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. 2018 Oct 11;2018(1):280. doi: 10.1186/s13660-018-1876-7

Subspace mixed rational time-frequency multiwindow Gabor frames and their Gabor duals

Yan Zhang 1,, Yun-Zhang Li 2
PMCID: PMC6182425  PMID: 30363791

Abstract

For a usual multiwindow Gabor system, all windows share common time-frequency shifts. A mixed multiwindow Gabor system is one of its generalizations, for which time-frequency shifts vary with the windows. This paper addresses subspace mixed multiwindow Gabor systems with rational time-frequency product lattices. It is a continuation of (Li and Zhang in Abstr. Appl. Anal. 2013:357242, 2013; Zhang and Li in J. Korean Math. Soc. 51:897–918, 2014). In (Li and Zhang in Abstr. Appl. Anal. 2013:357242, 2013) we dealt with discrete subspace mixed Gabor systems and in (Zhang and Li in J. Korean Math. Soc. 51:897–918, 2014) with L2(R) ones. In this paper, using a suitable Zak transform matrix method, we characterize subspace mixed multiwindow Gabor frames and their Gabor duals, obtain explicit expressions of Gabor duals, and characterize the uniqueness of Gabor duals. We also provide some examples, which show that there exist significant differences between mixed multiwindow Gabor frames and usual multiwindow Gabor frames.

Keywords: Gabor frame, Mixed multiwindow Gabor frame, Dual, Oblique dual, Gabor dual

Introduction

Let H be a separable Hilbert space. An at most countable sequence {hi}iI in H is called a frame for H if there exist constants 0<AB< such that

Af2iI|f,hi|2Bf2for fH, 1

where A and B are called frame bounds; it 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. In this case, the frame bounds are also called Riesz bounds. The fundamentals of frames can be found in [36]. For λR, define the modulation operator Eλ and translation operator Tλ on L2(R) respectively by

Eλf()=e2πiλf()andTλf()=f(λ)

for fL2(R). This paper addresses Gabor systems of the form

G(g,a,b)={EmblTnalgl:m,nZ,1lL}, 2

where L is a fixed positive integer, g={gl:1lL}L2(R), a=(a1,a2,,aL), and b=(b1,b2,,bL) with al,bl>0, 1lL. We denote by M(g,a,b) the closed linear span of G(g,a,b) in L2(R). A Gabor system G(g,a,b) is called a single-window Gabor system if L=1; it is called a mixed multiwindow Gabor system if L>1 and al (or bl) with 1lL are not all the same; it is called a multiwindow Gabor system if L>1, a1=a2==aL, and b1=b2==bL. Similarly, G(g,a,b) is called a subspace single-window Gabor frame if it is a frame for M(g,a,b) and L=1; it is called a subspace mixed multiwindow Gabor frame if it is a frame for M(g,a,b), L>1, and al (or bl) with 1lL are not all the same; it is called a subspace multiwindow Gabor frame if it is a frame for M(g,a,b), L>1, a1=a2==aL, and b1=b2==bL. In particular, when M(g,a,b)=L2(R), these frames are usual frames, which have been extensively studied [712]. To distinguish from subspace frames, we call them whole space frames.

For a Bessel sequence G(g,a,b) in L2(R), define the associated synthesis operator Tg:l2(Z2,CL)L2(R) by

Tgc=l=1LmZnZcl,m,nEmblTnalgl 3

for c=(c1,c2,,cL)l2(Z2,CL). Then it is a bounded operator, and its adjoint operator Tg (so-called analysis operator) is given by

Tgf=(c1(f),c2(f),,cL(f))for fL2(R),

where cl(f)={f,EmblTnalgl}m,nZ for 1lL. Similarly, for a Bessel sequence G(h,a,b) in L2(R) with h={h1,h2,,hL}, we associate it with Th. Define Sh,g=TgTh, that is,

Sh,gf=l=1LmZnZf,EmblTnalhlEmblTnalgl

for fL2(R). Let G(g,a,b) be a frame for M(g,a,b), and let G(h,a,b) be a Bessel sequence in L2(R). Then G(h,a,b) is called an oblique Gabor dual for G(g,a,b) if

Sh,gf=ffor fM(g,a,b).

Here we do not require that hM(g,a,b). In particular, an oblique Gabor dual G(h,a,b) for G(g,a,b) is said to be a Gabor dual of type I for G(g,a,b) if hM(g,a,b), and it is said to be a Gabor dual of type II for G(g,a,b) if range(Th)range(Tg). These notions are borrowed from [13] and [14]. Observe that a Gabor dual of type II is not required to be in M(g,a,b), but a moment containment relation is required.

For the whole space Gabor frames, single-window ones have been extensively studied in the past twenty years and more [4, 5, 7, 9, 15, 16]. Multiwindow frames were firstly studied by Zibulski and Zeevi [10] and Zeevi, Zibulski, and Porat [11]. By introduction of a Zak transform they developed a matrix (so-called Zibulski–Zeevi matrix) algebraic tool for multiwindow Gabor frames and applied it to image processing and computer vision. Since then, many researchers have studied multiwindow Gabor frames and related applications [2, 1720]. It was also pointed out in [12] that the Zibulski–Zeevi matrix method is not very efficient for mixed Gabor frames. In [2], with the help of a new Zak transform matrix, different from the Zibulski–Zeevi matrix, Zhang and Li investigated mixed rational time-frequency multiwindow Gabor frames (Riesz bases and orthonormal bases) and their Gabor duals in L2(R). For subspace Gabor frames, single-window ones have been considered by several papers [1, 13, 2126]. In [24, 26], and [27], a Zak transform matrix different from the Zibulski–Zeevi matrix was introduced and used effectively to study Gabor systems on periodic subsets of the real line, whereas the Zibulski–Zeevi matrix method does not work well for such Gabor systems. A variation of this method was applied to Gabor systems on discrete periodic sets [28, 29]. In [30], a density result for Gabor frames on periodic subsets of Rd is obtained via the Haar measure of the group generated by lattices. In [31], subspace multiwindow Gabor frames and their Gabor duals were characterized. All works mentioned, except [1] and [2], have not concerned real mixed multiwindow Gabor systems. Motivated by these observations, this paper is devoted to studying mixed multiwindow Gabor systems of the form (2). We work under the following assumptions:

Assumption 1

L is a positive integer;

Assumption 2

b1=b2==bL=b, and alb=plql with pl and ql being relatively prime positive integers for 1lL.

We denote by N the set of positive integers. As it is pointed out in Remark 1.1 of [2], if b1,b2, , bL in (2) are commensurable (there exist b>0, β1,β2, , and βLN such that b=βlbl for 1lL), then G(g,a,b) is a frame (a Riesz basis, an orthonormal basis) for M(g,a,b) if and only if

{e2πimbgl(τl)(nal):1lL,0τlβl1,m,nZ}

is a frame (a Riesz basis, an orthonormal basis) for M(g,a,b), where gl(τl)()=e2πiτlblgl(). It is well known that b1,b2,,bL are commensurable if they are all rational numbers or rational multiples of some fixed irrational number. We also remark that the restriction of “rational time-frequency” here is for using “finite-order” Zak transform matrix-valued functions. So Assumptions 1 and 2 are relatively general and reasonable to some extent.

Throughout this paper, p and q denote the least common multiple of pl and the greatest common divisor of ql with 1lL, respectively. It is easy to check that p and q are relatively prime and that pq is the least common multiple of plql with 1lL. So, for each 1lL, there exists a unique λlN such that

pq=λlplql. 4

This implies that λlal=pbq for 1lL by Assumption 2. We write

a=pbqandQ=ql=1Lλl, 5

and we denote by Nt the set

Nt={0,1,,t1}

and by It the t×t identity matrix for tN. Hereinafter we use I to denote the identity matrix when we need not specify its size. Given a measurable set S in R, a collection {Sk:kZ} of measurable sets in R is called a partition of S if

kZSk=SandSkSk=for kk in Z

up to a set of measure zero. For λ>0 and measurable sets S, SR, we say that S is λZ-congruent to S if there exists a partition {Sk:kZ} of S such that {Sk+λk:kZ} is a partition of S. In particular, only finitely many Sk among Sk, kZ, are nonempty if, in addition, both S and S are bounded. Obviously, S is also λZ-congruent to S if S is λZ-congruent to S. So, in this case, we usually say that S and S are λZ-congruent. For s, tN, we denote by Ms,t the set of all s×t complex matrices. Let MMs,t, which we consider as a linear mapping from Ct into Cs, and define the mapping : ker(M) range(M) by M˜x=Mx for x(ker(M)). Then is a bijection, and thus it has an inverse (M˜)1. We extend (M˜)1 to M: CsCt by defining

M(y+z)=(M˜)1y,yrange(M),z(range(M)).

The mapping M is called the pseudo-inverse of M.

The rest of this paper is organized as follows. In Sect. 2, using a suitable Zak transform matrix method, we characterize subspace mixed multiwindow Gabor frames, their Gabor duals of types I and II, and the uniqueness of Gabor duals and obtain explicit expressions of the Gabor duals. In Sect. 3, we give some examples and remarks. They show that there exist significant differences between mixed multiwindow Gabor frames and usual multiwindow Gabor frames. In particular, not every subspace mixed multiwindow Gabor frame G(g,a,b) admits an oblique Gabor dual. So there should be many challenging problems in this direction.

Frame and dual characterization

Let L, a, and b satisfy Assumptions 1 and 2. In this section, using a Zak transform matrix method, we characterize the Gabor systems G(g,a,b) that are frames for M(g,a,b) and Gabor systems G(h,a,b) that are duals of a frame G(g,a,b) of types I and II. We also characterize the uniqueness of Gabor duals.

For fL2(R), define the Zak transform Zaqf of f by

(Zaqf)(t,v):=Zf(t+aq)e2πiv 6

for a.e. (t,v)R2 and define

Zaqf(t,v)=(Zaqf(t,v)Zaqf(t+1b,v)Zaqf(t+p1b,v))

for a.e. (t,v)R2. It is easy to check that the Zak transform has quasi-periodicity:

(Zaqf)(t+kaq,v+)=e2πikv(Zaqf)(t,v)

for fL2(R), (k,)Z2 and a.e. (t,v)E2.

By Lemma 2.1 in [24], and by Lemma 2.1 in [2] we have the following:

Lemma 2.1

  • (i)

    Zaq(EmbTnaq+x0f)(t,v)=e2πimbte2πinvZaqf(tx0,v) for fL2(R), x0R and a.e. (t,v)R2.

  • (ii)

    Zaq is a unitary operator from L2(E) onto L2([0,1b)×[0,1),Cp), and Zaq is a unitary operator from L2(E) onto L2(S×[0,1)) for an arbitrary subset S of R thatis aqZ-congruent to [0,aq).

Definition 2.1

For g={g1,g2,,gL}L2(R), we associate it with a matrix-valued function G(t,v):R2MQ,p by

G(t,v)=(G1(t,v)G2(t,v)GL(t,v)), 7

where Gl(t,v) is a block matrix of the form

Gl(t,v)=(Gl(t,v)Gl(tal,v)Gl(t(λl1)al,v)) 8

with Gl(t,v):R2Mq,p for 1lL and

Gl(t,v)r,k=Zaqgl(tra+kb,v)

for rNq, kNp.

Remark 2.1

By the quasi-periodicity of Zaq, for an arbitrary fL2(R), Zaqf is uniquely determined by the values of (Zaqf)(,) on S×[0,1) with S being a set aqZ-congruent to [0,aq). So, by Lemma 2.1(ii) an arbitrary function FL2(S×[0,1)) determines a unique fL2(R) by

(Zaqf)(t,v)=F(t,v)for (t,v)S×[0,1).

Observe that [0,1bq)aNq+1bNp is aqZ-congruent to [0,aq). It is easy to prove that if a1=a2==aL, then Q=Lq, and an arbitrary function M(t,v):[0,1bq)×[0,1)MLq,p with all entries in L2([0,1bq)×[0,1)) determines a unique g={g1,g2,,gL}L2(R) by

G(t,v)=M(t,v)for a.e. (t,v)[0,1bq)×[0,1).

However, it is not the case if al, 1lL, are not all the same. By an argument similar to that in [1], we have Example 2.1, which provides us with a counterexample. Therefore, we must be careful when we define g by a function M(t,v):[0,1bq)×[0,1)MQ,p via

G(t,v)=M(t,v)for a.e. (t,v)[0,1bq)×[0,1)

if al, 1lL, are not all the same.

Example 2.1

Suppose al, 1lL, are not all the same. Then there exists 1lL such that λl>1. We may as well assume that λ1>1. Choose

M(t,v)=(M1(t,v)M2(t,v)ML(t,v)),Ml(t,v)=(Ml,0(t,v)Ml,1(t,v)Ml,λl1(t,v))

for 1lL and (t,v)[0,1bq)×[0,1) such that every entry of M(t,v) belongs to L2([0,1bq)×[0,1)) and

M1,0(t,v)0,M1,1(t,v)=0for (t,v)[0,1bq)×[0,1). 9

Suppose there exists g such that G(t,v)=M(t,v) for a.e. (t,v)[0,1bq)×[0,1). Then

G1(t,v)=M1,0(t,v), 10
G1(ta1,v)=M1,1(t,v) 11

for a.e. (t,v)[0,1bq)×[0,1), where

G1(t,v)r,k=Zaqg1(tra+kb,v),G1(ta1,v)r,k=Zaqg1(ta1ra+kb,v)

for (r,k)Nq×Np. Since the sets [0,1bq)aNq+1bNp and [0,1bq)aNq+1bNpa1 are both aqZ-congruent to [0,aq), by the quasi-periodicity of Zaq we have Zaqg1(t,v)0 for (t,v)E and some E[0,aq)×[0,1) with |E|>0 by (9), (10), whereas Zaqg1(t,v)=0 for a.e. (t,v)[0,aq)×[0,1) by (9) and (11). This is a contradiction.

Define the Fourier transform F: l2(Z2)L2([0,1b)×[0,1)) by

Fc(t,v)=bmZnZcm,ne2πimbte2πinv

for cl2(Z2) and a.e. (t,v)[0,1b)×[0,1), and define J: l2(Z2,CL)L2([0,1b)×[0,1),CQ) by

Jc(t,v)=C(t,v)

for c=(c1,c2,,cL)l2(Z2,CL), where

C(t,v)=(C1(t,v)C2(t,v)CL(t,v)),Cl(t,v)=(Cl(0)(t,v)Cl(1)(t,v)Cl(λl1)(t,v)), 12
Cl(βl)(t,v)=(mZnZcl,m,nqλl+βle2πimbte2πinvmZnZcl,m,(nq+1)λl+βle2πimbte2πinvmZnZcl,m,(nq+q1)λl+βle2πimbte2πinv) 13

for 1lL and βlNλl. Similarly, for an arbitrary dl2(Z2,CL), we associate it with D(t,v).

By a standard argument, we have

Lemma 2.2

The operators F and bJ are unitary operators from l2(Z2,CL) onto L2([0,1b)×[0,1)) and L2([0,1b)×[0,1),CQ), respectively.

By Lemmas 2.4 and 4.1 and by Remarks 2.6 and 2.7 in [31], we have following two lemmas.

Lemma 2.3

For g={g1,g2,,gL}L2(R), the following are equivalent:

  • (i)

    G(g,a,b) is a Bessel sequence in L2(R) with Bessel bound B.

  • (ii)

    G(t,v)G(t,v)bBI for a.e. (t,v)[0,1bq)×[0,1).

  • (iii)

    (G(t,v)G(t,v))2bBG(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

  • (iv)

    G(t,v)G(t,v)bBI for a.e. (t,v)[0,1bq)×[0,1).

  • (v)

    (G(t,v)G(t,v))2bBG(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

  • (vi)

    ZaqglL(R2) for 1lL.

Lemma 2.4

For g={g1,g2,,gL}L2(R), G(g,a,b) is complete in L2(R) if and only if

rank(G(t,v))=pffor a.e. (t,v)[0,1bq)×[0,1).

Lemma 2.5

For g={g1,g2,,gL}L2(R), we have

  • (i)
    f,EmbTnaq+ra+τalgl=01b01(Gl(tτal,v)Zaqf(t,v))re2πimbte2πinvdtdv
    for fL2(R), 1lL and (m,n,r,τ)Z×Z×Nq×Nλl.
  • (ii)
    G(t+nbq,v)=e2πimnvCnG(t,v)Dn for n=knq+(mnqrn)p with (kn,rn,mn)Np×Nq×Z and a.e. (t,v)R2, where
    Dn=(0e2πivIknIpkn0),
    Cn=diag(C1,n(v),C2,n(v),,CL,n(v)), Cl,n(v) denotes the block matrix (with λl blocks) of the form diag(Cn,Cn,,Cn) with
    Cn=(0Iqrne2πivIrn0).
  • (iii)

    ZaqTgc(t,v)=G(t,v)Jc(t,v) for a.e. (t,v)R2 whenever cl0(Z2,CL) (that is, c is finitely supported) or G(g,a,b) is a Bessel sequence in L2(R), where Tgc is as in (3).

  • (iv)
    If G(g,a,b) is a Bessel sequence in L2(R), then
    JTgf(t,v)=1bG(t,v)Zaqf(t,v)
    for fL2(R) and a.e. (t,v)R2.

Proof

(i), (ii), and (iii) are from Lemmas 2.2, 2.3, 2.5 in [2]. Next, we prove (iv). Write d=Tgf=(d1,d2,,dL). By Lemma 2.1 we have

dl,m,(nq+r)λl+βl=kNp01b01Zaqf(t+kb,v)Zaqgl(tβlalra+kb,v)e2πimbte2πinvdtdv=01b01(Gl(tβlal,v)Zaqf(t,v))re2πimbte2πinvdtdv 14

for (r,βl)Nq×Nλl, 1lL. When G(g,a,b) is a Bessel sequence, the integrand in (14) belongs to L2([0,1b)×[0,1)) by Lemma 2.3(vi). It follows that

D(βl)(t,v)=1bGl(tβlal,v)Zaqf(t,v)

for a.e (t,v)[0,1b)×[0,1). This leads to the lemma. □

Remark 2.2

By Lemma 2.5(ii),

rank(G(t+nbq,v))=rank(G(t,v)),
G(t+nbq,v)G(t+nbq,v)=DnG(t,v)G(t,v)Dn,
G(t+nbq,v)G(t+nbq,v)=CnG(t,v)G(t,v)Cn

for (t,v)R2 and nZ. It follows that the range of rank(G(t,v)) on [0,1bq)×[0,1) is that on R2, and the spectrum properties of G(t,v)G(t,v) and G(t,v)G(t,v) on [0,1bq)×[0,1) determine their spectrum properties on R2. For simplicity, all theorems further will be stated on [0,1bq)×[0,1).

By an argument similar to Lemmas 27, 28 in [1] and Lemmas 3.3, 3.4 in [31], we have the following two lemmas.

Lemma 2.6

For g={g1,g2,,gL}, h={h1,h2,,hL}L2(R), the following are equivalent:

  • (i)

    hM(g,a,b).

  • (ii)

    there exists a measurable function A:[0,1bq)×[0,1)MQ,Q such that H(t,v)=A(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

  • (iii)

    there exists a measurable function A:R2MQ,Q such that H(t,v)=A(t,v)G(t,v) for a.e. (t,v)R2.

Lemma 2.7

Given g={g1,g2,,gL}, h={h1,h2,,hL}L2(R), let G(g,a,b) and G(h,a,b) be Bessel sequences in L2(R). Then the following are equivalent:

  • (i)

    range(Th)range(Tg).

  • (ii)
    there exists a measurable function B:[0,1bq)×[0,1)Mp,p such that
    H(t,v)=G(t,v)B(t,v)for a.e. (t,v)[0,1bq)×[0,1).
  • (iii)
    there exists a measurable function B:R2Mp,p such that
    H(t,v)=G(t,v)B(t,v)for a.e. (t,v)R2.

Lemma 2.8

Given g={g1,g2,,gL}, h={h1,h2,,hL}L2(R), let G(g,a,b) and G(h,a,b) be Bessel sequences in L2(R). Then

Zaq(Sh,gf)(t,v)=1bG(t,v)H(t,v)Zaqf(t,v) 15

for fL2(R) and a.e. (t,v)R2.

Proof

Since Sh,g=TgTh, applying Lemma 2.5(iii), (iv) leads to the lemma. □

Lemma 2.9

Given g={g1,g2,,gL}, h={h1,h2,,hL}L2(R), let G(g,a,b) and G(h,a,b) be Bessel sequences in L2(R). Then the following are equivalent:

  • (i)

    G(h,a,b) is an oblique Gabor dual for G(g,a,b).

  • (ii)

    G(t,v)=1bG(t,v)H(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

  • (iii)

    G(t,v)=1bG(t,v)H(t,v)G(t,v) for a.e. (t,v)R2.

Proof

By Lemma 2.5(ii), (ii) and (iii) are equivalent. So, to prove the lemma, it suffices to prove the equivalence between (i) and the following equation:

G(t,v)=1bG(t,v)H(t,v)G(t,v)for a.e. (t,v)[0,1b)×[0,1). 16

Since range(Tg) is dense in M(g,a,b), (i) holds if and only if

Sh,gTgc=Tgcfor cl2(Z2,CL)

or, equivalently,

G(t,v)Jc(t,v)=1bG(t,v)H(t,v)G(t,v)Jc(t,v)for cl2(Z2,CL)

by Lemmas 2.1, 2.5, and 2.8, which is in turn equivalent to

G(t,v)d(t,v)=1bG(t,v)H(t,v)G(t,v)d(t,v) 17

for d(t,v)L2([0,1b)×[0,1),CQ) by Lemma 2.2. Obviously, (16) implies (17). Now suppose (17) holds. For arbitrary fixed xCQ, choose d(t,v) as

d(t,v)=xfor (t,v)[0,1b)×[0,1).

Then d(t,v)L2([0,1b)×[0,1),CQ), and thus

G(t,v)x=1bG(t,v)H(t,v)G(t,v)x

for a.e. (t,v)[0,1b)×[0,1) by (17). So (16) holds by the arbitrariness of x. The proof is completed. □

By the definition of pseudo-inverse, we have following two lemmas.

Lemma 2.10

For a d×d matrix A satisfying A=A, we have

AA=AA=Prange(A),

where Prange(A) denotes the orthogonal projection from Cd onto range(A).

Lemma 2.11

For an arbitrary s×t matrix A, we have

APrange(A)=A,APrange(A)=A,
range(AA)=range(A),range(AA)=range(A),

where Prange(A) denotes the orthogonal projection from Cs onto range(A).

Let us check the Gabor system G(g,a,b) under Assumptions 1 and 2. Since Z=λlZ+Nλl for each 1lL, we have

{EmbTnalgl:m,nZ}={EmbT(nλl+γl)algl:m,nZ,γlNλl}={EmbTnagl,γl:m,nZ,γlNλl},

where λl and a are as in (4) and (5), and gl,γl=Tγlalgl. So G(g,a,b)=G(g˜,a,b), where

G(g˜,a,b)={EmbTnagl,γl:m,nZ,γlNλl,1lL},g˜={g1,0,g1,1,,g1,λ11;g2,0,g2,1,,g2,λ21;;gL,0,gL,1,,gL,λL1}.

So the matrix-valued function G(t,v) in Definition 2.1 is exactly Ψg˜(t,v) by Definition 2.2 in [31]. It follows that G(g,a,b) and G(g˜,a,b) have the same frame properties. Therefore, using Theorems 2.9 and 2.14 and Remark 2.10 in [31], we have the following two theorems.

Theorem 2.1

For g={g1,g2,,gL}L2(R), the following are equivalent:

  • (i)

    G(g,a,b) is a frame for M(g,a,b) with frame bounds A and B.

  • (ii)

    bAG(t,v)G(t,v)(G(t,v)G(t,v))2bBG(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

  • (iii)

    bAG(t,v)G(t,v)(G(t,v)G(t,v))2bBG(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

Theorem 2.2

For g={g1,g2,,gL}L2(R), G(g,a,b) is a Riesz basis for M(g,a,b) with Riesz bounds A and B (an orthonormal basis) if and only if

bAIG(t,v)G(t,v)bBI(G(t,v)G(t,v)=bI)

for a.e. (t,v)[0,1bq)×[0,1).

By Lemmas 2.62.9, we have following theorem, which characterizes the Gabor duals of type I (resp., type II):

Theorem 2.3

Given g={g1,g2,,gL}L2(R), let G(g,a,b) be a frame for M(g,a,b). Then, for an arbitrary h={h1,h2,,hL}L2(R) with G(h,a,b) being a Bessel sequence in L2(R), G(h,a,b) is a Gabor dual of type I (type II) for G(g,a,b) if and only if the following hold:

  • (i)
    there exists a measurable function A:[0,1bq)×[0,1)MQ,Q (B:[0,1bq)×[0,1)Mp,p) such that
    H(t,v)=A(t,v)G(t,v)(H(t,v)=G(t,v)B(t,v));
  • (ii)

    G(t,v)=1bG(t,v)H(t,v)G(t,v) for a.e. (t,v)[0,1bq)×[0,1).

Theorem 2.4

Let G(g,a,b) be a frame for M(g,a,b). Then the following are equivalent:

  • (i)

    G(g,a,b) has a unique Gabor dual of type I (type II).

  • (ii)

    rank(G(t,v)){0,Q} (rank(G(t,v)){0,p}) for a.e. (t,v)[0,1bq)×[0,1).

  • (iii)

    rank(G(t,v)){0,Q} (rank(G(t,v)){0,p}) for a.e. (t,v)R2.

Proof

We only prove “type I” part. The other part can be proved similarly. By Lemma 2.5(ii) we only need to prove the equivalence between (i) and (ii). By Lemmas 2.6 and 2.9, (i) holds if and only if for a function A:[0,1bq)×[0,1)MQ,Q,

G(t,v)A(t,v)G(t,v)G(t,v)=0 18

implies

A(t,v)G(t,v)=0for a.e. [0,1bq)×[0,1). 19

Next, we prove the equivalence between (ii) and the above implication. Obviously, (ii) leads to this implication. Next, we prove that the implication fails if (ii) is violated and thus finish the proof. Suppose (ii) does not hold. Then rank(G(t,v)){0,Q} on some subset E0 of [0,1bq)×[0,1) with |E0|>0. Suppose {ei:1iQ} is the orthonormal basis for CQ, where ei is the vector with the ith component being 1 and others zeros. Since rank(G(t,v)){0,Q} for (t,v)E0, there exist 1rQ and E1E0 with |E1|>0 such that

G(t,v)er0andker(G(t,v)){0}for (t,v)E1.

By the same procedure as in Lemma 4.1 in [31], there exist 1kQ and E2E1 with |E2|>0 such that P(t,v)ek0 for (t,v)E2, where P(t,v) is the orthogonal projection of CQ onto ker(G(t,v)). By the argument of [8], p. 978, P(t,v) is measurable. Define

A(t,v)={P(t,v)ekerif (t,v)E2,0otherwise

for (t,v)[0,1bq)×[0,1). Then A(,) satisfies (18) but does not satisfy (19). The proof is completed. □

Theorem 2.5

Given g={g1,g2,,gL}L2(R), let G(g,a,b) be a frame for M(g,a,b). Then, for an arbitrary h={h1,h2,,hL}L2(R) with G(h,a,b) being a Bessel sequence in L2(R), we have

  • (i)
    G(h,a,b) is a Gabor dual of type I for G(g,a,b) if and only if there exists a measurable function A:[0,1bq)×[0,1)MQ,Q such that
    H(t,v)=b(G(t,v)G(t,v))G(t,v)[I1bG(t,v)A(t,v)G(t,v)]+A(t,v)G(t,v) 20
    for a.e. (t,v)[0,1bq)×[0,1).
  • (ii)
    G(h,a,b) is a Gabor dual of type II for G(g,a,b) if there exists a measurable function A:[0,1bq)×[0,1)Mp,p such that
    H(t,v)=bG(t,v)(G(t,v)G(t,v))[I1bG(t,v)G(t,v)A(t,v)]+G(t,v)A(t,v) 21
    f or a.e. (t,v)[0,1bq)×[0,1).
  • (iii)
    G(h,a,b) is an oblique dual of G(g,a,b) if one of the following conditions holds:
    1. H(t,v)=bG(t,v)(G(t,v)G(t,v))[I1bG(t,v)A(t,v)G(t,v)]+A(t,v)G(t,v)
      for some measurable function A:[0,1bq)×[0,1)MQ,Q and a.e. (t,v)[0,1bq)×[0,1);
    2. H(t,v)=b(G(t,v)G(t,v))G(t,v)[I1bG(t,v)G(t,v)A(t,v)]+G(t,v)A(t,v)
      for some measurable function A:[0,1bq)×[0,1)Mp,p and a.e. (t,v)[0,1bq)×[0,1).

Proof

We only prove (i); (ii) and (iii) can be proved similarly. First, we assume that (20) holds. Then, applying Lemmas 2.10 and 2.11, we have

1bG(t,v)H(t,v)G(t,v)=G(t,v),

and thus G(h,a,b) is a Gabor dual of type I for G(g,a,b) by Theorem 2.3.

Now we turn to the converse implication. Suppose G(h,a,b) is a Gabor dual of type I for G(g,a,b). Then there exists B(t,v):[0,1bq)×[0,1)MQ,Q such that

H(t,v)=B(t,v)G(t,v),G(t,v)=1bG(t,v)H(t,v)G(t,v) 22

for a.e. (t,v)[0,1bq)×[0,1) by Theorem 2.3. It follows that

b(G(t,v)G(t,v))G(t,v)G(t,v)=(G(t,v)G(t,v))G(t,v)G(t,v)H(t,v)G(t,v)

for a.e. (t,v)[0,1bq)×[0,1), and thus

b(G(t,v)G(t,v))G(t,v)=(G(t,v)G(t,v))G(t,v)G(t,v)H(t,v) 23

for a.e. (t,v)[0,1bq)×[0,1) since Cp=range(G(t,v))ker(G(t,v)). Put A(t,v)=B(t,v)(G(t,v)G(t,v)). Then (20) holds by Lemmas 2.10 and 2.11 and by a simple computation. The proof is completed. □

Some examples and remarks

This section is devoted to some examples and remarks. The ideas of this section are borrowed from [1]. Example 2.1 tells us that not every Q×p matrix-valued function with L2([0,1bq)×[0,1))-entries determines a g={g1,g2,,gL}L2(R) via the Zak transform matrix method if the time shift parameters a1,a2,,aL are not all the same. Example 3.2 shows that not every subspace mixed Gabor frame G(g,a,b) admits an oblique Gabor dual. Therefore, there exist significant differences between mixed multiwindow Gabor frames and usual multiwindow Gabor frames, and there should be many challenging problems in this direction.

Definition 3.1

Given g={g1,g2,,gL}L2(R), let G(g,a,b) be a Bessel sequence in L2(R). We say that G(g,a,b) has Riesz property if for cl2(Z2,CL), we must have c=0 whenever Tgc=0.

By Lemma 2.5(ii) and Theorem 2.1 in [2] we have the following:

Lemma 3.1

Given g={g1,g2,,gL}L2(R), let G(g,a,b) be a Bessel sequence in L2(R). Then G(g,a,b) has Riesz property if and only if rank(G(t,v))=Q for a.e. (t,v)[0,1bq)×[0,1).

Next we turn to examples of Theorems 2.1 and 2.2 with M(g,a,b)L2(R). See Examples 4.1, 4.2, 4.5, and 4.6 in [2], for examples with M(g,a,b)=L2(R). Suppose

a=(6,4)andb=(12,12). 24

Then ab=6 with p=6, q=1, and Q=5. For g={g1,g2} with g1, g2L2(R), we associate it with G as in Definition 2.1. Then

G(t,v)=(G1(t,v)G2(t,v)) 25

with

G1(t,v)=(G1(t,v)G1(t6,v))andG2(t,v)=(G2(t,v)G2(t4,v)G2(t8,v)),

where

Gl(t,v)=(Z12gl(t+2k,v))0,kM1,6 26

with kN6 and a.e. (t,v)[0,2)×[0,1) for l=1,2. By the quasi-periodicity of the Zak transform we have:

G1(t6,v)=(e2πivZ12g1(t+6,v),e2πivZ12g1(t+8,v),e2πivZ12g1(t+10,v),G1(t6,v)=(Z12g1(t,v),Z12g1(t+2,v),Z12g1(t+4,v)),G2(t4,v)=(e2πivZ12g2(t+8,v),e2πivZ12g2(t+10,v),Z12g2(t,v),G1(t6,v)=(Z12g2(t+2,v),Z12g2(t+4,v),Z12g2(t+6,v)),G2(t8,v)=(e2πivZ12g2(t+4,v),e2πivZ12g2(t+6,v),e2πivZ12g2(t+8,v),G1(t6,v)=(Z12g2(t+10,v),Z12g2(t,v),Z12g2(t+2,v)).

Thus, for a.e. (t,v)[0,2)×[0,1), G1(t6,v) and G2(t4,v), G2(t8,v) are uniquely determined by G1(t,v) and G2(t,v), respectively. Observe that [0,2)+2N6 is 12Z-congruent to [0,12). It follows that an arbitrary 5×6 matrix-valued function K(t,v) on [0,2)×[0,1) of the following form determines a unique g={g1,g2} by

G(t,v)=K(t,v)for  a.e. (t,v)[0,2]×[0,1]: 27
K(t,v)=(K1(t,v),K2(t,v)) 28

with K1(t,v),K2(t,v)M5,3 and

K1(t,v)=(a(t,v)λ(t,v)a(t,v)λ2(t,v)a(t,v)000λ(t,v)a(t,v)a(t,v)000λ(t,v)a(t,v)000),K2(t,v)=(000a(t,v)λ(t,v)a(t,v)λ2(t,v)a(t,v)000a(t,v)000λ(t,v)a(t,v)a(t,v))

for a.e. (t,v)[0,2)×[0,1).

Example 3.1

Let a and b be defined as in (24), and define g={g1,g2} by (27), where λ(,) and a(,) are continuous on [0,2]×[0,1]. Assume that

|λ(t,v)|<1,|a(t,v)|2(12|λ(t,v)|42|λ(t,v)|6|λ(t,v)|8)<1|λ(t,v)|3,|a(t,v)|2(1+2|λ(t,v)|3+2|λ(t,v)|5+|λ(t,v)|7)<1+|λ(t,v)|3

for (t,v)[0,2]×[0,1] satisfying a(t,v)0. Then G(g,a,b) is a frame for M(g,a,b), and M(g,a,b)L2(R). In particular, G(g,a,b) is a Riesz basis for M(g,a,b) if and only if, in addition,

a(t,v)λ(t,v)0for a.e. (t,v)[0,2)×[0,1). 29

Proof

Since G(t,v) is a 5×6 matrix, its rank cannot be p=6. So M(g,a,b)L2(R) by Lemma 2.4. Next, we prove that G(g,a,b) is a frame for M(g,a,b). By a simple computation we have

(G(t,v)G(t,v))2x,x=|a(t,v)|2l=15αl(t,v)|xl|2+β(t,v), 30
G(t,v)G(t,v)x,x=|a(t,v)|2l=15α˜l(t,v)|xl|2+β˜(t,v) 31

for (t,v)[0,2]×[0,1] and xC5, where

α1(t,v)=|λ(t,v)|3+(1+|λ(t,v)|2+|λ(t,v)|4)2|a(t,v)|2,α2(t,v)=1+(1+|λ(t,v)|2+|λ(t,v)|4)2|a(t,v)|2,α4(t,v)=1+|λ(t,v)|3+(1+|λ(t,v)|2)2|a(t,v)|2,α3(t,v)=α5(t,v)=(1+|λ(t,v)|2)2|a(t,v)|2,β(t,v)=2Re([λ(t,v)3(2+2|λ(t,v)|2+|λ(t,v)|4)a(t,v)2|a(t,v)|2]x¯1x4)β(t,v)=+2Re([(2+2|λ(t,v)|2+|λ(t,v)|4)a(t,v)2|a(t,v)|2]x¯2x4)β(t,v)=2Re(λ(t,v)3|a(t,v)|2x¯1x2),α˜1(t,v)=α˜2(t,v)=1+|λ(t,v)|2+|λ(t,v)|4,α˜3(t,v)=α˜4(t,v)=α˜5(t,v)=1+|λ(t,v)|2,β˜(t,v)=2Re(λ(t,v)3a(t,v)2x¯1x4)+2Re(a(t,v)2x¯2x4).

It is easy to check that

|β(t,v)||a(t,v)|2[β1(t,v)|x1|2+β2(t,v)|x2|2+β4(t,v)|x4|2],

w here

β1(t,v)=|λ(t,v)|3+|a(t,v)|2(2|λ(t,v)|3+2|λ(t,v)|5+|λ(t,v)|7),β2(t,v)=|λ(t,v)|3+|a(t,v)|2(2+2|λ(t,v)|2+|λ(t,v)|4),β4(t,v)=|a(t,v)|2(2+2|λ(t,v)|2+2|λ(t,v)|3+|λ(t,v)|4)β4(t,v)=+|a(t,v)|2(2|λ(t,v)|5+|λ(t,v)|7).

We have

|β˜(t,v)||a(t,v)|2[β˜1(t,v)|x1|2+β˜2(t,v)|x2|2+β˜4(t,v)|x4|2],

where

β˜1(t,v)=|λ(t,v)|3,β˜2(t,v)=1,β˜4(t,v)=1+|λ(t,v)|3.

Write

Cl(t,v)=αl(t,v)+βl(t,v),C˜l(t,v)=αl(t,v)βl(t,v)for l=1,2,4,C(t,v)=(1+|λ(t,v)|2)2|a(t,v)|2,Dl(t,v)=α˜l(t,v)+β˜l(t,v),D˜l(t,v)=α˜l(t,v)β˜l(t,v)for l=1,2,4,D(t,v)=|λ(t,v)|2+1.

Take

{C˜=min{C(t,v),C˜l(t,v):l=1,2,4,(t,v)[0,2]×[0,1]},C=max{C(t,v),Cl(t,v):l=1,2,4,(t,v)[0,2]×[0,1]},{D˜=min{D(t,v),D˜l(t,v):l=1,2,4,(t,v)[0,2]×[0,1]},D=max{D(t,v),Dl(t,v):l=1,2,4,(t,v)[0,2]×[0,1]}.

Then

C˜|a(t,v)|2x2(G(t,v)G(t,v))2x,xC|a(t,v)|2x2,D˜|a(t,v)|2x2G(t,v)G(t,v)x,xD|a(t,v)|2x2.

It follows that

12AG(t,v)G(t,v)x,x(G(t,v)G(t,v))2x,x12BG(t,v)G(t,v)x,x

for xC5and a.e. (t,v)[0,2)×[0,1), and thus

12AG(t,v)G(t,v)(G(t,v)G(t,v))212BG(t,v)G(t,v) 32

for a.e. (t,v)[0,2)×[0,1), where A=2C˜D, B=2CD˜. By Theorem 2.1, G(g,a,b) is a frame for M(g,a,b) with frame bounds A and B.

By simple computation, (29) holds if and only if rankG(t,v)=5 for a.e. (t,v)[0,2)×[0,1) or, equivalently, G(g,a,b) is a Riesz basis for M(g,a,b) by Lemma 3.1. □

Remark 3.1

Not every subspace mixed Gabor frame G(g,a,b) admits an oblique Gabor dual.

We show it by revisiting Example 3.1. Let us make the additional assumption that λ(t,v)=0 and a(t,v)0 for a.e. (t,v)[0,2)×[0,1]. Then G(g,a,b) is a frame but not a Riesz basis for M(g,a,b) by Example 3.1. Suppose G(h,a,b) with h={h1,h2} is an oblique Gabor dual for G(g,a,b). Then

G(t,v)=2G(t,v)H(t,v)G(t,v)for a.e. (t,v)[0,2)×[0,1) 33

by Lemma 2.9. Writing out (0,0)-, (3,1)-, and (5,4)-entries of both sides, we have

2(a(t,v))2Z12h1(t,v)=a(t,v), 34
2(a(t,v))2Z12h1(t,v)+2|a(t,v)|2Z12h2(t+2,v)=a(t,v), 35
2(a(t,v))2Z12h2(t+2,v)=a(t,v) 36

for a.e. (t,v)[0,2)×[0,1). By (35) and (36) we have 2a(t,v)Z12h1(t,v)=0 for a.e. (t,v)[0,2)×[0,1). This contradicts (34).

Observe that a1a2 in Remark 3.1 (a1=6 and a2=4). It is natural to ask:

Does every subspace Gabor frame G(g,a,b) admit no oblique Gabor dual whenever al, 1lL, are not all the same?

The following example gives a negative answer to this question.

Example 3.2

Let a=(1,2) and b=(13,13). Assume that

J(t,v)=(J1(t,v)J2(t,v))andE(t,v)=(E1(t,v)E2(t,v))

have the form

J1(t,v)=(a0,0(t,v)a0,1(t,v)a1,0(t,v)a1,1(t,v)a2,0(t,v)a2,1(t,v)a2,1(t,v)e2πiva2,0(t,v)e2πiva0,1(t,v)a0,0(t,v)e2πiva1,1(t,v)a1,0(t,v)),E1(t,v)=(c0,0(t,v)c0,1(t,v)c1,0(t,v)c1,1(t,v)c2,0(t,v)c2,1(t,v)c2,1(t,v)e2πivc2,0(t,v)e2πivc0,1(t,v)c0,0(t,v)e2πivc1,1(t,v)c1,0(t,v)),

and

J2(t,v)=(b0,0(t,v)b0,1(t,v)b1,0(t,v)b1,1(t,v)b2,0(t,v)b2,1(t,v)),E2(t,v)=(d0,0(t,v)d0,1(t,v)d1,0(t,v)d1,1(t,v)d2,0(t,v)d2,1(t,v))

for a.e. (t,v)[0,2)×[0,1) with all entries of J(t,v) and E(t,v) in L([0,2)×[0,1)), that (J1(t,v))E1(t,v) has the form

(A(t,v)00A(t,v))

and (J(t,v))J(t,v) has the form

(B(t,v)00B(t,v))

for (t,v)[0,2)×[0,1), and that

(J2(t,v))E2(t,v)=(13A(t,v)0013A(t,v))

for (t,v)[0,2)×[0,1) satisfying B(t,v)0. Define g={g1,g2} and h={h1,h2} by

G(t,v)=J(t,v)andH(t,v)=E(t,v)for (t,v)[0,2)×[0,1).

Then g and h are well defined by the quasi-periodicity of the Zak transform Z6 and 6Z-congruence between [0,1)2N3+3N2 and [0,6), and G(g,a,b) and G(h,a,b) are both Bessel sequences by the quasi-periodicity of the Zak transform and Lemma 2.3(vi). A simple computation shows that

G(t,v)G(t,v)=3G(t,v)H(t,v)G(t,v)G(t,v)

for (t,v)[0,2)×[0,1). This implies that G(t,v)=3G(t,v)H(t,v)G(t,v) for (t,v)[0,2)×[0,1) since C9=range(G(t,v))ker(G(t,v)). So G(h,a,b) is an oblique Gabor dual for G(h,a,b) by Lemma 2.9.

Conclusions

A mixed multiwindow Gabor system is one of generalizations of multiwindow Gabor systems, whose time-frequency shifts vary with the windows. This paper addresses subspace mixed multiwindow Gabor systems with rational time-frequency product lattices. Using a suitable Zak-transform matrix method, in this paper, we characterize subspace mixed multiwindow Gabor frames and their Gabor duals, obtain explicit expressions of Gabor duals, and characterize the uniqueness of Gabor duals. Some provided examples show that there exist significant differences between mixed multiwindow Gabor frames and usual multiwindow Gabor frames.

Acknowledgements

The authors would like to thank the reviewers for their comments.

Authors’ contributions

Both authors contributed equally to the writing of this paper. Both authors read and approved the final manuscript.

Funding

The article is supported by the National Natural Science Foundation of China (Grant No. 11501010, 11271037) and the Scientific Research Project of Ningxia Colleges and Universities (Grant No. NGY 2018-163).

Competing interests

The authors declare that they have no competing interests.

Footnotes

Publisher’s Note

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Contributor Information

Yan Zhang, Email: yzhangbun@163.com.

Yun-Zhang Li, Email: yzlee@bjut.edu.cn.

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