Common fixed points of Ćirić-type contractive mappings in two ordered generalized metric spaces
- M Abbas^{1},
- YJ Cho^{2} and
- T Nazir^{1}Email author
https://doi.org/10.1186/1687-1812-2012-139
© Abbas et al.; licensee Springer 2012
Received: 16 March 2012
Accepted: 22 August 2012
Published: 4 September 2012
Abstract
In this paper, using the setting of two ordered generalized metric spaces, a unique common fixed point of four mappings satisfying a generalized contractive condition is obtained. We also present an example to demonstrate the results presented herein.
MSC:54H25, 47H10, 54E50.
Keywords
1 Introduction and preliminaries
The study of a unique common fixed point of given mappings satisfying certain contractive conditions has been at the center of rigorous research activity. Mustafa and Sims [1] generalized the concept of a metric in which a real number is assigned to every triplet of an arbitrary set. Based on the notion of generalized metric spaces, Mustafa et al. [2–5] obtained some fixed point theorems for some mappings satisfying different contractive conditions. The existence of common fixed points in generalized metric spaces was initiated by Abbas and Rhoades [6] (see also [7] and [8]). For further study of common fixed points in generalized metric spaces, we refer to [9–12] and references mentioned therein. Abbas et al. [13] showed the existence of coupled common fixed points in two generalized metric spaces (for more results on couple fixed points, see also [14–21]).
The existence of fixed points in ordered metric spaces has been initiated in 2004 by Ran and Reurings [22] and further studied by Nieto and Lopez [23]. Subsequently, several interesting and valuable results have appeared in this direction [24–30].
The aim of this paper is to study common fixed point of four mappings that satisfy the generalized contractive condition in two ordered generalized metric spaces.
In the sequel, $\mathbb{R}$, ${\mathbb{R}}^{+}$ and $\mathbb{N}$ denote the set of real numbers, the set of nonnegative integers and the set of positive integers respectively. The usual order on $\mathbb{R}$ (respectively, on ${\mathbb{R}}^{+}$) will be indistinctly denoted by ≤ or by ≥.
In [1], Mustafa and Sims introduced the following definitions and results:
- (a)
$G(x,y,z)=0$ if $x=y=z$ for all $x,y,z\in X$;
- (b)
$0<G(x,y,z)$ for all $x,y,z\in X$ with $x\ne y$;
- (c)
$G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $y\ne z$;
- (d)
$G(x,y,z)=G(p\{x,y,z\})$, where p is a permutation of $x,y,z\in X$ (symmetry);
- (e)
$G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$.
Then G is called a G-metric on X and $(X,G)$ is called a G-metric space.
- (1)
a G-Cauchy sequence if, for any $\epsilon >0$, there exists ${n}_{0}\in N$ (the set of natural numbers) such that, for all $n,m,l\ge {n}_{0}$, $G({x}_{n},{x}_{m},{x}_{l})<\epsilon $;
- (2)
G-convergent if, for any $\epsilon >0$, there exist $x\in X$ and ${n}_{0}\in N$ such that, for all $n,m\ge {n}_{0}$, $G(x,{x}_{n},{x}_{m})<\epsilon $;
- (3)
A G-metric space X is said to be G-complete if every G-Cauchy sequence in X is G-convergent in X.
It is known that $\{{x}_{n}\}$ is G-convergent to a point $x\in X$ if and only if $G({x}_{m},{x}_{n},x)\to 0$ as $n,m\to \mathrm{\infty}$.
Proposition 1.3 [1]
- (1)
A sequence $\{{x}_{n}\}$ in X is G-convergent to a point $x\in X$;
- (2)
$G({x}_{n},{x}_{m},x)\to 0$ as $n,m\to \mathrm{\infty}$;
- (3)
$G({x}_{n},{x}_{n},x)\to 0$ as $n\to \mathrm{\infty}$;
- (4)
$G({x}_{n},x,x)\to 0$ as $n\to \mathrm{\infty}$.
Definition 1.4 A G-metric on X is said to be symmetric if $G(x,y,y)=G(y,x,x)$ for all $x,y\in X$.
for all $x,y\in X$.
for all $x,y\in X$.
Now, we give an example of a non-symmetric G-metric.
G -metric
(x,y,z) | G(x,y,z) |
---|---|
(1,1,1), (2,2,2) | 0 |
(1,1,2), (1,2,1), (2,1,1) | 0.5 |
(1,2,2), (2,1,2), (2,2,1) | 1.0 |
Note that G satisfies all the axioms of a generalized metric, but $G(x,x,y)\ne G(x,y,y)$ for two distinct points $x,y\in X$.
Definition 1.7 Let f and g be self-mappings on a set X. If $w=fx=gx$ for some $x\in X$, then the point x is called a coincidence point of f and g and w is called a point of coincidence of f and g.
Definition 1.8 [31]
Let f and g be self-mappings on a set X. Then f and g are said to be weakly compatible if they commute at every coincidence point.
Definition 1.9 [8]
Let X be a G-metric space and f, g be self-mappings on X. Then f and g are said to be R-weakly commuting if there exists a positive real number R such that $G(fgx,fgx,gfx)\le RG(fx,fx,gx)$ for all $x\in X$.
The maps f and g are R-weakly commuting on X if and only if they commute at their coincidence points.
- (a)
G is a generalized metric on X;
- (b)
⪯ is a partial order on X.
Definition 1.11 Let $(X,\u2aaf)$ be a partial ordered set. Then two points $x,y\in X$ are said to be comparable if $x\u2aafy$ or $y\u2aafx$.
Definition 1.12 [24]
Let $(X,\u2aaf)$ be a partially ordered set. A self-mapping f on X is said to be dominating if $x\u2aaffx$ for all $x\in X$.
Example 1.13 [24]
Let $X=[0,1]$ be endowed with usual ordering and $f:X\to X$ be a mapping defined by $fx=\sqrt[n]{x}$ for some $n\in \mathbb{N}$. Since $x\le {x}^{\frac{1}{n}}=fx$ for all $x\in X$, f is a dominating mapping.
Definition 1.14 Let $(X,\u2aaf)$ be a partially ordered set. A self-mapping f on X is said to be dominated if $fx\u2aafx$ for all $x\in X$.
Example 1.15 Let $X=[0,1]$ be endowed with usual ordering and $f:X\to X$ be a mapping defined by $fx={x}^{n}$ for some $n\in \mathbb{N}$. Since $fx={x}^{n}\le x$ for all $x\in X$, f is a dominated mapping.
Definition 1.16 A subset $\mathcal{K}$ of a partially ordered set X is said to be well-ordered if every two elements of $\mathcal{K}$ are comparable.
2 Common fixed point theorems
for all $x,y\in X$, where $a,b,c,e\ge 0$ with $a+b+c+2e<1$, then T has a unique fixed point provided that X is T-orbitally complete (for related definitions and results, we refer to [33]).
for all $x,y\in X$, where $0\le q<1$.
In this section, we show the existence of a unique common fixed point of four mappings satisfying Ćirić-type contractive condition in the framework of two ordered generalized metric spaces.
Now, we start with the following result:
- (a)
f, S are compatible, f or S is continuous and g, T are weakly compatible
- (b)
g, T are compatible, g or T is continuous and f, S are weakly compatible,
then f, g, S and T have a common fixed point. Moreover, the set of common fixed points of f, g, S and T is well-ordered if and only if f, g, S and T have one and only one common fixed point.
and so ${y}_{2k+1}={y}_{2k+2}$ since ${k}^{2}<1$.
Similarly, if $m=2k+1$, then one can easily obtain ${y}_{2k+2}={y}_{2k+3}$. Thus $\{{y}_{n}\}$ becomes a constant sequence and ${y}_{2n}$ serves as the common fixed point of f, g, S and T.
Suppose that ${G}_{1}({y}_{2n},{y}_{2n+1},{y}_{2n+1})>0$ for all $n\ge 0$.
where $h=\frac{k}{2-k}$. Obviously, $0\le h<1$.
Therefore, by using the above two inequalities, we have $fz=z$.
Thus (2.9) and (2.10) imply $fz=gv$. Since g and T are weakly compatible, we have $gz=gfz=gTv=Tgv=Tfz=Tz$, and so z is the coincidence point of g and T.
and so $z=gz$. Therefore, $fz=gz=Sz=Tz=z$. The proof is similar when f is continuous. Similarly, if (b) holds, then the result follows.
and hence $z=u$.
The converse follows immediately. This completes the proof. □
Two G -metrices
(x,y,z) | ${\mathit{G}}_{\mathbf{1}}\mathbf{(}\mathit{x},\mathit{y},\mathit{z}\mathbf{)}$ | ${\mathit{G}}_{\mathbf{2}}\mathbf{(}\mathit{x},\mathit{y},\mathit{z}\mathbf{)}$ |
---|---|---|
(0,0,0), (1,1,1), (2,2,2), (3,3,3), | 0 | 0 |
(0,0,2), (0,2,0), (2,0,0), (0,2,2), (2,0,2), (2,2,0), (0,0,1), (0,1,0), (1,0,0), (0,0,3), (0,3,0), (3,0,0), | 4 | 3 |
(0,1,1), (1,0,1), (1,1,0), (0,3,3), (3,0,3), (3,3,0), (1,1,2), (1,2,1), (2,1,1), (1,2,2), (2,1,2), (2,2,1), (1,1,3), (1,3,1), (3,1,1), (1,3,3), (3,1,3), (3,3,1), (2,2,3), (2,3,2), (3,2,2), (2,3,3), (3,2,3), (3,3,2), | 8 | 6 |
(0,1,2), (0,1,3), (0,2,1), (0,2,3), (0,3,1), (0,3,2), (1,0,2), (1,0,3), (1,2,0), (1,2,3), (1,3,0), (1,3,2), (2,0,1), (2,0,3), (2,1,0), (2,1,3), (2,3,0), (2,3,1), (3,0,1), (3,0,2), (3,1,0), (3,1,2), (3,2,0), (3,2,1), | 8 | 6 |
Self maps
x | f(x) | g(x) | S(x) | T(x) |
---|---|---|---|---|
0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 2 | 2 |
2 | 0 | 2 | 2 | 3 |
3 | 0 | 0 | 3 | 3 |
Dominated and dominating maps
x∈X | f is dominated | g is dominated | S is dominating | T is dominating |
---|---|---|---|---|
x = 0 | f(0)=0 | g(0)=0 | 0 = S(0) | 0 = T(0) |
x = 1 | f(1)=0<1 | g(1)=0<1 | 1<2 = S(1) | 1<2 = T(1) |
x = 2 | f(2)=0<2 | g(2)=2 | 2 = S(2) | 2<3 = T(2) |
x = 3 | f(3)=0<3 | g(3)=0<3 | 3 = S(3) | 3 = T(3) |
- (1)When $x=0$ and $y=2$, then $fx=0$, $gy=2$, $Sx=0$, $Ty=3$ and so$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& {G}_{1}(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_{2}(2,2,3)=\frac{3}{4}{G}_{2}(gy,gy,Ty)\\ \le & kmax\{{G}_{2}(Sx,Sx,Ty),{G}_{2}(fx,fx,Sx),{G}_{2}(gy,gy,Ty),\\ [{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)]/2\}\end{array}$
- (2)When $x=1$ and $y=2$, then $fx=0$, $gy=2$, $Sx=2$, $Ty=3$ and so$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& {G}_{1}(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_{2}(2,2,3)=\frac{3}{4}{G}_{2}(gy,gy,Ty)\\ \le & kmax\{{G}_{2}(Sx,Sx,Ty),{G}_{2}(fx,fx,Sx),{G}_{2}(gy,gy,Ty),\\ [{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)]/2\}\end{array}$
- (3)When $x=2$ and $y=2$, then $fx=0$, $gy=2$, $Sx=2$, $Ty=3$ and so$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& {G}_{1}(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_{2}(2,2,3)=\frac{3}{4}{G}_{2}(Sx,Sx,Ty)\\ \le & kmax\{{G}_{2}(Sx,Sx,Ty),{G}_{2}(fx,fx,Sx),{G}_{2}(gy,gy,Ty),\\ [{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)]/2\}\end{array}$
- (4)Finally, when $x=3$ and $y=2$, then $fx=0$, $gy=2$, $Sx=3$, $Ty=3$ and so$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& {G}_{1}(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_{2}(2,2,3)=\frac{3}{4}{G}_{2}(gy,gy,Ty)\\ \le & kmax\{{G}_{2}(Sx,Sx,Ty),{G}_{2}(fx,fx,Sx),{G}_{2}(gy,gy,Ty),\\ [{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)]/2\}\end{array}$
Thus, for all cases, the contractions (2.1) and (2.2) are satisfied. Hence all of the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique common fixed point of f, g, S and g.
for any $k\in [0,1)$. So corresponding results in ordinary metric spaces cannot be applied in this case.
Theorem 2.1 can be viewed as an extension of Theorem 2.1 of [8] to the case of two ordered G-metric spaces.
Since the class of weakly compatible mappings includes R-weakly commuting mappings, Theorem 2.1 generalizes the comparable results in [8].
- (a)
f, S are compatible, f or S is continuous and g, T are weakly compatible
- (b)
g, T are compatible, g or T is continuous and f, S are weakly compatible,
then f, g, S and T have a common fixed point in X. Moreover, the set of common fixed points of f, g, S and T is well-ordered if and only if f, g, S and T have one and only one common fixed point in X.
- (1)If $x,y\in [0,\frac{1}{2})$, then we have$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& \frac{1}{12}|x-3y|\le \frac{1}{12}(x+3y)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{9}{4}y\right)\\ =& {a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ \le & {a}_{1}{G}_{2}(Sx,Sx,Ty)+{a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ +{a}_{4}[{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)].\end{array}$
- (2)If $x\in [0,\frac{1}{2})$ and $y\in [\frac{1}{2},1]$, then we have$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& \frac{1}{12}|x-2y|\le \frac{1}{12}(x+2y)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{14}{6}y\right)\\ =& {a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ \le & {a}_{1}{G}_{2}(Sx,Sx,Ty)+{a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ +{a}_{4}[{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)].\end{array}$
- (3)If $y\in [0,\frac{1}{2})$ and $x\in [\frac{1}{2},1]$, then we have$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& \frac{1}{12}|x-3y|\le \frac{1}{12}(x+3y)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{9}{4}y\right)\\ =& {a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ \le & {a}_{1}{G}_{2}(Sx,Sx,Ty)+{a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ +{a}_{4}[{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)].\end{array}$
- (4)If $x,y\in [\frac{1}{2},1]$, then we obtain$\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& \frac{1}{12}|x-2y|\le \frac{1}{12}(x+2y)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{14}{6}y\right)\\ =& {a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ \le & {a}_{1}{G}_{2}(Sx,Sx,Ty)+{a}_{2}{G}_{2}(fx,fx,Sx)+{a}_{3}{G}_{2}(gy,gy,Ty)\\ +{a}_{4}[{G}_{2}(fx,fx,Ty)+{G}_{2}(gy,gy,Sx)].\end{array}$
Thus (2.13) is satisfied with ${a}_{1}={a}_{4}=\frac{1}{10}$ and ${a}_{2}={a}_{3}=\frac{3}{10}$, where ${a}_{1}+{a}_{2}+{a}_{3}+2{a}_{4}<1$. Similarly, (2.14) is satisfied. Thus all the conditions of Corollary 2.3 are satisfied. Moreover, 0 is the unique common fixed point of f and g.
3 Application
- (i)For each $s,t\in \mathrm{\Omega}$,${\int}_{\mathrm{\Omega}}{q}_{1}(t,s,u(s))\phantom{\rule{0.2em}{0ex}}ds\le u(s)$
- (ii)There exists $r:\mathrm{\Omega}\to \mathrm{\Omega}$ such that${\int}_{\mathrm{\Omega}}|{q}_{1}(t,s,u(t))-{q}_{2}(t,s,v(t))|\phantom{\rule{0.2em}{0ex}}dt\le r(t)|u(t)-v(t)|$
for each $s,t\in \mathrm{\Omega}$ with ${sup}_{t\in \mathrm{\Omega}}r(t)\le k$ where $k\in [0,1)$.
Then the integral equations (3.1) have a common solution in ${L}^{2}(\mathrm{\Omega})$.
is satisfied. Now we can apply Theorem 2.1 by taking S and T as identity maps to obtain the common solutions of integral equations (3.1) in ${L}^{2}(\mathrm{\Omega})$. □
- (5)
A G-metric naturally induces a metric ${d}_{G}$ given by ${d}_{G}(x,y)=G(x,y,y)+G(x,x,y)$. If the G-metric is not symmetric, then the inequalities (2.1), (2.2), (2.13) and (2.14) do not reduce to any metric inequality with the metric ${d}_{G}$. Hence our results do not reduce to fixed point problems in the corresponding metric space $(X,\u2aaf,{d}_{G})$.
Declarations
Acknowledgements
The authors thank the referees for their appreciation and suggestions regarding this work.
Authors’ Affiliations
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