Fixed Point Theorems in Complex-Valued Double Controlled Metric-Like Spaces

Abstract

In this chapter, we have discussed the concept of complex-valued double controlled metric-like spaces. These results extend the corresponding results about complex-valued double controlled metric-type spaces. As applications, we have proved some complex-valued fixed point theorems in this space and have proved the existence and uniqueness of the solution of a Fredholm type integral equation. Finally, some examples are also given in favor of our given results.

1. Introduction

Fixed point theory concept is a widely recognized and big subject of studies in engineering and mathematical sciences. This area is called the aggregate of evaluation including geometry, algebra, and topology. In this field, a big contribution has been done by Banach [1] by introducing notion of contraction mapping due to a complete metric space to find fixed point of the specified function. This classical theorem of Banach [1] has been studied and generalized by means of many researchers in diverse methods (see, for instance, [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]), as this principle is also considered as heart of fixed point theory. Classical definition of metric space was generalized by Harandi [21] by introducing the notion of metric-like space. Azam [22] was the first one to introduce the complex-valued metric space, which was generalized by Hosseni and Karizaki [23], who introduced complex-valued metric-like space. Bakhtin [2] and Czerwik [3] generalized metric space by giving the idea of b-metric spaces. Aslam et al. [4] introduced the notion of complex-valued controlled metric-type space. Abdeljawad [7] presented the idea of double controlled metric-type space (DCMLS) that was generalization of Kamran et al. [5] and Mlaiki et al. [6].

Further, let us recall some interesting definitions and results, useful in the introduction of our new concept.

Let $\mathbb{C}$ be the set of complex numbers and w₁, $w_2\in \mathbb{C}.$ Since we cannot compare in usual way two complex numbers let us add to the complex set $\mathbb{C}$ the following partial order $\precsim ,$ known in related literature as lexicographic order.

$w_1\precsim w_2$ if and only if $\mathit{Re}\left(w_1\right)\preccurlyeq \mathit{Re}\left(w_2\right)$ or (Re(w₁) = Re(w₂) and $\mathit{Im}\left(w_1\right)\preccurlyeq \mathit{Im}\left(w_2\right)).$.Taking into account the previous definition, we have that $w_1\precsim w_2$ if one of the next conditions is satisfied:

(P1) Re(w₁) < Re(w₂) and Im(w₁) < Im(w₂);

(P2) Re(w₁) < Re(w₂) and Im(w₁) = Im(w₂);

(P3) Re(w₁) < Re(w₂) and Im(w₁) > Im(w₂);

(P4) Re(w₁) = Re(w₂) and Im(w₁) < Im(w₂).

Let us recall the definition of complex-valued extended b-metric given by N. Ullah et al. in Ref. [9].

Definition [9]

Let $\mathbb{X}$ be a non-empty set and $\vartheta :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right).$ The function $h_{\mathit{eb}}:\mathbb{X}\times \mathbb{X}\to C$ is said to be complex-valued extended b-metric if the following conditions are satisfied:

$$\left({\mathit{CEB}}_1\right)0\precsim h_{\mathit{eb}}\left(p,q\right)\text{and}{h}_{\mathit{eb}}\left(p,q\right)=0\text{if and only if}p=q,$$

$$\left({\mathit{CEB}}_2\right)h_{\mathit{eb}}\left(p,q\right)=h_{\mathit{eb}}\left(q,p\right),$$

$$\left({\mathit{CEB}}_3\right)h_{\mathit{eb}}\left(p,r\right)\precsim \vartheta \left(p,r\right)\left[h_{\mathit{eb}}\left(p,q\right)+h_{\mathit{eb}}\left(q,r\right)\right],$$

for all p, q, $r\in \mathbb{X}.$ A pair $\left(\mathbb{X},h_{\mathit{eb}}\right)$ is called a complex-valued extended b-metric space.

Mlaiki et al. [6] generalized the notion of b-metric spaces as follows.

Definition [6]

Given $\mathrm{\varrho }:\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right),$ where $\mathbb{X}$ is nonempty. Let $h_c:\mathbb{X}\times \mathbb{X}\to \left[0,\infty \right).$ Suppose that

$$\left({\mathit{CMT}}_1\right)h_c\left(p,q\right)=0\text{if and only if}p=q,$$

$$\left({\mathit{CMT}}_2\right)h_c\left(p,q\right)=h_c\left(q,p\right),$$

$$\left({\mathit{CMT}}_3\right)h_c\left(p,q\right)\le \mathrm{\varrho }\left(p,r\right)h_c\left(p,r\right)+\mathrm{\varrho }\left(r,q\right)h_c\left(r,q\right),$$

for all p, q, $r\in \mathbb{X}.$ Then, h_c is called a controlled metric type and $\left(\mathbb{X},h_c\right)$ is called a controlled metric-type space.

Definition [13]

Given non-comparable functions $\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right),$ where $\mathbb{X}$ is nonempty.

If $h_{\mathit{dl}}:{\mathbb{X}}^2\to \left[0,\infty \right)$ satisfies

$$\left({\mathit{\text{DCML}}}_1\right)h_{\mathit{dl}}\left(p,q\right)=0\Rightarrow p=q,$$

$$\left({\mathit{\text{DCML}}}_2\right)h_{\mathit{dl}}\left(p,q\right)=h_{\mathit{dl}}\left(q,p\right),$$

$$\left({\mathit{\text{DCML}}}_3\right)h_{\mathit{dl}}\left(p,q\right)\le \mathrm{\varrho }\left(p,r\right)h_{\mathit{dl}}\left(p,r\right)+\varsigma \left(r,q\right)h_{\mathit{dl}}\left(r,q\right),$$

for all p, q, $r\in \mathbb{X}.$ Then, h_dl is called a double controlled metric like by $\mathrm{\varrho }$ and $\varsigma$ and $\left(\mathbb{X},h_{\mathit{dl}}\right)$ is called a double controlled metric-like space (DCMLS).

Definition [8]

Given non-comparable functions $\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right),$ where $\mathbb{X}$ is nonempty.

If $h_{\mathit{cdt}}:{\mathbb{X}}^2\to \left[0,\infty \right)$ satisfies

$$\left({\mathit{\text{CDCMT}}}_1\right)h_{\mathit{cdt}}\left(p,q\right)=0\iff p=q,$$

$$\left({\mathit{\text{CDCMT}}}_2\right)h_{\mathit{cdt}}\left(p,q\right)=h_{\mathit{cdt}}\left(q,p\right),$$

$$\left({\mathit{\text{CDCMT}}}_3\right)h_{\mathit{cdt}}\left(p,q\right)\precsim \mathrm{\varrho }\left(p,r\right)h_{\mathit{cdt}}\left(p,r\right)+\varsigma \left(r,q\right)h_{\mathit{cdt}}\left(r,q\right),$$

for all p, q, $r\in \mathbb{X}.$ Then, h_cdt is called a complex-valued double controlled metric type by $\mathrm{\varrho }$ and $\varsigma$ and $\left(\mathbb{X},h_{\mathit{cdt}}\right)$ is called a complex-valued double controlled metric-type space (CDCMTS).

Recently, Panda et al. [8] presented idea of complex-valued double controlled metric-type space (CDCMTS). Inspired by Panda et al. [8], in this chapter we will present the concept of complex-valued double controlled metric-like space (CDCMLS). Then, two fixed point theorems in CDCMLS are presented. One of them is the Banach contraction principle and the second one is the related Reich type result. The theorems are illustrated with the help of examples. Moreover, an application to prove the existence and the uniqueness of a Fredholm type integral equation is given.

2. Complex-valued double controlled metric-like spaces

In the section, we will introduce our generalization of complex-valued double controlled metric-like spaces (CDCMLS).

Definition

Given non-comparable functions $\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right),$ where $\mathbb{X}$ is nonempty.

If $h_{\mathit{cdl}}:{\mathbb{X}}^2\to \left[0,\infty \right)$ satisfies

$$\left({\mathit{\text{CDCML}}}_1\right)h_{\mathit{cdl}}\left(p,q\right)=0\Rightarrow p=q,$$

$$\left({\mathit{\text{CDCML}}}_2\right)h_{\mathit{cdl}}\left(p,q\right)=h_{\mathit{cdl}}\left(q,p\right),$$

$$\left({\mathit{\text{CDCML}}}_3\right)h_{\mathit{cdl}}\left(p,r\right)\precsim \mathrm{\varrho }\left(p,q\right)h_{\mathit{cdl}}\left(p,q\right)+\varsigma \left(q,r\right)h_{\mathit{cdl}}\left(q,r\right),$$

for all p, q, $r\in \mathbb{X}.$ Then, h_cdl is called a complex-valued double controlled metric like by $\mathrm{\varrho }$ and $\varsigma$ and $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ is called a complex-valued double controlled metric-like space (CDCMLS).

Remark

A complex-valued double controlled metric-type space is also complex-valued double controlled metric-like space in general. The converse is not true in general. Further this is also more generalized form than complex-valued extended b-metric-type space (see Example 2).

Let $\mathbb{X}=\left\{1,2,3\right\}.$ Consider the complex-valued double controlled metric like h = h_cdl defined by

$$\begin{array}{c}h\left(1,1\right)=h\left(2,2\right)=0\text{and}h\left(3,3\right)=\frac{i}{2},\\ h\left(1,2\right)=h\left(2,1\right)=2+4i,h\left(2,3\right)=h\left(3,2\right)=i,h\left(1,3\right)=h\left(3,1\right)=1-i.\end{array}$$

Take $\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right)$ to be symmetric and defined by

$$\begin{array}{c}\mathrm{\varrho }\left(1,1\right)=\mathrm{\varrho }\left(2,2\right)=\mathrm{\varrho }\left(3,3\right)=1,\mathrm{\varrho }\left(1,2\right)=\mathrm{\varrho }\left(2,1\right)=\frac{6}{5},\mathrm{\varrho }\left(2,3\right)=\mathrm{\varrho }\left(3,2\right)=\frac{8}{5},\\ \mathrm{\varrho }\left(3,1\right)=\mathrm{\varrho }\left(1,3\right)=\frac{151}{100}.\end{array}$$

and $\begin{array}{c}\varsigma \left(1,1\right)=\varsigma \left(2,2\right)=\varsigma \left(3,3\right)=1,\varsigma \left(1,2\right)=\varsigma \left(2,1\right)=\frac{6}{5},\varsigma \left(2,3\right)=\varsigma \left(3,2\right)=\frac{33}{20},\\ \varsigma \left(3,1\right)=\varsigma \left(1,3\right)=\frac{8}{3}.\end{array}$.

The conditions (CDCML₁) and (CDCML₂) hold. We will verify (CDCML₃).

Case 1: When p = q = r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,1\right)\mid =0\preccurlyeq 0=0+0=\mathrm{\varrho }\left(1,1\right)\mid h\left(1,1\right)\mid +\varsigma \left(1,1\right)\mid h\left(1,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 2: When p = 2, q = r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(2,1\right)\mid =\sqrt{20}\preccurlyeq \frac{6}{5}\sqrt{20}=\frac{6}{5}\sqrt{20}+0=\mathrm{\varrho }\left(2,1\right)\mid h\left(2,1\right)\mid +\varsigma \left(1,1\right)\mid h\left(1,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 3: When p = 3, q = r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(3,1\right)\mid =\sqrt{2}\preccurlyeq \frac{151}{100}\sqrt{2}=\frac{151}{100}\sqrt{20}+0=\mathrm{\varrho }\left(3,1\right)\mid h\left(3,1\right)\mid +\varsigma \left(1,1\right)\mid h\left(1,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 4: When p = r = 1, q = 2,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,1\right)\mid =0\preccurlyeq \frac{12}{5}\sqrt{20}=\frac{6}{5}\sqrt{20}+\frac{6}{5}\sqrt{20}=\mathrm{\varrho }\left(1,2\right)\mid h\left(1,2\right)\mid +\varsigma \left(2,1\right)\mid h\left(2,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 5: When p = r = 1, q = 3,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,1\right)\mid =0\preccurlyeq \frac{1253}{300}\sqrt{2}=\frac{151}{100}\sqrt{2}+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(1,3\right)\mid h\left(1,3\right)\mid +\varsigma \left(3,1\right)\mid h\left(3,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 6: When p = q = 1, r = 2,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,2\right)\mid =\sqrt{20}\preccurlyeq \frac{6}{5}\sqrt{20}=0+\frac{6}{5}\sqrt{20}=\mathrm{\varrho }\left(1,1\right)\mid h\left(1,1\right)\mid +\varsigma \left(1,2\right)\mid h\left(1,2\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 7: When p = q = 1, r = 3,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,3\right)\mid =\sqrt{2}\preccurlyeq \frac{8}{3}\sqrt{2}=0+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(1,1\right)\mid h\left(1,1\right)\mid +\varsigma \left(1,3\right)\mid h\left(1,3\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 8: When p = q = 2, r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(2,1\right)\mid =\sqrt{20}\preccurlyeq \frac{6}{5}\sqrt{20}=0+\frac{6}{5}\sqrt{20}=\mathrm{\varrho }\left(2,2\right)\mid h\left(2,2\right)\mid +\varsigma \left(2,1\right)\mid h\left(2,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 9: When p = 2, q = 3, r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(2,1\right)\mid =\sqrt{20}\preccurlyeq \frac{24+40\sqrt{2}}{15}=\frac{8}{5}+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(2,3\right)\mid h\left(2,3\right)\mid +\varsigma \left(3,1\right)\mid h\left(3,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 10: When p = 3, q = 2, r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(3,1\right)\mid =\sqrt{2}\preccurlyeq \frac{8+6\sqrt{20}}{5}=\frac{8}{5}+\frac{6}{5}\sqrt{20}=\mathrm{\varrho }\left(3,2\right)\mid h\left(3,2\right)\mid +\varsigma \left(2,1\right)\mid h\left(2,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 11: When p = q = 3, r = 1,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(3,1\right)\mid =\sqrt{2}\preccurlyeq \frac{3+16\sqrt{2}}{6}=\frac{1}{2}+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(3,3\right)\mid h\left(3,3\right)\mid +\varsigma \left(3,1\right)\mid h\left(3,1\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 12: When p = 1, q = r = 2,

$$\begin{array}{r}\mid h\left(p,r\right)\mid =\mid h\left(1,2\right)\mid =\sqrt{20}\preccurlyeq \frac{6}{5}\sqrt{20}=\frac{6}{5}\sqrt{2}+0=\mathrm{\varrho }\left(1,2\right)\mid h\left(1,2\right)\mid +\varsigma \left(2,2\right)\mid h\left(2,2\right)\mid \\ =\mathrm{\varrho }\left(p,q\right)\mid h\left(p,q\right)\mid +\varsigma \left(q,r\right)\mid h\left(q,r\right)\mid .\end{array}$$

Case 13: When p = 1, q = 3, r = 2,

$$\begin{array}{rcl}\left|h\left(p,r\right)\right|& =& \left|h\left(1,2\right)\right|=\sqrt{20}\preccurlyeq \frac{453\sqrt{2}+800}{300}=\frac{151}{100}\sqrt{2}+\frac{8}{3}=\mathrm{\varrho }\left(1,3\right)\left|h\left(1,3\right)\right|+\varsigma \left(3,2\right)\left|h\left(3,2\right)\right|\\ & =& \mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 14: When p = r = 2, q = 1,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,2\right)\right|=0\preccurlyeq \frac{12}{5}\sqrt{20}=\frac{6}{5}\sqrt{20}+\frac{6}{5}\sqrt{20}=\mathrm{\varrho }\left(2,1\right)\left|h\left(2,1\right)\right|+\varsigma \left(1,2\right)\left|h\left(1,2\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 15: When p = q = r = 2,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,2\right)\right|=0\preccurlyeq 0=0+0=\mathrm{\varrho }\left(2,2\right)\left|h\left(2,2\right)\right|+\varsigma \left(2,2\right)\left|h\left(2,2\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 16: When p = r = 2, q = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,2\right)\right|=0\preccurlyeq \frac{13}{4}=\frac{8}{5}+\frac{33}{20}=\mathrm{\varrho }\left(2,3\right)\left|h\left(2,3\right)\right|+\varsigma \left(3,2\right)\left|h\left(3,2\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 17: When p = 3, q = 1, r = 2,

$$\begin{array}{rcl}\left|h\left(p,r\right)\right|& =& \left|h\left(3,2\right)\right|=1\preccurlyeq \frac{151\sqrt{2}+120\sqrt{20}}{100}=\frac{151}{100}\sqrt{2}+\frac{6}{5}\sqrt{20}\\ & =& \mathrm{\varrho }\left(3,1\right)\left|h\left(3,1\right)\right|+\varsigma \left(1,2\right)\left|h\left(1,2\right)\right|=\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 18: When p = 3, q = r = 2,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(3,2\right)\right|=1\preccurlyeq \frac{8}{5}=\frac{8}{5}+0=\mathrm{\varrho }\left(3,2\right)\left|h\left(3,2\right)\right|+\varsigma \left(2,2\right)\left|h\left(2,2\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 19: When p = q = 3, r = 2,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(3,2\right)\right|=1\preccurlyeq \frac{43}{20}=\frac{1}{2}+\frac{33}{20}=\mathrm{\varrho }\left(3,3\right)\left|h\left(3,3\right)\right|+\varsigma \left(3,2\right)\left|h\left(3,2\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 20: When p = 1, q = 2, r = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(1,3\right)\right|=\sqrt{2}\preccurlyeq \frac{24\sqrt{20}+33}{20}=\frac{6}{5}\sqrt{20}+\frac{33}{20}=\mathrm{\varrho }\left(1,2\right)\left|h\left(1,2\right)\right|+\varsigma \left(2,3\right)\left|h\left(2,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 21: When p = 1, q = r = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(1,3\right)\right|=\sqrt{2}\preccurlyeq \frac{151\sqrt{2}+50}{100}=\frac{151}{100}\sqrt{2}+\frac{1}{2}=\mathrm{\varrho }\left(1,3\right)\left|h\left(1,3\right)\right|+\varsigma \left(3,3\right)\left|h\left(3,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 22: When p = 2, q = 1, r = 3,

$$\begin{array}{rcl}\left|h\left(p,r\right)\right|& =& \left|h\left(2,3\right)\right|=1\preccurlyeq \frac{18\sqrt{20}+40\sqrt{2}}{15}=\frac{6}{5}\sqrt{20}+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(2,1\right)\left|h\left(2,1\right)\right|+\varsigma \left(1,3\right)\left|h\left(1,3\right)\right|\\ & =& \mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 23: When p = q = 2, r = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,3\right)\right|=1\preccurlyeq \frac{33}{20}=0+\frac{33}{20}=\mathrm{\varrho }\left(2,2\right)\left|h\left(2,2\right)\right|+\varsigma \left(2,3\right)\left|h\left(2,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 24: When p = 2, q = r = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,3\right)\right|=1\preccurlyeq \frac{21}{10}=\frac{8}{5}+\frac{1}{2}=\mathrm{\varrho }\left(2,3\right)\left|h\left(2,3\right)\right|+\varsigma \left(3,3\right)\left|h\left(3,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 25: When p = r = 3, q = 1,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(3,3\right)\right|=\frac{1}{2}\preccurlyeq \frac{1253}{300}\sqrt{2}=\frac{151}{100}\sqrt{2}+\frac{8}{3}\sqrt{2}=\mathrm{\varrho }\left(3,1\right)\left|h\left(3,1\right)\right|+\varsigma \left(1,3\right)\left|h\left(1,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 26: When p = r = 3, q = 2,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(3,3\right)\right|=\frac{1}{2}\preccurlyeq \frac{13}{4}=\frac{8}{5}+\frac{33}{20}=\mathrm{\varrho }\left(3,2\right)\left|h\left(3,2\right)\right|+\varsigma \left(2,3\right)\left|h\left(2,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Case 27: When p = q = r = 3,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(3,3\right)\right|=\frac{1}{2}\preccurlyeq 1=\frac{1}{2}+\frac{1}{2}=\mathrm{\varrho }\left(3,3\right)\left|h\left(3,3\right)\right|+\varsigma \left(3,3\right)\left|h\left(3,3\right)\right|\\ =\mathrm{\varrho }\left(p,q\right)\left|h\left(p,q\right)\right|+\varsigma \left(q,r\right)\left|h\left(q,r\right)\right|.\end{array}$$

Thus, h = h_cdl is complex-valued double controlled metric-like space(CDCMLS).

But when p = 2, q = 3, r = 1,

$$\begin{array}{l}\left|h\left(p,r\right)\right|=\left|h\left(2,1\right)\right|=\sqrt{20}⋠\frac{6}{5}\left(1+\sqrt{2}\right)=\frac{6}{5}\left[1+\sqrt{2}\right]=\vartheta \left(2,1\right)\left[\left|h\left(2,3\right)\right|+\left|h\left(3,1\right)\right|\right]\\ =\vartheta \left(p,q\right)\left[\left|h\left(p,r\right)\right|+\left|h\left(q,r\right)\right|\right].\end{array}$$

Thus, h = h_cdl is not a complex-valued extended b-metric type for the function ϑ.

Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a complex-valued double controlled metric-like space (CDCMLS) by one or two functions.

1. The sequence {p_n} is convergent to some p in $\mathbb{X},$ if for each positive ε, there is some integer Z_ε such that $h_{\mathit{cdl}}\left(p_n,p\right)\prec \epsilon$ for each n ≥ Z_ε. It is written as lim_n → ∞ p_n = p.

2. The sequence {p_n} is said Cauchy, if for every ε > 0, $h_{\mathit{cdl}}\left(p_n,p_m\right)\prec \epsilon$ for all m, n ≥ Z_ε, where Z_ε is some integer.

3. $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ is said complete if every Cauchy sequence is convergent.

   Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a complex-valued double controlled metric-like space (CDCMLS) by either one function or two functions—for $p\in \mathbb{X}$ and l > 0.

   1. We define $\mathcal{B}\left(p,l\right)$ as

      $$\mathcal{B}\left(p,l\right)=\left\{y\in \mathbb{X},h_{\mathit{cdl}}\left(p,y\right)\prec l\right\}\cdot$$

   2. The self-map ϒ on $\mathbb{X}$ is said to be continuous at p in $\mathbb{X}$ if for all δ > 0, there exists l > 0 such that

      $$\mathrm{\Upsilon }\left(\mathcal{B}\left(p,l\right)\right)\subseteq \mathcal{B}\left(\mathrm{\Upsilon }p,\delta \right)\cdot$$

Note that if ϒ is continuous at p in $\left(\mathbb{X},h_{\mathit{cdl}}\right),$ then p_n → p implies that ϒp_n → ϒp when n tends to ∞.

One can prove the following lemmas for the specific case of CDCMLS, in a similar way as in [11]. Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS and assume a sequence {d_n} in $\mathbb{X}.$ Then {d_n} is Cauchy sequence $\Leftarrow \Rightarrow \left|h_{\mathit{cdl}}\left(d_m,d_n\right)\right|\to 0$ as m, n → ∞, where m, $n\in N.$

Suppose $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS and {d_n} be sequence in $\mathbb{X}.$ Then {d_n} converges to $d\Leftarrow \Rightarrow \left|h_{\mathit{cdl}}\left(d_n,d\right)\right|\to 0$ as n → ∞.

Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS. Then a sequence {d_n} in $\mathbb{X}$ is Cauchy sequence, such that d_m ≠ d_n, whenever m ≠ n. Then {d_n} converges to at most one point.

For a given complex-valued controlled space $\left(\mathbb{X},h_{\mathit{cdl}}\right),$ the complex-valued double controlled (c.v.dc) metric-like function $h_{\mathit{cdl}}:\mathbb{X}\times \mathbb{X}\to \mathbb{C}$ is continuous, with respect to the partial order ${}^{“}{\precsim }^{”}.$

Consider $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS. Limit of every convergent sequence in $\mathbb{X}$ is unique, if the functional $h_{\mathit{cdl}}:\mathbb{X}\times \mathbb{X}\to \mathbb{X}$ is continuous.

3. Fixed point theorems in CDCMLS

In our first theorem of this section, we prove the Banach contraction type theorem in CDCMLS.

Theorem 1.1 Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS by the functions $\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right).$ Suppose that $\mathrm{\Upsilon }:\mathbb{X}\times \mathbb{X}$ satisfies

$$h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\precsim {\mathit{lh}}_{\mathit{cdl}}\left(p,q\right)$$E1

for all $p,q\in \mathbb{X},$ where $l\in \left(0,1\right).$ For $p_0\in \mathbb{X},$ choose $p_n={\mathrm{\Upsilon }}^n{p}_0.$ Assume that

$$\underset{m\ge 1}{sup}\underset{\iota \to \infty }{lim}\frac{\mathrm{\varrho }\left(p_{\iota +1},p_{\iota +1}\right)}{\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)}\varsigma \left(p_{\iota +1},p_m\right)<\frac{1}{l}\cdot$$E2

In addition, for each $p\in \mathbb{X},$ suppose that

$$\underset{n\to \infty }{lim}\mathrm{\varrho }\left(p,p_n\right)\text{and}\underset{n\to \infty }{lim}\varsigma \left(p_n,p\right)\text{exist and}\mathrm{are}\text{finite}.$$E3

Then, ϒ has a unique fixed point.

Proof: Consider the sequence {p_n = ϒⁿp₀} in $\mathbb{X}$ that satisfies the hypothesis of the theorem. By using (1), we get

$$h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)\precsim l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)\text{for}\mathrm{all}n\ge 0.$$E4

Let n, m be integers such that n < m. We have

$$\begin{array}{l}{h}_{\mathit{cdl}}\left(p_n,p_m\right)\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)h_{\mathit{cdl}}\left(p_{n+1},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)\mathrm{\varrho }\left(p_{n+1},p_{n+2}\right)h_{\mathit{cdl}}\left(p_{n+1},p_{n+2}\right)\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)h_{\mathit{cdl}}\left(p_{n+2},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)\mathrm{\varrho }\left(p_{n+1},p_{n+2}\right)h_{\mathit{cdl}}\left(p_{n+1},p_{n+2}\right)\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)\mathrm{\varrho }\left(p_{n+2},p_{n+3}\right)h_{\mathit{cdl}}\left(p_{n+2},p_{n+3}\right)\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)\varsigma \left(p_{n+3},p_m\right)h_{\mathit{cdl}}\left(p_{n+3},p_m\right)\\ \precsim \cdots \end{array}$$

$$\begin{array}{l}\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\sum _{\iota =n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)h_{\mathit{cdl}}\left(p_{\iota },p_{\iota +1}\right)\\ +\prod _{k=n+1}^{m-1}\varsigma \left(p_k,p_m\right)h_{\mathit{cdl}}\left(p_{m-1},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ +\prod _{\iota =n+1}^{m-1}\varsigma \left(p_{\iota },p_m\right)l^{m-1}{h}_{\mathit{cdl}}\left(p_0,p_1\right)\end{array}$$

$$\begin{array}{l}\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ +\left(\prod _{\iota =n+1}^{m-1}\varsigma \left(p_{\iota },p_m\right)\right)l^{m-1}\mathrm{\varrho }\left(p_{m-1},p_m\right)h_{\mathit{cdl}}\left(p_0,p_1\right)\\ =\mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-1}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-1}\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^i{h}_{\mathit{cdl}}\left(p_0,p_1\right).\end{array}$$

We used (p, q) ≥ 1. Let

$$R_g+\sum _{\iota =0}^g\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }.$$

Hence, we have

$$h_{\mathit{cdl}}\left(p_n,p_m\right)\precsim h_{\mathit{cdl}}\left(p_0,p_1\right)\left[l^n\mathrm{\varrho }\left(p_n,p_{n+1}\right)+\left(R_{m-1},R_n\right)\right].$$E5

The ratio test together with (2) implies that the limit of the real number sequence {R_n} exits, and so {R_n} is Cauchy.

Indeed, the ration test is applied to the term $v_{\iota }=\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +\mathit{1}}\right).$.

Letting n, m tend to ∞ in (5) yields

$$\underset{n,m\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_n,p_m\right)=0,$$

so the sequence {p_n} is Cauchy. Since $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ is a complete double controlled metric-type space, there exists some $\kappa \in \mathbb{X}$ such that

$$\underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_n,\kappa \right)=0.$$

We claim that ϒκ = κ. By (DCML3), we have

$$h\left(\kappa ,p_{n+1}\right)\precsim \mathrm{\varrho }\left(\kappa ,p_n\right)h_{\mathit{cdl}}\left(\kappa ,p_n\right)+\varsigma \left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right).$$E6

Using (3) and (6), we get that

$$\underset{n\to \infty }{lim}h\left(\kappa ,p_{n+1}\right)=0.$$E7

By (1), we have

$$\begin{array}{r}h\left(\kappa ,\mathrm{\Upsilon }\kappa \right)\precsim \mathrm{\varrho }\left(\kappa ,p_{n+1}\right)h_{\mathit{cdl}}\left(\kappa ,p_{n+1}\right)+\varsigma \left(p_{n+1},\mathrm{\Upsilon }\kappa \right)h_{\mathit{cdl}}\left(p_{n+1},\mathrm{\Upsilon }\kappa \right)\\ \precsim \mathrm{\varrho }\left(p,p_{n+1}\right)h_{\mathit{cdl}}\left(\kappa ,p_{n+1}\right)+\mathit{l\varsigma }\left(p_{n+1},\mathrm{\Upsilon }\kappa \right)h_{\mathit{cdl}}\left(p_n,\kappa \right).\end{array}$$

Using (3) and (7), we get at the limit h_cdl (κ, ϒκ) = 0, that is, ϒκ = κ. Let ϖ in $\mathbb{X}$ be such that ϒη = ϖ and κ ≠ ϖ. We have

$$0\prec h_{\mathit{cdl}}\left(\kappa ,\varpi \right)=h_{\mathit{cdl}}\left(\mathrm{\Upsilon }\kappa ,\mathrm{\Upsilon }\kappa \right)\preccurlyeq {\mathit{lh}}_{\mathit{cdl}}\left(\kappa ,\varpi \right).$$

It is a contradiction, so κ = ϖ. Hence, κ is the unique fixed point of ϒ.

The assumption (3) in Theorem 1.1 above can be replaced by the assumptions that the mapping ϒ and the complex-valued double controlled metric h are continuous. Indeed, when p_n → κ, then ϒp_n → ϒκ and hence we have

$$\underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(\mathrm{\Upsilon }{p}_n,\mathrm{\Upsilon }\kappa \right)=0=\underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(\mathrm{\Upsilon }{p}_{n+1},\mathrm{\Upsilon }\kappa \right)=p\left(\kappa ,\mathrm{\Upsilon }\kappa \right),$$

and hence ϒκ = κ.

Theorem 1.1 is illustrated by the following examples.

We endow $\mathbb{X}=\left\{1,2,3\right\}$ by the following CDCMLS h = h_cdl

$$\begin{array}{c}h\left(1,1\right)=h\left(2,2\right)=0\text{and}h\left(3,3\right)=\frac{i}{2},\\ h\left(1,2\right)=h\left(2,1\right)=2+4i,h\left(2,3\right)=h\left(3,2\right)=i,h\left(3,1\right)=h\left(3,1\right)=1-i.\end{array}$$

Take $\left.\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to [1,\infty \right)$ to be symmetric and defined by

$$\begin{array}{c}\mathrm{\varrho }\left(1,1\right)=\mathrm{\varrho }\left(2,2\right)=\mathrm{\varrho }\left(3,3\right)=1,\mathrm{\varrho }\left(1,2\right)=\mathrm{\varrho }\left(2,1\right)=\frac{6}{5},\mathrm{\varrho }\left(2,3\right)=\mathrm{\varrho }\left(3,2\right)=\frac{8}{5},\\ \mathrm{\varrho }\left(3,1\right)=\mathrm{\varrho }\left(1,3\right)=\frac{151}{100}.\end{array}$$

and

$$\begin{array}{c}\varsigma \left(1,1\right)=\varsigma \left(2,2\right)=\varsigma \left(3,3\right)=1,\varsigma \left(1,2\right)=\varsigma \left(2,1\right)=\frac{6}{5},\varsigma \left(2,3\right)=\varsigma \left(3,2\right)=\frac{33}{20},\\ \varsigma \left(3,1\right)=\varsigma \left(1,3\right)=\frac{8}{3}.\end{array}$$

Now define the self-mapping ϒ on $\mathbb{X}$ as follows:

$$\mathrm{\Upsilon }1=\mathrm{\Upsilon }2=\mathrm{\Upsilon }3=2$$

Now we will verify the condition 1:

Case 1: When p = q = 1,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(1\right),\mathrm{\Upsilon }\left(1\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(1,1\right)\right|.$$

Case 2: When p = 1, q = 2,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(1\right),\mathrm{\Upsilon }\left(2\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(1,2\right)\right|.$$

Case 3: When p = 1, q = 3,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(1\right),\mathrm{\Upsilon }\left(3\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(1,3\right)\right|.$$

Case 4: When p = 2, q = 1,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(2\right),\mathrm{\Upsilon }\left(1\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(2,1\right)\right|.$$

Case 5: When p = q = 2,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(2\right),\mathrm{\Upsilon }\left(2\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(2,2\right)\right|.$$

Case 6: When p = 2, q = 3,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(2\right),\mathrm{\Upsilon }\left(3\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(2,3\right)\right|.$$

Case 7: When p = 3, q = 1,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(3\right),\mathrm{\Upsilon }\left(1\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(3,1\right)\right|.$$

Case 8: When p = 3, q = 2,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(3\right),\mathrm{\Upsilon }\left(2\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(3,2\right)\right|.$$

Case 9: When p = q = 3,

$$\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|=\left|h\left(\mathrm{\Upsilon }\left(3\right),\mathrm{\Upsilon }\left(3\right)\right)\right|=\left|h\left(2,2\right)\right|=0\precsim l\left|h\left(3,3\right)\right|.$$

For all $k\in \left(0,1\right),$ it is clear that the above conditions are satisfied, these conditions are also satisfied for ϒ1 = ϒ2 = ϒ3 = 1. For any $p_0\in \mathbb{X}$ condition (2) holds along with conditions of Theorem 1.1. Therefore, there exists a unique fixed point at 1.

Given $p_0\in \mathbb{X},$ the orbit $\mathbb{O}\left(u_0\right)$ of p₀ is defined as $\mathbb{O}\left(u_0\right)=\left\{p_0,\mathrm{\Upsilon }{p}_0,{\mathrm{\Upsilon }}^2{u}_0,\dots \right\},$ where ϒ is a self-map on the set $\mathbb{X}.$ The operator $\mathrm{\Gamma }:\mathbb{X}\to \mathrm{ℝ}$ is called ϒ-orbitally lower semi-continuous at $\varpi \in \mathbb{X}$ if when {p_n} in $\mathbb{O}\left(p_0\right)$ such that lim_n → ∞ h_cdl (p_n, ϖ) = 0, we get that $\mathrm{\Gamma }\left(\varpi \right)\preccurlyeq \underset{n\to \infty }{lim}inf\mathrm{\Gamma }\left(p_n\right).$

Proceeding similarly as [24] and using Definition 3, we have the following corollary generalizing the Theorem 1.1 in [13].

Let ϒ be a self-map on $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ a complete complex-valued double controlled metric-like space by two mappings $\mathrm{\varrho },\varsigma .$ Given $p_0\in \mathbb{X}.$ Let $l\in \left(0,1\right)$ be such that

$$h_{\mathit{cdl}}\left(\mathrm{\Upsilon }z,{\mathrm{\Upsilon }}^{2}z\right)\precsim {\mathit{lh}}_{\mathit{cdl}}\left(z,\mathrm{\Upsilon }z\right),\mathit{\text{foreachz}}\in \mathbb{O}\left(p_0\right).$$

Take p_n = ϒⁿp₀ and suppose that

$$\underset{m\ge 1}{sup}\underset{\iota \to \infty }{lim}\frac{\mathrm{\varrho }\left(p_{\iota +1},p_{\iota +2}\right)}{\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)}\varsigma \left(p_{\iota +1},p_m\right)<\frac{1}{l}.$$

Then, $\underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_n,\kappa \right)=0.$ We also we have that ϒκ = κ if and only if the operator $x\mapsto h_{\mathit{cdl}}\left(x,\mathrm{\Upsilon }x\right)$ is ϒ-orbitally lower semi-continuous at p.

Our next fixed point result involve a Reich type inequality, as follows.

Theorem 1.2 Let $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ be a CDCMLS by the functions $\left.\mathrm{\varrho },\varsigma :\mathbb{X}\times \mathbb{X}\to [1,\infty \right)$ and ϒ be a self mapping satisfying Reich condition. That is, ϒx satisfies

$$h_{\mathit{cdl}}\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\precsim \alpha h_{\mathit{cdl}}\left(p,q\right)+\beta \left(p,\mathrm{\Upsilon }p\right)+\gamma \left(q,\mathrm{\Upsilon }q\right)$$E8

for $\alpha ,\beta ,\gamma \in \left(0,1\right)$ with α + β + γ < 1 and $\gamma =\frac{\alpha +\beta }{1-\gamma }<1,$ for all $p,q\in \mathbb{X}$.

For $p_0\in \mathbb{X}$ we choose p_n = ϒⁿp₀. Assume that

$$\underset{m\ge 1}{sup}\underset{\iota \to \infty }{lim}\frac{\mathrm{\varrho }\left(p_{\iota +1},p_{\iota +2}\right)}{\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)}\varsigma \left(p_{\iota +1},p_m\right)<\frac{1}{l}.$$E9

$$\underset{n\to \infty }{lim}\mathrm{\varrho }\left(p,p_n\right)<\infty \text{exist and finite and}\underset{n\to \infty }{lim}\varsigma \left(p_n,p\right)<\frac{1}{\gamma }$$E10

Then, ϒ has a unique fixed point.

Proof: Let $p_0\in \mathbb{X}.$ Consider the sequence {p_n} with p_n + 1 = ϒp_n for all $n\in \mathbb{N}\cdot$ It is clear that if there exist n₀ for which $p_{n_0+1}=p_{n_0}$ then $\mathrm{\Upsilon }{p}_{n_0}=p_{n_0}.$ Then the proof is finished.

Thus, we suppose that $p_{n_o+1}\ne p_n$ for every $n\in \mathbb{N}.$ Therefore, we may assume that p_n + 1 = p_n for all $n\in \mathbb{N}.$ Now

$$\begin{array}{c}{h}_{\mathit{cdl}}\left(p_n,p_{n+1}\right)=h_{\mathit{cdl}}\left(\mathrm{\Upsilon }{p}_{n-1},\mathrm{\Upsilon }{p}_n\right)\precsim \alpha h_{\mathit{cdl}}\left(p_{n-1},p_n\right)+\beta h_{\mathit{cdl}}\left(p_{n-1},\mathrm{\Upsilon }{p}_{n-1}\right)+\gamma h_{\mathit{cdl}}\left(p_n,\mathrm{\Upsilon }{p}_n\right)\\ =\alpha h_{\mathit{cdl}}\left(p_{n-1},p_n\right)+\beta h_{\mathit{cdl}}\left(p_{n-1},\mathrm{\Upsilon }{p}_n\right)+\gamma h_{\mathit{cdl}}\left(\mathrm{\Upsilon }{p}_n,p_{n+1}\right).\end{array}$$E11

Therefore, we get

$$h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)\precsim \left(\frac{\alpha +\beta }{1-\gamma }\right)h_{\mathit{cdl}}\left(p_{n-1},p_n\right)={\mathit{lh}}_{\mathit{cdl}}\left(p_{n-1},p_n\right)$$E12

Thus, we obtain

$$h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)\precsim {\mathit{lh}}_{\mathit{cdl}}\left(p_{n-1},p_n\right)\precsim l^2{h}_{\mathit{cdl}}\left(\left(p_{n-2},p_{n-1}\precsim \dots \precsim l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)\right.\right..$$E13

for all $n,m\in \mathbb{N}$ with n < m

$$\begin{array}{c}{h}_{\mathit{cdl}}\left(p_n,p_m\right)\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)h_{\mathit{cdl}}\left(p_{n+1},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)\mathrm{\varrho }\left(p_{n+1},p_{n+2}\right)h_{\mathit{cdl}}\left(p_{n+1},p_{n+2}\right)\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)h_{\mathit{cdl}}\left(p_{n+2},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\varsigma \left(p_{n+1},p_m\right)\mathrm{\varrho }\left(p_{n+1},p_{n+2}\right)h_{\mathit{cdl}}\left(p_{n+1},p_{n+2}\right)\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)\mathrm{\varrho }\left(p_n\right.\\ +\varsigma \left(p_{n+1},p_m\right)\varsigma \left(p_{n+2},p_m\right)\varsigma \left(p_{n+3},p_m\right)h_{\mathit{cdl}}\left(p_{n+3},p_m\right)\\ \precsim \dots \end{array}$$

$$\begin{array}{l}\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\sum _{\iota =n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)h_{\mathit{cdl}}\left(p_{\iota },p_{\iota +1}\right)\\ +\prod _{k=n+1}^{m-1}\varsigma \left(p_k,p_m\right)h_{\mathit{cdl}}\left(p_{m-1},p_m\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ +\prod _{\iota =n+1}^{m-1}\varsigma \left(p_{\iota },p_m\right)l^{m-1}{h}_{\mathit{cdl}}\left(p_0,p_1\right)\end{array}$$

$$\begin{array}{l}\precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-2}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ +\left(\prod _{\iota =n+1}^{m-1}\varsigma \left(p_{\iota },p_m\right)\right)l^{m-1}\mathrm{\varrho }\left(p_{m-1},p_m\right)h_{\mathit{cdl}}\left(p_0,p_1\right)\\ =\mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-1}\left(\prod _{j=n+1}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right)\\ \precsim \mathrm{\varrho }\left(p_n,p_{n+1}\right)l^n{h}_{\mathit{cdl}}\left(p_0,p_1\right)+\sum _{\iota =n+1}^{m-1}\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_0,p_1\right).\\ R_n=\sum _{i=0}^n\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_1,p_0\right).\end{array}$$

Then we applying the ratio test, we have

$$\begin{array}{c}{g}_n=\left(\prod _{j=0}^{\iota }\varsigma \left(p_j,p_m\right)\right)\mathrm{\varrho }\left(p_{\iota },p_{\iota +1}\right)l^{\iota }{h}_{\mathit{cdl}}\left(p_1,p_0\right)\\ \frac{g_{n+1}}{g_n}=l\mathrm{\varrho }\left(p_{\iota +1},p_m\right)\frac{\varsigma \left(p_{\iota +1},p_{\iota +2}\right)}{\left(\varsigma \left(p_{\iota },p_{\iota +1}\right)\right.}\end{array}$$

Therefore under condition (9), the series $\sum _n{g}_n$ converges. Therefore, lim_n → ∞ R_n exist. So the real number sequence {R_n} is Cauchy.

Thus we obtained the inequality $h_{\mathit{cdl}}\left(p_n,p_m\right)\precsim h_{\mathit{cdl}}\left(p_1,p_0\right)\left[l^n\varsigma \left(p_n,p_{n+1}\right)+\left(R_{m-1}-R_n\right.\right]$.

Letting n, m → ∞, we get

$$\underset{n,m\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_n,p_m\right)=0,$$

so the sequence {p_n} is Cauchy. Since $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ is a complete CDCMLS, then there exists some $p_0^{\ast }\in \mathbb{X}$ such that

$$\underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_n,p_0^{\ast }\right)=0$$

Which means $p_n\to p_0^{\ast }$ and n → ∞.

Now, our claim is to show that $\mathrm{\Upsilon }{p}_0^{\ast }=p_0^{\ast }.$

$$\begin{array}{c}{h}_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\precsim \mathrm{\varrho }\left(p_0^{\ast },p_{n+1}\right)h_{\mathit{cdl}}\left(p_0^{\ast },p_{n+1}\right)+\varsigma \left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\\ =\mathrm{\varrho }\left(p_0^{\ast },p_{n+1}\right)h_{\mathit{cdl}}\left(p_0^{\ast },p_{n+1}\right)+\varsigma \left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)h_{\mathit{cdl}}\left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\\ \precsim \mathrm{\varrho }\left(p_0^{\ast },p_{n+1}\right)h_{\mathit{cdl}}\left(p_0^{\ast },p_{n+1}\right)+\varsigma \left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\left[\alpha h_{\mathit{cdl}}\left(p_n,\mathrm{\Upsilon }{p}_0^{\ast }\right)+\beta h_{\mathit{cdl}}\left(p_n,\mathrm{\Upsilon }{p}_n\right)+\gamma h_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\right]\\ =\varsigma \left(p_0^{\ast },p_{n+1}\right)h\left(p_0^{\ast },p_{n+1}\right)+\mathrm{\varrho }\left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\left[\alpha h_{\mathit{cdl}}\left(p_n,\mathrm{\Upsilon }{p}_0^{\ast }\right)+\beta h_{\mathit{cdl}}\left(p_n,p_{n+1}\right)+\gamma h_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\right]\end{array}$$

Using this facts in (10) and letting the limit as n → ∞ we obtain

$$h_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\precsim \mathrm{\varrho }\left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\left[\gamma \underset{n\to \infty }{lim}{h}_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }\left(p_0,\mathrm{\Upsilon }{p}_0^{\ast }\right)\right.\right].$$

Suppose that $\mathrm{\Upsilon }{p}_0^{\ast }\ne p_0^{\ast }.$ Since $\underset{n\to \infty }{lim}\mathrm{\varrho }\left(p_{n+1},\mathrm{\Upsilon }{p}_n\right)<\frac{1}{l}$ we have

$$\begin{array}{c}0\prec h_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\precsim \mathrm{\varrho }\left(p_{n+1},\mathrm{\Upsilon }{p}_0^{\ast }\right)\left[\gamma h_{\mathit{cdl}}\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right)\right]\\ \precsim \mathrm{\varrho }\left(p_0^{\ast },\mathrm{\Upsilon }{p}_0^{\ast }\right).\end{array}$$

It is a contradiction, which means $p_0^{\ast }=\mathrm{\Upsilon }{p}_0^{\ast }.$.

Finally, assume that ϒ has two fixed points, say p and q.

Then $h_{\mathit{cdl}}\left(p,q\right)=h_{\mathit{cdl}}\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\precsim \alpha h_{\mathit{cdl}}\left(p,q\right)+\beta h_{\mathit{cdl}}\left(p,\mathrm{\Upsilon }p\right)+\gamma h_{\mathit{cdl}}\left(q,\mathrm{\Upsilon }q\right)$ and so $h_{\mathit{cdl}}\left(p,q\right)\left(1-\alpha \right)\precsim 0.$ Since $\alpha \ne 1.$ We get h_cdl (p, q) = 0 which implies p = q.

This completes the proof.

Let $\mathbb{E}=\left\{1,2,3\right\}.$ Define $h=h_{\mathit{cdl}}:\mathbb{E}\times \mathbb{E}\to \mathbb{C}$ by

$$\begin{array}{c}h\left(1,1\right)=h\left(2,2\right)=0\text{and}h\left(3,3\right)=\frac{i}{2},\\ h\left(1,2\right)=h\left(2,1\right)=2+i,h\left(2,3\right)=h\left(3,2\right)=i,\\ h\left(1,3\right)=h\left(3,1\right)=1-i.\end{array}$$

Define $\mathrm{\varrho },\varsigma :\mathbb{E}\times \mathbb{E}\to \left[1,\infty \right)$ by

$$\begin{array}{c}\mathrm{\varrho }\left(1,1\right)=\mathrm{\varrho }\left(2,2\right)=\mathrm{\varrho }\left(3,3\right)=1,\\ \mathrm{\varrho }\left(1,2\right)=\mathrm{\varrho }\left(2,1\right)=1,\mathrm{\varrho }\left(2,3\right)=\mathrm{\varrho }\left(3,2\right)=\frac{8}{7},\mathrm{\varrho }\left(3,1\right)=\mathrm{\varrho }\left(1,3\right)=\frac{3}{2}.\end{array}$$

and

$$\begin{array}{c}\varsigma \left(1,1\right)=\varsigma \left(2,2\right)=\varsigma \left(3,3\right)=1,\\ \varsigma \left(1,2\right)=\varsigma \left(2,1\right)=\frac{7}{6},\varsigma \left(2,3\right)=\varsigma \left(3,2\right)=\frac{9}{2},\varsigma \left(3,1\right)=\varsigma \left(1,3\right)=1.\end{array}$$

Let

$$\mathrm{\Upsilon }\left(1\right)=2,\mathrm{\Upsilon }\left(2\right)=2,\mathrm{\Upsilon }\left(3\right)=2$$

Proof: $h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\precsim \mathit{\alpha h}\left(p,q\right)+\mathit{\beta h}\left(p,\mathrm{\Upsilon }p\right)+\mathit{\gamma h}\left(q,\mathrm{\Upsilon }q\right)$.

Case 1: When p = 1, q = 2

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim \frac{7\sqrt{5}}{12}=\frac{1}{3}\sqrt{5}=\alpha \mid h\left(1,2\right)\mid +\beta \mid h\left(1,2\right)\mid +\gamma \mid h\left(2,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

Case 2: When p = 1, q = 1

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim \frac{13\sqrt{5}}{36}=\frac{1}{3}\sqrt{5}+\frac{1}{9}\sqrt{5}=\alpha \mid h\left(1,1\right)\mid +\beta \mid h\left(1,2\right)\mid +\gamma \mid h\left(1,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

Case 3: When p = 2, q = 2

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim 0=0+0+0=\alpha \mid h\left(2,2\right)\mid +\beta \mid h\left(2,2\right)\mid +\gamma \mid h\left(2,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

Case 4: When p = 3, q = 3

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim \frac{12+13\sqrt{2}}{36\sqrt{2}}=\frac{1}{3\sqrt{2}}+\frac{1}{4}+\frac{1}{9}=\alpha \mid h\left(3,3\right)\mid +\beta \mid h\left(3,2\right)\mid +\gamma \mid h\left(3,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

Case 5: When p = 2, q = 1

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim \frac{4\sqrt{5}}{9}=\frac{1}{3}\sqrt{5}+0+\frac{1}{9}\sqrt{5}=\alpha \mid h\left(2,1\right)\mid +\beta \mid h\left(2,2\right)\mid +\gamma \mid h\left(1,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

Case 6: When p = 3, q = 1

$$\begin{array}{rcl}\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|& =& \left|h\left(2,2\right)\right|=0\precsim \frac{12\sqrt{2}+9+4\sqrt{5}}{36}=\frac{1}{3}\sqrt{2}+\frac{1}{4}+\frac{1}{9}\sqrt{5}\\ & =& \alpha \left|h\left(3,1\right)\right|+\beta \left|h\left(3,2\right)\right|+\gamma \left|h\left(1,2\right)\right|\\ & =& \alpha \left|h\left(p,q\right)\right|+\beta \left|h\left(p,\mathrm{\Upsilon }p\right)\right|+\gamma \left|h\left(q,\mathrm{\Upsilon }q\right)\right|.\end{array}$$

Case 7: When p = 1, q = 3

$$\begin{array}{rcl}\left|h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\right|& =& \left|h\left(2,2\right)\right|=0\precsim \frac{12\sqrt{2}+9\sqrt{5}+4}{36}=\frac{1}{3}\sqrt{2}+\frac{1}{4}\sqrt{5}+\frac{1}{9}\\ & =& \alpha \left|h\left(1,3\right)\right|+\beta \left|h\left(1,2\right)\right|+\gamma \left|h\left(3,2\right)\right|\\ & =& \alpha \left|h\left(p,q\right)\right|+\beta \left|h\left(p,\mathrm{\Upsilon }p\right)\right|+\gamma \left|h\left(q,\mathrm{\Upsilon }q\right)\right|.\end{array}$$

Case 8: When p = 2, q = 3

$$\begin{array}{r}\mid h\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)\mid =\mid h\left(2,2\right)\mid =0\precsim \frac{4}{9}=\frac{1}{3}+\frac{1}{4}\left(0\right)+\frac{1}{9}=\alpha \mid h\left(2,3\right)\mid +\beta \mid h\left(2,2\right)\mid +\gamma \mid h\left(3,2\right)\mid \\ =\alpha \mid h\left(p,q\right)\mid +\beta \mid h\left(p,\mathrm{\Upsilon }p\right)\mid +\gamma \mid h\left(q,\mathrm{\Upsilon }q\right)\mid .\end{array}$$

For all α, β, $\gamma \in \left(0,1\right)$ with α + β + γ < 1, it is clear that the above conditions are satisfied, these conditions are also satisfied for ϒ1 = ϒ2 = ϒ3 = 2. For any $p_0\in \mathbb{E}$ condition (9) holds along with conditions of theorem 1.2. Therefore, there exists a unique fixed point at 2.

4. Existence and uniqueness of the solution of a Fredholm type integral equation

During this section we suppose the following Fredholm integral equation

$$p\left(u\right)=f\left(u\right)+\underset{a}{\overset{b}{\int }}B\left(u,v,p\left(v\right)\right)\mathit{dv},u,v\in \left[a,b\right],p\left(u\right)\in \mathbb{X},$$E14

where $B\left(u,v,p\left(v\right)\right):\left[a,b\right]\times \left[a,b\right]\times \mathbb{C}\to \mathbb{C}$ and $f\left(u\right):\left[a,b\right]\to \mathbb{C}$ be two bounded and continuous functions.

To prove the existence of solution for integral eq. (14) we use Theorem 1.1. Then we give the following result.

Theorem 1.3 Let $\mathbb{X}=C\left(\left[a,b\right],\mathbb{C}\right)$ is the set of all continuous and complex valued functions which are defined on [a, b]. Also let $\mathrm{\Upsilon }:\mathbb{X}\to \mathbb{X}$ be an operator defined as:

$$p\left(u\right)=f\left(u\right)+{\int }_a^{b}B\left(u,v,p\left(v\right)\right)\mathit{dv},u,v\in \left[a,b\right].$$E15

Suppose the following conditions hold:

1. the functions $B\left(u,v,p\left(v\right)\right):\left[a,b\right]\times \left[a,b\right]\times \mathbb{C}\to \mathbb{C}$ and $f\left(u\right):\left[a,b\right]\to \mathbb{C}$ it’s a continuous function;

2. $\mid B\left(u,v,p\left(v\right)\right)-B\left(u,v,q\left(v\right)\right)\mid \precsim \frac{1}{\tau \sqrt{b-a}}\mid p\left(u\right)-q\left(u\right)\mid ,$ for all p, $q\in \mathbb{X}$ and $\omega \in \left(1,\frac{1}{\lambda }\right]$ with $\lambda \in \left(0,1\right).$

Then Eq. (14) has a unique solution.

Proof: Let $\mathbb{X}=C\left(\left[a,b\right],\mathbb{C}\right)$ and $h_{\mathit{cdl}}:\mathbb{X}\times \mathbb{X}\to \mathbb{C}$ such that,

$$h_{\mathit{cdl}}\left(p,q\right)={‖p-q‖}_{\infty }={\left|p\left(u\right)-q\left(u\right)\right|}^2{e}^{i{cos}^{-1}\tau },$$

where $\mid x\mid =\sqrt{{\alpha }^2+{\beta }^2},$ with α, $\beta \in \mathrm{ℝ},$ τ > 0 and $i=\sqrt{-1}\in \mathbb{C}.$

Let ϑ_u, ${\mathrm{\varrho }}_u:\mathbb{X}\times \mathbb{X}\to \left[1,\infty \right)$ be defined as

$${\mathrm{\varrho }}_u\left(p,q\right)=\left\{\begin{array}{ll}1,& \text{if}p,q\in \left[0,1\right],\\ max\left\{p\left(u\right),q\left(u\right)\right\},& \text{otherwise}.\end{array}\right.$$

$${\varsigma }_u\left(p,q\right)=\left\{\begin{array}{ll}1,& \text{if}p,q\in \left[0,1\right],\\ \frac{1+max\left\{p\left(u\right),q\left(u\right)\right\}}{min\left\{p\left(u\right),q\left(u\right)\right\}},& \text{otherwise}.\end{array}\right.$$

We observe that $\left(\mathbb{X},h_{\mathit{cdl}}\right)$ is a complete CDCMLS. Then the problem (14) can be translated to find a fixed point of the operator ϒ.

Then we have the next inequality

$$\begin{array}{l}{\left|\mathrm{\Upsilon }p\left(u\right)-\mathrm{\Upsilon }q\left(u\right)\right|}^2\precsim {\left|{\int }_a^{b}B(u,v,p\left(v\right))\mathit{dv}-{\int }_a^{b}B(u,v,q\left(v\right))\mathit{dv}\right|}^2\\ \precsim {\int }_a^b{\left|B(u,v,p\left(v\right))-B(u,v,q\left(v\right))\right|}^2\mathit{dv}\\ \precsim \frac{1}{{\tau }^2\left(b-a\right)}{\int }_a^b{\left|p\left(v\right)-q{\left(v\right)}^2\right|}^2\mathit{dv}\\ =\frac{e^{-i{cos}^{-1}\tau }}{{\tau }^2\left(b-a\right)}{\int }_a^b{\left|p\left(v\right)-q\left(v\right)\right|}^2{e}^{i{cos}^{-1}\tau }\mathit{dv}\\ =\frac{e^{-i{cos}^{-1}\tau }}{{\tau }^2\left(b-a\right)}{‖p-q‖}_{\infty }\left({\int }_a^b\mathit{dv}\right).\end{array}$$

Following the calculus we obtain

$${\left|\mathrm{\Upsilon }p\left(u\right)-\mathrm{\Upsilon }q\left(u\right)\right|}^2{e}^{i{cos}^{-1}\tau }={‖\mathrm{\Upsilon }p-\mathrm{\Upsilon }q‖}_{\infty }\precsim \frac{1}{{\tau }^2}{\left|p\left(u\right)-q\left(u\right)\right|}^2{e}^{i{cos}^{-1}\tau }=\frac{1}{{\tau }^2}{‖p-q‖}_{\infty }.$$

Using the hypothesis (ii) we have

$$h_{\mathit{cdl}}\left(\mathrm{\Upsilon }p,\mathrm{\Upsilon }q\right)={‖\mathrm{\Upsilon }p-\mathrm{\Upsilon }q‖}_{\infty }\precsim \frac{1}{{\tau }^2}{‖p-q‖}_{\infty }=\frac{1}{{\tau }^2}{h}_{\mathit{cdl}}\left(p,q\right).$$

It is easy to check that, for both cases of the expressions of ${\mathrm{\varrho }}_u\left(p,q\right)$ and ${\varsigma }_u\left(p,q\right),$ the conditions (2) and (3) are true.

Then, for $0<\delta =\frac{1}{{\tau }^2}<1,$ all the hypothesis of Theorem 1.1 holds. In this conditions, we get that Eq. (14) has a unique solution.

5. Conclusions

Continuing in the same way as in Refs. [12, 25], we have introduced the concept of complex-valued double controlled metric-like spaces(CDCMLS). Some fixed point results and supporting examples in this setting, the related Banach contraction principle, the Reich type fixed point results, are presented. Moreover, the last section is dedicated to apply our main result in order to prove the existence and uniqueness of the solution of a Fredholm type integral equation.

Conflict of interest

The authors declare no conflict of interest.