Fixed Point Theorems for 𝜓-Contractive Mappings in Ordered Metric Spaces

Abstract

We obtain some new fixed point theorems for 𝜓-contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., Harjani et al., 2011 and 2010).

1. Introduction and Preliminaries

Throughout this paper, by ℝ⁺, we denote the set of all real nonnegative 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,𝑑(𝑓𝑥,𝑓𝑦)≤𝛼⋅𝑑(𝑥,𝑦).(1.1)

The well-known Banach 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. It is well known that the Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, a mapping 𝑓∶𝑋→𝑋 is called a quasicontraction if there exists 𝑘<1 such that𝑑(𝑓𝑥,𝑓𝑦)≤𝑘⋅max{𝑑(𝑥,𝑦),𝑑(𝑥,𝑓𝑥),𝑑(𝑦,𝑓𝑦),𝑑(𝑥,𝑓𝑦),𝑑(𝑦,𝑓𝑥)},(1.2)

for any 𝑥,𝑦∈𝑋. In 1974, Ćirić [2] introduced these maps and proved an existence and uniqueness fixed point theorem.

In 1972, Chatterjea [3] introduced the following definition.

Definition 1.1. Let (𝑋,𝑑) be a metric space. A mapping 𝑓∶𝑋→𝑋 is said to be a 𝒞-contraction if there exists 𝛼∈(0,1/2) such that for all 𝑥,𝑦∈𝑋, the following inequality holds: 𝑑(𝑓𝑥,𝑓𝑦)≤𝛼⋅(𝑑(𝑥,𝑓𝑦)+𝑑(𝑦,𝑓𝑥)).(1.3)

Choudhury [4] introduced a generalization of 𝒞-contraction as follows.

Definition 1.2. Let (𝑋,𝑑) be a metric space. A mapping 𝑓∶𝑋→𝑋 is said to be a weakly 𝒞-contraction if for all 𝑥,𝑦∈𝑋, 𝑑(𝑓𝑥,𝑓𝑦)≤12(𝑑(𝑥,𝑓𝑦)+𝑑(𝑦,𝑓𝑥)−𝜙(𝑑(𝑥,𝑓𝑦),𝑑(𝑦,𝑓𝑥))),(1.4) where 𝜙∶ℝ⁺²→ℝ⁺ is a continuous function such that 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0.

In [3, 4], the authors proved some fixed point results for the 𝒞-contractions. In [5], Harjani et al. proved some fixed point results for weakly 𝒞-contractive mappings in a complete metric space endowed with a partial order.

In the following, we assume that the function 𝜓∶ℝ⁺⁵→ℝ⁺ satisfies the following conditions:(C1)𝜓 is a strictly increasing and continuous function in each coordinate, and(C2)for all 𝑡∈ℝ⁺⧵{0}, 𝜓(𝑡,𝑡,𝑡,0,2𝑡)<𝑡, 𝜓(𝑡,𝑡,𝑡,2𝑡,0)<𝑡, 𝜓(0,0,𝑡,𝑡,0)<𝑡, and 𝜓(𝑡,0,0,𝑡,𝑡)<𝑡.

Example 1.3. Let 𝜓∶ℝ⁺⁵→ℝ⁺ denote 𝜓𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅=𝑘⋅max𝑡₁,𝑡₂,𝑡₃,𝑡⁴2,𝑡2⁵,for𝑘∈(0,1).(1.5) Then, 𝜓 satisfies the above conditions (C1) and (C2).

Now, we define the following notion of a 𝜓-contractive mapping in metric spaces.

Definition 1.4. Let (𝑋,≤) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a metric space. The mapping 𝑓∶𝑋→𝑋 is said to be a 𝜓-contractive mapping, if 𝑑(𝑓𝑥,𝑓𝑦)≤𝜓(𝑑(𝑥,𝑦),𝑑(𝑥,𝑓𝑥),𝑑(𝑦,𝑓𝑦),𝑑(𝑥,𝑓𝑦),𝑑(𝑦,𝑓𝑥)),(∗) for 𝑥≥𝑦.

Using Example 1.3, it is easy to get the following examples of 𝜓-contractive mappings.

Example 1.5. Let 𝑋=ℝ⁺ endowed with usual ordering and with the metric 𝑑∶𝑋×𝑋→ℝ⁺ given by 𝑑(𝑥,𝑦)=||𝑥−𝑦||,for𝑥,𝑦∈𝑋.(1.6) Let 𝜓∶ℝ⁺⁵→ℝ⁺ denote 𝜓𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅=34⋅max𝑡¹,𝑡₂,𝑡₃,𝑡⁴2,𝑡2⁵,(1.7) where 𝑡₁=𝑑(𝑥,𝑦), 𝑡₂=𝑑(𝑥,𝑓𝑥), 𝑡₃=𝑑(𝑦,𝑓𝑦), 𝑡₄=𝑑(𝑥,𝑓𝑦),and 𝑡₅=𝑑(𝑦,𝑓𝑥), for all 𝑥,𝑦∈𝑋. Let 𝑓∶𝑋→𝑋 denote 𝑓(𝑥)=13𝑥.(1.8) Then, 𝑓 is a 𝜓-contractive mapping.

Example 1.6. Let 𝑋=ℝ⁺×ℝ⁺ endowed with the coordinate ordering (i.e., (𝑥,𝑦)≤(𝑧,𝑤)⇔𝑥≤𝑧 and 𝑦≤𝑤) and with the metric 𝑑∶𝑋×𝑋→ℝ⁺ given by 𝑑(𝑥,𝑦)=||𝑥₁−𝑦₁||+||𝑥₂−𝑦₂||,for𝑥=𝑥₁,𝑥₂,𝑦=𝑦₁,𝑦₂∈𝑋.(1.9) Let 𝜓∶ℝ⁺⁵→ℝ⁺ denote 𝜓𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅=34⋅max𝑡¹,𝑡₂,𝑡₃,𝑡⁴2,𝑡2⁵,(1.10) where 𝑡₁=𝑑(𝑥,𝑦), 𝑡₂=𝑑(𝑥,𝑓𝑥), 𝑡₃=𝑑(𝑦,𝑓𝑦), 𝑡₄=𝑑(𝑥,𝑓𝑦), and 𝑡₅=𝑑(𝑦,𝑓𝑥), for all 𝑥,𝑦∈𝑋. Let 𝑓∶𝑋→𝑋 denote 𝑓(𝑥)=13𝑥.(1.11) Then, 𝑓 is a 𝜓-contractive mapping.

In this paper, we obtain some new fixed point theorems for 𝜓-contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., [5, 6]).

2. Main Results

We start with the following definition.

Definition 2.1. Let (𝑋,≤) be a partially ordered set and 𝑓∶𝑋→𝑋. Then one says that 𝑓 is monotone nondecreasing if, for 𝑥,𝑦∈𝑋, 𝑥≤𝑦⟹𝑓𝑥≤𝑓𝑦.(2.1)

We now state the main fixed point theorem for 𝜓-contractive mappings in ordered metric spaces when the operator is nondecreasing, as follows.

Theorem 2.2. 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 continuous and nondecreasing 𝜓-contractive mapping. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑓𝑥₀, then 𝑓 has a fixed point in 𝑋.

Proof. If 𝑓(𝑥₀)=𝑥₀, then the proof is finished. Suppose that 𝑥₀<𝑓(𝑥₀). Since 𝑓 is nondecreasing mapping, by induction, we obtain that 𝑥₀<𝑓𝑥₀≤𝑓²𝑥₀≤𝑓³𝑥₀≤⋯≤𝑓^𝑛𝑥₀≤𝑓^𝑛+1𝑥₀≤⋯.(2.2) Put 𝑥_𝑛+1=𝑓𝑥_𝑛=𝑓^𝑛+1𝑥₀ for 𝑛∈ℕ∪{0}. Then, for each 𝑛∈ℕ, from (∗), and, as the elements 𝑥_𝑛 and 𝑥_𝑛−1 are comparable, we get 𝑑𝑥_𝑛+1,𝑥_𝑛≤≤𝜓𝜓𝑑𝑑𝑥𝑥_𝑛𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝑑𝑥_𝑛,𝑓𝑥_𝑛−1,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛≤𝜓𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛+1,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛,0,𝑑𝑥_𝑛−1,𝑥_𝑛+𝑑𝑥_𝑛,𝑥_𝑛+1,(2.3) and so we can deduce that, for each 𝑛∈ℕ, 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝑑𝑥_𝑛,𝑥_𝑛−1.(2.4) Let we denote 𝑐_𝑚=𝑑(𝑥_𝑚+1,𝑥_𝑚). Then, 𝑐_𝑚 is a nonincreasing sequence and bounded below. Thus, it must converge to some 𝑐≥0. If 𝑐>0, then by the above inequalities, we have 𝑐≤𝑐_𝑛+1≤𝜓𝑐_𝑛,𝑐_𝑛,𝑐_𝑛,0,2𝑐_𝑛.(2.5) Passing to the limit, as 𝑛→∞, we have 𝑐≤𝑐≤𝜓(𝑐,𝑐,𝑐,0,2𝑐)<𝑐,(2.6) which is a contradiction. So 𝑐=0.
We next claim that that the following result holds.
For each 𝛾>0, there is 𝑛₀(𝛾)∈ℕ such that for all 𝑚>𝑛>𝑛₀(𝛾),𝑑𝑥_𝑚,𝑥_𝑛<𝛾.(^∗) We will prove (∗) by contradiction. Suppose that (∗) is false. Then, there exists some 𝛾>0 such that for all 𝑘∈ℕ, there exist 𝑚_𝑘 and 𝑛_𝑘 with 𝑚_𝑘>𝑛_𝑘>𝑘 such that 𝑑𝑥_𝑚𝑘,𝑥_𝑛𝑘≥𝛾,𝑑𝑥_𝑚𝑘−1,𝑥_𝑛𝑘<𝛾.(2.7) Using the triangular inequality: 𝛾≤≤𝑑𝑑𝑥𝑥_{𝑚𝑘𝑚𝑘},𝑥_𝑛𝑘<𝛾+𝑑,𝑥𝑥_{𝑚𝑘−1𝑚𝑘}+𝑑𝑥_𝑚𝑘−1,𝑥_𝑛𝑘,𝑥_𝑚𝑘−1,(2.8) and letting 𝑘→∞, we have lim_𝑘→∞𝑑𝑥_𝑚𝑘,𝑥_𝑛𝑘=𝛾.(2.9)
Since 𝑓 is a 𝜓-contractive mapping, we also have𝛾≤≤𝜓𝑑𝑥𝑑_𝑚𝑘𝑥,𝑥_𝑛𝑘_𝑚𝑘−1=𝑑𝑓𝑥_𝑚𝑘−1,𝑓𝑥_𝑛𝑘−1≤𝜓𝑐_𝑚𝑘−1,𝑥_𝑛𝑘−1,𝑑𝑥_𝑚𝑘−1,𝑥_𝑚𝑘,𝑑𝑥_𝑛𝑘−1,𝑥_𝑛𝑘,𝑑𝑥_𝑚𝑘−1,𝑥_𝑛𝑘,𝑑𝑥_𝑛𝑘−1,𝑥_𝑚𝑘+𝑑𝑥_𝑚𝑘,𝑥_𝑛𝑘+𝑐_𝑛𝑘−1,𝑐_𝑚𝑘−1,𝑐_𝑛𝑘−1,𝑐_𝑚𝑘−1+𝑑𝑥_𝑚𝑘,𝑥_𝑛𝑘,𝑑𝑥_𝑚𝑘,𝑥_𝑛𝑘+𝑐_𝑛𝑘−1.(2.10) Letting 𝑘→∞. Then, we get 𝛾≤𝜓(𝛾,0,0,𝛾,𝛾)<𝛾,(2.11) a contradiction. It follows from (∗) that the sequence {𝑥_𝑛} must be a Cauchy sequence.
Similary, we also conclude that for each 𝑛∈ℕ, 𝑑𝑥_𝑛,𝑥_𝑛+1≤≤𝜓𝜓𝑑𝑑𝑥𝑥_{𝑛−1𝑛−1},𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑𝑥_𝑛−1,𝑓𝑥_𝑛,𝑑𝑥_𝑛,𝑓𝑥_𝑛−1≤𝜓𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑥_𝑛,𝑥_𝑛,𝑑𝑥_𝑛−1,𝑥_𝑛,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛+𝑑𝑥_𝑛,𝑥_𝑛+1,0,(2.12) and so we have that for each 𝑛∈ℕ, 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛−1,𝑥_𝑛.(2.13)
Let us denote 𝑏_𝑚=𝑑(𝑥_𝑚,𝑥_𝑚+1). Then, 𝑏_𝑚 is a nonincreasing sequence and bounded below. Thus, it must converge to some 𝑏≥0. If 𝑏>0, then by the above inequalities, we have 𝑏≤𝑏_𝑛+1≤𝜓𝑏_𝑛,𝑏_𝑛,𝑏_𝑛,2𝑏_𝑛,0.(2.14)Passing to the limit, as 𝑛→∞, we have 𝑏≤𝑏≤𝜓(𝑏,𝑏,𝑏,2𝑏,0)<𝑏,(2.15)which is a contradiction. So 𝑏=0. By the above argument, we also conclude that {𝑥_𝑛} is a Cauchy sequence.
Since 𝑋 is complete, there exists 𝜇∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝜇. Moreover, the continuity of 𝑓 implies that 𝜇=lim_𝑛→∞𝑥_𝑛+1=lim_𝑛→∞𝑓𝑥_𝑛=𝑓(𝜇).(2.16)
So we complete the proof.

In what follows, we prove that Theorem 2.2 is still valid for 𝑓 not necessarily continuous, assuming the following hypothesis in 𝑋(which appears in Theorem 1 of [7]).If {𝑥_𝑛} is a nondecreasing sequence in 𝑋, such that 𝑥_𝑛⟶𝑥,then𝑥_𝑛≤𝑥∀𝑛∈ℕ.(^∗∗)

Theorem 2.3. 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. Assume that 𝑋 satisfies (∗∗), and let 𝑓∶𝑋→𝑋 be a nondecreasing 𝜓-contractive mapping. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑓(𝑥₀), then 𝑓 has a fixed point in 𝑋.

Proof. Following the proof of Theorem 2.2, we only have to check that 𝑓(𝜇)=𝜇. As {𝑥_𝑛} is a nondecreasing sequence in 𝑋 and 𝑥_𝑛→𝜇, then the condition (∗∗) gives us that 𝑥_𝑛≤𝜇 for every 𝑛∈ℕ. Since 𝑓∶𝑋→𝑋 is a nondecreasing 𝜓-contractive mapping, we have 𝑑𝑥_𝑛+1,𝑓𝜇≤=𝜓𝑑𝑓𝑑𝑥_𝑛𝑥,𝑓_𝑛𝜇≤𝜓𝑑𝑥_𝑛,𝜇,𝑑𝑥_𝑛,𝑓𝑥_𝑛,𝑑(𝜇,𝑓𝜇),𝑑𝑥_𝑛,𝑓𝜇,𝑑𝜇,𝑓𝑥_𝑛,𝜇,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑(𝜇,𝑓𝜇),𝑑𝑥_𝑛,𝑓𝜇,𝑑𝜇,𝑥_𝑛+1.(2.17) Letting 𝑛→∞ and using the continuity of 𝜓, we have 𝑑(𝜇,𝑓𝜇)≤<𝑑𝜓((𝜇0,,𝑓0,𝜇𝑑),(𝜇,𝑓𝜇),𝑑(𝜇,𝑓𝜇),0)(2.18) and this is a contraction unless 𝑑(𝜇,𝑓𝜇)=0, or equivalently, 𝜇=𝑓𝜇.

In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition is the following and it appears in [8]:for𝑥,𝑦∈𝑋,thereexistsalowerboundoranupperbound.(2.19) In [7], it is proved that the above-mentioned condition is equivalent to the following: for𝑥,𝑦∈𝑋,thereexists𝑧∈𝑋whichiscomparableto𝑥and𝑦.(^∗∗∗)

Theorem 2.4. Adding condition (∗∗∗) to the hypothesis of Theorem 2.2 (or Theorem 2.3) and the condition for all 𝑡∈ℝ⁺, 𝜓(𝑡,0,2𝑡,𝑡,𝑡)<𝑡(or, 𝜓(𝑡,2𝑡,0,0,𝑡)<𝑡) to the function 𝜓, one obtains the uniqueness of the fixed point of 𝑓.

Proof. Suppose that there exist 𝜇,𝜈∈𝑋 which are fixed points of 𝑓. We distinguish two cases.
Case 1. If 𝜇 and 𝜈 are comparable and 𝜇≠𝜈, then 𝑓^𝑛𝜇=𝜇 is comparable to 𝑓^𝑛𝜈=𝜈 for all 𝑛∈ℕ, and 𝑑(=𝜇,𝑑𝜈)(𝑓_𝑛≤𝜓𝑑𝜇𝑓,𝑓^𝑛_𝑛−1𝜈)≤=<𝜓𝜓𝑑(((𝜇𝑑𝑑,((𝜈𝜇𝜇),,,𝜈𝜈)),,𝜇𝑑0,,𝑓(0𝜇,,𝑑𝜇^𝑛−1(𝜈)𝜇,,𝑑,𝜈𝑑()𝜈,,𝑑𝑓𝜈()𝜈,^𝑛−1,𝑑𝜇𝜇(,)𝜇)𝑓,^𝑛𝜈𝜇),𝑑,𝑑(𝜈,𝑓𝜇))^𝑛−1𝜈,𝑓^𝑛𝜈,𝑑𝑓^𝑛−1𝜇,𝑓^𝑛𝜈,𝑑𝑓^𝑛−1𝜈,𝑓^𝑛𝜇(2.20) and this is a contradiction unless 𝑑(𝜇,𝜈)=0, that is, 𝜇=𝜈.Case 2. If 𝜇 and 𝜈 are not comparable, then there exists 𝑥∈𝑋 comparable to 𝜇 and 𝜈. Monotonicity of 𝑓 implies that 𝑓^𝑛𝑥 is comparable to 𝑓^𝑛𝜇 and 𝑓^𝑛𝜈 for all 𝑛∈ℕ. We also distinguish two cases.Subcase 2.1. If there exists 𝑛₀∈ℕ with 𝑓^𝑛0𝑥=𝜇, then we have 𝑑(==𝜇,𝑑𝑑𝜈)(𝑓𝑓𝜇,𝑓_𝑛0+1𝜈)≤𝜓𝑑(𝑓𝑥_𝑛0,𝑓^𝑛0+1𝜈==<𝜓𝜓𝑑(((𝜇𝑑𝑑,((𝜈𝜇𝜇),,,𝜈𝜈𝑥)),,,𝑑0𝑓,(0^𝑛0𝜇𝜈,,𝑑)𝑓,(𝜇𝑑𝜇),,𝜈𝑓𝑑),(^𝑛0𝜈𝑥𝑑,,(𝜈𝑓𝜈),,𝜇𝑑)^𝑛0+1)(𝑥𝜇,,𝜈𝑑),𝑑𝑓(𝜈^𝑛0𝜈,𝑓,𝑓𝜇))^𝑛0+1𝜈,𝑑𝑓^𝑛0𝑥,𝑓^𝑛0+1𝜈,𝑑𝑓^𝑛0𝜈,𝑓^𝑛0+1𝑥(2.21) and this is a contradiction unless 𝑑(𝜇,𝜈)=0, that is, 𝜇=𝜈.Subcase 2.2. For all 𝑛∈ℕ with 𝑓^𝑛𝑥≠𝜇, since 𝑓 is a nondecreasing 𝜓-contractive mapping, we have 𝑑(=𝜇,𝑑𝑓^𝑛(𝑓𝑥)_𝑛≤𝜓𝑑𝜇𝑓,𝑓^𝑛_𝑛−1𝑥)≤𝜓𝑑𝜇,𝑓𝜇,_𝑛−1𝑓^𝑛−1𝑥,𝑑𝑓^𝑛−1𝜇,𝑓^𝑛𝜇,𝑑𝑓^𝑛−1𝑥,𝑓^𝑛𝑥,𝑑𝑓^𝑛−1𝜇,𝑓^𝑛𝑥,𝑑𝑓^𝑛−1𝑥,𝑓^𝑛𝜇≤𝜓𝑑𝜇,𝑓_𝑛−1𝑥,𝑑(𝜇,𝜇),𝑑𝑓^𝑛−1𝑥,𝑓^𝑛𝑥,𝑑(𝜇,𝑓^𝑛𝑥),𝑑𝑓^𝑛−1𝑥,𝜇𝑥,0,𝑑𝑓^𝑛−1𝑥,𝜇+𝑑(𝜇,𝑓^𝑛𝑥),𝑑(𝜇,𝑓^𝑛𝑥),𝑑𝑓^𝑛−1𝑥,𝜇.(2.22) Using the above inequality, we claim that for each 𝑛∈ℕ, 𝑑(𝜇,𝑓^𝑛𝑥)<𝑑𝜇,𝑓^𝑛−1𝑥.(2.23) If not, we assume that 𝑑(𝜇,𝑓^𝑛−1𝑥)≤𝑑(𝜇,𝑓^𝑛𝑥), then by the definition of 𝜓 and 𝜓(𝑡,0,2𝑡,𝑡,𝑡)<𝑡, we have 𝑑(𝜇,𝑓^𝑛𝑥)≤≤𝜓𝜓(𝑑𝑑(𝜇𝜇,,𝑓𝑓^𝑛−1_𝑛𝑥,0,𝑑𝑓^𝑛−1𝑥,𝜇+𝑑(𝜇,𝑓^𝑛𝑥),𝑑(𝜇,𝑓^𝑛𝑥),𝑑𝑓^𝑛−1𝑥,𝜇<𝑑(𝜇,𝑓_𝑛𝑥),0,2𝑑(𝑓^𝑛𝑥,𝜇),𝑑(𝜇,𝑓^𝑛𝑥),𝑑(𝑓^𝑛𝑥,𝜇))𝑥),(2.24) which implies a contradiction. Therefore, our claim is proved.
This proves that the nonnegative decreasing sequence {𝑑(𝜇,𝑓^𝑛𝑥)} is convergent. Put lim_𝑛→∞𝑑(𝜇,𝑓^𝑛𝑥)=𝜂, 𝜂≥0. We now claim that 𝜂=0. If 𝜂>0, then making 𝑛→∞, we get𝜂=lim_𝑛→∞𝑑(𝜇,𝑓^𝑛𝑥)≤𝜓(𝜂,0,2𝜂,𝜂,𝜂)<𝜂,(2.25)this is a contradiction. So 𝜂=0, that is, lim_𝑛→∞𝑑(𝜇,𝑓^𝑛𝑥)=0.
Analogously, it can be proved that lim_𝑛→∞𝑑(𝜈,𝑓^𝑛𝑥)=0.
Finally, the uniqueness of the limit gives us 𝜇=𝜈.
This finishes the proof.

In the following, we present a fixed point theorem for a 𝜓-contractive mapping when the operator 𝑓 is nonincreasing. We start with the following definition.

Definition 2.5. Let (𝑋,≤) be a partially ordered set and 𝑓∶𝑋→𝑋. Then one says that 𝑓 is monotone nonincreasing if, for 𝑥,𝑦∈𝑋, 𝑥≤𝑦⟹𝑓𝑥≥𝑓𝑦.(2.26)
Using a similar argument to that in the proof of Theorem 3.1 of [5], we get the following point results.

Theorem 2.6. Let (𝑋,≤) be a partially ordered set satisfying condition (∗∗∗) and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space, and let 𝑓 be a nonincreasing 𝜓-contractive mapping. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑓𝑥₀ or 𝑥₀≥𝑓𝑥₀, then inf{𝑑(𝑥,𝑓𝑥)∶𝑥∈𝑋}=0. Moreover, if in addition, 𝑋 is compact and 𝑓 is continuous, then 𝑓 has a unique fixed point in 𝑋.

Proof. If 𝑓𝑥₀=𝑥₀, then it is obvious that inf{𝑑(𝑥,𝑓𝑥)∶𝑥∈𝑋}=0. Suppose that 𝑥₀<𝑓𝑥₀ (the same argument serves for 𝑥₀<𝑓𝑥₀). Since 𝑓 is nonincreasing the consecutive terms of the sequence {𝑓^𝑛𝑥₀} are comparable, we have 𝑑≤𝑓𝜓^𝑛+1𝑥𝑑₀,𝑓𝑓^𝑛_𝑛𝑥₀≤𝜓𝑑𝑓_𝑛𝑥₀,𝑓^𝑛−1𝑥₀,𝑑𝑓^𝑛𝑥₀,𝑓^𝑛+1𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛+1𝑥₀,𝑑𝑓^𝑛𝑥₀,𝑓^𝑛𝑥₀≤𝜓𝑑𝑓_𝑛𝑥₀,𝑓^𝑛−1𝑥₀,𝑑𝑓^𝑛𝑥₀,𝑓^𝑛+1𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛+1𝑥₀,0𝑥₀,𝑓^𝑛−1𝑥₀,𝑑𝑓^𝑛𝑥₀,𝑓^𝑛+1𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛𝑥₀,𝑑𝑓^𝑛−1𝑥₀,𝑓^𝑛𝑥₀+𝑑𝑓^𝑛𝑥₀,𝑓^𝑛+1𝑥₀,(2.27) and so we conclude that for each 𝑛∈ℕ, 𝑑𝑓^𝑛+1𝑥₀,𝑓^𝑛𝑥₀<𝑑𝑓^𝑛𝑥₀,𝑓^𝑛−1𝑥₀.(2.28)
Thus, {𝑑(𝑓^𝑛+1𝑥₀,𝑓^𝑛𝑥₀)} is a decreasing sequence and bounded below, and it must converge to 𝜂≥0. We claim that 𝜂=0. If 𝜂>0, then by the above inequalities and the continuity of 𝜓, letting 𝑛→∞, we have 𝜂=≤<𝜓𝜂li,m(_𝑛→∞𝜂,𝑑𝜂,𝜂𝑓,2𝜂^𝑛+1𝑥,0₀),𝑓^𝑛𝑥₀(2.29) which is a contradiction. So 𝜂=0, that is, lim_𝑛→∞𝑑(𝑓^𝑛+1𝑥₀,𝑓^𝑛𝑥₀)=0. Consequently, inf{𝑑(𝑥,𝑓𝑥)∶𝑥∈𝑋}=0.
Further, since 𝑓 is continuous and 𝑋 is compact, we can find 𝜇∈𝑋 such that 𝑑(𝜇,𝑓𝜇)=inf{𝑑(𝑥,𝑓𝑥)∶𝑥∈𝑋}=0,(2.30)
and, therefore, 𝜇 is a fixed point of 𝑓.
The uniqueness of the fixed point is proved as in Theorem 2.4.

Acknowledgment

The authors would like to thank the referees whose helpful comments and suggestions led to much improvement in this paper.