# Discrete Subspace Multiwindow Gabor Frames and Their Duals

Authors: Yun-Zhang Li; Yan Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper addresses discrete subspace multiwindow Gabor analysis. Such a scenario can model many practical signals and has potential applications in signal processing. In this paper, using a suitable Zak transform matrix we characterize discrete subspace mixed multi-window Gabor frames (Riesz bases and orthonormal bases) and their duals with Gabor structure. From this characterization, we can easily obtain frames by designing Zak transform matrices. In particular, for usual multi-window Gabor frames (i.e., all windows have the same time-frequency shifts), we characterize the uniqueness of Gabor dual of type I (type II) and also give a class of examples of Gabor frames and an explicit expression of their Gabor duals of type I (type II).


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## Body

## 1. Introduction

Letℋ be a separable Hilbert space. An at most countable sequence {hi}i∈ℐ in ℋ is called a frame for ℋ if there exist 0<A≤B<∞ such that
(1)A∥f∥2≤∑i∈ℐ‍|〈f,hi〉|2≤B∥f∥2forf∈ℋ,
where A and B are called frame bounds. The sequence {hi}i∈ℐ is called a Bessel sequence in ℋ if the right-hand side inequality in (1) holds. In this case B is called a Bessel bound. A frame for ℋ is said to be a Riesz basis if it ceases to be a frame for ℋ whenever an arbitrary element is removed. And in this case, the frame bounds are also called Riesz bounds. The fundamentals of frames can be found in [1–4]. We denote by ℤ the set of integers, by ℕ the set of positive integers, by {ei} the vector in Euclidean spaces with the ith component being 1 and others being zero, and by ℕγ the set {0,1,…,γ-1} for γ∈ℕ. This paper addresses Gabor systems G(g,N,M) of the form
(2)G(g,N,M)={Em/MTnNlgl:1≤l≤L,m∈ℕM,n∈ℤ},
where L is a fixed positive integer, g=(g1,g2,…,gL) with each gl∈l2(ℤ), N=(N1,N2,…,NL) with each Nl∈ℕ, and M=(M1,M2,…,ML) with each Ml∈ℕ. Throughout this paper, we work under the following assumptions.Assumption 1.
L is a positive integer.Assumption 2.
M
1
=
M
2
=
⋯
=
M
L
=
M, and Nl/M=pl/ql with pl and ql being relatively prime positive integers for 1≤l≤L.We always denote byN the least common multiple of Nl with 1≤l≤L, by λl the positive integer satisfying N=λlNl for each 1≤l≤L, by p and q two relatively prime positive integers satisfying N/M=p/q, and by Q the number Q=q∑l=1Lλl. Obviously, they are all uniquely determined by Assumptions 1 and 2.Remark 3.
We do not lose generality by making Assumptions1 and 2. Let us check a general Gabor system of the form
(3)G(g,N,M)={Eml/MlTnNlgl:1≤l≤L,ml∈ℕMl,n∈ℤ}
with g=(g1,g2,…,gL) satisfying gl∈l2(ℤ), N=(N1,N2,…,NL), and M=(M1,M2,…,ML)(M1,M2,…,ML are not necessarily all the same). Suppose M is the greatest common divisor of M1,M2,…,ML, and Ml=αlM with αl∈ℕ for 1≤l≤L. Define gl(βl)(·)=e2πi(βl/Ml)·gl(·) for 1≤l≤L and βl∈ℕαl. Then, by a simple computation, G(g,N,M) is a frame for a closed subspace ℳ of l2(ℤ) if and only if the Gabor system
(4){Em/MTnNlgl(βl):1≤l≤L,βl∈ℕαl,m∈ℕM,n∈ℤ}
is a frame for ℳ. Therefore, the study of (3) is reduced to the study of (4).We denote byℳ(g,N,M) the closed linear span of G(g,N,M). A Gabor system G(g,N,M) of the form (2) is called a mixed multi-window Gabor system if L>1 since N1,N2,… and NL are not necessarily all the same. (In particular, this system is the usual multi-window Gabor system when N1=N2=⋯=NL). It is called a subspace mixed multi-window Gabor frame if it is a frame for ℳ(g,N,M) and L>1. Similarly, a Gabor system G(g,N,M) is called a single-window Gabor system if L=1 and called a single-window subspace Gabor frame if it is a frame for ℳ(g,N,M) and L=1.For a Bessel sequenceG(g,N,M) in l2(ℤ) of the form (2), we define the associated synthesis operator𝒯g:l2(ℕM×ℤ,ℂL)→l2(ℤ) by
(5)𝒯gc=∑l=1L‍∑m∈ℕM‍∑n∈ℤ‍cl,m,nEm/MTnNlgl
for c=(c1,c2,…,cL)∈l2(ℕM×ℤ,ℂL). Then it is a bounded operator, and its adjoint operator 𝒯g* (so-called analysis operator) is given by
(6)𝒯g*f=(c1(f),c2(f),…,cL(f))forf∈l2(ℤ),
where cl(f)={〈f,Em/MTnNlgl〉}m∈ℕM,n∈ℤ for each 1≤l≤L. For a Bessel sequence G(h,N,M) in l2(ℤ) with h=(h1,h2,…,hL), we associate it with 𝒯h similarly. Write 𝒮h,g=𝒯g𝒯h*; that is,
(7)𝒮h,gf=∑l=1L‍∑m∈ℕM‍∑n∈ℤ‍〈f,Em/MTnNlhl〉Em/MTnNlgl
for f∈l2(ℤ). Let G(g,N,M) be a frame for ℳ(g,N,M) and let G(h,N,M) be a Bessel sequence in l2(ℤ). G(h,N,M) is called an oblique Gabor dual for G(g,N,M) if
(8)𝒮h,gf=fforf∈ℳ(g,N,M).
Here it is not required that each hl belongs to ℳ(g,N,M). In particular, an oblique Gabor dual G(h,N,M) for G(g,N,M) is said to be a Gabor dual of typeI for G(g,N,M) if hl∈ℳ(g,N,M)  for each 1≤l≤L and is said to be a Gabor dual of type II for G(g,N,M) if range(𝒯h*)⊂range(𝒯g*). These notions of duals are a generalization of the ones in [5, 6]. They are borrowed from [7, 8] which dealt with Gabor frames in L2(ℝ). For a Gabor dual of type II, it is not required to be in ℳ(g,N,M), but a containment relation between the ranges of analysis operators is required. Observe that the canonical daulG(𝒮g,g-1g,N,M) for G(g,N,M)  belongs to any one of the three Gabor duals.In the past more than twenty years, the theory of frames has been growing rapidly. Gabor frames are a class of important frames among all kinds of frames. For continuous Gabor frames, single-window Gabor frames forL2(ℝ) have been studied extensively [2, 3, 9, 10]; multi-window Gabor frames for L2(ℝ) were firstly studied by Zibulski and Zeevi in [11], and then by others in [12–17]; single-window subspace (of L2(ℝ)) Gabor frames were studied in [7, 8, 18–23]. In digital signal processing, one usually encounters discrete signals instead of continuous ones. One can obtain discrete Gabor frames for l2(ℤ) via Gabor frames for L2(ℝ) through sampling under certain additional assumptions [24–26], though these assumptions are artificial and too technical. So it should be more reasonable to consider Gabor frames in l2(ℤ) without referring to frames in L2(ℝ). Intuitively, the general theory of discrete Gabor analysis is similar to the continuous case. Sometimes major differences occur. In 1989, Heil in [27] showed that while Gabor frames in the continuous case are bases only if they are generated by functions that are not smooth or have poor decay, it is possible in the discrete case to construct Gabor frames that are bases and are generated by sequences with good decay. The sampled Gaussian gives an example of such a signal. Also due to its good potential for digital signal processing, since then, Gabor analysis on l2(ℤ) has attracted many researchers (see [2, 3, 24–34] and the references therein). In [6, 35], single-window Gabor frames on discrete periodic subsets of ℤ were investigated. It is well known that a single-window Gabor expansion is not enough to analyze the dynamic time-frequency contents of signals that contain a wide range of spatial and frequency components. A multi-window Gabor expansion is capable of extracting local frequencies in an adaptive manner, in which wide windows are responsible for slow-changing components and narrow windows are designed to extract transient and rapid-changing components of a signal. See [15] and the references therein for details. In addition, Example 38 and the arguments before it in the last section show that, for Gabor duals, multi-window Gabor frames behave differently from single-window ones. Motivated by the above works, we in this paper study subspace Gabor frame of the form (2) under Assumptions 1 and 2.The rest of this paper is organized as follows. Section2 is an auxiliary one to following sections. In this section, we introduce the notion of Zak transform matrix associated with a Gabor system of the form (2) and investigate its properties. In terms of Zak transform matrices, we in Section 3 characterize subspace multi-window Gabor frames (Riesz bases, orthonormal bases) and in Section 4 Gabor duals of type I (type II) for subspace multi-window Gabor frames. Section 5 focuses on subspace multi-window Gabor frames with all the windows having the same time-frequency shifts. We characterize the uniqueness of Gabor duals of type I (type II) and obtain a class of examples of subspace multi-window Gabor frames (Riesz bases, orthonormal bases) and their Gabor duals of type I (type II) (see Theorems 36 and 37).


## 2. Zak Transforms and Zak Transform Matrices

This section focuses on some properties of Zak transforms and Zak transform matrices, which is an auxiliary one to following sections. We start with some notations and notions. Fors, t∈ℕ, we denote by ℳs,t the set of s×t complex matrices. Given γ∈ℕ, two subsets S1 and S2 of ℤ are said to be γℤcongruent if there exists a sequence of subsets {S1,k}k∈ℤ of S1 such that ⋃k∈ℤS1,k=S1, S1,k1∩S1,k2=∅ for k1≠k2 in ℤ and that S2=⋃k∈ℤ(S1,k+γk), (S1,k1+γk1)∩(S1,k2+γk2)=∅ for k1≠k2 in ℤ. If S1 or S2 is a finite set in addition, only finitely many S1,k in the above are nonempty, while the others are empty. For two vectors x and y in a Hilbert space ℋ, we always write their inner product as 〈x,y〉 instead of 〈x,y〉ℋ when it causes no confusions. Let E be a finite set in ℤ and let Γ∈ℕ. We denote by L2(E×[0,1),ℂΓ) the Hilbert space of vector-valued functions f satisfying f(j,·)∈L2([0,1),ℂΓ) for each j∈E endowed with the inner product
(9)〈f,f~〉=∑j∈E‍∫01‍〈f(j,v),f~(j,v)〉dv
for f, f~∈L2(E×[0,1),ℂΓ), where 〈f(j,v),f~(j,v)〉 denotes their inner product in ℂΓ; by l2(E×ℤ,ℂΓ) the Hilbert space of sequences c=(c1,c2,…,cΓ) satisfying cγ∈l2(E×ℤ) endowed with the inner product
(10)〈c,c~〉=∑γ=1Γ‍〈cγ,c~γ〉
for c, c~∈l2(E×ℤ,ℂΓ), where 〈cγ,c~γ〉 denotes their inner product in l2(E×ℤ) and by l0(E×ℤ,ℂΓ) the set of sequences c=(c1,c2,…,cΓ) with each cγ being a finitely supported sequence defined on E×ℤ. For simplicity, we write L2(E×[0,1)) for L2(E×[0,1),ℂΓ), l2(E×ℤ) for l2(E×ℤ,ℂΓ), and l0(E×ℤ) for l0(E×ℤ,ℂΓ) when Γ=1.Forf∈l2(ℤ), define the discrete Zak transform𝒵qNf of f by
(11)(𝒵qNf)(j,v)∶=∑ℓ∈ℤ‍f(j+ℓqN)e2πiℓv
for j∈ℤ and a.e. v∈ℝ. It is easy to check that 𝒵qN has quasi-periodicity:
(12)(𝒵qNf)(j+kqN,v+ℓ)=e-2πikv(𝒵qNf)(j,v)
for k,ℓ,j∈ℤ and a.e. v∈ℝ.Definition 4.
Forg=(g1,g2,…,gL) with each gl∈l2(ℤ), we associate it with a matrix-valued function Ψg(j,v):ℤ×ℝ→ℳQ,p by
(13)Ψg(j,v)=(G1(j,v)G2(j,v)⋮GL(j,v)),
where Q∶=q∑l=1Lλl and Gl(j,v) is a block matrix of the form
(14)Gl(j,v)=(𝒢l(j,v)𝒢l(j-Nl,v)⋮𝒢l(j-(λl-1)Nl,v))
with 𝒢l(j,v):ℤ×ℝ→ℳq,p for 1≤l≤L and
(15)𝒢l(j,v)r,k=𝒵qNgl(j-rN+kM,v)
for r∈ℕq, k∈ℕp.By the quasi-periodicity of𝒵qN, an arbitrary f∈l2(ℤ) is uniquely determined by the values of (𝒵qNf)(·,·) on S×[0,1) with S being a set qNℤ congruent to ℕqN. So an arbitrary function F∈L2(S×[0,1)) determines a unique f∈l2(ℤ) by
(16)(𝒵qNf)(j,v)=F(j,v)for(j,v)∈S×[0,1).
Let us make another assumption that N1=N2=⋯=NL in Definition 4. Then Ψg(j,v) is an Lq×p-matrix-valued function of the form
(17)Ψg(j,v)=(𝒢1(j,v)𝒢2(j,v)⋮𝒢L(j,v)),𝒢l(j,v)r,k=𝒵qNgl(j-rN+kM,v)
for 1≤l≤L, r∈ℕq, and k∈ℕp. Also observe that ℕM/q-Nℕq+Mℕp is qNℤ congruent to ℕqN. We have if N1=N2=⋯=NL, then an arbitrary function M(j,v):ℕM/q×[0,1)→ℳLq,p with all entries in L2(ℕM/q×[0,1)) determines a unique g=(g1,g2,…,gL) with each gl∈l2(ℤ) by
(18)Ψg(j,v)=M(j,v)forj∈ℕM/q,a.e.v∈[0,1).
But if Nl, 1≤l≤L, are not all the same, a function M(j,v):ℕM/q×[0,1)→ℳQ,p with all entries in L2(ℕM/q×[0,1)) does not necessarily determine a g. We show it by an example. In this case, there exists 1≤l≤L such that λl>1. We may as well assume that λ1>1. Choose
(19)M(j,v)=(M1(j,v)M2(j,v)⋮ML(j,v)),Ml(j,v)=(Ml,0(j,v)Ml,1(j,v)⋮Ml,λl-1(j,v))
for 1≤l≤L and (j,v)∈ℕM/q×[0,1) such that every entry of M(j,v) belongs to L2(ℕM/q×[0,1)) and that
(20)M1,0(j,v)≠0,M1,1(j,v)=0aaaaafor(j,v)∈ℕM/q×[0,1).
Suppose there exists g such that Ψg(j,v)=M(j,v) for j∈ℕM/q and a.e. v∈[0,1). Then
(21)𝒢1(j,v)=M1,0(j,v),(22)𝒢1(j-N1,v)=M1,1(j,v)
for j∈ℕM/q and a.e. v∈[0,1), where
(23)𝒢1(j,v)r,k=𝒵qNg1(j-rN+kM,v),𝒢1(j-N1,v)r,k=𝒵qNg1(j-N1-rN+kM,v)
for (r,k)∈ℕq×ℕp. Since the sets ℕM/q-Nℕq+Mℕp and ℕM/q-Nℕq+Mℕp-N1 are both qNℤ congruent to ℕqN, we have 𝒵qNg1(j0,v)≠0 for some j0∈ℤ and v∈E⊂[0,1) with |E|>0 by (20), (21), and the quasi-periodicity of 𝒵qN, while 𝒵qNg1(j,v)=0 for j∈ℤ and a.e. v∈[0,1) by (20), (22), and the quasi-periodicity of 𝒵qN. This is a contradiction. Therefore, we must be careful when we define g by a function M(j,v):ℕM/q×[0,1)→ℳQ,p if Nl, 1≤l≤L, are not all the same.Definition 5.
DefineZqN:l2(ℤ)→L2(ℕM×[0,1),ℂp) by
(24)ZqNf(j,v)=(𝒵qNf(j,v)𝒵qNf(j+M,v)⋮𝒵qNf(j+(p-1)M,v))aaaaaaaaaaaaaaaaaaafor(j,v)∈ℕM×[0,1).By the quasi-periodicity of𝒵qN and [35, Theorem 2.1], we have the following lemmas.Lemma 6.
Z
q
N is a unitary operator from l2(ℤ) onto L2(ℕM×[0,1),ℂp), and 𝒵qN is a unitary operator from l2(ℤ) onto L2(E×[0,1)) for an arbitrary subset E of ℤ which is qNℤ congruent to ℕqN.Definition 7.
Define theFourier transformℱ:l2(ℕM×ℤ)→L2(ℕM×[0,1)) by
(25)ℱc(j,v)=1M∑m∈ℕM‍∑n∈ℤ‍cm,ne2πi(m/M)je2πinv
for c∈l2(ℕM×ℤ), j∈ℕM, and a.e. v∈[0,1).Definition 8.
Define𝒥:l2(ℕM×ℤ,ℂL)→L2(ℕM×[0,1),ℂQ) by
(26)𝒥c(j,v)=𝒞(j,v)
for c=(c1,c2,…,cL)∈l2(ℕM×ℤ,ℂL), j∈ℕM, and a.e. v∈ℝ, where
(27)𝒞(j,v)=(𝒞1(j,v)𝒞2(j,v)⋮𝒞L(j,v)),𝒞l(j,v)=(𝒞l(0)(j,v)𝒞l(1)(j,v)⋮𝒞l(λl-1)(j,v)),𝒞l(τl)(j,v)=(∑m∈ℕM‍∑n∈ℤ‍cl,m,nqλl+τle2πi(m/M)je2πinv∑m∈ℕM‍∑n∈ℤ‍cl,m,(nq+1)λl+τle2πi(m/M)je2πinv⋮∑m∈ℕM‍∑n∈ℤ‍cl,m,(nq+q-1)λl+τle2πi(m/M)je2πinv)
for 1≤l≤L and τl∈ℕλl.Similarly, for an arbitraryd∈l2(ℕM×ℤ,ℂL), we associate it with 𝒟(j,v).Lemma 9.
(
i
)  ℱ is a unitary operator from l2(ℕM×ℤ) onto L2(ℕM×[0,1));
(
ii
)  (1/M)𝒥 is a unitary operator from l2(ℕM×ℤ,ℂL) onto L2(ℕM×[0,1),ℂQ).Proof.
Since{(1/M)e2πi(m/M)je2πinv:m∈ℕM,n∈ℤ} is an orthonormal basis for L2(ℕM×[0,1)), we have (i). (ii) is an immediate consequence of (i).Lemma 10.
Forg=(g1,g2,…,gL) with gl∈l2(ℤ) for 1≤l≤L, one has(i)
𝒵qN(Em/MTnqN+τlNl+rNgl)(j,v)=e2πi(m/M)je2πinv𝒵qNgl(j-τlNl-rN,v) for r∈ℕq;(ii)
(28)ZqNf(j,v)=Ψg*(j,v)-𝒥c(j,v)forf=𝒯gc with c∈l0(ℕM×ℤ,ℂL), where 𝒯gc is as in (5).Proof.
(i) can be obtained by a direct computation. Applying (i), we have(29)𝒵qNf(j,v)=𝒵qN(∑l=1L‍∑τl∈ℕλl‍∑r∈ℕq‍∑m∈ℕM‍∑n∈ℤ‍cl,m,(nq+r)λl+τl×Em/MTnqN+τlNl+rNgl∑l=1L‍∑τl∈ℕλl‍∑r∈ℕq‍∑m∈ℕM‍∑n∈ℤ‍cl,m,(nq+r)λl+τl)=∑l=1L‍∑τl∈ℕλl‍∑r∈ℕq‍𝒞l(τl)(j,v)r𝒵qNgl(j-τlNl-rN,v)
for (j,v)∈ℤ×ℝ. This implies (ii).Remark 11.
WhenG(g,N,M) in Lemma 10 is a Bessel sequence in l2(ℤ), by the same procedure as the above we can prove (28) holds for f=𝒯gc with c∈l2(ℕM×ℤ,ℂL).Lemma 12.
Forg=(g1,g2,…,gL) with gl∈l2(ℤ) for 1≤l≤L, one has
(30)Ψg(j+Mqn,v)=e-2πimnv𝒞n(v)Ψg(j,v)𝒟n(v)
for j∈ℤ, n=knq+(mnq-rn)p with (kn,rn,mn)∈ℕp×ℕq×ℤ, and a.e. v∈ℝ, where 𝒟n(v)=(0e-2πivIknIp-kn0), 𝒞n(v)=diag(𝒞1,n(v),𝒞2,n(v),…,𝒞L,n(v)), and 𝒞l,n(v) denotes the block matrix (with λl blocks) of the form diag(Cn(v),Cn(v),…,Cn(v)) with Cn(v)=(0Iq-rne2πivIrn0).Proof.
By [6, Lemma 5], we have
(31)𝒢l(j+Mqn,v)=e-2πimnvCn(v)𝒢l(j,v)𝒟n(v)
for 1≤l≤L, j∈ℤ, n=knq+(mnq-rn)p with (kn,rn,mn)∈ℕp×ℕq×ℤ, and a.e. v∈ℝ. This leads to the lemma.


## 3. Frame Characterization

This section is devoted to characterization of subspace Gabor framesG(g,N,M) of the form (2).Theorem 13.
For anyg=(g1,g2,…,gL) with each gl∈l2(ℤ), the following are equivalent.(i)
G(g,N,M) is a Bessel sequence in l2(ℤ) with Bessel bound B.(ii)
Ψg(j,v)Ψg*(j,v)≤(B/M)I for j∈ℕM/q and a.e. v∈[0,1).(iii)
Ψg(j,v)Ψg*(j,v)≤(B/M)I for j∈ℤ and a.e. v∈ℝ.Proof.
By Lemma12, conditions (ii) and (iii) are equivalent. So, to finish the proof, we only need to prove that (i) holds if and only if
(32)Ψg(j,v)Ψg*(j,v)≤BMIforj∈ℕM,a.e.v∈[0,1),
since G(g,N,M) is a Bessel sequence with Bessel bound B if and only if
(33)∥𝒯gc∥2≤B∥c∥2
for c∈l0(ℕM×ℤ,ℂL), which can be rewritten as
(34)∑j∈ℕM‍∫01‍∥Ψg*(j,v)-𝒥c(j,v)∥2dv≤BM∑j∈ℕM‍∫01‍∥𝒥c(j,v)∥2dv
by Lemmas 6–10.
Next we prove the equivalence between (32) and (34) to finish the proof. It is obvious that (32) implies (34). Now we turn to the converse implication. Suppose (34) holds. Since every entry in Ψg(j,·) belongs to L2([0,1)) and thus belongs to L1([0,1)) for j∈ℕM, almost every point in (0,1) is a Lebesgue point for every entry in Ψg*(j,v) and j∈ℕM. Let v0 be such a point and j0∈ℕM. Fix x∈ℂQ and ϵ>0 with (v0-ϵ,v0+ϵ)⊂(0,1). Define c∈l2(ℕM×ℤ,ℂL) by
(35)𝒥c(j,v)=12ϵχ{j0}×(v0-ϵ,v0+ϵ)(j,v)x-
for j∈ℕM, v∈[0,1). By Lemma 9 and the density of l0(ℕM×ℤ,ℂL) in l2(ℕM×ℤ,ℂL), (34) holds for c∈l2(ℕM×ℤ,ℂL). Substituting (35) into (34), we obtain that
(36)12ϵ∫(v0-ϵ,v0+ϵ)‍∥Ψg*(j0,v)x∥2dv≤BM∥x∥2.
Letting ϵ→0 leads to
(37)∥Ψg*(j0,v0)x∥2≤BM∥x∥2;
that is,
(38)〈Ψg(j0,v0)Ψg*(j0,v0)x,x〉≤BM∥x∥2.
This gives (32) by the arbitrariness of j0, v0, and x. The proof is completed.Lemma 14 ([1, Theorem A.6.5]).
LetU1, U2, U3 be self-adjoint operators on a Hilbert space ℋ. If U1≤U2, and U3≥0, and U3 commutes with U1 and U2, then U1U3≤U2U3.Remark 15.
The inequalityΨg(j,v)Ψg*(j,v)≤(B/M)I in Theorem 13 can be replaced by any one of the following inequalities:(i)
(Ψg(j,v)Ψg*(j,v))2≤(B/M)Ψg(j,v)Ψg*(j,v);(ii)
Ψg*(j,v)Ψg(j,v)≤(B/M)I;(iii)
(Ψg*(j,v)Ψg(j,v))2≤(B/M)Ψg*(j,v)Ψg(j,v).
Indeed, we have (i) by Lemma14 if Ψg(j,v)Ψg*(j,v)≤(B/M)I. From (i), we have
(39)〈Ψg*(j,v)Ψg(j,v)Ψg*(j,v)x,Ψg*(j,v)x〉≤BM〈Ψg*(j,v)x,Ψg*(j,v)x〉
for x∈ℂQ, and thus Ψg*(j,v)Ψg(j,v)≤(B/M)I on range(Ψg*(j,v)). This implies (ii) due to the fact that ℂp=range(Ψg*(j,v))⊕ker(Ψg(j,v)). It is obvious that (ii) implies (iii) by Lemma 14. By the same procedure as in “(i)⇒(ii)”, we can prove that (iii) implies Ψg(j,v)Ψg*(j,v)≤(B/M)I.Remark 16.
Condition (iii) in Theorem13 is equivalent to ∥Ψg(j,v)∥≤B/M for a.e. (j,v)∈ℤ×ℝ when Ψg(j,v) is understood as an operator from ℂp into ℂQ. However, such norm is equivalent to the one obtained by taking the maximum of the absolute values of all entries of a matrix. So G(g,N,M) is a Bessel sequence if and only if 𝒵qNgl∈L∞(ℤ×ℝ) for 1≤l≤L by Theorem 13.Lemma 17.
Forg=(g1,g2,…,gL) with each gl∈l2(ℤ), let G(g,N,M) be a Bessel sequence in l2(ℤ). For f∈l2(ℤ), write d(f)=𝒯g*f. Then
(40)𝒥d(f)(j,v)=MΨg(j,v)-ZqNf(j,v)
for (j,v)∈ℕM×ℝ when either G(g,N,M) is a Bessel sequence or f∈l0(ℤ).Proof.
By Lemmas6 and 10, we have
(41)dl,m,(nq+r)λl+τl(f)=∑k∈ℕp‍∑j∈ℕM‍∫01‍𝒵qNf(j+kM,v)×𝒵qNgl(j-τlNl-rN+kM,v)-×e-2πinvdve-2πi(m/M)j=∑j∈ℕM‍∫01(𝒢l(j-τlNl,v)-ZqNf(j,v))r×e-2πinvdve-2πi(m/M)j
for each r∈ℕq, τl∈ℕλl, and 1≤l≤L. When either G(g,N,M) is a Bessel sequence or f∈l0(ℤ), the integrand in (41) belongs to L2([0,1)) as a function about v by Remark 16. It follows that
(42)𝒟(τl)(j,v)=M𝒢l(j-τlNl,v)-ZqNf(j,v)
for (j,v)∈ℕM×ℝ. The lemma therefore follows.Theorem 18.
LetG(g,N,M) and G(h,N,M) be both Bessel sequences in l2(ℤ). Then
(43)ZqN𝒮h,gf(j,v)=MΨg*(j,v)Ψh(j,v)-ZqNf(j,v)
for j∈ℕM and a.e. v∈[0,1).Proof.
Writed=𝒯h*f. Then
(44)ZqN𝒮h,gf(j,v)=Ψg*(j,v)-𝒥d(j,v)
by Remark 11. Applying Lemma 17 to h, we have
(45)𝒥d(j,v)=MΨh*(j,v)-ZqNf(j,v),
and thus (43) holds by (44).Theorem 19.
Forg=(g1,g2,…,gL) with each gl∈l2(ℤ), the following are equivalent:(i)
G(g,N,M) is a frame for ℳ(g,N,M) with frame bounds A and B;(ii)
(A/M)Ψg(j,v)Ψg*(j,v)≤(Ψg(j,v)Ψg*(j,v))2≤(B/M)Ψg(j,v)Ψg*(j,v) for j∈ℕM/q and a.e. v∈[0,1);(iii)
(A/M)Ψg(j,v)Ψg*(j,v)≤(Ψg(j,v)Ψg*(j,v))2≤(B/M)Ψg(j,v)Ψg*(j,v) for j∈ℤ and a.e. v∈ℝ.Proof.
By Theorem13 and Remark 15, we may as well assume that G(g,N,M) is a Bessel sequence, and we need to treat the “lower frame bound” part under this assumption. By a similar argument to beginning proof of Theorem 13, we only need to prove that
(46)〈𝒮g,gf,f〉≥A∥f∥2forf∈ℳ(g,N,M)
if and only if
(47)(Ψg(j,v)Ψg*(j,v))2≥AMΨg(j,v)Ψg*(j,v)forj∈ℕM,v∈[0,1).
Since the linear span of G(g,N,M) is dense in ℳ(g,N,M), (46) holds if and only if
(48)〈𝒮g,gf,f〉≥A∥f∥2
for f=𝒯gc  with c∈l2(ℕM×ℤ,ℂL) by [1, Lemma 5.1.7]. By Lemma 10, Theorem 18, and Remark 11, (48) can be rewritten as
(49)M〈(Ψg(j,v)Ψg*(j,v))-2𝒥c(j,v),𝒥c(j,v)〉≥A〈Ψg(j,v)Ψg*(j,v)-𝒥c(j,v),𝒥c(j,v)〉,
equivalently,
(50)M〈(Ψg(j,v)Ψg*(j,v))2𝒥c(j,v)-,𝒥c(j,v)-〉≥A〈Ψg(j,v)Ψg*(j,v)𝒥c(j,v)-,𝒥c(j,v)-〉
for j∈ℕM, v∈[0,1), and c∈l2(ℕM×ℤ,ℂL). By the same procedure as in Theorem 13, we can prove the equivalence between (47) and (50). The proof is completed.Remark 20.
By an argument similar to Remark15, (Ψg(j,v)Ψg*(j,v))2 and Ψg(j,v)Ψg*(j,v) in Theorem 19 can be replaced by (Ψg*(j,v)Ψg(j,v))2 and Ψg*(j,v)Ψg(j,v), respectively.Definition 21.
Giveng=(g1,g2,…,gL) with each gl∈l2(ℤ), let G(g,N,M) be a Bessel sequence in l2(ℤ). We say that G(g,N,M) has Riesz property if, for c∈l2(ℕM×ℤ,ℂL), we must have c=0 whenever 𝒯gc=0.By an easy application of the spectral theorem of self-adjoint matrices, we have the following lemma (see also [36, page 978]).Lemma 22.
Given a measurable setE in ℝ with |E|>0, let 𝒜:E→ℳs,t be a matrix-valued measurable function. Define by 𝒫(v) the orthogonal projection of ℂt onto ker(𝒜(v)). Then
(51)𝒫(v)=limn→∞exp(-n𝒜*(v)𝒜(v))
for v∈E, and thus 𝒫(v) is measurable.Theorem 23.
Forg=(g1,g2,…,gL) with each gl∈l2(ℤ), the following are equivalent:(i)
G(g,N,M) has Riesz property;(ii)
rank(Ψg(j,v))=Q for j∈ℕM/q and a.e. v∈[0,1);(iii)
rank(Ψg(j,v))=Q for j∈ℤ and a.e. v∈ℝ.Proof.
By Lemma12, (ii) and (iii) are equivalent. So we only need to prove that c=0 is a unique solution to
(52)𝒯gc=0
in l2(ℕM×ℤ,ℂL) if and only if
(53)rank(Ψg(j,v))=Qforj∈ℕM,v∈[0,1).
By Remark 11, (52) can be rewritten as
(54)Ψg*(j,v)-𝒥c(j,v)=0forj∈ℕM,v∈[0,1).
So c=0 is a unique solution to (52) in l2(ℕM×ℤ,ℂL) if and only if d(j,v)=0 is a unique solution to
(55)Ψg*(j,v)-d(j,v)=0forj∈ℕM,a.e.v∈[0,1)
in L2(ℕM×[0,1),ℂQ) by Lemma 9. It is obvious that (53) implies that d(j,v)=0 is a unique solution to (55) in L2(ℕM×[0,1),ℂQ). Next we prove the converse implication. Suppose there exist j0∈ℕM and E⊂[0,1) with |E|>0 such that rank(Ψg(j0,v))<Q on E. Let 𝒫(j0,v) be the orthogonal projection of ℂQ onto ker(Ψg*(j0,v)). Then there exist i0∈ℕQ and E′⊂E with |E′|>0 such that 𝒫(j0,v)ei0≠0 for v∈E′. Define d(j,v)∈L2(ℕM×[0,1),ℂQ) by
(56)d(j,v)={𝒫(j,v)ei0,ifj=j0,v∈E′,0,otherwise
for j∈ℕM and v∈[0,1). Then d(j,v) is well defined, and ∥d(j,v)∥≤1 by Lemma 22. It follows that d(j,v) is a nonzero solution to (55) in L2(ℕM×[0,1),ℂQ). This is a contradiction. The proof is completed.Since a Riesz basis is exactly a frame having Riesz property, and an orthonormal basis is exactly a Riesz basis with Riesz bound1, we have the following theorem by Theorems 19 and 23.Theorem 24.
Forg=(g1,g2,…,gL) with each gl∈l2(ℤ), the following are equivalent:(i)
G(g,N,M) is a Riesz basis for G(g,N,M) with Riesz bounds A and B (an orthonormal basis);(ii)
(A/M)I≤Ψg(j,v)Ψg*(j,v)≤(B/M)I(Ψg(j,v)Ψg*(j,v)=(1/M)I) for j∈ℕM/q a.e. v∈[0,1);(iii)
(A/M)I≤Ψg(j,v)Ψg*(j,v)≤(B/M)I(Ψg(j,v)Ψg*(j,v)=(1/M)I) for j∈ℤ and a.e. v∈ℝ.Next we turn to examples of Theorems19 and 24. Suppose
(57)N=(6,4),M=(2,2).
Then N/M=6 with p=6, q=1, and Q=5. For g=(g1,g2) with g1,g2∈l2(ℤ), we associate it with Ψg as in Definition 4. Then
(58)Ψg(j,v)=(G1(j,v)G2(j,v))withG1(j,v)=(𝒢1(j,v)𝒢1(j-6,v)),G2(j,v)=(𝒢2(j,v)𝒢2(j-4,v)𝒢2(j-8,v)),
where
(59)𝒢l(j,v)=(𝒵12gl(j+2k,v))0,k∈ℳ1,6withj∈ℕ2,k∈ℕ6
for l=1,2. By the quasi-periodicity of Zak transform, we have
(60)𝒢1(j-6,v)=(e2πiv𝒵12g1(j+6,v),e2πiv𝒵12g1(j+8,v),e2πiv𝒵12g1(j+10,v),𝒵12g1(j,v),𝒵12g1(j+2,v),𝒵12g1(j+4,v)e2πiv),𝒢2(j-4,v)=(e2πiv𝒵12g2(j+8,v),e2πiv𝒵12g2(j+10,v),𝒵12g2(j,v),𝒵12g2(j+2,v),𝒵12g2(j+4,v),𝒵12g2(j+6,v)e2πiv),𝒢2(j-8,v)=(e2πiv𝒵12g2(j+4,v),e2πiv𝒵12g2(j+6,v),e2πiv𝒵12g2(j+8,v),𝒵12g2(j+10,v),𝒵12g2(j,v),𝒵12g2(j+2,v)e2πiv).
Thus for each j∈ℕ2 and a.e. v∈[0,1), 𝒢1(j-6,v), and 𝒢2(j-4,v), 𝒢2(j-8,v) are uniquely determined by 𝒢1(j,v) and 𝒢2(j,v), respectively. Observe that ℕ2+2ℕ6 is 12ℤ congruent to ℕ12. It follows that g is uniquely determined by the values of Ψg(j,v) for j∈ℕ2 and a.e. v∈[0,1). Therefore, an arbitrary matrix 5×6 matrix-valued function 𝒦(j,v) for j∈ℕ2 and a.e. v∈[0,1) with all entries being in L2(ℕ2×[0,1)) determines a unique g by
(61)Ψg(j,v)=𝒦(j,v)forj∈ℕ2,a.e.v∈[0,1).Let(62)𝒦(j,v)=(a(j,v)λ(j,v)a(j,v)λ2(j,v)a(j,v)000000a(j,v)λ(j,v)a(j,v)λ2(j,v)a(j,v)-λ(j,v)a(j,v)-a(j,v)-000000λ(j,v)a(j,v)-a(j,v)-000000-λ(j,v)a(j,v)-a(j,v)-)for j∈ℕ2 and a.e. v∈[0,1). Define g=(g1,g2) by
(63)Ψg(j,v)=𝒦(j,v)forj∈ℕ2,a.e.v∈[0,1].
Then we obtain the following example.Example 25.
LetN and M be defined as in (57) and define g=(g1,g2) by (63), where λ(j,·), a(j,·) are continuous on [0,1] for j∈{0,1}. Assume that
(64)|λ(j,v)|<1,|a(j,v)|2(1-2|λ(j,v)|4-2|λ(j,v)|6-|λ(j,v)|8)<1-|λ(j,v)|3,|a(j,v)|2(1+2|λ(j,v)|3+2|λ(j,v)|5+|λ(j,v)|7)<1+|λ(j,v)|3
for (j,v)∈{0,1}×[0,1] satisfying a(j,v)≠0. Then
(65)supp(g1)⊂ℕ6+12ℤ,supp(g2)⊂ℕ4+12ℤ,
and G(g,N,M) is a frame for ℳ(g,N,M).Proof.
Equation (65) is an immediate consequence of (63). Next we prove that G(g,N,M) is a frame for ℳ(g,N,M). By a simple computation, we have
(66)〈(Ψg(j,v)Ψg*(j,v))2x,x〉=|a(j,v)|2∑l=15‍αl(j,v)|xl|2+β(j,v),〈Ψg(j,v)Ψg*(j,v)x,x〉=|a(j,v)|2∑l=15‍α~l(j,v)|xl|2+β~(j,v)
for (j,v)∈{0,1}×[0,1] and x∈ℂ5, where
(67)α1(j,v)=|λ(j,v)|366666666+(1+|λ(j,v)|2+|λ(j,v)|4)2|a(j,v)|2,α2(j,v)=1+(1+|λ(j,v)|2+|λ(j,v)|4)266666666×|a(j,v)|2,α4(j,v)=1+|λ(j,v)|3+(1+|λ(j,v)|2)2|a(j,v)|2,α4(j,v)=1+|λ(j,v)|3+(1+|λ(j,v)|2)2|a(j,v)|2,α3(j,v)=α5(j,v)=(1+|λ(j,v)|2)2|a(j,v)|2,β(j,v)=2Re([λ(j,v)366666666666666×(2+2|λ(j,v)|2+|λ(j,v)|4)66666666666666×a(j,v)2|a(j,v)|2]x-1x4)66666666+2Re([(2+2|λ(j,v)|2+|λ(j,v)|4)666666666666666×a(j,v)2|a(j,v)|2]x-2x4)66666666-2Re(λ(j,v)3|a(j,v)|2x-1x2),α~1(j,v)=α~2(j,v)=1+|λ(j,v)|2+|λ(j,v)|4,α~3(j,v)=α~4(j,v)=α~5(j,v)=1+|λ(j,v)|2,β~(j,v)=-2Re(λ(j,v)3a(j,v)2x-1x4)66666666+2Re(a(j,v)2x-2x4).
It is easy to check that
(68)|β(j,v)|≤|a(j,v)|2[β1(j,v)|x1|2+β2(j,v)|x2|2+β4(j,v)|x4|2],
where
(69)β1(j,v)=|λ(j,v)|3+|a(j,v)|2×(2|λ(j,v)|3+2|λ(j,v)|5+|λ(j,v)|7),β2(j,v)=|λ(j,v)|3+|a(j,v)|2×(2+2|λ(j,v)|2+|λ(j,v)|4),β4(j,v)=|a(j,v)|2×(|λ(j,v)|72+2|λ(j,v)|2+2|λ(j,v)|3+|λ(j,v)|4+2|λ(j,v)|5+|λ(j,v)|7).|β~(j,v)|≤|a(j,v)|2×[β~1(j,v)|x1|2+β~2(j,v)|x2|2+β~4(j,v)|x4|2],
where
(70)β~1(j,v)=|λ(j,v)|3,β~2(j,v)=1,β~4(j,v)=1+|λ(j,v)|3.
Write
(71)Cl(j,v)=αl(j,v)+βl(j,v),C~l(j,v)=αl(j,v)-βl(j,v)forl=1,2,4,C(j,v)=(1+|λ(j,v)|2)2|a(j,v)|2,Dl(j,v)=α~l(j,v)+β~l(j,v),D~l(j,v)=α~l(j,v)-β~l(j,v)forl=1,2,4,D(j,v)=|λ(j,v)|2+1.
Take
(72)C~=min{C(j,v),C~l(j,v):l=1,2,4,j∈ℕ2,v∈[0,1]},C=max{C(j,v),Cl(j,v):l=1,2,4,j∈ℕ2,v∈[0,1]},D~=min{D(j,v),D~l(j,v):l=1,2,4,j∈ℕ2,v∈[0,1]},D=max{D(j,v),Dl(j,v):l=1,2,4,j∈ℕ2,v∈[0,1]}.
Then
(73)〈(Ψg(j,v)Ψg*(j,v))2x,x〉≤|a(j,v)|2[C1(j,v)|x1|2+C2(j,v)|x2|2666666666666666+C4(j,v)|x4|2+C(j,v)666666666666666×(|x3|2+|x5|2)]≤C|a(j,v)|2∥x∥2,〈(Ψg(j,v)Ψg*(j,v))2x,x〉≥|a(j,v)|2[C~1(j,v)|x1|2+C~2(j,v)|x2|2666666666666666+C~4(j,v)|x4|2+C(j,v)666666666666666×(|x3|2+|x5|2)C~1(j,v)|x1|2+C~2(j,v)|x2|2]≥C~|a(j,v)|2∥x∥2,〈Ψg(j,v)Ψg*(j,v)x,x〉≤|a(j,v)|2[D1(j,v)|x1|2+D2(j,v)|x2|2+D4(j,v)|x4|2+D(j,v)×(|x3|2+|x5|2)]≤D|a(j,v)|2∥x∥2,〈Ψg(j,v)Ψg*(j,v)x,x〉≥|a(j,v)|2[D~1(j,v)|x1|2+D~2(j,v)|x2|2+D~4(j,v)|x4|2+D(j,v)(|x3|2+|x5|2)D~1]≥D~|a(j,v)|2∥x∥2.
It follows that
(74)A2〈Ψg(j,v)Ψg*(j,v)x,x〉≤〈(Ψg(j,v)Ψg*(j,v))2x,x〉≤B2〈Ψg(j,v)Ψg*(j,v)x,x〉,
for x∈ℂ5, j∈ℕ2, and a.e. v∈[0,1), and thus
(75)A2Ψg(j,v)Ψg*(j,v)≤(Ψg(j,v)Ψg*(j,v))2≤B2Ψg(j,v)Ψg*(j,v)
for j∈ℕ2 and a.e. v∈[0,1), where A=2C~/D, B=2C/D~. By Theorem 19, G(g,N,M) is a frame for ℳ(g,N,M) with frame bounds A and B.Remark 26.
By a simple computation, we haverank(Ψg(j,v))=5 if and only if
(76)a(0,v)a(1,v)λ(0,v)λ(1,v)≠0fora.e.v∈[0,1).
Also observe that rank(Ψg(j,v))=5 is equivalent to Ψg(j,v)Ψg*(j,v) being invertible, and that (75) can be reduced to
(77)AI2≤Ψg(j,v)Ψg*(j,v)≤BI2forj∈ℕ2,a.e.v∈[0,1)
in this case. Therefore, G(g,N,M) is a Riesz basis for ℳ(g,N,M) with Riesz bounds A and B if and only if (76) holds.


## 4. Gabor Dual Characterization

LetG(g,N,M) be a frame for ℳ(g,N,M). In this section, we discuss three kinds of duals with Gabor structure. We establish characterizations of Gabor duals of type I and type II and obtain a sufficient condition for Gabor duals of type II and oblique Gabor duals.By an argument similar to [5, Lemma 2.6], we have the following lemmas.Lemma 27.
Forg=(g1,g2,…,gL) and h=(h1,h2,…,hL) with each gl, hl∈l2(ℤ), the following are equivalent:(i)
hl∈ℳ(g,N,M) for each 1≤l≤L;(ii)
there exists a functionA:ℕM/q×[0,1)→ℳQ,Q such that
(78)Ψh(j,v)=A(j,v)Ψg(j,v)forj∈ℕM/q and a.e. v∈[0,1);(iii)
there exists a functionA:ℤ×ℝ→ℳQ,Q such that
(79)Ψh(j,v)=A(j,v)Ψg(j,v)forj∈ℤ and a.e. v∈ℝ.Lemma 28.
Giveng=(g1,g2,…,gL) with each gl∈l2(ℤ), h=(h1,h2,…,hL) with each hl∈l2(ℤ), let G(g,N,M) and G(h,N,M) be both Bessel sequences in l2(ℤ). Then the following are equivalent:(i)
range(𝒯h*)⊂range(𝒯g*)-;(ii)
there exists a functionB:ℕM/q×[0,1)→ℳp,p such that
(80)Ψh(j,v)=Ψg(j,v)B(j,v)for(j,v)∈ℕM/q×[0,1);(iii)
there exists a functionB:ℤ×ℝ→ℳp,p such that
(81)Ψh(j,v)=Ψg(j,v)B(j,v)for(j,v)∈ℤ×ℝ.Proof.
If (ii) holds, we can obtain (iii) by choosingB:ℤ×ℝ→ℳp,p as
(82)B(j+Mqn,v)=𝒟n*(v)B(j,v)𝒟n(v)
for (j,v)∈ℕM/q×ℝ, where 𝒟n(v) is defined as in Lemma 12. So we only need to prove that (i) holds if and only if there exists a function B:ℕM×[0,1)→ℳp,p such that
(83)Ψh(j,v)=Ψg(j,v)B(j,v)for(j,v)∈ℕM×[0,1).
For c∈l2(ℕM×ℤ,ℂL), c is orthogonal to range(𝒯g*) (range(𝒯h*)) if and only if
(84)𝒥c(j,v)⊥Ψg(j,v)-ZqNf(j,v)(𝒥c(j,v)⊥Ψh(j,v)-ZqNf(j,v))
in L2(ℕM×[0,1),ℂQ)-inner product by Lemmas 9 and 17, equivalently,
(85)Ψg*(j,v)-𝒥c(j,v)⊥ZqNf(j,v)(Ψh*(j,v)-𝒥c(j,v)⊥ZqNf(j,v))forf∈l2(ℤ)
in L2(ℕM×[0,1),ℂp)-inner. Again by Lemma 6, (85) is equivalent to
(86)Ψg*(j,v)-𝒥c(j,v)=0(Ψh*(j,v)-𝒥c(j,v)=0)forj∈ℕM,a.e.v∈[0,1).
So we only need to prove that (83) holds if and only if, for 𝒥c(j,v)∈L2(ℕM×[0,1),ℂQ),
(87)Ψh*(j,v)-𝒥c(j,v)=0forj∈ℕM,a.e.v∈[0,1)
whenever
(88)Ψg*(j,v)-𝒥c(j,v)=0forj∈ℕM,a.e.v∈[0,1).
Obviously the necessity holds. Next we prove that a contradiction will occur if (83) is violated. Suppose (83) does not hold. Then there exist some j0∈ℕM and E⊂[0,1) with |E|>0, on which some r0th column hr0 of Ψh(j0,v)- satisfies
(89)hr0∉range(Ψg(j0,v)-)forv∈E.
Let 𝒫(j0,v) be the orthogonal projection of ℂQ onto ker(Ψg*(j0,v)-). Define 𝒥c(j,v)∈L2(ℕM×[0,1),ℂQ) by
(90)𝒥c(j,v)={𝒫(j0,v)hr0(j0,v),ifj=j0,v∈E,0,otherwise
for (j,v)∈ℕM×[0,1). Then 𝒥c(j,v) solves (88) and 𝒥c(j0,v)≠0 for v∈E. Also observe that the r0th component of Ψh*(j0,v)-𝒥c(j0,v) is exactly ∥𝒥c(j0,v)∥2 for v∈E. It follows that 𝒥c(j,v) fails to solve (87). This finishes the proof.Lemma 29.
Giveng=(g1,g2,…,gL), h=(h1,h2,…,hL) with each gl, hl∈l2(ℤ), let G(g,N,M) and G(h,N,M) be both Bessel sequences in l2(ℤ). Then
(91)𝒮h,gf=f
for f∈ℳ(g,N,M) if and only if
(92)Ψg*(j,v)=MΨg*(j,v)Ψh(j,v)Ψg*(j,v)
for j∈ℕM/q and a.e. v∈[0,1).Proof.
By Lemma12, (92) holds for j∈ℕM/q and a.e. v∈[0,1) if and only if it holds for j∈ℕM and a.e. v∈[0,1). Next we prove that (91) holds for f∈ℳ(g,N,M) if and only if (92) holds for j∈ℕM and a.e. v∈[0,1).
Sincerange(𝒯g) is dense in ℳ(g,N,M), (91) holds for f∈ℳ(g,N,M) if and only if it holds for f∈range(𝒯g), equivalently,
(93)Ψg*(j,v)𝒥c(j,v)=MΨg*(j,v)Ψh(j,v)Ψg*(j,v)𝒥c(j,v)
for c∈l2(ℕM×ℤ,ℂL) by Remark 11 and Theorem 18. This is also equivalent to
(94)Ψg*(j,v)d(j,v)=MΨg*(j,v)Ψh(j,v)Ψg*(j,v)d(j,v)
for d(j,v)∈L2(ℕM×[0,1),ℂQ), j∈ℕM, and a.e. v∈[0,1) by Lemma 9. It is obvious that (94) holds if (92) holds for j∈ℕM and a.e. v∈[0,1). Now suppose (94) holds. For an arbitrarily fixed x∈ℂQ, choose d(j,v) as
(95)d(j,v)=xfor(j,v)∈ℕM×[0,1).
Then d(j,v)∈L2(ℕM×[0,1),ℂQ), and thus Ψg*(j,v)x=MΨg*(j,v)Ψh(j,v)Ψg*(j,v)x for j∈ℕM and a.e. v∈[0,1) by (94). So (92) holds for j∈ℕM and a.e. v∈[0,1) by the arbitrariness of x. The proof is completed.By Lemmas27–29, we have the following theorem which characterizes the Gabor duals of type I (resp., type II).Theorem 30.
Giveng=(g1,g2,…,gL) with each gl∈l2(ℤ), let G(g,N,M) be a frame for ℳ(g,N,M). Then, for any Bessel sequence G(h,N,M) in l2(ℤ), G(h,N,M) is a Gabor dual of type I (type II) for G(g,N,M) if and only if the following hold:(i)
there existsA:ℕM/q×[0,1)→ℳQ,Q(B:ℕM/q×[0,1)→ℳp,p) such that
(96)Ψh(j,v)=A(j,v)Ψg(j,v)(Ψh(j,v)=Ψg(j,v)B(j,v));(ii)
Ψg*(j,v)=MΨg*(j,v)Ψh(j,v)Ψg*(j,v) for j∈ℕM/q and a.e. v∈[0,1).Theorem 31.
Giveng=(g1,g2,…,gL) with each gl∈l2(ℤ), let G(g,N,M) be a frame for ℳ(g,N,M). Then, for any Bessel sequence G(h,N,M) in l2(ℤ),(i)
G(h,N,M) is a Gabor dual of type I for G(g,N,M) if and only if there exists 𝒜:ℕM/q×[0,1)→ℳQ,Q such that
(97)Ψh(j,v)=1M(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)×[I-MΨg*(j,v)𝒜(j,v)Ψg(j,v)]+𝒜(j,v)Ψg(j,v)
for j∈ℕM/q and a.e. v∈[0,1);(ii)
G(h,N,M) is a Gabor dual of type II for G(g,N,M) if
(98)Ψh(j,v)=1MΨg(j,v)(Ψg*(j,v)Ψg(j,v))†×[I-MΨg*(j,v)Ψg(j,v)𝒜(j,v)]+Ψg(j,v)𝒜(j,v)
for j∈ℕM/q and a.e. v∈[0,1);(iii)
G(h,N,M) is an oblique dual of G(g,N,M) if one of the following conditions holds:
(a)
Ψh(j,v)=(1/M)Ψg(j,v)(Ψg*(j,v)Ψg(j,v))†[I-MΨg*(j,v)𝒜(j,v)Ψg(j,v)]+𝒜(j,v)Ψg(j,v) for j∈ℕM/q and a.e. v∈[0,1);(b)
Ψh(j,v)=(1/M)(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)[I-MΨg*(j,v)Ψg(j,v)𝒜(j,v)]+Ψg(j,v)𝒜(j,v) for j∈ℕM/q and a.e. v∈[0,1).Proof.
The items (ii) and (iii) can be proved similarly to item (i). Next we prove item (i). First we assume that (97) holds. Then
(99)MΨg*(j,v)Ψh(j,v)Ψg*(j,v)=Ψg*(j,v)(Ψg(j,v)Ψg*(j,v))†×Ψg(j,v)Ψg*(j,v)-MΨg*(j,v)(Ψg(j,v)Ψg*(j,v))†×Ψg(j,v)Ψg*(j,v)𝒜(j,v)×Ψg(j,v)Ψg*(j,v)+MΨg*(j,v)𝒜(j,v)Ψg(j,v)Ψg*(j,v)
for j∈ℕM/q and a.e. v∈[0,1). Also observe that (Ψg(j,v)Ψg*(j,v))†Ψg(j,v)Ψg*(j,v) is the projection from ℂQ onto range(Ψg(j,v)). It follows that MΨg*(j,v)Ψh(j,v)Ψg*(j,v)=Ψg*(j,v), and thus G(h,N,M) is a Gabor dual of type I for G(g,N,M) by Theorem 30.
Now we turn to the converse implication. SupposeG(h,N,M) is a Gabor dual of type I for G(g,N,M). Then there exists B(j,v):ℕM/q×[0,1)→ℳQ,Q such that
(100)Ψh(j,v)=B(j,v)Ψg(j,v),Ψg*(j,v)=MΨg*(j,v)Ψh(j,v)Ψg*(j,v)
for j∈ℕM/q and a.e. v∈[0,1). It follows that
(101)1M(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)Ψg*(j,v)=(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)×Ψg*(j,v)Ψh(j,v)Ψg*(j,v)
for j∈ℕM/q and a.e. v∈[0,1), and thus
(102)1M(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)=(Ψg(j,v)Ψg*(j,v))†×Ψg(j,v)Ψg*(j,v)Ψh(j,v)
for j∈ℕM/q and a.e. v∈[0,1) due to the fact that ℂp=range(Ψg*(j,v))⊕ker(Ψg(j,v)). Put 𝒜(j,v)=B(j,v)-(Ψg(j,v)Ψg*(j,v))†. Then the right-hand side of (97) equals
(103)Ψh(j,v)+(Ψg(j,v)Ψg*(j,v))†𝒫range(Ψg(j,v))Ψg(j,v)-(Ψg(j,v)Ψg*(j,v))†Ψg(j,v)
by (102), and this is exactly Ψh(j,v) for j∈ℕM/q and a.e. v∈[0,1) again by the fact that
(104)Ψg(j,v)Ψg*(j,v)(Ψg(j,v)Ψg*(j,v))†=𝒫range(Ψg(j,v)).


## 5. The Case ofN1=N2=⋯=NL

Theorems30 and 31 characterize duals with Gabor structure for general N=(N1,N2,…,NL). This section deals with the case N1=N2=⋯=NL. We start with a remark on the case of Nl, 1≤l≤L, being not all the same.Remark 32.
Not every subspace Gabor frameG(g,N,M) admits an oblique Gabor dual.We show it by revisiting Example25. Let us make an additional assumption that λ(0,v)=λ(1,v)=0 and a(0,v)a(1,v)≠0 for v∈[0,1]. Then G(g,N,M) is a frame but not a Riesz basis for ℳ(g,N,M) by Remark 26. Suppose G(h,N,M) with h=(h1,h2) is an oblique Gabor dual for G(g,N,M). Then
(105)Ψg*(j,v)=2Ψg*(j,v)Ψh(j,v)Ψg*(j,v)aaaaaaaaaaaforj∈ℕ2,a.e.v∈[0,1)
by Theorem 30. Writing out (0,0), (3,1), and (5,4) entries of both sides, we have
(106)2(a(j,v)-)2𝒵12h1(j,v)=a(j,v)-,(107)2(a(j,v)-)2𝒵12h1(j,v)+2|a(j,v)|2𝒵12h2(j+2,v)=a(j,v)-,(108)2(a(j,v))2𝒵12h2(j+2,v)=a(j,v)
for j∈ℕ2 and a.e. v∈[0,1). By (107) and (108), we have 2a(j,v)-𝒵12h1(j,v)=0 for j∈ℕ2 and a.e. v∈[0,1). This contradicts (106).Observe thatN1≠N2 in Remark 32 (N1=6 and N2=4). It is natural to ask the following question.Does every subspace Gabor frameG(g,N,M) admit no oblique Gabor dual whenever Nl, 1≤l≤L, are not all the same?The following example gives a negative answer to this question.Example 33.
LetN=(1,2) and M=(3,3). Assume that 𝒥(v)=(𝒥1(v)𝒥2(v)), ℰ(v)=(ℰ1(v)ℰ2(v)) have the form
(109)𝒥1(v)=(a0,0(v)a0,1(v)a1,0(v)a1,1(v)a2,0(v)a2,1(v)a2,1(v)e-2πiva2,0(v)e2πiva0,1(v)a0,0(v)e2πiva1,1(v)a1,0(v)),ℰ1(v)=(c0,0(v)c0,1(v)c1,0(v)c1,1(v)c2,0(v)c2,1(v)c2,1(v)e-2πivc2,0(v)e2πivc0,1(v)c0,0(v)e2πivc1,1(v)c1,0(v)),𝒥2(v)=(b0,0(v)b0,1(v)b1,0(v)b1,1(v)b2,0(v)b2,1(v)),ℰ2(v)=(d0,0(v)d0,1(v)d1,0(v)d1,1(v)d2,0(v)d2,1(v)),
for v∈[0,1) with all entries of 𝒥(v) and ℰ(v) in L∞([0,1)), that (𝒥1(v))*ℰ1(v) has the form (A(v)00A(v)) and (𝒥(v))*𝒥(v) has the form (B(v)00B(v)) for v∈[0,1), and that
(110)(𝒥2(v))*ℰ2(v)=(13-A(v)0013-A(v))
for v∈[0,1) satisfying B(v)≠0. Define g=(g1,g2) and h=(h1,h2) by
(111)Ψg(0,v)=𝒥(v),Ψh(0,v)=ℰ(v)forv∈[0,1).
Then g and h are well defined by the quasi-periodicity of the Zak transform 𝒵6 and 6ℤ congruence between {0}-2ℕ3+3ℕ2 and ℕ6, and G(g,N,M) and G(h,N,M) are both Bessel sequences by Remark 16. A simple computation shows that
(112)Ψg*(0,v)Ψg(0,v)=3Ψg*(0,v)Ψh(0,v)Ψg*(0,v)Ψg(0,v)
for v∈[0,1). This implies that Ψg*(0,v)=3Ψg*(0,v)Ψh(0,v)Ψg*(0,v) for v∈[0,1) due to the fact that ℂ9=range(Ψg(0,v))⊕ker(Ψg*(0,v)). So G(h,N,M) is an oblique Gabor dual for G(h,N,M) by Lemma 29.Remark32 and Example 33 show that the Gabor dual theory for subspace Gabor frames is complicated when Nl, 1≤l≤L, are not all the same. There are still many unresolved problems in this direction. Next we work under the following assumption.Assumption 34.
L is a positive integer, M=(M,M,…,M), and N=(N1,N2,…,NL) with N1=N2=⋯=NL=N.The following theorem characterizes the uniqueness of such duals.Theorem 35.
Giveng=(g1,g2,…,gL) with each gl∈l2(ℤ), let G(g,N,M) be a frame for ℳ(g,N,M). Then the following are equivalent:(i)
G(g,N,M) has a unique Gabor dual of type I (type II);(ii)
rank(Ψg(j,v))∈{0,Lq}(rank(Ψg(j,v))∈{0,p}) for a.e. (j,v)∈ℕM/q×[0,1);(iii)
rank(Ψg(j,v))∈{0,Lq}(rank(Ψg(j,v))∈{0,p}) for a.e. (j,v)∈ℤ×ℝ.Proof.
We first deal with the Gabor dual of type I. By Lemma12, (ii) and (iii) are equivalent. By Theorem 30 and Lemma 27, we only need to prove that (ii) holds if and only if a function A:ℕM×[0,1)→ℳLq,Lq solves
(113)A(j,v)Ψg(j,v)=0forj∈ℕM,a.e.v∈[0,1)
whenever it solves
(114)Ψg*(j,v)A(j,v)Ψg(j,v)Ψg*(j,v)=0forj∈ℕM,a.e.v∈[0,1).
Suppose (ii) holds and A:ℕM×[0,1)→ℳLq,Lq solves (114). If (j,v) is such that rank(Ψg(j,v))=0, then Ψg(j,v)=0, and thus A(j,v)Ψg(j,v)=0. If (j,v) is such that rank(Ψg(j,v))=Lq, then (114) implies that
(115)A(j,v)Ψg(j,v)Ψg*(j,v)=0,
which leads to A(j,v)=0 and thus A(j,v)Ψg(j,v)=0. Conversely, suppose (ii) does not hold. Then there exist some j0∈ℕM and ek, er such that
(116)𝒫(j0,v)ek≠0,er*Ψg(j0,v)≠0
on some E⊂[0,1) with |E|>0, where 𝒫(j0,v) is the orthogonal projection of ℂLq onto ker(Ψg*(j0,v)). Define
(117)A(j0,v)={𝒫(j0,v)eker*,ifj=j0,v∈E,0,otherwise
for (j,v)∈ℕM×[0,1). Then (114) holds, but (113) fails to hold.
Next we turn to the Gabor duals of type II. Similarly to the above arguments, by Theorem30 and Lemmas 27 and 28, we only need to prove that (ii) holds if and only if a function B:ℕM×[0,1)→ℳp,p solves
(118)Ψg(j,v)B(j,v)=0forj∈ℕM,a.e.v∈[0,1)
whenever it solves
(119)Ψg*(j,v)Ψg(j,v)B(j,v)Ψg*(j,v)=0forj∈ℕM,a.e.v∈[0,1).
The necessity can be proved similarly to the case of type I. Now we suppose (ii) does not hold. Then there exist some j0∈ℕM and ek, er such that
(120)𝒫(j0,v)ek≠0,Ψg(j0,v)er≠0
on some E⊂[0,1) with |E|>0, where 𝒫(j0,v) is the orthogonal projection of ℂp onto ker(Ψg(j0,v)). Define B(j0,v) by
(121)B*(j0,v)={𝒫(j0,v)eker*,ifj=j0,v∈E,0,otherwise
for (j,v)∈ℕM×[0,1). Then (119) holds, but (118) fails to hold. The proof is completed.It is well known that every scalark×n matrix M with rank r≥1 has a decomposition
(122)M=U(D000)V*,
where U and V are, respectively, k×k and n×n unitary matrices, (D000) is a k×n block matrix in which D is an r×r diagonal matrix with positive entries in the diagonal (see [1, Theorem 1.5.4]). Observe that rank(M) may change in variables if M is a matrix-valued function. Next we restrict ourselves to Gabor systems G(g,N,M) such that
(123)Ψg(j,v)=U(j,v)(D(j,v)000)V*(j,v)aaaaaaaaaaaaforj∈ℕM/q,a.e.v∈[0,1),
where U(j,v) and V(j,v) are, respectively, Lq×Lq and p×p unitary matrices for j∈ℕM/q and a.e. v∈[0,1), and (D(j,v)000) is an Lq×p block matrix in which D(j,v) is of the form
(124)D(j,v)=diag(λ1(j,v),λ2(j,v),…,λm(j,v))
with λi(j,v)≥0 for j∈ℕM/q and a.e. v∈[0,1) and ⋃j∈ℕM/q{v∈[0,1):λi(j,v)>0} being of positive measure for each 1≤i≤m.As an immediate consequence of Theorems19, 24, and 35, we have the following theorem.Theorem 36.
Let0≠g=(g1,g2,…,gL) be defined as in (123). Then(i)
G(g,N,M) is a frame for ℳ(g,N,M) with frame bounds A, B if and only if
(125)AM≤λi(j,v)≤BMon{(j,v)∈ℕM/q×[0,1):λi(j,v)>0}for1≤i≤m;(ii)
G(g,N,M) is a Riesz basis (an orthonormal basis) for ℳ(g,N,M) with Riesz bounds A, B if and only if
(126)m=Lq,AM≤λi(j,v)≤BM(λi(j,v)=1M)for1≤i≤m, j∈ℕM/q, and a.e. v∈[0,1);(iii)
G(g,N,M) is a frame for ℳ(g,N,M) with frame bounds A and B and simultaneously admits a unique Gabor dual of type I (type II) if and only if m=Lq (m=p), and for each j∈ℕM/q and a.e. v∈[0,1), either λi(j,v)=0 for each 1≤i≤m or A/M≤λi(j,v)≤B/M for each 1≤i≤m.Theorem 37.
Letg be defined as in (123) and let G(g,N,M) be a frame for ℳ(g,N,M). Then, for h=(hl,h2,…,hL) with each hl∈l2(ℤ), we have(i)
G(h,N,M) is a Gabor dual of type I for G(g,N,M) if and only if Ψh has the form
(127)Ψh(j,v)=1MU(j,v)C(j,v)(D(j,v)000)V*(j,v)forj∈ℕM/q,a.e.v∈[0,1),whereC:ℕM/q×[0,1)→ℳLq,Lq satisfies C(j,v)=(C1(j,v),0) with C1(j,v)∈ℳLq,m, the entries of the ith column of C1(j,v) are essentially bounded on {(j,v)∈ℕM/q×[0,1):λi(j,v)>0} for 1≤i≤m, and
(128)(C(j,v))i,n={0,i≠n,λi(j,v)λn(j,v)≠0,(λi(j,v))-2,i=n,λi(j,v)≠0for1≤i, n≤m;(ii)
G(h,N,M) is a Gabor dual of type II for G(g,N,M) if and only if Ψh has the form
(129)Ψh(j,v)=1MU(j,v)(D(j,v)000)C(j,v)V*(j,v)forj∈ℕM/q,a.e.v∈[0,1),where C:ℕM/q×[0,1)→ℳp,p satisfies C(j,v)=(C1(j,v)0) with C1(j,v)∈ℳm,p, the entries of the ith row of C1(j,v) are essentially bounded on {(j,v)∈ℕM/q×[0,1):λi(j,v)>0} for 1≤i≤m, and
(130)(C(j,v))i,n={0,i≠n,λi(j,v)λn(j,v)≠0,(λi(j,v))-2,i=n,λi(j,v)≠0
for 1≤i, n≤m.Proof.
We only prove item (i), and (ii) can be proved similarly. By Theorem30, G(h,N,M) is a Gabor dual of type I for G(g,N,M) if and only if there exists a function A:ℕM/q×[0,1)→ℳLq,Lq such that
(131)Ψh(j,v)=A(j,v)Ψg(j,v),(132)Ψg*(j,v)=MΨg*(j,v)A(j,v)Ψg(j,v)Ψg*(j,v)
for j∈ℕM/q and a.e. [0,1), and for each j∈ℕM/q the entries of Ψh(j,v) belong to L∞([0,1)). Write
(133)C(j,v)=MU*(j,v)A(j,v)U(j,v)=(C1(j,v),C2(j,v)),
where C1(j,v)∈ℳLq,m. A simple computation shows that (132) is equivalent to
(134)D(j,v)=D(j,v)C11(j,v)D2(j,v)forj∈ℕM/q,a.e.v∈[0,1),
where C1(j,v)=(C11(j,v)C21(j,v)), C11(j,v)∈ℳm,m. It is obvious that (134) is equivalent to (128), which implies that (132) is equivalent to (128). By (133), (132) can be rewritten as
(135)Ψh(j,v)=1MU(j,v)(C1(j,v)D(j,v),0)V*(j,v).
Since C2(j,v) does not appear in (135), we assume that C2(j,v)=0 without loss of generality. By the unitary properties of U(j,v) and V(j,v), for each j∈ℕM/q, the entries of Ψh(j,v) belong to L∞([0,1)) if and only if the entries of C1(j,v)D(j,v) belong to L∞([0,1)); that is, the entries of the ith column of C1(j,v) are essentially bounded on {(j,v)∈ℕM/q×[0,1):λi(j,v)>0} for 1≤i≤m. The proof is completed.Next we discuss a special case of Theorem36. Let N=1, M=2, 1→=(1,1), and 2→=(2,2) and write  V1(0,v)=V~(0,v)=1 and
(136)U1(0,v)=11+cos22πv(-1cos2πvcos2πv1),D1(0,v)=(1+cos22πv0),U~(0,v)=13(-1i-icos2πv-isin2πvcos2πvisin2πv-ii-1-i-isin2πvicos2πv-isin2πv-cos2πv-1-1)D~(0,v)=(3000)
for v∈[0,1). Define g  and  g~=(g~1,g~2) as
(137)Ψg(0,v)=U1(0,v)D1(0,v)V1*(0,v),Ψg~(0,v)=U~(0,v)D~(0,v)V~*(0,v)
by letting U=U1, D=D1, and V=V1 and U=U~, D=D~, and V=V~ in (123), respectively. Then
(138)Ψg(0,v)=(-1cos2πv),Ψg~(0,v)=(-1cos2πv-1-isin2πv)forv∈[0,1).
This implies that g=g~1=-χ{0}+(1/2)χ{-3,1}, g~2=-χ{0}-(1/2)χ{1}+(1/2)χ{-3} and that
(139)ℳ(g,1,2)=ℳ(g~,1→,2→)=l2(ℤ)
by [5, Theorem 3.1]. So G(g,1,2) and G(g~,1→,2→) are both frames for l2(ℤ), but neither is a Riesz basis by Theorem 36. Define h∈l2(ℤ) and h~=(h~1,h~2) with h~1,h~2∈l2(ℤ) by Ψh(0,v)=(1/2)Ψg(0,v)(Ψg*(0,v)Ψg(0,v))† and Ψh~(0,v)=(1/2)Ψg~(0,v)(Ψg~*(0,v)Ψg~(0,v))† for v∈[0,1). Then G(h,1,2) and G(h~,1→,2→) are, respectively, the canonical duals for G(g,1,2) and G(g~,1→,2→) by [5, Remark 5.5], and
(140)Ψh(0,v)=12(1+cos22πv)(-1cos2πv),(141)Ψh~(0,v)=16(-1cos2πv-1-isin2πv)
for v∈[0,1). It follows that
(142)h~=(-16χ{0}+112χ{-3,1},-16χ{0}-112χ{1}+112χ{-3}),
and h is infinitely supported. It is interesting that g~ and h~ are both finitely supported, but h is not although g is the first component of g~. Therefore, there exist significant differences in Gabor duals between multi-window Gabor frames and single-window ones. Now we conclude this paper by summarizing the above arguments as the following example.Example 38.
Let1→=(1,1) and 2→=(2,2),
(143)g=g~1=-χ{0}+12χ{-3,1},g~2=-χ{0}-12χ{1}+12χ{-3}.h and h~, as in (140) and (142) and write g~=(g~1,g~2). Then(i)
G(g,1,2) and G(g~,1→,2→) are both frames for l2(ℤ), but neither is a Riesz basis for l2(ℤ);(ii)
G(h,1,2) and G(h~,1→,2→) are, respectively, the canonical duals for G(g,1,2) and G(g~,1→,2→);(iii)
h~ is finitely supported, and h is not.



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*Source: 357242-2013-06-11.xml*