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

\AtBeginDocument{{\noindent\small {\em 
Electronic Journal of Differential Equations}, Vol. 2005(2005), 
No. 126, pp. 1--10.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/126\hfil Asymptotic stability results]
{Asymptotic stability results for certain integral equations}
\author[C. Avramescu, C. Vladimirescu\hfil EJDE-2005/126\hfilneg]
{Cezar Avramescu, Cristian Vladimirescu}  % in alphabetical order

\address{Cezar Avramescu \hfill\break
Department of Mathematics, University of Craiova\\
13 A.I. Cuza Str., Craiova RO 200585, Romania}
\email{zarce@central.ucv.ro}

\address{Cristian Vladimirescu \hfill\break
Department of Mathematics, University of Craiova\\
13 A.I. Cuza Str., Craiova RO 200585, Romania}
\email{vladimirescucris@yahoo.com}

\date{}
\thanks{Submitted October 17, 2005. Published November 14, 2005.}
\subjclass[2000]{47H10, 45D10} 
\keywords{Fixed points; integral equations}

\begin{abstract}
 This paper shows the existence of asymptotically
 stable solutions to an integral equation. This is done by using
 a fixed point theorem, and without requiring that the
 solutions be bounded.
\end{abstract}

\maketitle 
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction}

 Ban\'{a}s and Rzepka \cite{BR} study a very interesting
property for the solutions of some functional equations. The same
property was also studied  by Burton and Zhang in \cite{BZ}, in a
more general case. Let $F:BC(\mathbb{R}_{+})\to
BC(\mathbb{R}_{+})$ be an operator, where $BC(\mathbb{R}_{+})$
consists of bounded and continuous functions from $\mathbb{R}_{+}$ to
$\mathbb{R}^d$, $\mathbb{R}_{+}:=[0,\infty )$, $d\geq 1$. Let  
$|\cdot | $ be a norm  in $\mathbb{R}^{d}$.

The following definition is given in \cite{BR,BZ}, for solutions
$x\in BC(\mathbb{R}_{+})$ of the equation
\begin{equation}
x=Fx.  \label{1}
\end{equation}

\begin{definition} \label{Definition1} \rm
A function $x$ is said to be an {\bf asymptotically stable
solution} of \eqref{1} if for any $\varepsilon >0$ there exists
$T=T(\varepsilon )>0$ such that for every $t\geq T$ and for every
 solution $y$ of \eqref{1}, we have
\begin{equation}
| x(t)-y(t)| \leq \varepsilon . \label{2}
\end{equation}
\end{definition}

A sufficient condition for the existence of asymptotically stable
solutions is given by the following proposition.

\begin{proposition} \label{Proposition1}
Assume that there exist a constant $k\in [ 0,1)$ and a continuous
function $a:\mathbb{R}_{+}\to \mathbb{R}_{+}$ with $\lim_{t\to
\infty }a(t)=0$, such that
\begin{equation}
| (Fx)(t)-(Fy)(t) | \leq k| x(t)-y(t)| +a( t),\quad\forall t\in
\mathbb{R}_{+},\; \forall x,y\in BC(\mathbb{R}_{+}).  \label{3}
\end{equation}
Then every solution of \eqref{1} is asymptotically stable.
\end{proposition}

The proof of this proposition is immediate. Let us remark that
basically the property of the asymptotic stability is a property
of the fixed points of the operator $F$. Actually, in
\cite{AV,BZ}, the proof of the existence of an asymptotically
stable solution in done by applying a fixed point theorem, i.e.
Schauder's Theorem. It follows that it is sufficient to ask
Definition \ref{Definition1} to be fulfilled only on the closed,
bounded, and convex set on which Schauder's Theorem is applied.

  Another remark concerning Proposition
\ref{Proposition1} is that if \eqref{3} is fulfilled then {\bf
every} solution of \eqref{1}
 is asymptotically stable. Moreover, by \eqref{3} we
deduce that the result of Proposition \ref{Proposition1} is
appropriate for the case when $F=A+B$, where $A$ is contraction
and $\lim_{t\to \infty }(Bx)(t)=0$, for every $x$ belonging to the
set on which the fixed point theorem is applied. On the other
hand, the set of the fixed points of $F$ should be ``big'' enough
for Definition \ref{Definition1} to be consistent. In this
direction, in the case when Schauder's fixed point Theorem is used
an interesting result has been obtained by Zamfirescu in
\cite{Za}, stating that if $B_{\rho }$ is the closed ball of
radius $\rho >0$ from a Banach space and $F:B_{\rho}\to B_{\rho }$
is a compact operator, then for most functions $F$, the set of
solutions of \eqref{1} is homeomorphic to the Cantor set (``most''
means ``all'' except those in a first category set).

Finally, let us remark that in order to fulfil Definition
\ref{Definition1} it is not necessary that all the solutions of
\eqref{1} to be bounded on $\mathbb{R}_{+}$.
  The result obtained
by Burton and Zhang is contained in the following theorem.

\begin{theorem}
Assume that
\begin{itemize}
\item[(i)] $f:\mathbb{R}_{+}\times \mathbb{R}^{d}\to
\mathbb{R}^{d}$ is continuous and there exist a continuous
function $k:\mathbb{R}_{+}\to [ 0,1] $ with $0\leq k(t)<1$ for
$t>0$ and a constant $x_{0}\in \mathbb{R}^{d}$ such that
$x_{0}=f(0,x_{0})$ and
\[
\lim_{t\to 0^{+}}(1-k(t))^{-1}( f(t,x_{0})-f(0,x_{0}))=0;
\]

\item[(ii)] for each $t\in \mathbb{R}_{+}$ and $x$, $y\in
\mathbb{R}^{d}$,
\[
| f(t,x)-f(t,y)| \leq k(t) | x-y| ;
\]

\item[(iii)] $u:\mathbb{R}_{+}\times \mathbb{R}_{+}\times
\mathbb{R}^{d}\to \mathbb{R}^{d}$ is continuous and there are
continuous functions $a$, $b:\mathbb{R}_{+}\to \mathbb{R}_{+}$
such that $| u(t,s,x)| \leq a(t) b(s)$, for all $t$, $s\in
\mathbb{R}_{+}$ ($s\leq t$) and all $x\in \mathbb{R}^{d}$ with
\[
\lim_{t\to 0^{+}}\frac{a(t)}{1-k(t)} \int_{0}^{t}b(s)ds=0
\]
and
\[
\lim_{t\to \infty }\frac{a(t)}{1-k(t)} \int_{0}^{t}b(s)ds=0.
\]
\end{itemize}
Then \eqref{63} below has at least one solution, and every
solution of the equation
\begin{equation}
x(t)=f(t,x(t))+\int_{0}^{t}u( t,s,x(s))ds,\quad t\in
\mathbb{R}_{+} \label{63}
\end{equation}
is asymptotically stable and converges to the unique continuous
function
 $\psi :\mathbb{R}_{+}\to \mathbb{R}^{d}$ satisfying
\[
\psi (t)=f(t,\psi (t)),\quad t\geq 0.
\]
\end{theorem}

Note that hypothesis (i) is not necessary; in our note \cite{AV}
we prove a similar theorem without using this hypothesis.
  Let us remark that in  \eqref{63} one has
$F=A+B$, where $A $ is a contraction in $BC(\mathbb{R}_{+})$ and
$B$ is a compact operator which in the admitted hypotheses
fulfills the property
\begin{equation}
\lim_{t\to \infty }(Bx)(t)=0,  \label{4}
\end{equation}
the limit being uniform with respect to $x\in BC(\mathbb{R}_{+})$.
The second result in our Note \cite{AV} is obtained in the absence
of condition  \eqref{4}.

In the present paper we  prove the existence of the asymptotically
stable solutions to \eqref{1} when $F$ is a sum of three
operators, without requiring the boundedness of the solutions. The
general result that we present needs a more sophisticated argument
than the one used in \cite{AV}. To this aim, we consider the set
of continuous functions as the fundamental space
\[
X=C_{c}=C_{c}(\mathbb{R}_{+},\mathbb{R}^{d})
\]
which equipped with the numerable family of seminorms
\begin{equation}
| x| _{n}:=\sup_{t\in [ 0,n] }\{ | x( t)| \} ,\quad n\geq 1,
\label{5}
\end{equation}
becomes a Fr\'{e}chet space (i.e. a complete linear metrisable
space).
  We will use in addition another family of
seminorms,
\begin{equation}
\| x\| _{n}:=| x| _{\gamma _{n}}+| x| _{h_{n}},\quad n\geq 1,
\label{6}
\end{equation}
where
\[
| x| _{h_{n}}=\sup_{\gamma _{n}\leq t\leq n}\{ e^{-h_{n}(t-\gamma
_{n})}| x(t)| \} ,
\]
$\gamma _{n}\in (0,n)$ and $h_{n}>0$ are arbitrary numbers.

\begin{remark} \label{Remark1} \rm
The families \eqref{5} and \eqref{6} define  the same topology on
$X$, i.e. the topology of the uniform convergence on compact
subsets of $\mathbb{R}_{+}$. Consequently, a family in $X$ is
relatively compact if and only if it is equicontinuous and
uniformly bounded on compact subsets of $\mathbb{R}_{+}$.
\end{remark}

\subsection*{Notations and general hypotheses}
  We consider
the nonlinear integral equation
\begin{equation}
x(t)=q(t)+f(t,x(t)) +\int_{0}^{t}V(t,s)x(s)ds+\int_{0}^{t}G(
t,s,x(s))ds,\,t\in \mathbb{R}_{+},  \label{7}
\end{equation}
where $q:\mathbb{R}_{+}\to \mathbb{R}^{d}$,
$f:\mathbb{R}_{+}\times \mathbb{R}^{d}\to \mathbb{R}^{d}$,
$V:\Delta \to \mathcal{M}_{d}(\mathbb{R})$, $G:\Delta \times
\mathbb{R}^{d}\to \mathbb{R}^{d}$ are supposed to be continuous
and $\Delta =\{ (t,s)\in \mathbb{R}_{+}\times \mathbb{R}_{+},\quad
s\leq t\} $.

  In what follows we
 denote by $| \cdot | $ a vector norm and also a matrix norm,
such that for every vector $x\in \mathbb{R}^{d}$ and for every
real quadratic $d\times d$ matrix $Z\in
\mathcal{M}_{d}(\mathbb{R})$,
\[
| Zx| \leq | Z| | x| .
\]
We will use the following general hypotheses:
 \begin{itemize}
\item[(H1)] There is a constant $L\in [ 0,1)$ such that
\[
| f(t,x)-f(t,y)| \leq L| x-y| ,\mbox{ }\forall x,y\in
\mathbb{R}^{d},\;\forall t\in \mathbb{R}_{+};
\]

\item[(H2)] There are two continuous functions
$a,b:\mathbb{R}_{+}\to \mathbb{R}_{+}$, such that
\[
| V(t,s)| \leq a(t)b(s), \quad\forall (t,s)\in \Delta ;
\]

\item[(H3)] There is a continuous function $\omega : \Delta \to
\mathbb{R}_{+}$ such that
\[
| G(t,s,x)| \leq \omega (t,s),\quad%
\forall (t,s)\in \Delta ,\quad\forall x\in \mathbb{R}^{d}.
\]
\end{itemize}

\section{Preliminary result}

In $X$, we consider  the equation
\begin{equation}
x(t)=q(t)+f(t,x(t)),\quad t\in \mathbb{R}_{+}.  \label{8}
\end{equation}

\begin{lemma} \label{Lemma1}
Under assumptions (H1)-(H3), Equation \eqref{8} admits a unique
solution.
\end{lemma}

\begin{proof} We define the operator $\Phi :X\to X$ through
\begin{equation}
(\Phi x)(t)=q(t)+f(t,x(t) ),\quad x\in X, \quad t\in
\mathbb{R}_{+}. \label{9}
\end{equation}
By hypothesis (H1) and \eqref{9} it follows that
\[
| \Phi x-\Phi y| _{n}\leq L| x-y| _{n},\quad n\geq 1, \quad x,y\in
X.
\]
Let us define the sequence of iterates
\begin{gather*}
x_{0} \in X, \\
x_{m} =\Phi (x_{m-1}),\quad m\geq 1.
\end{gather*}
Straightforward estimates lead us to
\[
| x_{m+p}-x_{m}| _{n}\leq \frac{L^{m}}{1-L}| x_{1}-x_{0}|
,\quad\forall m,p\geq 1.
\]
Hence we obtain that for all $\varepsilon >0$ and all $n$ there
exists $N=N(\varepsilon ,n)$ such that
\[
| x_{m+p}-x_{m}| _{n}<\varepsilon , \quad\forall p\geq 1,\;
\forall m\geq N,
\]
which means that $\{ x_{m}\} _{m\geq 0}$ is a Cauchy sequence.
Since $X$ is complete, $\{ x_{m}\} _{m\geq 0}$ is convergent. Then
$\xi :=\lim_{m\to \infty }x_{m}$ is a fixed point of $\Phi $. The
uniqueness of $\xi $ is proved by  contradiction.
\end{proof}

\section{The associated equation}

In \eqref{7}, we make the transformation $x=y+\xi ( t)$, where
$\xi $ is the function defined by Lemma \ref{Lemma1}. Then
\eqref{7} becomes
\begin{equation}
y=Ay+By+Cy,  \label{12}
\end{equation}
where
\begin{gather*}
(Ay)(t)=q(t)+f(t,y(t)+\xi (t))-\xi (t), \\
(By)(t)=\int_{0}^{t}V(t,s)[y(s)+\xi (s)] ds, \\
(Cy)(t)=\int_{0}^{t}G(t,s,y(s) +\xi (s))ds.
\end{gather*}
Obviously, if $y$ is a solution of \eqref{12}, then $x=y+\xi (t)$
is a solution of \eqref{7}, and conversely. The operators $A,B,C$
satisfy the following properties
\begin{gather}
| (Ay_{1})(t)-(Ay_{2})( t)| \leq L| y_{1}(t)-y_{2}(t) | ,\quad A
0=0,  \label{13}
\\
| (By_{1})(t)-(By_{2})( t)| \leq a(t)\int_{0}^{t}b(s)|
y_{1}(s)-y_{2}(s)| ds,  \label{14}
\\
| (Cy)(t)| \leq \int_{0}^{t}\omega (t,s)ds.  \label{15}
\end{gather}
We set $D=A+B$. Then we can state and prove the following useful
lemma.

\begin{lemma} \label{Lemma2}
The operators $C$ and $D$ satisfy the following properties:
\begin{enumerate}
\item  $C:X\to X$ is compact operator; \item There exists a
numerable set $\{ \delta _{n}\} _{n}$ such that $\delta _{n}\in [
0,1)$, for all $n\geq 1$ and for all $x,y\in X$ and for all $n\geq
1$,
\begin{equation}
\| Dx-Dy\| _{n}\leq \delta _{n}\| x-y\|_{n}. \label{16}
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof} (1) First we prove that $C:X\to X$ is
continuous. Let $y_{m},y\in X$ be such that $y_{m}\to y$ in $X$,
i.e. for all $\varepsilon >0$ and all $n\geq 1$ there exists
$N=N(\varepsilon ,n)$ such that
\[
| y_{m}-y| _{n}<\varepsilon , \quad\forall m\geq N.
\]
Let $n\geq 1$ be fixed; we have
\[
| (Cy_{m})(t)-(Cy)( t)| \leq \int_{0}^{t}| G(t,s,y_{m}(s) +\xi
(s))-G(t,s,y(s)+\xi ( s))| ds,
\]
and so, for $t\in [ 0,n] $, we get
\[
| (Cy_{m})(t)-(Cy) (t)| \leq \int_{0}^{n}| G( t,s,y_{m}(s)+\xi
(s))-G( t,s,y(s)+\xi (s))| ds.
\]
But the convergence of $\{ y_{m}\} _{m}$ and the continuity of
$\xi $ implies that there is a number $L_{n}>0$ such that
\[
| y_{m}(t)+\xi (t)| \leq L_{n},\quad | y(t)+\xi (t)| \leq
L_{n},\quad \forall t\in [ 0,n] ,\quad n\geq 1.
\]
Since the function $G$ is uniformly continuous on the compact set
\[
\big\{ (t,s,x)\in \mathbb{R}_{+}\times \mathbb{R}_{+}\times
\mathbb{R}^{d},\quad t,s\in [ 0,n] ,\quad| x| \leq L_{n}\big\} ,
\]
it follows that
\[
| G(t,s,y_{m}(s)+\xi (s)) -G(t,s,y(s)+\xi (s))| \leq
\frac{\varepsilon }{n},\quad\forall m\geq N.
\]
Then
\[
| Cy_{m}-Cy| _{n}\leq \varepsilon ,\quad \forall m\geq N,
\]
and the continuity of $C$ is proved.

It remains to show that $C$ maps bounded sets into compact sets.
Let $\mathcal{S}\subset C_{c}$ be bounded. We have to prove that
for each $n\geq 1$ the family $\{ Cy\big|_{[ 0,n] }: y\in
\mathcal{S}\} $ is uniformly bounded and equicontinuous.

Recall that $\mathcal{S}\subset C_{c}$ is bounded if and only if
for all $n$, there exists $p_{n}>0$ such that for all $x\in
\mathcal{S}$, $| x| _{n}\leq p_{n}$. Let $n\geq 1$ be arbitrary
but fixed. For $t\in [ 0,n] $, $y\in \mathcal{S}$, we have
\[
| (Cy)(t)| \leq \int_{0}^{t}| G(t,s,y(s)+\xi ( s))| ds\leq
\int_{0}^{t}\omega ( t,s)ds\leq n\omega _{n},
\]
where
\[
\omega _{n}:=\sup_{(t,s)\in \Delta _{n}}\{ \omega ( t,s)\} ,
\]
\begin{equation}
\Delta _{n}:=\{ (t,s)\in [ 0,n] \times [ 0,n] ,\; s\leq t\} .
\label{17}
\end{equation}
Hence the family $\{ Cy\big| _{[ 0,n] }: y\in \mathcal{S}\} $ is
uniformly bounded.

Let $y\in \mathcal{S}$, $t\in [ 0,n]$; therefore $G(t,s,y(s)+\xi
(s))$ is continuous and so $(Cy)(t)$ is a continuous function of
$t$. Let $\xi _{n}:=\sup_{t\in [ 0,n] }\{| \xi (t) |\}$.
  Now, $G(t,s,x)$ is uniformly continuous on
\[
\Omega _{n}:=\{ (t,s,x),\quad 0\leq s\leq t\leq n,\; | x| \leq
p_{n}+\xi _{n}\} .
\]
Hence, for each $\varepsilon >0$, there is a $\delta =\delta
(\varepsilon )>0$, such that if $(t_{i},s_{i},x_{i})\in
\Omega_{n}$, $i=\overline{1,2}$, then
\[
| (t_{1},s_{1},x_{1})-(t_{2},s_{2},x_{2}) | <\delta
\]
implies that
\[
| G(t_{1},s_{1},x_{1})-G(t_{2},s_{2},x_{2}) | <\varepsilon.
\]
For $y\in \mathcal{S}$ and $t_{1}$, $t_{2}\in [ 0,n] $ with
$|t_{1}-t_{2}| <\delta $, since $\delta$ can be chosen such that
$\delta \leq \varepsilon$, we have successively
\begin{align*}
| (Cy)(t_{1})-(Cy)( t_{2})| &\leq \int_{0}^{t_{1}}\big|
G(t_{1},s,y(s)+\xi (s))-G(t_{2},s,y(s)
+\xi(s))\big| ds \\
&\quad +\Big| \int_{t_{2}}^{t_{1}}G(t_{2},s,y(s)+\xi (s))ds\Big| \\
&\leq \varepsilon n+\delta M_{n}\leq \varepsilon (n+M_{n}),
\end{align*}
where $M_{n}:=\sup_{(t,s,x)\in \Omega _{n}}\{ | G(t,s,x)| \}$.

Hence the set $\{ Cy\big|_{[ 0,n] }: y\in \mathcal{S}\} $ is
equicontinuous.
  By Remark \ref{Remark1} we deduce
that $C$ is compact operator. \smallskip

\noindent (2) Let $n\geq 1$ be arbitrary but fixed.   Let $t\in [
0,\gamma _{n}] $ be arbitrary. Then we have
\begin{align*}
| (Dx)(t)-(Dy)(t) | &\leq L| x(t)-y(t)| +a( t)\int_{0}^{t}b(s)|
x(s)-y(s)| ds \\
&\leq (L+\gamma _{n}c_{n})| x-y| _{\gamma _{n}},
\end{align*}
where $c_{n}:=\sup_{(t,s)\in \Delta _{n}}\{ a( t)b(s)\} $, and
$\Delta _{n}$ is given by \eqref{17}. Therefore,
\begin{equation}
| Dx-Dy| _{\gamma _{n}}\leq (L+\gamma _{n}c_{n}) | x-y| _{\gamma
_{n}}.  \label{18}
\end{equation}
Let $t\in [ \gamma _{n},n] $ be arbitrary. Then we have
\begin{align*}
| (Dx)(t)-(Dy)(t) | &\leq L| x(t)-y(t)| +a( t)\int_{0}^{\gamma
_{n}}b(s)| x(s)-y(s)| ds \\
&\quad +a(t)\int_{\gamma _{n}}^{t}b(s)| x( s)-y(s)| ds.
\end{align*}
After easy computations, it follows that
\begin{align*}
&| (Dx)(t)-(Dy)(t) | e^{-h_{n}(t-\gamma _{n})}\\
&< L| x(t) -y(t)| e^{-h_{n}(t-\gamma _{n})} +\gamma _{n}c_{n}|
x-y| _{\gamma _{n}}+\frac{c_{n}}{h_{n}} | x-y| _{h_{n}}
\end{align*}
and therefore
\begin{equation}
| Dx-Dy| _{h_{n}}\leq (L+\frac{c_{n}}{h_{n}})| x-y|
_{h_{n}}+\gamma _{n}c_{n}| x-y| _{\gamma _{n}}. \label{19}
\end{equation}
By \eqref{18} and \eqref{19} we obtain
\begin{equation}
\| Dx-Dy\| _{n}\leq (L+2\gamma _{n}c_{n})| x-y| _{\gamma
_{n}}+(L+\frac{c_{n}}{h_{n}})| x-y| _{h_{n}}.  \label{20}
\end{equation}
Since $L<1$, for $\gamma _{n}\in (0,\frac{1-L}{2c_{n}})$ we deduce
that $L+2\gamma _{n}c_{n}<1$ and for $h_{n}>\frac{c_{n}}{1-L}$ we
deduce that $L+\frac{c_{n}}{h_{n}}<1$. Let $\delta _{n}:=\max
\{L+2\gamma _{n}c_{n},L+\frac{c_{n}}{h_{n}}\} $. It follows that
$\delta _{n}\in [ 0,1)$ and, since \eqref{20},
\[
\| Dx-Dy\| _{n}\leq \delta _{n}\| x-y\| _{n},\quad \forall x,y\in
X.
\]
The proof of Lemma \ref{Lemma2} is now complete.
\end{proof}

\begin{remark} \label{Remark2} \rm
Obviously, each operator $D$ which fulfills $(\ref{16})$ with
$\delta _{n}>0$, $\forall n\geq 1$ is continuous on $X$; if, in
addition, $\delta _{n}<1$, $\forall n\geq 1$, then $I-D$ is
invertible and $(I-D)^{-1}$ is continuous ($I$ denotes the
identity operator). The proof of this assertion is immediate and
it follows the classical model when $X$ is a Banach space and $D$
is a contraction.
\end{remark}

\section{Some remarks on Krasnoselskii's Theorem}

A well known result in nonlinear analysis is Krasnoselskii's
Theorem, which states as follows.

\begin{theorem}[Krasnoselskii \cite{K}, \cite{Ze})] \label{Theorem2}
Let $M$ be a non-empty bounded closed convex subset of a Banach
space $U$.  Suppose that $P:M\to U$ is  a contraction
 and $Q:M\to U$ is a compact operator. If $H:=P+Q$ has
the property $H(M)\subset M$, then $H$ admits fixed points in $M$.
\end{theorem}

Burton \cite{B} remarks that in practice it is difficult to check
condition $H(M)\subset M$ and he proposes to replace it by the
condition
\[
(x=Px+Qy,\quad y\in M)\Longrightarrow (x\in M).
\]
In another paper, \cite{BK}, Burton and Kirk give another variant
of Krasnoselskii's Theorem:

\begin{theorem}[Burton and Kirk, \cite{BK}] \label{Theorem3}
 Let $U$ be a Banach space, $P$, $Q:U\to U$, $P$ a contraction
with $\alpha <1$ and $Q$ a compact operator. Then either
\begin{itemize}
\item[(a)] $x=\lambda P(\frac{x}{\lambda })+\lambda Qx$ has a
solution for $\lambda =1$  or

\item[(b)] the set $\{ x\in U: x=\lambda P(\frac{x}{\lambda })
+\lambda Qx,\; \lambda \in (0,1)\} $ is unbounded.
\end{itemize}
\end{theorem}
This result has been generalized in \cite{A1}, obtaining the
following proposition.

\begin{proposition}\label{Proposition2}
Let $X$ be a Fr\'{e}chet space, $C,D:X\to X$ two operators. Admit
that:
\begin{itemize}
\item[(a)] $C$ is compact operator on $X$; \item[(b)] $D$ fulfills
condition \eqref{16} for a family of seminorms $| \cdot | _{n}$,
$n\geq 1$; \item[(c)] The following set is bounded
\begin{equation}
\{ x\in X,\quad x=\lambda D(\frac{x}{\lambda })+\lambda
Cx,\quad\lambda \in (0,1)\}\,.  \label{21}
\end{equation}
\end{itemize}
Then the operator $C+D$ admits fixed points.
\end{proposition}

The proof of this proposition is a consequence of Schaefer's
Theorem (\cite {Ze}).

\section{Existence result}

One can state and prove now an existence theorem for  \eqref{12}
(and so for \eqref{7}).

\begin{theorem} \label{Theorem4}
If hypotheses (H1)--(H3) are fulfilled, then \eqref{12} admits
solutions.
\end{theorem}

\begin{proof} We will use Proposition \ref{Proposition2}. Taking into
 account Lemma \ref{Lemma2}, it will be
sufficient to show that the set $(\ref{21})$ is bounded. We recall
a general result stating that if a set is bounded with respect to
a family of seminorms, then it will be bounded with respect to
every other equivalent family of seminorms.
  So, let $y\in X$, $y=\lambda D(\frac{y}{\lambda })+\lambda Cy$,
$\lambda \in (0,1)$. Then, since $\lambda <1$, from \eqref {13}
and hypotheses (H2) and (H3), we deduce successively
\begin{align*}
| y(t)| &=\Big| \lambda A(\frac{y}{\lambda
})(t)+\int_{0}^{t}V(t,s)y(s)ds
 \\
&\quad +\lambda \int_{0}^{t}V(t,s)\xi (s) ds+\lambda
\int_{0}^{t}G(t,s,y(s)+\xi (s)
)ds\Big| \\
&\leq L| y(t)| +a(t) \int_{0}^{t}b(s)| y(s)| ds+a(
t)\int_{0}^{t}b(s)| \xi (s) |ds+\int_{0}^{t}\omega (t,s)ds
\end{align*}
and so
\begin{equation}
| y(t)| \leq \frac{a(t)}{1-L} \int_{0}^{t}b(s)| y(s)|
ds+\frac{a(t)}{1-L}\int_{0}^{t}b(s)|\xi (s)|ds
+\frac{1}{1-L}\int_{0}^{t}\omega (t,s)ds.  \label{22}
\end{equation}
Let us denote
\begin{equation}
c(t):=\frac{a(t)}{1-L}\int_{0}^{t}b(s) |\xi (s)|
ds+\frac{1}{1-L}\int_{0}^{t}\omega ( t,s)ds.  \label{23}
\end{equation}
Then \eqref{22} becomes
\begin{equation}
| y(t)| \leq \frac{a(t)}{1-L}%
\int_{0}^{t}b(s)| y(s)| ds+c( t).  \label{24}
\end{equation}
We set
\[
w(t)=\int_{0}^{t}b(s)| y(s) | ds
\]
and, since \eqref{24}, we obtain
\begin{equation}
w(0)=0,\quad w'(t)=b(t) | y(t)| \leq \frac{a(t)b(t)
}{1-L}w(t)+b(t)c(t).  \label{25}
\end{equation}
By \eqref{25}, classical estimates lead us to conclude
\begin{equation}
\begin{aligned}
| y(t)| &\leq \frac{a(t) }{1-L}e^{ \frac{1}{1-L}\int_{0}^{t}a(s)b(
s)ds}\cdot \int_{0}^{t}e^{-\frac{1}{1-L}\int_{0}^{s}a( u)b(u)
du}b(s)c(s)ds+c(t) \\
&=:h(t),\quad\forall t\in \mathbb{R}_{+}.
\end{aligned} \label{26}
\end{equation}
Since $h$ is a continuous function, by \eqref{26} it follows that
\[
| y| _{n}\leq \sup_{t\in [ 0,n] }\{ h( t)\} ,
\]
which allows us to conclude that the set \eqref{21} is bounded and
so the proof of Theorem \ref{Theorem4} is complete.
\end{proof}

\section{Main result}

\begin{theorem}\label{Theorem5}
Assume hypotheses (H1)--(H3). If
\begin{equation}
\lim_{t\to \infty }h(t)=0  \label{27}
\end{equation}
then every solution $x(t)$ to  \eqref{7} is asymptotically stable
and
\[
\lim_{t\to \infty }| x(t)-\xi (t) | =0.
\]
\end{theorem}

\begin{proof} Let $x_{1},x_{2}$ be two solutions to \eqref{7}.
Then $y_{i}=x_{i}+\xi $, $i\in \overline{1,2}$ are solutions to
\eqref{12}. Similar estimates as in the proof of the boundedness
of the set \eqref{21} in Theorem \ref{Theorem4}, allow us to
conclude that
\[
| y_{i}(t)| \leq h(t),\quad \forall t\in \mathbb{R}_{+},\; \forall
i\in \overline{1,2}.
\]
Then, from \eqref{26}, for every $t\in \mathbb{R}_{+}$ we have
\[
| x_{1}(t)-x_{2}(t)| =| y_{1}(t)-y_{2}(t)| \leq 2h(t).
\]
Finally, by \eqref{27}, the conclusion follows.
\end{proof}

Next, we present an example when condition \eqref{27} holds.

\begin{remark}\label{Remark3} \rm
Let the following assumptions be fulfilled:
\begin{enumerate}
\item $\lim_{t\to \infty }a(t)=0$; \item $\int_{0}^{\infty
}b(t)dt<\infty $; \item $\int_{0}^{\infty }a(t)b(t)dt<\infty $;
\item $\int_{0}^{t}b(s)| \xi (s)| ds <\infty$; \item $\lim_{t\to
\infty }\int_{0}^{t}\omega (t,s) ds=0$.
\end{enumerate}
Then \eqref{27} holds. Indeed, since (2)--(5) and
\begin{align*}
&\exp\big(-\frac{1}{1-L}\int_{0}^{t}a(u)b(u)du\big) b(t)c(t)\\
&\leq \frac{a(t)b(t)}{1-L}\int_{0}^{t}b(s)|\xi (s)|ds
+\frac{b(t)}{1-L}\int_{0}^{t}\omega (t,s)ds,\quad \forall t\in
\mathbb{R}_{+},
\end{align*}
it follows that
\begin{equation}
\int_{0}^{\infty }
\exp\big(-\frac{1}{1-L}\int_{0}^{s}a(u)b(u)du\big)
b(s)c(s)ds<\infty .  \label{28}
\end{equation}
Then, from (1), (3), (4), (5), and \eqref{28}, we deduce that
\[
\lim_{t\to \infty }h(t)=0.
\]
\end{remark}

\begin{remark} \label{Remark4} \rm
Unlike \cite{BR}, under assumptions (1)--(5), the mapping $\xi $
is not necessarily bounded (see also Remark 4 in \cite{BZ}).
\end{remark}

\begin{remark} \label{Remark5} \rm
 If the mapping $a$ is decreasing, then hypothesis (3) follows from
hypothesis (2).
\end{remark}

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\end{document}