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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 271, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/271\hfil Solutions of nonlinear integral equations]
{Existence and asymptotic behavior of solutions for some nonlinear integral
equations on an unbounded interval}

\author[B. \.Ilhan, \.I. \"Ozdem\.ir \hfil EJDE-2016/271\hfilneg]
{Bek\.ir \.Ilhan, \.Ismet \"Ozdem\.ir} % Bekir Ilhan, Ismet Ozdem

\address[Bek\.{\i}r \.Ilhan]
{Malatya Science High School,
44110, Malatya, Turkey}
\email{bekirilhan@gmail.com}

\address[\.Ismet \"Ozdem\.{\i}r]
{Faculty of Education, \.{I}n\"{o}n\"{u} University,
 44280, Malatya, Turkey}
\email{ismet.ozdemir@inonu.edu.tr}

\thanks{Submitted  May 30, 2016. Published October 10, 2016.}
\subjclass[2010]{45G10, 47H10, 47H08, 45M99}
\keywords{Nonlinear integral equation; measure of noncompactness;
\hfill\break\indent fixed-point theorem}

\begin{abstract}
 The goal in this paper is to prove an existence theorem for the
 solutions of a class of functional  integral equations which contain
 a number of classical nonlinear integral  equations as special cases.
 Our investigations will be carried out in the  space of continuous and
 bounded functions on an unbounded interval.
 The main tools are some techniques in analysis and the Schauder fixed
 point theorem via measures of noncompactness. Our results extend and improve
 some known results in the recent literature. Three nontrivial examples
 explain the generalizations and applications of our results.
\end{abstract}

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


\section{Introduction}

It is well known that the theory of nonlinear integral equations of various
types appears in many applications that arise in the fields of mathematical
analysis, nonlinear functional analysis, mathematical physics,
and engineering (see \cite{hak, 1, 2}). There has been a significant development in
ordinary and partial fractional differential and integral equations in recent years.

Agarwal and O'Regan \cite{5-18} gave the existence of solutions for the nonlinear
integral equation
\begin{equation} \label{5-8}
x(t)=\int^{\infty}_0k(t, s)f(s, x(s))ds,\quad t\in \mathbb{R}^{+},
\end{equation}
in the space of bounded and continuous functions
$\text{C}_{l}[0,\infty)$ which have limit at infinity.

 Meehan and O'Regan \cite{18-3, 10-1} discussed the existence of solutions
for the nonlinear integral equation
\begin{equation} \label{18-3}
x(t)=h(t)+ \mu\int^{\infty}_0k(t, s)f(s, x(s))ds,\quad
 t\in \mathbb{R}^{+},
\end{equation} in the space $\text{C}_{l}[0,\infty)$ and the existence
of solutions for the nonlinear integral equation
\begin{equation} \label{10-1}
x(t)=h(t)+\int^{\infty}_0k(t, s)[f(x(s))+g(x(s))]ds,\quad
 t\in \mathbb{R}^{+},
\end{equation}
in the space the space $BC(\mathbb{R}^{+},\mathbb{R} )$ of bounded and continuous
functions on $\mathbb{R}^{+}$. Later in \cite{10-11} they established the
 existence of at least one positive solution for the nonlinear integral equation
\begin{equation} \label{10-11}
x(t)=h(t)+ \int^{\infty}_0k(t, s)f(s, x(s))ds,\quad t\in \mathbb{R}^{+},
\end{equation}
in the space $L^{p}(\mathbb{R}^{+})$, in 2001.

In 2004, Bana\'{s} and Poludniak \cite{1-14} investigated the monotonic
solutions for the nonlinear integral equation
\begin{equation} \label{1-14}
x(t)=f(t)+ \int^{\infty}_0u(t, s, x(s))ds,\quad
 t\in \mathbb{R}^{+},
\end{equation}
in the space of Lebesque integrable functions on the halfaxis $\mathbb{R^{+}}$
by using the Darbo fixed point theorem and the measure
of noncompactness (both in strong and weak sense).

Bana\'{s} and Martin \cite{1-25} studied the existence and asymptotic stability
of solutions for the nonlinear integral equation
\begin{equation} \label{1-25}
x(t)=g(t)+ f(t, x(t))\int^{\infty}_0K(t, s)h(s, x(s))ds,\quad
 t\in \mathbb{R}^{+},
\end{equation}
in the Banach space $BC(\mathbb{R}^{+},\mathbb{R} )$, in 2006.

In 2004, Cabellaro and others \cite{7-8}, in 2008, Bana\'{s} and Olszowy \cite{2-4}
and more recently in 2013, Darwish and others \cite{yeni}
studied the existence of solutions for the Urysohn integral equation defined
on an unbounded interval
 \begin{equation} \label{2-4}
x(t)=a(t)+ f(t, x(t))\int^{\infty}_0u(t, s, x(s))ds,\quad
 t\in \mathbb{R}^{+},
\end{equation}
in the space $BC(\mathbb{R}^{+},\mathbb{R} )$.

In those papers, different conditions and various measures of noncompactness
were applied in proving existence theorems.

Olszowy \cite{8-2, 8-1, 8-3} studied \eqref{2-4} in the Fr\`echet space of
real functions being defined and continuous on $\mathbb{R}^{+}$ and has
 given results about monotonocity of the solutions of the integral equation
\eqref{2-4}.

In 2010, Karoui and others \cite{15-8} studied \eqref{2-4} in the space
$L^{p}(\mathbb{R}^{+})$ by means of Schauder's fixed point theorem.
Recently, Khosravi and others \cite{2015} studied the existence of solutions
for nonlinear functional integral
equations of convolution type
 \begin{equation} \label{2015}
x(t)= f(t, x(t))+\int^{\infty}_0k(t-s)(Qx)(s)ds,\quad t\in \mathbb{R}^{+},
\end{equation}
in the space $L^{p}(\mathbb{R}^{+})$, in 2015.

Motivated by recent researches in this field, we study the more general
nonlinear integral equation
 \begin{equation}\label{anadenk}
x(t)=(T_1x)(t)+ (T_2x)(t) \int^{\infty}_0u(t, s, x(s))ds,\quad
 t\in \mathbb{R}^{+},
\end{equation}
where the functions $u(t, s, x)$ and the operators $T_{i}x$, $(i=1,2) $ appearing
 in \eqref{anadenk} are given, while $x=x(t)$ is an unknown
 function. Using the technique of a suitable measure of noncompactness,
 we prove an existence theorem for \eqref{anadenk}. We give three
nontrivial examples that explain the generalizations and applications
of our results.
So our work improves directly results obtained in \cite{1-14,1-25,7-8,yeni,15-8}
and completes some results mentioned before.
 To the best of our knowledge, \eqref{anadenk} is more general than
those investigated up to now and includes \eqref{18-3}-\eqref{2015}
 as special cases.

\section{Auxiliary facts and notation}

In this section, we give a collection of auxiliary facts which will be
needed further on. Assume that $(E, \|\cdot\|)$ is a real Banach space with zero
element $\theta$.
Let $B(x, r)$ denote the closed ball centered at x and with radius $r$.
The symbol $B_{r}$ stands for the ball $B(\theta, r)$. If $X$ is a
subset of $E$, then $\overline{X}$ and $\operatorname{conv} X$ denote the closure
 and convex closure of $X$, respectively. With the symbols $\lambda X$ and $X+ Y$,
we denote
the standard algebraic operations on sets. Moreover, we denote by
$\mathfrak{M}_{E}$ the family of all nonempty and bounded subsets of $E$ and
$\mathfrak{N}_{E}$
its subfamily consisting of all relatively compact subsets.
The definition of the concept of a measure of noncompactness presented
below comes from \cite{w1}.

\begin{definition}\label{mnc} \rm
A function $\mu:\mathfrak{M}_{E}\to \mathbb {R^{+}}=[0,\infty)$
is said to be a measure of noncompactness in $E$ if it satisfies following
conditions:
\begin{itemize}
 \item[(1)] The family $ \ker \mu= \{X\in\mathfrak{M}_{E}: \mu(X)=0\}$ is nonempty
 and $\ker \mu\subset\mathfrak{N}_{E}$.
 \item[(2)] $ X\subset Y\Rightarrow \mu(X)\leq \mu(Y)$.
\item[(3)] $\mu (\overline{X})=\mu( \operatorname{conv} X)= \mu(X)$.
\item[(4)] $\mu(\lambda X+ (1-\lambda) Y)\leq \lambda\mu(X)+ (1-\lambda)\mu(Y)$,
for $\lambda \in [0,1]$.
\item[(5)] If $\{X_{n}\}$ is a sequence of nonempty, bounded, closed subsets
of the set $E$ such that $X_{n+1}\subset X_{n}, (n=1,2,\dots ) $ and
 $\lim_{n\to \infty} \mu (X_{n})=0$, then the set
$X_{\infty}=\cap^{\infty}_{n=1} X_{n}$ is nonempty.
\end{itemize}
\end{definition}

Notice that the intersection set $X_{\infty}$ from $5$ belongs to $ \ker \mu $.
In fact, from the inequality $\mu(X_{\infty})\leq \mu(X_{n})$
for any $n=1, 2, \dots $ we have that $\mu(X_{\infty})=0$.
This property of the set $X_{\infty}$ will be crucial later.

In the sequel, we will work in the Banach space $BC(\mathbb{R}^{+}, \mathbb{R})$
which is equipped with the standard norm
$\|x\|=\sup\{|x(t)|: t \in \mathbb{R}^{+}\}$.

 We will use a measure of noncompactness in the space
$BC(\mathbb{R}^{+}, \mathbb{R})$. To define this measure let us fix
a nonempty and bounded subset $X$ of $BC(\mathbb{R}^{+}, \mathbb{R})$.
For $x\in X$, $\varepsilon \geq 0$ and $L>0$, we denoted by
the modulus of continuity for function $x$, as
$$
w^{L}(x, \varepsilon)=\sup\{|x(s)-x(t)| : t,s \in [0, L] \text{ and }
 |t-s|\leq\varepsilon\}.
$$
Further let us put
\begin{gather}
w^{L}(X, \varepsilon)=\sup\{w^{L}(x, \varepsilon): x\in X\}, \nonumber\\
w^{L}_0(X)=\lim_{\varepsilon\to 0}w^{L}(X, \varepsilon), \nonumber\\
\label{omega}
w_0(X)=\lim_{L\to \infty}w^{L}_0(X).
\end{gather}
Next we define
\begin{gather}
\beta^{L}(x)=\sup\{|x(t)|:t\geq L\}, \nonumber\\
\beta^{L}(X)= \sup\{\beta^{L}(x):x\in X \}, \nonumber\\
\label{beta}
\beta(X)= \lim_{L\to\infty}\beta^{L}(X).
\end{gather}
Finally, let us define the function $\mu$ as
$$
\mu(X)=w_0(X)+\beta(X).
$$
It is shown in \cite{w1} that the function $\mu$ is a measure of noncompactness
in the space $BC(\mathbb{R}^{+}, \mathbb{R})$. Moreover, the kernel
$\ker \mu$ of this measure contains nonempty and bounded sets $X$
such that functions from $X$ are locally equicontinuous on $\mathbb{R}^{+}$
and tend to zero at infinity uniformly with respect to the set $X$; i.e.,
 for each $\varepsilon \geq0$ there exists $L > 0$
with the property that
\[
|x(t)| \leq\varepsilon; \quad \text{for all $x\in X$ and
 $t$ with $t \geq L$}.
\]
This property of $\ker \mu$ will be important in our further study.

\section{Main Result}

We shall study the existence of solutions to \eqref{anadenk}
 assuming that following conditions are satisfied:
 \begin{itemize}
 \item[(i)] The operators
$T_{i}:BC(\mathbb{R}^{+},\mathbb{R} ) \to BC(\mathbb{R}^{+},\mathbb{R} )$
are continuous and there exist continuous nondecreasing functions
$d_{i}:\mathbb{R}^{+}\to \mathbb{R}^{+}$ such that
 $$
|(T_{i}x)(t)|\leq d_{i}(\|x\|)
$$
for all $x\in BC(\mathbb{R}^{+},\mathbb{R} )$ and
$t\in \mathbb{R}^{+}$, $(i=1, 2)$.

\item[(ii)] $u:\mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R} \to \mathbb{R}$
is a continuous function and there exist a
 continuous function $g:\mathbb{R}^{+}\times \mathbb{R}^{+}\to \mathbb{R}^{+}$
and a continuous nondecreasing function
 $h:\mathbb{R}^{+}\to \mathbb{R}^{+}$ such that
\[
 |u(t, s, x)|\leq g(t, s)h(|x|)
\]
for all $t, s\in \mathbb{R}^{+}$ and $x \in \mathbb{R}$.

\item[(iii)] For every $t\geq 0$ the function $s\to g(t, s)$ is
integrable on $\mathbb{R}^{+}$ and the function
$t\to \int_0^{\infty}g(t, s)ds $ is bounded on $\mathbb{R}^{+}$.

\item[(iv)] The inequality $d_1(r)+ d_2(r)h(r)G\leq r$ has a positive
solution $r_0$, where $ G=\sup\{\int_0^{\infty}g(t, s)ds:
t\geq 0\}<\infty$.

\item[(v)]There exist the nonnegative constants $k_{i}$ and $m_{i}$ for
$r_0$ such that the inequalities
 \begin{gather*}
\omega_0(T_{i}X)\leq k_{i}\omega_0(X),\\
\beta(T_{i}X) \leq m_{i}\beta(X)
\end{gather*} hold for all nonempty and bounded subset $X$ of the ball
$B_{r_0}$, $(i=1, 2 )$, where $w_0$ and $\beta$ are defined by
\eqref{omega} and \eqref{beta}.

 \item[(vi)] $\max\big\{k_1+k_2Gh(r_0), m_1+ m_2G h(r_0)\big\}<1$.

\item[(vii)] $\lim_{L\to\infty}\big\{\sup\{ \int_{L}^{\infty}g(t, s)ds:
 t\in [0, L]\}\big\}=0$.

\end{itemize}
 Now we can formulate an existence result concerning the functional
integral equation \eqref{anadenk}.

\begin{theorem}\label{anateorem}
Under assumptions {\rm (i)--(vii)}, there exists at least one solution
$x=x(t)$ of \eqref{anadenk} in the space $BC(\mathbb{R}^{+},\mathbb{R})$
such that $x(t)\to 0$ as $t\to \infty$.
 \end{theorem}

\begin{proof}
 We define an operator $F$ on $BC(\mathbb{R}^{+}, \mathbb{R})$ as follows
 \[ % \label{ana denklem}
(Fx)(t)=(T_1x)(t)+ (T_2x)(t) \int^{\infty}_0u(t, s, x(s))ds.
\]
 Notice that in view of assumptions (i) and (ii) the function $t\to (Fx)(t)$
is well defined on the interval
$\mathbb{R}^{+}$. At first we show that the function $Fx$ is continuous on
$\mathbb{R}^{+}$. To do this fix arbitrarily $L>0$ and
$\varepsilon\geq 0$. Take arbitrary numbers $t, t_0\in [0, L]$ with
$|t-t_0|\leq \varepsilon$. Then in view of assumptions we obtain
\begin{equation} \label{sreklilik1}
\begin{aligned}
&|(Fx)(t)-(Fx)(t_0)| \\
&=\Big|(T_1x)(t)+ (T_2x)(t) \int^{\infty}_0u(t, s, x(s))ds -(T_1x)(t_0) \\
&\quad -(T_2x)(t_0) \int^{\infty}_0u(t_0, s, x(s))ds\Big| \\
 &\leq |(T_1x)(t)-(T_1x)(t_0)|+ |(T_2x)(t)-(T_2x)(t_0)|
\Big|\int^{\infty}_0u(t, s, x(s))ds \Big| \\
 &\quad + |(T_2x)(t_0)| \Big|\int^{\infty}_0
 [u(t, s, x(s))-u(t_0, s, x(s))]ds\Big| \\
 &\leq \omega^{L}(T_1x, \varepsilon)
 +\omega^{L}(T_2x, \varepsilon)\int^{\infty}_0|u(t, s, x(s))|ds \\
&\quad + d_2(\|x\|)\int^{\infty}_0|u(t, s, x(s))-u(t_0, s, x(s))|ds \\
 &\leq \omega^{L}(T_1x, \varepsilon)+\omega^{L}(T_2x, \varepsilon)
 \int^{\infty}_0g(t, s)h(|x(s)|)ds \\
 &\quad +d_2(\|x\|)\Big\{\int^{L}_0|u(t, s, x(s))-u(t_0, s, x(s))|ds \\
&\quad +\int^{\infty}_{L}|u(t, s,
 x(s))-u(t_0, s, x(s))|ds\Big\} \\
 &\leq \omega^{L}(T_1x, \varepsilon)+\omega^{L}(T_2x, \varepsilon)Gh(\|x\|)
+ d_2(\|x\|)L\omega^{L}_{\|x\|}(u, \varepsilon)\\
&\quad +2d_2(\|x\|) h(\|x\|)\sup\Big\{\int^{\infty}_{L}g(t,s)ds:t\in[0, L]\Big\},
\end{aligned}
\end{equation}
 where
 $$
\omega^{L}(T_{i}x, \varepsilon)=\sup\{|(T_{i}x)(s)-(T_{i}x)(t)| :
 t,s \in [0, L]\textrm{ and }|t-s|\leq\varepsilon\}
$$
for $i=1, 2$ and
 \begin{align*}
\omega^{L}_{\|x\|}(u, \varepsilon)
=\sup\big\{&|u(t, s, y)-u(t_0, s, y)| :
t, t_0, s \in [0, L], \\
& y\in[-\|x\|, \|x\|]\text{ and }|t-t_0|\leq\varepsilon\big\}.
\end{align*}
By the uniform continuity of the functions $T_{i}x$ on the set $[0,L]$ and $u$ on
the set $[0, L]\times[0, L]\times [-\|x\|, \|x\|] $, we deduce that
$\omega^{L}(T_{i}x, \varepsilon)\to 0$ and $\omega^{L}_{\|x\|}(u, \varepsilon)\to 0$
as $\varepsilon\to 0$.
 Further observe that by assumption (vii) we can choose a number
 $L$ so big that the last term of the estimate \eqref{sreklilik1}
is sufficiently small. Thus we infer that $Fx$ is continuous on the interval
$[0, L]$ for any $L>0$ big enough. This implies that $Fx$ is
continuous on the whole interval $\mathbb{R}^{+}$.
 Next we show that $Fx$ is bounded on $\mathbb{R}^{+}$. By our assumptions,
for arbitrarily fixed $t\in\mathbb{R}^{+}$ we have:
\begin{equation}\label{sahur3}
\begin{aligned}
|(Fx)(t)|&=\Big|(T_1x)(t)+(T_2x)(t) \int^{\infty}_0u
(t, s, x(s))ds\Big| \\
 &\leq |(T_1x)(t)|+|(T_2x)(t)| \int^{\infty}_0\left|u(t, s, x(s))\right|ds.
\end{aligned}
 \end{equation}
By the assumptions and from estimate \eqref{sahur3} we obtain
 \begin{align*}
 |(Fx)(t)|
&\leq d_1(\|x\|)+ d_2(\|x\|)\int^{\infty}_0g(t, s)h(|x(s)|)ds \\
&\leq d_1(\|x\|)+ d_2(\|x\|)h(\|x\|)G.
 \end{align*}
Hence, we obtain
 \begin{equation}\label{yuvar1}
\|Fx\|\leq d_1(\|x\|)+ d_2(\|x\|)h(\|x\|)G
\end{equation}
which implies that the function $Fx$ is bounded on $\mathbb{R}^{+}$.

Linking this fact with the continuity of the function $Fx$ on
$\mathbb{R}^{+}$ we conclude that the operator $F$ maps the
space $BC(\mathbb{R}^{+}, \mathbb{R})$ into itself. Furthermore, by estimate
\eqref{yuvar1} and assumption (iv) we infer that $F$ is a self mapping
of the ball $B_{r_0}$, where $r_0$ is mentioned in the assumption
(iv).

Next we take a nonempty subset $X$ of the ball
$B_{r_0}$. Fix $\varepsilon\geq0$ and $L>0$ and take an arbitrary function
$x\in X$. Then, using estimate \eqref{sreklilik1}, it follows that
 \begin{equation}\label{pazar}
\begin{aligned}
 \omega^{L}(Fx, \varepsilon)
&\leq \omega^{L}(T_1x, \varepsilon)+\omega^{L}(T_2x, \varepsilon)Gh(r_0)
 + d_2(r_0)L\omega^{L}_{r_0}(u,
 \varepsilon) \\
&\quad +2 d_2(r_0) h(r_0)\sup
\Big\{\int^{\infty}_{L}g(t,s)ds:t\in[0, L]\Big\}.
\end{aligned}
 \end{equation}
Hence by \eqref{pazar} we obtain
 \begin{equation}\label{gece}
\begin{aligned}
 \omega^{L}(FX, \varepsilon)
&\leq \omega^{L}(T_1X, \varepsilon)+\omega^{L}(T_2X, \varepsilon)Gh(r_0)
 + d_2(r_0)L\omega^{L}_{r_0}(u, \varepsilon) \\
& \quad +2 d_2(r_0) h(r_0)\sup\Big\{\int^{\infty}_{L}g(t,s)ds:t\in[0, L]\Big\}.
\end{aligned}
 \end{equation}
Now taking into account the properties of the component involved in the
estimate \eqref{gece}, we have
 \begin{equation}\label{sahur}
\begin{aligned}
 \omega^{L}_0(FX)
&\leq \omega^{L}_0(T_1X)+\omega^{L}_0(T_2X)Gh(r_0) \\
& \quad +2 d_2(r_0)
 h(r_0)\sup\Big\{\int^{\infty}_{L}g(t,s)ds:t\in[0, L]\Big\}.
\end{aligned}
 \end{equation}
Combining \eqref{sahur} with assumption (vii), we derive the estimate
 \begin{equation}\label{sahur1}
 \omega_0(FX)\leq [k_1+k_2Gh(r_0)]\omega_0(X).
 \end{equation}
Further taking $x\in X$ and choosing arbitrarily $L>0$, in view of the
estimate \eqref{sahur3} we obtain that
 \begin{equation}\label{bay}
\begin{aligned}
&\sup\{|(Fx)(t)|:t\geq L\}\\
&\leq \sup\{|(T_1x)(t)|:t\geq L\} \\
&\quad +\sup\{|(T_2x)(t)|:t\geq
 L\} h(r_0)\sup\Big\{\int_0^{\infty}g(t, s)ds:t\geq L\Big\}.
\end{aligned}
 \end{equation}
Hence by \eqref{bay} we obtain
 \begin{equation}\label{sahur4}
\begin{aligned}
 \beta(FX)&\leq \beta(T_1X)+ \beta(T_2X)h(r_0)
\sup\Big\{\int_0^{\infty}g(t, s)ds:t\geq L\Big\} \\
&\leq m_1\beta(X)+ m_2\beta(X)h(r_0)G \\
&\leq [ m_1+ m_2h(r_0)G]\beta(X).
\end{aligned}
 \end{equation}
Now, linking \eqref{sahur1} and \eqref{sahur4} we derive that
 \begin{equation}\label{darbo}
 \mu(FX)\leq k\mu(X),
 \end{equation}
where $k=\max\{k_1+k_2Gh(r_0),\,m_1+ m_2G h(r_0)\}$.

 Next, we consider the sequence of sets $(B_{r_0}^{n})$, where
$B^{1}_{r_0}=\operatorname{conv}F(B_{r_0})$,
 $B^{2}_{r_0}=\operatorname{conv}F(B_{r_0}^{1})$ and so on.
Observe that all sets of this sequence are nonempty, bounded,
closed and convex. Moreover, $B^{n+1}_{r_0} \subset B^{n}_{r_0}$ for
$n = 1, 2,\ldots$ Further, keeping in mind \eqref{darbo}
we obtain
\[
 \mu(B_{r_0}^{n})\leq k^{n}\mu(B_{r_0}).
\]
 Obviously in view of assumption (vi) we have that $k<1$.
Hence, from condition $5$ of Definition \ref{mnc}
we infer that the set $Y=\cap_{n=1}^{\infty}B_{r_0}^{n}$ is nonempty,
bounded, closed and convex. In fact, since
 $\mu(Y)\leq\mu(B_{r_0}^{n})$ for any n, we deduce that $\mu(Y)=0$ and
thus $Y\in \ker \mu$. It should be also noted that the operator $F$
maps the set $Y$ into itself. Now we show that $F$ is continuous on the set $Y$.
 To do this fix $\varepsilon\geq0$ and take functions $x, y\in Y$ such that
$\|x-y\|\leq\varepsilon$. Taking into account the fact that
$Y\in \ker \mu $
 and the description of sets belonging to $\ker \,\mu$ we can find a number
$L>0$ such that for each $z\in Y$ and $t\geq L$ the inequality
 $|z(t)|\leq \varepsilon$ is satisfied. Since $F:Y\to Y$, we have that
$Fx, Fy\in Y$. Thus, for $t\geq L$ we obtain that
\[
 |(Fx)(t)-(Fy)(t)|\leq |(Fx)(t)|+ |(Fy)(t)|\leq 2\varepsilon.
\]
 On the other hand, for $t\in [0, L]$ we obtain
 \begin{equation}\label{sr}
\begin{aligned}
&|(Fx)(t)-(Fy)(t)| \\
&=\Big|(T_1x)(t)+(T_2x)(t) \int^{\infty}_0u(t, s, x(s))ds-(T_1y)(t) \\
&\quad -(T_2y)(t) \int^{\infty}_0u(t, s, y(s))ds\Big| \\
&\leq |(T_1x)(t)-(T_1y)(t)|+|(T_2x)(t)-(T_2y)(t)|
\int^{\infty}_0 |u(t, s, x(s))|ds \\
&\quad +\left|(T_2y)(t)\right|\int^{\infty}_0\left |u(t, s, x(s))-u(t, s, y(s))
 \right|ds \\
&\leq \varepsilon+\varepsilon G h(r_0)+d_2(r_0)
 \Big\{\int^{L}_0\left|u(t, s, x(s))-u(t, s, y(s))\right|ds \\
&\quad +\int^{\infty}_{L}\left|u(t, s, x(s))-u(t, s, y(s))\right|ds\Big\} \\
&\leq \varepsilon+\varepsilon G h(r_0)+d_2(r_0)L\bar{\omega}_{r_0}^{L}
 (u, \varepsilon) \\
& \quad + 2d_2(r_0)h(r_0)\sup\Big\{\int^{\infty}_{L}g(t,s)ds:t\in[0, L]\Big\},
\end{aligned}
\end{equation}
 where
 $$
\bar{\omega}_{r_0}^{L}(u, \varepsilon)
=\sup\{|u(t, s, x)-u(t, s, y)|: t, s\in[0, L];\, x, y \in[-r_0, r_0] \textrm{ and }
 |x-y|\leq\varepsilon\}.
$$
 Observe that $\bar{\omega}_{r_0}^{L}(u, \varepsilon)\to 0$ as $\varepsilon\to 0$.
Moreover we can choose $L$ in such a way
 that the last term in estimate \eqref{sr} is small enough.
Taking into account the above facts we conclude that the operator $F$
 is continuous on the set $Y$.
 Finally, linking all above established properties of the set $Y$ and the
operator $F:Y\to Y$ and using the Schauder fixed
 point principle we infer that the operator $F$ has at least one fixed point
$x$ in the set $Y$. Moreover, keeping in mind that
 $Y\in\ker \mu$
 we obtain that $x(t)\to0$ as $t\to \infty$.
\end{proof}

\section{Examples}

 \begin{example} \label{examp4.1} \rm
 Let us consider the integral equation
 \begin{equation}\label{rnek}
 x(t)= \frac{t\sin x(t)}{3t+9}+ \frac{t^{2}x^{2}(t)}{3t^{2}+2}
\int_0^{\infty}\frac{t^{2}\exp( -t-s)\sqrt{|x(s)|}}{t^{2}+1}ds,
 \end{equation}
where $t\in \mathbb{R}^{+}$.

 If we put $(T_1x)(t)= t\sin x(t)/(3t+9)$,
 $(T_2x)(t)=t^{2}x^{2}(t)/(3t^{2}+2)$ and
 $u(t, s, x)=t^{2}\exp( -t-s)\sqrt{|x|}/(t^{2}+1)$, then  \eqref{rnek}
 is a special  case of  \eqref{anadenk}.
 It is easily verified that the assumptions of Theorem \ref{anateorem} are satisfied.
 Indeed, $T_1$ and $T_2$ are continuous operators on the space
 $BC(\mathbb{R}^{+}, \mathbb{R})$. Further for all
$t\in \mathbb{R}^{+}$ and $x\in BC(\mathbb{R}^{+}, \mathbb{R})$, we have
 \begin{gather*}
 |(T_1x)(t)|\leq \frac{1}{3}, \\
 |(T_2x)(t)|\leq \frac{ x^{2}(t)}{3}.
\end{gather*}
 Hence assumption (i) is satisfied with
 $d_1(x)=1/3$ and $d_2(x)= x^{2}/3$.
 Now note that the function $u$ is continuous on the set
$\mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R} $. Moreover
 we obtain
\[
 |u(t, s, x)|=\Big|\frac{t^{2}\exp( -t-s)\sqrt{|x|}}{t^{2}+1}\Big|
\leq\exp( -t-s)\sqrt{|x|}
\]
 and if we choose $g(t, s)=\exp( -t-s)$ and $h(x)=\sqrt{x}$, we see that
assumption (ii) is satisfied. To check that the assumption (iii) is satisfied
 let us observe that $s\to \exp( -t-s)$ is integrable on $\mathbb{R}^{+}$
and $t\to \int_0^{\infty}\exp( -t-s)ds $ is
 bounded on $\mathbb{R}^{+}$. Thus it is easily seen that
\[
G=\sup\Big\{\int_0^{\infty}\exp( -t-s)ds: t\in \mathbb{R^{+}}\Big\}
=\sup\{\exp( -t): t\in  \mathbb{R^{+}}\}=1.
\]
 Now notice that  the inequality in assumption (iv) has the form
 \begin{equation}\label{anaeitsizlik}
 \frac{1}{3}+ \frac{r^{2}\sqrt{r}}{3}\leq r.
 \end{equation}
 It can be easily verified that if $0.3590\leq r_0 < 1$.  Then
 $r_0$ is the solution of \eqref{anaeitsizlik}.
 Also for $\varepsilon\geq0$, $L>0$, $\|x\|\leq r_0$ and
$t, s\in[0, L] $ such that $|t-s|\leq\varepsilon$, we have
 \begin{equation}\label{ram}
\begin{aligned}
 \left|(T_1x)(t)-(T_1x)(s)\right|
 &= \left|\frac{t\sin x(t)}{3t+9}-\frac{s\sin x(s)}{3s+9}\right| \\
 &\leq \frac{t(s+3)|\sin x(t)-\sin x(s)|+3|\sin x(s)||t-s|}{3(t+3)(s+3)} \\
 &\leq \frac{t}{3(t+3)}| x(t)- x(s)|+\frac{\varepsilon |\sin x(s)|}{(t+3)(s+3)} \\
 &\leq \frac{1}{3}| x(t)- x(s)|+\frac{\varepsilon}{9}.
\end{aligned}
 \end{equation}
 Further, it can be seen that
 \begin{equation}\label{mar}
\begin{aligned}
|(T_2x)(t)-(T_2x)(s)|
&=\Big|\frac{t^{2}x^{2}(t)}{3t^{2}+2}-\frac{s^{2}x^{2}(s)}{3s^{2}+2}\Big| \\
&\leq \frac{2r_0t^{2}(3s^{2}+2)| x(t)-x(s)|+2r_0^{2}(t+s)|t-s|}{(3 t^{2}+2)
 (3s^{2}+2)} \\
&\leq \frac{2r_0t^{2}}{3t^{2}+2}| x(t)- x(s)|
 +\frac{2 r_0^{2}\varepsilon(t+s)}{(3 t^{2}+2)(3s^{2}+2)} \\
&\leq \frac{2r_0}{3}| x(t)- x(s)|+ r_0^{2}\varepsilon L.
\end{aligned}
 \end{equation}
From estimates \eqref{ram} and \eqref{mar} in view of the \eqref{mnc} we have
 \begin{gather*}
 \omega_0(T_1X)\leq\frac{1}{3}\omega_0(X), \\
 \omega_0(T_2X)\leq \frac{2r_0}{3}\omega_0(X),
 \end{gather*}
respectively. This implies that assumption (v) is satisfied with
$k_1=1/3$ and $k_2=2r_0/3$.
 Finally we obtain
 \begin{equation}\label{son}
\begin{aligned}
\sup\{ |(T_1x)(t)|: t\geq L\}
&= \sup\Big\{\Big|\frac{t\sin
 x(t)}{3t+9}\Big|:t\geq L\Big\} \\
 &= \frac{1}{3}\sup\{|\sin x(t)|:t\geq L\}  \\
 &\leq \frac{1}{3}\sup\{| x(t)|:t\geq L\}.
\end{aligned}
 \end{equation}
Thus from estimate \eqref{son} we have
\[
 \beta(T_1X) \leq \frac{1}{3}\beta(X).
\]
 Moreover, we derive that
 \begin{equation}\label{art5}
\begin{aligned}
\sup\{|(T_2x)(t)|: t\geq
 L\}
& =\sup\Big\{\Big|\frac{t^{2}x^{2}(t)}{3t^{2}+2}\Big|:t\geq L\Big\} \\
& = \frac{1}{3}\sup\{ |x(t)|^{2}:t\geq L\} \\
&\leq  \frac{r_0}{3}\sup\{| x(t)|:t\geq L\}.
\end{aligned}
\end{equation}
Thus from estimate \eqref{art5} we have
\[ % \label{art1}
 \beta(T_2X) \leq \frac{r_0}{3}\beta(X).
\]
 Therefore $m_1=1/3$ and $m_2=r_0/3$.

 Taking into account the above estimates we have
 $$
\max\{k_1+k_2Gh(r_0),\,m_1+ m_2G h(r_0)\}
= \max\Big\{\frac{1}{3}+\frac{2r_0}{3}\sqrt{r_0},\,
 \frac{1}{3}+\frac{r_0}{3}\sqrt{r_0}\Big\}<1.
$$
Hence  assumption (vi) is satisfied.
Finally, we have the following equality for the function $g(t,s)$
appearing in assumption (vii):
\begin{align*}
\sup\Big\{\int_{L}^{\infty}\exp( -t-s)ds:t\in[0, L]\Big\}
=\sup\Big\{\frac{1}{\exp(t+ L)}:t\in[0, L]\Big\}
=\frac{1}{\exp( L)}.
\end{align*}
Since $1/\exp( L)\to 0$ as $L\to \infty$, assumption (vii) is satisfied.

 Thus we showed that all assumptions
 of Theorem \ref{anateorem} are fulfilled. This yields that \eqref{rnek}
 has at least one solution $x=x(t)$ in the space
 $BC(\mathbb{R}^{+}, \mathbb{R})$ vanishing at infinity.
 \end{example}

 \begin{remark} \label{rmk4.1} \rm
Note that none of the existence theorems in
\cite{18-3, 10-1, 10-11, 1-14, 5-18, 1-25, 7-8, 2-4, yeni, 8-2, 8-1, 8-3,15-8} are
 applicable to \eqref{rnek}, since the integral equation \eqref{rnek}
 can not be derived from any of the integral equations handled in
 mentioned papers.
\end{remark}

 \begin{example} \label{examp4.2} \rm
Let us consider the following integral equation:
\begin{equation}\label{rnek1}
 x(t)=\frac{t}{10t+1}+\frac{ x^{2}(t)}{t+1}\int_0^{\infty}
\frac{\exp(-t)(e-1)x(s)}{(t+1)(s+e)(s+1)}ds,
 \end{equation}
where $t\in \mathbb{R}^{+}$. Observe that this equation has the form
of \eqref{anadenk} if we put
$(T_1x)(t)=t/(10t+1)$, $(T_2x)(t)=x^{2}(t)/(t+1)$ and
$u(t, s, x)=\exp(-t)(e-1)x/((t+1)(s+e)(s+1))$. It is clear that
$T_1$ and $T_2$ are continuous operators on the space
$BC(\mathbb{R}^{+},\mathbb{R} )$. Additionally for all $t\in \mathbb{R}^{+}$ and
$x\in BC(\mathbb{R}^{+}, \mathbb{R})$, we have
 \begin{gather*}
|(T_1x)(t)|\leq \Big|\frac{t}{10t+1}\Big|,\\
|(T_2x)(t)|\leq \Big|\frac{ x^{2}(t)}{t+1}\Big|.
\end{gather*}
Hence assumption (i) is satisfied with $d_1(x)=1/10$ and $d_2(x)= x^{2}$,
respectively.
 The function $u(t, s, x)$ is continuous on the set
$\mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R}$. Further we obtain
 \[ %\label{es15}
 |u(t, s, x)|=\Big|\frac{\exp(-t)(e-1)x}{(t+1)(s+e)(s+1)}\Big|
=\frac{\exp(-t)(e-1) |x|}{(t+1)(s+e)(s+1)}
 \]
for all $t, s \in \mathbb{R}^{+}$ and $x\in \mathbb{R}$.
 Thus the functions appearing in assumption (ii) have the form
$g(t, s)=\exp(-t)(e-1)/((t+1)(s+e)(s+1))$ and $h(x)=x$. Obviously
 $s\to \exp(-t)(e-1)/((t+1)(s+e)(s+1))$ is integrable on $\mathbb{R}^{+}$ and
$t\to \int_0^{\infty}\exp(-t)(e-1)ds/((t+1)(s+e)(s+1))$ is bounded on
$\mathbb{R}^{+}$.
Moreover, we have
\[
 G=\sup\Big\{\int_0^{\infty}\frac{\exp(-t)(e-1)}{(t+1)(s+e)(s+1)}ds: t\in
 \mathbb{R^{+}}\Big\}
=\sup\Big\{\frac{\exp(-t)}{t+1}: t\in \mathbb{R^{+}}\Big\}=1.
\]
Now note that  the inequality in assumption (iv) has the form:
 \begin{equation}\label{anaeitsizlik1}
 \frac{1}{10}+ r^{3}\leq r.
 \end{equation}
 It can be easily verified that if $0.1010\leq r_0 < 1/\sqrt{2}$, then
 $r_0$ is the solution of \eqref{anaeitsizlik1}.
 Also for $\varepsilon\geq 0$, $L>0$, $\|x\|\leq r_0$ and
$t, s\in[0, L]$ such that $|t-s|\leq\varepsilon$, we have that
 \begin{equation}\label{ak}
\begin{aligned}
|(T_1x)(t)-(T_1x)(s)|
&=\Big|\frac{t}{10t+1}-\frac{s}{10s+1}\Big| \\
&= \frac{|t-s|}{(10t+1)(10s+1)} \\
&\leq \frac{\varepsilon}{(10t+1)(10s+1)}
\leq \varepsilon.
\end{aligned}
 \end{equation}
Further, it can be seen that
 \begin{equation}\label{map}
\begin{aligned}
|(T_2x)(t)-(T_2x)(s)|
&=\Big|\frac{ x^{2}(t)}{t+1}-\frac{ x^{2}(s)}{s+1}\Big| \\
&\leq \frac{(t+1)|x^{2}(t)-x^{2}(s)|+|x^{2}(t)||s-t|}{(t+1)(s+1)} \\
&\leq \frac{2r_0|x(t)-x(s)|}{s+1}+\frac{r_0^{2}\varepsilon}{(t+1)(s+1)} \\
&\leq 2r_0|x(t)-x(s)|+r_0^{2}\varepsilon.
\end{aligned}
 \end{equation} From estimates \eqref{ak} and \eqref{map} in view of
\eqref{mnc}, we have
 \begin{gather*}
 \omega_0(T_1X)=0, \\
 \omega_0(T_2X)\leq 2r_0 \omega_0(X).
\end{gather*}
 This implies that assumption (v) is satisfied with $k_1=0 $ and $k_2=2r_0$.
 Further, we have
 \begin{equation}\label{son10}
\sup\{|(T_1x)(t)|: t\geq L\}=\sup\Big\{\Big|\frac{t}{10t+1}\Big|:t\geq
 L\Big\}=\frac{1}{10}.
 \end{equation}
 Moreover, we obtain
 \begin{equation}\label{art3}
\begin{aligned}
\sup\{|(T_2x)(t)|: t\geq L\}
&= \sup\Big\{\Big|\frac{ x^{2}(t)}{t+1}\Big|:t\geq L\Big\} \\
&=\frac{1}{L+1}\sup\{| x(t)|^{2}:t\geq
 L\} \\
 &\leq \frac{r_0}{L+1}\sup\{| x(t)|:t\geq L\}.
\end{aligned}
 \end{equation}
Thus from estimates \eqref{son10} and \eqref{art3} in view of \eqref{beta}
 we have that $\beta(T_1X)=1/10$ and $\beta(T_2X)=0$.
 Therefore we obtain $m_1=1/10$ and $m_2=0$.
Keeping in mind the above obtained constants $k_1,k_2, m_1 , m_2$ we obtain
$$
\max\Big\{2r_0^{2},\,\frac{1}{10}\Big\}<1
$$
and thus assumption (vi) is satisfied. Finally, we obtain
\begin{equation}\label{gts}
\begin{aligned}
&\sup\Big\{\int_{L}^{\infty}\frac{\exp(-t)(e-1)}{(t+1)
(s+e)(s+1)}ds:t\in[0, L]\Big\} \\
&= \sup\Big\{\frac{\exp(-t)}{(t+1)}\ln\Big(\frac{L+1}{L+e}\Big) :t\in[0, L]\Big\} \\
&= \frac{1}{\exp( L)(L+1)}\ln\Big(\frac{L+1}{L+e}\Big).
\end{aligned}
\end{equation}
Hence, we infer that  assumption (vii) is satisfied.

 Finally we conclude that the assumptions of Theorem
 \ref{anateorem} are satisfied.
This implies that the considered integral equation has a solution $x=x(t)$
belonging to the set $Y$. Moreover,  $x(t)\to 0$ as $t\to 0$.
 \end{example}


 \begin{remark} \label{rmk4.2} \rm
 Now we compare the result in Theorem \ref{anateorem} with the results in
\cite{2-4, 8-3,yeni}.

Notice that in \cite{2-4} it was assumed that $\lim_{t \to \infty}a(t)=0$.
Thus, the result given in  \cite{2-4} is inapplicable to\eqref{rnek1}.

 Now if we consider the assumption (iii) of  \cite[Theorem 8]{yeni}, we have
 $f(t, x)=x^{2}/(t+1)$ and it does not satisfy the Lipschitz condition
with respect to second variable for all $x\in\mathbb{R^{+}}$ and
 $t\in\mathbb{R^{+}}$. Hence, the existence theorem given in \cite{yeni}
is inapplicable to  \eqref{rnek1}.

 Further, in assumption (ii) of
\cite[Theorem 3.1]{8-3} states that $f(t, x)$ is nondecreasing with respect
to  both nonnegative  variables $t$ and  $x$.
Since we have $t\to f(t, x)=x^{2}/(t+1)$ is a
decreasing function in $t$, \cite[Theorem 3.1]{8-3} is not inapplicable to
\eqref{rnek1}.
 \end{remark}

 \begin{example} \label{examp4.3} \rm
 Consider the integral equation
 \begin{equation} \label{rnek3}
 x(t)=\frac{t}{\exp(t)}+\frac{\sqrt{x^{2}(t)+1}}{t+1}
\int_0^{\infty}\frac{\sqrt{1+|x(s)|}}{\exp(t+s+1)}ds,
 \end{equation}
where $t\in \mathbb{R}^{+}$. This equation is a special case of
\eqref{anadenk} if we put
$(T_1x)(t)=t/\exp(t)$, $(T_2x)(t)= \sqrt{x^{2}(t)+1}/(t+1)$ and
$u(t, s, x)=\sqrt{1+|x|}/\exp(t+s+1)$. Obviously, we have
that $T_1$ and $T_2$ are continuous operators on the space
$BC(\mathbb{R}^{+},\mathbb{R} )$. Moreover, for all $t\in \mathbb{R}^{+}$ and
$x\in BC(\mathbb{R}^{+}, \mathbb{R})$, we obtain
 \begin{gather*}
 |(T_1x)(t)|\leq \Big|\frac{t}{\exp(t)}\Big|\leq \frac{1}{e}, \\
 |(T_2x)(t)|\leq \Big|\frac{\sqrt{x^{2}(t)+1}}{t+1}\Big|
\leq\sqrt{\|x\|^{2}+1}.
\end{gather*}
 Thus, we conclude that assumption (i) is satisfied with $d_1(x)=1/e$
and $d_2(x)= \sqrt{x^{2}+1}$.
 Observe that the function $u(t, s, x)$ is continuous on the set
$\mathbb{R}^{+}\times \mathbb{R}^{+}\times \mathbb{R}$. Further, we obtain
\[
 |u(t, s, x)|=\Big|\frac{\sqrt{1+|x|}}{\exp(t+s+1)}\Big|
=\frac{\sqrt{1+|x|}}{\exp(t+s+1)}
\]
for all $t, s \in \mathbb{R}^{+}$ and $x\in \mathbb{R}$.
 Thus the function $u(t, s, x)$ satisfies assumption (ii) with
$g(t, s)=1/\exp(t+s+1)$ and $h(x)=\sqrt{1+|x|}$. Obviously,
 $s\to 1/\exp(t+s+1)$ is integrable on $\mathbb{R}^{+}$ and
$t\to \int_0^{\infty}ds/\exp(t+s+1) $ is
 bounded on $\mathbb{R}^{+}$. Further, we have that
\[
 G=\sup\Big\{\int_0^{\infty}\frac{1}{\exp(t+s+1)}ds: t\in \mathbb{R^{+}}\Big\}
=\sup\Big\{\frac{1}{\exp(t+1)}: t\in
 \mathbb{R^{+}}\Big\}=\frac{1}{e}.
\]
 Next, observe that the inequality from assumption (iv) has the form
 \begin{equation}\label{eşitsizlik19}
 \frac{1}{e}+ \sqrt{r_0^{2}+1}\sqrt{1+r_0}\frac{1}{e}\leq r_0.
 \end{equation}
It can be easily verified that if $1.2532\leq r_0 \leq 5.1357$, then
 $r_0$ is the solution of \eqref{eşitsizlik19}.
 Also for $\varepsilon\geq 0$, $L>0$, $\|x\|\leq r_0$ and
 $t, s\in[0, L]$ such that $|t-s|\leq\varepsilon$, we have that
 \begin{equation}\label{ak2}
|(T_1x)(t)-(T_1x)(s)|=\Big|\frac{t}{\exp(t)}-\frac{s}{\exp(s)}\Big|.
 \end{equation}
 Further, without loss of generality we can assume that $x(t)<x(s)$. Hence, we obtain
\begin{equation}\label{map2}
\begin{aligned}
|(T_2x)(t)-(T_2x)(s)|
&=\Big|\frac{\sqrt{ x^{2}(t)+1}}{t+1}-\frac{\sqrt{ x^{2}(s)+1}}{s+1}\Big| \\
 &\leq \frac{1}{t+1}\left|\sqrt{ x^{2}(t)+1}-\sqrt{ x^{2}(s)+1}\right|\\
&\quad +\sqrt{ x^{2}(s)+1}\Big|\frac{1}{t+1}-\frac{1}{s+1}\Big| \\
 &\leq \frac{|x(t)-x(s)||2\xi|}{2\sqrt{\xi^{2}+1}}
 +\frac{\sqrt{r_0^{2}+1}|s-t|}{(t+1)(s+1)} \\
 &\leq  |x(t)-x(s)|+\sqrt{r_0^{2}+1} \varepsilon,
\end{aligned}
 \end{equation}
where $\xi \in (x(t), x(s))$.
 In view of \eqref{mnc} and the uniform continuity of the function
$t\to t\exp(t)$ on the set
 $[0, L]$, we have by \eqref{ak2} and \eqref{map2} that
 \begin{gather*}
 \omega_0(T_1X)=0, \\
 \omega_0(T_2X)\leq \omega_0(X).
 \end{gather*}
This implies that assumption (v) is satisfied with $k_1=0$ and $k_2=1$.
 Further, we have
 \begin{equation}\label{son120}
\begin{aligned}
\sup\{|(T_1x)(t)|: t\geq  L\}
&= \sup\Big\{\Big|\frac{t}{\exp(t)}\Big|:t\geq L\Big\}\\
&= \begin{cases}
 1/e, & 0<L\leq 1\\
 L/\exp(L), & L>1.
 \end{cases}
\end{aligned}
\end{equation}
Moreover,
 \begin{equation}\label{art4}
\sup\{|(T_2x)(t)|: t\geq L\}
=\sup\Big\{\Big|\frac{\sqrt{ x^{2}(t)+1}}{t+1}\Big|:t\geq L\Big\}
\leq  \frac{\sqrt{r_0^{2}+1}}{L+1}.
 \end{equation}
Thus from estimates \eqref{son120} and \eqref{art4} in view of \eqref{beta}
 we have that $\beta(T_1X)=\beta(T_2X)=0$.
 Therefore we obtain $m_1=m_2=0$.
Keeping in mind the constants $k_1,k_2, m_1, m_2$ above obtained  we have
$$
\max\Big\{\frac{\sqrt{r_0+1}}{e},\,0\Big\}<1
$$
and assumption (vi) is satisfied.
Then, we obtain
\begin{align*}
&\sup\Big\{\int_{L}^{\infty}\frac{1}{\exp(t+s+1)}ds:t\in[0, L]\Big\} \\
&= \sup\Big\{\frac{1}{\exp(t+L+1)}:t\in[0, L]\Big\} \\
&= \frac{1}{\exp( L+1)}.
\end{align*}
 Hence, we infer that  assumption (vii) is satisfied.

Finally, we conclude that all of the assumptions of Theorem \ref{anateorem}
 are satisfied. This implies that the integral equation \eqref{rnek3}
 has a solution $x=x(t)$ belonging to the set $Y$. Moreover,
 $x(t)\to 0$ as $t\to 0$.
\end{example}

 \begin{remark} \label{rmk4.3} \rm
 Observe that if we put $a(t)=t\exp(-t)$, $f(t, x)=\sqrt{x^{2}+1}/(t+1)$ and
$u(t, s, x)=\sqrt{1+|x|}/\exp(t+s+1)$, then
 \eqref{rnek3} is a special case of \eqref{2-4} which is handled in
\cite{yeni}.
It is easily seen that $a \in  BC(\mathbb{R}^{+},\mathbb{R} )$ and
 $\|a\|=1/e$. $f(t, 0)\in BC(\mathbb{R}^{+},\mathbb{R} )$ and $f$
satisfies the Lipschitz condition with
 respect to the second variable for $k=1$.

 On the other hand we have $g(t, s)=1/\exp(t+s+1)$ and $h(r)=r$ which are
imposed in assumption (iv) of \cite[Theorem 8]{yeni} for the inequality
\[
 |u(t, s, x)-u(t, s, y)|\leq g(t,s)h(|x-y|)
\]
to be satisfied for all $t, s \in\mathbb{R^{+}}$ and $x, y \in\mathbb{R}$.

 Moreover we obtain $\bar{f}=1$, $\bar{g}=1/e$ and $\bar{u}=1/e$, where
\begin{gather*}
\bar{f}=\sup\{|f(t, 0)|: t\in \mathbb{R^{+}}\}, \quad
\bar{g}=\sup\Big\{\int_0^{\infty} g(t, s)ds: t\in \mathbb{R^{+}}\Big\}, \\
\bar{u}=\sup\Big\{\int_0^{\infty} |u(t, s, 0)|ds: t\in \mathbb{R^{+}}\Big\}.
\end{gather*}
 Thus assumptions (i)--(vi) of \cite[Theorem 8]{yeni} are fulfilled.

 Finally, let us note that the inequality of assumption (vii), given
in \cite{yeni}:
 $$
\|a\|+k\bar{g}rh(r)+k\bar{u}r+\bar{f}\bar{g}h(r)+\bar{f}\bar{u}\leq r
$$
takes the form
 \begin{equation}\label{344}
 \frac{r^{2}+2r+2}{e}\leq r.
 \end{equation}
It can be checked that \eqref{344} does not have a positive solution.
Therefore, \cite[Theorem 8]{yeni} is
 inapplicable to \eqref{rnek3},
\end{remark}

 \subsection*{Acknowledgement}
The authors are grateful to the editor and to the referees
for their valuable suggestions that helped us improve our manuscript.


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