STRONG CONVERGENCE TO COMMON FIXED POINTS OF NONEXPANSIVE MAPPINGS WITHOUT COMMUTATIVITY ASSUMPTION YONGHONG YAO, RUDONG CHEN, AND HAIYUN ZHOU Received 11 June 2006; Revised 27 July 2006; Accepted 2 August 2006 We introduce an iteration scheme for nonexpansive mappings in a Hilbert space and prove that the iteration converges strongly to common fixed points of the mappings without commutativity assumption. Copyright © 2006 Yonghong Yao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. A mapping T of C into itself is said to be nonexpansive if Tx − T y ≤ x − y , (1.1) for each x, y ∈ C. For a mapping T of C into itself, we denote by F(T) the set of fixed points of T. We also denote by N and R+ the set of positive integers and nonnegative real numbers, respectively. Baillon [1] proved the first nonlinear ergodic theorem. Let C be a nonempty bounded convex closed subset of a Hilbert space H and letT be a nonexpansive mapping of C into itself. Then, for an arbitrary x ∈ C, {(1/(n + 1)) ni=0 T i x}∞ n=0 converges weakly to a fixed point of T. Wittmann [9] studied the following iteration scheme, which has first been considered by Halpern [3]: x0 = x ∈ C, xn+1 = αn+1 x + 1 − αn+1 Txn , (1.2) n ≥ 0, ∞ where a sequence {αn } in [0,1] is chosen so that limn→∞ αn = 0, ∞ n=1 αn = ∞, and n =1 |αn+1 − αn | < ∞; see also Reich [7]. Wittmann proved that for any x ∈ C, the sequence Hindawi Publishing Corporation Fixed Point Theory and Applications Volume 2006, Article ID 89470, Pages 1–8 DOI 10.1155/FPTA/2006/89470 2 Nonexpansive mappings without commutativity assumption {xn } defined by (1.2) converges strongly to the unique element Px ∈ F(T), where P is the metric projection of H onto F(T). Recall that two mappings S and T of H into itself are called commutative if ST = TS, (1.3) for all x, y ∈ H. Recently, Shimizu and Takahashi [8] have first considered an iteration scheme for two commutative nonexpansive mappings S and T and proved that the iterations converge strongly to a common fixed point of S and T. They obtained the following result. Theorem 1.1 (see [8]). Let H be a Hilbert space, and let C be a nonempty closed convex subsetof H. Let S and T be nonexpansive mappings of C into itself such that ST = TS and F(S) F(T) is nonempty. Suppose that {αn }∞ n=0 ⊆ [0,1] satisfies (i) limn→∞ αn = 0, and (ii) ∞ n=0 αn = ∞. Then, for an arbitrary x ∈ C, the sequence {xn }∞ n=0 generated by x0 = x and xn+1 = αn x + 1 − αn n 2 Si T j xn , (n + 1)(n + 2) k=0 i+ j =k n ≥ 0, (1.4) converges strongly to a common fixed point Px of S and T, where P is the metric projection of H onto F(S) F(T). Remark 1.2. At this point, we note that the authors have imposed the commutativity on the mappings S and T. But there are many mappings, that do not satisfy ST = TS. For example, if X = [−1/2,1/2], and S and T of X into itself are defined by S = x2 , T = sinx, (1.5) then ST = sin2 x, whereas TS = sinx2 . In this paper, we deal with the strong convergence to common fixed points of two nonexpansive mappings in a Hilbert space. We consider an iteration scheme for nonexpansive mappings without commutativity assumption and prove that the iterations converge strongly to a common fixed point of the mappings Ti , i = 1,2. 2. Preliminaries Let C be a closed convex subset of a Hilbert space H and let S and T be nonexpansive mappings of C into itself. Then we consider the iteration scheme x0 = x ∈ C, xn+1 = αn x + 1 − αn y n = βn x n + 1 − βn n 2 Si T j yn , (n + 1)(n + 2) k=0 i+ j =k n 2 T i S j xn , (n + 1)(n + 2) k=0 i+ j =k n ≥ 0, (2.1) Yonghong Yao et al. 3 where {αn } and {βn } are two sequences in [0,1]. We know that a Hilbert space H satisfies Opial’s condition [6], that is, if a sequence {xn } in H converges weakly to an element y of H and y = z, then liminf xn − y < liminf xn − z. n→∞ n→∞ (2.2) In what follows, we will use PC to denote the metric projection from H onto C; that is, for each x ∈ H, PC is the only point in C with the property x − PC x = min u − x. u∈C (2.3) It is known that PC is nonexpansive and characterized by the following inequality: given x ∈ H and v ∈ H, then v = PC x if and only if x − v,v − y ≥ 0, y ∈ C. (2.4) Now, we introduce several lemmas for our main result in this paper. The first lemma can be found in [4, 5, 10]. Lemma 2.1. Assume {an } is a sequence of nonnegative real numbers such that an+1 ≤ 1 − γn an + δn , (2.5) γn } is a sequence in (0,1) and {δn } is a sequence such that where { (1) ∞ n =1 γ n = ∞ ; (2) limsupn→∞ δn /γn ≤ 0 or ∞ n=1 |δn | < ∞. Then limn→∞ an = 0. Lemma 2.2. Let C be a nonempty bounded closed convex subset of a Hilbert H, and let S,T be nonexpansive mappings of C into itself. For x ∈ C and n ∈ N ∪ {0}, put Gn (x) = n 2 Si T j x, (n + 1)(n + 2) k=0 i+ j =k n 2 Gn (x) = T i S j x. (n + 1)(n + 2) k=0 i+ j =k (2.6) Then lim sup Gn (x) − SGn (x) = 0, n → ∞ x ∈C lim sup Gn (x) − TGn (x) = 0. n → ∞ x ∈C (2.7) 4 Nonexpansive mappings without commutativity assumption Proof. We first prove limn→∞ supx∈C Gn (x) − SGn (x) = 0. n ∞ By an idea in [2], for {xi, j }∞ i+ j =k xi, j , z n = i, j =0 , {x i, j }i, j =0 ⊆ C and zn = (1/ln ) k =0 n (1/ln ) k=0 i+ j =k xi, j ∈ C, with ln = (n + 1)(n + 2)/2, we have n n zn − v 2 = 1 xi, j − v 2 − 1 xi, j − zn 2 ln k=0 i+ j =k ln k=0 i+ j =k (2.8) for each v ∈ H. For x ∈ C, put xi, j = Si T j x,xi, j = T i S j x and v = Szn ,v = Tzn . Then, we have n n i j i j Gn (x) − SGn (x)2 = 1 S T x − Szn 2 − 1 S T x − zn 2 ln k=0 i+ j =k ln k=0 i+ j =k n = n 1 T k x − Szn 2 + 1 Si T j x − Szn 2 l n k =0 ln k=1 i+ j =k,i≥1 n − 1 Si T j x − zn 2 ln k=0 i+ j =k n ≤ n 1 T k x − Szn 2 + 1 Si−1 T j x − zn 2 l n k =0 ln k=1 i+ j =k,i≥1 n − 1 Si T j x − zn 2 ln k=0 i+ j =k n = (2.9) n −1 1 T k x − Szn 2 + 1 Si T j x − zn 2 l n k =0 ln k=0 i+ j =k n − 1 Si T j x − zn 2 ln k=0 i+ j =k n = 1 T k x − Szn 2 − 1 Si T j x − zn 2 l n k =0 ln i+ j =n ≤ 2 1 T k x − Szn 2 ≤ 2 diam(C) , l n k =0 n+2 n where diam(C) is the diameter of C. So, we have, for each n ∈ N ∪ {0}, 2 sup Gn (x) − SGn (x) ≤ x ∈C 2 2 diam(C) , n+2 (2.10) and hence lim sup Gn (x) − SGn (x) = 0. n → ∞ x ∈C (2.11) Yonghong Yao et al. 5 Similarly, we have lim sup Gn (x) − TGn (x) = 0. (2.12) n → ∞ x ∈C 3. Convergence theorem Now we can prove a strong convergence theorem in a Hilbert space. Theorem 3.1. Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let S and T be nonexpansive mappings of C into itself such that F(S) F(T) is nonempty. ∞ Suppose that {αn }∞ n=0 and {βn }n=1 are two sequences in [0,1] satisfying the following conditions: (i) limn→∞ αn = 0, and (ii) ∞ n=0 αn = ∞. For an arbitrary x ∈ C, the sequence {xn }∞ n=0 is generated by x0 = x and xn+1 = αn x + 1 − αn n 2 Si T j yn , (n + 1)(n + 2) k=0 i+ j =k n 2 y n = βn x n + 1 − βn T i S j xn , (n + 1)(n + 2) k=0 i+ j =k (3.1) n ≥ 0. Let zn = n 2 Si T j yn , (n + 1)(n + 2) k=0 i+ j =k zn = n 2 T i S j xn , (n + 1)(n + 2) k=0 i+ j =k (3.2) ∞ ∞ ∞ for each n ∈ N ∪ {0}. If there exist subsequences {zni }∞ i=0 of {zn }n=0 and {z n j } j =0 of {z n }n=0 , respectively, which converge weakly to some common point z in some bounded subset D of C, then the sequence {xn }∞ n=0 defined by (3.1) converges strongly to PF(S)∩F(T) x. Proof. Let x ∈ C and w ∈ F(S) F(T). Putting r = x − w, then the set D = y ∈ H : y − w ≤ r ∩ C (3.3) is a nonempty bounded closed convex subset of C which is S- and T-invariant and contains x0 = x. So we may assume, without loss of generality, that S and T are the mappings of D into itself. Since P is the metric projection of H onto F(S) ∩ F(T), we have y − Px,x − Px ≤ 0 for each y ∈ F(S) F(T). (3.4) 6 Nonexpansive mappings without commutativity assumption From (3.4), we have limsup zn − Px,x − Px ≤ 0, limsup zn − Px,x − Px ≤ 0. n→∞ n→∞ (3.5) In fact, assume that, there exist two positive real numbers r0 and r1 such that limsup zn − Px,x − Px > r0 , limsup zn − Px,x − Px > r1 . n→∞ n→∞ (3.6) ∞ Since {zn }∞ n=0 and {z n }n=0 ⊆ D are bounded, from (3.6), there exist subsequences ∞ ∞ ∞ {zni }i=0 of {zn }n=0 and {zn j }∞ j =0 of {z n }n=0 , respectively, such that limsup zn − Px,x − Px = lim zni − Px,x − Px > r0 , i→∞ n→∞ limsup zn − Px,x − Px = lim zn j − Px,x − Px > r1 . (3.7) j →∞ n→∞ ∞ By the assumption, we know that {zni }∞ comi=0 and {z n j } j =0 converge weakly to some mon point z ∈ D. Thus from Lemma 2.2 and Opial’s condition, we have z ∈ F(S) F(T). In fact, if z = Sz, we have liminf zni − z < liminf zni − Sz i→∞ i→∞ ≤ liminf zni − Szni + Szni − Sz i→∞ (3.8) ≤ liminf zni − z. i→∞ This is a contradiction. Therefore, we have z = Sz. Similarly, we have z = Tz. So, we have z − Px,x − Px ≤ 0. (3.9) On the other hand, since {zni } converges weakly to z, we obtain z − Px,x − Px ≥ r0 . (3.10) This is a contradiction. Hence, we have limsup zn − Px,x − Px ≤ 0, n→∞ limsup zn − Px,x − Px ≤ 0. n→∞ (3.11) Yonghong Yao et al. 7 Since zn − Px ≤ ≤ 2 n 2 T i S j xn − Px (n + 1)(n + 2) k=0 i+ j =k 2 n 2 xn − Px (n + 1)(n + 2) k=0 i+ j =k 2 = xn − Px , yn − Px2 = βn xn + 1 − βn zn − Px2 2 = βn xn − Px + 1 − βn zn − Px 2 2 2 = βn2 xn − Px + 2βn 1 − βn xn − Px,zn − Px + 1 − βn zn − Px 2 2 2 xn − Px + zn − Px ≤ βn2 xn − Px + 2βn 1 − βn 2 2 2 2 + 1 − βn zn − Px ≤ xn − Px . (3.12) Then, we have xn+1 − Px2 = αn x + 1 − αn zn − Px2 2 2 = α2n x − Px2 + 1 − αn zn − Px + 2αn 1 − αn zn − Px,x − Px 2 ≤ 1 − αn 2 n 2 Si T j yn − Px (n + 1)(n + 2) k=0 i+ j =k + α2n x − Px2 + 2αn 1 − αn zn − Px,x − Px ≤ 1 − αn 2 2 n 2 yn − Px (n + 1)(n + 2) k=0 i+ j =k + α2n x − Px2 + 2αn 1 − αn zn − Px,x − Px 2 2 = 1 − αn yn − Px + α2n x − Px2 + 2αn 1 − αn zn − Px,x − Px 2 ≤ 1 − αn xn − Px + αn αn x − Px2 + 2 1 − αn zn − Px,x − Px . (3.13) Putting an = xn − Px2 , from (3.13), we have an+1 ≤ 1 − αn an + δn , where δn = αn {αn x − Px2 + 2(1 − αn ) zn − Px, x − Px}. (3.14) 8 Nonexpansive mappings without commutativity assumption It is easily seen that limsup δn /αn = limsup αn x − Px2 + 2 1 − αn zn − Px, x − Px n→∞ n→∞ ≤ 0. (3.15) Now applying Lemma 2.1 with (3.15) to (3.14) concludes that xn − Px → 0 as n → ∞. This completes the proof. References [1] J.-B. 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Yonghong Yao: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China E-mail address: yuyanrong@tjpu.edu.cn Rudong Chen: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China E-mail address: chenrd@tjpu.edu.cn Haiyun Zhou: Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang 050003, China E-mail address: witman66@yahoo.com.cn