Characterization of duals of continuous frames in Hilbert C*-modules

TL;DR

This paper explores the properties and construction methods of dual continuous frames in Hilbert C*-modules, establishing a correspondence with adjointable operators and analyzing dual sum conditions.

Contribution

It introduces a new characterization of duals, constructs dual families via fixed duals, and links duals to adjointable operators in Hilbert C*-modules.

Findings

Established a one-to-one correspondence between duals and adjointable operators.

Provided conditions for the sum of duals to also be a dual.

Presented methods for constructing duals from a fixed dual.

Abstract

In this paper, we investigate some characterizations of dual continuous frames and give some results about them. Also, we refer to the method of constructing a family of duals through a fixed dual and show there exists a one-to-one correspondence between duals of a continuous frame for Hilbert $C^*$-module $U$ and adjointable operators $K$ from $U$ to $L^{2}(\Omega ,\mathcal A)$. Then we check the conditions that the sum of two duals of a given continuous frame under the influence of adjointable mappings becomes a dual of it and state some results about them.

Equations

$$\Vert ⟨f,g⟩{\Vert }^2\le \Vert ⟨f,f⟩\Vert \Vert ⟨g,g⟩\Vert$$

$$⟨Tf,g⟩=⟨f,T^{\ast }g⟩$$

$$L^2(\Omega ,A)=\{\phi :\Omega \to A;\Vert {\int }_{\Omega }|(\phi (\omega ){)}^{\ast }{|}^{2}d\mu (\omega )\Vert <\infty \}$$

$$⟨\phi ,\psi ⟩={\int }_{\Omega }⟨\phi (\omega ),\psi (\omega )⟩d\mu (\omega ),$$

$$A⟨f,f⟩\le {\int }_{\Omega }⟨f,F(\omega )⟩⟨F(\omega ),f⟩d\mu (\omega )\le B⟨f,f⟩,(f\in U).$$

$$⟨T_F\phi ,f⟩={\int }_{\Omega }\phi (\omega )⟨F(\omega ),f⟩d\mu (\omega ),(f\in U).$$

$$(T_F^{\ast }f)(\omega )=⟨f,F(\omega )⟩(\omega \in \Omega ).$$

$$⟨S_{F}f,f⟩={\int }_{\Omega }⟨f,F(\omega )⟩⟨F(\omega ),f⟩d\mu (\omega ),(f\in U).$$

$$f={\int }_{\Omega }⟨f,G(\omega )⟩F(\omega )d\mu (\omega ),(f\in U)$$

$$⟨f,g⟩={\int }_{\Omega }⟨f,G(\omega )⟩⟨F(\omega ),g⟩d\mu (\omega ),(f,g\in U).$$

$$f={\int }_{\Omega }⟨f,F(\omega )⟩S^{-1}F(\omega )d\mu (\omega )={\int }_{\Omega }⟨f,S^{-1}F(\omega )⟩F(\omega )d\mu (\omega ).$$

$$G_{i+1}(\omega )=S_F^{-1}F(\omega )+S_F{G}_i(\omega )-F(\omega )$$

$${\int }_{\Omega }⟨f,F(\omega )⟩⟨G_2(\omega ),g⟩d\mu (\omega )$$

$$-{\int }_{\Omega }⟨f,F(\omega )⟩⟨F(\omega ),g⟩d\mu (\omega )$$

$$=⟨f,g⟩+⟨f,S_{F}g⟩-⟨S_{F}f,g⟩=⟨f,g⟩,$$

$${\int }_{\Omega }⟨f,F(\omega )⟩⟨G_{k+1}(\omega ),g⟩d\mu (\omega )$$

$$-{\int }_{\Omega }⟨f,F(\omega )⟩⟨F(\omega ),g⟩d\mu (\omega )$$

$$=⟨f,g⟩+⟨f,S_{F}g⟩-⟨S_{F}f,g⟩=⟨f,g⟩,$$

$$V_1(\omega )=S_F^{-1}F(\omega )+S_{F}G(\omega )-F(\omega ),$$

$$V_{i+1}(\omega )=S_F^{-1}F(\omega )+S_F^{i+1}G(\omega )-S_F^{i}F(\omega ),$$

$${\int }_{\Omega }⟨f,V_{i+1}F(\omega )⟩F(\omega )d\mu (\omega )$$

$$={\int }_{\Omega }⟨f,S_F^{-1}F(\omega )⟩F(\omega )d\mu (\omega )+{\int }_{\Omega }⟨f,S_F^{i+1}G(\omega )⟩F(\omega )d\mu (\omega )$$

$$-{\int }_{\Omega }⟨f,S^{i}F(\omega )⟩F(\omega )d\mu (\omega )$$

$$=f+S_F^{i+1}f-S_F(S_F^{i}f)=f,$$

$$G(\omega )=S_F^{-1}F(\omega )+L(\omega )(\omega \in \Omega ),$$

$${\int }_{\Omega }⟨f,F(\omega )⟩⟨L(\omega ),g⟩d\mu (\omega )$$

$$={\int }_{\Omega }⟨f,F(\omega )⟩⟨G(\omega ),g⟩d\mu (\omega )-{\int }_{\Omega }⟨f,F(\omega )⟩⟨S_F^{-1}F(\omega ),g⟩d\mu (\omega )$$

$$=⟨f,g⟩-⟨f,g⟩=0.$$

$${\int }_{\Omega }⟨f,F(\omega )⟩⟨G(\omega ),g⟩d\mu (\omega )$$

$$={\int }_{\Omega }⟨f,F(\omega )⟩⟨S_F^{-1}F(\omega ),g⟩d\mu (\omega )+{\int }_{\Omega }⟨f,F(\omega )⟩⟨L(\omega ),g⟩d\mu (\omega )$$

$$=⟨f,g⟩+⟨{\int }_{\Omega }⟨f,F(\omega )⟩L(\omega )d\mu (\omega ),g⟩$$

$$=⟨f,g⟩+0=⟨f,g⟩,$$

$$\begin{array}[]{ll}\langle.,.\rangle:\mathcal{A}\times\mathcal{A}\;\to\quad\mathcal{A}\\ \qquad\;(M,N)\longmapsto M(\overline{N})^{t}.\end{array}$$

$$S_F=\left(\begin{array}{cc}\frac{5}{3}& \frac{5}{3}\\ \frac{5}{3}& \frac{10}{3}\end{array}\right),S_F^{-1}=\left(\begin{array}{cc}\frac{6}{5}& \frac{-3}{5}\\ \frac{-3}{5}& \frac{3}{5}\end{array}\right),S_F^{-1}F(\omega )=\left(\begin{array}{cc}\frac{9}{5}\omega & \frac{-3}{5}\omega \\ \frac{-3}{5}\omega & \frac{6}{5}\omega \end{array}\right)$$

$$G(\omega )=\left(\begin{array}{cc}\frac{33}{10}\omega -1& \frac{9}{10}\omega -1\\ \frac{9}{10}\omega -1& \frac{27}{10}\omega -1\end{array}\right)$$

$$V_1(\omega )=\left(\begin{array}{cc}\frac{102\omega -50}{15}& \frac{66\omega -50}{15}\\ \frac{69\omega -50}{10}& \frac{87\omega -50}{10}\end{array}\right)$$

Full text

*The corresponding author: t.shateri@hsu.ac.ir; shateri@ualberta.ca (Tayebe Lal Shateri)

1. Introduction And Preliminaries

The concept of a generalization of frames to a family indexed by some locally compact space endowed with a Radon measure was proposed by G. Kaiser [10] and independently by Ali, Antoine and Gazeau [1]. These frames are known as continuous frames.

Frames for Hilbert spaces have natural generalizations in Hilbert $C^{*}$-modules that are generalizations of Hilbert spaces by allowing the inner product to take values in a $C^{*}$-algebra rather than $\mathbb{C}$. Note that there are many differences between Hilbert $C^{*}$-modules and Hilbert spaces. Unlike Hilbert space cases, not every closed submodule of a Hilbert $C^{*}$-module is complemented. Moreover, the well-known Riesz representation theorem for continuous functionals in Hilbert spaces does not hold in Hilbert $C^{*}$-modules, which implies that not all bounded linear operators on Hilbert $C^{*}$-modules are adjointable. There are many essential differences between Hilbert space frames and Hilbert $C^{*}$-module frames. In Hilbert spaces, every Riesz basis has a unique dual which is also a Riesz basis. But in Hilbert $C^{*}$-modules, due to the existence of zero-divisors, not all Riesz bases have unique duals, and not every dual is a Riesz basis.

Frank and Larson [4] presented a general approach to the frame theory in Hilbert $C^{*}$-modules. We refer the readers for a discussion of frames in Hilbert $C^{*}$-modules to Refs. [2, 8, 9, 14, 15].

The paper is organized as follows. First, we recall the basic definition of Hilbert $C^{*}$-modules, and we give some properties of them which we will use in the later sections. Also, we recall the notion of continuous frames in Hilbert $C^{*}$-modules. In Section 2, we describe the characterizations of the duals of a continuous frame and obtain some results about them. Finally, we give the conditions under which the sum of two duals of a continuous frame is also a dual.

We give only a brief introduction to the theory of Hilbert $C^{*}$-modules to make our explanations self-contained. For comprehensive accounts, we refer to [11, 13]. Throughout this paper, $\mathcal{A}$ shows a unital $C^{*}$-algebra.

Definition 1.1**.**

A pre-Hilbert module over unital $C^{*}$-algebra $\mathcal{A}$ is a complex vector space $U$ which is also a left $\mathcal{A}$-module equipped with an $\mathcal{A}$-valued inner product $\langle.,.\rangle:U\times U\to\mathcal{A}$ which is $\mathbb{C}$-linear and $\mathcal{A}$-linear in its first variable and satisfies the following conditions:

$(i)\;\langle f,f\rangle\geq 0$,

$(ii)\;\langle f,f\rangle=0$ iff $f=0$,

$(iii)\;\langle f,g\rangle^{*}=\langle g,f\rangle,$

$(iv)\;\langle af,g\rangle=a\langle f,g\rangle,$

for all $f,g\in U$ and $a\in\mathcal{A}$.

A pre-Hilbert $\mathcal{A}$-module $U$ is called Hilbert $\mathcal{A}$-module if $U$ is complete with respect to the topology determined by the norm $\|f\|=\|\langle f,f\rangle\|^{\frac{1}{2}}$.

The Cauchy-Schwartz inequality reconstructed in Hilbert $C^{*}$-modules as follow [13].

Lemma 1.2**.**

(Cauchy-Schwartz inequality) Let $U$ be a Hilbert $C^{*}$-modules over a unital $C^{*}$-algebra $\mathcal{A}$. Then

[TABLE]

for all $f,g\in U$.

For two Hilbert $C^{*}$-modules $U$ and $V$ over a unital $C^{*}$-algebra $\mathcal{A}$. A map $T:U\to V$ is said to be adjointable if there exists a map $T^{*}:V\to U$ satisfying

[TABLE]

for all $f\in U,g\in V$. Such a map $T^{*}$ is called the adjoint of $T$. By $End_{\mathcal{A}}^{*}(U)$ we denote the set of all adjointable maps on $U$. It is surprising that every adjointable operator is automatically linear and bounded. We need the following result in the next sections.

Theorem 1.3**.**

[12*, Theorem 2.1.4]** Let $U$ and $V$ be two Hilbert $C^{*}$-modules over a unital $C^{*}$-algebra $\mathcal{A}$ and $T\in End_{\mathcal{A}}^{*}(U,V)$. Then The following are equivalent:

(i) $T$ is bounded and $\mathcal{A}$-linear.

(ii) There exists $k>0$ such that $\langle Tf,Tf\rangle\leq k\langle f,f\rangle$, for all $f\in U$.*

Let $\mathcal{Y}$ be a Banach space, $(\mathcal{X},\mu)$ a measure space, and $f:\mathcal{X}\to\mathcal{Y}$ a measurable function. The integral of the Banach-valued function $f$ has been defined by Bochner and others. Most properties of this integral are similar to those of the integral of real-valued functions (see [3, 16]). Since every $C^{*}$-algebra and Hilbert $C^{*}$-module is a Banach space, hence we can use this integral in these spaces.In the sequel, we assume that $\mathcal{A}$ is a unital $C^{*}$-algebra and $U$ is a Hilbert $C^{*}$-module over $\mathcal{A}$ and $(\Omega,\mu)$ is a measure space. Define

[TABLE]

It was shown that $L^{2}(\Omega,A)$ is a Hilbert $\mathcal{A}$-module with the inner product

[TABLE]

and induced norm $\|\varphi\|=\|\langle\varphi,\varphi\rangle\|^{\frac{1}{2}}$, for any $\varphi,\psi\in L^{2}(\Omega,A)$. [11]

In the following, we recall the notion of continuous frames in Hilbert $C^{*}$-modules over a unital $C^{*}$-algebra $\mathcal{A}$, and mention some properties of these frames. For details, see [5, 6].

Definition 1.4**.**

A mapping $F:\Omega\to U$ is called a continuous frame for $U$ if

$(i)\;F$ is weakly-measurable, i.e., the mapping $\omega\longmapsto\langle f,F(\omega)\rangle$ is measurable on $\Omega$, for any $f\in U$.

$(ii)$ There exist constants $A,B>0$ such that

[TABLE]

The constants $A,B$ are called lower and upper frame bounds, respectively. The mapping $F$ is called Bessel if the right inequality in (1.1) holds and is called tight if $A=B$.

For a continuous frame $F:\Omega\to U$ the following operators were defined.

$(i)$ The synthesis operator or pre-frame operator $T_{F}:L^{2}(\Omega,A)\;\to U$ weakly defined by

[TABLE]

$(ii)$  The adjoint of $T,$ so called the analysis operator $T^{\ast}_{F}:U\;\to L^{2}(\Omega,A)$ is given by

[TABLE]

$(iii)$  The frame operator $S_{F}:U\;\to U$ is weakly defined by

[TABLE]

In [6] was proved the pre-frame operator $T_{F}:L^{2}(\Omega,A)\;\to U$ is well defined, surjective, adjointable $\mathcal{A}$-linear map and bounded with $\|T\|\leq\sqrt{B}$ . Moreover the analysis operator $T^{\ast}_{F}:U\;\to L^{2}(\Omega,A)$ is injective and has closed range. Also $S=TT^{\ast}$ is positive, adjointable, self-adjoint and invertible and $\|S\|\leq B$.

Now we recall the concept of duals of continuous frames in Hilbert $C^{\ast}$-modules [7].

Definition 1.5**.**

Let $F:\Omega\to U$ be a continuous Bessel mapping. A continuous Bessel mapping $G:\Omega\to U$ is called a dual for $F$ if

[TABLE]

or

[TABLE]

In this case $(F,G)$ is called a dual pair. If $T_{F}$ and $T_{G}$ denote the synthesis operators of $F$ and $G$, respectively, then (1.5) is equivalent to $T_{F}T^{*}_{G}=I_{U}$.

Remark 1.6*.*

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$. Then by reconstructin formula we have

[TABLE]

Therefore, $S^{-1}F$ is a dual for $F$, which is called the canonical dual. If a continuous frame $F$ has only one dual, it is called a Riesz-type frame.

We need the following theorem in the next section.

Theorem 1.7**.**

[6, Theorem 3.4]** Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$. Then $F$ is a Riesz-type frame if and only if the analysis operator $T^{*}_{F}:U\to L^{2}(\Omega,A)$ is onto.

2. Construction of dual continuous frames

In this section, we characterize the duals of a continuous frame and give some results about them. We refer to the method of constructing a family of duals through a fixed dual. We also examine the conditions so that a mapping becomes a continuous frame under the influence of a continuous frame.

In the following results, we show how we can construct a sequence of duals of a continuous frame from a given dual.

Theorem 2.1**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the continuous frame operator $S_{F}$ and also assume that $G_{1}:\Omega\to U$ is a dual of $F$. Then $\{G_{i}\}_{i\in\mathbb{N}}$ where

[TABLE]

is a sequence of duals of $F$.

Proof.

Clearly $G_{i}$ is a continuous Bessel mapping for $U$, for each $i\in\mathbb{N}$. Now we use induction to continue the proof. If $i=1$, then

[TABLE]

for every $f,g\in U$. Suppose that $G_{k}$ is a dual of $F$. We show that $G_{k+1}$ is also a dual of $F$.

[TABLE]

for every $f,g\in U$. ∎

Corollary 2.2**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the continuous frame operator $S_{F}$ and let $G:\Omega\to U$ be a dual of $F$. Then $\{V_{i}\}_{i\in\mathbb{N}}$ is a sequence of duals of $F$, where

[TABLE]

and

[TABLE]

for every $i\in\mathbb{N}$.

Proof.

By Theorem 2.1, the mapping $V_{1}$ is a dual of $F$. Also

[TABLE]

for every $f\in U$ and $i\in\mathbb{N}$. ∎

The following theorem shows that the difference between each arbitrary dual and the canonical dual of a continuous frame can be considered as a continuous Bessel mapping.

Theorem 2.3**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the frame operator $S_{F}$. Then a continuous Bessel mapping $G:\Omega\to U$ is a dual of $F$ if and only if there exists a continuous Bessel mapping $L:\Omega\to U$ such that

[TABLE]

where $\int_{\Omega}\langle f,F(\omega)\rangle L(\omega)d\mu(\omega)=0$ for each $f\in U$.

Proof.

Let $G$ be a dual of $F$.Then $f=\int_{\Omega}\langle f,GF(\omega)\rangle F(\omega)d\mu(\omega)$ for each $f\in U$. Suppose that $L=G-S^{-1}_{F}F$. Then for each $f,g\in U$,

[TABLE]

Hence $\langle\{\langle f,F(\omega)\rangle\}_{\omega\in\Omega},\{\langle g,L(\omega)\rangle\}_{\omega\in\Omega}\rangle=0$ and $\{\langle f,F(\omega)\rangle\}_{\omega\in\Omega}\in R(T^{*}_{L})^{\perp}=Ker(T_{L})$, where $T_{L}$ is the pre-frame operator of $L$. This shows that $\int_{\Omega}\langle f,F(\omega)\rangle L(\omega)d\mu(\omega)=0$.

Conversly, let $G=S^{-1}_{F}F+L$ where $\int_{\Omega}\langle f,F(\omega)\rangle L(\omega)d\mu(\omega)=0$ for each $f\in U$. Then,

[TABLE]

for each $f,g\in U$. Hence $G$ is a dual of $F$. ∎

Example 2.4**.**

Assume that \mathcal{A}=\Big{\{}\begin{pmatrix}a&b\\ c&d\end{pmatrix}:a,b,c,d\in\mathbb{C}\Big{\}} which is an unital $C^{*}$-algebra. We define the inner product

[TABLE]

It is not difficult to see that $\mathcal{A}$ with this inner product is a Hilbert $C^{*}$-module over itself. Suppose that $(\Omega,\mu)$ is a measure space where $\Omega=[0,1]$ and $\mu$ is the Lebesgue measure. Also the mapping $F:\Omega\to\mathcal{A}$ is defined by $F(\omega)=\begin{pmatrix}2\omega&\omega\\ \omega&3\omega\end{pmatrix}$, for any $\omega\in\Omega$.

It is easy to see that $F$ is a continuous frame for $\mathcal{A}$ with the frame bounds $\dfrac{1}{2},\dfrac{9}{2}$ and

[TABLE]

Also, the mapping $G:\Omega\to\mathcal{A}$ where

[TABLE]

is a dual of $F$ and by Theorem 2.2, $\{V_{i}\}_{i\in\mathbb{N}}$ is a sequence of duals of $F$, where

[TABLE]

and

[TABLE]

for all $i\in\mathbb{N}$.

Moreover, due to Theorem 2.3, each dual of $F$ is obtained as the mapping $G:\Omega\to\mathcal{A}$ defined by

[TABLE]

where $\alpha,\beta,\gamma,\delta\in\mathbb{R}$ are orbitrary.

It was shown in [17] that the difference between each arbitrary dual and the canonical dual of a continuous frame in Hilbert spaces can be considered as a bounded operator. In the following, we generalize it for Hilbert $C^{\ast}$-modules.

Theorem 2.5**.**

*Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the frame operator $S_{F}$ and the pre-frame operator $T_{F}$ and the frame bounds $A_{F},B_{F}>0$. Then there exists a one-to-one correspondence between duals of $F$ and adjointable operators $K:U\to L^{2}(\Omega,\mathcal{A})$ such that $T_{F}K=0$.

In this case, $\|K\|\leq\sqrt{D}+\dfrac{1}{\sqrt{A}}$ where $D$ is the upper frame dound of $G$.*

Proof.

Let $G$ be a dual of $F$. For each $f\in U$ , $\omega\in\Omega$ define

[TABLE]

where $(Kf)(\omega)=\langle f,G(\omega)-S_{F}^{-1}F(\omega)\rangle$.

It is obvious that $K$ is well-defined. Also for every $f\in U$ , $\varphi\in L^{2}(\Omega,\mathcal{A})$

[TABLE]

Hence $K$ is adjointable and $K^{*}=T_{G}-T_{S^{-1}F}$. Also

[TABLE]

i.e., $T_{F}K=0$.

Moreover, assume that $D$ is the upper frame bound of $G$. Then for $f\in U$ and $\varphi\in L^{2}(\Omega,\mathcal{A})$ where $\|\varphi\|\leq 1$ we have

[TABLE]

Conversly, let $K:U\to L^{2}(\Omega,\mathcal{A})$ be adjiontable and $T_{F}K=0$. Define

[TABLE]

where $\langle f,G(\omega)\rangle=(Kf)(\omega)+\langle f,S_{F}^{-1}F(\omega)\rangle$, for each $f\in U$ , $\omega\in\Omega$

Now we show that the mapping

[TABLE]

is well-defined and bounded. For $\varphi\in L^{2}(\Omega,\mathcal{A})$ and $f\in U$ which $\|f\|\leq 1$ we have

[TABLE]

This implies that $G$ is a continuous Bessel mapping. Also,

[TABLE]

Therefore $G$ is a dual of $F$. ∎

In the following, we give two interesting characterizations of the cononical dual of a continuous frame.

Theorem 2.6**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the continuous frame operator $S_{F}$ and also assume that $G:\Omega\to U$ is a dual of $F$. Then $G$ is the canonical dual of $F$ if and only if

[TABLE]

for all $\omega,\gamma\in\Omega$.

Proof.

Let $G$ be the canonical dual of $F$. Then

[TABLE]

for all $\omega,\gamma\in\Omega$.

Conversly, suppose that $\langle F(\omega),G(\gamma)\rangle=\langle G(\omega),F(\gamma)\rangle$, for all $\omega,\gamma\in\Omega$. Then

[TABLE]

for every $f\in U$. This shows that $F(\omega)=SG(\omega)$. Hence $G(\omega)=S^{-1}F(\omega)$, for all $\omega\in\Omega$.

∎

Theorem 2.7**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the continuous frame operator $S_{F}$ and also assume that $G:\Omega\to U$ is a dual of $F$. Then $G$ is the canonical dual of $F$ if and only if the inequality

[TABLE]

holds for any dual $D$ of $F$ and every $f\in U$, where $S_{G}$ and $S_{D}$ are the continuous frame operators of $G$ and $D$, respectively.

Proof.

Assume that $G=S^{-1}_{F}F$. Since both $S^{-1}_{F}F$ and $D$ are dual frames of $F$, so for each $f\in U$

[TABLE]

then

[TABLE]

and

[TABLE]

Hence

[TABLE]

Since each integral is a positive element of $C^{*}$-algebra $\mathcal{A}$, so

[TABLE]

holds for any dual $D$ of $F$ and every $f\in U$.

Conversly, for $D=S^{-1}_{F}F$, we have

[TABLE]

Also by first part of the proof,

[TABLE]

Therefore, $G=S^{-1}_{F}F$. ∎

3. Sum of dual continuous frames

In this section, we give the conditions under which the sum of two duals of a continuous frame is also a dual of original frame.

First, we investigate the conditions so that the sum of a continuous frame with its dual under the influence of adjointable mappings becomes a continuous frame.

Theorem 3.1**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with bounds $A_{F},B_{F}$ and $G:\Omega\to U$ is a dual of $F$ with bounds $A_{G},B_{G}$. Assume that $L_{1},L_{2}\in End^{*}(U)$ such that $L_{1}L_{2}^{*}=I_{U}$. Then $L_{1}F+L_{2}G$ is a continuous frame for $U$.

Proof.

For every $f\in U$, we have

[TABLE]

Then

[TABLE]

Also, since $L_{1},L_{2}$ are bounded and $\mathcal{A}$-linear so by Theorem 1.3, there exist $M_{1},M_{2}>0$ such that

[TABLE]

∎

Due to Theorem 3.1, the following results hold.

Corollary 3.2**.**

*Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ and $G:\Omega\to U$ is a dual of $F$. Assume that $L,L_{1},L_{2}\in End^{*}(U)$ such that $L_{1}L_{2}^{*}=I_{U}$. Then the following statements hold.

$(1)$  $(L_{1}F,L_{2}G)$ is a dual pair.

$(2)$  If $L$ is a unitary operator, then $LF+LG$ is a continuous frame for $U$.

$(3)$  $F+G$ is a continuous frame for $U$.

$(4)$  $L_{1}F+L_{2}G$ is a Riesz-type frame for $U$ if and only if $T^{*}_{F}L^{*}_{1}+T^{*}_{G}L^{*}_{2}$ is surjective, where $T_{F}$and $T_{G}$ are the pre-frames operators of $F$ and $G$, respectively.*

Proposition 3.3**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ and $G:\Omega\to U$ is a dual of $F$. If there exists $L\in End^{*}(U)$ such that $LG$ is a dual of $F$ then $L=I_{U}$.

Proof.

For each $f,g\in U$ we have

[TABLE]

Hence $L=I_{U}$. ∎

The following lemma is necessary to prove the next theorems.

Lemma 3.4**.**

Let $F:\Omega\to U$ be a continuous Bessel mapping for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the bound $B$ and $a\in\mathcal{A}$. Then $aF:\Omega\to U$ is a continuous Bessel mapping for $U$.

Proof.

[TABLE]

for all $f\in U$. ∎

Remark 3.5*.*

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ and $a\in\mathcal{A}$. It is easily to see that:

$(1)$  If $a$ is unitary, then $aF$ is a continuous frame for $U$ and $S_{aF}=S_{F}$.

$(2)$  If $a$ is unitary and $G:\Omega\to U$ is a dual of $F$, then $aG$ is a dual of $aF$.

$(3)$  If $ab=ba$, for all $b\in\mathcal{A}$, then $aF$ is a continuous frame for $U$ and $S_{aF}=|a|^{2}S_{F}$ and $T_{aF}=aT_{F}$.

In the following theorems, we check the conditions that the sum of two duals of a continuous frame becomes a dual of it.

Theorem 3.6**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ and $a_{1},a_{2}\in\mathcal{A}$ such that $a_{1}+a_{2}=1_{\mathcal{A}}$ and $a_{i}b=ba_{i}$, for $i=1,2$ and all $b\in\mathcal{A}$. If $G,K:\Omega\to U$ are duals of $F$, then $a_{1}G+a_{2}K$ is a dual of $F$.

Proof.

By Lemma 3.4, the mappings $a_{1}G$ and $a_{2}K$ are continuous Bessel mapping. Also

[TABLE]

for all $f,g\in U$. ∎

Theorem 3.7**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ with the pre-frame operator $T_{F}$. Also suppose that $G,K:\Omega\to U$ are two duals of $F$ and $V_{1},V_{2}\in End^{*}(U)$. Then $V_{1}G+V_{2}K$ is a dual of $F$ if and only if $V_{1}+V_{2}=I_{U}$.

Proof.

Assume that $T_{G}$ and $T_{K}$ are the pre-frame operators of $G$ and $K$ respectively, and $F^{\prime}=V_{1}G+V_{2}K$ is a dual of $F$ with the pre-frame operator $T_{F^{\prime}}$. Then $T_{F}T^{*}_{F^{\prime}}=I_{U}$ and

[TABLE]

Conversely, assume that $V_{1}+V_{2}=I_{U}$. Clearly $V_{1}G+V_{2}K$ is a continuous Bessel mapping for $U$. Then

[TABLE]

for all $f\in U$. ∎

Corollary 3.8**.**

Let $F:\Omega\to U$ be a continuous frame for Hilbert $C^{\ast}$-module $U$ over a unital $C^{*}$-algebra $\mathcal{A}$ and $\alpha,\beta\in\mathbb{C}$ such that $\alpha+\beta=1$. If $G,K:\Omega\to U$ are duals of $F$, then $\alpha G+\beta K$ is a dual of $F$.

Proof.

Applying Theorem 3.7, it is enough to set $V_{1}=\alpha I_{U}$ and $V_{2}=\beta I_{U}$. ∎

Conflict of interest statement and data availability:

We have no conflict of interest.