In this paper, we establish a new convergence theorem for best proximity of weak contractions in Branciari type generalized metric spaces under weak conditions

S, Arul Ravi

English

Turkish Journal of Computer and Mathematics Education (TURCOMAT)

Turkish Journal of Computer and Mathematics Education          Vol.12 No.6(2021), 321-326            Research Article 321 *Corresponding author: Arul Ravi S Assistant Professor, Department of Mathematics, St. Xavier’s College (Autonomous), Palayamkottai ammaarulravi@gamil.com    BEST PROXIMITY POINT THOREM FOR GENERALIZED (𝜓,)-WEAK CONTRACTIONS IN BRANCIARI TYPE GENERALIZED METRIC SPACES  Arul Ravi S  Assistant Professor, Department of Mathematics, St. Xavier’s College (Autonomous), Palayamkottai ammaarulravi@gamil.com  Article History:Received:11 november 2020; Accepted: 27 December 2020; Published online: 05 April 2021 ABSTRACT : In this paper, we establish a new convergence theorem for best proximity of weak contractions in Branciari type generalized metric spaces under weak conditions. Keywords: Branciari type generalized metric spaces, Best Proximity point, 𝑝 −property  1. Introduction and Preliminaries: The concept of generalized metric spaces, which is a generalization of netric spaces was first defined by Branciari [1] in 2000. The generalization is via the fact that the triangle inequality is replaced by rectangular inequality 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑢) + 𝑑(𝑢, 𝑣) + 𝑑(𝑣, 𝑦) for all pairwise distinct points 𝑥, 𝑦, 𝑢, 𝑣 ∈ 𝑋. Afterwards many authors studied and extended the existence of fixed point results in such spaces [1-18]. In thus paper, we are to generalize well known best proximity point theorems. For this we recall some basic definitions.  Definition:1.1 Let 𝑋 be nonempty set and 𝑑: 𝑋 × 𝑋 → [0,∞) be a mapping such that for all 𝑥, 𝑦 ∈  𝑋 and for all distinct points 𝑢, 𝑣 ∈ 𝑋 each of them different from 𝑥 and 𝑦 resoectively satisfying the following conditions: (i) 𝑑(𝑥, 𝑦) = 0 iff 𝑥 = 𝑦 (ii) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) (iii) 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑢) + 𝑑(𝑢, 𝑣) + 𝑑(𝑣, 𝑦) the rectangular inequality. Then (𝑋, 𝑑) is called a Branciari type generalized metric space. Remark:1.1 Every metric spaces is a Branciari type generalized metric space, but the converse is not true [2].  Definition:1.2 Let (𝑋, 𝑑) be a Branciari type generalized metric space and {𝑥𝑛} be a sequence in 𝑋 and 𝑥 ∈ 𝑋. We call that  (i) {𝑥𝑛} is convergent iff 𝑑(𝑥𝑛 , 𝑥) → 0 as 𝑛 → ∞ (denoted by 𝑥𝑛 → 𝑥) (ii) {𝑥𝑛} is a Cauchy sequence iff for each 𝜀 > 0 there exists a natural number 𝑁 such that 𝑑(𝑥𝑛, 𝑥𝑚) < 𝜀 for all 𝑛,𝑚 > 𝑁. (iii) 𝑋 is complete iff every Cauchy sequence is convergent in 𝑋. In 2012, Lakzian and Samet [4] obtained a fixed point theorem of the generalized metric spaces.  Theorem:1.1 Let (𝑋, 𝑑) be a Hausdorff and complete generalized metric space and Let 𝑇:𝑋 → 𝑋 be a self mapping satisfying 𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓(𝑑(𝑥, 𝑦)) − 𝜙(𝑑(𝑥, 𝑦)) for all 𝑥, 𝑦 ∈ 𝑋 where (i) 𝜓: [0,∞) → [0,∞) is a continuous and monotone nondecreasing function with 𝜓(𝑡) = 0 iff 𝑡 = 0. (ii) 𝜙: [0,∞) → [0,∞) is a continuous function with  𝜙(𝑡) = 0 iff 𝑡 = 0. Then  𝑇 has a unique fixed point. In 2013, Liu and Chai [8] gave a generalization of the above theorem.  Theorem:1.2 [5] Let (𝑋, 𝑑) be a Hausdorff and complete generalized metric space and Let 𝑇:𝑋 → 𝑋 be a self mapping satisfying 𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓(𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝜙(𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) +𝑎3𝑑(𝑦, 𝑇𝑦)) for all 𝑥, 𝑦 ∈ 𝑋 where (i) 𝜓: [0,∞) → [0,∞) is a continuous and monotone nondecreasing function with 𝜓(𝑡) = 0 iff 𝑡 = 0. (ii) 𝜙: [0,∞) → [0,∞) satisfying lim𝑡→𝑟𝜙(𝑡) > 0 for 𝑟 > 0 and lim𝑡→𝑟𝜙(𝑡) = 0 iff 𝑟 = 0  Arul Ravi S  322     (iii) 𝑎𝑖 ≥ 0(𝑖 = 1,2,3) with 𝑎1 + 𝑎2 + 𝑎3 ≤ 1. Then  𝑇 has a unique fixed point. Later in 2021, [22] Zhicum and Guiwen found a result for generalized (𝜓 − 𝜙)-weak contractions in Branciari type generalized metric spaces.  Theorem: 1.3 [22] Let Let (𝑋, 𝑑) be a Branciari type generalized metric space and Let 𝑇: 𝑋 → 𝑋 be a self mapping satisfying  𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓(𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝜙(𝑎1𝑑(𝑥, 𝑦) +𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) for all 𝑥, 𝑦 ∈ 𝑋 where 𝜓 ∈ 𝛹 and 𝜙 ∈ 𝛷 and 𝑎𝑖 ≥ 0(𝑖 = 1,2,3) with 𝑎1 + 𝑎2 + 𝑎3 ≤1. Then  𝑇 has a unique fixed point.  Definition:1.3 [21] 𝐴0 = {𝑥 ∈ 𝐴: 𝑑(𝑥, 𝑦) = 𝑑(𝐴, 𝐵), for 𝑦 ∈ 𝐵 𝐵0 = {𝑦 ∈ 𝐵: 𝑑(𝑥, 𝑦) = 𝑑(𝐴, 𝐵), for 𝑥 ∈ 𝐴 where 𝑑(𝐴, 𝐵) = 𝑖𝑛𝑓{𝑑(𝑥, 𝑦): 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵}  Definition: 1.4[20] Let (𝐴, 𝐵) be a pair of nonempty subsets of metric space (𝑋, 𝑑) with 𝐴0 ≠ 0. Then the pair (𝐴, 𝐵) is said to have 𝑝 −property iff for any 𝑥1, 𝑥2 ∈ 𝐴0 and 𝑦1, 𝑦2 ∈ 𝐵0, 𝑑(𝑥1, 𝑦1) = 𝑑(𝐴, 𝐵) = 𝑑(𝑥1, 𝑦2)  Remark: 1.2 [20] It is easy to that for any nonempty subsets 𝐴 of 𝑋, the pair (𝐴, 𝐴) has the 𝑝 −property.  2 Main Results: Theorem: 2.1 Let (𝐴, 𝐵) be a pair of nonempty subsets of a metric space such that 𝐴0 is nonempty. Let 𝑇: 𝐴 →𝐵 be a mapping satisfying 𝑇(𝐴0) ⊂ 𝐵0. Suppose 𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓((𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) +𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝑑(𝐴, 𝐵)) − 𝜙((𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝑑(𝐴, 𝐵)) for all 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 where 𝜓 ∈ 𝛹 and 𝜙 ∈ 𝛷 and 𝑎𝑖 ≥ 0(𝑖 = 1,2,3) with 𝑎1 + 𝑎2 + 𝑎3 ≤ 1. Proof: Choose 𝑥0 ∈ 𝐴. Since 𝑇𝑥0 ∈ 𝑇(𝐴0) ⊆ 𝐵0, there exists 𝑥1 ∈ 𝐴0 such that 𝑑(𝑥1, 𝑇𝑥0) = 𝑑(𝐴, 𝐵). Analogously, regarding the assumption, 𝑇𝑥1 ∈ 𝑇(𝐴0) ⊆ 𝐵0 ,  we determine 𝑥2 ∈ 𝐴0 such that 𝑑(𝑥2, 𝑇𝑥1) = 𝑑(𝐴, 𝐵).  Recursively, we obtain a sequence (𝑥𝑛) in 𝐴0 satisfying 𝑑(𝑥𝑛+1, 𝑇𝑥𝑛) = 𝑑(𝐴, 𝐵) for all 𝑛 ∈………..(2) Claim: 𝑑(𝑥𝑛 , 𝑥𝑛+1) → 0  If  𝑥𝑁 = 𝑥𝑁+1, then 𝑥𝑁 is a best proximity point. By the 𝑝-property, we have 𝑑(𝑥𝑛+1, 𝑥𝑛+2) = 𝑑(𝑇𝑥𝑛 , 𝑇𝑥𝑛+1)  Hence we assume that  𝑥𝑛 ≠ 𝑥𝑛+1 for all 𝑛 ∈ 𝑁.  Since 𝑑(𝑥𝑛+1, 𝑇𝑥𝑛) = 𝑑(𝐴, 𝐵),  from (1), we have for all 𝑛 ∈ 𝑁. 𝜓(𝑑(𝑥𝑛+1, 𝑥𝑛+2)) = 𝜓(𝑑(𝑇𝑥𝑛 , 𝑇𝑥𝑛+1))  ≤ 𝜓((𝑎1𝑑(𝑥𝑛 , 𝑥𝑛+1) + 𝑎2𝑑(𝑥𝑛 , 𝑇𝑥𝑛) + 𝑎3𝑑(𝑥𝑛+1, 𝑇𝑥𝑛+1)) − 𝑑(𝐴, 𝐵)) − 𝜙((𝑎1𝑑(𝑥𝑛 , 𝑥𝑛+1) + 𝑎2𝑑(𝑥𝑛 , 𝑇𝑥𝑛) +𝑎3𝑑(𝑥𝑛+1, 𝑇𝑥𝑛+1)) − 𝑑(𝐴, 𝐵))……......................…(3)                             = 𝜓((𝑎1 + 𝑎2 + 𝑎3)𝑑(𝑥𝑛 , 𝑇𝑥𝑛) −𝑑(𝐴, 𝐵)) − 𝜙((𝑎1 + 𝑎2 + 𝑎3)𝑑(𝑥𝑛 , 𝑇𝑥𝑛) − 𝑑(𝐴, 𝐵)) ≤ 𝜓((𝑑(𝑥𝑛, 𝑇𝑥𝑛) − 𝑑(𝐴, 𝐵)) − 𝜙((𝑎1 + 𝑎2 + 𝑎3)𝑑(𝑥𝑛 , 𝑇𝑥𝑛) − 𝑑(𝐴, 𝐵)) ie  𝜙((𝑎1 + 𝑎2 + 𝑎3)𝑑(𝑥𝑛 , 𝑇𝑥𝑛)) = 𝑑(𝐴, 𝐵) if ∑ 𝑎𝑖 ≠ 03𝑖=1 , we get 𝑑(𝑥𝑛 , 𝑥𝑛+1) = 0 a contradiction. If ∑ 𝑎𝑖 = 03𝑖=1  we get from (3) that  𝜓(𝑑(𝑥𝑛 , 𝑥𝑛+1)) = 0 𝑑(𝑥𝑛, 𝑥𝑛+1) = 0, contradicting our assumption Therefore 𝑑(𝑥𝑛+1, 𝑥𝑛+2) < 𝑑(𝑥𝑛 , 𝑥𝑛+1) for any 𝑛 ∈ 𝑁 and hence {𝑑(𝑥𝑛 , 𝑥𝑛+1)} is monotone decreasing sequence of nonnegative real numbers, hence there exists 𝑟 ≥ 0 such that lim𝑛→∞𝑑(𝑥𝑛 , 𝑥𝑛+1) = 𝑟. In the view of the fact from (2), for any 𝑛 ∈ 𝑁, we have 𝜓(𝑑(𝑥𝑛+1, 𝑥𝑛+2)) ≤ 𝜓(𝑑(𝑥𝑛 , 𝑥𝑛+1)) − 𝜙(𝑑(𝑥𝑛 , 𝑥𝑛+1)), Taking the limit as 𝑛 → ∞ in the above inequality, and using the conditions of  𝜓 and 𝜙 we have 𝜓(𝑟) ≤ 𝜓(𝑟) − 𝜙(𝑟) which implies 𝜙(𝑟) = 0 Hence lim𝑛→∞𝑑(𝑥𝑛 , 𝑥𝑛+1) = 0……………………………………………..………………...…(4) Next we show that (𝑥𝑛) is a Cauchy sequence. If otherwise there exists 𝜀 > 0, for which we can find two sequences of positive integers (𝑚𝑘) and (𝑛𝑘) such that for all positive integers 𝑚𝑘 > 𝑛𝑘  > 𝑘, 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) ≥ 𝜀 and 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘−1) < 𝜀. Now 𝜀 ≤  𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘)  ≤ 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘−1) + 𝑑(𝑥𝑛𝑘−1 , 𝑥𝑛𝑘), that is 𝜀 ≤  𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) < 𝜀 +  𝑑(𝑥𝑛𝑘−1 , 𝑥𝑛𝑘)  Taking the limit as 𝑘 → ∞ in the above inequality and using (4) we have  lim𝑛→∞ 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) = 𝜀………………………………………...………………………………(5) BEST PROXIMITY POINT THOREM FOR GENERALIZED (𝜓,)-WEAK CONTRACTIONS IN BRANCIARI TYPE GENERALIZED METRIC SPACES  323     Again 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) ≤  𝑑(𝑥𝑚𝑘 , 𝑥𝑚𝑘+1) +  𝑑(𝑥𝑚𝑘+1 , 𝑥𝑛𝑘+1) +  𝑑(𝑥𝑛𝑘+1 , 𝑥𝑛𝑘). Taking the limit as 𝑘 → ∞ in the above inequalities and using (4) and (5) we have  lim𝑘→∞ 𝑑(𝑥𝑚𝑘+1 , 𝑥𝑛𝑘+!) = 𝜀…………………………………..………………………………...(6) Again 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) ≤  𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘+1) +  𝑑(𝑥𝑛𝑘+1 , 𝑥𝑛𝑘)  ≤  𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) +  𝑑(𝑥𝑛𝑘 , 𝑥𝑛𝑘+1)  Letting  𝑘 → ∞ in the above inequalities and using (4) and (5) we have  lim𝑘→∞ 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘+1) = 𝜀…………………………...………………………………………….(7) lim𝑘→∞ 𝑑(𝑥𝑛𝑘 , 𝑥𝑚𝑘+1) = 𝜀 …………...…………………………………………………………(8) For 𝑥 = 𝑥𝑚𝑘 , 𝑦 = 𝑦𝑚𝑘 we have 𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘) − 𝑑(𝐴, 𝐵) ≤ 𝑑(𝑥𝑚𝑘 , 𝑥𝑚𝑘+1) +  𝑑(𝑥𝑚𝑘+1 , 𝑇𝑥𝑚𝑘) − 𝑑(𝐴, 𝐵)                                                                  = 𝑑(𝑥𝑚𝑘 , 𝑥𝑚𝑘+1)  Similarly 𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘) − 𝑑(𝐴, 𝐵) =  𝑑(𝑥𝑛𝑘 , 𝑥𝑛𝑘+1). Also 𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑛𝑘) − 𝑑(𝐴, 𝐵) = 𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘+1) and  𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑚𝑘) − 𝑑(𝐴, 𝐵) =  𝑑(𝑥𝑛𝑘 , 𝑥𝑚𝑘+1). From (1) we have  𝜓(𝑑(𝑥𝑚𝑘+1 , 𝑥𝑛𝑘+1)) = 𝜓(𝑑(𝑇𝑥𝑚𝑘 , 𝑇𝑥𝑛𝑘)) ≤ 𝜓((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) + 𝑎2𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘) + 𝑎3𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘)) − 𝑑(𝐴, 𝐵)) − 𝜙 ((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) +𝑎2𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘) + 𝑎3𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘)) − 𝑑(𝐴, 𝐵))         ≤ 𝜓((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) + 𝑎2𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘) + 𝑎3𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘))) − 𝜙 ((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) + 𝑎2𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘) +𝑎3𝑑(𝑥𝑛𝑘, 𝑇𝑥𝑛𝑘)))               It follows that  𝜓(𝑑(𝑇𝑥𝑚𝑘 , 𝑇𝑥𝑛𝑘))≤ 𝜓 ((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) + 𝑎2𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘+1) + 𝑎3𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘+1)))− 𝜙 ((𝑎1𝑑(𝑥𝑚𝑘 , 𝑥𝑛𝑘) + 𝑎2𝑑(𝑥𝑛𝑘 , 𝑇𝑥𝑛𝑘+1) + 𝑎3𝑑(𝑥𝑚𝑘 , 𝑇𝑥𝑚𝑘+1))) From (4), (5), (6) and (7) and letting 𝑘 → ∞ in the above inequalities and using the conditions of  𝜓 and 𝜙, we have  𝜓(𝜀) ≤ 𝜓(𝜀) − 𝜙(𝜀)  which is contradiction by virtue of property 𝜙. Hence {𝑥𝑛} is a Cauchy sequence. Since {𝑥𝑛}  ⊂ 𝐴 and  𝐴 is a closed subset of the complete metric space (𝑋, 𝑑), there exists 𝑥∗ in 𝐴 such that  𝑥𝑛 → 𝑥∗. Putting 𝑥 = 𝑥𝑛 and  𝑦 = 𝑥∗ and since  𝑑(𝑥𝑛, 𝑇𝑥∗) ≤ 𝑑(𝑥𝑛 , 𝑥∗) + 𝑑(𝑥∗, 𝑇𝑥𝑛) and 𝑑(𝑥∗, 𝑇𝑥𝑛) ≤ 𝑑(𝑥∗, 𝑇𝑥∗)+ 𝑑(𝑇𝑥∗, 𝑇𝑥𝑛)  We have 𝜓(𝑑(𝑥𝑛+1, 𝑇𝑥∗) − 𝑑(𝐴, 𝐵) ≤ 𝜓𝑑(𝑇𝑥𝑛 , 𝑇𝑥∗)) ≤ 𝜓((𝑎1𝑑(𝑥𝑛 , 𝑥∗) + 𝑎2𝑑(𝑥𝑛 , 𝑇𝑥𝑛) + 𝑎3𝑑(𝑥∗, 𝑇𝑥∗)) − 𝑑(𝐴, 𝐵)) − 𝜙 ((𝑎1𝑑(𝑥𝑛, 𝑥∗) + 𝑎2𝑑(𝑥𝑛 , 𝑇𝑥𝑛) +𝑎3𝑑(𝑥∗, 𝑇𝑥∗)) − 𝑑(𝐴, 𝐵))        Taking the limit as  𝑛 → ∞ in the above inequalities and using the conditions of  𝜓 and 𝜙,we have 𝜓(𝑑(𝑥∗, 𝑇𝑥∗) − 𝑑(𝐴, 𝐵)) ≤ 𝜓(𝑑(𝑥∗, 𝑇𝑥∗) − 𝑑(𝐴, 𝐵)) − 𝜙(𝑑(𝑥∗, 𝑇𝑥∗) − 𝑑(𝐴, 𝐵)) which implies that 𝑑(𝑥∗, 𝑇𝑥∗) = 𝑑(𝐴, 𝐵)   Hence  𝑥∗ is a best proximity point of  𝑇. For the uniqueness  Let 𝑝 and 𝑞 be two best proximity points  and suppose that 𝑝 ≠ 𝑞, Then putting 𝑥 = 𝑝 and 𝑦 = 𝑞 in (1) we obtain  𝜓 (𝑑(𝑇𝑝, 𝑇𝑞)) ≤ 𝜓 ((𝑎1𝑑(𝑝, 𝑞) + 𝑎2𝑑(𝑝, 𝑇𝑝) + 𝑎3𝑑(𝑞, 𝑇𝑞)) − 𝑑(𝐴, 𝐵)) − 𝜙 ((𝑎1𝑑(𝑝, 𝑞) + 𝑎2𝑑(𝑝, 𝑇𝑝) +𝑎3𝑑(𝑞, 𝑇𝑞)) − 𝑑(𝐴, 𝐵))  that is 𝜓(𝑑(𝑝, 𝑞)) ≤  𝜓(𝑑(𝑝, 𝑞)) −  𝜙(𝑑(𝑝, 𝑞)) which is contradiction by virtue of a property 𝜙. There 𝑝 = 𝑞 This completes the proof. Example:2.1 Let 𝑋 = 𝐴⋃𝐵,where 𝐴 = {12,13,14,16} and 𝐵 = {1,2}  Define the generalized metric space on 𝑋 as follows:  Arul Ravi S  324     𝑑(𝑥, 𝑦) ={      𝑑(𝑥, 𝑦), 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵   0,    𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 𝑤𝑖𝑡ℎ 𝑥 = 𝑦0.3, 𝑥 =12, 𝑦 =13 𝑜𝑟 𝑥 =14, 𝑦 =150.2, 𝑥 =12, 𝑦 =15𝑜𝑟 𝑥 =13, 𝑦 =140.6, 𝑥 =12, 𝑦 =14𝑜𝑟 𝑥 =15, 𝑦 =13|𝑥 − 𝑦|, 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵    Then (𝑋, 𝑑) is a Branciari type generalized metric space, but it is not metric space. In fact 0.6 = 𝑑 (12,14) > 𝑑 (12,13) + 𝑑 (13,14) = 0.5 Let 𝑇: 𝐴 → 𝐵 is defined by  𝑇𝑥 ={    15, 𝑥 ∈ [1,2]14, 𝑥 ∈ [12,13,14]13,              𝑥 =15  Define 𝜓(𝑡) = 𝑡, 𝜙(𝑡) =  𝑡5, 𝑡 ∈ [0,∞) then 𝑇 satisfies  𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓(𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝜙(𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) for all 𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 where 𝑎1 = 0.4, 𝑎2 = 0.4, 𝑎3 = 0.2. 𝜓(𝑑(𝑇𝑥, 𝑇𝑦)) ≤ 𝜓((𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) + 𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝑑(𝐴, 𝐵)) − 𝜙((𝑎1𝑑(𝑥, 𝑦) + 𝑎2𝑑(𝑥, 𝑇𝑥) +𝑎3𝑑(𝑦, 𝑇𝑦)) − 𝑑(𝐴, 𝐵))  Thus all the hypothesis of theorem are satisfied and 𝑇 has a best proximity point.  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