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CONTACT:  

Ahmad Khairul Umam            akumam@billfath.ac.id         Department of Mathematics, Universitas 
Billfath, Lamongan, East Java 62261 

 

       The article can be accessed here. https://doi.org/10.15642/mantik.2022.8.2.105-112  

Existence and Uniqueness of Fixed Points in Cone Metric Spaces 

for 𝝎-distance 
 

 

Ahmad Khairul Umam1, Nihaya Alivia C. Dewi1, 

Pukky Tetralian B. Ngastiti1 

  1Universitas Billfath, Lamongan, Indonesia 

 
Article history: 

Received Nov 19, 2021 

Revised May 14, 2022 

Accepted Dec 31, 2022 

 

Kata Kunci:  

Titik Tetap,  

Metrik Cone, 

Jarak-𝜔 

Abstrak. Di dalam penelitian ini, fungsi jarak yang digunakan 

adalah fungsi jarak-𝜔. Beberapa teorema, pembuktian teorema, dan 
contoh tentang ruang metrik cone dibahas pada penelitian ini. Jika 

diberikan ruang metrik cone lengkap dengan fungsi jarak-𝜔, cone 
normal 𝑃 di 𝑋, fungsi kontraksi 𝑓: 𝑋 → 𝑋 dengan bentuk 

𝜔(𝑓(𝑥), 𝑓(𝑦)) ≼ 𝛼𝜔(𝑥, 𝑦) + 𝛽[𝜔(𝑥, 𝑓(𝑥) + 𝜔(𝑦, 𝑓(𝑦)] +
𝛾[𝜔(𝑥, 𝑓(𝑦) + 𝜔(𝑦, 𝑓(𝑥)] untuk setiap 𝑥, 𝑦 ∈ 𝑋, dan 𝛼, 𝛽, 𝛾 adalah 
bilangan real tak negatif dimana 𝛼 + 2𝛽 + 2𝛾 < 1, maka fungsi 𝑓 
memiliki titik tetap tunggal di 𝑋. 
 

Keywords:  

Fixed Point, 

Cone Metric, 

𝜔-distance 

Abstract. In this study, the distance function used is the distance 

function-𝜔. Several theorems, proof of theorem, and example of 
cone metric space are discussed in this study. If given a complete 

cone metric space with distance function, the cone is normal at 

contraction function 𝑓: 𝑋 → 𝑋 with 𝜔(𝑓(𝑥), 𝑓(𝑦)) ≼ 𝛼𝜔(𝑥, 𝑦) +
𝛽[𝜔(𝑥, 𝑓(𝑥) + 𝜔(𝑦, 𝑓(𝑦)] + 𝛾[𝜔(𝑥, 𝑓(𝑦) + 𝜔(𝑦, 𝑓(𝑥)] for every 
𝑥, 𝑦 ∈ 𝑋, and 𝛼, 𝛽, 𝛾 is a non-negative real number where 𝛼 + 2𝛽 +
2𝛾 < 1, then function 𝑓 has unique fixed point at 𝑋. 
 

How to cite: 

A. K. Umam, N. H. A. Dewi, and P. T. B. Ngastiti, “Existence and Uniqueness of Fixed Points 

in Cone Metric Spaces for 𝝎-distance”, J. Mat. Mantik, vol. 8, no. 2, pp. 105-112, December 

2022. 

 
  

Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 105-112 

 

ISSN: 2527-3159 (print) 2527-3167 (online) 

mailto:akumam@billfath.ac.id
http://u.lipi.go.id/1458103791


Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 105-112 

 

 

106 

1. Introduction 
 

Fixed point has many useful for solving linear equation, ordinary differential 

equation, partial differential equation, intergral equation. The famous fixed point theorem 

is Banach fixed point theorem. According to [1], the Banach fixed point guarantee the 

existence and uniqueness of fixed point for function in complete space and contractive 

function. 

In this paper we discuss some fixed point theorems in cone metric space with 𝜔-
distance. According to [2], cone metric space is generalization of metric space. Range of 

cone metric space is Banach space. 

We use real Banach space and 𝜔-distance for mertic. According to [3], a 𝜔-distance 
is a function in metric spaces with three condition: symmetry, lower semicontinuous 

function, and relationship 𝜔-distance with metric itself. 
 

2. Preliminaries 
 

Definition 1. [4] Let 𝑉 is non-empty set with  addition and scalar (real number) 
multiplication operation. Addition operation: �̅�, �̅� ∈ 𝑉, �̅� + �̅� ∈ 𝑉; and scalar 
multiplication operation: 𝑘 ∈ ℝ, �̅� ∈ 𝑉, 𝑘�̅� ∈ 𝑉. 𝑉 is called vector space if satisfy 
V1.   ∀�̅�, �̅� ∈ 𝑉, �̅� + �̅� ∈ 𝑉; 
V2.   ∀�̅�, �̅�, �̅� ∈ 𝑉, (�̅� + �̅�) + �̅� = �̅� + (�̅� + �̅�); 
V3.   ∃0̅ ∈ 𝑉, ∀�̅� ∈ 𝑉, 0̅ + �̅� = �̅� + 0̅ = �̅�; 
V4.   ∀�̅� ∈ 𝑉, ∃ − �̅� ∈ 𝑉, �̅� + (−�̅�) = −�̅� + �̅� = 0̅; 
V5.   ∀�̅�, �̅� ∈ 𝑉, �̅� + �̅� = �̅� + �̅�; 
V6.   ∀𝑘 ∈ ℝ, ∀�̅� ∈ 𝑉, 𝑘�̅� ∈ 𝑉; 
V7.   ∀𝑘 ∈ ℝ, ∀�̅�, �̅� ∈ 𝑉, 𝑘(�̅� + �̅�) = 𝑘�̅� + 𝑘�̅�; 
V8.   ∀𝑘, ℎ ∈ ℝ, ∀�̅� ∈ 𝑉, (𝑘 + ℎ)�̅� = 𝑘�̅� + ℎ�̅�; 
V9.   ∀𝑘, ℎ ∈ ℝ, ∀�̅� ∈ 𝑉, 𝑘(ℎ�̅�) = 𝑘ℎ(�̅�); 
V10. ∃1 ∈ ℝ, ∀�̅� ∈ 𝑉, 1. �̅� = �̅�. 
 

Definition 2. [5] Let 𝑉 is vector space over the field 𝔽. Function ‖∙‖: 𝑉 → ℝ is called 
norm of 𝑉 if satisfy 
N1. ‖𝑥‖ ≥ 0 for all 𝑥 ∈ 𝑉; 
N2. If 𝑥 ∈ 𝑉 dan ‖𝑥‖ = 0 then 𝑥 = 0; 
N3. ‖𝛼𝑥‖ = |𝛼|‖𝑥‖ for all 𝑥 ∈ 𝑉 and 𝛼 ∈ 𝔽; 
N4. ‖𝑥 + 𝑦‖ ≤ ‖𝑥‖ + ‖𝑦‖ for all 𝑥, 𝑦 ∈ 𝑉. 
A normed space is a vector space 𝑉 together with a norm ‖∙‖. 
 

Definition 3. [6] All complete normed vector space is called Banach space. 

 

Definition 4. [7] A metric on a set 𝑋 is a function 𝑑: 𝑋 × 𝑋 → ℝ that satisfies the 
following properties: 

M1. 𝑑(𝑥, 𝑦) ≥ 0 for all 𝑥, 𝑦 ∈ 𝑋. (positivity) 
M2. 𝑑(𝑥, 𝑦) = 0 if and only if 𝑥 = 𝑦. (definiteness) 
M3. 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) for all 𝑥, 𝑦 ∈ 𝑋. (symmetry) 
M4. 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑧) + 𝑑(𝑧, 𝑦) for all 𝑥, 𝑦, 𝑧 ∈ 𝑋. (triangle inequality) 
A metric space (𝑋, 𝑑) is a set 𝑋 together with a metric 𝑑 on 𝑋. 
 

Definition 5. [8] Let 𝔼 is real Banach space and 𝑃 a subset of 𝔼. A set 𝑃 is called a cone 
if only if: 

i 𝑃 is closed, nonempty, and 𝑃 ≠ 0; 
ii 𝑎𝑥 + 𝑏𝑦 ∈ 𝑃 for all 𝑥, 𝑦 ∈ 𝑃 dan 𝑎, 𝑏 ∈ ℝ+ ∪ {0}; 



Ahmad Khairul Umam, Nihaya Alivia C. Dewi, and Pukky Tetralian B. N. 

Existence and Uniqueness of Fixed Points in Cone Metric Spaces for 𝝎-distance 

107 

iii 𝑃 ∩ (−𝑃) = 0.                                                                               

Further, if 𝑃 ⊆ 𝔼 is cone, then we define partial ordering “≼” with respect to 𝑃 by 𝑥 ≼ 𝑦 
if only if 𝑦 − 𝑥 ∈ 𝑃. And then 𝑥 < 𝑦 is interpreted 𝑥 ≼ 𝑦 and 𝑥 ≠ 𝑦. Whereas 𝑥 < 𝑦 is 
interpreted 𝑦 − 𝑥 ∈ 𝑖𝑛𝑡 𝑃 (interior of 𝑃). 
 

Definition 6. [9] The cone 𝑃 is called normal if there is a number 𝑀 > 0 such that for all 
𝑥, 𝑦 ∈ 𝔼, 0 ≤ 𝑥 ≤ 𝑦 implies 

‖𝑥‖ ≤ 𝑀‖𝑦‖.                                                             (1) 
The least positive 𝑀 satisfying (1) is called the normal constant of 𝑃. 
 

Definition 7. [2] A cone metric on non-empty set 𝑋 is a function 𝑑: 𝑋 × 𝑋 → 𝔼 that 
satisfies the following properties: 

C1. 0 < 𝑑(𝑥, 𝑦) for all 𝑥, 𝑦 ∈ 𝑋 and 𝑑(𝑥, 𝑦) = 0 if only if 𝑥 = 𝑦; 
C2. 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) for all 𝑥, 𝑦 ∈ 𝑋; 
C3. 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑧) + 𝑑(𝑧, 𝑦) for all 𝑥, 𝑦 dan 𝑧 ∈ 𝑋. 
A cone metric spaces (𝑋, 𝑑) is a set 𝑋 together with a metric 𝑑 on 𝑋. 
 

Definition 8. [10] Let (𝑋, 𝑑) be a cone metric space, 〈𝑥𝑛 〉 be a sequence in 𝑋. If for any 
𝑐 ∈ 𝔼 with 0 < 𝑐, there is 𝑁 such that for all 𝑛 > 𝑁, 𝑑(𝑥, 𝑥𝑛 ) < 𝑐, then 〈𝑥𝑛 〉 is called a 
convergent sequence to a point  𝑥 ∈ 𝑋. 
 

Definition 9. [2] Let (𝑋, 𝑑) be a cone metric space, 〈𝑥𝑛 〉 be a sequence in 𝑋. If for any 
𝑐 ∈ 𝔼 with 0 < 𝑐, there is 𝑁 such that for all 𝑚, 𝑛 > 𝑁, 𝑑(𝑥𝑛 , 𝑥𝑚 ) < 𝑐, then 〈𝑥𝑛 〉 is called 
a Cauchy sequence in 𝑋. 
 

Definition 10. [11] Let (𝑋, 𝑑) be a cone metric space, 〈𝑥𝑛 〉 be a sequence in 𝑋. (𝑋, 𝑑) is a 
complete cone metric space if every Cauchy sequence is convergent. 

 

Definition 11. [12] A function 𝜓 from a metric space (𝑋, 𝑑) to ℝ is called lower 
semicontinuous if for every 𝑦 ∈ 𝑋, 

𝑙𝑖𝑚
𝑥→𝑦

𝑖𝑛𝑓 𝜓(𝑥) ≥ 𝜓(𝑦). 

Example 1. Let a function 

𝑓(𝑥) = {
2, 𝑥 > 2
1, 𝑥 ≤ 2

. 

The function 𝑓(𝑥) is lower semicontinuous at 𝑥 = 2. 
       𝑦   

        2      

        1   

       

              -1   0       1    2     3    4     𝑥 

Figure 1. Example of lower semicontinuous function 

 

Definition 12. [13] Let (𝑋, 𝑑) be a metric space and a function 𝑓: 𝑋 → 𝑋. A point 𝑥 ∈ 𝑋 
is called a fixed point of function 𝑓 if 𝑥 = 𝑓(𝑥). 
 

Definition 13. [14] Let metric space (𝑋, 𝑑). Function 𝑓: 𝑋 → 𝑋 is said contraction 
function if there is a real number 𝑐 where 0 ≤ 𝑐 < 1 such that: 

𝑑(𝑓(𝑥), 𝑓(𝑦)) ≤ 𝑐𝑑(𝑥, 𝑦), ∀𝑥, 𝑦 ∈ 𝑋. 

Theorem 1. [15] (Banach’s Fixed Point). Let (𝑋, 𝑑) is complete metric space. If function 
𝑓: 𝑋 →  𝑋 is contraction function in 𝑋, then function 𝑓 has unique fixed point. 



Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 105-112 

 

 

108 

Defintion 14. [16] A function 𝜔: 𝑋 × 𝑋 → [0, ∞) is a 𝜔-distance on 𝑋 if it satisfies the 
following conditions for any 𝑥, 𝑦, 𝑧 ∈ 𝑋: 
ω1. 𝜔(𝑥, 𝑧) ≤ 𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑧) 
ω2. The function 𝜔(𝑥,⋅): 𝑋 → [0, ∞) is lower semicontinuous 
ω3. For any > 0, there exists 𝛿 > 0 such that 𝜔(𝑧, 𝑥) ≤ 𝛿 and 𝜔(𝑧, 𝑦) ≤ 𝛿 imply  
 𝑑(𝑥, 𝑦) ≤ . 

Of course, the metric 𝑑 is a 𝜔-distance on 𝑋. 
 

3. Result and Discussion 
 

Theorem 2. Let complete cone metric space (𝑋, 𝑑) with 𝜔-distance. Let normal cone 𝑃 
in 𝑋 and function 𝑓: 𝑋 → 𝑋 that satisfy 

𝜔(𝑓(𝑥), 𝑓(𝑦)) ≼ 𝛼𝜔(𝑥, 𝑦) + 𝛽[𝜔(𝑥, 𝑓(𝑥)) + 𝜔(𝑦, 𝑓(𝑦))] 

  +𝛾[𝜔(𝑥, 𝑓(𝑦)) + 𝜔(𝑦, 𝑓(𝑥))]                                         (2) 

for all 𝑥, 𝑦 ∈ 𝑋 where 𝛼, 𝛽, 𝛾 are non-negative real numbers such that 𝛼 + 2𝛽 + 2𝛾 < 1, 
then 𝑓 has unique fixed point in 𝑋. 
Proof: 

Let a sequence 〈𝑥𝑛 〉 in cone metric space (𝑋, 𝑑). A sequence 〈𝑥𝑛 〉 satisfies this 
property: 

𝑥𝑛 = 𝑓(𝑥𝑛−1) = 𝑓(𝑓(𝑥𝑛−2)) = 𝑓
2(𝑥𝑛−2) = ⋯ = 𝑓

𝑛(𝑥0) 

where 𝑛 ∈ ℕ. 
According (2), we get 

𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼ 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽[𝜔(𝑥0, 𝑓(𝑥0)) + 𝜔(𝑓(𝑥0), 𝑓

2(𝑥0))] 

+𝛾[𝜔(𝑥0, 𝑓
2(𝑥0)) + 𝜔(𝑓(𝑥0), 𝑓(𝑥0))] 

  = 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

+𝛾𝜔(𝑥0, 𝑓
2(𝑥0)) + 𝛾𝜔(𝑓(𝑥0), 𝑓(𝑥0)) 

  = 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

+𝛾𝜔(𝑥0, 𝑓
2(𝑥0)) + 0 

= 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

+𝛾𝜔(𝑥0, 𝑓
2(𝑥0)).                                                                          (3) 

We use triangle inequality and we get 

𝜔(𝑥0, 𝑓
2(𝑥0)) ≼ 𝜔(𝑥0, 𝑓(𝑥0)) + 𝜔(𝑓(𝑥0), 𝑓

2(𝑥0)) 

so that (3) become 

𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼ 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓

2(𝑥0)) 

+𝛾𝜔(𝑥0, 𝑓
2(𝑥0)) 

  ≼ 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

+𝛾[𝜔(𝑥0, 𝑓(𝑥0)) + 𝜔(𝑓(𝑥0), 𝑓
2(𝑥0))] 

  = 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

+𝛾𝜔(𝑥0, 𝑓(𝑥0)) + 𝛾𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

  = 𝛼𝜔(𝑥0, 𝑓(𝑥0)) + 𝛽𝜔(𝑥0, 𝑓(𝑥0)) + 𝛾𝜔(𝑥0, 𝑓(𝑥0)) 

+𝛽𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) + 𝛾𝜔(𝑓(𝑥0), 𝑓

2(𝑥0)) 

  = (𝛼 + 𝛽 + 𝛾)𝜔(𝑥0, 𝑓(𝑥0)) + (𝛽 + 𝛾)𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

or we can write 

    𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼ (𝛼 + 𝛽 + 𝛾)𝜔(𝑥0, 𝑓(𝑥0)) 

  +(𝛽 + 𝛾)𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) 

𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) − (𝛽 + 𝛾)𝜔(𝑓(𝑥0), 𝑓

2(𝑥0)) ≼ (𝛼 + 𝛽 + 𝛾)𝜔(𝑥0, 𝑓(𝑥0)) 

  [1 − (𝛽 + 𝛾)]𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼ (𝛼 + 𝛽 + 𝛾)𝜔(𝑥0, 𝑓(𝑥0)) 



Ahmad Khairul Umam, Nihaya Alivia C. Dewi, and Pukky Tetralian B. N. 

Existence and Uniqueness of Fixed Points in Cone Metric Spaces for 𝝎-distance 

109 

  𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼

𝛼 + 𝛽 + 𝛾

1 − (𝛽 + 𝛾)
𝜔(𝑥0, 𝑓(𝑥0)). 

Suppose 𝐾 =
𝛼+𝛽+𝛾

1−(𝛽+𝛾)
 according to definition 8 about contraction function, then 

𝜔(𝑓(𝑥0), 𝑓
2(𝑥0)) ≼ 𝐾𝜔(𝑥0, 𝑓(𝑥0))                                          (4) 

where 0 ≤ 𝐾 =
𝛼+𝛽+𝛾

1−(𝛽+𝛾)
< 1. 

According (4), we get 
𝜔(𝑓 𝑛(𝑥0), 𝑓

𝑛+1(𝑥0)) ≼ 𝐾𝜔(𝑓
𝑛−1(𝑥0), 𝑓

𝑛(𝑥0)) ≼ ⋯ ≼ 𝐾
𝑛 𝜔(𝑥0, 𝑓(𝑥0)) 

where 0 ≤ 𝐾 < 1. Let 𝑚 > 𝑛 ≥ 𝑁 ∈ ℕ, according to triangle inequality then 
𝜔(𝑓 𝑛(𝑥0), 𝑓

𝑚 (𝑥0)) ≼ 𝜔(𝑓
𝑛(𝑥0), 𝑓

𝑛+1(𝑥0)) + 𝜔(𝑓
𝑛+1(𝑥0), 𝑓

𝑛+2(𝑥0)) + ⋯ 

 +𝜔(𝑓 𝑚−2(𝑥0), 𝑓
𝑚−1(𝑥0)) + 𝜔(𝑓

𝑚−1(𝑥0), 𝑓
𝑚(𝑥0)) 

  ≼ 𝐾𝑛 𝜔(𝑥0, 𝑓(𝑥0)) + 𝐾
𝑛+1𝜔(𝑥0, 𝑓(𝑥0)) + ⋯ 

 +𝐾𝑚−2𝜔(𝑥0, 𝑓(𝑥0)) + 𝐾
𝑚−1𝜔(𝑥0, 𝑓(𝑥0)) 

  = (𝐾𝑛 + 𝐾𝑛+1 + ⋯ + 𝐾𝑚−2 + 𝐾𝑚−1)𝜔(𝑥0, 𝑓(𝑥0)) 

  = 𝐾𝑛 (1 + 𝐾 + 𝐾2 + ⋯ 𝐾𝑚−𝑛−1)𝜔(𝑥0, 𝑓(𝑥0)) 

  = 𝐾𝑛 ( ∑ 𝐾𝑖
𝑚−𝑛−1

𝑖=0

) 𝜔(𝑥0, 𝑓(𝑥0)) 

  ≼ 𝐾𝑛 (∑ 𝐾𝑖
∞

𝑖=0

) 𝜔(𝑥0, 𝑓(𝑥0)) 

  = 𝐾𝑛 (
1

1 − 𝐾
) 𝜔(𝑥0, 𝑓(𝑥0)) 

  = (
𝐾𝑛

1 − 𝐾
) 𝜔(𝑥0, 𝑓(𝑥0)). 

Suppose 𝜔(𝑥0, 𝑓(𝑥0)) = 𝑐, then 

𝜔(𝑓 𝑛 (𝑥0), 𝑓
𝑚(𝑥0)) ≼ (

𝐾𝑛

1 − 𝐾
) 𝜔(𝑥0, 𝑓(𝑥0)) =

𝑐𝐾𝑛

1 − 𝐾
 . 

Choose 𝑁 ∈ ℕ with 

𝑁 <  𝐾 log
(1 − 𝐾)

𝑐
 

such that for all 𝑚, 𝑛 ≥ 𝑁, we get  

𝜔(𝑓 𝑛(𝑥0), 𝑓
𝑚(𝑥0)) ≼

𝑐𝐾𝑛

1 − 𝐾
≼

𝑐𝐾𝑁

1 − 𝐾
<

𝑐

1 − 𝐾
 .

(1 − 𝐾)

𝑐
= . 

So, sequence 〈𝑥𝑛 〉 is Cauchy sequence. 
Because space (𝑋, 𝑑) is complete cone metric space, then sequence 〈𝑥𝑛 〉 is 

convergent to point 𝑥 ∈ 𝑋 or we can write 〈𝑥𝑛 〉 → 𝑥. And then, we will proof that point 𝑥 
is fixed point of function 𝑓. According to triangle inequality and (2), then 

𝜔(𝑓(𝑥), 𝑥) ≼ 𝜔(𝑓(𝑥), 𝑓(𝑥𝑛 )) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

   ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽[𝜔(𝑥, 𝑓(𝑥)) + 𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 ))] 

 +𝛾[𝜔(𝑥, 𝑓(𝑥𝑛 )) + 𝜔(𝑥𝑛 , 𝑓(𝑥))] + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

= 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥, 𝑓(𝑥)) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) 

 +𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 
or we can write  

    𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥, 𝑓(𝑥)) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) 

   +𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

    𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑓(𝑥), 𝑥) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) 

   +𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 



Jurnal Matematika MANTIK 

Vol. 8, No. 2, December 2022, pp. 105-112 

 

 

110 

𝜔(𝑓(𝑥), 𝑥) − 𝛽𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) 

   +𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

    (1 − 𝛽)𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) 

+𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

 (1 − 𝛽)𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) 

+𝛾𝜔(𝑥, 𝑓(𝑥𝑛 )) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 

 (1 − 𝛽)𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) 

+𝛾𝜔(𝑓(𝑥𝑛 ), 𝑥) + 𝜔(𝑓(𝑥𝑛 ), 𝑥) 
 (1 − 𝛽)𝜔(𝑓(𝑥), 𝑥) ≼ 𝛼𝜔(𝑥, 𝑥𝑛 ) + 𝛽𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) + 𝛾𝜔(𝑥𝑛 , 𝑓(𝑥)) 

+(𝛾 + 1)𝜔(𝑓(𝑥𝑛 ), 𝑥) 

    𝜔(𝑓(𝑥), 𝑥) ≼
𝛼

(1 − 𝛽)
𝜔(𝑥, 𝑥𝑛 ) +

𝛽

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑓(𝑥𝑛 )) 

+
𝛾

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑓(𝑥)) +

(𝛾 + 1)

(1 − 𝛽)
𝜔(𝑓(𝑥𝑛 ), 𝑥) 

 𝜔(𝑓(𝑥), 𝑥) ≼
𝛼

(1 − 𝛽)
𝜔(𝑥, 𝑥𝑛 ) +

𝛽

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑥𝑛+1) 

+
𝛾

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑓(𝑥)) +

(𝛾 + 1)

(1 − 𝛽)
𝜔(𝑥𝑛+1, 𝑥) 

because sequence 〈𝑥𝑛 〉 is convergent for 𝑛 → ∞, then: 

 𝜔(𝑓(𝑥), 𝑥) ≼
𝛼

(1 − 𝛽)
𝜔(𝑥, 𝑥𝑛 ) +

𝛽

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑥𝑛+1) 

+
𝛾

(1 − 𝛽)
𝜔(𝑥𝑛 , 𝑓(𝑥)) +

(𝛾 + 1)

(1 − 𝛽)
𝜔(𝑥𝑛+1, 𝑥) 

  =
𝛼

(1 − 𝛽)
(0) +

𝛽

(1 − 𝛽)
(0) +

𝛾

(1 − 𝛽)
(0) 

+
(𝛾 + 1)

(1 − 𝛽)
(0) 

  = 0.                                                                                      (5) 
Because 0 ≼ 𝜔(𝑓(𝑥), 𝑥), according (5) then 

𝜔(𝑓(𝑥), 𝑥) = 0, 
or we can write 𝑥 = 𝑓(𝑥). So, point 𝑥 is fixed point of function 𝑓. 

Furthermore, we will proof uniqueness of a fixed point. Suppose 𝑥 and 𝑦 are fixed 
points of function 𝑓, such that 

𝑥 = 𝑓(𝑥) and 𝑦 = 𝑓(𝑦) 
then 

𝜔(𝑥, 𝑦) = 𝜔(𝑓(𝑥), 𝑓(𝑦)) 

≼ 𝛼𝜔(𝑥, 𝑦) + 𝛽[𝜔(𝑥, 𝑓(𝑥)) + 𝜔(𝑦, 𝑓(𝑦))] + 𝛾[𝜔(𝑥, 𝑓(𝑦)) + 𝜔(𝑦, 𝑓(𝑥))] 

= 𝛼𝜔(𝑥, 𝑦) + 𝛽[𝜔(𝑥, 𝑥) + 𝜔(𝑦, 𝑦)] + 𝛾[𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑥)] 
= 𝛼𝜔(𝑥, 𝑦) + 𝛽[0 + 0] + 𝛾[𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑥)] 
= 𝛼𝜔(𝑥, 𝑦) + 𝛽[0] + 𝛾[𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑥)] 
= 𝛼𝜔(𝑥, 𝑦) + 0 + 𝛾[𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑥)] 
= 𝛼𝜔(𝑥, 𝑦) + 𝛾[𝜔(𝑥, 𝑦) + 𝜔(𝑦, 𝑥)] 
= 𝛼𝜔(𝑥, 𝑦) + 𝛾𝜔(𝑥, 𝑦) + 𝛾𝜔(𝑦, 𝑥) 
= 𝛼𝜔(𝑥, 𝑦) + 𝛾𝜔(𝑥, 𝑦) + 𝛾𝜔(𝑥, 𝑦) 
= 𝛼𝜔(𝑥, 𝑦) + 2𝛾𝜔(𝑥, 𝑦) 
= (𝛼 + 2𝛾)𝜔(𝑥, 𝑦) 

because 𝛼 + 2𝛽 + 2𝛾 < 1 and 𝛼 + 2𝛾 < 1, so that 
𝜔(𝑥, 𝑦) ≼ (𝛼 + 2𝛾)𝜔(𝑥, 𝑦) 

is contradiction. So, point 𝑥 is unique fixed point of function 𝑓 in 𝑋. 



Ahmad Khairul Umam, Nihaya Alivia C. Dewi, and Pukky Tetralian B. N. 

Existence and Uniqueness of Fixed Points in Cone Metric Spaces for 𝝎-distance 

111 

Example 2. Let set 𝔼 = ℝ2, 𝑃 = {(𝑥, 𝑦) ∈ 𝔼: 𝑥, 𝑦 ≥ 0} ⊂ ℝ2, 𝑋 = ℝ, and 𝑑: 𝑋 × 𝑋 → 𝔼 

such that 𝑑(𝑥, 𝑦) = (|𝑥 − 𝑦|, 𝛼|𝑥 − 𝑦|) where 𝛼 ≥ 0 is a constant, then (𝑋, 𝑑) is cone 

metric space.  

Example 3. Let metric space (𝑋, 𝑑) and function 𝜔: 𝑋 × 𝑋 → [0, ∞). Function 𝜔(𝑥, 𝑦) =

𝑐 for every 𝑥, 𝑦 ∈ 𝑋 is 𝜔-distance on 𝑋 (𝑐 is positive real number). Function 𝜔 is not 

metric space because 𝜔(𝑥, 𝑥) = 𝑐 ≠ 0 for any 𝑥 ∈ 𝑋. 

 

4. Conclusion 
 

Cone metric space is generalization of metric space. Range of cone metric in cone 

metric space is Banach space. In this research, we use 𝜔-distance for metrik. a 𝜔-distance 
is a function in metric spaces with three condition: symmetry, lower semicontinuous 

function, and relationship 𝜔-distance with metric itself. 
Fixed point has many useful for solving linear equation, ordinary differential 

equation, partial differential equation, integral equation. The famous fixed-point theorem 

is Banach fixed point theorem. The Banach fixed point guarantee the existence and 

uniqueness of fixed point for function in complete space and contractive function. Let 

complete cone metric space with 𝜔-distance. Let normal cone 𝑃 in 𝑋 and function 𝑓: 𝑋 →

𝑋 that satisfy 𝜔(𝑓(𝑥), 𝑓(𝑦)) ≼ 𝛼𝜔(𝑥, 𝑦) + 𝛽[𝜔(𝑥, 𝑓(𝑥)) + 𝜔(𝑦, 𝑓(𝑦))] +

𝛾[𝜔(𝑥, 𝑓(𝑦)) + 𝜔(𝑦, 𝑓(𝑥))] for all 𝑥, 𝑦 ∈ 𝑋 where 𝛼, 𝛽, 𝛾 are non negative real numbers 

such that 𝛼 + 2𝛽 + 2𝛾 < 1, then 𝑓 has unique fixed point in 𝑋. 
 

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