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. (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. [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. (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 [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:

  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

([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. (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

([26])

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. [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. (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. □

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]?