Fixed Point Theorems on Ordered Metric Spaces through a Rational Contraction

Abstract

Ran and Reurings (2004) established an interesting analogue of Banach Contraction Principle in a complete metric space equipped with a partial ordering and also utilized the same oneto discuss the existence of solutions to matrix equations. Motivated by this paper, we prove results on coincidence points for a pair of weakly increasing mappings satisfying a nonlinear contraction condition described by a rational expression on an ordered complete metric space. The uniqueness of common fixed point is also discussed. Some examples are furnished to demonstrate the validity of the hypotheses of our results. As an application, we derive an existence theorem for the solution of an integral equation.

1. Introduction with Preliminaries

A variety of generalizations of the Classical Banach Contraction Principle [1] are available in the existing literature of metric fixed point theory. The majority of these generalizations are obtained by improving the underlying contraction condition (e.g., [2]). Presently, there is vigorous research activity to prove existence results on complete metric spaces equipped with a partial ordering. In fact, various existence and uniqueness theorems on fixed and common fixed point for monotone mappings are of paramount importance in the study of nonlinear equations which generate natural interest to establish usable fixed point theorems in partial metric spaces (e.g., [1–24]).

Very recently, Harjani et al. [25] proved a fixed point theorem in partially ordered metric spaces satisfying a contractive condition of rational type due to Jaggi [26]. The aim of this paper is to prove some results of Harjani et al. [25] type for a pair of self-mappings. We accomplish this using the concept of weakly increasing property due to Nashine and Samet [14] (also see [4, 27, 28]). Some examples are also furnished to demonstrate the validity of the hypotheses of our results. As an application, we establish the existence of solution to an integral equation (also see [2, 23, 29, 30]).

Before presenting our results, we recall some notations, definitions, and examples required in our subsequent discussions.

Definition 1. Let 𝒳 be a nonempty set. Then (𝒳,𝑑,⪯) is called an ordered (partial) metric space if
(i) (𝒳,⪯) is a partially ordered set and (ii) (𝒳,𝑑) is a metric space.

Definition 2. Let (𝒳,⪯) be a partially ordered set. Then(a)elements 𝑥,𝑦∈𝒳 are called comparable with respect to “⪯” if either 𝑥⪯𝑦 or 𝑦⪯𝑥; (b)a mapping 𝒯∶𝒳→𝒳 is called nondecreasing with respect to “⪯” if 𝑥⪯𝑦 implies 𝒯𝑥⪯𝒯𝑦.

Let 𝒳 be a nonempty set and ℛ∶𝒳→𝒳 be 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}. (1)

Definition 3. Let (𝒳,⪯) be a partially ordered set and let 𝒯,ℛ∶𝒳→𝒳 be two mappings such that 𝒯𝒳⊆ℛ𝒳. We say that 𝒯 is weakly increasing with respect to ℛ if and only if for all 𝑥∈𝒳, one has: 𝒯𝑥⪯𝒯𝑦,∀𝑦∈ℛ⁻¹(𝒯𝑥). (2)

Remark 4. If ℛ∶𝒳→𝒳 is the identity mapping on 𝒳, then 𝒯 is weakly increasing with respect to ℛ if and only if 𝒯(𝑥)⪯𝒯𝒯(𝑥) for all 𝑥∈𝒳.

As mentioned earlier, the notion of weakly increasing mappings was introduced in [4] which is presently in use (e.g. [27, 28]). In what follows, we furnish a relatively new example to demonstrate the preceding definition.

Example 5. Consider 𝒳=[0,∞) endowed with the natural order ≤. Define two mappings 𝒯,ℛ∶𝒳→𝒳 as 𝑥𝒯𝑥=²𝑥+1,0≤𝑥<11,1≤𝑥,ℛ𝑥=³+1,0≤𝑥<11,1≤𝑥. (3)
In order to show that the mapping 𝒯 is weakly increasing with respect to mapping ℛ, we distinguish three cases.
Firstly, we consider the case 𝑥=0. Let 𝑦∈ℛ⁻¹(𝒯(0)), that is, ℛ(𝑦)=𝒯(0)=1 so that ℛ(𝑦)=1 and henceforth 𝑦=0 or 𝑦≥1. By the definitions of 𝒯, we have 1=𝒯(0)≤𝒯𝑦=1. (4)
Secondly, we argue the case 𝑥≥1. Let 𝑦∈ℛ⁻¹(𝒯(𝑥)), that is, ℛ(𝑦)=𝒯(𝑥)=1, which amounts to say (in view of definition 𝑅) that 𝑦=0 or 𝑦≥1, implying thereby 1=𝒯(𝑥)≤𝒯𝑦=1. (5)
Finally, we need to consider the case 0<𝑥<1. Let 𝑦∈ℛ⁻¹(𝒯(𝑥)), that is, ℛ(𝑦)=𝒯(𝑥). By the definition of 𝒯, we have 𝒯(𝑥)=𝑥²+1, so that ℛ(𝑦)=𝑥²+1. Now, in view of definition of ℛ, we have ℛ(𝑦)=𝑦³+1, so that 𝑦³+1=𝑥²+1, yielding thereby 𝑦=(𝑥)^2/3. Now, we have 𝑥²+1=𝒯(𝑥)≤𝒯𝑦=𝒯(𝑥)^2/3=(𝑥)^4/3+1. (6)
Thus we have shown that 𝒯 is weakly increasing with respect to ℛ.

Definition 6. Let (𝒳,⪯) be an 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 lim_𝑛→∞𝑧_𝑛=𝑧∈𝒳, then 𝑧_𝑛⪯𝑧 for all 𝑛∈ℕ.

Definition 7. A pair (ℛ,𝒯) of self-mappings of a metric space (𝒳,𝑑) is said to be compatible if and only if lim_𝑛→+∞𝑑(𝒯ℛ𝑥_𝑛,ℛ𝒯𝑥_𝑛)=0, whenever {𝑥_𝑛} is a sequence in 𝒳 such that lim_𝑛→+∞ℛ𝑥_𝑛=lim_𝑛→+∞𝒯𝑥_𝑛=𝑧∈𝒳.

Definition 8 (see [18]). A pair (ℛ,𝒯) of self-mappings of a metric space (𝒳,𝑑) is said to be reciprocally continuous if and only if lim_𝑛→∞ℛ𝒯𝑥_𝑛=ℛ𝑧 and lim_𝑛→∞𝒯ℛ𝑥_𝑛=𝒯𝑧 for every sequence {𝑥_𝑛} in 𝒳 satisfying lim_𝑛→∞ℛ𝑥_𝑛=lim_𝑛→∞𝒯𝑥_𝑛=𝑧 (7) for some 𝑧∈𝒳.
Notice that a pair of continuous mappings is always reciprocally continuous but not conversely as substantiated by examples in [18].

Definition 9 (see [19]). A pair (ℛ,𝒯) of self mappings of a metric space (𝒳,𝑑) is said to be weakly reciprocally continuous if and only if lim_𝑛→∞ℛ𝒯𝑥_𝑛=ℛ𝑧, for every sequence {𝑥_𝑛} in 𝒳 satisfying lim_𝑛→∞ℛ𝑥_𝑛=lim_𝑛→∞𝒯𝑥_𝑛=𝑧 (8) for some 𝑧∈𝒳.
Evidently, every pair of reciprocally continuous mappings is always weakly reciprocally but not conversely as demonstrated in Pant et al. [19].

2. Results

The main result of this paper runs as follows

Theorem 10. Let (𝒳,⪯) be a partially ordered set equipped with a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯,ℛ∶𝒳→𝒳 be two mappings satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝑑()𝒯𝑥,𝒯𝑦≤𝛼ℛ𝑥,𝒯𝑥⋅𝑑ℛ𝑦,𝒯𝑦,1+𝑑ℛ𝑥,ℛ𝑦+𝛽𝑑ℛ𝑥,ℛ𝑦 (9) where 𝛼, 𝛽 are nonnegative real numbers with 𝛼+𝛽<1. Suppose that (a)𝒳 is regular and 𝒯 is weakly increasing with ℛ, (b)the pair (ℛ,𝒯) is commuting as well as weakly reciprocally continuous. Then 𝒯 and ℛ have a coincidence point. That is, there exists 𝑢∈𝒳 such that ℛ𝑢=𝒯𝑢.

Proof. Let 𝑥₀ be an arbitrary point in 𝒳. Since 𝒯𝒳⊆ℛ𝒳, one can inductively construct a sequence {𝑥_𝑛} in 𝒳 defined by ℛ𝑥_𝑛=𝒯𝑥_𝑛−1,∀𝑛∈ℕ. (10)
As 𝑥₁∈ℛ⁻¹(𝒯𝑥₀) and 𝑥₂∈ℛ⁻¹(𝒯𝑥₁), using weakly increasing property of 𝒯 with respect to ℛ, we obtain ℛ𝑥₁=𝒯𝑥₀⪯𝒯𝑥₁=ℛ𝑥₂⪯𝒯𝑥₂=ℛ𝑥₃. (11)
Continuing this process indefinitely, we get ℛ𝑥₁⪯ℛ𝑥₂⪯ℛ𝑥₃⪯⋯⪯ℛ𝑥_𝑛⪯ℛ𝑥_𝑛+1⪯⋯. (12)
Now, we proceed to show that for 𝑛∈ℕ, 𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2≤𝛽1−𝛼^𝑛𝑑ℛ𝑥₁,ℛ𝑥₂. (13)
As ℛ(𝑥₂)≽ℛ(𝑥₁), using (9), we have 𝑑ℛ𝑥₂,ℛ𝑥₃=𝑑𝒯𝑥₁,𝒯𝑥₂𝑑ℛ𝑥≤𝛼¹,𝒯𝑥₁⋅𝑑ℛ𝑥₂,𝒯𝑥₂1+𝑑ℛ𝑥₁,ℛ𝑥₂+𝛽𝑑ℛ𝑥₁,ℛ𝑥₂,owingto𝑑ℛ𝑥₁,ℛ𝑥₂≤1+𝑑ℛ𝑥₁,ℛ𝑥₂≤𝛼𝑑ℛ𝑥₂,ℛ𝑥₃+𝛽𝑑ℛ𝑥₁,ℛ𝑥₂ (14) which implies that (1−𝛼)𝑑ℛ𝑥₂,ℛ𝑥₃≤𝛽𝑑ℛ𝑥₁,ℛ𝑥₂ (15) so that 𝑑ℛ𝑥₂,ℛ𝑥₃≤𝛽𝑑ℛ𝑥1−𝛼¹,ℛ𝑥₂. (16)
Let (13) holds for all 𝑛>0. As ℛ(𝑥_𝑛+1)≽ℛ(𝑥_𝑛), using (9), we have 𝑑ℛ𝑥_𝑛+2,ℛ𝑥_𝑛+3=𝑑𝒯𝑥_𝑛+1,𝒯𝑥_𝑛+2𝑑ℛ𝑥≤𝛼^𝑛+1,𝒯𝑥_𝑛+1𝑑ℛ𝑥_𝑛+2,𝒯𝑥_𝑛+21+𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2+𝛽𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2,owingto𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2≤1+𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2≤𝛼𝑑ℛ𝑥_𝑛+2,ℛ𝑥_𝑛+3+𝛽𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2, (17)
so that 𝑑ℛ𝑥_𝑛+2,ℛ𝑥_𝑛+3≤𝛽𝑑ℛ𝑥1−𝛼^𝑛+1,ℛ𝑥_𝑛+2≤𝛽1−𝛼^𝑛+1𝑑ℛ𝑥₁,ℛ𝑥₂. (18)
Now with ℎ=𝛽/(1−𝛼), for any 𝑚>𝑛, we have 𝑑ℛ𝑥_𝑛,ℛ𝑥_𝑚≤𝑑ℛ𝑥_𝑛,ℛ𝑥_𝑛+1+𝑑ℛ𝑥_𝑛+1,ℛ𝑥_𝑛+2+⋯𝑑ℛ𝑥_𝑚−1,ℛ𝑥_𝑚≤ℎ^𝑛+ℎ^𝑛+1+⋯ℎ^𝑚−1𝑑ℛ𝑥₁,ℛ𝑥₂≤ℎ^𝑛𝑑1−ℎℛ𝑥₁,ℛ𝑥₂. (19)
As 0<ℎ=𝛽/(1−𝛼)<1, we have lim_𝑛→+∞𝑑(ℛ(𝑥_𝑛),ℛ(𝑥_𝑚))=0, that is, {ℛ𝑥_𝑛} is Cauchy sequences in the complete metric space (𝒳,𝑑) and hence there exists some 𝑥∈𝒳 such that lim_𝑛→+∞𝒯𝑥_𝑛=lim_𝑛→+∞ℛ𝑥_𝑛=𝑥. (20)
Owing to commutativity of 𝒯 with ℛ and by using (10) (for each 𝑛≥1), we have ℛℛ𝑥_𝑛+1=ℛ𝒯𝑥_𝑛=𝒯ℛ𝑥_𝑛,thatisℛ⁻¹𝒯ℛ𝑥_𝑛=ℛ𝑥_𝑛+1. (21)
Since 𝒯 is weakly increasing with ℛ, we can write ℛℛ𝑥_𝑛+1=𝒯ℛ𝑥_𝑛⪯𝒯ℛ𝑥_𝑛+1=ℛℛ𝑥_𝑛+2 (22) so that ℛ(ℛ𝑥_𝑛) is nondecreasing. As the maps ℛ and 𝒯 are weakly reciprocally continuous, lim_𝑛→∞ℛ(ℛ𝑥_𝑛)=lim_𝑛→∞ℛ(𝒯𝑥_𝑛−1)=ℛ𝑥, which together with regularity of 𝒳 gives rise to ℛ(ℛ𝑥_𝑛)⪯ℛ𝑥; that is, ℛ(ℛ𝑥_𝑛) and ℛ𝑥 are comparable. On using condition (9), we have 𝑑(ℛ𝑥,𝒯𝑥)≤𝑑ℛ𝑥,ℛℛ𝑥_𝑛+1+𝑑ℛℛ𝑥_𝑛+1,𝒯𝑥=𝑑ℛ𝑥,ℛℛ𝑥_𝑛+1+𝑑ℛ𝒯𝑥_𝑛,𝒯𝑥=𝑑ℛ𝑥,ℛℛ𝑥_𝑛+1+𝑑𝒯ℛ𝑥_𝑛,𝒯𝑥=𝑑ℛ𝑥,ℛℛ𝑥_𝑛+1+𝑑𝒯𝑥,𝒯ℛ𝑥_𝑛≤𝑑ℛ𝑥,ℛℛ𝑥_𝑛+1()𝑑ℛ+𝛼𝑑ℛ𝑥,𝒯𝑥ℛ𝑥_𝑛,𝒯ℛ𝑥_𝑛1+𝑑ℛ𝑥,ℛℛ𝑥_𝑛+𝛽𝑑ℛ𝑥,ℛℛ𝑥_𝑛. (23)
On making 𝑛→∞ in the preceeding inequality, one gets 𝑑(ℛ𝑥,𝒯𝑥)=0, (24)
so that 𝒯𝑥=ℛ𝑥. Thus, we have shown that 𝒯 and ℛ have a coincidence point. This completes the proof.

Setting 𝛼=0 in Theorem 10, we deduce the following.

Corollary 11. Let (𝒳,⪯) be a partially ordered set on which there is a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯,ℛ∶𝒳→𝒳 be given mappings satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝒯𝑥,𝒯𝑦≤𝛽𝑑ℛ𝑥,ℛ𝑦, (25) where 𝛽 is a nonnegative real number with 𝛽<1. Suppose that (a)𝒳 is regular and 𝒯 is weakly increasing with ℛ, (b)the pair (ℛ,𝒯) is commuting as well as weakly reciprocally continuous. Then 𝒯 and ℛ have a coincidence point. That is, there exists 𝑢∈𝒳 such that ℛ𝑢=𝒯𝑢.

Theorem 12. Let (𝒳,⪯) be a partially ordered set equipped with a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯,ℛ∶𝒳→𝒳 be two mappings satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝑑()𝒯𝑥,𝒯𝑦≤𝛼ℛ𝑥,𝒯𝑥⋅𝑑ℛ𝑦,𝒯𝑦,1+𝑑ℛ𝑥,ℛ𝑦+𝛽𝑑ℛ𝑥,ℛ𝑦 (26) where 𝛼, 𝛽 are non-negative real numbers with 𝛼+𝛽<1. Suppose that (a)𝒯 is weakly increasing with ℛ, (b)the pair {𝒯,ℛ} is compatible and reciprocally continuous. Then 𝒯 and ℛ have a coincidence point. That is, there exists 𝑢∈𝒳 such that ℛ𝑢=𝒯𝑢.

Proof. Proceeding on the lines of the proof of Theorem 10, one can furnish a sequence {𝑥_𝑛} such that lim_𝑛→+∞𝒯𝑥_𝑛=lim_𝑛→+∞ℛ𝑥_𝑛=𝑥. (27)
Now, it remains to show that 𝑥 is a coincidence point of 𝒯 and ℛ. To accomplish this, as the pair {𝒯,ℛ} is compatible as well as reciprocally continuous, lim_𝑛→∞𝑑ℛ𝒯𝑥_𝑛,𝒯ℛ𝑥_𝑛ℛ(𝑥)=0,=lim_𝑛→∞ℛ𝒯𝑥_𝑛,𝒯(𝑥)=lim_𝑛→∞𝒯ℛ𝑥_𝑛, (28)(29) whenever lim_𝑛→+∞𝒯𝑥_𝑛=lim_𝑛→+∞ℛ𝑥_𝑛=𝑥. (30)
By using (29) in (28), we have 𝑑(𝒯𝑥,ℛ𝑥)=0, so that 𝒯𝑥=ℛ𝑥.

By appealing Theorems 10 and 12, one can also have the following natural theorem.

Theorem 13. Let (𝒳,⪯) be a partially ordered set on which there is a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯∶𝒳→𝒳 be nondecreasing mapping satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝑑()𝒯𝑥,𝒯𝑦≤𝛼𝑥,𝒯𝑥⋅𝑑𝑦,𝒯𝑦1+𝑑𝑥,𝑦+𝛽𝑑𝑥,𝑦, (31) where 𝛼, 𝛽 are non-negative real numbers such that 𝛼+𝛽<1. Suppose that (I)𝒯𝑥⪯𝒯(𝒯𝑥) for all 𝑥∈𝒳,(II)either 𝒯 is continuous or 𝒳 is regular. Then, 𝒯 has a fixed point.

Proof. The proof of this theorem can be outlined on the lines of the proof of Theorem 10 realizing ℛ to be the identity mapping on 𝒳.

Now, we introduce the following property which will be utilized in our next theorem.

Property (A). If ℛ(𝑥_𝑛) is a nondecreasing sequence in 𝑋 such that lim_𝑛→+∞ℛ(𝑥_𝑛)=𝑥, then ℛ(𝑥_𝑛) is comparable to ℛ(𝑥), for all 𝑛∈𝑁.

Theorem 14. Let (𝒳,⪯) be a partially ordered set on which there exists a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯,ℛ∶𝒳→𝒳 be given mappings satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝑑()𝒯𝑥,𝒯𝑦≤𝛼ℛ𝑥,𝒯𝑥⋅𝑑ℛ𝑦,𝒯𝑦,1+𝑑ℛ𝑥,ℛ𝑦+𝛽𝑑ℛ𝑥,ℛ𝑦 (32) where 𝛼, 𝛽 are non-negative real numbers with 𝛼+𝛽<1. Suppose that (a)𝒳 is regular and 𝒯 is weakly increasing with ℛ, (b)the pair (ℛ,𝒯) is commuting as well as weakly reciprocally continuous, (c)ℛ satisfies Property (A). Then ℛ and 𝒯 have a common fixed point.

Proof. Proceeding on the lines of the proof of Theorem 10, one can inductively construct nondecreasing sequence {ℛ𝑥_𝑛}, such that lim_𝑛→∞𝒯𝑥_𝑛=lim_𝑛→∞ℛ𝑥_𝑛+1=𝑥, and ℛ(𝑥)=𝒯(𝑥). Since ℛ(𝑥_𝑛) and ℛ(𝑥) are comparable (for all 𝑛∈𝑁), by using condition (32) we have 𝑑ℛ𝑥,ℛ𝑥_𝑛+1=𝑑𝒯𝑥,𝒯𝑥_𝑛()𝑑≤𝛼𝑑ℛ𝑥,𝒯𝑥ℛ𝑥_𝑛,𝒯𝑥_𝑛1+𝑑ℛ𝑥,ℛ𝑥_𝑛+𝛽𝑑ℛ𝑥,ℛ𝑥_𝑛. (33)
Taking the limit as 𝑛→∞, one gets 𝑥=ℛ𝑥=𝒯𝑥. (34) This completes the proof.

Theorem 15. Let (𝒳,⪯) be a partially ordered set on which there exists a metric 𝑑 on 𝒳 such that (𝒳,𝑑) is a complete metric space. Let 𝒯,ℛ∶𝒳→𝒳 be given mappings satisfying (for pairs (𝑥,𝑦)∈𝒳×𝒳 wherein ℛ𝑥 and ℛ𝑦 are comparable) 𝑑𝑑()𝒯𝑥,𝒯𝑦≤𝛼ℛ𝑥,𝒯𝑥⋅𝑑ℛ𝑦,𝒯𝑦,1+𝑑ℛ𝑥,ℛ𝑦+𝛽𝑑ℛ𝑥,ℛ𝑦 (35) where 𝛼, 𝛽 are non-negative real numbers with 𝛼+𝛽<1. Suppose that (a)𝒯 is weakly increasing with ℛ, (b)the pair {𝒯,ℛ} is compatible and reciprocally continuous, (c)ℛ satisfies Property (A).
Then ℛ and 𝒯 have a common fixed point.

Proof. Proof is obvious in view of Theorems 12 and 14.

3. Uniqueness Results

In what follows, we investigate the conditions under which Theorem 10 ensures the uniqueness of common fixed point.

Theorem 16. If, in addition to the hypotheses of Theorem 10, every (𝑥,𝑥^∗)∈𝒳×𝒳, there exists a 𝑢∈𝒳 such that 𝒯𝑢 is upper bound of 𝒯𝑥 and 𝒯𝑥^∗, then 𝒯 and ℛ have a unique common fixed point.

Proof. In view of Theorem 10, the set of coincidence points of the maps 𝒯 and ℛ is non-empty. If 𝑥 and 𝑥^∗ are two coincidence points of the maps 𝒯 and ℛ (i.e., ℛ𝑥=𝒯𝑥, ℛ𝑥^∗=𝒯𝑥^∗), then we proceed to show that ℛ𝑥=ℛ𝑥^∗. (36)
In view of the additional hypothesis of this theorem, there exists 𝑢∈𝒳 such that 𝒯𝑢 is upper bound of 𝒯𝑥 and 𝒯𝑥^∗. Put 𝑢₀=𝑢, and choose 𝑢₁∈𝒳 such that ℛ𝑢₁=𝒯𝑢₀. Now, proceeding on the lines of the proof of Theorem 10, one can inductively define sequence {ℛ𝑢_𝑛} such that ℛ𝑢_𝑛=𝒯𝑢_𝑛−1∀𝑛, (37) wherein ℛ𝑢₁⪯ℛ𝑢₂⪯ℛ𝑥₃⪯⋯⪯ℛ𝑢_𝑛⪯ℛ𝑢_𝑛+1⪯⋯. (38)
Further, setting 𝑥₀=𝑥, and 𝑥^∗₀=𝑥^∗, one can also define the sequences {ℛ𝑥_𝑛} and {ℛ𝑥^∗_𝑛} such that ℛ𝑥_𝑛=𝒯𝑥_𝑛−1,ℛ𝑥^∗_𝑛=𝒯𝑥^∗_𝑛−1. (39)
For every 𝑛≥1, we have ℛ𝑥_𝑛=𝒯𝑥,ℛ𝑥^∗_𝑛=𝒯𝑥^∗∀𝑛≥1. (40)
Since 𝒯𝑢=ℛ𝑢₁ is upper bound of 𝒯𝑥=ℛ𝑥₁ and 𝒯𝑥^∗=ℛ𝑥^∗₁, then ℛ𝑥₁=ℛ𝑥⪯ℛ𝑢₁,ℛ𝑥^∗₁=ℛ𝑥^∗⪯ℛ𝑢₁. (41)
It is easy to show that ℛ𝑥⪯ℛ𝑢_𝑛 and ℛ𝑥^∗⪯ℛ𝑢_𝑛 for all 𝑛≥1, then ℛ𝑥 and ℛ𝑥^∗ are comparable with ℛ𝑢_𝑛. On using (9), we have 𝑑ℛ𝑥,ℛ𝑢_𝑛+1=𝑑𝒯𝑥,𝒯𝑢_𝑛()𝑑≤𝛼𝑑ℛ𝑥,𝒯𝑥ℛ𝑢_𝑛,𝒯𝑢_𝑛1+𝑑ℛ𝑥,ℛ𝑢_𝑛+𝛽𝑑ℛ𝑥,ℛ𝑢_𝑛 (42) or 𝑑ℛ𝑥,ℛ𝑢_𝑛+1≤𝛽𝑑ℛ𝑥,ℛ𝑢_𝑛. (43)
Owing to (43), we can write 𝑑ℛ𝑥,ℛ𝑢_𝑛+1≤𝛽𝑑ℛ𝑥,ℛ𝑢_𝑛≤𝛽²𝑑ℛ𝑥,ℛ𝑢_𝑛−1≤⋯≤𝛽^𝑛+1𝑑ℛ𝑥,ℛ𝑢₀. (44)
Taking the limit as 𝑛→∞ in (44), we get lim_𝑛→∞𝑑ℛ𝑥,ℛ𝑢_𝑛=0, (45) as 0<𝛽<1.
Similarly, one can also show that lim_𝑛→∞𝑑ℛ𝑥^∗,ℛ𝑢_𝑛=0. (46)
On using (45) and (46), we can have 𝑑ℛ𝑥,ℛ𝑥^∗≤𝑑ℛ𝑥,ℛ𝑢_𝑛+1+𝑑ℛ𝑢_𝑛+1,ℛ𝑥^∗ (47) so that lim_𝑛→∞𝑑(ℛ𝑥,ℛ𝑥^∗)0, that is, ℛ𝑥=ℛ𝑥^∗. Thus, we have proved (36).
Since ℛ𝑥=𝒯𝑥, owing to commutativity of 𝒯 and ℛ, one can write ℛ(ℛ𝑥)=ℛ(𝒯𝑥)=𝒯(ℛ𝑥) (48) which on inserting ℛ𝑥=𝑧 gives rise to ℛ𝑧=𝒯𝑧. (49)
Thus, 𝑧 is another coincidence point of the pair. Now, due to ℛ𝑥=ℛ𝑥^∗ (for every coincidence point 𝑥 and 𝑥^∗) and owing to the fact that 𝑧 is coincidence point of the pair {ℛ,𝒯}, it follows that ℛ𝑧=ℛ𝑥, that is, ℛ𝑧=𝑧. (50)
On making use of (49) and (50), we can have 𝑧=ℛ𝑧=𝒯𝑧 (51) which shows that 𝑧 is a common fixed point of 𝒯 and ℛ.
To prove the uniqueness, let 𝑝 be another common fixed point of the pair {ℛ,𝒯}. Since 𝑝 and 𝑧 are coincidence points, then ℛ𝑝=ℛ𝑧.
Owing to that 𝑝 and 𝑧 are common fixed point of the pairs {ℛ,𝒯}, one can have 𝑝=ℛ𝑝=ℛ𝑧=𝑧. (52)
This completes the proof.

The following simple example demonstrates Theorem 16.

Example 17. Consider 𝒳={(1,0),(0,1)}⊂𝑅² equipped with natural order (𝑥,𝑦)≤(𝑧,𝑡)⇔𝑥≤𝑧 and 𝑦≤𝑡. Thus (𝒳,≤), is a partially ordered set wherein the two elements are not comparable to each other. Also, (𝒳,≤) is a complete metric space under Euclidean metric. Defined mappings 𝒯 and ℛ∶𝒳→𝒳 as 𝒯=(),=()(),=(),ℛ=(),=()(),=().𝑥,𝑦1,0𝑥,𝑦1,01,0𝑥,𝑦0,1𝑥,𝑦1,0𝑥,𝑦1,00,1𝑥,𝑦0,1 (53)(54)
As ℛ(𝑥,𝑦) and ℛ(𝑧,𝑡) in 𝒳 are merely comparable to themselves, inequality (9) is vacuously satisfied for every 𝛼,𝛽.
Notice that 𝒯(𝒳)=(1,0)⊂ℛ(𝒳)=𝒳. Also, for (𝑥,𝑦)=(1,0), ℛ(𝑧,𝑡)=𝒯𝑥,𝑦=𝒯(1,0)=(1,0)⟹(𝑧,𝑡)=(1,0) (55) so that 𝒯(1,0)=𝒯𝑥,𝑦≤𝒯(𝑧,𝑡)=𝒯(1,0). (56) Otherwise, for (𝑧,𝑡)=(0,1), ℛ(𝑧,𝑡)=𝒯𝑥,𝑦=𝒯(0,1)=(1,0)⟹(𝑧,𝑡)=(1,0) (57) so that (1,0)=𝒯(0,1)=𝒯𝑥,𝑦≤𝒯(𝑧,𝑡)=𝒯(1,0)=(1,0). (58) Thus 𝒯 is weakly increasing with respect to ℛ.
If {(𝑥_𝑛,𝑦_𝑛)}⊂𝒳 is a nondecreasing sequence converging to (𝑥,𝑦)∈𝒳, then necessarily {(𝑥_𝑛,𝑦_𝑛)} must be constant sequence, that is, (𝑥_𝑛,𝑦_𝑛)=(𝑥,𝑦), for all 𝑛∈𝑁, so that limit (𝑥,𝑦) is an upper bond for all the terms in the sequence which shows that 𝒳 is regular.
By using definitions of the maps 𝒯 and ℛ, we have 𝑥,𝑦=ℛ𝑥,𝑦=lim_𝑛→∞𝒯𝑥_𝑛,𝑦_𝑛=lim_𝑛→∞ℛ𝒯𝑥_𝑛,𝑦_𝑛 (59) wherein lim_𝑛→∞ℛ(𝑥_𝑛,𝑦_𝑛)=lim_𝑛→∞𝒯(𝑥_𝑛,𝑦_𝑛)=(𝑥,𝑦) so that ℛ𝑥,𝑦=lim_𝑛→∞ℛ𝒯𝑥_𝑛,𝑦_𝑛, (60) which shows that the pair {ℛ,𝒯} is weakly reciprocally continuous. Also, the pair {ℛ,𝒯} is clearly commuting.
Now, we show that for every 𝑥 and 𝑥^∗ in 𝒳, there exists a 𝑢∈𝒳 such that 𝒯𝑢 is upper bound of 𝒯𝑥 and 𝒯𝑥^∗. If 𝑥=𝑥^∗, then choice of 𝑢 is obvious. Otherwise, if 𝑥=(1,0) and 𝑥^∗=(0,1), we can choose 𝑢=(0,1) such that 𝒯𝑢=𝒯(0,1)=(1,0) is upper bound of 𝒯𝑥=(1,0) and 𝒯𝑥^∗=(1,0).
Thus, we have shown that all the conditions of Theorem 16 are satisfied and (𝑥,𝑦)=(1,0) is the unique common fixed point of 𝒯 and ℛ.

Theorem 18. If, in addition to the hypotheses of Theorem 12, for pairs (𝑥,𝑥^∗)∈𝒳×𝒳, there exists a 𝑢∈𝒳 such that 𝒯𝑢 is upper bound of 𝒯𝑥 and 𝒯𝑥^∗, then 𝒯 and ℛ have a unique common fixed point.

Proof. Proof is obvious in view of Theorems 12 and 16.

Corollary 19. In addition to the hypotheses of Theorem 16 (or Theorem 18), suppose that for every (𝑥,𝑥^∗)∈𝒳×𝒳 there exists a 𝑢∈𝒳 such that 𝒯𝑢 is upper bound of 𝒯𝑥 and 𝒯𝑥^∗. Then 𝒯 has a unique fixed point; that is, there exists a unique 𝑥∈𝒳 such that 𝑥=𝒯𝑥.

Proof. In Theorem 16, if ℛ=𝐼 (the identity mapping on 𝑋), we have the result.

The following example demonstrates Theorem 18.

Example 20. Consider 𝒳=[0,∞) equipped with the usual metric and natural order ≤. Define two mappings 𝒯,ℛ∶𝒳→𝒳 by 𝑥𝒯𝑥=³+2,0≤𝑥<13,1≤𝑥,ℛ𝑥=2𝑥³+1,0≤𝑥<13,1≤𝑥. (61) Then, evidently (𝒳,≤) is a partially ordered set and the maps 𝒯 and ℛ satisfy inequality (9) with 𝛽=1/2 and 𝛼=1/3, for 𝑥,𝑦∈[0,∞).
To show that 𝒯 is weakly increasing with respect to ℛ, notice that 𝒯(𝒳)=[2,3]⊂[1,3]=ℛ(𝒳). Firstly, we argue the case 𝑥=0. Let 𝑦∈ℛ⁻¹(𝑇0), that is,ℛ(𝑦)=𝒯(0)=2 so that 2𝑦³+1=2 or 𝑦=1/2^1/3. Using definitions of 𝒯 and ℛ, we have (0)𝑦12=𝒯≤𝒯=𝒯2^1/3=12+2. (62) Secondly, if 𝑥≥1, then we have ℛ(𝑦)=𝒯(𝑥)=3. By using definition of ℛ, we have 𝑦≥1, so that 3=𝒯(𝑥)≤𝒯𝑦=3. (63) Finally, we consider the case 0<𝑥<1. Let 𝑦∈ℛ⁻¹(𝒯(𝑥)), that is, ℛ(𝑦)=𝒯(𝑥). In view of definition of 𝒯, we have 𝒯(𝑥)=𝑥³+2, so that ℛ(𝑦)=𝑥³+2. By using the definition of ℛ, we have ℛ(𝑦)=2𝑦³+1, so that 2𝑦³+1=𝑥³+2 or 𝑦=((𝑥³+1)/2)^1/3. Thus, we have 𝑥³(𝑥)𝑦𝑥+2=𝒯≤𝒯=𝒯³+12^1/3=𝑥³+12+2. (64) Therefore 𝒯 is weakly increasing with respect to ℛ.
Since 𝒯 and ℛ, are continuous, therefore this pair of maps is reciprocally continuous.
Now, we show that maps 𝒯 and ℛ, are compatible. If lim_𝑛→∞𝑥_𝑛=𝑘, and 𝑘≥1, then lim_𝑛→∞𝒯𝑥_𝑛=lim_𝑛→∞ℛ𝑥_𝑛=3 and henceforth lim_𝑛→∞𝒯ℛ𝑥_𝑛=lim_𝑛→∞ℛ𝒯𝑥_𝑛=3 (65) implying thereby lim_𝑛→∞𝑑(ℛ𝒯𝑥_𝑛,𝒯ℛ𝑥_𝑛)=0 so that 𝒯 and ℛ, are compatible.
Thus, we have shown that all the conditions of Theorem 18 are satisfied and 𝑥=3 is the unique fixed point of 𝒯 and ℛ.

4. An Application

In this section, we present an application of Theorem 13 and used the idea of Ćirić et al. [29] to define a partial order and prove an existence theorem for the solution of an integral equation.

Theorem 21. Consider the integral equation 𝑥(𝑡)=^𝑇₀𝐾(𝑡,𝑠,𝑥(𝑠))𝑑𝑠+𝑔(𝑡),𝑡∈[0,𝑇] (66) with 𝑇>0 wherein(i)the functions 𝐾∶[0,𝑇]×[0,𝑇]×ℝ^𝑛→ℝ^𝑛 and 𝑔∶ℝ^𝑛→ℝ^𝑛are continuous,(ii)for each 𝑡,𝑠∈[0,𝑇], 𝐾(𝑡,𝑠,𝑥(𝑠))≪𝐾𝑡,𝑠,^𝑇₀𝐾(𝑠,𝜏,𝑥(𝜏))𝑑𝜏+𝑔(𝑠), (67)where ≪ denotes a partial order relation on ℝ^𝑛, (iii)there exists a continuous function 𝑝∶[0,𝑇]×[0,𝑇]→ℝ₊ such that |𝐾(𝑡,𝑠,𝑢)−𝐾(𝑡,𝑠,𝑣)|≤𝑝(𝑡,𝑠)|𝑢−𝑣| (68) for each 𝑡,𝑠∈[0,𝑇] and is also comparable 𝑢,𝑣∈ℝ^𝑛,(iv)sup_{𝑡∈[0,𝑇]}∫^𝑇₀𝑝(𝑡,𝑠)𝑑𝑠≤𝛽<1. Then the integral equation (66) has a unique solution 𝑥^∗ in 𝐶([0,𝑇],ℝ^𝑛).

Proof. Consider 𝒳∶=𝐶([0,𝑇],ℝ^𝑛) with the usual supremum norm, that is, ‖𝑥‖=max_{[]𝑡∈0,𝑇}|𝑥(𝑡)|, (69) for 𝑥,𝑦∈𝐶([0,𝑇]ℝ^𝑛). Define on 𝒳 a partial order as follows (𝑥,𝑦∈𝐶([0,𝑇],ℝ^𝑛)): 𝑥⪯𝑦iff𝑥(𝑡)≪𝑦(𝑡)forany𝑡∈[0,𝑇]. (70)
Then (𝒳,⪯) is a partially ordered set and (𝒳,‖⋅‖) is a complete metric space.
Moreover for any increasing sequence {𝑥_𝑛} in 𝒳 converging to 𝑥^∗∈𝒳, we have 𝑥_𝑛(𝑡)≪𝑥^∗(𝑡) for any 𝑡∈[0,𝑇]. Also for every 𝑥,𝑦∈𝒳 there exists 𝑐∈𝒳 which depends on 𝑥 and 𝑦 and is also comparable to 𝑥 and 𝑦 (cf. [21]).
Define 𝒯∶𝐶([0,𝑇])→𝐶([0,𝑇]) by 𝒯𝑢(𝑡)=^𝑇₀𝐾((𝑠(𝑡),[].𝑡,𝑠,𝑢))𝑑𝑠+𝑔𝑡∈0,𝑇,𝑢∈𝐶([0,𝑇]) (71)
Let us prove that 𝒯𝑢⪯𝒯(𝒯𝑢)∀𝑢∈𝐶([0,𝑇]). (72)
Let 𝑢∈𝐶([0,𝑇]). From (ii), for all 𝑡∈[0,𝑇], we have (𝑡)=𝒯𝑢_𝑇0𝐾((𝑠(𝑡)≪𝑡,𝑠,𝑢))𝑑𝑠+𝑔_𝑇0𝐾𝑡,𝑠,^𝑇₀𝐾((𝜏(𝑠)(𝑡)=𝑠,𝜏,𝑢))𝑑𝜏+𝑔𝑑𝑠+𝑔_𝑇0𝐾((𝑠(𝑡)(𝑡).𝑡,𝑠,𝒯𝑢))𝑑𝑠+𝑔=𝒯𝒯𝑢)( (73) Then (72) holds.
Now, for all 𝑥,𝑦∈𝐶([0,𝑇]) with 𝑦⪯𝑥, we have (by (iii)) (𝑡)(𝑡≤𝒯𝑥−𝒯𝑦)_𝑇0𝐾((𝑠(𝑠)≤𝑡,𝑠,𝑥))−𝐾𝑡,𝑠,𝑦𝑑𝑠_𝑇0𝑝()𝑥(𝑠)(𝑠≤𝑡,𝑠−𝑦)𝑑𝑠𝑥−𝑦^𝑇₀𝑝(𝑡,𝑠)𝑑𝑠. (74)
Hence (𝑡)(𝑡≤𝒯𝑥−𝒯𝑦)𝑥−𝑦sup_{[]𝑡∈0,𝑇}^𝑇₀𝑝().𝑡,𝑠𝑑𝑠≤𝛽𝑥−𝑦 (75)
On the other hand, it is demonstrated in [16] that condition of regularity is satisfied for 𝒳=𝐶([0,𝑇];ℝ^𝑛).
Thus, all the hypotheses of Theorem 13 are satisfied for 𝛼=0, and then 𝒯 has a fixed point 𝑢^∗∈𝐶([0,𝑇];ℝ^𝑛); that is, 𝑢^∗ is a solution of the integral equation (66).

Acknowledgments

The authors would like to thank the referees for careful reading and for providing valuable suggestions and comments for this paper. Also, Poom Kumam would like to thank the King Mongkuts University of Technology Thonburi (KMUTT) for the financial support and Dhananjay Gopal is thankful to CSIR, Government of India, Grant no. 25(0215)/13/EMR-II.