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\begin{document}
\title[Darbo type fixed and coupled fixed point...]{Fixed point theorem via
measure of noncompactness and application to Volterra integral equations in
Banach algebras}
\author[H. K. Nashine]{Hemant Kumar Nashine}
\address{Department of Mathematics\\
Texas A \& M University - Kingsville - 78363-8202, Texas, USA}
\email{hemant.nashine@tamuk.edu}
\author[J. R. Roshan]{Jamal Rezaei Roshan$^{\ast }$}
\address{Department of Mathematics\\
Qaemshahr Branch, Islamic Azad University, Qaemshahr, IRAN}
\email{jmlroshan@gmail.com, jml.roshan@qaemiau.ac.ir}
\subjclass{54H25, 47H10}
\keywords{measures of noncompactness; Darbo's fixed point theorem; coupled
fixed point; integral equations.\\
\indent$^{\ast }$Corresponding author.}

\begin{abstract}
We propose a new notion of contraction mappings for two class of functions
involving measure of noncompactness in Banach space and derive some basic
Darbo type fixed and coupled fixed point results. This work includes and
extends the results of Falset and Latrach[ Falset, J. G., Latrach, K. : On
Darbo-Sadovskii's fixed point theorems type for abstract measures of (weak)
noncompactness, Bull. Belg. Math. Soc. Simon Stevin 22 (2015), 797-812.] The
results are also correlated with the classical generalized Banach fixed
point theorems. Finally, we will discuss the applicability of obtained
results to the Volterra integral equations in Banach algebras with an
illustration.
\end{abstract}

\maketitle

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%\submitted{March 14, 2003}
%\received{March 16, 2000}
%\revised{June 1, 2000}
%\accepted{July 22, 2000}
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%\date{January 1, 2004}
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%\dedicatory{To my boss}
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%some Darbo type results existing in the literature.
%%% ----------------------------------------------------------------------
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\section{Introduction and preliminaries}

To understand the work in the underlying area, we start out with listing
some notations and preliminaries that we shall need to express our results. 
\newline
Throughout the paper,

$\mathbb{R}$ = the set of real numbers,

$\mathbb{N}$ the set of natural numbers,

$\mathbb{R}^+ = [0, + \infty)$ and $\mathbb{N}^*=\mathbb{N}\cup\{0\}$.%
\newline
Let $(E, \| . \|)$ be a real Banach space with zero element $0$. Let $%
\mathcal{B}(x, r)$ denote the closed ball centered at $x$ with radius $r$.
The symbol $\mathcal{B}_r$ stands for the ball $\mathcal{B}(0, r)$. For $X$,
a nonempty subset of $E$, we denote by $\overline{X}$ and $Conv X$ the
closure and the convex closure of $X$, respectively. Moreover, let us denote
by $\mathfrak{M}_E$ the family of nonempty bounded subsets of $E$ and by $%
\mathfrak{N}_E$ its subfamily consisting of all relatively compact sets.

We use the following definition of the measure of noncompactness (MNC) given
in \cite{BANAS4}.

\begin{definition}
\label{DEF1} A mapping $\mu : \mathfrak{M}_E \to \mathbb{R}^+$ is said to be
a MNC in $E$ if it satisfies the following conditions:

\begin{enumerate}
\item[($1^0$)] The family $ker \mu = \{X \in \mathfrak{M}_E : \mu(X) = 0\}$
is nonempty and $ker \mu \subset \mathfrak{N}_E$,

\item[($2^0$)] (Monotonicity) $X \subset Y \Rightarrow \mu(X) \leq \mu(Y)$,

\item[($3^0$)] (Invariance under closure) $\mu(\overline{X}) = \mu(X)$,

\item[($4^0$)] (invariance under passage to the convex hull) $\mu(Conv X) =
\mu(X)$,

\item[($5^0$)] (convexity) $\mu( \lambda X + (1 - \lambda)Y ) \leq \lambda
\mu(X) + (1 - \lambda) \mu(Y )$ for $\lambda \in [0, 1]$,

\item[($6^0$)] (Cantor's generalized intersection property) If $(X_n)$ is a
decreasing sequence of nonempty, closed sets in $\mathfrak{M}_E$ such that $%
X_{n+1} \subset X_n$ ($n =1, 2,\ldots )$ and if $\lim_{n \to \infty}
\mu(X_n) = 0$, then the set $X_{\infty} = \bigcap_{n=1}^{\infty} X_n$ is
compact.
\end{enumerate}
\end{definition}

The family $ker \mu$ defined in axiom ($1^0$) is called the kernel of the
MNC $\mu$.

One of the properties of the MNC is $X_{\infty} \in ker \mu$. Indeed, from
the inequality $\mu(X_{\infty}) \leq \mu(X_n)$ for $n = 1, 2, 3,\ldots,$ we
infer that $\mu(X_{\infty}) = 0$.

In 1930, Kuratowski \cite{KUR1} opened up a new direction of research with
the introduction of measure of noncompactness, denoted by $\alpha $

The Kuratowski MNC is the map $\alpha :\mathfrak{M}_{E}\rightarrow \mathbb{R}%
^{+}$ with 
\begin{equation}
\alpha (\mathcal{Q})=\inf \bigg\{\epsilon >0:\mathcal{Q}\subset
\bigcup_{k=1}^{n}S_{k},S_{k}\subset E,diam(S_{k})<\epsilon \,(k\in \mathbb{N}%
)\bigg\}.
\end{equation}

In 1955, Darbo \cite{DARBO1} used the notion of Kuratowski measure of
noncompactness $\alpha $ to prove fixed point theorem and generalized
topological Schauder fixed point theorem \cite{BANAS4} and classical Banach
fixed point theorem \cite{BANACH}.

\begin{theorem}
\label{T1}\cite{BANAS4} Let $\mathcal{X}$ be a closed, convex subset of a
Banach space $E$. Then every compact, continuous map $\mathcal{T} : \mathcal{%
X}\to \mathcal{X}$ has at least one fixed point.
\end{theorem}

\begin{theorem}
\cite{DARBO1} \label{T11} Let $\mathcal{X}$ be a nonempty, bounded, closed
and convex subset of a Banach space $E$, $\mu$ be the Kuratowski MNC on $E$.
Let $\mathcal{T} : \Omega \to \Omega$ be a continuous and $\mu$-set
contraction operator, that is, there exists a constant $k \in [0,1)$ with
\begin{equation*}
\mu(\mathcal{TM}) \leq k \mu(\mathcal{M})
\end{equation*}
for any nonempty subset $\mathcal{M}$ of $\mathcal{X}$. Then $\mathcal{T}$
has a fixed point.
\end{theorem}

Following this result, various authors proved several Darbo-type fixed point
and coupled theorems by using different types of control functions. Here we
mention the paper discussed in \cite%
{AGHAJANI-NOVI,AGHAJANI-JNCA,MUR-JCAM,DARBO1,FALSET1}.

With the above discussion in mind, we establish some results of Darbo's type
which generalizes and include work mentioned in \cite%
{AGHAJANI-NOVI,AGHAJANI-JNCA,MUR-JCAM,DARBO1,FALSET1} as well (see Remark %
\ref{rem1}). In the final section, we apply the obtained result to solve the
Volterra integral equations in Banach algebras and justify with an example.

\section{Fixed point theorems}

We start the section with the following notion:

\begin{defn}
\label{Z1} \cite{PIRI14} Let $\mathfrak{F}$ be the family of all functions $%
F : \mathbb{R}^+ \to \mathbb{R}$ with the following properties:

\begin{itemize}
\item[$(F_1)$] $F$ is continuous and strictly increasing;

\item[$(F_2)$] for each sequence $\{t_n\} \subseteq \mathbb{R}^+ $, $\lim_{n
\to \infty} t_n = 0 \text{ if and only if } \lim_{n \to \infty} F(t_n) =
-\infty. $
\end{itemize}
\end{defn}

The first main result is:

\begin{thm}
\label{MR1} Let $\mathcal{X}$ be a nonempty, bounded, closed, and convex
subset of a Banach space $E$, and $\mathcal{T}:\mathcal{X}\rightarrow 
\mathcal{X}$ is continuous operator. If there exist $\tau >0$, $F$
satisfying condition ($F_{2}$) and $\varphi :\mathbb{R}^{+}\rightarrow 
\mathbb{R}^{+}$ is continuous and strictly increasing mapping such that 
\begin{equation}
\mu (\mathcal{TM})>0\Rightarrow \tau +F(\mu (\mathcal{TM})+\varphi (\mu (%
\mathcal{TM})))\leq F(\mu (\mathcal{M})+\varphi (\mu (\mathcal{M}))),
\label{F1}
\end{equation}%
for all $\mathcal{M}\subseteq \mathcal{X}$, where $\mu $ is an arbitrary
MNC. Then $\mathcal{T}$ has at least one fixed point in $\mathcal{X}$.
\end{thm}

\begin{proof}
Starting with the assumption $\mathcal{X}_{0}=\mathcal{X}$, we define a
sequence $\{\mathcal{X}_{n}\}$ such that $\mathcal{X}_{n+1}=Conv(\mathcal{TX}%
_{n})$, for $n\in \mathbb{N}^{\ast }$. If $\mu (\mathcal{X}_{n_{0}})+\varphi
(\mu (\mathcal{X}_{n_{0}}))=0$, that is, $\mu (\mathcal{X}_{n_{0}})=0$ for
some natural number $n_{0}\in \mathbb{N}$, then $\mathcal{X}_{n_{0}}$ is
compact and since $\mathcal{T}(\mathcal{X}_{n_{0}})\subseteq Conv(\mathcal{TX%
}_{n_{0}})=\mathcal{X}_{{n_{0}}+1}\subseteq \mathcal{X}_{n_{0}}$. Thus we
conclude the result from Theorem \ref{T1}, hence we assume that $\mu (%
\mathcal{X}_{n})+\varphi (\mu (\mathcal{X}_{n}))>0$, for all $n\in \mathbb{N}%
^{\ast }$. From (\ref{F1}) and ($4^{0}$) of Definition \ref{DEF1}, we have 
\begin{eqnarray*}
F(\mu (\mathcal{X}_{n+1})+\varphi (\mu (\mathcal{X}_{n+1}))) &=&F(\mu (Conv(%
\mathcal{TX}_{n}))+\varphi (\mu (Conv(\mathcal{TX}_{n})))) \\
&=&F(\mu (\mathcal{TX}_{n})+\varphi (\mu (\mathcal{TX}_{n}))) \\
&\leq &F(\mu (\mathcal{X}_{n})+\varphi (\mu (\mathcal{X}_{n})))-\tau \\
&\leq &F(\mu (\mathcal{X}_{n-1})+\varphi (\mu (\mathcal{X}_{n-1})))-2\tau \\
&\vdots & \\
&\leq &F(\mu (\mathcal{X}_{0})+\varphi (\mu (\mathcal{X}_{0})))-(n+1)\tau ,
\end{eqnarray*}%
that is, 
\begin{equation}
F(\mu (\mathcal{X}_{n+1})+\varphi (\mu (\mathcal{X}_{n+1})))\leq F(\mu (%
\mathcal{X}_{0})+\varphi (\mu (\mathcal{X}_{0})))-(n+1)\tau ,\text{ for all }%
n\in \mathbb{N}.  \label{F21}
\end{equation}%
Therefore by \eqref{F21}, we get $F(\mu (\mathcal{X}_{n+1})+\varphi (\mu (%
\mathcal{X}_{n+1})))\rightarrow -\infty $ as ${n\rightarrow \infty }$.

Thus, from the property $(F_{2})$, we have 
\begin{equation*}
\lim_{n\rightarrow \infty }\mu (\mathcal{X}_{n})+\varphi (\mu (\mathcal{X}%
_{n}))=0,
\end{equation*}%
therefore 
\begin{equation*}
\lim_{n\rightarrow \infty }\mu (\mathcal{X}_{n})=0.
\end{equation*}%
Now from ($6^{0}$) of Definition \ref{DEF1}, we have $\mathcal{X}_{\infty
}=\bigcap_{n=1}^{\infty }\mathcal{X}_{n}$ is nonempty, closed, convex set
and $\mathcal{X}_{\infty }\subseteq \mathcal{X}_{n}$ for all $n\in \mathbb{N}
$. Also $\mathcal{T}(\mathcal{X}_{\infty })\subset \mathcal{X}_{\infty }$
and $\mathcal{X}_{\infty }\in ker\mu .$ Therefore, by Theorem \ref{T1}, $%
\mathcal{T}$ has a fixed point $u$ in the set $\mathcal{X}_{\infty }$ and
hence $u\in \mathcal{X}$.
\end{proof}

\begin{rem}
If $\varphi(t)=0$ in Theorem \ref{MR1}, then we get Theorem 3.2 \cite%
{FALSET1}.
\end{rem}

\begin{rem}
\label{rem1} Taking various concrete functions $F \in \mathfrak{F}$ in the
condition \eqref{F1} of Theorems \ref{MR1}, we can get various classes of $%
\mu$-set contractive conditions. We state just a few examples which include
results existing in the literature:

\begin{enumerate}
\item[(A1)] Taking $F(t)=\ln t$ ($t>0$), $\tau =ln(\frac{1}{\lambda })$
where $\lambda \in (0,1)$, we have condition 
\begin{equation*}
\mu (\mathcal{TM})>0\Rightarrow \mu (\mathcal{TM})+\varphi (\mu (\mathcal{TM}%
))\leq \lambda \lbrack \mu (\mathcal{M})+\varphi (\mu (\mathcal{M}))].
\end{equation*}

\item[(A2)] Taking $F(t)=\ln t+t$ ($t>0$), $\tau =ln(\frac{1}{\lambda })$
where $\lambda \in (0,1)$, we have condition 
\begin{equation*}
\mu (\mathcal{TM})>0\Rightarrow \lbrack \mu (\mathcal{TM})+\varphi (\mu (%
\mathcal{TM}))]{e^{\mu (\mathcal{TM})+\varphi (\mu (\mathcal{TM}))-[\mu (%
\mathcal{M})+\varphi (\mu (\mathcal{M}))]}}\leq \lambda \lbrack \mu (%
\mathcal{M})+\varphi (\mu (\mathcal{M}))].
\end{equation*}

\item[(A3)] Taking $F(t)=\ln t$ ($t>0$) and $\tau =\ln (\frac{1}{\lambda })$%
, $\varphi (t)=t$ where $\lambda \in (0,1)$, we have Darbo's $\mu $-set
contraction condition 
\begin{equation*}
\mu (\mathcal{TM})>0\Rightarrow \mu (\mathcal{TM})\leq \lambda \mu (\mathcal{%
M}).
\end{equation*}

\item[(A4)] Taking $F(t)=-\frac{1}{\sqrt{t}}$ ($t>0$), $\tau =\lambda $ ($%
\lambda >0$), the condition is 
\begin{equation*}
\mu (\mathcal{TM})>0\Rightarrow \mu (\mathcal{TM})+\varphi (\mu (\mathcal{TM}%
))\leq \frac{\mu (\mathcal{M})+\varphi (\mu (\mathcal{M}))}{[1+\lambda \sqrt{%
\mu (\mathcal{M})+\varphi (\mu (\mathcal{M}))}]^{2}}.
\end{equation*}
\end{enumerate}
\end{rem}

\begin{prop}
\label{MR2} Let $\mathcal{X}$ be a nonempty, bounded, closed, and convex
subset of a Banach space $E$ and $\mathcal{T}:\mathcal{X}\rightarrow 
\mathcal{X}$ is continuous operator. If there exist $\tau >0$, $F$
satisfying condition ($F_{2}$) and $\varphi :\mathbb{R}^{+}\rightarrow 
\mathbb{R}^{+}$ is continuous mapping such that 
\begin{equation}
\mu (\mathcal{TM})>0\Rightarrow \tau +F(diam(\mathcal{TM})+\varphi (diam(%
\mathcal{TM})))\leq F(diam(\mathcal{M})+\varphi (diam(\mathcal{M}))),
\label{P1}
\end{equation}%
for all $\mathcal{M}\subseteq \mathcal{X}$. Then $\mathcal{T}$ has a unique
fixed point in $\mathcal{X}$.
\end{prop}

\begin{proof}
Following Theorem \ref{MR1} and argument of Proposition 3.2 \cite{FALSET1}, $%
\mathcal{T}$ has a fixed point in $\mathcal{X}$. \newline
To prove uniqueness, we suppose that there exist two distinct fixed points $%
\zeta, \xi \in \mathcal{X}$, then we may define the set $\Upsilon :=
\{\zeta,\xi \}$. In this case $diam(\Upsilon) = diam(\mathcal{T}(\Upsilon))
= \|\xi-\zeta\| > 0$. Then using (\ref{P1}), we get 
\begin{equation*}
diam(\mathcal{T}(\Upsilon))>0 \Rightarrow \tau + F(diam(\mathcal{T}%
(\Upsilon))+ \varphi(diam(\mathcal{T}(\Upsilon)))) \leq
F(diam(\Upsilon)+\varphi(diam(\Upsilon))),
\end{equation*}
a contradiction and hence $\xi = \zeta$.
\end{proof}

Following is the generalized classical fixed point result derived from
Proposition \ref{MR2}.

\begin{cor}
\label{MR3} Let $\mathcal{X}$ be a nonempty, bounded, closed, and convex
subset of a Banach space $E$ and let $\mathcal{T}:\mathcal{X}\rightarrow 
\mathcal{X}$ be an operator. It there exist $\tau >0$, $F\in \mathfrak{F}$
and $\varphi :\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is continuous and
strictly increasing mapping such that 
\begin{equation}
\Vert \mathcal{T}u-\mathcal{T}v\Vert >0\Rightarrow \tau +F(\Vert \mathcal{T}%
u-\mathcal{T}v\Vert +\varphi (\Vert \mathcal{T}u-\mathcal{T}v\Vert ))\leq
F(\Vert u-v\Vert +\varphi (\Vert u-v\Vert ))  \label{P2}
\end{equation}%
for all $u,v\in \mathcal{X}$. Then $\mathcal{T}$ has a unique fixed point.
\end{cor}

\begin{proof}
Let $\mu :\mathfrak{M}_{E}\rightarrow \mathbb{R}^{+}$ be a set quantity
defined by the formula $\mu (\mathcal{X})=diam\mathcal{X}$, where $diam%
\mathcal{X}=\sup \{\Vert u-v\Vert :u,v\in \mathcal{X}\}$ stands for the
diameter of $\mathcal{X}$. It is easily seen that $\mu $ is a MNC in a space 
$E$ in the sense of Definition \ref{DEF1}. Therefore from (\ref{P2}) we have 
\begin{eqnarray*}
\sup_{u,v\in \mathcal{X}}\Vert \mathcal{T}u-\mathcal{T}v\Vert &>&0\Rightarrow
\\
\tau +F(\sup_{u,v\in \mathcal{X}}\Vert \mathcal{T}u-\mathcal{T}v\Vert
+\varphi (\sup_{u,v\in \mathcal{X}}\Vert \mathcal{T}u-\mathcal{T}v\Vert ))
&=&\tau +\sup_{u,v\in \mathcal{X}}F(\Vert \mathcal{T}u-\mathcal{T}v\Vert
+\varphi (\Vert \mathcal{T}u-\mathcal{T}v\Vert )) \\
&\leq &\sup_{u,v\in \mathcal{X}}F(\Vert u-v\Vert +\varphi (\Vert u-v\Vert ))
\\
&\leq &F(\sup_{u,v\in \mathcal{X}}\Vert u-v\Vert +\varphi (\sup_{u,v\in 
\mathcal{X}}\Vert u-v\Vert ))
\end{eqnarray*}%
which implies that 
\begin{equation*}
diam(\mathcal{T(X)})>0\Rightarrow \tau +F(diam(\mathcal{T(X)})+\varphi (diam(%
\mathcal{T(X)})))\leq F(diam(\mathcal{X})+\varphi (diam(\mathcal{X})).
\end{equation*}%
Thus following Proposition \ref{MR2}, $\mathcal{T}$ has an unique fixed
point.
\end{proof}

\begin{cor}
\label{MR4} Let $(E, \| \cdot \|)$ be a Banach space and let $\mathcal{X}$
be a closed convex subsets of $E$. Let $\mathcal{T}_1, \mathcal{T}_2: 
\mathcal{X} \rightarrow \mathcal{X} $ be two operators satisfying the
following conditions:

\begin{enumerate}
\item[(I)] $(\mathcal{T}_1 + \mathcal{T}_2)(\mathcal{X}) \subseteq \mathcal{X%
}$;

\item[(II)] there exist $\tau >0$, $F\in \mathfrak{F}$ satisfying condition (%
$F_{2}$) and $\varphi :\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is
continuous and increasing mapping such that 
\begin{equation}
\Vert \mathcal{T}_{1}u-\mathcal{T}_{1}v\Vert >0\Rightarrow \tau +F(\Vert 
\mathcal{T}_{1}u-\mathcal{T}_{1}v\Vert +\varphi (\Vert \mathcal{T}_{1}u-%
\mathcal{T}_{1}v)\Vert ))\leq F(\Vert u-v\Vert +\varphi (\Vert u-v\Vert ))
\label{P3}
\end{equation}

\item[(III)] $\mathcal{T}_{2}$ is a continuous and compact operator.
\end{enumerate}

Then $\mathcal{J} := \mathcal{T}_1 + \mathcal{T}_2: \mathcal{X} \rightarrow 
\mathcal{X}$ has a fixed point $u \in \mathcal{X}$.
\end{cor}

\begin{proof}
Suppose that $\mathcal{M}$ is a subset of $X$ with $\alpha (\mathcal{M})>0$.
By the notion of Kuratowski MNC, for each $n\in \mathbb{N}$, there exist $%
\mathcal{C}_{1},\ldots ,\mathcal{C}_{m(n)}$ bounded subsets such that $%
\mathcal{M}\subseteq \bigcup_{i=1}^{m(n)}\mathcal{C}_{i}$ and $diam(\mathcal{%
C}_{i})\leq \alpha (\mathcal{M})+\frac{1}{n}$. Suppose that $\alpha (%
\mathcal{T}_{1}(\mathcal{M}))>0$. Since $\mathcal{T}_{1}(\mathcal{M}%
)\subseteq \bigcup_{i=1}^{m(n)}\mathcal{T}_{1}(\mathcal{C}_{i})$, there
exists $i_{0}\in \{1,2,\ldots ,m(n)\}$ such that $\alpha (\mathcal{T}_{1}(%
\mathcal{M}))\leq diam(\mathcal{T}_{1}(\mathcal{C}_{i_{0}}))$. Using (\ref%
{P3}) condition of $\mathcal{T}_{1}$ with discussed arguments, we have 
\begin{eqnarray}
\tau +F(\alpha (\mathcal{T}_{1}(\mathcal{M}))+\varphi (\alpha (\mathcal{T}%
_{1}(\mathcal{M})))) &\leq &\tau +F(diam(\mathcal{T}_{1}(\mathcal{C}%
_{i_{0}}))+\varphi (diam(\mathcal{T}_{1}(\mathcal{C}_{i_{0}}))))  \notag
\label{P4} \\
&\leq &F(diam(\mathcal{C}_{i_{0}})+\varphi (diam(\mathcal{C}_{i_{0}})) 
\notag \\
&\leq &F\bigg(\alpha (\mathcal{M})+\frac{1}{n}+\varphi \bigg(\alpha (%
\mathcal{M})+\frac{1}{n}\bigg)\bigg).
\end{eqnarray}%
Passing to the limit in \eqref{P4} as $n\rightarrow \infty $, we get 
\begin{equation*}
\tau +F(\alpha (\mathcal{T}_{1}(\mathcal{M}))+\varphi (\alpha (\mathcal{T}%
_{1}(\mathcal{M}))))\leq F(\alpha (\mathcal{M})+\varphi (\alpha (\mathcal{M}%
))).
\end{equation*}%
Using (III) hypothesis, we have by the notion of $\alpha $ that 
\begin{eqnarray*}
\tau +F\left( \alpha (\mathcal{J}(\mathcal{M}))+\varphi \left( \alpha (%
\mathcal{J}(\mathcal{M}))\right) \right) &=&\tau +F(\alpha (\mathcal{T}_{1}(%
\mathcal{M})+\mathcal{T}_{2}(\mathcal{M}))+\varphi \left( \alpha \left( 
\mathcal{T}_{1}(\mathcal{M})+\mathcal{T}_{2}(\mathcal{M})\right) \right) ) \\
&\leq &\tau +F\left( \alpha (\mathcal{T}_{1}(\mathcal{M}))+\alpha (\mathcal{T%
}_{2}(\mathcal{M}))+\varphi \left( \alpha (\mathcal{T}_{1}(\mathcal{M}%
))+\alpha \left( \mathcal{T}_{2}(\mathcal{M})\right) \right) \right) \\
&=&\tau +F(\alpha (\mathcal{T}_{1}(\mathcal{M}))+\varphi \left( \alpha (%
\mathcal{T}_{1}(\mathcal{M}))\right) ) \\
&\leq &F(\alpha (\mathcal{M})+\varphi (\alpha (\mathcal{M}))).
\end{eqnarray*}%
Thus by Theorem \ref{MR1}, $\mathcal{J}$ has a fixed point $u\in \mathcal{X}$%
.
\end{proof}

\begin{rem}
If $\varphi(t)=0$ in Proposition \ref{MR2}-Corollary \ref{MR4}, we get
result given in \cite[Proposition 3.2-Corollary 3.4]{FALSET1}.
\end{rem}

\section{Coupled fixed point theorems}

This section is concern with the coupled fixed point theorems of Section 2.

\begin{defn}
\cite{GUO1} An element $(u, v) \in {E}^2$ is called a coupled fixed point of
a mapping $\mathcal{G}: {E}^2 \to {E}$ if $\mathcal{G}(u, v) = u$ and $%
\mathcal{G}(v, u) = v.$
\end{defn}

The first coupled fixed point result is:

\begin{thm}
\label{CT1} Let $\mathcal{X}$ be a nonempty, bounded, closed, and convex
subset of a Banach space $E$. Suppose that $\mathcal{G}:\mathcal{X}\times 
\mathcal{X}\rightarrow \mathcal{X}$ is continuous operator. If there exist $%
\tau >0$, $F\in \mathfrak{F}$ satisfying condition ($F_{2}$) and
sub-additive, and $\varphi :\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is
continuous, increasing and sub-additive mapping such that 
\begin{equation}
\left. 
\begin{array}{r}
\text{for }i,j\in \left\{ 1,2\right\} ,i\neq j,\text{ }\mu \left( \mathcal{G}%
\left( \mathcal{X}_{i}\times \mathcal{X}_{j}\right) \right) >0 \\ 
\Rightarrow \tau +F\left( \text{ }\mu \left( \mathcal{G}\left( \mathcal{X}%
_{i}\times \mathcal{X}_{j}\right) \right) +\varphi \left( \text{ }\mu \left( 
\mathcal{G}\left( \mathcal{X}_{i}\times \mathcal{X}_{j}\right) \right)
\right) \right) \\ 
\leq \dfrac{1}{2}F\left( \mu \left( \mathcal{X}_{i}\right) +\mu \left( 
\mathcal{X}_{j}\right) +\varphi \left( \mu \left( \mathcal{X}_{i}\right)
+\mu \left( \mathcal{X}_{j}\right) \right) \right)%
\end{array}%
\right\}  \label{C-1}
\end{equation}%
for all $\mathcal{X}_{i},\mathcal{X}_{j}\subseteq \mathcal{X}$, where $\mu $
is an arbitrary MNC. Then $\mathcal{G}$ has at least a coupled fixed point $%
(u,v)\in \mathcal{X}\times \mathcal{X}$.
\end{thm}

\begin{proof}
Consider the map $\widehat{{\mathcal{G}}}:\mathcal{X}\times \mathcal{X}%
\rightarrow \mathcal{X}\times \mathcal{X}$ defined by the formula 
\begin{equation*}
\widehat{\mathcal{G}}(u,v)=({\mathcal{G}}(u,v),{\mathcal{G}}(v,u)).
\end{equation*}%
$\widehat{\mathcal{G}}$ is continuous due to continuity of $\mathcal{G}$.
Following \cite{AGHAJANI-JNCA}, we define a new MNC in the space $\mathcal{X}%
\times \mathcal{X}$ as 
\begin{equation*}
\widehat{{\mu }}(\mathcal{M})=\mu (\mathcal{X}_{1})+\mu (\mathcal{X}_{2})
\end{equation*}%
where $\mathcal{X}_{i}$, $i=1,2$ denote the natural projections of $\mathcal{%
X}$. Without loss of generality, we can assume $\mathcal{M}$ is a nonempty
subset of $\mathcal{X}^{2}$. Hence, by the condition (\ref{C-1}) and using ($%
2^{0}$) of Definition \ref{DEF1} we conclude that 
\begin{eqnarray*}
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M})) &\leq &\widehat{\mu }({%
\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})\times {\mathcal{G}}%
(X_{2}\times \mathcal{X}_{1})) \\
&=&\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2}))+\mu ({\mathcal{%
G}}(\mathcal{X}_{2}\times \mathcal{X}_{1})),
\end{eqnarray*}%
therefore by the assumption, we have 
\begin{equation*}
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M}))>0,
\end{equation*}%
that implies 
\begin{eqnarray*}
&&\tau +F(\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M}))+\varphi (%
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M})))) \\
&\leq &\tau +F(\widehat{\mu }({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2})\times {\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))+\varphi (%
\widehat{\mu }({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})\times {%
\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1})))) \\
&\leq &\tau +F(\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2}))+\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))+\varphi
(\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})))+\varphi (\mu ({%
\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1})))) \\
&\leq &\tau +F(\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2}))+\varphi (\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2}))))+F(\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}%
_{1}))+\varphi (\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))))
\\
&\leq &\frac{1}{2}F(\mu (\mathcal{X}_{1})+\mu (\mathcal{X}_{2}))+\varphi
(\mu (\mathcal{X}_{1})+\mu (\mathcal{X}_{2})))+\frac{1}{2}F(\mu (\mathcal{X}%
_{2})+\mu (\mathcal{X}_{1})+\varphi (\mu (\mathcal{X}_{2})+\mu (\mathcal{X}%
_{1}))) \\
&=&F(\mu (\mathcal{X}_{1})+\mu (\mathcal{X}_{2})+\varphi (\mu (\mathcal{X}%
_{1})+\mu (\mathcal{X}_{2}))) \\
&=&F(\widehat{\mu }(\mathcal{M})+\varphi (\widehat{\mu }(\mathcal{M}))),
\end{eqnarray*}%
that is, 
\begin{equation*}
\widehat{\mu }(\widehat{\mathcal{G}}(\mathcal{M}))>0\Rightarrow \tau +F(%
\widehat{\mu }(\widehat{\mathcal{G}}(\mathcal{M}))+\varphi (\widehat{{\mu }}(%
\widehat{\mathcal{G}}(\mathcal{M}))))\leq F(\widehat{\mu }(\mathcal{M}%
)+\varphi (\widehat{\mu }(\mathcal{M}))).
\end{equation*}%
Therefore from Theorem \ref{MR1}, we get $\widehat{\mathcal{G}}$ has at
least one fixed point in $\mathcal{X}^{2}$, and hence $\mathcal{G}$ has a
coupled fixed point.
\end{proof}

The second result is as follows:

\begin{thm}
\label{CT2} Let $\mathcal{X}$ be a nonempty, bounded, closed, and convex
subset of a Banach space $E$. Suppose that $\mathcal{G}:\mathcal{X}\times 
\mathcal{X}\rightarrow \mathcal{X}$ is continuous operator. If there exist $%
\tau >0$, $F\in \mathfrak{F}$ satisfying condition ($F_{2}$), and $\varphi :%
\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is continuous and increasing
mapping such that 
\begin{equation}
\left. 
\begin{array}{r}
\text{for }i,j\in \left\{ 1,2\right\} ,i\neq j,\text{ }\mu \left( \mathcal{G}%
\left( \mathcal{X}_{i}\times \mathcal{X}_{j}\right) \right) >0 \\ 
\Rightarrow \tau +F\left( \text{ }\mu \left( \mathcal{G}\left( \mathcal{X}%
_{i}\times \mathcal{X}_{j}\right) \right) +\varphi \left( \text{ }\mu \left( 
\mathcal{G}\left( \mathcal{X}_{i}\times \mathcal{X}_{j}\right) \right)
\right) \right) \\ 
\leq F\left( \max \{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\}+\varphi
\left( \max \{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\}\right) \right)%
\end{array}%
\right\}  \label{C-3}
\end{equation}%
for all $\mathcal{X}_{i},\mathcal{X}_{j}\subseteq \mathcal{X}$, where $\mu $
is an arbitrary MNC. Then $\mathcal{G}$ has at least a coupled fixed point $%
(u,v)\in \mathcal{X}\times \mathcal{X}$.
\end{thm}

\begin{proof}
Consider the map $\widehat{\mathcal{G}}:\mathcal{X}\times \mathcal{X}%
\rightarrow \mathcal{X}\times \mathcal{X}$ defined by the formula 
\begin{equation*}
\widehat{\mathcal{G}}(u,v)=({\mathcal{G}}(u,v),{\mathcal{G}}(v,u)).
\end{equation*}%
$\widehat{\mathcal{G}}$ is continuous due to continuity of $\mathcal{G}$.
Following \cite{AGHAJANI-JNCA}, we define a new MNC in the space $\mathcal{X}%
\times \mathcal{X}$ as 
\begin{equation*}
\widehat{{\mu }}(\mathcal{M})=\max \{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}%
_{2})\}
\end{equation*}%
where $\mathcal{X}_{i}$, $i=1,2$ denote the natural projections of $\mathcal{%
M}$. Without loss of generality, we can assume $\mathcal{M}$ is a nonempty
subset of $\mathcal{X}\times \mathcal{X}$. Following previous theorem, we
have 
\begin{eqnarray*}
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M})) &\leq &\widehat{\mu }({%
\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})\times {\mathcal{G}}(%
\mathcal{X}_{2}\times \mathcal{X}_{1})) \\
&=&\max \{\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})),\mu ({%
\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))\},
\end{eqnarray*}%
which is, by the assumption, 
\begin{equation*}
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M}))>0.
\end{equation*}%
Hence, by the condition (\ref{C-3}) and using ($2^{0}$) of Definition \ref%
{DEF1} we conclude that 
\begin{eqnarray*}
&&\tau +F(\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M}))+\varphi (%
\widehat{{\mu }}(\widehat{\mathcal{G}}(\mathcal{M})))) \\
&\leq &\tau +F(\widehat{\mu }({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2})\times {\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))+\varphi (%
\widehat{\mu }({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})\times {%
\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1})))) \\
&=&\tau +F(\max \{\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2})),\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))\}+\varphi
(\max \{\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}_{2})),\mu ({%
\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}_{1}))\})) \\
&=&\tau +\max \{F(\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2}))+\varphi (\mu ({\mathcal{G}}(\mathcal{X}_{1}\times \mathcal{X}%
_{2})))),F(\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}%
_{1}))+\varphi (\mu ({\mathcal{G}}(\mathcal{X}_{2}\times \mathcal{X}%
_{1}))))\} \\
&\leq &\max \left\{ 
\begin{array}{cc}
F(\max \{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\}+\varphi (\max \{\mu (%
\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\})), &  \\ 
F(\max \{\mu (\mathcal{X}_{2}),\mu (\mathcal{X}_{1})\}+\varphi (\max \{\mu (%
\mathcal{X}_{2}),\mu (\mathcal{X}_{1})\})) & 
\end{array}%
\right\} \\
&=&F(\max \{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\}+\varphi (\max
\{\mu (\mathcal{X}_{1}),\mu (\mathcal{X}_{2})\})) \\
&=&F(\widehat{\mu }(\mathcal{M})+\varphi (\widehat{\mu }(\mathcal{M}))),
\end{eqnarray*}%
that is, 
\begin{equation*}
\widehat{\mu }(\widehat{\mathcal{G}}(\mathcal{M})>0\Rightarrow \tau +F(%
\widehat{\mu }(\widehat{\mathcal{G}}(\mathcal{M})))\leq F(\widehat{\mu }(%
\mathcal{M})+\varphi (\widehat{\mu }(\mathcal{M}))).
\end{equation*}%
Hence by Theorem \ref{MR1}, we reached that $\widehat{\mathcal{G}}$ has at
least one fixed point in $\mathcal{X}^{2}$, and thus $\mathcal{G}$ has a
coupled fixed point.
\end{proof}

\begin{rem}
\label{rem3} In view of the Remark \ref{rem1}(A1-(A3)), some new coupled
fixed point results can be derived from Theorems \ref{CT1} and Theorems \ref%
{CT2}.
\end{rem}

\section{Application to the Volterra integral equations in Banach algebras}

Let $\left( X,\left\Vert .\right\Vert \right) $ be a real Banach algebra and
the symbol $C\left( \left[ 0,T\right] ,X\right) $ stands for the space
consisting of all continuous mappings $f:\left[ 0,T\right] \rightarrow X,$
with $T>0.$ In this section inspired by Theorem 4.1 of J. Garcia-Falset 
\textit{et al}. \cite{FALSET1}, we will consider the existence of a solution 
$x\in C\left( \left[ 0,T\right] ,X\right) $ to the following nonlinear
Volterra integral equation

\begin{equation}
x\left( t\right) =f\left( t,x\left( t\right) \right) +Gx\left( t\right)
\int_{0}^{t}g\left( s,x\left( s\right) \right) Qx\left( s\right) ds.
\label{D-1}
\end{equation}

We will assume that the following conditions are satisfied:

\begin{itemize}
\item[(a)] $f:\left[ 0,T\right] \times X\rightarrow X$ is a continuous
mapping such that there exist a bijective, strictly increasing function $%
F:\left( 0,\infty \right) \rightarrow \left( -\infty ,0\right) $ and a \
nondecreasing function $\phi :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ and $\tau >0$ such that
\end{itemize}

\begin{equation}
\left\Vert f\left( t,x\right) -f\left( t,y\right) \right\Vert >0\implies
\tau +F\left( \left\Vert f\left( t,x\right) -f\left( t,y\right) \right\Vert
+\phi \left( \left\Vert f\left( t,x\right) -f\left( t,y\right) \right\Vert
\right) \right) \leq F\left( \left\Vert x-y\right\Vert +\phi \left(
\left\Vert x-y\right\Vert \right) \right) .  \label{F}
\end{equation}

\begin{itemize}
\item[(b)] $G$ and $Q$ are some operators acting continuously from the space 
$C\left( \left[ 0,T\right] ,X\right) $ into itself \ and there are
increasing functions $\varphi ,\psi :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ such that 
\begin{eqnarray*}
\left\Vert G\left( x\right) \right\Vert &\leq &\varphi \left( \left\Vert
x\right\Vert \right) \\
\left\Vert Q\left( x\right) \right\Vert &\leq &\psi \left( \left\Vert
x\right\Vert \right) ,
\end{eqnarray*}
\end{itemize}

for any $x\in C\left( \left[ 0,T\right] ,X\right) .$

\begin{itemize}
\item[(c)] $g:\left[ 0,T\right] \times X\rightarrow X$ is a continuous
mapping such that there exist an increasing function $u\in L^{1}\left( \left[
0,T\right] ,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\right) $ and an increasing continuous function $\theta :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ such that
\end{itemize}

\begin{equation*}
\left\Vert g\left( t,x\right) \right\Vert \leq u\left( t\right) \theta
\left( \left\Vert x\right\Vert \right) ,
\end{equation*}

for any $x\in X$ and a.e. $t\in \left[ 0,T\right].$ Moreover, for any fixed $%
x\in X$ the mapping $t\longrightarrow g\left( t,x\right) $ is measurable
over the interval $\left[ 0,T\right] $ and the mapping $x\longrightarrow
g\left( t,x\right) $ is continuous for a.e. $t\in \left[ 0,T\right] ,$

\begin{itemize}
\item[(d)] $\liminf_{\gamma \rightarrow \infty }\dfrac{\varphi \left( \gamma
\right) \psi \left( \gamma \right) \theta \left( \gamma \right) \left\Vert
u\right\Vert _{1}}{\gamma }<1.$
\end{itemize}

\begin{thm}
\label{Apl} Under assumptions (a)-(d), Eq.(\ref{D-1}) has at least one
solution in the space $x\in C\left( \left[ 0,T\right] ,X\right) .$
\end{thm}

\begin{proof}
Define an integral operator \ $\mathcal{J}:$ $C\left( \left[ 0,T\right]
,X\right) \rightarrow C\left( \left[ 0,T\right] ,X\right) $ by \ 
\begin{equation*}
\mathcal{J}x\left( t\right) =f\left( t,x\left( t\right) \right) +Gx\left(
t\right) \int_{0}^{t}g\left( s,x\left( s\right) \right) Qx\left( s\right) ds.
\end{equation*}%
We will show that the operator $\mathcal{J}$ has a one fixed point. To this
end we define the following two mappings $\mathcal{T}_{1},\mathcal{T}%
_{2}:C\left( \left[ 0,T\right] ,X\right) \rightarrow C\left( \left[ 0,T%
\right] ,X\right) $ by:%
\begin{eqnarray*}
\mathcal{T}_{1}x\left( t\right) &=&f\left( t,x\left( t\right) \right) , \\
\mathcal{T}_{2}x\left( t\right) &=&Gx\left( t\right) \int_{0}^{t}g\left(
s,x\left( s\right) \right) Qx\left( s\right) ds,
\end{eqnarray*}%
where $\mathcal{J=T}_{1}+\mathcal{T}_{2}.$

It is easy to see that \ $\mathcal{T}_{1}$ is well-defined. Now we show that 
$\mathcal{T}_{2}$ is well-defined, let $\varepsilon >0$ arbitrarily and $%
x\in C\left( \left[ 0,T\right] ,X\right) $ be given and fixed. Next let $%
M_{1}=\sup_{t\in \left[ 0,T\right] }\left\Vert g\left( t,x\left( t\right)
\right) \right\Vert ,$ since $Gx$ is uniformly continuous on $[0,T],$ there
exists $\delta _{1}\left( \varepsilon \right) >0$ such that for all $%
t_{1},t_{2}\in \lbrack 0,T],$ with $\left\vert t_{2}-t_{1}\right\vert
<\delta _{1}\left( \varepsilon \right) $ we have%
\begin{equation*}
\left\Vert Gx\left( t_{2}\right) -Gx\left( t_{1}\right) \right\Vert <\frac{%
\varepsilon }{2\left( 1+TM_{1}\left\Vert Qx\right\Vert _{\infty }\right) }.
\end{equation*}

Put $\delta \left( \varepsilon \right) =\min \left\{ \delta _{1}\left(
\varepsilon \right) ,\dfrac{\varepsilon }{2\left( 1+M_{1}\left\Vert
Gx\right\Vert _{\infty }\left\Vert Qx\right\Vert _{\infty }\right) }\right\}
.$ Without loss of generality we may assume that $t_{1}<t_{2}$ and $%
t_{2}-t_{1}<\delta \left( \varepsilon \right) $ and we obtain

\begin{eqnarray*}
\left\Vert \mathcal{T}_{2}x\left( t_{2}\right) -\mathcal{T}_{2}x\left(
t_{1}\right) \right\Vert &=&\left\Vert \left( Gx\left( t_{2}\right)
-Gx\left( t_{1}\right) \right) \int_{0}^{t_{1}}g\left( s,x\left( s\right)
\right) Qx\left( s\right) ds+Gx\left( t_{2}\right)
\int_{t_{1}}^{t_{2}}g\left( s,x\left( s\right) \right) Qx\left( s\right)
ds\right\Vert \\
&\leq &\left\Vert Gx\left( t_{2}\right) -Gx\left( t_{1}\right) \right\Vert
.\int_{0}^{t_{1}}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert
.\left\Vert Qx\left( s\right) \right\Vert ds+\left\Vert Gx\left(
t_{2}\right) \right\Vert . \\
&&\int_{t_{1}}^{t_{2}}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert .\left\Vert Qx\left( s\right) \right\Vert ds \\
&<&\frac{\varepsilon }{2\left( 1+TM_{1}\left\Vert Qx\right\Vert _{\infty
}\right) }TM_{1}\left\Vert Qx\right\Vert _{\infty }+M_{1}\left\Vert
Gx\right\Vert _{\infty }.\left\Vert Qx\right\Vert _{\infty }\left(
t_{2}-t_{1}\right) \\
&<&\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon .
\end{eqnarray*}

Next, we show that $\mathcal{T}_{2}$ is a continuous operator. Fix $y\in
C\left( \left[ 0,T\right] ,X\right) $ and $\ \varepsilon >0$ be given$,$
since $G$ and $Q$ are some operators acting continuously from the space $%
C\left( \left[ 0,T\right] ,X\right) $ into itself, so there exist $\delta
_{1}>0\ $and $\delta _{2}$ $>0,$ such that

\begin{eqnarray*}
\forall x &\in &C\left( \left[ 0,T\right] ,X\right) ,\text{ }\left(
\left\Vert x-y\right\Vert _{\infty }<\delta _{1}\Longrightarrow \left\Vert
Gx-Gy\right\Vert _{\infty }<\varepsilon _{1}\right) \\
\forall x &\in &C\left( \left[ 0,T\right] ,X\right) ,\text{ }\left(
\left\Vert x-y\right\Vert _{\infty }<\delta _{2}\Longrightarrow \left\Vert
Qx-Qy\right\Vert _{\infty }<\varepsilon _{2}\right) ,
\end{eqnarray*}

where $\varepsilon _{1}$ and $\varepsilon _{2}$ depend on $\varepsilon $ and
will be given.

On the other hand, since the mapping $x\longrightarrow g\left( t,x\right) $
is continuous for a.e. $t\in \left[ 0,T\right] ,$ there exits a $\delta _{3}$
$>0$ such that for a.e. $t\in \left[ 0,T\right] $ we have

\begin{equation*}
\forall x\in C\left( \left[ 0,T\right] ,X\right) ,\text{ }\left( \left\Vert
x\left( t\right) -y\left( t\right) \right\Vert <\delta _{3}\Longrightarrow
\left\Vert g\left( t,x\right) -g\left( t,y\right) \right\Vert <\varepsilon
_{3}\right) ,
\end{equation*}

where $\varepsilon _{3}$ is depends on $\varepsilon $ and will be given.

Now\ if we put $\delta =\min \left\{ \delta _{1},\delta _{2},\delta
_{3}\right\} ,$ then for any $x\in C\left( \left[ 0,T\right] ,X\right) $
that $\left\Vert x-y\right\Vert _{\infty }<\delta ,$ \ by the triangle
inequality we obtain%
\begin{eqnarray*}
\left\Vert \mathcal{T}_{2}x\left( t\right) -\mathcal{T}_{2}y\left( t\right)
\right\Vert &=&\left\Vert Gx\left( t\right) \int_{0}^{t}g\left( s,x\left(
s\right) \right) Qx\left( s\right) ds-Gy\left( t\right) \int_{0}^{t}g\left(
s,y\left( s\right) \right) Qy\left( s\right) ds\right\Vert \\
&\leq &\left\Vert Gx\left( t\right) -Gy\left( t\right) \right\Vert
.\int_{0}^{t}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert
.\left\Vert Qx\left( s\right) \right\Vert ds \\
&&+\left\Vert Qx-Qy\right\Vert _{\infty }.\left\Vert Gy\left( t\right)
\right\Vert .\int_{0}^{t}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert ds \\
&&+\left\Vert Gy\right\Vert _{\infty }.\left\Vert Qy\right\Vert _{\infty
}.\int_{0}^{t}\left\Vert g\left( s,x\left( s\right) \right) -g\left(
s,y\left( s\right) \right) \right\Vert ds \\
&\leq &\left\Vert Gx-Gy\right\Vert _{\infty }.\left\Vert Qx\right\Vert
_{\infty }.\int_{0}^{T}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert ds \\
&&+\left\Vert Qx-Qy\right\Vert _{\infty }.\left\Vert Gy\right\Vert _{\infty
}.\int_{0}^{T}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert ds \\
&&+\left\Vert Gy\right\Vert _{\infty }.\left\Vert Qy\right\Vert _{\infty
}.\int_{0}^{T}\left\Vert g\left( s,x\left( s\right) \right) -g\left(
s,y\left( s\right) \right) \right\Vert ds \\
&\leq &\varepsilon _{1}.\psi \left( \left\Vert x\right\Vert _{\infty
}\right) .\int_{0}^{T}u\left( s\right) \theta \left( \left\Vert x\left(
s\right) \right\Vert \right) ds+\varepsilon _{2}\varphi \left( \left\Vert
y\right\Vert _{\infty }\right) \int_{0}^{T}u\left( s\right) \theta \left(
\left\Vert x\left( s\right) \right\Vert \right) ds \\
&&+\varepsilon _{3}T\left\Vert Gy\right\Vert _{\infty }.\left\Vert
Qy\right\Vert _{\infty } \\
&\leq &\varepsilon _{1}.\psi \left( \left\Vert y\right\Vert _{\infty
}+\delta \right) .\left\Vert u\right\Vert _{1}.\theta \left( \left\Vert
y\right\Vert _{\infty }+\delta \right) +\varepsilon _{2}\varphi \left(
\left\Vert y\right\Vert _{\infty }\right) .\left\Vert u\right\Vert
_{1}.\theta \left( \left\Vert y\right\Vert _{\infty }+\delta \right) \\
&&+T\varepsilon _{3}\varphi \left( \left\Vert y\right\Vert _{\infty }\right)
.\psi \left( \left\Vert y\right\Vert _{\infty }\right) \\
&<&\frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}%
=\varepsilon ,
\end{eqnarray*}%
where%
\begin{eqnarray*}
\varepsilon _{1} &=&\dfrac{\varepsilon }{3\left( 1+\psi \left( \left\Vert
y\right\Vert _{\infty }+\delta \right) \left\Vert u\right\Vert _{1}\theta
\left( \left\Vert y\right\Vert _{\infty }+\delta \right) \right) }, \\
\varepsilon _{2} &=&\dfrac{\varepsilon }{3\left( 1+\varphi \left( \left\Vert
y\right\Vert _{\infty }\right) \left\Vert u\right\Vert _{1}\theta \left(
\left\Vert y\right\Vert _{\infty }+\delta \right) \right) }, \\
\varepsilon _{3} &=&\dfrac{\varepsilon }{3\left( 1+T\varepsilon _{3}\varphi
\left( \left\Vert y\right\Vert _{\infty }\right) .\psi \left( \left\Vert
y\right\Vert _{\infty }\right) \right) }.
\end{eqnarray*}

Now we show that $\mathcal{T}_{2}$ is a compact operator. If $B=\left\{ x\in
C\left( \left[ 0,T\right] ,X\right) :\left\Vert x\right\Vert _{\infty }<1%
\text{ }\right\} $ is the open unit ball of $C\left( \left[ 0,T\right]
,X\right) ,$ then we claim that $\overline{\mathcal{T}_{2}\left( B\right) }$
is a compact subset of $C\left( \left[ 0,T\right] ,X\right) .$ To see this,
by the Arzel\`{a}--Ascoli's theorem, we need only to show that $\mathcal{T}%
_{2}\left( B\right) $ is an uniformly bounded and equi-continuous subset of $%
C\left( \left[ 0,T\right] ,X\right) .$ First we show that $\mathcal{T}%
_{2}\left( B\right) =\left\{ \mathcal{T}_{2}x:x\in B\right\} $ is uniformly
bounded. By the conditions (a) and (b) for any $x\in B,$ we have the
following estimates%
\begin{eqnarray*}
\left\Vert \mathcal{T}_{2}x\left( t\right) \right\Vert &=&\left\Vert
Gx\left( t\right) \int_{0}^{t}g\left( s,x\left( s\right) \right) Qx\left(
s\right) ds\right\Vert \\
&\leq &\left\Vert Gx\left( t\right) \right\Vert .\left\Vert
\int_{0}^{t}g\left( s,x\left( s\right) \right) Qx\left( s\right)
ds\right\Vert \\
&\leq &\left\Vert Gx\right\Vert _{\infty }\int_{0}^{t}\left\Vert g\left(
s,x\left( s\right) \right) Qx\left( s\right) \right\Vert ds \\
&\leq &\left\Vert Gx\right\Vert _{\infty }\int_{0}^{t}\left\Vert g\left(
s,x\left( s\right) \right) \right\Vert .\left\Vert Qx\left( s\right)
\right\Vert ds \\
&\leq &\left\Vert Gx\right\Vert _{\infty }.\left\Vert Qx\right\Vert _{\infty
}\int_{0}^{T}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert ds \\
&\leq &\left\Vert Gx\right\Vert _{\infty }.\left\Vert Qx\right\Vert _{\infty
}.\int_{0}^{T}u\left( t\right) \theta \left( \left\Vert x\left( s\right)
\right\Vert \right) ds \\
&\leq &\varphi \left( \left\Vert x\right\Vert _{\infty }\right) .\varphi
\left( \left\Vert x\right\Vert _{\infty }\right) .\theta \left( \left\Vert
x\right\Vert _{\infty }\right) .\left\Vert u\right\Vert _{1} \\
&\leq &\varphi \left( 1\right) .\psi \left( 1\right) .\theta \left( 1\right)
.\left\Vert u\right\Vert _{1}.
\end{eqnarray*}

Hence, putting $M_{2}:=\varphi \left( 1\right) .\psi \left( 1\right) .\theta
\left( 1\right) .\left\Vert u\right\Vert _{1},$ we conclude that, $\mathcal{T%
}_{2}\left( B\right) $ is uniformly bounded. Now we show that, $\mathcal{T}%
_{2}\left( B\right) $ is an uniformly equi-continuous subset of $C\left( %
\left[ 0,T\right] ,X\right) .$ To see this, let $x\in B$ be arbitrary$,$ and
let $\varepsilon >0.$ Since $Gx$ is uniformly continuous, there exists some $%
\delta _{1}\left( \varepsilon \right) >0$ such that

\begin{equation*}
\forall t_{1},t_{2}\in \left[ 0,T\right] ,\left( \left\vert
t_{2}-t_{1}\right\vert <\delta _{1}\left( \varepsilon \right)
\Longrightarrow \left\Vert Gx\left( t_{2}\right) -Gx\left( t_{1}\right)
\right\Vert <\varepsilon _{1}\right),
\end{equation*}

where $\varepsilon _{1}$ is depends on $\varepsilon $ and will be given.
Without loss of generality we may assume that $t_{1}<t_{2}.$ Now by the Mean
Value Theorem for Integrals, there exists some $c_{x}\in \left(
t_{1},t_{2}\right) $ such that

\begin{equation*}
\int_{t_{1}}^{t_{2}}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert ds=\left( t_{2}-t_{1}\right) .\left\Vert g\left( c_{x},x\left(
c_{x}\right) \right) \right\Vert .
\end{equation*}

Let $\delta \left( \varepsilon \right) =\min \left\{ \delta _{1}\left(
\varepsilon \right) ,\varepsilon _{2}\right\} ,$ where $\varepsilon _{2}$ is
depends on $\varepsilon $ and will be given. Therefore, if $t_{1},t_{2}\in %
\left[ 0,T\right] $ satisfies $0<t_{2}-t_{1}<\delta \left( \varepsilon
\right) $ and $x\in B,$ then we have the following estimates

\begin{eqnarray*}
\left\Vert \mathcal{T}_{2}x\left( t_{2}\right) -\mathcal{T}_{2}x\left(
t_{1}\right) \right\Vert &=&\left\Vert Gx\left( t_{2}\right)
\int_{0}^{t_{2}}g\left( s,x\left( s\right) \right) Qx\left( s\right)
ds-Gx\left( t_{1}\right) \int_{0}^{t_{1}}g\left( s,x\left( s\right) \right)
Qx\left( s\right) ds\right\Vert \\
&\leq &\left\Vert Gx\left( t_{2}\right) -Gx\left( t_{1}\right) \right\Vert
.\int_{0}^{t_{1}}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert
.\left\Vert Qx\left( s\right) \right\Vert ds \\
&&+\left\Vert Gx\left( t_{2}\right) \right\Vert
.\int_{t_{1}}^{t_{2}}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert .\left\Vert Qx\left( s\right) \right\Vert ds \\
&\leq &\varepsilon _{1}\left\Vert Qx\right\Vert _{\infty
}.\int_{0}^{T}\left\Vert g\left( s,x\left( s\right) \right) \right\Vert
ds+\left\Vert Gx\right\Vert _{\infty }.\left\Vert Qx\right\Vert _{\infty
}.\int_{t_{1}}^{t_{2}}\left\Vert g\left( s,x\left( s\right) \right)
\right\Vert ds \\
&\leq &\varepsilon _{1}\psi \left( \left\Vert x\right\Vert _{\infty }\right)
.\int_{0}^{T}u\left( s\right) \theta \left( \left\Vert x\left( s\right)
\right\Vert \right) ds+\varphi \left( \left\Vert x\right\Vert _{\infty
}\right) .\psi \left( \left\Vert x\right\Vert _{\infty }\right) \left(
t_{2}-t_{1}\right) .\left\Vert g\left( c_{x},x\left( c_{x}\right) \right)
\right\Vert \\
&\leq &\varepsilon _{1}\psi \left( \left\Vert x\right\Vert _{\infty }\right)
.\left\Vert u\right\Vert _{1}.\theta \left( \left\Vert x\right\Vert _{\infty
}\right) +\varepsilon _{2}\varphi \left( \left\Vert x\right\Vert _{\infty
}\right) .\psi \left( \left\Vert x\right\Vert _{\infty }\right) u\left(
c_{x}\right) \theta \left( \left\Vert x\left( c_{x}\right) \right\Vert
\right) \\
&\leq &\varepsilon _{1}\psi \left( 1\right) .\theta \left( 1\right)
\left\Vert u\right\Vert _{1}+\varepsilon _{2}\varphi \left( 1\right) .\psi
\left( 1\right) u\left( T\right) \theta \left( 1\right) \\
&<&\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon,
\end{eqnarray*}
where%
\begin{eqnarray*}
\varepsilon _{1} &=&\dfrac{\varepsilon }{2\left( 1+\psi \left( 1\right)
.\theta \left( 1\right) .\left\Vert u\right\Vert _{1}\right) }, \\
\varepsilon _{2} &=&\dfrac{\varepsilon }{2\left( 1+\varphi \left( 1\right)
.\psi \left( 1\right) .u\left( T\right) .\theta \left( 1\right) \right) }.
\end{eqnarray*}

Therefore $\mathcal{T}_{2}$ is a compact operator.

Next, we show that $\mathcal{T}_{1}$ is a F-contraction. Let $x,y\in C\left( %
\left[ 0,T\right] ,X\right) ,$ and $\left\Vert \mathcal{T}_{1}x-\mathcal{T}%
_{1}y\right\Vert _{\infty }>0.$ By applying the fact that every continuous
function attains it's maximum on a compact set, there exists $t\in \left[ 0,T%
\right] $ such that $0<\left\Vert \mathcal{T}_{1}x-\mathcal{T}%
_{1}y\right\Vert _{\infty }=\left\Vert f\left( t,x\left( t\right) \right)
-f\left( t,y\left( t\right) \right) \right\Vert .$ By (a) and using the fact
that, $F$ and $\phi $ are strictly increasing functions we obtain

\begin{eqnarray*}
\tau +F\left( \left\Vert \mathcal{T}_{1}x-\mathcal{T}_{1}y\right\Vert
_{\infty }+\phi \left( \left\Vert \mathcal{T}_{1}x-\mathcal{T}%
_{1}y\right\Vert _{\infty }\right) \right) &=&\tau +F\left( \left\Vert
f\left( t,x\left( t\right) \right) -f\left( t,y\left( t\right) \right)
\right\Vert +\phi \left( \left\Vert f\left( t,x\left( t\right) \right)
-f\left( t,y\left( t\right) \right) \right\Vert \right) \right) \\
&\leq &F\left( \left\Vert x\left( t\right) -y\left( t\right) \right\Vert
+\phi \left( \left\Vert x\left( t\right) -y\left( t\right) \right\Vert
\right) \right) \\
&\leq &F\left( \left\Vert x-y\right\Vert _{\infty }+\phi \left( \left\Vert
x-y\right\Vert _{\infty }\right) \right) .
\end{eqnarray*}

Hence $\mathcal{T}_{1}$ is a F-contraction. Now we show that there exists
some $M_{3}>0$ such that $\left\Vert \mathcal{T}_{1}x\right\Vert _{\infty
}\leq M_{3}$ holds for each $x\in C\left( \left[ 0,T\right] ,X\right) .$
Since $F$ is bijective and strictly increasing we have

\begin{equation*}
\left\Vert \mathcal{T}_{1}x-\mathcal{T}_{1}y\right\Vert _{\infty }+\phi
\left( \left\Vert \mathcal{T}_{1}x-\mathcal{T}_{1}y\right\Vert _{\infty
}\right) \leq F^{-1}\left[ F\left( \left\Vert x-y\right\Vert _{\infty }+\phi
\left( \left\Vert x-y\right\Vert _{\infty }\right) \right) -\tau \right],
\end{equation*}

Let $0<\left\Vert x\right\Vert _{\infty }+\varphi \left( \left\Vert
x\right\Vert _{\infty }\right) ,$ since $F\left( \left\Vert x\right\Vert
_{\infty }+\varphi \left( \left\Vert x\right\Vert _{\infty }\right) \right)
<0,$ the above inequality implies that

\begin{eqnarray*}
\left\Vert \mathcal{T}_{1}x\right\Vert _{\infty } &\leq &\left\Vert \mathcal{%
T}_{1}x-\mathcal{T}_{1}0\right\Vert _{\infty }+\left\Vert \mathcal{T}%
_{1}0\right\Vert _{\infty } \\
&\leq &\left\Vert \mathcal{T}_{1}x-\mathcal{T}_{1}0\right\Vert _{\infty
}+\phi \left( \left\Vert \mathcal{T}_{1}x-\mathcal{T}_{1}0\right\Vert
_{\infty }\right) +\left\Vert \mathcal{T}_{1}0\right\Vert _{\infty } \\
&\leq &F^{-1}\left[ F\left( \left\Vert x\right\Vert _{\infty }+\phi \left(
\left\Vert x\right\Vert _{\infty }\right) \right) -\tau \right] +\left\Vert 
\mathcal{T}_{1}0\right\Vert _{\infty } \\
&\leq &F^{-1}\left[ -\tau \right] +\left\Vert \mathcal{T}_{1}0\right\Vert
_{\infty }.
\end{eqnarray*}

Therefore%
\begin{equation*}
\exists \text{ }M_{3}>0,\forall x\text{ }\left( x\in C\left( \left[ 0,T%
\right] ,X\right) \implies \left\Vert \mathcal{T}_{1}x\right\Vert _{\infty
}\leq M_{3}\right) ,
\end{equation*}

where, $M_{3}:=F^{-1}\left[ -\tau \right] +\left\Vert \mathcal{T}%
_{1}0\right\Vert _{\infty }.$ Finally, we claim that there exists some $r>0,$
such that $\mathcal{J}\left( B_{r}\left( 0\right) \right) \subseteq
B_{r}\left( 0\right) $ with $B_{r}\left( 0\right) =\left\{ x\in C\left( %
\left[ 0,T\right] ,X\right) :\left\Vert x\right\Vert _{\infty }\leq
r\right\} .$ On the contrary, for any $\gamma >0$ there exists some $%
x_{\gamma }\in B_{r}\left( 0\right) $ such that $\left\Vert \mathcal{J}%
\left( x_{\gamma }\right) \right\Vert >\gamma .$ This implies that $%
\liminf_{\gamma \rightarrow \infty }\frac{1}{\gamma }\left\Vert \mathcal{J}%
\left( x_{\gamma }\right) \right\Vert \geq 1.$ On the other hand, we have

\begin{eqnarray*}
\left\Vert \mathcal{J}x_{\gamma }\left( t\right) \right\Vert &=&\left\Vert
f\left( t,x_{\gamma }\left( t\right) \right) \right\Vert +\left\Vert
Gx_{\gamma }\left( t\right) \int_{0}^{t}g\left( s,x_{\gamma }\left( s\right)
\right) Qx_{\gamma }\left( s\right) ds\right\Vert \\
&\leq &\left\Vert \mathcal{T}_{1}x_{\gamma }\right\Vert _{\infty
}+\left\Vert Gx_{\gamma }\left( t\right) \right\Vert .\left\Vert
\int_{0}^{t}g\left( s,x_{\gamma }\left( s\right) \right) Qx_{\gamma }\left(
s\right) ds\right\Vert \\
&\leq &M_{3}+\left\Vert Gx_{\gamma }\right\Vert _{\infty }.\left\Vert
Qx_{\gamma }\right\Vert _{\infty }\int_{0}^{t}\left\Vert g\left( s,x_{\gamma
}\left( s\right) \right) \right\Vert ds \\
&\leq &M_{3}+\varphi \left( \left\Vert x_{\gamma }\right\Vert _{\infty
}\right) .\psi \left( \left\Vert x_{\gamma }\right\Vert _{\infty }\right)
.\int_{0}^{T}u\left( t\right) \theta \left( \left\Vert x_{\gamma }\left(
s\right) \right\Vert \right) ds \\
&\leq &M_{3}+\varphi \left( \gamma \right) .\psi \left( \gamma \right)
.\theta \left( \gamma \right) .\left\Vert u\right\Vert _{1}.
\end{eqnarray*}

Hence, by the above estimate and condition (d) we get

\begin{equation*}
\liminf_{\gamma \rightarrow \infty }\frac{1}{\gamma }\left\Vert \mathcal{J}%
\left( x_{\gamma }\right) \right\Vert _{\infty }\leq \liminf_{\gamma
\rightarrow \infty }\frac{\varphi \left( \gamma \right) .\psi \left( \gamma
\right) .\theta \left( \gamma \right) .\left\Vert u\right\Vert _{1}}{\gamma }%
<1
\end{equation*}%
which is a contradiction. Thus in view of the above discussions and
Corollary \ref{MR4} we conclude that Eq.(\ref{D-1}) has at least one
solution in $B_{r}\left( 0\right) \subseteq C\left( \left[ 0,T\right]
,X\right) .$
\end{proof}

\begin{example}
\label{Exam}Consider the Volterra integral equation of the form
\end{example}

\begin{gather}
x\left( t\right) =\arctan \left( e^{3t}\right) +\dfrac{\left\vert x\left(
t\right) \right\vert }{\left( 1+7\sqrt[5]{8\left\vert x\left( t\right)
\right\vert }\right) ^{5}}+\frac{1}{4}\frac{\sqrt[3]{\left\vert x\left(
t\right) \right\vert }}{2+\cos ^{4}x\left( t\right) }  \notag \\
\times \int_{0}^{t}\dfrac{s^{2}\sqrt[7]{\left\vert x\left( s\right)
\right\vert }e^{-x\left( s\right) }sinx\left( s\right) }{\left( 1+s\right)
\left( 1+x^{2}\left( s\right) \right) }\ln \left( 1+\frac{\sqrt[5]{%
\left\vert x\left( s\right) \right\vert }}{3}\right) ds,\text{ }t\in \left[
0,1\right] .  \label{D-2}
\end{gather}

Let $X:=%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,$ $\tau :=7$ and $T:=1.$ Now in order to investigate the conditions of
Theorem \ref{Apl} we have

\begin{itemize}
\item[(a)] Define the functions $f:\left[ 0,1\right] \times 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ given by $f\left( t,x\right) =\arctan \left( e^{3t}\right) +\dfrac{%
\left\vert x\left( t\right) \right\vert }{\left( 1+7\sqrt[5]{8\left\vert
x\left( t\right) \right\vert }\right) ^{5}},$ $f$ is continuous and
\end{itemize}

\begin{equation*}
\left\vert f\left( t,x\right) -f\left( t,y\right) \right\vert \leq
\left\vert \dfrac{\left\vert x\left( t\right) \right\vert }{\left( 1+7\sqrt[5%
]{8\left\vert x\left( t\right) \right\vert }\right) ^{5}}-\dfrac{\left\vert
y\left( t\right) \right\vert }{\left( 1+7\sqrt[5]{8\left\vert y\left(
t\right) \right\vert }\right) ^{5}}\right\vert .
\end{equation*}

Consider the function $h:\left[ 0,\infty \right) \rightarrow \left[ 0,\infty
\right) ,h\left( t\right) =\dfrac{t}{\left( 1+7\sqrt[5]{8t}\right) ^{5}}.$

Since $h^{\prime }\left( t\right) =\dfrac{1}{\left( 1+7\sqrt[5]{8t}\right)
^{6}}>0$ and \ $h^{\prime \prime }\left( t\right) =\dfrac{-336}{\sqrt[5]{%
\left( 8t\right) ^{4}}\left( 1+7\sqrt[5]{8t}\right) ^{7}}<0,$ so $h$ is
strictly increasing and concave. Moreover since $h\left( 0\right) =0$ and $h$
is concave, then $h\left( t+s\right) \leq h\left( t\right) +h\left( s\right)
.$

Without loss of generality, we can assume that $\left\vert x\right\vert \geq
\left\vert y\right\vert .$ In this case we obtain

\begin{eqnarray}
\left\vert f\left( t,x\right) -f\left( t,y\right) \right\vert &\leq &h\left(
\left\vert x\right\vert \right) -h\left( \left\vert y\right\vert \right) 
\notag \\
&\leq &h\left( \left\vert x\right\vert -\left\vert y\right\vert \right) \leq
h\left( \left\vert x-y\right\vert \right)  \notag \\
&=&\frac{\left\vert x-y\right\vert }{\left( 1+7\sqrt[5]{8\left\vert
x-y\right\vert }\right) ^{5}}.  \label{D-3}
\end{eqnarray}

Now, by choosing the function $F:\left[ 0,\infty \right) \rightarrow \left(
-\infty ,0\right) $ \ given by $F\left( t\right) =\dfrac{-1}{\sqrt[5]{t}},$
and the function $\phi :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ \ given by $\phi \left( t\right) =7t,$ it is easy to see that the
inequality (\ref{D-3}) implies that the condition (\ref{F}) holds.

Indeed we have

\begin{eqnarray*}
\tau +F\left( \left\vert f\left( t,x\right) -f\left( t,y\right) \right\vert
+\phi \left( \left\vert f\left( t,x\right) -f\left( t,y\right) \right\vert
\right) \right) &\leq &F\left( \left\vert x-y\right\vert +\phi \left(
\left\vert x-y\right\vert \right) \right) \\
&\Longleftrightarrow &7+F\left( 8\left\vert f\left( t,x\right) -f\left(
t,y\right) \right\vert \right) \leq F\left( 8\left\vert x-y\right\vert
\right) \\
&\Longleftrightarrow &7-\dfrac{1}{\sqrt[5]{8\left\vert f\left( t,x\right)
-f\left( t,y\right) \right\vert }}\leq \dfrac{-1}{\sqrt[5]{8\left\vert
x-y\right\vert }} \\
&\Longleftrightarrow &\frac{1+7\sqrt[5]{8\left\vert x-y\right\vert }}{\sqrt[5%
]{8\left\vert x-y\right\vert }}\leq \dfrac{1}{\sqrt[5]{8\left\vert f\left(
t,x\right) -f\left( t,y\right) \right\vert }} \\
&\Longleftrightarrow &\sqrt[5]{8\left\vert f\left( t,x\right) -f\left(
t,y\right) \right\vert }\leq \frac{\sqrt[5]{8\left\vert x-y\right\vert }}{1+7%
\sqrt[5]{8\left\vert x-y\right\vert }} \\
&\Longleftrightarrow &\left\vert f\left( t,x\right) -f\left( t,y\right)
\right\vert \leq \frac{\left\vert x-y\right\vert }{\left( 1+7\sqrt[5]{%
8\left\vert x-y\right\vert }\right) ^{5}}.
\end{eqnarray*}

\begin{itemize}
\item[(b)] Define the continuous operators $G$ $,$ $Q:C\left( \left[ 0,1%
\right] ,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right) \rightarrow C\left( \left[ 0,1\right] ,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right) $ given by
\end{itemize}

\begin{eqnarray*}
Gx &=&\frac{\sqrt[3]{\left\vert x\right\vert }}{4\left( 2+\cos ^{4}x\right) }%
, \\
Qx &=&\ln \left( 1+\frac{\sqrt[5]{\left\vert x\right\vert }}{3}\right) .
\end{eqnarray*}

\bigskip By choosing the strictly continuous functions $\varphi ,\psi :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ given by $\varphi \left( t\right) =\dfrac{\sqrt[3]{t}}{4}$ and $\psi
\left( t\right) =\dfrac{\sqrt[5]{t}}{3},$ we have

\begin{eqnarray*}
\left\vert Gx\right\vert &\leq &\varphi \left( \left\vert x\right\vert
\right) , \\
\left\vert Qx\right\vert &\leq &\psi \left( \left\vert x\right\vert \right) .
\end{eqnarray*}

\begin{itemize}
\item[(c)] Define the continuous function $g:\left[ 0,1\right] \times 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ given by
\end{itemize}

\begin{equation*}
g\left( t,x\right) =\frac{t^{2}\sqrt[7]{\left\vert x\right\vert }e^{-x}\sin x%
}{\left( 1+t\right) \left( 1+x^{2}\right) }.
\end{equation*}

Considering the increasing function $u\in L^{1}\left( \left[ 0,1\right] ,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\right) $ given by $u\left( t\right) =\dfrac{t^{2}}{1+t},$ and the
increasing and continuous function $\theta :%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{+}$ given by $\theta \left( x\right) =\sqrt[7]{x},$ and we have

\begin{equation*}
\left\vert g\left( t,x\right) \right\vert \leq \frac{t^{2}\sqrt[7]{%
\left\vert x\right\vert }}{\left( 1+t\right) }=u\left( t\right) .\theta
\left( \left\vert x\right\vert \right) .
\end{equation*}

\begin{itemize}
\item[(d)] Since $\int_{0}^{1}\dfrac{t^{2}}{1+t}dt=\ln 2-\dfrac{1}{2},$ so
we have
\end{itemize}

\begin{eqnarray*}
\lim_{\gamma \rightarrow \infty }\dfrac{\varphi \left( \gamma \right) \psi
\left( \gamma \right) \theta \left( \gamma \right) \left\Vert u\right\Vert
_{1}}{\gamma } &=&\lim_{\gamma \rightarrow \infty }\dfrac{\dfrac{\sqrt[3]{%
\gamma }}{4}\dfrac{\sqrt[5]{\gamma }}{3}\sqrt[7]{\gamma }\int_{0}^{1}\dfrac{%
t^{2}}{1+t}dt}{\gamma } \\
&=&\dfrac{1}{12}\left( \ln 2-\dfrac{1}{2}\right) \lim_{\gamma \rightarrow
\infty }\left( \dfrac{1}{\gamma }\right) ^{\frac{34}{105}}=0.
\end{eqnarray*}

So the all conditions of Theorem \ref{Apl} are satisfied and Eq.(\ref{D-2})
has at least one solution in $C\left( \left[ 0,1\right] ,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right) .$

\section*{Acknowledgement}

The first author is thankful to the United State-India Education Foundation,
New Delhi, India and IIE/CIES, Washington, DC, USA for Fulbright-Nehru PDF
Award (No. 2052/FNPDR/2015).

{\small \setlength{\parskip}{0pt} \setlength{\baselineskip}{13pt} \vspace{6pt%
} }

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\end{thebibliography}

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