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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 35, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/35\hfil Krasnoselskii-type fixed point theorems]
{Krasnoselskii-type fixed point theorems under weak topology
 settings and applications}

\author[T. Xiang, R. Yuan\hfil EJDE-2010/35\hfilneg]
{Tian Xiang, Rong Yuan}  % in alphabetical order

\address{Tian Xiang \newline
School of Mathematical Sciences,
Beijing Normal University,
Laboratory of Mathematics and Complex Systems,
Ministry of Education, Beijing 100875, China}
\email{tianx@mail.bnu.edu.cn}

\address{Rong Yuan \newline
School of Mathematical Sciences,
Beijing Normal University,
Laboratory of Mathematics and Complex Systems,
Ministry of Education, Beijing 100875, China}
\email{ryuan@bnu.edu.cn}

\thanks{Submitted February 9, 2009. Published March 9, 2010.}
\thanks{Supported by the National Natural Science Foundation of China}
\subjclass[2000]{37C25, 47H10, 31B10}
\keywords{Expansive mapping; fixed point; weakly
continuous; integral equation}

\begin{abstract}
 In this article, we establish  some  fixed point results
 of Krasnoselskii type for the sum $T+S$, where  $S$ is
 weakly continuous and $T$ may not be continuous.
 Some of the main results complement and encompass  the
 previous ones. As an application, we study the existence
 of solution to one parameter operator equations.
 Finally, our results are used to prove the existence of
 solution for integral equations in reflexive Banach spaces.
\end{abstract}

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


\section{Introduction}

 Recently, more and more authors are interested in the study of
the existence of solutions of nonlinear abstract operator equation
or the fixed point of a sum of two  operators of the form
 \begin{equation} \label{1.1}
Sx+Tx=x, \quad x\in K,
\end{equation}
 where $K$ is  a closed and convex subset of a Banach space $E$.
The reason is  that, on
the one hand,  varieties of problems arising from the fields of
natural science, when modelled under the mathematical viewpoint,
involve the study of solutions of \eqref{1.1};  on the other hand,
several analysis and topological situations in the theory and
applications of  nonlinear operator lead also  to the investigation
of fixed point of \eqref{1.1}.  Especially, many problems in
integral equations can be formulated in terms of \eqref{1.1}.
Krasnoselskii's fixed point theorem appeared as a prototyped for
solving such equations. Motivated by an observation
that inversion of a perturbed differential operator may yield the
sum of a  contraction and  a compact operator,  Krasnoselskii \cite{Kra, Kras}
proved the following theorem.

\begin{theorem} \label{thmA}
 Let $K$ be a
nonempty closed convex subset of a Banach space $E$. Suppose that
$T$ and $S$ map $K$ into $E$ such that
\begin{itemize}
    \item [(i)] $S$ is continuous and $S(K)$ is contained in a
 compact subset of E;
    \item [(ii)]$T$ is a contraction with constant $\alpha<1$;
    \item [(iii)] Any $x,y\in K$ imply $Tx+Sy\in K$.
\end{itemize}
Then there  exists $x^*\in K $  with $Sx^*+Tx^*=x^*$.
\end{theorem}

Since the above theorem  was published there have appeared a huge number
of papers contributing generalizations or modifications of  the
Krasnoselskii's fixed point theorem and their applications,  see
\cite{Burton, Barroso, Barroso2, Cain, Kras, D. O'regan,  Park,
Taou, Vij, Xiang} and  the references therein.  Meanwhile, a large
class of problems, for instance in integral equations and stability
theory, have been adapted by the Krasnoselskii's fixed point method.
Several improvements of Theorem \ref{thmA}  have been made in the literature
in the course of time by modifying  assumption (i), (ii) or (iii).
For example,  see \cite{Burton,  Barroso, Xiang}.  It has been
mentioned in \cite{Burton} that the condition (iii) is too strong
and hence the author, Burton, proposed the following improvement for
\begin{itemize}
\item[(iii)] If $ x=Tx+Sy $ with $ y\in K $, then $x\in K$.
\end{itemize}
Subsequently, if $T$ is a
 bounded linear operator on $E$, in \cite{Barroso},  Barroso introduced
the following asymptotic requirement for (iii):
\begin{quote}
If $\lambda\in (0, 1)$  and $x=\lambda Tx+Sy$
for some $y\in K$,  then  $x\in K$.
\end{quote}
More recently, in \cite{Xiang}, the
authors firstly considered that the   map  $T$ is expansive  rather
than contractive, and then relaxed the compactness of the operator
$S$ by a $k$-set contractive assumption.


 Based on the well known fact that infinite dimensional Banach
spaces are not locally compact and some practical equations of the
form \eqref{1.1} may encounter the problem that the operators
involved may not be continuous, inspired by  papers \cite{Barroso2,
Xiang} and the mentioned works,  in this paper, we continue to
 study \eqref{1.1} in the setting of locally convex  (weak) topology.
We  should  mention that other authors have already studied
\eqref{1.1} in locally convex spaces \cite{Barroso2, Cain, Taou, Vij}.
As  the condition (i)  involves continuity and compactness, thus, we
would like to replace the continuity and the compactness by weakly
continuity and weakly compactness, respectively.

 The condition (ii) is also, in some sense, a little limited and
artificial,  since, on the one hand, this condition implies
norm-continuous;  on the other hand, why $T$ can  not be other type?
In this article we  consider a more general condition besides
contraction, which includes  expansive mapping and other type ones.
For example, see Theorems \ref{thm2.2} and \ref{thm2.3}, Corollaries
\ref{cor2}, \ref{cor2.4}, \ref{cor2.5} and \ref{cor2.6}.  We also
investigate some modifications on the condition (iii). The point of
this paper is that we  replace the contractiveness of $T$ by the
expansiveness of $T$ in the setting of weak topology  and derive
some new fixed point results, some of which complement and encompass
the corresponding results of \cite{Barroso, Barroso2, Taou}. Finally, a new
existence criterion  for integral equations in reflexive Banach spaces is
obtained.

  The remainder of this paper is organized as follows. In section
2, we state the main results and show their proofs. In section 3, we
apply these fixed point results to study the existence of  a
solution to one parameter operator equations of the form
$$
\lambda Tx+Sx=x, \quad \lambda\geq 0, \quad x\in E.
$$
To illustrate the theories,  our final purpose is  to prove the
existence of a solution to the following nonlinear  integral
equations of the form
\begin{equation}\label{1.2}
u(t)=f(u)+\int_0^T g(s, u(s))ds, \quad t\in [0, T],
\end{equation}
where $u$ takes values in a
reflexive Banach space $E$. By imposing some conditions on $f$ and
$g$ (see section 4), we are able to establish the existence of a
solution to \eqref{1.2}.

\section{Fixed point theorems for the sum of operators}

 At the beginning  we recall several  basic definitions and concepts
used further on.
Let $(E,\|\cdot\|)$ be a Banach space. For a sequence $\{x_n\}\subset E$
and $x\in E$ we write $x_n\to x$ whenever the sequence $\{x_n\}$
converges to $x$ (in the norm $\|\cdot\|$). If $\{x_n\}$ converges
weakly to $x$  we will write that
$x_n\rightharpoonup x$.

 Next, let $X\subset E$ be a nonempty set. An operator
$T: X\to E$ is said to be sequentially weakly continuous on the
set $X$ if for every sequence $\{x_n\}\subset X$ and $x\in X$
such that $x_n\rightharpoonup x$ we have that $Tx_n\rightharpoonup Tx$.

 Due to the Eberlin-\v{S}mulian's theorem (\cite[Theorem 8.12.4]{Ed}),
it is well known that if a set $K$ is weak compact, then
each sequentially weakly continuous mapping $T: K\to E$  is weakly
continuous. Therefore, it may be  possible to solve  equations of
the form \eqref{1.1} in the weak topology setting by suitable fixed
points results. As a tool to this intention, we rely on the
following version of Schauder fixed point principle which was
obtained by Arino, Gautier and Penot \cite{Arino}.

\begin{lemma} \label{lem1}
Let $K$ be a weakly  compact convex
subset of a Banach space $E$. Then each  sequentially  weakly
continuous map  $T: K\to K$ has a fixed point in $K$.
\end{lemma}

 Before stating the main results we need some  definitions and
lemmas.

\begin{definition}\label{def1} \rm
Let $(X,d)$ be a metric space and $M$ be a subset of
$X$. The mapping $T:M\to{X} $ is said to be expansive, if
there exists a constant $h>1$ such that
 \begin{equation}\label{2.1}
 d(Tx,Ty)\geq h d(x,y),\quad \forall x, y\in M.
 \end{equation}
\end{definition}
In the sequel, we shall employ the following two lemmas which have
been established  in \cite{Xiang}.

\begin{lemma}\label{lem2}
Let $M$  be a closed subset of a complete metric space
$X$. Assume that the mapping $T:M\to{X} $ is expansive and
$T(M)\supset M$, then there exists a unique point $x^*\in M $ such
that $Tx^*=x^*$.
\end{lemma}

\begin{lemma} \label{lem2.3}
 Let $(X,\|.\|)$ be a linear normed space, $M\subset X$. Suppose
that the mapping $T:M\to X $
is expansive with constant $h>1$. Then the inverse of
$F:=I-T:M\to (I-T)(M)$ exists and \begin{equation}\label{2.2}
 \|F^{-1}x-F^{-1}y\|\leq \frac{1}{h-1}\|x-y\|, x, y\in F(M).
\end{equation}
\end{lemma}

\begin{definition} \label{def2} \rm
Let $M$,  $K$  be two subsets of a  linear normed space
$X$, $T: M\to X$ and $S: K \to X$ two mappings. We
denote by  $\mathbb{F}=\mathbb{F}(M, K; T, S)$ the set
$$
\mathbb{F}=\big\{x\in M: x=Tx+Sy \text{  for some } y\in K\big\}.
$$
\end{definition}

 We are now ready to state  and prove the first main result of this
article.

\begin{theorem}\label{thm1}
 Let $K\subset E$ be a nonempty closed convex subset.
Suppose that $T$ and $S$ map $K$ into $E$ such that
\begin{itemize}
    \item [(i)] $S$ is  sequentially weakly  continuous;
    \item [(ii)]$T$ is an expansive mapping;
    \item[(iii)] $z\in S(K)$ implies $T(K)+z\supset K$, where
$T(K)+z=\{y+z \mid\ y \in     T(K)\}$;
     \item[(iv)] If $\{x_n\}$ is a sequence in
$\mathbb{F}(K, K;  T, S)$ such that
     $x_n\rightharpoonup x$  and $Tx_n \rightharpoonup y$, then $y=Tx$;
      \item[(v)] The set $\mathbb{F}(K, K;  T, S)$ is relatively
weakly compact.
\end{itemize}
Then there exists a point $x^*\in K $ with $Sx^*+Tx^*=x^*$.
\end{theorem}

\begin{proof}
From (ii) and (iii), for each  $ y\in K $, we see that the
mapping $ T+Sy:K\to E $ satisfies the assumptions of Lemma
\ref{lem2}. Therefore, the equation
\begin{equation}\label{2.3}
 Tx+Sy=x
\end{equation}
  has a unique solution
 $ x=\tau(Sy)\in K $,   so that the mapping $\tau S: K\to K$ given
by $y\to \tau Sy$ is well-defined.
In view of Lemma \ref{lem2.3}, we obtain that $\tau Sy= (I-T)^{-1}
Sy$ for all $y\in K$. In addition, we observe that $\tau S
(K)\subset \mathbb{F}\subset K$. We claim that $\tau S$ is sequentially
weakly continuous in $K$. To see this, let $\{x_n\}$ be a sequence
in $K$ with $x_n\rightharpoonup x$ in $K$. Notice that
$\tau S(x_n)\in \mathbb{F}$. Thus, up to a
subsequence, we may assume by (v) that $\tau S(x_n)\rightharpoonup
y$ for some $y\in K$.  It follows from (i) that
$Sx_n\rightharpoonup Sx$.  From the equality
\begin{equation}\label{2.4}
\tau S x_n=T(\tau Sx_n)+Sx_n,
\end{equation}
passing the  weak limit in \eqref{2.4} yields
$$
 T(\tau Sx_n)\rightharpoonup y-Sx.
$$
The assumption (iv) now implies that $y-Sx=Ty$; i.e.,
$y=\tau Sx$ since  $x\in K$.  This
proves the assertion. Let the set  $C=\overline{co}( \mathbb{F} )$,
where $ \overline{co}(\mathbb{F})$ denotes the closed convex hull of
$\mathbb{F}$. Then $C\subset K$ and is a weakly compact set by the
Krein-\v{S}mulian theorem. Furthermore, it is straightforward
to see that $\tau S$ maps $C$ into $C$.
In virtue of Lemma \ref{lem1}, there
exists $x^*\in C$ such that $\tau Sx^*=x^*$.  From \eqref{2.3} we
deduce that
$$
T(\tau Sx^*)+Sx^*=\tau Sx^*;
$$
that is,
$Tx^*+Sx^*=x^*$.
The proof is complete.
\end{proof}

 \begin{remark} \label{rmk2.1}  \rm
We note that $T$ may not be continuous since it is only expansive.
If $T:K \to E$ is a contraction,
 then a similar
 result can be found in \cite{Barroso2}.
Hence Theorem \ref{thm1} complements \cite[Theorem 2.9]{Barroso2}.
 The proof presented here are analogous to the arguments in \cite{Barroso2}.
\end{remark}

\begin{corollary} \label{cor1}
Under the conditions of Theorem \ref{thm1}, if only the condition
(iii) of Theorem \ref{thm1} is replaced by that $T$ maps $K$ onto
$E$,  then there exists a point $x^*\in K $ with $Sx^*+Tx^*=x^*$.
\end{corollary}

 It is worthy of pointing out that the condition (iii) may be a
litter restrictive and the next result might be regarded as an
improvement of Theorem \ref{thm1}.
\begin{theorem}\label{thm2.2} Let $K\subset E$ be a nonempty closed convex subset. Suppose that $T:E\to E$  and $S:K\to E$ such that
\begin{itemize}
    \item [(i)]$S$ is  sequentially weakly continuous;
    \item [(ii)] $T$ is an expansive mapping;
    \item[(iii)] $S(K)\subset (I-T)(E)$ and $[x=Tx+Sy, y\in K]\Longrightarrow x\in K$
    (or  $S(K)\subset (I-T)(K)$ );
     \item[(iv)]  If $\{x_n\}$ is a sequence in  $\mathbb{F}(E, K;  T, S)$ such that
     $x_n\rightharpoonup x$  and $Tx_n \rightharpoonup y$, then $y=Tx$;
      \item[(v)] The set  $\mathbb{F}(E, K; T, S)$ is relatively weakly compact.
\end{itemize}
Then there exists a point $x^*\in K $ with $Sx^*+Tx^*=x^*$.
\end{theorem}

\begin{proof}
For each $y\in K$, by (iii), there exists $x\in E $ such that
$ x-Tx=Sy$.
 By Lemma \ref{lem2.3} and the second part of (iii), we have $ x=(I-T)^{-1}Sy\in K$.
As is shown in Theorem \ref{thm1},   one obtains that $(I-T)^{-1}S:
K\to K$ is sequentially weakly continuous and  there is  a
point $x^*\in K $ with $x^*=(I-T)^{-1}Sx^*$. This completes the
proof.
\end{proof}

  Let us now state some  consequences of Theorem \ref{thm2.2}.
First, the case when $E$ is a reflexive Banach space is considered,
so that a closed, convex and bounded set is weakly compact.
Rechecking the proof of Theorem \ref{thm1},
 we find that it is only required
$ \overline{co}(\mathbb{F})$ to be weakly compact.

\begin{corollary} \label{cor2.2}
Suppose that the conditions (i)-(iv) of Theorem \ref{thm2.2} for $T$
and $S$ are fulfilled.  If $\mathbb{F}(E, K; T, S)$ is a bounded
subset of a reflexive    Banach space $E$, then $T+S$ has at least
one  fixed point in $K$.
\end{corollary}

 The second consequence of Theorem \ref{thm2.2} is  concerned
the case when  $T$ is non-contractive on $M\subset
E$, i.e., $\|Tx-Ty\|\geq\|x-y\|$ for all $x,y\in M$.

\begin{corollary} \label{cor2}
Let $K\subset E$ be a nonempty  convex and weakly compact subset.
Suppose that $T:E\to E$  and $S:K\to E$ are sequentially weakly
continuous  such that
\begin{itemize}
    \item [(i)] $T$ is non-contractive on $E$ (or $K$);
    \item[(ii)] There is a sequence $\lambda_n>1$ with $\lambda_n\to 1$ such that
    $S(K)\subset (I-\lambda_nT)(E)$ and $[x=\lambda_nTx+Sy, y\in K]\Longrightarrow x\in K$
    (or  $S(K)\subset (I-\lambda_nT)(K)$ ).
\end{itemize}
Then $T+S$ has a fixed point in $K$.
\end{corollary}

\begin{proof}
Notice that  $\lambda_nT:E\to E$ is expansive with
constant $\lambda_n>1$.  By Theorem \ref{thm2.2}, there exists
$x_n^*\in K$ such that \begin{equation}\label{2.5} Sx_n^*+\lambda_nTx_n^*=x_n^*.
\end{equation} Up to a subsequence we may assume that $x_n^*\rightharpoonup
x^*$ in $K$ since $K$ is convex and weakly compact. Passing the weak
limit in \eqref{2.5} we complete the proof.
\end{proof}

 Given by Lemma \ref{lem2.3}, Theorem \ref{thm2.2} and
\cite[Theorem 2.9]{Barroso2},
the following weak type Krasnoselskii fixed point theorem may be
easily formulated,
which clearly contains, but not limited to (see Remarks \ref{rem2.2}
and \ref{rem2.3}),  Theorem \ref{thm2.2}
and \cite[Theorem 2.9]{Barroso2}.

\begin{theorem}\label{thm2.00}
 Let $K\subset E$ be a nonempty closed convex subset. Suppose
that $T:E\to E$  and $S:K\to E$ such that
\begin{itemize}
    \item [(i)]$S$ is  sequentially weakly continuous;
    \item [(ii)] $(I-T)$ is one-to-one;
    \item[(iii)] $S(K)\subset (I-T)(E)$ and $[x=Tx+Sy, y\in K]\Longrightarrow x\in K$
    (or  $S(K)\subset (I-T)(K)$ );
     \item[(iv)]  If $\{x_n\}$ is a sequence in  $\mathbb{F}(E, K;  T, S)$ such that
     $x_n\rightharpoonup x$  and $Tx_n \rightharpoonup y$, then $y=Tx$;
      \item[(v)] The set  $\mathbb{F}(E, K; T, S)$ is relatively weakly compact.
\end{itemize}
Then there exists a point $x^*\in K $ with $Sx^*+Tx^*=x^*$.
\end{theorem}

\begin{remark}\label{rem2.2} \rm
If $T:E\to E$ is a contraction mapping, then $(I-T)(E)=E$ and hence
$S(K)\subset (I-T)(E)$. It can be easily seen by (ii) and (iii)
that $\mathbb{F}(E, K;  T, S)=(I-T)^{-1}S(K)$. It has been shown
under the assumptions of \cite{Taou} that $\mathbb{F}$
is relatively weakly compact. Therefore, Theorem \ref{thm2.00}
also encompasses the main result of \cite[Theorem 2.1]{Taou}.
Moreover, the condition (iv) is weaker than the condition that
$T$ is sequentially weakly continuous.
\end{remark}

 For a given  $r>0$, let  $B_r$ denote the set
$\{x\in E: \|x\|\leq r\}$.  Taking advantage of the linearity
of the operator $T$, we derive the following result.

\begin{theorem} \label{thm2.3}
Let $E$ be a reflexive Banach space,  $T:
E\to E$ a linear operator and  $S:E \to E$  a
sequentially weakly continuous map. Assume that  the following
conditions are satisfied.
\begin{itemize}
    \item [(i)] $(I-T)$ is continuously invertible;
    \item[(ii)]There exists  $R>0$ such that
$S(B_R)\subset B_{\beta R}$, where
 $\beta\leq  \|(I-T)^{-1}\|^{-1}$;
     \item[(iii)] $S(B_R)\subset (I- T)(E)$.
\end{itemize}
Then $T+S$ possesses a fixed point in $B_R$.
\end{theorem}

\begin{proof} Let
$F=I-T:E\to (I-T)(E)$. By  (i),  one can easily see from
the fact that $T$ is linear and
$\beta\leq   \|(I-T)^{-1}\|^{-1}$ that
\begin{equation}\label{2.6}
 \|F^{-1}x-F^{-1}y\|\leq \frac{1}{\beta}\|x-y\|,\quad
\forall x,y \in F(E).
\end{equation}
It follows from \eqref{2.6}  that
$F^{-1}: F(E)\to E$ is continuous. Recall that $F^{-1}$ being linear
implies that $F^{-1}$ is weakly continuous. Consequently,  one knows
from (iii) that $F^{-1}S: B_R\to E$ is sequentially weakly
continuous. For any $x\in B_R$, one easily derive from \eqref{2.6}
and (ii) that $ \|F^{-1}Sx\|\leq R$. Hence, $F^{-1}S$ maps $B_R$
into itself.  Applying  Lemma
\ref{2.1}, we obtain that $F^{-1}S$ has a fixed point in $B_R$.
This completes  the proof.
\end{proof}

 Next, we shall present some concrete mappings which fulfil the
condition (i) of Theorem \ref{thm2.3}. Before stating the
consequences,  we introduce the following two  lemmas.  The first
one is known, its proof can be directly shown or founded in
\cite{Xiang}.

\begin{lemma} \label{lem3.1}
 Let $(X,\|.\|)$ be a linear normed space, $M\subset X$. Assume
that the mapping $T:M\to X $
is contractive with constant $\alpha<1$, then the inverse of
$F:=I-T:M\to (I-T)(M)$ exists and
\begin{equation}\label{2.14}
 \|F^{-1}x-F^{-1}y\|\leq \frac{1}{1-\alpha}\|x-y\|, x, y\in F(M).
\end{equation}
\end{lemma}

The second one is as follows,  we shall
provide all the details for the sake of convenience.

\begin{lemma} \label{lem2.5}
Let $E$ be a Banach space.
Assume that  $T:E\to E $ is linear and bounded  and $T^p$ is
a  contraction for some $p\in \mathbb{N}$. Then $(I-T)$ maps $E$
onto $E$, the inverse of $F:=I-T:E\to E$ exists and
\begin{equation}\label{2.15}
 \|F^{-1}x-F^{-1}y\|\leq \gamma_p\|x-y\|, x, y\in E,
\end{equation}
where
$$
\gamma_p= \begin{cases}
\frac{p}{1-\|T^p\|}, & \text{if } \|T\|=1,\\
 \frac{1}{1-\|T\|}, &  \text{if  } \|T\|< 1,\\
\frac{\|T\|^p-1}{(1-\|T^p\|)(\|T\|-1)}, & \text{if  } \|T\|>1.
\end{cases}
$$
\end{lemma}

\begin{proof}
Let  $y\in E$ be fixed and define the map $T_y: E\to E$ by
$$
T_y x= Tx+y.
$$
We first show that $T_y^p$ is a  contraction. To this end,
let $x_1, x_2\in E$. Notice that $T$ is linear. One has
$$
\|T_yx_1-T_yx_2\|=\|Tx_1-Tx_2\|.
$$
Again
$$
\|T_y^2x_1-T_y^2x_2\|=\|T^2x_1-T^2x_2\|.
$$
By induction,
$$
\|T_y^px_1-T_y^px_2\|=\|T^px_1-T^px_2\| \leq \|T^p\|\|x_1-x_2\|.
$$
So $T_y^p$ is a contraction on $E$.
Next, we claim that both $(I-T)$  and $(I-T^p)$ map $E$ onto $E$.
Indeed,  by Banach contraction mapping principle, there is a unique
$x^*\in E$ such that $T_y^px^*=x^*$. It then follows that $T_yx^*$ is
also a fixed point of $T_y^p$. In view of uniqueness, we obtain that
$T_yx^*=x^*$ and $x^*$ is the unique fixed point of $T_y$. Hence, we
have
$$
(I-T)x^*=y,
$$
which implies that $(I-T)$ maps $E$ onto $E$. It is clear that
$(I-T^p)$ maps $E$ onto $E$.  The claim is proved.  Next, for each
$x,y\in E$ and $x\neq y$, one easily obtain that
$$
\|(I-T^p)x-(I-T^p)y\|\geq (1-\|T^p\|)\|x-y\|>0,
$$
which shows that $(I-T^p)$ is one-to-one.  Summing the above
arguments, we derive that  $(I-T^p)^{-1}$ exists on $E$. Therefore,
we infer that $(I-T)^{-1}$ exists on $E$ due to  the fact that
\begin{equation}\label{2.9}
(I-T)^{-1}=(I-T^p)^{-1}\sum_{k=0}^{p-1} T^k.
\end{equation}
Since  $T^p$ is a  contraction,  we know from \eqref{2.14} that
\begin{equation}\label{2.10}
\|(I-T^p)^{-1}\|\leq \frac{1}{1-\|T^p\|}.
\end{equation}
We conclude from Lemma \ref{lem3.1}, \eqref{2.9} and \eqref{2.10} that
\begin{equation}\label{2.11}
\|(I-T)^{-1}\|\leq \begin{cases}
\frac{p}{1-\|T^p\|}, &  \text{if } \|T\|=1,\\
\quad \frac{1}{1-\|T\|}, &  \text{if  } \|T\|< 1,\\
\frac{\|T\|^p-1}{(1-\|T^p\|)(\|T\|-1)},
& \text{if  } \|T\|>1.
\end{cases}
\end{equation}
This proves the lemma.
\end{proof}

Together  Lemmas \ref{lem2.3}, \ref{lem3.1},   \ref{lem2.5}  and
Theorem \ref{2.3} immediately yield the following results.

\begin{corollary}\label{cor2.4}
Let $E, S$ be the same as Theorem \ref{thm2.3}.
Assume that  $T: E\to E$ is a linear expansion with constant
$h>1$ such that $S(B_R)\subset B_{(h-1)R}$ for some $R>0$ and
$S(B_R)\subset (I- T)(E)$.
Then fixed point for $T+S$ is achieved  in $B_R$.
\end{corollary}

\begin{corollary} \label{cor2.5}
Let $E, S$ be the same as Theorem \ref{thm2.3}.
Assume that  $T: E\to E$ is a linear  contraction  with
constant $\alpha<1$ such that $S(B_R)\subset  B_{(1-\alpha)R}$
for some $R>0$. Then the equation  $Tx+Sx=x$ has at least one
solution in $B_R$.
\end{corollary}

\begin{corollary} \label{cor2.6}
Let $E, S$ be the same as Theorem \ref{thm2.3}.
Assume that  $T: E\to E$  is linear and bounded  and $T^p$
is a contraction for some $p\in \mathbb{N}$ such that $S(B_R)\subset
    B_{\gamma_p^{-1}R}$ for some $R>0$, where $\gamma_p$ is given
in Lemma \ref{lem2.5}.
 Then the equation  $Tx+Sx=x$ has at least one  solution in $B_R$.
\end{corollary}

 \begin{remark}\label{rem2.3}\rm
Given by Lemma \ref{lem2.5}, it is easily verified that,
under the conditions in \cite[Theorem 2.1]{Barroso}, all
the assumptions of Theorem \ref{thm2.00} are fulfilled.
Furthermore,  when $T\in \mathcal{L}(E)$ and  $\|T^p\|\leq 1$ for
some $p\geq 1$, instead of requiring  $
[ x= Tx+Sy, y\in K]\Longrightarrow x\in K$,  we assume the following
condition holds in Theorem \ref{thm2.00}.
$$
[\lambda\in (0,1) \text{ and } x=\lambda Tx+Sy, y\in K]
\Longrightarrow x\in K.
$$
Then Theorem \ref{thm2.00} also covers the main result
\cite[Theorem 2.2]{Barroso}.
 However, it does not necessarily require
that $T$ is linear in Theorem \ref{thm2.00}.
 \end{remark}

 Finally, inspired by the work of Barroso \cite{Barroso3}, we give
the following asymptotic version of the
 Krasnoselskii fixed point theorem.

 \begin{theorem}\label{thm2.000}
Let $K,  E, S, T $  and  the conditions
 (ii), (iii) and (v)  for $S$ and $T$ be the same as
Theorem \ref{thm2.00}. In addition, assume that the
following hypotheses are fulfilled.
\begin{itemize}
    \item [(a)] $S$ is   demicontinuous, that is,
if $\{x_n\}\subset K$ and $x_n\to x$ then $Sx_n\rightharpoonup Sx$;
    \item [(b)] $T$ is sequentially weakly continuous and
$T\theta=\theta$;
\end{itemize}
Then there exists a sequence $\{u_n\}$ in $K$ so that
$(u_n-(S+T)u_n)_n$ converges weakly to zero.
\end{theorem}

\begin{proof} Keeping the conditions (a) and (b) in mind, using the
essentially same reasoning as in Theorem \ref{thm1}, one can show
easily that $(I-T)^{-1}S: C\to C$ is demicontinuous, where
$C=\overline{co}( \mathbb{F} )$. Due to \cite[Theorem 3.3]{Barroso3}
there is a sequence $\{u_n\}$ in $C$ such that
$u_n-(I-T)^{-1}Su_n\rightharpoonup \theta$, i.e.,
$(I-T)^{-1}[u_n-(S+T)u_n]\rightharpoonup \theta$. Invoking again the
item (b), one can readily deduce that $u_n-(S+T)u_n\rightharpoonup
\theta$. This ends the proof.
\end{proof}

\section{Fixed point results to one parameter operator equation}

 Throughout this section, $E$ will denote  a reflexive Banach
space. The main purpose of this section is to present some existence
results for the following nonlinear abstract operator equation in
Banach spaces.
\begin{equation}\label{3.1}
\lambda Tx+Sx=x,
 \end{equation}
where $T, S :E \to E$ and $\lambda\geq 0$ is a parameter.
The first result concerning about \eqref{3.1} is as
 follows.
\begin{theorem} \label{thm3.1}
Let  $T$ and  $S$ map $ E$ into $E$  being
sequentially weakly continuous operators. Suppose that there exists
$\lambda_0>0$ such that
\begin{itemize}
  \item[(i)] $T$ is  expansive  with constant $h>1$ and
$S(B_R)\subset (I-\lambda T)(E)$ for all $\lambda\geq   \lambda_0$;
  \item [(ii)] $S(B_R)\subset \{x\in E: \|x+\lambda
T\theta\|\leq (\lambda h-1)R\}$ for
    some $R>0$ and all $\lambda\geq \lambda_0$.
\end{itemize}
Then  \eqref{3.1} is solvable  for all $\lambda\geq \lambda_0$.
\end{theorem}

\begin{proof} For each $\lambda\geq \lambda_0$, it follows from  the
first part of (i) that
$$
\|(I-\lambda T)x-(I-\lambda T)y\|\geq(\lambda h-1)\|x-y\|,
$$
which implies that
 \begin{equation}\label{3.2}
\|(I-\lambda T)x+\lambda T\theta\|\geq (\lambda  h-1)\|x\|.
\end{equation}
Assume now that  $x=\lambda Tx+Sy$ with $y\in B_R$, then it follows
from \eqref{3.2} and (ii) that
$$
(\lambda h-1)\|x\|\leq\|(I-\lambda T)x+\lambda T\theta\|
= \|Sy+\lambda T\theta\|\leq (\lambda h-1)R,
$$
Thus $x\in B_R$, and hence, since  $B_R$ is weakly compact,
it follows from Corollary \ref{cor2.2}  that $\lambda T+S$
has a fixed point in $B_R$. This completes the proof.
\end{proof}

\begin{remark} \label{rem3000} \rm
Particularly, in Theorem \ref{thm3.1}, if $T\theta=\theta$,
then  condition (ii) can be replaced by that
$S(B_R)\subset \{x\in E: \|x\|\leq (\lambda_0 h-1)R\}$ for some $R>0$.
And the result of Theorem \ref{thm3.1} also holds.
\end{remark}

\begin{corollary} \label{cor3.2}
Assume that the condition (ii) of Theorem
\ref{thm3.1} holds. In addition,   if $T$  is expansive and onto,
then \eqref{3.1} is solvable for all $\lambda\geq \lambda_0$.
\end{corollary}

 Next, we can modify some assumptions to study \eqref{3.1}.
Before proceeding to the theorem, we shall give a needed definition.

\begin{definition}\label{def3.1}\rm
 Let $(X,d)$ be a metric space and $M$ be a subset of
$X$. A mapping $T:M\to{X} $ is said to be weakly
expansive, if there exists a constant $\beta>0$ such that
 \begin{equation}\label{3.3}
 d(Tx,Ty)\geq \beta  d(x,y),\quad \forall x, y\in M.
 \end{equation}
\end{definition}

\begin{remark} \label{rmk3.1} \rm
Clearly, if $\beta>1$, then weakly expansive map is just an
expansive one.  If $T$ is weakly expansive and satisfies
similar assumptions as (i), (ii) in Theorem \ref{thm3.1},
then there exists $\lambda_1\geq \lambda_0$
such that \eqref{3.1} has a solution for $\lambda\geq \lambda_1$.
\end{remark}

 Our second result of this section is as follows.

\begin{theorem} \label{thm3.2}
Suppose  that  $T$ and $ S: E\to E$ are
sequentially weakly continuous operators such that
\begin{itemize}
    \item [(i)] $T$ is  weakly expansive  with constant $h>0$ and onto;
    \item [(ii)] $S(B_R)\subset B_R $ and $\|T\theta\|<h R$  for
    some $R>0$.
\end{itemize}
Then there exists $\lambda_0>0$ such that \eqref{3.1} is
solvable for all $\lambda\geq \lambda_0$.
\end{theorem}

\begin{proof}
We first choose $\lambda_1,\epsilon>0$ such  that
$\lambda_1 h>1$ and
\begin{equation}\label{3.4}
\frac{\lambda h}{1+\epsilon}>1,  \quad \text{ for all }\lambda\geq
\lambda_1.
\end{equation}
In view of $\|T\theta\|<h R$, for such a small $\epsilon$ there
exists  $\lambda_2>0$ such that
\begin{equation}\label{3.5}
\lambda (hR-\|T\theta\|)\geq 2(1+\epsilon)R, \quad
\text{ for all }\quad \lambda\geq \lambda_2.
\end{equation}
We define $T',S': E\to E$ by
$$
T'x=\frac{\lambda Tx}{1+\epsilon} \quad    \text{and}
\quad     S'y=\frac{Sy+\epsilon y}{1+\epsilon}.
$$
Then, $T', S'$ are sequentially weakly continuous,  $S'$
maps $B_R$ into itself,  and it is easy to see from \eqref{3.3} and
\eqref{3.4} that $T'$ is expansive with constant
$\lambda h/(1+\epsilon)>1$ for $\lambda\geq \lambda_0$, where
$\lambda_0=\max\{\lambda_1, \lambda_2\}$.
Together with the expression of $T'$ and (i), Lemma \ref{lem2}
tells us that $I-T'$ maps $E$ onto $E$.
Therefore $S'(B_R)\subset (I-T')(E)$.  Now, if $x=T'x+S'y$
with $y\in B_R$, then
\begin{equation}\label{3.6}
(I-\lambda T)x=Sy+\epsilon y-\epsilon x.
\end{equation}
 From \eqref{3.2},  \eqref{3.5} and \eqref{3.6},
we deduce that
$$
 [\lambda h-(1+\epsilon)]\|x\|
\leq R+\lambda\| T\theta\|+\epsilon R
\leq [\lambda h-(1+\epsilon)]R, \quad \text{for } \lambda\geq
\lambda_0.
$$
Hence, $x\in B_R$. Applying Theorem \ref{thm2.2}, we know that
$T'+S'$ has a fixed point in $B_R$. This completes the
proof.
\end{proof}

\begin{theorem}\label{thm3.3}
Let $T: E\to E$ be a linear weakly   expansive
mapping with constant $h>0$ and  $S:E \to E$  a bounded and
sequentially weakly continuous operator. If there exist $R>0$ and
$\lambda_0\geq 0$ such that $S(B_R)\subset (I-\lambda T)(E)$ for all
$\lambda\geq \lambda_0$,  then there exists $\lambda_1\geq
\lambda_0$ such that the equation $Sx+\lambda Tx=x$ is solvable  in
$B_R$ for all $\lambda\geq \lambda_1$.
\end{theorem}

\begin{proof} Choose $\lambda_1'\geq \lambda_0$ so that
$\lambda_1' h>1$.  Thus $\lambda T: E\to E$ is  expansive with
constant $\lambda h>1$ for all  $\lambda\geq \lambda_1'$.
Let $F_\lambda=I-\lambda T$. By Lemma \ref{lem2.3}, we know
that the inverse of $F_\lambda:E\to F_\lambda(E)$ exists and
\begin{equation}\label{3.7}
\|F_\lambda^{-1}(x)-F_\lambda^{-1}(y)\|\leq
\frac{1}{\lambda h-1}\|x-y\|, \quad \forall  x, y\in F_\lambda(E).
\end{equation}
It follows from \eqref{3.7} that $F_\lambda^{-1}$ is linearly
bounded. So $F_\lambda^{-1}$ is weakly continuous. Consequently, one
knows from the assumptions that  $F_\lambda^{-1}S: B_R\to E$
is sequentially weakly continuous. There is
$\lambda_1\geq \lambda_1'$ such that $\|Sx\|\leq (\lambda h-1)R$
for all $x\in B_R$ and $\lambda\geq \lambda_1$ since $S$ is bounded.
Thus, we deduce from \eqref{3.7} that
\begin{equation}\label{3.8}
\|F_\lambda^{-1}Sx\|\leq R, \quad \text{for all } x\in B_R \quad
\text{and  }\lambda \geq \lambda_1.
\end{equation}
It follows from \eqref{3.8} that $F_\lambda^{-1}S$
maps $B_R$ into itself. Using Lemma \ref{2.1}, we obtain that
$F_\lambda^{-1}S$ has a fixed point
in $B_R$ for all $\lambda\geq \lambda_1$. The proof is complete.
\end{proof}

 As for the Lipschitzian  mapping, by analogous argument,
we derive the following result.

 \begin{theorem} \label{thm3.4}
Let $T: E\to E$ be a  bounded linear operator
  and  $S:E \to E$ a  sequentially weakly
continuous operator. Suppose that  there exist $R>0$ and
$\lambda_0\geq0$ such that
$S(B_R)\subset B_{(1-\lambda_0\|T\|)R}$.
Then there exists $\lambda_1\in [0, \lambda_0]$ such that
\eqref{3.1} has at least one solution in $B_R$ for all
$\lambda\in[0, \lambda_1]$.
\end{theorem}

 \begin{proof} For each $\lambda\in[0,
\lambda_0]$, we have  $\lambda \|T\|<1$ and hence $\lambda T:
E\to E$ is  a contraction   with constant $\lambda \|T\|<1$.
Let $F_\lambda=I-\lambda T$.  One easily know from Lemma
\ref{lem3.1} that the inverse of $F_\lambda:E\to E$ exists
and
\begin{equation}\label{3.9} \|F_\lambda^{-1}(x)-F_\lambda^{-1}(y)\|\leq
\frac{1}{1-\lambda \|T\|}\|x-y\|, \quad \forall  x, y\in E.
\end{equation}
One readily  sees from \eqref{3.9} that $F_\lambda^{-1}$ is weakly
continuous. Therefore, $F_\lambda^{-1}S: E\to E$ is
sequentially weakly continuous.
  One can obtain from the hypothesis  that
$\|Sx\|\leq (1-\lambda \|T\|)R$ for all $x\in B_R$ and
$\lambda\in[0, \lambda_0]$. We
derive from \eqref{3.9} that
\begin{equation}\label{3.10}\|F_\lambda^{-1}Sx\|\leq
R, \quad \text{for all } x\in B_R \quad \text{and  }\lambda\in[0,
\lambda_0].
\end{equation}
It follows from \eqref{3.10} that $F_\lambda^{-1}S$
maps $B_R$ into itself. Invoking Lemma \ref{2.1}, we infer  that
$F_\lambda^{-1}S$ has a fixed point in $B_R$ for $\lambda\in[0,
\lambda_1]$. The proof is complete.
\end{proof}

\begin{corollary} \label{coro3.2}
Let $T, S$ be the same as Theorem \ref{3.4}.
Suppose that for each $x\in E$, we have $\|Sx\|\leq
a\|x\|^p+b$, where $b\geq 0$, $0<p\leq 1$, $a\in [0, 1)$ if $p=1$;
$a\geq 0$ if $0<p<1$. Then there exists $\lambda_1\in [0,
\lambda_0]$ such that \eqref{3.1} is solvable in $B_R$ for all
$\lambda\in[0, \lambda_1]$.
\end{corollary}

\begin{proof}
For the  case that  $p=1$,  since $0\leq a<1$, there is
$\lambda_0\geq 0$ such that $a<1-\lambda_0 \|T\|$. Obviously, there
exists sufficiently large $R>0$ such that
\begin{equation}\label{3.11}
 \frac{b}{R}\leq (1-\lambda_0 \|T\|-a).
\end{equation}
It follows from  \eqref{3.11} and the hypothesis  that the
conditions  of
Theorem \ref{thm3.4} is satisfied.

Next, for the case that  $p\in (0, 1)$,  it suffices to choose
$\lambda_0\geq 0$ with $\lambda_0\|T\|<1$ and $R>0$ such that
$aR^p+b\leq (1-\lambda_0\|T\|)R$. This is obvious.
\end{proof}

\begin{remark} \label{rmk3.3}\rm
The fixed point results of section 2 can be applied to
study  the eigenvalue  problems  of  Krasnosel'skii-type in
the critical case, that is,  the map $T:M\subset E\to E$ is
non-expansive.
Their arguments are fully analogous to the discission presented in
this  section. Hence we omit it.
\end{remark}

\section{Application to integral equation }

 In this section, our aim is to present some existence results
for the  nonlinear integral equation
\begin{equation}\label{4.1}
u(t)=f(u)+\int_0^T g(s, u(s))ds, \quad u\in C(J, E),
\end{equation}
where $E$ is a reflexive Banach space and  $J=[0, T]$.
The integral in \eqref{4.1} is understood to be the Pettis integral.
To study \eqref{4.1}, we assume for the remained of this section  the
following  hypotheses are satisfied:
\begin{itemize}
    \item [(H1)] $f:E\to E$ is  sequentially weakly continuous and onto;

    \item [(H2)] $ \|f(x)-f(y)\|\geq h\|x-y\|$, ($h\geq 2$)
for all $x, y\in E$;  and $f$ maps relatively weakly compact
sets  into bounded sets and is uniformly continuous on
weakly compact sets;

    \item[(H3)] for any $t\in J$, the map $g_t=g(t, \cdot): E\to E$
is  sequentially weakly continuous;

     \item[(H4)] for each $x\in C(J, E)$, $g(\cdot, x(\cdot))$
is Pettis integrable on $[0, T]$;

      \item[(H5)] there exist  $\alpha\in L^1[0, T]$ and a nondecreasing
continuous function  $\phi$ from $[0, \infty)$ to $(0, \infty)$
such that $\|g(t, x)\|\leq \alpha(t)\phi(\|x\|)$ for
a.e. $t\in [0, T]$ and all $x\in E$. Further,   assume that
$\int_0^T\alpha(s)ds<\int_{\|f(\theta)\|}^\infty\frac{dr}{\phi(r)}$.
\end{itemize}
We now state and prove an existence principle  for \eqref{4.1}.

\begin{theorem} \label{thm4.1}
Suppose that the conditions (H1)-(H5) are fulfilled.
Then \eqref{4.1} has at least one solution $u\in C(J, E)$.
\end{theorem}

\begin{proof} Put
$$
\beta(t)=\int_{\|f(\theta)\|}^t\frac{dr}{\phi(r)} \quad
 \text{and} \quad
b(t)=(h-1)^{-1}\beta^{-1}\big(\int_0^t\alpha(s)ds\big).
$$
Then
\begin{equation}\label{4.00}
\int_{\|f(\theta)\|}^{(h-1)b(t)}\frac{dr}{\phi(r)}=\int_0^t\alpha(s)ds.
\end{equation}
It follows from \eqref{4.00} and the final part of (H5) that
$b(T)<\infty$.  We  define the set
$$
K=\big\{x\in C(J, E): \|x(t)\|\leq (h-1)b(t) \text{ for all } t\in J \big\}.
 $$
Then $K$ is a closed, convex and bounded subset of $C(J, E)$. Let us
now introduce the nonlinear operators $T$ and $S$ as follows:
\begin{gather*}
(Tx)(t)=f(x(t))-f(\theta), \\
(Sy)(t)=f(\theta)+\int_0^tg(s, y(s))ds.
\end{gather*}
The conditions  (H1) and (H4) imply that $T$ and $S$ are well
defined on $C(J,E)$, respectively.

  Our idea is to use Theorem \ref{thm2.2}  to find the
fixed point for the sum $T+S$ in $K$. The proof will be shown
 in several steps.

\noindent\textbf{Step 1:} Prove that $S$ maps $K$ into $K$,  $S(K)$ is
equicontinuous and relatively weakly compact.

 For any  $y\in K$, we shall show that $Sy\in K$. Let  $t\in J$
be fixed. Without loss of generality, we may assume that
$(Sy)(t)\neq 0$. In view of the Hahn-Banach theorem there exists
$y_t^*\in E^*$ with $\|y_t^*\|=1$ such that $\langle y_t^*,
(Sy)(t)\rangle=\|(Sy)(t)\|$. Thus, one can deduce  from  (H5) and
\eqref{4.00} that
\begin{equation}\label{4.2}
\begin{aligned}
\|(Sy)(t)\|&=\langle y_t^*,
f(\theta)\rangle+\int_0^t\langle y_t^*,g(s, y(s))\rangle ds\\
&\leq \|f(\theta)\|
+\int_0^t\alpha(s)\phi(\|y(s)\|)ds\\
&\leq \|f(\theta)\|+\int_0^t\alpha(s)\phi((h-1)b(s))ds\\
&= \|f(\theta)\|+(h-1)\int_0^tb'(s)ds=(h-1)b(t).
\end{aligned}
\end{equation}
It shows from \eqref{4.2} that $S(K)\subset K$ and hence is bounded.
This proves the first claim of Step 1. Next, let  $t, s\in J$ with
$s\neq t$. We may
assume that $(Sy)(t)-(Sy)(s)\neq0$. Then there exists $x_t^*\in E^*$
with $\|x_t^*\|=1$ and $\langle x_t^*,
(Sy)(t)-(Sy)(s)\rangle=\|(Sy)(t)-(Sy)(s)\|$. Consequently,
\begin{equation}\label{4.3}
\begin{aligned}
 \|(Sy)(t)-(Sy)(s)\|
&\leq \int_s^t\alpha(\tau)\phi(\|y(\tau)\|)d\tau\\
&\leq \int_s^t\alpha(\tau)\phi((h-1)b(\tau))d\tau \\
&\leq (h-1)\big|\int_s^tb'(\tau)d\tau\big|=(h-1)|b(t)-b(s)|.
\end{aligned}
\end{equation}
It follows from \eqref{4.3} that $S(K)$ is equicontinuous.
The reflexiveness of $E$ implies that $S(K)(t)$ is relatively weakly
compact for each $t\in J$, where $S(K)(t)=\{z(t): z\in S(K)\}$.
By a known result (see \cite{Ku,Mit}), one can easily get
that $S(K)$ is relatively weakly  compact in $C(J, E)$.
This  completes  Step 1.

\noindent\textbf{Step 2:}
 Prove that $S:K\to K$ is sequentially weakly continuous.
Let $\{x_n\}$ be a sequence in $K$ with $x_n\rightharpoonup x$
in $C(J, E)$, for some $x\in K$.  Then $x_n(t)\rightharpoonup x(t)$
in $E$ for all $t\in J$. Fix $t\in (0, T]$.  From the item (H3) one
sees that $g(t, x_n(t))\rightharpoonup g(t, x(t))$ in $E$. Together
with (H5) and the Lebesgue dominated convergence theorem for the
Pettis integral  yield for each $\varphi \in E^*$ that
$$
\langle \varphi, (Sx_n)(t)\rangle \to \langle \varphi, (Sx)(t)\rangle;
$$
i.e., $(Sx_n)(t)\rightharpoonup(Sx)(t)$ in $E$.
We can do this for each $t\in J$ and notice that $S(K)$ is
equicontinuous, and accordingly $Sx_n\rightharpoonup Sx$ by \cite{Mit}.
The Step 2 is proved.

\noindent\textbf{Step 3:} Prove that the conditions (ii) and
(iii) of Theorem \ref{thm2.2} hold.
 Since  $E$ is reflexive and $f$ is continuous on  weakly compact sets,
it shows that $T$ transforms $C(J,E)$ into itself. This,
in conjunction  with the first part of (H2), one easily gets
that $T: C(J, E)\to C(J, E)$ is expansive with constant $h\geq 2$.
  For all $x, y\in C(J, E)$, one can  see   from the first part
of (H2) that
$$
\|(I-T)x(t)-(I-T)y(t)\|\geq (h-1)\|x(t)-y(t)\|\geq \|x(t)-y(t)\|,
$$
where $I$ is identity map.  Thus, one has
\begin{equation}\label{4.4}
\|(I-T)x(t)\|\geq (h-1)\|x(t)\|\geq\|x(t)\|, \quad \forall x\in C(J,E).
\end{equation}
Assume now that $x=Tx+Sy $ for some $y\in K$. We conclude
from \eqref{4.2} and \eqref{4.4} that
$$
\|x(t)\|\leq\|(I-T)x(t)\|=\|(Sy)(t)\|\leq (h-1)b(t),
$$
which shows that $x\in K$.  Therefore, the second part of (iii) in
Theorem \ref{thm2.2} is fulfilled.  Next, for each $y\in C(J, E)$,
we define $T_y:C(J, E)\to C(J, E)$ by
$$
(T_yx)(t)=(Tx)(t)+y(t).
$$
Then $T_y$ is expansive with constant $h\geq 2$ and onto since $f$
maps $E$ onto $E$. By Lemma \ref{lem2}, we know there exists
$x^*\in C(J, E)$ such that $T_yx^*=x^*$, that is $ (I-T)x^*=y$. Hence
$S(K)\subset (I-T)(E)$.  This completes  Step 3.

\noindent\textbf{Step 4:} Prove that the condition (v) of
Theorem \ref{thm2.2} is satisfied.
 For each $x\in \mathbb{F}(E, K; T, S)$, then by the definition
of $\mathbb{F}$ and Lemma \ref{lem2.3} there exists $y\in K$ such that
\begin{equation}\label{9999}
x=(I-T)^{-1}Sy.
\end{equation}
Hence, for  $t, s\in J$, we obtain from Lemma \ref{lem2.3},
\eqref{9999} and \eqref{4.3}  that
$$
\|x(t)-x(s)\|\leq |b(t)-b(s)|,
$$
which illustrates that $\mathbb{F}(E, K; T, S)$ is equicontinuous in
$C(J, E)$. Let $\{x_n\}$ be a sequence in $\mathbb{F}$. Then $\{x_n\}$
is equicontinuous in $C(J, E)$ and there exists $\{y_n\}$ in $K$
with $x_n=Tx_n+Sy_n$. Thus, one has from \eqref{4.2} and \eqref{4.4}
that
$$
\|x_n(t)\|\leq \frac{1}{h-1}\|(Sy_n)(t)\|\leq b(t), \forall t\in J.
$$
It follows that, for each $t\in J$, the set $\{x_n(t)\}$ is
relatively weakly compact in $E$. The above discussion tells us  that $\{x_n: n\in \mathbb{N}\}$ is relatively
 weakly compact. The Eberlein-\v{S}mulian theorem implies
that $\mathbb{F}$ is relatively  weakly compact.  This achieves Step 4

\noindent\textbf{Step 5:} Prove that $T$ fulfils the condition (iv)
of Theorem \ref{2.2}.
 By  the second part of (H2) and the fact that $\mathbb{F}$ is
relatively weakly compact we obtain that $T(\mathbb{F})$ is bounded.
Again by the second part of (H2) and the fact that  $\mathbb{F}$ is
equicontinuous, one can readily deduce that  $T(\mathbb{F})$ is also
equicontinuous.  Now, let $\{x_n\}\subset \mathbb{F}$ with
$x_n\rightharpoonup x$ in  $C(J, E)$ for some $x\in K$. It follows
from (H1) that $(Tx_n)(t)\rightharpoonup (Tx)(t)$. Since $\{Tx_n:
n\in \mathbb{N}\}$ is
 equicontinuous in $C(J, E)$, as before, we conclude  that
$Tx_n\rightharpoonup Tx$ in $C(J, E)$. The Step 5 is proved.

 Now, invoking Theorem \ref{thm2.2} we obtain that there is
$x^*\in  K$ with $Tx^*+Sx^*=x^*$; i.e., $x^*$ is a solution to
\eqref{4.1}. This accomplishes the proof.
\end{proof}


\begin{remark}\label{rem4.1}\rm
 It is clearly seen that the following locally
``Lipscitizan'' type condition fulfills the second part of (H2): For
each bounded subset $U$ of $E$, there exists a continuous function
$\psi_U: \mathbb{R}^{+}\to \mathbb{R}^{+}$ with $\psi_U(0)=
0$, such that $\|f(x)-f(y)\|\leq \psi_U(\|x-y\|)$ for all $x, y \in
U$.  Although the proof of Theorem \ref{thm4.1} is analogous to that
of \cite[Theorem 5.1]{Barroso2}, it clarifies some vague points
made in \cite{Barroso2}. Moreover, it can be easily known that
Theorem \ref{thm4.1} does not contain  the corresponding result of
\cite[Theorem 5.1]{Barroso2}, vice versa. Therefore,  Theorem
\ref{thm4.1}  and \cite[Theorem 5.1]{Barroso2}  are
complementary.
\end{remark}

  We conclude this artcile by  presenting  a class of  maps
 which fulfil the assumptions (H1) and (H2) in Theorem
  \ref{thm4.1}. Assume  that  $f: \mathbb{R}^n\to  \mathbb{R}^n$
  is continuous and  is coercive, i.e., $f$ satisfies the
   inequality
  $$
(f(x)-f(y), x-y)\geq \alpha(\|x-y\|)\|x-y\|, \quad \forall x,
y\in\mathbb{R}^n,
  $$
where $\alpha(0)=0, \alpha(t)>0$ for all $t>0$;  $\lim_{t\to
\infty}\alpha(t)=\infty$.  Then it is well known that $f$ is
surjective. Particularly, if $\alpha(t)=ht$, $h>0$, then $f:
\mathbb{R}^n\to  \mathbb{R}^n$ is a homeomorphism. Specifically,
let us  consider   the function $f: \mathbb{R}\to \mathbb{R}$
defined by $f(x)= x^k+hx+x_0$, where $h\geq 2$, $k$ is a positive
odd number,  and $x_0$ is a given constant.
Then $f$ satisfies the assumptions (H1) and (H2) in
Theorem  \ref{thm4.1} for $E=\mathbb{R}$.


\subsection*{Acknowledgments}
The authors  are very grateful to the anonymous
referee who carefully read the manuscript, pointed out
a misinformation and several vague points, and gave us valuable
comments, and suggestions for improving the exposition.
In particular, the authors are thankful to the
referee for suggesting Theorems \ref{thm1} and \ref{thm2.000}
and for pointing out the references \cite{Barroso3} and \cite{Taou}.

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\end{document}