\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 22, pp. 1--12.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2008/22\hfil Compression-expansion theorem]
{Functional compression-expansion fixed point theorem}
\author[R. Avery, J. Henderson, D. O'regan\hfil EJDE-2008/22\hfilneg]
{Richard Avery, Johnny Henderson, Donal O'regan} % in alphabetical order
\address{Richard Avery \newline
College of Arts and Sciences, Dakota State University, Madison,
South Dakota 57042, USA}
\email{rich.avery@dsu.edu}
\address{Johnny Henderson \newline
Department of Mathematics, Baylor University,
Waco, Texas 76798, USA}
\email{Johnny\_Henderson@baylor.edu}
\address{Donal O'regan \newline
Department of Mathematics, National University of Ireland,
Galway, Ireland}
\email{donal.oregan@nuigalway.ie}
\thanks{Submitted January 31, 2008. Published February 22, 2008.}
\subjclass[2000]{47H10}
\keywords{Fixed-point theorems; cone; positive solutions; multivalued maps}
\begin{abstract}
This paper presents a generalization of the fixed point theorems
of compression and expansion of functional type. As an
application, the existence of a positive solution to a second
order conjugate boundary value problem is considered. We conclude
with an extension to multivalued maps.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\section{Introduction}
In this paper we provide a generalization of all the fixed point
theorems of compression and expansion involving functionals. The
generalization also gives an alternative, simpler argument of the
Sun-Zhang fixed point theorem of cone compression-expansion of
functional type which applies convex functionals. The use of
functionals with the fixed point index to yield positive solutions
can be traced back to Leggett and Williams \cite{leg}. Several
multiple fixed point theorems have employed the use of functionals
\cite{five,ah1,ah2,pete} and most recently they have been used
\cite{aa,sun} to verify the existence of at least one fixed point.
In \cite{aa} Anderson and Avery generalized the fixed point
theorem of Guo \cite{guo1} by replacing the norm in places by
sublinear functionals, and in \cite{sun} Sun and Zhang showed
that a certain set was a retract, thus replacing the norm from the
argument by a convex functional. In this paper we provide a
generalization of all of the compression-expansion arguments that
have utilized the norm and/or functionals (including
\cite{aa,guo1,guo2,kr,sun}) in verifying the existence of at least
one fixed point. Our result does not require sets to be invariant
under our operator and yet maintains the freedom gained by using
functionals that satisfy either Property A1 or Property A2,
properties defined in the next section. We will
follow the main result with an application and associated example,
and then we will conclude with an extension to multivalued maps.
\section{Preliminaries}
In this section we will state the definitions that are used in the
remainder of the paper.
\begin{definition} \label{def1} \rm
Let $E$ be a real Banach space. A nonempty closed
convex set $P \subset E$ is called a \emph{cone} if it satisfies the
following two conditions:
\begin{itemize}
\item[(i)] $x \in P, \lambda \geq 0$ implies $\lambda x \in P$;
\item[(ii)] $x \in P, -x \in P$ implies $x = 0$.
\end{itemize}
\end{definition}
Every cone $P \subset E$ induces an ordering in $E$ given by
$$
x \leq y \mbox{ if and only if } y - x \in P.
$$
\begin{definition} \label{def2} \rm
An operator is called completely continuous if it is
continuous and maps bounded sets into precompact sets.
\end{definition}
\begin{definition} \label{def3} \rm
A map $\alpha$ is said to be a nonnegative continuous
concave functional on a cone $P$ of a real Banach space $E$ if
$\alpha : P \to [0,\infty)$
is continuous and
$$
\alpha(tx + (1-t)y) \geq t\alpha(x) + (1-t)\alpha(y)
$$
for all $x,y \in P$ and $t \in [0,1]$. Similarly we say the map
$\beta$ is a nonnegative continuous convex functional on a cone $P$ of a
real Banach space $E$ if
$\beta : P \to [0,\infty)$
is continuous and
$$
\beta(tx + (1-t)y) \leq t\beta(x) + (1-t)\beta(y)
$$
for all $x,y \in P$ and $t \in [0,1]$.
We say the map $\psi$ is a sub-linear functional if
$$
\psi(tx)\leq t \psi(x) \;\; \hbox{for all} x \in P,\; t \in [0,1].
$$
\end{definition}
\subsection*{Property A1}
Let $P$ be a cone in a real Banach space $E$ and
$\Omega$ be a bounded open subset of $E$ with $0 \in \Omega$.
Then a continuous functional $\beta : P \to [0,\infty)$ is said
to satisfy Property A1 if one of the following conditions hold:
\begin{itemize}
\item[(i)] $\beta$ is convex, $\beta(0) = 0$, $\beta(x) \neq 0$ if $x \neq 0$, and $\displaystyle{\inf_{x \in P \cap \partial \Omega} \beta(x) > 0}$,
\item[(ii)] $\beta$ is sublinear, $\beta(0) = 0$, $\beta(x) \neq 0$ if $x \neq 0$, and $\displaystyle{\inf_{x \in P \cap \partial \Omega} \beta(x) > 0}$,
\item[(iii)] $\beta$ is concave and unbounded.
\end{itemize}
Note that if condition (i) of Property A1 is satisfied so
is condition (ii), since
$$
\beta(tx) = \beta(tx + (1-t)0) \geq t\beta(x) + (1-t)\beta(0) = t\beta(x).
$$
\subsection*{Property A2}
Let $P$ be a cone in a real Banach space $E$ and
$\Omega$ be a bounded open subset of $E$ with $0 \in \Omega$.
Then a continuous functional $\beta : P \to [0,\infty)$
is said to satisfy Property A2 if one of the following conditions hold:
\begin{itemize}
\item[(i)] $\beta$ is convex, $\beta(0) = 0$ and $\beta(x) \neq 0$ if
$x \neq 0$,
\item[(ii)] $\beta$ is sublinear, $\beta(0) = 0$ and $\beta(x) \neq 0$
if $x \neq 0$,
\item[(iii)] $\beta(x+y) \geq \beta(x) + \beta(y)$ for all $x,y \in P$,
$\beta(0) = 0$, $\beta(x) \neq 0$ if $x \neq 0$.
\end{itemize}
Note that he assumption $\beta(x+y) \geq \beta(x) + \beta(y)$
for all $x,y \in P$ in condition (iii) could be rephrased as $-\beta$
satisfying the triangle inequality on $P$. Also note that if
condition (i) of Property A2 is satisfied so is condition (ii)
(for the same reason as in the remark following Property A1).
\begin{definition} \label{def6} \rm
Let $D$ be a subset of a real Banach space $E$. If $\rho: E
\to D$ is continuous
with $\rho(x) = x$ for all $x \in D$, then $D$ is a \emph{retract} of $E$,
and the map $\rho$ is a \emph{retraction}.
The \emph{convex hull} of a subset $D$ of a real Banach space
$X$ is given by
$$
\mathop{\rm conv}(D) = \Big\{ \sum_{i = 1}^{n}\lambda_{i}x_{i} : x_{i} \in
D,\; \lambda_{i} \in [0,1],\; \sum_{i = 1}^{n}\lambda_{i} = 1,
\mbox{ and } n \in \mathbb{N} \Big\}.
$$
\end{definition}
The following theorem is due to Dugundji and a proof can
be found in \cite[p. 44]{deim}.
\begin{theorem}\label{dug}
For Banach spaces $X$ and $Y$, let $D \subset X$ be closed and let
$F: D \to Y$ be continuous. Then
$F$ has a continuous extension $\tilde{F}: X \to Y$ such
that $\tilde{F}(X) \subset \overline{\mathop{\rm conv}(F(D))}$.
\end{theorem}
\begin{corollary}\label{R2}
Every closed convex set of a Banach space is a
retract of that Banach space.
\end{corollary}
Note that for any positive real number $r$ and non-negative
continuous concave functional $\alpha$, $Q(\alpha,r) = \{x \in P :
r \leq \alpha(x)\}$ is a retract of $E$ by Corollary \ref{R2}.
Note also, if $r$ is a positive number and if $\alpha: P \to
[0,\infty)$ is a uniformly continuous convex functional with
$\alpha(0) = 0$ and $\alpha(x) > 0$ for $x \neq 0$, then
\cite[Theorem 2.1]{sun} guarantees that $Q(\alpha,r)$ is a retract
of $E$.
The following theorem, which establishes the existence and
uniqueness of the fixed point index, is from \cite[pp. 82-86]{gu};
an elementary proof can be found in \cite[pp. 58 \& 238]{deim}.
The proof of our main result in the next section will invoke the
properties of the fixed point index.
\begin{theorem}\label{index}
Let $X$ be a retract of a real Banach space $E$. Then, for every
bounded relatively open subset $U$ of $X$ and every completely
continuous operator $A: \overline{U} \to X$ which has no fixed
points on $\partial U$ (relative to $X$), there exists an integer
$i(A,U,X)$ satisfying the following conditions:
\begin{itemize}
\item[(G1)] Normality: $i(A,U,X) = 1$ if $Ax \equiv y_{0} \in U$ for
any $x \in \overline{U}$;
\item[(G2)] Additivity: $i(A,U,X) = i(A,U_{1},X) + i(A,U_{2},X)$
whenever $U_{1}$ and $U_{2}$ are disjoint open subsets of $U$ such
that $A$ has no fixed points on $\overline{U} - (U_{1} \cup U_{2})$;
\item[(G3)] Homotopy Invariance: $i(H(t,\cdot),U,X)$ is independent
of $t \in [0,1]$ whenever \newline $H: [0,1]\times\overline{U} \to X$
is completely continuous and $H(t,x)\neq x$ for any
$(t,x)\in [0,1]\times\partial U$;
\item[(G4)] Permanence: $i(A,U,X) = i(A,U \cap Y, Y)$ if
$Y$ is a retract of $X$ and $A(\overline{U}) \subset Y$;
\item[(G5)] Excision: $i(A,U,X) = i(A,U_{0},X)$ whenever $U_{0}$ is
an open subset of $U$ such that $A$ has no fixed points in
$\overline{U} - U_{0}$;
\item[(G6)] Solution: If $i(A,U,X) \neq 0$, then $A$ has at least one
fixed point in $U$.
\end{itemize}
Moreover, $i(A,U,X)$ is uniquely defined.
\end{theorem}
\section{Fixed Point Theorems}
The proof of the following fixed point results can be found
in \cite[pp. 88-89]{gu}.
\begin{lemma}\label{L1}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$ with $0 \in \Omega$, and
$A: P \cap \overline{\Omega} \to P$ a completely continuous operator.
If
$Ax \neq \mu x$
for all $x \in P \cap \partial \Omega$ and $\mu \geq 1$, then
$$
i(A, P \cap \Omega,P)=1.
$$
\end{lemma}
\begin{lemma}\label{L2}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$, and $A: P \cap \overline{\Omega}
\to P$ a completely continuous operator. If
\begin{itemize}
\item[(i)] $\inf_{x \in P \cap \partial \Omega} \|Ax\| > 0$; and
\item[(ii)] $Ax \neq \nu x$ for all $x \in P \cap \partial \Omega$ and
$\nu \in (0,1]$,
\end{itemize}
then
$i(A, P \cap \Omega,P)=0$.
\end{lemma}
\begin{lemma}\label{ML1}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$ with $0 \in \Omega$, and
$A: P \cap \overline{\Omega} \to P$ a completely continuous operator.
If the continuous functional $\alpha$ satisfies Property A1
and
$\alpha(Ax) \geq \alpha(x)$
for all $x \in P \cap \partial \Omega$, with $Ax \neq x$, for all $x \in P \cap \partial \Omega$, then
$$
i(A, P \cap \Omega,P)=0.
$$
\end{lemma}
\begin{proof}
Suppose $\alpha$ satisfies Property A1. Then at least one of the
conditions (i), (ii) or (iii) of Property A1 is satisfied. We
proceed in cases.
\smallskip
\noindent Case 1: Condition (i) of Property A1 is satisfied.
The result follows from the proof of the following case since if
condition (i) of Property A1 is satisfied then condition (ii) of
Property A1 is satisfied.
\smallskip
\noindent Case 2: Condition (ii) of Property A1 is satisfied.
Suppose that $\alpha$ is sublinear and $\alpha(0) = 0$,
$\alpha(x) \neq 0$ if $x \neq 0$, and ${\inf_{x \in P \cap
\partial \Omega} \alpha(x) > 0}$.
\smallskip
\noindent Claim: $Ax \neq \nu x$ for all $x \in P \cap \partial \Omega$
and $\nu \in (0,1]$.\\
Suppose to the contrary that there exists an $x_0 \in P \cap \partial
\Omega$ and $\nu_0 \in (0,1]$ such that
$$Ax_0 = \nu_0x_0$$
(since $Ax \neq x$ for $x \in P \cap \partial \Omega$ we have that
$\nu_0 \neq 1$). Then since $\alpha(x_0) > 0$ we have
$$
\alpha(Ax_0) = \alpha(\nu_0x_0) \leq \nu_0 \alpha(x_0) < \alpha(x_0),
$$
which is a contradiction.
Also,
$$
\inf_{x \in P \cap \partial \Omega} \alpha(Ax) \geq
\inf_{x \in P \cap \partial \Omega} \alpha(x) > 0.
$$
Now since $A$ is completely continuous
$\inf_{x\in P \cap \partial \Omega} \|Ax\| \geq 0$.
If $\inf_{x\in P \cap \partial \Omega} \|Ax\|=0$ then there
exists a sequence $x_n \in P \cap \partial \Omega$ with
$\|Ax_n\| \to 0$, i.e. $Ax_n \to 0$, as $n \to \infty$ and so $\alpha$
continuous with $\inf_{x\in P \cap \partial \Omega} \alpha(Ax)>0$ implies
$$
0=\alpha(0)=\alpha(\lim_{n\to \infty} Ax_n)
= \lim_{n\to \infty} \alpha(Ax_n) > 0.
$$
Therefore, $\inf_{x\in P \cap \partial \Omega} \|Ax\| > 0$
and hence by Lemma \ref{L2}
$$
i(A,P \cap \Omega,P)=0.
$$
\noindent Case 3: Condition (iii) of Property A1 is satisfied.
Let
$$
R_1 = \sup_{x\in P \cap \overline{\Omega}} \alpha(Ax)
\quad\text{and}\quad
R_2 = \sup_{x\in P \cap \overline{\Omega}} \alpha(x)
$$
and then define
$R=\max\{R_1,R_2\}+1$.
Let $x^* \in P(\alpha,R)=\{x\in P : \alpha(x) \geq R\}$
(which is nonempty since $\alpha$ satisfies condition (iii)
of Property A1), and
$$
H(t,x)= tAx+(1-t)x^*.
$$
Obviously, $H:[0,1]\times (P \cap \partial \Omega) \to P$
(note that $P \cap \partial \Omega = \partial (P \cap \Omega)$)
is completely continuous. \smallskip
\noindent Claim: $H(t,x) \neq x$ for all $(t,x) \in [0,1]
\times (P \cap \partial \Omega)$.\\
Suppose to the contrary, that is there is a
$(t_0,x_0) \in [0,1] \times (P \cap \partial \Omega)$ such that
$H(t_0,x_0) = x_0$. Note $t_0 \neq 0$ since $\alpha(x^{*}) \geq R$
and $\alpha(x_0) \leq R_2 t_0\alpha(Ax_0) + (1-t_0)\alpha(Ax_0)\\
& = \alpha(Ax_0)
\end{align*}
which is a contradiction. Thus by the homotopy invariance property
$$
i(A,P\cap \Omega, P) = i(x^*,P\cap \Omega, P)
$$
and $i(x^*,P\cap \Omega, P)=0$ since if
$i(x^*,P\cap \Omega, P) \neq 0$, then there would be an
$x_1\in P\cap \Omega$ such that
$x^*=x_1$ which is a contradiction since
$\alpha(x^*) \geq R > \alpha(x_1)$. Thus
$$
i(A,P\cap \Omega, P) = 0.
$$
Therefore, regardless which of the three conditions of Property A1
is satisfied we have that $i(A,P\cap \Omega, P) = 0$.
\end{proof}
\begin{lemma}\label{ML2}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$ with $0 \in \Omega$, and
$A: P \cap \overline{\Omega} \to P$ a completely continuous operator.
If the continuous functional $\alpha$ satisfies Property A2
and
$\alpha(Ax) \leq \alpha(x)$
for all $x \in P \cap \partial \Omega$ with $Ax \neq x$ for all
$x \in P \cap \partial \Omega$, then
$$
i(A, P \cap \Omega,P)=1.
$$
\end{lemma}
\begin{proof}
Suppose $\alpha$ satisfies Property A2. Then at least one of the
conditions (i), (ii) or (iii) of Property A2 is satisfied. We
proceed in cases.
\noindent Case 1: Condition (i) of Property A2 is satisfied.
The result follows from the proof of the following case since if
condition (i) of Property A2 is satisfied then condition (ii)
of Property A2 is satisfied.
\smallskip
\noindent Case 2: Condition (ii) of Property A2 is satisfied.
Suppose that $\alpha$ is sublinear, $\alpha(0)=0$ and $\alpha(x)
\neq 0$ if $x \neq 0$. \smallskip
\noindent Claim: $Ax \neq \lambda x$ for all $x \in P \cap \partial
\Omega$ and $\lambda \geq 1$. \\
Suppose to the contrary; that is, there exists an
$x_0 \in \partial
\Omega$ and $\lambda_0 \geq 1$ (since $Ax \neq x$ for all
$x \in P \cap \partial \Omega$, we have that $\lambda_0 \neq 1$), such that
$$Ax_0 = \lambda_0x_0.$$
Note, $\alpha(x_0) \neq 0$ and since $\lambda_0 > 1$, we have
$0<\frac{1}{\lambda_0}<1$, and
$x_0 = Ax_0/\lambda_0$.
Thus, by the sublinearity of $\alpha$
$$\alpha(x_0) = \alpha(\frac{Ax_0}{\lambda_0}) \leq \frac{\alpha(Ax_0)}{\lambda_0},$$
and thus
$$
\alpha(x_0) < \lambda_0 \alpha(x_0) \leq \alpha(Ax_0),
$$
which is a contradiction.
Note that $0 \in \Omega$ by assumption. Hence by Lemma \ref{L1}
$$
i(A,P\cap \Omega,P)=1.
$$
\noindent Case 3: Condition (iii) of Property A2 is satisfied.
Let $H(t,x)= tAx$. Obviously, $H:[0,1]\times (P \cap
\overline{\Omega}) \to P$ is completely continuous.
\smallskip
\noindent Claim: $H(t,x) \neq x$ for all $(t,x) \in [0,1] \times
(P \cap \partial{\Omega})$. \\
Suppose to the contrary; that is, there is a
$(t_0,x_0) \in [0,1] \times (P \cap \partial{\Omega})$ such that
$H(t_0,x_0) = x_0$. Note, $t_0 \neq 0$. Also, since $Ax_0 \neq x_0$
we have that $t_0 \neq 1$, and we have
$$
Ax_0 = t_0Ax_0 + (1-t_0)Ax_0 = x_0 + (1-t_0)Ax_0.
$$
Thus for $t_0 \in (0,1)$
\begin{align*}
\alpha(Ax_0) & = \alpha(x_0 + (1-t_0)Ax_0)\\
& \geq \alpha(x_0) + \alpha((1-t_0)Ax_0) \\
& > \alpha(x_0),
\end{align*}
since $(1-t_0)Ax_0 \neq 0$ (note, if $(1-t_0)Ax_0=0$ then
$Ax_0=x_0$ which contradicts $Ax_0 \neq x_0$), which is a contradiction.
Hence, for all $(t,x) \in [0,1] \times (P \cap \partial{\Omega})$,
$H(t,x) \neq x$,
and so by the homotopy invariance property of the fixed point index
$$
i(A,P\cap \Omega, P) = i(0,P\cap \Omega, P),
$$
and $i(0,P\cap \Omega, P)=1$ by the solution property of the fixed point
index. Therefore
$$
i(A,P\cap \Omega, P) = 1.
$$
\end{proof}
\begin{theorem}\label{main}
Let $\Omega_1$ and $\Omega_2$ be two bounded open sets in a Banach
Space $E$ such that $0 \in \Omega_1$ and
$\overline{\Omega_1} \subseteq \Omega_2$ and $P$ is a cone in $E$.
Suppose $A:P \cap (\overline{\Omega_2} - \Omega_1) \to P$ is completely
continuous, $\alpha$ and $\psi$ are nonnegative continuous
functionals on $P$, and one of the two conditions:
\begin{itemize}
\item[(K1)] $\alpha$ satisfies Property A1 with
$\alpha(Ax) \geq \alpha(x)$, for all $x\in P\cap \partial\Omega_1$,
and $\psi$ satisfies property {\rm (A2)} with $\psi(Ax) \leq \psi(x)$,
for all $x\in P\cap \partial\Omega_2$;
or
\item[(K2)] $\alpha$ satisfies Property A2 with
$\alpha(Ax) \leq \alpha(x)$, for all $x\in P\cap \partial\Omega_1$,
and $\psi$ satisfies Property A1 with $\psi(Ax) \geq \psi(x)$,
for all $x\in P\cap \partial\Omega_2$,
\end{itemize}
is satisfied. Then $A$ has at least one fixed point in
$P \cap (\overline{\Omega_2} - \Omega_1)$.
\end{theorem}
\begin{proof} If there exists an
$x \in P \cap \partial(\overline{\Omega_2} - \Omega_1)$ such
that $Ax = x$, then there is nothing to prove; thus suppose that
$Ax \neq x$ for all $x$ on the boundary of
$\overline{\Omega_2} - \Omega_1$. By Dugundji's Theorem
(Theorem \ref{dug}), $A$ has a completely continuous extension
(which we will also denote by $A$)
$$
A: P \cap \overline{\Omega_2} \to P.
$$
Suppose condition (K1) is satisfied; the proof when (K2) is satisfied is
nearly identical and will be omitted.
From Lemma \ref{ML1} we have that $i(A,P \cap \Omega_1,P)=0$, and
from Lemma \ref{ML2} we have that $i(A,P \cap \Omega_2,P)=1$, and since
$A$ has no fixed points on
$\Omega_2 - (\Omega_1 \cup (\Omega_2-\overline{\Omega_1}))$,
by the additivity property of the fixed point index, we have
$$
i(A,P \cap \Omega_2,P) = i(A,P \cap (\Omega_2-\overline{\Omega_1}),P)
+ i(A,P \cap \Omega_1,P),
$$
and hence $i(A,P \cap (\Omega_2-\overline{\Omega_1}),P)=1$.
By the solution property of the fixed point index, we have that $A$
has a fixed point in $\Omega_2-\overline{\Omega_1}$.
\end{proof}
\section{Application}
In this section, as an application of our main result, Theorem
\ref{main}, we are concerned with the existence of at least one
positive solution for the second order boundary value problem,
\begin{gather}
x''+f(x)=0,\qquad 0\leq t\leq 1,\label{e4.1}\\
x(0)=0=x(1),\label{e4.2}
\end{gather}
where $f: \mathbb{R} \to [0,\,\infty)$ is continuous. We look for
solutions $x\in C^{(2)}[0,1]$ of \eqref{e4.1}, \eqref{e4.2} which
are both nonnegative and concave on $[0,1]$. We will impose growth
conditions on $f$ which ensure the existence of at least one
symmetric positive solution of \eqref{e4.1}, \eqref{e4.2} by
applying Theorem \ref{main}. We will apply Theorem \ref{main} to a
completely continuous operator whose kernel $G(t,s)$ is the
Green's function for
\begin{equation}
-x''=0,\label{e4.3}
\end{equation}
satisfying \eqref{e4.2}. In particular,
\begin{equation}
G(t,s)=\begin{cases}
t(1-s),&0\leq t\leq s\leq 1,\\
s(1-t),&0\leq s\leq t\leq 1.\end{cases}\label{e4.4}
\end{equation}
We will make use of various properties of $G(t,s)$ which include
\begin{gather}
\int_0^1G(t,s) ds = \frac{t(1-t)}{2},\quad
0 \leq t \leq 1 ; \label{e4.5}\\
\int_t^{1-t} G(t,s) ds = \frac{t(1-2t)}{2}, \quad 0 \leq t \leq
\frac{1}{2}; \label{e4.6}\\
\max_{0\leq r\leq 1}\frac{G(\frac{1}{2},r)}{G(t,r)}=\frac{1}{2t},
\quad 0 < t \leq \frac{1}{2}. \label{e4.7}
\end{gather}
Let $E = C[0,1]$ be endowed with the maximum norm,
$$
\|x\|=\max_{0\leq t\leq 1}|x(t)|,
$$
and define the cone $P \subset E$ by
\begin{align*}
P = \Big\{& x \in E : x \text{ is concave, symmetric, nonnegative
valued on $[0,1]$, and}\\
&\min_{t \in [z,1-z]} x(t) \geq 2z \|x\|
\text{ for all $ z \in [0,1/2]$}\Big\}.
\end{align*}
Finally, let the nonnegative continuous functionals
$\alpha$ and $\psi$ be defined on the cone $P$ by
\begin{gather}
\alpha (x) = \min_{t \in [1/4,3/4]} x(t) =
x(1/4) , \label{e4.8} \\
\psi(x) = \max_{t \in [0,1]}x(t) = x(1/2). \label{e4.9}
\end{gather}
We observe here that, for each $x \in P$,
\begin{equation}
\|x\| = x(\frac{1}{2}) \leq 2 x(\frac{1}{4}) = 2\alpha(x)\label{e4.10}
\end{equation}
and that $x \in P$ is a solution of \eqref{e4.1}, \eqref{e4.2} if and
only if
\begin{equation}
x(t)=\int_0^1G(t,s)f(x(s))ds,\quad 0\leq t\leq 1.\label{e4.11}
\end{equation}
We now present an application of our main result.
\begin{theorem} \label{app}
Suppose there exists positive numbers $r$ and $R$ such that
$0 0$;
and
\item[(ii)] $\nu x \not \in Ax $ for all $x \in P \cap \partial \Omega$
and $\nu \in (0,1]$,
\end{itemize}
then $i(A, P \cap \Omega,P)=0.$
\end{lemma}
\begin{lemma}\label{MMML1}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$ with $0 \in \Omega$, and
$A: P \cap \overline{\Omega} \to CK(E)$ a completely continuous operator.
If the continuous functional $\alpha$ satisfies Property A1,
$$
\alpha(y) \geq \alpha(x)
$$
for all $x \in P \cap \partial \Omega$ and for all $y \in Ax$, and
$$x \not \in Ax$$
for all $x \in P \cap \partial \Omega$, then
$$i(A, P \cap \Omega,P)=0.$$
\end{lemma}
\begin{proof} Suppose $\alpha$ satisfies Property A1.
Then at least one of the conditions (i), (ii) or (iii)
of Property A1 is satisfied. The proofs when conditions (ii)
and (iii) of Property A1 are satisfied are nearly identical to
the arguments in the proof of Lemma \ref{ML1} and will therefore
be omitted. For completeness we provide the proof when condition (i)
of Property A1 is satisfied.
\smallskip
\noindent Claim:
$\nu x \not \in Ax$ for all $x \in P \cap \partial \Omega$ and
$\nu \in (0,1]$.\\
Suppose to the contrary, that is there is an
$x_0 \in P \cap \partial \Omega$ and $\nu \in (0,1)$ such that
$\nu x_0 = y_0 \in Ax_0$ (note we know that $\nu \neq 1$ since
$x_0 \not \in Ax_0$). Therefore, since $\alpha$ is convex,
\begin{align*}
\alpha(y_0) & = \alpha(\nu x_0)\\
&= \alpha(\nu x_0 + (1-\nu)0)\\
& \leq \nu \alpha(x_0)\\
&< \alpha(x_0)
\end{align*}
which is a contradiction. Also, since
$\alpha(y) \geq \alpha(x)$
for all $x \in P \cap \partial \Omega$ and for all $y \in Ax$,
we have that
$$
\inf_{y \in \{Ax \; : x \in P \cap \partial \Omega\}}
\alpha(y) \geq \inf_{x \in P \cap \partial \Omega} \alpha(x) > 0,
$$
and following the argument in Lemma \ref{ML1}, we have that
$$
\inf_{y \in \{Ax : x \in P \cap \partial \Omega\}} \|y\| > 0.
$$
Hence, by Lemma \ref{MML1} we have $i(A, P \cap \Omega,P)=0$.
\end{proof}
\begin{lemma}\label{MMML2}
Let $P$ be a cone in a real Banach space $E$,
$\Omega$ a bounded open subset of $E$ with $0 \in \Omega$, and
$A: P \cap \overline{\Omega} \to CK(E)$ a completely continuous operator.
If the continuous functional $\alpha$ satisfies Property A2,
$$
\alpha(y) \leq \alpha(x)$$
for all $x \in P \cap \partial \Omega$ and for all $y \in Ax$, and
$x \not \in Ax$
for all $x \in P \cap \partial \Omega$, then
$$
i(A, P \cap \Omega,P)=1.
$$
\end{lemma}
\begin{proof}
Suppose $\alpha$ satisfies Property A2. Then at least one of
the conditions (i), (ii) or (iii) of Property A2 is satisfied.
The proofs when conditions (ii) and (iii) of Property A2 are
satisfied are nearly identical to the arguments in the proof of
Lemma \ref{ML2} and will therefore be omitted. For completeness
we provide the proof when condition (i) of Property A2 is
satisfied.
\smallskip
\noindent Claim: $\mu x \not \in Ax$ for all $x \in P \cap
\partial \Omega$ and $\mu \geq 1$.\\
Suppose to the contrary, that is there is an $x_0 \in P \cap \partial \Omega$ and $\mu > 1$ such that
$\mu x_0 = y_0 \in Ax_0$ (note we know that $\mu\neq 1$ by assumption).
Therefore, since $\alpha$ is convex,
\begin{align*}
\alpha(x_0) & = \alpha(\frac{y_0}{\mu})\\
&= \alpha(\frac{y_0}{\mu}+ (1-\frac{1}{\mu})0)\\
& \leq \frac{\alpha(y_0)}{\mu}\\
&< \alpha(y_0),
\end{align*}
which is a contradiction. Hence, by Lemma \ref{MML2} we have
$i(A, P \cap \Omega,P)=1$.
\end{proof}
\begin{theorem}\label{MMmain}
Let $\Omega_1$ and $\Omega_2$ be two bounded open sets in a Banach
Space $E$ such that $0 \in \Omega_1$ and
$\overline{\Omega_1} \subseteq \Omega_2$ and $P$ is a cone in $E$.
Suppose $A:P \cap (\overline{\Omega_2} - \Omega_1) \to CK(E)$
is completely continuous,
$\alpha$ and $\psi$ are nonnegative continuous functionals on $P$,
and one of the two conditions:
\begin{itemize}
\item[(K1)] $\alpha$ satisfies Property A1 with
$\alpha(y) \geq \alpha(x)$ for all $x\in P\cap \partial\Omega_1$
and for all $y \in Ax$ and $\psi$ satisfies Property A2 with
$\psi(y) \leq \psi(x)$ for all $x\in P\cap \partial\Omega_2$ and
for all $y \in Ax$; or
\item[(K2)] $\alpha$ satisfies Property A2 with
$\alpha(y) \leq \alpha(x)$ for all $x\in P\cap \partial\Omega_1$
and for all $y \in Ax$ and $\psi$ satisfies Property A1 with
$\psi(y) \geq \psi(x)$ for all $x\in P\cap \partial\Omega_2$ and
for all $y \in Ax$
\end{itemize}
is satisfied. Then $A$ has at least one fixed point in
$P \cap (\overline{\Omega_2} - \Omega_1)$.
\end{theorem}
\begin{proof}
If there exists an $x \in P \cap \partial(\overline{\Omega_2} - \Omega_1)$
such that $x \in Ax$, then there is nothing to prove; thus suppose
that $x \not \in Ax $
for all $x$ on the boundary of $\overline{\Omega_2} - \Omega_1$.
By Ma's extension of Dugundji's Theorem
\cite{ma}, $A$ has a completely continuous extension (which we will
also denote by $A$)
$$
A: P \cap \overline{\Omega_2} \to CK(E).
$$
Suppose condition $(K1)$ is satisfied; the proof when $(K2)$ is
satisfied is nearly identical and will be omitted.
From Lemma \ref{MMML1} we have that $i(A,P \cap \Omega_1,P)=0$,
and from Lemma \ref{MMML2} we have that $i(A,P \cap \Omega_2,P)=1$,
and since $A$ has no fixed points on
$\Omega_2 - (\Omega_1 \cup (\Omega_2-\overline{\Omega_1}))$ by the
additivity property of the fixed point index, we have
$$
i(A,P \cap \Omega_2,P) = i(A,P \cap (\Omega_2-\overline{\Omega_1}),P)
+ i(A,P \cap \Omega_1,P).
$$
Hence $i(A,P \cap (\Omega_2-\overline{\Omega_1}),P)=1,$ and by the
solution property of the fixed point index, we have that $A$ has a
fixed point in $\Omega_2-\overline{\Omega_1}$.
\end{proof}
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\end{document}