The duals of $\ast$-operator frames for $End_{\mathcal{A}}^{\ast}(H)$

TL;DR

This paper introduces the concept of duals for $ ext{ extasterisk}$-operator frames within Hilbert $ ext{ extasterisk}$-modules, exploring their properties and establishing foundational results for their application.

Contribution

It defines dual $ ext{ extasterisk}$-operator frames in Hilbert $ ext{ extasterisk}$-modules and investigates their properties, providing new theoretical insights.

Findings

Established properties of dual $ ext{ extasterisk}$-operator frames.

Extended frame theory to Hilbert $ ext{ extasterisk}$-modules.

Provided foundational results for future applications.

Abstract

Frames play significant role in signal and image processing, which leads to many applications in differents fields. In this paper we define the dual of $\ast$-operator frames and we show their propreties obtained in Hilbert $\mathcal{A}$-modules and we establish some results.

Equations

$$A⟨x,x{⟩}_{\mathcal{A}}\le \sum _{i\in I}⟨T_{i}x,T_{i}x{⟩}_{\mathcal{A}}\le B⟨x,x{⟩}_{\mathcal{A}},\forall x\in \mathcal{H}.$$

$$A⟨x,x{⟩}_{\mathcal{A}}{A}^{\ast }\le \sum _{i\in I}⟨T_{i}x,T_{i}x{⟩}_{\mathcal{A}}\le B⟨x,x{⟩}_{\mathcal{A}}{B}^{\ast },\forall x\in \mathcal{H}.$$

$$Sx=\sum _{i\in I}{T}_i^{\ast }{T}_{i}x$$

$$x=\sum _{i\in I}{T}_i^{\ast }{T}_i{S}^{-1}x$$

$$x=\sum _{i\in I}{T}_i^{\ast }{\tilde{T}}_{i}x=R_T^{\ast }{R}_{\tilde{T}}x$$

$$⟨{\theta }^{\ast }\eta f,g⟩$$

$$=⟨\sum _{i\in I}{\Omega }_{i}f,\sum _{i\in I}{\Lambda }_{i}g⟩$$

$$=\sum _{i\in I}⟨{\Omega }_{i}f,{\Lambda }_{i}g⟩=\sum _{i\in I}⟨{\Lambda }^{\ast }{\Omega }_{i}f,g⟩$$

$$=⟨\sum _{i\in I}{\Lambda }^{\ast }{\Omega }_{i}f,g⟩=⟨f,g⟩$$

$$||({\eta }^{{}^{\prime }}{)}^{\ast }\eta {)}^{-1}|{|}^{-1}⟨f,f⟩\le ⟨{\eta }^{{}^{\prime }}f,\eta {}^{\prime }f⟩=\sum _{i\in I}⟨{\pi }_i\eta {}^{\prime }f,{\pi }_i\eta {}^{\prime }f⟩$$

$$S_{\Omega }$$

$$={\eta }^{\ast }\theta S^{-1}+{\eta }^{\ast }\eta -{\eta }^{\ast }\theta S^{-1}$$

$$={\eta }^{\ast }\theta S^{-1}+{\eta }^{\ast }\eta -{\eta }^{\ast }\theta S^{-1}{\theta }^{\ast }\eta$$

$$={\eta }^{\ast }\theta S^{-1}+{\eta }^{\ast }(I-\theta S^{-1}{\theta }^{\ast })\eta$$

$$\theta S^{-1}+(I-\theta S^{-1}{\theta }^{\ast }){\eta }^{{}^{\prime }}=\theta S^{-1}+\eta {}^{\prime }-\theta S^{-1}{\theta }^{\ast }{\eta }^{{}^{\prime }}=\theta S^{-1}+\eta {}^{\prime }-\theta S^{-1}I=\eta {}^{\prime }$$

$${\pi }_i\psi$$

$$={\Lambda }_i{S}^{-1}+{\Delta }_i-{\Lambda }_i{S}^{-1}\sum _{k\in I}{\Lambda }_k^{\ast }{\Delta }_k$$

$$={\Lambda }_i{S}^{-1}+{\pi }_i\eta -\sum _{k\in I}{\Lambda }_k^{\ast }{\Delta }_k$$

$$={\pi }_i\theta S^{-1}+{\pi }_i\eta -{\pi }_i\theta S^{-1}{\theta }^{\ast }\eta$$

$$={\pi }_i(\theta S^{-1}+\eta -\theta S^{-1}{\theta }^{\ast }\eta )$$

$$\psi =\theta S^{-1}+\eta -\theta S^{-1}{\theta }^{\ast }\eta$$

$$\sum _{j\in J}{\Lambda }_{g_j}^{\ast }{\Lambda }_{f_j}f=\sum _{j\in J}⟨f,f_j⟩g_j=\sum _{j\in J}⟨f,g_j⟩f_j=\sum _{j\in J}{\Lambda }_{f_j}^{\ast }{\Lambda }_{g_j}f.$$

$$\sum _{j\in J}⟨f,f_j⟩f_j=\sum _{j\in J}f{f}_j^{\ast }{f}_j=\sum _{j\in J}⟨⟨f,f_j⟩,f_j^{\ast }⟩=\sum _{j\in J}{\Lambda }_{f_j}^{\ast }{\Lambda }_{f_j}f.$$

$$g_j=S^{-1}{f}_j+h_j-\sum _{k\in J}⟨S^{-1}{f}_j,f_k⟩h_k$$

$${\Gamma }_j={\tilde{\Lambda }}_{f_j}+{\Lambda }_{h_j}+\sum _{k\in J}{\tilde{\Lambda }}_{f_j}{\Lambda }_{f_k}^{\ast }{\Lambda }_{h_k},\forall j\in J$$

$${\Gamma }_{j}f$$

$$={\Lambda }_{f_j}{S}^{-1}f+{\Lambda }_{h_j}f-\sum _{k\in J}{\Lambda }_{f_j}{S}^{-1}{\Lambda }_{f_k}^{\ast }⟨f,h_k⟩$$

$$=⟨S^{-1}f,f_j⟩+⟨f,h_j⟩-\sum _{k\in J}⟨S^{-1}{\Lambda }_{f_k}^{\ast }⟨f,h_k⟩,f_j⟩$$

$$=⟨S^{-1}f,f_j⟩+⟨f,h_j⟩-\sum _{k\in J}⟨S^{-1}{\Lambda }_{f_k}^{\ast }{\Lambda }_{h_k}f,f_j⟩$$

$$=⟨S^{-1}f,f_j⟩+⟨f,h_j⟩-\sum _{k\in J}⟨S^{-1}{\Lambda }_{h_k}f{f}_k,f_j⟩$$

$$=⟨S^{-1}f,f_j⟩+⟨f,h_j⟩-\sum _{k\in J}⟨S^{-1}f{h}_k^{\ast }{f}_k,f_j⟩$$

$$=⟨f,S^{-1}{f}_j⟩+⟨f,h_j⟩-\sum _{k\in J}⟨f{h}_k^{\ast }{f}_k,S^{-1}{f}_j⟩$$

$$=⟨f,S^{-1}{f}_j+h_j⟩-\sum _{k\in J}⟨f,f_k^{\ast }{h}_k{S}^{-1}{f}_j⟩$$

$$=⟨f,S^{-1}{f}_j+h_j-\sum _{k\in J}⟨S^{-1}{f}_j,f_k⟩h_k⟩$$

$$=⟨f,g_j⟩={\Lambda }_{g_j}f.$$

$${\tilde{\Lambda }}_{f_j}+{\Lambda }_{h_j}-\sum _{k\in J}{\tilde{\Lambda }}_{f_j}{\Lambda }_{f_k}^{\ast }{\Lambda }_{h_k}$$

$$\sum _{i\in I,j\in J}({\Lambda }_i\otimes {\Gamma }_j{)}^{\ast }(\widetilde{{\Lambda }_i}\otimes \widetilde{{\Gamma }_j})(x\otimes y)$$

$$=\sum _{i\in I,j\in J}({\Lambda }_i^{\ast }\widetilde{{\Lambda }_i}x\otimes {\Gamma }_j^{\ast }\widetilde{{\Gamma }_j}y)$$

$$=\sum _{i\in I}{\Lambda }_i^{\ast }\widetilde{{\Lambda }_i}x\otimes \sum _{j\in J}{\Gamma }_j^{\ast }\widetilde{{\Gamma }_j}y$$

$$=x\otimes y$$

$$\sum _{i\in I,j\in J}({\Lambda }_i\otimes {\Gamma }_j{)}^{\ast }(\widetilde{{\Lambda }_i}\otimes \widetilde{{\Gamma }_j})=I$$

Full text

Frame theory is recently an active research area in mathematics, computer science, and engineering with many exciting applications in a variety of different fields. They are generalizations of bases in Hilbert spaces. Frames for Hilbert spaces were first introduced in 1952 by Duffin and Schaeffer [5] for study of nonharmonic Fourier series. They were reintroduced and developed in 1986 by Daubechies, Grossmann and Meyer [4], and popularized from then on. Hilbert $C^{\ast}$-modules is a generalization of Hilbert spaces by allowing the inner product to take values in a $C^{\ast}$-algebra rather than in the field of complex numbers. The aim of this papers is to study the dual of $\ast$-operator frames.

The paper is organized as follows:

In section 2, we briefly recall the definitions and basic properties of operator frame and $\ast$-operator frame in Hilbert $C^{\ast}$-modules.

In section 3, we introduce the dual $\ast$-operator frame, the $\ast$-operator frame transform and the $\ast$-frame operator.

In section 4, we investigate tensor product of Hilbert $C^{\ast}$-modules, we show that tensor product of dual $\ast$-operator frames for Hilbert $C^{\ast}$-modules $\mathcal{H}$ and $\mathcal{K}$, present a dual $\ast$-operator frames for $\mathcal{H}\otimes\mathcal{K}$.

1. Preliminaries

Let $I$ be a countable index set. In this section we briefly recall the definitions and basic properties of $C^{\ast}$-algebra, Hilbert $C^{\ast}$-modules, frame, $\ast$-frame in Hilbert $C^{\ast}$-modules. For information about frames in Hilbert spaces we refer to [1]. Our reference for $C^{\ast}$-algebras is [3, 2]. For a $C^{\ast}$-algebra $\mathcal{A}$, an element $a\in\mathcal{A}$ is positive ($a\geq 0$) if $a=a^{\ast}$ and $sp(a)\subset\mathbf{R^{+}}$. $\mathcal{A}^{+}$ denotes the set of positive elements of $\mathcal{A}$.

Definition 1.1**.**

[6] .

A family of adjointable operators $\{T_{i}\}_{i\in I}$ on a Hilbert $\mathcal{A}$-module $\mathcal{H}$ over a unital $C^{\ast}$-algebra is said to be an operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$, if there exist positive constants $A,B>0$ such that

[TABLE]

The numbers $A$ and $B$ are called lower and upper bound of the operator frame, respectively. If $A=B=\lambda$, the operator frame is $\lambda$-tight. If $A=B=1$, it is called a normalized tight operator frame or a Parseval operator frame.

Definition 1.2**.**

[6].

A family of adjointable operators $\{T_{i}\}_{i\in I}$ on a Hilbert $\mathcal{A}$-module $\mathcal{H}$ over a unital $C^{\ast}$-algebra is said to be an $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$, if there exists two strictly nonzero elements $A$ and $B$ in $\mathcal{A}$ such that

[TABLE]

The elements $A$ and $B$ are called lower and upper bounds of the $\ast$-operator frame, respectively. If $A=B=\lambda$, the $\ast$-operator frame is $\lambda$-tight. If $A=B=1_{\mathcal{A}}$, it is called a normalized tight $\ast$-operator frame or a Parseval $\ast$-operator frame. If only upper inequality of hold, then $\{T_{i}\}_{i\in i}$ is called an $\ast$-operator Bessel sequence for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.

If the sum in the middle of (2.1) is convergent in norm, the operator frame is called standard. If only upper inequality of (2.1) hold, then $\{T_{i}\}_{i\in I}$ is called an operator Bessel sequence for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.

2. Dual of $\ast$-operator Frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$

We begin this section with the following definition.

Definition 2.1**.**

.

Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator frame for $\mathcal{H}$. If there exists an $\ast$-operator frame $\{\Lambda_{i}\}_{i\in I}$ such that $x=\sum_{i\in I}T^{\ast}_{i}\Lambda_{i}x$ for all $x\in\mathcal{H}$. then the $\ast$-operator frames $\{\Lambda_{i}\}_{i\in I}$ is called the duals $\ast$-operator frames of $\{T_{i}\}_{i\in I}$.

Example 2.2**.**

.

Let $\mathcal{A}$ be a Hilbert $\mathcal{A}$-module over itself, let $\{f_{j}\}_{j\in J}$ be an $\ast$-frame for $\mathcal{A}$.

We define the adjointable $\mathcal{A}$-module map $\Lambda_{f_{j}}:\mathcal{A}\to\mathcal{A}$ by $\Lambda_{f_{j}}f=\langle f,f_{j}\rangle$. Clearly, that $\{\Lambda_{f_{j}}\}_{j\in J}$ is an $\ast$-operator frame for $\mathcal{A}$.

Theorem 2.3**.**

.

Every $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ has a dual $\ast$-operator frame.

Proof.

.

Let $\{T_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ be an $\ast$-operator for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$, with $\ast$-operator $S$.

We see that $\{T_{i}S^{-1}\}_{i\in I}$ is an $\ast$-operator frame.

Or, $\forall x\in\mathcal{H}$ we have :

[TABLE]

then

[TABLE]

hence $\{T_{i}S^{-1}\}_{i\in I}$ is a dual $\ast$-operator frame of $\{T_{i}\}_{i\in I}$.

It is called the canonique dual $\ast$-operator frame of $\{T_{i}\}_{i\in I}$. ∎

Remark 2.4**.**

.

Assume that $T=\{T_{i}\}_{i\in I}$ is an $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ with analytic operator $R_{T}$ and $\tilde{T}=\{\tilde{T}_{i}\}_{i\in I}$ is a dual $\ast$-operator frame of T with analytic operator $R_{T}$, then for any $x\in\mathcal{H}$ we have:

[TABLE]

this show that every element of H can be reconstructed with a $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ and its dual.

Theorem 2.5**.**

.

Let $\{\Lambda_{i}\}_{i\in I}$ be an $\ast$-operator frame for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ with $\ast$-operator frame transform $\theta$, the $\ast$-operator frame $S$ and the canonical dual $\ast$-operator frames $\{\tilde{\Lambda}_{i}\}_{i\in\mathbb{J}}$.

Let $\{\Omega_{i}\}_{i\in I}$ be an arbitrary dual $\ast$-operator frame of $\{\Lambda_{i}\}_{i\in I}$ with the $\ast$-operator frame transform $\eta$; then the folowing statements are true:

• (1)

${\theta}^{\ast}\eta=I$.

• (2)

$\Omega_{i}=\Pi_{i}\eta$* for all $i\in I$.*

• (3)

If $\eta^{{}^{\prime}}:\mathcal{H}\longmapsto l^{2}(\mathcal{H})$ is any adjointable right inverse of $\theta^{\ast}$ then $\{\Pi_{i}{\eta^{{}^{\prime}}}\}_{i\in I}$ is a dual $\ast$-operator frame of $\{\Lambda\}_{i\in I}$ with the operator frames transform $\eta^{{}^{\prime}}$.

• (4)

The $\ast$-operator frame $S_{\Omega}$ of $\{\Omega_{i}\}_{i\in I}$ is equal to $S^{-1}+{\eta^{\ast}(I-\theta S^{-1}\theta^{\ast})\eta}$.

• (5)

Every adjointable right inverse $\eta^{{}^{\prime}}$ of $\theta^{\ast}$ is the forme :

$\eta^{{}^{\prime}}=\theta S^{-1}+(I-\theta S^{-1}\theta^{\ast})\psi$* for some adjointable map $\psi:\mathcal{H}\longmapsto l^{2}(\mathcal{H})$ and vice versa.*

• (6)

There exist a $\ast$-bessel operator $\{\Delta_{j}\}_{j\in J}\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$ $\{\Delta\}_{i\in\mathbb{I}}$ whose $\ast$-operator frame transform is $\eta$ and yields is $\eta$ and yields

$\Omega_{j}=\tilde{\Lambda}_{j}+\Delta_{j}-\sum_{k\in J}\tilde{\Lambda}_{j}\Lambda^{\ast}_{k}\Delta_{k},\forall j\in J$**

Proof.

.

• (1)

For $f,g\in\mathcal{H}$ we have :

[TABLE]

then ${\theta}^{\ast}\eta=I$.

• (2)

The proof is clear from the definition

• (3)

Since $\eta^{{}^{\prime}}$ is adjointable, it follows from prop3.1 that $\{\Pi_{i}{\eta^{{}^{\prime}}}\}_{i\in I}$ is a $\ast$-bessel sequence in $\mathcal{H}$.

Also, since $(\eta^{{}^{\prime}})^{\ast}\theta=I$; $(\eta^{{}^{\prime}})^{\ast}$ is surjective, by lemme 2.7, for $f\in\mathcal{H}$ we have:

[TABLE]

clearly, $\eta^{{}^{\prime}}$ is the pre-frame $\ast$-operator frame transform $\{\Pi_{i}{\eta^{{}^{\prime}}}\}_{i\in I}$

• (4)

[TABLE]

• (5)

If $\eta^{{}^{\prime}}$ is such a right inverse of $\theta$, then

[TABLE]

• (6)

Let $\{\Delta_{i}\}_{i\in I}$ be an $\ast$-operator bessel sequence for $End_{\mathcal{A}}^{\ast}(\mathcal{H})$ with the preframe operator $\eta$. For $i\in I$, let $\Omega_{i}=\tilde{\Lambda_{i}}+\Delta_{i}-\sum_{k\in I}\tilde{\Lambda_{i}}\Lambda_{k}^{\ast}\Delta_{k}$ Let $S$ and $\theta$ be the $\ast$-frame operator and the preframe operator of $\{\Delta_{i}\}_{i\in I}$ , resp. we define the linear operator $\psi:\mathcal{H}\longmapsto l^{2}(\mathcal{H})$ by $\psi f=(\Omega_{i}f)_{i\in I}$. clearly, $\psi$ is adjointable, for every $i\in I$, we have

[TABLE]

then

[TABLE]

by parts (3) and (5) of the theorem; $\{\Omega_{i}\}_{i\in I}$ becomes a dual of $\ast$-operator $\{\Lambda_{i}\}_{i\in I}$

∎

Example 2.6**.**

.

Let $\mathcal{A}$ be a Hilbert $\mathcal{A}$-module over itself, let $\{f_{j}\}_{j\in J}\subset\mathcal{A}$.

We define the adjointable $\mathcal{A}$-module map $\Lambda_{f_{j}}:\mathcal{A}\to\mathcal{A}$ with $\Lambda_{f_{j}}.f=\langle f,f_{j}\rangle$, clearly $\{f_{j}\}_{j\in J}$ is a $\ast$-frame in $\mathcal{A}$ if and only if $\{\Lambda_{f_{j}}\}_{j\in J}$ is a $\ast$-operator frame in $\mathcal{A}$.

In the folowing, we study the duals of such $\ast$-operator frame.

(a) Let $\{g_{j}\}_{j\in J}\subset\mathcal{A}$ for all $f\in\mathcal{A}$ :

[TABLE]

Therefore, $\{g_{j}\}_{j\in J}$ is a dual $\ast$-frame of $\{f_{j}\}_{j\in J}$ if and only if $\{\Lambda_{g_{j}}\}_{j\in J};$ is a dual $\ast$-operator of $\{\Lambda_{f_{j}}\}_{j\in J}$

(b) Let $S$ and $S_{\Lambda}$ be the $\ast$-frame operators of $\{f_{j}\}_{j\in J}$ and $\{\Lambda_{f_{j}},\mathcal{A}\}_{j\in J}$ respectively.

For all $f\in\mathcal{A}$ we have:

[TABLE]

It follows that $S=S_{\Lambda}$

(c) It is clearly to see that $\{h_{j}\}_{j\in J}\subset\mathcal{A}$ is an $\ast$-bessel sequence if and only if $\{\Lambda_{h_{j}},\mathcal{A}\}_{j\in J}$ is an $\ast$-bessel operator.

(d) for a $\ast$-bessel sequence $\{h_{j}\}_{j\in J}$ we define

[TABLE]

then the sequence $\{g_{j}\}_{j\in J}$ is a dual $\ast$-frame of $\{f_{j}\}_{j\in J}$.

By the last theorem, the sequence $\{\Gamma_{j}\}_{j\in J}$ is a dual $\ast$-operator frame of $\{\Lambda_{f_{j}}\}_{j\in J}$, where

[TABLE]

now we clain that $\Gamma_{j}=\Lambda_{g_{j}}$

In fact, $\forall f\in\mathcal{A}$ we have

[TABLE]

therefore, every $\ast$-operator frame of $\{\Lambda_{f_{j}}\}_{j\in J}$ has the form :

[TABLE]

where $\{h_{j}\}_{j\in J}$ is a $\ast$-bessel sequence in $\mathcal{A}$.

3. Tensor product

In this section, we study the tensor product of the duals $\ast$-operator frames.

Theorem 3.1**.**

.

Let $\mathcal{H}$ and $\mathcal{K}$ are two Hilbert $C^{\ast}$-modules over unitary $C^{\ast}$-Algebras $\mathcal{A}$ and $\mathcal{B}$ respectively, let $\{\Lambda_{i}\}_{i\in I}\subset End_{\mathcal{A}}^{\ast}(\mathcal{H})$ and $\{\Gamma_{j}\}_{j\in J}\subset End_{\mathcal{B}}^{\ast}(\mathcal{K})$ are an $\ast$-operators frames.

If $\{\tilde{\Lambda_{i}}\}_{i\in I}$ is a dual of $\{\Lambda_{i}\}_{i\in I}$ and $\{\tilde{\Gamma_{j}}\}_{j\in J}$ is a dual of $\{\Gamma_{j}\}_{j\in J}$

then $\{\tilde{\Lambda_{i}}\otimes\tilde{\Gamma_{j}}\}_{i\in I,j\in J}$ is a dual $\ast$-operator frame of $\{\Lambda_{i}\otimes\Gamma_{j}\}_{i\in I,j\in J}$.

Proof.

.

Let $x\in\mathcal{H}$ and $y\in\mathcal{K}$, we have :

[TABLE]

then

[TABLE]

hence $\{\tilde{\Lambda_{i}}\otimes\tilde{\Gamma_{j}}\}_{i\in I,j\in J}$ is a dual $\ast$-operator frames of $\{\Lambda_{i}\otimes\Gamma_{j}\}_{i\in I,j\in J}$. ∎

Corollary 3.2**.**

.

Let $(\Lambda_{ij})_{0\leq i\leq n;j\in J}$ be a family of $\ast$-operator and $(\tilde{\Lambda}_{ij})_{0\leq i\leq n;j\in J}$ its their dual, then $(\tilde{\Lambda}_{0j}\otimes\tilde{\Lambda}_{1j}\otimes...\otimes\tilde{\Lambda}_{nj})_{j\in J}$ is a dual of $(\Lambda_{0j}\otimes\Lambda_{1j}\otimes...\otimes\Lambda_{nj})_{j\in J}$.

Acknowledgment

The authors would like to thank from the anonymous reviewers for carefully reading of the manuscript and giving useful comments, which will help to improve the paper.