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

Abstract

Let $(X,⪯)$ 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.

1 Introduction and preliminaries

Let $(X,d)$ be a metric space, $P(X)$ be the class of all non-empty subsets of 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

$$D(A,B):=max\{\underset{b\in B}{sup}d(b,A),\underset{a\in A}{sup}d(a,B)\},$$

where

$$d(a,B):=\underset{b\in B}{inf}d(a,b).$$

D is called the Hausdorff metric induced by d.

Let $f:X\to X$ be any single-valued mapping. A mapping $F:X\to \mathit{CB}(X)$ is f-contractive if there exists a real number κ with $0<\kappa <1$ such that

$$D(Fx,Fy)\le \kappa d(fx,fy)\text{for all }x,y\in X.$$

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

$$D(Fx,Fy)\le \kappa max\{d(fx,fy),d(fx,Fx),d(fy,Fy),1/2[d(fx,Fy)+d(fy,Fx)]\}$$

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

$$d(fx,fy)\le \kappa [d(x,fx)+d(y,fy)]\mathit{\text{for all}}x,y\in X.$$

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. $x⪯x$ (reflexivity);

• 2. If $x⪯y$ and $y⪯x$, then $x=y$ (antisymmetry);

• 3. If $x⪯y$ and $y⪯z$, then $x⪯z$ (transitivity)

for all x, y and z in X.

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

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

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 $(X,⪯)$ 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

  $$d(f(x),f(y))\le \kappa d(x,y)\mathit{\text{for all}}x⪯y.$$

• 2. There exists $x_0\in X$ with $x_0⪯f{x}_0$ or $f{x}_0⪯x_0$.

Then f is a Picard operator (PO), that is, f has a unique fixed point $x^{\ast }\in X$ and for each $x\in X$,

$$\underset{n\to \mathrm{\infty }}{lim}{f}^{n}x=x^{\ast }.$$

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 $k-⪯$ 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 $ϵ>0$, there exists an element $b\in B$ such that $d(a,b)<D(A,B)+ϵ$.

Throughout the next two sections, we take $(X,⪯)$ 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 $t_2,t_3,\dots ,t_6$.

${\mathcal{T}}_2$: there exists a real number κ with $0<\kappa <1$ and $\epsilon >0$ such that the inequalities

$$u\le w+ϵ$$

and

$$T(w,v,v,u,u+v,0)\le 0$$

imply

$$w\le \kappa v.$$

${\mathcal{T}}_3$: $T(w,0,0,v,v,0)\le 0$ 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 $ϵ>0$ such that $\alpha u+ϵ<u$ (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 $u\le w+ϵ$. If $u\ge v$, then $u\le \alpha u+ϵ<u$. 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 $ϵ>0$ so that $\alpha u+ϵ<u$. Since $T(w,v,v,u,u+v,0)\le 0$, therefore $w-\alpha (max\{u,v\})\le 0$. Now let $u\le w+ϵ$. If $u\ge v$, then $u\le \alpha u+ϵ<u$. 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 $ϵ>0$ so that $(\alpha +2a(1-\alpha ))u+ϵ<u$ (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 $u\le w+ϵ$. If $u\ge v$, then $u\le [\alpha +2a(1-\alpha )]u+ϵ<u$, 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

$$T(D(Fx,Fy),d(fx,fy),d(fx,Fx),d(fy,Fy),d(fx,Fy),d(fy,Fx))\le 0$$

(A)

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 $x⪯q$;

• 3. If $y_n\in F{x}_n$ is such that $y_n\to y=fp$, then $x_n⪯p$ 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 $x_0⪯x_1$ 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

$$d(y_0,y_1)\le D(F{x}_0,F{x}_1)+\epsilon .$$

(2.1)

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 $x_1⪯x_2$.

Now using (A), we have

$$T(D(F{x}_0,F{x}_1),d(f{x}_0,f{x}_1),d(f{x}_0,F{x}_0),d(f{x}_1,F{x}_1),d(f{x}_0,F{x}_1),d(f{x}_1,F{x}_0))\le 0.$$

Using the facts that $d(f{x}_0,F{x}_0)\le d(f{x}_0,y_0)$, $d(f{x}_1,F{x}_1)\le d(y_0,y_1)$,

$$d(f{x}_0,F{x}_1)\le d(f{x}_0,y_1)\le d(f{x}_0,y_0)+d(y_0,y_1),d(f{x}_1,F(x_0))\le d(y_0,y_0),$$

and by ${\mathcal{T}}_1$, we have

$$T(D(F{x}_0,F{x}_1),d(f{x}_0,y_0),d(f{x}_0,y_0),d(y_0,y_1),d(f{x}_0,y_0)+d(y_0,y_1),0)\le 0,$$

that is,

$$T(w,v,v,u,u+v,0)\le 0,$$

where $w=D(F{x}_0,F{x}_1)$, $v=d(f{x}_0,y_0)$, $u=d(y_0,y_1)$. By using ${\mathcal{T}}_2$, we have ($w\le \kappa v$),

$$D(F{x}_0,F{x}_1)\le \kappa d(f{x}_0,y_0).$$

(2.2)

Using (2.2) in (2.1), we have

$$d(y_0,y_1)\le \kappa d(f{x}_0,y_0)+ϵ.$$

Since $y_1\in F{x}_1$, then for $\kappa >0$, from Lemma 1.7, there exists $y_2\in F{x}_2$ such that

$$d(y_1,y_2)\le D(F{x}_1,F{x}_2)+\kappa .$$

(2.3)

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 $x_2⪯x_3$.

Now, since $x_1⪯x_2$, by using (A), we have

$$T(D(F{x}_1,F{x}_2),d(f{x}_1,f{x}_2),d(f{x}_1,F{x}_1),d(f{x}_2,F{x}_2),d(f{x}_1,F{x}_2),d(f{x}_2,F{x}_1))\le 0,$$

by ${\mathcal{T}}_1$ we have

$$T(D(F{x}_1,F{x}_2),d(y_0,y_1),d(y_0,y_1),d(y_1,y_2),d(y_0,y_1)+d(y_1,y_2),0)\le 0,$$

that is,

$$T(w,v,v,u,u+v,0)\le 0,$$

where $w=D(F{x}_1,F{x}_2)$, $v=d(y_0,y_1)$, $u=d(y_1,y_2)$. Therefore, by using ${\mathcal{T}}_2$, we have ($w\le \kappa v$)

$$D(F{x}_1,F{x}_2)\le \kappa d(y_0,y_1).$$

(2.4)

And so from (2.3) and (2.4), we have

$$d(y_1,y_2)\le D(F{x}_1,F{x}_2)+\kappa \le \kappa d(y_0,y_1)+\kappa .$$

(2.5)

Again, since $y_2\in F{x}_2$, then for $\kappa >0$, from Lemma 1.7, there exists $y_3\in F{x}_3$ such that

$$d(y_2,y_3)\le D(F{x}_2,F{x}_3)+{\kappa }^2.$$

(2.6)

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 $x_3⪯x_4$.

Now, since $x_2⪯x_3$, by using (A) we have

$$T(D(F{x}_2,F{x}_3),d(f{x}_2,f{x}_3),d(f{x}_2,F{x}_2),d(f{x}_3,F{x}_3),d(f{x}_2,F{x}_3),d(f{x}_3,F{x}_2))\le 0,$$

by ${\mathcal{T}}_1$ we have

$$T(D(F{x}_2,F{x}_3),d(y_1,y_2),d(y_1,y_2),d(y_2,y_3),d(y_1,y_2)+d(y_2,y_3),0)\le 0,$$

that is,

$$T(w,v,v,u,u+v,0)\le 0,$$

where $w=D(F{x}_2,F{x}_3)$, $v=d(y_1,y_2)$, $u=d(y_2,y_3)$. Therefore, by using ${\mathcal{T}}_2$, we have ($w\le \kappa v$)

$$D(F{x}_2,F{x}_3)\le \kappa d(y_1,y_2).$$

(2.7)

Now, using (2.5), (2.6) and (2.7), we have

$$d(y_2,y_3)\le \kappa d(y_1,y_2)+{\kappa }^2\le {\kappa }^{2}d(y_0,y_1)+2{\kappa }^2.$$

(2.8)

Continuing in this way, we obtain a sequence $\{x_n\}$ with $x_n⪯x_{n+1}$ such that $y_n=f{x}_{n+1}\in F{x}_n$ for $n\ge 0$ and

(2.9)

(2.10)

Hence, $\{y_n\}$ is a Cauchy sequence. So, there exists a point y in the complete metric space X such that

$$\underset{n\to \mathrm{\infty }}{lim}{y}_n=\underset{n\to \mathrm{\infty }}{lim}f(x_{n+1})=y\in \underset{n\to \mathrm{\infty }}{lim}F(x_n).$$

Now, since $f(X)$ is closed, there exists $p\in X$ such that $y=fp\in f(X)$ and by assumption 3, $x_n⪯p$ for all n.

Now, using (A), we have

$$T(D(F{x}_n,Fp),d(f{x}_n,fp),d(f{x}_n,F{x}_n),d(fp,Fp),d(f{x}_n,Fp),d(fp,F{x}_n))\le 0.$$

Now, taking limit as $n\to \mathrm{\infty }$ and using ${\mathcal{T}}_2$, also the facts that $d(f{x}_n,fp)=d(y_{n-1},y)\to 0$, $d(f{x}_n,F{x}_n)\le d(y_{n-1},y_n)\to 0$, $d(fp,F{x}_n)\le d(y,y_n)\to 0$, we have

$$T(\underset{n\to \mathrm{\infty }}{lim}D(F{x}_n,Fp),0,0,d(y,Fp),d(y,Fp),0)\le 0,$$

that is,

$$T(w,0,0,v,v,0).$$

By using ${\mathcal{T}}_3$, we get

$$\underset{n\to \mathrm{\infty }}{lim}D(F{x}_n,Fp)\le \kappa (d(y,Fp)).$$

Next, since $y_n\in F{x}_n$,

$$d(y_n,Fv)\le D(F{x}_n,Fp),$$

taking limit as $n\to \mathrm{\infty }$, we obtain

$$d(y,Fp)\le \underset{n\to \mathrm{\infty }}{lim}D(F{x}_n,Fp)\le \kappa d(y,Fp).$$

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

$$D(Fx,Fy)\le \kappa d(fx,fy)$$

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 $x⪯q$.

• 3. If $y_n\in F{x}_n$ is such that $y_n\to y=fp$, then $x_n⪯p$ 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

$$D(Fx,Fy)\le \kappa max\{d(fx,fy),d(fx,Fx),d(fy,Fy),1/2[d(fx,Fy)+d(fy,Fx)]\}$$

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 $x⪯q$.

• 3. If $y_n\in F{x}_n$ is such that $y_n\to y=fp$, then $x_n⪯p$ 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

In this section, we define $K-⪯$ set-valued mappings on partially ordered metric spaces, and by using the definitions, we obtain the existence of common fixed points. Let A and B be two non-empty subsets of $(X,⪯)$, the relations between A and B are denoted and defined as follows:

$$A{\prec }_{1}B:\text{ if for every }a\in A\text{ there exists }b\in B\text{ such that }a⪯b.$$

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 $K-⪯$ 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 $u_x⪯u_y$ such that

$$d(u_x,u_y)\le \kappa [d(x,u_x)+d(y,u_y)]$$

for all $y\in M$ with $x⪯y$.

For $K-⪯$ 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 $K-⪯$ 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 $x_n⪯x$ 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 $x_0⪯x_1$. Now, since F is a $K-⪯$ set-valued mapping, there is $x_2\in F{x}_1$ with $x_1⪯x_2$ such that

$$d(x_1,x_2)\le \kappa [d(x_0,x_1)+d(x_1,x_2)],$$

which gives

$$(1-\kappa )d(x_1,x_2)\le \kappa d(x_0,x_1),$$

and consequently

$$d(x_1,x_2)\le \frac{\kappa }{1-\kappa }d(x_0,x_1).$$

Continuing in this way, we obtain a sequence $\{x_n\}$ whose consecutive terms are comparable and

$$d(x_n,x_{n+1})\le {\left[\frac{\kappa }{1-\kappa }\right]}^{n}d(x_0,x_1).$$

Take $0\le h=\frac{\kappa }{1-\kappa }<1$, then we have

$$d(x_n,x_{n+1})\le h^{n}d(x_0,x_1).$$

(3.1)

Next, we will show that $\{x_n\}$ is a Cauchy sequence in X. Let $m>n$. Then

$$\begin{array}{rcl}d(x_n,x_m)& \le & d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+d(x_{n+2},x_{n+3})+\cdots +d(x_{m-1},x_m)\\ \le & [h^n+h^{n+1}+h^{n+2}+\cdots +h^{m-1}]d(x_0,x_1)\\ =& h^n[1+h+h^2+\cdots +h^{m-n-1}]d(x_0,x_1)\\ =& h^n\frac{1-h^{m-n}}{1-h}d(x_0,x_1)\\ <& \frac{h^n}{1-h}d(x_0,x_1)\end{array}$$

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, $x_n⪯x$ for all n.

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

Since $x_n\in F{x}_{n-1}$ with $x_{n-1}⪯x_n$ also $x_n⪯x$ for all n and F is a $k-⪯$ set-valued mapping, there exists $u_n\in Fx$ with $x_n⪯u_n$ such that

$$d(x_n,u_n)\le \kappa [d(x_{n-1},x_n)+d(x,u_n)].$$

(3.2)

Now

$$d(x,u_n)\le d(x,x_n)+d(x_n,u_n),$$

and using (3.2), we have

$$d(x,u_n)\le \frac{1}{1-\kappa }[d(x,x_n)+\kappa d(x_{n-1},x_n)].$$

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:

$$d(x,y)=d((x_1,x_2),(y_1,y_2)):=max\{\mid x_1-y_1\mid ,\mid x_2-y_2\mid \}$$

for all $x,y\in X$, so that $(X,d)$ is a complete metric space. Define $F:M\to \mathit{Cl}(M)$ as follows:

$$F(x,y)=\{\begin{array}{cc}\{(0,0)\}\hfill & \text{if }x\ge y,\hfill \\ \{(0,0),(0,\frac{-1}{4})\}\hfill & \text{if }x<y,\hfill \end{array}$$

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

(B): 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 $u_x⪯u_y$ such that

$$d(u_x,u_y)\le \kappa [d(x,u_x)+d(y,u_y)]$$

for all $y\in M$ with $x⪯y$ 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 $x_n⪯x$ 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 $x_0⪯x_1$. Now, using (B), there is $x_2\in F_2{x}_1$ with $x_1⪯x_2$ such that

$$d(x_1,x_2)\le \kappa [d(x_0,x_1)+d(x_1,x_2)],$$

which gives

$$d(x_1,x_2)\le \frac{\kappa }{1-\kappa }d(x_0,x_1).$$

(3.3)

Next, for this $x_2$, there exists $x_3\in F_3{x}_2$ with $x_2⪯x_3$ such that

$$d(x_2,x_3)\le \kappa [d(x_1,x_2)+d(x_2,x_3)].$$

Using (3.3), we obtain

$$d(x_2,x_3)\le {\left[\frac{\kappa }{1-\kappa }\right]}^{2}d(x_0,x_1).$$

Continuing in this way, we obtain a sequence $x_n$ whose consecutive terms are comparable and

$$d(x_n,x_{n+1})\le {\left[\frac{\kappa }{1-\kappa }\right]}^{n}d(x_0,x_1).$$

Take $0\le h=\frac{\kappa }{1-\kappa }<1$, then we have

$$d(x_n,x_{n+1})\le h^{n}d(x_0,x_1).$$

(3.4)

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

Further, since 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 $x_{n-1}⪯x_n$. Also, $x_n⪯x$ for all n. By using (B), there exists $u_n\in F_{m}x$ such that

$$d(x_n,u_n)\le \kappa [d(x_{n-1},x_n)+d(x,u_n)].$$

(3.5)

Now

$$d(x,u_n)\le d(x,x_n)+d(x_n,u_n),$$

which gives

$$d(x,u_n)\le \frac{1}{1-\kappa }[d(x,x_n)+\kappa d(x_{n-1},x_n)].$$

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].