How artificial intelligence reduces human bias in diagnostics?

1.   Introduction and preliminaries

In 1922, S. Banach ^[15] provided the concept of Contraction theorem in the context of metric space. After, Nadler ^[28] introduced the concept of set-valued mapping in the module of Hausdroff metric space which is one of the potential generalizations of a Contraction theorem. Let $\left(X, d\right)$ is a complete metric space and a mapping $T:X\rightarrow CB\left(X\right)$ satisfying

for all $x, y\in X,$ where $0\leq \gamma < 1$, $H$ is a Hausdorff with respect to metric $d$ and $CB\left(X\right) = \left \{ \mathbf{S}\subseteq X:S\text{ is closed and bounded subset of }X\text{ equipped with a metric }d\right \}.$ Then $T$ has a fixed point in $X.$

In the recent past, Matthews ^[26] initiate the concept of partial metric spaces which is the classical extension of a metric space. After that, many researchers generalized some related results in the frame of partial metric spaces. Recently, Asadi et al. ^[4] introduced the notion of an $M$-metric space which is the one of interesting generalizations of a partial metric space. Later on, Samet et al. ^[33] introduced the class of mappings which known as $\left(\alpha, \psi \right)$-contractive mapping. The notion of $\left(\alpha, \psi \right)$ -contractive mapping has been generalized in metric spaces (see more ^{[10,12,14,17,19,25,29,30,32]}).

Throughout this manuscript, we denote the set of all positive integers by $\mathbb{N}$ and the set of real numbers by $\mathbb{R}.$ Let us recall some basic concept of an $M$-metric space as follows:

Definition 1.1. ^[4] Let $m:X\times X\rightarrow \mathbb{R} ^{+}$be a mapping on nonempty set $X$ is said to be an $M$-metric if for any $x, y, z$ in $X,$ the following conditions hold:

(i) $m\left(x, x\right) = m\left(y, y\right) = m\left(x, y\right)$ if and only if $x = y;$

(ii) $m_{xy}\leq m(x, y);$

(iii) $m\left(x, y\right) = m\left(y, x\right);$

(iv) $m\left(x, y\right) -m_{xy}\leq (m\left(x, z\right) -m_{xz})+(m\left(z, y\right) -m_{z, y})$ for all $x, y, z\in X.$ Then a pair $\left(X, m\right)$ is called $M$-metric space. Where

and

Remark 1.2. ^[4] For any $x, y, z$ in $M$-metric space $X$, we have

(i) $0\leq M_{xy}+m_{xy} = m\left(x, x\right) +m\left(y, y\right);$

(ii) $0\leq M_{xy}-m_{xy} = \left \vert m\left(x, x\right) -m\left(y, y\right) \right \vert;$

(iii) $M_{xy}-m_{xy}\leq \left(M_{xz}-m_{xz}\right) +\left(M_{zy}-m_{zy}\right).$

Example 1.3. ^[4] Let $\left(X, m\right)$ be an $M$-metric space. Define $m^{w},$ $m^{s}:X\times X\rightarrow \mathbb{R} ^{+}$ by:

(i)

(ii)

Then $m^{w}$ and $m^{s}$ are ordinary metrics. Note that, every metric is a partial metric and every partial metric is an $M$-metric. However, the converse does not hold in general. Clearly every $M$-metric on $X$ generates a $T_{0}$ topology $\tau _{m}$ on $X$ whose base is the family of open $M$ -balls

where

for all $x\in X$, $\varepsilon > 0.$ (see more ^[3,4,23]).

Definition 1.4. ^[4] Let $\left(X, m\right)$ be an $M$-metric space. Then,

(i) A sequence $\{x_{n}\}$ in $\left(X, m\right)$ is said to be converges to a point $x$ in $X$ with respect to $\tau _{m}$ if and only if

(ii) Furthermore, $\{x_{n}\}$ is said to be an $M$-Cauchy sequence in $\left(X, m\right)$ if and only if

exist (and are finite)$.$

(iii) An $M$-metric space $\left(X, m\right)$ is said to be complete if every $M$-Cauchy sequence $\{x_{n}\}$ in $\left(X, m\right)$ converges with respect to $\tau _{m}$ to a point $x\in X$ such that

Lemma 1.5. ^[4] Let $\left(X, m\right)$ be an $M$-metric space. Then:

(i) $\{x_{n}\}$ is an $M$-Cauchy sequence in $\left(X, m\right)$ if and only if $\{x_{n}\}$ is a Cauchy sequence in a metric space $\left(X, m^{w}\right).$

(ii) An $M$-metric space $\left(X, m\right)$ is complete if and only if the metric space $\left(X, m^{w}\right)$ is complete. Moreover,

Lemma 1.6. ^[4] Suppose that $\{x_{n}\}$ convergesto $x$ and $\left \{ y_{n}\right \}$ converges to $y$ as $n$ approaches to $\infty$ in $M$-metric space $\left(X, m\right).$ Then we have

Lemma 1.7. ^[4] Suppose that $\{x_{n}\}$ converges to $x$as $n$ approaches to $\infty$ in $M$-metric space $\left(X, m\right).$Then we have

Lemma 1.8. ^[4] Suppose that $\{x_{n}\}$ converges to $x$and $\{x_{n}\}$ converges to $y$ as $n$ approaches to $\infty$ in $M$-metric space $\left(X, m\right).$ Then $m\left(x, y\right) = m_{xy}$moreover if $m\left(x, x\right) =$ $m\left(y, y\right),$ then $x = y.$

Definition 1.9. Let $\alpha :X\times X\rightarrow \left[0, \infty \right).$ A mapping $T:X\rightarrow X$ is said to be an $\alpha$-admissible mapping if for all $x, y\in X$

Let $\Psi$ be the family of the $\left(c\right)$-comparison functions $\psi : \mathbb{R} ^{+}\cup \left \{ 0\right \} \rightarrow \mathbb{R} ^{+}\cup \left \{ 0\right \}$ which satisfy the following properties:

(i) $\psi$ is nondecreasing,

(ii) $\sum_{n = 0}^{\infty }\psi ^{n}\left(t\right) < \infty$ for all $t > 0,$ where $\psi ^{n}$ is the $n$-iterate of $\psi$ (see ^[7,8,10,11]).

Definition 1.10. ^[33] Let $\left(X, d\right)$ be a metric space and $\alpha :X\times X\rightarrow \left[0, \infty \right)$. A mapping $T:X\rightarrow X$ is called $(\alpha, \psi)$-contractive mapping if for all $x, y\in X,$ we have

where $\psi \in \Psi.$

A subset $K$ of an $M$-metric space $X$ is called bounded if for all $x\in K$, there exist $y\in X$ and $r > 0$ such that $x\in B_{m}\left(y, r\right).$ Let $\overline{K}$ denote the closure of $K.$ The set $K$ is closed in $X$ if and only if $\overline{K} = K.$

Definition 1.11. ^[31] Define $H_{m}:CB_{m}\left(X\right) \times CB_{m}\left(X\right) \rightarrow \lbrack 0, \infty)$ by

where

Lemma 1.12. ^[31] Let $F$ be any nonempty set in $M$-metric space $\left(X, m\right)$, then

Proposition 1.13. ^[31] Let $A, B, C\in CB_{m}\left(X\right)$, then

(i) $\nabla _{m}\left(A, A\right) = \sup_{x\in A}\left \{ \sup_{y\in A}m_{xy}\right \},$

(ii) $\left(\nabla _{m}\left(A, B\right) -\sup_{x\in A}\sup_{y\in B}m_{xy}\right) \leq \left(\nabla _{m}\left(A, C\right) -\inf_{x\in A}\inf_{z\in C}m_{xz}\right) +$

$\left(\nabla _{m}\left(C, B\right) -\inf_{z\in C}\inf_{y\in B}m_{zy}\right).$

Proposition 1.14. ^[31] Let $A, B, C\in CB_{m}\left(X\right)$ followingare hold

(i) $H_{m}\left(A, A\right) = \nabla _{m}\left(A, A\right) = \sup_{x\in A}\left \{ \sup_{y\in A}m_{xy}\right \},$

(ii) $H_{m}\left(A, B\right) = H_{m}\left(B, A\right),$

(iii) $H_{m}\left(A, B\right) -\sup_{x\in A}\sup_{y\in A}m_{xy})\leq H_{m}\left(A, C\right) +H_{m}\left(B, C\right) -\inf_{x\in A}\inf_{z\in C}m_{xz}-\inf_{z\in C}\inf_{y\in B}m_{zy}.$

Lemma 1.15. ^[31] Let $A, B\in CB_{m}\left(X\right)$ and $h > 1$.Then for each $x\in A,$ there exist at the least one $y\in B$ such that

Lemma 1.16. ^[31] Let $A, B\in CB_{m}\left(X\right)$ and $l > 0$.Then for each $x\in A,$ there exist at least one $y\in B$ such that

Theorem 1.17. ^[31] Let $\left(X, m\right)$ be a complete $M$-metric space and $T:X\rightarrow CB_{m}\left(X\right)$. Assume that there exist $h\in \left(0, 1\right)$ such that

for all $x, y\in X$. Then $T$ has a fixed point.

Proposition 1.18. ^[31] Let $T:X\rightarrow CB_{m}\left(X\right)$ be a set-valued mapping satisfying $\left(1.1\right)$ for all $x, y$ inan $M$-metric space $X$. If $z\in T\left(z\right)$ for some $z$ in $X$such that $m\left(x, x\right) = 0$ for $x\in T\left(z\right).$

2.   Main results

We start with the following definition:

Definition 2.1. Assume that $\Psi$ is a family of non-decreasing functions $\phi _{M}: \mathbb{R} ^{+}\rightarrow \mathbb{R} ^{+}$ such that

(i) $\sum_{n}^{+\infty }\phi _{M}^{n}\left(x\right) < \infty$ for every $x > 0$ where $\phi _{M}^{n}$ is a $n^{th}$-iterate of $\phi _{M},$

(ii) $\phi _{M}\left(x+y\right) \leq \phi _{M}\left(x\right) +\phi _{M}\left(y\right)$ for all $x, y\in \mathbb{R} ^{+},$

(iii) $\phi _{M}\left(x\right) < x,$ for each$\ x > 0.$

Remark 2.2. If $\sum \alpha _{n}|_{n = \infty }$ $= 0$ is a convergent series with positive terms then there exists a monotonic sequence $(\beta _{n})|_{n = \infty }$ such that $\beta _{n}|_{n = \infty } = \infty$ and $\sum \alpha _{n}\beta _{n}|_{n = \infty } = 0$ converges.

Definition 2.3. Let $\left(X, m\right)$ be an $M$-metric pace. A self mapping $T:X\rightarrow X$ is called $(\alpha _{\ast }, \phi _{M})$-contraction if there exist two functions $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right)$ and $\phi _{M}\in \Psi$ such that

for all $x, y\in X.$

Definition 2.4. Let $\left(X, m\right)$ be an $M$-metric space. A set-valued mapping $T:X\rightarrow CB_{m}\left(X\right)$ is said to be $(\alpha _{\ast }, \phi _{M})$-contraction if for all $x, y\in X,$ we have

where $\phi _{M}\in \Psi$ and $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right).$

A mapping $T$ is called $\alpha _{\ast }$-admissible if

for each $a_{1}\in T\left(x\right)$ and $b_{1}\in T\left(y\right).$

Theorem 2.5. Let $\left(X, m\right)$ be a complete $M$-metric space.Suppose that $\left(\alpha _{\ast }, \phi _{M}\right)$ contraction and $\alpha _{\ast }$-admissible mapping $T:X\rightarrow CB_{m}\left(X\right)$satisfies the following conditions:

(i) there exist $x_{0}\in X$ such that $\alpha _{\ast }\left(x_{0}, a_{1}\right) \geq 1$ for each $a_{1}\in T\left(x_{0}\right),$

(ii) if $\left \{ x_{n}\right \} \in X$ is a sequence such that $\alpha _{\ast }\left(x_{n}, x_{n+1}\right) \geq 1$ for all $n$ and $\left \{ x_{n}\right \} \rightarrow x\in X$ as $n\rightarrow \infty$, then $\alpha _{\ast }\left(x_{n}, x\right) \geq 1$ for all $n\in \mathbb{N}.$ Then $T$ has a fixed point.

Proof. Let $x_{1}\in T\left(x_{0}\right)$ then by the hypothesis $\left(i\right)$ $\alpha _{\ast }\left(x_{0}, x_{1}\right) \geq 1.$ From Lemma $1.16$, there exist $x_{2}\in T\left(x_{1}\right)$ such that

Similarly, there exist $x_{3}\in T\left(x_{2}\right)$ such that

Following the similar arguments, we obtain a sequence $\left \{ x_{n}\right \} \in X$ such that there exist $x_{n+1}\in T\left(x_{n}\right)$ satisfying the following inequality

Since $T$ is $\alpha _{\ast }$-admissible, therefore $\alpha _{\ast }\left(x_{0}, x_{1}\right) \geq 1\Rightarrow \alpha _{\ast }\left(x_{1}, x_{2}\right) \geq 1.$ Using mathematical induction, we get

By $\left(2.1\right)$ and $\left(2.2\right)$, we have

Let us assume that $\epsilon > 0,$ then there exist $n_{0}\in N$ such that

By the Remarks $\left(1.2\right)$ and $\left(2.2\right),$ we get

Using the above inequality and $\left(m_{2}\right)$, we deduce that

Owing to limit, we have $\lim_{n\rightarrow \infty }m\left(x_{n}, x_{n}\right) = 0,$

Now, we prove that $\left \{ x_{n}\right \}$ is $M$-Cauchy in $X.\,$ For $m, n$ in $N$ with $m > n$ and using the triangle inequality of an $M$-metric we get

$m\left(x_{n}, x_{m}\right) -m_{x_{n}x_{m}}\rightarrow 0,$ as $n\rightarrow \infty,$ we obtain $\lim_{m, n\rightarrow \infty }\left(M_{x_{n}x_{m}}-m_{x_{n}x_{m}}\right) = 0$. Thus $\left \{ x_{n}\right \}$ is a $M$-Cauchy sequence in $X.$ Since $\left(X, m\right)$ is $M$-complete, there exist $x^{\star }\in X$ such that

Also, $\lim_{n\rightarrow \infty }m\left(x_{n}, x_{n}\right) = 0$ gives that

which implies that $m\left(x^{\star }, x^{\star }\right) = 0$ and hence we obtain $m_{x^{\star }T(x^{\star })} = 0$. By using $\left(2.1\right)$ and $\left(2.3\right)$ with

Thus,

Now from $\left(2.3\right),$ $\left(2.4\right),$ and $x_{n+1}\in T\left(x_{n}\right),$ we have

Taking limit as $n\rightarrow \infty$ and using $\left(2.4\right),$ we obtain that

Since $m_{x_{n+1}T\left(x^{\star }\right) }\leq m\left(x_{n+1}, T\left(x^{\star }\right) \right)$ which gives

Using the condition $\left(m_{4}\right),$ we obtain

Applying limit as $n\rightarrow \infty$ and using $\left(2.3\right)$ and $\left(2.6\right),$ we have

From $\left(m_{2}\right),$ $m_{x^{\star }y}\leq m\left(x^{\star }y\right)$ for each $y\in T\left(x^{\star }\right)$ which implies that

Hence,

Then

Thus

Now, from $\left(2.7\right)$ and $\left(2.8\right)$, we obtain

Consequently, owing to Lemma $\left(1.12\right)$, we have $x^{\star }\in \overline{T\left(x^{\star }\right) } = T\left(x^{\star }\right).$

Corollary 2.6. Let $\left(X, m\right)$ be a complete $M$-metric space and anself mapping $T:X\rightarrow X$ an $\alpha _{\ast }$-admissible and $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction mapping. Assume that thefollowing properties hold:

(i) there exists $x_{0}\in X$ such that $\alpha _{\ast }\left(x_{0}, T\left(x_{0}\right) \right) \geq 1,$

(ii) either $T$ is continuous or for any sequence $\{x_{n}\} \in X$ with $\alpha _{\ast }\left(x_{n}, x_{n+1}\right) \geq 1$ for all $n\in \mathbb{N}$ and $\left \{ x_{n}\right \} \rightarrow x$ as $n$ $\rightarrow$ $\infty$, we have $\alpha _{\ast }\left(x_{n}, x\right) \geq 1$ for all $n\in \mathbb{N}.$ Then $T$ has a fixed point.

Some fixed point results in ordered $M$-metric space.

Definition 2.7. Let $\left(X, \preceq \right)$ be a partially ordered set. A sequence $\left \{ x_{n}\right \} \subset X$ is said to be non-decreasing if $x_{n}\preceq x_{n+1}$ for all $n$.

Definition 2.8. ^[16] Let $F$ and $G$ be two nonempty subsets of partially ordered set $\left(X, \preceq \right)$. The relation between $F$ and $G$ is defined as follows: $F\prec _{1}G$ if for every $x\in F,$ there exists $y\in G$ such that $x\preceq y.$

Definition 2.9. Let $\left(X, m, \preceq \right)$ be a partially ordered set on $M$-metric. A set-valued mapping $T:X\rightarrow CB_{m}\left(X\right)$ is said to be ordered $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction if for all $x, y\in X,$ with $x\preceq y$ we have

where $\phi _{M}\in \Psi.$ Suppose that $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right)$ is defined by

A mapping $T$ is called $\alpha _{\ast }$-admissible if

for each $a_{1}\in T\left(x\right)$ and $b_{1}\in T\left(y\right).$

Theorem 2.10. Let $\left(X, m, \preceq \right)$ be a partially orderedcomplete $M$-metric space and $T:X\rightarrow CB_{m}\left(X\right)$ an $\alpha _{\ast }$-admissible ordered $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction mapping satisfying the following conditions:

(i) there exist $x_{0}\in X$ such that $\left \{ x_{0}\right \} \prec _{1}\left \{ T\left(x_{0}\right) \right \},$ $\alpha _{\ast }\left(x_{0}, a_{1}\right) \geq 1$ for each $a_{1}\in T\left(x_{0}\right),$

(ii) for every $x, y\in X,$ $x\preceq y$ implies $T\left(x\right) \prec _{1}T\left(y\right),$

(iii) If $\left \{ x_{n}\right \} \in X$ is a non-decreasing sequence such that $x_{n}\preceq x_{n+1}$ for all $n$ and $\left \{ x_{n}\right \} \rightarrow x\in X$ as $n$ $\rightarrow \infty$ gives $x_{n}\preceq x$ for all $n\in \mathbb{N}.$ Then $T$ has a fixed point.

Proof. By assumption $\left(i\right)$ there exist $x_{1}\in T\left(x_{0}\right)$ such that $x_{0}\preceq x_{1}$ and $\alpha _{\ast }\left(x_{0}, x_{1}\right) \geq 1.$ By hypothesis $\left(ii\right),$ $T\left(x_{0}\right) \prec _{1}T\left(x_{1}\right).$ Let us assume that there exist $x_{2}\in T\left(x_{1}\right)$ such that $x_{1}\preceq x_{2}$ and we have the following

In the same way, there exist $x_{3}\in T\left(x_{2}\right)$ such that $x_{2}\preceq x_{3}$ and

Following the similar arguments, we have a sequence $\left \{ x_{n}\right \} \in X$ $\ $and $x_{n+1}\in T\left(x_{n}\right)$ for all $n\geq 0$ satisfying $x_{0}\preceq x_{1}\preceq x_{2}\preceq x_{3}\preceq...x_{n}\preceq x_{n+1}$. The proof is complete follows the arguments given in Theorem 2.5.

Example 2.11. Let $X = \left[\frac{1}{6}, 1\right]$ be endowed with an $M$ -metric given by $m\left(x, y\right) = \frac{x+y}{2}$. Define $T:X\rightarrow CB_{m}\left(X\right)$ by

Define a mapping $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right)$ by

Let $\phi _{M}: \mathbb{R} ^{+}\rightarrow \mathbb{R} ^{+}\ $be given by $\phi _{M}\left(t\right) = \frac{17}{10}$ where $\phi _{M}\in \Psi,$ for $x, y\in X$. If $x = \frac{1}{6},$ $y = \frac{1}{4}$ then $m\left(x, y\right) = \frac{5}{24}$, and

If $x = \frac{1}{3},$ $y = \frac{1}{2}$ then $m\left(x, y\right) = \frac{5}{12}$, and

If $x = \frac{1}{6},$ $y = 1,$ then $m\left(x, y\right) = \frac{7}{12}$ and

In all cases, $T$ is $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction mapping. If $x_{0} = \frac{1}{3},$ then $T\left(x_{0}\right) = \left \{ \frac{x }{2}, \frac{x}{3}\right \}.$Therefore $\alpha _{\ast }\left(x_{0}, a_{1}\right) \geq 1$ for every $a_{1}\in T\left(x_{0}\right).$ Let $x, y\in X$ be such that $\alpha _{\ast }\left(x, y\right) \geq 1,$ then $x, y\in \left[\frac{x}{2}, \frac{x}{3}\right]$ and $T\left(x\right) = \left \{ \frac{x}{2}, \frac{x}{3}\right \}$ and $T\left(y\right) =$ $\left \{ \frac{x}{2}, \frac{x}{3}\right \}$ which implies that $\alpha _{\ast }\left(a_{1}, b_{1}\right) \geq 1$ for every $a_{1}\in T\left(x\right)$ and $b_{1}\in T\left(x\right)$. Hence $T$ is $\alpha _{\ast }$-admissble.

Let $\left \{ x_{n}\right \} \in X$ be a sequence such that $\alpha _{\ast }\left(x_{n}, x_{n+1}\right) \geq 1$ for all $n$ in $\mathbb{N}$ and $x_{n}$ converges to $x$ as $n$ converges to $\infty,$ then $x_{n}\in \left[\frac{x}{2}, \frac{x}{3}\right].$ By definition of $\alpha _{\ast }$ -admissblity, therefore $x\in \left[\frac{x}{2}, \frac{x}{3}\right]$ and hence $\alpha _{\ast }\left(x_{n}, x\right) \geq 1.$ Thus all the conditions of Theorem 2.3 are satisfied. Moreover, $T$ has a fixed point$.$

Example 2.12. Let $X = \left \{ \left(0, 0\right), \left(0, -\frac{1}{5} \right), \left(-\frac{1}{8}, 0\right) \right \}$ be the subset of $\mathbb{R} ^{2}$ with order $\preceq$ defined as: For $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right) \in X,$ $\left(x_{1}, y_{1}\right) \preceq \left(x_{2}, y_{2}\right)$ if and only if $x_{1}\leq x_{2},$ $y_{1}\leq y_{2}.$ Let $m:X\times X\rightarrow \mathbb{R} ^{+}$ be defined by

Then $\left(X, m\right)$ is a complete $M$-metric space. Let $T:X\rightarrow CB_{m}\left(X\right)$ be defined by

Define a mapping $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right)$ by

Let $\phi _{M}: \mathbb{R} ^{+}\rightarrow \mathbb{R} ^{+}\ $be given by $\phi _{M}\left(t\right) = \frac{1}{2}$. Obviously, $\phi _{M}\in \Psi.$ For $x, y\in X,$

if $x = \left(0, -\frac{1}{5}\right)$ and $y = \left(0, 0\right),$ then $H_{m}\left(T\left(x\right), T\left(y\right) \right) = 0$ and $m\left(x, y\right) = \frac{1}{10}$ gives that

If $x = \left(-\frac{1}{8}, 0\right)$ and $y = \left(0, 0\right)$ then $H_{m}\left(T\left(x\right), T\left(y\right) \right) = 0,$ and $m\left(x, y\right) = \frac{1}{16}$ implies that

If $x = \left(0, 0\right)$ and $y = \left(0, 0\right)$ then $H_{m}\left(T\left(x\right), T\left(y\right) \right) = 0,$ and $m\left(x, y\right) = 0$ gives

If $x = \left(0, -\frac{1}{5}\right)$ and $y = \left(0, -\frac{1}{5}\right)$ then $H_{m}\left(T\left(x\right), T\left(y\right) \right) = 0,$ and $m\left(x, y\right) = \frac{1}{5}$ implies that

If $x = \left(0, -\frac{1}{8}\right)$ and $y = \left(0, -\frac{1}{8}\right)$ then $H_{m}\left(T\left(x\right), T\left(y\right) \right) = 0,$ and $m\left(x, y\right) = \frac{1}{8}$ gives that

Thus all the condition of Theorem 2.10 satisfied. Moreover, $\left(0, 0\right)$ is the fixed point of $T.$

3.   Application

In this section, we present an application of our result in homotopy theory. We use the fixed point theorem proved for set-valued $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction mapping in the previous section, to establish the result in homotopy theory. For further study in this direction, we refer to ^[6,35].

Theorem 3.1. Suppose that $\left(X, m\right)$ is a complete $M$-metricspace and $A$ and $B$ are closed and open subsets of $X$ respectively, suchthat $A\subset B$. For $a, b\in \mathbb{R}$, let $T:B\times \left[a, b\right] \rightarrow CB_{m}\left(X\right)$ be aset-valued mapping satisfying the following conditions:

$(i)$ $x\notin T\left(y, t\right)$ for each $y\in B/A$and $t\in \left[a, b\right],$

$(ii)$ there exist $\phi _{M}\in \Psi$ and $\alpha_{\ast }:X\times X\rightarrow \left[0, \infty \right)$ such that

for each pair $\left(x, y\right) \in B\times B$ and $t\in \left[a, b\right],$

$(iii)$ there exist a continuous function $\Omega :\left[a, b\right] \rightarrow \mathbb{R}$ such that for each $s, t\in \left[a, b\right]$ and $x\in B,$ we get

$(iv)$ if $x^{\star }\in T\left(x^{\star }, t\right),$then $T\left(x^{\star }, t\right) = \left \{ x^{\star }\right \},$

$(v)$ there exist $x_{0}$ in $X$ such that $x_{0}\in T\left(x_{0}, t\right),$

$(vi)$ a function $\Re :\left[0, \infty \right)\rightarrow \left[0, \infty \right)$ defined by $\Re \left(x\right) = x-\phi _{M}\left(x\right)$ is strictly increasing and continuous if $T\left(., t^{\intercal }\right)$ has a fixed point in $B$ for some $t^{\intercal }\in \left[a, b\right],$ then $T\left(., t\right)$ has afixed point in $A$ for all $t\in \left[a, b\right].$ Moreover, for a fixed $t\in \left[a, b\right]$, fixed point is unique provided that $\phi_{M}\left(t\right) = \frac{1}{2}t$ where $t > 0.$

Proof. Define a mapping $\alpha _{\ast }:X\times X\rightarrow \left[0, \infty \right)$ by

We show that $T$ is $\alpha _{\ast }$-admissible. Note that $\alpha _{\ast }\left(x, y\right) \geq 1$ implies that $x\in T\left(x, t\right)$ and $y\in T\left(y, t\right)$ for all $t\in \left[a, b\right]$. By hypothesis $\left(iv\right),$ $T\left(x, t\right) = \left \{ x\right \}$ and $T\left(y, t\right) = \left \{ y\right \}.$ It follows that $T$ is $\alpha _{\ast }$ -admissible. By hypothesis $\left(v\right),$ there exist $x_{0}\in X$ such that $x_{0}\in \left(x_{0}, t\right)$ for all $t$, that is $\alpha _{\ast }\left(x_{0}, x_{0}\right) \geq 1$. Suppose that $\alpha _{\ast }\left(x_{n}, x_{n+1}\right) \geq 1$ for all $n$ and $x_{n}$ converges to $q$ as $n$ approaches to $\infty$ and $x_{n}\in T\left(x_{n}, t\right)$ and $x_{n+1}\in T\left(x_{n+1}, t\right)$ for all $n$ and $t\in \left[a, b\right]$ which implies that $q\in T\left(q, t\right)$ and thus $\alpha _{\ast }\left(x_{n}, q\right) \geq 1.$ Set

So $T\left(., t^{\intercal }\right)$ has a fixed point in $B$ for some $t^{\intercal }\in \left[a, b\right]$, there exist $x\in B$ such that $x\in T\left(x, t\right).$ By hypothesis $\left(i\right)$ $x\in T\left(x, t\right)$ for $t\in \left[a, b\right]$ and $x\in A$ so $D\neq \phi$. Now we now prove that $D$ is open and close in $\left[a, b\right].$ Let $t_{0}\in D$ and $x_{0}\in A$ with $x_{0}\in T\left(x_{0}, t_{0}\right).$ Since $A$ is open subset of $X$, $\overline{B_{m}\left(x_{0}, r\right) } \subseteq A$ for some $r > 0$. For $\epsilon = r+m_{xx_{0}}-\phi \left(r+m_{xx_{0}}\right)$ and a continuous function $\Omega$ on $\left[a, b \right]$, there exist $\delta > 0$ such that

If $t\in \left(t_{0}-\delta, t_{0}+\delta \right)$ for $x\in B_{m}\left(x_{0}, r\right) = \left \{ x\in X:m\left(x_{0}\, , x\right) \leq m_{x_{0}x}+r\right \}$ and $l\in T\left(x, t\right),$ we obtain

Using the condition $\left(iii\right)$ of Proposition 1.13 and Proposition 1.18, we have

as $x\in T\left(x_{0}, t_{0}\right)$ and $x\in B_{m}\left(x_{0}, r\right) \subseteq A\subseteq B$, $t_{0}\in \left[a, b\right]$ with $\alpha _{\ast }\left(x_{0}, x_{0}\right) \geq 1.$ By hypothesis $\left(ii\right)$, $\left(iii\right)$ and $\left(2.9\right)$

Hence $l\in \overline{B_{m}\left(x_{0}, r\right) }$ and thus for each fixed $t\in \left(t_{0}-\delta, t_{0}+\delta \right),$ we obtain $T\left(x, t\right) \subset \overline{B_{m}\left(x_{0}, r\right) }$ therefore $T: \overline{B_{m}\left(x_{0}, r\right) }\rightarrow CB_{m}\left(\overline{ B_{m}\left(x_{0}, r\right) }\right)$ satisfies all the assumption of Theorem $\left(3.1\right)$ and $T\left(., t\right)$ has a fixed point $\overline{B_{m}\left(x_{0}, r\right) } = B_{m}\left(x_{0}, r\right) \subset B$. But by assumption of $\left(i\right)$ this fixed point belongs to $A$. So $\left(t_{0}-\delta, t_{0}+\delta \right) \subseteq D,$ thus $D$ is open in $\left[a, b\right].$ Next we prove that $D$ is closed. Let a sequence $\left \{ t_{n}\right \} \in D$ with $t_{n}$ converges to $t_{0}\in \left[a, b \right]$ as $n$ approaches to $\infty.$ We will prove that $t_{0}$ is in $D$.

Using the definition of $D,$ there exist $\left \{ t_{n}\right \}$ in $A$ such that $x_{n}\in T\left(x_{n}, t_{n}\right)$ for all $n$. Using Assumption $\left(iii\right)$–$\left(v\right),$ and the condition $\left(iii\right)$ of Proposition 1.13, and an outcome of the Proposition 1.18, we have

So, continuity of $\frac{1}{\Re },$ $\Re$ and convergence of $\left \{ t_{n}\right \},$ taking the limit as $m, n\rightarrow \infty$ in the last inequality, we obtain that

Sine $m_{x_{n}x_{m}}\leq m\left(x_{n}, x_{m}\right),$ therefore

Thus, we have $\lim_{n\rightarrow \infty }m\left(x_{n}, x_{n}\right) = 0 = \lim_{m\rightarrow \infty }m\left(x_{m}, x_{m}\right)$. Also,

Hence $\left \{ x_{n}\right \}$ is an $M$-Cauchy sequence. Using Definition 1.4, there exist $x^{\ast }$ in $X$ such that

As $\lim_{n\rightarrow \infty }m\left(x_{n}, x_{n}\right) = 0$, therefore

Thus, we have $m\left(x, x^{\ast }\right) = 0.$ We now show that $x^{\ast }\in T\left(x^{\ast }, t^{^{\ast }}\right).$ Note that

Applying the limit $n\rightarrow \infty$ in the above inequality, we have

Hence

Since $m\left(x^{\ast }, x^{\ast }\right) = 0,$ we obtain

From above two inequalities, we get

Thus using Lemma 1.12 we get $x^{\ast }\in T\left(x^{\ast }, t^{^{\ast }}\right).$ Hence $x^{\ast }\in A.$ Thus $x^{\ast }\in D$ and $D$ is closed in $\left[a, b\right],$ $D = \left[a, b\right]$ and $D$ is open and close in $\left[a, b\right].$ Thus $T\left(., t\right)$ has a fixed point in $A$ for all $t\in \left[a, b\right].$ For uniqueness, $t\in \left[a, b\right]$ is arbitrary fixed point, then there exist $x\in A$ such that $x\in T\left(x, t\right)$. Assume that $y$ is an other point of $T\left(x, t\right)$, then by applying condition 4, we obtain

For$\phi _{M}\left(t\right) = \frac{1}{2}t,$ where $t > 0,$ the uniqueness follows.

4.   Application to integral equation

In this section we will apply the previous theoretical results to show the existence of solution for some integral equation. For related results (see ^[13,20]). We see for non-negative solution of $\left(3.1\right)$ in $X = C\left(\left[0, \delta \right], \mathbb{R} \right).$ Let $X = C\left(\left[0, \delta \right], \mathbb{R} \right)$ be a set of continuous real valued functions defined on $\left[0, \delta \right]$ which is endowed with a complete $M$-metric given by

Consider an integral equation

Define $g:X\rightarrow X$ by

where

(i) for $\delta > 0$, $\ \left(a\right)$ $J:\left[0, \delta \right] \times \mathbb{R} \rightarrow \mathbb{R},$ $\left(b\right)$ $h:\left[0, \delta \right] \times \left[0, \delta \right] \rightarrow \left[0, \infty \right),$ $\left(c\right)$ $\rho : \left[0, \delta \right] \rightarrow \mathbb{R}$ are all continuous functions

(ii) Assume that $\sigma :X\times X\rightarrow \mathbb{R}$ is a function with the following properties,

(iii) $\sigma \left(x, y\right) \geq 0$ implies that $\sigma \left(T\left(x\right), T\left(y\right) \right) \geq 0,$

(iv) there exist $x_{0}\in X$ such that $\sigma \left(x_{0}, T\left(x_{0}\right) \right) \geq 0,$

(v) if $\left \{ x_{n}\right \} \in X$ is a sequence such that $\sigma \left(x_{n}, x_{n+1}\right) \geq 0$ for all $n\in \mathbb{N}$ and $x_{n}\rightarrow x$ as $n\rightarrow \infty,$ then $\sigma \left(x, T\left(x\right) \right) \geq 0$

(vi)

where $t\in \left[0, \delta \right]$, $s\in \mathbb{R},$

$\left(vii\right)$ there exist $\phi _{M}\in \Psi$, $\sigma \left(y, T\left(y\right) \right) \geq 1$ and $\sigma \left(x, T\left(x\right) \right) \geq 1$ such that for each $t\in \left[0, \delta \right],$ we have

Theorem 4.1. Under the assumptions $\left(i\right) -\left(vii\right)$ theintegral Eq $\left(3.1\right)$ has a solution in $\left \{ X = C\left(\left[0, \delta \right], \mathbb{R} \right) \mathit{\text{ for all }}t\in \left[0, \delta \right] \right \}.$

Proof. Using the condition $\left(vii\right)$, we obtain that

Define $\alpha _{\ast }:X\times X\rightarrow \left[0, +\infty \right)$ by

which implies that

Hence all the assumption of the Corollary 2.6 are satisfied, the mapping $g$ has a fixed point in $X = C\left(\left[0, \delta \right], \mathbb{R} \right)$ which is the solution of integral Eq $\left(3.1\right).$

5.   Conclusions

In this study we develop some set-valued fixed point results based on $\left(\alpha _{\ast }, \phi _{M}\right)$-contraction mappings in the context of $M$-metric space and ordered $M$-metric space. Also, we give examples and applications to the existence of solution of functional equations and homotopy theory.

Conflict of interest

The authors declare that they have no competing interests.