Continuous $\ast$-K-g-Frame in Hilbert $C^{\ast}$-Modules

TL;DR

This paper introduces the concept of Continuous *-K-g-Frames within Hilbert C*-Modules, exploring their properties and laying foundational groundwork for future research in this mathematical framework.

Contribution

It defines and analyzes the properties of Continuous *-K-g-Frames in Hilbert C*-Modules, a novel concept in this area.

Findings

Established the definition of Continuous *-K-g-Frames

Derived key properties of these frames

Provided foundational insights for further research

Abstract

In this paper, we introduce the concept of Continuous $\ast$-K-g-Frame in Hilbert $C^{\ast}$-Modules and we give some properties.

Equations

$$⟨Tx,Tx⟩\le \Vert T{\Vert }^2⟨x,x⟩,\forall x\in \mathcal{H}.$$

$$\Vert (T^{\ast }T{)}^{-1}{\Vert }^{-1}\le T^{\ast }T\le \Vert T{\Vert }^2.$$

$$\Vert (T{T}^{\ast }{)}^{-1}{\Vert }^{-1}\le T{T}^{\ast }\le \Vert T{\Vert }^2.$$

$${\oplus }_{w\in \Omega }{V}_w=\left\{x=\{x_w\}:x_w\in V_w,\Vert {\int }_{\Omega }|x_w{|}^{2}d\mu (w)\Vert <\infty \right\}.$$

$$A⟨K^{\ast }x,K^{\ast }x⟩A^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}x,{\Lambda }_{w}x⟩d\mu (w)\le B⟨x,x⟩B^{\ast },\forall x\in U.$$

$$uv=\{u_j{v}_j{\}}_{j\in \mathbf{N}},u^{\ast }=\{\overline{u_j}{\}}_{j\in \mathbf{N}},\Vert u\Vert =\underset{j\in \mathbf{N}}{\sup }|u_j|.$$

$$⟨u,v⟩=u{v}^{\ast }=\{u_j\overline{u_j}{\}}_{j\in \mathbf{N}}.$$

$$\sum _{j\in \mathbf{N}}⟨{\Lambda }_{j}x,{\Lambda }_{j}x⟩=\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbf{N}}^{\ast }}⟨x,x⟩\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbf{N}}^{\ast }}.$$

$$⟨K^{\ast }x,K^{\ast }x{⟩}_{\mathcal{A}}\le \sum _{j\in \mathbf{N}}⟨{\Lambda }_{j}x,{\Lambda }_{j}x⟩=\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbf{N}}^{\ast }}⟨x,x⟩\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbf{N}}^{\ast }}.$$

$$\Lambda_{w}(x)=\frac{1}{\mu(\Omega_{k})}\langle x,f_{k}\rangle h_{\omega},{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}..}\forall x\in H$$

$${\int }_{\Omega }⟨{\Lambda }_{\omega }x,{\Lambda }_{\omega }x⟩d\mu (\omega )=\sum _{j\in \mathbb{N}}⟨x,f_j⟩⟨f_j,x⟩$$

$$⟨K^{\ast }x,K^{\ast }x⟩\le {\int }_{\Omega }⟨{\Lambda }_{\omega }x,{\Lambda }_{\omega }x⟩d\mu (\omega )$$

$$=\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbb{N}}^{\ast }}⟨x,x⟩\{\frac{1}{2}+\frac{1}{i}{\}}_{i\in {\mathbb{N}}^{\ast }}$$

$$A⟨x,x⟩A^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}x,{\Lambda }_{w}x⟩d\mu (w)\le B⟨x,x⟩B^{\ast },\forall x\in U.$$

$$⟨K^{\ast }x,K^{\ast }x⟩\le ||K|{|}^2⟨x,x⟩,\forall x\in U.$$

$$\begin{gathered} (||K|{|}^{-1}A)⟨K^{\ast }x,K^{\ast }x⟩(||K|{|}^{-1}A{)}^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}x,{\Lambda }_{w}x⟩d\mu (w) \\ \le B⟨x,x⟩B^{\ast } \end{gathered}$$

$$m⟨x,x⟩\le ⟨K^{\ast }x,K^{\ast }x⟩$$

$$(A\sqrt{m})⟨x,x⟩(A\sqrt{m}{)}^{\ast }\le A⟨K^{\ast }x,K^{\ast }x⟩A^{\ast }$$

$$(A\sqrt{m})⟨x,x⟩(A\sqrt{m}{)}^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}x,{\Lambda }_{w}x⟩d\mu (w)\le B⟨x,x⟩B^{\ast }$$

$$⟨Sx,y⟩$$

$$={\int }_{\Omega }⟨{\Lambda }_w^{\ast }{\Lambda }_{w}x,y⟩d\mu (w)$$

$$={\int }_{\Omega }⟨x,{\Lambda }_w^{\ast }{\Lambda }_{w}y⟩d\mu (w)$$

$$=⟨x,{\int }_{\Omega }{\Lambda }_w^{\ast }{\Lambda }_{w}yd\mu (w)⟩$$

$$=⟨x,Sy⟩.$$

$$A⟨K^{\ast }x,K^{\ast }x⟩A^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}x,{\Lambda }_{w}x⟩d\mu (w)\le B⟨x,x⟩B^{\ast }.$$

$$A⟨K^{\ast }x,K^{\ast }x⟩A^{\ast }\le ⟨Sx,x⟩\le B⟨x,x⟩B^{\ast }.$$

$$\Vert A^{-1}{\Vert }^{-2}\Vert ⟨K{K}^{\ast }x,x⟩\Vert \le \Vert ⟨Sx,x⟩\Vert \le \Vert B{\Vert }^2\Vert ⟨x,x⟩\Vert .$$

$$\Vert A^{-1}{\Vert }^{-2}\Vert K{\Vert }^2\le \Vert S\Vert \le \Vert B{\Vert }^2.$$

$$A⟨K^{\ast }Tx,K^{\ast }Tx⟩A^{\ast }\le {\int }_{\Omega }⟨{\Lambda }_{w}Tx,{\Lambda }_{w}Tx⟩d\mu (w)\le B⟨Tx,Tx⟩B^{\ast },\forall x\in U.$$

$$m⟨Tx,Tx⟩\le ⟨K^{\ast }Tx,K^{\ast }Tx⟩$$

$$m\Vert (T^{\ast }T{)}^{-1}{\Vert }^{-1}⟨x,x⟩\le ⟨K^{\ast }Tx,K^{\ast }Tx⟩$$

$$m\Vert (T^{\ast }T{)}^{-1}{\Vert }^{-1}A⟨x,x⟩A^{\ast }\le A⟨K^{\ast }Tx,K^{\ast }Tx⟩A^{\ast }$$

$$(\Vert T^{-1}{\Vert }^{-1}\sqrt{m}A)⟨x,x⟩(\Vert T^{-1}{\Vert }^{-1}\sqrt{m}A{)}^{\ast }\le A⟨K^{\ast }Tx,K^{\ast }Tx⟩A^{\ast },\forall x\in U.$$

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1. Introduction and preliminaries

The concept of frames in Hilbert spaces has been introduced by Duffin and Schaeffer [7] in 1952 to study some deep problems in nonharmonic Fourier series, after the fundamental paper [6] by Daubechies, Grossman and Meyer, frame theory began to be widely used, particularly in the more specialized context of wavelet frames and Gabor frames [9].

Traditionally, frames have been used in signal processing, image processing, data compression and sampling theory. A discreet frame is a countable family of elements in a separable Hilbert space which allows for a stable, not necessarily unique, decomposition of an arbitrary element into an expansion of the frame elements. 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 [11] and independently by Ali, Antoine and Gazeau [1]. These frames are known as continuous frames. Gabardo and Han in [10] called these frames associated with measurable spaces, Askari-Hemmat, Dehghan and Radjabalipour in [2] called them generalized frames and in mathematical physics they are referred to as coherent states [1].

In this paper, we introduce the notion of Continuous $\ast$-K-g-Frame which are generalization of $\ast$-K-g-Frame in Hilbert $C^{\ast}$-Modules introduced by M. Rossafi and S. Kabbaj [12] and we establish some new results.

The paper is organized as follows, we continue this introductory section we briefly recall the definitions and basic properties of $C^{\ast}$-algebra, Hilbert $C^{\ast}$-modules. In Section 2, we introduce the Continuous $\ast$-K-g-Frame, the Continuous pre-$\ast$-K-g-frame operator and the Continuous $\ast$-K-g-frame operator, also we establish here properties.

In the following we briefly recall the definitions and basic properties of $C^{\ast}$-algebra, Hilbert $\mathcal{A}$-modules. Our reference for $C^{\ast}$-algebras is [8, 5]. For a $C^{\ast}$-algebra $\mathcal{A}$ if $a\in\mathcal{A}$ is positive we write $a\geq 0$ and $\mathcal{A}^{+}$ denotes the set of positive elements of $\mathcal{A}$.

Definition 1.1**.**

[13].

Let $\mathcal{A}$ be a unital $C^{\ast}$-algebra and $\mathcal{H}$ be a left $\mathcal{A}$-module, such that the linear structures of $\mathcal{A}$ and $\mathcal{H}$ are compatible. $\mathcal{H}$ is a pre-Hilbert $\mathcal{A}$-module if $\mathcal{H}$ is equipped with an $\mathcal{A}$-valued inner product $\langle.,.\rangle_{\mathcal{A}}:\mathcal{H}\times\mathcal{H}\rightarrow\mathcal{A}$, such that is sesquilinear, positive definite and respects the module action. In the other words,

• (i)

$\langle x,x\rangle_{\mathcal{A}}\geq 0$ for all $x\in\mathcal{H}$ and $\langle x,x\rangle_{\mathcal{A}}=0$ if and only if $x=0$.

• (ii)

$\langle ax+y,z\rangle_{\mathcal{A}}=a\langle x,y\rangle_{\mathcal{A}}+\langle y,z\rangle_{\mathcal{A}}$ for all $a\in\mathcal{A}$ and $x,y,z\in\mathcal{H}$.

• (iii)

$\langle x,y\rangle_{\mathcal{A}}=\langle y,x\rangle_{\mathcal{A}}^{\ast}$ for all $x,y\in\mathcal{H}$.

For $x\in\mathcal{H},$ we define $||x||=||\langle x,x\rangle_{\mathcal{A}}||^{\frac{1}{2}}$. If $\mathcal{H}$ is complete with $||.||$, it is called a Hilbert $\mathcal{A}$-module or a Hilbert $C^{\ast}$-module over $\mathcal{A}$. For every $a$ in $C^{\ast}$-algebra $\mathcal{A}$, we have $|a|=(a^{\ast}a)^{\frac{1}{2}}$ and the $\mathcal{A}$-valued norm on $\mathcal{H}$ is defined by $|x|=\langle x,x\rangle_{\mathcal{A}}^{\frac{1}{2}}$ for $x\in\mathcal{H}$.

Let $\mathcal{H}$ and $\mathcal{K}$ be two Hilbert $\mathcal{A}$-modules, A map $T:\mathcal{H}\rightarrow\mathcal{K}$ is said to be adjointable if there exists a map $T^{\ast}:\mathcal{K}\rightarrow\mathcal{H}$ such that $\langle Tx,y\rangle_{\mathcal{A}}=\langle x,T^{\ast}y\rangle_{\mathcal{A}}$ for all $x\in\mathcal{H}$ and $y\in\mathcal{K}$.

We reserve the notation $End_{\mathcal{A}}^{\ast}(\mathcal{H},\mathcal{K})$ for the set of all adjointable operators from $\mathcal{H}$ to $\mathcal{K}$ and $End_{\mathcal{A}}^{\ast}(\mathcal{H},\mathcal{H})$ is abbreviated to $End_{\mathcal{A}}^{\ast}(\mathcal{H})$.

The following lemmas will be used to prove our mains results

Lemma 1.2**.**

[13]**. Let $\mathcal{H}$ be Hilbert $\mathcal{A}$-module. If $T\in End_{\mathcal{A}}^{\ast}(\mathcal{H})$, then

[TABLE]

Lemma 1.3**.**

[3]**. Let $\mathcal{H}$ and $\mathcal{K}$ two Hilbert $\mathcal{A}$-modules and $T\in End^{\ast}(\mathcal{H},\mathcal{K})$. Then the following statements are equivalent:

• (i)

$T$* is surjective.*

• (ii)

$T^{\ast}$* is bounded below with respect to norm, i.e., there is $m>0$ such that $\|T^{\ast}x\|\geq m\|x\|$ for all $x\in\mathcal{K}$.*

• (iii)

$T^{\ast}$* is bounded below with respect to the inner product, i.e., there is $m^{\prime}>0$ such that $\langle T^{\ast}x,T^{\ast}x\rangle\geq m^{\prime}\langle x,x\rangle$ for all $x\in\mathcal{K}$.*

Lemma 1.4**.**

[4]**. Let $\mathcal{H}$ and $\mathcal{K}$ two Hilbert $\mathcal{A}$-modules and $T\in End^{\ast}(\mathcal{H},\mathcal{K})$. Then:

• (i)

If $T$ is injective and $T$ has closed range, then the adjointable map $T^{\ast}T$ is invertible and

[TABLE]

• (ii)

If $T$ is surjective, then the adjointable map $TT^{\ast}$ is invertible and

[TABLE]

2. Continuous $\ast$-K-g-Frame in Hilbert $C^{\ast}$-Modules

Let $X$ be a Banach space, $(\Omega,\mu)$ a measure space, and function $f:\Omega\to X$ a measurable function. Integral of the Banach-valued function $f$ has defined Bochner and others. Most properties of this integral are similar to those of the integral of real-valued functions. Because every $C^{\ast}$-algebra and Hilbert $C^{\ast}$-module is a Banach space thus we can use this integral and its properties.

Let $(\Omega,\mu)$ be a measure space, let $U$ and $V$ be two Hilbert $C^{\ast}$-modules, $\{V_{w}:w\in\Omega\}$ is a sequence of subspaces of V, and $End_{\mathcal{A}}^{\ast}(U,V_{w})$ is the collection of all adjointable $\mathcal{A}$-linear maps from $U$ into $V_{w}$. We define

[TABLE]

For any $x=\{x_{w}:w\in\Omega\}$ and $y=\{y_{w}:w\in\Omega\}$, if the $\mathcal{A}$-valued inner product is defined by $\langle x,y\rangle=\int_{\Omega}\langle x_{w},y_{w}\rangle d\mu(w)$, the norm is defined by $\|x\|=\|\langle x,x\rangle\|^{\frac{1}{2}}$, the $\oplus_{w\in\Omega}V_{w}$ is a Hilbert $C^{\ast}$-module.

Definition 2.1**.**

.

Let $K\in End_{\mathcal{A}}^{\ast}(U)$, We call $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ a continuous $\ast$-K-g-frame for Hilbert $C^{\ast}$-module $U$ with respect to $\{V_{w}:w\in\Omega\}$ if:

• •

for any $x\in U$, the function $\tilde{x}:\Omega\rightarrow V_{w}$ defined by $\tilde{x}(w)=\Lambda_{w}x$ is measurable;

• •

there exist two strictly nonzero elements $A$ and $B$ in $\mathcal{A}$ such that

[TABLE]

The elements $A$ and $B$ are called continuous $\ast$-K-g-frame bounds.

If $A=B$ we call this continuous $\ast$-K-g-frame a continuous tight $\ast$-K-g-frame, and if $A=B=1_{\mathcal{A}}$ it is called a continuous Parseval $\ast$-K-g-frame. If only the right-hand inequality of (2.1) is satisfied, we call $\{\Lambda_{w}:w\in\Omega\}$ a continuous $\ast$-K-g-Bessel for $U$ with respect to $\{\Lambda_{w}:w\in\Omega\}$ with Bessel bound $B$.

Example 2.2**.**

. Let $l^{\infty}$ be the set of all bounded complex-valued sequences. For any $u=\{u_{j}\}_{j\in\mathbf{N}},v=\{v_{j}\}_{j\in\mathbf{N}}\in l^{\infty}$, we define

[TABLE]

Then $\mathcal{A}=\{l^{\infty},\|.\|\}$ is a $\mathbb{C}^{\ast}$-algebra.

Let $\mathcal{H}=C_{0}$ be the set of all sequences converging to zero. For any $u,v\in\mathcal{H}$ we define

[TABLE]

Then $\mathcal{H}$ is a Hilbert $\mathcal{A}$-module.

Define $f_{j}=\{f_{i}^{j}\}_{i\in\mathbf{N}^{\ast}}$ by $f_{i}^{j}=\frac{1}{2}+\frac{1}{i}$ if $i=j$ and $f_{i}^{j}=0$ if $i\neq j$ $\forall j\in\mathbf{N}^{\ast}$.

Now define the adjointable operator $\Lambda_{j}:\mathcal{H}\to\mathcal{A},\;\;\Lambda_{j}x=\langle x,f_{j}\rangle$.

then for every $x\in\mathcal{H}$ we have

[TABLE]

So $\{\Lambda_{j}\}_{j}$ is a $\{\frac{1}{2}+\frac{1}{i}\}_{i\in\mathbf{N}^{\ast}}$-tight $\ast$-g-frame.

Let $K:\mathcal{H}\to\mathcal{H}$ defined by $Kx=\{\frac{x_{i}}{i}\}_{i\in\mathbf{N}^{\ast}}$.

Then for every $x\in\mathcal{H}$ we have

[TABLE]

Now, let $(\Omega,\mu)$ be a $\sigma$-finite measure space with infinite measure and $\{H_{\omega}\}_{\omega\in\Omega}$ be a family of Hilbert $A$-module (H_{\omega}=C_{0},{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}..}\forall w\in\Omega).

Since $\Omega$ is a $\sigma$-finite, it can be written as a disjoint union $\Omega=\bigcup\Omega_{\omega}$ of countably many subsets $\Omega_{\omega}\subseteq\Omega$, such that \mu(\Omega_{k})<\infty,{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}..}\forall k\in\mathbb{N}. Without less of generality, assume that \mu(\Omega_{k})>0{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}..}\forall k\in\mathbb{N}.

For each $\omega\in\Omega$, define the operator : $\Lambda_{\omega}:H\rightarrow H_{w}$ by :

[TABLE]

where $k$ is such that $w\in\Omega_{\omega}$ and $h_{\omega}$ is an arbitrary element of $H_{\omega}$, such that $\|h_{\omega}\|=1$.

For each $x\in H$, $\{\Lambda_{\omega}x\}_{\omega\in\Omega}$ is strongly measurable (since $h_{\omega}$ are fixed) and

[TABLE]

So, therefore

[TABLE]

So $\{\Lambda_{\omega}\}_{\omega\in\Omega}$ is a continuous $\ast$-K-g-frame.

Remark 2.3**.**

.

• •

Every continuous $\ast$-g-frame is a continuous $\ast$-K-g-frame

indeed:

Let $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ a continuous $\ast$-g-frame for Hilbert $C^{\ast}$-module $U$ with respect to $\{V_{w}:w\in\Omega\}$, then

[TABLE]

or

[TABLE]

then

[TABLE]

so $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ be a continuous $\ast$-K-g-frame with lower and upper bounds $||K||^{-1}A$ and $B$, respectively.

• •

If $K\in End_{\mathcal{A}}^{\ast}(H)$ is a surjectif operator, then every continuous $\ast$-K-g-frame for $H$ with respect to $\{V_{w}:w\in\Omega\}$ is a continuous $\ast$-g-frame.

indeed:

If $K$ is surjectif there exists $m>0$ such that

[TABLE]

then

[TABLE]

or $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ be a continuous $\ast$-K-g-frame, we have

[TABLE]

hence $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ be a continuous $\ast$-g-frame for $U$ with lower and upper bounds $A\sqrt{m}$ and $B$, respectively

Let $K\in End_{\mathcal{A}}^{\ast}(U)$, and $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ a continuous $\ast$-K-g-frame for Hilbert $C^{\ast}$-module $U$ with respect to $\{V_{w}:w\in\Omega\}$.

We define an operator $T:U\rightarrow\oplus_{w\in\Omega}V_{w}$ by:

$Tx=\{\Lambda_{w}x:w\in\Omega\}\forall x\in U$,

then $T$ is called the continuous $\ast$-K-g-frame transform.

So its adjoint operator is: $T^{\ast}:\oplus_{w\in\Omega}V_{w}\rightarrow U$ given by:

$T^{\ast}(\{x_{\omega}\}_{\omega\in\Omega})=\int_{\Omega}\Lambda^{\ast}_{\omega}x_{\omega}d\mu(w)$

By composing $T$ and $T^{\ast}$, the frame operator $S=T^{\ast}T$ given by:

$Sx=\int_{\Omega}\Lambda^{\ast}_{\omega}\Lambda_{\omega}xd\mu(w)$, S is called continuous $\ast$-K-g frame operator

Theorem 2.4**.**

.

The continuous $\ast$-K-g frame operator $S$ is a bounded, positive, selfadjoint and $||A^{-1}||^{-2}\|K\|^{2}\leq||S||\leq||B||^{2}$

Proof.

First we show, $S$ is a selfadjoint operator. By definition we have $\forall x,y\in U$

[TABLE]

Then $S$ is a selfadjoint.

Clearly $S$ is positive.

By definition of a continuous $\ast$-K-g-frame we have

[TABLE]

So

[TABLE]

This give

[TABLE]

If we take supremum on all $x\in U$, where $\|x\|\leq 1$, we have

[TABLE]

∎

Theorem 2.5**.**

.

*Let $K\in End_{\mathcal{A}}^{\ast}(H)$ be surjective and $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ be a continuous $\ast$-K-g-frame for $U$, with lower and upper bounds $A$ and $B$, respectively and with the continuous $\ast$-K-g-frame operator $S$.

Let $T\in End_{\mathcal{A}}^{\ast}(U)$ be invertible, then $\{\Lambda_{w}T:w\in\Omega\}$ is a continuous $\ast$-K-g-frame for $U$ with continuous $\ast$-K-g-frame operator $T^{\ast}ST$.*

Proof.

.

We have

[TABLE]

Using Lemma (1.3) , we have $\|(T^{\ast}T)^{-1}\|^{-1}\langle x,x\rangle\leq\langle Tx,Tx\rangle$, $\forall x\in U$.

$K$ is surjectif, then there exist $m$ such that:

[TABLE]

then

[TABLE]

so

[TABLE]

Or $\|T^{-1}\|^{-2}\leq\|(T^{\ast}T)^{-1}\|^{-1}$, this implies:

[TABLE]

And we know that $\langle Tx,Tx\rangle\leq\|T\|^{2}\langle x,x\rangle$, $\forall x\in U$. This implies that:

[TABLE]

Using (2.2), (2.3), (2.4) we have:

[TABLE]

So $\{\Lambda_{w}T:w\in\Omega\}$ is a continuous $\ast$-K-g-frame for $U$.

Moreover for every $x\in U$, we have

[TABLE]

This completes the proof. ∎

Corollary 2.6**.**

.

Let $\{\Lambda_{w}\in End_{\mathcal{A}}^{\ast}(U,V_{w}):w\in\Omega\}$ be a continuous $\ast$-K-g-frame for $U$ and $K\in End_{\mathcal{A}}^{\ast}(U)$ be surjective, with continuous $\ast$-K-g-frame operator $S$. Then $\{\Lambda_{w}S^{-1}:w\in\Omega\}$ is a continuous $\ast$-K-g-frame for $U$.

Proof.

.

Result from the last theorem by taking $T=S^{-1}$ ∎

the following lemma caracterize a continuous $\ast$-K-g-frame by its frame operator

Theorem 2.7**.**

.

Let $\{\Lambda_{\omega}\}_{\omega\in\Omega}$ be a continuous $\ast$-g-Bessel for $H$ with respect $\{H_{\omega}\}_{\omega\in\Omega}$.

then $\{\Lambda_{\omega}\}_{\omega\in\Omega}$ is a continuous $\ast$-K-g-frame for $H$ with respect to $\{H_{\omega}\}_{\omega\in\Omega}$ if and only if there exist a constant $A>0$ such that $S\geq AKK^{\ast}$ where $S$ is the frame operator for $\{\Lambda_{\omega}\}_{\omega\in\Omega}$.

Proof.

.

We know $\{\Lambda_{\omega}\}_{\omega\in\Omega}$ is a continuous $\ast$-K-g-frame for $H$ with bounded $A$ et $B$ if and only if

[TABLE]

If and only if

[TABLE]

If and only if

[TABLE]

Where $S$ is the continuous $\ast$-K-g frame operator for $\{\Lambda_{\omega}\}_{\omega\in\Omega}$.

Therefore, the conclusuin holds. ∎