1 Introduction
In metric fixed point theory, the classical Banach contraction principle [11] remains a vital instrument which ensures the existence and uniqueness of fixed points of contraction maps in the setting of complete metric spaces. However, many researchers generalized and extended the Banach contraction principle in numerous ways by improving contraction conditions, using auxiliary mappings, and enlarging the class of metric spaces for this kind of results. One may recall the existing notions, namely of partial metric space [16], partial symmetric space [9], partial JS-metric space [7], metric like space [1], b-metric space [14], rectangular metric space [8, 12], cone metric space [15], M-metric space [5], \(M_{b}\)-metric space [18], rectangular M-metric space [25], and several others. Very recently, Asim et al. [10] introduced the class of rectangular \(M_{rb}\)-metric spaces to enlarge the classes of \(M_{b}\)-metric spaces and rectangular M-metric spaces wherein the newly refined ideas are utilized to prove some fixed point results.
In 2000, Branciari [12] enlarged the class of metric spaces by introducing an interesting class of ν-generalized metric spaces wherein the triangular inequality is replaced by a more general inequality, often called polygonal inequality (namely, involving \(x,u_{1},u_{2},\ldots,u_{\nu },y\) points instead of three). In [12], Branciari proved a generalization of Banach contraction principle whose proof was erroneous (see [28, 29]). However, one is required to be careful while proving results involving ν-generalized metric spaces because such spaces need not have a compatible topology (see [30]).
Anzeige
In 2014, Asadi et al. [5] extended the partial metric spaces (see [17]) by introducing M-metric spaces and utilized the same to prove fixed point results, which were extended in many ways (see [2‐4, 6, 13, 19‐27]). Thereafter, Özgür [25] extended both the rectangular metric spaces and M-metric spaces by introducing rectangular \(M_{r}\)-metric spaces which were used to prove fixed point results.
Inspired by the concepts of M-metric spaces and ν-generalized metric spaces, we introduce the notion of an \(M_{\nu }\)-metric space and utilize the same approach to prove an analogue of the Banach contraction principle in such a space. Also, we adopt an example to establish the genuineness of our main result. Finally, as an application of our main result, we prove a result establishing the existence and uniqueness of solution for a Fredholm integral equation.
2 Preliminaries
In this section, we begin with some notions and definitions which are needed in our subsequent discussions.
Notation 1
([5])
Anzeige
The following notations will be utilized in our presentation:
(1)
\(m_{x,y}=\min \{m(x,x),m(y,y)\}\),
(2)
\(M_{x,y}=\max \{m(x,x),m(y,y)\}\).
(3)
\(m_{r_{x,y}}=\min \{m_{r}(x,x),m_{r}(y,y)\}\),
(4)
\(M_{r_{x,y}}=\max \{m_{r}(x,x),m_{r}(y,y)\}\).
In 2014, Asadi et al. [5] introduced the notion of an M-metric spaces as follows:
Definition 2.1
([5])
Let X be a nonempty set. A mapping \(m:X\times X\to \mathbb{R}_{+}\) is said to be an M-metric, if m satisfies the following (for all \(x,y,z\in X\)):
Then the pair \((X,m)\) is said to be an M-metric space.
(1m)
\(m(x,x)=m(x,y)=m(y,y)\) if and only if \(x=y\),
(2m)
\(m_{x,y}\leq m(x,y)\),
(3m)
\(m(x,y)=m(y,x)\),
(4m)
\((m(x,y)-m_{x,y})\leq (m(x,z)-m_{x,z})+(m(z,y)-m_{z,y})\).
In 2018, Özgür [25] introduced the notion of rectangular \(M_{r}\)-metric spaces as follows:
Definition 2.2
([25])
Let X be a nonempty set. A mapping \(m_{r}:X\times X \to \mathbb{R}_{+}\) is said to be a rectangular \(M_{r}\)-metric, if \(m_{r}\) satisfies the following (for all \(x,y\in X\) and all distinct \(u,v\in X\setminus \{x,y\}\)):
Then the pair \((X,m_{r})\) is said to be a rectangular \(M_{r}\)-metric space.
(\(1m_{r}\))
\(m_{r}(x,x)=m_{r}(x,y)=m_{r}(y,y)\) if and only if \(x=y\),
(\(2m_{r}\))
\(m_{r_{x,y}}\leq m_{r}(x,y)\),
(\(3m_{r}\))
\(m_{r}(x,y)=m_{r}(y,x)\),
(\(4m_{r}\))
\((m_{r}(x,y)-m_{r_{x,y}})\leq (m_{r}(x,u)-m_{r_{x,u}})+(m_{r}(u,v)-m _{r_{u,v}})+(m_{r}(v,y)-m_{r_{v,y}})\).
In 2000, Branciari [12] introduced the notion of rectangular metric spaces as follows:
Definition 2.3
([12])
Let X be a nonempty set. A mapping \(r:X\times X \to \mathbb{R}^{+}\) is said to be a rectangular metric on X, if r satisfies the following (for all \(x,y\in X\) and all distinct \(u,v\in X\setminus \{x,y\}\)):
Then the pair \((X,r)\) is said to be a rectangular metric space.
(1r)
\(r(x,y)=0\) if and only if \(x=y\),
(2r)
\(r(x,y)=r(y,x)\),
(3r)
\(r(x,y)\leq r(x,u)+r(u,v)+r(v,y)\).
In 2000, Branciari [12] introduced the following very interesting metric.
Definition 2.4
([12])
Let X be a nonempty set. A mapping \(r_{\nu }:X\times X \to \mathbb{R}^{+}\) is said to be a ν-generalized metric on X, if \(r_{\nu }\) satisfies the following (for all distinct \(x,u_{1},u _{2},\ldots,u_{\nu },y\in X\)):
Then the pair \((X,r_{\nu })\) is said to be a ν-generalized metric space.
(\(1r_{\nu }\))
\(r_{\nu }(x,y)=0\) if and only if \(x=y\),
(\(2r_{\nu }\))
\(r_{\nu }(x,y)=r_{v}(y,x)\),
(\(3r_{\nu }\))
\(r_{\nu }(x,y)\leq r_{\nu }(x,u_{1})+r_{\nu }(u_{1},u_{2})+\cdots +r _{\nu }(u_{\nu },y)\).
Remark 2.1
Observe that when \(\nu =1\) the ν-generalized metric space coincides with a metric space whereas for \(\nu =2\) the same space coincides with a rectangular metric space.
3 Main results
In this section, we introduce the notion of an \(M_{\nu }\)-metric space (for any fixed \(\nu \in \mathbb{N}\)) and utilize it to prove a fixed point theorem besides deriving some lemmas, propositions, and corollaries. Some natural examples are also furnished. The following notations will be utilized in the sequel.
Notation 2
(1)
\(m_{\nu _{x,y}}=\min \{m_{\nu }(x,x),m_{\nu }(y,y)\}\),
(2)
\(M_{\nu _{x,y}}=\max \{m_{\nu }(x,x),m_{\nu }(y,y)\}\).
Definition 3.1
Let X be a nonempty set. A mapping \(m_{\nu }:X\times X\to \mathbb{R}_{+}\) is said to be an \(M_{\nu }\)-metric, if \(m_{\nu }\) satisfies the following (for all \(x,u_{1},u_{2},\ldots,u_{\nu },y\in X\)):
Then the pair \((X,m_{\nu })\) is said to be an \(M_{\nu }\)-metric space.
(\(1m_{\nu }\))
\(m_{\nu }(x,x)=m_{\nu }(x,y)=m_{\nu }(y,y)\) if and only if \(x=y\),
(\(2m_{\nu }\))
\(m_{\nu _{x,y}}\leq m_{\nu }(x,y)\),
(\(3m_{\nu }\))
\(m_{\nu }(x,y)=m_{\nu }(y,x)\),
(\(4m_{\nu }\))
\((m_{\nu }(x,y)-m_{\nu _{x,y}})\leq (m_{\nu }(x,u_{1})-m_{\nu _{x,u_{1}}})+(m _{\nu }(u_{1},u_{2})-m_{\nu _{u_{1},u_{2}}})+\cdots +(m_{\nu }(u_{ \nu },y)-m_{\nu _{u_{\nu },y}})\) such that \(x,u_{1},u_{2},\ldots,u_{ \nu },y\) are distinct.
Notice that \((X,m_{\nu })\) is an M-metric space if and only if \((X,m_{\nu })\) is an \(M_{1}\)-metric space and a rectangular \(M_{r}\)-metric space if and only if \((X,m_{\nu })\) is an \(M_{2}\)-metric space.
Now, we adopt an example in support of Definition 3.1 which is as follows:
Example 3.1
Let \(X=\mathbb{R}\). Define \(m_{\nu }:X\times X\to \mathbb{R}_{+}\) by
Here, one can easily check that conditions \((1m_{\nu })\)-\((3m_{ \nu })\) are trivially satisfied. Now, we merely need to show that condition \((4m_{\nu })\) holds. In doing so, we distinguish the following six cases:
$$ m_{\nu }(x,y)=\frac{ \vert x \vert + \vert y \vert }{2},\quad \text{for all }x,y\in X. $$
Case 1. Firstly, assume that \(\vert u_{1} \vert \leq \vert u_{2} \vert \leq \cdots \leq \vert u_{\nu } \vert \leq \vert x \vert \leq \vert y \vert \). Hence, \(m_{\nu _{x,y}}= \vert x \vert \), \(m_{\nu _{x,u_{1}}}= \vert u_{1} \vert \), \(m_{\nu _{u_{1},u_{2}}}= \vert u_{1} \vert , m_{ \nu _{u_{2},u_{3}}}= \vert u_{2} \vert ,\ldots,m_{\nu _{u_{\nu },y}}= \vert u_{\nu } \vert \). Then \((4m_{\nu })\) can be written as
and since \(\vert u_{1} \vert \leq \vert x \vert \), the above inequality is correct.
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert u_{1} \vert +\frac{ \vert u _{1} \vert + \vert u_{2} \vert }{2}- \vert u_{1} \vert + \cdots +\frac{ \vert u_{\nu } \vert + \vert y \vert }{2}- \vert u_{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert u_{1} \vert + \vert u_{1} \vert +\cdots + \vert u_{\nu } \vert - \vert u_{1} \vert - \cdots - \vert u_{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert u_{1} \vert , \end{aligned}$$
Case 2. Next, assume that \(\vert u_{1} \vert \leq \cdots \leq \vert u_{r} \vert \leq \vert x \vert \leq \vert u_{r+1} \vert \leq \cdots \leq \vert u_{\nu } \vert \leq \vert y \vert \), for some \(1< r< v\). Then, \((4m_{\nu })\) can be written as
and since \(\vert u_{\nu } \vert \leq \vert x \vert \), then above inequality is correct.
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert u_{1} \vert +\frac{ \vert u _{1} \vert + \vert u_{2} \vert }{2}- \vert u_{1} \vert + \cdots +\frac{ \vert u_{r} \vert + \vert x \vert }{2}- \vert u_{r} \vert \\ & {} +\frac{ \vert x \vert + \vert u_{r+1} \vert }{2}- \vert x \vert +\cdots +\frac{ \vert u_{\nu } \vert + \vert y \vert }{2}- \vert u _{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert u_{1} \vert + \vert u_{1} \vert +\cdots + \vert u_{\nu } \vert - \vert u_{1} \vert - \cdots - \vert u_{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert u_{1} \vert , \end{aligned}$$
Case 3. Now, assume that \(\vert u_{1} \vert \leq \cdots \leq \vert u_{r} \vert \leq \vert x \vert \leq \vert y \vert \leq \vert u_{r+1} \vert \leq \cdots \leq \vert u_{\nu } \vert \), for some \(1< r< v\). Then, \((4m_{\nu })\) can be written as
and since \(\vert x \vert + \vert y \vert - \vert u_{\nu } \vert \leq \vert x \vert \), then above inequality is correct.
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert u_{1} \vert +\frac{ \vert u _{1} \vert + \vert u_{2} \vert }{2}- \vert u_{1} \vert + \cdots +\frac{ \vert u_{r} \vert + \vert x \vert }{2}- \vert u_{r} \vert \\ & {} +\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert +\frac{ \vert y \vert + \vert u_{r+1} \vert }{2}- \vert y \vert +\cdots +\frac{ \vert u _{\nu } \vert + \vert y \vert }{2}- \vert y \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert u_{1} \vert + \vert u_{1} \vert +\cdots+ \vert u_{\nu } \vert - \vert u_{1} \vert -\cdots - \vert y \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}-\bigl( \vert x \vert + \vert y \vert - \vert u_{\nu } \vert \bigr), \end{aligned}$$
Case 4. Now, assume that \(\vert x \vert \leq \vert u_{1} \vert \leq \vert u_{2} \vert \leq \cdots \leq \vert u_{\nu } \vert \leq \vert y \vert \). Then, \((4m_{\nu })\) can be written as
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert x \vert +\frac{ \vert u_{1} \vert + \vert u _{2} \vert }{2}- \vert u_{1} \vert +\cdots \frac{ \vert u_{\nu } \vert + \vert y \vert }{2}- \vert u_{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert + \vert u_{1} \vert + \cdots + \vert u_{\nu } \vert - \vert u_{1} \vert - \cdots - \vert u _{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert . \end{aligned}$$
Case 5. Now, assume that \(\vert x \vert \leq \vert u_{1} \vert \leq \cdots \leq \vert u_{r} \vert \leq \vert y \vert \leq \vert u_{r+1} \vert \leq \cdots \leq \vert u_{\nu } \vert \), for some \(1< r< v\). Then, \((4m_{\nu })\) can be written as:
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert x \vert +\frac{ \vert u_{1} \vert + \vert u _{2} \vert }{2}- \vert u_{1} \vert +\cdots +\frac{ \vert u_{r} \vert + \vert y \vert }{2}- \vert u_{r} \vert \\ & {} +\frac{ \vert y \vert + \vert u_{r+1} \vert }{2}- \vert y \vert +\cdots +\frac{ \vert u_{\nu } \vert + \vert y \vert }{2}- \vert u _{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert + \vert u_{1} \vert + \cdots + \vert u_{\nu } \vert - \vert u_{1} \vert - \cdots - \vert u _{\nu } \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert . \end{aligned}$$
Case 6. Finally, assume that \(\vert x \vert \leq \vert y \vert \leq \vert u_{1} \vert \leq \vert u_{2} \vert \leq \cdots \leq \vert u_{\nu } \vert \). Then, \((4m_{\nu })\) can be written as:
and since \(\vert x \vert + \vert y \vert - \vert u_{\nu } \vert \leq \vert x \vert \), then above inequality is correct.
$$\begin{aligned} \frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert \leq &\frac{ \vert x \vert + \vert u_{1} \vert }{2}- \vert x \vert +\frac{ \vert u_{1} \vert + \vert u _{2} \vert }{2}- \vert u_{1} \vert +\cdots +\frac{ \vert u_{\nu } \vert + \vert y \vert }{2}- \vert y \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}- \vert x \vert + \vert u_{1} \vert + \cdots + \vert u_{\nu } \vert - \vert u_{1} \vert - \cdots - \vert y \vert \\ =&\frac{ \vert x \vert + \vert y \vert }{2}-\bigl( \vert x \vert + \vert y \vert - \vert u_{\nu } \vert \bigr), \end{aligned}$$
Now, we furnish two examples by which one can obtain a ν-generalized metric space from an \(M_{\nu }\)-metric space.
Example 3.2
Let \((X,m_{\nu })\) be an \(M_{\nu }\)-metric space. Define a function \(m_{\nu }^{*}:X\times X\to \mathbb{R}_{+}\) by (for all \(x,y\in X\))
Then \(m_{\nu }^{*}\) is a ν-generalized metric and the pair \((X,m_{\nu }^{*})\) is ν-generalized metric space.
$$ m_{\nu }^{*}(x,y)=m_{\nu }(x,y)-2m_{\nu _{x,y}}+M_{\nu _{x,y}}. $$
(3.1)
Proof
To verify condition (\(1r_{\nu }\)), for any \(x,y\in X\), we have
Also,
Now, for condition (\(2r_{\nu }\)), for any \(x,y\in X\), we have
Finally, we show that condition (\(3r_{\nu }\)) holds. Observe that for all distinct \(x,u_{1},u_{2},\ldots,u_{\nu },y\in X\), we have
Thus, \((X,m_{\nu }^{*})\) is a ν-generalized metric space. □
$$\begin{aligned}& m_{\nu }^{*}(x,y) = 0 \\ & \quad \Longleftrightarrow\quad m_{\nu }(x,y)-2m_{\nu _{x,y}}+M_{\nu _{x,y}}=0 \\ & \quad \Longleftrightarrow \quad m_{\nu }(x,y)=2m_{\nu _{x,y}}-M_{\nu _{x,y}}. \end{aligned}$$
$$\begin{aligned}& m_{\nu _{x,y}} \leq 2m_{\nu _{x,y}}-M_{\nu _{x,y}} \\ & \quad \Longleftrightarrow \quad M_{\nu _{x,y}}\leq m_{\nu _{x,y}} \\ & \quad \Longleftrightarrow \quad m_{\nu }(x,y)=m_{\nu _{x,y}}=M_{\nu _{x,y}} \\ & \quad \Longleftrightarrow \quad x=y. \end{aligned}$$
$$\begin{aligned} m_{\nu }^{*}(x,y) =&m_{\nu }(x,y)-2m_{\nu _{x,y}}+M_{\nu _{x,y}} \\ =&m_{\nu }(y,x)-2m_{\nu _{y,x}}+M_{\nu _{y,x}} \\ =&m_{\nu }^{*}(y,x). \end{aligned}$$
$$\begin{aligned} m_{\nu }^{*}(x,y) =&m_{\nu }(x,y)-2m_{\nu _{x,y}}+M_{\nu _{x,y}} \\ =&\bigl(m_{\nu }(x,y)-m_{\nu _{x,y}}\bigr)+(M_{\nu _{x,y}}-m_{\nu _{x,y}}) \\ \leq & \bigl[\bigl(m_{\nu }(x,u_{1})-m_{\nu _{x,u_{1}}}\bigr)+ \bigl(m_{\nu }(u_{1},u _{2})-m_{\nu _{u_{1},u_{2}}}\bigr) \\ & {} +\cdots +\bigl(m_{\nu }(u_{\nu },y)-m_{\nu _{u_{\nu },y}}\bigr) \bigr]+ \bigl[(M _{\nu _{x,u_{1}}}-m_{\nu _{x,u_{1}}}) \\ & {} +(M_{\nu _{u_{1},u_{2}}}-m_{\nu _{u_{1},u_{2}}})+\cdots +(M_{ \nu _{u_{\nu },y}}-m_{\nu _{u_{\nu },y}}) \bigr] \\ =&m_{\nu }^{*}(x,u_{1})+m_{\nu }^{*}(u_{1},u_{2})+ \cdots +m_{\nu } ^{*}(u_{\nu },y). \end{aligned}$$
Example 3.3
Let \((X,m_{\nu })\) be an \(M_{\nu }\)-metric space. Define a function \(m_{\nu }^{**}:X\times X\to \mathbb{R}_{+}\) by (for all \(x,y\in X\))
Then \(m_{\nu }^{**}\) is a ν-generalized metric and the pair \((X,m_{\nu }^{**})\) is ν-generalized metric space.
$$ m_{\nu }^{**}(x,y)=m_{\nu }(x,y)-m_{\nu _{x,y}}. $$
(3.2)
Proof
By similar arguments as in Example 3.2, one can easily show that \(m_{\nu }^{**}\) is a ν-generalized metric. □
With a view to discuss topology corresponding to new \(M_{\nu }\)-metric, let \((X,m_{\nu })\) be an \(M_{\nu }\)-metric space. Then, for all \(x\in X\) and \(\epsilon >0\), the open ball with center x and radius ϵ is defined by
Observe that \(x\in B_{m_{\nu }}(x,\epsilon )\) for each \(\epsilon >0\). Indeed, we have
Similarly, for all \(x\in X\) and \(\epsilon >0\), the closed ball with center x and radius ϵ is defined by
$$ B_{m_{\nu }}(x,\epsilon )=\bigl\{ y\in X:m_{\nu }(x,y)< m_{\nu _{x,y}}+ \epsilon \bigr\} . $$
$$ m_{\nu }(x,x)-m_{\nu _{x,x}}=m_{\nu }(x,x)-m_{\nu }(x,x)=0< \epsilon . $$
$$ B_{m_{\nu }}[x,\epsilon ]=\bigl\{ y\in X:m_{\nu }(x,y)\leq m_{\nu _{x,y}}+ \epsilon \bigr\} . $$
Lemma 3.1
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space. Then the collection of all open balls on X,
forms a basis on X.
$$ \mathcal{U}_{m_{\nu }}=\bigl\{ B_{m_{\nu }}(x,\epsilon ): x\in X,\epsilon >0 \bigr\} , $$
Proof
Let \(u_{0}\in B_{m_{\nu }}(x,\epsilon )\). Then by the definition of \(B_{m_{\nu }}(x,\epsilon )\), we have
Let \(\delta =\epsilon +m_{\nu _{x,u_{0}}}-m_{\nu }(x,u_{0})>0\). We claim that
Let \(u_{1}\in B_{m_{\nu }}(u_{0},\delta )\). Then by the definition, we have
Again let \(\delta _{1}=\delta +m_{\nu _{u_{1},u_{0}}}-m_{\nu }(u_{1},u _{0})\). Inductively, let \(u_{\nu }\in B_{m_{\nu }}(u_{\nu -1}, \delta _{\nu -1})\), for any finite \(\nu \geq 2\). Then
Let us choose \(\delta _{\nu }>0\) such that
Now, from condition (\(4m_{\nu }\)), we have
Hence, \(B_{m_{\nu }}(u_{0},\delta )\subseteq B_{m_{\nu }}(x,\epsilon )\). Therefore, \(\mathcal{U}_{m_{\nu }}\) forms a basis on X. □
$$ m_{\nu }(x,u_{0})< m_{\nu _{x,u_{0}}}+\epsilon . $$
$$ B_{m_{\nu }}(u_{0},\delta )\subseteq B_{m_{\nu }}(x,\epsilon ). $$
$$ m_{\nu }(u_{1},u_{0})< m_{\nu _{u_{1},u_{0}}}+\delta . $$
$$ m_{\nu }(u_{\nu },u_{\nu -1})< m_{\nu _{u_{\nu },u_{\nu -1}}}+ \delta _{\nu -1}. $$
$$ \delta _{\nu }=\delta _{\nu -1}+m_{\nu _{u_{\nu },u_{\nu -1}}}-m_{\nu }(u _{\nu },u_{\nu -1}). $$
$$\begin{aligned} \bigl(m_{\nu }(x,u_{\nu })-m_{\nu _{x,u_{\nu }}}\bigr) \leq & \bigl(m_{\nu }(x,u_{0})-m _{\nu _{x,u_{0}}}\bigr)+ \bigl(m_{\nu }(u_{0},u_{1})-m_{\nu _{u_{0},u_{1}}}\bigr) \\ & {} +\cdots +\bigl(m_{\nu }(u_{\nu -1},u_{\nu })-m_{\nu _{u_{\nu -1},u_{ \nu }}} \bigr) \\ < &(\epsilon -\delta )+(\delta -\delta _{1})+(\delta _{1}- \delta _{2}) \\ & {} +\cdots +(\delta _{\nu -2}-\delta _{\nu -1})+(\delta _{\nu -1}- \delta _{\nu }) \\ < &(\epsilon -\delta _{\nu }). \end{aligned}$$
Definition 3.2
Let \((X,m_{\nu })\) be an \(M_{\nu }\)-metric space and \(\tau _{m_{\nu }}\) a topology generated by the open balls \(B_{m_{\nu }}(x,\epsilon )\). Then the pair \((X,\tau _{m_{\nu }})\) is called an \(M_{\nu }\)-space.
Proposition 3.1
An
\(M_{\nu }\)-space is a
\(T_{0}\)-space.
Proof
Let \((X,\tau _{m_{\nu }})\) be an \(M_{\nu }\)-metric space and \(x,y\in X\) are two distinct points. Then from condition \((2m_{\nu })\), we have
that is,
Firstly, assume that \(m_{\nu }(x,x)=m_{\nu }(y,y)\). Then we have
yielding
If we choose \(\epsilon >0\) such that \(m_{\nu }(x,y)- m_{\nu }(x,x)= \epsilon \), then \(m_{\nu }(x,y)=m_{\nu _{x,y}}+\epsilon \), so that \(y\notin B_{m_{\nu }}(x,\epsilon )\).
$$ m_{\nu _{x,y}}\leq m_{\nu }(x,y)\quad \Rightarrow \quad \min \bigl\{ m_{ \nu }(x,x),m_{\nu }(y,y)\bigr\} \leq m_{\nu }(x,y), $$
$$ m_{\nu }(x,x)\leq m_{\nu }(x,y) \quad \text{or}\quad m_{\nu }(y,y) \leq m_{\nu }(x,y). $$
$$ m_{\nu _{x,y}}=m_{\nu }(x,x)=m_{\nu }(y,y)< m_{\nu }(x,y), $$
$$ m_{\nu }(x,y)-m_{\nu _{x,y}}=m_{\nu }(x,y)-m_{\nu }(x,x)>0. $$
Next, assume that \(m_{\nu }(x,x)< m_{\nu }(y,y)\). Then
implying
Again, if we choose \(\epsilon >0\) such that \(m_{\nu }(x,y)- m_{\nu }(x,x)= \epsilon \), then
so that \(y\notin B_{m_{\nu }}(x,\epsilon )\).
$$ m_{\nu _{x,y}}=m_{\nu }(x,x)< m_{\nu }(x,y), $$
$$ m_{\nu }(x,y)-m_{\nu _{x,y}}=m_{\nu }(x,y)-m_{\nu }(x,x)>0. $$
$$ m_{\nu }(x,y)=m_{\nu _{x,y}}+\epsilon $$
Similarly, for \(m_{\nu }(x,x)>m_{\nu }(y,y)\), one can easily show that \(x\in B_{m_{\nu }}(x,\epsilon )\) and \(y\notin B_{m_{\nu }}(x,\epsilon )\).
Therefore, for any two distinct points in \(x,y\in X\), there is a ball containing one and not containing the other point. Hence, \((X,m_{ \nu })\) is a \(T_{0}\)-space. □
In an \(M_{\nu }\)-metric space, the concepts of basic topological notions, namely of \(m_{\nu }\)-Cauchy sequence, \(m_{\nu }\)-convergent sequence, and \(m_{\nu }\)-complete \(M_{\nu }\)-metric space can be easily adopted as follows.
Definition 3.3
A sequence \(\{x_{n}\}\) in \((X,m_{\nu })\) is said to be \(m_{\nu }\)-convergent to \(x\in X\) if and only if
$$ \lim_{n\to \infty }\bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr)=0. $$
Definition 3.4
A sequence \(\{x_{n}\}\) in \((X,m_{\nu })\) is said to be \(m_{\nu }\)-convergent uniquely to \(x\in X\) if and only if \(\lim_{n\to \infty }(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}})=0\) holds and \(\lim_{n\to \infty }(m_{\nu }(x_{n},y)-m_{\nu _{x_{n},y}})=0\) does not hold for \(y\in X\setminus \{x\}\).
Definition 3.5
A sequence \(\{x_{n}\}\) in \((X,m_{\nu })\) is said to be \(m_{\nu }\)-Cauchy if and only if
exist and are finite.
$$ \lim_{n,m\to \infty } \bigl(m_{\nu }(x_{n},x_{m})-m_{ \nu _{x_{n},x_{m}}} \bigr)\quad \text{and}\quad \lim_{n,m\to \infty } (M_{\nu _{x_{n},x_{m}}}-m_{\nu _{x_{n},x_{m}}} ) $$
Definition 3.6
An \(M_{\nu }\)-metric space \((X,m_{\nu })\) is said to be \(m_{\nu }\)-complete if every \(m_{\nu }\)-Cauchy sequence in X is \(m_{\nu }\)-convergent to a point \(x\in X\) such that
$$ \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr)=0\quad \text{and}\quad \lim_{n\to \infty } (M_{ \nu _{x_{n},x}}-m_{\nu _{x_{n},x}} )=0. $$
Definition 3.7
A self-mapping f on \((X,m_{\nu })\) is said to be sequentially \(m_{\nu }\)-continuous if and only if the fact that \(\{x_{n}\}\)
\(m_{\nu }\)-converges to x implies that \(\{fx_{n}\}\)
\(m_{\nu }\)-converges to fx.
Definition 3.8
A sequence \(\{x_{n}\}\) in \((X,m_{\nu })\) is said to be \(m_{\nu }\)-κ-Cauchy if and only if
exist and are finite with \(\kappa \in \mathbb{N}\) and \(j\in \mathbb{N}_{0}\).
$$ \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x_{n+1+j\kappa })-m_{ \nu _{x_{n},x_{n+1+j\kappa }}} \bigr)\quad \text{and}\quad \lim_{n\to \infty } (M_{\nu _{x_{n},x_{n+1+j\kappa }}}-m_{ \nu _{x_{n},x_{n+1+j\kappa }}} ) $$
Definition 3.9
An \(M_{\nu }\)-metric space \((X,m_{\nu })\) is said to be an \(m_{\nu }\)-κ-complete if every \(m_{\nu }\)-κ-Cauchy in X is \(m_{\nu }\)-convergent to a point \(x\in X\) such that
$$ \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr)=0 \quad \text{and}\quad \lim_{n\to \infty } (M_{ \nu _{x_{n},x}}-m_{\nu _{x_{n},x}} )=0. $$
Remark 3.1
We prefer to write “\(m_{\nu }\)-Cauchy” instead of “\(m_{\nu }\)-1-Cauchy” and “\(m_{\nu }\)-complete” instead of “\(m _{\nu }\)-1-complete”.
Lemma 3.2
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space. Then, we have
(i)
A sequence
\(\{x_{n}\}\)
in
\((X,m_{\nu })\)
is
\(m_{\nu }\)-Cauchy in
\((X,m_{\nu })\)
if and only if
\(\{x_{n}\}\)
in
\((X,m_{\nu })\)
is
\(m_{\nu }\)-Cauchy in
\((X,m_{\nu }^{*})\) (resp. \((X,m_{\nu }^{**})\)).
(ii)
\((X,m_{\nu })\)
is
\(m_{\nu }\)-complete if and only if
\((X,m_{\nu }^{*})\) (resp. \((X,m_{\nu }^{**})\)) is
\(m_{\nu }\)-complete. Moreover,
$$ \lim_{n\to \infty }m_{\nu }^{*}(x_{n},x)=0 \quad \Longleftrightarrow \quad \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr)=0=\lim_{n\to \infty } (M_{\nu _{x_{n},x}}-m_{\nu _{x _{n},x}} ). $$
Proposition 3.2
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space and
\(\kappa ,\lambda \in \mathbb{N}\)
such that
κ
divides λ. Then
(i)
Every
\(m_{\nu }\)-κ-Cauchy sequence is
\(m_{\nu }\)-λ-Cauchy.
(ii)
If
X
is
\(m_{\nu }\)-κ-complete, then
X
is
\(m_{\nu }\)-λ-complete.
Proof
(i) Let \(\{x_{n}\}\) be an \(m_{\nu }\)-κ-Cauchy sequence in X. By the definition of an \(m_{\nu }\)-κ-Cauchy sequence, we have that
exist and are finite with \(j\in \mathbb{N}_{0}\). Since, \(\kappa , \lambda \in \mathbb{N}\) are such that κ divides λ, one can find a \(j\in \mathbb{N}_{0}\) such that \(\lambda =j\kappa \). Thus,
exist and are finite with \(j\in \mathbb{N}_{0}\). Therefore, \(\{x_{n}\}\) is an \(m_{\nu }\)-λ-Cauchy.
$$ \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x_{n+1+j\kappa })-m_{ \nu _{x_{n},x_{n+1+j\kappa }}} \bigr)\quad \text{and}\quad \lim_{n\to \infty } (M_{\nu _{x_{n},x_{n+1+j\kappa }}}-m_{ \nu _{x_{n},x_{n+1+j\kappa }}} ) $$
$$ \lim_{n\to \infty }\bigl(m_{\nu }(x_{n},x_{n+1+j\lambda })-m_{ \nu _{x_{n},x_{n+1+j\lambda }}} \bigr) \quad \text{and}\quad \lim_{n\to \infty }(M_{\nu _{x_{n},x_{n+1+j\lambda }}}-m_{ \nu _{x_{n},x_{n+1+j\lambda }}}) $$
(ii) Let \((X,m_{\nu })\) be \(m_{\nu }\)-κ-complete. Then every \(m_{\nu }\)-κ-Cauchy sequence is also \(m_{\nu }\)-λ-Cauchy which is \(m_{\nu }\)-convergent to some point in X. Hence, \((X,m_{\nu })\) is \(m_{\nu }\)-λ-complete. □
Now, we prove the following lemma which is used in our subsequent discussion:
Lemma 3.3
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space. Let
\(\{x_{n}\}\)
be a sequence in
X
such that all
\(x_{n}\)s are distinct and
\(\sum_{n=1} ^{\infty } (m_{\nu }(x_{n},x_{n+1})-m_{\nu _{x_{n},x_{n+1}}} )< \infty \). Then
\(\{x_{n}\}\)
is
\(m_{\nu }\)-ν-Cauchy.
Proof
Fix \(\epsilon >0\), then there exists \(N\in \mathbb{N}\) such that \(\sum_{i=N}^{\infty } (m_{\nu }(x_{i},x_{i+1})-m_{\nu _{x_{i},x_{i+1}}} )<\epsilon \). Fix \(n\in \mathbb{N}\) with \(n\geq N\). We will show that
For \(j=0\), (3.3) trivially holds. Now, from \((4m_{\nu })\), we have (for some \(j\in \mathbb{N}\))
Then (3.5) holds for \(k=k+1\). Thus, by mathematical induction, (3.3) holds for any \(j\in \mathbb{N}_{0}\). Hence,
Also, by condition \((2m_{\nu })\) and recalling our notation, we have
Therefore, \(\{x_{n}\}\) is an \(m_{\nu }\)-ν-Cauchy sequence. □
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{\nu _{x_{n},x_{n+1+j\nu }}} \bigr)\leq \sum_{i=n}^{n+j\nu } \bigl(m_{\nu }(x_{i},x_{i+1})-m_{ \nu _{x_{i},x_{i+1}}} \bigr). $$
(3.3)
$$\begin{aligned} \bigl(m_{\nu }(x_{n},x_{n+1+(j+1)\nu })-m_{\nu _{x_{n},x_{n+1+(j+1) \nu }}} \bigr) \leq & \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{ \nu _{x_{n},x_{n+1+j\nu }}} \bigr) \\ &{}+\sum_{i=n+1+j\nu }^{n+(j+1)\nu } \bigl(m_{\nu }(x_{i},x_{i+1})-m_{ \nu _{x_{i},x_{i+1}}} \bigr) \\ \leq &\sum_{i=n}^{n+(j+1)\nu } \bigl(m_{\nu }(x_{i},x_{i+1})-m_{ \nu _{x_{i},x_{i+1}}} \bigr). \end{aligned}$$
$$\begin{aligned} \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{\nu _{x_{n},x_{n+1+j\nu }}} \bigr) \leq &\sum_{i=n}^{n+j\nu } \bigl(m_{\nu }(x_{i},x_{i+1})-m_{ \nu _{x_{i},x_{i+1}}} \bigr) \\ \leq &\sum_{i=N}^{\infty } \bigl(m_{\nu }(x_{i},x_{i+1})-m_{ \nu _{x_{i},x_{i+1}}} \bigr)< \epsilon . \end{aligned}$$
$$ (M_{\nu _{x_{n},x_{n+1+j\nu }}}-m_{\nu _{x_{n},x_{n+1+j\nu }}} )\leq \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{\nu _{x_{n},x_{n+1+j \nu }}} \bigr)< \epsilon . $$
Proposition 3.3
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space where
ν
is odd. Let
\(\{x_{n}\}\)
be an
\(m_{\nu }\)-ν-Cauchy sequence such that all
\(x_{n}\)
are distinct. Then
\(\{x_{n}\}\)
is an
\(m_{\nu }\)-Cauchy sequence.
Proof
We first note that if \(\nu =1\), then from Remark 3.1 the conclusion clearly holds. Now, we assume that \(\nu \geq 3\). Fix \(\epsilon >0\), then there exists \(N\in \mathbb{N}\) such that
Next, we fix \(j\in \mathbb{N}_{0}\) with \(n\geq N\). Now, we first show that
If \(k=0\), then (3.5) trivially holds by (3.4). So, let \(0< k\leq (\nu -1)/2\). Now using (\(4m_{\nu }\)), we have
Then, (3.5) holds for \(k=k+1\). Thus, by mathematical induction, (3.5) holds for every k, which implies
for \(j\in \mathbb{N}_{0}\), \(k=0,1,\ldots,(\nu -1)/2\) with \(n\geq N\). Therefore, we have
for \(j\in \mathbb{N}_{0}\), \(k=0,1,\ldots,(\nu -3)/2\) with \(n\geq N\). Also, by condition \((2m_{\nu })\) and our notation, we have
Therefore, \(\{x_{n}\}\) is an \(m_{\nu }\)-Cauchy sequence. □
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{\nu _{x_{n},x_{n+1+j\nu }}} \bigr)< \epsilon , \quad \text{for }n\geq N\text{ and }j\in \mathbb{N} _{0}. $$
(3.4)
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-m_{\nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr)< (k\nu +1)\epsilon , \quad \text{for }k=0,1,\ldots,(\nu -1)/2. $$
(3.5)
$$\begin{aligned}& \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2(k+1)}) - m_{ \nu _{x_{n},x_{n+1+j\nu +2(k+1)}}} \bigr) \\& \quad \leq \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-m_{ \nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr) \\& \quad\quad{} + \bigl(m_{\nu }(x_{n+1+j\nu +2k},x_{n+2+(j+1)\nu +2k})-m_{ \nu _{x_{n+1+j\nu +2k},x_{n+2+(j+1)\nu +2k}}} \bigr) \\& \quad\quad{} + \sum_{i=n+1+j\nu +2(k+1)}^{n+1+(j+1)\nu +2k} \bigl(m_{\nu }(x_{i},x _{i+1})-m_{\nu _{x_{i},x_{i+1}}} \bigr) \\& \quad \leq (k\nu +1)\epsilon +\epsilon +(\nu -1)\epsilon \\& \quad = \bigl((k+1)\nu +1\bigr)\epsilon . \end{aligned}$$
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-m_{\nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr)< \biggl(\frac{\nu ^{2}}{2}-\frac{\nu }{2}+1 \biggr), $$
$$\begin{aligned}& \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k+1}) - m_{ \nu _{x_{n},x_{n+1+j\nu +2k+1}}} \bigr) \\& \quad \leq \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-m_{ \nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr) \\& \quad \quad{} + \bigl(m_{\nu }(x_{n+1+j\nu +2k},x_{n+2+j\nu +2k+\nu -1})-m_{ \nu _{x_{n+1+j\nu +2k},x_{n+2(j\nu +2k+\nu -1}}} \bigr) \\& \quad \quad{} + \sum_{i=n+2+j\nu +2k}^{n+(j+1)\nu +2k} \bigl(m_{\nu }(x_{i},x_{i+1})-m _{\nu _{x_{i},x_{i+1}}} \bigr) \\& \quad \leq 2 \biggl(\frac{\nu ^{2}}{2}-\frac{\nu }{2}+1 \biggr)\epsilon +( \nu -1) \epsilon \\& \quad = \bigl(\nu ^{2}+1\bigr)\epsilon \end{aligned}$$
$$ (M_{\nu _{x_{n},x_{n+1+j\nu +2k}}}-m_{\nu _{x_{n},x_{n+1+j\nu +2k}}} )\leq \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-_{ \nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr)< \epsilon . $$
Proposition 3.4
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space where
ν
is even. Let
\(\{x_{n}\}\)
be an
\(m_{\nu }\)-ν-Cauchy sequence such that all
\(x_{n}\)
are distinct. Then
\(\{x_{n}\}\)
is an
\(m_{\nu }\)-2-Cauchy sequence.
Proof
Fix \(\epsilon >0\), then there exists \(N\in \mathbb{N}\) such that
Next, we fix \(j\in \mathbb{N}_{0}\) with \(n\geq N\). Then by mathematical induction as in the proof of Proposition 3.3, one can show
Therefore, we have
Furthermore, by using condition \((2m_{\nu })\) and our notation, we obtain
for any \(j\in \mathbb{N}_{0}\). So that \(\{x_{n}\}\) is \(m_{\nu }\)-2-Cauchy. □
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu })-m_{\nu _{x_{n},x_{n+1+j\nu }}} \bigr)< \epsilon , \quad \text{for }n\geq N\text{ and }j\in \mathbb{N} _{0}. $$
(3.6)
$$ \bigl(m_{\nu }(x_{n},x_{n+1+j\nu +2k})-m_{\nu _{x_{n},x_{n+1+j\nu +2k}}} \bigr)< (k\nu +1)\epsilon ,\quad \text{for }k=0,1,\ldots,\nu /2-1. $$
(3.7)
$$ \bigl(m_{\nu }(x_{n},x_{n+1+2j})-m_{\nu _{x_{n},x_{n+1+2j}}} \bigr)< \biggl(\frac{\nu ^{2}}{2}-\nu +1 \biggr)\epsilon , \quad \text{for any }j \in \mathbb{N}_{0}. $$
$$ (M_{\nu _{x_{n},x_{n+1+2j}}}-m_{\nu _{x_{n},x_{n+1+2j}}} )\leq \bigl(m_{\nu }(x_{n},x_{n+1+2j})-m_{\nu _{x_{n},x_{n+1+2j}}} \bigr)< \biggl(\frac{\nu ^{2}}{2}-\nu +1 \biggr)\epsilon $$
Remark 3.2
Observe that every \(m_{\nu }\)-Cauchy sequence is \(m_{\nu }\)-2-Cauchy but the converse is not true in general. For converse part, we prove the following lemma.
Lemma 3.4
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space. Let
\(\{x_{n}\}\)
be an
\(m_{\nu }\)-2-Cauchy sequence in
X
such that all
\(x_{n}\)s are distinct and
Then
\(\{x_{n}\}\)
is an
\(m_{\nu }\)-Cauchy sequence.
$$ \lim_{n\to \infty } \bigl(m_{\nu }(x_{n},x_{n+2})-m_{ \nu _{x_{n},x_{n+2}}} \bigr)=0. $$
Proof
Since \(\{x_{n}\}\) is an \(m_{\nu }\)-2-Cauchy, for every \(\epsilon >0\), there exists \(N\in \mathbb{N}\) such that
for any \(j\in \mathbb{N}_{0}\) with \(n\geq N\). Fix \(j\in \mathbb{N} _{0}\) and for \(n\geq N\). Then for \(\nu =1\), we have
Now, we take the case \(\nu \geq 2\), and then have
By using condition \((2m_{\nu })\) and our notation, we obtain
Therefore, \(\{x_{n}\}\) is an \(m_{\nu }\)-Cauchy sequence. □
$$ \bigl(m_{\nu }(x_{n},x_{n+1+2j})-m_{\nu _{x_{n},x_{n+1+2j}}} \bigr)< \epsilon \quad \text{and}\quad \bigl(m_{\nu }(x_{n},x_{n+2})-m_{ \nu _{x_{n},x_{n+2}}} \bigr)< \epsilon $$
$$\begin{aligned} \bigl(m_{\nu }(x_{n},x_{n+2+2j})-m_{\nu _{x_{n},x_{n+2+2j}}} \bigr) \leq & \bigl(m_{\nu }(x_{n},x_{n+1+2j})-m_{\nu _{x_{n},x_{n+1+2j}}} \bigr) \\ &{}+ \bigl(m_{\nu }(x_{n+1+2j},x_{n+2+2j})-m_{\nu _{x_{n+1+2j},x_{n+2+2j}}} \bigr) \\ < &\epsilon . \end{aligned}$$
$$\begin{aligned} \bigl(m_{\nu }(x_{n},x_{n+2+2j})-m_{\nu _{x_{n},x_{n+2+2j}}} \bigr) \leq & \bigl(m_{\nu }(x_{n},x_{n+1+2j})-m_{\nu _{x_{n},x_{n+1+2j}}} \bigr) \\ &{}+ \bigl(m_{\nu }(x_{n+1+2j},x_{n+2j+2\nu })-m_{ \nu _{x_{n+1+2j},x_{n+2j+2\nu }}} \bigr) \\ &{}+\sum_{i=n+2+2j}^{n+2(\nu -1)+2j} \bigl(m_{\nu }(x_{i},x_{i+2})-m_{ \nu _{x_{i},x_{i+2}}} \bigr) \\ < &(\nu +1)\epsilon . \end{aligned}$$
$$ (M_{\nu _{x_{n},x_{n+2+2j}}}-m_{\nu _{x_{n},x_{n+2+2j}}} )\leq \bigl(m_{\nu }(x_{n},x_{n+2+2j})-m_{\nu _{x_{n},x_{n+2+2j}}} \bigr)< ( \nu +1)\epsilon . $$
Next, we present the following lemma required in the sequel.
Lemma 3.5
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space and
\(f:X\to X\)
a self-mapping on
X
such that
for some
\(\lambda \in [0,1)\). Consider the sequence
\(\{x_{n}\}\)
defined by
\(x_{n+1}=fx_{n}\). If
\(x_{n}\to x\)
as
\(n\to \infty \), then
\(fx_{n}\to fx\)
as
\(n\to \infty \).
$$ m_{\nu }(fx,fy)\leq \lambda m_{\nu }(x,y) $$
(3.8)
Proof
Assume that \(m_{\nu }(fx_{n},fx)=0\), then \(m_{{\nu }_{fx_{n},fx}} \leq m_{\nu }(fx_{n},fx)=0\), so that \(m_{\nu }(fx_{n},fx)-m_{{\nu } _{fx_{n},fx}}\to 0\) as \(n\to \infty \) and \(fx_{n}\to fx\) as \(n\to \infty \).
On the other hand, assume that \(m_{\nu }(fx_{n},fx)>0\). By (3.8) we have \(m_{\nu }(fx_{n},fx)\leq \lambda m_{\nu }(x_{n},x)\). Here, we distinguish two cases as follows:
Firstly, assume that \(m_{\nu }(x,x)\leq m_{\nu }(x_{n},x_{n})\). Then, by using (3.8), we have
By taking limit as \(n\to \infty \), we get
Since \(m_{\nu }(fx,fx)< m_{\nu }(x,x)=0\), we obtain that \(m_{\nu }(fx,fx)= \lambda m_{\nu }(x,x)=0\) (for \(\lambda \in [0,1)\)). Then, by the definition of \(m_{\nu }\)-convergence of a sequence \(x_{n}\), which converges to x, we have
Since, \(m_{\nu _{x_{n},x}}=\min \{m_{\nu }(x_{n},x_{n}),m_{\nu }(x,x) \}\) and hence \(m_{\nu _{x_{n},x}}\to 0\) as \(n\to \infty \) so that \(m_{\nu }(x_{n}, x)\to 0\), \(n\to \infty \). Hence, we obtain \(m_{\nu }(fx _{n},fx)< m_{v}(x_{n},x)\to 0\). Therefore, \(m_{\nu }(fx_{n},fx)- m _{{\nu }{fx_{n},fx}}\to 0\) so that \(fx_{n}\to fx\).
$$ m_{\nu }(x_{n},x_{n})=m_{\nu }(fx_{n-1},fx_{n-1}) \leq \lambda m_{ \nu }(x_{n-1},x_{n-1})\leq \cdots \leq \lambda ^{n-1}m_{\nu }(x_{0},x _{0}). $$
$$ \lim_{n\to \infty }m_{\nu }(x_{n},x_{n})=0 \quad \Longrightarrow \quad m_{\nu }(x,x)=0. $$
$$ \lim_{n\to \infty }\bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr)=0. $$
Secondly, assume that \(m_{\nu }(x,x)\geq m_{\nu }(x_{n},x_{n})\). Similarly, one can show that
Hence, \(m_{\nu }(x_{n},x)\to 0\). Since \(m_{\nu }(fx_{n},fx)< m_{\nu }(x _{n},x)\to 0\), we have \(m_{\nu }(fx_{n},fx)- m_{{\nu }{fx_{n},fx}} \to 0\) so that \(fx_{n}\to fx\). This finishes the proof. □
$$ \lim_{n\to \infty }m_{\nu }(x_{n},x_{n})=0 \quad \Longrightarrow \quad m_{{\nu }{x_{n},x}}\to 0. $$
Now, we are equipped to prove our main result as follows:
Theorem 3.1
Let
\((X,m_{\nu })\)
be an
\(M_{\nu }\)-metric space and
\(f:X\to X\). Assume that the following conditions are satisfied:
Then
f
has a unique fixed point
\(x\in X\)
such that
\(m_{\nu }(x,x)=0\).
(i)
there exists
\(\lambda \in [0,1)\)
such that (for all
\(x,y\in X\))
$$ m_{\nu }(fx,fy)\leq \lambda m_{\nu }(x,y) $$
(3.9)
(ii)
\((X,m_{\nu })\)
is
\(m_{\nu }\)-complete.
Proof
Let \(x_{0}\in X\). Construct an iterative sequence \(\{x_{n}\}\) by:
Now, we assert that \(\lim_{n\to \infty }m_{\nu }(x_{n},x_{n+1})=0\). On setting \(x=x_{n}\) and \(y=x_{n+1}\) in (3.9), we have
which, letting \(n\to \infty \), gives rise to
Now, by taking \(x=x_{n}\) and \(y=x_{n+2}\) in (3.9), we obtain
and, taking limit as \(n\to \infty \), we have
Similarly, from condition (3.9), we get
By taking limit as \(n\to \infty \), we get
Also, we have
and
Therefore, from equations (3.11), (3.12) and by recalling our notation, we obtain
Firstly, we show that \(x_{n}\neq x_{m}\) for any \(n\neq m\). Let on the contrary \(x_{n}=x_{m}\) for some \(n>m\), then we have \(x_{n+1}=fx_{n}=fx _{m}=x_{m+1}\). Then from (3.9), we get
a contradiction. Thus, in what follows, we can assume that \(x_{n} \neq x_{m}\) for all \(n\neq m\).
$$ x_{1}=fx_{0}, \qquad x_{2}=f^{2}x_{0}, \qquad x_{3}=f^{3}x_{0}, \qquad \ldots, \qquad x_{n}=f^{n}x_{0}, \qquad \dots . $$
$$\begin{aligned} m_{\nu }(x_{n},x_{n+1}) =&m_{\nu }(fx_{n-1},fx_{n}) \\ \leq &\lambda m_{\nu }(x_{n-1},x_{n}) \\ \leq &\lambda ^{n}m_{\nu }(x_{0},x_{1}), \end{aligned}$$
$$ \lim_{n\to \infty }m_{\nu }(x_{n},x_{n+1})=0. $$
$$\begin{aligned} m_{\nu }(x_{n},x_{n+2}) =&m_{\nu }(fx_{n-1},fx_{n+1}) \\ \leq &\lambda m_{\nu }(x_{n-1},x_{n+1}) \\ \leq &\lambda ^{n-1}m_{\nu }(x_{0},x_{2}), \end{aligned}$$
$$ \lim_{n\to \infty }m_{\nu }(x_{n},x_{n+2})=0. $$
$$ m_{\nu }(x_{n},x_{n})=m_{\nu }(fx_{n-1},fx_{n-1}) \leq \lambda m_{ \nu }(x_{n-1},x_{n-1})\leq \cdots \leq \lambda ^{n-1}m_{\nu }(x_{0},x _{0}). $$
$$ \lim_{n\to \infty }m_{\nu }(x_{n},x_{n})=0. $$
(3.10)
$$ \sum_{n=1}^{\infty }m_{\nu }(x_{n},x_{n+1}) \leq \sum_{n=1}^{\infty } \lambda ^{n}m_{\nu }(x_{0},x_{1})< \infty $$
(3.11)
$$ \sum_{n=1}^{\infty }m_{\nu }(x_{n},x_{n}) \leq \sum_{n=1}^{\infty } \lambda ^{n}m_{\nu }(x_{0},x_{0})< \infty . $$
(3.12)
$$ \sum_{n=1}^{\infty } \bigl(m_{\nu }(x_{n},x_{n+1})-m_{\nu _{x_{n},x_{n+1}}} \bigr)< \infty . $$
$$ m_{\nu }(x_{m},x_{m+1})=m_{\nu }(x_{n},x_{n+1})< m_{\nu }(x_{n-1},x _{n})< \cdots < m_{\nu }(x_{m},x_{m+1}), $$
Now, we assert that \(\{x_{n}\}\) is an \(m_{\nu }\)-Cauchy sequence in \((X,m_{\nu })\). By Lemma 3.3, \(\{x_{n}\}\) is an \(m_{\nu }\)-ν-Cauchy sequence. By Propositions 3.3 and 3.4, \(\{x_{n}\}\) is an \(m_{\nu }\)-Cauchy sequence. Thus, we have
Since X is \(m_{\nu }\)-complete, there exists \(x\in X\) such that \(x_{n}\to x\). Now, we show that \(fx=x\). By Lemma 3.5 we have
so that \(m_{\nu }(fx,x)=m_{\nu _{x,fx}}\). Since \(m_{\nu _{x,fx}}=\min \{m_{\nu }(x,x),m_{\nu }(fx,fx)\}\), therefore \(m_{\nu _{x,fx}}=m_{ \nu }(x,x)\) or \(m_{\nu _{x,fx}}=m_{\nu }(fx,fx)\) which amounts to saying that \(fx=x\).
$$ \lim_{n,m\to \infty }\bigl(m_{\nu }(x_{n},x_{m})-m_{\nu _{x_{n},x _{m}}} \bigr)=0 \quad \text{and}\quad \lim_{n,m\to \infty }(M_{ \nu _{x_{n},x_{m}}-m_{\nu _{x_{n},x_{m}}}})=0. $$
$$\begin{aligned} \lim_{n\to \infty }\bigl(m_{\nu }(x_{n},x)-m_{\nu _{x_{n},x}} \bigr) =&0 \\ =&\lim_{n\to \infty }\bigl(m_{\nu }(x_{n+1},x)-m_{\nu _{x_{n+1},x}} \bigr) \\ =&\lim_{n\to \infty }\bigl(m_{\nu }(fx_{n},x)-m_{\nu _{fx_{n},x}} \bigr) \\ =&\lim_{n\to \infty }\bigl(m_{\nu }(fx,x)-m_{\nu _{fx,x}} \bigr) \end{aligned}$$
Now, we show the uniqueness of the fixed point x. Suppose on the contrary that f has two fixed points \(x,y\in X\), that is, \(fx=x\) and \(fy=y\). Thus
which implies that \(m_{\nu }(x,y)=0\) and hence, \(x=y\). Finally, we show that if x is a fixed point, then \(m_{\nu }(x,x)=0\). Assume that x is a fixed point of f. Observe that
yielding \(m_{\nu }(x,x)=0\). This completes the proof. □
$$ m_{\nu }(x,y)=m_{\nu }(fx,fy)\leq \lambda m_{\nu }(x,y)< m_{\nu }(x,y), $$
$$ m_{\nu }(x,x)=m_{\nu }(fx,fx)\leq \lambda m_{\nu }(x,x) < m_{\nu }(x,x), $$
Now, we present an example which demonstrates the utility of our newly proved result:
Example 3.4
Consider \(X=[0,1]\) and an \(M_{\nu }\)-metric \(m_{\nu }:X\times X\to \mathbb{R}_{+}\) defined by
Then \((X,m_{\nu })\) is an \(m_{\nu }\)-complete \(M_{\nu }\)-metric space. Define a self-mapping f on X by
Observe that, for all \(x,y\in X\), we obtain
Thus, all conditions of Theorem 3.1 are satisfied and f has a unique fixed point (namely \(x=0\)).
$$ m_{\nu }(x,y)=\frac{x+y}{2},\quad \text{for all }x,y\in X. $$
$$ fx=\frac{3x}{5},\quad \text{for all }x\in X. $$
$$\begin{aligned} m_{\nu }(fx,fy) =&\frac{fx+fy}{2}=\frac{\frac{3x}{5}+\frac{3y}{5}}{2} \\ \leq &\frac{3}{5} \biggl(\frac{x+y}{2} \biggr)= \frac{3}{5}m_{\nu }(x,y). \end{aligned}$$
The following corollary is due to Asadi et al. [5].
Corollary 3.1
Let
\((X,m)\)
be an
M-metric space and
\(f:X\to X\). Assume that the following conditions are satisfied:
Then
f
has a unique fixed point
x
such that
\(m(x,x)=0\).
(i)
there exists
\(\lambda \in [0,1)\)
such that (for all
\(x,y\in X\))
$$ m(fx,fy)\leq \lambda m(x,y) $$
(ii)
\((X,m)\)
is
m-complete.
Proof
By choosing \(\nu =1\) in Theorem 3.1, the above result is immediate. □
The following corollary is due to Özgür et al. [25].
Corollary 3.2
Let
\((X,m_{r})\)
be a rectangular
\(M_{r}\)-metric space and
\(f:X\to X\). Assume that the following conditions are satisfied:
Then
f
has a unique fixed point
x
such that
\(m_{r}(x,x)=0\).
(i)
there exists
\(\lambda \in [0,1)\)
such that (for all
\(x,y\in X\))
$$ m_{r}(fx,fy)\leq \lambda m_{r}(x,y) $$
(ii)
\((X,m_{r})\)
is
\(m_{r}\)-complete.
Proof
The above result is immediate from Theorem 3.1 by choosing \(\nu =2\). □
4 An application to an integral equation
In this section, we endeavor to apply Theorem 3.1 to investigate the existence and uniqueness of solution of the Fredholm integral equation.
Let \(X=C([0,1],\mathbb{R})\) be the set of continuous real-valued functions defined on \([0,1]\). Now, we consider the following Fredholm type integral equation:
where \(G\in C([0,1],\mathbb{R})\). Define \(m_{\nu }:X\times X\to \mathbb{R}^{+}\) as in Example 3.1, that is,
Then \((X,m_{\nu })\) is an \(m_{\nu }\)-complete \(M_{\nu }\)-metric space.
$$ x(t)= \int _{0}^{1}G\bigl(t,s,x(t)\bigr)\,ds,\quad \text{for }t,s\in [0,1], $$
(4.1)
$$ m_{\nu }\bigl(x(t),y(t)\bigr)=\sup_{t\in [a,b]} \biggl( \frac{ \vert x(t) \vert + \vert y(t) \vert }{2} \biggr), \quad \text{for all }x,y\in X. $$
Now, we are equipped to state and prove our result as follows:
Theorem 4.1
Assume that (for all
\(x,y\in C([0,1],\mathbb{R})\))
where
\(\lambda \in [0,1)\). Then the integral equation (4.1) has a unique solution.
$$ \bigl\vert G\bigl(t,s,x(t)\bigr)+G\bigl(t,s,y(t)\bigr) \bigr\vert \leq \lambda \bigl\vert x(t)+y(t) \bigr\vert ,\quad \textit{for all }t,s \in [0,1], $$
(4.2)
Proof
Define \(f:X\to X\) by
Observe that existence of a fixed point of the operator f is equivalent to the existence of a solution of the integral equation (4.1). Now, for all \(x,y\in X\), we have
Thus, condition (3.9) is satisfied. Therefore, all conditions of Theorem 3.1 are satisfied. Hence, operator f has a unique fixed point, which means that the Fredholm integral equation (4.1) has a unique solution. This completes the proof. □
$$ fx(t)= \int _{0}^{1}G\bigl(t,s,x(t)\bigr)\,ds,\quad \text{for all }t,s\in [0,1]. $$
$$\begin{aligned} m_{\nu }(fx,fy) = &\biggl\vert \frac{fx(t)+fy(t)}{2} \biggr\vert = \biggl\vert \int _{0} ^{1} \biggl(\frac{G(t,s,x(t))+G(t,s,y(t))}{2} \biggr) \,ds \biggr\vert \\ \leq & \int _{0}^{1} \biggl\vert \frac{G(t,s,x(t))+G(t,s,y(t))}{2} \biggr\vert \,ds \\ \leq &\lambda \int _{0}^{1} \biggl\vert \frac{x(t)+y(t)}{2} \biggr\vert \,ds \\ \leq &\lambda \int _{0}^{1} \biggl(\frac{ \vert x(t) \vert + \vert y(t) \vert }{2} \biggr)\,ds \\ \leq & \lambda \sup_{t\in [a,b]} \biggl(\frac{ \vert x(t) \vert + \vert y(t) \vert }{2} \biggr) \int _{0}^{1}\,ds \\ \leq &\lambda m_{\nu }(x,y). \end{aligned}$$
Acknowledgements
The authors are thankful to learned referees for their critical and important suggestions.
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