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Some Fixed Point Results for (𝜶, 𝝍, 𝝋)- Geraghty Contraction 

Mappings in Bipolar Metric Space 

 

Diksha1, Manoj Kumar2,*  
1, 2Department of Mathematics, Baba Mastnath University, Asthal Bohar, Rohtak-124021, Haryana, India 

1Department of Mathematics, Government College, Matanhail, Jhajjar-124106, Haryana, India 

2Department of Mathematics, Maharishi Markandeshwar (Deemed to be University), Mullana, Ambala-133207, 

Haryana, India 

Email addresses- dikshaahlawat14@gmail.com, manojantil18@gmail.com 

(*Corresponding Author) 

 

Article History: 

Received: 22-10-2024 

Revised: 06-12-2024 

Accepted: 13-12-2024 

Abstract:  

In this paper, we explore a generalization of (𝛼, 𝜓)- Geraghty contractions and investigate 

the existence and uniqueness of fixed points for mappings satisfying this condition. The 

study extends, improves, and generalizes some earlier results in the literature on this topic. 

We consider the interplay of three parameters: 𝛼, 𝜓 and 𝜑, which play a crucial role in 

defining the contraction properties. Our goal is to establish fixed point theorems that 

encompass various scenarios within bipolar metric spaces. 

Keywords: Fixed point, (𝛼, 𝜓, 𝜑)- Geraghty contraction mappings, covariant and 

contravariant mappings, bipolar metric space. 

2020 MSC: 47H10, 54H25 

1. Introduction 

The concept of fixed points is fundamental in mathematics, and it arises in diverse fields such as 

differential equations and optimization. In 1922, Banach [2] introduced Banach contraction principle 

as the first constructive method to get a fixed point for a self map on a complete metric space. 

Continuation of this, in 1973, Geraghty [8] gives an extension of the Banach contraction mapping 

principle, provides a powerful tool for proving the existence of fixed points. Specifically, Geraghty’s 

result ensures a unique fixed point under certain contractive conditions. Many authors generalized his 

work, see [1,4,6,7,13].  

In 2012, Samet et al. [14] introduced the concepts of 𝛼-contractive and 𝛼-admissible mappings and 

proved various fixed point theorems of 𝛼- admissible contractive mappings in complete metric spaces. 

Recently, in 2015, Chandok [4] introduced the concept of (𝛼, 𝛽)- admissible Geraghty type contractive 

mappings and proved some fixed point theorems of such kind of mappings in complete metric spaces. 

Some researcher extended their work in various spaces        [9-16].  

In 2019, Karapinar et al. [7] introduced the notion 𝜑-Geraghty and Ciric type 𝜑-Geraghty contractive 

mappings in complete metric space and proved some fixed points theorems and uniqueness of fixed 

points. 

To get a new approach for fixed point results in 2016, Mutlu and Gürdal [10] introduced the concept 

of bipolar metric space. The major difference between the previously defined spaces and bipolar is of 

mailto:dikshaahlawat14@gmail.com
mailto:manojantil18@gmail.com


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distance function. In bipolar metric space, the distance function is from the cartesian product of two 

different sets to non-negative real numbers. Since then, many authors have proved several fixed point 

results in bipolar metric space see [5], [9-11], [13].  

Motivated by the work of Abduletif et al. [1], the main objective of this manuscript is to prove some 

fixed points results and their uniqueness for (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping in complete 

bipolar metric spaces. Furthermore, we offer illustrations to support our essential findings. 

2. Preliminaries 

We need to introduce some new notations and terminology and provide some fundamental definitions 

which is used for the fixed point theorems for  (𝛼, 𝜓, 𝜑)- Geraghty contraction mappings in bipolar 

metric spaces. 

Definition 2.1. In 2016, Mutlu and Gürdal [10] introduced the concept of bipolar metric space. 

Let 𝑋 and 𝑌 are two non-empty sets and 𝑑 ∶ 𝑋 × 𝑌 → [0, ∞) be a function satisfying the following 

conditions: 

(BP1) 𝑑(𝑥, 𝑦) = 0  if and only if 𝑥 =  𝑦, where (𝑥, 𝑦) ∈ 𝑋 × 𝑌, 

(BP2) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) for all 𝑥, 𝑦 ∈ 𝑋 ∩ 𝑌, 

(BP3) 𝑑(𝑥1, 𝑦2) ≤ 𝑑(𝑥1, 𝑦1) + 𝑑(𝑥2, 𝑦1) + 𝑑(𝑥2, 𝑦2) for all 𝑥1, 𝑥2 ∈ 𝑋 and 𝑦1, 𝑦2 ∈ 𝑌. 

Then 𝑑 is called bipolar metric and (𝑋, 𝑌, 𝑑) is called bipolar metric space. 

If  𝑋 ∩ 𝑌 = ∅,  then space is called disjoint otherwise joint. The set 𝑋 is called left pole and 𝑌 is called 

right pole of bipolar metric space (𝑋, 𝑌, 𝑑) and any element of left pole (𝑋), right pole (𝑌)  and  𝑋 ∩ 𝑌 

is called left element, right element and central element respectively. 

Definition 2.2. Let (𝑋, 𝑌, 𝑑)   be a bipolar metric space. Then any sequence (𝑥𝑛) ⊆ 𝑋  is called left 

sequence and is said to be convergent to right element say ‘𝑦’ if 𝑑(𝑥𝑛, 𝑦) → 0  as  𝑛 → ∞. Similarly, a 

right sequence  (𝑦𝑛) ⊆ 𝑌  is  said to be convergent to a left element say ‘𝑥’ if  𝑑(𝑥, 𝑦𝑛) → 0  as 𝑛 → ∞. 

Definition 2.3. Let (𝑋1, 𝑌1, 𝑑1) and (𝑋2, 𝑌2, 𝑑2)  be two bipolar metric spaces.  

Let 𝑇 ∶  𝑋1 ∪ 𝑌1  →  𝑋2 ∪ 𝑌2  be a function such that 

(i)If 𝑇(𝑋1) ⊆ 𝑋2 and 𝑇(𝑌1)  ⊆ 𝑌2 , then 𝑇 is called covariant map and is denoted by                         𝑇 ∶

(𝑋1, 𝑌1, 𝑑1)   ⇉ (𝑋2, 𝑌2, 𝑑2).  

(ii)If  𝑇(𝑋1)  ⊆ 𝑌2 and 𝑇(𝑌1) ⊆ 𝑋2, then 𝑇 is called contravariant map and is denoted by        𝑇 ∶

(𝑋1, 𝑌1, 𝑑1)   ⤨ (𝑋2, 𝑌2, 𝑑2).  

Definition 2.4. Let (𝑋1, 𝑌1, 𝑑1) and (𝑋2, 𝑌2, 𝑑2)  be two bipolar metric spaces. 

(i)A map 𝑇 ∶  (𝑋1, 𝑌1, 𝑑1)   ⇉ (𝑋2, 𝑌2, 𝑑2) is called left continuous at a point 𝑥0 ∈ 𝑋1 if for every 𝜖 > 0 

there exists 𝛿 > 0 such that 𝑑2(𝑇𝑥0, 𝑇𝑦) < 휀 whenever 𝑑1(𝑥0, 𝑦) < 𝛿. 

(ii)A map 𝑇 ∶  (𝑋1, 𝑌1, 𝑑1)   ⇉ (𝑋2, 𝑌2, 𝑑2) is called right continuous at a point 𝑦0 ∈ 𝑌1 if for every 𝜖 > 0 

there exists 𝛿 > 0 such that 𝑑2(𝑇𝑥, 𝑇𝑦0) < 휀 whenever 𝑑1(𝑥, 𝑦0) < 𝛿. 



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(iii)A map 𝑇 is called continuous, if it is left continuous at each 𝑥0 ∈ 𝑋1  and right continuous at each 𝑦0 ∈

𝑌1.   

(iv)A contravariant map 𝑇 ∶  (𝑋1, 𝑌1, 𝑑1) ⤨ (𝑋2, 𝑌2, 𝑑2) is continuous if and only if it is continuous as a 

covariant map 𝑇 ∶  (𝑋1, 𝑌1, 𝑑1)   ⇉ (𝑋2, 𝑌2, 𝑑2).   

Definition 2.5. Let  (𝑋, 𝑌, 𝑑) be a bipolar metric space. 

(i)A sequence {(𝑥𝑛, 𝑦𝑛)} on the set 𝑋 × 𝑌  is called a bisequence on  (𝑋, 𝑌, 𝑑). 

(ii)If both the sequences (𝑥𝑛) and (𝑦𝑛) converge, then bisequence {(𝑥𝑛, 𝑦𝑛)} is said to be convergent. If 

both the sequences (𝑥𝑛) and (𝑦𝑛) converge to same point 𝑣 and 𝑣 ∈  𝑋 ∩ 𝑌, then this bisequence is 

said to be biconvergent. 

(iii)A bisequence  {(𝑥𝑛, 𝑦𝑛)}  on  (𝑋, 𝑌, 𝑑)  is said to be Cauchy bisequence, if for each 𝜖 > 0 there exists 

a positive integer 𝑁 ∈ ℕ such that 𝑑(𝑥𝑛, 𝑦𝑚) < 𝜖 for all 𝑛, 𝑚 ≥ 𝑁.  

(iv)A bipolar metric space is said to be complete if every Cauchy bisequence is convergent in this space. 

Definition 2.6. [11] Let 𝑋 and 𝑌 be two non-empty sets. Let 𝑇 ∶  (𝑋, 𝑌)  ⇉ (𝑋, 𝑌) and                              𝛼 ∶

𝑋 × 𝑌 → [0, +∞). Then 𝑇 is called 𝛼-admissible (covariant) if 

𝛼(𝑥, 𝑦) ≥ 1 ⇒ 𝛼(𝑇𝑥, 𝑇𝑦) ≥ 1 

for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌. 

Definition 2.7. [11] Let 𝑋 and 𝑌 be two non-empty sets. Let 𝑇 ∶  (𝑋, 𝑌) ⤨  (𝑋, 𝑌) and                      𝛼 ∶

𝑋 × 𝑌 → [0, +∞). Then 𝑇 is called 𝛼-admissible (contravariant) if 

𝛼(𝑥, 𝑦) ≥ 1 ⇒ 𝛼(𝑇𝑦, 𝑇𝑥) ≥ 1 

for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌. 

Let Ψ be the family of functions 𝜓 ∶ [0, ∞) → [0, ∞) which satisfying the following conditions: 

(i) 𝜓 is continuous, 

(ii) 𝜓 is strictly increasing, 

(iii) 𝜓(0) = 0. 

Consider Θ be the family of functions 𝜃 ∶ [0, ∞) → [0,1) such that for any bounded sequence {𝑡𝑛} of 

positive reals, 𝜃(𝑡𝑛) → 1 implies that 𝑡𝑛 → 0, as 𝑛 → ∞. 

Let Θ𝑡 be the family of functions 𝜃 ∶ [0, ∞) → [0,1) such that for any bounded sequence {𝑡𝑛} of 

positive reals, lim 𝑠𝑢𝑝 𝜃(𝑡𝑛) → 1 implies that 𝑡𝑛 → 0, as 𝑛 → ∞. 

3. Main Results 

In this section, we will introduce new notations for 𝜑-Geraghty contraction mappings and prove 

various fixed point theorems for such type of mappings in complete bipolar metric spaces. 

Definition 3.1. Let 𝑋  and  𝑌 be two non-empty sets. Consider  (𝑋, 𝑌, 𝑑)  be a bipolar metric space, 

𝑇 ∶  (𝑋, 𝑌)  ⇉  (𝑋, 𝑌) is called Geraghty contraction if there exist a function 𝜃 ∈ Θ which satisfies the 

following condition:  

                                                 𝑑(𝑇𝑥, 𝑇𝑦) ≤ 𝜃( 𝑑(𝑥, 𝑦)) 𝑑(𝑥, 𝑦) for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌.  



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 Definition 3.2. Let 𝑋 and 𝑌 be two non-empty sets and (𝑋, 𝑌, 𝑑) be a bipolar metric space,           𝑇 ∶

 (𝑋, 𝑌)  ⇉  (𝑋, 𝑌). Suppose that 𝜑 ∶  ℝ+ → ℝ+ is function and 𝜃 ∈ Θ and 𝑇 is called                𝜑- 

Geraghty contraction if it satisfies the following condition: 

(i) 𝜑(𝑡) < 𝑡 for any 𝑡 ∈ (0, ∞),                                                                                           

(ii) For any 휀 > 0, there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀,                 

(iii)𝑑(𝑇𝑥, 𝑇𝑦) ≤ 𝜃( 𝑑(𝑥, 𝑦)) 𝜑(𝑑(𝑥, 𝑦)) for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌.                                 

Definition 3.3. Let 𝑋 and 𝑌 be two non-empty sets and (𝑋, 𝑌, 𝑑) be a bipolar metric space,             𝑇 ∶

 (𝑋, 𝑌)  ⇉  (𝑋, 𝑌) is a self map and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). A mapping 𝑇 is said to be       (𝛼, 𝜓, 𝜑)- 

Geraghty contraction mapping if there exist 𝜑 ∶  ℝ+ → ℝ+ , 𝜓 ∈ Ψ and 𝜃 ∈ Θ satisfies the following 

condition: 

(i) 𝜑(𝑡) < 𝑡 for any 𝑡 ∈ (0, ∞),                                                                                                (3.1) 

(ii) For any 휀 > 0, there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀,                            (3.2) 

(iii)𝛼(𝑥, 𝑦)𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜃(𝜓( 𝑑(𝑥, 𝑦))) 𝜑(𝜓(𝑑(𝑥, 𝑦))), for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌.       (3.3)                              

Theorem 3.4. Let (𝑋, 𝑌, 𝑑) be a complete bipolar metric space, 𝑇 ∶  (𝑋, 𝑌) ⇉  (𝑋, 𝑌) is a covariant 

mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). Suppose that the following conditions hold: 

(i) 𝑇 is 𝛼- admissible mapping, 

(ii) 𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping, 

(iii) there exist 𝑥0 ∈ 𝑋and 𝑦0 ∈ 𝑌 such that 𝛼(𝑥0 , 𝑦0 ) ≥ 1 and 𝛼(𝑥0 , 𝑇𝑦0 ) ≥ 1. 

Then 𝑇 has fixed point. 

Proof: Let 𝑥0 ∈ 𝑋and 𝑦0 ∈ 𝑌 such that 𝛼(𝑥0 , 𝑦0 ) ≥ 1 and 𝛼(𝑥0 , 𝑇𝑦0 ) ≥ 1. Now we define a 

bisequence  {(𝑥𝑛, 𝑦𝑛)} in (𝑋, 𝑌) by 

                                              𝑇𝑥𝑛 = 𝑥𝑛+1 and 𝑇𝑦𝑛 = 𝑦𝑛+1 for all 𝑛 ∈ ℕ ∪ {0}. 

Since 𝑇 is  𝛼- admissible mapping.  

So, 𝛼(𝑥0 , 𝑦0 ) ≥ 1 ⇒ 𝛼(𝑇𝑥0 , 𝑇𝑦0 ) ≥ 1,  

      𝛼(𝑥0, 𝑦1 ) = 𝛼(𝑥0 , 𝑇𝑦0 ) ≥ 1, 

      𝛼(𝑥1 , 𝑦1 ) = 𝛼(𝑇𝑥0 , 𝑇𝑦0 ) ≥ 1, 

Using mathematical induction, we get 

𝛼(𝑥𝑛 , 𝑦𝑛+1 ) ≥ 1 and 𝛼(𝑥𝑛 , 𝑦𝑛 ) ≥ 1 for all 𝑛 ∈ ℕ ∪ {0}.                                                                   (3.4) 

Putting 𝑥 = 𝑥𝑛+1  and 𝑦 = 𝑦𝑛+2  in equation (3.3), using equations (3.1), (3.2) and by the properties of  

𝜓 and 𝜃, we have 

𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+2 )) = 𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑦𝑛+1 ))  

                                 ≤  𝛼(𝑥𝑛 , 𝑦𝑛+1 )𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑦𝑛+1 ))  

                                 ≤ 𝜃( 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛+1 )))𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛+1 ))) 



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                                 ≤ 𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛+1 ))) 

                                < 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛+1 )) .                                                                                          (3.5) 

Hence, 𝜓 is strictly increasing function so, we get 

                                𝑑(𝑥𝑛+1 , 𝑦𝑛+2 ) < 𝑑(𝑥𝑛 , 𝑦𝑛+1 ) for all 𝑛 ≥ 0.                                                      (3.6)                                                

Similarly, putting 𝑥 = 𝑥𝑛+1 and 𝑦 = 𝑦𝑛+1  in equation (3.3), using equations (3.1), (3.2) and by the 

properties of  𝜓 and 𝜃, we have the following 

𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) = 𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑦𝑛 ))  

                                 ≤  𝛼(𝑥𝑛 , 𝑦𝑛)𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑦𝑛 ))  

                                 ≤ 𝜃( 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 )))𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 ))) 

                                 ≤ 𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 ))) 

                                < 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 )) .                                                                                              (3.7) 

Hence, 𝜓 is strictly increasing function so, we obtain 

                              𝑑(𝑥𝑛+1 , 𝑦𝑛+1 ) < 𝑑(𝑥𝑛 , 𝑦𝑛 ) for all 𝑛 ≥ 0.                                                         (3.8) 

From the above, we conclude that the sequences {𝑑(𝑥𝑛 , 𝑦𝑛+1 )} and {𝑑(𝑥𝑛, 𝑦𝑛 )} are monotonically 

decreasing and for the non-negative monotonically decreasing sequences  {𝑑(𝑥𝑛 , 𝑦𝑛+1 )} and 

{𝑑(𝑥𝑛 , 𝑦𝑛 )}, there exist some 𝑟1 ≥ 0 and 𝑟2 ≥ 0, such that 

        𝑑(𝑥𝑛 , 𝑦𝑛+1 ) →  𝑟1 , 𝑑(𝑥𝑛 , 𝑦𝑛 ) →  𝑟2  as 𝑛 → ∞ .                                                                           (3.9) 

We suppose on the contrary that  𝑟1 > 0.   

Hence, we have 0 < 𝑟1 <  𝑑(𝑥𝑛 , 𝑦𝑛+1 ) for all 𝑛 ≥ 0. Set 휀 = 𝑟1 . 

From equation (3.2), there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀. 

On the other hand, by the definition of 휀, we can choose 𝑛0 ∈ ℕ such that 

                             휀 <  𝑑(𝑥𝑛0 , 𝑦𝑛0+1) < 휀 + 𝛿  

By the properties of 𝜓,   𝜃,using equations (3.1), (3.2) and (3.3), we have 

                          𝜓(휀) < 𝜓( 𝑑(𝑥𝑛0 , 𝑦𝑛0+1)) < 𝜓(휀 + 𝛿) = 𝜓(휀) + 𝜓(𝛿) 

this implies that   

                                   𝜑(𝜓( 𝑑(𝑥𝑛0 , 𝑦𝑛0+1))) ≤  𝜓(휀).                                                                     (3.10) 

We have also       

                              휀 <  𝑑(𝑥𝑛0+2, 𝑦𝑛0+3) <  𝑑(𝑥𝑛0+1, 𝑦𝑛0+2) =  𝑑(𝑇𝑥𝑛0
, 𝑇𝑦𝑛0+1), 

which implies that     

                              𝜓(휀) < 𝜓(𝑑(𝑥𝑛0+2, 𝑦𝑛0+3)) 



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< 𝜓(𝑑(𝑥𝑛0+1, 𝑦𝑛0+2)) =  𝜓(𝑑(𝑇𝑥𝑛0
, 𝑇𝑦𝑛0+1)) 

                                       ≤ 𝛼(𝑥𝑛0
, 𝑦𝑛0+1) 𝜓(𝑑(𝑇𝑥𝑛0

, 𝑇𝑦𝑛0+1)) 

                                       ≤ 𝜃( 𝜓 (𝑑(𝑥𝑛0
, 𝑦𝑛0+1))) 𝜑(𝜓 (𝑑(𝑥𝑛0

, 𝑦𝑛0+1))) 

                                       < 𝜑(𝜓 (𝑑(𝑥𝑛0
, 𝑦𝑛0+1))) 

                                       ≤  𝜓( 휀), 

which is a contradiction. Hence  

                     lim
𝑛→∞

 𝑑(𝑥𝑛 , 𝑦𝑛+1 ) = 𝑟1 = 0.                                                                                 (3.11) 

Similarly,     lim
𝑛→∞

 𝑑(𝑥𝑛 , 𝑦𝑛 ) = 𝑟2 = 0.                                                                                     (3.12) 

Now, we shall prove that {(𝑥𝑛 , 𝑦𝑛 )} is a Cauchy bisequence. 

We fix 휀1 > 0, then by (3.2) there exists 𝛿1 > 0 such that 

                      𝑡 < 휀1 + 𝛿1 ⇒ 𝜑(𝑡) ≤ 휀1.                                                                                     (3.13) 

Without loss of generality, we assume 𝛿1 < 휀1. Due to (3.11), there exist 𝑛0 ∈ ℕ such that 

                       𝑑(𝑥𝑛 , 𝑦𝑛+1 ) <  𝛿1, for all 𝑛 ≥  𝑛0,                                                                     (3.14) 

which implies that  

                       𝜓( 𝑑(𝑥𝑛 , 𝑦𝑛+1 )) <  𝜓(𝛿1). 

By mathematical induction, we show that for any fixed 𝑘 ≥  𝑛0 

                       𝑑(𝑥𝑘 , 𝑦𝑘+𝑙 ) < 휀1 + 𝛿1, for all 𝑙 ∈ ℕ.                                                                 (3.15) 

For 𝑙 = 1, this inequality trivially holds by (3.14). 

Now, assume that (3.15) is satisfied for some 𝑗 ∈ ℕ and we have to show that it holds for                𝑙 =

𝑗 + 1. 

From the triangle inequality (BP3), properties of 𝜓 ,𝜃, equations (3.1), (3.2) and (3.3)          

𝜓(𝑑(𝑥𝑘 , 𝑦𝑘+𝑗+1 )) ≤  𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 ) +   𝑑(𝑥𝑘+1 , 𝑦𝑘+1 ) +  𝑑(𝑥𝑘+1 , 𝑦𝑘+𝑗+1 )) 

       ≤ 𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) + 𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+𝑗+1 )) 

       ≤ 𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) + 𝜓( 𝑑(𝑇𝑥𝑘 , 𝑇𝑦𝑘+𝑗 )) 

       ≤ 𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) 

            +𝛼(𝑥𝑘, 𝑦𝑘+𝑗 )𝜓( 𝑑(𝑇𝑥𝑘, 𝑇𝑦𝑘+𝑗 ))    ≤ 𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 ))    

            +𝜃(𝜓(𝑑(𝑥𝑘, 𝑦𝑘+𝑗 )))𝜑(𝜓( 𝑑(𝑥𝑘, 𝑦𝑘+𝑗 ))) 

        <  𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) +  𝜑(𝜓( 𝑑(𝑥𝑘, 𝑦𝑘+𝑗 ))). 

Using equations (3.14) and (3.15), we get 



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               𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+𝑗+1 )) ≤  𝜓(  𝛿1) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) +  𝜓( 휀1). 

By letting 𝑘 → ∞, using equation (3.12), we obtain 

        𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+𝑗+1 )) ≤  𝜓( 𝛿1) +  𝜓(휀1) 

                                        = 𝜓(휀1 + 𝛿1). 

By the property of 𝜓 , we get 

                        𝑑(𝑥𝑘 , 𝑦𝑘+𝑗+1 ) < 휀1 + 𝛿1. 

So, equation (3.15) is holds for 𝑙 = 𝑗 + 1. 

Hence, by induction we prove that 𝑑(𝑥𝑘 , 𝑦𝑘+𝑗+1 ) < 휀1 + 𝛿1 for all 𝑘 ≥  𝑛0 and 𝑙 ≥ 1. 

Since 휀1 is arbitrary, we conclude that 

                        lim
𝑚,𝑛→∞

 𝑑(𝑥𝑛 , 𝑦𝑚) = 0. 

Hence, {(𝑥𝑛 , 𝑦𝑛)} is a Cauchy bisequence and (𝑋, 𝑌, 𝑑) is a complete bipolar metric space. So, 

{(𝑥𝑛 , 𝑦𝑛 )} is convergent and in fact biconvergent. So, there exists 𝑢 ∈ 𝑋 ∩ 𝑌 such that (𝑥𝑛 ) → 𝑢, 

 (𝑦𝑛 ) → 𝑢 as 𝑛 → ∞.  

We claim that 𝑇𝑢 = 𝑢. 

Let, if possible, 𝑇𝑢 ≠ 𝑢. Then there exist 𝑟′ > 0 such that 𝑑(𝑢, 𝑇𝑢) = 𝑟′ > 0. 

Since (𝑥𝑛 ) → 𝑢, (𝑦𝑛 ) → 𝑢 as 𝑛 → ∞, we can choose 𝑛0 ∈ ℕ such that 

𝑑(𝑥𝑛 , 𝑢) <
𝑟′

2
 , for all 𝑛 ≥  𝑛0 and 𝑑(𝑦𝑛 , 𝑢) <

𝑟′

2
 , for all 𝑛 ≥  𝑛0.                                               (3.16) 

From the triangle inequality (BP3), properties of 𝜓, 𝜃, equations (3.1), (3.2) and (3.3), 

                  𝜓(𝑟 ′) = 𝜓(𝑑(𝑢, 𝑇𝑢)) 

                            ≤  𝜓(𝑑(𝑢, 𝑦𝑛+1 ) + 𝑑(𝑥𝑛+1 , 𝑦𝑛+1 ) + 𝑑(𝑥𝑛+1 , 𝑇𝑢)) 

                            ≤  𝜓(𝑑(𝑢, 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑇𝑢)) 

                            ≤  𝜓(𝑑(𝑢, 𝑦𝑛+1)) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) +  𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑢)) 

                            ≤  𝜓(𝑑(𝑢, 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝛼(𝑥𝑛, 𝑢) 𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑢)) 

                            ≤  𝜓(𝑑(𝑢, 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝜃(𝜓((𝑥𝑛, 𝑢)) 𝜑( 𝜓(𝑑(𝑥𝑛 , 𝑢))) 

                            <  𝜓(𝑑(𝑢, 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) +  𝜑( 𝜓(𝑑(𝑥𝑛 , 𝑢))) 

                            <  𝜓(𝑑(𝑢, 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1)) +  𝜓(𝑑(𝑥𝑛 , 𝑢)). 

Using equations (3.12) and (3.16), we get 

𝜓(𝑟 ′) <  𝜓 (
𝑟 ′

2
) +  𝜓 (

𝑟 ′

2
) 



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                                                                      =  𝜓 (
𝑟′

2
+

𝑟′

2
) =  𝜓(𝑟 ′), 

which is a contradiction. 

Thus 𝑇𝑢 = 𝑢. i.e., 𝑢 is the fixed point of 𝑇. 

Example 3.5. Let 𝑋 = [0, + ∞) and 𝑌 = [−1,1] and let 𝑑 ∶ 𝑋 × 𝑌 → [0, +∞) be a function such that 

𝑑(𝑥, 𝑦) = |𝑥2 − 𝑦2| for all (𝑥, 𝑦) ∈ 𝑋 × 𝑌. 

Then, clearly (𝑋, 𝑌, 𝑑) be a complete bipolar metric space. 

Define 𝑇 ∶  (𝑋, 𝑌)  ⇉ (𝑋, 𝑌) such that 𝑇𝑥 =
𝑥

2
 is a mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞) such that 𝛼(𝑥, 𝑦) =

3

2
  for (𝑥, 𝑦) ∈ 𝑋 × 𝑌.  

Clearly, 𝑇 is 𝛼-admissible mapping and there exist  (𝑥0 , 𝑦0) ∈ 𝑋 × 𝑌  such  that 𝛼(𝑥0 , 𝑇𝑦0 ) ≥ 1, 𝑋 ∩

𝑌 = {0} and 𝑇0 = 0. 

Taking 𝜓(𝑡) =
𝑡

4
, 𝜑(𝑡) =

𝑡

2
  and 𝜃(𝑡) =

3

4
 . 

Left hand side of equation (3.3) becomes 

 𝛼(𝑥, 𝑦) 𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) =  
3

2

|𝑥2−𝑦2|

16
. 

Right hand side of equation (3.3) becomes 

𝜃( 𝜓(𝑑(𝑥, 𝑦))) 𝜑(𝜓(𝑑(𝑥, 𝑦))) =  
3

2

|𝑥2−𝑦2|

16
, for all (𝑥, 𝑦) ∈ 𝑋 × 𝑌, 

which implies equation (3.3) holds. 

Hence, 𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping. 

All the conditions of Theorem 3.4. are satisfied.  

So, 𝑇 has a fixed point and 𝑥 = 0 is the fixed point of 𝑇. 

Theorem 3.6. Let (𝑋, 𝑌, 𝑑) be a complete bipolar metric space, 𝑇 ∶  (𝑋, 𝑌) ⇉  (𝑋, 𝑌) is a covariant 

mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). Suppose that the following conditions hold: 

(i) 𝑇 is 𝛼-admissible mapping, 

(ii) 𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping, 

(iii)there exist 𝑥0 ∈ 𝑋and 𝑦0 ∈ 𝑌 such that 𝛼(𝑥0 , 𝑦0 ) ≥ 1 and 𝛼(𝑥0 , 𝑇𝑦0 ) ≥ 1. 

Then 𝑇 has a unique fixed point. 

Proof: Following the proof of Theorem 3.4. 𝑇 has fixed point. To prove the uniqueness of fixed point 

of covariant mapping 𝑇 in complete bipolar metric space, let us assume, if possible, 𝑢 and 𝑣 are two 

distinct fixed point of 𝑇. i.e. 𝑇𝑢 = 𝑢 and 𝑇𝑣 = 𝑣. 

By using the properties of 𝜓, 𝜃, equations (3.1), (3.2) and (3.3), 

                                           𝜓(𝑑(𝑢, 𝑣)) = 𝜓(𝑑(𝑇𝑢, 𝑇𝑣)) 



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                                                              ≤ 𝛼(𝑢, 𝑣)𝜓(𝑑(𝑇𝑢, 𝑇𝑣)) 

                                                              ≤ 𝜃(𝜓(𝑑(𝑢, 𝑣)))𝜑(𝜓(𝑑(𝑢, 𝑣))) 

                                                              ≤  𝜑(𝜓(𝑑(𝑢, 𝑣))) 

                                                              <  𝜓(𝑑(𝑢, 𝑣)). 

This implies that 𝑑(𝑢, 𝑣) < 𝑑(𝑢, 𝑣), which is a contradiction. Thus, 𝑢 is the unique fixed point of 𝑇. 

Example 3.7. In the Example 3.5, we can easily say that 𝑇 satisfies all the conditions of Theorem 3.6. 

So, 𝑇 has a unique fixed point. 

Clearly, ‘0’ is unique fixed point of 𝑇. 

Definition 3.8. Let 𝑋 and 𝑌 be two non-empty sets. Consider (𝑋, 𝑌, 𝑑) be a bipolar metric space, a 

contravariant mapping 𝑇 ∶  (𝑋, 𝑌)  ⤨  (𝑋, 𝑌)  is called Geraghty contraction if there exist a function 

𝜃 ∈ Θ which satisfies the following condition:  

                                              𝑑(𝑇𝑦, 𝑇𝑥) ≤ 𝜃( 𝑑(𝑥, 𝑦)) 𝑑(𝑥, 𝑦) for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌.  

Definition 3.9. Let  (𝑋, 𝑌, 𝑑) be a bipolar metric space and  𝑇 ∶  (𝑋, 𝑌)  ⤨  (𝑋, 𝑌) is a contravariant 

mapping where 𝑋 and 𝑌 are two non-empty sets. Suppose that 𝜑 ∶  ℝ+ → ℝ+ is function and     𝜃 ∈ Θ 

and 𝑇 is called 𝜑- Geraghty contraction if it satisfies the following condition: 

(i)𝜑(𝑡) < 𝑡 for any 𝑡 ∈ (0, ∞),                                                                                          

(ii)For any 휀 > 0, there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀,                  

(iii)𝑑(𝑇𝑦, 𝑇𝑥) ≤ 𝜃( 𝑑(𝑥, 𝑦)) 𝜑(𝑑(𝑥, 𝑦)) for all 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌.                                 

Definition 3.10. Let 𝑋 and 𝑌 be two non-empty sets and (𝑋, 𝑌, 𝑑) be a bipolar metric space,          𝑇 ∶

 (𝑋, 𝑌) ⤨  (𝑋, 𝑌) is a contravariant self map and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). A mapping 𝑇 is said to be 

(𝛼, 𝜓, 𝜑)- Geraghty contraction mapping if there exist 𝜑 ∶  ℝ+ → ℝ+ , 𝜓 ∈ Ψ and 𝜃 ∈ Θ satisfies the 

following condition: 

(i)𝜑(𝑡) < 𝑡 for any 𝑡 ∈ (0, ∞),                                                                                      

(ii)For any 휀 > 0, there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀,                   

(iii)𝛼(𝑥, 𝑦)𝜓(𝑑(𝑇𝑦, 𝑇𝑥)) ≤ 𝜃(𝜓( 𝑑(𝑥, 𝑦))) 𝜑(𝜓(𝑑(𝑥, 𝑦))),∀ 𝑥 ∈ 𝑋 and 𝑦 ∈ 𝑌             (3.17)           

Theorem 3.11. Let (𝑋, 𝑌, 𝑑) be a complete bipolar metric space, 𝑇 ∶  (𝑋, 𝑌) ⤨  (𝑋, 𝑌) is a 

contravariant mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). Suppose that the following conditions hold: 

(i)𝑇 is 𝛼- admissible mapping, 

(ii)𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping, 

(iii) there exist 𝑥0 ∈ 𝑋 such that  𝛼(𝑥0 , 𝑇𝑥0 ) ≥ 1. 

Then 𝑇 has fixed point. 

Proof: Let 𝑥0 ∈ 𝑋and 𝑦0 ∈ 𝑌 such that 𝛼(𝑥0 , 𝑦0 ) ≥ 1 and 𝛼(𝑥0 , 𝑇𝑥0 ) ≥ 1. Now we define a 

bisequence  {(𝑥𝑛, 𝑦𝑛)} in (𝑋, 𝑌) by 𝑇𝑥𝑛 = 𝑦𝑛 and 𝑇𝑦𝑛 = 𝑥𝑛+1 for all 𝑛 ∈ ℕ ∪ {0}. Since 𝑇 is       𝛼-

admissible mapping. 



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So, 𝛼(𝑥0 , 𝑦0 ) = 𝛼(𝑥0 , 𝑇𝑥0 ) ≥ 1,  

      𝛼(𝑥1, 𝑦0 ) = 𝛼(𝑇𝑦0 , 𝑇𝑥0 ) ≥ 1,  

      𝛼(𝑥1, 𝑦1 ) = 𝛼(𝑥1 , 𝑇𝑥1) ≥ 1. 

Using mathematical induction, we get 

𝛼(𝑥𝑛+1 , 𝑦𝑛 ) ≥ 1 and 𝛼(𝑥𝑛 , 𝑦𝑛 ) ≥ 1 for all 𝑛 ∈ ℕ ∪ {0}.                                                               (3.18)  

Putting 𝑥 = 𝑥𝑛+1  and 𝑦 = 𝑦𝑛  in equation (3.17), using equations (3.1), (3.2) and by the properties of  

𝜓 and 𝜃, we have the following 

         𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 )) = 𝜓(𝑑(𝑇𝑦𝑛 , 𝑇𝑥𝑛))  

                                      ≤  𝛼(𝑥𝑛 , 𝑦𝑛 )𝜓(𝑑(𝑇𝑦𝑛 , 𝑇𝑥𝑛))  

                                     ≤ 𝜃( 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 )))𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 ))) 

                                     ≤ 𝜑(𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 ))) 

        𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 )) < 𝜓(𝑑(𝑥𝑛 , 𝑦𝑛 )) .                                                                                             (3.19) 

Hence, 𝜓 is strictly increasing function so, we get 

                                𝑑(𝑥𝑛+1 , 𝑦𝑛 ) < 𝑑(𝑥𝑛 , 𝑦𝑛 ) for all 𝑛 ≥ 0.                                                             (3.20)                                                

Similarly, putting 𝑥 = 𝑥𝑛+1 and 𝑦 = 𝑦𝑛+1  in equation (3.17), using equations (3.1), (3.2) and by the 

properties of  𝜓 and 𝜃, we have the following 

𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) = 𝜓(𝑑(𝑇𝑦𝑛 , 𝑇𝑥𝑛+1 ))  

                                 ≤  𝛼(𝑥𝑛+1 , 𝑦𝑛)𝜓(𝑑(𝑇𝑦𝑛 , 𝑇𝑥𝑛+1 ))  

                                 ≤ 𝜃( 𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 )))𝜑(𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 ))) 

                                 ≤ 𝜑(𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 ))) 

                                < 𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 )) .                                                                                          (3.21) 

Hence, 𝜓 is strictly increasing function so, we obtain 

                              𝑑(𝑥𝑛+1 , 𝑦𝑛+1 ) < 𝑑(𝑥𝑛+1 , 𝑦𝑛 ) for all 𝑛 ≥ 0.                                                   (3.22) 

From the above, we conclude that the sequences {𝑑(𝑥𝑛+1 , 𝑦𝑛 )} and {𝑑(𝑥𝑛+1, 𝑦𝑛+1 )} are monotonically 

decreasing and for the non-negative monotonically decreasing sequences  {𝑑(𝑥𝑛+1 , 𝑦𝑛 )} and 

{𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )}, there exist some 𝑟1 ≥ 0 and 𝑟2 ≥ 0, such that 

        𝑑(𝑥𝑛+1 , 𝑦𝑛 ) →  𝑟1 , 𝑑(𝑥𝑛+1 , 𝑦𝑛+1 ) →  𝑟2  as 𝑛 → ∞.                                                                  (3.23) 

We suppose on the contrary that  𝑟1 > 0.   

Hence, we have 0 < 𝑟1 <  𝑑(𝑥𝑛+1 , 𝑦𝑛 ) for all 𝑛 ≥ 0. Set 휀 = 𝑟1 . 

From equation (3.2), there exist 𝛿 > 0 such that 휀 < 𝑡 < 휀 + 𝛿 ⇒ 𝜑(𝑡) ≤ 휀. 



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On the other hand, by the definition of 휀, we can choose 𝑛0 ∈ ℕ such that        

                                  휀 <  𝑑(𝑥𝑛0+1 , 𝑦𝑛0
) < 휀 + 𝛿.  

By the properties of 𝜓, 𝜃,using equations (3.1), (3.2) and (3.17), we have 

                             𝜓(휀) < 𝜓( 𝑑(𝑥𝑛0+1 , 𝑦𝑛0
)) < 𝜓(휀 + 𝛿) = 𝜓(휀) + 𝜓(𝛿). 

This implies that   

                                   𝜑(𝜓( 𝑑(𝑥𝑛0+1 , 𝑦𝑛0
))) ≤  𝜓(휀).                                                                   (3.24) 

We have also       

                              휀 <  𝑑(𝑥𝑛0+2, 𝑦𝑛0+1) <  𝑑(𝑥𝑛0+1, 𝑦𝑛0+1) =  𝑑(𝑇𝑦𝑛0
, 𝑇𝑥𝑛0+1), 

which implies that     

                             𝜓(휀) < 𝜓(𝑑(𝑥𝑛0+2, 𝑦𝑛0+1)) 

                                      < 𝜓(𝑑(𝑥𝑛0+1, 𝑦𝑛0+1)) =  𝜓(𝑑(𝑇𝑦𝑛0
, 𝑇𝑥𝑛0+1)) 

                                      ≤ 𝛼(𝑥𝑛0+1, 𝑦𝑛0
) 𝜓(𝑑(𝑇𝑦𝑛0

, 𝑇𝑥𝑛0+1)) 

                                      ≤ 𝜃( 𝜓 (𝑑(𝑥𝑛0+1, 𝑦𝑛0
))) 𝜑(𝜓 (𝑑(𝑥𝑛0+1, 𝑦𝑛0

))) 

                                      < 𝜑(𝜓 (𝑑(𝑥𝑛0+1, 𝑦𝑛0
))) 

                                       ≤  𝜓( 휀), 

which is a contradiction. Hence  

                     lim
𝑛→∞

 𝑑(𝑥𝑛+1 , 𝑦𝑛 ) = 𝑟1 = 0.                                                                                 (3.25) 

Similarly,      lim
𝑛→∞

 𝑑(𝑥𝑛 , 𝑦𝑛 ) = 𝑟2 = 0.                                                                                    (3.26) 

Now, we shall prove that {(𝑥𝑛 , 𝑦𝑛 )} is a Cauchy bisequence. 

We fix 휀1 > 0, then by (3.2) there exists 𝛿1 > 0 such that 

                      𝑡 < 휀1 + 𝛿1 ⇒ 𝜑(𝑡) ≤ 휀1.                                                                                      (3.27) 

Without loss of generality, we assume 𝛿1 < 휀1. Due to (3.25), there exist 𝑛0 ∈ ℕ such that 

                       𝑑(𝑥𝑛+1 , 𝑦𝑛 ) <  𝛿1, for all 𝑛 ≥  𝑛0,                                                                       (3.28) 

which implies that  

                       𝜓( 𝑑(𝑥𝑛+1 , 𝑦𝑛 )) <  𝜓(𝛿1). 

By mathematical induction, we show that for any fixed 𝑘 ≥  𝑛0 

                       𝑑(𝑥𝑘+𝑙 , 𝑦𝑘) < 휀1 + 𝛿1, for all 𝑙 ∈ ℕ.                                                                    (3.29) 

For 𝑙 = 1, this inequality trivially holds by (3.28). 



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Let us assume that (3.29) is satisfied for some 𝑗 ∈ ℕ and we will show that it holds for 𝑙 = 𝑗 + 1. 

From the triangle inequality (BP3), properties of 𝜓 ,𝜃, equations (3.1), (3.2) and (3.3) 

     𝜓( 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘 )) ≤  𝜓( 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘+1) +   𝑑(𝑥𝑘+1 , 𝑦𝑘+1 ) +  𝑑(𝑥𝑘+1 , 𝑦𝑘 )) 

                                    ≤ 𝜓( 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) + 𝜓( 𝑑(𝑥𝑘+ 1 , 𝑦𝑘 )) 

                                    ≤ 𝜓( 𝑑(𝑇𝑦𝑘+𝑗 , 𝑇𝑥𝑘+1 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) + 𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘)) 

                                    ≤ 𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) 

                                        +𝛼(𝑥𝑘+1, 𝑦𝑘+𝑗 )𝜓( 𝑑(𝑇𝑦𝑘+𝑗, 𝑇𝑥𝑘 )) 

                                    ≤ 𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) 

                                        +𝜃(𝜓(𝑑(𝑥𝑘, 𝑦𝑘+𝑗 )))𝜑(𝜓( 𝑑(𝑥𝑘, 𝑦𝑘+𝑗 ))) 

                                    <  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘 )) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) +  𝜑(𝜓( 𝑑(𝑥𝑘 , 𝑦𝑘+𝑗 ))). 

Using equations (3.28) and (3.29), we get 

       𝜓( 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘 )) ≤  𝜓(  𝛿1) +  𝜓( 𝑑(𝑥𝑘+1 , 𝑦𝑘+1 )) +  𝜓( 휀1). 

By letting 𝑘 → ∞, using equation (3.26), we obtain 

       𝜓( 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘 )) ≤  𝜓( 𝛿1) +  𝜓(휀1), 

                                       = 𝜓(휀1 + 𝛿1). 

By the property of 𝜓, we get 

                        𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘 ) < 휀1 + 𝛿1. 

So, equation (3.29) is holds for 𝑙 = 𝑗 + 1. 

Hence, by induction we prove that 𝑑(𝑥𝑘+𝑗+1 , 𝑦𝑘 ) < 휀1 + 𝛿1 for all 𝑘 ≥  𝑛0 and 𝑙 ≥ 1. 

Since 휀1 is arbitrary, we conclude that 

                        lim
𝑚,𝑛→∞

 𝑑(𝑥𝑛 , 𝑦𝑚) = 0. 

Hence, {(𝑥𝑛 , 𝑦𝑛)} is a Cauchy bisequence and (𝑋, 𝑌, 𝑑) is a complete bipolar metric space. So, 

{(𝑥𝑛 , 𝑦𝑛 )}  is convergent and in fact biconvergent. So, there exists 𝑢 ∈ 𝑋 ∩ 𝑌such that (𝑥𝑛 ) → 𝑢, 

(𝑦𝑛 ) → 𝑢 as 𝑛 → ∞. 

We claim that 𝑇𝑢 = 𝑢.  

Let, if possible, 𝑇𝑢 ≠ 𝑢. Then there exist 𝑟′ > 0 such that 𝑑(𝑇𝑢, 𝑢) = 𝑟′ > 0. 

Since (𝑥𝑛 ) → 𝑢, (𝑦𝑛 ) → 𝑢 as 𝑛 → ∞, we can choose 𝑛0 ∈ ℕ such that 

𝑑(𝑥𝑛 , 𝑢) <
𝑟′

2
 , for all 𝑛 ≥  𝑛0 and 𝑑(𝑦𝑛 , 𝑢) <

𝑟′

2
 , for all 𝑛 ≥  𝑛0.                                             (3.30) 

From the triangle inequality (BP3), properties of 𝜓, 𝜃, equations (3.1), (3.2) and (3.17) 



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                  𝜓(𝑟 ′) = 𝜓(𝑑(𝑇𝑢, 𝑢)), 

                            ≤  𝜓(𝑑(𝑇𝑢, 𝑦𝑛) + 𝑑(𝑥𝑛+1 , 𝑦𝑛 ) + 𝑑(𝑥𝑛+1 , 𝑢)) 

                            ≤  𝜓(𝑑(𝑇𝑢, 𝑦𝑛 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑢)) 

                            ≤  𝜓(𝑑(𝑇𝑢, 𝑇𝑥𝑛)) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑢)) 

                            ≤ 𝛼(𝑥𝑛, 𝑢) 𝜓(𝑑(𝑇𝑢, 𝑇𝑥𝑛 )) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝜓(𝑑(𝑥𝑛+1 , 𝑢)) 

                            ≤  𝜃(𝜓((𝑥𝑛, 𝑢)) 𝜑( 𝜓(𝑑(𝑥𝑛 , 𝑢))) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝜓(𝑑(𝑥𝑛+1 , 𝑢)) 

                            <   𝜑( 𝜓(𝑑(𝑥𝑛 , 𝑢))) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝜓(𝑑(𝑥𝑛+1 , 𝑢)) 

                            <  𝜓(𝑑(𝑥𝑛 , 𝑢)) +  𝜓(𝑑(𝑥𝑛+1 , 𝑦𝑛+1 )) + 𝜓(𝑑(𝑥𝑛+1 , 𝑢)). 

Using equations (3.26) and (3.30), we get  

𝜓(𝑟 ′) <  𝜓 (
𝑟 ′

2
) +  𝜓 (

𝑟 ′

2
) 

                                                                  =  𝜓 (
𝑟′

2
+

𝑟′

2
) =  𝜓(𝑟 ′), 

which is a contradiction. 

Thus 𝑇𝑢 = 𝑢. i.e., 𝑢 is the fixed point of 𝑇. 

Example 3.12. Let 𝑋 = [0, + ∞) and 𝑌 = [−1,1] and let 𝑑 ∶ 𝑋 × 𝑌 → [0, +∞) be a function such that 

𝑑(𝑥, 𝑦) = |𝑥2 − 𝑦2| for all (𝑥, 𝑦) ∈ 𝑋 × 𝑌. 

Then, clearly (𝑋, 𝑌, 𝑑) be a complete bipolar metric space. 

Define 𝑇 ∶  (𝑋, 𝑌) ⤨  (𝑋, 𝑌) such that 𝑇𝑥 =
−𝑥

2
 is a mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞) such that 

𝛼(𝑥, 𝑦) =
3

2
  for (𝑥, 𝑦) ∈ 𝑋 × 𝑌.  

Clearly, 𝑇 is 𝛼-admissible mapping and there exist 𝑥0 ∈ 𝑋 such that 𝛼(𝑥0 , 𝑇𝑥0 ) ≥ 1 and            𝑋 ∩

𝑌 = {0} and 𝑇0 = 0. 

Taking 𝜓(𝑡) =
𝑡

4
, 𝜑(𝑡) =

𝑡

2
  and 𝜃(𝑡) =

3

4
 . 

Left hand side of equation (3.17) becomes 

 𝛼(𝑥, 𝑦) 𝜓(𝑑(𝑇𝑦, 𝑇𝑥)) =  
3

2

|𝑥2−𝑦2|

16
.  

Right hand side becomes 

𝜃( 𝜓(𝑑(𝑥, 𝑦))) 𝜑(𝜓(𝑑(𝑥, 𝑦))) =  
3

2

|𝑥2−𝑦2|

16
, for all (𝑥, 𝑦) ∈ 𝑋 × 𝑌, 

which implies equation (3.17) holds. 

Hence, 𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping. 



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All the conditions of Theorem 3.11. are satisfied. So, 𝑇 has a fixed point and 𝑥 = 0 is the fixed point 

of 𝑇. 

Theorem 3.13. Let (𝑋, 𝑌, 𝑑) be a complete bipolar metric space, 𝑇 ∶  (𝑋, 𝑌) ⤨  (𝑋, 𝑌) is a 

contravariant mapping and 𝛼 ∶ 𝑋 × 𝑌 → [0, ∞). Suppose that the following conditions hold: 

(i)𝑇 is 𝛼- admissible mapping, 

(ii)𝑇 is an  (𝛼, 𝜓, 𝜑)- Geraghty contraction mapping, 

(iii)there exist 𝑥0 ∈ 𝑋 such that 𝛼(𝑥0 , 𝑇𝑥0 ) ≥ 1. 

Then 𝑇 has a unique fixed point. 

Proof: Following the proof of Theorem 3.13. 𝑇 has fixed point. To prove the uniqueness of fixed point 

of contravariant mapping 𝑇 in complete bipolar metric space, let if possible, 𝑢 and 𝑣 are two distinct 

fixed point of 𝑇. i.e., 𝑇𝑢 = 𝑢 and 𝑇𝑣 = 𝑣. 

By using the properties of 𝜓 ,𝜃, equations (3.1), (3.2) and (3.17), 

                                           𝜓(𝑑(𝑢, 𝑣)) = 𝜓(𝑑(𝑇𝑢, 𝑇𝑣)) 

                                                              ≤ 𝛼(𝑢, 𝑣)𝜓(𝑑(𝑇𝑢, 𝑇𝑣)) 

                                                              ≤ 𝜃(𝜓(𝑑(𝑢, 𝑣)))𝜑(𝜓(𝑑(𝑢, 𝑣))) 

                                                              ≤  𝜑(𝜓(𝑑(𝑢, 𝑣))) 

                                                              <  𝜓(𝑑(𝑢, 𝑣)). 

This implies that 𝑑(𝑢, 𝑣) < 𝑑(𝑢, 𝑣), which is a contradiction. 

Hence, 𝑇 has a unique fixed point. 

Example 3.14. In the Example 3.12, we can easily say that 𝑇 satisfies all the conditions of Theorem 

3.13. So, 𝑇 has a unique fixed point. 

Clearly, ‘0’ is unique fixed point of 𝑇.                  

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