# Douglas’ factorization theorem and atomic system in Hilbert
pro$-C^{\ast}-$module

Mohamed Rossafi1∗, Roumaissae Eljazzar2 and Ram Mohapatra3 1LaSMA Laboratory
Department of Mathematics Faculty of Sciences, Dhar El Mahraz University Sidi
Mohamed Ben Abdellah, Fez, Morocco<EMAIL_ADDRESS>2Laboratory of
Partial Differential Equations, Spectral Algebra and Geometry, Department of
Mathematics, Faculty Of Sciences, University of Ibn Tofail, Kenitra, Morocco
<EMAIL_ADDRESS>3Department of Mathematics, University of
Central Florida, Orlando, FL., 32816, USA<EMAIL_ADDRESS>
###### Abstract.

In the present paper we introduce the generalized inverse operators which have
an interesting role in operator theory. We establish Douglas’ factorization
theorem type for Hilbert pro-$C^{\ast}$-module. We introduce the notion of
atomic system and of $K$-frame in Hilbert pro-$C^{\ast}$-module and we study
the relationship between them. We also demonstrat some properties of $K$-frame
by using Douglas’ factorization theorem.Finally we demonstrate that the sum of
two $K$-frames in a Hilbert pro-$C^{\ast}$-module with certain conditions is
once again a $K$-frame.

###### Key words and phrases:

Douglas majorization, atomic system, Hilbert pro-$C^{\ast}$-modules.
∗Corresponding author: Mohamed Rossafi (rossafimohamed@gmail.com).

###### 2010 Mathematics Subject Classification:

42C15, 46L05, 47A05.

## INTRODUCTION

Douglas [2] has studied the equation $AX=B$ with a view to finding a solution
for bounded linear operators on Hilbert spaces. A generalization of Douglas
theorem for Hilbert $C^{\ast}$-module was given in [4] and [5]. Those authors
have extended the Douglas factor decomposition for closed densely defined
operators on Hilbert spaces in the context of regular operators in Hilbert
$C^{\ast}$-modules.

Frames, introduced by Duffin and Schaefer [3] in 1952 to analyse some deep
problems in nonharmonic Fourier series by abstracting the fundamental notion
introduced by Gabor [8] for signal processing. Today, frame theory is an
exciting, dynamic and fast paced subject with applications to a wide variety
of areas in mathematics and engineering, including sampling theory, operator
theory, harmonic analysis, nonlinear sparse approximation, pseudodifferential
operators, wavelet theory, wireless communication, data transmission with
erasures, filter banks, signal processing, image processing, geophysics,
quantum computing, sensor networks, and more.

In 2008, Joita [12] extended the theory of frames in Hilbert modules over
pro-$C^{\ast}$-algebras. The concept of $K$-frame was first introduced by
Laura Găvruţa [9] in order to study atomic systems for a given bounded linear
operator $K$ in a separable Hilbert space. It is well known that K-frames
present a genralization of ordinary frames, which permits to reconstruct the
elements in the domain of a linear and bounded operator in a Hilbert space.

This paper is devoted to establishing Douglas’ Factorization theorem type
results for pro-$C^{\ast}$-modules. Moreover, we define the atomic system in
the framework of pro-$C^{\ast}$-modules and mention some properties. Also, we
present $K$-frame in Hilbert pro-$C^{\ast}$-modules and establish some
results.

This article will be organized as follows: In section 2, we briefly recall
some definitions and basic properties of pro-$C^{\ast}$-algebra. Section 3 is
devoted to introduce the concept of generalized inverse which will be used to
prove a Douglas type theorem on a Hilbert pro-$C^{\ast}$-module. In Section 4
we define the atomic system and show that every Bessel sequence is an atomic
system for the frame operator. Finaly we show that the sum of two $K$-frames
under certain conditions is again a $K$-frame.

##  1\. Preliminaries

The basic information about pro-$C^{\ast}$-algebras can be found in the works
[6, 7, 14, 10, 15, 16].

A $C^{\ast}$-algebra whose topology is induced by a family of continuous
$C^{\ast}-$seminorms instead of a $C^{\ast}$-norm is called
pro-$C^{\ast}$-algebra. Hilbert pro-$C^{\ast}$-modules are generalizations of
Hilbert spaces the inner product takes values in a pro-$C^{\ast}$-algebra
rather than in the field of complex numbers.

Pro-$C^{\ast}$-algebra is defined as a complete Hausdorff complex topological
$\ast$-algebra $\mathcal{A}$ whose topology is determined by its continuous
$C^{\ast}$-seminorms in the sens that a net $\\{a_{\alpha}\\}$ converges to
$0$ if and only if $p(a_{\alpha})$ converges to $0$ for all continuous
$C^{\ast}$-seminorms $p$ on $\mathcal{A}$ (see [10, 13, 16]), and we have:

  1. 1)

$p(ab)\leq p(a)p(b)$

  2. 2)

$p(a^{\ast}a)=p(a)^{2}$

for all $a,b\in\mathcal{A}$. If the topology of pro-$C^{*}$-algebra is
determined by only countably many $C^{*}$-seminorms, then it is called a
$\sigma$-$C^{*}$-algebra. We denote by $sp(a)$ the spectrum of $a$ such that:
$sp(a)=\left\\{\lambda\in\mathbb{C}:\lambda 1_{\mathcal{A}}-a\right.$ is not
invertible $\\}$ for all $a\in\mathcal{A}$, where $\mathcal{A}$ is unital
pro-$C^{*}$-algebra with unity $1_{\mathcal{A}}$.

Let the set of all continuous $C^{\ast}$-seminorms on $\mathcal{A}$ be denoted
by $S(\mathcal{A})$. If $\mathcal{A}^{+}$ denotes the set of all positive
elements of $\mathcal{A}$, then $\mathcal{A}^{+}$ is a closed convex
$C^{*}$-seminorms on $\mathcal{A}.$

We also denote by $\mathcal{H}_{\mathcal{A}}$ the set of all sequences (
$\left.a_{n}\right)_{n}$ with $a_{n}\in\mathcal{A}$ such that
$\sum_{n}a_{n}^{*}a_{n}$ converges in $\mathcal{A}$.

######  Example 1.1.

Every $C^{*}$-algebra is a pro-$C^{*}$-algebra.

######  Proposition 1.2.

[10] Let $\mathcal{A}$ be a unital pro-$C^{*}$-algebra with an identity
$1_{\mathcal{A}}.$ Then for any $p\in S(\mathcal{A}),$ we have:

  1. (1)

$p(a)=p(a^{*})$ for all $a\in A$

  2. (2)

$p\left(1_{\mathcal{A}}\right)=1$

  3. (3)

If $1_{\mathcal{A}}\leq b$, then $b$ is invertible and $b^{-1}\leq
1_{\mathcal{A}}$

  4. (4)

If $a,b\in\mathcal{A}^{+}$ are invertible and $0\leq a\leq b$, then $0\leq
b^{-1}\leq a^{-1}$

  5. (5)

If $a,b\in\mathcal{A}^{+}$ and $a^{2}\leq b^{2}$, then $0\leq a\leq b$

######  Definition 1.3.

[16] A pre-Hilbert module over pro-$C^{\ast}$-algebra $\mathcal{A}$, is a
complex vector space $E$ which is also a left $\mathcal{A}$-module compatible
with the complex algebra structure, equipped with an $\mathcal{A}$-valued
inner product $\langle.,.\rangle$ $E\times E\rightarrow\mathcal{A}$ which is
$\mathbb{C}$-and $\mathcal{A}$-linear in its first variable and satisfies the
following conditions:

  1. 1)

$\langle\xi,\eta\rangle^{*}=\langle\eta,\xi\rangle$ for every $\xi,\eta\in E$

  2. 2)

$\langle\xi,\xi\rangle\geq 0$ for every $\xi\in E$

  3. 3)

$\langle\xi,\xi\rangle=0$ if and only if $\xi=0$

for every $\xi,\eta\in E.$ We say $E$ is a Hilbert $\mathcal{A}$-module (or
Hilbert pro-$C^{\ast}$-module over $\mathcal{A}$ ). If $E$ is complete with
respect to the topology determined by the family of seminorms

$\bar{p}_{E}(\xi)=\sqrt{p(\langle\xi,\xi\rangle)}\quad\xi\in E,p\in
S(\mathcal{A})$

Let $\mathcal{A}$ be a pro-$C^{\ast}$-algebra and let $\mathcal{X}$ and
$\mathcal{Y}$ be Hilbert $\mathcal{A}$-modules and assume that I and J be
countable index sets. A bounded $\mathcal{A}$-module map from $\mathcal{X}$ to
$\mathcal{Y}$ is called an operators from $\mathcal{X}$ to $\mathcal{Y}$. We
denote the set of all operator from $\mathcal{X}$ to $\mathcal{Y}$ by
$Hom_{\mathcal{A}}(\mathcal{X},\mathcal{Y})$.

######  Definition 1.4.

An $\mathcal{A}$-module map $T:\mathcal{X}\longrightarrow\mathcal{Y}$ is
adjointable if there is a map $T^{\ast}:\mathcal{Y}\longrightarrow\mathcal{X}$
such that $\langle T\xi,\eta\rangle=\left\langle\xi,T^{\ast}\eta\right\rangle$
for all $\xi\in\mathcal{X},\eta\in\mathcal{Y}$, and is called bounded if for
all $p\in S(\mathcal{A})$, there is $M_{p}>0$ such that
$\bar{p}_{\mathcal{Y}}(T\xi)\leq M_{p}\bar{p}_{\mathcal{X}}(\xi)$ for all
$\xi\in\mathcal{X}$.

We denote by $Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$, the set of
all adjointable operator from $\mathcal{X}$ to $\mathcal{Y}$ and
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})=Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{X})$

######  Definition 1.5.

Let $\mathcal{A}$ be a pro-$C^{\ast}$-algebra and $\mathcal{X},\mathcal{Y}$ be
two Hilbert $\mathcal{A}$-modules. The operator
$T:\mathcal{X}\rightarrow\mathcal{Y}$ is called uniformly bounded below, if
there exists $C>0$ such that for each $p\in S(\mathcal{A})$,

$\bar{p}_{\mathcal{Y}}(T\xi)\leqslant C\bar{p}_{\mathcal{X}}(\xi),\quad\text{
for all }\xi\in\mathcal{X}$

and is called uniformly bounded above if there exists $C^{\prime}>0$ such that
for each $p\in S(\mathcal{A})$,

$\bar{p}_{\mathcal{Y}}(T\xi)\geqslant
C^{\prime}\bar{p}_{\mathcal{X}}(\xi),\quad\text{ for all }\xi\in\mathcal{X}$
$\|T\|_{\infty}=\inf\\{M:M\text{ is an upper bound for }T\\}$
$\hat{p}_{\mathcal{Y}}(T)=\sup\left\\{\bar{p}_{\mathcal{Y}}(T(x)):\xi\in\mathcal{X},\quad\bar{p}_{\mathcal{X}}(\xi)\leqslant
1\right\\}$

It’s clear to see that, $\hat{p}(T)\leqslant\|T\|_{\infty}$ for all $p\in
S(\mathcal{A})$.

######  Definition 1.6.

[12] A sequence $\left\\{\xi_{i}\right\\}_{i}$ in $M(\mathcal{X})$ is a
standard frame of multipliers in $\mathcal{X}$ if for each
$\xi\in\mathcal{X},\sum_{i}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}$
converges in $\mathcal{A}$, and there are two positive constants $C$ and $D$
such that

$C\langle\xi,\xi\rangle_{\mathcal{X}}\leqslant\sum_{i}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leqslant
D\langle\xi,\xi\rangle_{\mathcal{X}}$

for all $\xi\in\mathcal{X}$. If $D=C=1$ we say that
$\left\\{\xi_{i}\right\\}_{i}$ is a standard normalized frame of multipliers.

Particularly, if the right inequality

$\sum_{i}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leqslant
D\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in\mathcal{X}$

holds true, we call $\left\\{\xi_{i}\right\\}_{i\in I}$ a Bessel sequence.

######  Definition 1.7.

Let $\left\\{\xi_{i}\right\\}_{i}$ be a standard frame of multipliers in
$\mathcal{X}$. The module morphism
$U:\mathcal{X}\rightarrow\mathcal{H}_{\mathcal{A}}$ defined by $U(x)=$
$\left(\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)_{i}$ is
called the frame transform for $\left\\{\xi_{i}\right\\}_{i}$.

######  Definition 1.8.

Let $\mathcal{X}$ be a Hilbert module over a pro-$C^{*}$-algebra $\mathcal{A}$
and $\left\\{\xi_{i}\right\\}_{i}$ be a standard frame of multipliers for
$\mathcal{X}$. The invertible positive and bounded element $L$ in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$, such that
$\sum_{i}\xi_{i}\cdot\left\langle\xi_{i},L(\xi)\right\rangle_{M(\mathcal{X})}=\xi$
for all $\xi\in$ is called the frame operator associated with the standard
frame of multipliers $\left\\{\xi_{i}\right\\}_{i}$.

######  Proposition 1.9.

[1]. Let $\mathcal{X}$ be a Hilbert module over pro-$C^{*}$-algebra
$\mathcal{A}$ and $T$ be an invertible element in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ such that both are uniformly bounded.
Then for each $\xi\in\mathcal{X}$,

$\left\|T^{-1}\right\|_{\infty}^{-2}\langle\xi,\xi\rangle\leq\langle
T\xi,T\xi\rangle\leq\|T\|_{\infty}^{2}\langle\xi,\xi\rangle.$

######  Theorem 1.10.

[11] Let $T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$. If $T$ has
closed range then:

  1. (1)

$Ker(T)$ is a complemented submodule of $\mathcal{X}$;

  2. (2)

$Ran(T)$, the range of $T$, is a complemented submodule of $\mathcal{Y}$.

######  Proposition 1.11.

[17] Let $T:\mathcal{X}\rightarrow\mathcal{X}$ be an operator, then the
following statements are equivalent:

  1. (1)

$T$ is a positive element in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

  2. (2)

for any element $\xi\in\mathcal{X}$ the inequality $\langle
T\xi,\xi\rangle\geq 0$ holds, i.e. this element is positive in $\mathcal{A}$.

######  Lemma 1.12.

[10] Let $\mathcal{A}$ be a pro-$C^{*}$-algebra, suppose
$\alpha,\beta\in\mathcal{A}^{+}$be such that $\alpha\leq\beta$. Then
$p(\alpha)\leq p(\beta)$ and
$\lambda^{*}\alpha\lambda\leq\lambda^{*}\beta\lambda$ for each
$\lambda\in\mathcal{A}$.

##  2\. Douglas’ factorization theorem in Hilbert pro$-C^{\ast}-$module

We start by defining the bounded generalized inverse module map. We give an
equivalente characterization of a bounded generalized inverse
$\mathcal{A}$-module map, which is the main tool to obtain Douglas’ type
factor decomposition theorem of some uniformly bounded module maps.

######  Definition 2.1.

Let $\mathcal{X}$ and $\mathcal{Y}$ be two Hilbert pro-$C^{*}$-modules over a
pro-$C^{*}$-algebra $\mathcal{A}$, $T\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$. If there exists a
$T^{\dagger}\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$ such that

  1. (1)

$TT^{\dagger}T=T$;

  2. (2)

$T^{\dagger}TT^{\dagger}=T^{\dagger}$;

  3. (3)

$\left(TT^{\dagger}\right)^{*}=TT^{\dagger};$

  4. (4)

$\left(T^{\dagger}T\right)^{*}=T^{\dagger}T$,

then $T^{\dagger}$ is called a bounded generalized inverse module map of $T$.

######  Remark 2.2.

According to the conditions of the previous definition, we can easily deduce
that $T^{\dagger}$ is unique. In fact suppose that $T$ has another generalized
inverse which we denote by $T^{\prime}$. The definition 2.1 implies
$Ran\left(TT^{\dagger}\right)=Ran(T)=Ran\left(TT^{\prime}\right)$ and
$Ran\left(T^{\dagger}T\right)=Ran\left(T^{*}\right)=Ran\left(T^{\prime}T\right)$.
Note that $TT^{\dagger},T^{\dagger}T,TT^{\prime},T^{\prime}T$ are all
projections, hence $TT^{\dagger}=TT^{\prime}$ and $T^{\dagger}T=T^{\prime}T$.
As a consequence,,
$T^{\dagger}=T^{\dagger}TT^{\dagger}=T^{\dagger}TT^{\prime}=T^{\prime}TT^{\prime}=T^{\prime}$.

######  Theorem 2.3.

Let $T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$, and let
$\mathcal{X},\mathcal{Y}$ be Hilbert pro-$C^{\ast}$-modules over a
pro-$C^{*}$-algebra $\mathcal{A}$. Then the following statements are
equivalent:

  1. (1)

$T$ has a generalized inverse module map in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$,

  2. (2)

$Ran(T)$ is a closed submodule in $\mathcal{Y}$.

###### Proof.

$(1)\Rightarrow(2)$ is easy to prove, see the Remark 2.2.

$(2)\Rightarrow(1)$. Let $Ran(T)$ be closed. By Theorem 1.10, $Ker(T)$ and
$Ran(T)$ are both complemented submodules in $\mathcal{X}$ and in
$\mathcal{Y}$, respectively, then $\mathcal{X}=Ker(T)\oplus
Ran\left(T^{*}\right)$ and $\mathcal{Y}=Ker\left(T^{*}\right)\oplus Ran(T)$.
Define a map $T^{\dagger}:\mathcal{Y}\mapsto\mathcal{X}$ which is linear by

$T^{\dagger}\xi=\begin{cases}\left(T\mid Ker(T)^{\perp}\right)^{-1}\xi&\xi\in
Ran(T)\\\ 0&\xi\in Ker\left(T^{*}\right)\end{cases}$

and define a linear map
$\left(T^{\dagger}\right)^{*}:\mathcal{X}\mapsto\mathcal{Y}$ by

$\left(T^{\dagger}\right)^{*}\xi=\begin{cases}\left(T^{*}\mid
Ker\left(T^{*}\right)^{\perp}\right)^{-1}\xi&\xi\in Ran\left(T^{*}\right)\\\
0&\xi\in Ker(T).\end{cases}$

To prove that $T^{\dagger}$is the generalized inverse module map of $T$, we
should prove that $T^{\dagger}\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$, equivalently,
$\left\langle
T^{\dagger}\xi,\eta\right\rangle=\left\langle\xi,\left(T^{\dagger}\right)^{*}\eta\right\rangle,\xi\in\mathcal{Y}$,
$\eta\in\mathcal{X}$. The verification of this identity is uncomplicated
utilizing the orthogonal direct sum decompositions. It’s also easy to verify
the conditions $(1)-(4)$ of Definition 2.1 ∎

######  Theorem 2.4.

Let $\mathcal{X}$ be a Hilbert $\mathcal{A}$-module over a pro-$C^{*}$-algebra
$\mathcal{A}.$ Let $T,L\in Hom_{\mathcal{A}}^{*}(\mathcal{X})$. If $Ran(L)$ is
closed, then the following statements are equivalent:

  1. (1)

$Ran(T)\subseteq Ran(L)$.

  2. (2)

$TT^{*}\leq\alpha^{2}LL^{*}$ for some $\alpha\geq 0$.

  3. (3)

There exists $U\in Hom_{\mathcal{A}}^{*}(\mathcal{X})$ uniformly bounded such
that $T=LU$.

###### Proof.

$(3)\Longrightarrow(1)$ is obvious.

$(1)\Longrightarrow(3)$ Suppose that $(1)$ holds. For every
$\xi\in\mathcal{X},$ we have $T(\xi)\in Ran(L)$. Since $L$ has closed range,
then by Theorem 1.10 $Ker(L)$ is a complemented submodule, which it results
that there exists a unique $\eta\in KerL^{\perp}$ such tat $T(\xi)=L(\eta)$.
Let’s define the map $U$ as follow $U:U(\xi)=\eta$. Observe the fact that then
$U(\xi a)=\eta a$ indicates that $U$ is a module map, for each
$a\in\mathcal{A}$. The construction of $U$ gives $T=LU$. We will next show
that $U\in Hom_{\mathcal{A}}^{*}(\mathcal{X})$. By Theorem 2.3 $L$ has the
generalized inverse module map $L^{\dagger}\in
Hom_{\mathcal{A}}^{*}(\mathcal{X})$. Since $T=LU$ then
$L^{\dagger}T=L^{\dagger}LU$. Note that $kerL^{\perp}=Ran\left(L^{*}\right)$,
hence $L^{\dagger}LU(\xi)=L^{\dagger}L(\eta)=\eta=U(\xi)$, for every
$\xi\in\mathcal{X}$, that is, $L^{\dagger}LU=U$. According to the above proof
we have $U=L^{\dagger}T\in Hom_{\mathcal{A}}^{*}(\mathcal{X})$. In fact,
$U^{*}=T^{*}\left(L^{\dagger}\right)^{*}$.

$(3)\Longrightarrow(2)$ Suppose that (3) holds. then there exists $U\in
Hom_{\mathcal{A}}^{*}(\mathcal{X})$ such that $T=LU$,
$TT^{*}=LUU^{*}U^{*}\leq\|U\|_{\infty}^{2}LL^{*}$, so
$TT^{*}\leq\lambda^{2}LL^{*}$, where $\lambda=\|U\|_{\infty}$.

$(2)\Longrightarrow(3)$ Let be $U_{1}$ a module map defined as follow
$U_{1}:L^{*}\mathcal{X}\longrightarrow T^{*}\mathcal{X}$ such that
$U_{1}L^{*}(\xi)=T^{*}(\xi),\forall\xi\in\mathcal{X}.$ By $(2)$ we have that
$\bar{p}_{\mathcal{X}}(U_{1}\left(L^{*}\xi\right))^{2}=\bar{p}_{\mathcal{X}}(T^{*})^{2}=p(\sqrt{\left\langle
TT^{*}\xi,\xi\right\rangle})\leq p(\sqrt{\lambda^{2}\left\langle
LL^{*}\xi,\xi\right\rangle})=\lambda^{2}\bar{p}_{\mathcal{X}}(L^{*}\xi)^{2}$.
As well as $U_{1}$ is well defined and uniformly bounded. Since $R(L)$ is
closed, and also $R\left(L^{*}\right)$ [11, Theorem 3.2.4]. Out of the
observation above, setting
$U_{1}\xi=0,\xi\in\left(L^{*}\mathcal{X}\right)^{\perp}$, it is simple to
verify that $U_{1}$ is bounded in $\mathcal{X}$, and from the construction of
$U_{1}$ we have $U_{1}L^{*}=T^{*}$. Notify that
$L^{*}\left(L^{*}\right)^{\dagger}L^{*}=L^{*}$, so $U_{1}L^{*}(\xi)=$
$U_{1}L^{*}\left(\left(L^{*}\right)^{\dagger}L^{*}(\xi)\right)=T^{*}\left(\left(L^{*}\right)^{\dagger}L^{*}(\xi)\right)$,
from which it results that $T^{*}=T^{*}\left(L^{*}\right)^{\dagger}L^{*}$,
i.e, $T=LL^{\dagger}T$ (in this case by using the fact
$\left(L^{\dagger}\right)^{*}=\left(L^{*}\right)^{\dagger}$), from which
$T=LU$, where $U=L^{\dagger}T$.

∎

######  Theorem 2.5.

Let $T,L\in Hom_{\mathcal{A}}^{*}(\mathcal{X})$ be two uniformly bounded
operators, then

$Ran(T)+Ran(L)=Ran(\sqrt{TT^{\ast}+LL^{*}}).$

###### Proof.

Let $M=\left(\begin{array}[]{cc}T&-L\\\ 0&0\end{array}\right)$ on
$\mathcal{X}\oplus\mathcal{X}$. Then we have

$\displaystyle(Ran(T)+Ran(L))\oplus\\{0\\}$ $\displaystyle=Ran(M)$
$\displaystyle=Ran(MM^{\ast})^{1/2}$
$\displaystyle=Ran\left(\begin{array}[]{cc}\left(TT^{*}+LL^{*}\right)^{1/2}&0\\\
0&0\end{array}\right)$
$\displaystyle=Ran\left(\sqrt{TT^{*}+LL^{*}}\right)\oplus\\{0\\}.$

Which gives

$Ran(T)+Ran(L)=Ran(\sqrt{TT^{\ast}+LL^{*}}).$

∎

######  Corollary 2.6.

Let $T,L_{1},L_{2}\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. The statements
below are equivalent:

  1. (1)

$Ran(T)\subset Ran\left(L_{1}\right)+Ran\left(L_{2}\right)$.

  2. (2)

$TT^{*}\leqslant\alpha^{2}\left(L_{1}L_{1}^{\ast}+L_{2}L_{2}^{*}\right)$ for
some constant $\alpha>0$.

  3. (3)

There are two uniformly bounded operators $U$ and $V$ such that
$T=L_{1}U+L_{2}V$.

###### Proof.

By using Theorems 2.3 and 2.5 we observe that the conditions $(1)$ and $(2)$
are equivalent.

Since it is evident that the condition $(3)$ implies $(1)$, it is enough to
prove that the statement $(1)$ implies $(3)$.

Now if we assume that $M=\left(\begin{array}[]{ll}T&0\\\
0&0\end{array}\right)$ and $N=\left(\begin{array}[]{ll}L_{1}&L_{2}\\\
0&0\end{array}\right)$, the assertion $(1)$ implies $Ran(M)\subset Ran(N)$.
Therefore $M=NU$ for some $2\times 2$ matrix $U$ and then Theorem 2.3 implies
$(3)$.

∎

######  Lemma 2.7.

Let $\mathcal{A}$ be a pro-$C^{*}$-algebra and
$\alpha,\beta\in\mathcal{A}^{+}$be such that

$p(\alpha\lambda)\leq p(\beta\lambda)\text{ for all
}\lambda\in\mathcal{A}^{+}.$

Then $\alpha^{2}\leq\beta^{2}$.

###### Proof.

Without loss of generality, assume that $p(\alpha)\leq 1$ and $p(\beta)\leq 1$
. Suppose that the inequality $\alpha^{2}\leq\beta^{2}$ is not true.
Then there exists $x_{0}\in\operatorname{sp}\left(\alpha^{2}-\beta^{2}\right)$
such that $x_{0}>0$, where
$\operatorname{sp}\left(\alpha^{2}-\beta^{2}\right)$ denotes the spectrum of
$\alpha^{2}-\beta^{2}$. It follows that

$m=\max\left\\{x:x\in\operatorname{sp}\left(\alpha^{2}-\beta^{2}\right)\right\\}>0$

Let $f$ be any continuous real-valued function defined on the real line such
that

$\left\\{\begin{array}[]{ll}&0\leq f(t)\leq 1\text{ for all }t\in\mathbb{R}\\\
&f(t)=0\text{ for all }t\in\left(-\infty,\frac{m}{2}\right]\\\ &f(t)=1\text{
for all }t\in[m,+\infty)\end{array}\right.$

Choose $\lambda=f\left(\alpha^{2}-\beta^{2}\right).$ Then
$\lambda\in\mathcal{A}^{+}$since $f$ is non-negative. By use of the functional
calculus, we have

$p(\lambda\left(\alpha^{2}-\beta^{2}\right)\lambda)=\max\left\\{xf(x)^{2}:x\in\operatorname{sp}\left(\alpha^{2}-\beta^{2}\right)\right\\}=m>0$
(2.1)

Since $x>\frac{m}{2}$ whenever $f(x)\neq 0$, it follows again by use of the
functional calculus that

$\lambda\left(\alpha^{2}-\beta^{2}\right)\lambda\geq\frac{m}{2}\lambda^{2}.$
(2.2)

Note that $\lambda\beta^{2}\lambda$ is self-adjoint, so there exists a state
$\rho$ acting on $\mathcal{A}$ such that
$\phi\left(\lambda\beta^{2}\lambda\right)=p(\lambda\beta^{2}\lambda)$. Then by
2.2 and the assumption $p(\beta)\leq 1$, we have

$\displaystyle
p(\beta\alpha^{2}\beta)\geq\phi\left(\lambda\alpha^{2}\lambda\right)$
$\displaystyle\geq\phi\left(\lambda\beta^{2}\lambda\right)+\frac{m}{2}\phi\left(\lambda^{2}\right)$
$\displaystyle\geq\frac{m+2}{2}\phi\left(\lambda\beta^{2}\lambda\right)=\frac{m+2}{2}p(\lambda\beta^{2}\lambda)$

This shows that if $\lambda\beta^{2}\lambda\neq 0$, then
$p(\lambda\alpha^{2}\lambda)>p(\lambda\beta^{2}\lambda).$ On the other hand,
if $\lambda\beta^{2}\lambda=0$, then it follows from 2.1 that
$p(\lambda\alpha^{2}\lambda)>0$. So in either case, we have

$p(\alpha\lambda)^{2}=p(\lambda\alpha^{2}\lambda)>p(cb^{2}c)=p(\beta\lambda)^{2},$

in contradiction to the assumption that $p(\alpha\lambda)\leq
p(\beta\lambda)$.

∎

######  Theorem 2.8.

Let $\mathcal{X},\mathcal{Y}$ be two Hilbert $\mathcal{A}$-modules, and let
$T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y})$ and $L\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X},\mathcal{Y}).$ Then the following two
statements are equivalent:

  1. (i)

$TT^{*}\leq LL^{*}$;

  2. (ii)

$\bar{p}_{\mathcal{X}}(T^{*}\xi)\leq\bar{p}_{\mathcal{X}}(L^{*}\xi)$ for all
$\xi\in\mathcal{Y}$.

###### Proof.

$(i)\Rightarrow(ii)$ follows from Proposition 1.11 and Lemma 1.12.

$(ii)\Rightarrow(i)$ let $\xi\in\mathcal{Y}$ and $\alpha=\langle
TT^{\ast}\xi,\xi\rangle$ and $\beta=\langle LL^{\ast}\xi,\xi\rangle$. Then
$\alpha,\beta\in{\mathcal{A}}^{+}$, for any $\lambda\in\mathcal{A}^{+}$ it is
established that

$\displaystyle p(\alpha^{\frac{1}{2}}\lambda)^{2}$
$\displaystyle=p((\alpha^{\frac{1}{2}}\lambda)^{\ast}\alpha^{\frac{1}{2}}\lambda)=p(\lambda^{\ast}\alpha\lambda)=p(\lambda\alpha\lambda)=p(\lambda\langle
TT^{*}\xi,\xi\rangle\lambda)$ $\displaystyle=p(\langle
TT^{*}(\xi\lambda),(\xi\lambda)\rangle)\leq p(\langle
LL^{*}(\xi\lambda),(\xi\lambda)\rangle)=p(\beta^{\frac{1}{2}}\lambda)^{2}$

As a result, by Lemma 2.7, we have $\alpha\leq\beta$ and so

$\left\langle TT^{*}\xi,\xi\right\rangle\leq\left\langle
LL^{*}\xi,\xi\right\rangle$

Therefore by Proposition 1.11, that $TT^{*}\leq LL^{*}$.

∎

##  3\. Atomic system in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$

Next, we introduce the concept of atomic systems for operators in Hilbert
pro-$C^{\ast}$-modules and we establish some results.

######  Definition 3.1.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$, we say that
$\\{\xi_{i}\\}_{i=1}^{\infty}$ is an atomic system for $K$ if :

  1. i)

The serie $\sum_{i}m_{i}\xi_{i}$ converges for all
$m=\left(m_{i}\right)\in\mathcal{H}_{\mathcal{A}}$;

  2. ii)

There exists $C>0$ such that for every $x\in\mathcal{X}$ there exists
$m_{x}=\left(m_{i}\right)\in\mathcal{H}_{\mathcal{A}}$ such that $\langle
m_{x},m_{x}\rangle_{\mathcal{H}_{\mathcal{A}}}\leqslant C\langle
x,x\rangle_{\mathcal{X}}$ and $Kx=\sum_{i}m_{i}\xi_{i}$.

######  Proposition 3.2.

Suppose that $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is a Bessel sequence in
$\mathcal{X}$. Then $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is an atomic
system for $L$, where $L$ is the frame operator.

###### Proof.

We have the module morphism

$\theta:\mathcal{X}\rightarrow{\mathcal{H}_{\mathcal{A}}}$ defined by $\theta
x=\\{\left\langle x,\xi_{i}\right\rangle\\}_{i=1}^{\infty}$,

and its adjoint

$\theta:{\mathcal{H}_{\mathcal{A}}}\rightarrow\mathcal{X}$ defined by
$\theta\left(\left\\{c_{i}\right\\}_{i=1}^{\infty}\right)=\sum_{i=1}^{\infty}c_{i}\xi_{i}$.

Let $L=\theta^{\ast}\theta$. Then

$L:\mathcal{X}\rightarrow\mathcal{X},\quad Lx=\sum_{i=1}^{\infty}\left\langle
x,\xi_{i}\right\rangle\xi_{i}$

Let
$a_{\xi}=\\{a_{i}\\}=\\{\langle\xi,\xi_{i}\rangle\\}_{i=1}^{\infty}\in{\mathcal{H}_{\mathcal{A}}}$

Now

$\displaystyle\bar{p}_{\mathcal{X}}(a_{\xi})^{2}=$ $\displaystyle p(\langle
a_{\xi},a_{\xi}\rangle)$ $\displaystyle=$ $\displaystyle p\left(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle\leq$ $\displaystyle D\bar{p}_{\mathcal{X}}(\xi)^{2}$

Then, $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is an atomic system for $L$.

∎

######  Lemma 3.3.

Let $\mathcal{A}$ be a unital pro-$C^{*}$-algebra and
$\left\\{\xi_{i}\right\\}_{j\in I}$ be a sequence of a finitely or countably
generated Hilbert $\mathcal{A}$-module $\mathcal{X}$ over $\mathcal{A}$. Then
$\left\\{\xi_{i}\right\\}_{i\in I}$ is a Bessel sequence with bound $D$ if and
only if

$p(\sum_{i\in I}\left\langle
x,\xi_{i}\right\rangle\left\langle\xi_{i},x\right\rangle)\leq
D\bar{p}_{\mathcal{X}}(\xi)^{2}$

for all $\xi\in\mathcal{X}.$

###### Proof.

$"\Rightarrow"$ Evident

$"\Leftarrow"$ Let $T$ be a linear operator defined as follow
$T:\mathcal{X}\rightarrow\mathcal{H}_{\mathcal{A}}$

$T\xi=\sum_{i\in I}\left\langle\xi,\xi_{i}\right\rangle
e_{i},\quad\forall\xi\in\mathcal{X}.$

Then

$\bar{p}_{\mathcal{X}}(T\xi)^{2}=p(\langle T\xi,T\xi\rangle)=p(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle\left\langle\xi_{i},\xi\right\rangle)\leq
D\bar{p}_{\mathcal{X}}(\xi)^{2},$

g which implies that $T$ is bounded.

It is clear that $T$ is $\mathcal{A}$-linear. Then we have

$\langle T\xi,T\xi\rangle\leq D\langle\xi,\xi\rangle$

equivalently, $\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle\left\langle\xi_{i},\xi\right\rangle\leq
D\langle\xi,\xi\rangle$, as desired.

∎

######  Theorem 3.4.

If $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$, then there exists an atomic
system for the operator $K$.

###### Proof.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a standard normalized tight frame
for $\mathcal{X}$. Since

$\xi=\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle\xi_{i},\quad\xi\in\mathcal{X},$

we have

$K\xi=\sum_{i\in I}\left\langle\xi,\xi_{i}\right\rangle
K\xi_{i},\quad\xi\in\mathcal{X}.$

For $\xi\in\mathcal{H}$, putting
$a_{i,\xi}=\left\langle\xi,\xi_{i}\right\rangle$ and $f_{i}=K\xi_{i}$ for all
$i\in I$, we get

$\displaystyle\sum_{i\in I}\left\langle\xi,f_{i}\right\rangle\left\langle
f_{i},\xi\right\rangle$ $\displaystyle=\sum_{i\in
I}\left\langle\xi,K\xi_{i}\right\rangle\left\langle K\xi_{i},\xi\right\rangle$
$\displaystyle=\sum_{i\in I}\left\langle
K^{*}\xi,\xi_{i}\right\rangle\left\langle\xi_{i},K^{*}\xi\right\rangle=\left\langle
K^{*}\xi,K^{*}\xi\right\rangle$
$\displaystyle\leqslant\left\|K^{*}\right\|_{\infty}^{2}\langle\xi,\xi\rangle.$

Therefore $\left\\{f_{i}\right\\}_{i\in I}$ is a Bessel sequence for
$\mathcal{H}$ and we conclude that the series $\sum_{i\in I}c_{i}f_{i}$
converges for all $c=\left\\{c_{i}\right\\}_{i\in I}\in\ell^{2}(A)$ by Lemma
3.3. We also have

$\sum_{i\in I}a_{i,\xi}a_{i,\xi}^{*}=\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle\left\langle\xi_{i},\xi\right\rangle=\langle\xi,\xi\rangle,$

which completes the proof.

∎

Our next result deals with Bessel sequences.

######  Theorem 3.5.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a Bessel sequence for $\mathcal{X}$
and $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. Suppose that $T\in
Hom_{\mathcal{A}}^{\ast}\left(\mathcal{X},\mathcal{H}_{\mathcal{A}}\right)$ is
given by $T(\xi)=\left\\{\left\langle\xi,\xi_{i}\right\rangle\right\\}_{i\in
I}$ and $\overline{Ran(T)}$ is orthogonally complemented. Then the following
statements are equivalent:

  1. (1)

$\left\\{\xi_{i}\right\\}_{i\in I}$ is an atomic system for $K$;

  2. (2)

there exists two positives values $C$ and $D$ such that

$\displaystyle
C^{-1}\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}\leq
p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle\leq D\left(\bar{p}_{\mathcal{X}}(\xi)\right)^{2};$

  3. (3)

There exists $Q\in
Hom_{\mathcal{A}}^{\ast}\left(\mathcal{X},\mathcal{H}_{\mathcal{A}}\right)$
such that $K=T^{*}Q$.

###### Proof.

$(1)\Rightarrow(2)$ For each $\xi\in\mathcal{X}$

$\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}=\left(\sup\left\\{p(\langle\xi,Ky\rangle):\bar{p}_{\mathcal{X}}(y)\leq
1\right\\}\right)^{2}$

And $Ky=\sum_{i}m_{i}\xi_{i}$. Then

$\displaystyle\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}$
$\displaystyle=\left(\sup\left\\{p\left(\left\langle\xi,\sum_{i\in
l}\xi_{i}m_{i}\right\rangle\right):\bar{p}_{\mathcal{X}}(y)\leq
1\right\\}\right)^{2}$ $\displaystyle=\left(\sup\left\\{p\left(\sum_{i\in
I}m_{i}^{*}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right):\bar{p}_{\mathcal{X}}(y)\leq
1\right\\}\right)^{2}$ $\displaystyle\leq\sup_{\bar{p}_{\mathcal{X}}(y)\leq
1}\left\\{p\left(\sum_{i\in I}m_{i}^{*}m_{i}\right)p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)\right\\}$
$\displaystyle\leq Cp\left(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$

Then

$C^{-1}\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}\leq
p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$

Since $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is Bessel sequence, then there
exists positive value $D$ such that

$p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)\leq
D\left(\bar{p}_{\mathcal{X}}(\xi)\right)^{2}$

Then

$\displaystyle
C^{-1}\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}\leq
p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle\leq D\left(\bar{p}_{\mathcal{X}}(\xi)\right)^{2}.$

$(2)\Rightarrow(3)$ Since $\left\\{\xi_{i}\right\\}_{i\in I}$ is a Bessel
sequence, we get $T^{*}e_{i}=\xi_{i}$, where $\left\\{e_{i}\right\\}_{i\in I}$
is the standard orthonormal basis for $\mathcal{H}_{\mathcal{A}}$. Therefore

$\displaystyle
C^{-1}\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}\leq
p\left(\sum_{i\in
l}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle=p\left(\sum_{i\in
I}\left\langle\xi,T^{*}e_{i}\right\rangle\left\langle
T^{*}e_{i},\xi\right\rangle\right)$ $\displaystyle=p\left(\sum_{i\in
I}\left\langle T\xi,e_{i}\right\rangle\left\langle
e_{i},T\xi\right\rangle\right)=\bar{p}_{\mathcal{X}}(T\xi)^{2},\quad\xi\in\mathcal{X}$

Theorem 2.4 yields that there exists $Q\in
Hom_{\mathcal{A}}^{\ast}\left(\mathcal{X},\mathcal{H}_{\mathcal{A}}\right)$
such that $K=T^{*}Q$.

$(3)\Rightarrow(1)$ For every $\xi\in\mathcal{X}$, we have

$Q\xi=\sum_{i\in I}\left\langle Q\xi,e_{i}\right\rangle e_{i}$

Thus

$T^{*}Q\xi=\sum_{i\in I}\left\langle Q\xi,e_{i}\right\rangle
T^{*}e_{i},\quad\xi\in\mathcal{X}$

Let $c_{i}=\left\langle Q\xi,e_{i}\right\rangle$, so for all
$\xi\in\mathcal{X}$ we get

$\sum_{i\in I}c_{i}c_{i}^{*}=\sum_{i\in I}\left\langle
Q\xi,e_{i}\right\rangle\left\langle e_{i},Q\xi\right\rangle=\langle
Q\xi,Q\xi\rangle\leqslant\|Q\|_{\infty}^{2}\langle\xi,\xi\rangle$

Since $\left\\{\xi_{i}\right\\}_{i\in I}$ is a Bessel sequence for
$\mathcal{X}$, we get that $\left\\{\xi_{i}\right\\}_{i\in I}$ is an atomic
system for $K$.

∎

######  Corollary 3.6.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a frame of multiplier for
$\mathcal{X}$ with bounds $C,D>0$ and $K\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. Then $\left\\{\xi_{i}\right\\}_{i\in
I}$ is an atomic system for $K$ with bounds $\frac{1}{C^{-1}\|K\|^{2}}$ and
$D$.

###### Proof.

Let $S$ be the frame operator of $\left\\{\xi_{i}\right\\}_{i\in I}$.We show
that condition $(2)$ of the Theorem 3.5 is verified. Since
$\left\\{S^{-1}\xi_{i}\right\\}_{i\in I}$ is a frame for $\mathcal{X}$ with
bounds $D^{-1},C^{-1}>0$ and $x=\sum_{n\in J}\left\langle x,f_{n}\right\rangle
S^{-1}f_{n}$ for all $x\in\mathcal{H}$, we get

$\left(\bar{p}_{\mathcal{X}}\left(K^{*}\xi\right)\right)^{2}=\left(\sup\left\\{p(\langle\xi,Ky\rangle):\bar{p}_{\mathcal{X}}(y)\leq
1\right\\}\right)^{2}$

And $Ky=\sum_{i}m_{i}\xi_{i}$.

$\displaystyle\bar{p}_{\mathcal{X}}(K^{*}x)^{2}$
$\displaystyle=\left(\sup\left\\{p(\langle\xi,Ky\rangle):\bar{p}_{\mathcal{X}}(y)\leq
1\right\\}\right)^{2}$ $\displaystyle=(\sup\\{p\left\langle(\sum_{n\in
J}\langle x,f_{n}\rangle
K^{*}S^{-1}f_{n},y)\right\rangle:\bar{p}_{\mathcal{X}}(y)\leq 1\\})^{2}$
$\displaystyle=(\sup\\{p(\sum_{n\in J}\left\langle
x,f_{n}\right\rangle\left\langle
K^{*}S^{-1}f_{n},y\right\rangle:\bar{p}_{\mathcal{X}}(y)\leq 1\\})^{2}$
$\displaystyle\leq\sup_{\bar{p}_{\mathcal{X}}(y)\leq
1}\left\\{p\left(\sum_{n\in J}\left\langle
Ky,S^{-1}f_{n}\right\rangle\left\langle
S^{-1}f_{n},Ky\right\rangle\right)p\left(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)\right\\}$
$\displaystyle\leq\sup_{\bar{p}_{\mathcal{X}}(y)\leq 1}C^{-1}p\left(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right){\bar{p}_{\mathcal{X}}(Ky)}^{2}$
$\displaystyle\leq C^{-1}{\bar{p}_{\mathcal{X}}(K)^{2}}p\left(\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\right)$

Hence $\left\\{\xi_{i}\right\\}_{i\in I}$ is an atomic system for $K$.

∎

######  Theorem 3.7.

Let $K_{1},K_{2}\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$. If
$\left\\{\xi_{i}\right\\}_{i\in I}$ is an atomic system for $K_{1}$ and
$K_{2}$, and $\alpha,\beta$ are scalars, then
$\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is an atomic system for $\lambda
K_{1}+\gamma K_{2}$ and $K_{1}K_{2}$.

###### Proof.

$\\{\xi\\}_{i=1}^{\infty}$ is an atomic system for $K_{1}$ and $K_{2}$, then
there are two positives constantes $0<\lambda_{n}\leq\gamma_{n}<\infty$
$(n=1,2)$ such that

$\lambda_{n}\left(\bar{p}_{\mathcal{X}}\left(K_{n}^{*}x\right)\right)^{2}\leq
p\left(\sum_{i\in l}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)\leq\gamma_{n}\left(\bar{p}_{\mathcal{X}}(x)\right)^{2},\text{for
all}\;x\in\mathcal{X}.$ (3.1)

We have

$\displaystyle\left(\bar{p}_{\mathcal{X}}\left(\alpha
K_{1}+K_{2}\right)^{*}x\right)^{2}$ $\displaystyle=p(\langle\left(\alpha
K_{1}+K_{2}\right)^{*}x,\left(\alpha K_{1}+K_{2}\right)^{*}x\rangle)$
$\displaystyle=p(\langle\left(\alpha K_{1}\right)^{*}x,\left(\alpha
K_{1}\right)^{*}x\rangle+\langle\left(\beta K_{2}\right)^{*}x,\left(\beta
K_{2}\right)^{*}x\rangle)$ $\displaystyle\leq p(\alpha^{2})Cp\left(\sum_{i\in
I}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle+p(\beta^{2})Dp\left(\sum_{i\in I}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)$
$\displaystyle\leq\max(p(\alpha^{2})C,p(\beta^{2})D)p\left(\sum_{i\in
I}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)$

by setting $M=\max(p(\alpha^{2})C,p(\beta^{2})D)$ It follows that

$M^{-1}\left(\bar{p}_{\mathcal{X}}\left(\alpha K_{1}+\beta
K_{2}\right)^{*}x\right)^{2}\leq p\left(\sum_{i\in I}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)$

Hence $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ satisfies the lower frame
condition. And from inequalities 3.1 , we get

$p\left(\sum_{i\in l}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}\right)\leq\frac{\gamma_{1}+\gamma_{2}}{2}\left(\bar{p}_{\mathcal{X}}(x)\right)^{2},\text{for
all}\;x\in\mathcal{X}$

Hence $\\{\xi_{i}\\}_{i=0}^{\infty}$ is an atomic system for $\alpha
K_{1}+\beta K_{2}$.

∎

##  4\. $K$-frames in Hilbert pro-$C^{\ast}$-module

Now we define $K$-frame in Hilbert pro-$C^{\ast}$-modules and we show that
under some conditions every ordinary frame of multiplier is a $K$-frame. Then,
we use Douglas’ factorization theorem to demonstrat some properties of
$K$-frames. We finally show the relationship between Bessel sequences and
$K$-frame in pro-$C^{\ast}$-modules.

######  Definition 4.1.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.
$\\{\xi_{i}\\}_{i=0}^{\infty}$ is a sequence in $\mathcal{X}$,
$\\{\xi_{i}\\}_{i=0}^{\infty}$ is called $K$-frame for $\mathcal{X}$ if there
existe two positive constants $A$ and $B$ such that

$A\langle K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in\mathcal{X}$

$A$ and $B$ are respectively called lower and upper bounds for $K$-frame
$\\{\xi_{i}\\}_{i\in I}$

######  Definition 4.2.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be a bounded operator. A
sequence $\left\\{\xi_{i}\right\\}_{i\in I}$ in $\mathcal{X}$ is said to be a
tight $K$-frame with bound $A$ if

$A\langle K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}=\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\text{
for all }\xi\in\mathcal{X}$

When $A=1$, it is called a Parseval $K$-frame .

######  Theorem 4.3.

Let $K$ be an invertible element in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$
such that both are uniformly bounded with $\|K\|_{\infty}^{2}$. Then every
ordinary frame of multiplier is $K$-frame for $\mathcal{X}$.

###### Proof.

Suppose that $\\{\xi_{i}\\}_{i=0}^{\infty}$ is frame for $\mathcal{X}$ then
there two constants $A$ and $B$ such that

$A\langle\xi,\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in\mathcal{X}$ (4.1)

For $K$ an invertible element in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$, we
have $\langle
K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq\|K\|_{\infty}^{2}\langle\xi,\xi\rangle_{\mathcal{X}}$.
Then $\frac{1}{\|K\|_{\infty}^{2}}\langle
K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq\langle\xi,\xi\rangle_{\mathcal{X}}$,
for all $\xi\in\mathcal{X}$. From the inequality 4.1, we have

$A\frac{1}{\|K\|_{\infty}^{2}}\langle
K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq
A\langle\xi,\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}$

Hence $\\{\xi_{i}\\}_{i=0}^{\infty}$ is $K$-frame for $\mathcal{X}$.

∎

######  Proposition 4.4.

Let $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ be a $K$-frame for
$\mathcal{X}.$ Let $L\in\mathcal{X}$ be a bounded uniformly operator with
$Ran(L)\subseteq$ $Ran(K)$. Then $\left\\{f_{i}\right\\}_{i=1}^{\infty}$ is a
$L$-frame for $\mathcal{X}$.

###### Proof.

Let $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ be $K$-frame for $\mathcal{X}$.
Then there are positive constants $A$ and $B$ such that

$A\langle K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in\mathcal{X}$ (4.2)

Since $Ran(L)\subseteq Ran(K)$, by Douglas’ theorem 2.4, there exists
$\alpha>0$ such that $LL^{*}\leq\alpha^{2}KK^{*}$. From the inequality (4.2),
we have

$\displaystyle\frac{A}{\alpha^{2}}\langle
L^{\ast}\xi,L^{\ast}\xi\rangle_{\mathcal{X}}\leq A\langle
K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}$ $\displaystyle\leq\sum_{i\in
I}\left\langle
x,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},x\right\rangle_{M(\mathcal{X})}$
$\displaystyle\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in\mathcal{X}.$

Therefore $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is a $L$-frame for
$\mathcal{X}$.

∎

######  Theorem 4.5.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and
$\left\\{\xi_{i}\right\\}_{i\in I}$ be a Bessel sequence for $\mathcal{X}$ .
Suppose that $T\in
Hom_{\mathcal{A}}^{\ast}\left(\mathcal{X},\mathcal{H}_{\mathcal{A}}\right)$ is
given by $T(\xi)=\left\\{\left\langle\xi,\xi_{i}\right\rangle\right\\}_{i\in
I}$ and $\overline{Ran(T)}$ is orthogonally complemented. Then
$\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$ if and
only if there exists a linear bounded operator
$L:\mathcal{H}_{\mathcal{A}}\rightarrow\mathcal{X}$ such that $Le_{i}=\xi_{i}$
and $Ran(K)\subseteq Ran(L)$, where $\left\\{e_{i}\right\\}_{i}$ is the
orthonormal basis for $\mathcal{H}_{\mathcal{A}}$.

###### Proof.

Suppose that $\left\\{\xi_{i}\right\\}_{i\in I}$ is $K$-frame. Then
$A\bar{p}_{\mathcal{X}}(K^{*}\xi)^{2}\leqslant\bar{p}_{\mathcal{X}}(T\xi)^{2}$
for all $\xi\in\mathcal{X}$. By Theorem 2.8, there is $\alpha>0$ such that

$KK^{*}\leqslant\alpha T^{*}T.$

Setting $T^{*}=L$, we get $KK^{*}\leqslant\alpha LL^{*}$ and therefore
$Ran(K)\subseteq Ran(L)$. Conversely, since $Ran(K)\subseteq Ran(L)$, by
Theorem 2.4 there exists $\alpha>0$ such that $KK^{*}\leqslant\alpha LL^{*}$.
Therefore

$\frac{1}{\alpha}\left\langle
K^{*}\xi,K^{*}\xi\right\rangle\leqslant\left\langle
L^{*}\xi,L^{*}\xi\right\rangle=\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle\left\langle\xi_{i},\xi\right\rangle,\quad\xi\in\mathcal{X}$

Then $\left\\{\xi_{i}\right\\}_{i\in I}$ is $K$-frame for $\mathcal{X}$.

∎

######  Proposition 4.6.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a Bessel sequence of $\mathcal{X}$
, $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame with bounds $A,B>0$ if
and only if $L\geqslant AKK^{*}$, where $L$ is the frame operator for
$\left\\{\xi_{i}\right\\}_{i\in I}.$

###### Proof.

A sequence $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$ if and only if

$\left\langle AKK^{*}\xi,\xi\right\rangle=A\left\langle
K^{*}\xi,K^{*}\xi\right\rangle\leqslant\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}=\langle
L\xi,\xi\rangle\leqslant B\langle\xi,\xi\rangle,$

Then $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame if and only if
$L\geqslant AKK^{*}$.

∎

######  Theorem 4.7.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a Bessel sequence in $\mathcal{X}$.
Then $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$ if
and only if $K=L^{1/2}L$, for some $U\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.

###### Proof.

Suppose $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame, by Proposition 4.6
there exist two positives constantes $A$ and $B$

$AKK^{*}\leq L^{1/2}L^{1/2^{*}}$

Therefore by Douglas’ theorem 2.4, $K=L^{1/2}U$, for some $L$ bounded in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X}).$

Conversly, let $K=L^{1/2}U$, for some $L$ bounded in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X}).$ Then by Douglas’ factorization
theorem, $L^{1/2}$ majorizes $K^{*}$. Then there is a positive number $A$ such
that

$L\geq A^{2}KK^{*}$. Therefore by Proposition 4.6
$\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$.

∎

######  Example 4.8.

Suppose that $\left\\{u_{i}\right\\}_{i=1}^{\infty}$ is an orthonormal basis
in $\mathcal{H}_{\mathcal{A}}$. Define operators $L$ and $K$ on
$\mathcal{H}_{\mathcal{A}}$ by $Lu_{i}=u_{i-1}$ for $i>1$ and $Lu_{1}=0$ and
$Ku_{i}=u_{i+1}$ respectively. It is clear that
$\left\\{Ku_{i}\right\\}_{i=1}^{\infty}$ is a $K$-frame for
$\mathcal{H}_{\mathcal{A}}$. Suppose $\left\\{Ku_{i}\right\\}_{i\in I}$ is a
$L$-frame. Then by Proposition 4.6, there exists $A>0$ such that $KK^{*}\geq
ALL^{*}$. Hence by Douglas, theorem, $Ran(L)\subseteq Ran(K)$. But this is
contradiction to $Ran(L)\nsubseteq Ran(K)$, since $u_{1}\in Ran(L)$ but
$u_{1}\notin Ran(K)$.

######  Theorem 4.9.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be an uniformly bounded
operator such that $Ran(K)$ is closed. The frame operator of a $K$-frame is
invertible on the subspace $Ran(K)$ of $\mathcal{X}$.

###### Proof.

Suppose $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$.
Then there exists $A>0$ such that

$A\langle K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}$
(4.3)

Since $Ran(K)$ is closed, then $KK^{\dagger}\xi=\xi$, for all $\xi\in Ran(K)$.
That is,

$\left.KK^{\dagger}\right|_{Ran(K)}=I_{Ran(K)}$

we have
$I_{Ran(K)}^{*}=\left(\left.K^{\dagger}\right|_{Ran(K)}\right)^{*}K^{*}$. For
any $\xi\in Ran(K)$, we obtain

$\langle\xi,\xi\rangle=\langle\left(\left.K^{\dagger}\right|_{R(K)}\right)^{*}K^{*}\xi,\left(\left.K^{\dagger}\right|_{R(K)}\right)^{*}K^{*}\xi\rangle\leq\left\|K^{\dagger}\right\|_{\infty}^{2}\cdot\langle
K^{*}\xi,K^{*}\xi\rangle$

Therefore

$\langle
K^{*}\xi,K^{*}\xi\rangle\geq\left\|K^{\dagger}\right\|_{\infty}^{-2}\cdot\langle\xi,\xi\rangle$

In combination with 4.3, we obtain

$\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\geq
A\langle K^{\ast}\xi,K^{\ast}\xi\rangle_{\mathcal{X}}\geq
A\left\|K^{\dagger}\right\|_{\infty}^{-2}\cdot\langle\xi,\xi\rangle\;\;\text{for
all}\;\xi\in Ran(K).$

Hence, by the definition of $K$-frame, we get

$A\left\|K^{\dagger}\right\|_{\infty}^{-2}\cdot\langle\xi,\xi\rangle_{\mathcal{X}}\leq\sum_{i\in
I}\left\langle\xi,\xi_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i},\xi\right\rangle_{M(\mathcal{X})}\leq
B\langle\xi,\xi\rangle_{\mathcal{X}}\;\;\forall\xi\in Ran(K)$

Therefore

$A\left\|K^{\dagger}\right\|_{\infty}^{-2}\cdot\langle\xi,\xi\rangle_{\mathcal{X}}\leq\langle
S\xi,S\xi\rangle_{\mathcal{X}}\leq B\langle\xi,\xi\rangle_{\mathcal{X}}$

And so $S:Ran(K)\rightarrow Ran(S)$ is a bounded linear operator and
invertible on $Ran(K)$.

∎

######  Theorem 4.10.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded such
that $Ran(K)$ is dense. Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame
and $T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded such
that $Ran(T)$ is closed. If $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a
$K$-frame for $\mathcal{X}$, then $T$ is surjective.

###### Proof.

Let’s assume that $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$ with frame bounds $A$ and $B$. Then for any $\xi\in\mathcal{X}$,

$A\langle K^{*}\xi,K^{*}\xi\rangle\leq\sum_{i\in
I}\langle\xi,T\xi_{i}\rangle_{M(\mathcal{X})}\langle
T\xi_{i},\xi\rangle_{M(\mathcal{X})}\leq B\langle\xi,\xi\rangle.$ (4.4)

Since $Ran(K)$ is dense, $\mathcal{X}=\overline{Ran(K)}$, so $K^{*}$ is
injective. Then from 4.4, $T^{*}$ is injective since
$Ker\left(T^{*}\right)\subseteq Ker\left(K^{*}\right)$. Moreover,
$Ran(T)=Ker\left(T^{*}\right)^{\perp}=\mathcal{X}$. Thus $T$ is surjective.

∎

######  Theorem 4.11.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded and let
$\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame for $\mathcal{X}$. If $T\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ is uniformly bounded and has a closed
range with $TK=KT$, then $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a $K$-frame
for $Ran(T)$.

###### Proof.

As $T$ has a closed range, it has the generalized inverse $T^{\dagger}$ such
that $TT^{\dagger}=I$. Now $I=I^{*}=T^{\dagger^{*}}T^{*}$. Then for each
$\xi\in Ran(T),K^{*}\xi=T^{\dagger^{*}}T^{*}K^{*}\xi$, so we have

$\langle K^{*}\xi,K^{*}\xi\rangle=\langle
T^{\dagger^{*}}T^{*}K^{*}\xi,T^{\dagger^{*}}T^{*}K^{*}\xi\rangle\leq\left\|T^{\dagger^{*}}\right\|_{\infty}^{2}\langle
T^{*}K^{*}\xi,T^{*}K^{*}\xi\rangle$

Therefore

$\left\|T^{\dagger^{*}}\right\|_{\infty}^{-2}\langle
K^{*}\xi,K^{*}\xi\rangle\leq\langle T^{*}K^{*}\xi,T^{*}K^{*}\xi\rangle.$

Now for each $\xi\in Ran(T)$,

$\displaystyle\sum_{i\in I}\langle\xi,T\xi_{i}\rangle_{\mathcal{X}}\langle
T\xi_{i},\xi\rangle_{\mathcal{X}}$ $\displaystyle=\sum_{i\in I}\langle
T^{\ast}\xi,\xi_{i}\rangle_{\mathcal{X}}\langle\xi_{i},T^{\ast}\xi\rangle_{\mathcal{X}}\geq
A\langle K^{*}T^{*}\xi,K^{*}T^{*}\xi\rangle$ $\displaystyle=A\langle
T^{*}K^{*}\xi,T^{*}K^{*}\xi\rangle$ $\displaystyle\geq
A\left\|T^{\dagger^{*}}\right\|_{\infty}^{-2}\langle
K^{*}\xi,K^{*}\xi\rangle.$

Since $\left\\{\xi_{i}\right\\}_{i\in I}$ is a Bessel sequence with bound $B$,
for each $\xi\in Ran(T)$, we have

$\sum_{i\in I}\langle\xi,T\xi_{i}\rangle_{M(\mathcal{X})}\langle
T\xi_{i},\xi\rangle_{M(\mathcal{X})}=\sum_{i\in I}\langle
T^{\ast}\xi,\xi_{i}\rangle_{M(\mathcal{X})}\langle\xi_{i},T^{\ast}\xi\rangle_{M(\mathcal{X})}\leq
B\langle T^{*}\xi,T^{*}\xi\rangle\leq
B\|T\|_{\infty}^{2}\langle\xi,\xi\rangle.$

Therefore $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $Ran(T)$.

∎

######  Theorem 4.12.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded such
that $Ran(K)$ is dense. Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame
and let $T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded such
that $Ran(T)$ is closed. If $\left\\{T\xi_{i}\right\\}_{i\in I}$ and
$\left\\{T^{*}\xi_{i}\right\\}_{i\in I}$ are $K$-frames for $\mathcal{X}$,
then $T$ is invertible.

###### Proof.

Suppose that $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$ with frame bounds $A_{1}$ and $B_{1}$. Then for any
$\xi\in\mathcal{X}$,

$A_{1}\langle K^{*}\xi,K^{*}\xi\rangle\leq\sum_{i\in
I}\langle\xi,T\xi_{i}\rangle_{M(\mathcal{X})}\langle
T\xi_{i},\xi\rangle_{M(\mathcal{X})}\leq B_{1}\langle\xi,\xi\rangle$ (4.5)

As $Ran(K)$ is dense, $K^{*}$ is injective. Then from 4.5, $T^{*}$ is
injective since $Ker\left(T^{*}\right)\subseteq Ker\left(K^{*}\right)$.
Moreover $Ran(T)=Ker\left(T^{*}\right)^{\perp}=\mathcal{X}$. Then $T$ is
surjective.

Suppose $\left\\{T^{*}f_{i}\right\\}_{i=1}^{\infty}$ is a $K$-frame for
$\mathcal{X}$ with frame bounds $A_{2}$ and $B_{2}$. Then for any
$\xi\in\mathcal{X}$,

$A_{2}\langle K^{*}\xi,K^{*}\xi\rangle\leq\sum_{i\in
I}\langle\xi,T^{\ast}\xi_{i}\rangle_{M(\mathcal{X})}\langle
T^{\ast}\xi_{i},\xi\rangle_{M(\mathcal{X})}\leq B_{2}\langle\xi,\xi\rangle$
(4.6)

As $K$ has a dense range, $K^{*}$ is injective. Then from 4.6, $T$ is
injective since $Ker(T)\subseteq Ker\left(K^{*}\right)$. Therefore $T$ is
bijective. Using the Bounded Inverse Theorem, $T$ is invertible.

∎

######  Theorem 4.13.

Let $K\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ and let
$\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame for $\mathcal{X}$ and let
$T\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ be uniformly bounded and be a co-
isometry with $TK=KT$. Then $\left\\{T\xi_{i}\right\\}_{i=1}^{\infty}$ is a
$K$-frame for $\mathcal{X}$.

###### Proof.

Suppose $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$.
For every $\xi\in\mathcal{X}$

$\displaystyle\sum_{i\in I}\langle\xi,T\xi_{i}\rangle_{\mathcal{X}}\langle
T\xi_{i},\xi\rangle_{\mathcal{X}}$ $\displaystyle=\sum_{i\in I}\langle
T^{\ast}\xi,\xi_{i}\rangle_{\mathcal{X}}\langle\xi_{i},T^{\ast}\xi\rangle_{\mathcal{X}}\geq
A\langle K^{*}T^{*}\xi,K^{*}T^{*}\xi\rangle$ $\displaystyle=A\langle
T^{*}K^{*}\xi,T^{*}K^{*}\xi\rangle$ $\displaystyle=A\langle
K^{*}\xi,K^{*}\xi\rangle$

It is obvious that $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a Bessel sequence.
Since $\left\\{\xi_{i}\right\\}_{i\in I}$ is a Bessel sequence, for each
$\xi\in\mathcal{X}$ we have

$\sum_{i\in I}\langle\xi,T\xi_{i}\rangle_{\mathcal{X}}\langle
T\xi_{i},\xi\rangle_{\mathcal{X}}=\sum_{i\in I}\langle
T^{\ast}\xi,\xi_{i}\rangle_{\mathcal{X}}\langle\xi_{i},T^{\ast}\xi\rangle_{\mathcal{X}}\leq
B\|T\|_{\infty}^{2}\langle\xi,\xi\rangle.$

Therefore $\left\\{T\xi_{i}\right\\}_{i\in I}$ is a K-frame for $\mathcal{X}$.

∎

##  5\. Sums of $K$-Frames

In the following section we show that the sums of $K$-frames under some
conditions is again a $K$-frame in Hilbet pro-$C^{\ast}$-modules.

######  Theorem 5.1.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ and $\left\\{\eta_{i}\right\\}_{i\in
I}$ be two $K$-frames for $\mathcal{X}$ and let $L_{1}$ and $L_{2}$ be
respectively their corresponding operators in Proposition 4.6. If
$L_{1}L_{2}^{*}$ and $L_{2}L_{1}^{*}$ are positive operators and
$Ran\left(L_{1}\right)+Ran\left(L_{2}\right)$ is closed, then
$\left\\{\xi_{i}+\eta_{i}\right\\}_{i\in I}$ is a $K$-frame for $\mathcal{X}$.

###### Proof.

Suppose that $\left\\{\xi_{i}\right\\}_{i\in I}$ and
$\left\\{\eta_{i}\right\\}_{i\in I}$ are two $K$-frames for $\mathcal{X}$. By
the assumption, there are two bounded operators $L_{1}$ and $L_{2}$ such that
$L_{1}u_{i}=\xi_{i},$ $L_{2}u_{i}=\eta_{i}$ and $Ran(K)\subseteq
Ran\left(L_{1}\right),Ran(K)\subseteq Ran\left(L_{2}\right)$, we denote by
$\left\\{u_{i}\right\\}_{i\in I}$ is an orthonormal basis for
$\mathcal{H}_{\mathcal{A}}$

So $Ran(K)\subseteq Ran\left(L_{1}\right)+Ran\left(L_{2}\right)$, by Corollary
2.6

$KK^{*}\leq\alpha^{2}\left(L_{1}L_{1}^{*}+L_{2}L_{2}^{*}\right),\text{ for
some }\alpha>0$

For every $\xi\in\mathcal{X}$, we have

$\displaystyle\sum_{i\in
I}\langle\xi,\xi_{i}+\eta_{i}\rangle\langle\xi_{i}+\eta_{i},\xi\rangle$
$\displaystyle=\sum_{i\in
I}\langle(L_{1}^{*}+L_{2}^{*})\xi,u_{i}\rangle\langle
u_{i},(L_{1}{}^{*}+L_{2}^{*})\xi\rangle$ $\displaystyle=\sum_{i\in
I}\langle(L_{1}+L_{2})^{*}\xi,u_{i}\rangle\langle
u_{i},(L_{1}+L_{2})^{*}\xi\rangle$
$\displaystyle=\langle(L_{1}+L_{2})^{*}\xi,(L_{1}+L_{2})^{*}\xi\rangle$
$\displaystyle=\langle L_{1}{}^{*}\xi,L_{1}{}^{*}\xi\rangle+\langle
L_{1}{}^{*}\xi,L_{2}{}^{*}\xi\rangle$ $\displaystyle+\langle
L_{2}{}^{*}\xi,L_{1}{}^{*}\xi\rangle+\langle
L_{2}{}^{*}\xi,L_{2}{}^{*}\xi\rangle$
$\displaystyle\geqslant\langle(L_{1}L_{1}{}^{*}+L_{2}L_{2}{}^{*})\xi,\xi\rangle$
$\displaystyle\geqslant\frac{1}{\alpha^{2}}\left(\left\langle
KK^{*}\xi,\xi\right\rangle\right).$
$\displaystyle\geqslant\frac{1}{\alpha^{2}}\left(\left\langle
K^{*}\xi,K^{*}\xi\right\rangle.\right.$

Therefore $\left\\{\xi_{i}+\eta_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$.

∎

######  Corollary 5.2.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ and $\left\\{\eta_{i}\right\\}_{i\in
I}$ be $K$-frames for $\mathcal{X}$ with frame operators $L_{1}$ and $L_{2}$
respectively. Then $K=L_{1}^{1/2}T_{1}+L_{2}^{1/2}T_{2}$, for some bounded
operators $T_{1}$ and $T_{2}$ in $Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$.

###### Proof.

Since $\left\\{\xi_{i}\right\\}_{i\in I}$ and $\left\\{\eta_{i}\right\\}_{i\in
I}$ are $K$-frames for $\mathcal{X}$, by Proposition 4.6, there are positive
constants $A_{1}$ and $A_{2}$ such that

$L_{1}\geq A_{1}KK^{*}\text{ and }L_{2}\geq A_{2}KK^{*}$

Then by Douglas’ theorem, we have

$Ran(K)\subseteq Ran\left(L_{1}^{1/2}\right)$

and

$Ran(K)\subseteq Ran\left(L_{2}^{1/2}\right).$

Hence $Ran(K)\subseteq
Ran\left(S_{1}^{1/2}\right)+Ran\left(S_{2}^{1/2}\right).$ Hence by Corollary
2.6, there exists two bounded operators $T_{1},T_{2}$ in
$Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ such that
$K=L_{1}^{1/2}T_{1}+L_{2}^{1/2}T_{2}$.

∎

######  Theorem 5.3.

Suppose that $\left\\{\xi_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$ such that $L$ is its frame operator and let $T$ be a positive
operator. Then $\left\\{\xi_{i}+T\xi_{i}\right\\}_{i\in I}$ is a $K$-frame.
Furthermore $\left\\{\xi_{i}+T^{n}f_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$ for any natural number $n$.

###### Proof.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame for $\mathcal{X}$. Then
by Proposition 4.6, there is $A>0$ such that $L\geq AKK^{*}$.
$(I+T)L(I+T)^{*}$ is the frame operator for
$\left\\{\xi_{i}+T\xi_{i}\right\\}_{i\in I}$ for the raison that for each
$\xi\in\mathcal{X}$,

$\displaystyle\sum_{i\in
I}\left\langle\xi,\left(\xi_{i}+T\xi_{i}\right)\right\rangle\left(\xi_{i}+T\xi_{i}\right)$
$\displaystyle=(I+T)\sum_{i\in
I}\left\langle\xi,(I+T)\xi_{i}\right\rangle\xi_{i}$
$\displaystyle=(I+T)L(I+T)^{*}\xi$

In addition, we have

$(I+T)L(I+T)^{*}=L+LT^{*}+TL+TLT^{*}\geq L\geq AKK^{*}$

By Propostion 4.6, we can say that $\left\\{\xi_{i}+T\xi_{i}\right\\}_{i\in
I}$ is a $K$-frame for $\mathcal{X}$. For every given natural number $n$, the
frame operator for $\\{\xi_{i}+T^{n}\xi_{i}\\}_{i\in I}$ is
$(I+T^{n})L(I+T^{n})^{*}\geq L$. Therefore
$\left\\{\xi_{i}+T^{n}\xi_{i}\right\\}_{i=1}^{\infty}$ is a $K$-frame for
$\mathcal{X}$.

∎

######  Corollary 5.4.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ be a $K$-frame for $\mathcal{X}$ and
let $L$ be its frame operator. Suppose that $\left\\{I_{1},I_{2}\right\\}$ is
a partition of $\mathbb{N}$. For $j=1,2$, let $L_{j}$ be the frame operator
for the Bessel sequence $\left\\{\xi_{i}\right\\}_{i\in I_{j}}.$ Then

$\left\\{\xi_{i}L_{1}^{m}\xi_{i}\right\\}_{i\in
I_{1}}\cup\left\\{\xi_{i}+L_{2}^{n}\xi_{i}\right\\}_{i\in I_{2}}$

is a $K$-frame for $\mathcal{X}$ for every $m,n\in\mathbb{N}$.

###### Proof.

For every $m\in\mathbb{N}$, $L^{m}$ can be defined as follow

$L^{m}\xi=\sum_{i\in
I_{j}}\left\langle\xi,L^{\frac{m-1}{2}}\xi_{i}\right\rangle
L^{\frac{m-1}{2}}\xi_{i}$

For each $\xi\in\mathcal{X}$,

$\displaystyle\sum_{i\in
I_{1}}\left\langle\xi,\xi_{i}+L_{1}^{m}\xi_{i}\right\rangle\left(\xi_{i}+L_{1}^{m}\xi_{i}\right)$
$\displaystyle=\left(I+L_{1}^{m}\right)\sum_{i\in
I_{1}}\left\langle\xi,\xi_{i}+L_{1}^{m}\xi_{i}\right\rangle\xi_{i}$
$\displaystyle=\left(I+L_{1}^{m}\right)\sum_{i\in
I_{1}}\left\langle\xi,\left(I+L_{1}^{m}\right)\xi_{i}\right\rangle\xi_{i}$
$\displaystyle=\left(I+L_{1}^{m}\right)L_{1}\left(I+L_{1}^{m}\right)^{*}\xi$
$\displaystyle=\left(I+L_{1}^{m}\right)\left(L_{1}+L_{1}^{(1+m)}\right)\xi$
$\displaystyle=\left(L_{1}+2L_{1}^{(1+m)}+L_{1}^{(1+2m)}\right)\xi.$

Therefore $L_{1}+2L_{1}^{(1+m)}+$ $L_{1}^{(1+2m)}$ and
$L_{2}+2L_{2}^{(1+n)}+L_{2}^{(1+2n)}$ are the frame operators for
$\left\\{\xi_{i}+L_{1}^{m}\xi_{i}\right\\}_{i\in I_{1}}$ and
$\left\\{\xi_{i}+L_{2}^{n}\xi_{i}\right\\}_{i\in I_{2}}$ respectively.
Consider that $L_{0}$ is the frame operator for
$\left\\{\xi_{i}+L_{1}^{m}\xi_{i}\right\\}_{i\in
I_{1}}\cup\left\\{\xi_{i}+L_{2}^{n}\xi_{i}\right\\}_{i\in I_{2}}.$

As $\left\\{\xi_{i}\right\\}_{i=1}^{\infty}$ is a $K$-frame for $\mathcal{X}$,
then there exists $A>0$ such that $L\geq AKK^{*}$ and $L_{0}\geq
L_{1}+L_{2}=L\geq AKK^{*}$. From which
$\left\\{\xi_{i}+L_{1}^{m}\xi_{i}\right\\}_{i\in
I_{1}}\cup\left\\{\xi_{i}+L_{2}^{n}\xi_{i}\right\\}_{i\in I_{2}}$ is a
$K$-frame for $\mathcal{X}$.

∎

######  Theorem 5.5.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ and $\left\\{\eta_{i}\right\\}_{i\in
I}$ be Parseval $K$-frames for $\mathcal{X}$, with synthesis operators $L_{1}$
and $L_{2}$ respectively. If $L_{1}L_{2}^{*}=0$ then
$\left\\{\xi_{i}+\eta_{i}\right\\}_{i\in I}$ is a $2$-tight $K$-frame for
$\mathcal{X}$.

###### Proof.

Suppose $\left\\{\xi_{i}\right\\}_{i\in I}$ and
$\left\\{\eta_{i}\right\\}_{i\in I}$ are two Parseval $K$-frames for
$\mathcal{X}$. Then there are transform operators $L_{1},L_{2}\in
Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ such that $L_{1}e_{i}=\xi_{i}$ and
$L_{2}e_{i}=\eta_{i}$ with
$Ran(K)=Ran\left(L_{1}\right),Ran(K)=Ran\left(L_{2}\right)$ respectively. For
each $\xi\in\mathcal{X}$, we have

$\displaystyle\sum_{i\in
I}\left\langle\xi,\xi_{i}+\eta_{i}\right\rangle_{M(\mathcal{X})}\left\langle\xi_{i}+\eta_{i},\xi\right\rangle_{M(\mathcal{X})}$
$\displaystyle=\langle\left(L_{1}+L_{2}\right)^{*}\xi,\left(L_{1}+L_{2}\right)^{*}\xi\rangle_{\mathcal{X}}$
$\displaystyle=\langle
L_{1}^{*}\xi,L_{1}^{*}\xi\rangle_{\mathcal{X}}+\left\langle
L_{2}L_{1}^{*}\xi,\xi\right\rangle_{\mathcal{X}}$ $\displaystyle+\left\langle
L_{1}L_{2}^{*}\xi,L_{1}L_{2}^{*}\xi\right\rangle_{\mathcal{X}}+\langle
L_{2}^{*}\xi,L_{2}^{*}\xi\rangle_{\mathcal{X}}$ $\displaystyle=\langle
L_{1}^{*}\xi,L_{1}^{*}\xi\rangle_{\mathcal{X}}+\langle
L_{2}^{*}\xi,L_{2}^{*}\xi\rangle_{\mathcal{X}}$ $\displaystyle=2\langle
K^{*}\xi,K^{*}\xi\rangle_{\mathcal{X}}$

∎

######  Theorem 5.6.

Let $\left\\{\xi_{i}\right\\}_{i\in I}$ and $\left\\{\eta_{i}\right\\}_{i\in
I}$ be $K$-frames for $\mathcal{X}$, and let $L_{1}$ and $L_{2}$ be synthesis
operators for sequences $\left\\{\xi_{i}\right\\}_{i\in I}$ and
$\left\\{\eta_{i}\right\\}_{i\in I}$ respectively, such that
$L_{1}L_{2}^{*}=0$ and let $T_{j}\in Hom_{\mathcal{A}}^{\ast}(\mathcal{X})$ an
operator uniformly bounded with $R\left(L_{j}\right)\subseteq
R\left(T_{j}L_{j}\right)$, for $j=1,2$. Then
$\left\\{T_{1}\xi_{i}+T_{2}\eta_{i}\right\\}_{i\in I}$ is a $K$-frame for
$\mathcal{X}$.

###### Proof.

Suppose that $\left\\{\xi_{i}\right\\}_{i\in I}$ and
$\left\\{\eta_{i}\right\\}_{i\in I}$ are two $K$-frames for $\mathcal{X}$.
Then by Theorem 4.5, there exists an orthonormal basis
$\left\\{e_{i}\right\\}_{i=1}^{\infty}$ in $\mathcal{H}_{\mathcal{A}}$ such
that $L_{1}e_{i}=\xi_{i},L_{2}e_{i}=\eta_{i}$ and $R(K)\subseteq
R\left(L_{1}\right),R(K)\subseteq R\left(L_{2}\right)$. For each
$\xi\in\mathcal{X}$

$\displaystyle\sum_{i\in
I}\langle\xi,T_{1}\xi_{i}+T_{2}\eta_{i}\rangle_{M(\mathcal{X})}\langle\xi,T_{1}\xi_{i}+T_{2}\eta_{i}\rangle_{M(\mathcal{X})}$
$\displaystyle=\sum_{i\in
I}\langle\xi,T_{1}T_{1}L_{1}e_{i}+T_{2}L_{2}e_{i}\rangle_{M(\mathcal{X})}\langle\xi,T_{1}L_{1}e_{i}+T_{2}L_{2}e_{i}\rangle_{M(\mathcal{X})}$
$\displaystyle=\langle\left(T_{1}L_{1}+T_{2}L_{2}\right)^{*}\xi,\left(T_{1}L_{1}+T_{2}L_{2}\right)^{*}\xi\rangle$
$\displaystyle=\langle\left(T_{1}L_{1}\right)^{*}\xi,\left(T_{1}L_{1}\right)^{*}\xi\rangle+\left\langle
T_{2}L_{2}L_{1}^{*}T_{1}^{*}\xi,\xi\right\rangle$ $\displaystyle+\left\langle
T_{1}L_{1}L_{2}^{*}T_{2}^{*}\xi,\xi\right\rangle+\langle\left(T_{2}L_{2}\right)^{*}\xi,\left(T_{2}L_{2}\right)^{*}\xi\rangle$
$\displaystyle=\langle\left(T_{1}L_{1}\right)^{*}\xi,\left(T_{1}L_{1}\right)^{*}\xi\rangle+\langle\left(T_{2}L_{2}\right)^{*}\xi,\left(T_{2}L_{2}\right)^{*}\xi\rangle$

We have that $Ran(K)\subseteq Ran\left(L_{j}\right)\subseteq
Ran\left(T_{j}L_{j}\right)$ for $j=1,2$. So by Douglas’ factorization theorem,
for each $j=1,2$, there exists $\alpha_{j}>0$ such that

$KK^{*}\leq\alpha_{j}\left(T_{j}L_{j}\right)\left(T_{j}L_{j}\right)^{*}$

Then from the above inequality, for each $\xi\in\mathcal{X}$

$\displaystyle\sum_{i\in
I}\langle\xi,T_{1}\xi_{i}+T_{2}\eta_{i}\rangle_{M(\mathcal{X})}\langle\xi,T_{1}\xi_{i}+T_{2}\eta_{i}\rangle_{M(\mathcal{X})}$
$\displaystyle=\langle\left(T_{1}L_{1}\right)^{*}\xi,\left(T_{1}L_{1}\right)^{*}\xi\rangle+\langle\left(T_{2}L_{2}\right)^{*}\xi,\left(T_{2}L_{2}\right)^{*}\xi\rangle$
$\displaystyle\geq\left(\frac{1}{\alpha_{1}}+\frac{1}{\alpha_{2}}\right)\langle
K^{*}\xi,K^{*}\xi\rangle$

Hence $\left\\{T_{1}\xi_{i}+T_{2}\eta_{i}\right\\}_{i\in I}$ is a $K$-frame
for $\mathcal{X}$.

∎

## Declarations

Availablity of data and materials
Not applicable.

Competing interest
The authors declare that they have no competing interests.

Fundings
Authors declare that there is no funding available for this article.

Authors’ contributions
The authors equally conceived of the study, participated in its design and
coordination, drafted the manuscript, participated in the sequence alignment,
and read and approved the final manuscript.

Acknowledgements
The authors are thankful to the area editor and referees for giving valuable
comments and suggestions

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