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. [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 G-metric on X and $(X,G)$ is called a G-metric space.

Definition 1.2 A sequence $\{x_n\}$ in a G-metric space X is called:

• (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]

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

• (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 $x,y\in X$.

Proposition 1.5 Every G-metric on X defines a metric $d_G$ on X by

$$d_G(x,y)=G(x,y,y)+G(y,x,x)$$

(1.1)

for all $x,y\in X$.

For a symmetric G-metric, we have

$$d_G(x,y)=2G(x,y,y)$$

(1.2)

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

$$\frac{3}{2}G(x,y,y)\le d_G(x,y)\le 3G(x,y,y)$$

(1.3)

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

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

for all $x,y\in X$.

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

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

Table 1 G -metric

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.

Recall that two mappings f and g on a G-metric 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 G-convergent to some $t\in X$,

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

Definition 1.10 Let X be a nonempty set. Then $(X,⪯,G)$ 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 $(X,⪯)$ 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 $(X,⪯)$ 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 $X=[0,1]$ 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 $(X,⪯)$ 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 $X=[0,1]$ 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

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

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

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

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

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:

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

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:

Theorem 2.1 Let $(X,⪯)$ be a partially ordered set and $G_1$, $G_2$ be two G-metrics 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 self-mappings on X satisfying the following properties:

$$\begin{array}{rcl}{G}_1(fx,fx,gy)& \le & kmax\{G_2(Sx,Sx,Ty),G_2(fx,fx,Sx),G_2(gy,gy,Ty),\\ [G_2(fx,fx,Ty)+G_2(gy,gy,Sx)]/2\}\end{array}$$

(2.1)

and

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& \le & kmax\{G_2(Sx,Ty,Ty),G_2(fx,Sx,Sx),G_2(gy,Ty,Ty),\\ [G_2(fx,Ty,Ty)+G_2(gy,Sx,Sx)]/2\}\end{array}$$

(2.2)

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 $y_n⪯x_n$ for all $n\in \mathbb{N}$, $y_n\to u$ implies that $u⪯x_n$ 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 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 $f(X)\subseteq T(X)$ and $g(X)\subseteq S(X)$, we can define the sequences $\{x_n\}$ and $\{y_n\}$ in X by

$$y_{2n}=g{x}_{2n}=S{x}_{2n+1},y_{2n+1}=f{x}_{2n+1}=T{x}_{2n+2}$$

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 $G_1(y_{2n},y_{2n+1},y_{2n+1})>0$ 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

(2.3)

Again, from (2.2), it follows that

(2.4)

Thus (2.3) and (2.4) imply that

$$G_1(y_{2k+2},y_{2k+1},y_{2k+1})\le k^2{G}_1(y_{2k+2},y_{2k+1},y_{2k+1})$$

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

$$G_1(y_{n+1},y_{n+1},y_n)\le k{G}_1(y_n,y_n,y_{n-1}).$$

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,

$$G_1(y_{n+1},y_{n+1},y_n)\le k{G}_1(y_n,y_n,y_{n-1})$$

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

$$G_1(y_{n+1},y_{n+1},y_n)\le k^n{G}_1(y_1,y_1,y_0)$$

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

$$\begin{array}{c}{G}_1(y_m,y_m,y_n)\hfill \\ \le G_1(y_n,y_{n+1},y_{n+1})+G_1(y_{n+1},y_{n+2},y_{n+2})+\cdots +G_1(y_{m-1},y_m,y_m)\hfill \\ \le k^n{G}_1(y_0,y_1,y_1)+k^{n+1}{G}_1(y_0,y_1,y_1)+\cdots +k^{m-1}{G}_1(y_0,y_1,y_1)\hfill \\ =k^n{G}_1(y_0,y_1,y_1)\sum _{i=0}^{m-n-1}{k}^i\hfill \\ \le \frac{k^n}{1-k}{G}_1(y_0,y_1,y_1)\hfill \end{array}$$

and so $G_1(y_n,y_m,y_m)\to 0$ as $m,n\to \mathrm{\infty }$. Hence $\{y_n\}$ 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 $\{f,S\}$ 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

$$\begin{array}{rcl}{G}_1(Sz,Sz,z)& \le & kmax\{G_2(Sz,Sz,z),G_2(Sz,Sz,Sz),G_2(z,z,z),\\ [G_2(Sz,Sz,z)+G_2(z,z,Sz)]/2\}\\ \le & kmax\{G_1(Sz,Sz,z),[G_1(Sz,Sz,z)+G_1(z,z,Sz)]/2\}\\ =& \frac{k}{2}[G_1(Sz,Sz,z)+G_1(z,z,Sz)],\end{array}$$

which further implies that

$$G_1(Sz,Sz,z)\le h{G}_1(z,z,Sz),$$

(2.5)

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

Similarly, we obtain

$$G_1(Sz,z,z)\le h{G}_1(z,Sz,Sz).$$

(2.6)

From (2.5) and (2.6), we have

$$G_1(Sz,Sz,z)\le h^2{G}_1(z,Sz,Sz)$$

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

$$\begin{array}{c}{G}_1(fz,fz,g{x}_{2n})\hfill \\ \le kmax\{G_2(Sz,Sz,T{x}_{2n}),G_2(fz,fz,Sz),G_2(g{x}_{2n},g{x}_{2n},T{x}_{2n}),\hfill \\ [G_2(fz,fz,T{x}_{2n})+G_2(g{x}_{2n},g{x}_{2n},Sz)]/2\}\hfill \\ =kmax\{G_2(z,z,T{x}_{2n}),G_2(fz,fz,z),G_2(g{x}_{2n},g{x}_{2n},T{x}_{2n}),\hfill \\ [G_2(fz,fz,T{x}_{2n})+G_2(g{x}_{2n},g{x}_{2n},z)]/2\},\hfill \end{array}$$

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

$$\begin{array}{rcl}{G}_1(fz,fz,z)& \le & kmax\{G_2(z,z,z),G_2(fz,fz,z),G_2(z,z,z),\\ [G_2(fz,fz,z)+G_2(z,z,z)]/2\}\\ \le & k{G}_1(fz,fz,z).\end{array}$$

(2.7)

Similarly, we obtain

$$G_1(fz,z,z)\le k{G}_1(z,fz,fz).$$

(2.8)

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 $v⪯Tv=fz⪯z$, it follows from (2.1) that

$$\begin{array}{rcl}{G}_1(fz,fz,gv)& \le & kmax\{G_2(Sz,Sz,Tv),G_2(fz,fz,Sz),G_2(gv,gv,Tv),\\ [G_2(fz,fz,Tv)+G_2(gv,gv,Sz)]/2\}\\ =& kmax\{G_2(fz,fz,fz),G_2(fz,fz,fz),G_2(gv,gv,fz),\\ [G_2(fz,fz,fz)+G_2(gv,gv,fz)]/2\}\\ \le & k{G}_1(fz,gv,gv).\end{array}$$

(2.9)

Similarly, we get

$$G_1(fz,gv,gv)\le k{G}_1(fz,fz,gv).$$

(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

$$\begin{array}{rcl}{G}_1(z,z,gz)& =& G_1(fz,fz,gz)\\ \le & kmax\{G_2(Sz,Sz,Tz),G_2(fz,fz,Sz),G_2(gz,gz,Tz),\\ [G_2(fz,fz,Tz)+G_2(gz,gz,Sz)]/2\}\\ =& kmax\{G_2(z,z,gz),G_2(z,z,z),G_2(gz,gz,gz),\\ [G_2(z,z,gz)+G_2(gz,gz,z)]/2\}\\ =& kmax\{G_2(z,z,gz),[G_2(z,z,gz)+G_2(gz,gz,z)]/2\}\\ \le & \frac{k}{2}[G_1(z,z,gz)+G_1(gz,gz,z)],\end{array}$$

that is,

$$G_1(z,z,gz)\le h{G}_1(gz,gz,z),$$

(2.11)

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

$$G_1(z,gz,gz)\le h{G}_1(z,z,gz).$$

(2.12)

Combining the above two inequalities, we get

$$G_1(z,z,gz)\le h^2{G}_1(z,z,gz)$$

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

$$\begin{array}{rcl}{G}_1(z,z,u)& =& G_1(fz,fz,gu)\\ \le & kmax\{G_2(Sz,Sz,Tu),G_2(fz,fz,Sz),G_2(gu,gu,Tu),\\ [G_2(fz,fz,Tu)+G_2(gu,gu,Sz)]/2\}\\ =& kmax\{G_2(z,z,u),G_2(z,z,z),G_2(u,u,u),\\ [G_2(z,z,u)+G_2(u,u,z)]/2\}\\ =& \frac{k}{2}[G_2(z,z,u)+G_2(u,u,z)]\\ \le & \frac{1}{2}{G}_1(z,z,u)+\frac{k}{2}{G}_1(u,u,z),\end{array}$$

that is,

$$G_1(z,z,u)\le k{G}_1(z,u,u).$$

Similarly, using (2.2), we obtain

$$G_1(z,u,u)\le k{G}_1(z,z,u).$$

Combining the above two inequalities, we get

$$G_1(z,z,u)\le k^2{G}_1(z,z,u)$$

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 G-metrics on X defined by Table 2. Then $G_1$ and $G_2$ are non-symmetric 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.

Table 2 Two G -metrices

Table 3 Self maps

Table 4 Dominated and dominating maps

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

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& =& G_1(0,2,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,3,3)=\frac{3}{4}{G}_2(gy,Ty,Ty)\\ \le & kmax\{G_2(Sx,Ty,Ty),G_2(fx,Sx,Sx),G_2(gy,Ty,Ty),\\ [G_2(fx,Ty,Ty)+G_2(gy,Sx,Sx)]/2\}.\end{array}$$

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

  $$\begin{array}{rcl}{G}_1(fx,fx,gy)& =& G_1(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,2,3)=\frac{3}{4}{G}_2(gy,gy,Ty)\\ \le & kmax\{G_2(Sx,Sx,Ty),G_2(fx,fx,Sx),G_2(gy,gy,Ty),\\ [G_2(fx,fx,Ty)+G_2(gy,gy,Sx)]/2\}\end{array}$$

and

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& =& G_1(0,2,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,3,3)=\frac{3}{4}{G}_2(gy,Ty,Ty)\\ \le & kmax\{G_2(Sx,Ty,Ty),G_2(fx,Sx,Sx),G_2(gy,Ty,Ty),\\ [G_2(fx,Ty,Ty)+G_2(gy,Sx,Sx)]/2\}.\end{array}$$

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

  $$\begin{array}{rcl}{G}_1(fx,fx,gy)& =& G_1(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,2,3)=\frac{3}{4}{G}_2(Sx,Sx,Ty)\\ \le & kmax\{G_2(Sx,Sx,Ty),G_2(fx,fx,Sx),G_2(gy,gy,Ty),\\ [G_2(fx,fx,Ty)+G_2(gy,gy,Sx)]/2\}\end{array}$$

and

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& =& G_1(0,2,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,3,3)=\frac{3}{4}{G}_2(Sx,Ty,Ty)\\ \le & kmax\{G_2(Sx,Ty,Ty),G_2(fx,Sx,Sx),G_2(gy,Ty,Ty),\\ [G_2(fx,Ty,Ty)+G_2(gy,Sx,Sx)]/2\}.\end{array}$$

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

  $$\begin{array}{rcl}{G}_1(fx,fx,gy)& =& G_1(0,0,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,2,3)=\frac{3}{4}{G}_2(gy,gy,Ty)\\ \le & kmax\{G_2(Sx,Sx,Ty),G_2(fx,fx,Sx),G_2(gy,gy,Ty),\\ [G_2(fx,fx,Ty)+G_2(gy,gy,Sx)]/2\}\end{array}$$

and

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& =& G_1(0,2,2)=4\\ <& \frac{3}{4}(6)=\frac{3}{4}{G}_2(2,3,3)=\frac{3}{4}{G}_2(gy,Ty,Ty)\\ \le & kmax\{G_2(Sx,Ty,Ty),G_2(fx,Sx,Sx),G_2(gy,Ty,Ty),\\ [G_2(fx,Ty,Ty)+G_2(gy,Sx,Sx)]/2\}.\end{array}$$

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.

If we consider the same set equipped with two metrics given by $d_1(x,y)=|x-y|$ and $d_2(x,y)=\frac{1}{2}|x-y|$ for all $x,y\in X$, then for $x=1$ and $y=2$, we have

$$\begin{array}{rcl}{d}_1(fx,gy)& =& d_1(0,2)=2\nleqq 2k\\ \le & kmax\{d_2(2,3),d_2(0,2),d_2(2,3),[d_2(0,3)+d_2(2,2)]/2\}\\ =& kmax\{d_2(Sx,Ty),d_2(fx,Sx),d_2(gy,Ty),[d_2(fx,Ty)+d_2(gy,Sx)]/2\}\end{array}$$

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].

Corollary 2.3 Let $(X,⪯)$ be a partially ordered set and $G_1$, $G_2$ be two G-metrics 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 self-mappings on X satisfying the following properties:

$$\begin{array}{rcl}{G}_1(fx,fx,gy)& \le & a_1{G}_2(Sx,Sx,Ty)+a_2{G}_2(Sx,Sx,fx)+a_3{G}_2(Ty,Ty,gy)\\ +a_4[G_2(Sx,Sx,gy)+G_2(Ty,Ty,fx)]\end{array}$$

(2.13)

and

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& \le & a_1{G}_2(Sx,Ty,Ty)+a_2{G}_2(Sx,fx,fx)+a_3{G}_2(Ty,gy,gy)\\ +a_4[G_2(Sx,gy,gy)+G_2(Ty,fx,fx)]\end{array}$$

(2.14)

for all comparable $x,y\in X$, where $a_1+a_2+a_3+2a_4<1$. 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 $y_n⪯x_n$ for all $n\in \mathbb{N}$, $y_n\to u$ implies that $u⪯x_n$ 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 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 $X=[0,1]$ be endowed with the usual ordering and $G_1$, $G_2$ be two G-metrics on X given in [13]:

$$\begin{array}{c}{G}_1(a,b,c)=|a-b|+|b-c|+|c-a|,\hfill \\ G_2(a,b,c)=\frac{1}{2}[|a-b|+|b-c|+|c-a|].\hfill \end{array}$$

Define the mappings $f,g,S,T:X\to X$ as

$$fx=\frac{x}{12},gx=\{\begin{array}{cc}\frac{x}{4}\hfill & \text{if }x\in [0,\frac{1}{2}),\hfill \\ \frac{x}{6}\hfill & \text{if }x\in [\frac{1}{2},1),\hfill \end{array}S(x)=\frac{3x}{2},T(x)=\frac{5x}{2}$$

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

  $$\begin{array}{rcl}{G}_1(fx,fx,gy)& =& \frac{1}{12}|x-3y|\le \frac{1}{12}(x+3y)\\ \le & \frac{3}{10}\left(\frac{17}{12}x\right)+\frac{3}{10}\left(\frac{9}{4}y\right)\\ =& a_2{G}_2(fx,fx,Sx)+a_3{G}_2(gy,gy,Ty)\\ \le & a_1{G}_2(Sx,Sx,Ty)+a_2{G}_2(fx,fx,Sx)+a_3{G}_2(gy,gy,Ty)\\ +a_4[G_2(fx,fx,Ty)+G_2(gy,gy,Sx)].\end{array}$$

• (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 $a_1+a_2+a_3+2a_4<1$. 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: $x,y\in X$, $x⪯y\iff x(t)\le y(t)$, for all $t\in \mathrm{\Omega }$. We consider the integral equations

$$\begin{array}{r}x(t)={\int }_{\mathrm{\Omega }}{q}_1(t,s,x(s))ds-c(t),\\ y(t)={\int }_{\mathrm{\Omega }}{q}_2(t,s,y(s))ds-c(t),\end{array}$$

(3.1)

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

$$\begin{array}{c}{G}_1(x,y,z)=\underset{t\in \mathrm{\Omega }}{sup}|x(t)-y(t)|+\underset{t\in \mathrm{\Omega }}{sup}|y(t)-z(t)|+\underset{t\in \mathrm{\Omega }}{sup}|z(t)-x(t)|,\hfill \\ G_2(x,y,z)=\frac{1}{2}\left[\underset{t\in \mathrm{\Omega }}{sup}\right|x(t)-y(t)|+\underset{t\in \mathrm{\Omega }}{sup}|y(t)-z(t)|+\underset{t\in \mathrm{\Omega }}{sup}|z(t)-x(t)\left|\right].\hfill \end{array}$$

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

  $${\int }_{\mathrm{\Omega }}{q}_1(t,s,u(s))ds\le u(s)$$

and

$${\int }_{\mathrm{\Omega }}{q}_2(t,s,u(s))ds\le u(s)$$

hold.

• (ii) There exists $r:\mathrm{\Omega }\to \mathrm{\Omega }$ such that

  $${\int }_{\mathrm{\Omega }}|q_1(t,s,u(t))-q_2(t,s,v(t))|dt\le r(t)|u(t)-v(t)|$$

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 $fx(t)={\int }_{\mathrm{\Omega }}{q}_1(t,s,x(t))dt-c(t)$ and $gx(t)={\int }_{\mathrm{\Omega }}{q}_2(t,s,x(t))dt-c(t)$. As $fx(t)\le x(t)$ and $gx(t)\le x(t)$, so f and g are dominated maps. Now, for all comparable $x,y\in X$,

$$\begin{array}{rcl}{G}_1(fx,fx,gy)& =& 2\underset{t\in \mathrm{\Omega }}{sup}|fx(t)-gy(t)|\\ =& 2\underset{t\in \mathrm{\Omega }}{sup}|{\int }_{\mathrm{\Omega }}{q}_1(t,s,x(t))dt-{\int }_{\mathrm{\Omega }}{q}_2(t,s,y(t))dt|\\ \le & 2\underset{t\in \mathrm{\Omega }}{sup}{\int }_{\mathrm{\Omega }}|q_1(t,s,x(t))-q_2(t,s,y(t))|dt\\ \le & 2\underset{t\in \mathrm{\Omega }}{sup}r(t)|x(t)-y(t)|\\ \le & 2k\underset{t\in \mathrm{\Omega }}{sup}|x(t)-y(t)|\\ =& k{G}_2(x,y,y)\\ \le & kmax\{G_2(x,x,y),G_2(fx,fx,x),G_2(gy,gy,y),\\ [G_2(fx,fx,y)+G_2(gy,gy,x)]/2\}.\end{array}$$

Similarly,

$$\begin{array}{rcl}{G}_1(fx,gy,gy)& \le & kmax\{G_2(x,y,y),G_2(fx,x,x),G_2(gy,y,y),\\ [G_2(fx,y,y)+G_2(gy,x,x)]/2\}\end{array}$$

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 $L^2(\mathrm{\Omega })$. □

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 G-metric 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 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 $(X,⪯,d_G)$.