A new generalization of metric spaces: rectangular M-metric spaces

Abstract

In this paper, we introduce the concept of the rectangular M-metric spaces, along with its topology and we prove some fixed-point theorems under different contraction principles with various techniques. The obtained results generalize some classical fixed-point results such as the Banach’s contraction principle, the Kannan’s fixed-point theorem and the Chatterjea’s fixed-point theorem. Also we give an application to the fixed-circle problem.

Introduction

The well-known Banach contraction principle has been studied and generalized in many different directions such as generalizing the used metric spaces. Recently, new generalized metric spaces have been presented for this purpose. For example, M-metric spaces, rectangular metric spaces, partial rectangular metric spaces have been introduced and studied (see [2, 3, 7]). Branciari in [3] defined rectangular metric spaces as follows:

Definition 1.1

[3] (Rectangular metric space (Branciari metric space)) Let X be a nonempty set. A mapping $d:X\times X\rightarrow {\mathbb {R}}^{+}$ is said to be a rectangular metric on X if for any $x,y\in X$ and all distinct points $u,v\in X\setminus \{x,y\}$, it satisfies the following conditions: 

$(R_{1})$ :

$x=y$ if and only if $d(x,y)=0;$

$(R_{2})$ :

$d(x,y)=d(y,x);$

$(R_{3})$ :

$d(x,y)\le d(x,u)+d(u,v)+d(v,y)$ (rectangular inequality).

In this case the pair (X, d) is called a rectangular metric space.

Inspired by the work of Branciari, Shukla in [7] defined rectangular partial metric spaces which are generalizations of rectangular metric spaces.

Definition 1.2

[7] (Partial rectangular metric space) Let X be a nonempty set. A mapping $\rho :X\times X\rightarrow {\mathbb {R}}^{+}$ is said to be a partial rectangular metric on X if for any $x,y\in X$ and all distinct points $u,v\in X\setminus \{x,y\}$, it satisfies the following conditions : 

$(RP_{1})$ :

$x=y$ if and only if $\rho (x,y)=\rho (x,x)=\rho (y,y);$

$(RP_{2})$ :

$\rho (x,x)\le \rho (x,y);$

$(RP_{3})$ :

$\rho (x,y)=\rho (y,x);$

$(RP_{4})$ :

$\rho (x,y)\le \rho (x,u)+\rho (u,v)+\rho (v,y)-\rho (u,u)-\rho (v,v).$

In this case, the pair $(X,\rho )$ is called a partial rectangular metric space.

Asadi et al. in [2] gave an extension to the partial metric spaces, called M-metric spaces, defined as follows.

Notation 1.3

[2]

• 1. $m_{x,y}:={\mathrm{{min}}}\{m(x,x),m(y,y)\}$.

• 2. $M_{x,y}:= {\mathrm{{max}}} \{ m(x,x),m(y,y) \}$.

Definition 1.4

[2] Let X be a nonempty set. If the function $m:X\times X\rightarrow {\mathbb {R}}^{+}$ satisfies the following conditions for all $x,y,z\in X$

$(M_{1})$ :

$m(x,x)=m(y,y)=m(x,y)$ if and only if $x=y,$

$(M_{2})$ :

$m_{x,y}\le m(x,y),$

$(M_{3})$ :

$m(x,y)=m(y,x),$

$(M_{4})$ :

$(m(x,y)-m_{x,y})\le (m(x,z)-m_{x,z})+(m(z,y)-m_{z,y}),$

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

On these new spaces, some generalized fixed-point results have been obtained (see [1,2,3, 6, 7]). In this paper, we introduce the concept of a rectangular M-metric space, along with proving some fixed-point theorems for self-mappings in rectangular M-metric spaces. In Sect. 2, we define the notion of a rectangular M-metric space and investigate some basic properties of this new space. In Sect. 3, we present some topological concepts about open balls and convergence in rectangular M -metric spaces. In Sect. 4, we prove new generalizations of classical fixed-point results such as the Banach’s contraction principle, the Kannan’s fixed-point theorem and the Chatterjea’s fixed-point theorem. In Sect. 5, we define the notions of a circle and a fixed circle. Using these concepts, we present an application to fixed-circle problem.

Rectangular M-metric spaces

At first, we need to present the following notation.

Notation 2.1

• 1. $m_{{r}_{x,y}}:={\mathrm{{min}}}\{m_{r}(x,x),m_{r}(y,y)\}$.

• 2. $M_{{r}_{x,y}}:={\mathrm{{max}}} \{ m_{r}(x,x),m_{r}(y,y) \}$.

Definition 2.2

Let X be a nonempty set and $m_{r}:X\times X\rightarrow [0,\infty )$ be a function. If the following conditions are satisfied for all x, y in X

$(RM_{1})$ :

$m_{r}(x,y)=m_{{r}_{x,y}}=M_{{r}_{x,y}}\Longleftrightarrow x=y,$

$(RM_{2})$ :

$m_{{r}_{x,y}}\le m_{r}(x,y),$

$(RM_{3})$ :

$m_{r}(x,y)=m_{r}(y,x),$

$(RM_{4})$ :

$m_{r}(x,y)-m_{r_{x,y}}\le m_{r}(x,u)-m_{r_{x,u}}+m_{r}(u,v)-m_{r_{u,v}}+m_{r}(v,y)-m_{r_{v,y}}$ for all $u,v\in X\setminus \{x,y\},$

then the pair $(X,m_{r})$ is called a rectangular M-metric space.

Notice that every M-metric is also a rectangular M-metric.

Remark 2.3

Let $(X,m_{r})$ be a rectangular M-metric space. Clearly, we have

• (1) $0\le M_{{r}_{x,y}}+m_{r_{x,y}}=m_{r}(x,x)+m_{r}(y,y),$

• (2) $0\le M_{{r}_{x,y}}-m_{r_{x,y}}=\left| m_{r}(x,x)-m_{r}(y,y)\right|$

  for every $x,y\in X$.

  Also it can be easily verified the following inequality under some cases:

• (3) $M_{{r}_{x,y}}-m_{r_{x,y}}\le \left( M_{{r}_{{x,u}}}-m_{{r}_{{x,u} }}\right) +\left( M_{{r}_{{u}{,v}}}-m_{{r}_{{u}{,v}}}\right) +\left( M_{{r}_{ {v}{,y}}}-m_{{r}_{{v}{,y}}}\right).$

For example, if we consider the case

$$m_{r}(x,x)\ge m_{r}(u,u)\ge m_{r}(v,v)\ge m_{r}(y,y),$$

then we get

$$\begin{aligned} M_{{r}_{x,y}}-m_{r_{x,y}}= & {} \,m_{r}(x,x)-m_{r}(y,y) \\= & {} \,m_{r}(x,x)-m_{r}(y,y)+m_{r}(u,u)-m_{r}(u,u)+m_{r}(v,v)-m_{r}(v,v) \\= & {}\, \left( m_{r}(x,x)-m_{r}(u,u)\right) +\left( m_{r}(u,u)-m_{r}(v,v)\right) +\left( m_{r}(v,v)-m_{r}(y,y)\right) \\= & {}\, \left( M_{{r}_{{x,u}}}-m_{{r}_{{x,u}}}\right) +\left( M_{{r}_{{u}{,v} }}-m_{{r}_{{u}{,v}}}\right) +\left( M_{{r}_{{v}{,y}}}-m_{{r}_{{v}{,y} }}\right) . \end{aligned}$$

Now we give some examples.

Example 2.4

Let ${\mathbb {C}}$ be the set of all complex numbers, and consider the set $X_{\theta }=\left\{ z\in {\mathbb {C}}:\arg (z)=\theta \right\} \cup \left\{ 0\right\}$ for a fixed $\theta ,0\le \theta <2\pi$. If we define the self-mapping $m_{r}$ on $X_{\theta }$ given by $m_{r}(x,y)=\displaystyle \frac{\left| x\right| +\left| y\right| }{2}$ for all $x,y\in X_{\theta }$, then $(X_{\theta },m_{r})$ is a rectangular M-metric space.

We will only show that the following triangular inequality holds since the other conditions of the metric are satisfied (easy to check).

$$m_{r}(x,y)-m_{r_{x,y}}\le m_{r}(x,u)-m_{r_{x,u}}+m_{r}(u,v)-m_{r_{u,v}}+m_{r}(v,y)-m_{r_{v,y}}.$$

(2.1)

Let $x,y,u,v\in X_{\theta }$. We suppose without loss of generality that $\left| x\right| \le \left| y\right|$. Then, $m_{r}(x,y)= \displaystyle \frac{\left| x\right| +\left| y\right| }{2}$, $m_{r}(x,u)=\displaystyle \frac{\left| x\right| +\left| u\right| }{2}$, $m_{r}(u,v)=\displaystyle \frac{\left| u\right| +\left| v\right| }{2}$. We need to consider the following cases:

Case 1::

$\left| u\right| \le \left| v\right| \le \left| x\right| \le \left| y\right|$:

We have $m_{r_{x,y}}=\left| x\right|$, $m_{r_{x,u}}=\left| u\right|$, $m_{r_{u,v}}=\left| u\right|$ and $m_{r_{v,y}}=\left| v\right|$. Therefore, (2.1) holds. Indeed, (2.1) can be written as follows:

$$\begin{aligned} \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -\left| x\right|&\le \displaystyle \frac{\left| x\right| +\left| u\right| }{2}-\left| u\right| +\displaystyle \frac{ \left| u\right| +\left| v\right| }{2}-\left| u\right| +\displaystyle \frac{\left| v\right| +\left| y\right| }{2}-\left| v\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} +\left| u\right| +\left| v\right| -\left| u\right| -\left| u\right| -\left| v\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -\left| u\right| , \end{aligned}$$

which is correct since $\left| u\right| \le \left| x\right|$.

Case 2::

$\left| u\right| \le \left| x\right| \le \left| v\right| \le \left| y\right| :$ It follows as in the previous case.

Case 3::

$\left| u\right| \le \left| x\right| \le \left| y\right| \le \left| v\right| :$

We obtain

$$\begin{aligned} \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -\left| x\right|&\le \displaystyle \frac{\left| x\right| +\left| u\right| }{2}-\left| u\right| +\displaystyle \frac{ \left| u\right| +\left| v\right| }{2}-\left| u\right| +\displaystyle \frac{\left| v\right| +\left| y\right| }{2}-\left| y\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -(\left| u\right| +\left| y\right| -\left| v\right| ), \end{aligned}$$

which is correct given that $\left| u\right| +\left| y\right| -\left| v\right| <\left| x\right|$.

Case 4::

$\left| x\right| \le \left| u\right| \le \left| v\right| \le \left| y\right| :$

$$\begin{aligned} \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -\left| x\right|&\le \displaystyle \frac{\left| x\right| +\left| u\right| }{2}-\left| u\right| +\displaystyle \frac{ \left| u\right| +\left| v\right| }{2}-\left| u\right| +\displaystyle \frac{\left| v\right| +\left| y\right| }{2}-\left| v\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} +\left| u\right| +\left| v\right| -\left| u\right| -\left| u\right| -\left| v\right| \\&\le \displaystyle \frac{\left| x\right| +\left| y\right| }{ 2}-\left| x\right| . \end{aligned}$$

Case 5::

$\left| x\right| \le \left| u\right| \le \left| y\right| \le \left| v\right| :$

$$\begin{aligned} \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -\left| x\right|&\le \displaystyle \frac{\left| x\right| +\left| u\right| }{2}-\left| x\right| +\displaystyle \frac{ \left| u\right| +\left| v\right| }{2}-\left| u\right| +\displaystyle \frac{\left| v\right| +\left| y\right| }{2}-\left| y\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} +\left| u\right| +\left| v\right| -\left| x\right| -\left| u\right| -\left| y\right| \\&= \displaystyle \frac{\left| x\right| +\left| y\right| }{2} -(\left| x\right| +\left| y\right| -\left| v\right| ). \end{aligned}$$

Since $\left| x\right| +\left| y\right| -\left| v\right| \le \left| x\right|$, therefore the inequality holds.

Case 6::

$\left| x\right| \le \left| y\right| \le \left| u\right| \le \left| v\right| :$ It follows as in the previous case.

We note that if we permute u and v in all the precedent cases, (2.1) is still valid. Hence, $(X_{\theta },m_{r})$ is a rectangular M -metric space.

Proposition 2.5

Let (X, d) be a rectangular metric space and a function $\xi : \left[ 0,\infty \right) \rightarrow \left[ \alpha ,\infty \right)$ be a one-to-one and nondecreasing function with $\xi (0)=\alpha$ such that

$$\xi (x+y+z)\le \xi (x)+\xi (y)+\xi (z)-2\alpha {,}$$

for all $x,y,z\in \left[ 0,\infty \right)$. Then, the function $m_{r}:X\times X\rightarrow [0,\infty )$ is defined as

$$m_{r}(x,y)=\xi (d(x,y)){,}$$

for all $x,y\in X$ is a rectangular M-metric.

Proof

From the hypothesis, it can be easily checked that the conditions $(RM_{1})$ , $(RM_{2})$ and $(RM_{3})$ are satisfied. Now we show that the condition $(RM_{4})$ is satisfied. Using the condition $(R_{3})$, we obtain

$$\begin{aligned} \xi (d(x,y))\le & {} \xi (d(x,u)+d(u,v)+d(v,y)) \\\le & {} \xi (d(x,u))+\xi (d(u,v))+\xi (d(v,y))-2\alpha \end{aligned}$$

and

$$\xi (d(x,y))-\alpha \le \left( \xi (d(x,u))-\alpha \right) +\left( \xi (d(u,v))-\alpha \right) +\left( \xi (d(v,y))-\alpha \right) .$$

Therefore, we get

$$m_{r}(x,y)-m_{r_{x,y}}\le m_{r}(x,u)-m_{r_{x,u}}+m_{r}(u,v)-m_{r_{u,v}}+m_{r}(v,y)-m_{r_{v,y}}.$$

Consequently, $m_{r}$ is a rectangular M-metric. $\square$

Example 2.6

Let (X, d) be a rectangular metric space and a function $\xi : \left[ 0,\infty \right) \rightarrow \left[ \alpha ,\infty \right)$ be defined as

$$\xi (t)=mt+n{,}$$

with $\xi (0)=\alpha$ for all $t\in \left[ 0,\infty \right)$. From Proposition 2.5, the function $m_{r}(x,y)=md(x,y)+n$ is a rectangular M-metric.

Note that we can obtain a rectangular metric space from a rectangular M -metric space as seen in the following examples.

Example 2.7

Let $(X,m_{r})$ be a rectangular M-metric space and $m_{r}^{w}:X\times X\rightarrow \left[ 0,\infty \right)$ be a function defined as

$$m_{r}^{w}(x,y)=m_{r}(x,y)-2m_{{r}_{{x,y}}}+M_{{r}_{{x,y}}},$$

for all $x,y\in X$. Then, $m_{r}^{w}$ is a rectangular metric and the pair $(X,m_{r}^{w})$ is a rectangular metric space.

Now we show that the conditions $(R_{1})$, $(R_{2})$ and $(R_{3})$ are satisfied as follows: 

$(R_{1})$ Using the conditions $(RM_{1})$ and $(RM_{2})$, we get

$$\begin{aligned} m_{r}^{w}(x,y)= & {} 0\Leftrightarrow m_{r}(x,y)-2m_{{r}_{{x,y}}}+M_{{r}_{{x,y} }}=0 \\\Leftrightarrow & {} \,m_{r}(x,y)=2m_{{r}_{{x,y}}}-M_{{r}_{{x,y}}} \end{aligned}$$

and

$$\begin{aligned} m_{{r}_{{x,y}}}\le & {} m_{r}(x,y)=2m_{{r}_{{x,y}}}-M_{{r}_{{x,y} }}\Leftrightarrow M_{{r}_{{x,y}}}\le m_{{r}_{{x,y}}}\Leftrightarrow M_{{r}_{ {x,y}}}=m_{{r}_{{x,y}}} \\\Leftrightarrow & {} m_{r}(x,y)=m_{r}(x,x)=m_{r}(y,y)\Leftrightarrow x=y{,} \end{aligned}$$

for all $x,y\in X$.

$(R_{2})$ Using the conditions $(RM_{3})$, we have

$$\begin{aligned} m_{r}^{w}(x,y)= & {}\, m_{r}(x,y)-2m_{{r}_{{x,y}}}+M_{{r}_{{x,y}}} \\= & {}\, m_{r}(y,x)-2m_{{r}_{{y}{,x}}}+M_{{r}_{{y}{,x}}}=m_{r}^{w}(y,x), \end{aligned}$$

for all $x,y\in X$.

$(R_{3})$ Using the conditions $(RM_{4})$ and the inequality (3) given in Remark 2.3, we get

$$\begin{aligned} m_{r}^{w}(x,y)= & {}\, m_{r}(x,y)-2m_{{r}_{{x,y}}}+M_{{r}_{{x,y}}} \\= & {}\, \left( m_{r}(x,y)-m_{{r}_{{x,y}}}\right) +\left( M_{{r}_{{x,y}}}-m_{{r}_{{ x,y}}}\right) \\\le & {}\, \left[ m_{r}(x,u)-m_{r_{x,u}}+m_{r}(u,v)-m_{r_{u,v}}+m_{r}(v,y)-m_{r_{v,y}}\right] \\&+\left[ \left( M_{{r}_{{x,u}}}-m_{{r}_{{x,u}}}\right) +\left( M_{{r}_{{u}{ ,v}}}-m_{{r}_{{u}{,v}}}\right) +\left( M_{{r}_{{v}{,y}}}-m_{{r}_{{v}{,y} }}\right) \right] \\= & {}\, m_{r}^{w}(x,u)+m_{r}^{w}(u,v)+m_{r}^{w}(v,y){,} \end{aligned}$$

for all $u,v\in X\setminus \{x,y\}$. Consequently, $(X,m_{r}^{w})$ is a rectangular metric space.

Example 2.8

Let $(X,m_{r})$ be a rectangular M-metric space and $m_{r}^{s}:X\times X\rightarrow \left[ 0,\infty \right)$ be a function defined as

$$m_{r}^{s}(x,y)=m_{r}(x,y)-m_{{r}_{{x,y}}},$$

for all $x,y\in X$such that if$\ m_{r}^{s}(x,y)=0$then$\ x=y$. Then, $m_{r}^{s}$ is a rectangular metric and the pair $(X,m_{r}^{s})$ is a rectangular metric space.

Now we show that the conditions $(R_{1})$, $(R_{2})$ and $(R_{3})$ are satisfied as follows : 

$(R_{1})$ Using the hypothesis and the definition of $m_{r}^{s}$, we get

$$\begin{aligned} x= & {}\, y\Rightarrow m_{r}^{s}(x,x)=m_{r}(x,x)-m_{r_{x,x}} \\= & \,{} m_{r}(x,x)-\min \left\{ m_{r}(x,x),m_{r}(x,x)\right\} =0 \end{aligned}$$

and

$$m_{r}^{s}(x,y)=0\Rightarrow x=y,$$

for all $x,y\in X$.

$(R_{2})$ Using the condition $(RM_{3})$, we have

$$m_{r}^{s}(x,y)=m_{r}(x,y)-m_{{r}_{{x,y}}}=m_{r}(y,x)-m_{{r}_{{y}{,x} }}=m_{r}^{s}(y,x),$$

for all $x,y\in X$.

$(R_{3})$ Using the condition $(RM_{4})$, we obtain

$$\begin{aligned} m_{r}^{s}(x,y)= & {}\, m_{r}(x,y)-m_{{r}_{{x,y}}} \\\le & {}\, m_{r}(x,u)-m_{r_{_{x,u}}}+m_{r}(u,v)-m_{r_{_{u,v}}}+m_{r}(v,y)-m_{r_{_{v,y}}} \\= & {}\, m_{r}^{s}(x,u)+m_{r}^{s}(u,v)+m_{r}^{s}(v,y){,} \end{aligned}$$

for all $u,v\in X\setminus \{x,y\}$. Consequently, $(X,m_{r}^{s})$ is a rectangular metric space.

In the following proposition, we see the relationship between a rectangular partial metric and a rectangular M-metric.

Proposition 2.9

Every partial rectangular metric is a rectangular M-metric.

Proof

Let $m_{r}$ be a partial rectangular metric. Let us consider the following cases:

• (1) $m_{r}(x,x)=m_{r}(y,y)=m_{r}(u,u)=m_{r}(v,v)$,

• (2) $m_{r}(x,x)<m_{r}(y,y)<m_{r}(u,u)<m_{r}(v,v)$,

• (3) $m_{r}(x,x)=m_{r}(y,y)=m_{r}(u,u)<m_{r}(v,v)$,

• (4) $m_{r}(x,x)=m_{r}(y,y)<m_{r}(u,u)<m_{r}(v,v)$,

• (5) $m_{r}(x,x)=m_{r}(y,y)<m_{r}(u,u)<m_{r}(v,v)$,

• (6) $m_{r}(x,x)<m_{r}(y,y)<m_{r}(u,u)=m_{r}(v,v)$,

• (7) $m_{r}(x,x)<m_{r}(y,y)=m_{r}(u,u)<m_{r}(v,v)$,

• (8) $m_{r}(x,x)<m_{r}(y,y)=m_{r}(u,u)=m_{r}(v,v)$,

• (9) $m_{r}(x,x)>m_{r}(y,y)>m_{r}(u,u)>m_{r}(v,v)$,

• (10) $m_{r}(x,x)=m_{r}(y,y)=m_{r}(u,u)>m_{r}(v,v)$,

• (11) $m_{r}(x,x)=m_{r}(y,y)>m_{r}(u,u)=m_{r}(v,v)$,

• (12) $m_{r}(x,x)=m_{r}(y,y)>m_{r}(u,u)>m_{r}(v,v)$,

• (13) $m_{r}(x,x)>m_{r}(y,y)>m_{r}(u,u)=m_{r}(v,v)$,

• (14) $m_{r}(x,x)>m_{r}(y,y)=m_{r}(u,u)>m_{r}(v,v)$,

• (15) $m_{r}(x,x)>m_{r}(y,y)=m_{r}(u,u)=m_{r}(v,v)$.

Under the above cases, the condition $(RM_{4})$ is satisfied. For example, if we consider case (2), then we get

$$m_{r}(x,y)\le m_{r}(x,u)+m_{r}(u,v)+m_{r}(v,y)-m_{r}(u,u)-m_{r}(u,v)$$

and so

$$\begin{aligned} m_{r}(x,y)-m_{rx,y}= & {}\, m_{r}(x,y)-m_{r}(x,x) \\\le & {}\, m_{r}(x,u)+m_{r}(u,v)+m_{r}(v,y)-m_{r}(u,u)-m_{r}(v,v)-m_{r}(x,x) \\\le & {}\, \left[ m_{r}(x,u)-m_{r}(x,x)\right] +\left[ m_{r}(u,v)-m_{r}(u,u) \right] +\left[ m_{r}(v,y)-m_{r}(y,y)\right] \\= & {}\, m_{r}(x,y)-m_{r_{x,y}}+m_{r}(u,v)-m_{r_{u,v}}+m_{r}(v,y)-m_{r_{v,y}}{ ,} \end{aligned}$$

for all $u,v\in X\setminus \{x,y\}$. Using the similar arguments, it can be easily seen that the condition $(RM_{4})$ is satisfied under the other cases. Therefore, the partial rectangular metric $m_{r}$ is a rectangular M -metric. $\square$

The converse statement of Proposition 2.9 is not always true as seen in the following example.

Example 2.10

Let $X=\left\{ 1,2,3,4\right\}$ and the function $m_{r}:X\times X\rightarrow [0,\infty )$ be defined by

$$\begin{aligned} \begin{array}{c} m_{r}(1,1)=m_{r}(2,2)=m_{r}(3,3)=1\text { and }m_{r}(4,4)=8, \\ m_{r}(1,2)=m_{r}(2,1)=4, \\ m_{r}(1,3)=m_{r}(3,1)=4, \\ m_{r}(1,4)=m_{r}(4,1)=4, \\ m_{r}(2,3)=m_{r}(3,2)=5, \\ m_{r}(2,4)=m_{r}(4,2)=6, \\ m_{r}(3,4)=m_{r}(4,3)=7, \end{array} \end{aligned}$$

for all $x,y\in X$. Then, $m_{r}$ is a rectangular M-metric, but it is not a rectangular partial metric on X. Indeed, for $x=4$, $y=3$, we have

$$m_{r}(4,4)=8\le m_{r}(4,3)=7{,}$$

which is a contradiction. Therefore, the condition $(RP_{2})$ is not satisfied.

It is known that every metric space is a rectangular metric space (see [4]) and that every rectangular metric space is a partial rectangular metric space with zero self-distance (see [7]). Also every metric space is a partial metric space and every partial metric space is an M -metric space (see [2, 5]). Consequently, we can give the following diagram. Here, arrows stand for inclusions.

Some topological notions of rectangular M-metric spaces

In this section, we investigate some topological properties of rectangular M-metric spaces.

Convergence in rectangular M-metric spaces

Definition 3.1

Let $(X,m_{r})$ be a rectangular M-metric space. Then, we have

• (1) A sequence $\{x_{n}\}$ in X converges to a point x if and only if

  $$\lim _{n\rightarrow \infty }(m_{r}(x_{n},x)-m_{{r}_{{x_{n},x}}})=0.$$

• (2) A sequence $\{x_{n}\}$ in X is said to be $m_{r}$-Cauchy sequence if and only if

  $$\begin{aligned} \lim _{n,m\rightarrow \infty }(m_{r}(x_{n},x_{m})-m_{{r}_{{x_{n},x_{m}}}}) \text { and }\lim _{n,m\rightarrow \infty }(M_{{r}_{{x_{n},x_{m}}}}-m_{{r}_{{ x_{n},x_{m}}}}) \end{aligned}$$

  exist and finite.

• (3) A rectangular M-metric space is said to be $m_{r}$-complete if every $m_{r}$-Cauchy sequence $\{x_{n}\}$ converges to a point x such that

  $$\lim _{n\rightarrow \infty }(m_{r}(x_{n},x)-m_{{r}_{x_{n},x}})=0\text { and } \lim _{n\rightarrow \infty }(M_{{r}_{x_{n},x}}-m_{{r}_{x_{n},x}})=0.$$

Lemma 3.2

Assume that $x_{n}\longrightarrow x$ and $y_{n} \longrightarrow y$ as $n\longrightarrow \infty$ in a rectangular M-metric space $(X,m_{r})$. Then,

$$\displaystyle \lim _{n\rightarrow \infty }(m_{r}(x_{n},y_{n})-m_{{r}_{{ x_{n},y_{n}}}})=m_{r}(x,y)-m_{{r}_{{x,y}}}.$$

(3.1)

Proof

Using the triangular inequality of the rectangular M-metric, we obtain

$$m_{r}(x_{n},y_{n})-m_{{r}_{{x_{n},y_{n}}}}\le m_{r}(x_{n},x)-m_{{r}_{{ x_{n},x}}}+m_{r}(x,y)-m_{{r}_{{x,y}}}+m_{r}(y,y_{n})-m_{{r}_{{y,y_{n}}}}.$$

Then,

$$m_{r}(x_{n},y_{n})-m_{{r}_{{x_{n},y_{n}}}}-m_{r}(x,y)+m_{{r}_{{x,y}}}\le m_{r}(x_{n},x)-m_{{r}_{{x_{n},x}}}+m_{r}(y,y_{n})-m_{{r}_{{y,y_{n}}}}.$$

(3.2)

Knowing that $(x_{n})$ converges to x and $(y_{n})$ converges to y, we obtain the result from (3.2), that is,

$$m_{r}(x,y)-m_{r_{x,y}}\le m_{r}(x,x_{n})-m_{r_{x,x_{n}}}+m_{r}(x_{n},y_{n})-m_{r_{x_{n},y_{n}}}+m_{r}(y_{n},y)-m_{r_{y_{n},y}}$$

and then

$$m_{r}(x,y)-m_{r_{x,y}}\le \underset{n\rightarrow \infty }{\lim }\left( m_{r}(x_{n},y_{n})-m_{r_{x_{n},y_{n}}}\right) {.}$$

$\square$

From Lemma 3.2, we can deduce the following lemma.

Lemma 3.3

Assume that $x_{n}\longrightarrow x$ as $n\longrightarrow \infty$ in a rectangular M-metric space $(X,m_{r})$. Then

$$\displaystyle \lim _{n\rightarrow \infty }(m_{r}(x_{n},y)-m_{{r}_{{x_{n},y} }})=m_{r}(x,y)-m_{{r}_{{x,y}}}\ \text{for all}\ y\in X.$$

(3.3)

Lemma 3.4

Assume that $x_{n}\longrightarrow x$ and $y_{n} \longrightarrow y$ as $n\longrightarrow \infty$ in a rectangular M-metric space $(X,m_{r})$. Then, $m_{r}(x,y)=m_{{r}_{{x,y}}}$. Further if $m_{r}(x,x)=m_{r}(y,y)$, then $x=y$.

Proof

From Lemma 3.2, we have

$$0=\displaystyle \lim _{n\rightarrow \infty }(m_{r}(x_{n},y_{n})-m_{{r}_{{ x_{n},y_{n}}}})=m_{r}(x,y)-m_{{r}_{{x,y}}}$$

and then

$$m_{r}(x,y)=m_{{r}_{{x,y}}}{.}$$

From the condition $(RM_{1})$ and the hypothesis $m_{r}(x,x)=m_{r}(y,y)$, we get $x=y$. $\square$

Lemma 3.5

Let $\{x_{n}\}$ be a sequence in a rectangular M-metric space $(X,m_{r})$, such that

$${\textit{there} \, \textit{exists}\, }r\in [0,1){ \,\textit{such} \, \textit{that} \,} m_{r}(x_{n+1},x_{n})\le rm_{r}(x_{n},x_{n-1})\ {\,\textit{for} \, \textit{all} \,}n\in {\mathbb { N}}.$$

(3.4)

Then,

• (A) $\displaystyle \lim _{n \rightarrow \infty }m_r(x_n,x_{n-1})=0,$

• (B) $\displaystyle \lim _{n \rightarrow \infty }m_r(x_n,x_{n})=0,$

• (C) $\displaystyle \lim _{n,m\rightarrow \infty }m_{{r}_{{x_{n},x_{m}} }}=0,$

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

Proof

Using the definition of convergence and inequality (3.4), the proof of the condition (A) follows easily. From the condition $(RM_{2})$ and the condition (A), we get

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\min \left\{ m_{r}(x_{n},x_{n}),m_{r}(x_{n-1},x_{n-1})\right\}&= \underset{n\rightarrow \infty }{\lim }m_{{r}_{{x_{n},x_{n-1}}}} \\ & \quad \le \lim _{n\rightarrow \infty }m_{r}(x_{n},x_{n-1})=0{.} \end{aligned}$$

Therefore, the condition (B) holds. Since $\lim _{n\rightarrow \infty }m_{r}(x_{n},x_{n})=0$, the condition (C) holds. Using the previous conditions and Definition 3.1, we see that the condition (D) holds. $\square$

Lemma 3.6

Let $(X,m_{r})$ be a rectangular M-metric space. Then, we get

• (1) $\{x_{n}\}$ is an $m_{r}$-Cauchy sequence in $(X,m_{r})$ if and only if $\{x_{n}\}$ is a Cauchy sequence in $(X,m_{r}^{w})$ (resp. $(X,m_{r}^{s}))$.

• (2) $(X,m_{r})$ is $m_{r}$-complete if and only if $(X,m_{r}^{w})$ ( resp. $(X,m_{r}^{s}))$ is complete.

Proof

Using Examples 2.7 and 2.8, the proof follows easily. $\square$

Topology of rectangular M-metric spaces

Let $m_{r}$ be a rectangular M-metric on X. For all $x\in X$ and $\varepsilon >0$, the open ball with the center x and the radius $\varepsilon$ is

$$B(x,\varepsilon )=\left\{ y\in X:m_{r}(x,y)-m_{r_{x,y}}<\varepsilon \right\} {.}$$

Notice that we have $x\in B(x,\varepsilon )$ for all $\varepsilon >0$. Indeed, we get

$$m_{r}(x,x)-m_{r_{x,x}}=m_{r}(x,x)-m_{r}(x,x)=0<\varepsilon {.}$$

Similarly, the closed ball with the center x and the radius $\varepsilon$ is

$$B[x,\varepsilon ]=\left\{ y\in X:m_{r}(x,y)-m_{r_{x,y}}\le \varepsilon \right\} {.}$$

Lemma 3.7

Let $m_{r}$ be a rectangular M-metric on X. The collection of all open balls on X

$${\mathcal {B}}_{m_{r}}=\left\{ B(x,\varepsilon )\right\} _{x\in X}^{\varepsilon >0}{,}$$

forms a basis on X.

Proof

Let $y\in B(x,\varepsilon )$. Then, we have

$$m_{r}(x,y)-m_{r_{x,y}}<\varepsilon {,}$$

for all $x\in X$ and $\varepsilon >0$. If we take

$$\delta =\varepsilon -m_{r}(x,y)+m_{r_{x,y}}{,}$$

(3.5)

then we get $\delta >0$. Now we show that

$$B(y,\delta )\subseteq B(x,\varepsilon ){.}$$

Let $z\in B(y,\delta )$. Then, we obtain

$$m_{r}(y,z)-m_{r_{y,z}}<\delta {.}$$

(3.6)

From the conditions $(RM_{4})$, (3.5) and (3.6), we get

$$\begin{aligned}&m_{r}(x,z)-m_{r_{x,z}} \le m_{r}(x,y)-m_{r_{x,y}}+m_{r}(y,y)-m_{r_{y,y}}+m_{r}(y,z)-m_{r_{y,z}} \\< & {} \varepsilon -\delta +\delta =\varepsilon {.} \end{aligned}$$

Consequently, we find $B(y,\delta )\subseteq B(x,\varepsilon )$ and ${\mathcal {B}}_{m_{r}}$ is a basis on X. $\square$

Definition 3.8

• a) Let $m_{r}$ be a rectangular M-metric on X and $\tau _{m_{r}}$ be the topology generated by the open balls $B(x,\varepsilon )$. Then, the pair $\left( X,\tau _{m_{r}}\right)$ is called a rectangular M -space.

• b) Let $(X,\tau _{m_{r}})$ be a rectangular M- space. $(X,\tau _{m_{r}})$ is called a $T_{0}$-space if for any distinct pair of points $x,y\in X$, there exists an open ball containing x but not y or an open ball containing y but not x.

Theorem 3.9

A rectangular M-space is a $T_{0}$-space.

Proof

Let $\left( X,\tau _{m_{r}}\right)$ be a rectangular M-space and $x,y\in X$ with $x\ne y$. Without loss of generality, let us consider the following cases:

Case 1 If $m_{r}(x,x)=m_{r}(y,y)$, then using the hypothesis, the conditions ($RM_{1}$) and ($RM_{2}$), we get

$$m_{r_{x,y}}=m_{r}(x,x)=m_{r}(y,y)<m_{r}(x,y)$$

and

$$m_{r}(x,y)-m_{r_{x,y}}=m_{r}(x,y)-m_{r}(x,x)>0{.}$$

Hence, if we take $\varepsilon =m_{r}(x,y)-m_{r}(x,x),$ then we obtain $y\notin B(x,\varepsilon )$.

Case 2 If $m_{r}(x,x)<m_{r}(y,y)$, then using the conditions ($RM_{1}$) and ($RM_{2}$), we get

$$m_{r}(x,y)-m_{r_{x,y}}>0$$

and

$$m_{r}(x,y)-m_{r_{x,y}}=m_{r}(x,y)-m_{r}(x,x)>0{.}$$

Hence, if we take $\varepsilon =m_{r}(x,y)-m_{r}(x,x),$ then we obtain $y\notin B(x,\varepsilon )$.

Consequently, $\left( X,\tau _{m_{r}}\right)$ is a $T_{0}$-space. $\square$

Some fixed-point results

At first, we prove the following useful lemma.

Lemma 4.1

Let $(X,m_{r})$ be a rectangular M-metric space and T be a self-mapping on X. If there exists $k\in [0,1)$ such that

$$m_{r}(Tx,Ty)\le km_{r}(x,y)\ {\,\textit{for} \, \textit{all}\,}\ x,y\in X$$

(4.1)

and consider the sequence $\{x_{n}\}_{n\ge 0}$ defined by $x_{n+1}=Tx_{n}.$ If $x_{n}\rightarrow u$ as $n\rightarrow \infty ,$ then $Tx_{n}\rightarrow Tu$ as $n\rightarrow \infty$.

Proof

First, note that if $m_{r}(Tx_{n},Tu)=0,$ then $m_{{r}_{{Tx_{n},Tu}}}=0$ and that is due to the fact that $m_{{r}_{{Tx_{n},Tu}}}\le m_{r}(Tx_{n},Tu),$ which implies that

$$\begin{aligned} m_{r}(Tx_{n},Tu)-m_{{r}_{{Tx_{n},Tu}}}\rightarrow 0\ \text {as}\ n\rightarrow \infty \ \text {and that is}\ Tx_{n}\rightarrow Tu\ \text {as}\ n\rightarrow \infty . \end{aligned}$$

So, we may assume that $m_{r}(Tx_{n},Tu)>0,$ since by (4.1) we have $m_{r}(Tx_{n},Tu)<m_{r}(x_{n},u),$ then we have the following two cases:

If $m_{r}(u,u)\le m_{r}(x_{n},x_{n}),$ then it is easy to see that $m_{r}(x_{n},x_{n})\rightarrow 0,$ which implies that $m_{r}(u,u)=0,$ and since $m_{r}(Tu,Tu)<m_{r}(u,u)=0,$ we deduce that $m_{r}(Tu,Tu)=m_{r}(u,u)=0,$ and $m_{r}(x_{n},u)\rightarrow 0;$ on the other, we have $m_{r}(Tx_{n},Tu) \le m_{r}(x_{n},u)\rightarrow 0.$

Hence, $m_{r}(Tx_{n},Tu)-m_{{r}_{{Tx_{n},Tu}}}\rightarrow 0$ and thus $Tx_{n}\rightarrow Tu.$

If $m_{r}(u,u)\ge m_{r}(x_{n},x_{n}),$ and once again it is easy to see that $m_{r}(x_{n},x_{n})\rightarrow 0,$ which implies that $m_{{r}_{{x_{n},u} }}\rightarrow 0.$ Hence, $m_{r}(x_{n},u)\rightarrow 0$

and since $m_{r}(Tx_{n},Tu)<m_{r}(x_{n},u)\rightarrow 0,$ we have $m_{r}(Tx_{n},Tu)-m_{{r}_{{Tx_{n},Tu}}}\rightarrow 0$ and thus $Tx_{n}\rightarrow Tu$ as desired. $\square$

Now we give some fixed point theorems.

Theorem 4.2

Let $(X,m_{r})$ be a complete rectangular M-metric space and T a self-mapping on X. If there exists $0<k<1$ such that

$$m_{r}(Tx,Ty)\le km_{r}(x,y)\ \text { for \,all }\ x,y\in X,$$

(4.2)

then T has a unique fixed point u in X,  where $m_{r}(u,u)=0.$

Proof

Let x in X be arbitrary. Using (4.2), we have

$$m_r\big (T^nx,T^{n+1}x\big )\le k m_r\big (T^{n-1}x,T^{n}x\big )\le \cdots \le k^n m_r(x,Tx),$$

(4.3)

for all $n\ge 1$. We distinguish two cases.

Case 1 Let $T^nx=T^mx$ for some integers $n\ne m$. For example, take $m>n$ . We have $T^{m-n}(T^nx)=T^nx$. Choose $y=T^nx$ and $p=m-n$. Then,

$$T^py=y,$$

that is, y is a periodic point of T. By (4.2) and (4.3), we have

$$m_r(y,Ty)=m_r\big (T^py,T^{p+1}y\big )\le k^p m_r(y,Ty).$$

Since $k\in (0,1)$, we get $m_r(y,Ty)=0.$ On the other hand, we have

$$m_r(y,y)=m_r(T^py,T^py)\le km_r(T^{p-1}y,T^{p-1}y)\le \cdots< m_r(Ty,Ty)\le km_r(y,y)<m_r(y,y).$$

Thus,

$$m_r(y,y)=m_r(Ty,Ty)=0.$$

Hence, $y=Ty$, that is, y is a fixed point of T.

Case 2 Suppose that $T^nx\ne T^mx$ for all integers $n\ne m$.

We rewrite (4.3) as

$$m_r\big (T^nx,T^{n+1}x\big )\le k^n m_r(x,Tx)\le \frac{k^{n}}{1-k}m_r(x,Tx).$$

(4.4)

Similarly, by (4.2), we have

$$m_r\big (T^nx,T^{n+2}x\big )\le k m_r\big (T^{n-1}x,T^{n+1}x\big )\le \cdots \le k^n m_r\big (x,T^2x\big )\le \frac{k^{n}}{1-k}m_r\big (x,T^2x\big ).$$

(4.5)

Now, if $m>2$ is odd, then consider $m=2p+1$ with $p\ge 1$. By (4.2) and (4.4), we have

$$\begin{aligned} m_{r}\big (T^{n}x,T^{n+m}x\big )&\le m_{r}\big (T^{n}x,T^{n+1}x\big )+m_{r}\big (T^{n+1}x,T^{n+2}x\big )+ \cdots +m_{r}\big (T^{n+2p}x,T^{n+2p+1}x\big ) \\&\le k^{n}m_r(x,Tx)+k^{n+1}m_r(x,Tx)+\cdots +k^{n+2p}m_r(x,Tx) \\&= k^{n}m_r(x,Tx)\big [1+k+k^{2}+\cdots +k^{2p}\big ] \\&\le \frac{k^{n}}{1-k}m_r(x,Tx). \end{aligned}$$

On the other hand, if $m>2$ is even, then consider $m=2p$ with $p\ge 2$. Again, by (4.2), (4.4) and (4.5),

$$\begin{aligned} m_{r}\big (T^{n}x,T^{n+m}x\big )&\le m_{r}\big (T^{n}x,T^{n+2}x\big )+m_{r}\big (T^{n+2}x,T^{n+3}x\big )+ \cdots +m_{r}\big (T^{n+2p-1}x,T^{n+2p}x\big ) \\&\le k^{n}m_r\big (x,T^2x\big )+k^{n+2}m_r(x,Tx)+k^{n+3}m_r(x,Tx)+\cdots +k^{n+2p-1}m_r(x,Tx) \\&\le k^{n}m_r\big (x,T^2x\big )+\frac{k^{n+2}}{1-k}m_r(x,Tx)\le k^{n}m_r\big (x,T^2x\big )+ \frac{k^{n}}{1-k}m_r(x,Tx). \end{aligned}$$

We deduce from all cases that

$$\begin{aligned} m_{r}\big (T^{n}x,T^{n+m}x\big )\le k^{n}m_r\big (x,T^2x\big )+\frac{k^{n}}{1-k} m_r(x,Tx)\,\,\,\,\text{for all }\,\,n, m\ge 0. \end{aligned}$$

(4.6)

The right-hand side tends to 0 as $n\rightarrow \infty$, and since

$$m_{r}\big (T^{n}x,T^{n+m}x\big )-m_{r_{T^{n}x,T^{n+m}x}}\le m_{r}\big (T^{n}x,T^{n+m}x\big ),$$

we deduce that the sequence $\{T^nx\}$ is $m_{r}$-Cauchy in the $m_{r}$-complete rectangular M-metric space $(X,m_r)$. Hence, there exists some $u\in X$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }m_{r}\big (T^{n}x,u\big )=\lim _{n,m\rightarrow \infty }m_{r}\big (T^{n}x,T^{m}x\big )=m_{r}(u,u). \end{aligned}$$

In view (4.6), we get

$$m_{r}(u,u)=\lim _{n\rightarrow \infty }m_{r}(T^{n}x,u)=\lim _{n,m\rightarrow \infty }m_{r}(T^{n}x,T^{m}x)=0.$$

(4.7)

We shall prove that $Tu=u$. Mention that we are still in case 2, that is, $T^nx\ne T^mx$ for all integers $n\ne m$. Now, we distinguish three subcases.

Subcase 1 If for all $n\ge 0$, $T^nx\not \in \{u, Tu\}$, the rectangular inequality implies that

$$\begin{aligned} m_r(u,Tu)&\le m_r(u,T^nx)+m_r(T^nx,T^{n+1}x)+m_r(T^{n+1}x,Tu) \\&\le m_r(u,T^nx)+m_r(T^nx,T^{n+1}x)+k m_r(T^{n}x,u). \end{aligned}$$

Taking limit as $n\rightarrow \infty$ and using (4.4) and (4.7 ), we get $m_{r}(u,Tu)=0$ that is, $Tu=u.$

Subcase 2 If there exists an integer N such that $T^Nx=u$. Due to case 2, $T^nx \ne u$ for all $n>N$. Similarly, $T^nx \ne Tu$ for all $n>N$. We reach subcase 1, so u is a fixed point of T.

Subcase 3 If there exists an integer N such that $T^Nx=Tu$. Again, necessarily $T^nx\ne u$ and $T^nx\ne Tu$ for all $n>N$. Similarly, we get $Tu=u$.

We deduce that u is a fixed point of T. To show the uniqueness of the fixed point u, assume that T has another fixed point v. By (4.2),

$$m_{r}(u,v)=m_{r}(Tu,Tv)\le k m_{r}(u,v),$$

which holds unless $m_{r}(u,v)=0,$ so $u=v$. $\square$

Example 4.3

Taking $\theta =0$, we consider the rectangular M-metric space $(X_{0},m_{r})$ introduced in Example 2.4 where $X_{0}=[0,\infty )$ and $m_{r}(x,y)=\dfrac{x+y}{2}$ for all $x,y\in X_{0}$. Define the mapping T,

$$\begin{aligned} \begin{array}{ccccc} T &{} : &{} X_{0} &{} \rightarrow &{} X_{0} \\ &{} &{} x &{} \mapsto &{} \displaystyle \frac{x}{2}. \\ &{} &{} &{} &{} \end{array} \end{aligned}$$

Let $x,y\in [0,\infty )$, we have

$$\begin{aligned} m_{r}(Tx,Ty)=m_{r}\left( x/2,y/2\right) =\left( x/2+y/2\right) /2=\dfrac{x+y }{4}\le \frac{2}{3}\dfrac{x+y}{2}. \end{aligned}$$

Then, T satisfies $m_{r}(Tx,Ty)\le km_{r}(x,y)$ with $0<k=\frac{2}{3}<1$.

Finally, all the conditions of Theorem 4.2 are satisfied. Therefore, T has $u=0$ as a fixed point in $X_{0}$.

Theorem 4.4

Let $(X,m_{r})$ be a complete rectangular M-metric space and T be a self-mapping on X. If there exists $0<k<1$ such that

$$m_{r}(Tx,Ty)\le k\max \{m_{r}(x,y),m_{r}(x,Tx),m_{r}(y,Ty)\}\ \text { for all }\ x,y\in X,$$

(4.8)

then T has a unique fixed point u in X,  where $m_{r}(u,u)=0.$

Proof

Let $x_{0}\in X$ and the sequence $\{x_{n}\}$ be defined as in the proof of Theorem 4.2. So, we may assume that $x_{n}\ne x_{n+1}$ for all n.

For all natural number n, we have

$$\begin{aligned} m_{n}=m_{r}(x_{n},x_{n+1})=m_{r}(Tx_{n-1},Tx_{n})\le k\max \{m_{r}(x_{n},x_{n+1}),m_{r}(x_{n-1},x_{n})\}.\ \ \end{aligned}$$

(4.9)

Hence, if $\max \{m_{r}(x_{n},x_{n+1}),m_{r}(x_{n-1},x_{n})\}=m_{r}(x_{n},x_{n+1}),$ then inequality (8) implies

$$m_{r}(x_{n},x_{n+1})<m_{r}(x_{n},x_{n+1})$$

which leads to a contradiction. Therefore,

$$\max \{m_{r}(x_{n},x_{n+1}),m_{r}(x_{n-1},x_{n})\}=m_{r}(x_{n-1},x_{n})\ \text {for \,all}\ n.$$

(4.10)

Thus, the sequence $\{x_{n}\}$ satisfies the hypothesis of Theorem 4.2. So, similarly to the proof of Theorem 4.2, we can easily deduce that T has a unique fixed point u in X,  where $m_{r}(u,u)=0.$ $\square$

Theorem 4.5

Let $(X,m_{r})$ be a complete rectangular M-metric space and T be a self-mapping on X. If there exists $0\le k<\frac{1}{2}$ such that

$$m_{r}(Tx,Ty)\le k\left[ m_{r}(x,Tx)+m_{r}(y,Ty)\right] \ \text { for \,all }\ x,y\in X,$$

(4.11)

then T has a unique fixed point u in X,  where $m_{r}(u,u)=0.$

Proof

Let $x_{0}\in X$ and define the sequence $\{x_{n}\}$ by

$$\begin{aligned} x_{n}=Tx_{n-1}\quad \text {for \,all }\ n=1,2,\ldots \end{aligned}$$

If there exists a natural number k such that $x_{k}=x_{k+1}$, then $x_{k}$ is a fixed point of T. Indeed, we have

$$x_{k}=Tx_{k-1}=x_{k+1}=Tx_{k}$$

and $x_{k}$ is the desired point. Therefore, we can assume that $x_{n}\ne x_{n+1}$ for all n. By (4.11), we have

$$m_{r}(x_{n},x_{n+1})=m_{r}(Tx_{n-1},Tx_{n})\le k\left[ m_{r}(x_{n-1},x_{n})+m_{r}(x_{n},x_{n+1})\right]$$

and so

$$m_{r}(x_{n},x_{n+1})\le \frac{k}{1-k} m_{r}(x_{n-1},x_{n})=rm_{r}(x_{n-1},x_{n}){,}$$

where $0\le r=\frac{k}{1-k}<1$. Then, by the completeness of X and Lemma 3.5, we obtain $x_{n}\rightarrow x$ for some $x\in X$. Hence, we find

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\left[ m_{r}(x_{n},x)-m_{r_{x_{n},x}} \right] =0\,\, \text { and }\,\, \underset{n\rightarrow \infty }{\lim }\left[ M_{r_{x_{n},x}}-m_{r_{x_{n},x}}\right] =0 \end{aligned}$$

and since $m_{r_{x_{n},x}}\rightarrow 0$ we have $m_{r}(x_{n},x)\rightarrow 0$ and $M_{r_{x_{n},x}}\rightarrow 0$. By Remark 2.3, we get $m_{r}(x,x)=0=m_{r_{x,Tx}}$ and by (4.11)

$$m_{r}(x_{n+1},Tx)=m_{r}(Tx_{n},Tx)\le k\left[ m_{r}(x_{n},x_{n+1})+m_{r}(x,Tx)\right] {.}$$

Using the fact $m_{r}(x_{n},x_{n+1})\rightarrow 0$, we get

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim \sup }m_{r}(x_{n+1},Tx)=\underset{ n\rightarrow \infty }{\lim \sup }m_{r}(Tx_{n},Tx)\le km_{r}(x,Tx){.} \end{aligned}$$

On the other hand,

$$m_{r}(x,Tx)-m_{r_{x,Tx}}\le m_{r}(x,x_{n})+m_{r}(x_{n},Tx)$$

implies that

$$\begin{aligned} m_{r}(x,Tx)\le \underset{n\rightarrow \infty }{\lim \sup }\left[ m_{r}(x,x_{n})+m_{r}(x_{n},Tx)\right] \le km_{r}(x,Tx) \end{aligned}$$

since $m_{r_{x,Tx}}=0$ and $m_{r}(x_{n},x)\rightarrow 0$. Consequently, $m_{r}(x,Tx)=0$. By contradiction (4.11), we have

$$m_{r}(Tx,Tx)\le k\left[ m_{r}(x,Tx)+m_{r}(x,Tx)\right] =2km_{r}(x,Tx)$$

and so

$$m_{r}(Tx,Tx)=0=m_{r}(x,x)=m_{r}(x,Tx){.}$$

This shows that $x=Tx$ by the condition ($RM_{1}$). Uniqueness of the fixed point follows by (4.12). Assume that T has two fixed points u, v. We have

$$m_{r}(u,v)=m_{r}(Tu,Tv)\le k\left[ m_{r}(u,Tu)+m_{r}(v,Tv)\right] =0$$

and

$$m_{r}(u,v)=0{.}$$

Using the fact $m_{r}(u,u)=0=m_{r}(v,v)$, we get $u=v$ as required. $\square$

Theorem 4.6

Let $(X,m_{r})$ be a complete rectangular M-metric space and T be a self-mapping on X. If there exists $0\le k<\frac{\sqrt{3}-1}{2}$ such that

$$m_{r}(Tx,Ty)\le k\left[ m_{r}(x,Ty)+m_{r}(y,Tx)\right] ,$$

(4.12)

for all $x,y\in X$, then T has a unique fixed point u in X,  where $m_{r}(u,u)=0.$

Proof

Suppose that $x_{0}\in X$ and $T^{n}x_{0}=x_{n}$. Now we show that

$$m_{r}(x_{n},x_{n+1})\rightarrow 0\ \text { as }\ n\rightarrow \infty {.}$$

Using inequality (4.12), we get

$$m_{r}(x_{n},x_{n})=m_{r}(Tx_{n-1},Tx_{n-1})\le 2km_{r}(x_{n-1},x_{n}){, }$$

(4.13)

for all $n\in {\mathbb {N}}$. From the inequality (4.13) and the condition ($RM_{4}$), we obtain

$$\begin{aligned} m_{r}(x_{n},x_{n+1})&= m_{r}(Tx_{n-1},Tx_{n})\le k\left[ m_{r}(x_{n-1},x_{n+1})+m_{r}(x_{n},x_{n})\right] \\\le k\left[ \begin{array}{c} m_{r}(x_{n-1},x_{n})-m_{r_{x_{n-1},x_{n}}}+m_{r}(x_{n},x_{n})-m_{r_{x_{n},x_{n}}} \\ +m_{r}(x_{n},x_{n+1})-m_{r_{x_{n},x_{n+1}}}+m_{r_{x_{n-1},x_{n+1}}}+m_{r}(x_{n},x_{n}) \end{array} \right] {.} \end{aligned}$$

If we take

$$R_{n}=m_{r}(x_{n},x_{n})-m_{r_{x_{n-1},x_{n}}}-m_{r_{x_{n},x_{n+1}}}+m_{r_{x_{n-1},x_{n+1}}},$$

then we get the following cases:

Case 1 Let us consider

$$m_{r}(x_{n-1},x_{n-1})\le m_{r}(x_{n},x_{n})\le m_{r}(x_{n+1},x_{n+1}) {.}$$

Then, we get

$$\begin{aligned} m_{r_{x_{n},x_{n+1}}}=m_{r}(x_{n},x_{n}),\text { } m_{r_{x_{n-1},x_{n}}}=m_{r}(x_{n-1},x_{n-1}),\text { } m_{r_{x_{n-1},x_{n+1}}}=m_{r}(x_{n-1},x_{n-1}) \end{aligned}$$

and so

$$R_{n}=0{.}$$

Case 2 Let us consider

$$\begin{aligned} m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n},x_{n})\le m_{r}(x_{n-1},x_{n-1}) {.} \end{aligned}$$

By the similar arguments used in Case 1, we get

$$R_{n}=0{.}$$

Case 3 Let us consider

$$m_{r}(x_{n},x_{n})\le m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n-1},x_{n-1}) {.}$$

Then, we get

$$\begin{aligned} m_{r_{x_{n},x_{n+1}}}=m_{r}(x_{n},x_{n}),\text { } m_{r_{x_{n-1},x_{n}}}=m_{r}(x_{n},x_{n}),\text { } m_{r_{x_{n-1},x_{n+1}}}=m_{r}(x_{n+1},x_{n+1}) \end{aligned}$$

and so

$$\begin{aligned} R_{n}=m_{r}(x_{n+1},x_{n+1})-m_{r}(x_{n},x_{n})\le m_{r}(x_{n+1},x_{n+1}) {.} \end{aligned}$$

Case 4 Let us consider

$$\begin{aligned} m_{r}(x_{n-1},x_{n-1})\le m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n},x_{n}) {.} \end{aligned}$$

By the similar arguments used in Case 3, we get

$$\begin{aligned} R_{n}=m_{r}(x_{n},x_{n})-m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n},x_{n}){. } \end{aligned}$$

Case 5 Let us consider

$$\begin{aligned} m_{r}(x_{n},x_{n})\le m_{r}(x_{n-1},x_{n-1})\le m_{r}(x_{n+1},x_{n+1}) {.} \end{aligned}$$

By the similar arguments used in Case 3, we get

$$\begin{aligned} R_{n}=m_{r}(x_{n-1},x_{n-1})-m_{r}(x_{n},x_{n})\le m_{r}(x_{n+1},x_{n+1}) {.} \end{aligned}$$

Case 6 Let us consider

$$\begin{aligned} m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n-1},x_{n-1})\le m_{r}(x_{n},x_{n}) {.} \end{aligned}$$

By the similar arguments used in Case 3, we get

$$\begin{aligned} R_{n}=m_{r}(x_{n},x_{n})-m_{r}(x_{n+1},x_{n+1})\le m_{r}(x_{n},x_{n}){. } \end{aligned}$$

If $R_{n}=0,$ then we obtain

$$\begin{aligned} m_{r}(x_{n},x_{n+1})\le k\left[ m_{r}(x_{n-1},x_{n})+m_{r}(x_{n},x_{n+1}) \right] \end{aligned}$$

and so

$$\begin{aligned} m_{r}(x_{n},x_{n+1})\le \frac{k}{1-k}m_{r}(x_{n-1},x_{n}){.} \end{aligned}$$

If $R_{n}<m_{r}(x_{n},x_{n}),$ then using inequality (4.13), we obtain

$$R_{n}<2km_{r}(x_{n-1},x_{n})$$

and so

$$\begin{aligned} m_{r}(x_{n},x_{n+1})\le \frac{k(2k+1)}{1-k}m_{r}(x_{n-1},x_{n}){.} \end{aligned}$$

If $R_{n}<m_{r}(x_{n+1},x_{n+1}),$ then using inequality (4.13), we obtain

$$R_{n}<2km_{r}(x_{n},x_{n+1})$$

and so

$$\begin{aligned} m_{r}(x_{n},x_{n+1})\le \frac{k}{1-k(2k+1)}m_{r}(x_{n-1},x_{n}){.} \end{aligned}$$

Since $0\le k<\frac{\sqrt{3}-1}{2}$, then we get $\frac{k}{1-k}<1$, $\frac{ k(2k+1)}{1-k}<1$, $\frac{k}{1-k(2k+1)}<1$ and so using Lemma 3.5, we have

$$\begin{aligned} m_{r}(x_{n},x_{n+1})\rightarrow 0{,} \end{aligned}$$

as $n\rightarrow \infty$. Using the completeness hypothesis, we obtain $x_{n}\rightarrow u$ for some $u\in X$ and so

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\left[ m_{r}(x_{n},u)-m_{r_{x_{n},u}} \right] =0\text { and }\underset{n\rightarrow \infty }{\lim }\left[ M_{r_{x_{n},u}}-m_{r_{x_{n},u}}\right] =0{.} \end{aligned}$$

Since $m_{r_{x_{n},u}}\rightarrow 0,$ we have $m_{r}(x_{n},u)\rightarrow 0$ and $M_{r_{x_{n},u}}\rightarrow 0$. By Remark 2.3, we get

$$m_{r}(u,u)=0=m_{r_{u,Tu}}{.}$$

From inequality (4.12), we obtain

$$\begin{aligned} m_{r}(u,Tu)\le & {} \underset{n\rightarrow \infty }{\lim \sup }m_{r}(u,x_{n})+ \underset{n\rightarrow \infty }{\lim \sup }m_{r}(x_{n},Tu) \\= & {}\, \underset{n\rightarrow \infty }{\lim \sup }m_{r}(x_{n},Tu) \\\le & {} \underset{n\rightarrow \infty }{\lim \sup }\left( k\left[ m_{r}(x_{n-1},Tu)+m_{r}(x_{n},u)\right] \right) \\\le & {} \underset{n\rightarrow \infty }{\lim \sup }km_{r}(x_{n-1},Tu)+ \underset{n\rightarrow \infty }{\lim \sup }km_{r}(x_{n},u) \\\le & {} \underset{n\rightarrow \infty }{\lim \sup }k\left[ m_{r}(x_{n-1},u)-m_{r_{x_{n-1},u}}+m_{r}(u,u)-m_{r_{u,u}}+m_{r}(u,Tu)-m_{r_{u,Tu}} \right] \\\le & {} km_{r}(u,Tu){,} \end{aligned}$$

which implies $m_{r}(u,Tu)=0$ since $0\le k<\frac{\sqrt{3}-1}{2}$. Using inequality (4.13), we get

$$\begin{aligned} 0\le m_{r}(Tu,Tu)\le 2km_{r}(u,Tu)=0 \end{aligned}$$

and so

$$\begin{aligned} m_{r}(Tu,Tu)=m_{r}(u,Tu)=m_{r}(u,u){.} \end{aligned}$$

Using condition ($RM_{1}$), we get $u=Tu$. Now we prove that u is a unique fixed point of T. Let us consider $u,v\in X$ with $u\ne v$, $Tu=u$ and $Tv=v$. Using the condition (4.12), we have

$$\begin{aligned} 0< m_{r}(u,v)&=m_{r}(Tu,Tv)\le k\left[ m_{r}(u,Tv)+m_{r}(v,Tu)\right] \\&= 2km_{r}(u,v)<m_{r}(u,v){,} \end{aligned}$$

which is a contradiction. Therefore, $u=v$ and T has a unique fixed point u in X. $\square$

An application to fixed-circle problem

The notions of a circle and of a fixed circle on a rectangular M-metric space are defined as follows:

Let $r>0$ and $x_{0}\in X$. The circle $C_{x_{0},r}^{m_{r}}$ with the center $x_{0}$ and the radius r is defined by

$$\begin{aligned} C_{x_{0},r}^{m_{r}}=\left\{ x\in X:m_{r}(x,x_{0})-m_{r_{x,x_{0}}}=r\right\} {.} \end{aligned}$$

Let $C_{x_{0},r}^{m_{r}}$ be a circle and $T:X\rightarrow X$ be a self-mapping. If $Tx=x$ for any $x\in C_{x_{0},r}^{m_{r}}$ then the circle $C_{x_{0},r}^{m_{r}}$ is called as the fixed circle of T.

Now we give the following fixed-circle result.

Theorem 5.1

Let $(X,m_{r})$ be a rectangular M-metric space and $C_{x_{0},r}^{m_{r}}$ be any circle on X. Let us define the mapping

$$\begin{aligned} \varphi :X\rightarrow \left[ 0,\infty \right) {, }\varphi (x)=m_{r}(x,x_{0})-m_{r_{x,x_{0}}}{,} \end{aligned}$$

for all $x\in X$. If there exists a self-mapping $T:X\rightarrow X$ satisfying

$(CM_{1})$ :

$m_{r}(x,Tx)-m_{r_{x,Tx}}\le \varphi (x)-\varphi (Tx)$,

$(CM_{2})$ :

$m_{r}(Tx,x_{0})-m_{r_{Tx,x_{0}}}\ge r$,

$(CM_{3})$ :

$m_{r_{x,Tx}}=M_{r_{x,Tx}}$,

for each $x\in C_{x_{0},r}^{m_{r}}$, then the circle $C_{x_{0},r}^{m_{r}}$ is a fixed circle of T.

Proof

Let $x\in C_{x_{0},r}^{m_{r}}$. Then, we have $m_{r}(x,x_{0})-m_{r_{x,x_{0}}}=r$. Now we prove that $Tx=x$ whenever $x\in C_{x_{0},r}^{m_{r}}$. From the condition $(CM_{1})$, we obtain

$$\begin{aligned} m_{r}(x,Tx)-m_{r_{x,Tx}}\le & {} \varphi (x)-\varphi (Tx) \nonumber \\= & {} \left[ m_{r}(x,x_{0})-m_{r_{x,x_{0}}}\right] -\left[ m_{r}(Tx,x_{0})-m_{r_{Tx,x_{0}}}\right] {.} \end{aligned}$$

(5.1)

Using the conditions $(RM_{2})$ and $(CM_{2})$, we get

$$\begin{aligned} m_{r}(Tx,x_{0})-m_{r_{Tx,x_{0}}}=r \end{aligned}$$

and $Tx\in C_{x_{0},r}^{m_{r}}$. Using inequality (5.1), we obtain

$$\begin{aligned} m_{r}(x,Tx)-m_{r_{x,Tx}}=0\Rightarrow m_{r}(x,Tx)=m_{r_{x,Tx}}{.} \end{aligned}$$

(5.2)

From conditions (5.2), $(CM_{3})$ and $(RM_{1})$, we find

$$\begin{aligned} m_{r}(x,Tx)=m_{r_{x,Tx}}=M_{r_{x,Tx}}\Rightarrow x=Tx{.} \end{aligned}$$

Consequently, the circle $C_{x_{0},r}^{m_{r}}$ is a fixed circle of T. $\square$

Now we give an illustrative example.

Example 5.2

Let us consider the rectangular M-metric space $\left( X_{ \frac{\pi }{2}},m_{r}\right)$ introduced in Example 2.4, the circle $C_{2i,1}^{m_{r}}$ on $X_{\frac{\pi }{2}}$ and define the self-mapping $T:X_{ \frac{\pi }{2}}\rightarrow X_{\frac{\pi }{2}}$ as

$$\begin{aligned} Tz=\left\{ \begin{array}{ccc} iz &{} ; &{} \left| z\right| <4 \\ z &{} ; &{} \left| z\right| \ge 4 \end{array} \right. {,} \end{aligned}$$

for all $z\in X_{\frac{\pi }{2}}$. Then, the self-mapping T satisfies the conditions $(CM_{1})$, $(CM_{2})$ and $(CM_{3})$ for $x\in C_{2i,1}^{m_{r}}$ such that

$$\begin{aligned} C_{2i,1}^{m_{r}}=\left\{ x\in X_{\frac{\pi }{2}}:\frac{\left| x\right| +\left| 2i\right| }{2}-m_{rx,2i}=1\right\} =\left\{ 0,4i\right\} {.} \end{aligned}$$

Clearly $C_{2i,1}^{m_{r}}$ is a fixed circle of T.

Conclusion

Let us consider Example 2.6 and the contractive condition given in Theorem 4.2. Then, we have

$$\begin{aligned} m_{r}(Tx,Ty)\le km_{r}(x,y){,} \end{aligned}$$

for all $x,y\in X$ and $k\in (0,1)$. Using the definition of $m_{r}$ defined in Example 2.6, we get

$$\begin{aligned} m_{r}(Tx,Ty)& = md(Tx,Ty)+n\le k\left[ md(x,y)+n\right] =kmd(x,y)+kn \nonumber \\\Rightarrow & {} d(Tx,Ty)\le kd(x,y)+\frac{n(k-1)}{m}. \end{aligned}$$

(6.1)

Inequality (6.1) does not satisfy the Banach contraction principle

$$\begin{aligned} d(Tx,Ty)\le kd(x,y){,} \end{aligned}$$

for all $x,y\in X$ and $k\in (0,1)$ on a rectangular metric space. Therefore, it is important to study fixed-point theorems using different contractive conditions on a rectangular M-metric space even if a rectangular M-metric and a rectangular metric generate same topology. Furthermore, in the last section, we have given an introduction to the fixed-circle problem [8]. On this new space, it is possible to study some fixed-circle results by various aspects.