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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 137, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/137\hfil Existence of solutions]
{Existence of solutions for mixed Volterra-Fredholm
integral equations}

\author[A. Aghajani, Y. Jalilian, K. Sadarangani \hfil EJDE-2012/137\hfilneg]
{Asadollah Aghajani, Yaghoub Jalilian, Kishin Sadarangani}  % in alphabetical order

\address{Asadollah Aghajani \newline
School of Mathematics, Iran University of Science and Technology,
Narmak, Tehran 16846-13114, Iran}
\email{aghajani@iust.ac.ir}

\address{Yaghoub Jalilian \newline
Department of Mathematics, Razi University, Kermanshah 67149, Iran}
\email{yajalilian@iust.ac.ir}

\address{Kishin Sadarangani \newline
Department of Mathematics, University of Las Palmas de Gran Canaria,
35017 Las Palmas de Gran Canaria, Spain}
\email{ksadaran@dma.ulpgc.es}

\thanks{Submitted May 11, 2012. Published August 19, 2012.}
\subjclass[2000]{45G10, 47H08}
\keywords{Functional integral equation; fixed-point; Nemytski operator;
\hfill\break\indent measure of weak noncompactness;  Carath\'eodory conditions}

\begin{abstract}
 In this article, we give some results concerning the continuity of
 the nonlinear Volterra and Fredholm integral operators on the
 space $L^{1}[0,\infty)$. Then by using the concept of measure of
 weak noncompactness, we prove an existence result for a functional
 integral equation which includes several classes of nonlinear
 integral equations. Our results extend some previous works.
\end{abstract}

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

\section{Introduction} 

Integral Equations occur in mechanics and many related fields of
engineering and mathematical physics \cite{b3,b4,b5,b8,b9,b10,c1,d2,k2,m2,m3,t1,z1}.
They also form one of useful mathematical tools in many branches of
pure analysis such as  functional analysis \cite{k1,t1}.
Recently many papers have been devoted to the  existence
of solutions of nonlinear functional integral equations
\cite{a1,a2,b1,b2,b5,b8}.
Our main  purpose is to prove an existence theorem
for a class of functional integral equations which contains many
integral or functional integral equations. For example, we can mention
the nonlinear Volterra integral equations, mixed Volterra-Fredholm  integral
equations and Fredholm  integral equations on the unbounded interval
$[0,\infty)$.

The concept of  measure of weak noncompactness was developed by De
 Blasie  \cite{d1}. Bana\'s and Knap \cite{b3} introduced a measure of weak
noncompactness in the space of real Lebesgue integrable functions
on an interval which is convenient for our purpose.
In the proof of main result we will use a
 measure of weak noncompactness given by Bana\'s and Knap to find a special
subset of $L^{1}[0,\infty)$ and also by  applying the Schauder fixed point theorem
 on this set, the existence result which generalizes
 several previous
 works \cite{a3,b4,b5,b6,b8,b10,d2,z1} will be proven.

Organization of this article: Section 2 gives some definitions and
preliminary results about continuous operator on
$L^{1}(\mathbb{R}_{+})$, Section 3 describes the concept of measure
of weak noncompactness and weakly compact sets in
$L^{1}(\mathbb{R}_{+})$ and finally in Section 4 we give our main
result and some examples.

\section{Notations and auxiliary results}

In this paper, $\mathbb{R}_{+}$ indicates the interval $[0,\infty)$
and for the Lebesgue measurable subset $D$ of $\mathbb{R}$, $m(D)$
implies the Lebesgue measure of $D$. Also, let $L^{1}(D)$ be the
space of all Lebesgue integrable functions $y$ on $D$ equipped
with the standard norm $\|y\|_{L^{1}(D)}=\int_{D}|y(x)|dx$.

\begin{lemma}[\cite{d5}] \label{lem2.1} 
Let $\Omega$ be a Lebesgue measurable subset of $\mathbb{R}$ and 
$1\leq p\leq\infty$. 
 If $\{f_{n}\}$ is convergent to $f\in L^{p}(\Omega)$ in the
$L^{p}$-norm, then there is a subsequence $\{f_{n_{k}}\}$ which
converges to $f$ a.e., and there is $g\in L^{p}(\Omega)$, 
$g\geq 0$, such that
\[
 |f_{n_{k}}(x)|\leq g(x),\quad \text{a.e. } x\in \Omega.
\]
\end{lemma}

 Let $I\subset\mathbb{R}$ be an interval. A function
$f:I\times\mathbb{R}\to\mathbb{R}$ is said to have the
Carath\'{e}odory property if
\begin{itemize}
\item[(M)] for all $x\in\mathbb{R}$ the function $t\mapsto f(t,x)$ is 
Lebesgue measurable on $I$;

\item[(C)] for  almost all $t\in I$ the function $x\mapsto f(t,x)$ is continuous on
  $\mathbb{R}$.
\end{itemize}

  One of the most important nonlinear mappings is the so-called
Nemytski operator which is also called the substitution (or
superposition) operator \cite{b3,b5,d5,z1}. 
By  substituting the function $x:I\to\mathbb{R}$ into the
function $f$ the Nemytski operator $F:x\to f(.,x(.))$ has
been obtained which acts on a space containing functions
$x$.
 Krasnosel'skii \cite{k2} and Appell and
Zabreiko \cite{a3} have proven the following assertion when $I$ is
a bounded and an unbounded domain  respectively.

\begin{theorem} \label{thm2.1}
 The superposition operator $F$ generated
by  function $f(t, x)$ maps the space $L^{1}(I)$ continuously
into itself  if and only if $|f (t,x)|\leq g(t)+c|x|$ for all $t$ in an interval $I$, 
and $x\in\mathbb{R}$, where
 $g$ is a function from the space $L^{1}(I)$ and $c$ is a nonnegative
constant. 
\end{theorem}

 \begin{remark} \label{rmk2.1}\rm
The Carath\'{e}odory property can be generalized to functions
$f:\Omega\times\mathbb{R}^{m}\to\mathbb{R}$ where $\Omega$ is a
measurable subset of $\mathbb{R}^{n}$. 
Theorem \ref{thm2.1} holds similarly if
and only if there exist $c\in\mathbb{R}$ and $g\in L^{1}(\Omega)$
such that
\begin{equation}\label{}
|f(x,y)|\leq g(x)+c\sum_{i=1}^{m}|y_i|,
\end{equation}
for almost all $x\in\Omega$ and all
$y=(y_1,\dots ,y_{m})\in\mathbb{R}^{m}$.
\end{remark}

Now we are going to review a theorem from \cite{b3} about the
continuity of the linear Volterra integral operator on the space
$L^{1}=L^{1}(\mathbb{R}_{+})$. 
Let $\Delta=\{(t,s):0\leq s\leq t\}$ and $k:\bigtriangleup\to \mathbb{R}$
 be a measurable function with respect to both variables. Consider
\begin{eqnarray*}
(Kx)(t)=\int_0^{t}k(t,s)x(s)ds,\quad t\in\mathbb{R}_{+},\; x\in L^{1}(\mathbb{R}_{+}).
\end{eqnarray*}
We notice that $K$ is a linear Volterra integral operator
generated by $k$.

\begin{theorem} \label{thm2.2}
 Let $k$ be measurable on $\Delta$ and such that
\begin{equation}\label{e}
\operatorname{ess\,sup}_{s\geq0}\int_{s}^{\infty}|k(t,s)|dt<\infty.
\end{equation}
Then the Volterra integral operator $K$ generated by $k$ maps
(continuously) the space $L^{1}(\mathbb{R}_{+})$ into itself and
the norm $\|K\|$ of this operator is majorized by the number
$\operatorname{ess\,sup}_{s\geq0}\int_{s}^{\infty}|k(t,s)|dt$.
\end{theorem}

Assume that $A$ is a measurable subset of $\mathbb{R}_{+}$, we denote
by $\|K\|_{A}$ the norm of linear Volterra operator
$K:L^{1}(A)\to L^{1}(A)$.

Now we give a result concerning the continuity of the
nonlinear Volterra operator on $L^{1}(\mathbb{R}_{+})$. In what
follows we suppose that
$u:\mathbb{R}_{+}\times\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$ is
a function which satisfies:
\begin{itemize}
\item[(a)]  $t\to u(t,s,x)$ is measurable for all
$s\in \mathbb{R}_{+}$ and $x\in\mathbb{R}$;
\item[(b)] $(s,x)\to u(t,s,x)$ is continuous for
almost all $t\in\mathbb{R}_{+}$.
\end{itemize}

\begin{theorem} \label{thm2.3}
Let $u:\mathbb{R}_{+}\times\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$
 be a function such that
\begin{equation} \label{eq4}
 |u(t,s,x)|\leq k_1(t,s)+k_2(t,s)|x|,\quad t,s\in
 \mathbb{R}_{+}\; x\in\mathbb{R},
\end{equation}
where
$k_i:\mathbb{R}_{+}\times\mathbb{R}_{+}\to\mathbb{R}_{+}$
(i=1,2) are measurable functions. Moreover, the
integral operator $K_2$ generated by $k_2$ is a continuous map
from $L^{1}(\mathbb{R}_{+})$ into itself and
$\int_0^{t}k_1(t,s)ds\in L^{1}(\mathbb{R}_{+})$. Then the
operator
\[
 (Ux)(t)=\int_0^{t}u(t,s,x(s))ds,
\]
maps $L^{1}(\mathbb{R}_{+})$ continuously into itself
\end{theorem}

 \begin{proof} 
Let $\{x_{n}\}$ be an arbitrary sequence in
 $L^{1}=L^{1}(\mathbb{R}_{+})$ which converges to $x\in L^{1}$
in the $L^{1}$-norm. By using Lemma \ref{lem2.1} there is a subsequence
$\{x_{n_{k}}\}$ which converges to $x$ a.e., and there is 
$g\in L^{1}$, $g\geq 0$, such that
\begin{equation} \label{eq6}
 |x_{n_{k}}(s)|\leq g(s),\quad \text{a.e. on } \mathbb{R}_{+}.
\end{equation}
Since $x_{n_{k}}\to x$ almost everywhere in $\mathbb{R}_{+}$,
it readily follows from (b) that
\begin{equation}\label{eq7}
u(t,s,x_{n_{k}}(s))\to
u(t,s,x(s)),\quad \text{for almost all } s,t\in \mathbb{R}_{+}.
\end{equation}
From inequalities \eqref{eq4} and \eqref{eq6}, we infer that
\begin{equation} \label{eq8}
 |u(t,s,x_{n_{k}}(s))|\leq k_1(t,s)+k_2(t,s)g(s),\quad ~
  \text{for almost all }s, t\in \mathbb{R}_{+}.
\end{equation}
As a consequence of the Lebesgue's Dominated Convergence Theorem,
\eqref{eq7} and \eqref{eq8} yield
\[
\int_0^{t}
u(t,s,x_{n_{k}}(s))ds\to\int_0^{t}u(t,s,x(s))ds,
\]
 for almost all $t\in\mathbb{R}_{+}$. Inequality \eqref{eq8} implies that
\begin{equation} \label{eq88}
 |(Ux_{n_{k}})(t)|\leq\int_0^{t}|u(t,s,x_{n_{k}}(s))|ds
\leq \int_0^{t}k_1(t,s)ds+
 \int_0^{t}k_2(t,s)g(s)ds,
\end{equation}
for almost all $t\in \mathbb{R}_{+}$. Regarding the assumptions on
$k_1$ and $k_2,$ we obtain
\begin{equation} \label{eq888}
\int_0^{\infty} \int_0^{t}k_1(t,s)\,ds\,dt+\int_0^{\infty}
 \int_0^{t}k_2(t,s)g(s)\,ds\,dt<\infty.
\end{equation}
Then inequalities \eqref{eq88}-\eqref{eq888} and  the Lebesgue's
Dominated Convergence Theorem imply
\begin{eqnarray*}
\|Ux_{n_{k}}-Ux\|_{L^{1}}\to0.
\end{eqnarray*}
Since any sequence $\{x_{n}\}$ converging to $x$ in $L^{1}$ has a
subsequence $\{x_{n_{k}}\}$ such that $Ux_{n_{k}}\to Ux$
in $L^{1}$, we can conclude that $U:L^{1}(\mathbb{R}_{+})\to
L^{1}(\mathbb{R}_{+})$ is a continuous operator.
\end{proof}

Similar to the above theorem, we can prove the following theorem
for the nonlinear Fredholm integral operators.

 \begin{theorem} \label{thm2.4}
Let $v:\mathbb{R}_{+}\times\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$
be a function satisfying {\rm (a)--(b)} such that
\begin{equation} \label{e.29}
 |v(t,s,x)|\leq k_1(t,s)+k_2(t,s)|x|,\quad t,s\in  \mathbb{R}_{+}\; x\in\mathbb{R},
\end{equation}
 where $k_i:\mathbb{R}_{+}\times\mathbb{R}_{+}\to\mathbb{R}_{+}$
($i=1,2$) are measurable functions. Moreover, the
integral operator
$(K_2x)(t)=\int_0^{\infty} k_2(t,s)|x(s)|ds$ maps $L^{1}(\mathbb{R}_{+})$
continuously into itself and $k_1(t,s)\in L^{1}(\mathbb{R}_{+}\times\mathbb{R}_{+})$.
Then the operator
\begin{eqnarray*}
 (Vx)(t)=\int_0^{\infty}v(t,s,x(s))ds,
\end{eqnarray*}
maps $L^{1}(\mathbb{R}_{+})$ continuously into itself.
\end{theorem}

\section{Measure of weak noncompactness}

 Let $(E,\|\cdot\|)$ be an infinite dimensional Banach
space with zero element $\theta$. We write $B(x,r)$ to denote the
closed ball centered at $x$ with radius $r$ and $\operatorname{conv}X$ to denote
the closed convex hull of $X$. Further let:

  $\textbf{m}_{E}$ be the family of all nonempty bounded subsets of $E$,
 $\textbf{n}_{E}^{w}:$ the subfamily of $\textbf{m}_{E}$ consisting
of all relatively weakly compact sets, and
$\overline{X}^{w}:$ the weak closure of a set $X$.

In this paper, we use the following definition of the measure of
weak noncompactness \cite{b6}.

\begin{definition} \label{def3.1}
A mapping $\mu :\textbf{m}_{E}\to \mathbb{R}_{+}$ is said to be a measure of
weak noncompactness if it satisfies the following conditions:
\begin{itemize}
\item[(1)] The family $\ker \mu=\{X\in \textbf{m}_{E}:\mu(X)=0\}$ is nonempty
 and $\ker \mu\subset \textbf{n}_{E}^{w}$,
\item[(2)] $X\subset Y\Rightarrow \mu(X)\leq\mu(Y)$,
\item[(3)] $\mu (\operatorname{conv}X)=\mu(X)$,
\item[(4)] $\mu(\lambda X+(1-\lambda)Y)\leq\lambda
\mu(X)+(1-\lambda)\mu(Y)$ for $\lambda\in[0,1]$,
\item[(5)] If $X_{n}\in \textbf{m}_{E}$,
$X_{n}=\overline{{X}}_{n}^{w}$ for $n=1,2,\dots $ and if
$\lim_{n\to\infty} \mu(X_{n})=0,$ then the intersection set
$X_{\infty}=\bigcap_{n=1}^{\infty} X_{n}$ is nonempty.
\end{itemize}
\end{definition}

In the sequel, we will use a measure of weak noncompactness
represented by a convenient formula in the space
$L^{1}(\mathbb{R}_{+})$ \cite{b7}. 
For $X\in\textbf{m}_{L^{1}(\mathbb{R}_{+})}$ define:
\begin{gather*}
c(X)=\lim_{\varepsilon\to 0}\{\sup_{x\in
X}\{\sup[\int_{D}|x(t)|dt:D\subset\mathbb{R}_{+},~m(D)\leq\varepsilon]\}\},\\
d(X)=\lim_{T\to\infty}\{\sup[\int_{T}^{\infty}|x(t)|dt:x\in
X]\},
\\
\mu(X)=c(X)+d(X).
\end{gather*}
In \cite{b7}, it is shown that $\mu$ is a measure of weak
noncompactness on $L^{1}(\mathbb{R}_{+})$. By using the following
theorem \cite{d4}, we can infer that
$\ker\mu=\textbf{n}_{L^{1}(\mathbb{R}_{+})}^{w}$.

 \begin{theorem} \label{thm3.1}
 A bounded set $X$ is relatively weakly compact in
$L^{1}(\mathbb{R}_{+})$ if and only if the following two conditions
are satisfied:
\begin{itemize}
\item[(1)] for any $\varepsilon>0$ there exists $\delta>0$ such that if
$m(D)\leq \delta$ then $\int_{D}|x(t)|dt\leq\varepsilon$ for all
$x\in X$,
\item[(2)] for any $\varepsilon>0$ there exists $T>0$ such that
$\int_{T}^{\infty}|x(t)|dt\leq\varepsilon$ for any $x\in X$.
\end{itemize}
\end{theorem}

\section{Main results}

In this section, we study the existence of solutions to the
functional integral equation 
\begin{equation}\label{eq}
    x(t)=f\Big(t,\int_0^{t}u(t,s,x(s))ds,\int_0^{\infty}a_2(t)v(s,x(s))ds\Big),
    \quad t\geq0.
\end{equation}
This equation  is a general form of many integral
equations, such as the mixed
Volterra-Fredholm integral equation
\begin{equation}\label{eq1}
    x(t)=g(t)+\int_0^{t}k(t,s)u(s,x(s))ds+a(t)\int_0^{\infty}v(s,x(s))ds,
\quad t\geq0.
\end{equation}
Equations like \eqref{eq1} have been considered by many authors;
see for example \cite{b9,c2,d3,m1,m3}
and references cited therein.
Moreover, \eqref{eq} contains the nonlinear Volterra
and Fredholm integral equations on $\mathbb{R}_{+}$ such as:
\begin{gather*}
x(t)=g(t)+\int_0^{t}u(t,s,x(s))ds,\quad t\geq0, \\
x(t)=f(t)+a(t)\int_0^{\infty}v(s,x(s))ds,\quad t\geq0.
\end{gather*}
We consider  equation \eqref{eq} under the following
assumptions:
\begin{itemize}
\item[(i)] The function
$f:\mathbb{R}_{+}\times\mathbb{R}^{2}\to\mathbb{R}$ satisfies
Carath\'{e}dory conditions and there exist constant
$B\in\mathbb{R}_{+}$ and function $a_1\in L^{1}(\mathbb{R}_{+})$ such
that
\begin{equation} \label{eq2}
|f(t,x,y)|\leq
a_1(t)+B(|x|+|y|),\quad t\in\mathbb{R}_{+},\; x,y\in\mathbb{R}.
\end{equation}

\item[(ii)] $u:\mathbb{R}_{+}\times\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$
 satisfies (a)--(b) and $|u(t,s,x)|\leq k_1(t,s)+k_2(t,s)|x|$ for
$(t,s,x)\in \mathbb{R}_{+}\times\mathbb{R}_{+}\times\mathbb{R}$, where
$k_i:\mathbb{R}_{+}\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ (i=1,2)
satisfies Carath\'{e}odory conditions. Moreover, the integral
operator $K_2$ generated by $k_2$ i.e.
\begin{equation}\label{eq3}
    (K_2x)(t)=\int_0^{t}k_2(t,s)x(s)ds,
\end{equation}
is a continuous map from $L^{1}(\mathbb{R}_{+})$ into itself and
${\int_0^{t}k_1(t,s)ds\in L^{1}(\mathbb{R}_{+})}$.

\item[(iii)] $v:\mathbb{R}_{+}\times\mathbb{R}\to\mathbb{R}$
 satisfies Carath\'{e}odory conditions and $|v(s,x)|\leq n(s)+b|x|$ for
 $(s,x)\in\mathbb{R}_{+}\times\mathbb{R}$ where $n\in L^{1}(\mathbb{R}_{+})$
 and $b$ is a positive constant.
\item[(iv)] $a_2:\mathbb{R}_{+}\to\mathbb{R}$ is a function in
$L^{1}(\mathbb{R}_{+})$.
\item[(v)] $B(b\|a_2\|+\|K_2\|)<1$,
where $\|K_2\|$ denotes the norm of operator $K_2$.

\end{itemize}
To prove the main result of this paper, we need the next lemma.

Let  $X$ be a nonempty, closed, convex, bounded and weakly compact subset of
$L^{1}=L^{1}(\mathbb{R}_{+})$ and $I=[0,a]$ where $a>0$. Moreover, we
define the operator $F$ on $L^{1}=L^{1}(\mathbb{R}_{+})$ as follows:
\begin{equation} \label{f}
(Fx)(t)=f\Big(t,\int_0^{t}u(t,s,x(s))ds,\int_0^{\infty}a_2(t)v(s,x(s))ds\Big).
\end{equation}

\begin{lemma} \label{lem4.1} 
 Suppose that assumptions
(i)--(iv) hold and the operator
$F$ takes $X$ into itself. Then for any $\varepsilon>0$ there
exists $D_{\varepsilon}\subset I$ with $m(I \setminus
D_{\varepsilon})\leq\varepsilon$ such that $F(\operatorname{conv} FX)$ on
$D_{\varepsilon}$ is a relatively compact subset of
$C(D_{\varepsilon})$. 
\end{lemma}

\begin{proof} Consider an arbitrary  but fixed $\varepsilon>0$. Then
using Lusin theorem and generalized version of Scorza-Dragoni
theorem \cite{c1} we can find the closed set
$D_{\varepsilon}\subset I$ with 
$m(I \setminus D_{\varepsilon})\leq\varepsilon$, such that the functions
$a_i\big|_{D_{\varepsilon}},$
$k\big|_{D_{\varepsilon}\times\mathbb{R}_{+}}$,
$u\big|_{D_{\varepsilon}\times\mathbb{R}_{+}\times\mathbb{R}}$ and
$f\big|_{D_{\varepsilon}\times\mathbb{R}_{+}\times\mathbb{R}}$ are
continuous. Let us take an arbitrary $x\in X$. Then for 
$t\in D_{\epsilon}$ we have
\begin{equation} \label{bu}
\begin{split}
\big|\int_0^{t}u(t,s,x(s))ds\big|
& \leq \int_0^{t}k_1(t,s)+ \int_0^{t}k_2(t,s)|x(s)|ds\\
&\leq \bar{k}_1a+\bar{k}_2\|x\|\leq\bar{k}_1a+\bar{k}_2\|X\|=:U_{\varepsilon},
\end{split}
\end{equation}
and
\begin{equation} \label{bv}
\big|\int_0^{\infty}a_2(t)v(s,x(s))ds\big|
\leq \bar{a}_2 (\|n\|+b\|x\|)\leq \bar{a}_2 (\|n\|+b\|X\|)=:V_{\varepsilon},
\end{equation}
where $\|X\|=\sup\{\|x\|:x\in X\}$,
$\bar{a}_i=\sup\{|a_i(t)|:t\in D_{\varepsilon}\}$ and
$\bar{k}_i=\sup\{|k_i(t,s)|:(t,s)\in D_{\varepsilon}\times
I\}$ for $i=1,2$. Now let $y\in FX$. Then there exists $x\in X$
such that $y=Fx$. Using the inequalities \eqref{bu} and \eqref{bv}
for $t\in D_{\varepsilon}$ we obtain
\begin{equation} \label{by}
\begin{split}
|y(t)|=|(Fx)(t)|
&\leq a_1(t)+|\int_0^{t}u(t,s,x(s))ds|+|\int_0^{\infty}a_2(t)v(s,x(s))ds|\\
&\leq \bar{a}_1+U_{\varepsilon}+V_{\varepsilon}=:Y_{\varepsilon}.
\end{split}
\end{equation}
We can easily deduce that the inequality \eqref{by} is true, for
any $y\in Y=\operatorname{conv} FX$. Now assume that $\{y_{n}\}$ is a sequence in
$Y$ and let $t_1,t_2\in D_{\varepsilon}$. Without loss of
generality we can assume that $t_1\leq t_2$. Relatively weakly
compactness of the set $\{y_{n}\}$ implies that for
$\varepsilon_0=t_2-t_1$ there exists
$0<\delta_0\leq\varepsilon_0$ such that for any measurable
subset $D$ of $[0,t_1]$ with $m(D)\leq \delta_0$, we have:
\begin{equation} \label{wy}
\int_{D}|y_{n}(t)|dt\leq \varepsilon_0\quad \text{for } n=1,2,\dots .
\end{equation}
We see that the estimate \eqref{by} does not depend on the choice of
$\varepsilon$. Thus for $\varepsilon=\delta_0$ there exists a
closed set $D_{\delta_0}\subset [0,t_1]$ with $m([0,t_1]
\setminus D_{\delta_0})\leq \delta_0$ such that
\begin{equation} \label{e4.10}
|y_{n}(t)|\leq Y_{\delta_0}\quad \text{for }t\in
D_{\delta_0},\; n=1,2,\dots .
\end{equation}
Hence from \eqref{wy} and uniform continuity of
$u\big|_{D_{\varepsilon}\times
D_{\delta_0}\times[-Y_{\delta_0},Y_{\delta_0}]}$ and
$k_i\big|_{D_{\varepsilon}\times[0,a]}$ $(i=1,2)$ we infer that
\begin{equation} \label{eu}
\begin{split}
&\int_0^{t_1}|u(t_1,s,y_{n}(s))-u(t_2,s,y_{n}(s))|ds\\
&\leq \int_{D_{\delta_0}}|u(t_1,s,y_{n}(s))-u(t_2,s,y_{n}(s))|ds\\
&\quad + \int_{[0,t_1] \setminus D_{\delta_0}}|u(t_1,s,y_{n}(s))-u(t_2,s,y_{n}(s))|ds \\
&\leq O(|t_1-t_2|)+2\bar{k}_1m([0,t_1] \setminus
D_{\delta_0})
+2\bar{k}_2\int_{[0,t_1] \setminus D_{\delta_0}} |y_{n}(t)|dt \\
&\leq O(|t_1-t_2|)+2(\bar{k}_1+\bar{k}_2)|t_1-t_2|.
\end{split}
\end{equation}
Here $O$ is a function which $O(\eta)\to0$ as
$\eta\to0$. Thus from \eqref{eu} we have:
\begin{equation} \label{1a}
\begin{split}
&|\int_0^{t_1}u(t_1,s,y_{n}(s))ds-\int_0^{t_2}u(t_2,s,y_{n}(s))ds|\\
&\leq \int_0^{t_1}|u(t_1,s,y_{n}(s))-u(t_2,s,y_{n}(s))|ds
 +|\int_{t_1}^{t_2}u(t_2,s,y_{n}(s))ds| \\
&\leq O(|t_1-t_2|)+2(\bar{k}_1+\bar{k}_2)|t_1-t_2|
 +\int_{t_1}^{t_2}k_1(t_2,s)ds
 +\int_{t_1}^{t_2}k_2(t_2,s)|y_{n}(s)|ds \\
&\leq O(|t_1-t_2|)+2(\bar{k}_1+\bar{k}_2)|t_1-t_2|
 +\bar{k}_1|t_1-t_2|+\bar{k}_2\int_{t_1}^{t_2}|y_{n}(s)|ds.
\end{split}
\end{equation}
Weakly compactness of the set $\{y_{n}\}$ implies that
$\int_{t_1}^{t_2}|y_{n}(s)| ds$ is arbitrary small uniformly with
respect to $n\in\mathbb{N}$ if $t_2-t_1$ is small enough. Then
from \eqref{1a} and \eqref{bu} the sequence $\{Uy_{n}\}$ which
\[
(Uy_{n})(t)=\int_0^{t}u(t,s,y_{n}(s))ds,
\]
is equibounded and equicontinuous on the set $D_{\varepsilon}$.
Obviously from assumption (iii) and inequality \eqref{bv} we can
easily infer that the sequence $\{Vy_{n}\}$ is equibounded and
equicontinuous on $D_{\varepsilon}$ where
\[
(Vy_{n})(t)=\int_0^{\infty}a_1(t)v(s,y_{n}(s))ds.
\]
Hence, uniform continuity of
$f\big|_{D_{\varepsilon}\times[-U_{\varepsilon},U_{\varepsilon}]\times[-V_{\varepsilon},V_{\varepsilon}]}$
implies that the sequence $\{Fy_{n}\}$ is equibounded and
equicontinuous on $D_{\varepsilon}$. Then, by Ascoli theorem the
sequence $\{Fy_{n}\}$ has a convergent subsequence in the norm
$C(D_{\varepsilon})$. Therefore, $F(\operatorname{conv} FX)$ is a relatively
compact subset of $C(D_{\varepsilon})$.
\end{proof}
Now we present our main result.


\begin{theorem} \label{thm4.1}  Under assumptions
{\rm (i)--(v)}, the functional
integral equation \eqref{eq} has at least one
solution $x\in L^{1}(\mathbb{R}_{+})$.
\end{theorem}

\begin{proof} At first we define the operator $F$ on
$L^{1}=L^{1}(\mathbb{R}_{+})$ by
\[
(Fx)(t)=f\Big(t,\int_0^{t}u(t,s,x(s))ds,\int_0^{\infty}a_2(t)v(s,x(s))ds\Big).
\]
We prove the theorem in the following steps.

 Step 1. $F:L^{1}(\mathbb{R}_{+})\to
L^{1}(\mathbb{R}_{+})$ is a continuous operator. \\
Using Theorems \ref{thm2.3} and \ref{thm2.4}, the operators
\[
 (Ux)(t)=\int_0^{t}u(t,s,x(s))ds,\quad 
 (Vx)(t)=\int_0^{\infty}v(t,s,x(s))ds,
\]
map $L^{1}(\mathbb{R}_{+})$ continuously into itself. Also by
assumptions (i)--(iv) and Remark \ref{rmk2.1}, the Nemytski operator
generated by $f$ is a continuous operator from
$L^{1}(\mathbb{R}_{+})$ into
$L^{1}(\mathbb{R}_{+})$. Thus the operator $F$ is continuous.

Step 2. There exists a positive number $r$ such that the
operator $F$ takes the ball $B(\theta,r)$ into itself.
Let $x\in L^{1}(\mathbb{R}_{+})$. Then
\begin{equation} \label{iq}
\begin{split}
\|Fx\|&=\int_0^{\infty}|(Fx)(t)|dt\\
&=\Big |\int_0^{\infty}f\big(t,\int_0^{t}u(t,s,x(s))ds,
\int_0^{\infty}a_2(t)v(s,x(s))ds\big)dt\Big| \\
&\leq \|a_1\|
+B\int_0^{\infty}\Big(\int_0^{t}|u(t,s,x(s))|ds
 +\int_0^{\infty}|a_2(t)v(s,x(s))|ds\Big)dt \\
&\leq \|a_1\|+B(K_1+\|K_2\|\|x\|)+B\|a_2\|(\|n\|+b\|x\|),
\end{split}
\end{equation}
where
$$
K_1=\int_0^{\infty}\int_0^{t}k_1(t,s)\,ds\,dt.
$$
From \eqref{iq} and assumption (v), one can deduce that for
${r=\frac{\|a_1\|+B(K_1+\|a_2\|\|n\|)}{1-B(\|K_2\|+\|a_2\|b)}}$,
the operator $F$ takes $B_{r}=B(\theta,r)$ into itself.

Step 3. There exists a weakly compact subset $Y$ such that the
operator $F$ maps $Y$ into itself.
 Let $X$ be a nonempty subset of $B_{r}$. Let
$\varepsilon>0$ be an arbitrary  number and $D\subset\mathbb{R}_{+}$
be a measurable subset with $m(D)\leq\varepsilon$. Then for $x\in
X$ we have:
\begin{equation} \label{c1}
\begin{split}
&\int_{D}|(Fx)(t)|dt\\
&\leq \int_{D}a_1(t)dt+B\int_{D}\Big(\int_0^{t}|u(t,s,x(s))|ds
 +\int_0^{\infty}|a_2(t)v(s,x(s))|ds\Big)dt \\
&\leq \int_{D}a_1(t)dt + B\int_{D}\int_0^{t} k_1(t,s)\,ds\,dt\\
&\quad +B \int_{D}|(K_2x)(t)|dt+B(\|n\|+br)\int_{D}|a_2(t)|dt \\
&\leq \int_{D}a_1(t)dt + B\int_{D}\int_0^{t}k_1(t,s)\,ds\,dt\\
&\quad +B \|K_2\|_{D}\int_{D}|x(t)|dt+B(\|n\|+br)\int_{D}|a_2(t)|dt
\end{split}
\end{equation}
Further, as a simple consequence of the fact that a single set in
$L^{1}(\mathbb{R}_{+})$ is weakly compact, for
$\gamma(t)=a_1(t),{\int_0^{t}k_1(t,s)ds}$  or $a_2(t)$, we
have:
\begin{itemize}
\item[(C1)] $\lim_{\varepsilon\to 0}
\{\sup[\int_{D}|\gamma(t)|dt:D\subset\mathbb{R}_{+},~m(D)\leq\varepsilon]\}=0$,

\item[(C2)] $\lim_{T\to\infty}\int_{T}^{\infty}|\gamma(t)|dt=0$.
\end{itemize}

Then from \eqref{c1} and (C1) we conclude that
\begin{equation} \label{c3}
c(FX)\leq B\|K_2\| c(X).
\end{equation}
By similar calculations we obtain:
\begin{equation} \label{c2}
\begin{split}
\int_{T}^{\infty}|(Fx)(t)|dt
&\leq \int_{T}^{\infty}a_1(t)dt+B
\int_{T}^{\infty} \int_0^{t}k_1(t,s)ds\,dt\\
&\quad +B\|K_2\|\int_{T}^{\infty}|x(t)|dt+B(\|n\|+br)
\int_{T}^{\infty}|a_2(t)|dt.
\end{split}
\end{equation}
Therefore, from (C2), we have
\begin{equation} \label{c4}
d(FX)\leq B\|K_2\| d(X).
\end{equation}
Hence by adding \eqref{c3} and \eqref{c4} we obtain
\begin{equation} \label{c5}
\mu(FX)\leq B\|K_2\| \mu(X).
\end{equation}
 Assumption (v) implies that $B\|K_2\|< 1$. Thus inequality
\eqref{c5} yields that there exists a closed, convex and weakly
compact set $X_{\infty}\subset B_{r}$ such that
$FX_{\infty}\subset X_{\infty} $.
 Let $Y=\operatorname{conv} FX_{\infty}$. Obviously $FY\subset Y\subset
 X_{\infty}$. Thus $FY$ and $ Y$ are relatively weakly compact.

Step 4. The set $FY$ obtained in the Step 3 is a
  relatively compact subset of  $L^{1}(\mathbb{R}_{+})$.
Suppose $\{y_{n}\}\subset Y$, and fix an arbitrary $\varepsilon>0$.
Applying Theorem \ref{thm3.1} for relatively weakly compact set $FY$ implies that
there exists $T>0$ such that for $m,n\in\mathbb{N}$
\begin{equation} \label{c6}
\int_{T}^{\infty}|(Fy_{n})(t)-(Fy_{m})(t)|dt\leq
\frac{\varepsilon}{2}.
\end{equation}
Further, by using Lemma \ref{lem4.1} for any $k\in\mathbb{N}$  there exists a
closed set $D_{k}\subset[0,T]$ with $m([0,T]\setminus D_{k})\leq
\frac{1}{k}$ such that $\{Fy_{n}\}$ is a relatively compact subset
of $C(D_{k})$. So for any $k\in\mathbb{N}$ there exists a subsequence
$\{y_{k,n}\}$ of $\{y_{n}\}$ which is a Cauchy sequence in
$C(D_{k})$. Also these subsequences  can be chosen such that
$\{y_{k+1,n}\}\subseteq \{y_{k,n}\}$. Consequently the subsequence
$\{y_{n,n}\}$ is a Cauchy sequence in each space $C(D_{k})$ for
any $k\in\mathbb{N}$ which for simplicity we call it again
$\{y_{n}\}$.

 From the relatively weakly compactness of $\{Fy_{n}\}$ we can find
$\delta>0$ such that for each closed subset $D_{\delta}$ with
$m([0,T]\setminus D_{\delta})\leq\delta$ we obtain:
\begin{equation} \label{c7}
\int_{[0,T]\setminus D_{\delta}}|(Fy_{n})(t)-(Fy_{m})(t)|dt\leq
\frac{\varepsilon}{4},\quad ~m,n\in\mathbb{N}.
\end{equation}
Considering the fact $\{Fy_{n}\}$ is Cauchy in $C(D_{k})$ for each
$k\in\mathbb{N}$ one can find $k_0$ such that $m([0,T]\setminus
D_{k_0})\leq\delta$ and for $m,n\geq k_0$
\begin{equation} \label{c8}
\|(Fy_{n})-(Fy_{m})\|_{C(D_{k_0})}\leq
\frac{\varepsilon}{4(m(D_{k_0})+1)},
\end{equation}
consequently \eqref{c7} and \eqref{c8} imply that
\begin{equation} \label{c9}
\begin{split}
&\int_0^{T}|(Fy_{n})(t)-(Fy_{m})(t)|dt\\
&= \int_{D_{k_0}}|(Fy_{n})(t)-(Fy_{m})(t)|dt
 +\int_{[0,T]\setminus D_{k_0}}|(Fy_{n})(t)-(Fy_{m})(t)|dt\\
&\leq \frac{\varepsilon}{2},
\end{split}
\end{equation}
for $m,n\geq k_0$. Now by considering \eqref{c6} and \eqref{c9}
for $m,n\geq k_0$ we obtain the inequality
\begin{equation} \label{c10}
\|(Fy_{n})-(Fy_{m})\|_{L^{1}}=\int_0^{\infty}|(Fy_{n})(t)-(Fy_{m})(t)|dt\leq
\varepsilon,
\end{equation}
which shows that the sequence $\{Fy_{n}\}$ is a Cauchy sequence in
the Banach space $L^{1}({\mathbb{R}_{+}})$. Then $\{Fy_{n}\}$ has a
convergent subsequence which implies that $FY$ is a relatively
compact subset of $L^{1}({\mathbb{R}_{+}})$.

Step 5. By the Step 4 there exists a bounded, closed, convex set
$Y\subset L^{1}({\mathbb{R}_{+}})$ such that the operator $F:Y\to Y$
is continuous and compact. Then Schauder fixed point theorem
 completes the proof.
\end{proof}

Next, by applying our theorem we prove the existence of
solutions for some integral equations.

\begin{example} \label{examp4.1}\rm
 Consider the  Fredholm integral equation
\begin{equation}\label{w}
 x(t)=
\frac{t^{2/3}}{t^{3}+1}+\int_0^{\infty}
a_2(t)\tanh(\frac{s+|x(s)|}{(1+s^{2})^{2}})ds,\quad t\geq0,
\end{equation}
where
$$
a_2(t)=\frac{t\pi}{8}\chi_{[0,1]}+\frac{1}{4(1+t^{2})}\chi_{(1,\infty)},
$$
in which for $A\subset \mathbb{R}_{+}$ and
$\chi_{A}(x)=\begin{cases}
    1, & x\in A, \\
   0, & x\in \mathbb{R}_{+} \setminus A.
\end{cases}
$
Put
\begin{gather*}
f(t,x,y)=\frac{t^{2/3}}{1+t^{3}}+y, \quad u(t,s,x)=0,\\
v(s,x)=\tanh(\frac{s+|x|}{(1+s^{2})^{2}}),\quad a_1(t)=\frac{t^{2/3}}{1+t^{3}},\\
n(s)=\frac{s}{(1+s^{2})^{2}},\quad B=1,\; b=1.
\end{gather*}
We know that $\tanh(\alpha)\leq \alpha$, for $\alpha>0$. Then
\begin{gather*}
 |v(s,x)|\leq n(s)+b|x|,\\
 |f(t,x,y)|\leq\frac{t^{2/3}}{1+t^{3}}+B(|x|+|y|),
\end{gather*}
Further
\[
\|a_2\|=\int_0^{\infty}|a_2(t)|dt=\frac{\pi}{8}.
\]
 Since $u=0$ we can choose $k_1=k_2=0$ and then $\|K_2\|=0$. 
Thus, $B(b\|a_2\|+\|K_2\|)=\frac{\pi}{8}<1$.
 It is easy to see that assumptions (i)-(v) are
fulfilled. Consequently Theorem \ref{thm4.1} ensures that the equation
\eqref{w} has at least one solution in $L^{1}(\mathbb{R}_{+})$.
\end{example}

\begin{example} \label{examp4.2} \rm 
Consider the  mixed Volterra-Fredholm integral equation
\begin{equation}\label{w1}
\begin{split}
x(t)
&=\frac{1+t^{2}}{\cosh(t)}+\int_0^{t}\frac{\lfloor
t+s^{2}\rfloor}{2}\exp(-t)\sin( x(s))ds\\
&\quad +\int_0^{\infty}\frac{
t\ln(1+sx^{2}(s))}{3(1+t^{2})^{2}(s+1)}ds,\quad t\geq0,
\end{split}
\end{equation}
where the symbol $\lfloor z \rfloor$  means the largest integer 
less than or equal to $z$. 
Let
\begin{gather*}
f(t,x,y)=\frac{1+t^{2}}{\cosh(t)}+x+y,\\
u(t,s,x)=\frac{\lfloor t+s^{2}\rfloor}{2}\exp(-t)\sin( x),\\
v(s,x)=\frac{\ln(1+sx^{2})}{s+1},\quad a_2(t)=\frac{t}{3(1+t^{2})^{2}},\\
k_2(t,s)=\frac{\lfloor t+s^{2}\rfloor}{2}\exp(-t),\quad B=1,\quad b=1.
\end{gather*}
We know that $\ln(1+\alpha^{2})\leq|\alpha|$ for
$\alpha\in\mathbb{R}$. Then
\begin{gather*}
|f(t,x,y)|\leq\frac{1+t^{2}}{\cosh(t)}+B(|x|+|y|),\\
|u(t,s,x)|\leq k_2(t,s)|x|,\quad |v(s,x)|\leq b|x|,\\
\|a_2\|=\int_0^{\infty}|a_2(t)|dt=\int_0^{\infty}\frac{t}{3(1+t^{2})^{2}}dt
=\frac{1}{6},\\ 
\int_{s}^{\infty}|k_2(t,s)|dt
=\int_{s}^{\infty}\frac{\lfloor t+s^{2}\rfloor}{2}\exp(-t)dt
\leq \frac{(s^{2}+s+1)}{2}\exp(-s)\leq\frac{3}{2}\exp(-1),
\end{gather*}
for $s,t\in\mathbb{R}_{+}$ and $x\in\mathbb{R}$. Therefore, from 
Theorem \ref{thm2.2}, we have that $B\|K_2\|\leq\frac{3}{2}\exp(-1)$ and then
$B(b\|a_2\|+\|K_2\|)\leq \frac{1}{6}+ \frac{3}{2}\exp(-1)<1$.
Using Theorem \ref{thm4.1} we deduce that the equation \eqref{w1} has at
least one solution in $L^{1}(\mathbb{R}_{+})$.
\end{example}

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