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Common fixed point and coincidence point of generalized contractions in ordered metric spaces
Fixed Point Theory and Applications volume 2012, Article number: 229 (2012)
Abstract
Let (X,\u2aaf) be a partially ordered set and d be a complete metric on X. The notion of fcontractive for a setvalued 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 nonempty subsets of X, \mathit{Cl}(X) be the class of all nonempty closed subsets of X and \mathit{CB}(X) be the class of all nonempty closed and bounded subsets of X. For A,B\in \mathit{CB}(X), let
where
D is called the Hausdorff metric induced by d.
Let f:X\to X be any singlevalued mapping. A mapping F:X\to \mathit{CB}(X) is fcontractive if there exists a real number κ with 0<\kappa <1 such that
By introducing the notion of fcontractiveness for setvalued 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 fcontractive 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 Kmapping 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 orderpreserving or orderreversing) 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 fcontractiveness through an implicit relation. This implicit relation is then used to obtain the existence of coincidence and common fixed points for a pair of singlevalued mapping and setvalued 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 setvalued 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 realvalued functions T:{R}_{+}^{6}\to R satisfying the following conditions:
{\mathcal{T}}_{1}: T({t}_{1},{t}_{2},\dots ,{t}_{6}) is nonincreasing 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
and
imply
{\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}.
Now using (A), we have
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}),
and by {\mathcal{T}}_{1}, we have
that is,
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),
Using (2.2) in (2.1), we have
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
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}.
Now, since {x}_{1}\u2aaf{x}_{2}, by using (A), we have
by {\mathcal{T}}_{1} we have
that is,
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)
And so from (2.3) and (2.4), we have
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
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}.
Now, since {x}_{2}\u2aaf{x}_{3}, by using (A) we have
by {\mathcal{T}}_{1} we have
that is,
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)
Now, using (2.5), (2.6) and (2.7), we have
Continuing in this way, we obtain a sequence \{{x}_{n}\} with {x}_{n}\u2aaf{x}_{n+1} such that {y}_{n}=f{x}_{n+1}\in F{x}_{n} for n\ge 0 and
Hence, \{{y}_{n}\} is a Cauchy sequence. So, there exists a point 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.
Now, using (A), we have
Now, taking limit as n\to \mathrm{\infty} and using {\mathcal{T}}_{2}, also the facts that d(f{x}_{n},fp)=d({y}_{n1},y)\to 0, d(f{x}_{n},F{x}_{n})\le d({y}_{n1},{y}_{n})\to 0, d(fp,F{x}_{n})\le d(y,{y}_{n})\to 0, we have
that is,
By using {\mathcal{T}}_{3}, we get
Next, since {y}_{n}\in F{x}_{n},
taking limit as n\to \mathrm{\infty}, we obtain
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
In this section, we define K\u2aaf setvalued 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 nonempty subsets of (X,\u2aaf), the relations between A and B are denoted and defined as follows:
Definition 3.1 Let M be a nonempty subset of X and F:M\to P(X). A mapping F is said to be K\u2aaf setvalued 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 setvalued mappings, we just required comparability of the elements, but order does not matter.
Theorem 3.2 Let M be a nonempty closed subset of X and F:M\to \mathit{Cl}(M) be a K\u2aaf setvalued 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 setvalued mapping, there is {x}_{2}\in F{x}_{1} with {x}_{1}\u2aaf{x}_{2} such that
which gives
and consequently
Continuing in this way, we obtain a sequence \{{x}_{n}\} whose consecutive terms are comparable and
Take 0\le h=\frac{\kappa}{1\kappa}<1, then we have
Next, we will show that \{{x}_{n}\} is a Cauchy sequence in X. Let m>n. Then
because h\in (0,1), 1{h}^{mn}<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.
Since {x}_{n}\in F{x}_{n1} with {x}_{n1}\u2aaf{x}_{n} also {x}_{n}\u2aafx for all n and F is a k\u2aaf setvalued mapping, there exists {u}_{n}\in Fx with {x}_{n}\u2aaf{u}_{n} such that
Now
and using (3.2), we have
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:
for all x,y\in X, so that (X,d) is a complete metric space. Define F:M\to \mathit{Cl}(M) 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 setvalued 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 nonempty 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}\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
which gives
Next, for this {x}_{2}, there exists {x}_{3}\in {F}_{3}{x}_{2} with {x}_{2}\u2aaf{x}_{3} such that
Using (3.3), we obtain
Continuing in this way, we obtain a sequence {x}_{n} whose consecutive terms are comparable and
Take 0\le h=\frac{\kappa}{1\kappa}<1, then we have
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.
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}_{n1} with {x}_{n1}\u2aaf{x}_{n}. Also, {x}_{n}\u2aafx for all n. By using (B), there exists {u}_{n}\in {F}_{m}x such that
Now
which gives
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].
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IB gave the idea and ARB wrote the initial draft. Both authors read and agreed upon the draft and finalized the manuscript. Correspondence was mainly done by IB. All authors read and approved the final manuscript.
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Beg, I., Butt, A.R. Common fixed point and coincidence point of generalized contractions in ordered metric spaces. Fixed Point Theory Appl 2012, 229 (2012). https://doi.org/10.1186/168718122012229
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DOI: https://doi.org/10.1186/168718122012229
Keywords
 fixed point
 coincidence point
 partially ordered set
 metric space
 setvalued mapping