\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{B. Mohammadi, F. Golkarmanesh and V. Parvaneh} \fancyhead[CO]{Some fixed point theorems for nonexpansive... } %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} {\noindent Journal of Mathematical Extension \\ Vol. XX, No. XX, (2014), pp-pp (Will be inserted by layout editor)}\\ ISSN: 1735-8299\\ URL: http://www.ijmex.com\\ \vspace*{9mm} \begin{center} {\Large \bf Some fixed point theorems for nonexpansive self-mappings and multi-valued mappings in $b$-metric spaces\\} {\bf \\ } \let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)} {\bf B. Mohammadi}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Marand Branch, Islamic Azad University, Marand, Iran.} \vspace{2mm} {\bf F. Golkarmanesh}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Sanandaj Branch, Islamic Azad University, Sanandaj, Iran.} \vspace{2mm} {\bf V. Parvaneh$^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\ \vspace{2mm} {\small Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, Gilan-E-Gharb, Iran.} \vspace{2mm} \end{center} \vspace{4mm} {\footnotesize \begin{quotation} {\noindent \bf Abstract.} In this paper, motivated by [F. Vetro, Filomat, 29:9 (2015), 2011-2020] we present some fixed point results for a class of nonexpansive self-mappings and multi-valued mappings in the framework of $b$-metric spaces. Our results generalize and improve the consequences of [ Khojasteh et al. Abstract and Applied Analysis, vol. 2014, Article ID 325840, 5 pages, 2014.] and [F. Vetro, Filomat, 29:9 (2015), 2011-2020]. Some examples are provided to illustrate our results. \end{quotation} \begin{quotation} \noindent{\bf AMS Subject Classification:} MSC 47H04; 47H10 \noindent{\bf Keywords and Phrases:} fixed point, multi-valued mappings, nonexpansive self-mapping, $b$-metric spaces \end{quotation}} \section{Introduction} \label{intro} % It is advised to give each section and subsection a unique label. Assume that $(X,d)$ be a metric space and $f:X\to X$ be a single-valued mapping. Then, $f$ is called a $k$-Lipscitz mapping if $d(fx,fy)\leq kd(x,y)$ for all $x,y\in X$, where $k\geq 0$. If $k\in [0,1)$, then it is called a contractive mapping and if $k=1$, $f$ is called a nonexpansive mapping. In \cite{Banach} Banach proved a very important fixed point result for a contractive self-mapping. Obviously, any contractive mapping is nonexpansive, but the reverse is not true in general. Several researchers investigated fixed point theory for nonexpansive mappings such as \cite{Edelstein, Khamsi}. In \cite{Khojasteh} Khojasteh et al. investigated fixed point results for a new type of self-mappings and multivalued mappings. Then, Vetro in \cite{Vetro} extended their results for nonexpansive mappings using a binary relation on $X$, called $f$-invariant. Also, he established fixed point results for nonexpansive multi-valued mappings with a new type of contraction. On the other hand, fixed point theory for mappings on $b$-metric spaces as a generalization of metric spaces has attracted many researchers for many years (see \cite{Bakhtin, Boriceanu1, Boriceanu2, Czerwik1, Czerwik2, Mustafa, Nieto, Pacurar, Parvaneh, Ran, Roshan1, 1, 2, 3, Suzana3, Suzana4}). In this paper, motivated by Vetro \cite{Vetro} we give some fixed point results for a class of nonexpansive self-mappings and multi-valued mappings on $b$-metric spaces. Our results generalize and improve the results of Khojasteh et al. \cite{ Khojasteh} and Vetro \cite{Vetro}. Some examples are given to illustrate our results. \section{Preliminaries} In this section we give some notions and results that will be needed in the sequel. \begin{definition}\cite{Bakhtin} Let $X$ be a nonempty set and $s\geq 1$ be a constant real number. The function $d:X\times X\rightarrow [0,\infty)$ is called a $b$-metric on $X$ if the following conditions hold: \begin{itemize} \item[(i)] $d(x,y)=0$ iff $x=y$ for all $x,y\in X$; \item[(ii)] $d(x,y)=d(y,x)$ for all $x,y\in X$; \item[(iii)] $d(x,y)\leq s[d(x,z)+d(z,y)]$ for all $x,y,z\in X$. \end{itemize} In this case, $(X,d)$ is called a $b$-metric space with parameter $s$. \end{definition} Denote by $CB(X)$ the set of all nonempty closed bounded subsets of $X$. Assume that $H$ be the Pompeiu-Hausdorff metric on $CB(X)$ defined by $$H(A,B)=\max\{\sup_{x\in A}d(x,B),\sup_{y\in B}d(y,A)\},$$ for all $A, B\in CB(X)$, where $d(x,B)=\inf_{y\in B}d(x,y)$. An element $x\in X$ is said to be a fixed point of a multi-valued mapping $T:X\to CB(X)$ whenever $x\in Tx$. It is said that $T:X\to CB(X)$ is contractive whenever $H(Tx,Ty)\leq kd(x,y)$ for all $x,y\in X$, where $k\in [0,1)$. If $k=1$, then $T$ is called a nonexpansive multivalued mapping. Nadler \cite{Nadler} proved the existence of fixed point for contractive multivalued mappings. \begin{lemma} \cite{Vetro} If $\{a_n\}$ be a nonincreasing sequence of nonnegative real numbers, then the sequence $\{\frac{a_n + a_{n+1}}{a_n + a_{n+1} + 1}\}$ is nonincreasing too. \end{lemma} \begin{cor}\label{c2.2} Let $(X,d)$ be a $b$-metric space and $f:X\to X$ be a nonexpansive mapping. If $x_0\in X$ and $\{x_n\}$ be a Picard sequence starting with $x_0$, that is, $x_n=fx_{n-1}$ for all $n\in \mathbb{N}$, then, the sequence $$\{\frac{d(x_{n-1},x_n) +d(x_n,x_{n+1})}{d(x_{n-1},x_n) +d(x_n,x_{n+1}) + 1}\}$$ is nonincreasing too. \end{cor} \section{Fixed point results for nonexpansive self-mappings} We prove some results for single-valued mappings defined on a $b$-metric space endowed with an arbitrary binary relation. Let $X$ be a nonempty set, $f:X\to X$ be a mapping and $\mathcal{R} $ be a binary relation on $ X$, that is, $\mathcal{R}$ is a subset of $ X \times X $. Then, $ \mathcal{R} $ is Banach $f$-invariant if $(f x, f^2x) \in \mathcal{R} $, whenever $ (x, f x) \in \mathcal{R}$. Also, a subset $ Y $ of $ X $ is well ordered with respect to $\mathcal{R}$ if for all $x, y\in Y$ we have $(x,y)\in \mathcal{R}$ or $(y, x)\in \mathcal{R}$. Let $Fix( f ) =\{x\in X : x = f x\}$ denotes the set of all fixed points of $ f $ on $ X $. \begin{theorem}\label{t3.1} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ endowed with a binary relation $ \mathcal{R} $ on $ X $ and $f : X\to X $ be a nonexpansive mapping such that \begin{equation}\label{e1} d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y) \end{equation} for all $ (x, y)\in \mathcal{R} $, where $k\in [0, 1)$. Also assume that \begin{itemize} \item[(a)] $ \mathcal{R} $ is Banach $ f $ -invariant, \item[(b)] if $\{x_n\}$ is a sequence in $ X $ such that $ (x_{n-1}, x_n) \in \mathcal{R} $ for all $ n\in \mathbb{N}$ and $ x_n\to z \in X $ as $ n\to \infty $, then $ (x_{n-1}, z)\in \mathcal{R} $, for all $ n\in \mathbb{N} $; \item[(c)] $ Fix( f ) $ is well ordered with respect to $ \mathcal{R} $. \end{itemize} Let there exists $ x_0\in X $ such that $(x_0, f x_0)\in \mathcal{R}$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0; f^2x_0)}{d(x_0, f x_0) + d( f x_0; f^2x_0) + 1} + k .$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence of initial point $ x_0\in X $ converges to a fixed point of $ f $, \item[(iii)] if $ z,w\in X $ are two distinct fixed points of $ f $, then $ d(z, w)\geq \frac{s- k}{2} $. \end{itemize} \end{theorem} \begin{proof} Let $ x_0\in X $ be such that $ (x_0, fx_0)\in \mathcal{R}$, $\lambda s<1$ and let $\{x_n\}$ be a Picard sequence with initial point $ x_0 $. If $ x_{n-1}=x_n $ for some $ n\in \mathbb{N} $, then $ x_{n-1} $ is a fixed point of $ f $ and the existence of a fixed point is proved. Now, we suppose that $ x_{n-1}\neq x_n $ for all $ n\in \mathbb{N} $. From $ (x_0, x_1) = (x_0, f x_0)\in \mathcal{R} $, since $ \mathcal{R} $ is Banach $ f $-invariant, we deduce that $(x_1, x_2) = ( f x_0, f^2x_0)\in \mathcal{R}$. This implies that $ (x_{n-1}, x_n) = ( f^{n-1}x_0, f^nx_0)\in \mathcal{R} $ for all $ n\in \mathbb{N}$. Using the contractive condition (\ref{e1}) with $ x = x_{n-1} $ and $ y = x_n $, we get \begin{equation}\label{e2} \begin{array}{lll} d(x_{n}, x_{n+1})=d(f x_{n-1}, f x_n) &\leq& (\frac{d(x_{n-1}, f x_n) + d(x_n, f x_{n-1})}{s[d(x_{n-1}, f x_{n-1}) + d(x_n, fx_n) + 1]}+ k)d(x_{n-1}, x_n)\\ &=&(\frac{d(x_{n-1}, x_{n+1})}{s[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n)\\ &\leq&(\frac{d(x_{n-1}, x_{n})+d(x_{n}, x_{n+1})}{[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n) \end{array} \end{equation} for all $ n\in \mathbb{N} $. From (\ref{e2}), by Corollary \ref{c2.2}, we get \begin{equation}\label{e3} \begin{array}{lll} d(x_{n}, x_{n+1})&\leq& (\frac{d(x_{n-1}, x_{n})+d(x_{n}, x_{n+1})}{[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n)\\ &\leq& (\frac{d(x_0, x_1)+d(x_1, x_2)}{[d(x_0, x_1) + d(x_1, x_2) + 1]}+ k)d(x_{n-1}, x_n)\\ &=&\lambda d(x_{n-1}, x_n), \end{array} \end{equation} for all $ n\in \mathbb{N}$. Thus, for any $m,n\in \mathbb{N}$ with $m>n$, we have \begin{equation}\label{e4} \begin{array}{lll} d(x_n, x_m)&\leq& s[d(x_n, x_{n+1})+d(x_{n+1}, x_m)]\\ &\leq& sd(x_n, x_{n+1})+s^2d(x_{n+1}, x_{n+2})+...+s^{m-n-1}d(x_{m-1}, x_m)\\ &=&\sum_{i=1}^{m-n-1}s^id(x_{n+i-1}, x_{n+i})\leq \sum_{i=1}^{m-n-1}s^i\lambda^{i+n-1}d(x_0, x_{1})\\ &=&\lambda^{n-1}d(x_0, x_{1}) \sum_{i=1}^{m-n-1}(\lambda s)^{i}\\ &=&\lambda^{n-1}d(x_0, x_{1})(\lambda s)^{1-n} \sum_{i=n}^{m-1}(\lambda s)^{i}\\ &=&d(x_0, x_{1})s^{1-n} \sum_{i=n}^{m-1}(\lambda s)^{i}. \end{array} \end{equation} As $ \lambda s<1 $ and $s>1$, the last term in the above tends to zero, as $m,n\to \infty$. Thus, $\{x_n\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists $z\in X$ such that $x_n\to z$. Now we show that $z$ is a fixed point of $f$. By assumption (b), we deduce that $(x_n,z)\in \mathcal \mathcal{R}$. So, by (\ref{e1}), we have \begin{equation}\label{e5} \begin{array}{lll} d(x_{n+1}, fz)=d(f x_{n}, fz)&\leq& (\frac{d(x_{n}, f z) + d(z, f x_{n})}{s[d(x_{n}, f x_{n}) + d(z, fz) + 1]}+ k)d(x_{n}, z)\\ &=& (\frac{d(x_{n}, f z) + d(z, x_{n+1})}{s[d(x_{n}, x_{n+1}) + d(z, fz) + 1]}+ k)d(x_{n}, z). \end{array} \end{equation} Taking limit as $n\to \infty$, in the above inequality, we get $d(z,fz)\leq 0$. Thus, $d(z,fz)=0$, that is $z=fz$. Thus (i) and (ii) hold. Now, let $z,w$ are two distinct fixed points of $f$. Then, we have \[d(z,w)=d(fz,fw)\leq (\frac{d(z,fw)+d(w,fz)+k}{s})d(z,w),\] which implies that $d(z,w)\geq \frac{s-k}{2}$. \end{proof} Also, we can prove the following result with a weaker contractive condition. \begin{theorem}\label{t3.2} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ endowed with a binary relation $ \mathcal{R} $ on $ X $ and $f : X\to X $ be a nonexpansive mapping such that \begin{equation}\label{e6} d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y)+Ld(y,fx) \end{equation} for all $ (x, y)\in \mathcal{R} $, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Also, assume that \begin{itemize} \item[(a)] $ \mathcal{R} $ is Banach $f$-invariant, \item[(b)] if $\{x_n\}$ is a sequence in $X$ such that $(x_{n-1}, x_n)\in \mathcal{R}$ for all $n\in \mathbb{N}$ and $x_n\to z \in X$ as $ n\to \infty$, then $(x_{n-1},z)\in \mathcal{R}$, for all $n\in \mathbb{N}$; \item[(c)] $Fix( f )$ is well ordered with respect to $\mathcal{R}$. \end{itemize} Let there exists $ x_0\in X $ such that $(x_0, f x_0)\in \mathcal{R}$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0; f^2x_0)}{d(x_0, f x_0) + d( f x_0; f^2x_0) + 1} + k.$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence with initial point $ x_0\in X $ converges to a fixed point of $ f $, \item[(iii)] if $ z,w\in X $ are two distinct fixed points of $ f $, then $d(z, w)\geq \max\{\frac{s(1-L)- k}{2},0\}$. \end{itemize} \end{theorem} \begin{proof} Let $ x_0\in X $ be such that $ (x_0, fx_0)\in \mathcal{R}$, $\lambda s<1$ and let $\{x_n\}$ be a Picard sequence with initial point $ x_0 $. If $ x_{n-1}=x_n $ for some $ n\in \mathbb{N} $, then $ x_{n-1} $ is a fixed point of $ f $ and the existence of a fixed point is proved. Now, we suppose that $ x_{n-1}\neq x_n $ for all $ n\in \mathbb{N}$. From $ (x_0, x_1) = (x_0, f x_0)\in \mathcal{R} $, since $ \mathcal{R} $ is Banach $ f $-invariant, we deduce $(x_1, x_2) = ( f x_0, f^2x_0)\in \mathcal{R}$. This implies that $ (x_{n-1}, x_n) = ( f^{n-1}x_0, f^nx_0)\in \mathcal{R} $ for all $ n\in \mathbb{N}$. Using the contractive condition (\ref{e6}) with $ x = x_{n-1} $ and $ y = x_n $, we get \[ \begin{array}{lll} d(x_{n}, x_{n+1})=d(f x_{n-1}, f x_n)&\leq& (\frac{d(x_{n-1}, f x_n) + d(x_n, f x_{n-1})} {s[d(x_{n-1}, f x_{n-1}) + d(x_n, fx_n) + 1]}+ k)d(x_{n-1}, x_n)+Ld(x_n, x_n)\\ &=&(\frac{d(x_{n-1}, x_{n+1})}{s[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n)\\ &\leq&(\frac{d(x_{n-1}, x_{n})+d(x_{n}, x_{n+1})}{[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n) \end{array} \] for all $ n\in \mathbb{N}$. As in the proof of Theorem \ref{t3.1}, $\{x_n\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists $z\in X$ such that $x_n\to z$. Now we show that $z$ is a fixed point of $f$. By assumption (b), we deduce that $(x_n,z)\in \mathcal{R}$. So, by (\ref{e6}), we have \begin{equation}\label{e7} \begin{array}{lll} d(x_{n+1}, fz)=d(f x_{n}, fz)&\leq& (\frac{d(x_{n}, f z) + d(z, f x_{n})} {s[d(x_{n}, f x_{n}) + d(z, fz) + 1]}+ k)d(x_{n}, z)+Ld(x_{n}, z)\\ &=& (\frac{d(x_{n}, f z) + d(z, x_{n+1})}{s[d(x_{n}, x_{n+1}) + d(z, fz) + 1]}+ k)d(x_{n}, z)+Ld(x_{n}, z). \end{array} \end{equation} Taking limit as $n\to \infty$ in the above inequality, we get $d(z,fz)\leq 0$. Thus, $d(z,fz)=0$, that is $z=fz$. Thus, (i) and (ii) hold. Now, let $z,w$ are two distinct fixed points of $f$. Then, we have \[d(z,w)=d(fz,fw)\leq (\frac{d(z,fw)+d(w,fz)+k}{s})d(z,w)+Ld(z,w),\] which implies that $d(z,w)\geq \frac{s(1-L)-k}{2}$. Thus, (iii) holds. \end{proof} Putting $\mathcal{R}=X\times X$ in Theorems \ref{t3.1} and \ref{t3.2}, we obtain the following results in $b$-metric spaces: \begin{theorem}\label{t3.3} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ and let $f : X\to X $ be a nonexpansive mapping such that \[d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y)\] for all $ x, y\in X $, where $k\in [0, 1)$. Assume that there exists $ x_0\in X $ such that $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0; f^2x_0)}{d(x_0, f x_0) + d( f x_0; f^2x_0) + 1} + k.$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence with initial point $x_0\in X$ converges to a fixed point of $f$, \item[(iii)] if $z,w\in X$ are two distinct fixed points of $f$, then $d(z, w)\geq \frac{s- k}{2}$. \end{itemize} \end{theorem} \begin{theorem}\label{t3.4} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ and let $f : X\to X $ be a nonexpansive mapping such that \[ d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y)+Ld(y,fx) \] for all $ (x, y)\in X$, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Let there exists $ x_0\in X $ such that $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0, f^2x_0)}{d(x_0, f x_0) + d( f x_0, f^2x_0) + 1} + k.$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence with initial point $ x_0\in X $ converges to a fixed point of $ f $, \item[(iii)] if $ z,w\in X $ are two distinct fixed points of $ f $, then $d(z, w)\geq \max\{\frac{s(1-L)- k}{2},0\}$. \end{itemize} \end{theorem} \begin{example} Let $X=[0,1]\cup [\frac{5}{2},\infty)$ and $d:X\times X:\to [0,\infty)$ be defined by $d(x,y)=(x-y)^2$. Define $f:X\to X$ by $ fx= \Bigg\{\begin{array}{lll} \frac{1}{2}+\frac{1}{2}x\ & if\ x\in [0,1],\\ \frac{5}{4}+\frac{1}{2}x\ & if\ x\in [ \frac{5}{2},\infty). \end{array} $ It is clear that $(X,d)$ is a complete $b$-metric space with parameter $s=2$ and $f$ is nonexpansive. Also, if $x,y\in [0,1]$ or $x,y\in [\frac{5}{2},\infty)$, then \[d(f x, f y)=\frac{1}{4}(x-y)^2\leq \frac{1}{2}(x-y)^2=\frac{1}{2}d(x,y).\] If $x\in [0,1]$ and $y\in [\frac{5}{2},\infty)$, then \begin{equation}\label{e8} \frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}\geq \frac{3}{4}. \end{equation} In fact, \begin{equation}\label{eq1} \begin{array}{lll} &&\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}\geq \frac{3}{4}\\ &&\Longleftrightarrow \frac{(x-\frac{5}{4}-\frac{1}{2}y)^2 + (y-\frac{1}{2}-\frac{1}{2}x)^2}{[(x-\frac{1}{2}-\frac{1}{2}x)^2 + (y-\frac{5}{4}-\frac{1}{2}y)^2 + 1]} \geq \frac{3}{2} \\ &&\Longleftrightarrow 2x^2+2(\frac{5}{4}+\frac{1}{2}y)^2-4x(\frac{5}{4} +\frac{1}{2}y)+2y^2+2(\frac{1}{2}+\frac{1}{2}x)^2-4y(\frac{1}{2}+\frac{1}{2}x) \\ &&\geq 3x^2+3(\frac{1}{2}+\frac{1}{2}x)^2-6x(\frac{1}{2}+\frac{1}{2}x) +3y^2+3(\frac{5}{4}+\frac{1}{2}y)^2-6y(\frac{5}{4}+\frac{1}{2}y)+3\\ &&\Longleftrightarrow 7x^2+7y^2-16xy-10x+17y\geq \frac{77}{4}\\ &&\Longleftrightarrow 7(y-x)^2+17y\geq \frac{63}{4}+\frac{14}{4}+2x(y+5). \end{array} \end{equation} Since $(y-x)\geq \frac{3}{2}$, thus $7(y-x)^2\geq \frac{63}{4}$. It is sufficient to show that $17y\geq \frac{14}{4}+2x(y+5)$. Now, since $x\leq 1$, it is sufficient to show $17y\geq \frac{14}{4}+2(y+5)$ or equally $y\geq \frac{54}{60}$ which is desired. Thus, (\ref{e7}) holds. Now \begin{equation}\label{eq1} \begin{array}{lll} d(fx,fy)&=&(\frac{5}{4}+\frac{1}{2}y-\frac{1}{2}-\frac{1}{2}x)^2\\ &=&(\frac{3}{4}+\frac{1}{2}(y-x))^2\leq (\frac{3}{4}+\frac{1}{4})(y-x)^2\\ &\leq & (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+\frac{1}{4})d(x,y). \end{array} \end{equation} Also, for $k=\frac{1}{4}$ and $x_0=\frac{1}{8}$, we have $$\lambda=\frac{d(x_0, f x_0) + d( fx_0, f^2x_0)}{d(x_0, f x_0) + d( f x_0, f^2x_0) + 1} + k=\frac{245}{1269}+\frac{1}{4}.$$ Therefore, $\lambda s=(\frac{245}{1269}+\frac{1}{4})2=\frac{490}{1269}+\frac{1}{2}<1.$ Thus, all of the conditions of Theorem \ref{t3.3} are satisfied and so $T$ has a fixed point. Here, $z=1$ and $w=\frac{5}{2}$ are two fixed points of $f$. Also, \[d(z,w)=(1-\frac{5}{2})^2=\frac{9}{4}\geq \frac{7}{8}=\frac{2-\frac{1}{4}}{2}=\frac{s-k}{2}.\] Note that $f$ is not a contraction. In fact, $d(f1,f\frac{5}{2})=d(1,\frac{5}{2})$. \end{example} Let $(X,d)$ be a $b$-metric space and $\preceq$ be an order on $X$. Then, the triple $(X,d,\preceq)$ is called a partial ordered $b$-metric space. Then, two elements $x,y\in X$ are called comparable if $x\preceq y$ or $y\preceq x$. Also, $(X,d,\preceq)$ is called regular if for any nondecreasing sequence $\{x_n\}$ in $ X $ such that $ x_n\to z \in X $ as $ n\to \infty $, then $ x_n\preceq z $, for all $ n\in \mathbb{N} $. Note that $\mathcal{R}=\{(x,y):x\preceq y\}$ is a binary relation on $X$. Also, if $f:X\to X$ be nondecreasing, then $\mathcal{R}$ is Banach $f$-invariant. \begin{theorem}\label{t3.5} Let $(X,d,\preceq)$ be a complete ordered $b$-metric space with parameter $s> 1$ and $f : X\to X $ be a nonexpansive nondecreasing mapping such that \[ d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y) \] for all comparable elements $ x, y\in X $, where $k\in [0, 1)$. Also assume that \begin{itemize} \item[(a)] $(X,d,\preceq)$ is regular, \item[(b)] $ Fix( f ) $ is well ordered with respect to $ \preceq $. \end{itemize} Let there exists $ x_0\in X $ such that $x_0\preceq f x_0$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0; f^2x_0)}{d(x_0, f x_0) + d( f x_0; f^2x_0) + 1} + k.$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence with initial point $ x_0\in X $ converges to a fixed point of $ f $, \item[(iii)] if $ z,w\in X $ are two distinct fixed points of $ f $, then $ d(z, w)\geq \frac{s- k}{2} $. \end{itemize} \end{theorem} \begin{theorem}\label{t3.6} Let $(X,d,\preceq)$ be a complete ordered $b$-metric space with parameter $s> 1$ and $f : X\to X $ be a nonexpansive nondecreasing mapping such that \[ d(f x, f y) \leq (\frac{d(x, f y) + d(y, f x)}{s[d(x, f x) + d(y, f y) + 1]}+ k)d(x, y)+Ld(x, y) \] for all comparable elements $ x, y\in X $, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Also assume that \begin{itemize} \item[(a)] $(X,d,\preceq)$ is regular, \item[(b)] $ Fix( f ) $ is well ordered with respect to $ \preceq $. \end{itemize} Let there exists $ x_0\in X $ such that $x_0\preceq f x_0$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, f x_0) + d( fx_0; f^2x_0)}{d(x_0, f x_0) + d( f x_0; f^2x_0) + 1} + k.$$ Then, \begin{itemize} \item[(i)] $ f $ has at least one fixed point $ z\in X $, \item[(ii)] the Picard sequence with initial point $ x_0\in X $ converges to a fixed point of $ f $, \item[(iii)] if $ z,w\in X $ are two distinct fixed points of $ f $, then $d(z,w)\geq \frac{s(1-L)-k}{2}$. \end{itemize} \end{theorem} \section{Fixed point results for nonexpansive multi-valued mappings} In this section, we give some fixed point results for nonexpansive multi-valued mappings in $b$-metric spaces. Let $(X,d)$ be a $b$-metric space and $K(X)$ be the set of all nonempty compact subsets of $X$. \begin{definition}\label{d4.1} Let $ \mathcal{R} $ be a binary relation on $ X $. Then, $ \mathcal{R} $ is called Banach $T$-invariant if for any $x\in X$ and $y\in Tx$ with $(x,y)\in \mathcal{R}$, then we have $(y,z)\in \mathcal{R}$ for all $z\in Ty$. \end{definition} Note that if $T$ be a single-valued mapping, then Definition \ref{d4.1} reduces to the definition of $f$-invariant for single-valued mappings. \begin{theorem}\label{t4.1} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ endowed with a binary relation $ \mathcal{R} $ on $ X $ and $T:X\to K(X)$ be a nonexpansive multi-valued mapping such that \begin{equation}\label{e9} H(Tx, Ty) \leq (\frac{d(x, Ty) + d(y, Tx)}{s[d(x, Tx) + d(y, Ty) + 1]}+ k)d(x, y) \end{equation} for all $ (x, y)\in \mathcal{R} $, where $k\in [0, 1)$. Also, assume that \begin{itemize} \item[(a)] $ \mathcal{R} $ is Banach $T$-invariant, \item[(b)] if $\{x_n\}$ is a sequence in $ X $ such that $ (x_{n-1}, x_n) \in \mathcal{R} $ for all $ n\in \mathbb{N} $ and $ x_n\to z \in X $ as $ n\to \infty $, then $ (x_{n-1}, z)\in \mathcal{R} $, for all $ n\in \mathbb{N} $; \end{itemize} Let there exists $ x_0\in X $ and $x_1\in Tx_0$ such that $(x_0, x_1)\in \mathcal{R}$, $d(x_0, x_1)=d(x_0, Tx_0)$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d(x_1, Tx_1) + 1} + k.$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} \begin{proof} Let $ x_0\in X $ and $x_1\in Tx_0$ be such that $(x_0, x_1)\in \mathcal{R}$, $d(x_0, x_1)=d(x_0, Tx_0)$ and $\lambda s<1$. Since $Tx_1$ is compact, there exists $x_2\in Tx_1$ such that $d(x_1, x_2)=d(x_1, Tx_1)$. From $ (x_0, x_1)\in \mathcal{R}$, since $ \mathcal{R} $ is Banach $ T $-invariant, we get $(x_1, x_2)\in \mathcal{R}$. Continuing this process, we have a sequence $\{x_n\}$ in $X$, such that $(x_{n-1}, x_n)\in \mathcal{R}$, $x_n\in Tx_{n-1}$ and $d(x_{n-1}, x_n)=d(x_{n-1}, Tx_{n-1})$ for all $n\in \mathbb{N}$. Since $T$ is nonexpansine, we have $d(x_{n}, x_{n+1})=d(x_{n}, Tx_{n})\leq H(Tx_{n-1}, Tx_{n})\leq d(x_{n-1}, x_n)$. Thus, by Corollary \ref{c2.2}, $$\{\frac{d(x_{n-1},x_n) +d(x_n,x_{n+1})}{d(x_{n-1},x_n) +d(x_n,x_{n+1}) + 1}\}$$ is nonincreasing. Using the contractive condition (\ref{e9}) with $ x = x_{n-1} $ and $ y = x_n $, we get \begin{equation} \begin{array}{lll} d(x_{n}, x_{n+1})&=&d(x_{n},T x_{n})\leq H(T x_{n-1}, T x_n)\\ &\leq& (\frac{d(x_{n-1}, T x_n) + d(x_n, T x_{n-1})}{s[d(x_{n-1}, T x_{n-1}) + d(x_n, Tx_n) + 1]}+ k)d(x_{n-1}, x_n)\\ &=&(\frac{d(x_{n-1}, x_{n+1})}{s[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n)\\ &\leq&(\frac{d(x_{n-1}, x_{n})+d(x_{n}, x_{n+1})}{[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]}+ k)d(x_{n-1}, x_n) \end{array} \end{equation} for all $ n\in \mathbb{N} $. Thus, we get \begin{equation}\label{e10} \begin{array}{lll} d(x_{n}, x_{n+1})&\leq& (\frac{d(x_{n-1}, x_{n})+d(x_{n}, x_{n+1})}{[d(x_{n-1}, x_{n}) + d(x_n, x_{n+1}) + 1]} + k)d(x_{n-1}, x_n)\\ &\leq& (\frac{d(x_0, x_1)+d(x_1, x_2)}{[d(x_0, x_1) + d(x_1, x_2) + 1]}+ k)d(x_{n-1}, x_n)\\ &=&\lambda d(x_{n-1}, x_n), \end{array} \end{equation} for all $ n\in \mathbb{N}$. Thus, as in Theorem \ref{t3.1}, $\{x_n\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists $z\in X$ such that $x_n\to z$. Now, we show that $z$ is a fixed point of $T$. By assumption (b), we deduce that $(x_n,z)\in \mathcal{R}$. So, by (\ref{e9}), we have \begin{equation}\label{e11} \begin{array}{lll} d(z,Tz)&\leq& s[d(z,x_{n+1})+d(x_{n+1}, Tz)]\\ &\leq& sd(z,x_{n+1})+sH(Tx_{n}, Tz)\\ &\leq& sd(z,x_{n+1})+(\frac{d(x_{n}, Tz) + d(z, Tx_{n})}{[d(x_{n}, Tx_{n}) + d(z, Tz) + 1]}+ k)d(x_{n}, z)\\ &=&sd(z,x_{n+1}+(\frac{d(x_{n}, Tz) + d(z, x_{n+1})}{[d(x_{n}, x_{n+1}) + d(z, Tz) + 1]}+ k)d(x_{n}, z). \end{array} \end{equation} Taking limit as $n\to \infty$, in the above inequality, we get $d(z,Tz)\leq 0$. Thus, $d(z,Tz)=0$, that is, $z\in Tz$. \end{proof} Also, we can prove the following result for nonexpansive multi-valued mapping with a weaker contractive condition. \begin{theorem}\label{t4.2} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ endowed with a binary relation $ \mathcal{R} $ on $ X $ and $T:X\to K(X)$ be a nonexpansive multi-valued mapping such that \begin{equation}\label{e12} H(Tx, Ty) \leq (\frac{d(x, Ty) + d(y, Tx)}{s[d(x, Tx) + d(y, Ty) + 1]}+ k)d(x, y)+Ld(x,y) \end{equation} for all $ (x, y)\in \mathcal{R} $, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Also, assume that \begin{itemize} \item[(a)] $ \mathcal{R} $ is Banach $T$-invariant, \item[(b)] if $\{x_n\}$ is a sequence in $ X $ such that $ (x_{n-1}, x_n) \in \mathcal{R} $ for all $ n\in \mathbb{N} $ and $ x_n\to z \in X $ as $ n\to \infty $, then $ (x_{n-1}, z)\in \mathcal{R} $, for all $ n\in \mathbb{N} $; \end{itemize} Let there exists $ x_0\in X $, $x_1\in Tx_0$ such that $(x_0, x_1)\in \mathcal{R}$, $d(x_0, x_1)=d(x_0, Tx_0)$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d(x_1, Tx_1) + 1}+ k.$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} Putting $\mathcal{R} =X\times X$ in Theorems \ref{t4.1} and \ref{t4.2}, we obtain the following results in $b$-metric spaces: \begin{theorem}\label{t4.3} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ and $T: X\to K(X) $ be a nonexpansive multi-valued mapping such that \[ H(T x, T y) \leq (\frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}+ k)d(x, y) \] for all $ x, y\in X $, where $k\in [0, 1)$. Assume that there exists $ x_0\in X $ and $x_1\in Tx_0$ such that $d(x_0, x_1)=d(x_0, Tx_0)$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d(x_1, Tx_1) + 1}+ k.$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} \begin{theorem}\label{t4.4} Let $(X,d)$ be a complete $b$-metric space with parameter $s> 1$ and $T: X\to K(X) $ be a nonexpansive multi-valued mapping such that \[ H(T x, T y) \leq (\frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}+ k)d(x, y)+Ld(x, y) \] for all $ x, y\in X $, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Assume that there exists $ x_0\in X $ and $x_1\in Tx_0$ such that $d(x_0, x_1)=d(x_0, Tx_0)$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d(x_1, Tx_1) + 1}+ k .$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} \begin{example} Let $X=[0,1]\cup [\frac{5}{2},\infty)$ and let $d:X\times X:\to [0,\infty)$ be defined by $d(x,y)=(x-y)^2$. Define $T:X\to K(X)$ by $$ Tx=\left\{ \begin{array}{ll} \left[ \frac{1}{2}+\frac{1}{2}x,1\right] , \hbox{if}\,\, x\in [ 0,1] ,\\ \left[\frac{5}{2},\frac{5}{4}+\frac{1}{2}x\right] , \hbox{if}\,\, x\in [ \frac{5}{2},\infty). \end{array} \right. $$ It is clear that $(X,d)$ is a complete $b$-metric space with parameter $s=2$ and $T$ is nonexpansive. Also, if $x,y\in [0,1]$ or $x,y\in [\frac{5}{2},\infty)$, then \[H(Tx, Ty)=\frac{1}{4}(x-y)^2\leq \frac{1}{2}(x-y)^2=\frac{1}{2}d(x,y).\] If $x\in [0,1]$ and $y\in [\frac{5}{2},\infty)$, then \[ \frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}\geq \frac{3}{4}. \] Now, \[ \begin{array}{lll} H(Tx,Ty)&=&\max\{\frac{5}{2}-\frac{1}{2}-\frac{1}{2}x,1+\frac{1}{2}y-1\})^2\\ &=&\max\{2-\frac{1}{2}x,\frac{1}{2}y\})^2\leq (\frac{3}{4}+\frac{1}{4})(y-x)^2\\ &\leq& (\frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}+\frac{1}{4})d(x,y). \end{array} \] Also, for $k=\frac{1}{4}$ and $x_0=\frac{1}{8}$, $d(x_0,Tx_0)=d(\frac{1}{8},\frac{3}{16})$. Thus, with $x_1=\frac{3}{16}$ we have $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d( x_1, Tx_1) + 1} + k=\frac{245}{1269}+\frac{1}{4}.$$ Therefore, $\lambda s=(\frac{245}{1269}+\frac{1}{4})2=\frac{490}{1269}+\frac{1}{2}<1.$ Thus, all of the conditions of Theorem \ref{t4.3} are satisfied and so $T$ has a fixed point. Here, $z=1$ and $w=\frac{5}{2}$ are two fixed points of $T$. Also, \[d(z,w)=(1-\frac{5}{2})^2=\frac{9}{4}\geq \frac{7}{8}=\frac{2-\frac{1}{4}}{2}=\frac{s-k}{2}.\] Note that $T$ is not a contraction. In fact, $H(T1,T\frac{5}{2})=d(1,\frac{5}{2})$. \end{example} \begin{theorem}\label{t4.5} Let $(X,d,\preceq)$ be a complete ordered $b$-metric space with parameter $s> 1$ and $T : X\to K(X) $ be a nonexpansive multi-valued mapping such that \[ H(T x, T y) \leq (\frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}+ k)d(x, y) \] for all comparable elements $ x, y\in X $, where $k\in [0, 1)$. Also, assume that \begin{itemize} \item[(a)] for any $x\in X$ and $y\in Tx$ with $x\preceq y$, then we have $y\preceq z$ for all $z\in Ty$. \item[(b)] $(X,d,\preceq)$ is regular, \end{itemize} Let there exists $ x_0\in X $ and $x_1\in Tx_0$ such that $d( x_0,Tx_0)=d(x_0,x_1)$, $x_0\preceq x_1$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d( x_1, Tx_1) + 1} + k.$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} \begin{theorem} Let $(X,d,\preceq)$ be a complete ordered $b$-metric space with parameter $s> 1$ and $T : X\to K(X) $ be a nonexpansive multi-valued mapping such that \[ H(T x, T y) \leq (\frac{d(x, T y) + d(y, T x)}{s[d(x, T x) + d(y, T y) + 1]}+ k)d(x, y)+Ld(x,y) \] for all comparable elements $ x, y\in X $, where $k\in [0, 1)$ and $L$ is a nonnegative real number. Also, assume that \begin{itemize} \item[(a)] for any $x\in X$ and $y\in Tx$ with $x\preceq y$, then we have $y\preceq z$ for all $z\in Ty$, \item[(b)] $(X,d,\preceq)$ is regular. \end{itemize} Let there exists $ x_0\in X $ and $x_1\in Tx_0$ such that $d( x_0,Tx_0)=d(x_0,x_1)$, $x_0\preceq x_1$ and $\lambda s<1$, where $$\lambda=\frac{d(x_0, x_1) + d( x_1, Tx_1)}{d(x_0, x_1) + d( x_1, Tx_1) + 1} + k.$$ Then, $ T $ has at least one fixed point $ z\in X $. \end{theorem} % BibTeX users please use one of %\bibliographystyle{spbasic} % basic style, author-year citations %\bibliographystyle{spmpsci} % mathematics and physical sciences %\bibliographystyle{spphys} % APS-like style for physics %\bibliography{} % name your BibTeX data base % Non-BibTeX users please use \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. % \bibitem{Bakhtin} I. A. 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(2018) 20:147, https://doi.org/10.1007/s11784-018-0626-2. \bibitem{Suzana4} Suzana Aleksi\'{c}, Tatjana Do.enovi\'{c}, Zoran D.Mitrovi\'{c} and Stojan Radenovi\'{c}, Remarks on common fixed point results for generalized $\alpha$-$\psi$-contraction multivalued mappings in b-metric spaces, to appear in Adv. Fixed Point Theory. \end{thebibliography} \end{center} {\small \noindent{\bf B. Mohammadi} \noindent Department of Mathematics \noindent Assistant Professor of Mathematics \noindent Department of Mathematics, Marand Branch, Islamic Azad University \noindent Marand, Iran \noindent E-mail: bmohammadi@marandiau.ac.ir}\\ {\small \noindent{\bf F. Golkarmanesh } \noindent Department of Mathematics \noindent Assistant Professor of Mathematics \noindent Department of Mathematics, Sanandaj Branch, Islamic Azad University \noindent Sanandaj, Iran \noindent E-mail: Fgolkarmanesh@yahoo.com}\\ {\small \noindent{\bf V. Parvaneh } \noindent Department of Mathematics \noindent Assistant Professor of Mathematics \noindent Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University \noindent Gilan-E-Gharb, Iran \noindent E-mail: zam.dalahoo@gmail.com}\\ \end{document}