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Fixed point theorems for weak Ccontractions in partially ordered 2metric spaces
Fixed Point Theory and Applicationsvolume 2013, Article number: 161 (2013)
Abstract
The aim of this paper is to state some fixed point results for weak Ccontractions in a partially ordered 2metric space. Examples are given to illustrate the results.
1 Introduction and preliminaries
Chatterjea in [1] introduced the notion of a Ccontraction.
Definition 1.1 [1]
Let $(X,d)$ be a metric space and $T:X\to X$ be a map. Then T is called a Ccontraction if there exists $\alpha \in (0,\frac{1}{2})$ such that for all $x,y\in X$, $d(Tx,Ty)\le \alpha [d(x,Ty)+d(y,Tx)]$.
This notion was generalized to a weak Ccontraction by Choudhury in [2].
Definition 1.2 ([2], Definition 1.3)
Let $(X,d)$ be a metric space and $T:X\to X$ be a map. Then T is called a weak Ccontraction if there exists $\psi :{[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty})$ which is continuous, and $\psi (s,t)=0$ if and only if $s=t=0$ such that
for all $x,y\in X$.
In [2], Choudhury proved that if X is a complete metric space, then every weak Ccontraction has a unique fixed point; see [[2], Theorem 2.1]. This result was generalized to a complete, partially ordered metric space in [3]; see [[3], Theorems 2.1, 2.3 and 3.1].
There were some generalizations of a metric such as a 2metric, a Dmetric, a Gmetric, a cone metric, and a complexvalued metric. The notion of a 2metric has been introduced by Gähler in [4]. Note that a 2metric is not a continuous function of its variables, whereas an ordinary metric is. This led Dhage to introduce the notion of a Dmetric in [5]. But in [6] Mustafa and Sims showed that most of topological properties of Dmetric were not correct. In [7] Mustafa and Sims introduced the notion of a Gmetric to overcome flaws of a Dmetric. After that, many fixed point theorems on Gmetric spaces have been stated. However, it was shown in [8] and [9] that in several situations fixed point results in Gmetric spaces can be in fact deduced from fixed point theorems in metric or quasimetric spaces.
In [10] Huang and Zhang defined the notion of a cone metric. After that, many authors extended some fixed point theorems on metric spaces to cone metric spaces. However, it was shown later by various authors that in several cases the fixed point results in cone metric spaces can be obtained by reducing them to their standard metric counterparts; for example, see [11–14]. In [15] Azam, Fisher and Khan have introduced the notion of a complexvalued metric and some fixed point theorems have been stated. But in [16] Sastry, Naidu and Bekeshie showed that some fixed point theorems recently generalized to complexvalued metric spaces are consequences of their counterparts in the setting of metric spaces and hence are redundant.
Note that in the above generalizations, only a 2metric space has not been known to be topologically equivalent to an ordinary metric. Then there was no easy relationship between results obtained in 2metric spaces and metric spaces. In particular, the fixed point theorems on 2metric spaces and metric spaces may be unrelated easily. For the fixed point theorems on 2metric spaces, the readers may refer to [17–26].
The aim of this paper is to state some fixed point results for weak Ccontractions in a partially ordered 2metric space. Examples are given to illustrate the results.
First we recall some notions and lemmas which will be useful in what follows.
Definition 1.3 [4]
Let X be a nonempty set and let $d:X\times X\times X\to \mathbb{R}$ be a map satisfying the following conditions:

1.
For every pair of distinct points $x,y\in X$, there exists a point $z\in X$ such that $d(x,y,z)\ne 0$.

2.
If at least two of three points $x,y,z$ are the same, then $d(x,y,z)=0$.

3.
The symmetry: $d(x,y,z)=d(x,z,y)=d(y,x,z)=d(y,z,x)=d(z,x,y)=d(z,y,x)$ for all $x,y,z\in X$.

4.
The rectangle inequality: $d(x,y,z)\le d(x,y,t)+d(y,z,t)+d(z,x,t)$ for all $x,y,z,t\in X$.
Then d is called a 2metric on X and $(X,d)$ is called a 2metric space which will be sometimes denoted by X if there is no confusion. Every member $x\in X$ is called a point in X.
Definition 1.4 [4]
Let $(X,d)$ be a 2metric space and $a,b\in X$, $r\ge 0$. The set
is called a 2ball centered at a and b with radius r. The topology generated by the collection of all 2balls as a subbasis is called a 2metric topology on X.
Definition 1.5 [22]
Let $\{{x}_{n}\}$ be a sequence in a 2metric space $(X,d)$.

1.
$\{{x}_{n}\}$ is said to be convergent to x in $(X,d)$, written ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$, if for all $a\in X$, ${lim}_{n\to \mathrm{\infty}}d({x}_{n},x,a)=0$.

2.
$\{{x}_{n}\}$ is said to be Cauchy in X if for all $a\in X$, ${lim}_{n,m\to \mathrm{\infty}}d({x}_{n},{x}_{m},a)=0$, that is, for each $\epsilon >0$, there exists ${n}_{0}$ such that $d({x}_{n},{x}_{m},a)<\epsilon $ for all $n,m\ge {n}_{0}$.

3.
$(X,d)$ is said to be complete if every Cauchy sequence is a convergent sequence.
Definition 1.6 ([24], Definition 8)
A 2metric space $(X,d)$ is said to be compact if every sequence in X has a convergent subsequence.
Lemma 1.7 ([24], Lemma 3)
Every 2metric space is a ${T}_{1}$space.
Lemma 1.8 ([24], Lemma 4)
${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in a 2metric space $(X,d)$ if and only if ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in the 2metric topological space X.
Lemma 1.9 ([24], Lemma 5)
If $T:X\to Y$ is a continuous map from a 2metric space X to a 2metric space Y, then ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$ in X implies ${lim}_{n\to \mathrm{\infty}}T{x}_{n}=Tx$ in Y.
Remark 1.10

1.
It is straightforward from Definition 1.3 that every 2metric is nonnegative and every 2metric space contains at least three distinct points.

2.
A 2metric $d(x,y,z)$ is sequentially continuous in one argument. Moreover, if a 2metric $d(x,y,z)$ is sequentially continuous in two arguments, then it is sequentially continuous in all three arguments, see [[27], p.975].

3.
A convergent sequence in a 2metric space need not be a Cauchy sequence, see [[27], Remark 01 and Example 01].

4.
In a 2metric space $(X,d)$, every convergent sequence is a Cauchy sequence if d is continuous, see [[27], Remark 02].

5.
There exists a 2metric space $(X,d)$ such that every convergent sequence is a Cauchy sequence but d is not continuous, see [[27], Remark 02 and Example 02].
2 Main results
First, we introduce the notion of a weak Ccontraction on a partially ordered 2metric space.
Definition 2.1 Let $(X,\u2aaf,d)$ be a partially ordered 2metric space and $T:X\to X$ be a map. Then T is called a weak Ccontraction if there exists $\psi :{[0,\mathrm{\infty})}^{2}\to [0,\mathrm{\infty})$ which is continuous, and $\psi (s,t)=0$ if and only if $s=t=0$ such that
for all $x,y,a\in X$ and $x\u2aafy$ or $y\u2aafx$.
The following example gives some examples of ψ in Definition 2.1. Note that in [[2], Example 2.1], Choudhury considered the function $\psi (a,b)=\frac{1}{2}min\{a,b\}$ for all $a,b\in [0,\mathrm{\infty})$. Unfortunately, for this function, we have $\psi (0,1)=0$, which is a contradiction to the condition that $\psi (s,t)=0$ if and only if $s=t=0$ in [[2], Definition 1.3], also in Definition 2.1.
Example 2.2

1.
$\psi (a,b)=\frac{a+b}{2}$ for all $a,b\in [0,\mathrm{\infty})$.

2.
$\psi (a,b)=\frac{max\{a,b\}}{4}$ for all $a,b\in [0,\mathrm{\infty})$.
The first result is a sufficient condition for the existence of a fixed point of a weak Ccontraction on a 2metric space. For the preceding one in metric spaces, see [[3], Theorem 2.1].
Theorem 2.3 Let $(X,\u2aaf,d)$ be a complete, partially ordered 2metric space and $T:X\to X$ be a weak Ccontraction such that:

1.
T is continuous and nondecreasing.

2.
There exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$.
Then T has a fixed point.
Proof If ${x}_{0}=T{x}_{0}$, then the proof is finished. Suppose now that ${x}_{0}\prec T{x}_{0}$. Since T is a nondecreasing map, we have ${x}_{0}\prec T{x}_{0}\u2aaf{T}^{2}{x}_{0}\u2aaf\cdots \u2aaf{T}^{n}{x}_{0}\u2aaf\cdots $ . Put ${x}_{n+1}=T{x}_{n}$. Then, for all $n\ge 1$, from (2.1) and noting that ${x}_{n1}$ and ${x}_{n}$ are comparable, we get
for all $a\in X$. By choosing $a={x}_{n1}$ in (2.2), we obtain $d({x}_{n+1},{x}_{n},{x}_{n1})\le 0$, that is,
It follows from (2.2) and (2.3) that
It implies that
Thus $\{d({x}_{n},{x}_{n+1},a)\}$ is a decreasing sequence of nonnegative real numbers and hence it is convergent. Let
Taking the limit as $n\to \mathrm{\infty}$ in (2.4) and using (2.6), we get
That is,
Taking the limit as $n\to \mathrm{\infty}$ in (2.2) and using (2.6), (2.7), we get $r\le \frac{1}{2}2r\psi (0,2r)\le \frac{1}{2}2r=r$. It implies that $\psi (0,2r)=0$, that is, $r=0$. Then (2.6) becomes
From (2.5), we have if $d({x}_{n1},{x}_{n},a)=0$, then $d({x}_{n},{x}_{n+1},a)=0$. Since $d({x}_{0},{x}_{1},{x}_{0})=0$, we have $d({x}_{n},{x}_{n+1},{x}_{0})=0$ for all $n\in \mathbb{N}$. Since $d({x}_{m1},{x}_{m},{x}_{m})=0$, we have
for all $n\ge m1$. For $0\le n<m1$, noting that $m1\ge n+1$, from (2.9) we have
It implies that
Since $d({x}_{n},{x}_{n+1},{x}_{n+1})=0$, from (2.10) we have
for all $0\le n<m1$. From (2.9) and (2.11), we have $d({x}_{n},{x}_{n+1},{x}_{m})=0$ for all $n,m\in \mathbb{N}$.
Now, for all $i,j,k\in \mathbb{N}$ with $i<j$, we have $d({x}_{j1},{x}_{j},{x}_{i})=d({x}_{j1},{x}_{j},{x}_{k})=0$. Therefore,
This proves that for all $i,j,k\in \mathbb{N}$
In what follows, we will prove that $\{{x}_{n}\}$ is a Cauchy sequence. Suppose to the contrary that $\{{x}_{n}\}$ is not a Cauchy sequence. Then there exists $\epsilon >0$ for which we can find subsequences $\{{x}_{m(k)}\}$ and $\{{x}_{n(k)}\}$ where $n(k)$ is the smallest integer such that $n(k)>m(k)>k$ and
for all $k\in \mathbb{N}$. Therefore,
By using (2.12), (2.13) and (2.14), we have
Taking the limit as $k\to \mathrm{\infty}$ in (2.15) and using (2.8), we have
Also, from (2.12), we have
and
Taking the limit as $k\to \mathrm{\infty}$ in (2.17), (2.18) and using (2.8), (2.16), we obtain
Since $n(k)>m(k)$ and ${x}_{n(k)1}$, ${x}_{m(k)1}$ are comparable, by using (2.1), we have
Taking the limit as $k\to \mathrm{\infty}$ in (2.20) and using (2.16), (2.19) and the continuity of ψ, we have
This proves that $\psi (\epsilon ,\epsilon )=0$, that is, $\epsilon =0$. It is a contradiction. This proves that $\{{x}_{n}\}$ is a Cauchy sequence. Since X is complete, there exists $z\in X$ such that ${lim}_{n\to \mathrm{\infty}}{x}_{n}=z$. It follows from the continuity of T that $z={lim}_{n\to \mathrm{\infty}}{x}_{n+1}={lim}_{n\to \mathrm{\infty}}T{x}_{n}=Tz$. Then z is a fixed point of T. □
The next result is another one for the existence of the fixed point of a weak Ccontraction on a 2metric space. For the preceding one in metric spaces, see [[3], Theorem 2.2].
Theorem 2.4 Let $(X,\u2aaf,d)$ be a complete, partially ordered 2metric space and $T:X\to X$ be a weak Ccontraction such that:

1.
T is nondecreasing.

2.
If $\{{x}_{n}\}$ is nondecreasing such that ${lim}_{n\to \mathrm{\infty}}{x}_{n}=x$, then ${x}_{n}\u2aafx$ for all $n\in \mathbb{N}$.

3.
There exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$.
Then T has a fixed point.
Proof As in the proof of Theorem 2.3, we have a Cauchy sequence $\{{x}_{n}\}$ with ${lim}_{n\to \mathrm{\infty}}{x}_{n}=z$ in X. We only have to prove that $Tz=z$. Since $\{{x}_{n}\}$ is nondecreasing and ${lim}_{n\to \mathrm{\infty}}{x}_{n}=z$, we have ${x}_{n}\u2aafz$ for all $n\in \mathbb{N}$. It follows from (2.1) that
Taking the limit as $n\to \mathrm{\infty}$ in (2.21), we have
It implies that $d(z,Tz,a)=0$ for all $a\in X$, that is, $Tz=z$. □
In what follows, we prove a sufficient condition for the uniqueness of the fixed point in Theorem 2.3 and Theorem 2.4.
Theorem 2.5 Suppose that:

1.
Either hypotheses of Theorem 2.3 or hypotheses of Theorem 2.4 hold.

2.
For each $x,y\in X$, there exists $z\in X$ that is comparable to x and y.
Then T has a unique fixed point.
Proof As in the proofs of Theorem 2.3 and Theorem 2.4, we see that T has a fixed point.
Now we prove the uniqueness of the fixed point of T. Let x, y be two fixed points of T. We consider the following two cases.
Case 1. x is comparable to y. Then ${T}^{n}x$ is comparable to ${T}^{n}y$ for all $n\in \mathbb{N}$. For all $a\in X$, we have
This proves that $\psi (d(x,y,a),d(y,x,a))=0$, that is, $d(x,y,a)=0$ for all $a\in X$. Then $x=y$.
Case 2. x is not comparable to y. Then there exists $z\in X$ that is comparable to x and y. It implies that ${T}^{n}z$ is comparable to ${T}^{n}x=x$ and ${T}^{n}y=y$. For all $n\in \mathbb{N}$ and $a\in X$, we have
It implies that $d(x,{T}^{n}z,a)\le d(x,{T}^{n1}z,a)$. Then there exists ${lim}_{n\to \mathrm{\infty}}d(x,{T}^{n}z,a)=l$. Taking the limit as $n\to \mathrm{\infty}$ in (2.22) and noting that ψ is continuous, we have $l\le \frac{1}{2}(l+l)\psi (l,l)\le l$. This proves that $\psi (l,l)=0$. Then $l=0$, that is, ${lim}_{n\to \mathrm{\infty}}{T}^{n}z=x$. Similarly, ${lim}_{n\to \mathrm{\infty}}{T}^{n}z=y$. By Lemma 1.7, we get $x=y$. □
Remark 2.6 Note that if $(X,\u2aaf)$ is totally ordered, then the condition (2) in Theorem 2.5 is always satisfied.
The following result is an analogue of [[3], Theorem 3.1].
Theorem 2.7 Let $(X,\u2aaf,d)$ be a complete, partially ordered 2metric space and $T:X\to X$ be a weak Ccontraction such that:

1.
For all $x,y\in X$, if $x\u2aafy$ then $Tx\u2ab0Ty$.

2.
For each $x,y\in X$, there exists $z\in X$ that is comparable to x and y.

3.
There exists ${x}_{0}\in X$ with ${x}_{0}\u2aafT{x}_{0}$ or ${x}_{0}\u2ab0T{x}_{0}$.
Then, for all $a\in X$, $inf\{d(x,Tx,a):x\in X\setminus \{a\}\}=0$. In particular, $inf\{d(x,Tx,a):x\in X\}=0$.
Proof We consider the following two cases.
Case 1. ${x}_{0}\u2aafT{x}_{0}$. By the hypothesis (1), consecutive terms of the sequence $\{{T}^{n}{x}_{0}\}$ are comparable. It follows from (2.1) that for all $a\in X$,
As in the proof of (2.12) of Theorem 2.3, we have $d({x}_{i},{x}_{j},{x}_{k})=0$ for all $i,j,k\in \mathbb{N}$. Then (2.23) implies
That is, $d({T}^{n+1}{x}_{0},{T}^{n}{x}_{0},a)\le d({T}^{n}{x}_{0},{T}^{n1}{x}_{0},a)$. Then there exists ${lim}_{n\to \mathrm{\infty}}d({T}^{n+1}{x}_{0},{T}^{n}{x}_{0},a)=r$. As in the proof of Theorem 2.3, we get $r=0$. Then ${lim}_{n\to \mathrm{\infty}}d({T}^{n+1}{x}_{0},{T}^{n}{x}_{0},a)=0$. That is, $inf\{d(x,Tx,a):x\in X\}=0$.
Case 2. ${x}_{0}\succ T{x}_{0}$. The same as in Case 1. □
For each $a\in X$, if ${d}_{a}(x,y)=d(x,y,a)$ for all $x,y\in X$ is a metric on X, then the formula (2.1) becomes (1.1). Also, the above proofs may be similar to the method used in [2] and [3]. The following example guarantees that this fact is not true in general.
Example 2.8 There exists a 2metric space $(X,d)$ such that for each $a\in X$, the formula ${d}_{a}(x,y)=d(x,y,a)$ for all $x,y\in X$ is not a metric on X.
Proof Let $X=[0,+\mathrm{\infty})$ and
Then $(X,d)$ is a 2metric space. For each $a\in X$, we have
If $a=0$, then we have
If $a\ne 0$, then we have
This proves that ${d}_{a}$ is not a metric on X for all $a\in X$. □
The following example shows that hypotheses in Theorem 2.3 and Theorem 2.4 do not guarantee the uniqueness of the fixed point.
Example 2.9 Let $X=\{(1,0),(0,1),(2,2)\}\subset {\mathbb{R}}^{2}$ with the order
Define a 2metric d on X as follows:
Then $(X,\u2aaf,d)$ is a partially ordered, complete 2metric space whose different elements are not comparable. The identity map $T(x,y)=(x,y)$ for all $(x,y)\in X$ is continuous, nondecreasing, and contraction conditions in Theorem 2.3 and Theorem 2.4 are satisfied. Moreover, $(1,0)\u2aafT(1,0)$ and T has more than one fixed point.
The following example is an illustration of Theorem 2.4 and Theorem 2.7.
Example 2.10 Let $X=\{a,b,c\}$ with the order $x\u2aafy$ if and only if $x=y$ for all $x,y\in X$. Let d be a 2metric on X defined by the symmetry of all three variables and
Let $T:X\to X$ be defined by $Ta=b$, $Tb=a$, $Tc=c$. It is easy to see that Theorem 2.4 and Theorem 2.7 are applicable to T and c is a unique fixed point of T. Moreover, the condition (2) in Theorem 2.5 does not hold, then it is not a necessary condition of the uniqueness of the fixed point.
Note that, in [[3], Theorem 3.1], if X is a compact metric space and T is continuous, then T has a unique fixed point. The following example shows that, in Theorem 2.7, if X is a compact 2metric space and T is continuous, then T may not have a unique fixed point.
Example 2.11 Let $(X,d)$ be a 2metric space as in Example 2.10. Let $T:X\to X$ be defined by
We see that all assumptions in Theorem 2.7 are satisfied but T has more than one fixed point.
Finally, Example 2.12 and Example 2.13 show that the above results can be used to prove the existence of a fixed point when standard arguments in metric spaces in [2] and [3] fail, even for trivial maps.
Example 2.12 Let X be the 2metric space $(R,\sigma )$ on [[4], p.145] with the usual order and $Tx=0$ for all $x\in X$. Then we have:

1.
X is a complete, totally ordered 2metric space.

2.
X is not metrizable.

3.
T is a Cweak contraction on the 2metric space X.
Proof (1) and (2) See [[4], p.145].
(3) By choosing $\psi (a,b)=\frac{a+b}{2}$ for all $a,b\in [0,+\mathrm{\infty})$, then the condition (2.1) holds. This proves that T is a Cweak contraction on the 2metric space X. □
Example 2.13 Let $X=\{0,1,2,\dots ,n,\dots \}$ with the usual order,
and $Tx=0$ for all $x\in X$. Then we have:

1.
$(X,d)$ is a complete, totally ordered 2metric space.

2.
$(X,d)$ is not completely metrizable, that is, there does not exist any metric ρ on X such that the metric topology and the completeness on $(X,\rho )$ are coincident with the 2metric topology and the completeness on $(X,d)$, respectively.

3.
T is a Cweak contraction on the 2metric space $(X,d)$.
Proof (1) It is easy to see that $(X,d)$ is a partially ordered 2metric space.
Let $\{{x}_{n}\}$ be a Cauchy sequence in $(X,d)$. We have ${lim}_{n,m\to \mathrm{\infty}}d({x}_{n},{x}_{m},a)=0$ for all $a\in X$. Then, for each $a\in X$, there exists ${n}_{0}(a)$ such that $d({x}_{n},{x}_{m},a)=0$ for all $n,m\ge {n}_{0}(a)$. We consider the following three cases.
Case 1. For all $n,m\ge {n}_{0}(a)$, ${x}_{n}={x}_{m}$. This proves that $\{{x}_{n}\}$ is convergent.
Case 2. For all $n\ge {n}_{0}(a)$, ${x}_{n}=a$. It is a contradiction because a is an arbitrary point of X.
Case 3. For all $n,m\ge {n}_{0}(a)$ and all $k\ge 1$, ${x}_{n}\ne {x}_{m}\ne a$ and $\{k,k+1\}\not\subset \{{x}_{n},{x}_{m},a\}$. Then $d({x}_{n},0,a)=0$ for all $n\ge {n}_{0}(a)$. This proves that $\{{x}_{n}\}$ is convergent.
By the above three cases, the Cauchy sequence $\{{x}_{n}\}$ is convergent in $(X,d)$. This proves that $(X,d)$ is complete.
(2) Since ${lim}_{n\to \mathrm{\infty}}d(n,0,a)=0$, we have $\{n\}$ is a convergent sequence in $(X,d)$. On the other hand, ${lim}_{n\to \mathrm{\infty}}d(n,n+1,a)=1$, then $\{n\}$ is not a Cauchy sequence in $(X,d)$. This proves that $(X,d)$ is not completely metrizable.
(3) By choosing $\psi (a,b)=\frac{a+b}{2}$, the condition (2.1) holds. This proves that T is a Cweak contraction on the 2metric space $(X,d)$. □
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Keywords
 Point Theorem
 Distinct Point
 Cauchy Sequence
 Convergent Subsequence
 Convergent Sequence