Fixed Points of Contractive Mappings in 𝑏-Metric-Like Spaces

Abstract

We discuss topological structure of 𝑏-metric-like spaces and demonstrate a fundamental lemma for the convergence of sequences. As an application we prove certain fixed point results in the setup of such spaces for different types of contractive mappings. Finally, some periodic point results in 𝑏-metric-like spaces are obtained. Two examples are presented in order to verify the effectiveness and applicability of our main results.

1. Introduction

There are a lot of generalizations of the concept of metric space. The concepts of 𝑏-metric space and partial metric space were introduced by Czerwik [1] and Matthews [2], respectively. Combining these two notions, Shukla [3] introduced another generalization which is called a partial 𝑏-metric space.

On the other hand, Amini-Harandi [4] introduced a new extension of the concept of partial metric space, called a metric-like space. The concept of 𝑏-metric-like space which generalizes the notions of partial metric space, metric-like space, and 𝑏-metric space was introduced by Alghamdi et al. in [5]. They established the existence and uniqueness of fixed points in a 𝑏-metric-like space as well as in a partially ordered 𝑏-metric-like space. In addition, as an application, they derived some new fixed point and coupled fixed point results in partial metric spaces, metric-like spaces, and 𝑏-metric spaces (see also [6–14]).

The aim of this paper is to examine more closely the topological structure of these spaces. In this context, we demonstrate a fundamental lemma for the convergence of sequences in 𝑏-metric-like spaces and by using it we prove some fixed point results in the setup of such spaces. Finally, some periodic point results in 𝑏-metric-like spaces are obtained. Two examples are presented in order to verify the effectiveness and applicability of our main results.

2. Preliminaries

2.1. Definitions and Basic Properties of Certain Types of Spaces

To clarify the issue, we first recall definitions of 𝑏-metric, partial metric, partial 𝑏-metric, and metric-like spaces.

Definition 1 (see [1]). Let 𝑋 be a nonempty set and 𝑠≥1 a given real number. A function 𝑑∶𝑋×𝑋→𝑅⁺ is a 𝑏-metric if, for all 𝑥,𝑦,𝑧∈𝑋, the following conditions are satisfied:(b₁)𝑑(𝑥,𝑦)=0 if and only if 𝑥=𝑦,(b₂)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥),(b₃)𝑑(𝑥,𝑦)≤𝑠[𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)].
The pair (𝑋,𝑑) is called a 𝑏-metric space with coefficient 𝑠.

Definition 2 (see [2]). A partial metric on a nonempty set 𝑋 is a mapping 𝑝∶𝑋×𝑋→ℝ⁺ such that for all 𝑥,𝑦,𝑧∈𝑋(p₁)𝑥=𝑦 if and only if 𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦),(p₂)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦),(p₃)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥),(p₄)𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑧)+𝑝(𝑧,𝑦)−𝑝(𝑧,𝑧).
A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is a nonempty set and 𝑝 is a partial metric on 𝑋.

Definition 3 (see [3]). A partial 𝑏-metric on a nonempty set 𝑋 is a mapping 𝑝_𝑏∶𝑋×𝑋→ℝ⁺ such that for some real number 𝑠≥1 and all 𝑥,𝑦,𝑧∈𝑋(p_𝑏1)𝑥=𝑦 if and only if 𝑝_𝑏(𝑥,𝑥)=𝑝_𝑏(𝑥,𝑦)=𝑝_𝑏(𝑦,𝑦),(p_𝑏2)𝑝_𝑏(𝑥,𝑥)≤𝑝_𝑏(𝑥,𝑦),(p_𝑏3)𝑝_𝑏(𝑥,𝑦)=𝑝_𝑏(𝑦,𝑥),(p_𝑏4)𝑝_𝑏(𝑥,𝑦)≤𝑠[𝑝_𝑏(𝑥,𝑧)+𝑝_𝑏(𝑧,𝑦)]−𝑝_𝑏(𝑧,𝑧).
A partial 𝑏-metric space is a pair (𝑋,𝑝_𝑏) such that 𝑋 is a nonempty set and 𝑝_𝑏 is a partial 𝑏-metric on 𝑋. The number 𝑠 is called the coefficient of (𝑋,𝑝_𝑏).

Definition 4 (see [4]). A metric-like on a nonempty set 𝑋 is a mapping 𝜎∶𝑋×𝑋→ℝ⁺ such that for all 𝑥,𝑦,𝑧∈𝑋(𝜎1)𝜎(𝑥,𝑦)=0 implies 𝑥=𝑦,(𝜎2)𝜎(𝑥,𝑦)=𝜎(𝑦,𝑥),(𝜎3)𝜎(𝑥,𝑦)≤𝜎(𝑥,𝑧)+𝜎(𝑧,𝑦).
The pair (𝑋,𝜎) is called a metric-like space.

Every partial metric space is a metric-like space. Below we give some other examples of metric-like spaces.

Example 5 (see [15]). Let 𝑋=[0,1]. Then the mapping 𝜎₁∶𝑋×𝑋→ℝ⁺ defined by 𝜎₁(𝑥,𝑦)=𝑥+𝑦−𝑥𝑦 is a metric-like on 𝑋.

Example 6 (see [15]). Let 𝑋=ℝ; then the mappings 𝜎_𝑖∶𝑋×𝑋→ℝ⁺ (𝑖∈{2,3,4}), defined by 𝜎₂=|𝑥|+𝑦𝜎𝑥,𝑦+𝑎,₃=||+,𝜎𝑥,𝑦𝑥−𝑏𝑦−𝑏₄𝑥,𝑦=𝑥²+𝑦², (1) are metric-like on 𝑋, where 𝑎≥0 and 𝑏∈ℝ.

Definition 7 (see [5]). Let 𝑋 be a nonempty set and 𝑠≥1 a given real number. A function 𝜎_𝑏∶𝑋×𝑋→ℝ⁺ is 𝑏-metric-like if, for all 𝑥,𝑦,𝑧∈𝑋, the following conditions are satisfied:(𝜎_𝑏1)𝜎_𝑏(𝑥,𝑦)=0 implies 𝑥=𝑦,(𝜎_𝑏2)𝜎_𝑏(𝑥,𝑦)=𝜎_𝑏(𝑦,𝑥),(𝜎_𝑏3)𝜎_𝑏(𝑥,𝑦)≤𝑠[𝜎_𝑏(𝑥,𝑧)+𝜎_𝑏(𝑧,𝑦)].
A 𝑏-metric-like space is a pair (𝑋,𝜎_𝑏) such that 𝑋 is a nonempty set and 𝜎_𝑏 is 𝑏-metric-like on 𝑋. The number 𝑠 is called the coefficient of (𝑋,𝜎_𝑏).

In a 𝑏-metric-like space (𝑋,𝜎_𝑏) if 𝑥,𝑦∈𝑋 and 𝜎_𝑏(𝑥,𝑦)=0, then 𝑥=𝑦, but the converse may not be true and 𝜎_𝑏(𝑥,𝑥) may be positive for 𝑥∈𝑋. It is clear that every partial 𝑏-metric space is a 𝑏-metric-like space with the same coefficient 𝑠 and every 𝑏-metric space is also a 𝑏-metric-like space with the same coefficient 𝑠. However, the converses of these facts need not hold.

Example 8. Let 𝑋=ℝ⁺, 𝑝>1 a constant, and 𝜎_𝑏∶𝑋×𝑋→ℝ⁺ be defined by 𝜎_𝑏𝑥,𝑦=𝑥+𝑦^𝑝,∀𝑥,𝑦∈𝑋. (2) Then (𝑋,𝜎_𝑏) is a 𝑏-metric-like space with coefficient 𝑠=2^𝑝−1, but it is not a partial 𝑏-metric space. Indeed, for any 0<𝑦<𝑥 we have 𝜎_𝑏(𝑥,𝑥)=(𝑥+𝑥)^𝑝>(𝑥+𝑦)^𝑝=𝜎_𝑏(𝑥,𝑦), so (p_𝑏2) of Definition 3 is not satisfied.

The following propositions help us to construct some more examples of 𝑏-metric-like spaces.

Proposition 9. Let (𝑋,𝜎) be a metric-like space and 𝜎_𝑏(𝑥,𝑦)=[𝜎(𝑥,𝑦)]^𝑝, where 𝑝>1 is a real number. Then 𝜎_𝑏 is 𝑏-metric-like with coefficient 𝑠=2^𝑝−1.

Proof. The proof follows from the fact that (𝑎+𝑏)^𝑝≤2^𝑝−1(𝑎^𝑝+𝑏^𝑝), where 𝑎,𝑏∈ℝ⁺.

From the above proposition and Examples 5 and 6 we have the following examples of 𝑏-metric-like spaces.

Example 10. Let 𝑋=[0,1]. Then the mapping 𝜎_𝑏1∶𝑋×𝑋→ℝ⁺ defined by 𝜎_𝑏1(𝑥,𝑦)=(𝑥+𝑦−𝑥𝑦)^𝑝, where 𝑝>1 is a real number, is 𝑏-metric-like on 𝑋 with coefficient 𝑠=2^𝑝−1.

Example 11. Let 𝑋=ℝ. Then the mappings 𝜎_𝑏𝑖∶𝑋×𝑋→ℝ⁺ (𝑖∈{2,3,4}), defined by 𝜎_𝑏2𝑥,𝑦=|𝑥|+𝑦+𝑎^𝑝,𝜎_𝑏3𝑥,𝑦=|𝑥−𝑏|+𝑦−𝑏^𝑝,𝜎_𝑏4𝑥,𝑦=𝑥²+𝑦^2𝑝, (3) are 𝑏-metric-like on 𝑋, where 𝑝>1,𝑎≥0, and 𝑏∈ℝ.

Proposition 12. Let 𝑋 be a nonempty set such that 𝑑 and 𝑝_𝑏 are 𝑏-metric and partial 𝑏-metric, respectively, 𝑠>1, and 𝜎 is a metric-like on 𝑋. Then the mappings 𝜎_𝑏𝑖∶𝑋×𝑋→ℝ⁺ (𝑖∈{5,6,7}), defined by 𝜎_𝑏5𝑥,𝑦=𝑝_𝑏,𝜎𝑥,𝑦+𝑑𝑥,𝑦_𝑏6𝑥,𝑦=𝜎𝑥,𝑦+𝑝_𝑏,𝜎𝑥,𝑦_𝑏7𝑥,𝑦=𝜎𝑥,𝑦+𝑑𝑥,𝑦, (4) for all 𝑥,𝑦∈𝑋 are 𝑏-metric-like on 𝑋.

Proof. Let (𝑋,𝑝_𝑏) be a partial 𝑏-metric space and (𝑋,𝑑) a 𝑏-metric space with 𝑠>1. Then conditions (𝜎_𝑏1), (𝜎_𝑏2), and (𝜎_𝑏3) are obvious for the function 𝜎_𝑏5. For instance, if 𝑥,𝑦,𝑧∈𝑋 are arbitrary then, as 𝑝_𝑏 is partial 𝑏-metric and 𝑑 is 𝑏-metric on 𝑋, we have 𝜎_𝑏5𝑥,𝑦=𝑝_𝑏𝑝𝑥,𝑦+𝑑𝑥,𝑦≤𝑠_𝑏(𝑥,𝑧)+𝑝_𝑏𝑧,𝑦−𝑝_𝑏()𝑑()𝑝𝑧,𝑧+𝑠𝑥,𝑧+𝑑𝑧,𝑦≤𝑠_𝑏(𝑥,𝑧)+𝑑(𝑥,𝑧)+𝑝_𝑏𝜎𝑧,𝑦+𝑑𝑧,𝑦=𝑠_𝑏5(𝑥,𝑧)+𝜎_𝑏5𝑧,𝑦. (5) Therefore, (𝜎_𝑏3) is satisfied and so (𝑋,𝜎_𝑏5) is a 𝑏-metric-like space. Similarly, one can show that (𝑋,𝜎_𝑏6) and (𝑋,𝜎_𝑏7) are 𝑏-metric-like spaces.

From the above proposition and Examples 5 and 6 we have the following examples.

Example 13. Let 𝑋=[0,1]. Then the mapping 𝜎_𝑏7∶𝑋×𝑋→ℝ⁺ defined by 𝜎_𝑏7(𝑥,𝑦)=𝑥+𝑦−𝑥𝑦+|𝑥−𝑦|^𝑝, where 𝑝>1 is a real number, is 𝑏-metric-like on 𝑋 with coefficient 𝑠=2^𝑝−1.

Example 14. Let 𝑋=ℝ. Then the mappings 𝜎_𝑏𝑖∶𝑋×𝑋→ℝ⁺ (𝑖∈{8,9,10}), defined by 𝜎_𝑏8𝑥,𝑦=|𝑥|+𝑦+𝑎+𝑥−𝑦^𝑝,𝜎_𝑏9𝑥,𝑦=|𝑥−𝑏|+𝑦−𝑏+𝑥−𝑦^𝑝,𝜎_𝑏10𝑥,𝑦=𝑥²+𝑦²+𝑥−𝑦^𝑝, (6) are 𝑏-metric-like on 𝑋 with coefficient 𝑠=2^𝑝−1, where 𝑝>1, 𝑎≥0, and 𝑏∈ℝ.

Each 𝑏-metric-like 𝜎_𝑏 on 𝑋 generates a topology 𝜏_𝜎𝑏 on 𝑋 whose base is the family of all open 𝜎_𝑏-balls {𝐵_𝜎𝑏(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝜎𝑏(𝑥,𝜀)={𝑦∈𝑋∶|𝜎_𝑏(𝑥,𝑦)−𝜎_𝑏(𝑥,𝑥)|<𝜀} for all 𝑥∈𝑋 and 𝜀>0.

Now, we define the concepts of Cauchy sequence and convergent sequence in a 𝑏-metric-like space.

Definition 15 (see [5]). Let (𝑋,𝜎_𝑏) be a 𝑏-metric-like space with coefficient 𝑠, and let {𝑥_𝑛} be any sequence in 𝑋 and 𝑥∈𝑋. Then
(i)the sequence {𝑥_𝑛} is said to be convergent to 𝑥 with respect to 𝜏_𝜎𝑏, if lim_𝑛→∞𝜎_𝑏(𝑥_𝑛,𝑥)=𝜎_𝑏(𝑥,𝑥);(ii)the sequence {𝑥_𝑛} is said to be a Cauchy sequence in (𝑋,𝜎_𝑏) if lim_{𝑛,𝑚→∞}𝜎_𝑏(𝑥_𝑛,𝑥_𝑚) exists and is finite;(iii)(𝑋,𝜎_𝑏) is said to be a complete 𝑏-metric-like space if for every Cauchy sequence {𝑥_𝑛} in 𝑋 there exists 𝑥∈𝑋 such that lim_{𝑛,𝑚→∞}𝜎_𝑏𝑥_𝑛,𝑥_𝑚=lim_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑥=𝜎_𝑏(𝑥,𝑥). (7)

It is clear that the limit of a sequence in a 𝑏-metric-like space is usually not unique (since already partial metric spaces share this property).

We start our work by proving the following crucial lemma.

Lemma 16. Let (𝑋,𝜎_𝑏) be a 𝑏-metric-like space with coefficient 𝑠>1, and suppose that {𝑥_𝑛} and {𝑦_𝑛} are convergent to 𝑥 and 𝑦, respectively. Then one has 1𝑠²𝜎_𝑏−1𝑥,𝑦𝑠𝜎_𝑏(𝑥,𝑥)−𝜎_𝑏𝑦,𝑦≤liminf_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑦_𝑛≤limsup_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑦_𝑛≤𝑠𝜎_𝑏(𝑥,𝑥)+𝑠²𝜎_𝑏𝑦,𝑦+𝑠²𝜎_𝑏𝑥,𝑦. (8) In particular, if 𝜎_𝑏(𝑥,𝑦)=0, then one has lim_𝑛→∞𝜎_𝑏(𝑥_𝑛,𝑦_𝑛)=0.
Moreover, for each 𝑧∈𝑋 one has 1𝑠𝜎_𝑏(𝑥,𝑧)−𝜎_𝑏(𝑥,𝑥)≤liminf_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑧≤limsup_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑧≤𝑠𝜎_𝑏(𝑥,𝑧)+𝑠𝜎_𝑏(𝑥,𝑥). (9) In particular, if 𝜎_𝑏(𝑥,𝑥)=0, then 1𝑠𝜎_𝑏(𝑥,𝑧)≤liminf_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑧≤limsup_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑧≤𝑠𝜎_𝑏(𝑥,𝑧). (10)

Proof. Using the triangle inequality in a 𝑏-metric-like space it is easy to see that 𝜎_𝑏𝑥,𝑦≤𝑠𝜎_𝑏𝑥,𝑥_𝑛+𝑠²𝜎_𝑏𝑥_𝑛,𝑦_𝑛+𝑠²𝜎_𝑏𝑦_𝑛,𝜎,𝑦_𝑏𝑥_𝑛,𝑦_𝑛≤𝑠𝜎_𝑏𝑥_𝑛,𝑥+𝑠²𝜎_𝑏𝑥,𝑦+𝑠²𝜎_𝑏𝑦,𝑦_𝑛. (11) Taking the lower limit as 𝑛→∞ in the first inequality and the upper limit as 𝑛→∞ in the second inequality we obtain the first desired result. If 𝜎_𝑏(𝑥,𝑦)=0, then by the triangle inequality we get 𝜎_𝑏(𝑥,𝑥)=0 and 𝜎_𝑏(𝑦,𝑦)=0. Therefore, we have lim_𝑛→∞𝜎_𝑏(𝑥_𝑛,𝑦_𝑛)=0. Similarly, using again the triangle inequality the other assertions follow.

2.2. Contraction Conditions and Fixed Point Results

It is well known that a self-map 𝑓 on a metric space (𝑋,𝑑) is said to be a Banach contraction mapping, if there exists a number 𝑘∈[0,1) such that 𝑑𝑓𝑥,𝑓𝑦≤𝑘𝑑𝑥,𝑦 (12) for all 𝑥,𝑦∈𝑋. A mapping 𝑓∶𝑋→𝑋 is called a quasicontraction if for some constant 𝛼∈[0,1) and for every 𝑥,𝑦∈𝑋𝑑𝑑,𝑑.𝑓𝑥,𝑓𝑦≤𝛼max𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦𝑥,𝑓𝑦,𝑑𝑦,𝑓𝑥 (13) This concept was introduced and studied by Ćirić in 1974 [16]. A result of Ćirić shows that every quasicontraction in a complete metric space 𝑓 has a unique fixed point.

The existence of fixed points in partially ordered metric spaces was first investigated in 2004 by Ran and Reurings [17] and then by Nieto and Rodríguez-López [18].

In this paper, we establish some fixed point theorems for quasicontractive type mappings in a partially ordered complete 𝑏-metric-like space. We investigate also the so-called 𝑃-property for mappings in such spaces.

3. Main Results

3.1. Fixed Points of Quasicontraction-Type Mappings

Throughout this paper, let (𝑋,⪯) be a partially ordered set, and let (𝑋,𝜎_𝑏) be a 𝑏-metric-like space (we will say, for short, that (𝑋,⪯,𝜎_𝑏) is a partially ordered 𝑏-metric-like space). Further, let 𝐹(𝑓)={𝑥∈𝑋∶𝑓𝑥=𝑥} be the fixed point set of 𝑓, (𝐿𝐹)_𝑓={𝑥∈𝑋∶𝑥⪯𝑓𝑥} the lower fixed point set of 𝑓, and 𝑀𝑥,𝑦=max𝜎_𝑏𝑥,𝑦,𝜎_𝑏𝑥,𝑓𝑥,𝜎_𝑏,𝜎𝑦,𝑓𝑦_𝑏𝑥,𝑓𝑦,𝜎_𝑏𝑦,𝑓𝑥. (14) In this section, we obtain some fixed point results for quasicontractions defined on a partially ordered complete 𝑏-metric-like space.

We will also make use of the following notion.

Definition 17. An ordered 𝑏-metric-like space (𝑋,⪯,𝜎_𝑏) is said to have the sequential limit comparison property if for every nondecreasing sequence (nonincreasing sequence) {𝑥_𝑛}_𝑛∈ℕ in 𝑋, 𝑥_𝑛→𝑥 implies that 𝑥_𝑛⪯𝑥(𝑥⪯𝑥_𝑛), for all 𝑛∈ℕ.

Theorem 18. Let (𝑋,⪯,𝜎_𝑏) be a complete partially ordered 𝑏-metric-like space. If 𝑓∶𝑋→𝑋 is a nondecreasing map such that, for all elements 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, 𝜎_𝑏𝑓𝑥,𝑓𝑦≤𝛼𝑀𝑥,𝑦, (15) where 𝛼∈[0,1/2𝑠²), then 𝐹(𝑓)≠∅ provided that there exists an 𝑥₀∈(𝐿𝐹)_𝑓 and one of the following two conditions is satisfied:(a)𝑓 is a continuous self-map on 𝑋,(b)(𝑋,⪯,𝜎_𝑏) has the sequential limit comparison property.
Moreover, 𝑓 has a unique fixed point if and only if each pair of fixed points of 𝑓 is comparable.

Proof. Since 𝑥₀∈(𝐿𝐹)_𝑓 and 𝑓 is nondecreasing, therefore 𝑓^𝑛𝑥₀⪯𝑓^𝑛+1𝑥₀ for each 𝑛∈ℕ. Define a sequence {𝑥_𝑛} in 𝑋 with 𝑥_𝑛=𝑓^𝑛𝑥₀ and so 𝑥_𝑛+1=𝑓𝑥_𝑛 for all 𝑛∈ℕ. If there exists a positive integer 𝑛 such that 𝑥_𝑛=𝑥_𝑛+1, then 𝑓^𝑛𝑥₀=𝑓^𝑛+1𝑥₀=𝑓𝑓^𝑛𝑥₀ which implies that 𝑓^𝑛𝑥₀ is a fixed point of 𝑓. Assume that 𝑥_𝑛≠𝑥_𝑛+1 for every positive integer 𝑛. Since 𝑥_𝑛−1⪯𝑥_𝑛, therefore by replacing 𝑥 by 𝑥_𝑛−1 and 𝑦 by 𝑥_𝑛 in (15), we have 𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1=𝜎_𝑏𝑓𝑥_𝑛−1,𝑓𝑥_𝑛≤𝛼max𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛,𝜎_𝑏𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝜎_𝑏𝑥_𝑛,𝑓𝑥_𝑛,𝜎_𝑏𝑥_𝑛−1,𝑓𝑥_𝑛,𝜎_𝑏𝑥_𝑛,𝑓𝑥_𝑛−1=𝛼max𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛,𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛,𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1,𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+1,𝜎_𝑏𝑥_𝑛,𝑥_𝑛≤𝛼max𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛,𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1,𝑠𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1,2𝑠𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛. (16) If, for some 𝑛, 𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)>𝜎_𝑏(𝑥_𝑛−1,𝑥_𝑛)>0, then according to the above inequality 𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)≤2𝑠𝛼𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)<𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1) which is a contradiction.
Hence, for all 𝑛, 𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)≤𝜎_𝑏(𝑥_𝑛−1,𝑥_𝑛) and therefore 𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛≤ℎ𝜎_𝑏𝑥_𝑛,𝑥_𝑛−1, (17) where ℎ=2𝛼𝑠. Obviously, 0≤ℎ<1. Repeating the above process, we get 𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛≤ℎ𝜎_𝑏𝑥_𝑛,𝑥_𝑛−1≤⋯≤ℎ^𝑛𝜎_𝑏𝑥₁,𝑥₀, (18) for all 𝑛≥1, and so, for 𝑚>𝑛, we have 𝜎_𝑏𝑥_𝑛,𝑥_𝑚≤𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1+𝑠²𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛+2+⋯+𝑠^𝑚−𝑛𝜎_𝑏𝑥_𝑚−1,𝑥_𝑚≤𝑠ℎ^𝑛𝜎_𝑏𝑥₀,𝑥₁+𝑠²ℎ^𝑛+1𝜎_𝑏𝑥₀,𝑥₁+⋯+𝑠^𝑚−𝑛ℎ^𝑚−1𝜎_𝑏𝑥₀,𝑥₁=𝑠ℎ^𝑛1+𝑠ℎ+⋯+𝑠^{𝑚−𝑛−1}ℎ^{𝑚−𝑛−1}𝜎_𝑏𝑥₀,𝑥₁≤𝑠ℎ^𝑛𝜎1−𝑠ℎ^𝑏𝑥₀,𝑥₁. (19) Since, by assumption, ℎ<1/𝑠, it follows that lim_{𝑚,𝑛→∞}𝜎_𝑏(𝑥_𝑛,𝑥_𝑚)=0. Since 𝑋 is complete, there exists an element 𝑢∈𝑋 such that 0=lim_{𝑛,𝑚→∞}𝜎_𝑏𝑥_𝑛,𝑥_𝑚=lim_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑢=𝜎_𝑏(𝑢,𝑢). (20) If 𝑓 is a continuous self-map on 𝑋, then 𝑓𝑢=𝑢. Indeed, by the triangle inequality we have 𝜎_𝑏𝑓𝑢,𝑢≤𝑠𝜎_𝑏𝑓𝑢,𝑥_𝑛+1+𝜎_𝑏𝑥_𝑛+1𝜎,𝑢=𝑠_𝑏𝑓𝑢,𝑓𝑥_𝑛+𝜎_𝑏𝑥_𝑛+1,𝑢. (21) Taking the limit as 𝑛→∞ in the above inequality, the desired result is obtained.
If condition (b) is fulfilled then 𝑥_𝑛=𝑓^𝑛𝑥₀⪯𝑢 for all 𝑛∈ℕ, and it follows that 𝜎_𝑏𝑓𝑥_𝑛,𝑓𝑢≤𝛼max𝜎_𝑏𝑥_𝑛,𝑢,𝜎_𝑏𝑥_𝑛,𝑓𝑥_𝑛,𝜎_𝑏,𝜎𝑢,𝑓𝑢_𝑏𝑥_𝑛,𝑓𝑢,𝜎_𝑏𝑢,𝑓𝑥_𝑛=𝛼max𝜎_𝑏𝑥_𝑛,𝑢,𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1,𝜎_𝑏,𝜎𝑢,𝑓𝑢_𝑏𝑥_𝑛,𝑓𝑢,𝜎_𝑏𝑢,𝑥_𝑛+1. (22) On taking the upper limit as 𝑛→∞, (using Lemma 16, as 𝜎_𝑏(𝑢,𝑢)=0) it follows that 1𝑠𝜎_𝑏𝑢,𝑓𝑢≤𝛼𝑠𝜎_𝑏𝑢,𝑓𝑢, (23) and hence 𝜎_𝑏(𝑢,𝑓𝑢)≤𝛼𝑠²𝜎_𝑏(𝑢,𝑓𝑢), or equivalently, 𝑓𝑢=𝑢.
Suppose that the fixed points of 𝑓 are comparable. Let 𝑤 be another fixed point of 𝑓 such that 𝑤≠𝑢. Assume, for example, that 𝑢⪯𝑤. Using (15), we obtain that 𝜎_𝑏(𝑢,𝑤)=𝜎_𝑏𝜎𝑓𝑢,𝑓𝑤≤𝛼max_𝑏(𝑢,𝑤),𝜎_𝑏𝑢,𝑓𝑢,𝜎_𝑏,𝜎𝑤,𝑓𝑤_𝑏𝑢,𝑓𝑤,𝜎_𝑏𝜎𝑤,𝑓𝑢=𝛼max_𝑏(𝑢,𝑤),𝜎_𝑏(𝑢,𝑢),𝜎_𝑏(),𝜎𝑤,𝑤_𝑏(𝑢,𝑤),𝜎_𝑏(𝑤,𝑢)≤2𝑠𝛼𝜎_𝑏(𝑢,𝑤), (24) which further implies that 𝑢=𝑤. The converse is trivial.

Example 19. Let 𝑋=[0,1] be endowed with the usual order. Define 𝜎_𝑏∶𝑋×𝑋→ℝ⁺ by 𝜎_𝑏𝑥,𝑦=𝑥²+𝑦²+𝑥−𝑦² (25) for all 𝑥,𝑦∈𝑋. Then (𝑋,𝜎_𝑏) is a complete 𝑏-metric-like space with coefficient 𝑠=2 (see Example 14).
Let 𝑓∶𝑋→𝑋 be defined by 𝑓𝑥=ln(1+𝑥/4). It is easy to see that 𝑓 is a nondecreasing and continuous self-map on 𝑋, and 0∈(𝐿𝐹)_𝑓. Using the Mean Value Theorem for any 𝑥,𝑦∈𝑋 with 𝑥≤𝑦 and that 𝑓𝑥≤𝑥/4, we have 𝜎_𝑏𝑓𝑥,𝑓𝑦=𝑓²𝑥+𝑓²𝑦+𝑓𝑦−𝑓𝑥²=𝑥ln1+4²+𝑦ln1+4²+𝑦ln1+4𝑥−ln1+4²≤𝑥4²+𝑦4²+𝑦4−𝑥4²=1𝑥16²+𝑦²+𝑥−𝑦²=1𝜎16^𝑏≤1𝑥,𝑦𝑀.16𝑥,𝑦 (26) Thus (15) is satisfied with 𝛼=1/16∈[0,1/8). Thus all conditions of Theorem 18 are satisfied. Moreover, 0 is the unique fixed point of 𝑓.

Theorem 20. Let (𝑋,⪯,𝜎_𝑏) be a partially ordered complete 𝑏-metric-like space with coefficient 𝑠>1. If a nondecreasing map 𝑓∶𝑋→𝑋 satisfies 𝑠𝜎_𝑏𝑓𝑥,𝑓𝑦≤𝑀𝑥,𝑦−𝜑𝑚𝑥,𝑦, (27) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, where 𝑀=𝜎𝑥,𝑦^𝑏𝑥,𝑓𝑦+𝜎_𝑏𝑦,𝑓𝑥,𝑚𝜎4𝑠𝑥,𝑦=min_𝑏𝑥,𝑓𝑦,𝜎_𝑏𝑦,𝑓𝑥, (28) and 𝜑∶[0,∞)→[0,∞) is a continuous and nondecreasing function such that 𝜑(𝑡)>0 for all 𝑡∈(0,∞) and 𝜑(0)=0, then 𝐹(𝑓)≠∅ provided that there exists an 𝑥₀∈(𝐿𝐹)_𝑓 and one of the following two conditions is satisfied:(a)𝑓 is a continuous self-map on 𝑋.(b)(𝑋,⪯,𝜎_𝑏) has the sequential limit comparison property.
Moreover, 𝑓 has a unique fixed point if and only if the fixed points of 𝑓 are comparable.

Proof. Define the sequence {𝑥_𝑛} as given in the proof of Theorem 18. If there exists a positive integer 𝑛 such that 𝑥_𝑛=𝑥_𝑛+1, then 𝑥_𝑛 is a fixed point of 𝑓. Assume that 𝑥_𝑛≠𝑥_𝑛+1 for every positive integer 𝑛. Since 𝑥_𝑛−1⪯𝑥_𝑛, therefore by replacing 𝑥 by 𝑥_𝑛−1 and 𝑦 by 𝑥_𝑛 in (27), we obtain 𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1=𝑠𝜎_𝑏𝑓𝑥_𝑛−1,𝑓𝑥_𝑛≤𝑀𝑥_𝑛−1,𝑥_𝑛−𝜑𝑚𝑥_𝑛−1,𝑥_𝑛≤𝜎_𝑏𝑥_𝑛−1,𝑓𝑥_𝑛+𝜎_𝑏𝑥_𝑛,𝑓𝑥_𝑛−1𝑚𝑥4𝑠−𝜑_𝑛−1,𝑥_𝑛=𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+1+𝜎_𝑏𝑥_𝑛,𝑥_𝑛𝑚𝑥4𝑠−𝜑_𝑛−1,𝑥_𝑛≤𝑠𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1+2𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1𝑚𝑥4𝑠−𝜑_𝑛−1,𝑥_𝑛≤𝑠𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1+2𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1.4𝑠 (29) Therefore, for all 𝑛≥0, 𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)≤𝜎_𝑏(𝑥_𝑛,𝑥_𝑛−1) and {𝜎_𝑏(𝑥_𝑛,𝑥_𝑛+1)} is nonincreasing and bounded from below. Hence, there exists 𝑟≥0 such that lim_𝑛→∞𝜎_𝑏(𝑥_𝑛+1,𝑥_𝑛)=𝑟.
From the above argument we have 𝑠𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛≤𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+1+𝜎_𝑏𝑥_𝑛,𝑥_𝑛≤4𝑠𝑠𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1+2𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1.4𝑠 (30) If 𝑛→∞, we get 𝑠𝑟≤lim_𝑛→∞𝜎_𝑏𝑥_𝑛−1,𝑥_𝑛+1+𝜎_𝑏𝑥_𝑛,𝑥_𝑛4𝑠≤𝑟, (31) which is, because of 𝑠>1, possible only if 𝑟=0. So we have lim_𝑛→∞𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛=0. (32)
Next, we show that {𝑥_𝑛} is a Cauchy sequence. If not, then there exists 𝜀>0 for which we can find subsequences {𝑥_𝑚𝑘} and {𝑥_𝑛𝑘} of the sequence {𝑥_𝑛} where 𝑛_𝑘 is the smallest index for which 𝑛_𝑘>𝑚_𝑘>𝑘 with 𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≥𝜀. (33) Then, 𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘−1<𝜀. (34) From (33) and (34), we obtain 𝜀≤𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝑠𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘−1+𝑠𝜎_𝑏𝑥_𝑛𝑘−1,𝑥_𝑛𝑘<𝑠𝜀+𝑠𝜎_𝑏𝑥_𝑛𝑘−1,𝑥_𝑛𝑘. (35) Taking the upper and lower limits as 𝑘→∞, from (32) we conclude that 𝜀≤liminf_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤limsup_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝑠𝜀. (36) Note that 𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1≤𝑠𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑚𝑘+𝑠𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘−1. (37) Using (34) and (32), we get limsup_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1≤𝑠𝜀. (38) On the other hand, 𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝑠𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝑠²𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1+𝑠²𝜎_𝑏𝑥_𝑛𝑘−1,𝑥_𝑛𝑘. (39) Using (36) and (32), we get 𝜀𝑠²≤liminf_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1. (40) From (38) and (40), we have 𝜀𝑠²≤liminf_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1≤limsup_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘−1≤𝑠𝜀. (41) Also, 𝜀≤𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝑠𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝑠𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘. (42) Taking the upper limit as 𝑘→∞, we conclude that 𝜀≤limsup_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘≤limsup_𝑘→∞𝑠𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘, (43) or equivalently, 𝜀𝑠≤limsup_𝑘→∞𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘. (44) As {𝑥_𝑛} is nondecreasing and 𝑚_𝑘<𝑛_𝑘, from (27) we have 𝑠𝜎_𝑏𝑥_𝑚𝑘+1,𝑥_𝑛𝑘=𝑠𝜎_𝑏𝑓𝑥_𝑚𝑘,𝑓𝑥_𝑛𝑘−1≤𝑀𝑥_𝑚𝑘,𝑥_𝑛𝑘−1−𝜑𝑚𝑥_𝑚𝑘,𝑥_𝑛𝑘−1, (45) where 𝑀𝑥_𝑚𝑘,𝑥_𝑛𝑘−1=𝜎_𝑏𝑥_𝑚𝑘,𝑓𝑥_𝑛𝑘−1+𝜎_𝑏𝑥_𝑛𝑘−1,𝑓𝑥_𝑚𝑘=𝜎4𝑠_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘+𝜎_𝑏𝑥_𝑛𝑘−1,𝑥_𝑚𝑘+1,𝑚𝑥4𝑠_𝑚𝑘,𝑥_𝑛𝑘−1=min𝜎_𝑏𝑥_𝑚𝑘,𝑥_𝑛𝑘,𝜎_𝑏𝑥_𝑛𝑘−1,𝑥_𝑚𝑘+1, (46) wherefrom on taking the upper limit as 𝑘→∞, from (36) and (41), we have limsup_𝑘→∞𝑀𝑥_𝑚𝑘,𝑥_𝑛𝑘−1≤𝜀2,liminf_𝑘→∞𝑚𝑥_𝑚𝑘,𝑥_𝑛𝑘−1𝜀≥min𝜀,𝑠²=𝜀𝑠². (47) Thus, from (44) and (45), we have 𝜀𝜀≤2𝜀−𝜑𝑠², (48) which gives 𝜀=0, a contradiction. Hence {𝑥_𝑛} is a Cauchy sequence in 𝑋. Since 𝑋 is complete, there exists an element 𝑢∈𝑋 such that 0=lim_{𝑛,𝑚→∞}𝜎_𝑏𝑥_𝑛,𝑥_𝑚=lim_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑢=𝜎_𝑏(𝑢,𝑢). (49)
If 𝑓 is a continuous self-map on 𝑋, then 𝑓𝑢=𝑢. If assumption (b) is satisfied then 𝑥_𝑛=𝑓^𝑛𝑥₀⪯𝑢 for all 𝑛∈ℕ; it follows that 𝑠𝜎_𝑏𝑓𝑥_𝑛,𝑓𝑢≤𝑀𝑥_𝑛,𝑢−𝜑𝑚𝑥_𝑛,𝑢, (50) where 𝑀𝑥_𝑛=𝜎,𝑢^𝑏𝑥_𝑛,𝑓𝑢+𝜎_𝑏𝑢,𝑥_𝑛+1,𝑚𝑥4𝑠_𝑛,𝑢=min𝜎_𝑏𝑥_𝑛,𝑓𝑢,𝜎_𝑏𝑢,𝑥_𝑛+1. (51) On taking the upper limit as 𝑛→∞ in (50), from Lemma 16 we obtain 1𝑠⋅𝑠𝜎_𝑏𝑢,𝑓𝑢≤𝑠𝜎_𝑏𝑢,𝑓𝑢(0),4𝑠−𝜑 (52) and hence 𝜎_𝑏(𝑢,𝑓𝑢)=0, implying that 𝑢=𝑓𝑢.
Now, suppose that the fixed points of 𝑓 are comparable. Let 𝑤 be another fixed point of 𝑓. We will show that 𝑤=𝑢. If not, then without loss of generality, we assume that 𝑢⪯𝑤. Using (27), we obtain 𝜎_𝑏(𝑢,𝑤)=𝜎_𝑏𝑓𝑢,𝑓𝑤≤𝑀(𝑢,𝑤)−𝜑(𝑚(𝑢,𝑤)), (53) where 𝑀()=𝜎𝑢,𝑤^𝑏𝑢,𝑓𝑤+𝜎_𝑏𝑤,𝑓𝑢=𝜎4𝑠_𝑏(𝑢,𝑤)+𝜎_𝑏(𝑤,𝑢)=𝜎4𝑠^𝑏(𝑢,𝑤).2𝑠 (54) Thus, 𝜎_𝑏(𝑢,𝑤)=𝜎_𝑏≤𝜎𝑓𝑢,𝑓𝑤^𝑏(𝑢,𝑤)𝜎2𝑠−𝜑_𝑏(𝑢,𝑤), (55) which implies that 𝜎_𝑏(𝑢,𝑤)=0, which yields that 𝑢=𝑤. The converse is trivial.

Example 21. Let 𝑋=[0,∞) be endowed with the usual order ≤. Define 𝜎_𝑏∶𝑋×𝑋→ℝ⁺ by 𝜎_𝑏𝑥,𝑦=𝑥²+𝑦²² (56) for all 𝑥,𝑦∈𝑋. Then (𝑋,≤,𝜎_𝑏) is a complete ordered 𝑏-metric-like space with coefficient 𝑠=2 (see Example 11).
Let 𝑓∶𝑋→𝑋 be defined by 𝑓𝑥=(1/4)ln((𝑥²/3)+1). It is easy to see that 𝑓 is a nondecreasing and continuous self-map on 𝑋, and 0∈(𝐿𝐹)_𝑓. Using the Mean Value Theorem for the function 𝑓²𝑥=(1/16)ln(𝑥²/3+1) for any 𝑥,𝑦∈𝑋 with 𝑥≤𝑦, we have 𝑓²𝑦−𝑓²1𝑥=𝑦16ln²3𝑥+1−ln²3≤1+1𝑦48²−𝑥²≤𝑦²−𝑥², (57) which implies that 𝑓²𝑦+𝑥²≤𝑓²𝑥+𝑦². (58) Hence 𝑚𝑥,𝑦=min𝜎_𝑏𝑓𝑦,𝑥,𝜎_𝑏𝑓𝑥,𝑦=𝜎_𝑏𝑓𝑦,𝑥. (59) Now we define 𝜑∶[0,∞)→[0,∞) by 𝜑(𝑡)=𝑡/8, and, for any 𝑥,𝑦∈𝑋 with 𝑥≤𝑦, we have 𝑠𝜎_𝑏𝑓𝑥,𝑓𝑦≤𝑀𝑥,𝑦−𝜑𝑚𝑥,𝑦⟺2𝜎_𝑏≤𝜎𝑓𝑥,𝑓𝑦^𝑏𝑓𝑦,𝑥+𝜎_𝑏𝑓𝑥,𝑦8−𝜎_𝑏𝑓𝑦,𝑥8⟺16𝜎_𝑏𝑓𝑥,𝑓𝑦≤𝜎_𝑏𝑓𝑓𝑥,𝑦⟺16²𝑥+𝑓²𝑦²≤𝑓²𝑥+𝑦²²⟺4𝑓²𝑥+𝑓²𝑦≤𝑓²𝑥+𝑦²⟺3𝑓²𝑥+4𝑓²𝑦≤𝑦²⟸7𝑓²𝑦≤𝑦²⟺𝑓²𝑦𝑦≤²7⟸1𝑦16ln²3≤𝑦+1²,48 (60) since 𝑓²𝑥=(1/16)ln(𝑥²/3+1) is an increasing function and 𝑓²𝑥≤𝑥²/48.
Thus, all conditions of Theorem 20 are satisfied. Moreover, 0 is the unique fixed point of 𝑓.

3.2. Coincidence Points of Four Mappings under Generalized Weakly Contractive Conditions

Definition 22. Let 𝑓 and 𝑔 be two self-maps on partially ordered set 𝑋. A pair (𝑓,𝑔) is said to be(i)weakly increasing if 𝑓𝑥⪯𝑔𝑓𝑥 and 𝑔𝑥⪯𝑓𝑔𝑥, for all 𝑥∈𝑋 [19, 20],(ii)partially weakly increasing if 𝑓𝑥⪯𝑔𝑓𝑥, for all 𝑥∈𝑋 [21].

Let 𝑋 be a nonempty set and 𝑓∶𝑋→𝑋 a given mapping. For every 𝑥∈𝑋, let 𝑓⁻¹(𝑥)={𝑢∈𝑋∶𝑓𝑢=𝑥}.

Definition 23. Let (𝑋,⪯) be a partially ordered set and 𝑓,𝑔,ℎ∶𝑋→𝑋 mappings such that 𝑓𝑋⊆ℎ𝑋 and 𝑔𝑋⊆ℎ𝑋. The ordered pair (𝑓,𝑔) is said to be(a)weakly increasing with respect to ℎ if and only if, for all 𝑥∈𝑋, 𝑓𝑥⪯𝑔𝑦 for all 𝑦∈ℎ⁻¹(𝑓𝑥), and 𝑔𝑥⪯𝑓𝑦 for all 𝑦∈ℎ⁻¹(𝑔𝑥) [22],(b)partially weakly increasing with respect to ℎ if 𝑓𝑥⪯𝑔𝑦, for all 𝑦∈ℎ⁻¹(𝑓𝑥) [23].

Remark 24. In the above definition (i) if 𝑓=𝑔, we say that 𝑓 is weakly increasing (partially weakly increasing) with respect to ℎ and (ii) if ℎ=𝐼 (the identity mapping on 𝑋), then the above definition reduces to the weakly increasing (partially weakly increasing) mapping (see [22, 24]).

The study of unique common fixed points of mappings satisfying weakly contractive conditions has been at the center of vigorous research activity. Motivated by the work in [21, 23–30], we prove some coincidence point results for nonlinear generalized (𝜓,𝜑)-weakly contractive mappings in partially ordered 𝑏-metric-like spaces.

Recall [31] that an altering distance function is a mapping 𝜓∶[0,∞)→[0,∞) which satisfies that(𝜓₁)𝜓(𝑡) is increasing and continuous,(𝜓₂)𝜓(𝑡)=0 if and only if 𝑡=0.

Let (𝑋,⪯,𝜎_𝑏) be an ordered 𝑏-metric-like space and 𝑓,𝑔,𝑅,𝑆∶𝑋→𝑋 four self-mappings. Throughout this subsection, unless otherwise stated, for all 𝑥,𝑦∈𝑋, let 𝑀𝑥,𝑦=max𝜎_𝑏𝑆𝑥,𝑅𝑦,𝜎_𝑏𝑆𝑥,𝑓𝑥,𝜎_𝑏,𝜎𝑅𝑦,𝑔𝑦_𝑏𝑆𝑥,𝑔𝑦+𝜎_𝑏𝑅𝑦,𝑓𝑥.4𝑠 (61) Suppose that 𝑓(𝑋)⊆𝑅(𝑋) and 𝑔(𝑋)⊆𝑆(𝑋); let 𝑥₀ be an arbitrary point of 𝑋. Choose 𝑥₁∈𝑋 such that 𝑓𝑥₀=𝑅𝑥₁ and 𝑥₂∈𝑋 such that 𝑔𝑥₁=𝑆𝑥₂. Continuing in this way, construct a sequence {𝑧_𝑛} defined by 𝑧_2𝑛+1=𝑅𝑥_2𝑛+1=𝑓𝑥_2𝑛 and 𝑧_2𝑛+2=𝑆𝑥_2𝑛+2=𝑔𝑥_2𝑛+1, for all 𝑛≥0. The sequence {𝑧_𝑛} in 𝑋 is said to be a Jungck-type iterative sequence with initial guess 𝑥₀.

Theorem 25. Let (𝑋,⪯,𝜎_𝑏) be a partially ordered 𝑏-metric-like space with sequential limit comparison property, 𝑓,𝑔,𝑅,𝑆∶𝑋→𝑋 four mappings such that 𝑓(𝑋)⊆𝑅(𝑋) and 𝑔(𝑋)⊆𝑆(𝑋), and 𝑅𝑋 and 𝑆𝑋𝜎_𝑏-complete subsets of 𝑋. Suppose that, for comparable elements 𝑆𝑥,𝑅𝑦∈𝑋, one has 𝜓𝑠²𝜎_𝑏𝑓𝑥,𝑔𝑦≤𝜓𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (62) where 𝜓,𝜑∶[0,∞)→[0,∞) are altering distance functions. Then, the pairs (𝑓,𝑆) and (𝑔,𝑅) have a coincidence point 𝑧^∗ in 𝑋 provided that the pairs (𝑓,𝑆) and (𝑔,𝑅) are weakly compatible, and the pairs (𝑓,𝑔) and (𝑔,𝑓) are partially weakly increasing with respect to 𝑅 and 𝑆, respectively.

Proof. Let {𝑧_𝑛} be a Jungck-type iterative sequence with initial guess 𝑥₀ in 𝑋; that is, 𝑧_2𝑛+1=𝑅𝑥_2𝑛+1=𝑓𝑥_2𝑛, 𝑧_2𝑛+2=𝑆𝑥_2𝑛+2=𝑔𝑥_2𝑛+1 for all 𝑛≥0.
As 𝑥₁∈𝑅⁻¹(𝑓𝑥₀) and 𝑥₂∈𝑆⁻¹(𝑔𝑥₁), and the pairs (𝑓,𝑔) and (𝑔,𝑓) are partially weakly increasing with respect to 𝑅 and 𝑆, so we have 𝑅𝑥₁=𝑓𝑥₀⪯𝑔𝑥₁=𝑆𝑥₂. (63) Continuing this process, we obtain 𝑅𝑥_2𝑛+1⪯𝑆𝑥_2𝑛+2, for 𝑛≥0. We will complete the proof in three steps.
Step  I. We will prove that lim_𝑘→∞𝜎_𝑏(𝑧_𝑘,𝑧_𝑘+1)=0.
Define 𝜎_𝑏𝑘=𝜎_𝑏(𝑧_𝑘,𝑧_𝑘+1). Suppose 𝜎_𝑏𝑘0=0, for some 𝑘₀. Then, 𝑧_𝑘0=𝑧_𝑘0+1. If 𝑘₀=2𝑛, then 𝑧_2𝑛=𝑧_2𝑛+1 gives 𝑧_2𝑛+1=𝑧_2𝑛+2. Indeed, 𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2=𝜓𝜎_𝑏𝑓𝑥_2𝑛,𝑔𝑥_2𝑛+1≤𝜓𝑠²𝜎_𝑏𝑓𝑥_2𝑛,𝑔𝑥_2𝑛+1≤𝜓𝑀𝑥_2𝑛,𝑥_2𝑛+1−𝜑𝑀𝑥_2𝑛,𝑥_2𝑛+1, (64) where 𝑀𝑥_2𝑛,𝑥_2𝑛+1=max𝜎_𝑏𝑆𝑥_2𝑛,𝑅𝑥_2𝑛+1,𝜎_𝑏𝑆𝑥_2𝑛,𝑓𝑥_2𝑛,𝜎_𝑏𝑅𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝜎_𝑏𝑆𝑥_2𝑛,𝑔𝑥_2𝑛+1+𝜎_𝑏𝑅𝑥_2𝑛+1,𝑓𝑥_2𝑛𝜎4𝑠=max_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+2+𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+1𝜎4𝑠≤max_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,𝑠𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1+𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2+2𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2×(4𝑠)⁻¹=max0,0,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,0+3𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2𝜎4𝑠=max_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,3𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+24=𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2. (65) Thus, 𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2≤𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2−𝜑𝑀𝑧_2𝑛,𝑧_2𝑛+1, (66) which implies that 𝜑(𝑀(𝑧_2𝑛,𝑧_2𝑛+1))=0; that is, 𝑧_2𝑛+1=𝑧_2𝑛+2. Similarly, if 𝑘₀=2𝑛+1, then 𝑧_2𝑛+1=𝑧_2𝑛+2 gives 𝑧_2𝑛+2=𝑧_2𝑛+3. Consequently, the sequence {𝑧_𝑘} becomes constant for 𝑘≥𝑘₀ and hence lim_𝑘→∞𝜎_𝑏(𝑧_𝑘,𝑧_𝑘+1)=0.
Suppose that 𝜎_𝑏𝑘=𝜎_𝑏𝑧_𝑘,𝑧_𝑘+1>0, (67) for each 𝑘. We now claim that the following inequality holds: 𝜎_𝑏𝑧_𝑘+1,𝑧_𝑘+2≤𝜎_𝑏𝑧_𝑘,𝑧_𝑘+1=𝑀𝑥_𝑘,𝑥_𝑘+1 (68) for each 𝑘=1,2,….
Let 𝑘=2𝑛, and for an 𝑛≥0, 𝜎_𝑏(𝑧_2𝑛+1,𝑧_2𝑛+2)>𝜎_𝑏(𝑧_2𝑛,𝑧_2𝑛+1)>0. Then, as 𝑆𝑥_2𝑛⪯𝑅𝑥_2𝑛+1, using (62) we obtain that 𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2≤𝜓𝑠²𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2=𝜓𝑠²𝜎_𝑏𝑓𝑥_2𝑛,𝑔𝑥_2𝑛+1≤𝜓𝑀𝑥_2𝑛,𝑥_2𝑛+1−𝜑𝑀𝑥_2𝑛,𝑥_2𝑛+1, (69) where 𝑀𝑥_2𝑛,𝑥_2𝑛+1=max𝜎_𝑏𝑆𝑥_2𝑛,𝑅𝑥_2𝑛+1,𝜎_𝑏𝑆𝑥_2𝑛,𝑓𝑥_2𝑛,𝜎_𝑏𝑅𝑥_2𝑛+1,𝑔𝑥_2𝑛+1,𝜎_𝑏𝑆𝑥_2𝑛,𝑔𝑥_2𝑛+1+𝜎_𝑏𝑅𝑥_2𝑛+1,𝑓𝑥_2𝑛𝜎4𝑠=max_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+2+𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+1𝜎4𝑠≤max_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,𝑠𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1+𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2+2𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2×(4𝑠)⁻¹=max𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2,𝑠𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1+3𝑠𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2𝜎4𝑠=max_𝑏𝑧_2𝑛,𝑧_2𝑛+1,𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2. (70) If, for some 𝑛, 𝜎_𝑏(𝑧_2𝑛+1,𝑧_2𝑛+2)>𝜎_𝑏(𝑧_2𝑛,𝑧_2𝑛+1)>0, (69) implies that 𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2≤𝜓𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2−𝜑𝑀𝑥_2𝑛,𝑥_2𝑛+1, (71) which is possible only if 𝑀(𝑥_2𝑛,𝑥_2𝑛+1)=0; that is, 𝜎_𝑏(𝑧_2𝑛,𝑧_2𝑛+1)=0, a contradiction to (67). Hence, 𝜎_𝑏(𝑧_2𝑛+1,𝑧_2𝑛+2)≤𝜎_𝑏(𝑧_2𝑛,𝑧_2𝑛+1) and 𝑀𝑥_2𝑛,𝑥_2𝑛+1=𝜎_𝑏𝑧_2𝑛,𝑧_2𝑛+1. (72) Therefore, (68) is proved for 𝑘=2𝑛.
Similarly, it can be shown that 𝜎_𝑏𝑧_2𝑛+2,𝑧_2𝑛+3≤𝜎_𝑏𝑧_2𝑛+1,𝑧_2𝑛+2=𝑀𝑥_2𝑛+1,𝑥_2𝑛+2. (73) Hence, {𝜎_𝑏(𝑧_𝑘,𝑧_𝑘+1)} is a nonincreasing sequence of nonnegative real numbers. Therefore, there is an 𝑟≥0 such that lim_𝑘→∞𝜎_𝑏𝑧_𝑘,𝑧_𝑘+1=𝑟. (74) Taking the limit as 𝑘→∞ in (68), we obtain lim_𝑘→∞𝑀𝑥_𝑘,𝑥_𝑘+1=𝑟. (75) Taking the limit as 𝑛→∞ in (69), using (74), (75), and the continuity of 𝜓 and 𝜑, we have 𝜓(𝑟)≤𝜓(𝑟)−𝜑(𝑟). Therefore 𝜑(𝑟)=0. Hence, lim_𝑘→∞𝜎_𝑏𝑧_𝑘,𝑧_𝑘+1=0, (76) from our assumptions about 𝜑.
Step  II. We now show that {𝑧_𝑛} is a 𝜎_𝑏-Cauchy sequence in 𝑋. Because of (76), it is sufficient to show that {𝑧_2𝑛} is 𝜎_𝑏-Cauchy.
We assume on the contrary that there exists 𝜀>0 for which we can find subsequences {𝑧_2𝑚(𝑘)} and {𝑧_2𝑛(𝑘)} of {𝑧_2𝑛} such that 𝑛(𝑘)>𝑚(𝑘)≥𝑘 and 𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_2𝑛(𝑘)≥𝜀, (77) and 𝑛(𝑘) is the smallest number such that the above statement holds; that is, 𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−2}<𝜀. (78) From triangle inequality, we have 𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_2𝑛(𝑘)≤𝑠𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑚(𝑘)+1}+𝑠𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_2𝑛(𝑘). (79) Taking the upper limit as 𝑘→∞ in (79), from (77) and (76) we obtain that 𝜀𝑠≤limsup_𝑘→∞𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_2𝑛(𝑘). (80) Using triangle inequality, we have 𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−1}≤𝑠𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−2}+𝑠𝜎_𝑏𝑧_{2𝑛(𝑘)−2},𝑧_{2𝑛(𝑘)−1}. (81) Taking the upper limit as 𝑘→∞ in (81), from (78) and (76) we have limsup_𝑘→∞𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−1}≤𝑠𝜀. (82) Also, 𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_2𝑛(𝑘)≤𝑠𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−2}+𝑠𝜎_𝑏𝑧_{2𝑛(𝑘)−2},𝑧_2𝑛(𝑘). (83) Taking the upper limit as 𝑘→∞ in (83) and using (76) and (78) we have limsup_𝑘→∞𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_2𝑛(𝑘)≤𝑠𝜀. (84) Consider 𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_{2𝑛(𝑘)−1}≤𝑠𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_2𝑚(𝑘)+𝑠𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−1}. (85) Taking the limit as 𝑘→∞ and using (76) and (82), we have limsup_𝑘→∞𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_{2𝑛(𝑘)−1}≤𝑠²𝜀. (86) As 𝑆𝑥_2𝑚(𝑘)⪯𝑅𝑥_{2𝑛(𝑘)−1}, so from (62), we have 𝜓𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_2𝑛(𝑘)=𝜓𝜎_𝑏𝑓𝑥_2𝑚(𝑘),𝑔𝑥_{2𝑛(𝑘)−1}≤𝜓𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}−𝜑𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}, (87) where 𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}=max𝜎_𝑏𝑆𝑥_2𝑚(𝑘),𝑅𝑥_{2𝑛(𝑘)−1},𝜎_𝑏𝑆𝑥_2𝑚(𝑘),𝑓𝑥_2𝑚(𝑘),𝜎_𝑏𝑅𝑥_{2𝑛(𝑘)−1},𝑔𝑥_{2𝑛(𝑘)−1},𝜎_𝑏𝑆𝑥_2𝑚(𝑘),𝑔𝑥_{2𝑛(𝑘)−1}+𝜎_𝑏𝑅𝑥_{2𝑛(𝑘)−1},𝑓𝑥_2𝑚(𝑘)𝜎4𝑠=max_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑛(𝑘)−1},𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_{2𝑚(𝑘)+1},𝜎_𝑏𝑧_{2𝑛(𝑘)−1},𝑧_2𝑛(𝑘),𝜎_𝑏𝑧_2𝑚(𝑘),𝑧_2𝑛(𝑘)+𝜎_𝑏𝑧_{2𝑛(𝑘)−1},𝑧_{2𝑚(𝑘)+1}.4𝑠 (88) Taking the upper limit in the above and using (76), (82), (84), and (86), we get limsup_𝑘→∞𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}≤𝑠𝜀. (89) Now, taking the upper limit as 𝑘→∞ in (87) and using (80) and (89) we have 𝜓(𝑠𝜀)=𝜓𝑠²𝜀𝑠≤𝜓𝑠²limsup_𝑘→∞𝜎_𝑏𝑧_{2𝑚(𝑘)+1},𝑧_2𝑛(𝑘)≤𝜓limsup_𝑘→∞𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}−𝜑liminf_𝑘→∞𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}≤𝜓(𝑠𝜀)−𝜑liminf_𝑘→∞𝑀𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}, (90) which implies that 𝜑(liminf_𝑘→∞𝑀(𝑥_2𝑚(𝑘),𝑥_{2𝑛(𝑘)−1}))=0. Hence, liminf_𝑘→∞𝜎_𝑏𝑥_2𝑚(𝑘),𝑥_2𝑛(𝑘)=0, (91) which is in contradiction with (77). Hence, {𝑧_𝑛} is a 𝜎_𝑏-Cauchy sequence.
Step  III. We will show that 𝑓, 𝑔, 𝑅, and 𝑆 have a coincidence point.
Since {𝑧_𝑛} is a 𝜎_𝑏-Cauchy sequence and 𝑅(𝑋) and 𝑆(𝑋) are 𝜎_𝑏-complete 𝜎_𝑏-metric spaces, there exists 𝑧^∗∈𝑅(𝑋)∩𝑆(𝑋) such that lim_𝑘→∞𝜎_𝑏𝑧_𝑘,𝑧^∗=𝜎_𝑏𝑧^∗,𝑧^∗=lim_{𝑚,𝑛→∞}𝜎_𝑏𝑧_𝑛,𝑧_𝑚=0. (92) There exists 𝑢∈𝑋 such that 𝑧^∗=𝑅𝑢 and lim_𝑛→∞𝜎_𝑏𝑧_2𝑛+1,𝑅𝑢=lim_𝑛→∞𝜎_𝑏𝑅𝑥_2𝑛+1,𝑅𝑢=𝜎_𝑏(𝑅𝑢,𝑅𝑢). (93) Similarly, there exists 𝑣∈𝑋 such that 𝑧^∗=𝑆𝑣 and lim_𝑛→∞𝜎_𝑏𝑧_2𝑛+2,𝑆𝑣=lim_𝑛→∞𝜎_𝑏𝑆𝑥_2𝑛+2,𝑆𝑣=𝜎_𝑏(𝑆𝑣,𝑆𝑣). (94) Now, we prove that 𝑧^∗ is a coincidence point of 𝑔 and 𝑅. For this purpose, we show that 𝑅𝑢=𝑔𝑢. Since 𝑆𝑥_2𝑛+2→𝑧^∗=𝑅𝑢 as 𝑛→∞, so 𝑆𝑥_2𝑛+2⪯𝑅𝑢.
Therefore, from (62), we have 𝜓𝑠²𝜎_𝑏𝑓𝑥_2𝑛+2,𝑔𝑢≤𝜓𝑀𝑥_2𝑛+2𝑀𝑥,𝑢−𝜑_2𝑛+2,𝑢, (95) where 𝑀𝑥_2𝑛+2,𝑢=max𝜎_𝑏𝑆𝑥_2𝑛+2,𝑅𝑢,𝜎_𝑏𝑆𝑥_2𝑛+2,𝑓𝑥_2𝑛+2,𝜎_𝑏,𝜎𝑅𝑢,𝑔𝑢_𝑏𝑆𝑥_2𝑛+2,𝑔𝑢+𝜎_𝑏𝑅𝑢,𝑓𝑥_2𝑛+2.4𝑠 (96) Taking the limit as 𝑛→∞ in (95), as 𝜎_𝑏(𝑧^∗,𝑧^∗)=0, and using Lemma 16 we obtain that 𝜓𝑠𝜎_𝑏𝑧^∗𝑠,𝑔𝑢=𝜓²1𝑠𝜎_𝑏𝑧^∗,𝑔𝑢≤𝜓max𝑠𝜎_𝑏𝑧^∗,𝑅𝑢,0,𝜎_𝑏,𝜎𝑅𝑢,𝑔𝑢_𝑏𝑧^∗,𝑔𝑢+𝜎_𝑏𝑧^∗,𝑅𝑢4𝑠−𝜑max𝑠𝜎_𝑏𝑧^∗,𝑅𝑢,0,𝜎_𝑏,𝜎𝑅𝑢,𝑔𝑢_𝑏𝑧^∗,𝑔𝑢+𝜎_𝑏𝑧^∗,𝑅𝑢𝜎4𝑠=𝜓_𝑏𝑧^∗,𝑔𝑢−𝜑𝜎_𝑏𝑧^∗,𝑔𝑢≤𝜓𝑠𝜎_𝑏𝑧^∗,𝑔𝑢−𝜑𝜎_𝑏𝑧^∗,𝑔𝑢 (97) which implies that 𝑅𝑢=𝑧^∗=𝑔𝑢.
As 𝑔 and 𝑅 are weakly compatible, we have 𝑔𝑧^∗=𝑔𝑅𝑢=𝑅𝑔𝑢=𝑅𝑧^∗. Thus 𝑧^∗ is a coincidence point of 𝑔 and 𝑅.
Similarly it can be shown that 𝑧^∗ is a coincidence point of the pair (𝑓,𝑆)

Theorem 26. Let (𝑋,⪯,𝜎_𝑏) be a partially ordered 𝜎_𝑏-complete 𝜎_𝑏-metric-like space. Let 𝑓,𝑔∶𝑋→𝑋 be two mappings. Suppose that for every two comparable elements 𝑥,𝑦∈𝑋 one has 𝜓2𝑠𝜎_𝑏𝑓𝑥,𝑔𝑦≤𝜓𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (98) where 𝜓,𝜑∶[0,∞)→[0,∞) are altering distance functions and 𝑀𝑥,𝑦=max𝜎_𝑏𝑥,𝑦,𝜎_𝑏𝑥,𝑓𝑥,𝜎_𝑏,𝜎𝑦,𝑔𝑦_𝑏𝑥,𝑔𝑦+𝜎_𝑏𝑦,𝑓𝑥.4𝑠 (99) Let 𝑓 and 𝑔 be continuous and let the pair (𝑓,𝑔) be weakly increasing. Then, 𝑓 and 𝑔 have a common fixed point 𝑧^∗ in 𝑋.

Proof. Let 𝑥₀∈𝑋. Let {𝑥_𝑛} in 𝑋 be constructed such that 𝑥_2𝑛+1=𝑓𝑥_2𝑛 and 𝑥_2𝑛+2=𝑔𝑥_2𝑛+1, for all nonnegative integers 𝑛. As 𝑓 and 𝑔 are weakly increasing, we have 𝑥₁=𝑓𝑥₀⪯𝑔𝑓𝑥₀=𝑥₂=𝑔𝑥₁⪯𝑓𝑔𝑥₁=𝑥₃⪯⋯⪯𝑥_2𝑛+1=𝑓𝑥_2𝑛⪯𝑔𝑓𝑥_2𝑛=𝑥_2𝑛+2⪯…. (100) Following the proof of the above theorem there exists 𝑧^∗∈𝑋 such that lim_𝑛→∞𝜎_𝑏𝑥_2𝑛+1,𝑧^∗=lim_𝑛→∞𝜎_𝑏𝑓𝑥_2𝑛,𝑧^∗=0,lim_𝑛→∞𝜎_𝑏𝑥_2𝑛+2,𝑧^∗=lim_𝑛→∞𝜎_𝑏𝑔𝑥_2𝑛+1,𝑧^∗=0. (101) Using the triangular inequality, we get 𝜎_𝑏𝑧^∗,𝑓𝑧^∗≤𝑠𝜎_𝑏𝑧^∗,𝑓𝑥_2𝑛+𝑠𝜎_𝑏𝑓𝑥_2𝑛,𝑓𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗≤𝑠𝜎_𝑏𝑧^∗,𝑔𝑥_2𝑛+1+𝑠𝜎_𝑏𝑔𝑥_2𝑛+1,𝑔𝑧^∗. (102) Letting 𝑛→∞ and using continuity of 𝑓 and 𝑔, we get 𝜎_𝑏𝑧^∗,𝑓𝑧^∗≤𝑠lim_𝑛→∞𝜎_𝑏𝑧^∗,𝑓𝑥_2𝑛+𝑠lim_𝑛→∞𝜎_𝑏𝑓𝑥_2𝑛,𝑓𝑧^∗=𝑠𝜎_𝑏𝑓𝑧^∗,𝑓𝑧^∗≤2𝑠𝜎_𝑏𝑓𝑧^∗,𝑔𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗≤𝑠lim_𝑛→∞𝜎_𝑏𝑧^∗,𝑔𝑥_2𝑛+1+𝑠lim_𝑛→∞𝜎_𝑏𝑔𝑥_2𝑛+1,𝑔𝑧^∗=𝑠𝜎_𝑏𝑔𝑧^∗,𝑔𝑧^∗≤2𝑠𝜎_𝑏𝑓𝑧^∗,𝑔𝑧^∗. (103) Therefore, max𝜎_𝑏𝑧^∗,𝑓𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗≤max𝑠𝜎_𝑏𝑓𝑧^∗,𝑓𝑧^∗,𝑠𝜎_𝑏𝑔𝑧^∗,𝑔𝑧^∗≤2𝑠𝜎_𝑏𝑔𝑧^∗,𝑓𝑧^∗. (104) From (98), we have 𝜓2𝑠𝜎_𝑏𝑓𝑧^∗,𝑔𝑧^∗≤𝜓𝑀𝑧^∗,𝑧^∗−𝜑𝑀𝑧^∗,𝑧^∗, (105) where 𝑀𝑧^∗,𝑧^∗=max𝜎_𝑏𝑧^∗,𝑧^∗,𝜎_𝑏𝑧^∗,𝑓𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗+𝜎_𝑏𝑧^∗,𝑓𝑧^∗𝜎4𝑠=max_𝑏𝑧^∗,𝑓𝑧^∗,𝜎_𝑏𝑧^∗,𝑔𝑧^∗. (106) Hence, by (104) and (105), we get that 𝜑(𝑀(𝑧^∗,𝑧^∗))=0. Thus, we have 𝑓𝑧^∗=𝑔𝑧^∗=𝑧^∗.

3.3. Common Fixed Points of Dominated Maps on Closed Balls

Motivated by [32, Theorem 2.1] we have the following result.

Theorem 27. Let (𝑋,⪯,𝜎_𝑏) be a complete ordered 𝑏-metric-like space, 𝑆,𝑇∶𝑋→𝑋 dominated maps, and 𝑥₀ an arbitrary point in 𝑋. Suppose that for 𝑘∈[0,1/𝑠) and for 𝑆≠𝑇 one has 𝜎_𝑏𝑆𝑥,𝑇𝑦≤𝑘𝜎_𝑏𝑥,𝑦𝑓𝑜𝑟𝑎𝑙𝑙𝑐𝑜𝑚𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠𝑥,𝑦𝑖𝑛𝐵𝑥₀,𝑟𝑠,𝜎_𝑏𝑥₀,𝑆𝑥₀≤(1−𝑘𝑠)𝑟. (107)(108) Let, for each nonincreasing sequence {𝑥_𝑛} in 𝐵(𝑥₀,𝑟𝑠), {𝑥_𝑛}→𝑢 imply that 𝑢⪯𝑥_𝑛. Then there exists 𝑥^∗∈𝐵(𝑥₀,𝑟𝑠) such that 𝜎_𝑏(𝑥^∗,𝑥^∗)=0 and 𝑥^∗=𝑆𝑥^∗=𝑇𝑥^∗. Also, if for any two points 𝑥,𝑦 in 𝐵(𝑥₀,𝑟𝑠) there exists a point 𝑧∈𝐵(𝑥₀,𝑟𝑠) such that 𝑧⪯𝑥 and 𝑧⪯𝑦, that is, every pair of elements has a lower bound, then 𝑥^∗ is a unique common fixed point in 𝐵(𝑥₀,𝑟𝑠).

Proof. Choose a point 𝑥₁ in 𝑋 such that 𝑥₁=𝑆𝑥₀. As 𝑆𝑥₀⪯𝑥₀, so 𝑥₁⪯𝑥₀ and let 𝑥₂=𝑇𝑥₁. Now 𝑇𝑥₁⪯𝑥₁ gives 𝑥₂⪯𝑥₁, and continuing this process, we construct a sequence 𝑥_𝑛 of points in 𝑋 such that 𝑥_2𝑖+1=𝑆𝑥_2𝑖,𝑥_2𝑖+2=𝑇𝑥_2𝑖+1,𝑥_2𝑖+1=𝑆𝑥_2𝑖⪯𝑥_2𝑖where𝑖=0,1,2,…. (109) First we show that 𝑥_𝑛∈𝐵(𝑥₀,𝑟𝑠) for all 𝑛∈𝑁. Using inequality (108), we have 𝜎_𝑏𝑥₀,𝑥₁≤(1−𝑘𝑠)𝑟≤𝑟𝑠. (110) It follows that 𝑥₁∈𝐵(𝑥₀,𝑟𝑠).
Let 𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟𝑠) for some 𝑗∈𝑁. If 𝑗=2𝑖+1, then 𝑥_2𝑖+1⪯𝑥_2𝑖, where 𝑖=0,1,2,…,(𝑗−1)/2 so, using inequality (107), we obtain 𝜎_𝑏𝑥_2𝑖+1,𝑥_2𝑖+2=𝜎_𝑏𝑆𝑥_2𝑖,𝑇𝑥_2𝑖+1≤𝑘𝜎_𝑏𝑥_2𝑖,𝑥_2𝑖+1≤𝑘²𝜎_𝑏𝑥_2𝑖−1,𝑥_2𝑖≤⋯≤𝑘^2𝑖+1𝜎_𝑏𝑥₀,𝑥₁. (111) If 𝑗=2𝑖+2, then as 𝑥₁,𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟𝑠) and 𝑥_2𝑖+2⪯𝑥_2𝑖+1 (𝑖=0,1,2,…,(𝑗−2)/2) we obtain 𝜎_𝑏𝑥_2𝑖+2,𝑥_2𝑖+3≤𝑘^2(𝑖+1)𝜎_𝑏𝑥₀,𝑥₁. (112) Thus from inequalities (111) and (112), we have 𝜎_𝑏𝑥_𝑗,𝑥_𝑗+1≤𝑘^𝑗𝜎_𝑏𝑥₀,𝑥₁. (113) Now 𝜎_𝑏𝑥₀,𝑥_𝑗+1≤𝑠𝜎_𝑏𝑥₀,𝑥₁+𝑠²𝜎_𝑏𝑥₁,𝑥₂+⋯+𝑠^𝑗𝜎_𝑏𝑥_𝑗−1,𝑥_𝑗+𝑠^𝑗𝜎_𝑏𝑥_𝑗,𝑥_𝑗+1≤𝑠𝜎_𝑏𝑥₀,𝑥₁+𝑘𝑠²𝜎_𝑏𝑥₀,𝑥₁+⋯+𝑘^𝑗−1𝑠^𝑗𝜎_𝑏𝑥₀,𝑥₁+𝑘^𝑗𝑠^𝑗𝜎_𝑏𝑥₀,𝑥₁by(113)≤𝑠𝜎_𝑏𝑥₀,𝑥₁1+𝑘𝑠+⋯+𝑘^𝑗−1𝑠^𝑗−1+𝑘^𝑗𝑠^𝑗≤𝑠1−𝑘^𝑗+1𝑠^𝑗+1𝜎1−𝑘𝑠^𝑏𝑥₀,𝑥₁≤𝑠1−𝑘^𝑗+1𝑠^𝑗+1()𝑟1−𝑘𝑠1−𝑘𝑠by(108)≤𝑠1−𝑘^𝑗+1𝑠^𝑗+1𝑟≤𝑟𝑠. (114) Thus, 𝑥_𝑗+1∈𝐵(𝑥₀,𝑟𝑠). Hence 𝑥_𝑛∈𝐵(𝑥₀,𝑟𝑠) for all 𝑛∈𝑁. This implies that 𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1≤𝑘^𝑛𝜎_𝑏𝑥₀,𝑥₁,∀𝑛∈𝑁. (115) It follows that 𝜎_𝑏𝑥_𝑛,𝑥_𝑛+𝑖≤𝑠𝜎_𝑏𝑥_𝑛,𝑥_𝑛+1+𝑠²𝜎_𝑏𝑥_𝑛+1,𝑥_𝑛+2+⋯+𝑠^𝑖−1𝜎_𝑏𝑥_{𝑛+𝑖−2},𝑥_{𝑛+𝑖−1}+𝑠^𝑖−1𝜎_𝑏𝑥_{𝑛+𝑖−1},𝑥_𝑛+𝑖≤𝑠𝑘^𝑛𝜎_𝑏𝑥₀,𝑥₁+𝑠²𝑘^𝑛+1𝜎_𝑏𝑥₀,𝑥₁+⋯+𝑠^𝑖−1𝑘^𝑖−2𝜎_𝑏𝑥₀,𝑥₁+𝑠^𝑖−1𝑘^{𝑛+𝑖−1}𝜎_𝑏𝑥₀,𝑥₁by(115)≤𝑠𝑘^𝑛𝜎_𝑏𝑥₀,𝑥₁1+𝑘𝑠+⋯+𝑘^𝑖−2𝑠^𝑖−2+𝑘^𝑖−1𝑠^𝑖−2𝑘≤𝑠^𝑛1−𝑘^𝑖𝑠^𝑖𝜎1−𝑘𝑠^𝑏𝑥₀,𝑥₁⟶as𝑛⟶∞. (116) Notice that the sequence {𝑥_𝑛} is a Cauchy sequence in (𝐵(𝑥₀,𝑟𝑠),𝜎_𝑏). Therefore there exists a point 𝑥^∗∈𝐵(𝑥₀,𝑟𝑠) with lim_𝑛→∞𝑥_𝑛=𝑥^∗. Also, lim_𝑛→∞𝜎_𝑏𝑥_𝑛,𝑥^∗=0. (117) Now, 𝜎_𝑏𝑥^∗,𝑆𝑥^∗≤𝑠𝜎_𝑏𝑥^∗,𝑥_2𝑛+2+𝑠𝜎_𝑏𝑥_2𝑛+2,𝑆𝑥^∗. (118) On taking the limit as 𝑛→∞ and using the fact that 𝑥^∗⪯𝑥_𝑛 when 𝑥_𝑛→𝑥^∗, we have 𝜎_𝑏𝑥^∗,𝑆𝑥^∗≤𝑠lim_𝑛→∞𝜎_𝑏𝑥^∗,𝑥_2𝑛+2+𝑘𝜎_𝑏𝑥_2𝑛+1,𝑥^∗. (119) By (117) we obtain 𝜎_𝑏𝑥^∗,𝑆𝑥^∗≤0, (120) and hence 𝑥^∗=𝑆𝑥^∗. Similarly, by using 𝜎_𝑏𝑥^∗,𝑇𝑥^∗≤𝑠𝜎_𝑏𝑥^∗,𝑥_2𝑛+1+𝑠𝜎_𝑏𝑥_2𝑛+1,𝑇𝑥^∗, (121) we can show that 𝑥^∗=𝑇𝑥^∗. Hence 𝑆 and 𝑇 have a common fixed point in 𝐵(𝑥₀,𝑟𝑠). Now, 𝜎_𝑏𝑥^∗,𝑥^∗=𝜎_𝑏𝑆𝑥^∗,𝑇𝑥^∗≤𝑘𝜎_𝑏𝑥^∗,𝑥^∗. (122) This implies that 𝜎_𝑏𝑥^∗,𝑥^∗=0. (123)
For the uniqueness, assume that 𝑦 is another fixed point of 𝑇 and 𝑆 in 𝐵(𝑥₀,𝑟𝑠). If 𝑥^∗ and 𝑦 are comparable then 𝜎_𝑏𝑥^∗,𝑦=𝜎_𝑏𝑆𝑥^∗,𝑇𝑦≤𝑘𝜎_𝑏𝑥^∗,𝑦. (124) This shows that 𝑥^∗=𝑦. Now if 𝑥^∗ and 𝑦 are not comparable then there exists a point 𝑧₀∈𝐵(𝑥₀,𝑟𝑠) such that 𝑧₀⪯𝑥^∗ and 𝑧₀⪯𝑦. Choose a point 𝑧₁ in 𝑋 such that 𝑧₁=𝑇𝑧₀. As 𝑇𝑧₀⪯𝑧₀, so 𝑧₁⪯𝑧₀; let 𝑧₂=𝑆𝑧₁. Now 𝑆𝑧₁⪯𝑧₁ gives 𝑧₂⪯𝑧₁, and continuing this process choose 𝑧_𝑛 in 𝑋 such that 𝑧_2𝑖+1=𝑇𝑧_2𝑖,𝑧_2𝑖+2=𝑆𝑧_2𝑖+1,𝑧_2𝑖+1=𝑇𝑧_2𝑖⪯𝑧_2𝑖where𝑖=0,1,2,…. (125) It follows that 𝑧_𝑛+1⪯𝑧_𝑛⪯⋯⪯𝑧₀⪯𝑥^∗. As 𝑧₀⪯𝑥^∗ and 𝑧₀⪯𝑦, it follows that 𝑧_𝑛⪯𝑇𝑥^∗ and 𝑧_𝑛⪯𝑇𝑦 for all 𝑛∈𝑁. We will prove that 𝑧_𝑛∈𝐵(𝑥₀,𝑟𝑠) for all 𝑛∈𝑁 by using mathematical induction. For 𝑛=1, 𝜎_𝑏𝑥₀,𝑧₁≤𝑠𝜎_𝑏𝑥₀,𝑥₁+𝑠𝜎_𝑏𝑥₁,𝑧₁≤𝑠(1−𝑘𝑠)𝑟+𝑘𝑠𝜎_𝑏𝑥₀,𝑧₀≤𝑠(1−𝑘𝑠)𝑟+𝑘𝑠²𝑟=𝑠𝑟. (126) It follows that 𝑧₁∈𝐵(𝑥₀,𝑟𝑠). Let 𝑧₂,𝑧₃…,𝑧_𝑗∈𝐵(𝑥₀,𝑟𝑠) for some 𝑗∈𝑁. Note that if 𝑗 is odd then 𝜎_𝑏𝑥_𝑗+1,𝑧_𝑗+1=𝜎_𝑏𝑇𝑥_𝑗,𝑆𝑧_𝑗≤𝑘𝜎_𝑏𝑥_𝑗,𝑧_𝑗≤⋯≤𝑘^𝑗+1𝜎_𝑏𝑥₀,𝑧₀, (127) and if 𝑗 is even then 𝜎_𝑏𝑥_𝑗+1,𝑧_𝑗+1=𝜎_𝑏𝑆𝑥_𝑗,𝑇𝑧_𝑗≤𝑘𝜎_𝑏𝑥_𝑗,𝑧_𝑗≤⋯≤𝑘^𝑗+1𝜎_𝑏𝑥₀,𝑧₀. (128) Now, 𝜎_𝑏𝑥₀,𝑧_𝑗+1≤𝑠𝜎_𝑏𝑥₀,𝑥₁+𝑠²𝜎_𝑏𝑥₁,𝑥₂+⋯+𝑠^𝑗𝜎_𝑏𝑥_𝑗,𝑥_𝑗+1+𝑠^𝑗𝜎_𝑏𝑥_𝑗+1,𝑧_𝑗+1≤𝑠𝜎_𝑏𝑥₀,𝑥₁+𝑘𝑠²𝜎_𝑏𝑥₀,𝑥₁+⋯+𝑘^𝑗𝑠^𝑗𝜎_𝑏𝑥₀,𝑥₁+𝑘^𝑗+1𝑠^𝑗𝜎_𝑏𝑥₀,𝑧₀≤𝑠𝜎_𝑏𝑥₀,𝑥₁1+𝑘𝑠+⋯+𝑘^𝑗𝑠^𝑗−1+𝑘^𝑗+1𝑠^𝑗+1𝑟≤𝑠𝜎_𝑏𝑥₀,𝑥₁1+𝑘𝑠+⋯+𝑘^𝑗𝑠^𝑗+𝑘^𝑗+1𝑠^𝑗+1𝑟≤𝑠(1−𝑘𝑠)𝑟1−𝑘^𝑗+1𝑠^𝑗+11−𝑘𝑠+𝑘^𝑗+1𝑠^𝑗+1𝑟=𝑠𝑟−𝑘^𝑗+1𝑠^𝑗+2𝑟+𝑘^𝑗+1𝑠^𝑗+1𝑟≤𝑠𝑟−𝑘^𝑗+1𝑠^𝑗+2𝑟+𝑘^𝑗+1𝑠^𝑗+2𝑟=𝑠𝑟, (129) and this implies that 𝜎_𝑏𝑥₀,𝑧_𝑗+1≤𝑠𝑟. (130) Thus 𝑧_𝑗+1∈𝐵(𝑥₀,𝑟𝑠). Hence 𝑧_𝑛∈𝐵(𝑥₀,𝑟𝑠) for all 𝑛∈𝑁. As 𝑧₀⪯𝑥^∗ and 𝑧₀⪯𝑦, it follows that 𝑧_𝑛⪯𝑇^𝑛𝑥^∗, 𝑧_𝑛⪯𝑆^𝑛𝑥^∗, 𝑧_𝑛⪯𝑆^𝑛𝑦, and 𝑧_𝑛⪯𝑇^𝑛𝑦 for all 𝑛∈𝑁 as 𝑆^𝑛𝑥^∗=𝑇^𝑛𝑥^∗=𝑥^∗ and 𝑆^𝑛𝑦=𝑇^𝑛𝑦=𝑦 for all 𝑛∈𝑁. If 𝑛 is odd then 𝜎_𝑏𝑥^∗,𝑦=𝜎_𝑏𝑇^𝑛𝑥^∗,𝑇^𝑛𝑦≤𝑠𝜎_𝑏𝑇^𝑛𝑥^∗,𝑆𝑧_𝑛+𝑠𝜎_𝑏𝑆𝑧_𝑛,𝑇^𝑛𝑦≤𝑘𝑠𝜎_𝑏𝑇^𝑛−1𝑥^∗,𝑧_𝑛+𝑘𝑠𝜎_𝑏𝑧_𝑛,𝑇^𝑛−1𝑦=𝑘𝑠𝜎_𝑏𝑆^𝑛−1𝑥^∗,𝑇𝑧_𝑛−1+𝑘𝑠𝜎_𝑏𝑇𝑧_𝑛−1,𝑆^𝑛−1𝑦≤𝑘²𝑠𝜎_𝑏𝑆^𝑛−2𝑥^∗,𝑧_𝑛−1+𝑘²𝑠𝜎_𝑏𝑧_𝑛−1,𝑆^𝑛−2𝑦⋮≤𝑘^𝑛+1𝑠𝜎_𝑏𝑥^∗,𝑧₀+𝑘^𝑛+1𝑠𝜎_𝑏𝑧₀,𝑦⟶0as𝑛⟶∞. (131) Hence, 𝑥^∗=𝑦. Similarly, we can show that 𝑥^∗=𝑦 if 𝑛 is even.
Thus, 𝑥^∗ is a unique common fixed point of 𝑇 and 𝑆 in 𝐵(𝑥₀,𝑟𝑠).

3.4. Periodic Point Results

Clearly, a fixed point of 𝑓 is also a fixed point of 𝑓^𝑛 for every 𝑛∈ℕ; that is, 𝐹(𝑓)⊂𝐹(𝑓^𝑛). However, the converse is false. For example, the mapping 𝑓∶ℝ→ℝ, defined by 𝑓𝑥=1/2−𝑥, has a unique fixed point 1/4 but every 𝑥∈ℝ is a fixed point of 𝑓². If 𝐹(𝑓)=𝐹(𝑓^𝑛) for every 𝑛∈ℕ, then 𝑓 is said to have property 𝑃. For more details, we refer to [33, 34] and the references mentioned therein.

Recently, the study of periodic points for contraction mappings has been considered by many authors; for instance, every quasicontraction 𝑓∶𝑋→𝑋 with the constant 𝛼∈[0,1/2), where 𝑋 is a cone metric space, has the property 𝑃 [35, Theorem 3.1] and if (𝑋,𝑑) is a cone metric space and 𝑇-Hardy-Rogers contraction 𝑓∶𝑋→𝑋 satisfies some appropriate conditions, then 𝑓 has property 𝑃 [36, Corollary 3.3].

Definition 28 (see [21]). Let (𝑋,⪯) be a partially ordered set. A mapping 𝑓 is called dominating on 𝑋 if 𝑥⪯𝑓𝑥 for each 𝑥 in 𝑋.

Example 29 (see [21]). Let 𝑋=[0,∞) be endowed with the usual ordering. Let 𝑓∶𝑋→𝑋 be defined by 𝑓𝑥=√^𝑛𝑥 for 𝑥∈[0,1) and 𝑓𝑥=𝑥^𝑛 for 𝑥∈[1,∞), for any 𝑛∈ℕ. Then, for all 𝑥∈𝑋, 𝑥≤𝑓𝑥; that is, 𝑓 is a dominating map.

We have the following results.

Theorem 30. Let (𝑋,⪯,𝜎_𝑏) be a partially ordered complete 𝑏-metric-like space. Let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that for all 𝑥∈𝑋 with 𝑥⪯𝑓𝑥 one has 𝜎_𝑏𝑓𝑥,𝑓²𝑥≤𝜆𝜎_𝑏𝑥,𝑓𝑥, (132) where 𝜆∈[0,1). Then 𝑓 has the property 𝑃 provided that 𝐹(𝑓) is nonempty and 𝑓 is dominating on 𝐹(𝑓^𝑛).

Proof. Let 𝑢∈𝐹(𝑓^𝑛) for some 𝑛>1. We will show that 𝑢=𝑓𝑢. Since 𝑓 is dominating on 𝐹(𝑓^𝑛), therefore 𝑢⪯𝑓𝑢 which implies that 𝑓^𝑛−1𝑢⪯𝑓^𝑛𝑢 as 𝑓 is nondecreasing. Using (132), we obtain that 𝜎_𝑏𝑢,𝑓𝑢=𝜎_𝑏𝑓𝑓^𝑛−1𝑢,𝑓²𝑓^𝑛−1𝑢≤𝜆𝜎_𝑏𝑓^𝑛−1𝑢,𝑓^𝑛𝑢=𝜆𝜎_𝑏𝑓𝑓^𝑛−2𝑢,𝑓²𝑓^𝑛−2𝑢. (133) Repeating the above process, we get 𝜎_𝑏𝑢,𝑓𝑢≤𝜆^𝑛𝜎_𝑏𝑢,𝑓𝑢, (134) which, on taking the limit as 𝑛→∞, implies that 𝜎_𝑏(𝑢,𝑓𝑢)=0, implying that 𝑢=𝑓𝑢.

Theorem 31. Let 𝑋 and 𝑓 be as in Theorem 18. If 𝑓 is dominating on 𝑋, then 𝑓 satisfies property 𝑃.

Proof. From Theorem 18, 𝐹(𝑓)≠∅. We will prove that (132) is satisfied for all 𝑥⪯𝑓𝑥.
If 𝜎_𝑏(𝑓𝑥,𝑓²𝑥)=0, then it is easy to see that (132) is satisfied. Now we suppose that 𝜎_𝑏(𝑓𝑥,𝑓²𝑥)>0. Since 𝑓 is dominating, we have 𝑥⪯𝑓𝑥. Also, 𝑓𝑥⪯𝑓²𝑥 as 𝑓 is nondecreasing. Using (15), we have 𝜎_𝑏𝑓𝑥,𝑓²𝑥=𝜎_𝑏𝜎𝑓𝑥,𝑓𝑓𝑥≤𝛼max_𝑏𝑥,𝑓𝑥,𝜎_𝑏𝑥,𝑓𝑥,𝜎_𝑏𝑓𝑥,𝑓²𝑥,𝜎_𝑏𝑥,𝑓²𝑥,𝜎_𝑏𝜎𝑓𝑥,𝑓𝑥=𝛼max_𝑏𝑥,𝑓𝑥,𝜎_𝑏𝑓𝑥,𝑓²𝑥,𝜎_𝑏𝑥,𝑓²𝑥,𝜎_𝑏𝜎𝑓𝑥,𝑓𝑥≤𝛼max_𝑏𝑥,𝑓𝑥,𝜎_𝑏𝑓𝑥,𝑓²𝑥,𝑠𝜎_𝑏𝑥,𝑓𝑥+𝑠𝜎_𝑏𝑓𝑥,𝑓²𝑥,2𝑠𝜎_𝑏𝑥,𝑓𝑥. (135) We will show that 𝜎_𝑏(𝑓𝑥,𝑓²𝑥)≤𝜎_𝑏(𝑥,𝑓𝑥). On the contrary, if 𝜎_𝑏(𝑥,𝑓𝑥)<𝜎_𝑏(𝑓𝑥,𝑓²𝑥), then from the above inequality we have 𝜎_𝑏𝑓𝑥,𝑓²𝑥<2𝑠𝛼𝜎_𝑏𝑓𝑥,𝑓²𝑥, (136) which implies that 1<2𝑠𝛼<2𝑠⋅(1/2𝑠²)<1, a contradiction.
So we have 𝜎_𝑏𝑓𝑥,𝑓²𝑥≤𝜆𝜎_𝑏𝑥,𝑓𝑥, (137) where 𝜆=2𝑠𝛼. Obviously, 𝜆∈[0,1). By Theorem 30, 𝑓 has property 𝑃.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Authors’ Contribution

All authors contributed equally and significantly in writing this paper. All authors read and approved the final paper.

Acknowledgments

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, N. Hussain acknowledges with thanks DSR, KAU, for financial support. Z. Kadelburg is thankful to the Ministry of Education, Science and Technological Development of Serbia.