Fixed Point Results in Modular b-Metric-like Spaces with an Application

Abstract

In this study, we introduce a new space called the modular b-metric-like space. We investigate some properties of this new concept and define notions of $\xi$-convergence, $\xi$-Cauchy sequence, $\xi$-completeness and $\xi$-contraction. The existence and uniqueness of fixed points in the modular b-metric-like space are handled. Moreover, we give some examples and an application to an integral equation to illustrate the usability of the obtained results.

1. Introduction

Metric space theory was established by Fréchet [1] and Hausdorff [2]. Banach’s fixed point theorem (also known as Banach Contraction Principle), in which the concept of metric space is used, is the cornerstone of fixed point theory. Banach [3] introduced this theorem in 1922, and it has since become one of the most effective theorems in mathematics due to its wide applicability and simplicity.

Czerwik [4] presented generalization of some fixed point theorems of the Banach type, using the idea that some problems, especially the problem of convergence of measurable functions, lead to a generalization of the concept of the metric. This generalization of the concept of metric is called the b-metric by Czerwik [4]. For some fixed point results for a multivalued generalized contraction on a set with two b-metrics, see [5].

Amini-Harandi [6] first introduced a new space called a metric-like space. In this new concept, X is a nonempty set and $\sigma :X\times X\to I{R}^{+}$ satisfies all conditions of a metric except that $\sigma (x,x)$ may be different from zero for $x\in X$. Then, Amini-Harandi [6] established the fixed point theory in metric-like spaces by giving some fixed point results in such spaces. For several concepts related to metric-like spaces, such as equal-like points, cluster points, completely separate points, distance between a point and a subset of a metric-like space and distance between two subsets of a metric-like space, see [7].

The concept of b-metric-like space which is generalization of the concepts of metric-like space and b-metric space was presented by Alghamdi et al. [8]. They also investigated the existence of fixed points in a b-metric-like space and provided examples and applications to integral equations.

Nakano [9] introduced the concept of modular spaces. The concept of modular spaces was also studied by Orlicz [10]. Concepts of metric modular and modular metric spaces were introduced by Chistyakov [11,12] who constructed the theory of this structure. According to Chistyakov [13], while the metric on a set represents the non-negative finite distances between any two points of set, the purpose of a metric modular is to represent non-negative (possibly infinite valued) velocities. Some results achieved by Chistyakov are available in [14]. For fixed point results obtained by Chistyakov and applications of them, see [13,15,16,17]. Chistyakov compiled many of the works on metric modular spaces in [18]. Mongkolkeha et al. [19] obtained some results on the existence of fixed points by proving the fixed point theorems for contraction mappings in modular metric spaces.

Ege and Alaca [20] defined the notion of modular b-metric, which is the generalization of the metric modular, and introduced definitions to prove the Banach contraction principle in this new structure. Then, they gave an application of this principle to the system of linear equations.

Rasham et al. [21] introduced the concept of modular-like metric space. Then, they achieved some fixed point results for two families of set-valued mappings, satisfying a contraction in modular-like metric spaces. In [21], some results in graph theory were improved by using multigraph-dominated functions in modular-like metric spaces. Moreover, applications of fixed point theorems on the existence and uniqueness of the solution of integral equations have been investigated in [21,22,23,24]. For more fixed point results in modular-like metric spaces, see [25].

The theory of fixed points also has very proper applications in geometry besides integral equations, systems of linear equations and differential equations. For example, fixed points of principal $E_6$-bundles over a compact algebraic curve and of the automorphisms of the vector bundle moduli space over a compact Riemann surface were introduced by Antón-Sancho [26,27]. Furthermore, Antón-Sancho [28] presented the notion of an $\alpha$-trialitarian G-bundle to describe the fixed points of the automorphism of moduli space.

Despite the important generality of the theory of modular spaces over linear spaces, due to problems arising from multivalued analysis, such as the definition of metric functional spaces, selection principles, and the existence of regular selections of multifunctions, the concepts of modular and corresponding modular linear space are very restrictive. For this reason, Chistyakov introduced a new concept of modular space on an arbitrary set that is consistent with the classical concept. With this paper, our aim is to modulate the b-metric-like space in order not to face these restrictions, allowing us to present a more general form of a metric modular and a modular extension of the concepts in b-metric-like spaces.

In this study, we present the concept of “modular b-metric-like space” by using the approaches in [8,14] and investigate fixed point theorems for contractive mappings in modular b-metric-like space. We give the concepts of a $\xi$-open ($\xi$-closed) set, $\xi$-convergence, $\xi$-Cauchy sequence, $\xi$-completeness and $\xi$-contraction with the help of intelligible examples. Furthermore, we demonstrate the existence of a solution of integral equations to support our results.

2. Preliminaries

This section presents fundamental definitions and concepts to facilitate the comprehension of the primary results. Throughout this paper, IR and IN will be used to denote the set of all real numbers and the set of all positive integer numbers, respectively.

We see that the structure of space changes with the change of axioms and then the concept of modular is combined to these structures below.

Definition 1

([4]). Let $X\ne \varnothing$ and $K\ge 1$ be a real number. A mapping $d:X\times X\to [0,\infty )$ is called b-metric on X if the following hold for each $x,y,z\in X$:

(bM1) $d(x,y)=0\iff x=y$;

(bM2) $d(x,y)=d(y,x)$;

(bM3) $d(x,y)\le K[d(x,z)+d(z,y)]$.

The pair $(X,d)$ is called a b-metric space.

Definition 2

([6]). Let $X\ne \varnothing$. A mapping $\sigma :X\times X\to I{R}^{+}$ is called metric-like on X if the following hold for each $x,y,z\in X$:

(ML1) $\sigma (x,y)=0\Rightarrow x=y$;

(ML2) $\sigma (x,y)=\sigma (y,x)$;

(ML3) $\sigma (x,z)\le \sigma (x,y)+\sigma (y,z)$.

The pair $(X,\sigma )$ is called a metric-like space.

Definition 3

([8]). Let $X\ne \varnothing$ and $K\ge 1$ be a real number. A mapping $\rho :X\times X\to [0,\infty )$ is called b-metric-like on X if the following hold for each $x,y,z\in X$:

(bML1) $\rho (x,y)=0\Rightarrow x=y$;

(bML2) $\rho (x,y)=\rho (y,x)$;

(bML3) $\rho (x,z)\le K[\rho (x,y)+\rho (y,z)]$.

The pair $(X,\rho )$ is called a b-metric-like space.

Example 1

([8]). Let $X=[0,\infty )$. Define the function $\rho :X\times X\to [0,\infty )$ by $\rho (x,y)={(x+y)}^2$. Then, $(X,\rho )$ is a b-metric-like space with constant $K=2$.

Example 2

([8]). Let $X=[0,\infty )$. Define the function $\rho :X\times X\to [0,\infty )$ by $\rho (x,y)={\left(\max \{x,y\}\right)}^2$. Then, $(X,\rho )$ is a b-metric-like space with constant $K=2$.

Definition 4

([10]). Let X be a real linear space. A functional $\delta :X\to [0,\infty ]$ is called classical modular on X if the following hold for each $x,y\in X$:

(CM1) $\delta \left(0\right)=0$;

(CM2) If $\delta \left(\alpha x\right)=0$ for all $\alpha >0$, then $x=0$;

(CM3) $\delta (-x)=\delta \left(x\right)$;

(CM4) $\delta (\alpha x+\beta y)\le \delta \left(x\right)+\delta \left(y\right)$ for all $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$.

Definition 5

([14]). Let $X\ne \varnothing$. A mapping $v:(0,\infty )\times X\times X\to [0,\infty ]$ is called metric modular on X if the following hold for each $x,y,z\in X$:

(MM1) $v_{\lambda }(x,y)=0\iff x=y$, for all $\lambda >0$;

(MM2) $v_{\lambda }(x,y)=v_{\lambda }(y,x)$, for all $\lambda >0$;

(MM3) $v_{\lambda +\mu }(x,z)\le v_{\lambda }(x,y)+v_{\mu }(y,z)$, for all $\lambda ,\mu >0$.

Definition 6

([20]). Let $X\ne \varnothing$ and $K\ge 1$ be a real number. A mapping $u:(0,\infty )\times X\times X\to [0,\infty ]$ is called modular b-metric on X if the following hold for each $x,y,z\in X$:

(MbM1) $u_{\lambda }(x,y)=0\iff x=y$, for all $\lambda >0$;

(MbM2) $u_{\lambda }(x,y)=u_{\lambda }(y,x)$, for all $\lambda >0$;

(MbM3) $u_{\lambda +\mu }(x,z)\le K[u_{\lambda }(x,y)+u_{\mu }(y,z)]$, for all $\lambda ,\mu >0$.

Then, we say that $(X,u)$ is a modular b-metric space.

Definition 7

([21]). Let $X\ne \varnothing$. A mapping $w:(0,\infty )\times X\times X\to [0,\infty )$ is called modular-like metric on X if the following hold for each $x,y,z\in X$:

(MLM1) $w_{\lambda }(x,y)=0\Rightarrow x=y$, for all $\lambda >0$;

(MLM2) $w_{\lambda }(x,y)=w_{\lambda }(y,x)$, for all $\lambda >0$;

(MLM3) $w_{\lambda +\mu }(x,z)\le w_{\lambda }(x,y)+w_{\mu }(y,z)$, for all $\lambda ,\mu >0$.

Then, $(X,w)$ is called a modular-like metric space.

3. Modular b-Metric-like Space

In this section, we start with the introduction of a modular b-metric-like space and give some properties of this concept besides useful examples to support the structure.

Definition 8.

Let $X\ne \varnothing$ and $s\ge 1$ be a real number. A function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$ is called modular b-metric-like on X if it satisfies the following three conditions for each $x,y,z\in X$:

(MbML1) ${\xi }_{\lambda }(x,y)=0\Rightarrow x=y$, for all $\lambda >0$,

(MbML2) ${\xi }_{\lambda }(x,y)={\xi }_{\lambda }(y,x)$, for all $\lambda >0$,

(MbML3) ${\xi }_{\lambda +\mu }(x,y)\le s[{\xi }_{\lambda }(x,z)+{\xi }_{\mu }(z,y)]$, for all $\lambda ,\mu >0$.

Then, the triplet $(X,\xi ,s)$ is called modular b-metric-like space.

If we replace (MbML1) with ${\xi }_{\lambda }(x,y)=0\iff x=y$, then ξ becomes a modular b-metric on X.

In the rest of this paper, for all $\lambda >0$ and $x,y\in X$, ${\xi }_{\lambda }(x,y)=\xi (\lambda ,x,y)$ denotes the map $\xi :(0,\infty )\times X\times X\to [0,\infty ]$.

Example 3.

Let $X=[0,\infty )$. Define the function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$

by ${\xi }_{\lambda }(x,y)={\displaystyle \frac{{(x+y)}^2}{\lambda }}$ for all $\lambda >0$ and $x,y\in X=[0,\infty )$. Then, $(X,\xi ,2)$ is a modular b-metric-like space.

It is clear that the conditions (MbML1) and (MbML2) hold. For this reason, only the condition (MbML3) will be shown:

(MbML3) Since ${(x+y)}^2\le 2[{(x+z)}^2+{(z+y)}^2]$ for all $x,y,z\in X$, we have

$$\begin{array}{cc}{\displaystyle \frac{{(x+y)}^2}{\lambda +\mu }}\hfill & \le {\displaystyle \frac{2}{\lambda +\mu }}[{(x+z)}^2+{(z+y)}^2]\hfill \\ & =2[{\displaystyle \frac{{(x+z)}^2}{\lambda +\mu }}+{\displaystyle \frac{{(z+y)}^2}{\lambda +\mu }}]\hfill \\ & \le 2[{\displaystyle \frac{{(x+z)}^2}{\lambda }}+{\displaystyle \frac{{(z+y)}^2}{\mu }}]\hfill \end{array}$$

for all $x,y,z\in X$ and all $\lambda ,\mu >0$.

That means that ${\xi }_{\lambda +\mu }(x,y)\le 2[{\xi }_{\lambda }(x,z)+{\xi }_{\mu }(z,y)].$ Thus, $(X,\xi ,2)$ is a modular b-metric-like space.

The graphical behavior of the function ξ defined as ${\xi }_{\lambda }(x,y)={\displaystyle \frac{{(x+y)}^2}{\lambda }}$ on the set $[1,10]\times [1,10]$ for the values $\lambda =1,2,3,4,5$ is given in the Figure 1. Thus, we get a visual idea about how the function ξ changes on the set $[1,10]\times [1,10]$ with the change of the value of λ from 1 to 5.

Example 4.

Let $X=[0,\infty )$. Define the function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$ by

${\xi }_{\lambda }(x,y)={\displaystyle \frac{{\left(\max \{x,y\}\right)}^2}{\lambda }}$ for all $\lambda >0$ and $x,y\in X=[0,\infty )$. Then, $(X,\xi ,2)$ is a modular b-metric-like space.

It is clear that the conditions (MbML1) and (MbML2) hold. For this reason, only the condition (MbML3) will be shown:

(MbML3) Since ${\left(\max \{x,y\}\right)}^2\le 2[{\left(\max \{x,z\}\right)}^2+{\left(\max \{z,y\}\right)}^2]$ for all $x,y,z\in X$, we have

$$\begin{array}{cc}{\displaystyle \frac{{\left(\max \{x,y\}\right)}^2}{\lambda +\mu }}\hfill & \le {\displaystyle \frac{2}{\lambda +\mu }}[{\left(\max \{x,z\}\right)}^2+{\left(\max \{z,y\}\right)}^2]\hfill \\ & =2[{\displaystyle \frac{{\left(\max \{x,z\}\right)}^2}{\lambda +\mu }}+{\displaystyle \frac{{\left(\max \{z,y\}\right)}^2}{\lambda +\mu }}]\hfill \\ & \le 2[{\displaystyle \frac{{\left(\max \{x,z\}\right)}^2}{\lambda }}+{\displaystyle \frac{{\left(\max \{z,y\}\right)}^2}{\mu }}]\hfill \end{array}$$

for all $x,y,z\in X$ and all $\lambda ,\mu >0$. That means that ${\xi }_{\lambda +\mu }(x,y)\le 2[{\xi }_{\lambda }(x,z)+{\xi }_{\mu }(z,y)].$ Thus, $(X,\xi ,2)$ is a modular b-metric-like space.

Example 5.

Let $\aleph =C[0,L]$ be the set of all continuous real-valued functions defined on $[0,L]$, where $L>0$. Define the function $\xi :(0,\infty )\times \aleph \times \aleph \to [0,\infty ]$ by

${\xi }_{\lambda }(\varpi \left(s\right),\phi \left(s\right))={\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda }}$ for all $\lambda >0$ and all $\varpi ,\phi \in \aleph$. Then, $(\aleph ,\xi ,2)$ is a modular b-metric-like space.

It is clear that the conditions (MbML1) and (MbML2) hold. For this reason, only the condition (MbML3) will be shown:

Since ${\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2\le 2[{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\kappa \left(s\right)|)}^2+{\max }_{s\in [0,L]}(|\kappa \left(s\right)|+{|\phi \left(s\right)|)}^2]$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda +\mu }}& \le & {\displaystyle \frac{2}{\lambda +\mu }}[\underset{s\in [0,L]}{\max }(|\varpi \left(s\right)|+{|\kappa \left(s\right)|)}^2+\underset{s\in [0,L]}{\max }(|\kappa \left(s\right)|+{|\phi \left(s\right)|)}^2]\hfill \\ & =& 2[{\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\kappa \left(s\right)|)}^2}{\lambda +\mu }}+{\displaystyle \frac{{\max }_{s\in [0,L]}(|\kappa \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda +\mu }}]\hfill \\ & \le & 2[{\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\kappa \left(s\right)|)}^2}{\lambda }}+{\displaystyle \frac{{\max }_{s\in [0,L]}(|\kappa \left(s\right)|+{|\phi \left(s\right)|)}^2}{\mu }}]\hfill \end{array}$$

for all $\varpi ,\phi ,\kappa \in \aleph$ and all $\lambda ,\mu >0$. That means that

$${\xi }_{\lambda +\mu }(\varpi ,\phi )\le 2[{\xi }_{\lambda }(\varpi ,\kappa )+{\xi }_{\mu }(\kappa ,\phi )].$$

Thus, $(\aleph ,\xi ,2)$ is a modular b-metric-like space.

Proposition 1.

Let $X=[0,\infty )$, and let $(X,d)$ be a b-metric-like space with constant $s\ge 1$. Define the function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$ by ${\xi }_{\lambda }(x,y)={\displaystyle \frac{d(x,y)}{\lambda }}$ for all $\lambda >0$ such that $x,y\in X=[0,\infty )$. Then, $(X,\xi ,s)$ is a modular b-metric-like space.

Proof. 

($MbML1$) ${\xi }_{\lambda }(x,y)={\displaystyle \frac{d(x,y)}{\lambda }}=0\Rightarrow d(x,y)=0$ for all $\lambda >0$. Hence, we have $x=y$, since d is b-metric-like.

$\left(MbML2\right)$${\xi }_{\lambda }(x,y)={\displaystyle \frac{d(x,y)}{\lambda }}={\displaystyle \frac{d(y,x)}{\lambda }}={\xi }_{\lambda }(y,x)$ for all $\lambda >0$.

$\left(MbML3\right)$ Since $(X,d)$ is a b-metric-like space with constant s, we have $d(x,y)\le s[d(x,z)+d(z,y)]$ for all $x,y,z\in X.$ It follows that ${\displaystyle \frac{d(x,y)}{\lambda +\mu }}\le {\displaystyle \frac{s}{\lambda +\mu }}[d(x,z)+d(z,y)]$ $=s[{\displaystyle \frac{d(x,z)}{\lambda +\mu }}+{\displaystyle \frac{d(z,y)}{\lambda +\mu }}]$ $\le s[{\displaystyle \frac{d(x,z)}{\lambda }}+{\displaystyle \frac{d(z,y)}{\mu }}]$ for all $x,y,z\in X$ and all $\lambda ,\mu >0$. That means that ${\xi }_{\lambda +\mu }(x,y)\le s[{\xi }_{\lambda }(x,z)+{\xi }_{\mu }(z,y)].$ Thus, $(X,\xi ,s)$ is a modular b-metric-like space. □

Definition 9.

Let ξ be a modular b-metric-like on X, and let $x_0$ be an arbitrary element in X.

Define set ${X_{\xi }}^{fin}$ by ${X_{\xi }}^{fin}\equiv {X_{\xi }}^{fin}\left(x_0\right)=\{x\in X:{\xi }_{\lambda }(x,x_0)<\infty$ for all $\lambda >0$}.

Definition 10.

Let $(X,\xi ,s)$ be a modular b-metric-like space. Let $x\in X$, $r>0$ and $\lambda >0$. Then, set $B_{{\xi }_{\lambda }}(x,r)=\{y\in X:|{\xi }_{\lambda }(x,y)-{\xi }_{\lambda }(x,x)|<r\}$ is called a $\xi -open$ ball relative to λ with center x and radius $r>0$.

Definition 11.

Let $(X,\xi ,s)$ be a modular b-metric-like space and U be a subset of X. If there exists $r_0>0$ such that $B_{{\xi }_{{\lambda }_0}}(x,r_0)$ $\subset U$ for all $x\in U$ and some ${\lambda }_0>0$, then U is called a $\xi -open$ subset of X.

If $X\setminus U$ is a $\xi -open$ set, then U is called a $\xi -closed$ set.

Definition 12.

Let $(X,\xi ,s)$ be a modular b-metric-like space, ${\left\{x_n\right\}}_{n\in IN}\subset {X_{\xi }}^{fin}$ and $x\in {X_{\xi }}^{fin}$.

$\left(i\right)$x is called $\xi -limit$ of the sequence ${\left\{x_n\right\}}_{n\in N}$ if ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x)={\xi }_{\lambda }(x,x)$ for all $\lambda >0$; moreover, we say that the sequence ${\left\{x_n\right\}}_{n\in N}$ is $\xi -convergent$ to x and we denote it by $x_n{\to }^{\xi }x$.

$\left(ii\right)$ Sequence ${\left\{x_n\right\}}_{n\in N}$ is called $\xi -Cauchy$ if ${lim}_{n,m\to \infty }{\xi }_{\lambda }(x_n,x_m)$ exists and is finite for all $\lambda >0$.

$\left(iii\right)$ Modular b-metric-like space ${X_{\xi }}^{fin}$ is called $\xi -complete$ if every $\xi -Cauchy$ sequence ${\left\{x_n\right\}}_{n\in N}$ is $\xi -convergent$ to any x such that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x)={\xi }_{\lambda }(x,x)={lim}_{n,m\to \infty }{\xi }_{\lambda }$ $(x_n,x_m)$ for all $\lambda >0$.

Proposition 2.

Let $(X,\xi ,s)$ be a modular b-metric-like space, and let V be a subset of X. V is ξ-closed if and only if for any sequence $\left\{x_n\right\}\subset V$, which is ξ-convergent to $x\in X$, we have $x\in V$.

Proof. 

Suppose that V is a $\xi$-closed set, $\left\{x_n\right\}\subset V$, $x\in X$, $x_n{\to }^{\xi }x$. Let $x\notin V$. By Definition 11, $X\setminus V$ is a $\xi$-open set. Since $x\in X\setminus V$, there exists $r_0>0$ such that $B_{{\xi }_{{\lambda }_0}}(x,r_0)\subset X\setminus V$ for some ${\lambda }_0>0$. Since $x_n{\to }^{\xi }x$, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x)={\xi }_{\lambda }(x,x)$ for all $\lambda >0$. In other words, ${lim}_{n\to \infty }|{\xi }_{\lambda }(x_n,x)-{\xi }_{\lambda }(x,x)|=0$ for all $\lambda >0$. Hence, for all $\lambda >0$, there exists $n_0\in IN$ such that $|{\xi }_{\lambda }(x_n,x)-{\xi }_{\lambda }(x,x)|<r_0$ for all $n\ge n_0$. Especially for $\lambda ={\lambda }_0$, we have $|{\xi }_{{\lambda }_0}(x_n,x)-{\xi }_{{\lambda }_0}(x,x)|<r_0$ for all $n\ge n_0$. Thus, $x_n\subset B_{{\xi }_{{\lambda }_0}}(x,r_0)\subset X\setminus V$ for all $n\ge n_0$, which is a contradiction. Hence, $x\in V$.

Conversely, assume that for any sequence $\left\{x_n\right\}\subset V$, which is $\xi$-convergent to $x\in X$, we have $x\in V$. Let $y\in X\setminus V$. We need to show that there exists $r_0>0$ such that $B_{{\xi }_{{\lambda }_0}}(y,r_0)\cap V=\varnothing$ for some ${\lambda }_0$. Suppose that for all $\lambda >0$ and $r>0$, we have

$B_{{\xi }_{\lambda }}(y,r_0)\cap V\ne \varnothing$. Then, for all $n\in IN$ and $\lambda >0$, choose $x_n\in B_{{\xi }_{\lambda }}(y,{\displaystyle \frac{1}{n}})\cap V\ne \varnothing$. Hence, $|{\xi }_{\lambda }(x_n,y)-{\xi }_{\lambda }(y,y)|<{\displaystyle \frac{1}{n}}$ for all $\lambda >0$ and $n\in IN$. Then, $0\le {lim}_{n\to \infty }|{\xi }_{\lambda }(x_n,y)-{\xi }_{\lambda }(y,y)|<{lim}_{n\to \infty }{\displaystyle \frac{1}{n}}$ and we obtain ${lim}_{n\to \infty }|{\xi }_{\lambda }(x_n,y)-{\xi }_{\lambda }(y,y)|=0$. Therefore, ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,y)={\xi }_{\lambda }(y,y)$ for all $\lambda >0$ and we get $x_n{\to }^{\xi }x$. Since $\left\{x_n\right\}\subset V$, we have $y\in V$ from our assumption, which is a contradiction. Then, for all $y\notin V$, there exists $r_0>0$ such that $B_{{\xi }_{{\lambda }_0}}(y,r_0)\subset X\setminus V$ for some ${\lambda }_0>0$. Thus, $X\setminus V$ is a $\xi$-open set. So, V is a $\xi$-closed set. □

Proposition 3.

Let $(X,\xi ,s)$ be a modular b-metric-like space, and let $\left\{x_n\right\}$ be a sequence in X such that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x)=0$ for all $\lambda >0$. Then, x is unique.

Proof. 

Suppose that there exists $y\in X$ such that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,y)=0$ for all $\lambda >0$. Then, for all $\lambda >0$,

$0\le {\xi }_{\lambda }\left(x_{,}y\right)\le s[{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x,x_n)+{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,y)]$.

⇒$0\le {lim}_{n\to \infty }{\xi }_{\lambda }\left(x_{,}y\right)\le s[{lim}_{n\to \infty }{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x,x_n)+{lim}_{n\to \infty }{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,y)]$

⇒$0\le {\xi }_{\lambda }\left(x_{,}y\right)\le 0$.

Hence, ${\xi }_{\lambda }\left(x_{,}y\right)=0$ for all $\lambda >0$, and $x=y$. □

Remark 1.

In a modular b-metric-like space, the $\xi -limit$ of the $\xi -convergent$ sequence $\left\{x_n\right\}$ may not be unique. Let $X=[0,\infty )$. Define the function $d:X\times X\to [0,\infty )$ by $d(x,y)=\max \{x,y\}$. Then, we know that $(X,d)$ is a b-metric-like space with any constant $s\ge 1$. Consider Proposition 1 and define a sequence $\left\{x_n\right\}\subset {X_{\xi }}^{fin}$ by $\left\{x_n\right\}=\{1+{\displaystyle \frac{1}{n}}\}$.

If $x\ge 2$, then ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x)={lim}_{n\to \infty }{\displaystyle \frac{d(x_n,x)}{\lambda }}={lim}_{n\to \infty }{\displaystyle \frac{\max \{x_n,x\}}{\lambda }}={lim}_{n\to \infty }{\displaystyle \frac{x}{\lambda }}={\displaystyle \frac{\max \{x,x\}}{\lambda }}={\xi }_{\lambda }(x,x)$ for all $\lambda >0$. Hence, the sequence $\left\{x_n\right\}$ is $\xi -convergent$ to all $x\in {X_{\xi }}^{fin}$ with $x\ge 2$.

4. Fixed Point Results

We prove some related fixed point theorems and give examples to support these theorems in this part.

Definition 13.

Let ξ be a modular b-metric-like on X, and let $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ be a mapping. If for every $x,y\in {X_{\xi }}^{fin}$ and all $\lambda >0$ there exists $0<k<1$ such that ${\xi }_{\lambda }(Tx,Ty)\le k{\xi }_{\lambda }(x,y)$, then the mapping T is called $\xi -contraction$.

Theorem 1.

Let $(X,\xi ,s)$ be a modular b-metric-like space such that ${X_{\xi }}^{fin}$ is ξ-complete. Let $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ be a ξ-contraction with restriction $0<k<1$. Then, for the sequence defined as $x_n=T{x}_{n-1}=T^n{x}_0$ where $x_0\in {X_{\xi }}^{fin}$, there exists an element $\overline{x}\in {X_{\xi }}^{fin}$ such that $\left\{x_n\right\}$ is ξ-convergent to $\overline{x}$ and $\overline{x}$ is a unique fixed point of T.

Proof. 

Let $x_0\in {X_{\xi }}^{fin}$ and $\left\{x_n\right\}\subset {X_{\xi }}^{fin}$ be defined by $x_n=T{x}_{n-1}=T^n{x}_0$. Since T is a $\xi$-contraction, we obtain

$${\xi }_{\lambda }(T^2{x}_0,T^2{x}_1)\le k{\xi }_{\lambda }(T{x}_0,T{x}_1)\le k^2{\xi }_{\lambda }(x_0,x_1).$$

If this procedure is iterated, we get

$${\xi }_{\lambda }(T^n{x}_0,T^n{x}_1)\le k^n{\xi }_{\lambda }(x_0,x_1),$$

for all $\lambda >0$ and $n\in IN.$

Since $T^n{x}_1=T^n\left(T{x}_0\right)=T^{n+1}{x}_0=x_{n+1}$ and $T^n{x}_0=x_n$, for all $\lambda >0$ and $n\in IN$, we have

$${\xi }_{\lambda }(x_n,x_{n+1})\le k^n{\xi }_{\lambda }(x_0,x_1).$$

Taking the limit as $n\to \infty$ in the above inequality, we get ${lim}_{n\to \infty }\left(k^n{\xi }_{\lambda }(x_0,T{x}_0)\right)={lim}_{n\to \infty }\left(k^n{\xi }_{\lambda }(x_0,x_1)\right)=0$ because of $k\in (0,1)$ by the definition of the $\xi$-contraction and ${\xi }_{\lambda }(x,Tx)<\infty$ for all $\lambda >0$ and all $x\in {X_{\xi }}^{fin}$.

Then, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x_{n+1})=0$ for all $\lambda >0$. Hence, for all $\lambda >0$ and $ϵ>0$, there exists $n_0\in IN$ such that ${\xi }_{\lambda }(x_n,x_{n+1})<ϵ$ for all $n\ge n_0$. Without loss of generality, suppose $m,n\in N$ and $m>n$. Observe that, for ${\displaystyle \frac{\lambda }{m-n}}>0$, there exists $n_{{\displaystyle \frac{\lambda }{m-n}}}\in N$ such that ${\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})<{\displaystyle \frac{ϵ}{{\sum }_{p=1}^{m-n}{s}^p}}$, for all $n\ge n_{{\displaystyle \frac{\lambda }{m-n}}}$.

Now, we have

$$\begin{array}{ccc}\hfill {\xi }_{\lambda }(x_n,x_m)& \le & s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_{n+1},x_{n+2})+...+s^{m-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_{m-1},x_m)\hfill \\ & =& s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(T{x}_n,T{x}_{n+1})+...+s^{m-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(T{x}_{m-2},T{x}_{m-1})\hfill \\ & \le & s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^{2}k{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^3{k}^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+...+s^{m-n}{k}^{m-1-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & \le & s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+...+s^{m-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & =& [s+s^2+s^3+...+s^{m-n}]{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & =& \sum _{p=1}^{m-n}{s}^p{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & \le & \sum _{p=1}^{m-n}{s}^p{\displaystyle \frac{ϵ}{{\sum }_{p=1}^{m-n}{s}^p}}\hfill \\ & =& ϵ.\hfill \end{array}$$

for all $m>n$ and all $n\ge n_{{\displaystyle \frac{\lambda }{m-n}}}$.

Therefore, we have ${lim}_{n,m\to \infty }{\xi }_{\lambda }(x_n,x_m)=0$; hence, $\left\{x_n\right\}\subset {X_{\xi }}^{fin}$ is a $\xi$-Cauchy sequence. Since ${X_{\xi }}^{fin}$ is a $\xi$-complete set, there exists $\overline{x}\in {X_{\xi }}^{fin}$ such that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,\overline{x})={\xi }_{\lambda }(\overline{x},\overline{x})={lim}_{n,m\to \infty }{\xi }_{\lambda }(x_n,x_m)$ for all $\lambda >0$. Since ${lim}_{n,m\to \infty }{\xi }_{\lambda }(x_n,x_m)=0$ for all $\lambda >0$, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,\overline{x})={\xi }_{\lambda }(\overline{x},\overline{x})=0$ for all $\lambda >0$.

It follows that

$$\begin{array}{ccc}\hfill {\xi }_{\lambda }(T\overline{x},\overline{x})& \le & s[{\xi }_{{\displaystyle \frac{\lambda }{2}}}(T\overline{x},x_n)+{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,\overline{x})]\hfill \\ & =& s[{\xi }_{{\displaystyle \frac{\lambda }{2}}}(T\overline{x},T{x}_{n-1})+{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,\overline{x})]\hfill \\ & \le & s[k{\xi }_{{\displaystyle \frac{\lambda }{2}}}(\overline{x},x_{n-1})+{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,\overline{x})]\hfill \end{array}$$

for all $\lambda >0$ and all $n\in IN$.

Taking the limit as $n\to \infty$ in the above inequality, we get

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{\xi }_{\lambda }(T\overline{x},\overline{x})& \le & \underset{n\to \infty }{lim}(s[k{\xi }_{{\displaystyle \frac{\lambda }{2}}}(\overline{x},x_{n-1})+{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,\overline{x})])\hfill \\ & =& s[k\underset{n\to \infty }{lim}{\xi }_{{\displaystyle \frac{\lambda }{2}}}(\overline{x},x_{n-1})+\underset{n\to \infty }{lim}{\xi }_{{\displaystyle \frac{\lambda }{2}}}(x_n,\overline{x})]\hfill \\ & =& 0\hfill \end{array}$$

for all $\lambda >0$.

It follows that ${\xi }_{\lambda }(T\overline{x},\overline{x})=0$ for all $\lambda >0$. Hence, we have $T\overline{x}=\overline{x}$ from condition $\left(MbML1\right)$. Thus, $\overline{x}$ is a fixed point of T. Next, we prove that this fixed point $\overline{x}$ is unique.

Suppose that y is another fixed point of T such that $\overline{x}\ne y$. Therefore, we have $Ty=y$. Since T is a $\xi$-contraction, we have ${\xi }_{\lambda }(\overline{x},y)={\xi }_{\lambda }(T\overline{x},Ty)\le k{\xi }_{\lambda }(\overline{x},y)$ for all $\lambda >0$.

It follows that $(1-k){\xi }_{\lambda }(\overline{x},y)\le 0$. Hence, we have ${\xi }_{\lambda }(\overline{x},y)=0$ for all $\lambda >0$. Thus, we get $\overline{x}=y$ from condition $\left(MbML1\right)$. □

Example 6.

Let $X=[0,\infty )$. Define the function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$ by ${\xi }_{\lambda }(x,y)={\displaystyle \frac{d(x,y)}{\lambda }}$ for all $\lambda >0$ such that $d(x,y)={(x+y)}^2$ and $x,y\in X=[0,\infty )$. Then, $(X,\xi ,2)$ is a modular b-metric-like space such that ${X_{\xi }}^{fin}$ is ξ-complete since ${X_{\xi }}^{fin}=X$.

Define the map $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ by $Tx=\alpha x$ such that $\alpha \in (0,1)$. Then, we have ${\xi }_{\lambda }(Tx,Ty)={\xi }_{\lambda }(\alpha x,\alpha y)={\displaystyle \frac{{(\alpha x+\alpha y)}^2}{\lambda }}={\displaystyle \frac{{\alpha }^2{(x+y)}^2}{\lambda }}={\alpha }^2{\displaystyle \frac{d(x,y)}{\lambda }}={\alpha }^2{\xi }_{\lambda }(x,y)$ for all $\lambda >0$. Since $\alpha \in (0,1)$, we have $k={\alpha }^2\in (0,1)$. Thus, the mapping T is a ξ-contraction with constant $k={\alpha }^2$. Then, by Theorem 1, there exists a unique fixed point $\overline{x}=0\in {X_{\xi }}^{fin}$ such that $x_n=T{x}_{n-1}=T^n{x}_0$ is ξ-convergent to $\overline{x}=0$.

Indeed, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,0)={lim}_{n\to \infty }{\displaystyle \frac{d(x_n,0)}{\lambda }}={lim}_{n\to \infty }{\displaystyle \frac{{(x_n+0)}^2}{\lambda }}={lim}_{n\to \infty }{\displaystyle \frac{{\left(x_n\right)}^2}{\lambda }}$ $={\displaystyle \frac{1}{\lambda }}\left({lim}_{n\to \infty }{x}_n{lim}_{n\to \infty }{x}_n\right)$. Then, it follows that $x_n=T^n{x}_0=T^{n-1}\left(T{x}_0\right)=T^{n-1}\left(\alpha x_0\right)=T^{n-2}\left(T\left(\alpha x_0\right)\right)=T^{n-2}\left({\alpha }^2{x}_0\right)=···=T^{n-(n-1)}\left({\alpha }^{n-1}{x}_0\right)=T\left({\alpha }^{n-1}{x}_0\right)={\alpha }^n{x}_0$, and this means that $x_n={\alpha }^n{x}_0$.

Since $x_0\in {X_{\xi }}^{fin}$, we have $x_0<\infty$. Then, ${lim}_{n\to \infty }{x}_n={lim}_{n\to \infty }{\alpha }^n{x}_0=0$ since $\alpha \in (0,1)$ and $x_0<\infty$. Therefore, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,0)={\displaystyle \frac{1}{\lambda }}\left({lim}_{n\to \infty }{x}_n{lim}_{n\to \infty }{x}_n\right)=0={\displaystyle \frac{{(0+0)}^2}{\lambda }}={\xi }_{\lambda }(0,0)$, and this means that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,0)={\xi }_{\lambda }(0,0)$ for all $\lambda >0$. Also, since $T\left(0\right)=\alpha 0=0$ holds, 0 is a unique fixed point of T.

Remark 2.

Let $(X,\xi ,s)$ be a modular b-metric-like space. Define ${{\xi }_{\lambda }}^z:X^2\to [0,\infty )$ by ${{\xi }_{\lambda }}^z(x,y)=|2{\xi }_{\lambda }(x,y)-{\xi }_{\lambda }(x,x)-{\xi }_{\lambda }(y,y)|$. Clearly, ${{\xi }_{\lambda }}^z(x,x)=0$ for all $x\in X$.

Theorem 2.

Let $(X,\xi ,s)$ be a modular b-metric-like space such that ${X_{\xi }}^{fin}$ is ξ-complete. Suppose that the mapping $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ is onto and satisfies

$${\xi }_{\lambda }(Tx,Ty)\ge [r+lmin\{{{\xi }_{\lambda }}^z(x,Tx),{{\xi }_{\lambda }}^z(y,Ty),{{\xi }_{\lambda }}^z(x,Ty),{{\xi }_{\lambda }}^z(y,Tx)\}]{\xi }_{\lambda }(x,y)$$

(1)

for all $x,y\in {X_{\xi }}^{fin}$ and all $\lambda >0$, where $r>s$, $l\ge 0$. Then, T has a unique fixed point.

Proof. 

Let $x_0\in {X_{\xi }}^{fin}$. Since T is onto mapping, there exists $x_1\in {X_{\xi }}^{fin}$ such that $x_0=T{x}_1$. By continuing this process, we get $x_n=T{x}_{n+1}$ for all $n\in IN$. In case $x_{n_0}=x_{n_0+1}$ for some $n_0\in IN$, we have $T{x}_{n_0+1}=x_{n_0+1}$ since $T{x}_{n_0+1}=x_{n_0}$. Thus, $x_{n_0+1}$ is a fixed point of T. Now assume that $x_n\ne x_{n+1}$ for all n. From (1) with $x=x_n$ and $y=x_{n+1}$, we get

$${\xi }_{\lambda }(T{x}_n,T{x}_{n+1})\ge [r+lmin\{{{\xi }_{\lambda }}^z(x_n,T{x}_n),{{\xi }_{\lambda }}^z(x_{n+1},T{x}_{n+1}),{{\xi }_{\lambda }}^z(x_n,T{x}_{n+1}),{{\xi }_{\lambda }}^z(x_{n+1},T{x}_n)\}]{\xi }_{\lambda }(x_n,x_{n+1})$$

for all $\lambda >0$.

It follows that

$$\begin{array}{ccc}\hfill {\xi }_{\lambda }(x_{n-1},x_n)& \ge & [r+lmin\{{{\xi }_{\lambda }}^z(x_n,x_{n-1}),{{\xi }_{\lambda }}^z(x_{n+1},x_n),{{\xi }_{\lambda }}^z(x_n,x_n),{{\xi }_{\lambda }}^z(x_{n+1},x_{n-1})\}]{\xi }_{\lambda }(x_n,x_{n+1})\hfill \\ & =& r{\xi }_{\lambda }(x_n,x_{n+1})\hfill \end{array}$$

for all $\lambda >0$ since $T{x}_{n+1}=x_n$ for all $n\in IN$, which implies ${\xi }_{\lambda }(x_{n-1},x_n)\ge r{\xi }_{\lambda }(x_n,x_{n+1})$. Hence, ${\xi }_{\lambda }(x_n,x_{n+1})\le {\displaystyle \frac{1}{r}}{\xi }_{\lambda }(x_{n-1},x_n)$, and so we have ${\xi }_{\lambda }(x_n,x_{n+1})\le h{\xi }_{\lambda }(x_{n-1},x_n)$ where $h={\displaystyle \frac{1}{r}}<{\displaystyle \frac{1}{s}}$ since $r>s$. Now, we will show that $x_n$ is a $\xi$-Cauchy sequence.

Since ${\xi }_{\lambda }(x_n,x_{n+1})\le h{\xi }_{\lambda }(x_{n-1},x_n)$ for all $n\in IN$ and all $\lambda >0$, we have

$$h{\xi }_{\lambda }(x_{n-1},x_n)\le h\left(h{\xi }_{\lambda }(x_{n-2},x_{n-1})\right)=h^2{\xi }_{\lambda }(x_{n-2},x_{n-1})\le h^3{\xi }_{\lambda }(x_{n-3},x_{n-2})\le \cdots \le h^n{\xi }_{\lambda }(x_0,x_1)$$

which implies ${\xi }_{\lambda }(x_n,x_{n+1})\le h^n{\xi }_{\lambda }(x_0,x_1)$ for all $n\in IN$ and all $\lambda >0$.

We have

$$\underset{n\to \infty }{lim}\left(h^n{\xi }_{\lambda }(T{x}_1,x_1)\right)=\underset{n\to \infty }{lim}\left(h^n{\xi }_{\lambda }(x_0,x_1)\right)=0$$

since $h<{\displaystyle \frac{1}{s}}\le 1$ and ${\xi }_{\lambda }(Tx,x)<\infty$ for all $\lambda >0$ and all $x\in {X_{\xi }}^{fin}$. It follows that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x_{n+1})=0$ for all $\lambda >0$. So, for all $\lambda >0$, we have that for all $\epsilon >0$ there exists $n_0\in IN$ such that ${\xi }_{\lambda }(x_n,x_{n+1})<\epsilon$ for all $n\in IN$ with $n\ge n_0$. Without loss of generality, suppose $m,n\in IN$ and $m>n$. Observe that, for ${\displaystyle \frac{\lambda }{m-n}}>0$, there exists $n_{{\displaystyle \frac{\lambda }{m-n}}}\in IN$ such that

$${\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})<{\displaystyle \frac{ϵ}{{\sum }_{p=1}^{m-n}{s}^p}}$$

for all $n\ge n_{{\displaystyle \frac{\lambda }{m-n}}}$.

Now, we have

$$\begin{array}{ccc}\hfill {\xi }_{\lambda }(x_n,x_m)& \le & s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_{n+1},x_{n+2})+\cdots +s^{m-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_{m-1},x_m)\hfill \\ & \le & s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^{2}h{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^3{h}^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+\cdots +s^{m-n}{h}^{m-1-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & <& s{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+s^2{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})+\cdots +s^{m-n}{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & =& [s+s^2+s^3+\cdots +s^{m-n}]{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & =& \sum _{p=1}^{m-n}{s}^p{\xi }_{{\displaystyle \frac{\lambda }{m-n}}}(x_n,x_{n+1})\hfill \\ & \le & \sum _{p=1}^{m-n}{s}^p{\displaystyle \frac{ϵ}{{\sum }_{p=1}^{m-n}{s}^p}}\hfill \\ & =& ϵ\hfill \end{array}$$

for all $m>n$ and all $n\ge n_{{\displaystyle \frac{\lambda }{m-n}}}$.

Thus, we have ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x_m)=0$. Since ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,x_m)=0$ exists and is finite, $\left\{x_n\right\}$ is a $\xi$-Cauchy sequence. Since $({X_{\xi }}^{fin},\xi ,s)$ is $\xi$-complete, the sequence $\left\{x_n\right\}$ in ${X_{\xi }}^{fin}$ is $\xi$-convergent to $z_0\in {X_{\xi }}^{fin}$ such that

$$\underset{n\to \infty }{lim}{\xi }_{\lambda }(x_n,z_0)={\xi }_{\lambda }(z_0,z_0)=\underset{n,m\to \infty }{lim}{\xi }_{\lambda }(x_n,x_m)$$

for all $\lambda >0$.

Since T is onto mapping, there exists $v\in {X_{\xi }}^{fin}$ such that $Tv=z_0$. From (1) and since $T{x}_{n+1}=x_n$, we have

$${\xi }_{\lambda }(x_n,z_0)={\xi }_{\lambda }(T{x}_{n+1},Tv)\ge [r+lmin\{{{\xi }_{\lambda }}^z(x_{n+1},T{x}_{n+1}),{{\xi }_{\lambda }}^z(v,Tv),{{\xi }_{\lambda }}^z(x_{n+1},Tv),{{\xi }_{\lambda }}^z(v,T{x}_{n+1})\}]{\xi }_{\lambda }(x_{n+1},v)$$

for all $\lambda >0$ and all $n\in IN$.

By taking the limit as $n\to \infty$ in the above inequality, we get

$$\underset{n\to \infty }{lim}\left([r+lmin\{{{\xi }_{\lambda }}^z(x_{n+1},T{x}_{n+1}),{{\xi }_{\lambda }}^z(v,Tv),{{\xi }_{\lambda }}^z(x_{n+1},Tv),{{\xi }_{\lambda }}^z(v,T{x}_{n+1})\}]{\xi }_{\lambda }(x_{n+1},v)\right)\le \underset{n\to \infty }{lim}{\xi }_{\lambda }(x_n,z_0)$$

for all $\lambda >0$.

It follows that

$$\begin{array}{c}\underset{n\to \infty }{lim}[r+lmin\{{{\xi }_{\lambda }}^z(x_{n+1},T{x}_{n+1}),{{\xi }_{\lambda }}^z(v,Tv),{{\xi }_{\lambda }}^z(x_{n+1},Tv),{{\xi }_{\lambda }}^z(v,T{x}_{n+1})\}]\underset{n\to \infty }{lim}{\xi }_{\lambda }(x_{n+1},v)\hfill \\ \le \underset{n\to \infty }{lim}{\xi }_{\lambda }(x_n,z_0)\hfill \end{array}$$

for all $\lambda >0$.

Thus, we have $0\le r{lim}_{n\to \infty }{\xi }_{\lambda }(x_{n+1},v)\le {lim}_{n\to \infty }{\xi }_{\lambda }(x_n,z_0)=0$ for all $\lambda >0$ since ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,z_0)={\xi }_{\lambda }(z_0,z_0)={lim}_{n,m\to \infty }{\xi }_{\lambda }(x_n,x_m)=0$ for all $\lambda >0$, which implies $r{lim}_{n\to \infty }{\xi }_{\lambda }(x_{n+1},v)=0$ for all $\lambda >0$. It follows that ${lim}_{n\to \infty }{\xi }_{\lambda }(x_{n+1},v)=0$ for all $\lambda >0$ since $r>s\ge 1$. By Proposition 3, v is unique. Also, since ${lim}_{n\to \infty }{\xi }_{\lambda }(x_n,z_0)=0$, $z_0$ is unique again from Proposition 3, that is why we have $v=z_0$. It follows that $T{z}_0=z_0$ since $Tv=z_0$. Thus, $z_0$ is a fixed point of T. Next, we prove that this fixed point $z_0$ is unique.

Suppose that $y_0$ is another fixed point of T such that $z_0\ne y_0$. Therefore, we have $T{y}_0=y_0$. Thus, from (1), we have

$${\xi }_{\lambda }(T{z}_0,T{y}_0)\ge r{\xi }_{\lambda }(z_0,y_0)$$

for all $\lambda >0$, which implies ${\xi }_{\lambda }(z_0,y_0)\ge r{\xi }_{\lambda }(z_0,y_0)$ for all $\lambda >0$.

It follows that $0\ge (r-1){\xi }_{\lambda }(z_0,y_0)$ for all $\lambda >0$.

Since $r>s\ge 1$, which implies $r>1$, we have $r-1>0$. That is why we have ${\xi }_{\lambda }(z_0,y_0)=0$ for all $\lambda >0$. Thus, we get $z_0=y_0$ from condition $\left(MbML1\right)$.

If we take $l=0$ in Theorem 2, then we deduce the following corollary. □

Corollary 1.

Let $(X,\xi ,s)$ be a modular b-metric-like space such that ${X_{\xi }}^{fin}$ is ξ-complete. Suppose that the mapping $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ is onto and satisfies ${\xi }_{\lambda }(Tx,Ty)\ge r{\xi }_{\lambda }(x,y)$ for all $x,y\in {X_{\xi }}^{fin}$ and all $\lambda >0$, where $r>s$. Then, T has a unique fixed point.

Example 7.

Let $X=[0,\infty )$. Define the function $\xi :(0,\infty )\times X\times X\to [0,\infty ]$ by ${\xi }_{\lambda }(x,y)={\displaystyle \frac{d(x,y)}{\lambda }}$ for all $\lambda >0$ such that $d(x,y)={(x+y)}^2$ and $x,y\in X=[0,\infty )$. Then, $(X,\xi ,2)$ is a modular b-metric-like space such that ${X_{\xi }}^{fin}$ is ξ-complete since ${X_{\xi }}^{fin}=X$. Let $T:{X_{\xi }}^{fin}\to {X_{\xi }}^{fin}$ be defined by

$${\displaystyle Tx=\left\{\begin{array}{cc}4x,\hfill & x\in [0,1),\hfill \\ 3x+2,\hfill & x\in [1,2),\hfill \\ 6x+1,\hfill & x\in [2,\infty ).\hfill \end{array}\right.}$$

Clearly, T is an onto mapping. Now, we consider the following cases:

∗ Let $x,y\in [0,1)$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(4x,4y)}{\lambda }}={\displaystyle \frac{{(4x+4y)}^2}{\lambda }}={\displaystyle \frac{16{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

∗ Let $x,y\in [1,2)$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(3x+2,3y+2)}{\lambda }}={\displaystyle \frac{{(3x+2+3y+2)}^2}{\lambda }}={\displaystyle \frac{{(3x+3y+4)}^2}{\lambda }}\ge {\displaystyle \frac{{(3x+3y)}^2}{\lambda }}={\displaystyle \frac{9{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

∗ Let $x,y\in [2,\infty )$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(6x+1,6y+1)}{\lambda }}={\displaystyle \frac{{(6x+1+6y+1)}^2}{\lambda }}={\displaystyle \frac{{(6x+6y+2)}^2}{\lambda }}\ge {\displaystyle \frac{{(6x+6y)}^2}{\lambda }}={\displaystyle \frac{36{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

∗ Let $x\in [0,1)$ and $y\in [1,2)$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(4x,3y+2)}{\lambda }}={\displaystyle \frac{{(4x+3y+2)}^2}{\lambda }}\ge {\displaystyle \frac{{(3x+3y)}^2}{\lambda }}={\displaystyle \frac{9{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

∗ Let $x\in [0,1)$ and $y\in [2,\infty )$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(4x,6y+1)}{\lambda }}={\displaystyle \frac{{(4x+6y+1)}^2}{\lambda }}\ge {\displaystyle \frac{{(4x+4y)}^2}{\lambda }}={\displaystyle \frac{16{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

∗ Let $x\in [1,2)$ and $y\in [2,\infty )$. Then, ${\xi }_{\lambda }(Tx,Ty)={\displaystyle \frac{d(3x+2,6y+1)}{\lambda }}={\displaystyle \frac{{(3x+2+6y+1)}^2}{\lambda }}={\displaystyle \frac{{(3x+6y+3)}^2}{\lambda }}\ge {\displaystyle \frac{{(3x+3y)}^2}{\lambda }}={\displaystyle \frac{9{(x+y)}^2}{\lambda }}\ge {\displaystyle \frac{3{(x+y)}^2}{\lambda }}=3{\xi }_{\lambda }(x,y)$ for all $\lambda >0$.

That is, ${\xi }_{\lambda }(Tx,Ty)\ge r{\xi }_{\lambda }(x,y)$ for all $x,y\in {X_{\xi }}^{fin}$ and all $\lambda >0$, where $r=3>2=s$. The conditions of Corollary 1 are satisfied, and T has a unique fixed point $x_0=0$.

5. An Application to an Integral Equation

In this section, we investigate the existence of a solution for an integral equation by using Theorem 1.

Consider the following integral equation:

$$\varpi \left(s\right)={\int }_0^L\varsigma (s,q,\varpi \left(q\right))dq,$$

(2)

where $L>0$ and $\varsigma :[0,L]\times [0,L]\times IR\to IR$.

Let $\aleph =C[0,L]$ be the set of all continuous real-valued functions defined on $[0,L]$. Consider the modular b-metric-like given as ${\xi }_{\lambda }(\varpi \left(s\right),\phi \left(s\right))={\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda }}$ for all $\lambda >0$ and all $\varpi ,\phi \in \aleph$. Clearly, $(\aleph ,\xi ,2)$ is modular b-metric-like space such that ${{\aleph }_{\xi }}^{fin}$ is $\xi$-complete since $\aleph ={{\aleph }_{\xi }}^{fin}$.

Let $\mathsf{\Psi }\varpi \left(s\right)={\int }_0^L\varsigma (s,q,\varpi \left(q\right))dq$ for all $\varpi \in \aleph$ and $s\in [0,L]$. Observe that the existence of a solution of (2) is equivalent to the existence of a fixed point of $\mathsf{\Psi }$.

Theorem 3.

Suppose that the following conditions hold. Then, considering the above, the Integral Equation $\left(2\right)$ has a unique solution:

(1) $\varsigma :[0,L]\times [0,L]\times IR\to IR$ is continuous.

(2) There is a continuous function $\delta :[0,L]\times [0,L]\to I{R}^{+}$ for all $s,q\in [0,L]$ such that

$|\varsigma (s,q,\varpi \left(q\right))|+|\varsigma (s,q,\phi \left(q\right))|\le {\vartheta }^{{\displaystyle \frac{1}{2}}}\delta (s,q)(|\varpi \left(s\right)|+|\phi \left(s\right)|)$ where $\vartheta \in (0,1)$.

(3) ${sup}_{s\in [0,L]}{\int }_0^L\delta (s,q)dq\le 1$.

Proof. 

For all $s\in [0,L]$, we have

$$\begin{array}{ccc}\hfill {\displaystyle \frac{(|\mathsf{\Psi }\varpi (s)|+{|\mathsf{\Psi }\phi (s)|)}^2}{\lambda }}& =& {\displaystyle \frac{(|{\int }_0^L\varsigma (s,q,\varpi (q))dq|+|{\int }_0^L\varsigma (s,q,\phi (q))dq{|)}^2}{\lambda }}\hfill \\ & \le & {\displaystyle \frac{({\int }_0^L|\varsigma (s,q,\varpi (q))|dq+{\int }_0^L{|\varsigma (s,q,\phi (q))|dq)}^2}{\lambda }}\hfill \\ & =& {\displaystyle \frac{({\int }_0^L|\varsigma (s,q,\varpi (q))|+{|\varsigma (s,q,\phi (q))|dq)}^2}{\lambda }}\hfill \\ & \le & {\displaystyle \frac{({\int }_0^L{\vartheta }^{\frac{1}{2}}\delta (s,q)(|\varpi (s)|+{|\phi (s)|)dq)}^2}{\lambda }}\hfill \\ & =& {\displaystyle \frac{({\int }_0^L{\vartheta }^{\frac{1}{2}}\delta (s,q)((|\varpi (s)|+{|\phi (s)|)}^2{{)}^{\frac{1}{2}}dq)}^2}{\lambda }}\hfill \\ & =& {\displaystyle \frac{\vartheta (|\varpi (s)|+{|\phi (s)|)}^2{({\int }_0^L\delta (s,q)dq)}^2}{\lambda }}\hfill \\ & \le & {\displaystyle \frac{\vartheta (|\varpi (s)|+{|\phi (s)|)}^2{({sup}_{s\in [0,L]}{\int }_0^L\delta (s,q)dq)}^2}{\lambda }}\hfill \\ & \le & \vartheta {\displaystyle \frac{(|\varpi (s)|+{|\phi (s)|)}^2}{\lambda }}.\hfill \end{array}$$

Then, for all $s\in [0,L]$, we have

$${\displaystyle \frac{(|\mathsf{\Psi }\varpi \left(s\right)|+{|\mathsf{\Psi }\phi \left(s\right)|)}^2}{\lambda }}\le \underset{s\in [0,L]}{\max }\vartheta {\displaystyle \frac{(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda }}.$$

It follows that

$$\underset{s\in [0,L]}{\max }{\displaystyle \frac{(|\mathsf{\Psi }\varpi \left(s\right)|+{|\mathsf{\Psi }\phi \left(s\right)|)}^2}{\lambda }}\le \underset{s\in [0,L]}{\max }\vartheta {\displaystyle \frac{(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda }}.$$

Hence,

$${\displaystyle \frac{{\max }_{s\in [0,L]}(|\mathsf{\Psi }\varpi \left(s\right)|+{|\mathsf{\Psi }\phi \left(s\right)|)}^2}{\lambda }}\le \vartheta {\displaystyle \frac{{\max }_{s\in [0,L]}(|\varpi \left(s\right)|+{|\phi \left(s\right)|)}^2}{\lambda }}.$$

Thus, we have

$${\xi }_{\lambda }(\mathsf{\Psi }\varpi \left(s\right),\mathsf{\Psi }\phi \left(s\right))\le \vartheta {\xi }_{\lambda }(\varpi \left(s\right),\phi \left(s\right)).$$

Also, observe that all conditions of Theorem 1 are satisfied. Therefore, the operator $\mathsf{\Psi }$ has a unique fixed point. This means that the Integral Equation $\left(2\right)$ has a unique solution. □

Example 8.

Consider the integral equation below.

$$\varpi \left(s\right)={\displaystyle \frac{1}{3}}{\int }_0^{1}q\varpi \left(q\right)dq$$

(3)

Then, it has a solution in ℵ.

Let $\mathsf{\Psi }:\aleph \to \aleph$ be defined by $\mathsf{\Psi }\varpi \left(s\right)={\displaystyle \frac{1}{3}}{\int }_0^{1}q\varpi \left(q\right)dq$. By setting $\varsigma (s,q,\varpi \left(q\right))={\displaystyle \frac{1}{3}}q\varpi \left(q\right)$ in Theorem 3, we get

(1) $\varsigma :[0,1]\times [0,1]\times IR\to IR$ is continuous.

(2) There is a continuous function $\delta (s,q)=q$ for all $s,q\in [0,1]$ such that

$$\begin{array}{ccc}\hfill |\varsigma (s,q,\varpi \left(q\right))|+|\varsigma (s,q,\phi \left(q\right))|& =& |{\displaystyle \frac{1}{3}}q\varpi \left(q\right)|+|{\displaystyle \frac{1}{3}}q\phi \left(q\right)|\hfill \\ & =& {\displaystyle \frac{1}{3}}q(|\varpi \left(q\right)|+|\phi \left(q\right)|)\hfill \\ & \le & {\displaystyle \frac{1}{2}}q(|\varpi \left(q\right)|+|\phi \left(q\right)|)\hfill \\ & =& {\left({\displaystyle \frac{1}{4}}\right)}^{{\displaystyle \frac{1}{2}}}q(|\varpi \left(q\right)|+|\phi \left(q\right)|)\hfill \\ & =& {\vartheta }^{{\displaystyle \frac{1}{2}}}\delta (s,q)(|\varpi \left(q\right)|+|\phi \left(q\right)|)\hfill \end{array}$$

where $\vartheta ={\displaystyle \frac{1}{4}}\in (0,1)$.

(3) ${sup}_{s\in [0,1]}{\int }_0^1\delta (s,q)dq={sup}_{s\in [0,1]}{\int }_0^{1}qdq\le 1$.

Hence, all conditions of Theorem 3 are satisfied. Therefore, the problem $\left(3\right)$ has a solution in ℵ.

6. Conclusions

Fixed point results are important to solve many mathematical problems, such as differential equations, integral equations, and systems of linear equations. That is why we provided some fixed point results on a new space called a modular b-metric-like space and an application of these results to an integral equation. Our work is useful from a theoretical and applied perspective, as the result of this paper enables the further development of fixed point theory and its application. Also, our results may provide motivation for researchers to improve fixed point theory by working in this new space. New contraction mappings can be defined on this new space; thus, different application areas can be found.