Generalization of Some Coupled Fixed Point Results on Partial Metric Spaces

Abstract

Using the setting of partial metric spaces, we prove some coupled fixed point results. Our results generalize several well-known comparable results of H. Aydi (2011). Also, we introduce an example to support our results.

1. Introduction and Preliminaries

The notion of coupled fixed point of a mapping 𝐹∶𝑋×𝑋→𝑋 was introduced by Gnana Bhaskar and Lakshmikantham in [1]. Later on, many authors investigated many coupled fixed point results in different spaces such as usual metric spaces, fuzzy metric spaces, generalized metric spaces, partial metric spaces, and partially ordered metric spaces (see [1–20]).

Definition 1.1 (see [1]). An element (𝑥,𝑦)∈𝑋×𝑋 is called a coupled fixed point of a mapping 𝐹∶𝑋×𝑋→𝑋 if 𝐹𝑥,𝑦=𝑥,𝐹𝑦,𝑥=𝑦. (1.1)

Matthews [21] in 1994 introduced the notion of partial metric spaces in such a way that each object does not necessarily have to have a zero distance from itself. Consistent with Matthews [21], the following definitions and results will be needed in the sequel.

Definition 1.2 (see [21]). A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→ℝ⁺ such that for all 𝑥,𝑦,𝑧∈𝑋:(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 nonempty set and 𝑝 is a partial metric on 𝑋.

Each partial metric 𝑝 on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋. The set {𝐵_𝑝(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝑝(𝑥,𝜀)={𝑦∈𝑋∶𝑝(𝑥,𝑦)<𝑝(𝑥,𝑥)+𝜀} for all 𝑥∈𝑋 and 𝜀>0, forms the base of 𝜏_𝑝.

If 𝑝 is a partial metric on 𝑋, then the function 𝑝^𝑠∶𝑋×𝑋→ℝ⁺ given by 𝑝^𝑠𝑥,𝑦=2𝑝𝑥,𝑦−𝑝(𝑥,𝑥)−𝑝𝑦,𝑦 (1.2) is a metric on 𝑋.

Definition 1.3 (see [21]). Let (𝑋,𝑝) be a partial metric space. Then:(1)a sequence (𝑥_𝑛) in a partial metric space (𝑋,𝑝) converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥,𝑥_𝑛),(2)a sequence (𝑥_𝑛) in a partial metric space (𝑋,𝑝) is called a Cauchy sequence if there exists (and is finite) lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚),(3)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 that 𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).

Lemma 1.4 (see [21]). Let (𝑋,𝑝) be a partial metric space.(1)(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).(2)A partial metric space (𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑝^𝑠) is complete. 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. (1.3)

Abdeljawad et al. [22–24], Altun et al. [25], Karapinar and Erhan [26–28], Oltra and Valero [29] and Romaguera [30] studied fixed point theorems in partial metric spaces. For more works in partial metric spaces, we refer the reader to [31–40].

Aydi [2] proved the following coupled fixed point theorems in partial metric spaces.

Theorem 1.5. Let (𝑋,𝑝) be a complete partial metric space. Suppose that the mapping 𝐹∶𝑋×𝑋→𝑋 satisfies the following contractive condition for all 𝑥,𝑦,𝑢,𝑣∈𝑋: 𝑝𝐹𝑥,𝑦,𝐹(𝑢,𝑣)≤𝑘𝑝(𝑥,𝑢)+𝑙𝑝𝑦,𝑣, (1.4) where 𝑘,𝑙 are nonnegative constants with 𝑘+𝑙<1. Then 𝐹 has a unique coupled fixed point.

Theorem 1.6. Let (𝑋,𝑝) be a complete partial metric space. Suppose that the mapping 𝐹∶𝑋×𝑋→𝑋 satisfies the following contractive condition for all 𝑥,𝑦,𝑢,𝑣∈𝑋: 𝑝𝐹𝑥,𝑦,𝐹(𝑢,𝑣)≤𝑘𝑝𝐹𝑥,𝑦,𝑥+𝑙𝑝(𝐹(𝑢,𝑣),𝑢), (1.5) where 𝑘,𝑙 are nonnegative constants with 𝑘+𝑙<1. Then 𝐹 has a unique coupled fixed point.

In this paper, we prove some coupled fixed point results. Our results generalize Theorems 1.5 and 1.6. Also, we introduce an example to support our results.

2. The Main Result

Theorem 2.1. Let (𝑋,𝑝) be a complete partial metric space. Suppose that the mapping 𝐹∶𝑋×𝑋→𝑋 satisfies 𝑝𝐹𝑥,𝑦,𝐹(𝑢,𝑣)≤𝑟max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝𝐹𝑥,𝑦,𝑥,𝑝(𝐹(𝑢,𝑣),𝑢), (2.1) for all 𝑥,𝑦,𝑢,𝑣∈𝑋. If 𝑟∈[0,1), then 𝐹 has a unique coupled fixed point.

Proof. Choose 𝑥₀,𝑦₀∈𝑋. Let 𝑥₁=𝐹(𝑥₀,𝑦₀) and 𝑦₁=𝐹(𝑦₀,𝑥₀). Again let 𝑥₂=𝐹(𝑥₁,𝑦₁) and 𝑦₂=𝐹(𝑦₁,𝑥₁). By continuing in the same way, we construct two sequences (𝑥_𝑛) and (𝑦_𝑛) in 𝑋 such that 𝑥_𝑛+1=𝐹𝑥_𝑛,𝑦_𝑛𝑦,𝑛=0,1,2,3,…,_𝑛+1=𝐹𝑦_𝑛,𝑥_𝑛,𝑛=0,1,2,3,…. (2.2) Then by (2.1), we have 𝑝𝑥_𝑛+1,𝑥_𝑛+2=𝑝𝐹𝑥_𝑛,𝑦_𝑛,𝐹𝑥_𝑛+1,𝑦_𝑛+1≤𝑟max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝐹𝑥_𝑛,𝑦_𝑛,𝑥_𝑛,𝑝𝐹𝑥_𝑛+1,𝑦_𝑛+1,𝑥_𝑛+1≤𝑟max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝑥_𝑛+1,𝑥_𝑛,𝑝𝑥_𝑛+2,𝑥_𝑛+1≤𝑟max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1𝑝𝑦_𝑛+1,𝑦_𝑛+2=𝑝𝐹𝑦_𝑛,𝑥_𝑛,𝐹𝑦_𝑛+1,𝑥_𝑛+1≤𝑟max𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝐹𝑦_𝑛,𝑥_𝑛,𝑦_𝑛,𝑝𝐹𝑦_𝑛+1,𝑥_𝑛+1,𝑦_𝑛+1}≤𝑟max𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛+1,𝑦_𝑛,𝑝𝑦_𝑛+2,𝑦_𝑛+1≤𝑟max𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝑥_𝑛,𝑥_𝑛+1. (2.3) Thus from (2.3), we have max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1≤𝑟max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑦_𝑛−1,𝑦_𝑛. (2.4) By repeating (2.4) 𝑛-times, we get that max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1≤𝑟max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑦_𝑛−1,𝑦_𝑛≤𝑟²max𝑝𝑥_𝑛−2,𝑥_𝑛−1,𝑝𝑦_𝑛−2,𝑦_𝑛−1⋮≤𝑟^𝑛max𝑝𝑥₀,𝑥₁,𝑝𝑦₀,𝑦₁. (2.5) Letting 𝑛→+∞ in (2.5), we get that lim_𝑛→+∞max𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑦_𝑛,𝑦_𝑛+1=0. (2.6) Therefore, we have lim_𝑛→+∞𝑝𝑥_𝑛,𝑥_𝑛+1=0,lim_𝑛→+∞𝑝𝑦_𝑛,𝑦_𝑛+1=0. (2.7) For 𝑛,𝑚∈ℕ with 𝑚>𝑛, we have 𝑝𝑥_𝑛,𝑥_𝑚≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑚−𝑝𝑥_𝑛+1,𝑥_𝑛+1≤𝑝𝑥_𝑛,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥_𝑛+2+𝑝𝑥_𝑛+2,𝑥_𝑚−𝑝𝑥_𝑛+1,𝑥_𝑛+1−𝑝𝑥_𝑛+2,𝑥_𝑛+2⋮≤^𝑚−1_𝑖=𝑛𝑝𝑥_𝑖,𝑥_𝑖+1−^𝑚−2_𝑖=𝑛𝑝𝑥_𝑖+1,𝑥_𝑖+1≤^𝑚−1_𝑖=𝑛𝑝𝑥_𝑖,𝑥_𝑖+1. (2.8) By (2.5) and (2.8), we have 𝑝𝑥_𝑛,𝑥_𝑚≤^𝑚−1_𝑖=𝑛𝑟^𝑖max𝑝𝑥₀,𝑥₁,𝑝𝑦₀,𝑦₁≤^+∞_𝑖=𝑛𝑟^𝑖max𝑝𝑥₀,𝑥₁,𝑝𝑦₀,𝑦₁=𝑟^𝑛𝑝𝑥1−𝑟max₀,𝑥₁,𝑝𝑦₀,𝑦₁. (2.9) Letting 𝑛,𝑚→+∞ in (2.9), we get that lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚=0. (2.10) Thus lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists and is finite. Hence (𝑥_𝑛) is a Cauchy sequence in (𝑋,𝑝). Similarly, we may show that lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚=0, (2.11) and hence (𝑦_𝑛) is a Cauchy sequence in (𝑋,𝑝). By Lemma 1.4 there exist 𝑥,𝑦∈𝑋 such that lim_𝑛→∞𝑝^𝑠(𝑥_𝑛,𝑥)=0 (resp., lim_𝑛→∞𝑝^𝑠(𝑦_𝑛,𝑦)=0) if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝𝑥_𝑛,𝑥=lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚=0,resp.,𝑝𝑦,𝑦=lim_𝑛→∞𝑝𝑦_𝑛,𝑦=lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚=0. (2.12) Now, we prove that 𝑥=𝐹(𝑥,𝑦). By (2.1), we have 𝑝𝐹𝑥,𝑦,𝑥≤𝑝𝐹𝑥,𝑦,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑥−𝑝𝑥_𝑛+1,𝑥_𝑛+1≤𝑝𝐹𝑥,𝑦,𝑥_𝑛+1+𝑝𝑥_𝑛+1𝐹𝑥,𝑥≤𝑝𝑥,𝑦,𝐹_𝑛,𝑦_𝑛+𝑝𝑥_𝑛+1𝑝,𝑥≤𝑟max𝑥,𝑥_𝑛,𝑝𝑦,𝑦_𝑛,𝑝𝐹𝑥,𝑦,𝑥,𝑝𝐹𝑥_𝑛,𝑦_𝑛,𝑥_𝑛+𝑝𝑥_𝑛+1𝑝,𝑥=𝑟max𝑥,𝑥_𝑛,𝑝𝑦,𝑦_𝑛,𝑝𝐹𝑥,𝑦,𝑥,𝑝𝑥_𝑛+1,𝑥_𝑛+𝑝𝑥_𝑛+1,𝑥. (2.13) Letting 𝑛→∞ in the above inequality and using (2.12), we get that 𝑝𝐹𝑥,𝑦,𝑥≤𝑟𝑝𝐹𝑥,𝑦,𝑥. (2.14)
Since 𝑟∈[0,1), we conclude that 𝑝(𝐹(𝑥,𝑦),𝑥)=0. By (p₁) and (p₂), we have 𝐹(𝑥,𝑦)=𝑥. Similarly, we may show that 𝐹(𝑦,𝑥)=𝑦. Thus (𝑥,𝑦) is a coupled fixed point of 𝐹. To prove the uniqueness of the fixed point, we let (𝑢,𝑣) be a coupled fixed point of 𝐹. We will show that 𝑥=𝑢 and 𝑦=𝑣. By (2.1), we have 𝑝()𝐹()𝑝()𝐹(𝐹()𝑝()()(.𝑥,𝑢=𝑝𝑥,𝑦,𝐹𝑢,𝑣≤𝑟max𝑥,𝑢,𝑝𝑥,𝑦,𝑥,𝑝𝑦,𝑣,𝑝𝑢,𝑣,𝑢)=𝑟max𝑥,𝑢,𝑝𝑦,𝑣,𝑝𝑥,𝑥,𝑝𝑢,𝑢) (2.15) Since 𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑢) and 𝑝(𝑢,𝑢)≤𝑝(𝑥,𝑢), we have 𝑝(𝑥,𝑢)≤𝑟max𝑝(𝑥,𝑢),𝑝𝑦,𝑣. (2.16) Also, from (2.1), we have 𝑝𝐹()𝑝𝐹()(𝐹()𝑝()(.𝑦,𝑣=𝑝𝑦,𝑥,𝐹𝑣,𝑢≤𝑟max𝑦,𝑣,𝑝𝑦,𝑥,𝑦,𝑝𝑥,𝑢,𝑝𝑣,𝑢,𝑣)=𝑟max𝑥,𝑢,𝑝𝑦,𝑣,𝑝𝑦,𝑦,𝑝𝑣,𝑣) (2.17) Since 𝑝(𝑦,𝑦)≤𝑝(𝑦,𝑣) and 𝑝(𝑣,𝑣)≤𝑝(𝑦,𝑣), we have 𝑝𝑦,𝑣≤𝑟max𝑝(𝑥,𝑢),𝑝𝑦,𝑣. (2.18) From (2.16) and (2.18), we have max𝑝(𝑥,𝑢),𝑝𝑦,𝑣≤𝑟max𝑝(𝑥,𝑢),𝑝𝑦,𝑣. (2.19) Since 𝑟<1, we have max{𝑝(𝑥,𝑢),𝑝(𝑦,𝑣)}=0. Hence 𝑝(𝑥,𝑢)=0 and 𝑝(𝑦,𝑣)=0. By (p₁) and (p₂), we have 𝑥=𝑢 and 𝑦=𝑣.

Corollary 2.2. Let (𝑋,𝑝) be a complete partial metric space. Suppose that there are 𝑎,𝑏,𝑐,𝑑∈[0,1) with 𝑎+𝑏+𝑐+𝑑<1 such that the mapping 𝐹∶𝑋×𝑋→𝑋 satisfies 𝑝𝐹𝑥,𝑦,𝐹(𝑢,𝑣)≤𝑎𝑝(𝑥,𝑢)+𝑏𝑝𝑦,𝑣+𝑐𝑝𝐹𝑥,𝑦,𝑥+𝑑𝑝(𝐹(𝑢,𝑣),𝑢) (2.20) for all 𝑥,𝑦,𝑢,𝑣∈𝑋. Then 𝐹 has a unique coupled fixed point.

Proof. The proof follows from Theorem 2.1 by noting that: ()𝐹(𝐹())≤()𝑝()𝐹(𝐹().𝑎𝑝𝑥,𝑢+𝑏𝑝𝑦,𝑣+𝑐𝑝𝑥,𝑦,𝑥+𝑑𝑝𝑢,𝑣,𝑢𝑎+𝑏+𝑐+𝑑max𝑥,𝑢,𝑝𝑦,𝑣,𝑝𝑥,𝑦,𝑥,𝑝𝑢,𝑣,𝑢) (2.21)

Remarks.(1)Theorem 1.5 [2, Theorem 2.1] is a special case of Corollary 2.2.(2)[2, Corollary 2.2] is a special case of Corollary 2.2.(3)Theorem 1.6 [2, Theorem 2.4] is a special case of Corollary 2.2.(4)[2, Corollary 2.6] is a special case of Corollary 2.2.Now, we introduce an example satisfying the hypotheses of Theorem 2.1 but not the hypotheses of Theorems 2.1 and 2.4 of [2].

Example 2.3. Define 𝑝∶[0,1]×[0,1]→[0,1] by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. Then ([0,1],𝑝) is a complete partial metric space. Let 𝐹∶[0,1]×[0,1]→[0,1] be the mapping defined by 𝐹𝑥,𝑦=𝑥−𝑦2. (2.22) Then,(a)𝑝(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤(1/2)max{𝑝(𝑥,𝑢),𝑝(𝑦,𝑣),𝑝(𝐹(𝑥,𝑦),𝑥),𝑝(𝐹(𝑢,𝑣),𝑢)} for all 𝑥,𝑦,𝑢,𝑣∈[0,1].(b) There are no 𝑎,𝑏∈[0,1) with 𝑎+𝑏<1 such that 𝑝(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤𝑎𝑝(𝑥,𝑢)+𝑏𝑝(𝑦,𝑣) for all 𝑥,𝑦,𝑢,𝑣∈[0,1].(c) There are no 𝑎,𝑏∈[0,1) with 𝑎+𝑏<1 such that 𝑝(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤𝑎𝑝(𝐹(𝑥,𝑦),𝑥)+𝑏𝑝(𝐹(𝑢,𝑣),𝑢) for all 𝑥,𝑦,𝑢,𝑣∈[0,1].

Proof. To prove part (a), given 𝑥,𝑦,𝑣,𝑢∈[0,1]. Then: 𝑝𝐹𝑥,𝑦,𝐹(𝑢,𝑣)=max𝑥−𝑦2,|𝑢−𝑣|2=12,|=1max𝑥−𝑦𝑢−𝑣|2≤1max𝑥−𝑦,𝑦−𝑥,𝑢−𝑣,𝑣−𝑢2=1max𝑥,𝑦,𝑢,𝑣2𝑝()≤1max𝑥,𝑢,𝑝𝑦,𝑣2max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝𝐹𝑥,𝑦,𝑥,𝑝(𝐹(𝑢,𝑣),𝑢). (2.23) To prove part (b), suppose that there are 𝑎,𝑏∈[0,1) with 𝑎+𝑏<1 such that 𝑝(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤𝑎𝑝(𝑥,𝑢)+𝑏𝑝(𝑦,𝑣) for all 𝑥,𝑦,𝑢,𝑣∈[0,1].
Since 𝑝(𝐹()(11,0,𝐹0,0))=𝑝2=1,02()()𝑝(𝐹()(1≤𝑎𝑝1,0+𝑏𝑝0,0=𝑎,0,1,𝐹0,0))=𝑝2=1,02≤𝑎𝑝(0,0)+𝑏𝑝(1,0)=𝑏, (2.24) we have 𝑎+𝑏≥1, which is a contradiction.
To prove part (c), suppose that there are 𝑎,𝑏∈[0,1) with 𝑎+𝑏<1 such that 𝑝(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))≤𝑎𝑝(𝐹(𝑥,𝑦),𝑥)+𝑏𝑝(𝐹(𝑢,𝑣),𝑢) for all 𝑥,𝑦,𝑢,𝑣∈[0,1].
Since 𝑝(𝐹()(11,0,𝐹0,0))=𝑝2=1,02(𝐹())(𝐹())1≤𝑎𝑝1,0,1+𝑏𝑝0,0,0=𝑎𝑝2()𝑝(𝐹()(1,1+𝑏𝑝0,0=𝑎0,0,𝐹1,0))=𝑝0,2=12(𝐹())(𝐹())()1≤𝑎𝑝0,0,0+𝑏𝑝1,0,1=𝑎𝑝0,0+𝑏𝑝2,1=𝑏, (2.25) we have 𝑎+𝑏≥1, which is a contradiction.

Thus by Theorem 2.1, 𝐹 has a unique coupled fixed point. Here, (0,0) is the unique fixed point of 𝐹.

Acknowledgments

The authors thank the editor and the referees for their valuable comments and suggestions.