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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 57, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2009/57\hfil Volterra integral equation with deviating arguments]
{On solutions of a Volterra integral equation with deviating arguments}

\author[M. D. Julie, K. Balachandran \hfil EJDE-2009/57\hfilneg]
{M. Diana Julie, Krishnan Balachandran} 

\address{M. Diana Julie \newline
 Department of Mathematics, Bharathiar University, 
 Coimbatore-641 046, India}
\email{diana.stephen@rediffmail.com}

\address{Krishnan Balachandran \newline
Department of Mathematics, Bharathiar University, 
Coimbatore-641 046, India}
\email{kbkb1956@yahoo.com}

\thanks{Submitted October 14, 2008. Published April 23, 2009.}
\subjclass[2000]{45D05, 45M05}
\keywords{Volterra integral equation; deviating arguments;
asymptotic behavior; \hfill\break\indent fixed point theorem}

\begin{abstract}
 In this article,  we establish the existence and asymptotic
 characterization of solutions to a nonlinear Volterra integral
 equation with deviating arguments. Our proof is based on
 measure of noncompactness and the Schauder fixed point theorem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

Several authors have studied nonlinear Volterra
integral equations with deviating arguments and functional integral
equations; see for example \cite{b1,b4,b5,b11, d1,e1}. Bana\'s \cite{b3}
proved an existence theorem
for functional integral equation  while  Balachandran and Illamaran
\cite{b1} proved an existence theorem for Volterra integral equation with
a deviating argument. Existence of solutions to  nonlinear
integral equations, which contain particular cases of
important integral and functional equations such as
nonlinear Volterra integral equation, Urysohn integral equation and
integral equations of Chandrasekhar type,  have been considered in many
papers and monographs \cite{b8,c1,h1}.

In this paper we study the nonlinear Volterrra integral equation
\begin{equation}
x(t) = g\big(t,x(h_1(t)),\dots , x(h_n(t))\big)+\int_0^t k\big(t,s,x(H_1(s)),
\dots , x(H_m(s))\big) ds, \label{e1.1}
\end{equation}
where $t\geq 0$. In particular, we prove the existence and asymptotic
stability of solutions for this equation.
The investigation is done on the space of
continuous and tempered functions on $\mathbb{R}^n$. The main tool used in
our considerations is the  measure of noncompactness and
the Schauder fixed point theorem. The results obtained in this paper
generalize several results presented in \cite{b6,b9,b10,l1}.

\section{Notation and auxiliary results}

Suppose $E$ is a real Banach space with the norm $\|\cdot\|$ and
zero element $\theta$. Denote by $B(x, r )$ the closed ball centered
at $x$ and with radius $r$. We write $B_r$ for the ball
$B(\theta, r)$. If $X$ is a subset of $E$ then the symbols
$\overline{X}$ and $\mathop{\rm Conv} X$ stand for the closure and
convex closure of $X$, respectively.
Further, let $\mathcal{M}_E$ denote the family of all
nonempty and bounded subsets of $E$ and $\mathcal{N}_E$ its
subfamily consisting of all relatively compact sets.

We define the the following notion of measure of noncompactness.

\begin{definition} \label{def2.1}\rm
A mapping $\mu:\mathcal{M }_E \to \mathbb{R}_+$ is said to be a measure
of noncompactness in the space $E$ if
it satisfies the following conditions:
\begin{itemize}
\item [(i)] The family $\ker \mu=\{X\in \mathcal{M }_E  :\mu(X) = 0\}$
is nonempty and $\ker \mu \subset \mathcal{N }_E $;
\item [(ii)] $X \subset Y\Rightarrow \mu(X) \leq\mu(Y)$;
\item [(iii)] $\mu(\overline X)$= $\mu(\mathop{\rm Conv }X)= \mu(X)$;
\item [(iv)] $\mu(\lambda X+(1-\lambda)Y)\leq \lambda\mu(X)
+(1-\lambda)\mu(Y)$ for $ \lambda \in [0,1]$;
\item [(v)] If $(X_n)$ is a sequence of closed sets from
$\mathcal{M }_E $ such that $X_{n+1} \subset X_n$
for $n = 1, 2, 3\dots $ and if $ \lim_{n \to \infty}\mu(X_n) = 0$,
then the set $X_\infty = \cap_{n=1}^\infty X_n $ is nonempty.
\end{itemize}
\end{definition}

The family $\ker  \mu $ defined in axiom (i) is called the kernel
of the measure of noncompactness $\mu$.

 Let us remark that the intersection set
$X_\infty$ from (v) is a member of the kernel of the measure of
noncompactness $\mu$. Indeed, from the inequality
$\mu(X_\infty)\leq \mu(X_n)$ for $n = 1, 2, \dots$, we infer
that $\mu(X_\infty)$ = 0,
so $X_\infty \in$ $\ker \mu$. This property of the intersection set
$X_\infty$ will be crucial in our study.

Further facts concerning measures of noncompactness and their
properties may be found in \cite{b2,b7}.

Now, let us assume that $p = p(t)$ is a given function defined and
continuous on the interval $\mathbb{R}_+$ with real positive values. We will
denote by $C\big(\mathbb{R}_+, p(t); \mathbb{R}^n\big)=C_p$, the Banach space
consisting of all continuous functions from $\mathbb{R}_+$ into
$\mathbb{R}^n$ such that
\[
\|x\| =\sup\{|x(t)|p(t) : t \geq 0 \}< \infty.
\]

Now we recall the definition of the measure of noncompactness in the
space $C_p$ which will be used in the sequel \cite{b2,b7}. Let $X$ be a
nonempty and bounded subset of the space $C_p$. Fix a positive
number $T $. For $x \in X$ and $\epsilon > 0$ denote by
$\omega^T(x,\epsilon)$ the modulus of continuity of the function
$x$(tempered by the function $p$) on the interval $[0, T ]$, i.e.
\[
\omega^T(x,\epsilon) = \sup\{|x(t)p(t) - x(s)p(s)| : t, s \in [0, T ],
\ |t - s|\leq \epsilon \}.
\]
 Further, let us put
\begin{gather*}
\omega^T(X,\epsilon)= \sup\{\omega^T(x,\epsilon): x\in X \},\quad
\omega_0^T(X)= \lim_{\epsilon \to 0} \omega^T(X,\epsilon),\\
\omega_0(X) =  \lim_{T \to \infty} \omega_0^T(X).
\end{gather*}
Moreover, we put
\[
\alpha(X) =  \lim_{T \to \infty}\{ \sup_{x \in X}\{\sup\{|x(t)|p(t)
: t \geq T \}\}\}.
\]
 Finally, let us the define the function $\mu$ on the family
$\mathcal{M}_{C_p}$ by formula
$\mu(X) = \omega_0(X) + \alpha(X)$.

 It may be shown that the function $\mu$ is the measure of
noncompactness in the space $C_p$ \cite{b2,b7}. The kernel $\ker \mu$ is the
family of all nonempty and bounded sets $X$ such that functions
belonging to $X$ are locally equicontinuous on $\mathbb{R}_+$ and such that
$ \lim_{t \to \infty} x(t)p(t)= 0$ uniformly with
respect to the set $X$, i.e. for each $\epsilon > 0$ there exists
$T> 0$ with the property that $|x(t)|p(t)\leq \epsilon$ for $t\geq
T$ and for $x\in X$.

Finally, let us assume that $x\in C_p$. For $T > 0$ and denote by
$\nu^T (x,\epsilon)$ the usual modulus of continuity of the function
$x$ on the interval $[0, T ]$:
\[
\nu^T (x, \epsilon) = \sup\{|x(t) - x(s)| : t, s \in [0, T ],\
|t - s|\leq \epsilon\}.
\]

\section{Main Result}

We assume the following conditions:
\begin{itemize}
\item [(H1)] The function $g:\mathbb{R}_+ \times R^{n^2} \to \mathbb{R}^n$
is continuous and there exists a constant $K\geq 0$ such that
$$
|g(t,x_1,\dots ,x_n)-g(t,y_1,\dots ,y_n)|\leq K\sum_{i=1}^n|x_i-y_i|
$$
for all $t\in \mathbb{R}_+$ and $(x_1,x_2,\dots ,x_n)$,
$(y_1,y_2,\dots ,y_n)\in R^{n^2}$;

\item [(H2)] There exists a continuous function
$a:\mathbb{R}_+ \to \mathbb{R}_+$ such that
$|g(t,0,\dots ,0)|\leq a(t)$
for all $t \in \mathbb{R}_+$;

\item [(H3)] Let $\Delta=\{(t,s): 0\leq s\leq t<\infty \}$.
The kernel $k : \Delta \times R^{nm} \to \mathbb{R}^n $ is continuous
and there exists a continuous function $L_0 : \mathbb{R}_+ \to \mathbb{R}_+$,
a continuous nonincreasing function $ L_1: \mathbb{R}_+ \to R$ and a
continuous nondecreasing function $q:\mathbb{R}_+\to \mathbb{R}_+$
such that
$$
|k(t,s,x_1,\dots ,x_m)|\leq L_0(s)+\exp (L_1(t))q(s)\sum_{i=1}^m |x_i|
$$
for all $t, s \in \mathbb{R}_+$ and $(x_1,x_2,\dots ,x_m) \in R^{nm}$;

\item [(H4)] For $i=1, 2,\dots n$, $j=1,\ 2,\dots m$  the functions
$h_i,H_j : \mathbb{R}_+  \to \mathbb{R}_+$ are continuous and
$h_i(t) \leq t$, $H_j(t) \leq t$;

\item [(H5)]$ \lim_{t \to \infty}t q(t)\exp L_1(t) = 0 $ and
$ \lim_{t \to \infty}[a(t)+b(t)]\exp(-Mb(t))= 0$ where
$b(t) =  \int_0^t L_0(s) ds$;
\end{itemize}

Let us define the function $L(t)$ as
$$
L(t) =  \int_0^t[ L_0(s)+q(s)\exp L_1(s)]ds.
$$
Fix a constant $M>1$ and denote by
$C_L=C\big(\mathbb{R}_+, \exp(-ML(t));\mathbb{R}^n\big)$.
It is obvious that
the function $L(t)$ is nondecreasing and continuous on $\mathbb{R}_+$.

\begin{theorem} \label{thm3.1}
Under assumptions {\rm (H1)-(H5)}, Equation \eqref{e1.1}
has at least one solution $x\in C_L$ such that
$x(t) =o(\exp(ML(t))$ as $t\to \infty$, provided $(Kn+m/M)<1$.
\end{theorem}

\begin{proof}
 Consider the operator $F$ defined on the space $C_L$ by the formula
\[
(Fx)(t) = g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)
+\int_0^t k\big(t,s,x(H_1(s)),\dots ,x(H_m(s))\big) ds,
\]
where $t\geq 0$.
Obviously the function $Fx$ is continuous on the interval $\mathbb{R}_+$.
Moreover, in view of our assumptions,
for arbitrarily fixed $x\in C_L$ and $t \in \mathbb{R}_+$, we obtain
\begin{align*}
&|(Fx)(t)|\exp(-ML(t))\\
&\leq  \Big[ \big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)\big|\\
&\quad +\int_0^t\big|k\big(t,s,x(H_1(s)),\dots ,x(H_m(s))\big)\big|ds\Big]
 \exp(-ML(t))\\
&\leq  \Big[\Big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)-g(t,0\dots ,0)\Big|
 + |g(t,0\dots ,0)|\Big]\exp(-ML(t))\\
&\quad +\Big[\int_0^t L_0(s)ds + \exp(L_1(t)) \int_0^t q(s)
 \sum_{i=1}^{m}|x(H_i(s))|ds \Big]\exp(-ML(t))\\
&\leq  K\sum_{i=1}^n|x(h_i(t))|\exp(-ML(t))+[a(t)+b(t)]\exp(-ML(t))\\
&\quad +\exp(L_1(t))\int_0^tq(s)\|x\|\sum_{i=1}^m \exp(ML(H_i(s)))ds\cdot
 \exp(-ML(t))\\
&\leq  K\sum_{i=1}^n|x(h_i(t))|\exp(-ML(h_i(t)))+[a(t)+b(t)]\exp(-ML(t))\\
&\quad +m\|x\|\int_0^t[L_0(s)+ q(s)\exp(L_1(s))]\exp(ML(s))ds\cdot
\exp(-ML(t))\\
&\leq  Kn\|x\|+[a(t)+b(t)]\exp(-Mb(t))+\big(\frac{m}{M}\big)\|x\|\\
&\leq  \big[Kn+\frac{m}{M}\big]\|x\|+D,
\end{align*}
where $D = \sup\{[a(t)+b(t)]\exp(-Mb(t)): t\in \mathbb{R}_+ \}$.
 Obviously $D<\infty$ in view of the hypothesis (H5).
The obtained estimate shows that $Fx$ is bounded on $\mathbb{R}_+$.
This fact with the continuity of $Fx$ on $\mathbb{R}_+$ yields that $F$
transforms the space $C_L$ into itself. Moreover for $r = D/(1-Kn-m/M)$,
we have that $F$ maps the ball $B_r$ into itself.

 Let us take an arbitrary nonempty subset $A$ of the ball $B_r$.
Fix $T>0$ and $\epsilon> 0$. Next, take arbitrary $t, s \in [0,T]$
with $|t-s|\leq \epsilon$. Then, for arbitrarily fixed $x\in A$, we get
 \begin{align}
&|(Fx)(t)-(Fx)(s)| \notag \\
& \leq  \big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)
 -g\big(s,x(h_1(s)),\dots ,x(h_n(s))\big)\big|  \notag\\
&\quad +\big|\int_0^t k\big(t,\tau,x(H_1(\tau)),\dots ,
x(H_m(\tau))\big)d\tau   \notag\\
&\quad - \int_0^s k\big(s,\tau,x(H_1(\tau)),\dots ,
x(H_m(\tau))\big)d\tau \big|   \notag\\
&\leq  \big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)
 -g\big(t,x(h_1(s)),\dots ,x(h_n(s))\big)\big|   \notag\\
&\quad +\Big|g\big(t,x(h_1(s)),\dots ,x(h_n(s))\big)
 -g\big(s,x(h_1(s)),\dots ,x(h_n(s))\big)\Big|   \notag\\
&\quad +\int_0^t \Big|k\big(t,\tau,x(H_1(\tau)),\dots ,x(H_m(\tau))\big)
 -k\big(s,\tau,x(H_1(\tau)),\dots ,x(H_m(\tau))\big)\Big|d\tau \notag \\
&\quad +\int_s^t\Big|k\big(s,\tau,x(H_1(\tau)),\dots ,x(H_m(\tau))\big)
  \Big|d\tau \label{e3.1} \\
&\leq  K\sum_{i=1}^n|x(h_i(t))-x(h_i(s))|+\nu_1^T(g,\epsilon)
 +T\nu_1^T(k,\epsilon) \notag\\
&\quad +\epsilon \sup\Big\{\big|k\big(s,\tau,x(H_1(\tau)),\dots ,
x(H_m(\tau))\big)\big|: s,\tau \in[0,T], \notag\\
&\quad  |x(H_i(\tau))|\leq r\exp(ML(T)),\;  i=1, 2,\dots ,m \Big\}, \notag
\end{align}
where
\begin{align*}
\nu_1^T(g,\epsilon)
&= \sup\Big\{|g(t,x_1,\dots ,x_n)-g(s,x_1,\dots ,x_n)|:
 t, s \in[0,T],\\
&\quad |t-s|\leq \epsilon,\; |x_i|\leq r\exp(ML(T)),\;
 i=1, 2,\dots ,n \Big\},\\
\nu_1^T(k,\epsilon)
&= \sup\Big\{|k(t,\tau,x_1,\dots ,x_m)-k(s,\tau,x _1,\dots ,x_m)|:
 t, s, \tau \in[0,T],\\
&\quad |t-s|\leq \epsilon, \;
  |x_i|\leq r\exp(ML(T)),\;  i=1, 2,\dots ,m \Big\}.
\end{align*}
Now let us denote
\begin{align*}
\beta(\epsilon) &= \nu_1^T(g,\epsilon)+T\nu_1^T(k,\epsilon)
+\epsilon \sup\Big\{\big|k\big(s,\tau,x(H_1(\tau)),\dots ,
x(H_m(\tau))\big)\big|: \\
&\quad s, \tau \in[0,T],\;
 |x(H_i(\tau))|\leq r\exp(ML(T)),\;  i=1, 2,\dots ,m  \Big\}.
\end{align*}
Keeping in mind, the uniform continuity of
$g(t,x_1,\dots ,x_n)$ and $k(t,s,x_1,\dots ,x_m)$ on compact subsets
 of $\mathbb{R}_+ \times R^{n^2}$ and $\Delta \times R^{nm}$ respectively,
 we deduce that $\beta(\epsilon) \to 0$ as $\epsilon \to 0$.
Furthermore, from \eqref{e3.1}, we have
\begin{align*}
&|(Fx)(t)\exp(-ML(t))-(Fx)(s)\exp(-ML(s))|\\
&\leq  |(Fx)(t)\exp(-ML(t))-(Fx)(s)\exp(-ML(t))|\\
&\quad +|(Fx)(s)|  |\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K\sum_{i=1}^n|x(h_i(t))-x(h_i(s))|\exp(-ML(t))
 +\beta(\epsilon)\exp(-ML(t))\\
&\quad +|(Fx)(s)||\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K\sum_{i=1}^n|x(h_i(t))\exp(-ML(h_i(t))-x(h_i(s))\exp(-ML(h_i(s))|\\
&\quad +K\sum  _{i=1}^n|x(h_i(s))\exp(-ML(h_i(s))-x(h_i(s))
 \exp(-ML(h_i(t))|\\
&\quad +\beta(\epsilon)\exp(-ML(t))+|(Fx)(s)||\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K\sum_{i=1}^n\omega^T(x,\nu^T(h_i,\epsilon))+Kr\sum_{i=1}^n
 \exp(ML(h_i(s)))\nu^T\big(\exp(-ML(h_i(t))),\epsilon \big)\\
&\quad +\beta(\epsilon)\exp(-ML(t))+r\exp(ML(T))\nu^T
 \big(\exp(-ML(t)),\epsilon \big).
\end{align*}

The above estimate and the fact that
$\exp(-ML(t))$  and  $\exp(-ML(h_i(t)))$ are uniformly
continuous on $[0,T]$, yields the  inequality
$$
\omega_0^T(FA)\leq Kn\omega_0^T(A).
$$
Hence,
\begin{equation} \label{e3.2}
\omega_0(FA)\leq Kn\omega_0(A).
\end{equation}
Next, let us assume that $t\geq T$. Then, by virtue of our assumptions
we obtain
\begin{align*}
& |(Fx)(t)|\exp(-ML(t)) \\
&\leq  \Big[\big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)-g(t,0\dots ,0)\big|
+ \big|g(t,0\dots ,0)\big|\Big]\exp(-ML(t)) \\
&\quad +\Big[\int_0^t L_0(s)ds
+ \exp(L_1(t)) \int_0^t q(s)\sum_{i=1}^m|x(H_i(s))|ds \Big]\exp(-ML(t)) \\
&\leq  K\sum_{i=1}^n|x(h_i(t))|\exp(-ML(t))+[a(t)+b(t)]\exp(-ML(t)) \\
&\quad +\exp(L_1(t))\int_0^t q(s)\|x\|\sum_{i=1}^m \exp(ML(H_i(s)))ds\cdot
 \exp(-ML(t)) \\
&\leq  K\sum_{i=1}^n|x(h_i(t))|\exp(-ML(h_i(t)))+[a(t)+b(t)]\exp(-ML(t)) \\
&\quad +m\|x\|\exp(L_1(t))\int_0^t q(s)\exp(ML(s))ds\cdot \exp(-ML(t)) \\
&\leq  K\sum_{i=1}^n|x(h_i(t))|\exp(-ML(h_i(t)))+[a(t)+b(t)]\exp(-Mb(t)) \\
&\quad +mrtq(t)\exp(L_1(t))
\end{align*}
Now, taking into account  (H5), from the above estimate
we obtain
\begin{equation} \label{e3.4}
\alpha(FA)\leq Kn \alpha(A).
\end{equation}
where $\alpha(A)$ was defined previously. Next, from \eqref{e3.2} and \eqref{e3.4} we have
\begin{equation} \label{e3.5}
\mu(FA)\leq C \mu(A)
\end{equation}
where $C = Kn >0$ is a constant and $\mu$ denotes the measure of
noncompactness defined earlier.

Furthermore, let us consider the sequence of sets $(B_r^n)$, where
$B_r^1 =\mathop{\rm  Conv} F(B_r)$,
$B_r^2 = \mathop{\rm Conv} F(B_r^1)$ and so on. Observe that all sets
of this sequence are nonempty, bounded, closed and convex. Moreover,
$B_r^{n+1}\subset B_r^n $ for $n=1, 2, 3,\dots $.
Thus in virtue of \eqref{e3.5} it is easily seen that
\[
\mu(B_r^n)\leq C^n\mu(B_r)
\]
This inequality and the fact that $C < 1$ yields that
$ \lim_{n \to \infty}\mu(B_r^n) = 0$. Thus from  (v) in
Definition \ref{def2.1}, we deduce that the set $B=\cap_{n=1}^\infty B_r^n$ is
nonempty, bounded, closed and convex. Moreover, in view of the remark
we have that $B \in$ $\ker \mu$. It should be also noted that the
operator $F$ maps the set $B$ into itself.

Next we show that $F$ is continuous on the set $B$. To do this fix
$\epsilon > 0$ and take arbitrary functions $x, y \in B$ such that
$\|x-y\|\leq \epsilon$. Taking into account the fact that $B \in \ker \mu$
and the description of sets belonging to $\ker \mu$  we can find $T > 0$
such that for each $z \in B$ and $t \geq T$ the inequality
$|z(t)| \exp(-ML(t)) \leq \epsilon/2$ is satisfied.

Let us observe that in view of our assumptions, for an arbitrarily fixed
$t \in \mathbb{R}_+$, we obtain
\begin{equation} \label{e3.6}
\begin{aligned}
& |(Fx)(t)-(Fy)(t)|\exp(-ML(t)) \\
&\leq  \Big|g\big(t,x(h_1(t)),\dots ,x(h_n(t))\big)
 -g\big(t,y(h_1(t)),\dots ,y(h_n(t))\big)\Big|\exp(-ML(t)) \\
&\quad + \int_0^t \Big|k\big(t,\tau,x(H_1(\tau)),\dots ,x(H_m(\tau))\big) \\
&\quad -k\big(t,\tau,y(H_1(\tau)),\dots ,y(H_m(\tau))\big)\Big|d\tau \cdot
 \exp(-ML(t)) \\
&\leq K\sum_{i=1}^n|x(h_i(t))-y(h_i(t))|\exp(-ML(t))
 +T\nu^T(k,\epsilon)\exp(-ML(t)).
\end{aligned}
\end{equation}
Now let $t \in [0,T]$, then we have
\begin{equation} \label{e3.7}
|(Fx)(t)-(Fy)(t)|\exp(-ML(t)) \leq  C \epsilon+T\nu^T(k,\epsilon)\exp(-ML(t))
\end{equation}
where the quantity $\nu^T(k,\epsilon)$ is defined as follows
\begin{align*}
\nu^T(k,\epsilon) &= \sup\Big\{|k(t,\tau,x_1,\dots ,x_m)-k(t,\tau,y _1,
\dots ,y_m)|: t, \tau \in[0,T],\\
&\quad |x_i|,\ |y_i|\leq r\exp(ML(T)),\; |x_i-y_i|\leq \epsilon \exp(ML(T))
,\; i=1, 2,\dots ,m \Big\}.
\end{align*}
 From the definition of the function $k(t,s,x_1,\dots ,x_m)$, we conclude
that $\nu^T(k,\epsilon) \to$ 0 as $\epsilon \to 0$.

Further, let us assume that $t>T$. Then, taking into account that
$x, y \in B$ and that $F$ transforms the set $B$ into itself we have
that $Fx$, $Fy \in B$. Hence by the characterization of the set $B$
given above, we get
\begin{equation} \label{e3.8}
\begin{aligned}
&|(Fx)(t)-(Fy)(t)|\exp(-ML(t))\\
&\leq  |(Fx)(t)|\exp(-ML(t))+|(Fy)(t)|\exp(-ML(t))
\leq \epsilon.
\end{aligned}
\end{equation}
Now from \eqref{e3.6}-\eqref{e3.8}, we deduce that the operator
 $F$ is continuous on the set $B$.

Finally, taking into account all facts concerning set $B$ and the
operator $F:B\to B$ and applying the classical Schauder
fixed point principle we infer that $F$ has at least one fixed point
$x$ in the set $B$. Obviously, the function $x=x(t)$ is a solution
of \eqref{e1.1}. Moreover, keeping in mind that $B\in$ $\ker \mu$, we
obtain that $x(t)= o(\exp(ML(t)))$ as $t\to \infty$.
\end{proof}

\section{Example}

Consider the following Volterra integral equation with deviating arguments
\begin{equation} \label{e4.1}
 x(t)=\frac{1+\arctan \big(x(t/3)\big)}{4+t^2}+
 \int_0^t \big[s\cos2t+3s^2e^{-2t}\ln\big(1+|x(s/2)|\big)\big]ds,
\end{equation}
where $t\geq 0$.
This equation is clearly of the form  \eqref{e1.1} with
$g(t,x) =  \displaystyle \frac{1+\arctan (x)}{4+t^2}$
and $k(t,s,x) =  s\cos(2t)+3s^2e^{-2t}\ln\big(1+|x|\big)$ where
$m = n = 1$, $h_1(t)= t/3$ and $H_1(t) =  t/2$.

It is easily seen that for \eqref{e4.1}, the assumptions (H1) and (H2)
are satisfied with $K=  1/4$ and $a(t) =  \displaystyle \frac{1}{4+t^2}$.
Let us observe that
\[
|k(t,s,x)|\leq s+3s^2e^{-2t}|x|.
\]
Note that the assumption (H3) is satisfied with $L_0(t) = t$,
$L_1(t) = -2t$ and $q(s) = 3s^2$.
On the other hand, \eqref{e4.1} satisfies assumption (H5) with
$b(t) = t^2/2$. Now, we get $(Kn+m/M) < 1$ for $M>1$.

Thus, in view of Theorem \ref{thm3.1}  we conclude that problem \eqref{e4.1}
has a solution $x=x(t)$ such that
\[
x(t)=o\Big(\exp\big(M\big[\frac{t^2}{2}-\frac{3e^{-2t}}{2}
(t^2+t+\frac{1}{2})+\frac{3}{4}\big]\big)\Big)
\]
as $t\to \infty$, where $M>1$ is a constant.

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\end{document}