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).

1. Introduction

Let be a separable Hilbert space. An at most countable sequence in is called a frame for if there exist such that where and are called frame bounds. The sequence is called a Bessel sequence in if the right-hand side inequality in (1) holds. In this case is called a Bessel bound. A frame for is said to be a Riesz basis if it ceases to be a frame for 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 [14]. We denote by the set of integers, by the set of positive integers, by the vector in Euclidean spaces with the th component being and others being zero, and by the set for . This paper addresses Gabor systems of the form where is a fixed positive integer, with each , with each , and with each . Throughout this paper, we work under the following assumptions.

Assumption 1. is a positive integer.

Assumption 2. , and with and being relatively prime positive integers for .

We always denote by the least common multiple of with , by the positive integer satisfying for each , by and two relatively prime positive integers satisfying , and by the number . Obviously, they are all uniquely determined by Assumptions 1 and 2.

Remark 3. We do not lose generality by making Assumptions 1 and 2. Let us check a general Gabor system of the form with satisfying , , and are not necessarily all the same). Suppose is the greatest common divisor of , and with for . Define for and . Then, by a simple computation, is a frame for a closed subspace of if and only if the Gabor system is a frame for . Therefore, the study of (3) is reduced to the study of (4).

We denote by the closed linear span of . A Gabor system of the form (2) is called a mixed multi-window Gabor system if since and are not necessarily all the same. (In particular, this system is the usual multi-window Gabor system when ). It is called a subspace mixed multi-window Gabor frame if it is a frame for and . Similarly, a Gabor system is called a single-window Gabor system if and called a single-window subspace Gabor frame if it is a frame for and .

For a Bessel sequence in of the form (2), we define the associated synthesis operator by for . Then it is a bounded operator, and its adjoint operator (so-called analysis operator) is given by where for each . For a Bessel sequence in with , we associate it with similarly. Write ; that is, for . Let be a frame for and let be a Bessel sequence in . is called an oblique Gabor dual for if Here it is not required that each belongs to . In particular, an oblique Gabor dual for is said to be a Gabor dual of type for if for each and is said to be a Gabor dual of type II for if . 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 . For a Gabor dual of type II, it is not required to be in , but a containment relation between the ranges of analysis operators is required. Observe that the canonical daul for 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 for have been studied extensively [2, 3, 9, 10]; multi-window Gabor frames for were firstly studied by Zibulski and Zeevi in [11], and then by others in [1217]; single-window subspace (of ) Gabor frames were studied in [7, 8, 1823]. In digital signal processing, one usually encounters discrete signals instead of continuous ones. One can obtain discrete Gabor frames for via Gabor frames for through sampling under certain additional assumptions [2426], though these assumptions are artificial and too technical. So it should be more reasonable to consider Gabor frames in without referring to frames in . 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 has attracted many researchers (see [2, 3, 2434] and the references therein). In [6, 35], single-window Gabor frames on discrete periodic subsets of 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. Section 2 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. For , , we denote by the set of complex matrices. Given , two subsets and of are said to be congruent if there exists a sequence of subsets of such that , for in and that , for in . If or is a finite set in addition, only finitely many in the above are nonempty, while the others are empty. For two vectors and in a Hilbert space , we always write their inner product as instead of when it causes no confusions. Let be a finite set in and let . We denote by the Hilbert space of vector-valued functions satisfying for each endowed with the inner product for , , where denotes their inner product in ; by the Hilbert space of sequences satisfying endowed with the inner product for , , where denotes their inner product in and by the set of sequences with each being a finitely supported sequence defined on . For simplicity, we write for , for , and for when .

For , define the discrete Zak transform of by for and a.e. . It is easy to check that has quasi-periodicity: for and a.e. .

Definition 4. For with each , we associate it with a matrix-valued function by where and is a block matrix of the form with for and for , .

By the quasi-periodicity of , an arbitrary is uniquely determined by the values of on with being a set congruent to . So an arbitrary function determines a unique by Let us make another assumption that in Definition 4. Then is an -matrix-valued function of the form for , , and . Also observe that is congruent to . We have if , then an arbitrary function with all entries in determines a unique with each by But if , , are not all the same, a function with all entries in does not necessarily determine a . We show it by an example. In this case, there exists such that . We may as well assume that . Choose for and such that every entry of belongs to and that Suppose there exists such that for and a.e. . Then for and a.e. , where for . Since the sets and are both congruent to , we have for some and with by (20), (21), and the quasi-periodicity of , while for and a.e. by (20), (22), and the quasi-periodicity of . This is a contradiction. Therefore, we must be careful when we define by a function if , , are not all the same.

Definition 5. Define by

By the quasi-periodicity of and [35, Theorem 2.1], we have the following lemmas.

Lemma 6. is a unitary operator from onto , and is a unitary operator from onto for an arbitrary subset of which is congruent to .

Definition 7. Define the Fourier transform by for , , and a.e. .

Definition 8. Define by for , , and a.e. , where for and .

Similarly, for an arbitrary , we associate it with .

Lemma 9.    is a unitary operator from onto ;
   is a unitary operator from onto .

Proof. Since is an orthonormal basis for , we have (i). (ii) is an immediate consequence of (i).

Lemma 10. For with for , one has(i) for ;(ii)for with , where is as in (5).

Proof. (i) can be obtained by a direct computation. Applying (i), we have for . This implies (ii).

Remark 11. When in Lemma 10 is a Bessel sequence in , by the same procedure as the above we can prove (28) holds for with .

Lemma 12. For with for , one has for , with , and a.e. , where , , and denotes the block matrix (with blocks) of the form with .

Proof. By [6, Lemma 5], we have for , , with , and a.e. . This leads to the lemma.

3. Frame Characterization

This section is devoted to characterization of subspace Gabor frames of the form (2).

Theorem 13. For any with each , the following are equivalent.(i) is a Bessel sequence in with Bessel bound .(ii) for and a.e. .(iii) for and a.e. .

Proof. By Lemma 12, conditions (ii) and (iii) are equivalent. So, to finish the proof, we only need to prove that (i) holds if and only if since is a Bessel sequence with Bessel bound if and only if for , which can be rewritten as by Lemmas 610.
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 belongs to and thus belongs to for , almost every point in is a Lebesgue point for every entry in and . Let be such a point and . Fix and with . Define by for , . By Lemma 9 and the density of in , (34) holds for . Substituting (35) into (34), we obtain that Letting leads to that is, This gives (32) by the arbitrariness of , , and . The proof is completed.

Lemma 14 ([1, Theorem A.6.5]). Let , , be self-adjoint operators on a Hilbert space . If , and , and commutes with and , then .

Remark 15. The inequality in Theorem 13 can be replaced by any one of the following inequalities:(i);(ii);(iii).
Indeed, we have (i) by Lemma 14 if . From (i), we have for , and thus on . This implies (ii) due to the fact that . It is obvious that (ii) implies (iii) by Lemma 14. By the same procedure as in “”, we can prove that (iii) implies .

Remark 16. Condition (iii) in Theorem 13 is equivalent to for a.e. when is understood as an operator from into . However, such norm is equivalent to the one obtained by taking the maximum of the absolute values of all entries of a matrix. So is a Bessel sequence if and only if for by Theorem 13.

Lemma 17. For with each , let be a Bessel sequence in . For , write . Then for when either is a Bessel sequence or .

Proof. By Lemmas 6 and 10, we have for each , , and . When either is a Bessel sequence or , the integrand in (41) belongs to as a function about by Remark 16. It follows that for . The lemma therefore follows.

Theorem 18. Let and be both Bessel sequences in . Then for and a.e. .

Proof. Write . Then by Remark 11. Applying Lemma 17 to , we have and thus (43) holds by (44).

Theorem 19. For with each , the following are equivalent:(i) is a frame for with frame bounds and ;(ii) for and a.e. ;(iii) for and a.e. .

Proof. By Theorem 13 and Remark 15, we may as well assume that 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 if and only if Since the linear span of is dense in , (46) holds if and only if for with by [1, Lemma ]. By Lemma 10, Theorem 18, and Remark 11, (48) can be rewritten as equivalently, for , , and . 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 Remark 15, and in Theorem 19 can be replaced by and , respectively.

Definition 21. Given with each , let be a Bessel sequence in . We say that has Riesz property if, for , we must have whenever .

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 set in with , let be a matrix-valued measurable function. Define by the orthogonal projection of onto . Then for , and thus is measurable.

Theorem 23. For with each , the following are equivalent:(i) has Riesz property;(ii) for and a.e. ;(iii) for and a.e. .

Proof. By Lemma 12, (ii) and (iii) are equivalent. So we only need to prove that is a unique solution to in if and only if By Remark 11, (52) can be rewritten as So is a unique solution to (52) in if and only if is a unique solution to in by Lemma 9. It is obvious that (53) implies that is a unique solution to (55) in . Next we prove the converse implication. Suppose there exist and with such that on . Let be the orthogonal projection of onto . Then there exist and with such that for . Define by for and . Then is well defined, and by Lemma 22. It follows that is a nonzero solution to (55) in . 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 bound , we have the following theorem by Theorems 19 and 23.

Theorem 24. For with each , the following are equivalent:(i) is a Riesz basis for with Riesz bounds and (an orthonormal basis);(ii) for a.e. ;(iii) for and a.e. .

Next we turn to examples of Theorems 19 and 24. Suppose Then with , , and . For with , we associate it with as in Definition 4. Then where for . By the quasi-periodicity of Zak transform, we have Thus for each and a.e. , , and , are uniquely determined by and , respectively. Observe that is congruent to . It follows that is uniquely determined by the values of for and a.e. . Therefore, an arbitrary matrix matrix-valued function for and a.e. with all entries being in determines a unique by

Letfor and a.e. . Define by Then we obtain the following example.

Example 25. Let and be defined as in (57) and define by (63), where , are continuous on for . Assume that for satisfying . Then and is a frame for .

Proof. Equation (65) is an immediate consequence of (63). Next we prove that is a frame for . By a simple computation, we have for and , where It is easy to check that where where Write Take Then It follows that for , , and a.e. , and thus for and a.e. , where , . By Theorem 19, is a frame for with frame bounds and .

Remark 26. By a simple computation, we have if and only if Also observe that is equivalent to being invertible, and that (75) can be reduced to in this case. Therefore, is a Riesz basis for with Riesz bounds and if and only if (76) holds.

4. Gabor Dual Characterization

Let be a frame for . 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. For and with each , , the following are equivalent:(i) for each ;(ii)there exists a function such that for and a.e. ;(iii) there exists a function such that for and a.e. .

Lemma 28. Given with each , with each , let and be both Bessel sequences in . Then the following are equivalent:(i); (ii)there exists a function such that (iii)there exists a function such that

Proof. If (ii) holds, we can obtain (iii) by choosing as for , where is defined as in Lemma 12. So we only need to prove that (i) holds if and only if there exists a function such that For , is orthogonal to () if and only if in -inner product by Lemmas 9 and 17, equivalently, in -inner. Again by Lemma 6, (85) is equivalent to So we only need to prove that (83) holds if and only if, for , whenever 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 and with , on which some th column of satisfies Let be the orthogonal projection of onto . Define by for . Then solves (88) and for . Also observe that the th component of is exactly for . It follows that fails to solve (87). This finishes the proof.

Lemma 29. Given , with each , , let and be both Bessel sequences in . Then for if and only if for and a.e. .

Proof. By Lemma 12, (92) holds for and a.e. if and only if it holds for and a.e. . Next we prove that (91) holds for if and only if (92) holds for and a.e. .
Since is dense in , (91) holds for if and only if it holds for , equivalently, for by Remark 11 and Theorem 18. This is also equivalent to for , , and a.e. by Lemma 9. It is obvious that (94) holds if (92) holds for and a.e. . Now suppose (94) holds. For an arbitrarily fixed , choose as Then , and thus for and a.e. by (94). So (92) holds for and a.e. by the arbitrariness of . The proof is completed.

By Lemmas 2729, we have the following theorem which characterizes the Gabor duals of type I (resp., type II).

Theorem 30. Given with each , let be a frame for . Then, for any Bessel sequence in , is a Gabor dual of type I (type II) for if and only if the following hold:(i) there exists such that (ii) for and a.e. .

Theorem 31. Given with each , let be a frame for . Then, for any Bessel sequence in ,(i)  is a Gabor dual of type I for if and only if there exists such that for and a.e. ;(ii)  is a Gabor dual of type II for if for and a.e. ;(iii) is an oblique dual of if one of the following conditions holds:(a) for and a.e. ;(b) for and a.e. .

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 for and a.e. . Also observe that is the projection from onto . It follows that , and thus is a Gabor dual of type I for by Theorem 30.
Now we turn to the converse implication. Suppose is a Gabor dual of type I for . Then there exists such that for and a.e. . It follows that for and a.e. , and thus for and a.e. due to the fact that . Put . Then the right-hand side of (97) equals by (102), and this is exactly for and a.e. again by the fact that

5. The Case of

Theorems 30 and 31 characterize duals with Gabor structure for general . This section deals with the case . We start with a remark on the case of , , being not all the same.

Remark 32. Not every subspace Gabor frame admits an oblique Gabor dual.

We show it by revisiting Example 25. Let us make an additional assumption that and for . Then is a frame but not a Riesz basis for by Remark 26. Suppose with is an oblique Gabor dual for . Then by Theorem 30. Writing out , , and entries of both sides, we have for and a.e. . By (107) and (108), we have for and a.e. . This contradicts (106).

Observe that in Remark 32 ( and ). It is natural to ask the following question.

Does every subspace Gabor frame admit no oblique Gabor dual whenever , , are not all the same?

The following example gives a negative answer to this question.

Example 33. Let and . Assume that , have the form for with all entries of and in , that has the form and has the form for , and that for satisfying . Define and by Then and are well defined by the quasi-periodicity of the Zak transform and congruence between and , and and are both Bessel sequences by Remark 16. A simple computation shows that for . This implies that for due to the fact that . So is an oblique Gabor dual for by Lemma 29.

Remark 32 and Example 33 show that the Gabor dual theory for subspace Gabor frames is complicated when , , are not all the same. There are still many unresolved problems in this direction. Next we work under the following assumption.

Assumption 34. is a positive integer, , and with .

The following theorem characterizes the uniqueness of such duals.

Theorem 35. Given with each , let be a frame for . Then the following are equivalent:(i) has a unique Gabor dual of type I (type II);(ii) for a.e. ;(iii) for a.e. .

Proof. We first deal with the Gabor dual of type I. By Lemma 12, (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 solves whenever it solves Suppose (ii) holds and solves (114). If is such that , then , and thus . If is such that , then (114) implies that which leads to and thus . Conversely, suppose (ii) does not hold. Then there exist some and , such that on some with , where is the orthogonal projection of onto . Define for . Then (114) holds, but (113) fails to hold.
Next we turn to the Gabor duals of type II. Similarly to the above arguments, by Theorem 30 and Lemmas 27 and 28, we only need to prove that (ii) holds if and only if a function solves whenever it solves The necessity can be proved similarly to the case of type I. Now we suppose (ii) does not hold. Then there exist some and , such that on some with , where is the orthogonal projection of onto . Define by for . Then (119) holds, but (118) fails to hold. The proof is completed.

It is well known that every scalar matrix with rank has a decomposition where and are, respectively, and unitary matrices, is a block matrix in which is an diagonal matrix with positive entries in the diagonal (see [1, Theorem ]). Observe that may change in variables if is a matrix-valued function. Next we restrict ourselves to Gabor systems such that where and are, respectively, and unitary matrices for and a.e. , and is an block matrix in which is of the form with for and a.e. and being of positive measure for each .

As an immediate consequence of Theorems 19, 24, and 35, we have the following theorem.

Theorem 36. Let be defined as in (123). Then(i) is a frame for with frame bounds , if and only if for ;(ii) is a Riesz basis (an orthonormal basis) for with Riesz bounds , if and only if for , , and a.e. ;(iii) is a frame for with frame bounds and and simultaneously admits a unique Gabor dual of type I (type II) if and only if (), and for each and a.e. , either for each or for each .

Theorem 37. Let be defined as in (123) and let be a frame for . Then, for with each , we have(i) is a Gabor dual of type I for if and only if has the form where satisfies with , the entries of the th column of are essentially bounded on for , and for , ;(ii) is a Gabor dual of type II for if and only if has the form where satisfies with , the entries of the th row of are essentially bounded on for , and for , .

Proof. We only prove item (i), and (ii) can be proved similarly. By Theorem 30, is a Gabor dual of type I for if and only if there exists a function such that for and a.e. , and for each the entries of belong to . Write where . A simple computation shows that (132) is equivalent to where , . It is obvious that (134) is equivalent to (128), which implies that (132) is equivalent to (128). By (133), (132) can be rewritten as Since does not appear in (135), we assume that without loss of generality. By the unitary properties of and , for each , the entries of belong to if and only if the entries of belong to ; that is, the entries of the th column of are essentially bounded on for . The proof is completed.

Next we discuss a special case of Theorem 36. Let , , , and and write and for . Define and as by letting , , and and , , and in (123), respectively. Then This implies that , and that by [5, Theorem 3.1]. So and are both frames for , but neither is a Riesz basis by Theorem 36. Define and with by and for . Then and are, respectively, the canonical duals for and by [5, Remark 5.5], and for . It follows that and is infinitely supported. It is interesting that and are both finitely supported, but is not although is the first component of . 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. Let and , and , as in (140) and (142) and write . Then(i) and are both frames for , but neither is a Riesz basis for ;(ii) and are, respectively, the canonical duals for and ;(iii) is finitely supported, and is not.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (Grant no. 11271037), Beijing Natural Science Foundation (Grant no. 1122008), and the Scientific Research Common Program of Beijing Municipal Commission of Education (Grant no. KM201110005030).