\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 $$x=Fx. \label{1}$$ \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 $$| x(t)-y(t)| \leq \varepsilon . \label{2}$$ \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 $$| (Fx)(t)-(Fy)(t) | \leq k| x(t)-y(t)| +a( t),\quad\forall t\in \mathbb{R}_{+},\; \forall x,y\in BC(\mathbb{R}_{+}). \label{3}$$ 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 $$x(t)=f(t,x(t))+\int_{0}^{t}u( t,s,x(s))ds,\quad t\in \mathbb{R}_{+} \label{63}$$ 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 $$\lim_{t\to \infty }(Bx)(t)=0, \label{4}$$ 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 $$| x| _{n}:=\sup_{t\in [ 0,n] }\{ | x( t)| \} ,\quad n\geq 1, \label{5}$$ becomes a Fr\'{e}chet space (i.e. a complete linear metrisable space). We will use in addition another family of seminorms, $$\| x\| _{n}:=| x| _{\gamma _{n}}+| x| _{h_{n}},\quad n\geq 1, \label{6}$$ 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 $$x(t)=q(t)+f(t,x(t)) +\int_{0}^{t}V(t,s)x(s)ds+\int_{0}^{t}G( t,s,x(s))ds,\,t\in \mathbb{R}_{+}, \label{7}$$ 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 $$x(t)=q(t)+f(t,x(t)),\quad t\in \mathbb{R}_{+}. \label{8}$$ \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 $$(\Phi x)(t)=q(t)+f(t,x(t) ),\quad x\in X, \quad t\in \mathbb{R}_{+}. \label{9}$$ 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 $$y=Ay+By+Cy, \label{12}$$ 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 $n\geq 1$, $$\| Dx-Dy\| _{n}\leq \delta _{n}\| x-y\|_{n}. \label{16}$$ \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 $\omega _{n}:=\sup_{(t,s)\in \Delta _{n}}\{ \omega ( t,s)\} ,$ $$\Delta _{n}:=\{ (t,s)\in [ 0,n] \times [ 0,n] ,\; s\leq t\} . \label{17}$$ 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, $$| Dx-Dy| _{\gamma _{n}}\leq (L+\gamma _{n}c_{n}) | x-y| _{\gamma _{n}}. \label{18}$$ 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 $$| Dx-Dy| _{h_{n}}\leq (L+\frac{c_{n}}{h_{n}})| x-y| _{h_{n}}+\gamma _{n}c_{n}| x-y| _{\gamma _{n}}. \label{19}$$ By \eqref{18} and \eqref{19} we obtain $$\| Dx-Dy\| _{n}\leq (L+2\gamma _{n}c_{n})| x-y| _{\gamma _{n}}+(L+\frac{c_{n}}{h_{n}})| x-y| _{h_{n}}. \label{20}$$ 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 $$\{ x\in X,\quad x=\lambda D(\frac{x}{\lambda })+\lambda Cx,\quad\lambda \in (0,1)\}\,. \label{21}$$ \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 $$| y(t)| \leq \frac{a(t)}{1-L} \int_{0}^{t}b(s)| y(s)| ds+\frac{a(t)}{1-L}\int_{0}^{t}b(s)|\xi (s)|ds +\frac{1}{1-L}\int_{0}^{t}\omega (t,s)ds. \label{22}$$ Let us denote $$c(t):=\frac{a(t)}{1-L}\int_{0}^{t}b(s) |\xi (s)| ds+\frac{1}{1-L}\int_{0}^{t}\omega ( t,s)ds. \label{23}$$ Then \eqref{22} becomes $$| y(t)| \leq \frac{a(t)}{1-L}% \int_{0}^{t}b(s)| y(s)| ds+c( t). \label{24}$$ We set $w(t)=\int_{0}^{t}b(s)| y(s) | ds$ and, since \eqref{24}, we obtain $$w(0)=0,\quad w'(t)=b(t) | y(t)| \leq \frac{a(t)b(t) }{1-L}w(t)+b(t)c(t). \label{25}$$ By \eqref{25}, classical estimates lead us to conclude \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} 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 $$\lim_{t\to \infty }h(t)=0 \label{27}$$ 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 $$\int_{0}^{\infty } \exp\big(-\frac{1}{1-L}\int_{0}^{s}a(u)b(u)du\big) b(s)c(s)ds<\infty . \label{28}$$ 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} \begin{thebibliography}{9} \bibitem{A1} C. 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