# Regular and Irregular Sampling Theorem for Multiwavelet Subspaces

Authors: Liu Zhanwei; Hu Guoen; Wu Guochang
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We study the sampling theorem for frames in multiwavelet subspaces. Firstly,
a sufficient condition under which the regular sampling theorem holds is established. Then, notice that
irregular sampling is also useful in practice; we consider the general cases of the irregular sampling
and establish a general irregular sampling theorem for multiwavelet subspaces. Finally, using this
generalized irregular sampling theorem, we obtain an estimate for the perturbations of regular sampling
in shift-invariant spaces.


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## Body

## 1. Introduction

At the present time the sampling theorem plays a crucial role in signal processing and communication, as it establishes an equivalence between discrete signals and analogue (continuous) signals. For a band-limited signal, the classical Shannon sampling theorem provides an exact representation by its uniform samples with a sampling rate higher than its Nyquist rate. But there exist several problems. Firstly, real-world signals or images are never exactly band-limited. Secondly, there is no such device as an ideal (antialiasing or reconstruction) low-pass filter. Thirdly, Shannon reconstruction formula is rarely used in practice (especially with images) because of the slow decay of the sinc function. Therefore, this classical Shannon sampling theorem has been generalized to many other forms.Extensions of Shannon sampling theorem to scalar wavelets can be found in [1–5], but a scalar wavelet cannot have the orthogonality, compact support, and symmetry at the same time (except the Haar wavelet). It is a disadvantage for signal processing. Meanwhile, multiwavelets have attracted much attention in the research community, since multiwavelet has more desired properties than any scalar wavelet function, such as orthogonality, short compact support, symmetry, high approximation order, and so on. The first orthogonal multiwavelet with symmetry, approximation order, and compact support was presented by Geronimo et al. [6]. In addition, the sampling theorems for multiwavelet subspaces were studied in [7–10]. The authors of [7, 9] presented the construction of compactly supported orthogonal multiscaling functions that are continuously differentiable and cardinal. The scaling functions thereby support a Shannon-like sampling theorem. However, the multiwavelets of [7, 9] do not have symmetry. It is not good for digital signal processing and image compression. They also did not study the sampling theorem for frame in multiwavelet subspaces, which is very important in application.A reconstruction from more general sets of points is necessary if the measurements cannot be made at uniform points. Hence, irregular sampling is also useful in practice. In [11], the authors obtained a Feichtinger-Grochenig iterative algorithm based on the quasiinterpolation projection procedure to recover signals from irregular samples for multiwavelet subspaces. However, the maximal allowable gap between two sampling points needed for reconstructing a function from its samples was not obtained, which was supposed to exist in theory. The authors of [10] generalized the multiwavelet sampling theorem by reproducing a kernel and derived an estimate for the perturbations of uniform noninteger sampling in shift-invariant spaces, but their results just based on the Riesz basis.In our paper, we will show a sufficient condition for regular sampling theorem to hold in multiwavelet subspaces for frames. Notice that a reconstruction from more general sets of points is necessary if the measurements cannot be made an uniform; we establish the irregular sampling theorems in multiwavelet subspaces. Finally, an estimate for the perturbations of regular sampling in shift-invariant spaces is derived.This paper is organized as follows. Section2 contains some definitions in this correspondence. Also, we review some relative notations. In Section 3, we discuss general uniform noninteger sampling and obtain a sufficient condition for uniform noninteger sampling theorem to hold. In Section 4, an irregular sampling theorem in general multiwavelet subspaces is established. Finally, by applying the result in Section 4, we estimate the perturbations of uniform noninteger sampling in shift-invariant spaces.


## 2. Preliminary

We now introduce some notations used in this correspondence.C(R) is the space of continuous function.The shift-invariant closed subspaceV0 generated by {ϕ1,ϕ2,…,ϕr}(2.1)V0=span{ϕi(·-k):1≤i≤r,k∈Z}¯⊂L2(R).For a functionf∈L2(R), we denote by 〈f〉 the minimal closed shift invariant subspace that contains f.Let(2.2)f=[f1,f2,…,fr]T
denote vector (we denote vectors and matrices in this paper in boldface). The integration ∫Rf(x)dx is defined as (2.3)∫Rf(x)dx=[∫Rf1(x)dx,∫Rf2(x)dx,…,∫Rfr(x)dx]T.The Fourier transform of vectorf is defined by (2.4)f̂(ω)=∫Rf(x)e-ixωdx.The inverse Fourier transform of vectorf is written by (2.5)f̌(x)=12π∫Rf(ω)eixωdω.Zf(x,ω)=∑n∈Zf(x+n)e-inω is the Zak transform of function f. The Zak transform of vector f is defined by (2.6)Zf(x,ω)=[Zf1(x,ω),Zf2(x,ω),…,Zfr(x,ω)]T.A collection of elements{ϕi(·-k):1≤i≤r,k∈Z} in a Hilbert space H is called a frame if there exist constants A and B, 0<A≤B<∞, such that (2.7)A∥f∥2≤∑k∈Z∑i=1r|〈f,ϕi(·-k)〉|2≤B∥f∥2,∀f∈H.If{ϕi(·-k):1≤i≤r,k∈Z} is a frame for H, then there exists a dual frame {ϕ̃k}k∈Z for {ϕi(·-k):1≤i≤r,k∈Z} [12, Theorem 5.6.5].Forf∈H,we can write(2.8)Ef={ω∈R∣Gf(ω)>0},Gf(ω)=∑k∈Z|f̂(ω+2πk)|2,χEf={1,t∈Ef,0,t∉Ef.Let(2.9)Φ=[ϕ1,ϕ2,…,ϕr]T,Φ̃=[ϕ̃1,ϕ̃2,…,ϕ̃r]T.
The reproducing kernel is defined as (2.10)q(t,·)=∑l=1r∑k∈Zϕl(t-k)¯ϕ̃l(·-k)=∑k∈ZΦ*(t-k)Φ̃(·-k),
where * denotes the Hermitian conjugate. Put t=n+am. Then (2.11)q(n+am,·)=∑l=1r∑k∈Zϕl(n+am-k)¯ϕ̃l(·-k)=∑k∈ZΦ*(n+am-k)Φ̃(·-k),
where 1≤m≤r, n∈Z, am∈[0,1) is constant, and ai≠aj for i≠j, i,j=1,2,…,r.LetA⊂R, Max{A} means the largest number in subset A, and Min{A} means the smallest number in subset A.Forf,g∈L2(R), let bracket function [f,g] be the function defined a.e. by [f,g](ξ)=∑k∈Zf(ξ+k)g(ξ+k)¯.


## 3. General Uniform Sampling Theorem

The main purpose of this section is to study the regular sampling theorem for frame in multiwavelet subspace.Firstly, we start with some useful lemmas.Lemma 3.1.
Letf,g∈L2(R). If  〈f〉⊥〈g〉 and  |h|,|s|∈〈f〉⊕〈g〉 for all  s∈〈f〉,h∈〈g〉, then  [s,h]=0a.e.Proof.
Suppose that{f(·-k)} is a frame sequence in L2(R). Let s∈〈f〉, then there exists a sequence {ci}i∈Z∈l2 such that s=∑i∈Zcif(·-i).
Define(3.1)χ(s)={1,s≥0,-1,s<0.
We have
(3.2)|s|=χ(s)s=χ(s)∑i∈Zcif(·-i)=∑i∈Zχ(s)cif(·-i).
It is easy to check that {χ(s)ci}i∈Z∈l2 and |s|⊥〈g〉; thus |s|∈〈f〉. Similar to the above argument, forallh∈〈g〉, we get |h|∈〈g〉.
Suppose that〈f〉⊥〈g〉 holds; by |s|∈〈f〉 and |h|∈〈g〉, we can obtain(3.3)0≤∫T|∑k∈Zs(x+k)·h(x+k)¯|dx≤∫T∑k∈Z|s(x+k)·h(x+k)¯|dx=∫R|s(x)·h(x)¯|dx≤〈|s|,|h|〉=0.
It implies that [s,h]=0a.e. The proof is completed.Lemma 3.2.
Let{fi}i∈{1,2,…,r}⊂L2(R), then the following two assertions are equivalent:(a)
for any{cki}k∈Z∈l2, i∈{1,2,…,r}, ∑i=1r∑k∈Zckifi(x-k) converges pointwise to a continuous function;(b)
{fi}i∈{1,2,…,r}⊂C(R) and supx∑i=1r∑k∈Z|fi(x-k)|2<+∞.Proof.
(a)⇒(b). It is easy to see that foralli∈{1,2,…,r},fi∈C(R). For each x∈R, since ∑i=1r∑k∈Zckifi(x-k) is convergent for each {cki}∈l2, clearly, ∑k∈Z|fi(x-k)|2<∞. For each x∈[0,1], define(3.4)Txc=∑i=1r∑k∈Zckifi(x-k),∀c={cki}∈l2.
Then Tx is a bounded linear functional on l2 with the norm ∥Tx∥=(∑i=1r∑k∈Z|fi(x-k)|2)1/2. For any {cki}∈l2, define g(t)=∑i=1r∑k∈Zckifi(t-k). Since g(t) is continuous on R, we have
(3.5)supx∈[0,1]|Txc|=supx∈[0,1]|g(x)|<+∞.
By the Banach-Steinhaus theorem [13], supx∈[0,1]∥Tx∥<+∞, that is, ∑i=1r∑k∈Z|fi(x-k)|2 is bounded on R.
(a)⇒(b). By the Cauchy inequality, ∑i=1r∑k∈Zckifi(x-k) in convergent uniformly on R, so the limit function is continuous.Lemma 3.3.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for  V0 with bounds A and B. If  〈ϕi〉⊥〈ϕj〉 for all  i,j∈{1,2,…,r},  i≠j, then  {ϕi(·-k)}k∈Z is a frame for the subspace  〈ϕi〉.Proof.
Suppose that{ϕi(·-k):1≤i≤r,k∈Z} is a frame for the subspace V0, then there exist constants 0<A≤B<∞ such that
(3.6)A∥f∥2≤∑i=1r∑k∈Z|〈f,ϕi(x-k)〉|2≤B∥f∥2.
Notice that 〈ϕi〉⊥〈ϕj〉 for all i,j∈{1,2,…,r}, i≠j; if f∈〈ϕi〉, then 〈f,ϕj(x-k)〉=0. Hence,
(3.7)A∥f∥2≤∑k∈Z|〈f,ϕi(x-k)〉|2≤B∥f∥2.
By the definition of Frame, {ϕi(·-k)}k∈Z is a frame for the closed subspace 〈ϕi〉. Then we get the desired result.Lemma 3.4.
Letf∈L2(R) and  f∈C(R), then there exists a set  E∈R,  |R∖E|=0 such that for any  x∈E and  forall{ck}k∈Z∈l2,  ∑k∈Zckf(x-k) converges pointwise to a continuous function.Proof.
Byf∈L2(R), then ∫T∑k∈Z|f(x-k)|2dx<∞. Hence there exists a set E∈R,|R∖E|=0 such that supx∈E∑k∈Z|f(x-k)|2<∞. Notice that f∈C(R); using Lemma 3.2, clearly, for any {ck}k∈Z∈l2, ∑kckf(x-k) converges pointwise to a continuous function in set E.From the above lemmas, we have the following result.Theorem 3.5.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for V0. Suppose that ϕi, 1≤i≤r are continuous functions, 〈ϕi〉⊥〈ϕj〉 for all i≠j, i,j∈{1,2,…,r}, and
(3.8)supx∈R∑i=1r∑k∈Z|ϕi(x-k)|2<+∞.
Then
(3.9)f(t)=∫Rf(s)q(t,s)ds
holds for each f∈V0 and t∈R.Proof.
Suppose that{ϕi(·-k):1≤i≤r,k∈Z} is a frame for V0 and 〈ϕi〉⊥〈ϕj〉, i≠j, then, by [14, Proposition 3.1] and Lemma 3.3, there exist constants 0<Ai≤Bi<∞ such that(3.10)AiχEϕi(ω)≤∑n∈Z|ϕ̂i(ω+2nπ)|2≤BiχEϕi(ω),a.e.
From Lemma 4 in [15], we have the dual frame ϕ̃i for {ϕi(·-k):1≤i≤r,k∈Z}, where ϕ̃i  is  defined by
(3.11)ϕ̃i(ω)={ϕ̂i(ω)Gϕi(ω),ω∈Eϕi,0,ω∉Eϕi.
Hence, for all1≤i≤r,(3.12)ϕ̂i(ω)Bi≤ϕ̃î(ω)≤ϕ̂i(ω)Ai
holds.
So, it follows that(3.13)ϕĩ(x)=∫ω∈Rϕĩ̂(ω)eiωxdω≤1Ai∫ω∈Rϕî(ω)eiωxdω=1Aiϕî(x),ϕĩ(x)=∫ω∈Rϕĩ̂(ω)eiωxdω≥1Bi∫ω∈Rϕî(ω)eiωxdω=1Biϕî(x).
From above results, obviously, if(3.14)supx∈R∑i=1r∑k∈Z|ϕi(x-k)|2<+∞,
then
(3.15)supx∈R∑i=1r∑k∈Z|ϕ̃i(x-k)|2<+∞.
Hence,
(3.16)q(t,s)=∑l=1r∑k∈Zϕl(t-k)¯,ϕ̃l(s-k)=∑k∈ZΦ*(t-k)Φ̃(s-k)
converges to q(t,·) in L2(R) for any f∈V0 and t∈R.
By〈ϕi〉⊥〈ϕj〉 for all i,j∈{1,2,…,r}, i≠j, it is easy to check that {ϕ̃i(·-k):1≤i≤r,k∈Z} is a dual frame for {ϕi(·-k):1≤i≤r,k∈Z}; then we have(3.17)f(t)=∑l=1r∑k∈Z〈f(s),ϕ̃l(s-k)〉ϕl(t-k)=〈f(s),∑l=1r∑k∈Zϕl(t-k)¯,ϕ̃l(s-k)〉=〈f(s),∑k∈ZΦ*(t-k)Φ̃(s-k)〉=∫Rf(s)q(t,s)ds
for any f∈V0 and t∈R. So we get the desired result.Based on these facts, the following sampling theorem is established.Theorem 3.6.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for V0. Suppose that 〈ϕi〉⊥〈ϕj〉 for all i,j∈{1,2,…,r}, i≠j, and ϕi, 1≤i≤r are continuous such that
(3.18)supt∈R∑i=1r∑k∈Z|ϕi(t-k)|2<+∞.
If there exist constants Ai and Bi, 0<Ai≤Bi<+∞ such that
(3.19)AiχEϕi(ω)≤|ϕ̂i*(ω)|≤BiχEϕi(ω),a.e.ω∈R,
where ϕ̂i*(ω)=∑n∈Zϕi(n)e-inω, then there exists a frame {si(·-k):1≤i≤r,k∈Z} of V0, for i,j∈{1,2,…,r}, i≠j(3.20)f(t)=∑n∈Z∑m=1rf(n+am)sm(t-n),f∈V0,
holds, where the convergence is in L2(R).Proof.
Define the functionqi(t,s) by qi(t,s)=∑k∈Zϕi(t-k)¯ϕ̃i(s-k), where ϕ̃i is defined by (3.11). From Lemma 3.3, it is easy to see that {ϕi(·-k)}k∈Z and {ϕ̃i(·-k)}k∈Z are the frames for the subspace 〈ϕi〉.
Notice that(3.21)q̂i(n,ω)=ϕ̂i*(ω)¯ϕ̃î(ω),AiχEϕi(ω)≤|ϕ̂i*(ω)|≤BiχEϕi(ω),a.e.ω∈R.
From the proof process of Proposition 3.1 in [14], {qi(n,·)}n∈Z={qi(0,·-n)}n∈Z is a frame for the subspace 〈ϕi〉. Similar to the above argument, obviously, for 1≤m≤r, {qi(n+am,·)}n∈Z={qi(0,·-n-am)}n∈Z is the frame for the subspace 〈ϕi〉. Hence, forallfi∈Vi,forallm∈{1,2,…,r}, there exist constants 0<Cim≤Dim<∞ such that
(3.22)Cim∥fi∥2≤∑n∈Z|〈fi,qi(n+am,·)〉|2≤Dim∥fi∥2.
Thus, we have
(3.23)fi(s)=∑n∈Z〈fi(s),qi(n+am,s)〉q̃i(n+am,s),
where q̃i(n+am,s) is the dual frame for qi(n+am,s).
From above results, we get(3.24)∑i=1rCim∥fi∥2≤∑i=1r∑n∈Z|〈fi,qi(n+am,s)〉|2≤∑i=1rDim∥fi∥2,
where m∈{1,2,…,r}.
Notice that{ϕi(·-k)}k∈Z, {qi(0,·-n-am)}n∈Z,m∈{1,2,…,r} and {qi(0,·-n)}n∈Z are the frame for subspace 〈ϕi〉; by [16, Lemma 3], we have Eϕi=Eqi(0,·-n-am)=Eqi(0,·-n) (except a null measurable set).
By Lemma3.3, it follows that for all f∈V0, f=∑i=1rfi holds, where fi∈Vi. Again by Lemma 3.3, we have Vi⊥Vl, 1≤i≠l≤r, then,(3.25)∥f∥2=〈∑i=1rfi,∑l=1rfl¯〉=∑i=1r∑l=1r〈fi,fl¯〉=∑i=1r〈fi,fi¯〉=∑i=1r∥fi∥2.
From (3.23) and (3.25), forallm∈{1,2,…,r}, we get
(3.26)Min{Cim}i=1r∥f∥2=Min{Cim}i=1r∑i=1r∥fi∥2≤∑i=1rCim∥fi∥2≤∑i=1r∑n∈Z|〈fi,qi(n+am,)〉|2,∑i=1r∑n∈Z|〈fi,qi(n+am,·)〉|2≤∑i=1rDim∥fi∥2≤Max{Dim}i=1r∑i=1r∥fi∥2=Max{Dim}i=1r∥f∥2.
For(3.27)〈f(t),qi(n+ai,t)〉=∫R∑l=1rfl(t)qi(n+ai,t)¯dt=∑l=1r∫Rfl(t)qi(n+ai,t)¯dt=∫Rfi(t)qi(n+ai,t)¯dt=〈fi(t),qi(n+ai,t)〉,
notice (3.26),  we have
(3.28)Min{Cii}i=1r∥f∥2=Min{Cii}i=1r∑i=1r∥fi∥2≤∑i=1rCii∥fi∥2≤∑n∈Z∑i=1r|〈fi,qi(n+ai,·)〉|2=∑n∈Z∑i=1r|〈f,qi(n+ai,·)〉|2,∑n∈Z∑i=1r|〈f,qi(n+ai,·)〉|2=∑n∈Z∑i=1r|〈fi,qi(n+ai,·)〉|2≤∑i=1rDii∥fi∥2≤Max{Dii}i=1r∑i=1r∥fi∥2=Max{Dii}i=1r∥f∥2.
LetC=Min{Cii}i=1r,D=Max{Dii}i=1r, then forallf∈V0, there exist constants number 0<C≤D<∞ such that(3.29)C∥f∥2≤∑i=1r∑n∈Z|〈f,qi(n+ai,·)〉|2≤D∥f∥2.
Hence, according to the definition of frame, {qi(n+ai,s):1≤i≤r,n∈Z} is the frame for the subspace V0. Let q(n+ai,s)=∑l=1rql(n+ai,s); by Theorem 3.5, clearly, 〈f(s),q(n+ai,s)〉=f(n+ai).
Notice that(3.30)|〈f(s),∑l=1rql(n+ai,s)〉|2=|∑l=1r〈f(s),ql(n+ai,s)〉|2≤∑l=1r|〈f(s),ql(n+ai,s)〉|2.
Using (3.27) and (3.22), then
(3.31)∑i=1r∑n∈Z|〈f(s),q(n+ai,s)〉|2=∑i=1r∑n∈Z|〈f(s),∑l=1rql(n+ai,s)〉|2≤∑i=1r∑n∈Z∑l=1r|〈f(s),ql(n+ai,s)〉|2=∑i=1r∑n∈Z∑l=1r|〈fl(s),ql(n+ai,s)〉|2≤∑l=1r(∑i=1rDli∥fl∥2)≤∑l=1rMax{Dli:i∈{1,2,…,r}}∥fl∥2≤Max{Max{Dli:i∈{1,2,…,r}}:l∈{1,2,…,r}}∥f∥2=D'∥f∥2.
Forfl∈〈ϕl〉,fp∈〈ϕp〉, by Lemmas 3.1 and 3.4, notice that for all l∈{1,2,…,r}, {fl(n+ai)}n∈Z∈l2, then ∑n∈Zfl(n+ai)fp(n+ai)¯=0, for all i∈{1,2,…,r}, l≠p. Hence,(3.32)∑i=1r∑n∈Z|〈f,q(n+ai,·)〉|2=∑i=1r∑n∈Z(〈f,q(n+ai,·)〉)(〈f,q(n+ai,·)〉)¯=∑i=1r∑n∈Z(∑l=1r〈fl,ql(n+ai,·)〉)(∑p=1r〈fp,qp(n+ai,·)〉¯)=∑i=1r∑n∈Z∑l=1r|〈fl,ql(n+ai,·)〉|2≥Min{Min{Cli:i∈{1,2,…,r}}:l∈{1,2,…,r}}∥f∥2≥C'∥f∥2
holds. From (3.31) and (3.32), we get that {q(n+ai,s):i=1,2,…,r,n∈Z} is the frame for the subspace V0. Then there exists a dual frame {si(·-k):1≤i≤r,k∈Z} of {q(n+ai,s):i=1,2,…,r,n∈Z} such that
(3.33)f=∑i=1r∑n∈Z〈f,q(n+ai,·)〉si(·-k)=∑i=1r∑n∈Zf(ai+n)si(·-k),∀f∈V0.
The proof is completed.


## 4. General Irregular Sampling Theorem

In many realistic situations, we try to reconstruct a continuous signalf∈V0 by using its discrete samples {f(n+am+δmn):1≤m≤r,n∈Z}. Obviously, the samples cannot be arbitrary; we must discuss the values δmn of deviation from the uniformly noninteger points. The following theorem studies this problem.Based on the results in Section3, we get the irregular sampling theorems.Theorem 4.1.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for  V0 with bounds A and B. Suppose that  〈ϕi〉⊥〈ϕj〉 for all  i,j∈{1,2,…,r},  i≠j, and  ϕi,  1≤i≤r are continuous such that
(4.1)supt∈R∑i=1r∑k∈Z|ϕi(t-k)|2<+∞.
Moreover, suppose that there exist constants Ai and Bi, 0<Ai≤Bi<+∞ such that
(4.2)AiχEϕi(ω)≤|ϕ̂i*(ω)|≤BiχEϕi(ω),a.e.ω∈R.
If there exists a sequence {δmn:n∈Z,m=1,2,…,r} and two constants C and D, 0<C≤D<∞ such that
(4.3)C∑m=1r∑n∈Z|f(n+am)|2≤∑m=1r∑n∈Z|f(n+am+δmn)-f(n+am)|2≤D∑m=1r∑n∈Z|f(n+am)|2
holds for any f∈V0. Then, there exists a frame {sm,n(t):1≤m≤r,n∈Z} for V0 such that for any f∈V0,
(4.4)f(t)=∑n∈Z∑m=1rf(n+am+δmn)sm,n(t)
holds, where the convergence is both in L2(R) and uniform on R.Proof.
From the argument of Theorem3.6, it is easy to see that {q(a1,·-n),q(a2,·-n),…,q(ar,·-n)}n∈Z={q(n+a1,·),q(n+a2,·),…,q(n+ar,·)}n∈Z is a frame for the closed subspace V0 of L2(R). Hence, there exist two constants C1 and C2, 0<C1≤C2<∞ such that
(4.5)C1∥f∥2≤∑n∈Z∑k=lr|〈f,q(n+ak,·)〉|2=∑n∈Z∑k=lr|f(n+ak)|2≤C2∥f∥2,∀f∈V0.
By (4.3), we have(4.6)CC1∥f∥2≤∑n∈Z∑k=lr|〈f,q(n+ak+δkn,·)〉|2=∑n∈Z∑k=lr|f(n+ak+δkn)|2≤DC2∥f∥2.
Then,{q(n+a1+δ1n,·),q(n+a2+δ2n,·),…,q(n+ar+δrn,·)}n is also a frame for the closed subspace V0 of L2(R). So, there exists a dual frame {si(·-k):1≤i≤r,k∈Z} of {q(n+ai+δin,·):1≤i≤r,k∈Z} such that
(4.7)f(t)=∑n∈Z∑m=1rf(n+am+δmn)sm(t-n)
for any f∈V0.Theorem 4.2.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for  V0 with bounds A and B. Suppose that  〈ϕi〉⊥〈ϕj〉 for all  i,j∈{1,2,…,r},  i≠j, and  ϕi,  1≤i≤r are continuous such that
(4.8)supt∈R∑i=1r∑k∈Z|ϕi(t-k)|2<+∞.
Moreover, suppose that there exist constants Ai and Bi, 0<Ai≤Bi<+∞ such that
(4.9)AiχEϕi(ω)≤|ϕ̂i*(ω)|≤BiχEϕi(ω),a.e.ω∈R.
If there exists a sequence {δmn:n∈Z,m=1,2,…,r} and a constant 0<θ<1 such that
(4.10)∑m=1r∑n∈Z|f(n+am+δmn)-f(n+am)|2≤θ2∑m=1r∑n∈Z|f(n+am)|2
holds for any f∈V0, then, there exists a frame {sm,n(t):1≤m≤r,n∈Z} for V0 such that for any f∈V0(4.11)f(t)=∑n∈Z∑m=1rf(n+am+δmn)sm,n(t)
holds, where the convergence is both in L2(R) and uniform on R.Proof.
If(4.12)∑m=1r∑n∈Z|f(n+am+δmn)-f(n+am)|2≤θ2∑m=1r∑n∈Z|f(n+am)|2,
then
(4.13)(1-θ2)∑m=1r∑n∈Z|f(n+am)|2≤∑m=1r∑n∈Z|f(n+am+δmn)|2≤(1+θ2)∑m=1r∑n∈Z|f(n+am)|2.
By Theorem 4.1, we get the desired result.


## 5.  Perturbation of Uniform Noninteger Sampling in Shift-Invariant Spaces

Our objective is to find explicit formulas or algorithms to calculate the ranges of the perturbations of uniform noninteger sampling in shift-invariant spaces. In order to establish the algorithm for perturbations of uniform noninteger sampling for frame in shift-invariant spaces, we need to introduce the function classLamλ[a,b](λ>0,σ∈[0,1),0∈[a,b]⊂[-1,1]) defined and used in [10].Definition 5.1.
Lamλ[a,b](λ>0 and σ∈[0,1),0∈[a,b]⊂[-1,1]) consists of all the measurable functions f, for which the norm
(5.1)∥f∥Lamλ[a,b]=sup{rk}k⊂[a,b]∑k|f(k+σ+rk)-f(k+σ)|supk|rk|λ<∞.Theorem 5.2.
Let{ϕi(·-k):1≤i≤r,k∈Z} be a frame for  V0 with bounds A and B. Suppose that there exist constants  Ai and  Bi,  0<Ai≤Bi<+∞ such that
(5.2)AiχEϕi(ω)≤|ϕ̂i*(ω)|≤BiχEϕi(ω),a.e.ω∈R.
Moreover, suppose that 〈ϕi〉⊥〈ϕj〉 for all i,j∈{1,2,…,r}, i≠j, and ϕi, 1≤i≤r are continuous such that
(5.3)supt∈R∑i=1r∑k∈Z|ϕi(t-k)|2<+∞.
Then, for any {δmn}⊂[-θΦ,θΦ]∩[a,b], 1≤m≤r, n∈Z, there is a frame {Sm,n} of V0 such that
(5.4)f(t)=∑n∈Z∑m=1rf(n+am+δmn)sm(t-n)
holds in L2(R) for any f∈V0 if
(5.5)δΦ<(C1/2∑m=1r∥ϕm∥Lamλ[a,b])1/λ.Proof.
Notice that{ϕi(·-k):1≤i≤r,k∈Z} is a frame for V0, then, forallf∈V0, there exists a square summable sequence {cmn} such that f=∑n∈Z∑m=1rcmnϕm(·-n) holds in L2(R).
Similar to the argument of Theorem5 in [10], let(5.6)Δ=∑n∈Z∑m=1r|f(n+am+δmn)-f(n+am)|2=∑n∈Z∑m=1r|∑k∈Z∑i=1rcik(ϕi(n+am+δmn-k)-ϕi(n+am-k))|2=∑n∈Z∑m=1r∑k∈Z∑i=1r∑l∈Z∑j=1rcikcjl(ϕi(n+am+δmn-k)-ϕi(n+am-k))×(ϕj(n+am+δmn-k)-ϕj(n+am-k))=∑k∈Z∑i=1r∑l∈Z∑j=1rcikcjl∑n∈Z∑m=1r(ϕi(n+am+δmn-k)-ϕi(n+am-k))×(ϕj(n+am+δmn-k)-ϕj(n+am-k)).
Take
(5.7)ak,i;l,j=∑n∈Z∑m=1r(ϕi(n+am+δmn-k)-ϕi(n+am-k))×(ϕj(n+am+δmn-k)-ϕj(n+am-k)).
Then, ak,i;l,j=al,j;k,i holds for any k,l∈Z, 1≤i,j≤r, and
(5.8)Δ=∑k∈Z∑i=1r∑l∈Z∑j=1rcikcjlak,i;l,j≤∑k∈Z∑i=1r∑l∈Z∑j=1r|ak,i;l,j|((cik)2+(cjl)2)2=12(∑k∈Z∑i=1r(∑l∈Z∑j=1r|ak,i;l,j|)(cik)2+∑l∈Z∑j=1r(∑k∈Z∑i=1r|ak,i;l,j|)(cjl)2)=∑k∈Z∑i=1r(∑l∈Z∑j=1r|ak,i;l,j|)(cik)2≤(supk∈Z,1≤i≤r∑l∈Z∑j=1r|ak,i;l,j|)∑k∈Z∑i=1r(cik)2.
Moreover
(5.9)supk∈Z,1≤i≤r∑l∈Z∑j=1r|ak,i;l,j|≤supk∈Z,1≤i≤r∑l∈Z∑j=1r∑n∈Z∑m=1r|ϕi(n+am+δmn-k)-ϕi(n+am-k)|×|ϕj(n+am+δmn-l)-ϕj(n+am-l)|=supk∈Z,1≤i≤r∑n∈Z∑m=1r|ϕi(n+am+δmn-k)-ϕi(n+am-k)|×∑l∈Z∑j=1r|ϕj(n+am+δmn-l)-ϕj(n+am-l)|≤(∑m=1r∥ϕm∥Lamλ[a,b])2(supn,m|δmn|2).
Hence
(5.10)Δ≤∑k∈Z∑i=1r(cik)2(∑m=1r∥ϕm∥Lamλ[a,b])2(supn,m|δmn|λ)2.
From the argument in Theorem3.6, {q(n+ai,·):n∈Z,1≤i≤r} is a frame for the closed subspace V0. By (3.25), then(5.11)∑i=1r∑n∈Z|f(n+ai)|2=∑i=1r∑n∈Z|〈f,q(n+ai)〉|2≥C∥f∥2≥C∑i=1r∥f∥i.
Again from the proof process in Theorem 3.6, obviously,
(5.12)∥fi∥2≥∑k∈Z|〈fi,qi(k+ai,·)〉|2=∑k∈Z|fi(k)|2=12π∥∑k∈Z∑l∈Zcilϕi(k-l)e-ikω∥*2=12π∥ϕ̂i*∑k∈Zcik∥*2≥Ai2∑k∈Z(cik)2
holds, where ∥f∥*  is defined by
(5.13)∥f∥*=(∫02π|f(t)|2)1/2,
then
(5.14)∑i=1r∑n∈Z|f(n+ai)|2≥CMin{A12,A22,…,Ar2}∑i=1r∑k∈Z(cik)2.
By Theorem 4.2, we only need to show that
(5.15)(∑m=1r∥ϕm∥Lamλ[a,b])2(supn,m|δmn|λ)2≤θ2C.
This is exactly implied by (5.5). We get the desired result.


## 6. Conclusion

Multiwavelets have orthogonality, regularity, short compact support, symmetry, and high approximation order. It is not possible in the scalar wavelet case. In our paper, we study the sampling theorem for frames in multiwavelet subspaces and obtain the sufficient condition under which the regular sampling theorem holds. Notice that the measurements may not be made at uniform points; we establish an irregular sampling theorem for frame in multiwavelet subspaces. Finally, an estimate for the perturbations of regular sampling in shift-invariant spaces is derived.



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*Source: 757039-2010-09-20.xml*