Frames and generalized operator orbits

Abstract

This short note is concerned with various operator representations of frames in a Hilbert space ${{{\mathcal {H}}}}.$ While it is known that only very special frames can be represented in the form $\{ T^n \varphi \}_{n=0}^\infty$ for a bounded operator T,  we prove that every frame (actually, every Bessel sequence) has a representation of the form $\{UT^k \varphi \}_{k=0}^\infty$ for certain bounded operators U and T. We also provide a lifting procedure that allows to represent any given Bessel sequence in the form $\{PT^k \varphi \}_{k=0}^\infty ,$ where T is a bounded operator on an ambient Hilbert space and P denotes the orthogonal projection onto the given Hilbert space ${{{\mathcal {H}}}}.$ In particular, this implies that for any frame, the frame coefficients of $f\in {{{\mathcal {H}}}}$ can be calculated as inner products between f and a system of functions of the form $\{T^k \varphi \}_{k=0}^\infty ,$ for a bounded operator T on an ambient Hilbert space.

1 Introduction

By now, dynamical sampling has been a very active research area for almost a decade. In its pure functional analytic formulation, it aims at construction and characterization of bounded operators T on a Hilbert space ${{{\mathcal {H}}}}$ such that $\{T^k \varphi \}_{k=0}^\infty$ is a frame for ${{{\mathcal {H}}}},$ for some $\varphi \in {{{\mathcal {H}}}}$ [1,2,3]. The dual question, namely construction of a bounded operator T such that a given frame $\{f_k\}_{k=1}^\infty$ has a representation $\{f_k\}_{k=1}^\infty = \{T^k f_1\}_{k=0}^\infty ,$ has been analysed, e.g., in the papers [4,5,6,7].

It turns out that explicit construction of a bounded operator T such that $\{T^k \varphi \}_{k=0}^\infty$ is a frame, is a very complicated task. Indeed, the literature mainly contains negative results, with the Carleson frame [1] being the prominent exception. Concerning the dual question, the only additional class of frames that is known to have such a representation with a bounded operator T,  is the class of Riesz sequences.

The aim of the current paper is to provide a number of alternative operator representations of frames. We prove that every frame (actually, every Bessel sequence) has a representation of the form $\{UT^k \varphi \}_{k=0}^\infty$ for certain bounded operators U and T. We also provide a lifting procedure (reminiscent of the Naimark’s Theorem) that allows to represent any given Bessel sequence in the form $\{PT^k \varphi \}_{k=0}^\infty ,$ where T is a bounded operator on an ambient Hilbert space ${{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ and P denotes the orthogonal projection onto ${{{\mathcal {H}}}}.$ The impact of these observations on the frame representations will be discussed as well. In particular, we show that for any frame, the frame coefficients of $f\in {{{\mathcal {H}}}}$ can be calculated as inner products between f and a system of functions of the form $\{T^k \varphi \}_{k=0}^\infty ,$ for a bounded operator T on an ambient Hilbert space.

The paper is organized as follows. In the rest of this introduction we set the stage by stating a number of basic definitions. In Sect. 2 we prove that every frame has a representation of the form $\{UT^k \varphi \}_{k=0}^\infty$ for certain bounded operators U and T. As a consequence of the result we obtain a characterization of the frames that can be represented via iterates of a bounded operator in terms of a certain intertwining property. The lifting procedure and some of its consequences are discussed in Sect. 3. Finally, Sect. 4 collects a number of results related to operator representations and duality.

In the entire paper ${{{\mathcal {H}}}}$ and ${{{\mathcal {K}}}}$ will denote infinite-dimensional separable Hilbert spaces. We will use the following terminology. A sequence $\{f_k\}_{k=1}^\infty$ in ${{{\mathcal {H}}}}$ is a frame for ${{{\mathcal {H}}}}$ if there exist constants $A,B>0$ such that

$$\begin{aligned} A \, ||f||^2 \le \sum _{k=1}^\infty | \langle f, f_k\rangle |^2 \le B \, ||f||^2 \quad \forall f\in {{{\mathcal {H}}}}; \end{aligned}$$

it is a frame sequence if the stated inequalities hold for all $f\in \overline{\text {span}}\{f_k\}_{k=1}^\infty .$ The sequence $\{f_k\}_{k=1}^\infty$ is a Bessel sequence if at least the upper frame condition holds. Also, $\{f_k\}_{k=1}^\infty$ is called a Riesz sequence if there exist constants $A,B>0$ such that

$$\begin{aligned} A \sum |c_k|^2 \le \left| \left| \sum c_k f_k \right| \right| ^2 \le B\sum |c_k|^2 \end{aligned}$$

for all finite scalar sequences $\{c_k\}_{k=1}^\infty .$ A Riesz basis is a Riesz sequence $\{f_k\}_{k=1}^\infty$ for which $\overline{\text {span}}\{f_k\}_{k=1}^\infty = {{{\mathcal {H}}}}.$

Definition 1.1

Let W denote an operator on ${{{\mathcal {H}}}}.$

• (i) A family of the form $\{W^k\varphi \}_{k=0}^\infty$ for some $\varphi \in {{{\mathcal {H}}}}$ is called an orbit of the operator W.

• (ii) A family of the form $\{VW^k\varphi \}_{k=0}^\infty$ for some $\varphi \in {{{\mathcal {H}}}}$ and some operator V on ${{{\mathcal {H}}}}$ is called a generalized orbit of the operator W.

The shift-operator ${{{\mathcal {T}}}},$ introduced next, will play a key role throughout the paper.

Definition 1.2

Let $\{e_k\}_{k=1}^\infty$ denote an orthonormal basis for ${{{\mathcal {H}}}}.$ The shift-operator with respect to $\{e_k\}_{k=1}^\infty$ is the bounded linear operator ${{{\mathcal {T}}}}: {{{\mathcal {H}}}}\rightarrow {{{\mathcal {H}}}}$ defined by ${{{\mathcal {T}}}}e_k:=e_{k+1}, \, k\in {\mathbb {N}}.$

That the operator ${{{\mathcal {T}}}}$ indeed is bounded, follows by a standard calculation.

Recall that if $\{f_k\}_{k=1}^\infty$ is a Bessel sequence, the bounded operator

$$\begin{aligned} U: {\ell }^2({\mathbb {N}})\rightarrow {{{\mathcal {H}}}}, U\{c_k\}_{k=1}^\infty = \sum _{k=1}^\infty c_kf_k \end{aligned}$$

is called the synthesis operator. For the purpose of the current paper it is convenient to use a slightly different terminology.

Definition 1.3

Consider a Bessel sequence $\{f_k\}_{k=1}^\infty$ and fix an orthonormal basis $\{e_k\}_{k=1}^\infty$ for ${{{\mathcal {H}}}}.$ Identifying the canonical orthonormal basis for ${\ell }^2({\mathbb {N}})$ with $\{e_k\}_{k=1}^\infty ,$ we can consider the synthesis operator for $\{f_k\}_{k=1}^\infty$ as a bounded operator from ${{{\mathcal {H}}}}$ to ${{{\mathcal {H}}}}.$ This operator is called the synthesis operator for $\{f_k\}_{k=1}^\infty$ with respect to $\{e_k\}_{k=1}^\infty .$

2 Frames and operator representations

In [4] it was proved that every Riesz basis for a Hilbert space ${{{\mathcal {H}}}}$ has a representation $\{T^n \varphi \}_{n=0}^\infty$ for a bounded operator $T: {{{\mathcal {H}}}}\rightarrow {{{\mathcal {H}}}}.$ We now extend this result by providing an explicit representation for the operator T.

Proposition 2.1

Let $\{e_k\}_{k=1}^\infty$ be an orthonormal basis for ${{{\mathcal {H}}}},$ with associated shift-operator ${{{\mathcal {T}}}}.$ Then the following hold:

• (i) $\{e_k\}_{k=1}^\infty = \{{{{\mathcal {T}}}}^k e_1\}_{k=0}^\infty .$

• (ii) Let $\{f_k\}_{k=1}^\infty$ denote a Riesz sequence in ${{{\mathcal {H}}}},$ having synthesis operator U with respect to $\{e_k\}_{k=1}^\infty .$ Then, denoting the pseudo-inverse of U by $U^\dagger ,$

  $$\begin{aligned} \{f_k\}_{k=1}^\infty = \{(U{{{\mathcal {T}}}}U^{\dagger })^k f_1\}_{k=0}^\infty . \end{aligned}$$

  In particular, if $\{f_k\}_{k=1}^\infty$ is a Riesz basis, we obtain the representation

  $$\begin{aligned} \{f_k\}_{k=1}^\infty = \{(U{{{\mathcal {T}}}}U^{-1})^k f_1\}_{k=0}^\infty . \end{aligned}$$

Proof

(i) follows directly from Definition 1.2. For (ii), note that the synthesis operator associated with a Riesz sequence is an injective operator with closed range $R(U)=\overline{\text {span}}\{f_k\}_{k=1}^\infty$. Therefore its pseudo inverse $U^\dagger : {{{\mathcal {H}}}}\rightarrow {{{\mathcal {H}}}}$ is bounded and surjective, and $U^\dagger U f = f$ for all $f\in {{{\mathcal {H}}}}$. Thus

$$\begin{aligned} f_k= Ue_k = U{{{\mathcal {T}}}}^{k-1}e_1 = \left( U{{{\mathcal {T}}}}U^{\dagger }\right) ^{k-1} Ue_1=\left( U{{{\mathcal {T}}}}U^{\dagger }\right) ^{k-1} f_1, \end{aligned}$$

as claimed. If $\{f_k\}_{k=1}^\infty$ is a Riesz basis, then U is invertible and $U^\dagger =U^{-1}$. $\square$

As already discussed, only frames with very particular properties can have a representation of the form $\{T^k \varphi \}_{k=0}^\infty$ for a bounded operator T. We now prove that every frame (indeed, every Bessel sequence) has a representation of the form $\{ UT^k \varphi \}_{k=0}^\infty$ for some bounded operators T, U.

Proposition 2.2

Let $\{f_k\}_{k=1}^\infty$ be a Bessel sequence in ${{{\mathcal {H}}}},$ having synthesis operator U with respect to a fixed orthonormal basis $\{e_k\}_{k=1}^\infty .$ Then

$$\begin{aligned} \{f_k\}_{k=1}^\infty = \{ U {{{\mathcal {T}}}}^k e_1\}_{k=0}^\infty . \end{aligned}$$

(2.1)

Furthermore, the following hold:

• (i) The operator U is bounded.

• (ii) If $\{f_k\}_{k=1}^\infty$ is a frame, the operator U is bounded and surjective.

• (iii) If $\{f_k\}_{k=1}^\infty$ is a Riesz basis, the operator U is bounded and bijective.

Proof

It is well known that if $\{f_k\}_{k=1}^\infty$ is a Bessel sequence, then $f_k=Ue_k.$ Thus (2.1) follows directly from Proposition 2.1 (i). The rest is standard. $\square$

As a consequence of the above results we can provide a characterization of Bessel sequences having a representation of the form $\{ T^k \varphi \}_{k=0}^\infty$ in terms of a certain intertwining property:

Corollary 2.3

Let $\{f_k\}_{k=1}^\infty$ be a Bessel sequence having synthesis operator U with respect to a fixed orthonormal basis $\{e_k\}_{k=1}^\infty$, and consider a bounded operator $T:{{{\mathcal {H}}}}\rightarrow {{{\mathcal {H}}}}.$ Then the following are equivalent:

• (i) $\{f_k\}_{k=1}^\infty = \{ T^k \varphi \}_{k=0}^\infty$ for some $\varphi \in {{{\mathcal {H}}}}.$

• (ii) U satisfies the intertwining relation

  $$\begin{aligned} U{{{\mathcal {T}}}}= T U. \end{aligned}$$

  (2.2)

Proof

(i) $\Rightarrow$ (ii). If (i) holds, then by (2.1), we have

$$\begin{aligned} \{ T^k \varphi \}_{k=0}^\infty = \{f_k\}_{k=1}^\infty = \{ U {{{\mathcal {T}}}}^k e_1\}_{k=0}^\infty . \end{aligned}$$

That means for every $k\in {\mathbb {N}}$,

$$\begin{aligned} T Ue_k = T f_k = f_{k+1} = U e_{k+1} = U{{{\mathcal {T}}}}e_k. \end{aligned}$$

Since $\{e_k\}_{k=1}^\infty$ is an orthonormal basis for ${{{\mathcal {H}}}}$, it follows that (2.2) holds.

(ii) $\Rightarrow$ (i). Assume that (2.2) holds. Then, letting $\varphi := Ue_1,$ (2.1) implies that

$$\begin{aligned} \{f_k\}_{k=1}^\infty = \{ U {{{\mathcal {T}}}}^k e_1\}_{k=0}^\infty = \{ T^k Ue_1\}_{k=0}^\infty = \{ T^k \varphi \}_{k=0}^\infty , \end{aligned}$$

as desired. $\square$

3 The lifting procedure

It is well-known (Naimark’s Theorem) that every tight frame can be considered as the projection of an orthogonal sequence in an ambient Hilbert space; and that every frame is a projection of a Riesz sequence in an ambient Hilbert space. We will now provide a certain lifting procedure, which yields an explicit version of this result, valid for any Bessel sequence $\{f_k\}_{k=1}^\infty$ in the given Hilbert space.

Proposition 3.1

Let $\{f_k\}_{k=1}^\infty$ be a Bessel sequence in ${{{\mathcal {H}}}},$ and fix an orthonormal basis $\{e_k\}_{k=1}^\infty$ for an infinite-dimensional Hilbert space ${{{\mathcal {K}}}}.$ Then the following hold:

• (i) The sequence $\{(f_k, e_k)\}_{k=1}^\infty \subset {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ is a Riesz sequence in the Hilbert space ${{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$.

• (ii) There exists a bounded operator $T: {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}\rightarrow {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ such that

  $$\begin{aligned} \{(f_k, e_k)\}_{k=1}^\infty = \{T^k (f_1, e_1)\}_{k=0}^\infty . \end{aligned}$$

• (iii) Letting P denote the orthogonal projection of ${{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ onto ${{{\mathcal {H}}}},$

  $$\begin{aligned} \{f_k\}_{k=1}^\infty = \{PT^k (f_1, e_1)\}_{k=0}^\infty . \end{aligned}$$

Proof

For any finite scalar-sequence $\{c_k\},$ we have that

$$\begin{aligned} \left| \left| \sum c_k(f_k, e_k) \right| \right| _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}}^2 = \left| \left| \sum c_kf_k \right| \right| _{{{\mathcal {H}}}}^2 + \left| \left| \sum c_k e_k \right| \right| _{{{\mathcal {K}}}}^2. \end{aligned}$$

It follows from here that $\{(f_k, e_k)\}_{k=1}^\infty$ is a Bessel sequence, and also that

$$\begin{aligned} \left| \left| \sum c_k(f_k, e_k) \right| \right| _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}}^2 \ge \sum |c_k|^2. \end{aligned}$$

This proves that $\{(f_k, e_k)\}_{k=1}^\infty$ is a Riesz sequence. (ii) now follows from Proposition 2.1, and (iii) follows from (ii). $\square$

As a consequence of the lifting procedure, we now prove that for any frame, we can provide a frame decomposition of $f\in {{{\mathcal {H}}}}$ where the frame coefficients are obtained by taking inner products of f and an orbit of a bounded operator on an ambient Hilbert space. The proof idea is inspired by the theory for pseudoduals due to Li [8].

Theorem 3.2

Let ${{{\mathcal {H}}}}$ and ${{{\mathcal {K}}}}$ be infinite-dimensional Hilbert spaces and assume that $\{f_k\}_{k=1}^\infty$ is a frame for ${{{\mathcal {H}}}}.$ Then there exists a bounded operator $T: {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}\rightarrow {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ and $\varphi \in {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ such that

$$\begin{aligned} f= \sum _{k=1}^\infty \langle (f,0), T^{k-1}\varphi \rangle _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}} f_k, \, \forall f \in {{{\mathcal {H}}}}. \end{aligned}$$

Proof

Let $\{g_k \}_{k=1}^\infty$ denote a dual frame of $\{f_k\}_{k=1}^\infty .$ Applying the lifting procedure in Proposition 3.1, we can extend $\{g_k \}_{k=1}^\infty$ to a Riesz sequence $\{(g_k, e_k)\}_{k=1}^\infty$ in ${{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ and write $\{(g_k, e_k)\}_{k=1}^\infty =\{T^{k-1}\varphi \}_{k=1}^\infty$ for a certain bounded operator $T: {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}\rightarrow {{{\mathcal {H}}}}\times {{{\mathcal {K}}}}$ and $\varphi :=(g_1,e_1).$ Note that for $f\in {{{\mathcal {H}}}},$

$$\begin{aligned} \langle f, g_k\rangle = \langle (f,0), (g_k,e_k)\rangle _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}} = \langle (f,0), T^{k-1}\varphi \rangle _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}}. \end{aligned}$$

It follows that

$$\begin{aligned} f= \sum _{k=1}^\infty \langle f, g_k\rangle f_k= \sum _{k=1}^\infty \langle (f,0), T^{k-1}\varphi \rangle _{{{{\mathcal {H}}}}\times {{{\mathcal {K}}}}} f_k, \end{aligned}$$

as claimed. $\square$

Theorem 3.2 has a natural and immediate application to the case of a frame sequence spanning a subspace of ${{{\mathcal {H}}}}$ having infinite codimension:

Corollary 3.3

Let $\{f_k\}_{k=1}^\infty$ denote a frame sequence in ${{{\mathcal {H}}}},$ and assume that $\text{ span }{\{f_k\}_{k=1}^\infty }^{\bot }=\infty .$ Then there exist $\varphi \in {{{\mathcal {H}}}}$ and a bounded operator $V: {{{\mathcal {H}}}}\rightarrow {{{\mathcal {H}}}}$ such that

$$\begin{aligned} f= \sum _{k=1}^\infty \langle f, V^{k-1} \varphi \rangle f_k, \, \forall f\in \overline{\text {span}}\{f_k\}_{k=1}^\infty . \end{aligned}$$

Proof

The result follows by applying Theorem 3.2 with the Hilbert space ${{{\mathcal {H}}}}$ replaced by $\overline{\text {span}}\{f_k\}_{k=1}^\infty$ and ${{{\mathcal {K}}}}$ replaced by $\overline{\text {span}}{\{f_k\}_{k=1}^\infty }^{\bot }.$ $\square$

4 Duality and frame decompositions

Assuming that a given frame is an orbit of a bounded operator, as well the canonical dual frame as the associated tight frame are orbits of bounded operators:

Proposition 4.1

Consider a frame of the form $\{T^k \varphi \}_{k=0}^\infty ,$ and denote the associated frame operator by S. Then the following hold:

• (i) The canonical dual frame is given as the orbit $\{ (S^{-1}T S)^k S^{-1}\varphi \}_{k=0}^\infty .$

• (ii) The tight frame associated with $\{T^k \varphi \}_{k=0}^\infty$ is given as the orbit $\{ (S^{-1/2} T S^{1/2})^k S^{-1/2} \varphi \}_{k=0}^\infty .$

Finally, we note that if a frame $\{T^k \varphi \}_{k=0}^\infty$ has an approximate dual frame $\{{\widetilde{T}}^k {\widetilde{\varphi }}\}_{k=0}^\infty$, it actually has a (exact) dual frame of that form.

Proposition 4.2

Assume that $\{T^k \varphi \}_{k=0}^\infty$ and $\{{\widetilde{T}}^k {\widetilde{\varphi }}\}_{k=0}^\infty$ are approximate dual frames, with synthesis operators U,  respectively, ${\widetilde{U}}.$ Then $\{T^k \varphi \}_{k=0}^\infty$ and $\{ (U{\widetilde{U}}^*)^{-1}T^k \varphi \}_{k=0}^\infty$ are dual frames.

Proof

By definition of approximate dual frames, we have that for some $0<\epsilon <1,$   $||I-U{\widetilde{U}}^*|| \le \epsilon .$ Thus the operator $U{\widetilde{U}}^*$ is invertible, and

$$\begin{aligned} f= (U{\widetilde{U}}^*)^{-1}U{\widetilde{U}}^*f=(U{\widetilde{U}}^*)^{-1} \sum _{k=0}^\infty \langle f, {\widetilde{T}}^k {\widetilde{\varphi }} \rangle T^k \varphi = \sum _{k=0}^\infty \langle f, {\widetilde{T}}^k {\widetilde{\varphi }} \rangle (U{\widetilde{U}}^*)^{-1}T^k \varphi . \end{aligned}$$

Since both sequences $\{{\widetilde{T}}^k {\widetilde{\varphi }}\}_{k=0}^\infty$ and $\{ (U{\widetilde{U}}^*)^{-1}T^k \varphi \}_{k=0}^\infty$ are Bessel sequences, they are dual frames. Finally, a direct calculation shows that $(U{\widetilde{U}}^*)^{-1}T^k \varphi = (( U{\widetilde{U}}^*)^{-1}T U{\widetilde{U}}^*)^k (U{\widetilde{U}}^*)^{-1} \varphi .$ $\square$