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Common fixed points of Ćirićtype contractive mappings in two ordered generalized metric spaces
Fixed Point Theory and Applications volume 2012, Article number: 139 (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.
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:
Definition 1.1 Let X be a nonempty set. Suppose that a mapping G:X\times X\times X\to {\mathbb{R}}^{+} satisfies the following conditions:

(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 Gmetric on X and (X,G) is called a Gmetric space.
Definition 1.2 A sequence \{{x}_{n}\} in a Gmetric space X is called:

(1)
a GCauchy 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)
Gconvergent 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 Gmetric space X is said to be Gcomplete if every GCauchy sequence in X is Gconvergent in X.
It is known that \{{x}_{n}\} is Gconvergent 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]
Let X be a Gmetric space. Then the following items are equivalent:

(1)
A sequence \{{x}_{n}\} in X is Gconvergent 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 Gmetric on X is said to be symmetric if G(x,y,y)=G(y,x,x) for all x,y\in X.
Proposition 1.5 Every Gmetric on X defines a metric {d}_{G} on X by
for all x,y\in X.
For a symmetric Gmetric, we have
for all x,y\in X. However, if G is nonsymmetric, then the following inequality holds:
for all x,y\in X. It is obvious that
for all x,y\in X.
Now, we give an example of a nonsymmetric Gmetric.
Example 1.6 Let X=\{1,2\} and G:X\times X\times X\to {\mathbb{R}}^{+} be a mapping defined by Table 1.
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 selfmappings 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 selfmappings 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 Gmetric space and f, g be selfmappings on X. Then f and g are said to be Rweakly 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 Rweakly commuting on X if and only if they commute at their coincidence points.
Recall that two mappings f and g on a Gmetric space X are said to be compatible if, for a sequence \{{x}_{n}\} in X such that \{f{x}_{n}\} and \{g{x}_{n}\} are Gconvergent to some t\in X,
Definition 1.10 Let X be a nonempty set. Then (X,\u2aaf,G) is called an ordered generalized metric space if the following conditions hold:

(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 selfmapping 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 selfmapping 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 wellordered if every two elements of \mathcal{K} are comparable.
2 Common fixed point theorems
In [32], Kannan proved a fixed point theorem for a single valued selfmapping T on a metric space X satisfying the following property:
for all x,y\in X, where h\in [0,\frac{1}{2}). If a selfmapping T on a metric space X satisfies the following property:
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 Torbitally complete (for related definitions and results, we refer to [33]).
Afterwards, Ćirić [34] obtained a fixed point result for a mapping satisfying the following property:
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 Ćirićtype contractive condition in the framework of two ordered generalized metric spaces.
Now, we start with the following result:
Theorem 2.1 Let (X,\u2aaf) be a partially ordered set and {G}_{1}, {G}_{2} be two Gmetrics on X such that {G}_{2}(x,y,z)\le {G}_{1}(x,y,z) for all x,y,z\in X with a complete metric {G}_{1} on X. Suppose that f, g, S and T are selfmappings on X satisfying the following properties:
and
for all comparable x,y\in X, where k\in [0,1). Suppose that f(X)\subseteq T(X) and g(X)\subseteq S(X), f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence \{{x}_{n}\} in X with {y}_{n}\u2aaf{x}_{n} for all n\in \mathbb{N}, {y}_{n}\to u implies that u\u2aaf{x}_{n} and either

(a)
f, S are compatible, f or S is continuous and g, T are weakly compatible
or

(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 wellordered if and only if f, g, S and T have one and only one common fixed point.
Proof Let {x}_{0} be an arbitrary point in X. Since f(X)\subseteq T(X) and g(X)\subseteq S(X), we can define the sequences \{{x}_{n}\} and \{{y}_{n}\} in X by
for all n\ge 0. By the given assumptions, we have
Thus, for all n\ge 0, we have {x}_{n+1}\u2aaf{x}_{n}. Suppose that {G}_{1}({y}_{2n},{y}_{2n+1},{y}_{2n+1})>0 for all n\ge 0. If not, then, for some m\ge 0, {y}_{m}={y}_{m+1}. Indeed, if m=2k, then {y}_{2k}={y}_{2k+1} and from (2.1), it follows that
Again, from (2.2), it follows that
Thus (2.3) and (2.4) imply that
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.
If n\in \mathbb{N} is even, then n=2k for some k\in \mathbb{N}; then it follows from (2.1) that
which implies that
If n\in \mathbb{N} is odd, then n=2k+1 for some k\in \mathbb{N}. Again, it follows from (2.1) that
that is,
for all n\in \mathbb{N}. Continuing the above process, we have
for all n\in \mathbb{N}. Thus, for all n,m\in \mathbb{N} with m>n, we have
and so {G}_{1}({y}_{n},{y}_{m},{y}_{m})\to 0 as m,n\to \mathrm{\infty}. Hence \{{y}_{n}\} is a GCauchy sequence in X. Since X is {G}_{1}complete, there exists a point z\in X such that {lim}_{n\to \mathrm{\infty}}{y}_{n}=z. Consequently, we have
and
If S is continuous and \{f,S\} is compatible, then
Since S{x}_{2n+1}=g{x}_{2n}\u2aaf{x}_{2n}, (2.1) gives
Taking the limit as n\to \mathrm{\infty}, we obtain
which further implies that
where h=\frac{k}{2k}. Obviously, 0\le h<1.
Similarly, we obtain
From (2.5) and (2.6), we have
and so Sz=z since 0\le {h}^{2}<1. Since g{x}_{2n}\u2aaf{x}_{2n} and g{x}_{2n}\to z as n\to \mathrm{\infty} implies z\u2aaf{x}_{2n}, it follows from (2.1) that
which, taking the limit as n\to \mathrm{\infty}, gives
Similarly, we obtain
Therefore, by using the above two inequalities, we have fz=z.
Since f(X)\subseteq T(X), there exists a point v\in X such that fz=Tv. Since v\u2aafTv=fz\u2aafz, it follows from (2.1) that
Similarly, we get
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.
Now, from (2.1), we have
that is,
where h=\frac{k}{2k}. Obviously, 0\le h<1. Using (2.2), we have
Combining the above two inequalities, we get
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.
Now, suppose that the set of common fixed points of f, g, S and T is well ordered. We show that a common fixed point of f, g, S and T is unique. Let u be another common fixed point of f, g, S and T. Then, from (2.1), we have
that is,
Similarly, using (2.2), we obtain
Combining the above two inequalities, we get
and hence z=u.
The converse follows immediately. This completes the proof. □
Example 2.2 Let X=\{0,1,2,3\} be endowed with the usual ordering and {G}_{1}, {G}_{2} be two Gmetrics on X defined by Table 2. Then {G}_{1} and {G}_{2} are nonsymmetric since {G}_{1}(1,1,0)\ne {G}_{1}(1,0,0) and {G}_{2}(1,1,0)\ne {G}_{2}(1,0,0) with {G}_{2}(x,y,z)\le {G}_{1}(x,y,z) for all x,y,z\in X. Let f,g,S,T:X\to X be the mappings defined by Table 3. Clearly, f(X)\subseteq T(X), g(X)\subseteq S(X), f, g are dominated mappings and S, T are dominating mappings, see Table 4.
Now, we shall show that for all comparable x,y\in X, (2.1) and (2.2) are satisfied with k=\frac{3}{4}\in [0,1). Note that for all x,y\in \{0,1,3\}, G(fx,fx,gy)=G(fx,gy,gy)=0 and (2.1), (2.2) are satisfied obviously.

(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}
and

(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}
and

(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}
and

(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}
and
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.
If we consider the same set equipped with two metrics given by {d}_{1}(x,y)=xy and {d}_{2}(x,y)=\frac{1}{2}xy for all x,y\in X, then for x=1 and y=2, we have
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 Gmetric spaces.
Since the class of weakly compatible mappings includes Rweakly commuting mappings, Theorem 2.1 generalizes the comparable results in [8].
Corollary 2.3 Let (X,\u2aaf) be a partially ordered set and {G}_{1}, {G}_{2} be two Gmetrics on X such that {G}_{2}(x,y,z)\le {G}_{1}(x,y,z) for all x,y,z\in X with a complete metric {G}_{1} on X. Suppose that f, g, S and T are selfmappings on X satisfying the following properties:
and
for all comparable x,y\in X, where {a}_{1}+{a}_{2}+{a}_{3}+2{a}_{4}<1. Suppose that f(X)\subseteq T(X), g(X)\subseteq S(X) and f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence \{{x}_{n}\} with {y}_{n}\u2aaf{x}_{n} for all n\in \mathbb{N}, {y}_{n}\to u implies that u\u2aaf{x}_{n} and either

(a)
f, S are compatible, f or S is continuous and g, T are weakly compatible
or

(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 wellordered if and only if f, g, S and T have one and only one common fixed point in X.
Example 2.4 Let X=[0,1] be endowed with the usual ordering and {G}_{1}, {G}_{2} be two Gmetrics on X given in [13]:
Define the mappings f,g,S,T:X\to X as
for all x\in X. Clearly, f, g are dominated mappings and S, T are dominating mappings with f(X)\subseteq T(X) and g(X)\subseteq S(X). Also, f, S are compatible, f is continuous and g, T are weakly compatible. Now, for all comparable x,y\in X, we check the following cases:

(1)
If x,y\in [0,\frac{1}{2}), then we have
\begin{array}{rcl}{G}_{1}(fx,fx,gy)& =& \frac{1}{12}x3y\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}x2y\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}x3y\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}x2y\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
Let X={L}^{2}(\mathrm{\Omega}), the set of comparable functions on Ω whose square is integrable on Ω where \mathrm{\Omega}=[0,1], be a bounded set in \mathbb{R}. We endow X with the partial ordered ⪯ given by: x,y\in X, x\u2aafy\iff x(t)\le y(t), for all t\in \mathrm{\Omega}. We consider the integral equations
where {q}_{1},{q}_{2}:\mathrm{\Omega}\times \mathrm{\Omega}\times \mathbb{R}\to \mathbb{R} and c:\mathrm{\Omega}\to {\mathbb{R}}^{+}, to be given continuous mappings. Recently, Abbas et al. [35] obtained a common solution of integral equations (3.1) as an application of their results in the setup of ordered generalized metric spaces. Here we study a sufficient condition for the existence of a common solution of integral equations in the framework of two generalized metric spaces. Define {G}_{1},{G}_{2}:X\times X\times X\to {\mathbb{R}}^{+} by
Obviously, {G}_{2}(x,y,z)\le {G}_{1}(x,y,z) for all x,y,z\in X. Suppose that the following hypotheses hold:

(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)
and
hold.

(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}).
Proof Define fx(t)={\int}_{\mathrm{\Omega}}{q}_{1}(t,s,x(t))\phantom{\rule{0.2em}{0ex}}dtc(t) and gx(t)={\int}_{\mathrm{\Omega}}{q}_{2}(t,s,x(t))\phantom{\rule{0.2em}{0ex}}dtc(t). As fx(t)\le x(t) and gx(t)\le x(t), so f and g are dominated maps. Now, for all comparable x,y\in X,
Similarly,
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}). □
Remarks (1) If we take f=g in Theorem 2.1, then it generalizes Corollary 2.3 in [8] to a more general class of commuting mappings in the setup of two ordered Gmetric spaces.

(2)
If we take S=T in Theorem 2.1, then Corollary 2.4 in [8] is a special case of Theorem 2.1.

(3)
If S=T={I}_{X} (: the identity mapping on X) in Theorem 2.1, then we obtain Corollary 2.5 in [8] in a more general setup.
(4) Corollary 2.6 of [8] becomes a special case of Theorem 2.1 if we take f=g and S=T={I}_{X}.

(5)
A Gmetric naturally induces a metric {d}_{G} given by {d}_{G}(x,y)=G(x,y,y)+G(x,x,y). If the Gmetric 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}).
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Abbas, M., Cho, Y. & Nazir, T. Common fixed points of Ćirićtype contractive mappings in two ordered generalized metric spaces. Fixed Point Theory Appl 2012, 139 (2012). https://doi.org/10.1186/168718122012139
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DOI: https://doi.org/10.1186/168718122012139