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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 253, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/253\hfil Extension of compression-expansion]
{An extension of the compression-expansion fixed point theorem of functional type}

\author[R. I. Avery, D. R. Anderson, J. Henderson \hfil EJDE-2016/253\hfilneg]
{Richard I. Avery, Douglas R. Anderson, Johnny Henderson}

\address{Richard I. Avery \newline
College of Arts and Sciences,
 Dakota State University,
Madison, South Dakota 57042, USA}
\email{rich.avery@dsu.edu}

\address{Douglas R. Anderson \newline
Department of Mathematics,
Concordia College,
Moorhead, MN 56562, USA}
\email{andersod@cord.edu}

\address{Johnny Henderson \newline
Department of Mathematics,
Baylor University,
 Waco, TX  76798, USA }
\email{Johnny\_Henderson@baylor.edu}

\thanks{Submitted July 11, 2016. Published September 21, 2016.}
\subjclass[2010]{47H10}
\keywords{Fixed-point theorem; k-contractive; expansion; compression}

\begin{abstract}
 In this article we use an interval of functional type as the underlying
 set in our compression-expansion fixed point theorem argument which can
 be used to exploit properties of the operator to improve conditions that
 will guarantee the existence of a fixed point in applications.
 An example is provided to demonstrate how intervals of functional type
 can improve conditions in applications to boundary value problems.
 We also show how one can use suitable $k$-contractive conditions to prove
 that a fixed point in a functional-type interval is unique.
\end{abstract}

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

\section{Introduction}

The compression-expansion fixed point theorems of functional type, 
see for example \cite{five,aah}, have relied on functional frustrums of a 
cone which are sets of the form
$$
P(\beta,b,\alpha,a) = \{x \in P : a < \alpha(x)  \text{ and }
 \beta(x) < b\},
$$
and if compression-expansion conditions are met, then one concludes 
that there is a fixed point for an operator $T$ in this set.  
In this paper we show that the underlying set can be generalized by using 
functional-type intervals which are subsets of $P(\beta,b,\alpha,a)$, that is,
$$
A(\beta,b,\alpha,a) = \{x \in A : a < \alpha(x)  \text{ and }  \beta(x) < b\},
$$
where $A$ is an open subset of the cone $P$ and most importantly
$$
\partial A \cap \overline{A(\beta,b,\alpha,a)} = \emptyset.
$$
In the spirit of the Leggett-Williams fixed point theorem \cite{leg}, 
which many of the compression-expansion fixed point theorems of functional 
type have generalized, we do not know that $T$ is invariant on 
$A(\beta,b,\alpha,a)$.  However, if suitable $k$-contractive conditions 
are met we can use similar arguments as presented in the Banach 
fixed point theorem (see \cite[{\em p} 17]{z} for a presentation of 
these concepts, and one can also see these techniques in the work of
 Petryshyn \cite{petr}), to prove that the fixed point of $T$ in  
$A(\beta,b,\alpha,a)$ is unique.  The uniqueness argument does not require
 iterates to converge nor does it require the operator to be invariant on 
the functional interval.

\section{Preliminaries}

 For completeness we provide the following definitions and theorems, 
which are nearly identical to the presentation in other compression-expansion 
fixed point papers, in particular \cite{aah}.

\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 $y - x \in P$.


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

In Theorem \ref{result} we will show how one can use functional-type 
intervals instead of functional frustrums of a cone 
(sets of the form $P(\beta,b,\alpha,a)$).  The definition of a 
functional type interval appears next.


\begin{definition} \label{def4} \rm 
Let $A$ be a relatively open subset of a cone $P$, $a$ and $b$ be nonnegative
 numbers, $\alpha$ be a concave functional on $P$, and $\beta$ be a convex 
functional on $P$. Then the set
$$
A(\beta,b,\alpha,a) = \{x \in A : a < \alpha(x) \text{ and } \beta(x) < b\}
$$
is an interval of functional type.
\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}.
\end{definition}

\begin{remark} \label{rmk} \rm 
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{remark}

  The following theorem is due to Dugundji and its proof can
be found in \cite[{\em 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 $\tilde{F}(X) \subset \overline{\operatorname{conv}(F(D))}$.
\end{theorem}

\begin{corollary} \label{R2}
Every closed convex set of a Banach space is a
retract of the Banach space.
\end{corollary}

The following theorem is developed from topological degree theory 
and the abstract (axiomatic) form that appears below can be found 
in \cite[p 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[(1)] Normality: $i(A,U,X) = 1$ if $Ax  \equiv y_{0} \in U$ for any 
 $x \in \overline{U}$;
 \item[(2)] 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[(3)] 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[(4)] 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}

There are many functional fixed point theorems in the literature, 
for example see the following papers 
\cite{aa,five, aah,guo1,guo2,kr,leg, pet,petr,sun} for foundational 
arguments related to functional fixed point theorems.  
The following functional fixed point theorem extends the conclusions 
of the fixed point theorems of functional type in the literature by 
providing a uniqueness condition and a generalization of the underlying 
set in which the fixed point is known to exist.  
The underlying set in our arguments is called a functional type interval 
and provides a mechanism for properties of the operator to be introduced 
into applications of the theorem which is illustrated at the conclusion 
of this paper.

\begin{theorem}\label{result}
Suppose $P$ is a cone in a real Banach space $E$, $A$ is a relatively open subset 
of $P$, $\alpha$ and $\psi$ are nonnegative continuous concave functionals on 
$P$,  $\beta$ and $\theta$  are nonnegative continuous convex functionals on $P$, 
and $T: P \to P$ is a completely continuous operator.  If there exist nonnegative 
numbers $a, b, c, $ and $d$ such that
\begin{enumerate}
\item[(A1)] $A(\beta,b,\alpha,a)$ is bounded, 
 $A(\beta,b,\alpha,a) \cap A(\theta,c,\psi,d) \neq \emptyset$, and \\
 $\partial A \cap \overline{A(\beta,b,\alpha,a)} = \emptyset$;

\item[(A2)] if $x \in \partial A(\beta,b,\alpha,a)$ with $\alpha(x) =a$ and either 
 $\theta(x)\leq c$ or $\theta(Tx) > c$, then $\alpha (Tx) > a$;

\item[(A3)] if $x \in \partial A(\beta,b,\alpha,a)$ with  $\beta(x) =b$ and either 
$\psi(Tx) < d$ or $\psi(x)\geq d$, then $\beta (Tx) < b$;
\end{enumerate}
then $T$ has a fixed point $x^* \in A(\beta,b,\alpha,a)$.  Moreover, if for all 
$x \in A(\beta,b,\alpha,a)$ there exists a $k \in [0,1)$ such that
$$
 \|Tx - x^*\| \leq k\|x - x^*\|
$$
then $x^*$ is the unique fixed point of $T$ in $A(\beta,b,\alpha,a)$.
\end{theorem}


\begin{proof}  
By Corollary \ref{R2}, $P$ is a retract of the Banach space $E$ since it is 
convex  and closed.
\smallskip

\noindent  \textbf{Claim 1:}
 $Tx \neq x$ for all $x\in \partial A(\beta,b,\alpha,a)$.
  The functional interval $A(\beta,b,\alpha,a) = A \cap P(\beta,b,\alpha,a)$, hence
\begin{align*}
&\partial A(\beta,b,\alpha,a) \\
& =  \partial (A \cap P(\beta,b,\alpha,a)) \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap
  \overline{(P - (A \cap P(\beta,b,\alpha,a)))} \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap 
 \overline{(P - A) \cup (P-P(\beta,b,\alpha,a))} \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap 
 (\overline{(P - A)} \cup \overline{(P-P(\beta,b,\alpha,a))})\\
& \subseteq  (\overline{A}\cap \overline{ P(\beta,b,\alpha,a)}) 
 \cap (\overline{(P - A)} \cup \overline{(P-P(\beta,b,\alpha,a))}) \\
& = (\overline{A}\cap \overline{ P(\beta,b,\alpha,a)} \cap 
 \overline{(P - A)}) \cup (\overline{A}\cap 
 \overline{ P(\beta,b,\alpha,a)} \cap \overline{(P-P(\beta,b,\alpha,a))}) \\
& = (\partial A \cap \overline{ P(\beta,b,\alpha,a)}) \cup 
 (\overline{A}\cap \partial P(\beta,b,\alpha,a)) \\
& = \overline{A}\cap \partial P(\beta,b,\alpha,a) 
\end{align*}
since $\partial A \cap \overline{P(\beta,b,\alpha,a)} 
= \partial A \cap \overline{A(\beta,b,\alpha,a)} = \emptyset$. 
Thus, if $z_0 \in \partial A(\beta,b,\alpha,a)$ then 
$z_0 \in \partial P(\beta,b,\alpha,a)$ so either $\beta (z_0) = b$ or 
$\alpha(z_0) = a$.  We want to show that $z_0$ is not a fixed point of $T$;
 so, suppose to the contrary that $T(z_0)= z_0$.
\smallskip

\noindent\textbf{Case 1.1: $\beta (z_0) = b$.}
If $\psi(Tz_0) < d$ or $\psi(z_0) = \psi(Tz_0) \geq d$, then 
$\beta(Tz_0) < b$ by condition (A3).  Hence we have that $Tz_0 \neq z_0$.
\smallskip

\noindent\textbf{Case 1.2: $\alpha (z_0) = a$.}
If $\theta(Tz_0) > c$ or $\theta(Tz_0) = \theta(z_0) \leq c$, then 
$\alpha(Tz_0) > a$ by condition (A2).  Hence we have that $Tz_0 \neq z_0$.

Therefore, $T$ does not have any fixed points on $\partial A(\beta,b,\alpha,a)$.

  Let $w_0 \in A(\beta,b,\alpha,a) \cap A(\theta,c,\psi,d)$  
(see condition (A1)), and define
$H: [0,1] \times \overline{A(\beta,b,\alpha,a)} \to P$
by
$$
H(t,x)=(1-t)Tx + t w_0.
$$
Clearly, $H$ is continuous and  $H([0,1] \times \overline{A(\beta,b,\alpha,a))}$ 
is precompact.
\smallskip

\noindent\textbf{Claim 2:}  $H(t,x) \neq x$ for all $(t,x)
 \in [0,1] \times \partial A(\beta,b,\alpha,a)$.
Suppose not; that is, suppose there exists
 $(t_0,x_0) \in [0,1] \times \partial A(\beta,b,\alpha,a)$ such that 
$$
H(t_0,x_0)=x_0.
$$ 
Since $x_0 \in \partial A(\beta,b,\alpha,a)$, we have that $\beta(x_0) = b$ 
or $\alpha(x_0) = a$ since $\partial A \cap \overline{A(\beta,b,\alpha,a)} 
= \emptyset$.  Also, since $T$ has no fixed point on $\partial A(\beta,b,\alpha,a)$, 
we have that $t_0 \neq 0$.
\smallskip

\noindent\textbf{Case 2.1: $\beta (x_0) = b$.}
 Either $\psi(Tx_0) < d$ or $\psi(Tx_0) \geq d$.
\smallskip

\noindent\textbf{Subcase 2.1.1: $\psi(Tx_0) < d$.}
 By condition (A3) we have $\beta(Tx_0) < b$, thus it follows that
$$
b    =   \beta(x_0) = \beta((1-t_0)Tx_0 + t_0 w_0)
            \leq  (1-t_0)\beta(Tx_0) + t_0\beta(w_0) < b,$$
which is a contradiction.
\smallskip

\noindent\textbf{Subcase 2.1.2: $\psi(Tx_0) \geq d$.} 
We have that $\psi(x_0) \geq d$ since
$$
\psi(x_0)    =   \psi((1-t_0)Tx_0 + t_0w_0)
            \geq  (1-t_0)\psi(Tx_0) + t_0\psi(w_0)
             \geq d,
$$
and thus by condition (A3) we have $\beta(Tx_0) < b$,
which is the same contradiction we arrived at in the previous subcase.
\smallskip

\noindent\textbf{Case 2.2: $\alpha (x_0) = a$.}
 Either $\theta(Tx_0) \leq c$ or $\theta(Tx_0) > c$.
\smallskip

\noindent\textbf{Subcase 2.2.1: $\theta(Tx_0) > c$.} 
By condition (A2) we have $\alpha(Tx_0) > a$, thus we have
$$
a    =   \alpha(x_0) = \alpha((1-t_0)Tx_0 + t_0w_0)
            \geq  (1-t_0)\alpha(Tx_0) + t_0\alpha(w_0) > a,$$
which is a contradiction.
\smallskip

\noindent\textbf{Subase 2.2.2: $\theta(Tx_0) \leq c$.} 
We have that $\theta(x_0) \leq c$ since
$$
\theta(x_0)    =   \theta((1-t_0)Tx_0 + t_0w_0)
            \leq  (1-t_0)\theta(Tx_0) + t_0\theta(w_0)
             \leq  c,
$$
and thus by condition (A2) we have $\alpha(Tx_0) > a$,
which is the same contradiction we arrived at in the previous case.

  Therefore, we have shown that  $H(t,x) \neq x$ for all 
$(t,x) \in [0,1] \times \partial A(\beta,b,\alpha,a)$, and thus by the 
homotopy invariance property of the fixed point index
$$
i(T,A(\beta,b,\alpha,a),P) = i(w_0,A(\beta,b,\alpha,a),P),
$$
and by the normality property  of the fixed point index
$$
i(T,A(\beta,b,\alpha,a),P) = i(w_0,A(\beta,b,\alpha,a),P)=1.
$$
Consequently by the solution property of the fixed point index, $T$ 
has a fixed point $x^* \in A(\beta,b,\alpha,a)$.

  Furthermore, if for all $x \in A(\beta,b,\alpha,a)$  there exists a $k \in [0,1)$ 
such that
$$
\|Tx - x^*\| \leq k\|x - x^*\|,
$$
then for any fixed point $z^* \in A(\beta,b,\alpha,a)$ we have that
$$
\|z^* - x^*\| =\|Tz^* - x^*\| \leq k \|z^* - x^*\|.
$$
Therefore $\|z^* - x^*\| = 0$ as $k<1$, and we have verified that under this 
condition $T$ has a unique fixed point in $A(\beta,b,\alpha,a)$.
\end{proof}

\section{Application}

In this section, using a non-standard functional technique 
(using an evaluation of the derivative as one of the functionals) and an open 
subset $A$ of the cone $P$ (the set $A$ is not bounded nor is it a cone), 
we will illustrate the key techniques for verifying the existence and uniqueness 
of a positive solution for a right focal boundary value problem in our interval 
of functional type using our main result. Note that the resulting conditions 
for a fixed point to exist in our functional-type interval 
(conditions $(a)$ and $(b)$ of Theorem \ref{main} below) do not force the 
operator $T$ to be invariant on our functional-type interval. 
 We consider the classical right 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.
It is well known that 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
$$
G(t,s)=\min\{t,s\}, \quad (t,s)\in[0,1]\times[0,1],
$$
then $x$ is a solution of the boundary value problem \eqref{b1}, \eqref{b2}.

 Define the cone $P \subset E=C^1[0,1]$ by
$$ 
P := \{ x \in E : \text{$x$ is nonnegative, nondecreasing, concave, and }
 x(0)=0  \}. 
$$
For $x \in P$, define the convex functional $\beta$ on $P$ by
$$ 
\beta(x):= \max_{t\in[0,1]}  x(t) =  x(1), 
$$
the (concave) functional $\psi$ on  $P$ by
$$
\psi(x):=x'\big(\frac14\big),
$$
and the concave functional $\alpha$ on $P$ by
$$ 
\alpha(x):= \min_{t\in[\frac14,1]}  x(t) =  x\big(\frac14\big). 
$$
We are now ready to prove the existence of a unique positive solution to
 \eqref{b1}, \eqref{b2} in our functional-type interval, 
if the conditions in the following theorem are satisfied.

\begin{theorem} \label{main}
If $b > 0$, and  $f: [0,20b] \to [0,\infty)$ is a continuous differentiable 
function such that
\begin{itemize}
\item[(a)] $\frac{256b}{11} < f(w) < 80b$ for $w \in [0,8b]$,
\item[(b)] $\frac{256b}{11} < f(w) < \frac{368b}{11}$ for $w \in [8b,20b]$,
\item[(c)] $|f'(w)| < 16$ for $w \in [0,5b]$, and
\item[(d)] $|f'(w)| < 1$ for $w \in [5b,20b]$,
\end{itemize}
then the right focal problem  \eqref{b1}, \eqref{b2} has a unique positive 
solution 
\[
x^* \in A(\beta,20b,\alpha,5b).
\]
\end{theorem}

\begin{proof}
Let
\begin{gather*}
A = \big\{x \in P : x\big(\frac38\big) - x\big(\frac14\big) > 2b \big\},\\
A(\beta,20b,\alpha,5b) = \big\{x \in A : 5b < x(t) < 20b  \text{ for all } 
 \frac14 \leq t \leq 1\big\}.
\end{gather*}
Thus $A$ is an open subset of $P$ and $A(\beta,20b,\alpha,5b)$  is  an open,
 nonempty, bounded, subset of $P$.  By properties of Green's function we have
$$
(Tx)''(t) = -f(x(t)) \text{ and } Tx(0)= 0 = (Tx)'(1);
$$
that is, fixed points of the operator $T$ are solutions of the boundary 
value problem  \eqref{b1}, \eqref{b2}.  
Applying the Arzela-Ascoli Theorem we have
$T: \overline{A(\beta,20b,\alpha,5b)} \to P$ is a completely continuous operator, 
and applying Dugunji's Theorem, there is a continuous extension, which we 
will again denote by $T$, such that  $T:P \to P$ (the extension is necessary 
to extend the domain of $T$ from $\overline{A(\beta,20b,\alpha,5b)}$ to $P$ 
since $f$ is only defined on $[0,20b]$).

  Letting $x \in \overline{A(\beta,20b,\alpha,5b)}$ we see that 
$\alpha(x) = x(\frac14) \geq 5b$;
therefore
\begin{align*}
(Tx)\big(\frac38\big) - (Tx)\big(\frac14\big) 
& = \int_{1/4}^{3/8} \big(s-\frac14\big)  
f(x(s)) \,ds +\big(\frac18\big)\int_{3/8}^1   f(x(s)) \,ds \\
& > \int_{1/4}^{3/8} \big(s-\frac14\big)  \frac{256b}{11} \,ds
+\big(\frac18\big)\int_{3/8}^1  \frac{256b}{11} \,ds =2b.
\end{align*}
Hence, $x \not \in \partial A$ thus
 $\partial A \cap \overline{A(\beta,20b,\alpha,5b)} = \emptyset$.
Let
$$
w_0(t) = \int_0^1 G(t,s)  36b \,ds
$$
thus
\begin{gather*}
\beta(w_0) = \theta(w_0) = w_0(1) = \int_0^1 G(1,s)  36b \,ds = 18b < 20b,\\
\alpha(w_0) = w_0\big(\frac14\big) = \int_0^1 G\left(\frac14,s\right)  36b \,ds
 = \frac{63b}{8} > 5b,\\
\psi(w_0) = w_0'\big(\frac14\big) = \int_{1/4}^{1} 35b \,ds
= \frac{105b}{4} >24b
\end{gather*}
hence 
$$
w_0 \in A(\beta,20b,\alpha,5b) \cap  A(\beta,20b,\psi,24b).
$$
\smallskip

\noindent\textbf{Claim 1:}
 If $x\in \partial A(\beta,20b,\alpha,5b)$ with   $\beta(x) =20b$ 
and $\psi(x)\geq 24b$, then $\beta (Tx) < 20b$.

Since $\psi(x)\geq 24b$ and $x$ is increasing, 
$x\big(\frac14\big) \geq 6b$ since
$$
6b \leq \frac{\psi(x)}{4} = \int_0^{1/4}  x'\big(\frac14\big) \,ds 
\leq \int_0^{1/4}   x'(s) \,ds
= x\big(\frac14\big).
$$
Also, since $x \in A$, $x\big(\frac38\big) \geq x\big(\frac14\big) + 2b \geq 8b$.
Therefore,
\begin{align*}
\beta(Tx) 
& = \int_0^1  G(1,s)  f(x(s)) \,ds  = \int_0^1  s  f(x(s)) \,ds \\
& = \int_0^{3/8}  s  f(x(s)) \,ds  + \int_{3/8}^1  s  f(x(s)) \,ds \\
& < \int_0^{3/8}  80bs  \,ds  + \int_{3/8}^1  \frac{368bs}{11}  \,ds \\
& =  80b\big(\frac{9}{128}\big) + \big(\frac{368b}{11}\big)
\big(\frac{55}{128}\big) = 20b.
\end{align*}
\smallskip

\noindent\textbf{Claim 2:}
 If $x\in \partial A(\beta,20b,\alpha,5b)$ with  $\beta(x) =20b$ and 
$\psi(Tx) < 24b$, then $\beta (Tx) < 20b$.

Let $x\in \partial A(\beta,20b,\alpha,5b)$ with $\beta(x) =20b$ and 
$\psi(Tx) < 24b$. Then
\begin{align*}
24b   &  >   \psi(Tx) = (Tx)'\big(\frac14\big)  \\
& =  \int_{1/4}^1  f(x(s)) \,ds \\
& =  \int_{1/4}^1  s  f(x(s)) \,ds + \int_{1/4}^1  (1-s)  f(x(s)) \,ds,
\end{align*}
and since we have that
$$
\int_{1/4}^1  (1-s)  f(x(s)) \,ds \geq \int_{1/4}^1 \frac{256b(1-s)}{11}
\,ds > \frac{13b}{2},
$$
we also have that
$$
\int_{1/4}^1  s  f(x(s)) \,ds < \frac{35b}{2}.
$$
Thus,
\begin{align*}
\beta(Tx) 
& = \int_0^1 \; G(1,s)  f(x(s)) \,ds   \\
& = \int_0^{1/4}  s  f(x(s)) \,ds  + \int_{1/4}^1  s  f(x(s)) \,ds \\
& < \int_0^{1/4}  80bs  \,ds  + \frac{35b}{2}\\
& =  \frac{5b}{2} +  \frac{35b}{2} =20b.
\end{align*}
Hence, the previous two claims have verified condition (A3)
 of Theorem \ref{result} is satisfied.
\smallskip

\noindent\textbf{Claim 3:}
 If $x\in \partial A(\beta,20b,\alpha,5b)$ with  $\alpha(x) =5b$, then 
$\alpha(Tx) > 5b$.

  Let $x\in \partial A(\beta,20b,\alpha,5b)$ with $\alpha(x) =5b$. Then
\begin{align*}
\alpha(Tx) 
& = \ Tx\big(\frac14\big) 
= \int_0^1  G\big(\frac14,s\big)  f(x(s)) \,ds   \\
& = \int_0^{1/4}   s  f(x(s)) \,ds
 + \int_{1/4}^1   \frac{f(x(s))}{4} \,ds \\
& >  \int_0^{1/4}   s  \frac{256b}{11} \,ds
 + \int_{1/4}^1   \frac{256b}{44} \,ds > 5b.
\end{align*}
Therefore, condition (A2)  of Theorem \ref{result} is satisfied since 
we have shown that
\[
\text{if $x\in \partial A(\beta,20b,\alpha,5b)$ with $\alpha(x) =5b$ implies
 $\alpha(Tx) > 5b$}
\]
which guarantees that
\begin{quote}
if $x \in \partial A(\beta,b,\alpha,a)$ with $\alpha(x) =a$ and either
$\theta(x)\leq c$ or $\theta(Tx) > c $ implies $\alpha (Tx) > a$.
\end{quote}
Therefore, by Theorem \ref{result}, the operator $T$ has a fixed point 
$x^* \in A(\beta,20b,\alpha,5b)$
which is a desired solution of \eqref{b1}, \eqref{b2}. Furthermore, 
if $x \in A(\beta,20b,\alpha,5b)$, then
\begin{align*}
\|Tx-Tx^*\| 
& = \max_{t\in [0,1]} \big|\int_0^1  G(t,s)  (f(x(s))-f(x^*(s))) \,ds\big|   \\
& \leq \max_{t\in [0,1]} \int_0^1 \; G(t,s)  | f(x(s))-f(x^*(s))| \,ds   \\
& =  \int_0^1  s  | f(x(s))-f(x^*(s))| \,ds   \\
& =  \int_0^{1/4}  s  | f(x(s))-f(x^*(s))| \,ds
 + \int_{1/4}^1  s | f(x(s))-f(x^*(s))| \,ds  \\
& \leq  \int_0^{1/4}  16s  \|x-x^*\| \,ds
+ \int_{1/4}^1  s \|x-x^*\| \,ds  \\
& =  \big(\frac{31}{32}\big)\|x-x^*\|.
\end{align*}
Thus
$$
\|Tx-x^*\| = \|Tx-Tx^*\|  \le \big(\frac{31}{32}\big)\|x-x^*\|,
$$
hence $x^*$ is the unique fixed point of $T$ in $ A(\beta,20b,\alpha,5b)$.
\end{proof}


\section*{Acknowledgements}
The authors would like to thank the anonymous referee for the careful
 reading of the manuscript and pertinent comments; 
the referee's constructive suggestions substantially improved the
 quality of this work.


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\section{Addendum posted on October 6, 2016}

The authors would like to modify Theorem \ref{result} (A1) to provide 
a more general fixed point theorem that we can use to correct the mistake 
in the application, Theorem \ref{main} (Theorem \ref{result} is correct, 
we are just slightly generalizing a condition).  
Below is the modified statement of the extension of the compression-expansion 
fixed point theorem of functional type which is presented as 
Theorem \ref{result2}, as well as the only component of the proof of 
Theorem \ref{result} that is modified.  Following Theorem \ref{result2} 
we'll outline the modifications necessary in the proof of the application 
to fix the boundedness issue.


\begin{theorem}\label{result2}
Suppose $P$ is a cone in a real Banach space $E$, $A$ is a relatively open subset 
of $P$, $\alpha$ and $\psi$ are nonnegative continuous concave functionals on 
$P$,  $\beta$ and $\theta$  are nonnegative continuous convex functionals on $P$, 
and $T: P \to P$ is a completely continuous operator.  If there exist nonnegative 
numbers $a, b, c, $ and $d$ such that
\begin{enumerate}
\item[(A1)] $A(\beta,b,\alpha,a)$ is bounded, 
 $A(\beta,b,\alpha,a) \cap A(\theta,c,\psi,d) \neq \emptyset$, and \\
  if $x \in \partial A \cap \overline{P(\beta,b,\alpha,a)}$ then $Tx \neq x$;

\item[(A2)] if $x \in \partial A(\beta,b,\alpha,a)$ with $\alpha(x) =a$ and either 
 $\theta(x)\leq c$ or $\theta(Tx) > c$, then $\alpha (Tx) > a$;

\item[(A3)] if $x \in \partial A(\beta,b,\alpha,a)$ with  $\beta(x) =b$ and either 
$\psi(Tx) < d$ or $\psi(x)\geq d$, then $\beta (Tx) < b$;
\end{enumerate}
then $T$ has a fixed point $x^* \in A(\beta,b,\alpha,a)$.  Moreover, if for all 
$x \in A(\beta,b,\alpha,a)$ there exists a $k \in [0,1)$ such that
$$
 \|Tx - x^*\| \leq k\|x - x^*\|
$$
then $x^*$ is the unique fixed point of $T$ in $A(\beta,b,\alpha,a)$.
\end{theorem}


\begin{proof}  
By Corollary \ref{R2}, $P$ is a retract of the Banach space $E$ since it is 
convex  and closed.
\smallskip

\noindent  \textbf{Claim 1:}
 $Tx \neq x$ for all $x\in \partial A(\beta,b,\alpha,a)$.
  The functional interval $A(\beta,b,\alpha,a) = A \cap P(\beta,b,\alpha,a)$, hence
\begin{align*}
&\partial A(\beta,b,\alpha,a) \\
& =  \partial (A \cap P(\beta,b,\alpha,a)) \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap
  \overline{(P - (A \cap P(\beta,b,\alpha,a)))} \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap 
 \overline{(P - A) \cup (P-P(\beta,b,\alpha,a))} \\
& =  \overline{(A \cap P(\beta,b,\alpha,a))} \cap 
 (\overline{(P - A)} \cup \overline{(P-P(\beta,b,\alpha,a))})\\
& \subseteq  (\overline{A}\cap \overline{ P(\beta,b,\alpha,a)}) 
 \cap (\overline{(P - A)} \cup \overline{(P-P(\beta,b,\alpha,a))}) \\
& = (\overline{A}\cap \overline{ P(\beta,b,\alpha,a)} \cap 
 \overline{(P - A)}) \cup (\overline{A}\cap 
 \overline{ P(\beta,b,\alpha,a)} \cap \overline{(P-P(\beta,b,\alpha,a))}) \\
& = (\partial A \cap \overline{ P(\beta,b,\alpha,a)}) \cup 
 (\overline{A}\cap \partial P(\beta,b,\alpha,a)). 
\end{align*}
If $z_0 \in \partial A(\beta,b,\alpha,a)$ then 
$z_0 \in \partial P(\beta,b,\alpha,a)$ so either $\beta (z_0) = b$ or 
$\alpha(z_0) = a$.  We want to show that $z_0$ is not a fixed point of $T$;
 so, suppose to the contrary that $T(z_0)= z_0$.
\smallskip

\noindent\textbf{Case 1.1: $\beta (z_0) = b$.}
If $\psi(Tz_0) < d$ or $\psi(z_0) = \psi(Tz_0) \geq d$, then 
$\beta(Tz_0) < b$ by condition (A3).  Hence we have that $Tz_0 \neq z_0$.
\smallskip

\noindent\textbf{Case 1.2: $\alpha (z_0) = a$.}
If $\theta(Tz_0) > c$ or $\theta(Tz_0) = \theta(z_0) \leq c$, then 
$\alpha(Tz_0) > a$ by condition (A2).  Hence we have that $Tz_0 \neq z_0$.

Therefore, $T$ does not have any fixed points on $\partial A(\beta,b,\alpha,a)$ 
since we just verified that $T$ has no fixed points on
 $\overline{A}\cap \partial P(\beta,b,\alpha,a)$ and we assumed $T$ did not 
have any fixed points on $\partial A \cap \overline{ P(\beta,b,\alpha,a)}$. 
\smallskip

\noindent There are no other changes to the proof of Theorem \ref{result}.
\end{proof}


In the application, Theorem \ref{main}, we need to modify the underlying 
set so that it is bounded in the $C^1[0,1]$ norm.  Below are the modifications 
to the beginning of the proof that fixes this issue, there are no changes 
needed in the statement nor the remainder of the proof of the application.

\begin{proof}
Let
\begin{gather*}
A = \big\{x \in P : x\big(\frac38\big) - x\big(\frac14\big) > 2b \;\; \mbox{and} \;\; x'(0) < 80b \big\},\\
A(\beta,20b,\alpha,5b) = \big\{x \in A : 5b < x(t) < 20b  \text{ for all } 
 \frac14 \leq t \leq 1\big\}.
\end{gather*}
By properties of Green's function we have
$$
(Tx)''(t) = -f(x(t)) \text{ and } Tx(0)= 0 = (Tx)'(1);
$$
that is, fixed points of the operator $T$ are solutions of the boundary 
value problem  \eqref{b1}, \eqref{b2}.  
Applying the Arzela-Ascoli Theorem we have
$T: \overline{A(\beta,20b,\alpha,5b)} \to P$ is a completely continuous operator, 
and applying Dugunji's Theorem, there is a continuous extension, which we 
will again denote by $T$, such that  $T:P \to P$ (the extension is necessary 
to extend the domain of $T$ from $\overline{A(\beta,20b,\alpha,5b)}$ to $P$ 
since $f$ is only defined on $[0,20b]$).  
Also note that if $x \in A(\beta,20b,\alpha,5b)$ that
$$
\|x\| = \sup_{t \in [0,1]}|x(t)| + \sup_{t \in [0,1]}|x'(t)| = x(1)+x'(0) 
\leq 20b + 80b = 100b
$$
thus $A(\beta,20b,\alpha,5b)$ is a bounded, open subset of $P$.

  Letting $x \in \overline{A(\beta,20b,\alpha,5b)}$ we see that 
$\alpha(x) = x(\frac14) \geq 5b$;
therefore
\begin{align*}
(Tx)\big(\frac38\big) - (Tx)\big(\frac14\big) 
& = \int_{1/4}^{3/8} \big(s-\frac14\big)  
f(x(s)) \,ds +\big(\frac18\big)\int_{3/8}^1   f(x(s)) \,ds \\
& > \int_{1/4}^{3/8} \big(s-\frac14\big)  \frac{256b}{11} \,ds
+\big(\frac18\big)\int_{3/8}^1  \frac{256b}{11} \,ds =2b.
\end{align*}
Also,
$$
(Tx)'(0)  =  \int_{0}^1 \; f(x(s)) \; ds < \int_{0}^1 \; 80b \; ds = 80b.
$$
Hence, $Tx\neq x$ if $x \in \partial A \cap \overline{P(\beta,20b,\alpha,5b)}$. 
Let
$$
w_0(t) = \int_0^1 G(t,s)  36b \,ds
$$
thus
\begin{gather*}
\beta(w_0) = \theta(w_0) = w_0(1) = \int_0^1 G(1,s)  36b \,ds = 18b < 20b,\\
\alpha(w_0) = w_0\big(\frac14\big) = \int_0^1 G\left(\frac14,s\right)  36b \,ds
 = \frac{63b}{8} > 5b,\\
\psi(w_0) = w_0'\big(\frac14\big) = \int_{1/4}^{1} 36b \,ds
= 27b >24b
\end{gather*}
hence 
$$
w_0 \in A(\beta,20b,\alpha,5b) \cap  A(\beta,20b,\psi,24b).
$$
\smallskip

\noindent There are no other changes to the proof of Theorem \ref{main}.
\end{proof}


\subsection*{Acknowledgements}
The authors would like to thank Professor Jeff Webb for pointing out 
the application error and for his thorough review and comments of our results.

End of addendum.

\end{document}