Suzuki-Edelstein Type Contractions via Auxiliary Functions

Abstract

We prove the existence and uniqueness of a fixed point of certain type mapping, extension of Suzuki-Edelstein mapping, in a partially ordered complete metric space. Our results extend, improve, and generalize the existence results on the topic in the literature. We state some examples to illustrate our results.

1. Introduction

Fixed point theory lies in the center of nonlinear functional analysis because it has a broad range of applications in fields such as economics, computer science, and many others (see, e.g., [1–7]). In particular, fixed point theory is quite useful in finding the solutions of inverse problems and structural optimizations in science and engineering [8–10]. Banach contraction mapping principle [11] is considered to be the fundamental result in this theory. It states that each contraction defined on a complete metric space has a unique fixed point. The strength of Banach's result in applications comes from two remarkable implications of this principle. The first one is that it guarantees the existence and uniqueness of a fixed point of a contraction. The second and most effective one is that it provides a technique to evaluate the fixed point. The importance of these two properties of Banach contraction mapping principle evidently has attracted many prominent mathematicians (see, e.g., [12–16]) interested in the fixed point theory and its applications as the use of this principle has widened considerably since its first appearance [17–21]. In particular, in this paper, we will focus on one of the most remarkable generalizations of Banach contraction mapping principle developed by Edelstein [22]. This theorem can be stated as follows.

Theorem 1. Let (𝑋,𝑑) be a compact metric space and let 𝑇 be a mapping on 𝑋. Assume 𝑑(𝑇𝑥,𝑇𝑦)<𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋 with 𝑥≠𝑦. Then 𝑇 has a unique fixed point.

Inspired by Edelstein's theorem, Suzuki [23, 24] further improved Banach's principle through the two theorems below.

Theorem 2. Define a nonincreasing function 𝜃 from [0,1) onto (1/2,1] by 𝜃(𝑟)=√1𝑖𝑓0≤𝑟≤5−12,(1−𝑟)𝑟⁻²√𝑖𝑓5−12√≤𝑟≤22,(1+𝑟)⁻¹√𝑖𝑓22≤𝑟<1. (1) Then for a metric space (𝑋,𝑑), the following is equivalent. (1)𝑋is complete. (2)Every mapping 𝑇 on 𝑋 satisfying the following has a fixed point: there exists 𝑟∈[0,1) such that 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑(𝑥,𝑦) implies 𝑑(𝑇𝑥,𝑇𝑦)≤𝑟𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋.

Theorem 3. Let (𝑋,𝑑) be a compact metric space and let 𝑇 be a mapping on 𝑋. Assume that (1/2)𝑑(𝑥,𝑇𝑥)<𝑑(𝑥,𝑦) implies 𝑑(𝑇𝑥,𝑇𝑦)<𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋. Then 𝑇 has a unique fixed point.

Theorems 2 and 3 are extensively studied by many authors (see, e.g., [25–29]). It is worth to point out that the studies mentioned above can be classified as the extensions of Banach's principle on compact/complete metric spaces, which are totally ordered.

On the other hand, Turinici [30], Ran and Reurings [31], and Nieto and Rodríguez-López [32] proved an analog of the desired Banach's principle in the context of partially ordered complete metric spaces for a certain class of maps. In particular, Ran and Reurings [31] applied their result to solve a matrix equation and Nieto and Rodríguez-López [32] applied their theorem to guarantee the existence and uniqueness of the solution of some boundary value problems.

Motivated by all these developments, we shall prove new fixed point theorems extending Edelstein-Suzuki type contraction results in the setting of partially ordered complete metric spaces.

2. Main Results

We denote by Φ_𝜑 the set of functions 𝜑∶[0,+∞)→[0,+∞) satisfying the following condition: ()𝜑(𝑡)≤1𝜃12𝑡∀𝑡≥0. (2)

We denote by Φ the set of nondecreasing functions 𝜙∶[0,+∞)→[0,+∞) such that lim_𝑛→+∞𝜙^𝑛(𝑡)=0∀𝑡>0. (3)

We have the following lemma.

Lemma 4 (see [33, 34]). If 𝜓∈Φ, then 𝜙(𝑡)<𝑡 for all 𝑡>0.

Definition 5 (see [35]). Let 𝑓∶𝑋→𝑋 and 𝛼∶𝑋→ℝ⁺. We say that 𝑓 is 𝛼-admissible if 𝑥,𝑦∈𝑋𝛼𝑥,𝑦≥1⇒𝛼𝑓𝑥,𝑓𝑦≥1. (4)

Theorem 6. Let (𝑋,𝑑,⪯) be an ordered metric space such that (𝑋,𝑑) is complete. Let 𝑓∶𝑋→𝑋 be a self-mapping. Assume that there exists 𝜑∈Φ_𝜑 such that 𝜑𝑑𝑑𝛼𝑑𝑑𝑥,𝑓𝑥≤𝛼𝑥,𝑦𝑥,𝑦𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑥,𝑦𝑓𝑥,𝑓𝑦≤𝜙𝑥,𝑦 (5) for all comparable 𝑥,𝑦∈𝑋, where 𝜙∈Φ. If the following conditions hold: (i)𝑓 is non-decreasing, continuous, and 𝛼-admissible, (ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑓𝑥₀ and 𝛼(𝑥₀,𝑓𝑥₀)≥1, then 𝑓 has a fixed point in 𝑋.

Proof. Let 𝑥₀⪯𝑓𝑥₀. If 𝑥₀=𝑓𝑥₀, then the theorem follows. Hence we suppose that 𝑥₀≺𝑓𝑥₀. Define a sequence {𝑥_𝑛} by 𝑥_𝑛=𝑓^𝑛𝑥₀=𝑓𝑥_𝑛−1 for all 𝑛∈ℕ. Since 𝑓 is non-decreasing and 𝑥₀≺𝑓𝑥₀ then 𝑥₀≺𝑥₁⪯𝑥₂⪯⋯, (6) and hence {𝑥_𝑛} is an increasing sequence. If 𝑥_𝑛=𝑥_𝑛+1=𝑓𝑥_𝑛 for some 𝑛∈ℕ, then the result is proved as 𝑥_𝑛 is a fixed point of 𝑓. In what follows, we will suppose that 𝑥_𝑛≠𝑥_𝑛+1. Hence, {𝑥_𝑛} is a strictly increasing sequence. Since 𝑓 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑓𝑥₀)≥1, we find 𝛼(𝑥₁,𝑥₂)=𝛼(𝑓𝑥₀,𝑓𝑥₁)≥1. By continuing this process, we get 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ∪{0}. Now we obtain that 𝜑𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤12𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤𝛼𝑥_𝑛−1,𝑓𝑥_𝑛−1𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1 (7) and 𝑥_𝑛−1 and 𝑓𝑥_𝑛−1 are comparable for all 𝑛∈ℕ. Then by (5), we have 𝛼𝑥_𝑛−1,𝑓𝑥_𝑛−1𝑑𝑓𝑥_𝑛−1,𝑓²𝑥_𝑛−1≤𝜙𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1. (8)
Then 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝛼𝑥_𝑛−1,𝑥_𝑛𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜙𝑑𝑥_𝑛−1,𝑥_𝑛. (9) By induction, we have 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜙^𝑛𝑑𝑥₀,𝑥₁. (10)
By taking the limit as 𝑛→+∞ in the inequality above, we deduce lim_𝑛→+∞𝑑𝑥_𝑛,𝑥_𝑛+1=0. (11) For a fixed 𝜖>0, there exists 𝑁∈ℕ such that 𝑑𝑥_𝑛,𝑥_𝑛+1<𝜖−𝜙(𝜖)∀𝑛∈ℕ. (12) As 𝜙 is nondecreasing, we get 𝜙𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜙𝜖−𝜙(𝜖)≤𝜙(𝜖)∀𝑛∈ℕ. (13) Then 𝑑𝑥_𝑛,𝑥_𝑛+2≤𝑑𝑥_𝑛,𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑥_𝑛+2<𝜖−𝜙(𝜖)+𝜙𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜖−𝜙(𝜖)+𝜙(𝜖)=𝜖. (14) By continuing this process, we get 𝑑𝑥_𝑛,𝑥_𝑛+𝑘<𝜖 (15) for all 𝑛≥𝑁 and 𝑘∈ℕ. Consequently lim_{𝑚,𝑛,→+∞}𝑑(𝑥_𝑛,𝑥_𝑚)=0. Hence {𝑥_𝑛} is a Cauchy sequence. Since 𝑋 is complete, there is 𝑧∈𝑋 such that 𝑥_𝑛→𝑧. Now, since 𝑓 is continuous, we have 𝑓𝑧=lim_𝑛→∞𝑓𝑥_𝑛=lim_𝑛→∞𝑥_𝑛=𝑧. (16) So 𝑧 is a fixed point of 𝑓.

Example 7. Let 𝑋=[0,∞). Define the metric 𝑑 on 𝑋 by 𝑑(𝑥,𝑦)=|𝑥−𝑦|. Define 𝑓∶𝑋→𝑋 by 𝑓(𝑥)=121𝑥−4𝑥²[],1,if𝑥∈0,1ln𝑥+2if𝑥∈(1,∞) (17) and 𝛼∶𝑋²→ℝ⁺ by 𝛼=[],𝑥,𝑦1,if𝑥,𝑦∈0,10otherwise. (18)
Let 𝜓(𝑡)=(1/2)𝑡. Let 𝑥⪯𝑦 if and only if 𝑥≤𝑦. Hence the conditions of Theorem 6 hold; that is, 𝑓 has a fixed point.

Proof. Let 𝑥,𝑦∈𝑋. First we assume that 𝛼(𝑥,𝑦)≥1. Then 𝑥,𝑦∈[0,1]. On the other hand, for all 𝑥,𝑦∈[0,1], we have 𝑓𝑥≤1/2. Hence 𝛼(𝑓𝑥,𝑓𝑦)≥1. That is, 𝑓 is 𝛼-admissible. Clearly, 𝛼(0,𝑓0)≥1 and 0⪯𝑓0. Let 𝑥≽𝑦, and 𝑥,𝑦∈[0,1). Then we derive 𝛼𝑥,𝑦𝑑𝑓𝑥,𝑓𝑦=𝑥−𝑦2−𝑥²−𝑦²4≤𝑥−𝑦2=𝜓𝑑𝑥,𝑦. (19) Next we assume that 𝛼(𝑥,𝑦)=0. As a result we find 0=𝛼𝑥,𝑦𝑑𝑓𝑥,𝑓𝑦≤𝜓𝑑𝑥,𝑦. (20) Then the conditions of Theorem 6 hold; therefore, 𝑓 has a fixed point.

Theorem 8. Let (𝑋,𝑑,⪯) be an ordered metric space such that (𝑋,𝑑) is complete. Let 𝑓∶𝑋→𝑋 be a self-mapping. Assume that there exists 𝜑∈Φ_𝜑 such that 𝜑𝑑𝑑𝛼𝑑𝑑𝑥,𝑓𝑥≤𝛼𝑥,𝑦𝑥,𝑦implies𝑥,𝑦𝑓𝑥,𝑓𝑦≤𝜙𝑥,𝑦 (21) for all comparable 𝑥,𝑦∈𝑋, where 𝜙∈Φ. If the following conditions hold: (i)𝑓 is non-decreasing and 𝛼-admissible, (ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑓𝑥₀ and 𝛼(𝑥₀,𝑓𝑥₀)≥1, (iii)for a strictly increasing sequence {𝑥_𝑛}⊂𝑋, converging to 𝑥∈𝑋 one has 𝑥_𝑛≺𝑥 for all 𝑛∈ℕ, (iv)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥 as 𝑛→∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛, then 𝑓 has a fixed point.

Proof. Proceeding as in the proof of Theorem 6, we know that there is a point 𝑧∈𝑋 such that 𝑥_𝑛→𝑧 and 𝑥_𝑛≺𝑧 by condition (iii). We prove that 𝑧=𝑓𝑧. From (iv), we have 𝛼(𝑥_𝑛,𝑧)≥1 for all 𝑛∈ℕ. Now, we suppose that the following inequalities hold 12𝑑𝑥_𝑛,𝑥_𝑛+1≥𝜑𝑑𝑥_𝑛,𝑥_𝑛+1>𝛼𝑥_𝑛,𝑧𝑑𝑥_𝑛,𝑧≥𝑑𝑥_𝑛,1,𝑧2𝑑𝑥_𝑛+1,𝑥_𝑛+2≥𝜑𝑑𝑥_𝑛+1,𝑥_𝑛+2>𝛼𝑥_𝑛+1,𝑧𝑑𝑥_𝑛+1,𝑧≥𝑑𝑥_𝑛+1,𝑧 (22) for some 𝑛∈ℕ.
Since, 𝑥_𝑛≠𝑥_𝑛+1 for all 𝑛, we get 𝑑(𝑥_𝑛,𝑥_𝑛+1)>0 for all 𝑛. Now by Lemma 4 and (9), we have 𝑑(𝑥_𝑛+1,𝑥_𝑛+2)<𝑑(𝑥_𝑛,𝑥_𝑛+1). As a result, we obtain 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛,𝑧+𝑑𝑥_𝑛+1<1,𝑧2𝑑𝑥_𝑛,𝑥_𝑛+1+12𝑑𝑥_𝑛+1,𝑥_𝑛+2<12𝑑𝑥_𝑛,𝑥_𝑛+1+12𝑑𝑥_𝑛,𝑥_𝑛+1=𝑑𝑥_𝑛,𝑥_𝑛+1. (23) This is a contradiction. Thus, for all 𝑛∈ℕ, either 𝜑𝑑𝑥_𝑛,𝑥_𝑛+1≤𝛼𝑥_𝑛,𝑧𝑑𝑥_𝑛,𝑧 (24) or 𝜑𝑑𝑥_𝑛+1,𝑥_𝑛+2≤𝛼𝑥_𝑛+1,𝑧𝑑𝑥_𝑛+1,𝑧 (25) holds. By (21), we have 𝑑𝑓𝑥_𝑛,𝑓𝑧≤𝛼𝑥_𝑛,𝑧𝑑𝑓𝑥_𝑛,𝑓𝑧≤𝜙𝑑𝑥_𝑛,𝑧 (26) or 𝑑𝑓𝑥_𝑛+1,𝑓𝑧≤𝛼𝑥_𝑛+1,𝑧𝑑𝑓𝑥_𝑛+1,𝑓𝑧≤𝜙𝑑𝑥_𝑛+1,𝑧. (27)
Now since 𝑥_𝑛≺𝑧, we have 𝑑(𝑥_𝑛,𝑧)>0 for all 𝑛∈ℕ, that is, 𝑑𝑥_𝑛+1,𝑓𝑧<𝑑𝑥_𝑛,𝑧or𝑑𝑥_𝑛+2,𝑓𝑧<𝑑𝑥_𝑛+1,𝑧. (28)
Consequently, there exists an infinite subset 𝐽⊂ℕ such that one of these inequalities holds for every 𝑛∈𝐽. If we take the limit as 𝑛→+∞ and 𝑛∈𝐽, then we get 𝑑(𝑧,𝑓𝑧)=0, that is, 𝑧=𝑓𝑧, which is a contradiction. Hence it is absurd to suppose 𝑧≠𝑓𝑧. So 𝑧 is a fixed point of 𝑓.

Example 9. Let 𝑋=[0,∞). Define the metric 𝑑 on 𝑋 by 𝑑(𝑥,𝑦)=|𝑥−𝑦|. Define 𝑓∶𝑋→𝑋 by 𝑓(𝑥)=18𝑥²,if𝑥∈[0,1],4if𝑥=3,𝑥+1if𝑥∈ℝ⁺⧵[0,1]∪{3} (29) and 𝛼∶𝑋²→ℝ⁺ by 𝛼=[],𝑥,𝑦11,if𝑥,𝑦∈0,14,if𝑥=3,𝑦=1,0,otherwise. (30)
Let 𝜑(𝑡)=(1/2)𝑡, and 𝜙(𝑡)=(1/4)𝑡. Let 𝑥⪯𝑦 if and only if 𝑥≤𝑦. Hence the conditions of Theorem 8 hold. That is, 𝑓 has a fixed point.

Proof. Let 𝑥,𝑦∈𝑋. If 𝛼(𝑥,𝑦)≥1, then 𝑥,𝑦∈[0,1]. On the other hand, for all 𝑥,𝑦∈[0,1], we derive 𝑓𝑥≤1/2. Hence 𝛼(𝑓𝑥,𝑓𝑦)≥1. That is, 𝑓 is 𝛼-admissible.
Clearly, there exists 𝑥₀∈[0,1] such that 𝛼(𝑥₀,𝑓𝑥₀)≥1.
If {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥 as 𝑛→∞. Then {𝑥_𝑛}⊆[0,1] and 𝑥∈[0,1]. That is, 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛. We need to consider the following cases.(i)If 𝑥,𝑦∈[0,1] and 𝑥≥𝑦 then 𝛼𝑑==1𝑥,𝑦𝑓𝑥,𝑓𝑦𝑓𝑥−𝑓𝑦8𝑥²−𝑦²≤14𝑥−𝑦. (31)(ii)If 𝑥=3 and 𝑦=1, then 𝛼()𝑑()=3,1𝑓3,𝑓1=𝛼3,1𝑓3−𝑓131>1322=14|3−1|=𝜙(𝑑(2,3)). (32)But we see that 𝜑𝑑=1𝑥,𝑓𝑥213−𝑓3=1>2=𝛼(3,1)|3−1|=𝛼(3,1)𝑑(3,1). (33)
If it is the case that 𝛼(𝑥,𝑦)=0, then 𝛼(𝑥,𝑦)𝑑(𝑓𝑥,𝑓𝑦)=0≤𝜙(𝑑(𝑥,𝑦)). Hence, 𝜑𝑑𝑑𝛼𝑑𝑑.𝑥,𝑓𝑥≤𝛼𝑥,𝑦𝑥,𝑦implies𝑥,𝑦𝑓𝑥,𝑓𝑦≤𝜙𝑥,𝑦 (34)
Then the conditions of Theorem 8 hold. Consequently, 𝑓 has a fixed point.

The following notions will be used to show the uniqueness of a fixed point. (A)For all 𝑥,𝑦∈𝑋 that are not comparable, there exists 𝑣∈𝑋 comparable with 𝑥 and 𝑦 such that 𝛼(𝑥,𝑣)≥1 and 𝛼(𝑦,𝑣)≥1. (B)For all 𝑥,𝑦∈𝑋 that are comparable, there exists 𝑤∈𝑋 such that 𝛼(𝑥,𝑣)≥1 and 𝛼(𝑦,𝑣)≥1.

Theorem 10. Adding conditions (A) and (B) to the hypotheses of Theorem 8 one obtains uniqueness of the fixed point of 𝑓.

Proof. Suppose that 𝑧 and 𝑧^∗ are two fixed points of 𝑓 such that 𝑧≠𝑧^∗. If 𝑧 and 𝑧^∗ are not comparable, by condition (A), there exists 𝑣∈𝑋 comparable with 𝑧 and 𝑧^∗, such that 𝛼(𝑧,𝑣)≥1 and 𝛼(𝑧^∗,𝑣)≥1. Since 𝑓 is 𝛼-admissible, we have 𝛼(𝑓^𝑛𝑧,𝑓^𝑛𝑣)≥1 and 𝛼(𝑓^𝑛𝑧^∗,𝑓^𝑛𝑣)≥1. Since we have 𝜑𝑑𝑓^𝑛−1𝑧,𝑓^𝑛𝑧≤12𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣≤𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣≤𝛼𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣 (35) and 𝑓^𝑛−1𝑧 and 𝑓^𝑛−1𝑣 are comparable, by (21), we derive 𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑣≤𝛼𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑣≤𝜙𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣. (36) By induction, we get 𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑣≤𝜙^𝑛(𝑑(𝑧,𝑣)). (37) If 𝑑(𝑧,𝑣)=0, then 𝑧=𝑣. That is, 𝑧 and 𝑧^∗ are comparable, which is a contradiction. Hence we suppose that 𝑑(𝑧,𝑣)>0.
Taking the lim as 𝑘→+∞ in the first of the inequalities above, we have lim_𝑛→+∞𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑥=0. (38) Similarly, we obtain that lim_𝑛→+∞𝑑𝑓^𝑛𝑧^∗,𝑓^𝑛𝑣=0. (39) From the inequality 𝑑𝑧,𝑧^∗=𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑧^∗≤𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑣+𝑑𝑓^𝑛𝑣,𝑓^𝑛𝑧^∗, (40) we get 𝑑(𝑧,𝑧^∗)=0 by taking the limit as 𝑛→+∞, that is, 𝑧=𝑧^∗.
Similarly, if 𝑧 and 𝑧^∗ are comparable, then from (21) and condition (B), it follows that 𝑧=𝑧^∗.

We denote by Θ_𝛾 the set of functions 𝛾∶[0,+∞)→[0,+∞) satisfying the following condition: (𝜃2)𝛾(𝑡)≥𝑡∀𝑡≥0. (41)

Theorem 11. Let (𝑋,𝑑,⪯) be an ordered metric space such that (𝑋,𝑑) is complete. Let 𝑓∶𝑋→𝑋 be a self-mapping. Assume that there exist 𝜑∈Φ_𝜑 and 𝛾∈Θ_𝛾 such that 𝜑𝑑𝑑𝜓𝑑𝑑𝑑𝑥,𝑓𝑥≤𝛾𝑥,𝑦𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑓𝑥,𝑓𝑦≤𝜓𝑥,𝑦−𝜙𝑥,𝑦 (42) for all comparable 𝑥,𝑦∈𝑋, where 𝜙∈Φ and 𝜓∈Ψ. If the following conditions hold: (i)𝑓 is non-decreasing, (ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑓𝑥₀, (iii)if a non-decreasing sequence {𝑥_𝑛} is such that 𝑥_𝑛→𝑥 as 𝑛→∞, then 𝑥_𝑛⪯𝑥∀𝑛∈ℕ, (43) then 𝑓 has a fixed point. Moreover, 𝑓 has a unique fixed point if: (iv)for all 𝑥,𝑦∈𝑋 that are not comparable there exists 𝑣∈𝑋 comparable with 𝑥 and 𝑦.

Proof. Let 𝑥₀⪯𝑓𝑥₀. If 𝑥₀=𝑓𝑥₀, then the conclusion follows. Hence we suppose that 𝑥₀≺𝑓𝑥₀. Define a sequence {𝑥_𝑛} by 𝑥_𝑛=𝑓^𝑛𝑥₀=𝑓𝑥_𝑛−1 for all 𝑛∈ℕ. Since 𝑓 is non-decreasing and 𝑥₀≺𝑓𝑥₀, we have 𝑥₀≺𝑥₁⪯𝑥₂⪯⋯. (44) Hence {𝑥_𝑛} is a non-decreasing sequence. Since 𝜑𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤12𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤𝛾𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1 (45) and 𝑥_𝑛−1 and 𝑓𝑥_𝑛−1 are comparable for all 𝑛∈ℕ, by (42), we have 𝜓𝑑𝑓𝑥_𝑛−1,𝑓²𝑥_𝑛−1≤𝜓𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1−𝜙𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1≤𝜓𝑑𝑥_𝑛−1,𝑓𝑥_𝑛−1. (46)
Then we find 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛−1,𝑥_𝑛 (47) for all 𝑛∈ℕ. This implies that the sequence {𝑑(𝑥_𝑛−1,𝑥_𝑛)} is decreasing and so there is a real number 𝑠≥0 such that lim_𝑛→+∞𝑑𝑥_𝑛,𝑥_𝑛+1=𝑠. (48)
Now, we show that 𝑠 must be equal to 0. Let 𝑠>0. Then by taking the limit as 𝑛→∞ in (46), we get 𝜓(𝑠)≤𝜓(𝑠)−𝜙(𝑠) (49) which is a contradiction. Hence we conclude that lim_𝑛→+∞𝑑𝑥_𝑛,𝑥_𝑛+1=0. (50)
Next, we prove that {𝑥_𝑛} is a Cauchy sequence. Suppose, to the contrary, that {𝑥_𝑛} is not a Cauchy sequence. Then there exists an 𝜀>0 such that for all positive integer 𝑘, 𝑛(𝑘)>𝑚(𝑘)>𝑘,𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)𝑑𝑥≥𝜀,_𝑛(𝑘),𝑥_{𝑚(𝑘)−1}<𝜀 (51) for some sequences {𝑚(𝑘)} and {𝑛(𝑘)}. For all 𝑘∈ℕ, we have 𝜀≤𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)≤𝑑𝑥_𝑛(𝑘),𝑥_{𝑚(𝑘)−1}+𝑑𝑥_{𝑚(𝑘)−1},𝑥_𝑚(𝑘)<𝜀+𝑑𝑥_{𝑚(𝑘)−1},𝑥_𝑚(𝑘). (52) Taking the limit as 𝑘→+∞ in the inequality above and using (50), we get lim_𝑘→+∞𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)=𝜀. (53) Again, from the expressions 𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)≤𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1+𝑑𝑥_𝑚(𝑘)+1,𝑥_𝑛(𝑘)+1+𝑑𝑥_𝑛(𝑘)+1,𝑥_𝑛(𝑘),𝑑𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1≤𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1+𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+𝑑𝑥_𝑛(𝑘)+1,𝑥_𝑛(𝑘), (54) (50), and (53), we deduce lim_𝑘→+∞𝑑𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1=𝜀 (55) by taking the limit as 𝑘→+∞. From the facts lim_𝑘→+∞𝑑(𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1)=0 and 𝑑(𝑥_𝑚(𝑘),𝑥_𝑛(𝑘))≥𝜀>0, we deduce that there is 𝑁∈ℕ such that for all 𝑘≥𝑁𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1≤𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘). (56)
Since 𝜑𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1≤12𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1≤12𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝛾𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘) (57) and 𝑥_𝑚(𝑘) and 𝑥_𝑛(𝑘) are comparable, by (5), we get 𝜓𝑑𝑓𝑥_𝑚(𝑘),𝑓𝑥_𝑛(𝑘)≤𝜓𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)−𝜙𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘). (58) Taking the limit as 𝑘→+∞ in the inequality above, we have 𝜓(𝜀)≤𝜓(𝜀)−𝜙(𝜀), (59) which is a contradiction. Hence {𝑥_𝑛} is a Cauchy sequence. Since 𝑋 is complete, there is a point 𝑧∈𝑋 such that 𝑥_𝑛→𝑧. Furthermore, by condition (iii), 𝑥_𝑛⪯𝑧.
Now, we suppose that the following inequalities hold: 12𝑑𝑥_𝑛,𝑥_𝑛+1≥𝜑𝑑𝑥_𝑛,𝑥_𝑛+1>𝛾𝑑𝑥_𝑛𝑥,𝑧≥𝑑_𝑛,1,𝑧2𝑑𝑥_𝑛+1,𝑥_𝑛+2≥𝜑𝑑𝑥_𝑛+1,𝑥_𝑛+2>𝛾𝑑𝑥_𝑛+1,𝑧≥𝑑𝑥_𝑛+1,𝑧 (60) for some 𝑛∈ℕ. By (47), we have 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛,𝑧+𝑑𝑥_𝑛+1<1,𝑧2𝑑𝑥_𝑛,𝑥_𝑛+1+12𝑑𝑥_𝑛+1,𝑥_𝑛+2<12𝑑𝑥_𝑛,𝑥_𝑛+1+12𝑑𝑥_𝑛,𝑥_𝑛+1=𝑑𝑥_𝑛,𝑥_𝑛+1, (61) a contradiction. Thus, for all 𝑛∈ℕ, either 𝜑𝑑𝑥_𝑛,𝑥_𝑛+1≤𝛾𝑑𝑥_𝑛,𝑧 (62) or 𝜑𝑑𝑥_𝑛+1,𝑥_𝑛+2≤𝛾𝑑𝑥_𝑛+1,𝑧 (63) holds. By (42), it follows that either 𝜓𝑑𝑓𝑥_𝑛,𝑓𝑧≤𝜓𝑑𝑥_𝑛,𝑧−𝜙𝑑𝑥_𝑛,𝑧 (64) or 𝜓𝑑𝑓𝑥_𝑛+1,𝑓𝑧≤𝜓𝑑𝑥_𝑛+1,𝑧−𝜙𝑑𝑥_𝑛+1,𝑧. (65)
Consequently, there exists an infinite subset 𝐽⊂ℕ such that one of these inequalities holds for every 𝑛∈𝐽. If we take the limit as 𝑛→+∞ and 𝑛∈𝐽, then we get that 𝑑(𝑧,𝑓𝑧)=0, that is, 𝑧=𝑓𝑧 which contradicts with the assumption 𝑧≠𝑓𝑧. Hence 𝑧 is a fixed point of 𝑓.
Now, assume that condition (iv) holds. To prove the uniqueness of 𝑧, suppose that 𝑧^∗ is another fixed point of 𝑓. It is easy to see from (5) that 𝑧=𝑧^∗, if 𝑧 and 𝑧^∗ are comparable. If 𝑧 and 𝑧^∗ are not comparable, by condition (iv), there exists an 𝑥 comparable with 𝑧 and 𝑧^∗. Now we suppose that lim_𝑛→+∞𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑥=𝑠₁>0,lim_𝑛→+∞𝑑𝑓^𝑛𝑧^∗,𝑓^𝑛𝑥=𝑠₂>0. (66) First we note that for each 𝑥∈𝑋 comparable with 𝑧, by (5) with 𝑥=𝑓^𝑛−1𝑧 and 𝑦=𝑓^𝑛−1𝑥, we have 𝜓𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑥≤𝜓𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑥−𝜙𝑑𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑥. (67) Taking the limsup as 𝑘→+∞ in the inequalities above, we have 𝜓𝑠₁≤𝜓𝑠₁−𝜙𝑠₁, (68) which is a contradiction. Hence lim_𝑛→+∞𝑑(𝑓^𝑛𝑧,𝑓^𝑛𝑥)=0. Similarly, we obtain that lim_𝑛→+∞𝑑𝑓^𝑛𝑧^∗,𝑓^𝑛𝑥=0. (69) From the inequality 𝑑𝑧,𝑧^∗=𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑧^∗≤𝑑𝑓^𝑛𝑧,𝑓^𝑛𝑥+𝑑𝑓^𝑛𝑥,𝑓^𝑛𝑧^∗, (70) we get 𝑑(𝑧,𝑧^∗)=0, that is, 𝑧=𝑧^∗, by taking the limit as 𝑛→+∞.

Example 12. Let 𝑋={(0,0),(4,0),(0,4),(4,5),(5,4)}. We define a partial order ⊑ by 𝑥,𝑦⊑(𝑢,𝑣)⟺𝑥≤𝑢,𝑣≤𝑦. (71)
Define the metric 𝑑 on 𝑋 by 𝑑((𝑥₁,𝑥₂),(𝑦₁,𝑦₂))=|𝑥₁−𝑦₁|+|𝑥₂−𝑦₂|. Clearly, (𝑋,𝑑) is a complete metric space. Also define 𝑓∶𝑋→𝑋 by 𝑓𝑥₁,𝑥₂=𝑥₁,0,If𝑥₁≤𝑥₂,0,𝑥₂If𝑥₁>𝑥₂. (72)
Let 𝜑(𝑡)=(1/2)𝑡, 𝛾(𝑡)=𝑡, 𝜓(𝑡)=𝑡, and 𝜙(𝑡)=(1−𝑟)𝑡, where 𝑟∈[0,1). Hence the conditions of Theorem 11 hold; that is, 𝑓 has a unique fixed point. But Theorem2.1 in [36] cannot be applied to 𝑓 in this example.

Proof. Let (𝑥,𝑦),(𝑢,𝑣)∈𝑋, with (𝑥,𝑦)⊑(𝑢,𝑣). Then for all elements of 𝑋 we have (0,0)⊑(4,0),(0,4)⊑(5,4),(4,5)⊑(5,4). (73) We examine each of the following cases: (i)for (0,0)⊑(4,0), we have; 𝑑𝑓()(40,0,𝑓4,0)=0≤5𝑑((0,0),(4,0)); (74)(ii)for (0,4)⊑(5,4), we have 𝑑𝑓()(40,4,𝑓5,4)=4≤4=5𝑑((0,4),(5,4)); (75)(iii)for (4,5)⊑(5,4), we have 𝑑𝑓()(84,5,𝑓5,4)=8>5=45𝑑((4,5),(5,4)). (76)That is, in [36, Theorem  2.1] cannot be applied to 𝑓. But for (4,5)⊑(5,4), 12𝑑()(=54,5,𝑓4,5)2>2=𝑑((4,5),(5,4)). (77) Hence the conditions of Theorem 11 hold; that is, 𝑓 has a unique fixed point in 𝑋.

3. Application to Existence of Solutions of Integral Equations

Let 𝑋=𝐶([0,𝑇],ℝ) be the set of real continuous functions defined on [0,𝑇] and 𝑑_𝛼∶𝑋×𝑋→[0,+∞) be defined by 𝑑(𝑥,𝑦)=sup_{𝑡∈[0,𝑇]}|𝑥(𝑡)−𝑦(𝑡)| for all 𝑥,𝑦∈𝑋. Let ⪯ be the partial order on 𝑋 defined by 𝑥⪯𝑦 if and only if 𝑥(𝑡)≤𝑦(𝑡) for all 𝑡∈[0,𝑇]. Then (𝑋,𝑑,⪯) is a complete partially ordered metric space.

Consider the following integral equation: 𝑥(𝑡)=𝑝(𝑡)+^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠, (78) where(A)𝑓∶[0,𝑇]×ℝ→ℝ is continuous, (B)𝑝∶[0,𝑇]→ℝ is continuous, (C)𝑆∶[0,𝑇]×[0,𝑇]→[0,+∞) is continuous and sup_{𝑡∈[0,𝑇]}^𝑇₀𝑆(𝑡,𝑠)𝑑𝑠≤1, (79)(D)there exists a nonempty and closed subset 𝑈⊆𝑋 such that for all 𝑥,𝑦∈𝑈 with 𝑥⪯𝑦, we have 𝑓(𝑠)((𝑠𝑦(𝑠)(𝑠)[],𝑠,𝑦−𝑓𝑠,𝑥))≤𝜓−𝑥∀𝑠∈0,𝑇 (80) where 𝜓∈Ψ and for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, we have 0≤𝑓𝑠,𝑦(𝑠)−𝑓(𝑠,𝑥(𝑠)), (81)(E)for all 𝑥∈𝑈, we have 𝑥(𝑡)≤𝑝(𝑡)+^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠∈𝑈, (82)(F)for a strictly increasing sequence {𝑥_𝑛}⊂𝑋 converging to 𝑥∈𝑋, we have 𝑥_𝑛≺𝑥 for all 𝑛∈ℕ. We have the following result of existence of solutions for integral equations.

Theorem 13. Under assumptions (A)–(E), the integral equation (78) has a solution in 𝑋=𝐶([0,𝑇],𝑅).

Proof. Let 𝐻∶𝑋→𝑋 be defined by 𝐻𝑥(𝑡)=𝑝(𝑡)+^𝑇₀𝑆()𝑓((𝑠[]𝑡,𝑠𝑠,𝑥))𝑑𝑠𝑡∈0,𝑇,∀𝑥∈𝑋. (83)
First, we will prove that 𝐻 is a nondecreasing mapping with respect to ⪯. Let 𝑥⪯𝑦. By (D), we have 0≤𝑓(𝑠,𝑦(𝑠))−𝑓(𝑠,𝑥(𝑠)) for all 𝑠∈[0,𝑇]. On the other hand by definition of 𝐻, we have 𝐻𝑦−𝐻𝑥=^𝑇₀𝑆(𝑓(𝑠)((𝑠∀[].𝑡,𝑠)𝑠,𝑦−𝑓𝑠,𝑥))𝑑𝑠≥0𝑡∈0,𝑇 (84)
Then 𝐻𝑥⪯𝐻𝑦, that is, 𝐻 is a non-decreasing mapping with respect to ⪯. Now suppose that 𝑥,𝑦∈𝑈 with 𝑥⪯𝑦. Then by (C), (D), and the definition of 𝐻, we get 𝑑𝐻𝑥,𝐻𝑦=sup_{[]𝑡∈0,𝑇}𝐻𝑥(𝑡)−𝐻𝑦(𝑡)=sup_{𝑡∈[0,𝑇]}^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝑑𝑠≤sup_{[]𝑡∈0,𝑇}^𝑇₀𝑆(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝑑𝑠≤sup_{[]𝑡∈0,𝑇}^𝑇₀𝑆()𝜓𝑦(𝑠)(𝑠)≤𝑡,𝑠−𝑥𝑑𝑠sup_{[]𝑡∈0,𝑇}^𝑇₀𝑆(𝑡,𝑠)𝑑𝑠×𝜓sup_{[]𝑠∈0,𝑇}𝑦(𝑠)(𝑠𝑑.−𝑥)≤𝜓𝑥,𝑦 (85)
Define 𝛼∶𝑋²→ℝ₊ by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈𝑈with𝑥⪯𝑦,0,otherwise. (86)
Clearly, 𝛼(𝑥,𝑦)𝑑(𝐻𝑥,𝐻𝑦)≤𝜓(𝑑(𝑥,𝑦)) holds for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦. If 𝛼(𝑥,𝑦)≥1 then 𝑥,𝑦∈𝑈 and 𝑥⪯𝑦. Since, 𝐻 is increasing, we obtain 𝐻𝑥⪯𝐻𝑦. Similarly, since 𝐻𝑥,𝐻𝑦∈𝑈, we get 𝛼(𝐻𝑥,𝐻𝑦)≥1. Therefore, 𝐻 is an 𝛼-admissible mapping. Also, assume that {𝑥_𝑛} is a sequence such that 𝑥_𝑛→𝑥 as 𝑛→∞ and 𝛼(𝑥_𝑛,𝑥)≥1. Then {𝑥_𝑛}⊆𝑈. Since 𝑈 is a closed set, we have 𝑥∈𝑈. That is, 𝛼(𝑥_𝑛,𝑥)≥1. Then conditions of Theorem 8 hold and 𝐻 has a solution in 𝑋.

Acknowledgment

The first author is supported by Islamic Azad University, Astara Branch.