 Research
 Open Access
Further fixed point results on Gmetric spaces
 Erdal Karapınar^{1}Email author and
 Ravi P Agarwal^{2, 3}
https://doi.org/10.1186/168718122013154
© Karapınar and Agarwal; licensee Springer. 2013
 Received: 26 February 2013
 Accepted: 28 May 2013
 Published: 17 June 2013
Abstract
Very recently, Samet et al. (Int. J. Anal. 2013:917158, 2013) and JleliSamet (Fixed Point Theory Appl. 2012:210, 2012) noticed that some fixed point theorems in the context of a Gmetric space can be deduced by some wellknown results in the literature in the setting of a usual (quasi) metric space. In this paper, we note that the approach of Samet et al. (Int. J. Anal. 2013:917158, 2013) and JleliSamet (Fixed Point Theory Appl. 2012:210, 2012) is inapplicable unless the contraction condition in the statement of the theorem can be reduced into two variables. For this purpose, we modify some existing results to suggest new fixed point theorems that fit with the nature of a Gmetric space. The expressions in our result, the contraction condition, cannot be expressed in two variables, therefore the techniques used in (Int. J. Anal. 2013:917158, 2013; Fixed Point Theory Appl. 2012:210, 2012) are not applicable.
MSC:47H10, 54H25.
Keywords
 coupled fixed point
 fixed point
 ordered set
 metric space
1 Introduction
The concept of Gmetric space was introduced by Mustafa and Sims [1] in order to extend and generalize the notion of metric space. In this paper, the authors characterized the Banach contraction mapping principle [2] in the context of a Gmetric space. Following this initial report, a number of authors have characterized many wellknown fixed point theorems in the setting of Gmetric space (see, e.g., [1, 3–36]). Since one is adapted from the other, there is a close relation between a usual metric space and a Gmetric space (see, e.g., [1, 23–27]). In fact, the nature of a Gmetric space is to understand the geometry of three points instead of two points via perimeter of a triangle. However, most of the published papers dealing with a Gmetric space did not give much importance to these details. Consequently, a great majority of results were obtained by transforming the contraction conditions from the usual metric space context to a Gmetric space without carrying enough of the characteristics of the Gmetric.
Very recently, Samet et al. [37] and JleliSamet [38] observed that some fixed point theorems in the context of a Gmetric space in the literature can be concluded by some existing results in the setting of a (quasi)metric space. In fact, if the contraction condition of the fixed point theorem on a Gmetric space can be reduced to two variables instead of three variables, then one can construct an equivalent fixed point theorem in the setup of a usual metric space. More precisely, in [37, 38], the authors noticed that $d(x,y)=G(x,y,y)$ forms a quasimetric. Hence, if one can transform the contraction condition of existence results in a Gmetric space in such terms, $G(x,y,y)$, then the related fixed point results become the known fixed point results in the context of a quasimetric space.
In this paper, we notice that the techniques used in [37, 38] are valid if the contraction condition in the statement of the theorem can be expressed in two variables. Furthermore, we prove some fixed point theorems in the context of a Gmetric space for which the techniques in [37, 38] are inapplicable.
2 Preliminaries
In this section we recollect basic definitions and a detailed overview of the fundamental results. Throughout this paper, ℕ is the set of nonnegative integers, and ${\mathbb{N}}^{\ast}$ is the set of positive integers.
Definition 2.1 (See [1])
 (G1)
$G(x,y,z)=0$ if $x=y=z$,
 (G2)
$0<G(x,x,y)$ for all $x,y\in X$ with $x\ne y$,
 (G3)
$G(x,x,y)\le G(x,y,z)$ for all $x,y,z\in X$ with $y\ne z$,
 (G4)
$G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots $ (symmetry in all three variables),
 (G5)
$G(x,y,z)\le G(x,a,a)+G(a,y,z)$ for all $x,y,z,a\in X$ (rectangle inequality).
Then the function G is called a generalized metric or, more specifically, a Gmetric on X, and the pair $(X,G)$ is called a Gmetric space.
for all $x,y,z\in X$, is a Gmetric on X.
Definition 2.2 (See [1])
that is, for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $G(x,{x}_{n},{x}_{m})<\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 2.1 (See [1])
 (1)
$\{{x}_{n}\}$ is Gconvergent to x,
 (2)
$G({x}_{n},{x}_{n},x)\to 0$ as $n\to +\mathrm{\infty}$,
 (3)
$G({x}_{n},x,x)\to 0$ as $n\to +\mathrm{\infty}$,
 (4)
$G({x}_{n},{x}_{m},x)\to 0$ as $n,m\to +\mathrm{\infty}$.
Definition 2.3 (See [1])
Let $(X,G)$ be a Gmetric space. A sequence $\{{x}_{n}\}$ is called a GCauchy sequence if, for any $\epsilon >0$, there is $N\in \mathbb{N}$ such that $G({x}_{n},{x}_{m},{x}_{l})<\epsilon $ for all $m,n,l\ge N$, that is, $G({x}_{n},{x}_{m},{x}_{l})\to 0$ as $n,m,l\to +\mathrm{\infty}$.
Proposition 2.2 (See [1])
 (1)
the sequence $\{{x}_{n}\}$ is GCauchy,
 (2)
for any $\epsilon >0$, there exists $N\in \mathbb{N}$ such that $G({x}_{n},{x}_{m},{x}_{m})<\epsilon $ for all $m,n\ge N$.
Definition 2.4 (See [1])
A Gmetric space $(X,G)$ is called Gcomplete if every GCauchy sequence is Gconvergent in $(X,G)$.
We will use the following result which can be easily derived from the definition of a Gmetric space (see, e.g., [1]).
Definition 2.5 (See [1])
Let $(X,G)$ be a Gmetric space. A mapping $T:X\to X$ is said to be Gcontinuous if $\{T({x}_{n})\}$ is Gconvergent to $T(x)$ where $\{{x}_{n}\}$ is any Gconvergent sequence converging to x.
In [22], Mustafa characterized the wellknown Banach contraction mapping principle in the context of Gmetric spaces in the following ways.
Theorem 2.1 (See [22])
where $k\in [0,1)$. Then T has a unique fixed point.
Theorem 2.2 (See [22])
where $k\in [0,1)$. Then T has a unique fixed point.
Remark 2.1 The condition (2) implies the condition (3). The converse is true only if $k\in [0,\frac{1}{2})$. For details, see [22].
Theorem 2.3 (See [26])
for all x, y, z, where a, b, c, d are positive constants such that $k=a+b+c+d<1$. Then there is a unique $x\in X$ such that $Tx=x$.
Theorem 2.4 (See [27])
for all x, y, z, where $k\in [0,\frac{1}{3})$. Then there is a unique $x\in X$ such that $Tx=x$.
Theorem 2.5 (See [26])
for all x, y, z, where a, b are positive constants such that $k=a+b<1$. Then there is a unique $x\in X$ such that $Tx=x$.
Theorem 2.6 (See [26])
for all x, y, z, where a, b are positive constants such that $k=a+b<1$. Then there is a unique $x\in X$ such that $Tx=x$.
Theorem 2.7 (See [25])
for all x, y, z, where $k\in [0,\frac{1}{2})$. Then there is a unique $x\in X$ such that $Tx=x$.
Theorem 2.8 (See, e.g., [38])
for all $x,y\in X$, where $\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ is continuous with ${\phi}^{1}(\{0\})=0$. Then there is a unique $x\in X$ such that $Tx=x$.
Definition 2.6 (See, e.g., [38])
A quasimetric on a nonempty set X is a mapping $p:X\times X\to [0,\mathrm{\infty})$ such that
(p_{1}) $x=y$ if and only if $p(x,y)=0$,
(p_{2}) $p(x,y)\le p(x,z)+p(z,y)$,
for all $x,y,z\in X$. A pair $(X,p)$ is said to be a quasimetric space.
is a metric on X.
for all $x\in X$, where q is a constant such that $q\in [0,1)$. Then T has a unique fixed point.
Samet et al. [37] proved that Theorem 2.4Theorem 2.7 are the consequences of Theorem 2.9 by using the following proposition.
 (A)
If $(X,G)$ is a complete Gmetric space, then $(X,d)$ is a complete metric space.
 (B)
If $(X,G)$ is a sequentially Gcompact Gmetric space, then $(X,d)$ is a compact metric space.
3 Main results
We first state the following theorem about the existence and uniqueness of a common fixed point, which is a generalization of Theorem 2.7. Furthermore, the techniques of the papers [37, 38] are not applicable to this theorem.
Then there is a unique $x\in X$ such that $Tx=x$.
a contradiction since $0\le k<\frac{1}{2}$.
where $h=\frac{k}{1k}<1$ since $0\le k<\frac{1}{2}$.
where $0\le k<\frac{1}{2}$.
by the rectangular property (G5). Since $0\le k<\frac{1}{2}$, the inequalities above yield contradictions. Hence we have $G({x}^{\ast},T{x}^{\ast},T{x}^{\ast})=0$, that is, ${x}^{\ast}=T{x}^{\ast}$.
Since $0\le k<\frac{1}{2}$, the expressions (26) and (27) yield contradictions. Thus, ${x}^{\ast}$ is the unique fixed point of T. □
In Theorem 3.1, the interval of constant of the contractive condition can be extended to the interval $[0,1)$ by eliminating the same terms. Since the proof is the mimic of Theorem 3.1, we omit it.
Then there is a unique $x\in X$ such that $Tx=x$.
Remark 3.1 Theorem 2.1Theorem 2.6 are the consequences of Theorem 3.1 and Theorem 3.2.
Inspired by Theorem 2.8, we state the following theorem for which the methods in [37, 38] are not applicable.
for all $x,y\in X$, where $\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ is continuous with ${\phi}^{1}(\{0\})=0$. Then there is a unique $x\in X$ such that $Tx=x$.
a contradiction. Hence, T has a unique fixed point.
a contradiction. Hence, $\{{x}_{n}\}$ is a GCauchy sequence. Since $(X,G)$ is Gcomplete, there is $z\in X$ such that ${x}_{n}\to z$.
Hence $G(z,z,Tz)=0$, that is, $Tz=z$. □
for all $x,y\in X$. It is easy to see that both mappings p and q do not satisfy the conditions of Definition 2.6. Hence, Theorem 3.1 and Theorem 3.3 cannot be characterized in the context of quasimetric as it is suggested in [37, 38].
and $\phi (t)=\frac{3}{4}t$ for all $t\in [0,+\mathrm{\infty})$.
Proof For the proof the Example 3.1, we examine the following cases:

Let $0\le x,y<1/3$. Then$\begin{array}{rcl}G(Tx,{T}^{2}x,Ty)& =& max\{\frac{1}{4}x,\frac{1}{16}x,\frac{1}{4}y\}\le \frac{1}{4}max\{x,\frac{1}{4}x,y\}\\ =& G(x,Tx,y)\phi (G(x,Tx,y)).\end{array}$

Let $1/3\le x,y<1$. Then$\begin{array}{rcl}G(Tx,{T}^{2}x,Ty)& =& max\{\frac{1}{8}{x}^{4},\frac{1}{64}{x}^{16},\frac{1}{8}{y}^{4}\}\le \frac{1}{4}max\{x,\frac{1}{8}{x}^{4},y\}\\ =& G(x,Tx,y)\phi (G(x,Tx,y)).\end{array}$

Let $0\le x<1/3\le y<1$. Then$\begin{array}{rcl}G(Tx,{T}^{2}x,Ty)& =& max\{\frac{1}{4}x,\frac{1}{16}x,\frac{1}{8}{y}^{4}\}\le \frac{1}{4}max\{x,\frac{1}{4}x,y\}\\ =& G(x,Tx,y)\phi (G(x,Tx,y)).\end{array}$

Let $0\le y<1/3\le x<1$. Then$\begin{array}{rcl}G(Tx,{T}^{2}x,Ty)& =& max\{\frac{1}{8}{x}^{4},\frac{1}{64}{x}^{16},\frac{1}{4}y\}\le \frac{1}{4}max\{x,\frac{1}{8}{x}^{4},y\}\\ =& G(x,Tx,y)\phi (G(x,Tx,y)).\end{array}$
Then the conditions of Theorem 3.3 hold and T has a unique fixed point. Notice that $(0,0,0)$ is the desired fixed point of T. □
Declarations
Acknowledgements
The authors express their gratitude to the anonymous referees for constructive and useful remarks, comments and suggestions. The first author thank to student Peyman Salimi for his help for Example 3.1.
Authors’ Affiliations
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