Suzuki-Type Generalization of Chatterjea Contraction Mappings on Complete Partial Metric Spaces

Abstract

Motivated by Suzuki (2008), we prove a Suzuki-type fixed point theorem employing Chatterjea contraction on partial metric spaces.

1. Introduction and Preliminaries

Historically, the idea of a complete metric space has interesting and important applications in classical analysis especially in existence and uniqueness theories on one hand while on the other hand Banach fixed point theorem [1] is one of the most useful results in nonlinear analysis. In the recent past, many authors considered the equivalence of existence results on fixed points (of mappings) by proving suitable equivalence theorems ascertaining the completeness of the underlying metric space. Kirk [2] proved that a metric space (𝑋,𝑑) is complete if and only if every Caristi type mapping has a unique fixed point. Similarly, Subrahmanyam [3] showed that a metric space (𝑋,𝑑) is complete if and only if every Kannan mapping has a unique fixed point. However, Banach contraction condition does not characterize the metric completeness of the underlying space. Indeed, Connell [4] gave an example of an incomplete metric space on which every contraction map has a fixed point. Despite this fact, Suzuki obtained Banach contraction principle that characterizes the metric completeness of the space using a different type of contraction. Thereafter, many authors proved different generalizations by proving different types of fixed point theorems in complete metric space, for example, Suzuki [5] proved Kannan version of a Suzuki-type generalized result wherein authors discussed contraction mappings and Kannan mappings from a different point of view while Popescu [6] attempted Chatterjea version on complete metric space. In this continuation, Altun and Erduran proved a Suzuki-type fixed point theorem using an implicit function on complete metric space [7] wherein authors obtained unified and generalized results of Suzuki-type. Finally, Kikkawa and Suzuki proved multivalued version of Suzuki-type results which generalize classical results of Markin [8] and Nadler [9]. For further details on this theme, one can be referred to [7, 10–15].

Evidently, contractions are always continuous, but Kannan mappings are not necessarily continuous. Also, one may notice that a contraction mapping is not necessarily Kannan mapping, and a Kannan mapping is not essentially a contraction mapping so that both conditions are not essentially comparable but Kannan contractions are relatively stronger than Banach contraction in certain sense. However, Chatterjea contraction is obtained by interchanging the roles of variables 𝑥, 𝑦 in Kannan contraction. Therefore, Chatterjea contraction is similar to Kannan contraction enabling one to infer that the mappings satisfying a Kannan contraction also satisfy the corresponding Chatterjea contraction.

In 1992, Matthews [16] introduced partial metric spaces wherein the distance of a point from itself may not always be zero. After introducing the idea of partial metric spaces, Matthews proved the partial metric version of Banach fixed point theorem. Thereafter, many authors further studied partial metric spaces and their topological properties (e.g., [17–22]) which are followed by Valero [23], Oltra and Valero [24], and Altun et al. [25] wherein authors obtained some generalizations of the core result of Matthews [16]. In a noted paper, Romaguera [26] extended Kirk's characterization of metric completeness [2] to partial metric spaces.

Recently, many authors proved several fixed point results in partial metric spaces. Out of such results, Suzuki-type (cf. [5]) generalizations are noted ones. Very recently, Paesano and Vetro [27] proved an analogous fixed point result for a self-mapping defined on a partial metric space (also on a partially ordered metric space) which generalizes certain noted results of Ran and Reurings [26] besides offering a characterization of partial metric 0-completeness in terms of fixed points. This result can be viewed as an extension of Suzuki-type (cf. [5]) characterization of metric completeness.

Inspired by Paesano and Vetro [27], we prove Suzuki-type fixed point result for Chatterjea contraction mappings in 0-complete partial metric spaces wherein we show that constant 𝜓(𝑟) is the best for every 𝑟 which substantiates the genuineness and utility of our result.

In what follows, we give some relevant definitions and results from the existing literature which are relevant to our subsequent discussion.

Definition 1 (see [16]). A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→𝑅⁺ such that for all 𝑥, 𝑦, 𝑧∈𝑋: (i) 𝑥=𝑦⇔𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦), (ii) 𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦), (iii) 𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥), (iv) 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 whose base is the family of open 𝑝-balls {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.

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) is a metric on 𝑋.

Definition 2. A mapping 𝑓∶𝑋→𝑋 is said to be continuous at 𝑥₀∈𝑋, if for every 𝜀>0, there exists 𝛿>0 such that 𝑓(𝐵_𝑝(𝑥₀,𝛿))⊂𝐵_𝑝(𝑓(𝑥₀),𝜀).

Definition 3 (see [16]). Let (𝑋,𝑝) be a partial metric space and {𝑥_𝑛} a sequence in 𝑋. Then,(i){𝑥_𝑛} converges to a point 𝑥∈𝑋 if 𝑝(𝑥,𝑥)=lim_𝑛→+∞𝑝(𝑥,𝑥_𝑛);(ii){𝑥_𝑛} is called a Cauchy sequence if lim_{𝑛,𝑚→+∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists and is finite.

Moreover, if lim_{𝑚,𝑛→+∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0, then {𝑥_𝑛} is said to be a 0-Cauchy sequence in (𝑋,𝑝).

Definition 4. 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).

A partial metric space (𝑋,𝑝) is called 0-complete [26] if every 0-Cauchy sequence {𝑥_𝑛} in 𝑋 converges to some 𝑥∈𝑋 (with respect to 𝜏_𝑝) such that 𝑝(𝑥,𝑥)=0. Therefore, (𝑋,𝑝) is 0-complete if and only if every 0-Cauchy sequence converges with respect to 𝜏_𝑝𝑠. It is clear that every 0-Cauchy sequence in (𝑋,𝑝) is a Cauchy sequence in (𝑋,𝑝). Therefore, if (𝑋,𝑝) is complete, then it is 0-complete but not conversely (e.g., [26]).

In general, 𝑝(𝑥,𝑦) is not a continuous function in two variables, in the sense that 𝑥_𝑛→𝑥 and 𝑦_𝑛→𝑦 (in 𝜏_𝑝) imply that 𝑝(𝑥_𝑛,𝑦_𝑛)→𝑝(𝑥,𝑦), as 𝑛→+∞. However, the following holds:

Lemma 5. Let (𝑋,𝑝) be a partial metric space and {𝑥_𝑛}⊂𝑋. If 𝑥_𝑛→𝑧 as 𝑛→∞ and 𝑝(𝑧,𝑧)=0, then lim_𝑛→∞𝑝𝑥_𝑛,𝑥=𝑝(𝑧,𝑥). (2)

Lemma 6 (see [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)(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. (3)

Before proving our main result (i.e., Theorem 8), we state the following lemma available in [5, 13] for standard metric spaces. As the proof of the lemma is identical for the partial metrices, therefore, we merely state the lemma (for partial metric spaces).

Lemma 7. Let (𝑋,𝑝) be a partial metric space and let 𝑇 be mapping on 𝑋. Let 𝑥∈𝑋 satisfies 𝑝(𝑇𝑥,𝑇²𝑥)≤𝑟𝑝(𝑥,𝑇𝑥) for some 𝑟∈[0,1). Then for 𝑦∈𝑋, either 1𝑝()11+𝑟𝑥,𝑇𝑥≤𝑝𝑥,𝑦or𝑝1+𝑟𝑇𝑥,𝑇²𝑥≤𝑝𝑇𝑥,𝑦 (4) holds.

Now, we prove our main result as follows.

Theorem 8. Define a nonincreasing function 𝜓 from [0,1) to (1/2,1] by 𝜓(𝑟)=11,0≤𝑟<√2,1,11+𝑟√2≤𝑟<1. (5) Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇 a self-mapping defined on 𝑋. Let 𝛾∈[0,1/2) and put 𝜆∶=𝛾/(1−𝛾)∈[0,1). If 𝜓(𝑟)𝑝()𝑝𝑥,𝑇𝑥≤𝑝𝑥,𝑦⟹𝑝𝑇𝑥,𝑇𝑦≤𝛾𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥 (6) for all 𝑥,𝑦∈𝑋, then 𝑇 has a unique fixed point.

Proof. Since 𝜓(𝑟)≤1, 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)≤𝑝(𝑥,𝑇𝑥) holds. In view of one of the hypotheses, we can write the following: 𝑝𝑇𝑥,𝑇²𝑥≤𝛾𝑝𝑥,𝑇²𝑥(𝑝()+𝑝𝑇𝑥,𝑇𝑥)≤𝛾𝑥,𝑇𝑥+𝑝𝑇𝑥,𝑇²𝑥 (7) so that 𝑝𝑇𝑥,𝑇²𝑥≤𝛾𝑝()1−𝛾𝑥,𝑇𝑥 (8) for 𝑥∈𝑋. Now, fix 𝑢₀∈𝑋 and define a sequence {𝑢_𝑛} in 𝑋 by 𝑢_𝑛=𝑇^𝑛𝑢₀=𝑇𝑢_𝑛−1. On using (8), we can have the following: 𝑝𝑢_𝑛,𝑢_𝑛+1=𝑝𝑇^𝑛𝑢₀,𝑇^𝑛+1𝑢₀≤𝜆𝑝𝑇^𝑛−1𝑢₀,𝑇^𝑛𝑢₀⋮≤𝜆^𝑛𝑝𝑢₀,𝑇𝑢₀ (9) so that 𝑝𝑢_𝑛,𝑢_𝑚≤𝜆^𝑛𝑝𝑢1−𝜆⁰,𝑇𝑢₀()⟶0as𝑛⟶∞since0≤𝜆<1 (10) for all 𝑛,𝑚∈ℕ and so {𝑢_𝑛} is a 0-Cauchy sequence. Since (𝑋,𝑝) is 0-complete, {𝑢_𝑛} converges to some point 𝑢∈𝑋; that is, lim_𝑛→∞𝑝𝑢_𝑛,𝑢=𝑝(𝑢,𝑢). (11) Now, we show that 𝑝(𝑢,𝑇𝑥)≤𝜆𝑝(𝑥,𝑇𝑥)∀𝑥∈𝑋⧵{𝑢}. (12) Since 𝑢_𝑛→𝑢, there exists 𝑛₀∈ℕ such that 𝑝(𝑢_𝑛,𝑢)≤𝑝(𝑢,𝑥)/3 for all 𝑛≥𝑛₀. Owing to 𝑝(𝑥,𝑢)≤𝑝(𝑥,𝑢_𝑛)+𝑝(𝑢_𝑛,𝑢) and 𝑝(𝑢_𝑛,𝑢_𝑛+1)≤𝑝(𝑢_𝑛,𝑢)+𝑝(𝑢_𝑛+1,𝑢), for all 𝑛≥𝑛₀, we obtain the following: 𝜓(𝑟)𝑝𝑢_𝑛,𝑇𝑢_𝑛≤𝑝𝑢_𝑛,𝑇𝑢_𝑛=𝑝𝑢_𝑛,𝑢_𝑛+1≤𝑝𝑢_𝑛,𝑢+𝑝𝑢_𝑛+1≤2,𝑢3𝑝()()−𝑝()𝑢,𝑥=𝑝𝑢,𝑥𝑢,𝑥3≤𝑝(𝑢,𝑥)−𝑝𝑢_𝑛𝑢,𝑢≤𝑝_𝑛,𝑥, (13) and henceforth 𝑝𝑇𝑢_𝑛,𝑇𝑥≤𝛾𝑝𝑢_𝑛,𝑇𝑥+𝑝𝑥,𝑇𝑢_𝑛∀𝑛≥𝑛₀. (14) Now, 𝑝(𝑢,𝑇𝑥)≤lim_𝑛→∞𝑝𝑢_𝑛+1,𝑇𝑥=lim_𝑛→∞𝑝𝑇𝑢_𝑛,𝑇𝑥≤lim_𝑛→∞𝛾𝑝𝑢_𝑛,𝑇𝑥+𝑝𝑥,𝑇𝑢_𝑛=𝜆𝑝(𝑥,𝑇𝑥)owingtoLemma5 (15) for 𝑥∈𝑋 with 𝑥≠𝑢. Let us prove that 𝑢 is a fixed point of 𝑇. In case 0≤𝑟<1/√2, let on contrary that 𝑇𝑢≠𝑢. Then owing to (12), 𝑝𝑢,𝑇²𝑢≤𝜆𝑝𝑇𝑢,𝑇²𝑢≤𝛾𝜆𝑝(𝑢,𝑇𝑢) (16) so that 𝑝(𝑢,𝑇𝑢)≤𝑝𝑢,𝑇²𝑢+𝑝𝑇²()()≤𝑢,𝑇𝑢≤𝛾𝜆𝑝𝑢,𝑇𝑢+𝛾𝑝𝑢,𝑇𝑢𝛾+2𝛾²𝑝()<1+𝛾𝑢,𝑇𝑢1+𝛾𝑝()(),𝛾+1𝑢,𝑇𝑢<𝑝𝑢,𝑇𝑢 (17) which is a contradiction. Therefore, we have 𝑇𝑢=𝑢. In case 1/√2≤𝑟<1, using Lemma 7, we have the following: 𝜓(𝑟)𝑝𝑢_2𝑛,𝑢_2𝑛+1≤𝑝𝑢_2𝑛(𝑟)𝑝𝑢,𝑢or𝜓_2𝑛+1,𝑢_2𝑛+2≤𝑝𝑢_2𝑛+1,𝑢 (18) for 𝑛∈ℕ. Thus, there exists a subsequence {𝑛_𝑗} of {𝑛} such that 𝜓(𝑟)𝑝𝑢_𝑛𝑗,𝑢_𝑛𝑗+1≤𝑝𝑢_𝑛𝑗,𝑢 (19) for 𝑗∈ℕ. On using one of our assumption, we have the following: 𝑝(𝑢,𝑇𝑢)=lim_𝑗→∞𝑝𝑢_𝑛𝑗+1,𝑇𝑢≤lim_𝑗→∞𝛾𝑝𝑢_𝑛𝑗,𝑇𝑢+𝑝𝑢,𝑇𝑢_𝑛𝑗=𝛾𝑝(𝑢,𝑇𝑢). (20) As 𝛾<1/2, we have 𝑇𝑢=𝑢. The uniqueness of fixed point is obvious in view of (12). This concludes the proof.

Notice that, in the course of proving Theorem 9 below, we have shown the following: (i) (𝑋,𝑝) is a complete partial metric space, (ii) 𝑇 satisfies condition (21) but has no fixed point.

The following theorem shows that 𝜓(𝑟) is the best constant for every value of 𝑟.

Theorem 9. Define a function 𝜓 as in Theorem 8. For every 𝛾∈[0,1/2), setting 𝜆∶=𝛾/(1−𝛾), there exists a 0-complete partial metric space (𝑋,𝑝) and a mapping 𝑇 on 𝑋 such that 𝑇 has no fixed point but 𝜓(𝑟)𝑝()𝑝𝑥,𝑇𝑥<𝑝𝑥,𝑦⟹𝑝𝑇𝑥,𝑇𝑦≤𝛾𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥 (21) for all 𝑥,𝑦∈𝑋.

Proof. Let 𝑋=[0,1]∪[2,3] and define a complete partial metric 𝑝 on 𝑋 by the following: 𝑝=,⊂[],,∩[]𝑥,𝑦𝑥−𝑦𝑥,𝑦0,1max𝑥,𝑦𝑥,𝑦2,3≠∅. (22) Then (𝑋,𝑝) is obviously 0-complete. Define a mapping 𝑇 on 𝑋 by the following: 3𝑇𝑥=1,0≤𝑥<1,41,𝑥=1,4,2≤𝑥≤3. (23) Notice that 𝑇 has no fixed point.
Firstly if 0≤𝑟<1/√2, then the following cases arise.(i) If 𝑥=𝑦=0, 𝑥=𝑦=1, 𝑥=𝑦=2, and 𝑥=𝑦=3, then 𝑇𝑥=𝑇𝑦. Hence, the right-hand side of (21) trivially holds so that implication is also verified.(ii) If 𝑥=1, 𝑦=0, then 𝑇𝑥=3/4, 𝑇𝑦=1 and hence the left-hand side of (21) holds good as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=1/4<1=𝑝(𝑥,𝑦). Also, the right-hand side of (21) is satisfied in view of the following: 𝑝()=1𝑇𝑥,𝑇𝑦=𝑝𝑇1,𝑇04𝑝=3≤𝛾𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥4𝛾 (24) as there exists such a 𝛾∈[0,1/√2).(iii) If 𝑥=0, 𝑦=1, then 𝑇𝑥=1, 𝑇𝑦=3/4 and hence the right-hand side of (21) trivially holds so that implication is also realized.(iv) If {𝑥,𝑦}⊂(0,1), then 𝑇𝑥=𝑇𝑦=1 so that the left-hand side of (21) holds good as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=|𝑥−1|<|𝑥−𝑦|=𝑝(𝑥,𝑦) (as 𝑦<1). The right-hand side of (21) also remains true as 𝑝(𝑇𝑥,𝑇𝑦)=0≤𝛾[𝑝(𝑥,𝑇𝑦)+𝑝(𝑦,𝑇𝑥)].(v) If {𝑥,𝑦}∩(2,3)≠∅, then 𝑇𝑥=𝑇𝑦=1/4, wherein we examine two cases. If 𝑥≥𝑦, then the right-hand side of (21) trivially holds and hence so does the implication. If 𝑥<𝑦, then the left-hand side of (21) remains true as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=𝑥<𝑝(𝑥,𝑦). The right-hand side of (21) also holds good as 𝑝𝑇𝑥,𝑇𝑦=0≤𝛾𝑝𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥. (25)
Secondly, if 1/√2≤𝑟<1, then we distinguish the following cases.(i) If 𝑥=𝑦=0, 𝑥=𝑦=1, 𝑥=𝑦=2, and 𝑥=𝑦=3, then 𝑇𝑥=𝑇𝑦. Hence, the right-hand side of (21) as well as implication remains true.(ii) If {𝑥,𝑦}⊂(0,1), then 𝑇𝑥=𝑇𝑦=1,  𝑝(𝑥,𝑇𝑥)=|𝑥−1|. The left-hand side of (21) holds good as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=(1/(𝑟+1))|𝑥−1|<|𝑥−𝑦|. The right-hand side of (21) remains true as 𝑝𝑇𝑥,𝑇𝑦≤𝛾𝑝𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥. (26)(iii) If 𝑥=0, 𝑦=1, then 𝑇𝑥=1, 𝑇𝑦=3/4. The left-hand side of (21) remains true as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=1/(𝑟+1)<1=𝑝(𝑥,𝑦), The right-hand side of (21) is satisfied as follows: 𝑝=1𝑇𝑥,𝑇𝑦43≤𝛾4≤𝛾𝑝𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥 (27) if we choose 𝛾≥1/3.(iv) If 𝑥=1, 𝑦=0, then 𝑇𝑥=3/4, 𝑇𝑦=1. The left-hand side of (21) holds good as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=(1/(𝑟+1))(1/4)<1=𝑝(𝑥,𝑦). The right-hand side of (21) is also satisfied as follows: 𝑝=1𝑇𝑥,𝑇𝑦43≤𝛾4≤𝛾𝑝𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥 (28) if we can choose 𝛾≥1/3.(v) If {𝑥,𝑦}∩(2,3)≠∅, then 𝑇𝑥=𝑇𝑦=1/4. If 𝑥<𝑦, then the left-hand side of (21) remains true as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=(1/(1+𝑟))𝑥<𝑦=𝑝(𝑥,𝑦). The right-hand side of (21) is also verified as follows: 𝑝𝑇𝑥,𝑇𝑦=0≤𝛾𝑝(𝑥,𝑇𝑥)+𝑝𝑦,𝑇𝑦. (29) If 𝑥≥𝑦, then left-hand side of (21) holds good as 𝜓(𝑟)𝑝(𝑥,𝑇𝑥)=(1/(1+𝑟))𝑥<𝑥=𝑝(𝑥,𝑦). The right-hand side of (21) remains true as 𝑝𝑇𝑥,𝑇𝑦=0≤𝛾𝑝(𝑥,𝑇𝑥)+𝑝𝑦,𝑇𝑦. (30)
Thus, in all possible cases the condition (21) is satisfied.

Acknowledgment

Both the authors are grateful to both the learned referees for their fruitful comments towards the improvement of this paper.