Common Fixed Point Theorem for Six Selfmaps of a Complete G-Metric Space

J. Niranjan Goud¹, M. Rangamma²

¹Department of Mathematics, Government College for Men, Kurnool, India.
²Department of Mathematics, Osmania University, Hyderabad, India.

DOI: 10.4236/apm.2017.73015                

Abstract

By using weakly compatible conditions of selfmapping pairs, we prove a com-mon fixed point theorem for six mappings in generalized complete metric spaces. An example is provided to support our result.

Keywords

G-Metric Space, Weakly Compatible Mappings, Fixed Point, Associated Sequence of a Point Relative to Six Selfmaps

1. Introduction

The study of fixed point theory has been at the centre of vigorous activity and it has a wide range of applications in applied mathematics and sciences. Over the past two decades, a considerable amount of research work for the development of fixed point theory have executed by several authors.

In 1963, Gahler [1] [2] introduced 2-metric spaces and claimed them as generalizations of metric spaces. But many researchers proved that there was no relation between these two spaces. These considerations led Dhage [3] to initiate a study of general metric spaces called D-metric spaces. As a probable modification to D-metric spaces, Shaban Sedghi, Nabi Shobe and Haiyun Zhou [4] have introduced D*-metric spaces. In 2006, Zead Mustafa and Brailey Sims [5] initiated $G$ -metric spaces. Several researchers proved many common fixed point theorems on $G$ -metric spaces.

The purpose of this paper is to prove a common fixed point theorem for six weakly compatible selfmaps of a complete $G$ -metric space. Now we recall some basic definitions and results on $G$ -metric space.

2. Preliminaries

We begin with

Definition 2.1: ( [5] , Definition 3) Let $X$ be a non-empty set and $G:X^3\to \left[0,\infty \right)$ be a function satisfying:

(G1) $G\left(x,y,z\right)=0$ if $x=y=z.$

(G2) $0<G\left(x,x,y\right)$ for all $x,y\in X$ with $x\ne y.$

(G3) $G\left(x,x,y\right)<G\left(x,y,z\right)$ for all $x,y,z\in X$ with $y\ne z.$

(G4) $G\left(x,y,z\right)=G\left(\sigma \left(x,y,z\right)\right)$ for all $x,y,z\in X$ , where $\sigma \left(x,y,z\right)$ is a permutation of the set $\left\{x,y,z\right\}.$

And

(G5) $G\left(x,y,z\right)<G\left(x,w,w\right)+G\left(w,y,z\right)$ for all $x,y,z,w\in X.$

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

Definition 2.2: ( [5] , Definition 4) A G-metric Space $\left(X,G\right)$ is said to be symmetric if

(G6) $G\left(x,y,y\right)=G\left(x,x,y\right)$ for all $x,y\in X.$

The example given below is a non-symmetric G-metric space.

Example 2.3: ( [5] , Example 1): Let $X=\left\{a,b\right\}$ Define $G:X^3\to \left[0,\infty \right)$ by

$G\left(a,a,a\right)=G\left(b,b,b\right)=0;$ $G\left(a,a,b\right)=1,G\left(a,b,b\right)=2$ and extend $G$ to all of $X^3$ by using (G4).

Then it is easy to verify that $\left(X,G\right)$ is a G-metric space. Since $G\left(a,a,b\right)\ne G\left(a,b,b\right)$ , the space $\left(X,G\right)$ is non-symmetric, in view of (G6).

Example 2.4: Let $\left(X,d\right)$ be a metric space. Define $G_s^d:X^3\to \left[0,\infty \right)$ by

$G_s^d\left(x,y,z\right)=\frac{1}{3}\left[d\left(x,y\right)+d\left(y,z\right)+d\left(z,x\right)\right]$ for $x,y,z\in X$ .Then $\left(X,G_s^d\right)$ is a G-metric Space.

Lemma (2.5): ( [5] , p. 292) If $\left(X,G\right)$ is a G-metric space then $G\left(x,y,y\right)\le 2G\left(y,x,x\right)$ for all $x,y\in X$ .

Definition 2.6: Let $\left(X,G\right)$ be a G-metric Space. A sequence $\left\{x_n\right\}$ in $X$ is said to be G-convergent if there is a $x_0\in X$ such that to each $\epsilon >0$ there is a natural number $N$ for which $G\left(x_n,x_n,x_0\right)<\epsilon$ for all $n\ge N$ .

Lemma 2.7: ( [5] , Proposition 6) Let $\left(X,G\right)$ be a G-metric Space, then for a sequence $\left\{x_n\right\}\subseteq X$ and point $x\in X$ the following are equivalent.

(1) $\left\{x_n\right\}$ is G- convergent to $x$ .

(2) $d_G\left(x_n,x\right)\to 0$ as $n\to \infty$ (that is $\left\{x_n\right\}$ converges to $x$ relative to the metric $d_G$ ).

(3) $G\left(x_n,x_n,x\right)\to 0$ as $n\to \infty .$

(4) $G\left(x_n,x,x\right)\to 0$ as $n\to \infty .$

(5) $G\left(x_m,x_n,x\right)\to 0$ as $m,n\to \infty .$

Definition 2.8: ( [5] , Definition 8) Let $\left(X,G\right)$ be a G-metric space, then a sequence $\left\{x_n\right\}\subseteq X$ is said to be G-Cauchy if for each $\epsilon >0$ , there exists a natural number N such that $G\left(x_n,x_m,x_l\right)<\epsilon$ for all $n,m,l\ge N$ .

Note that every G-convergent sequence in a G-metric space $\left(X,G\right)$ is G- Cauchy.

Definition 2.9: ( [5] , Definition 9) A G-metric space $\left(X,G\right)$ is said to be G- complete if every G -Cauchy sequence in $\left(X,G\right)$ is G-convergent in $\left(X,G\right)$ .

Gerald Jungck [6] initiated the notion of weakly compatible mappings, as a generalization of commuting maps. We now give the definition of weakly compatibility in a G-metric space.

Definition 2.10: [7] Suppose f and g are selfmaps of a G-metric space $\left(X,G\right)$ . The pair $\left(f,g\right)$ is said to be weakly compatible if $G\left(fgx,gfx,gfx\right)=0$ whenever $G\left(fx,gx,gx\right)=0.$

3. Main Theorem

Theorem 3.1: Suppose $f,g,h,p,Q$ and $R$ are six selfmaps of a complete $G$ -metric space $\left(X,G\right)$ satisfying the following conditions.

(3.1.1) $fg\left(X\right)\subseteq R\left(X\right)$ and $hp\left(X\right)\subseteq Q\left(X\right)$ ,

(3.1.2)

$\begin{array}{c}G\left(hpx,fgy,fgy\right)\le \alpha G\left(Rx,Qy,Qy\right)+\beta \left[G\left(Rx,hpx,hpx\right)+G\left(Qy,fgy,fgy\right)\right]\\ \text{+}\gamma \left[G\left(Rx,fgy,fgy\right)+G\left(hpx,Qy,Qy\right)\right]\end{array}$

for all $x,y\in X$ and $\alpha ,\beta ,\gamma$ are non-negative real numbers such that $\alpha +2\beta +2\gamma <1$ ,

(3.1.3) one of $R\left(X\right),Q\left(X\right)$ is closed sub subset of $X$ ,

(3.1.4) $\left(fg,Q\right)$ and $\left(hp,R\right)$ are weakly compatible pairs,

(3.1.5) The pairs $\left(h,p\right),\left(h,R\right),\left(f,g\right),$ and $\left(f,Q\right)$ are commuting.

Then $f,g,h,p,Q$ and $R$ have a unique common fixed point in $X$ .

Proof: Let $x_0\in X$ be an arbitrary point. Since $fg\left(X\right)\subseteq R\left(X\right)$ and $hp\left(X\right)\subseteq Q\left(X\right)$ there exists $x_1,x_2\in X$ such that $hp{x}_0=Q{x}_1$ and $fg{x}_1=R{x}_2$ again there exists $x_3,x_4\in X$ such that $hp{x}_2=Q{x}_3$ and $fg{x}_3=R{x}_4,$ continuing in the same manner for each $n\ge 0,$ we obtain a sequence $\left\{x_n\right\}$ in X such that

$y_{2n}=hp{x}_{2n}=Q{x}_{2n+1},y_{2n+1}=fg{x}_{2n+1}=R{x}_{2n+2}\text{for}n\ge 0.$ (3.1.6)

From condition (3.1.2), we have

$\begin{array}{c}G\left(y_{2n},y_{2n+1},y_{2n+1}\right)=G\left(hp{x}_{2n},fg{x}_{2n+1},fg{x}_{2n+1}\right)\\ \le \alpha G\left(R{x}_{2n},Q{x}_{2n+1},Q{x}_{2n+1}\right)+\beta [G\left(R{x}_{2n},hp{x}_{2n},hp{x}_{2n}\right)+G\left(Q{x}_{2n+1},fg{x}_{2n+1},fg{x}_{2n+1}\right)]\\ +\gamma \left[G\left(R{x}_{2n},fg{x}_{2n+1},fg{x}_{2n+1}\right)+G\left(hp{x}_{2n},Q{x}_{2n+1},Q{x}_{2n+1}\right)\right]\\ =\alpha G\left(y_{2n-1},y_{2n},y_{2n}\right)+\beta \left[G\left(y_{2n-1},y_{2n},y_{2n}\right)+G\left(y_{2n},y_{2n+1},y_{2n+1}\right)\right]\\ +\gamma \left[G\left(y_{2n-1},y_{2n+1},y_{2n+1}\right)+G\left(y_{2n},y_{2n},y_{2n}\right)\right]\\ \le \left(\alpha +\beta +\gamma \right)G\left(y_{2n-1},y_{2n},y_{2n}\right)+\left(\beta +\gamma \right)G\left(y_{2n},y_{2n+1},y_{2n+1}\right).\end{array}$

Therefore

$\begin{array}{c}\left(1-\beta -\gamma \right)G\left(y_{2n},y_{2n+1},y_{2n+1}\right)\le \left(\alpha +\beta +\gamma \right)G\left(y_{2n-1},y_{2n},y_{2n}\right)\\ G\left(y_{2n},y_{2n+1},y_{2n+1}\right)\le \frac{\left(\alpha +\beta +\gamma \right)}{\left(1-\beta -\gamma \right)}G\left(y_{2n-1},y_{2n},y_{2n}\right)\\ G\left(y_{2n},y_{2n+1},y_{2n+1}\right)\le kG\left(y_{2n-1},y_{2n},y_{2n}\right)\end{array}$ (3.1.7)

where $k=\frac{\left(\alpha +\beta +\gamma \right)}{\left(1-\beta -\gamma \right)}<1$ .

Similarly, we can show that

$G\left(y_{2n+1},y_{2n+2},y_{2n+2}\right)\le kG\left(y_{2n},y_{2n+1},y_{2n+1}\right).$ (3.18)

From (3.1.7) and (3.1.8) we have

$G\left(y_n,y_{n+1},y_{n+1}\right)\le kG\left(y_{n-1},y_n,y_n\right)\le \cdots \le k^{n}G\left(y_0,y_1,y_1\right).$

Now for every $n,m\in N$ such that $m>n$ we have

$\begin{array}{c}G\left(y_n,y_m,y_m\right)\le G\left(y_n,y_{n+1},y_{n+1}\right)+G\left(y_{n+1},y_{n+2},y_{n+2}\right)+\cdots +G\left(y_{m-1},y_m,y_m\right)\\ \le k^{n}G\left(y_0,y_1,y_1\right)+k^{n+1}G\left(y_0,y_1,y_1\right)+\cdots +k^{m-1}G\left(y_0,y_1,y_1\right)\\ \le k^n\left(1+k+k^2+\cdots +k^{m-n+1}\right)G\left(y_0,y_1,y_1\right)\\ \le k^n\frac{\left(1-k^{m-n}\right)}{1-k}G\left(h{x}_0,h{x}_1,h{x}_1\right)\to 0\text{as}n\to \infty .\end{array}$

Since $k<1.$

Therefore, $\left\{y_n\right\}$ is a Cauchy sequence in $X$ . Since $X$ is a complete G-metric space, then there exists a point $z\in X$ such that

$\underset{n\to \infty }{\lim }hp{x}_{2n}=\underset{n\to \infty }{\lim }Q{x}_{2n+1}=\underset{n\to \infty }{\lim }fg{x}_{2n+1}=\underset{n\to \infty }{\lim }R{x}_{2n+2}=z.$ (3.1.9)

If $R\left(X\right)$ is a closed subset of $X$ , then there exists a point $u\in X$ such that $z=Ru$ .

Now from (3.1.2), we have

$\begin{array}{c}G\left(hpu,fg{x}_{2n+1},fg{x}_{2n+1}\right)\le \alpha G\left(Ru,Q{x}_{2n+1},Q{x}_{2n+1}\right)+\beta \left[G\left(Ru,hpu,hpu\right)+G\left(Q{x}_{2n+1},fg{x}_{2n+1},fg{x}_{2n+1}\right)\right]\\ +\gamma \left[G\left(Ru,fg{x}_{2n+1},fg{x}_{2n+1}\right)+G\left(hpu,Q{x}_{2n+1},Q{x}_{2n+1}\right)\right].\end{array}$ (3.1.10)

Letting $n\to \infty$ in (3.1.10) and by the continuity of G we have

$\begin{array}{c}G\left(hpu,z,z\right)\le \alpha G\left(z,z,z\right)+\beta \left[G\left(z,hpu,hpu\right)+G\left(z,z,z\right)\right]\\ +\gamma \left[G\left(z,z,z\right)+G\left(hpu,z,z\right)\right]\\ \le \left(2\beta +\gamma \right)G\left(hpu,z,z\right),\end{array}$

which leads to a contradiction as $2\beta +\gamma <1$ .

Hence $G\left(hpu,z,z\right)=0,$ which implies $hpu=z.$

Therefore,

$hpu=Ru=z.$ (3.1.11)

Now since $hp\left(X\right)\subseteq Q\left(X\right)$ then there exists a point $v\in X$ such that $z=Qv.$

Then we have by (3.1.2)

$\begin{array}{c}G\left(hpu,fgv,fgv\right)\le \alpha G\left(Ru,Qv,Qv\right)+\beta \left[G\left(Ru,hpu,hpu\right)+G\left(Qv,fgv,fgv\right)\right]\\ +\gamma \left[G\left(Ru,fgv,fgv\right)+G\left(hpu,Qv,Qv\right)\right]\end{array}$ (3.1.12)

$\begin{array}{c}G\left(z,fgv,fgv\right)\le \alpha G\left(z,z,z\right)+\beta \left[G\left(z,z,z\right)+G\left(z,fgv,fgv\right)\right]\\ +\gamma \left[G\left(z,fgv,fgv\right)+G\left(z,z,z\right)\right]\\ \le \left(\beta +\gamma \right)G\left(z,fgv,fgv\right),\end{array}$

which leads to a contradiction, since $\beta +\gamma <1$ . Hence $fgv=z.$

Therefore,

$fgv=Qv=z.$ (3.1.13)

From (3.1.11) and (3.1.13) we have $Ru=hpu=fgv=Qv=z.$

Since the pair $\left(fg,Q\right)$ is weakly compatible then $fgQv=Qfgv$ which gives $fgz=Qz.$

Now (3.1.2) we have

$\begin{array}{c}G\left(z,fgz,fgz\right)=G\left(hpu,fgz,fgz\right)\\ \le \alpha G\left(Ru,Qz,Qz\right)+\beta \left[G\left(Ru,hpu,hpu\right)+G\left(Qz,fgz,fgz\right)\right]\\ +\gamma \left[G\left(Ru,fgz,fgz\right)+G\left(hpu,Qz,Qz\right)\right]\\ =\alpha G\left(z,fgz,fgz\right)+\beta \left[G\left(z,z,z\right)+G\left(fgz,fgz,fgz\right)\right]\\ +\gamma \left[G\left(z,fgz,fgz\right)+G\left(z,fgz,fgz\right)\right]\\ =\left(\alpha +2\gamma \right)G\left(z,fgz,fgz\right)\end{array}$

which is a contradiction, since $\alpha +2\gamma <1.$ Hence $G\left(z,fgz,fgz\right)=0$ thus $fgz=z.$

Showing that $z$ is a common fixed point of $fg$ and $Q.$

Since the pair $\left(hp,R\right)$ is weakly compatible then $hpRu=Rhpu$ which gives $hpz=Rz.$

Then we have by (3.1.2)

$\begin{array}{c}G\left(hpz,z,z\right)=G\left(hpz,fgz,fgz\right)\\ \le \alpha G\left(Rz,Qz,Qz\right)+\beta \left[G\left(Rz,hpz,hpz\right)+G\left(Qz,fgz,fgz\right)\right]\\ +\gamma \left[G\left(Rz,fgz,fgz\right)+G\left(hpz,Qz,Qz\right)\right]\\ =\alpha G\left(hpz,z,z\right)+\beta \left[G\left(hpz,hpz,hpz\right)+G\left(z,z,z\right)\right]\\ +\gamma \left[G\left(hpz,z,z\right)+G\left(hpz,z,z\right)\right]\\ =\left(\alpha +2\gamma \right)G\left(hpz,z,z\right),\end{array}$

which is a contradiction, since $\alpha +2\gamma <1$ . Hence $G\left(hpz,z,z\right)=0$ thus $hpz=z.$

Showing that $z$ is a common fixed point of $hp$ and $R.$

Therefore, $z$ is a common fixed point of $fg$ , $hp$ , $R$ and $Q.$

By commuting conditions of the pairs in (3.1.5), we have

$fz=f\left(fgz\right)=f\left(gfz\right)=fg\left(fz\right),fz=f\left(Qz\right)=Q\left(fz\right).$

And

$hz=h\left(hpz\right)=h\left(phz\right)=hp\left(hz\right),hz=h\left(Rz\right)=R\left(hz\right).$

From (3.1.2)

$\begin{array}{c}G\left(z,fz,fz\right)=G\left(hpz,fgfz,fgfz\right)\\ \le \alpha G\left(Rz,Qfz,Qfz\right)+\beta \left[G\left(Rz,hpz,hpz\right)+G\left(Qfz,fgfz,fgfz\right)\right]\\ +\gamma \left[G\left(Rz,fgfz,fgfz\right)+G\left(hpz,Qfz,Qfz\right)\right]\\ =\alpha G\left(z,fz,fz\right)+\beta \left[G\left(z,z,z\right)+G\left(fz,fz,fz\right)\right]\\ +\gamma \left[G\left(z,fz,fz\right)+G\left(z,fz,fz\right)\right]\\ =\left(\alpha +2\gamma \right)G\left(z,fz,fz\right).\end{array}$

Since $\alpha +2\gamma <1,$ we have $G\left(z,fz,fz\right)=0$ thus $fz=z.$

Also $gz=gfz=fgz=z.$

Therefore, we have $fz=gz=Rz=fgz=z.$

Similarly, we have $hz=pz=Qz=hpz=z.$

Therefore, $z$ is a common fixed point of $f,g,h,p,Q$ and $R.$

The proof is similar in case if $Q\left(X\right)$ is a closed subset of $X.$

We now prove the uniqueness of the common fixed point.

If possible, assume that $w$ is another common fixed point of $f,g,h,p,Q$ and $R$ .

By condition (3.1.2) we have

$\begin{array}{c}G\left(z,w,w\right)=G\left(hpz,fgw,fgw\right)\\ \le \alpha G\left(Rz,Qw,Qw\right)+\beta \left[G\left(Rz,hpz,hpz\right)+G\left(Qw,fgw,fgw\right)\right]\\ +\gamma \left[G\left(Rz,fgw,fgw\right)+G\left(hpz,Qw,Qw\right)\right]\\ =\alpha G\left(z,w,w\right)+\beta \left[G\left(z,z,z\right)+G\left(w,w,w\right)\right]+\gamma \left[G\left(z,w,w\right))+G\left(z,w,w\right)\right]\\ =\left(\alpha +2\gamma \right)G\left(z,w,w\right),\end{array}$

which is a contradiction, since $\alpha +2\gamma <1$ .

Hence $G\left(z,w,w\right)=0$ which gives $z=w$ .

Therefore, $z$ is a unique common fixed point of $f,g,h,p,Q$ and $R$ .

As an example, we have the following.

3.1. Example

Let $X=\left[0,1\right]$ with $G\left(x,y,z\right)=\left|x-y\right|+\left|y-z\right|+\left|z-x\right|$ for $x,y,z\in X$ . Then G is a G-metric on $X$ .

Define

$f:X\to X,g:X\to X,h:X\to X,p:X\to X,Q:X\to X,R:X\to X$

by

$\begin{array}{l}fx=hx=\frac{x+1}{3},\forall x\in X,\\ gx=px=\frac{3x+1}{5},\forall x\in X,\\ Qx=Rx=x,\forall x\in X.\end{array}$

$fgx=f\left(\frac{3x+1}{5}\right)=\frac{x+2}{5},hpx=h\left(\frac{3x+1}{5}\right)=\frac{x+2}{5},$

$fgX=\left[\frac{2}{5},\frac{3}{5}\right],hpX=\left[\frac{2}{5},\frac{3}{5}\right],RX=\left[0,1\right],QX=\left[0,1\right]$

$fgX\subseteq RX,hpX\subseteq QX.$

Proving the condition (3.1.1) of the Theorem (3.1).

$RX$ and $QX$ are closed subsets of $X$ . Proving the condition (3.1.3) of the Theorem (3.1).

Since $fg\left(\frac{1}{2}\right)=\frac{1}{2}$ and $Q\left(\frac{1}{2}\right)=\frac{1}{2}$ then $fgQ\left(\frac{1}{2}\right)=Qfg\left(\frac{1}{2}\right),$ showing that the pair $\left(fg,Q\right)$ is weakly compatible.

Also, the pair $\left(hp,R\right)$ is weakly compatible.

Proving the condition (3.1.4) of the Theorem (3.1).

$\begin{array}{l}hp\left(x\right)=\frac{x+2}{5}=ph\left(x\right),hR\left(x\right)=h\left(x\right)=Rh\left(x\right),\\ fg\left(x\right)=\frac{x+2}{5}=gf\left(x\right),fQ\left(x\right)=f(x)=Qf\left(x\right),\end{array}$

showing that $\left(h,R\right),\left(f,Q\right),\left(h,p\right)$ and $\left(f,g\right)$ are commuting pairs.

Proving the condition (3.1.5) of the Theorem (3.1).

Now we prove the condition (3.1.2) of the Theorem (3.1).

On taking $\alpha =\frac{1}{10},\beta =\frac{1}{8},\gamma =\frac{1}{12}$ then $\alpha +2\beta +2\gamma =\frac{31}{60}<1.$

Now $G\left(hpx,fgy,fgy\right)=2\left|hpx-fgy\right|=\frac{2}{5}\left|x-y\right|$

$\begin{array}{l}G\left(Rx,Qy,Qy\right)=2\left|Rx-Qy\right|=2\left|x-y\right|,\\ G\left(Rx,hpx,hpx\right)=2\left|Rx-hpx\right|=\frac{4}{5}\left|2x-1\right|,\\ G\left(Qy,fgy,fgy\right)=2\left|fgy-Qy\right|=\frac{4}{5}\left|1-2y\right|,\\ G\left(Rx,fgy,fgy\right)=2\left|Rx-fgy\right|=\frac{2}{5}\left|5x-y-2\right|,\\ G\left(hpx,Qy,Qy\right)=2\left|hpx-Qy\right|=\frac{2}{5}\left|x+2-5y\right|\end{array}$

$\begin{array}{l}\alpha G\left(Rx,Qy,Qy\right)+\beta \left[G\left(Rx,hpx,hpx\right)+G\left(Qy,fgy,fgy\right)\right]\\ +\gamma \left[G\left(Rx,fgy,fgy\right)+G\left(hpx,Qy,Qy\right)\right]\\ =2\alpha \left|x-y\right|+\frac{4}{5}\beta \left(\left|2x-1\right|+\left|1-2y\right|\right)+\frac{2}{5}\gamma \left(\left|5x-y-2\right|+\left|x-5y-2\right|\right)\\ \ge 2\alpha \left|x-y\right|+\frac{4}{5}\beta \left|2x-2y\right|+\frac{2}{5}\gamma \left|6x-6y\right|\\ =\left(2\alpha +\frac{8\beta }{5}+\frac{12}{5}\gamma \right)\left|x-y\right|\\ =\frac{3}{5}\left|x-y\right|\ge \frac{2}{5}\left|x-y\right|=G\left(fgx,hpy,hpy\right).\end{array}$

Therefore,

$\begin{array}{c}G\left(hpx,fgy,fgy\right)\le \alpha G\left(Rx,Qy,Qy\right)+\beta \left[G\left(Rx,hpx,hpx\right)+G\left(Qy,fgy,fgy\right)\right]\\ +\gamma \left[G\left(Rx,fgy,fgy\right)+G\left(hpx,Qy,Qy\right)\right].\end{array}$

Proving the condition (3.1.2) of the Theorem (3.1).

Hence all the conditions of the Theorem (3.1) are satisfied.

Therefore, $\frac{1}{2}$ is a unique common fixed point of $f,g,h,p,Q$ and $R$ .

3.2. Corollary

Suppose $f,p,Q$ and $R$ are four selfmaps of a complete $G$ -metric space $\left(X,G\right)$ satisfying the following conditions:

(3.1.1) $f\left(X\right)\subseteq R\left(X\right)$ and $p\left(X\right)\subseteq Q\left(X\right)$ ,

(3.1.2) $\begin{array}{c}G\left(px,fy,fy\right)\le \alpha G\left(Rx,Qy,Qy\right)+\beta \left[G\left(Rx,px,px\right)+G\left(Qy,fy,fy\right)\right]\\ +\gamma \left[G\left(Rx,fy,fy\right)+G\left(px,Qy,Qy\right)\right]\end{array}$

for all $x,y\in X$ and $\alpha ,\beta ,\gamma$ are non-negative real numbers such that $\alpha +2\beta +2\gamma <1$ ,

(3.1.3) One of $R\left(X\right),Q\left(X\right)$ is closed sub subset of X,

(3.1.4) $\left(p,R\right)$ and $\left(f,Q\right)$ are weakly compatible pairs,

Then $f,p,Q$ and $R$ have a unique common fixed point in $X$ .

Proof: Follows from the Theorem (3.1) if $g=h=I$ the identity map.

3.3. Corollary

Suppose $f,p$ and $R$ are three selfmaps of a complete $G$ -metric space $\left(X,G\right)$ satisfying the following conditions:

(3.1.1) $f\left(X\right)\subseteq R\left(X\right)$ and $p\left(X\right)\subseteq R\left(X\right)$ ,

(3.1.2) $\begin{array}{c}G\left(px,fy,fy\right)\le \alpha G\left(Rx,Ry,Ry\right)+\beta \left[G\left(Rx,px,px\right)+G\left(Ry,fy,fy\right)\right]\\ +\gamma \left[G\left(Rx,fy,fy\right)+G\left(px,Ry,Ry\right)\right]\end{array}$

for all $x,y\in X$ and $\alpha ,\beta ,\gamma$ are non-negative real numbers such that $\alpha +2\beta +2\gamma <1$ ,

(3.1.3) $R\left(X\right)$ is closed sub subset of X,

(3.1.4) $\left(p,R\right)$ and $\left(f,R\right)$ are weakly compatible pairs.

Then $f,p$ and $R$ have a unique common fixed point in $X$ .

Proof: Follows from the Theorem (3.1) if $g=h=I$ the identity map, and $Q=R$ .

Conflicts of Interest

The authors declare no conflicts of interest.

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