Theory and Modern Applications

# Common fixed point results for three maps in generalized metric space

## Abstract

Mustafa and Sims [Fixed Point Theory Appl. 2009, Article ID 917175, 10, (2009)] generalized a concept of a metric space and proved fixed point theorems for mappings satisfying different contractive conditions. In this article, we extend and generalize the results obtained by Mustafa and Sims and prove common fixed point theorems for three maps in these spaces. It is worth mentioning that our results do not rely on continuity and commutativity of any mappings involved therein. We also introduce the notation of a generalized probabilistic metric space and obtain common fixed point theorem in the frame work of such spaces.

2000 Mathematics Subject Classification: 47H10.

## 1. Introduction and Preliminaries

The study of fixed points of mappings satisfying certain contractive conditions has been at the center of vigorous research activity. Mustafa and Sims [1] generalized the concept of a metric space. Based on the notion of generalized metric spaces, Mustafa et al. [25] obtained some fixed point theorems for mappings satisfying different contractive conditions. Abbas and Rhoades [6] motivated the study of a common fixed point theory in generalized metric spaces. Recently, Saadati et al. [7] proved some fixed point results for contractive mappings in partially ordered G-metric spaces.

The purpose of this article is to initiate the study of common fixed point for three mappings in complete G-metric space. It is worth mentioning that our results do not rely on the notion of continuity, weakly commuting, or compatibility of mappings involved therein. We generalize various results of Mustafa et al. [3, 5].

Consistent with Mustafa and Sims [1], the following definitions and results will be needed in the sequel.

Definition 1.1. Let X be a nonempty set. Suppose that a mapping G : X × X × XR + satisfies:

1. (a)

G(x, y, z) = 0 if and only if x = y = z,

2. (b)

0 < G(x, y, z) for all x, y X, with xy,

3. (c)

G(x, x, y) ≤ G(x, y, z) for all x, y, z X, with zy,

4. (d)

G(x, y, z) = G(x, z, y) = G(y, z, x) = (symmetry in all three variables), and

5. (e)

G(x, y, z) ≤ G(x, a, a) + G(a, y, z) for all x, y, z, a X.

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

Definition 1.2. A G-metric is said to be symmetric if G(x, y, y) = G(y, x, x) for all x, y X.

Definition 1.3. Let (X, G) be a G-metric space. We say that {x n } is

1. (i)

a G-Cauchy sequence if, for any ε > 0, there is an n 0 N (the set of all positive integers) such that for all n, m, ln 0, G(x n , x m , x l ) < ε;

2. (ii)

a G-Convergent sequence if, for any ε > 0, there is an x X and an n 0 N, such that for all n, mn 0, G(x, x n , x m ) < ε.

A G-metric space X is said to be complete if every G-Cauchy sequence in X is convergent in X. It is known that {x n } converges to x (X, G) if and only if G(x m , x n , x) → 0 as n, m → ∞.

Proposition 1.4. Every G-metric space (X, G) will define a metric space (X, d G ) by

$\begin{array}{cc}\hfill {d}_{G}\left(x,y\right)=G\left(x,y,y\right)+G\left(y,x,x\right),\hfill & \hfill \forall \hfill \end{array}x,y\in X.$

Definition 1.5. Let (X, G) and (X′, G′) be G-metric spaces and let f : (X, G) → (X′, G′) be a function, then f is said to be G-continuous at a point a X if and only if, given ε > 0, there exists δ > 0 such that x, y X; and G(a, x, y) < δ implies G′(f(a), f(x), f(y)) < ε. A function f is G-continuous at X if and only if it is G-continuous at all a X.

## 2. Common Fixed Point Theorems

In this section, we obtain common fixed point theorems for three mappings defined on a generalized metric space. We begin with the following theorem which generalize [[5], Theorem 1].

Theorem 2.1. Let f, g, and h be self maps on a complete G-metric space X satisfying

$G\left(fx,gy,hz\right)\le kU\left(x,y,z\right)$
(2.1)

where $k\in \left[0,\frac{1}{2}\right)$ and

$\begin{array}{ccc}\hfill U\left(x,y,z\right)\hfill & \hfill =\hfill & \hfill max\left\{G\left(x,y,z\right),G\left(fx,fx,x\right),G\left(y,gy,gy\right),G\left(z,hz,hz\right),\hfill \\ \hfill G\left(x,gy,gy\right),G\left(y,hz,hz\right),G\left(z,fx,fx\right)\hfill \end{array}$

for all x, y, z X. Then f, g, and h have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. Suppose x 0 is an arbitrary point in X. Define {x n } by x 3n+1= fx 3n , x 3n+2= gx 3n+1, x 3n+3= hx 3n+2for n ≥ 0. We have

$\begin{array}{ll}\hfill G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left(f{x}_{3n},g{x}_{3n+1},h{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kU\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\end{array}$

for n = 0, 1, 2, ..., where

$\begin{array}{l}U\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left(f{x}_{3n},f{x}_{3n},{x}_{3n}\right),G\left({x}_{3n+1},g{x}_{3n+1},g{x}_{3n+1}\right),\\ G\left({x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right),G\left({x}_{3n},g{x}_{3n+1},g{x}_{3n+1}\right),\\ G\left({x}_{3n+1},h{x}_{3n+2},h{x}_{3n+2}\right),G\left({x}_{3n+2},f{x}_{3n},f{x}_{3n}\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left({x}_{3n+1},{x}_{3n+1},{x}_{3n}\right),G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+2}\right),\\ G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right),G\left({x}_{3n},{x}_{3n+2},{x}_{3n+2}\right),\\ G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right),G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right)\right\}\\ \le \phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),\\ G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right),G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),\\ G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right),\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\right\}\end{array}$

In case max{G(x 3n , x 3n+1, x 3n+2), G(x 3n+1, x 3n+2, x 3n+3)} = G(x 3n , x 3n+1, x 3n+2), we obtain that

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

If max{G(x 3n , x 3n+1, x 3n+2), G(x 3n+1, x 3n+2, x 3n+3)} = G(x 3n+1, x 3n+2, x 3n+3), then

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right),$

which implies that G(x 3n+1, x 3n+2, x 3n+3) = 0, and x 3n+1= x 3n+2= x 3n+3and the result follows immediately.

Hence,

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Similarly it can be shown that

$G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right)\le kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)$

and

$G\left({x}_{3n+3},{x}_{3n+4},{x}_{3n+5}\right)\le kG\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right).$

Therefore, for all n,

$\begin{array}{ll}\hfill G\left({x}_{n+1},{x}_{n+2},{x}_{n+3}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}kG\left({x}_{n},{x}_{n+1},{x}_{n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\cdots \le {k}^{n+1}G\left({x}_{0},{x}_{1},{x}_{2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Now, for any l, m, n with l > m > n,

$\begin{array}{ll}\hfill G\left({x}_{n},{x}_{m},{x}_{l}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}G\left({x}_{n},{x}_{n+1},{x}_{n+1}\right)+G\left({x}_{n+1},{x}_{n+1},{x}_{n+2}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\cdots +G\left({x}_{l-1},{x}_{l-1},{x}_{l}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}G\left({x}_{n},{x}_{n+1},{x}_{n+2}\right)+G\left({x}_{n},{x}_{n+1},{x}_{n+2}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\cdots +G\left({x}_{l-2},{x}_{l-1},{x}_{l}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\left[{k}^{n}+{k}^{n+1}+\cdots +{k}^{l}\right]G\left({x}_{0},{x}_{1},{x}_{2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{{k}^{n}}{1-k}G\left({x}_{0},{x}_{1},{x}_{2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

The same holds if l = m > n and if l > m = n we have

$G\left({x}_{n},{x}_{m},{x}_{l}\right)\le \frac{{k}^{n-1}}{1-k}G\left({x}_{0},{x}_{1},{x}_{2}\right).$

Consequently G(x n , x m , x l ) → 0 as n, m, l → ∞. Hence {x n } is a G-Cauchy sequence. By G-completeness of X, there exists u X such that {x n } converges to u as n → ∞. We claim that fu = u. If not, then consider

$G\left(fu,{x}_{3n+2},{x}_{3n+3}\right)=G\left(fu,g{x}_{3n+1},h{x}_{3n+2}\right)\le kU\left(u,{x}_{3n+1},{x}_{3n+2}\right),$

where

$\begin{array}{l}U\left(u,{x}_{3n+1},{x}_{3n+2}\right)\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,{x}_{3n+1},{x}_{3n+2}\right),G\left(fu,fu,u\right),G\left({x}_{3n+1},g{x}_{3n+1},g{x}_{3n+1}\right),\\ G\left({x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right),G\left(u,g{x}_{3n+1},g{x}_{3n+2}\right),\\ G\left({x}_{3n+1},h{x}_{3n+2},h{x}_{3n+2}\right),G\left({x}_{3n+2},fu,fu\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,{x}_{3n+1},{x}_{3n+2}\right),G\left(fu,fu,u\right),G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+2}\right),\\ G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right),G\left(u,{x}_{3n+2},{x}_{3n+2}\right),\\ G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right),G\left({x}_{3n+2},fu,fu\right)\right\}.\end{array}$

On taking limit n → ∞, we obtain that

$G\left(fu,u,u\right)\le kU\left(u,u,u\right),$

where

$\begin{array}{c}U\left(u,u,u\right)\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,u,u\right),G\left(fu,fu,u\right),G\left(u,u,u\right),G\left(u,u,u\right)\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(u,u,u\right),G\left(u,u,u\right),G\left(u,fu,fu\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}G\left(fu,fu,u\right).\end{array}$

Thus

$G\left(fu,u,u\right)\le kG\left(fu,fu,u\right)\le 2kG\left(fu,u,u\right),$

a contradiction. Hence, fu = u. Similarly it can be shown that gu = u and hu = u. To prove the uniqueness, suppose that if v is another common fixed point of f, g, and h, then

$G\left(u,v,v\right)=G\left(fu,gv,hv\right)\le kU\left(u,v,v\right),$

where

$\begin{array}{c}U\left(u,v,v\right)\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,v,v\right),G\left(fu,fu,u\right),G\left(v,gv,gv\right),G\left(v,hv,hv\right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(u,gv,gv\right),G\left(v,hv,hv\right),G\left(v,fu,fu\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,v,v\right),G\left(u,u,u\right),G\left(v,v,v\right),G\left(v,v,v\right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(u,v,v\right),G\left(v,v,v\right),G\left(v,u,u\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(u,v,v\right),G\left(v,u,u\right)\right\}\end{array}$

If U(u, v, v) = G(u, v, v), then

$G\left(u,v,v\right)\le kG\left(u,v,v\right),$

which gives that G(u, v, v) = 0, and u = v. Also for U(u, v, v) = G(v, u, u) we obtain

$G\left(u,v,v\right)\le kG\left(v,u,u\right)\le 2kG\left(u,v,v\right),$

which gives that G(u, v, v) = 0 and u = v. Hence, u is a unique common fixed point of f, g, and h.

Now suppose that for some p in X, we have f(p) = p. We claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$G\left(p,gp,hp\right)=G\left(fp,gp,hp\right)\le kU\left(p,p,p\right),$

where

$\begin{array}{c}U\left(p,p,p\right)\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(p,p,p\right),G\left(fp,fp,p\right),G\left(p,gp,gp\right),G\left(p,hp,hp\right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(p,gp,gp\right),G\left(p,hp,hp\right),G\left(p,fp,fp\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left(p,gp,gp\right),G\left(p,hp,hp\right)\right\}.\end{array}$

Now U(p, p, p) = G(p, gp, gp) gives

$G\left(p,gp,hp\right)\le kG\left(p,gp,gp\right)\le kG\left(p,gp,hp\right),$

a contradiction. For U(p, p, p) = G(p, hp, hp), we obtain

$G\left(p,gp,hp\right)\le kG\left(p,hp,hp\right)\le kG\left(p,gp,hp\right),$

a contradiction. Similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence, in all cases, we conclude that p = gp = hp. The same conclusion holds if p = gp or p = hp.   □

Example 2.2. Let X = {0, 1, 2, 3} be a set equipped with G-metric defined by

$\begin{array}{cc}\hfill \left(x,y,z\right)\hfill & \hfill G\left(x,y,z\right)\hfill \\ \hfill \left(0,0,0\right),\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right),\hfill & \hfill 0\hfill \\ \hfill \left(0,0,2\right),\left(0,2,0\right),\left(2,0,0\right),\left(0,2,2\right),\left(2,0,2\right),\left(2,2,0\right),\hfill & \hfill 1\hfill \\ \hfill \begin{array}{c}\hfill \left(0,0,1\right),\left(0,1,0\right),\left(1,0,0\right),\left(0,1,1\right),\left(1,0,1\right),\left(1,1,0\right),\hfill \\ \hfill \left(0,0,3\right),\left(0,3,0\right),\left(3,0,0\right),\left(0,3,3\right),\left(3,0,3\right),\left(3,3,0\right),\hfill \\ \hfill \left(1,1,2\right),\left(1,2,1\right),\left(2,1,1\right),\left(1,2,2\right),\left(2,1,2\right),\left(2,2,1\right),\hfill \\ \hfill \left(1,1,3\right),\left(1,3,1\right),\left(3,1,1\right),\left(1,3,3\right),\left(3,1,3\right),\left(3,3,1\right),\hfill \\ \hfill \left(2,2,3\right),\left(2,3,2\right),\left(3,2,2\right),\left(2,3,3\right),\left(3,2,3\right),\left(3,3,2\right),\hfill \end{array}\hfill & \hfill 3\hfill \\ \hfill \begin{array}{c}\hfill \left(0,1,2\right),\left(0,1,3\right),\left(0,2,1\right),\left(0,2,3\right),\left(0,3,1\right),\left(0,3,2\right),\hfill \\ \hfill \left(1,0,2\right),\left(1,0,3\right),\left(1,2,0\right),\left(1,2,3\right),\left(1,3,0\right),\left(1,3,2\right),\hfill \\ \hfill \left(2,0,1\right),\left(2,0,3\right),\left(2,1,0\right),\left(2,1,3\right),\left(2,3,0\right),\left(2,3,1\right),\hfill \\ \hfill \left(3,0,1\right),\left(3,0,2\right),\left(3,1,0\right),\left(3,1,2\right),\left(3,2,0\right),\left(3,2,1\right),\hfill \end{array}\hfill & \hfill 3\hfill \end{array}$

and f, g, h : XX be defined by

$\begin{array}{cccc}\hfill x\hfill & \hfill f\left(x\right)\hfill & \hfill g\left(x\right)\hfill & \hfill h\left(x\right)\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}$

It may be verified that the mappings satisfy contractive condition (2.1) with contractivity factor equal to $\frac{1}{3}$. Moreover, 0 is a common fixed point of mappings f, g, and h.

Corollary 2.3. Let f, g, and h be self maps on a complete G-metric space X satisfying

$\begin{array}{c}G\left({f}^{m}x,{g}^{m}y,{h}^{m}z\right)\phantom{\rule{0.5em}{0ex}}\le \phantom{\rule{0.5em}{0ex}}k\mathrm{max}\left\{G\left(x,y,z\right),G\left({f}^{m}x,{f}^{m}x,x\right),G\left(y,{g}^{m}y,{g}^{m}y\right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(z,{h}^{m}z,{h}^{m}z\right),G\left(x,{g}^{m}y,{g}^{m}y\right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}G\left(y,{h}^{m}z,{h}^{m}z\right),G\left(z,{f}^{m}x,{f}^{m}x\right)\right\}\end{array}$
(2.2)

for all x, y, z X, where $k\in \left[0,\frac{1}{2}\right)$ . Then f, g, and h have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. It follows from Theorem 2.1, that f m , g m and h m have a unique common fixed point p. Now f(p) = f(f m (p)) = f m+1(p) = f m (f(p)), g(p) = g(g m (p)) = g m+1(p) = g m (g(p)) and h(p) = h(h m (p)) = h m+1(p) = h m (h(p)) implies that f(p), g(p) and h(p) are also fixed points for f m , g m and h m . Now we claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left({f}^{m}p,{g}^{m}\left(gp\right),{h}^{m}\left(hp\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kmax\left\{G\left(p,gp,hp\right),G\left({f}^{m}p,{f}^{m}p,p\right),G\left(gp,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(hp,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right),G\left(p,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(gp,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right),G\left(hp,{f}^{m}p,{f}^{m}p\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(p,gp,hp\right),G\left(p,p,p\right),G\left(gp,gp,gp\right),G\left(hp,hp,hp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,gp,gp\right),G\left(gp,hp,hp\right),G\left(hp,p,p\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(p,gp,hp\right),G\left(gp,hp,hp\right),G\left(hp,p,p\right)\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kG\left(p,gp,hp\right),\phantom{\rule{2em}{0ex}}\end{array}$

which is a contradiction. Similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence in all cases, we conclude that, f(p) = g(p) = h(p) = p. It is obvious that every fixed point of f is a fixed point of g and h and conversely.   □

Theorem 2.4. Let f, g, and h be self maps on a complete G-metric space X satisfying

$G\left(fx,gy,hz\right)\le kU\left(x,y,z\right),$
(2.3)

where $k\in \left[0,\frac{1}{3}\right)$ and

$\begin{array}{ll}\hfill U\left(x,y,z\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(y,fx,fx\right)+G\left(x,gy,gy\right),G\left(z,gy,gy\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+G\left(y,hz,hz\right),G\left(z,fx,fx\right)+G\left(x,hz,hz\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, y, z X. Then f, g, and h have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. Suppose x 0 is an arbitrary point in X. Define {x n } by x 3n+1= fx 3n , x 3n+2= gx 3n+1, x 3n+3= hx 3n+2. We have

$\begin{array}{ll}\hfill G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left(f{x}_{3n},g{x}_{3n+1},h{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kU\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\end{array}$

for n = 0, 1, 2, ..., where

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}U\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left({x}_{3n+1},f{x}_{3n},f{x}_{3n}\right)+G\left({x}_{3n},g{x}_{3n+1},g{x}_{3n+1}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+2},g{x}_{3n+1},g{x}_{3n+1}\right)+G\left({x}_{3n+1},h{x}_{3n+2},h{x}_{3n+2}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+2},f{x}_{3n},f{x}_{3n}\right)+G\left({x}_{3n},h{x}_{3n+2},h{x}_{3n+2}\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left({x}_{3n+1},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n},{x}_{3n+2},{x}_{3n+2}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+2},{x}_{3n+2},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)\right\}\phantom{\rule{2em}{0ex}}\\ \le max\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

Now if U(x 3n , x 3n+1, x 3n+2) = G(x 3n , x 3n+1, x 3n+2), then

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Also if U(x 3n , x 3n+1, x 3n+2) = G(x 3n+1, x 3n+2, x 3n+3), then

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right),$

which implies that G(x 3n+1, x 3n+2, x 3n+3) = 0, and x 3n+1= x 3n+2= x 3n+3and the result follows immediately.

Finally U(x 3n , x 3n+1, x 3n+2) = G(x 3n+2, x 3n+1, x 3n+1) + G(x 3n , x 3n+3, x 3n+3), implies

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{2em}{0ex}}\end{array}$

which further implies that

$\left(1-k\right)G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le 2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Thus,

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le \lambda G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),$

where $\lambda =\frac{2k}{1-k}$. Obviously 0 < λ < 1.

Hence,

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Similarly it can be shown that

$G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right)\le kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)$

and

$G\left({x}_{3n+3},{x}_{3n+4},{x}_{3n+5}\right)\le kG\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right).$

Therefore, for all n,

$\begin{array}{ll}\hfill G\left({x}_{n+1},{x}_{n+2},{x}_{n+3}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}kG\left({x}_{n},{x}_{n+1},{x}_{n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\cdots \le {k}^{n+1}G\left({x}_{0},{x}_{1},{x}_{2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Following similar arguments to those given in Theorem 2.1, G(x n , x m , x l ) → 0 as n, m, l → ∞. Hence, {x n } is a G-Cauchy sequence. By G-completeness of X, there exists u X such that {x n } converges to u as n → ∞. We claim that fu = u. If not, then consider

$G\left(fu,{x}_{3n+2},{x}_{3n+3}\right)=G\left(fu,g{x}_{3n+1},h{x}_{3n+2}\right)\le kU\left(u,{x}_{3n+1},{x}_{3n+2}\right),$

where

$\begin{array}{c}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}U\left(u,{x}_{3n+1},{x}_{3n+2}\right)\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n+1},fu,fu\right)+G\left(u,g{x}_{n+1},g{x}_{n+1}\right),G\left({x}_{3n+2},g{x}_{3n+1},g{x}_{3n+1}\right)\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}+G\left({x}_{3n+1},h{x}_{3n+2},h{x}_{3n+2}\right),G\left({x}_{3n+2},fu,fu\right)+G\left(u,h{x}_{3n+2},h{x}_{3n+2}\right)\right\}\\ =\phantom{\rule{0.5em}{0ex}}\mathrm{max}\left\{G\left({x}_{3n+1},fu,fu\right)+G\left(u,{x}_{n+2},{x}_{n+2}\right),G\left({x}_{3n+2},{x}_{3n+2},{x}_{3n+2}\right)\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right),G\left({x}_{3n+2},fu,fu\right)+G\left(u,{x}_{3n+3},{x}_{3n+3}\right)\right\}\end{array}$

On taking limit n → ∞, we obtain that

$G\left(fu,u,u\right)\le kU\left(u,u,u\right),$

where

$\begin{array}{ll}\hfill U\left(u,u,u\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(u,fu,fu\right)+G\left(u,u,u\right),G\left(u,u,u\right)+G\left(u,u,u\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,fu,fu\right)+G\left(u,u,u\right)\right\}=G\left(fu,fu,u\right).\phantom{\rule{2em}{0ex}}\end{array}$

Thus

$G\left(fu,u,u\right)\le kG\left(fu,fu,u\right)\le 2kG\left(fu,u,u\right),$

gives a contradiction. Hence, fu = u. Similarly it can be shown that gu = u and hu = u. To prove the uniqueness, suppose that if v is another common fixed point of f, g, and h, then

$G\left(u,v,v\right)=G\left(fu,gv,hv\right)\le kU\left(u,v,v\right),$

where

$\begin{array}{ll}\hfill U\left(u,v,v\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(v,fu,fu\right)+G\left(u,gv,gv\right),G\left(v,gv,gv\right)+G\left(v,hv,hv\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(v,fu,fu\right)+G\left(u,hv,hv\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left(v,u,u\right)+G\left(u,v,v\right),G\left(v,v,v\right)+G\left(v,v,v\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(v,u,u\right)+G\left(u,v,v\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}G\left(v,u,u\right)+G\left(u,v,v\right).\phantom{\rule{2em}{0ex}}\end{array}$

Hence,

$G\left(u,v,v\right)\le k\left[G\left(v,u,u\right)+G\left(u,v,v\right)\right]\le 3kG\left(u,v,v\right),$

which gives that G(u, v, v) = 0, and u = v. Therefore, u is a unique common fixed point of f, g, and h.

Now suppose that for some p in X, we have f(p) = p. We claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$G\left(p,gp,hp\right)=G\left(fp,gp,hp\right)\le kU\left(p,p,p\right),$

where

$\begin{array}{ll}\hfill U\left(p,p,p\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(p,fp,fp\right)+G\left(p,gp,gp\right),G\left(p,gp,gp\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+G\left(p,hp,hp\right),G\left(p,fp,fp\right)+G\left(p,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left(p,p,p\right)+G\left(p,gp,gp\right),G\left(p,gp,gp\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+G\left(p,hp,hp\right),G\left(p,p,p\right)+G\left(p,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left(p,gp,gp\right),G\left(p,gp,gp\right)+G\left(p,hp,hp\right),G\left(p,hp,hp\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

If U(p, p, p) = G(p, gp, gp), then

$G\left(p,gp,hp\right)\le kG\left(p,gp,gp\right)\le kG\left(p,gp,hp\right),$

Also for U(p, p, p) = G(p, gp, gp) + G(p, hp, hp), we obtain

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left(p,gp,gp\right)+G\left(p,hp,hp\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left(p,gp,hp\right),\phantom{\rule{2em}{0ex}}\end{array}$

a contradiction. If U(p, p, p) = G(p, hp, hp), then

$G\left(p,gp,hp\right)\le kG\left(p,hp,hp\right)\le kG\left(p,gp,hp\right),$

a contradiction. Similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence, in all cases, we conclude that p = gp = hp.   □

Corollary 2.5. Let f, g, and h be self maps on a complete G-metric space X satisfying

$G\left({f}^{m}x,{g}^{m}y,{h}^{m}z\right)\le kU\left(x,y,z\right),$
(2.4)

where $k\in \left[0,\frac{1}{3}\right)$ and

$\begin{array}{ll}\hfill U\left(x,y,z\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(y,{f}^{m}x,{f}^{m}x\right)+G\left(x,{g}^{m}y,{g}^{m}y\right),G\left(z,{g}^{m}y,{g}^{m}y\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+G\left(y,{h}^{m}z,{h}^{m}z\right),G\left(z,{f}^{m}x,{f}^{m}x\right)+G\left(x,{h}^{m}z,{h}^{m}z\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, y, z X. Then f, g, and h have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. It follows from Theorem 2.4 that f m , g m , and h m have a unique common fixed point p. Now f(p) = f(f m (p)) = f m+1(p) = f m (f(p)), g(p) = g(g m (p)) = g m+1(p) = g m (g(p)) and h(p) = h(h m (p)) = h m+1(p) = h m (h(p)) implies that f(p), g(p) and h(p) are also fixed points for f m , g m and h m .

We claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left({f}^{m}p,{g}^{m}\left(gp\right),{h}^{m}\left(hp\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kU\left(p,gp,hp\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(gp,{f}^{m}p,{f}^{m}p\right)+G\left(p,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(hp,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right)+G\left(gp,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(hp,{f}^{m}p,{f}^{m}p\right)+G\left(p,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(gp,p,p\right)+G\left(p,gp,gp\right),G\left(hp,gp,gp\right)+G\left(gp,hp,hp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(hp,p,p\right)+G\left(p,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left(p,gp,hp\right).\phantom{\rule{2em}{0ex}}\end{array}$

a contradiction. Similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence, in all cases, we conclude that, f(p) = g(p) = h(p) = p.   □

Theorem 2.6. Let f, g, and h be self maps on a complete G-metric space X satisfying

$G\left(fx,gy,hz\right)\le kU\left(x,y,z\right),$
(2.5)

where $k\in \left[0,\frac{1}{3}\right)$ and

$\begin{array}{ll}\hfill U\left(x,y,z\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(x,gy,gy\right)+G\left(y,gy,gy\right)+G\left(z,gy,gy\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, y, z X. Then f, g, and h have a common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. Suppose x 0 is an arbitrary point in X. Define {x n } by x 3n+1= fx 3n , x 3n+2= gx 3n+1, x 3n+3= hx 3n+2. We have

$\begin{array}{ll}\hfill G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left(f{x}_{3n},g{x}_{3n+1},h{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kU\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\end{array}$

for n = 0, 1, 2, ..., where

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}U\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left({x}_{3n},f{x}_{3n},f{x}_{3n}\right)+G\left({x}_{3n+1},f{x}_{3n},f{x}_{3n}\right)+G\left({x}_{3n+2},f{x}_{3n},f{x}_{3n}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},g{x}_{3n+1},g{x}_{3n+1}\right)+G\left({x}_{3n+1},g{x}_{3n+1},g{x}_{3n+1}\right)+G\left({x}_{3n+2},g{x}_{3n+1},g{x}_{3n+1}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},h{x}_{3n+2},h{x}_{3n+2}\right)+G\left({x}_{3n+1},h{x}_{3n+2},h{x}_{3n+2}\right)+G\left({x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+1},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},{x}_{3n+2},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+2}\right)+G\left({x}_{3n+2},{x}_{3n+2},{x}_{3n+2}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left({x}_{3n},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},{x}_{3n+2},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+2}\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

Now if U(x 3n , x 3n+1, x 3n+2) = G(x 3n , x 3n+1, x 3n+1) + G(x 3n+2, x 3n+1, x 3n+1), then

$\begin{array}{ll}\hfill G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+2},{x}_{3n+1},{x}_{3n+1}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Also if U(x 3n , x 3n+1, x 3n+2) = G(x 3n , x 3n+2, x 3n+2) + G(x 3n+1, x 3n+2, x 3n+2), then

$\begin{array}{ll}\hfill G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+2},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+2}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Finally for U(x 3n , x 3n+1, x 3n+2) = G(x 3n , x 3n+3, x 3n+3) + G(x 3n+1, x 3n+3, x 3n+3) + G(x 3n+2, x 3n+3, x 3n+3), implies

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[2G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+1}\right)+G\left({x}_{3n+1},{x}_{3n+3},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left[G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right)+kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\right]\phantom{\rule{2em}{0ex}}\end{array}$

implies that

$\left(1-k\right)G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le 2kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Thus,

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le \lambda G\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right),$

where $\lambda =\frac{2k}{1-k}$. Obviously 0 < λ < 1.

Hence,

$G\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)\le kG\left({x}_{3n},{x}_{3n+1},{x}_{3n+2}\right).$

Similarly it can be shown that

$G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right)\le kG\left({x}_{3n+1},{x}_{3n+2},{x}_{3n+3}\right)$

and

$G\left({x}_{3n+3},{x}_{3n+4},{x}_{3n+5}\right)\le kG\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+4}\right).$

Therefore, for all n,

$\begin{array}{ll}\hfill G\left({x}_{n+1},{x}_{n+2},{x}_{n+3}\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}kG\left({x}_{n},{x}_{n+1},{x}_{n+2}\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\cdots \le {k}^{n+1}G\left({x}_{0},{x}_{1},{x}_{2}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Following similar arguments to those given in Theorem 2.1, G(x n , x m , x l ) → 0 as n, m, l → ∞. Hence, {x n } is a G-Cauchy sequence. By G-completeness of X, there exists u X such that {x n } converges to u as n → ∞. We claim that fu = gu = u. If not, then consider

$G\left(fu,gu,{x}_{3n+3}\right)=G\left(fu,gu,h{x}_{3n+2}\right)\le kU\left(u,u,{x}_{3n+2}\right),$

where

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}U\left(u,{x}_{3n+1},{x}_{3n+2}\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left(u,fu,fu\right)+G\left(u,fu,fu\right)+G\left({x}_{3n+2},fu,fu\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,gu,gu\right)+G\left(u,gu,gu\right)+G\left({x}_{3n+2,}gu,gu\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,h{x}_{3n+2},h{x}_{3n+2}\right)+G\left(u,h{x}_{3n+2},h{x}_{3n+2}\right)+G\left({x}_{3n+2},h{x}_{3n+2},h{x}_{3n+2}\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{2G\left(u,fu,fu\right)+G\left({x}_{3n+2},fu,fu\right),2G\left(u,gu,gu\right)+G\left({x}_{3n+2},gu,gu\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}2G\left(u,{x}_{3n+3},{x}_{3n+3}\right)+G\left({x}_{3n+2},{x}_{3n+3},{x}_{3n+3}\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

On taking limit as n → ∞, we obtain that

$G\left(fu,gu,u\right)\le kU\left(u,u,u\right),$

where

$\begin{array}{ll}\hfill U\left(u,u,u\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{2G\left(u,fu,fu\right)+G\left(u,fu,fu\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}2G\left(u,gu,gu\right)+G\left(u,gu,gu\right),2G\left(u,u,u\right)+G\left(u,u,u\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{3G\left(u,fu,fu\right),3G\left(u,gu,gu\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

Now for U(u, u, u) = 3G(fu, fu, fu), then

$G\left(fu,gu,u\right)\le 3kG\left(fu,fu,u\right)\le 3kG\left(fu,gu,u\right),$

a contradiction. Hence, fu = gu = u. Also for U(u, u, u) = 3G(u, gu, gu),

$G\left(fu,gu,u\right)\le 3kG\left(u,gu,gu\right)\le 3kG\left(fu,gu,u\right),$

a contradiction. Hence, fu = gu = u. Similarly it can be shown that gu = u and hu = u.

Now suppose that for some p in X, we have f(p) = p. We claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$G\left(p,gp,hp\right)=G\left(fp,gp,hp\right)\le kU\left(p,p,p\right),$

where

$\begin{array}{ll}\hfill U\left(p,p,p\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(p,fp,fp\right)+G\left(p,fp,fp\right),G\left(p,fp,fp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,gp,gp\right)+G\left(p,gp,gp\right)+G\left(p,gp,gp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,hp,hp\right)+G\left(p,hp,hp\right)+G\left(p,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{3G\left(p,p,p\right),3G\left(p,gp,gp\right),3G\left(p,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{3G\left(p,gp,gp\right),3G\left(p,hp,hp\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

If U(p, p, p) = 3G(p, gp, gp), then

$G\left(p,gp,hp\right)\le 3kG\left(p,gp,gp\right)\le 3kG\left(p,gp,hp\right),$

a contradiction. Also, U(p, p, p) = 3G(p, hp, hp) gives

$G\left(p,gp,hp\right)\le 3kG\left(p,hp,hp\right)\le 3kG\left(p,gp,hp\right),$

a contradiction. Similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence in all cases, we conclude that p = gp = hp.   □

Remark 2.7. Let f, g, and h be self maps on a complete G-metric space X satisfying (2.5). Then f, g and h have a unique common fixed point in X provided that $0\le k<\frac{1}{4}$.

Proof. Existence of common fixed points of f, g, and h follows from Theorem 2.6. To prove the uniqueness, suppose that if v is another common fixed point of f, g, and h, then

$G\left(u,v,v\right)=G\left(fu,gv,hv\right)\le kU\left(u,v,v\right),$

where

$\begin{array}{ll}\hfill U\left(u,v,v\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(u,fu,fu\right)+G\left(v,fu,fu\right)+G\left(v,fu,fu\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,gv,gv\right),G\left(v,gv,gv\right)+G\left(v,gv,gv\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,hv,hv\right)+G\left(v,hv,hv\right)+G\left(v,hv,hv\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{G\left(u,u,u\right)+G\left(v,u,u\right)+G\left(v,u,u\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,v,v\right)+G\left(v,v,v\right)+G\left(v,v,v\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(u,v,v\right)+G\left(v,v,v\right)+G\left(v,v,v\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}max\left\{2G\left(v,u,u\right),G\left(u,v,v\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

U(u, v, v) = 2G(v, u, u), implies that

$G\left(u,v,v\right)\le 2kG\left(v,u,u\right)\le 4kG\left(u,v,v\right),$

which gives u = v. And U(u, v, v) = G(u, v, v), gives

$G\left(u,v,v\right)\le kG\left(u,v,v\right),$

U = v. Hence, u is a unique common fixed point of f, g, and h.   □

Corollary 2.8. Let f, g, and h be self maps on a complete G-metric space X satisfying

$G\left({f}^{m}x,{g}^{m}y,{h}^{m}z\right)\le kU\left(x,y,z\right),$
(2.6)

where $k\in \left[0,\frac{1}{4}\right)$ and

$\begin{array}{ll}\hfill U\left(x,y,z\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{G\left(x,{f}^{m}x,{f}^{m}x\right)+G\left(y,{f}^{m}x,{f}^{m}x\right)+G\left(z,{f}^{m}x,{f}^{m}x\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(x,{g}^{m}y,{g}^{m}y\right)+G\left(y,{g}^{m}y,{g}^{m}y\right)+G\left(z,{g}^{m}y,{g}^{m}y\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(x,{h}^{m}z,{h}^{m}z\right)+G\left(y,{h}^{m}z,{h}^{m}z\right)+G\left(z,{h}^{m}z,{h}^{m}z\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, y, z X. Then f, g and h have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point g and h and conversely.

Proof. It follows from Theorem 2.6, that f m , g m , and h m have a unique common fixed point p. Now f(p) = f(f m (p)) = f m+1(p) = f m (f(p)), g(p) = g(g m (p)) = g m+1(p) = g m (g(p)) and h(p) = h(h m (p)) = h m+1(p) = h m (h(p)) implies that f(p), g(p) and h(p) are also fixed points for f m , g m , and h m . Now we claim that p = g(p) = h(p), if not then in case when pg(p) and ph(p), we obtain

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}G\left({f}^{m}p,{g}^{m}\left(gp\right),{h}^{m}\left(hp\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}kU\left(p,gp,hp\right)\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(p,{f}^{m}p,{f}^{m}p\right)+G\left(gp,{f}^{m}p,{f}^{m}p\right)+G\left(hp,{f}^{m}p,{f}^{m}p\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right)+G\left(gp,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right)+G\left(hp,{g}^{m}\left(gp\right),{g}^{m}\left(gp\right)\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right),G\left(gp,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right)+G\left(hp,{h}^{m}\left(hp\right),{h}^{m}\left(hp\right)\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(p,p,p\right)+G\left(gp,p,p\right)+G\left(hp,p,p\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,gp,gp\right)+G\left(gp,gp,gp\right)+G\left(hp,gp,gp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,hp,hp\right)\right),G\left(gp,hp,hp\right)+G\left(hp,hp,hp\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}kmax\left\{G\left(gp,p,p\right)+G\left(hp,p,p\right),G\left(p,gp,gp\right)+G\left(hp,gp,gp\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}G\left(p,hp,hp\right)\right)+G\left(gp,hp,hp\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

Now if U(p, gp, hp) = G(gp, p, p) + G(hp, p, p), then

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left(gp,p,p\right)+G\left(hp,p,p\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left(p,gp,hp\right),\phantom{\rule{2em}{0ex}}\end{array}$

a contradiction. Also if U(p, gp, hp) = G(p, gp, gp) + G(hp, gp, gp), then

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left(p,gp,gp\right)+G\left(hp,gp,gp\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left(p,gp,hp\right),\phantom{\rule{2em}{0ex}}\end{array}$

a contradiction. Finally, if U(p, gp, hp) = G(p, hp, hp) + G(gp, hp, hp), then

$\begin{array}{ll}\hfill G\left(p,gp,hp\right)\phantom{\rule{1em}{0ex}}& \le \phantom{\rule{1em}{0ex}}k\left[G\left(p,hp,hp\right)+G\left(gp,hp,hp\right)\right]\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}2kG\left(p,gp,hp\right),\phantom{\rule{2em}{0ex}}\end{array}$

Also similarly when pg(p) and p = h(p) or when ph(p) and p = g(p), we arrive at a contradiction following the similar arguments to those given above. Hence, in all cases, we conclude that f(p) = g(p) = h(p) = p

Example 2.9. Let X = [0, 1] and G(x, y, z) = max{|x - y|, |y - z|, |z - x|} be a G-metric on X. Define f, g, h : XX by

$\begin{array}{ll}\hfill f\left(x\right)& =\left\{\begin{array}{cc}\hfill \frac{x}{12}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\frac{1}{2}\right)\hfill \\ \hfill \frac{x}{10}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[\frac{1}{2},1\right],\hfill \end{array}\right\\phantom{\rule{2em}{0ex}}\\ \hfill g\left(x\right)& =\left\{\begin{array}{cc}\hfill \frac{x}{8}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\frac{1}{2}\right)\hfill \\ \hfill \frac{x}{6}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[\frac{1}{2},1\right],\hfill \end{array}\right\\phantom{\rule{2em}{0ex}}\end{array}$

and

$h\left(x\right)=\left\{\begin{array}{cc}\hfill \frac{x}{5}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[0,\frac{1}{2}\right)\hfill \\ \hfill \frac{x}{3}\hfill & \hfill \mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[\frac{1}{2},1\right].\hfill \end{array}\right\$

Note that f, g and h are discontinuous maps. Also $fg\left(\frac{1}{2}\right)=f\left(\frac{1}{12}\right)=\frac{1}{144}$, $gf\left(\frac{1}{2}\right)=g\left(\frac{1}{20}\right)=\frac{1}{160}$, $gh\left(\frac{1}{2}\right)=g\left(\frac{1}{6}\right)=\frac{1}{48}$, $hg\left(\frac{1}{2}\right)=h\left(\frac{1}{12}\right)=\frac{1}{60}$, and $fh\left(\frac{1}{2}\right)=f\left(\frac{1}{6}\right)=\frac{1}{72}$, $hf\left(\frac{1}{2}\right)=h\left(\frac{1}{20}\right)=\frac{1}{100}$, which shows that f, g and h does not commute with each other.

Note that for $x,y,z\in \left[0,\frac{1}{2}\right)$,

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right]=\frac{11x}{12}+\left|y-\frac{x}{12}\right|+\left|z-\frac{x}{12}\right|,\\ \left[G\left(x,gy,gy\right)+G\left(y,gy,gy\right)+G\left(z,gy,gy\right)\right]=\left|x-\frac{y}{8}\right|+\frac{7y}{8}+\left|z-\frac{y}{8}\right|,\end{array}$

and

$\left[G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)\right]=\left|x-\frac{z}{5}\right|+\left|y-\frac{z}{5}\right|+\frac{4z}{5}.$

Now

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{\left|\frac{x}{12}-\frac{y}{8}\right|,\left|\frac{y}{8}-\frac{z}{5}\right|,\left|\frac{z}{5}-\frac{x}{12}\right|\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{8}max\left\{\left|\frac{2x}{3}-y\right|,\left|y-\frac{8z}{5}\right|,\left|\frac{8z}{5}-\frac{2x}{3}\right|\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

For U(x, y, z) = G(x, fx, fx) + G(y, fx, fx) + G(z, fx, fx), we obtain

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{8}max\left\{\left|\frac{2x}{3}-y\right|,\left|y-\frac{8z}{5}\right|,\left|\frac{8z}{5}-\frac{2x}{3}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{8}\left[\frac{11x}{12}+\left|y-\frac{x}{12}\right|+\left|z-\frac{x}{12}\right|\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{8}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

In case U(x, y, z) = G(x, gy, gy) + G(y, gy, gy) + G(z, gy, gy), then

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{8}max\left\{\left|\frac{2x}{3}-y\right|,\left|y-\frac{8z}{5}\right|,\left|\frac{8z}{5}-\frac{2x}{3}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\left|x-\frac{y}{8}\right|+\frac{7y}{8}+\left|z-\frac{y}{8}\right|\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

And for U(x, y, z) = G(x, hz, hz) + G(y, hz, hz) + G(z, hz, hz), we have

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{8}max\left\{\left|\frac{2x}{3}-y\right|,\left|y-\frac{8z}{5}\right|,\left|\frac{8z}{5}-\frac{2x}{3}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\left|x-\frac{z}{5}\right|+\left|y-\frac{z}{5}\right|+\frac{4z}{5}\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

Thus, (2.5) is satisfied for $k=\frac{1}{4}<\frac{1}{3}$.

For $x,y,z\in \left[\frac{1}{2},1\right]$

$\begin{array}{c}G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)=\frac{9x}{10}+\left|y-\frac{x}{10}\right|+\left|z-\frac{x}{10}\right|,\\ \phantom{\rule{1em}{0ex}}G\left(x,gy,gy\right)+G\left(y,gy,gy\right)+G\left(z,gy,gy\right)=\left|x-\frac{y}{6}\right|+\frac{5y}{6}+\left|z-\frac{y}{6}\right|,\end{array}$

and

$G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)=\left|x-\frac{z}{3}\right|+\left|y-\frac{z}{3}\right|+\frac{2z}{3}.$

Now,

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{\left|\frac{x}{10}-\frac{y}{6}\right|,\left|\frac{y}{6}-\frac{z}{3}\right|,\left|\frac{z}{3}-\frac{x}{10}\right|\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|\frac{3x}{5}-y\right|,\left|2z-y\right|,\left|2z-\frac{3x}{5}\right|\right\},\phantom{\rule{2em}{0ex}}\end{array}$

For U(x, y, z) = G(x, fx, fx) + G(y, fx, fx) + G(z, fx, fx), we obtain

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|\frac{3x}{5}-y\right|,\left|2z-y\right|,\left|2z-\frac{3x}{5}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\frac{9x}{10}+\left|y-\frac{x}{10}\right|+\left|z-\frac{x}{10}\right|\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

In case, U(x, y, z) = G(x, gy, gy) + G(y, gy, gy) + G(z, gy, gy), then

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|\frac{3x}{5}-y\right|,\left|2z-y\right|,\left|2z-\frac{3x}{5}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\left|x-\frac{y}{6}\right|+\frac{5y}{6}+\left|z-\frac{y}{6}\right|\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

And U(x, y, z) = G(x, hz, hz) + G(y, hz, hz) + G(z, hz, hz) gives that

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|\frac{3x}{5}-y\right|,\left|2z-y\right|,\left|2z-\frac{3x}{5}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\left|x-\frac{z}{3}\right|+\left|y-\frac{z}{3}\right|+\frac{2z}{3}\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

Hence (2.5) is satisfied for $k=\frac{1}{4}<\frac{1}{3}$.

Now for $x\in \left[0,\frac{1}{2}\right)$, $y,z\in \left[\frac{1}{2},1\right]$,

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)\right]=\frac{11x}{12}+\left|y-\frac{x}{12}\right|+\left|z-\frac{x}{12}\right|,\\ \phantom{\rule{1em}{0ex}}\left[G\left(x,gy,gy\right)+G\left(y,gy,gy\right)+G\left(z,gy,gy\right)\right]=\left|x-\frac{y}{6}\right|+\frac{5y}{6}+\left|z-\frac{y}{6}\right|,\end{array}$

and

$\left[G\left(x,hz,hz\right)+G\left(y,hz,hz\right)+G\left(z,hz,hz\right)\right]=\left|x-\frac{z}{3}\right|+\left|y-\frac{z}{3}\right|+\frac{2z}{3}.$

Also

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}max\left\{\left|\frac{x}{12}-\frac{y}{6}\right|,\left|\frac{y}{6}-\frac{z}{3}\right|,\left|\frac{z}{3}-\frac{x}{12}\right|\right\}\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|y-\frac{x}{2}\right|,\left|2z-y\right|,\left|2z-\frac{x}{2}\right|\right\}.\phantom{\rule{2em}{0ex}}\end{array}$

Now for U(x, y, z) = G(x, fx, fx) + G(y, fx, fx) + G(z, fx, fx), then

$\begin{array}{ll}\hfill G\left(fx,gy,hz\right)\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\frac{1}{6}max\left\{\left|y-\frac{x}{2}\right|,\left|2z-y\right|,\left|2z-\frac{x}{2}\right|\right\}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\frac{1}{4}\left[\frac{11x}{12}+\left|y-\frac{x}{12}\right|+\left|z-\frac{x}{12}\right|\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{4}\left[G\left(x,fx,fx\right)+G\left(y,fx,fx\right)+G\left(z,fx,fx\right)