# Common fixed points of Ćirić-type contractive mappings in two ordered generalized metric spaces

## 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. [25] 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 [912] 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 [1421]).

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 [2430].

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 $G:X×X×X\to {\mathbb{R}}^{+}$ satisfies the following conditions:

1. (a)

$G\left(x,y,z\right)=0$ if $x=y=z$ for all $x,y,z\in X$;

2. (b)

$0 for all $x,y,z\in X$ with $x\ne y$;

3. (c)

$G\left(x,x,y\right)\le G\left(x,y,z\right)$ for all $x,y,z\in X$ with $y\ne z$;

4. (d)

$G\left(x,y,z\right)=G\left(p\left\{x,y,z\right\}\right)$, where p is a permutation of $x,y,z\in X$ (symmetry);

5. (e)

$G\left(x,y,z\right)\le G\left(x,a,a\right)+G\left(a,y,z\right)$ for all $x,y,z,a\in X$.

Then G is called a G-metric on X and $\left(X,G\right)$ is called a G-metric space.

Definition 1.2 A sequence $\left\{{x}_{n}\right\}$ in a G-metric space X is called:

1. (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 $n,m,l\ge {n}_{0}$, $G\left({x}_{n},{x}_{m},{x}_{l}\right)<\epsilon$;

2. (2)

G-convergent if, for any $\epsilon >0$, there exist $x\in X$ and ${n}_{0}\in N$ such that, for all $n,m\ge {n}_{0}$, $G\left(x,{x}_{n},{x}_{m}\right)<\epsilon$;

3. (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 $\left\{{x}_{n}\right\}$ is G-convergent to a point $x\in X$ if and only if $G\left({x}_{m},{x}_{n},x\right)\to 0$ as $n,m\to \mathrm{\infty }$.

Proposition 1.3 [1]

Let X be a G-metric space. Then the following items are equivalent:

1. (1)

A sequence $\left\{{x}_{n}\right\}$ in X is G-convergent to a point $x\in X$;

2. (2)

$G\left({x}_{n},{x}_{m},x\right)\to 0$ as $n,m\to \mathrm{\infty }$;

3. (3)

$G\left({x}_{n},{x}_{n},x\right)\to 0$ as $n\to \mathrm{\infty }$;

4. (4)

$G\left({x}_{n},x,x\right)\to 0$ as $n\to \mathrm{\infty }$.

Definition 1.4 A G-metric on X is said to be symmetric if $G\left(x,y,y\right)=G\left(y,x,x\right)$ for all $x,y\in X$.

Proposition 1.5 Every G-metric on X defines a metric ${d}_{G}$ on X by

${d}_{G}\left(x,y\right)=G\left(x,y,y\right)+G\left(y,x,x\right)$
(1.1)

for all $x,y\in X$.

For a symmetric G-metric, we have

${d}_{G}\left(x,y\right)=2G\left(x,y,y\right)$
(1.2)

for all $x,y\in X$. However, if G is non-symmetric, then the following inequality holds:

$\frac{3}{2}G\left(x,y,y\right)\le {d}_{G}\left(x,y\right)\le 3G\left(x,y,y\right)$
(1.3)

for all $x,y\in X$. It is obvious that

$G\left(x,x,y\right)\le 2G\left(x,y,y\right)$

for all $x,y\in X$.

Now, we give an example of a non-symmetric G-metric.

Example 1.6 Let $X=\left\{1,2\right\}$ and $G:X×X×X\to {\mathbb{R}}^{+}$ be a mapping defined by Table 1.

Note that G satisfies all the axioms of a generalized metric, but $G\left(x,x,y\right)\ne G\left(x,y,y\right)$ 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 $G\left(fgx,fgx,gfx\right)\le RG\left(fx,fx,gx\right)$ 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.

Recall that two mappings f and g on a G-metric space X are said to be compatible if, for a sequence $\left\{{x}_{n}\right\}$ in X such that $\left\{f{x}_{n}\right\}$ and $\left\{g{x}_{n}\right\}$ are G-convergent to some $t\in X$,

$\underset{n\to \mathrm{\infty }}{lim}G\left(fg{x}_{n},fg{x}_{n},gf{x}_{n}\right)=0.$

Definition 1.10 Let X be a nonempty set. Then $\left(X,⪯,G\right)$ is called an ordered generalized metric space if the following conditions hold:

1. (a)

G is a generalized metric on X;

2. (b)

is a partial order on X.

Definition 1.11 Let $\left(X,⪯\right)$ be a partial ordered set. Then two points $x,y\in X$ are said to be comparable if $x⪯y$ or $y⪯x$.

Definition 1.12 [24]

Let $\left(X,⪯\right)$ be a partially ordered set. A self-mapping f on X is said to be dominating if $x⪯fx$ for all $x\in X$.

Example 1.13 [24]

Let $X=\left[0,1\right]$ 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 $\left(X,⪯\right)$ be a partially ordered set. A self-mapping f on X is said to be dominated if $fx⪯x$ for all $x\in X$.

Example 1.15 Let $X=\left[0,1\right]$ 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

In [32], Kannan proved a fixed point theorem for a single valued self-mapping T on a metric space X satisfying the following property:

$d\left(Tx,Ty\right)\le h\left\{d\left(x,Tx\right)+d\left(y,Ty\right)\right\}$

for all $x,y\in X$, where $h\in \left[0,\frac{1}{2}\right)$. If a self-mapping T on a metric space X satisfies the following property:

$d\left(Tx,Ty\right)\le ad\left(x,y\right)+bd\left(x,Tx\right)+cd\left(y,Ty\right)+e\left[d\left(x,Ty\right)+d\left(y,Tx\right)\right]$

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]).

Afterwards, Ćirić [34] obtained a fixed point result for a mapping satisfying the following property:

$d\left(Tx,Ty\right)\le qmax\left\{d\left(x,y\right),d\left(x,Tx\right),d\left(y,Ty\right),\frac{d\left(x,Ty\right)+d\left(y,Tx\right)}{2}\right\}$

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.

Theorem 2.1 Let $\left(X,⪯\right)$ be a partially ordered set and ${G}_{1}$, ${G}_{2}$ be two G-metrics on X such that ${G}_{2}\left(x,y,z\right)\le {G}_{1}\left(x,y,z\right)$ for all $x,y,z\in X$ with a complete metric ${G}_{1}$ on X. Suppose that f, g, S and T are self-mappings on X satisfying the following properties:

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& \le & kmax\left\{{G}_{2}\left(Sx,Sx,Ty\right),{G}_{2}\left(fx,fx,Sx\right),{G}_{2}\left(gy,gy,Ty\right),\\ \left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right]/2\right\}\end{array}$
(2.1)

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& \le & kmax\left\{{G}_{2}\left(Sx,Ty,Ty\right),{G}_{2}\left(fx,Sx,Sx\right),{G}_{2}\left(gy,Ty,Ty\right),\\ \left[{G}_{2}\left(fx,Ty,Ty\right)+{G}_{2}\left(gy,Sx,Sx\right)\right]/2\right\}\end{array}$
(2.2)

for all comparable $x,y\in X$, where $k\in \left[0,1\right)$. Suppose that $f\left(X\right)\subseteq T\left(X\right)$ and $g\left(X\right)\subseteq S\left(X\right)$, f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence $\left\{{x}_{n}\right\}$ in X with ${y}_{n}⪯{x}_{n}$ for all $n\in \mathbb{N}$, ${y}_{n}\to u$ implies that $u⪯{x}_{n}$ and either

1. (a)

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

or

1. (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.

Proof Let ${x}_{0}$ be an arbitrary point in X. Since $f\left(X\right)\subseteq T\left(X\right)$ and $g\left(X\right)\subseteq S\left(X\right)$, we can define the sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X by

${y}_{2n}=g{x}_{2n}=S{x}_{2n+1},\phantom{\rule{1em}{0ex}}{y}_{2n+1}=f{x}_{2n+1}=T{x}_{2n+2}$

for all $n\ge 0$. By the given assumptions, we have

$\begin{array}{c}{x}_{2n+2}⪯T{x}_{2n+2}=f{x}_{2n+1}⪯{x}_{2n+1},\hfill \\ {x}_{2n+1}⪯S{x}_{2n+1}=g{x}_{2n}⪯{x}_{2n}.\hfill \end{array}$

Thus, for all $n\ge 0$, we have ${x}_{n+1}⪯{x}_{n}$. Suppose that ${G}_{1}\left({y}_{2n},{y}_{2n+1},{y}_{2n+1}\right)>0$ 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

(2.3)

Again, from (2.2), it follows that

(2.4)

Thus (2.3) and (2.4) imply that

${G}_{1}\left({y}_{2k+2},{y}_{2k+1},{y}_{2k+1}\right)\le {k}^{2}{G}_{1}\left({y}_{2k+2},{y}_{2k+1},{y}_{2k+1}\right)$

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 $\left\{{y}_{n}\right\}$ becomes a constant sequence and ${y}_{2n}$ serves as the common fixed point of f, g, S and T.

Suppose that ${G}_{1}\left({y}_{2n},{y}_{2n+1},{y}_{2n+1}\right)>0$ 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

${G}_{1}\left({y}_{n+1},{y}_{n+1},{y}_{n}\right)\le k{G}_{1}\left({y}_{n},{y}_{n},{y}_{n-1}\right).$

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,

${G}_{1}\left({y}_{n+1},{y}_{n+1},{y}_{n}\right)\le k{G}_{1}\left({y}_{n},{y}_{n},{y}_{n-1}\right)$

for all $n\in \mathbb{N}$. Continuing the above process, we have

${G}_{1}\left({y}_{n+1},{y}_{n+1},{y}_{n}\right)\le {k}^{n}{G}_{1}\left({y}_{1},{y}_{1},{y}_{0}\right)$

for all $n\in \mathbb{N}$. Thus, for all $n,m\in \mathbb{N}$ with $m>n$, we have

$\begin{array}{c}{G}_{1}\left({y}_{m},{y}_{m},{y}_{n}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {G}_{1}\left({y}_{n},{y}_{n+1},{y}_{n+1}\right)+{G}_{1}\left({y}_{n+1},{y}_{n+2},{y}_{n+2}\right)+\cdots +{G}_{1}\left({y}_{m-1},{y}_{m},{y}_{m}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {k}^{n}{G}_{1}\left({y}_{0},{y}_{1},{y}_{1}\right)+{k}^{n+1}{G}_{1}\left({y}_{0},{y}_{1},{y}_{1}\right)+\cdots +{k}^{m-1}{G}_{1}\left({y}_{0},{y}_{1},{y}_{1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={k}^{n}{G}_{1}\left({y}_{0},{y}_{1},{y}_{1}\right)\sum _{i=0}^{m-n-1}{k}^{i}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{k}^{n}}{1-k}{G}_{1}\left({y}_{0},{y}_{1},{y}_{1}\right)\hfill \end{array}$

and so ${G}_{1}\left({y}_{n},{y}_{m},{y}_{m}\right)\to 0$ as $m,n\to \mathrm{\infty }$. Hence $\left\{{y}_{n}\right\}$ is a G-Cauchy 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

$\underset{n\to \mathrm{\infty }}{lim}{y}_{2n+1}=\underset{n\to \mathrm{\infty }}{lim}f{x}_{2n+1}=\underset{n\to \mathrm{\infty }}{lim}T{x}_{2n+2}=z$

and

$\underset{n\to \mathrm{\infty }}{lim}{y}_{2n}=\underset{n\to \mathrm{\infty }}{lim}g{x}_{2n}=\underset{n\to \mathrm{\infty }}{lim}S{x}_{2n+1}=z.$

If S is continuous and $\left\{f,S\right\}$ is compatible, then

$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{S}^{2}{x}_{2n+1}=Sz,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}fS{x}_{2n+1}=\underset{n\to \mathrm{\infty }}{lim}Sf{x}_{2n+1}=Sz.\hfill \end{array}$

Since $S{x}_{2n+1}=g{x}_{2n}⪯{x}_{2n}$, (2.1) gives

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

$\begin{array}{rcl}{G}_{1}\left(Sz,Sz,z\right)& \le & kmax\left\{{G}_{2}\left(Sz,Sz,z\right),{G}_{2}\left(Sz,Sz,Sz\right),{G}_{2}\left(z,z,z\right),\\ \left[{G}_{2}\left(Sz,Sz,z\right)+{G}_{2}\left(z,z,Sz\right)\right]/2\right\}\\ \le & kmax\left\{{G}_{1}\left(Sz,Sz,z\right),\left[{G}_{1}\left(Sz,Sz,z\right)+{G}_{1}\left(z,z,Sz\right)\right]/2\right\}\\ =& \frac{k}{2}\left[{G}_{1}\left(Sz,Sz,z\right)+{G}_{1}\left(z,z,Sz\right)\right],\end{array}$

which further implies that

${G}_{1}\left(Sz,Sz,z\right)\le h{G}_{1}\left(z,z,Sz\right),$
(2.5)

where $h=\frac{k}{2-k}$. Obviously, $0\le h<1$.

Similarly, we obtain

${G}_{1}\left(Sz,z,z\right)\le h{G}_{1}\left(z,Sz,Sz\right).$
(2.6)

From (2.5) and (2.6), we have

${G}_{1}\left(Sz,Sz,z\right)\le {h}^{2}{G}_{1}\left(z,Sz,Sz\right)$

and so $Sz=z$ since $0\le {h}^{2}<1$. Since $g{x}_{2n}⪯{x}_{2n}$ and $g{x}_{2n}\to z$ as $n\to \mathrm{\infty }$ implies $z⪯{x}_{2n}$, it follows from (2.1) that

$\begin{array}{c}{G}_{1}\left(fz,fz,g{x}_{2n}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le kmax\left\{{G}_{2}\left(Sz,Sz,T{x}_{2n}\right),{G}_{2}\left(fz,fz,Sz\right),{G}_{2}\left(g{x}_{2n},g{x}_{2n},T{x}_{2n}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\left[{G}_{2}\left(fz,fz,T{x}_{2n}\right)+{G}_{2}\left(g{x}_{2n},g{x}_{2n},Sz\right)\right]/2\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=kmax\left\{{G}_{2}\left(z,z,T{x}_{2n}\right),{G}_{2}\left(fz,fz,z\right),{G}_{2}\left(g{x}_{2n},g{x}_{2n},T{x}_{2n}\right),\hfill \\ \phantom{\rule{2em}{0ex}}\left[{G}_{2}\left(fz,fz,T{x}_{2n}\right)+{G}_{2}\left(g{x}_{2n},g{x}_{2n},z\right)\right]/2\right\},\hfill \end{array}$

which, taking the limit as $n\to \mathrm{\infty }$, gives

$\begin{array}{rcl}{G}_{1}\left(fz,fz,z\right)& \le & kmax\left\{{G}_{2}\left(z,z,z\right),{G}_{2}\left(fz,fz,z\right),{G}_{2}\left(z,z,z\right),\\ \left[{G}_{2}\left(fz,fz,z\right)+{G}_{2}\left(z,z,z\right)\right]/2\right\}\\ \le & k{G}_{1}\left(fz,fz,z\right).\end{array}$
(2.7)

Similarly, we obtain

${G}_{1}\left(fz,z,z\right)\le k{G}_{1}\left(z,fz,fz\right).$
(2.8)

Therefore, by using the above two inequalities, we have $fz=z$.

Since $f\left(X\right)\subseteq T\left(X\right)$, there exists a point $v\in X$ such that $fz=Tv$. Since $v⪯Tv=fz⪯z$, it follows from (2.1) that

$\begin{array}{rcl}{G}_{1}\left(fz,fz,gv\right)& \le & kmax\left\{{G}_{2}\left(Sz,Sz,Tv\right),{G}_{2}\left(fz,fz,Sz\right),{G}_{2}\left(gv,gv,Tv\right),\\ \left[{G}_{2}\left(fz,fz,Tv\right)+{G}_{2}\left(gv,gv,Sz\right)\right]/2\right\}\\ =& kmax\left\{{G}_{2}\left(fz,fz,fz\right),{G}_{2}\left(fz,fz,fz\right),{G}_{2}\left(gv,gv,fz\right),\\ \left[{G}_{2}\left(fz,fz,fz\right)+{G}_{2}\left(gv,gv,fz\right)\right]/2\right\}\\ \le & k{G}_{1}\left(fz,gv,gv\right).\end{array}$
(2.9)

Similarly, we get

${G}_{1}\left(fz,gv,gv\right)\le k{G}_{1}\left(fz,fz,gv\right).$
(2.10)

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

$\begin{array}{rcl}{G}_{1}\left(z,z,gz\right)& =& {G}_{1}\left(fz,fz,gz\right)\\ \le & kmax\left\{{G}_{2}\left(Sz,Sz,Tz\right),{G}_{2}\left(fz,fz,Sz\right),{G}_{2}\left(gz,gz,Tz\right),\\ \left[{G}_{2}\left(fz,fz,Tz\right)+{G}_{2}\left(gz,gz,Sz\right)\right]/2\right\}\\ =& kmax\left\{{G}_{2}\left(z,z,gz\right),{G}_{2}\left(z,z,z\right),{G}_{2}\left(gz,gz,gz\right),\\ \left[{G}_{2}\left(z,z,gz\right)+{G}_{2}\left(gz,gz,z\right)\right]/2\right\}\\ =& kmax\left\{{G}_{2}\left(z,z,gz\right),\left[{G}_{2}\left(z,z,gz\right)+{G}_{2}\left(gz,gz,z\right)\right]/2\right\}\\ \le & \frac{k}{2}\left[{G}_{1}\left(z,z,gz\right)+{G}_{1}\left(gz,gz,z\right)\right],\end{array}$

that is,

${G}_{1}\left(z,z,gz\right)\le h{G}_{1}\left(gz,gz,z\right),$
(2.11)

where $h=\frac{k}{2-k}$. Obviously, $0\le h<1$. Using (2.2), we have

${G}_{1}\left(z,gz,gz\right)\le h{G}_{1}\left(z,z,gz\right).$
(2.12)

Combining the above two inequalities, we get

${G}_{1}\left(z,z,gz\right)\le {h}^{2}{G}_{1}\left(z,z,gz\right)$

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

$\begin{array}{rcl}{G}_{1}\left(z,z,u\right)& =& {G}_{1}\left(fz,fz,gu\right)\\ \le & kmax\left\{{G}_{2}\left(Sz,Sz,Tu\right),{G}_{2}\left(fz,fz,Sz\right),{G}_{2}\left(gu,gu,Tu\right),\\ \left[{G}_{2}\left(fz,fz,Tu\right)+{G}_{2}\left(gu,gu,Sz\right)\right]/2\right\}\\ =& kmax\left\{{G}_{2}\left(z,z,u\right),{G}_{2}\left(z,z,z\right),{G}_{2}\left(u,u,u\right),\\ \left[{G}_{2}\left(z,z,u\right)+{G}_{2}\left(u,u,z\right)\right]/2\right\}\\ =& \frac{k}{2}\left[{G}_{2}\left(z,z,u\right)+{G}_{2}\left(u,u,z\right)\right]\\ \le & \frac{1}{2}{G}_{1}\left(z,z,u\right)+\frac{k}{2}{G}_{1}\left(u,u,z\right),\end{array}$

that is,

${G}_{1}\left(z,z,u\right)\le k{G}_{1}\left(z,u,u\right).$

Similarly, using (2.2), we obtain

${G}_{1}\left(z,u,u\right)\le k{G}_{1}\left(z,z,u\right).$

Combining the above two inequalities, we get

${G}_{1}\left(z,z,u\right)\le {k}^{2}{G}_{1}\left(z,z,u\right)$

and hence $z=u$.

The converse follows immediately. This completes the proof. □

Example 2.2 Let $X=\left\{0,1,2,3\right\}$ be endowed with the usual ordering and ${G}_{1}$, ${G}_{2}$ be two G-metrics on X defined by Table 2. Then ${G}_{1}$ and ${G}_{2}$ are non-symmetric since ${G}_{1}\left(1,1,0\right)\ne {G}_{1}\left(1,0,0\right)$ and ${G}_{2}\left(1,1,0\right)\ne {G}_{2}\left(1,0,0\right)$ with ${G}_{2}\left(x,y,z\right)\le {G}_{1}\left(x,y,z\right)$ for all $x,y,z\in X$. Let $f,g,S,T:X\to X$ be the mappings defined by Table 3. Clearly, $f\left(X\right)\subseteq T\left(X\right)$, $g\left(X\right)\subseteq S\left(X\right)$, 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 $k=\frac{3}{4}\in \left[0,1\right)$. Note that for all $x,y\in \left\{0,1,3\right\}$, $G\left(fx,fx,gy\right)=G\left(fx,gy,gy\right)=0$ and (2.1), (2.2) are satisfied obviously.

1. (1)

When $x=0$ and $y=2$, then $fx=0$, $gy=2$, $Sx=0$, $Ty=3$ and so

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& {G}_{1}\left(0,0,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,2,3\right)=\frac{3}{4}{G}_{2}\left(gy,gy,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Sx,Ty\right),{G}_{2}\left(fx,fx,Sx\right),{G}_{2}\left(gy,gy,Ty\right),\\ \left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right]/2\right\}\end{array}$

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& =& {G}_{1}\left(0,2,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,3,3\right)=\frac{3}{4}{G}_{2}\left(gy,Ty,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Ty,Ty\right),{G}_{2}\left(fx,Sx,Sx\right),{G}_{2}\left(gy,Ty,Ty\right),\\ \left[{G}_{2}\left(fx,Ty,Ty\right)+{G}_{2}\left(gy,Sx,Sx\right)\right]/2\right\}.\end{array}$
1. (2)

When $x=1$ and $y=2$, then $fx=0$, $gy=2$, $Sx=2$, $Ty=3$ and so

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& {G}_{1}\left(0,0,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,2,3\right)=\frac{3}{4}{G}_{2}\left(gy,gy,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Sx,Ty\right),{G}_{2}\left(fx,fx,Sx\right),{G}_{2}\left(gy,gy,Ty\right),\\ \left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right]/2\right\}\end{array}$

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& =& {G}_{1}\left(0,2,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,3,3\right)=\frac{3}{4}{G}_{2}\left(gy,Ty,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Ty,Ty\right),{G}_{2}\left(fx,Sx,Sx\right),{G}_{2}\left(gy,Ty,Ty\right),\\ \left[{G}_{2}\left(fx,Ty,Ty\right)+{G}_{2}\left(gy,Sx,Sx\right)\right]/2\right\}.\end{array}$
1. (3)

When $x=2$ and $y=2$, then $fx=0$, $gy=2$, $Sx=2$, $Ty=3$ and so

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& {G}_{1}\left(0,0,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,2,3\right)=\frac{3}{4}{G}_{2}\left(Sx,Sx,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Sx,Ty\right),{G}_{2}\left(fx,fx,Sx\right),{G}_{2}\left(gy,gy,Ty\right),\\ \left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right]/2\right\}\end{array}$

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& =& {G}_{1}\left(0,2,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,3,3\right)=\frac{3}{4}{G}_{2}\left(Sx,Ty,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Ty,Ty\right),{G}_{2}\left(fx,Sx,Sx\right),{G}_{2}\left(gy,Ty,Ty\right),\\ \left[{G}_{2}\left(fx,Ty,Ty\right)+{G}_{2}\left(gy,Sx,Sx\right)\right]/2\right\}.\end{array}$
1. (4)

Finally, when $x=3$ and $y=2$, then $fx=0$, $gy=2$, $Sx=3$, $Ty=3$ and so

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& {G}_{1}\left(0,0,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,2,3\right)=\frac{3}{4}{G}_{2}\left(gy,gy,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Sx,Ty\right),{G}_{2}\left(fx,fx,Sx\right),{G}_{2}\left(gy,gy,Ty\right),\\ \left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right]/2\right\}\end{array}$

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& =& {G}_{1}\left(0,2,2\right)=4\\ <& \frac{3}{4}\left(6\right)=\frac{3}{4}{G}_{2}\left(2,3,3\right)=\frac{3}{4}{G}_{2}\left(gy,Ty,Ty\right)\\ \le & kmax\left\{{G}_{2}\left(Sx,Ty,Ty\right),{G}_{2}\left(fx,Sx,Sx\right),{G}_{2}\left(gy,Ty,Ty\right),\\ \left[{G}_{2}\left(fx,Ty,Ty\right)+{G}_{2}\left(gy,Sx,Sx\right)\right]/2\right\}.\end{array}$

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.

If we consider the same set equipped with two metrics given by ${d}_{1}\left(x,y\right)=|x-y|$ and ${d}_{2}\left(x,y\right)=\frac{1}{2}|x-y|$ for all $x,y\in X$, then for $x=1$ and $y=2$, we have

$\begin{array}{rcl}{d}_{1}\left(fx,gy\right)& =& {d}_{1}\left(0,2\right)=2\nleqq 2k\\ \le & kmax\left\{{d}_{2}\left(2,3\right),{d}_{2}\left(0,2\right),{d}_{2}\left(2,3\right),\left[{d}_{2}\left(0,3\right)+{d}_{2}\left(2,2\right)\right]/2\right\}\\ =& kmax\left\{{d}_{2}\left(Sx,Ty\right),{d}_{2}\left(fx,Sx\right),{d}_{2}\left(gy,Ty\right),\left[{d}_{2}\left(fx,Ty\right)+{d}_{2}\left(gy,Sx\right)\right]/2\right\}\end{array}$

for any $k\in \left[0,1\right)$. 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].

Corollary 2.3 Let $\left(X,⪯\right)$ be a partially ordered set and ${G}_{1}$, ${G}_{2}$ be two G-metrics on X such that ${G}_{2}\left(x,y,z\right)\le {G}_{1}\left(x,y,z\right)$ for all $x,y,z\in X$ with a complete metric ${G}_{1}$ on X. Suppose that f, g, S and T are self-mappings on X satisfying the following properties:

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& \le & {a}_{1}{G}_{2}\left(Sx,Sx,Ty\right)+{a}_{2}{G}_{2}\left(Sx,Sx,fx\right)+{a}_{3}{G}_{2}\left(Ty,Ty,gy\right)\\ +{a}_{4}\left[{G}_{2}\left(Sx,Sx,gy\right)+{G}_{2}\left(Ty,Ty,fx\right)\right]\end{array}$
(2.13)

and

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& \le & {a}_{1}{G}_{2}\left(Sx,Ty,Ty\right)+{a}_{2}{G}_{2}\left(Sx,fx,fx\right)+{a}_{3}{G}_{2}\left(Ty,gy,gy\right)\\ +{a}_{4}\left[{G}_{2}\left(Sx,gy,gy\right)+{G}_{2}\left(Ty,fx,fx\right)\right]\end{array}$
(2.14)

for all comparable $x,y\in X$, where ${a}_{1}+{a}_{2}+{a}_{3}+2{a}_{4}<1$. Suppose that $f\left(X\right)\subseteq T\left(X\right)$, $g\left(X\right)\subseteq S\left(X\right)$ and f, g are dominated mappings and S, T are dominating mappings. If, for any nonincreasing sequence $\left\{{x}_{n}\right\}$ with ${y}_{n}⪯{x}_{n}$ for all $n\in \mathbb{N}$, ${y}_{n}\to u$ implies that $u⪯{x}_{n}$ and either

1. (a)

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

or

1. (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.

Example 2.4 Let $X=\left[0,1\right]$ be endowed with the usual ordering and ${G}_{1}$, ${G}_{2}$ be two G-metrics on X given in [13]:

$\begin{array}{c}{G}_{1}\left(a,b,c\right)=|a-b|+|b-c|+|c-a|,\hfill \\ {G}_{2}\left(a,b,c\right)=\frac{1}{2}\left[|a-b|+|b-c|+|c-a|\right].\hfill \end{array}$

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 $f\left(X\right)\subseteq T\left(X\right)$ and $g\left(X\right)\subseteq S\left(X\right)$. 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. (1)

If $x,y\in \left[0,\frac{1}{2}\right)$, then we have

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& \frac{1}{12}|x-3y|\le \frac{1}{12}\left(x+3y\right)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{9}{4}y\right)\\ =& {a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ \le & {a}_{1}{G}_{2}\left(Sx,Sx,Ty\right)+{a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ +{a}_{4}\left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right].\end{array}$
2. (2)

If $x\in \left[0,\frac{1}{2}\right)$ and $y\in \left[\frac{1}{2},1\right]$, then we have

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& \frac{1}{12}|x-2y|\le \frac{1}{12}\left(x+2y\right)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{14}{6}y\right)\\ =& {a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ \le & {a}_{1}{G}_{2}\left(Sx,Sx,Ty\right)+{a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ +{a}_{4}\left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right].\end{array}$
3. (3)

If $y\in \left[0,\frac{1}{2}\right)$ and $x\in \left[\frac{1}{2},1\right]$, then we have

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& \frac{1}{12}|x-3y|\le \frac{1}{12}\left(x+3y\right)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{9}{4}y\right)\\ =& {a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ \le & {a}_{1}{G}_{2}\left(Sx,Sx,Ty\right)+{a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ +{a}_{4}\left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right].\end{array}$
4. (4)

If $x,y\in \left[\frac{1}{2},1\right]$, then we obtain

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& \frac{1}{12}|x-2y|\le \frac{1}{12}\left(x+2y\right)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{14}{6}y\right)\\ =& {a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ \le & {a}_{1}{G}_{2}\left(Sx,Sx,Ty\right)+{a}_{2}{G}_{2}\left(fx,fx,Sx\right)+{a}_{3}{G}_{2}\left(gy,gy,Ty\right)\\ +{a}_{4}\left[{G}_{2}\left(fx,fx,Ty\right)+{G}_{2}\left(gy,gy,Sx\right)\right].\end{array}$

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 $X={L}^{2}\left(\mathrm{\Omega }\right)$, the set of comparable functions on Ω whose square is integrable on Ω where $\mathrm{\Omega }=\left[0,1\right]$, be a bounded set in $\mathbb{R}$. We endow X with the partial ordered given by: $x,y\in X$, $x⪯y⇔x\left(t\right)\le y\left(t\right)$, for all $t\in \mathrm{\Omega }$. We consider the integral equations

$\begin{array}{r}x\left(t\right)={\int }_{\mathrm{\Omega }}{q}_{1}\left(t,s,x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds-c\left(t\right),\\ y\left(t\right)={\int }_{\mathrm{\Omega }}{q}_{2}\left(t,s,y\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds-c\left(t\right),\end{array}$
(3.1)

where ${q}_{1},{q}_{2}:\mathrm{\Omega }×\mathrm{\Omega }×\mathbb{R}\to \mathbb{R}$ 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 ${G}_{1},{G}_{2}:X×X×X\to {\mathbb{R}}^{+}$ by

$\begin{array}{c}{G}_{1}\left(x,y,z\right)=\underset{t\in \mathrm{\Omega }}{sup}|x\left(t\right)-y\left(t\right)|+\underset{t\in \mathrm{\Omega }}{sup}|y\left(t\right)-z\left(t\right)|+\underset{t\in \mathrm{\Omega }}{sup}|z\left(t\right)-x\left(t\right)|,\hfill \\ {G}_{2}\left(x,y,z\right)=\frac{1}{2}\left[\underset{t\in \mathrm{\Omega }}{sup}|x\left(t\right)-y\left(t\right)|+\underset{t\in \mathrm{\Omega }}{sup}|y\left(t\right)-z\left(t\right)|+\underset{t\in \mathrm{\Omega }}{sup}|z\left(t\right)-x\left(t\right)|\right].\hfill \end{array}$

Obviously, ${G}_{2}\left(x,y,z\right)\le {G}_{1}\left(x,y,z\right)$ for all $x,y,z\in X$. Suppose that the following hypotheses hold:

1. (i)

For each $s,t\in \mathrm{\Omega }$,

${\int }_{\mathrm{\Omega }}{q}_{1}\left(t,s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\le u\left(s\right)$

and

${\int }_{\mathrm{\Omega }}{q}_{2}\left(t,s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\le u\left(s\right)$

hold.

1. (ii)

There exists $r:\mathrm{\Omega }\to \mathrm{\Omega }$ such that

${\int }_{\mathrm{\Omega }}|{q}_{1}\left(t,s,u\left(t\right)\right)-{q}_{2}\left(t,s,v\left(t\right)\right)|\phantom{\rule{0.2em}{0ex}}dt\le r\left(t\right)|u\left(t\right)-v\left(t\right)|$

for each $s,t\in \mathrm{\Omega }$ with ${sup}_{t\in \mathrm{\Omega }}r\left(t\right)\le k$ where $k\in \left[0,1\right)$.

Then the integral equations (3.1) have a common solution in ${L}^{2}\left(\mathrm{\Omega }\right)$.

Proof Define $fx\left(t\right)={\int }_{\mathrm{\Omega }}{q}_{1}\left(t,s,x\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-c\left(t\right)$ and $gx\left(t\right)={\int }_{\mathrm{\Omega }}{q}_{2}\left(t,s,x\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-c\left(t\right)$. As $fx\left(t\right)\le x\left(t\right)$ and $gx\left(t\right)\le x\left(t\right)$, so f and g are dominated maps. Now, for all comparable $x,y\in X$,

$\begin{array}{rcl}{G}_{1}\left(fx,fx,gy\right)& =& 2\underset{t\in \mathrm{\Omega }}{sup}|fx\left(t\right)-gy\left(t\right)|\\ =& 2\underset{t\in \mathrm{\Omega }}{sup}|{\int }_{\mathrm{\Omega }}{q}_{1}\left(t,s,x\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt-{\int }_{\mathrm{\Omega }}{q}_{2}\left(t,s,y\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt|\\ \le & 2\underset{t\in \mathrm{\Omega }}{sup}{\int }_{\mathrm{\Omega }}|{q}_{1}\left(t,s,x\left(t\right)\right)-{q}_{2}\left(t,s,y\left(t\right)\right)|\phantom{\rule{0.2em}{0ex}}dt\\ \le & 2\underset{t\in \mathrm{\Omega }}{sup}r\left(t\right)|x\left(t\right)-y\left(t\right)|\\ \le & 2k\underset{t\in \mathrm{\Omega }}{sup}|x\left(t\right)-y\left(t\right)|\\ =& k{G}_{2}\left(x,y,y\right)\\ \le & kmax\left\{{G}_{2}\left(x,x,y\right),{G}_{2}\left(fx,fx,x\right),{G}_{2}\left(gy,gy,y\right),\\ \left[{G}_{2}\left(fx,fx,y\right)+{G}_{2}\left(gy,gy,x\right)\right]/2\right\}.\end{array}$

Similarly,

$\begin{array}{rcl}{G}_{1}\left(fx,gy,gy\right)& \le & kmax\left\{{G}_{2}\left(x,y,y\right),{G}_{2}\left(fx,x,x\right),{G}_{2}\left(gy,y,y\right),\\ \left[{G}_{2}\left(fx,y,y\right)+{G}_{2}\left(gy,x,x\right)\right]/2\right\}\end{array}$

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 ${L}^{2}\left(\mathrm{\Omega }\right)$. □

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 G-metric spaces.

1. (2)

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

2. (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 $S=T={I}_{X}$.

1. (5)

A G-metric naturally induces a metric ${d}_{G}$ given by ${d}_{G}\left(x,y\right)=G\left(x,y,y\right)+G\left(x,x,y\right)$. 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 $\left(X,⪯,{d}_{G}\right)$.

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## Acknowledgements

The authors thank the referees for their appreciation and suggestions regarding this work.

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Correspondence to T Nazir.

### Competing interests

The authors declare that they have no competing interests.

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Abbas, M., Cho, Y. & Nazir, T. Common fixed points of Ćirić-type contractive mappings in two ordered generalized metric spaces. Fixed Point Theory Appl 2012, 139 (2012). https://doi.org/10.1186/1687-1812-2012-139