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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 63, pp. 1--9.\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-2010/63\hfil Functional compression-expansion]
{Functional expansion - compression fixed point theorem of
Leggett-Williams type}

\author[D. R. Anderson, R. I. Avery, J. Henderson\hfil EJDE-2010/63\hfilneg]
{Douglas R. Anderson, Richard I. Avery, Johnny Henderson}  % 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}

\thanks{Submitted November 19, 2009. Published May 5, 2010.}
\subjclass[2000]{47H10}
\keywords{Fixed-point theorems; Leggett-Williams; expansion;
 compression; \hfill\break\indent positive solutions}

\begin{abstract}
 This paper presents a  fixed point theorem of compression
 and expansion of functional type in the spirit of the original
 fixed point work of Leggett-Williams.  Neither the entire
 lower nor the entire upper boundary is required to be mapped
 inward or outward.
\end{abstract}

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

\section{Introduction}

The spirit of the original Leggett-Williams fixed point theorem
\cite{leg} is to take a subset of the elements in the  cone in
which $\alpha(x) = a$ and map these outward in the sense that
$\alpha(Tx) \geq a$, where $\alpha$ is a concave positive
functional defined on the cone.  The subset that Leggett-Williams
considered can be thought of as the set of all elements of the
cone in which $\|x\| \leq b$ and $\alpha(x) = a$.  There were  no
outward conditions on the operator $T$ in the Leggett-Williams
fixed point theorem concerning those elements with $\|x\| > b$ and
$\alpha(x) = a$, and hence they avoided any invariance-like
conditions with respect to one boundary. The entire upper boundary
was mapped inward (Leggett-Williams had invariance-like conditions
with respect to only the outer boundary).  That is, all of the
elements in the cone for which $\|x\| = c$ were mapped inward  in
the sense that $\|Tx\| \leq c$.  Leggett-Williams created only a
compression result;  Leggett-Williams did not create an expansion
result.

In this paper we use techniques similar to those of
Leggett-Williams that will require only subsets of both boundaries
to be mapped inward and outward, respectively. We thus provide
more general results than those obtained by using the
Krasnosel'skii fixed point theorem \cite{kr}, prior functional
compression-expansion results which mapped at least one boundary
inward or outward \cite{aa,aho,guo1,guo2,leg,sun}, or the
topological generalizations of fixed point theorems introduced by
Kwong \cite{ko} which require both boundaries to be mapped inward
or outward (invariance-like conditions).  Moreover, conditions
involving the norm in the original Leggett-Williams fixed point
theorem are replaced by more general conditions on a convex
functional.

\section{Preliminaries}


 In this section we will state the definitions that are used in the
remainder of the paper.

\begin{definition} \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
$$
x \leq y  \quad  \text{if and only if} \quad  y - x \in P.
$$

\begin{definition} \rm
An operator is called completely continuous if it is
continuous and maps bounded sets into precompact sets.
\end{definition}

\begin{definition} \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 $t \in [0,1]$.
\end{definition}

 Let $\psi$ and  $\delta$ be nonnegative continuous
functionals on $P$; then, for positive real numbers $a$ and $b$,
we define the sets:
\begin{gather}\label{Q1}
 P(\psi,b) = \{x \in P :  \psi(x) \leq b  \},\\
\label{Q2}
 P(\psi,\delta,a,b) = \{x \in P : a \leq \psi (x)\text{ and } \delta
(x) \leq b\}.
\end{gather}

\begin{definition} \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
$$
\mathop{\rm conv}(D) = \Big\{ \sum_{i = 1}^{n}\lambda_{i}x_{i} :
x_{i} \in D,\; \lambda_{i} \in [0,1],\; \sum_{i =
1}^{n}\lambda_{i} = 1, \text{ and } n \in \mathbb{N} \Big\}.
$$
\end{definition}

 The next theorem is due to Dugundji and its 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 the Banach space.
\end{corollary}

\section{Fixed point index}

 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{Main Result}

\begin{theorem}\label{main}
Suppose $P$ is a cone in a real Banach space $E$,
 $\alpha$ is a nonnegative continuous concave functional on $P$,  $\beta$  is a nonnegative continuous convex functional on $P$ and $T: P \to P$
is a completely continuous operator.  If there exists nonnegative numbers $a$, $b$, $c$ and $d$ such that
\begin{itemize}
\item[(A1)] $\{x\in  P : a < \alpha (x) \text{ and } \beta (x) < b\} \neq \emptyset$;
\item[(A2)] if $x\in P$ with $\beta(x) =b$ and $\alpha(x)\geq a$, then $\beta (Tx) < b$;
\item[(A3)] if $x\in P$ with $\beta(x) =b$ and $\alpha(Tx) < a$ ,then $\beta (Tx) < b$;
\item[(A4)] $\{x\in  P : c < \alpha (x) \text{ and } \beta (x) < d\} \neq \emptyset$;
\item[(A5)] if $x\in P$ with $\alpha(x) =c$ and $\beta(x)\leq d$, then $\alpha(Tx) > c$;
\item[(A6)] if $x\in P$ with $\alpha(x) =c$ and $\beta(Tx) > d$, then $\alpha (Tx) > c$;
\end{itemize}
and if
\begin{itemize}
\item[(H1)] $a<c$, $b<d$, $\{x\in  P : b < \beta (x) \text{ and } \alpha (x) < c\} \neq \emptyset$, $P(\beta,b) \subset P(\alpha,c)$, \newline and $P(\alpha,c)$ is bounded then $T$ has a fixed point $x^*$ in $P(\beta,\alpha,b,c)$;
\item[(H2)] $c<a$, $d<b$, $\{x\in  P : a < \alpha (x) \text{ and } \beta (x) < d\} \neq \emptyset$, $P(\alpha,a) \subset P(\beta,d)$,\newline  and $P(\beta,d)$ is bounded then $T$ has a fixed point $x^*$ in $P(\alpha,\beta,a,d)$.
\end{itemize}
\end{theorem}

\begin{proof}
We will prove the expansion result (H1).  The proof of the compression
result (H2) is nearly identical; moreover, a topological proof can
be found in \cite{aah} for the compression result.
If we let
\begin{gather*}
U = \{x \in P : \beta (x) < b \},\\
V = \{x \in P : \alpha (x) < c\},
\end{gather*}
then the interior of $V-U$ is given by $W = (V-U)^{\circ} = \{x
\in V : b < \beta (x) \text{ and } \alpha(x) < c\}$.  Thus
$U$, $V$ and $W$ are bounded (they are subsets of $V$ which is
bounded by condition (H1)), non-empty (by conditions (A1),
(A4) and (H1)) and open subsets of $P$.  To prove the
existence of a fixed point for our operator $T$ in
$P(\beta,\alpha,b,c)$, it is enough for us to show that $i(T,W,P)
\neq 0$ since $W$ is the interior of $P(\beta,\alpha,b,c)$.  By
Corollary \ref{R2}, $P$ is a retract of the Banach space $E$ since
it is convex  and closed.


\noindent\textbf{Claim $1$:} $Tx \neq x$ for all $x \in \partial U$.

Let $z_0 \in \partial U$, then $\beta (z_0) = b$.
We want to show that $z_0$ is not a fixed point of $T$;
so suppose to the contrary that $T(z_0)= z_0$.
If $\alpha(Tz_0) < a$ then $\beta(Tz_0) < b$ by condition $(A3)$,
and if $\alpha(z_0) = \alpha(Tz_0) \geq a$ then $\beta(Tz_0) < b$
by condition (A2).  Hence in either case we have that $Tz_0 \neq z_0$,
 thus $T$ does not have any fixed points on $\partial U$.


\noindent\textbf{Claim $2$:} $Tx \neq x$ for all $x \in \partial V$.

Let $z_1 \in \partial V$, then $\alpha (z_1) = c$.  We want to
show  that $z_1$ is not a fixed point of $T$; so suppose to the
contrary that $T(z_1)= z_1$.  If $\beta(Tz_1) > d$ then
$\alpha(Tz_1) > c$ by condition $(A6)$, and if $\beta(z_1) =
\beta(Tz_1) \leq d$  then $\alpha(Tz_1) > c$ by condition $(A5)$.
Hence in either case we have that $Tz_1 \neq z_1$, thus $T$ does
not have any fixed points on $\partial V$.



 Let $w_1 \in \{x\in  P  : a < \alpha (x) \text{ and } \beta (x) < b\}$  (see condition $(A1)$) and let
$H_1: [0,1] \times \overline{U} \to P$
be defined by
$$
H_1(t,x)=(1-t)Tx + t w_1.
$$
Clearly, $H_1$ is
continuous and  $H_1([0,1] \times \overline{U})$ is relatively
compact.

\noindent\textbf{Claim $3$:}
$H_1(t,x) \neq x$ for all $(t,x) \in [0,1] \times \partial U$.

Suppose not; that is, there exists
$(t_1,x_1) \in [0,1] \times \partial U$ such that $H(t_1,x_1)=x_1$.
Since $x_1 \in \partial U$ we have that $\beta(x_1) = b$.
Either $\alpha(Tx_1) < a$ or $\alpha(Tx_1) \geq a$.

 Case $1$: $\alpha(Tx_1) < a$.
By condition $(A3)$ we have $\beta(Tx_1) < b$,
which is a contradiction since
\begin{align*}
b &=  \beta(x_1) = \beta((1-t_1)Tx_1 + t_1 w_1)\\
  &\leq  (1-t_1)\beta(Tx_1) + t_1\beta(w_1) < b.
\end{align*}

 Case $2$: $\alpha(Tx_1) \geq a$.
We have that $\alpha(x_1) \geq a$ since
\begin{align*}
\alpha(x_1) &=  \alpha((1-t_1)Tx_1 + t_1w_1)\\
           &\geq  (1-t_1)\alpha(Tx_1) + t_1\alpha(w_1)
           \geq  a,
\end{align*}
and thus by condition (A2) we have $\beta(Tx_1) < b$,
which is the same contradiction we arrived at in the previous
case.

Therefore, we have shown that  $H_1(t,x) \neq x$ for all
$(t,x) \in [0,1] \times \partial U$, and thus by the
homotopy invariance property $(G3)$ of the fixed point index
$$
i(T,U,P) = i(w_1,U,P),
$$
and by the normality property (G1) of the fixed point index
$$
i(T,U,P) = i(w_1,U,P)=1.
$$

 Let $w_2 \in \{x\in  P : c < \alpha (x) \text{ and } \beta (x) < d\}$
(see condition (A4)) and let
$$
H_2: [0,1] \times \overline{V} \to P
$$
be defined by
$$
H_2(t,x)=(1-t)Tx + t w_2.
$$
Clearly, $H_2$ is continuous and
$H_2([0,1] \times \overline{V})$ is relatively compact.



\noindent\textbf{Claim $4$:}  $H_2(t,x) \neq x$ for all
$(t,x) \in [0,1] \times \partial V$.

Suppose not; that is, there exists
$(t_2,x_2) \in [0,1] \times \partial V$ such that $H(t_2,x_2)=x_2$.
 Since $x_2 \in \partial V$ we have that $\alpha(x_2) = c$.
Either $\beta(Tx_2) \leq d$ or $\beta(Tx_2) > d$.

 Case $1$: $\beta(Tx_2) > d$.
By condition $(A6)$ we have $\alpha(Tx_2) > c$,
which is a contradiction since
\begin{align*}
c &=  \alpha(x_2) = \alpha((1-t_2)Tx_2 + t_2w_2)\\
           &\geq  (1-t_2)\alpha(Tx_2) + t_2\alpha(w_2) > c.
\end{align*}


 Case $2$ : $\beta(Tx_2) \leq d$.
We have that $\beta(x_2) \leq d$ since
\begin{align*}
\beta(x_2) &=  \beta((1-t_2)Tx_2 + t_2w_2)\\
           &\leq  (1-t_2)\beta(Tx_2) + t_2\beta(w_2)
           \leq  d,
\end{align*}
and thus by condition (A5) we have $\alpha(Tx_2) > c$,
which is the same contradiction we arrived at in the previous case.

 Therefore, we have shown that  $H_2(t,x) \neq x$ for all
$(t,x) \in [0,1] \times \partial V$ and thus by the homotopy
invariance property $(G3)$ of the fixed point index
$$
i(T,V,P) = i(w_2,V,P),
$$
and by the solution property (G6) of the fixed point index
(since $w_2 \not\in V$ the index cannot be nonzero) we have
$$
i(T,V,P) = i(w_2,V,P)=0.
$$


 Since $U$ and $W$ are disjoint open subsets of $V$ and $T$ has
no fixed points in $\overline{V} - (U \cup W)$
(by claims $1$ and $2$), by the additivity property (G2)
 of the fixed point index
$$
i(T,V,P) = i(T,U,P) + i(T,W,P).
$$
Consequently, we have
$$
i(T,W,P)= -1,
$$
and thus by the solution property (G6) of the fixed point index
the operator $T$ has a fixed point
$x^* \in W \subset P(\beta,\alpha,b,c)$.
\end{proof}


\section{Application}

In this section we will illustrate the key techniques for verifying
the existence of a positive solution for a boundary value problem
using our main result. In particular, under the expansion condition (H1)
we apply the properties of a Green's function, bound the
nonlinearity by constants over some intervals, and use concavity
to deal with a singularity.  To proceed, consider the second-order
nonlinear focal boundary value problem
\begin{gather}\label{b1}
  x''(t) + f(x(t)) = 0, \quad t\in(0,1), \\
\label{b2}
  x(0)= 0 = x'(1),
\end{gather}
where $f: \mathbb{R} \to [0,\infty)$ is continuous.  If $x$ is a fixed
point of the operator $T$ defined by
$$
Tx(t) := \int_{0}^{1} G(t,s) f(x(s)) ds,
$$
where
$$
G(t,s)=\min\{t,s\}, \quad (t,s)\in[0,1]\times[0,1]
$$
is the Green's function for the operator $L$ defined by
$Lx(t) := -x''$
with right-focal boundary conditions
$x(0)= 0 = x'(1)$,
then it is well known that $x$ is a solution of the boundary value
problem \eqref{b1}, \eqref{b2}.
Throughout this section of the paper we will use the facts
that $G(t,s)$ is nonnegative, and for each fixed $s\in[0,1]$,
the Green's function is nondecreasing in $t$.

 Let $\tau \in (0,1)$ and define the cone $P \subset E=C[0,1]$ by
$$
P := \left\{ x \in E : \text{$x$ is nonnegative, nondecreasing,
concave and} \; x(\tau) \geq \tau x(1)  \right\};
$$
for $x \in P$, define the concave functional $\alpha$ on $P$  by
$$
 \alpha(x):=\min_{t\in[\tau,1]}  x(t) =  x(\tau)
$$
and the convex functional $\beta$ on $P$ by
$$
 \beta(x):= \max_{t\in[0,1]}  x(t) =  x(1).
$$
In the following theorem, we demonstrate how to apply the expansive
condition of Theorem $\ref{main}$ to prove the existence of at
least one positive solution to \eqref{b1}, \eqref{b2}.

\begin{theorem} \label{main2}
If $\tau \in (0,1)$ is fixed, $b$ and $c$ are positive real
numbers with $3b \leq c$, and  $f: [0,\infty) \to [0,\infty)$ is a
continuous function such that
\begin{itemize}
\item[(a)] $f(w) > \frac{c}{\tau (1-\tau)}$ for
 $w \in \left[c, \frac{c}{\tau}\right]$,
\item[(b)] $f(w)$ is decreasing for $w \in [0,b\tau]$ with
 $f(b\tau) \geq f(w)$ for $w \in [b\tau,b]$, and
\item[(c)] $\int_0^{\tau}  s  f(bs) \,ds <
\frac{2b - f(b\tau)(1-\tau^2)}{2}$,
\end{itemize}
 then the focal problem  \eqref{b1}, \eqref{b2} has at least one
positive solution $x^* \in P(\beta,\alpha,b,c)$.
\end{theorem}


\begin{proof}
If we let  $a = b \tau$ and $d = c/\tau$, then we have that $a<c$
and $b<d$ since $3b \leq c$. For $x \in P(\beta,\alpha,b,c)$,
if $t \in (0,1)$, then by the properties of the Green's function
$(Tx)''(t) = -f(x(t))$
and $Tx(0)= 0 = (Tx)'(1)$.
For any $y,w \in [0,1]$ with $y \leq w$ we have the following
important property of the Green's function,
\begin{equation}\label{G}
  \min_{s \in [0,1]} \frac{G(y,s)}{G(w,s)} \geq \frac{y}{w};
\end{equation}
thus for any $x \in P$ we have that
\begin{align*}
\alpha(Tx)
&=  Tx(\tau) = \int_0^1 G(\tau,s) \;  f(x(s)) \,ds \\
&\geq  \int_0^1 \tau  G(1,s)   f(x(s)) \,ds
 = \tau Tx(1) = \tau \beta(Tx).
\end{align*}
Therefore we have that $T:P \to P$. By the Arzela-Ascoli Theorem
it is a standard exercise to show that $T$ is a completely
continuous operator using the properties of $G$ and $f$. We also
point out that $P(\alpha,c)$ is a bounded subset of the cone $P$,
since if $x \in P(\alpha,c)$, then
$$
\tau \beta(x) \leq \alpha(x) \leq c,
$$
and so
$$
\|x\| = \beta(x) \leq \frac{\alpha(x)}{\tau} \leq \frac{c}{\tau}.
$$
Also, if $x \in P(\beta,b)$, then
$$
\alpha(x) \leq \beta(x) \leq b <c,
$$
and hence $P(\beta,b) \subset P (\alpha,c)$.

For any $M \in (2b,c)$ the function $x_M$ defined by
$$
x_M(t)\equiv  \int_{0}^{1} M G(t,s) ds = \frac{Mt(2-t)}{2}
\in P(\beta,\alpha,b,c),
$$
since
$$
\alpha(x_M) = x_M(\tau) = \frac{M\tau(2-\tau)}{2}
< \frac{c \tau(2-\tau)}{2} \leq c
$$
and
$$
\beta(x_M) = x_M(1) = \frac{M}{2} >  b.
$$
Consequently we have that $\{x\in  P : b < \beta (x)
\text{ and } \alpha (x) < c\} \neq \emptyset$.

Similarly, for any $L \in \big(\frac{2b}{2-\tau},2b\big)$
the function $x_L$ defined by
$$
x_L(t)\equiv  \int_{0}^{1} L G(t,s) ds
= \frac{Lt(2-t)}{2} \in \{x\in  P : a < \alpha (x) \text{ and }
\beta (x) < b\},
$$
since
$$
\alpha(x_L) = x_L(\tau) = \frac{L\tau(2-\tau)}{2} > b \tau = a
$$
and
$$
\beta(x_L) = x_L(1) = \frac{L}{2} <  b.
$$
Likewise, for any
$J \in \big(\frac{2c}{\tau(2-\tau)},\frac{2c}{\tau}\big)$,
 the function $x_J$ defined by
$$
x_J(t)\equiv  \int_{0}^{1} J G(t,s) ds = \frac{Jt(2-t)}{2}
\in \{x\in  P : c < \alpha (x) \text{ and } \beta (x) < d\},
$$
since
$$
\alpha(x_J) = x_J(\tau) = \frac{J\tau(2-\tau)}{2} > c
$$
and
$$
\beta(x_J) = x_J(1) = \frac{J}{2} <   \frac{c}{\tau}  =d.
$$
We have that both
$$
\{x\in  P : a < \alpha (x) \text{ and } \beta (x) < b\} \neq \emptyset,
$$
and
$$
\{x\in  P : c < \alpha (x) \text{ and }\beta (x) < d\} \neq \emptyset,
$$
 and hence conditions (A1) and (A4) of Theorem \ref{main} are satisfied.


\noindent\textbf{Claim $1$:} $\beta(Tx) < b$ for all $x \in P$
with $\beta(x) =b$ and $\alpha(x) \geq a$.
 Let $x \in P$ with $\beta(x) =b$ and $\alpha(x) \geq a$.
By the concavity of $x$, for $s \in [0,\tau]$  we have
$$
x(s) \geq \Big(\frac{x(\tau)}{\tau}\Big)  s \geq bs,
$$
and for all $s \in [\tau,1]$, we have
$b \tau \leq x(s) \leq b$.
Hence by properties (b) and (c), it follows that
\begin{align*}
\beta(Tx)
&= \int_0^1 \; G(1,s) \; f(x(s)) \,ds
= \int_0^1  s  f(x(s)) \,ds \\
&= \int_0^{\tau}  s  f(x(s)) \,ds  + \int_{\tau}^1  s  f(x(s)) \,ds \\
&\leq \int_0^{\tau}  s  f(bs) \,ds  + f(b\tau) \int_{\tau}^1  s  \,ds \\
&<  \frac{2b - f(b\tau)(1-\tau^2)}{2}
+  \frac{f(b\tau)(1-\tau^2)}{2} = b.
\end{align*}


\noindent\textbf{Claim $2$:}
  If $x\in P$ and $\alpha(Tx) < a$, then $\beta (Tx) < b$.
 Let  $x \in P$ with $\alpha(Tx)< a$.  Thus by the properties
of $G(t,s)$ given in \eqref{G},
\begin{align*}
\beta(Tx)
&=  \int_0^1 \; G(1,s) \; f(x(s)) \,ds\\
&\leq   \big(\frac{1}{\tau}\big)\int_0^1 G(\tau,s)  f(x(s)) \,ds \\
&=  \big(\frac{1}{\tau}\big) \alpha(Tx)
<  \big(\frac{a}{\tau}\big) =b.
\end{align*}

\noindent\textbf{Claim $3$:}
  $\alpha(Tx) > c$ for all $x \in P$ with $\alpha(x) =c$ and
$\beta(x) \leq d$.
 Let $x \in P$ with $\alpha(x) =c$ and $\beta(x) \leq d$.
Then for $s \in [\tau,1]$  we have
$$
c \leq x(s) \leq d = \frac{c}{\tau}.
$$
Hence by property (a),
\begin{align*}
\alpha(Tx)
&= \int_0^1 \; G(\tau,s)  f(x(s)) \,ds
\geq \int_{\tau}^{1}  G(\tau,s) \; f(x(s)) \,ds  \\
&= \int_{\tau}^{1}  \tau  f(x(s)) \,ds
 >  \int_{\tau}^{1}  \frac{c}{1-\tau} \,ds =c.
\end{align*}


\noindent\textbf{Claim $4$:} If $x\in P$ and $\beta(Tx) > d$,
then $\alpha (Tx) > c$.
Let $x \in P$ with $\beta(Tx) > d$. Again by the properties of $G$
given in \eqref{G},
\begin{align*}
\alpha(Tx)
&= \int_0^1  G(\tau,s)  f(x(s)) \,ds  \\
&\geq  \tau  \int_0^1  G(1,s)  f(x(s)) \,ds \\
&= \tau  \beta(Tx) > \tau d  = c.
\end{align*}
Therefore, the expansion hypotheses of Theorem \ref{main} have
been satisfied; thus the operator $T$ has at least one fixed
point $x^* \in P(\beta,\alpha,b,c)$, which is a desired solution
of \eqref{b1}, \eqref{b2}.
\end{proof}

\subsection*{Example} Let
$b=1$ $c=5$, and $\tau = 1/2$.
Then the boundary value problem
\[
  x'' + \frac{1}{\sqrt{x}}+ e^{x-2}=0,
\]
with right-focal boundary conditions
\[
  x(0)= 0 = x'(1),
\]
 has at least one positive solution $x^*$ which can be verified
by the above theorem, with
$1\le x^*(1)$ and $x^*(\tau)\le 5$.

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\end{document}