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\fancyhead[RE]{ A. Karamian and R. Lashkaripour} 
\fancyhead[LO]{Fixed point results for increasing mapping and  the relationship between (relative) algebraic interior and topological interior }



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{\noindent Journal of Mathematical Extension \\
Vol. XX, No. XX, (2017), pp-pp (Will be inserted by layout editor)}
\vspace*{9mm}

\begin{center}

{\Large \bf 
Fixed point results for increasing mapping  and  the relationship between (relative) algebraic interior and topological interior\\}

\let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)}
{\bf  A. Karamian$^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\
\vspace{2mm} {\small   University 
of Sistan and Baluchestan} \vspace{2mm}

{\bf R. Lashkaripour}\vspace*{-2mm}\\
\vspace{2mm} {\small   University 
of Sistan and Baluchestan} \vspace{2mm}


\end{center}

\vspace{4mm}


{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.}  In this paper, we show that the relative algebraic interior is a suitable replacement  for both of the topological interior and the algebraic interior for the cases where these are empty. Also, we presente some properties of (relative) algebraic interior and  some fixed point theorems for increasing mapping. The results obtained can be viewed as an extension and improvement of the known corresponding results. Some examples are provided here to support our conclusions.
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification:} 47H10

\noindent{\bf Keywords and Phrases:} fixed point; nonlinear scalarization mapping; algebraic interior; relative algebraic interior.
\end{quotation}}

\section{Introduction and preliminaries}
\label{intro} % It is advised to give each section and subsection a unique label.
In this section we recall some definitions and facts to set up our results in the sequel.

\begin{definition} \cite{A. Karamian}
 Let $X$ be a real vector space with the zero vector $\theta_{X}$. A nonempty, nontrivial(i.e., $ P\neq 
\lbrace \theta_{X}\rbrace$), closed subset $P$ of $X$ is called cone if  the following conditions are satisfied;
\begin{itemize}
\item [(i)] $2P\subseteq P$;
\item [(ii)] $tx+(1-t)y\in P, ~\forall(x,y,t) \in P \times P \times[0,1]$;
\item [(iii)] $P\cap (-P)=\lbrace\theta_{X}\rbrace.$
\end{itemize}
 
\end{definition}


A cone $P \subseteq X$ defines an ordering $\preceq_{P}$  on $X$ with respect to $P$ by letting $x\preceq_{P} y$ whenever $y-x\in P$. We use the notation  $x\prec _{p}y$ for $y-x\in P\smallsetminus \lbrace\theta_{X}\rbrace$.
 Also, if $X$ is a topological vector space and $P$ is a solid cone of $X$, i.e., $intP\neq \emptyset$, then we can define a pre-order(it is not reflexive) on $X$ by $$x \prec\prec_{p} y\Longleftrightarrow y-x\in intP.$$
 If the cone $P$ is known, for simplicity, we replace $\preceq_{P}$,  $\prec _{p}$ and $\prec \prec_{int P}$ by $\preceq$, $\prec $  and $\prec\prec$; respectively.
 Note that trough the paper we reserve the symbol $\leq$ and its obvious variants, for the usual order on $\mathbb{R}$.\\
 The pair $(X, P)$ consisting of a real normed space $X$ and a cone $P$ of $X$ is called a
partially ordered  normed space.

\begin{definition}\cite{core}\label{core} Let $S$ be a nonempty subset of a real linear space $X$. The algebraic interior of $S$, denoted by cor$(S)$, and the relative algebraic interior of $P$, denoted by icr$(P)$, are defined as follows:\\
cor$(S) :=\lbrace x \in S : \forall x^{\prime}\in X,  \exists \lambda^{\prime  }>0; \forall \lambda \in [0,\lambda^{\prime  }],  x+\lambda x^{\prime }\in S \rbrace $,\\
icr$(S) :=\lbrace x \in S : \forall x^{\prime}\in L(S), \exists \lambda^{\prime  }>0 ; \forall \lambda \in [0,\lambda^{\prime  }],  x+\lambda x^{\prime }\in S \rbrace $,\\
where $L(S) = span(S - S)$, and $S-S=\lbrace s_{1}-s_{2} : s_{i}\in S, i=1,2\rbrace$. If $S$ is a subset of a topological vector space, then int$S\subseteq $cor$(S)\subseteq$ icr$(S)\subseteq S$.
\end{definition}

\begin{definition}\cite{ZFV}\label{normal}
  Let $(X,P)$ be a partially ordered normed space. A cone $P$ is said to be normal, if there exists a constant $k>0$  such that $\theta_{X} \preceq x \preceq y$ implies $\parallel x\parallel \leq k \parallel y \parallel $ for all $x, y \in X$. The least positive constant $k$ satisfying the above inequality is called the normal constant of $P$.
\end{definition}
\begin{definition}\cite{ZFV} Let $(X,P)$ be a partially ordered normed space. A cone
$P$ is said to be (sequentially) regular, if every sequence in $X$
which is increasing and ordered bounded above must be
convergent in $X$. \\ This means that, if $\lbrace a_{n}\rbrace$ is a
sequence in $X$ such that $a_{1}\preceq a_{2} \preceq... \preceq a_{n}
\preceq... \preceq m$ for some $m \in X$, then there exists $a \in X$ such
that $\Vert a_{n}-a\Vert \rightarrow 0$ as $n \rightarrow \infty$.
Equivalently, if every sequence in $X$ which is decreasing and
bounded below is convergent in $X.$\\
It is well known that any regular cone is normal and the converse is true if the  space $X$ is reflexive (see \cite{A. Karamian}).
\end{definition}
\begin{definition}\label{2.6}
The mapping $T:X \to Y$ acting in partially ordered real vector linear
spaces $X$ and $Y$ is called increasing if $x \preceq y$ implies $T(x) \preceq T(y)$.
Note that if we take $X = Y = \mathbb{R}$, then Definition \ref{2.6} collapses to the usual definition of an
increasing mapping.
\end{definition}


The next definition was firstly introduced in \cite{Chr.} (for more details, see \cite{faraj}) in order to apply
in optimization theory and then was used for equilibrium problems in \cite{Chen}.
\begin{definition} Let $(X,P)$ be a partially ordered  topological vector space (t.v.s.) with a solid cone
$P$ and $e\in intP.$ 
The nonlinear scalarization function
$\xi_e:X\to  \mathbb{R} $ is defined as
follows:
$$\xi_e(y)=\inf \{r\in \mathbb{R} \,:\, y\in re- P\}=\min \{r\in  \mathbb{R}:\, y\preceq re\}.$$
\end{definition}
\vspace{2mm}
The following lemma characterizes some of the important properties
of the nonlinear scalarization function, which are used in the
sequel.\vspace{2mm}
\begin{lemma}\cite{thai}\label{pzeta} Let $(X,P)$ be a partially ordered normed space with a solid cone
$P$. For each $ r\in \mathbb{R} $ and $y\in X$, the following statements are satisfied:
\begin{itemize}
\item [(i)] $ \xi_{e} (re)=r$, particulary $\xi_{e} (\theta_{X})=0$;
\item [(ii)] if $y_{2} \preceq y_{1}$, then $\xi_{e}(y_{2}) \leq \xi_{e}(y_{1})$ for any $y_{1},y_{2} \in X$;
\item [(iii)] if $y_{2} \prec \prec y_{1}$, then $\xi_{e}(y_{2}) < \xi_{e}(y_{1})$ for any $y_{1},y_{2} \in X$;
\item [(iv)] $\xi_{e}(y) \leq r \Longleftrightarrow y \in re-P $;
\item [(v)] $\xi_{e}(y) > r \Longleftrightarrow y \notin re-P $;
\item [(vi)] $\xi_{e}(y) < r \Longleftrightarrow y \in re-int P $;
\item [(vii)] $\xi_{e}(y) \geq r \Longleftrightarrow y \notin re-int P  $;
\item [(viii)] $\xi_{e}$ is subadditive on $X$, i.e., $\xi_{e}(x+y) \leq \xi_{e}(x) + \xi_{e}(y)$ for all $x,y \in X$;
\item [(ix)]  $\xi_{e}$ is positively homogeneous on $X$, i.e., $\xi_{e}(\beta x)=\beta \xi_{e}(x)$ for every $x \in X$ and a positive real number $\beta$;
\item [(10)]  $\xi_{e}$ is continuous on $X$.
\end{itemize}
\end{lemma}


The following lemma guarantees the existence of two points $u_{0}$ and $ v_{0}\in (
u_{-}, u_{+})_{c}\subseteq[
u_{-}, u_{+}]_{o}$ such that $u_{0} \preceq Tu_{0}, Tv_{0}\preceq v_{0}$ which plays a
crucial role reaching to one of the main goal in this paper(i.e. Theorem \ref{2part}), where 
$$[u_{-}, u_{+}]_{o}=\lbrace x\in X: u_{-} \preceq x \preceq u_{+}\rbrace,$$
$$(u_{-}, u_{+})_{c}=\lbrace x\in X: x=tu_{+}+(1-t)u_{-}; 0<t<1\rbrace.$$
\begin{lemma}\label{u*}\cite{A. Karamian}
Let $(X,P)$ be a partially ordered normed space with a solid cone
$P$. Assume that there are two points $u_{-}$ and $u_{+}$ in $X$ and an increasing continuous mapping $T:[u_{-},u_{+}]_{c} \rightarrow X$ such that  $u_{-} \preceq u_{+}$, $u_{+} \prec \prec Tu_{+}$, and $Tu_{-}\prec \prec u_{-}$. Then there exist
$u_{0}$ and $v_{0} \in (u_{-},u_{+} )_{c}$ such that  $u_{0}\preceq Tu_{0},$ $Tv_{0}\preceq v_{0}$, where $[u_{-},u_{+} ]_{c}=\lbrace x\in X: x=tu_{+}+(1-t)u_{-}; 0 \leq t \leq1\rbrace.$
\end{lemma}


\section{Main results}
Now, we are in a position to give our main results.\\

In 2012, Kostrykin and Oleynik  presented Theorem 1 in \cite{Kost}  which is an
extension of a key lemma (Lemma 2.1) of \cite{Kost2} that plays a key role in \cite{Kost2}. Moreover, it can be considered as an important
existence result of the unstable bumps in neural, Integral equations and operator theory (see, for
instance, \cite{Kost2,Li,Liao,Wu}).\\

In the following we show that Theorem 1 in \cite{Kost} is not true in general because in page 3 (\cite{Kost}), the authors claimed that for operator $\hat{T}:[ u_{-}, u_{+}]_{o}\rightarrow X$ defined by $\hat{T}u:=sup\lbrace inf\lbrace T u, u_{+}\rbrace, u_{-}\rbrace$ every fixed point $u_{*}$ of operator $\hat{T}$ satisfying $ u_{-}\prec u_{*} \prec u_{+}$ is a fixed point of the operator $T$. This assertion is not true  in general because $inf\lbrace T u_{*}, u_{+}\rbrace$ does not exists in general. Note that $Tu_{*} $ and $u_{+}$ are not comparable in general. Thus the results in \cite{Kost2,Liao,Wu} which are based upon  on Theorem 1 can not be true. In the sequel we give two corrected versions of this theorem.
\begin{remark}
It is worth noting  that there exists some other papers exist whose authors have made the same mistake as mentioned above. For example, Theorem \ref{1wang}  and Theorem \ref{2wang} (the following two theorems) together with Corollaries 3.1, 3.2, 3.3 and 3.4 in \cite{wang} are not true. It is not hard to see that  in the proof of Theorem \ref{1wang} (that is Theorem 2.1 in \cite{wang}) the relation (2.7), i.e., $u_0\preceq Av_0\preceq Au_0\preceq v_0$ does not hold. Similarly, in the proof of Theorem  \ref{2wang} (that is Theorem 2.2 in \cite{wang}) the relation (2.14), i.e., $u_0\preceq Av_0\preceq Au_0\preceq v_0$ dose not hold. Note that for any $x \in D$, $inf\lbrace Ax,x\rbrace=u_0$ and $sup\lbrace Ax,x \rbrace=v_0$ depend on a variable $x$. Thus $u_0$ and $Av_0=Ax$ (resp. $v_0$ and $Au_0=Ax$) are not comparable in general. 
\end{remark}

\begin{theorem}\label{1wang}\cite{wang}
 Let $E$ be an ordered Banach space with lattice structure, $D \subseteq E$ be bounded,
and $A:D \longrightarrow D$ be a decreasing and condensing operator. Then the operator $A$ has a fixed
point in $D.$
\end{theorem}


\begin{theorem}\label{2wang}\cite{wang}
 Let $E$ be an ordered Banach space with lattice structure, $P \subseteq E$ be a normal
cone, and $A:E \longrightarrow E$ be a decreasing and condensing operator. Then the operator $A$ has a fixed
point in $E.$
\end{theorem}
The following counter example shows that Theorem  \ref{1wang} does not hold in general, even in the special case $E=\mathbb{R}$. In the same way, one can show that  Theorem  \ref{2wang} is not true in general.
\begin{example}\label{}
Let $E =\mathbb{R}$ be endowed by the norm 
$$\Vert a\Vert=\vert a\vert, $$
$$P=\lbrace x\in \mathbb{R}: x \geq 0 \rbrace,$$ 
$$D=[-1,0] \smallsetminus \lbrace \frac{-1}{2}\rbrace,$$
and define $A:D\longrightarrow D$ by $A(x)=-x-1$. One can check that $A$ satisfies all the assumptions of Theorem 2.4 while $A$ doese not have a fixed point.
\end{example}




In the following we  show that the  relative algebraic interior is a suitable replacement of the topological interior for the case where it is empty. Moreover, the authors establishe Theorem \ref{NewSemnan} in real Banach space which is an improvement version of Theorem 1 in \cite{Kost} by relaxing minihedrality on cone and replacing the toplological interior of the cone by the relative algebric interior.\\


Now, we present the following lemma which shows that algebraic interior is not a suitable replacement of the topological interior.

\begin{lemma}\label{cor=int} If $S$ is a nonempty, closed, convex  subset of a Banach space  $X$, then cor$(S)=intS$. 

\begin{proof}It is clear that the core of $S$ is contained in its interior, i.e., int$S\subseteq$cor$(S)$.
Conversely, Let $\overline{x}\in corS$, from Definition \ref{core}, we have $S \smallsetminus \lbrace \overline{x} \rbrace $ is absorbing. Since $S$ is a convex closed subset of a Banach space  $X$, then $\theta _{X}\ \in int(S \smallsetminus \lbrace\overline{x} \rbrace)  $, which implies that  $\overline{x}\in intS $. This complets the proof.
\end{proof}
\end{lemma}
To illustrate Lemma \ref{cor=int} we provide some examples.
\begin{example}
Assume that $X = C[0,1]$ denotes the Banach space of all real-valued continuous mapping defined on $[0,1]$ endowed by the $\Vert f \Vert =\displaystyle\max_{0 \leq t \leq 1}\vert f(t)\vert $. Let  $S =\lbrace f\in X:f(x)\geq 0,\forall  x\in [0,1]\rbrace.$ It is easy to check that $S$ is closed, convex and int$S=\lbrace f\in X:f(x)>0,\forall  x\in [0,1]\rbrace$. We show that int$S$=cor$(S)$. To see this, let $f\in S$ and $f>0$. Set $m=\displaystyle\inf_{x\in [0,1]} f(x)$. If $g\in C[0,1] $ is an orbitrary mapping, then there exists $M_{g}>0$ such that $-M_{g}\leq g(x) \leq M_{g}, \forall x\in [0,1]. $ Therefore there exists $\lambda^{\prime}=\dfrac{m}{M_{g}}>0$ such that $f+\lambda g>0, \forall \lambda \in[0,\lambda^{\prime}] $. Thus cor$(S)=$int$S$.
\end{example}
 We know that  $l_{\infty}$ (that is the space consisting of all bounded sequences) with the sup norm, i.e., $\Vert x\Vert_{\infty}=\displaystyle \sup_{n\in \mathbb{N}} \vert x_{n}\vert$, is a Banach space. Now, we give the following lemma which shows that there exists a Banach space which has no any cone, namely $P$, satisfying $intP \neq cor(P) $ . In other words, for each cone in $l_{\infty}$ the algebraic interior of the cone coincide with  interior of the cone. 

\begin{lemma}\label{corL=intL}  If  $S$ is a nonempty closed, convex  subset of the Banach space $l_{\infty}$ , then  
$cor(S)=intS$.

\begin{proof}If $cor(S)= \emptyset$, then the proof is clear. Suppose that $cor(S)\neq \emptyset$. Let $\overline{x}\in cor(S), x\in l_{\infty} $ and $B=\lbrace x\in l_{\infty} : \Vert x\Vert_{\infty} \leq 1\rbrace$. Define the real valued mapping $G: B\longrightarrow \mathbb{R}^{+}$ by $$G(x)=sup \lbrace \lambda>0 :\overline{x}+\lambda x\in S \rbrace.$$ Put $\delta^{\ast}:=\displaystyle\inf_{x\in B}G(x)>0$. So there exists $x_{\ast} \in B$ such that $\delta^{\ast}=G(x_{\ast})$. Thus, $\overline{x}+\lambda x\in S $ for all $x\in B$, $0<\lambda \leq \delta^{\ast}$, and so $\overline{x}\in intS$. The proof is complete.
\end{proof}
\end{lemma}
In the following example we preasent a cone of a real vector space endowed with a topology which is not a topological  vector space(t.v.s.). Moreover, the relative algebraic interior of this cone coincide with  interior of it but its algebraic interior is empty. This example shows that in Lemma \ref{cor=int} it is essential that the space to be topological vector space.

\begin{example}
Let  $R_{1}=\mathbb{R}$ and $R_{2}=\mathbb{R}$ be two topological spaces equipped with standard topology on the real line (a topology generated by the collection of all open intervals in the real line) and discrete topology(the collection of all subsets of $\mathbb{R}$), respectively. We consider the  topological space   $X=R_{1}\times R_{2}$ and the cone $P=[0,\infty)\times \lbrace 0\rbrace\subseteq R_{1}\times R_{2}$. It is easy to show that $intP=icr(P)=(0,\infty)\times \lbrace 0\rbrace$ and $cor(P)=\emptyset$. Here, the scaler multiplication is not continuous, i.e, the mapping $\centerdot:\mathbb{R}\times X\rightarrow X$ definded by $\centerdot(\alpha ,x)\rightarrow \alpha x$  is not continuous. Because the set $\lbrace 1\rbrace $ is a closed set in $X$, but $~\centerdot^{-1} \lbrace 1 \rbrace $ is not closed in $\mathbb{R}\times X$. Thus $X$ is not a  topological vector space.
\end{example}

In the following lemma we establish several characterizations which are used in Theorem \ref{NewSemnan}. For the sake of the reader we give the proof.


\begin{lemma}\label{icr}
Let $P$ be a cone in a vector space $X$ with a nonempty  relative algebraic
interior. Then
\begin{itemize}

\item [(i)] $\theta_{X} \notin icr(P)$; 
\item [(ii)]  if $x\in icr(P)$ and $y\in P $, then $[x,y)_{c}\in icr(P).$ In particular, the set icr(P) is convex;
\item [(iii)]$ icr(P)\cup \lbrace \theta_{X} \rbrace$ is a cone;
\item [(iv)] $icr(icr(P))=icr(P).$
\end{itemize} 
\begin{proof}
 (i) On the contrary, suppose that  $\theta_{X} \in icr(P)$. Thus $\forall x^{\prime}\in L(P), \exists  \lambda^{\prime  }>0 ; \forall \lambda \in [0,\lambda^{\prime  }],  \lambda x^{\prime }\in P  $. Therefore $ L(P)=P$ which is contradicted by $P\cap(-P)=\lbrace \theta_{X} \rbrace$.\\
(ii) Let $0\leq t <1$, $z=(1-t)x+t y,  \nu\in L(P) $. Take $\lambda>0$ such that $x+\lambda\nu\in P.$  Since $z+(1-t)\lambda\nu=(1-t)(x+\lambda\nu)+t y$, then $z+(1-t)\lambda\nu\in P$. Now, the assertion easily follows.\\
(iii) It is clear that $\emptyset=icr(P) \cap -icr(P)\subseteq P\cap (-P)=\lbrace\theta_{X}\rbrace$. Now, applying (i) and (ii) we have (iii).\\
(iv) It is clear that $icr(icr(P)) \subseteq icr(P)$. Conversly, let $x\in$ icr$(P)$, $x^{\prime}\in L(P) $. There exists $\lambda^{\prime}>0$ such that $x+\lambda x^{\prime}\in P, \forall \lambda\in [0,\lambda^{\prime}]$.  We can write $t(x+\lambda x^{\prime})+(1-t)x \in icr(P), \forall ~0\leq t <1$, applying (ii). Thus $x+t\lambda x^{\prime} \in icr(P), ~\forall ~ t\lambda \in [0, t\lambda^{\prime}]$ and hence $x\in icr(icr(P)).$
\end{proof}
\end{lemma}
 The following example shows that it is possible that int$P=$cor$(P)=\emptyset ,$ and $icr(P)\neq \emptyset$. Moreover, it shows that the  relative algebraic interior is a suitable replacement of the topological interior and algebraic interior for the cases where these are empty.
\begin{example}Let  $l_{\infty}$ be as given in Lemma 2.7 and $$P=\lbrace (x_{1},x_{2},...,x_{n},0,0,0,...) : x_{i}\geq 0 ,1\leq i \leq n\rbrace.$$
One can check that $P$ is a cone, $intP=cor(P)=\emptyset$, and $$ icr(P)=\lbrace (x_{1},x_{2},...,x_{n},0,0,0,...) : x_{i}>0 ,1\leq i \leq n\rbrace.$$

\end{example}
It is easy to check that the following theorem is an  improvement of Theorem 1 in \cite{Kost} by using relative algebraic interior of cone. 
\begin{theorem}\label{NewSemnan}
Suppose that  $X$ is a real Banach space and let $P$ be a normal cone with nonempty  relative algebraic
interior (i.e., $icr(P)\neq \phi$). Assume that $K = icr(P)\cup \lbrace \theta_{X} \rbrace  $, there are two points $u_{-}$ and $u_{+}$ in $X$, where  $u_{-} \prec _{ P}u_{+},$
and an increasing  mapping $T : [u_{-},u_{+}]_{o}\longrightarrow X$. Let $h_{0}=u_{+}-u_{-}$. If one of the following assumptions holds: \\
\begin{itemize}
\item [(i)] $T$ is convex , $Tu_{+}\prec_{K} u_{+}$ ,$u_{-}\preceq_{P} Tu_{-}$;
\item [(ii)]  $T$ is concave, $u_{-}\preceq_{K}Tu_{-} $, $Tu_{+}\prec_{P}u_{+}$,
\end{itemize} 
then $T$ has a unique fixed point $x_{\ast}\in[u_{-},u_{+}]_{o}$. Moreover, each iteration $Tx_{n}= x_{n-1}$ for all $n = 1, 2,3,... $ with
$x_{0}\in[u_{-},u_{+}]_{o}$ converges to $x_{\ast}$ and there exist $M>0$ and $r\in(0,1)$ such that $$\Vert x_{n}-x_{\star}\Vert \leq Mr^{n}.$$
\begin{proof}
Assume that (i) holds(the proof is similar if condition (ii) holds). Applying Lemma \ref{icr} (iii), the set $K = icr(P)\cup \lbrace \theta_{X} \rbrace $ is a cone and by Lemma \ref{icr} (iv), we have$$u_{+}-Tu_{+}\in K \smallsetminus \lbrace \theta_{X} \rbrace=icr(P)=icr(icr(P)).$$ By the assumption $u_{-} \prec _{ P}u_{+}$, we get $h_{0}=u_{+}-u_{-}\in P\subseteq L(P),$ where $L(P)$ is the smallest subspace containing $P$. So $ (u_{-}-u_{+})\in L(P)$ and there exists $\lambda^{\prime} >0$ such that $$(u_{+}-Tu_{+})+\lambda (u_{-}-u_{+})\in K=icr(P)\cup\lbrace \theta_{X} \rbrace\subseteq P, \forall \lambda\in [0, \lambda^{\prime}].$$ Since $\lambda^{\prime} >0$, we can choose $\varepsilon\in (0,1)$ such that $$Tu_{+}\preceq_{P}u_{+}-\varepsilon (u_{+}-u_{-}).$$
This means that $Tu_{+}\preceq_{P}u_{+}-\varepsilon h_{0}.$ Now, applying the next theorem and the inequality $$u_{-}\preceq_{P} Tu_{-},$$ complete the proof.
\end {proof}
\end{theorem}
\begin{theorem} \cite{Y.Du}\label{Yihong DU} 
Suppose that $X$ is a real Banach space, $P$ is a normal cone, and $u_{-},u_{+} \in X$ with $u_{0} \prec_{P} v_{0}$. Moreover, $T : [u_{-},u_{+}]_{o}\longrightarrow X$ is an increasing mapping. Let $h_{0}=u_{+}-u_{-}$. If one of the following assumptions holds:
\begin{itemize}
\item [(i)] $T$ is convex mapping, $Tu_{+}\preceq_{P} u_{+}-\varepsilon h_{0}$, $u_{-}\preceq_{P} Tu_{-}$  where $\varepsilon \in (0,1)$ is a constant; 
\item [(ii)]  $T$ is concave mapping, $u_{-}+\varepsilon h_{0} \preceq_{P}Tu_{-} $, $Tu_{+}\prec_{P}u_{+}$ where $\varepsilon \in (0,1)$ is a constant,
\end{itemize} 
then $T$ has a unique fixed point  $x_{\ast}\in[u_{-},u_{+}]_{o}$. Moreover, for any $x_{0}\in[u_{-},u_{+}]_{o}$, the iterative sequnce $\lbrace x_{n} \rbrace$ given by $x_{n}=Tx_{n-1} (n=1, 2, ...)$ satisfying that $$\Vert x_{n}-x_{\ast}\Vert \rightarrow 0 (n\rightarrow \infty),$$
$$\quad \quad \quad \quad \quad \Vert x_{n}-x_{\ast}\Vert \leq M(1-r)^{n} (n=1, 2, ...),$$
with $M$ a positive constant independent of $x_{0}$.
\end{theorem}

Now, we are ready to give  another corrected version of Theorem 1 in \cite{Kost} by relaxing some assumptions of it (such as minihedrality on cone and the compactness on the mapping $T$) and an extension of it in general spaces. The following theorem  can be considered as repairment  an improvement of Theorem 1 in \cite{Kost}. 
\begin{theorem}\label{2part}
Let $X$ be a normed spac, $P$ be a regular cone with nonempty interior.
Assume that there are two points $u_{-}$ and $u_{+}$ in $X$ such that $u_{-}\prec\prec u_{+}$ and an increasing and continuous mapping $T:[u_{-},u_{+}]_{o}\longrightarrow X$. If $Tu_{-}\prec\prec u_{-}$ and $u_{+}\prec\prec Tu_{+}$, then we have at least one of the following: \\
(i) There exist $u_{0}$ and $v_{0}\in (u_{-},u_{+})_{c} $ such that $T$ has a maximal fixed point $v^{\ast}$ and minimal fixed point $u_{\ast}$ in $ [u_{0},v_{0}]_{o}$ and $u_{\ast}=\displaystyle\lim_{n\rightarrow \infty} u_{n},$  $v^{\ast}=\displaystyle\lim_{n\rightarrow \infty} v_{n}$,
where $u_{n}=T^{n}u_{0}$, $v_{n}=T^{n}v_{0} (n=1,2,3,...),$ and
\begin{equation}
u _{0}\preceq u_{1}\preceq...\preceq u_{n}\preceq...\preceq u_{\ast}\preceq v^{\ast}\preceq...\preceq v_{n}\preceq...\preceq v_{1}\preceq v_{0}.
\end{equation}\label{e2}
$(ii)$ For each $n\in \mathbb{N}$, there exist $u_{n} $ and $ v_{n}\in [v_{n-1},u_{n-1}]_{c}$ such that $Tv_{n}\preceq v_{n}$ and $u_{n}\preceq Tu_{n}$. Moreovere, $Tv_{\ast}\preceq v_{\ast}=\displaystyle\lim_{n\rightarrow\infty} v_{n}$, $u^{\ast}=\displaystyle\lim_{n\rightarrow\infty} u_{n} \preceq Tu^{\ast} $ and $ u_{-} \preceq v _{0}\preceq v_{1}\preceq...\preceq v_{n}\preceq...\preceq v_{\ast}\preceq u^{\ast}\preceq ... \preceq u_{n}\preceq...\preceq u_{1}\preceq u_{0}\preceq u_{+}$.\\
\begin{proof}
Applying Lemma \ref{u*}, there exist $u_{0} , v_{0}\in (u_{-},u_{+})_{c} $ such that  
\begin{equation}\label{e3}
 u_{0}\preceq Tu_{0}\quad and \quad
 Tv_{0}\preceq v_{0}.
 \end{equation}
Now, we have one of the following cases:
$$ u_{0}=v_{0},\quad u_{0}\prec v_{0} ,\quad and \quad v_{0} \prec u_{0}.$$
Case I: $u_{0}=v_{0}$.\\
In this case, $u_{\ast}=u_{0}=v_{0}=v^{\ast}$ is a fixed point of $T$.\\\\
Case II: $u_{0}\prec v_{0}$.\\
In this case, set $u_{n}=T^{n}u_{0}=Tu_{n-1} $ and $v_{n}=T^{n}v_{0}=Tv_{n-1}$.\\
Since $T$ is increasing, applying (2.2) and $u_{0}\prec v_{0}$, we have
\begin{equation}
u _{0}\preceq u_{1}\preceq...\preceq u_{n}\preceq...\preceq v_{n}\preceq...\preceq v_{1}\preceq v_{0}.
\end{equation}\label{e4}
By the regularity of $P$ and (2.3), $u_{n}\rightarrow u_{\ast}$ as $n\rightarrow \infty$ and $u_{n}\preceq u_{\ast} \preceq v_{n} (n=0,1,2,...).$
Since  $T$ is continuous, we have  $u_{n+1}=Tu_{n}\rightarrow Tu_{\ast}$. So $Tu_{\ast}= u_{\ast}$ . Similarly, we can show that $v_{n}\rightarrow v^{\ast}$ as $n\rightarrow \infty$, $Tv^{\ast}=v^{\ast}$ and
\begin{equation}
 u_{n}\preceq u_{\ast}\preceq v^{\ast}\preceq v_{n}\quad\quad\quad\quad(n=1,2,...).
\end{equation}\label{e5}
Hence, applying (2.4) and (2.3), (2.1) holds.
Finally, one can check that $u_{\ast}$ and $v^{\ast}$ are the minimal fixed point and the maximal fixed point of $T$ in $[u_{0},v_{0}]_{c}$, respectively. \\\\
Case III: $ v_{0} \prec u_{0}$.\\
In this case, we have $T:[v_{0},u_{0}]_{o} \longrightarrow  [Tv_{0},Tu_{0}]_{o}(\supseteq [v_{0},u_{0}]_{o})$. Define $T^{\ast}:[v_{0},u_{0}]_{c} \longrightarrow  \Bbb R$ by $T^{\ast}t=\xi_{e}(t-Tt)$. Since $ u_{0}-Tu_{0}\preceq \theta$ and $\theta\preceq v_{0}-Tv_{0}$, applying Lemma \ref{pzeta} (parts (ii) and (ii)), we have $T^{\ast}u_{0}\leq \xi_{e}(\theta)=0$ and $\xi_{e}(\theta)=0 \leq T^{\ast}v_{0}$.
Applying  Intermediate Value Theorem (for more details see Lemma \ref{u*}), there exists $u_{1}\in [v_{0},u_{0}]_{c}$ such that  $T^{\ast}u_{1}=0$. Soapplying Lemma \ref{pzeta}( parts (iv) and (vii)) we get $u_{1}-Tu_{1}\in -P\quad \backslash -intP$,  and so $u_{1}\preceq Tu_{1}$. Similarly, if we put $T^{\ast}t=\xi_{e}(Tt-t)$, there exists $v_{1}\in [v_{0},u_{0}]_{c}$ such that $Tv_{1}\preceq v_{1}.$\\
Now, we have one of the following cases:
$$u_{1}=v_{1} \quad or \quad u_{1}\prec v_{1}\quad or \quad v_{1} \prec u_{1}.$$
If  $u_{1}=v_{1}$ or $ u_{1}\prec v_{1}$, then $T$ has a maximal fixed point and  a minimal fixed point by repeating the process in case  I or case II.
If $v_{1} \prec u_{1}$, applying case III, there exist $v_{2}$ and $u_{2}$ such that  $u_{2}, v_{2}\in  [v_{1},u_{1}]_{c}$, and $u_{2}\preceq Tu_{2} ,Tv_{2} \preceq v_{2}$. By repeating this process and the regularity of cone, one can complet the proof.
 \end{proof}
 \end{theorem}

\vskip 5mm
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{\small

\noindent{\bf Ardeshir Karamian}

\noindent Department of Mathematics, Faculty of  Mathematics

\noindent Ph.D Student of Mathematics

\noindent  University 
of Sistan and Baluchestan,

\noindent  Zahedan, Iran

\noindent E-mail: ar$_{-}$karamian1979@yahoo.com}\\

{\small
\noindent{\bf  Rahmatollah Lashkaripour }

\noindent  \noindent Department of Mathematics, Faculty of  Mathematics

\noindent  Professor of Mathematics

\noindent  University 
of Sistan and Baluchestan,


\noindent Zahedan, Iran

\noindent E-mail:   lashkari@hamoon.usb.ac.ir}\\



\end{document}
{\normalfont (See \cite{RefJ2})}