# Common fixed point and coincidence point of generalized contractions in ordered metric spaces

- Ismat Beg
^{1}Email author and - Asma Rashid Butt
^{2}

**2012**:229

https://doi.org/10.1186/1687-1812-2012-229

© Beg and Butt; licensee Springer. 2012

**Received: **26 July 2012

**Accepted: **29 November 2012

**Published: **18 December 2012

## Abstract

Let $(X,\u2aaf)$ be a partially ordered set and *d* be a complete metric on *X*. The notion of *f*-contractive for a set-valued mapping due to Latif and Beg is extended through an implicit relation. Coincidence and fixed point results are obtained for mappings satisfying generalized contractions in a partially ordered metric space *X*. Our results improve and extend several known results in the existing literature.

**MSC:**47H10, 47H04, 47H07.

### Keywords

fixed point coincidence point partially ordered set metric space set-valued mapping## 1 Introduction and preliminaries

*X*, $\mathit{Cl}(X)$ be the class of all non-empty closed subsets of

*X*and $\mathit{CB}(X)$ be the class of all non-empty closed and bounded subsets of

*X*. For $A,B\in \mathit{CB}(X)$, let

*D* is called the *Hausdorff metric* induced by *d*.

*f*-contractive if there exists a real number

*κ*with $0<\kappa <1$ such that

By introducing the notion of *f*-contractiveness for set-valued mappings, Kaneko [1] obtained a result which gives the existence of a coincidence point in a metric space. An improved version of Kaneko’s result was obtained by Latif and Beg [2]. They proved the following.

**Theorem 1.1** [[2], Theorem 2.6]

*Let* $f:X\to X$ *be a continuous map with* $f(X)$ *complete*. *Suppose that* $F:X\to \mathit{CB}(X)$ *is a* *f*-*contractive map such that* $F(X)\subseteq f(X)$. *Then there exists* ${x}_{0}\in X$ *such that* $f{x}_{0}\in F{x}_{0}$.

In [3], Kaneko and Sessa proved the following coincidence point result.

**Theorem 1.2** [3]

*Let*$(X,d)$

*be a complete metric space*$f:X\to X$

*and*$F:X\to \mathit{CB}(X)$

*be compatible continuous mappings such that*$F(X)\subseteq f(X)$

*and*

*for all* $x,y\in X$, *where* $0\le \kappa <1$.

*Then there exists* ${x}_{0}\in X$ *such that* $f{x}_{0}\in F{x}_{0}$.

*A mapping*$f:X\to X$

*is called*

*K*-

*mapping if there exists a real number*

*κ*

*with*$0\le \kappa <\frac{1}{2}$

*such that*

**Definition 1.3** Let $F:X\to P(X)$ be a mapping. A point $x\in X$ is said to be a *fixed point* of *F* if $x\in Fx$.

**Definition 1.4** Let $F:X\to P(X)$ and $f:X\to X$ be mappings. A point $x\in X$ is said to be a *coincidence point* of *F* and *f* if $fx\in Fx$.

**Definition 1.5**A

*partial order*is a binary relation ⪯ over a set

*X*which satisfies the following conditions:

- 1.
$x\u2aafx$ (reflexivity);

- 2.
If $x\u2aafy$ and $y\u2aafx$, then $x=y$ (antisymmetry);

- 3.
If $x\u2aafy$ and $y\u2aafz$, then $x\u2aafz$ (transitivity)

for all *x*, *y* and *z* in *X*.

A set with a partial order ⪯ is called a *partially ordered set*.

Let $(X,\u2aaf)$ be a partially ordered set and $x,y\in X$. Elements *x* and *y* are said to be *comparable elements* of *X* if either $x\u2aafy$ or $y\u2aafx$.

Recently, there have been so many exciting developments in the field of existence of a fixed point in partially ordered sets (see [4–20] and the references cited therein). This trend was started by Ran and Reurings in [11] where they extended the Banach contraction principle in partially ordered sets with some application to a matrix equation. Ran and Reurings [11] proved the following seminal result.

**Theorem 1.6** [11]

*Let*$(X,\u2aaf)$

*be a partially ordered set such that every pair*$x,y\in X$

*has an upper and lower bound*.

*Let*

*d*

*be a complete metric on*

*X*

*and*$f:X\to X$

*be a continuous monotone*(

*either order*-

*preserving or order*-

*reversing*)

*mapping*.

*Suppose that the following conditions hold*:

- 1.
*There exists*$\kappa \in (0,1)$*with*$d(f(x),f(y))\le \kappa d(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}x\u2aafy.$ - 2.
*There exists*${x}_{0}\in X$*with*${x}_{0}\u2aaff{x}_{0}$*or*$f{x}_{0}\u2aaf{x}_{0}$.

*Then*

*f*

*is a Picard operator*(

*PO*),

*that is*,

*f*

*has a unique fixed point*${x}^{\ast}\in X$

*and for each*$x\in X$,

Theorem 1.6 was further extended and refined in [7–9, 12, 15, 19]. These results are hybrid of the two fundamental classical theorems: Banach’s fixed point theorem (see [21]) and Tarski’s [22] fixed point theorem. Our aim in this paper is to introduce a generalization of *f*-contractiveness through an implicit relation. This implicit relation is then used to obtain the existence of coincidence and common fixed points for a pair of single-valued mapping and set-valued mapping on a partially ordered metric space. In Section 2, we prove a coincidence point theorem where we use an implicit relation only for comparable elements of a partially ordered set *X*. Our result generalizes/extends [1–3, 23, 24] work to partial ordered sets. Section 3 deals with the existence of a common fixed point by using the notion of $k-\u2aaf$ set-valued mapping, which improves the results of Latif and Beg [2] to partially ordered sets.

We will make use of the following lemma in the proof of our result in the next section.

**Lemma 1.7** [25]

*Let* $A,B\in \mathit{CB}(X)$ *and* $a\in A$. *Then for* $\u03f5>0$, *there exists an element* $b\in B$ *such that* $d(a,b)<D(A,B)+\u03f5$.

Throughout the next two sections, we take $(X,\u2aaf)$ as a partially ordered set with a complete metric *d*.

## 2 Implicit relation and coincidence points

Implicit relations in metric spaces have been considered by several authors in connection with solving nonlinear functional equations (see, for instance, [16–18, 26] and the references cited therein). First, we give some implicit relations for subsequent use.

Let ${R}_{+}$ be the set of nonnegative real numbers and $\mathcal{T}$ be the set of continuous real-valued functions $T:{R}_{+}^{6}\to R$ satisfying the following conditions:

${\mathcal{T}}_{1}$: $T({t}_{1},{t}_{2},\dots ,{t}_{6})$ is non-increasing in ${t}_{2},{t}_{3},\dots ,{t}_{6}$.

*κ*with $0<\kappa <1$ and $\epsilon >0$ such that the inequalities

${\mathcal{T}}_{3}$: $T(w,0,0,v,v,0)\le 0$ implies $w\le \kappa v$.

Next, we give some examples for *T* satisfying ${\mathcal{T}}_{1}$-${\mathcal{T}}_{3}$.

**Example 2.1** $T({t}_{1},\dots ,{t}_{6})={t}_{1}-\alpha (max\{{t}_{2},{t}_{3},{t}_{4},1/2({t}_{5}+{t}_{6})\})$, where $0<\alpha <1$.

${\mathcal{T}}_{1}$: is obvious.

${\mathcal{T}}_{2}$: Let $u>0$, then choose $\u03f5>0$ such that $\alpha u+\u03f5<u$ (this is possible since $0<\alpha <1$). As $T(w,v,v,u,u+v,0)\le 0$, therefore $w-\alpha (max\{u,v\})\le 0$. Now let $u\le w+\u03f5$. If $u\ge v$, then $u\le \alpha u+\u03f5<u$. Hence a contradiction. Thus $u<v$ and $w\le \alpha v$. If $u=0$, then $w\le \alpha v$. Thus ${\mathcal{T}}_{2}$ is satisfied.

${\mathcal{T}}_{3}$: Since $T(w,0,0,v,v,0)\le 0$, therefore $w-\alpha v\le 0$. It further implies that $w\le \alpha v$.

**Example 2.2** $T({t}_{1},\dots ,{t}_{6})={t}_{1}-\alpha (max\{{t}_{2},\frac{{t}_{3}+{t}_{4}}{2},\frac{{t}_{5}+{t}_{6}}{2}\})$, where $0<\alpha <1$.

${\mathcal{T}}_{1}$: is obvious.

${\mathcal{T}}_{2}$: Let $u>0$, then choose $\u03f5>0$ so that $\alpha u+\u03f5<u$. Since $T(w,v,v,u,u+v,0)\le 0$, therefore $w-\alpha (max\{u,v\})\le 0$. Now let $u\le w+\u03f5$. If $u\ge v$, then $u\le \alpha u+\u03f5<u$. Hence a contradiction. Thus $u<v$ and $w\le \alpha v$. If $u=0$, then $w\le \alpha v$. Thus ${\mathcal{T}}_{2}$ is satisfied.

${\mathcal{T}}_{3}$: Since $T(w,0,0,v,v,0)\le 0$, therefore $w-\frac{\alpha}{2}v\le 0$. It further implies that $w\le \frac{\alpha}{2}v\le \kappa v$.

**Example 2.3** $T({t}_{1},\dots ,{t}_{6})={t}_{1}-\alpha max\{{t}_{2},{t}_{3},{t}_{4}\}-(1-\alpha )(a{t}_{5}+b{t}_{6})$, where $0<\alpha <1$ , $0\le a,b<1/2$.

${\mathcal{T}}_{1}$: is obvious.

${\mathcal{T}}_{2}$: Let $u>0$, then choose $\u03f5>0$ so that $(\alpha +2a(1-\alpha ))u+\u03f5<u$ (this is possible since $0<\alpha +2a(1-\alpha )<1$). Since $T(w,v,v,u,u+v,0)\le 0$, therefore $w-\alpha max\{u,v\}-(1-\alpha )a(u+v)\le 0$. Now let $u\le w+\u03f5$. If $u\ge v$, then $u\le [\alpha +2a(1-\alpha )]u+\u03f5<u$, hence a contradiction. Thus $u<v$ and $w\le (\alpha +2a(1-\alpha ))v$. Thus ${\mathcal{T}}_{2}$ is satisfied with $\kappa =\alpha +2a(1-\alpha )$.

${\mathcal{T}}_{3}$: Since $T(w,0,0,v,v,0)\le 0$, therefore $w-\alpha v-(1-\alpha )av\le 0$. It further implies that $w\le [\alpha +a(1-\alpha )]v\le \kappa v$.

**Theorem 2.4**

*Let*$f:X\to X$

*and*$F:X\to \mathit{CB}(X)$

*satisfying*

*for all comparable elements*

*x*,

*y*

*of*

*X*

*and for some*$T\in \mathcal{T}$.

*If the following conditions are satisfied*:

- 1.
$F(X)\subseteq f(X)$

*and*$f(X)$*is closed*; - 2.
*If*$fq\in Fx$,*then*$x\u2aafq$; - 3.
*If*${y}_{n}\in F{x}_{n}$*is such that*${y}_{n}\to y=fp$,*then*${x}_{n}\u2aafp$*for all**n*,

*then there exists* *p* *with* $fp\in Fp$.

*Proof*Let ${x}_{0}\in X$, then by using assumptions 1 and 2, we can choose ${x}_{1}\in X$ with ${x}_{0}\u2aaf{x}_{1}$ such that ${y}_{0}=f{x}_{1}\in F{x}_{0}$. Since ${y}_{0}\in F{x}_{0}$, then for any $\epsilon >0$, from Lemma 1.7, there exists ${y}_{1}\in F{x}_{1}$ such that

Using assumptions 1 and 2, since ${y}_{1}\in F{x}_{1}\subseteq f(X)$, there exists ${x}_{2}\in X$ such that ${y}_{1}=f{x}_{2}\in F{x}_{1}$ and so ${x}_{1}\u2aaf{x}_{2}$.

Using assumptions 1 and 2, since ${y}_{2}\in F{x}_{2}\subseteq f(X)$, there exists ${x}_{3}\in X$ such that ${y}_{2}=f{x}_{3}\in F{x}_{2}$ and so ${x}_{2}\u2aaf{x}_{3}$.

Using assumptions 1 and 2, since ${y}_{3}\in F{x}_{3}\subseteq f(X)$, there exists ${x}_{4}\in X$ such that ${y}_{3}=f{x}_{4}\in F{x}_{3}$ and so ${x}_{3}\u2aaf{x}_{4}$.

*y*in the complete metric space

*X*such that

Now, since $f(X)$ is closed, there exists $p\in X$ such that $y=fp\in f(X)$ and by assumption 3, ${x}_{n}\u2aafp$ for all *n*.

From above, we have $\kappa >1$, a contradiction. So, $d(y,Fp)=0$. Therefore, $d(y,Fp)=0$ and $fp=y\in \overline{Fp}=Fp$. □

**Remark 2.5** In assumptions 2 and 3 of Theorem 2.4, we need only comparability of the elements. Theorem 2.4 with Example 2.2 partially improve [[27], Theorem 3.10].

**Corollary 2.6**

*Let*$f:X\to X$

*and*$F:X\to \mathit{CB}(X)$

*satisfy*

*for some* *κ* *with* $0<\kappa <1$ *and for all comparable elements* *x*, *y* *of* *X*.

*Also*,

*assume that the following conditions are satisfied*:

- 1.
$F(X)\subseteq f(X)$

*and*$f(X)$*is closed*. - 2.
*If*$fq\in Fx$,*then*$x\u2aafq$. - 3.
*If*${y}_{n}\in F{x}_{n}$*is such that*${y}_{n}\to y=fp$,*then*${x}_{n}\u2aafp$*for all**n*.

*Then there exists* *p* *such that* $fp\in Fp$.

*Proof* Let $T({t}_{1},\dots ,{t}_{6}):={t}_{1}-\kappa {t}_{2}$, then it is obvious that $T\in \mathcal{T}$. Therefore, the proof is complete from Theorem 2.4. □

**Corollary 2.7**

*Let*$f:X\to X$

*and*$F:X\to \mathit{CB}(X)$

*satisfy*

*for some*

*κ*

*with*$0<\kappa <1$

*and for all comparable elements*

*x*,

*y*

*of*

*X*.

*Also*,

*assume that the following conditions are satisfied*:

- 1.
$F(X)\subseteq f(X)$

*and*$f(X)$*is closed*. - 2.
*If*$fq\in Fx$,*then*$x\u2aafq$. - 3.
*If*${y}_{n}\in F{x}_{n}$*is such that*${y}_{n}\to y=fp$,*then*${x}_{n}\u2aafp$*for all**n*.

*Then* $fv\in Fv$.

**Remark 2.8** Corollary 2.6 extends Latif and Beg [[2], Theorem 2.6], the result of Kaneko and Sessa [1] in a partially ordered set. Corollary 2.7 also extends the results of [1–3, 24] to a partially ordered set.

## 3 Common fixed points

*A*and

*B*be two non-empty subsets of $(X,\u2aaf)$, the relations between

*A*and

*B*are denoted and defined as follows:

**Definition 3.1**Let

*M*be a non-empty subset of

*X*and $F:M\to P(X)$. A mapping

*F*is said to be $K-\u2aaf$ set-valued if there exists $0\le \kappa <\frac{1}{2}$, and for any $x\in M$, ${u}_{x}\in Fx$ there exists a ${u}_{y}\in Fy$ with ${u}_{x}\u2aaf{u}_{y}$ such that

for all $y\in M$ with $x\u2aafy$.

For $K-\u2aaf$ set-valued mappings, we just required comparability of the elements, but order does not matter.

**Theorem 3.2**

*Let*

*M*

*be a non*-

*empty closed subset of*

*X*

*and*$F:M\to \mathit{Cl}(M)$

*be a*$K-\u2aaf$

*set*-

*valued mapping satisfying*:

- 1.
*There exists*${x}_{0}$*in**M**such that*$\{{x}_{0}\}{\prec}_{1}F{x}_{0}$; - 2.
*If*${x}_{n}\to x$*is a sequence in**M**whose consecutive terms are comparable*,*then*${x}_{n}\u2aafx$*for all**n*.

*Then there exists* $x\in M$ *with* $x\in Fx$.

*Proof*Let ${x}_{0}\in M$. Then by assumption 1, there exists ${x}_{1}\in F{x}_{0}$ such that ${x}_{0}\u2aaf{x}_{1}$. Now, since

*F*is a $K-\u2aaf$ set-valued mapping, there is ${x}_{2}\in F{x}_{1}$ with ${x}_{1}\u2aaf{x}_{2}$ such that

*X*. Let $m>n$. Then

because $h\in (0,1)$, $1-{h}^{m-n}<1$.

Therefore, $d({x}_{n},{x}_{m})\to 0$ as $n\to \mathrm{\infty}$, which further implies that $\{{x}_{n}\}$ is a Cauchy sequence. So, there exists some point (say) *x* in the complete metric space *X* such that ${x}_{n}\to x$. By using assumption 2, ${x}_{n}\u2aafx$ for all *n*.

Further, since *M* is closed, $x\in M$. Now we want to show that $x\in Fx$.

*n*and F is a $k-\u2aaf$ set-valued mapping, there exists ${u}_{n}\in Fx$ with ${x}_{n}\u2aaf{u}_{n}$ such that

Letting $n\to \mathrm{\infty}$, we obtain ${u}_{n}\to x$.

As ${u}_{n}\in Fx$ and *Fx* is closed, so $x\in Fx$. □

**Example 3.3**Let $M=\{(0,0),(0,\frac{1}{4}),(0,\frac{-1}{4}),(\frac{-1}{4},\frac{1}{4})\}$ be a subset of $X={R}^{2}$ with usual order defined as follows: for $(u,v),(x,y)\in X$, $(u,v)\le (x,y)$ if and only if $u\le x,y\le v$. Let

*d*be a metric on

*X*defined as follows:

$(0,\frac{-1}{4})\le (0,0)\le (0,\frac{1}{4})$ and $(\frac{-1}{4},\frac{1}{4})\le (0,\frac{1}{4})$.

Consider $x=(0,\frac{-1}{4})\le (0,0)=y$ for ${u}_{x}=(0,0)\in Fx$, there exists ${u}_{y}=(0,0)\in Fy$ such that ${u}_{x}\le {u}_{y}$.

Next $0=d((0,0),(0,0))\le \kappa [d((0,\frac{-1}{4}),(0,0))+d((0,0),(0,0))]=\frac{\kappa}{4}$.

Similarly, for other comparable elements of *M*, one can see that *F* is a $K-\le $ set-valued mapping.

Further $(0,0)\in M$ is such that $\{0,0\}{\prec}_{1}F(0,0)$. Also, assumption 2 of Theorem 3.2 is satisfied and $(0,0)$ is the fixed point of *F*.

**Theorem 3.4** *Let* *M* *be a non*-*empty closed subset of* *X* *and* ${F}_{n}:M\to \mathit{Cl}(M)$ *be a sequence of mappings satisfying the following*:

*For any two mappings*${F}_{i}$, ${F}_{j}$

*and for any*$x\in M$, ${u}_{x}\in {F}_{i}x$,

*there exists a*${u}_{y}\in {F}_{j}y$

*with*${u}_{x}\u2aaf{u}_{y}$

*such that*

*for all* $y\in M$ *with* $x\u2aafy$ *and for some* $0\le \kappa <\frac{1}{2}$.

*Assume that the following conditions also hold*:

- 1.
*For each*${x}_{0}\in M$, $\{{x}_{0}\}{\prec}_{1}{F}_{1}{x}_{0}$. - 2.
*If*${x}_{n}\to x$*is a sequence in**X**whose consecutive terms are comparable*,*then*${x}_{n}\u2aafx$*for all**n*.

*Then there exists* $x\in M$ *with* $x\in \cap {F}_{n}x$.

*Proof*Let ${x}_{0}\in M$. Then by assumption 1, there exists ${x}_{1}\in {F}_{1}{x}_{0}$ such that ${x}_{0}\u2aaf{x}_{1}$. Now, using (B), there is ${x}_{2}\in {F}_{2}{x}_{1}$ with ${x}_{1}\u2aaf{x}_{2}$ such that

Therefore, $\{{x}_{n}\}$ is a Cauchy sequence in a complete metric space *X*, so ${x}_{n}\to x$. By using assumption 2, ${x}_{n}\u2aafx$ for all *n*.

*M*is closed, $x\in M$. Let ${F}_{m}$ be any arbitrary member of ${F}_{n}$. Now, since ${x}_{n}\in {F}_{n}{x}_{n-1}$ with ${x}_{n-1}\u2aaf{x}_{n}$. Also, ${x}_{n}\u2aafx$ for all

*n*. By using (B), there exists ${u}_{n}\in {F}_{m}x$ such that

Letting $n\to \mathrm{\infty}$, we have ${u}_{n}\to x$.

As ${u}_{n}\in {F}_{m}x$ and ${F}_{m}x$ is closed, so $x\in {F}_{m}x$, *i.e.*, $x\in {F}_{n}x$. □

**Remark 3.5** Theorems 3.2 and 3.4 improve/extend [[2], Theorems 4.1 and 4.2].

## Declarations

### Acknowledgements

The present version of the paper owes much to the precise and kind remarks of the learned referees.

## Authors’ Affiliations

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