A Simple Method for Obtaining Coupled Fixed Points of 𝛼-𝜓-Contractive Type Mappings

Abstract

In 2012, the notion of 𝛼-𝜓-contractive type mappings was introduced by Samet, C. Vetro, and P. Vetro. By using a simple method, we give some coupled fixed point results for 𝛼-𝜓-contractive type mappings.

1. Introduction

In 1987, Guo and Lakshmikantham introduced the notion of coupled fixed points [1]. Then some authors proved some coupled fixed point results via some applications in the last decade of the previous century [2–6]. Later, this field was completed by some researchers by using different sights (see for example [7–18]). In 2012, Samet et al. introduced the notion of 𝛼-𝜓-contractive type mappings [19]. Also, Amini-Harandi has provided a method for obtaining coupled fixed point results [20]. The aim of this paper is to provide a simple method for obtaining some coupled fixed point results for 𝛼-𝜓-contractive type mappings.

Denote by Ψ the family of nondecreasing functions 𝜓∶[0,∞)→[0,∞) such that ∑^∞_𝑛=1𝜓^𝑛(𝑡)<∞ for all 𝑡>0, where 𝜓^𝑛 is the 𝑛th iterate of 𝜓. It is known that 𝜓(𝑡)<𝑡 for all 𝑡>0 and 𝜓∈Ψ [19]. Let (𝑋,𝑑) be a metric space and 𝑇 a self-map on 𝑋. Then 𝑇 is called a 𝛼-𝜓-contraction mapping whenever there exist 𝜓∈Ψ and 𝛼∶𝑋×𝑋→[0,∞) such that 𝛼(𝑥,𝑦)𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑑(𝑥,𝑦)) for all 𝑥,𝑦∈𝑋 [19]. Also, we say that 𝑇 is 𝛼-admissible whenever 𝛼(𝑥,𝑦)≥1 implies 𝛼(𝑇𝑥,𝑇𝑦)≥1 [19]. Also, we say that 𝑋 has the property (𝐵) if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛≥1 and 𝑥_𝑛→𝑥, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛≥1 [19]. Let (𝑋,𝑑) be a complete metric space and 𝑇 a 𝛼-admissible 𝛼-𝜓-contractive mapping on 𝑋. Suppose that there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. If 𝑇 is continuous or 𝑋 has the property (𝐵), then 𝑇 has a fixed point (see [19]; Theorems 2.1 and 2.2). Also, we say that 𝑋 has the property (𝐻) whenever for each 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1. If 𝑋 has the property (𝐻) in the Theorems 2.1 and 2.2, then 𝑋 has a unique fixed point ([19]; Theorem 2.3). It is considerable that the results of Samet et al. generalize similar ordered results in the literature (see the results of third section in [19]). Let 𝐹∶𝑋×𝑋→𝑋 be a mapping, where (𝑋,𝑑) is a metric space. We say that (𝑥₀,𝑦₀)∈𝑋×𝑋 is a coupled fixed point of 𝐹 whenever 𝐹(𝑥₀,𝑦₀)=𝑥₀ and 𝐹(𝑦₀,𝑥₀)=𝑦₀. Define 𝑇_𝐹∶𝑋×𝑋→𝑋×𝑋 by 𝑇_𝐹(𝑥,𝑦)=(𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) for all (𝑥,𝑦)∈𝑋×𝑋. Then, it is easy to check that (𝑥₀,𝑦₀) is a coupled fixed point of 𝐹 if and only if (𝑥₀,𝑦₀) is a fixed point of 𝑇_𝑓.

2. Main Results

In this section, we define 𝛿((𝑥,𝑦),(𝑢,𝑣))=𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣) and 𝑚,()𝛿,(),𝐹,𝛿),(𝐹()(,1𝑥,𝑦𝑢,𝑣=max𝑥,𝑦𝑢,𝑣,𝛿𝑥,𝑦𝑥,𝑦,𝐹𝑦,𝑥((𝑢,𝑣𝑢,𝑣,𝐹𝑣,𝑢)))2𝛿,(𝐹()((),𝐹𝑥,𝑦𝑢,𝑣,𝐹𝑣,𝑢))+𝛿𝑢,𝑣𝑥,𝑦,𝐹𝑦,𝑥 (1) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋, where (𝑋,𝑑) is a metric space and 𝐹∶𝑋×𝑋→𝑋 is a mapping. Now, we are ready to state and prove our main results.

Lemma 1. Let (𝑋,𝑑) be a complete metric space, 𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ and 𝑇 a self-map on 𝑋 such that 𝛼𝑑𝑑(),1𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜓max𝑥,𝑦,𝑑𝑥,𝑇𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥 (2) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. If 𝑋 has the property (𝐵), then 𝑇 has a fixed point.

Proof. Take 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 and define the sequence {𝑥_𝑛} in 𝑋 by 𝑥_𝑛+1=𝑇𝑥_𝑛 for all 𝑛≥0. If 𝑥_𝑛=𝑥_𝑛+1 for some 𝑛, then 𝑥^∗=𝑥_𝑛 is a fixed point of 𝑇. Assume that 𝑥_𝑛≠𝑥_𝑛+1 for all 𝑛. Since 𝑇 is 𝛼-admissible, it is easy to check that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛≥1. Thus, for each natural number 𝑛 one has 𝑑𝑥_𝑛,𝑥_𝑛+1=𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜓max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛−1,𝑥_𝑛,12𝑑𝑥_𝑛,𝑥_𝑛+𝑑𝑥_𝑛−1,𝑥_𝑛+1≤𝜓max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1,12𝑑𝑥_𝑛,𝑥_𝑛−1+𝑑𝑥_𝑛,𝑥_𝑛+1=𝜓max𝑑𝑥_𝑛,𝑥_𝑛−1,𝑑𝑥_𝑛,𝑥_𝑛+1. (3) If max{𝑑(𝑥_𝑛,𝑥_𝑛−1),𝑑(𝑥_𝑛,𝑥_𝑛+1)}=𝑑(𝑥_𝑛,𝑥_𝑛+1), then 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝜓𝑑𝑥_𝑛,𝑥_𝑛+1<𝑑𝑥_𝑛+1,𝑥_𝑛 (4) which is contradiction. Thus, max{𝑑(𝑥_𝑛,𝑥_𝑛−1), 𝑑(𝑥_𝑛,𝑥_𝑛+1)}=𝑑(𝑥_𝑛,𝑥_𝑛−1) for all 𝑛. Hence, 𝑑(𝑥_𝑛+1,𝑥_𝑛)≤𝜓(𝑑(𝑥_𝑛,𝑥_𝑛−1)) and so 𝑑(𝑥_𝑛+1,𝑥_𝑛)≤𝜓^𝑛(𝑑(𝑥₁,𝑥₀)) for all 𝑛. It is easy to check that {𝑥_𝑛} is a Cauchy sequence. Thus, there exists 𝑥^∗∈𝑋 such that 𝑥_𝑛→𝑥^∗. By using the assumption, we have 𝛼(𝑥_𝑛,𝑥^∗)≥1 for all 𝑛. Thus, 𝑑𝑇𝑥^∗,𝑥^∗≤𝑑𝑇𝑥^∗,𝑇𝑥_𝑛+𝑑𝑥_𝑛+1,𝑥^∗≤𝛼𝑥_𝑛,𝑥^∗𝑑𝑇𝑥^∗,𝑇𝑥_𝑛+𝑑𝑥_𝑛+1,𝑥^∗≤𝜓max𝑑𝑥_𝑛,𝑥^∗,𝑑𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥^∗,𝑇𝑥^∗,12𝑑𝑥_𝑛,𝑇𝑥^∗+𝑑𝑥^∗,𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑥^∗=𝜓max𝑑𝑥_𝑛,𝑥^∗,𝑑𝑥_𝑛,𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑥^∗ (5) for all 𝑛. Hence, 𝑑(𝑇𝑥^∗,𝑥^∗)=0 and so 𝑇𝑥^∗=𝑥^∗.

By using a similar argument we can prove the following results.

Lemma 2. Let (𝑋,𝑑) be a complete metric space, 𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ and 𝑇 a continuous self-map on 𝑋 such that 𝛼𝑑𝑑(),1𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜓max𝑥,𝑦,𝑑𝑥,𝑇𝑥,𝑑𝑦,𝑇𝑦2𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥 (6) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Then 𝑇 has a fixed point.

Lemma 3. Let (𝑋,𝑑) be a complete metric space, 𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ and 𝑇 a self-map on 𝑋 such that 𝛼𝑥,𝑦𝑑𝑇𝑥,𝑇𝑦≤𝜓𝑑(𝑥,𝑢)+𝑑𝑦,𝑣 (7) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. If 𝑋 has the property (𝐵), then 𝑇 has a fixed point.

Lemma 4. Let (𝑋,𝑑) be a complete metric space, 𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ and 𝑇 a continuous self-map on 𝑋 such that 𝛼𝑥,𝑦𝑑𝑇𝑥,𝑇𝑦≤𝜓𝑑(𝑥,𝑢)+𝑑𝑦,𝑣 (8) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Then 𝑇 has a fixed point.

Theorem 5. Let (𝑋,𝑑) be a complete metric space and 𝐹∶𝑋×𝑋→𝑋 a mapping. Suppose that there exist a mapping 𝛼∶𝑋²×𝑋²→[0,+∞) and 𝜓∈Ψ such that 𝛼((𝑥,𝑦),(𝑢,𝑣))𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝑚((𝑥,𝑦),(𝑢,𝑣))) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋, 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼𝐹𝑥,𝑦,𝐹𝑦,𝑥,(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢))≥1 (9) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 and there exists (𝑥₀,𝑦₀)∈𝑋×𝑋 such that 𝛼𝑥₀,𝑦₀,𝐹𝑥₀,𝑦₀,𝐹𝑦₀,𝑥₀≥1 (10) and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1. Also, suppose that for each convergent sequences {𝑥_𝑛} and {𝑦_𝑛} with 𝑥_𝑛→𝑥, 𝑦_𝑛→𝑦, 𝛼((𝑥_𝑛,𝑦_𝑛),(𝑥_𝑛+1,𝑦_𝑛+1))≥1 and 𝛼((𝑦_𝑛+1,𝑥_𝑛+1),(𝑦_𝑛,𝑥_𝑛))≥1 for all 𝑛, we have 𝛼((𝑥_𝑛,𝑦_𝑛),(𝑥,𝑦))≥1 and 𝛼((𝑦,𝑥),(𝑦_𝑛,𝑥_𝑛))≥1 for all 𝑛. Then 𝐹 has a coupled fixed point.

Proof. It is easy to check that 𝑚((𝑥,𝑦),(𝑢,𝑣))=𝑚((𝑣,𝑢),(𝑦,𝑥)) for all (𝑥,𝑦),(𝑢,𝑣) in 𝑋×𝑋, the metric space (𝑋×𝑋,𝛿) is complete and 𝛽(𝜉,𝜂)𝛿(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝑚(𝜉,𝜂)) for all 𝜉=(𝜉₁,𝜉₂), 𝜂=(𝜂₁,𝜂₂)∈𝑋×𝑋, where 𝛽𝜉₁,𝜉₂,𝜂₁,𝜂₂=min𝛼𝜉₁,𝜉₂,𝜂₁,𝜂₂,𝛼𝜂₂,𝜂₁,𝜉₂,𝜉₁. (11) Thus, 𝑇_𝐹 is a 𝛽-𝜓-contractive mapping. It is easy to check that 𝑇_𝐹 is 𝛽-admissible and 𝛽((𝑥₀,𝑦₀),𝑇_𝐹(𝑥₀,𝑦₀))≥1. Since (𝑋×𝑋,𝛿) has the property (𝐵), 𝑇_𝐹 has a fixed point by using Lemma 1 and so 𝐹 has a coupled fixed point.

Theorem 6. Let (𝑋,𝑑) be a complete metric space and 𝐹∶𝑋×𝑋→𝑋 a continuous mapping. Suppose that there exist a mapping 𝛼∶𝑋²×𝑋²→[0,+∞) and 𝜓∈Ψ such that 𝛼((𝑥,𝑦),(𝑢,𝑣))𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝑚((𝑥,𝑦),(𝑢,𝑣))) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋, 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼𝐹𝑥,𝑦,𝐹𝑦,𝑥,(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢))≥1 (12) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 and there exists (𝑥₀,𝑦₀)∈𝑋×𝑋 such that 𝛼𝑥₀,𝑦₀,𝐹𝑥₀,𝑦₀,𝐹𝑦₀,𝑥₀≥1 (13) and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1. Then 𝐹 has a coupled fixed point.

Proof. The metric space (𝑋×𝑋,𝛿) is complete and 𝛽(𝜉,𝜂)𝛿(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝑚(𝜉,𝜂)) for all 𝜉=(𝜉₁,𝜉₂),𝜂=(𝜂₁,𝜂₂)∈𝑋×𝑋, where 𝛽𝜉₁,𝜉₂,𝜂₁,𝜂₂=min𝛼𝜉₁,𝜉₂,𝜂₁,𝜂₂,𝛼𝜂₂,𝜂₁,𝜉₂,𝜉₁. (14) Thus, 𝑇_𝐹 is a continuous 𝛽-𝜓-contractive mapping. Also, it is easy to check that 𝑇_𝐹 is 𝛽-admissible and 𝛽((𝑥₀,𝑦₀),𝑇_𝐹(𝑥₀,𝑦₀))≥1. Now by using Lemma 2, 𝑇_𝐹 has a fixed point and so 𝐹 has a coupled fixed point.

Example 7. Let 𝑋=[0,+∞) and 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋. Define the mapping 𝐹∶𝑋×𝑋→𝑋 by 𝐹(𝑥,𝑦)=(𝑥−𝑦)/4 whenever 𝑥≥𝑦 and 𝐹(𝑥,𝑦)=0 whenever 𝑥<𝑦. Then 𝐹 is continuous mapping. Also, define 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼𝑥,𝑦,(𝑢,𝑣)=1,𝑥≥𝑦,𝑢≥𝑣,0,otherwise. (15) Thus, for each (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 we have 𝛼,()𝑑𝐹()≤𝐹(=𝑥,𝑦𝑢,𝑣𝑥,𝑦,𝐹𝑢,𝑣𝑥,𝑦−𝐹𝑢,𝑣)𝑥−𝑦4−𝑢−𝑣4=14)+≤1(𝑥−𝑢𝑣−𝑦4||+=1𝑥−𝑢𝑣−𝑦4𝑑()=1𝑥,𝑢+𝑑𝑦,𝑣4𝛿,()≤1𝑥,𝑦𝑢,𝑣2𝜓𝑚x,𝑦,(𝑢,𝑣), (16) where 𝜓(𝑡)=𝑡/2 for all 𝑡≥0. Also, it is easy to check that 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼((𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)),(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)))≥1 for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋. Also, 𝛼((𝑥₀,𝑦₀),(𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)))≥1 and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1, where (𝑥₀,𝑦₀)=(1,1). Thus by using Theorem 6, 𝐹 has a coupled fixed point.

Corollary 8. Let (𝑋,𝑑) be a complete metric space, 𝐹∶𝑋×𝑋→𝑋 a mapping and ⪯ an order on 𝑋×𝑋. Suppose that there exist (𝑥₀,𝑦₀)∈𝑋×𝑋 such that (𝑥₀,𝑦₀) and (𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)) are comparable and also (𝑦₀,𝑥𝑦₀) and (𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)) are comparable, and a mapping 𝜓∈Ψ such that 𝑑𝐹(≤1𝑥,𝑦,𝐹𝑢,𝑣)2𝜓𝑚𝑥,𝑦,(𝑢,𝑣) (17) for all comparable elements (𝑥,𝑦),(𝑢,𝑣) in 𝑋×𝑋. Suppose that (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) is comparable with (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) whenever (𝑥,𝑦) is comparable with (𝑢,𝑣). Also, suppose that for each convergent sequences {𝑥_𝑛} and {𝑦_𝑛} with 𝑥_𝑛→𝑥, 𝑦_𝑛→𝑦, (𝑥_𝑛,𝑦_𝑛) is comparable with (𝑥_𝑛+1,𝑦_𝑛+1) and (𝑦_𝑛+1,𝑥_𝑛+1) is comparable with (𝑦_𝑛,𝑥_𝑛) for all 𝑛, one gets that (𝑥_𝑛,𝑦_𝑛) is comparable with (𝑥,𝑦) and (𝑦,𝑥) is comparable with (𝑦_𝑛,𝑥_𝑛) for all 𝑛. Then 𝐹 has a coupled fixed point.

Proof. Define the mapping 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼((𝑥,𝑦),(𝑢,𝑣))=1 whenever (𝑥,𝑦) and (𝑢,𝑣) are comparable and 𝛼((𝑥,𝑦),(𝑢,𝑣))=0 otherwise. Then by using Theorem 5, 𝐹 has a coupled fixed point.

Corollary 9. Let (𝑋,𝑑) be a complete metric space, 𝐹∶𝑋×𝑋→𝑋 a continuous mapping, (𝑥^∗,𝑦^∗)∈𝑋×𝑋 a fixed element and ⪯ an order on 𝑋×𝑋. Suppose that there exist (𝑥₀,𝑦₀)∈𝑋×𝑋 such that (𝑥₀,𝑦₀), (𝑦₀,𝑥₀), (𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)) and (𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)) are comparable with (𝑥^∗,𝑦^∗), and a mapping 𝜓∈Ψ such that 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝑚((𝑥,𝑦),(𝑢,𝑣))) for all (𝑥,𝑦) and (𝑢,𝑣) in 𝑋×𝑋 which are comparable with (𝑥^∗,𝑦^∗). Assume that (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) and (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) are comparable with (𝑥^∗,𝑦^∗) whenever (𝑥,𝑦) and (𝑢,𝑣) so are. Then 𝐹 has a coupled fixed point.

Proof. Define the mapping 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼((𝑥,𝑦),(𝑢,𝑣))=1 whenever (𝑥,𝑦) and (𝑢,𝑣) are comparable with (𝑥^∗,𝑦^∗) and 𝛼((𝑥,𝑦),(𝑢,𝑣))=0 otherwise. Then by using Theorem 6, 𝐹 has a coupled fixed point.

Theorem 10. Let (𝑋,𝑑) be a complete metric space and 𝐹∶𝑋×𝑋→𝑋 a mapping. Suppose that there exist a mapping 𝛼∶𝑋²×𝑋²→[0,+∞) and 𝜓∈Ψ such that 𝛼,()𝑑𝐹()≤1𝑥,𝑦𝑢,𝑣𝑥,𝑦,𝐹𝑢,𝑣2𝜓𝛿𝑥,𝑦,(𝑢,𝑣) (18) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋, 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼𝐹𝑥,𝑦,𝐹𝑦,𝑥,(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢))≥1 (19) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 and there exists (𝑥₀,𝑦₀)∈𝑋×𝑋 such that 𝛼𝑥₀,𝑦₀,𝐹𝑥₀,𝑦₀,𝐹𝑦₀,𝑥₀≥1 (20) and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1. Also, suppose that for each convergent sequences {𝑥_𝑛} and {𝑦_𝑛} with 𝑥_𝑛→𝑥, 𝑦_𝑛→𝑦, 𝛼((𝑥_𝑛,𝑦_𝑛),(𝑥_𝑛+1,𝑦_𝑛+1))≥1 and 𝛼((𝑦_𝑛+1,𝑥_𝑛+1),(𝑦_𝑛,𝑥_𝑛))≥1 for all 𝑛, one has 𝛼((𝑥_𝑛,𝑦_𝑛),(𝑥,𝑦))≥1 and 𝛼((𝑦,𝑥),(𝑦_𝑛,𝑥_𝑛))≥1 for all 𝑛. Then 𝐹 has a coupled fixed point.

Proof. The metric space (𝑋×𝑋,𝛿) is complete and 𝛽(𝜉,𝜂)𝛿(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝛿(𝜉,𝜂)) for all 𝜉=(𝜉₁,𝜉₂),𝜂=(𝜂₁,𝜂₂)∈𝑋×𝑋, where 𝛽𝜉₁,𝜉₂,𝜂₁,𝜂₂=min𝛼𝜉₁,𝜉₂,𝜂₁,𝜂₂,𝛼𝜂₂,𝜂₁,𝜉₂,𝜉₁. (21) Thus, 𝛽(𝜉,𝜂)𝑑(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝛿(𝜉,𝜂)) for all 𝜉,𝜂∈𝑋×𝑋. Also, it is easy to check that 𝑇_𝐹 is 𝛽-admissible and 𝛽((𝑥₀,𝑦₀),𝑇_𝐹(𝑥₀,𝑦₀))≥1. Since (𝑋×𝑋,𝛿) has the property (𝐵), 𝑇_𝐹 has a fixed point by using Lemma 3, and so 𝐹 has a coupled fixed point.

Theorem 11. Let (𝑋,𝑑) be a complete metric space and 𝐹∶𝑋×𝑋→𝑋 a continuous mapping. Suppose that there exist a mapping 𝛼∶𝑋²×𝑋²→[0,+∞) and 𝜓∈Ψ such that 𝛼((𝑥,𝑦),(𝑢,𝑣))𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝛿((𝑥,𝑦),(𝑢,𝑣))) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋, 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼𝐹𝑥,𝑦,𝐹𝑦,𝑥,(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢))≥1 (22) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 and there exists (𝑥₀,𝑦₀)∈𝑋×𝑋 such that 𝛼𝑥₀,𝑦₀,𝐹𝑥₀,𝑦₀,𝐹𝑦₀,𝑥₀≥1 (23) and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1. Then 𝐹 has a coupled fixed point.

Proof. The metric space (𝑋×𝑋,𝛿) is complete and 𝛽(𝜉,𝜂)𝛿(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝛿(𝜉,𝜂)) for all 𝜉=(𝜉₁,𝜉₂),𝜂=(𝜂₁,𝜂₂)∈𝑋×𝑋, where 𝛽𝜉₁,𝜉₂,𝜂₁,𝜂₂=min𝛼𝜉₁,𝜉₂,𝜂₁,𝜂₂,𝛼𝜂₂,𝜂₁,𝜉₂,𝜉₁. (24) Thus, 𝛽(𝜉,𝜂)𝑑(𝑇_𝐹𝜉,𝑇_𝐹𝜂)≤𝜓(𝛿(𝜉,𝜂)) for all 𝜉,𝜂∈𝑋×𝑋. Also, it is easy to check that 𝑇_𝐹 is a continuous 𝛽-admissible mapping and 𝛽((𝑥₀,𝑦₀),𝑇_𝐹(𝑥₀,𝑦₀))≥1. Now by using Lemma 4, 𝑇_𝐹 has a fixed point and so 𝐹 has a coupled fixed point.

Example 12. Let 𝑋=ℝ and 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋. Define the mapping 𝐹∶𝑋×𝑋→𝑋 by 𝐹(𝑥,𝑦)=(𝑥−𝑦)/3+1 whenever 𝑥≥𝑦 and 𝐹(𝑥,𝑦)=0 whenever 𝑥<𝑦. Then 𝐹 is a discontinuous mapping. Define 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼𝑥,𝑦,(𝑢,𝑣)=1,𝑥≥𝑦,𝑢≥𝑣,0,otherwise. (25) Thus, for each (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 we have 𝛼,()𝑑𝐹()≤𝐹(=𝑥,𝑦𝑢,𝑣𝑥,𝑦,𝐹𝑢,𝑣𝑥,𝑦−𝐹𝑢,𝑣)𝑥−𝑦3−𝑢−𝑣3=13)+≤1(𝑥−𝑢𝑣−𝑦3+=1|(𝑥−𝑢)|𝑣−𝑦3𝑑()=1𝑥,𝑢+𝑑𝑦,𝑣3𝛿,(=1𝑥,𝑦𝑢,𝑣)2𝜓𝛿𝑥,𝑦,(𝑢,𝑣), (26) where 𝜓(𝑡)=2𝑡/3 for all 𝑡≥0. Also, 𝛼((𝑥,𝑦),(𝑢,𝑣))≥1 implies 𝛼𝐹𝑥,𝑦,𝐹𝑦,𝑥,(𝐹(𝑢,𝑣),𝐹(𝑣,𝑢))≥1 (27) for all (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋. Finally, 𝛼((𝑥₀,𝑦₀),(𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)))≥1 and 𝛼((𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)),(𝑦₀,𝑥₀))≥1, where (𝑥₀,𝑦₀)=(2,2). Thus by using Theorem 10, 𝐹 has a coupled fixed point.

Corollary 13. Let (𝑋,𝑑) be a complete metric space, 𝐹∶𝑋×𝑋→𝑋 a mapping and ⪯ an order on 𝑋×𝑋. Suppose that there exists a mapping 𝜓∈Ψ such that 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝛿((𝑥,𝑦),(𝑢,𝑣))) for all comparable elements (𝑥,𝑦) and (𝑢,𝑣) in 𝑋×𝑋. Suppose that there exists (𝑥₀,𝑦₀)∈𝑋×𝑋 such that (𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)) and (𝑥₀,𝑦₀) are comparable and also (𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)) and (𝑦₀,𝑥₀) are comparable, (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) is comparable with (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) whenever (𝑥,𝑦) is comparable with (𝑢,𝑣) and for each sequences {𝑥_𝑛} and {𝑦_𝑛} in 𝑋 such that 𝑥_𝑛→𝑥, 𝑦_𝑛→𝑦, (𝑥_𝑛,𝑦_𝑛) and (𝑥_𝑛+1,𝑦_𝑛+1) are comparable elements of 𝑋×𝑋 for all 𝑛, (𝑥_𝑛,𝑦_𝑛) and (𝑥,𝑦) are comparable elements for all 𝑛. Then 𝐹 has a coupled fixed point.

Proof. Define the mapping 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼((𝑥,𝑦),(𝑢,𝑣))=1 whenever (𝑥,𝑦) and (𝑢,𝑣) are comparable and 𝛼((𝑥,𝑦),(𝑢,𝑣))=0 otherwise. Then by using Theorem 10, 𝐹 has a coupled fixed point.

Corollary 14. Let (𝑋,𝑑) be a complete metric space, 𝐹∶𝑋×𝑋→𝑋 a continuous mapping, (𝑥^∗,𝑦^∗)∈𝑋×𝑋 a fixed element and ⪯ an order on 𝑋×𝑋. Suppose that there exist (𝑥₀,𝑦₀)∈𝑋×𝑋 such that (𝑥₀,𝑦₀), (𝑦₀,𝑥₀), (𝐹(𝑥₀,𝑦₀),𝐹(𝑦₀,𝑥₀)) and (𝐹(𝑦₀,𝑥₀),𝐹(𝑥₀,𝑦₀)) are comparable with (𝑥^∗,𝑦^∗), a mapping 𝜓∈Ψ such that 𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)𝜓(𝛿((𝑥,𝑦),(𝑢,𝑣))) for all (𝑥,𝑦) and (𝑢,𝑣) in 𝑋×𝑋 which are comparable with (𝑥^∗,𝑦^∗). Assume that (𝐹(𝑥,𝑦),𝐹(𝑦,𝑥)) and (𝐹(𝑢,𝑣),𝐹(𝑣,𝑢)) are comparable with (𝑥^∗,𝑦^∗) whenever (𝑥,𝑦) and (𝑢,𝑣) so are. Then 𝐹 has a coupled fixed point.

Proof. Define the mapping 𝛼∶𝑋²×𝑋²→[0,+∞) by 𝛼((𝑥,𝑦),(𝑢,𝑣))=1 whenever (𝑥,𝑦) and (𝑢,𝑣) are comparable with (𝑥^∗,𝑦^∗) and 𝛼((𝑥,𝑦),(𝑢,𝑣))=0 otherwise. Then by using Theorem 11, 𝐹 has a coupled fixed point.

Now, we give an application of Lemma 1. In this way, we study the nonlinear fractional differential equation 𝐷^𝛼0𝑥(𝑡)+𝐷^𝛽0𝑥(𝑡)=𝑓₀(𝑡,𝑥(𝑡)) for 𝑡∈[0,1] via the two-point boundary value condition 𝑥(0)=𝑥(1)=0, where 𝑓₀∶[0,1]×ℝ→ℝ is a continuous function and 0<𝛽₀<𝛼₀<1. Recall that the Green function associated to the equation is given by 𝐺(𝑡)=𝑡^𝛼0−1𝐸_{𝛼0−𝛽0,𝛼0}(−𝑡^{𝛼0−𝛽0}), where 𝐸_𝛼0,𝛽0 is the two-parametric Mittag-Leffler function defined by 𝐸_{(𝛼0,𝛽0)}(𝑧)=^∞_𝑘=0𝑧^𝑘Γ𝑘𝛼₀+𝛽₀ (28) for 𝛼₀>0 and 𝛽₀>0 (see [21, 22]). Let 𝑋=𝐶_ℝ([0,1]) and 𝑑𝑓,𝑔=sup_[]𝑥∈0,1𝑓(𝑥)−𝑔(𝑥). (29) Now by considering some conditions, we give the following result about existence of solution of the nonlinear fractional differential equation.

Theorem 15. Suppose that (i)there exist a function 𝜉∶ℝ²→ℝ and 𝜓∈Ψ such that 𝑓₀(𝑡,𝑎)−𝑓₀(𝑡,𝑏)≤𝛼₀𝜓(𝑑(𝑎,𝑏)) (30) for all 𝑡∈[0,1] and 𝑎,𝑏∈ℝ with 𝜉(𝑎,𝑏)≥0, (ii)there exist 𝑥₀∈𝑋 such that 𝜉(𝑥₀(𝑡),∫^𝑡₀𝐺(𝑡−𝑠)𝑓₀(𝑠,𝑥₀(𝑠))𝑑𝑠)≥0 for all 𝑡∈[0,1], for each 𝑡∈[0,1] and 𝑥,𝑦∈𝑋, 𝜉(𝑥(𝑡),𝑦(𝑡))≥0 implies 𝜉^𝑡₀𝐺(𝑡−𝑠)𝑓₀(𝑠,𝑥(𝑠)),^𝑡₀𝐺(𝑡−𝑠)𝑓₀𝑠,𝑦(𝑠)𝑑𝑠≥0, (31)(iii) if {𝑥_𝑛} is a sequence in 𝑋 with 𝑥_𝑛→𝑥 and 𝜉(𝑥_𝑛,𝑥_𝑛+1)≥0 for all 𝑛, then 𝜉(𝑥_𝑛,𝑥)≥0 for all 𝑛.Then the nonlinear fractional differential equation has at least one solution.

Proof. It is well known that 𝑥∈𝑋 is a solution of the nonlinear fractional differential equation if and only if is a solution of the integral equation 𝑥(𝑡)=^𝑡₀𝐺(𝑡−𝑠)𝑓₀(𝑠,𝑥(𝑠))𝑑𝑠 (32) for all 𝑡∈[0,1]. Define the operator 𝐹∶𝑋→𝑋 by 𝐹𝑥(𝑡)=∫^𝑡₀𝐺(𝑡−𝑠)𝑓₀(𝑠,𝑥(𝑠))𝑑𝑠 for all 𝑡∈[0,1]. Thus, for finding a solution of the the nonlinear fractional differential equation it is sufficient we find a fixed point of the continuous operator 𝐹. Let 𝑥,𝑦∈X be such that 𝜉(𝑥(𝑡),𝑦(𝑡))≥0 for all 𝑡∈[0,1]. By using (i), we get (𝑡)(𝑡=𝐹𝑥−𝐹𝑦)_𝑡0𝐺(𝑡−𝑠)𝑓₀(𝑠,𝑥(𝑠))−𝑓₀(𝑠)≤𝑠,𝑦𝑑𝑠_𝑡0|𝐺(𝑡−𝑠)|𝑓₀(𝑠,𝑥(𝑠))−𝑓₀(𝑠)≤𝑠,𝑦𝑑𝑠_𝑡0|𝐺(𝑡−𝑠)|𝛼₀𝜓𝑥(𝑠)−𝑦(s)𝑑𝑠≤𝛼₀𝜓𝑥−𝑦_∞sup_𝑡∈𝐼^𝑡₀|𝐺(𝑡−𝑠)|𝑑𝑠≤𝜓𝑥−𝑦_∞. (33) Note that, 𝐺(𝑡)=𝑡^𝛼0−1𝐸_{𝛼0−𝛽0,𝛼0}(−𝑡^{𝛼0−𝛽0})≤𝑡^𝛼0−1(1/(1+|−𝑡^{𝛼0−𝛽0}|))≤𝑡^𝛼0−1 for all 𝑡∈[0,1]. Thus, sup_𝑡∈[0,1]∫^𝑡₀𝐺(𝑡−𝑠)𝑑𝑠≤1/𝛼₀. Now, define 𝛼∶𝑋×𝑋→[0,∞) by 𝛼(𝑥,𝑦)=1 whenever 𝜉(𝑥(𝑡),𝑦(𝑡))≥0 for all 𝑡∈[0,1] and 𝛼(𝑥,𝑦)=0 otherwise. Hence, 𝛼(𝑥,𝑦)‖𝐹𝑥−𝐹𝑦‖_∞≤𝜓(‖𝑥−𝑦‖_∞) and so 𝛼𝑑𝑑(),1𝑥,𝑦𝐹𝑥,𝐹𝑦≤𝜓max𝑥,𝑦,𝑑𝑥,𝐹𝑥,𝑑𝑦,𝐹𝑦2𝑑𝑥,𝐹𝑦+𝑑𝑦,𝐹𝑥 (34) for all 𝑥,𝑦∈𝑋. Thus, 𝐹 is an 𝛼-𝜓-contractive mapping. By using (iii), 𝛼(𝑥,𝑦)≥1 implies 𝛼(𝐹𝑥,𝐹𝑦)≥1. Therefore, 𝐹 is 𝛼-admissible. From (ii), there exists 𝑥₀∈𝐶(𝐼) such that 𝛼(𝑥₀,𝐹𝑥₀)≥1. Now by using (iv) and Lemma 1, there exists 𝑥^∗∈𝑋 such that 𝐹𝑥^∗=𝑥^∗.

Similar to the work of Samet et al. [19], we say that the space 𝑋 has the property (𝐻^∗) whenever for each (𝑥,𝑦),(𝑢,𝑣)∈𝑋×𝑋 there exists (𝑧₁,𝑧₂)∈𝑋×𝑋 such that 𝛼((𝑥,𝑦),(𝑧₁,𝑧₂))≥1, 𝛼((𝑧₂,𝑧₁),(𝑦,𝑥))≥1, 𝛼((𝑢,𝑣),(𝑧₁,𝑧₂))≥1 and 𝛼((𝑧₂,𝑧₁),(𝑣,𝑢))≥1. It is easy to check that if 𝑋 has the property (𝐻^∗) in before results, then the mapping has a unique coupled fixed point.

If there is an order ≤ on 𝑋, then one can construct the order ≼ on 𝑋×𝑋 by (𝑥,𝑦)≼(𝑢,𝑣) whenever 𝑥≤𝑢 and 𝑣≤𝑦. By using this idea, it has been provided some coupled fixed point results for mixed monotone mappings on ordered metric spaces [7]. Let (𝑋,≼) be a partially ordered set and 𝐹∶𝑋×𝑋→𝑋 a mapping. We say that 𝐹 has the mixed monotone property whenever 𝑥₁≼𝑥₂ implies 𝐹(𝑥₁,𝑦)≼𝐹(𝑥₂,𝑦) and 𝑦₁≼𝑦₂ implies 𝐹(𝑥,𝑦₂)≼𝐹(𝑥,𝑦₁) for all 𝑥,𝑦∈𝑋 [7]. Thus by considering the provided corollaries and the explanations in the text-body, we can obtain some similar coupled fixed point results for mixed monotone mappings as special case of above results. Finally, it has been published interesting fixed point results on metric spaces with a graph (see e.g., [23–26]). There is a connection between fixed point results on ordered metric spaces and fixed point results on metric spaces with a graph [23, 27]. It is notable that one can get some similar coupled fixed point results on metric spaces with a graph.

Acknowledgments

The authors express their gratitude to the referees for their helpful suggestions which improved final version of this paper.