Fixed Point Theory of Weak Contractions in Partially Ordered Metric Spaces

Abstract

We prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.

1. Introduction and Preliminaries

Throughout this paper, by ℝ⁺, we denote the set of all nonnegative real numbers, while ℕ is the set of all natural numbers. Let (𝑋,𝑑) be a metric space, 𝐷 a subset of 𝑋, and 𝑓∶𝐷→𝑋 a map. We say 𝑓 is contractive if there exists 𝛼∈[0,1) such that, for all 𝑥,𝑦∈𝐷,𝑑𝑓𝑥,𝑓𝑦≤𝛼⋅𝑑𝑥,𝑦. (1) The well-known Banach's fixed point theorem asserts that if 𝐷=𝑋, 𝑓 is contractive and (𝑋,𝑑) is complete, then 𝑓 has a unique fixed point in 𝑋. In nonlinear analysis, the study of fixed points of given mappings satisfying certain contractive conditions in various abstract spaces has been investigated deeply. The Banach contraction principle [1] is one of the initial and crucial results in this direction. Also, this principle has many generalizations. For instance, Alber and Guerre-Delabriere in [2] suggested a generalization of the Banach contraction mapping principle by introducing the concept of weak contraction in Hilbert spaces. In [2], the authors also proved that the result of Eslamian and Abkar [3] is equivalent to the result of Dutta and Choudhury [4]. Later, weakly contractive mappings and mappings satisfying other weak contractive inequalities have been discussed in several works, some of which are noted in [4–16].

In 2008, Dutta and Choudhury proved the following theorem.

Theorem 1 (see [4]). Let (𝑋,𝑑) be a complete metric space, and let 𝑓∶𝑋→𝑋 be such that 𝜓𝑑𝑓𝑥,𝑓𝑦≤𝜓𝑑𝑥,𝑦−𝜙𝑑𝑥,𝑦,𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑥,𝑦∈𝑋, (2) where 𝜓,𝜙∶ℝ⁺→ℝ⁺ are continuous and nondecreasing, and 𝜓(𝑡)=𝜙(𝑡)=0 if and only if 𝑡=0. Then 𝑓 has a fixed point in 𝑋.

Recently, Eslamian and Abkar [3] proved the following theorem.

Theorem 2 (see [3]). Let (𝑋,𝑑) be a complete metric space, and let 𝑓∶𝑋→𝑋 be such that 𝜓𝑑𝑓𝑥,𝑓𝑦≤𝛼𝑑𝑥,𝑦−𝛽𝑑𝑥,𝑦,𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑥,𝑦∈𝑋, (3) where 𝜓,𝛼,𝛽∶ℝ⁺→ℝ⁺ are such that 𝜓 is continuous and nondecreasing, 𝛼 is continuous, 𝛽 is lower semicontinuous, and 𝜓(𝑡)(𝑡)(𝑡)𝜓(𝑡)(0)(0)−𝛼+𝛽>0∀𝑡>0,=0𝑖𝑓𝑎𝑛𝑑𝑜𝑛𝑙𝑦𝑖𝑓𝑡=0,𝛼=𝛽=0. (4) Then 𝑓 has a fixed point in 𝑋.

In the recent, fixed point theory has developed rapidly in partially ordered metric spaces (e.g., [17–22]).

In 2012, Choudhury and Kundu [23] proved the following fixed point theorem as a generalization of Theorem 2.

Theorem 3 (see [23]). Let (𝑋,⊑) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space and let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that 𝜓𝑑𝑓𝑥,𝑓𝑦≤𝛼𝑑𝑥,𝑦−𝛽𝑑𝑥,𝑦,𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑥,𝑦∈𝑋𝑠𝑢𝑐ℎ𝑡ℎ𝑎𝑡𝑥⊑𝑦, (5) where 𝜓,𝛼,𝛽∶ℝ⁺→ℝ⁺ are such that 𝜓 is continuous and nondecreasing, 𝛼 is continuous, 𝛽 is lower semicontinuous, and 𝜓(𝑡)(𝑡)(𝑡)𝜓(𝑡)(0)(0)−𝛼+𝛽>0∀𝑡>0,=0𝑖𝑓𝑎𝑛𝑑𝑜𝑛𝑙𝑦𝑖𝑓𝑡=0,𝛼=𝛽=0. (6) Also, if any nondecreasing sequence {𝑥_𝑛} in 𝑋 converges to 𝜈, then one assumes that 𝑥_𝑛⊑𝜈∀𝑛∈ℕ. (7) If there exists 𝑥₀∈𝑋 with 𝑥₀⊑𝑓𝑥₀, then 𝑓 and 𝑔 have a coincidence point in 𝑋.

In this paper, we prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.

2. Fixed Point Results (I)

We start with the following definition.

Definition 4. Let (𝑋,⊑) be a partially ordered set and 𝑓∶𝑋→𝑋. Then 𝑓 is said to be monotone nondecreasing if, for 𝑥,𝑦∈𝑋, 𝑥⊑𝑦⟹𝑓𝑥⊑𝑓𝑦. (8)

Let (𝑋,⊑) be a partially ordered set. 𝑥,𝑦∈𝑋 are said to be comparable if either 𝑥⊑𝑦 or 𝑦⊑𝑥 holds.

In the section, we denote by Ψ the class of functions 𝜓∶ℝ⁺³→ℝ⁺ satisfying the following conditions: (𝜓₁)𝜓 is an increasing, continuous function in each coordinate;(𝜓₂) for all 𝑡∈ℝ⁺, 𝜓(𝑡,𝑡,𝑡)≤𝑡, 𝜓(0,0,𝑡)≤𝑡 and 𝜓(𝑡,0,0)≤𝑡.

Next, we denote by Φ the class of functions 𝜙∶ℝ⁺→ℝ⁺ satisfying the following conditions: (𝜙₁)𝜙 is a continuous, nondecreasing function; (𝜙₂)𝜙(𝑡)>0 for 𝑡>0 and 𝜙(0)=0; (𝜙₃)𝜙 is subadditive; that is, 𝜙(𝑡₁+𝑡₂)≤𝜙(𝑡₁)+𝜙(𝑡₂) for all 𝑡₁,𝑡₂>0.

And we denote the following sets of functions: Θ=𝜑∶ℝ⁺→ℝ⁺suchthat𝜑iscontinuous,Ξ=𝜉∶ℝ⁺→ℝ⁺suchthat𝜉islowercontinuous. (9)

Let 𝑋 be a nonempty set, and let (𝑋,⊑) be a partially ordered set endowed with a metric 𝑑. Then, the triple (𝑋,⊑,𝑑) is called a partially ordered complete metric space.

We now state the main fixed point theorem for (𝜑,𝜓,𝜙,𝜉)-contractions in partially ordered metric spaces, as follows.

Theorem 5. Let (𝑋,⊑,𝑑) be a partially ordered complete metric space. Let 𝑓∶𝑋→𝑋 be monotone nondecreasing, and 𝜑𝑑𝜙𝑑𝑑𝑑𝑑,𝑓𝑥,𝑓𝑦≤𝜓𝑥,𝑦,𝜙𝑥,𝑓𝑥,𝜙𝑦,𝑓𝑦−𝜉max𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦 (10) for all comparable 𝑥,𝑦∈𝑋, where 𝜑∈Θ, 𝜓∈Ψ, 𝜙∈Φ, and 𝜉∈Ξ, and 𝜑(𝑡)(𝑡)(𝑡)𝜑(𝑡)(0)(0)−𝜙+𝜉>0∀𝑡>0,=0𝑖𝑓𝑎𝑛𝑑𝑜𝑛𝑙𝑦𝑖𝑓𝑡=0,𝜙=𝜉=0. (11) Suppose that either (a)  𝑓 is continuous or (b)  if any nondecreasing sequence {𝑥_𝑛} in 𝑋 converges to 𝜈, then one assumes that 𝑥_𝑛⊑𝜈∀𝑛∈ℕ. (12)If there exists 𝑥₀∈𝑋 with 𝑥₀⊑𝑓𝑥₀, then 𝑓 has a fixed point in 𝑋.

Proof. Since 𝑓 is nondecreasing, by induction, we construct the sequence {𝑥_𝑛} recursively as 𝑥_𝑛=𝑓^𝑛𝑥₀=𝑓𝑥_𝑛−1∀𝑛∈ℕ. (13) Thus, we also conclude that 𝑥₀⊏𝑥₁=𝑓𝑥₀⊑𝑥₂=𝑓𝑥₁⊑⋯⊑𝑥_𝑛=𝑓𝑥_𝑛−1⊑⋯. (14) If any two consecutive terms in (14) are equal, then the 𝑓 has a fixed point, and hence the proof is completed. So we may assume that 𝑑𝑥_𝑛−1,𝑥_𝑛≠0,∀𝑛∈ℕ. (15) Now, we claim that 𝑑(𝑥_𝑛,𝑥_𝑛+1)≤𝑑(𝑥_𝑛−1,𝑥_𝑛) for all 𝑛∈ℕ. If not, we assume that 𝑑(𝑥_𝑛−1,𝑥_𝑛)<𝑑(𝑥_𝑛,𝑥_𝑛+1) for some 𝑛∈ℕ; substituting 𝑥=𝑥_𝑛 and 𝑦=𝑥_𝑛+1 in (10) and using the definition of the function 𝜓, we have 𝜓𝜙𝑑𝑥_𝑛,𝑥_𝑛+1,𝜙𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝜙𝑑𝑥_𝑛+1,𝑓𝑥_𝑛+1=𝜓𝜙𝑑𝑥_𝑛,𝑥_𝑛+1,𝜙𝑑𝑥_𝑛,𝑥_𝑛+1,𝜙𝑑𝑥_𝑛+1,𝑥_𝑛+2≤𝜙𝑑𝑥_𝑛+1,𝑥_𝑛+2,𝜉max𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝑥_𝑛+1,𝑓𝑥_𝑛+1=𝜉max𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛+1,𝑥_𝑛+2=𝜉𝑑𝑥_𝑛+1,𝑥_𝑛+2, (16) and hence 𝜑𝑑𝑥_𝑛+1,𝑥_𝑛+2=𝜑𝑑𝑓𝑥_𝑛,𝑓𝑥_𝑛+1≤𝜙𝑑𝑥_𝑛+1,𝑥_𝑛+2−𝜉𝑑𝑥_𝑛+1,𝑥_𝑛+2. (17) Since 𝜑(𝑡)−𝜙(𝑡)+𝜉(𝑡)>0 for all 𝑡>0, we have that 𝑑(𝑥_𝑛+1,𝑥_𝑛+2)=0, which contradicts (15). Therefore, we conclude that 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛−1,𝑥_𝑛∀𝑛∈ℕ. (18) From the previous argument, we also have that for each 𝑛∈ℕ𝜑𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜙𝑑𝑥_𝑛−1,𝑥_𝑛−𝜉𝑑𝑥_𝑛−1,𝑥_𝑛. (19) It follows in (18) that the sequence {𝑑(𝑥_𝑛,𝑥_𝑛+1)} is monotone decreasing; it must converge to some 𝜂≥0. Taking limit as 𝑛→∞ in (19) and using the continuities of 𝜑 and 𝜙 and the lower semicontinuous of 𝜉, we get 𝜑𝜂≤𝜙𝜂−𝜉𝜂, (20) which implies that 𝜂=0. So we conclude that lim_𝑛→∞𝑑𝑥_𝑛,𝑥_𝑛+1=0. (21)

We next claim that {𝑥_𝑛} is a Cauchy sequence; that is, for every 𝜀>0, there exists 𝑛∈ℕ such that if 𝑝,𝑞≥𝑛, then 𝑑(𝑥_𝑝,𝑥_𝑞)<𝜀.

Suppose, on the contrary, that there exists 𝜖>0 such that, for any 𝑛∈ℕ, there are 𝑝_𝑛,𝑞_𝑛∈ℕ with 𝑝_𝑛>𝑞_𝑛≥𝑛 satisfying 𝑑𝑥_𝑞𝑛,𝑥_𝑝𝑛≥𝜖. (22) Further, corresponding to 𝑞_𝑛≥𝑛, we can choose 𝑝_𝑛 in such a way that it the smallest integer with 𝑝_𝑛>𝑞_𝑛≥𝑛 and 𝑑(𝑥_𝑞𝑛,𝑥_𝑝𝑛)≥𝜖. Therefore 𝑑(𝑥_𝑞𝑛,𝑥_𝑝𝑛−1)<𝜖. Now we have that for all 𝑛∈ℕ𝜖≤𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛≤𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛<𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝜖. (23)

By letting 𝑛→∞. we get thatlim_𝑛→∞𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛=𝜖. (24) On the other hand, we have 𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛≤𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1+𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛,𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1≤𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛+𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛+𝑑𝑥_𝑞𝑛,𝑥_𝑞𝑛−1. (25) Letting 𝑛→∞, then we get lim_𝑛→∞𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=𝜖. (26) By (14), we have that the elements 𝑥_𝑝𝑛 and 𝑥_𝑞𝑛 are comparable. Substituting 𝑥=𝑥_𝑝𝑛−1 and 𝑦=𝑥_𝑞𝑛−1 in (10), we have that, for all 𝑛∈ℕ, 𝜓𝜙𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝜙𝑑𝑥_𝑝𝑛−1,𝑓𝑥_𝑝𝑛−1,𝜙𝑑𝑥_𝑞𝑛−1,𝑓𝑥_𝑞𝑛−1≤𝜓𝜙𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝜙𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛,𝜙𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛,𝑀𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=max𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝑑𝑥_𝑝𝑛−1,𝑓𝑥_𝑝𝑛−1,𝑑𝑥_𝑞𝑛−1,𝑓𝑥_𝑞𝑛−1=max𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛,𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛. (27) By the previous argument and using inequality (10), we can conclude that 𝜑(𝜖)𝜙(𝜖)(𝜖)(𝜖)(𝜖),≤𝜓,0,0−𝜉≤𝜙−𝜉 (28) which implies that 𝜖=0, a contradiction. Therefore, the sequence {𝑥_𝑛} is a Cauchy sequence.

Since 𝑋 is complete, there exists 𝜈∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝜈. (29) Suppose that (a) holds. Then 𝜈=lim_𝑛→∞𝑥_𝑛+1=lim_𝑛→∞𝑓𝑥_𝑛=𝑓𝜈. (30)

Thus, 𝜈 is a fixed point in 𝑋.

Suppose that (b) holds; that is, 𝑥_𝑛⊑𝜈 for all 𝑛∈ℕ. Substituting 𝑥=𝑥_𝑛 and 𝑦=𝜈 in (10), we have that 𝜑𝑑𝑥_𝑛+1𝑑,𝑓𝜈=𝜑𝑓𝑥_𝑛𝜙𝑑𝑥,𝑓𝜈≤𝜓_𝑛,𝜈,𝜙𝑑𝑥_𝑛,𝑓𝑥_𝑛𝑑𝑑𝑥,𝜙𝜈,𝑓𝜈−𝜉max_𝑛,𝜈,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝜈,𝑓𝜈. (31) Taking limit as 𝑛→∞ in equality (31), we have 𝜑𝑑𝜙(0)(𝑜)𝑑𝑑𝑑𝑑,𝜈,𝑓𝜈≤𝜓,𝜙,𝜙𝜈,𝑓𝜈−𝜉𝜈,𝑓𝜈≤𝜙𝜈,𝑓𝜈−𝜉𝜈,𝑓𝜈 (32) which implies that 𝑑(𝜈,𝑓𝜈)=0; that is 𝜈=𝑓𝜈. So we complete the proof.

If we let 𝜓𝜙𝑑𝑑𝑑𝜙𝑑𝑑𝑑,𝑥,𝑦,𝜙𝑥,𝑓𝑥,𝜙𝑦,𝑓𝑦=max𝑥,𝑦,𝜙𝑥,𝑓𝑥,𝜙𝑦,𝑓𝑦 (33)it is easy to get the following theorem.

Theorem 6. Let (𝑋,⊑,𝑑) be a partially ordered complete metric space. Let 𝑓∶𝑋→𝑋 be monotone nondecreasing, and 𝜑𝑑𝜙𝑑𝑑𝑑𝑑,𝑓𝑥,𝑓𝑦≤max𝑥,𝑦,𝜙𝑥,𝑓𝑥,𝜙𝑦,𝑓𝑦−𝜉max𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦 (34) for all comparable 𝑥,𝑦∈𝑋, where 𝜑∈Θ, 𝜙∈Φ, and 𝜉∈Ξ, and 𝜑(𝑡)(𝑡)(𝑡)𝜑(𝑡)(0)(0)−𝜙+𝜉>0∀𝑡>0,=0𝑖𝑓𝑎𝑛𝑑𝑜𝑛𝑙𝑦𝑖𝑓𝑡=0,𝜙=𝜉=0. (35) Suppose that either(a)  𝑓 is continuous or (b)  if any nondecreasing sequence {𝑥_𝑛} in 𝑋 converges to 𝜈, then one assumes that 𝑥_𝑛⊑𝜈∀𝑛∈ℕ. (36) If there exists 𝑥₀∈𝑋 with 𝑥₀⊑𝑓𝑥₀, then 𝑓 has a fixed point in 𝑋.

3. Fixed Point Results (II)

In the section, we denote by Ψ the class of functions 𝜓∶ℝ⁺³→ℝ⁺ satisfying the following conditions: (𝜓₁)𝜓 is an increasing and continuous function in each coordinate; (𝜓₂) for 𝑡∈ℝ⁺, 𝜙(𝑡,𝑡,𝑡)≤𝑡, 𝜙(𝑡,0,0)≤𝑡, and 𝜙(0,0,𝑡)≤𝑡.

Next, we denote by Θ the class of functions 𝜑∶ℝ⁺→ℝ⁺ satisfying the following conditions: (𝜑₁)𝜑 is continuous and nondecreasing; (𝜑₂) for 𝑡>0, 𝜑(𝑡)>0 and 𝜑(0)=0.

And we denote by Φ the class of functions 𝜙∶ℝ⁺→ℝ⁺ satisfying the following conditions.(𝜙₁)𝜙 is continuous; (𝜙₂) for 𝑡>0, 𝜙(𝑡)>0 and 𝜙(0)=0.

We now state the main fixed point theorem for the (𝜓,𝜑,𝜙)-contractions in partially ordered metric spaces, as follows.

Theorem 7. Let (𝑋,⊑,𝑑) be a partially ordered complete metric space, and let 𝑓∶𝑋→𝑋 be monotone nondecreasing, and 𝜑𝑑𝜑𝑑𝑑𝑑𝑀,𝑓𝑥,𝑓𝑦≤𝜓𝑥,𝑦,𝜑𝑥,𝑓𝑥,𝜑𝑦,𝑓𝑦−𝜙𝑥,𝑦+𝐿⋅𝑚𝑥,𝑦 (37) for all comparable 𝑥,𝑦∈𝑋 and 𝜓∈Ψ, 𝜑∈Θ, 𝜙∈Φ, where 𝐿>0 and 𝑀𝑑,𝑚𝑑,𝑑.𝑥,𝑦=max𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦𝑥,𝑦=min𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦,𝑑𝑥,𝑓𝑦𝑦,𝑓𝑥 (38) Suppose that either(a)  𝑓 is continuous or(b)  if any nondecreasing sequence {𝑥_𝑛} in 𝑋 converges to 𝜈, then one assumes that 𝑥_𝑛⊑𝜈∀𝑛∈ℕ. (39)If there exists 𝑥₀∈𝑋 with 𝑥₀⊑𝑓𝑥₀, then 𝑓 has a fixed point in 𝑋.

Proof. If 𝑓𝑥₀=𝑥₀, then the proof is finished. Suppose that 𝑥₀⊏𝑓𝑥₀. Since 𝑓 is nondecreasing, by induction, we construct the sequence {𝑥_𝑛} recursively as 𝑥_𝑛=𝑓^𝑛𝑥₀=𝑓𝑥_𝑛−1∀𝑛∈ℕ. (40) Thus, we also conclude that 𝑥₀⊏𝑥₁=𝑓𝑥₀⊑𝑥₂=𝑓𝑥₁⊑⋯⊑𝑥_𝑛=𝑓𝑥_𝑛−1⊑⋯. (41) We now claim that lim_𝑛→∞𝑑𝑥_𝑛,𝑥_𝑛+1=0. (42) Put 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛 in (37). Note that 𝑚𝑥_𝑛−1,𝑥_𝑛=min𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛,𝑑𝑥_𝑛,𝑓𝑥_𝑛−1=min𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑥_𝑛=0. (43) So, we obtain that 𝜑𝑑𝑥_𝑛,𝑥_𝑛+1=𝜑𝑑𝑓𝑥_𝑛−1,𝑓𝑥_𝑛≤𝜓𝜑𝑑𝑥_𝑛−1,𝑥_𝑛,𝜑𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝜑𝑑𝑥_𝑛,𝑓𝑥_𝑛−𝜙𝑀𝑥_𝑛−1,𝑥_𝑛≤𝜓𝜑𝑑𝑥_𝑛−1,𝑥_𝑛,𝜑𝑑𝑥_𝑛−1,𝑥_𝑛,𝜑𝑑𝑥_𝑛,𝑥_𝑛+1−𝜙𝑀𝑥_𝑛−1,𝑥_𝑛, (44) where 𝑀𝑥_𝑛−1,𝑥_𝑛=max𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝑑𝑥_𝑛,𝑓𝑥_𝑛=max𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛+1. (45) We now claim that 𝑑𝑥_𝑛,𝑥_𝑛+1<𝑑𝑥_𝑛−1,𝑥_𝑛,∀𝑛∈ℕ. (46) If not, we assume that 𝑑(𝑥_𝑛−1,𝑥_𝑛)≤𝑑(𝑥_𝑛,𝑥_𝑛+1); then 𝜑(𝑑(𝑥_𝑛−1,𝑥_𝑛))≤𝜑(𝑑(𝑥_𝑛,𝑥_𝑛+1)), since 𝜑 is non-decreasing. Using inequality (44) and the conditions of the function 𝜓, we have that, for each 𝑛∈ℕ, 𝜑𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜑𝑑𝑥_𝑛,𝑥_𝑛+1−𝜙𝑑𝑥_𝑛,𝑥_𝑛+1, (47) which implies that 𝜙(𝑑(𝑥_𝑛,𝑥_𝑛+1))=0, and hence 𝑑(𝑥_𝑛,𝑥_𝑛+1)=0. This contradicts our initial assumption.
From the previous argument, we have that, for each 𝑛∈ℕ, 𝜑𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜑𝑑𝑥_𝑛−1,𝑥_𝑛−𝜙𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛+1<𝑑𝑥_𝑛−1,𝑥_𝑛. (48) And since the sequence {𝑑(𝑥_𝑛,𝑥_𝑛+1)} is decreasing, it must converge to some 𝜂≥0. Taking limit as 𝑛→∞ in (48) and by the continuity of 𝜑 and 𝜙, we get 𝜑𝜂≤𝜑𝜂−𝜙𝜂, (49) and so we conclude that 𝜙(𝜂)=0 and 𝜂=0.
We next claim that {𝑥_𝑛} is Cauchy; that is, for every 𝜀>0, there exists 𝑛∈ℕ such that if 𝑝,𝑞≥𝑛, then 𝑑(𝑥_𝑝,𝑥_𝑞)<𝜀.
Suppose, on the contrary, that there exists 𝜖>0 such that, for any 𝑛∈ℕ, there are 𝑝_𝑛,𝑞_𝑛∈ℕ with 𝑝_𝑛>𝑞_𝑛≥𝑛 satisfying 𝑑𝑥_𝑞𝑛,𝑥_𝑝𝑛≥𝜖. (50) Further, corresponding to 𝑞_𝑛≥𝑛, we can choose 𝑝_𝑛 in such a way that it the smallest integer with 𝑝_𝑛>𝑞_𝑛≥𝑛 and 𝑑(𝑥_𝑞𝑛,𝑥_𝑝𝑛)≥𝜖. Therefore 𝑑(𝑥_𝑞𝑛,𝑥_𝑝𝑛−1)<𝜖. By the rectangular inequality, we have 𝜖≤𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛≤𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛<𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝜖. (51) Letting 𝑛→∞, then we get lim_𝑛→∞𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛=𝜖. (52) On the other hand, we have 𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛≤𝑑𝑥_𝑝𝑛,𝑥_𝑝𝑛−1+𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1+𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛,𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1≤𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛+𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛+𝑑𝑥_𝑞𝑛,𝑥_𝑞𝑛−1. (53) By letting 𝑛→∞, we get that lim_𝑛→∞𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=𝜖. (54) Using inequalities (37), (52), and (54) and putting 𝑥=𝑥_𝑝𝑛−1 and 𝑦=𝑥_𝑞𝑛−1, we have that 𝜑𝑑𝑥_𝑝𝑛,𝑥_𝑞𝑛=𝜑𝑑𝑓𝑥_𝑝𝑛−1,𝑓𝑥_𝑞𝑛−1≤𝜓𝜑𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝜑𝑑𝑥_𝑝𝑛−1,𝑓𝑥_𝑝𝑛−1,𝜑𝑑𝑥_𝑞𝑛−1,𝑓𝑥_𝑞𝑛−1−𝜙𝑀𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1+𝐿⋅𝑚𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=𝜓𝜑𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝜑𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛,𝜑𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛−𝜙𝑀𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1+𝐿⋅𝑚𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1, (55) where 𝑀𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=max𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛,𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛,𝑚𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=min𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1,𝑑𝑥_𝑝𝑛−1,𝑥_𝑝𝑛,𝑑𝑥_𝑞𝑛−1,𝑥_𝑞𝑛,𝑑𝑥_𝑝𝑛−1,𝑥_𝑞𝑛,𝑑𝑥_𝑞𝑛−1,𝑥_𝑝𝑛. (56) Letting 𝑛→∞, then we obtain that lim_𝑛→∞𝑀𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1=𝜖,lim_𝑛→∞𝑚𝑥_𝑝𝑛−1,𝑥_𝑞𝑛−1𝜑(𝜖)𝜑(𝜖)(𝜖)(𝜖)(𝜖).=0,≤𝜓,0,0−𝜙≤𝜑−𝜙 (57) This implies that 𝜙(𝜖)=0, and hence 𝜖=0. So we get a contraction. Therefore {𝑥_𝑛} is a Cauchy sequence.
Since 𝑋 is complete, there exists 𝜈∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝜈. (58)
Suppose that (a) holds. Then 𝜈=lim_𝑛→∞𝑥_𝑛+1=lim_𝑛→∞𝑓𝑥_𝑛=𝑓𝜈. (59) Thus, 𝜈 is a fixed point in 𝑋
Suppose that (b) holds; that is, 𝑥_𝑛⊑𝜈 for all 𝑛∈ℕ. Substituting 𝑥=𝑥_𝑛 and 𝑦=𝜈 in (37), we have that 𝜑𝑑𝑥_𝑛+1𝑑,𝑓𝜈=𝜑𝑓𝑥_𝑛𝜑𝑑𝑥,𝑓𝜈≤𝜓_𝑛,𝜈,𝜑𝑑𝑥_𝑛,𝑓𝑥_𝑛𝑑𝑀𝑥,𝜑𝜈,𝑓𝜈−𝜙_𝑛,𝜈+𝐿⋅𝑚𝑥_𝑛,𝜈, (60) where 𝑀𝑥_𝑛,𝜈=max𝑑𝑥_𝑛,𝜈,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑚𝑥,𝑑𝜈,𝑓𝜈_𝑛𝑑𝑥,𝜈=min_𝑛,𝜈,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝜈,𝑓𝜈,𝑑𝑥_𝑛𝑑,𝑓𝜈𝜈,𝑓𝑥_𝑛. (61) Letting 𝑛→∞, then we obtain that 𝑀𝑥_𝑛,𝜈⟶𝑑𝜈,𝑓𝜈,𝑚𝑥_𝑛𝜑𝑑𝜑(0)(0)𝑑𝑑𝑑𝑑,,𝜈⟶0,𝜈,𝑓𝜈≤𝜓,𝜑,𝜑𝜈,𝑓𝜈−𝜙𝜈,𝑓𝜈≤𝜑𝜈,𝑓𝜈−𝜙𝜈,𝑓𝜈 (62) which implies that 𝑑(𝜈,𝑓𝜈)=0; that is, 𝜈=𝑓𝜈. So we complete the proof.

If we let 𝜓𝜑𝑑𝑑𝑑𝜑𝑑𝑑𝑑,𝑥,𝑦,𝜑𝑥,𝑓𝑥,𝜑𝑦,𝑓𝑦=max𝑥,𝑦,𝜑𝑥,𝑓𝑥,𝜑𝑦,𝑓𝑦 (63)it is easy to get the following theorem.

Theorem 8. Let (𝑋,⊑,𝑑) be a partially ordered complete metric space, and let 𝑓∶𝑋→𝑋 be monotone nondecreasing, and 𝜑𝑑𝜑𝑑𝑑𝑑𝑀,𝑓𝑥,𝑓𝑦≤max𝑥,𝑦,𝜑𝑥,𝑓𝑥,𝜑𝑦,𝑓𝑦−𝜙𝑥,𝑦+𝐿⋅𝑚𝑥,𝑦 (64) for all comparable 𝑥,𝑦∈𝑋 and 𝜑∈Θ, 𝜙∈Φ, where 𝐿>0 and 𝑀𝑑,𝑚𝑑,𝑑.𝑥,𝑦=max𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦𝑥,𝑦=min𝑥,𝑦,𝑑𝑥,𝑓𝑥,𝑑𝑦,𝑓𝑦,𝑑𝑥,𝑓𝑦𝑦,𝑓𝑥 (65) Suppose that either(a)  𝑓 is continuous or (b)  if any nondecreasing sequence {𝑥_𝑛} in 𝑋 converges to 𝜈, then one assumes that 𝑥_𝑛⊑𝜈∀𝑛∈ℕ. (66)If there exists 𝑥₀∈𝑋 with 𝑥₀⊑𝑓𝑥₀, then 𝑓 has a fixed point in 𝑋.

Acknowledgment

The authors would like to thank the referee(s) for the many useful comments and suggestions for the improvement of the paper.