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.

1 Introduction

One of the most interesting applications of fixed point theory is solving integral and differential equations; see, for example, [1]. The Banach contraction principle was generalized many times to extend its application. As an example of these generalizations, b-spaces (see [2]) are an extension of the regular metric spaces; see [3–15]. Lately, Kamran [16] 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 [17–23]. First, we start by reminding the reader the definition of extended b-metric spaces.

Definition 1.1

([16])

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

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

• (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. [24] gave an extension to this type of metric spaces as follows.

Definition 1.2

([24])

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

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

• (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 [25], the so-called controlled metric-like spaces.

Definition 1.3

([25])

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:

• (CML1) $d_{c}(s,r)=0$ ⇒ $s=r$;

• (CML2) $d_{c}(s,r)=d_{c}(r,s)$;

• (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

([25])

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

([25])

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

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

• (3) 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

([26])

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

• (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. [27] (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

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

2 Main results

Our first main result corresponds to a nonlinear Banach-type result on CMLS, which is also an extension of the results in [28].

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 [24], 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 [24], 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 $u=\frac{1}{2}$.

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

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

4 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 [26]?