Research Article | Open Access

H. R. Fernández-Morales, A. G. García, M. A. Hernández-Medina, M. J. Muñoz-Bouzo, "On Some Sampling-Related Frames in -Invariant Spaces", *Abstract and Applied Analysis*, vol. 2013, Article ID 761620, 14 pages, 2013. https://doi.org/10.1155/2013/761620

# On Some Sampling-Related Frames in -Invariant Spaces

**Academic Editor:**Patricia J. Y. Wong

#### Abstract

This paper is concerned with the characterization as frames of some sequences in -invariant spaces of a separable Hilbert space where denotes an unitary operator defined on ; besides, the dual frames having the same form are also found. This general setting includes, in particular, shift-invariant or modulation-invariant subspaces in , where these frames are intimately related to the generalized sampling problem. We also deal with some related perturbation problems. In doing so, we need the unitary operator to belong to a continuous group of unitary operators.

#### 1. Introduction

This paper is concerned with the study of some special frames in -invariant spaces. Given an unitary operator on a separable Hilbert space , we consider closed subspaces having the form , where denotes some fixed element in . In case that the sequence is a Riesz basis for , we have

Recall that a *Riesz basis* in a separable Hilbert space is the image of an orthonormal basis by means of a bounded invertible operator. Any Riesz basis has a unique biorthogonal (dual) Riesz basis , that is, , such that the expansions
hold for every . We state the definition by considering the integers set as the index set since throughout the paper most of sequences are indexed in . A *Riesz sequence* in is a Riesz basis for its closed span. A necessary and sufficient condition in order for the sequence to be a Riesz sequence in is given in Theorem 3 infra.

Given elements , , in , a challenging problem is to characterize the sequence as a frame (Riesz basis) in , where denotes a positive integer. Besides, another interesting problem is to look for those dual frames having the same form for some , , so that, for any the expansion holds.

At this point, we give an explanation about the expression *sampling-related frames* appearing in the title. Namely, -invariant subspaces in are natural generalizations of shift-invariant or modulation-invariant subspaces of ; there, the unitary involved operators are, respectively, the translation operator or the modulation operator . In the shift-invariant subspace generated by , for any , the inner products are
where for each . Thus, the above inner products are nothing but samples of some filtered versions of the function itself; this is precisely the generalized sampling problem in the shift-invariant space . Mathematically, it consists of the stable recovery of any from the above sequence of samples, that is, to obtain sampling formulas in having the following form:
such that the sequence of reconstruction functions is a frame for the shift-invariant space . As a consequence, expansions (3) and (5) have the same nature. Recall that a sequence is a *frame* for a separable Hilbert space if there exist constants (frame bounds), such that

A sequence in satisfying only the right hand inequality above is said to be a *Bessel sequence* for . Given a frame for , the representation property of any vector as a series is retained, but, unlike the case of Riesz bases *(exact frames)*, the uniqueness of this representation (for *overcomplete frames*) is sacrificed. Suitable frame coefficients which depend continuously and linearly on are obtained by using the dual frames of ; that is, is another frame for such that for each . For more details on the frame theory see [1].

Sampling in shift-invariant spaces of has been profusely treated in the mathematical literature (see, for instance, [2–13]).

The existence of expansions like (3) in -invariant subspaces was treated for the first time in [14]; see also [15, 16]. Following similar techniques to those in [14], we give, in Section 2, a complete characterization in of sequences having the form where for each . In other words, we carry out the study of the completeness, Bessel, frame, or Riesz basis properties of the sequence . Whenever it is a frame for , we find a family of dual frames having the same form for some , . In Section 4, we also discuss the case where some ; although the sequence is not contained in , something can be said in the light of the theory of pseudoframes (see [17, 18]).

All the obtained results in Section 2 involve the discrete group of unitary operators which is completely determined by . If we want to deal with something similar to the time-jitter error version of (5), that is, the recovery of any from a perturbed sequence of samples with errors , then the availability of a continuous group of unitary operators containing, in particular, the operator (say for instance ) becomes essential. In Section 3, after a brief on groups of unitary operators, we deal with two types of perturbation problems. The first one concerns the study of sequences as in , for small enough error sequence -norm, the sequence is a Riesz sequence in . The second one goes into the recovery of any from the perturbed sequence of inner products , for small enough errors , there exists a frame expansion for having the inner products as coefficients.

#### 2. -Invariant Subspaces

In a Hilbert space , the -invariant subspaces are intimately related to stationary sequences.

##### 2.1. Some Preliminaries on Stationary Sequences

A sequence in a separable Hilbert space is said to be *stationary* if

The function , for every , is called the autocovariance function of the sequence . Moreover, two stationary sequences and are said to be stationary correlated if
and , for every defines the corresponding *cross-covariance* function. The following result is a well-known characterization of stationary sequences (see [19]).

Lemma 1. *To every stationary sequence in a Hilbert space there exists a unique unitary operator and such that for all . Conversely, every pair of a unitary operator and an defines by , , a stationary sequence in .**Moreover, two stationary sequence and are stationary correlated if and only if they are generated by the same unitary operator ; that is, and for some . *

The autocovariance and the cross-covariance functions admit a spectral representation which is related to the integral representation of the unitary operator (see [19]).

Theorem 2. *For every stationary sequence in a Hilbert space , the autocovariance function admits a spectral representation
**
in the form of an integral with respect to a (positive) spectral measure . *

*For every two stationary correlated sequences , in a Hilbert space , the cross-covariance function admits a spectral representation*

*in the form of an integral with respect to a (complex) spectral measure .*

###### 2.1.1. The -Invariant Subspace

For a fixed , we consider the subspace of given by . In case that the sequence is a Riesz sequence in , we have

A necessary and sufficient condition in order for the sequence to be a Riesz sequence in can be stated in terms of the Lebesgue decomposition of the spectral measure into an absolute and a singular part as .

Theorem 3. *Let be a sequence obtained from an unitary operator in a separable Hilbert space with spectral measure , and let be the closed subspace spanned by . Then, is a Riesz basis for if and only if and
**
where .*

*Proof. *For a fixed -sequence , we have
if is not absolutely continuous with respect to Lebesgue measure , Lemma 4 below implies that there exists a bounded sequence such that tends to infinity with , so cannot be a Bessel sequence, therefore, not a Riesz basis. Assume now that and is a Riesz basis, this implies that there exists such that
for all . Let , assume that , this implies that for all . Now, (13), (14), and Parseval's theorem imply that
for all . Introducing the Fourier expansion of the characteristic function in (15), we obtain for all , which contradicts ; thus, the assumption is false. In a similar way, it can be proved that .

For the sufficient condition, assume now that ; then, (13) implies that
if, in addition, condition (12) is satisfied, we get
which implies that
Therefore, the sequence is a Riesz basis for .

Lemma 4. *Let be a finite positive measure on which is not absolutely continuous with respect to Lebesgue measure . Then, there exists a bounded sequence such that
*

*Proof. *If the measure is not absolutely continuous with respect to Lebesgue measure, then for a Lebesgue measurable set of Lebesgue measure zero. Thus, there exists a Borel set of Lebesgue measure zero such that . In fact, the set is an intersection of a countable collection of open sets (see [20, page 63]). Therefore, and . On the other hand, every finite Borel measure on is inner regular (see [20, page 340]); that is,
then there exists a compact set such that

For any , there exists a sequence of disjoint open intervals such that
(see [20, pages 58 and 42]). Since is compact, we may take the sequence to be finite. Hence, for every , there exist open disjoint intervals in such that
Besides, .

Consider the function , where , that satisfies

We modify and extend each to obtain a -periodic function such that and its derivative are continuous on , and for every . Let be the Fourier series of . First, by using Parseval's identity, we have
so that is a bounded sequence in . Besides, the regularity of each ensures that each Fourier series converges uniformly to . Therefore, each series converges to in and consequently,
from which we obtain the desired result.

The proof of Theorem 3 is similar to that of Theorem 6 in [16], except we do not exclude the case in which the singular measure is atomless. Recently, we have been aware that Theorem 3 was exposed in [21].

##### 2.2. Studying the Sequence in

For , , consider the sequence . For every , the spectral measure in the integral representation of the cross-covariance function of the sequences and has no singular part. Indeed, according to Theorem 3, the spectral measure associated with the autocovariance function of the sequence has no singular part; then by using the Cauchy-Schwarz type inequality in [22, page 125] we get the result. In the sequel, we will use the abridged notation ; our goal in this section is to study the sequence in in terms of an matrix which is introduced below. For the sake of completeness, we include some needed calculations which appear in [14].

First of all, we have where stands for the cross spectral density of the stationary correlated sequences and . Define

In what follows, we will use the *left-shift operator * defined as
or equivalently, by , where denotes the unidimensional torus. Also, we will consider the *decimation operator *, is a positive integer, defined as
which can equivalently be written as

For each , set the matrix of functions on the torus as follows: and define the matrix of functions on the torus as follows:

It is worth to mention that the matrix was explicitly computed in [14] for the translation and modulation cases in .

Next, for any , we obtain an expression for the inner products , and . Indeed, writing where , we have that is, where .

Now, for , define the sequence . Thus, we can write where .

After some easy calculations, we get

Defining , (38) implies that which can be written in matrix form as where denotes the multiplication operator by and

By (resp., ), we denote the product space for times (resp., times). Besides,

The above calculations let us prove the following result.

Theorem 5. *Let for and let be the associated matrix given in (33). Then, the following results hold. *(a)*The sequence is a complete system in if and only if the rank of the matrix is a.e. in . *(b)*The sequence is a Bessel sequence for if and only if there exists a constant such that
*(c)*The sequence is a frame for if and only if there exist constants such that
*(d)*The sequence is a Riesz basis for if and only if it is a frame and . *

*Proof. *To prove (a), assume that there exists a set with positive measure such that for each . Then, there exists a measurable function , , such that and in . This function can be constructed as in [23, Lemma 2.4]. Define such that if , and if . Hence, from (40), we obtain that the system is not complete. Conversely, if the system is not complete, by using (40), we obtain a different from 0 in a set with positive measure such that . Thus, on a set with positive measure.

To prove (b), we keep in mind that is a Riesz basis for , the mapping , given by is bijective and there exist two constants such that

Assume first that (43) is satisfied. It follows from (40) and (42) that

By construction and , using (45), it follows from (46) that
Conversely, assume that is a Bessel sequence for , then there exists such that

Using (45), this implies that
for all . Inserting the right hand side of (42) for , it is straightforward to see that (43) holds with . The proof of (c) is completed proceeding as in (b).

To prove (d), consider the following mapping:

According to (40), the mapping is isometric equivalent to , and assuming that is a frame, it is a Riesz basis if and only if is surjective.

First, if is a Riesz basis, then it is a frame and is surjective. Applying (a) yields that is bijective, and therefore is bijective. Hence, is for almost every in , so
and finally .

Conversely, if is a frame and , (a) implies that is invertible for almost every in , which implies that is surjective; then is surjective and is a Riesz basis.

The following lemma will allow us to restate Theorem 5.

Lemma 6. *Let be an matrix with entries in , and consider the following constants:
**
where (resp., ) denotes the smallest (resp., the largest) eigenvalue of the positive semidefinite matrix . Then, *(a)*the matrix has essentially bounded entries on if and only if ;*(b)*there exist constants such that , a.e., if and only if . *

*Proof. *The first part of lemma follows from that , and
where denotes the spectral norm of the matrix (see, for instance, [24]).

Now, we prove the second part of the lemma. Since means that for all , in particular, taking an eigenvector associated with the largest eigenvalue of such that , one has that . Hence, . In a similar way, implies that .

Conversely, Rayleigh-Ritz theorem [24, page 176] yields that
Thus, implies that

In other words, ; analogously, .

It is easy to deduce from the proof that and are the optimal constants and satisfying the inequalities , a.e., .

As a consequence of Lemma 6, statements (b) and (c) in Theorem 5 can be restated in terms of the following constants: as follows.

Theorem 7. *Let for , and let be the associated matrix given in (33) and its related constants (56). Then, the following results hold.*(i)*The sequence is a Bessel sequence for if and only if the constant . *(ii)*The sequence is a frame for if and only if the constants and satisfy . In this case, and are the optimal frame bounds for . *

##### 2.3. The Frame Expansion

Define the matrix of functions on the torus as follows: The following expansion involving the inner products of holds.

Lemma 8. *Assume that the matrix has essentially bounded entries on . For any , one has
**
where denotes the th Fourier coefficient of the function defined in (41), and the sequence is given in (35). *

*Proof. *Indeed,

At this point, we are ready to prove the following expansion result.

Theorem 9. *Let for , and assume that the associated matrix given in (33) has essentially bounded entries on , that is, . The following statements are equivalent. *(i)*The constant . *(ii)*There exist , , such that the sequence is a frame for , yielding, for any , the following expansion:
**In case the equivalent conditions hold, and form a pair of dual frames in .*

*Proof. *First, we prove that (i) implies (ii). Observe that can be written as where . Next,
For and define , where , and . Then, (61) implies

In order to be allowed to use [1, Lemma 5.6.2], we have to prove that the above constructed sequence is a Bessel sequence for . To this end, we compute the corresponding matrix for . Setting , we obtain

Now,
where . Hence, we have deduced that

Therefore, for , we have
and consequently, the matrix can be written as
where

As a consequence of Theorem 7, the proof ends if we prove that the matrix has essentially bounded entries. Clearly, the decimation operator sends bounded functions into bounded functions; Theorem 3 implies that is bounded so, taking into account (67), it remains to check that the matrix has essentially bounded entries.

Now, , the lower bound condition (c) in Theorem 5 and Lemma 6 imply that has bounded entries, and therefore the matrix has bounded entries. We have shown that has bounded entries; then Theorem 7, part (a) and Lemma 6 guarantee that the sequence is a Bessel sequence; then, the sequences and form a pair of dual frames in (see [1, Lemma 5.6.2]).

Finally, condition (ii) implies condition (i). According to [1, Lemma 5.6.2], the sequence is a frame for since it is a Bessel sequence and the expansion in (ii) holds. By using Theorem 7 we obtain that .

It is worth to observe that the analysis done in Theorem 9 provides a whole family of dual frames for the sequence . In fact, everything works if we replace in (57) by any matrix of the following form: