Fixed Points of Geraghty-Type Mappings in Various Generalized Metric Spaces

Abstract

Fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in the frame of partial metric spaces, ordered partial metric spaces, and metric-type spaces. Examples are given showing that these results are proper extensions of the existing ones.

1. Introduction

Let 𝒮 denote the class of real functions 𝛽∶[0,+∞)→[0,1) satisfying the condition 𝛽𝑡_𝑛⟶1implies𝑡_𝑛⟶0. (1.1) An example of a function in 𝒮 may be given by 𝛽(𝑡)=𝑒^−2𝑡 for 𝑡>0 and 𝛽(0)∈[0,1). In an attempt to generalize the Banach contraction principle, M. Geraghty proved in 1973 the following.

Theorem 1.1 (see [1]). Let (𝑋,𝑑) be a complete metric space, and let 𝑓∶𝑋→𝑋 be a self-map. Suppose that there exists 𝛽∈𝒮 such that 𝑑𝑓𝑥,𝑓𝑦≤𝛽𝑑𝑥,𝑦𝑑𝑥,𝑦 (1.2) holds for all 𝑥,𝑦∈𝑋. Then 𝑓 has a unique fixed point 𝑧∈𝑋 and for each 𝑥∈𝑋 the Picard sequence {𝑓^𝑛𝑥} converges to 𝑧 when 𝑛→∞.

Recently, A. Amini-Harandi and H. Emami extended this result to partially ordered metric spaces as follows.

Theorem 1.2 (see [2]). Let (𝑋,𝑑,⪯) be a complete partially ordered metric space. Let 𝑓∶𝑋→𝑋 be an increasing self-map such that there exists 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀. Suppose that there exists 𝛽∈𝒮 such that (1.2) holds for all 𝑥,𝑦∈𝑋 with 𝑥≽𝑦. Assume that either 𝑓 is continuous or 𝑋 is such that 𝑖𝑓𝑎𝑛𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒𝑥_𝑛𝑖𝑛𝑋𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠𝑡𝑜𝑥∈𝑋,𝑡ℎ𝑒𝑛𝑥_𝑛⪯𝑥∀𝑛. (1.3) Then, 𝑓 has a fixed point in 𝑋. If, moreover, 𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑥,𝑦∈𝑋𝑡ℎ𝑒𝑟𝑒𝑒𝑥𝑖𝑠𝑡𝑠𝑧∈𝑋𝑐𝑜𝑚𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑤𝑖𝑡ℎ𝑥,𝑦, (1.4) then the fixed point of 𝑓 is unique.

Similar results were also obtained in [3, 4].

In recent years several authors have worked on domain theory in order to equip semantics domain with a notion of distance. In particular, Matthews [5] 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 principle can be generalized to the partial metric context for applications in program verification. Subsequently, several authors (see, e.g., [6, 7]) studied fixed point theorems in partial metric spaces, as well as ordered partial metric spaces (see, e.g., [8, 9]).

Huang and Zhang introduced cone metric spaces in [10], replacing the set of real numbers by an ordered Banach space as the codomain for a metric. Cone metric spaces over normal cones inspired another generalization of metric spaces that were called metric-type spaces by Khamsi [11] (see also [12]; note that, in fact, spaces of this kind were used earlier under the name of 𝑏-spaces by Czerwik [13]).

In the present paper, we extend Theorems 1.1 and 1.2 to the frame of partial metric spaces, ordered partial metric spaces, and metric type spaces. Examples are given to distinguish new results from the existing ones.

2. Notation and Preliminary Results

2.1. Partial Metric Spaces

The following definitions and details can be seen in [5–9, 14, 15].

Definition 2.1. 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.

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.

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 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 (2.1) 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. (2.2)

Example 2.2. (1) A basic example of a partial metric space is the pair (ℝ⁺,𝑝), where 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈ℝ⁺. The corresponding metric is 𝑝^𝑠𝑥,𝑦=2max𝑥,𝑦−𝑥−𝑦=𝑥−𝑦. (2.3)(2) If (𝑋,𝑑) is a metric space and 𝑐≥0 is arbitrary, then 𝑝𝑥,𝑦=𝑑𝑥,𝑦+𝑐 (2.4) defines a partial metric on 𝑋 and the corresponding metric is 𝑝^𝑠(𝑥,𝑦)=2𝑑(𝑥,𝑦).

Other examples of partial metric spaces which are interesting from a computational point of view may be found in [5, 15].

Remark 2.3. Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the function 𝑝(⋅,⋅) need not be continuous in the sense that 𝑥_𝑛→𝑥 and 𝑦_𝑛→𝑦 implies 𝑝(𝑥_𝑛,𝑦_𝑛)→𝑝(𝑥,𝑦). 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→∞.

Definition 2.4 (see [8]). Let  (𝑋,𝑝)  be a partial metric space. Then one has 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).

Lemma 2.5 (see [5, 6]). 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.

Definition 2.6. Let  𝑋  be a nonempty set. Then  (𝑋,𝑝,⪯)  is called an ordered partial metric space if:(i)(𝑋,𝑝) is a partial metric space and (ii) (𝑋,⪯)  is a partially ordered set.
The space (𝑋,𝑝,⪯) is called regular if the following holds: if {𝑥_𝑛} is a nondecreasing sequence in 𝑋 with respect to ⪯ such that 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ.

2.2. Some Auxiliary Results

Assertions similar to the following lemma (see, e.g., [16]) were used (and proved) in the course of proofs of several fixed point results in various papers.

Lemma 2.7. Let (𝑋,𝑑) be a metric space, and let {𝑥_𝑛} be a sequence in 𝑋 such that lim_𝑛→∞𝑑𝑥_𝑛+1,𝑥_𝑛=0. (2.5) If {𝑥_2𝑛} is not a Cauchy sequence, then there exist 𝜀>0 and two sequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the following four sequences tend to 𝜀 when 𝑘→∞: 𝑑𝑥_2𝑚𝑘,𝑥_2𝑛𝑘,𝑑𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1,𝑑𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘,𝑑𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘+1. (2.6)

As a corollary we obtain the following.

Lemma 2.8. Let (𝑋,𝑝) be a partial metric space, and let {𝑥_𝑛} be a sequence in 𝑋 such that lim_𝑛→∞𝑝𝑥_𝑛+1,𝑥_𝑛=0. (2.7) If {𝑥_2𝑛} is not a Cauchy sequence in (𝑋,𝑝), then there exist 𝜀>0 and two sequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the following four sequences tend to 𝜀 when 𝑘→∞: 𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘,𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1,𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘,𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘+1. (2.8)

Proof. Suppose that {𝑥_𝑛} is a sequence in (𝑋,𝑝) satisfying (2.7) such that {𝑥_2𝑛} is not Cauchy. According to Lemma 2.5, it is not a Cauchy sequence in the metric space (𝑋,𝑝^𝑠), either. Applying Lemma 2.7 we get the sequences 𝑝^𝑠𝑥_2𝑚𝑘,𝑥_2𝑛𝑘,𝑝^𝑠𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1,𝑝^𝑠𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘,𝑝^𝑠𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘+1 (2.9) tending to some 2𝜀>0 when 𝑘→∞. Using definition (2.1) of the associated metric and (2.7) (which by (p₂) implies that also lim_𝑛→∞𝑝(𝑥_𝑛,𝑥_𝑛)=0), we get that the sequences (2.8) tend to 𝜀 when 𝑘→∞.

2.3. Property (P)

Let 𝑋 be a nonempty set and 𝑓∶𝑋→𝑋 a self-map. As usual, we denote by 𝐹(𝑓) the set of fixed points of 𝑓. Following Jeong and Rhoades [17], we say that the map 𝑓 has property (𝑃) if it satisfies 𝐹(𝑓)=𝐹(𝑓^𝑛) for each 𝑛∈ℕ. The proof of the following lemma is the same as in the metric case [17, Theorem  1.1].

Lemma 2.9. Let (𝑋,𝑝) be a partial metric space, and let 𝑓∶𝑋→𝑋 be a selfmap such that 𝐹(𝑓)≠∅. Then 𝑓 has property (𝑃) if 𝑝𝑓𝑥,𝑓²𝑥≤𝜆𝑝𝑥,𝑓𝑥 (2.10) holds for some 𝜆∈(0,1) and either (i) for all 𝑥∈𝑋  or (ii) for all 𝑥≠𝑓𝑥.

2.4. Metric Type Spaces

Definition 2.10 (see [11]). Let  𝑋  be a nonempty set,  𝐾≥1  a real number, and let a function  𝐷∶𝑋×𝑋→ℝsatisfy the following properties:(a)𝐷(𝑥,𝑦)=0 if and only if 𝑥=𝑦;(b)𝐷(𝑥,𝑦)=𝐷(𝑦,𝑥) for all 𝑥,𝑦∈𝑋;(c)𝐷(𝑥,𝑧)≤𝐾(𝐷(𝑥,𝑦)+𝐷(𝑦,𝑧)) for all 𝑥,𝑦,𝑧∈𝑋. Then (𝑋,𝐷,𝐾) is called a metric type space.

Obviously, for 𝐾=1, metric type space is simply a metric space.

The notions such as convergent sequence, Cauchy sequence, and complete space are defined in an obvious way.

A metric type space may satisfy some of the following additional properties:(d)𝐷(𝑥,𝑧)≤𝐾(𝐷(𝑥,𝑦₁)+𝐷(𝑦₁,𝑦₂)+⋯+𝐷(𝑦_𝑛,𝑧)) for arbitrary points 𝑥,𝑦₁,𝑦₂,…,𝑦_𝑛,𝑧∈𝑋;(e)function 𝐷 is continuous in two variables, that is, 𝑥_𝑛⟶𝑥and𝑦_𝑛⟶𝑦(in(𝑋,𝐷,𝐾))implies𝐷𝑥_𝑛,𝑦_𝑛⟶𝐷𝑥,𝑦. (2.11) (The last condition is in the theory of symmetric spaces usually called “property (𝐻_𝐸)’’.)

Condition (d) was used instead of (c) in the original definition of a metric type space by Khamsi [11].

Note that weaker version of property (e):(e′)𝑥_𝑛→𝑥 and 𝑦_𝑛→𝑥 (in (𝑋,𝐷,𝐾)) implies that 𝐷(𝑥_𝑛,𝑦_𝑛)→0is satisfied in an arbitrary metric type space. It can also be proved easily that the limit of a sequence in a metric type space is unique. Indeed, if 𝑥_𝑛→𝑥 and 𝑥_𝑛→𝑦 (in (𝑋,𝐷,𝐾)) and 𝐷(𝑥,𝑦)=𝜀>0, then 0≤𝐷𝑥,𝑦≤𝐾𝐷𝑥,𝑥_𝑛+𝐷𝑥_𝑛𝜀,𝑦<𝐾+𝜀2𝐾2𝐾=𝜀 (2.12) for sufficiently large 𝑛, which is impossible.

3. Results

3.1. Results in Partial Metric Spaces

Theorem 3.1. Let (𝑋,𝑝) be a complete partial metric space, and let 𝑓∶𝑋→𝑋 be a self-map. Suppose that there exists 𝛽∈𝒮 such that 𝑝𝑓𝑥,𝑓𝑦≤𝛽𝑝𝑥,𝑦𝑝𝑥,𝑦 (3.1) holds for all 𝑥,𝑦∈𝑋. Then 𝑓 has a unique fixed point 𝑧∈𝑋 and for each 𝑥∈𝑋 the Picard sequence {𝑓^𝑛𝑥} converges to 𝑧 when 𝑛→∞.

Proof. Let 𝑥₁∈𝑋 be arbitrary, and let 𝑥_𝑛+1=𝑓𝑥_𝑛 for 𝑛∈ℕ. Consider the following two cases:(1)𝑝(𝑥_𝑛0+1,𝑥_𝑛0)=0 for some 𝑛₀∈ℕ;(2)𝑝(𝑥_𝑛+1,𝑥_𝑛)>0 for each 𝑛∈ℕ.Case 1. Under this assumption we get that 𝑝𝑥_𝑛0+2,𝑥_𝑛0+1=𝑝𝑓𝑥_𝑛0+1,𝑓𝑥_𝑛0≤𝛽𝑝𝑥_𝑛0+1,𝑥_𝑛0𝑝𝑥_𝑛0+1,𝑥_𝑛0=𝛽(0)⋅0=0, (3.2) and it follows that 𝑝(𝑥_𝑛0+2,𝑥_𝑛0+1)=0. By induction, we obtain that 𝑝(𝑥_𝑛+1,𝑥_𝑛)=0 for all 𝑛≥𝑛₀ and so 𝑥_𝑛=𝑥_𝑛0 for all 𝑛≥𝑛₀. Hence, {𝑥_𝑛} is a Cauchy sequence, converging to 𝑥_𝑛0 which is a fixed point of 𝑓.Case 2. We will prove first that in this case the sequence 𝑝(𝑥_𝑛+1,𝑥_𝑛) is decreasing and tends to 0 as 𝑛→∞.
For each 𝑛∈ℕ we have that 0<𝑝𝑥_𝑛+2,𝑥_𝑛+1=𝑝𝑓𝑥_𝑛+1,𝑓𝑥_𝑛≤𝛽𝑝𝑥_𝑛+1,𝑥_𝑛𝑝𝑥_𝑛+1,𝑥_𝑛<𝑝𝑥_𝑛+1,𝑥_𝑛. (3.3) Hence, 𝑝(𝑥_𝑛+1,𝑥_𝑛) is decreasing and bounded from below, thus converging to some 𝑞≥0. Suppose that 𝑞>0. Then, it follows from (3.3) that 𝑝𝑥_𝑛+2,𝑥_𝑛+1𝑝𝑥_𝑛+1,𝑥_𝑛≤𝛽𝑝𝑥_𝑛+1,𝑥_𝑛<1, (3.4) where from, passing to the limit when 𝑛→∞, we get that lim_𝑛→∞𝛽(𝑝(𝑥_𝑛+1,𝑥_𝑛))=1. Using property (1.1) of the function 𝛽, we conclude that lim_𝑛→∞𝑝(𝑥_𝑛+1,𝑥_𝑛)=0, that is, 𝑞=0, a contradiction. Hence, lim_𝑛→∞𝑝(𝑥_𝑛+1,𝑥_𝑛)=0 is proved.
In order to prove that {𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝), suppose the contrary. As was already proved, 𝑝(𝑥_𝑛+1,𝑥_𝑛)→0 as 𝑛→∞, and so, using (p₂), 𝑝(𝑥_𝑛,𝑥_𝑛)→0 as 𝑛→∞. Hence, using (2.1), we get that 𝑝^𝑠(𝑥_𝑛+1,𝑥_𝑛)→0 as 𝑛→∞. Using Lemma 2.8, we obtain that there exist 𝜀>0 and two sequences {𝑚_𝑘} and {𝑛_𝑘} of positive integers such that the following four sequences tend to 𝜀 when 𝑘→∞: 𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘,𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1,𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘,𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘+1. (3.5) Putting in the contractive condition 𝑥=𝑥_{2𝑚𝑘−1} and 𝑦=𝑥_2𝑛𝑘, it follows that 𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1≤𝛽𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘<𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘. (3.6) Hence, 𝑝𝑥_2𝑚𝑘,𝑥_2𝑛𝑘+1𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘≤𝛽𝑝𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘<1 (3.7) and lim_𝑘→∞𝛽(𝑝(𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘))=1. Since 𝛽∈𝒮, it follows that lim_𝑘→∞𝑝(𝑥_{2𝑚𝑘−1},𝑥_2𝑛𝑘)=0, which is in contradiction with 𝜀>0.
Thus {𝑥_𝑛} is a Cauchy sequence, both in (𝑋,𝑝) and in (𝑋,𝑝^𝑠). Since these spaces are complete, it follows that sequence {𝑥_𝑛} converges in the metric space (𝑋,𝑝^𝑠), say lim_𝑛→∞𝑝^𝑠(𝑥_𝑛,𝑧)=0. Again from Lemma 2.5, we have 𝑝(𝑧,𝑧)=lim_𝑛→∞𝑝𝑥_𝑛,𝑧=lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚. (3.8) Moreover since {𝑥_𝑛} is a Cauchy sequence in the metric space (𝑋,𝑝^𝑠), we have lim_{𝑛,𝑚→∞}𝑝^𝑠(𝑥_𝑛,𝑥_𝑚)=0 and so, by the definition of 𝑝^𝑠, we have lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚)=0. Then (3.8) implies that 𝑝(𝑧,𝑧)=0 and lim_𝑛→∞𝑝𝑥_𝑛,𝑧=𝑝(𝑧,𝑧)=0. (3.9) We will prove that 𝑧 is a fixed point of 𝑓.
By (p₄), and using the contractive condition, we get that 𝑝𝑧,𝑓𝑧≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝑓𝑧−𝑝𝑥_𝑛+1,𝑥_𝑛+1≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑓𝑥_𝑛,𝑓𝑧≤𝑝𝑧,𝑥_𝑛+1+𝛽𝑝𝑥_𝑛,𝑧𝑝𝑥_𝑛,𝑧≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑥_𝑛,𝑧⟶0+0=0. (3.10) Thus, 𝑝(𝑧,𝑓𝑧)=0 and 𝑓𝑧=𝑧.
Assume that 𝑢≠𝑣 are two fixed points of 𝑓. Then 0<𝑝(𝑢,𝑣)=𝑝𝑓𝑢,𝑓𝑣≤𝛽𝑝(𝑢,𝑣)𝑝(𝑢,𝑣)<𝑝(𝑢,𝑣), (3.11) a contradiction. Hence the fixed point of 𝑓 is unique. The theorem is proved.

Remark 3.2. It follows from Lemma  1, (viii)⇔(x) of the paper [18] of Jachymski, that under conditions of Theorem 3.1 there exists a continuous and nondecreasing function 𝜑∶[0,+∞)→[0,+∞) such that 𝜑(𝑡)<𝑡 for all 𝑡>0 and 𝑝(𝑓𝑥,𝑓𝑦)≤𝜑(𝑝(𝑥,𝑦)) for all 𝑥,𝑦∈𝑋.
On the other hand, Romaguera [19] recently obtained a partial metric extension of the celebrated Boyd and Wong fixed point theorem, from which it follows that if (𝑋,𝑝) is a complete partial metric space and 𝑓∶𝑋→𝑋 is a map satisfying  𝑝(𝑓𝑥,𝑓𝑦)≤𝜑(𝑝(𝑥,𝑦)) for all 𝑥,𝑦∈𝑋, with a function 𝜑 with the aforementioned properties, then 𝑓 has a unique fixed point. Hence, combining Jachymski’s and Romaguera’s results, an alternative proof of Theorem 3.1 is obtained.

Theorem 3.3. If 𝑓∶𝑋→𝑋 satisfies conditions of Theorem 3.1, then it has property (P).

Proof. By Theorem 3.1, the set of fixed points of 𝑓 is a singleton, 𝐹(𝑓)={𝑧}. Then also 𝑧∈𝐹(𝑓^𝑛) for all 𝑛∈ℕ. Let 𝑣∈𝐹(𝑓^𝑛) for some 𝑛>1, and suppose that 𝑧≠𝑣, that is, 𝑝(𝑧,𝑣)>0. Then 0<𝑝(𝑧,𝑣)=𝑝𝑓𝑓^𝑛−1𝑧,𝑓𝑓^𝑛−1𝑣≤𝛽𝑝𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣𝑝𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣<𝑝𝑓^𝑛−1𝑧,𝑓^𝑛−1𝑣. (3.12) We have that 𝑓^𝑛−1𝑧≠𝑓^𝑛−1𝑣 (otherwise 𝑧=𝑓^𝑛𝑧=𝑓^𝑛𝑣=𝑣, which is excluded). It follows that 0<𝑝(𝑧,𝑣)<𝑝𝑓𝑓^𝑛−2𝑧,𝑓𝑓^𝑛−2𝑣≤𝛽𝑝𝑓^𝑛−2𝑧,𝑓^𝑛−2𝑣𝑝𝑓^𝑛−2𝑧,𝑓^𝑛−2𝑣<𝑝𝑓^𝑛−2𝑧,𝑓^𝑛−2𝑣. (3.13) Continuing, we obtain that 0<𝑝(𝑧,𝑣)<𝑝𝑓^𝑛−1𝑥,𝑓^𝑛−1𝑣<⋯<𝑝(𝑧,𝑣), (3.14) a contradiction. Hence, 𝑝(𝑧,𝑣)=0 and 𝑧=𝑣, that is, 𝐹(𝑓)=𝐹(𝑓^𝑛) for each 𝑛∈ℕ.

Example 3.4. Let 𝑋=[0,1], 𝑑(𝑥,𝑦)=2|𝑥−𝑦|, 𝑝(𝑥,𝑦)=max{𝑥,𝑦}, 𝛽(𝑡)=𝑒^−𝑡/(𝑡+1) for 𝑡>0 and 𝛽(0)∈[0,1). The mapping 𝑓∶(𝑋,𝑑)→(𝑋,𝑑) defined by 𝑓𝑥=(1/6)𝑥 does not satisfy conditions of Theorem 1.1. Indeed, take 𝑥=1, 𝑦=0 and obtain that 𝑑1𝑓1,𝑓0=2𝑑61,0=26=1−03,𝛽(𝑑(𝑑()(2)𝑒1,0))1,0=𝛽⋅2=⁻²2+1⋅2=2𝑒⁻²3<13. (3.15) On the other hand, take 𝑥,𝑦∈𝑋 with, for example, 𝑥≥𝑦. Then 𝑝1𝑓𝑥,𝑓𝑦=𝑝61𝑥,6𝑦=16𝛽𝑝𝑝(𝑥)𝑒𝑥,𝑥,𝑦𝑥,𝑦=𝛽⋅𝑥=^−𝑥1𝑥+1⋅𝑥≥6𝑥, (3.16) since 𝑒^−𝑥/(𝑥+1)≥1/2𝑒>1/6 for 𝑥∈[0,1]. Hence, 𝑓 satisfies conditions of Theorem 3.1 and thus has a unique fixed point (𝑧=0).

3.2. Results in Ordered Partial Metric Spaces

Theorem 3.5. Let (𝑋,𝑝,⪯) be a complete ordered partial metric space. Let 𝑓∶𝑋→𝑋 be an increasing self-map (with respect to. ⪯) such that there exists 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀. Suppose that there exists 𝛽∈𝒮 such that (3.1) holds for all comparable 𝑥,𝑦∈𝑋. Assume that either 𝑓 is continuous or 𝑋 is regular. Then, 𝑓 has a fixed point in 𝑋. The set 𝐹(𝑓) of fixed points of 𝑓 is a singleton if and only if it is well ordered.

Proof. Take 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀ and, using monotonicity of 𝑓, form the sequence 𝑥_𝑛=𝑓𝑥_𝑛−1 with 𝑥₀⪯𝑥₁⪯𝑥₂⪯⋯⪯𝑥_𝑛⪯⋯. (3.17) Since 𝑥_𝑛−1 and 𝑥_𝑛 are comparable we can apply contractive condition to obtain 𝑝𝑥_𝑛+1,𝑥_𝑛=𝑝𝑓𝑥_𝑛,𝑓𝑥_𝑛−1≤𝛽𝑝𝑥_𝑛−1,𝑥_𝑛𝑝𝑥_𝑛−1,𝑥_𝑛≤𝑝𝑥_𝑛−1,𝑥_𝑛. (3.18) Proceeding as in the proof of Theorem 3.1 we obtain that lim_𝑛→∞𝑝(𝑥_𝑛+1,𝑥_𝑛)=0, that {𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝) (and in (𝑋,𝑝^𝑠)). Thus, it converges (in 𝑝 and in 𝑝^𝑠) to a point 𝑧∈𝑋 such that 𝑝(𝑧,𝑧)=lim_𝑛→∞𝑝𝑥_𝑛,𝑧=lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚. (3.19) Also, it follows as in the proof of Theorem 3.1 that lim_𝑛→∞𝑝𝑥_𝑛,𝑧=𝑝(𝑧,𝑧)=0. (3.20) We will prove that 𝑧 is a fixed point of 𝑓.(i)Suppose that 𝑓∶(𝑋,𝑝)→(𝑋,𝑝) is continuous. We have, by (p₄), 𝑝𝑧,𝑓𝑧≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑓𝑥_𝑛,𝑓𝑧. (3.21) Passing to the limit when 𝑛→∞ and using continuity of 𝑓 we get that 𝑝𝑧,𝑓𝑧≤𝑝(𝑧,𝑧)+𝑝𝑓𝑧,𝑓𝑧=𝑝𝑓𝑧,𝑓𝑧≤𝑝𝑧,𝑓𝑧byp₂. (3.22) It follows that 𝑝(𝑧,𝑓𝑧)=𝑝(𝑓𝑧,𝑓𝑧). Since 𝑧⪯𝑧, using contractive condition, we get that 𝑝𝑓𝑧,𝑓𝑧≤𝛽𝑝(𝑧,𝑧)𝑝(𝑧,𝑧)=0 (3.23) and so 𝑝(𝑧,𝑓𝑧)=0 and 𝑓𝑧=𝑧.(ii)If (𝑋,𝑝) is regular, since {𝑥_𝑛} is an increasing sequence tending to 𝑧, we have that 𝑥_𝑛⪯𝑧 for each 𝑛∈ℕ. So we can apply (p₄) and contractive condition to obtain 𝑝𝑧,𝑓𝑧≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑓𝑥_𝑛,𝑓𝑧≤𝑝𝑧,𝑥_𝑛+1+𝛽𝑝𝑥_𝑛,𝑧𝑝𝑥_𝑛,𝑧≤𝑝𝑧,𝑥_𝑛+1+𝑝𝑥_𝑛,𝑧. (3.24) Letting 𝑛→∞ we get 𝑝𝑧,𝑓𝑧≤𝑝(𝑧,𝑧)+𝑝(𝑧,𝑧)=0. (3.25) Hence, we again obtain that 𝑓𝑧=𝑧.
Let the set 𝐹(𝑓) of fixed points of 𝑓 be well ordered, and suppose that there exist two distinct points 𝑢,𝑣∈𝐹(𝑓). Then these points are comparable, and we can apply the contractive condition to obtain 0<𝑝(𝑢,𝑣)=𝑝𝑓𝑢,𝑓𝑣≤𝛽𝑝(𝑢,𝑣)𝑝(𝑢,𝑣)<𝑝(𝑢,𝑣), (3.26) a contradiction. Hence, the set 𝐹(𝑓) is a singleton. The converse is trivial.

Example 3.6. Let 𝑋={1,2,3}, and define the partial order ⪯ on 𝑋 by ⪯={(1,1),(2,2),(3,3),(1,3),(3,2),(1,2)}. (3.27) Consider the function 𝑓∶𝑋→𝑋 given as 𝑓=¹²³₃₂₂, which is increasing with respect to ⪯.
Define first the metric 𝑑 on 𝑋 by 𝑑(1,2)=𝑑(1,3)=1, 𝑑(2,3)=1/2, 𝑑(𝑥,𝑥)=0 for 𝑥∈𝑋 and 𝑑(𝑦,𝑥)=𝑑(𝑥,𝑦) for 𝑥,𝑦∈𝑋. Then (𝑋,𝑑) is a complete partially ordered metric space. The function 𝛽∶[0,+∞)→[0,1), defined by 𝛽(𝑡)=𝑒^−𝑡, 𝑡>0, and 𝛽(0)∈[0,1), belongs to the class 𝒮.
Take 𝑥=1 and 𝑦=3. Then 𝑑()=1𝑓1,𝑓3=𝑑3,22>1𝑒=𝛽(1)⋅1=𝛽(𝑑(1,3))𝑑(1,3). (3.28) Hence, conditions of Theorem 1.2 are not fulfilled and this theorem cannot be used to prove the existence of a fixed point of 𝑓.
Define now the partial metric 𝑝 on 𝑋 by 𝑝(1,2)=𝑝(1,3)=1, 𝑝(2,3)=1/9, 𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥) for 𝑥,𝑦∈𝑋; further 𝑝(1,1)=1, 𝑝(2,2)=0, and 𝑝(3,3)=1/10 (it is easy to check conditions (p₁)–(p₄)). Let us check contractive condition (3.1) of Theorem 3.5: 𝑝()=1𝑓1,𝑓1=𝑝3,3<110𝑒(1)𝑝(𝑝(),𝑝()=1=𝛽⋅1=𝛽1,1)1,1𝑓1,𝑓2=𝑝3,29<1𝑒(1)𝑝()𝑝(),𝑝()=1=𝛽⋅1=𝛽1,21,2𝑓1,𝑓3=𝑝3,29<1𝑒(1)𝑝()𝑝(),𝑝()(0)𝑝()𝑝(),𝑝()1=𝛽⋅1=𝛽1,31,3𝑓2,𝑓2=𝑝2,2=0=0=𝛽⋅0=𝛽2,22,2𝑓2,𝑓3=𝑝2,2=0<9𝑒^−1/91=𝛽9⋅19𝑝()𝑝(),𝑝()1=𝛽2,32,3𝑓3,𝑓3=𝑝2,2=0<𝑒10^−1/101=𝛽⋅110𝑝()𝑝().10=𝛽3,33,3 (3.29) Hence, we can apply Theorem 3.5 to conclude that there is a unique fixed point of 𝑓 (which is 𝑧=2).

A variant of Theorem 1.2 which uses an altering function was obtained in [3, Theorems  2.2, 2.3]. Recall that 𝜓∶[0,+∞)→[0,+∞) is called an altering function if it is continuous, increasing and 𝜓⁻¹(0)={0}. We state a partial metric version of this result. The proof is omitted since it is similar to the previous one.

Theorem 3.7. Let (𝑋,𝑝,⪯) be a complete ordered partial metric space. Let 𝑓∶𝑋→𝑋 be an increasing self-map (w.r.t. ⪯) such that there exists 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀. Suppose that there exist 𝛽∈𝒮 and an altering function 𝜓 such that 𝜓𝑝𝑓𝑥,𝑓𝑦≤𝛽𝑝𝑥,𝑦𝜓𝑝𝑥,𝑦 (3.30) holds for all comparable 𝑥,𝑦∈𝑋. Assume that either 𝑓 is continuous or 𝑋 is regular. Then, 𝑓 has a fixed point in 𝑋. The set 𝐹(𝑓) of fixed points of 𝑓 is a singleton if and only if it is well ordered.

3.3. Results in Metric Type Spaces

For the use in metric type spaces (with the given 𝐾>1) we will consider the class of functions 𝒮_𝐾, where 𝛽∈𝒮_𝐾 if 𝛽∶[0,+∞)→[0,1/𝐾) and has the property𝛽𝑡_𝑛⟶1𝐾implies𝑡_𝑛⟶0. (3.31) An example of a function in 𝒮_𝐾 is given by 𝛽(𝑡)=(1/𝐾)𝑒^−𝑡 for 𝑡>0 and 𝛽(0)∈[0,1/𝐾).

Theorem 3.8. Let 𝐾>1, and let (𝑋,𝐷,𝐾) be a complete metric type space. Suppose that a mapping 𝑓∶𝑋→𝑋 satisfies the condition 𝐷𝑓𝑥,𝑓𝑦≤𝛽𝐷𝑥,𝑦𝐷𝑥,𝑦 (3.32) for all 𝑥,𝑦∈𝑋 and some 𝛽∈𝒮_𝐾. Then 𝑓 has a unique fixed point 𝑧∈𝑋, and for each 𝑥∈𝑋 the Picard sequence {𝑓^𝑛𝑥} converges to 𝑧 in (𝑋,𝐷,𝐾).

Proof. Using condition (3.32) it is easy to show that the fixed point of 𝑓 in (𝑋,𝐷,𝐾) is unique (if it exists) and that 𝑓 is 𝐷-continuous in the sense that 𝑥_𝑛→𝑥 implies that 𝑓𝑥_𝑛→𝑓𝑥 in (𝑋,𝐷,𝐾) (for details see [12]).
Let 𝑥₀∈𝑋 be arbitrary and 𝑥_𝑛=𝑓𝑥_𝑛−1 for 𝑛∈ℕ. If 𝑥_𝑛0+1=𝑥_𝑛0 for some 𝑛₀, then it is easy to show that 𝑥_𝑛=𝑥_𝑛0 for 𝑛≥𝑛₀, and the proof is complete. Suppose that 𝑥_𝑛≠𝑥_𝑛+1 for all 𝑛≥0. Then, using (3.32), we get that 𝐷𝑥_𝑛+1,𝑥_𝑛=𝐷𝑓𝑥_𝑛,𝑓𝑥_𝑛−1≤𝛽𝐷𝑥_𝑛,𝑥_𝑛−1𝐷𝑥_𝑛,𝑥_𝑛−1<1𝐾𝐷𝑥_𝑛,𝑥_𝑛−1. (3.33) By [12, Lemma  3.1], {𝑥_𝑛} is a Cauchy sequence in (𝑋,𝐷,𝐾). As this space is complete, {𝑥_𝑛} converges to some 𝑧∈𝑋 as 𝑛→∞. Obviously, also 𝑓𝑥_𝑛−1=𝑥_𝑛→𝑧 and continuity of 𝑓 implies that 𝑓𝑥_𝑛−1→𝑓𝑧. Since the limit of a sequence in a metric type space is unique, it follows that 𝑓𝑧=𝑧.

Example 3.9. Let 𝑋={0,1,3} be equipped with the metric type function 𝐷 given by 𝐷(𝑥,𝑦)=(𝑥−𝑦)² with 𝐾=2. Consider the mapping 𝑓∶𝑋→𝑋 defined by 𝑓(0)=1, 𝑓1=1, 𝑓3=0, and the function 𝛽∈𝒮_𝐾 given by 𝛽(𝑡)=(1/2)𝑒^−𝑡/9, 𝑡>0, and 𝛽(0)∈[0,1/2). Then 𝐷()1𝑓0,𝑓1=𝐷1,1=0<2𝑒^−1/9(1)(𝐷(𝐷(),𝐷()9=𝛽⋅1=𝛽0,1))0,1𝑓0,𝑓3=𝐷1,0=1<(9)(𝐷(𝐷(),𝐷()2𝑒=𝛽⋅9=𝛽0,3))0,3𝑓1,𝑓3=𝐷1,0=1<2𝑒^−4/9=𝛽(4)⋅4=𝛽(𝐷(1,3))𝐷(1,3). (3.34) Hence, 𝑓 satisfies all the assumptions of Theorem 3.8 and thus it has a unique fixed point (which is 𝑧=1).

Acknowledgments

The authors are thankful to the referees for very useful suggestions that helped to improve the paper, and they are thankful to the Ministry of Science and Technological Development of Serbia.