# A note on some coupled fixed-point theorems on G-metric spaces

## Abstract

The purpose of this paper is to extend some recent coupled fixed-point theorems in the context of G-metric space by essentially different and more natural way. We state some examples to illustrate our results.

MSC:46N40, 47H10, 54H25, 46T99.

## 1 Introduction

In nonlinear functional analysis, one of the most productive tools is the fixed-point theory, which has numerous applications in many quantitative disciplines such as biology, chemistry, computer science, and additionally in many branches of engineering. In this theory, the Banach contraction principle can be considered as a cornerstone pioneering result which in elementary terms states that each contraction has a unique fixed point in a complete metric space. Due to its potential of applications in the fields above mentioned and many more, the fixed-point theory, in particular, the Banach contraction principle, attracts considerable attention from many authors (see, e.g., ). Especially, it is considered very natural and curious to investigate the existence and uniqueness of a fixed point for several contraction type mappings in various abstract spaces. A major example in this direction is the work of Mustafa and Sims  in which they introduced the concept of G-metric spaces as a generalization of (usual) metric spaces in 2004. After this remarkable paper, a number of papers have appeared on this topic in the literature (see, e.g., [18, 10, 12, 1829]).

For the sake of completeness, we recall some basic definitions and elementary results from the literature. Throughout this paper, $\mathbb{N}$ is the set of nonnegative integers, and ${\mathbb{N}}^{\ast }$ is the set of positive integers.

Definition 1 (See )

Let X be a nonempty set, $G:X×X×X\to {\mathbb{R}}^{+}$ be a function satisfying the following properties:

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

(G2) $0 for all $x,y\in X$ with $x\ne y$,

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

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

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

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

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

(1.1)

Example 2 Let $\left(X,d\right)$ be a metric space. The function $G:X×X×X\to \left[0,+\mathrm{\infty }\right)$, defined by

$G\left(x,y,z\right)=max\left\{d\left(x,y\right),d\left(y,z\right),d\left(z,x\right)\right\},$

or

$G\left(x,y,z\right)=d\left(x,y\right)+d\left(y,z\right)+d\left(z,x\right),$

for all $x,y,z\in X$, is a G-metric on X.

Definition 3 (See )

Let $\left(X,G\right)$ be a G-metric space, and let $\left\{{x}_{n}\right\}$ be a sequence of points of X, therefore, we say that $\left({x}_{n}\right)$ is G-convergent to $x\in X$ if ${lim}_{n,m\to +\mathrm{\infty }}G\left(x,{x}_{n},{x}_{m}\right)=0$, that is, for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $G\left(x,{x}_{n},{x}_{m}\right)<\epsilon$, for all $n,m\ge N$. We call x the limit of the sequence and write ${x}_{n}\to x$ or ${lim}_{n\to +\mathrm{\infty }}{x}_{n}=x$.

Proposition 4 (See )

Let$\left(X,G\right)$be a G-metric space. The following are equivalent:

1. (1)

$\left\{{x}_{n}\right\}$ is G-convergent to x,

2. (2)

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

3. (3)

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

4. (4)

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

Definition 5 (See )

Let $\left(X,G\right)$ be a G-metric space. A sequence $\left\{{x}_{n}\right\}$ is called a G-Cauchy sequence if, for any $\epsilon >0$, there is $N\in \mathbb{N}$ such that $G\left({x}_{n},{x}_{m},{x}_{l}\right)<\epsilon$ for all $m,n,l\ge N$, that is, $G\left({x}_{n},{x}_{m},{x}_{l}\right)\to 0$ as $n,m,l\to +\mathrm{\infty }$.

Proposition 6 (See )

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

1. (1)

the sequence $\left\{{x}_{n}\right\}$ is G-Cauchy,

2. (2)

for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $G\left({x}_{n},{x}_{m},{x}_{m}\right)<\epsilon$, for all $m,n\ge N$.

Definition 7 (See )

A G-metric space $\left(X,G\right)$ is called G-complete if every G-Cauchy sequence is G-convergent in $\left(X,G\right)$.

Definition 8 Let $\left(X,G\right)$ be a G-metric space. A mapping $F:X×X×X\to X$ is said to be continuous if for any three G-convergent sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ converging to x, y, and z, respectively, $\left\{F\left({x}_{n},{y}_{n},{z}_{n}\right)\right\}$ is G-convergent to $F\left(x,y,z\right)$.

Definition 9 Let $F:X×X\to X$ and $g:X\to X$ be mappings. The mappings F and g are said to commute if

In , Shatanawi proved the following theorems.

Theorem 10 Let$\left(X,G\right)$be a G-metric space. Let$F:X×X\to X$and$g:X\to X$be two mappings such that

(1.2)

Assume that F and g satisfy the following conditions:

1. (1)

$F\left(X×X\right)\subset g\left(X\right)$,

2. (2)

$g\left(X\right)$ is G-complete,

3. (3)

g is G-continuous and commutes with F.

If$k\in \left[0,\frac{1}{2}\right)$, then there is a unique$x\in X$such that$gx=F\left(x,x\right)=x$.

Corollary 11 Let$\left(X,G\right)$be a complete G-metric space. Let$F:X×X\to X$be a mapping such that

(1.3)

If$k\in \left[0,\frac{1}{2}\right)$, then there is a unique$x\in X$such that$F\left(x,x\right)=x$.

In this paper, we aim to extend the above coupled fixed-point results.

## 2 Main results

We start with an example to show the weakness of Theorem 10.

Example 12 Let $X=\left[0,1\right]$. Define $G:X×X×X\to \left[0,+\mathrm{\infty }\right)$ by

$G\left(x,y,z\right)=|x-y|+|x-z|+|y-z|$

for all $x,y,z\in X$. Then $\left(X,G\right)$ is a G-metric space. Define a map $F:X×X\to X$ by $F\left(x,y\right)=\frac{1}{3}x+\frac{1}{8}y$ and $g:X\to X$ by $g\left(x\right)=\frac{x}{2}$ for all $x,y\in X$. Then, for all $x,y,u,v,z,w\in X$ with $y=v=w$, we have

$\begin{array}{rcl}G\left(F\left(x,y\right),F\left(u,v\right),F\left(z,w\right)\right)& =& G\left(\frac{1}{3}x+\frac{1}{8}y,\frac{1}{3}u+\frac{1}{8}v,\frac{1}{3}z+\frac{1}{8}w\right)\\ =& \frac{|x-u|+|x-z|+|u-z|}{3}\end{array}$

and

$\begin{array}{rcl}G\left(gx,gu,gz\right)+G\left(gy,gv,gw\right)& =& G\left(\frac{x}{2},\frac{u}{2},\frac{z}{2}\right)+G\left(\frac{y}{2},\frac{v}{2},\frac{w}{2}\right)\\ =& \frac{|x-u|+|x-z|+|u-z|}{2}.\end{array}$

Then it is easy to that there is no $k\in \left[0,\frac{1}{2}\right)$ such that

$G\left(F\left(x,y\right),F\left(u,v\right),F\left(z,w\right)\right)\le k\left[G\left(gx,gu,gz\right)+G\left(gy,gv,gw\right)\right]$

for all $x,y,u,v,z,w\in X$. Thus, Theorem 10 cannot be applied to this example. However, it is easy to see that 0 is the unique point $x\in X$ such that $x=gx=F\left(x,x\right)$.

We now state our first result which successively guarantee a coupled fixed point.

Theorem 13 Let$\left(X,G\right)$be a G-metric space. Let$F:X×X\to X$and$g:X\to X$be two mappings such that (2.1)

for all$x,y,u,v\in X$. Assume that F and g satisfy the following conditions:

1. (1)

$F\left(X×X\right)\subset g\left(X\right)$,

2. (2)

$g\left(X\right)$ is G-complete,

3. (3)

g is G-continuous and commutes with F.

If$k\in \left[0,1\right)$, then there is a unique$x\in X$such that$gx=F\left(x,x\right)=x$.

Proof Take ${x}_{0},{y}_{0}\in X$. Noting that $F\left(X×X\right)\subset g\left(X\right)$, we can construct two sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in X such that

$g{x}_{n+1}=F\left({x}_{n},{y}_{n}\right),\phantom{\rule{2em}{0ex}}g{y}_{n+1}=F\left({y}_{n},{x}_{n}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$

Let

${M}_{n}=G\left(g{x}_{n},g{x}_{n+1},g{x}_{n+1}\right)+G\left(g{y}_{n},g{y}_{n+1},g{y}_{n+1}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$

Then, by using (2.1), for each $n\in {\mathbb{N}}^{\ast }$, we have

$\begin{array}{rcl}{M}_{n}& =& G\left(g{x}_{n},g{x}_{n+1},g{x}_{n+1}\right)+G\left(g{y}_{n},g{y}_{n+1},g{y}_{n+1}\right)\\ =& G\left(F\left({x}_{n-1},{y}_{n-1}\right),F\left({x}_{n},{y}_{n}\right),F\left({x}_{n},{y}_{n}\right)\right)+G\left(F\left({y}_{n-1},{x}_{n-1}\right),F\left({y}_{n},{x}_{n}\right),F\left({y}_{n},{x}_{n}\right)\right)\\ \le & k\left[G\left(g{x}_{n-1},g{x}_{n},g{x}_{n}\right)+G\left(g{y}_{n-1},g{y}_{n},g{y}_{n}\right)\right]\\ =& k{M}_{n-1},\end{array}$

which yields that

${M}_{n}\le {k}^{n}{M}_{0},\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$
(2.2)

Now, for all $m,n\in \mathbb{N}$ with $m>n$, by using rectangle inequality of G-metric and (2.2), we get

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

which yields that

$\underset{n,m\to +\mathrm{\infty }}{lim}G\left(g{x}_{n},g{x}_{m},g{x}_{m}\right)+G\left(g{y}_{n},g{y}_{m},g{y}_{m}\right)=0.$

Then, by Proposition 6, we conclude that the sequences $\left\{g{x}_{n}\right\}$ and $\left\{g{y}_{n}\right\}$ are G-Cauchy.

Noting that $g\left(X\right)$ is G-complete, there exist $x,y\in g\left(X\right)$ such that $\left\{g{x}_{n}\right\}$ and $\left\{g{y}_{n}\right\}$ are G-convergent to x and y, respectively, i.e.,

$\underset{n\to +\mathrm{\infty }}{lim}G\left(g{x}_{n},x,x\right)=0,\phantom{\rule{2em}{0ex}}\underset{n\to +\mathrm{\infty }}{lim}G\left(g{y}_{n},y,y\right)=0.$

Also, since g is G-continuous, we get

$\underset{n\to +\mathrm{\infty }}{lim}G\left(gg{x}_{n},gx,gx\right)=0,\phantom{\rule{2em}{0ex}}\underset{n\to +\mathrm{\infty }}{lim}G\left(gg{y}_{n},gy,gy\right)=0.$
(2.3)

In addition, by (2.1) and the fact g commutes with F, we get

$\begin{array}{c}G\left(gg{x}_{n+1},F\left(x,y\right),F\left(x,y\right)\right)+G\left(gg{y}_{n+1},F\left(y,x\right),F\left(y,x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=G\left(g\left(F\left({x}_{n},{y}_{n}\right)\right),F\left(x,y\right),F\left(x,y\right)\right)+G\left(g\left(F\left({y}_{n},{x}_{n}\right)\right),F\left(y,x\right),F\left(y,x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=G\left(F\left(g{x}_{n},g{y}_{n}\right),F\left(x,y\right),F\left(x,y\right)\right)+G\left(F\left(g{y}_{n},g{x}_{n}\right),F\left(y,x\right),F\left(y,x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le k\left[G\left(gg{x}_{n},gx,gx\right)+G\left(gg{y}_{n},gy,gy\right)\right].\hfill \end{array}$

Combining this with (2.3), we get

$G\left(gg{x}_{n+1},F\left(x,y\right),F\left(x,y\right)\right)+G\left(gg{y}_{n+1},F\left(y,x\right),F\left(y,x\right)\right)\to 0,\phantom{\rule{1em}{0ex}}n\to +\mathrm{\infty }.$

On the other hand, by the fact that G is continuous on its variables (cf. ), we have

$\begin{array}{c}G\left(gg{x}_{n+1},F\left(x,y\right),F\left(x,y\right)\right)+G\left(gg{y}_{n+1},F\left(y,x\right),F\left(y,x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\to G\left(gx,F\left(x,y\right),F\left(x,y\right)\right)+G\left(gy,F\left(y,x\right),F\left(y,x\right)\right),\phantom{\rule{1em}{0ex}}n\to +\mathrm{\infty }.\hfill \end{array}$

Thus, we conclude that

$G\left(gx,F\left(x,y\right),F\left(x,y\right)\right)+G\left(gy,F\left(y,x\right),F\left(y,x\right)\right)=0,$

i.e.,

$G\left(gx,F\left(x,y\right),F\left(x,y\right)\right)=G\left(gy,F\left(y,x\right),F\left(y,x\right)\right)=0,$

which yields that

$gx=F\left(x,y\right),\phantom{\rule{2em}{0ex}}gy=F\left(y,x\right).$

Moreover, it follows from

$\begin{array}{c}G\left(gx,gy,gy\right)+G\left(gy,gx,gx\right)\hfill \\ \phantom{\rule{1em}{0ex}}=G\left(F\left(x,y\right),F\left(y,x\right),F\left(y,x\right)\right)+G\left(F\left(y,x\right),F\left(x,y\right),F\left(x,y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le k\left[G\left(gx,gy,gy\right)+G\left(gy,gx,gx\right)\right]\hfill \end{array}$

that $G\left(gx,gy,gy\right)+G\left(gy,gx,gx\right)=0$. Thus, $G\left(gx,gy,gy\right)=0$, i.e., $gx=gy$.

Next, let us show that $gx=F\left(x,x\right)=x$. By using rectangle inequality of G-metric and (2.1), we have

$\begin{array}{c}G\left(x,gx,gx\right)+G\left(y,gy,gy\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le G\left(x,g{x}_{n+1},g{x}_{n+1}\right)+G\left(g{x}_{n+1},gx,gx\right)+G\left(y,g{y}_{n+1},g{y}_{n+1}\right)+G\left(g{y}_{n+1},gy,gy\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[G\left(x,g{x}_{n+1},g{x}_{n+1}\right)+G\left(y,g{y}_{n+1},g{y}_{n+1}\right)\right]\hfill \\ \phantom{\rule{2em}{0ex}}+\left[G\left(F\left({x}_{n},{y}_{n}\right),F\left(x,y\right),F\left(x,y\right)\right)+G\left(F\left({y}_{n},{x}_{n}\right),F\left(y,x\right),F\left(y,x\right)\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[G\left(x,g{x}_{n+1},g{x}_{n+1}\right)+G\left(y,g{y}_{n+1},g{y}_{n+1}\right)\right]+k\left[G\left(g{x}_{n},gx,gx\right)+G\left(g{y}_{n},gy,gy\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[G\left(x,g{x}_{n+1},g{x}_{n+1}\right)+G\left(y,g{y}_{n+1},g{y}_{n+1}\right)\right]\hfill \\ \phantom{\rule{2em}{0ex}}+k\left[G\left(x,gx,gx\right)+G\left(y,gy,gy\right)\right]+k\left[G\left(g{x}_{n},x,x\right)+G\left(g{y}_{n},y,y\right)\right],\hfill \end{array}$

which gives that

$\begin{array}{c}G\left(x,gx,gx\right)+G\left(y,gy,gy\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{G\left(x,g{x}_{n+1},g{x}_{n+1}\right)+G\left(y,g{y}_{n+1},g{y}_{n+1}\right)+k\left[G\left(g{x}_{n},x,x\right)+G\left(g{y}_{n},y,y\right)\right]}{1-k}.\hfill \end{array}$

Combing this with the fact that $\left\{g{x}_{n}\right\}$ and $\left\{g{y}_{n}\right\}$ are G-convergent to x and y, respectively, we conclude that

$G\left(x,gx,gx\right)+G\left(y,gy,gy\right)=0,$

which yields that

$x=gx,\phantom{\rule{2em}{0ex}}y=gy.$

Recalling that $gx=gy$ and $gx=F\left(x,y\right)$, we get $x=y$ and $x=gx=F\left(x,x\right)$.

It remains to show the uniqueness. Let $u\in X$ be such that $u=gu=F\left(u,u\right)$. Then we have

$\begin{array}{rcl}2G\left(u,x,x\right)& =& G\left(F\left(u,u\right),F\left(x,x\right),F\left(x,x\right)\right)+G\left(F\left(u,u\right),F\left(x,x\right),F\left(x,x\right)\right)\\ \le & k\left[G\left(gu,gx,gx\right)+G\left(gu,gx,gx\right)\right]\\ \le & 2kG\left(u,x,x\right),\end{array}$

which yields that $\left(2-2k\right)G\left(u,x,x\right)\le 0$. Thus, $G\left(u,x,x\right)=0$, which means $u=x$. This completes the proof. □

Remark 14 It is easy to see that Theorem 10, appearing in , is a direct corollary of Theorem 13. On the other hand, Theorem 13 can deal with some cases, which Theorem 10 cannot be applied. For this, let us reconsider Example 12. In fact, for all $x,y,u,v\in X$, we have

$\begin{array}{c}G\left(F\left(x,y\right),F\left(u,v\right),F\left(u,v\right)\right)+G\left(F\left(y,x\right),F\left(v,u\right),F\left(v,u\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=G\left(\frac{1}{3}x+\frac{1}{8}y,\frac{1}{3}u+\frac{1}{8}v,\frac{1}{3}u+\frac{1}{8}v\right)+G\left(\frac{1}{3}y+\frac{1}{8}x,\frac{1}{3}v+\frac{1}{8}u,\frac{1}{3}v+\frac{1}{8}u\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{11\left(|x-u|+|y-v|\right)}{12}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{11}{12}\left[G\left(gx,gu,gu\right)+G\left(gy,gv,gv\right)\right],\hfill \end{array}$

i.e., (2.1) holds. Other assumptions of Theorem 13 are easy to verify. So, by Theorem 13, there exists a unique $x\in X$ such that $gx=F\left(x,x\right)=x$.

Letting $g=I$, we can get the following result.

Corollary 15 Let$\left(X,G\right)$be a complete G-metric space. Let$F:X×X\to X$be a mapping such that (2.4)

for all$x,y,u,v\in X$. If$k\in \left[0,1\right)$, then there is a unique$x\in X$such that$F\left(x,x\right)=x$.

Example 16 Let $\left(X,G\right)$ be the same as in Example 12. Then $\left(X,G\right)$ is a G-metric space. Also, it is not difficult to verify that $\left(X,G\right)$ is G-complete. Define a map $F:X×X\to X$ by $F\left(x,y\right)=1-\frac{1}{16}{x}^{2}-\frac{5}{16}{y}^{2}$ for all $x,y\in X$. Then, for all $x,y,u,v\in X$, we have

$\begin{array}{c}G\left(F\left(x,y\right),F\left(u,v\right),F\left(u,v\right)\right)+G\left(F\left(y,x\right),F\left(v,u\right),F\left(v,u\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=G\left(1-\frac{1}{16}{x}^{2}-\frac{5}{16}{y}^{2},1-\frac{1}{16}{u}^{2}-\frac{5}{16}{v}^{2},1-\frac{1}{16}{u}^{2}-\frac{5}{16}{v}^{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+G\left(1-\frac{1}{16}{y}^{2}-\frac{5}{16}{x}^{2},1-\frac{1}{16}{v}^{2}-\frac{5}{16}{u}^{2},1-\frac{1}{16}{v}^{2}-\frac{5}{16}{u}^{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{8}|{u}^{2}-{x}^{2}|+\frac{5}{8}|{v}^{2}-{y}^{2}|+\frac{1}{8}|{v}^{2}-{y}^{2}|+\frac{5}{8}|{u}^{2}-{x}^{2}|\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{3}{4}|{u}^{2}-{x}^{2}|+\frac{3}{4}|{v}^{2}-{y}^{2}|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{3}{2}|u-x|+\frac{3}{2}|v-y|\hfill \end{array}$

and

$\begin{array}{c}G\left(x,u,u\right)+G\left(y,v,v\right)\hfill \\ \phantom{\rule{1em}{0ex}}=2\left(|x-u|+|y-v|\right).\hfill \end{array}$

Thus, the statement (2.4) of Corollary 15 is satisfied for any $k\in \left[\frac{3}{4},1\right)$. Thus, there is a unique $x\in X$ such that $F\left(x,x\right)=x$.

Remark 17 Corollary 11 cannot be applied to Example 16 since (1.3) does not hold. In fact, if (1.3) holds for some $k\in \left[0,\frac{1}{2}\right)$, then

$\begin{array}{rcl}\frac{9}{40}& =& G\left(\frac{11}{16},\frac{4}{5},\frac{4}{5}\right)=G\left(F\left(0,1\right),F\left(0,\frac{4}{5}\right),F\left(0,\frac{4}{5}\right)\right)\\ \le & k\left[G\left(0,0,0\right)+G\left(1,\frac{4}{5},\frac{4}{5}\right)\right]\\ =& \frac{2k}{5}\le \frac{1}{5},\end{array}$

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## Acknowledgements

The authors are indebted to the referees for their careful reading of the manuscript and valuable suggestions. Hui-Sheng Ding acknowledges support from the NSF of China (11101192), the Key Project of Chinese Ministry of Education (211090), the NSF of Jiangxi Province (20114BAB211002), the Jiangxi Provincial Education Department (GJJ12173), and the Program for Cultivating Youths of Outstanding Ability in Jiangxi Normal University.

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Correspondence to Hui-Sheng Ding.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Ding, HS., Karapınar, E. A note on some coupled fixed-point theorems on G-metric spaces. J Inequal Appl 2012, 170 (2012). https://doi.org/10.1186/1029-242X-2012-170

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• DOI: https://doi.org/10.1186/1029-242X-2012-170

### Keywords

• coupled fixed point
• coincidence point
• mixed g-monotone property
• ordered set
• G-metric space 