then the following are equivalent:

(i) rank((ω+ 2))kZd+1is a constant for anyω∈Rd+1.

(ii) There exist some positive constantsC1andC2such that

C1[,](ω)≤[,](ω)[,](ω)TC2[,](ω), ω[–π,π]d+1.

Proposition 2.2([1, Lemma 2]) Let∈(L2)(r)satisfyrank((ξ+ 2kπ))

k∈Zd+1=k0≥1for allξ∈Rd+1.Then there exists a finite index set,η

λ∈[–π,π]d+1, 0 <δλ< 1/4,nonsingular

2π-periodic r×r matrix Pλ(ξ)with all entries in the Wiener class and Kλ⊂Zd+1 with

cardinality() =k0for allλ,having the following properties:

(i)

[–π,π]d+1

λ

B(δλ,δλ/2),

(4)

(ii)

(ξ)(ξ) =

1,λ(ξ)

2,λ(ξ)

, ξ∈Rd+1andλ,

where1,λand2,λare functions fromRd+1toCk0andCr–k0,respectively,satisfying

rank1,λ(ξ+ 2πk) kK

λ=k0, ∀ξB(δλ,δλ/2)

and

2,λ(ξ) = 0, ∀ξB(δλ, 8δλ/5) + 2πZd+1.

Furthermore,there exist2π-periodicC∞functions hλ(ξ),λ,onRd+1such that

λ

(ξ) = 1, ∀ξ∈Rd+1

and

supp(ξ)⊂B(δλ,δλ/2) + 2πZd+1.

The following lemma can be proved similarly to [7, Theorem 3.4]. And we leave the details to the interested reader.

Lemma 2.3 Assume that fLp,q(1 <p,q<∞)and gL∞,∞.Then

RRdf(x1,x2)g(x1k1,x2k2)dx1dx2:k1∈Z,k2∈Z

d

p,q

fLp,qgL∞,∞.

Lemma 2.4 Let c1.Then one has:

(i) IfθLp,q(1 <p,q<),then

c∗sdθLp,qc1θLp,q.

(ii) IfθL∞,∞,then

c∗sdθL∞,∞≤ c1θL∞,∞.

Proof (i) By Young’s inequality and the triangle inequality, one has

c∗sdθLp,q =

n∈Z

[0,1]d

l∈Zd

c∗sdθ(·+n,y+l)

q dy

1/q

Lp[0,1]

=

n∈Z

[0,1]d

l∈Zd

n∈Z

l∈Zd

cn,lθ

·+nn,y+ll

q dy

1/q

Lp[0,1]

n∈Z

[0,1]d

n∈Z

l∈Zd

l∈Zd

cn,lθ·+nn,y+ll

q dy

1/q

(5)

n∈Z

[0,1]d

n∈Z

l∈Zd

|cn,l| l∈Zd

θ·+nn,y+l

q

dy

1/q

Lp[0,1]

n∈Z

n∈Z

l∈Zd

|cn,l|

[0,1]d

l∈Zd

θ·+nn,y+l

q dy

1/q

Lp[0,1]

n∈Z

l∈Zd

|cn,l|

n∈Z

[0,1]d

l∈Zd

θ·+nn,y+l

q dy

1/q

Lp[0,1]

n∈Z

l∈Zd

|cn,l| n∈Z

[0,1]d

l∈Zd

θ·+nn,y+l

q dy

1/q

Lp[0,1]

n∈Z

l∈Zd

|cn,l| n∈Z

[0,1]d

l∈Zd

θ·+n,y+l

q dy

1/q

Lp[0,1]

=c1θLp,q.

The desired result (i) in Lemma2.4is obtained.

(ii) The desired result (ii) in Lemma2.4can be found in [8, Lemma 2.4].

Lemma 2.5 Assume thatθLp,q (1 <p,q<)and

k∈Zd+1θ(·–k) = 0.Then for any function h onRd+1satisfying

h(x)≤D1 +|x| –d–2 and h(x) –h(y)≤D|xy|1 +min|x|,|y| –d–2, (2.1)

one has

lim n→∞2

–n(d+1)

k∈Zd+1

h2–nk θ(·–k)

Lp,q

= 0.

Here D in(2.1)is a positive constant.

Proof SinceθLp,q, for anyε> 0, there isN02 satisfying

|l|≥N0

[0,1]d

k∈Zd

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

< (2.2)

and

l∈Z

[0,1]d

kEdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

<, (2.3)

whereEdN0={(k1, . . . ,kd) : there exists some 1≤i0dsuch that|ki0|>N0}. Set

θ1(x1, . . . ,xd+1) =θ(x1, . . . ,xd+1)χON0(x1, . . . ,xd+1)

+

(k1,...,kd+1)∈EdN+10

(6)

whereON0=|ki|≤N0,1id+1[(k1, . . . ,kd+1) + [0, 1]d+1] andχ

Sis the characteristic function

ofS.

ThuskZd+1θ1(·–k) =kZd+1θ(·–k) = 0 andθ1–θLp,q< 5. In fact

θ1–θLp,q

=

l∈Z

[0,1]d

k∈Zd

(θ1–θ)(·+l,x+k)

q dx

1/q

Lp[0,1]

[0,1]d

k∈Zd

(θ1–θ)(·,x+k)

q dx

1/q

Lp[0,1]

+

l=0

[0,1]d

k∈Zd

(θ1–θ)(·+l,x+k)

q dx

1/q

Lp[0,1]

=I1+I2.

First of all, one treatsI1: by (2.2) and (2.3), one has

I1

[0,1]d

(θ1–θ)(·,x) q dx 1/q + [0,1]d

k=0

(θ1–θ)(·,x+k)

q dx

1/q

Lp[0,1]

[0,1]d

(θ1–θ)(·,x) q

dx

1/q

Lp[0,1]

+

[0,1]d

k=0

(θ1–θ)(·,x+k)

q dx

1/q

Lp[0,1]

[0,1]d

(k1,...,kd+1)∈EdN+10

θ(·+k1, . . . ,xd+1+kd+1)

q dx

1/q

Lp[0,1]

+

[0,1]d

kEdN 0

θ(·,x+k)

q dx

1/q

Lp[0,1]

[0,1]d

|l|>N0,k∈Zd

+

l∈Z,k∈EdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

l∈Z

[0,1]d

kEdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

[0,1]d

|l|>N0,k∈Zd

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

[0,1]d

l∈Z,k∈EdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

(7)

≤ |l|>N0

[0,1]d

k∈Zd

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

l∈Z

[0,1]d

kEdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

<++= 3. Next, one treatsI2:

I2

|l|>N0

[0,1]d

k∈Zd

(θ1–θ)(·+l,x+k)

q dx

1/q

Lp[0,1]

+

|l|≤N0,l=0

[0,1]d

k∈Zd

(θ1–θ)(·+l,x+k)

q dx

1/q

Lp[0,1]

=

|l|>N0

[0,1]d

k∈Zd

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

|l|≤N0,l=0

[0,1]d

kEdN 0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

|l|>N0

[0,1]d

k∈Zd

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

+

l∈Z

[0,1]d

kEd N0

θ(·+l,x+k)

q dx

1/q

Lp[0,1]

<+= 2.

Therefore, one hasθ1–θLp,q< 5.

Using Lemma2.4and (2.1), there exists some positive constantCsuch that

2–n(d+1)

k∈Zd+1

h2–nk φ(·–k) –φ1(·–k)

Lp,q

≤2–n(d+1)

k∈Zd+1

h2–nk φ1–φLp,qC.

Thus

2–n(d+1)

k∈Zd+1

h2–nk θ1(·–k)

Lp,q

= 2–n(d+1)

j1∈Z

[0,1]d

j2∈Zd

k1∈Z,k2∈Zd

h2–nk1, 2–nk2

×θ1(·+j1k1,x2+j2k2)

qdx2

1/q

Lp[0,1]

= 2–n(d+1)

j1∈Z

[0,1]d

j2∈Zd

k1∈Z,k2∈Zd

h2–nk1, 2–nk2 –h

2–nj1, 2–nj2

(8)

×θ1(·+j1k1,x2+j2k2)

qdx2

1/q

Lp[0,1]

≤2–n(d+2)C1(N0)

j1∈Z

[0,1]d

j2∈Zd

k1∈Z,k2∈Zd

1 + 2–n(k1,k2) –(d+2)

×θ1(·+j1–k1,x2+j2–k2)

q dx2

1/q

Lp[0,1]

= 2–n(d+2)C1(N0)

j1∈Z

[0,1]d

k1∈Z

k2∈Zd

1 + 2–n(k1,k2) –(d+2)

×

j2∈Zd

θ1(·+j1k1,x2+j2)

q dx2

1/q

Lp[0,1]

≤2–n(d+2)C1(N0)

k1∈Z,k2∈Zd

1 + 2–n(k1,k2) –(d+2)

×

j1∈Z

[0,1]d

j2∈Zd

θ1(·+j1k1,x2+j2)

q dx2

1/q

Lp[0,1]

≤2–nC2(N0)

j1∈Z

[0,1]d

j2∈Zd

θ1(·+j1,x2+j2)

q dx2

1/q

Lp[0,1]

= 2–nC2(N0)θ

1Lp,q≤2–nC2(N0)θLp,q+ 5 .

HereCi(N0) (i= 1, 2) are positive constants depending only onN0andd. This completes

the proof.

3 Proof of Theorem1.5

In this section, we give the proof of Theorem1.5. The main steps of the proof are as follows: (iv)⇒(iii)⇒(i)⇒(ii)⇒(iv).

(iv)⇒(iii):

Letf =rj=1kZd+1f,ψj(·–k)θj(·–k). Then, by Lemma2.3, one has

inf f=rj=1cj∗sdφj

r

j=1

cjp,q

r

j=1

f,ψj(·–k1,·–k2)

:k1∈Z,k2∈Zdp,q

r

j=1

fLp,qψjL∞,∞=fLp,q r

j=1

ψjL∞,∞.

Conversely, iff =rj=1cj∗sdθj, then, by Proposition1.3and the triangle inequality

fLp,q =

r

j=1 cj∗sdθj

Lp,q

r

j=1

cj∗sdθjLp,q

r

j=1

cjp,jLp,q≤max

1≤jrθjL

p,q

r

j=1

(9)

Taking the infimum for (3.1), one gets

fLp,q≤max

1≤jrθjL

p,q inf

f=rj=1cj∗sdθj

r

j=1

cjp,q.

LetB1= 1/max1≤jrθjLp,q andB2=r

j=1ψjL∞,∞. Then one has

B1fLp,q≤ inf

f=rj=1cj∗sdφj

r

j=1

cjp,qB2fLp,q, ∀fVp,q().

(iii)⇒(i):

For convenience, letT: (p,q)(r)V

p,q() be a mapping which is defined by

TC=

r

j=1

cj∗sdθj, C= (c1,c2, . . . ,cr)T

p,q (r),

and letfinf=inff=rj=1cj∗sdθj

r

j=1cjp,q. Then, obviously, · infis a norm. Assumefn⊂ Ran(T) (n≥1) is a Cauchy sequence. HereRan(T) denotes the range ofT. Without loss of generality, letfnfn–1inf< 2–n. Using the definition of · inf, there isCn∈(p,q)(r)

(n≥2) such thatTCn=fnfn–1andCn(p,q)(r)< 2–nfor anyn≥2. By the completeness

of (p,q)(r)and

n=2Cn(p,q)(r)<∞, one hasZ=∞n=2Cn∈(p,q)(r)andf1+TZ∈Ran(T).

Note thatTCinf≤ C(p,q)(r)for anyC∈(p,q)(r). One has

fnf1TZinf=

T

k=n+1 Ck

inf

k=n+1 Ck

(p,q)(r)

≤ ∞

k=n+1

Ck(p,q)(r)→0,

whenn→ ∞. Therefore,Ran(T) is closed. SinceVp,q() =Ran(T), one sees thatVp,q() is closed.

(i)⇒(ii):

Similarly to [1, Proof of (i)⇒(iii)], one can prove (i)⇒(ii) by usingL∞,∞L, and

substitutingLp,q,L∞,∞, Proposition2.1and Lemma2.5forLp,L, Lemma 1 and Lemma 3

in [1], respectively. (ii)⇒(iv):

Assume that(ω),(ω) and1,λ(ω) are as in Proposition2.2. Define

(ω) =(ω)T

[1,λ,1,λ](ω)–1 0

0 I

(ω)(ω). (3.2)

Here(ω) is a function with period 2πwhich satisfiessuppB(ηλ,δλ) + 2πZd+1and

(ω) = 1 onsupp. Thus∈(WC)(r×r). Let= (ψ1,ψ2, . . . ,ψr)Tbe defined by

(ω) =

λ

(ω)(ξ)(ω). (3.3)

Then, by Lemma 2.4, one has L∞,∞. For any fVp,q(), using the definition of

(10)

A(ω)T(ω). Putting

g=

r

j=1

k∈Zd+1

f,ψj(·–k)

θj(·–k).

By the periodicity of(ω) and(ω), (3.2), (3.3) and Proposition2.2, one has ˆ

g(ω) =A(ω)T[,](ω)(ω) =

λ

A(ω)TPλ(ω)–1

×

[1,λ,1,λ](ω) 0

0 0

[1,λ,1,λ](ω)–1 0

0 I

1,λ(ω)

0

(ω)

=

λ

A(ω)TPλ(ω)–1

1,λ(ω)

0

(ω)

=

λ

A(ω)TPλ(ω)–1(ω)(ω)(ω)

=

λ

A(ω)T(ω)(ω)

=A(ω)T(ω) =ˆf(ω).

Thusfˆ(ω) =gˆ(ω). Thereforef=g, namely

f =

r

j=1

k∈Zd

f,ψj(·–k)

θj(·–k).

Similar arguments show that

f =

r

j=1

k∈Zd

f,θj(·–k)

ψj(·–k).

4 Concluding remarks

In this paper, we study the closedness of shift invariant subspaces inLp,q(Rd+1). We first

define the shift invariant subspaces generated by the shifts of finite functions inLp,q(Rd+1).

Then we give some necessary and sufficient conditions for the shift invariant subspaces in

Lp,q(Rd+1) to be closed.

However, in this paper, we only consider the closedness of shift invariant subspace of Lp,q(Rd+1). Studying theLp,q-frames in a shift invariant subspace of mixed Lebesgue Lp,q(Rd) is the goal of future work.

Funding

(11)

Competing interests

The author declares that he has no competing interests.

Authors’ contributions

QZ provided the questions and gave the proof for the main result. He read and approved the manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 27 February 2018 Accepted: 23 May 2018

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