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Common fixed point results for mappings under nonlinear contraction of cyclic form in ordered metric spaces
 Wasfi Shatanawi^{1} and
 Mihai Postolache^{2}Email author
https://doi.org/10.1186/16871812201360
© Shatanawi and Postolache; licensee Springer 2013
 Received: 21 November 2012
 Accepted: 26 February 2013
 Published: 18 March 2013
Abstract
In this paper, we introduce the notion of a cyclic $(\psi ,A,B)$contraction for the pair $(f,T)$ of selfmappings on the set X. We utilize our definition to introduce some common fixed point theorems for the two mappings f and T under a set of conditions. Also, we introduce an example to support the validity of our results. As application of our results, we derive some common fixed point theorems of integral type.
MSC:54H25, 47H10.
Keywords
 metric spaces
 common fixed point
 altering distance function
 almost contraction
 ordered metric spaces
 weakly increasing mappings
1 Introduction
In recent years many authors established interesting results in fixed point theory in (ordered) metric spaces. One of the popular topics in the fixed point theory is the cyclic contraction. Kirk et al. [1] established the first result in this interesting area. Meantime, other authors obtained important results in this area (see [1–12]).
We begin with the definition of a cyclic map.
Definition 1.1 Let A and B be nonempty subsets of a metric space $(X,d)$ and $T:A\cup B\to A\cup B$. Then T is called a cyclic map if $T(A)\subseteq B$ and $T(B)\subseteq A$.
In 2003, Kirk et al. [1] gave the following interesting theorem in fixed point theory for a cyclic map.
Theorem 1.1 ([1])
If $k\in [0,1)$, then T has a unique fixed point in $A\cap B$.
Recently, several authors proved many results in fixed point theory for cyclic mappings, satisfying various (nonlinear) contractive conditions (see [1–12]). Some of contractive conditions are based on functions called control functions which alter the distance between two points in a metric space. Such functions were introduced by Khan et al. [13].
Definition 1.2 (altering distance function, [13])
 (1)
ϕ is continuous and nondecreasing;
 (2)
$\varphi (t)=0$ if and only if $t=0$.
For some fixed point theorems based on an altering distance function, we refer the reader to [14–20].
Let X be a nonempty set. Then $(X,d,\u2aaf)$ is called an ordered metric space if and only if $(X,d)$ is a metric space and $(X,\u2aaf)$ is a partially ordered set. Two elements $x,y\in X$ are called comparable if $x\u2aafy$ or $y\u2aafx$.
Altun et al. [21, 22] introduced the notion of weakly increasing mappings and proved some existing theorems. For some works in the theory of weakly increasing mappings, we refer the reader to [23, 24].
Definition 1.3 ([21])
Let $(X,\u2aaf)$ be a partially ordered set. Two mappings $F,G:X\to X$ are said to be weakly increasing if $Fx\u2aafGFx$ and $Gx\u2aafFGx$ for all $x\in X$.
The purpose of this paper is to obtain common fixed point results for mappings satisfying nonlinear contractive conditions of a cyclic form based on the notion of an altering distance function.
2 Main result
We start with the following definition.
 (1)
ψ is an altering distance function;
 (2)
$A\cup B$ has a cyclic representation w.r.t. the pair $(f,T)$; that is, $fA\subseteq B$, $TB\subseteq A$ and $X=A\cup B$;
 (3)
Definition 2.2 Let $(X,\u2aaf)$ be a partially ordered set and A, B be closed subsets of X with $X=A\cup B$. Let $f,T:X\to X$ be two mappings. The pair $(f,T)$ is said to be $(A,B)$weakly increasing if $fx\u2aafTfx$ for all $x\in A$ and $Tx\u2aaffTx$ for all $x\in B$.
From now on, by ψ we mean altering distance functions unless otherwise stated.
In the rest of this paper, ℕ stands for the set of nonnegative integer numbers.
 (1)
The pair $(f,T)$ is a cyclic $(\psi ,A,B)$contraction;
 (2)
f or T is continuous.
Then f and T have a common fixed point.
Proof Choose ${x}_{0}\in A$. Let ${x}_{1}=f{x}_{0}$. Since $fA\subseteq B$, we have ${x}_{1}\in B$. Also, let ${x}_{2}=T{x}_{1}$. Since $TB\subseteq A$, we have ${x}_{2}\in A$. Continuing this process, we can construct a sequence $\{{x}_{n}\}$ in X such ${x}_{2n+1}=f{x}_{2n}$, ${x}_{2n+2}=T{x}_{2n+1}$, ${x}_{2n}\in A$ and ${x}_{2n+1}\in B$.
We divide our proof into the following steps.
Step 1: We will show that $\{{x}_{n}\}$ is a Cauchy sequence in $(X,d)$.
Since $\delta <1$, we have $\psi (d({x}_{2n+1},{x}_{2n+2}))=0$ and hence ${x}_{2n+2}={x}_{2n+1}$. Similarly, we may show that ${x}_{2n+3}={x}_{2n+2}$. Hence $\{{x}_{n}\}$ is a constant sequence in X, so it is a Cauchy sequence in $(X,d)$.
Since $\delta <1$, we have $\psi (\u03f5)=0$ and hence $\u03f5=0$, a contradiction. Thus $\{{x}_{n}\}$ is a Cauchy sequence in $(X,d)$.
Step 2: Existence of a common fixed point.
Since $\delta <1$, we get that $d(u,Tu)=0$ and hence $u=Tu$. □
 (P)
If $({x}_{n})$ is a nondecreasing sequence in X with ${x}_{n}\to x$, then ${x}_{n}\u2aafx$.
Now, we state and prove the following result.
 (1)
The pair $(f,T)$ is a cyclic $(\psi ,A,B)$contraction;
 (2)
X satisfies property (P).
Then f and T have a common fixed point.
Letting $n\to +\mathrm{\infty}$ in the above inequality, we get $\psi (d(u,Tu))\le \delta \psi (d(u,Tu))$. Since $\delta <1$, we get $d(u,Tu)=0$, hence $u=Tu$. Similarly, we may show that $u=fu$. Thus u is a common fixed point of f and T. □
Taking $\psi ={I}_{[0,+\mathrm{\infty})}$ (the identity function) in Theorem 2.1, we have the following result.
If f or T is continuous, then f and T have a common fixed point.
The continuity of f or T in Corollary 2.1 can be dropped.
If X satisfies property (P), then f and T have a common fixed point.
By taking $f=T$ in Theorem 2.1, we have the following result.
 (1)
f is a cyclic map;
 (2)
f is continuous.
Then f has a fixed point.
The continuity of f in Corollary 2.3 can be dropped.
 (1)
f is a cyclic map;
 (2)
X satisfies property (P).
Then f has a fixed point.
Taking $A=B=X$ in Theorem 2.1, we have the following result.
If f or T is continuous, then f and T have a common fixed point.
The continuity of f or T in Corollary 2.5 can be dropped.
If X satisfies property (P), then f and T have a common fixed point.
To support the validity of our results, we introduce the following nontrivial example.
 (1)
$(X,d,\u2aaf)$ is a complete ordered metric space;
 (2)
$A\cup B$ has a cyclic representation with respect to the pair $(f,T)$;
 (3)
The pair $(f,T)$ is weakly $(A,B)$increasing;
 (4)
X satisfies property (P);
 (5)For every two comparable elements $x,y\in X$ with $x\in A$ and $y\in B$, we have$\psi (d(fx,Ty))\le {e}^{1}\psi (max\{d(x,y),d(x,fx),d(y,Ty),\frac{1}{2}(d(x,Ty)+d(fx,y))\}).$
Proof The proof of part (1) is clear. Since $fA=\{0,1,3,5\dots \}\subseteq B$ and $TB=\{0,2,4,\dots \}\subseteq A$, we conclude that $A\cup B$ has a cyclic representation with respect to the pair $(f,T)$. To prove part (3), given $x\in A$. If $x\in \{0,1,2,3,4\}$, then $T(fx)=0$. Thus $Tfx\le fx$ and hence $fx\u2aafT(fx)$. If $x\ge 5$, then $fx=x1$ and $T(fx)=T(x1)=x4$. Thus $T(fx)\le fx$ and hence $fx\u2aafT(fx)$. Therefore $fx\u2aafT(fx)$ for all $x\in A$. Similarly, we may show that $Tx\u2aaff(Tx)$ for all $x\in B$. So, the pair $(f,T)$ is weakly $(A,B)$increasing. To prove part (4), let $\{{x}_{n}\}$ be a nondecreasing sequence such that ${x}_{n}\to x\in X$. Then $d({x}_{n},x)\to d(x,x)=0$. So, ${x}_{n}=x$ for all n except for finitely many. Since $({x}_{n})$ is a nondecreasing with respect to ⪯, we have ${x}_{1}\ge {x}_{2}\ge {x}_{3}\cdots $ . Since ${x}_{n}=x$ for all but finitely many, then there exists $k\in \mathbb{N}$ such that ${x}_{1}\ge \cdots \ge {x}_{k1}\ge {x}_{n}=x$ for all $n\ge k$. So, ${x}_{n}\ge x$ for all $n\in \mathbb{N}$ and hence ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$. Thus X satisfies property (P). To prove part (5), given two comparable elements $x,y\in X$ with $x\in A$ and $y\in B$. We divide the proof into the following cases:

Case one: $x=0$ and $y\in \{0,1,3\}$. Here, we have $fx=Ty=0$ and hence $\psi (d(fx,Ty))=0$. Thus$\psi (d(fx,Ty))\le {e}^{1}\psi (max\{d(x,y),d(x,fx),d(y,Ty),\frac{1}{2}(d(x,Ty)+d(fx,y))\}).$

Case two: $x\ge 2$ and $y\ge 5$. Here $fx=x1$ and $Ty=y3$. Since $x\in A$ and $y\in B$, then $x=2t$ and $y=2n+1$ for some $t,n\in \mathbb{N}$.
If $fx=Ty$, then $x1=y3$ and hence $2t1=2n2$. Thus $2t=2n1$, which is impossible.

Case three: $x=0$ and $y\ge 5$. Here $fx=0$ and $Ty=y3$. Thus$\begin{array}{rcl}\psi (d(fx,Ty))& =& \psi (y3)\\ =& (y3){e}^{y3}\\ \le & {e}^{1}y{e}^{y}\\ =& {e}^{1}\psi (y)\\ =& {e}^{1}\psi (d(y,Ty))\\ \le & {e}^{1}\psi (max\{d(x,y),d(x,fx),d(y,Ty),\frac{1}{2}(d(x,Ty)+d(fx,y))\}).\end{array}$

Case four: $x\ge 2$ and $y\in \{0,1,3\}$. Here $fx=x1$ and $Ty=0$.$\begin{array}{rcl}\psi (d(fx,Ty))& =& \psi (d(x1,0))=\psi (x1)\\ =& (x1){e}^{x1}\\ \le & {e}^{1}x{e}^{x}\\ =& {e}^{1}\psi (x)\\ =& {e}^{1}\psi (d(x,fx))\\ \le & {e}^{1}\psi (max\{d(x,y),d(x,fx),d(y,Ty),\frac{1}{2}(d(x,Ty)+d(fx,y))\}).\end{array}$
Note that f and T satisfy all the hypotheses of Theorem 2.1. Hence f and T have a fixed point. Here 0 is the fixed point of f and T. □
3 Applications
 (h1)
μ is a Lebesgueintegrable mapping on each compact of $[0,+\mathrm{\infty})$;
 (h2)For every $\u03f5>0$, we have${\int}_{0}^{\u03f5}\mu (t)\phantom{\rule{0.2em}{0ex}}dt>0.$
If f or T is continuous, then f and T have a common fixed point.
Proof Follows from Theorem 2.1 by defining $\psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ via $\psi (t)={\int}_{0}^{t}\mu (s)\phantom{\rule{0.2em}{0ex}}ds$ and noting that ψ is an altering distance function. □
The continuity of f or T in Theorem 3.1 can be dropped.
If X satisfies property (P), then f and T have a common fixed point.
By taking $A=B=X$ in Theorems 3.1 and 3.2, we have the following results.
If f or T is continuous, then f and T have a common fixed point.
If X satisfies property (P), then f and T have a common fixed point.
Declarations
Acknowledgements
The authors thank the editor and the referees for their valuable comments and suggestions regarding the initial version of our article.
Authors’ Affiliations
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