Theory and Modern Applications

# Fredholm-type integral equation in controlled metric-like spaces

## Abstract

In this article we make an improvement in the Banach contraction using a controlled function in controlled metric like spaces, which generalizes many results in the literature. Moreover, we present an application on Fredholm type integral equation.

## Introduction

One of the most interesting applications of fixed point theory is solving integral and differential equations; see, for example, . The Banach contraction principle was generalized many times to extend its application. As an example of these generalizations, b-spaces (see ) are an extension of the regular metric spaces; see . Lately, Kamran  introduced what the so-called extended b-metric spaces by adding a control function $$\theta (\mathfrak{p},\mathfrak{q})$$ in the triangle inequality. For more on b-metric spaces and its extensions, we refer the reader to . First, we start by reminding the reader the definition of extended b-metric spaces.

### Definition 1.1

()

Consider the set $$X\neq \emptyset$$ and $$\theta : X\times X\rightarrow [1,\infty )$$. Let $$d_{e}:X\times X\rightarrow \mathbb{[}0,\infty )$$ be such that for all $$\mathfrak{p},\mathfrak{q},z \in X$$,

1. (1)

$$d_{e}(\mathfrak{p},\mathfrak{q})=0$$ if and only if $$\mathfrak{p}=\mathfrak{q}$$;

2. (2)

$$d_{e}(\mathfrak{p},\mathfrak{q}) = d_{e}(\mathfrak{q},\mathfrak{p})$$;

3. (3)

$$d_{e}(\mathfrak{p},\mathfrak{q}) \leq \theta (\mathfrak{p}, \mathfrak{q}) [d_{e}(\mathfrak{p},z) + d_{e}(z,\mathfrak{q})]$$.

Then $$(X,d_{e})$$ is called an extended b-metric space.

Mlaiki et al.  gave an extension to this type of metric spaces as follows.

### Definition 1.2

()

Consider the set $$X\neq \emptyset$$ and $$\varrho : X\times X\rightarrow [1,\infty )$$. Suppose that a function $$d_{c}: X\times X\rightarrow [0,\infty )$$ satisfies the following:

1. (1)

$$d_{c}(\mathfrak{p},\mathfrak{q})=0$$ if and only if $$\mathfrak{p}=\mathfrak{q}$$;

2. (2)

$$d_{c}(\mathfrak{p},\mathfrak{q})=d_{c}(\mathfrak{q},\mathfrak{p})$$;

3. (3)

$$d_{c}(\mathfrak{p},\mathfrak{q})\leq \varrho (\mathfrak{p},z) d_{c}( \mathfrak{p},z)+\varrho (z,\mathfrak{q}) d_{c}(z,\mathfrak{q})$$ for all $$\mathfrak{p},\mathfrak{q},z\in X$$.

Then $$(X,d_{c})$$ is called a controlled metric-type space.

In 2021, a new generalization of the b-metric spaces introduced in , the so-called controlled metric-like spaces.

### Definition 1.3

()

Consider the set $$X\neq \emptyset$$ and $$\varrho : X\times X\rightarrow [1,\infty )$$. Suppose that a function $$d_{c}: X\times X\rightarrow [0,\infty )$$ satisfies the following:

1. (CML1)

$$d_{c}(s,r)=0$$ $$s=r$$;

2. (CML2)

$$d_{c}(s,r)=d_{c}(r,s)$$;

3. (CML3)

$$d_{c}(s,r)\leq \varrho (s,z) d_{c}(s,z)+\varrho (z,r) d_{c}(z,r)$$ for all $$s,r,z\in X$$.

Then $$(X,d_{c})$$ is called a controlled metric-like space.

### Example 1.4

()

Let $$X=\{0,1,2\}$$. Define the function $$d_{c}$$ by

$$d_{c}(0,0)=d_{c}(1,1)=0,\qquad d_{c}(2,2)= \frac{1}{10}$$

and

$$d_{c}(0,1)=d_{c}(1,0)=1, \qquad d_{c}(0,2)=d_{c}(2,0)= \frac{1}{2},\qquad d_{c}(1,2)=d_{c}(2,1)= \frac{2}{5}.$$

Let $$\varrho : X\times X\rightarrow [1,\infty )$$ a symmetric function defined by

$$\varrho (0,0)=\varrho (1,1)=\varrho (2,2)=\varrho (0,2)=1,\qquad \varrho (1,2)= \frac{5}{4}, \qquad \varrho (0,1)=\frac{11}{10}.$$

Here $$d_{c}$$ is a controlled metric-like on X.

We have $$d_{c}(2,2)=\frac{1}{10}\neq 0$$, which implies that $$(X,d_{c})$$ is not a controlled metric-type space.

### Definition 1.5

()

Let $$(X,d_{c})$$ be a controlled metric-like space, and let $$\{s_{n}\}_{n\ge 0}$$ be a sequence in X.

1. (1)

$$\{s_{n}\}$$ converges to s in X if and only if

$$\lim_{n\rightarrow \infty }d_{c}(s_{n},s)=d_{c}(s,s).$$

Then we write $$\lim_{n \rightarrow \infty }{s_{n}=s}$$.

2. (2)

$$\{s_{n}\}$$ is a Cauchy sequence if and only if $$\lim_{n,m\rightarrow \infty }d_{c}(s_{n},s_{m})$$ exists and is finite.

3. (3)

We say that $$(X,d_{c})$$ is complete if for every Cauchy sequence $$\{s_{n}\}$$, there is $$s\in \chi$$ such that

$$\lim_{n\rightarrow \infty }d_{c}(s_{n},s)=d_{c}(s,s)= \lim_{n,m \rightarrow \infty }d_{c}(s_{n},s_{m}).$$

### Definition 1.6

()

Let $$(X,d_{c})$$ be a controlled metric-like space. Let $$s\in X$$ and $$\tau >0$$.

1. (i)

The open ball $$B(s,\tau )$$ is

$$B(s,\tau )=\bigl\{ y\in X, \bigl\vert d_{c}(s,r)-d_{c}(s,s) \bigr\vert < \tau \bigr\} .$$

We denote controlled metric-like spaces by CMLS.

Note that if $$\mathfrak{T}$$ is continuous at $$\mathfrak{p}$$ in the CMLS $$(X,d_{c})$$, then $$\mathfrak{p}_{n}\rightarrow \mathfrak{p}$$ implies that $$\mathfrak{T}\mathfrak{p}_{n}\rightarrow \mathfrak{T}\mathfrak{p}$$ as $$n\rightarrow \infty$$.

Now let $$(X,d_{c})$$ be a controlled metric-like space, and let $$\mathfrak{T}:X\rightarrow X$$. The following control functions were introduced by Sintunavarat et al.  (in this paper, we will exclude zero from their range):

$$\mathrm{A}= \bigl\{ \vartheta :X\rightarrow (0,1), \vartheta ( \mathfrak{T} \mathfrak{p})\leq \vartheta (\mathfrak{p}) \mbox{ for all } \mathfrak{p}\in X \bigr\} .$$

and

$$\mathrm{B}= \bigl\{ \vartheta :X\rightarrow (0,1/2), \vartheta ( \mathfrak{T} \mathfrak{p})\leq \vartheta (\mathfrak{p}) \mbox{ for all } \mathfrak{p}\in X \bigr\} .$$

## Main results

Our first main result corresponds to a nonlinear Banach-type result on CMLS, which is also an extension of the results in .

### Theorem 2.1

Let $$(X,d_{c})$$ be a complete CMLS. Consider the mapping $$\mathfrak{T}\colon X\rightarrow X$$ such that

$$d_{c}(\mathfrak{T}\mathfrak{p},\mathfrak{T} \mathfrak{q})\leq \vartheta (\mathfrak{p}) d_{c}(\mathfrak{p}, \mathfrak{q})$$
(2.1)

for all $$\mathfrak{p},\mathfrak{q}\in X$$, where $$\vartheta \in \mathrm{A}$$. For $$\mathfrak{p}_{0}\in X$$, take $$\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}$$. Suppose that

$$\sup_{m\geq 1}\lim_{i\rightarrow \infty } \frac{\varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{i+2})}{\varrho (\mathfrak{p}_{i},\mathfrak{p}_{i+1})} \varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{m})< \frac{1}{\vartheta (\mathfrak{p}_{0})}.$$
(2.2)

Also, assume that for every $$\mathfrak{p}\in X$$, we have

$$\lim_{n\rightarrow \infty }\varrho (\mathfrak{p}_{n}, \mathfrak{p}) \quad \textit{and}\quad \lim_{n\rightarrow \infty }\varrho ( \mathfrak{p}, \mathfrak{p}_{n})\quad \textit{exist and are finite}.$$
(2.3)

Then $$\mathfrak{T}$$ has a unique fixed point.

### Proof

Consider the sequence $$\{\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}\}$$. By (2.1) we get

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \le \vartheta (\mathfrak{p}_{n-1})d_{c}( \mathfrak{p}_{n-1},\mathfrak{p}_{n}) \quad \mbox{for all } n\geq 1.$$

Since $$\vartheta \in \mathrm{A}$$, we have

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \le \vartheta (\mathfrak{p}_{0})d_{c}( \mathfrak{p}_{n-1},\mathfrak{p}_{n}) \quad \mbox{for all } n\geq 1.$$

By induction,

$$d_{c}(\mathfrak{p}_{n}, \mathfrak{p}_{n+1})\le \bigl[\vartheta ( \mathfrak{p}_{0}) \bigr]^{n} d_{c}(\mathfrak{p}_{0}, \mathfrak{p}_{1})\quad \mbox{for all } n\geq 0.$$
(2.4)

Choose $$k=:\vartheta (\mathfrak{p}_{0})\in (0,1)$$. For all natural numbers $$n< m$$, as in , we have

\begin{aligned} d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{m})& \le \varrho (\mathfrak{p}_{n}, \mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+ \varrho ( \mathfrak{p}_{n+1},\mathfrak{p}_{m})d_{c}( \mathfrak{p}_{n+1}, \mathfrak{p}_{m}) \\ &\leq \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1})k^{n}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1})+\sum _{i=n+1}^{m-1} \Biggl(\prod _{j=n+1}^{i} \varrho (\mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho ( \mathfrak{p}_{i}, \mathfrak{p}_{i+1}) k^{i}d_{c}( \mathfrak{p}_{0}, \mathfrak{p}_{1}) \\ &\leq \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1})k^{n}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1})+\sum _{i=n+1}^{m-1} \Biggl(\prod _{j=0}^{i} \varrho (\mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho ( \mathfrak{p}_{i}, \mathfrak{p}_{i+1}) k^{i}d_{c}( \mathfrak{p}_{0}, \mathfrak{p}_{1}). \end{aligned}

Let

$$S_{p}= \sum_{i=0}^{p} \Biggl(\prod_{j=0}^{i}\varrho ( \mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho ( \mathfrak{p}_{i},\mathfrak{p}_{i+1}) k^{i}.$$

Hence we have

$$d_{c}(\mathfrak{p}_{n}, \mathfrak{p}_{m})\leq d_{c}(\mathfrak{p}_{0}, \mathfrak{p}_{1}) \bigl[ k^{n} \varrho ( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+(S_{m-1}-S_{n}) \bigr].$$
(2.5)

Now by condition (2.2) and the ratio test, we deduce that $$\lim_{n\rightarrow \infty }S_{n}$$ exists, and therefore $$\{S_{n}\}$$ is a Cauchy sequence. Taking the limit in (2.5), we obtain

$$\lim_{n,m\rightarrow \infty }d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{m})=0.$$
(2.6)

Hence $$\{\mathfrak{p}_{n}\}$$ is a Cauchy sequence. Since $$(X,d_{c})$$ is complete, we deduce that $$\{\mathfrak{p}_{n}\}$$ converges to some $$u\in X$$. We claim that u is a fixed point of $$\mathfrak{T}$$. To prove this claim, we start by applying the triangle inequality of the CMLS as follows:

$$d_{c}(u,\mathfrak{p}_{n+1})\leq \varrho (u, \mathfrak{p}_{n})d_{c}(u, \mathfrak{p}_{n})+ \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1}).$$

By (2.2), (2.3), and (2.6) we conclude that

$$\lim_{n\rightarrow \infty }d_{c}(u, \mathfrak{p}_{n+1})=0.$$
(2.7)

Thus

\begin{aligned} d_{c}(u,\mathfrak{T}u)&\leq \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)d_{c}( \mathfrak{p}_{n+1}, \mathfrak{T}u) \\ &\le \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \vartheta (\mathfrak{p}_{n})\varrho ( \mathfrak{p}_{n+1},\mathfrak{T}u)d_{c}( \mathfrak{p}_{n}, u) \\ &\le \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \vartheta (\mathfrak{p}_{0})\varrho ( \mathfrak{p}_{n+1},\mathfrak{T}u)d_{c}( \mathfrak{p}_{n}, u). \end{aligned}

Note that, as $$n\rightarrow \infty$$ in (2.3) and (2.7), we have $$d_{c}(u,\mathfrak{T}u)=0$$, that is, $$\mathfrak{T}u=u$$. Now we may assume that $$\mathfrak{T}$$ has fixed points u and v. Hence

$$d_{c}(u,v)=d_{c}(\mathfrak{T}u,\mathfrak{T}v)\le \vartheta (u) d_{c}(u,v)< d_{c}(u,v) ,$$

which leads us to a contradiction. Thereby $$d_{c}(u,v)=0$$, which implies $$u=v$$, as desired. □

Next, we present the following example.

### Example 2.2

Let $$X=[0,1]$$. Consider the CMLS $$(X,d_{c})$$ defined by

$$d_{c}(\mathfrak{p},\mathfrak{q})= \vert \mathfrak{p}-\mathfrak{q} \vert ^{2},$$

where $$\varrho (\mathfrak{p},\mathfrak{q})=\mathfrak{p}\mathfrak{q}+1$$ for $$\mathfrak{p},\mathfrak{q}\in X$$. Take $$\mathfrak{T}\mathfrak{p}=\frac{\mathfrak{p}^{2}}{4}$$. Choose $$\vartheta :X\rightarrow [0,1)$$ as $$\vartheta (\mathfrak{p})=\frac{\mathfrak{p}+1}{4}$$. Then $$\vartheta \in \mathrm{A}$$. Take $$\mathfrak{p}_{0}=0$$, so (2.2) is satisfied. Let $$\mathfrak{p},\mathfrak{q}\in X$$. Then

\begin{aligned} d_{c}(\mathfrak{T}\mathfrak{p}, \mathfrak{T}\mathfrak{q})&=\frac{(\mathfrak{p}^{2}-\mathfrak{q}^{2})^{2}}{16}= \frac{1}{16} \vert \mathfrak{p}-\mathfrak{q} \vert ^{2}(\mathfrak{p}+ \mathfrak{q})^{2} \\ &\leq \frac{1}{4} \vert \mathfrak{p}-\mathfrak{q} \vert ^{2} \\ &\leq \frac{\mathfrak{p}+1}{4} \vert \mathfrak{p}-\mathfrak{q} \vert ^{2} \\ &=\vartheta (\mathfrak{p})d_{c}(\mathfrak{p},\mathfrak{q}). \end{aligned}

Note that all the hypotheses of Theorem 2.1 are satisfied. Thus there exists an element $$u\in X$$ such that $$\mathfrak{T}u=u$$, which is $$u=0$$.

In the following theorem, we propose a fixed point result using the nonlinear Kannan-type contraction via the auxiliary function $$\vartheta \in \mathrm{B}$$.

### Theorem 2.3

Let $$(X,d_{c})$$ be a complete CMLS by the function $$\varrho :X\times X\rightarrow [1,\infty )$$. Let $$\mathfrak{T}\colon X\rightarrow X$$ where

$$d_{c}(\mathfrak{T}\mathfrak{p},\mathfrak{T} \mathfrak{q})\leq \vartheta (\mathfrak{p}) \bigl[d_{c}(\mathfrak{p},\mathfrak{T} \mathfrak{p})+d_{c}( \mathfrak{q},\mathfrak{T}\mathfrak{q})\bigr]$$
(2.8)

for all $$\mathfrak{p},\mathfrak{q}\in X$$, where $$\vartheta \in \mathrm{B}$$. For $$\mathfrak{p}_{0}\in X$$, take $$\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}$$. Suppose that

$$\sup_{m\geq 1}\lim_{i\rightarrow \infty } \frac{\varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{i+2})}{\varrho (\mathfrak{p}_{i},\mathfrak{p}_{i+1})} \varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{m})< \frac{1-\vartheta (\mathfrak{p}_{0})}{\vartheta (\mathfrak{p}_{0})}.$$
(2.9)

Also, assume that for every $$\mathfrak{p}\in X$$, we have

$$\lim_{n\rightarrow \infty }\varrho (\mathfrak{p}, \mathfrak{p}_{n}) \quad \textit{exists, is finite and}\quad \lim _{n \rightarrow \infty }\varrho (\mathfrak{p}_{n},\mathfrak{p})< \frac{1}{\vartheta (\mathfrak{p}_{0})}.$$
(2.10)

Then there exists a unique fixed point of $$\mathfrak{T}$$.

### Proof

Let $$\{\mathfrak{p}_{n}=\mathfrak{T}\mathfrak{p}_{n-1}\}$$ be a sequence in X satisfying hypotheses (2.9) and (2.10). From (2.8) we obtain

\begin{aligned} d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) = & d_{c}(\mathfrak{T} \mathfrak{p}_{n-1},\mathfrak{T} \mathfrak{p}_{n}) \\ \leq & \vartheta (\mathfrak{p}_{n-1})\bigl[d_{c}( \mathfrak{p}_{n-1}, \mathfrak{T}\mathfrak{p}_{n-1})+d_{c}( \mathfrak{p}_{n},\mathfrak{T} \mathfrak{p}_{n})\bigr] \\ \leq & \vartheta (\mathfrak{p}_{0})\bigl[d_{c}( \mathfrak{p}_{n-1}, \mathfrak{p}_{n})+d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})\bigr]. \end{aligned}

Consider $$a=\vartheta (\mathfrak{p}_{0})$$. Then $$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \leq \frac{a}{1-a}d_{c}( \mathfrak{p}_{n-1},\mathfrak{p}_{n})$$. By induction we get

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \leq \biggl(\frac{a}{1-a}\biggr)^{n}d_{c}( \mathfrak{p}_{1},\mathfrak{p}_{0}),\quad \forall n\geq 0 .$$
(2.11)

For all natural numbers n, m, we have

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{m}) \leq \varrho (\mathfrak{p}_{n}, \mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+ \varrho ( \mathfrak{p}_{n+1},\mathfrak{p}_{m})d_{c}( \mathfrak{p}_{n+1}, \mathfrak{p}_{m}).$$

Following the steps of the proof of Theorem 2.1, we deduce

\begin{aligned} d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{m}) & \leq \varrho (\mathfrak{p}_{n}, \mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+\sum _{i=n+1}^{m-2} \Biggl(\prod _{j=n+1}^{i}\varrho (\mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho (\mathfrak{p}_{i}, \mathfrak{p}_{i+1}) d_{c}( \mathfrak{p}_{i}, \mathfrak{p}_{i+1}) \\ &\quad {} + \prod_{k=n+1}^{m-1} \varrho ( \mathfrak{p}_{k},\mathfrak{p}_{m})d_{c}( \mathfrak{p}_{m-1},\mathfrak{p}_{m}) \\ & \leq \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \biggl(\frac{a}{1-a}\biggr)^{n}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1}) \\ &\quad {}+\sum _{i=n+1}^{m-2} \Biggl(\prod _{j=n+1}^{i} \varrho (\mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho ( \mathfrak{p}_{i}, \mathfrak{p}_{i+1}) \biggl(\frac{a}{1-a}\biggr)^{i}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1}) \\ &\quad {}+\prod_{i=n+1}^{m-1} \varrho ( \mathfrak{p}_{i},\mathfrak{p}_{m}) \biggl( \frac{a}{1-a}\biggr)^{m-1}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1}). \end{aligned}

Since $$0\leq a<\frac{1}{2}$$, we have $$\frac{a}{1-a}\in (0,1)$$. Therefore $$\{\mathfrak{p}_{n}\}$$ is a Cauchy sequence, and since $$(X,d_{c})$$ is a complete CMLS, $$\{\mathfrak{p}_{n}\}$$ converges to some $$u\in X$$. Suppose that $$\mathfrak{T}u\neq u$$. Then

\begin{aligned} 0 < & d_{c}(u,\mathfrak{T}u)\leq \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)d_{c}( \mathfrak{p}_{n+1}, \mathfrak{T}u) \\ \leq & \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)\vartheta (\mathfrak{p}_{n})\bigl[d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+d_{c}(u, \mathfrak{T}u)\bigr] \\ \leq & \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)\vartheta (\mathfrak{p}_{0})\bigl[d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+d_{c}(u, \mathfrak{T}u)\bigr]. \end{aligned}
(2.12)

As $$n\rightarrow \infty$$ in (2.12) and by (2.10), we conclude that $$0< d_{c}(u,\mathfrak{T}u)< d_{c}(u,\mathfrak{T}u)$$, which leads us to a contradiction. Thereby $$\mathfrak{T}u=u$$. Now we may assume that $$\mathfrak{T}$$ has fixed points u and v. Thus

\begin{aligned} d_{c}(u,v) =&d_{c}(\mathfrak{T}u,\mathfrak{T}v) \leq \vartheta (u)\bigl[d_{c}(u, \mathfrak{T}u)+d_{c}(v, \mathfrak{T}v)\bigr] \\ = & \vartheta (u)\bigl[d_{c}(u,u)+d_{c}(v,v) \bigr]=0. \end{aligned}

Hence $$u=v$$. Therefore the fixed point is unique, as required. □

### Example 2.4

Consider $$X=\{0,1,2\}$$. Take the controlled metric-like $$d_{c}$$ defined as

$$d_{c}(0,1)=\frac{1}{2},\qquad d_{c}(0,2)= \frac{11}{20}, \qquad d_{c}(1,2)= \frac{3}{20}.$$

Let $$\varrho : X\times X\rightarrow [1,\infty )$$ be defined by

\begin{aligned}& \varrho (0,0)=\varrho (1,1)=\varrho (2,2)=\varrho (1,2)=\varrho (2,1)=1, \\& \varrho (0,2)=\varrho (2, 0)=2,\qquad \varrho (0,1)=\varrho (1,0)= \frac{3}{2}. \end{aligned}

Let $$\mathfrak{T}: X\rightarrow X$$ be given by

$$\mathfrak{T}0=2\quad \mbox{and}\quad \mathfrak{T}1=\mathfrak{T}2=1.$$

Let $$\vartheta :X\rightarrow [0,\frac{1}{2})$$ be given by $$\vartheta (0)=\frac{99}{200}$$, $$\vartheta (1)=\frac{3}{10}$$, and $$\vartheta (2)=\frac{49}{100}$$. Then $$\vartheta \in \mathrm{B}$$. Take $$\mathfrak{p}_{0}=0$$, so that (2.9) is satisfied.

Also, it is easy to see that (2.8) holds. By Theorem 2.3 there exists a unique u such that $$\mathfrak{T}u=u$$, that is, $$u=1$$.

Now,we again give a response to an open question in , which is a study of a nonlinear Chatterjea-type contraction via an auxiliary function $$\vartheta \in \mathrm{B}$$.

### Theorem 2.5

Let $$(X,d_{c})$$ be a complete CMLS by the function

\begin{aligned}& \varrho :X\times X\rightarrow [1,\infty ). \\& \textit{Let } \mathfrak{T}\colon X\rightarrow X \textit{ be such that } d_{c}(\mathfrak{T}\mathfrak{p},\mathfrak{T} \mathfrak{q})\leq \vartheta (\mathfrak{p}) \bigl[d_{c}(\mathfrak{p}, \mathfrak{T} \mathfrak{q})+d_{c}(\mathfrak{q},\mathfrak{T}\mathfrak{p})\bigr] \end{aligned}
(2.13)

for all $$\mathfrak{p},\mathfrak{q}\in X$$, where $$\vartheta \in \mathrm{B}$$. For $$\mathfrak{p}_{0}\in X$$, take $$\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}$$. Suppose that

\begin{aligned}& \sup_{i\geq 1}\varrho (\mathfrak{p}_{i-1}, \mathfrak{p}_{i})=\beta\quad (\textit{exists and is finite}), \end{aligned}
(2.14)
\begin{aligned}& 0< \vartheta (\mathfrak{p}_{0})< \frac{1}{2\beta }, \end{aligned}
(2.15)

and

$$\sup_{m\geq 1}\lim_{i\rightarrow \infty } \frac{\varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{i+2})}{\varrho (\mathfrak{p}_{i},\mathfrak{p}_{i+1})} \varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{m})< \frac{\beta \vartheta (\mathfrak{p}_{0})}{1-\beta \vartheta (\mathfrak{p}_{0})}.$$
(2.16)

Also, assume that $$d_{c}$$ is continuous with respect to the first variable and that for every $$\mathfrak{p}\in X$$,

$$\lim_{n\rightarrow \infty }\varrho (\mathfrak{p}, \mathfrak{p}_{n})\quad \textit{exists, is finite, and} \quad \lim _{n \rightarrow \infty }\varrho (\mathfrak{p}_{n},\mathfrak{p})< \frac{1}{\vartheta (\mathfrak{p}_{0})}.$$
(2.17)

Then $$\mathfrak{T}$$ possesses a unique fixed point in X.

### Proof

Consider the sequence $$\{\mathfrak{p}_{n}=\mathfrak{T}\mathfrak{p}_{n-1}\}$$ in X satisfying hypotheses (2.14), (2.15), (2.16), and (2.17). From (2.13) and (2.14) we obtain

\begin{aligned} d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) = & d_{c}(\mathfrak{T} \mathfrak{p}_{n-1},\mathfrak{T} \mathfrak{p}_{n}) \\ \leq & \vartheta (\mathfrak{p}_{n-1})\bigl[d_{c}( \mathfrak{p}_{n-1}, \mathfrak{T}\mathfrak{p}_{n})+d_{c}( \mathfrak{p}_{n},\mathfrak{T} \mathfrak{p}_{n-1})\bigr] \\ = & \vartheta (\mathfrak{p}_{n-1})d_{c}( \mathfrak{p}_{n-1}, \mathfrak{p}_{n+1}) \\ \leq & \vartheta (\mathfrak{p}_{0})\bigl[\varrho ( \mathfrak{p}_{n-1}, \mathfrak{p}_{n})d_{c}( \mathfrak{p}_{n-1},\mathfrak{p}_{n})+\varrho ( \mathfrak{p}_{n},\mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n}, \mathfrak{p}_{n+1})\bigr] \\ \leq & \beta \vartheta (\mathfrak{p}_{0})\bigl[d_{c}( \mathfrak{p}_{n-1}, \mathfrak{p}_{n})+d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})\bigr]. \end{aligned}

Let $$b= \frac{\beta \vartheta (\mathfrak{p}_{0})}{1-\beta \vartheta (\mathfrak{p}_{0})}$$. By (2.15) we have $$b\in (0,1)$$. Then $$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \leq b d_{c}(\mathfrak{p}_{n-1}, \mathfrak{p}_{n})$$. By induction we get

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{n+1}) \leq b^{n} d_{c}( \mathfrak{p}_{0}, \mathfrak{p}_{1}),\quad \forall n\geq 0 .$$
(2.18)

For all natural numbers n, m, we have

$$d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{m}) \leq \varrho (\mathfrak{p}_{n}, \mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+ \varrho ( \mathfrak{p}_{n+1},\mathfrak{p}_{m})d_{c}( \mathfrak{p}_{n+1}, \mathfrak{p}_{m}).$$

Following the steps of the proof of Theorem 2.1, we get

\begin{aligned} d_{c}(\mathfrak{p}_{n},\mathfrak{p}_{m}) & \leq \varrho (\mathfrak{p}_{n}, \mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1})+\sum _{i=n+1}^{m-2} \Biggl(\prod _{j=n+1}^{i}\varrho (\mathfrak{p}_{j}, \mathfrak{p}_{m}) \Biggr)\varrho (\mathfrak{p}_{i}, \mathfrak{p}_{i+1}) d_{c}( \mathfrak{p}_{i}, \mathfrak{p}_{i+1}) \\ &\quad {} + \prod_{k=n+1}^{m-1} \varrho ( \mathfrak{p}_{k},\mathfrak{p}_{m})d_{c}( \mathfrak{p}_{m-1},\mathfrak{p}_{m}) \\ & \leq \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1}) (b^{n}d_{c}( \mathfrak{p}_{0}, \mathfrak{p}_{1})+\sum_{i=n+1}^{m-2} \Biggl(\prod_{j=n+1}^{i} \varrho ( \mathfrak{p}_{j},\mathfrak{p}_{m}) \Biggr)\varrho ( \mathfrak{p}_{i},\mathfrak{p}_{i+1}) b^{i}d_{c}( \mathfrak{p}_{0}, \mathfrak{p}_{1}) \\ &\quad {}+\prod_{i=n+1}^{m-1} \varrho ( \mathfrak{p}_{i},\mathfrak{p}_{m})b^{m-1}d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1}). \end{aligned}

This implies that $$\{\mathfrak{p}_{n}\}$$ is a Cauchy sequence CMLS $$(X,d_{c})$$. Since the space is complete, the sequence $$\{\mathfrak{p}_{n}\}$$ converges to some $$u\in X$$. Now suppose that $$\mathfrak{T}u\neq u$$. Then

\begin{aligned} 0 < & d_{c}(u,\mathfrak{T}u)\leq \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)d_{c}( \mathfrak{p}_{n+1}, \mathfrak{T}u) \\ \leq & \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)\vartheta (\mathfrak{p}_{n})\bigl[d_{c}( \mathfrak{p}_{n},\mathfrak{T}u)+d_{c}(u, \mathfrak{p}_{n+1})\bigr] \\ \leq & \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ \varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)\vartheta (\mathfrak{p}_{0})\bigl[d_{c}( \mathfrak{p}_{n},\mathfrak{T}u)+d_{c}(u, \mathfrak{p}_{n+1})\bigr]. \end{aligned}
(2.19)

As $$n\rightarrow \infty$$ in (2.19), by (2.17) and using the continuity of $$d_{c}$$ with respect to its first variable, we deduce that $$0< d_{c}(u,\mathfrak{T}u)< d_{c}(u,\mathfrak{T}u)$$, which leads us to a contradiction. Thus $$\mathfrak{T}u=u$$.

Now let us assume that $$\mathfrak{T}$$ has fixed points u and v. Then

\begin{aligned} d_{c}(u,v) =&d_{c}(\mathfrak{T}u,\mathfrak{T}v) \leq \vartheta (u)\bigl[d_{c}(u, \mathfrak{T}v)+d_{c}(v, \mathfrak{T}u)\bigr] \\ = & \vartheta (u)\bigl[d_{c}(u,u)+d_{c}(v,v) \bigr]=0. \end{aligned}

Therefore $$u=v$$, and thus the fixed point of $$\mathfrak{T}$$ is unique. □

Now we introduce cyclical orbital contractions in the class of CMLS.

### Definition 2.6

Let U and V be two nonempty subsets of a CMLS $$(X,d_{c})$$. Let $$\mathfrak{T}:U\cup V\rightarrow U\cup V$$ be a cyclic mapping (i.e., $$\mathfrak{T}(U)\subseteq V$$ and $$\mathfrak{T}V\subseteq U$$) such that for some $$\mathfrak{p}\in U$$, there exists $$k_{\mathfrak{p}}\in (0, 1)$$ such that

$$d_{c}\bigl(\mathfrak{T}^{2n} \mathfrak{p},\mathfrak{T}\mathfrak{q}\bigr)\leq k_{\mathfrak{p}} d_{c}\bigl(\mathfrak{T}^{2n-1}\mathfrak{p},\mathfrak{q} \bigr),$$
(2.20)

where $$n=1,2,\ldots$$ and $$\mathfrak{q}\in U$$. Then $$\mathfrak{T}$$ is called a controlled cyclic orbital contraction mapping.

Finally, we prove the following result.

### Theorem 2.7

Let U and V be two nonempty closed subsets of a complete CMLS $$(X,d_{c})$$. Let $$\mathfrak{T}\colon X\rightarrow X$$ be a controlled cyclic orbital contraction mapping. For $$\mathfrak{p}_{0}\in U$$, take $$\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}$$. Suppose that

$$\sup_{m\geq 1}\lim_{i\rightarrow \infty } \frac{\varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{i+2})}{\varrho (\mathfrak{p}_{i},\mathfrak{p}_{i+1})} \varrho (\mathfrak{p}_{i+1},\mathfrak{p}_{m})< \frac{1}{k_{\mathfrak{p}_{0}}}.$$
(2.21)

Also, assume that for every $$\mathfrak{p}\in X$$,

$$\lim_{n\rightarrow \infty }\varrho (\mathfrak{p}_{n}, \mathfrak{p})\quad \textit{and}\quad \lim_{n\rightarrow \infty }\varrho ( \mathfrak{p}, \mathfrak{p}_{n}) \quad \textit{exist and are finite}.$$
(2.22)

Then $$U\cap V$$ is nonempty, and $$\mathfrak{T}$$ has a unique fixed point.

### Proof

Suppose there exists $$\mathfrak{p}$$ (say $$\mathfrak{p}_{0}$$) in U satisfying (2.20). Define the iterative sequence $$\{\mathfrak{p}_{n}=\mathfrak{T}^{n}\mathfrak{p}_{0}\}$$. Since $$\mathfrak{p}_{0}\in U$$ and $$\mathfrak{T}$$ is cyclic, we have

$$\mathfrak{p}_{2n}\in U \quad \mbox{and}\quad \mathfrak{p}_{2n+1}\in V\quad \mbox{for all } n\geq 0.$$
(2.23)

By (2.20) we get

$$d_{c}\bigl(\mathfrak{T}^{2}\mathfrak{p},\mathfrak{T} \mathfrak{p}\bigr)\leq k_{\mathfrak{p}} d_{c}(\mathfrak{T} \mathfrak{p},\mathfrak{p}).$$

Again,

$$d_{c}\bigl(\mathfrak{T}^{3}\mathfrak{p}, \mathfrak{T}^{2}\mathfrak{p}\bigr)=d_{c}\bigl( \mathfrak{T}^{2}\mathfrak{p},\mathfrak{T}\bigl(\mathfrak{T}^{2} \mathfrak{p}\bigr)\bigr)\leq k_{\mathfrak{p}} d_{c}\bigl( \mathfrak{T}\mathfrak{p}, \mathfrak{T}^{2}\mathfrak{p}\bigr)\leq (k_{\mathfrak{p}})^{2} d_{c}( \mathfrak{T} \mathfrak{p},\mathfrak{p}).$$

By induction we obtain that

$$d_{c}(\mathfrak{p}_{n}, \mathfrak{p}_{n+1})\le [k_{\mathfrak{p}}]^{n} d_{c}( \mathfrak{p}_{0},\mathfrak{p}_{1}) \quad \mbox{for all } n\geq 0.$$
(2.24)

Similarly to the proof of Theorem 2.1, we can easily deduce that

$$\lim_{n,m\rightarrow \infty }d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{m})=0,$$
(2.25)

that is, $$\{\mathfrak{p}_{n}\}$$ is a Cauchy sequence in the complete CMLS $$(X,d_{c})$$, so $$\{\mathfrak{p}_{n}\}$$ converges to some $$u\in X$$. Since $$\{\mathfrak{T}^{2n}\mathfrak{p}\}$$ is in U and U is closed, the limit u is in $$S_{1}$$. Similarly, $$\{\mathfrak{T}^{2n-1}\mathfrak{p}\}$$ is in the closed subset V, so $$u\in V$$, that is, $$u\in U\cap V$$, and hence $$U\cap V$$ is not empty. Let us prove that u is a fixed point of $$\mathfrak{T}$$. We have

$$d_{c}(u,\mathfrak{p}_{n+1})\leq \varrho (u, \mathfrak{p}_{n})d_{c}(u, \mathfrak{p}_{n})+ \varrho (\mathfrak{p}_{n},\mathfrak{p}_{n+1})d_{c}( \mathfrak{p}_{n},\mathfrak{p}_{n+1}).$$

Using (2.21), (2.22), and (2.25), we get that

$$\lim_{n\rightarrow \infty }d_{c}(u, \mathfrak{p}_{n+1})=0.$$
(2.26)

By (2.20) we deduce

\begin{aligned} d_{c}(u,\mathfrak{T}u)&\leq \varrho \bigl(u,\mathfrak{T}^{2n} \mathfrak{p}\bigr)d_{c}\bigl(u, \mathfrak{T}^{2n} \mathfrak{p}\bigr)+ \varrho \bigl(\mathfrak{T}^{2n} \mathfrak{p}, \mathfrak{T}u\bigr)d_{c}\bigl(\mathfrak{T}^{2n} \mathfrak{p}, \mathfrak{T}u\bigr) \\ &\le \varrho \bigl(u,\mathfrak{T}^{2n}\mathfrak{p} \bigr)d_{c}\bigl(u,\mathfrak{T}^{2n} \mathfrak{p}\bigr)+ k_{\mathfrak{p}}\varrho \bigl(\mathfrak{T}^{2n}\mathfrak{p}, \mathfrak{T}u\bigr)d_{c}\bigl(\mathfrak{T}^{2n-1} \mathfrak{p}, u\bigr) \\ &= \varrho (u,\mathfrak{p}_{n+1})d_{c}(u, \mathfrak{p}_{n+1})+ k_{\mathfrak{p}}\varrho (\mathfrak{p}_{n+1}, \mathfrak{T}u)d_{c}( \mathfrak{p}_{2n-1}, u). \end{aligned}

Taking the limit as $$n\rightarrow \infty$$ and using (2.22) and (2.26), we deduce that $$d_{c}(u,\mathfrak{T}u)=0$$, that is, $$\mathfrak{T}u=u$$. Finally, assume that $$\mathfrak{T}$$ has two fixed points, say u and v (they are in U). Then

$$d_{c}(u,v)=d_{c}(\mathfrak{T}u,\mathfrak{T}v)=d_{c} \bigl(\mathfrak{T}^{2n}u, \mathfrak{T}v\bigr)\le k_{u} d_{c}\bigl(\mathfrak{T}^{2n-1}u,v\bigr)=k_{u} d_{c}(u,v),$$

which holds unless $$d_{c}(u,v)=0$$, so $$u=v$$. Hence $$\mathfrak{T}$$ has a unique fixed point. □

The following example illustrates Theorem 2.7.

### Example 2.8

Let $$X=U\cup V$$, where $$U=[\frac{1}{4},\frac{1}{2}] \text{and} V=[\frac{1}{2},1]$$. Consider the controlled metric-like $$d_{c}$$ defined as

$$d_{c}(\mathfrak{p},\mathfrak{q})= \vert \mathfrak{p}-\mathfrak{q} \vert ^{2},$$

where $$\varrho (\mathfrak{p},\mathfrak{q})=\mathfrak{p}\mathfrak{q}+1$$ for $$\mathfrak{p},\mathfrak{q}\in X$$. Take $$\mathfrak{T}\mathfrak{p}=\frac{1}{2}$$ if $$\mathfrak{p} \in U$$ and $$\mathfrak{T}\mathfrak{p}=\frac{\mathfrak{p}}{2}$$ if $$\mathfrak{p}\in V\setminus \{\frac{1}{2}\}$$. Now let $$k_{\mathfrak{p}}:X\rightarrow [0,1]$$ be defined as $$k_{\mathfrak{p}}=\frac{\mathfrak{p}+1}{2}$$. Note that for all $$\mathfrak{p}\in U$$, we have

$$\mathfrak{T}\mathfrak{p}=\frac{1}{2}, \qquad \mathfrak{T}^{2} \mathfrak{p}= \frac{1}{2},\qquad \dots ,\qquad \mathfrak{T}^{2n-1} \mathfrak{p}=\frac{1}{2},\qquad \mathfrak{T}^{2n}\mathfrak{p}= \frac{1}{2},\qquad \dots .$$

For all $$\mathfrak{q}\in U$$, using the fact that

$$d_{c}\bigl(\mathfrak{T}^{2n}\mathfrak{p},\mathfrak{T} \mathfrak{q}\bigr)=d_{c}\biggl( \frac{1}{2}, \frac{1}{2}\biggr)=0,$$

we deduce that

$$d_{c}\bigl(\mathfrak{T}^{2n}\mathfrak{p},\mathfrak{T} \mathfrak{q}\bigr)\leq k_{\mathfrak{p}} d_{c}\bigl( \mathfrak{T}^{2n-1}\mathfrak{p},\mathfrak{q}\bigr).$$

It is not difficult to see that $$\mathfrak{T}$$ satisfies all the hypotheses of Theorem 2.7. Therefore $$\mathfrak{T}$$ has a unique fixed point $$u=\frac{1}{2}$$.

## Fredholm-type integral equation

Consider the set $$X = C([0,1], (-\infty ,\infty ))$$ and the following Fredholm-type integral equation:

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

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

\begin{aligned} d_{c}:{}& X \times X \longrightarrow \mathbb{R}^{+} \\ & (\mathfrak{p},\mathfrak{q}) \mapsto \sup_{t \in [0,1]}\biggl( \frac{ \vert \mathfrak{p}'(t) \vert + \vert \mathfrak{q}(t) \vert }{2}\biggr). \end{aligned}

Note that $$(X, d_{c})$$ is a complete CMLS, where

$$\varrho (\mathfrak{p},\mathfrak{q})=2.$$

### Theorem 3.1

Assume that for all $$\mathfrak{p},\mathfrak{q} \in X$$,

1. (1)

$$|\mathbb{S}(t,s,\mathfrak{p}'(t))| + |\mathbb{S}(t,s,\mathfrak{q}(t))| \leq \vartheta (\sup_{t \in [0,1]}(|\mathfrak{p}'(t)|+|\mathfrak{q}(t)|)) (|\mathfrak{p}'(t)| +|\mathfrak{q}(t)|)$$ for some $$\vartheta \in \mathrm{B}$$.

2. (2)

$$\mathbb{S}(t,s, \int _{0}^{1}\mathbb{S}(t,s, \mathfrak{p}'(t))\,ds ) < \mathbb{S}(t,s, \mathfrak{p}'(t) )$$ for all t, s.

Then the integral equation (3.1) has a unique solution.

### Proof

Let $$\mho : X \longrightarrow X$$ be defined by $$\mho \mathfrak{p}'(t) = \int _{0}^{1}\mathbb{S}(t,s, \mathfrak{p}'(t))\,ds$$. Then

$$d_{c}(\mho \mathfrak{p}',\mho \mathfrak{q})= \sup_{t \in [0,1]}\biggl( \frac{|\mho \mathfrak{p}'(t)|+|\mho \mathfrak{q}(t)|}{2}\biggr).$$

Now we have

\begin{aligned} d_{c}\bigl(\mho \mathfrak{p}'(t),\mho \mathfrak{q}(t)\bigr)&= \frac{ \vert \mho \mathfrak{p}'(t) \vert + \vert \mho \mathfrak{q}(t) \vert }{2} \\ & = \frac{ \vert \int _{0}^{1}\mathbb{S}(t,s, \mathfrak{p}'(t))\,ds \vert + \vert \int _{0}^{1}\mathbb{S}(t,s, \mathfrak{q}(t))\,ds \vert }{2} \\ & \le \frac{\int _{0}^{1} \vert \mathbb{S}(t,s, \mathfrak{p}'(t)) \vert \,ds+\int _{0}^{1} \vert \mathbb{S}(t,s, \mathfrak{q}(t)) \vert \,ds}{2} \\ & = \frac{\int _{0}^{1}( \vert \mathbb{S}(t,s, \mathfrak{p}'(t)) \vert + \vert \mathbb{S}(t,s, \mathfrak{q}(t)) \vert )\,ds}{2} \\ & \le \frac{\int _{0}^{1}\vartheta (\sup_{t \in [0,1]}( \vert \mathfrak{p}'(t) \vert + \vert \mathfrak{q}(t) \vert )) ( \vert \mathfrak{p}'(t) \vert + \vert \mathfrak{q}(t) \vert )\,ds}{2} \\ & \le \vartheta \Bigl(\sup_{t \in [0,1]}\bigl( \bigl\vert \mathfrak{p}'(t) \bigr\vert + \bigl\vert \mathfrak{q}(t) \bigr\vert \bigr)\Bigr) d_{c}\bigl(\mathfrak{p}'(t), \mathfrak{q}(t)\bigr). \end{aligned}

Thus $$d_{c}(\mho \mathfrak{p}',\mho \mathfrak{q}) \le \vartheta ( \sup_{t \in [0,1]}(|\mathfrak{p}'(t)|+|\mathfrak{q}(t)|)) d_{c}( \mathfrak{p}',\mathfrak{q})$$. Also, notice that

$$\varrho (\mathfrak{p},\mathfrak{q})< \frac{1}{\vartheta (\sup_{t \in [0,1]}( \vert \mathfrak{p}'(t) \vert + \vert \mathfrak{q}(t) \vert ))}.$$

Therefore all the hypotheses of Theorem 2.1 are satisfied, and hence equation (3.1) has a unique solution. □

## Conclusion

We have proved the existence and uniqueness of a fixed point for a self-mapping in controlled metric-like spaces under different nonlinear contractions with a control function. Also, we present an application of our results to Fredholm-type integral equations. Moreover, we would like to bring the reader’s attention to the following question.

### Question 4.1

Under what conditions we can obtain the same results for a self-mapping in double controlled metric-like spaces ?

Not applicable.

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## Acknowledgements

The first and the second 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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