Journal of Applied Mathematics

Journal of Applied Mathematics / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 756453 | 10 pages | https://doi.org/10.1155/2012/756453

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

Academic Editor: Yansheng Liu
Received19 Sep 2011
Accepted25 Nov 2011
Published18 Jan 2012

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 ğ’ž-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 ğ’ž-contraction as follows.

Definition 1.2. Let (𝑋,𝑑) be a metric space. A mapping 𝑓∶𝑋→𝑋 is said to be a weakly ğ’ž-contraction if for all 𝑥,𝑦∈𝑋, 1𝑑(𝑓𝑥,𝑓𝑦)≤2(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 ğ’ž-contractions. In [5], Harjani et al. proved some fixed point results for weakly ğ’ž-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𝑡=𝑘⋅max1,𝑡2,𝑡3,𝑡42,𝑡52,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−𝑦for𝑥,𝑦∈𝑋.(1.6) Let 𝜓∶ℝ+5→ℝ+ denote 𝜓𝑡1,𝑡2,𝑡3,𝑡4,𝑡5=34𝑡⋅max1,𝑡2,𝑡3,𝑡42,𝑡52,(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 1𝑓(𝑥)=3𝑥.(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(𝑥,𝑦)=1−𝑦1||+||𝑥2−𝑦2||,for𝑥𝑥=1,𝑥2𝑦,𝑦=1,𝑦2∈𝑋.(1.9) Let 𝜓∶ℝ+5→ℝ+ denote 𝜓𝑡1,𝑡2,𝑡3,𝑡4,𝑡5=34𝑡⋅max1,𝑡2,𝑡3,𝑡42,𝑡52,(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 1𝑓(𝑥)=3𝑥.(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≤⋯≤𝑓𝑛𝑥0≤𝑓𝑛+1𝑥0≤⋯.(2.2) Put 𝑥𝑛+1=𝑓𝑥𝑛=𝑓𝑛+1𝑥0 for 𝑛∈ℕ∪{0}. Then, for each 𝑛∈ℕ, from (∗), and, as the elements 𝑥𝑛 and 𝑥𝑛−1 are comparable, we get 𝑑𝑥𝑛+1,𝑥𝑛𝑑𝑥≤𝜓𝑛,𝑥𝑛−1𝑥,𝑑𝑛,𝑓𝑥𝑛𝑥,𝑑𝑛−1,𝑓𝑥𝑛−1𝑥,𝑑𝑛,𝑓𝑥𝑛−1𝑥,𝑑𝑛−1,𝑓𝑥𝑛𝑑𝑥≤𝜓𝑛,𝑥𝑛−1𝑥,𝑑𝑛,𝑥𝑛+1𝑥,𝑑𝑛−1,𝑥𝑛𝑥,𝑑𝑛,𝑥𝑛𝑥,𝑑𝑛−1,𝑥𝑛+1𝑑𝑥≤𝜓𝑛,𝑥𝑛−1𝑥,𝑑𝑛,𝑥𝑛+1𝑥,𝑑𝑛−1,𝑥𝑛𝑥,0,𝑑𝑛−1,𝑥𝑛𝑥+𝑑𝑛,𝑥𝑛+1,(2.3) and so we can deduce that, for each 𝑛∈ℕ, 𝑑𝑥𝑛+1,𝑥𝑛𝑥≤𝑑𝑛,𝑥𝑛−1.(2.4) Let we denote 𝑐𝑚=𝑑(𝑥𝑚+1,𝑥𝑚). Then, 𝑐𝑚 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≤𝑐𝑛+1𝑐≤𝜓𝑛,𝑐𝑛,𝑐𝑛,0,2𝑐𝑛.(2.5) Passing to the limit, as ğ‘›â†’âˆž, 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𝑥𝑚,𝑥𝑛<𝛾.(∗) We will prove (∗) by contradiction. Suppose that (∗) is false. Then, there exists some 𝛾>0 such that for all 𝑘∈ℕ, there exist 𝑚𝑘 and 𝑛𝑘 with 𝑚𝑘>𝑛𝑘>𝑘 such that 𝑑𝑥𝑚𝑘,𝑥𝑛𝑘𝑥≥𝛾,𝑑𝑚𝑘−1,𝑥𝑛𝑘<𝛾.(2.7) Using the triangular inequality: 𝑥𝛾≤𝑑𝑚𝑘,𝑥𝑛𝑘𝑥≤𝑑𝑚𝑘,𝑥𝑚𝑘−1𝑥+𝑑𝑚𝑘−1,𝑥𝑛𝑘𝑥<𝛾+𝑑𝑚𝑘,𝑥𝑚𝑘−1,(2.8) and letting ğ‘˜â†’âˆž, we have limğ‘˜â†’âˆžğ‘‘î€·ğ‘¥ğ‘šğ‘˜,𝑥𝑛𝑘=𝛾.(2.9)
Since 𝑓 is a 𝜓-contractive mapping, we also have𝑥𝛾≤𝑑𝑚𝑘,𝑥𝑛𝑘=𝑑𝑓𝑥𝑚𝑘−1,𝑓𝑥𝑛𝑘−1𝑑𝑥≤𝜓𝑚𝑘−1,𝑥𝑛𝑘−1𝑥,𝑑𝑚𝑘−1,𝑥𝑚𝑘𝑥,𝑑𝑛𝑘−1,𝑥𝑛𝑘𝑥,𝑑𝑚𝑘−1,𝑥𝑛𝑘𝑥,𝑑𝑛𝑘−1,𝑥𝑚𝑘𝑐≤𝜓𝑚𝑘−1𝑥+𝑑𝑚𝑘,𝑥𝑛𝑘+𝑐𝑛𝑘−1,𝑐𝑚𝑘−1,𝑐𝑛𝑘−1,𝑐𝑚𝑘−1𝑥+𝑑𝑚𝑘,𝑥𝑛𝑘𝑥,𝑑𝑚𝑘,𝑥𝑛𝑘+𝑐𝑛𝑘−1.(2.10) Letting ğ‘˜â†’âˆž. Then, we get 𝛾≤𝜓(𝛾,0,0,𝛾,𝛾)<𝛾,(2.11) a contradiction. It follows from (∗) that the sequence {𝑥𝑛} must be a Cauchy sequence.
Similary, we also conclude that for each 𝑛∈ℕ, 𝑑𝑥𝑛,𝑥𝑛+1𝑑𝑥≤𝜓𝑛−1,𝑥𝑛𝑥,𝑑𝑛−1,𝑓𝑥𝑛−1𝑥,𝑑𝑛,𝑓𝑥𝑛𝑥,𝑑𝑛−1,𝑓𝑥𝑛𝑥,𝑑𝑛,𝑓𝑥𝑛−1𝑑𝑥≤𝜓𝑛−1,𝑥𝑛𝑥,𝑑𝑛−1,𝑥𝑛𝑥,𝑑𝑛,𝑥𝑛+1𝑥,𝑑𝑛−1,𝑥𝑛+1𝑥,𝑑𝑛,𝑥𝑛𝑑𝑥≤𝜓𝑛−1,𝑥𝑛𝑥,𝑑𝑛−1,𝑥𝑛𝑥,𝑑𝑛,𝑥𝑛+1𝑥,𝑑𝑛−1,𝑥𝑛𝑥+𝑑𝑛,𝑥𝑛+1,,0(2.12) and so we have that for each 𝑛∈ℕ, 𝑑𝑥𝑛,𝑥𝑛+1𝑥≤𝑑𝑛−1,𝑥𝑛.(2.13)
Let us denote 𝑏𝑚=𝑑(𝑥𝑚,𝑥𝑚+1). Then, 𝑏𝑚 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≤𝑏𝑛+1𝑏≤𝜓𝑛,𝑏𝑛,𝑏𝑛,2𝑏𝑛.,0(2.14)Passing to the limit, as ğ‘›â†’âˆž, 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 {𝑥𝑛} is a Cauchy sequence.
Since 𝑋 is complete, there exists 𝜇∈𝑋 such that limğ‘›â†’âˆžğ‘¥ğ‘›=𝜇. Moreover, the continuity of 𝑓 implies that 𝜇=limğ‘›â†’âˆžğ‘¥ğ‘›+1=limğ‘›â†’âˆžğ‘“î€·ğ‘¥ğ‘›î€¸=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 {𝑥𝑛} is a nondecreasing sequence in 𝑋, such that 𝑥𝑛⟶𝑥,then𝑥𝑛≤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 {𝑥𝑛} is a nondecreasing sequence in 𝑋 and 𝑥𝑛→𝜇, then the condition (∗∗) gives us that 𝑥𝑛≤𝜇 for every 𝑛∈ℕ. Since 𝑓∶𝑋→𝑋 is a nondecreasing 𝜓-contractive mapping, we have 𝑑𝑥𝑛+1,𝑓𝜇=𝑑𝑓𝑥𝑛𝑑𝑥,𝑓𝜇≤𝜓𝑛𝑥,𝜇,𝑑𝑛,𝑓𝑥𝑛𝑥,𝑑(𝜇,𝑓𝜇),𝑑𝑛,𝑓𝜇,𝑑𝜇,𝑓𝑥𝑛𝑑𝑥≤𝜓𝑛𝑥,𝜇,𝑑𝑛,𝑥𝑛+1𝑥,𝑑(𝜇,𝑓𝜇),𝑑𝑛,𝑓𝜇,𝑑𝜇,𝑥𝑛+1.(2.17) Letting ğ‘›â†’âˆž and using the continuity of 𝜓, we have 𝑑(𝜇,𝑓𝜇)≤𝜓(0,0,𝑑(𝜇,𝑓𝜇),𝑑(𝜇,𝑓𝜇),0)<𝑑(𝜇,𝑓𝜇),(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 𝑓𝑛𝜇=𝜇 is comparable to 𝑓𝑛𝜈=𝜈 for all 𝑛∈ℕ, and 𝑑(𝜇,𝜈)=𝑑(𝑓𝑛𝜇,𝑓𝑛𝑑𝑓𝜈)≤𝜓𝑛−1𝜇,𝑓𝑛−1𝜈𝑓,𝑑𝑛−1𝜇,𝑓𝑛𝜇𝑓,𝑑𝑛−1𝜈,𝑓𝑛𝜈𝑓,𝑑𝑛−1𝜇,𝑓𝑛𝜈𝑓,𝑑𝑛−1𝜈,𝑓𝑛𝜇≤𝜓(𝑑(𝜇,𝜈),𝑑(𝜇,𝜇),𝑑(𝜈,𝜈),𝑑(𝜇,𝜈),𝑑(𝜈,𝜇))=𝜓(𝑑(𝜇,𝜈),0,0,𝑑(𝜇,𝜈),𝑑(𝜈,𝜇))<𝑑(𝜇,𝜈),(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 𝑓𝑛𝑥 is comparable to 𝑓𝑛𝜇 and 𝑓𝑛𝜈 for all 𝑛∈ℕ. We also distinguish two cases.Subcase 2.1. If there exists 𝑛0∈ℕ with 𝑓𝑛0𝑥=𝜇, then we have 𝑓𝑑(𝜇,𝜈)=𝑑(𝑓𝜇,𝑓𝜈)=𝑑𝑛0+1𝑥,𝑓𝑛0+1𝜈𝑑≤𝜓(𝑓𝑛0𝑥,𝑓𝑛0𝜈𝑓),𝑑𝑛0𝑥,𝑓𝑛0+1𝑥𝑓,𝑑𝑛0𝜈,𝑓𝑛0+1𝜈𝑓,𝑑𝑛0𝑥,𝑓𝑛0+1𝜈𝑓,𝑑𝑛0𝜈,𝑓𝑛0+1𝑥=𝜓(𝑑(𝜇,𝜈),𝑑(𝜇,𝑓𝜇),𝑑(𝜈,𝜈),𝑑(𝜇,𝜈),𝑑(𝜈,𝑓𝜇))=𝜓(𝑑(𝜇,𝜈),0,0,𝑑(𝜇,𝜈),𝑑(𝜈,𝜇))<𝑑(𝜇,𝜈),(2.21) and this is a contradiction unless 𝑑(𝜇,𝜈)=0, that is, 𝜇=𝜈.Subcase 2.2. For all 𝑛∈ℕ with 𝑓𝑛𝑥≠𝜇, since 𝑓 is a nondecreasing 𝜓-contractive mapping, we have 𝑑(𝜇,𝑓𝑛𝑥)=𝑑(𝑓𝑛𝜇,𝑓𝑛𝑑𝑓𝑥)≤𝜓𝑛−1𝜇,𝑓𝑛−1𝑥𝑓,𝑑𝑛−1𝜇,𝑓𝑛𝜇𝑓,𝑑𝑛−1𝑥,𝑓𝑛𝑥𝑓,𝑑𝑛−1𝜇,𝑓𝑛𝑥𝑓,𝑑𝑛−1𝑥,𝑓𝑛𝜇𝑑≤𝜓𝜇,𝑓𝑛−1𝑥𝑓,𝑑(𝜇,𝜇),𝑑𝑛−1𝑥,𝑓𝑛𝑥,𝑑(𝜇,𝑓𝑛𝑓𝑥),𝑑𝑛−1𝑑𝑥,𝜇≤𝜓𝜇,𝑓𝑛−1𝑥𝑓,0,𝑑𝑛−1𝑥,𝜇+𝑑(𝜇,𝑓𝑛𝑥),𝑑(𝜇,𝑓𝑛𝑓𝑥),𝑑𝑛−1.𝑥,𝜇(2.22) Using the above inequality, we claim that for each 𝑛∈ℕ, 𝑑(𝜇,𝑓𝑛𝑥)<𝑑𝜇,𝑓𝑛−1𝑥.(2.23) If not, we assume that 𝑑(𝜇,𝑓𝑛−1𝑥)≤𝑑(𝜇,𝑓𝑛𝑥), then by the definition of 𝜓 and 𝜓(𝑡,0,2𝑡,𝑡,𝑡)<𝑡, we have 𝑑(𝜇,𝑓𝑛𝑥𝑑)≤𝜓𝜇,𝑓𝑛−1𝑥𝑓,0,𝑑𝑛−1𝑥,𝜇+𝑑(𝜇,𝑓𝑛𝑥),𝑑(𝜇,𝑓𝑛𝑥𝑓),𝑑𝑛−1𝑥,𝜇≤𝜓(𝑑(𝜇,𝑓𝑛𝑥),0,2𝑑(𝑓𝑛𝑥,𝜇),𝑑(𝜇,𝑓𝑛𝑥),𝑑(𝑓𝑛𝑥,𝜇))<𝑑(𝜇,𝑓𝑛𝑥),(2.24) which implies a contradiction. Therefore, our claim is proved.
This proves that the nonnegative decreasing sequence {𝑑(𝜇,𝑓𝑛𝑥)} is convergent. Put limğ‘›â†’âˆžğ‘‘(𝜇,𝑓𝑛𝑥)=𝜂, 𝜂≥0. We now claim that 𝜂=0. If 𝜂>0, then making ğ‘›â†’âˆž, we get𝜂=limğ‘›â†’âˆžğ‘‘(𝜇,𝑓𝑛𝑥)≤𝜓(𝜂,0,2𝜂,𝜂,𝜂)<𝜂,(2.25)this is a contradiction. So 𝜂=0, that is, limğ‘›â†’âˆžğ‘‘(𝜇,𝑓𝑛𝑥)=0.
Analogously, it can be proved that limğ‘›â†’âˆžğ‘‘(𝜈,𝑓𝑛𝑥)=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 {𝑓𝑛𝑥0} are comparable, we have 𝑑𝑓𝑛+1𝑥0,𝑓𝑛𝑥0𝑑𝑓≤𝜓𝑛𝑥0,𝑓𝑛−1𝑥0𝑓,𝑑𝑛𝑥0,𝑓𝑛+1𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛+1𝑥0𝑓,𝑑𝑛𝑥0,𝑓𝑛𝑥0𝑑𝑓≤𝜓𝑛𝑥0,𝑓𝑛−1𝑥0𝑓,𝑑𝑛𝑥0,𝑓𝑛+1𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛+1𝑥0𝑑𝑓,0≤𝜓𝑛𝑥0,𝑓𝑛−1𝑥0𝑓,𝑑𝑛𝑥0,𝑓𝑛+1𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛𝑥0𝑓,𝑑𝑛−1𝑥0,𝑓𝑛𝑥0𝑓+𝑑𝑛𝑥0,𝑓𝑛+1𝑥0,(2.27) and so we conclude that for each 𝑛∈ℕ, 𝑑𝑓𝑛+1𝑥0,𝑓𝑛𝑥0𝑓<𝑑𝑛𝑥0,𝑓𝑛−1𝑥0.(2.28)
Thus, {𝑑(𝑓𝑛+1𝑥0,𝑓𝑛𝑥0)} 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 ğ‘›â†’âˆž, we have 𝜂=limğ‘›â†’âˆžğ‘‘î€·ğ‘“ğ‘›+1𝑥0,𝑓𝑛𝑥0≤𝜓(𝜂,𝜂,𝜂,2𝜂,0)<𝜂,(2.29) which is a contradiction. So 𝜂=0, that is, limğ‘›â†’âˆžğ‘‘(𝑓𝑛+1𝑥0,𝑓𝑛𝑥0)=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.

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Copyright © 2012 Chi-Ming Chen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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