Double controlled $M$-metric spaces and some fixed point results

1.   Introduction

Fixed point results is a well-known and established concept in mathematical analysis and also has a firm utilization in many mathematical fields. Fixed point results are also used to solve differential equations, integral equations ^{[3,4,5,6,12,14,15]}.

In 1994, Matthews ^[7] developed the concept of partial metric space. In 2014, $m$-metric space was presented by Asadi et al. ^[2] which is the extended form of a partial metric space. In 2016, Mlaiki et al. ^[10] developed the notion of an $m_{b}$-metric space which extends $m$-metric space and also which is the generalize form of $b$ -metric space (see ^[13]). In 2018, Mlaiki et al. ^[9] established the idea of extended $m_{b}$-metric spaces. In 2018, Mlaiki et al. ^[8] presented the concept of controlled metric space. In 2018, Abdeljawad et al. ^[1] developed the idea of double controlled metric space by using two control functions. More details can be found in ^[16].

In this article, we establish a new double controlled $M$-metric space, by employing two control functions $\alpha, \beta :E\times E\rightarrow \lbrack 1, \infty)$ with the following new double controlled $M$-metric type triangle inequality:

Further we provide some important examples to validate our work. At last, we present an application related to our main results for Fredholm type integral equation.

2.   Preliminaries

In this part, we discuss some important definitions, which would be useful in understanding this work.

Definition 2.1. ^[7] Let $E\neq \phi$. A function $g:E\times E\rightarrow \lbrack 0, \infty)$ is said to be a partial metric if the following conditions hold, for all $u, v, w\in E$,

$1)$ $u = \nu$ if and only if $g(u, u) = g(u, \nu) = g(\nu, \nu);$

$2)$ $g(u, \nu) = g(\nu, u);$

$3)$ $g(u, u)\leq g(u, \nu);$

$4)$ $g(u, \nu)\leq g(u, w)+g(w, \nu)-g(w, w).$

Example 2.2. ^[7] Let $E = [0, \infty)$ with $g(u, v) = max\{u, v\}$. Then $(E, g)$ is a partial metric space.

We will use the following notations given in ^[7].

$1)$ $m_{u, \nu } = m_{{b}_{u, \nu}} = \min \{m(u, u), m(\nu, \nu)\}$;

$2)$ $M_{u, \nu } = M_{{b}_{u, \nu}} = \max \{m(u, u), m(\nu, \nu)\}$.

Definition 2.3. ^[2] Let $E\neq \phi$. A function $m:E\times E\rightarrow \left[ 0, \infty \right)$ is said to be an $m$-metric if the following conditions hold, for all $u, v, w\in E$,

$1)$ $u = \nu$ $\Leftrightarrow m\left(u, u\right) = m\left(u, \nu \right) = m\left(\nu, \nu \right);$

$2)$ $m_{u, \nu }\leq m\left(u, \nu \right);$

$3)$ $m\left(u, \nu \right) = m\left(\nu, u\right);$

$4)$ $\left(m\left(u, \nu \right) -m_{u, \nu }\right) \leq \left(m\left(u, w\right) -m_{u, w}\right) +\left(m\left(w, \nu \right) -m_{w, \nu }\right).$

Example 2.4. ^[2] Let $E = [0, \infty)$. Then $m(u, v) = u+v$ on $E$ is an $m$-metric.

Definition 2.5. ^[11] Let $E\neq \phi$. A function $m_{b}:E\times E\rightarrow [0, \infty)$ is said be an $m_{b}$-metric with coefficient $s\geq 1$ if the following conditions hold, for all $u, v, w\in E$,

$1)$ $u = \nu$ if and only if $m_{b}\left(u, u\right) = m_{b}\left(u, \nu \right) = m_{b}\left(\nu, \nu \right);$

$2)$ $m_{b\; {\rm{ }}\; u, \nu }\leq m_{b}\left(u, \nu \right)$;

$3)$ $m_{b}\left(u, \nu \right) = m_{b}\left(\nu, u\right)$;

$4)$ $\left(m_{b}\left(u, \nu \right) -m_{b\; {\rm{ }}\; u, \nu }\right) \leq s \left[ \left(m_{b}\left(u, w\right) -m_{b\; {\rm{ }}\; u, w}\right) +\left(m_{b}\left(w, \nu \right) -m_{b\; {\rm{ }}\; w, \nu }\right) \right] -m_{b}\left(w, w\right).$

Definition 2.6. ^[9] Let $E\neq \phi$ and $\alpha :E^{2}\rightarrow \left[ 1, \; {\rm{ }}\; \infty \right)$ be a function. A function $m_{b}:E^{2}\rightarrow \mathbb{R} ^{+}$ is said be an extended $m_{b}$-metric if the following conditions hold, for all $u, v, w\in E$,

$1)$ $u = \nu$ if and only if $m_{b}\left(u, u\right) = m_{b}\left(u, \nu \right) = m_{b}\left(\nu, \nu \right);$

$2)$ $m_{b\; {\rm{ }}\; u, \nu }\leq m_{b}\left(u, \nu \right)$;

$3)$ $m_{b}\left(u, \nu \right) = m_{b}\left(\nu, u\right)$;

$4)$ $\left(m_{b}\left(u, \nu \right) -m_{bu, \nu }\right) \leq \alpha \left(u, \; {\rm{ }}\; \nu \right) \left[ \left(m_{b}\left(u, w\right) -m_{u, w}\right) +\left(m_{b}\left(w, \nu \right) -m_{bw, \nu }\right) \right].$

The pair $(E, m_{b})$ is said to be an extended $m_{b}$-metric space.

Example 2.7. ^[9] Let $E = C([a, b], \mathbb{R})$ be the set of all continuous real valued functions on $[a, b]$. We define the functions

by

Then $(E, m_{b})$ is an extended $m_{b}$-metric space with the function $\alpha$.

3.   Main results

Definition 3.1. Let $E\neq \phi$ and $\alpha,$ $\beta :E\times E\rightarrow \left[ 1, \; {\rm{ }}\; \infty \right)$ be functions. A mapping $M:E\times E\rightarrow \left[ 0, \infty \right)$ is said be a double controlled $M$-metric if the following conditions hold, for all $u, v, w\in E,$

$(1)$ $u = \nu$ if and only if $M\left(u, u\right) = M\left(u, \nu \right) = M\left(\nu, \nu \right);$

$(2)$ $M_{u, \nu }\leq M\left(u, \nu \right);$

$(3)$ $M\left(u, \nu \right) = M\left(\nu, u\right);$

$(4)$ $\left(M\left(u, \nu \right) -M_{u, \nu }\right) \leq \alpha \left(u, w\right) \left(M\left(u, w\right) -M_{u, w}\right) +\beta \left(w, \nu \right) \left(M\left(w, \nu \right) -M_{w, \nu }\right)$.

Example 3.2. Let $E = [0, 1]$ and for distinct $u, v$ and $w\in E$ we take

Define $\alpha, \beta :E\times E\rightarrow \left[ 1, \; {\rm{ }}\; \infty \right)$ by

and

Clearly, (1)–(3) are satisfied. Now for $\left( 4\right)$

Clearly $M$ is not an $m$-metric space, but if we take

then

Hence the space is not an $m$-metric space, but it is a double controlled $M$ -metric space

Example 3.3. Let $E = \left \{ 1, \; {\rm{ }}\; 2, \; {\rm{ }}\; 3\right \}$ and for distinct $u, v$ and $w\in E$ we take

Define $\alpha, \beta :E\times E\rightarrow \left[ 1, \; {\rm{ }}\; \infty \right)$ by

and

Clearly, (1)–(3) $are satisfied. Now for$ \left( 4\right) $

if we take

Also clearly $M$ is not an $m$-metric space, since

Hence the space is not an $m$-metric space, but it is a double controlled $M$ -metric space.

Remark 3.4. Observe that if $\alpha \left(u, \; {\rm{ }}\; v\right) = \beta \left(u, \; {\rm{ }}\; v\right)$, then $M$ is an extended $m_{b}$-metric but if $\alpha \left(u, \; {\rm{ }}\; v\right) = \beta \left(u, \; {\rm{ }}\; v\right) = 1$, then $M$ becomes an $m$-metric.

Example 3.5. Let $E = \left[ 0, \; {\rm{ }}\; \infty \right)$ and define

and for distinct $u,$ $v$ and $w\in E$

Also, define $\alpha,$ $\beta :E^{2}\rightarrow \left[ 1, \; {\rm{ }}\; \infty \right)$ by

and

Then clearly $E$ is not an $m$-metric and extended $m_{b}$-metric space but it is a double controlled $M$-metric space.

Definition 3.6. Let $(E, M)$ be a double controlled $M$-metric space.

$\left(1\right)$ A sequence $\left \{ u_{n}\right \}$ in $E$ converges at a point $u$ if

$\left(2\right)$ A sequence $\left \{ u_{n}\right \}$ in $E$ is said to be an $M$-Cauchy sequence if

$\left(3\right) \ $A double controlled $M$-metric space is said to be $M$ -complete if every $M$-Cauchy sequence $\left \{ u_{n}\right \}$ converges to a point $u$, i.e.,

Theorem 3.7. Let $(E, M)$ be a complete double controlled $M$-metric space byfunctions $\alpha, \mu :E\times E\rightarrow \lbrack 1, \infty)$. Supposethat a continuous mapping $R:E\rightarrow E$ satisfies

for all $u,$ $v\in E$ where $K\in (0, 1).$ For $u_{0}\in E$, let $u_{n} = R^{n}u_{0}$. Assume that

In addition, for each $u\in E$, suppose that

Then the mapping $R$ has a unique fixed point.

Proof. Let $\{u_{n} = R^{n}u_{0}\}$ be a sequence in $E$ satisfying the hypothesis of the theorem. By (3.1), we get

for all $n\geq 0$. For $n, m\in \mathbb{Z}$ with $n\leq m$, we get

and so we have

Letting

we have

By (3.2), the limit of the real sequences exists and so $\{s_{n}\}$ is Cauchy. Indeed, the ratio test is applied to the term

Letting $n, m$ tend to $\infty$ in (3.4), we get

which implies that $\{u_{n}\}$ is a Cauchy sequence and by using the completeness of $M$ there exists $u\in E$ such that

and

By the continuity of $R$ and taking the limit, we obtain

Now let

Since $K\in (0, 1),$ by applying limit, we get

Finally, since

and

$Ru = u$.

Now suppose $R$ has two fixed points, i.e., $Ra = a$ and $Rb = b$. Then

Since $M(a, b) = M_{a, b}$, we obtain

which implies that $a = b$.

Example 3.8. Let $E = [0, \infty)$. Consider the double controlled $M$-metric type defined by $M(u, v) = u+v$ for all $u, v\in E$ and the functions $\alpha, \mu$ given by

and

Letting $Ru = 1$ for all $u\in E$ and $u_{0} = 1$ and $k = \frac{1}{4}$, we have

that is, (3.2) holds. In addition to that for every $u\in \lbrack 0, \infty)$ we have

that is, (3.3) holds. All the conditions of the above theorem hold and $u = 1$ is the unique fixed point.

Theorem 3.9. Let $(E, M)$ be a complete double controlled $M$-metric type and $R:E\rightarrow E$ be a continues mapping. Suppose that there exist $\breve{o}, \check{c}\in \lbrack 0, \infty)$ with

where $u_{n} = R^{n}u_{0}$ and for any $u\in E$

If

then $R$ has a unique fixed point.

Proof. Let $u_{o}\in E$ and define $\{u_{n}\}$ as follow: $u_{1} = Ru_{0}$, $u_{2} = Ru_{1} = R^{2}u_{0}$ $u_{n} = R^{n}u_{0}$. We first prove that

To prove this, let $n\in {\mathbb{N}}$. Then

and so

Hence

Since

by the ratio test,

converges. This implies that $M(u_{n}, u_{n+1})$ converges to $0$. Further for $n, m\in \mathbb{N}$

Using the observation of the above inequality, we obtain that $M(u_{n}, u_{m})$ converges to $0$. Since

we conclude that

Also suppose that

Then

Taking the limit $n\rightarrow \infty$, we obtain

Thus $\{u_{n}\}$ is an $M$-Cauchy sequence and by using the completeness of $M$ we obtain that $\{u_{n}\}$ converges to some $u\in E$ and so $\{Ru_{n} = u_{n+1}\}$ converges to $u\in E$. Furthermore by the hypothesis of the theorem, i.e, $R:(E, M)\rightarrow (E, M)$ is continues, it is not difficult to show that $\{Ru_{n}\}$ converges to $R\bar{e}$. By [9,Lemma 3.3], we have

and so

Hence

By using the same observation given in the above equality, we get

Since $(E, M)$ is complete, it follows that

Thus we obtain that

is a fixed point of $R.$

Next we will show the uniqueness. Suppose there exists $v\in E$ such that

By $\left(1.6\right)$, we obtain

Hence

Since

and so

Using (3.6), we obtain

So the mapping $R$ has a unique fixed point.

Example 3.10. Let $E = \{1, 2, 3\}$ and $\alpha, \mu$ be defined in Example 3.3. Let

and $R:E\rightarrow E$ be defined by

Then clearly

Letting $Ru = 1$ for $u_{o} = 1$, we have

That is, (3.5) holds. Since

all the conditions of Theorem 3.9 are satisfied and hence $u = 1$ is a fixed point.

Theorem 3.11. Let $(E, M)$ be a complete double controlled $M$-metrc space by functions $\alpha, \mu :E\times E\rightarrow \lbrack 1, \infty)$. Suppose that acontinuous mapping $R$ satisfies Bianchini type condition

for all $u, v\in E$ where $0 < \hat{h} < 1$. For $u_{0}\in E$, choose $u_{n} = R^{n}u_{0}$. Suppose that

In addition, for each $u\in E$ suppose that

Then $R$ has a unique fixed point.

Proof. Let $u_{o}$ and $u_{1\text{ }}$ be points as in Theorem 3.7. If for some $m$, $u_{m} = u_{m+1} = Ru_{m\text{ }}$, then clearly $u_{m\text{ }}$is a fixed point of $R$.

Now suppose $u_{n\text{ }}\neq u_{n+1\text{ }}$ for all $n$. By (3.7), we have

If $\max \left \{ M(u_{n}, u_{n+1}), M(u_{n-1}, u_{n})\right \} = M(u_{n}, u_{n+1})$, then

which is a contradiction since $0 < \hat{h} < 1.$

If $\max \left \{ M(u_{n}, u_{n+1}), M(u_{n-1}, u_{n})\right \} = M(u_{n-1}, u_{n})$, then

If we proceed it continually, then we come to the conclusion that for each $n\geq 0$ we obtain

Taking the limit on both sides, we get

which implies that $\{u_{n}\}$ is a Cauchy sequence. By the same procedure used in Theorem 3.7, for all $m\geq n$, we may get

By using the assumption used in (3.8) and the ratio test to the series derived in the above inequality as in Theorem 3.7, the sequence $\{u_{n}\}$ is Cauchy. By the completeness of double controlled $M$-metric space, there exists $u\in E$ such that

Thus $u$ is a fixed point of $R$, which follows from the same procedure as we done in Theorem 3.7.

Example 3.12. Let $E = [0, \infty)$ and $R:E\rightarrow E$ be defined by $R(u) = \frac{u}{2+2u}$ and $M(u, v) = \frac{(u+v)^{2}}{2}$. Then $M(u, Rv) = M(Ru, v) = (u+v)^{2}$. If either one of the elements in the form of $Ru$ or $Rv$, we get

Define $\alpha (u, v) = \mu (u, v) = 1+u+v$. Now by induction it is not difficult to deduce that

for all $n\in N$. Thus

On the other hand,

Hence all the hypothesis of Theorem 3.11 hold. Therefore, $f$ has a unique fixed point in $E$.

4.   Application

Let $E = C([0, 1], \mathbb{R})$ and

be a Fredholm type integral equation, where $G(p, q, u^{^{\prime }}(p))$ is a continuous function from $[0, 1]\times \lbrack 0, 1]\rightarrow$ $\mathbb{R}$. Now define

Note that $(E, M)$ is a double controlled $M$-metric type space as already shown in Example 3.2.

Theorem 4.1. Suppose that for all $u, v\in E$

(1)

where $k\in \lbrack 0, \frac{1}{(1+sup_{p, q}|G(p, q, u^{^{\prime}}(p))||G(p, q, v(p))|))^{2}}]$ and for some $p\in \lbrack 0, 1]$

(2)

for all $p, q$. Then the above integral integral has a unique solution.

Proof. Let $R:E\rightarrow E$ be defined by $Ru^{^{\prime }}(p) = \int_{0}^{1}G(p, q, u^{^{\prime }}(p))$. Then

So we get

Furthermore, let $n\in N$ and $u\in E$. Then

Thus for all $p\in [0, 1]$ we find that the sequences $\{R^{n}u^{^{\prime }}(p)\}$ is bounded below and strictly decreasing and so it converges to some $l$. Since $\{R^{n}\}_{n}$ is monotonic, by using Denis theorem, the sup value converges to some point $l$. Observe that $M(R(u), R(v)) = \sup_{p\in \lbrack 0, 1]}\left(\frac{|Ru(p)+Rv(p)|}{2}\right)$ converges to $l$. So all the hypothesis of Theorem 3.7 are satisfied and the Eq (4.1) has a unique solution.

5.   Conclusions

This paper dealt with the achievement of introducing the notion of double controlled $m$-metric spaces as a generalization of extended $m$-$b$-metric space and studiedg some of its properties and results. Moreover, some fixed points have been investigated for mapping satisfying different conditions in this new frame work. This idea can be applied for further investigations in studying fixed points for other structures in metric spaces.

Conflict of interest

The authors declare that they have no competing interests.