Comparison Functions and Fixed Point Results in Partial Metric Spaces

Abstract

The aim of this article is to clearly formulate various possible assumptions for a comparison function in contractive conditions and to deduce respective (common) fixed point results in partial metric spaces. Since standard metric spaces are special cases, these results also apply for them. We will show by examples that there exist situations when a partial metric result can be applied, while the standard metric one cannot.

1. Introduction

In recent years many authors have worked on domain theory in order to equip semantics domain with a notion of distance. In particular, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks and obtained, among other results, a nice relationship between partial metric spaces and so-called weightable quasimetric spaces. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [2–11]) proved fixed point theorems in partial metric spaces.

Contractive conditions with the so-called comparison function 𝜑 of the form 𝑑𝑇𝑥,𝑇𝑦≤𝜑𝑀𝑥,𝑦 (1.1) have been used for obtaining (common) fixed point results of mappings in metric spaces since the celebrated result of Boyd and Wong [12]. In various articles, different assumptions for function 𝜑 were made. Sometimes these assumptions were not clearly stated, and sometimes the assumptions were stronger than needed. This includes some recent fixed point results in partial metric spaces.

The aim of this paper is to clearly formulate various possible conditions for a comparison function and to deduce respective (common) fixed point results in partial metric spaces. Since standard metric spaces are special cases, these results also apply for them. We will show by examples that there exist situations when a partial metric result can be applied, while the standard metric one cannot.

2. Preliminaries

The following definitions and details can be seen, for example, in [1, 3, 6, 9, 11].

Definition 2.1. A partial metric space is a pair (𝑋,𝑝) of a nonempty set 𝑋 and a partial metric 𝑝 on 𝑋, that 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).

It is clear that if 𝑝(𝑥,𝑦)=0, then, from (𝑝₁) and (𝑝₂), it follows that 𝑥=𝑦. But 𝑝(𝑥,𝑥) may not be 0.

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∶𝑝(𝑥,𝑦)<𝑝(𝑥,𝑥)+𝜀} for all 𝑥∈𝑋 and 𝜀>0. A sequence {𝑥_𝑛} in (𝑋,𝑝) converges to a point 𝑥∈𝑋, with respect to 𝜏_𝑝, if lim_𝑛→∞𝑝(𝑥,𝑥_𝑛)=𝑝(𝑥,𝑥). This will be denoted as 𝑥_𝑛→𝑥, 𝑛→∞ or lim_𝑛→∞𝑥_𝑛=𝑥. If 𝑇∶𝑋→𝑋 is continuous at 𝑥₀∈𝑋 (in 𝜏_𝑝), then, for each sequence {𝑥_𝑛} in 𝑋, 𝑥_𝑛⟶𝑥₀⟹𝑇𝑥_𝑛⟶𝑇𝑥₀. (2.1)

Remark 2.2. Clearly, a limit of a sequence in a partial metric space needs not to be unique. For example, if 𝑋=[0,+∞) and 𝑝(𝑥,𝑦)=max{𝑥,𝑦}for 𝑥,𝑦∈𝑋, then for {𝑥_𝑛}={1},𝑝(𝑥_𝑛,𝑥)=𝑥=𝑝(𝑥,𝑥) for each 𝑥≥1 and so, for example, 𝑥_𝑛→2 and 𝑥_𝑛→3 when 𝑛→∞. Moreover, the function 𝑝(⋅,⋅) needs not to be continuous in the sense that 𝑥_𝑛→𝑥 and 𝑦_𝑛→𝑦 imply 𝑝(𝑥_𝑛,𝑦_𝑛)→𝑝(𝑥,𝑦).

Definition 2.3. Let (𝑋,𝑝) be a partial metric space. Then we have the following.(1)A sequence {𝑥_𝑛} in (𝑋,𝑝) is called a Cauchy sequence if lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists (and is finite).(2)The 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).

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

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 (2.2) 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. (2.3)

Lemma 2.4. Let (𝑋,𝑝) be a partial metric space: (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)the space (𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑝^𝑠) is complete.

3. Auxiliary Results

We will consider the following properties of functions 𝜑∶[0,+∞)→[0,+∞). 𝜑^𝑛 will denote the 𝑛th iteration of 𝜑:(I)𝜑(𝑡)<𝑡 for each 𝑡>0 and 𝜑^𝑛(𝑡)→0, 𝑛→∞ for each 𝑡≥0,(II)𝜑 is nondecreasing and 𝜑^𝑛(𝑡)→0, 𝑛→∞ for each 𝑡≥0,(III)𝜑 is right-continuous, and 𝜑(𝑡)<𝑡 for each 𝑡>0,(IV)𝜑 is nondecreasing and ∑_𝑛≥1𝜑^𝑛(𝑡)<+∞ for each 𝑡≥0.

Lemma 3.1. (1) (II) ⇒ (I).
(2) (III) +𝜑 is nondecreasing ⇒ (II).
(3) (IV) ⇒ (II).
(4) (III) and (IV) are not comparable (even if 𝜑 is nondecreasing).

Proof. (1) (see [13]). Suppose that (II) holds and that there is some 𝑡₀∈(0,+∞) such that 𝜑(𝑡₀)≥𝑡₀. Then monotonicity of 𝜑 implies that 𝜑²(𝑡₀)≥𝜑(𝑡₀)≥𝑡₀. Continuing by induction we get that 𝜑^𝑛(𝑡₀)≥𝑡₀ and so 𝜑^𝑛(𝑡₀)→0, 𝑛→∞ is impossible.
(2) Let (III) hold and let 𝜑 be nondecreasing. Monotonicity of 𝜑 implies that, for each fixed 𝑡≥0, the sequence {𝜑^𝑛(𝑡)} is nonincreasing (and nonnegative); hence, there exists lim_𝑛→∞𝜑^𝑛(𝑡)=𝛼≥0. Suppose that 𝛼>0. Then it follows by (III) that 0<𝛼≤lim_𝑛→∞𝜑^𝑛+1(𝑡)=lim_{𝜑𝑛(𝑡)→𝛼+}𝜑𝜑^𝑛(𝑡)=𝜑lim_{𝜑𝑛(𝑡)→𝛼+}𝜑^𝑛(𝑡)=𝜑(𝛼), (3.1) which is a contradiction with 𝜑(𝑡)<𝑡.
(3) Obvious.
(4) It is demonstrated in the following example.

Example 3.2. (1) The function 𝜑(𝑡)=131𝑡,0≤𝑡<1,22,𝑡=1,3𝑡,𝑡>1, (3.2) satisfies (IV) but not (III).
(2) The function 𝜑(𝑡)=𝑡/(1+𝑡) is nondecreasing and satisfies (III) but not (IV), since 𝜑^𝑛(𝑡)=𝑡/(1+𝑛𝑡).

Assertions similar to the following lemma (see, e.g., [14]) were used (and proved) in the course of proofs of several fixed point results in various papers.

Lemma 3.3. Let (𝑋,𝑑) be a metric space and let {𝑦_𝑛} be a sequence in 𝑋 such that {𝑑(𝑦_𝑛+1,𝑦_𝑛)} is nonincreasing and lim_𝑛→∞𝑑𝑦_𝑛+1,𝑦_𝑛=0. (3.3) If {𝑦_2𝑛} is not a Cauchy sequence, then there exist 𝜀>0 and two subsequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the following four sequences tend to 𝜀+0 when 𝑘→∞: 𝑑𝑦_2𝑚𝑘,𝑦_2𝑛𝑘,𝑑𝑦_2𝑚𝑘,𝑦_2𝑛𝑘+1,𝑑𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘,𝑑𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘+1. (3.4)

As a consequence we obtain the following.

Lemma 3.4. Let (𝑋,𝑝) be a partial metric space and let {𝑦_𝑛} be a sequence in 𝑋 such that {𝑝(𝑦_𝑛+1,𝑦_𝑛)} is nonincreasing and lim_𝑛→∞𝑝𝑦_𝑛+1,𝑦_𝑛=0. (3.5) If {𝑦_2𝑛} is not a Cauchy sequence in (𝑋,𝑝), then there exist 𝜀>0 and two subsequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the following four sequences tend to 𝜀+0 when 𝑘→∞: 𝑝𝑦_2𝑚𝑘,𝑦_2𝑛𝑘,𝑝𝑦_2𝑚𝑘,𝑦_2𝑛𝑘+1,𝑝𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘,𝑝𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘+1. (3.6)

Proof. Suppose that {𝑦_𝑛} is a sequence in (𝑋,𝑝) satisfying (3.5) such that {𝑦_2𝑛} is not Cauchy. According to Lemma 2.4, it is not a Cauchy sequence in the metric space (𝑋,𝑝^𝑠), either. Applying Lemma 3.3 we get the sequences 𝑝^𝑠𝑦_2𝑚𝑘,𝑦_2𝑛𝑘,𝑝^𝑠𝑦_2𝑚𝑘,𝑦_2𝑛𝑘+1,𝑝^𝑠𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘,𝑝^𝑠𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘+1 (3.7) tending (from above) to some 2𝜀>0 when 𝑘→∞. Using definition (2.2) of the associated metric and (3.5) (which implies that also lim_𝑛→∞𝑝(𝑦_𝑛,𝑦_𝑛)=0), we get that the sequences (3.6) tend to 𝜀+0 when 𝑘→∞.

4. Common Fixed Point Results for Four Mappings

In this section we prove two common fixed point results for four mappings in partial metric spaces, using two distinct properties of a comparison function mentioned in Section 3.

Theorem 4.1. Let (𝑋,𝑝) be a complete partial metric space, and let 𝐴,𝐵,𝑆,𝑇∶𝑋→𝑋. Suppose that 𝐴𝑋⊂𝑇𝑋, 𝐵𝑋⊂𝑆𝑋 and one of these four subsets of 𝑋 is closed. Let further 𝑝𝐴𝑥,𝐵𝑦≤𝜑𝑀𝑥,𝑦 (4.1) hold for all 𝑥,𝑦∈𝑋, where 𝑀𝑝(),1𝑥,𝑦=max𝑆𝑥,𝑇𝑦,𝑝𝐴𝑥,𝑆𝑥,𝑝𝐵𝑦,𝑇𝑦2𝑝𝑆𝑥,𝐵𝑦+𝑝𝐴𝑥,𝑇𝑦, (4.2) and 𝜑∶[0,+∞)→[0,+∞) satisfies property (III), that is, 𝜑 is right-continuous, and 𝜑(𝑡)<𝑡 for each 𝑡>0. Then 𝐴 and 𝑆, as well as 𝐵 and 𝑇, have a coincidence point. If, moreover, pairs (𝐴,𝑆) and (𝐵,𝑇) are weakly compatible, then 𝐴,𝐵,𝑆,𝑇 have a unique common fixed point.

Recall that a point 𝑦∈𝑋 is called a point of coincidence for mappings 𝑓,𝑔∶𝑋→𝑋 if there exists 𝑥∈𝑋 such that 𝑓(𝑥)=𝑔(𝑥)=𝑦. Then, 𝑥 is called a coincidence point. Mappings 𝑓 and 𝑔 are called weakly compatible if they commute at their coincidence points.

Proof. Starting from arbitrary 𝑥₀∈𝑋 and using that 𝐴𝑋⊂𝑇𝑋, 𝐵𝑋⊂𝑆𝑋 construct a Jungck sequence {𝑦_𝑛} by 𝑦_2𝑛=𝐴𝑥_2𝑛=𝑇𝑥_2𝑛+1,𝑦_2𝑛+1=𝐵𝑥_2𝑛+1=𝑆𝑥_2𝑛+2,𝑛=0,1,2,…. (4.3) Consider two possible cases.
(1)𝑝(𝑦_𝑛,𝑦_𝑛+1)=0 (and so 𝑦_𝑛=𝑦_𝑛+1) for some 𝑛∈ℕ. Let, for example, 𝑛=2𝑘−1, 𝑘∈ℕ. Then 𝑝𝑦_2𝑘,𝑦_2𝑘+1=𝑝𝐴𝑥_2𝑘,𝐵𝑥_2𝑘+1≤𝜑𝑀𝑥_2𝑘,𝑥_2𝑘+1, (4.4) where 𝑀𝑥_2𝑘,𝑥_2𝑘+1=max𝑝𝑦_2𝑘−1,𝑦_2𝑘,𝑝𝑦_2𝑘,𝑦_2𝑘−1,𝑝𝑦_2𝑘+1,𝑦_2𝑘,12𝑝𝑦_2𝑘−1,𝑦_2𝑘+1+𝑝𝑦_2𝑘,𝑦_2𝑘=𝑝𝑦_2𝑘,𝑦_2𝑘+1byp₄. (4.5) It follows that 𝑝(𝑦_2𝑘,𝑦_2𝑘+1)≤𝜑(𝑝(𝑦_2𝑘,𝑦_2𝑘+1))<𝑝(𝑦_2𝑘,𝑦_2𝑘+1), which is impossible, unless 𝑝(𝑦_2𝑘,𝑦_2𝑘+1)=0 and 𝑦_2𝑘=𝑦_2𝑘+1. In a similar way, if 𝑛=2𝑘, 𝑘∈ℕ, it follows that also 𝑦_2𝑘+1=𝑦_2𝑘+2. Hence, in both cases we obtain that the sequence {𝑦_𝑛} is eventually constant, and so a Cauchy one.
(2) Suppose that 𝑝(𝑦_𝑛,𝑦_𝑛+1)>0 for all 𝑛∈ℕ. Then, as above, 𝑝𝑦_2𝑛,𝑦_2𝑛+1=𝑝𝐴𝑥_2𝑛,𝐵𝑥_2𝑛+1≤𝜑𝑀𝑥_2𝑛,𝑥_2𝑛+1, (4.6) where 𝑀𝑥_2𝑛,𝑥_2𝑛+1=max𝑝𝑦_2𝑛−1,𝑦_2𝑛,𝑝𝑦_2𝑛,𝑦_2𝑛−1,𝑝𝑦_2𝑛+1,𝑦_2𝑛,12𝑝𝑦_2𝑛−1,𝑦_2𝑛+1+𝑝𝑦_2𝑛,𝑦_2𝑛=max𝑝𝑦_2𝑛−1,𝑦_2𝑛,𝑝𝑦_2𝑛,𝑦_2𝑛+1byp₄. (4.7) Similarly, 𝑝(𝑦_2𝑛+1,𝑦_2𝑛+2)≤𝜑(𝑀(𝑥_2𝑛+1,𝑥_2𝑛+2)),𝑀(𝑥_2𝑛+1,𝑥_2𝑛+2)=max{𝑝(𝑦_2𝑛,𝑦_2𝑛+1),𝑝(𝑦_2𝑛+1,𝑦_2𝑛+2)}, that is 𝑝𝑦_𝑛,𝑦_𝑛+1≤𝜑max𝑝𝑦_𝑛−1,𝑦_𝑛,𝑝𝑦_𝑛,𝑦_𝑛+1,𝑛=1,2,…. (4.8) If 𝑝(𝑦_𝑛,𝑦_𝑛+1)≥𝑝(𝑦_𝑛−1,𝑦_𝑛), then 𝑝(𝑦_𝑛,𝑦_𝑛+1)≤𝜑(𝑝(𝑦_𝑛,𝑦_𝑛+1))<𝑝(𝑦_𝑛,𝑦_𝑛+1), a contradiction. It follows that𝑝𝑦_𝑛,𝑦_𝑛+1<𝑝𝑦_𝑛−1,𝑦_𝑛,𝑛=1,2,…. (4.9) Thus, in this case {𝑝(𝑦_𝑛,𝑦_𝑛+1)} is a decreasing sequence of positive numbers. Denote lim_𝑛→∞𝑝(𝑦_𝑛,𝑦_𝑛+1)=𝑟≥0. In fact, 𝑟=0. Indeed, if 𝑟>0, then passing to the limit when 𝑛→∞ in 𝑝(𝑦_𝑛,𝑦_𝑛+1)≤𝜑(𝑝(𝑦_𝑛−1,𝑦_𝑛)), and using properties (III) of 𝜑, we get that 𝑟≤𝜑(𝑟)<𝑟, (4.10) a contradiction.
We have proved that lim_𝑛→∞𝑝(𝑦_𝑛,𝑦_𝑛+1)=0. Suppose that {𝑦_2𝑛} is not a Cauchy sequence. Then, Lemma 3.4 implies that there exist 𝜀>0 and two subsequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the sequences (3.6) tend to 𝜀 (from above) when 𝑘→∞. Now, using (4.1) with 𝑥=𝑥_2𝑚𝑘 and 𝑦=𝑥_2𝑛𝑘+1, we get that𝑝𝑦_2𝑚𝑘,𝑦_2𝑛𝑘+1≤𝜑𝑀𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1, (4.11) where 𝑀𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1=max𝑝𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘,𝑝𝑦_2𝑚𝑘,𝑦_{2𝑚𝑘−1},𝑝𝑦_2𝑛𝑘+1,𝑦_2𝑛𝑘,12𝑝𝑦_{2𝑚𝑘−1},𝑦_2𝑛𝑘+1+𝑝𝑦_2𝑚𝑘,𝑦_2𝑛𝑘⟶𝜀+0,𝑘⟶∞. (4.12) Using properties (III) of 𝜑, we obtain a contradiction 𝜀≤𝜑(𝜀)<𝜀, since 𝜀>0.
Thus {𝑦_2𝑛} and so also {𝑦_𝑛} is a Cauchy sequence, both in (𝑋,𝑝) and in (𝑋,𝑝^𝑠). Suppose that, for example, 𝑆𝑋 is closed in 𝑋, and hence complete. It follows that sequence {𝑦_𝑛} converges in the metric space (𝑋,𝑝^𝑠), say lim_𝑛→∞𝑝^𝑠(𝑦_𝑛,𝑧)=0, where 𝑧=𝑆𝑢 for some 𝑢∈𝑋. Again from Lemma 2.4, we have 𝑝(𝑧,𝑧)=lim_𝑛→∞𝑝𝑦_𝑛,𝑧=lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚. (4.13)
Moreover, since {𝑦_𝑛} is a Cauchy sequence in the metric space (𝑋,𝑝^𝑠), we have lim_{𝑛,𝑚→∞}𝑝^𝑠(𝑦_𝑛,𝑦_𝑚)=0 and so, by the definition of 𝑝^𝑠, we have lim_{𝑛,𝑚→∞}𝑝(𝑦_𝑛,𝑦_𝑚)=0. Then (4.13) implies that 𝑝(𝑧,𝑧)=0 and lim_𝑛→∞𝑝𝑦_𝑛,𝑧=𝑝(𝑧,𝑧)=0. (4.14) We will prove that 𝑧 is a point of coincidence of 𝐴 and 𝑆, that is, 𝐴𝑢=𝑧.
Suppose that 𝑝(𝐴𝑢,𝑧)>0. Then, using (𝑝₄),𝑝(𝐴𝑢,𝑧)≤𝑝𝐴𝑢,𝐵𝑥_2𝑛+1+𝑝𝐵𝑥_2𝑛+1,𝑧−𝑝𝐵𝑥_2𝑛+1,𝐵𝑥_2𝑛+1≤𝜑max𝑝𝑧,𝑦_2𝑛,𝑝(𝐴𝑢,𝑧),𝑝𝑦_2𝑛+1,𝑦_2𝑛,12𝑝𝑧,𝑦_2𝑛+1+𝑝𝐴𝑢,𝑦_2𝑛+𝑝𝑦_2𝑛+1𝑝,𝑧≤𝜑max𝑧,𝑦_2𝑛,𝑝(𝐴𝑢,𝑧),𝑝𝑦_2𝑛+1,𝑦_2𝑛,12𝑝𝑧,𝑦_2𝑛+1+𝑝(𝐴𝑢,𝑧)+𝑝𝑧,𝑦_2𝑛+𝑝𝑦_2𝑛+1,𝑧. (4.15) All terms in the previous set {⋯}, which depend on 𝑛, tend to 0 when 𝑛→∞ and they are smaller, for 𝑛 large enough, than, say, (1/2)𝑝(𝐴𝑢,𝑧). It follows that 𝑝(𝐴𝑢,𝑧)≤𝜑𝑝(𝐴𝑢,𝑧)+𝑝𝑦_2𝑛+1,𝑧. (4.16) Letting 𝑛→∞, and using that 𝜑(𝑡)<𝑡, we get that 𝑝(𝐴𝑢,𝑧)≤𝜑(𝑝(𝐴𝑢,𝑧))<𝑝(𝐴𝑢,𝑧), a contradiction. Hence, 𝑝(𝐴𝑢,𝑧)=0 and 𝐴𝑢=𝑧.
Now 𝐴𝑋⊂𝑇𝑋 implies that 𝑧=𝐴𝑢∈𝑇𝑋 and so there exists 𝑣∈𝑋 such that 𝑇𝑣=𝑧. If 𝑝(𝑧,𝐵𝑣)>0, then 𝑝()()𝑝()()(),1𝑧,𝐵𝑣=𝑝𝐴𝑢,𝐵𝑣≤𝜑max𝑆𝑢,𝑇𝑣,𝑝𝐴𝑢,𝑆𝑢,𝑝𝐵𝑣,𝑇𝑣2𝑝()()𝑝()(),𝑆𝑢,𝐵𝑣+𝑝𝐴𝑢,𝑇𝑣≤𝜑𝑧,𝐵𝑣<𝑝𝑧,𝐵𝑣 (4.17)
a contradiction. Hence, 𝑝(𝑧,𝐵𝑣)=0 and 𝐵𝑣=𝑧, so 𝑧 is a point of coincidence, both for (𝐴,𝑆) and for (𝐵,𝑇). It order to show that this point of coincidence is unique, one has only to use that 𝜑(𝑡)<𝑡 for 𝑡>0 and property (𝑝₂) of partial metric. Hence, if these pairs of mappings are weakly compatible, it follows by a well-known result that 𝑧 is a unique common fixed point of 𝐴,𝐵,𝑆,𝑇. The theorem is proved.

Theorem 4.2. Let (𝑋,𝑝) be a complete partial metric space, and let 𝐴,𝐵,𝑆,𝑇∶𝑋→𝑋. Suppose that 𝐴𝑋⊂𝑇𝑋, 𝐵𝑋⊂𝑆𝑋 and one of these four subsets of 𝑋 is closed. Let further 𝑝𝐴𝑥,𝐵𝑦≤𝜑𝑀𝑥,𝑦 (4.18) hold for all 𝑥,𝑦∈𝑋, where 𝑀𝑝(),1𝑥,𝑦=max𝑆𝑥,𝑇𝑦,𝑝𝐴𝑥,𝑆𝑥,𝑝𝐵𝑦,𝑇𝑦2𝑝𝑆𝑥,𝐵𝑦+𝑝𝐴𝑥,𝑇𝑦, (4.19) and 𝜑∶[0,+∞)→[0,+∞) satisfies property (IV), that is, 𝜑 is nondecreasing and ∑_𝑛≥1𝜑^𝑛(𝑡)<+∞ for each 𝑡≥0. Then 𝐴 and 𝑆, as well as 𝐵 and 𝑇, have a coincidence point. If, moreover, pairs (𝐴,𝑆) and (𝐵,𝑇) are weakly compatible, then 𝐴,𝐵,𝑆,𝑇 have a unique common fixed point.

Proof. Construct a Jungck sequence {𝑦_𝑛} as in the proof of Theorem 4.1. By Lemma 3.1, conditions (IV) imply that 𝜑^𝑛(𝑡)→0, 𝑛→∞ for each 𝑡∈[0,+∞). Since 𝜑 is nondecreasing, it further implies that 𝜑(𝑡)<𝑡 for each 𝑡>0. Condition (4.9) follows as in the proof of Theorem 4.1. It further implies that 𝑝𝑦_𝑛,𝑦_𝑛+1<𝜑^𝑛𝑝𝑦₀,𝑦₁,𝑛∈ℕ. (4.20) By the definition of associated metric (2.2), we get that 𝑝^𝑠𝑦_𝑛,𝑦_𝑛+1<4𝜑^𝑛𝑝𝑦₀,𝑦₁,𝑛∈ℕ. (4.21) Now, for arbitrary 𝑘∈ℕ, 𝑝^𝑠𝑦_𝑛,𝑦_𝑛+𝑘≤𝑝^𝑠𝑦_𝑛,𝑦_𝑛+1+𝑝^𝑠𝑦_𝑛+1,𝑦_𝑛+2+⋯+𝑝^𝑠𝑦_{𝑛+𝑘−1},𝑦_𝑛+𝑘≤4^{𝑛+𝑘−1}_𝑖=𝑛𝜑^𝑖𝑝𝑦₀,𝑦₁≤4^∞_𝑖=𝑛𝜑^𝑖𝑝𝑦₀,𝑦₁⟶0 (4.22) as 𝑛→∞, by conditions (IV). Hence {𝑦_𝑛} is a Cauchy sequence in 𝑋.
The rest of the proof is the same as in Theorem 4.1, since it uses only the contractive condition and that 𝜑(𝑡)<𝑡 for 𝑡>0.

Remark 4.3. (1) It follows from the proofs of previous theorems that, under the respective hypotheses, an arbitrary Jungck sequence {𝑦_𝑛} converges to the (unique) common fixed point 𝑧 of 𝐴,𝐵,𝑆,𝑇.
(2) Taking 𝜑(𝑡)=𝑘𝑡, 𝑘∈[0,1) in Theorems 4.1 and 4.2, we obtain an extension of Fisher’s theorem for four mappings [15] to the setting of partial metric spaces.
(3) Taking 𝐴=𝐵=𝑆=𝑇, Theorems 3 and 4 from [8] are obtained, and so also extensions of [3, Theorem 1] and [2, Theorem 1]. For [4, Theorem 2.1], precise assumptions for the function 𝜑 are stated.
(4) Our results are far extensions of the classical Boyd-Wong result [12] to the setting of partial metric spaces.
(5) A related result for so-called Geraghty-type mappings was recently obtained in [5].

We illustrate the results of this section with an example.

Example 4.4. Let 𝑋=[0,1] be equipped with the partial metric 𝑝(𝑥,𝑦)=max{𝑥,𝑦} and let 𝜑(𝑡)=(𝑡/(𝑡+1))𝑒^−𝑡 for 𝑡≥0 (𝜑 satisfies condition (III)). The mappings 1𝐴𝑥=6𝑥²1,𝐵𝑥=8𝑥²1,𝑇𝑥=2𝑥²,𝑆𝑥=𝑥² (4.23) are such that 𝐴𝑋⊂𝑇𝑋, 𝐵𝑋⊂𝑆𝑋, and (𝐴,𝑆) as well as (𝐵,𝑇) are weakly compatible. In order to check condition (4.1), take arbitrary 𝑥,𝑦∈𝑋 with, first, 𝑥≥𝑦. Then, 𝑝1𝐴𝑥,𝐵𝑦=𝑝6𝑥²,18𝑦²=16𝑥²,𝑀𝑥,𝑦=max𝑝𝑥²,12𝑦²1,𝑝6𝑥²,𝑥²1,𝑝8𝑦²,12𝑦²,12𝑝𝑥²,18𝑦²1+𝑝6𝑥²,12𝑦²=max𝑥²,𝑥²,12𝑥²,12𝑥²1+max6𝑥²,12𝑦²=𝑥²,𝜑𝑀=𝑥𝑥,𝑦²𝑥²𝑒+1^−𝑥2≥16𝑥², (4.24) since 𝑒^−𝑥2/(𝑥²+1)≥1/2𝑒>1/6 for 𝑥∈[0,1]. In the case 𝑥<𝑦, the same can be checked after careful calculations. Hence, 𝐴,𝐵,𝑆,𝑇 satisfy all the conditions of Theorem 4.1 and they have a unique common fixed point (𝑧=0).

5. Common Fixed Points for Two Mappings under Weaker Condition for the Comparison Function

In the next theorem we consider weaker condition (II) for the comparison function 𝜑. As a compensation, we assume a bit stronger contractive condition.

Theorem 5.1. Let (𝑋,𝑝) be a complete partial metric space and let 𝐴,𝑇∶𝑋→𝑋 be two self-maps. Let 𝐴𝑋⊂𝑇𝑋 and at least one of these subsets of 𝑋 is closed. Suppose that there is a function 𝜑∶[0,+∞)→[0,+∞) satisfying property (II), such that 𝑝𝑝(),1𝐴𝑥,𝐴𝑦≤𝜑max𝑇𝑥,𝑇𝑦,𝑝𝐴𝑥,𝑇𝑥,𝑝𝐴𝑦,𝑇𝑦2𝑝𝑇𝑥,𝐴𝑦 (5.1) holds for all 𝑥,𝑦∈𝑋. Then 𝐴 and 𝑇 have a unique point of coincidence. If 𝐴 and 𝑇 are weakly compatible, then they have a unique common fixed point 𝑢∈𝑋.

Proof. We will construct a Jungck sequence in the usual way. Take arbitrary 𝑥₀∈𝑋 and using that 𝐴𝑋⊂𝑇𝑋, choose 𝑥_𝑛∈𝑋 such that 𝑦_𝑛=𝐴𝑥_𝑛=𝑇𝑥_𝑛+1, 𝑛=0,1,2,…. First, suppose that there exists an 𝑛 such that 𝑝(𝑦_𝑛,𝑦_𝑛−1)=0. Then the sequence {𝑦_𝑛} is eventually constant. Indeed, from (5.1) it follows that 𝑝𝑦_𝑛+1,𝑦_𝑛=𝑝𝐴𝑥_𝑛+1,𝐴𝑥_𝑛≤𝜑max𝑝𝑇𝑥_𝑛+1,𝑇𝑥_𝑛,𝑝𝐴𝑥_𝑛+1,𝑇𝑥_𝑛+1,𝑝𝐴𝑥_𝑛,𝑇𝑥_𝑛,12𝑝𝑇𝑥_𝑛+1,𝐴𝑥_𝑛=𝜑max𝑝𝑦_𝑛,𝑦_𝑛−1,𝑝𝑦_𝑛,𝑦_𝑛+1,𝑝𝑦_𝑛−1,𝑦_𝑛,12𝑝𝑦_𝑛,𝑦_𝑛=𝜑𝑝𝑦_𝑛+1,𝑦_𝑛, (5.2) since (1/2)𝑝(𝑦_𝑛,𝑦_𝑛)≤𝑝(𝑦_𝑛,𝑦_𝑛)≤𝑝(𝑦_𝑛+1,𝑦_𝑛) by (𝑝₂). This is a contradiction with 𝜑(𝑡)<𝑡 (which follows from assumption (I)), unless 𝑝(𝑦_𝑛+1,𝑦_𝑛)=0. Hence, 𝑦_𝑛+1=𝑦_𝑛=𝑦_𝑛−1. Continuing this process, we obtain that {𝑦_𝑛} is an eventually constant sequence and hence a Cauchy one. Moreover, from 𝑦_𝑛=𝐴𝑥_𝑛=𝑇𝑥_𝑛+1=𝑦_𝑛+1=𝐴𝑥_𝑛+1=𝑇𝑥_𝑛+1 it follows that 𝐴𝑥_𝑛+1=𝑇𝑥_𝑛+1 and 𝐴 and 𝑇 have a coincidence point.
Suppose now that 𝑝(𝑦_𝑛,𝑦_𝑛−1)>0 for all 𝑛∈ℕ. Then, similarly as above, we get 𝑝𝑦_𝑛+1,𝑦_𝑛≤𝜑max𝑝𝑦_𝑛,𝑦_𝑛−1,𝑝𝑦_𝑛+1,𝑦_𝑛. (5.3)
Using property (I) of function 𝜑 it follows that 𝑝𝑦_𝑛+1,𝑦_𝑛≤𝜑𝑝𝑦_𝑛,𝑦_𝑛−1<𝑝𝑦_𝑛,𝑦_𝑛−1. (5.4) Hence, 𝑝(𝑦_𝑛+1,𝑦_𝑛)<𝜑^𝑛(𝑝(𝑦₀,𝑦₁))→0 when 𝑛→∞ and {𝑝(𝑦_𝑛+1,𝑦_𝑛)} is a decreasing sequence, tending to 0.
Now, using mathematical induction, we prove that {𝑦_𝑛} is a Cauchy sequence. Since 𝑝(𝑦_𝑛,𝑦_𝑛+1)→0, for each 𝜀>0, there exists 𝑛(𝜀) such that 𝑝(𝑦_𝑛,𝑦_𝑛+1)<𝜀−𝜑(𝜀) for 𝑛>𝑛(𝜀). Let, for some 𝑘∈ℕ, 𝑝(𝑦_𝑛,𝑦_𝑛+𝑘)<𝜀. Then we have 𝑝𝑦_𝑛,𝑦_𝑛+𝑘+1≤𝑝𝑦_𝑛,𝑦_𝑛+1+𝑝𝑦_𝑛+1,𝑦_𝑛+𝑘+1, (5.5)
where 𝑝𝑦_𝑛+1,𝑦_𝑛+𝑘+1=𝑝𝐴𝑥_𝑛+1,𝐴𝑥_𝑛+𝑘+1≤𝜑max𝑝𝑇𝑥_𝑛+1,𝑇𝑥_𝑛+𝑘+1,𝑝𝐴𝑥_𝑛+1,𝑇𝑥_𝑛+1,𝑝𝐴𝑥_𝑛+𝑘+1,𝑇𝑥_𝑛+𝑘+1,12𝑝𝑇𝑥_𝑛+1,𝐴𝑥_𝑛+𝑘+1=𝜑max𝑝𝑦_𝑛,𝑦_𝑛+𝑘,𝑝𝑦_𝑛+1,𝑦_𝑛,𝑝𝑦_𝑛+𝑘+1,𝑦_𝑛+𝑘,12𝑝𝑦_𝑛,𝑦_𝑛+𝑘+1. (5.6) The first three members of the last set are smaller than 𝜀. Concerning the fourth one, we have that 12𝑝𝑦_𝑛,𝑦_𝑛+𝑘+1≤12𝑝𝑦_𝑛,𝑦_𝑛+𝑘+12𝑝𝑦_𝑛+𝑘,𝑦_𝑛+𝑘+1<121𝜀+2𝜀−𝜑(𝜀)<𝜀, (5.7) and it follows that 𝑝𝑦_𝑛+1,𝑦_𝑛+𝑘+1<𝜑(𝜀). (5.8) Hence, 𝑝𝑦_𝑛,𝑦_𝑛+𝑘+1<𝜀−𝜑(𝜀)+𝜑(𝜀)=𝜀, (5.9) and the inductive proof is over.
Using the definition of the associated metric 𝑝^𝑠, we get that 𝑝^𝑠𝑦_𝑛,𝑦_𝑛+𝑘≤4𝑝𝑦_𝑛,𝑦_𝑛+𝑘<4𝜀, (5.10)
for each 𝑛>𝑛(𝜀) and 𝑘∈ℕ. Thus, {𝑦_𝑛} is also a 𝑝^𝑠-Cauchy sequence, and so there exists 𝑦∈𝑋 such that 𝑝𝑦,𝑦=lim_𝑛→∞𝑝𝑦_𝑛,𝑦=lim_{𝑚,𝑛→∞}𝑝𝑦_𝑛,𝑦_𝑚. (5.11) Suppose that, for example, 𝑇𝑋 is closed. Then 𝑦=𝑇𝑢 for some 𝑢∈𝑋. We will prove that also 𝐴𝑢=𝑦. Suppose that 𝑝(𝐴𝑢,𝑦)>0. Then, using (5.1), we get 𝑝𝑦,𝐴𝑢≤𝑝𝑦,𝐴𝑥_𝑛+𝑝𝐴𝑥_𝑛,𝐴𝑢≤𝑝𝑦,𝐴𝑥_𝑛+𝜑max𝑝𝑇𝑥_𝑛,𝑇𝑢,𝑝𝐴𝑥_𝑛,𝑇𝑥_𝑛(),1,𝑝𝐴𝑢,𝑇𝑢2𝑝𝑇𝑢,𝐴𝑥_𝑛=𝑝𝑦,𝐴𝑥_𝑛+𝜑max𝑝𝑦_𝑛−1,𝑇𝑢,𝑝𝑦_𝑛,𝑦_𝑛−1,1,𝑝𝐴𝑢,𝑦2𝑝𝑇𝑢,𝑦_𝑛. (5.12) Since 𝑦_𝑛→𝑦=𝑇𝑢, there exists 𝑛₀ such that 𝑝(𝑦_𝑛−1,𝑇𝑢)<𝑝(𝐴𝑢,𝑦) and 𝑝(𝑦_𝑛,𝑦_𝑛−1)<𝑝(𝐴𝑢,𝑦) whenever 𝑛>𝑛₀. It follows that 𝑝𝑦,𝐴𝑢≤𝑝𝑦,𝐴𝑥_𝑛+𝜑𝑝𝐴𝑢,𝑦 (5.13) and, passing to the limit when 𝑛→∞, we obtain that 𝑝𝑦,𝐴𝑢≤0+𝜑𝑝𝐴𝑢,𝑦<𝑝𝐴𝑢,𝑦, (5.14) which is a contradiction. Hence, 𝑝(𝐴𝑢,𝑦)=0 and 𝐴𝑢=𝑦=𝑇𝑢 is a point of coincidence of 𝐴 and 𝑇.
Note that continuity of 𝜑 was not needed in the previous conclusions.
In order to prove that the point of coincidence is unique, suppose that there exists 𝑦₁≠𝑦 (and so 𝑝(𝑦₁,𝑦)>0) such that 𝐴𝑢₁=𝑇𝑢₁=𝑦₁ for some 𝑢₁∈𝑋. Then (5.1) implies that 𝑝𝑦₁,𝑦=𝑝𝐴𝑢₁,𝐴𝑢≤𝜑max𝑝𝑇𝑢₁,𝑇𝑢,𝑝𝐴𝑢₁,𝑇𝑢₁(),1,𝑝𝐴𝑢,𝑇𝑢2𝑝𝑇𝑢₁𝑝𝑦,𝐴𝑢=𝜑max₁,𝑦,𝑝𝑦₁,𝑦₁,1,𝑝𝑦,𝑦2𝑝𝑦₁,𝑦≤𝜑𝑝𝑦₁,𝑦<𝑝𝑦₁,𝑦, (5.15)since 𝑝(𝑦,𝑦)≤𝑝(𝑦,𝑦₁) and 𝑝(𝑦₁,𝑦₁)≤𝑝(𝑦,𝑦₁). This contradiction shows that 𝑝(𝑦₁,𝑦)=0 and 𝑦₁=𝑦.
The last assertion of the theorem follows easily from the definition of weak compatibility and the uniqueness of the coincidence point.

We present an example where the existence of a common fixed point can be proved using Theorem 5.1 (and conditions formulated in terms of a partial metric) but cannot be obtained using respective conditions in the associated (standard) metric.

Example 5.2. Consider on 𝑋={0,1,2} both the partial metric 𝑝(𝑥,𝑦)=max{𝑥,𝑦} and the associated metric 𝑑(𝑥,𝑦)=|𝑥−𝑦|. Let 𝐴∶𝑋→𝑋 be given by 𝐴0=𝐴1=0, 𝐴2=1, let 𝑇=𝑖_𝑋 (identity map), and let 𝜑(𝑡)=(3/4)𝑡, 𝑡∈[0,+∞). Then 𝑝()()𝑝()()(2)𝑝()()()(,𝑝()()(2)𝑝()()()(,𝐴0,𝐴1=𝑝0,0=0,𝐴0,𝐴2=𝑝0,1=1<3/2=𝜑=𝜑max0,2,𝑝0,0,𝑝2,1,1/2𝑝0,1)𝐴1,𝐴2=𝑝0,1=1<3/2=𝜑=𝜑max1,2,𝑝1,0,𝑝2,1,1/2𝑝1,1) (5.16) and condition (5.1) is satisfied. The respective condition in the standard metric does not hold since 𝑑(𝐴1,𝐴2)=𝑑(0,1)=1>3/4=𝜑(1)=𝜑(max{𝑑(1,2),𝑑(1,0),𝑑(2,1),1/2𝑑(1,1)}). (5.17)

6. Meir Keeler-Type Result in Partial Metric Spaces

It is well-known that the celebrated Meir-Keeler fixed point result [16] can also be formulated in a form with a comparison function. We present a partial metric version of this result. The function 𝜑∶[0,+∞)→[0,+∞) will be called a Meir-Keeler function if ∀𝜀>0∃𝛿>0∀𝑡𝜀≤𝑡<𝜀+𝛿⟹𝜑(𝑡)<𝜀. (6.1)

Theorem 6.1. Let (𝑋,𝑝) be a complete partial metric space and let a mapping 𝑓∶𝑋→𝑋 satisfies the following condition ∀𝑥,𝑦∈𝑋𝑝𝑓𝑥,𝑓𝑦≤𝜑𝑝𝑥,𝑦, (6.2) for a Meir-Keeler function 𝜑. Then 𝑓 has a unique fixed point, say 𝑢, and for each 𝑥∈𝑋, the Picard sequence {𝑓^𝑛𝑥} converges to 𝑢, satisfying 𝑝(𝑢,𝑢)=0.

Proof. We prove first the uniqueness. Let 𝑢,𝑣 be two fixed points of 𝑓 and let 𝑝(𝑢,𝑣)>0. Then, taking 𝜀=𝑝(𝑢,𝑣) in (9), it follows that there exists 𝛿>0 such that 𝑝(𝑢,𝑣)≤𝑡<𝑝(𝑢,𝑣)+𝛿 implies that 𝜑(𝑡)<𝑝(𝑢,𝑣). In particular, 𝜑(𝑝(𝑢,𝑣))<𝑝(𝑢,𝑣). But then, by (6.2), 𝑝(𝑢,𝑣)=𝑝(𝑓𝑢,𝑓𝑣)≤𝜑(𝑝(𝑢,𝑣))<𝑝(𝑢,𝑣), a contradiction. Hence, 𝑝(𝑢,𝑣)=0 and 𝑢=𝑣.
Take now arbitrary 𝑥∈𝑋 and form the Picard sequence {𝑓^𝑛𝑥}. If, for some 𝑛, 𝑝(𝑓^𝑛𝑥,𝑓^𝑛+1𝑥)=0, then 𝑓𝑓^𝑛𝑥=𝑓^𝑛𝑥 and 𝑓^𝑛𝑥 is a (unique) fixed point of 𝑓. Suppose that 𝑝(𝑓^𝑛𝑥,𝑓^𝑛+1𝑥)>0 for each 𝑛=0,1,2,…. Taking (for fixed 𝑛) 𝜀=𝑝(𝑓^𝑛𝑥,𝑓^𝑛+1𝑥) in (6.1), there exists 𝛿>0 such that 𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥≤𝑡<𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥+𝛿⟹𝜑(𝑡)<𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥. (6.3)
Then, by (6.2), 𝑝𝑓^𝑛+1𝑥,𝑓^𝑛+2𝑥=𝑝𝑓𝑓^𝑛𝑥,𝑓𝑓^𝑛+1𝑥≤𝜑𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥<𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥. (6.4) It follows that {𝑝(𝑓^𝑛𝑥,𝑓^𝑛+1𝑥)} is a decreasing sequence of positive numbers, tending to some 𝑟≥0. If 𝑟>0 then, again by (6.1) and (6.2), we can find 𝛿 such that 𝑟≤𝑝𝑓^𝑛𝑥,𝑓^𝑛+1𝑥<𝑟+𝛿⟹𝑝𝑓^𝑛+1𝑥,𝑓^𝑛+2𝑥<𝑟. (6.5) This is a contradiction since 𝑝(𝑓^𝑛+1𝑥,𝑓^𝑛+2𝑥)≥𝑟. We conclude that 𝑝(𝑓^𝑛𝑥,𝑓^𝑛+1𝑥)→0 when 𝑛→∞.
In order to prove that {𝑓^𝑛𝑥} is a Cauchy sequence, we use again Lemma 3.4. If we suppose the contrary, then there exist 𝜀>0 and two sequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that 𝑝(𝑓^𝑛𝑘𝑥,𝑓^𝑚𝑘𝑥)→𝜀⁺ and also 𝑝(𝑓^𝑛𝑘+1𝑥,𝑓^𝑚𝑘+1𝑥)→𝜀⁺. But, (6.1) and (6.2) imply that there exists 𝛿>0 such that 𝜀≤𝑝(𝑓^𝑛𝑘𝑥,𝑓^𝑚𝑘𝑥)<𝜀+𝛿 implies 𝑝(𝑓^𝑛𝑘+1𝑥,𝑓^𝑚𝑘+1𝑥)<𝜀. This is a contradiction since 𝑝(𝑓^𝑛𝑘+1𝑥,𝑓^𝑚𝑘+1𝑥)≥𝜀 for large 𝑘.
Thus, {𝑓^𝑛𝑥} is a Cauchy sequence, both in (𝑋,𝑝) and in (𝑋,𝑝^𝑠) (Lemma 2.4) and there exists an element 𝑢 in the complete (partial) metric space 𝑋 such that lim_𝑛→∞𝑝^𝑠(𝑓^𝑛𝑥,𝑢)=0, wherefrom 𝑝(𝑢,𝑢)=lim_𝑛→∞𝑝𝑓^𝑛𝑥,𝑢=lim_{𝑚,𝑛→∞}𝑝𝑓^𝑛𝑥,𝑓^𝑚𝑥. (6.6)
By the definition (2.2) of metric 𝑝^𝑠, and lim_𝑛→∞𝑝𝑓^𝑛𝑥,𝑓^𝑛𝑥=lim_𝑚→∞𝑝𝑓^𝑚𝑥,𝑓^𝑚𝑥=0, (6.7) it follows that 𝑝(𝑢,𝑢)=lim_𝑛→∞𝑝𝑓^𝑛𝑥,𝑢=lim_{𝑚,𝑛→∞}𝑝𝑓^𝑛𝑥,𝑓^𝑚𝑥=0. (6.8) We have proved that 0=𝑝(𝑢,𝑢)=lim_𝑛→∞𝑝𝑓^𝑛𝑥,𝑢. (6.9) It remains to show that 𝑓𝑢=𝑢. Take 𝑓^𝑛𝑥 instead of 𝑥 and 𝑢 instead of 𝑦 in (6.2). Then, using (6.1) we get that 𝑝𝑓^𝑛+1𝑥,𝑓𝑢<𝑝𝑓^𝑛,𝑝𝑥,𝑢𝑢,𝑓𝑢≤𝑝𝑢,𝑓^𝑛+1𝑥+𝑝𝑓^𝑛+1𝑥,𝑓𝑢−𝑝𝑓^𝑛+1𝑥,𝑓^𝑛+1𝑥≤𝑝𝑢,𝑓^𝑛+1𝑥+𝑝𝑓^𝑛+1𝑥,𝑓𝑢<𝑝𝑢,𝑓^𝑛+1𝑥+𝑝𝑓^𝑛()()𝑥,𝑢⟶𝑝𝑢,𝑢+𝑝𝑢,𝑢=0as𝑛⟶∞. (6.10) If follows that 𝑝(𝑢,𝑓𝑢)=0 and 𝑓𝑢=𝑢.

Example 6.2. Let 𝑋, 𝑝 and 𝑑 be as in Example 5.2. Consider mapping 𝑓∶𝑋→𝑋 and function 𝜑∶[0,+∞)→[0,+∞) given by (𝑡)=3𝑓=012001,𝜑4𝑡. (6.11) Then 𝜑 is a Meir-Keeler function. Indeed, for arbitrary 𝜀>0 choose 𝛿=(1/3)𝜀 and 𝜀≤𝑡<𝜀+𝛿=(4/3)𝜀 implies that 𝜑(𝑡)<𝜀. We will check that 𝑓 satisfies condition (6.2) of Theorem 6.1.
In the cases 𝑥=𝑦=0; 𝑥=𝑦=1; and 𝑥=0,𝑦=1, the left-hand side of (6.2) is equal to zero. In all other cases (𝑥=𝑦=2; 𝑥=0,𝑦=2; and 𝑥=1,𝑦=2), it is 𝑝(𝑓𝑥,𝑓𝑦)=1 and 𝜑(𝑝(𝑥,𝑦))=𝜑(2)=3/2. Hence, condition (6.2) always holds true, and mapping 𝑓 has a unique fixed point (𝑢=0).
Note again that in the case when standard metric 𝑑 is used instead of partial metric 𝑝, this conclusion cannot be obtained. Indeed, for 𝑥=1, 𝑦=2 we have that 𝑑()3𝑓1,𝑓2=𝑑0,1=1>4=𝜑(1)=𝜑(𝑑(1,2)). (6.12)

Acknowledgments

The authors are highly indebted to the referees for their comments that undoubtedly helped to improve the text. The first author gratefully acknowledges the support from the Deanship of Scientific Research (DSR) at King Abdulaziz University during this research. The fourth author acknowledges the financial support provided by University of Tabuk through the project of international cooperation. The second and third authors are thankful to the Ministry of Science and Technological Development of Serbia.