\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 170, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2010/170\hfil Krasnoselskii-Schaefer theorem]
{Periodic solutions of neutral delay integral equations of advanced type}
\author[M. N. Islam, N. Sultana, J. Booth\hfil EJDE-2010/170\hfilneg]
{Muhammad N. Islam, Nasrin Sultana, James Booth}
\address{Muhammad N. Islam \newline
Department of Mathematics\\
University of Dayton\\
Dayton, OH 45469-2316 USA}
\email{muhammad.islam@notes.udayton.edu}
\address{Nasrin Sultana \newline
Department of Mathematics\\
University of Dayton\\
Dayton, OH 45469-2316 USA}
\email{sultannz@notes.udayton.edu}
\address{James Booth \newline
Department of Mathematics\\
University of Dayton\\
Dayton, OH 45469-2316 USA}
\email{boothjaa@notes.udayton.edu}
\thanks{Submitted October 1, 2010. Published November 29, 2010.}
\subjclass[2000]{45D05, 45J05}
\keywords{Volterra integral equation;
neutral delay integral equation; \hfill\break\indent
periodic solution; Krasnosel'sii's fixed point theorem;
Schaefer's fixed point theorem; \hfill\break\indent
Liapunov's method}
\begin{abstract}
We study the existence of continuous periodic solutions
of a neutral delay integral equation of advanced type.
In the analysis we employ three fixed point theorems:
Banach, Krasnosel'skii, and Krasnosel'skii-Schaefer.
Krasnosel'skii-Schaefer fixed point theorem requires an
\emph{a priori} bound on all solutions. We employ a Liapunov
type method to obtain such bound.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{lemma}[theorem]{Lemma}
\section{Introduction}
In this article we use three fixed point theorems to obtain
continuous periodic solutions to the integral equation
\begin{equation} \label{e1}
x(t)=f(t,x(t),x(t-h))-\int_t^{\infty}c(t,s)g(s,x(s),x(s-h))ds,
\end{equation}
where $f,g:\mathbb{R}\times \mathbb{R}\times \mathbb{R}\to
\mathbb{R}$ and $c:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$
are continuous functions, and the delay $h$ is a positive constant.
Equations such as \eqref{e1} are known as neutral delay integral equations
of advanced type \cite[p. 295]{B2}.
Several applications in physical sciences lead to neutral functional
differential equations such as
\begin{equation}\label{e2}
x'(t)=ax(t)+\alpha x'(t-h)-q(x(t),x(t-h))+r(t)\,;
\end{equation}
see for example \cite{B2,BH,GO,GZ,HL,K1} and their references.
By integrating the above equation
we obtain the integral equation
\begin{equation}\label{e3}
x(t)=\alpha x(t-h)-\int_t^{\infty}[q(x(s),x(s-h))
-a \alpha x(s-h)]e^{a(t-s)}ds + p(t)\,,
\end{equation}
which is a particular case of the equation to be studied here.
Although integrals from $t$ to $\infty$ are not common in these
type of problems, they are found when studying
unstable manifolds for ordinary differential equations,
Coddington and Levinson \cite[1955, p. 331]{CL}.
Studies of integrals of this and other types can be found
in \cite{B2,BH,C,GLS} and their references.
The three fixed point theorems:
Banach, Krasnosel'skii, and Krasnosel'skii-Schaefer will be
used in this article.
The latter theorem is a combination of Krasnosel'skii
theorem and Schaefer fixed point theorem, as explained in
Burton and Kirk \cite{BC}.
Advantages and disadvantages of each method will be stated
at the end of their corresponding sections.
Krasnosel'skii-Schaefer's theorem requires \emph{a priori}
bounds for solutions of an auxiliary equation.
To obtain such bounds, we construct a Lyapunov-like functional,
assuming certain sign conditions on $c$ and its derivatives.
Since one might define various Liapunov functionals,
and obtain the required bounds under various conditions,
the Krasnosel'skii-Schaefer theorem has a potential for
providing results better than ours.
The same three theorems were used by the first author in \cite{I1} for
studying \eqref{e1} with $\int_{t-h}^t$ instead of
$\int_t^\infty$. There, the Lyapunov functional and some
assumptions are different from the ones here.
We start by stating the classical theorems.
\begin{theorem}[Krasnosel'skii Theorem \cite{S}] \label{thmkrasno}
Let $M$ be a closed convex non-empty subset of a Banach
space $(S,\|\cdot\|)$. Suppose $A$ and $B$ map $M$ into $S$ such that
\begin{itemize}
\item[(i)] $A\Phi + B\Psi \in M$ for all $\Phi ,\Psi \in M$;
\item[(ii)] $A$ is continuous and $AM$ is contained in a compact set;
\item[(iii)] $B$ is a contraction.
\end{itemize}
Then there exists a $\Phi \in M$ with $\Phi=A\Phi+B\Phi$.
\end{theorem}
\begin{theorem}[Schaefer Theorem \cite{BC}] \label{thmsch}
Let $(S,\|\cdot\|)$ be a normed space, and let $H$ a continuous mapping
from $S$ into $S$ which is compact on each bounded subset $X$ of $S$.
Then either
\begin{itemize}
\item[(i)] the equation $x=\lambda Hx$ has a solution for
$\lambda=1$, or
\item[(ii)] the set of all such solutions, $0<\lambda <1$,
is unbounded.
\end{itemize}
\end{theorem}
\begin{theorem}[Krasnosel'skii-Schaefer Theorem \cite{BC}]
\label{thmschaefer}
Let $(S,\|\cdot\|)$ be a Banach space. Suppose $B:S\to S$
is a contraction map, and $A:S\to S$ is continuous and maps
bounded sets into compact sets. Then either
\begin{itemize}
\item[(i)] $x=\lambda B(\frac{x}{\lambda})+\lambda Ax$
has a solution in $S$ for $\lambda=1$, or
\item[(ii)] the set of all such solutions, $0<\lambda <1$,
is unbounded.
\end{itemize}
\end{theorem}
As general assumptions, we have the following:
\begin{itemize}
\item[(H1)] For a positive constant $T$, we assume
\begin{gather*}
f(t+T,x,y)=f(t,x,y), \quad
g(t+T,x,y)=g(t,x,y),\\
c(t+T,s+T)=c(t,s).
\end{gather*}
\end{itemize}
Note that $f$ and $g$ are periodic in the variable $t$,
with the same period ($T$-periodic for short). Using this
property, we define
\begin{equation}
f_0:=\sup_{0\leq t\leq T}|f(t,0,0)|,\quad
g_0:=\sup_{0\leq t\leq T}|g(t,0,0)|. \label{ef0}
\end{equation}
Also note that $c$ is not necessarily periodic in $t$ or in $s$;
for example $c(t,s)=\phi(t-s)$ satisfies the above condition,
without being periodic.
However, $\int_t^\infty c(t,s)\,ds$ is $T$-periodic, which is used in
the definition
\begin{equation} \label{ebarc}
\bar{c}:=\sup_{0\leq t\leq T}\int_t^{\infty} |c(t,s)|\,ds\,.
\end{equation}
\begin{itemize}
\item[(H2)] The kernel $c$ satisfies
$\int_t^{\infty} |c(t,s)|\,ds<\infty$ for $t\geq 0$, and
\begin{equation} \label{intcont}
\lim_{u\to t} \int_{\max\{u,t\}}^{\infty} |c(u,s)-c(t,s)|\,ds=0\,.
\end{equation}
This condition is satisfied for example by $c(t,s)=\exp(t-s)$.
\item[(H3)] There are constants $f_1,f_2$ such that
\begin{equation}
|f(t,x_1,y_1)-f(t,x_2,y_2)|\leq f_1|x_1-x_2|+f_2|y_1-y_2|,
\label{flips}
\end{equation}
for $t, x_1, x_2, y_1, y_2\in \mathbb{R}$.
\item[(H4)] There are constants $g_1,g_2$ such that
\begin{equation}
|g(t,x_1,y_1)-g(t,x_2,y_2)|\leq g_1|x_1-x_2|+g_2|y_1-y_2|
\label{glips}
\end{equation}
for $ t, x_1, x_2, y_1, y_2\in \mathbb{R}$.
\end{itemize}
Let $(P_T,\|\cdot\|)$ denote the space of continuous
$T$-periodic functions with the norm
$\|x\|=\sup_{0\leq t\leq T}|x(t)|$.
On this space, we define the operators
\begin{equation} \label{eAB}
Ax(t)=-\int^{\infty}_t c(t,s)g(s,x(s),x(s-h))\,ds,\quad
Bx(t)=f(t,x(t),x(t-h))\,.
\end{equation}
All three fixed point theorems in this article use $B$ as
a contraction, which requires \eqref{flips} with additional
conditions on $f_1$ and $f_2$.
\section{Using Banach's Fixed point theorem}
\begin{theorem} \label{thm1}
Assume {\rm (H1)-(H4)} and
\begin{equation} \label{banach}
f_1+f_2+(g_1+g_2)\bar{c}<1\,.
\end{equation}
Then \eqref{e1} has a unique continuous T-periodic solution.
\end{theorem}
\begin{proof}
First note that every fixed point of $(A+B)$ in $P_T$ is a periodic
solution of \eqref{e1}.
Let $x$ be a function in $P_T$. Then the continuity and periodicity of $Bx$
follows from the continuity and periodicity of $f$ and $x$.
Next we show that $Ax$ is $T$-periodic.
Using the change of variable $u=s-T$, the equality
$c(t+T,u+T)=c(t,u)$, and that $g(s,x(s),x(s-h))$ is periodic in $s$,
we have
\begin{align*}
Ax(t+T)
&=\int_{t+T}^\infty c(t+T,s)g(s,x(s),x(s-h))\,ds\\
&=\int_t^\infty c(t+T,u+T)g(u+T,x(u+T),x(u+T-h))\,du\\
&=\int_t^\infty c(t,u)g(u,x(u),x(u-h))\,du
=Ax(t)\,.
\end{align*}
Next we show that $Ax$ is continuous.
Using that $g(s,x(s),x(s-h))$ is continuous and periodic in $s$,
we define $\bar{g}_x:=\sup_{0\leq s\leq T}|g(s,x(s),x(s-h))|<\infty$.
Then for $u\geq t$, we have
\begin{align*}
&|Ax(u)-Ax(t)|\\
&=\big|\int_u^\infty c(u,s)g(s,x(s),x(s-h))\,ds
-\int_t^\infty c(t,s)g(s,x(s),x(s-h))\,ds\big|\\
&\leq \bar{g}_x \int_t^u |c(t,s)|\,ds
+\bar{g}_x \int_u^\infty |c(u,s)- c(t,s)|\,ds\,.
\end{align*}
Similarly for $t\geq u$, we have
\[
|Ax(u)-Ax(t)| \leq \bar{g}_x \int_u^t |c(u,s)|\,ds
+\bar{g}_x \int_t^\infty |c(u,s)- c(t,s)|\,ds\,.
\]
Taking the limit as $u\to t$, by \eqref{intcont}, the right-hand side approaches
zero. This shows the continuity of $Ax$.
Furthermore, when considering $x$ in a bounded set $M\subset P_T$,
the constant $\bar{g}_x$ can be made independent of $x$.
Thus $AM$ becomes a family of equi-continuous functions.
Now we show that $A+B$ is a contraction. Let $x,y$ be functions in
$P_T$. By \eqref{flips},
\begin{equation} \label{Bcontract}
|Bx(t)-By(t)|=|f(t,x(t),x(t-h))-f(t,y(t),y(t-h))|
\leq (f_1+f_2)\|x-y\|.
\end{equation}
By \eqref{glips} and \eqref{ebarc},
\begin{align*}
|Ax(t)-Ay(t)|
&=\big|\int_t^\infty c(t,s)\big(g(s,x(s),x(s-h))-g(s,y(s),y(s-h))\big)\,ds\big|\\
&\leq (g_1+g_2)\|x-y\| \int_t^\infty |c(t,s)|\\
&\leq (g_1+g_2)\|x-y\| \bar{c}\,.
\end{align*}
From the two inequalities above and \eqref{banach}, the operator
$(A+B)$ is a contracting mapping in the Banach space $P_T$.
Therefore, there is a unique fixed point which is a solution of \eqref{e1}.
This completes the proof.
\end{proof}
\section{Using Krasnosel'skii's Fixed point theorem}
We look for periodic solutions of \eqref{e1} in
\[
M_r=\{x\in P_T:\|x\|\leq r\}\,,
\]
which is a closed and convex subset of $P_T$.
\begin{theorem} \label{thm2}
Under the hypotheses of Theorem \ref{thm1}, Krasnosel'skii's
Theorem also provides continuous periodic solutions to \eqref{e1}, and
at least one solution satisfies
\[
\|x\|\leq \frac{f_0+g_0\bar{c}}{1-(f_1+f_2+(g_1+g_2)\bar{c})}\,.
\]
\end{theorem}
\begin{proof}
We will show that items (i)--(iii) of Theorem \ref{thmkrasno}
are satisfied.
Part (i): Let $x,y$ be functions in $M_r$.
That $Bx$ and $Ax$ are continuous and periodic
follows from the arguments in the proof of Theorem \ref{thm1}.
We need to show that $\|Bx+Ay\|\leq r$,
for certain value $r$.
From \eqref{ef0}, \eqref{glips} and \eqref{ebarc}, we have
\begin{align*}
|Ay(t)|
&=\big|\int_t^\infty c(t,s)
\big(g(s,0,0)+g(s,y(s),y(s-h))-g(s,0,0)\big)\,ds\big|\\
&\leq \big(g_0+(g_1+g_2)r\big)\bar{c}\,.
\end{align*}
From \eqref{ef0} and \eqref{flips},
\[
|Bx(t)|=\big|f(t,0,0) + f(t,x(t),x(t-h))-f(t,0,0)\big|
\leq f_0 + (f_1+f_2)r
\]
Combining the two inequalities above,
\[
\frac{|Ay(t)+ Bx(t)|}{r} \leq
\frac{f_0+g_0\bar{c}}{r}+ f_1+f_2+(g_1+g_2)\bar{c}\,.
\]
As $r\to\infty$, by \eqref{banach}, the right-hand side on
the above inequality approaches $(f_1+f_2+(g_1+g_2)\bar{c})$ which
is less than 1. Then, there exists $r_0$ such that $r\geq r_0$ implies
$|Ay(t)+ Bx(t)|\leq r$; i, e., $Ay(t)+ Bx(t)\in M_r$.
Using \eqref{banach}, we can select
\[
r_0=\frac{f_0+g_0\bar{c}}{1-(f_1+f_2+(g_1+g_2)\bar{c})}\,.
\]
Part (ii):
First we show that $A$ is a continuous mapping on $M_r$.
Let $x,y\in M_r$. By \eqref{glips} and \eqref{ebarc}, we have
\begin{align*}
|Ax(t)-Ay(t)|
&\leq\int_t^{\infty}|c(t,s)|\,|g(s,x(s),x(s-h))-g(s,y(s),y(s-h))|\,ds\\
&\leq \int_t^{\infty}|c(t,s)|(g_1+g_2)\|x-y\|\,ds\\
&\leq \bar{c} (g_1+g_2) \|x-y\|\,.
\end{align*}
This shows that $A$ is a continuous mapping on $M_r$.
Now, we show that $AM_r$ is a compact set, by means of the
Arzela-Ascoli's theorem. Since $AM_r\subset M_r$,
the functions in $AM_r$ are unformly bounded by $r$.
In the proof of Theorem \ref{thm1} we showed that when $M_r$ is bounded,
the functions in $AM_r$ are equi-continuous.
Part (iii):
That $B$ maps $M_r$ into $M_r$ and that $B$ is a contracting operator
follows from the proof Theorem \ref{thm1}, \eqref{Bcontract},
and \eqref{banach}.
From Theorem \ref{thmkrasno}, the operator $(A+B)$ has a fixed point
in $M_r\subset P_T$.
This is a solution of \eqref{e1}, which is periodic and bounded by $r$.
\end{proof}
\begin{remark} \label{rmk1} \rm
In our setting, the Banach and Kasnosels'kii Theorems require
the same assumptions. Banach fixed point theorem provides the uniqueness of the solution,
and Kasnosels'kii's theorem provides an estimate on the size of at least one solution.
\end{remark}
\section{Using Krasnosel'skii-Schaefer Fixed point theorem}
In this section, we omit (H4), but use the following conditions:
\begin{itemize}
\item[(H5)] The kernel $c(t,s)$ is twice differentiable,
$c_{st}(t,s)\leq 0$ for $t\leq s$, and
\begin{equation} \label{stc}
\lim_{s\to \infty} (s-t)c(t,s)=0\,.
\end{equation}
\item[(H6)]
\begin{equation} \label{Lsup}
L:=\sup_{0\leq t\leq T} \int_t^{\infty}|c_s(t,s)|(1+\frac{s-t}{T})ds
<\infty.
\end{equation}
\item[(H7)] $x g(t,x,y)\geq 0$, and for $f_0,f_1,f_2$ as defined above,
there exist positive constants $K,\beta$ such that
\[
f_2|y g(t,x,y)|+(f_0+\beta)|g(t,x,y)| \leq K +(1-f_1)xg(t,x,y)\,,
\]
for $t,x,y\in \mathbb{R}$.
\end{itemize}
Note that (H7) implies that
for each fixed $x$, $\lim_{|y|\to\infty}|y g(t,x,y)|<\infty$.
For the parameter $\lambda$ in Krasnosel'ksii-Schaefer Theorem,
we define the auxiliary equation
\begin{equation}\label{e8}
x(t)=\lambda f\Big(t,\frac{x(t)}{\lambda},
\frac {x(t-h)}{\lambda}\Big)-\lambda\int_t^{\infty}
c(t,s)g(s,x(s),x(s-h))ds.
\end{equation}
With $A$ and $B$ as defined in \eqref{eAB}, we write the above equation
as
\begin{equation}\label{e10}
x(t)=\lambda B(\frac{x}{\lambda})(t)+\lambda (Ax)(t).
\end{equation}
Notice that this equation becomes \eqref{e1} when $\lambda=1$.
\begin{theorem} \label{thm3}
Assume {\rm (H1)--(H3), (H5)-(H7)} and that
\begin{equation} \label{fschaefer}
f_1+f_2<1\,.
\end{equation}
Then \eqref{e1} has a continuous T-periodic solution.
\end{theorem}
\begin{proof}
Let $x\in P_T$. Then by the same argument as in
the proof of Theorem \ref{thm1}, $Ax$ and $Bx$ are also in $P_T$.
That $B$ is a contraction, also follows from the proof
of Theorem \ref{thm1}, assuming \eqref{fschaefer}.
To show that \eqref{e10} has a solution in $P_T$ when $\lambda=1$,
we show that all solutions of \eqref{e10} are bounded by a
constant independent of the solution when $0<\lambda\leq 1$.
Note that when $x$ is $T$-periodic, so is $|g(u,x(u),x(u-h))|$, and
it is bounded by a constant $\bar{g}_x$.
Therefore, by \eqref{stc},
\[
\lim_{s\to\infty}|c(t,s)\int_t^s g(u,.,.)\,du|
\leq \lim_{s\to\infty} \bar{g}_x |s-t|\,|c(t,s)|=0\,,
\]
where $g(u,.,.):=g(u,x(u),x(u-h))$. Using this limit,
integrating by parts in \eqref{e8}, we have
\begin{equation} \label{e11}
x(t)= \lambda f\big(t,\frac{x(t)}{\lambda},
\frac {x(t-h)}{\lambda}\big)
+\lambda \int_t^{\infty} c_s(t,s)\int_t^s g(u,.,.)\,du\,ds\,.
\end{equation}
The first term in the right-hand side is estimated using \eqref{ef0},
\eqref{flips}, and
\begin{equation} \label{trianglef}
\begin{aligned}
|\lambda f\big(t,\frac{x(t)}{\lambda},\frac {x(t-h)}{\lambda}\big)|
&\leq |\lambda f(t,0,0)+
\lambda f\big(t,\frac{x(t)}{\lambda},
\frac {x(t-h)}{\lambda}\big)-\lambda f(t,0,0)|\\
&\leq \lambda f_0+ f_1|x(t)|+f_2|x(t-h)|\,.
\end{aligned}
\end{equation}
To estimate the integral in \eqref{e11},
we define the Lyapunov-like functional
\begin{equation}
V(t)=V(t,x(t))=\frac{\lambda}{2}
\int_t^{\infty} c_s(t,s)
\Big(\int_t^s g(u,x(u),x(u-h))\,du\Big)^2 ds.
\end{equation}
Differentiating with respect to $t$,
\begin{align*}
V'(t)
&=\frac{\lambda}{2} \int_t^{\infty}\frac{d}{dt}\big[c_s(t,s)
\Big(\int_t^s g(u,.,.)\,du\Big)^2 \big]\,ds\\
&= \frac{\lambda}{2} \int_t^{\infty} c_{st}(t,s)
\Big(\int_t^s g(u,.,.)\,du\Big)^2
- c_s(t,s)2 \Big(\int_t^s g(u,.,.)\,du \Big)g(t,.,.) \,ds\,.
\end{align*}
Using that $c_{st}(t,s)\leq 0$ by (H5), we have
\[
V'(t)\leq -\lambda g(t,.,.) \int_t^\infty
c_s(t,s) \int_t^s g(u,.,.)\,du \,ds\,.
\]
Integrating by parts and using \eqref{stc} and that $x$ is a solution
of \eqref{e8}, we have
\begin{align*}
V'(t)&\leq g(t,.,.) \Big(\lambda\int_t^\infty c(t,s)g(s,.,.)\,ds\Big)\\
&= g(t,.,.) \Big(\lambda f\big(t,\frac{x(t)}{\lambda},
\frac {x(t-h)}{\lambda}\big)-x(t)\Big)\\
&=g(t,.,.)\lambda f\big(t,\frac{x(t)}{\lambda},
\frac {x(t-h)}{\lambda}\big)-x(t)g(t,.,.) \,.
\end{align*}
By \eqref{trianglef}, and that $0\leq x(t)g(t,.,.)$, we have
\begin{align*}
V'(t)
&\leq |g(t,.,.)|
\big(\lambda f_0+ f_1|x(t)|+f_2|x(t-h)|\big)-x(t)g(t,.,.)\\
&=(f_1-1)x(t)g(t,.,.) +f_2|x(t-h)|\,|g(t,.,.)|+ \lambda f_0|g(t,.,.)|\,.
\end{align*}
Using that $\lambda\leq 1$, by (H7) there exists positive constants
$K$ and $\beta$, such that
\begin{equation} \label{e12}
V'(t) \leq K -\beta |g(t,x(t),x(t-h))|\,.
\end{equation}
Note that $V(t)$ is $T$-periodic which can be shown as follows:
First show that $c_s(t,s)=c_s(t+T,s+T)$ and that
$\int_t^s g(u,x(u),x(u-h))\,du$ is $T$-periodic.
Then show that $V(t)$ is $T$-periodic as is done for $Ax$ in
Theorem \ref{thm1}.
By the periodicity of $V$,
\[
0=V(T)-V(0)\leq KT-\beta \int_0^T |g(t,x(t),x(t-h))|\,dt\,,
\]
which implies
\[
\int_0^T |g(t,x(t),x(t-h))|\,dt\leq \frac{KT}{\beta}\,.
\]
With this inequality, and using that $|g(t,x(t),x(t-h))|$ is
$T$-periodic, we have
\[
\int_t^s |g(u,.,.)|\,du
\leq \big(1+\frac{s-t}{T}\big)\int_0^T |g(u,.,.)|\,du
\leq \big(1+\frac{s-t}{T}\big)\frac{KT}{\beta}\,.
\]
From \eqref{e11}, \eqref{trianglef}, and (H6), we have
\begin{align*}
|x(t)|
&\leq \lambda f_0 +f_1|x(t)| +f_2|x(t-h)|+
\int_t^{\infty} |c_s(t,s)|\int_t^s |g(u,.,.)|\,du\,ds\\
&\leq \lambda f_0 +f_1|x(t)| +f_2|x(t-h)|+
\frac{LKT}{\beta}\,.
\end{align*}
Computing the supremum over $[0,T]$ and using that $\lambda\leq 1$,
we have
\[
\|x(t)\|\leq \frac{f_0+\frac{LKT}{\beta}}{1-(f_1+f_2)}\,.
\]
Note that the right-hand side is independent of the solution;
Thus the set of solutions to \eqref{e10} is bounded
for all $\lambda\in (0,1]$.
The proof is complete.
\end{proof}
\begin{remark} \label{rmk2} \rm
When applying Krasnosel'skii, (H4) requires $g(t,x,y)$ to satisfy
a Lipschitz condition at every value of $x$ and $y$.
When applying Krasnosel'skii-Schaefer, (H7) replaces these conditions
by $yg(t,x,y)$ being bounded.
There are functions that satisfy (H4) but not (H7), for example
$g(t,x,y)=\sin(t)(x/5+y/5)$.
There are also functions that satisfy (H7) but not (H4), for example
$g(t,x,y)=\sin^2(t)x^{1/3}/(1+y^2)$.
Note that Krasnosel'skii-Schaefer's theorem imposes additional
restrictions on the kernel $c(t,s)$, (H5)-(H6).
\end{remark}
\begin{remark} \label{rmk3} \rm
In this article, the operator $B$ is a contraction.
However, our results remain valid if the
function $f$ of \eqref{e1} defines a ``large contraction'' or a
``separate contraction''.
Burton \cite{B2} introduced a concept of large contraction.
Then Liu and Li \cite{LL} defined a separate contraction,
and showed that a large contraction is also a separate contraction.
They also proved that Krasnosel'skii's theorem holds for
separate contractions.
\end{remark}
\subsection*{Acknowledgments}
The authors thank Professor Julio G. Dix for his inputs, and for many
suggestions in the write-up of the manuscript.
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\end{document}