Fixed Points for Weak 𝛼-𝜓-Contractions in Partial Metric Spaces

Abstract

Recently, Samet et al. (2012) introduced the notion of 𝛼-𝜓-contractive mappings and established some fixed point results in the setting of complete metric spaces. In this paper, we introduce the notion of weak 𝛼-𝜓-contractive mappings and give fixed point results for this class of mappings in the setting of partial metric spaces. Also, we deduce fixed point results in ordered partial metric spaces. Our results extend and generalize the results of Samet et al.

1. Introduction

The notion of partial metric is one of the most useful and interesting generalizations of the classical concept of metric. The partial metric spaces were introduced in 1994 by Matthews [1] as a part of the study of denotational semantics of data for networks, showing that the contraction mapping principle can be generalized to the partial metric context for applications in program verification. Later on, many authors studied the existence of several connections between partial metrics and topological aspects of domain theory (see [2–8] and the references therein). On the other hand, some researchers [9, 10] investigated the characterization of partial metric 0-completeness in terms of fixed point theory, extending the characterization of metric completeness [11–14].

Recently, Samet et al. [15] introduced the notion of 𝛼-𝜓-contractive mappings and established some fixed point results in the setting of complete metric spaces. In this paper, we introduce the notion of weak 𝛼-𝜓-contractive mappings and give fixed point results for this class of mappings in the setting of partial metric spaces. Also, we deduce fixed point results in ordered partial metric spaces. Our results extend and generalize Theorems 2.1–2.3 of [15] and many others. An application to ordinary differential equations concludes the paper.

2. Preliminaries

In this section, we recall some definitions and some properties of partial metric spaces that can be found in [1, 5, 10, 16, 17]. A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→[0,+∞) such that, for all 𝑥,𝑦,𝑧∈𝑋, we have(𝑝₁)𝑥=𝑦⇔𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦), (𝑝₂)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦), (𝑝₃)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥), (𝑝₄)𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑧)+𝑝(𝑧,𝑦)−𝑝(𝑧,𝑧).

A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is a non-empty set and 𝑝 is a partial metric on 𝑋. It is clear that if 𝑝(𝑥,𝑦)=0, then from (𝑝₁) and (𝑝₂) it follows that 𝑥=𝑦. But if 𝑥=𝑦, 𝑝(𝑥,𝑦) may not be 0. A basic example of a partial metric space is the pair ([0,+∞),𝑝), where 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈[0,+∞). Other examples of partial metric spaces which are interesting from a computational point of view can be found in [1].

Each partial metric 𝑝 on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has as a base the family of open 𝑝-balls {𝐵_𝑝(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝑝(𝑥,𝜀)=𝑦∈𝑋∶𝑝𝑥,𝑦<𝑝(𝑥,𝑥)+𝜀 (1) for all 𝑥∈𝑋 and 𝜀>0.

Let (𝑋,𝑝) be a partial metric space. A sequence {𝑥_𝑛} in (𝑋,𝑝) converges to a point 𝑥∈𝑋 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→+∞𝑝(𝑥,𝑥_𝑛).

A sequence {𝑥_𝑛} in (𝑋,𝑝) is called a Cauchy sequence if there exists (and is finite) lim_{𝑛,𝑚→+∞}𝑝(𝑥_𝑛,𝑥_𝑚).

A partial metric space (𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→+∞}𝑝(𝑥_𝑛,𝑥_𝑚).

A sequence {𝑥_𝑛} in (𝑋,𝑝) is called 0-Cauchy if lim_{𝑛,𝑚→+∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0. We say that (𝑋,𝑝) is 0-complete if every 0-Cauchy sequence in 𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=0.

On the other hand, the partial metric space (ℚ∩[0,+∞),𝑝), where ℚ denotes the set of rational numbers and the partial metric 𝑝 is given by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, provides an example of a 0-complete partial metric space which is not complete.

It is easy to see that every closed subset of a complete partial metric space is complete.

Notice that if 𝑝 is a partial metric on 𝑋, then the function 𝑝^𝑠∶𝑋×𝑋→[0,+∞) given by 𝑝^𝑠𝑥,𝑦=2𝑝𝑥,𝑦−𝑝(𝑥,𝑥)−𝑝𝑦,𝑦 (2) is a metric on 𝑋. Furthermore, lim_𝑛→+∞𝑝^𝑠(𝑥_𝑛,𝑥)=0 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→+∞𝑝𝑥_𝑛,𝑥=lim_{𝑛,𝑚→+∞}𝑝𝑥_𝑛,𝑥_𝑚. (3)

Lemma 1 (see [1, 16]). Let (𝑋,𝑝) be a partial metric space. Then (a){𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝) if and only if it is a Cauchy sequence in the metric space (𝑋,𝑝^𝑠), (b)a partial metric space (𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑝^𝑠) is complete.

Let 𝑋 be a non-empty set. If (𝑋,𝑝) is a partial metric space and (𝑋,⪯) is a partially ordered set, then (𝑋,𝑝,⪯) is called an ordered partial metric space. Then 𝑥,𝑦∈𝑋 are called comparable if 𝑥⪯𝑦 or 𝑦⪯𝑥 holds. Let (𝑋,⪯) be a partially ordered set, and let 𝑇∶𝑋→𝑋 be a mapping. 𝑇 is a non-decreasing mapping if 𝑇𝑥⪯𝑇𝑦 whenever 𝑥⪯𝑦 for all 𝑥,𝑦∈𝑋.

Definition 2 (see [15]). Let 𝑇∶𝑋→𝑋 and 𝛼∶𝑋×𝑋→[0,+∞). One says that 𝑇 is 𝛼-admissible if 𝑥,𝑦∈𝑋,𝛼𝑥,𝑦≥1⟹𝛼𝑇𝑥,𝑇𝑦≥1. (4)

Example 3. Let 𝑋=[0,+∞), and define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=𝑒𝑥,𝑦^𝑥−𝑦if0𝑥≥𝑦,if𝑥<𝑦. (5) Then, every non-decreasing mapping 𝑇∶𝑋→𝑋 is 𝛼-admissible. For example the mappings defined by 𝑇𝑥=ln(1+𝑥) and 𝑇𝑥=𝑥/(1+𝑥) for all 𝑥∈𝑋 are 𝛼-admissible.

3. Main Results

Throughout this paper, the standard notations and terminologies in nonlinear analysis are used. We start the main section by presenting the new notion of weak 𝛼-𝜓-contractive mappings.

Denote by Ψ the family of non-decreasing functions 𝜓∶[0,+∞)→[0,+∞) such that 𝜓(𝑡)>0 and lim_𝑛→+∞𝜓^𝑛(𝑡)=0 for each 𝑡>0, where 𝜓^𝑛 is the 𝑛th iterate of 𝜓.

Remark 4. Notice that the family Ψ used in this paper is larger (less restrictive) than the corresponding family of functions defined in [15], see also next Examples 12–13.

Lemma 5. For every function 𝜓∈Ψ, one has 𝜓(𝑡)<𝑡 for each 𝑡>0.

Definition 6. Let (𝑋,𝑝) be a partial metric space, and let 𝑇∶𝑋→𝑋 be a given mapping. We say that 𝑇 is a weak 𝛼-𝜓-contractive mapping if there exist two functions 𝛼∶𝑋×𝑋→[0,+∞) and 𝜓∈Ψ such that 𝛼𝑝𝑝(),𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜓max𝑥,𝑦,𝑝𝑥,𝑇𝑥,𝑝𝑦,𝑇𝑦 (6) for all 𝑥,𝑦∈𝑋. If 𝛼𝑥,𝑦𝑝𝑇𝑥,𝑇𝑦≤𝜓𝑝𝑥,𝑦, (7) for all 𝑥,𝑦∈𝑋, then 𝑇 is an 𝛼-𝜓-contractive mapping.

Remark 7. If 𝑇∶𝑋→𝑋 satisfies the contraction mapping principle, then 𝑇 is a weak 𝛼-𝜓-contractive mapping, where 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋 and 𝜓(𝑡)=𝑘𝑡 for all 𝑡≥0 and some 𝑘∈[0,1).

In the sequel, we consider the following property of regularity. Let (𝑋,𝑝) be a partial metric space, and let 𝛼∶𝑋×𝑋→[0,+∞) be a function. Then(r)𝑋 is 𝛼-regular if for each sequence {𝑥_𝑛}⊂𝑋, such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and 𝑥_𝑛→𝑥, we have that 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ, (c)𝑋 has the property (C) with respect to 𝛼 if for each sequence {𝑥_𝑛}⊂𝑋, such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ, there exists 𝑛₀∈ℕ such that 𝛼(𝑥_𝑚,𝑥_𝑛)≥1 for all 𝑛>𝑚≥𝑛₀.

Remark 8. Let 𝑋 be a non-empty set, and let 𝛼∶𝑋×𝑋→[0,+∞) be a function. Denote ℛ∶=𝑥,𝑦∶𝛼𝑥,𝑦≥1. (8) If ℛ is a transitive relation on 𝑋, then 𝑋 has the property (C) with respect to 𝛼.

In fact, if {𝑥_𝑛}⊂𝑋 is a sequence such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ, then (𝑥_𝑛,𝑥_𝑛+1)∈ℛ for all 𝑛∈ℕ. Now, fix 𝑚≥1 and show that 𝛼𝑥_𝑚,𝑥_𝑛≥1∀𝑛>𝑚. (9)

Obviously, (9) holds if 𝑛=𝑚+1. Assume that (9) holds for some 𝑛>𝑚. From (𝑥_𝑚,𝑥_𝑛),(𝑥_𝑛,𝑥_𝑛+1)∈ℛ, since ℛ is transitive, we get (𝑥_𝑚,𝑥_𝑛+1)∈ℛ. This implies that 𝛼(𝑥_𝑚,𝑥_𝑛+1)≥1, and so 𝛼(𝑥_𝑚,𝑥_𝑛)≥1 for all 𝑛>𝑚; that is, 𝑋 has the property (C) with respect to 𝛼.

Remark 9. Let (𝑋,𝑝,⪯) be an ordered partial metric space, and let 𝛼∶𝑋×𝑋→[0,+∞) be a function defined by 𝛼=1𝑥,𝑦if0𝑥⪯𝑦,otherwise. (10) Then 𝑋 has the property (C) with respect to 𝛼. Moreover, if for each sequence {𝑥_𝑛}, such that 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ convergent to some 𝑥∈𝑋, we have 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ, and then 𝑋 is 𝛼-regular.

By Remark 8, 𝑋 has the property (C) with respect to 𝛼. Now, let {𝑥_𝑛} be a sequence such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ convergent to some 𝑥∈𝑋, and then 𝑥_𝑛⪯𝑥_𝑛+1 for all 𝑛∈ℕ, and hence 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ. This implies that 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛∈ℕ, and so 𝑋 is 𝛼-regular.

Our first result is the following theorem that generalizes Theorem 2.1 of [15].

Theorem 10. Let (𝑋,𝑝) be a complete partial metric space, and let 𝑇∶𝑋→𝑋 be a weak 𝛼-𝜓-contractive mapping satisfying the following conditions: (i)𝑇 is 𝛼-admissible,(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, (iii)𝑋 has the property (C) with respect to 𝛼,(iv)𝑇 is continuous on (𝑋,𝑝^𝑠).Then, 𝑇 has a fixed point, that is; there exists 𝑥^∗∈𝑋 such that 𝑇𝑥^∗=𝑥^∗.

Proof. Let 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Define the sequence {𝑥_𝑛} in 𝑋 by 𝑥_𝑛+1=𝑇𝑥_𝑛,∀𝑛∈ℕ. (11) If 𝑥_𝑛=𝑥_𝑛+1 for some 𝑛∈ℕ, then 𝑥^∗=𝑥_𝑛 is a fixed point for 𝑇. Assume that 𝑥_𝑛≠𝑥_𝑛+1 for all 𝑛∈ℕ. Since 𝑇 is 𝛼-admissible, we have 𝛼𝑥₀,𝑥₁=𝛼𝑥₀,𝑇𝑥₀≥1⟹𝛼𝑇𝑥₀,𝑇𝑥₁=𝛼𝑥₁,𝑥₂≥1. (12) By induction, we get 𝛼𝑥_𝑛,𝑥_𝑛+1≥1,∀𝑛∈ℕ. (13) Applying inequality (6) with 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛 and using (13), we obtain 𝑝𝑥_𝑛,𝑥_𝑛+1=𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜓max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝑝𝑥_𝑛,𝑇𝑥_𝑛=𝜓max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1. (14) If max{𝑝(𝑥_𝑛−1,𝑥_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1)}=𝑝(𝑥_𝑛,𝑥_𝑛+1), from 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓𝑝𝑥_𝑛,𝑥_𝑛+1<𝑝𝑥_𝑛,𝑥_𝑛+1, (15) we obtain a contradiction; therefore, max{𝑝(𝑥_𝑛−1,𝑥_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1)}=𝑝(𝑥_𝑛−1,𝑥_𝑛), and hence 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓𝑝𝑥_𝑛−1,𝑥_𝑛,∀𝑛∈ℕ. (16) By induction, we get 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜓^𝑛𝑝𝑥₀,𝑥₁,∀𝑛∈ℕ. (17) This implies that lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛+1=0. (18) Fix 𝜀>0, and let 𝑛(𝜀)∈ℕ such that 𝑝𝑥_𝑚,𝑥_𝑚+1<𝜀−𝜓(𝜀),∀𝑚≥𝑛(𝜀). (19) Since 𝑋 has the property (C) with respect to 𝛼, there exists 𝑛₀∈ℕ such that 𝛼(𝑥_𝑚,𝑥_𝑛)≥1 for all 𝑛>𝑚≥𝑛₀. Let 𝑚∈ℕ with 𝑚≥max{𝑛₀,𝑛(𝜀)}, and we show that 𝑝𝑥_𝑚,𝑥_𝑛+1<𝜀,∀𝑛≥𝑚. (20) Note that (20) holds for 𝑛=𝑚. Assume that (20) holds for some 𝑛>𝑚, then 𝑝𝑥_𝑚,𝑥_𝑛+2≤𝑝𝑥_𝑚,𝑥_𝑚+1+𝑝𝑥_𝑚+1,𝑥_𝑛+2−𝑝𝑥_𝑚+1,𝑥_𝑚+1≤𝑝𝑥_𝑚,𝑥_𝑚+1+𝑝𝑇𝑥_𝑚,𝑇𝑥_𝑛+1≤𝑝𝑥_𝑚,𝑥_𝑚+1+𝛼𝑥_𝑚,𝑥_𝑛+1×𝑝𝑇𝑥_𝑚,𝑇𝑥_𝑛+1≤𝑝𝑥_𝑚,𝑥_𝑚+1+𝜓max𝑝𝑥_𝑚,𝑥_𝑛+1,𝑝𝑥_𝑚,𝑥_𝑚+1,𝑝𝑥_𝑛+1,𝑥_𝑛+2<𝜀−𝜓(𝜀)+𝜓(𝜀)=𝜀. (21) This implies that (20) holds for 𝑛≥𝑚, and hence lim_{𝑚,𝑛→+∞}𝑝𝑥_𝑚,𝑥_𝑛=0. (22) Thus, we proved that {𝑥_𝑛} is a Cauchy sequence in the partial metric space (𝑋,𝑝) and hence, by Lemma 1, in the metric space (𝑋,𝑝^𝑠). Since (𝑋,𝑝) is complete, by Lemma 1, also (𝑋,𝑝^𝑠) is complete. This implies that there exists 𝑥^∗∈𝑋 such that 𝑝^𝑠(𝑥_𝑛,𝑥^∗)→0 as 𝑛→+∞; that is, 𝑝𝑥^∗,𝑥^∗=lim_𝑛→+∞𝑝𝑥^∗,𝑥_𝑛=lim_{𝑚,𝑛→+∞}𝑝𝑥_𝑚,𝑥_𝑛=0. (23) From the continuity of 𝑇 on (𝑋,𝑝^𝑠), it follows that 𝑥_𝑛+1=𝑇𝑥_𝑛→𝑇𝑥^∗ as 𝑛→+∞. By the uniqueness of the limit, we get 𝑥^∗=𝑇𝑥^∗; that is, 𝑥^∗ is a fixed point of 𝑇.

In the next theorem, which is a proper generalization of Theorem 2.2 in [15], we omit the continuity hypothesis of 𝑇. Moreover, we assume 0-completeness of the space.

Theorem 11. Let (𝑋,𝑝) be a 0-complete partial metric space, and let 𝑇∶𝑋→𝑋 be a weak 𝛼-𝜓-contractive mapping satisfying the following conditions: (i)𝑇 is 𝛼-admissible,(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1,(iii)𝑋 has the property (C) with respect to 𝛼,(iv)𝑋 is 𝛼-regular.Then, 𝑇 has a fixed point.

Proof. Let 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Define the sequence {𝑥_𝑛} in 𝑋 by 𝑥_𝑛+1=𝑇𝑥_𝑛, for all 𝑛∈ℕ. Following the proof of Theorem 10, we know that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and that {𝑥_𝑛} is a 0-Cauchy sequence in the 0-complete partial metric space (𝑋,𝑝). Consequently, there exists 𝑥^∗∈𝑋 such that 𝑝𝑥^∗,𝑥^∗=lim_𝑛→+∞𝑝𝑥^∗,𝑥_𝑛=lim_{𝑚,𝑛→+∞}𝑝𝑥_𝑚,𝑥_𝑛=0. (24)
On the other hand, from 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ and the hypothesis (iv), we have 𝛼𝑥_𝑛,𝑥^∗≥1,∀𝑛∈ℕ. (25) Now, using the triangular inequality, (6) and (25), we get 𝑝𝑇𝑥^∗,𝑥^∗≤𝑝𝑇𝑥^∗,𝑇𝑥_𝑛+𝑝𝑥_𝑛+1,𝑥^∗−𝑝𝑥_𝑛+1,𝑥_𝑛+1≤𝛼𝑥_𝑛,𝑥^∗𝑝𝑇𝑥_𝑛,𝑇𝑥^∗+𝑝𝑥_𝑛+1,𝑥^∗≤𝜓max𝑝𝑥_𝑛,𝑥^∗,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥^∗,𝑇𝑥^∗+𝑝𝑥_𝑛+1,𝑥^∗. (26)
Since 𝑝(𝑥_𝑛,𝑥^∗),𝑝(𝑥_𝑛,𝑥_𝑛+1)→0 as 𝑛→+∞, for 𝑛 great enough, we have max𝑝𝑥_𝑛,𝑥^∗,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥^∗,𝑇𝑥^∗=𝑝𝑥^∗,𝑇𝑥^∗, (27) and hence 𝑝𝑇𝑥^∗,𝑥^∗≤𝜓𝑝𝑥^∗,𝑇𝑥^∗<𝑝𝑥^∗,𝑇𝑥^∗. (28) This is a contradiction, and so we obtain 𝑝(𝑇𝑥^∗,𝑥^∗)=0; that is, 𝑇𝑥^∗=𝑥^∗.

The following example illustrates the usefulness of Theorem 10.

Example 12. Let 𝑋=[0,+∞) and 𝑝∶𝑋×𝑋→[0,+∞) be defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Clearly, (𝑋,𝑝) is a complete partial metric space. Define the mapping 𝑇∶𝑋→𝑋 by 3𝑇𝑥=2𝑥−2if𝑥𝑥>1,1+𝑥if0≤𝑥≤1. (29) At first, we observe that we cannot find 𝑘∈[0,1) such that 𝑝𝑇𝑥,𝑇𝑦≤𝑘max𝑝𝑥,𝑦,𝑝(𝑥,𝑇𝑥),𝑝𝑦,𝑇𝑦 (30) for all 𝑥,𝑦∈𝑋, since we have 𝑝()1𝑇1,𝑇2=max2,52=525>𝑘2{}1=𝑘maxmax1,2,max1,2,5max2,2 (31) for all 𝑘∈[0,1). Now, we define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=1𝑥,𝑦if[],0𝑥,𝑦∈0,1otherwise. (32) Clearly 𝑇 is a weak 𝛼-𝜓-contractive mapping with 𝜓(𝑡)=𝑡/(1+𝑡) for all 𝑡≥0. In fact, for all 𝑥,𝑦∈𝑋, we have 𝛼𝑝𝑥𝑥,𝑦𝑇𝑥,𝑇𝑦=max,𝑦1+𝑥𝑝𝑝(),𝑝.1+𝑦=𝜓𝑥,𝑦≤𝜓max𝑥,𝑦,𝑝𝑥,𝑇𝑥𝑦,𝑇𝑦 (33) Moreover, there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. In fact, for 𝑥₀=1, we have 𝛼()11,𝑇1=𝛼1,2=1. (34) Obviously, 𝑇 is continuous on (𝑋,𝑝^𝑠) since 𝑝^𝑠(𝑥,𝑦)=|𝑥−𝑦|, and so we have to show that 𝑇 is 𝛼-admissible. In doing so, let 𝑥,𝑦∈𝑋 such that 𝛼(𝑥,𝑦)≥1. This implies that 𝑥,𝑦∈[0,1], and by the definitions of 𝑇 and 𝛼, we have 𝑥𝑇𝑥=∈[]𝑦1+𝑥0,1,𝑇𝑦=∈[],𝛼1+𝑦0,1𝑇𝑥,𝑇𝑦=1. (35) Then, 𝑇 is 𝛼-admissible. Moreover, if {𝑥_𝑛} is a sequence such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛∈ℕ, then 𝑥_𝑛∈[0,1] for all 𝑛∈ℕ, and hence 𝛼(𝑥_𝑚,𝑥_𝑛)≥1 for all 𝑛>𝑚≥1. Thus, 𝑋 has the property (C) with respect to 𝛼.
Now, all the hypotheses of Theorem 10 are satisfied, and so 𝑇 has a fixed point. Notice that Theorem 10 (also Theorem 11) guarantees only the existence of a fixed point but not the uniqueness. In this example, 0 and 3/2 are two fixed points of 𝑇.
Moreover, ∑^+∞_𝑛=1𝜓^𝑛(𝑡)=∑^+∞_𝑛=1(𝑡/(1+𝑛𝑡))≮+∞, and so 𝑇 is not an 𝛼-𝜓-contractive mapping in the sense of [15] with respect to the complete metric space (𝑋,𝑝^𝑠); that is, Theorem 2.1 of [15] cannot be applied in this case.

Now, we give an example involving a mapping 𝑇 that is not continuous. Also, this example shows that our Theorem 11 is a proper generalization of Theorem 2.2 in [15].

Example 13. Let 𝑋=ℚ∩[0,+∞) and 𝑝 as in Example 12. Clearly, (𝑋,𝑝) is a 0-complete partial metric space which is not complete. Then, Theorem 10 is not applicable in this case. Define the mapping 𝑇∶𝑋→𝑋 by 5𝑇𝑥=2𝑥−2if𝑥𝑥>2,1+𝑥if0≤𝑥≤2. (36) It is clear that 𝑇 is not continuous at 𝑥=2 with respect to the metric 𝑝^𝑠. Define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=1𝑥,𝑦if[],0𝑥,𝑦∈0,2otherwise. (37) Clearly 𝑇 is a weak 𝛼-𝜓-contractive mapping with 𝜓(𝑡)=𝑡/(1+𝑡) for all 𝑡≥0. In fact, for all 𝑥,𝑦∈𝑋, we have 𝛼𝑝𝑝𝑝,𝑝().𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜓𝑥,𝑦≤𝜓max𝑥,𝑦𝑥,𝑇𝑥,𝑝𝑦,𝑇𝑦 (38)
Proceeding as in Example 12, the reader can show that all the required hypotheses of Theorem 11 are satisfied, and so 𝑇 has a fixed point. Here, 0 and 5/2 are two fixed points of 𝑇.

Moreover, since (𝑋,𝑝^𝑠) is not complete, where 𝑝^𝑠(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋, we conclude that neither Theorem 2.1 nor Theorem 2.2 of [15] can be applied to cover this case, also because ∑^+∞_𝑛=1𝜓^𝑛(𝑡)≮+∞.

To ensure the uniqueness of the fixed point, we will consider the following hypothesis: (H)for all 𝑥,𝑦∈𝑋 with 𝛼(𝑥,𝑦)<1, there exists 𝑧∈𝑋 such that 𝛼(𝑥,𝑧)≥1, 𝛼(𝑦,𝑧)≥1, and lim_𝑛→+∞𝑝(𝑇^𝑛−1𝑧,𝑇^𝑛𝑧)=0.

Theorem 14. Adding condition (𝐻) to the hypotheses of Theorem 10 (resp., Theorem 11), one obtain the uniqueness of the fixed point of 𝑇.

Proof. Suppose that 𝑥^∗ and 𝑦^∗ are two fixed points of 𝑇 with 𝑥^∗≠𝑦^∗. If 𝛼(𝑥^∗,𝑦^∗)≥1, using (6), we get 𝑝𝑥^∗,𝑦^∗≤𝛼𝑥^∗,𝑦^∗𝑝𝑇𝑥^∗,𝑇𝑦^∗=𝜓𝑝𝑥^∗,𝑦^∗<𝑝𝑥^∗,𝑦^∗, (39) which is a contradiction, and so 𝑥^∗=𝑦^∗. If 𝛼(𝑥^∗,𝑦^∗)<1 by (H), there exists 𝑧∈𝑋 such that 𝛼𝑥^∗,𝑧≥1,𝛼𝑦^∗,𝑧≥1. (40) Since 𝑇 is 𝛼-admissible, from (40), we get 𝛼𝑥^∗,𝑇^𝑛𝑧≥1,𝛼𝑦^∗,𝑇^𝑛𝑧≥1,∀𝑛∈ℕ. (41) Let 𝑧_𝑛=𝑇^𝑛𝑧 for all 𝑛∈ℕ. Using (41) and (6), we have 𝑝𝑥^∗,𝑧_𝑛=𝑝𝑇𝑥^∗,𝑇𝑧_𝑛−1≤𝛼𝑥^∗,𝑧_𝑛−1𝑝𝑇𝑥^∗,𝑇𝑧_𝑛−1≤𝜓max𝑝𝑥^∗,𝑧_𝑛−1,𝑝𝑥^∗,𝑇𝑥^∗,𝑝𝑧_𝑛−1,𝑇𝑧_𝑛−1=𝜓max𝑝𝑥^∗,𝑧_𝑛−1,𝑝𝑧_𝑛−1,𝑧_𝑛. (42) Now, let 𝐽={𝑛∈ℕ∶max{𝑝(𝑥^∗,𝑧_𝑛−1),𝑝(𝑧_𝑛−1,𝑧_𝑛)}=𝑝(𝑧_𝑛−1,𝑧_𝑛). If 𝐽 is an infinite subset of ℕ, then 𝑝𝑥^∗,𝑧_𝑛≤𝜓𝑝𝑧_𝑛−1,𝑧_𝑛<𝑝𝑧_𝑛−1,𝑧_𝑛∀𝑛∈𝐽. (43) Then, letting 𝑛→+∞ with 𝑛∈𝐽 in the previous inequality, we get lim_𝑛→+∞𝑝𝑥^∗,𝑧_𝑛=0. (44)
If 𝐽 is a finite subset of ℕ, then there exists 𝑛₀∈ℕ such that max𝑝𝑥^∗,𝑧_𝑛−1,𝑝𝑧_𝑛−1,𝑧_𝑛=𝑝𝑥^∗,𝑧_𝑛−1∀𝑛>𝑛₀. (45)
This implies that 𝑝𝑥^∗,𝑧_𝑛≤𝜓^𝑛−𝑛0𝑝𝑥^∗,𝑧_𝑛0,∀𝑛>𝑛₀. (46)
Then, letting 𝑛→+∞, we get lim_𝑛→+∞𝑝𝑥^∗,𝑧_𝑛=0. (47)
Similarly, using (41) and (6), we get lim_𝑛→+∞𝑝𝑦^∗,𝑧_𝑛=0. (48) Since 𝑝^𝑠(𝑥,𝑦)≤2𝑝(𝑥,𝑦), using (47) and (48), we deduce that lim_𝑛→+∞𝑝^𝑠𝑥^∗,𝑧_𝑛=lim_𝑛→+∞𝑝^𝑠𝑦^∗,𝑧_𝑛=0. (49) Now, the uniqueness of the limit gives us 𝑥^∗=𝑦^∗. This finishes the proof.

From Theorems 10 and 11, we obtain the following corollaries.

Corollary 15. Let (𝑋,𝑝) be a complete partial metric space, and let 𝑇∶𝑋→𝑋 be an 𝛼-𝜓-contractive mapping satisfying the following conditions: (i)𝑇 is 𝛼-admissible, (ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1, (iii)𝑋 has the property (𝐶) with respect to 𝛼, (iv)𝑇 is continuous on (𝑋,𝑝^𝑠).Then, 𝑇 has a fixed point.

Corollary 16. Let (𝑋,𝑝) be a 0-complete partial metric space, and let 𝑇∶𝑋→𝑋 be an 𝛼-𝜓-contractive mapping satisfying the following conditions: (i)T is 𝛼-admissible,(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1,(iii)𝑋 has the property (𝐶) with respect to 𝛼,(iv)𝑋 is 𝛼-regular.Then, 𝑇 has a fixed point.

From the proof of Theorem 14, we deduce the following corollaries.

Corollary 17. One adds to the hypotheses of Corollary 15 (resp., Corollary 16) the following condition: (HC) for all 𝑥,𝑦∈𝑋 with 𝛼(𝑥,𝑦)<1, there exists 𝑧∈𝑋 such that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1, and one obtains the uniqueness of the fixed point of 𝑇.

4. Consequences

Now, we show that many existing results in the literature can be deduced easily from our theorems.

4.1. Contraction Mapping Principle

Theorem 18 (Matthews [1]). Let (𝑋,𝑝) be a 0-complete partial metric space, and let 𝑇∶𝑋→𝑋 be a given mapping satisfying 𝑝𝑇𝑥,𝑇𝑦≤𝑘𝑝𝑥,𝑦 (50) for all 𝑥,𝑦∈𝑋, where 𝑘∈[0,1). Then 𝑇 has a unique fixed point.

Proof. Let 𝛼∶𝑋×𝑋→[0,+∞) be defined by 𝛼(𝑥,𝑦)=1, for all 𝑥,𝑦∈𝑋, and let 𝜓∶[0,+∞)→[0,+∞) be defined by 𝜓(𝑡)=𝑘𝑡. Then 𝑇 is an 𝛼-𝜓-contractive mapping. It is easy to show that all the hypotheses of Corollaries 16 and 17 are satisfied. Consequently, 𝑇 has a unique fixed point.

Remark 19. In Example 12, Theorem 18 cannot be applied since 𝑝(𝑇1,𝑇2)>𝑝(2,1). However, using our Corollary 15, we obtain the existence of a fixed point of 𝑇.

4.2. Fixed Point Results in Ordered Metric Spaces

The existence of fixed points in partially ordered sets has been considered in [18]. Later on, some generalizations of [18] are given in [19–24]. Several applications of these results to matrix equations are presented in [18]; some applications to periodic boundary value problems and particular problems are given in [22, 23], respectively.

In this section, we will show that many fixed point results in ordered metric spaces can be deduced easily from our presented theorems.

4.2.1. Ran and Reurings Type Fixed Point Theorem

In 2004, Ran and Reurings proved the following theorem.

Theorem 20 (Ran and Reurings [18]). Let (𝑋,⪯) be a partially ordered set, and suppose that there exists a metric 𝑑 in 𝑋 such that the metric space (𝑋,𝑑) is complete. Let 𝑇∶𝑋→𝑋 be a continuous and non-decreasing mapping with respect to ⪯. Suppose that the following two assertions hold: (i)there exists 𝑘∈[0,1) such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝑘𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦,(ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀,(iii)𝑇 is continuous. Then, 𝑇 has a fixed point.

From Theorem 10, we deduce the following generalization and extension of the Ran and Reurings theorem in the framework of ordered complete partial metric spaces.

Theorem 21. Let (𝑋,𝑝,⪯) be an ordered complete partial metric space, and let 𝑇∶𝑋→𝑋 be a non-decreasing mapping with respect to ⪯. Suppose that the following assertions hold: (i)there exists 𝜓∈Ψ such that 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(max{𝑝(𝑥,𝑦),𝑝(𝑥,𝑇𝑥),𝑝(𝑦,𝑇𝑦)}) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦,(ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀,(iii)𝑇 is continuous on (𝑋,𝑝^𝑠).Then, 𝑇 has a fixed point.

Proof. Define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=1𝑥,𝑦if0𝑥⪯𝑦,otherwise. (51) From (i), we have 𝛼𝑝𝑝,𝑝(),𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜓max𝑥,𝑦𝑥,𝑇𝑥,𝑝𝑦,𝑇𝑦∀𝑥,𝑦∈𝑋. (52) Then, 𝑇 is a weak 𝛼-𝜓-contractive mapping. Now, let 𝑥,𝑦∈𝑋 such that 𝛼(𝑥,𝑦)≥1. By the definition of 𝛼, this implies that 𝑥⪯𝑦. Since 𝑇 is a non-decreasing mapping with respect to ⪯, we have 𝑇𝑥⪯𝑇𝑦, which gives us that 𝛼(𝑇𝑥,𝑇𝑦)=1. Then 𝑇 is 𝛼-admissible. From (ii), there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, and so 𝛼(𝑥₀,𝑇𝑥₀)=1. Moreover, by Remark 9, 𝑋 has the property (C) with respect to 𝛼.
Therefore, all the hypotheses of Theorem 10 are satisfied, and so 𝑇 has a fixed point.

Example 22. Let 𝑋=[0,+∞) and 𝑝∶𝑋×𝑋→[0,+∞) be defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Clearly, (𝑋,𝑝) is a complete partial metric space. Define the mapping 𝑇∶𝑋→𝑋 by 𝑇𝑥=2𝑥,∀𝑥∈𝑋. (53) Clearly 𝑇 is a continuous mapping with respect to the metric 𝑝^𝑠. We endow 𝑋 with the usual order of real numbers. Now, condition (𝑖) of Theorem 21 is not satisfied for 𝑥=1≤3=𝑦. In fact, if we assume the contrary, then 𝑝(𝑇1,𝑇3)=6≤𝜓𝑝(1,3)=𝜓(3)<3, (54) which is a contradiction. Then, we cannot apply Theorem 21 to prove the existence of a fixed point of 𝑇.

Define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=1𝑥,𝑦4if≠(),1𝑥,𝑦0,0if𝑥,𝑦=(0,0). (55) It is clear that 𝛼𝑝≤1𝑥,𝑦𝑇𝑥,𝑇𝑦2𝑝𝑥,𝑦,∀𝑥,𝑦∈𝑋. (56) Then, 𝑇 is a weak 𝛼-𝜓-contractive mapping with 𝜓(𝑡)=𝑡/2 for all 𝑡≥0. Now, let 𝑥,𝑦∈𝑋 such that 𝛼(𝑥,𝑦)≥1. By the definition of 𝛼, this implies that 𝑥=𝑦=0. Then we have 𝛼(𝑇𝑥,𝑇𝑦)=𝛼(0,0)=1, and so 𝑇 is 𝛼-admissible. Also, for 𝑥₀=0, we have 𝛼(𝑥₀,𝑇𝑥₀)=1. Consequently, all the hypotheses of Theorem 10 are satisfied, then we deduce the existence of a fixed point of 𝑇. Here 0 is a fixed point of 𝑇.

4.2.2. Nieto and Rodríguez-López Type Fixed Point Theorem

In 2005, Nieto and Rodríguez-López proved the following theorem.

Theorem 23 (Nieto and Rodríguez-López [22]). Let (𝑋,⪯) be a partially ordered set, and suppose that there exists a metric 𝑑 in 𝑋 such that the metric space (𝑋,𝑑) is complete. Let 𝑇∶𝑋→𝑋 be a non-decreasing mapping with respect to ⪯. Suppose that the following assertions hold: (i)there exists 𝑘∈[0,1) such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝑘𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦,(ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀,(iii)if {𝑥_𝑛} is a non-decreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥∈𝑋 as 𝑛→+∞, then 𝑥_𝑛⪯𝑥 for all 𝑛.Then, 𝑇 has a fixed point.

From Theorem 11, we deduce the following generalization and extension of the Nieto and Rodríguez-López theorem in the framework of ordered 0-complete partial metric spaces.

Theorem 24. Let (𝑋,𝑝,⪯) be an ordered 0-complete partial metric space, and let 𝑇∶𝑋→𝑋 be a non-decreasing mapping with respect to ⪯. Suppose that the following assertions hold: (i)there exists 𝜓∈Ψ such that 𝑝(𝑇𝑥,𝑇𝑦)≤𝜓(max{𝑝(𝑥,𝑦),𝑝(𝑥,𝑇𝑥),𝑝(𝑦,𝑇𝑦)}) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, (ii)there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, (iii)if {𝑥_𝑛} is a non-decreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥∈𝑋 as 𝑛→+∞, then 𝑥_𝑛⪯𝑥 for all 𝑛. Then, 𝑇 has a fixed point.

Proof. Define the function 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼=1𝑥,𝑦if0𝑥⪯𝑦,otherwise. (57) The reader can show easily that 𝑇 is a weak 𝛼-𝜓-contractive and 𝛼-admissible mapping. Now, by Remark 9, 𝑋 has the property (C) with respect to 𝛼 and is 𝛼-regular. Thus all the hypotheses of Theorem 11 are satisfied, and 𝑇 has a fixed point.

Remark 25. In, Example 22, also Theorem 24 cannot be applied since condition (𝑖) is not satisfied.

Remark 26. To establish the uniqueness of the fixed point, Ran and Reurings, Nieto and Rodríguez-López [18, 22] considered the following hypothesis: (u) for all 𝑥,𝑦∈𝑋, there exists 𝑧∈𝑋 such that 𝑥⪯𝑧 and 𝑦⪯𝑧. Notice that in establishing the uniqueness it is enough to assume that (u) holds for all 𝑥,𝑦∈𝑋 that are not comparable. This result is also a particular case of Corollary 17. Precisely, if 𝑥,𝑦∈𝑋 are not comparable, then there exists 𝑧∈𝑋 such that 𝑥⪯𝑧 and 𝑦⪯𝑧. This implies that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1, and here, we consider the same function 𝛼 used in the previous proof. Then, hypothesis (HC) of Corollary 17 is satisfied, and so we deduce the uniqueness of the fixed point. For establishing the uniqueness of the fixed point in Theorems 21 and 24, we consider the following hypothesis: (U) for all 𝑥,𝑦∈𝑋 that are not comparable, there exists 𝑧∈𝑋 such that 𝑥⪯𝑧, 𝑦⪯𝑧, and lim_𝑛→+∞𝑝(𝑇^𝑛−1𝑧,𝑇^𝑛𝑧)=0.

5. Application to Ordinary Differential Equations

In this section, we present a typical application of fixed point results to ordinary differential equations. In fact, in the literature there are many papers focusing on the solution of differential problems approached via fixed point theory (see, e.g., [15, 25, 26] and the references therein). For such a case, even without any additional problem structure, the optimal strategy can be obtained by finding the fixed point of an operator 𝑇 which satisfies a contractive condition in certain spaces.

Here, we consider the following two-point boundary value problem for second order differential equation: −𝑑²𝑥𝑑𝑡²((𝑡[]𝑥(0)(1)=𝑓𝑡,𝑥)),𝑡∈0,1=𝑥=0, (58) where 𝑓∶[0,1]×ℝ→ℝ is a continuous function. Recall that the Green's function associated to (58) is given by 𝐺()=𝑡()𝑠()𝑡,𝑠1−𝑠0≤𝑡≤𝑠≤1,1−𝑡0≤𝑠≤𝑡≤1. (59)

Let 𝐶(𝐼) (𝐼=[0,1]) be the space of all continuous functions defined on 𝐼. It is well known that such a space with the metric given by 𝑑𝑥,𝑦=𝑥−𝑦_∞=max_𝑡∈𝐼𝑥(𝑡)−𝑦(𝑡) (60) is a complete metric space.

Now, we consider the following conditions:(i)for all 𝑡∈𝐼, for all 𝑎,𝑏∈ℝ with |𝑎|,|𝑏|≤1, we have 𝑓(𝑡,𝑎)−𝑓(𝑡,𝑏)≤8𝜓(|𝑎−𝑏|), (61) where 𝜓∈Ψ, (ii)there exists 𝑥₀∈𝐶(𝐼) such that ‖𝑥₀‖_∞≤1, (iii)for all 𝑥∈𝐶(𝐼), ‖𝑥‖_∞≤1⟹¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠_∞≤1. (62)

Theorem 27. Suppose that conditions (𝑖)–(𝑖𝑖𝑖) hold. Then (58) has at least one solution 𝑥^∗∈𝐶²(𝐼).

Proof. Consider 𝐶(𝐼) endowed with the partial metric given by 𝑝𝑥,𝑦=𝑥−𝑦_∞if‖𝑥‖_∞,𝑦_∞≤1,𝑥−𝑦_∞+𝜌otherwise, (63) where 𝜌>0. It is easy to show that (𝐶(𝐼),𝑝) is 0-complete but is not complete. In fact, 𝑝^𝑠=2𝑥,𝑦𝑥−𝑦_∞if‖𝑥‖_∞,𝑦_∞≤1or‖𝑥‖_∞,𝑦_∞,2>1𝑥−𝑦_∞+𝜌otherwise, (64) and consequently (𝐶(𝐼),𝑝^𝑠) is not complete.
On the other hand, it is well known that 𝑥∈𝐶(𝐼), and is a solution of (58), is equivalent to 𝑥∈𝐶(𝐼) is a solution of the integral equation 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠,∀𝑡∈𝐼. (65) Define the operator 𝑇∶𝐶(𝐼)→𝐶(𝐼) by 𝑇𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠,∀𝑡∈𝐼. (66)
Then solving problem (58) is equivalent to finding 𝑥^∗∈𝐶(𝐼) that is a fixed point of 𝑇. Now, let 𝑥,𝑦∈𝐶(𝐼) such that ‖𝑥‖_∞,‖𝑦‖_∞≤1. From (i), we have (𝑡)(𝑡=𝑇𝑥−𝑇𝑦)₁₀𝐺()×𝑓((𝑠(𝑠)≤𝑡,𝑠𝑠,𝑥))−𝑓𝑠,𝑦𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑥(𝑠))−𝑓𝑠,𝑦(𝑠)𝑑𝑠≤8¹₀𝐺(𝑡,𝑠)𝜓𝑥(𝑠)−𝑦(𝑠)𝑑𝑠≤8sup_𝑡∈𝐼¹₀𝐺(𝑡,𝑠)𝑑𝑠×𝜓𝑥−𝑦_∞≤𝜓𝑥−𝑦_∞. (67)
Note that for all 𝑡∈𝐼, ∫¹₀𝐺(𝑡,𝑠)𝑑𝑠=(−𝑡²/2)+(𝑡/2), which implies that sup_𝑡∈𝐼¹₀𝐺()1𝑡,𝑠𝑑𝑠=8. (68)
Then, for all 𝑥,𝑦∈𝐶(𝐼) such that ‖𝑥‖_∞,‖𝑦‖_∞≤1, we have 𝑇𝑥−𝑇𝑦_∞≤𝜓𝑥−𝑦_∞. (69)
Define the function 𝛼∶𝐶(𝐼)×𝐶(𝐼)→[0,+∞) by 𝛼=1𝑥,𝑦if‖𝑥‖_∞,𝑦_∞0≤1,otherwise. (70) For all 𝑥,𝑦∈𝐶(𝐼), we have 𝛼𝑥,𝑦𝑇𝑥−𝑇𝑦_∞≤𝜓𝑥−𝑦_∞. (71) Then, 𝑇 is an 𝛼-𝜓-contractive mapping. From condition (iii), for all 𝑥,𝑦∈𝐶(𝐼), we get 𝛼𝑥,𝑦≥1⟹‖𝑥‖_∞,𝑦_∞⟹‖‖≤1𝑇𝑥_∞,𝑇𝑦_∞≤1⟹𝛼𝑇𝑥,𝑇𝑦≥1. (72) Then, 𝑇 is 𝛼-admissible. From conditions (ii) and (iii), there exists 𝑥₀∈𝐶(𝐼) such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Thus, all the conditions of Corollary 16 are satisfied, and hence we deduce the existence of 𝑥^∗∈𝐶(𝐼) such that 𝑥^∗=𝑇𝑥^∗; that is, 𝑥^∗ is a solution to (58).

Acknowledgments

This work was supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission (under Grant no. NRU56000508).