Generalized Contractions of Rational Type in Ordered Partial Metric Spaces

Abstract

Without the continuity and nondecreasing property of the comparison function, we in this paper prove some fixed point theorems of generalized contractions of rational type in ordered partial metric spaces, which generalize and improve the corresponding results of Luong and Thuan. An example is given to support the usability of our results.

1. Introduction and Preliminaries

Throughout this paper, ℝ₊ and ℕ will denote the set of nonnegative real numbers and the set of all positive integer numbers.

It is well known that the Banach contraction principle is one of the pivotal results of analysis. Generalizations of this principle have been obtained in several directions. In [1], Jaggi introduced a contraction of rational type in a metric space and proved the unique existence of fixed point as the contraction is continuous and the metric space is complete, which was then extended to the case of ordered metric spaces by Harjani et al. [2]. In [3], Luong and Thuan considered weak contractions of rational type in ordered metric spaces and proved the following fixed point theorem.

Theorem 1 (see [3]). Let (𝑋,𝑑,⪯) be a complete ordered metric space and 𝑇∶𝑋→𝑋 a nondecreasing mapping such that there exists a function 𝜂∶ℝ₊→ℝ₊ with 𝜂⁻¹({0})={0} such that 𝑑𝑇𝑥,𝑇𝑦≤𝑀_𝑑⋅𝑟𝑥,𝑦−𝜂𝑀_𝑑⋅𝑟𝑥,𝑦, (1) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦 and 𝑥≠𝑦, where 𝑀_𝑑⋅𝑟(𝑥,𝑦)=max{𝑑(𝑥,𝑦),((𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦))/𝑑(𝑥,𝑦))}. Assume that 𝜂 is lower semicontinuous, and 𝑋 has the following property:(A1) if {𝑥_𝑛} is a nondecreasing sequence in 𝑋 such that 𝑥→−_𝑛^𝑑𝑥, then 𝑥=sup{𝑥_𝑛}.If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point.

In [4], Matthews introduced the partial metric space and extended the Banach contraction principle to the case of partial metric spaces, which was then improved by Oltra and Valero [5]. In [6–8], the authors studied the unique existence of fixed point of generalized contractions in partial metric spaces. Recall that a mapping 𝑇∶𝑋→𝑋 is called a 𝑀^𝑖_𝑝-generalized contraction (𝑀^𝑖_𝑝-GC) if there exists a comparison function 𝜑∶ℝ₊→ℝ₊ such that 𝑝𝑇𝑥,𝑇𝑦≤𝜑𝑀^𝑖_𝑝𝑥,𝑦, (2) where (𝑋,𝑝) is a partial metric space, 𝑀¹_𝑝(𝑥,𝑦)=max{𝑝(𝑥,𝑦),𝑝(𝑥,𝑇𝑥),𝑝(𝑦,𝑇𝑦),(𝑝(𝑥,𝑇𝑦)+𝑝(𝑦,𝑇𝑥))/2}, and 𝑀²_𝑝(𝑥,𝑦)=max{𝑝(𝑥,𝑦),𝑝(𝑥,𝑇𝑥),𝑝(𝑦,𝑇𝑦)}. Under several different assumptions made on 𝜑, they obtained the following fixed point results.

Theorem 2 (see Theorem 1 of [6], Theorem 1 of [7], and Theorems 3 and 4 of [8]). Let (𝑋,𝑝) be a complete partial metric space. Assume the following.

(i) 𝑇∶𝑋→𝑋 is a 𝑀¹_𝑝-GC and one of the following conditions is satisfied:(H1)𝜑 is continuous and nondecreasing, and 𝜑(𝑡)<𝑡 for all 𝑡>0;(H2)𝜑 is nondecreasing, and the series ∑^∞_𝑛=0𝜑^𝑛(𝑡) is convergent for all 𝑡>0 (𝜑^𝑛 denotes the 𝑛th iterate of 𝜑);(H3)𝜑 is upper semicontinuous from the right, and 𝜑(𝑡)<𝑡 for all 𝑡>0.

Or (ii) 𝑇∶𝑋→𝑋 is a 𝑀²_𝑝-GC and the following condition is satisfied:(H4)𝜑 is nondecreasing, and lim_𝑛→∞𝜑^𝑛(𝑡)=0 for all 𝑡>0.

Then 𝑇 has a unique fixed point.

For other references concerned with various fixed point results and common fixed point results for contractions in the setting of metric-like, partial metric, and ordered partial metric spaces, we refer the readers to [9–24].

In this paper, we establish some fixed point theorems for generalized contractions of rational type in ordered partial metric spaces, which generalize and improve Theorems 1 and 2. An example is given to support the usability of our results. Even in the setting of metric spaces, the results presented in this paper are still new since the comparison function 𝜑 is not necessarily assumed to be upper semicontinuous from the right or nondecreasing.

Following [4, 5], a partial metric on a set 𝑋 is a function 𝑝∶𝑋×𝑋→ℝ₊ such that, for each 𝑥,𝑦,𝑧∈𝑋,(p1)𝑝(𝑥,𝑦)=𝑝(𝑥,𝑥)=𝑝(𝑦,𝑦) if and only if 𝑥=𝑦;(p2)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥); (p3)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦);(p4)𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑧)+𝑝(𝑧,𝑦)−𝑝(𝑧,𝑧).Observe that, if 𝑝(𝑥,𝑦)=0, then 𝑥=𝑦. A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is a set and 𝑝 is a partial metric on 𝑋. Each partial metric 𝑝 on 𝑋 induces a 𝑇₀ topology 𝜏_𝑝 on 𝑋 which has as a base of the family of open balls {𝐵_𝑝(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝑝(𝑥,𝜀)={𝑦∈𝑋∶𝑝(𝑥,𝑦)<𝜀+𝑝(𝑥,𝑥)} for all 𝑥∈𝑋 and 𝜀>0.

Let (𝑋,𝑝) be a partial metric space and {𝑥_𝑛} a sequence of 𝑋. The sequence {𝑥_𝑛} converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 (denoted by 𝑥−→_𝑛^𝑝𝑥) if lim_𝑛→∞𝑝(𝑥_𝑛,𝑥)=𝑝(𝑥,𝑥). Define a function 𝑝^𝑠∶𝑋×𝑋→ℝ₊ by 𝑝^𝑠(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦). Then 𝑝^𝑠 is a metric on 𝑋. The sequence {𝑥_𝑛} converges, with respect to 𝜏_𝑝𝑠, to a point 𝑥∈𝑋 (denoted by 𝑥−−→_𝑛^𝑝𝑠𝑥) if and only if lim_𝑛→∞𝑝𝑥_𝑛,𝑥=lim_{𝑚,𝑛→∞}𝑝𝑥_𝑚,𝑥_𝑛=𝑝(𝑥,𝑥). (3) The sequence {𝑥_𝑛} is called a Cauchy sequence if lim_{𝑚,𝑛→∞}𝑝(𝑥_𝑚,𝑥_𝑛) exists and is finite; (𝑋,𝑝) is called complete if every Cauchy sequence {𝑥_𝑛}⊂𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=lim_{𝑚,𝑛→∞}𝑝(𝑥_𝑚,𝑥_𝑛). In particular, {𝑥_𝑛} is called a 0-Cauchy sequence if lim_{𝑚,𝑛→∞}𝑝(𝑥_𝑚,𝑥_𝑛)=0;  (𝑋,𝑝) is called 0-complete if every 0-Cauchy sequence {𝑥_𝑛}⊂𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋 such that 𝑝(𝑥,𝑥)=0. Every complete partial metric space (𝑋,𝑝) is 0-complete, but the converse may not be true; see [17].

Remark 3 (see [4, 5]). A partial metric space (𝑋,𝑝) is complete if and only if (𝑋,𝑝^𝑠) is complete.

2. Fixed Point Theorems

Let (𝑋,𝑝,⪯) be an ordered partial metric space and 𝑇∶𝑋→𝑋. For all 𝑥,𝑦∈𝑋 with 𝑝(𝑥,𝑦)>0, set 𝑀¹_𝑝⋅𝑟𝑝(),𝑝()𝑥,𝑦=max𝑥,𝑦,𝑝𝑥,𝑇𝑥,𝑝𝑦,𝑇𝑦𝑥,𝑇𝑥⋅𝑝𝑦,𝑇𝑦𝑝,𝑝𝑥,𝑦𝑥,𝑇𝑦+𝑝𝑦,𝑇𝑥2,𝑀²_𝑝⋅𝑟𝑝(),𝑝()𝑥,𝑦=max𝑥,𝑦,𝑝𝑥,𝑇𝑥,𝑝𝑦,𝑇𝑦𝑥,𝑇𝑥⋅𝑝𝑦,𝑇𝑦𝑝.𝑥,𝑦 (4) A mapping 𝑇∶𝑋→𝑋 is said to be a 𝑀^𝑖_𝑝.𝑟-generalized contraction of rational type (𝑀^𝑖_𝑝.𝑟-GCRT), if there exists a comparison function 𝜑∶ℝ₊→ℝ₊ such that 𝑝𝑇𝑥,𝑇𝑦≤𝜑𝑀^𝑖_𝑝.𝑟𝑥,𝑦, (5) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦 and 𝑝(𝑥,𝑦)>0.

Lemma 4. Let (𝑋,𝑝,⪯) be an ordered partial metric space and 𝑇∶𝑋→𝑋 a nondecreasing 𝑀¹_𝑝.𝑟-GCRT. Assume that (H5)𝜑(𝑠)<𝑠 and limsup_𝑡→𝑠+(𝜑(𝑡)/𝑡)<1 for all 𝑠>0.For each 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, let 𝑥_𝑛=𝑇^𝑛𝑥₀ for all 𝑛∈ℕ. If 𝑥_𝑛≠𝑥_𝑛−1 for all 𝑛∈ℕ, then lim_{𝑚,𝑛→∞}𝑝𝑥_𝑚,𝑥_𝑛=0. (6)

Proof. Note that 𝑇 is nondecreasing and 𝑥₀⪯𝑇𝑥₀; then {𝑥_𝑛} is nondecreasing. Since 𝑥_𝑛≠𝑥_𝑛−1 for all 𝑛∈ℕ, then 𝑝𝑥_𝑛−1,𝑥_𝑛>0, (7) for all 𝑛∈ℕ. By (4) and (p4), for all 𝑛∈ℕ, we have max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1≤max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥_𝑛−1,𝑥_𝑛+1+𝑝𝑥_𝑛,𝑥_𝑛2=𝑀¹_𝑝⋅𝑟𝑥_𝑛−1,𝑥_𝑛≤max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥_𝑛−1,𝑥_𝑛+𝑝𝑥_𝑛,𝑥_𝑛+12=max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1. (8) Thus by the nondecreasing property of {𝑥_𝑛}, (7), and (8), 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜑𝑀¹_𝑝⋅𝑟𝑥_𝑛−1,𝑥_𝑛=𝜑max𝑝𝑥_𝑛−1,𝑥_𝑛,𝑝𝑥_𝑛,𝑥_𝑛+1, (9) for all 𝑛∈ℕ. Now, we claim that, for all 𝑛∈ℕ, 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝑝𝑥_𝑛−1,𝑥_𝑛. (10) Suppose, on the contrary, that there exists some 𝑛∈ℕ such that 𝑝(𝑥_𝑛,𝑥_𝑛+1)>𝑝(𝑥_𝑛−1,𝑥_𝑛). Then by (8), (9), and 𝜑(𝑠)<𝑠 for all 𝑠>0, we have 𝑝𝑥_𝑛,𝑥_𝑛+1≤𝜑𝑝𝑥_𝑛,𝑥_𝑛+1<𝑝𝑥_𝑛,𝑥_𝑛+1. (11) This is a contradiction and hence (10) is true. Consequently, {𝑝(𝑥_𝑛,𝑥_𝑛+1)} is a decreasing sequence of positive real numbers. This yields that there exists 𝛼≥0 such that 𝑝𝑥_𝑛,𝑥_𝑛+1⟶𝛼⁺. (12) In the following, we will show that 𝛼=0. Suppose, on the contrary, that 𝛼>0. It follows from (8), (9), and (10) that, for all 𝑛∈ℕ, 𝑝𝑥_𝑛,𝑥_𝑛+1𝑝𝑥_𝑛−1,𝑥_𝑛≤𝜑𝑝𝑥_𝑛−1,𝑥_𝑛𝑝𝑥_𝑛−1,𝑥_𝑛. (13) Letting 𝑛→∞ in (13), by (12), (H5), and 𝛼>0, we get a contradiction 1≤limsup_𝑡→𝛼+(𝜑(𝑡)/𝑡)<1. Hence 𝛼=0, and consequently by (12) lim_𝑛→∞𝑝𝑥_𝑛,𝑥_𝑛+1=0. (14) Now, we show that conclusion (6) is true. If otherwise, there exist some 𝛽>0 and subsequences {𝑚_𝑘}_𝑘∈ℕ and {𝑛_𝑘}_𝑘∈ℕ in ℕ, with 𝑚_𝑘>𝑛_𝑘≥𝑘, such that, for all 𝑘∈ℕ, 𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘≥𝛽. (15) From (15) we may assume that, without loss of generality, for all 𝑘∈ℕ, 𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘−1<𝛽. (16) Then by (p4), for all 𝑘∈ℕ, we have 𝛽≤𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘−1+𝑝𝑥_𝑚𝑘,𝑥_𝑚𝑘−1<𝛽+𝑝𝑥_𝑚𝑘,𝑥_𝑚𝑘−1. (17) Letting 𝑘→∞ in the above inequality, by (14), we get lim_𝑘→∞𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘=𝛽. (18) Also by (p4), for all 𝑘∈ℕ, we have 𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘−𝑝𝑥_𝑛𝑘,𝑥_𝑛𝑘+1−𝑝𝑥_𝑚𝑘,𝑥_𝑚𝑘+1≤𝑝𝑥_𝑛𝑘+1,𝑥_𝑚𝑘+1≤𝑝𝑥_𝑛𝑘,𝑥_𝑛𝑘+1+𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘+𝑝𝑥_𝑚𝑘,𝑥_𝑚𝑘+1. (19) Letting 𝑘→∞ in the above inequality, by (14) and (18), we get lim_𝑘→∞𝑝𝑥_𝑛𝑘+1,𝑥_𝑚𝑘+1=𝛽. (20) Similarly, we can obtain lim_𝑘→∞𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘+1=lim_𝑘→∞𝑝𝑥_𝑛𝑘+1,𝑥_𝑚𝑘=𝛽. (21) By the nondecreasing property of {𝑥_𝑛}, (5), and (15), we get, for all 𝑘∈ℕ, 𝑝𝑥_𝑛𝑘+1,𝑥_𝑚𝑘+1=𝑝𝑇𝑥_𝑛𝑘,𝑇𝑥_𝑚𝑘≤𝜑𝑀¹_𝑝.𝑟𝑥_𝑛𝑘,𝑥_𝑚𝑘, (22) where 𝑀¹_𝑝.𝑟(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=max{𝑝(𝑥_𝑛𝑘,𝑥_𝑚𝑘),𝑝(𝑥_𝑛𝑘,𝑥_𝑛𝑘+1), 𝑝(𝑥_𝑚𝑘,𝑥_𝑚𝑘+1),(𝑝((𝑥_𝑛𝑘,𝑥_𝑛𝑘+1)⋅𝑝(𝑥_𝑚𝑘,𝑥_𝑚𝑘+1))/𝑝(𝑥_𝑛𝑘,𝑥_𝑚𝑘)), (𝑝(𝑥_𝑛𝑘,𝑥_𝑚𝑘+1)+𝑝(𝑥_𝑛𝑘+1,𝑥_𝑚𝑘)/2)}. By (14) and (18), we have lim_𝑘→∞𝑝𝑥_𝑛𝑘,𝑥_𝑛𝑘+1⋅𝑝𝑥_𝑚𝑘,𝑥_𝑚𝑘+1𝑝𝑥_𝑛𝑘,𝑥_𝑚𝑘=0, (23) which together with (14), (18), and (21) implies that there exists 𝑘₁∈ℕ such that, for all 𝑘≥𝑘₁, 𝑀¹_𝑝⋅𝑟𝑥_𝑛𝑘,𝑥_𝑚𝑘=𝑀₁𝑥_𝑛𝑘,𝑥_𝑚𝑘, (24) where 𝑀₁(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=max{𝑝(𝑥_𝑛𝑘,𝑥_𝑚𝑘),(𝑝(𝑥_𝑛𝑘,𝑥_𝑚𝑘+1)+𝑝(𝑥_𝑛𝑘+1,𝑥_𝑚𝑘))/2}. It follows from (15), (18), and (21) that 𝑀₁𝑥_𝑛𝑘,𝑥_𝑚𝑘⟶𝛽⁺, (25) By (15), (22), and (24), we have, for all 𝑘≥𝑘₁, 𝑝𝑥_𝑛𝑘+1,𝑥_𝑚𝑘+1𝑀₁𝑥_𝑛𝑘,𝑥_𝑚𝑘≤𝜑𝑀₁𝑥_𝑛𝑘,𝑥_𝑚𝑘𝑀₁𝑥_𝑛𝑘,𝑥_𝑚𝑘. (26) Letting 𝑘→∞ in (26), by (20), (25), (H5), and 𝛽>0, we get a contradiction 1≤limsup_𝑡→𝛽⁺(𝜑(𝑡)/𝑡)<1. Hence conclusion (6) is true. The proof is complete.

Remark 5. It is easy to see that Lemma 4 is still valid for 𝑀²_𝑝⋅𝑟-GCRT and 𝑀¹_𝑝-GC.

Theorem 6. Let (𝑋,𝑝,⪯) be a 0-complete ordered partial metric space and 𝑇∶𝑋→𝑋 a nondecreasing 𝑀¹_𝑝⋅𝑟-GCRT. Assume that(H6)𝜑(𝑠)<𝑠 and limsup_𝑡→𝑠(𝜑(𝑡)/𝑡)<1,  for all 𝑠>0, and 𝑋 has the following property:(A2) if {𝑥_𝑛} is a nondecreasing sequence in 𝑋 such that 𝑥−−→_𝑛^𝑝𝑠𝑥, then 𝑥=sup{𝑥_𝑛}.If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. Let 𝑥_𝑛=𝑇^𝑛𝑥₀ for all 𝑛∈ℕ. If there exists some 𝑛∈ℕ such that 𝑇𝑥_𝑛−1=𝑥_𝑛=𝑥_𝑛−1, then 𝑥_𝑛−1 is a fixed point and hence the proof is complete. Therefore, we may assume that 𝑥_𝑛≠𝑥_𝑛−1 for all 𝑛∈ℕ. By Lemma 4, {𝑥_𝑛} is 0-Cauchy. So by the 0-completeness of 𝑋, there exists 𝑥_∗∈𝑋 such that 𝑥−→_𝑛^𝑝𝑥_∗, and consider lim_𝑛→∞𝑝𝑥_𝑛,𝑥_∗=𝑝𝑥_∗,𝑥_∗=lim_{𝑚,𝑛→∞}𝑝𝑥_𝑚,𝑥_𝑛=0. (27) That is, 𝑥−−→_𝑛^𝑝𝑠𝑥_∗. By (A2), sup{𝑥_𝑛}=𝑥^∗, and so, for all 𝑛∈ℕ, 𝑥_𝑛⪯𝑥_∗. (28) Now, we claim that, for all 𝑛∈ℕ, 𝑥_𝑛≠𝑥_∗. (29) If otherwise, there exists some 𝑛_∗ such that 𝑥_𝑛∗=𝑥_∗; then 𝑥_𝑛=𝑥_∗ for all 𝑛≥𝑛_∗ since {𝑥_𝑛} is nondecreasing. This contradicts with the assumption 𝑥_𝑛≠𝑥_𝑛−1 for all 𝑛∈ℕ, and hence (29) is true. Since 𝑇 is nondecreasing, then 𝑥_𝑛+1=𝑇𝑥_𝑛⪯𝑇𝑥_∗, for all 𝑛∈ℕ, and hence by (A2) 𝑥_∗=sup𝑥_𝑛⪯𝑇𝑥_∗. (30) Let 𝑦_𝑛=𝑇^𝑛𝑥_∗ for all 𝑛∈ℕ. Then {𝑦_𝑛} is a nondecreasing sequence since 𝑇 is nondecreasing. We may assume that, for all 𝑛∈ℕ, 𝑦_𝑛≠𝑦_𝑛−1. (31) If otherwise, there exists some 𝑛∈ℕ such that 𝑇𝑦_𝑛−1=𝑦_𝑛=𝑦_𝑛−1; then 𝑦_𝑛−1 is a fixed point and hence the proof is complete. By Lemma 4 and the 0-completeness of 𝑋, there exists 𝑦_∗∈𝑋 such that 𝑦−→_𝑛^𝑝𝑦_∗, and consider lim_𝑛→∞𝑝𝑦_𝑛,𝑦_∗=𝑝𝑦_∗,𝑦_∗=lim_{𝑚,𝑛→∞}𝑝𝑦_𝑚,𝑦_𝑛=0. (32) That is, 𝑦−−→_𝑛^𝑝𝑠𝑦_∗. By (A2), we have, for all 𝑛∈ℕ, 𝑦_𝑛⪯𝑦_∗. (33) It follows from 𝑥−−→_𝑛^𝑝𝑠𝑥_∗, 𝑦−−→_𝑛^𝑝𝑠𝑦_∗, and the continuity of the metric 𝑝^𝑠 that lim_𝑛→∞𝑝^𝑠𝑥_𝑛,𝑦_𝑛=𝑝^𝑠𝑥_∗,𝑦_∗. (34) That is, lim_𝑛→∞2𝑝𝑥_𝑛,𝑦_𝑛−𝑝𝑥_𝑛,𝑥_𝑛−𝑝𝑦_𝑛,𝑦_𝑛=2𝑝𝑥_∗,𝑦_∗, (35) which together with (27) and (32) implies that lim_𝑛→∞𝑝𝑥_𝑛,𝑦_𝑛=𝑝𝑥_∗,𝑦_∗. (36) By (p4), we have, 𝑛∈ℕ, 𝑝𝑥_𝑛,𝑦_𝑛−𝑝𝑥_𝑛,𝑥_𝑛+1≤𝑝𝑥_𝑛+1,𝑦_𝑛≤𝑝𝑥_𝑛,𝑦_𝑛+𝑝𝑥_𝑛,𝑥_𝑛+1,𝑝𝑥_𝑛,𝑦_𝑛−𝑝𝑦_𝑛,𝑦_𝑛+1≤𝑝𝑥_𝑛,𝑦_𝑛+1≤𝑝𝑥_𝑛,𝑦_𝑛+𝑝𝑦_𝑛,𝑦_𝑛+1. (37) Letting 𝑛→∞ in the above two inequalities, by (27), (32), and (36), we get lim_𝑛→∞𝑝𝑥_𝑛+1,𝑦_𝑛=lim_𝑛→∞𝑝𝑥_𝑛,𝑦_𝑛+1=𝑝𝑥_∗,𝑦_∗. (38) In what follows, we will show 𝑥_∗=𝑦_∗. Suppose, on the contrary, that 𝑥_∗≠𝑦_∗; then 𝑝(𝑥_∗,𝑦_∗)>0. Clearly, 𝑥_𝑛⪯𝑦_𝑛 and 𝑥_𝑛≠𝑦_𝑛 for all 𝑛∈ℕ by (29), (30), and (31). Consequently, 𝑝𝑥_𝑛,𝑦_𝑛>0, (39) for all 𝑛∈ℕ. Thus by (5), we have, for all 𝑛∈ℕ, 𝑝𝑥_𝑛+1,𝑦_𝑛+1=𝑝𝑇𝑥_𝑛,𝑇𝑦_𝑛≤𝜑𝑀¹_𝑝.𝑟𝑥_𝑛,𝑦_𝑛, (40) where 𝑀¹_𝑝⋅𝑟(𝑥_𝑛,𝑦_𝑛)=max{𝑝(𝑥_𝑛,𝑦_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1),𝑝(𝑦_𝑛,𝑦_𝑛+1), (𝑝(𝑥_𝑛,𝑥_𝑛+1)⋅𝑝(𝑦_𝑛,𝑦_𝑛+1))/𝑝(𝑥_𝑛,𝑦_𝑛), (𝑝(𝑥_𝑛,𝑦_𝑛+1)+𝑝(𝑦_𝑛,𝑥_𝑛+1))/2}. By (27), (32), and (36), we have lim_𝑛→∞𝑝𝑥_𝑛,𝑥_𝑛+1⋅𝑝𝑦_𝑛,𝑦_𝑛+1𝑝𝑥_𝑛,𝑦_𝑛=0, (41) which together with (27), (32), (36), and (38) implies that there exists 𝑛₁∈ℕ such that, for all 𝑛≥𝑛₁, 𝑀¹_𝑝.𝑟𝑥_𝑛,𝑦_𝑛=𝑀₂𝑥_𝑛,𝑦_𝑛, (42) where 𝑀₂(𝑥_𝑛,𝑦_𝑛)=max{𝑝(𝑥_𝑛,𝑦_𝑛),(𝑝(𝑥_𝑛,𝑦_𝑛+1)+𝑝(𝑦_𝑛,𝑥_𝑛+1))/2}. It follows from (36) and (38) that 𝑀₂𝑥_𝑛,𝑦_𝑛⟶𝑝𝑥_∗,𝑦_∗. (43) By (39), (40), and (42), we have, for all 𝑛≥𝑛₁, 𝑝𝑥_𝑛+1,𝑦_𝑛+1𝑀₂𝑥_𝑛,𝑦_𝑛≤𝜑𝑀₂𝑥_𝑛,𝑦_𝑛𝑀₂𝑥_𝑛,𝑦_𝑛. (44) Letting 𝑛→∞ in (44), by (36), (43), (H6), and 𝑝(𝑥_∗,𝑦_∗)>0, we get a contradiction 1≤limsup_{𝑡→𝑝(𝑥∗,𝑦∗)}(𝜑(𝑡)/𝑡)<1. Hence 𝑝(𝑥_∗,𝑦_∗)=0, and consequently 𝑥_∗=𝑦_∗. From (30) and (33), it follows that 𝑥_∗⪯𝑇𝑥_∗=𝑦₁⪯𝑦_∗. This together with 𝑥_∗=𝑦_∗ yields that 𝑥_∗=𝑇𝑥_∗. The proof is complete.

In particular when 𝑇 is a 𝑀²_𝑝⋅𝑟-GCRT, condition (H6) could be weakened, and we have the following result.

Theorem 7. Let (𝑋,𝑝,⪯) be a 0-complete ordered partial metric space and 𝑇∶𝑋→𝑋 a nondecreasing 𝑀²_𝑝⋅𝑟-GCRT. Assume that (H5) is satisfied and 𝑋 has property (A2). If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. Following the proof of Theorem 6, we find that (27)–(36) and (39) still hold by Remark 5. Instead of (40), by (5), (6), (39), and 𝑥_𝑛⪯𝑦_𝑛, we have, for all 𝑛∈ℕ, 𝑝𝑥_𝑛+1,𝑦_𝑛+1=𝑝𝑇𝑥_𝑛,𝑇𝑦_𝑛≤𝜑𝑀²_𝑝.𝑟𝑥_𝑛,𝑦_𝑛<𝑀²_𝑝.𝑟𝑥_𝑛,𝑦_𝑛, (45) where 𝑀²_𝑝⋅𝑟(𝑥_𝑛,𝑦_𝑛)=max{𝑝(𝑥_𝑛,𝑦_𝑛),𝑝(𝑥_𝑛,𝑥_𝑛+1),𝑝(𝑦_𝑛,𝑦_𝑛+1), (𝑝(𝑥_𝑛,𝑥_𝑛+1)⋅𝑝(𝑦_𝑛,𝑦_𝑛+1))/𝑝(𝑥_𝑛,𝑦_𝑛)}. By (28), (32), (36), and 𝑝(𝑥_∗,𝑦_∗)>0, there exists 𝑛₂∈ℕ such that, for all 𝑛≥𝑛₂, 𝑀²_𝑝.𝑟𝑥_𝑛,y_𝑛=𝑝𝑥_𝑛,𝑦_𝑛, (46) which together with (45) implies that, for all 𝑛≥𝑛₂, 𝑝𝑥_𝑛+1,𝑦_𝑛+1<𝑝𝑥_𝑛,𝑦_𝑛. (47) That is, {𝑝(𝑥_𝑛,𝑦_𝑛)}^∞_𝑛=𝑛2 is decreasing. Moreover by (36), we have 𝑝𝑥_𝑛,𝑦_𝑛⟶𝑝𝑥_∗,𝑦_∗⁺. (48) Instead of (44), by (39), (45), and (46), for all 𝑛≥𝑛₂, 𝑝𝑥_𝑛+1,𝑦_𝑛+1𝑝𝑥_𝑛,𝑦_𝑛≤𝜑𝑝𝑥_𝑛,𝑦_𝑛𝑝𝑥_𝑛,𝑦_𝑛. (49) Letting 𝑛→∞ in (49), by (48), (H5), and 𝑝(𝑥_∗,𝑦_∗)>0, we get a contradiction 1≤lim_{𝑡→𝑝(𝑥∗,𝑦∗)}⁺(𝜑(𝑡)/𝑡)<1. Hence 𝑝(𝑥_∗,𝑦_∗)=0, and consequently 𝑥_∗=𝑦_∗. From (30) and (33), it follows that 𝑥_∗⪯𝑇𝑥_∗=𝑦₁⪯𝑦_∗. This together with 𝑥_∗=𝑦_∗ yields that 𝑥_∗=𝑇𝑥_∗. The proof is complete.

Corollary 8. Let (𝑋,𝑝,⪯) be a 0-complete ordered partial metric space and 𝑇∶𝑋→𝑋 a nondecreasing mapping such that there exists a function 𝜂∶ℝ₊→ℝ₊ with 𝜂⁻¹({0})={0} such that 𝑝𝑇𝑥,𝑇𝑦≤𝑀²_𝑝⋅𝑟𝑥,𝑦−𝜂𝑀²_𝑝⋅𝑟𝑥,𝑦, (50) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦 and 𝑝(𝑥,𝑦)>0. Assume that, for all 𝑠>0, liminf_𝑡→𝑠+𝜂(𝑡)𝑡>0, (51) and 𝑋 has property (A2). If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. Let 𝜑(𝑡)=𝑡−𝜂(𝑡). Clearly, 𝜑(𝑡)<𝑡 for all 𝑡>0 by 𝜂⁻¹({0})={0}. For all 𝑠>0, it follows from (51) that limsup_𝑡→𝑠+𝜑(𝑡)𝑡=limsup_𝑡→𝑠+𝑡−𝜂(𝑡)𝑡≤1−liminf_𝑡→𝑠+𝜂(𝑡)𝑡<1. (52) Then the conclusion follows immediately from Theorem 7. The proof is complete.

Remark 9. If 𝜂 is lower semicontinuous from the right and 𝜂⁻¹({0})={0}, then, for all 𝑠>0, liminf_𝑡→𝑠+𝜂(𝑡)𝑡≥𝜂(𝑠)𝑠>0. (53) Therefore Theorem 1 follows immediately from Corollary 8.

Now we illustrate Theorems 6 and 7 by the following example.

Example 10. Let 𝑋=ℝ₊ with the partial metric 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. Clearly, 𝑝^𝑠(𝑥,𝑦)=2𝑝(𝑥,𝑦)−𝑝(𝑥,𝑥)−𝑝(𝑦,𝑦)=|𝑥−𝑦| and hence, by Remark 3, (𝑋,𝑝) is a complete partial metric space since (𝑋,𝑝^𝑠) is a complete metric space. Define a partial order ⪯ on 𝑋 by 0}),{0}𝑥⪯𝑦⟺𝑥=𝑦,𝑥,𝑦∈𝑋⧵({∪𝐵𝑥≤𝑦,𝑥,𝑦∈∪𝐵, (54) where 𝐵={(1/𝑛)∶𝑛=2,3,…} and ≤ is the usual order of reals. Let 𝑇𝑥=𝑥0,𝑥=𝑋⧵𝐵,11+𝑥,𝑥=𝑛𝜑(𝑡)=,𝑛=2,4,…,𝑡−𝑡³[)𝑡,𝑡∈0,1⧵𝐵,11+𝑡,𝑡=𝑛1,𝑛=2,3,…,𝑡−2,𝑡≥1. (55)(56) It is easy to see that 𝑇 is nondecreasing, 𝑋 has property (A2), and 𝜑(𝑡)<𝑡 for all 𝑡>0. Direct calculations give that limsup_𝑡→𝑠𝜑(𝑡)𝑡=1−𝑠²(),1<1,𝑠∈0,11−2𝑠<1,𝑠≥1. (57) That is, (H6) is satisfied. In particular, (H5) is satisfied. For each 𝑥,𝑦∈𝑋⧵𝐵 with 𝑥⪯𝑦 and 𝑝(𝑥,𝑦)>0, we must have 𝑝(𝑇𝑥,𝑇𝑦)=0 and hence 𝑝(𝑇𝑥,𝑇𝑦)≤𝜑(𝑀^𝑖_𝑝.𝑟(𝑥,𝑦)),𝑖=1,2. For each 𝑥,𝑦∈{0}∪𝐵 with 𝑥⪯𝑦 and 𝑝(𝑥,𝑦)>0, we have two cases:(i)if 𝑥=0 and𝑦=(1/𝑛),𝑛≥2, then 𝑝(𝑇𝑥,𝑇𝑦)=1/(𝑛+1)≤1/𝑛=𝑝(𝑥,𝑦)=𝑀^𝑖_𝑝.𝑟(𝑥,𝑦),𝑖=1,2;(ii)if 𝑦=1/𝑛 and 𝑥=1/𝑚,𝑚≥𝑛≥3, then 𝑝(𝑇𝑥,𝑇𝑦)=1/(𝑛+1)≤1/𝑛=𝑝(𝑥,𝑦)=𝑀^𝑖_𝑝.𝑟(𝑥,𝑦),𝑖=1,2.
Thus by (56), 𝑝(𝑇𝑥,𝑇𝑦)≤𝑀^𝑖_𝑝.𝑟(𝑥,𝑦)=𝜑(𝑀^𝑖_𝑝⋅𝑟(𝑥,𝑦)),𝑖=1,2. This shows that 𝑇 is a 𝑀^𝑖_𝑝⋅𝑟-GCRT(𝑖=1,2), and consequently the existence of fixed point follows immediately from Theorem 6 or Theorem 7. In fact, 0 is a fixed point of 𝑇.

In what follows, we will extend Theorem 2 to the case of ordered partial metric space under weaker conditions.

Theorem 11. Let (𝑋,𝑝,⪯) be a 0-complete ordered partial metric space and 𝑇∶𝑋→𝑋 a nondecreasing mapping such that, for each 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦, (2) is satisfied for 𝑀¹_𝑝. Assume that either (H4) or (H5) is satisfied and 𝑋 has the following property:(A3) if {𝑥_𝑛} is a nondecreasing sequence in 𝑋 such that 𝑥−−→_𝑛^𝑝𝑠𝑥, then 𝑥_𝑛⪯𝑥 for all 𝑛∈ℕ.
If there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀, then 𝑇 has a fixed point.

Proof.   
Case 1 ((H5) is satisfied). Following the proof of Theorem 6, we get (27) and (28) by Remark 5 and (A3). By (p4), for all 𝑛∈ℕ, 𝑝𝑥_∗,𝑇𝑥_∗−𝑝𝑥_𝑛+1,𝑥_∗≤𝑝𝑥_𝑛+1,𝑇𝑥_∗≤𝑝𝑥_∗,𝑇𝑥_∗+𝑝𝑥_𝑛+1,𝑥_∗. (58) Letting 𝑛→∞ in (58), by (27), we get lim_𝑛→∞𝑝𝑥_𝑛+1,𝑇𝑥_∗=𝑝𝑥_∗,𝑇𝑥_∗. (59) Now, we show that 𝑥_∗ is a fixed point of 𝑇. If otherwise, then 𝑝(𝑥_∗,𝑇𝑥_∗)>0. By (p4), (2), and (28), we have, for all 𝑛∈ℕ, 𝑝𝑥_∗,𝑇𝑥_∗≤𝑝𝑥_𝑛+1,𝑥_∗+𝑝𝑇𝑥_𝑛,𝑇𝑥_∗≤𝑝𝑥_𝑛+1,𝑥_∗+𝜑𝑀¹_𝑝𝑥_𝑛,𝑥_∗, (60) where 𝑀¹_𝑝(𝑥_𝑛,𝑥_∗)=max{𝑝(𝑥_𝑛,𝑥_∗),𝑝(𝑥_𝑛,𝑥_𝑛+1),𝑝(𝑥_∗,𝑇𝑥_∗), (𝑝(𝑥_𝑛,𝑇𝑥_∗)+𝑝(𝑥_∗,𝑥_𝑛+1))/2}. It follows from (27) and (59) that lim_𝑛→∞𝑝𝑥_𝑛,𝑇𝑥_∗+𝑝𝑥_∗,𝑥_𝑛+12=𝑝𝑥_∗,𝑇𝑥_∗2, (61) which together with (27) implies that there exist 𝛿₀∈(0,𝑝(𝑥_∗,𝑇𝑥_∗)/2] and 𝑛₀∈ℕ such that, for all 𝑛≥𝑛₀, 𝑝𝑥_𝑛,𝑇𝑥_∗+𝑝𝑥_∗,𝑥_𝑛+12≤𝑝𝑥_∗,𝑇𝑥_∗2+𝛿₀≤𝑝𝑥_∗,𝑇𝑥_∗,𝑝𝑥_𝑛,𝑥_∗≤𝛿₀,𝑝𝑥_𝑛,𝑥_𝑛+1≤𝛿₀. (62) Thus, for all 𝑛≥𝑛₀, 𝑀¹_𝑝𝑥_𝑛,𝑥_∗=𝑝𝑥_∗,𝑇𝑥_∗, (63) and hence, by (60), 𝑝𝑥_∗,𝑇𝑥_∗≤𝑝𝑥_𝑛+1,𝑥_∗+𝜑𝑝𝑥_∗,𝑇𝑥_∗. (64) Letting 𝑛→∞ in (64), by (27), we get a contradiction 𝑝𝑥_∗,𝑇𝑥_∗≤𝜑𝑝𝑥_∗,𝑇𝑥_∗<𝑝𝑥_∗,𝑇𝑥_∗, (65) since 𝜑(𝑠)<𝑠 for all 𝑠>0. Hence 𝑥_∗ is a fixed point of 𝑇.
Case 2 ((H4) is satisfied). By the same method used in Lemma 4 and [8, Theorem 4], we can show that (27) and (59) still hold. Then by analogy to Case 1, 𝑥_∗ is a fixed point of 𝑇. The proof is complete.

Remark 12. If 𝑇 is a generalized contraction of rational type, then the method used in Theorem 11 fails to work since there exists 𝑛₀∈ℕ such that 𝑀^𝑖_𝑝.𝑟(𝑥_𝑛,𝑥_∗)=𝑝(𝑥_∗,𝑇𝑥_∗) for all 𝑛≥𝑛₀ unless lim_𝑛→∞𝑝(𝑥_𝑛,𝑥_𝑛+1)/𝑝(𝑥_𝑛,𝑥_∗)≤1, which could not be done. Therefore the existence of fixed point of generalized contractions of rational type could not be obtained under the weaker condition (A3).

By the methods used in Theorem 11 and [8, Theorem 4], we have the following fixed point result.

Theorem 13. Let (𝑋,𝑝) be a 0-complete partial metric space and 𝑇∶𝑋→𝑋 a 𝑀¹_𝑝-GC. If either (H4) or (H5) is satisfied, then 𝑇 has a unique fixed point.

Remark 14. (i) In the case that (H5) is satisfied, the continuity and nondecreasing property of 𝜑 necessarily assumed in Theorem 2 is removed in Theorem 13. Note that (H3) implies (H5); then Theorem 13 improves Theorem1 of [6], Theorem 3 of [8], and Theorem 2.3 of [23].
(ii) In the case that (H4) is satisfied, Theorem 13 is generalization of Theorem 4 of [8] to the case of 𝑀¹_𝑝-GC and hence improves Theorem 1 of [7] since (H2) implies (H4).

Example 15. Let 𝑋={0}∪{1/𝑛∶𝑛=2,3,…} with the partial metric 𝑝(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈𝑋. By analogy to Example 10, we find from Remark 3 that (𝑋,𝑝) is a complete partial metric space. Let 𝑇𝑥=𝑥0,𝑥=0,11+𝑥,𝑥=𝑛,𝑛=2,3,…, (66) and 𝜑 is given by (56). It is easy to check that 𝑇 is a 𝑀¹_𝑝-GC and (H5) is satisfied. Therefore the unique existence of fixed point follows immediately from Theorem 13. In fact, 0 is the unique fixed point of 𝑇.
For each 𝑠=1/𝑛,𝑛≥2, we have limsup_𝑡→𝑠+𝜑(𝑡)=(𝑛²−1)/𝑛³>1/(𝑛+1)=𝜑(𝑠). This implies that 𝜑 is not upper semicontinuous from the right at 𝑠=1/𝑛,𝑛≥2. Meanwhile, it is clear that 𝜑 is decreasing on (1/3,1). Therefore, we cannot invoke Theorem 2 to show the unique existence of fixed point.

Remark 16. It is worth mentioning that, even in the setting of metric spaces, the main results in this paper are still new since the continuity and nondecreasing property of the comparison function 𝜑 necessarily assumed in [24] and other relating references is removed.

Acknowledgments

This work was supported by the Natural Science Foundation of China (11161022), the Natural Science Foundation of Jiangxi Province (20114BAB211006, 20122BAB201015), the Educational Department of Jiangxi Province (GJJ12280, GJJ13297), and the Program for Excellent Youth Talents of JXUFE(201201).