Some fixed point theorems on non-convex sets

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1 @ Appl Gen Topol 18, no 017), doi:104995/agt c AGT, UPV, 017 Some fixed point theorems on non-convex sets M Radhakrishnan a, S Rajesh b and Sushama Agrawal a a Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai , India radhariasm@gmailcom,sushamamdu@gmailcom) b Department of Mathematics, Indian Institute of Technology, Tirupati , India srajeshiitmdt@gmailcom) 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, T : K K is a nonexpansive map satisfying x+tx K for all x K and if is 3 uniformly convex or has the Opial property, then T has a fixed point in K 010 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 A mapping T : K K is said to be nonexpansive if Tx Ty x y for all x,y K The following theorem was proved independently by Browder [] and Göhde [8] in the setting of uniformly convex Banach spaces Theorem 11 []) Let K be a nonempty weakly compact convex subset of a uniformly convex Banach space 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 11 Received 8 March 017 Accepted 08 June 017

2 M Radhakrishnan, S Rajesh and Sushama Agrawal Theorem 1 [10]) Let K be a nonempty weakly compact convex subset having normal structure in a Banach space 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 = [, 1] [1,] R and T is a self map on K defined by Tx = 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 11 and Theorem 1, Veeramani [0] introduced the notion of T regular set as follows: Let T be a self map on a nonempty subset K of a Banach space Then K is said to be a T regular set if x+tx 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 setsee Example 3) Further, Veeramani [0] proved the following fixed point theorem Theorem 13 [0]) Let K be a nonempty weakly compact subset of a uniformly convex Banach space and T : K K be a nonexpansive map Further, 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 existence 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 l p,1 N) [1] for 1 < p < is UC but not 1 UC where l p,1 N) is the l p N) space with suitable renorm Motivated by Theorem 1, Theorem 13 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+tx K for all x K? In this paper, we give an affirmative answer to the above question, if is a 3 UC Banach space For the proof of this result, Lemma 33 and Lemma 34 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, every nonempty weakly compact convex set in a Banach space which satisfy c AGT, UPV, 017 Appl Gen Topol 18, no 378

3 Some fixed point theorems on non-convex sets Opial s condition has fixed point property for a nonexpansive mapping Recently, 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 property for all such maps In this paper, we prove that if K is a nonempty weakly compact set in a Banach space 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 [1] iterated sequence {x n } where x n+1 = xn+txn for all n N and x 1 K weakly converges to a fixed point Preliminaries Now, we give some basic definitions and results which are used in this paper Let be a Banach space For a nonempty subset A of, let { n } n coa) = λ ix i : x i A,λ i 0, for i = 1,,,n and λ i = 1,n N i=1 { n affa) = λ ix i : x i A,λ i R, for i = 1,,,n and i=1 i=1 } n λ i = 1,n N The sets coa) and affa) are called the convex hull and the affine hull of A respectively A set A is affine if A = affa) Every affine set is a translation of a subspace 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 coa) is the dimension of affa) Sliverman [17] introduced the notion of volume of k+1 vectors, denoted by Vx 1,x,,x k+1 ), as follows: Given x 1,x,,x k+1, f 1x x 1) f 1x k+1 x 1) Vx 1,x,,x k+1 ) = 1 k! sup f x x 1) f x k+1 x 1) : f 1,,f k B f k x x 1) f k x k+1 x 1) By the consequences of Hahn-Banach theorem, Vx 1,x ) = x 1 x for any x 1,x Note that Vx 1,x,,x k+1 ) = 0 iff the dimension of the convex hull of {x 1,x,,x k+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) = sup{vx 1,,x k+1 ) : x 1,,x k+1 B } i=1 c AGT, UPV, 017 Appl Gen Topol 18, no 379

4 M Radhakrishnan, S Rajesh and Sushama Agrawal Definition 1 [18]) The modulus of k convexity is defined as { } δ k) ) = inf 1 1 k+1 x i : x 1,,x k+1 B and Vx 1,,x k+1 ) k +1 where [0,µ k) ) i=1 A Banach space is said to be k uniformly convex if δ k) ) > 0 for every 0 < < µ k) 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) of k convexity using the following inequality Theorem [13]) Let be a Banach space and k be any positive integer For every 0 < 1 < c < < µ k), δ k) c) δk) 1) c 1 1 k 1/k 1/k 1 ) 1 1/k 1 Corollary 3 [13]) Let be a Banach space Then δ k) ) is continuous on [0,µ k) ) Definition 4 [16]) A Banach space is said to have the Opial property if {x n } is a weakly convergent sequence in with limit z, then for all y with y z lim inf n x n z < liminf n x n y It is known that [5] Hilbert spaces, finite dimensional Banach spaces and l p N) 1 < p < ) have the Opial property Edelstein [3] introduced the notion of asymptotic center as follows: Definition 5 [3]) Let K be a nonempty subset of a Banach space and {x n } be a bounded sequence in For each x, define rx) = limsup x n x n The number r = inf rx) and the set AK,{x n}) = {x K : rx) = r} x K are called the asymptotic radius and asymptotic center of {x n } with respect to K respectively We use the next lemma in the sequel, which is proved by Goebel and Kirk [4] Lemma 6 [4]) Let {z n } and {w n } be bounded sequences in a Banach space and let λ 0,1) Suppose that z n+1 = λw n +1 λ)z n and w n+1 w n z n+1 z n for all n N Then lim n w n z n = 0 c AGT, UPV, 017 Appl Gen Topol 18, no 380

5 Some fixed point theorems on non-convex sets 3 Main results 31 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 Banachspace Also, we provethe existence offixed points for a nonexpansive map T defined on a weakly compact T regular set in a 3 UC Banach space Theorem 31 Let K be a nonempty compact subset of a Banach space and T : K K be a nonexpansive map Further, assume that K is T regular Define a sequence {x n } in K by x n+1 = xn+txn for n N and x 1 K Then T has a fixed point in K and {x n } strongly converges to a fixed point of T Proof Since x n+1 = xn+txn for n N, by Lemma 6, we have lim x n n Tx n = 0 Since K is compact and {x n } K, there exists a subsequence {x nk } of {x n } and z K such that {x nk } converges to z Now, by the continuity of T, {Tx nk } converges to Tz But, note that lim x n k Tx nk = 0 Hence {x nk } also converges to Tz This implies that Tz = z Also, note that { x n z } is a decreasing sequence For, x n+1 z 1 x n z + 1 Tx n z x n z, for all n N Therefore {x n } converges to z, as {x nk } converges to z in norm Example3 LetK = {x,0, 1 ),0,y, 1 1 n ),x,x, n ),x,0,0),0,y,0),x,x,0) : n 0 x,y 1 and n N} be a subset of R 3, ) Define a map T : K K by Tx,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 = x 1,y 1,z 1 ),y = x,y,z ) K Then Tx Ty = y 1 y,x 1 x,0) x 1 x,y 1 y,z 1 z ) = x y ByTheorem31,T hasafixedpointink,sincek iscompactandt regular Also, note that FixT) = {x,x,0) : 0 x 1} Lemma 33 Let K be a nonempty weakly compact subset of a Banach space and T : K K be a nonexpansive map Further, assume that K is T regular Define a sequence {x n } in K by x n+1 = xn+txn for n N and x 1 K Then the asymptotic center AK,{x n }) of {x n } 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 AK,{x n }) = K Proof Since rx) = lim sup x x n is a weakly lower semicontinuous function n on and K is weakly compact, AK,{x n }) = {x K : rx) = inf ry) = r} y K is nonempty c AGT, UPV, 017 Appl Gen Topol 18, no 381

6 M Radhakrishnan, S Rajesh and Sushama Agrawal Also {x : rx) inf ry)} is a weakly closed set, this implies that y K ry)} K is a weakly closed set AK,{x n }) = {x : rx) inf y K Moreover, since T is nonexpansive and lim x n Tx n = 0, AK,{x n }) is n T invariant Now, it is claimed that AK,{x n }) is a T regular set Let x AK,{x n }) Then Tx AK,{x n }) and This implies that x+tx x n 1 x x n + 1 Tx x n lim sup n x+tx x n = r Therefore x+tx AK,{x n }) Hence AK,{x n }) is a nonempty weakly compact T regular subset of K Suppose that K is a nonempty minimal weakly compact T regular set Then AK,{x n }) = K, as AK,{x n }) K is also a nonempty weakly compact T regular set Lemma 34 Let be a k UC Banach space, for some k N and x 1,x,, x k+1 B such that Vx 1,x,,x k+1 ) = > 0 Then t 1 x 1 +t x + +t k+1 x k+1 1 k +1)min{t 1,t,,t k+1 }δ k) ), where k+1 i=1 t i = 1, t i 0 for i = 1,,,k +1 Proof Withoutlossofgenerality,wecanassumethatt 1 = min{t 1,t,,t k+1 } t 1 x 1 +t x + +t k+1 x k+1 = t 1 x 1 + +x k+1 )+t t 1 )x +t 3 t 1 )x 3 + +t k+1 t 1 )x k+1 x 1 +x + +x k+1 k +1)t 1 k +1 +t t 1 ) x +t 3 t 1 ) x 3 + +t k+1 t 1 ) x k+1 k +1)t 1 1 δ k) +t +t 3 + +t k+1 kt 1 = k +1)t 1 k +1)t 1 δ k) )+1 k +1)t 1 = 1 k +1)t 1 δ k) ) Hence t 1 x 1 +t x + +t k+1 x k+1 1 k+1)min{t 1,t,,t k+1 }δ k) ) Remark 35 Now from Lemma 34, we have: 1) If k = and t 1 = t = 1 4, then x x 4 + x δ) ) c AGT, UPV, 017 Appl Gen Topol 18, no 38

7 Some fixed point theorems on non-convex sets ) If k = 3 and t 1 = t = 1 8,t 3 = 1 4 then x x 8 + x x δ3) ) 3) If k = 3 and t 1 +t +t 3 = 1, then t 1x 1 +t x +t 3 x x 4 1 4min{t 1,t,t 3 }δ 3) ) We obtain the intuitive and geometric idea for the proof of our main result Theorem 37 from the proof technique of the following theorem Theorem 36 Let K be a nonempty weakly compact subset of a uniformly convex Banach space 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 x 1 K and define x k+1 = x k+tx k K, for k N By Lemma 33, we have AK,{x k }) = K ie, rx) = limsup x x k = r, for all x K Note that r = 0 if and only if K is singleton For, if r = 0, then limsup x x k = 0, for all x K This gives {x k } 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 {x k } K We claim that r = 0 Suppose that r > 0 This implies that x Tx, for all x K It is claimed that Tx n aff{x 1,Tx 1 } for all n N Suppose that there exists n N such that Tx n / aff{x 1,Tx 1 } Without loss of generality, we can assume that Tx / aff{x 1,Tx 1 } Thisgives{x 1,Tx 1,Tx }isaffinelyindependentanddimco{x 1,Tx 1,Tx } = Hence Vx 1,Tx 1,Tx ) = for some > 0 Since is UC and δ ) is continuous, we have limr +ρ) 1 3 ρ 0 4 δ) r+ρ) This implies that there is a ρ 0 > 0 such that = r δ) r +ρ 0 ) δ) r +ρ 0 ) < r r < r c AGT, UPV, 017 Appl Gen Topol 18, no 383

8 M Radhakrishnan, S Rajesh and Sushama Agrawal Since AK,{x k }) = K and for this ρ 0 > 0, there exists N N such that for k N, we have As is UC, we have x 1 +Tx 1 +Tx x k r+ρ 0 ) 3 x 1 x k r +ρ 0 Tx 1 x k r +ρ 0 Tx x k r +ρ 0 1 δ ) r+ρ 0 ), for k N Note that x 3 = x1 4 + Tx1 4 + Tx co{x 1,Tx 1,Tx } and by Lemma 34, we get x 3 x k = x Tx Tx x k r +ρ 0 ) δ) r+ρ 0 ), for k N This implies that rx 3 ) = limsup r+ρ 0 ) x 3 x k δ) r +ρ 0 ) This gives a contradiction to AK,{x k }) = K Therefore Tx n aff{x 1,Tx 1 }, for all n N This implies that {x n } aff{x 1,Tx 1 } Since {x n } is a bounded sequence and dimaff{x 1,Tx 1 }) = 1, so it has a convergent subsequence say {x nj } of {x n } and z K such that x nj z as j Since lim x n j Tx nj = 0 and T is nonexpansive, Tz = z Hence j 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 < r Theorem 37 Let K be a nonempty weakly compact subset of a 3 uniformly convex Banach space 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 x 1 K and define x k+1 = x k+tx k K, for k N By Lemma 33, we have AK,{x k }) = K ie, rx) = limsup x x k = r, for all x K We claim that r = 0 Suppose that r > 0 This implies that x Tx, for all x K c AGT, UPV, 017 Appl Gen Topol 18, no 384

9 Some fixed point theorems on non-convex sets Suppose that for every n N, Tx n aff{x 1,Tx 1 } Then {x n } is a bounded sequence in aff{x 1,Tx 1 }, as K is bounded Hence {x n } has a convergent subsequence This implies that T has a fixed point in K Suppose that there exists n N such that Tx n aff{x 1,Tx 1 } Without loss of generality, we can assume that Tx aff{x 1,Tx 1 } It is claimed that Tx n aff{x 1,Tx 1,Tx }, for all n N We use mathematical induction to prove our claim Case 1 It is claimed that Tx 3 aff{x 1,Tx 1,Tx } Suppose that Tx 3 aff{x 1,Tx 1,Tx } This gives {x 1,Tx 1,Tx,Tx 3 } is affinely independent and dimco{x 1,Tx 1, Tx,Tx 3 }) = 3 Hence Vx 1,Tx 1,Tx,Tx 3 ) =, for some > 0 Since is 3 UC and δ 3) is continuous, there is a ρ 0 > 0 such that r +ρ 0 ) 1 1 δ3) r +ρ 0 ) 3 < r Since AK,{x k }) = K, there exists N N such that for k N, we have x 1 x k r +ρ 0 Tx 1 x k r +ρ 0 Tx x k r +ρ 0 Tx 3 x k r +ρ 0 As is 3 UC, we have for k N x 1 +Tx 1 +Tx +Tx 3 x k r+ρ 0 ) 4 1 δ 3) Notethatx 4 = x3+tx3 Now, by Lemma 34, we get x 4 x k = x Tx Tx 4 + Tx 3 x k r +ρ 0 ) 1 1 δ3) r+ρ 0 ) 3 This implies that = x+tx 4 + Tx3 = x1 8 +Tx1 8 +Tx 4 +Tx3 rx 4 ) = limsup r+ρ 0 ) x 4 x k 1 1 δ3) r +ρ 0 ) 3 r+ρ 0 ) 3 co{x 1,Tx 1,Tx,Tx 3 },for k N < r This gives a contradiction to AK,{x k }) = K Hence Tx 3 aff{x 1,Tx 1,Tx } Case It is claimed that Tx 4 aff{x 1,Tx 1,Tx } Suppose that Tx 4 / aff{x 1,Tx 1,Tx } This gives {x 1,Tx 1,Tx,Tx 4 } is affinely independent and dimco{x 1,Tx 1, Tx,Tx 4 }) = 3 Since Tx 3 aff{x 1,Tx 1,Tx }, we have the following cases: c AGT, UPV, 017 Appl Gen Topol 18, no 385

10 M Radhakrishnan, S Rajesh and Sushama Agrawal a) Tx 3 aff{x,tx } b) Tx 3 aff{x,tx } Subcase a) Suppose that Tx 3 aff{x,tx } Then Tx 3 = 1 µ 3 )x + µ 3 Tx, for some µ 3 R By the nonexpansiveness of T, we have 1 Tx x = x 3 x Tx 3 Tx = 1 µ 3 Tx x This gives 1 µ 3 3 Note that µ 3 1 For, if µ 3 = 1, then Tx 3 = x 3 Now x 4 = x 3 +Tx 3 = 1 ) x +Tx +Tx 3 = x ) Tx3 1 µ 3 )x + Tx 3 4 µ 3 ) ) µ3 1 µ3 +1 = x + Tx 3 4µ 3 4µ 3 ) ) ) µ3 1 µ3 1 µ3 +1 = x 1 + Tx 1 + Tx 3 8µ 3 8µ 3 4µ 3 = t 1 x 1 +t 1 Tx 1 +1 t 1 )Tx 3 where t 1 = µ 3 1 8µ 3 Since µ 3 > 1, we have t 1 > 0 and 1 t 1 > 0 This gives x 4 lies in the interior of co{x 1,Tx 1,Tx 3 } Since{x 1,Tx 1,Tx,Tx 4 }isaffinelyindependentandtx 3 aff{x,tx },we have{x 1,Tx 1,Tx 3,Tx 4 }isaffinelyindependentanddimco{x 1,Tx 1,Tx 3,Tx 4 }) = 3 Hence Vx 1,Tx 1,Tx 3,Tx 4 ) = for some > 0 Since δ 3) is continuous and is 3 UC, there is a ρ 0 > 0 such that r+ρ 0 ) 1 min{t 1,1 t 1 }δ 3) r +ρ 0 ) 3 < r As AK,{x k }) = K, there exist N N such that for k N, we have x 1 +Tx 1 +Tx 3 +Tx 4 x k r+ρ 0 ) 1 δ 3) 4 r+ρ 0 ) 3 Note that x 5 = x4+tx4 = 1 t 1x 1 +t 1 Tx 1 +1 t 1 )Tx 3 +Tx 4 ) This implies that x 5 lies in the interior of co{x 1,Tx 1,Tx 3,Tx 4 } Now, by Lemma 34, for k N we have x 5 x k = 1 This implies that rx 5 ) = limsup t 1x 1 +t 1 Tx 1 +1 t 1 )Tx 3 +Tx 4 ) x k r+ρ 0 ) 1 min{t 1,1 t 1 }δ 3) r +ρ 0 ) 3 r +ρ 0 ) x 5 x k 1 min{t 1,1 t 1 }δ 3) r+ρ 0 ) 3 < r c AGT, UPV, 017 Appl Gen Topol 18, no 386

11 Some fixed point theorems on non-convex sets This gives a contradiction to AK,{x k }) = K Hence Tx 4 aff{x 1,Tx 1,Tx } Subcase b) Suppose that Tx 3 / aff{x,tx } Then {x,tx,tx 3 } is affinely independent and dimco{x,tx,tx 3 }) = Since Tx 3 aff{x 1,Tx 1,Tx } and Tx 3 aff{x,tx }, we have Tx 3 = ax 1 +btx 1 +1 a+btx, for a,b R with a b Since {x 1,Tx 1,Tx,Tx 4 } is affinely independent and Tx 3 = ax 1 +btx a + btx, we have {x,tx,tx 3,Tx 4 } is affinely independent and dimco{x,tx, Tx 3,Tx 4 }) = 3 This implies that Vx,Tx,Tx 3,Tx 4 ) =, for some > 0 Therefore by case 1, we get rx 5 ) < r ThisgivesacontradictiontoAK,{x k }) = KHenceTx 4 aff{x 1,Tx 1,Tx } Case 3 Now, we assume that Tx n aff{x 1,Tx 1,Tx }, for 1 n m 1 To prove that Tx m aff{x 1,Tx 1,Tx } Suppose not Then {x 1,Tx 1,Tx,Tx m } is affinely independent Since Tx k aff{x 1,Tx 1,Tx } for 3 k m 1, we have the following cases: a) Tx k aff{x,tx } for k = 3,4,,m 1 b) Tx k aff{x,tx } for some k {3,4,,m 1} Subcase 3a) Suppose that Tx k aff{x,tx } for 3 k m 1 Then x k aff{x,tx } for 3 k m, as x k = x k 1+Tx k 1 Let x k = 1 λ k )x + λ k Tx for some λ k R, k m and Tx k = 1 µ k )x +µ k Tx for some µ k R, k m 1 Note that λ k+1 = λ k+µ k, for k m 1, as x k+1 = x k+tx k Hence λ 3 = 1, as λ = 0, µ = 1 Since we work with the aff{x,tx }, we can identify the aff{x,tx } with the real line R by assuming x = 0 and Tx = 1 In this way, we get that x k = λ k and Tx k = µ k for k m 1 As Tx k x k, we have λ k µ k and λ k λ k+1 for k m 1 Note that, from case a), we have λ 3 < µ 3 This implies that λ 3 < λ 4 < µ 3, as λ k+1 = λ k+µ k It is claimed that λ k < λ k+1 and λ k < µ k, for 4 k m 1 Since T is nonexpansive, we have µ 4 µ 3 x Tx = Tx 3 Tx 4 x 3 x 4 = λ 4 λ 3 ) x Tx This implies that λ 4 +λ 3 µ 4 µ 3 λ 4 λ 3 Now, since λ 4 = λ3+µ3, we have λ 4 < µ 4 This gives λ 4 < λ 5 < µ 4 Continuing in this way, we get λ k < λ k+1 < µ k for 3 k m 1 Hence 0 = λ < λ 3 < λ 4 < < λ m 1 < λ m < µ m 1 This implies that λ k lies in the interior of co{λ,µ m 1 } for 3 k m Hence x k lies in the interior of co{x,tx m 1 } for 3 k m This implies that x m lies in the interior of co{x 1,Tx 1,Tx m 1 }, as x = x 1+Tx 1 Now,sinceaff{x 1,Tx 1,Tx } =aff{x 1,Tx 1,Tx m 1 }andtx m aff{x 1,Tx 1,Tx }, we have {x 1,Tx 1,Tx m 1,Tx m } is affinely independent and dimco{x 1,Tx 1, Tx m 1,Tx m }) = 3 c AGT, UPV, 017 Appl Gen Topol 18, no 387

12 M Radhakrishnan, S Rajesh and Sushama Agrawal Hence x m+1 lies in the interior of co{x 1,Tx 1,Tx m 1,Tx m }, as x m+1 = x m+tx m Now, by using the arguments as in case a), it is easy to see that rx m+1 ) = lim sup x m+1 x k < r ThisgivesacontradictiontoAK,{x k }) = KHenceTx m aff{x 1,Tx 1,Tx } Subcase 3b) Suppose that there exists k N such that 3 k m 1 and Tx k aff{x,tx } Let k 0 be the least integer satisfying Tx k0 aff{x,tx } This implies Tx 3,Tx 4,, Tx k0 1 aff{x,tx } Then {x k0 1,Tx k0 1,Tx k0 } is affinely independent and aff{x k0 1,Tx k0 1, Tx k0 } = aff{x 1,Tx 1,Tx } Now, we consider the set {x k0 1,Tx k0 1,Tx k0 } Suppose that Tx k aff{x k0,tx k0 } for k 0 +1 k m 1 Then using the arguments as in case 3a), it is easy to see that x m+1 lies in theinteriorofco{x k0 1,Tx k0 1,Tx m 1,Tx m }and{x k0 1,Tx k0 1,Tx m 1,Tx m } is affinely independent Now, it is apparent that rx m+1 ) < r, as is 3 UC ThisgivesacontradictiontoAK,{x k }) = KHenceTx m aff{x 1,Tx 1,Tx } Suppose that there exists k N such that k 0 +1 k m 1 and Tx k aff{x k0,tx k0 } Let k 1 be the least integer satisfying Tx k1 aff{x k0,tx k0 } This implies that Tx k0+1,tx k0+,,tx k1 1 aff{x k0,tx k0 } Then {x k1 1,Tx k1 1,Tx k1 } is affinely independent and aff{x k1 1,Tx k1 1, Tx k1 } = aff{x k0 1,Tx k0 1,Tx k0 } Now, we consider the set {x k1 1,Tx k1 1,Tx k1 } Continuing in this way, we can find n 0 is the largest integer such that k 1 n 0 m 1 and Tx n0 aff{x n0 1,Tx n0 1} This implies that Tx n aff{x n0,tx n0 } for n 0 n m 1 Then using the arguments as in case 3a), it is easy to see that x m+1 lies in the interior of co{x n0 1,Tx n0 1,Tx m 1,Tx m } and {x n0 1,Tx n0 1, Tx m 1,Tx m } is affinely independent Now, it is apparent that rx m+1 ) < r, as is 3 UC ThisgivesacontradictiontoAK,{x k }) = KHenceTx m aff{x 1,Tx 1,Tx } Hence, by mathematical indution Tx n aff{x 1,Tx 1,Tx }, for all n N This implies that {x n } aff{x 1,Tx 1,Tx } Since {x n } is a bounded sequence and dimaff{x 1,Tx 1,Tx }) =, so it has a convergent subsequence ie, there exists a subsequence {x nj } of {x n } and z K such that x nj z as j Since lim x n j Tx nj = 0 and T is nonexpansive, we have Tz = z Hence j r = 0 This implies that K is singleton and T has a fixed point in K Remark 38 In the light of Theorem 36 and Theorem 37, it is natural to expect that if K is a nonempty weakly compact subset of a k UC Banach space, for k > 3 and if T : K K is a nonexpansive map satisfying x+tx K for all x K, then T has a fixed point in K c AGT, UPV, 017 Appl Gen Topol 18, no 388

13 Some fixed point theorems on non-convex sets 3 Banach space with Opial property Theorem 39 Let K be a nonempty weakly compact subset of a Banach space having the Opial property and T : K K be a nonexpansive map Further, assume that K is T regular Define a sequence {x n } in K by x n+1 = xn+txn for n N and x 1 K Then T has a fixed point in K and {x n } converges weakly to a fixed point of T Proof By Lemma 6, we have lim n x n Tx n = 0 Since K is weakly compact, there exists a subsequence {x nk } of {x n } and z K such that {x nk } converges weakly to z Also, we have Hence x nk Tz x nk Tx nk + Tx nk Tz, for all k N lim inf x n k Tz liminf x n k z Since has the Opial property, we obtain Tz = z Also note that, { x n z } is a decreasing sequence It is claimed that {x n } converges weakly to z Suppose that {x n } does not converge weakly to z Then there exists a subsequence {x nj } of {x n } which does not converge weakly to z Since K is weakly compact and {x nj } K, there exists a subsequence of {x nj } whose weak limit is w K and z w Without loss of generality, we can assume that {x nj } converges weakly to w It is easy to see that Tw = w, as lim j x n j Tx nj = 0 Also, it is apparent that { x n w } is a decreasing sequence, as Tw = w Since has the Opial property, {x nj } converges weakly to w and {x nk } converges weakly to z, we have lim x n z = lim x n n k z < lim x n k w = lim x n w n = lim x nj w < lim j j x n j z = lim x n z n This is a contradiction Hence {x n } weakly converges to z Acknowledgements The authors thank the anonymous reviewer for the comments and suggestions Also, the authors thank Prof P Veeramani, Department of Mathematics, IIT Madras India) for the fruitful discussions regarding the subject matter of this paper The first author thank the University Grants Commission India), for providing financial support to carry out this research work in the form of Project fellow through Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai c AGT, UPV, 017 Appl Gen Topol 18, no 389

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