Common Fixed Point Results Using Generalized Altering Distances on Orbitally Complete Ordered Metric Spaces

Abstract

We prove the existence of common fixed points for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space using orbital continuity of one of the involved maps. We furnish a suitable example to demonstrate the validity of the hypotheses of our results.

1. Introduction and Preliminaries

Browder and Petryshyn introduced the concept of asymptotic regularity of a self-map at a point in a metric space.

Definition 1.1 (see [1]). A self-map 𝒯 on a metric space (𝒳,𝑑) is said to be asymptotically regular at a point 𝑥∈𝒳 if lim_𝑛→∞𝑑(𝒯^𝑛𝑥,𝒯^𝑛+1𝑥)=0.

Recall that the set 𝒪(𝑥₀;𝒯)={𝒯^𝑛𝑥₀∶𝑛=0,1,2,…} is called the orbit of the self-map 𝒯 at the point 𝑥₀∈𝒳.

Definition 1.2 (see [2]). A metric space (𝒳,𝑑) is said to be 𝒯-orbitally complete if every Cauchy sequence contained in 𝒪(𝑥;𝒯) (for some 𝑥 in 𝒳) converges in 𝒳.

Here, it can be pointed out that every complete metric space is 𝒯-orbitally complete for any 𝒯, but a 𝒯-orbitally complete metric space need not be complete.

Definition 1.3 (see [1]). A self-map 𝒯 defined on a metric space (𝒳,𝑑) is said to be orbitally continuous at a point 𝑧 in 𝒳 if for any sequence {𝑥_𝑛}⊂𝒪(𝑥;𝒯) (for some 𝑥∈𝒳), 𝑥_𝑛→𝑧 as 𝑛→∞ implies 𝒯𝑥_𝑛→𝒯𝑧 as 𝑛→∞.

Clearly, every continuous self-mapping of a metric space is orbitally continuous, but not conversely.

Sastry et al. [3] extended the above concepts to two and three mappings and employed them to prove common fixed point results for commuting mappings. In what follows, we collect such definitions for three maps.

Definition 1.4 (see [3]). Let 𝒮,𝒯,ℛ be three self-mappings defined on a metric space (𝒳,𝑑).

(1)If for a point 𝑥₀∈𝒳, there exists a sequence {𝑥_𝑛} in 𝒳 such that ℛ𝑥_2𝑛+1=𝒮𝑥_2𝑛, ℛ𝑥_2𝑛+2=𝒯𝑥_2𝑛+1, 𝑛=0,1,2,…, then the set 𝒪(𝑥₀;𝒮,𝒯,ℛ)={ℛ𝑥_𝑛∶𝑛=1,2,…} is called the orbit of (𝒮,𝒯,ℛ) at 𝑥₀.(2)The space (𝒳,𝑑) is said to be (𝒮,𝒯,ℛ)-orbitally complete at 𝑥₀ if every Cauchy sequence in 𝒪(𝑥₀;𝒮,𝒯,ℛ) converges in 𝒳.(3)The map ℛ is said to be orbitally continuous at 𝑥₀ if it is continuous on 𝒪(𝑥₀;𝒮,𝒯,ℛ).(4)The pair (𝒮,𝒯) is said to be asymptotically regular (in short a.r.) with respect to ℛ at 𝑥₀ if there exists a sequence {𝑥_𝑛} in 𝒳 such that ℛ𝑥_2𝑛+1=𝒮𝑥_2𝑛, ℛ𝑥_2𝑛+2=𝒯𝑥_2𝑛+1, 𝑛=0,1,2,…, and 𝑑(ℛ𝑥_𝑛,ℛ𝑥_𝑛+1)→0 as 𝑛→∞.

On the other side, Khan et al. [4] introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metric space. This notion has been used by several authors to establish fixed point results in a number of subsequent works, some of which are noted in [5–9]. In [5], Choudhury introduced the concept of a generalized altering distance function in three variables which was further generalized by Rao et al. [10] to four variables and is defined as follows.

Definition 1.5 (see [10]). A function 𝜓∶[0,+∞)⁴→[0,+∞) is said to be a generalized altering distance function if(i)𝜓 is continuous,(ii)𝜓 is nondecreasing in each variable,(iii)𝜓(𝑡₁,𝑡₂,𝑡₃,𝑡₄)=0⇔𝑡₁=𝑡₂=𝑡₃=𝑡₄=0.ℱ₄ will denote the set of all functions 𝜓 satisfying conditions (i)–(iii).

Simple examples of generalized altering distance functions with four variables are𝜓𝑡₁,𝑡₂,𝑡₃,𝑡₄=𝑘max𝑡₁,𝑡₂,𝑡₃,𝑡₄𝜓𝑡,𝑘>0;₁,𝑡₂,𝑡₃,𝑡₄=max𝑡₁,𝑡₂,𝑡₃,𝑡₄1+max𝑡₁,𝑡₂,𝑡₃,𝑡₄;𝜓𝑡₁,𝑡₂,𝑡₃,𝑡₄=𝑡^𝑝₁+𝑡^𝑞₂+𝑡^𝑟₃+𝑡^𝑠₄,𝑝,𝑞,𝑟,𝑠≥1. (1.1)

On the other hand, fixed point theory has developed rapidly in metric spaces endowed with a partial ordering. The first result in this direction was given by Ran and Reurings [11] who presented its applications to matrix equations. Subsequently, Nieto and Rodríguez-López [12] extended this result for nondecreasing mappings and applied it to obtain a unique solution for a first order ordinary differential equation with periodic boundary conditions. Thereafter, several authors obtained many fixed point theorems in ordered metric spaces. For more details see [13–20] and the references cited therein.

In this paper, an attempt has been made to derive some common fixed point theorems for three relatively asymptotically regular mappings defined on an orbitally complete ordered metric space, using orbital continuity of one of the involved maps and conditions involving a generalized altering distance function. The presented theorems generalize, extend, and improve some recent results given in [7, 14, 21, 22]. In the hypotheses, we have considered the space as not necessarily complete, the maps ℛ,𝒮, and 𝒯 as not necessarily continuous and the range of 𝒮 and 𝒯 may not be contained in the range of ℛ.

2. Results

2.1. Notations and Definitions

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

If (𝒳,≼) is a partially ordered set, then 𝑥,𝑦∈𝒳 are called comparable if 𝑥≼𝑦 or 𝑦≼𝑥 holds. A subset 𝒦 of 𝒳 is said to be well ordered if every two elements of 𝒦 are comparable. If 𝒯∶𝒳→𝒳 is such that, for 𝑥,𝑦∈𝒳, 𝑥≼𝑦 implies 𝒯𝑥≼𝒯𝑦, then the mapping 𝒯 is said to be nondecreasing.

Definition 2.1. Let (𝒳,≼) be a partially ordered set and 𝒮,𝒯∶𝒳→𝒳. (1)[23] The pair (𝒮,𝒯) is called weakly increasing if 𝒮𝑥≼𝒯𝒮𝑥 and 𝒯𝑥≼𝒮𝒯𝑥 for all 𝑥∈𝒳.(2)[24] The pair (𝒮,𝒯) is called partially weakly increasing if 𝒮𝑥≼𝒯𝒮𝑥 for all 𝑥∈𝒳.(3)[24] The mapping 𝒮 is called a weak annihilator of 𝒯 if 𝒮𝒯𝑥≼𝑥 for all 𝑥∈𝒳.(4)[24] The mapping 𝒮 is called dominating if 𝑥≼𝒮𝑥 for each 𝑥∈𝒳.

Note that none of two weakly increasing mappings need to be nondecreasing. There exist some examples to illustrate this fact in [23]. Obviously, the pair (𝒮,𝒯) is weakly increasing if and only if the ordered pairs (𝒮,𝒯) and (𝒯,𝒮) are partially weakly increasing. Following is an example of an ordered pair (𝒮,𝒯) which is partially weakly increasing but not weakly increasing.

Example 2.2 (see [24]). Let 𝒳=[0,1] be endowed with usual ordering.

(1)Let 𝒮,𝒯∶𝒳→𝒳 be defined by 𝒮𝑥=𝑥² and 𝒯𝑥=√𝑥. Clearly, (𝒮,𝒯) is partially weakly increasing. But 𝒯𝑥=√𝑥≰𝑥=𝒮𝒯𝑥 for 𝑥∈(0,1) implies that (𝒯,𝒮) is not partially weakly increasing.(2)Let 𝒮,𝒯∶𝒳→𝒳 be defined by 𝒮𝑥=𝑥² and 𝒯𝑥=𝑥³. Obviously, 𝒮𝒯𝑥=𝑥⁶≤𝑥 for all 𝑥∈𝒳. Thus 𝒮 is a weak annihilator of 𝒯.(3)Let 𝒮∶𝒳→𝒳 be defined by 𝒮𝑥=√^𝑛𝑥. Since 𝑥≤√^𝑛𝑥=𝒮𝑥 for all 𝑥∈𝒳, 𝒮 is a dominating map.

Definition 2.3 (see [25, 26]). Let (𝒳,𝑑) be a metric space and 𝑓,𝑔∶𝒳→𝒳. The mappings 𝑓 and 𝑔 are said to be compatible 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.

Definition 2.4. Let 𝒳 be a nonempty set. Then (𝒳,𝑑,≼) is called an ordered metric space if(i)(𝒳,𝑑) is a metric space,(ii)(𝒳,≼) is a partially ordered set.

2.2. Main Results

The first main result is as follows.

Theorem 2.5. Let (𝒳,𝑑,≼) be an ordered metric space. Let 𝒮,𝒯,ℛ∶𝒳→𝒳 be given mappings satisfying Ψ₁𝑑𝒮𝑥,𝒯𝑦≤𝜓₁𝑀[𝒮,𝒯,ℛ]𝑥,𝑦−𝜓₂𝑀[𝒮,𝒯,ℛ]𝑥,𝑦, (2.1) for all 𝑥,𝑦∈𝒪(𝑥₀;𝒮,𝒯,ℛ) (for some 𝑥₀) such that ℛ𝑥 and ℛ𝑦 are comparable, where 𝑀[=𝑑(),1𝒮,𝒯,ℛ]𝑥,𝑦ℛ𝑥,ℛ𝑦,𝑑ℛ𝑥,𝒮𝑥,𝑑ℛ𝑦,𝒯𝑦2𝑑ℛ𝑥,𝒯𝑦+𝑑ℛ𝑦,𝒮𝑥, (2.2) and  𝜓₁ and 𝜓₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(𝑡)=𝜓₁(𝑡,𝑡,𝑡,𝑡). We assume the following hypotheses:(i)(𝒮,𝒯) is a.r. with respect to ℛ at 𝑥₀∈𝒳;(ii)𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at 𝑥₀;(iii)(ℛ,𝒮) and (ℛ,𝒯) are partially weakly increasing;(iv)𝒮 and 𝒯 are dominating maps;(v)𝒮 and 𝒯 are weak annihilators of ℛ;(vi)for a nondecreasing sequence {𝑥_𝑛}, 𝑥_𝑛≼𝑦_𝑛 for all 𝑛 and 𝑦_𝑛→𝑢 as 𝑛→∞ imply that 𝑥_𝑛≼𝑢 for all 𝑛∈ℕ. Assume either(a)𝒮 and ℛ are compatible; 𝒮 or ℛ is orbitally continuous at 𝑥₀ or(b)𝒯 and ℛ are compatible; 𝒯 or ℛ is orbitally continuous at 𝑥₀.Then 𝒮,𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(𝑥₀;𝒮,𝒯,ℛ) is well ordered if and only if it is a singleton.

Proof. Since (𝒮,𝒯) is a.r. with respect to ℛ at 𝑥₀ in 𝒳, there exists a sequence {𝑥_𝑛} in 𝒳 such that 𝒮𝑥_2𝑛−2=ℛ𝑥_2𝑛−1,𝒯𝑥_2𝑛−1=ℛ𝑥_2𝑛,∀𝑛∈ℕ. (2.3) By the given assumptions, 𝑥_2𝑛−2≼𝒮𝑥_2𝑛−2=ℛ𝑥_2𝑛−1≼𝒮ℛ𝑥_2𝑛−1≼𝑥_2𝑛−1, and 𝑥_2𝑛−1≼𝒯𝑥_2𝑛−1=ℛ𝑥_2𝑛≼𝒯ℛ𝑥_2𝑛≼𝑥_2𝑛. Thus, for all 𝑛≥1, we have 𝑥_𝑛≼𝑥_𝑛+1. (2.4) In view of (i), we have lim_𝑛→∞𝑑ℛ𝑥_𝑛,ℛ𝑥_𝑛+1=0. (2.5) Now, we assert that {ℛ𝑥_𝑛} is a Cauchy sequence in the metric space 𝒪(𝑥₀;𝒮,𝒯,ℛ).
From (2.5), 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 integers 𝑖, 𝑛(𝑖)>𝑚(𝑖)>𝑖,𝑑ℛ𝑥_2𝑚(𝑖),ℛ𝑥_2𝑛(𝑖)≥𝜀,𝑑ℛ𝑥_2𝑚(𝑖),ℛ𝑥_{2𝑛(𝑖)−2}<𝜀. (2.6) From (2.6) and using the 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). (2.7) Letting 𝑖→∞ in the above inequality and using (2.5), we obtain lim_𝑖→∞𝑑ℛ𝑥_2𝑚(𝑖),ℛ𝑥_2𝑛(𝑖)=𝜀. (2.8) Again, the triangular inequality gives us 𝑑ℛ𝑥_2𝑛(𝑖),ℛ𝑥_{2𝑚(𝑖)−1}−𝑑ℛ𝑥_2𝑛(𝑖),ℛ𝑥_2𝑚(𝑖)≤𝑑ℛ𝑥_{2𝑚(𝑖)−1},ℛ𝑥_2𝑚(𝑖). (2.9) Letting 𝑖→∞ in the above inequality and using (2.5) and (2.8), we get lim_𝑖→∞𝑑ℛ𝑥_2𝑛(𝑖),ℛ𝑥_{2𝑚(𝑖)−1}=𝜀. (2.10) Similarly, we have lim_𝑖→∞𝑑ℛ𝑥_{2𝑛(𝑖)+1},ℛ𝑥_{2𝑚(𝑖)−1}=𝜀. (2.11) 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}. (2.12) Then, from (2.5), (2.8), and the continuity of Ψ₁, we get by letting 𝑖→∞ in (2.1) Ψ₁(𝜀)≤lim_𝑖→∞Ψ₁𝑑𝒮𝑥_{2𝑚(𝑖)−1},𝒯𝑥_2𝑛(𝑖). (2.13) Now, using the considered contractive condition (2.1) for 𝑥=𝑥_{2𝑚(𝑖)−1} and 𝑦=𝑥_2𝑛(𝑖), we have Ψ₁𝑑𝒮𝑥_{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). (2.14) Then, from (2.5), (2.10), (2.11), and the continuity of 𝜓₁ and 𝜓₂, we get by letting 𝑖→∞ in the above inequality lim_𝑖→∞Ψ₁𝑑𝒮𝑥_{2𝑚(𝑖)−1},𝒯𝑥_2𝑛(𝑖)≤𝜓₁(𝜀,0,0,𝜖)−𝜓₂(𝜀,0,0,𝜖)≤Ψ₁(𝜀)−𝜓₂(𝜀,0,0,𝜖). (2.15) Now, combining (2.13) with the above inequality, we get Ψ₁(𝜀)≤Ψ₁(𝜀)−𝜓₂(𝜀,0,0,𝜖), (2.16) which implies that 𝜓₂(𝜀,0,0,𝜖)=0, which is a contradiction since 𝜀>0. Hence {ℛ𝑥_𝑛} is a Cauchy sequence in 𝒪(𝑥₀;𝒮,𝒯,ℛ). Since 𝒳 is (𝒮,𝒯,ℛ)-orbitally complete at 𝑥₀, there exists some 𝑧∈𝒳 such that ℛ𝑥_𝑛→𝑧 as 𝑛→∞.
Finally, we prove the existence of a common fixed point of the three mappings 𝒮,𝒯, and ℛ.
We have ℛ𝑥_2𝑛+1=𝒮𝑥_2𝑛⟶𝑧as𝑛⟶∞,ℛ𝑥_2𝑛+2=𝒯𝑥_2𝑛+1⟶𝑧as𝑛⟶∞. (2.17) Suppose that (a) holds. Since {𝒮,ℛ} are compatible, we have lim_𝑛→∞𝒮ℛ𝑥_2𝑛+2=lim_𝑛→∞ℛ𝒮𝑥_2𝑛+2=ℛ𝑧. (2.18) Also, 𝑥_2𝑛+1≼𝒯𝑥_2𝑛+1=ℛ𝑥_2𝑛+2. Now Ψ₁𝑑𝒮ℛ𝑥_2𝑛+2,𝒯𝑥_2𝑛+1≤𝜓₁𝑑ℛℛ𝑥_2𝑛+2,ℛ𝑥_2𝑛+1,𝑑ℛℛ𝑥_2𝑛+2,𝒮ℛ𝑥_2𝑛+2,𝑑ℛ𝑥_2𝑛+1,𝒯𝑥_2𝑛+1,12𝑑ℛℛ𝑥_2𝑛+2,𝒯𝑥_2𝑛+1+𝑑𝒮ℛ𝑥_2𝑛+2,ℛ𝑥_2𝑛+1−𝜓₂𝑑ℛℛ𝑥_2𝑛+2,ℛ𝑥_2𝑛+1,𝑑ℛℛ𝑥_2𝑛+2,𝒮ℛ𝑥_2𝑛+2,𝑑ℛ𝑥_2𝑛+1,𝒯𝑥_2𝑛+1,12𝑑ℛℛ𝑥_2𝑛+2,𝒯𝑥_2𝑛+1+𝑑𝒮ℛ𝑥_2𝑛+2,ℛ𝑥_2𝑛+1. (2.19) Assume that ℛ is orbitally continuous. Passing to the limit as 𝑛→∞, we obtain Ψ₁(𝑑(ℛ𝑧,𝑧))≤𝜓₁(𝑑(ℛ𝑧,𝑧),0,0,𝑑(ℛ𝑧,𝑧))−𝜓₂(𝑑(ℛ𝑧,𝑧),0,0,𝑑(ℛ𝑧,𝑧))≤Ψ₁(𝑑(ℛ𝑧,𝑧))−𝜓₂(𝑑(ℛ𝑧,𝑧),0,0,𝑑(ℛ𝑧,𝑧)), (2.20) so 𝜓₂(𝑑(ℛ𝑧,𝑧),0,0,𝑑(ℛ𝑧,𝑧))=0, which implies that ℛ𝑧=𝑧. (2.21) Now, 𝑥_2𝑛+1≼𝒯𝑥_2𝑛+1 and 𝒯𝑥_2𝑛+1→𝑧 as 𝑛→∞, so by the assumption we have 𝑥_2𝑛+1≼𝑧 and (2.1) becomes Ψ₁𝑑𝒮𝑧,𝒯𝑥_2𝑛+1≤𝜓₁𝑑ℛ𝑧,ℛ𝑥_2𝑛+1,𝑑(𝒮𝑧,ℛ𝑧),𝑑𝒯𝑥_2𝑛+1,ℛ𝑥_2𝑛+1,12𝑑ℛ𝑧,𝒯𝑥_2𝑛+1+𝑑𝒮𝑧,ℛ𝑥_2𝑛+1−𝜓₂𝑑ℛ𝑧,ℛ𝑥_2𝑛+1,𝑑(𝒮𝑧,ℛ𝑧),𝑑𝒯𝑥_2𝑛+1,ℛ𝑥_2𝑛+1,12𝑑ℛ𝑧,𝒯𝑥_2𝑛+1+𝑑𝒮𝑧,ℛ𝑥_2𝑛+1. (2.22) Passing to the limit as 𝑛→∞ in the above inequality and using (2.21), it follows that Ψ₁(𝑑(𝒮𝑧,𝑧))≤𝜓₁()10,𝑑𝒮𝑧,𝑧,0,2𝑑(𝒮𝑧,𝑧)−𝜓₂()10,𝑑𝒮𝑧,𝑧,0,2𝑑(𝒮𝑧,𝑧)≤Ψ₁(𝑑(𝒮𝑧,𝑧))−𝜓₂()10,𝑑𝒮𝑧,𝑧,0,2𝑑(𝒮𝑧,𝑧), (2.23) which holds unless 𝜓₂(0,𝑑(𝒮𝑧,𝑧),0,(1/2)𝑑(𝒮𝑧,𝑧))=0, so 𝒮𝑧=𝑧. (2.24) Now, since 𝑥_2𝑛≼𝒮𝑥_2𝑛 and 𝒮𝑥_2𝑛→𝑧 as 𝑛→∞ implies that 𝑥_2𝑛≼𝑧, from (2.1) Ψ₁𝑑𝒮𝑥_2𝑛,𝒯𝑧≤𝜓₁𝑑ℛ𝑥_2𝑛,ℛ𝑧,𝑑ℛ𝑥_2𝑛,𝒮𝑥_2𝑛(),1,𝑑ℛ𝑧,𝒯𝑧2𝑑ℛ𝑥_2𝑛,𝒯𝑧+𝑑𝒮𝑥_2𝑛,ℛ𝑧−𝜓₂𝑑ℛ𝑥_2𝑛,ℛ𝑧,𝑑ℛ𝑥_2𝑛,𝒮𝑥_2𝑛(),1,𝑑ℛ𝑧,𝒯𝑧2𝑑ℛ𝑥_2𝑛,𝒯𝑧+𝑑𝒮𝑥_2𝑛,ℛ𝑧. (2.25) Passing to the limit as 𝑛→∞, we have Ψ₁(𝑑(𝑧,𝒯𝑧))≤𝜓₁(0,0,𝑑(𝑧,𝒯𝑧),𝑑(𝑧,𝒯𝑧))−𝜓₂(𝑑(𝑧,𝒯𝑧),0,0,𝑑(𝑧,𝒯𝑧))≤Ψ₁(𝑑(𝑧,𝒯𝑧))−𝜓₂(0,0,𝑑(𝑧,𝒯𝑧),𝑑(𝑧,𝒯𝑧)), (2.26) which gives that 𝑧=𝒯𝑧. (2.27) Therefore, 𝒮𝑧=𝒯𝑧=ℛ𝑧=𝑧, hence 𝑧 is a common fixed point of ℛ,𝒮, and 𝒯. The proof is similar when 𝒮 is orbitally continuous.
Similarly, the result follows when condition (b) holds.
Now, suppose that the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(𝑥₀;𝒮,𝒯,ℛ) is well ordered. We claim that it cannot contain more than one point. Assume to the contrary that 𝒮𝑢=𝒯𝑢=ℛ𝑢=𝑢 and 𝒮𝑣=𝒯𝑣=ℛ𝑣=𝑣 but 𝑢≠𝑣. By supposition, we can replace 𝑥 by 𝑢 and 𝑦 by 𝑣 in (2.1) to obtain Ψ₁(𝑑(𝑢,𝑣))=Ψ₁(𝑑(𝒮𝑢,𝒯𝑣))≤𝜓₁𝑑()()(),1ℛ𝑢,ℛ𝑣,𝑑ℛ𝑢,𝒮𝑢,𝑑ℛ𝑣,𝒯𝑣2[𝑑(ℛ𝑢,𝒯𝑣)+𝑑(𝒮𝑢,ℛ𝑣)]−𝜓₂𝑑()()(),1ℛ𝑢,ℛ𝑣,𝑑ℛ𝑢,𝒮𝑢,𝑑ℛ𝑣,𝒯𝑣2[𝑑(ℛ𝑢,𝒯𝑣)+𝑑(𝒮𝑢,ℛ𝑣)]=𝜓₁(𝑑(𝑢,𝑣),0,0,𝑑(𝑢,𝑣))−𝜓₂(𝑑(𝑢,𝑣),0,0,𝑑(𝑢,𝑣))<Ψ₁(𝑑(𝑢,𝑣)), (2.28) a contradiction. Hence, 𝑢=𝑣. The converse is trivial.

Now, it is easy to state a corollary of Theorem 2.5 involving a contraction of integral type.

Corollary 2.6. Let 𝒮,𝒯, and ℛ satisfy the conditions of Theorem 2.5, except that condition (2.1) 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 ^Ψ1₀^{(𝑑(𝒮𝑥,𝒯𝑦))}𝑢(𝑡)𝑑𝑡≤^𝜓1₀^{(𝑀[𝒮,𝒯,ℛ](𝑥,𝑦))}𝑢(𝑡)𝑑𝑡−^𝜓2₀^{(𝑀[𝒮,𝒯,ℛ](𝑥,𝑦))}𝑢(𝑡)𝑑𝑡. (2.29) Then, 𝒮,𝒯, and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒮,𝒯, and ℛ in 𝒪(𝑥₀;𝒮,𝒯,ℛ) is well ordered if and only if it is a singleton.

Remark 2.7. If we take 𝜓₁𝑡₁,𝑡₂,𝑡₃,𝑡₄=max𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝜓₂𝑡₁,𝑡₂,𝑡₃,𝑡₄=(1−𝑘)max𝑡₁,𝑡₂,𝑡₃,𝑡₄, (2.30) for 𝑘∈(0,1), then Ψ₁(𝑡)=𝑡 for all 𝑡≥0, and the contractive condition (2.1) becomes 𝑑𝑑(),1𝒮𝑥,𝒯𝑦≤𝑘maxℛ𝑥,ℛ𝑦,𝑑ℛ𝑥,𝒮𝑥,𝑑ℛ𝑦,𝒯𝑦2𝑑ℛ𝑥,𝒯𝑦+𝑑ℛ𝑦,𝒮𝑥, (2.31) which corresponds to the contraction given by Theorem  2.1 in [24] by taking 𝜓(𝑡)=𝑡 and 𝜑(𝑡)=(1−𝑘)𝑡. Hence, the result of Abbas et al. [24] is covered by Theorem 2.5 for three maps.
Other results could be derived for other choices of 𝜓₁ and 𝜓₂.

As consequences of Theorem 2.5, we may state the following corollaries.

Corollary 2.8. Let (𝒳,𝑑,≼) be an ordered metric space. Let 𝒯,ℛ∶𝒳→𝒳 be given mappings satisfying Ψ₁𝑑𝒯𝑥,𝒯𝑦≤𝜓₁𝑑(),1ℛ𝑥,ℛ𝑦,𝑑ℛ𝑥,𝒯𝑥,𝑑ℛ𝑦,𝒯𝑦2𝑑ℛ𝑥,𝒯𝑦+𝑑ℛ𝑦,𝒯𝑥−𝜓₂𝑑(),1ℛ𝑥,ℛ𝑦,𝑑ℛ𝑥,𝒯𝑥,𝑑ℛ𝑦,𝒯𝑦2𝑑ℛ𝑥,𝒯𝑦+𝑑ℛ𝑦,𝒯𝑥, (2.32) for all 𝑥,𝑦∈𝒪(𝑥₀;𝒯,𝒯,ℛ) such that ℛ𝑥 and ℛ𝑦 are comparable, where 𝜓₁ and 𝜓₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(𝑡)=𝜓₁(𝑡,𝑡,𝑡,𝑡). We assume the following hypotheses:(i)𝒯 is a.r. with respect to ℛ at 𝑥₀;(ii)𝒳 is (𝒯,ℛ)-orbitally complete at 𝑥₀;(iii)𝒯 or ℛ is orbitally continuous at 𝑥₀;(iv)(𝒯,ℛ) is partially weakly increasing;(v)𝒯 is a dominating map;(vi)𝒯 is a weak annihilator of ℛ;(vii)𝒯 and ℛ are compatible.Let for a nondecreasing sequence {𝑥_𝑛} with 𝑥_𝑛≼𝑦_𝑛  for all 𝑛, 𝑦_𝑛→𝑢 as 𝑛→∞ imply that 𝑥_𝑛≼𝑢 for all 𝑛∈ℕ.
Then 𝒯 and ℛ have a common fixed point. Moreover, the set of common fixed points of 𝒯 and ℛ in 𝒪(𝑥₀;𝒯,𝒯,ℛ) is well ordered if and only if it is a singleton.

Corollary 2.9. Let (𝒳,𝑑,≼) be an ordered metric space. Let 𝒯∶𝒳→𝒳 be a mapping satisfying Ψ₁𝑑𝒯𝑥,𝒯𝑦≤𝜓₁𝑑(),1𝑥,𝑦,𝑑𝑥,𝒯𝑥,𝑑𝑦,𝒯𝑦2𝑑𝑥,𝒯𝑦+𝑑𝑦,𝒯𝑥−𝜓₂𝑑(),1𝑥,𝑦,𝑑𝑥,𝒯𝑥,𝑑𝑦,𝒯𝑦2𝑑𝑥,𝒯𝑦+𝑑𝑦,𝒯𝑥, (2.33) for all (𝑥,𝑦)∈𝒪(𝑥₀;𝒯) such that 𝑥 and 𝑦 are comparable, where 𝜓₁ and 𝜓₂ are generalized altering distance functions (in ℱ₄) and Ψ₁(𝑡)=𝜓₁(𝑡,𝑡,𝑡,𝑡). We assume the following hypotheses:(i)𝒯 is a.r. at some point 𝑥₀ of 𝒳;(ii)𝒳 is 𝒯-orbitally complete at 𝑥₀;(iii)𝒯 is a dominating map. Let for a nondecreasing sequence {𝑥_𝑛} with 𝑥_𝑛≼𝑦_𝑛  for all 𝑛, 𝑦_𝑛→𝑢 as 𝑛→∞ imply that 𝑥_𝑛≼𝑢 for all 𝑛∈ℕ.
Then 𝒯 has a fixed point. Moreover, the set of fixed points of 𝒯 in 𝒪(𝑥₀;𝒯) is well ordered if and only if it is a singleton.

We present an example showing the usage of our results.

Example 2.10. Let the set 𝒳=[0,+∞) be equipped with the usual metric 𝑑 and the order defined by 𝑥≼𝑦⟺𝑥≥𝑦. (2.34) Consider the following self-mappings on 𝒳: 1ℛ𝑥=6𝑥,𝒮𝑥=21𝑥,0≤𝑥≤2,1𝑥,𝑥>2,1𝒯𝑥=31𝑥,0≤𝑥≤3,1𝑥,𝑥>3. (2.35) Take 𝑥₀=1/2. Then it is easy to show that 𝒪𝑥₀⊂1;𝒮,𝒯,ℛ2^𝑘⋅3^𝑙∶𝑘,𝑙∈ℕ,𝒪𝑥₀;𝒮,𝒯,ℛ=𝒪𝑥₀;𝒮,𝒯,ℛ∪{0} (2.36) and all the conditions (i)–(vi) and (a)-(b) of Theorem 2.5 are fulfilled. Take 𝜓₁(𝑡₁,𝑡₂,𝑡₃,𝑡₄)=max{𝑡₁,𝑡₂,𝑡₃,𝑡₄} and 𝜓₂(𝑡₁,𝑡₂,𝑡₃,𝑡₄)=(5/6)max{𝑡₁,𝑡₂,𝑡₃,𝑡₄}. Then contractive condition (2.1) takes the form 121𝑥−3𝑦≤16max6𝑥−6𝑦,112𝑥,1731𝑦,216𝑥−3𝑦+16𝑦−2𝑥, (2.37) for 𝑥,𝑦∈𝒪(𝑥₀;𝒮,𝒯,ℛ). Using substitution 𝑦=𝑡𝑥, 𝑡>0, the last inequality reduces to |3−2𝑡|≤max6|1−𝑡|,112,1731𝑡,216−3𝑡+16𝑡−2, (2.38) and can be checked by discussion on possible values for 𝑡>0. Hence, all the conditions of Theorem 2.5 are satisfied and 𝒮,𝒯,ℛ have a common fixed point (which is 0).

Remark 2.11. It was shown by examples in [22] that (in similar situations)(1)if the contractive condition is satisfied just on 𝒪(𝑥₀;𝒮,𝒯,ℛ), there might not exist a (common) fixed point;(2)under the given hypotheses (common), fixed point might not be unique in the whole space 𝒳.

Acknowledgment

The authors thank the referees for their careful reading of the text and for suggestions that helped to improve the exposition of the paper. The second and third authors are thankful to the Ministry of Science and Technological Development of Serbia.