# Generalized form of fixed-point theorems in generalized Banach algebra
relative to the weak topology with an application

Aref Jeribi (1)(i), Najib Kaddachi (1)(ii) and Zahra Laouar(1)(i)

_(1) Department of Mathematics. University of Sfax. Faculty of Sciences of
Sfax.
Soukra Road Km 3.5 B.P. 1171, 3000, Sfax, Tunisia_.

e-mail : (i<EMAIL_ADDRESS>(ii<EMAIL_ADDRESS>
Abstract. In this paper, a general hybrid fixed point theorem for the
contractive mappings in generalized Banach spaces is proved via measure of
weak non-compactness and it is further applied to fractional integral
equations for proving the existence results for the solutions under mixed
Lipschitz and weakly sequentially continuous conditions. Finally, an example
is given to illustrate the result.

Keywords: Banach space, weakly compact, Measure of weak non-compactness, Fixed
point theory, Integral equations.

Mathematics Subject Classification: 46E15, 46E40, 47H10, 37C25, 47N20.

##  1\. Introduction

Many nonlinear problems involve the study of nonlinear equations of the form

(1.1) $\displaystyle x=Ax\cdot Bx+Cx,\ \ x\in S$

where $S$ is a nonempty, closed, and convex subset of a Banach algebra $X,$
see for example [6, 7, 9, 10, 11] and the references therein. These studies
were mainly based on the Schauder fixed point theorem, and properties of the
operators $A,B$ and $C$ (cf. completely continuous, k-set contractive,
condensing, and the potential tool of the axiomatic measure of
noncompactness,$\ldots).$ Various attempts have been made in the literature to
extend this study to a weak topology, see for example [2, 4, 15]. In recent
years, the authors Ben Amar et al. have established in [2] some fixed point
theorems for the operator equation $(\ref{abc})$ in Banach algebra under the
weak topology, their results were based on the class of weak sequential
continuity, weakly compact and weakly condensing conditions. In other
direction, the classical Banach contraction principle was extended for
contractive maps on spaces endowed with vector-valued metrics by Perov [19,
20]. Schauder’s fixed point theorem has been extended from Banach spaces to
generalized Banach spaces by Viorel [25] and Krasnosel’skii’s fixed point
theorem has been extended by Petru and ouahab in [18, 21]. More recently, the
authors Nieton et al. [17] have also established some new variants of fixed-
point theorems for operator equation $(\ref{abc})$ in vector-valued metrics
endowed with an internal composition law $(\cdot).$ These studies were mainly
based on the convexity and the closure of the bounded domain and properties of
the operators $A,B$ and $C.$ Their analysis was carried out via arguments of
strong topology.

This paper is centered around the following question: under which conditions
on its entries, the operator equation $(\ref{abc})$ acting on a generalized
Banach algebras with respect to the weak topology, has a solution?
Our main results are applied to investigate the existence of solutions for the
following coupled system of quadratic integral equations of fractional order

$x_{i}(t)=f_{i}(t,x_{1}(t),x_{2}(t))\cdot\displaystyle\int_{0}^{t}\displaystyle\frac{(t-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{i})}g_{i}(s,x_{1}(s),x_{2}(s))ds+\displaystyle\sum_{k=1}^{m}I^{\beta_{i}^{k}}h_{i}^{k}(t,x_{1}(t),x_{2}(t)),\
\ i=1,2$

where $k\in\\{1,...,m\\},$ $t,\alpha_{i}\in(0,1),$ $I^{\beta_{i}^{k}}$ is the
fractional Pettis integral of order $\beta_{i}^{k}>0,$ $\Gamma(\cdot)$ is the
Gamma function and the functions $f_{i},g_{i},h_{i}^{k}$ are given functions,
whereas $x_{1}=x_{1}(t)$ and $x_{2}=x_{2}(t)$ are unknown functions.
The present paper is arranged as follows. The next section, we give some
preliminary results for future use. Moreover, we shall extend the result of B.
Amar, Jeribi and Mnif [[5], Theorem $2.5$]. In addition, we establish the
fixed point theorem by using the concept of the measure of weak non
compactness in generalized Banach space $($see Theorem 2.3$)$. Note our result
$($Theorem 2.3$)$ improves and generalizes Theorem $3.2$ in [3]. Apart from
that we introduce a class of generalized Banach algebras satisfying certain
sequential conditions called here the condition $(\mathcal{GP})$ $($see
Definition 2.8 $)$. In section $3,$ we present a collection of new fixed point
theorems in generalized Banach algebras satisfying condition $(\mathcal{GP})$.
Our results extend and improve well-known results in [2, 4, 17]. In the last
section of this manuscript, we apply Theorem 3.2 to discuss the existence of
solutions for a system of fractional integral equations $(\ref{Sy3})$ and an
example is given to explain the applicability of the results.

##  2\.  Preliminaries and Results

Let $(X,\|\cdot\|)$ be a generalized Banach space in the sense of Perov such
that the vector-valued norm $\|\cdot\|:X\longrightarrow\mathbb{R}^{n}_{+}$ is
given by

$\|x\|=\left(\begin{array}[]{ccc}\|x\|_{1}\\\ \vdots\\\ \|x\|_{n}\\\
\end{array}\right),\ \ x\in X$

with $\|\cdot\|_{i},i=1,...,n$ define $n$ norm on $X.$ We denote by $\theta$
the zero element of $X$ and ${B}_{r}=B(\theta,r)$ the closed ball centered at
$\theta$ with radius $r\in\mathbb{R}^{n}_{+}.$ For more details, the reader
may consult the monograph of John R. Graef, Johnny Henderson and Abdelghani
Ouahab [13].
Let $(Y_{i},\tau_{i})_{i\in I}$ be a family of topological spaces and let
$f_{i}:X\longrightarrow Y_{i},\ \ i\in I$ be a linear and continuous mappings.
We define the smallest topology on $X$ such that all the mappings $f_{i}$
remain continuous with respect to this topology. Its basic open sets are of
the form $\displaystyle\cap_{i\in J}f_{i}^{-1}u_{i},$ with $J$ a finite subset
of $I$ and $u_{i}\in\tau_{i}$ for each $i.$ This topology is called the weak
topology on $X$ generated by the $(f_{i})_{i\in I}$ and we denote it by
$\sigma(X,(f_{i})_{i\in I}).$ So that a sequence $(x_{n})_{n}$ in $X$
converges to $x$ in $\sigma(X,(f_{i})_{i\in I})$ if and only if
$(f_{i}(x_{n}))_{n}$ converges to $f_{i}(x),$ for all $i\in I.$ In generalized
Banach spaces, weakly open subset, weakly closed subset and weak compactness,
are similar to those for usual Banach spaces. We denote by $\mathcal{B}(X)$
the collection of all nonempty bounded subsets of X and $\mathcal{W}(X)$ is a
subset of $\mathcal{B}(X)$ consisting of all weakly compact subset of $X.$ If
$x,y\in\mathbb{R}^{n},$ $x=(x_{1},...,x_{n}),\ \ y=(y_{1},...,y_{n})$ by
$x\leq y$ we means $x_{i}\leq y_{i}$ for all $i=1,...,n.$

######  Definition 2.1.

A square matrix of real numbers $M$ is said to be convergent to zero if and
only if its spectral radius $\rho(M)$ is strictly less than $1.$ In other
words, $|\lambda|<1,$ for every $\lambda\in\mathbb{C},$ with $det(M-\lambda
I)=0,$ where $I$ denotes the unit matrix of $\mathcal{M}_{n\times
n}(\mathbb{R}_{+}).$ $\diamondsuit$

######  Lemma 2.1.

[24] Let $M$ The following assertions are equivalent:
$(i)$ $M$ is a matrix convergent to zero,
$(ii)$ $M^{k}\rightarrow 0$ as $k\rightarrow\infty,$
$(iii)$ The matrix $(I-M)$ is nonsingular and

$(I-M)^{-1}=I+M+M^{2}+...+M^{k}+...,$

$(iv)$ The matrix $(I-M)$ is nonsingular and $(I-M)^{-1}$ has nonnegative
elements. $\diamondsuit$

######  Definition 2.2.

[23] Let $X$ be a generalized Banach space. An operator $T:X\longrightarrow X$
is said to be contractive if there exists a matrix $M\in\mathcal{M}_{n\times
n}(\mathbb{R}_{+})$ convergent to zero such that

$\|Tx-Ty\|\leq M\|x-y\|,\ \ \text{ for all }x,y\\\ \text{ in }X.$

For $n=1$, we recover the classical Banach’s contraction fixed point result.
$\diamondsuit$

######  Definition 2.3.

An operator $T:X\longrightarrow X$ is called weakly sequentially continuous on
$X,$ if for every sequence $(x_{n})_{n\in\mathbb{N}}$ with
$x_{n}\rightharpoonup x\ \ \text{ we have }\ \ Tx_{n}\rightharpoonup Tx,$ here
$\rightharpoonup$ denotes weak convergence. $\diamondsuit$

######  Definition 2.4.

An operator $T:X\longrightarrow X$ is said to be weakly compact if $T(V)$ is
relatively weakly compact for every bounded subset $V\subseteq X.$
$\diamondsuit$

######  Theorem 2.1.

[25, 14] Let $X$ be a generalized Banach space, $S$ be a nonempty, compact and
convex subset of $X$ and $A:S\longrightarrow S$ be a continuous operator. Then
$A$ has at least a fixed point in $S.$ $\diamondsuit$

Now, we will prove the following theorem.

######  Theorem 2.2.

Let $S$ be a nonempty, closed and convex subset of a generalized Banach space
$X.$ Let $T:S\longrightarrow S$ be a weakly sequentially continuous mapping.
If $T(S)$ is relatively weakly compact, then $T$ has a fixed point.
$\diamondsuit$

Proof. Let $M=\overline{co}(T(S))$ the closed convex hull of $T(S).$ Because
$T(S)$ is relatively weakly compact, then $M$ is weakly compact convex subset
of $X.$ On the other hand, $T(M)\subset T(S)\subset\overline{co}(T(S)=M$ ie :
$T$ maps $M$ into itself. Since $T$ is weakly sequentially continuous on $M,$
we infer that $T$ is weakly continue on $M.$ (consider
$X=(X,\sigma(X,(f_{i})_{i\in{I}}))$ the space endowed with the weak topology
and note that $T:M\longrightarrow M$ is continuous with M is compact). The use
of Theorem 2.1 achieves the proof Q.E.D.

######  Definition 2.5.

the measure of weak non-compactness of the generalized Banach space $X$ is the
map $\mu:\mathcal{B}{(X)}\longrightarrow\mathbb{R}_{+}^{n}$ defined in the
following way

$\mu(S)=\inf\big{\\{}r\in\mathbb{R}^{n}_{+},\ \ \text{ there exists
}K\in\mathcal{W}(X)\text{ such that }S\subseteq K+{B}_{r}\big{\\}},$

for all $S\in\mathcal{B}(X).$ $\diamondsuit$

######  Lemma 2.2.

Let $S_{1}$, $S_{2}$ be two elements of $\mathcal{B}(X).$ Then the functional
$\mu$ has the properties:
$(i)$ $S_{1}\subseteq S_{2}$ implies $\mu(S_{1})\leq\mu(S_{2}),$
$(ii)$ $\mu(\overline{S_{1}^{w}})=\mu(S_{1}),$
$(iii)$ $\mu(S_{1})=0,$ if and only if,
$\overline{S_{1}^{w}}\in\mathcal{W}{(X)},$ where $\overline{S_{1}^{w}}$ is the
weak closure of $S_{1},$
$(iv)$ $\mu(co(S_{1}))=\mu(S_{1}),$ where $co(S_{1})$ is the convex hull of
$S_{1},$
$(v)$
$\mu(S_{1}+S_{2})\leq\mu(S_{1})+\mu(S_{2}),\mu(\\{a\\}+S_{1})\leq\mu(S_{1}),$
$(vi)$ $\mu(S_{1}\cup S_{2})=\max\\{\mu(S_{1}),\mu(S_{2})\\},$
$(vii)$ $\mu(\lambda S_{1})=|\lambda|\mu(S_{1}),$ for all
$\lambda\in\mathbb{R}.$
$\diamondsuit$

Proof. The statements $(i)$ and $(iv)$ \- $(viii)$ are immediate consequences
of the definition of $\mu$ . Let us prove $(ii)$. From the definition of $\mu$
there exists a subset $K_{1}\in\mathcal{W}(X)$ and
$r_{1}\in\mathbb{R}^{n}_{+}\setminus\\{0_{\mathbb{R}^{n}}\\}$ such that
$S_{1}\subseteq K_{1}+{B}(\theta,r_{1}+\mu(S_{1})),$ then
$S_{1}\subseteq\overline{co}{K_{1}}+{B}(\theta,r_{1}+\mu(S)).$ By the
Kerein-S̆mulian theorem $\overline{co}{K_{1}}$ is weakly compact. Since
${B}(\theta,r_{1}+\mu(S_{1}))$ is weakly closed, we infer that
$\overline{co}{K_{1}}+{B}(\theta,r_{1}+\mu(S_{1}))$ is weakly closed. So
$\overline{S}_{1}^{w}\subseteq\overline{co}{K_{1}}+{B}(\theta,r_{1}+\mu(S_{1}))$
implies that $\mu(\overline{S}_{1}^{w})\leq r_{1}+\mu(S_{1}).$ Letting
$r_{1}\to 0_{\mathbb{R}^{n}}$ in the above inequality, we get
$\mu(\overline{S}_{1}^{w})\leq\mu(S_{1}).$ The reverse inequality is obvious .
The proof of the ”only if” part of $(iii).$ By the definition of $\mu,$ there
exists a subset $K_{1}\in\mathcal{W}(X)$ and
$r_{1}\in\mathbb{R}^{n}_{+}\setminus\\{0_{\mathbb{R}^{n}}\\}$ such that

$\overline{S}_{1}^{w}\subseteq\overline{K}_{1}^{w}+{B}(\theta,r_{1}).$

Let $\\{x_{n}\\}_{n\in\mathbb{N}}$ be a sequence on $\overline{S}_{1}^{w}.$ So
there existe sebsequences $\\{y_{n}\\}_{n\in\mathbb{N}}$ and
$\\{z_{n}\\}_{n\in\mathbb{N}}$ of $\overline{K}_{1}^{w}$ and
${B}(\theta,r_{1})$ respectively, such that $x_{n}=y_{n}+z_{n}.$ If
$r_{1}\longrightarrow 0_{\mathbb{R}^{n}},$ we deduce that $\\{x_{n}\\}$ has a
weakly convergence subsequence, So $\overline{S}_{1}^{w}$ is weakly compact,
while the ”if” part is trivial.
Q.E.D.

For $n=1$, we recover the classical De Blasi measure of weak non-compactness.

######  Definition 2.6.

Let $S$ be a nonempty subset of a generalized Banach space $X$ and
$M\in\mathcal{M}_{n\times n}(\mathbb{R}_{+})$ is a matrix convergent to zero.
If $T$ maps $S$ into $X,$ we say that
$(i)$ $T$ is $M$-set contraction with respect to $\mu$ if $T$ is bounded and
for any bounded subset $V\text{ in }S$ and $\mu(T(V))\leq M\mu(V),$
$(ii)$ $T$ is condensing with $\mu$ if $T$ is bounded and $\mu(T(V))<\mu(V)$
for all bounded subsets $V$ of $S$ with $\mu(V)>0_{\mathbb{R}^{n}}.$
$\diamondsuit$

######  Remark 2.1.

If we assume that $T$ is $M$-set contraction, then $T$ is condensing. Indeed,
let $V$ be a bounded subset on $S$ with $\mu(V)>0_{\mathbb{R}^{n}}.$ We claim
that $M\mu(V)<\mu(V).$ If not we obtain $(I-M)\mu(V)\leq 0_{\mathbb{R}^{n}}.$
Since $(I-M)^{-1}$ has nonnegative elements , it follows that
$\mu(V)=0_{\mathbb{R}^{n}}.$ which is a contradiction with
$\mu(V)>0_{\mathbb{R}^{n}}.$ $\diamondsuit$

By using the concept of a measure of weak noncompactness in vector-valued
Banach spaces, we obtain the following fixed point theorem.

######  Theorem 2.3.

Let $S$ be a nonempty closed convex subset of a generalized Banach space $X.$
Let $T:S\longrightarrow S$ be a weakly seqeuntially continuous operator and
condensing with respect to $\mu.$ If $T(S)$ is bounded, then $T$ has a fixed
point in $S.$ $\diamondsuit$

Proof. let $x_{0}$ be fixed in $S$ and define the set

$\Sigma=\\{K:\ \ x_{0}\in K\subseteq S,\ \ K\text{ is closed, convex, bonded
and }T(K)\subseteq K\\}.$

Clearly, $\Sigma\neq\emptyset$ since
$\overline{co}(T(S)\cup\\{x_{0}\\})\subseteq S$ and we have

$T(\overline{co}(T(S)\cup\\{x_{0}\\})\subseteq
T(S)\subseteq\overline{co}(T(S)\cup\\{x_{0}\\}).$

Which shows that $\overline{co}(T(S)\cup\\{x_{0}\\})\in\Sigma.$ If we consider
$M=\displaystyle\cap_{K\in\Sigma}K,$ then $x_{0}\in M\subseteq S,M$ is also a
closed convex subset and $T(M)\subseteq M.$ Therefore, we have that
$M\in\Sigma.$ We will prove that $M$ is weakly compact. Denoting by
$M_{0}=\overline{co}(T(M)\cup{x_{0}})$, we have $M_{0}\subseteq M$, which
implies that $T(M_{0})\subseteq T(M)\subseteq M_{0}$. Therefore
$M_{0}\in\Sigma$, $M\subseteq M_{0}$. Hence $M=M_{0}.$ Since $M$ is weakly
closed, it suffices to show that $M$ is relatively weakly compact. If
$\mu(M)>0_{\mathbb{R}^{n}},$ we obtain

$\mu(M)=\mu(\overline{co}(T(M)\cup\\{x_{0}\\}))<\mu(M)$

which is a contradiction. Hence, $\mu(M)=0_{\mathbb{R}^{n}}$ and so $M$ is
relatively weakly compact. Now, $T$ is weakly continuous on $M.$ Consider
$X=(X,\sigma(X,(f_{i})_{i\in{I}}))$ the space endowed with the weak topology.
Hence, an application of Theorem 2.1 shows that $T$ has at least one fixed
point in $M.$ Q.E.D.

######  Definition 2.7.

[17] A generalized normed algebra $X$ is an algebra endowed with a norm
satisfying the following property

$\text{ for all }x,y\in X\ \ \|x.y\|\leq\|x\|\|y\|,$

where

$\|x.y\|=\left(\begin{array}[]{ccc}\|x.y\|_{1}\\\ \vdots\\\ \|x.y\|_{n}\\\
\end{array}\right)$

and

$\|x\|\|y\|=\left(\begin{array}[]{ccc}\|x\|_{1}\|y\|_{1}\\\ \vdots\\\
\|x\|_{n}\|y\|_{n}\\\ \end{array}\right).$

A complete generalized normed algebra is called a generalized Banach algebra.
$\diamondsuit$

Because the product of two sequentially weakly continuous functions in
generalized Banach algaebra is not necessarily sequentially weakly continuous,
we will introduce:

######  Definition 2.8.

we will say that the generalized Banach algebra $X$ satisfies conditions
$(\mathcal{GP})$ if
$(\mathcal{GP})\left\\{\begin{array}[]{ll}\mbox{\leavevmode\nobreak\ For any
sequences \leavevmode\nobreak\ }\\{x_{n}\\}\mbox{\leavevmode\nobreak\
and\leavevmode\nobreak\ }\\{y_{n}\\}\mbox{\leavevmode\nobreak\
of\leavevmode\nobreak\ }X\mbox{\leavevmode\nobreak\ such
that\leavevmode\nobreak\ }x_{n}\rightharpoonup x\mbox{\leavevmode\nobreak\
and\leavevmode\nobreak\ }y_{n}\rightharpoonup y,&\hbox{}\\\
\mbox{\leavevmode\nobreak\ then\leavevmode\nobreak\ }x_{n}\cdot
y_{n}\rightharpoonup x\cdot y;\mbox{\leavevmode\nobreak\
here\leavevmode\nobreak\ }\rightharpoonup\mbox{\leavevmode\nobreak\ denotes
weak convergence\leavevmode\nobreak\ }.&\hbox{}\end{array}\right.$
$\diamondsuit$

If $X$ is Banach algebra, we recover the classical sequential condition
$(\mathcal{P})$ [[4],Definition $3.1$]

######  Lemma 2.3.

If $K$ and $K^{\prime}$ are weakly compact subsets of generalized Banach
algebra $X$ satisfying the condition $(\mathcal{GP}),$ then $K\cdot
K^{\prime}$ is weakly compact. $\diamondsuit$

Proof. We will show that $K\cdot K^{\prime}$ is weakly sequentially compact.
For that, let $\\{z_{n}\\}_{n}$ be any sequence of $K\cdot K^{\prime}.$ So,
there exist sequences $\\{x_{n}\\}_{n}$ and $\\{x^{\prime}_{n}\\}_{n}$ of $K$
and $K^{\prime}$ respectively. By hypothesis, there is a renamed subsequence
$\\{x_{n}\\}_{n}$ such that $x_{n}\rightharpoonup x\in K.$ Again, there is a
renamed subsequence $\\{x^{\prime}_{n}\\}_{n}$ of $K^{\prime}$ such that
$x^{\prime}_{n}\rightharpoonup x^{\prime}\in K.$ This, together with condition
$(\mathcal{GP})$ yields that

$z_{n}\rightharpoonup z=x\cdot x^{\prime}.$

This implies that $K\cdot K^{\prime}$ is weakly sequentially compact. Hence,
an application of the Eberlein-S̆mulian’s theorem yields that $K\cdot
K^{\prime}$ is weakly compact. Q.E.D.

##  3\.  Existence solutions of Equation (1.1)

In this section, we are prepared to state our first fixed point theorems in
generalized Banach algebra in order to found the existence of solutions for
the operator equation $(\ref{abc})$ under weak topology in the case where
$A,B$ and $C$ are weakly sequentially continuous operators.

######  Theorem 3.1.

Let $S$ be a nonempty, closed, and convex subset of a generalized Banach
algebra $X$ satisfying the sequential condition $(\mathcal{GP}).$ Assume that
$A,C:X\longrightarrow X$ and $B:S\longrightarrow X$ are three weakly
sequentially continuous operators satisfying the following conditions:
$(i)$ $A$ and $C$ are contractive with Lipschitz matrix $M_{A}$ and $M_{C}$
respectively,
$(ii)$ $A$ is regular on $X,$ $($i.e. $A$ maps $X$ into the set of all
invertible elements of $X),$
$(iii)$ $B(S)$ is bounded,
$(iv)$ $x=Ax\cdot By+Cx,\ \ y\in S\Rightarrow x\in S,$ and
$(v)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ is weakly compact.
Then, the operator equation $($1.1$)$ has, at least, a solution in $S$ as soon
as $\|B(S)\|M_{A}+M_{C}\in\mathcal{M}_{n\times n}(\mathbb{R}_{+})$ is a matrix
converging to zero. $\diamondsuit$

######  Remark 3.1.

If $M_{A}=(a_{ij})_{1\leq i,j\leq n},\ \ M_{C}=(\overline{a}_{ij})_{1\leq
i,j\leq n}$ and $\|B(S)\|=(b_{j})_{1\leq j\leq n}$ then

$\|B(S)\|M_{A}+M_{C}=\left(\begin{array}[]{ccc}b_{1}a_{11}+\overline{a}_{11}&\ldots&b_{n}a_{1n}+\overline{a}_{1n}\\\
\vdots&\ddots&\vdots\\\
b_{1}a_{n1}+\overline{a}_{n1}&\ldots&b_{n}a_{nn}+\overline{a}_{nn},\\\
\end{array}\right).$

$\diamondsuit$

Proof of Theorem 3.1.  It is easy to check that the vector $x\in S$ is a
solution for the operator equation $x=Ax\cdot Bx+Cx,$ if and only if $x$ is a
fixed point for the inverse operator
$T:=\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}B.$ The use of assumption
$(i)$ as well as perov’s theorem [22] leads to for each $y\in S,$ there is a
unique $x_{y}\in X$ such that $x_{y}=Ty.$ Indeed, let $y\in S$ be arbitrary
and let us define the mapping $\varphi_{y}:X\longrightarrow X$ by the formula

$\varphi_{y}(x)=Ax\cdot By+Cx.$

Notice that $\varphi_{y}$ is contractive with a Lipschitz matrix
$\|B(S)\|M_{A}+M_{C}.$ Applying Perov’s theorem [22], we obtain that
$\varphi_{y}$ has a unique fixed point in $S,$ say $x_{y}.$ From assumption
$(ii),$ it follows that the operator
$\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ is well defined on $B(S).$
Let $\\{x_{n},\ n\in\mathbb{N}\\}$ be a weakly convergent sequence of $S$ to a
point $x\in S$ and let $y_{n}=Tx_{n}.$ Then the relation $y_{n}=Ay_{n}\cdot
Bx_{n}+Cy_{n}$ holds and, therefore $\\{y_{n},\ n\in\mathbb{N}\\}\subset S$ in
view of assumption $(iv).$ Since $T(S)$ is relatively weakly compact, there is
a subsequence $(x_{n_{k}})$ of $\\{x_{n},\ n\in\mathbb{N}\\}$ such that
$y_{n_{k}}=Tx_{n_{k}}\rightharpoonup y,$ for some $y\in X.$ The weak
sequential closedness of $S$ gives $y\in S.$ Making use of the condition
$(\mathcal{GP}),$ together with the assumptions on $A,B$ and $C,$ enables us
to have

$y_{n_{k}}=Ay_{n_{k}}\cdot Bx_{n_{k}}+Cy_{n_{k}}\rightharpoonup Ay\cdot
Bx+Cy.$

This implies that $y=Ay\cdot Bx+Cy.$ Consequently $Tx_{n_{k}}\rightharpoonup
Tx$ in light of assumption $(ii).$ Now a standard argument shows that
$Tx_{n}\rightharpoonup Tx.$ Suppose the contrary, then there exists a weakly
neighborhood $V^{w}$ of $Tx$ and a subsequence $(x_{n_{j}})_{j}$ of
$\\{x_{n},\ n\in\mathbb{N}\\}$ such that $Tx_{n_{j}}\notin V^{w},$ for all
$j\geq 1.$ Arguing as before, we may extract a subsequence
$(x_{n_{j_{k}}})_{k\in\mathbb{N}}$ of $\\{x_{n_{j}},\ j\in\mathbb{N}\\}$
verifying $Tx_{n_{j_{k}}}\rightharpoonup y,$ which is a contradiction and
consequently $T$ is weakly sequentially continuous. Hence, $T$ has, at least,
one fixed point $x$ in $S$ in view of Theorem 2.2. Q.E.D.

An interesting fixed point result of Theorem 3.1 is

######  Corollary 3.1.

Let $S$ be a nonempty, closed, and convex subset of a generalized Banach
algebra $X$ satisfying the sequential condition $(\mathcal{GP}).$ Assume that
$A,C:X\longrightarrow X$ and $B:S\longrightarrow X$ are three weakly
sequentially continuous operators satisfying the following conditions:
$(i)$ $A$ and $C$ are contractive with Lipschitz matrix $M_{A}$ and $M_{C}$
respectively,
$(ii)$ $A$ is regular on $X,$
$(iii)$ $A(S),B(S)$ and $C(S)$ are relatively weakly compact,
$(iv)$ $x=Ax{\cdot}By+Cx,\ \ y\in S\Rightarrow x\in S.$
Then, the operator equation $($1.1$)$ has, at least, a solution in $S$ as soon
as $\|B(S)\|M_{A}+M_{C}$ is a matrix converging to zero. $\diamondsuit$

######  Theorem 3.2.

Let $S$ be a nonempty, bounded, closed, and convex subset of a generalized
Banach algebra $X$ satisfying the sequential condition $(\mathcal{GP}).$
Assume that $A,C:X\longrightarrow X$ and $B:S\longrightarrow X$ are three
weakly sequentially continuous operators satisfying the following conditions:
$(i)$ $A$ and $C$ are contractive with Lipschitz matrix $M_{A}$ and $M_{C}$
respectively,
$(ii)$ $A$ is regular on $X$,
$(iii)$ $B$ is weakly compact,
$(iv)$ $x=Ax\cdot By+Cx,\ \ y\in S\Rightarrow x\in S.$
Then, the operator equation $($1.1$)$ has, at least, a solution in $S$ as soon
as $\|B(S)\|M_{A}+M_{C}$ is a matrix converging to zero. $\diamondsuit$

Before proving the theorem, we need to establish two lemmas.

######  Lemma 3.1.

Let $X$ be a generalized Banach algebra. If $S\in\mathcal{B}(X)$ and
$K\in\mathcal{W}(X),$ then $\mu\left(S\cdot K\right)\leq\|K\|\mu(S).$
$\diamondsuit$

Proof. Assume that $\|K\|>0_{\mathbb{R}^{n}}.$ By the definition of $\mu,$
there exists a subset $K^{\prime}$ of $\mathcal{W}(X)$ and
$r\in{\mathbb{R}^{n}_{+}}\setminus\\{0_{\mathbb{R}^{n}}\\}$ such that

$S\cdot K\subseteq K^{\prime}\cdot
K+{B}\left(\theta,\mu(S)+r\|K^{-1}\|\right)\cdot K,$

where

$r\|K^{-1}\|:=\left(\begin{array}[]{ccc}\dfrac{r_{1}}{\|K\|_{1}}\\\ \vdots\\\
\dfrac{r_{n}}{\|K\|}_{n}\\\ \end{array}\right).$

So

$S\cdot K\subseteq K^{\prime}\cdot K+{B}\left(\theta,\|K\|\mu(S)+r\right).$

Keeping in mind the subadditivity of the measure of weak noncompactness and
using lemma 2.3, we get

$\begin{array}[]{rcl}\displaystyle\mu(S\cdot
K)&\leq&\displaystyle\mu\left(K\cdot
K^{\prime}\right)+\mu\left({B}\left(0,\|K\|\mu(S)+r\right)\right)\\\ \\\
&\leq&\|K\|\mu(S)+r.\end{array}$

Since $\varepsilon$ is arbitrary, we deduce that

$\mu(S\cdot K)\leq\|K\|\mu(S).$

Q.E.D.

######  Lemma 3.2.

Let $X$ be a generalized Banach algebra. Assume that $T:X\longrightarrow X$ is
weakly sequentially continuous. If $T$ is contractive with Lipschitz matrix
$M\in\mathcal{M}_{n\times n}(\mathbb{R}_{+}),$ then for any bounded subset $S$
of $X,$ one has

$\mu\left(T(S)\right)\leq M\mu\left(S\right).$

$\diamondsuit$

Proof. Let $S$ be a bounded subset of $X$ and $\varepsilon\in\mathbb{R}^{n}$
such that $\varepsilon>\mu(S)$ and let $M=(a_{ij})_{1\leq i,j\leq n}.$ By the
definition of $\mu,$ we have there exists $r\in\mathbb{R}^{n}$ and a weakly
compact subset $K$ of $X$ such that $0<r<\varepsilon$ and $S\subseteq
K+{B}(\theta,{r}).$ Let $y\in T\left(K+{B}(\theta,{r})\right),$ then there
exists $x\in K+{B}(\theta,{r})$ such that $y=Tx.$ Since $x\in
K+{B}(\theta,{r}),$ there are $k\in K$ and $b\in{B}\left(\theta,{r})\right)$
such that $x=k+b,$ and so

$\|y-Tk\|_{i}=\|Tx-
Tk\|_{i}\leq\sum_{j=1}^{n}a_{ij}\|x-k\|_{j}=\sum_{j=1}^{n}a_{ij}\|b\|_{j}\leq\sum_{j=1}^{n}a_{ij}r_{j}.$

This means that

$\|y-Tk\|\leq\left(\begin{array}[]{c}\displaystyle\sum_{j=1}^{n}a_{1j}r_{j}\\\
\displaystyle\sum_{j=1}^{n}a_{2j}r_{j}\\\ \vdots\\\
\displaystyle\sum_{j=1}^{n}a_{nj}r_{j}\\\ \end{array}\right)=Mr.$

That is, $y\in TK+{B}(\theta,{Mr})$ and consequently $TS\subset
TK+{B}(\theta,{Mr}).$ Moreover, since $T$ is sequentially weakly continuous we
have $\overline{TK}^{w}\in\mathcal{W}(X).$ Accordingly,

$\mu(TS)\leq Mr.$

Letting $\varepsilon\to\mu(S)$ in the above inequality, we get $\mu(TS)\leq
M\mu(S).$ Q.E.D.

Proof of Theorem 3.2. Following the same procedures as in the proof of Theorem
3.1, it can be proved that the inverse operator
$T:=\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}B$ exists on $S.$ Now, we
claim that $T(S)$ is a relatively compact subset of $X.$ If this is not the
case, then $\mu(TS)>0_{\mathbb{R}^{n}}.$ Keeping in mind the subadditivity of
the De Blasi’s measure of weak noncompactness and using the equality:

(3.1) $\displaystyle\displaystyle T=AT\cdot B+CT$

we obtain

$\displaystyle\mu(T(S))$ $\displaystyle\leq\mu\left(AT(S)\cdot
B(S)\right)+\mu\left(CT(S)\right).$

The use of Lemma 3.1 as well as Lemma 3.2 leads to

$\displaystyle\mu(TS)$
$\displaystyle\leq\left(\|B(S)\|M_{A}+M_{C}\right)\mu(TS).$

Since $\|B(S)\|M_{A}+M_{C}$ is a matrix converging to zero, we get a
contradiction and consequently the claim is approved. Q.E.D.

As easy consequences of Theorem 3.2 we obtain the following result.

######  Corollary 3.2.

Let $S$ be a nonempty, bounded, closed and convex subset of a generalized
Banach algebra $X$ satisfying the sequential condition $(\mathcal{GP}).$
Assume that $A,C:X\longrightarrow X$ and $B:S\longrightarrow X$ are three
weakly sequentially continuous operators satisfying the following conditions:
$(i)$ $A$ is regular and is contractive with a Lipschitz matrix $M,$
$(ii)$ $B$ and $C$ are weakly compact,
$(iii)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ exists on $B(S),$
$(iv)$ $x=Ax\cdot By+Cx,\ \ y\in S\Rightarrow x\in S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S$ as
soon as $\|B(S)\|M$ is a matrix converging to zero. $\diamondsuit$

In the following result, we will consider that $A$ and $B$ are weakly compact
operators.

######  Theorem 3.3.

Let $S$ be a nonempty, bounded, closed and convex subset of a generalized
Banach algebra $X$ satisfying the condition $(\mathcal{GP}).$ Assume that
$A,B,C:S\longrightarrow X$ are three weakly sequentially continuous operators
satisfying the following conditions:
$(i)$ $A$ is regular,
$(ii)$ $A$ and $B$ are weakly compact,
$(iii)$ If $(I-C)x_{n}\rightharpoonup y,$ then there exists a weakly
convergent subsequence of $(x_{n})_{n},$
$(iv)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ exists on $B(S)$, and
$(v)$ $x=Ax\cdot By+Cx\in S,\ \ y\in S$ $\Rightarrow$ $x\in S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S.$
$\diamondsuit$

Proof. Let $\\{y_{n},\ n\in\mathbb{N}\\}\subset T(S),$ there is a sequence
$\\{x_{n},\ n\in\mathbb{N}\\}\subset S$ such that

$y_{n}=Tx_{n}=\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}Bx_{n}.$

Or equivalently, $y_{n}=Ay_{n}\cdot Bx_{n}+Cy_{n}.$ Taking into account the
weak compactness of the weak closure of $A(S)$ and $B(S),$ we infer that

$ATx_{n_{k}}\rightharpoonup x\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \textrm{ and }\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
Bx_{n_{k_{j}}}\rightharpoonup x^{\prime},\textrm{ for some }x,x^{\prime}\in
S,$

where $(x_{n_{k}})$ is a subsequence of $\\{x_{n},\ n\in\mathbb{N}\\}$ and
$(x_{n_{k_{j}}})$ is a subsequence of $\\{x_{n_{k}},\ k>n\\}.$ Using the
condition $(\mathcal{GP}),$ we get

$Ay_{n_{k_{j}}}\cdot Bx_{n_{k_{j}}}=(I-C)y_{n_{k_{j}}}\rightharpoonup x\cdot
x^{\prime}\ \ \text{ in }X.$

Based on assumption $(iii),$ it follows that there exists a weakly convergent
subsequence of $(y_{n_{k_{j}}})$ and consequently $T(S)$ is a relatively
weakly compact subset of $X.$ The use of Theorem 3.1 achieves the proof.
Q.E.D.

######  Remark 3.2.

If we assume that $C$ is sequentially weakly continuous and contractive with a
Lipschitz Matrix $M,$ then $C$ satisfies the condition $(iii)$ of Theorem 3.3.
In fact, let $\\{x_{n},\ n\in\mathbb{N}\\}$ be a sequence in $S$ such that
$(I-C)x_{n}\rightharpoonup y,$ for some $y\in X.$ Based on the subadditivity
of the De Blasi’s measure of weak non-compactness it is shown that

$\beta(\\{x_{n},\ n\in\mathbb{N}\\})\leq\beta(\\{(I-C)x_{n},\
n\in\mathbb{N}\\})+\beta(\\{Cx_{n},\ n\in\mathbb{N}\\}).$

If we consider the weak compactness of the weak closure of $\\{(I-C)x_{n},\
n\in\mathbb{N}\\}$ and deploy Lemma 3.2, we get

$\beta(\\{x_{n},\ n\in\mathbb{N}\\})\leq M\beta(\\{x_{n},\
n\in\mathbb{N}\\}).$

Since $M$ is a matrix converging to zero, then there is a weakly convergent
subsequence of $\\{x_{n},\ n\in\mathbb{N}\\}.$ $\diamondsuit$

A consequence of Theorem 3.3 is

######  Corollary 3.3.

Let $S$ be a nonempty, bounded, closed and convex subset of a generalized
Banach algebra $X$ satisfying the condition $(\mathcal{GP}).$ Assume that
$A,B,C:S\longrightarrow X$ are three weakly sequentially continuous operators
satisfying the following conditions:
$(i)$ $A$ is regular and $B$ is weakly compact,
$(ii)$ If $\left(\displaystyle\dfrac{I-C}{A}\right)x_{n}\rightharpoonup y,$
then there exists a weakly convergent subsequence of $(x_{n})_{n},$
$(iii)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ exists and
$\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}B(S)\subseteq S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S.$
$\diamondsuit$

Let us study the case where the operators $B$ and
$\left(I-\displaystyle\dfrac{I-C}{A}\right)$ are $M_{1}$-$\mu$-contraction and
$M_{2}$-$\mu$-contraction respectively.

######  Theorem 3.4.

Let $S$ be a nonempty, bounded, closed and convex subset of a generalized
Banach algebra $X.$ Assume that $A,C:X\longrightarrow X$ and
$B:S\longrightarrow X$ are three operators satisfying the following
conditions:
$(i)$ $A$ is regular,
$(ii)$ $B$ and $\left(I-\displaystyle\dfrac{I-C}{A}\right)$ are
$M_{1}$-$\mu$-contraction and $M_{2}$-$\mu$-contraction respectively,
$(iii)$ $B$ and $\left(\displaystyle\dfrac{I-C}{A}\right)$ are weakly
sequentially continuous,
$(iv)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ exists on $B(S),$ and
$(v)$ $x=Ax\cdot By+Cx,\ \ y\in S$ $\Rightarrow$ $x\in S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S$ as
soon as $(I-M_{2})^{-1}M_{1}$ is a matrix converging to zero. $\diamondsuit$

Proof. It is easy to see that the operator
$T:=\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}B:S\longrightarrow S$ is well
defined. Let $\\{x_{n},\ \ n\in\mathbb{N}\\}$ be a weakly convergent sequence
of $S$ to a point $x\in S.$ Keeping in mind the weak sequential continuity of
the operator $B$ and using the equality

(3.2) $\displaystyle T=B+\left(I-\dfrac{I-C}{A}\right)T,$

we obtain

$\begin{array}[]{rcl}\mu\left(\\{Tx_{n},\ \
n\in\mathbb{N}\\}\right)&\leq&\displaystyle\mu\left(Bx_{n},\ \
n\in\mathbb{N}\right)+\mu\left(\left(I-\dfrac{I-C}{A}\right)(\\{Tx_{n},\ \
n\in\mathbb{N}\\})\right)\\\ \\\ &\leq&M_{2}\mu\left(\\{Tx_{n},\ \
n\in\mathbb{N}\\}\right).\end{array}$

This inequality means, in particular, that $\\{Tx_{n},\ \ n\in\mathbb{N}\\}$
is relatively weakly compact. Consequently, there is a subsequence
$(x_{n_{k}})_{k}$ of $\\{Tx_{n},\ \ n\in\mathbb{N}\\}$ such that
$Tx_{n_{k}}\rightharpoonup y,$ fore some $y\in S.$ Making use of equality
$(\ref{mu}),$ together with the assumptions on $B$ and
$\left(\frac{I-C}{A}\right)$; enables us to have $y=Tx.$ Now a standard
argument shows that $Tx_{n}\rightharpoonup Tx.$ Suppose the contrary, then
there exists a weakly neighborhood $V^{w}$ of $Tx$ and a subsequence
$(x_{n_{j}})_{j}$ of $\\{x_{n},\ n\in\mathbb{N}\\}$ such that
$Tx_{n_{j}}\notin V^{w},$ for all $j\geq 1.$ Arguing as before, we may extract
a subsequence $(x_{n_{j_{k}}})_{k\in\mathbb{N}}$ of $\\{x_{n_{j}},\
j\in\mathbb{N}\\}$ verifying $Tx_{n_{j_{k}}}\rightharpoonup y,$ which is a
contradiction and consequently $T$ is sequentially weakly continuous. Next,
$T$ is $\mu$-condensing. In fact, let $V$ be a bounded subset of $S$ with
$\mu(V)>0.$ Using the subadditivity of the De Blasi’s measure of weak
noncompactness, we get

$\begin{array}[]{rcl}\mu\left(T(V)\right)&\leq&\mu\left(B(V)\right)+\mu\left(\left(I-\dfrac{I-C}{A}\right)T(V)\right)\\\
\\\ &\leq&M_{1}\mu(V)+M_{2}\mu(T(V)).\end{array}$

This implies that

$\mu\left(T(V)\right)\leq(I-M_{2})^{-1}M_{1}\mu\left(V\right).$

Hence, $T$ has, at least, one fixed point $x$ in $S$ in view of Theorem 2.3 .
Q.E.D.

If we take $A=1_{X}$ in the above result, where $1_{X}$ is the unit element of
the generalized Banach algebra $X$, we obtain the following Corollary.

######  Corollary 3.4.

Let $S$ be a nonempty, bounded, closed and convex subset of a generalized
Banach algebra $X.$ Assume that $C:X\longrightarrow X$ and $B:S\longrightarrow
X$ are two sequentially weakly continuous operators satisfying the following
conditions:
$(i)$ $B$ and $C$ are $M_{1}$-$\mu$-contraction and $M_{2}$-$\mu$-contraction
respectively,
$(ii)$ $\left(I-C\right)^{-1}$ exists on $B(S),$
$(iii)$ $x=By+Cx,\ \ y\in S$ $\Rightarrow$ $x\in S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S$ as
soon as $(I-M_{2})^{-1}M_{1}$ is a matrix converging to zero. $\diamondsuit$

######  Remark 3.3.

Note that condition $(iii)$ in Theorem 3.4 may be replaced by ”$A,B$ and $C$
are weakly sequentially continuous”, but the generalized Banach algebra must
satisfy condition $(\mathcal{GP}).$ Now, we can study the folllwing result.
$\diamondsuit$

######  Theorem 3.5.

Let $S$ be a nonempty, bounded, closed, and convex subset of a generalized
Banach algebra $X$ satisfying the sequential condition $(\mathcal{GP}).$
Assume that $A,C:X\longrightarrow X$ and $B:S\longrightarrow X$ are three
weakly sequentially continuous operators satisfying the following conditions:
$(i)$ $A$ is regular and weakly compact,
$(ii)$ $B$ and $C$ are $M_{1}$-$\mu$-contraction and $M_{2}$-$\mu$-contraction
respectively,
$(iii)$ $\left(\displaystyle\dfrac{I-C}{A}\right)^{-1}$ exists on $B(S),$ and
$(iv)$ $x=Ax\cdot By+Cx,\ \ y\in S$ $\Rightarrow$ $x\in S.$
Then, the operator equation $($1.1$)$ has, at least, one solution in $S$ as
soon as $(I-M_{2})^{-1}\|A(S)\|M_{1}$ is a matrix converging to zero.
$\diamondsuit$

##  4\. Integral Equations of Fractional Order

Let $(X,\|\cdot\|)$ be a reflexive Banach and let ${C}(J,{X})$ be the Banach
algebra of all $X$-valued continuous functions defined on $J=[0,1]$, endowed
with the norm $\|f\|_{\infty}=\displaystyle\sup_{t\in J}\|f(t)\|.$ We will use
Theorem 3.2 to examine the existence of solutions to the coupled system of
quadratic integral equations of fractional order $(\ref{Sy3}).$ We need the
following definition and proposition in the sequel.

######  Definition 4.1.

[1] Let $f:[0,T]\rightarrow X$ be a function. The fractional Pettis integral
of the function $f$ of order $\alpha\in\mathbb{R_{+}}$ is defined by

$I^{\alpha}f(t)=\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s)ds,$

where the sign $"\int"$ denotes the Pettis integral.

######  Proposition 4.1.

[1] If $f:[0,T]\rightarrow X$ is Riemann integrable on $[0,T],$ then
$I^{\alpha}f$ exists on $[0,T]$ and fractional Pettis integral.

Let us now introduce the following assumptions:
($\mathcal{H}_{0}$) The function $f_{i}:J\times X\times X\longrightarrow
X,i=1,2$ is such that:
$($a$)$ The partial function $x\longrightarrow f_{1}(t,x,y)$ is regular on $X$
$($b$)$ The partial function $y\longrightarrow f_{2}(t,x,y)$ is regular on $X$
$($c$)$ The partial function $t\mapsto f_{i}(t,x,y)$ is continuous,
$($d$)$ The partial function $(x,y)\mapsto f_{i}(t,x,y)$ is weakly
sequentially continuous,
$($e$)$ There is nonnegative real numbers $a_{i1}$ and $a_{i2},$ $i=1,2$ such
that

$\|f_{i}(t,x,y)-f_{i}(t,\tilde{x},\tilde{y})\|\leq
a_{i1}\|x-\tilde{x}\|+a_{i2}\|y-\tilde{y}\|.$

($\mathcal{H}_{1}$) The function $g_{i}:J\times X\times X\longrightarrow
X,i=1,2$ is such that:
$($a$)$The partial function $t\mapsto g_{i}(t,x,y)$ is Riemann integrable,
$($b$)$ The partial function $(x,y)\mapsto g_{i}(t,x,y)$ is weakly
sequentially continuous.
($\mathcal{H}_{2}$) The function $h_{i}^{k}:J\times X\times X\longrightarrow
X,$ $k={1,\ldots,m}$ is such that:
$($a$)$The partial function $t\mapsto h_{i}^{k}(t,x,y)$ is Riemann integrable,
$($b$)$ The partial function $(x,y)\mapsto h_{i}^{k}(t,x,y)$ is weakly
sequentially continuous,
$($c$)$ There is nonnegative real numbers ${b}_{i1}^{k}$ and ${b}_{i2}^{k}$
such that

$\|h_{i}^{k}(t,x,y)-h_{i}^{k}(t,\tilde{x},\tilde{y})\|\leq{b}_{i1}^{k}\|x-\tilde{x}\|+{b}_{i2}^{k}\|y-\tilde{y}\|.$

######  Theorem 4.1.

Suppose that the assumptions $(\mathcal{H}_{0})-(\mathcal{H}_{3})$ are
satisfied. Moreover, assume that there exists a real number $r_{0}>0$ and
$P\in\mathbb{R}^{\ast}_{+}$ such that

(4.6)
$\displaystyle\displaystyle\left\\{\begin{array}[]{lll}\|g_{i}(t,s,x,y)\|\leq
P\ \ \text{ with }\|x\|\leq r_{0}\text{ and }\|y\|\leq r_{0}\\\ \\\
\rho\big{[}P\big{(}\frac{T^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}+\frac{T^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}\big{)}+\sum_{k=1}^{m}\big{(}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}+\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}\big{)}\big{]}<1,\textrm{
and }\\\ \\\ M_{A},M_{C}\textrm{ and }\|B(S)\|M_{A}+M_{C}\textrm{ are three
matrices converging to zero, where }\end{array}\right.$
$\rho=\max\\{a_{11},a_{12},a_{21},{a_{22}},{b}_{11}^{k},{b}_{12}^{k},{b}_{21}^{K},{b}_{21}^{k},{b}_{22}^{k};\
\ k=1,\ldots,m\\},$ $M_{A}=\left(\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\\\ \end{array}\right),\ \
M_{C}=\left(\begin{array}[]{cc}\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{11}^{k}}&\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{12}^{k}}\\\
\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{21}^{k}}&\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{22}^{k}}\\\
\end{array}\right)\texttt{ and }$ $\ \
\|B(S)\|M_{A}+M_{C}=\left(\begin{array}[]{cc}\dfrac{PT^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}a_{11}+\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{11}^{k}}&\dfrac{PT^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}a_{12}+\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{12}^{k}}\\\
\dfrac{PT^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}a_{21}+\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{21}^{k}}&\dfrac{PT^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}a_{22}+\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{22}^{k}}\\\
\end{array}\right).$

Then the problem $(\ref{Sy3})$ has a solution in $C(J,X)\times C(J,X).$
$\diamondsuit$

Proof. We recall that the problem $(\ref{Sy3}$) is equivalent to the operator
equation

$(x,y)=(A_{1}(x,y),A_{2}(x,y))(B_{1}(x,y),B_{2}(x,y))+(C_{1}(x,y),C_{2}(x,y)),$

where the operators $A_{i},B_{i}$ and $C_{i},\ \ i=1,2$ are defined by

$\left\\{\begin{array}[]{ll}A_{i}(x,y)(t)=f_{i}(t,x(t),y(t))\\\ \\\
B_{i}(x,y)(t)=\int_{0}^{t}\frac{(t-s)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})}g_{i}(s,x(s),y(s))ds\\\
\\\
C_{i}(x,y)(t)=\sum_{k=1}^{m}I^{\beta_{1}^{i}}h_{i}^{k}(t,x(t),y(t)).\end{array}\right.$

Let us define the subset $S$ of $C(J,X)\times C(J,X)$ by

$S=\left\\{(x,y)\in C(J,X)\times C(J,X),\ \
\|(x,y)\|=\left(\begin{array}[]{cc}\|x\|_{\infty}\\\ \|y\|_{\infty}\\\
\end{array}\right)\leq\left(\begin{array}[]{cc}r_{0}\\\ r_{0}\\\
\end{array}\right)\right\\},$

where

$r_{0}\geq\frac{F_{0}P\big{(}\frac{T^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}+\frac{T^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}\big{)}+H_{0}\sum_{k=1}^{m}\big{(}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}+\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}\big{)}}{1-\rho\big{[}P\big{(}\frac{T^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}+\frac{T^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}\big{)}+\sum_{k=1}^{m}\big{(}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}+\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}\big{)}\big{]}},$

with

$\left\\{\begin{array}[]{ll}F_{0}=\max\
\big{\\{}\|f_{1}(\cdot,0,0)\|,\|f_{2}(\cdot,0,0)\|\big{\\}}\\\ H_{0}=\max\
\big{\\{}\|h_{1}^{k}(\cdot,0,0)\|,\|h_{2}^{k}(\cdot,0,0)\|;\ \
k=1,\ldots,m\big{\\}}\\\ \end{array}\right.$

Our strategy is to apply Theorem 3.2 to prove the existence of a fixed point
for the nonlinear equation $(\ref{Sy3})$ in $S.$ Then, we need to verify the
following steps:
Claim 1: We start by showing that the operators $A,C:C(J,X)\times
C(J,X)\longrightarrow C(J,X)\times C(J,X)$ and $B:S\longrightarrow
C(J,X)\times C(J,X)$ are weakly sequentially continuous. Firstly, we verify
that the operator $A_{i}(x,y),\text{ for }i=1,2$ is continuous on $J$ for all
$(x,y)\in C(J,X)\times C(J,X).$ To see this, let $\\{t_{n},\ \
n\in\mathbb{N}\\}$ be any sequence in J converging to a point $t$ in $J.$
Then,

$\begin{array}[]{rcl}\|A_{i}(x,y)(t_{n})-A_{i}(x,y)(t)\|&=&\|f_{i}(t_{n},x(t_{n}),y(t_{n}))-f_{i}(t,x(t),y(t))\|\\\
\\\ &\leq&a_{i1}\|x(t_{n})-x(t)\|+a_{i2}\|y(t_{n})-y(t)\|\\\ \\\
&+&\|f_{i}(t_{n},x(t),y(t)-f_{i}(t,x(t),y(t)\|.\end{array}$

The continuity of $x,y$ and $t\mapsto f_{i}(t,x,y)$ on $[0,1]$ implies that
the function $A_{i}(x,y)$ is continuous. Let
$\\{(x_{n},y_{n}),n\in\mathbb{N}\\}$ be a weakly convergent sequence of
$C(J,X)\times C(J,X)$ to a point $(x,y).$ In this case, the set
$\\{(x_{n},y_{n}),n\in\mathbb{N}\\}$ is bounded and so, we can apply the
Dobrakov’s theorem [12] in order to get

$(x_{n}(t),y_{n}(t))\rightharpoonup(x(t),y(t))\text{ in }X\times X.$

Based on assumption $(\mathcal{H}_{0})(d),$ it is shown that
$A_{i}(x_{n},y_{n})(t)\rightharpoonup A_{i}(x,y)(t)$ and then, we can again
apply the Dobrakov’s theorem to obtain the weak sequential continuity of the
operator $A.$ Besides, the use of assumption $(\mathcal{H}_{1})(b)$ and
assumption $(\mathcal{H}_{2})(b)$ as well as the Dobrakov’s theorem [12, page
36] leads to the two maps $B$ and $C$ are weakly sequentially continuous.
Claim 2: The operators $A$ and $C$ are contractive. The claim regarding the
operator $A$ is immediate, from assumption $(\mathcal{H}_{0})(e).$ Let us fix
arbitrary $(x,y),(\tilde{x},\tilde{y})\in C(J,X)\times C(J,X).$ If we take an
arbitrary $t\in[0,1],$ then we get

$\begin{array}[]{rcl}\|C_{i}(x,y)(t)-C_{i}(\tilde{x},\tilde{y})(t)\|&=&\left\|\displaystyle\sum_{k=1}^{m}I^{\beta_{i}^{k}}h_{i}^{k}(t,x(t),y(t)-\displaystyle\sum_{k=1}^{m}I^{\beta_{i}^{k}}h_{i}^{k}(t,\tilde{x}(t),\tilde{y}(t))\right\|\\\
\\\
&\leq&\displaystyle\sum_{k=1}^{m}I^{\beta_{i}^{k}}\|h_{i}^{k}(t,x(t),y(t)-h_{i}^{k}(t,\tilde{x}(t),\tilde{y}(t))\|\\\
\\\
&\leq&\displaystyle\sum_{k=1}^{m}I^{\beta_{i}^{k}}\left(b_{i1}^{k}\|x(t)-\tilde{x}(t)\|+b_{i1}^{k}\|y(t)-\tilde{y}(t)\|\right)\\\
\\\
&\leq&\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{i}^{k}}}{\Gamma(\beta_{i}^{k}+1)}\left(b_{i1}^{k}\|x(t)-\tilde{x}(t)\|+b_{i1}^{k}\|y(t)-\tilde{y}(t)\|\right).\end{array}$

This implies that $\left\|C(x,y)-C(\tilde{x},\tilde{y})\right\|\leq
M_{C}\left\|(x,y)-(\tilde{x},\tilde{y})\right\|,$ where

$M_{C}=\left(\begin{array}[]{cc}\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{11}^{k}}&\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{12}^{k}}\\\
\\\
\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{21}^{k}}&\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{22}^{k}}\end{array}\right).$

Claim 3: Let $\varepsilon>0,\ \ (x,y)\in S$ and $t,t^{\prime}\in J$ such that
$|t^{\prime}-t|<\varepsilon.$ From the Hahn-Banach theorem there exists a
linear function $\phi\in X^{\ast}$ with $\|\phi\|=1$ such that

$\left\|B_{i}(x,y)(t^{\prime})-B_{i}(x,y)(t)\right\|=\phi\left(B_{i}(x,y)(t)-B_{i}(x,y)(t)\right),\
\ i=1,2.$

Based on the first inequality in (4.6) it is shown that

$\begin{array}[]{rcl}\left\|B_{i}(x,y)(t^{\prime})-B_{i}(x,y)(t)\right\|&=&\phi\left(\displaystyle\int_{0}^{t^{\prime}}\frac{(t-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{j})}g_{i}(s,x(s),y(s))ds-\displaystyle\int_{0}^{t}\frac{(t-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{i})}g_{i}(s,x(s),y(s))\right)\\\
\\\
&\leq&\displaystyle\int_{0}^{t}\frac{|(t^{\prime}-s)^{\alpha_{i}-1}-(t-s)^{\alpha_{i}-1}|}{\Gamma(\alpha_{i})}\phi(g_{i}(s,x(s),y(s)))ds\\\
\\\
&+&\displaystyle\int_{t}^{t^{\prime}}\frac{(t^{\prime}-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{i})}\phi(g_{i}(s,x(s),y(s)))ds\\\
\\\
&\leq&P\displaystyle\int_{0}^{t}\frac{|(t^{\prime}-s)^{\alpha_{i}-1}-(t-s)^{\alpha_{i}-1}|}{\Gamma(\alpha_{i})}ds+P\displaystyle\int_{t}^{t^{\prime}}\frac{(t^{\prime}-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{i})}ds.\end{array}$

This implies that $\left\|B_{i}(x,y)(t^{\prime})-B_{i}(x,y)(t)\right\|\to
0\text{ as }\varepsilon\to 0,$ and consequently $B(S)$ is a weakly equi-
continuous subset. Let now $\\{(x_{n},y_{n}),n\in\mathbb{N}\\}$ be any
sequence in $S.$ From the first inequality in (4.6), it follows that

$\begin{array}[]{rcl}\left\|B_{i}(t,x_{n}(t),y_{n}(t))\right\|&\leq&\displaystyle\int_{0}^{t}\frac{(t-s)^{\alpha_{i}-1}}{\Gamma(\alpha_{i})}\left\|g_{i}(s,x_{n}(s),y_{n}(s))\right\|ds\\\
\\\
&\leq&\displaystyle\frac{PT^{\alpha_{i}}}{\Gamma(\alpha_{i}+1)},\end{array}$

for all $t\in[0,1].$ This demonstrate that $\\{B_{i}(x_{n},y_{n}),\ \
n\in\mathbb{N}\\}$ is a uniformly bounded sequence in $B(S)$ and so, $B(S)(t)$
is sequentially relatively weakly compact. Hence, $B(S)$ is sequentially
relatively weakly compact in light of the Arzelà-Ascoli’s theorem [26]. An
application of Eberlein-S̆mulian’s theorem [8] yields that $B(S)$ is
relatively weakly compact. Claim 4: The operators $A,B$ and $C$ satisfy
assumption $(iv)$ of Theorem $\ref{3}.$ To see this, let $(x,y)\in
C(J,X)\times C(J,X)$ and $(u,v)\in S$ with $(x,y)=A(x,y)\cdot B(u,v)+C(x,y).$
We shall show that $(x,y)\in S.$ For all $t\in[0,1],$ we have

$\begin{array}[]{rcl}\|x(t)\|&\leq&\left\|f_{1}(t,x(t),y(t))\right\|\displaystyle\int_{0}^{t}\displaystyle\frac{(t-s)^{\alpha_{1}-1}}{\Gamma(\alpha_{1})}\left\|g_{1}(s,u(s),v(s))\right\|ds+\displaystyle\sum_{i=1}^{m}I^{\beta_{1}^{i}}\|h_{1}^{i}(t,x(t),y(t))\|\\\
\\\
&\leq&\left[\rho(\|x\|_{\infty}+\|y\|_{\infty})+F_{0}\right]\displaystyle\frac{PT^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}+\displaystyle\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}\left[\rho(\|x\|_{\infty}+\|y\|_{\infty})+H_{0}\right]\end{array}$

This implies that

$\begin{array}[]{rcl}\|x(t)\|+\|y(t)\|&\leq&\big{[}\rho(\|x\|_{\infty}+\|y\|_{\infty})+F_{0}\big{]}P\left(\frac{T^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}+\frac{T^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}\right)\\\
&+&\sum_{k=1}^{m}\left(\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}+\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}\right)\big{[}\rho(\|x\|_{\infty}+\|y\|_{\infty})+H_{0}\big{]}\end{array}$

Consequently,

$\|x\|_{\infty}\leq r_{0}\text{ and }\|y\|_{\infty}\leq r_{0}.$

To end the proof, we apply Theorem $(\ref{mple})$, we deduce that the problem
$(\ref{Sy3})$ has, at least, one solution in $C(J,X)\times C(J,X).$
Example : Let $C(J,\mathbb{R})$ be the Banach algebra of all continuous
functions from $J$ to $\mathbb{R}$ endowed with the sup-norm $\|\cdot\|$
defined by $\|x\|_{\infty}=\sup\limits_{0\leq t\leq T}|x(t)|,$ for each $x\in
C(J,\mathbb{R}).$

(4.7)
$\left\\{\begin{array}[]{lll}D^{\frac{1}{2}}\big{[}\frac{x(t)-\sum_{k=1}^{2}I^{\beta_{1}^{k}}h_{1}^{k}(t,x(t),y(t))}{f_{1}(t,x(t),y(t))}\big{]}=\frac{3}{35(13-t^{2})}(7|x(t)|+15|y(t)|),&\hbox{}\\\
\\\
D^{\frac{1}{2}}\big{[}\frac{y(t)-\sum_{k=1}^{2}I^{\beta_{2}^{k}}h_{2}^{k}(t,x(t),y(t))}{f_{2}(t,x(t),y(t))}\big{]}=\frac{3}{35(13-t^{2})}(7|x(t)|+15|y(t)|),&\hbox{}\\\
\\\ x(0)=y(0)=0,\ \ \hfill{t\in[0,1]}.\end{array}\right.$

Note that this problem may be transformed into the fixed point problem
$(\ref{Sy3})$ in view of lemma $2.5$ in [16], where

$\sum_{k=1}^{2}I^{\beta_{1}^{k}}h_{1}^{k}(t,x(t),y(t))=I^{1/3}\frac{2te^{-3t}}{15(3+t)}\left(\frac{x^{2}(t)+9|x(t)|}{|x(t)|+5}+\frac{12e^{3t}}{5}\right)+I^{10/3}\frac{2t\sin\pi
t}{14+t^{2}}\left(\frac{x^{2}(t)+5|x(t)|}{|x(t)|+8}+\frac{1}{3}\right),$
$\sum_{k=1}^{2}I^{\beta_{2}^{k}}h_{2}^{k}(t,x(t),y(t))=I^{7/4}\frac{t\sin
t}{7(4+e^{t})}\left(\frac{y^{2}(t)+4|y(t)|}{|y(t)|+3}+\cos
t\right)+I^{29/6}\frac{3t\cos
t}{10(4-t^{2})}\left(\frac{y^{2}(t)+5|y(t)|}{|y(t)|+4}+\frac{t}{t+2}\right),$

and

$f_{1}(t,x(t),y(t))=\dfrac{3cos\pi
t+2t}{5(2+10t^{2})(|x(t)|+3)},f_{2}(t,x(t),y(t))=\dfrac{4cos\pi
t+3t}{7(3+8t^{2})(|y(t)|+6)},$

and

$g_{1}(t,x(t),y(t))=g_{2}(t,x(t),y(t))=\frac{3}{35(13-t^{2})}(7|x(t)|+15|y(t)|,$

here
$\alpha_{1}=\alpha_{2}=\dfrac{1}{2},T=1,m=2,\beta_{1}^{1}=\dfrac{1}{3},\beta_{1}^{2}=\dfrac{10}{3},\beta_{2}^{1}=\dfrac{7}{4}$
and $\beta_{2}^{2}=\dfrac{29}{6}.$
We can show that

$\displaystyle|f_{1}(t,x,y)-f_{1}(t,\tilde{x},\tilde{y})|\leq\frac{3+2t}{5(2+10t^{2})}|x-\tilde{x}|,$
$\displaystyle|f_{2}(t,x,y)-f_{2}(t,\tilde{x},\tilde{y})|\leq\frac{4+3t}{7(1+5t^{2})}|y-\tilde{y}|,$

and

$\displaystyle|h_{1}^{1}(t,x,y)-h_{1}^{1}(t,\tilde{x},\tilde{y})|\leq\frac{18t}{75(3+t)}|x-\tilde{x}|,$
$\displaystyle|h_{1}^{2}(t,x,y)-h_{1}^{2}(t,\tilde{x},\tilde{y})|\leq\frac{10t}{8(14+t^{2})}|x-\tilde{x}|,$
$\displaystyle|h_{2}^{1}(t,x,y)-h_{2}^{1}(t,\tilde{x},\tilde{y})|\leq\frac{4t}{21(4+e^{t})}|y-\tilde{y}|,$
$\displaystyle|h_{2}^{2}(t,x,y)-h_{2}^{2}(t,x,\tilde{y})|\leq\frac{3t}{8(4-t^{2})}|y-\tilde{y}|.$

It follows that $a_{11}=\dfrac{1}{12},\ \ a_{21}=0,\ \ a_{12}=0,\ \
a_{22}=\dfrac{1}{6},\ \ b^{1}_{11}=\dfrac{3}{50},\ \ b^{1}_{12}=0,\ \
b_{21}^{1}=0,\ \ b_{22}^{1}=\dfrac{4}{21(4+e)},\ \ b_{11}^{2}=\dfrac{1}{12},\
\ b_{12}^{2}=0,\ \ b_{21}^{2}=0,\ \ b_{22}^{2}=\dfrac{1}{8}.$ Now

$|g_{1}(t,x,y)|=|g_{2}(t,x,y)|\leq\dfrac{1}{4}$

It is easy to verify that $\rho=\dfrac{1}{6},\ \ F_{0}=\dfrac{1}{36},\ \
H_{0}=\dfrac{2}{25}.$ We see that condition $(\mathcal{H}_{3})$ is followed
with a number $r_{0}=2.$ Moreover,
$\dfrac{PT^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}a_{11}+\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{11}^{k}}\simeq
0.0990,\ \
\dfrac{PT^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}a_{12}+\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{12}^{k}}=\dfrac{PT^{\alpha_{1}}}{\Gamma(\alpha_{1}+1)}a_{21}+\sum_{k=1}^{m}\frac{T^{\beta_{1}^{k}}}{\Gamma(\beta_{1}^{k}+1)}b{{}_{21}^{k}}=0,\
\
\dfrac{PT^{\alpha_{2}}}{\Gamma(\alpha_{2}+1)}a_{22}+\sum_{k=1}^{m}\frac{T^{\beta_{2}^{k}}}{\Gamma(\beta_{2}^{k}+1)}b{{}_{22}^{k}}\simeq
0.0653.$ Theorem $(\ref{mple})$ proves the existence of a solution to system
$(\ref{exe})$ . Q.E.D.

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