Generalized Altering Distances and Common Fixed Points in Ordered Metric Spaces

Abstract

Coincidence point and common fixed point results with the concept of generalized altering distance functions in complete ordered metric spaces are derived. These results generalize the existing fixed point results in the literature. To illustrate our results and to distinguish them from the existing ones, we equip the paper with examples. As an application, we study the existence of a common solution to a system of integral equations.

1. Introduction and Preliminaries

The study of common fixed points of mappings satisfying certain contractive conditions has been researched extensively by many mathematicians since fixed point theory plays a major role in mathematics and applied sciences (see [1–40] and others). A new category of contractive fixed point problems was addressed by Khan et al. [1]. In this work, they introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metric space.

Definition 1.1 (see [1]). A function 𝜑∶[0,+∞)→[0,+∞) is called an altering distance function if and only if(i)𝜑is continuous,(ii)𝜑 is nondecreasing,(iii)𝜑(𝑡)=0⇔𝑡=0.
Khan et al. [1] proved the following result.

Theorem 1.2 (see [1]). Let (𝑋,𝑑) be a complete metric space, 𝜑∶[0,+∞)→[0,+∞) an altering distance function, and 𝑇∶𝑋→𝑋 a self-mapping which satisfies the following inequality: 𝜑𝑑𝑇𝑥,𝑇𝑦≤𝑐𝜑𝑑𝑥,𝑦 (1.1) for all 𝑥,𝑦∈𝑋 and for some 0<𝑐<1. Then 𝑇 has a unique fixed point.

Putting 𝜑(𝑡)=𝑡 in Theorem 1.2, we retrieve immediately the Banach contraction principle.

In 1997, Alber and Guerre-Delabriere [2] introduced the concept of weak contractions in Hilbert spaces.

This concept was extended to metric spaces by Rhoades in [3].

Definition 1.3. A mapping 𝑇∶𝑋→𝑋, where (𝑋,𝑑) is a metric space, is said to be weakly contractive if and only if 𝑑𝑇𝑥,𝑇𝑦≤𝑑𝑥,𝑦−𝜑𝑑𝑥,𝑦,∀𝑥,𝑦∈𝑋, (1.2) where 𝜑∶[0,+∞)→[0,+∞) is an altering distance function.

Theorem 1.4 (see [3]). Let (𝑋,𝑑) be a complete metric space and 𝑇∶𝑋→𝑋 a weakly contractive map. Then, 𝑇 admits a unique fixed point.

Weak inequalities of the above type have been used to establish fixed point results in a number of subsequent works, some of which are noted in [4–7, 20]. In [5], Choudhury introduced the concept of a generalized altering distance function for three variables.

Definition 1.5 (see [5]). A function 𝜑∶[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞) is said to be a generalized altering distance function if and only if(i)𝜑 is continuous,(ii)𝜑 is nondecreasing in all the three variables,(iii)𝜑(𝑥,𝑦,𝑧)=0⇔𝑥=𝑦=𝑧=0.

In [5], Choudhury proved the following common fixed point theorem using altering distances for three variables.

Theorem 1.6 (see [5]). Let (𝑋,𝑑) be a complete metric space and 𝑆,𝑇∶𝑋→𝑋 two self mappings such that the following inequality is satisfied: Φ₁𝑑𝑆𝑥,𝑇𝑦≤𝜓₁𝑑𝑥,𝑦,𝑑(𝑥,𝑆𝑥),𝑑𝑦,𝑇𝑦−𝜓₂𝑑𝑥,𝑦,𝑑(𝑥,𝑆𝑥),𝑑𝑦,𝑇𝑦 (1.3) for all 𝑥,𝑦∈𝑋, where 𝜓₁ and 𝜓₂ are generalized altering distance functions and Φ₁(𝑥)=𝜓₁(𝑥,𝑥,𝑥). Then 𝑆 and 𝑇 have a common fixed point.

In [31], Rao et al. generalized Theorem 1.6 for four mappings satisfying a generalized contractive condition and used the following generalized altering distance function for four variables.

Definition 1.7 (see [31]). A function 𝜑∶[0,+∞)×[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞) is said to be a generalized altering distance function if and only if(i)𝜑 is continuous,(ii)𝜑 is nondecreasing in all the three variables,(iii)𝜑(𝑡₁,𝑡₂,𝑡₃,𝑡₄)=0⇔𝑡₁=𝑡₂=𝑡₃=𝑡₄=0.
Let Ψ₄ denote the set of all functions 𝜑 satisfying (i)–(iii) in Definition 1.7.

Recently, there have been so many exciting developments in the field of existence of fixed point in partially ordered sets (see [8–19, 21–36] and the references cited therein). The first result in this direction was given by Turinici [36], where he extended the Banach contraction principle in partially ordered sets. Ran and Reurings [33] presented some applications of Turinici’s theorem to matrix equations. The obtained result by Turinici was further extended and refined in [28–32]. Subsequently, Harjani and Sadarangani [19] generalized their own results [18] by considering pair of altering functions (𝜓,𝜑). Nashine and Altun [21] and Nashine and Altun [22] generalized the results of Harjani and Sadarangani [18, 19]. Also, Nashine and Altun [22] and Shatanawi and Samet [41] worked for a pair (𝑇,𝑆) of weakly increasing mappings with respect to a third mapping 𝑅. In another paper, Nashine et al. [24] prove coincidence point and common fixed point results for mappings satisfying a contractive inequality which involves two generalized altering distance functions for three variables in ordered complete metric spaces. As application, they studied the existence of a common solution to a system of integral equations.

The aim of this paper is to generalize the results of Nashine et al. [24] in the sense of four variables. We obtain coincidence point and common fixed point theorems in complete ordered metric spaces for mappings satisfying a contractive condition which involves two generalized altering distance functions in four variables. Presented theorems are ordered version of Theorem  2.1 of Rao et al. [31] for three mappings. In addition, an application to the study of the existence of a common solution to a system of integral equations is given.

2. Main Results

First we introduce some notations and definitions that will be used later.

2.1. Notations and Definitions

The following definition was introduced by Jungck in [37].

Definition 2.1 (see [37]). Let (𝑋,𝑑) be a metric space and 𝑓,𝑔∶𝑋→𝑋. If 𝑤=𝑓𝑥=𝑔𝑥, for some 𝑥∈𝑋, then 𝑥 is called a coincidence point of 𝑓 and 𝑔, and 𝑤 is called a point of coincidence of 𝑓 and 𝑔. The pair {𝑓,𝑔} is said to be compatible if and only if lim_𝑛→+∞𝑑(𝑓𝑔𝑥_𝑛,𝑔𝑓𝑥_𝑛)=0, whenever {𝑥_𝑛} is a sequence in 𝑋 such that lim_𝑛→+∞𝑓𝑥_𝑛=lim_𝑛→+∞𝑔𝑥_𝑛=𝑡 for some 𝑡∈𝑋.

Let 𝑋 be a nonempty set and 𝑅∶𝑋→𝑋 a given mapping. For every 𝑥∈𝑋, we denote by 𝑅⁻¹(𝑥) the subset of 𝑋 defined by 𝑅⁻¹(𝑥)∶={𝑢∈𝑋∣𝑅𝑢=𝑥}. (2.1)

Definition 2.2 (see [23]). Let (𝑋,⪯) be a partially ordered set and 𝑇,𝑆,𝑅∶𝑋→𝑋 are given mappings such that 𝑇𝑋⊆𝑅𝑋 and 𝑆𝑋⊆𝑅𝑋. We say that 𝑆 and 𝑇 are weakly increasing with respect to 𝑅 if and only if, for all 𝑥∈𝑋, we have 𝑇𝑥⪯𝑆𝑦,∀𝑦∈𝑅⁻¹(𝑇𝑥),𝑆𝑥⪯𝑇𝑦,∀𝑦∈𝑅⁻¹(𝑆𝑥). (2.2)

Remark 2.3. If 𝑅∶𝑋→𝑋 is the identity mapping (𝑅𝑥=𝑥 for all 𝑥∈𝑋), then 𝑆 and 𝑇 are weakly increasing with respect to 𝑅whichimplies that 𝑆 and 𝑇 are weakly increasing mappings. Note that the notion of weakly increasing mappings was introduced in [9] (also see [17, 38]).

Example 2.4. Let 𝑋={1,2,3} be endowed with the partial order ⪯ given by ⪯∶={(1,1),(2,2),(3,3),(2,3),(3,1),(2,1)}. (2.3) Define the mappings 𝑇,𝑆,𝑅∶𝑋→𝑋 by 𝑇1=𝑇3=1,𝑇2=3,𝑆1=𝑆2=𝑆3=1,𝑅1=1,𝑅2=𝑅3=2. (2.4) We will show that the mappings 𝑆 and 𝑇 are weakly increasing with respect to 𝑅.
Let 𝑥,𝑦∈𝑋 such that 𝑦∈𝑅⁻¹(𝑇𝑥). By the definition of 𝑆, we have 𝑆𝑦=1. On the other hand, 𝑇𝑥∈{1,3} and (1,1),(3,1)∈⪯. Thus, we have 𝑇𝑥⪯𝑆𝑦 for all 𝑦∈𝑅⁻¹(𝑇𝑥).
Let 𝑥,𝑦∈𝑋 such that 𝑦∈𝑅⁻¹(𝑆𝑥). By the definitions of 𝑆 and 𝑅, we have 𝑅⁻¹(𝑆𝑥)=𝑅⁻¹(1)={1}. Then we have 𝑦=1. On the other hand, 1=𝑆𝑥⪯𝑇𝑦=𝑇1=1. Then, 𝑆𝑥⪯𝑇𝑦 for all 𝑦∈𝑅⁻¹(𝑆𝑥).
Thus, we proved that 𝑆 and 𝑇 are weakly increasing with respect to 𝑅.

Definition 2.5. Let 𝑋 be a nonempty set. Then (𝑋,𝑑,⪯) is called an ordered metric space if and only if:(i)(𝑋,𝑑) is a metric space,(ii)(𝑋,⪯) is a partially ordered set.

2.2. Results

Our first result is the following.

Theorem 2.6. Let (𝑋,𝑑,⪯) be an ordered complete metric space. Let 𝑇,𝑆,𝑅∶𝑋→𝑋 be given mappings satisfying for every pair (𝑥,𝑦)∈𝑋×𝑋 such that 𝑅𝑥 and 𝑅𝑦 are comparable, Φ₁𝑑𝑆𝑥,𝑇𝑦≤𝜓₁𝑑(),1𝑅𝑥,𝑅𝑦,𝑑𝑅𝑥,𝑆𝑥,𝑑𝑅𝑦,𝑇𝑦2𝑑𝑅𝑥,𝑇𝑦+𝑑𝑅𝑦,𝑆𝑥−𝜓₂𝑑(),1𝑅𝑥,𝑅𝑦,𝑑𝑅𝑥,𝑆𝑥,𝑑𝑅𝑦,𝑇𝑦2𝑑𝑅𝑥,𝑇𝑦+𝑑𝑅𝑦,𝑆𝑥, (2.5) where 𝜓₁ and 𝜓₂ are generalized altering distance functions (in Ψ₄) and Φ₁(𝑥)=𝜓₁(𝑥,𝑥,𝑥,𝑥). One assumes the following hypotheses:(i)𝑇, 𝑆, and 𝑅 are continuous,(ii)𝑇𝑋⊆𝑅𝑋, 𝑆𝑋⊆𝑅𝑋,(iii)𝑇 and 𝑆 are weakly increasing with respect to 𝑅,(iv) the pairs {𝑇,𝑅} and {𝑆,𝑅} are compatible.
Then, 𝑇, 𝑆, and 𝑅 have a coincidence point, that is, there exists 𝑢∈𝑋 such that 𝑅𝑢=𝑇𝑢=𝑆𝑢.

Proof. Let 𝑥₀∈𝑋 be an arbitrary point. Since 𝑇𝑋⊆𝑅𝑋, there exists 𝑥₁∈𝑋 such that 𝑅𝑥₁=𝑇𝑥₀. Since 𝑆𝑋⊆𝑅𝑋, there exists 𝑥₂∈𝑋 such that 𝑅𝑥₂=𝑆𝑥₁. Continuing this process, we can construct a sequence {𝑅𝑥_𝑛} in 𝑋 defined by 𝑅𝑥_2𝑛+1=𝑇𝑥_2𝑛,𝑅𝑥_2𝑛+2=𝑆𝑥_2𝑛+1,∀𝑛∈ℕ. (2.6) We claim that 𝑅𝑥_𝑛⪯𝑅𝑥_𝑛+1,∀𝑛∈ℕ. (2.7) To this aim, we will use the increasing property of the mappings 𝑆 and 𝑇 with respect to 𝑅. From (2.6), we have 𝑅𝑥₁=𝑇𝑥₀⪯𝑆𝑦,∀𝑦∈𝑅⁻¹𝑇𝑥₀. (2.8) Since 𝑅𝑥₁=𝑇𝑥₀, then 𝑥₁∈𝑅⁻¹(𝑇𝑥₀), and we get 𝑅𝑥₁=𝑇𝑥₀⪯𝑆𝑥₁=𝑅𝑥₂. (2.9) Again, 𝑅𝑥₂=𝑆𝑥₁⪯𝑇𝑦,∀𝑦∈𝑅⁻¹𝑆𝑥₁. (2.10) Since 𝑥₂∈𝑅⁻¹(𝑆𝑥₁), we get 𝑅𝑥₂=𝑆𝑥₁⪯𝑇𝑥₂=𝑅𝑥₃. (2.11) Hence, by induction, (2.7) holds.
Without loss of the generality, we can assume that 𝑅𝑥_𝑛≠𝑅𝑥_𝑛+1,∀𝑛∈ℕ. (2.12)
Now, we will prove our result on three steps.
Step  1. We claim that lim_𝑛→+∞𝑑𝑅𝑥_𝑛+1,𝑅𝑥_𝑛+2=0. (2.13) Putting 𝑥=𝑥_2𝑛+1 and 𝑦=𝑥_2𝑛, from (2.7) and the considered contraction (2.5), we have Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+1=Φ₁𝑑𝑆𝑥_2𝑛+1,𝑇𝑥_2𝑛≤𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑆𝑥_2𝑛+1,𝑑𝑅𝑥_2𝑛,𝑇𝑥_2𝑛,12𝑑𝑅𝑥_2𝑛+1,𝑇𝑥_2𝑛+𝑑𝑅𝑥_2𝑛,𝑆𝑥_2𝑛+1−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑆𝑥_2𝑛+1,𝑑𝑅𝑥_2𝑛,𝑇𝑥_2𝑛,12𝑑𝑅𝑥_2𝑛+1,𝑇𝑥_2𝑛+𝑑𝑅𝑥_2𝑛,𝑆𝑥_2𝑛+1=𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2. (2.14) Suppose, for some 𝑛∈ℕ, that 𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2>𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1. (2.15) Using (2.15) and a triangular inequality, we have 12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2≤12(𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1+𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2<𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2. (2.16) Using this and (2.15) together with a property of the generalized altering function 𝜓₁, we get 𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2≤Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+1. (2.17) Hence, we obtain Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+1≤Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+1−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2. (2.18) This implies that 𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2=0, (2.19) which yields that 𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛=0. (2.20) Hence, we obtain a contradiction with (2.12). We deduce that 𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2≤𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,∀𝑛∈ℕ. (2.21) Similarly, putting 𝑥=𝑥_2𝑛+1 and 𝑦=𝑥_2𝑛+2, from (2.7) and the considered contraction (2.5), we have Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3≤𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3,12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+3−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3,12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+3. (2.22) Suppose, for some 𝑛∈ℕ, that 𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3>𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2. (2.23) Then, by a triangular inequality, we have 12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+3≤12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2+𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3<𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3. (2.24) Hence, from this, (2.22), and (2.23), we obtain Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3≤Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3,12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+3. (2.25) This implies that 𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3,12𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+3=0, (2.26) which leads to 𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2=0. (2.27) Hence, we obtain a contradiction with (2.12). We deduce that 𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2≥𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+3,∀𝑛∈ℕ. (2.28) Combining (2.21) and (2.28), we obtain 𝑑𝑅𝑥_𝑛+1,𝑅𝑥_𝑛+2≥𝑑𝑅𝑥_𝑛+2,𝑅𝑥_𝑛+3,∀𝑛∈ℕ. (2.29) Then, {𝑑(𝑅𝑥_𝑛+1,𝑅𝑥_𝑛+2)} is a nonincreasing sequence of positive real numbers. This implies that there exists 𝑟≥0 such that lim_𝑛→+∞𝑑𝑅𝑥_𝑛+1,𝑅𝑥_𝑛+2=𝑟. (2.30) By (2.14), we have Φ₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑥_2𝑛+1≤𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+2≤𝜓₁𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,0=Φ₁𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1−𝜓₂𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2,d𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,0. (2.31) Letting 𝑛→+∞ in (2.31) and using the continuities of Φ₁ and 𝜓₂, we obtain Φ₁(𝑟)≤Φ₁(𝑟)−𝜓₂(𝑟,𝑟,𝑟,0), (2.32) which implies that 𝜓₂(𝑟,𝑟,𝑟,0)=0, so 𝑟=0. Hence lim_𝑛→+∞𝑑𝑅𝑥_𝑛+1,𝑅𝑥_𝑛+2=0. (2.33) Hence, (2.13) is proved.
Step  2. We claim that {𝑅𝑥_𝑛} is a Cauchy sequence.
From (2.13), it will be sufficient to prove that {𝑅𝑥_2𝑛} is a Cauchy sequence. We proceed by negation and suppose that {𝑅𝑥_2𝑛} is not a Cauchy sequence. Then, there exists 𝜀>0 for which we can find two sequences of positive integers {𝑚(𝑖)} and {𝑛(𝑖)} such that, for all positive integer 𝑖, 𝑛(𝑖)>𝑚(𝑖)>𝑖,𝑑𝑅𝑥_2𝑚(𝑖),𝑅𝑥_2𝑛(𝑖)≥𝜀,𝑑𝑅𝑥_2𝑚(𝑖),𝑅𝑥_{2𝑛(𝑖)−2}<𝜀. (2.34) From (2.34) and using a triangular inequality, we get 𝜀≤𝑑𝑅𝑥_2𝑚(𝑖),𝑅𝑥_2𝑛(𝑖)≤𝑑𝑅𝑥_2𝑚(𝑖),𝑅𝑥_{2𝑛(𝑖)−2}+𝑑𝑅𝑥_{2𝑛(𝑖)−2},𝑅𝑥_{2𝑛(𝑖)−1}+𝑑𝑅𝑥_{2𝑛(𝑖)−1},𝑅𝑥_2𝑛(𝑖)<𝜀+𝑑𝑅𝑥_{2𝑛(𝑖)−2},𝑅𝑥_{2𝑛(𝑖)−1}+𝑑𝑅𝑥_{2𝑛(𝑖)−1},𝑅𝑥_2𝑛(𝑖). (2.35) Letting 𝑖→+∞ in the previous inequality and using (2.13), we obtain lim_𝑖→+∞𝑑𝑅𝑥_2𝑚(𝑖),𝑅𝑥_2𝑛(𝑖)=𝜀. (2.36) Again, a triangular inequality gives us 𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑚(𝑖)−1}−𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_2𝑚(𝑖)≤𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_2𝑚(𝑖). (2.37) Letting 𝑖→+∞ in the above inequality and using (2.13) and (2.36), we get lim_𝑖→+∞𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑚(𝑖)−1}=𝜀. (2.38) On the other hand, we have 𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_2𝑚(𝑖)≤𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑛(𝑖)+1}+𝑑𝑅𝑥_{2𝑛(𝑖)+1},𝑅𝑥_2𝑚(𝑖)=𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑛(𝑖)+1}+𝑑𝑇𝑥_2𝑛(𝑖),𝑆𝑥_{2𝑚(𝑖)−1}. (2.39) Then, from (2.13), (2.36), and the continuity of Φ₁, we get by letting 𝑖→+∞ in the above inequality Φ₁(𝜀)≤lim_𝑖→+∞Φ₁𝑑𝑆𝑥_{2𝑚(𝑖)−1},𝑇𝑥_2𝑛(𝑖). (2.40) Now, using the considered contractive condition (2.5) for 𝑥=𝑥_{2𝑚(𝑖)−1} and 𝑦=𝑥_2𝑛(𝑖), we have Φ₁𝑑𝑆𝑥_{2𝑚(𝑖)−1},𝑇𝑥_2𝑛(𝑖)≤𝜓₁𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_2𝑛(𝑖),𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_2𝑚(𝑖),𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑛(𝑖)+1},12𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_{2𝑛(𝑖)+1}+𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_2𝑚(𝑖)−𝜓₂𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_2𝑛(𝑖),𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_2𝑚(𝑖),𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_{2𝑛(𝑖)+1},12𝑑𝑅𝑥_{2𝑚(𝑖)−1},𝑅𝑥_{2𝑛(𝑖)+1}+𝑑𝑅𝑥_2𝑛(𝑖),𝑅𝑥_2𝑚(𝑖). (2.41) Then, from (2.13), (2.38), and the continuities of 𝜓₁ and 𝜓₂, we get by letting 𝑖→+∞ in the above inequality lim_𝑖→+∞Φ₁𝑑𝑆𝑥_{2𝑚(𝑖)−1},𝑇𝑥_2𝑛(𝑖)≤𝜓₁(𝜀,0,0,𝜖)−𝜓₂(𝜀,0,0,𝜖)≤Φ₁(𝜀)−𝜓₂(𝜀,0,0,𝜖). (2.42) Now, combining (2.40) with the previous inequality, we get Φ₁(𝜀)≤Φ₁(𝜀)−𝜓₂(𝜀,0,0,𝜖), (2.43) which implies that 𝜓₂(𝜀,0,0,𝜖)=0, that is a contradiction since 𝜀>0. We deduce that {𝑅𝑥_𝑛} is a Cauchy sequence.
Step  3. We claimexistence of a coincidence point.
Since {𝑅𝑥_𝑛} is a Cauchy sequence in the complete metric space (𝑋,𝑑), there exists 𝑢∈𝑋 such that lim_𝑛→+∞𝑅𝑥_𝑛=𝑢. (2.44) From (2.44) and the continuity of 𝑅, we get lim_𝑛→+∞𝑅𝑅𝑥_𝑛=𝑅𝑢. (2.45) By the triangular inequality, we have 𝑑(𝑅𝑢,𝑇𝑢)≤𝑑𝑅𝑢,𝑅𝑅𝑥_2𝑛+1+𝑑𝑅𝑇𝑥_2𝑛,𝑇𝑅𝑥_2𝑛+𝑑𝑇𝑅𝑥_2𝑛,𝑇𝑢. (2.46) On the other hand, we have 𝑅𝑥_2𝑛⟶𝑢,𝑇𝑥_2𝑛⟶𝑢as𝑛→+∞. (2.47) Since 𝑅 and 𝑇 are compatible mappings, this implies that lim_𝑛→+∞𝑑𝑅𝑇𝑥_2𝑛,𝑇𝑅𝑥_2𝑛=0. (2.48) Now, from the continuity of 𝑇 and (2.44), we have lim_𝑛→+∞𝑑𝑇𝑅𝑥_2𝑛,𝑇𝑢=0. (2.49) Combining (2.45), (2.48), and (2.49) and letting 𝑛→+∞ in (2.46), we obtain 𝑑(𝑅𝑢,𝑇𝑢)≤0, (2.50) that is, 𝑅𝑢=𝑇𝑢. (2.51) Again, by a triangular inequality, we have 𝑑(𝑅𝑢,𝑆𝑢)≤𝑑𝑅𝑢,𝑅𝑅𝑥_2𝑛+2+𝑑𝑅𝑆𝑥_2𝑛+1,𝑆𝑅𝑥_2𝑛+1+𝑑𝑆𝑅𝑥_2𝑛+1,𝑆𝑢. (2.52) On the other hand, we have 𝑅𝑥_2𝑛+1⟶𝑢,𝑆𝑥_2𝑛+1⟶𝑢as𝑛⟶+∞. (2.53) Since 𝑅 and 𝑆 are compatible mappings, this implies that lim_𝑛→+∞𝑑𝑅𝑆𝑥_2𝑛+1,𝑆𝑅𝑥_2𝑛+1=0. (2.54) Now, from the continuity of 𝑆 and (2.44), we have lim_𝑛→+∞𝑑𝑆𝑅𝑥_2𝑛+1,𝑆𝑢=0. (2.55) Combining (2.45), (2.54), and (2.55) and letting 𝑛→+∞ in (2.52), we obtain 𝑑(𝑅𝑢,𝑆𝑢)≤0, (2.56) that is, 𝑅𝑢=𝑆𝑢. (2.57) Finally, from (2.51) and (2.57), we have 𝑇𝑢=𝑅𝑢=𝑆𝑢, (2.58) that is, 𝑢 is a coincidence point of 𝑇, 𝑆, and 𝑅. This makes end to the proof.

In the next theorem, we omit the continuity hypotheses satisfied by 𝑇, 𝑆, and 𝑅.

Definition 2.7. Let (𝑋,𝑑,⪯) be a partially ordered metric space. We say that 𝑋 is regular if and only if the following hypothesis holds: if {𝑧_𝑛} is a nondecreasing sequence in 𝑋 with respect to ⪯such that 𝑧_𝑛→𝑧∈𝑋 as 𝑛→+∞, then 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ.

Now, our second result is the following.

Theorem 2.8. Let (𝑋,𝑑,⪯) be an ordered complete metric space. Let 𝑇,𝑆,𝑅∶𝑋→𝑋 be given mappings satisfying for every pair (𝑥,𝑦)∈𝑋×𝑋 such that 𝑅𝑥 and 𝑅𝑦 are comparable, Φ₁𝑑𝑆𝑥,𝑇𝑦≤𝜓₁𝑑(),1𝑅𝑥,𝑅𝑦,𝑑𝑅𝑥,𝑆𝑥,𝑑𝑅𝑦,𝑇𝑦2𝑑𝑅𝑥,𝑇𝑦+𝑑𝑅𝑦,𝑆𝑥−𝜓₂𝑑(),1𝑅𝑥,𝑅𝑦,𝑑𝑅𝑥,𝑆𝑥,𝑑𝑅𝑦,𝑇𝑦2𝑑𝑅𝑥,𝑇𝑦+𝑑𝑅𝑦,𝑆𝑥, (2.59) where 𝜓₁ and 𝜓₂ are generalized altering distance functions and Φ₁(𝑥)=𝜓₁(𝑥,𝑥,𝑥,𝑥). We assume the following hypotheses:(i)𝑋 is regular,(ii)𝑇 and 𝑆 are weakly increasing with respect to 𝑅,(iii)𝑅𝑋 is a complete subspace of (𝑋,𝑑),(iv)𝑇𝑋⊆𝑅𝑋, 𝑆𝑋⊆𝑅𝑋.
Then, 𝑇, 𝑆, and 𝑅 have a coincidence point.

Proof. Following the proof of Theorem 2.6, we have {𝑅𝑥_𝑛} is a Cauchy sequence in (𝑅𝑋,𝑑). Since 𝑅𝑋 is a complete, there exists 𝑢=𝑅𝑣, 𝑣∈𝑋 such that lim_𝑛→+∞𝑅𝑥_𝑛=𝑢=𝑅𝑣. (2.60) Since {𝑅𝑥_𝑛} is a nondecreasing sequence and 𝑋 is regular, it follows from (2.60) that 𝑅𝑥_𝑛⪯𝑅𝑣 for all 𝑛∈ℕ. Hence, we can apply the contractive condition (2.5). Then, for 𝑥=𝑣 and 𝑦=𝑥_2𝑛, we obtain Φ₁𝑑𝑆𝑣,𝑅𝑥_2𝑛+1=Φ₁𝑑𝑆𝑣,𝑇𝑥_2𝑛≤𝜓₁𝑑𝑅𝑣,𝑅𝑥_2𝑛,𝑑(𝑅𝑣,𝑆𝑣),𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑣,𝑇𝑥_2𝑛+𝑑𝑅𝑥_2𝑛,𝑆𝑣−𝜓₂𝑑𝑅𝑣,𝑅𝑥_2𝑛,𝑑(𝑅𝑣,𝑆𝑣),𝑑𝑅𝑥_2𝑛,𝑅𝑥_2𝑛+1,12𝑑𝑅𝑣,𝑇𝑥_2𝑛+𝑑𝑅𝑥_2𝑛,𝑆𝑣. (2.61) Letting 𝑛→+∞ in the above inequality and using (2.13), (2.60), and the properties of 𝜓₁ and 𝜓₂, we obtain Φ₁(𝑑(𝑆𝑣,𝑅𝑣))≤𝜓₁()10,𝑑𝑅𝑣,𝑆𝑣,0,2𝑑(𝑅𝑣,𝑆𝑣)−𝜓₂()10,𝑑𝑅𝑣,𝑆𝑣,0,2𝑑(𝑅𝑣,𝑆𝑣)≤Φ₁(𝑑(𝑆𝑣,𝑅𝑣))−𝜓₂()10,𝑑𝑅𝑣,𝑆𝑣,0,2𝑑(𝑅𝑣,𝑆𝑣). (2.62) This implies that 𝜓₂(0,𝑑(𝑅𝑣,𝑆𝑣),0,(1/2)𝑑(𝑅𝑣,𝑆𝑣))=0, which gives us that 𝑑(𝑅𝑣,𝑆𝑣)=0, that is, 𝑅𝑣=𝑆𝑣. (2.63) Similarly, for 𝑥=𝑥_2𝑛+1 and 𝑦=𝑣, we obtain Φ₁𝑑𝑅𝑥_2𝑛+2,𝑇𝑣=Φ₁𝑑𝑆𝑥_2𝑛+1,𝑇𝑣≤𝜓₁𝑑𝑅𝑥_2𝑛+2,𝑅𝑣,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2(),1,𝑑𝑅𝑣,𝑇𝑣2𝑑𝑅𝑥_2𝑛+1,𝑇𝑣+𝑑𝑅𝑣,𝑆𝑥_2𝑛+1−𝜓₂𝑑𝑅𝑥_2𝑛+2,𝑅𝑣,𝑑𝑅𝑥_2𝑛+1,𝑅𝑥_2𝑛+2(),1,𝑑𝑅𝑣,𝑇𝑣2𝑑𝑅𝑥_2𝑛+1,𝑇𝑣+𝑑𝑅𝑣,𝑆𝑥_2𝑛+1. (2.64) Letting 𝑛→+∞ in the above inequality, we get Φ₁(𝑑(𝑅𝑣,𝑇𝑣))≤𝜓₁(),10,0,𝑑𝑅𝑣,𝑇𝑣2𝑑(𝑅𝑣,𝑇𝑣)−𝜓₂(),10,0,𝑑𝑅𝑣,𝑇𝑣2𝑑(𝑅𝑣,𝑇𝑣)≤Φ₁(𝑑(𝑅𝑣,𝑇𝑣))−𝜓₂(),10,0,𝑑𝑅𝑣,𝑇𝑣2𝑑(𝑅𝑣,𝑇𝑣). (2.65) This implies that 𝜓₂(0,0,𝑑(𝑅𝑣,𝑇𝑣),(1/2)𝑑(𝑅𝑣,𝑇𝑣))=0, and then, 𝑅𝑣=𝑇𝑣. (2.66) Now, combining (2.63) and (2.66), we obtain 𝑅𝑣=𝑇𝑣=𝑆𝑣. (2.67) Hence, 𝑣 is a coincidence point of 𝑇, 𝑆, and 𝑅. This makes end to the proof.

Now, we give a sufficient condition that assures the uniqueness of the common coincidence point of {𝑇,𝑅} and {𝑆,𝑅}.

Theorem 2.9. Under the hypotheses of Theorem 2.6 (resp., Theorem 2.8) and suppose that (𝑆𝑋,⪯) is a totally ordered set and 𝑆 is one-to-one mapping, then one obtains a unique common coincidence point of {𝑇,𝑅} and {𝑆,𝑅}.

Proof. Following the proof of Theorem 2.6 (resp., Theorem 2.8), the set of common coincidence points of {𝑇,𝑅} and {𝑆,𝑅} is nonempty. Let 𝜇,𝜈∈𝑋be two common coincidence points of {𝑇,𝑅} and {𝑆,𝑅}, that is, 𝑇𝜇=𝑅𝜇=𝑆𝜇,𝑇𝜈=𝑅𝜈=𝑆𝜈. (2.68) This implies that 𝑅𝜇,𝑅𝜈∈𝑆𝑋, and then, 𝑅𝜇 and 𝑅𝜈 are comparable with respect to ⪯. Then, we can apply the contractive condition (2.5). We have Φ₁𝑑𝑆𝜇,𝑇𝜈≤𝜓₁𝑑(),1𝑅𝜇,𝑅𝜈,𝑑𝑅𝜇,𝑆𝜇,𝑑𝑅𝜈,𝑇𝜈2𝑑𝑅𝜇,𝑇𝜈+𝑑𝑅𝜈,𝑆𝜇−𝜓₂𝑑(),1𝑅𝜇,𝑅𝜈,𝑑𝑅𝜇,𝑆𝜇,𝑑𝑅𝜈,𝑇𝜈2𝑑𝑅𝜇,𝑇𝜈+𝑑𝑅𝜈,𝑆𝜇=𝜓₁𝑑𝑆𝜇,𝑇𝜈,0,0,𝑑𝑆𝜇,𝑇𝜈−𝜓₂𝑑𝑆𝜇,𝑇𝜈,0,0,𝑑𝑆𝜇,𝑇𝜈≤Φ₁𝑑𝑆𝜇,𝑇𝜈−𝜓₂𝑑𝑆𝜇,𝑇𝜈,0,0,𝑑𝑆𝜇,𝑇𝜈. (2.69) This implies that 𝜓₂(𝑑(𝑆𝜇,𝑇𝜈),0,0,𝑑(𝑆𝜇,𝑇𝜈)), which gives us that 𝑑(𝑆𝜇,𝑇𝜈)=0, that is, 𝑆𝜇=𝑇𝜈. From (2.68), we get 𝑆𝜇=𝑆𝜈. Since 𝑆 is one-to-one, we have 𝜇=𝜈. This makes end to the proof.

Now, it is easy to state a corollary of Theorem 2.6 or Theorem 2.8 involving contractions of integral type.

Corollary 2.10. Let 𝑅,𝑆 and 𝑇 satisfy the conditions of Theorem 2.6 or Theorem 2.8, except that condition (2.5) is replaced by the following: there exists a positive Lebesgue integrable function 𝑢 on ℝ₊ such that ∫^𝜀₀𝑢(𝑡)𝑑𝑡>0 for each 𝜀>0 and that ^Φ1₀^{(𝑑(𝑆𝑥,𝑇𝑦))}𝑢(𝑡)𝑑𝑡≤^𝜓1₀^{(𝑑(𝑅𝑥,𝑅𝑦),𝑑(𝑅𝑥,𝑆𝑥),𝑑(𝑅𝑦,𝑇𝑦),(1/2)[𝑑(𝑅𝑥,𝑇𝑦)+𝑑(𝑅𝑦,𝑆𝑥)])}𝑢(𝑡)−𝑑𝑡_{𝜓20(𝑑(𝑅𝑥,𝑅𝑦),𝑑(𝑅𝑥,𝑆𝑥),𝑑(𝑅𝑦,𝑇𝑦),(1/2)[𝑑(𝑅𝑥,𝑇𝑦)+𝑑(𝑅𝑦,𝑆𝑥)])}𝑢(𝑡)𝑑𝑡. (2.70) Then, 𝑇, 𝑆, and 𝑅 have a coincidence point.

If 𝑅∶𝑋→𝑋 is the identity mapping, we can deduce easily the following common fixed point results.

The next result is an immediate consequence of Theorem 2.6.

Corollary 2.11. Let (𝑋,𝑑,⪯) be an ordered complete metric space. Let 𝑇,𝑆∶𝑋→𝑋 be given mappings satisfying for every pair (𝑥,𝑦)∈𝑋×𝑋 such that 𝑥 and 𝑦 are comparable, Φ₁𝑑𝑆𝑥,𝑇𝑦≤𝜓₁𝑑(),1𝑥,𝑦,𝑑𝑥,𝑆𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑆𝑥−𝜓₂𝑑(),1𝑥,𝑦,𝑑𝑥,𝑆𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑆𝑥, (2.71) where 𝜓₁ and 𝜓₂ are generalized altering distance functions and Φ₁(𝑥)=𝜓₁(𝑥,𝑥,𝑥,𝑥). One assumes the following hypotheses:(i)𝑇 and 𝑆 are continuous,(ii)𝑇 and 𝑆 are weakly increasing.
Then, 𝑇 and 𝑆 have a common fixed point, that is, there exists 𝑢∈𝑋 such that 𝑢=𝑇𝑢=𝑆𝑢.

The following result is an immediate consequence of Theorem 2.8.

Corollary 2.12. Let (𝑋,𝑑,⪯) be an ordered complete metric space. Let 𝑇,𝑆∶𝑋→𝑋 be given mappings satisfying for every pair (𝑥,𝑦)∈𝑋×𝑋 such that 𝑥 and 𝑦 are comparable, Φ₁𝑑𝑆𝑥,𝑇𝑦≤𝜓₁𝑑(),1𝑥,𝑦,𝑑𝑥,𝑆𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑆𝑥−𝜓₂𝑑(),1𝑥,𝑦,𝑑𝑥,𝑆𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑆𝑥, (2.72) where 𝜓₁ and 𝜓₂ are generalized altering distance functions and Φ₁(𝑥)=𝜓₁(𝑥,𝑥,𝑥,𝑥). One assume the following hypotheses:(i)𝑋 is regular,(ii)𝑇 and 𝑆 are weakly increasing.
Then, 𝑇 and 𝑆 have a common fixed point.

Now, we give some examples to support our results.

Example 2.13. Let 𝑋={4,5,6} be endowed with the usual metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋, and ⪯∶={(4,4),(5,5),(6,6),(6,4)}. Consider the mappings 𝑆=𝑇=456464,𝑅=456456. (2.73) We define the functions 𝜓₁,𝜓₂∶[0,+∞)⁴→[0,+∞) by 𝜓₁𝑡₁,𝑡₂,𝑡₃,𝑡₄=14𝑡₁+𝑡₂+𝑡₃+𝑡₄,𝜓₂𝑡₁,𝑡₂,𝑡₃,𝑡₄=1𝑡16¹+𝑡₂+𝑡₃+𝑡₄. (2.74) Clearly, Φ₁(𝑡)=𝑡 for all 𝑡≥0. Now, we will check that all the hypotheses required by Theorem 2.8 are satisfied.
(i) 𝑋 is regular.
Let {𝑧_𝑛} be a nondecreasing sequence in 𝑋 with respect to ⪯ such that 𝑧_𝑛→𝑧∈𝑋 as 𝑛→+∞. We have 𝑧_𝑛⪯𝑧_𝑛+1 for all 𝑛∈ℕ.(a)If 𝑧₀=4, then 𝑧₀=4⪯𝑧₁. From the definition of ⪯, we have 𝑧₁=4. By induction, we get 𝑧_𝑛=4 for all 𝑛∈ℕ and 𝑧=4. Then, 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ.(b)If 𝑧₀=5, then 𝑧₀=5⪯𝑧₁. From the definition of ⪯, we have 𝑧₁=5. By induction, we get 𝑧_𝑛=5 for all 𝑛∈ℕ and 𝑧=5. Then, 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ.(c)If 𝑧₀=6, then 𝑧₀=6⪯𝑧₁. From the definition of ⪯, we have 𝑧₁∈{6,4}. By induction, we get 𝑧_𝑛∈{6,4} for all 𝑛∈ℕ. Suppose that there exists 𝑝≥1 such that 𝑧_𝑝=4. From the definition of ⪯, we get 𝑧_𝑛=𝑧_𝑝=4 for all 𝑛≥𝑝. Thus, we have 𝑧=4 and 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ. Now, suppose that 𝑧_𝑛=6 for all 𝑛∈ℕ. In this case, we get 𝑧=6 and 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ.
Thus, we proved that, in all cases, we have 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ. Then, 𝑋 is regular.
(ii) 𝑇 and 𝑆 are weakly increasing.
Since 𝑆=𝑇, we have to check that 𝑇𝑥⪯𝑇(𝑇𝑥) for all 𝑥∈𝑋.
For 𝑥=4, we have 𝑇4=4⪯𝑇(𝑇4)=𝑇4=4. (2.75) For 𝑥=5, we have 𝑇5=6⪯𝑇(𝑇5)=𝑇6=4. (2.76) For 𝑥=6, we have 𝑇6=4⪯𝑇(𝑇6)=𝑇4=4. (2.77) Thus, we proved that 𝑇 and 𝑆 are weakly increasing.
On the other hand, it is very easy to show that (2.5) is satisfied for all 𝑥,𝑦∈𝑋 such that 𝑅𝑥⪯𝑅𝑦.
Now, all the hypotheses of Theorem 2.8 are satisfied. Then 𝑅, 𝑆, and 𝑇 have a coincidence point 𝑢=4.
Note that inequality (2.5) is not satisfied for 𝑥=4 and 𝑦=5. Indeed, 𝜓₁𝑑()()(),1𝑅4,𝑅5,𝑑𝑅4,𝑆4,𝑑𝑅5,𝑇52[𝑑(𝑅4,𝑇5)+𝑑(𝑅5,𝑆4)]−𝜓₂𝑑()()(),1𝑅4,𝑅5,𝑑𝑅4,𝑆4,𝑑𝑅5,𝑇52[𝑑(𝑅4,𝑇5)+𝑑(𝑅5,𝑆4)]=𝜓₁31,0,1,2−𝜓₂31,0,1,2=2132<Φ₁(𝑑(𝑆4,𝑇5))=Φ₁(2)=2. (2.78) Then, Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.

Example 2.14. Let 𝑋={(0,1),(1,0),(1,1)}⊂ℝ² with the Euclidean distance 𝑑₂. (𝑋,𝑑₂) is, obviously, a complete metric space. Moreover, we consider the order ⪯ in 𝑋 given by ℛ={(𝑥,𝑥),𝑥∈𝑋}. Notice that the elements in 𝑋 are only comparable to themselves, so (𝑋,⪯) is regular. Also we consider 𝑅,𝑆,𝑇∶𝑋→𝑋 given by 𝑇()()()=(),𝑇()()=()()=(),𝑇()()()=().1,0=𝑆1,0=𝑅1,01,00,1=𝑆0,11,0,𝑅0,10,11,1=𝑆1,1=𝑅1,11,1 (2.79) It is easy that, for all 𝑥∈𝑋, 𝑇𝑆𝑥⪯𝑆𝑥, and 𝑆𝑇𝑥⪯𝑇𝑥, so the pair {𝑆,𝑇} is weakly increasing.
Define 𝜓₁,𝜓₂∶[0,+∞)⁴→[0,+∞) by 𝜓₁𝑡₁,𝑡₂,𝑡₃,𝑡₄=max𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝜓₂𝑡₁,𝑡₂,𝑡₃,𝑡₄=12max𝑡₁,𝑡₂,𝑡₃,𝑡₄. (2.80) Then, 𝜓₁(𝑡)=𝑡 for all 𝑡≥0.
As the elements in 𝑋 are only comparable to themselves, condition (2.5) appearing in Theorem 2.8 is, obviously, satisfied. Now, all the hypotheses of Theorem 2.8 are satisfied. (1,1) and (1,0) are the coincidence points of the mappings 𝑅, 𝑆, and 𝑇.
On the other hand, the inequality (2.5) is not satisfied for 𝑥=(1,0) and 𝑦=(1,1). Indeed, 𝜓₁𝑑₂(𝑅(1,0),𝑅(1,1)),𝑑₂(𝑅(1,0),𝑆(1,0)),𝑑₂(𝑅()(,11,1,𝑇1,1))2𝑑₂(𝑅(1,0),𝑇(1,1))+𝑑₂(𝑅(1,1),𝑆(1,0))−𝜓₂𝑑₂(𝑅(1,0),𝑅(1,1)),𝑑₂(𝑅(1,0),𝑆(1,0)),𝑑₂(𝑅()(,11,1,𝑇1,1))2𝑑₂(𝑅(1,0),𝑇(1,1))+𝑑₂(𝑅(1,1),𝑆(1,0))=𝜓₁(1,0,0,1)−𝜓₂()=11,0,0,12<Φ₁𝑑₂(𝑆(1,0),𝑇(1,1))=Φ₁(1)=1. (2.81) Then, again Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.

A number of fixed point results may be obtained by assuming different forms for the functions 𝜓₁ and 𝜓₂. In particular, fixed point results under various contractive conditions may be derived from the above theorems. For example, if we consider 𝜓₁𝑥,𝑦,𝑧,𝑡=𝑘₁𝑥^𝑠+𝑘₂𝑦^𝑠+𝑘₃𝑧^𝑠+𝑘₄𝑡⁴,𝜓₂𝑥,𝑦,𝑧,𝑡=(1−𝑘)𝑘₁𝑥^𝑠+𝑘₂𝑦^𝑠+𝑘₃𝑧^𝑠+𝑘₄𝑡⁴, (2.82) where 𝑠>0 and 0<𝑘=𝑘₁+𝑘₂+𝑘₃+𝑘₄<1, we obtain the following results.

The next result is an immediate consequence of Corollaries 2.11 and 2.12.

Corollary 2.15. Let (𝑋,𝑑,⪯) be an ordered complete metric space. Let 𝑇,𝑆∶𝑋→𝑋 be given mappings satisfying for every pair (𝑥,𝑦)∈𝑋×𝑋 such that 𝑥 and 𝑦 are comparable, 𝑑𝑆𝑥,𝑇𝑦^𝑠≤𝑘₁𝑑𝑥,𝑦^𝑠+𝑘₂[𝑑(𝑥,𝑆𝑥)]^𝑠+𝑘₃𝑑𝑦,𝑇𝑦^𝑠+𝑘₄12𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑆𝑥^𝑠, (2.83) where 𝑠>0 and 0<𝑘=𝑘₁+𝑘₂+𝑘₃+𝑘₄<1. One assumes the following hypotheses:(i)𝑇 and 𝑆 are continuous or 𝑋 is regular,(ii)𝑇 and 𝑆 are weakly increasing.
Then, 𝑇 and 𝑆 have a common fixed point, that is, there exists 𝑢∈𝑋 such that 𝑢=𝑇𝑢=𝑆𝑢.

Remark 2.16. Other fixed point results may also be obtained under specific choices of 𝜓₁ and 𝜓₂.

3. Application

Consider the integral equations: 𝑢(𝑡)=^𝑇₀𝐾₁((𝑠(𝑡)[],𝑢(𝑡)=𝑡,𝑠,𝑢))𝑑𝑠+𝑔,𝑡∈0,𝑇_𝑇0𝐾₂(𝑡,𝑠,𝑢(𝑠))𝑑𝑠+𝑔(𝑡),𝑡∈[0,𝑇], (3.1) where 𝑇>0.

The purpose of this section is to give an existence theorem for common solution of (3.1) using Corollary 2.15. This application is inspired by [9].

Previously, we consider the space 𝑋=𝐶(𝐼)(𝐼=[0,𝑇]) of continuous functions defined on 𝐼. Obviously, this space with the metric given by 𝑑𝑥,𝑦=sup_𝑡∈𝐼𝑥(𝑡)−𝑦(𝑡),∀𝑥,𝑦∈𝐶(𝐼), (3.2) is a complete metric space. 𝐶(𝐼) can also be equipped with the partial order ⪯ given by 𝑥,𝑦∈𝐶(𝐼),𝑥⪯𝑦⟺𝑥(𝑡)≤𝑦(𝑡),∀𝑡∈𝐼. (3.3) Moreover, in [28], it is proved that (𝐶(𝐼),⪯) is regular.

Now, we will prove the following result.

Theorem 3.1. Suppose that the following hypotheses hold:
(i)𝐾₁,𝐾₂∶𝐼×𝐼×ℝ→ℝ and 𝑔∶ℝ→ℝ are continuous,
(ii) for all 𝑡,𝑠∈𝐼, 𝐾₁(𝑡,𝑠,𝑢(𝑡))≤𝐾₂𝑡,𝑠,^𝑇₀𝐾₁((𝜏(𝑠),𝐾𝑠,𝜏,𝑢))𝑑𝜏+𝑔₂(𝑡,𝑠,𝑢(𝑡))≤𝐾₁𝑡,𝑠,^𝑇₀𝐾₂(𝑠,𝜏,𝑢(𝜏))𝑑𝜏+𝑔(𝑠), (3.4)
(iii) there exist 𝑘₁,𝑘₂,𝑘₃≥0 such that 𝐺_𝑥(𝑡)−𝐹_𝑦(𝑡)≤𝑘₁𝑦(𝑡)−𝑥(𝑡)+𝑘₂𝑥(𝑡)−𝐺_𝑥(𝑡)−𝑔(𝑡)+𝑘₃𝑦(𝑡)−𝐹_𝑦(𝑡)−𝑔(𝑡), (3.5) where 𝐹_𝑥(𝑡)=^𝑇₀𝐾₁(𝑡,𝑠,𝑥(𝑠))𝑑𝑠,𝑡∈𝐼,𝐺_𝑥(𝑡)=^𝑇₀𝐾₂(𝑡,𝑠,𝑥(𝑠))𝑑𝑠,𝑡∈𝐼, (3.6)
and 𝑘₁+𝑘₂+𝑘₃<1, for every 𝑥,𝑦∈𝑋 and 𝑥⪯𝑦 and 𝑡∈𝐼.
Then, the integral equations (3.1) have a solution 𝑢^∗∈𝐶(𝐼).

Proof. Define 𝑇,𝑆∶𝐶(𝐼)→𝐶(𝐼) by 𝑇𝑥(𝑡)=𝐹_𝑥(𝑡)(𝑡)(𝑡)+𝑔,𝑡∈𝐼,𝑆𝑥=𝐺_𝑥(𝑡)+𝑔(𝑡),𝑡∈𝐼. (3.7) Now, we will prove that 𝑇 and 𝑆 are weakly increasing. From (ii), for all 𝑡∈𝐼, we have 𝑇𝑥(𝑡)=^𝑇₀𝐾₁((𝑠(𝑡)≤𝑡,𝑠,𝑥))𝑑𝑠+𝑔_𝑇0𝐾₂𝑡,𝑠,^𝑇₀𝐾₁((𝜏(𝑠)(𝑡)=𝑠,𝜏,𝑥))𝑑𝜏+𝑔𝑑𝑠+𝑔_𝑇0𝐾₂((𝑠(𝑡)(𝑡).𝑡,𝑠,𝑇𝑥))𝑑𝑠+𝑔=𝑆𝑇𝑥 (3.8) Similarly, 𝑆𝑥(𝑡)=^𝑇₀𝐾₂((𝑠(𝑡)≤𝑡,𝑠,𝑥))𝑑𝑠+𝑔_𝑇0𝐾₁𝑡,𝑠,^𝑇₀𝐾₂((𝜏(𝑠)(𝑡)=𝑠,𝜏,𝑥))𝑑𝜏+𝑔𝑑𝑠+𝑔_𝑇0𝐾₁((𝑠(𝑡)(𝑡).𝑡,𝑠,𝑆𝑥))𝑑𝑠+𝑔=𝑇𝑆𝑥 (3.9) Then, we have 𝑇𝑥⪯𝑆𝑇𝑥 and 𝑆𝑥⪯𝑇𝑆𝑥 for all 𝑥∈𝐶(𝐼). This implies that 𝑇 and 𝑆 are weakly increasing.
Now, for all 𝑥,𝑦∈𝐶(𝐼) such that 𝑥⪯𝑦, by (iii), we have 𝑆𝑥(𝑡)−𝑇𝑦(𝑡)=𝐺_𝑥(𝑡)−𝐹_𝑦(𝑡)≤𝑘₁𝑦(𝑡)−𝑥(𝑡)+𝑘₂𝑥(𝑡)−𝐺_𝑥(𝑡)−𝑔(𝑡)+𝑘₃𝑦(𝑡)−𝐹_𝑦(𝑡)−𝑔(𝑡). (3.10) Hence 𝑑𝑆𝑥,𝑇𝑦=sup_{[]𝑡∈0,𝑇}𝑆𝑥(𝑡)−𝑇𝑦(𝑡)≤𝑘₁sup_{[]𝑡∈0,𝑇}𝑥(𝑡)−𝑦(𝑡)+𝑘₂sup_{[]𝑡∈0,𝑇}𝑥(𝑡)−𝐺_𝑥(𝑡)−𝑔(𝑡)+𝑘₃sup_{[]𝑡∈0,𝑇}𝑦(𝑡)−𝐹_𝑦(𝑡)−𝑔(𝑡)=𝑘₁sup_{[]𝑡∈0,𝑇}𝑥(𝑡)−𝑦(𝑡)+𝑘₂sup_{[]𝑡∈0,𝑇}|𝑥(𝑡)−𝑆𝑥(𝑡)|+𝑘₃sup_{[]𝑡∈0,𝑇}𝑦(𝑡)−𝑇𝑦(𝑡). (3.11) Then 𝑑𝑆𝑥,𝑇𝑦≤𝑘₁𝑑𝑥,𝑦+𝑘₂𝑑(𝑥,𝑆𝑥)+𝑘₃𝑑𝑦,𝑇𝑦 (3.12) for all 𝑥,𝑦∈𝑋 such that 𝑦⪯𝑥.
This implies that, for all 𝑥,𝑦∈𝐶(𝐼) such that 𝑥⪯𝑦, 𝑑𝑆𝑥,𝑇𝑦≤𝑘₁𝑑𝑥,𝑦+𝑘₂𝑑(𝑥,𝑆𝑥)+𝑘₃𝑑𝑦,𝑇𝑦. (3.13) Hence the contractive condition required by Corollary 2.15 is satisfied with 𝑠=1, 𝑘₄=0, and 𝑘₁+𝑘₂+𝑘₃<1.
Now, all the required hypotheses of Corollary 2.15 are satisfied. Then, there exists 𝑢^∗∈𝐶(𝐼), a common fixed point of 𝑇 and 𝑆, that is, 𝑢^∗ is a solution to (3.1).