International Journal of Analysis and Applications ISSN 2291-8639 Volume 12, Number 1 (2016), 1-9 http://www.etamaths.com COMMON BEST PROXIMITY POINTS FOR CYCLIC ϕ−CONTRACTION MAPS M. AHMADI BASERI1,∗, H. MAZAHERI1 AND T. D. NARANG2 Abstract. The purpose of this paper is to introduce new types of contraction condition for a pair of maps (S, T ) in metric spaces. We give convergence and existence results of best proximity points of such maps in the setting of uniformly convex Banach spaces. Moreover, we obtain existence theorems of best proximity points for such contraction pairs in reflexive Banach spaces. Our results generalize, extend and improve results on the topic in the literature. 1. Introduction and preliminaries Let A and B be nonempty subsets of a metric space X := (X,d) and T a cyclic map on A∪B i.e. T(A) ⊆ B, T(B) ⊆ A. An element x ∈ A ∪ B is called a best proximity point of the mapping T if d(x,Tx) = d(A,B), where d(A,B) is distance of A and B i.e. d(A,B) = inf{d(x,y) : x ∈ A,y ∈ B}. The map T is called a cyclic contraction [2] if d(Tx,Ty) ≤ kd(x,y) + (1 −k)d(A,B), for some k ∈ (0, 1) and for all x ∈ A and y ∈ B. If ϕ : [0,∞) → [0,∞) is a strictly increasing map then the cyclic map T is called cyclic ϕ−contraction map [1] if d(Tx,Ty) ≤ d(x,y) −ϕ(d(x,y)) + ϕ(d(A,B)), for every x ∈ A and y ∈ B. Given two self maps S and T on A∪B, a common best proximity point of the pair (S,T) is a point x ∈ A∪B satisfying d(x,Sx) = d(x,Tx) = d(A,B). The pair (S,T) is called a semi-cyclic contraction [3] if: (i) S(A) ⊆ B, T(B) ⊆ A (ii) ∃α ∈ (0, 1), such that d(Sx,Ty) ≤ αd(x,y) + (1 −α)d(A,B), for every x ∈ A and y ∈ B. Let S(A) ⊆ B, T(B) ⊆ A. If ϕ : [0,∞) → [0,∞) is a strictly increasing map and d(Sx,Ty) ≤ d(x,y) −ϕ(d(x,y)) + ϕ(d(A,B)), for every x ∈ A and y ∈ B, then the pair (S,T) is called semi-cyclic ϕ−contraction [8]. Clearly, if S = T then a semi-cyclic ϕ−contraction pair reduces to a cyclic ϕ− contraction map. The best proximity point theorems emerge as a natural generalization of fixed point theorems, because a best proximity point reduces to a fixed point if A∩B 6= ∅. A fundamental result in fixed point theory is the Banach contraction principle. One of the interesting extentions of this result was given by Kirk, Srinivasan and Veermani [5]. Eldred and Veeramani [2] gave existence and convergence results of best proximity point for cyclic contraction maps in uniformly convex Banach spaces and metric spaces, to include the case A ∩ B = ∅. Al-Thagafi and Shahzad [1] obtained some such results for cyclic ϕ- contraction maps . Also Rezapour, Derafshpour and Shahzad [7] gave best proximity point of cyclic ϕ- contraction map on reflexive Banach spaces. In 2011, Gabeleh and Abkar [3] proved theorems on the existence and convergence of best proximity point for semi-cyclic contraction pair in Banach spaces. In 2014, Thakur and Sharma [8] obtaind best 2010 Mathematics Subject Classification. 41A65, 41A52, 46N10. Key words and phrases. best proximity point; generalized semi-cyclic ϕ-contraction pair; proximal property. c©2016 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 1 2 BASERI, MAZAHERI AND NARANG proximity point for semi-cyclic ϕ−contraction pair in uniformly convex Banach spaces. Inspired by these results, we introduce a generalized semi-cyclic ϕ-contraction pair in metric spaces, which contain the general contractive pair of maps, and prove the existence and convergence of best proximity point for such pair of maps in metric spaces, uniformly convex Banach spaces and in reflexive Banach spaces. The following, definitions are needed for our resuls. Definition 1.1. [1] Let A and B be nonempty subsets of a normed linear space X, T : A∪B → A∪B, T(A) ⊆ B and T(B) ⊆ A. We say that T satisfies the proximal property if xn w→ x ∈ A∪B, ‖xn −Txn‖→ d(A,B) =⇒‖x−Tx‖ = d(A,B), for {xn}n≥0 ∈ A∪B. Definition 1.2. A Banach space X is said to be (i) uniformly convex if there exists a stricly increasing function δ : (0, 2] → [0, 1] such that the following implication holds for all x1,x2,p ∈ X,R > 0 and r ∈ [0, 2R] : ‖xi −p‖≤ R, i = 1, 2 and ‖x1 −x2‖≥ r ⇒‖(x1 + x2)/2 −p‖≤ (1 − δ(r/R))R (ii) strictly convex if the following impilication holds for all x1,x2,p ∈ X and R > 0 ‖xi −p‖≤ R, i = 1, 2 and x1 6= x2 ⇒‖(x1 + x2)/2 −p‖ < R. 2. Main results We introduce generalized semi-cyclic ϕ−contraction pair in metric spaces as under: Definition 2.1. Let A and B be nonempty subsets of a metric space X and S,T : A ∪ B → A ∪ B be such that S(A) ⊆ B and T(B) ⊆ A. Then the pair (S,T) is said to be generalized semi-cyclic ϕ-contraction if ϕ : [0, +∞) → [0, +∞) is a strictly increasing map and d(Sx,Ty) ≤ (1/3){d(x,y) + d(Sx,x) + d(Ty,y)} − ϕ(d(x,y) + d(Sx,x) + d(Ty,y)) + ϕ(3d(A,B)), for all x ∈ A and y ∈ B. Example 2.1. Take ϕ(t) = (1 −k)(t/3) for t ≥ 0 and 0 < k < 1, we obtain d(Sx,Ty) ≤ (k/3){d(x,y) + d(x,Sx) + d(y,Ty)} + (1 −k)d(A,B), for all x ∈ A and y ∈ B, which is generalization of semi-cyclic contraction. Note that, S = T then we obtain generalized cyclic contraction map [4]. Example 2.2. Let X = R with the usual metric. For A = [0, 1], B = [−1, 0], define S,T : A∪B → A∪B by S(x) =   −x 2 x ∈ A x 2 x ∈ B, T(x) =   x 2 x ∈ A −x 2 x ∈ B. Clearly S(A) ⊆ B and T(B) ⊆ A. With a ∈ A, b ∈ B and ϕ(t) = t 2 1+8t for t ≥ 0, (S,T) is a generalized semi-cyclic ϕ-contraction. Let (S,T) be a generalized semi-cyclic ϕ-contraction. Consider x0 ∈ A, then Sx0 ∈ B, so there exists y0 ∈ B such that y0 = Sx0. Now Ty0 ∈ A, so there exists x1 ∈ A such that x1 = Ty0. Inductively, we define sequences {xn} and {yn} in A and B, respectively by (1) xn+1 = Tyn, yn = Sxn. Lemma 2.1. Let A and B be nonempty subsets of a metric space X and S,T : A ∪ B → A ∪ B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction . For x0 ∈ A∪B, the sequences {xn} and {yn} are generated by (1) then for all x ∈ A, y ∈ B, and n ≥ 1, we have (a) −ϕ(d(x,y) + d(x,Sx) + d(y,Ty)) + ϕ(3d(A,B)) ≤ 0, (b) d(Sx,Ty) ≤ (1/3){d(x,y) + d(x,Sx) + d(y,Ty)}, (c) d(xn,Sxn) ≤ d(xn−1,Sxn−1), (d) d(xn+1,yn) ≤ d(yn,Tyn−1), COMMON BEST PROXIMITY POINTS 3 (e) d(yn+1,Tyn) ≤ d(yn,Tyn−1). Proof. We have 3d(A,B) ≤ d(x,y) + d(Tx,x) + d(Ty,y). Hence ϕ is a strictly increasing map, (a) and (b) are obtained. Since d(xn,Sxn) ≤ (1/3){d(yn−1,xn) + d(xn,Sxn) + d(yn−1,xn)}, so (2) d(xn,Sxn) ≤ d(yn−1,xn). Also, since d(yn−1,xn) ≤ (1/3){d(yn−1,xn−1) + d(xn−1,Sxn−1) + d(yn−1,xn)}, we have (3) d(yn−1,xn) ≤ d(xn−1,Sxn−1). From (2) and (3), inequality (c) is obtained. Since d(xn+1,yn) ≤ (1/3){d(xn,yn) + d(xn+1,yn) + d(xn,yn)}, so d(xn+1,yn) ≤ d(yn,Tyn−1), that is inequlity (d). Now, since d(yn+1,Tyn) ≤ (1/3){d(xn+1,yn) + d(yn+1,Tyn) + d(xn+1,yn)}, we have d(yn+1,Tyn) ≤ d(xn+1,yn). By using (d), inequality (e) is obtained. Following result will be needed in what follows. Proposition 2.1. Let A and B be nonempty subsets of a metric space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction map. For x0 ∈ A∪B, the sequences {xn} and {yn} are generated by (1). Then d(xn,Sxn) → d(A,B) and d(yn,Tyn−1) → d(A,B). Proof. Let dn = d(xn,Sxn). It follows from Lemma 2.1(c), that {dn} is decreasing and bounded. So limn→∞dn = t0. Since (S,T) is a generalized semi-cyclic ϕ-contraction pair, we obtain dn+1 ≤ d(Sxn,Tyn) ≤ (1/3){2dn + d(yn,Tyn)}−ϕ(2dn + d(yn,Tyn)) + ϕ(3d(A,B)) ≤ dn −ϕ(2dn + d(yn,Tyn)) + ϕ(3d(A,B)). Hence, ϕ(3d(A,B)) ≤ ϕ(2dn + d(yn,Tyn)) ≤ dn −dn+1 + ϕ(3d(A,B)). Thus (4) lim n→∞ ϕ(2dn + d(yn,Tyn)) = ϕ(3d(A,B)). Since ϕ is strictly increasing and dn ≥ d(yn,Tyn) ≥ dn+1 ≥ t0 ≥ d(A,B), we have lim n→∞ ϕ(2dn + d(yn,Tyn)) ≥ ϕ(3t0) ≥ ϕ(3d(A,B)). From (4), ϕ(3t0) = ϕ(3d(A,B)). As ϕ is strictly increasing, we have t0 = d(A,B). Theorem 2.1. Let A and B be nonempty subsets of a metric space X and S,T : A∪B → A∪B be such that the pair (S,T) is a generalized semi-cyclic ϕ-contraction map. For x0 ∈ A∪B, the sequences {xn} and {yn} are generated by (1). If {xn} and {yn} have convergent subsequences in A and B, then there exists x ∈ A and y ∈ B such that d(x,Sx) = d(A,B) = d(y,Ty). 4 BASERI, MAZAHERI AND NARANG Proof. Let {ynk} be a subsequence of {yn} such that ynk → y. The relation d(A,B) ≤ d(Tynk,y) ≤ d(ynk,y) + d(ynk,Tynk ) holds for each k ≥ 1. Letting k →∞, by Proposion 2.1 and Lemma 2.1(d), we obtain lim k→∞ d(Tynk,y) = d(A,B). From Lemma2.1(b), d(Ty,ynk ) ≤ (1/3){d(y,xnk ) + d(y,Ty) + d(xnk,Sxnk )} ≤ (1/3){d(y,ynk ) + d(ynk,xnk ) + d(y,ynk ) + d(Ty,ynk ) + d(xnk,Sxnk )}. Letting k →∞, by Proposion 2.1, we get (2/3)d(A,B) ≤ (2/3) lim k→∞ d(Ty,ynk ) ≤ (2/3)d(A,B). So d(Ty,y) = d(A,B). Similary, it can be proved that d(x,Sx) = d(A,B). Proposition 2.2. Let A and B be nonempty subsets of a metric space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction.Then the sequences {xn} and {yn} generated by (1) are bounded. Proof. By Proposition 2.1, we have d(xn,Sxn) → d(A,B) as n → ∞. It is sufficient to show that {Sxn} is bounded. For the unbounded map ϕ, take M > 0 such that ϕ(M) > (4/3)d(x0,Sx0) + ϕ(3d(A,B)). If {Sxn} is not bounded, then there exists a natural number N ∈ N, such that d(x1,SxN ) > M, d(x1,SxN−1) ≤ M. Then M < d(x1,SxN ) ≤ d(y0,xN ) ≤ (1/3){d(x0,yN−1) + d(x0,Sx0) + d(yN−1,TyN−1)} −ϕ(d(x0,yN−1) + d(x0,Sx0) + d(yN−1,TyN−1)) + ϕ(3d(A,B)) ≤ (1/3){d(x0,x1) + d(x1,yN−1) + d(x0,Sx0) + d(xN−1,yN−1)} −ϕ(d(x0,yN−1)) + ϕ(3d(A,B)) ≤ (1/3){d(x0,y0) + d(y0,x1) + M + d(x0,Sx0) + d(xN−1,yN−2)} −ϕ(d(x0,yN−1)) + ϕ(3d(A,B)) ≤ (1/3){3d(x0,Sx0) + M + d(xN−2,yN−2)} −ϕ(d(x0,yN−1)) + ϕ(3d(A,B)) + ϕ(3d(A,B)) < (4/3)d(x0,Sx0) + M −ϕ(d(x0,yN−1)) + ϕ(3d(A,B)). Hence ϕ(d(x0,yN−1)) < (4/3)d(x0,Sx0) + ϕ(3d(A,B)). Therefore ϕ(M) < ϕ(d(x1,SxN )) ≤ ϕ(d(y0,xN )) ≤ ϕ(d(x0,yN−1)) < (4/3)d(x0,Sx0) + ϕ(3d(A,B)), which is a contradiction. Hence {Sxn} is bounded, therefore {xn} is bounded. Corollary 2.1. Let A and B be nonempty subsets of a metric space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. If A and B are boundedly compact then there exists x ∈ A and y ∈ B such that d(x,Sx) = d(A,B) = d(y,Ty). Proof. The result is an immediately consequence of Proposition 2.2 and Theorem 2.1. Now, define a sequence {zn} in A∪B as: (5) zn = { Tyk n = 2k Sxk n = 2k − 1. COMMON BEST PROXIMITY POINTS 5 In the following, we consider a uniformly convex Banach space X and give best proximity point for generalized semi-cyclic ϕ-contraction pair (S,T). Lemma 2.2. Let A and B be nonempty convex subsets of a uniformly convex Banach space X and S,T : A ∪ B → A ∪ B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. For x0 ∈ A ∪ B, if the sequences {xn} and {yn} are generated by (1) and sequence {zn} is generated by (5), then ‖z2n+2 −z2n‖→ 0 and ‖z2n+3 −z2n+1‖→ 0 as n →∞. Proof. To show ‖z2n+2 −z2n‖→ 0 as n →∞, assume the contrary. Then there exists �0 > 0 such that for each k ≥ 1, there exists nk ≥ k such that ‖z2nk+2 −z2nk‖≥ �0. (6) Choose � > 0 such that ( 1 − δ ( �0 d(A,B)+� )) (d(A,B) + �) < d(A,B). By Proposition 2.1, there exists N1 such that ‖z2nk+2 −z2nk+1‖≤ d(A,B) + �, (7) for every nk ≥ N1. Also, ‖z2nk −z2nk+1‖≤‖ynk −xnk+1‖≤‖xnk −ynk‖→ d(A,B), so, there exists N2 such that ‖z2nk −z2nk+1‖≤ d(A,B) + �, (8) for all nk ≥ N2. Let N = max{N1,N2}. From (6)-(8) and the uniform convex- ity of X, we get∥∥∥z2nk+2+z2nk2 −z2nk+1∥∥∥ ≤ (1 − δ( �0d(A,B)+�)) (d(A,B) + �), for all nk ≥ N. As (z2nk+2 + z2nk )/2 ∈ A, the choice of � implies that∥∥∥z2nk+2+z2nk2 −z2nk+1∥∥∥ < d(A,B), for all nk ≥ N, a contradiction. By a similar argument we can show that ‖z2n+3 − z2n+1‖ → 0 as n →∞. Proposition 2.3. Let A and B be nonempty convex subsets of a uniformly convex Banach space X and S,T : A ∪ B → A ∪ B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. For x0 ∈ A ∪ B, if the sequences {xn} and {yn} are generated by (1) and sequence {zn} is generated by (5), then for each � > 0, there exists a positive integer N0 such that for all m > n ≥ N0, ‖z2m −z2n+1‖ < d(A,B) + �. Proof. Suppose the contrary. So there exists �0 > 0 such that for each k ≥ 1, there is mk > nk ≥ k satisfying ‖z2mk −z2nk+1‖≥ d(A,B) + �0 (9) and ‖z2(mk−1) −z2nk+1‖ < d(A,B) + �0. (10) Now from (9) and (10), we get d(A,B) + �0 ≤‖z2mk −z2nk+1‖≤‖z2mk −z2(mk−1)‖ + ‖z2(mk−1) −z2nk+1‖ < ‖z2mk −z2(mk−1)‖ + d(A,B) + �0. Letting k →∞, Lemma 2.2 implies lim k→∞ ‖z2mk −z2nk+1‖ = d(A,B) + �0. (11) By Lemma 2.1(b) and (d), 6 BASERI, MAZAHERI AND NARANG ‖z2mk −z2nk+1‖≤‖z2mk −z2mk+2‖ + ‖z2mk+2 −z2nk+3‖ + ‖z2nk+3 −z2nk+1‖ ≤‖z2mk −z2mk+2‖ + (1/3){‖ymk+1 −xnk+2‖ +‖ymk+1 −xmk+2‖ + ‖xnk+2 −ynk+2‖} + ‖z2nk+3 −z2nk+1‖ ≤‖z2mk −z2mk+2‖ + (1/9){‖xmk+1 −ynk+1‖ +‖xmk+1 −ymk+1‖ + ‖ynk+1 −xnk+1‖} +(1/3){‖ymk+1 −xmk+1‖ + ‖xnk+2 −ynk+2‖} + ‖z2nk+3 −z2nk+1‖. Letting k →∞, from (11), Lemma 2.2 and Proposition 2.1 we get d(A,B) + �0 ≤ (1/9)(d(A,B) + �0) + (2/9)d(A,B) + (2/3)d(A,B), so d(A,B) + �0 ≤ d(A,B) + (1/9)�0, this is a contradiction. Theorem 2.2. Let A and B be nonempty closed and convex subsets of a uniformly convex Banach space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. For x0 ∈ A∪B, if the sequences {xn} and {yn} are generated by (1) and sequence {zn} is generated by (5), then there exist unique x ∈ A and y ∈ B such that z2n → x, z2n+1 → y and ‖x − Sx‖ = d(A,B) = ‖y −Ty‖. Proof. First, we show that {z2n} is a Cauchy sequence in A. If d(A,B) = 0, then let �0 > 0 be given. By Lemma 2.1(d) and Proposition 2.1, ‖z2n −z2n+1‖ = ‖Tyn −Sxn+1‖≤‖xn −Sxn‖→ d(A,B) = 0. So, there exists a positive integer N1 such that ‖z2n −z2n+1‖ < �, for every n ≥ N1. By Proposition 2.2, there exists a positive integer N2 such that ‖z2m −z2n+1‖ < �, for every m > n ≥ N2. Let N = max{N1,N2}. It follows that ‖z2m −z2n‖≤‖z2m −z2n+1‖ + ‖z2n −z2n+1‖ < 2�, for all m > n ≥ N. Therefore {z2n} is a Cauchy sequence in A. Now, we assume that d(A,B) > 0. To show that {z2n} is a Cauchy sequence in A, we assume the contrary. Then there exists �0 > 0 such that for each k ≥ 1 there exists mk > nk ≥ k so that ‖z2mk −z2nk‖≥ �0. (12) Choose � > 0 such that ( 1 − δ ( �0 d(A,B) + � )) (d(A,B) + �) < d(A,B). By Lemma 2.1(d) and Proposition 2.1, ‖z2nk −z2nk+1‖ = ‖Tynk −Sxnk+1‖≤‖xnk −Sxnk‖→ d(A,B). Hence, there exists a positive integer N1 such that ‖z2nk −z2nk+1‖≤ d(A,B) + �, (13) for all nk ≥ N1. By Proposition 2.2 there exists a positive integer N2 such that ‖z2mk −z2nk+1‖≤ d(A,B) + �, (14) for all mk > nk ≥ N2. Let N = max{N1,N2}. From (12)-(14) and the uniform convexity of X, we get∥∥∥∥z2mk + z2nk2 −z2nk+1 ∥∥∥∥ ≤ ( 1 −δ ( �0 d(A,B) + � )) (d(A,B) + �), for all mk > nk ≥ N. As (z2mk + z2nk )/2 ∈ A, the choice of � implies that∥∥∥∥z2mk + z2nk2 −z2nk+1 ∥∥∥∥ < d(A,B), COMMON BEST PROXIMITY POINTS 7 for all mk > nk ≥ N, a contradiction. Thus {z2n} Cauchy sequence in A. By a similar argument, we can show that {z2n+1} is a Cauchy sequence in B. The completeness of X and the closedness of A implies that z2n → x as n →∞. By Theorem 2.1, ‖x−Sx‖ = d(A,B). Also, it follows from closedness of B and Theorem 2.1 that ‖y−Ty‖ = d(A,B). To prove uniqueness, assume that there is a ∈ A such that a 6= x and ‖a−Sa‖ = d(A,B). By Lemma 2.1 (b), ‖TSx−Sx‖≤ (1/3){2‖Sx−x‖ + ‖TSx−Sx‖}, hence (2/3)d(A,B) ≤ (2/3)‖TSx−Sx‖≤ (2/3)‖Sx−x‖ = (2/3)d(A,B). Therefore, ‖TSx−Sx‖ = d(A,B), it follows that TSx = x. Now ‖Sx−a‖ = ‖Sx−TSa‖≤ (1/3){‖Sa−x‖ + ‖x−Sx‖ + ‖Sa−a‖} ≤ (1/9){‖Sx−a‖ + ‖Sa−a‖ + ‖x−Sx‖} +(1/3){‖x−Sx‖ + ‖Sa−a‖}. Hence (8/9)d(A,B) ≤ (8/9)‖Sx−a‖≤ (8/9)d(A,B), which implies that, ‖Sx−a‖ = d(A,B). From convexity of A and strict convexity of X, we get∥∥∥∥x + a2 −Sx ∥∥∥∥ = ∥∥∥∥x−Sx2 + a−Sx2 ∥∥∥∥ < d(A,B), a contradiction. Thus x = a. Similarly, we show the uniqueness of y ∈ B. Now, we show the existence of a best proximity point for generalized semi-cyclic ϕ-contraction pair (S,T) in reflexive Banach spaces. First, we prove the following theorem. Theorem 2.3. Let A and B be nonempty weakly closed subsets of a reflexive Banach space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. Then there exists (x,y) ∈ A×B such that ‖x−y‖ = d(A,B). Proof. If d(A,B) = 0, the result follows by Theorem 3.1(i) of [6]. So we assume that d(A,B) > 0. For x0 ∈ A, if the sequences {xn} and {yn} are generated by (1) and sequence {zn} is generated by (5) then from Proposition 2.3, the sequences {z2n} and {z2n+1} are bounded. Since X is reflexive and A is weakly closed, the sequence {z2n} has a subsequence {z2nk} such that z2nk w→ x ∈ A. Also B is weakly closed, hence z2nk+1 w→ y ∈ B as k → ∞. Since z2nk − z2nk+1 w→ x−y 6= 0 as k → ∞, there exists a bounded liner functional f : X → [0,∞) such that ‖f‖ = 1 and f(x−y) = ‖x−y‖. For all k ≥ 1, we have |f(z2nk −z2nk+1)| ≤ ‖f‖‖z2nk −z2nk+1‖ = ‖z2nk −z2nk+1‖. Since limk→∞ |f(z2nk −z2nk+1)| = ‖x−y‖, by Lemma 2.1(d) and Proposition 2.1, we get ‖x−y‖ = limk→∞ |f(z2nk −z2nk+1)| ≤ limk→∞‖z2nk −z2nk+1‖ ≤ limk→∞‖xnk −Sxnk‖ = d(A,B). So, ‖x−y‖ = d(A,B). Theorem 2.4. Let A and B be nonempty weakly closed subsets of a reflexive Banach space X and S,T : A∪B → A∪B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. Then there exists x ∈ A and y ∈ B such that ‖x−Sx‖ = d(A,B) = ‖Ty −y‖, provided that one of the following conditions is satisfied (i) S is weakly continuous on A and T is weakly continuous on B. (ii) T, S satisfy the proximal property. 8 BASERI, MAZAHERI AND NARANG Proof. If d(A,B) = 0, the result follows from Theorem 3.1(i) of [6]. So we assume that d(A,B) > 0. For x0 ∈ A, if the sequences {xn} and {yn} are generated by (1) and sequence {zn} is generated by (5) then by Proposition 2.3 the sequences {z2n} and {z2n+1} are bounded. Since A is weakly closed, the sequence {z2n} has a subsequence {z2nk} such that z2nk w→ x ∈ A. From (i), z2nk+1 w→ Sx ∈ B as k → ∞. So z2nk − z2nk+1 w→ x − Sx 6= 0 as k → ∞. Now the proof continues similar to that of Theorem 2.3. Also B is weakly closed, so z2nk+1 w→ y ∈ B as k → ∞. Since, (i) holds, z2nk+2 w→ Ty, as k →∞. Next the proof continues similar to that of Theorem 2.3. From (ii), by Lemma 2.1(d) and Proposition 2.1, ‖z2nk+1 − Tz2nk+1‖ → d(A,B) as k → ∞. So ‖y −Ty‖ = d(A,B). Also, ‖z2nk −Sz2nk‖→ d(A,B) as k →∞. Thus ‖x−Sx‖ = d(A,B). Next, we consider reflexive and strictly convex Banach spaces and give best proximity point for gen- eralized semi-cyclic ϕ-contraction pair. Theorem 2.5. Let A and B be nonempty closed and convex subsets of a reflexive and strictly convex Banach space X and S,T : A ∪ B → A ∪ B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. If (A−A) ∩ (B−B) = {0}, then there exists a unique x ∈ A and a unique y ∈ B such that ‖x−Sx‖ = d(A,B) = ‖Ty −y‖. Proof. If d(A,B) = 0, the result follows from Theorem 3.1(i) of [6]. So we assume that d(A,B) > 0. Since A and B are closed and convex, they are weakly closed. By Theorem 2.3, there exists (x,y) ∈ A×B with ‖x−y‖ = d(A,B). Suppose that there exists (a,b) ∈ A×B with ‖a−b‖ = d(A,B). Since (A−A) ∩ (B −B) = {0}, x−y 6= a− b. By the strict convexity of X, and convexity of A and B, we have ‖(x + a)/2 − (y + b)/2‖ = ‖(x−y)/2 + (a− b)/2‖ < d(A,B), which is a contraction. This shows that (x,y) is unique. Theorem 2.6. Let A and B be nonempty closed and convex subsets of a reflexive and strictly convex Banach space X and S,T : A ∪ B → A ∪ B be such that the pair (S,T) is generalized semi-cyclic ϕ-contraction. Then there exist unique x ∈ A and y ∈ B such that ‖x−Sx‖ = d(A,B) = ‖Ty −y‖, provided that one of the following conditions is satisfied (i) S is weakly continuous on A and T is weakly continuous on B. (ii) T, S satisfy the proximal property. Proof. If d(A,B) = 0, the result follows from Theorem 3.1(i) of [6]. So we assume that d(A,B) > 0. Since A and B are closed and convex, they are weakly closed. By Theorem 2.4, there exists x ∈ A and y ∈ B such that ‖x−Sx‖ = d(A,B) = ‖Ty −y‖. For the uniqueness of x, suppose that there exists a ∈ A such that ‖a−Sa‖ = d(A,B). By the strict convexity of X, and convexity of A and B, we have ‖(x + a)/2 − (Sx + Sa)/2‖ = ‖(x−Sx)/2 + (a−Sa)/2‖ < d(A,B), which is a contraction. Now, for uniqueness of y, suppose that there exists b ∈ B such that ‖Tb− b‖ = d(A,B). Since ‖(y + b)/2 − (Ty + Tb)/2‖ = ‖(y −Ty)/2 + (b−Tb)/2‖ < d(A,B), which is a contraction. References [1] M. A. Al-Thagafi, N. Shahzad, Convergence and existence result for best proximity points, Nonlinear Analysis, Theory, Methods and Applications, 70(2009), 3665-3671. [2] A. Anthony Eldred, P. Veeramani, Existence and convergence of best proximity points, J. Math. Anal. Appl, 323(2006), 1001-1006. COMMON BEST PROXIMITY POINTS 9 [3] M. Gabeleh, A. Abkar, Best proximity points for semi-cyclic contractive pairs in Banach spaces, Int. Math. Forum, 6(2011), 2179 - 2186. [4] E. Karapinar, Best proximity points of cyclic mappings, Appl. Math. Letters, 25(2012), 1761-1766. [5] W. A. Kirk, P. S. Srinivasan and P. Veeramani, Fixed points for mapping cyclic contractions, Fixed Point Theory, 4(2003), 79-89. [6] B. Prasad, A Best proximity theorem for some general contractive pair of maps, Proc. of Int. Conf. on Emerging Trends in Engineering and Technology, 2013. [7] Sh. Rezapour, M. Derafshpour and N. Shahzad, Best proximty points of cyclic ϕ− contractions on reflexive Banach spaces, Fixed Point Theory and Application, 2010(2010), Art. ID 946178. [8] B. S. Thakur, A. Sharma, Existence and convergence of best proximity points for semi-cyclic contraction pairs, International Journal of Analysis and Applications, 5(2014), 33-44. 1Faculty of Mathematics, Yazd University, Yazd, Iran 2Guru Nanak Dev University, Amristar, India ∗Corresponding author: m.ahmadi@stu.yazd.ac.ir