Some unique fixed point theorems for rational contractions in partially ordered metric spaces
 Muhammad Arshad^{1}Email author,
 Erdal Karapınar^{2} and
 Jamshaid Ahmad^{3}
https://doi.org/10.1186/1029242X2013248
© Arshad et al.; licensee Springer 2013
Received: 6 February 2013
Accepted: 2 May 2013
Published: 17 May 2013
Abstract
In this paper, we prove some unique fixed point results for an operator T satisfying certain rational contraction condition in a partially ordered metric space. Our results generalize the main result of Jaggi (Indian J. Pure Appl. Math. 8(2):223230, 1977). We give several examples to show that our results are proper generalization of the existing one.
MSC:47H10, 54H25, 46J10, 46J15.
Keywords
1 Introduction
Fixed point theory is one of the famous and traditional theories in mathematics and has a broad set of applications. In this theory, contraction is one of the main tools to prove the existence and uniqueness of a fixed point. Banach’s contraction principle, which gives an answer on the existence and uniqueness of a solution of an operator equation $Tx=x$, is the most widely used fixed point theorem in all of analysis. This principle is constructive in nature and is one of the most useful tools in the study of nonlinear equations. There are many generalizations of Banach’s contraction mapping principle in the literature [1–6]. These generalizations were made either by using the contractive condition or by imposing some additional conditions on an ambient space. There have been a number of generalizations of metric spaces such as rectangular metric spaces, pseudo metric spaces, fuzzy metric spaces, quasi metric spaces, quasi semimetric spaces, probabilistic metric spaces, Dmetric spaces and cone metric spaces
The basic topological properties of ordered sets were discussed by Wolk [7] and Monjardet [8]. The existence of fixed points in partially ordered metric spaces was considered by Ran and Reurings [9]. After this paper, Nieto et al. [10–12] published some new results. Recently, many papers have been reported on partially ordered metric spaces (see, e.g., [9–19] and also [8, 20–33]).
The triple $(X,d,\u2aaf)$ is called partially ordered metric spaces (POMS) if $(X,\u2aaf)$ is a partially ordered set and $(X,d)$ is a metric space. Further, if $(X,d)$ is a complete metric space, the triple $(X,d,\le )$ is called partially ordered complete metric spaces (POCMS). Throughout the manuscript, we assume that $X\ne \mathrm{\varnothing}$. A partially ordered metric space $(X,d,\u2aaf)$ is called ordered complete (OC) if for each convergent sequence ${\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}\subset X$, the following condition holds: either

if $\{{x}_{n}\}$ is a nonincreasing sequence in X such that ${x}_{n}\to {x}^{\ast}$ implies ${x}^{\ast}\u2aaf{x}_{n}$$\mathrm{\forall}n\in \mathbb{N}$, that is, ${x}^{\ast}=inf\{{x}_{n}\}$, or

if $\{{x}_{n}\}$ is a nondecreasing sequence in X such that ${x}_{n}\to {x}^{\ast}$ implies ${x}_{n}\u2aaf{x}^{\ast}$$\mathrm{\forall}n\in \mathbb{N}$, that is, ${x}^{\ast}=sup\{{x}_{n}\}$.
In this manuscript, we prove that an operator T satisfying certain rational contraction condition has a fixed point in a partially ordered metric space. Our results generalize the main result of Jaggi [34].
2 Main results
We start this section with the following definition.
for any distinct $x,y\in X$ with $x\u2aafy$, where $L\ge 0$ and $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$.
Theorem 2 Let $(X,d,\u2aaf)$ be a complete partially ordered metric space. Suppose that a selfmapping T is an almost Jaggi contraction, continuous and nondecreasing. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. Then T has a unique fixed point.
a contradiction. Hence u is a unique fixed point of T in X. □
Then T is continuous and nondecreasing. Take $\beta =\frac{1}{4}$. Then, for any $\alpha \in [0,1)$ with $\alpha +\beta <1$, we have the result. Let us examine in detail. Without loss of generality, we assume that $y\u2aafx$.
holds for any $L\ge 1$ and any $\alpha \in [0,1)$ with $\alpha +\beta <1$. Thus, all the conditions of Theorem 2 are satisfied.
holds for any $L\ge 0$ and any $\alpha \in [0,1)$ with $\alpha +\beta <1$. Hence, all the conditions of Theorem 2 are satisfied.
Notice that $0\in X$ is the fixed point of T.
for any distinct $x,y\in X$ with $x\u2aafy$, where $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$.
Corollary 5 Let $(X,d,\u2aaf)$ be a complete partially ordered metric space. Suppose that a selfmapping T is a Jaggi contraction, continuous and nondecreasing. Suppose that there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. Then T has a fixed point.
Proof Set $L=0$ in Theorem 2. □
Also, $x\u2aafy$ iff $x\le y$. Clearly, T is an increasing and continuous selfmapping on X. We shall prove that conditions of Corollary 5 hold and T has a fixed point.
Proof For the proof of this example, we have the following cases.

Let $0\le x<y\le 2$. Then$\begin{array}{rl}d(Tx,Ty)& =max\{Tx,Ty\}=\frac{y}{8(1+y)}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =\frac{1}{4}[\frac{max\{x,\frac{x}{8(1+x)}\}max\{y,\frac{y}{8(1+y)}\}}{max\{x,y\}}+max\{x,y\}]\\ =\frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}$

Let $2<x<y$. Then$\begin{array}{rcl}d(Tx,Ty)& =& max\{Tx,Ty\}=\frac{y}{12}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =& \frac{1}{4}[\frac{max\{x,\frac{x}{12}\}max\{y,\frac{y}{12}\}}{max\{x,y\}}+max\{x,y\}]\\ =& \frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}$

Let $0\le x\le 2$ and $2<y$. Then$\begin{array}{rcl}d(Tx,Ty)& =& max\{Tx,Ty\}=\frac{y}{12}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =& \frac{1}{4}[\frac{max\{x,\frac{x}{8(1+x)}\}max\{y,\frac{y}{12}\}}{max\{x,y\}}+max\{x,y\}]\\ =& \frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}$
Then conditions of Corollary 5 hold and T has a fixed point (here, $x=0$ is a fixed point of T). □
In the next theorem, we establish the existence of a unique fixed point of a map T by assuming only the continuity of some iteration of T.
Theorem 7 Let $(X,d,\u2aaf)$ be a complete partially ordered metric space. Suppose that a selfmapping T is nondecreasing and an almost Jaggi contraction. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. If the operator ${T}^{p}$ is continuous for some positive integer p, then T has a unique fixed point.
a contradiction. Hence $Tz=z$. The uniqueness of z follows as in Theorem 2. □
Corollary 8 Let $(X,d,\u2aaf)$ be a complete partially ordered metric space. Suppose that a selfmapping T is nondecreasing and a Jaggi contraction. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. If the operator ${T}^{p}$ is continuous for some positive integer p, then T has a unique fixed point.
Proof Set $L=0$ in Theorem 7. □
The following theorem generalizes Theorem 2.
for any distinct $x,y\in X$ with $x\u2aafy$ and for some $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$ and $L\ge 0$. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aaf{T}^{m}{x}_{0}$. If ${T}^{m}$ is continuous, then T has a unique fixed point.
Thus, Tz is also a fixed point of ${T}^{m}$. But, by Theorem 2, we know that ${T}^{m}$ has a unique fixed point z. It follows that $z=Tz$. Hence, z is the unique fixed point of T. □
for all distinct $x,y\in X$ and for some $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aaf{T}^{m}{x}_{0}$. If ${T}^{m}$ is continuous, then T has a unique fixed point.
Proof Set $L=0$ in Theorem 9. □
Now, we give the following example.
It can be easily seen that T is discontinuous and does not satisfy (1) for any $\alpha ,\beta \in [0,1)$ with $\alpha +\beta \prec 1$ when $x=\frac{1}{3}$, $y=1$. Now ${T}^{2}(x)=0$ for all $x\in [0,1]$. It can be verified that ${T}^{2}$ satisfies the conditions of Theorem 9 and 0 is a unique fixed point of ${T}^{2}$.
for any points $x,y\in X$ with $x\u2aafy$, and for some $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$ and $L\ge 0$. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. Then T has a fixed point.
where $k=\frac{\beta}{1\alpha}<1$. Letting $n\to \mathrm{\infty}$, then the righthand side of the inequality (9) tends to 0. Thus, the sequence $\{{x}_{n}\}$ is Cauchy.
which is possible only if $d(Tz,z)=0$. Thus, $Tz=z$.
a contradiction. Hence z is a unique fixed point of T in X. □
for any points $x,y\in X$ with $x\u2aafy$, and for some $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$. Then T has a fixed point.
Proof Set $L=0$ in Theorem 12. □
Also, $x\u2aafy$ iff $x\le y$. Suppose that $\beta =\frac{1}{3}$ and $\alpha \in [0,1)$ such that $\alpha +\beta <1$. Clearly, T is an injective, continuous and sequentially convergent mapping on X. We shall prove that conditions of Corollary 8 hold and T has a fixed point.
Proof For the proof of this example, we have the following cases.
Hence, conditions of Corollary 8 hold and T has a fixed point (here $x=0$ is a fixed point of T). □
3 Further results
for all $x,y\in X$ with $y\u2aafx$, where $A=d(y,Tx)+d(x,Ty)$ and λ, μ are nonnegative reals such that $\lambda +\mu <1$. If there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$, then T has a fixed point.
by iteration. Put ${x}_{n+1}=T{x}_{n}$. If there exists ${n}_{0}\in N$ such that ${x}_{{n}_{0}}={x}_{{n}_{0}+1}$, then from ${x}_{{n}_{0}}={x}_{{n}_{0}+1}=T{x}_{{n}_{0}}$, we get ${x}_{{n}_{0}}$ is a fixed point, and the proof is finished. Suppose that ${x}_{n}\ne {x}_{n+1}$ for $n\in N$. Since the points ${x}_{n}$ and ${x}_{n1}$ are comparable for all $n\in N$ due to (14), we have the following two cases.
and this proves that $d({x}_{m},{x}_{n})\to 0$ as $m,n\to \mathrm{\infty}$.
Thus z is a fixed point.
Case 2. If $A=d({x}_{n1},T{x}_{n})+d({x}_{n},T{x}_{n1})=0$, then $d({x}_{n+1},{x}_{n})=0$. This implies that ${x}_{n}={x}_{n+1}$, a contradiction. Thus there exists a fixed point z of T. □
is satisfied. This shows that conditions of Theorem 15 hold and T has a fixed point $\frac{1}{2}\in [0,1]$.
We may remove the continuity criteria on T in Theorem 15 as follows.
for all $x,y\in X$ with $y\u2aafx$, where $A=d(x,Tx)+d(y,Ty)$ and λ, μ are nonnegative reals with $\lambda +\mu <1$. And also suppose that X has the (OC) property. If there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$, then T has a fixed point.
Proof We only have to check that $z=Tz$. As $\{{x}_{n}\}\subset X$ is a nondecreasing sequence and ${x}_{n}\to z\in X$, then $z=sup\{{x}_{n}\}$ for all $n\in N$. Since T is a nondecreasing mapping, then $T{x}_{n}\u2aafTz$ for all $n\in \mathbb{N}$ or, equivalently, ${x}_{n+1}\u2aafTz$ for all $n\in N$. Moreover, as ${x}_{0}\prec {x}_{1}\u2aafTz$ and $z=sup\{{x}_{n}\}$, we get $z\u2aafTz$. Suppose that $z\prec Tz$. Using a similar argument as that in the proof of Theorem 15 for ${x}_{0}\u2aafT{x}_{0}$, we obtain that $\{{T}^{n}z\}$ is a nondecreasing sequence and ${lim}_{n\to \mathrm{\infty}}{T}^{n}z=y$ for certain $y\in X$. Again, using (OC), we have that $y=sup\{{T}^{n}z\}$. Moreover, from ${x}_{0}\u2aafz$, we get ${x}_{n}={T}^{n}{x}_{0}\u2aaf{T}^{n}z$ for $n\ge 1$ and ${x}_{n}\prec {T}^{n}z$ for $n\ge 1$ because ${x}_{n}\u2aafz\prec Tz\u2aaf{T}^{n}z$ for $n\ge 1$ as ${x}_{n}$ and ${T}^{n}z$ are comparable and distinct for $n\ge 1$.
As $\lambda <1$, $d(x,z)=0$, thus $z=y$. Particularly, $z=y=sup\{{T}^{n}z\}$ and consequently, $Tz\u2aafz$ which is a contradiction. Hence, we conclude that $Tz=z$.
Case 2. If $A=d({T}^{n}z,T{x}_{n})+d({x}_{n},{T}^{n+1}z)=0$, then $d({x}_{n+1},{T}^{n+1}z)=0$. Taking the limit as $n\to \mathrm{\infty}$, we get $d(z,y)=0$. Then $z=y=sup\{{T}^{n}z\}$, which implies that $Tz\u2aafz$, a contradiction. Thus $Tz=z$. □
Now we prove the sufficient condition for the uniqueness of the fixed point in Theorem 15 and Theorem 17, that is,
U: for any $y,z\in X$, there exists $x\in X$ which is comparable to y and z.
Theorem 18 Adding the above mentioned condition to the hypothesis of Theorem 15 (or Theorem 17), one obtains the uniqueness of the fixed point of T.
Proof We distinguish two cases.
 (i)If $d(z,Ty)+d(y,Tz)\ne 0$, then using the contractive condition, we have$\begin{array}{rcl}d(y,z)& =& d(Ty,Tz)\le \lambda d(y,z)+\mu \frac{d(y,Ty)d(y,Tz)+d(z,Ty)d(z,Tz)}{d(z,Ty)+d(y,Tz)}\\ \le & \lambda d(y,z)+\mu \frac{d(y,y)d(y,z)+d(z,y)d(z,z)}{d(z,y)+d(y,z)}=\lambda d(y,z).\end{array}$
 (ii)
If $d(z,Ty)+d(y,Tz)=0$, then $d(y,z)=0$, a contradiction. Thus $y=z$.
Case 2. If y and z are not comparable, then by a given condition there exists $x\in X$ comparable to y and z. Monotonicity implies that ${T}^{n}x$ is comparable to ${T}^{n}y=y$ and ${T}^{n}z=z$ for $n=0,1,2,\dots $ .
 (i)If $d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)\ne 0$, then using the contractive condition, we obtain, for $n\ge 2$,$\begin{array}{rcl}d({T}^{n}x,y)& =& d({T}^{n}x,{T}^{n}y)\\ \le & \lambda d({T}^{n1}x,{T}^{n1}y)\\ +\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,{T}^{n}y)+d({T}^{n1}y,{T}^{n}x)d({T}^{n1}y,{T}^{n}y)}{d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)}\\ \le & \lambda d({T}^{n1}x,y)+\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,y)+d(Ty,{T}^{n}x)d(y,y)}{d(y,{T}^{n}x)+d({T}^{n1}x,y)}\\ \le & \lambda d({T}^{n1}x,y)+\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,y)}{d(y,{T}^{n}x)+d({T}^{n1}x,y)}.\end{array}$
 (ii)If $d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)=0$, then $d({T}^{n}x,y)=0$. Then$\underset{n\to \mathrm{\infty}}{lim}{T}^{n}x=y.$
Now, the uniqueness of the limit gives that $y=z$. This completes the proof. □
Remark 19 If in Theorem 15Theorem 18 $\mu =0$, then we obtain Theorem 2.1Theorem 2.3 of [10].
We get the following fixed point theorem in partially ordered metric spaces if we take $\lambda =0$ in the theorems of Section 3.
for all $x,y\in X$ with $y\u2aafx$, where $A=d(y,Tx)+d(x,Ty)$ and μ is a nonnegative real with $0\le \mu <1$. Suppose also that either T is continuous or X satisfies the condition (OC). If there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$, then T has a fixed point.
If $(X,\u2aaf)$ satisfies the condition used in Theorem 18, then the uniqueness of a fixed point can be proved.
4 Applications
In this section we state some applications of the main results. The first result is the consequence of Theorem 2.
for all distinct $x,y\in X$ with $x\u2aafy$ and for $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$, where $L\ge 0$. If there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$ then T has at least one fixed point.
Similarly, the following result is the consequence of Corollary 5.
for any distinct $x,y\in X$ with $x\u2aafy$, where $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$. Suppose there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. Then T has a fixed point.
The following result is the consequence of Theorem 12.
for any $x,y\in X$ with $x\u2aafy$, and for some $\alpha ,\beta \in [0,1)$ with $\alpha +\beta <1$ and $L\ge 0$. Suppose that there exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$. Then T has a fixed point.
Declarations
Acknowledgements
The authors express their gratitude to the anonymous referees for constructive and useful remarks, comments and suggestions.
Authors’ Affiliations
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