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On Fixed Point for Nonexpansive Mappings in Partial Metric Spaces 

 

Arta Ekayanti1, Jumadi2, Erika Eka Santi3 

Department of Mathematics Education, Universitas Muhammadiyah Ponorogo, East Java, Indonesia 

arta_ekayanti@umpo.ac.id1, jumadi@umpo.ac.id2, erika_ekasanti@umpo.ac.id3 

 

Article History: 

Received: 12-04-2024 

Revised: 19-05-2024 

Accepted: 02-06-2024 

Abstract:  

In this paper, we establish some fixed points theorem for nonexpansive mappings in partial 

metric spaces. Our result generalizes Vetro’s results (2015) in the setting of partial metric 

spaces. This work proves and generalizes some results of Aydi (2017). Suitable example is 

provided to illustrate the usability of our results. 

Keywords: Fixed points, nonexpansive mapping, partial metric spaces. 

2020 Mathematics Subject Classification: 47H10, 47H09 

 

1. Introduction 

The Banach Contraction Principle is a fundamental topic in mathematics, especially in the focus of 

fixed point theory. S. Banach introduced the Banach Contraction Principle in 1922 [2]. The Banach 

Contraction Principle guarantees the existence of fixed points from a contraction mapping. In its 

development, studies regarding the existence of fixed points from contraction mapping have attracted 

much interest from researchers. Various studies were carried out in an effort to develop fixed point 

theory, including by providing a new definition of contraction mapping in various applications [11, 

12,14] as a generalization of the Banach Contraction Principle [3]. 

Contraction mapping is a special case of Lipschitz-𝜆 mapping [16]. Note that if we are given a metric 

space (𝑋, 𝑑) and given a mapping 𝑓: 𝑋 → 𝑋, then the mapping 𝑓 is said to be a Lipschitz-𝜆 mapping 

if 

𝑑(𝑓(𝑥), 𝑓(𝑦)) ≤ 𝜆𝑑(𝑥, 𝑦) 

for every 𝑥, 𝑦 ∈ 𝑋 with 𝜆 ≥ 0. In this case, if 𝜆 ∈ [0,1), then the mapping 𝑓 is said to be a contraction 

mapping, whereas if 𝜆 = 1, the mapping 𝑓 is said to be a non-expansive mapping. Like contraction 

mapping widely studied in relation to fixed points, non-expansive mapping also has an equally 

important role in fixed point studies [4,8,13,15]. One of them was carried out by Vetro [16], and the 

existence of fixed points from non-expansive mapping in metric space was successfully demonstrated. 

Furthermore, Aydi [1] extend a fixed point theorem for 𝛼-nonexpansive mappings on partial metric 

spaces. Motivated by Vetro [16] and Aydi [1], we will prove the fixed point theorem for nonexpansive 

mapping in partial metric spaces. And we also prove a more general theorem. 

2. Preliminaries 

In 1992, Matthews introduced a new concept as a generalization of standard metrics, namely partial 

metrics. Let we consider the following definitions. 



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Definition 2.1 [9,10] Let X be any nonempty set. A partial metric on 𝑋 is a mapping 𝑝: 𝑋 × 𝑋 → [0, ∞) 

which satisfies following conditions: 

(P1) 𝑝(𝑥, 𝑦) = 𝑝(𝑦, 𝑥), 

(P2) If 𝑝(𝑥, 𝑥) = 𝑝(𝑥, 𝑦) = 𝑝(𝑦, 𝑦) then 𝑥 = 𝑦, 

(P3) 𝑝(𝑥, 𝑥) ≤ 𝑝(𝑥, 𝑦), 

(P4) 𝑝(𝑥, 𝑧) + 𝑝(𝑦, 𝑦) ≤ 𝑝(𝑥, 𝑦) + 𝑝(𝑦, 𝑧), 

for all 𝑥, 𝑦, 𝑧 ∈ 𝑋. Then pair (𝑋, 𝑝) is called a partial metric space. 

Partial metric 𝑝 will be a metric if 𝑝(𝑥, 𝑥) = 0. Several properties of partial metric spaces are given as 

follow [5,6,7,9,10]. 

Definition 2.2. Let (𝑋, 𝑝) be a partial metric space. A sequence (𝑥𝑛) is said to converges to a point 

𝑥 ∈ 𝑋 if and only if 

𝑙𝑖𝑚𝑛→∞ 𝑝(𝑥𝑛, 𝑥) = 𝑝(𝑥, 𝑥). 

Definition 2.3. Let (𝑋, 𝑝) be a partial metric space. A sequence (𝑥𝑛) is called Cauchy sequence if and 

only if  

𝑙𝑖𝑚𝑛→∞ 𝑝(𝑥𝑛, 𝑥𝑚) 

is finite. 

Definition 2.4. Let (𝑋, 𝑝) be a partial metric space. If every Cauchy sequence (𝑥𝑛) converges to a 

point 𝑥 ∈ 𝑋 such that 

𝑙𝑖𝑚𝑛→∞ 𝑝(𝑥𝑛, 𝑥𝑚) = 𝑝(𝑥, 𝑥), 

then (𝑋, 𝑝) is known as complete partial metric space. 

For 𝑝 metric spaces on 𝑋, the mapping 𝑝𝑠: 𝑋 × 𝑋 → [0, ∞) defined by 

𝑝𝑠(𝑥, 𝑦) = 2𝑝(𝑥, 𝑦) − 𝑝(𝑥, 𝑥) − 𝑝(𝑦, 𝑦) 

for each 𝑥, 𝑦 ∈ 𝑋 is a metric on 𝑋. A sequence (𝑥𝑛) is Cauchy in partial metric spaces (𝑋, 𝑝) if and 

only if (𝑥𝑛) is Cauchy sequence in metric space (𝑋, 𝑝𝑠). It implies, a partial metric space (𝑋, 𝑝) is 

complete if and only if metric spaces (𝑋, 𝑝𝑠) is complete. Therefore, for (𝑥𝑛) is sequence in partial 

metric spaces (𝑋, 𝑝) and 𝑥 ∈ 𝑋, we have lim
𝑛→∞

𝑝𝑠(𝑥𝑛, 𝑥) = 0 if and only if  

𝑝(𝑥, 𝑥) = lim
𝑛→∞

𝑝(𝑥𝑛, 𝑥) = lim
𝑛,𝑚→∞

𝑝(𝑥𝑛, 𝑥𝑚) 

Definition 2.5. Let (𝑋, 𝑝) be a partial metric space. Mapping 𝑓: 𝑋 → 𝑋 is (sequentially) continuous if 

𝑝(𝑥𝑛, 𝑥) → 𝑝(𝑥, 𝑥) then for 𝑛 → ∞ we have 

𝑝(𝑓(𝑥𝑛), 𝑓(𝑥)) → 𝑝(𝑓(𝑥), 𝑓(𝑥)). 

Lemma 2.6. Suppose that (𝑋, 𝑝) be a partial metric space, then 

1. If 𝑝(𝑥, 𝑦) = 0 then 𝑥 = 𝑦 

2. If 𝑥 ≠ 𝑦 then 𝑝(𝑥, 𝑦) > 0 

Lemma 2.7. [16] If (𝑎𝑛) is nonincreasing sequences of nonnegative real numbers, then the sequence 



Communications on Applied Nonlinear Analysis  

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(
𝑎𝑛 + 𝑎𝑛+1

𝑎𝑛 + 𝑎𝑛+1 + 1
) 

is nonincreasing too. 

We extend Corollary 2.5 in [16] to partial metric spaces. 

Corollary 2.8. Let (𝑋, 𝑝) be a partial metric space. Suppose that 𝑓: 𝑋 → 𝑋 be a nonexpansive mapping 

and 𝑥0 ∈ 𝑋. If (𝑥𝑛) is a Picard sequence of initial point 𝑥0, then the sequence 

(
𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
) (2.1) 

is nonincreasing. 

Proof:  

Since (𝑥𝑛) is a Picard sequence of initial point 𝑥0 then 𝑥𝑛 = 𝑓(𝑥𝑛−1) = 𝑓𝑛(𝑥0) for all 𝑛 ∈ ℕ. 

Furthermore, since 𝑓 is nonexpansive mapping then  

𝑝(𝑥𝑛, 𝑥𝑛+1) = 𝑝(𝑓(𝑥𝑛−1), 𝑓(𝑥𝑛)) ≤ 𝑝(𝑥𝑛−1, 𝑥𝑛) 

for all 𝑛 ∈ ℕ. Hence by Lemma 2.7, the statement (2.1) holds. This completes the proof. 

3. Main Results 

Theorem 3.1. Let (𝑋, 𝑝) be a complete partial metric spaces endowed with a binary relation ℜ on 𝑋. 

Suppose that 𝑓: 𝑋 → 𝑋 be a nonexpansive mappings such that  

𝑝(𝑓(𝑥), 𝑓(𝑦)) ≤ (
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) (3.1) 

for each 𝑥, 𝑦 ∈ ℜ, where 𝑘 ∈ [0,1). Assume that: 

1. 𝑓 is preserving mapping 

2. 𝑓 is continuous mapping 

3. 𝐹𝑖𝑥 (𝑓) is well ordered with respect to ℜ. 

If there exist 𝑥0 ∈ 𝑋 such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and 

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘 < 1 (3.2) 

then there exist 𝑧 ∈ 𝑋 such that 𝑝(𝑧, 𝑧) = 0. Furthermore 

a. There exists 𝑧 ∈ 𝑋 fixed point of 𝑓 

b. The Picard sequences of initial point 𝑥0 ∈ 𝑋 converges to fixed point of 𝑓 

c. If 𝑧 and 𝑤 are fixed point of 𝑓 where 𝑧 ≠ 𝑤 then  

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
≥

1 − 𝑘

2
 

 



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Proof: 

Let 𝑥0 ∈ 𝑋 be such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and (3.2) holds. Suppose that (𝑥𝑛) be a Picard sequence of 

initial point 𝑥0 such that  

𝑥𝑛 = 𝑓(𝑥𝑛−1) = 𝑓𝑛(𝑥0) 

for all 𝑛 ∈ ℕ. Let we consider the following cases: 

Case 1. Suppose that 𝑥𝑛−1 = 𝑥𝑛 for some 𝑛 ∈ ℕ then  

𝑥𝑛−1 = 𝑓(𝑥𝑛−1) 

It means 𝑥𝑛−1 is fixed point of 𝑓. Hence, the existence of a fixed point of 𝑓 is proved. 

Case 2. Suppose that 𝑥𝑛−1 ≠ 𝑥𝑛 for all 𝑛 ∈ ℕ. Let we consider that (𝑥0, 𝑥1) = (𝑥0, 𝑓(𝑥0)) ∈ ℜ. Since 

𝑓 is preserving mapping then we have  

(𝑓(𝑥0), 𝑓2(𝑥0)) ∈ ℜ 

By induction, we have 

(𝑥𝑛−1, 𝑥𝑛) = (𝑓𝑛−1(𝑥0), 𝑓𝑛(𝑥0)) ∈ ℜ 

for all 𝑛 ∈ ℕ. Since (𝑥𝑛−1, 𝑥𝑛) ∈ ℜ then by (3.1) we obtain 

𝑝(𝑥𝑛, 𝑥𝑛+1) = 𝑝(𝑓(𝑥𝑛−1), 𝑓(𝑥𝑛))

⬚ ≤ (
𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛−1))

𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛−1)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛)) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ = (
𝑝(𝑥𝑛−1, 𝑥𝑛+1) + 𝑝(𝑥𝑛, 𝑥𝑛)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤ (
𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) − 𝑝(𝑥𝑛, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ = (
𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

 

Since 𝑓 in nonexpansive mapping, then by using Corollary 2.8 we obtain 

𝑝(𝑥𝑛, 𝑥𝑛+1) ≤ (
𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤ (
𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2)

𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

 

By (3.2) then we have 

𝑝(𝑥𝑛, 𝑥𝑛+1) ≤ (
𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2)

𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤ (
𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ = 𝛼𝑝(𝑥𝑛−1, 𝑥𝑛)

 



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where 𝛼 < 1. Hence (𝑥𝑛) is Cauchy sequences in 𝑋 since 𝑝(𝑥𝑛, 𝑥𝑛+1) ≤ 𝛼𝑝(𝑥𝑛−1, 𝑥𝑛), 𝛼 < 1. 

Therefor we have  

lim
𝑛,𝑚→∞

𝑝(𝑥𝑛, 𝑥𝑚) = 0 

Since 𝑋 is complete partial metric spaces, then sequences (𝑥𝑛) is converges, namely 𝑥𝑛 → 𝑧 ∈ 𝑋. It 

implies 

lim
𝑛→∞

𝑝(𝑥𝑛, 𝑧) = 𝑝(𝑧, 𝑧) = lim
𝑛,𝑚→∞

𝑝(𝑥𝑛, 𝑥𝑚) = 0. 

Therefor, 𝑝(𝑧, 𝑧) = 0. Furthermore, we will prove that 𝑧 is fixed point of 𝑓. Since 𝑓 is nonexpansive 

mapping then 

𝑝(𝑓(𝑧), 𝑓(𝑧)) ≤ 𝑝(𝑧, 𝑧) = 0 

Hence 𝑝(𝑓(𝑧), 𝑓(𝑧)) = 0. By hypothesis (2), the continuity of 𝑓 then by (3.1)  

𝑝(𝑧, 𝑓(𝑧)) = lim
𝑛→∞

𝑝(𝑥𝑛+1, 𝑓(𝑧))

⬚ = lim
𝑛→∞

𝑝(𝑓(𝑥𝑛), 𝑓(𝑧))

⬚ = 𝑝(𝑧, 𝑧)

 

Since 𝑝(𝑧, 𝑧) = 0, it implies 𝑝(𝑧, 𝑓(𝑧)) = 0, i.e. 𝑧 ∈ 𝑓(𝑧). It means 𝑧 is fixed point of 𝑓. Suppose that 

𝑤 is another fixed point of 𝑓 where 𝑧 ≠ 𝑤. By hypothesis (3) we obtain (𝑧, 𝑤) ∈ ℜ, then (3.1) we have 

𝑝(𝑧, 𝑤) = 𝑝(𝑓(𝑧), 𝑓(𝑤))

⬚ ≤ (
𝑝(𝑧, 𝑓(𝑤)) + 𝑝(𝑤, 𝑓(𝑧))

𝑝(𝑧, 𝑓(𝑧)) + 𝑝(𝑤, 𝑓(𝑤)) + 1
+ 𝑘) 𝑝(𝑧, 𝑤)

⬚ = (
𝑝(𝑧, 𝑤) + 𝑝(𝑤, 𝑧)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
+ 𝑘) 𝑝(𝑧, 𝑤)

⬚ = (
2𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
+ 𝑘) 𝑝(𝑧, 𝑤)

 

Furthermore, 

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
≥

1 − 𝑘

2
 

This completes the proof. 

We can replace the continuity hypothesis in Theorem 3.1 with the convergence hypothesis as the 

following theorem. 

Theorem 3.2. Let (𝑋, 𝑝) be a complete partial metric spaces endowed with a binary relation ℜ on 𝑋. 

Suppose that 𝑓: 𝑋 → 𝑋 be a nonexpansive mappings such that  

𝑝(𝑓(𝑥), 𝑓(𝑦)) ≤ (
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) (3.3) 

 



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for each 𝑥, 𝑦 ∈ ℜ, where 𝑘 ∈ [0,1). Assume that: 

1. 𝑓 is preserving mapping 

2. If 𝑥𝑛 sequences in 𝑋 such that (𝑥𝑛−1, 𝑥𝑛) ∈ ℜ for each 𝑛 ∈ ℕ and 𝑥𝑛 → 𝑧 ∈ 𝑋 as 𝑛 → ∞, then 

(𝑥𝑛−1, 𝑧) ∈ ℜ for all 𝑛 ∈ ℕ 

3. 𝐹𝑖𝑥 (𝑓) is well ordered with respect to ℜ. 

If there exist 𝑥0 ∈ 𝑋 such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and 

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘 < 1 (3.4) 

then there exist 𝑧 ∈ 𝑋 such that 𝑝(𝑧, 𝑧) = 0. Furthermore 

a. There exists 𝑧 ∈ 𝑋 fixed point of 𝑓 

b. The Picard sequences of initial point 𝑥0 ∈ 𝑋 converges to fixed point of 𝑓 

c. If 𝑧 and 𝑤 are fixed point of 𝑓 where 𝑧 ≠ 𝑤 then  

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
≥

1 − 𝑘

2
 

Proof: 

The proof is following Theorem 3.1’s proof. We only have to check that 𝑧 is fixed point of 𝑓. By 

hypothesis (2) we have (𝑥𝑛, 𝑧) ∈ ℜ for all 𝑛 ∈ ℕ. Therefor by (3.3) we obtain 

𝑝(𝑥𝑛+1, 𝑓(𝑧)) = 𝑝(𝑓(𝑥𝑛), 𝑓(𝑧))

⬚ ≤ (
𝑝(𝑥𝑛, 𝑓(𝑧)) + 𝑝(𝑧, 𝑓(𝑥𝑛))

𝑝(𝑥𝑛, 𝑓(𝑥𝑛)) + 𝑝(𝑧, 𝑓(𝑧)) + 1
+ 𝑘) 𝑝(𝑥𝑛, 𝑧)

⬚ = (
𝑝(𝑥𝑛, 𝑓(𝑧)) + 𝑝(𝑧, 𝑥𝑛+1)

𝑝(𝑥𝑛, 𝑥𝑛+1) + 𝑝(𝑧, 𝑓(𝑧)) + 1
+ 𝑘) 𝑝(𝑥𝑛, 𝑧)

 

Taking 𝑛 → ∞ then we obtain 

𝑝(𝑧, 𝑓(𝑧)) ≤ (
𝑝(𝑧, 𝑓(𝑧)) + 𝑝(𝑧, 𝑧)

𝑝(𝑧, 𝑧) + 𝑝(𝑧, 𝑓(𝑧)) + 1
+ 𝑘) 𝑝(𝑧, 𝑧) 

Since 𝑝(𝑧, 𝑧) = 0, it implies 𝑝(𝑧, 𝑓(𝑧)) ≤ 0. Hence 𝑝(𝑧, 𝑓(𝑧)) = 0, i.e. 𝑧 ∈ 𝑓(𝑧). It means 𝑧 is fixed 

point of 𝑓. For the following proof on the similar line with Theorem 3.1’s proof. This completes the 

proof. 

Let we consider, if 𝑝(𝑧, 𝑧) and 𝑝(𝑤, 𝑤) in Theorem 3.1 and 3.2 are zero, then we have  

𝑝(𝑧, 𝑤) ≥
1 − 𝑘

2
 

Furthermore, we provide the following example motivated by Aydi [1]. 

Example 3.3. Let 𝑋 = [0,1] and 𝑝(𝑥, 𝑦) = max {𝑥, 𝑦}. Define 𝑓: 𝑋 → 𝑋 by 𝑓(𝑥) = 𝑥2 for all 𝑥 ∈ 𝑋. 

Clearly, (𝑋, 𝑝) is complete partial metric spaces and 𝑓 is continuous mappings. Suppose that partial 

metric spaces (𝑋, 𝑝) is endowed with binary relations ℜ where (𝑥, 𝑦) ∈ ℜ it means 𝑥 ≥ 𝑦. 



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Without loss of generality, take 𝑥 ≥ 𝑦 > 0. Let we consider that 𝑓 is nonexpasive mapping since 

𝑝(𝑓(𝑥), 𝑓(𝑦)) = max{𝑥2, 𝑦2} = 𝑥2 ≤ 𝑥 = max{𝑥, 𝑦} = 𝑝(𝑥, 𝑦). 

Take 𝑘 =
1

2
 and let we consider the following cases: 

Case 1: For 𝑥2 ≥ 𝑦 we have 

𝑝(𝑓(𝑥), 𝑓(𝑦)) = 𝑥2, and 

(
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) = (

𝑥 + 𝑥2

𝑥 + 𝑦 + 1
+

1

2
) 𝑥

⬚ =
2𝑥2 + 2𝑥3 + (1 + 𝑥 + 𝑦)𝑥

2(𝑥 + 𝑦 + 1)

⬚ =
2𝑥2(1 + 𝑥) + 2𝑥2𝑦 − 2𝑥2𝑦 + (1 + 𝑥 + 𝑦)𝑥

2(𝑥 + 𝑦 + 1)

⬚ = 𝑥2 +
(1 + 𝑥 + 𝑦 − 2𝑥𝑦)𝑥

2(𝑥 + 𝑦 + 1)

 

Since 2𝑥𝑦 ≤ 𝑥2 + 𝑦2 ≤ 𝑥 + 𝑦 then 

(1 + 𝑥 + 𝑦 − 2𝑥𝑦)𝑥

2(𝑥 + 𝑦 + 1)
> 0 

It implies 

(
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) ≥ 𝑥2 = 𝑝(𝑓(𝑥), 𝑓(𝑦)) = 𝑥2 

Case 2: For 𝑥2 < 𝑦 we have 

𝑝(𝑓(𝑥), 𝑓(𝑦)) = 𝑥2, and 

(
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) = (

𝑥 + 𝑦

𝑥 + 𝑦 + 1
+

1

2
) 𝑥

⬚ =
2𝑥2 + 2𝑥𝑦 + (1 + 𝑥 + 𝑦)𝑥

2(𝑥 + 𝑦 + 1)

⬚ =
2𝑥2(1 + 𝑦) + 2𝑥3 − 2𝑥3 + (1 + 𝑥 + 𝑦)𝑥

2(𝑥 + 𝑦 + 1)

⬚ = 𝑥2 +
(1 + 𝑥 + 𝑦 − 2𝑥2)𝑥

2(𝑥 + 𝑦 + 1)

 

Since 𝑥 + 𝑦 ≤ 𝑥2 + 𝑦2 ≤ 2𝑥2 then 

(1 + 𝑥 + 𝑦 − 2𝑥2)𝑥

2(𝑥 + 𝑦 + 1)
> 0 

 

 



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Hence  

(
𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝑝(𝑥, 𝑦) ≥ 𝑥2 = 𝑝(𝑓(𝑥), 𝑓(𝑦)) 

This condition shows that (3.1) in Theorem 3.1 (resp. (3.3) in Theorem 3.2) is verified for all 𝑥, 𝑦 ∈ 𝑋. 

Furthermore, 𝑓 is preserving mappings since (𝑥, 𝑓(𝑥)) ∈ ℜ, i.e. 𝑥 ≥ 𝑥2 then we have 𝑥2 ≥ 𝑥3 or 

(𝑓(𝑥), 𝑓2(𝑥)) ∈ ℜ. For 𝑥0 =
1

2
 we have (𝑥0, 𝑓(𝑥0)) ∈ ℜ since 

1

2
>

1

4
= (

1

2
)

2

 and 

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘 =

1
2 +

1
4

1
2 +

1
4 + 1

+
1

2
=

13

14
< 1. 

Therefore, all hypotheses of Theorem 3.1 are satisfied. In this case 𝑓 has two fixed points which are 

𝑧 = 0 and 𝑤 = 1, and 

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
=

𝑝(0,1)

𝑝(0,0) + 𝑝(1,1) + 1
=

1

2
>

1

4
=

1 − 𝑘

2
. 

 

In the next results, we replace 𝑝(𝑥, 𝑦) in (3.1) by 𝒩(𝑥, 𝑦) as on the following theorem. This result 

generalizes Theorem 3.1. 

Theorem 3.4. Let (𝑋, 𝑝) be a complete partial metric spaces endowed with a binary relation ℜ on 𝑋. 

Suppose that 𝑓: 𝑋 → 𝑋 be a nonexpansive mappings such that  

𝑝(𝑓(𝑥), 𝑓(𝑦)) ≤
1

2
(

𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝒩(𝑥, 𝑦) (3.5) 

for each 𝑥, 𝑦 ∈ ℜ, where 𝑘 ∈ [0,1) and 

 

𝒩(𝑥, 𝑦) = max {𝑝(𝑥, 𝑦), 𝑝(𝑥, 𝑓(𝑥)), 𝑝(𝑦, 𝑓(𝑦)), 𝑝(𝑥, 𝑓(𝑦)), 𝑝(𝑦, 𝑓(𝑥))} (3.6) 

Assume that: 

1. 𝑓 is preserving mapping 

2. 𝑓 is continuous mapping 

3. 𝐹𝑖𝑥 (𝑓) is well ordered with respect to ℜ. 

If there exist 𝑥0 ∈ 𝑋 such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and 

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘 < 1 (3.7) 

then there exist 𝑧 ∈ 𝑋 such that 𝑝(𝑧, 𝑧) = 0. Furthermore 

a. There exists 𝑧 ∈ 𝑋 fixed point of 𝑓 

b. The Picard sequences of initial point 𝑥0 ∈ 𝑋 converges to fixed point of 𝑓 



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c. If 𝑧 and 𝑤 are fixed point of 𝑓 where 𝑧 ≠ 𝑤 then  

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
≥

1 − 𝑘

2
 

Proof: 

Suppose that 𝑥0 ∈ 𝑋 such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and (3.5) holds. Let (𝑥𝑛) be a Picard sequence of initial 

point 𝑥0 such that  

𝑥𝑛 = 𝑓(𝑥𝑛−1) = 𝑓𝑛(𝑥0) 

for all 𝑛 ∈ ℕ. Let we consider, if 𝑥𝑛−1 = 𝑥𝑛 for some 𝑛 ∈ ℕ then 𝑥𝑛−1 = 𝑓(𝑥𝑛−1). It means 𝑥𝑛−1 is 

fixed point of 𝑓. Hence, the existence of a fixed point of 𝑓 is proved. Another condition, suppose that 

𝑥𝑛−1 ≠ 𝑥𝑛 for all 𝑛 ∈ ℕ. Let we consider that (𝑥0, 𝑥1) = (𝑥0, 𝑓(𝑥0)) ∈ ℜ. Since 𝑓 is preserving 

mapping then we have  

(𝑓(𝑥0), 𝑓2(𝑥0)) ∈ ℜ 

By induction, we have 

(𝑥𝑛−1, 𝑥𝑛) = (𝑓𝑛−1(𝑥0), 𝑓𝑛(𝑥0)) ∈ ℜ 

for all 𝑛 ∈ ℕ. Since (𝑥𝑛−1, 𝑥𝑛) ∈ ℜ then by (3.5) we obtain 

𝑝(𝑥𝑛, 𝑥𝑛+1) = 𝑝(𝑓(𝑥𝑛−1), 𝑓(𝑥𝑛))

⬚ ≤
1

2
(

𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛−1))

𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛−1)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛)) + 1
+ 𝑘) 𝒩(𝑥𝑛−1, 𝑥𝑛)

 

Let we consider 

𝒩(𝑥𝑛−1, 𝑥𝑛) = max {𝑝(𝑥𝑛−1, 𝑥𝑛), 𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛−1)), 𝑝(𝑥𝑛, 𝑓(𝑥𝑛)), 𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛)), 𝑝(𝑥𝑛, 𝑓(𝑥𝑛−1))} 

 = max{𝑝(𝑥𝑛−1, 𝑥𝑛), 𝑝(𝑥𝑛, 𝑥𝑛+1), 𝑝(𝑥𝑛−1, 𝑥𝑛+1), 𝑝(𝑥𝑛, 𝑥𝑛)} 

 = max{𝑝(𝑥𝑛−1, 𝑥𝑛), 𝑝(𝑥𝑛, 𝑥𝑛+1), 𝑝(𝑥𝑛−1, 𝑥𝑛+1)} 

Since 𝑓 is nonexpansive mapping then  

𝑝(𝑥𝑛−1, 𝑥𝑛) = 𝑝(𝑓(𝑥𝑛), 𝑓(𝑥𝑛+1)) ≤ 𝑝(𝑥𝑛, 𝑥𝑛+1) 

Therefore 

𝒩(𝑥𝑛−1, 𝑥𝑛) = max {𝑝(𝑥𝑛−1, 𝑥𝑛), 𝑝(𝑥𝑛−1, 𝑥𝑛+1)} 

Let we consider 

𝑝(𝑥𝑛−1, 𝑥𝑛+1) ≤ 𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) − 𝑝(𝑥𝑛, 𝑥𝑛)

⬚ ≤ 𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1)

⬚ ≤ 𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ = 2𝑝(𝑥𝑛−1, 𝑥𝑛)

 

Thus 

𝒩(𝑥𝑛−1, 𝑥𝑛) ≤ max{𝑝(𝑥𝑛−1, 𝑥𝑛), 2𝑝(𝑥𝑛−1, 𝑥𝑛)} = 2𝑝(𝑥𝑛−1, 𝑥𝑛) 



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Hence 

𝑝(𝑥𝑛, 𝑥𝑛+1) = 𝑝(𝑓(𝑥𝑛−1), 𝑓(𝑥𝑛))

⬚ ≤
1

2
(

𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛−1))

𝑝(𝑥𝑛−1, 𝑓(𝑥𝑛−1)) + 𝑝(𝑥𝑛, 𝑓(𝑥𝑛)) + 1
+ 𝑘) 2𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ =
1

2
(

𝑝(𝑥𝑛−1, 𝑥𝑛+1) + 𝑝(𝑥𝑛, 𝑥𝑛)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 2𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤
1

2
(

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) − 𝑝(𝑥𝑛, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 2𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ =
1

2
(

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1)

𝑝(𝑥𝑛−1, 𝑥𝑛) + 𝑝(𝑥𝑛, 𝑥𝑛+1) + 1
+ 𝑘) 2𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤
1

2
(

𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2)

𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2) + 1
+ 𝑘) 2𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ = (
𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2)

𝑝(𝑥0, 𝑥1) + 𝑝(𝑥1, 𝑥2) + 1
+ 𝑘) 𝑝(𝑥𝑛−1, 𝑥𝑛)

⬚ ≤ 𝛼𝑝(𝑥𝑛−1, 𝑥𝑛)

 

where 𝛼 < 1. Hence (𝑥𝑛) is Cauchy sequences in 𝑋 since 𝑝(𝑥𝑛, 𝑥𝑛+1) ≤ 𝛼𝑝(𝑥𝑛−1, 𝑥𝑛), 𝛼 < 1. The 

next proof is the same as the proof of Theorem 3.1. This completes the proof. 

Analogous to Theorem 3.4, we can generalize theorem 3.2 as follows. 

Theorem 3.5. Let (𝑋, 𝑝) be a complete partial metric spaces endowed with a binary relation ℜ on 𝑋. 

Suppose that 𝑓: 𝑋 → 𝑋 be a nonexpansive mappings such that  

𝑝(𝑓(𝑥), 𝑓(𝑦)) ≤
1

2
(

𝑝(𝑥, 𝑓(𝑦)) + 𝑝(𝑦, 𝑓(𝑥))

𝑝(𝑥, 𝑓(𝑥)) + 𝑝(𝑦, 𝑓(𝑦)) + 1
+ 𝑘) 𝒩(𝑥, 𝑦) 

for each 𝑥, 𝑦 ∈ ℜ, where 𝑘 ∈ [0,1) and 

𝒩(𝑥, 𝑦) = max {𝑝(𝑥, 𝑦), 𝑝(𝑥, 𝑓(𝑥)), 𝑝(𝑦, 𝑓(𝑦)), 𝑝(𝑥, 𝑓(𝑦)), 𝑝(𝑦, 𝑓(𝑥))} 

Assume that: 

1. 𝑓 is preserving mapping 

2. If 𝑥𝑛 sequences in 𝑋 such that (𝑥𝑛−1, 𝑥𝑛) ∈ ℜ for each 𝑛 ∈ ℕ and 𝑥𝑛 → 𝑧 ∈ 𝑋 as 𝑛 → ∞, then 

(𝑥𝑛−1, 𝑧) ∈ ℜ for all 𝑛 ∈ ℕ 

3. 𝐹𝑖𝑥 (𝑓) is well ordered with respect to ℜ. 

If there exist 𝑥0 ∈ 𝑋 such that (𝑥0, 𝑓(𝑥0)) ∈ ℜ and 

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0))

𝑝(𝑥0, 𝑓(𝑥0)) + 𝑝(𝑓(𝑥0), 𝑓2(𝑥0)) + 1
+ 𝑘 < 1 

then there exist 𝑧 ∈ 𝑋 such that 𝑝(𝑧, 𝑧) = 0. Furthermore 

a. There exists 𝑧 ∈ 𝑋 fixed point of 𝑓 

b. The Picard sequences of initial point 𝑥0 ∈ 𝑋 converges to fixed point of 𝑓 



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c. If 𝑧 and 𝑤 are fixed point of 𝑓 where 𝑧 ≠ 𝑤 then  

𝑝(𝑧, 𝑤)

𝑝(𝑧, 𝑧) + 𝑝(𝑤, 𝑤) + 1
≥

1 − 𝑘

2
 

Proof: 

The proof of this theorem is following Theorem 3.4 and Theorem 3.2’s proof. 

4. Conclusions 

In this article, we have extended the fixed point theorem to non-expansive mappings resulting from 

Vetro in the setting of partial metric spaces. These results also provide proof of one of Aydi's results. 

Furthermore, we generalize our previous result by using the 𝒩 function. We also give examples to 

illustrate our results. 

5. Acknowledgement 

This research was funded by the Institute for Research and Community Service, Universitas 

Muhammadiyah Ponorogo, Indonesia. 

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