The closedness of shift invariant subspaces in Lp,q

Zhang Journal of Inequalities and Applications (2018) 2018:166 https://doi.org/10.1186/s13660-018-1755-2

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The closedness of shift invariant subspaces in Lp,q(Rd+1) Qingyue Zhang1* *

Correspondence: [email protected] College of Science, Tianjin University of Technology, Tianjin, China 1

Abstract In this paper, we consider the closedness of shift invariant subspaces in Lp,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 in Lp,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 spaces Lp,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 in Lp,q (Rd+1 ) [6–8]. In this environment, we study the closedness of shift invariant subspaces in Lp,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 in L2 (Rd ) [9–11]. And Jia, Micchelli, 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 of Lp,q (Rd+1 ) is as follows. Definition 1.1 For 1 < p, q < +∞. Lp,q = Lp,q (Rd+1 ) is made up of all functions f satisfying   f 

Lp,q

=

R

Rd

 pq  p1   f (x, y)q dy dx < +∞.

© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Zhang Journal of Inequalities and Applications (2018) 2018:166

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



p,q

=

p,q



Z

d+1



  p 1  q q p c(n, l) = c : cp,q = < +∞ . n∈Z

l∈Zd

Given a function f , define f Lp,q

 

:=

k1 ∈Z



[0,1]d

k2 ∈Zd

1/q 

  q

f (· + k1 , x2 + k2 ) dx2

.

Lp [0,1]

For 1 ≤ p, q ≤ ∞, let Lp,q = Lp,q (Rd+1 ) be the linear space of all functions f for which f Lp,q < ∞. The norms are defined above and with usual modification in the case of p or q = ∞. Lp,q is a generalization of Lp (the definition of Lp see [14, Sect. 1]). Clearly, for 1 ≤ p, q ≤ ∞, one has L∞,∞ ⊂ L∞ and L∞,∞ ⊂ Lp,q ⊂ L1,1 . Let fˆ (ω) denote the Fourier transform of f ∈ L1 (Rd+1 ): fˆ (ω) =

 Rd+1

f (x)e–iωx dx.

 For a given sequence c and a function φ, c ∗sd φ = k∈Zd+1 c(k)φ(· – k) is called semiconvolution of c and φ. Assume that B is a Banach space. (B )(r) denotes r copies B × B × · · · × B of B . If C =  (c1 , c2 , . . . , cr )T ∈ (B )(r) , then one defines the norm of C by C(B)(r) = rj=1 cj B . WC p,q (1 ≤ p, q ≤ ∞) consists of all distributions whose Fourier coefficients belong to p,q  . When p = q = 1, WC 1,1 becomes the Wiener class WC . Suppose that  = (θ1 , θ2 , . . . , θr )T and  = (ψ1 , ψ2 , . . . , ψs )T are two vector functions j (ω) (1 ≤ j ≤ r, 1 ≤ j ≤ s) are integrable. One defines which satisfy  θj (ω)ψ ,   ](ω) = [

 

 j (ω + 2kπ) θj (ω + 2kπ)ψ

k∈Zd+1

 1≤j≤r,1≤j ≤s

.

,   ](ω) ∈ WC for any ,  ∈ Remark 1.2 By [14, Theorem 3.1 and Theorem 3.2], [ ∞,∞ ∞ 2 ∞,∞ ,   ](ω) and L ⊂ L ⊂ L . Therefore, for any  ∈ L , using the continuity of [    rank[, ](ω) = rank((ω + 2kπ))k∈Zd+1 , one obtains, for any n ≥ 0, the set n = {ω :  (ω + 2kπ))k∈Zd+1 > n} is open. rank( The following proposition shows that the shift invariant subspaces in Lp,q (1 < p, q < ∞) are well defined. Proposition 1.3 ([8, Lemma 2.2]) Let θ ∈ Lp,q , where 1 < p, q < ∞. Then, for any c ∈ p,q , c ∗sd θ Lp,q ≤ cp,q θ Lp,q . Definition 1.4 For  = (θ1 , θ2 , . . . , θr )T ∈ (L∞,∞ )(r) , the multiply generated shift invariant subspace in the mixed Lebesgue spaces Lp,q is defined by Vp,q () =

 r   j=1 k∈Zd+1

   d+1 p,q ∈  ,1 ≤ j ≤ r . cj (k)θj (· – k) : cj = cj (k) : k ∈ Z

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The following is our main result. Theorem 1.5 Assume  = (θ1 , θ2 , . . . , θr )T ∈ (L∞,∞ )(r) and 1 < p, q < ∞. Then the following four conditions are equivalent. (i) Vp,q () is closed in Lp,q . (ii) There exist some positive constants C1 and C2 satisfying ,   ](ω) ≤ [ ,  ](ω)[ ,   ](ω)T ≤ C2 [ ,   ](ω), C 1 [

∀ω ∈ [–π, π]d+1 .

(iii) There exist constants B1 , B2 > 0 satisfying B1 f 

Lp,q

≤

 inf f = rj=1 cj ∗sd φj

r 

cj p,q ≤ B2 f Lp,q ,

∀f ∈ Vp,q ().

j=1

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

f =

r     f , ψj (· – k) θj (· – k) j=1 k∈Zd+1

=

r     f , θj (· – k) ψj (· – k),

∀f ∈ Vp,q ().

j=1 k∈Zd+1

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 Theorem 1.5. 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 Theorem 1.5. Proposition 2.1 ([1, Lemma 1]) Let  ∈ (L2 )(r) . Then the following are equivalent:  (ω + 2kπ))k∈Zd+1 is a constant for any ω ∈ Rd+1 . (i) rank( (ii) There exist some positive constants C1 and C2 such that ,   ](ω) ≤ [ ,  ](ω)[ ,   ](ω)T ≤ C2 [ ,   ](ω), C 1 [

∀ω ∈ [–π, π]d+1 .

(ξ + 2kπ))k∈Zd+1 = k0 ≥ 1 for Proposition 2.2 ([1, Lemma 2]) Let  ∈ (L2 )(r) satisfy rank( all ξ ∈ Rd+1 . Then there exists a finite index set , ηλ ∈ [–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(Kλ ) = k0 for all λ ∈ , having the following properties: (i) [–π, π]d+1 ⊂



B(δλ , δλ /2),

λ∈

where B(x0 , δ) denotes the open ball in Rd+1 with center x0 and radius δ;

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(ii)  1,λ (ξ )  (ξ ) = , Pλ (ξ ) 2,λ (ξ )  

ξ ∈ Rd+1 and λ ∈ ,

where 1,λ and 2,λ are functions from Rd+1 to Ck0 and Cr–k0 , respectively, satisfying  1,λ (ξ + 2πk) rank  = k0 , k∈K λ

∀ξ ∈ B(δλ , δλ /2)

and 2,λ (ξ ) = 0, 

∀ξ ∈ B(δλ , 8δλ /5) + 2πZd+1 .

Furthermore, there exist 2π -periodic C∞ functions hλ (ξ ), λ ∈ , on Rd+1 such that 

hλ (ξ ) = 1,

∀ξ ∈ Rd+1

λ∈

and supp hλ (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 f ∈ Lp,q (1 < p, q < ∞) and g ∈ L∞,∞ . Then

 

R



f (x1 , x2 )g(x1 – k1 , x2 – k2 ) dx1 dx2 : k1 ∈ Z, k2 ∈ Zd

d

R

p,q

≤ f Lp,q gL∞,∞ .

Lemma 2.4 Let c ∈ 1 . Then one has: (i) If θ ∈ Lp,q (1 < p, q < ∞), then c ∗sd θ Lp,q ≤ c1 θ Lp,q . (ii) If θ ∈ L∞,∞ , then c ∗sd θ L∞,∞ ≤ c1 θ L∞,∞ . Proof (i) By Young’s inequality and the triangle inequality, one has



c ∗sd θ Lp,q =

n∈Z



=

n∈Z



≤

n∈Z

 [0,1]d

 1/q   q

c ∗sd θ (· + n, y + l) dy

l∈Zd

Lp [0,1]

[0,1]

   1/q 

 q     

cn ,l θ · + n – n , y + l – l  dy 

d

[0,1]

    1/q 

  q    

cn ,l θ · + n – n , y + l – l  dy 

d

l∈Zd n ∈Z l ∈Zd

Lp [0,1]

n ∈Z l∈Zd l ∈Zd

Lp [0,1]

Zhang Journal of Inequalities and Applications (2018) 2018:166



≤

n∈Z

Page 5 of 11

  [0,1]d

n ∈Z l∈Zd

  1/q  

 q θ · + n – n , y + l 

|cn ,l | dy

   

 ≤ |cn ,l |



  

 ,l | ≤ |c n



[0,1]d

n∈Z n ∈Z l∈Zd

n ∈Z l∈Zd

≤

 n ∈Z l∈Zd

≤

 n ∈Z l∈Zd

[0,1]d

n∈Z



 |cn ,l |

n∈Z



 |cn ,l |

n∈Z

l ∈Zd

l ∈Zd



[0,1]d

l ∈Zd



[0,1]d

Lp [0,1]

l ∈Zd

l ∈Zd

 1/q

   q

θ · + n – n , y + l  dy

Lp [0,1]

 1/q

   q

θ · + n – n , y + l  dy

Lp [0,1]

   θ · + n – n , y + l 

q

1/q

dy

Lp [0,1]

 1/q

   q

θ · + n, y + l  dy

Lp [0,1]

= c1 θ Lp,q . The desired result (i) in Lemma 2.4 is obtained. (ii) The desired result (ii) in Lemma 2.4 can be found in [8, Lemma 2.4]. Lemma 2.5 Assume that θ ∈ Lp,q (1 < p, q < ∞) and function h on Rd+1 satisfying    h(x) ≤ D 1 + |x| –d–2

 k∈Zd+1



θ (· – k) = 0. Then for any

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

(2.1)

one has

  –n

lim 2–n(d+1) h 2 k θ (· – k)

n→∞

= 0.

Lp,q

k∈Zd+1

Here D in (2.1) is a positive constant. Proof Since θ ∈ Lp,q , for any ε > 0, there is N0 ≥ 2 satisfying

 



 1/q

  q

θ (· + l, x + k) dx

N0 }. Set θ1 (x1 , . . . , xd+1 ) = θ (x1 , . . . , xd+1 )χON0 (x1 , . . . , xd+1 )  + θ (x1 + k1 , . . . , xd+1 + kd+1 )χ[0,1]d+1 (x1 , . . . , xd+1 ), d+1 (k1 ,...,kd+1 )∈EN 0

Zhang Journal of Inequalities and Applications (2018) 2018:166

where ON0 = of S. Thus





Page 6 of 11

|ki |≤N0 ,1≤i≤d+1 [(k1 , . . . , kd+1 ) + [0, 1]

k∈Zd+1 θ1 (· – k) =

θ1 – θ Lp,q



=

l∈Z



 [0,1]d



≤

Lp [0,1]

 1/q

 q 

(θ1 – θ )(·, x + k) dx

Lp [0,1]



[0,1]d

l =0

θ (· – k) = 0 and θ1 – θ Lp,q < 5. In fact

 1/q  q 

(θ1 – θ )(· + l, x + k) dx

k∈Zd



+

] and χS is the characteristic function

k∈Zd



[0,1]d

k∈Zd+1

d+1

 1/q

 q 

(θ1 – θ )(· + l, x + k) dx

Lp [0,1]

k∈Zd

= I1 + I2 .

First of all, one treats I1 : by (2.2) and (2.3), one has



I1 ≤

[0,1]d

  (θ1 – θ )(·, x) q dx

 +

[0,1]d



≤

[0,1]d



+



≤



≤

 1/q    q

(θ1 – θ )(·, x + k) dx

  (θ1 – θ )(·, x) q dx

[0,1]d

[0,1]d

Lp [0,1]

Lp [0,1]

k =0

 1/q   q

θ (· + k1 , . . . , xd+1 + kd+1 ) dx

Lp [0,1]

d+1 (k1 ,...,kd+1 )∈EN 0

[0,1]d

1/q     q

θ (·, x + k) dx

d k∈EN

 [0,1]d

l∈Z

[0,1]d

Lp [0,1]

0



+

|l|>N0 ,k∈Zd

[0,1]d





+

1/q



 1/q    q

(θ1 – θ )(·, x + k) dx





+



≤

Lp [0,1]

k =0





+

1/q

d l∈Z,k∈EN

Lp [0,1]

0

  1/q  q 

θ (· + l, x + k) dx

d k∈EN

 |l|>N0 ,k∈Zd

[0,1]d

1/q     q

θ (· + l, x + k) dx

Lp [0,1]

0

 1/q

  q

θ (· + l, x + k) dx

Lp [0,1]

   1/q  q 

θ (· + l, x + k) dx

d l∈Z,k∈EN

Lp [0,1]

0

+

Zhang Journal of Inequalities and Applications (2018) 2018:166

 

≤

|l|>N0

 1/q  q 

θ (· + l, x + k) dx



[0,1]d

Lp [0,1]

k∈Zd

  1/q

  q

θ (· + l, x + k) dx



+

[0,1]d

l∈Z

Page 7 of 11

d k∈EN

+

Lp [0,1]

0

<  +  +  = 3. Next, one treats I2 :

 

I2 ≤

|l|>N0

[0,1]d

 

+

|l|≤N0 ,l =0

 

=

|l|>N0

[0,1]



 1/q   q

(θ1 – θ )(· + l, x + k) dx

k∈Zd

Lp [0,1]

 1/q    q

θ (· + l, x + k) dx

d

 

≤

1/q     q

θ (· + l, x + k) dx

d k∈EN



[0,1]d



+

[0,1]d

Lp [0,1]

k∈Zd

[0,1]d

|l|≤N0 ,l =0

l∈Z

Lp [0,1]

k∈Zd

[0,1]d

 

+

|l|>N0

 1/q   q

(θ1 – θ )(· + l, x + k) dx



Lp [0,1]

0

 1/q  q 

θ (· + l, x + k) dx

Lp [0,1]

k∈Zd

 1/q 

  q

θ (· + l, x + k) dx

d k∈EN

Lp [0,1]

0

<  +  = 2. Therefore, one has θ1 – θ Lp,q < 5. Using Lemma 2.4 and (2.1), there exists some positive constant C such that

–n(d+1)   –n 

2 h 2 k φ(· – k) – φ (· – k) 1

Lp,q

k∈Zd+1

≤ 2–n(d+1)

    h 2–n k φ1 – φLp,q ≤ C.

k∈Zd+1

Thus

–n(d+1)   –n

2 h 2 k θ1 (· – k)

Lp,q

k∈Zd+1



–n(d+1) =2

j1 ∈Z

[0,1]

    d



j2 ∈Zd k1 ∈Z,k2 ∈Zd

 h 2–n k1 , 2–n k2

q 1/q



 × θ1 (· + j1 – k1 , x2 + j2 – k2 ) dx2

p L [0,1]

  

   –n   h 2 k1 , 2–n k2 – h 2–n j1 , 2–n j2 = 2–n(d+1)

 j1 ∈Z

[0,1]d

j2 ∈Zd k1 ∈Z,k2 ∈Zd

Zhang Journal of Inequalities and Applications (2018) 2018:166

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q 1/q



× θ1 (· + j1 – k1 , x2 + j2 – k2 ) dx2



≤ 2–n(d+2) C1 (N0 )



[0,1]d

j1 ∈Z

Lp [0,1]



  –(d+2)  1 + 2–n (k1 , k2 )

j2 ∈Zd k1 ∈Z,k2 ∈Zd

1/q 

  q

  × θ1 (· + j1 – k1 , x2 + j2 – k2 ) dx2

Lp [0,1]



–n(d+2) =2 C1 (N0 )

[0,1]d

j1 ∈Z

    –(d+2)  1 + 2–n (k1 , k2 ) k1 ∈Z k2 ∈Zd

 1/q   q

 

× dx2 θ1 (· + j1 – k1 , x2 + j2 )

Lp [0,1]

j2 ∈Zd



≤ 2–n(d+2) C1 (N0 )

  –(d+2) 1 + 2–n (k1 , k2 )

k1 ∈Z,k2 ∈Zd



 ×



[0,1]d

j1 ∈Z



j2 ∈Zd

1/q 

  q

θ1 (· + j1 – k1 , x2 + j2 ) dx2



≤ 2–n C2 (N0 )

Lp [0,1]



[0,1]d

j1 ∈Z

j2 ∈Zd

1/q 

  q

θ1 (· + j1 , x2 + j2 ) dx2

Lp [0,1]

 = 2–n C2 (N0 )θ1 Lp,q ≤ 2–n C2 (N0 ) θ Lp,q + 5 . Here Ci (N0 ) (i = 1, 2) are positive constants depending only on N0 and d. This completes 

the proof.

3 Proof of Theorem 1.5 In this section, we give the proof of Theorem 1.5. The main steps of the proof are as follows: (iv) ⇒ (iii) ⇒ (i) ⇒ (ii) ⇒ (iv). (iv) ⇒ (iii):   Let f = rj=1 k∈Zd+1 f , ψj (· – k)θj (· – k). Then, by Lemma 2.3, one has r 

rinf

j=1 cj ∗sd φj j=1

f=

cj p,q ≤

r 

  

f , ψj (· – k1 , · – k2 ) : k1 ∈ Z, k2 ∈ Zd p,q  j=1

≤

r 

f Lp,q ψj L∞,∞ = f Lp,q

r 

j=1

Conversely, if f =

f 

Lp,q

r

j=1 cj

∗sd θj , then, by Proposition 1.3 and the triangle inequality

r



= cj ∗sd θj

j=1

≤

r  j=1

ψj L∞,∞ .

j=1

Lp,q

≤

r 

cj ∗sd θj Lp,q

j=1

cj p,q θj Lp,q ≤ max θj Lp,q 1≤j≤r

r  j=1

cj p,q .

(3.1)

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Taking the infimum for (3.1), one gets f Lp,q ≤ max θj Lp,q 1≤j≤r

rinf

f=

j=1 cj ∗sd θj j=1

Let B1 = 1/ max1≤j≤r θj Lp,q and B2 = B1 f Lp,q ≤

 inf f = rj=1 cj ∗sd φj

r 

r 

cj p,q .

r

j=1 ψj L∞,∞ .

Then one has

cj p,q ≤ B2 f Lp,q ,

∀f ∈ Vp,q ().

j=1

(iii) ⇒ (i): For convenience, let T : (p,q )(r) → Vp,q () be a mapping which is defined by

TC =

r 

cj ∗sd θj ,

 (r) C = (c1 , c2 , . . . , cr )T ∈ p,q ,

j=1

 and let f inf = inff =rj=1 cj ∗sd θj rj=1 cj p,q . Then, obviously,  · inf is a norm. Assume fn ⊂ Ran(T) (n ≥ 1) is a Cauchy sequence. Here Ran(T) denotes the range of T. Without loss of generality, let fn – fn–1 inf < 2–n . Using the definition of  · inf , there is Cn ∈ (p,q )(r) (n ≥ 2) such that TCn = fn – fn–1 and Cn (p,q )(r) < 2–n for any n ≥ 2. By the completeness  ∞ p,q (r) of (p,q )(r) and ∞ n=2 Cn (p,q )(r) < ∞, one has Z = n=2 Cn ∈ ( ) and f1 + TZ ∈ Ran(T). p,q (r) Note that TCinf ≤ C(p,q )(r) for any C ∈ ( ) . One has fn – f1 – TZinf

 ∞ 



= T Ck

k=n+1

inf

∞



≤ Ck

k=n+1

(p,q )(r)

≤

∞ 

Ck (p,q )(r) → 0,

k=n+1

when n → ∞. Therefore, Ran(T) is closed. Since Vp,q () = Ran(T), one sees that Vp,q () is closed. (i) ⇒ (ii): Similarly to [1, Proof of (i) ⇒ (iii)], one can prove (i) ⇒ (ii) by using L∞,∞ ⊂ L∞ , and substituting Lp,q , L∞,∞ , Proposition 2.1 and Lemma 2.5 for Lp , L∞ , Lemma 1 and Lemma 3 in [1], respectively. (ii) ⇒ (iv): 1,λ (ω) are as in Proposition 2.2. Define Assume that hλ (ω), Pλ (ω) and 

Dλ (ω) = Pλ

(ω)T

 1,λ ](ω)–1 1,λ ,  [ 0

 0 Pλ (ω)Hλ (ω). I

(3.2)

Here Hλ (ω) is a function with period 2π which satisfies supp Hλ ⊂ B(ηλ , δλ ) + 2πZd+1 and Hλ (ω) = 1 on supp hλ . Thus Dλ ∈ (WC )(r×r) . Let  = (ψ1 , ψ2 , . . . , ψr )T be defined by  (ω) = 



(ω). hλ (ω)Dλ (ξ )

(3.3)

λ∈

Then, by Lemma 2.4, one has  ∈ L∞,∞ . For any f ∈ Vp,q (), using the definition of Vp,q (), there exists a distribution A(ω) ∈ (WC p,q )(r) with period 2π which satisfies fˆ (ω) =

Zhang Journal of Inequalities and Applications (2018) 2018:166

Page 10 of 11

 (ω). Putting A(ω)T 

g=

r     f , ψj (· – k) θj (· – k). j=1 k∈Zd+1

By the periodicity of hλ (ω) and Dλ (ω), (3.2), (3.3) and Proposition 2.2, one has ,   ](ω)  (ω) gˆ (ω) = A(ω)T [  A(ω)T Pλ (ω)–1 = λ∈

 1,λ ](ω) 0 1,λ ,  1,λ ](ω)–1 1,λ ,  [ [ × 0 0 0    1,λ (ω)  A(ω)T Pλ (ω)–1 hλ (ω) = 0 λ∈   (ω)hλ (ω) A(ω)T Pλ (ω)–1 Pλ (ω) = 

0 I



 1,λ (ω)  hλ (ω) 0

λ∈

=



(ω)hλ (ω) A(ω)T 

λ∈

 (ω) = A(ω)T  = fˆ (ω). Thus fˆ (ω) = gˆ (ω). Therefore f = g, namely r     f , ψj (· – k) θj (· – k). f= j=1 k∈Zd

Similar arguments show that

f=

r     f , θj (· – k) ψj (· – k). j=1 k∈Zd

4 Concluding remarks In this paper, we study the closedness of shift invariant subspaces in Lp,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 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 the Lp,q -frames in a shift invariant subspace of mixed Lebesgue Lp,q (Rd ) is the goal of future work. Funding This work was supported partially by the National Natural Science Foundation of China under Grants Nos. 11371200, 11326094 and 11401435. This work was also partially supported by the Program for Visiting Scholars at the Chern Institute of Mathematics.

Zhang Journal of Inequalities and Applications (2018) 2018:166

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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.

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