On Coupled Fixed Point Theorems for Nonlinear Contractions in Partially Ordered 𝐺-Metric Spaces

Abstract

Two concepts—one of the coupled fixed point and the other of the generalized metric space—play a very active role in recent research on the fixed point theory. The definition of coupled fixed point was introduced by Bhaskar and Lakshmikantham (2006) while the generalized metric space was introduced by Mustafa and Sims (2006). In this work, we determine some coupled fixed point theorems for mixed monotone mapping satisfying nonlinear contraction in the framework of generalized metric space endowed with partial order. We also prove the uniqueness of the coupled fixed point for such mappings in this setup.

1. Introduction

Fixed point theory is a very useful tool in solving variety of problems in the control theory, economic theory, nonlinear analysis and, global analysis. The Banach contraction principle [1] is the most famous, most simplest, and one of the most versatile elementary results in the fixed point theory. A huge amount of literature is witnessed on applications, generalizations, and extensions of this principle carried out by several authors in different directions, for example, by weakening the hypothesis, using different setups, and considering different mappings.

Recently, the idea of generalized metric spaces was introduced and studied by Mustafa and Sims [2] originated from the concept of metric spaces. Some fixed point theorem in this setup was first determined by Mustafa et al. [3]; particularly, the Banach contraction principle was established in this work. Since then several fixed point, coupled fixed point, and triple fixed point theorems in the framework of generalized metric spaces have been investigated in [4–11].

In the recent past, many authors obtained important fixed point theorems in partially ordered metric spaces, that is, metric spaces endowed with a partial ordering (see [12–24]).

The aim of this paper is to determine some coupled fixed point theorems for nonlinear contractions in the framework of partially ordered generalized metric spaces.

2. Definitions and Preliminary Results

We will assume throughout this paper that the symbol ℝ and ℕ will denote the set of real and natural numbers, respectively. In this section, we recall some definitions and preliminary results which we will use throughout the paper. Mustafa and Sims [2] have recently introduced the concept of generalized metric space as follows.

Let 𝑋 be a nonempty set and a mapping 𝐺∶𝑋×𝑋×𝑋→ℝ. Then 𝐺 is called a generalized metric (for short, 𝐺-metric) on 𝑋 and (𝑋,𝐺) a generalized metric space or simply 𝐺-metric space if the following conditions are satisfied: (i)𝐺(𝑥,𝑦,𝑧)=0 if 𝑥=𝑦=𝑧, (ii)𝐺(𝑥,𝑥,𝑦)>0, for all 𝑥,𝑦∈𝑋 and 𝑥≠𝑦, (iii)𝐺(𝑥,𝑥,𝑦)≤𝐺(𝑥,𝑦,𝑧), for all 𝑥,𝑦,𝑧∈𝑋 and 𝑦≠𝑧, (iv)𝐺(𝑥,𝑦,𝑧)=𝐺(𝑥,𝑧,𝑦)=𝐺(𝑦,𝑧,𝑥)=⋯ (symmetry in all three variables), (v)𝐺(𝑥,𝑦,𝑧)≤𝐺(𝑥,𝑎,𝑎)+𝐺(𝑎,𝑦,𝑧), for all 𝑥,𝑦,𝑧,𝑎∈𝑋 (rectangle inequality).

We remark that every 𝐺-metric on 𝑋 defines a metric 𝑑_𝐺 on 𝑋 by 𝑑_𝐺(𝑥,𝑦)=𝐺(𝑥,𝑦,𝑦)+𝐺(𝑦,𝑥,𝑥), for all 𝑥,𝑦∈𝑋.

Example 2.1 (see [2]). Let (𝑋,𝑑) be a metric space. The function 𝐺∶𝑋×𝑋×𝑋→[0,∞) defined by 𝐺𝑥,𝑦,𝑧=max𝑑𝑥,𝑦,𝑑𝑦,𝑧,𝑑(𝑧,𝑥), (2.1) or 𝐺𝑥,𝑦,𝑧=𝑑𝑥,𝑦+𝑑𝑦,𝑧+𝑑(𝑧,𝑥), (2.2) for all 𝑥,𝑦,𝑧∈𝑋, is a 𝐺-metric on 𝑋.
The concepts of convergence and Cauchy sequences and continuous functions in 𝐺-metric space are studied in [2].
Let (𝑋,𝐺) be a 𝐺-metric space. Then, a sequence (𝑥_𝑛) is said to be convergent in (𝑋,𝐺) or simply 𝐺-convergent to 𝑥∈𝑋 if for every 𝜖>0 there exists 𝑁∈ℕ such that 𝐺(𝑥_𝑛,𝑥_𝑚,𝑥)<𝜖, for all 𝑛,𝑚≥𝑁.
Let (𝑋,𝐺) be a 𝐺-metric space. Then, (𝑥_𝑛) is said to be Cauchy in (𝑋,𝐺) or simply 𝐺-Cauchy if for every 𝜖>0 there exists 𝑁∈ℕ such that 𝐺(𝑥_𝑛,𝑥_𝑚,𝑥_𝑘)<𝜖, for all 𝑛,𝑚,𝑘≥𝑁. A 𝐺-metric space (𝑋,𝐺) is said to be complete if every 𝐺-Cauchy sequence is 𝐺-convergent.
Let (𝑋,𝐺) be a 𝐺-metric space and 𝑓∶𝑋→𝑋 a mapping. Then, 𝑓 is said to G-continuous at a point 𝑥∈𝑋 if and only if it is 𝐺-sequentially continuous at 𝑥; that is, whenever (𝑥_𝑛) is 𝐺-convergent to 𝑥, we have (𝑓(𝑥_𝑛))𝐺-convergent to 𝑓(𝑥).

Proposition 2.2 (see [2]). Let (𝑋,𝐺) be a 𝐺-metric space and (𝑥_𝑛) a sequence in 𝑋. Then, for all 𝑥∈𝑋, the following statements are equivalent: (i)(𝑥_𝑛) is 𝐺-convergent to 𝑥,(ii)𝐺(𝑥_𝑛,𝑥_𝑛,𝑥)→0 as 𝑛→∞,(iii)𝐺(𝑥_𝑛,𝑥,𝑥)→0 as 𝑛→∞,(iv)𝐺(𝑥_𝑛,𝑥_𝑚,𝑥)→0 as 𝑛,𝑚→∞.

Proposition 2.3 (see [2]). Let (𝑋,𝐺) be a 𝐺-metric space and (𝑥_𝑛) a sequence in 𝑋. Then, the following statements are equivalent: (i)(𝑥_𝑛) is 𝐺-Cauchy, (ii)For every 𝜖>0 there exists 𝑁∈ℕ such that 𝐺(𝑥_𝑛,𝑥_𝑚,𝑥_𝑚)<𝜖, for all 𝑛,𝑚≥𝑁.

Lemma 2.4 (see [2]). If (𝑋,𝐺) is a 𝐺-metric space then 𝐺(𝑥,𝑦,𝑦)≤2𝐺(𝑦,𝑥,𝑥) for all 𝑥,𝑦∈𝑋.

Let (𝑋,𝐺) be a 𝐺-metric space and 𝐹∶𝑋×𝑋→𝑋 a mapping. Then, a map 𝐹 is said to be continuous [10] in (𝑋,𝐺) if for every 𝐺-convergent sequences 𝑥_𝑛→𝑥 and 𝑦_𝑛→𝑦, (𝐹(𝑥_𝑛,𝑦_𝑛)) is 𝐺-convergent to 𝐹(𝑥,𝑦).

Quite recently, Bhaskar and Lakshmikantham [14] defined and studied the concepts of mixed monotone property and coupled fixed point in partially ordered metric space.

Let (𝑋,≤) be a partially ordered set and 𝐹∶𝑋×𝑋→𝑋 a mapping. Then, a map 𝐹 is said to have mixed monotone property if 𝐹(𝑥,𝑦) is monotone nondecreasing in 𝑥 and is monotone nonincreasing in 𝑦; that is, for any 𝑥,𝑦∈𝑋, 𝑥₁,𝑥₂∈𝑋,𝑥₁≤𝑥₂implies𝐹𝑥₁,𝑦≤𝐹𝑥₂,𝑦,𝑦₁,𝑦₂∈𝑋,𝑦₁≤𝑦₂implies𝐹𝑥,𝑦₁≥𝐹𝑥,𝑦₂. (2.3)

An element (𝑥,𝑦)∈𝑋×𝑋 is said to be a coupled fixed point of the mapping 𝐹∶𝑋×𝑋→𝑋 if 𝐹𝑥,𝑦=𝑥,𝐹𝑦,𝑥=𝑦. (2.4)

The following class of functions are considered in [25]. Denote with Φ the set of all functions 𝜑∶[0,∞)→[0,∞) which satisfy that(i)𝜑 is continuous and nondecreasing, (ii)𝜑(𝑡)=0 if and only if 𝑡=0, (iii)𝜑(𝑡+𝑠)≤𝜑(𝑡)+𝜑(𝑠), for all 𝑡,𝑠∈[0,∞).

By Ψ we denote the set of all functions 𝜓∶[0,∞)→(0,∞) which satisfy lim_𝑡→𝑟𝜓(𝑡)>0, for all 𝑟>0 and lim_𝑡→0⁺𝜓(𝑡)=0.

For example, functions 𝜑₁,𝜑₂,𝜑₃,𝜑₃∈Φ, where 𝜑₁(𝑡)=𝑘𝑡(𝑘>0), 𝜑₂(𝑡)=𝑡/𝑡+1, 𝜑₃(𝑡)=ln(𝑡+1), and 𝜑₄(𝑡)=min{𝑡,1}, and the functions 𝜓₁,𝜓₂,𝜓₃∈Ψ, where 𝜓₁(𝑡)=𝑘𝑡, 𝜓₂(𝑡)=ln(2𝑡+1)/2, and 𝜓₃(𝑡)=1,if𝑡𝑡=0,𝑡=1,,𝑡+1if𝑡0<𝑡<1,2,if𝑡>1. (2.5)

3. Main Results

In this section, we establish some coupled fixed point results by considering a map on generalized metric spaces endowed with a partial order.

Theorem 3.1. Let (𝑋,⪯) be a partially ordered set, and let 𝐺 be a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space. Suppose that there exist 𝜑∈Φ, 𝜓∈Ψ, and a mapping 𝐹∶𝑋×𝑋→𝑋 such that 𝜑𝐺𝐹()()≤1𝑥,𝑦,𝐹𝑢,𝑣,𝐹𝑠,𝑡2𝜑𝐺()𝐺()𝑥,𝑢,𝑠+𝐺𝑦,𝑣,𝑡−𝜓𝑥,𝑢,𝑠+𝐺𝑦,𝑣,𝑡2, (3.1) for all 𝑥,𝑦,𝑢,𝑣,𝑠,𝑡∈𝑋 with 𝑥≽𝑢≽𝑠 and 𝑦⪯𝑣⪯𝑡 where either 𝑢≠𝑠 or 𝑣≠𝑡. Suppose 𝐹 has a mixed monotone property and also suppose that either (a)𝐹 is continuous or(b)𝑋 has the following property:(i)if a nondecreasing sequence (𝑥_𝑛) is 𝐺-convergent to 𝑥, then 𝑥_𝑛⪯𝑥, for all 𝑛,(ii)if a nonincreasing sequence (𝑦_𝑛) is 𝐺-convergent to 𝑦, then 𝑦_𝑛≽𝑦, for all 𝑛.If there exist 𝑥₀,𝑦₀∈𝑋 such that 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀), then 𝐹 has a coupled point; that is, there exist 𝑥,𝑦∈𝑋 such that 𝐹(𝑥,𝑦)=𝑥 and 𝐹(𝑦,𝑥)=𝑦.

Proof. Let 𝑥₀,𝑦₀∈𝑋 be such that 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀). We can choose 𝑥₁,𝑦₁∈𝑋 such that 𝑥₁=𝐹(𝑥₀,𝑦₀) and 𝑦₁=𝐹(𝑦₀,𝑥₀). Write 𝑥_𝑛+1=𝐹𝑥_𝑛,𝑦_𝑛,𝑦_𝑛+1=𝐹𝑦_𝑛,𝑥_𝑛, (3.2) for all 𝑛≥1. Due to the mixed monotone property of 𝐹, we can find 𝑥₂≽𝑥₁≽𝑥₀ and 𝑦₂⪯𝑦₁⪯𝑦₀. By straightforward calculation, we obtain 𝑥₀⪯𝑥₁⪯𝑥₂⪯⋯⪯𝑥_𝑛+1𝑦⪯⋯,₀≽𝑦₁≽𝑦₂≽⋯≽𝑦_𝑛+1≽⋯. (3.3) Using (3.1) and (3.2), we obtain 𝜑𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛=𝜑𝐺𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛−1,𝑦_𝑛−1≤12𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1𝐺𝑥−𝜓^𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−12, (3.4) and similarly 𝜑𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛=𝜑𝐺𝐹𝑦_𝑛,𝑥_𝑛,𝐹𝑦_𝑛,𝑥_𝑛,𝐹𝑦_𝑛−1,𝑥_𝑛−1≤12𝜑𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1+𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1𝐺𝑦−𝜓^𝑛,𝑦_𝑛,𝑦_𝑛−1+𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−12. (3.5) Adding (3.4) and (3.5), we get 𝜑𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝜑𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛≤𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1𝐺𝑥−2𝜓^𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−12. (3.6) Using the property 𝜑(𝑡+𝑠)≤𝜑(𝑡)+𝜑(𝑠) for all 𝑡,𝑠∈[0,∞), we get 𝜑𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛≤𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1𝐺𝑥−2𝜓^𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−12, (3.7) which implies that 𝜑𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛≤𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1. (3.8) Since 𝜑 is nondecreasing, we have 𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛≤𝐺𝑥_𝑛,𝑥_𝑛,𝑥_𝑛−1+𝐺𝑦_𝑛,𝑦_𝑛,𝑦_𝑛−1. (3.9) For all 𝑛∈ℕ, set 𝑠_𝑛=𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛, (3.10) then a sequence (𝑠_𝑛) is decreasing. Therefore, there exists some 𝑠≥0 such that lim_𝑛→∞𝑠_𝑛=lim_𝑛→∞𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛=𝑠. (3.11) Now we have to show that 𝑠=0. On the contrary, suppose that 𝑠>0. Letting 𝑛→∞ in (3.7) (equivalently, 𝑠_𝑛 is 𝐺-convergent to 𝑠) and using the property of 𝜑 and 𝜓, we get 𝜑(𝑠)=lim_𝑛→∞𝜑𝑠_𝑛≤lim_𝑛→∞𝜑𝑠_𝑛−1𝑠−2𝜓^𝑛−12=𝜑(𝑠)−2lim_{𝑠𝑛−1→𝑠}𝜓𝑠_𝑛−12<𝜑(𝑠), (3.12) which is a contradiction. Thus 𝑠=0; from (3.11), we have lim_𝑛→∞𝑠_𝑛=lim_𝑛→∞𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦_𝑛=0. (3.13) Again, we have to show that (𝑥_𝑛) and (𝑦_𝑛) are Cauchy sequences in the 𝐺-metric space (𝑋,𝐺). On the contrary, suppose that at least one of (𝑥_𝑛) or (𝑦_𝑛) is not a Cauchy sequence in (𝑋,𝐺). Then there exists 𝜖>0, for which we can find subsequences (𝑥_𝑘(𝑗)), (𝑥_𝑙(𝑗)) and (𝑦_𝑘(𝑗)), (𝑦_𝑙(𝑗)) of the sequences (𝑥_𝑛) and (𝑦_𝑛), respectively, with 𝑘(𝑗)>𝑙(𝑗)≥𝑗, for all 𝑗∈ℕ such that 𝛼_𝑗=𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)≥𝜖. (3.14) We may also assume that 𝐺𝑥_{𝑘(𝑗)−1},𝑥_{𝑘(𝑗)−1},𝑥_𝑙(𝑗)+𝐺𝑦_{𝑘(𝑗)−1},𝑦_{𝑘(𝑗)−1},𝑦_𝑙(𝑗)<𝜖, (3.15) by choosing 𝑘(𝑗) to be the smallest number exceeding 𝑙(𝑗), for which (3.14) holds. From (3.14) and (3.15) and using the rectangle inequality, we obtain 𝜖≤𝛼_𝑗=𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)≤𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_{𝑘(𝑗)−1}+𝐺𝑥_{𝑘(𝑗)−1},𝑥_{𝑘(𝑗)−1},𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_{𝑘(𝑗)−1}+𝐺𝑦_{𝑘(𝑗)−1},𝑦_{𝑘(𝑗)−1},𝑦_𝑙(𝑗)<𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_{𝑘(𝑗)−1}+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_{𝑘(𝑗)−1}+𝜖. (3.16) Letting 𝑗→∞ in the above inequality and using (3.13), we get lim_𝑗→∞𝛼_𝑗=lim_𝑗→∞𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)=𝜖. (3.17) Again, by using rectangle inequality, we obtain 𝛼_𝑗=𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)≤𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑘(𝑗)+1+𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝐺𝑥_𝑙(𝑗)+1,𝑥_𝑙(𝑗)+1,𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑘(𝑗)+1+𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1+𝐺𝑦_𝑙(𝑗)+1,𝑦_𝑙(𝑗)+1,𝑦_𝑙(𝑗)=𝑠_𝑙(𝑗)+𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑘(𝑗)+1+𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑘(𝑗)+1+𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1. (3.18) By using Lemma 2.4, the above inequality becomes 𝛼_𝑗≤𝑠_𝑙(𝑗)+2𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+2𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1; (3.19) this implies that 𝛼_𝑗≤𝑠_𝑙(𝑗)+2𝑠_𝑘(𝑗)+𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1. (3.20) Operating 𝜑 on both sides of the above inequality, 𝜑𝛼_𝑗≤𝜑𝑠_𝑙(𝑗)+2𝑠_𝑘(𝑗)+𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1=𝜑𝑠_𝑙(𝑗)+2𝑠_𝑘(𝑗)+𝜑𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝜑𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1. (3.21) Now we find the expressions 𝜑(𝐺(𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1)) and 𝜑(𝐺(𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1)) in terms of 𝜑 and 𝜓 by using (3.1) and (3.2); that is, 𝜑𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1=𝜑𝐺𝐹𝑥_𝑘(𝑗),𝑦_𝑘(𝑗),𝐹𝑥_𝑘(𝑗),𝑦_𝑘(𝑗),𝐹𝑥_𝑙(𝑗),𝑦_𝑙(𝑗)≤12𝜑𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)𝐺𝑥−𝜓^𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)+𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)2,𝜑𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1=𝜑𝐺𝐹𝑦_𝑘(𝑗),𝑥_𝑘(𝑗),𝐹𝑦_𝑘(𝑗),𝑥_𝑘(𝑗),𝐹𝑦_𝑙(𝑗),𝑥_𝑙(𝑗)≤12𝜑𝐺𝑦_𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)+𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)𝐺𝑦−𝜓^𝑘(𝑗),𝑦_𝑘(𝑗),𝑦_𝑙(𝑗)+𝐺𝑥_𝑘(𝑗),𝑥_𝑘(𝑗),𝑥_𝑙(𝑗)2. (3.22)(3.23) Adding (3.22) and (3.23), we get 𝜑𝐺𝑥_𝑘(𝑗)+1,𝑥_𝑘(𝑗)+1,𝑥_𝑙(𝑗)+1+𝜑𝐺𝑦_𝑘(𝑗)+1,𝑦_𝑘(𝑗)+1,𝑦_𝑙(𝑗)+1≤𝜑𝛼_𝑗𝛼−2𝜓^𝑗2. (3.24) From (3.21) and (3.24), we obtain 𝜑𝛼_𝑗≤𝜑𝑠_𝑙(𝑗)+2𝑠_𝑘(𝑗)+𝜑𝛼_𝑗𝛼−2𝜓^𝑗2. (3.25) Taking limit as 𝑗→∞ on both sides of the above inequality, we get 𝜑(𝜖)≤𝜑(0)+𝜑(𝜖)−2lim_𝑗→∞𝜓𝛼_𝑗2=𝜑(𝜖)−2lim_{𝛼𝑗→𝜖}𝜓𝛼_𝑗2<𝜑(𝜖), (3.26) which is a contradiction, and hence (𝑥_𝑛) and (𝑦_𝑛) are Cauchy sequences in the 𝐺-metric space (𝑋,𝐺). Since (𝑋,𝐺) is complete 𝐺-metric space, hence (𝑥_𝑛) and (𝑦_𝑛) are 𝐺-convergent. Then, there exist 𝑥,𝑦∈𝑋 such that (𝑥_𝑛) and (𝑦_𝑛) are 𝐺-convergent to 𝑥 and 𝑦, respectively. Suppose that condition (a) holds. Letting 𝑛→∞ in (3.2), we get 𝑥=𝐹(𝑥,𝑦) and 𝑦=𝐹(𝑦,𝑥). Lastly, suppose that assumption (b) holds. Since a sequence (𝑥_𝑛) is nondecreasing and 𝐺-convergent to 𝑥 and also (𝑦_𝑛) is nonincreasing sequence and 𝐺-convergent to 𝑦, by assumption (b), we have 𝑥_𝑛⪯𝑥 and 𝑦_𝑛≽𝑦 for all 𝑛. Using the rectangle inequality, write 𝐺𝑥,𝑥,𝐹𝑥,𝑦≤𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝐹𝑥,𝑦=𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥,𝑦. (3.27) Applying the function 𝜑 on both sides of the above equation and using (3.1), we have 𝜑𝐺𝑥,𝑥,𝐹𝑥,𝑦≤𝜑𝐺𝑥,𝑥,𝑥_𝑛+1+𝜑𝐺𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛,𝑦_𝑛𝐺,𝐹𝑥,𝑦≤𝜑𝑥,𝑥,𝑥_𝑛+1+12𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑥+𝐺𝑦_𝑛,𝑦_𝑛𝐺𝑥,𝑦−𝜓^𝑛,𝑥_𝑛,𝑥+𝐺𝑦_𝑛,𝑦_𝑛,𝑦2. (3.28) Letting 𝑛→∞, we get 𝐺(𝑥,𝑥,𝐹(𝑥,𝑦))=0. Hence 𝑥=𝐹(𝑥,𝑦). Similarly we obtain 𝑦=𝐹(𝑦,𝑥). Thus, we conclude that 𝐹 has a coupled fixed point.

Corollary 3.2. Let (𝑋,⪯) be a partially ordered set, and let 𝐺 be a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space. Suppose that 𝐹∶𝑋×𝑋→𝑋 is a mapping having mixed monotone property. Assume that there exists 𝜓∈Ψ such that for all 𝑥,𝑦,𝑢,𝑣,𝑠,𝑡∈𝑋, 𝐺𝐹()()≤𝐺()𝑥,𝑦,𝐹𝑢,𝑣,𝐹𝑠,𝑡𝑥,𝑢,𝑠+𝐺𝑦,𝑣,𝑡2𝐺()−𝜓𝑥,𝑢,𝑠+𝐺𝑦,𝑣,𝑡2, (3.29) with 𝑥≽𝑢≽𝑠 and 𝑦⪯𝑣⪯𝑡 where either 𝑢≠𝑠 or 𝑣≠𝑡. Suppose that either (a)𝐹 is continuous or (b)𝑋 has the following property: (i)if a nondecreasing sequence (𝑥_𝑛) is 𝐺-convergent to 𝑥, then 𝑥_𝑛⪯𝑥, for all 𝑛,(ii)if a nonincreasing sequence (𝑦_𝑛) is 𝐺-convergent to 𝑦, then 𝑦_𝑛≽𝑦, for all 𝑛.If there exist 𝑥₀,𝑦₀∈𝑋 such that 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀), then there exist 𝑥,𝑦∈𝑋 such that 𝐹(𝑥,𝑦)=𝑥 and 𝐹(𝑦,𝑥)=𝑦; that is, 𝐹 has a coupled point in 𝑋.

Proof. Taking 𝜑(𝑡)=𝑡 in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.

Corollary 3.3. Let (𝑋,⪯) be a partially ordered set, and let 𝐺 be a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space. Suppose 𝐹∶𝑋×𝑋→𝑋 is a mapping having mixed monotone property and assume that there exists 𝑘∈[0,1) such that 𝐺𝐹()()≤𝑘𝑥,𝑦,𝐹𝑢,𝑣,𝐹𝑠,𝑡2𝐺(𝑥,𝑢,𝑠)+𝐺𝑦,𝑣,𝑡, (3.30) for all 𝑥,𝑦,𝑢,𝑣,𝑠,𝑡∈𝑋 with 𝑥≽𝑢≽𝑠 and 𝑦⪯𝑣⪯𝑡 where either 𝑢≠𝑠 or 𝑣≠𝑡. Suppose that either (a)𝐹 is continuous or (b)𝑋 has the following property: (i)if a nondecreasing sequence (𝑥_𝑛) is 𝐺-convergent to 𝑥, then 𝑥_𝑛⪯𝑥, for all 𝑛,(ii)if a nonincreasing sequence (𝑦_𝑛) is 𝐺-convergent to 𝑦, then 𝑦_𝑛≽𝑦, for all 𝑛.If there exist 𝑥₀,𝑦₀∈𝑋 such that 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀), then 𝐹 has a coupled point in 𝑋.

Proof. Taking 𝜑(𝑡)=𝑡 and 𝜓(𝑡)=((1−𝑘)/2)𝑡 in Theorem 3.1 and proceeding the same lines as in this theorem, we get the desired result.

Remark 3.4. To assure the uniqueness of a coupled fixed point, we shall consider the following condition. If (𝑌,⪯) is a partially ordered set, we endowed the product 𝑌×𝑌 with 𝑥,𝑦⪯(𝑢,𝑣)iff𝑥⪯𝑢,𝑦≽𝑣, (3.31) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑌×𝑌.

Theorem 3.5. In addition to the hypothesis of Theorem 3.1, suppose that for all (𝑥,𝑦),(𝑠,𝑡)∈𝑋×𝑋, there exists (𝑢,𝑣)∈𝑋×𝑋 such that (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) is comparable with (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) and (𝐹(𝑠,𝑡),𝐹(𝑡,𝑠)). Then, 𝐹 has a unique coupled fixed point.

Proof. It follows from Theorem 3.1 that the set of coupled fixed points is nonempty. Suppose (𝑥,𝑦) and (𝑠,𝑡) are coupled fixed points of the mappings 𝐹∶𝑋×𝑋→𝑋; that is, 𝑥=𝐹(𝑥,𝑦), 𝑦=𝐹(𝑦,𝑥), and 𝑠=𝐹(𝑠,𝑡), 𝑡=𝐹(𝑡,𝑠). By assumption there exists (𝑢,𝑣) in 𝑋×𝑋 such that (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) is comparable to (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) and (𝐹(𝑠,𝑡),𝐹(𝑡,𝑠)). Put 𝑢=𝑢₀ and 𝑣=𝑣₀ and choose 𝑢₁,𝑣₁∈𝑋 such that 𝑢₁=𝐹(𝑢₁,𝑣₁) and 𝑣₁=𝐹(𝑣₁,𝑢₁). Thus, we can define two sequences (𝑢_𝑛) and (𝑣_𝑛) as 𝑢_𝑛+1=𝐹𝑢_𝑛,𝑣_𝑛,𝑣_𝑛+1=𝐹𝑣_𝑛,𝑢_𝑛. (3.32) Since (𝑢,𝑣) is comparable to (𝑥,𝑦), we can assume that (𝑥,𝑦)≽(𝑢,𝑣)=(𝑢₀,𝑣₀). Then it is easy to show that (𝑢_𝑛,𝑣_𝑛) and (𝑥,𝑦) are comparable; that is, (𝑥,𝑦)≽(𝑢_𝑛,𝑣_𝑛), for all 𝑛. Thus, from (3.1), we have 𝜑𝐺𝑢_𝑢+1,𝑥,𝑥=𝜑𝐺𝐹𝑢_𝑢,𝑣_𝑛≤1,𝐹𝑥,𝑦,𝐹𝑥,𝑦2𝜑𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛𝐺𝑢,𝑦,𝑦−𝜓^𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦2,𝜑𝐺𝑦,𝑦,𝑣_𝑢+1=𝜑𝐺𝐹𝑦,𝑥,𝐹𝑦,𝑥,𝐹𝑣_𝑢,𝑢_𝑛≤12𝜑𝐺𝑦,𝑦,𝑣_𝑛+𝐺𝑥,𝑥,𝑢_𝑛𝐺−𝜓𝑦,𝑦,𝑣_𝑛+𝐺𝑥,𝑥,𝑢_𝑛2. (3.33)(3.34) Using the property of 𝜑 and adding (3.33) and (3.34), we get 𝜑𝐺𝑢_𝑢+1,𝑥,𝑥+𝐺𝑣_𝑢+1,𝑦,𝑦≤𝜑𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛𝐺𝑢,𝑦,𝑦−2𝜓^𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦2, (3.35) which implies that 𝜑𝐺𝑢_𝑢+1,𝑥,𝑥+𝐺𝑣_𝑢+1,𝑦,𝑦≤𝜑𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦. (3.36) Therefore, 𝐺𝑢_𝑢+1,𝑥,𝑥+𝐺𝑣_𝑢+1,𝑦,𝑦≤𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦. (3.37) We see that the sequence (𝐺(𝑢_𝑛,𝑥,𝑥)+𝐺(𝑣_𝑛,𝑦,𝑦)) is decreasing; there exists some 𝜉≥0 such that lim_𝑛→∞𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦=𝜉. (3.38) Now we have to show that 𝜉=0. On the contrary, suppose that 𝜉>0. Letting 𝑛→∞ in (3.35), we get 𝜑(𝜉)≤𝜑(𝜉)−2𝜓lim_𝑛→∞𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦2<𝜑(𝜉), (3.39) which is not possible. Hence 𝜉=0. Therefore, (3.38) becomes lim_𝑛→∞𝐺𝑢_𝑛,𝑥,𝑥+𝐺𝑣_𝑛,𝑦,𝑦=0, (3.40) which implies lim_𝑛→∞𝐺𝑢_𝑛,𝑥,𝑥=0=lim_𝑛→∞𝐺𝑣_𝑛,𝑦,𝑦. (3.41) Similarly, we can show that lim_𝑛→∞𝐺(𝑢_𝑛,𝑠,𝑠)=0=lim_𝑛→∞𝐺(𝑣_𝑛,𝑡,𝑡). We conclude that 𝑥=𝑠 and 𝑦=𝑡. Thus, 𝐹 has a unique coupled fixed point.

Theorem 3.6. In addition to the hypothesis of Theorem 3.1, suppose that 𝑥₀ and 𝑦₀ are comparable. Then, 𝐹 has a unique fixed point.

Proof. Proceeding the same lines as in the proof of Theorem 3.1, we know that a mapping 𝐹∶𝑋×𝑋→𝑋 has a coupled fixed point (𝑥,𝑦). Now we need to show that 𝑥=𝑦. Since 𝑥₀ and 𝑦₀ are comparable, we can assume that 𝑥₀≽𝑦₀. It is easy to show that 𝑥_𝑛≽𝑦_𝑛 for all 𝑛≥0, where 𝑥_𝑛+1=𝐹(𝑥_𝑛,𝑦_𝑛) and 𝑦_𝑛+1=𝐹(𝑦_𝑛,𝑥_𝑛). Suppose that 𝐺(𝑥,𝑥,𝑦)>0, for all 𝑥,𝑦∈𝑋 with 𝑥≠𝑦. Using the rectangle inequality, write 𝐺𝑥,𝑥,𝑦≤𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝑥_𝑛+1,𝑥_𝑛+1,𝑦_𝑛+1+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦=𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑦_𝑛,𝑥_𝑛+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦. (3.42) Operating 𝜑 on both sides of the above inequality, we get 𝜑𝐺𝑥,𝑥,𝑦≤𝜑𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦+𝜑𝐺𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑦_𝑛,𝑥_𝑛. (3.43) From (3.1), we have 𝜑𝐺𝑥,𝑥,𝑦≤𝜑𝐺𝑥,𝑥,𝑥_𝑛+1+𝐺𝑦_𝑛+1,𝑦_𝑛+1+1,𝑦2𝜑𝐺𝑥_𝑛,𝑥_𝑛,𝑦_𝑛+𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛𝐺𝑥−𝜓^𝑛,𝑥_𝑛,𝑦_𝑛+𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛2. (3.44) Similarly, we obtain 𝜑𝐺𝑦,𝑦,𝑥≤𝜑𝐺𝑦,𝑦,𝑦_𝑛+1+𝐺𝑥_𝑛+1,𝑥_𝑛+1+1,𝑥2𝜑𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛+𝐺𝑥_𝑛,𝑥_𝑛,𝑦_𝑛𝐺𝑦−𝜓^𝑛,𝑦_𝑛,𝑥_𝑛+𝐺𝑥_𝑛,𝑥_𝑛,𝑦_𝑛2. (3.45) Adding (3.44) and (3.45) and using the property of 𝜑, we get 𝜑𝐺𝐺𝑥,𝑥,𝑦+𝐺𝑦,𝑦,𝑥≤𝜑𝑥,𝑥,𝑥_𝑛+1+𝐺𝑦_𝑛+1,𝑦_𝑛+1,𝑦+𝜑𝐺𝑦,𝑦,𝑦_𝑛+1+𝐺𝑥_𝑛+1,𝑥_𝑛+1𝐺𝑥,𝑥+𝜑_𝑛,𝑥_𝑛,𝑦_𝑛+𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛𝐺𝑥−2𝜓^𝑛,𝑥_𝑛,𝑦_𝑛+𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛2. (3.46) Letting 𝑛→∞ in the above inequality, we get 𝜑𝐺𝑥,𝑥,𝑦+𝐺𝑦,𝑦,𝑥≤𝜑(0)+𝜑(0)+𝜑𝐺𝑥,𝑥,𝑦+𝐺𝑦,𝑦,𝑥−2lim_𝑛→∞𝜓𝐺𝑥_𝑛,𝑥_𝑛,𝑦_𝑛+𝐺𝑦_𝑛,𝑦_𝑛,𝑥_𝑛2, (3.47) which implies that 𝜑𝐺𝑥,𝑥,𝑦+𝐺𝑦,𝑦,𝑥<𝜑𝐺𝑥,𝑥,𝑦+𝐺𝑦,𝑦,𝑥, (3.48) which is not possible, and hence 𝐺(𝑥,𝑥,𝑦)=0. Thus 𝑥=𝑦, whence the result.

Theorem 3.7. Let (𝑋,⪯) be a partially ordered set, and let 𝐺 be a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space. Let 𝐹∶𝑋×𝑋→𝑋 be a mapping such that 𝐹 has a mixed monotone property and 𝐹(𝑥,𝑦)⪯𝐹(𝑦,𝑥) whenever 𝑥⪯𝑦. Suppose that there exist 𝜑∈Φ and 𝜓∈Ψ such that for all 𝑥,𝑦,𝑢,𝑣,𝑠,𝑡∈𝑋, (3.1) holds with 𝑥≽𝑢≽𝑠, 𝑦⪯𝑣⪯𝑡 and 𝑥≺𝑦 where either 𝑢≠𝑠 or 𝑣≠𝑡. Assume that either (a)𝐹 is continuous or (b)𝑋 has the following property: (i)if a nondecreasing sequence (𝑥_𝑛) is 𝐺-convergent to 𝑥, then 𝑥_𝑛⪯𝑥, for all 𝑛,(ii)if a nonincreasing sequence (𝑦_𝑛) is 𝐺-convergent to 𝑦, then 𝑦_𝑛≽𝑦, for all 𝑛.If there exist 𝑥₀,𝑦₀∈𝑋 such that 𝑥₀⪯𝑦₀, 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀), then 𝐹 has a coupled point; that is, there exist 𝑥,𝑦∈𝑋 such that 𝐹(𝑥,𝑦)=𝑥 and 𝐹(𝑦,𝑥)=𝑦.

Proof. Let 𝑥₀,𝑦₀∈𝑋 be such that 𝑥₀⪯𝐹(𝑥₀,𝑦₀) and 𝑦₀≽𝐹(𝑦₀,𝑥₀). We can choose 𝑥₁,𝑦₁∈𝑋 such that 𝑥₁=𝐹(𝑥₀,𝑦₀)≽𝑥₀ and 𝑦₁=𝐹(𝑦₀,𝑥₀)⪯𝑦₀. Since 𝑥₀⪯𝑦₀, we have 𝐹(𝑥₀,𝑦₀)⪯𝐹(𝑦₀,𝑥₀). Accordingly, 𝑥₀⪯𝑥₁=𝐹𝑥₀,𝑦₀⪯𝐹𝑦₀,𝑥₀=𝑦₁⪯𝑦₀. (3.49) Continuing this process, we can construct two sequences (𝑥_𝑛) and (𝑦_𝑛) in 𝑋 such that 𝑥_𝑛⪯𝐹𝑥_𝑛,𝑦_𝑛=𝑥_𝑛+1⪯𝑦_𝑛+1=𝐹𝑦_𝑛,𝑥_𝑛⪯𝑦_𝑛, (3.50) for all 𝑛≥0. Therefore, 𝑥₀⪯𝑥₁⪯𝑥₂⪯⋯⪯𝑥_𝑛⪯𝑥_𝑛+1⋯⪯𝑦_𝑛+1⪯𝑦_𝑛⪯⋯⪯𝑦₂⪯𝑦₁⪯𝑦₀. (3.51) The rest of the proof can be done on the same lines as in Theorem 3.1.

4. Example and the Concluding Remark

In the following, we construct an example of a 𝐺-metric space involving the idea of coupled fixed point to see the applicability of our results.

Let 𝑋=ℕ∪{0}. Define a mapping 𝐺 from 𝑋³ to ℝ by 𝐺𝑥,𝑦,𝑧=𝑥+𝑦+𝑧,if𝑥,𝑦,𝑧arealldistinctanddifferentfromzero,𝑥+𝑧,if𝑥=𝑦≠𝑧andallaredifferentfromzero,𝑦+𝑧+1,if𝑥=0,𝑦≠𝑧and𝑦,𝑧aredifferentfromzero,𝑦+2,if𝑥=0,𝑦=𝑧≠0,𝑧+1,if𝑥=0=𝑦,𝑧≠0,0,if𝑥=𝑦=𝑧. (4.1) Then (𝑋,𝐺) is a complete 𝐺-metric space [10]. Let us consider a partial order ⪯ on 𝑋 be such that 𝑥⪯𝑦 holds if 𝑥>𝑦, 3 divides (𝑥−𝑦), and 3⪯1 and 0⪯1 hold, for all 𝑥,𝑦∈𝑋. Consider a mapping 𝐹∶𝑋×𝑋→𝑋 defined by 𝐹𝑥,𝑦=1,if𝑥≺𝑦,0,otherwise. (4.2) Suppose that 𝑠⪯𝑢⪯𝑥≺𝑦⪯𝑣⪯𝑡 holds. Therefore, we have 𝐹(𝑥,𝑦)=1=𝐹(𝑢,𝑣)=𝐹(𝑠,𝑡). It follows that 𝐺(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣),𝐹(𝑠,𝑡))=0.

Now, applying the function 𝜑 to this equality and then using the hypothesis of this function, we see that (3.1) is satisfied since the left-hand side of (3.1) becomes 0. For 𝑥₀=81 and 𝑦₀=0, Theorem 3.7 is applicable. In this case, the coupled fixed point is not unique. Hence (0,0) and (1,0) are two coupled fixed points of 𝐹.

We remark that inequality (3.1) is not satisfied when 𝑠=𝑢=𝑥=𝑦=3, 𝑣=0, and 𝑡=1, and hence Theorem 3.1 does not work for this example. We know that a 𝐺-metric naturally induces a metric 𝑑_𝐺 given by 𝑑_𝐺(𝑥,𝑦)=𝐺(𝑥,𝑦,𝑦)+𝐺(𝑥,𝑥,𝑦) [2], but inequality (3.1) does not reduce to any metric inequality with the metric 𝑑_𝐺 due to the condition that either 𝑢≠𝑠 or 𝑣≠𝑡. Hence our theorems are more general, different from the classical results, and do not reduce to fixed point problems in the corresponding metric space (𝑋,𝑑_𝐺).

Acknowledgment

The authors have benefited from the reports of the anonymous referees and they are thankful for their valuable comments on the first draft of this paper which improved the presentation and readability.