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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 77, pp. 1--9.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/77\hfil Asymptotic behavior of solutions]
{Asymptotic behavior of solutions to functional integral equation
with deviating arguments}

\author[M. D. Julie, K. Balachandran \hfil EJDE-2008/77\hfilneg]
{M. Diana Julie, Krishnan Balachandran} % not in alphabetical order

\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{balachandran\_k@lycos.com}

\thanks{Submitted March 14, 2008. Published May 27, 2008.}
\subjclass[2000]{45D05, 45M05}
\keywords{Functional integral equation; deviating argument;\hfill\break\indent
asymptotic behavior; measures of noncompactness; fixed point
theorem}

\begin{abstract}
 This article presents results on the existence and asymptotic
 behavior of solutions of a functional integral equation with deviating
 arguments. The proof of our main result uses  the classical Schauder
 fixed point theorem and the technique of measures of noncompactness.
\end{abstract}

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

\section{Introduction}

The theory of functional integral equations with deviating argument
is very important and significant branch of nonlinear analysis.
It is worthwhile mentioning that these theories find numerous
applications in physics, mechanics, control theory, biology, ecology,
economics, theory of nuclear reactors, engineering, natural sciences
and so on \cite{b6,c1,d1}.
One of the basic problems considered in the theory of functional
integral equations with deviating arguments is to establish
convenient conditions guaranteeing the existence of solutions of
those equations. It is well known that existence of solutions of
equations of such a type depends strongly on the size of the delay
arguments involved in those equations.

In this article we will examine the  functional integral equation
\begin{equation} \label{e1.1}
x(t)= f\Big(t,x(\sigma_1(t)),\int_0^t k(t,s)g\big(s,x(\sigma_2(s))\big)ds
\Big),\quad t\geq 0.
\end{equation}

The functional integral equation of the above form contains a lot
of  special types of functional integral equations. The
differential equations with transformed argument or differential
equations of neutral type can also be transformed to functional
integral equations. Such type of equations were investigated in
lots of papers \cite{b4,e1}.

The aim of this paper is to investigate the existence and
asymptotic  behavior of solutions of \eqref{e1.1}. The main
tools used in our considerations are the concept of a measure of
noncompactness and the classical Schauder fixed point principle.
The investigations of the paper are placed in the space of
continuous and tempered functions on the real line. The result
obtained here generalizes several ones obtained earlier by many
authors \cite{a1,c1,c2,d1,d2}.


\section{Notation
 and auxiliary results}

Let $E$ be a real Banach space with the norm $\|\cdot\|$ and the
zero element $\theta$. Denote by $B(x, r )$ the closed ball
centered at $x$ and with radius $r$. The ball $B(\theta, r )$ will
be denoted by $B_r$.

If $X$ is a subset of $E$ then  $\overline{X}$ denotes the
closure, and $\mathop{\rm Conv}X$  denotes the convex closure of $X$.
We use the standard notation $X + Y$ , $\lambda X$
to denote the usual algebraic operations on subsets $X$, $Y$ of
the space $E$. 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 will accept the following definition of  measure of
 noncompactness \cite{b5}.

\begin{definition} \label{def2.1} \rm
A mapping $\mu:{\mathcal M }_E \to \mathbb{R}_+$ = $[0,+\infty)$ 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)] $X \subset Y$ implies $\mu(X) \leq\mu(Y)$;
\item[(iii)] $\mu(\mathop{\rm Conv }X)= \mu(X)$;
\item[(iv)] $\mu(\overline X)= \mu(X)$;
\item[(v)] $\mu(\lambda X+(1-\lambda)Y)\leq
 \lambda\mu(X)+(1-\lambda)\mu(Y)$ for $ \lambda \in [0,1]$;
\item[{(vi)}] If $(X_n)$ is a sequence of sets from ${\mathcal M }_E $
such that $X_{n+1} \subset X_n$, $\overline X_n = X_n$
($n= 1, 2, 3,\dots )$ and if $\lim_{n \to \infty}\mu(X_n) = 0$,
then the intersection $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 $\mu$.

\begin{remark} \label{rmk2.1} \rm
Note that the intersection set $X_\infty$ described in axiom (vi)
is a member of the kernel of the measure of noncompactness $\mu$.
In fact, the inequality $\mu(X_\infty)\leq \mu(X_n)$ for
$n = 1, 2, \dots$ implies that $\mu(X_\infty)$ = 0.
Hence $X_\infty \in$ $\ker \mu$. This property of the set
$X_\infty$ will be very important in our investigations.
\end{remark}

Now, let us assume that $p = p(t)$ is a given function defined and
continuous on the interval $\mathbb{R}_+$ with real positive values.
Denote by $C\big(\mathbb{R}_+, p(t)\big)$ = $C_p$ the Banach space
consisting of all real functions $x = x(t)$ defined and continuous
on $\mathbb{R}_+$ and such that
$$
\sup\{|x(t)|p(t) : t \geq 0\} < \infty.
$$
The space $C_p$ is furnished with the standard norm
$$
\|x\| =\sup\{|x(t)|p(t) : t \geq 0 \}.
$$
Further we recall the definition of the measure of noncompactness in the
space $C_p$ which will be used in our considerations \cite{b1,b5}. Let $X$ be
a nonempty and bounded subset of the space $C_p$. Fix positive number $T > 0$.
For an arbitrary function $x \in X$ 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 \},\\
\omega_0^T(X)= \lim_{\epsilon \to 0} \omega^T(X,\epsilon),\quad
\omega_0(X) =   \lim_{T \to \infty} \omega_0^T(X).
\end{gather*}
Also, we put
\[
b(X) = \lim_{T \to \infty}\Big\{ \sup_{x \in X}\big\{\sup\{|x(t)|p(t)
: t \geq T \}\big\}\Big\}.
\]
Finally, we define the function $\mu$ on the family $\mathcal{M}_{C_p}$
by putting $\mu(X) = \omega_0(X) + b(X)$.

It may be shown that the function $\mu$ is the measure of noncompactness
in the space $C_p$ \cite{b1}. 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 $\lim_{t \to \infty} x(t)p(t)= 0$
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 $x\in X$. This property will be crucial in our further
study.

Finally, let us assume that $x$ is a real function defined and continuous
on $\mathbb{R}_+$. Fix $T > 0$ 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 will consider the nonlinear functional-integral equation \eqref{e1.1}
under the following assumptions:
\begin{itemize}
\item[(H1)] $f:\mathbb{R}_+ \times R \times \mathbb{R} \to R $ is a continuous function and
there exists a constant $K\geq 0$ such that
$$
|f(t,x_1,y)-f(t,x_2,y)|\leq K|x_1-x_2|
$$
for all $t\in \mathbb{R}_+$, $x_1$, $x_2$, $y \in\mathbb{R}$;

\item[(H2)] There exists a continuous function $L_0: \mathbb{R}_+ \to \mathbb{R}_+$ and
a continuous nonincreasing function $L_1: \mathbb{R}_+ \to \mathbb{R}$  such that
$$
|f(t,0,y)|\leq L_0(t)+|y|\exp L_1(t)
$$
for all $t\in \mathbb{R}_+$ and $y\in\mathbb{R}$;

\item[(H3)] $k:\mathbb{R}_+ \times \mathbb{R}_+ \to \mathbb{R}$
is a continuous function and there exists a continuous function
$a:\mathbb{R}_+ \to \mathbb{R}_+$ and a continuous
nondecreasing function $b:\mathbb{R}_+ \to \mathbb{R}_+$  such that
$$
|k(t,s)|\leq a(t)b(s) \quad  \text{for all } t, s \in \mathbb{R}_+;
$$

\item[(H4)] $g:\mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}$ is a
continuous function and there exists a continuous nondecreasing
function $p:\mathbb{R}_+ \to \mathbb{R}_+$ such that
$$
|g(s,x)|\leq p(s)|x| \quad  \text{for all }
 s \in \mathbb{R}_+  \text{ and } x\in\mathbb{R};
$$

\item[(H5)] $\sigma_1, \sigma_2: \mathbb{R}_+ \to \mathbb{R}_+ $ are
continuous functions such that $\sigma_1(t)\leq t$ and $\sigma_2(t)\leq t$;

\item[(H6)] $\lim_{t \to \infty}t a(t)b(t)p(t) = 0$ and
 $\lim_{t \to \infty}L_0(t)\exp\Big(-\int_0^t L_0(s)ds\Big) = 0$;

\end{itemize}
Let $A= \sup\{a(t):\ t\in \mathbb{R}_+ \}$ and
$$
L(t)= \int_0^t [L_0(s)+b(s)p(s)\exp L_1(s)]ds.
$$
Obviously the function $L(t)$ is nondecreasing and continuous on
$\mathbb{R}_+$. Denote by $C_L$, the space
$C\big(\mathbb{R}_+, \exp(-ML(t))\big)$, where $M > 1$ is an arbitrarily
fixed constant.

\begin{theorem} \label{thm3.1}
 Assume {\rm(H1)-(H6)} and that $(K+A/M)<1$. Then  \eqref{e1.1} has
at least one solution $x\in C_L$ such that $x(t) = o(\exp(ML(t))$ as
$t\to \infty$.
\end{theorem}

\begin{proof}
Consider the operator $F$ defined on the space $C_L$ by the formula
\[
(Fx)(t) = f\Big(t,x(\sigma_1(t)),\int_0^t k(t,s)g
\big(s,x(\sigma_2(s))\big)ds\Big),\quad t\geq 0.
\]
Observe that the operator $F$ is well defined on the space $C_L$ and
 the function $Fx$ is continuous on $\mathbb{R}_+$.
Next, in view of our assumptions, for arbitrarily fixed $x\in C_L$
and $t \in \mathbb{R}_+$, we get
\begin{align*}
&|(Fx)(t)|\exp(-ML(t))\\
&\leq  \big|f\Big(t,x(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big)\\
&\quad -f\Big(t,0,\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t))\\
&\quad +\big|f\Big(t,0,\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t))\\
&\leq  K|x(\sigma_1(t))|\exp(-ML(t))+L_0(t)\exp(-ML(t))\\
&\quad +\exp L_1(t)\int_0^t
\big|k(t,s)\big|\ \big|g\Big(s,x(\sigma_2(s))\Big)\big|ds
\exp(-ML(t))\\
&\leq  K\|x\|\exp(M(L(\sigma_1(t)-L(t))))+L_0(t)\exp(-ML(t))\\
&\quad +\exp L_1(t)\int_0^t
a(t)b(s)p(s)|x(\sigma_2(s))|ds \exp(-ML(t)) \\
&\leq  K\|x\|+L_0(t)\exp(-ML(t))\\&\quad +a(t)\|x\|\int_0^t
b(s)p(s)\exp L_1(s)\exp (ML(s))ds \exp(-ML(t))\\
&\leq  K\|x\|+L_0(t)\exp\Big(-  \int_0^t L_0(s)ds\Big)\\
&\quad +a(t)\|x\|\int_0^t \left[L_0(s)+
b(s)p(s)\exp L_1(s)\right]\exp (ML(s))ds  \exp(-ML(t))\\
&\leq  K\|x\|+\frac{A}{M}\|x\|+B,
\end{align*}
where
\[
B = \sup\big\{L_0(t)\exp\Big(-  \int_0^t L_0(s)ds\Big):
t\in \mathbb{R}_+ \big\}.
\]
Obviously $B<\infty$ by virtue of assumption (H6).

The above obtained estimate shows that $Fx$ is bounded on
$\mathbb{R}_+$. This implies that the operator $F$ is a self mapping
of the space $C_L$. Moreover for $r = B/(1-K-A/M)$, the operator
$F$ transforms the ball $B_r$ into itself.

Let us take an arbitrary nonempty subset $X$ of the ball $B_r$.
Fix $T>0$ and take an arbitrary function $x\in X$.
Then evaluating similarly as before, for a fixed $t$, $t\geq T$, we get
\begin{align*}
&|(Fx)(t)|\exp(-ML(t))\\
 &\leq  K|x(\sigma_1(t))|\exp(-ML(t))+L_0(t)\exp(-ML(t))\\
&\quad +\exp L_1(t)\int_0^t
a(t)b(s)p(s)|x(\sigma_2(s))|ds\cdot \exp(-ML(t))\\
&\leq  K|x(\sigma_1(t))|\exp(-ML(\sigma_1(t)))+L_0(t)
\exp(-ML(t))\\&\quad +a(t)\exp L_1(t)\|x\|
\int_0^t b(s)p(s)\exp(ML(s))ds \exp(-ML(t))\\
&\leq  K|x(\sigma_1(t))|\exp(-ML(\sigma_1(t)))
 +L_0(t)\exp(-ML(t))\\&\quad +rta(t)b(t)p(t)\exp L_1(t)\\
&\leq  K|x(\sigma_1(t))|\exp(-ML(\sigma_1(t)))+L_0(t)
 \exp\Big(-\int_0^t L_0(s)ds \Big)\\
&\quad +rta(t)b(t)p(t)\exp L_1(0).
\end{align*}
Hence in view of the assumption (H6), we infer that
\begin{equation}
b(FX)\leq K b(X). \label{e3.1}
\end{equation}
where $b(X)$ was defined previously.

Next, let us fix $T>0$ and $\epsilon >0$.
Take arbitrary $t,\ s \in [0,T]$ with $|t-s|\leq \epsilon$.
Then we derive the following chain of inequalities.
\begin{equation}
\begin{aligned}
|(Fx)(t)-(Fx)(s)|
& \leq  \big|f\Big(t,x(\sigma_1(t)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big) \\
&\quad - f\Big(t,x(\sigma_1(s)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big)\big| \\
&\quad + \big|f\Big(t,x(\sigma_1(s)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big) \\
&\quad -f\Big(s,x(\sigma_1(s)),\int_0^s k(s,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big)\big| \\
&\leq  K |x(\sigma_1(t))-x(\sigma_1(s))| \\
&\quad + \big|f\Big(t,x(\sigma_1(s)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big) \\
&\quad -f\Big(s,x(\sigma_1(s)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big)\big| \\
&\quad +\big|f\Big(s,x(\sigma_1(s)),\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau
\Big) \\
&\quad -f\Big(s,x(\sigma_1(s)),\int_0^s k(s,\tau)g\Big(\tau,x(\sigma_2(\tau)
)\Big)d\tau \Big) \big| \\
&\leq  K |x(\sigma_1(t))-x(\sigma_1(s))|+\nu_1^T(f,\epsilon)
+\nu_2^T(f,\epsilon),
\end{aligned} \label{e3.2}
\end{equation}
where
\begin{align*}
\nu_1^T(f,\epsilon) = \sup\big\{&|f(t,x,y)-f(s,x,y)|: t,\, s \in [0,T],\,
 |t-s|\leq \epsilon,\\
& |x|\leq r\exp(ML(T)),\, |y|\leq N_1 \big\}, \\
\nu_2^T(f,\epsilon) = \sup\big\{&|f(t,x,y_1)-f(t,x,y_2)|: t \in [0,T],\,
|x|\leq r\exp(ML(T)),\\
&|y_1|, |y_2|\leq N_1,\, |y_1-y_2|\leq N_2 \big\}
\end{align*}
while the constants $N_1$ and $N_2$ appearing above are defined in the
following way
\[
N_1 = r\sup \big\{a(t)\int_0^t b(s)p(s)\exp(ML(s))ds:
 s,\ t \in[0,T] \},
\]
\begin{align*}
N_2 &\leq  \big|\int_0^t k(t,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau -\int_0^t k(s,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau \big|\\&\quad +\big|\int_0^t k(s,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau-\int_0^s k(s,\tau)g\big(\tau,x(\sigma_2(\tau))\big)d\tau\big|\\
&\leq  \int_0^t \big|k(t,\tau)- k(s,\tau)\big|
 \big|g\big(\tau,x(\sigma_2(\tau))\big)\big|d\tau \\
&\quad +\epsilon \sup\big\{\big|k(t,\tau)\big|
 \big|g\big(\tau,x(\sigma_2(\tau)\big)\big|: t,\ \tau \in[0,T]\big\}\\
&\leq  \nu^T(k, \epsilon)T \sup\big\{\big|g\big(\tau,x(\sigma_2(\tau)\big)
\big|: \tau \in[0,T]\big\}\\
&\quad +\epsilon \sup\big\{\big|k(t,\tau)\big|\
\big|g\big(\tau,x(\sigma_2(\tau)\big)\big|:\ t,\ \tau \in[0,T]\big\}.
\end{align*}
Now, let us denote
$$
q(\epsilon) = \nu_1^T(f,\epsilon)+ \nu_2^T(f,\epsilon).
$$
 From the uniform continuity of the function $f(t,x,y)$ on compact
subsets of $\mathbb{R}_+ \times \mathbb{R} \times \mathbb{R}$, we deduce
that $q(\epsilon)\to 0 $ as $\epsilon \to 0$.
Further, from \eqref{e3.2}, 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))-(Fx)(s)\exp(-ML(s))|\\
&\leq  |(Fx)(t)-(Fx)(s)|\exp(-ML(t))\\
&\quad +|(Fx)(s)|\ |\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K |x(\sigma_1(t))-x(\sigma_1(s))|\exp(-ML(t))+q(\epsilon)
 \exp(-ML(t))\\
&\quad +|(Fx)(s)|\ |\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K |x(\sigma_1(t))\exp(-ML(t))-x(\sigma_1(s))\exp(-ML(s))|\\
&\quad +K|x(\sigma_1(s))\exp(-ML(s))-x(\sigma_1(s))\exp(-ML(t))|\\
&\quad +q(\epsilon)\exp(-ML(t))+|(Fx)(s)|\ |\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K|x(\sigma_1(t))\exp(-ML(\sigma_1(t)))-x(\sigma_1(s))
 \exp(-ML(\sigma_1(s)))|\\
&\quad +K|x(\sigma_1(s))|\ |\exp(-ML(\sigma_1(s)))
 -\exp(-ML(\sigma_1(t)))|\\&\quad +q(\epsilon)\exp(-ML(t))+|(Fx)(s)|
 |\exp(-ML(t))-\exp(-ML(s))|\\
&\leq  K\omega^T\Big(x,\nu^T(\sigma_1,\epsilon)\Big)+Kr
 \exp(ML(\sigma_1(T)))\\&\quad  \nu^T\Big(\exp(-ML(\sigma_1(t))),
 \epsilon \Big)+q(\epsilon)\exp(-ML(t))\\&\quad +r\nu^T
 \Big(\exp(-ML(t)),\epsilon \Big)\Big[Kr\exp(ML(T))+\sup\left\{L_0(t):
 t \in [0,T]\right\}\\&\quad +r\sup\left\{t\exp(L_1(t))a(t)b(t)p(t)
 \exp(ML(t)): t\in [0,T]\right\}\Big].
\end{align*}
Keeping in mind, the uniform continuity of the functions $t\to \exp(-ML(t))$
 and $t\to \exp(-ML(\sigma_1(t)))$ on the interval $[0,T]$, from the above
estimate we infer that
\[
\omega_0^T(FX)\leq K\omega_0^T(X).
\]
Consequently,
\begin{equation}
\omega_0(FX)\leq K\omega_0(X). \label{e3.3}
\end{equation}
Now linking \eqref{e3.1} and \eqref{e3.3}, we get
\begin{equation}
\mu(FX)\leq K\mu(X) \label{e3.4}
\end{equation}
where $\mu$ denotes the measure of noncompactness defined earlier.

Further let us consider the sequence $(B_r^n)$, where
$B_r^1 = \mathop{\rm Conv} F(B_r)$, $B_r^2 = \mathop{\rm Conv }F(B_r^1)$,
\dots.  Obviously all sets of this sequence are nonempty, bounded,
convex and closed. Apart from this we have that
$B_r^{n+1}\subset B_r^n \subset B_r $ for $n=1, 2, 3,\dots$.
Thus, keeping in mind that $K < 1$ and taking into account of \eqref{e3.4},
we infer that $\lim_{n \to \infty}\mu(B_r^n) = 0$. Hence, in view of
the axiom (vi) of Definition \ref{def2.1}, we deduce that the set
$Y=\cap_{n=1}^\infty B_r^n$ is nonempty, bounded, convex
and closed. Moreover, in the light of Remark \ref{rmk2.1} we have that
$Y \in \ker \mu$. Let us also observe that the operator $F$
maps the set $Y$ into itself.

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

Observe that based on our assumptions, for an arbitrarily
fixed $t \in \mathbb{R}_+$, we have
\begin{equation}
\begin{aligned}
&|(Fx)(t)-(Fy)(t)|\exp(-ML(t)) \\
&\leq  \big| f\Big(t,x(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big) \\
&\quad - f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t)) \\
&\quad +\big|f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds \
\Big) \\
&\quad -f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,y(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t)) \\
&\leq  K\epsilon + \big|f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big) \\
&\quad -f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,y(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t)).
\end{aligned} \label{e3.5}
\end{equation}
Now, let us assume that $t\in [0,T]$, where $T$ is chosen as above.
Then we obtain
\begin{equation}
\begin{aligned}
&\big|f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,x(\sigma_2(s))\Big)ds\
\Big) \\
&-f\Big(t,y(\sigma_1(t)),\int_0^t k(t,s)g\Big(s,y(\sigma_2(s))\Big)ds\
\Big)\big|\exp(-ML(t)) \\
&\leq  \nu^T(f,\epsilon)\exp(-ML(t)) \\
&\leq  \nu^T(f,\epsilon)
\end{aligned} \label{e3.6}
\end{equation}
where $\nu^T(f,\epsilon)$ is defined as
\begin{align*}
\nu^T(f,\epsilon) = \sup\big\{&|f(t,v,x)-f(t,v,y)|: t\in[0,T],\,
 |v|\leq r\exp(ML(T)),\\
& |x|,\, |y| \leq N_1,\, |x-y|\leq N_3\big\}
\end{align*}
while the constant $N_3$ appearing above is defined by the formula
\begin{align*}
N_3 &= \sup\big\{\int_0^t k(t,s)\big|g\big(s,x(\sigma_2(s))\big)
-g\big(s,y(\sigma_2(s))\big)\big|ds: t, s \in [0,T]\big\}\\
 &=T\nu^T(g,\epsilon)\sup\big\{|k(t,s)|: t, s\in [0,T]\big\}.
\end{align*}
Observe that in view of the uniform continuity of the function $f(t,v,x)$
on the set $[0,T]\times [-r\exp(ML(T)),r\exp(ML(T))]\times [-N_1,N_1]$,
we conclude that $\nu^T(f,\epsilon)\to 0$ as $\epsilon \to 0$.

Next, we assume that $t\geq T$. Then keeping in mind that $x, y \in Y$
and $F:Y\to Y$, we derive easily the  estimate
\begin{equation}
\begin{aligned}
&|(Fx)(t)-(Fy)(t)|\exp(-ML(t)) \\
&\leq  |(Fx)(t)|\exp(-ML(t))+|(Fy)(t)|\exp(-ML(t))
\leq \epsilon.
\end{aligned} \label{e3.7}
\end{equation}

Now linking \eqref{e3.5}-\eqref{e3.7}, we conclude that the operator
$F$ is continuous on the set $Y$.

Finally, taking into account the properties of the set $Y$ and the
operator $F:Y\to Y$ established above and applying the classical
Schauder fixed point theorem we infer that the operator $F$
has at least one fixed point $x=x(t)$ in $Y$.
Obviously, the function $x(t)$ is a solution of  \eqref{e1.1}.

Moreover, keeping in mind that $Y\in$ $\ker \mu$,
we obtain that $x(t)= o(\exp(ML(t))$ as $t\to \infty$.
\end{proof}

\section{Example}

Consider the  functional integral equation, with deviating arguments,
\begin{equation}
\begin{aligned}
x(t)&=t^2+ \arctan\Big[\frac{x(t/3)}{4+t^2}+  \frac{1}{(1+t^3)}
\int_0^t \exp(-t)s^2x(s-\exp(-s))\\
&\quad \times \cos\big( x^2(s-\exp(-s))\big)ds\Big], \quad t\geq 0.
\end{aligned} \label{e4.1}
\end{equation}
Note that the above equation represents a special case
 of \eqref{e1.1} where
$$
f(t,x,y)= t^2+\arctan\big[  \frac{x}{4+t^2}+\frac{y}{(1+t^3)}\big],
$$
$\sigma_1(t)= t/3$ and $\sigma_2(t)= t-\exp(-t)$.
Moreover, the functions $k(t,s)$ and $g(s,x)$ take the form
$k(t,s) = s\exp(-t)$ and $g(s,x)= s\ x\cos x^2$.

It is easily seen that for \eqref{e4.1}, the assumption (H2) of
Theorem \ref{thm3.1} is satisfied with $L_0(t) = t^2$ and $L_1(t) = -\ln(1+t^3)$.
In fact,  observe that
\[
|f(t,0,y)|\leq t^2+\big|\arctan\big(\frac{y}{1+t^3}\big)\big|
\leq t^2+\frac{|y|}{1+t^3}.
\]
Also, note that the function $f(t,x,y)$ is continuous on the set
$\mathbb{R}_+\times \mathbb{R}\times \mathbb{R}$ and for arbitrary
$t\in \mathbb{R}_+$, $x_1, x_2, y \in\mathbb{R}$, we obtain
\begin{align*}
|f(t,x_1,y)-f(t,x_2,y)|\leq \frac{1}{4+t^2}|x_1-x_2|
\leq\frac{1}{4}|x_1-x_2|.
\end{align*}
This means the function $f(t,x,y)$ satisfies the Lipschitz condition
with respect to $x$ with the constant $K=\frac{1}{4}$.

Moreover the assumptions (H3) and (H4) of Theorem \ref{thm3.1}
are satisfied with $a(t)= \exp(-t)$, $b(t)=p(t)=t$ and hence
 $A$=1. Now, we get $(K+A/M)<1$ for $M>1$.

Similarly we can verify other details concerning the assumptions
of Theorem \ref{thm3.1}.
Taking into account the above established facts and applying
Theorem \ref{thm3.1}, we infer that \eqref{e4.1} has at least one solution
$x=x(t)$ such that $x\in C_L$, where
\[
L(t)=\frac{t^3}{3}+\frac{1}{3}\ln(1+t^3).
\]
Apart from this, we have that
\[
x(t)=o\Big(\exp\Big(M\big(\frac{t^3}{3}+\frac{1}{3}\ln(1+t^3)\big)\Big)\Big)
\]
as $t\to \infty$, where $M>1$ is a constant.

\begin{thebibliography}{00}

\bibitem{a1} J. Appell and P. P. Zabrejko;
Nonlinear Superposition Operators, in:
Cambridge Tracts in Mathematics, vol. 95,
{\it Cambridge University Press}, 1990.

\bibitem{b1} J. Banas;
Measures of noncompactness in the space of continuous tempered functions,
{\it Demonstratio Math.} 14 (1981), 127-133.

\bibitem{b2} J. Banas;
The existence of solutions of some functional-differential and
functional-integral equations, {\it Funct. Approx.} 13 (1982), 3-12.

\bibitem{b3} J. Banas and I. J. Cabrera;
On existence and asymptotic behavior of solutions of a functional
integral equation, {\it Nonlinear Anal.} 66 (2007), 2246-2254.

\bibitem{b4} J. Banas and I. J. Cabrera;
On solutions of a neutral differential equation with deviating argument,
 {\it Math. Comput. Modeling} 44 (2006), 1080-1088.

\bibitem{b5} J. Banas and K. Goebel;
 Measures of Noncompactness in Banach Spaces, {\it Marcel Dekker},
New York, 1980.

\bibitem{b6} T. A. Burton;
Volterra Integral and Differential Equations, {\it Academic Press},
 Newyork, 1983.

\bibitem{c1} C. Corduneanu;
Integral Equations and Applications, {\it Cambridge University Press},
Cambridge, 1991.

\bibitem{c2} S. Czerwik;
The existence of global solutions of a functional-differential equations,
{\it Colloq. Math.} 36 (1976), 121-125.

\bibitem{d1} K. Deimling;
 Nonlinear Functional Analysis, {\it Springer Verlag}, Berlin, 1985.

\bibitem{d2} B. C. Dhage and S. K. Ntouyas;
 Existence results for nonlinear functional integral equations via
a fixed point theorem of Krasnoselskii-Schaefer type,
{\it Nonlinear Stud.} 9 (2002), 307-317.

\bibitem{e1} W. G. El-Sayed;
Solvability of a differential equation with deviated argument;
{\it J. Math. Anal. Appl.} 327 (2007), 342-350.

\end{thebibliography}



\end{document}