# On Some Sampling-Related Frames inU-Invariant Spaces

Authors: H. R. Fernández-Morales; A. G. García; M. A. Hernández-Medina; M. J. Muñoz-Bouzo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the characterization as frames of some sequences inU-invariant spaces of a separable Hilbert space ℋ where U denotes an unitary operator defined on ℋ; besides, the dual frames having the same form are also found. This general setting includes, in particular, shift-invariant or modulation-invariant subspaces in L2ℝ, 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 U to belong to a continuous group of unitary operators.


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## Body

## 1. Introduction

This paper is concerned with the study of some special frames inU-invariant spaces. Given an unitary operator U on a separable Hilbert space ℋ, we consider closed subspaces having the form 𝒜a≔span¯{Una:n∈ℤ}, where a denotes some fixed element in ℋ. In case that the sequence {Una}n∈ℤ is a Riesz basis for 𝒜a, we have
(1)𝒜a={∑n∈ℤ‍αnUna:{αn}∈ℓ2(ℤ)}.Recall that aRiesz basis in a separable Hilbert space is the image of an orthonormal basis by means of a bounded invertible operator. Any Riesz basis {xn}n∈ℤ has a unique biorthogonal (dual) Riesz basis {yn}n∈ℤ, that is, 〈xn,ym〉ℋ=δn,m, such that the expansions
(2)x=∑n∈ℤ‍〈x,yn〉ℋxn=∑n∈ℤ‍〈x,xn〉ℋyn
hold for every x∈ℋ. We state the definition by considering the integers set ℤ as the index set since throughout the paper most of sequences are indexed in ℤ. A Riesz sequence in ℋ is a Riesz basis for its closed span. A necessary and sufficient condition in order for the sequence {Una}n∈ℤ to be a Riesz sequence in ℋ is given in Theorem 3 infra.Givens elements bj, j=1,2,…,s, in 𝒜a, a challenging problem is to characterize the sequence {Urkbj}k∈ℤ;j=1,2,…,s as a frame (Riesz basis) in 𝒜a, where r≥1 denotes a positive integer. Besides, another interesting problem is to look for those dual frames having the same form {Urkcj}k∈ℤ;j=1,2,…,s for some cj∈𝒜a, j=1,2,…,s, so that, for any x∈𝒜a the expansion
(3)x=∑j=1s∑k∈ℤ〈x,Urkbj〉Urkcjinℋ
holds.At this point, we give an explanation about the expressionsampling-related frames appearing in the title. Namely, U-invariant subspaces in ℋ are natural generalizations of shift-invariant or modulation-invariant subspaces of L2(ℝ); there, the unitary involved operators are, respectively, the translation operator T:f(t)↦f(t-1) or the modulation operator M:f(t)↦e2πitf(t). In the shift-invariant subspace Vφ2={∑n‍αnφ(t-n):{αn}∈ℓ2(ℤ)}⊂L2(ℝ) generated by φ∈L2(ℝ), for any f∈Vφ2, the inner products {〈x,Urkbj〉}k∈ℤ;j=1,2,…,s are
(4)∫-∞∞‍f(t)bj(t-rk)¯dt=(f*hj)(rk),0000000000000k∈ℤ,j=1,2,…,s,
where hj(t)≔bj(-t)¯ for each j=1,2,…,s. Thus, the above inner products are nothing but samples of some filtered versions f*hj of the function f itself; this is precisely the generalized sampling problem in the shift-invariant space Vφ2. Mathematically, it consists of the stable recovery of any f∈Vφ2 from the above sequence of samples, that is, to obtain sampling formulas in Vφ2 having the following form:
(5)f(t)=∑j=1s‍∑k∈ℤ‍(f*hj)(rk)Sj(t-rk),t∈ℝ,
such that the sequence of reconstruction functions {Sj(·-rk)}k∈ℤ;j=1,2,…,s is a frame for the shift-invariant space Vφ2. As a consequence, expansions (3) and (5) have the same nature. Recall that a sequence {xn}n∈ℤ is a frame for a separable Hilbert space ℋ if there exist constants A,B>0 (frame bounds), such that
(6)A∥x∥2≤∑n∈ℤ‍|〈x,xn〉|2≤B∥x∥2∀x∈ℋ.A sequence{xn}n∈ℤ in ℋ satisfying only the right hand inequality above is said to be a Bessel sequence for ℋ. Given a frame {xn}n∈ℤ for ℋ, the representation property of any vector x∈ℋ as a series x=∑n∈ℤ‍cnxn is retained, but, unlike the case of Riesz bases (exact frames), the uniqueness of this representation (for overcomplete frames) is sacrificed. Suitable frame coefficients cn which depend continuously and linearly on x are obtained by using the dual frames {yn}n∈ℤ of {xn}n∈ℤ; that is, {yn}n∈ℤ is another frame for ℋ such that x=∑n∈ℤ‍〈x,yn〉xn=∑n∈ℤ‍〈x,xn〉yn for each x∈ℋ. For more details on the frame theory see [1].Sampling in shift-invariant spaces ofL2(ℝ) has been profusely treated in the mathematical literature (see, for instance, [2–13]).The existence of expansions like (3) in U-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 𝒜a of sequences having the form {Urkbj}k∈ℤ;j=1,2,…,s where bj∈𝒜a for each j=1,2,…,s. In other words, we carry out the study of the completeness, Bessel, frame, or Riesz basis properties of the sequence {Urkbj}k∈ℤ;j=1,2,…,s. Whenever it is a frame for 𝒜a, we find a family of dual frames having the same form {Urkcj}k∈ℤ;j=1,2,…,s for some cj∈𝒜a, j=1,2,…,s. In Section 4, we also discuss the case where some bj∉𝒜a; although the sequence {Urkbj}k∈ℤ;j=1,2,…,s is not contained in 𝒜a, something can be said in the light of the theory of pseudoframes (see [17, 18]).All the obtained results in Section2 involve the discrete group of unitary operators {Un}n∈ℤ which is completely determined by U. If we want to deal with something similar to the time-jitter error version of (5), that is, the recovery of any f∈Vφ2 from a perturbed sequence of samples {(f*hj)(rk+ϵkj)}k∈ℤ;j=1,2,…,s with errors ϵkj∈ℝ, then the availability of a continuous group of unitary operators {Ut}t∈ℝ containing, in particular, the operator U (say for instance U=U1) 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 {Urk+ϵkjbj}k∈ℤ;j=1,2,…,s in ℋ, for small enough error sequence ϵ≔{ϵkj}ℓ2-norm, the sequence is a Riesz sequence in ℋ. The second one goes into the recovery of any x∈𝒜a from the perturbed sequence of inner products {〈x,Urk+ϵkjbj〉ℋ}k∈ℤ;j=1,2,…,s, for small enough errors ϵmj, there exists a frame expansion for x∈𝒜a having the inner products {〈x,Urk+ϵkjbj〉ℋ}k∈ℤ;j=1,2,…,s as coefficients.


## 2.U-Invariant Subspaces

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


### 2.1. Some Preliminaries on Stationary Sequences

A sequences={sk}k∈ℤ in a separable Hilbert space ℋ is said to be stationary if
(7)〈sm+k,sn+k〉ℋ=〈sm,sn〉ℋ∀m,n,k∈ℤ.The functionRs(k)≔〈sk,s0〉ℋ, for every k∈ℤ, is called the autocovariance function of the sequence s. Moreover, two stationary sequences s={sk}k∈ℤ and w={wk}k∈ℤ are said to be stationary correlated if
(8)〈sm+k,wn+k〉ℋ=〈sm,wn〉ℋ∀m,n,k∈ℤ,
and Rs,w(k)≔〈sk,w0〉ℋ, for every k∈ℤ defines the corresponding cross-covariance function. The following result is a well-known characterization of stationary sequences (see [19]).Lemma 1.
To every stationary sequences={sn}n∈ℤ in a Hilbert space ℋ there exists a unique unitary operator U:ℋ→ℋ and s∈ℋ such that sn=Uns for all n∈ℤ. Conversely, every pair (U,s) of a unitary operator U and an s∈ℋ defines by sn=Uns, n∈ℤ, a stationary sequence s={sn}n∈ℤ in ℋ.
Moreover, two stationary sequences and w are stationary correlated if and only if they are generated by the same unitary operator U; that is, sn=Uns and wn=Unw for some s,w∈ℋ.The autocovarianceRs and the cross-covariance Rs,w functions admit a spectral representation which is related to the integral representation of the unitary operator U (see [19]).Theorem 2.
For every stationary sequences={sn}n∈ℤ in a Hilbert space ℋ, the autocovariance function Rs admits a spectral representation
(9)Rs(k)=〈sk,s0〉ℋ=12π∫-ππeikθdμs(θ),k∈ℤ,
in the form of an integral with respect to a (positive) spectral measureμs.
For every two stationary correlated sequencess={sn}n∈ℤ, w={wn}n∈ℤ in a Hilbert space ℋ, the cross-covariance function admits a spectral representation
(10)Rs,w(k)=〈sk,w0〉ℋ=12π∫-ππeikθdμs,w(θ),k∈ℤ,
in the form of an integral with respect to a (complex) spectral measure μs,w.


#### 2.1.1. TheU-Invariant Subspace 𝒜a

For a fixeda∈ℋ, we consider the subspace of ℋ given by 𝒜a≔span¯{Una,n∈ℤ}. In case that the sequence {Una}n∈ℤ is a Riesz sequence in ℋ, we have
(11)𝒜a={∑n∈ℤ‍αnUna:{αn}∈ℓ2(ℤ)}.A necessary and sufficient condition in order for the sequencea={Una}n∈ℤ to be a Riesz sequence in ℋ can be stated in terms of the Lebesgue decomposition of the spectral measure μa into an absolute and a singular part as dμa(θ)=ϕa(eiθ)dθ+dμas(θ).Theorem 3.
Leta≔{Una}n∈ℤ be a sequence obtained from an unitary operator in a separable Hilbert space ℋ with spectral measure dμa(θ)=ϕa(eiθ)dθ+dμas(θ), and let 𝒜a be the closed subspace spanned by a. Then, a is a Riesz basis for 𝒜a if and only if μas≡0 and
(12)0<Aa≔ess infζ∈𝕋ϕa(ζ)≤Ba≔ess supζ∈𝕋ϕa(ζ)<∞,
where 𝕋≔{eiθ:θ∈[-π,π)}.Proof.
For a fixedℓ2-sequence c≔{cn}n∈ℤ, we have
(13)∥∑k∈ℤ‍ckUka∥2=∑m∈ℤ‍∑n∈ℤ‍cmcn¯〈Uma,Una〉=∑m∈ℤ‍∑n∈ℤ‍cmcn¯12π∫-ππ‍ei(m-n)θdμa(θ)=12π∫-ππ‍∑m∈ℤ‍∑n∈ℤ‍cmcn¯ei(m-n)θdμa(θ)=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dμa(θ),
if μa is not absolutely continuous with respect to Lebesgue measure λ, Lemma 4 below implies that there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that ∥∑k∈ℤ‍ckNUka∥2 tends to infinity with N, so a cannot be a Bessel sequence, therefore, not a Riesz basis. Assume now that μas≡0 and a is a Riesz basis, this implies that there exists B<∞ such that
(14)∥∑k∈ℤ‍ckUka∥2≤B∥c∥ℓ22
for all c={ck}k∈ℤ∈ℓ2(ℤ). Let Ωn≔{ζ∈𝕋:ϕa(ζ)≥n}, assume that Ba=∞, this implies that λ(Ωn)>0 for all n. Now, (13), (14), and Parseval's theorem imply that
(15)12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ≤B2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ
for all c={ck}k∈ℤ∈ℓ2(ℤ). Introducing the Fourier expansion of the characteristic function χΩn∈L2(-π,π) in (15), we obtain n≤B for all n, which contradicts B<∞; thus, the assumption Ba=∞ is false. In a similar way, it can be proved that 0<Aa.
For the sufficient condition, assume now thatμas≡0; then, (13) implies that
(16)∥∑k∈ℤ‍ckUka∥2=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ,
if, in addition, condition (12) is satisfied, we get
(17)Aa2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ≤∥∑k∈ℤ‍ckUka∥2≤Ba2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ,
which implies that
(18)Aa∥c∥ℓ22≤∥∑k∈ℤ‍ckUka∥2≤Ba∥c∥ℓ22for  eachc∈ℓ2(ℤ).
Therefore, the sequence a is a Riesz basis for 𝒜a.Lemma 4.
Letμ be a finite positive measure on (-π,π) which is not absolutely continuous with respect to Lebesgue measure λ. Then, there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that
(19)limN→∞∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∞.Proof.
If the measureμ is not absolutely continuous with respect to Lebesgue measure, then μ(M)>0 for a Lebesgue measurable set M⊂(-π,π) of Lebesgue measure zero. Thus, there exists a Borel set B of Lebesgue measure zero such that M⊂B⊂(-π,π). In fact, the set B is an intersection of a countable collection of open sets (see [20, page 63]). Therefore, μ(B)>0 and λ(B)=0. On the other hand, every finite Borel measure on (-π,π) is inner regular (see [20, page 340]); that is,
(20)μ(B)=sup{μ(C):C⊂B,Ccompact},
then there exists a compact set C⊂(-π,π) such that
(21)μ(C)>0,λ(C)=0.
For anyɛ>0, there exists a sequence of disjoint open intervals Ij⊂(-π,π) such that
(22)C⊂⋃j=1∞‍Ij,∑j=1∞‍λ(Ij)≤λ(C)+ɛ=ɛ,
(see [20, pages 58 and 42]). Since C is compact, we may take the sequence to be finite. Hence, for every N∈ℕ, there exist open disjoint intervals I1N,I2N,…,IiNN in (-π,π) such that
(23)C⊂⋃i=1iN‍IiN,∑i=1iN‍λ(IiN)≤13N.
Besides, ∑i=1iN‍μ(IiN)≥μ(C).
Consider the functiongN:(-π,π)→ℝ, where gN=2N/2χ⋃i=1iN‍IiN, that satisfies
(24)∥gN∥22=2N∑i=1iN‍λ(IiN)≤2N3N<1.
We modify and extend eachgN to obtain a 2π-periodic function fN:ℝ→ℝ such that fN and its derivative are continuous on ℝ, ∥fN∥22≤1 and fN(θ)=gN(θ) for every θ∈⋃i=1iNIiN. Let ∑kckNeikθ be the Fourier series of fN. First, by using Parseval's identity, we have
(25)∥ckN∥ℓ22=12π∥fN∥22≤12πfor  everyN∈ℕ,
so that {cN}N=1∞ is a bounded sequence in ℓ2(ℤ). Besides, the regularity of each fN ensures that each Fourier series converges uniformly to fN. Therefore, each series ∑k‍ckNeikθ converges to fN in Lμ2(-π,π) and consequently,
(26)∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∫-ππ‍|fN|2dμ≥∫-ππ‍|gN|2dμ=2N∑i=1iN‍μ(IiN)≥2Nμ(C),
from which we obtain the desired result.The proof of Theorem3 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{Urkbj}k∈ℤ;j=1,2,…,s in 𝒜a

Forbj∈𝒜a, j=1,2,…,s, consider the sequence {Urkbj}k∈ℤ;j=1,2,…,s. For every j=1,2,…,s, the spectral measure μa,bj in the integral representation of the cross-covariance function of the sequences a≔{Uka}k∈ℤ and bj≔{Ukbj}k∈ℤ has no singular part. Indeed, according to Theorem 3, the spectral measure associated with the autocovariance function of the sequence {Uka}k∈ℤ 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 bk,j≔Urkbj; our goal in this section is to study the sequence {bk,j}k∈ℤ;j=1,2,…,s in 𝒜a in terms of an s×r matrix Ψa,b(eiθ) which is introduced below. For the sake of completeness, we include some needed calculations which appear in [14].First of all, we have(27)〈Uka,bn,j〉=12π∫-ππ‍ei(k-rn)θϕa,bj(eiθ)dθ,
where ϕa,bj stands for the cross spectral density of the stationary correlated sequences a≔{Uka}k∈ℤ and bj≔{Ukbj}k∈ℤ. Define
(28)Φa,b(eiθ)≔(ϕa,b1(eiθ),ϕa,b2(eiθ),…,ϕa,bs(eiθ))⊤.In what follows, we will use theleft-shift operatorS defined as
(29)S:L2(𝕋)⟶L2(𝕋)∑k∈ℤ‍akeikθ⟼∑k∈ℤ‍ak+1eikθ,
or equivalently, by (Sf)(eiθ)=f(eiθ)e-iθ, where 𝕋≔{eiθ:θ∈[-π,π)} denotes the unidimensional torus. Also, we will consider the decimation operatorDr, r is a positive integer, defined as
(30)Dr:L2(𝕋)⟶L2(𝕋)∑k∈ℤ‍akeikθ⟼∑k∈ℤ‍arkeikθ,
which can equivalently be written as
(31)(Drf)(eiθ)=1r∑k=0r-1‍f(ei((θ+2kπ)/r)).For eachl=0,1,…,r-1, set the s×1 matrix of functions on the torus 𝕋 as follows:
(32)Ψa,bl(eiθ)≔(DrS-lΦa,b)(eiθ),
and define the s×r matrix of functions on the torus 𝕋 as follows:
(33)Ψa,b(eiθ)≔(Ψa,b0(eiθ)Ψa,b1(eiθ)⋯Ψa,br-1(eiθ)).It is worth to mention that the matrixΨa,b was explicitly computed in [14] for the translation and modulation cases in L2(ℝ).Next, for anyx∈𝒜a, we obtain an expression for the inner products αn,j≔〈x,Urnbj〉, n∈ℤ and j=1,2,…,s. Indeed, writing x=∑k∈ℤ‍xkUka where {xk}k∈ℤ∈ℓ2(ℤ), we have
(34)αn,j=〈x,Urnbj〉=∑k∈ℤ‍xk〈Uka,Urnbj〉=∑k∈ℤ‍xk12π∫-ππ‍ei(k-rn)θϕa,bj(eiθ)dθ=12π∫-ππ‍∑k∈ℤ‍xkeikθϕa,bj(eiθ)e-irnθdθ,
that is,
(35)αn≔(αn,1,αn,2,…,αn,s)⊤αn=12π∫-ππ‍Φa,b(eiθ)X(eiθ)e-irnθdθ,
where X(eiθ)≔∑k∈ℤ‍xkeikθ.Now, forl=0,1,…,r-1, define the sequence x(l)≔{xk(l)≔xkr+l}k∈ℤ. Thus, we can write
(36)X(eiθ)=∑l=0r-1‍∑k∈ℤ‍xkr+lei(kr+l)θ=∑l=0r-1‍X(l)(eirθ)eilθ,
where X(l)(eiθ)=∑k∈ℤ‍xk(l)eikθ.Using (36) in (35), we obtain
(37)αn=∑l=0r-1‍12π∫-ππ‍Φa,b(eiθ)X(l)(eirθ)eilθe-irnθdθ.After some easy calculations, we get(38)αn=∑l=0r-1‍12π∫-ππ‍S-lΦa,b(eiθ)X(l)(eirθ)e-irnθdθ=∑l=0r-1‍12π∫-rπrπ‍S-lΦa,b(ei(θ/r))rX(l)(eiθ)e-inθdθ=∑l=0r-1‍∑k=0r-1‍∫2πk2π(k+1)‍S-lΦa,b(ei(θ/r))2πrX(l)(eiθ)e-inθdθ=∫02π‍∑l=0r-1‍∑k=0r-1‍S-lΦa,b(ei((θ+2πk)/r))2πrX(l)(eiθ)e-inθdθ=12π∫-ππ‍∑l=0r-1‍(DrS-lΦa,b)(eiθ)X(l)(eiθ)e-inθdθ.DefiningC(eiθ)≔∑k∈ℤ‍αkeikθ, (38) implies that
(39)C(eiθ)=∑l=0r-1‍(DrS-lΦa,b)(eiθ)X(l)(eiθ),
which can be written in matrix form as
(40)C(eiθ)=(∑k∈ℤ‍αk,1eikθ,∑k∈ℤ‍αk,2eikθ,…,∑k∈ℤ‍αk,seikθ)⊤=Ψa,b(eiθ)(X(0)(eiθ),X(1)(eiθ),…,X(r-1)(eiθ))⊤=Ψa,b(eiθ)X~(eiθ)=(LΨa,bX~)(eiθ),
where LΨa,b:Lr2(𝕋)→Ls2(𝕋) denotes the multiplication operator by Ψa,b and
(41)X~(eiθ)≔(X(0)(eiθ),X(1)(eiθ),…,X(r-1)(eiθ))⊤.ByLr2(𝕋) (resp., Ls2(𝕋)), we denote the product space L2(𝕋)×⋯×L2(𝕋) for r times (resp., s times). Besides,
(42)∥Ψa,bX~∥Ls2(𝕋)2=12π∫-ππ‍〈Ψa,b(eiθ)X~(eiθ),Ψa,b(eiθ)X~(eiθ)〉ℂrdθ=12π∫-ππ‍〈Ψa,b*(eiθ)Ψa,b(eiθ)X~(eiθ),X~(eiθ)〉ℂrdθ.The above calculations let us prove the following result.Theorem 5.
Letbj∈𝒜a for j=1,2,…,s and let Ψa,b be the associated matrix given in (33). Then, the following results hold. (a)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a complete system in 𝒜a if and only if the rank of the matrix Ψa,b(ζ) is r a.e. ζ in 𝕋.(b)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a if and only if there exists a constant B<∞ such that
(43)Ψa,b*(ζ)Ψa,b(ζ)≤B𝕀ra.e.ζin𝕋.(c)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a if and only if there exist constants 0<A≤B<∞ such that
(44)A𝕀r≤Ψa,b*(ζ)Ψa,b(ζ)≤B𝕀ra.e.ζin𝕋.(d)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Riesz basis for 𝒜a if and only if it is a frame and s=r.Proof.
To prove (a), assume that there exists a setΩ⊆𝕋 with positive measure such that rank[Ψa,b(ζ)]<r for each ζ∈Ω. Then, there exists a measurable function v(ζ), ζ∈Ω, such that Ψa,b(ζ)v(ζ)=0 and ∥v(ζ)∥Lr2(𝕋)=1 in Ω. This function can be constructed as in [23, Lemma 2.4]. Define V~∈Lr2(𝕋) such that V~(ζ)=v(ζ) if ζ∈Ω, and V~(ζ)=0 if ζ∈𝕋∖Ω. Hence, from (40), we obtain that the system is not complete. Conversely, if the system is not complete, by using (40), we obtain a V~(ζ) different from 0 in a set with positive measure such that Ψa,b(ζ)V~(ζ)=0. Thus, rankΨa,b(ζ)<r on a set with positive measure.
To prove (b), we keep in mind that{Uka}k∈ℤ is a Riesz basis for 𝒜a, the mapping T:ℓ2(ℤ)→𝒜a, given by T{xk}k∈ℤ=x=∑k∈ℤ‍xkUka is bijective and there exist two constants 0<ma≤Ma<∞ such that
(45)ma∥{xk}∥ℓ22≤∥T{xk}∥ℋ2≤Ma∥{xk}∥ℓ22.
Assume first that (43) is satisfied. It follows from (40) and (42) that
(46)∥Ψa,bX~∥Ls2(𝕋)2≤B∥X~∥Lr2(𝕋)2.
By construction∥Ψa,bX~∥Ls2(𝕋)2=∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2 and ∥X~∥Lr2(𝕋)2=∥{xk}k∈ℤ∥ℓ22, using (45), it follows from (46) that
(47)∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2≤Bma∥x∥ℋ2.
Conversely, assume that {bkj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a, then there exists 0<B′<∞ such that
(48)∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2≤B′∥x∥ℋ2.
Using (45), this implies that
(49)∥Ψa,bX~∥Ls2(𝕋)2≤B′Ma∥X~∥Lr2(𝕋)2
for all X~∈Lr2(𝕋). Inserting the right hand side of (42) for ∥Ψa,bX~∥Ls2(𝕋)2, it is straightforward to see that (43) holds with B=B′Ma. The proof of (c) is completed proceeding as in (b).
To prove (d), consider the following mapping:(50)S:𝒜a⟶ℓs2(ℤ)x⟼{〈x,bk,j〉}k∈ℤ;j=1,2,…,s.
According to (40), the mapping S is isometric equivalent to LΨa,b, and assuming that {bk,j}k∈ℤ;j=1,2,…,s is a frame, it is a Riesz basis if and only if S is surjective.
First, if{bk,j}k∈ℤ;j=1,2,…,s is a Riesz basis, then it is a frame and S is surjective. Applying (a) yields that LΨa,b is bijective, and therefore LΨa,b*=LΨa,b* is bijective. Hence, rank[Ψa,b(ζ)Ψa,b*(ζ)] is s for almost every ζ in 𝕋, so
(51)r=rank[Ψa,b*(ζ)Ψa,b(ζ)]=rank[Ψa,b(ζ)Ψa,b*(ζ)]=s,
and finally s=r.
Conversely, if{bk,j}k∈ℤ;j=1,2,…,s is a frame and s=r, (a) implies that Ψa,b(ζ) is invertible for almost every ζ in 𝕋, which implies that LΨa,b is surjective; then S is surjective and {bk,j}k∈ℤ;j=1,2,…,s is a Riesz basis.The following lemma will allow us to restate Theorem5.Lemma 6.
LetG(ζ) be an s×r matrix with entries in L2(𝕋), and consider the following constants:
(52)AG≔ess infζ∈𝕋λmin[G*(ζ)G(ζ)],BG≔ess supζ∈𝕋λmax[G*(ζ)G(ζ)],
where λmin (resp., λmax) denotes the smallest (resp., the largest) eigenvalue of the positive semidefinite matrix G*(ζ)G(ζ). Then, (a)
the matrixG(ζ) has essentially bounded entries on 𝕋 if and only if BG<∞;(b)
there exist constants0<A≤B<∞ such that A𝕀r≤G*(ζ)G(ζ)≤B𝕀r, a.e., ζ∈𝕋 if and only if 0<AG≤BG<∞.Proof.
The first part of lemma follows from thatλmax[G*(ζ)G(ζ)]=∥G(ζ)∥22, and
(53)maxi,j|aij|≤∥A∥2≤mnmaxi,j|aij|.for  any  matrixA=[aij]i=1,2…,mj=1,2…,n,
where ∥A∥2 denotes the spectral norm of the matrix A (see, for instance, [24]).
Now, we prove the second part of the lemma. SinceG*(ζ)G(ζ)≤B𝕀r means that 〈Bx-G*(ζ)G(ζ)x,x〉≥0 for all x∈ℂr, in particular, taking an eigenvector x associated with the largest eigenvalue λmax of G*(ζ)G(ζ) such that ∥x∥=1, one has that B≥λmax(G*(ζ)G(ζ)). Hence, B≥ess supζ∈𝕋λmax[G*(ζ)G(ζ)]. In a similar way, A𝕀r≤G*(ζ)G(ζ) implies that A≤ess infζ∈𝕋λmin[G*(ζ)G(ζ)].
Conversely, Rayleigh-Ritz theorem [24, page 176] yields that
(54)λmax[G*(ζ)G(ζ)]=maxx∈ℂrx*G*(ζ)G(ζ)x*x=maxx∈ℂr〈G*(ζ)G(ζ)x,x〉〈x,x〉.
Thus, ess supζ∈𝕋λmax[G*(ζ)G(ζ)]=BG implies that
(55)maxx∈ℂr〈G*(ζ)G(ζ)x,x〉〈x,x〉≤BG,a.e.ζ∈𝕋.
In other words,BG𝕀r≥G*(ζ)G(ζ); analogously, G*(ζ)G(ζ)≥AG𝕀r.
It is easy to deduce from the proof thatAG and BG are the optimal constants A>0 and B<∞ satisfying the inequalities A𝕀r≤G*(ζ)G(ζ)≤B𝕀r, a.e., ζ∈𝕋.As a consequence of Lemma6, statements (b) and (c) in Theorem 5 can be restated in terms of the following constants:
(56)AΨ≔ess infζ∈𝕋λmin[Ψa,b*(ζ)Ψa,b(ζ)];BΨ≔ess supζ∈𝕋λmax[Ψa,b*(ζ)Ψa,b(ζ)],
as follows.Theorem 7.
Letbj∈𝒜a for j=1,2,…,s, and let Ψa,b be the associated matrix given in (33) and its related constants (56). Then, the following results hold.(i)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a if and only if the constant BΨ<∞.(ii)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a if and only if the constants AΨ and BΨ satisfy 0<AΨ≤BΨ<∞. In this case, AΨ and BΨ are the optimal frame bounds for {Urkbj}k∈ℤ;j=1,2,…,s.


### 2.3. The Frame Expansion

Define ther×s matrix of functions on the torus 𝕋 as follows:
(57)Γ(eiθ)≔∑k∈ℤΓkeikθ=[Ψa,b*(eiθ)Ψa,b(eiθ)]-1Ψa,b*(eiθ).
The following expansion involving the inner products αn,j=〈x,Urnbj〉 of x∈𝒜a holds.Lemma 8.
Assume that the matrixΨa,b(ζ) has essentially bounded entries on 𝕋. For any x=∑k∈ℤxkUka∈𝒜a, one has
(58)x~n=∑k∈ℤ‍Γkαn-k,
where x~n denotes the nth Fourier coefficient of the function X~(eiθ) defined in (41), and the sequence {αn}n∈ℤ is given in (35).Proof.
Indeed,(59)x~n=12π∫-ππ‍X~(eiθ)e-inθdθ=12π∫-ππ‍(∑k∈ℤ‍Γkeikθ)Ψa,b(eiθ)X~(eiθ)e-inθdθ=∑k∈ℤ‍Γk12π∫-ππ‍Ψa,b(eiθ)X~(eiθ)e-i(n-k)θdθ=∑k∈ℤ‍Γkαn-k.At this point, we are ready to prove the following expansion result.Theorem 9.
Letbj∈𝒜a for j=1,2,…,s, and assume that the associated matrix Ψa,b given in (33) has essentially bounded entries on 𝕋, that is, BΨ<∞. The following statements are equivalent. (i)
The constantAΨ>0.(ii)
There existcj∈𝒜a, j=1,2,…,s, such that the sequence {Urkcj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a, yielding, for any x∈𝒜a, the following expansion:
(60)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Urkcjinℋ.
In case the equivalent conditions hold,{Urkbj}k∈ℤ;j=1,2,…,s and {Urkcj}k∈ℤ;j=1,2,…,s form a pair of dual frames in 𝒜a.Proof.
First, we prove that (i) implies (ii). Observe thatx=∑k∈ℤxkUka can be written as ∑n∈ℤ‍x~n⊤a~n where a~n=(Unra,Unr+1a,…,Unr+r-1a)⊤. Next,
(61)x=∑n∈ℤ‍x~n⊤a~n=∑n∈ℤ‍(∑k∈ℤ‍Γkαn-k)⊤a~n=∑n∈ℤ‍∑k∈ℤ‍αn-k⊤Γk⊤a~n=∑n∈ℤ‍∑k∈ℤ‍αn⊤Γk⊤a~n+k=∑n∈ℤ‍αn⊤(∑k∈ℤ‍Γk⊤a~n+k).
For l∈ℤ and j=1,2,…,s define cl,j≔Urlcj, where (c1,c2,…,cs)⊤=∑k∈ℤ‍Γk⊤a~k, and bl,j≔Urlbj. Then, (61) implies
(62)x=∑n∈ℤ‍αn⊤(∑k∈ℤ‍Γk⊤a~n+k)=∑n∈ℤ‍αn⊤Unr(∑k∈ℤ‍Γk⊤a~k)=∑l=1s‍∑n∈ℤ‍〈x,bn,l〉cn,linℋ.
In order to be allowed to use [1, Lemma 5.6.2], we have to prove that the above constructed sequence {ck,j}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a. To this end, we compute the corresponding Ψa,c matrix for c≔{ck,j}k∈ℤ;j=1,2,…,s. Setting [Γk⊤]ij=aijk, we obtain
(63)〈Uka,cn,j〉=∑l∈ℤ‍∑i=1r‍〈Uka,Unr(ajilUlr+ir+i-1a)〉=∑l∈ℤ‍∑i=1r‍a¯jil〈Uk-nr-lr-i+1a,a〉=∑l∈ℤ‍∑i=1r‍a¯jil12π∫-ππ‍ei(k-nr-lr-i+1)θϕa(eiθ)dθ.
Now,(64)(〈Uka,cn,1〉〈Uka,cn,2〉⋮〈Uka,cn,s〉)=∑l∈ℤ‍Γ¯l⊤(12π∫-ππ‍ei(k-nr-lr)θϕa(eiθ)dθ12π∫-ππ‍ei(k-nr-lr-1)θϕa(eiθ)dθ⋮12π∫-ππ‍ei(k-nr-lr-r+1)θϕa(eiθ)dθ)=12π∫-ππ‍∑l∈ℤ‍Γ¯l⊤e-ilrθ(ei(k-nr)θei(k-nr-1)θ⋮ei(k-nr-r+1)θ)ϕa(eiθ)dθ=12π∫-ππ‍ei(k-nr)θΓ*(eirθ)e~(eiθ)ϕa(eiθ)dθ,
where e~(eiθ)≔(1,e-iθ,…,e-i(r-1)θ)⊤. Hence, we have deduced that
(65)Φa,c(eiθ)=Γ*(eirθ)e~(eiθ)ϕa(eiθ).
Therefore, forl=0,1,…,r-1, we have
(66)Ψa,cl(eiθ)≔DrS-l[Γ*(eirθ)e~(eiθ)ϕa(eiθ)],
and consequently, the s×r matrix Ψa,c(eiθ)≔(Ψa,c0(eiθ),Ψa,c1(eiθ),…,Ψa,cr-1(eiθ)) can be written as
(67)Ψa,c(eiθ)=Dr[ϕa(eiθ)Γ*(eirθ)E~(eiθ)],
where
(68)E~(eiθ)≔(1eiθ⋯ei(r-1)θe-iθ1⋯ei(r-2)θ⋮⋮⋱⋮e-i(r-1)θe-i(r-2)θ⋯1).
As a consequence of Theorem7, the proof ends if we prove that the matrix Ψa,c(eiθ) has essentially bounded entries. Clearly, the decimation operator Dr sends bounded functions into bounded functions; Theorem 3 implies that ϕa is bounded so, taking into account (67), it remains to check that the matrix Γ*(eirθ) has essentially bounded entries.
Now,Γ*(eirθ)=Ψa,b(eirθ)[Ψa,b*(eirθ)Ψa,b(eirθ)]-1, the lower bound condition (c) in Theorem 5 and Lemma 6 imply that [Ψa,b*(eirθ)Ψa,b(eirθ)]-1 has bounded entries, and therefore the matrix Γ*(eirθ) has bounded entries. We have shown that Ψa,c(eiθ) has bounded entries; then Theorem 7, part (a) and Lemma 6 guarantee that the sequence {ck,j}k∈ℤ;j=1,2,…,s is a Bessel sequence; then, the sequences {bk,j}k∈ℤ,j=1,2,…,s and {ck,j}k∈ℤ,j=1,2,…,s form a pair of dual frames in 𝒜a (see [1, Lemma 5.6.2]).
Finally, condition (ii) implies condition (i). According to [1, Lemma 5.6.2], the sequence {Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a since it is a Bessel sequence and the expansion in (ii) holds. By using Theorem 7 we obtain that AΨ>0.It is worth to observe that the analysis done in Theorem9 provides a whole family of dual frames for the sequence {Urkbj}k∈ℤ;j=1,2,…,s. In fact, everything works if we replace Γ(eiθ) in (57) by any matrix of the following form:
(69)Γ𝕌(eiθ)≔Ψa,b†(eiθ)+𝕌(eiθ)[𝕀s-Ψa,b(eiθ)Ψa,b†(eiθ)],
where 𝕌(eiθ) is any r×s matrix with entries in L∞(𝕋), and Ψa,b† denotes the Moore-Penrose left-inverse Ψa,b†(eiθ)≔[Ψa,b*(eiθ)Ψa,b(eiθ)]-1Ψa,b*(eiθ). Note that we need essentially bounded entries in the matrix Γ𝕌(eiθ) since the multiplication operator MF:f↦Ff in L2(𝕋) is well-defined (and consequently bounded) if and only if F∈L∞(𝕋).Notice that ifs=r, we have Ψa,b†=Ψa,b-1 which implies a unique Γ𝕌, and we are in the presence of a pair of dual Riesz bases. In fact, the following result holds.Corollary 10.
Letbj∈𝒜a for j=1,2,…,r; that is, r=s in Theorem 9. Assume that the square matrix Ψa,b given in (33) has entries essentially bounded on 𝕋, that is, BΨ<∞. The following statements are equivalent.(a)
The constantAΨ>0.(b)
There exists a Riesz basis{Ck,j}k∈ℤ;j=1,2,…,s such that for any x∈𝒜a, the expansion
(70)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Ck,jinℋholds.
In case the equivalent conditions are satisfied, necessarily, there existcj∈𝒜a, j=1,2,…,r, such that Ck,j=Urkcj for k∈ℤ and j=1,2,…,r. Moreover, the sequences {Urkcj}k∈ℤ;j=1,2,…,s and {Urkbj}k∈ℤ;j=1,2,…,s are dual Riesz bases in 𝒜a, and the interpolation property 〈cj,Urkbj′〉=δj,j′δk,0, where k∈ℤ and j,j′=1,2,…,r, holds.Proof.
To prove(a)⇒(b), we use Theorem 9; whenever 0<AΨ≤BΨ<∞, there exist cj∈𝒜a, j=1,2,…,s, such that the sequence {Urkcj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a and for any x∈𝒜a, the expansion
(71)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Urkcjinℋ,
holds. Actually, from Theorem 5, we get that r=s implies that {Urkbj}k∈ℤ;j=1,2,…,s is a Riesz basis, and consequently, {Urkcj}k∈ℤ;j=1,2,…,s is indeed its dual Riesz basis.
The converse follows easily from the fact that if{Ck,j}k∈ℤ;j=1,2,…,s is a Riesz basis, then (b) implies that {Urkbj}k∈ℤ;j=1,2,…,s is its dual Riesz basis; hence, Theorem 5 provides AΨ>0. The interpolation property comes out from the biorthogonal condition of a pair of dual Riesz bases.Closing this section it is worth to mention that the results stated and proved in Sections2.2 and 2.3 mathematically enrich some of the remarkable results concerning regular sampling contained in the interesting [14]. Here, we have assumed only one generator a∈ℋ and that bj∈𝒜a for all j=1,2,…,s. If bj∉𝒜a for some j, see the additional remarks in Section 4. The case of several generators al∈ℋ, l=1,2,…,L, can be essentially treated in the same way.


## 2.1. Some Preliminaries on Stationary Sequences

A sequences={sk}k∈ℤ in a separable Hilbert space ℋ is said to be stationary if
(7)〈sm+k,sn+k〉ℋ=〈sm,sn〉ℋ∀m,n,k∈ℤ.The functionRs(k)≔〈sk,s0〉ℋ, for every k∈ℤ, is called the autocovariance function of the sequence s. Moreover, two stationary sequences s={sk}k∈ℤ and w={wk}k∈ℤ are said to be stationary correlated if
(8)〈sm+k,wn+k〉ℋ=〈sm,wn〉ℋ∀m,n,k∈ℤ,
and Rs,w(k)≔〈sk,w0〉ℋ, for every k∈ℤ defines the corresponding cross-covariance function. The following result is a well-known characterization of stationary sequences (see [19]).Lemma 1.
To every stationary sequences={sn}n∈ℤ in a Hilbert space ℋ there exists a unique unitary operator U:ℋ→ℋ and s∈ℋ such that sn=Uns for all n∈ℤ. Conversely, every pair (U,s) of a unitary operator U and an s∈ℋ defines by sn=Uns, n∈ℤ, a stationary sequence s={sn}n∈ℤ in ℋ.
Moreover, two stationary sequences and w are stationary correlated if and only if they are generated by the same unitary operator U; that is, sn=Uns and wn=Unw for some s,w∈ℋ.The autocovarianceRs and the cross-covariance Rs,w functions admit a spectral representation which is related to the integral representation of the unitary operator U (see [19]).Theorem 2.
For every stationary sequences={sn}n∈ℤ in a Hilbert space ℋ, the autocovariance function Rs admits a spectral representation
(9)Rs(k)=〈sk,s0〉ℋ=12π∫-ππeikθdμs(θ),k∈ℤ,
in the form of an integral with respect to a (positive) spectral measureμs.
For every two stationary correlated sequencess={sn}n∈ℤ, w={wn}n∈ℤ in a Hilbert space ℋ, the cross-covariance function admits a spectral representation
(10)Rs,w(k)=〈sk,w0〉ℋ=12π∫-ππeikθdμs,w(θ),k∈ℤ,
in the form of an integral with respect to a (complex) spectral measure μs,w.


### 2.1.1. TheU-Invariant Subspace 𝒜a

For a fixeda∈ℋ, we consider the subspace of ℋ given by 𝒜a≔span¯{Una,n∈ℤ}. In case that the sequence {Una}n∈ℤ is a Riesz sequence in ℋ, we have
(11)𝒜a={∑n∈ℤ‍αnUna:{αn}∈ℓ2(ℤ)}.A necessary and sufficient condition in order for the sequencea={Una}n∈ℤ to be a Riesz sequence in ℋ can be stated in terms of the Lebesgue decomposition of the spectral measure μa into an absolute and a singular part as dμa(θ)=ϕa(eiθ)dθ+dμas(θ).Theorem 3.
Leta≔{Una}n∈ℤ be a sequence obtained from an unitary operator in a separable Hilbert space ℋ with spectral measure dμa(θ)=ϕa(eiθ)dθ+dμas(θ), and let 𝒜a be the closed subspace spanned by a. Then, a is a Riesz basis for 𝒜a if and only if μas≡0 and
(12)0<Aa≔ess infζ∈𝕋ϕa(ζ)≤Ba≔ess supζ∈𝕋ϕa(ζ)<∞,
where 𝕋≔{eiθ:θ∈[-π,π)}.Proof.
For a fixedℓ2-sequence c≔{cn}n∈ℤ, we have
(13)∥∑k∈ℤ‍ckUka∥2=∑m∈ℤ‍∑n∈ℤ‍cmcn¯〈Uma,Una〉=∑m∈ℤ‍∑n∈ℤ‍cmcn¯12π∫-ππ‍ei(m-n)θdμa(θ)=12π∫-ππ‍∑m∈ℤ‍∑n∈ℤ‍cmcn¯ei(m-n)θdμa(θ)=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dμa(θ),
if μa is not absolutely continuous with respect to Lebesgue measure λ, Lemma 4 below implies that there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that ∥∑k∈ℤ‍ckNUka∥2 tends to infinity with N, so a cannot be a Bessel sequence, therefore, not a Riesz basis. Assume now that μas≡0 and a is a Riesz basis, this implies that there exists B<∞ such that
(14)∥∑k∈ℤ‍ckUka∥2≤B∥c∥ℓ22
for all c={ck}k∈ℤ∈ℓ2(ℤ). Let Ωn≔{ζ∈𝕋:ϕa(ζ)≥n}, assume that Ba=∞, this implies that λ(Ωn)>0 for all n. Now, (13), (14), and Parseval's theorem imply that
(15)12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ≤B2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ
for all c={ck}k∈ℤ∈ℓ2(ℤ). Introducing the Fourier expansion of the characteristic function χΩn∈L2(-π,π) in (15), we obtain n≤B for all n, which contradicts B<∞; thus, the assumption Ba=∞ is false. In a similar way, it can be proved that 0<Aa.
For the sufficient condition, assume now thatμas≡0; then, (13) implies that
(16)∥∑k∈ℤ‍ckUka∥2=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ,
if, in addition, condition (12) is satisfied, we get
(17)Aa2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ≤∥∑k∈ℤ‍ckUka∥2≤Ba2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ,
which implies that
(18)Aa∥c∥ℓ22≤∥∑k∈ℤ‍ckUka∥2≤Ba∥c∥ℓ22for  eachc∈ℓ2(ℤ).
Therefore, the sequence a is a Riesz basis for 𝒜a.Lemma 4.
Letμ be a finite positive measure on (-π,π) which is not absolutely continuous with respect to Lebesgue measure λ. Then, there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that
(19)limN→∞∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∞.Proof.
If the measureμ is not absolutely continuous with respect to Lebesgue measure, then μ(M)>0 for a Lebesgue measurable set M⊂(-π,π) of Lebesgue measure zero. Thus, there exists a Borel set B of Lebesgue measure zero such that M⊂B⊂(-π,π). In fact, the set B is an intersection of a countable collection of open sets (see [20, page 63]). Therefore, μ(B)>0 and λ(B)=0. On the other hand, every finite Borel measure on (-π,π) is inner regular (see [20, page 340]); that is,
(20)μ(B)=sup{μ(C):C⊂B,Ccompact},
then there exists a compact set C⊂(-π,π) such that
(21)μ(C)>0,λ(C)=0.
For anyɛ>0, there exists a sequence of disjoint open intervals Ij⊂(-π,π) such that
(22)C⊂⋃j=1∞‍Ij,∑j=1∞‍λ(Ij)≤λ(C)+ɛ=ɛ,
(see [20, pages 58 and 42]). Since C is compact, we may take the sequence to be finite. Hence, for every N∈ℕ, there exist open disjoint intervals I1N,I2N,…,IiNN in (-π,π) such that
(23)C⊂⋃i=1iN‍IiN,∑i=1iN‍λ(IiN)≤13N.
Besides, ∑i=1iN‍μ(IiN)≥μ(C).
Consider the functiongN:(-π,π)→ℝ, where gN=2N/2χ⋃i=1iN‍IiN, that satisfies
(24)∥gN∥22=2N∑i=1iN‍λ(IiN)≤2N3N<1.
We modify and extend eachgN to obtain a 2π-periodic function fN:ℝ→ℝ such that fN and its derivative are continuous on ℝ, ∥fN∥22≤1 and fN(θ)=gN(θ) for every θ∈⋃i=1iNIiN. Let ∑kckNeikθ be the Fourier series of fN. First, by using Parseval's identity, we have
(25)∥ckN∥ℓ22=12π∥fN∥22≤12πfor  everyN∈ℕ,
so that {cN}N=1∞ is a bounded sequence in ℓ2(ℤ). Besides, the regularity of each fN ensures that each Fourier series converges uniformly to fN. Therefore, each series ∑k‍ckNeikθ converges to fN in Lμ2(-π,π) and consequently,
(26)∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∫-ππ‍|fN|2dμ≥∫-ππ‍|gN|2dμ=2N∑i=1iN‍μ(IiN)≥2Nμ(C),
from which we obtain the desired result.The proof of Theorem3 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].


## 2.1.1. TheU-Invariant Subspace 𝒜a

For a fixeda∈ℋ, we consider the subspace of ℋ given by 𝒜a≔span¯{Una,n∈ℤ}. In case that the sequence {Una}n∈ℤ is a Riesz sequence in ℋ, we have
(11)𝒜a={∑n∈ℤ‍αnUna:{αn}∈ℓ2(ℤ)}.A necessary and sufficient condition in order for the sequencea={Una}n∈ℤ to be a Riesz sequence in ℋ can be stated in terms of the Lebesgue decomposition of the spectral measure μa into an absolute and a singular part as dμa(θ)=ϕa(eiθ)dθ+dμas(θ).Theorem 3.
Leta≔{Una}n∈ℤ be a sequence obtained from an unitary operator in a separable Hilbert space ℋ with spectral measure dμa(θ)=ϕa(eiθ)dθ+dμas(θ), and let 𝒜a be the closed subspace spanned by a. Then, a is a Riesz basis for 𝒜a if and only if μas≡0 and
(12)0<Aa≔ess infζ∈𝕋ϕa(ζ)≤Ba≔ess supζ∈𝕋ϕa(ζ)<∞,
where 𝕋≔{eiθ:θ∈[-π,π)}.Proof.
For a fixedℓ2-sequence c≔{cn}n∈ℤ, we have
(13)∥∑k∈ℤ‍ckUka∥2=∑m∈ℤ‍∑n∈ℤ‍cmcn¯〈Uma,Una〉=∑m∈ℤ‍∑n∈ℤ‍cmcn¯12π∫-ππ‍ei(m-n)θdμa(θ)=12π∫-ππ‍∑m∈ℤ‍∑n∈ℤ‍cmcn¯ei(m-n)θdμa(θ)=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dμa(θ),
if μa is not absolutely continuous with respect to Lebesgue measure λ, Lemma 4 below implies that there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that ∥∑k∈ℤ‍ckNUka∥2 tends to infinity with N, so a cannot be a Bessel sequence, therefore, not a Riesz basis. Assume now that μas≡0 and a is a Riesz basis, this implies that there exists B<∞ such that
(14)∥∑k∈ℤ‍ckUka∥2≤B∥c∥ℓ22
for all c={ck}k∈ℤ∈ℓ2(ℤ). Let Ωn≔{ζ∈𝕋:ϕa(ζ)≥n}, assume that Ba=∞, this implies that λ(Ωn)>0 for all n. Now, (13), (14), and Parseval's theorem imply that
(15)12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ≤B2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ
for all c={ck}k∈ℤ∈ℓ2(ℤ). Introducing the Fourier expansion of the characteristic function χΩn∈L2(-π,π) in (15), we obtain n≤B for all n, which contradicts B<∞; thus, the assumption Ba=∞ is false. In a similar way, it can be proved that 0<Aa.
For the sufficient condition, assume now thatμas≡0; then, (13) implies that
(16)∥∑k∈ℤ‍ckUka∥2=12π∫-ππ‍|∑k∈ℤ‍ckeikθ|2ϕa(eiθ)dθ,
if, in addition, condition (12) is satisfied, we get
(17)Aa2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ≤∥∑k∈ℤ‍ckUka∥2≤Ba2π∫-ππ‍|∑k∈ℤ‍ckeikθ|2dθ,
which implies that
(18)Aa∥c∥ℓ22≤∥∑k∈ℤ‍ckUka∥2≤Ba∥c∥ℓ22for  eachc∈ℓ2(ℤ).
Therefore, the sequence a is a Riesz basis for 𝒜a.Lemma 4.
Letμ be a finite positive measure on (-π,π) which is not absolutely continuous with respect to Lebesgue measure λ. Then, there exists a bounded sequence {cN}N=1∞⊂ℓ2(ℤ) such that
(19)limN→∞∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∞.Proof.
If the measureμ is not absolutely continuous with respect to Lebesgue measure, then μ(M)>0 for a Lebesgue measurable set M⊂(-π,π) of Lebesgue measure zero. Thus, there exists a Borel set B of Lebesgue measure zero such that M⊂B⊂(-π,π). In fact, the set B is an intersection of a countable collection of open sets (see [20, page 63]). Therefore, μ(B)>0 and λ(B)=0. On the other hand, every finite Borel measure on (-π,π) is inner regular (see [20, page 340]); that is,
(20)μ(B)=sup{μ(C):C⊂B,Ccompact},
then there exists a compact set C⊂(-π,π) such that
(21)μ(C)>0,λ(C)=0.
For anyɛ>0, there exists a sequence of disjoint open intervals Ij⊂(-π,π) such that
(22)C⊂⋃j=1∞‍Ij,∑j=1∞‍λ(Ij)≤λ(C)+ɛ=ɛ,
(see [20, pages 58 and 42]). Since C is compact, we may take the sequence to be finite. Hence, for every N∈ℕ, there exist open disjoint intervals I1N,I2N,…,IiNN in (-π,π) such that
(23)C⊂⋃i=1iN‍IiN,∑i=1iN‍λ(IiN)≤13N.
Besides, ∑i=1iN‍μ(IiN)≥μ(C).
Consider the functiongN:(-π,π)→ℝ, where gN=2N/2χ⋃i=1iN‍IiN, that satisfies
(24)∥gN∥22=2N∑i=1iN‍λ(IiN)≤2N3N<1.
We modify and extend eachgN to obtain a 2π-periodic function fN:ℝ→ℝ such that fN and its derivative are continuous on ℝ, ∥fN∥22≤1 and fN(θ)=gN(θ) for every θ∈⋃i=1iNIiN. Let ∑kckNeikθ be the Fourier series of fN. First, by using Parseval's identity, we have
(25)∥ckN∥ℓ22=12π∥fN∥22≤12πfor  everyN∈ℕ,
so that {cN}N=1∞ is a bounded sequence in ℓ2(ℤ). Besides, the regularity of each fN ensures that each Fourier series converges uniformly to fN. Therefore, each series ∑k‍ckNeikθ converges to fN in Lμ2(-π,π) and consequently,
(26)∥∑k‍ckNeikθ∥Lμ2(-π,π)2=∫-ππ‍|fN|2dμ≥∫-ππ‍|gN|2dμ=2N∑i=1iN‍μ(IiN)≥2Nμ(C),
from which we obtain the desired result.The proof of Theorem3 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{Urkbj}k∈ℤ;j=1,2,…,s in 𝒜a

Forbj∈𝒜a, j=1,2,…,s, consider the sequence {Urkbj}k∈ℤ;j=1,2,…,s. For every j=1,2,…,s, the spectral measure μa,bj in the integral representation of the cross-covariance function of the sequences a≔{Uka}k∈ℤ and bj≔{Ukbj}k∈ℤ has no singular part. Indeed, according to Theorem 3, the spectral measure associated with the autocovariance function of the sequence {Uka}k∈ℤ 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 bk,j≔Urkbj; our goal in this section is to study the sequence {bk,j}k∈ℤ;j=1,2,…,s in 𝒜a in terms of an s×r matrix Ψa,b(eiθ) which is introduced below. For the sake of completeness, we include some needed calculations which appear in [14].First of all, we have(27)〈Uka,bn,j〉=12π∫-ππ‍ei(k-rn)θϕa,bj(eiθ)dθ,
where ϕa,bj stands for the cross spectral density of the stationary correlated sequences a≔{Uka}k∈ℤ and bj≔{Ukbj}k∈ℤ. Define
(28)Φa,b(eiθ)≔(ϕa,b1(eiθ),ϕa,b2(eiθ),…,ϕa,bs(eiθ))⊤.In what follows, we will use theleft-shift operatorS defined as
(29)S:L2(𝕋)⟶L2(𝕋)∑k∈ℤ‍akeikθ⟼∑k∈ℤ‍ak+1eikθ,
or equivalently, by (Sf)(eiθ)=f(eiθ)e-iθ, where 𝕋≔{eiθ:θ∈[-π,π)} denotes the unidimensional torus. Also, we will consider the decimation operatorDr, r is a positive integer, defined as
(30)Dr:L2(𝕋)⟶L2(𝕋)∑k∈ℤ‍akeikθ⟼∑k∈ℤ‍arkeikθ,
which can equivalently be written as
(31)(Drf)(eiθ)=1r∑k=0r-1‍f(ei((θ+2kπ)/r)).For eachl=0,1,…,r-1, set the s×1 matrix of functions on the torus 𝕋 as follows:
(32)Ψa,bl(eiθ)≔(DrS-lΦa,b)(eiθ),
and define the s×r matrix of functions on the torus 𝕋 as follows:
(33)Ψa,b(eiθ)≔(Ψa,b0(eiθ)Ψa,b1(eiθ)⋯Ψa,br-1(eiθ)).It is worth to mention that the matrixΨa,b was explicitly computed in [14] for the translation and modulation cases in L2(ℝ).Next, for anyx∈𝒜a, we obtain an expression for the inner products αn,j≔〈x,Urnbj〉, n∈ℤ and j=1,2,…,s. Indeed, writing x=∑k∈ℤ‍xkUka where {xk}k∈ℤ∈ℓ2(ℤ), we have
(34)αn,j=〈x,Urnbj〉=∑k∈ℤ‍xk〈Uka,Urnbj〉=∑k∈ℤ‍xk12π∫-ππ‍ei(k-rn)θϕa,bj(eiθ)dθ=12π∫-ππ‍∑k∈ℤ‍xkeikθϕa,bj(eiθ)e-irnθdθ,
that is,
(35)αn≔(αn,1,αn,2,…,αn,s)⊤αn=12π∫-ππ‍Φa,b(eiθ)X(eiθ)e-irnθdθ,
where X(eiθ)≔∑k∈ℤ‍xkeikθ.Now, forl=0,1,…,r-1, define the sequence x(l)≔{xk(l)≔xkr+l}k∈ℤ. Thus, we can write
(36)X(eiθ)=∑l=0r-1‍∑k∈ℤ‍xkr+lei(kr+l)θ=∑l=0r-1‍X(l)(eirθ)eilθ,
where X(l)(eiθ)=∑k∈ℤ‍xk(l)eikθ.Using (36) in (35), we obtain
(37)αn=∑l=0r-1‍12π∫-ππ‍Φa,b(eiθ)X(l)(eirθ)eilθe-irnθdθ.After some easy calculations, we get(38)αn=∑l=0r-1‍12π∫-ππ‍S-lΦa,b(eiθ)X(l)(eirθ)e-irnθdθ=∑l=0r-1‍12π∫-rπrπ‍S-lΦa,b(ei(θ/r))rX(l)(eiθ)e-inθdθ=∑l=0r-1‍∑k=0r-1‍∫2πk2π(k+1)‍S-lΦa,b(ei(θ/r))2πrX(l)(eiθ)e-inθdθ=∫02π‍∑l=0r-1‍∑k=0r-1‍S-lΦa,b(ei((θ+2πk)/r))2πrX(l)(eiθ)e-inθdθ=12π∫-ππ‍∑l=0r-1‍(DrS-lΦa,b)(eiθ)X(l)(eiθ)e-inθdθ.DefiningC(eiθ)≔∑k∈ℤ‍αkeikθ, (38) implies that
(39)C(eiθ)=∑l=0r-1‍(DrS-lΦa,b)(eiθ)X(l)(eiθ),
which can be written in matrix form as
(40)C(eiθ)=(∑k∈ℤ‍αk,1eikθ,∑k∈ℤ‍αk,2eikθ,…,∑k∈ℤ‍αk,seikθ)⊤=Ψa,b(eiθ)(X(0)(eiθ),X(1)(eiθ),…,X(r-1)(eiθ))⊤=Ψa,b(eiθ)X~(eiθ)=(LΨa,bX~)(eiθ),
where LΨa,b:Lr2(𝕋)→Ls2(𝕋) denotes the multiplication operator by Ψa,b and
(41)X~(eiθ)≔(X(0)(eiθ),X(1)(eiθ),…,X(r-1)(eiθ))⊤.ByLr2(𝕋) (resp., Ls2(𝕋)), we denote the product space L2(𝕋)×⋯×L2(𝕋) for r times (resp., s times). Besides,
(42)∥Ψa,bX~∥Ls2(𝕋)2=12π∫-ππ‍〈Ψa,b(eiθ)X~(eiθ),Ψa,b(eiθ)X~(eiθ)〉ℂrdθ=12π∫-ππ‍〈Ψa,b*(eiθ)Ψa,b(eiθ)X~(eiθ),X~(eiθ)〉ℂrdθ.The above calculations let us prove the following result.Theorem 5.
Letbj∈𝒜a for j=1,2,…,s and let Ψa,b be the associated matrix given in (33). Then, the following results hold. (a)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a complete system in 𝒜a if and only if the rank of the matrix Ψa,b(ζ) is r a.e. ζ in 𝕋.(b)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a if and only if there exists a constant B<∞ such that
(43)Ψa,b*(ζ)Ψa,b(ζ)≤B𝕀ra.e.ζin𝕋.(c)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a if and only if there exist constants 0<A≤B<∞ such that
(44)A𝕀r≤Ψa,b*(ζ)Ψa,b(ζ)≤B𝕀ra.e.ζin𝕋.(d)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Riesz basis for 𝒜a if and only if it is a frame and s=r.Proof.
To prove (a), assume that there exists a setΩ⊆𝕋 with positive measure such that rank[Ψa,b(ζ)]<r for each ζ∈Ω. Then, there exists a measurable function v(ζ), ζ∈Ω, such that Ψa,b(ζ)v(ζ)=0 and ∥v(ζ)∥Lr2(𝕋)=1 in Ω. This function can be constructed as in [23, Lemma 2.4]. Define V~∈Lr2(𝕋) such that V~(ζ)=v(ζ) if ζ∈Ω, and V~(ζ)=0 if ζ∈𝕋∖Ω. Hence, from (40), we obtain that the system is not complete. Conversely, if the system is not complete, by using (40), we obtain a V~(ζ) different from 0 in a set with positive measure such that Ψa,b(ζ)V~(ζ)=0. Thus, rankΨa,b(ζ)<r on a set with positive measure.
To prove (b), we keep in mind that{Uka}k∈ℤ is a Riesz basis for 𝒜a, the mapping T:ℓ2(ℤ)→𝒜a, given by T{xk}k∈ℤ=x=∑k∈ℤ‍xkUka is bijective and there exist two constants 0<ma≤Ma<∞ such that
(45)ma∥{xk}∥ℓ22≤∥T{xk}∥ℋ2≤Ma∥{xk}∥ℓ22.
Assume first that (43) is satisfied. It follows from (40) and (42) that
(46)∥Ψa,bX~∥Ls2(𝕋)2≤B∥X~∥Lr2(𝕋)2.
By construction∥Ψa,bX~∥Ls2(𝕋)2=∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2 and ∥X~∥Lr2(𝕋)2=∥{xk}k∈ℤ∥ℓ22, using (45), it follows from (46) that
(47)∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2≤Bma∥x∥ℋ2.
Conversely, assume that {bkj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a, then there exists 0<B′<∞ such that
(48)∑j=1s‍∑k∈ℤ‍|〈x,bk,j〉|2≤B′∥x∥ℋ2.
Using (45), this implies that
(49)∥Ψa,bX~∥Ls2(𝕋)2≤B′Ma∥X~∥Lr2(𝕋)2
for all X~∈Lr2(𝕋). Inserting the right hand side of (42) for ∥Ψa,bX~∥Ls2(𝕋)2, it is straightforward to see that (43) holds with B=B′Ma. The proof of (c) is completed proceeding as in (b).
To prove (d), consider the following mapping:(50)S:𝒜a⟶ℓs2(ℤ)x⟼{〈x,bk,j〉}k∈ℤ;j=1,2,…,s.
According to (40), the mapping S is isometric equivalent to LΨa,b, and assuming that {bk,j}k∈ℤ;j=1,2,…,s is a frame, it is a Riesz basis if and only if S is surjective.
First, if{bk,j}k∈ℤ;j=1,2,…,s is a Riesz basis, then it is a frame and S is surjective. Applying (a) yields that LΨa,b is bijective, and therefore LΨa,b*=LΨa,b* is bijective. Hence, rank[Ψa,b(ζ)Ψa,b*(ζ)] is s for almost every ζ in 𝕋, so
(51)r=rank[Ψa,b*(ζ)Ψa,b(ζ)]=rank[Ψa,b(ζ)Ψa,b*(ζ)]=s,
and finally s=r.
Conversely, if{bk,j}k∈ℤ;j=1,2,…,s is a frame and s=r, (a) implies that Ψa,b(ζ) is invertible for almost every ζ in 𝕋, which implies that LΨa,b is surjective; then S is surjective and {bk,j}k∈ℤ;j=1,2,…,s is a Riesz basis.The following lemma will allow us to restate Theorem5.Lemma 6.
LetG(ζ) be an s×r matrix with entries in L2(𝕋), and consider the following constants:
(52)AG≔ess infζ∈𝕋λmin[G*(ζ)G(ζ)],BG≔ess supζ∈𝕋λmax[G*(ζ)G(ζ)],
where λmin (resp., λmax) denotes the smallest (resp., the largest) eigenvalue of the positive semidefinite matrix G*(ζ)G(ζ). Then, (a)
the matrixG(ζ) has essentially bounded entries on 𝕋 if and only if BG<∞;(b)
there exist constants0<A≤B<∞ such that A𝕀r≤G*(ζ)G(ζ)≤B𝕀r, a.e., ζ∈𝕋 if and only if 0<AG≤BG<∞.Proof.
The first part of lemma follows from thatλmax[G*(ζ)G(ζ)]=∥G(ζ)∥22, and
(53)maxi,j|aij|≤∥A∥2≤mnmaxi,j|aij|.for  any  matrixA=[aij]i=1,2…,mj=1,2…,n,
where ∥A∥2 denotes the spectral norm of the matrix A (see, for instance, [24]).
Now, we prove the second part of the lemma. SinceG*(ζ)G(ζ)≤B𝕀r means that 〈Bx-G*(ζ)G(ζ)x,x〉≥0 for all x∈ℂr, in particular, taking an eigenvector x associated with the largest eigenvalue λmax of G*(ζ)G(ζ) such that ∥x∥=1, one has that B≥λmax(G*(ζ)G(ζ)). Hence, B≥ess supζ∈𝕋λmax[G*(ζ)G(ζ)]. In a similar way, A𝕀r≤G*(ζ)G(ζ) implies that A≤ess infζ∈𝕋λmin[G*(ζ)G(ζ)].
Conversely, Rayleigh-Ritz theorem [24, page 176] yields that
(54)λmax[G*(ζ)G(ζ)]=maxx∈ℂrx*G*(ζ)G(ζ)x*x=maxx∈ℂr〈G*(ζ)G(ζ)x,x〉〈x,x〉.
Thus, ess supζ∈𝕋λmax[G*(ζ)G(ζ)]=BG implies that
(55)maxx∈ℂr〈G*(ζ)G(ζ)x,x〉〈x,x〉≤BG,a.e.ζ∈𝕋.
In other words,BG𝕀r≥G*(ζ)G(ζ); analogously, G*(ζ)G(ζ)≥AG𝕀r.
It is easy to deduce from the proof thatAG and BG are the optimal constants A>0 and B<∞ satisfying the inequalities A𝕀r≤G*(ζ)G(ζ)≤B𝕀r, a.e., ζ∈𝕋.As a consequence of Lemma6, statements (b) and (c) in Theorem 5 can be restated in terms of the following constants:
(56)AΨ≔ess infζ∈𝕋λmin[Ψa,b*(ζ)Ψa,b(ζ)];BΨ≔ess supζ∈𝕋λmax[Ψa,b*(ζ)Ψa,b(ζ)],
as follows.Theorem 7.
Letbj∈𝒜a for j=1,2,…,s, and let Ψa,b be the associated matrix given in (33) and its related constants (56). Then, the following results hold.(i)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a if and only if the constant BΨ<∞.(ii)
The sequence{Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a if and only if the constants AΨ and BΨ satisfy 0<AΨ≤BΨ<∞. In this case, AΨ and BΨ are the optimal frame bounds for {Urkbj}k∈ℤ;j=1,2,…,s.


## 2.3. The Frame Expansion

Define ther×s matrix of functions on the torus 𝕋 as follows:
(57)Γ(eiθ)≔∑k∈ℤΓkeikθ=[Ψa,b*(eiθ)Ψa,b(eiθ)]-1Ψa,b*(eiθ).
The following expansion involving the inner products αn,j=〈x,Urnbj〉 of x∈𝒜a holds.Lemma 8.
Assume that the matrixΨa,b(ζ) has essentially bounded entries on 𝕋. For any x=∑k∈ℤxkUka∈𝒜a, one has
(58)x~n=∑k∈ℤ‍Γkαn-k,
where x~n denotes the nth Fourier coefficient of the function X~(eiθ) defined in (41), and the sequence {αn}n∈ℤ is given in (35).Proof.
Indeed,(59)x~n=12π∫-ππ‍X~(eiθ)e-inθdθ=12π∫-ππ‍(∑k∈ℤ‍Γkeikθ)Ψa,b(eiθ)X~(eiθ)e-inθdθ=∑k∈ℤ‍Γk12π∫-ππ‍Ψa,b(eiθ)X~(eiθ)e-i(n-k)θdθ=∑k∈ℤ‍Γkαn-k.At this point, we are ready to prove the following expansion result.Theorem 9.
Letbj∈𝒜a for j=1,2,…,s, and assume that the associated matrix Ψa,b given in (33) has essentially bounded entries on 𝕋, that is, BΨ<∞. The following statements are equivalent. (i)
The constantAΨ>0.(ii)
There existcj∈𝒜a, j=1,2,…,s, such that the sequence {Urkcj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a, yielding, for any x∈𝒜a, the following expansion:
(60)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Urkcjinℋ.
In case the equivalent conditions hold,{Urkbj}k∈ℤ;j=1,2,…,s and {Urkcj}k∈ℤ;j=1,2,…,s form a pair of dual frames in 𝒜a.Proof.
First, we prove that (i) implies (ii). Observe thatx=∑k∈ℤxkUka can be written as ∑n∈ℤ‍x~n⊤a~n where a~n=(Unra,Unr+1a,…,Unr+r-1a)⊤. Next,
(61)x=∑n∈ℤ‍x~n⊤a~n=∑n∈ℤ‍(∑k∈ℤ‍Γkαn-k)⊤a~n=∑n∈ℤ‍∑k∈ℤ‍αn-k⊤Γk⊤a~n=∑n∈ℤ‍∑k∈ℤ‍αn⊤Γk⊤a~n+k=∑n∈ℤ‍αn⊤(∑k∈ℤ‍Γk⊤a~n+k).
For l∈ℤ and j=1,2,…,s define cl,j≔Urlcj, where (c1,c2,…,cs)⊤=∑k∈ℤ‍Γk⊤a~k, and bl,j≔Urlbj. Then, (61) implies
(62)x=∑n∈ℤ‍αn⊤(∑k∈ℤ‍Γk⊤a~n+k)=∑n∈ℤ‍αn⊤Unr(∑k∈ℤ‍Γk⊤a~k)=∑l=1s‍∑n∈ℤ‍〈x,bn,l〉cn,linℋ.
In order to be allowed to use [1, Lemma 5.6.2], we have to prove that the above constructed sequence {ck,j}k∈ℤ;j=1,2,…,s is a Bessel sequence for 𝒜a. To this end, we compute the corresponding Ψa,c matrix for c≔{ck,j}k∈ℤ;j=1,2,…,s. Setting [Γk⊤]ij=aijk, we obtain
(63)〈Uka,cn,j〉=∑l∈ℤ‍∑i=1r‍〈Uka,Unr(ajilUlr+ir+i-1a)〉=∑l∈ℤ‍∑i=1r‍a¯jil〈Uk-nr-lr-i+1a,a〉=∑l∈ℤ‍∑i=1r‍a¯jil12π∫-ππ‍ei(k-nr-lr-i+1)θϕa(eiθ)dθ.
Now,(64)(〈Uka,cn,1〉〈Uka,cn,2〉⋮〈Uka,cn,s〉)=∑l∈ℤ‍Γ¯l⊤(12π∫-ππ‍ei(k-nr-lr)θϕa(eiθ)dθ12π∫-ππ‍ei(k-nr-lr-1)θϕa(eiθ)dθ⋮12π∫-ππ‍ei(k-nr-lr-r+1)θϕa(eiθ)dθ)=12π∫-ππ‍∑l∈ℤ‍Γ¯l⊤e-ilrθ(ei(k-nr)θei(k-nr-1)θ⋮ei(k-nr-r+1)θ)ϕa(eiθ)dθ=12π∫-ππ‍ei(k-nr)θΓ*(eirθ)e~(eiθ)ϕa(eiθ)dθ,
where e~(eiθ)≔(1,e-iθ,…,e-i(r-1)θ)⊤. Hence, we have deduced that
(65)Φa,c(eiθ)=Γ*(eirθ)e~(eiθ)ϕa(eiθ).
Therefore, forl=0,1,…,r-1, we have
(66)Ψa,cl(eiθ)≔DrS-l[Γ*(eirθ)e~(eiθ)ϕa(eiθ)],
and consequently, the s×r matrix Ψa,c(eiθ)≔(Ψa,c0(eiθ),Ψa,c1(eiθ),…,Ψa,cr-1(eiθ)) can be written as
(67)Ψa,c(eiθ)=Dr[ϕa(eiθ)Γ*(eirθ)E~(eiθ)],
where
(68)E~(eiθ)≔(1eiθ⋯ei(r-1)θe-iθ1⋯ei(r-2)θ⋮⋮⋱⋮e-i(r-1)θe-i(r-2)θ⋯1).
As a consequence of Theorem7, the proof ends if we prove that the matrix Ψa,c(eiθ) has essentially bounded entries. Clearly, the decimation operator Dr sends bounded functions into bounded functions; Theorem 3 implies that ϕa is bounded so, taking into account (67), it remains to check that the matrix Γ*(eirθ) has essentially bounded entries.
Now,Γ*(eirθ)=Ψa,b(eirθ)[Ψa,b*(eirθ)Ψa,b(eirθ)]-1, the lower bound condition (c) in Theorem 5 and Lemma 6 imply that [Ψa,b*(eirθ)Ψa,b(eirθ)]-1 has bounded entries, and therefore the matrix Γ*(eirθ) has bounded entries. We have shown that Ψa,c(eiθ) has bounded entries; then Theorem 7, part (a) and Lemma 6 guarantee that the sequence {ck,j}k∈ℤ;j=1,2,…,s is a Bessel sequence; then, the sequences {bk,j}k∈ℤ,j=1,2,…,s and {ck,j}k∈ℤ,j=1,2,…,s form a pair of dual frames in 𝒜a (see [1, Lemma 5.6.2]).
Finally, condition (ii) implies condition (i). According to [1, Lemma 5.6.2], the sequence {Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a since it is a Bessel sequence and the expansion in (ii) holds. By using Theorem 7 we obtain that AΨ>0.It is worth to observe that the analysis done in Theorem9 provides a whole family of dual frames for the sequence {Urkbj}k∈ℤ;j=1,2,…,s. In fact, everything works if we replace Γ(eiθ) in (57) by any matrix of the following form:
(69)Γ𝕌(eiθ)≔Ψa,b†(eiθ)+𝕌(eiθ)[𝕀s-Ψa,b(eiθ)Ψa,b†(eiθ)],
where 𝕌(eiθ) is any r×s matrix with entries in L∞(𝕋), and Ψa,b† denotes the Moore-Penrose left-inverse Ψa,b†(eiθ)≔[Ψa,b*(eiθ)Ψa,b(eiθ)]-1Ψa,b*(eiθ). Note that we need essentially bounded entries in the matrix Γ𝕌(eiθ) since the multiplication operator MF:f↦Ff in L2(𝕋) is well-defined (and consequently bounded) if and only if F∈L∞(𝕋).Notice that ifs=r, we have Ψa,b†=Ψa,b-1 which implies a unique Γ𝕌, and we are in the presence of a pair of dual Riesz bases. In fact, the following result holds.Corollary 10.
Letbj∈𝒜a for j=1,2,…,r; that is, r=s in Theorem 9. Assume that the square matrix Ψa,b given in (33) has entries essentially bounded on 𝕋, that is, BΨ<∞. The following statements are equivalent.(a)
The constantAΨ>0.(b)
There exists a Riesz basis{Ck,j}k∈ℤ;j=1,2,…,s such that for any x∈𝒜a, the expansion
(70)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Ck,jinℋholds.
In case the equivalent conditions are satisfied, necessarily, there existcj∈𝒜a, j=1,2,…,r, such that Ck,j=Urkcj for k∈ℤ and j=1,2,…,r. Moreover, the sequences {Urkcj}k∈ℤ;j=1,2,…,s and {Urkbj}k∈ℤ;j=1,2,…,s are dual Riesz bases in 𝒜a, and the interpolation property 〈cj,Urkbj′〉=δj,j′δk,0, where k∈ℤ and j,j′=1,2,…,r, holds.Proof.
To prove(a)⇒(b), we use Theorem 9; whenever 0<AΨ≤BΨ<∞, there exist cj∈𝒜a, j=1,2,…,s, such that the sequence {Urkcj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a and for any x∈𝒜a, the expansion
(71)x=∑j=1s‍∑k∈ℤ‍〈x,Urkbj〉Urkcjinℋ,
holds. Actually, from Theorem 5, we get that r=s implies that {Urkbj}k∈ℤ;j=1,2,…,s is a Riesz basis, and consequently, {Urkcj}k∈ℤ;j=1,2,…,s is indeed its dual Riesz basis.
The converse follows easily from the fact that if{Ck,j}k∈ℤ;j=1,2,…,s is a Riesz basis, then (b) implies that {Urkbj}k∈ℤ;j=1,2,…,s is its dual Riesz basis; hence, Theorem 5 provides AΨ>0. The interpolation property comes out from the biorthogonal condition of a pair of dual Riesz bases.Closing this section it is worth to mention that the results stated and proved in Sections2.2 and 2.3 mathematically enrich some of the remarkable results concerning regular sampling contained in the interesting [14]. Here, we have assumed only one generator a∈ℋ and that bj∈𝒜a for all j=1,2,…,s. If bj∉𝒜a for some j, see the additional remarks in Section 4. The case of several generators al∈ℋ, l=1,2,…,L, can be essentially treated in the same way.


## 3. Some Perturbation Results

In Section2, we have only used the discrete group of unitary operators {Un}n∈ℤ which is completely determined by the unitary operator U. In order to carry out the study of some perturbation results stated below, we need the availability of a continuous group of unitary operators {Ut}t∈ℝ which includes the unitary operator U, say, for instance, that U≔U1.


### 3.1. On Continuous Groups of Unitary Operators

Let{Ut}t∈ℝ denote a continuous group of unitary operators in ℋ such that U≔U1. Recall that {Ut}t∈ℝ is a family of unitary operators in ℋ satisfying (see [25, vol. 2; page 29]) the following:(1)
UtUt′=Ut+t′,(2)
U0=Iℋ,(3)
〈Utx,y〉ℋ is a continuous function of t for any x,y∈ℋ.Note that(Ut)-1=U-t, and since (Ut)*=(Ut)-1, we have (Ut)*=U-t.Classical Stone's theorem [26] assures us the existence of a self-adjoint operator T (possibly unbounded) such that Ut≡eitT. This self-adjoint operator T, defined on the dense domain of ℋ as follows:
(72)DT≔{x∈ℋsuch  that∫-∞∞‍w2d∥Ewx∥2<∞},
admits the spectral representationT=∫-∞∞‍wdEw which means that
(73)〈Tx,y〉=∫-∞∞wd〈Ewx,y〉for  anyx∈DT,y∈ℋ,
where {Ew}w∈ℝ is the corresponding resolution of the identity, that is, a one-parameter family of projection operators Ew in ℋ such that(i)
E-∞≔limw→-∞Ew=Oℋ, E∞≔limw→∞Ew=Iℋ,(ii)
Ew-=Ew for every -∞<w<∞,(iii)
EuEv=Ew where w=min{u,v}.Recall that∥Ewx∥2 and 〈Ewx,y〉, as functions of w, have bounded variation and define, respectively, a positive and a complex Borel measure on ℝ.Furthermore, for anyx∈DT, we have that limt→0((Utx-x)/t)=iTx and the operator iT is said to be the infinitesimal generator of the group {Ut}t∈ℝ. For each x∈DT, Utx is a continuous differentiable function of t. Notice that, whenever the self-adjoint operator T is bounded, DT=ℋ and eitT can be defined as the usual exponential series; in any case, Ut≡eitT means that
(74)〈Utx,y〉=∫-∞∞‍eiwtd〈Ewx,y〉,t∈ℝ,
where x∈DT and y∈ℋ. A comment on the continuity of a group of unitary operators is in order: The group is said to be strongly continuous if, for each x∈ℋ and t0∈ℝ, Utx→Ut0x as t→t0. If ℋ is a separable Hilbert space, strong continuity can be deduced from continuity and even from weak measurability; that is, 〈Utx,y〉ℋ is a Lebesgue measurable function of t for any x,y∈ℋ.The following result taken from [25, vol. 2; page 24] will be used later. For x∈DT and y∈ℋ, the inequality
(75)|∫-∞∞‍φ(w)d〈Ewx,y〉|≤∥y∥∫-∞∞‍|φ(w)|2d〈Ewx,x〉,
holds, where φ is a real or complex function which is continuous in ℝ with the possible exception of a finite number of points.For the details on the theory of continuous groups of unitary operators, see [22, 25, 27, 28].


### 3.2.  Studying the Perturbed Sequence{Urk+ϵkjbj}k∈ℤ;j=1,2,…s

Given an error sequenceϵ≔{ϵkj}k∈ℤ;j=1,2,…,s, consider the corresponding perturbed sequence {Urk+ϵkjbj}k∈ℤ;j=1,2,…,s. Consider the following perturbation result (see [1, page 354] for the proof).Lemma 11.
Let{xn}n=1∞ be a frame for the Hilbert space ℋ with frame bounds A,B, and let {yn}n=1∞ be a sequence in ℋ. If there exists a constant R<A such that
(76)∑n=1∞‍|〈xn-yn,x〉|2≤R∥x∥2foreachx∈ℋ,
then, the sequence {yn}n=1∞ is also a frame for ℋ with bounds A(1-R/A)2 and B(1+R/B)2. If {xn}n=1∞ is a Riesz basis, then {yn}n=1∞ is a Riesz basis.Note that it cannot be directly applied to the sequences{Urkbj}k∈ℤ;j=1,2,…,s and {Urk+ϵkjbj}k∈ℤ;j=1,2,…,s since the first one is not a frame for the entire Hilbert space ℋ, and its perturbed sequence does not necessarily belong to the subspace 𝒜a. However, something can be said in case {Urkbj}k∈ℤ;j=1,2,…r is a Riesz sequence in ℋ.Theorem 12.
Assume that for somebj∈DT; that is, ∫-∞∞‍w2d∥Ewbj∥2<∞ for each 1≤j≤r, the sequence {Ukrbj}k∈ℤ;j=1,2,…,r is a Riesz basis for 𝒜a with Riesz bounds 0<AΨ≤BΨ<∞. For a sequence ϵ≔{ϵkj}k∈ℤ,j=1,2,…,r of errors, let R be the constant given by
(77)R≔∥ϵ∥2maxj=1,2,…,r{∫-∞∞‍w2d∥Ewbj∥2},
where ∥ϵ∥ denotes the ℓr2-norm of the sequence ϵ.
IfR<AΨ, then the sequence {Ukr+ϵkjbj}k∈ℤ;j=1,2,…,r is a Riesz sequence in ℋ with Riesz bounds AΨ(1-R/AΨ)2 and BΨ(1+R/BΨ)2.Proof.
By using (75), we have
(78)|〈x,Ukrbj-Ukr+ϵkjbj〉|=|∫-∞∞‍e-ikrwd〈Ewx,bj〉-∫-∞∞‍e-ikrw-iϵkjwd〈Ewx,bj〉|=|∫-∞∞‍e-ikrw(1-e-iϵkjw)d〈Ewx,bj〉|=|∫-∞∞‍eikrw(1-eiϵkjw)d〈Ewbj,x〉|≤∥x∥∫-∞∞‍|1-eiϵkjw|2d∥Ewbj∥2≤∥x∥∫-∞∞‍w2|ϵkj|2d∥Ewbj∥2=|ϵkj|∥x∥∫-∞∞‍w2d∥Ewbj∥2.
Hence,(79)∑j=1r‍∑k∈ℤ‍|〈x,Ukrbj-Ukr+ϵkjbj〉|2≤∥x∥2∑j=1r‍∑k∈ℤ‍(∫-∞∞‍w2d∥Ewbj∥2)|ϵkj|2≤∥x∥2maxj=1,2,…,r{∫-∞∞‍w2d∥Ewbj∥2}∑j=1r‍∑k∈ℤ‍|ϵkj|2.
Hence, Lemma 11 and Theorem 15.3.2 in [1, page 356] give the desired results.


### 3.3. On the Perturbed Frame Expansion

Next, we deal with the problem of the recovery of anyx∈𝒜a in a stable way from the following perturbed sequence:
(80){〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s,
where ϵ≔{ϵmj}m∈ℤ;j=1,2,…,s denotes a sequence of real errors. In order to face this problem, we propose a possible strategy. Let 𝒯U,a:L2(0,1)→𝒜a be the isomorphism which maps the orthonormal basis {e2πinw}n∈ℤ onto the Riesz basis {Una}n∈ℤ for 𝒜a. In other words,
(81)𝒯U,a:L2(0,1)⟶𝒜a∑n∈ℤ‍αne2πinw⟼x=∑n∈ℤ‍αnUna.Thus,(82)〈x,Utbj〉ℋ=〈∑n∈ℤ‍αnUna,Utbj〉ℋ=∑n∈ℤ‍αn〈Utbj,Una〉¯ℋ=〈F,∑n∈ℤ‍〈Utbj,Una〉ℋe2πinw〉L2(0,1)=〈F,Kjt〉L2(0,1),
where 𝒯U,aF=x, and the function Kjt(w)≔∑n∈ℤ‍〈Utbj,Una〉ℋe2πinw belongs to L2(0,1) since the sequence {〈Utb,Una〉ℋ}n∈ℤ belongs to ℓ2(ℤ) for each t∈ℝ.Hence, for anyx∈𝒜a, we have the following expressions:
(83)〈x,Urmbj〉ℋ=〈F,gj(w)¯e2πirmw〉L2(0,1),〈x,Urm+ϵmjbj〉ℋ=〈F,gm,j(w)¯e2πirmw〉L2(0,1),
where the functions
(84)gj(w)≔∑k∈ℤ‍〈a,Ukbj〉ℋe2πikw,gm,j(w)≔∑k∈ℤ‍〈a,Uk+ϵmjbj〉ℋe2πikw
belong to L2(0,1). Therefore, we can see the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s as a perturbation of the sequence {gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s. From [6, Lemma 3], we know that this sequence is a frame for L2(0,1) if and only if 0<α𝔾≤β𝔾<∞ where the constants α𝔾 and β𝔾 are given by
(85)α𝔾≔ess infw∈(0,1/r)λmin[𝔾*(w)𝔾(w)],β𝔾≔ess supw∈(0,1/r)λmax[𝔾*(w)𝔾(w)],
and 𝔾(w) is the s×r matrix
(86)𝔾(w)≔[gj(w+k-1r)]j=1,2,…,sk=1,2,…,r.
Besides, the optimal frame bounds for {gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s are α𝔾/r and β𝔾/r.Given an error sequenceϵ≔{ϵmj}m∈ℤ;j=1,2,…,s, we define
(87)dm,k(j)≔〈a,Urm-k+ϵmjbj〉-〈a,Urm-kbj〉.
For any sequence c={ck}k∈ℤ∈ℓ2(ℤ), we have
(88)∑j=1s‍∑m∈ℤ‍|∑k∈ℤ‍dm,k(j)ck|2≤∑j=1s‍∑m∈ℤ‍∑l,k∈ℤ‍|dm,l(j)cld¯m,k(j)c¯k|=∑j=1s‍∑l,k∈ℤ‍|cl||ck|∑m∈ℤ‍|dm,l(j)dm,k(j)|≤∑j=1s‍∑l,k∈ℤ‍|cl|2+|ck|22∑m∈ℤ‍|dm,l(j)dm,k(j)|=∑j=1s‍∑l∈ℤ|cl|2∑k,m∈ℤ‍|dm,l(j)dm,k(j)|.Now, for|γ|<1/2, define the following functions:
(89)Ma,bj(γ)≔∑k∈ℤ‍maxt∈[-γ,γ]|〈a,Uk+tbj〉-〈a,Ukbj〉|,Na,bj(γ)≔maxk=0,1,…,r-1∑m∈ℤ‍maxt∈[-γ,γ]|〈a,Urm+k+tbj〉-〈a,Urm+kbj〉|.Notice thatNa,bj(γ)≤Ma,bj(γ) and for r=1, the equality holds. Moreover, assuming that the continuous functions φj(t)≔〈a,Utbj〉, j=1,2,…,s, satisfy a decay condition as φj(t)=O(|t|-(1+ηj)) when |t|→∞ for some ηj>0, we may deduce that the functions Na,bj(γ) and Ma,bj(γ) are continuous near to 0.Theorem 13.
Assume that for the functionsgj, j=1,2,…,s, given in (84) one has 0<α𝔾≤β𝔾<∞. For an error sequence ϵ≔{ϵmj}m∈ℤ;j=1,…,s, define the constant γj≔supm∈ℤ|ϵmj| for each j=1,2,…,s. Then the following condition:
(90)∑j=1sMa,bj(γj)Na,bj(γj)<α𝔾r
implies that there exists a frame {Cm,jϵ}m∈ℤ;j=1,2,…,s for 𝒜a such that, for any x∈𝒜a, the following sampling expansion:
(91)x=∑j=1s∑m∈ℤ‍〈x,Urm+ϵmjbj〉ℋCm,jϵinℋ
holds. Moreover, when r=s, the sequence {Cm,jϵ}m∈ℤ;j=1,2,…,s is a Riesz basis for 𝒜a, and the interpolation property 〈Cn,lϵ,Urm+ϵmjbj〉ℋ=δl,jδn,m holds.Proof.
The sequence{gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame (a Riesz basis if r=s) for L2(0,1) with frame (Riesz) bounds α𝔾 and β𝔾. For any F(w)=∑l∈ℤ‍ale2πilw in L2(0,1), we have
(92)∑m∈ℤ‍∑j=1s‍|〈gm,j(·)¯e2πirm·-gj(·)¯e2πirm·,F(·)〉L2(0,1)|2=∑m∈ℤ‍∑j=1s‍|-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯000000000000000-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯|2=∑m∈ℤ‍∑j=1s‍|-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Urm-k+ϵmjbj〉¯000000000000000-〈a,Urm-kbj〉¯〈a,Uk+ϵmjbj〉¯)e2πik·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯|2=∑m∈ℤ‍∑j=1s‍|∑k∈ℤ‍(〈a,Urm-k+ϵmjbj〉¯-〈a,Urm-kbj〉¯)a¯k|2=∑j=1s‍∑m∈ℤ‍|∑k∈ℤ‍dm,k(j)ak|2.
From (88) and the definition of the functions Ma,bj and Na,bj, we obtain
(93)∑m∈ℤ‍∑j=1s‍|〈gm,j(·)¯e2πirm·-gj(·)¯e2πirm·,F(·)〉L2(0,1)|2≤∑j=1s‍Ma,bj(γj)Na,bj(γj)∥{al}l∈ℤ∥2≤∑j=1s‍Ma,bj(γj)Na,bj(γj)∥F∥L2(0,1)2.
By using Lemma 11, we obtain that the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame for L2(0,1) (a Riesz basis if r=s). Let {hm,jϵ}m∈ℤ;j=1,2,…,s be its canonical dual frame. Hence, for any F∈L2(0,1),
(94)F=∑m∈ℤ‍∑j=1s‍〈F(·),gm,j(·)¯e2πirm·〉L2(0,1)hm,jϵ=∑m∈ℤ‍∑j=1s‍〈x,Urm+ϵmjbj〉ℋhm,jϵ.
Applying the isomorphism𝒯U,a, one gets (91), where Cm,jϵ=𝒯U,a(hm,jϵ). Since 𝒯U,a is an isomorphism between L2(0,1) and 𝒜a; the sequence {Cm,jϵ}m∈ℤ;j=1,2,…,s is a frame for 𝒜a (a Riesz basis if r=s). The interpolatory property in the case r=s follows from the uniqueness of the coefficients with respect to a Riesz basis.


### 3.4. A Frame Algorithm inℓ2(ℤ)

Sampling formula (91) is useless from a practical point of view, it is impossible to determine the involved frame {Cm,jϵ}m∈ℤ;j=1,2,…,s. As a consequence, in order to recover x∈𝒜a from the sequence of inner products {〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s, we should implement a frame algorithm in ℓ2(ℤ). Another possibility is given in the recent reference [29].Next we are going to implement a frame algorithm in theℓ2(ℤ) setting. To this end, consider the canonical isometry
(95)𝒰~:ℓ2(ℤ)⟶L2(0,1)such  that𝒰~c≔∑k∈ℤcke2πikw00000000000000000000000.forc={ck}k∈ℤ∈ℓ2(ℤ).Forx=∑n∈ℤ‍cnUna∈𝒜a, denote by 𝔽 the following sequence:
(96)𝔽≔𝒰~-1𝒯U,a-1x={cn}n∈ℤ∈ℓ2(ℤ).The inner products{〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s can be written as
(97)〈x,Urm+ϵmjbj〉ℋ=〈𝒯U,a-1x,gm,j(w)¯e2πirmw〉L2(0,1)=〈𝔽,𝕃m,j〉ℓ2(ℤ),
where, for j=1,2,…,s and m∈ℤ,
(98)𝕃m,j≔𝒰~-1(gm,j(·)¯e2πirm·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Uk+ϵmjbj〉ℋe2πik·¯e2πirm·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Uk+ϵmjbj〉¯ℋe2πi(rm-k)·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Urm-k+ϵmjbj〉¯ℋe2πik·)𝕃m,j={〈a,Urm-k+ϵmjbj〉¯ℋ}k∈ℤ.The sequence{𝕃m,j}m∈ℤ;j=1,2,…,s is a frame for ℓ2(ℤ). Indeed, assume that the error sequence ϵ≔{ϵm,j}m∈ℤ;j=1,…,s satisfies the hypothesis of Theorem 13; that is,
(99)Kϵ≔∑j=1s‍Ma,bj(γj)Na,bj(γj)<α𝔾r,
where γj≔supm∈ℤ|ϵmj| for each j=1,2,…,s. As a consequence, the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame for L2(0,1) with bounds
(100)Aϵ≔α𝔾r(1-rKϵα𝔾)2,Bϵ≔β𝔾r(1-rKϵβ𝔾)2.Since𝒰~-1 is an isometry, the sequence {𝕃m,j}m∈ℤ;j=1,2,…,s is a frame for ℓ2(ℤ) with the same bounds. Hence, the recovery of the element x=𝒯U,a(𝒰~𝔽)∈𝒜a from the samples {〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s is reduced to recover 𝔽∈ℓ2(ℤ) from the following sequence:
(101){〈𝔽,𝕃m,j〉ℓ2(ℤ)}m∈ℤ;j=1,2,…,s.In doing so, the classical frame algorithm reads (see, for instance, [1]), let 𝒮 be the frame operator in ℓ2(ℤ) of {𝕃m,j}m∈ℤ;j=1,2,…,s, and define
(102)𝔽0≔2𝒮Aϵ+Bϵ𝔽𝔽0=2Aϵ+Bϵ∑j=1s‍∑m∈ℤ‍〈𝔽,𝕃m,j〉𝕃m,j𝔽0=2Aϵ+Bϵ∑j=1s‍∑m∈ℤ‍〈x,Urm+ϵmjbj〉𝕃m,j,
and recursively,
(103)𝔽k+1=𝔽k+2𝒮Aϵ+Bϵ(𝔽-𝔽k)for  eachk∈ℕ.Then, the sequence{xn}n∈ℕ in 𝒜a given by xn≔∑k∈ℤ(𝔽n)kUka, satisfies
(104)∥x-xn∥ℋ≤∥𝒯U,a∥∥𝔽-𝔽k∥ℓ2(ℤ)≤∥𝒯U,a∥γϵk+1∥𝔽∥ℓ2(ℤ)≤∥𝒯U,a∥∥𝒯U,a-1∥γϵk+1∥x∥ℋ,
where γϵ≔(Bϵ-Aϵ)/(Bϵ+Aϵ).


## 3.1. On Continuous Groups of Unitary Operators

Let{Ut}t∈ℝ denote a continuous group of unitary operators in ℋ such that U≔U1. Recall that {Ut}t∈ℝ is a family of unitary operators in ℋ satisfying (see [25, vol. 2; page 29]) the following:(1)
UtUt′=Ut+t′,(2)
U0=Iℋ,(3)
〈Utx,y〉ℋ is a continuous function of t for any x,y∈ℋ.Note that(Ut)-1=U-t, and since (Ut)*=(Ut)-1, we have (Ut)*=U-t.Classical Stone's theorem [26] assures us the existence of a self-adjoint operator T (possibly unbounded) such that Ut≡eitT. This self-adjoint operator T, defined on the dense domain of ℋ as follows:
(72)DT≔{x∈ℋsuch  that∫-∞∞‍w2d∥Ewx∥2<∞},
admits the spectral representationT=∫-∞∞‍wdEw which means that
(73)〈Tx,y〉=∫-∞∞wd〈Ewx,y〉for  anyx∈DT,y∈ℋ,
where {Ew}w∈ℝ is the corresponding resolution of the identity, that is, a one-parameter family of projection operators Ew in ℋ such that(i)
E-∞≔limw→-∞Ew=Oℋ, E∞≔limw→∞Ew=Iℋ,(ii)
Ew-=Ew for every -∞<w<∞,(iii)
EuEv=Ew where w=min{u,v}.Recall that∥Ewx∥2 and 〈Ewx,y〉, as functions of w, have bounded variation and define, respectively, a positive and a complex Borel measure on ℝ.Furthermore, for anyx∈DT, we have that limt→0((Utx-x)/t)=iTx and the operator iT is said to be the infinitesimal generator of the group {Ut}t∈ℝ. For each x∈DT, Utx is a continuous differentiable function of t. Notice that, whenever the self-adjoint operator T is bounded, DT=ℋ and eitT can be defined as the usual exponential series; in any case, Ut≡eitT means that
(74)〈Utx,y〉=∫-∞∞‍eiwtd〈Ewx,y〉,t∈ℝ,
where x∈DT and y∈ℋ. A comment on the continuity of a group of unitary operators is in order: The group is said to be strongly continuous if, for each x∈ℋ and t0∈ℝ, Utx→Ut0x as t→t0. If ℋ is a separable Hilbert space, strong continuity can be deduced from continuity and even from weak measurability; that is, 〈Utx,y〉ℋ is a Lebesgue measurable function of t for any x,y∈ℋ.The following result taken from [25, vol. 2; page 24] will be used later. For x∈DT and y∈ℋ, the inequality
(75)|∫-∞∞‍φ(w)d〈Ewx,y〉|≤∥y∥∫-∞∞‍|φ(w)|2d〈Ewx,x〉,
holds, where φ is a real or complex function which is continuous in ℝ with the possible exception of a finite number of points.For the details on the theory of continuous groups of unitary operators, see [22, 25, 27, 28].


## 3.2.  Studying the Perturbed Sequence{Urk+ϵkjbj}k∈ℤ;j=1,2,…s

Given an error sequenceϵ≔{ϵkj}k∈ℤ;j=1,2,…,s, consider the corresponding perturbed sequence {Urk+ϵkjbj}k∈ℤ;j=1,2,…,s. Consider the following perturbation result (see [1, page 354] for the proof).Lemma 11.
Let{xn}n=1∞ be a frame for the Hilbert space ℋ with frame bounds A,B, and let {yn}n=1∞ be a sequence in ℋ. If there exists a constant R<A such that
(76)∑n=1∞‍|〈xn-yn,x〉|2≤R∥x∥2foreachx∈ℋ,
then, the sequence {yn}n=1∞ is also a frame for ℋ with bounds A(1-R/A)2 and B(1+R/B)2. If {xn}n=1∞ is a Riesz basis, then {yn}n=1∞ is a Riesz basis.Note that it cannot be directly applied to the sequences{Urkbj}k∈ℤ;j=1,2,…,s and {Urk+ϵkjbj}k∈ℤ;j=1,2,…,s since the first one is not a frame for the entire Hilbert space ℋ, and its perturbed sequence does not necessarily belong to the subspace 𝒜a. However, something can be said in case {Urkbj}k∈ℤ;j=1,2,…r is a Riesz sequence in ℋ.Theorem 12.
Assume that for somebj∈DT; that is, ∫-∞∞‍w2d∥Ewbj∥2<∞ for each 1≤j≤r, the sequence {Ukrbj}k∈ℤ;j=1,2,…,r is a Riesz basis for 𝒜a with Riesz bounds 0<AΨ≤BΨ<∞. For a sequence ϵ≔{ϵkj}k∈ℤ,j=1,2,…,r of errors, let R be the constant given by
(77)R≔∥ϵ∥2maxj=1,2,…,r{∫-∞∞‍w2d∥Ewbj∥2},
where ∥ϵ∥ denotes the ℓr2-norm of the sequence ϵ.
IfR<AΨ, then the sequence {Ukr+ϵkjbj}k∈ℤ;j=1,2,…,r is a Riesz sequence in ℋ with Riesz bounds AΨ(1-R/AΨ)2 and BΨ(1+R/BΨ)2.Proof.
By using (75), we have
(78)|〈x,Ukrbj-Ukr+ϵkjbj〉|=|∫-∞∞‍e-ikrwd〈Ewx,bj〉-∫-∞∞‍e-ikrw-iϵkjwd〈Ewx,bj〉|=|∫-∞∞‍e-ikrw(1-e-iϵkjw)d〈Ewx,bj〉|=|∫-∞∞‍eikrw(1-eiϵkjw)d〈Ewbj,x〉|≤∥x∥∫-∞∞‍|1-eiϵkjw|2d∥Ewbj∥2≤∥x∥∫-∞∞‍w2|ϵkj|2d∥Ewbj∥2=|ϵkj|∥x∥∫-∞∞‍w2d∥Ewbj∥2.
Hence,(79)∑j=1r‍∑k∈ℤ‍|〈x,Ukrbj-Ukr+ϵkjbj〉|2≤∥x∥2∑j=1r‍∑k∈ℤ‍(∫-∞∞‍w2d∥Ewbj∥2)|ϵkj|2≤∥x∥2maxj=1,2,…,r{∫-∞∞‍w2d∥Ewbj∥2}∑j=1r‍∑k∈ℤ‍|ϵkj|2.
Hence, Lemma 11 and Theorem 15.3.2 in [1, page 356] give the desired results.


## 3.3. On the Perturbed Frame Expansion

Next, we deal with the problem of the recovery of anyx∈𝒜a in a stable way from the following perturbed sequence:
(80){〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s,
where ϵ≔{ϵmj}m∈ℤ;j=1,2,…,s denotes a sequence of real errors. In order to face this problem, we propose a possible strategy. Let 𝒯U,a:L2(0,1)→𝒜a be the isomorphism which maps the orthonormal basis {e2πinw}n∈ℤ onto the Riesz basis {Una}n∈ℤ for 𝒜a. In other words,
(81)𝒯U,a:L2(0,1)⟶𝒜a∑n∈ℤ‍αne2πinw⟼x=∑n∈ℤ‍αnUna.Thus,(82)〈x,Utbj〉ℋ=〈∑n∈ℤ‍αnUna,Utbj〉ℋ=∑n∈ℤ‍αn〈Utbj,Una〉¯ℋ=〈F,∑n∈ℤ‍〈Utbj,Una〉ℋe2πinw〉L2(0,1)=〈F,Kjt〉L2(0,1),
where 𝒯U,aF=x, and the function Kjt(w)≔∑n∈ℤ‍〈Utbj,Una〉ℋe2πinw belongs to L2(0,1) since the sequence {〈Utb,Una〉ℋ}n∈ℤ belongs to ℓ2(ℤ) for each t∈ℝ.Hence, for anyx∈𝒜a, we have the following expressions:
(83)〈x,Urmbj〉ℋ=〈F,gj(w)¯e2πirmw〉L2(0,1),〈x,Urm+ϵmjbj〉ℋ=〈F,gm,j(w)¯e2πirmw〉L2(0,1),
where the functions
(84)gj(w)≔∑k∈ℤ‍〈a,Ukbj〉ℋe2πikw,gm,j(w)≔∑k∈ℤ‍〈a,Uk+ϵmjbj〉ℋe2πikw
belong to L2(0,1). Therefore, we can see the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s as a perturbation of the sequence {gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s. From [6, Lemma 3], we know that this sequence is a frame for L2(0,1) if and only if 0<α𝔾≤β𝔾<∞ where the constants α𝔾 and β𝔾 are given by
(85)α𝔾≔ess infw∈(0,1/r)λmin[𝔾*(w)𝔾(w)],β𝔾≔ess supw∈(0,1/r)λmax[𝔾*(w)𝔾(w)],
and 𝔾(w) is the s×r matrix
(86)𝔾(w)≔[gj(w+k-1r)]j=1,2,…,sk=1,2,…,r.
Besides, the optimal frame bounds for {gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s are α𝔾/r and β𝔾/r.Given an error sequenceϵ≔{ϵmj}m∈ℤ;j=1,2,…,s, we define
(87)dm,k(j)≔〈a,Urm-k+ϵmjbj〉-〈a,Urm-kbj〉.
For any sequence c={ck}k∈ℤ∈ℓ2(ℤ), we have
(88)∑j=1s‍∑m∈ℤ‍|∑k∈ℤ‍dm,k(j)ck|2≤∑j=1s‍∑m∈ℤ‍∑l,k∈ℤ‍|dm,l(j)cld¯m,k(j)c¯k|=∑j=1s‍∑l,k∈ℤ‍|cl||ck|∑m∈ℤ‍|dm,l(j)dm,k(j)|≤∑j=1s‍∑l,k∈ℤ‍|cl|2+|ck|22∑m∈ℤ‍|dm,l(j)dm,k(j)|=∑j=1s‍∑l∈ℤ|cl|2∑k,m∈ℤ‍|dm,l(j)dm,k(j)|.Now, for|γ|<1/2, define the following functions:
(89)Ma,bj(γ)≔∑k∈ℤ‍maxt∈[-γ,γ]|〈a,Uk+tbj〉-〈a,Ukbj〉|,Na,bj(γ)≔maxk=0,1,…,r-1∑m∈ℤ‍maxt∈[-γ,γ]|〈a,Urm+k+tbj〉-〈a,Urm+kbj〉|.Notice thatNa,bj(γ)≤Ma,bj(γ) and for r=1, the equality holds. Moreover, assuming that the continuous functions φj(t)≔〈a,Utbj〉, j=1,2,…,s, satisfy a decay condition as φj(t)=O(|t|-(1+ηj)) when |t|→∞ for some ηj>0, we may deduce that the functions Na,bj(γ) and Ma,bj(γ) are continuous near to 0.Theorem 13.
Assume that for the functionsgj, j=1,2,…,s, given in (84) one has 0<α𝔾≤β𝔾<∞. For an error sequence ϵ≔{ϵmj}m∈ℤ;j=1,…,s, define the constant γj≔supm∈ℤ|ϵmj| for each j=1,2,…,s. Then the following condition:
(90)∑j=1sMa,bj(γj)Na,bj(γj)<α𝔾r
implies that there exists a frame {Cm,jϵ}m∈ℤ;j=1,2,…,s for 𝒜a such that, for any x∈𝒜a, the following sampling expansion:
(91)x=∑j=1s∑m∈ℤ‍〈x,Urm+ϵmjbj〉ℋCm,jϵinℋ
holds. Moreover, when r=s, the sequence {Cm,jϵ}m∈ℤ;j=1,2,…,s is a Riesz basis for 𝒜a, and the interpolation property 〈Cn,lϵ,Urm+ϵmjbj〉ℋ=δl,jδn,m holds.Proof.
The sequence{gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame (a Riesz basis if r=s) for L2(0,1) with frame (Riesz) bounds α𝔾 and β𝔾. For any F(w)=∑l∈ℤ‍ale2πilw in L2(0,1), we have
(92)∑m∈ℤ‍∑j=1s‍|〈gm,j(·)¯e2πirm·-gj(·)¯e2πirm·,F(·)〉L2(0,1)|2=∑m∈ℤ‍∑j=1s‍|-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯000000000000000-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯|2=∑m∈ℤ‍∑j=1s‍|-〈a,Ukbj〉¯〈a,Uk+ϵmjbj〉¯)e2πi(rm-k)·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Urm-k+ϵmjbj〉¯000000000000000-〈a,Urm-kbj〉¯〈a,Uk+ϵmjbj〉¯)e2πik·,F(·)∑k∈ℤ(〈a,Uk+ϵmjbj〉¯〉L2(0,1)〈∑k∈ℤ(〈a,Uk+ϵmjbj〉¯|2=∑m∈ℤ‍∑j=1s‍|∑k∈ℤ‍(〈a,Urm-k+ϵmjbj〉¯-〈a,Urm-kbj〉¯)a¯k|2=∑j=1s‍∑m∈ℤ‍|∑k∈ℤ‍dm,k(j)ak|2.
From (88) and the definition of the functions Ma,bj and Na,bj, we obtain
(93)∑m∈ℤ‍∑j=1s‍|〈gm,j(·)¯e2πirm·-gj(·)¯e2πirm·,F(·)〉L2(0,1)|2≤∑j=1s‍Ma,bj(γj)Na,bj(γj)∥{al}l∈ℤ∥2≤∑j=1s‍Ma,bj(γj)Na,bj(γj)∥F∥L2(0,1)2.
By using Lemma 11, we obtain that the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame for L2(0,1) (a Riesz basis if r=s). Let {hm,jϵ}m∈ℤ;j=1,2,…,s be its canonical dual frame. Hence, for any F∈L2(0,1),
(94)F=∑m∈ℤ‍∑j=1s‍〈F(·),gm,j(·)¯e2πirm·〉L2(0,1)hm,jϵ=∑m∈ℤ‍∑j=1s‍〈x,Urm+ϵmjbj〉ℋhm,jϵ.
Applying the isomorphism𝒯U,a, one gets (91), where Cm,jϵ=𝒯U,a(hm,jϵ). Since 𝒯U,a is an isomorphism between L2(0,1) and 𝒜a; the sequence {Cm,jϵ}m∈ℤ;j=1,2,…,s is a frame for 𝒜a (a Riesz basis if r=s). The interpolatory property in the case r=s follows from the uniqueness of the coefficients with respect to a Riesz basis.


## 3.4. A Frame Algorithm inℓ2(ℤ)

Sampling formula (91) is useless from a practical point of view, it is impossible to determine the involved frame {Cm,jϵ}m∈ℤ;j=1,2,…,s. As a consequence, in order to recover x∈𝒜a from the sequence of inner products {〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s, we should implement a frame algorithm in ℓ2(ℤ). Another possibility is given in the recent reference [29].Next we are going to implement a frame algorithm in theℓ2(ℤ) setting. To this end, consider the canonical isometry
(95)𝒰~:ℓ2(ℤ)⟶L2(0,1)such  that𝒰~c≔∑k∈ℤcke2πikw00000000000000000000000.forc={ck}k∈ℤ∈ℓ2(ℤ).Forx=∑n∈ℤ‍cnUna∈𝒜a, denote by 𝔽 the following sequence:
(96)𝔽≔𝒰~-1𝒯U,a-1x={cn}n∈ℤ∈ℓ2(ℤ).The inner products{〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s can be written as
(97)〈x,Urm+ϵmjbj〉ℋ=〈𝒯U,a-1x,gm,j(w)¯e2πirmw〉L2(0,1)=〈𝔽,𝕃m,j〉ℓ2(ℤ),
where, for j=1,2,…,s and m∈ℤ,
(98)𝕃m,j≔𝒰~-1(gm,j(·)¯e2πirm·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Uk+ϵmjbj〉ℋe2πik·¯e2πirm·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Uk+ϵmjbj〉¯ℋe2πi(rm-k)·)𝕃m,j=𝒰~-1(∑k∈ℤ‍〈a,Urm-k+ϵmjbj〉¯ℋe2πik·)𝕃m,j={〈a,Urm-k+ϵmjbj〉¯ℋ}k∈ℤ.The sequence{𝕃m,j}m∈ℤ;j=1,2,…,s is a frame for ℓ2(ℤ). Indeed, assume that the error sequence ϵ≔{ϵm,j}m∈ℤ;j=1,…,s satisfies the hypothesis of Theorem 13; that is,
(99)Kϵ≔∑j=1s‍Ma,bj(γj)Na,bj(γj)<α𝔾r,
where γj≔supm∈ℤ|ϵmj| for each j=1,2,…,s. As a consequence, the sequence {gm,j(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame for L2(0,1) with bounds
(100)Aϵ≔α𝔾r(1-rKϵα𝔾)2,Bϵ≔β𝔾r(1-rKϵβ𝔾)2.Since𝒰~-1 is an isometry, the sequence {𝕃m,j}m∈ℤ;j=1,2,…,s is a frame for ℓ2(ℤ) with the same bounds. Hence, the recovery of the element x=𝒯U,a(𝒰~𝔽)∈𝒜a from the samples {〈x,Urm+ϵmjbj〉ℋ}m∈ℤ;j=1,2,…,s is reduced to recover 𝔽∈ℓ2(ℤ) from the following sequence:
(101){〈𝔽,𝕃m,j〉ℓ2(ℤ)}m∈ℤ;j=1,2,…,s.In doing so, the classical frame algorithm reads (see, for instance, [1]), let 𝒮 be the frame operator in ℓ2(ℤ) of {𝕃m,j}m∈ℤ;j=1,2,…,s, and define
(102)𝔽0≔2𝒮Aϵ+Bϵ𝔽𝔽0=2Aϵ+Bϵ∑j=1s‍∑m∈ℤ‍〈𝔽,𝕃m,j〉𝕃m,j𝔽0=2Aϵ+Bϵ∑j=1s‍∑m∈ℤ‍〈x,Urm+ϵmjbj〉𝕃m,j,
and recursively,
(103)𝔽k+1=𝔽k+2𝒮Aϵ+Bϵ(𝔽-𝔽k)for  eachk∈ℕ.Then, the sequence{xn}n∈ℕ in 𝒜a given by xn≔∑k∈ℤ(𝔽n)kUka, satisfies
(104)∥x-xn∥ℋ≤∥𝒯U,a∥∥𝔽-𝔽k∥ℓ2(ℤ)≤∥𝒯U,a∥γϵk+1∥𝔽∥ℓ2(ℤ)≤∥𝒯U,a∥∥𝒯U,a-1∥γϵk+1∥x∥ℋ,
where γϵ≔(Bϵ-Aϵ)/(Bϵ+Aϵ).


## 4. Some Additional Remarks

Givens vectors bj∈𝒜a, j=1,2,…,s with s≥r, we have proved in Section 2 that the sequence {Urkbj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a if and only if the constants AΨ and BΨ defined in (56) satisfy 0<AΨ≤BΨ<∞. Furthermore, we have obtained a family of dual frames having the same form. As it was mentioned in the introduction, now we deal with the case that some bj∉𝒜a.(i)
We have assumed in Theorems5 and 7 that bj belongs to 𝒜a for each j=1,2,…,s since we required the sequence {Urkbj}k∈ℤ;j=1,2,…,s to be contained in 𝒜a. In case that some bj∉𝒜a, the sequence {Urkbj}k∈ℤ;j=1,2,…,s is not necessarily contained in 𝒜a. However, a close look into the proof of Theorem 5 shows that whenever 0<AΨ≤BΨ<∞, the following inequalities:
(105)AΨ∥x∥2≤∑j=1s‍∑k∈ℤ‍|〈x,Urkbj〉|2≤BΨ∥x∥2∀x∈𝒜ahold, and conversely. Hence, the sequence{Urkbj}k∈ℤ;j=1,2,…,s is a pseudoframe for 𝒜a (see [17, 18]).Denoting byP𝒜a the orthogonal projection onto 𝒜a, since for each x∈𝒜a,
(106)〈x,Urkbj〉=〈x,P𝒜a(Urkbj)〉,k∈ℤ,j=1,2,…,s.Theorems5 and 7 can be reformulated in terms of {P𝒜a(Urkbj)}k∈ℤ;j=1,2,…,s which is a sequence in 𝒜a.(ii)
An analysis of the proof of Theorem9 shows that, even if not all of the bj belong to 𝒜a, there exist cj∈𝒜a, j=1,2,…,s, such that the sequence {Urkcj}k∈ℤ;j=1,2,…,s is a frame for 𝒜a, and the expansion (60) holds for each x∈𝒜a. Therefore, in case that some bj∉𝒜a, the sequence {Urkbj}k∈ℤ;j=1,2,…,s is a pseudodual frame of the frame {Urkcj}k∈ℤ;j=1,2,…,s for 𝒜a (see [17]).(iii)
In Section3.3, having in mind the isomorphism 𝒯U,a, for x=𝒯U,aF∈𝒜a, we have obtained the following expressions:
(107)〈x,Urmbj〉ℋ=〈F,gj(w)¯e2πirmw〉L2(0,1),000000000.wherem∈ℤ,j=1,2,…,s.Furthermore, we know that the sequence{gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s is a frame for L2(0,1) if and only if the constants α𝔾 and β𝔾 given in (85) satisfy 0<α𝔾≤β𝔾<∞; besides, the optimal frame bounds for {gj(w)¯e2πirmw}m∈ℤ;j=1,2,…,s are α𝔾/r and β𝔾/r. Hence, we obtain
(108)α𝔾r∥𝒯U,a∥-2∥x∥2≤∑j=1s‍∑k∈ℤ‍|〈x,Urkbj〉|2≤β𝔾r∥𝒯U,a-1∥2∥x∥2,x∈𝒜a.As a consequence, since we are dealing with optimal frame bounds, from (105) and (108), we derive the following equalities:
(109)AΨ=α𝔾r∥𝒯U,a∥-2,BΨ=β𝔾r∥𝒯U,a-1∥2.



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