Common fixed point results under a new contractive condition without using continuity

Abstract

In this paper, using the concept of the common $(E.A)$ property, we prove a common fixed point theorem for a class of twice power type weakly compatible mappings in generalized metric space. Our results do not rely on any commuting or continuity condition of the mappings. We also state some examples to illustrate our new results in symmetric and nonsymmetric generalized metric spaces. It should be pointed out that this is the first time to use common $(E.A)$ properties to discuss common fixed point problems of contractive mappings for twice power type in generalized metric spaces.

1 Introduction and preliminaries

In 2006, Mustafa and Sims [1] introduced a new structure of generalized metric space which is called a G-metric. Based on the notion of generalized metric spaces, Mustafa et al. [2–5], Obiedat and Mustafa [6], Aydi et al. [7, 8], Gajić and Stojaković [9], Zhou and Gu [10], Shatanawi [11] obtained some fixed point results for mappings satisfying different contractive conditions. Chugh et al. [12] obtained some fixed point results for maps satisfying property P in G-metric spaces. The study of common fixed point problems in G-metric spaces was initiated by Abbas and Rhoades [13]. Subsequently, many authors obtained many common fixed point theorems for the mappings satisfying different contractive conditions; see [14–31] for more details. Recently, some authors using $(E.A)$ property in generalized metric space to prove common fixed point, such as Abbas et al. [32], Mustafa et al. [33], Long et al. [34], Gu and Yin [35], Gu and Shatanawi [36].

Recently, Jleli and Samet [37] and Samet et al. [38] observed that some fixed point theorems in the context of a G-metric space can be proved (by simple transformation) using related existing results in the setting of a (quasi-) metric space. Namely, if the contraction condition of the fixed point theorem on G-metric space can be reduced to two variables, then one can construct an equivalent fixed point theorem in setting of usual metric space. This idea is not completely new, but it was not successfully used before (see [39]).

Very recently, Karapınar and Agarwal suggest new contraction conditions in G-metric space in a way that the techniques in [37, 38] are not applicable. In this approach [40], contraction conditions cannot be expressed in two variables. So, in some cases, as is noticed even in Jleli-Samet’s paper [37], when the contraction condition is of nonlinear type, this strategy cannot be always successfully used. This is exactly the case in our paper.

The purpose of this paper is to use the concept of the common $(E.A)$ property and weakly compatible mappings to discuss common fixed point problem for a class of twice power type contractive mappings in the framework of a generalized metric space. Our results do not rely on any commuting or continuity condition of the mappings. We also state some examples to illustrate our new results in the framework of symmetric and nonsymmetric generalized metric spaces.

As far as we know, this is the first time to use common $(E.A)$ properties to discuss common fixed point problems of contractive mappings for twice power type in generalized metric spaces.

Now we give preliminaries and basic definitions which are used throughout the paper.

Definition 1.1 [1]

Let X be a nonempty set, and let $G:X\times X\times X\to R^{+}$ be a function satisfying the following axioms:

• (G1) $G(x,y,z)=0$ if $x=y=z$;

• (G2) $0<G(x,y,z)$ for all $x,y\in X$ with $x\ne y$;

• (G3) $G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $z\ne y$;

• (G4) $G(x,y,z)=G(x,z,y)=G(y,z,x)\cdots$ (symmetry in all three variables);

• (G5) $G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$ (rectangle inequality).

Then the function G is called a generalized metric or a G-metric on X, and the pair $(X,G)$ is called a G-metric space.

Definition 1.2 [1]

Let $(X,G)$ be a G-metric space, and let $\{x_n\}$ a sequence of points in X, a point x in X is said to be the limit of the sequence $\{x_n\}$, ${lim}_{n\to \mathrm{\infty }}G(x,x_n,x_m)=0$, and one says that sequence $\{x_n\}$ is G-convergent to x.

Thus, if $x_n\to x$ or ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ in a G-metric space $(X,G)$, then if for each $ϵ>0$, there exists a positive integer N such that $G(x,x_n,x_m)<ϵ$ for all $n,m\ge N$.

Proposition 1.1 [1]

Let $(X,G)$ be a G-metric space. Then the following are equivalent:

• (1) $\{x_n\}$ is G-convergent to x.

• (2) $G(x_n,x_n,x)\to 0$ as $n\to \mathrm{\infty }$.

• (3) $G(x_n,x,x)\to 0$ as $n\to \mathrm{\infty }$.

• (4) $G(x_m,x_n,x)\to 0$ as $m,n\to \mathrm{\infty }$.

Definition 1.3 [1]

Let $(X,G)$ be a G-metric space. A sequence $\{x_n\}$ is called G-Cauchy if, for each $ϵ>0$, there exists a positive integer N such that $G(x_m,x_n,x)<ϵ$ for all $n,m,l\ge N$; i.e. if $G(x_m,x_n,x_l)\to 0$ as $m,n,l\to \mathrm{\infty }$.

Proposition 1.2 [1]

If $(X,G)$ is a G-metric space then the following are equivalent:

• (1) The sequence $\{x_n\}$ is G-Cauchy.

• (2) For each $ϵ>0$, there exists a positive integer N such that $G(x_m,x_n,x)<ϵ$ for all $n,m,l\ge N$.

Proposition 1.3 [1]

Let $(X,G)$ be a G-metric space. Then the function $G(x,y,z)$ is jointly continuous in all three of its variables.

Definition 1.4 [1]

A G-metric space $(X,G)$ is said to be G-complete if every G-Cauchy sequence in $(X,G)$ is G-convergent in X.

Definition 1.5 [41]

Let f and g be self-maps of a set X. If $w=fx=gx$ for some x in X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g.

Definition 1.6 [41]

Two self-mappings f and g on X are said to be weakly compatible if they commute at coincidence points.

Definition 1.7 [32]

Let X be a G-metric space. Self-maps f and g on X are said to satisfy the G-$(E.A)$ property if there exists a sequence $\{x_n\}$ in X such that $\{f{x}_n\}$ and $\{g{x}_n\}$ are G-convergent to some $t\in X$.

Definition 1.8 [32]

Let $(X,d)$ be a G-metric space and A, B, S, and T be four self-maps on X. The pairs $(A,S)$ and $(B,T)$ are said to satisfy the common $(E.A)$ property if there exist two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$\underset{n\to \mathrm{\infty }}{lim}A{x}_n=\underset{n\to \mathrm{\infty }}{lim}S{x}_n=\underset{n\to \mathrm{\infty }}{lim}B{y}_n=\underset{n\to \mathrm{\infty }}{lim}T{y}_n=t$$

for some $t\in X$.

Definition 1.9 [17]

Self-mappings f and g of a G-metric space $(X,G)$ are said to be compatible if ${lim}_{n\to \mathrm{\infty }}G(fg{x}_n,gf{x}_n,gf{x}_n)=0$ and ${lim}_{n\to \mathrm{\infty }}G(gf{x}_n,fg{x}_n,fg{x}_n)=0$, whenever $\{x_n\}$ is a sequence in X such that

$$\underset{n\to \mathrm{\infty }}{lim}f{x}_n=\underset{n\to \mathrm{\infty }}{lim}g{x}_n=t$$

for some $t\in X$.

2 Main results

Theorem 2.1 Let $(X,G)$ be a G-metric space. Suppose mappings $f,g,h,R,S,T:X\to X$ satisfying the following conditions:

$$G^2(fx,gy,hz)\le kmax\left\{\begin{array}{c}G(Rx,Sy,Tz)G(fx,Rx,Rx),\\ G(gy,Sy,Sy)G(hz,Tz,Tz),\\ G(fx,Sy,Tz)G(Rx,gy,Tz),\\ G(Rx,Sy,hz)G(fx,gy,Tz),\\ G(fx,Sy,hz)G(Rx,gy,hz)\end{array}\right\}$$

(2.1)

for all $x,y,z\in X$, $0\le k<1$. If one of the following conditions is satisfied, then the pairs $(f,R)$, $(g,S)$, and $(h,T)$ have a common point of coincidence in X.

• (i) The subspace RX is closed in X, $fX\subseteq SX$, $gX\subseteq TX$, and the two pairs of $(f,R)$ and $(g,S)$ satisfy the common $(E.A)$ property.

• (ii) The subspace SX is closed in X, $gX\subseteq TX$, $hX\subseteq RX$, and the two pairs of $(g,S)$ and $(h,T)$ satisfy the common $(E.A)$ property.

• (iii) The subspace TX is closed in X, $fX\subseteq SX$, $hX\subseteq RX$, and the two pairs of $(f,R)$ and $(h,T)$ satisfy the common $(E.A)$ property.

Further, if the pairs $(f,R)$, $(g,S)$, and $(h,T)$ are weakly compatible, then f, g, h, R, S, and T have a unique common fixed point in X.

Proof First we suppose that RX is closed in X, $fX\subseteq SX$, $gX\subseteq TX$, and the two pairs of $(f,R)$ and $(g,S)$ satisfy the common $(E.A)$ property, then by Definition 1.8 we know that there exist two sequences $\{x_n\}$ and $\{y_n\}$ in X such that

$$\underset{n\to \mathrm{\infty }}{lim}f{x}_n=\underset{n\to \mathrm{\infty }}{lim}R{x}_n=\underset{n\to \mathrm{\infty }}{lim}g{y}_n=\underset{n\to \mathrm{\infty }}{lim}S{y}_n=t$$

for some $t\in X$.

Since $gX\subseteq TX$, there exists a sequence $\{z_n\}$ in X such that $g{y}_n=T{z}_n$. So we get ${lim}_{n\to \mathrm{\infty }}T{z}_n={lim}_{n\to \mathrm{\infty }}g{y}_n=t$. By the condition (2.1) we have

$$G^2(f{x}_n,g{y}_n,h{z}_n)\le kmax\left\{\begin{array}{c}G(R{x}_n,S{y}_n,T{z}_n)G(f{x}_n,R{x}_n,R{x}_n),\\ G(g{y}_n,S{y}_n,S{y}_n)G(h{z}_n,T{z}_n,T{z}_n),\\ G(f{x}_n,S{y}_n,T{z}_n)G(R{x}_n,g{y}_n,T{z}_n),\\ G(R{x}_n,S{y}_n,h{z}_n)G(f{x}_n,g{y}_n,T{z}_n),\\ G(f{x}_n,S{y}_n,h{z}_n)G(R{x}_n,g{y}_n,h{z}_n)\end{array}\right\}.$$

Letting $n\to \mathrm{\infty }$, we have

$$G^2(t,t,\underset{n\to \mathrm{\infty }}{lim}h{z}_n)\le k{G}^2(t,t,\underset{n\to \mathrm{\infty }}{lim}h{z}_n);$$

this gives $G^2(t,t,{lim}_{n\to \mathrm{\infty }}h{z}_n)=0$, since $0\le k<1$. Hence ${lim}_{n\to \mathrm{\infty }}h{z}_n=t$.

Since RX is a closed subspace of X, and ${lim}_{n\to \mathrm{\infty }}R{x}_n=t$, there exists a point $u\in X$ such that $Ru=t$. By the condition (2.1) we have

$$G^2(fu,g{y}_n,h{z}_n)\le kmax\left\{\begin{array}{c}G(Ru,S{y}_n,T{z}_n)G(fu,Ru,Ru),\\ G(g{y}_n,S{y}_n,S{y}_n)G(h{z}_n,T{z}_n,T{z}_n),\\ G(fu,S{y}_n,T{z}_n)G(Ru,g{y}_n,T{z}_n),\\ G(Ru,S{y}_n,h{z}_n)G(fu,g{y}_n,T{z}_n),\\ G(fu,S{y}_n,h{z}_n)G(Ru,g{y}_n,h{z}_n)\end{array}\right\}.$$

Letting $n\to \mathrm{\infty }$, we have $G^2(fu,t,t)\le 0$, hence $fu=t$. Thus $Ru=fu=t$, so u is the coincidence point of the pair $(f,R)$.

Since $fX\subseteq SX$ and $fu=t$, there exists a point $v\in X$ such that $Sv=fu=t$. By the condition (2.1) we have

$$G^2(fu,gv,h{z}_n)\le kmax\left\{\begin{array}{c}G(Ru,Sv,T{z}_n)G(fu,Ru,Ru),\\ G(gv,Sv,Sv)G(h{z}_n,T{z}_n,T{z}_n),\\ G(fu,Sv,T{z}_n)G(Ru,gv,T{z}_n),\\ G(Ru,Sv,h{z}_n)G(fu,gv,T{z}_n),\\ G(fu,Sv,h{z}_n)G(Ru,gv,h{z}_n)\end{array}\right\}.$$

Letting $n\to \mathrm{\infty }$, we have $G^2(t,gv,t)\le 0$, hence $gv=t$. Thus $Sv=gv=t$, so v is the coincidence point of the pair $(g,S)$.

Since $gX\subseteq TX$ and $gv=t$, there exists a point $w\in X$ such that $Tw=gv=t$. By the condition (2.1) we have

$$\begin{array}{rl}{G}^2(t,t,hw)& =G^2(fu,gv,hw)\\ \le kmax\left\{\begin{array}{c}G(Ru,Sv,Tw)G(fu,Ru,Ru),\\ G(gv,Sv,Sv)G(hw,Tw,Tw),\\ G(fu,Sv,Tw)G(Ru,gv,Tw),\\ G(Ru,Sv,hw)G(fu,gv,Tw),\\ G(fu,Sv,hw)G(Ru,gv,hw)\end{array}\right\}\\ =k{G}^2(t,t,hw),\end{array}$$

hence $hw=t$, since $0\le k<1$. Thus $Tw=hw=t$, so w is the coincidence point of the pair $(h,T)$.

In the above proof we get $fu=Ru=gv=Sv=hw=Tw=t$. Then we get $ft=Rt$, $gt=St$, and $ht=Tt$, since the pairs $(f,R)$, $(g,S)$, and $(h,T)$ are weakly compatible. By the condition (2.1), we have

$$\begin{array}{rl}{G}^2(ft,t,t)& =G^2(ft,gv,hw)\\ \le kmax\left\{\begin{array}{c}G(Rt,Sv,Tw)G(ft,Rt,Rt),\\ G(gv,Sv,Sv)G(hw,Tw,Tw),\\ G(ft,Sv,Tw)G(Rt,gv,Tw),\\ G(Rt,Sv,hw)G(ft,gv,Tw),\\ G(ft,Sv,hw)G(Rt,gv,hw)\end{array}\right\}\\ =k{G}^2(ft,t,t),\end{array}$$

hence $G^2(ft,t,t)=0$, since $0\le k<1$. Thus $ft=t=Rt$. Similarly, it can be shown that $gt=St=t$ and $ht=Tt=t$, which means that t is a common fixed point of f, g, h, R, S, and T.

Now we prove the uniqueness of the common fixed point t.

Let t and p be two common fixed point of f, g, h, R, S, and T, then using the condition (2.1), we have

$$\begin{array}{rl}{G}^2(p,t,t)& =G^2(fp,gt,ht)\\ \le kmax\left\{\begin{array}{c}G(Rp,St,Tt)G(fp,Rp,Rp),\\ G(gt,St,St)G(ht,Tt,Tt),\\ G(fp,St,Tt)G(Rp,gt,Tt),\\ G(Rp,St,ht)G(fp,gt,Tt),\\ G(fp,St,ht)G(Rp,gt,ht)\end{array}\right\}\\ =k{G}^2(p,t,t),\end{array}$$

hence $G^2(p,t,t)=0$, since $0\le k<1$. Thus $p=t$. So common fixed point is unique. □

Example 2.1 Let $X=[0,1]$ be a G-metric space with

$$G(x,y,z)=|x-y|+|y-z|+|z-x|.$$

We define mappings f, g, h, R, S, and T on X by

$$\begin{array}{c}fx=\{\begin{array}{ll}1,& x\in [0,\frac{1}{2}],\\ \frac{4}{5},& x\in (\frac{1}{2},1],\end{array}gx=\{\begin{array}{ll}\frac{5}{6},& x\in [0,\frac{1}{2}],\\ \frac{4}{5},& x\in (\frac{1}{2},1],\end{array}hx=\{\begin{array}{ll}\frac{6}{7},& x\in [0,\frac{1}{2}],\\ \frac{4}{5},& x\in (\frac{1}{2},1],\end{array}\hfill \\ Rx=\{\begin{array}{ll}0,& x\in [0,\frac{1}{2}],\\ \frac{4}{5},& x\in (\frac{1}{2},1),\\ \frac{6}{7},& x=1,\end{array}Sx=\{\begin{array}{ll}1,& x\in [0,\frac{1}{2}]\\ \frac{4}{5},& x\in (\frac{1}{2},1),\\ 0,& x=1,\end{array}Tx=\{\begin{array}{ll}0,& x\in [0,\frac{1}{2}],\\ \frac{4}{5},& x\in (\frac{1}{2},1),\\ \frac{5}{6},& x=1.\end{array}\hfill \end{array}$$

Clearly, from the above functions we know that the subspace RX is closed in X, $fX\subseteq SX$, $gX\subseteq TX$, $hX\subseteq RX$ and the pairs $(f,R)$, $(g,S)$, $(h,T)$ be weakly compatible. The pairs $(f,R)$ and $(g,S)$ satisfy the common $(E.A)$ property, let $x_n=\frac{6}{7}$ and $y_n=\frac{5}{6}$ for each $n\in N$ be the required sequences.

Now we prove that the mappings f, g, h, R, S, and T are satisfying the condition (2.1) of Theorem 2.1 with $k=\frac{1}{4}\in [0,1)$. Let

$$M(x,y,z)=max\left\{\begin{array}{c}G(Rx,Sy,Tz)G(fx,Rx,Rx),\\ G(gy,Sy,Sy)G(hz,Tz,Tz),\\ G(fx,Sy,Tz)G(Rx,gy,Tz),\\ G(Rx,Sy,hz)G(fx,gy,Tz),\\ G(fx,Sy,hz)G(Rx,gy,hz)\end{array}\right\}.$$

Case (1) If $x,y,z\in [0,\frac{1}{2}]$, then we have

$$\begin{array}{c}{G}^2(fx,gy,hz)=G^2(1,\frac{5}{6},\frac{6}{7})=\frac{1}{9},\hfill \\ G(Rx,Sy,Tz)G(fx,Rx,Rx)=G(0,1,0)G(1,0,0)=4.\hfill \end{array}$$

Thus we have

$$G^2(fx,gy,hz)=\frac{1}{9}<\frac{1}{4}\cdot 4=kG(Rx,Sy,Tz)G(fx,Rx,Rx)\le kM(x,y,z).$$

Case (2) If $x,y\in [0,\frac{1}{2}]$, $z\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(1,\frac{5}{6},\frac{4}{5})=\frac{4}{25}.$$

If $z=1$, then

$$G(Rx,Sy,Tz)G(fx,Rx,Rx)=G(0,1,\frac{5}{6})G(1,0,0)=4.$$

If $z\in (\frac{1}{2},1)$, then

$$G(Rx,Sy,Tz)G(fx,Rx,Rx)=G(0,1,\frac{4}{5})G(1,0,0)=4.$$

So we know $G(Rx,Sy,Tz)G(fx,Rx,Rx)=4$. Thus we have

$$G^2(fx,gy,hz)=\frac{4}{25}<\frac{1}{4}\cdot 4=kG(Rx,Sy,Tz)G(fx,Rx,Rx)\le kM(x,y,z).$$

Case (3) If $x,z\in [0,\frac{1}{2}]$, $y\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(1,\frac{4}{5},\frac{6}{7})=\frac{4}{25}.$$

If $y=1$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(1,0,0)G(0,\frac{4}{5},0)=\frac{16}{5}.$$

If $y\in (\frac{1}{2},1)$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(1,\frac{4}{5},0)G(0,\frac{4}{5},0)=\frac{16}{5}.$$

So we know $G(Rx,Sy,Tz)G(fx,Rx,Rx)=\frac{16}{5}$. Thus we have

$$G^2(fx,gy,hz)=\frac{4}{25}<\frac{1}{4}\cdot \frac{16}{5}=kG(fx,Sy,Tz)G(Rx,gy,Tz)\le kM(x,y,z).$$

Case (4) If $y,z\in [0,\frac{1}{2}]$, $x\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(\frac{4}{5},\frac{5}{6},\frac{6}{7})=\frac{16}{1,225}.$$

If $x=1$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},1,0)G(\frac{6}{7},\frac{5}{6},0)=\frac{24}{7}.$$

If $x\in (\frac{1}{2},1)$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},1,0)G(\frac{4}{5},\frac{5}{6},0)=\frac{10}{3}.$$

So we know $G(fx,Sy,Tz)G(Rx,gy,Tz)\ge \frac{10}{3}$. Thus we have

$$G^2(fx,gy,hz)=\frac{16}{1,225}<\frac{1}{4}\cdot \frac{10}{3}\le kG(fx,Sy,Tz)G(Rx,gy,Tz)\le kM(x,y,z).$$

Case (5) $x\in [0,\frac{1}{2}]$, $y,z\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(1,\frac{4}{5},\frac{4}{5})=\frac{4}{25}.$$

If $y=1$, then

$$G(fx,Sy,hz)G(Rx,gy,hz)=G(1,0,\frac{4}{5})G(0,\frac{4}{5},\frac{4}{5})=\frac{16}{5}.$$

If $y\in (\frac{1}{2},1)$, then

$$G(fx,Sy,hz)G(Rx,gy,hz)=G(1,\frac{4}{5},\frac{4}{5})G(0,\frac{4}{5},\frac{4}{5})=\frac{16}{25}.$$

So we know $G(fx,Sy,hz)G(Rx,gy,hz)\ge \frac{16}{25}$. Thus we have

$$G^2(fx,gy,hz)=\frac{4}{25}=\frac{1}{4}\cdot \frac{16}{25}\le kG(fx,Sy,hz)G(Rx,gy,hz)\le kM(x,y,z).$$

Case (6) $y\in [0,\frac{1}{2}]$, $x,z\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(\frac{4}{5},\frac{5}{6},\frac{4}{5})=\frac{1}{225}.$$

If $x=1$, then

$$G(fx,Sy,hz)G(Rx,gy,hz)=G(\frac{4}{5},1,\frac{4}{5})G(\frac{6}{7},\frac{5}{6},\frac{4}{5})=\frac{8}{175}.$$

If $x\in (\frac{1}{2},1)$, then

$$G(fx,Sy,hz)G(Rx,gy,hz)=G(\frac{4}{5},1,\frac{4}{5})G(\frac{4}{5},\frac{5}{6},\frac{4}{5})=\frac{2}{75}.$$

So we know $G(fx,Sy,hz)G(Rx,gy,hz)\ge \frac{2}{75}$. Thus we have

$$G^2(fx,gy,hz)=\frac{1}{225}<\frac{1}{4}\cdot \frac{2}{75}\le kG(fx,Sy,hz)G(Rx,gy,hz)\le kM(x,y,z).$$

Case (7) $z\in [0,\frac{1}{2}]$, $x,y\in (\frac{1}{2},1]$, then we have

$$G^2(fx,gy,hz)=G^2(\frac{4}{5},\frac{4}{5},\frac{6}{7})=\frac{16}{1,225}.$$

If $x=1$, $y=1$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},0,0)G(\frac{6}{7},\frac{4}{5},0)=\frac{96}{35}.$$

If $x=1$, $y\in (\frac{1}{2},1)$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},\frac{4}{5},0)G(\frac{6}{7},\frac{4}{5},0)=\frac{96}{35}.$$

If $y=1$, $x\in (\frac{1}{2},1)$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},0,0)G(\frac{4}{5},\frac{4}{5},0)=\frac{64}{25}.$$

If $x,y\in (\frac{1}{2},1)$, then

$$G(fx,Sy,Tz)G(Rx,gy,Tz)=G(\frac{4}{5},\frac{4}{5},0)G(\frac{4}{5},\frac{4}{5},0)=\frac{64}{25}.$$

So we know $G(fx,Sy,Tz)G(Rx,gy,Tz)\ge \frac{64}{25}$. Thus we have

$$G^2(fx,gy,hz)=\frac{16}{1,225}<\frac{1}{4}\cdot \frac{64}{25}\le kG(fx,Sy,Tz)G(Rx,gy,Tz)\le kM(x,y,z).$$

Case (8) If $x,y,z\in (\frac{1}{2},1]$, then

$$G^2(fx,gy,hz)=G^2(\frac{4}{5},\frac{4}{5},\frac{4}{5})=0\le \frac{1}{4}M(x,y,z)=kM(x,y,z).$$

Then in all the above cases, the mappings f, g, h, R, T, and S are satisfying the condition (2.1) of Theorem 2.1 with $k=\frac{1}{4}$, so that all the conditions of Theorem 2.1 are satisfied. Moreover, $\frac{4}{5}$ is the unique common fixed point of f, g, h, R, T, and S.

The following example supports the usability of our results for nonsymmetric generalized metric spaces.

Example 2.2 Let $X=\{0,1,2\}$ be a set with G-metric defined by Table 1. It is easy to see that $(X,G)$ is a nonsymmetric generalized metric space. Let the maps $f,g,h,R,S,T:X\to X$ be defined by Table 2.

Table 1 The definition of G -metric on X

Table 2 The definition of maps f , g , h , R , S and T on X

Clearly, the subspace RX, SX, and TX are closed in X, $fX\subseteq SX$, $gX\subseteq TX$, and $hX\subseteq RX$ with the pairs $(f,R)$, $(g,S)$, and $(h,T)$ being weakly compatible. Also two pairs $(f,R)$ and $(g,S)$ satisfy the common $(E.A)$ property, indeed, $x_n=0$ and $y_n=1$ for each $n\in \mathbb{N}$ are the required sequences.

To check the contractive condition (2.1) for all $x,y\in X$, we consider the following cases.

Note that for Case (1) $x=y=z=0$, (2) $x=y=0$, $z=2$, (3) $x=z=0$, $y=1$, (4) $x=0$, $y=1$, $z=2$, (5) $x=1$, $y=z=0$, (6) $x=1$, $y=0$, $z=2$, (7) $x=y=1$, $z=0$, (8) $x=y=1$, $z=2$, (9) $x=2$, $y=z=0$, (10) $x=z=2$, $y=0$, (11) $x=2$, $y=1$, $z=0$, and (12) $x=z=2$, $y=1$.

We have $G(fx,gy,hz)=G(0,0,0)=0$, and hence (2.1) is obviously satisfied.

Case (13) If $x=y=0$, $z=1$, then $fx=gy=0$, $hz=1$, $Rx=Sy=0$, $Tz=2$, hence we have

$$\begin{array}{rcl}{G}^2(f0,g0,h1)& =& G^2(0,0,1)=1\\ <& \frac{1}{3}\times 9=\frac{1}{3}G(0,0,2)G(0,0,2)\\ =& \frac{1}{3}G(f0,S0,T1)G(R0,g0,T1)\\ \le & \frac{1}{3}M(0,0,1).\end{array}$$

Case (14) If $x=0$, $y=z=1$, then $fx=gy=0$, $hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}{G}^2(f0,g1,h1)& =& G^2(0,0,1)=1\\ <& \frac{1}{3}\times 9=\frac{1}{3}G(0,0,2)G(0,0,2)\\ =& \frac{1}{3}G(f0,S1,T1)G(R0,g1,T1)\\ \le & \frac{1}{3}M(0,1,1).\end{array}$$

Case (15) If $x=z=0$, $y=2$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}{G}^2(f0,g2,h0)& =& G^2(0,1,0)=1\\ =& \frac{1}{3}\times 3=\frac{1}{3}G(0,2,0)G(0,1,0)\\ =& \frac{1}{3}G(R0,S2,h0)G(f0,g2,T0)\\ \le & \frac{1}{3}M(0,2,0).\end{array}$$

Case (16) If $x=0$, $y=2$, $z=1$, then $fx=0$, $gy=hz=1$, $Rx=0$, $Sy=Tz=2$, hence we have

$$\begin{array}{rcl}{G}^2(f0,g2,h1)& =& G^2(0,1,1)=1\\ <& \frac{1}{3}\times 16=\frac{1}{3}G(0,2,1)G(0,1,2)\\ =& \frac{1}{3}G(R0,S2,h1)G(f0,g2,T1)\\ \le & \frac{1}{3}M(0,2,1).\end{array}$$

Case (17) If $x=0$, $y=z=2$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}{G}^2(f0,g2,h2)& =& G^2(0,1,0)=1\\ =& \frac{1}{3}\times 3=\frac{1}{3}G(0,2,0)G(0,1,1)\\ =& \frac{1}{3}G(R0,S2,h2)G(f0,g2,T2)\\ \le & \frac{1}{3}M(0,2,2).\end{array}$$

Case (18) If $x=z=1$, $y=0$, then $fx=gy=0$, $hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}{G}^2(f1,g0,h1)& =& G^2(0,0,1)=1\\ <& \frac{1}{3}\times 12=\frac{1}{3}G(0,0,2)G(1,0,2)\\ =& \frac{1}{3}G(f1,S0,T1)G(R1,g0,T1)\\ \le & \frac{1}{3}M(1,0,1).\end{array}$$

Case (19) $x=y=z=1$, then $fx=gy=0$, $hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}{G}^2(f1,g1,h1)& =& G^2(0,0,1)=1\\ <& \frac{1}{3}\times 12=\frac{1}{3}G(0,0,2)G(1,0,2)\\ =& \frac{1}{3}G(f1,S1,T1)G(R1,g1,T1)\\ \le & \frac{1}{3}M(1,1,1).\end{array}$$

Case (20) If $x=1$, $y=2$, $z=0$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}{G}^2(f1,g2,h0)& =& G^2(0,1,0)=1\\ <& \frac{1}{3}\times 4=\frac{1}{3}G(1,2,0)G(0,1,0)\\ =& \frac{1}{3}G(R1,S2,h0)G(f1,g2,T0)\\ \le & \frac{1}{3}M(1,2,0).\end{array}$$

Case (21) If $x=z=1$, $y=2$, then $fx=0$, $gy=hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,1,1)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(hz,Tz,Tz)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (22) If $x=1$, $y=z=2$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,1,0)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(gy,Sy,Sy)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (23) If $x=2$, $y=0$, $z=1$, then $fx=gy=0$, $hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,0,1)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(hz,Tz,Tz)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (24) If $x=2$, $y=z=1$, then $fx=gy=0$, $hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,0,1)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(hz,Tz,Tz)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (25) $x=y=2$, $z=0$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,1,0)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(gy,Sy,Sy)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (26) $x=y=2$, $z=1$, then $fx=0$, $gy=hz=1$, $Tz=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,1,1)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(hz,Tz,Tz)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Case (27) If $x=y=z=2$, then $fx=hz=0$, $gy=1$, $Sy=2$, hence we have

$$\begin{array}{rcl}\psi (G(fx,gy,hz))& =& 3G(0,1,0)=3\\ <& \frac{11}{4}\times 2=\frac{11}{4}G(1,2,2)\\ =& \frac{11}{4}G(gy,Sy,Sy)\le \frac{11}{4}M(x,y,z)\\ =& \psi (M(x,y,z))-\varphi (M(x,y,z)).\end{array}$$

Hence, all of the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique common fixed point of f, g, h, R, S, and T.

Corollary 2.1 Let $(X,G)$ be a G-metric space. Suppose mappings $f,g,h,R,S,T:X\to X$ satisfying the following conditions:

$$\begin{array}{rcl}{G}^2(fx,gy,hz)& \le & a_{1}G(Rx,Sy,Tz)G(fx,Rx,Rx)+a_{2}G(gy,Sy,Sy)G(hz,Tz,Tz)\\ +a_{3}G(fx,Sy,Tz)G(Rx,gy,Tz)+a_{4}G(Rx,Sy,hz)G(fx,gy,Tz)\\ +a_{5}G(fx,Sy,hz)G(Rx,gy,hz)\end{array}$$

(2.2)

for all $x,y,z\in X$. Here $a_i\ge 0$ ($i=1,2,3,4,5$) and $0\le a_1+a_2+a_3+a_4+a_5<1$. If one of the following conditions is satisfied, then the pairs $(f,R)$, $(g,S)$, and $(h,T)$ have a common point of coincidence in X.

• (i) The subspace RX is closed in X, $fX\subseteq SX$, $gX\subseteq TX$, and the two pairs of $(f,R)$ and $(g,S)$ satisfy the common $(E.A)$ property.

• (ii) The subspace SX is closed in X, $gX\subseteq TX$, $hX\subseteq RX$, and the two pairs of $(g,S)$ and $(h,T)$ satisfy the common $(E.A)$ property.

• (iii) The subspace TX is closed in X, $fX\subseteq SX$, $hX\subseteq RX$, and the two pairs of $(f,R)$ and $(h,T)$ satisfy the common $(E.A)$ property.

Further, if the pairs $(f,R)$, $(g,S)$, and $(h,T)$ are weakly compatible, then f, g, h, R, S, and T have a unique common fixed point in X.

Proof Suppose that

$$M(x,y,z)=max\left\{\begin{array}{c}G(Rx,Sy,Tz)G(fx,Rx,Rx),\\ G(gy,Sy,Sy)G(hz,Tz,Tz),\\ G(fx,Sy,Tz)G(Rx,gy,Tz),\\ G(Rx,Sy,hz)G(fx,gy,Tz),\\ G(fx,Sy,hz)G(Rx,gy,hz)\end{array}\right\}.$$

Then

$$\begin{array}{r}{a}_{1}G(Rx,Sy,Tz)G(fx,Rx,Rx)+a_{2}G(gy,Sy,Sy)G(hz,Tz,Tz)\\ +a_{3}G(fx,Sy,Tz)G(Rx,gy,Tz)+a_{4}G(Rx,Sy,hz)G(fx,gy,Tz)\\ +a_{5}G(fx,Sy,hz)G(Rx,gy,hz)\\ \le (a_1+a_2+a_3+a_4+a_5)M(x,y,z).\end{array}$$

So, if the condition (2.2) holds, then $G^2(fx,gy,hz)\le (a_1+a_2+a_3+a_4+a_5)M(x,y,z)$. Taking $k=a_1+a_2+a_3+a_4+a_5$ in Theorem 2.1, the conclusion of Corollary 2.1 can be obtained from Theorem 2.1, since $0\le a_1+a_2+a_3+a_4+a_5<1$. □