\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 105, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/105\hfil fix-point theorem]
{Fixed point theorem and its application to perturbed integral
equations in modular function spaces}
\author[A. Hajji, E. Hanebaly\hfil EJDE-2005/105\hfilneg]
{Ahmed Hajji, Ela\"idi Hanebaly }  % in alphabetical order

\address{Ahmed Hajji\hfill\break
 Department of Mathematics And Informatic,
 Mohammed V University,
 BP. 1014, Rabat, Morocco}
\email{hajid2@yahoo.fr}

\address{Ela\"idi Hanebaly\hfill\break
 Boulevard Mohammed El Yazidi. S 12 C6 Hay Riad, Rabat, Morocco}
\email{hanebaly@hotmail.com}

\date{}
\thanks{Submitted October 14, 2004. Published October 3, 2005.}
\subjclass[2000]{46A80, 47H10, 45G10} 
\keywords{Modular space; fixed point; integral equation}

\begin{abstract}
 In this paper, we present a modular version of Krasnoselskii's
 fixed point theorem. Then this result is applied to the
 existence of solutions to perturbed integral equations
 in modular function spaces.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

 \section{Introduction}

Using the same argument as in \cite{ai}, we present a modular
version of Krasnoselskii's fixed point theorem, result that is well
known in Banach spaces. The modular $\rho$ considered here is
convex, satisfies the Fatou property, and satisfies the $\Delta_2
$-condition. We are interested in the existence of a fixed point for
the  application $S: B\to B$; where $B$ is a convex, closed, and
bounded subset of $X_\rho $; $S = T+U $ with $ T: B \to B $ that
satisfies a contraction type hypothesis (see \cite{ai}); and  $U: B
\to B $ is $\rho$-completely continuous.

 Since $\rho $ satisfies the $\Delta_2 $-condition,
$U$ being  $\rho$-completely continuous is equivalent to
the condition $U,\;\|\cdot\|_\rho $-completely continuous,
where $\|\cdot\|_\rho $ is the Luxemburg norm. On the other hand if $T$
is $\rho$-contraction, then $T$ is not necessarily
$\|\cdot\|_\rho $-contraction (see counterexample in
\cite[page 945, Ex. 2.15]{krk}).

 We apply our main theorem to the study of solutions to the perturbed
integral equation
\begin{equation}
u(t) = \exp {(-t)} f_0 + \int_0 ^t  \exp{(s-t)} ( T + h) u(s)ds
\label{eI}
\end{equation}
in the modular space $C^\varphi =C ([0,b], L^\varphi )$,
where $L^\varphi $ is the Musielak-Orlicz space, $f_0 $ is a fixed
element in $L^\varphi $. Some hypotheses on the operators $T$
and $h$ are stated below. Also, we present an example of this
class of equations.

For more details about  modular spaces, we refer the reader
to the books edited by Musielak \cite{jm} and by Kozlowski \cite{rh}.
Now recall some definitions.

 Let $X$ be an arbitrary vector space over  $K$
 ($K=\mathbb{R}$ or  $K=\mathbb{C}$).

\noindent(a) A functional $\rho: X \to [0, +\infty]$ is called  modular if
\begin{itemize}
\item[(i)] $\rho(x)= 0$ implies $x=0$.
\item[(ii)] $ \rho(-x)= \rho(x)$ for all $x$ in $X$ in the case of
$X$ being real.
 $\rho(e^{it} x)= \rho(x)$ for any real $t$ in the case of $X$ being complex.
\item[(iii)] $ \rho( \alpha x+ \beta y )\leq \rho(x) + \rho (y)$ for
$ \alpha , \beta \geq 0$ and $ \alpha +\beta=1$.
\end{itemize}
If in place of (iii) there holds
\begin{itemize}
\item[(iii')] $\rho( \alpha x+ \beta y )\leq {\alpha} \rho(x)
+{\beta} \rho (y)$ for $ \alpha , \beta \geq 0 $ and
$  {\alpha} +{\beta}=1$,
\end{itemize}
then the modular $\rho$ is called convex.

\noindent (b) If $\rho$ is a modular in $X$, then the set
$ X_{\rho} =\{x\in X : \rho ( \lambda x)\to0 \mbox{ as }\lambda \to 0 \} $
 is called a modular space.

\noindent (c) (i) If $\rho $ is a modular in $X$, then $|x|_\rho =
\inf  \{ u> 0 ,  \rho (\frac{x}{u} )\leq u \}$ is a $F$-norm.\\
 (ii)
If $\rho$ is a convex modular, then $ \|x \|_{\rho}  = Inf \{u
>0 , \rho( \frac{x}{ u  }) \leq 1 \} $ is called the Luxemburg
norm.
\smallskip

 Let $X_\rho $ be a modular space.
\noindent (a) A sequence $(x_n )_{n\in \mathbb{N} } $ in $X_\rho $ is said to be\\
(i) $\rho$-convergent to $x$, denoted by $ x_n \stackrel{\rho}\to x $,
if $\rho (x_n -x ) \to 0 $ as $n \to +\infty $.\\
(ii) $\rho$-Cauchy if $\rho (x_n -x_m ) \to 0 $ as $n,m \to +\infty $.

\noindent(b) $X_\rho $ is $\rho $-complete if any $\rho$-Cauchy sequence
is $\rho$-convergent.

\noindent(c) A subset $B$ of $X_\rho $ is said to be $\rho$-closed if
 for any sequence $(x_n )_{n\in \mathbb{N}} \subset B $, such that
$ x_n \overset{\rho}{\to} x $, then $x\in B$.
Here ${\overline{B}}^\rho $ denotes the closure of $B$ in the sense
of $\rho $.

We say that the subset $A$ of $ X_\rho $ is $\rho $-bounded if: \\
$\sup_{x,y \in A} \rho (x-y) < +\infty $, and let the
$\rho$-diameter of $A$, denoted by $\delta_\rho (A)$, to be
$$
\delta_\rho (A)= \sup_{x,y \in A} \rho (x-y).
$$

 Recall also that $\rho$ has the Fatou property if
$ \rho (x-y) \leq \liminf \rho(x_n - y_n ) $,  whenever
 $ x_n \overset{\rho}{\to} x $ and $ y_n \overset{\rho}{\to} y$.

We say that $ \rho $  satisfies the $ \Delta_2 $-condition if:\\
$ \rho (2x_n  ) \to 0$ as $n \to+\infty$ whenever
$\rho (x_n  ) \to 0$ as $n \to+\infty$, for any sequence
$ (x_n )_{{n}{\in \mathbb{N} }}$ in $ X_ \rho$.

\section{Main result}

\begin {theorem} \label{thm2.1}
Let $\rho $ be a convex modular that satisfies the $\Delta_2 $-condition,
$X_\rho $ be a $\rho$-complete modular space and $B$ be a convex,
$\rho$-closed, $\rho$-bounded subset of $X_\rho $.
Assume that $U$ and $T$ are two applications from $B$ into $B$ such that
$U$ is $\rho$-completely continuous and there exist real numbers $k>0 $,
 and $ c>\max (1,k)$ that satisfy
 $ \rho (c(Tx -Ty )) \leq k\rho (x-y) $ for any $x,y $ in  $B$.
 And $T(B)  + U(B) \subset B$.
Then the operator $S= T+U $ has a fixed point.
\end{theorem}

\begin{remark} \label{rmk2.0} \rm
Since an operator $\rho $-Lipschitz is not necessarily $\|.\|_\rho
$-Lipschitz (see counterexample in \cite[page 945, Ex. 2.15]{krk}),
then the result above  gives a modular version of Krasnoselskii's
fixed point
 theorem.
\end{remark}

We need the following lemma for proving Theorem \ref{thm2.1}.

\begin{lemma} \label{lem2.1}
Let $\rho$ be a convex modular and $X_\rho $ be a modular space.
If a subset $B$ of $X_\rho $ is $\rho$-bounded then $B$ is
$\|.\|_\rho $-bounded.
\end{lemma}

\begin{proof}
Suppose that $B$ is not $\|.\|_\rho $-bounded. So there exist
sequences $(x_n )_{n \in \mathbb{N} } $ and $(y_n )_{n \in \mathbb{N} } $
in $B$ such that $ \|x_n - y_n \|_\rho \to +\infty $ as $n \to +\infty$.
Hence  for any $ A > 1$ there exists  $ N \in \mathbb{N}$, such that if $n>N$,
then $\|x_n - y_n \|_\rho  > A $ i.e. $\|\frac{x_n - y_n }{A} \|_\rho > 1 $
whenever $n>N$. This implies
$\rho(\frac{ x_n - y_n}{A}) \geq \|\frac{ x_n - y_n }{A} \|_\rho   > 1 $
(see \cite[p.8]{jm}). Hence,
$1< \rho(\frac{ x_n - y_n}{A})\leq \frac{1}{A} \rho( x_n - y_n)$
whenever $n>N $. So $ A<\rho( x_n - y_n)$ for any $n>N $.
This shows that $B$ is not $\rho$-bounded. Hence, the lemma is established.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
 Firstly, we show that the operator $I-T $ is a bijection  from $B$
into $U(B)$ (where $I$ is the identity function).
Let $x$ in $B$, and consider the following sequence defined by
 $ y_{n+1}= Ty_n + Ux$,  with $y_0 $  a fixed element in $B$.
Then the sequence $(y_n )_{n\in \mathbb{N}} $ is Cauchy.
 Indeed,
\begin{align*}
\rho (y_{n+m} - y_n ) &= \rho ( Ty_{m+n-1} - Ty_{n-1} )\\
&= \rho ( \frac{1}{c} (c (Ty_{n+m-1} - Ty_{n-1} )))\\
&\leq \frac{k}{c}\rho (y_{m+n-1} -y_{n-1} ),
\end{align*}
 by induction, we have
$$
\rho (y_{m+n} -y_n ) \leq (\frac{k}{c})^n \rho (y_m -y_0 )
$$
and by  hypothesis, $B$ is $\rho$-bounded, then we have
$\rho (y_m-y_0 ) \leq \delta_\rho (B) <\infty $ for any $m\in \mathbb{N} $,
which implies
$$
\rho (y_{m+n} -y_n ) \leq (\frac{k}{c})^n \delta_\rho (B) ,
$$
and by hypothesis  $c> \max (1,k) $ we have $(\frac{k}{c})^n \to 0$ as
$n \to +\infty $. Therefore,
$\rho (y_{m+n} -y_n ) \to 0 $ as $ n,m \to +\infty $. Which implies
that the sequence $(y_n )_{n\in \mathbb{N}} $ is $\rho$-Cauchy. Since $X_\rho $
is $\rho$-complete, $B$ is closed and $T$ is continuous then the
sequence $(y_n)_{n\in \mathbb{N}}$ is  convergent to an element $y\in B$ and
$ y= Ty + Ux $. Indeed,
\begin{align*}
\rho  (\frac{ y -Ty - U(x)}{2})
&= \rho  (\frac{y- y_n + y_n -Ty -U(x) }{2})\\
&= \rho  (\frac{y- y_n +Ty_{n-1}-Ty}{2})\\
&\leq \rho (y-y_n ) + \rho (Ty_{n-1} -Ty ),
\end{align*}
which implies that $y- Ty = U(x)$.

 Then it follows that for any  $x\in B $, there exists $y\in B $ such
   that $(I-T) y = Ux $. Therefore, we get that
$ (I-T) (B) \subset U(B) $ (Indeed, if we suppose that  there exists
$y\in B$ such that $y-Ty \notin  U(B)$ i.e., for any $x\in B$,
 we have $y-Ty \ne U(x) $ which is absurd), and  $I-T$ is a
surjective operator from  $B$ into $U(B)$.

Let $y_1, y_2 $ in $B$ such that $(I-T) y_1 = (I-T)y_2 $,
 then $ y_1 -y_2 = Ty_1 -Ty_2 $; therefore,
  $\rho (y_1 -y_2 ) \leq \frac{k}{c}\rho (y_1 -y_2 )$, and since
   $ c> \max (1,k) $ it follows that $\rho (y_1 -y_2 ) =0 $ and $y_1 =y_2$.
Which shows that $I-T$ is injective operator. Therefore,
$I-T $ is a bijection operator from $B$ into $U(B) $.

 Secondly, we show  that $ (I-T)^{-1} $ is continuous.
Let $(x_n )_{n\in \mathbb{N} } \subset U(B) $ be a convergent sequence to
$x\in U(B)$, and consider the sequence defined by
$z_n = (I-T)^{-1}(x_n)$, then $(z_n)_{n\in \mathbb{N}}$ is $\rho$-Cauchy. Indeed,
\begin{align*}
z_{n+m} -z_n
&= z_{m+n} - T z_{m+n} + Tz_{m+n} - Tz_n + Tz_n - z_n \\
&= x_{m+n} + Tz_{m+n} - Tz_n -x_n \\
&= x_{m+n} - x_n + T z_{m+n} - Tz_n ;
\end{align*}
 therefore, if we take $\alpha$ such that $\frac{1}{\alpha} + \frac{1}{c} =1$,
then
\begin{align*}
\rho (z_{m+n} -z_n )
&= \rho (\frac{1}{c}(c(Tz_{m+n} -Tz_n )) + \frac{1}{\alpha} \alpha (x_{m+n} -x_n ))\\
&\leq \frac{k}{c} \rho (z_{m+n} - z_n )+ \frac{1}{\alpha}\rho (\alpha (x_{m+n} -x_n )).
\end{align*}
Then,
$$
\rho( z_{m+n} -z_n ) \leq \frac{c}{c-k} \frac{1}{\alpha}
\rho (\alpha (x_{m+n} -x_n )).
$$
And since  $\rho (x_{m+n} -x_n ) \to 0 $ as $m, n \to +\infty $, then by the
 $\Delta_2 $-condition $\rho (\alpha(x_{m+n} -x_n )) \to 0 $ as
$m, n \to +\infty $.
Therefore, $\rho (z_{m+n} - z_n ) \to 0 $ as $m, n \to +\infty $, and
by hypothesis $X_\rho$ is $\rho$-complete, then the sequence
$(z_n)_{n\in \mathbb{N}}$ is convergent to an element $z\in B$.
On the other hand, $x_n = z_n -T(z_n )$ is convergent to  $ x= z-T(z)$.
Indeed,
$$
\rho ( \frac{z_n -T(z_n ) - z+ T(z)}{2}) \leq \rho(z_n -z )
+ \rho (T(z_n) - T(z)).
$$
Since $\rho (z_n -z) \to 0 $ as  $n\to +\infty $ and  $T$ is continuous,
 $ \rho ( \frac{z_n -T(z_n ) - z+ T(z)}{2}) \to 0 $ as
 $n \to + \infty $, and by  $\Delta_2 $-condition we have
$ \rho (z_n - T(z_n ) -( z- T(z)) \to 0 $ as
 $n \to + \infty $. Therefore, $ (I-T)^{-1} (x_n ) $ converges
to $ (I-T)^{-1} (x)$,   which implies that $(I-T)^{-1} $ is continuous.

 Finally, we consider the function $f$ defined by
$$ f(x) = (I-T )^{-1}U(x).$$
Since $U$ is $\rho$-completely continuous and $(I-T)^{-1} $ is
 $\rho$-continuous, it follows by the $\Delta_2$-condition
that $U$ is $\|\cdot\|_\rho$- completely
 continuous and $(I-T)^{-1} $ is $\|.\|_\rho $-continuous.
Which implies that $f$ is   $\|.\|_\rho$-completely continuous from
$B$ into $B$. By the $\Delta_2$-condition, $B$ is
  $\| .\|_\rho $-closed. Then, using Lemma \ref{lem2.1} and  Schauder's
   fixed point theorem, $f$ has a fixed point.
    Let $x_0 $ be such that  $f(x_0 ) =x_0 $, then we have
     $x_0 = f(x_0) = (I-T)^{-1} U(x_0) $ which implies that
$ x_0 = (T+U) (x_0)$. Therefore, $S$ has a fixed point , which
completes the proof.
\end{proof}

The next section presents an application of Theorem \ref{thm2.1}.
We study the existence of solutions in the modular space
$ C^\varphi = C ([0,b], L^\varphi )$.
For details about the spaces $C^\varphi $ and $L^\varphi $, we refer
the reader to \cite{ai} and to books edited by Musielak \cite{jm}
and Kozlowski \cite{rh}.

\section{Perturbed integral equations}

In this section, we study the existence of solutions to perturbed integral
equations on the Musielak-Orlicz space $L^\varphi $.
For this, we begin by setting the functional framework of this integral
equation.
\subsection*{Functional framework}
 Let $L^\varphi $ be the Musielak-Orlicz space. Then both the modular $\rho$
and its associated F-norm  satisfy the Fatou property.
Hence forth, we assume that $\rho$ is convex and satisfies
 the $\Delta_2 $-condition (the $F$-norm becomes the Luxemburg norm
\cite{ka}). Therefore, we have
$$
 \|x_n -x \|_\rho \to 0 \Longleftrightarrow \rho (x_n -x) \to 0
$$
as $n \to +\infty $ on $L^\varphi $. This implies that the topologies
generated  by $\|.\|_\rho $ and $\rho $ are
equivalent. Note that, under such conditions on $\rho$,
$(L^\varphi (\Omega ), \|.\|_\rho ) $ is a Banach space, where
$\Omega =[0,b]$ .

We denote by $ C^\varphi = C([0,b], L^\varphi )$ the space of all
 $\rho$-continuous functions from $[0,b]$ to $L^\varphi $, endowed with
 the modular $\rho_a $ defined by
  $ \rho_a (u) = \sup_{t \in [0,b]}  \exp{(-at)} \rho (u(t)) $,
where $a\geq 0$.
  On the space $C^\varphi$ one can consider the three topologies associated
with the modular $\rho_a$ (see \cite{jm}  and \cite{ha}),
the Luxemburg norm $\|.\|_{\rho_a } $, and  the norm $|.|_\infty $
  defined by $|u|_\infty =\sup_{t \in [0,b]}  \|u(t) \|_\rho $.

We note that the  three topologies above are
equivalent in the following sense $ \rho_a (x_n -x) \to 0
\Leftrightarrow \| x_n -x\|_{\rho_a}\to 0 \Leftrightarrow |x_n
-x|\infty \to 0$ as $ n\to +\infty $.
Indeed, let $(x_n)_{n\in \mathbb{N} }$ be a sequence in $C^\varphi$ such
that $| x_n -x |_\infty \to 0 $ as $ n\to +\infty $ and with
$x\in C^\varphi $, hence for all $0<\epsilon <1 $ there exists
$N\in \mathbb{N} $ such that for any $n> N $ we have
$$
 \sup_{t \in [0,b]} \| x_n (t) - x(t) \|_\rho \leq \epsilon <1 .
$$
On the other hand, $\| x_n (t) -x(t) \|_\rho \leq \epsilon <1 $ for all
$t \in [0,b]$ implies $\rho (x_n (t)-x(t) ) \leq \epsilon <1 $
for all $t \in [0,b]$. Then
$$
\sup_{t \in [0,b]}  \exp{(-at)} \rho (x_n (t)-x(t))\leq \epsilon
$$
for all $n\geq N$. This  implies $\rho_a (x_n -x ) \to 0 $ as
$n\to +\infty $. By the  $\Delta_2 $-condition we have
 $\| x_n -x \|_{\rho_a} \to 0$ as $ n\to +\infty $.

Conversely, by letting  $ u>0$ be such that $ \sup_{t \in
[0,b]} \exp{(-at)} \rho (\frac{x_n (t) -x(t)}{u})\leq 1 $, we
have
$$
e^{-ab} \rho (\frac{x_n (t) -x(t) }{u}) \leq  e^{-at}
 \rho (\frac{x_n (t) -x(t)}{u}) \leq 1
$$
for all $t\in [0,b]$. This implies
$$
e^{-ab} \rho (\frac{x_n (t) -x(t) }{u}) \leq \sup_{t \in [0,b]}
  \exp{(-at)} \rho (\frac{x_n (t) -x(t)}{u})\leq 1.
$$
Therefore,
\begin{align*}
A&:= \{ u>0 ; \sup_{t \in [0,b]}  \exp{(-at)} \rho (\frac{x_n (t) -x(t)}{u})
\leq 1 \} \\
&\subset B:= e^{-ab} \{ u>0 ; \rho (\frac{x_n (t) -x(t)}{u})\leq 1 \} .
\end{align*}
Hence,
$ \inf (A) \geq \inf (B) $,
which implies
$$
 \| x_n -x \|_{\rho_a } \geq e^{-ab} \| x_n (t) -x(t) \|_\rho
$$
for all $t\in [0,b] $. Hence,
$$
e^{ab} \| x_n -x \|_{\rho_a } \geq  \sup_{t \in [0,b]}
\| x_n (t) -x(t) \|_\rho = | x_n -x |_\infty .
$$
Therefore, $|x_n -x|_\infty \to 0 $ as $n\to +\infty $ is equivalent
to $\|x_n -x \|_{\rho_a} \to 0 $ as $n\to +\infty $.

To study the integral equation \eqref{eI}.
we set the following hypotheses:
\begin{itemize}
\item[(H1)] Let $B$ be a convex, $\rho$-closed,  $ \rho$-bounded subset
 of  $L^\varphi$,   and $0\in B$.

\item[(H2)] Let $T : B \to B $ be an application for which there exists a
real number $k> 0$  such that $ \rho (Tx -Ty) \leq k \rho (x - y ) $
for all $x,y \in B $.
 Also let $ h: B \to B $ be an application $ \rho$-completely continuous
such that   $T(B)+h(B) \subseteq B$.

\item[(H3)] Let $f_0 $ be a fixed element of $B$.
\end{itemize}

\begin{theorem} \label{thm3.1}
Under these hypotheses and for any $b>0$, the integral equation \eqref{eI}
 has a solution $ u \in C^\varphi =C ([0,b],L^\varphi ) $.
\end{theorem}

When we restrict our attention to the Banach space
 $( L^\varphi , \|. \|_\rho )$, Equation \eqref{eI} can be written as
$$
 u'(t) + (I-(T+h))u(t) =0.
$$
When $h\equiv 0$, Equation  \eqref{eI} becomes
$$
u(t) = \exp {(-t)} f_0 + \int_0 ^t  \exp{(s-t)} T u(s)ds .
 $$
The  equation above has been studied in \cite{ai} and \cite{m1}. The
proof of Theorem \ref{thm3.1} is based on Lemma \ref{lem2.1} and the
next lemma.

\begin{lemma} \label{lem3.1}
If a family $M \subset C^\varphi $ is equicontinuous in the sense
of $\|.\|_\rho$, then $M$ is equicontinuous in the sense of
$\rho$.\end{lemma}

\begin{proof}
Recall that if $\|x\|_\rho <1 $, then $\rho(x) \leq \|x\|_\rho$
(see \cite[p.2]{jm}).
Let $0< \epsilon<1$, there exists $\delta >0$ such that if
$|t-\overline{t}|< \delta $ then
$\|f(t) -f(\overline{t}) \|_\rho \leq \epsilon <1$ for all $f \in M$.
 Hence,
$\rho (f(t) -f(\overline{t}) ) \leq \|f(t) -f(\overline{t})
\|_\rho \leq \epsilon $ for any $f \in M$  whenever
$|t-\overline{t} |< \delta$. This implies that $M$ is $\rho$-equicontinuous
 and the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
Let $a > 0$ and $\rho_a $ be a modular in  $D = C ( [0,b], B)$
defined by $\rho_a (u ) = \sup_{t \in [0,b]}
\exp{(-at)} \rho (u(t)) $ for $u \in D $ (see \cite{ai}).

 By \cite[Prop. 2.1 (3)]{ai}, $D$ is convex,
$\rho_a$-closed and since $B$ is $\rho$-bounded, then $D$ is
$\rho_a$-bounded .

\noindent \textbf{Claim:} $D$ is invariant under the operator
$S$ given by
 $$
Su(t) = \exp {(-t)} f_0 + \int_0 ^t  \exp{(s-t)} (T+h)u(s)ds.
$$
First, we prove that $Su$ is continuous from $[0,b]$ into
$(L^\varphi , \|.\|_\rho )$. Let $t_n$,  $t_0 \in [0,b]$ such
 that $t_n \to t_0 $ as $n\to +\infty$. Since $T $ and $h$ are
$\rho$-continuous, then $(T+h)u$ is $\rho$-continuous at $t_0$.
Indeed,
\begin{align*}
&\rho ((T+h)u(t_n ) - (T+h)u(t_0 ))\\
& \leq \frac{1}{2} \rho (2(Tu(t_n ) -Tu(t_0 ))
+ \frac{1}{2} \rho (2(hu (t_n ) - hu (t_0))).
\end{align*}
By the $\Delta_2 $-condition, we have
$\rho ((T+h)u(t_n ) - (T+h)u(t_0 )) \to 0 $ as $ n \to +\infty $.
Again by $\Delta_2 $-condition, $ (T+h)u $ is $\|.\|_\rho$-continuous
at $t_0$. Hence $Su$ is $\|.\|_\rho$-continuous at $t_0$.

Next, we prove  that $Su(t) \in B $, for any $t\in [0,b]$. It is
well known that in Banach space $(L^\varphi , \|.\|_\rho )$,
\begin{align*}
&\int_0 ^t  \exp{(s-t)} (T+h)u(s)ds \\
&\in (\int_0 ^t  \exp{(s-t)}ds)  {\overline {\mathop{\rm co}}}^{\|.\|_\rho}
\{(T+ h)u(s), \quad 0 \leq s \leq t \},
\end{align*}
 where ${\overline{\mathop{\rm co}}^{\|.\|_\rho}}$ denotes the closure
of the convex hull  in the sense of $\|.\|_\rho$.
Since $ (T+h)(B) \subseteq B$,
 $\int_0 ^t  \exp{(s-t)} (T+h)u(s)ds \in (1- \exp (-t))
{\overline {\mathop{\rm co}}}^{\|.\|_\rho} (B)$.
 But $B$ is convex and $\rho$-closed.
Thus ${\overline {\mathop{\rm co}}}^{\|.\|_\rho} (B)
=  {\overline {B}}^{\|.\|_\rho}  \subset {\overline {B}}^{\rho}= B$.
Therefore, $Su(t) \in \exp (-t) B + (1-\exp (-t) ) B \subseteq B $
for all $t \in [0,b]$. Hence, $D$ is invariant by $S$.

Now consider the operators:
$T_1 u(t) = \exp {(-t)} f_0 + \int_0 ^t  \exp{(s-t)} Tu(s)ds$ and
$h_1 u(t)=  \int_0 ^t  \exp{(s-t)}  hu(s)ds$. Observe that $S=T_1 +h_1 $.
 Next, we show that $T_1 $ and $h_1 $ satisfy the hypotheses of Theorem
\ref{thm2.1}.

\noindent (1) We note that, by the same argument in the proof of
fixed point theorem (see \cite{ai}), we show that $D$ is invariant
under $h_1$ and $T_1 $ and there exists $c>\max (1,k_0 )$ such that
$$
\rho_a (c(T_1 u -T_1 v )) \leq k_0 \rho_a (u-v), \quad \forall u,v
\in D,
$$
where $ 1< c\leq \frac{e^b}{e ^b -1}$, $k_0 =c\frac{k}{1+a}$ and $a\geq k$.
The same techniques used in the proof of $S(D) \subset D$ are used
to establish  $ T_1 (D)+ h_1 (D) \subset D$: By taking the hypothesis
$T(B) + h(B) \subset B$, which gives
 $ T_1 u(t) + h_1 v(t) \in \exp (-t) B + (1- \exp
(-t)) B \subset B$ for any $t \in [0,b] $ and $u,v \in  D$.

\noindent (2) Claim:  $h_1 $ is $\rho_a $-completely continuous. Let
$M\subset D$, then $h_1 (M)$ is equicontinuous in the sense of
$\|.\|_{\rho} $. Indeed, let $u \in M $, we have
\begin{align*}
& h_1 u(t) -h_1 u( \overline{t})\\
&= \int_0 ^t \exp {(s-t)} hu(s)ds - \int_0 ^{\overline{t}}
\exp {(s-\overline{t})} hu(s) ds \\
&= e^{-t} \int_0 ^t  e^s  hu(s)ds - e^{-\overline{t} }
 \int_0 ^{\overline{t}} e^s hu(s) ds \\
&= e^{-t} \int_0 ^t e^s hu(s)ds - e^{-\overline{t} } \int_0 ^t e^s hu(s) ds
+ e^{-\overline{t} } \int_0 ^t e^s hu(s) ds - e^{-\overline{t} }
\int_0 ^{\overline{t}} e^s hu(s) ds \\
&= (e^{-t} - e^{-\overline{t}} ) \int_0 ^t e^s hu(s) ds
+ e^{-\overline{t} } \int_{\overline{t}}  ^t e^s hu(s) ds.
\end{align*}
Hence,
\begin{align*}
\| h_1 u(t) -h_1 u(\overline {t}) \|_\rho
&\leq | e^{-t} - e^{-\overline {t}} | b e^b \delta_{\|.\|_\rho } (B)
+ \delta_{\|.\|_\rho } (B) | \int_{{\overline{t}}} ^{t}  e^s ds |\\
&\leq | e^{-t} - e^{-\overline {t}} | b e^b \delta_{\|.\|_\rho } (B)
+ \delta_{\|.\|_\rho } (B) | e^{t} - e^{\overline {t}} |
\end{align*}
On the other hand, the functions $ t \mapsto e^{-t} $ and $ t \mapsto e^{t}$
are  uniformly continuous on the compact $[0,b]$. Hence for
  $\epsilon >0 $, there exists  $ \eta_1 >0 $ such that if
$| t- \overline{t} | < \eta_1 $ then
  $| e^{-t} - e^{-\overline {t}} |\leq \frac{\epsilon}{2b e^b
\delta_{\|.\|_\rho } (B) }$,
and there exists $\eta_2 >0 $ such that if $| t- \overline{t} | < \eta_2 $
then $| e^{t} - e^{\overline {t}} |\leq \frac{\epsilon}{2 \delta_{\|.\|_\rho }
 (B) }$.

Hence, there exists $\eta = \min (\eta_1 , \eta_2 ) $ such that if
$| t- \overline{t} | < \eta $ then
 $\| h_1 u(t) -h_1 u(\overline {t}) \|_\rho \leq \epsilon $ for any
$u \in M $.
Therefore, $h_1 (M) $ is equicontinuous in the sense of $\|.\|_\rho $,
and by Lemma \ref{lem3.1}, $h_1 (M)$ is  $\rho$-equicontinuous. Otherwise,
\begin{align*}
h_1 u(t)=  \int_0 ^t \exp{(s-t)} hu(s)ds
&\in (1- \exp (-t))  {\overline {\mathop{\rm co}}}^{\|.\|_\rho} \{ hu(s),
\; 0 \leq s \leq t \}\\
&\subset (1- \exp (-t)) {\overline {\mathop{\rm co}}}^{\|.\|_\rho} (h(B)).
\end{align*}
Hence $ h_1 (M(t)) \subset (1-\exp (-t)){\overline {\mathop{\rm co}}}
^{\|.\|_\rho} ( h(B))$ for all  $t\in [0,b]$. But $h(B)$ is $\rho$-compact
and by $\Delta_2$-condition $h(B)$ is  $\|.\|_\rho $  compact, which
implies that ${\overline {\mathop{\rm co}}}^{\|.\|_\rho} (h(B))$
 is compact. Therefore, ${\overline{h_1 (M(t))}}$ is $\|.\|_\rho $
compact for all $t\in [0,b]$, and by Ascoli's theorem
${\overline{h_1 (M)}}^{|.|_\infty}$ is  compact.
  Hence, by the equivalence of three topologies considered in functional
framework,  ${\overline { h_1 (M)}} $ is $\rho_a $-compact.
 Using the standard techniques
\cite[proof of the Theorem 3 page 103]{ro}, we show that $h_1 $ is
$\|.\|_{\rho_a} $-continuous then  $h_1 $ is $\rho_a $-continuous.
Hence,  $ h_1$ is $\rho_a$-completely continuous. It then follows
from Theorem \ref{thm2.1} that $S $ has a fixed point which is a
solution of the equation \eqref{eI}.
\end{proof}

\subsection{Example of equation \eqref{eI}}

In this example,  we study  the existence of a solution of the integral
equation
\begin{equation} \label{eII}
\begin{aligned}
u(t) &= \exp {(-t)} f_0 + \int_0 ^t  \exp{(s-t)}
 (\int_0 ^b \exp (-\xi ) g_2 ( s, \xi , u(\xi))d\xi ) ds \\
&\quad + \int_0 ^t \exp (s-t) (\int_0 ^b \exp (-\xi )
g_1 (s, \xi , u(\xi ))d\xi ) ds/
\end{aligned}
\end{equation}
under the hypotheses stated below.
 Let $X_\rho $ be a finite dimensional vector subspace of
$L^\varphi $, and $\rho $ be a convex modular on $L^\varphi $,
satisfying the $\Delta_2 $-condition.  Let $B$ be a convex,
$\rho$-closed,   $ \rho$-bounded subset of $X_\rho $ and $0\in B$.
Let $b>0$ very small,  $g_1 , \ g_2 $ be functions from $ [0,b]
\times [0,b] \times B$ into $B $,  $ \gamma : [0,b]\times
[0,b]\times [0,b] \to \mathbb{R}^+ $ and
 $ \beta:  [0,b]\times [0,b] \to \mathbb{R}^+ $ be measurable functions  
 such that:
\begin{itemize}
\item[(H1')] (i) $g_i (t,.,x): s \mapsto g_i (t,s,x) $ where $i\in
\{1,2\}$ are measurable
functions on $[0,b]$ for each $x\in B$ and for almost all $t\in [0,b]$.
\\
(ii) $g_i (t,s,.): x \mapsto g_i (t,s,x) $, where $i\in \{1,2\}$, are
$\rho$-continuous on $B$ for almost all $t,s \in [0,b]$.

\item[(H2')] For any $i\in \{1,2\}$,
 $\rho ( g_i (t,s,x) - g_i (\tau , s,x ) ) \leq \gamma  (t,\tau , s) $
for all  $(t,s,x)$ and $(\tau , s,x )$  in $[0,b] \times [0,b] \times B $
  and $ \lim_{t\to \tau } \int_0 ^{b} \gamma (t,\tau ,s ) ds =0 $
 uniformly for $\tau \in [0,b] $.

\item[(H3')] $\rho (g_2 (t,s,x) -g_2 (t,s,y))  \leq \rho (x-y) $ for all
$(t,s,x)$ and $(t , s,y )$  in $[0,b] \times [0,b] \times B $.
\end{itemize}
These hypotheses have been used by Martin \cite{rm}.

Now, assume that $f_0$ is a fixed element of $B$, and that
$h,\,T$ are the Uryshon operators  on $C([0,b], B ) $  defined by:
\begin{gather*}
[h u ] (t) = \int_0 ^{b} \exp (-s) g_1 (t,s,u (s)) ds, \\
[T u ] (t) = \int_0 ^{b}  \exp (-s) g_2 (t,s,u (s)) ds,
\end{gather*}
for  $t\in [0,b]$ and $u \in ( C([0,b], B ), \rho_a )$ with $(a>0)$.

\begin{proposition} \label{prop3.1}
(1) Under the hypotheses (H1')--(H3'), the operator  $T$ is
$\rho_a $-Lipschitz from $C([0,b], B )$ into $C([0,b], B )$.

\noindent(2) Under the hypotheses (H1')--(H2'), the operator
 $h$ is $\rho_a $-completely continuous from $C([0,b], B )$
into $C([0,b], B )$.
\end{proposition}

\begin{proof} (1) We show that $C([0,b], B ) $ is invariant by $T$.
(i) Note that $(X_\rho , \|.\|_\rho ) $ is a Banach space with finite
dimension.  By hypothesis (H1')(i),
$g_2 (t,.,u(.) ) : s \mapsto g_2 (t,s,u(s)) $ is measurable,
and since $ B$ is $\rho$-bounded,
 $g_2 (t,.,u(.) ) : s \mapsto g_2 (t,s,u(s)) $ is an integrable function
from   $[0,b]$ into $(X_\rho , \|.\|_\rho ) $. Then for $u\in C([0,b], B )$,
 we have
\begin{align*}
&[Tu ] (t) \in \int_0 ^{b} \exp (-s) ds
  {\overline {\mathop{\rm co}}}^{\|.\|_\rho} \{ g_2 (t,s,u (s)) ,
s\in [0,b] \} \\
&\subset (1-\exp(-b) ) {\overline {\mathop{\rm co}}}^{\|.\|_\rho} (B).
\end{align*}
But $B$ is convex and $\rho$-closed thus
${\overline {\mathop{\rm co}}}^{\|.\|_\rho} (B)
=  {\overline {B}}^{\|.\|_\rho} \subset {\overline {B}}^{\rho}= B$.
Since $0\in B $ and $0< 1-\exp (-b) <1 $, we have
$ [Tu ] (t) \in B $ for all $t\in [0,b]$.

\noindent (ii) Let $u \in C([0,b], B ) $ then $Tu$ is continuous from
$[0,b]$ into $(B , \rho )$. Indeed, let $(t_n )_{n\in \mathbb{N} } $ be a
 sequence and $r $ in $[0,b]$  such that $ t_n \to r $ as $n\to +\infty $
and we have
$$
 [Tu ] (t_n ) - [Tu ] (r ) = \int_0 ^b \exp(-s) (g_2 (t_n ,s,u(s))
 -g_2 (r ,s,u(s))) ds .
$$
Let $K= \{ s_0 ,s_1 ,\dots, s_m \}$ be a subdivision of $[0,b]$.
Then $ \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) e^{-s_i}x(s_i )$ is
$\|.\|_\rho $-convergent. Thus $\rho$-converges to
$\int_0 ^{b} \exp (-s) x(s) ds $ in $X_\rho $ when
 $|K| = sup \{|s_{i+1} - s_i |, i=0,\dots,m-1\} \to 0 $ as
$m \to +\infty $. Since
\begin{align*}
 & \int_0 ^{b} \exp (-s) ( g_2 (t,s,u(s) ) - g_2 (\tau ,s,u (s) )) ds\\
 &= \lim \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i )
( g_2 (t,s_i ,u (s_i ) )- g_2 (\tau ,s_i ,u (s_i ) )),
\end{align*}
 and $\sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i ) \leq  \int_0 ^{b}
  \exp (-s) ds = 1- \exp (-b)  < 1 $, then  by the Fatou property we have:
\begin{align*}
&\rho ([Tu ](t_n) -[Tu ](r ) ) \\
&\leq  \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i ) 
\rho (g_2 (t_n ,s_i ,u (s_i ) ) - g_2 (r ,s_i ,u (s_i ) )) \\
&\leq \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i ) \gamma (t_n ,r , s_i ) \\
&\leq \int_0 ^{b} \exp (-s)  \gamma (t_n ,r , s ) ds \\
&\leq \int_0 ^{b}  \gamma (t_n ,r , s ) ds
\end{align*}
Hence by hypothesis (H2') $Tu$ is $\rho$-continuous at $r $.

\noindent (2) We show that $T$ is $\rho_a $-Lipschitz. Let $u,v $ in
 $C([0,b],B)$, we have.
\begin{align*}
&\rho ([Tu ](t) -[Tv ](t ) ) \\
&\leq  \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i )( \exp (-s_i ))
 \rho (g_2 (t,s_i ,u (s_i ) ) - g_2 (t ,s_i ,v(s_i ) )) \\
&\leq \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i )
\rho (u (s_i ) -v (s_i ) ) \\
&\leq \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (as_i ) \rho_a
(u  - v  ) .
\end{align*}
Therefore,
\begin{align*}
\exp (-at) \rho ([Tu ](t) -[Tv ](t ) )
&\leq \exp(-at)  ( \int_0 ^{b} \exp(as) ds ) \ \ \rho_a (u  -v  ) \\
&\leq   \frac{e^{b a} -1}{a}  \rho_a (u  -v  )  .
\end{align*}
Hence,
$$
\rho_a ([Tu ] -[Tv ] ) \leq   \frac{e^{b a} -1}{a}  \rho_a (u  -v  )  .
$$
(3) Using the same argument of (1), we show that $C([0,b],B)$ is invariant
by $h$.

\noindent(4) Now, we claim that $ h (C([0,b], B )) $ is equicontinuous
in the sense of $\rho $, and $\rho_a $-compact.
We have:
$$
[hu ](t) - [hu ] (\tau ) = \int_0 ^{b} \exp (-s) ( g_1 (t,s,u (s) )
- g_1 (\tau ,s,u (s) )) ds .
$$
We easily obtain
$$
\rho ([hu ](t) - [hu ] (\tau )) \leq \int_0 ^{b} \gamma (t,\tau ,s ) ds,
$$
by using again the same argument in (1). And since,
 $ \lim_{t\to \tau } \int_0 ^{b}  \gamma (t,\tau ,s ) ds =0 $
uniformly for   $\tau \in [0,b]$, then $h( C([0,b], B ))$ is
$\rho$-equicontinuous.
On the other hand, since $B$ is $\rho$-bounded then, $h(C([0,b], B ))$
is $\rho_a $-bounded  subset of $C([0,b], B ) $. Indeed,
let $u,v$ in $C([0,b],B )$, we have
$$
[hu ](t) - [hv ] (t ) = \int_0 ^b \exp(-s) (g_1 (t,s,u(s)) -g_1 (t,s,v(s))) ds.
 $$
 Again from (1), we obtain
\begin{align*}
&\rho ([hu ](t) -[hv ](t ) )\\
&\leq  \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i )
\rho (g_1 (t ,s_i ,u (s_i ) ) - g_1 (t ,s_i ,v (s_i ) )) \\
&\leq \liminf \sum_{i=0} ^{m-1} (s_{i+1} -s_i ) \exp (-s_i )
 \delta_\rho (B) \\
&\leq (\int_0 ^{b} \exp (-s)ds )\ \  \delta_\rho (B).
\end{align*}
Hence,
$$
\rho_a ([hu ] - [hv ]) \leq  (1-e^{-b} ) \delta_\rho (B) < \infty
$$
Therefore, $h(C([0,b], B ))$ is a $\rho_a $-bounded subset of
$C([0,b], B )$ and by Lemma \ref{lem2.1}, it is $\|.\|_{\rho_a } $-bounded
subset of $C([0,b], B ) $. On the other hand, since
$(X_\rho ,\|.\|_\rho ) $ is a Banach space with finite dimensional ,
 then for each $t\in [0,b]$ we have
 $\overline{h(C([0,b],B))(t)} $ is $\|.\|_\rho $-compact.
Thus, by Ascoli's theorem we have $\overline{h(C([0,b],B))} $
is $\|.\|_{\rho_a} $-compact, then $\overline{h(C([0,b],B))} $
is $\rho_a $-compact. Hence  for any $M\subset C([0,b],B) $, we have
$\overline{h(M)} $ is $\rho_a $-compact.
Using the standard techniques \cite[Theorem 3 page 103]{ro}, we
that $h $ is $\|.\|_{\rho_a} $-continuous then $h $ is
$\rho_a $-continuous. So $h$ is $\rho_a $-completely continuous.
\end{proof}

\subsection*{Acknowledgment}
 The authors would like to thank the anonymous
referees for their suggestions and interesting remarks.


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\end{document}