@ Appl. Gen. Topol. 16, no. 2(2015), 119-126doi:10.4995/agt.2015.3242 c© AGT, UPV, 2015 Best proximity points for cyclical contractive mappings J. Maria Felicit and A. Anthony Eldred Department of Mathematics, St. Joseph’s College (Autonomous), Tiruchirappalli, Tamil Nadu, India (malarfelicit@gmail.com, anthonyeldred@yahoo.co.in) Abstract We consider p-cyclic mappings and prove an analogous result to Edel- stien contractive theorem for best proximity points. Also we give similar results satisfying Boyd-Wong and Geraghty contractive conditions. 2010 MSC: 47H09; 47H10. Keywords: best proximity point; p-cyclic mapping; cyclical contractive mapping; cyclical proximal property. 1. Introduction Best proximity theorems has evoked considerable interest in recent years following the results of [1], where the authors investigate the existence of an element x satisfying d(x,Tx) = d(A,B) = inf{d(x,y)/x ∈ A,y ∈ B} for the map T : A ∪ B → A ∪ B satisfying T(A) ⊂ B and T(B) ⊂ A. In [1] the authors proved a Banach contraction type result in a uniformly convex Banach space setting, which was extended by Di Bari et. al. [4] for cyclic Meir-Keeler contractions. Karpagam et. al. [7] and Vetro [3] considered p-cyclic mappings T : ∪ p i=1Ai → ∪ p i=1Ai satisfying T(Ai) ⊂ Ai+1, for 1 ≤ i ≤ p and Ap+i = Ai and they explored the existence of the best proximity point x ∈ Ai satisfying d(x,Tx) = d(Ai,Ai+1). In fact, p-cyclic mappings were first considered by Kirk et. al. [8] in which they discussed fixed point theorems for mappings satisfying the contraction condition. They have also considered extensions of fixed point theorems of Edelstien [5], Boyd-Wong [2] and Geraghty [6]. Received 25 August 2014 – Accepted 11 April 2015 http://dx.doi.org/10.4995/agt.2015.3242 J. Maria Felicit and A. Anthony Eldred In this paper we give analogous results to the above fixed point theorems using cyclical contractive conditions which does not force ∩ p i=1Ai 6= ∅ as in [7] and thereby we investigate the existence of best proximity point x ∈ Ai satisfying d(x,Tx) = d(Ai,Ai+1). The contractive conditions given in this pa- per behave differently from the ones used in [7] and [3], in the sense that the nonexpansive implication is nontrivial as we shall see in section 3. 2. Basic definitions and results In this section we give some basic concepts related to our results. Given two nonempty subsets A and B of a metric space X, the following notations and definitions are used in the sequel. d(A,B) = inf{d(x,y) : x ∈ A,y ∈ B}; d(x,A) = inf{d(x,y) : y ∈ A} A0 = {x ∈ A : d(x,y ′) = d(A,B) for some y′ ∈ B}; B0 = {y ∈ B : d(x ′,y) = d(A,B) for some x′ ∈ A}; PA(x) = {y ∈ A : d(x,y) = d(x,A)}. A Banach space X is said to be (a) uniformly convex if there exists a strictly increasing function δ : (0,2] → [0,1] such that for all x,y,p ∈ X,R > 0 and r ∈ [0,2R] : ‖x − p‖ ≤ R,‖y − p‖ ≤ R,‖x − y‖ ≥ r ⇒ ∥ ∥ ∥ x + y 2 − p ∥ ∥ ∥ ≤ ( 1 − δ ( r R )) R; (b) strictly convex if for all x,y,p ∈ X and R > 0 : ‖x − p‖ ≤ R,‖y − p‖ ≤ R,x 6= y ⇒ ∥ ∥ ∥ x + y 2 − p ∥ ∥ ∥ < R. Definition 2.1 ([7]). Let A1,A2, . . . ,Ap be nonempty subsets of a metric space X. Then T : ∪ p i=1Ai → ∪ p i=1Ai is called p-cyclic mapping if T(Ai) ⊂ Ai+1 for i = 1,2, . . . ,p, where Ap+i = Ai. A point x ∈ ∪ p i=1Ai is said to be a best proximity point if d(x,Tx) = d(Ai,Ai+1). Definition 2.2 ([1]). Let A1,A2, . . . ,Ap be nonempty subsets of a metric space X. A p-cyclic map T on ∪ p i=1Ai is a p-cyclic contraction mapping if for some k ∈ (0,1), (2.1) d(Tx,Ty) ≤ kd(x,y) + (1 − k)d(Ai,Ai+1) for all x ∈ Ai,y ∈ Ai+1, i = 1,2, . . . ,p. Remark 2.3. Note that Definition 2.2 implies that T satisfies d(Tx,Ty) ≤ d(x,y), for all x ∈ Ai,y ∈ Ai+1, moreover, the inequality (2.1) can be written as d(Tx,Ty) − d(Ai,Ai+1) ≤ k[d(x,y) − d(Ai,Ai+1)] for all x ∈ Ai,y ∈ Ai+1. Definition 2.4 ([7]). Let A1,A2, . . . ,Ap be nonempty subsets of a metric space X. Then a p-cyclic mapping T : ∪ p i=1Ai → ∪ p i=1Ai is called a p-cyclic nonexpansive mapping if d(Tx,Ty) ≤ d(x,y) for all x ∈ Ai,y ∈ Ai+1, i = 1,2, . . . ,p. c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 120 Best proximity points for cyclical contractive mappings The nonexpansive condition ensures the equality of distance between con- secutive sets. Lemma 2.5 ([7]). Let (X,d) be a metric space and let A1,A2, . . . ,Ap be nonempty subsets of X. If T : ∪ p i=1Ai → ∪ p i=1Ai is a p-cyclic nonexpansive mapping then d(Ai,Ai+1) = d(Ai+1,Ai+2) = · · · = d(A1,A2), i = 1,2, . . . ,p−1. Lemma 2.6 ([1]). Let A be a nonempty closed and convex subset and B be a nonempty closed subset of a uniformly convex Banach space . Let {xn} and {zn} be a sequences in A , and let {yn} be a sequence in B satisfying (i) ‖zn − yn‖ → d(A,B), (ii) for every ǫ > 0, there exists N0 ∈ N, such that for all m > n > N0,‖xm − yn‖ ≤ d(A,B) + ǫ. Then, for every ∈> 0, there exists N1 ∈ N, such that for all m > n > N1,‖xm− zn‖ ≤ ǫ. Lemma 2.7 ([1]). Let A be a nonempty closed convex subset and B be a nonempty closed subset of a uniformly convex Banach space. Let{xn} and{zn} be a sequences in A and let {yn} be a sequence in B satisfying (i) ‖xn − yn‖ → d(A,B), (ii) ‖zn − yn‖ → d(A,B). Then ‖xn − zn‖ converges to zero. Theorem 2.8 ([7]). Let A1,A2, . . . ,Ap be nonempty subsets of a metric space X and let T : ∪ p i=1Ai → ∪ p i=1Ai be a p-cyclic mapping. If for some x ∈ Ai, the sequence {T pnx} ∈ Ai contains a convergent subsequence {T pnjx} converging to ξ ∈ Ai, then ξ is a best proximity point in Ai. Definition 2.9. Let A1,A2, . . . ,Ap be nonempty subsets of a metric space X. A p-cyclic mapping T on ⋃p i=1 Ai is said to be a p-cyclic contractive map if d(Tx,Ty) < d(x,y), for all x ∈ Ai,y ∈ Ai+1 satisfying d(x,y) > d(Ai,Ai+1), for all i = 1, . . . ,p. Definition 2.10. The nonempty subsets A1,A2, . . . ,Ap of a metric space X are said to satisfy cyclical proximal property if there exists xi ∈ Ai for all 1 ≤ i ≤ p such that xi = xi+p for all i = 1, . . . ,p whenever ‖xi − xi+1‖ = d(Ai,Ai+1). 3. Main Results The following lemma shows that any p-cyclic contractive mapping is also p-cyclic non-expansive. Lemma 3.1. Let A1,A2, . . . ,Ap be nonempty closed and convex subsets of a uniformly convex Banach space X. Let T : ⋃p i=1 Ai → ⋃p i=1 Ai be such that (i) T(Ai) ⊂ Ai+1, i = 1,2, . . . ,p, where Ap+i = Ai, (ii) ‖Tx − Ty‖ < ‖x − y‖, for all x ∈ Ai, y ∈ Ai+1 and ‖x − y‖ 6= d(Ai,Ai+1). Then ‖Tx − Ty‖ ≤ ‖x − y‖, for all x ∈ Ai, y ∈ Ai+1. c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 121 J. Maria Felicit and A. Anthony Eldred Proof. It is easy to observe that d(Ai,Ai+1) = d(Ai+1,Ai+2), for all i = 1, . . . ,p− 1. We shall prove that ‖Tx−Ty‖ = d(Ai,Ai+1), whenever ‖x−y‖ = d(Ai,Ai+1). Assume that ‖x − y‖ = d(Ai,Ai+1), then it is possible to choose sequences {xn} ∈ Ai and {yn} ∈ Ai+1 such that ‖xn − yn‖ > d(Ai,Ai+1) and ‖xn − yn‖ → d(Ai,Ai+1) with xn 6= x,yn 6= y. Since d(Ai,Ai+1) ≤ ‖Txn − Ty‖ < ‖xn − y‖, ‖Txn − Ty‖ → d(Ai,Ai+1). Similar argument as- serts that ‖Tyn − Tx‖ → d(Ai,Ai+1). Since ‖PAi+1Ty − Ty‖ ≤ ‖Txn − Ty‖, Txn → PAi+1Ty and Tyn → PAi+2Tx. As ‖Txn − Tyn‖ → d(Ai,Ai+1), we have ‖PAi+1Ty−PAi+2Tx‖ = d(Ai,Ai+1). By uniqueness of the proximal point, Ty = PAi+2Tx, Tx = PAi+1Ty. Hence the lemma. � It is necessary to ensure the non-expansive condition as it may not be ex- plicitly given in the contractive condition for example Theorem 3.4, whereas the conditions used in Theorem 3.6 directly imply the non-expansive condition. Theorem 3.2. Let A1,A2, . . . ,Ap be nonempty closed and convex subsets of a strictly convex Banach space X satisfying cyclical proximal property. Further, assume one of the subsets is compact. Let T : ⋃p i=1 Ai → ⋃p i=1 Ai be a p- cyclic mapping such that ‖Tx − Ty‖ < ‖x − y‖ for all x ∈ Ai, y ∈ Ai+1 and ‖x − y‖ 6= d(Ai,Ai+1), then for each i, 1 ≤ i ≤ p, there exists a unique best proximity point such that, for any x0 ∈ Ai0 (with respect to Ai+1), the sequence {xpn} converges to the best proximity point. Proof. Assume Ai is compact. Define φ : Ai0 → R + by φ(y) = d(y,Ty) for all y ∈ Ai0. From the Lemma 2.7 it is easy to observe that T is continuous on Ai0. In general, T m is continuous on any Ai, i = 1, . . . ,p, where m is pos- itive integer. So φ is continuous and hence there exists y0 ∈ Ai0, such that d(y0,Ty0) = φ(y0) = infy∈Ai0 d(y,Ty). Suppose d(y0,Ty0) > d(Ai,Ai+1), then d(T py0,T p+1y0) < d(y0,Ty0) which is a contradiction. Hence d(y0,Ty0) = d(Ai,Ai+1). Assume that x0 ∈ Ai0, and {xpn} ∈ Ai0, for all n = 1,2, . . . . Suppose for some n, xpn = y0, then xpn+1 = Txpn = Ty0, Assume xpn 6= y0 for any n. Since ‖T ny0 − T n+1y0‖ = d(Ai,Ai+1) and T py0 = y0, by cyclical proximal property. d(xpn,PAi+1(y0)) = d(T pxpn−p,T p+1y0) ≤ d(xpn−p,Ty0) = d(xp(n−1),PAi+1(y0)). Therefore d(xpn,PAi+1(y0)) is a decreasing sequences converging to some r ≥ 0. Since Ai is compact, it follows that the sequence {xpn} has a subsequence {xpnk} converging to some z ∈ Ai. If d(z,PAi+1(y0)) ≤ d(Ai,Ai+1), then there is nothing to prove. Assume that d(z,PAi+1(y0)) > d(Ai,Ai+1), then d(z,PAi+1(y0)) = lim n→∞ d(xpn,PAi+1(y0)) = lim n→∞ d(T pxpn,PAi+1(y0)) = lim k→∞ d(T pxpnk,PAi+1(y0)) = d(T pz,T p+1y0) (Since T p is continuous on Ai0) < d(z,Ty0) = d(z,PAi+1(y0)), c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 122 Best proximity points for cyclical contractive mappings which is a contradiction. Therefore z = y0. Since any convergent subsequence of {xpn} converges to y0, {xpn} itself converges to y0 which is the best proximity point. For uniqueness, suppose there exists z ∈ Ai with z 6= y0 such that ‖z−Tz‖ = d(Ai,Ai+1), by cyclical proximal property T py0 = y0,T pz = z. If ‖y0 − Tz‖ − d(Ai,Ai+1) > 0 then ‖Ty0 − T 2z‖ − d(Ai,Ai+1) < ‖y0 − Tz‖ − d(Ai,Ai+1) = ‖T py0 − T p+1z‖ − d(Ai,Ai+1) ≤ ‖Ty0 − T 2z‖ − d(Ai,Ai+1). which is a contradiction. � Example 3.3. Let A1 = {(0,0,x) ∈ R 3/x ≥ 1}, A2 = {(0,1,x) ∈ R 3/x ≥ 1}, A3 = {(1,1,x) ∈ R 3/x ≥ 1}, and A4 = {(1,0,x) ∈ R 3/x ≥ 1} be subsets in the space R3 with euclidean norm. Clearly A1,A2,A3 and A4 satisfy cyclical proximal property. Define T on ∪4i=1Ai as T(0,0,x) = ( 0,1,x + 1 x ) , for (0,0,x) ∈ A1, T(0,1,x) = ( 1,1,x + 1 x ) , for (0,1,x) ∈ A2 T(1,1,x) = ( 1,0,x + 1 x ) , for (1,1,x) ∈ A3, T(1,0,x) = ( 0,0,x + 1 x ) , for (1,0,x) ∈ A4. For any (0,0,x) ∈ A1, and (0,1,y) ∈ A2. If ‖(0,0,x)−(0,1,y)‖ > d(A1,A2) = 1, then x 6= y. Also ‖T(0,0,x) − T(0,1,y)‖ = ‖ ( 0,1,x + 1 x ) − ( 1,1,y + 1 y ) ‖ < (1 + (x − y)2) 1 2 = ‖(0,0,x) − (0,1,y)‖ Hence T is a cyclic contractive map. Also for any (0,0,x) ∈ A1, ‖(0,0,x) − T(0,0,x)‖ = ‖(0,0,x) − ( 0,1,x + 1 x ) ‖ = ( 1 + (1 x )2) 1 2 > 1 = d(A1,A2). Here T does not admit any best proximity point as none of the sets are compact. Next we consider two of the famous extensions of Banach contraction theo- rem due to Boyd-Wong and Gregathy. Theorem 3.4. Let A1,A2, . . . ,Ap be nonempty closed subsets of a complete metric space (X,d). Let T : ∪ p i=1Ai → ∪ p i=1Ai be a p-cyclic mapping . Suppose d(Tx,Ty) ≤ ψ(d(x,y) − d(Ai,Ai+1)) + d(Ai,Ai+1) for all x ∈ Ai, y ∈ Ai+1, where ψ : [0,∞) → [0,∞) is upper semi-continuous from the right and satisfies 0 ≤ ψ(t) < t for all t > 0. Then c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 123 J. Maria Felicit and A. Anthony Eldred (i) d(T pnx,T pn+1y) → d(Ai,Ai+1) as n → ∞ (ii) d(T p(n+1)x,T pn+1y) → d(Ai,Ai+1) as n → ∞ Note: The contractive condition here does not directly guarantee the non- expansive condition and hence the importance of Lemma 3.1. Proof. (i) Choose x0 ∈ Ai, set sn = d(T pnx0,T pn+1x0) − d(Ai,Ai+1). Given ψ(t) < t for all t > 0, from the Lemma 3.1, it follows that d(T p(n+1)x0,T p(n+1)+1x0) ≤ d(T pnx0,T pn+1x0). Therefore {sn} is a decreasing sequence and hence converges . Let r be the limit of sn. Then r ≥ 0. Suppose r > 0. Then d(T p(n+1)x0,T p(n+1)+1x0) − d(Ai,Ai+1) ≤ d(T p(n+1)−1x0,T p(n+1)x0) ≤ d(T p(n+1)−2x0,T p(n+1)−1x0) ≤ . . . ≤ d(T pn+1x0,T pn+2x0) ≤ ψ(d(T pnx0,T pn+1x0) −d(Ai,Ai+1)). Taking lim sup on both sides, lim supd(T p(n+1)x0,T p(n+1)+1x0) − d(Ai,Ai+1) ≤ lim supψ(d(T pnx0,T pn+1x0) − d(Ai,Ai+1)). We obtain r ≤ ψ(r), which is a contradiction. Hence d(T pnx0,T pn+1x0) → d(Ai,Ai+1) as n → ∞. Similar argument shows that d(T p(n+1)x,T pn+1y) → d(Ai,Ai+1) as n → ∞. � Theorem 3.5. Let A1,A2, . . . ,Ap be nonempty closed and convex subsets of a uniformly convex Banach space X. Let T : ⋃p i=1 Ai → ⋃p i=1 Ai be a p-cyclic mapping such that d(Tx,Ty) ≤ ψ(d(x,y) − d(Ai,Ai+1)) + d(Ai,Ai+1) for all x ∈ Ai, y ∈ Ai+1, where ψ : [0,∞) → [0,∞) is upper semi-continuous from the right and satisfies 0 ≤ ψ(t) < t for all t > 0 and ψ(0) = 0. Then for each i,1 ≤ i ≤ p,there exists a unique best proximity point such that, for any x0 ∈ Ai, {T pnx0} converges to the best proximity point. Proof. Choose x0 ∈ Ai. Suppose d(Ai,Ai+1) = 0, then T has a unique fixed point x ∈ ∩ p i=1Ai, see in [8]. Assume that d(Ai,Ai+1) 6= 0, then by Theorem 3.4 it follows that ‖T pnx0−T pn+1x0‖ → d(Ai,Ai+1) and ‖T p(n+1)x0−T pn+1x0‖ → d(Ai,Ai+1). By Lemma 2.7, it follows that ‖T pnx0 − T p(n+1)x0‖ → 0. Sim- ilarly ‖T pn+1x0 − T p(n+1)+1x0‖ → 0. To complete the proof, we have to show that for every ǫ > 0, there exists N0, such that for all m > n ≥ N0, ‖T pmx0 − T pn+1x0‖ ≤ d(Ai,Ai+1) + ǫ. Suppose not, then there exists ǫ > 0, such that for all k ∈ N there exists mk > nk ≥ k for which ‖T pmkx0 − T pnk+1x0‖ ≥ d(Ai,Ai+1) + ǫ. This mk can be chosen such that it is the least integer greater than nk to satisfy the above inequality and ‖T p(mk−1)x0 − c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 124 Best proximity points for cyclical contractive mappings T pnk+1x0‖ < d(Ai,Ai+1) +ǫ. Consequently ‖T pnx0 −T pn+1x0‖ → d(Ai,Ai+1) and ‖T p(n+1)x0 − T pn+1x0‖ → d(Ai,Ai+1). By Lemma2.7, it follows that ‖T pnx0 − T p(n+1)x0‖ → 0. Similarly ‖T pn+1x0 − T p(n+1)+1x0‖ → 0. d(Ai,Ai+1) + ǫ ≤ ‖T pmkx0 − T pnk+1x0‖ ≤ ‖T pmkx0 − T p(mk−1)x0‖ + ‖T p(mk−1)x0 − T pnk+1x0‖ ≤ ‖T pmkx0 − T p(mk−1)x0‖ + d(Ai,Ai+1) + ǫ. This implies that limk→∞‖T pmkx0 − T pnk+1x0‖ = d(Ai,Ai+1) + ǫ. Since ‖T p(mk+1)x0 − T p(nk+1)+1x0‖ ≤ ‖T pmk+1x0 − T pnk+2x0‖, ‖T pmkx0 − T pnk+1x0‖ ≤ ‖T pmkx0 − T p(mk+1)x0‖ + ‖T p(mk+1)x0 −T p(nk+1)+1x0‖ + ‖T p(nk+1)+1x0 − T pnk+1x0‖ ≤ ‖T pmkx0 − T p(mk+1)x0‖ +‖T pmk+1x0 − T pnk+2x0‖ +‖T p(nk+1)+1x0 − T pnk+1x0‖ ≤ ‖T pmkx0 − T p(mk+1)x0‖ +ψ(‖T pmkx0 − T pnk+1x0‖ −d(Ai,Ai+1)) + d(Ai,Ai+1) +‖T p(nk+1)+1x0 − T pnk+1x0‖, which yields that ‖T pmkx0 − T pnk+1x0‖ − d(Ai,Ai+1) ≤ ‖T pmkx0 − T p(mk+1)x0‖ + ψ(‖T pmkx0 − T pnk+1x0‖ −d(Ai,Ai+1)) + ‖T p(nk+1)+1x0 − T pnk+1x0‖. Therefore lim supk ‖T pmkx0−T pnk+1x0‖−d(Ai,Ai+1) ≤ lim supk ψ(‖T pmkx0− T pnk+1x0‖−d(Ai,Ai+1)), as ‖T pmkx0−T p(mk+1)x0‖ → 0 and ‖T p(nk+1)+1x0 − T pnk+1x0‖ → 0. Hence ǫ ≤ ψ(ǫ), a contradiction. By Lemma 2.6, {T pnx0} is a cauchy sequence and converges to x ∈ Ai. From Theorem 2.8, it follows that ‖x − Tx‖ = d(Ai,Ai+1). To see that T px = x, we note that ‖x − T p+1x‖ = lim n→∞ ‖T pnx0 − T p+1x‖ ≤ lim n→∞ ‖T p(n−1)x0 − Tx‖ = ‖x − Tx‖ = d(Ai,Ai+1). Since Ai+1 is convex set and X is uniformly convex Banach space, Tx = T p+1x. Consequently ‖T px − Tx‖ = ‖T px − T p+1x‖ ≤ ‖x − Tx‖ = d(Ai,Ai+1). Hence T px = x. Uniqueness follows as in Theorem 3.2. � c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 2 125 J. Maria Felicit and A. Anthony Eldred The following result on Geraghty contractive condition can be proved in a similar fashion. Theorem 3.6. Let A1,A2, . . . ,Ap be nonempty closed and convex subsets of a uniformly convex Banach space X and let S = {α : R+ → [0,1) : α(tn) → 1 ⇒ tn → 0}. Let T : ⋃p i=1 Ai → ⋃p i=1 Ai be a p-cyclic mapping such that ‖Tx − Ty‖ ≤ α(‖x − y‖)(‖x − y‖) + (1 − α(‖x − y‖))d(Ai,Ai+1) for all x ∈ Ai, y ∈ Ai+1, where α ∈ S. Then for each i,1 ≤ i ≤ p, there exists a unique best proximity point such that, for any x0 ∈ Ai, {T pnx0} converges to the best proximity point. Acknowledgements. The authors are very grateful to the reviewers for their comments which have been very useful when improving the manuscript. 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