Fixed Point of 𝑇-Hardy-Rogers Contractive Mappings in Partially Ordered Partial Metric Spaces

Abstract

We prove some fixed point theorems for a T-Hardy-Rogers contraction in the setting of partially ordered partial metric spaces. We apply our results to study periodic point problems for such mappings. We also provide examples to illustrate the results presented herein.

1. Introduction and Preliminaries

The notion of a partial metric space was introduced by Matthews in [1]. In partial metric spaces, the distance of a point in the self may not be zero. After the definition of a partial metric space, Matthews proved the partial metric version of Banach fixed point theorem. A motivation behind introducing the concept of a partial metric was to obtain appropriate mathematical models in the theory of computation and, in particular, to give a modified version of the Banach contraction principle, more suitable in this context [1]. Subsequently, several authors studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions (e.g., [2–21], [22]). Existence of fixed points in partially ordered metric spaces has been initiated in 2004 by Ran and Reurings [23]. Subsequently, several interesting and valuable results have appeared in this direction [14]. The aim of this paper is to study the necessary conditions for existence of fixed point of mapping satisfying 𝑇-Hardy-Rogers conditions in the framework of partially ordered partial metric spaces. Our results extend and strengthen various known results [8, 24]. In the sequel, the letters ℝ, ℝ⁺, and ℕ will denote the set of real numbers, the set of nonnegative real numbers, and the set of nonnegative integer numbers, respectively. The usual order on ℝ (resp., on ℝ⁺) will be indistinctly denoted by ≤ or by ≥.

Consistent with [1, 8] (see [25–29]) the following definitions and results will be needed in the sequel.

Definition 1.1 (see [1]). A partial metric on a nonempty set 𝑋 is a mapping 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. If 𝑝(𝑥,𝑦)=0, then (𝑝₁) and (𝑝₂) imply that 𝑥=𝑦. But converse does not hold always. A trivial example of a partial metric space is the pair (ℝ⁺,𝑝), where 𝑝∶ℝ⁺×ℝ⁺→ℝ⁺ is defined as 𝑝(𝑥,𝑦)=max{𝑥,𝑦}. Each partial metric 𝑝 on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has as a base the family 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}.

On a partial metric space the concepts of convergence, Cauchy sequence, completeness, and continuity are defined as follows.

Definition 1.2 (see [1]). Let (𝑋,𝑝) be a partial metric space and let {𝑥_𝑛} be a sequence in 𝑋. Then (i) {𝑥_𝑛} converges to a point 𝑥∈𝑋 if and only if 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝(𝑥,𝑥_𝑛) (we may still write this as lim_𝑛→∞𝑥_𝑛=𝑣 or 𝑥_𝑛→𝑣); (ii) {𝑥_𝑛} is called a Cauchy sequence if there exists (and is finite) lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚).

Definition 1.3 (see [1]). A partial metric space (𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges to a point 𝑥∈𝑋, such that 𝑝(𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚). If 𝑝 is a partial metric on 𝑋, then the function 𝑝^𝑠∶𝑋×𝑋→𝑅⁺ given by 𝑝^𝑠(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦) is a metric on 𝑋.

Lemma 1.4 (see [1, 20]). 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. (1.1)

Remark 1.5. (1) (see [19]) Clearly, a limit of a sequence in a partial metric space does not need to be unique. Moreover, the function 𝑝(⋅,⋅) does not need to 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→∞.

(2) (see [7]) However, if 𝑝(𝑥_𝑛,𝑥)→𝑝(𝑥,𝑥)=0 then 𝑝(𝑥_𝑛,𝑦)→𝑝(𝑥,𝑦) for all 𝑦∈𝑋.

Definition 1.6 (see [30]). Suppose that (𝑋,𝑝) is a partial metric space. Denote 𝜏(𝑝) its topology. We say 𝑇∶(𝑋,𝑝)→(𝑋,𝑝) is continuous if both 𝑇∶(𝑋,𝜏(𝑝))→(𝑋,𝜏(𝑝)) and 𝑇∶(𝑋,𝜏(𝑝^𝑠))→(𝑋₂,𝜏(𝑝^𝑠)) are continuous.

Remark 1.7. It is worth to notice that the notions 𝑝-continuous and 𝑝^𝑠-continuous of any function in the context of partial metric spaces are incomparable, in general. Indeed, if 𝑋=[0,+∞),  𝑝(𝑥,𝑦)=max{𝑥,𝑦},  𝑝^𝑠(𝑥,𝑦)=|𝑥−𝑦|,  𝑓0=1, and 𝑓𝑥=𝑥² for all 𝑥>0 and 𝑔𝑥=|sin𝑥|, then 𝑓 is a 𝑝-continuous and 𝑝^𝑠-discontinuous at point 𝑥=0; while 𝑔 is a 𝑝-discontinuous and 𝑝^𝑠-continuous at 𝑥=𝜋.

According to [31], we state the following definition.

Definition 1.8. Let (𝑋,𝑝) be a partial metric space. A mapping 𝑇∶𝑋→𝑋 is said to be(i)sequentially convergent if for any sequence {𝑦_𝑛} in 𝑋 such that {𝑇𝑦_𝑛} is convergent in (𝑋,𝑝^𝑠) implies that {𝑦_𝑛} is convergent in (𝑋,𝑝^𝑠),(ii)subsequentially convergent if for any sequence {𝑦_𝑛} in 𝑋 such that {𝑇𝑦_𝑛} is convergent in (𝑋,𝑝^𝑠) implies that {𝑦_𝑛} has a convergent subsequence in (𝑋,𝑝^𝑠).

Consistent with [24, 31] we define a 𝑇-Hardy-Rogers contraction in the framework of partial metric spaces.

Definition 1.9. Let (𝑋,𝑝) be a partial metric space and 𝑇,𝑓∶𝑋→𝑋 be two mappings. A mapping 𝑓 is said to be a 𝑇-Hardy-Rogers contraction if there exist 𝑎_𝑖≥0, 𝑖=1,…,5 with 𝑎₁+𝑎₂+𝑎₃+𝑎₄+𝑎₅<1 such that for all 𝑥,𝑦∈𝑋𝑝𝑇𝑓𝑥,𝑇𝑓𝑦≤𝑎₁𝑝𝑇𝑥,𝑇𝑦+𝑎₂𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₃𝑝𝑇𝑦,𝑇𝑓𝑦+𝑎₄𝑝𝑇𝑥,𝑇𝑓𝑦+𝑎₅𝑝𝑇𝑦,𝑇𝑓𝑥. (1.2)

Putting 𝑎₁=𝑎₄=𝑎₅=0 and 𝑎₂=𝑎₃≠0, (resp., 𝑎₁=𝑎₂=𝑎₃=0 and 𝑎₄=𝑎₅≠0) in the previous definition, then the inequality (1.2) is said a 𝑇-Kannan (resp., 𝑇-Chatterjea) type contraction. Also, if 𝑎₄=𝑎₅=0 and 𝑎₁,𝑎₂,𝑎₃≠0, (1.2) is said the 𝑇-Reich type contraction.

Definition 1.10. Let 𝑋 be a nonempty set. Then (𝑋,𝑝,⪯) is called a partially ordered partial metric space if and only if (i) 𝑝 is a partial metric on 𝑋 and (ii) ⪯ is a partial order on 𝑋.
Let (𝑋,𝑝) be a partial metric space endowed with a partial order ⪯ and let 𝑓∶𝑋→𝑋 be a given mapping. We define sets Δ,Δ₁⊂𝑋×𝑋 by Δ=𝑥,𝑦∈𝑋×𝑋∶𝑥⪯𝑦or,Δ𝑦⪯𝑥₁=(𝑥,𝑥)∈𝑋×𝑋∶𝑥⪯𝑓𝑥or𝑓𝑥⪯𝑥. (1.3)

A point 𝑥∈𝑋 is called a fixed point of mapping 𝑓∶𝑋→𝑋 if 𝑥=𝑓𝑥. The set of all fixed points of the mapping 𝑓 is denoted by 𝐹_𝑓.

2. Fixed Point Results

In this section, we obtain fixed point results for a mapping satisfying a 𝑇-Hardy-Rogers contractive condition defined on a partially ordered partial metric space which is complete.

We start with the following result.

Theorem 2.1. Let (𝑋,⪯,𝑝) be an partially ordered partial metric space which is complete. Let 𝑇∶𝑋→𝑋 be a continuous, injective mapping and 𝑓∶𝑋→𝑋 a nondecreasing 𝑇-Hardy-Rogers contraction for all (𝑥,𝑦)∈Δ. If there exists 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀, and one of the following two conditions is satisfied(a)𝑓 is a continuous self-map on 𝑋;(b)for any nondecreasing sequence {𝑥_𝑛} in (𝑋,⪯) with lim_𝑛→∞𝑝^𝑠(𝑧,𝑥_𝑛)=0 it follows 𝑥_𝑛⪯𝑧 for all 𝑛∈ℕ;
then 𝐹_𝑓≠𝜙 provided that 𝑇 is subsequentially or sequentially convergent. Moreover, 𝑓 has a unique fixed point if 𝐹_𝑓×𝐹_𝑓⊂Δ.

Proof. As 𝑓 is nondecreasing, therefore by given assumption, we have 𝑥₁=𝑓𝑥₀⪯𝑓²𝑥₀⪯⋯⪯𝑓^𝑛𝑥₀⪯𝑓^𝑛+1𝑥₀⪯⋯ (2.1) Define a sequence {𝑥_𝑛} in 𝑋 with 𝑥_𝑛=𝑓^𝑛𝑥₀ and so 𝑥_𝑛+1=𝑓𝑥_𝑛 for 𝑛∈ℕ. Since (𝑥_𝑛−1,𝑥_𝑛)∈Δ therefore by replacing 𝑥 by 𝑥_𝑛−1 and 𝑦 by 𝑥_𝑛 in (1.2), we have 𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1=𝑝𝑇𝑓𝑥_𝑛−1,𝑇𝑓𝑥_𝑛≤𝑎₁𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₂𝑝𝑇𝑥_𝑛−1,𝑇𝑓𝑥_𝑛−1+𝑎₃𝑝𝑇𝑥_𝑛,𝑇𝑓𝑥_𝑛+𝑎₄𝑝𝑇𝑥_𝑛−1,𝑇𝑓𝑥_𝑛+𝑎₅𝑝𝑇𝑥_𝑛,𝑇𝑓𝑥_𝑛−1=𝑎₁𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₂𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₃𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1+𝑎₄𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+1+𝑎₅𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛≤𝑎₁+𝑎₂𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₃𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1+𝑎₄𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1−𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+𝑎₅𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛=𝑎₁+𝑎₂+𝑎₄𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₃+𝑎₄𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1+𝑎₅−𝑎₄𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛, (2.2) that is, 1−𝑎₃−𝑎₄𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1≤𝑎₁+𝑎₂+𝑎₄𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₅−𝑎₄𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛. (2.3) Similarly, replacing 𝑥 by 𝑥_𝑛 and 𝑦 by 𝑥_𝑛−1 in (1.2), we obtain 1−𝑎₂−𝑎₅𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1≤𝑎₁+𝑎₃+𝑎₅𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛+𝑎₄−𝑎₅𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛. (2.4) Summing (2.3) and (2.4), we obtain 𝑝(𝑇𝑥_𝑛,𝑇𝑥_𝑛+1)≤𝛿𝑝(𝑇𝑥_𝑛−1,𝑇𝑥_𝑛), where 𝛿=(2𝑎₁+𝑎₂+𝑎₃+𝑎₄+𝑎₅)/(2−𝑎₂−𝑎₃−𝑎₄−𝑎₅). Obviously 0≤𝛿<1. Therefore, for all 𝑛≥1, 𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1≤𝛿𝑝𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤⋯≤𝛿^𝑛𝑝𝑇𝑥₀,𝑇𝑥₁. (2.5) Now, for any 𝑚∈𝑁 with 𝑚>𝑛, we have 𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑚≤𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑛+1+𝑝𝑇𝑥_𝑛+1,𝑇𝑥_𝑛+2+⋯+𝑝𝑇𝑥_𝑚−1,𝑇𝑥_𝑚≤𝛿^𝑛+𝛿^𝑛+1+⋯+𝛿^𝑚−1𝑝𝑇𝑥₀,𝑇𝑥₁≤𝛿^𝑛𝑝1−𝛿𝑇𝑥₀,𝑇𝑥₁, (2.6) which implies that 𝑝(𝑇𝑥_𝑛,𝑇𝑥_𝑚)→0 as 𝑛,𝑚→∞. Hence {𝑇𝑥_𝑛} is a Cauchy sequence in (𝑋,𝑝) and in (𝑋,𝑝^𝑠). Since (𝑋,𝑝) is complete, therefore from Lemma 1.4, (𝑋,𝑝^𝑠) is a complete metric space. Hence {𝑇𝑥_𝑛} converges to some 𝑣∈𝑋 with respect to the metric 𝑝^𝑠, that is, lim_𝑛→∞𝑝^𝑠𝑇𝑥_𝑛,𝑣=0, (2.7) or equivalently, 𝑝(𝑣,𝑣)=lim_𝑛→∞𝑝𝑇𝑥_𝑛,𝑣=lim_{𝑛,𝑚→∞}𝑝𝑇𝑥_𝑛,𝑇𝑥_𝑚=0. (2.8) Suppose that 𝑇 is subsequentially convergent, therefore convergence of {𝑇𝑥_𝑛} in (𝑋,𝑝^𝑠) implies that {𝑥_𝑛} has a convergent subsequence {𝑥_𝑛𝑖} in (𝑋,𝑝^𝑠). So lim_𝑖→∞𝑝^𝑠𝑥_𝑛𝑖,𝑢=0, (2.9) for some 𝑢∈𝑋. As 𝑇 is continuous, so (2.9) and Definition 1.6. imply that lim_𝑖→∞𝑝^𝑠(𝑇𝑥_𝑛𝑖,𝑇𝑢)=0. From (2.7) and by the uniqueness of the limit in metric space (𝑋,𝑝^𝑠), we obtain 𝑇𝑢=𝑣. Consequently, 0=𝑝(𝑇𝑢,𝑇𝑢)=lim_𝑖→∞𝑝𝑇𝑥_𝑛𝑖,𝑇𝑢=lim_{𝑖,𝑗→∞}𝑝𝑇𝑥_𝑛𝑖,𝑇𝑥_𝑛𝑗. (2.10)(1^∘) If 𝑓 is a continuous self-map on 𝑋, then 𝑓𝑥_𝑛𝑖→𝑓𝑢 and 𝑇𝑓𝑥_𝑛𝑖→𝑇𝑓𝑢 as 𝑖→∞. Since 𝑇𝑥_𝑛𝑖→𝑇𝑢 as 𝑖→∞, we obtain that 𝑇𝑓𝑢=𝑇𝑢. As 𝑇 is injective, so we have 𝑓𝑢=𝑢.(2^∘) If 𝑓 is not continuous then by given assumption we have 𝑥_𝑛⪯𝑢 for all 𝑛∈ℕ. Thus for a subsequence {𝑥_𝑛𝑖} of {𝑥_𝑛} we have 𝑥_𝑛𝑖⪯𝑢 and (𝑥_𝑛𝑖,𝑢)∈Δ. Now, 𝑝𝑇𝑓𝑢,𝑇𝑢≤𝑝𝑇𝑓𝑢,𝑇𝑓𝑥_𝑛𝑖+𝑝𝑇𝑓𝑥_𝑛𝑖,𝑇𝑢−𝑝𝑇𝑓𝑥_𝑛𝑖,𝑇𝑓𝑥_𝑛𝑖≤𝑎₁𝑝𝑇𝑥_𝑛𝑖,𝑇𝑢+𝑎₂𝑝𝑇𝑥_𝑛𝑖,𝑇𝑓𝑥_𝑛𝑖+𝑎₃𝑝𝑇𝑢,𝑇𝑓𝑢+𝑎₄𝑝𝑇𝑥_𝑛𝑖,𝑇𝑓𝑢+𝑎₅𝑝𝑇𝑢,𝑇𝑓𝑥_𝑛𝑖+𝑝𝑇𝑓𝑥_𝑛𝑖,𝑇𝑢−𝑝𝑇𝑓𝑥_𝑛𝑖,𝑇𝑓𝑥_𝑛𝑖=𝑎₁𝑝𝑇𝑥_𝑛𝑖,𝑇𝑢+𝑎₂𝑝𝑇𝑥_𝑛𝑖,𝑇𝑥_𝑛𝑖+1+𝑎₃𝑝𝑇𝑢,𝑇𝑓𝑢+𝑎₄𝑝𝑇𝑥_𝑛𝑖,𝑇𝑓𝑢+𝑎₅𝑝𝑇𝑢,𝑇𝑥_𝑛𝑖+1+𝑝𝑇𝑥_𝑛𝑖+1,𝑇𝑢−𝑝𝑇𝑥_𝑛𝑖+1,𝑇𝑥_𝑛𝑖+1≤𝑎₁𝑝𝑇𝑥_𝑛𝑖,𝑇𝑢+𝑎₂𝑝𝑇𝑥_𝑛𝑖,𝑇𝑥_𝑛𝑖+1+𝑎₃𝑝𝑇𝑢,𝑇𝑓𝑢+𝑎₄𝑝𝑇𝑥_𝑛𝑖,𝑇𝑢+𝑝𝑇𝑢,𝑇𝑓𝑢−𝑝(𝑇𝑢,𝑇𝑢)+𝑎₅𝑝𝑇𝑢,𝑇𝑥_𝑛𝑖+1+𝑝𝑇𝑥_𝑛𝑖+1,𝑇𝑢−𝑝𝑇𝑥_𝑛𝑖+1,𝑇𝑥_𝑛𝑖+1. (2.11)On taking limit as 𝑖→∞ and applying Remark 1.5. (2) we get 𝑝𝑇𝑢,𝑇𝑓𝑢≤𝑎₁𝑝(𝑇𝑢,𝑇𝑢)+𝑎₂𝑝(𝑇𝑢,𝑇𝑢)+𝑎₃𝑝𝑇𝑢,𝑇𝑓𝑢+𝑎₄𝑝(𝑇𝑢,𝑇𝑢)+𝑝𝑇𝑢,𝑇𝑓𝑢−𝑝(𝑇𝑢,𝑇𝑢)+𝑎₅𝑝()()()=𝑎𝑇𝑢,𝑇𝑢+𝑝𝑇𝑢,𝑇𝑢−𝑝𝑇𝑢,𝑇𝑢₁+𝑎₂+𝑎₅𝑝(𝑇𝑢,𝑇𝑢)+𝑎₃+𝑎₄𝑝≤𝑎𝑇𝑢,𝑇𝑓𝑢₁+𝑎₂+𝑎₃+𝑎₄+𝑎₅𝑝,𝑇𝑢,𝑇𝑓𝑢<𝑝𝑇𝑢,𝑇𝑓𝑢 (2.12) which implies that 𝑝(𝑇𝑢,𝑇𝑓𝑢)=0, and so 𝑇𝑢=𝑇𝑓𝑢. Now injectivity of 𝑇 gives 𝑢=𝑓𝑢. Following similar arguments to those given above, the result holds when 𝑇 is sequentially convergent.
Suppose that 𝐹_𝑓×𝐹_𝑓⊂Δ. Let 𝑤 be a fixed point of 𝑓. As 𝐹_𝑓×𝐹_𝑓⊂Δ, therefore (𝑢,𝑤)∈Δ. From (1.2), we have 𝑝(𝑇𝑢,𝑇𝑤)=𝑝𝑇𝑓𝑢,𝑇𝑓𝑤≤𝑎₁𝑝(𝑇𝑢,𝑇𝑤)+𝑎₂𝑝𝑇𝑢,𝑇𝑓𝑢+𝑎₃𝑝𝑇𝑤,𝑇𝑓𝑤+𝑎₄𝑝𝑇𝑢,𝑇𝑓𝑤+𝑎₅𝑝𝑇𝑤,𝑇𝑓𝑢=𝑎₁𝑝(𝑇𝑢,𝑇𝑤)+𝑎₂𝑝(𝑇𝑢,𝑇𝑢)+𝑎₃𝑝(𝑇𝑤,𝑇𝑤)+𝑎₄𝑝(𝑇𝑢,𝑇𝑤)+𝑎₅𝑝()≤𝑎𝑇𝑤,𝑇𝑢₁+𝑎₂+𝑎₃+𝑎₃+𝑎₄+𝑎₅𝑝()(),𝑇𝑢,𝑇𝑤<𝑝𝑇𝑢,𝑇𝑤 (2.13) and hence 𝑝(𝑇𝑢,𝑇𝑤)=0, which further implies that 𝑢=𝑤 as 𝑇 is injective.

Example 2.2. Let 𝑋=[0,1] be endowed with usual order and let 𝑝 be the complete partial metric on 𝑋 defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Let 𝑇,𝑓∶𝑋→𝑋 be defined by 𝑇𝑥=4𝑥/5 and 𝑓𝑥=𝑥/4. Note that Δ=𝑋×𝑋. For any (𝑥,𝑦)∈Δ, we have 𝑝𝑥𝑇𝑓𝑥,𝑇𝑓𝑦=max5,𝑦5=15max𝑥,𝑦≤12≤235max𝑥,𝑦7max4𝑥5,4𝑦5+17max4𝑥5,𝑥5+17max4𝑦5,𝑦5+17max4𝑥5,𝑦5+17max4𝑦5,𝑥5=𝑎₁𝑝𝑇𝑥,𝑇𝑦+𝑎₂𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₃𝑝𝑇𝑦,𝑇𝑓𝑦+𝑎₄𝑝𝑇𝑥,𝑇𝑓𝑦+𝑎₅𝑝𝑇𝑦,𝑇𝑓𝑥. (2.14) Therefore, 𝑓 is a 𝑇-Hardly-Rogers contraction with 𝑎₁=2/7,   𝑎₂=𝑎₃=𝑎₄=𝑎₅=1/7. Obviously, 𝑇 is continuous and sequentially convergent. Thus, all the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique fixed point of 𝑓.

Example 2.3. Let 𝑋=[0,∞) be endowed with usual order and let 𝑝 be a partial metric on 𝑋 defined by 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Define 𝑇,𝑓∶𝑋→𝑋 by 𝑇𝑥=𝑥²and𝑓𝑥=𝑥/3. Note that Δ=𝑋×𝑋. For any (𝑥,𝑦)∈Δ, we have 𝑝𝑥𝑇𝑓𝑥,𝑇𝑓𝑦=max²9,𝑦²9=19max𝑥²,𝑦²≤16max𝑥²,𝑦²=𝑎₁𝑝𝑇𝑥,𝑇𝑦≤𝑎₁𝑝𝑇𝑥,𝑇𝑦+𝑎₂𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₃𝑝𝑇𝑦,𝑇𝑓𝑦+𝑎₄𝑝𝑇𝑥,𝑇𝑓𝑦+𝑎₅𝑝𝑇𝑦,𝑇𝑓𝑥. (2.15) Therefore, 𝑓 is a 𝑇-Hardly-Rogers contraction with 𝑎₁=𝑎₂=𝑎₃=𝑎₄=𝑎₅=1/6. Also, 𝑇 is continuous and sequentially convergent. Thus all the conditions of Theorem 2.1 are satisfied. Moreover, 0 is the unique fixed point of 𝑓.

Taking 𝑇𝑥=𝑥 in (1.2) and Theorem 2.1, we get the Hardy-Rogers type [32] (and so the Kannan, Chatterjea, and Reich) fixed point theorem on partially ordered partial metric spaces.

Corollary 2.4. Let (𝑋,⪯,𝑝) be a partially ordered partial metric space which is complete. Let 𝑓∶𝑋→𝑋 be a nondecreasing mapping such that for all (𝑥,𝑦)∈Δ, we have 𝑝𝑓𝑥,𝑓𝑦≤𝑎₁𝑝𝑥,𝑦+𝑎₂𝑝𝑥,𝑓𝑥+𝑎₃𝑝𝑦,𝑓𝑦+𝑎₄𝑝𝑥,𝑓𝑦+𝑎₅𝑝𝑦,𝑓𝑥, (2.16) where 𝑎_𝑖≥0, 𝑖=1,…,5 with 𝑎₁+𝑎₂+𝑎₃+𝑎₄+𝑎₅<1. If there exists 𝑥₀∈𝑋 with 𝑥₀⪯𝑓𝑥₀, and one of the following two conditions is satisfied.(a)𝑓 is a continuous self map on 𝑋;(b)for any nondecreasing sequence {𝑥_𝑛} in (𝑋,⪯) with lim_𝑛→∞𝑝^𝑠(𝑧,𝑥_𝑛)=0 it follows 𝑥_𝑛⪯𝑧 for all 𝑛∈ℕ; then 𝐹_𝑓≠𝜙. Moreover, 𝑓 has a unique fixed point if 𝐹_𝑓×𝐹_𝑓⊂Δ.

Remark 2.5. Corollary 2.4 corresponds to Theorem 2 of Altun et al. [8] in partially ordered partial metric spaces. For particular choices of the coefficients (𝑎_𝑖)_𝑖=1,…,5 in Theorem 2.1, we obtain the 𝑇-Kannan, 𝑇-Chatterjea, and 𝑇-Reich type fixed point theorems. Also, Theorem 2.1 is an extension of Theorem 2.1 of Filipović et al. [24] from the cone metric spaces to partial metric spaces.

3. Periodic Point Results

Let 𝑓∶𝑋→𝑋. If the map 𝑓 satisfies 𝐹_𝑓=𝐹_𝑓^𝑛 for each 𝑛∈ℕ, then it is said to have the property 𝑃, for more details see [33].

Definition 3.1. Let (𝑋,⪯) be a partially ordered set. A mapping 𝑓 is called (1) a dominating map on 𝑋 if 𝑥⪯𝑓𝑥 for each 𝑥 in 𝑋 and (2) a dominated map on 𝑋 if 𝑓𝑥⪯𝑥 for each 𝑥 in 𝑋.

Example 3.2. Let 𝑋=[0,1] be endowed with usual ordering. Let 𝑓∶𝑋→𝑋 defined by 𝑓𝑥=𝑥^1/3, then 𝑥≤𝑥^1/3=𝑓𝑥 for all 𝑥∈𝑋. Thus 𝑓 is a dominating map.

Example 3.3. Let 𝑋=[0,∞) be endowed with usual ordering. Let 𝑓∶𝑋→𝑋 defined by 𝑓𝑥=√^𝑛𝑥 for 𝑥∈[0,1) and let 𝑓𝑥=𝑥^𝑛 for 𝑥∈[1,∞), for any 𝑛∈ℕ, then for all 𝑥∈𝑋,   𝑥≤𝑓𝑥 that is 𝑓 is the dominating map. Note that Δ₁≠𝜙 if 𝑓 is a dominating or a dominated mapping.
We have the following result.

Theorem 3.4. Let (𝑋,⪯,𝑝) be a partially ordered partial metric space which is complete. Let 𝑇∶𝑋→𝑋 be an injective mapping and 𝑓∶𝑋→𝑋 a nondecreasing such that for all (𝑥,𝑥)∈Δ₁, we have 𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥≤𝜆𝑝𝑇𝑥,𝑇𝑓𝑥, (3.1) for some 𝜆∈[0,1) and for all 𝑥∈𝑋, 𝑥≠𝑓𝑥. Then 𝑓 has the property 𝑃 provided that 𝐹_𝑓 is nonempty and 𝑓 is a dominating map on 𝐹_𝑓^𝑛.

Proof. Let 𝑢∈𝐹_𝑓^𝑛 for some 𝑛>1. Now we show that 𝑢=𝑓𝑢. Since 𝑓 is dominating on 𝐹_𝑓^𝑛, therefore 𝑢⪯𝑓𝑢 which further implies that 𝑓^𝑛−1𝑢⪯𝑓^𝑛𝑢 as 𝑓 is nondecreasing. Hence (𝑓^𝑛−1𝑢,𝑓^𝑛−1𝑢)∈Δ₁. Now by using (3.1), we have 𝑝𝑇𝑢,𝑇𝑓𝑢=𝑝𝑇𝑓𝑓^𝑛−1𝑢,𝑇𝑓²𝑓^𝑛−1𝑢≤𝜆𝑝𝑇𝑓^𝑛−1𝑢,𝑇𝑓^𝑛𝑢=𝜆𝑝𝑇𝑓𝑓^𝑛−2𝑢,𝑇𝑓²𝑓^𝑛−2𝑢. (3.2) Repeating the above process, we get 𝑝𝑇𝑢,𝑇𝑓𝑢≤𝜆^𝑛𝑝𝑇𝑢,𝑇𝑓𝑢. (3.3) Taking limit as 𝑛→∞, we obtain 𝑝(𝑇𝑢,𝑇𝑓𝑢)=0 and 𝑇𝑢=𝑇𝑓𝑢. As 𝑇 is injective, so 𝑢=𝑓𝑢, that is, 𝑢∈𝐹_𝑓.

Theorem 3.5. Let (𝑋,⪯,𝑝) be a partially ordered partial metric space which is complete. let 𝑇,𝑓∶𝑋→𝑋 be mappings satisfy the condition of Theorem 2.1. If 𝑓 is dominating on 𝑋, then 𝑓 has the property 𝑃.

Proof. From Theorem 2.1, 𝐹_𝑓≠∅. We will prove that (3.1) is satisfied for all (𝑥,𝑥)∈Δ₁. Indeed, 𝑓 is a dominating map so that 𝑥⪯𝑓𝑥 and also 𝑓 is nondecreasing so that 𝑓𝑥⪯𝑓²𝑥 and hence (𝑥,𝑓𝑥)∈Δ. Now from (1.2), 𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥=𝑝𝑇𝑓𝑥,𝑇𝑓𝑓𝑥≤𝑎₁𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₂𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₃𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥+𝑎₄𝑝𝑇𝑥,𝑇𝑓²𝑥+𝑎₅𝑝≤𝑎𝑇𝑓𝑥,𝑇𝑓𝑥₁+𝑎₂𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₃𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥+𝑎₄𝑝𝑇𝑥,𝑇𝑓𝑥+𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥−𝑝𝑇𝑓𝑥,𝑇𝑓𝑥+𝑎₅𝑝𝑇𝑓𝑥,𝑇𝑓𝑥, (3.4) that is, 1−𝑎₃−𝑎₄𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥≤𝑎₁+𝑎₂+𝑎₄𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₅−𝑎₄𝑝𝑇𝑓𝑥,𝑇𝑓𝑥. (3.5) Again by using (1.2), we have 𝑝𝑇𝑓²𝑥,𝑇𝑓𝑥=𝑝𝑇𝑓𝑓𝑥,𝑇𝑓𝑥≤𝑎₁𝑝𝑇𝑓𝑥,𝑇𝑥+𝑎₂𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥+𝑎₃𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₄𝑝𝑇𝑓𝑥,𝑇𝑓𝑥+𝑎₅𝑝𝑇𝑥,𝑇𝑓²𝑥≤𝑎₁+𝑎₃𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₂𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥+𝑎₄𝑝𝑇𝑓𝑥,𝑇𝑓𝑥+𝑎₅𝑝𝑇𝑥,𝑇𝑓𝑥+𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥−𝑝𝑇𝑓𝑥,𝑇𝑓𝑥, (3.6) which implies that 1−𝑎₂−𝑎₅𝑝𝑇𝑓𝑥,𝑇𝑓²𝑥≤𝑎₁+𝑎₃+𝑎₅𝑝𝑇𝑥,𝑇𝑓𝑥+𝑎₄−𝑎₅𝑝𝑇𝑓𝑥,𝑇𝑓𝑥. (3.7) Summing (3.5) and (3.7) implies 𝑝(𝑇𝑓𝑥,𝑇𝑓²𝑥)≤𝜆𝑝(𝑇𝑥,𝑇𝑓𝑥),𝜆=(2𝑎₁+𝑎₂+𝑎₃+𝑎₄+𝑎₅)/(2−𝑎₂−𝑎₃−𝑎₄−𝑎₅). Obviously, 𝜆∈[0,1). By Theorem 3.4, 𝑓 has the property 𝑃.

Acknowledgment

S. Radenović is thankful to The Ministry of Science and Technology Development of Serbia.