\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 42, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2011/42\hfil Fixed point theorem]
{Operator type expansion-compression fixed point theorem}
\author[D. R. Anderson, R. I. Avery, J. Henderson, X. Liu\hfil
EJDE-2011/42\hfilneg]
{Douglas R. Anderson, Richard I. Avery, Johnny Henderson, Xueyan Liu}
% in alphabetical order
\address{Douglas R. Anderson \newline
Department of Mathematics and Computer Science,
Concordia College, Moorhead, MN 56562, USA}
\email{andersod@cord.edu}
\address{Richard I. 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, TX 76798, USA }
\email{Johnny\_Henderson@baylor.edu}
\address{Xueyan (Sherry) Liu \newline
Department of Mathematics, Baylor University,
Waco, TX 76798, USA}
\email{Xueyan\_Liu@baylor.edu}
\thanks{Submitted October 27, 2010. Published March 25, 2011.}
\subjclass[2000]{47H10}
\keywords{Fixed-point theorems; Leggett-Williams;
expansion; compression; \hfill\break\indent positive solutions}
\begin{abstract}
This article presents an alternative to the compression
and expansion fixed point theorems of functional type by
using operators and functions to replace the functionals
and constants that are used in functional compression
and expansion fixed point theorems. Only portions of
the boundaries are required to be mapped outward or
inward in the spirit of the original work of Leggett-Williams.
We conclude with an application verifying the existence of
a positive solution to a second-order boundary-value problem.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\section{Introduction}
Mavridis \cite{mav} published the first extension to the work of
Leggett-Williams \cite{leg} that replaced the arguments involving
functionals with arguments involving operators. An invariance
condition was a key component of the arguments in that paper, that
is, $T(K_{A,B}(u,v) \cap K_c) \subset K_{A,B}(u,v)$ (condition (i)
of Theorem 2.8, the main result therein). A similar approach was
taken in the topological generalizations of fixed point theorems
presented by Kwong \cite{ko} which required boundaries to be
mapped inward or outward (invariance-like conditions). The spirit
of the Leggett-Williams fixed point theorems \cite{leg} and the
functional extensions by Avery \cite{five},
Anderson-Avery-Henderson \cite{aah2}, and Sun-Zhang \cite{sun}, to
mention a few, is that at least one of the boundaries is void of
any invariance like conditions. Anderson-Avery-Henderson
\cite{aah2} recently published the first results of this nature
that do not require either boundary to have invariance like
conditions.
The difficulty in replacing the arguments involving functionals
with arguments involving operators lies in the ability to compare
the output of an operator to a function using the comparison
generated by an underlying cone $P$. That is, for an operator $R$
and a specified function $x_R$, one needs to be able to say, for
any $y \in P$, that either $R(y) < x_R$ or $x_R \leq R(y)$. In
this paper we accomplish this by restricting our attention to a
cone $P$ of a real Banach space $E$ which is a subset of $F(K)$,
the set of real valued functions defined on a set $K\subset \mathbb{R}$.
We introduce what it means to say that an operator $R$ is
comparable to a function $x_R$ on a cone $P$ relative to $J_R$
which is a subset of $K$. This allows us to maintain the spirit
of the original work of Leggett-Williams and the extensions to the
outer boundary by Anderson-Avery-Henderson by avoiding any
invariance-like conditions in our arguments.
The proof of the main results hinge on the comparability criteria.
The Operator Type Expansion-Compression Fixed Point Theorem can be
used to verify the existence of positive solutions to boundary
value problems such as $x''+g(t)f(x,x')=0$ for $t\in[0,1]$ with
$x(0)=x'(1)=0$ (see Section 4). In the following example we
illustrate the comparability criteria which will formally be
defined in the next section. For $x$ in the cone $P$ of
increasing, nonnegative functions of $C^1[0,1]$, define the
operator
$$
(Ax)(t)=x'(0)t
$$
which is a continuous linear operator mapping $P$ to $P$.
Let $b\in(0,\infty)$, $\tau \in (0,1)$, $J_A = [\tau,1]$,
and $x_A(t)=bt$. Then for all $x \in P$, either
$$
x'(0) < b \quad\text{or} \quad b \leq x'(0).
$$
Hence for all $t \in J_A$, either
$$
(Ax)(t) = x'(0)t< bt = x_A(t) \quad\text{or} \quad
x_A(t)=bt \leq x'(0)t = (Ax)(t).
$$
Therefore, for any $x \in P$, either
$$
Ax(t) < x_A(t) \quad\text{or} \quad
x_A(t) \leq Ax(t) \quad \forall\,t\in J_A,
$$
which we will denote as
$$
Ax<_{J_A} x_A \quad\text{or} \quad x_A \leq_{J_A} Ax,
$$
and we say the operator $A$ is comparable to $x_A$ on $P$
relative to $J_A$. The operator $A$ is also an example
of both a convex and a concave operator which we will
formally define in the next section.
\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, for all
$x \in P$ and $\lambda \geq 0$, $\lambda x \in P$,
and if $x, -x \in P$ then $x = 0$.
\end{definition}
Every cone $P \subset E$ induces an ordering in $E$ given by
$x \leq y$ if and only if $y - x \in P$, and we say that $x < y$
whenever $x \leq y$ and $x \neq y$.
Let $K$ be a subset of real numbers and $F(K)$ the set of all
real valued functions defined over $K$.
If $J \subset K$ and $x,y \in F(K)$ we will say that:
$$
x <_J y \quad\text{if and only if } x(t) < y(t)
\text{ for all $t \in J$},
$$
and
$$ x \leq_J y \quad\text{if and only if } x(t) \leq y(t)
\text{ for all $t \in J$}.
$$
Furthermore, we will say that
$$
x \leqq_J y \quad\text{if and only if } x \leq_J y \quad
\text{and there exists a $t_0 \in J$ such that } x(t_0)=y(t_0).
$$
\begin{definition} \label{def2} \rm
An operator is called \emph{completely continuous} if it is
continuous and maps bounded sets into precompact sets.
\end{definition}
\begin{definition} \label{def3} \rm
Let $P$ be a cone in a real Banach space $E$. Then we say
that $A:P \to P$ is a \emph{continuous concave operator} on $P$ if
$A : P \to P$
is continuous and
$$
tA(x) + (1-t)A(y) \leq A(tx + (1-t)y)
$$
for all $x,y \in P$ and $t \in [0,1]$. Similarly we say that
$B:P \to P$ is a \emph{continuous convex operator} on $P$ if
$B : P \to P$
is continuous and
$$
B(tx + (1-t)y) \leq tB(x) + (1-t)B(y)
$$
for all $x,y \in P$ and $t \in [0,1]$.
\end{definition}
Let $R$ and $S$ be operators on a cone $P$ of a real
Banach space $E$ which is a subset of $F(K)$, the set
of real valued functions defined on a set $K$.
For $J_R, J_S \subset K$ and $x_R, x_S \in E$ we define
the sets,
\begin{gather*}
P_{J_R}(R,x_R) = \{y \in P : R(y) <_{J_R} x_R\},\\
P(R,S,x_R,x_S,J_R,J_S) = P_{J_S}(S,x_S) - \overline{P_{J_R}(R,x_R)}.
\end{gather*}
\begin{definition} \label{def4} \rm
Let $R$ be an operator on a cone $P$ of a real Banach space $E$
which is a subset of $F(K)$, the set of real valued functions
defined on a set $K$. For $J_R \subset K$ and $x_R \in E$,
we say that \emph{$R$ is comparable to $x_R$ on $P$ relative to $J_R$}
if, given any $y \in P$, either
$$
R(y) <_{J_R} x_R \quad\text{or} \quad x_R \leq_{J_R} R(y).
$$
\end{definition}
\begin{definition} \label{def5} \rm
Let $D$ be a subset of a real Banach space $E$.
If $r: E \to D$ is continuous
with $r(x) = x$ for all $x \in D$, then $D$ is a \emph{retract} of $E$,
and the map $r$ is a \emph{retraction}.
The \emph{convex hull} of a subset $D$ of a real Banach space
$X$ is given by
$$
\operatorname{conv}(D) = \big\{ \sum_{i = 1}^{n}\lambda_{i}x_{i} : x_{i} \in
D,\; \lambda_{i} \in [0,1],\; \sum_{i = 1}^{n}\lambda_{i} = 1,
\text{ and } n \in \mathbb{N} \big\}.
$$
\end{definition}
The following theorem is due to Dugundji and its proof can be
found in \cite[ p. 22]{schthomp}.
\begin{theorem}\label{scth}
Let $E$ and $X$ be Banach spaces and let $f:C \to K$ be a continuous
mapping, where $C$ is closed in $E$ and $K$ is convex in $X$.
there exists a continuous mapping $\tilde{f}: E \to K$ such that
$\tilde{f} (u) = f(u)$, $u \in C$.
\end{theorem}
Yet in establishing our main results, we will use the following
form of Dugundji's theorem \cite[p. 44]{deim}.
\begin{corollary}\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 $\tilde{F}(X) \subset \overline{\operatorname{conv}(F(D))}$.
\end{corollary}
\begin{corollary}\label{R2}
Every closed convex set in a Banach space is a
retract of the Banach space.
\end{corollary}
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
$H: [0,1]\times\overline{U} \to X$ is completely continuous
and $H(t,x)\neq x$ for any $(t,x)\in[0,1]\times\partial U$;
\item[(G4)] 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{Main Results}
Anderson, Avery, and Henderson \cite{aah2} proved an
expansion-compression fixed point theorem of Leggett-Williams type;
imbedded in the proof were two lemmas which were the primary
means of generalizing the fixed point theorems of Leggett-Williams.
The two lemmas in this section are the operator versions of those
lemmas and they will be the essential components in our
expansion-compression fixed point theorem of operator type.
The key to Leggett-Williams type arguments is in using concavity
to remove an invariance condition (an inward condition).
Note that if one of $A(y) <_{J_A} x_A$ or $x_A \leq_{J_A} A(y)$
holds for each $y \in P$ (this is the condition that $A$ is comparable
to $x_A$ on $P$ relative to $J_A$), then an invariance condition
of the form
\begin{itemize}
\item[(F0)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$, then
$B(Ty) <_{J_B} x_B$
\end{itemize}
is equivalent to the following two conditions
\begin{itemize}
\item[(F1)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$ x_A \leq_{J_A} A(y)$, then $B(Ty) <_{J_B} x_B.$
\item[(F2)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$ A(y) <_{J_A}x_A$, then $B(Ty) <_{J_B} x_B$.
\end{itemize}
In the spirit of the original work of Leggett-Williams, using
the properties of a concave operator $A$, one can replace (F2) with
\begin{itemize}
\item[(F2$^\prime$)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$ A(Ty) <_{J_A}x_A$, then $B(Ty) <_{J_B} x_B$.
\end{itemize}
These inward conditions (F1) and (F2$^\prime$) are used in the next
Lemma \ref{i=1}. A similar technique was employed by Anderson,
Avery and Henderson to remove outward conditions in
the expansion-compression arguments in \cite{aah2} and
the operator version appears below as Lemma \ref{i=0}.
Although the technique appears unmotivated, the beauty comes
in applications when a clever choice of the operator $A$
effortlessly verifies (F2$^\prime$). Then verification of (F0)
is replaced by verification of (F1) in our fixed point arguments.
Condition (F1) requires fewer $y \in P$ to be checked compared
to (F0), therefore it is easier to use in applications.
The operator $A$ is only a tool for obtaining a fixed point of $T$,
consequently it is not part of the conclusion of any lemma or theorem.
\begin{lemma}\label{i=1}
Let $F(K)$ be the set of real valued functions defined on
$K \subset \mathbb{R}$, $J_A$ and $J_B$ be subsets of $K$ with
$J_B$ being compact, and $P$ be a cone of non-negative functions
in a real Banach space $E$ which is a subset of $F(K)$. Suppose that
$A$ is a concave operator on $P$, $B$ is a continuous convex
operator on $P$, and $T: P \to P$ is a completely continuous operator.
Suppose there exist $x_A, x_B \in E$ such that
\begin{itemize}
\item[(B0)] $A$ is comparable to $x_A$ on $P$ relative to $J_A$;
\item[(B1)] $\{y\in P : x_A <_{J_A} A(y) \text{ and }
B(y) <_{J_B} x_B\} \neq \emptyset$;
\item[(B2)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$ x_A \leq_{J_A} A(y)$, then $B(Ty) <_{J_B} x_B$;
\item[(B3)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$A(Ty) <_{J_A} x_A$, then $B(Ty) <_{J_B} x_B$.
\end{itemize}
If $\overline{P_{J_B}(B,x_B)}$ is bounded, then
$i(T,P_{J_B}(B,x_B),P) = 1$.
\end{lemma}
\begin{proof}
By Corollary \ref{R2}, $P$ is a retract of the Banach space
$E$ since it is convex and closed.
\medskip
\noindent Claim $1$: $Ty \neq y$ for all
$y \in \partial P_{J_B}(B,x_B)$.
Let $z_0 \in \partial P_{J_B}(B,x_B)$. By the continuity of $B$
and the compactness of $J_B$, $B(z_0) \leqq_{J_B} x_B$.
We want to show that $z_0$ is not a fixed point of $T$; so
suppose to the contrary that $T(z_0)= z_0$. Since $A$ is
comparable to $x_A$ on $P$ relative to $J_A$, either
$A(Tz_0) <_{J_A} x_A$ or
$x_A \leq_{J_A} A(Tz_0)$. If $A(Tz_0) <_{J_A} x_A$,
then $B(Tz_0) <_{J_B} x_B$ by condition (B3), and if
$x_A \leq_{J_A} A(Tz_0) = A(z_0)$, then $B(Tz_0) <_{J_B} x_B$
by condition (B2). Hence, in either case we have that
$B(Tz_0) <_{J_B} x_B$ and $B(z_0) \leqq_{J_B} x_B$.
Thus $Tz_0 \neq z_0$ and we have verified that $T$ does not
have any fixed points on $\partial P_{J_B}(B,x_B)$.
Let $z_1 \in \{y\in P : x_A <_{J_A} A(y) \text{ and }
B(y) <_{J_B} x_B\}$ (see condition (B1)), and let
$H_1: [0,1] \times \overline{P_{J_B}(B,x_B)} \to P$
be defined by $H_1(t,y)=(1-t)Ty + t z_1$.
Clearly, $H_1$ is continuous and
$H_1([0,1] \times \overline{P_{J_B}(B,x_B)})$ is relatively compact.
\medskip
\noindent Claim $2$:
$H_1(t,y) \neq y$ for all $(t,y) \in [0,1] \times
\partial P_{J_B}(B,x_B)$.
Suppose not; that is, suppose there exists
$(t_1,y_1) \in [0,1] \times \partial P_{J_B}(B,x_B)$
such that $H(t_1,y_1)=y_1$.
Since $y_1 \in \partial P_{J_B}(B,x_B)$ we have that
$B(y_1) \leqq_{J_B} x_B$. Again, since $A$ is comparable
to $x_A$ on $P$ relative to $J_A$, either $A(Ty_1) <_{J_A} x_A$
or $x_A \leq_{J_A} A(Ty_1)$.
\indent Case $1$: $A(Ty_1) <_{J_A} x_A$.
By condition (B3), we have $B(Ty_1) <_{J_B} x_B$.
Since $B$ is convex on $P$, then
$$
B(y_1) = B((1-t_1)Ty_1 + t_1 z_1)
\leq (1-t_1)B(Ty_1) + t_1B(z_1)<_{J_B} x_B,
$$
which contradicts $B(y_1) \leqq_{J_B} x_B$.
\indent Case $2$: $x_A \leq_{J_A} A(Ty_1)$.
Since $A$ is concave on $P$,
$$
(1-t_1)A(Ty_1) + t_1A(z_1) \leq A((1-t_1)Ty_1 + t_1z_1)=A(y_1).
$$
Since $P$ is a cone of non-negative functions,
$$
x_A\leq_{J_A} (1-t_1)A(Ty_1) + t_1A(z_1)\leq_{J_A} A(y_1),
$$
and thus by condition (B2) we have $B(Ty_1) <_{J_B} x_B$.
This is the same contradiction we reached in the previous case.
Therefore, we have shown that $H_1(t,y) \neq y$ for all
$(t,y) \in [0,1] \times \partial P_{J_B}(B,x_B)$, and
thus by the homotopy invariance property $(G3)$ of the fixed
point index,
$i(T,P_{J_B}(B,x_B),P) = i(z_1,P_{J_B}(B,x_B),P)$.
And by the normality property $(G1)$ of the fixed point index,
$i(T,P_{J_B}(B,x_B),P) = i(z_1,P_{J_B}(B,x_B),P)=1$.
\end{proof}
\begin{lemma}\label{i=0}
Let $F(K)$ be the set of real valued functions defined on
$K \subset \mathbb{R}$, $J_C$ and $J_D$ be subsets of $K$
with $J_C$ being compact, and $P$ be a cone of non-negative
functions in a real Banach space $E$ which is a subset of $F(K)$.
Suppose that $C$ is a continuous concave operator on $P$,
$D$ is a convex operator on $P$, and $T: P \to P$
is a completely continuous operator. Suppose there exist
$x_C, x_D \in E$ such that
\begin{itemize}
\item[(A0)] $D$ is comparable to $x_D$ on $P$ relative to $J_D$;
\item[(A1)] $\{y\in P : x_C <_{J_C} C(y) \text{ and }
D(y) <_{J_D} x_D\} \neq \emptyset$;
\item[(A2)] if $y\in P$ with $C(y) \leqq_{J_C} x_C$ and
$D(y)\leq_{J_D} x_D$, then $x_C <_{J_C} C(Ty)$;
\item[(A3)] if $y\in P$ with $C(y) \leqq_{J_C} x_C$ and
$x_D <_{J_D} D(Ty)$, then $x_C <_{J_C} C(Ty)$.
\end{itemize}
If $\overline{P_{J_C}(C,x_C)}$ is bounded, then
$i(T,P_{J_C}(C,x_C),P) = 0$.
\end{lemma}
\begin{proof}
By Corollary \ref{R2}, $P$ is a retract of the Banach space $E$
since it is convex and closed.
\medskip
\noindent Claim $1$:
$Ty \neq y$ for all $y \in \partial P_{J_C}(C,x_C)$.
Let $w_0 \in \partial P_{J_C}(C,x_C)$. By the continuity of $C$
and the compactness of $J_C$, $C(w_0) \leqq_{J_C} x_C$.
We want to show that $w_0$ is not a fixed point of $T$; so
suppose to the contrary that $T(w_0)= w_0$. Since $D$ is
comparable to $x_D$ on $P$ relative to ${J_D}$, either
$x_D <_{J_D} D(Tw_0)$ or $D(Tw_0) \leq_{J_D} x_D$.
If $x_D <_{J_D} D(Tw_0)$, then
$x_C <_{J_C} C(Tw_0)$ by condition (A3), and if
$D(w_0) = D(Tw_0) \leq_{J_D} x_D$, then
$x_C <_{J_C} C(Tw_0)$ by condition (A2). Hence, in either case,
we have that $x_C <_{J_C} C(Tw_0)$ and $C(w_0) \leqq_{J_C} x_C$.
Thus $Tw_0 \neq w_0$ and we have verified that $T$ does not
have any fixed points on $\partial P_{J_C}(C,x_C)$.
Let $w_1 \in \{y\in P : x_C <_{J_C} C(y) \text{ and } D(y)
<_{J_D} x_D\}$ (see condition (A1)), and let
$H_2: [0,1] \times \overline{P_{J_C}(C,x_C)} \to P$
be defined by
$H_2(t,y)=(1-t)Ty + t w_1$.
Clearly, $H_2$ is continuous and
$H_2([0,1] \times \overline{P_{J_C}(C,x_C)})$ is relatively compact.
\medskip
\noindent Claim $2$:
$H_2(t,y) \neq y$ for all $(t,y) \in [0,1] \times
\partial P_{J_C}(C,x_C)$.
Suppose not; that is, there exists
$(t_2,y_2) \in [0,1] \times \partial P_{J_C}(C,x_C)$ such
that $H_2(t_2,y_2)=y_2$. Since $y_2 \in \partial P_{J_C}(C,x_C)$,
we have that $C(y_2) \leqq_{J_C} x_C$. Also, since $D$ is comparable
to $x_D$ on $P$ relative to ${J_D}$, either $x_D <_{J_D} D(Ty_2) $
or $D(Ty_2) \leq_{J_D} x_D$.
\indent Case $1$: $ x_D <_{J_D} D(Ty_2)$.
By condition (A3) we have $x_C <_{J_C} C(Ty_2)$,
which is a contradiction, since
$$
x_C <_{J_C} (1-t_2)C(Ty_2) + t_2C(w_1) \leq_{J_C} C((1-t_2)
Ty_2 + t_2w_1) = C(y_2) \leqq_{J_C} x_C.
$$
\indent Case $2$: $D(Ty_2) \leq_{J_D} x_D$.
We have that $D(y_2) \leq_{J_D} x_D$, since
$$
D(y_2) = D((1-t_2)Ty_2 + t_2w_1)
\leq_{J_D} (1-t_2)D(Ty_2) + t_2D(w_1)
\leq_{J_D} x_D,
$$
where $D((1-t_2)Ty_2 + t_2w_1)
\leq_{J_D} (1-t_2)D(Ty_2) + t_2D(w_1)$
is guaranteed by the fact that $D$ is convex on $P$ and $P$ is a
cone of non-negative functions. Thus by condition (A2),
we have $x_C <_{J_C} C(Ty_2)$. This is the same contradiction
$x_C <_{J_C} x_C$ we reached in the previous case.
Therefore, we have shown that $H_2(t,y) \neq y$ for all
$(t,y) \in [0,1] \times \partial P_{J_C}(C,x_C)$, and thus by
the homotopy invariance property (G3) of the fixed point index,
$i(T,P_{J_C}(C,x_C),P) = i(w_1,P_{J_C}(C,x_C),P)$.
And by the solution property (G4) of the fixed point index
(since $w_1 \not\in P_{J_C}(C,x_C)$ the index cannot be nonzero),
we have $i(T,P_{J_C}(C,x_C),P) = i(w_1,P_{J_C}(C,x_C),P)=0$.
\end{proof}
\begin{theorem}\label{operator}
Let $F(K)$ be the set of real valued functions defined on
$K \subset \mathbb{R}$, let $J_A$, $J_B$, $J_C$ and $J_D$ be subsets
of $K$ such that $J_B$ and $J_C$ are compact, and let $P$ be
a cone of non-negative functions in a real Banach space $E$ that
is a subset of $F(K)$. Suppose that $A$ and $C$ are concave
operators on $P$ and that $B$ and $D$ are convex operators on
$P$ such that $B$ and $C$ are continuous, and that $T: P \to P$
is a completely continuous operator. For $J \subset K$ and
$x,y \in F(K)$ let $x <_J y$ $(x \leq_J y)$ if and only if
$x(t) < y(t)$ $(x(t) \leq y(t))$ for all $t \in J$, whereas
$x \leqq_J y$ if and only if $x \leq_J y$ and there exists
a $t_0 \in J$ such that $x(t_0)=y(t_0)$.
Suppose there exist $x_A, x_B, x_C, x_D \in E$ such that
\begin{itemize}
\item[(D1)] $T$ is LW-outward with respect to $P_{J_C}(C,x_C):=\{y \in P: C(y) <_{J_C} x_C\}$, that is, the following conditions are satisfied:
\begin{itemize}
\item[(A0)] either $D(y) <_{J_D} x_D$ or $x_D \leq_{J_D} D(y)$
for any $y \in P$;
\item[(A1)] $\{y\in P : x_C <_{J_C} C(y) \text{ and }
D(y) <_{J_D} x_D\} \neq \emptyset$;
\item[(A2)] if $y\in P$ with $C(y) \leqq_{J_C} x_C$ and
$D(y)\leq_{J_D} x_D$, then $x_C <_{J_C} C(Ty)$;
\item[(A3)] if $y\in P$ with $C(y) \leqq_{J_C} x_C$ and
$x_D <_{J_D} D(Ty)$, then $x_C <_{J_C} C(Ty)$;
\end{itemize}
and $\operatorname{closure}\{y \in P: C(y) <_{J_C} x_C\}$ is bounded.
\item[(D2)] $T$ is LW-inward with respect to
$P_{J_B}(B,x_B):=\{y \in P: B(y) <_{J_B} x_B\}$,
that is, the following conditions are satisfied:
\begin{itemize}
\item[(B0)] either $A(y) <_{J_A} x_A$ or $x_A \leq_{J_A} A(y)$
for any $y \in P$;
\item[(B1)] $\{y\in P : x_A <_{J_A} A(y) \text{ and }
B(y) <_{J_B} x_B\} \neq \emptyset$;
\item[(B2)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$ x_A \leq_{J_A} A(y)$, then $B(Ty) <_{J_B} x_B$;
\item[(B3)] if $y\in P$ with $B(y) \leqq_{J_B} x_B$ and
$A(Ty) <_{J_A} x_A$, then $B(Ty) <_{J_B} x_B$;
\end{itemize}
and $\operatorname{closure}\{y \in P: B(y) <_{J_B} x_B\}$ is bounded.
\end{itemize}
If
\begin{itemize}
\item[(H1)] $\operatorname{closure}\{y \in P: B(y) <_{J_B} x_B\} \subsetneq \{y \in P: C(y) <_{J_C} x_C\}$, then $T$ has a fixed point $y\in P$ such that
$C(y) <_{J_C} x_C$ with $y\notin\operatorname{closure}\{u \in P: B(u) <_{J_B} x_B\}$,
\end{itemize}
whereas, if
\begin{itemize}
\item[(H2)] $\operatorname{closure}\{y \in P: C(y) <_{J_C} x_C\} \subsetneq \{y \in P: B(y) <_{J_B} x_B\}$, then $T$ has a fixed point $y\in P$ such that
$B(y) <_{J_B} x_B$ with $y\notin\operatorname{closure}\{u \in P: C(u) <_{J_C} x_C\}$.
\end{itemize}
\end{theorem}
\begin{proof}
We will prove the expansive result (H1), as the proof of the
compressive result (H2) is nearly identical. First, define the sets
$$
P_{J_R}(R,x_R) := \{y \in P : R(y) <_{J_R} x_R\}
$$
and
$$
P(B,C,x_B,x_C,J_B,J_C) := P_{J_C}(C,x_C) - \overline{P_{J_B}(B,x_B)}.
$$
To prove the existence of a fixed point for our operator $T$
in $P(B,C,x_B,x_C,J_B,J_C)$, it is enough for us to show
that $i(T,P(B,C,x_B,x_C,J_B,J_C),P) \neq 0$.
Since $T$ is LW-inward with respect to $P_{J_B}(B,x_B)$, we have
by Lemma \ref{i=1} that
$i(T,P_{J_B}(B,x_B),P) = 1$,
and since $T$ is LW-outward with respect to $P_{J_C}(C,x_C)$, we have by Lemma \ref{i=0} that
$i(T,P_{J_C}(C,x_C),P) = 0$.
$T$ has no fixed points in
$\overline{P_{J_C}(C,x_C)} - (P_{J_B}(B,x_B) \cup
P(B,C,x_B,x_C,J_B,J_C))$, since if
$y \in \overline{P_{J_C}(C,x_C)} - (P_{J_B}(B,x_B) \cup
P(B,C,x_B,x_C,J_B,J_C))$, then either \\
$B(y) \leqq_{J_B} x_B$ or $C(y) \leqq_{J_C} x_C$.
Now, if $B(y) \leqq_{J_B} x_B$, then we showed
in Lemma \ref{i=1} that $y$ was not a fixed point of $T$,
and if $C(y) \leqq_{J_C} x_C$, then
we showed in Lemma \ref{i=0} that $y$ was not a fixed point of $T$.
Also, the sets $P_{J_B}(B,x_B)$ and $ P(B,C,x_B,x_C,J_B,J_C)$
are nonempty, disjoint, open subsets of $\overline{P_{J_C}(C,x_C)}$,
since $\overline{P_{J_B}(B,x_B)} \subsetneq P_{J_C}(C,x_C)$
implies that $ P(B,C,x_B,x_C,J_B,J_C)
= P_{J_C}(C,x_C) - \overline{P_{J_B}(B,x_B)} \neq \emptyset$.
Therefore, by the additivity property $(G2)$ of the fixed point index
$$
i(T,P_{J_C}(C,x_C),P) = i(T,P_{J_B}(B,x_B),P)
+ i(T,P(B,C,x_B,x_C,J_B,J_C),P).
$$
Consequently, we have
$i(T,P(B,C,x_B,x_C,J_B,J_C),P)= -1$,
and thus by the solution property $(G4)$ of the fixed point index,
the operator $T$ has a fixed point $y \in P(B,C,x_B,x_C,J_B,J_C)$.
\end{proof}
\section{Application}
As an application of our main results, we consider the following
second order nonlinear right focal boundary value problem,
\begin{gather}\label{e1}
x'' +g(t)f(x,x')=0,\quad t\in[0,1],\\
\label{e2}
x(0)=x'(1)=0,
\end{gather}
where $g: [0,1] \to [0,\infty)$ and
$f: \mathbb{R}^2 \to [0,\infty)$ are continuous.
Let the Banach space $E=C^1[0,1]$ with the norm of
$\|x\|=\max_{t\in[0,1]}|x(t)|+\max_{t\in[0,1]}|x'(t)|$, and
define the cone $P \subset E$ by
$$
P:=\{x\in E: x(t)\geq 0,\,x'(t)\geq 0,\text{ for }
t\in[0,1],\,x \text{ is concave, and }x(0)=0\}.
$$
Then for any $x\in P$,
we have $\|x\|=x(1)+x'(0)$. And from the concavity of any $x\in P$,
we have that $x(t)\geq tx(1)$ and $x(t)\leq x'(0)t$ for $t\in[0,1]$.
It is well known that the Green's function for $-x''=0$ and
satisfying \eqref{e2} is given by
$$
G(t,s)=\min\{t,s\},\quad (t,s)\in[0,1]\times[0,1].
$$
We note that, for any $s\in[0,1]$, $G(t,s)\geq tG(1,s)$ and
$G(t,s)$ is nondecreasing in $t$.
By using properties of the Green's function, solutions of \eqref{e1},
\eqref{e2} are the fixed points of the
completely continuous operator $T: P \to E$ defined by
$$
Tx(t)=\int_0^1G(t,s)g(s)f(x(s),x'(s))ds.
$$
Since $(Tx)''(t)=-g(t)f(x,x')\leq0$ on $[0,1]$ and
$(Tx)(0)=(Tx)'(1)=0$, we have $T: P\to P$.
Let $\tau\in(0,1)$. For $x\in P$, we define the following operators:
\[
(Ax)(t)=(Cx)(t)=x'(0)t,\quad
(Bx)(t)=\Big(\frac{x'(0)+x(1)}{2}\Big)t,\qquad
(Dx)(t)=\Big(\frac{x(\tau)}{\tau}\Big)t.
\]
All the above operators are continuous linear operators mapping
$P$ to $P$, and are convex or concave continuous operators as well.
In the following theorem, we demonstrate how to apply the
expansive condition of Theorem $\ref{operator}$ to prove
the existence of at least one positive solution to
\eqref{e1}, \eqref{e2}.
\begin{theorem} \label{application}
Suppose there is some $\tau\in(0,1)$ and $0 \frac{d}{\int_0^\tau g(s)ds}$,
for $(u_1,u_2)\in[0,d\tau]\times[0,d]$,
\item[(b)] $f(u_1,u_2) < \frac{2b}{\int_0^1(1+s)g(s)ds}$,
for $(u_1,u_2)\in[0,b]\times[0,2b)$.
\end{itemize}
Then the right focal problem
\eqref{e1}, \eqref{e2} has at least one positive solution
$y \in P$ with ${y}'(0)>d$ and ${y}'(0)+{y}(1)<2b$.
\end{theorem}
\begin{proof}
We choose $x_A(t)=bt$, $x_B(t)=b$, $x_C(t)=dt$, $x_D(t)=d\tau$
defined on $[0,1]$, and $J_A=J_C=[\tau,1]$, $J_B=\{1\}$ and
$J_D=\{\tau\}$. Then it is easy to see that $J_A,J_B,J_C,J_D$
are compact subsets of $[0,1]$ and $x_A,x_B,x_C,x_D\in E$.
\medskip
\noindent Claim $1$: $T$ is LW-inward with respect
to $P_{J_B}(B,x_B)$.
$A$ is comparable to $x_A$ on $P$ relative to $J_A$, since for
any $y\in P$, we have $(Ay)(t)=y'(0)t\geq x_A(t)=bt$, or
$(Ay)(t)=y'(0)tbt=x_A(t)$, for $t\in[\tau,1]=J_A$, and
$(By_0)(1)=\frac{y_0'(0)+y_0(1)}{2}=\frac{2a+a}{2}**dt=x_C(t)$, for $t\in[\tau,1]=J_C$, and
$(Dy_0)(\tau)=a\tau(2-\tau) \frac{d}{\int_0^\tau g(s)ds},\quad t\in[0,\tau],
$$
and so for $t\in[\tau,1]$,
\begin{align*}
(CTy)(t)
&=(Ty)'(0)t =\int_{0}^1g(s)f(y(s),y'(s))ds\cdot t\\
&\geq \int_{0}^\tau g(s)f(y(s),y'(s))ds\cdot t\\
&> \int_{0}^\tau g(s)ds\cdot \frac{dt}{\int_0^\tau g(s)ds}
=dt=x_C(t);
\end{align*}
i.e., $x_C <_{J_C} C(Ty)$.
\medskip
\noindent Subclaim $2.2$: If $y\in P$ with $C(y) \leqq_{J_C} x_C$
and $x_D <_{J_D} D(Ty)$, then $x_C <_{J_C} C(Ty)$.
Let $y\in P$ with $C(y) \leqq_{J_C} x_C$ and $x_D <_{J_D} D(Ty)$.
From $x_D <_{J_D} D(Ty)$, we have that $(Ty)(\tau)>d\tau$.
Hence, $(Ty)'(0)\geq \frac{(Ty)(\tau)}{\tau}>d$. Therefore,
$(CTy)(t)=(Ty)'(0)t>dt=x_C(t)$, for $t\in[\tau,1]$; i.e.,
$x_C <_{J_C}C(Tx)$.
It is easy to see that $\overline{P_{J_C}(C,x_C)}$ is bounded,
thus $T$ is LW-outward with respect to $P_{J_C}(C,x_C)$.
\medskip
\noindent Claim $3$:
$\overline{P_{J_C}(C,x_C)}\subset P_{J_B}(B,x_B)$ and
$P(C,B,x_C,x_B,J_C,J_B)\neq\emptyset$.
Let $y\in\overline{P_{J_C}(C,x_C)}$. Then, $y'(0)\leq d$.
From $y(1)\leq y'(0)\leq d$, we have
$$
(By)(1)=\frac{y'(0)+y(1)}{2}\leq \frac{d+d}{2}=d****dt=x_C(t)$,
for $t\in[\tau,1]$, and
$(By_0)(1)=\frac{y_0'(0)+y_0(1)}{2}=\frac{3a}{2}****0.3$
and $x'(0)+x(1)<1.2$.
\subsection*{Acknowledgements}
The authors are indebted to the anonymous referee for each of the
in-depth and extensive suggestions. This has led to a significantly
improved article.
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**