\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 $$u(t) = \exp {(-t)} f_0 + \int_0 ^t \exp{(s-t)} ( T + h) u(s)ds \label{eI}$$ 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 \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} 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. \begin{thebibliography}{99} \bibitem{ai} Taleb A. Ait, E. Hanebaly; \emph{A fixed point theorem and its application to integral equations in modular function spaces}, Proc. Amer. Math. Soc.128 , no 2, 419-426 (2000). \bibitem{ha}A. Hajji; \emph{Forme Equivalente \a la Condition $\Delta_2$ et Certains r\'esultats de s\'eparations dans les Espaces Modulaires}, Math. FA/0509482 v1. \bibitem{m1} M. A. 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