() @ Appl. Gen. Topol. 18, no. 2 (2017), 377-390doi:10.4995/agt.2017.7452 c© AGT, UPV, 2017 Some fixed point theorems on non-convex sets M. Radhakrishnana, S. Rajeshb and Sushama Agrawala a Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai 600 005, India. (radhariasm@gmail.com,sushamamdu@gmail.com) b Department of Mathematics, Indian Institute of Technology, Tirupati 517 506, India. (srajeshiitmdt@gmail.com) Communicated by E. A. Sánchez-Pérez Abstract In this paper, we prove that if K is a nonempty weakly compact set in a Banach space X, T : K → K is a nonexpansive map satisfying x+T x 2 ∈ K for all x ∈ K and if X is 3−uniformly convex or X has the Opial property, then T has a fixed point in K. 2010 MSC: 47H09; 47H10. Keywords: fixed points; nonexpansive mappings; T −regular sets; k−uniform convex Banach spaces; Opial property. 1. Introduction Let K be a nonempty subset of a Banach space X. A mapping T : K → K is said to be nonexpansive if ‖T x − T y‖ ≤ ‖x − y‖ for all x, y ∈ K. The following theorem was proved independently by Browder [2] and Göhde [8] in the setting of uniformly convex Banach spaces. Theorem 1.1 ([2]). Let K be a nonempty weakly compact convex subset of a uniformly convex Banach space X and T : K → K be a nonexpansive map. Then T has a fixed point in K. Using the notion of normal structure, Kirk [10] proved the following theorem which is more general than Theorem 1.1. Received 28 March 2017 – Accepted 08 June 2017 http://dx.doi.org/10.4995/agt.2017.7452 M. Radhakrishnan, S. Rajesh and Sushama Agrawal Theorem 1.2 ([10]). Let K be a nonempty weakly compact convex subset hav- ing normal structure in a Banach space X and T : K → K be a nonexpansive map. Then T has a fixed point in K. The convexity assumption cannot be dispense in the above theorems as can be seen from the following simple example. Let K = [−2, −1]∪ [1, 2] ⊆ R and T is a self map on K defined by T x = −x for all x ∈ K. Then T is nonexpansive, but T has no fixed points in K. This implies that nonexpansive map on a non-convex set in a Banach space need not have a fixed point. Motivated by Theorem 1.1 and Theorem 1.2, Veeramani [20] introduced the notion of T −regular set as follows: Let T be a self map on a nonempty subset K of a Banach space X. Then K is said to be a T −regular set if x+T x 2 ∈ K for all x ∈ K. Clearly, if K is a convex set and T : K → K, then K is T −regular. But a T −regular set need not be a convex set(see Example 3.2). Further, Veeramani [20] proved the following fixed point theorem. Theorem 1.3 ([20]). Let K be a nonempty weakly compact subset of a uni- formly convex Banach space X and T : K → K be a nonexpansive map. Fur- ther, assume that K is T −regular. Then T has a fixed point in K. Khan and Hussain [9] used the notion of T −regular sets to prove the ex- istence of fixed points for nonexpansive mappings in the setting of metrizable topological vector space. Also, Goebel and Schöneberg [6] proved the existence of fixed point for a nonexpansive map on certain nonconvex sets in a Hilbert space. Sullivan [18] introduced the concept of k−uniform convexity, k−UC in short, where k is any positive integer and proved that every k−uniformly convex Banach space has normal structure. Note that for k = 1, it is uniformly convex. Sullivan [18] observed that every k−UC Banach space is a (k + 1)−UC. But the converse is not true. For example, the Banach space lp,1(N) [1] for 1 < p < ∞ is 2−UC but not 1−UC where lp,1(N) is the lp(N) space with suitable renorm. Motivated by Theorem 1.2, Theorem 1.3 and the fact that k−UC Banach spaces have normal structure [18], we raise the following question: Does a nonexpansive map T on a nonempty weakly compact set K in a k−UC Banach space have a fixed point if x+T x 2 ∈ K for all x ∈ K? In this paper, we give an affirmative answer to the above question, if X is a 3−UC Banach space. For the proof of this result, Lemma 3.3 and Lemma 3.4 (the geometric inequality on k−UC Banach space) are crucial. In another direction, Opial [16] introduced a class of spaces for which the asymptotic center of a weakly convergent sequence coincides with the weak limit point of the sequence. Gossez and Lami Dozo [7] have observed that all such spaces have normal structure. Hence, in view of Kirk’s theorem, ev- ery nonempty weakly compact convex set in a Banach space which satisfy c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 378 Some fixed point theorems on non-convex sets Opial’s condition has fixed point property for a nonexpansive mapping. Re- cently, Suzuki [19] introduced a new class of mappings which also includes nonexpansive maps and proved that every nonempty weakly compact convex set in a Banach space which satisfy Opial’s condition also has fixed point prop- erty for all such maps. In this paper, we prove that if K is a nonempty weakly compact set in a Banach space X having the Opial property, T : K → K is a nonexpansive map and if K is T −regular set, then T has a fixed point point in K. Moreover, the Krasnoseleskii’s [12] iterated sequence {xn} where xn+1 = xn+T xn 2 for all n ∈ N and x1 ∈ K weakly converges to a fixed point. 2. Preliminaries Now, we give some basic definitions and results which are used in this paper. Let X be a Banach space. For a nonempty subset A of X, let co(A) = { n ∑ i=1 λixi : xi ∈ A, λi ≥ 0, for i = 1, 2, . . . , n and n ∑ i=1 λi = 1, n ∈ N } aff(A) = { n ∑ i=1 λixi : xi ∈ A, λi ∈ R, for i = 1, 2, . . . , n and n ∑ i=1 λi = 1, n ∈ N } The sets co(A) and aff(A) are called the convex hull and the affine hull of A respectively. A set A is affine if A = aff(A). Every affine set is a translation of a sub- space and the subspace is uniquely defined by the affine set. The dimension of an affine set is the dimension of the corresponding subspace. Further, the dimension of a convex set A is defined as the dimension of the smallest affine set which contains A. This shows that the dimension of co(A) is the dimension of aff(A). Sliverman [17] introduced the notion of volume of k + 1 vectors, denoted by V (x1, x2, . . . , xk+1), as follows: Given x1, x2, . . . , xk+1 ∈ X, V (x1, x2, . . . , xk+1) = 1 k! sup          ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ f1(x2 − x1) . . . f1(xk+1 − x1) f2(x2 − x1) . . . f2(xk+1 − x1) . .. . .. . .. fk(x2 − x1) . . . fk(xk+1 − x1) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ : f1, . . . , fk ∈ BX∗          By the consequences of Hahn-Banach theorem, V (x1, x2) = ‖x1 − x2‖ for any x1, x2 ∈ X. Note that V (x1, x2, . . . , xk+1) = 0 iff the dimension of the convex hull of {x1, x2, . . . , xk+1} does not exceed k − 1. Using the notion of volume of k+1 vectors, Sullivan [18] defined the concept of k−uniform convexity. We put µ (k) X = sup{V (x1, . . . , xk+1) : x1, . . . , xk+1 ∈ BX}. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 379 M. Radhakrishnan, S. Rajesh and Sushama Agrawal Definition 2.1 ([18]). The modulus of k−convexity is defined as δ (k) X (ǫ) = inf { 1 − 1 k + 1 ∥ ∥ ∥ ∥ ∥ k+1 ∑ i=1 xi ∥ ∥ ∥ ∥ ∥ : x1, . . . , xk+1 ∈ BX and V (x1, . . . , xk+1) ≥ ǫ } where ǫ ∈ [0, µ (k) X ). A Banach space X is said to be k−uniformly convex if δ (k) X (ǫ) > 0 for every 0 < ǫ < µ (k) X . Note that all Banach spaces of dimension less than k + 1 are k−UC. For more information on k−UC, one can refer to [11, 14, 15]. Lim [13] proved the continuity of modulus δ (k) X of k−convexity using the following inequality. Theorem 2.2 ([13]). Let X be a Banach space and k be any positive integer. For every 0 < ǫ1 < c < ǫ2 < µ (k) X , δ (k) X (c) − δ (k) X (ǫ1) c − ǫ1 ≤ 1 k(ǫ 1/k 2 − ǫ 1/k 1 )ǫ 1−1/k 1 Corollary 2.3 ([13]). Let X be a Banach space. Then δ (k) X (ǫ) is continuous on [0, µ (k) X ). Definition 2.4 ([16]). A Banach space X is said to have the Opial property if {xn} is a weakly convergent sequence in X with limit z, then lim inf n→∞ ‖xn − z‖ < lim inf n→∞ ‖xn − y‖ for all y ∈ X with y 6= z. It is known that [5] Hilbert spaces, finite dimensional Banach spaces and lp(N) (1 < p < ∞) have the Opial property. Edelstein [3] introduced the notion of asymptotic center as follows: Definition 2.5 ([3]). Let K be a nonempty subset of a Banach space X and {xn} be a bounded sequence in X. For each x ∈ X, define r(x) = lim sup n→∞ ‖x − xn‖. The number r = inf x∈K r(x) and the set A(K, {xn}) = {x ∈ K : r(x) = r} are called the asymptotic radius and asymptotic center of {xn} with respect to K respectively. We use the next lemma in the sequel, which is proved by Goebel and Kirk [4]. Lemma 2.6 ([4]). Let {zn} and {wn} be bounded sequences in a Banach space X and let λ ∈ (0, 1). Suppose that zn+1 = λwn + (1 − λ)zn and ‖wn+1 − wn‖ ≤ ‖zn+1 − zn‖ for all n ∈ N. Then lim n→∞ ‖wn − zn‖ = 0. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 380 Some fixed point theorems on non-convex sets 3. Main results 3.1. 3−UC Banach spaces. In this section, we first give the convergence theorem for a nonexpansive map T defined on a compact T −regular set in a Banach space X. Also, we prove the existence of fixed points for a nonexpansive map T defined on a weakly compact T −regular set in a 3−UC Banach space X. Theorem 3.1. Let K be a nonempty compact subset of a Banach space X and T : K → K be a nonexpansive map. Further, assume that K is T −regular. Define a sequence {xn} in K by xn+1 = xn+T xn 2 for n ∈ N and x1 ∈ K. Then T has a fixed point in K and {xn} strongly converges to a fixed point of T. Proof. Since xn+1 = xn+T xn 2 for n ∈ N, by Lemma 2.6, we have lim n→∞ ‖xn − T xn‖ = 0. Since K is compact and {xn} ⊆ K, there exists a subsequence {xnk} of {xn} and z ∈ K such that {xnk } converges to z. Now, by the continuity of T , {T xnk} converges to T z. But, note that lim k→∞ ‖xnk − T xnk‖ = 0. Hence {xnk} also converges to T z. This implies that T z = z. Also, note that {‖xn − z‖} is a decreasing sequence. For, ‖xn+1 − z‖ ≤ 1 2 ‖xn − z‖ + 1 2 ‖T xn − z‖ ≤ ‖xn − z‖, for all n ∈ N Therefore {xn} converges to z, as {xnk} converges to z in norm. � Example 3.2. Let K = {(x, 0, 1 2n ), (0, y, 1 2n ), (x, x, 1 2n ), (x, 0, 0), (0, y, 0), (x, x, 0) : 0 ≤ x, y ≤ 1 and n ∈ N} be a subset of (R3, ‖.‖2). Define a map T : K → K by T (x, y, z) = (y, x, 0) for all (x, y, z) ∈ K. It is easy to see that K is T −regular. Also, note that T is nonexpansive. For, let x = (x1, y1, z1), y = (x2, y2, z2) ∈ K. Then ‖T x − T y‖2 = ‖(y1 − y2, x1 − x2, 0)‖2 ≤ ‖(x1 − x2, y1 − y2, z1 − z2)‖2 = ‖x − y‖2 By Theorem 3.1, T has a fixed point in K, since K is compact and T −regular. Also, note that Fix(T ) = {(x, x, 0) : 0 ≤ x ≤ 1}. Lemma 3.3. Let K be a nonempty weakly compact subset of a Banach space X and T : K → K be a nonexpansive map. Further, assume that K is T −regular. Define a sequence {xn} in K by xn+1 = xn+T xn 2 for n ∈ N and x1 ∈ K. Then the asymptotic center A(K, {xn}) of {xn} with respect to K is also a nonempty weakly compact T −regular subset of K. Moreover, if K is a minimal weakly compact T −regular set, then A(K, {xn}) = K. Proof. Since r(x) = lim sup n→∞ ‖x−xn‖ is a weakly lower semicontinuous function on X and K is weakly compact, A(K, {xn}) = {x ∈ K : r(x) = inf y∈K r(y) = r} is nonempty. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 381 M. Radhakrishnan, S. Rajesh and Sushama Agrawal Also {x ∈ X : r(x) ≤ inf y∈K r(y)} is a weakly closed set, this implies that A(K, {xn}) = {x ∈ X : r(x) ≤ inf y∈K r(y)} ∩ K is a weakly closed set. Moreover, since T is nonexpansive and lim n→∞ ‖xn − T xn‖ = 0, A(K, {xn}) is T −invariant. Now, it is claimed that A(K, {xn}) is a T −regular set. Let x ∈ A(K, {xn}). Then T x ∈ A(K, {xn}) and ∥ ∥ ∥ ∥ x + T x 2 − xn ∥ ∥ ∥ ∥ ≤ 1 2 ‖x − xn‖ + 1 2 ‖T x − xn‖. This implies that lim sup n→∞ ∥ ∥ ∥ ∥ x + T x 2 − xn ∥ ∥ ∥ ∥ = r. Therefore x+T x 2 ∈ A(K, {xn}). Hence A(K, {xn}) is a nonempty weakly com- pact T −regular subset of K. Suppose that K is a nonempty minimal weakly compact T −regular set. Then A(K, {xn}) = K, as A(K, {xn}) ⊆ K is also a nonempty weakly compact T −regular set. � Lemma 3.4. Let X be a k−UC Banach space, for some k ∈ N and x1, x2, . . . , xk+1 ∈ BX such that V (x1, x2, . . . , xk+1) = ǫ > 0. Then ‖t1x1 + t2x2 + · · · + tk+1xk+1‖ ≤ 1 − (k + 1) min{t1, t2, . . . , tk+1}δ (k) X (ǫ), where k+1 ∑ i=1 ti = 1, ti ≥ 0 for i = 1, 2, . . . , k + 1. Proof. Without loss of generality, we can assume that t1 = min{t1, t2, . . . , tk+1}. ‖t1x1 + t2x2 + · · · + tk+1xk+1‖ = ‖t1(x1 + · · · + xk+1) + (t2 − t1)x2 + (t3 − t1)x3 + · · · + (tk+1 − t1)xk+1‖ ≤ (k + 1)t1 ∥ ∥ ∥ ∥ x1 + x2 + · · · + xk+1 k + 1 ∥ ∥ ∥ ∥ + (t2 − t1)‖x2‖ +(t3 − t1)‖x3‖ + · · · + (tk+1 − t1)‖xk+1‖ ≤ (k + 1)t1(1 − δ (k) X (ǫ)) + t2 + t3 + · · · + tk+1 − kt1 = (k + 1)t1 − (k + 1)t1δ (k) X (ǫ) + 1 − (k + 1)t1 = 1 − (k + 1)t1δ (k) X (ǫ) Hence ‖t1x1+t2x2+· · ·+tk+1xk+1‖ ≤ 1−(k+1) min{t1, t2, . . . , tk+1}δ (k) X (ǫ). � Remark 3.5. Now from Lemma 3.4, we have: (1) If k = 2 and t1 = t2 = 1 4 , then ∥ ∥ ∥ x1 4 + x2 4 + x3 2 ∥ ∥ ∥ ≤ 1 − 3 4 δ (2) X (ǫ). c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 382 Some fixed point theorems on non-convex sets (2) If k = 3 and t1 = t2 = 1 8 , t3 = 1 4 then ∥ ∥ ∥ x1 8 + x2 8 + x3 4 + x4 2 ∥ ∥ ∥ ≤ 1 − 1 2 δ (3) X (ǫ). (3) If k = 3 and t1 + t2 + t3 = 1 2 , then ∥ ∥ ∥ ∥ t1x1 + t2x2 + t3x3 + 1 2 x4 ∥ ∥ ∥ ∥ ≤ 1 − 4 min{t1, t2, t3}δ (3) X (ǫ). We obtain the intuitive and geometric idea for the proof of our main result Theorem 3.7 from the proof technique of the following theorem. Theorem 3.6. Let K be a nonempty weakly compact subset of a 2−uniformly convex Banach space X and T : K → K be a nonexpansive map. Further, assume that K is T −regular. Then T has a fixed point in K. Proof. Define F = {F ⊆ K : F is nonempty weakly compact T −regular set} . It is easy to see that the set inclusion ⊆, defines a partial order relation on F. By Zorn’s lemma, we get a minimal element in F. Without loss of generality, we can assume that K is minimal in F. Let x1 ∈ K and define xk+1 = xk+T xk 2 ∈ K, for k ∈ N. By Lemma 3.3, we have A(K, {xk}) = K i.e., r(x) = lim sup k→∞ ‖x − xk‖ = r, for all x ∈ K. Note that r = 0 if and only if K is singleton. For, if r = 0, then lim sup k→∞ ‖x − xk‖ = 0, for all x ∈ K. This gives {xk} converges to every point in K. Hence K is singleton. Conversely, suppose that K is singleton. Then it is easy to see that r = 0, as {xk} ⊆ K. We claim that r = 0. Suppose that r > 0. This implies that x 6= T x, for all x ∈ K. It is claimed that T xn ∈ aff{x1, T x1} for all n ∈ N. Suppose that there exists n ∈ N such that T xn /∈ aff{x1, T x1}. Without loss of generality, we can assume that T x2 /∈ aff{x1, T x1}. This gives {x1, T x1, T x2} is affinely independent and dim(co{x1, T x1, T x2} = 2. Hence V (x1, T x1, T x2) = ǫ for some ǫ > 0. Since X is 2−UC and δ (2) X is continuous, we have lim ρ→0 (r + ρ) ( 1 − 3 4 δ (2) X ( ǫ (r + ρ)2 )) = r ( 1 − 3 4 δ (2) X ( ǫ r2 ) ) < r This implies that there is a ρ0 > 0 such that (r + ρ0) ( 1 − 3 4 δ (2) X ( ǫ (r + ρ0)2 )) < r. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 383 M. Radhakrishnan, S. Rajesh and Sushama Agrawal Since A(K, {xk}) = K and for this ρ0 > 0, there exists N ∈ N such that for k ≥ N, we have ‖x1 − xk‖ ≤ r + ρ0 ‖T x1 − xk‖ ≤ r + ρ0 ‖T x2 − xk‖ ≤ r + ρ0 As X is 2−UC, we have ∥ ∥ ∥ ∥ x1 + T x1 + T x2 3 − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − δ (2) X ( ǫ (r + ρ0)2 )) , for k ≥ N. Note that x3 = x1 4 + T x1 4 + T x2 2 ∈ co{x1, T x1, T x2} and by Lemma 3.4, we get ‖x3 − xk‖ = ∥ ∥ ∥ ∥ x1 4 + T x1 4 + T x2 2 − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − 3 4 δ (2) X ( ǫ (r + ρ0)2 )) , for k ≥ N. This implies that r(x3) = lim sup k→∞ ‖x3 − xk‖ ≤ (r + ρ0) ( 1 − 3 4 δ (2) X ( ǫ (r + ρ0)2 )) < r. This gives a contradiction to A(K, {xk}) = K. Therefore T xn ∈ aff{x1, T x1}, for all n ∈ N. This implies that {xn} ⊆ aff{x1, T x1}. Since {xn} is a bounded sequence and dim(aff{x1, T x1}) = 1, so it has a convergent subsequence say {xnj } of {xn} and z ∈ K such that xnj → z as j → ∞. Since lim j→∞ ‖xnj − T xnj ‖ = 0 and T is nonexpansive, T z = z. Hence r = 0. This implies that K is singleton and T has a fixed point in K. � Next we prove the main result of this paper. Theorem 3.7. Let K be a nonempty weakly compact subset of a 3−uniformly convex Banach space X and T : K → K be a nonexpansive map. Further, assume that K is T −regular. Then T has a fixed point in K. Proof. Note that by using Zorn’s lemma, we get a nonempty minimal weakly compact T −regular subset of K. Without loss of generality, we can assume that K is a nonempty minimal weakly compact T −regular set. Let x1 ∈ K and define xk+1 = xk+T xk 2 ∈ K, for k ∈ N. By Lemma 3.3, we have A(K, {xk}) = K i.e., r(x) = lim sup k→∞ ‖x − xk‖ = r, for all x ∈ K. We claim that r = 0. Suppose that r > 0. This implies that x 6= T x, for all x ∈ K. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 384 Some fixed point theorems on non-convex sets Suppose that for every n ∈ N, T xn ∈ aff{x1, T x1}. Then {xn} is a bounded sequence in aff{x1, T x1}, as K is bounded. Hence {xn} has a convergent subsequence. This implies that T has a fixed point in K. Suppose that there exists n ∈ N such that T xn 6∈ aff{x1, T x1}. Without loss of generality, we can assume that T x2 6∈ aff{x1, T x1}. It is claimed that T xn ∈ aff{x1, T x1, T x2}, for all n ∈ N. We use mathematical induction to prove our claim. Case 1. It is claimed that T x3 ∈ aff{x1, T x1, T x2}. Suppose that T x3 6∈ aff{x1, T x1, T x2}. This gives {x1, T x1, T x2, T x3} is affinely independent and dim(co{x1, T x1, T x2, T x3}) = 3. Hence V (x1, T x1, T x2, T x3) = ǫ, for some ǫ > 0. Since X is 3−UC and δ (3) X is continuous, there is a ρ0 > 0 such that (r + ρ0) ( 1 − 1 2 δ (3) X ( ǫ (r + ρ0)3 )) < r. Since A(K, {xk}) = K, there exists N ∈ N such that for k ≥ N, we have ‖x1 − xk‖ ≤ r + ρ0 ‖T x1 − xk‖ ≤ r + ρ0 ‖T x2 − xk‖ ≤ r + ρ0 ‖T x3 − xk‖ ≤ r + ρ0 As X is 3−UC, we have for k ≥ N ∥ ∥ ∥ ∥ x1 + T x1 + T x2 + T x3 4 − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − δ (3) X ( ǫ (r + ρ0)3 )) . Note that x4 = x3+T x3 2 = x2+T x2 4 +T x3 2 = x1 8 +T x1 8 +T x2 4 +T x3 2 ∈ co{x1, T x1, T x2, T x3}. Now, by Lemma 3.4, we get ‖x4 − xk‖ = ∥ ∥ ∥ ∥ x1 8 + T x1 8 + T x2 4 + T x3 2 − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − 1 2 δ (3) X ( ǫ (r + ρ0)3 )) , for k ≥ N. This implies that r(x4) = lim sup k→∞ ‖x4 − xk‖ ≤ (r + ρ0) ( 1 − 1 2 δ (3) X ( ǫ (r + ρ0)3 )) < r. This gives a contradiction to A(K, {xk}) = K. Hence T x3 ∈ aff{x1, T x1, T x2}. Case 2. It is claimed that T x4 ∈ aff{x1, T x1, T x2}. Suppose that T x4 /∈ aff{x1, T x1, T x2}. This gives {x1, T x1, T x2, T x4} is affinely independent and dim(co{x1, T x1, T x2, T x4}) = 3. Since T x3 ∈ aff{x1, T x1, T x2}, we have the following cases: c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 385 M. Radhakrishnan, S. Rajesh and Sushama Agrawal (a). T x3 ∈ aff{x2, T x2} (b). T x3 6∈ aff{x2, T x2}. Subcase 2(a). Suppose that T x3 ∈ aff{x2, T x2}. Then T x3 = (1 − µ3)x2 + µ3T x2, for some µ3 ∈ R. By the nonexpansiveness of T, we have 1 2 ‖T x2 − x2‖ = ‖x3 − x2‖ ≥ ‖T x3 − T x2‖ = |1 − µ3|‖T x2 − x2‖. This gives 1 2 ≤ µ3 ≤ 3 2 . Note that µ3 6= 1 2 . For, if µ3 = 1 2 , then T x3 = x3. Now x4 = x3 + T x3 2 = 1 2 ( x2 + T x2 2 + T x3 ) = x2 4 + 1 4 ( T x3 − (1 − µ3)x2 µ3 ) + T x3 2 = ( 2µ3 − 1 4µ3 ) x2 + ( 2µ3 + 1 4µ3 ) T x3 = ( 2µ3 − 1 8µ3 ) x1 + ( 2µ3 − 1 8µ3 ) T x1 + ( 2µ3 + 1 4µ3 ) T x3 = t1x1 + t1T x1 + (1 − 2t1)T x3 where t1 = 2µ3 − 1 8µ3 . Since µ3 > 1 2 , we have t1 > 0 and 1 − 2t1 > 0. This gives x4 lies in the interior of co{x1, T x1, T x3}. Since {x1, T x1, T x2, T x4} is affinely independent and T x3 ∈ aff{x2, T x2}, we have {x1, T x1, T x3, T x4} is affinely independent and dim(co{x1, T x1, T x3, T x4}) = 3. Hence V (x1, T x1, T x3, T x4) = ǫ for some ǫ > 0. Since δ (3) X is continuous and X is 3−UC, there is a ρ0 > 0 such that (r + ρ0) ( 1 − 2 min{t1, 1 − 2t1}δ (3) X ( ǫ (r + ρ0)3 )) < r As A(K, {xk}) = K, there exist N ∈ N such that for k ≥ N, we have ∥ ∥ ∥ ∥ x1 + T x1 + T x3 + T x4 4 − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − δ (3) X ( ǫ (r + ρ0)3 )) . Note that x5 = x4+T x4 2 = 1 2 (t1x1 + t1T x1 + (1 − 2t1)T x3 + T x4) . This implies that x5 lies in the interior of co{x1, T x1, T x3, T x4}. Now, by Lemma 3.4, for k ≥ N we have ‖x5 − xk‖ = ∥ ∥ ∥ ∥ 1 2 (t1x1 + t1T x1 + (1 − 2t1)T x3 + T x4) − xk ∥ ∥ ∥ ∥ ≤ (r + ρ0) ( 1 − 2 min{t1, 1 − 2t1}δ (3) X ( ǫ (r + ρ0)3 )) . This implies that r(x5) = lim sup k→∞ ‖x5 − xk‖ ≤ (r + ρ0) ( 1 − 2 min{t1, 1 − 2t1}δ (3) X ( ǫ (r + ρ0)3 )) < r. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 386 Some fixed point theorems on non-convex sets This gives a contradiction to A(K, {xk}) = K. Hence T x4 ∈ aff{x1, T x1, T x2}. Subcase 2(b). Suppose that T x3 /∈ aff{x2, T x2}. Then {x2, T x2, T x3} is affinely independent and dim(co{x2, T x2, T x3}) = 2. Since T x3 ∈ aff{x1, T x1, T x2} and T x3 6∈ aff{x2, T x2}, we have T x3 = ax1 + bT x1 + (1 − (a + b))T x2, for a, b ∈ R with a 6= b. Since {x1, T x1, T x2, T x4} is affinely independent and T x3 = ax1 + bT x1 + (1 − (a + b))T x2, we have {x2, T x2, T x3, T x4} is affinely independent and dim(co{x2, T x2, T x3, T x4}) = 3. This implies that V (x2, T x2, T x3, T x4) = ǫ, for some ǫ > 0. Therefore by case 1, we get r(x5) < r. This gives a contradiction to A(K, {xk}) = K. Hence T x4 ∈ aff{x1, T x1, T x2}. Case 3. Now, we assume that T xn ∈ aff{x1, T x1, T x2}, for 1 ≤ n ≤ m − 1. To prove that T xm ∈ aff{x1, T x1, T x2}. Suppose not. Then {x1, T x1, T x2, T xm} is affinely independent. Since T xk ∈ aff{x1, T x1, T x2} for 3 ≤ k ≤ m − 1, we have the following cases: (a). T xk ∈ aff{x2, T x2} for k = 3, 4, . . . , m − 1 (b). T xk 6∈ aff{x2, T x2} for some k ∈ {3, 4, . . . , m − 1}. Subcase 3(a). Suppose that T xk ∈ aff{x2, T x2} for 3 ≤ k ≤ m − 1. Then xk ∈ aff{x2, T x2} for 3 ≤ k ≤ m, as xk = xk−1+T xk−1 2 . Let xk = (1 − λk)x2 + λkT x2 for some λk ∈ R, 2 ≤ k ≤ m and T xk = (1 − µk)x2 + µkT x2 for some µk ∈ R, 2 ≤ k ≤ m − 1. Note that λk+1 = λk+µk 2 , for 2 ≤ k ≤ m − 1, as xk+1 = xk+T xk 2 . Hence λ3 = 1 2 , as λ2 = 0, µ2 = 1. Since we work with the aff{x2, T x2}, we can identify the aff{x2, T x2} with the real line R by assuming x2 = 0 and T x2 = 1. In this way, we get that xk = λk and T xk = µk for 2 ≤ k ≤ m − 1. As T xk 6= xk, we have λk 6= µk and λk 6= λk+1 for 2 ≤ k ≤ m − 1. Note that, from case 2(a), we have λ3 < µ3. This implies that λ3 < λ4 < µ3, as λk+1 = λk+µk 2 . It is claimed that λk < λk+1 and λk < µk, for 4 ≤ k ≤ m − 1. Since T is nonexpansive, we have |µ4 − µ3|‖x2 − T x2‖ = ‖T x3 − T x4‖ ≤ ‖x3 − x4‖ = (λ4 − λ3)‖x2 − T x2‖. This implies that −λ4 + λ3 ≤ µ4 − µ3 ≤ λ4 − λ3. Now, since λ4 = λ3+µ3 2 , we have λ4 < µ4. This gives λ4 < λ5 < µ4. Continuing in this way, we get λk < λk+1 < µk for 3 ≤ k ≤ m − 1. Hence 0 = λ2 < λ3 < λ4 < · · · < λm−1 < λm < µm−1. This implies that λk lies in the interior of co{λ2, µm−1} for 3 ≤ k ≤ m. Hence xk lies in the interior of co{x2, T xm−1} for 3 ≤ k ≤ m. This implies that xm lies in the interior of co{x1, T x1, T xm−1}, as x2 = x1+T x1 2 . Now, since aff{x1, T x1, T x2} =aff{x1, T x1, T xm−1} and T xm 6∈ aff{x1, T x1, T x2}, we have {x1, T x1, T xm−1, T xm} is affinely independent and dim(co{x1, T x1, T xm−1, T xm}) = 3. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 387 M. Radhakrishnan, S. Rajesh and Sushama Agrawal Hence xm+1 lies in the interior of co{x1, T x1, T xm−1, T xm}, as xm+1 = xm+T xm 2 . Now, by using the arguments as in case 2(a), it is easy to see that r(xm+1) = lim sup k→∞ ‖xm+1 − xk‖ < r. This gives a contradiction to A(K, {xk}) = K. Hence T xm ∈ aff{x1, T x1, T x2}. Subcase 3(b). Suppose that there exists k ∈ N such that 3 ≤ k ≤ m − 1 and T xk 6∈ aff{x2, T x2}. Let k0 be the least integer satisfying T xk0 6∈ aff{x2, T x2}. This implies T x3, T x4, . . . , T xk0−1 ∈ aff{x2, T x2}. Then {xk0−1, T xk0−1, T xk0} is affinely independent and aff{xk0−1, T xk0−1, T xk0} = aff{x1, T x1, T x2}. Now, we consider the set {xk0−1, T xk0−1, T xk0}. Suppose that T xk ∈ aff{xk0, T xk0} for k0 + 1 ≤ k ≤ m − 1. Then using the arguments as in case 3(a), it is easy to see that xm+1 lies in the interior of co{xk0−1, T xk0−1, T xm−1, T xm} and {xk0−1, T xk0−1, T xm−1, T xm} is affinely independent. Now, it is apparent that r(xm+1) < r, as X is 3−UC. This gives a contradiction to A(K, {xk}) = K. Hence T xm ∈ aff{x1, T x1, T x2}. Suppose that there exists k ∈ N such that k0 + 1 ≤ k ≤ m − 1 and T xk 6∈ aff{xk0, T xk0}. Let k1 be the least integer satisfying T xk1 6∈ aff{xk0, T xk0}. This implies that T xk0+1, T xk0+2, . . . , T xk1−1 ∈ aff{xk0, T xk0}. Then {xk1−1, T xk1−1, T xk1} is affinely independent and aff{xk1−1, T xk1−1, T xk1} = aff{xk0−1, T xk0−1, T xk0}. Now, we consider the set {xk1−1, T xk1−1, T xk1}. Continuing in this way, we can find n0 is the largest integer such that k1 ≤ n0 ≤ m − 1 and T xn0 6∈ aff{xn0−1, T xn0−1}. This implies that T xn ∈ aff{xn0, T xn0} for n0 ≤ n ≤ m − 1. Then using the arguments as in case 3(a), it is easy to see that xm+1 lies in the interior of co{xn0−1, T xn0−1, T xm−1, T xm} and {xn0−1, T xn0−1, T xm−1, T xm} is affinely independent. Now, it is apparent that r(xm+1) < r, as X is 3−UC. This gives a contradiction to A(K, {xk}) = K. Hence T xm ∈ aff{x1, T x1, T x2}. Hence, by mathematical indution T xn ∈ aff{x1, T x1, T x2}, for all n ∈ N. This implies that {xn} ⊆ aff{x1, T x1, T x2}. Since {xn} is a bounded sequence and dim(aff{x1, T x1, T x2}) = 2, so it has a convergent subsequence i.e., there exists a subsequence {xnj } of {xn} and z ∈ K such that xnj → z as j → ∞. Since lim j→∞ ‖xnj − T xnj ‖ = 0 and T is nonexpansive, we have T z = z. Hence r = 0. This implies that K is singleton and T has a fixed point in K. � Remark 3.8. In the light of Theorem 3.6 and Theorem 3.7, it is natural to expect that if K is a nonempty weakly compact subset of a k−UC Banach space X, for k > 3 and if T : K → K is a nonexpansive map satisfying x+T x 2 ∈ K for all x ∈ K, then T has a fixed point in K. c© AGT, UPV, 2017 Appl. Gen. Topol. 18, no. 2 388 Some fixed point theorems on non-convex sets 3.2. Banach space with Opial property. Theorem 3.9. Let K be a nonempty weakly compact subset of a Banach space X having the Opial property and T : K → K be a nonexpansive map. Further, assume that K is T −regular. Define a sequence {xn} in K by xn+1 = xn+T xn 2 for n ∈ N and x1 ∈ K. Then T has a fixed point in K and {xn} converges weakly to a fixed point of T. Proof. By Lemma 2.6, we have lim n→∞ ‖xn − T xn‖ = 0. Since K is weakly com- pact, there exists a subsequence {xnk} of {xn} and z ∈ K such that {xnk } converges weakly to z. Also, we have ‖xnk − T z‖ ≤ ‖xnk − T xnk‖ + ‖T xnk − T z‖, for all k ∈ N. Hence lim inf k→∞ ‖xnk − T z‖ ≤ lim inf k→∞ ‖xnk − z‖. Since X has the Opial property, we obtain T z = z. Also note that, {‖xn − z‖} is a decreasing sequence. It is claimed that {xn} converges weakly to z. Suppose that {xn} does not converge weakly to z. Then there exists a subsequence {xn̂j } of {xn} which does not converge weakly to z. Since K is weakly compact and {xn̂j } ⊆ K, there exists a subse- quence of {xn̂j } whose weak limit is w ∈ K and z 6= w. Without loss of generality, we can assume that {xn̂j } converges weakly to w. It is easy to see that T w = w, as lim j→∞ ‖xn̂j −T xn̂j ‖ = 0. Also, it is apparent that {‖xn − w‖} is a decreasing sequence, as T w = w. Since X has the Opial property, {xn̂j } converges weakly to w and {xnk } converges weakly to z, we have lim n→∞ ‖xn − z‖ = lim k→∞ ‖xnk − z‖ < lim k→∞ ‖xnk − w‖ = lim n→∞ ‖xn − w‖ = lim j→∞ ‖xn̂j − w‖ < lim j→∞ ‖xn̂j − z‖ = lim n→∞ ‖xn − z‖. 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