A Generalization of Prešić Type Mappings in 0-Complete Ordered Partial Metric Spaces

Abstract

The purpose of this paper is to prove some coincidence and common fixed point results for mappings satisfying Prešić type contraction condition in 0-complete ordered partial metric spaces. The results proved in this paper generalize and extend the results of Prešić and Matthews in 0-complete ordered partial metric spaces. Some examples are included which show that the generalization is proper.

1. Introduction

Banach contraction mapping principle is one of the most interesting and useful tools in applied mathematics. In recent years many generalizations of Banach contraction mapping principle have appeared. In 1965 Prešić [1, 2] extended Banach contraction mapping principle to mappings defined on product spaces and proved the following theorem.

Theorem 1. Let (𝑋,𝑑) be a complete metric space, 𝑘 a positive integer, and 𝑇∶𝑋^𝑘→𝑋 a mapping satisfying the following contractive type condition: 𝑑𝑇𝑥₁,𝑥₂,…,𝑥_𝑘,𝑇𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝑞_𝑖𝑑𝑥_𝑖,𝑥_𝑖+1, (1) for every 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋, where 𝑞₁,𝑞₂,…,𝑞_𝑘 are nonnegative constants such that 𝑞₁+𝑞₂+⋯+𝑞_𝑘<1. Then there exists a unique point 𝑥∈𝑋 such that 𝑇(𝑥,𝑥,…,𝑥)=𝑥. Moreover if 𝑥₁,𝑥₂,…,𝑥_𝑘 are arbitrary points in 𝑋 and for 𝑛∈ℕ,𝑥_𝑛+𝑘=𝑇(𝑥_𝑛,𝑥_𝑛+1,…,𝑥_{𝑛+𝑘−1}), then the sequence {𝑥_𝑛} is convergent and lim𝑥_𝑛=𝑇(lim𝑥_𝑛,lim𝑥_𝑛,…,lim𝑥_𝑛).

Note that condition (1) in the case 𝑘=1 reduces to the well-known Banach contraction mapping principle. So, Theorem 1 is a generalization of the Banach fixed point theorem. Some generalization of Theorem 1 can be seen in [1–11].

On the other hand in 1994 Matthews [12] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, showing that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. In partial metric space the usual metric was replaced by partial metric, with a property that the self-distance of any point may not be zero. Result of Matthews is generalized by several authors in different directions (see, e.g., [13–29]). Romaguera [30] introduced the notion of 0-Cauchy sequence and 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness. Some results on 0-complete partial metric spaces can be seen in [30–32].

The existence of fixed point in partially ordered sets was investigated by Ran and Reurings [33] and then by Nieto and Rodríguez-López [34, 35]. Fixed point results in ordered partial metric spaces were obtained by several authors (see, e.g., [14, 16–19, 27, 29]). Very recently, in [7] (see also [36]) authors introduced the ordered Prešić type contraction and generalized the result of Prešić and proved some fixed point theorems for such mappings. In this paper, we generalize and extend the result of Prešić [1, 2] in 0-complete ordered partial metric spaces. A generalization of result of Prešić in 0-complete partial metric spaces is also established. Some examples are included which show that the generalization is proper.

First we recall some definitions and properties of partial metric space [12, 30, 32, 37, 38].

Definition 2. A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→ℝ⁺ (ℝ⁺ stands for nonnegative reals) 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.

It is clear that if 𝑝(𝑥,𝑦)=0, then from (P1) and (P2) 𝑥=𝑦. But if 𝑥=𝑦,𝑝(𝑥,𝑦) may not be 0. Also every metric space is a partial metric space, with zero self-distance.

Example 3. If 𝑝∶ℝ⁺×ℝ⁺→ℝ⁺ is defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, for all 𝑥,𝑦∈ℝ⁺, then (ℝ⁺,𝑝) is a partial metric space.

Each partial metric on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has a base 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.

Theorem 4 (see [12]). For each partial metric 𝑝∶𝑋×𝑋→ℝ⁺ the pair (𝑋,𝑑), where 𝑑(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦) for all 𝑥,𝑦∈𝑋, is a metric space.

Here (𝑋,𝑑) is called induced metric space and 𝑑 is induced metric. In further discussion until unless specified (𝑋,𝑑) will represent induced metric space.

Let (𝑋,𝑝) be a partial metric space. (1)A sequence {𝑥_𝑛} in (𝑋,𝑝) converges to a point 𝑥∈𝑋 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑥).(2)A sequence {𝑥_𝑛} in (𝑋,𝑝) is called Cauchy sequence if there exists (and is finite) lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(3)(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).(4)A sequence {𝑥_𝑛} in (𝑋,𝑝) is called 0-Cauchy sequence if lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0. The space (𝑋,𝑝) is said to be 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.

Lemma 5 (see [12, 30, 32, 37]). Let (𝑋,𝑝) be a partial metric space and {𝑥_𝑛} any sequence in 𝑋.(i){𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝) if and only if it is a Cauchy sequence in metric space (𝑋,𝑑).(ii)(𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑑) is complete. Furthermore, lim_𝑛→∞𝑑(𝑥_𝑛,𝑥)=0 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥_𝑛,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).(iii)Every 0-Cauchy sequence in (𝑋,𝑝) is Cauchy in (𝑋,𝑑).(iv)If (𝑋,𝑝) is complete, then it is 0-complete.

The converse assertions of (iii) and (iv) do not hold. Indeed 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 easy example of a 0-complete partial metric space which is not complete. Also, it is easy to see that every closed subset of a 0-complete partial metric space is 0-complete.

Definition 6. Let 𝑋 be a nonempty set, 𝑘 a positive integer, and 𝑓∶𝑋^𝑘→𝑋 a mapping. If 𝑓(𝑥,𝑥,…,𝑥)=𝑥, then 𝑥∈𝑋 is called a fixed point of 𝑓.

Definition 7. Let 𝑋 be a nonempty set, 𝑘 a positive integer, and 𝑓∶𝑋^𝑘→𝑋 and 𝑔∶𝑋→𝑋 mappings. (a)An element 𝑥∈𝑋is said to be a coincidence point of 𝑓 and 𝑔 if 𝑔𝑥=𝑓(𝑥,…,𝑥). (b)If 𝑤=𝑔𝑥=𝑓(𝑥,…,𝑥), then 𝑤 is called a point of coincidence of 𝑓 and 𝑔. (c)If 𝑥=𝑔𝑥=𝑓(𝑥,…,𝑥), then 𝑥 is called a common fixed point of 𝑓 and 𝑔. (d)Mappings 𝑓 and 𝑔 are said to be commuting if 𝑔(𝑓(𝑥,…,𝑥))=𝑓(𝑔𝑥,…,𝑔𝑥) for all 𝑥∈𝑋. (e)Mappings 𝑓 and 𝑔 are said to be weakly compatible if they commute at their coincidence points.

Lemma 8 (see [8]). Let 𝑋 be a nonempty set, 𝑘 a positive integer, and 𝑓∶𝑋^𝑘→𝑋,𝑔∶𝑋→𝑋 two weakly compatible mappings. If 𝑓 and 𝑔 have a unique point of coincidence 𝑦=𝑓(𝑥,𝑥,…,𝑥)=𝑔(𝑥), then 𝑦 is the unique common fixed point of 𝑓 and 𝑔.

Remark that the previous definition in the case 𝑘=1 reduces to the usual definitions of commuting and weakly compatible mappings in the sense of [39] (for details, see Introduction from [39]).

Definition 9. Let a nonempty set 𝑋 be equipped with a partial order “⊑” such that (𝑋,𝑝) is a partial metric space; then (𝑋,⊑,𝑝) is called an ordered partial metric space. A sequence {𝑥_𝑛} in 𝑋 is said to be nondecreasing with respect to “⊑” if 𝑥₁⊑𝑥₂⊑⋯⊑𝑥_𝑛⊑⋯. Let 𝑘 be a positive integer and 𝑓∶𝑋^𝑘→𝑋 a mapping; then 𝑓 is said to be nondecreasing with respect to “⊑” if for any finite nondecreasing sequence {𝑥_𝑛}^𝑘+1_𝑛=1 we have 𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘)⊑𝑓(𝑥₂,𝑥₃,…,𝑥_𝑘+1). Let 𝑔∶𝑋→𝑋 be a mapping. 𝑓 is said to be 𝑔-nondecreasing with respect to “⊑” if for any finite nondecreasing sequence {𝑔𝑥_𝑛}^𝑘+1_𝑛=1 we have 𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘)⊑𝑓(𝑥₂,𝑥₃,…,𝑥_𝑘+1).

Remark 10. For 𝑘=1 previous definitions reduce to usual definitions of fixed point and nondecreasing mapping in partial metric space.

Definition 11. Let 𝑋 be a nonempty set equipped with partial order “⊑”, and let 𝑔∶𝑋→𝑋 be a mapping. A nonempty subset 𝒜 of 𝑋 is said to be well ordered if every two elements of 𝒜 are comparable. Elements 𝑎,𝑏∈𝒜 are called 𝑔-comparable if 𝑔𝑎 and 𝑔𝑏 are comparable. 𝒜 is called 𝑔-well ordered if for all 𝑎,𝑏∈𝒜,𝑎 and 𝑏 are 𝑔-comparable; that is, 𝑔𝑎 and 𝑔𝑏 are comparable.

Example 12. Let 𝑋={0,1,2,3}, “⊑” a partial order relation on 𝑋 defined by ⊑={(0,0),(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. Let 𝒜={0,1,3} and 𝑔∶𝑋→𝑋 be defined by 𝑔0=1,𝑔1=2,𝑔2=3,𝑔3=3. Then it is clear that 𝒜 is not well ordered but it is 𝑔-well ordered.

Let (𝑋,⊑,𝑝) be an ordered partial metric space, 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋 be a mapping; 𝑓 is called ordered Prešić contraction if 𝑝𝑓𝑥₁,𝑥₂…,𝑥_𝑘,𝑓𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑥_𝑖,𝑥_𝑖+1, (2) for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋 with 𝑥₁⊑𝑥₂⊑⋯⊑𝑥_𝑘+1, where 𝛼_𝑖 are nonnegative constant such that ∑^𝑘_𝑖=1𝛼_𝑖<1.

If (2) is satisfied for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋, then 𝑓 is called Prešić contraction.

Note that in ordered partial metric spaces a Prešić contraction is necessarily an ordered Prešić contraction, but converse may not be true (see Example 21 of this paper).

Remark 13. Since every metric space is a partial metric space and there exist mappings which are Prešić contraction in partial metric spaces but not in metric spaces (see Example 20 of this paper), therefore the class of Prešić contraction in partial metric spaces is wider than that in metric spaces.
The following lemma will be useful in sequel.

Lemma 14. Let (𝑋,⊑,𝑝) be any ordered partial metric space, 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋,𝑔∶𝑋→𝑋 be two mappings satisfying 𝑝𝑓𝑥₁,𝑥₂…,𝑥_𝑘,𝑓𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑔𝑥_𝑖,𝑔𝑥_𝑖+1, (3) for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋 with 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘+1 where 𝛼_𝑖 are nonnegative constant such that ∑^𝑘_𝑖=1𝛼_𝑖<1. If 𝑓 and 𝑔 have a point of coincidence 𝑣, then 𝑝(𝑣,𝑣)=0.

Proof. Suppose that 𝑣 is a point of coincidence of 𝑓 and 𝑔 with coincidence point 𝑢∈𝑋, that is, 𝑔𝑢=𝑓(𝑢,…,𝑢)=𝑣. As 𝑔𝑢⊑𝑔𝑢, using (3) we obtain 𝑝()𝑓()()≤𝑣,𝑣=𝑝𝑢,…,𝑢,𝑓𝑢,…,𝑢_𝑘𝑖=1𝛼_𝑖𝑝=𝑔𝑢,𝑔𝑢_𝑘𝑖=1𝛼_𝑖𝑝(𝑣,𝑣); (4) as ∑^𝑘_𝑖=1𝛼_𝑖<1, we obtain 𝑝(𝑣,𝑣)=0.

Now we can state our main results.

2. Main Results

Theorem 15. Let (𝑋,⊑,𝑝) be any 0-complete ordered partial metric space and 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋,𝑔∶𝑋→𝑋 be two mappings such that 𝑓(𝑋^𝑘)⊂𝑔(𝑋) and 𝑔(𝑋) is a closed subspace of 𝑋. Suppose the following conditions hold: (I)𝑝𝑓𝑥₁,𝑥₂…,𝑥_𝑘,𝑓𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑔𝑥_𝑖,𝑔𝑥_𝑖+1, (5) for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋 with 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘+1, where 𝛼_𝑖 are nonnegative constant such that ∑^𝑘_𝑖=1𝛼_𝑖<1; (II) there exist 𝑥₁,𝑥₂,…,𝑥_𝑘∈𝑋 such that 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘⊑𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘); (III)𝑓 is 𝑔-nondecreasing; (IV) if a nondecreasing sequence {𝑔𝑥_𝑛} converges to 𝑔𝑢∈𝑋, then 𝑔𝑥_𝑛⊑𝑔𝑢 for all 𝑛∈ℕ and 𝑔𝑢⊑𝑔𝑔𝑢.
Then 𝑓 and 𝑔 have a point of coincidence. If in addition 𝑓 and 𝑔 are weakly compatible, then 𝑓 and 𝑔 have a common fixed point 𝑣∈𝑋 with 𝑝(𝑣,𝑣)=0. Moreover, the set of common fixed points of 𝑓 and 𝑔 is 𝑔-well ordered if and only if 𝑓 and 𝑔 have a unique common fixed point.

Proof. Starting with given 𝑥₁,𝑥₂,…,𝑥_𝑘∈𝑋 we define a sequence {𝑦_𝑛} as follows: let 𝑦_𝑛=𝑔𝑥_𝑛,𝑛=1,2,…,𝑘. As 𝑓(𝑋^𝑘)⊂𝑔(𝑋) and 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘⊑𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘), define 𝑦_𝑛+𝑘=𝑔𝑥_𝑛+𝑘=𝑓(𝑥_𝑛,𝑥_𝑛+1,…,𝑥_{𝑛+𝑘−1}),𝑛=1,2,…. Then 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘⊑𝑔𝑥_𝑘+1 and 𝑓 is 𝑔-nondecreasing, so 𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘)⊑𝑓(𝑥₂,𝑥₃,…,𝑥_𝑘+1), that is, 𝑔𝑥_𝑘+1⊑𝑔𝑥_𝑘+2. Continuing this procedure, we obtain 𝑔𝑥₁⊑𝑔𝑥₂⊑⋯⊑𝑔𝑥_𝑘⊑𝑔𝑥_𝑘+1⊑⋯, (6) that is, 𝑦₁⊑𝑦₂⊑⋯⊑𝑦_𝑘⊑𝑦_𝑘+1⊑⋯. (7)
Thus {𝑦_𝑛}={𝑔𝑥_𝑛} is a nondecreasing sequence with respect to “⊑.”
For simplicity set 𝑝_𝑛=𝑝(𝑦_𝑛,𝑦_𝑛+1)=𝑝(𝑔𝑥_𝑛,𝑔𝑥_𝑛+1),𝜇=max{𝑝₁/𝜃,𝑝₂/𝜃²,…,𝑝_𝑘/𝜃^𝑘}, where 𝜃=[∑^𝑘_𝑖=1𝛼_𝑖]^1/𝑘.
By mathematical induction we will show that 𝑝_𝑛≤𝜇𝜃^𝑛,∀𝑛∈ℕ. (8)
According to the definition of 𝜇 it is clear that (8) is true for 𝑛=1,2,…,𝑘. Let the following 𝑘 inequalities 𝑝_𝑛≤𝜇𝜃^𝑛,𝑝_𝑛+1≤𝜇𝜃^𝑛+1,…,𝑝_{𝑛+𝑘−1}≤𝜇𝜃^{𝑛+𝑘−1} (9) be the induction hypothesis.
As 𝑔𝑥_𝑛⊑𝑔𝑥_𝑛+1⊑⋯⊑𝑔𝑥_𝑛+𝑘, so using (5) we obtain 𝑝_𝑛+𝑘=𝑝𝑦_𝑛+𝑘,𝑦_𝑛+𝑘+1=𝑝𝑓𝑥_𝑛,…,𝑥_{𝑛+𝑘−1},𝑓𝑥_𝑛+1,…,𝑥_𝑛+𝑘≤𝛼₁𝑝𝑔𝑥_𝑛,𝑔𝑥_𝑛+1+𝛼₂𝑝𝑔𝑥_𝑛+1,𝑔𝑥_𝑛+2+⋯+𝛼_𝑘𝑝𝑔𝑥_{𝑛+𝑘−1},𝑔𝑥_𝑛+𝑘≤𝛼₁𝑝_𝑛+𝛼₂𝑝_𝑛+1+⋯+𝛼_𝑘𝑝_{𝑛+𝑘−1}≤𝛼₁𝜇𝜃^𝑛+𝛼₂𝜇𝜃^𝑛+1+⋯+𝛼_𝑘𝜇𝜃^{𝑛+𝑘−1}≤𝜇𝜃^𝑛+𝛼₂𝜇𝜃^𝑛+⋯+𝛼_𝑘𝜇𝜃^𝑛as𝜃=^𝑘_𝑖=1𝛼_𝑖^1/𝑘=<1_𝑘𝑖=1𝛼_𝑖𝜇𝜃^𝑛=𝜇𝜃^𝑛+𝑘, (10) and the inductive proof of (8) is complete.
Let 𝑛,𝑚∈ℕ with 𝑚>𝑛; then 𝑝𝑦_𝑛,𝑦_𝑚≤𝑝𝑦_𝑛,𝑦_𝑛+1+𝑝𝑦_𝑛+1,𝑦_𝑛+2+⋯+𝑝𝑦_𝑚−1,𝑦_𝑚−𝑝𝑦_𝑛+1,𝑦_𝑛+1+𝑝𝑦_𝑛+2,𝑦_𝑛+2+⋯+𝑝𝑦_𝑚−1,𝑦_𝑚−1≤𝑝_𝑛+𝑝_𝑛+1+⋯+𝑝_𝑚−1≤𝜇𝜃^𝑛+𝜇𝜃^𝑛+1+⋯+𝜇𝜃^𝑚−1≤1+𝜃+𝜃²+⋯𝜇𝜃^𝑛=𝜇𝜃^𝑛.1−𝜃 (11)
As 𝜃<1, previous inequality implies that lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚=0. (12)
Thus {𝑦_𝑛}={𝑔𝑥_𝑛} is a 0-Cauchy sequence in (𝑋,𝑝). Since 𝑔(𝑋) is 0-complete as it is closed subspace, it follows that {𝑦_𝑛}={𝑔𝑥_𝑛} converges in (𝑔(𝑋),𝑝). Therefore lim_𝑛→∞𝑝𝑦_𝑛,𝑣=lim_𝑛→∞𝑝𝑦_𝑛,𝑔𝑢=𝑝(𝑣,𝑣)=𝑝𝑔𝑢,𝑔𝑢=0, (13) where 𝑣=𝑔𝑢∈𝑔(𝑋) for some 𝑢∈𝑋.
We will show that 𝑢 is coincidence point of and 𝑓 and 𝑔.
Then for any integer 𝑛, we have 𝑝𝑔𝑢,𝑓(𝑢,…,𝑢)≤𝑝𝑔𝑢,𝑔𝑥_𝑛+𝑘+𝑝𝑔𝑥_𝑛+𝑘,𝑓(𝑢,…,𝑢)−𝑝𝑔𝑥_𝑛+𝑘,𝑔𝑥_𝑛+𝑘≤𝑝𝑔𝑢,𝑔𝑥_𝑛+𝑘+𝑝𝑓𝑥_𝑛,…,𝑥_{𝑛+𝑘−1},𝑓(𝑢,…,𝑢)≤𝑝𝑔𝑢,𝑔𝑥_𝑛+𝑘+𝑝𝑓𝑥_𝑛,…,𝑥_{𝑛+𝑘−1},𝑓𝑥_𝑛+1,…,𝑥_{𝑛+𝑘−1}𝑓𝑥,𝑢+𝑝_𝑛+1,…,𝑥_{𝑛+𝑘−1},𝑢,𝑓𝑥_𝑛+2,…,𝑥_{𝑛+𝑘−1}𝑓𝑥,𝑢,𝑢+⋯+𝑝_{𝑛+𝑘−1},𝑢,…,𝑢,𝑓(𝑢,…,𝑢). (14) By assumption (IV) it follows that 𝑔𝑥_𝑛⊑𝑔𝑢 for all 𝑛∈ℕ and 𝑔𝑢⊑𝑔𝑔𝑢, so using (5), previous inequality implies that 𝑝𝑔𝑢,𝑓(𝑢,…,𝑢)≤𝑝𝑔𝑢,𝑦_𝑛+𝑘+𝛼₁𝑝_𝑛+⋯+𝛼_𝑘−1𝑝_{𝑛+𝑘−2}+𝛼_𝑘𝑝𝑦_{𝑛+𝑘−1},𝑔𝑢+𝛼₁𝑝_𝑛+1+⋯+𝛼_𝑘−2𝑝_{𝑛+𝑘−2}+𝛼_𝑘−1𝑝𝑦_{𝑛+𝑘−1},𝑔𝑢+𝛼_𝑘𝑝𝑔𝑢,𝑔𝑢+⋯+𝛼₁𝑝𝑦_{𝑛+𝑘−1}+𝛼,𝑔𝑢₂+⋯+𝛼_𝑘𝑝𝑔𝑢,𝑔𝑢. (15) Using (13) and the fact that lim_𝑛→∞𝑝_𝑛=0, previous inequality implies that 𝑝𝑔𝑢,𝑓(𝑢,…,𝑢)=0thatis𝑔𝑢=𝑓(𝑢,…,𝑢)=𝑣. (16)
Thus 𝑣 is point of coincidence, and 𝑢 is coincidence point of 𝑓 and 𝑔.
Suppose 𝑓 and 𝑔 are weakly compatible, so 𝑓()𝑓();𝑣,…,𝑣=𝑓𝑔𝑢,…,𝑔𝑢=𝑔𝑢,…,𝑢=𝑔𝑣=𝑤say (17) so 𝑤 is a point of coincidence of 𝑓 and 𝑔, and by Lemma 14, we have 𝑝(𝑤,𝑤)=𝑝(𝑔𝑣,𝑔𝑣)=0. Again since 𝑔𝑢⊑𝑔𝑔𝑢=𝑔𝑣 we obtain from (5) that 𝑝()𝑓()()𝑓()()𝑓()()𝑓()()≤𝛼𝑣,𝑓𝑣,…,𝑣=𝑝𝑢,…,𝑢,𝑓𝑣,…,𝑣≤𝑝𝑢,…,𝑢,𝑓𝑢,…,𝑢,𝑣+𝑝𝑢,…,𝑢,𝑣,𝑓𝑢,…,𝑢,𝑣,𝑣+⋯+𝑝𝑢,𝑣,…,𝑣,𝑓𝑣,…,𝑣₁+⋯+𝛼_𝑘−1𝑝𝑔𝑢,𝑔𝑢+𝛼_𝑘𝑝+𝛼𝑔𝑢,𝑔𝑣₁+⋯+𝛼_𝑘−2𝑝𝑔𝑢,𝑔𝑢+𝛼_𝑘−1𝑝𝑔𝑢,𝑔𝑣+𝛼_𝑘𝑝𝑔𝑣,𝑔𝑣+⋯+𝛼₁𝑝+𝛼𝑔𝑢,𝑔𝑣₂+⋯+𝛼_𝑘𝑝𝑔𝑣,𝑔𝑣. (18) As 𝑝(𝑔𝑢,𝑔𝑢)=𝑝(𝑔𝑣,𝑔𝑣)=0,𝑔𝑢=𝑣, and 𝑔𝑣=𝑓(𝑣,…,𝑣) it follows from previous inequality that 𝑝𝑣,𝑓(𝑣,…,𝑣)≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑣,𝑓(𝑣,…,𝑣); (19) as ∑^𝑘_𝑖=1𝛼_𝑖<1, we obtain 𝑝𝑣,𝑓(𝑣,…,𝑣)=0thatis𝑣=𝑓(𝑣,…,𝑣)=𝑔𝑣. (20) Thus 𝑣 is common fixed point of 𝑓 and 𝑔.
Suppose that the set of common fixed points is 𝑔-well ordered. We will show that common fixed point is unique. Assume on the contrary that 𝑣^′ is another common fixed point of 𝑓 and 𝑔, that is, 𝑣^′=𝑓(𝑣^′,…,𝑣^′)=𝑔𝑣^′ and 𝑣≠𝑣^′. As 𝑣 and 𝑣^′ are 𝑔-comparable let, for example, 𝑔𝑣⊑𝑔𝑣^′. From (5), it follows that 𝑝𝑣,𝑣^′=𝑝𝑓(𝑣,…,𝑣),𝑓𝑣^′,…,𝑣^′≤𝑝𝑓(𝑣,…,𝑣),𝑓𝑣,…,𝑣,𝑣^′+𝑝𝑓𝑣,…,𝑣,𝑣^′,𝑓𝑣,…,𝑣^′,𝑣^′+⋯+𝑝𝑓𝑣,𝑣^′,…,𝑣^′,𝑓𝑣^′,…,𝑣^′≤𝛼₁+⋯+𝛼_𝑘−1𝑝𝑔𝑣,𝑔𝑣+𝛼_𝑘𝑝𝑔𝑣,𝑔𝑣^′+𝛼₁+⋯+𝛼_𝑘−2𝑝𝑔𝑣,𝑔𝑣+𝛼_𝑘−1𝑝𝑔𝑣,𝑔𝑣^′+𝛼_𝑘𝑝𝑔𝑣^′,𝑔𝑣^′+⋯+𝛼₁𝑝𝑔𝑣,𝑔𝑣^′+𝛼₂+⋯+𝛼_𝑘𝑝𝑔𝑣^′,𝑔𝑣^′. (21)
As 𝑝(𝑔𝑣,𝑔𝑣)=𝑝(𝑣,𝑣)=0 and from Lemma 14, 𝑝(𝑣^′,𝑣^′)=𝑝(𝑔𝑣^′,𝑔𝑣^′)=0, it follows from previous inequality that 𝑝𝑣,𝑣^′≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑔𝑣,𝑔𝑣^′=^𝑘_𝑖=1𝛼_𝑖𝑝𝑣,𝑣^′<𝑝𝑣,𝑣^′as^𝑘_𝑖=1𝛼_𝑖<1, (22) a contradiction. Therefore common fixed point is unique. For converse, if common fixed point of 𝑓 and 𝑔 is unique, then the set of common fixed points of 𝑓 and 𝑔is singleton therefore 𝑔-well ordered.

Remark 16. For 𝑘=1 previous theorem is a generalization and extension of result of Matthews [12] in ordered partial metric spaces.

Taking 𝑔=𝐼_𝑋, that is, identity mapping of 𝑋, we get the following fixed point result for ordered Prešić contraction in partial metric spaces.

Corollary 17. Let (𝑋,⊑,𝑝) be any 0-complete ordered partial metric space and 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋 be a mapping and the following conditions hold: (I)𝑓 is an ordered Prešić contraction; (II)there exist 𝑥₁,𝑥₂,…,𝑥_𝑘∈𝑋 such that 𝑥₁⊑𝑥₂⊑⋯⊑𝑥_𝑘⊑𝑓(𝑥₁,𝑥₂,…,𝑥_𝑘); (III)𝑓 is nondecreasing; (IV)if a nondecreasing sequence {𝑥_𝑛} converges to 𝑢∈𝑋, then 𝑥_𝑛⊑𝑢 for all 𝑛∈ℕ.
Then 𝑓 has a fixed point 𝑣∈𝑋 with 𝑝(𝑣,𝑣)=0. Moreover, the set of fixed points of 𝑓 is well ordered if and only if 𝑓 has a unique fixed point.

Following theorem generalizes and extends the result of Prešić in partial metric spaces.

Theorem 18. Let (𝑋,𝑝) be any 0-complete partial metric space and 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋,𝑔∶𝑋→𝑋 be two mappings such that 𝑓(𝑋^𝑘)⊂𝑔(𝑋), 𝑔(𝑋) is a closed subspace of 𝑋, and 𝑝𝑓𝑥₁,𝑥₂…,𝑥_𝑘,𝑓𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑔𝑥_𝑖,𝑔𝑥_𝑖+1, (23) for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋, where 𝛼_𝑖 are nonnegative constant such that ∑^𝑘_𝑖=1𝛼_𝑖<1. Then 𝑓 and 𝑔 have a point of coincidence. Furthermore, if 𝑓 and 𝑔 are weakly compatible, then 𝑓 and 𝑔 have a unique common fixed point 𝑣∈𝑋 with 𝑝(𝑣,𝑣)=0.

Proof. We note that the inequality (23) is true for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋; therefore the proof of theorem follows from similar process as used in the proof of Theorem 15.

Taking 𝑔=𝐼_𝑋 that is, identity mapping in the previous theorem, we obtain the following corollary.

Corollary 19 (Prešić type). Let (𝑋,𝑝) be any 0-complete partial metric space and 𝑘 a positive integer. Let 𝑓∶𝑋^𝑘→𝑋 be a mapping satisfying 𝑝𝑓𝑥₁,𝑥₂…,𝑥_𝑘,𝑓𝑥₂,𝑥₃,…,𝑥_𝑘+1≤^𝑘_𝑖=1𝛼_𝑖𝑝𝑥_𝑖,𝑥_𝑖+1, (24) for all 𝑥₁,𝑥₂,…,𝑥_𝑘+1∈𝑋, where 𝛼_𝑖 are nonnegative constant such that ∑^𝑘_𝑖=1𝛼_𝑖<1. Then 𝑓 has a unique fixed point 𝑣∈𝑋 with 𝑝(𝑣,𝑣)=0.

Following is an example which illustrates that a Prešić type contraction in partial metric space need not to be a Prešić type contraction in metric space.

Example 20. Let 𝑋=[0,1] and defined 𝑝∶𝑋×𝑋→ℝ⁺ by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Then (𝑋,𝑝) is a 0-complete partial metric space. For 𝑘=2, suppose 𝑓∶𝑋²→𝑋 be defined by 𝑓𝑥,𝑦=0if𝑥=𝑦=1,𝑥+𝑦5otherwise. (25) Then 𝑓 is Prešić type contraction in partial metric space (𝑋,𝑝), that is, it satisfies (24) with 𝛼₁=𝛼₂=1/5, while it is not a Prešić type contraction in usual metric space (𝑋,𝑑) where 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋 (note that 𝑑 is also the induced metric).

Proof. For 𝑘=2 and 𝛼₁=𝛼₂=𝛼 condition (24) becomes 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃≤𝛼𝑝𝑥₁,𝑥₂+𝑝𝑥₂,𝑥₃, (26) for all 𝑥₁,𝑥₂,𝑥₃∈𝑋, where 𝛼 is a nonnegative constant such that 𝛼<1/2. Note that if 𝑥₁=𝑥₂=𝑥₃=1, then (26) is satisfied trivially.
If 𝑥₁,𝑥₂,𝑥₃∈[0,1) and 𝑥₁≤𝑥₂≤𝑥₃, then (26) is valid for 𝛼∈[1/5,1/2). Indeed since 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃𝑥=𝑝¹+𝑥₂5,𝑥₂+𝑥₃5𝑥=max¹+𝑥₂5,𝑥₂+𝑥₃5=𝑥₂+𝑥₃5,𝛼𝑝𝑥₁,𝑥₂+𝑝𝑥₂,𝑥₃=𝛼max𝑥₁,𝑥₂+max𝑥₂,𝑥₃=𝛼𝑥₂+𝑥₃, (27) so result follows.
If any two of 𝑥₁,𝑥₂,𝑥₃ are equal to 1, for example, if 𝑥₁=𝑥₂=1 and 𝑥₃∈[0,1), then (26) is valid. In this case 𝑝𝑓(1,1),𝑓1,𝑥₃=𝑝0,1+𝑥₃5=max0,1+𝑥₃5=1+𝑥₃5,𝛼𝑝(1,1)+𝑝1,𝑥₃=𝛼max{1,1}+max1,𝑥₃=2𝛼. (28)
Hence, if 𝛼∈[1/5,1/2), then 𝑝(𝑓(1,1),𝑓(1,𝑥₃))≤𝛼[𝑝(1,1)+𝑝(1,𝑥₃)].
Similarly in all possible cases (26) is satisfied for 𝛼=1/5. Note that all other conditions of Corollary 19 are satisfied and 𝑓 has a unique fixed point 0, that is, 𝑓(0,0)=0, with 𝑝(0,0)=0.
Again for 𝑘=2,𝑓 will be Prešić contraction in (𝑋,𝑑) if 𝑑𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃≤𝛼₁𝑑𝑥₁,𝑥₂+𝛼₂𝑑𝑥₂,𝑥₃, (29) for all 𝑥₀,𝑥₁,𝑥₂∈𝑋, where 𝛼₁,𝛼₂ are nonnegative constant such that 𝛼₁+𝛼₂<1. We will show that (29) is not satisfied for certain points in [0,1]. Let 𝑥₁=10/11,𝑥₂=𝑥₃=1; then 𝑑𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃=𝑑𝑓10(11,1,𝑓1,1)=𝑑21=55,021,𝛼55₁𝑑𝑥₁,𝑥₂+𝛼₂𝑑𝑥₂,𝑥₃=𝛼₁𝑑1011,1+𝛼₂𝑑()=11,1𝛼11¹ (30) but 𝛼₁+𝛼₂<1; therefore for these values of 𝑥₁,𝑥₂, and 𝑥₃ (29) does not hold. Thus 𝑓 is not a Prešić contraction in (𝑋,𝑑), and so we cannot apply the result of Prešić to conclude the existence of fixed point of 𝑓.

Following example illustrates that an ordered Prešić contraction need not to be a Prešić contraction in ordered partial metric spaces, and the fixed point of ordered Prešić contraction may not be unique (when set of fixed points of 𝑓 is not well ordered).

Example 21. Let 𝑋=[0,2] and let order relation “⊑” be defined by [)∪[)∪;⊑=𝑥,𝑦∶𝑥,𝑦∈0,1with𝑦≤𝑥𝑥,𝑦∶𝑥,𝑦∈1,2with𝑦≤𝑥{(2,2)} (31) suppose 𝑝∶𝑋×𝑋→ℝ⁺ be defined by 𝑝𝑥,𝑦=0if𝑥=𝑦=2,max𝑥,𝑦otherwise. (32)
Then (𝑋,⊑,𝑝) is an ordered complete partial metric space. For 𝑘=2, suppose 𝑓∶𝑋²→𝑋 be a mapping defined by 𝑓=∈[)×[]∪[]×[),𝑥,𝑦𝑥+𝑦−1if𝑥,𝑦0,11,21,20,12if𝑥=𝑦=2,𝑥+𝑦4otherwise. (33)
Then(a)𝑓 is an ordered Prešić contraction that, satisfies (2) with 𝛼₁=𝛼₂=1/4,(b)𝑓 is not a Prešić contraction; that is, 𝑓 does not satisfy (2) for all 𝑥₁,𝑥₂∈𝑋.

Proof. (a) For 𝑘=2 and 𝛼₁=𝛼₂=𝛼, (2) becomes 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃≤𝛼𝑝𝑥₁,𝑥₂+𝑝𝑥₂,𝑥₃, (34) for all 𝑥₁,𝑥₂,𝑥₃∈𝑋 with 𝑥₁⊑𝑥₂⊑𝑥₃, where 𝛼 is nonnegative constant such that 𝛼<1/2. Note that if 𝑥₁=𝑥₂=𝑥₃=2, then (34) is satisfied trivially.
Now we have to check the validity of inequality (34) only for 𝑥₁,𝑥₂,𝑥₃∈[0,1) and 𝑥₁,𝑥₂,𝑥₃∈[1,2).
Let 𝑥₁,𝑥₂,𝑥₃∈[0,1) with 𝑥₁⊑𝑥₂⊑𝑥₃, that is, 𝑥₃≤𝑥₂≤𝑥₁; then (34) is valid for 𝛼∈[1/4,1/2). Indeed since 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃𝑥=𝑝¹+𝑥₂4,𝑥₂+𝑥₃4𝑥=max¹+𝑥₂4,𝑥₂+𝑥₃4=𝑥₁+𝑥₂4,𝛼𝑝𝑥₁,𝑥₂+𝑝𝑥₂,𝑥₃=𝛼max𝑥₁,𝑥₂+max𝑥₂,𝑥₃=𝛼𝑥₁+𝑥₂, (35) so result follows.
Similarly in all possible cases (34) is satisfied for 𝛼=1/4.
Note that all other conditions of Corollary 17 are satisfied (except that the set of fixed points of 𝑓 is well ordered) and 𝑓 has two fixed points 0 and 2, that is, 𝑓(0,0)=0,𝑓(2,2)=2, with 𝑝(0,0)=𝑝(2,2)=0. Note that set of fixed points of 𝑓,𝒜={0,2} is not well ordered as (0,2),(2,0)∉⊑.
(b) For 𝑘=2, (2) becomes 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃≤𝛼₁𝑝𝑥₁,𝑥₂+𝛼₂𝑝𝑥₂,𝑥₃. (36)
We will show that (36) is not satisfied for all 𝑥₁,𝑥₂,𝑥₃∈𝑋, with 𝛼₁+𝛼₂<1, in particular, let 𝑥₁=2,𝑥₂=2 and 𝑥₃=0; then 𝑝𝑓𝑥₁,𝑥₂,𝑓𝑥₂,𝑥₃𝑓()()(){}𝛼=𝑝2,2,𝑓2,0=𝑝2,2+0−1=max2,1=2,₁𝑝𝑥₁,𝑥₂+𝛼₂𝑝𝑥₂,𝑥₃=𝛼₁𝑝(2,2)+𝛼₂𝑝(2,0)=𝛼₂max{2,0}=2𝛼₂, (37) but 𝛼₁+𝛼₂<1; therefore for these values of 𝑥₁,𝑥₂, and 𝑥₃ (36) does not hold. Thus 𝑓 is not Prešić contraction in (𝑋,𝑝).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.