New extension of p-metric spaces with some fixed-point results on M-metric spaces

Abstract

In this paper, we extend the p-metric space to an M-metric space, and we shall show that the definition we give is a real generalization of the p-metric by presenting some examples. In the sequel we prove some of the main theorems by generalized contractions for getting fixed points and common fixed points for mappings.

1 Introduction and preliminaries

In 1994, in [1] Matthews introduced the notion of a partial metric space and proved the contraction principle of Banach in this new framework. Next, many fixed-point theorems in partial metric spaces have been given by several mathematicians. Recently Haghi et al. published [2] a paper which stated that we should ‘be careful on partial metric fixed point results’ along with giving some results. They showed that fixed-point generalizations to partial metric spaces can be obtained from the corresponding results in metric spaces.

In this paper, we extend the p-metric space to an M-metric space, and we shall show that our definition is a real generalization of the p-metric by presenting some examples. In the sequel we prove some of the main theorems by generalized contractions for getting fixed points and common fixed points for mappings.

Definition 1.1 ([1], [[3], Definition 1.1])

A partial metric on a nonempty set X is a function $p:X\times X\to {\mathbb{R}}^{+}$ such that for all $x,y,z\in X$:

• (p1) $p(x,x)=p(y,y)=p(x,y)\iff x=y$,

• (p2) $p(x,x)\le p(x,y)$,

• (p3) $p(x,y)=p(y,x)$,

• (p4) $p(x,y)\le p(x,z)+p(z,y)-p(z,z)$.

A partial metric space is a pair $(X,p)$ such that X is a nonempty set and p is a partial metric on X.

Notation The following notation is useful in the sequel.

• 1. $m_{xy}:=min\{m(x,x),m(y,y)\}$,

• 2. $M_{xy}:=max\{m(x,x),m(y,y)\}$.

Now we want to extend Definition 1.1 as follows.

Definition 1.2 Let X be a nonempty set. A function $m:X\times X\to {\mathbb{R}}^{+}$ is called an m-metric if the following conditions are satisfied:

• (m1) $m(x,x)=m(y,y)=m(x,y)\iff x=y$,

• (m2) $m_{xy}\le m(x,y)$,

• (m3) $m(x,y)=m(y,x)$,

• (m4) $(m(x,y)-m_{xy})\le (m(x,z)-m_{xz})+(m(z,y)-m_{zy})$.

Then the pair $(X,m)$ is called an M-metric space.

According to the above definition the condition (p1) in the definition of [1] changes to (m1), and (p2) is expressed for $p(x,x)$ where $p(y,y)=0$ may become $p(y,y)\ne 0$. Thus we improve that condition by replacing it by $min\{p(x,x),p(y,y)\}\le p(x,y)$, and also we improve the condition (p4) extending it to the form of (m4). In the sequel we present an example that holds for the m-metric but not for the p-metric.

Remark 1.1 For every $x,y\in X$

• 1. $0\le M_{xy}+m_{xy}=m(x,x)+m(y,y)$,

• 2. $0\le M_{xy}-m_{xy}=|m(x,x)-m(y,y)|$,

• 3. $M_{xy}-m_{xy}\le (M_{xz}-m_{xz})+(M_{zy}-m_{zy})$.

The next examples show that $m^s$ and $m^w$ are ordinary metrics.

Example 1.1 Let $X:=[0,\mathrm{\infty })$. Then $m(x,y)=\frac{x+y}{2}$ on X is an m-metric.

Example 1.2 Let m be an m-metric. Put

• 1. $m^w(x,y)=m(x,y)-2m_{xy}+M_{xy}$,

• 2. $m^s(x,y)=m(x,y)-m_{xy}$ when $x\ne y$ and $m^s(x,y)=0$ if $x=y$.

Then $m^w$ and $m^s$ are ordinary metrics.

Proof If $m^w(x,y)=0$, then

$$m(x,y)=2m_{xy}-M_{xy}.$$

(1)

But from equation (1) and $m_{xy}\le m(x,y)$ we get $m_{xy}=M_{xy}=m(x,x)=m(y,y)$, so by equation (1) we obtain $m(x,y)=m(x,x)=m(y,y)$ and therefore $x=y$. For the triangle inequality it is enough that we consider Remark 1.1 and (m4). □

Remark 1.2 For every $x,y\in X$

• 1. $m(x,y)-M_{xy}\le m^w(x,y)\le m(x,y)+M_{xy}$,

• 2. $(m(x,y)-M_{xy})\le m^s(x,y)\le m(x,y)$.

In other words

$$|m^w(x,y)-m(x,y)|\le M_{xy},|m^s(x,y)-m(x,y)|\le M_{xy}.$$

In the following example we present an example of an m-metric which is not a p-metric.

Example 1.3 Let $X=\{1,2,3\}$; define

$$\begin{array}{r}m(1,1)=1,m(2,2)=9,m(3,3)=5,\\ m(1,2)=m(2,1)=10,m(1,3)=m(3,1)=7,m(3,2)=m(2,3)=7.\end{array}$$

So m is an m-metric, but it is not p-metric.

Example 1.4 Let $(X,d)$ be a metric space. Let $\varphi :[0,\mathrm{\infty })\to [\varphi (0),\mathrm{\infty })$ be a one to one and nondecreasing or strictly increasing mapping, with $\varphi (0)$ defined such that

$$\varphi (x+y)\le \varphi (x)+\varphi (y)-\varphi (0),\mathrm{\forall }x,y\ge 0.$$

Then $m(x,y)=\varphi (d(x,y))$ is an m-metric.

Proof (m1), (m2), and (m3) are clear. For (m4) we have

$$\begin{array}{c}\begin{array}{rl}\varphi (d(x,y))& \le \varphi (d(x,z)+d(z,y))\\ \le \varphi (d(x,z))+\varphi (d(z,y))-\varphi (0),\end{array}\hfill \\ (\varphi (d(x,y))-\varphi (0))\le (\varphi (d(x,z))-\varphi (0))+(\varphi (d(z,y))-\varphi (0)),\hfill \\ (m(x,y)-m_{xy})\le (m(x,z)-m_{xz})+(m(z,y)-m_{zy}).\hfill \end{array}$$

 □

Example 1.5 Let $(X,d)$ be a metric space. Then $m(x,y)=ad(x,y)+b$ where $a,b>0$ is an m-metric, because we can put $\varphi (t)=at+b$.

Remark 1.3 According to Example 1.5, by the Banach contraction

$$\mathrm{\exists }k\in [0,1),m(Tx,Ty)\le km(x,y),\text{for all }x,y\in X,$$

we have

$$m(Tx,Ty)=ad(Tx,Ty)+b\le kad(x,y)+kb\Rightarrow d(Tx,Ty)\le kd(x,y)+\frac{b(k-1)}{a},$$

which does not imply the ordinary Banach contraction

$$\mathrm{\exists }k\in [0,1),d(Tx,Ty)\le kd(x,y),\text{for all }x,y\in X,$$

for all self-maps T on X. Thus, this states that even if the m-metric m and the ordinary metric d have the same topology, the Banach contraction of the m-metric does not imply the Banach contraction of the ordinary metric d.

Lemma 1.1 Every p-metric is an m-metric.

Proof Let m be a p-metric. It is enough that we consider the following cases:

• 1. $m(x,x)=m(y,y)=m(z,z)$,

• 2. $m(x,x)<m(y,y)<m(z,z)$,

• 3. $m(x,x)=m(y,y)<m(z,z)$,

• 4. $m(x,x)=m(y,y)>m(z,z)$,

• 5. $m(x,x)<m(y,y)=m(z,z)$,

• 6. $m(x,x)>m(y,y)=m(z,z)$.

For example, to prove (2), we have

$$\begin{array}{c}m(x,y)\le m(x,z)+m(z,y)-m(z,z),\hfill \\ m(x,y)\le m(x,z)+m(z,y)-m(y,y),\hfill \\ m(x,y)-m(x,x)\le m(x,z)-m(x,x)+m(z,y)-m(y,y),\hfill \\ m(x,y)-m_{x,y}\le m(x,z)-m_{x,z}+m(z,y)-m_{z,y}.\hfill \end{array}$$

 □

2 Topology for M-metric space

It is clear that each m-metric p on X generates a $T_0$ topology ${\tau }_m$ on X. The set

$$\{B_m(x,\epsilon ):x\in X,\epsilon >0\},$$

where

$$B_m(x,\epsilon )=\{y\in X:m(x,y)<m_{x,y}+\epsilon \},$$

for all $x\in X$ and $\epsilon >0$, forms a base of ${\tau }_m$.

Definition 2.1 Let $(X,m)$ be a m-metric space. Then:

• 1. A sequence $\{x_n\}$ in a M-metric space $(X,m)$ converges to a point $x\in X$ if and only if

  $$\underset{n\to \mathrm{\infty }}{lim}(m(x_n,x)-m_{x_n,x})=0.$$

  (2)

• 2. A sequence $\{x_n\}$ in a M-metric space $(X,m)$ is called an m-Cauchy sequence if

  $$\underset{n,m\to \mathrm{\infty }}{lim}(m(x_n,x_m)-m_{x_n,x_m}),\underset{n,m\to \mathrm{\infty }}{lim}(M_{x_n,x_m}-m_{x_n,x_m})$$

  (3)

  exist (and are finite).

• 3. An M-metric space $(X,m)$ is said to be complete if every m-Cauchy sequence $\{x_n\}$ in X converges, with respect to ${\tau }_m$, to a point $x\in X$ such that

  $$(\underset{n\to \mathrm{\infty }}{lim}(m(x_n,x)-m_{x_n,x})=0\mathrm{\&}\underset{n\to \mathrm{\infty }}{lim}(M_{x_n,x}-m_{x_n,x})=0).$$

Lemma 2.1 Let $(X,m)$ be a m-metric space. Then:

• 1. $\{x_n\}$ is an m-Cauchy sequence in $(X,m)$ if and only if it is a Cauchy sequence in the metric space $(X,m^w)$.

• 2. An M-metric space $(X,m)$ is complete if and only if the metric space $(X,m^w)$ is complete. Furthermore,

  $$\underset{n\to \mathrm{\infty }}{lim}{m}^w(x_n,x)=0\iff (\underset{n\to \mathrm{\infty }}{lim}(m(x_n,x)-m_{x_n,x})=0,\underset{n\to \mathrm{\infty }}{lim}(M_{x_n,x}-m_{x_n,x})=0).$$

Likewise the above definition holds also for $m^s$.

Lemma 2.2 Assume that $x_n\to x$ and $y_n\to y$ as $n\to \mathrm{\infty }$ in an M-metric space $(X,m)$. Then

$$\underset{n\to \mathrm{\infty }}{lim}(m(x_n,y_n)-m_{x_n,y_n})=m(x,y)-m_{xy}.$$

Proof We have

$$|(m(x_n,y_n)-m_{x_n,y_n})-(m(x,y)-m_{x,y})|\le (m(x_n,x)-m_{x_n,x})+(m(y,y_n)-m_{y,y_n}).$$

 □

From Lemma 2.2 we deduce the following lemma.

Lemma 2.3 Assume that $x_n\to x$ as $n\to \mathrm{\infty }$ in an M-metric space $(X,m)$. Then

$$\underset{n\to \mathrm{\infty }}{lim}(m(x_n,y)-m_{x_n,y})=m(x,y)-m_{x,y},$$

for all $y\in X$.

Lemma 2.4 Assume that $x_n\to x$ and $x_n\to y$ as $n\to \mathrm{\infty }$ in an M-metric space $(X,m)$. Then $m(x,y)=m_{xy}$. Furthermore, if $m(x,x)=m(y,y)$, then $x=y$.

Proof By Lemma 2.2 we have

$$0=\underset{n\to \mathrm{\infty }}{lim}(m(x_n,x_n)-m_{x_n,x_n})=m(x,y)-m_{xy}.$$

 □

Lemma 2.5 Let $\{x_n\}$ be a sequence in an m-metric space $(X,m)$, such that

$$\mathrm{\exists }r\in [0,1),m(x_{n+1},x_n)\le rm(x_n,x_{n-1}),\mathrm{\forall }n\in \mathbb{N}.$$

(4)

Then

• (A) ${lim}_{n\to \mathrm{\infty }}m(x_n,x_{n-1})=0$,

• (B) ${lim}_{n\to \mathrm{\infty }}m(x_n,x_n)=0$,

• (C) ${lim}_{m,n\to \mathrm{\infty }}{m}_{x_m{x}_n}=0$,

• (D) $\{x_n\}$ is an m-Cauchy sequence.

Proof From equation (4) we have

$$m(x_n,x_{n-1})\le rm(x_{n-1},x_{n-2})\le r^{2}m(x_{n-2},x_{n-3})\le \cdots \le r^{n}m(x_0,x_1),$$

thus,

$$\underset{n\to \mathrm{\infty }}{lim}m(x_n,x_{n-1})=0,$$

which implies that (A) holds.

From (m2) and (A) we have

$$\underset{n\to \mathrm{\infty }}{lim}min\{m(x_n,x_n),m(x_{n-1},x_{n-1})\}=\underset{n\to \mathrm{\infty }}{lim}{m}_{x_n{x}_{n-1}}\le \underset{n\to \mathrm{\infty }}{lim}m(x_n,x_{n-1})=0.$$

That is, (B) holds.

Clearly, (C) holds, since ${lim}_{n\to \mathrm{\infty }}m(x_n,x_n)=0$. □

Theorem 2.1 The topology ${\tau }_m$ is not Hausdorff.

Proof Let $x,y,z\in X$ be such that

$$a:=m(x,x)<m(z,z)=\frac{a+b}{2}<b:=m(y,y)$$

with

$$\frac{b}{2}<\frac{k}{2}<m(x,y)<M_{x,y}=b,r:=2m(x,y)-a-b>0$$

and

$$max\{m(x,z),m(z,y)\}\le (2m(x,y)-k)\frac{\epsilon }{r};$$

without loss of generality we assume that for each $\epsilon >0$ we have $\epsilon <r$. We want to show that the intersection of the following neighborhoods is not empty:

$$U_x=\{z\in X:m(x,z)-m_{xz}<\epsilon \},V_y=\{z\in X:m(y,z)-m_{yz}<\epsilon \}.$$

To prove $z\in U_x$, we have

$$\begin{array}{c}m(x,z)<(2m(x,y)-k)\frac{\epsilon }{r},\hfill \\ \begin{array}{rl}m(x,z)-m_{xz}& <(2m(x,y)-k)\frac{\epsilon }{r}-a\\ <(2m(x,y)-k-a)\frac{\epsilon }{r}\\ <(2m(x,y)-a-b)\frac{\epsilon }{r}=\epsilon \end{array}\hfill \end{array}$$

and for $z\in V_y$

$$\begin{array}{c}m(y,z)<(2m(x,y)-k)\frac{\epsilon }{r},\hfill \\ \begin{array}{rl}m(x,z)-m_{yz}& <(2m(x,y)-k)\frac{\epsilon }{r}-\frac{a+b}{2}\\ <(2m(x,y)-k)\frac{\epsilon }{r}-\frac{a+b}{2}\frac{\epsilon }{r}\\ <(2m(x,y)-k-\frac{a+b}{2})\frac{\epsilon }{r}\\ <(2m(x,y)-a-b)\frac{\epsilon }{r}=\epsilon ,\end{array}\hfill \end{array}$$

so we can find $x,y\in X$ such that for all nonempty neighborhoods $U_x$ of x and $V_y$ of y we have $U_x\cap V_y\ne \mathrm{\varnothing }$. □

3 Fixed point results on M-metric space

Theorem 3.1 Let $(X,m)$ be a complete M-metric space and let $T:X\to X$ be a mapping satisfying the following condition:

$$\mathrm{\exists }k\in [0,1)\mathit{\text{ such that }}m(Tx,Ty)\le km(x,y)\mathit{\text{for all }}x,y\in X.$$

(5)

Then T has a unique fixed point.

Proof Let $x_0\in X$ and $x_n:=T{x}_{n-1}$, so we have

$$m(x_n,x_{n-1})=m(T{x}_{n-1},T{x}_{n-2})\le km(x_{n-1},x_{n-2})$$

(6)

and so (A), (B), (C), and (D) of Lemma 2.5 hold. By completeness of X we get $x_n\to x$ for some $x\in X$. Thus by equation (5) $m(T{x}_n,Tx)\le km(x_n,x)\to 0$. Hence by (m2) $m_{T{x}_n,Tx}\le m(T{x}_n,Tx)\to 0$ so by equation (2) $T{x}_n\to Tx$.

Contraction (5) implies that $m(x_n,T{x}_n)\to 0$ and $m(Tx,Tx)<m(x,x)$. Since $m_{x_n,T{x}_n}\to 0$, by Lemma 2.2, we get $m(x,Tx)=m_{x,Tx}=m(Tx,Tx)$.

On the other hand, by Lemma 2.2 and $x_n=T{x}_{n-1}\to x$,

$$0=\underset{n\to \mathrm{\infty }}{lim}(m(x_n,T{x}_n)-m_{x_n,T{x}_n})=\underset{n\to \mathrm{\infty }}{lim}(m(x_n,x_{n-1})-m_{x_n,T{x}_n})=m(x,x)-m_{x,Tx},$$

thus $m(x,x)=m(x,Tx)$. Since $m(x,Tx)=m_{x,Tx}=m(Tx,Tx)$ now by (m1) $x=Tx$. Uniqueness by the contraction (5) is clear. □

Theorem 3.2 Let $(X,m)$ be a complete M-metric space and let $T:X\to X$ be a mapping satisfying the following condition:

$$\mathrm{\exists }k\in [0,\frac{1}{2})\mathit{\text{ such that }}m(Tx,Ty)\le k(m(x,Tx)+m(y,Ty))\mathit{\text{for all }}x,y\in X.$$

(7)

Then T has an unique fixed point.

Proof Let $x_0\in X$ and $x_n:=T{x}_{n-1}$, so we have

$$\begin{array}{rcl}m(x_n,x_{n-1})& =& m(T{x}_{n-1},T{x}_{n-2})\\ \le & k(m(x_{n-1},x_n)+m(x_{n-2},x_{n-1})).\end{array}$$

So

$$m(x_n,x_{n-1})\le rm(x_{n-2},x_{n-1}),$$

where $0\le r=\frac{k}{1-k}<1$.

By Lemma 2.5 and completeness of X, $x_n\to x$ for some $x\in X$. So

$$m(x_n,x)-m_{x_n,x}\to 0,M_{x_n,x}-m_{x_n,x}\to 0,$$

and since $m_{x_n,x}\to 0$, we have $m(x_n,x)\to 0$ and $M_{x_n,x}\to 0$. Therefore by Remark 1.1, $m(x,x)=0=m_{x,Tx}$;

$$m(x_{n+1},Tx)=m(T{x}_n,Tx)\le k(m(x_n,x_{n+1})+m(x,Tx)),$$

hence by $m(x_n,x_{n+1})\to 0$

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}m(x_{n+1},Tx)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}m(T{x}_n,Tx)\le km(x,Tx).$$

On the other hand

$$m(x,Tx)-m_{x,Tx}\le m(x,x_n)+m(x_n,Tx)$$

implies that

$$m(x,Tx)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(m(x,x_n)+m(x_n,Tx))\le km(x,Tx),$$

because $m_{x,Tx}=0$ and $m(x_n,x)\to 0$. So $m(x,Tx)=0$. Now by contraction (7) we have $m(Tx,Tx)\le 2km(x,Tx)=0$, so $m(Tx,Tx)=0=m(x,x)=m(x,Tx)$, thus $x=Tx$ by (m1). □

The next theorem is still open.

Theorem 3.3 Let $(X,m)$ be a complete M-metric space and let $T:X\to X$ be a mapping satisfying the following condition:

$$\mathrm{\exists }k\in [0,\frac{1}{2})\mathit{\text{ such that }}m(Tx,Ty)\le k(m(x,Ty)+m(Tx,y))\mathit{\text{for all }}x,y\in X.$$

(8)

Then T has a unique fixed point.