# Fredholm type integral equation in extended $$M_{b}$$-metric spaces

## Abstract

In this manuscript, we prove some new fixed point results for a self-mapping on extended $$M_{b}$$-metric spaces, under some new types of contractions, which generalizes many results in the literature. Also, we present some interesting examples to illustrate our work.

## Introduction and preliminaries

Lately, fixed point theory has become the focus of many researchers and that is due to its applications in many fields, such as engineering and computer sciences. Also fixed point theory can be used to solve differential equations along with integral equations . M-metric spaces were introduced by Asadi, Karapinar and Salimi, in , they are an extension of a partial metric space. Then some relationships between a partial metric and an M-metric were investigated in . So, first we remind the reader of the definition of a partial metric space and an M-metric space along with some other notations.

### Definition 1.1

([17, 18])

A partial metric on a nonempty set X is a function $$p_{i}:X^{2}\rightarrow [ 0,+\infty )$$ such that for all $$\lambda ,\epsilon ,z\in X$$

($$p_{i}1$$):

$$p_{i}(\lambda ,\lambda )=p_{i}(\epsilon ,\epsilon )=p_{i}(\lambda , \epsilon )$$ if and only if $$\lambda =\epsilon$$,

($$p_{i}2$$):

$$p_{i}(\lambda ,\lambda )\leq p_{i}(\lambda ,\epsilon )$$,

($$p_{i}3$$):

$$p_{i}(\lambda ,\epsilon )=p_{i}(\epsilon ,\lambda )$$,

($$p_{i}4$$):

$$p_{i}(\lambda ,\epsilon )\leq p_{i}(\lambda ,z)+p_{i}(z,\epsilon )-p_{i}(z,z)$$.

A partial metric space is a pair $$(X,p_{i})$$ such that X is a nonempty set and $$p_{i}$$ is a partial metric on X.

### Notation 1.2

()

1. 1.

$$k_{\lambda ,\epsilon }:= \min \{ K(\lambda ,\lambda ),K(\epsilon , \epsilon ) \}$$.

2. 2.

$$M_{\lambda ,\epsilon }:= \max \{ K(\lambda ,\lambda ),K(\epsilon , \epsilon ) \}$$.

### Definition 1.3

()

Let X be a nonempty set. If the function $$K:X^{2}\rightarrow [ 0,+\infty )$$ satisfies the following conditions:

1. (1)

$$K(\lambda ,\lambda )=K(\epsilon ,\epsilon )=K(\lambda ,\epsilon )$$ if and only if $$\lambda =\epsilon$$,

2. (2)

$$k_{\lambda ,\epsilon } \le K(\lambda ,\epsilon )$$,

3. (3)

$$K(\lambda ,\epsilon )=K(\epsilon ,\lambda )$$,

4. (4)

$$(K(\lambda ,\epsilon )-k_{\lambda ,\epsilon })\leq (K(\lambda ,z)-k_{ \lambda ,z})+(K(z,\epsilon )-k_{z,\epsilon })$$,

for all $$\lambda ,\epsilon ,z\in X$$, then the pair $$(X,K)$$ is called an M-metric space.

Recently, Mlaiki et al. , developed the concept of an $$M_{b}$$-metric space which extends an M-metric space, and some fixed point theorems are established. Also, $$M_{b}$$-metric spaces are a generalization of b-metric spaces; see . Now, we remind the reader of some definitions and notations of $$M_{b}$$-metric spaces.

### Notation 1.4

()

1. 1.

$$k_{{b}{\lambda ,\epsilon }}:= \min \{ K_{b}(\lambda ,\lambda ),K_{b}( \epsilon ,\epsilon ) \}$$.

2. 2.

$$M_{{b}{\lambda ,\epsilon }}:= ma\lambda \{ K_{b}(\lambda ,\lambda ),K_{b}( \epsilon ,\epsilon ) \}$$.

### Definition 1.5

()

An $$M_{b}$$-metric on a nonempty set X is a function $$K_{b}:X^{2}\rightarrow [ 0,+\infty )$$ that satisfies the following conditions:

1. (1)

$$K_{b}(\lambda ,\lambda )=K_{b}(\epsilon ,\epsilon )=K_{b}(\lambda , \epsilon )$$ if and only if $$\lambda =\epsilon$$,

2. (2)

$$k_{{b}{\lambda ,\epsilon }} \le K_{b}(\lambda ,\epsilon )$$,

3. (3)

$$K_{b}(\lambda ,\epsilon )=K_{b}(\epsilon ,\lambda )$$,

4. (4)

there exists a real number $$s\geq 1$$ such that for all $$\lambda ,\epsilon ,z\in X$$ we have

$$\bigl(K_{b}(\lambda ,\epsilon )-k_{{b}{\lambda ,\epsilon }} \bigr)\leq s \bigl[ \bigl(K_{b}( \lambda ,z)-k_{{b}{\lambda ,z}} \bigr)+ \bigl(K_{b}(z,\epsilon )-k_{{b}{z,\epsilon }} \bigr) \bigr]-K_{b}(z,z),$$

for all $$\lambda ,\epsilon ,z\in X$$. Then the pair $$(X,K_{b})$$ is called an $$M_{b}$$-metric space and the number s is called the coefficient of the $$M_{b}$$-metric space $$(X,K_{b})$$.

Note that the condition (4) given in Definition 1.5 is equivalent to the following condition:

(4)′:

There exists a real number $$s\geq 1$$ such that for all $$\lambda ,\epsilon ,z\in X$$ we have

$$\bigl(K_{b}(\lambda ,\epsilon )-k_{{b}{\lambda ,\epsilon }} \bigr)\leq s \bigl[ \bigl(K_{b}( \lambda ,z)-k_{{b}{\lambda ,z}} \bigr)+ \bigl(K_{b}(z,\epsilon )-k_{{b}{z,\epsilon }} \bigr) \bigr],$$
for all $$\lambda ,\epsilon ,z\in X$$.

Indeed, if we take $$\lambda =r$$ under the condition (4) then we get

$$K_{b}(\lambda ,\lambda )-k_{{b}{\lambda ,\lambda }}=K_{b}( \lambda , \lambda )-\min \bigl\{ K_{b}(\lambda ,\lambda ),K_{b}(\lambda , \lambda ) \bigr\} =0$$

and so we have

$$0\leq s \bigl[ \bigl(K_{b}(\lambda ,\lambda )-k_{{b}{\lambda ,\lambda }} \bigr)+ \bigl(K_{b}( \lambda ,\lambda )-k_{{b\lambda }{,\lambda }} \bigr) \bigr]-K_{b}(\lambda , \lambda )\leq -K_{b}(\lambda , \lambda )\text{,}$$

for $$z=\lambda$$. Therefore we get $$K_{b}(\lambda ,\lambda )=0$$ for all $$\lambda \in X$$ since $$K_{b}(\lambda ,\lambda )\in [ 0,+\infty )$$.

The concept of extended $$M_{b}$$-metric spaces was introduced in , which is a generalization of an $$M_{b}$$-metric space which also generalizes extended b-metric spaces . We give basic properties of this new space and its relation with some known metric spaces.

First, we give the following notation.

### Notation 1.6

1. (1)

$$k_{{\alpha }{\lambda ,\epsilon }}:=\min \{K_{\alpha }(\lambda , \lambda ),K_{\alpha }(\epsilon ,\epsilon )\}$$.

2. (2)

$$M_{{\alpha }{\lambda ,\epsilon }}:=\max \{K_{\alpha }(\lambda , \lambda ),K_{\alpha }(\epsilon ,\epsilon )\}$$.

### Definition 1.7

Let $$\alpha :X^{2}\rightarrow {}[ 1,+\infty )$$ be a function. An extended $$M_{b}$$-metric on a nonempty set X is a function $$K_{\alpha }:X^{2}\rightarrow {}[ 0,+\infty )$$ satisfying the following conditions:

1. (1)

$$K_{\alpha }(\lambda ,\lambda )=K_{\alpha }(\epsilon ,\epsilon )=K_{ \alpha }(\lambda ,\epsilon )$$ if and only if $$\lambda =\epsilon$$,

2. (2)

$$k_{{\alpha }{\lambda ,\epsilon }} \le K_{\alpha }(\lambda ,\epsilon )$$,

3. (3)

$$K_{\alpha }(\lambda ,\epsilon )=K_{\alpha }(\epsilon ,\lambda )$$,

4. (4)

$$(K_{\alpha }(\lambda ,\epsilon )-k_{{\alpha }{\lambda ,\epsilon }}) \leq \alpha (\lambda ,\epsilon )[(K_{\alpha }(\lambda ,z)-k_{{\alpha }{ \lambda ,z}})+(K_{\alpha }(z,\epsilon )-k_{{\alpha }{z,\epsilon }})]$$,

for all $$\lambda ,\epsilon ,z\in X$$. Then the pair $$(X,K_{\alpha })$$ is called an extended $$M_{b}$$-metric space.

We note that if $$\alpha (\lambda ,\epsilon )=s$$ for $$s\geq 1$$, then we get the definition of an $$M_{b}$$-metric space.

### Example 1.8

Let $$X=C([a,d],\mathbb{R})$$ be the set of all continuous real valued functions on $$[a,b]$$. We define the functions $$K_{\alpha }:X^{2}\rightarrow {}[ 0,+\infty )$$ and $$\alpha :X^{2}\rightarrow {}[ 1,+\infty )$$ by

$$K_{\alpha } \bigl(\lambda (t),\epsilon (t) \bigr)=\sup _{t\in {}[ a,b]} \bigl\vert \lambda (t)-y(t) \bigr\vert ^{2}$$

and

$$\alpha \bigl(\lambda (t),\epsilon (t) \bigr)= \bigl\vert \lambda (t) \bigr\vert + \bigl\vert y(t) \bigr\vert +2.$$

Then $$(X,K_{\alpha })$$ is an extended $$M_{b}$$-metric space with the function α.

Now we give the following proposition.

### Proposition 1.9

Let$$(X,K_{\alpha })$$be an extended$$M_{b}$$-metric space and$$\lambda ,\epsilon ,z\in X$$. Then we have

1. (1)

$$M_{\alpha \lambda ,\epsilon }+k_{\alpha \lambda ,\epsilon }=K_{ \alpha }(\lambda ,\lambda )+K_{\alpha }(\epsilon ,\epsilon )\geq 0$$,

2. (2)

$$M_{\alpha \lambda ,\epsilon }-k_{\alpha \lambda ,\epsilon }= \vert K_{\alpha }(\lambda ,\lambda )-K_{\alpha }(\epsilon ,\epsilon ) \vert \geq 0$$,

3. (3)

$$M_{\alpha \lambda ,\epsilon }-k_{\alpha \lambda ,\epsilon }\leq \alpha (\lambda ,\epsilon ) [ ( M_{\alpha \lambda ,z}-k_{ \alpha \lambda ,z} ) + ( M_{\alpha z,\epsilon }-k_{\alpha z, \epsilon } ) ]$$.

In this section, we give some topological notions on an extended $$M_{b}$$-metric space.

### Definition 1.10

Let $$(X,K_{\alpha })$$ be an extended $$M_{b}$$-metric space. Then:

1. (1)

A sequence $$\{\lambda _{n}\}$$ in X converges to a point λ if and only if

$$\lim_{n\rightarrow +\infty } \bigl(K_{\alpha }(\lambda _{n}, \lambda )-k_{{ \alpha }{\lambda _{n},\lambda }} \bigr)=0.$$
2. (2)

A sequence $$\{\lambda _{n}\}$$ in X is said to be a $$K_{\alpha }$$-Cauchy sequence if

$$\lim_{n,m\rightarrow +\infty } \bigl(K_{\alpha }(\lambda _{n}, \lambda _{m})-k_{{ \alpha }{\lambda _{n},\lambda _{m}}} \bigr)$$

and

$$\lim_{n\rightarrow +\infty }(M_{{\alpha }{\lambda _{n},\lambda _{m}}}-k_{{ \alpha }{\lambda _{n},\lambda _{m}}})$$

exist and are finite.

3. (3)

An extended $$M_{b}$$-metric space is said to be $$K_{\alpha }$$-complete if every $$K_{\alpha }$$-Cauchy sequence $$\{\lambda _{n}\}$$ converges to a point λ such that

$$\lim_{n\rightarrow +\infty } \bigl(K_{\alpha }(\lambda _{n}, \lambda )-k_{{ \alpha }{\lambda _{n},\lambda }} \bigr)=0$$

and

$$\lim_{n\rightarrow +\infty }(M_{{\alpha }{\lambda _{n}, \lambda }}-k_{{\alpha }{\lambda _{n},\lambda }})=0.$$

### Remark 1.11

If we consider Example 1.8, then it is not difficult to see that $$(X,K_{\alpha })$$ is a complete extended $$M_{b}$$-metric space.

### Lemma 1.12

Let$$(X,K_{\alpha })$$be an extended$$M_{b}$$-metric space. Then we get:

1. (1)

$$\{ \lambda _{n} \}$$is an$$K_{\alpha }$$-Cauchy sequence in$$(X,K_{\alpha })$$if and only if$$\{ \lambda _{n} \}$$is a Cauchy sequence in$$(X,K_{\alpha }^{b})$$.

2. (2)

$$(X,K_{\alpha })$$is complete if and only if$$(X,K_{\alpha }^{b})$$is complete.

## Main result

First, we start this section by proving the following theorem, which we consider our main result.

### Theorem 2.1

Let$$(X,K_{\alpha })$$be a complete extended$$M_{b}$$-metric space andfbe a continuous self-mapping onX. Suppose that there exists$$p \in [0,1)$$such that for all$$\lambda ,\epsilon \in X$$we have

$$K_{\alpha }(f\lambda ,f\epsilon ) \leq p \alpha (\lambda ,\epsilon ) K_{ \alpha }(\lambda ,\epsilon ) .$$
(2.1)

Also, fix$$\lambda _{0} \in X$$and define the sequence$$(\lambda _{n})$$defined by$$\lambda _{i} = f\lambda _{i-1}$$. If

$$\sup_{m \geq 1} \lim_{n} \alpha (\lambda _{m},\lambda _{n}) \alpha ( \lambda _{n}, \lambda _{n +1}) < \frac{1}{p},$$

and for every$$\lambda \in X$$we have$$\{ \alpha (\lambda ,\lambda _{n})\}_{n}$$and$$\{\alpha (\lambda _{n},\lambda )\}_{n}$$are bounded. Thenfhas a fixed point onX. Moreover, if for every two fixed points$$r,s \in X$$we have$$\alpha (r,s)<\frac{1}{p}$$, then the fixed point is unique.

### Proof

Using the sequence as defined in the hypotheses of the theorem and (2.1)

\begin{aligned} K_{\alpha }(\lambda _{n}, \lambda _{n+1})&= K_{\alpha }(f\lambda _{n-1}, f\lambda _{n}) \\ & \leq p \alpha (\lambda _{n-1}, \lambda _{n})K_{\alpha }( \lambda _{n-1}, \lambda _{n}) \\ & =p \alpha (\lambda _{n-1}, \lambda _{n})K_{\alpha }(f \lambda _{n-2}, \lambda _{n-1}) \\ & \leq p^{2} \alpha (\lambda _{n-1}, \lambda _{n}) \alpha (\lambda _{n-2}, \lambda _{n-1}) K_{\alpha }(\lambda _{n-2}, \lambda _{n-1}) \\ &\quad \vdots \\ & \leq p^{n}\prod_{i=1}^{n} \alpha (\lambda _{i-1},\lambda _{i})K_{ \alpha }( \lambda _{0},\lambda _{1}). \end{aligned}

Now, consider $$n,m \in \mathbb{N}$$ where $$m>n$$. Then

\begin{aligned} K_{\alpha }(\lambda _{n}, \lambda _{m})- k_{\alpha \lambda _{n}, \lambda _{m}} & \leq \alpha (\lambda _{n}, \lambda _{m}) \bigl[ \bigl(K_{\alpha }( \lambda _{n}, \lambda _{n+1})- k_{\alpha \lambda _{n}, \lambda _{n+1}} \bigr) \\ &\quad {}+ \bigl(K_{\alpha }(\lambda _{n +1}, \lambda _{m})- k_{\alpha \lambda _{n +1}, \lambda _{m}} \bigr) \bigr] \\ & \leq \alpha (\lambda _{n},\lambda _{m}) \bigl[K_{\alpha }(\lambda _{n}, \lambda _{n+1})- k_{\alpha \lambda _{n}, \lambda _{n+1}} \bigr] \\ &\quad {} +\alpha (\lambda _{n},\lambda _{m}) \alpha ( \lambda _{n+1},\lambda _{m}) \bigl[ \bigl(K_{\alpha }( \lambda _{n+1}, \lambda _{n+2})- k_{\alpha \lambda _{n+1}, \lambda _{n+2}} \bigr) \\ &\quad {}+ \bigl(K_{\alpha }(\lambda _{n +2}, \lambda _{m})- k_{\alpha \lambda _{n +2}, \lambda _{m}} \bigr) \bigr] \\ &\quad \vdots \\ & \leq \sum_{i=n}^{m-1}\prod _{j=n}^{i}\alpha (\lambda _{j}, \lambda _{m}) \bigl[K_{ \alpha }(\lambda _{i}, \lambda _{i+1})- k_{\alpha \lambda _{i}, \lambda _{i+1}} \bigr] \\ & \leq \sum_{i=n}^{m-1}\prod _{j=n}^{i}\alpha (\lambda _{j}, \lambda _{m}) \bigl[K_{ \alpha }(\lambda _{i}, \lambda _{i+1}) \bigr] \\ & \leq \sum_{i=n}^{m-1}\prod _{j=n}^{i}\alpha (\lambda _{j}, \lambda _{m})p^{i} \prod_{s=1}^{i} \alpha (\lambda _{s-1},\lambda _{s})K_{\alpha }( \lambda _{0},\lambda _{1}). \end{aligned}

Now, let

$$\varGamma _{i} : = \prod_{j=n}^{i} \alpha (\lambda _{j},\lambda _{m})p^{i} \prod_{s=1}^{i}\alpha (\lambda _{s-1},\lambda _{s})K_{\alpha }( \lambda _{0},\lambda _{1}),$$

then

$$\varGamma _{i + 1} : = \prod_{j=n}^{i+1} \alpha (\lambda _{j},\lambda _{m})p^{i} \prod_{s=1}^{i +1 }\alpha (\lambda _{s-1},\lambda _{s})K_{\alpha }( \lambda _{0},\lambda _{1}).$$

Thus,

$$\frac{\varGamma _{i + 1}}{\varGamma _{i }}= \alpha (\lambda _{i+1},\lambda _{m}) \alpha (\lambda _{i+1},\lambda _{i})p .$$

Therefore

$$\sup_{m \geq 1} \lim_{i} \frac{\varGamma _{i + 1}}{\varGamma _{i }}= p \sup_{m \geq 1} \lim_{i} \alpha (\lambda _{m},\lambda _{i}) \alpha ( \lambda _{i},\lambda _{i +1}) < 1,$$

which leads us to conclude that $$(\lambda _{n})$$ is $$K_{\alpha }$$-Cauchy sequence. Since $$(X,K_{\alpha })$$ is a complete extended $$M_{b}$$-metric space, we deduce that $$(\lambda _{n})$$ is convergent in X to some $$u \in X$$. Note that $$k_{\alpha u, fu} \leq K_{\alpha }(u, fu)$$ and

$$K_{\alpha }(u, fu)- k_{\alpha u, fu} \leq \alpha (u,fu) \bigl[ \bigl(K_{\alpha }(u, \lambda _{n})- k_{\alpha u, \lambda _{n}} \bigr)+ \bigl(K_{\alpha }(\lambda _{n },fu) - k_{\alpha \lambda _{n}, fu} \bigr) \bigr].$$

Since f is continuous and taking the limit in the above inequality we deduce that

$$K_{\alpha }(u, fu)= k_{\alpha u, fu}.$$

Now, without loss of generality we can suppose that $$M_{\alpha u, fu}= K_{\alpha } (u, u)$$.

\begin{aligned} M_{\alpha \lambda _{n}, f\lambda _{n}}= K_{\alpha }(\lambda _{n}, \lambda _{n}) & \leq p \alpha (\lambda _{n-1}, \lambda _{n-1})K_{ \alpha }(\lambda _{n-1}, \lambda _{n-1}) \\ & \leq p^{2} \alpha (\lambda _{n-1}, \lambda _{n-1}) \alpha (\lambda _{n-2}, \lambda _{n-2}) K_{\alpha }(\lambda _{n-2}, \lambda _{n-2}) \\ & \quad \vdots \\ & \leq p^{n}\prod_{i=0}^{n-1} \alpha (\lambda _{i},\lambda _{i})K_{ \alpha }( \lambda _{0},\lambda _{0}). \end{aligned}

Taking the limit on both sides as $$n \rightarrow +\infty$$ we have

$$M_{\alpha u, fu}=0.$$

Finally, since $$K_{\alpha }(u, fu)= k_{\alpha u, fu} \leq M_{\alpha u, fu}=0$$ and since $$K_{\alpha }(fu, fu)= k_{\alpha u, fu}$$, it is easy to conclude that $$fu=u$$. That is, f has a fixed point. Now,assume that f has two fixed points say $$s,r \in X$$, that is, $$fs=s$$ and $$fr=r$$. Thus,

\begin{aligned} K_{\alpha }(s,r) &= K_{\alpha }(fs,fr) \\ & \le p \alpha (s,r) K_{\alpha }(s,r) \\ & < p.\frac{1}{p}K_{\alpha }(s,r) \\ &=K_{\alpha }(s,r), \end{aligned}

which implies that $$K_{\alpha }(s,r)=0$$, therefore $$K_{\alpha }(s,r)=k_{\alpha _{r,s}}=0$$. Now, we may assume that $$M_{\alpha _{r,s}}=K_{\alpha }(s,s)$$, hence $$K_{\alpha }(s,s)=K_{\alpha }(fs,fs)\le p \alpha (s,s) K_{\alpha }(s,s)< K_{ \alpha }(s,s)$$. Hence, $$K_{\alpha }(s,s)=0$$, which leads us to conclude that

$$K_{\alpha }(s,r)=k_{\alpha _{r,s}}=M_{\alpha _{r,s}}=0$$

and that $$r=s$$ as required. □

### Example 2.2

Let $$X=[0,1]$$ and let f: XX defined by

$$f(\lambda )= \frac{\lambda }{2 + 2\lambda }.$$

Then f has a unique fixed point.

### Proof

For all $$\lambda ,\epsilon \in X$$, let $$K_{\alpha }(\lambda ,\epsilon )= \frac{(\lambda +\epsilon )^{2}}{2}$$ and $$\alpha (\lambda ,\epsilon )= 1+\lambda +\epsilon$$. An easy argument shows that $$(X, K_{\alpha } )$$ is a $$K_{\alpha }$$-complete extended $$M_{b}$$-metric space. Also we have

\begin{aligned} K_{\alpha }(f\lambda ,f\epsilon )&= \frac{(f\lambda + f\epsilon )^{2}}{2 } = \frac{(\frac{\lambda }{2 + 2\lambda } + \frac{\epsilon }{2 + 2\epsilon })^{2}}{2 } \\ &\le \frac{1}{4} \frac{(\lambda +\epsilon )^{2}}{2} \\ &\le \frac{1}{4}(1 + \lambda + \epsilon ) \frac{(\lambda +\epsilon )^{2}}{2} \\ &= \frac{1}{4} \alpha (\lambda ,\epsilon ) K_{\alpha }(\lambda , \epsilon ). \end{aligned}

Hence,

$$K_{\alpha }(f\lambda ,f\epsilon )\le p \alpha (\lambda ,\epsilon ) K_{ \alpha }(\lambda ,\epsilon ),\quad \text{where } p= \frac{1}{4}.$$

Now, by induction it is not difficult to deduce that

$$\lambda _{n}= f^{n}(\lambda )= \frac{\lambda }{2^{n} + (\sum_{k=1}^{n}2^{k})\lambda }$$

for all $$n \in \mathbb{N}$$. Thus,

$$\lim_{n\rightarrow +\infty } \alpha (\lambda ,\lambda _{n})=\lim _{n \rightarrow +\infty } \alpha (\lambda _{n},\lambda ) = 1+\lambda .$$

On the other hand,

\begin{aligned} \sup_{m\geq 1}\lim_{n \rightarrow +\infty }\alpha (\lambda _{n}, \lambda _{m})\alpha (\lambda _{n+1}, \lambda _{n}) = &\sup_{m\geq 1} \biggl(1+ \frac{\lambda }{2^{m} + (\sum_{k=1}^{m}2^{k})\lambda } \biggr) \\ =& 1 + \frac{\lambda }{2 + 2\lambda } \\ \le &2< 4=\frac{1}{p}. \end{aligned}

It is not difficult to check that f: $$(X, K_{\alpha })$$$$(X, K_{\alpha })$$ is continuous. Finally, note that f satisfies all the hypotheses of Theorem 2.1. Therefore, f has a unique fixed point in X. □

### Theorem 2.3

Let$$(X,K_{\alpha })$$be a complete extended$$M_{b}$$-metric space, and letfbe a continuous self-mapping onX. Assume that there exist$$a,b \in [0, + \infty )$$with

$$\lim_{n} \frac{a\alpha (\lambda _{n},\lambda _{n -1})}{1-b\alpha (\lambda _{n},\lambda _{n +1})} < 1$$

and

$$\alpha (\lambda _{n},\lambda _{n +1}) < \frac{1}{a+b},\quad \textit{where } \lambda _{n}= f^{n} \lambda _{0}.$$

If$$K_{\alpha }(f\lambda ,fy)\le a \alpha (\lambda ,f\lambda K_{\alpha }( \lambda ,f\lambda + b \alpha (\epsilon ,fy) K_{\alpha }(\epsilon ,fy)$$, thenfhas a unique fixed point inX.

### Proof

Let $$\lambda _{0}\in X$$ and define the sequence $$\{\lambda _{n}\}$$ as follows:

$$\lambda _{1}=f\lambda _{0},\qquad \lambda _{2}=f \lambda _{1}=f^{2}\lambda _{0}, \qquad \ldots ,\qquad \lambda _{n}=f\lambda _{n-1}=f^{n}\lambda _{0},\qquad \ldots.$$

We prove first that

$$K_{\alpha }(\lambda _{n},\lambda _{n+1}) \leq a^{n}\prod_{i=1}^{n} \biggl[ \frac{\alpha (\lambda _{i},\lambda _{i -1})}{1-b\alpha (\lambda _{i},\lambda _{i +1})} \biggr]K_{ \alpha }(\lambda _{0},\lambda _{1}).$$

To this end, let $$n \in \mathbb{N}^{\star }$$, then

\begin{aligned} K_{\alpha }(\lambda _{n},\lambda _{n+1})&= K_{\alpha }(f\lambda _{n-1},f \lambda _{n}) \\ &\le a \alpha (\lambda _{n-1},f\lambda _{n - 1}) K_{\alpha }(\lambda _{n-1},f \lambda _{n-1})+ b \alpha (\lambda _{n},f\lambda _{n}) K_{\alpha }( \lambda _{n},f\lambda _{n}) \\ &= a \alpha (\lambda _{n-1},\lambda _{n}) K_{\alpha }(\lambda _{n-1}, \lambda _{n})+ b \alpha (\lambda _{n},\lambda _{n+1}) K_{\alpha }( \lambda _{n},\lambda _{n+ 1}). \end{aligned}

Hence,

\begin{aligned} K_{\alpha }(\lambda _{n},\lambda _{n+1})& \leq a \frac{\alpha (\lambda _{n},\lambda _{n -1})}{1-b\alpha (\lambda _{n},\lambda _{n +1})}K_{ \alpha }(\lambda _{n},\lambda _{n-1}) \\ & = a \frac{\alpha (\lambda _{n},\lambda _{n -1})}{1-b\alpha (\lambda _{n},\lambda _{n+1})}K_{ \alpha }(f\lambda _{n - 1},f \lambda _{n-2}) \\ &\leq a^{2} \frac{\alpha (\lambda _{n},\lambda _{n -1}\alpha (\lambda _{n-1},\lambda _{n -2})}{(1-b\alpha (\lambda _{n},\lambda _{n +1}))(1-b\alpha (\lambda _{n-1},\lambda _{n}))}K_{ \alpha }(\lambda _{n-1},\lambda _{n-2}) \\ &\le \cdots \\ & \le a^{n}\prod_{i=1}^{n} \biggl[ \frac{\alpha (\lambda _{i},\lambda _{i -1})}{1-b\alpha (\lambda _{i},\lambda _{i +1})} \biggr]K_{ \alpha }(\lambda _{0}, \lambda _{1}). \end{aligned}

Since $$\lim_{n} \frac{a\alpha (\lambda _{n},\lambda _{n -1})}{1-b\alpha (\lambda _{n},\lambda _{n +1})} < 1$$, it follows from the ratio test that

$$\sum_{n=1}^{\infty }a^{n}\prod _{i=1}^{n} \frac{\alpha (\lambda _{i},\lambda _{i -1})}{1-b\alpha (\lambda _{i},\lambda _{i +1})}$$

converges, which implies that $$K_{\alpha }(\lambda _{n},\lambda _{n+1}) \text{converges to 0}$$.

Next, let $$n, m \in \mathbb{N}^{\star }$$, then

\begin{aligned} K_{\alpha }(\lambda _{n},\lambda _{m})&= K_{\alpha }(f\lambda _{n-1},f \lambda _{m-1}) \\ & \leq a \alpha (\lambda _{n-1},\lambda _{n}) K_{\alpha }(\lambda _{n-1}, \lambda _{n})+ b \alpha (\lambda _{m-1},f\lambda _{m}) K_{\alpha }( \lambda _{m-1},\lambda _{m}). \end{aligned}

By the above inequality, we deduce that $$K_{\alpha }(\lambda _{n},\lambda _{m}) \text{converges to 0}$$. Since

$$k_{\alpha \lambda _{m},\lambda _{n}}:= \min \bigl(K_{\alpha }(\lambda _{n}, \lambda _{n}), K_{\alpha }(\lambda _{m},\lambda _{m}) \bigr) \leq K_{ \alpha }(\lambda _{n}, \lambda _{m}),$$

we conclude that

$$\lim_{n,m \rightarrow + \infty } \bigl(K_{\alpha }(\lambda _{n},\lambda _{m})- k_{\alpha \lambda _{m},\lambda _{n}} \bigr) = 0.$$

Now, without loss of generality we may assume that

$$M_{\alpha \lambda _{m},\lambda _{n}}:= \max \bigl(K_{\alpha }(\lambda _{n}, \lambda _{n}), K_{\alpha }(\lambda _{m},\lambda _{m}) \bigr)= K_{\alpha }( \lambda _{n},\lambda _{n}).$$

Hence, we obtain

\begin{aligned} M_{\alpha \lambda _{m},\lambda _{n}}-k_{\alpha \lambda _{m},\lambda _{n}} & \leq M_{\alpha \lambda _{m},\lambda _{n}} \\ & = K_{\alpha }(\lambda _{n},\lambda _{n}) \\ &\leq (a+b)\alpha (\lambda _{n-1},\lambda _{n})K_{\alpha }( \lambda _{n-1}, \lambda _{n}) . \end{aligned}

Taking the limit of the above inequality as $$n \rightarrow +\infty$$ we deduce that

$$\lim_{n,m \rightarrow + \infty } (M_{\alpha \lambda _{m},\lambda _{n}}-k_{ \alpha \lambda _{m},\lambda _{n}}) = 0.$$

Thus, the sequence $$\{\lambda _{n}\}$$ is a $$K_{\alpha }$$-Cauchy sequence. Since $$(X, K_{\alpha })$$ is a $$K_{\alpha }$$-complete extended b-metric space, we conclude that $$\{\lambda _{n}\}$$ converges to some $$\omega \in X$$, and so $$\{f\lambda _{n} = \lambda _{n+1}\}$$ converges to $$\omega \in X$$. On the other hand, by the hypotheses of the theorem ($$f: (X, K_{\alpha })\rightarrow (X, K_{\alpha })$$ is continuous) it is not difficult to conclude that $$\{f\lambda _{n}\}$$ converges to $$f\omega \in X$$. From Lemma 3.3 in , we have

$$K_{\alpha }(\omega ,f\omega )- k_{\alpha \omega ,f\omega } = 0.$$

Then

\begin{aligned} K_{\alpha }(\omega ,f\omega )&= K_{\alpha }( f\omega ,f\omega ) \\ & \leq a \alpha (\omega ,f\omega ) K_{\alpha }(\omega ,f\omega )+ b \alpha (\omega ,f\omega ) K_{\alpha }(\omega ,f\omega ) \\ &=(a+b)\alpha (\omega ,f\omega ) K_{\alpha }(\omega ,f\omega ) \\ & < K_{\alpha }(\omega ,f\omega ) . \end{aligned}

Hence

$$K_{\alpha }(\omega ,f\omega )= K_{\alpha }( f\omega ,f\omega )=0.$$

Similarly to the above we have

$$K_{\alpha } \bigl(f\omega ,f^{2}\omega \bigr)= K_{\alpha } \bigl( f^{2}\omega ,f^{2} \omega \bigr)=0.$$
(2.2)

Since $$(X, K_{\alpha })$$ is an extended b-metric space, it follows that

$$ff\omega =f\omega .$$

We deduce that $$\omega '= f\omega$$ is a fixed point of f. Finally, to show uniqueness assume that there exists another fixed point of f, say u. By the contractive property of f we have

\begin{aligned} K_{\alpha } \bigl(u,\omega ' \bigr)&=K_{\alpha } \bigl(fu,f\omega ' \bigr) \\ & \leq a \alpha (u,fu) K_{\alpha }(u,fu)+ b \alpha \bigl(\omega ',f\omega ' \bigr) K_{\alpha } \bigl(\omega ',f\omega ' \bigr). \end{aligned}

From (2.2) we get

$$K_{\alpha } \bigl(u,\omega ' \bigr) \leq a \alpha (u,fu) K_{\alpha }(u,fu).$$

Hence,

\begin{aligned} K_{\alpha } \bigl(u,\omega ' \bigr) & \leq a \alpha (u,fu) K_{\alpha }(u,fu) \\ & =a \alpha (u,fu) K_{\alpha }(fu,fu) \\ & \leq a \alpha (u,fu) \bigl[(a+b)\alpha (u,fu) \bigr] K_{\alpha }(u,fu) \\ &\le \cdots \\ & \le a \alpha (u,fu) \bigl[(a+b)\alpha (u,fu) \bigr]^{n} K_{\alpha }(u,fu). \end{aligned}

Since $$(a+b)\alpha (u,fu) = (a+b)\alpha (fu,f^{2}u)< 1$$, it follows that $$[(a+b)\alpha (u,fu)]^{n}$$ converges to 0. So $$K_{\alpha }(u,\omega ')= K_{\alpha }(u,u)=0$$. By (2.2) we have

$$K_{\alpha } \bigl(u,\omega ' \bigr)= K_{\alpha }(u,u)= K_{\alpha } \bigl(\omega ', \omega ' \bigr)= 0.$$

Thus, f has a unique fixed point as required. □

## Application

Consider the set $$X= C([0,1], \mathbb{R})$$ and the following Fredholm type integral equation:

$$x'(t)= \int _{0}^{1}\mathbb{G} \bigl(t,s, x'(t) \bigr)\,ds, \quad \text{for } t,s \in [0,1] ,$$
(3.1)

where $$\mathbb{G}(t,s,x'(t))$$ is a continuous function from $$[0,1]^{2}$$ into $$\mathbb{R}$$. Now, define

\begin{aligned} K_{\alpha }:{} &X \times X \longrightarrow \mathbb{R} \\ & (x,y) \mapsto \sup_{t \in [0,1]} \biggl(\frac{ \vert x'(t) \vert + \vert y(t) \vert }{2} \biggr). \end{aligned}

Note that $$(X, K_{\alpha })$$ is a $$K_{\alpha }$$-complete extended $$M_{b}$$-metric space, where

$$\alpha (x,y)=1 + \sup_{t \in [0,1]} \bigl( \bigl\vert x'(t) \bigr\vert \bigl\vert y(t) \bigr\vert \bigr).$$

### Theorem 3.1

Assume that for all$$x,y \in X$$:

1. (1)

$$|\mathbb{G}(t,s,x'(t))| + |\mathbb{G}(t,s,y(t))| \leq p (1 + \sup_{t \in [0,1]}\{|x'(t)||y(t)|\}) (|x'(t)| +|y(t)|)$$, for some$$p \in [0, \frac{1}{(1 + \sup_{t,s}|\mathbb{G}(t,s, x'(t))||\mathbb{G}(t,s, y(t))|)^{2}})$$.

2. (2)

$$\mathbb{G}(t,s, \int _{0}^{1}\mathbb{G}(t,s, x'(t))\,ds ) < \mathbb{G}(t,s, x'(t) )$$for allt, s.

Then the above integral equation has a unique solution.

### Proof

Let $$f: X \longrightarrow X$$ be defined by $$fx'(t) = \int _{0}^{1}\mathbb{G}(t,s, x'(t))\,ds$$, then

$$K_{\alpha }(fx,fy)= \sup_{t \in [0,1]}\biggl(\frac{|fx'(t)|+|fy(t)|}{2}\biggr).$$

Now we have

\begin{aligned} \frac{ \vert fx'(t) \vert + \vert fy(t) \vert }{2} & = \frac{ \vert \int _{0}^{1}\mathbb{G}(t,s, x'(t))\,ds \vert + \vert \int _{0}^{1}\mathbb{G}(t,s, y(t))\,ds \vert }{2} \\ & \leq \frac{\int _{0}^{1} \vert \mathbb{G}(t,s, x'(t)) \vert \,ds+\int _{0}^{1} \vert \mathbb{G}(t,s, y(t)) \vert \,ds}{2} \\ & = \frac{\int _{0}^{1}( \vert \mathbb{G}(t,s, x'(t)) \vert + \vert \mathbb{G}(t,s, y(t)) \vert )\,ds}{2} \\ & \leq \frac{\int _{0}^{1}p(1 + \sup_{t \in [0,1]}\{ \vert x'(t) \vert \vert y(t) \vert \})( \vert x'(t) \vert + \vert y(t) \vert )\,ds}{2} \\ & \leq p \alpha (x,y)K_{\alpha }(x,y). \end{aligned}

Consequently, $$K_{\alpha }(fx,fy) \leq \alpha (x,y)K_{\alpha }(x,y)$$. On the other hand, let $$n \in \mathbb{N}^{\star }$$ and $$x \in X$$, then

\begin{aligned} \bigl(f^{n}x \bigr) (t) &= f \bigl(f^{n-1}x'(t) \bigr) = \int _{0}^{1}\mathbb{G} \bigl(t,s, f^{n-1}x'(t) \bigr)\,ds \\ & = \int _{0}^{1}\mathbb{G} \bigl(t,s, f \bigl(f^{n-2}x \bigr) (t) \bigr)\,ds \\ & = \int _{0}^{1}\mathbb{G} \biggl(t,s, \int _{0}^{1}\mathbb{G} \bigl(t,s, \bigl(f^{n-2}x'(t) \bigr) \bigr) \biggr)\,ds \\ & < \int _{0}^{1}\mathbb{G} \bigl(t,s, \bigl(f^{n-2}x'(t) \bigr) \bigr)\,ds= \bigl(f^{n-1}x'(t) \bigr). \end{aligned}

Thus, for all $$t \in [0,1]$$ we find that $$(f^{n}x'(t))_{n}$$ is strictly decreasing and a sequence bounded below, and so it converges to some l. Since $$(f_{n})_{n}$$ is a monotone sequence, it follows from the Dini theorem that $$\sup_{t}|f^{n}x'(t)|$$ converges to some $$l' \leq \sup_{t,s}|\mathbb{G}(t,s, x'(t))|$$. Observe that $$\alpha (f^{n}x, f^{m}(x))= 1+ \sup_{t}|f^{n}x'(t)| |f^{m}x'(t)|$$ converges to $$1+l^{\prime 2} \leq 1+ (\sup_{t,s}|\mathbb{G}(t,s, x'(t))|)^{2}$$. So

$$\sup_{m} \lim_{n} \alpha \bigl(f^{n}x, f^{m}(x) \bigr)\alpha \bigl(f^{n}x, f^{n+1}x'(t) \bigr) \leq \Bigl(1+ \Bigl(\sup _{t,s} \bigl\vert \mathbb{G} \bigl(t,s, x'(t) \bigr) \bigr\vert \Bigr)^{2} \Bigr)^{2} < \frac{1}{p}.$$

Now, note that all the hypotheses of Theorem 2.1, are satisfied and thus Eq. (3.1) has a unique solution. □

## Conclusion

In closing, note that in this manuscript we proved fixed point results for mappings that satisfy more general contractions, which generalizes many results obtained for mapping satisfying Banach contraction and by taking $$\alpha (\lambda ,\epsilon )= 1$$ for all $$\lambda , \epsilon \in X$$ in Theorem 2.1.

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### Acknowledgements

The first and thrid authors would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.

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