Viscosity Iterative Approximating the Common Fixed Points of Non-expansive Semigroups in Banach Spaces

Viscosity Iterative Approximating the Common Fixed Points of Non-expansive Semigroups in Banach Spaces YUAN-HENG WANG Zhejiang Normal University Department of Mathematics Yingbing Road 688, 321004 Jinhua CHINA wangyuanheng@yahoo.com.cn YAN LI Nanyang Industrial School The School Office Wenhua Avenue 168, 47300 Nanyang CHINA yhwang@zjnu.cn Abstract: Let C be a closed convex subset of a reflexive strictly convex Banach space E F = {T (t); t > 0} be a non-expansive semigroup on the C with the nonempty set of their common fixed points. The purpose of this paper is to study a new viscosity iterative method for a non-expansive semigroup weakly contraction mappings. And it is proved that the new iterative approximate sequences converge strongly to the solution of a certain variational inequality. These results improve extend some recent results of the other authors. Key Words: Non-expansive semigroup, Common fixed point, Uniformly Gâteaux differentiable norm, Weakly contraction, Iterative approximation, Strong convergence 1 Introduction Let C be a closed convex subset of Hilbert space H T be a nonexpansive mapping from C into itself. We denote by F (T ) the set of fixed points of T. Let F (T ) be nonempty u be an element of C. In 1967, Halpern [1] firstly introduced the following explicit iterative scheme (1) in Hilbert space, x n+1 = u + (1 )T x n, (1) where { } is a real sequence [0, 1]. He pointed out that the control conditions (C 1 ) lim = 0 (C 2 ) = are necessary for the convergence of the iterative scheme (1) to a fixed point of T. In 1992, Wittman [2] showed that the strong convergence of the iteration scheme (1) under the control conditions (C 1 ), (C 2 ) (C 3 ) +1 < in the Hilbert space. After that, Shioji Takahashi [3] extended Wittman s results to a uniformly convex Banach space with a uniformly Gâteaux differentiable norm. In 2004, H. K. Xu [4] proposed the following viscosity iterative process {x n }: x n+1 = f(x n ) + (1 )T x n, (2) where 0 1, T : C C is a nonexpansive mapping with F (T ), f : C C is a fixed contractive mapping. He showed that {x n } strongly converges to a fixed point q of T in a uniformly smooth Banach space. Recently, Chen Song [5] introduced the following implicit explicit viscosity iteration processes defined by (3) (4) to nonexpansive semigroup case, x n = f(x n ) +(1 ) 1 t x n+1 = f(x n ) +(1 ) 1 t t 0 t 0 T (s)xds, n 1, (3) T (s)xds, n 1, (4) showed that {x n } converges to a same point of t>0 F ix(t (t)) in a uniformly convex Banach space with a uniformly Gâteaux differentiable norm. Note however that their iterate x n at step n is constructed through the average of the semigroup over the interval (0, t). Suzuki [6] was the first to introduce again in a Hilbert space the following implicit iteration process: x n = u + (1 )T (t n )x n, n 1, (5) E-ISSN: 2224-2880 85 Issue 1, Volume 12, January 2013

for the nonexpansive semigroup case. Benavides, Aceda Xu [7] proved that if F = {T (t) : t > 0} satisfies an asymptotic regularity condition fulfills the control conditions (C 1 ) (C 2 ) (C 4 ) lim = 1 +1 in a uniformly smooth Banach space, then both the implicit iteration process (5) the explicit iteration (6) converge to a same point of F ix(f), x n+1 = u + (1 )T (t n )x n, n 1. (6) Song Xu [8] introduced the following implicit explicit viscosity iterative schemes, respectively: x n = f(x n ) + (1 )T (t n )x n, n 1, (7) x n+1 = f(x n ) + (1 )T (t n )x n, n 1. (8) They proved that the two iteration processes strongly converges to a same point q of F ix(f) which is a solution of certain variational inequality in a reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm. Motivated inspired by the above results, in this paper, we study the strong convergence of the viscosity iterative processes {z m } {x n } by respectively equations (9) (10). We consider the case T (t)(t > 0) is a noexpansive semigroup with t>0 F (T (t)), f : C C is a weakly contractive self-mapping, define the implicit viscosity iterative method explicit viscosity iterative method as follows z m = α m f(x m ) + (1 α m )T (t m )z m, m 1, (9) x n+1 = f(x n ) + β n x n +(1 β n )T (t n )y n, y n = γ n x n + (1 γ n )T (t n )x n, n 1. (10) where { }, {β n } are two sequences in (0,1) with + β n 1(n 1), {α m }, {γ n } are two sequences in [0,1]. In a reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm, we will prove that {z m } {x n } strongly converge to some point p t>0 F (T (t)), where p is a solution to the following variational inequality: (f I)p, j(x p) 0, x t>0 F (T (t)). So, our results extend improve some related results considered by Song Xu [8], Xu [9], Wu, Chang Yuan[10] the other authors. 2 Preliminaries Throughout this paper, let E be a reflexive strictly convex Banach space C be a closed convex subset of E. Let J denote the normalized duality mapping from E into 2 E given by J(x) = {f E, x, f = x f, x = f }, x E, where E denotes the dual space of E.,. denotes the generalized duality pairing. We shall denote the single-valued duality mapping by j. When {x n } is a sequence in E, then x n x (respectively x n x, x n x) will denote strong (respectively weak, weak ) convergence of the sequence {x n } to x. A Banach space E is said to be strictly convex if for x + y 2 < 1 x = y = 1, x y; the function δ : [0, 2] [0, 1] is said to be the modulus of convexity of Banach space E, where δ ε = inf{1 x y /2 : x 1, y 1, x y ε}. E is said to be uniformly convex if for each δ ε > 0. Let S(E) = {x E : x = 1}. The norm of Banach space E is said to be Gâteaux differentiable, if the x + ty x lim t 0 t exists for each x, y S(E). Moreover, if for each y S(E), the limit exists uniformly for x S(E), we say that the norm of E is uniformly Gâteaux differentiable. It is well known that each uniformly convex Banach space E is reflexive strictly convex if E is reflexive smooth, then the duality mapping J is single valued (see [11-13]). E-ISSN: 2224-2880 86 Issue 1, Volume 12, January 2013

Definition 1 Let C be a nonempty subset of a Banach space E T : C C a mapping. T is called a Lipschitzian mapping if there exists a constant L > 0 such that T x T y L x y for all x, y C, L is called Lipschitz constant of T. T is called nonexpansive mapping if L = 1, T is called contraction mapping if L [0, 1). Definition 2 [12] An operator T with domain D(T ) rang R(T ) in a Banach space E is said to be weakly contraction, if T x T y x y ψ( x y ), x, y C, where ψ : [0, ) [0, ) is a continuous nondecreasing function such that ψ(0) = 0, ψ(t) > 0 for all t > 0 lim t ψ(t) =. Remark 3 If ψ(t) = kt for all t 0, where k (0, 1), then T is a contraction with Lipschitz constant 1 k. It is obvious that the class of contraction mappings is a subclass of the class of weakly contraction. Definition 4 A family F = {T (t) : t 0} of mapping of C into itself is called nonexpansive semigroup of C, if it satisfies the following conditions: (1) T (t 1 +t 2 )x = T (t 1 )T (t 2 )x, for each t 1, t 2 0 x C; (2) T (0)x = x, for each x C; (3) lim t 0 T (t)x = x, for x C; (4) for each t > 0, T (t) is nonexpansive, that is, T (t)x T (t)y x y, x, y C. We shall denote by F the common fixed point set of F, that is, F := F ix(f) = {x C : T (t)x = x, t > 0} = t>0 F ix(t (t)), where F ix(t ) = {x C : T x = x} is the set of fixed points of a mapping T. Definition 5 F is said to be uniformly asymptotically regular (in short,u.a.r) on C if for all h 0 any bounded subset K of C, lim sup T (h)(t (t)x) T (t)x = 0. t,x K Let µ be a continuous linear functional on l let (a 0, a 1, ) l, we use µ m (a m ) instead of µ((a 0, a 1, )), we call µ a Banach limit when µ satisfies µ = µ m (1) = 1 µ m (a m+1 ) = µ m (a m ) for each (a 0, a 1, ) l. Lemma 6 [14] Let C be a nonempty closed convex subset of a Banach space E with a uniformly Gâteaux differentiable norm, {x m } a bounded sequence of E, let z 0 be a element of C µ be a Banach limit. Then if only if µ m x m z 0 2 = min y C µ m x m y 2, µ m y z 0, j(x m z 0 ) 0, y C. Lemma 7 [15] Let {x n } {y n } be bounded sequences in a Banach space E let {β n } be a sequence in [0,1] with Suppose 0 < lim inf β n lim sup β n < 1. x n+1 = β n x n + (1 β n )y n for all integers n 0 Then lim sup( y n+1 y n x n+1 x n ) 0. lim y n x n = 0. Lemma 8 [12] Let {u n } {v n } be two sequences v of nonnegative real numbers such that lim n un = 0 u n =. Let {λ n } be a sequence of nonnegative real numbers satisfying the recursive inequality: λ n+1 λ n u n ϕ(λ n ) + v n, n N, where ϕ : [0, ) [0, ) is a continuous nondecreasing function such that ϕ(0) = 0 ϕ(t) > 0 for all t > 0. Then {λ n } converges to zero. 3 Main Results Theorem 9 Let E be a real reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm, C a nonempty closed convex subset of E, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a weakly contractive mapping with function ψ. Suppose lim t m = α m [0, 1] such that lim α m = 0. If {z m } is defined by z m = α m f(z m ) + (1 α m )T (t m )z m, m 1. E-ISSN: 2224-2880 87 Issue 1, Volume 12, January 2013

Let z 1 C. Then as m, {z m } converges strongly to some common fixed point p of F such that p is the unique solution in F to the following variational inequality: f(p) p, j(x p) 0, x F. (11) Proof: We first show that the uniqueness of solution to the variational inequality (11) in F. In fact, suppose p, q F satisfy (11), we have that f(p) p, j(q p) 0 f(q) q, j(p q) 0. Combining the above two inequalities, we have Thus, q p 2 f(p) f(q) q p ( p q ψ( p q )) q p. q p p q ψ( p q ), we can obtain that p q = 0, or p = q. Next we show the boundedness of {z m }. Indeed, for any fixed y F, we have z m y = α m f(z m ) + (1 α m )T (t m )z m y α m f(z m ) y + (1 α m ) T (t m )z m y α m f(z m ) f(y) + α m f(y) y +(1 α m ) z m y α m z m y α m ψ( z m y ) +α m f(y) y + (1 α m ) z m y = z m y α m ψ( z m y ) +α m f(y) y. So, we obtain that ψ( z m y ) f(y) y. Suppose {z m y} is not bounded. Then there exists a sequence {m k } in (0, ) with m k as k such that z mk y > k, k N. (12) Since ψ is nondecreasing lim t ψ(t) =, it follows from (12) that ψ(k) < ψ( z mk y ) f(y) y, a contraction. Thus {z m } is bounded, so are {T (t m )z m } {f(z m )}. This implies that lim z m T (t m )z m = lim α m T (t m )z m f(z m ) = 0. Since {T (t)} is u.a.r nonexpansive semigroup lim t m =, then for all h > 0, lim T (h)t (t m )z m T (t m )z m lim sup T (h)t (t m )x T (t m )x = 0, x K where K is any bounded subset of C containing {z m }. Hence, z m That is, for all h > 0, T (h)z m z m T (t m )z m + T (t m )z m T (h)t (t m )z m + T (h)t (t m )z m T (h)z m 2 z m T (t m )z m + T (t m )z m T (h)t (t m )z m 0, m. lim z m T (h)z m = 0. (13) We claim that the set {z m } is sequentially compact. Define the function φ : C R by Since E is reflexive, φ(x) := µ m z m x 2, x C. lim φ( x ) =, x φ is continuous convex function, we have that the set M := {y C : φ(y) = inf φ(x)}, (14) x C which is nonempty closed convex bounded. Furthermore, M is invariant under T (t) (for all t > 0). In fact, for each y M, we have φ(t (t)y) = µ m z m T (t)y 2 µ m T (t)z m T (t)y 2 µ m z m y 2 = φ(y). Hence, T (t)y M. As y is arbitrary, then T (t)(m) M. Let u F, since every nonempty closed convex subset of a strictly convex reflexive Banach space is a Chebyshev set (see [13]), there exists an unique p M such that u p = inf u x, x M since T (t)u = u T (t)p M, u T (t)p = T (t)u T (t)p u p. E-ISSN: 2224-2880 88 Issue 1, Volume 12, January 2013

Hence T (t)p = p by the uniqueness of p M. Since t is arbitrary, it follows that p F. Using Lemma 6 together with p M, we obtain that In particular µ m z p, j(z m p) 0, z C. µ m f(p) p, j(z m p) 0. (15) Since f is weakly contraction, we have z m p 2 = z m f(z m ), j(z m p) + f(z m ) f(p), j(z m p) + f(p) p, j(z m p) z m f(z m ), j(z m p) + f(z m ) f(p) z m p + f(p) p, j(z m p) z m f(z m ), j(z m p) + z m p 2 z m p ψ( z m p ) + f(p) p, j(z m p), z m p ψ( z m p ) z m f(z m ), j(z m p) + f(p) p, j(z m p), (16) z m f(z m ), j(z m p) = (1 α m ) T (t m )z m f(z m ), j(z m p) = (1 α m ) T (t m )z m z m +z m f(z m ), j(z m p), z m f(z m ), j(z m p) 1 α m α m T (t m )z m z m, j(z m p) 1 α m α m T (t m )z m T (t m )p + p z m, j(z m p) 0. Hence, we get z m f(z m ), j(z m p) 0, m N. (17) Together with above inequalities (15), (16), (17), we obtain that µ m z m p ψ( z m p ) 0, therefore, there exists a subsequence {z mi } of {z m } such that z mi p(i ). Next we show that p is a solution in F to the variational inequality(11). Since the duality map j is a single-valued norm topology to weak topology uniformly continuous on bounded subset of E, z mi p, (i ), we have (I f)z m (I f)p 0, (i ), for all x F, we observe that z mi f(z mi ), j(z mi x) p f(p), j(p x) = z mi f(z mi ) (p f(p)), j(z mi x) + p f(p), j(z mi x) j(p x) z mi f(z mi ) (p f(p)) z mi x + p f(p), j(z mi x) j(p x) 0, i. It follows from (17) that f(p) p, j(x p) = lim i f(z mi ) z mi, j(x z mi ) 0. That is, p F is a solution of (11). Hence p = q by uniqueness. In a similar way, it can be show that each cluster point of the sequence {z m } is equal to q. Therefore, z m p as m. Corollary 10 Let E be a real reflexive strictly convex Banach space with a uniformly Gâteaux differentiable norm, C a nonempty closed convex subset of E, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a fixed contractive mapping with contractive coefficient β [0, 1). Suppose lim t m = α m [0, 1] such that lim α m = 0. If {z m } is defined by z m = α m f(z m ) + (1 α m )T (t m )z m, m 1. Let z 1 C. Then as m, {z m } converges strongly to some common fixed point p of F such that p is the unique solution in F to the variational inequality(11). Theorem 11 Let C be a nonempty closed convex subset of a real reflexive strictly convex Banach space E with a uniformly Gâteaux differentiable norm, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a weakly contractive mapping with function ψ. Suppose lim t n =. Let { }, {β n } be two sequences in (0,1) with + β n 1(n 1), {γ n } a sequence in [0,1]. The sequence {x n } is given by (10). Let x 1 C assume that { }, {β n }, {γ n }, ψ satisfy the following conditions: (i) lim = 0; (ii) = ; (iii) 0 < lim inf β n lim sup β n < 1; (iv) inf{ψ( x n q )/ x n q : x n q, n N} = δ > 0 for q F ; (v) lim γ n+1 γ n = 0 lim inf γ n > 0. Then as n, {x n } defined by (10) converges strongly to some common fixed point p of F such that p is the unique solution in F to the variational inequality (11). E-ISSN: 2224-2880 89 Issue 1, Volume 12, January 2013

Proof: The proof is divided into five steps. Step 1. We show that {x n } is bounded. Take q F. It follows that x n+1 q f(x n ) q + β n x n q +(1 β n ) T (t n )y n q f(x n ) f(q) + f(q) q +β n x n q + (1 β n ) y n q x n q ψ( x n q ) + f(q) q + β n x n q +(1 β n ) y n q = ( + β n ) x n q ψ( x n q ) + f(q) q + (1 β n ) y n q, y n q = γ n (x n q) +(1 γ n )(T (t n )x n q) γ n x n q + (1 γ n ) x n q = x n q. Since 0 < δ = inf{ψ( x n q )/ x n q : x n q, n N}, together with the above two inequalities, we have x n+1 By induction, q x n q δ x n q + f(q) q = (1 δ) x n q + f(q) q. x n q max{ x 1 q, 1 f(q) q }, n 1, δ consequently, {x n } is bounded, so are {y n }, {T (t n )x n }, {T (t n )y n } {f(x n )}. Step 2. We show that lim x n+1 x n = 0. Indeed, define a sequence {z n } by we have x n+1 = β n x n + (1 β n )z n, n 1, z n+1 z n = x n+2 β n+1 x n+1 x n+x n = +1f(x n+1 )+(1 +1 β n+1 )T (t n+1 )y n+1 f(x n )+(1 β n )T (t n )y n = +1 f(x n ) f(x n+1 ) + 1 + T (t n+1 )y n+1 T (t n )y n 1 β n = +1 f(x n+1 ) f(x n ) + 1 + (T (t n+1 )y n+1 T (t n+1 )y n ) + 1 + (T (t n+1 )y n T (t n )y n ) +( 1 + 1 αn βn )T (t n )y n, y n+1 y n = γ n+1 x n+1 + (1 γ n+1 )T (t n+1 )x n+1 γ n x n (1 γ n )T (t n )x n γ n+1 x n+1 x n + γ n+1 γ n x n +(1 γ n+1 ) T (t n+1 )x n+1 T (t n+1 )x n +(1 γ n+1 ) T (t n+1 )x n T (t n )x n + γ n+1 γ n T (t n )x n x n+1 x n + γ n+1 γ n x n +(1 γ n+1 ) T (t n+1 )x n T (t n )x n + γ n+1 γ n T (t n )x n. Together with the above two inequalities, we obtain that z n+1 z n x n+1 x n +1 f(x n+1 ) + f(x n ) T (t n+1 )y n+1 T (t n+1 )y n + 1 + + 1 + + 1 + x n+1 x n +1 T (t n+1 )y n T (t n )y n 1 β n T (t n )y n f(x n+1 ) + αn f(x n ) + 1 + [ x n+1 x n + γ n+1 γ n ( x n + T (t n )x n ) +(1 γ n+1 ) T (t n+1 )x n T (t n )x n ] T (t n+1 )y n T (t n )y n + 1 + + 1 + x n+1 x n = +1 1 αn βn T (t n )y n f(x n+1 ) + f(x n ) 1) x n+1 x n +( 1 + + 1 + [ γ n+1 γ n ( x n + T (t n )x n ) +(1 γ n+1 ) T (t n+1 )x n T (t n )x n ] + 1 + n+1 + 1 + T (t n+1 )y n T (t n )y n 1 β n T (t n )y n. If t n+1 > t n, by (u.a.r), we have T (t n+1 )x n T (t n )x n = T (t n+1 t n )T (t n )x n T (t n )x n 0. (18) If t n+1 < t n,interchange t n+1 t n, we also can obtain similarly, we get T (t n+1 )x n T (t n )x n 0, T (t n+1 )y n T (t n )y n 0. (19) E-ISSN: 2224-2880 90 Issue 1, Volume 12, January 2013

Thus it follows from the conditions (i), (iii), (v) (18), (19), we obtain that lim sup( z n+1 z n x n+1 x n ) 0. Hence, by Lemma 7, we have which imply that lim z n x n = 0, lim x n+1 x n = 0. (20) Step 3. For each t (0, ), T (t)x n x n 0. Indeed, we have that x n+1 T (t n )x n x n+1 T (t n )y n + T (t n )y n T (t n )x n x n+1 T (t n )y n + y n x n = x n+1 T (t n )y n + (1 γ n ) x n T (t n )x n x n+1 T (t n )y n + (1 γ n ) x n+1 x n +(1 γ n ) x n+1 T (t n )x n. So And as x n+1 T (t n )x n 1 γ n x n+1 T (t n )y n + 1 γ n γ n x n+1 x n. x n+1 T (t n )y n = f(x n ) + β n x n +(1 β n )T (t n )y n T (t n )y n f(x n ) T (t n )y n + β n x n T (t n )y n f(x n ) T (t n )y n +β n x n x n+1 +β n x n+1 T (t n )y n, (1 β n ) x n+1 T (t n )y n f(x n ) T (t n )y n + β n x n x n+1 x n+1 T (t n )y n f(x n ) T (t n )y n + βn x n x n+1, by (i), (iii), (v), (20) together with above inequalities, we get x n T (t n )x n 0, (n ). (21) Let K be any bounded subset of C which contains the sequence {x n }. It follows that So we have T (t)x n x n 0, (n ). (22) Step 4. We show that Let lim sup (I f)p, j(p x n+1 ) 0. (23) z m = α m f(z m ) + (1 α m )T (t m )z m, where t m α m satisfies the conditions of Theorem 9. Then we have that lim z m = p. From the definition of ψ, we know that Since lim ψ( z m p ) = ψ(0) = 0. z m x n+1 2 = α m f(z m ) x n+1, j(z m x n+1 ) +(1 α m ) T (t m )z m x n+1, j(z m x n+1 ) = (1 α m ) T (t m )z m T (t m )x n+1, j(z m x n+1 ) +(1 α m ) T (t m )x n+1 x n+1, j(z m x n+1 ) +α m f(z m ) f(p) + f(p) + z m z m + p p x n+1, j(z m x n+1 ) (1 α m ) z m x n+1 2 +(1 α m ) T (t m )x n+1 x n+1, j(z m x n+1 ) +α m f(z m ) f(p) z m + p, j(z m x n+1 ) +α m f(p) p, j(z m x n+1 ) +α m z m x n+1, j(z m x n+1 ) z m x n+1 2 +(1 α m ) T (t m )x n+1 x n+1, j(z m x n+1 ) +α m f(p) p, j(z m x n+1 ) +α m ( f(z m ) f(p) + z m p ) z m x n+1, so, we can obtain that f(p) p, j(x n+1 z m ) 1 α m α m T (t m )x n+1 x n+1 z m x n+1 +(2 z m p ψ( z m p )) z m x n+1 1 α M m 1 α m T (t m )x n+1 x n+1 +2M 1 z m p ψ( z m p )M 1 M 1 α m T (t m )x n+1 x n+1 +2M 1 z m p ψ( z m p )M 1, T (t)x n x n T (t)x n T (t)t (t n )x n + T (t)t (t n )x n T (t n )x n + T (t n )x n x n 2 x n T (t n )x n + sup x K T (t)t (t n )x T (t n )x. where M 1 is a constant such that x n+1 z m M 1. E-ISSN: 2224-2880 91 Issue 1, Volume 12, January 2013

Firstly, we take limit as n, then as m in above inequality (using (22)) lim sup lim sup f(p) p, j(x n+1 z m ) 0. On the other h, since J is single-valued norm topology to weak topology uniformly continuous on bounded set of E we get Therefore, we have lim z m = p, lim (x n+1 z m ) = x n+1 p. f(p) p, j(x n+1 z m ) f(p) p, j(x n+1 p). Thus, given ε > 0, there exists N 1, such that if m > N, for all n, we have f(p) p, j(x n+1 p) < f(p) p, j(x n+1 z m ) + ε. (24) Therefore, by taking upper limit as n firstly, then as m in both sides of (24) lim sup f(p) p, j(x n+1 p) lim sup lim sup f(p) p, j(x n+1 z m ) + ε. Since ε is arbitrary, we obtain(23). Thus, there exists a sequence {ε n } in (0, ) which lim ε n = 0 such that (I f)p, j(p x n+1 ) ε n, n N. Step 5. lim x n p = 0. Indeed, we have x n+1 p 2 = (f(x n ) p) + β n (x n p) +(1 β n )(T (t n )y n p), j(x n+1 p) = (f(x n ) f(p)) + β n (x n p) +(1 β n )(T (t n )y n p), j(x n+1 p) + f(p) p, j(x n+1 p) (f(x n ) f(p)) + β n (x n p) +(1 β n )(T (t n )y n p) x n+1 p + ε n [ f(x n ) f(p) + β n x n p +(1 β n ) T (t n )y n p ] x n+1 p + ε n So, [ x n p ψ( x n p ) +β n x n p +(1 β n ) y n p ] x n+1 p + ε n [( + β n ) x n p ψ( x n p ) +(1 β n ) γ n (x n p) +(1 γ n )(T (t n )x n p) ] x n+1 p + ε n [ x n p ψ( x n p )] x n+1 p + ε n 1 2 [ x n p ψ( x n p )] 2 + 1 2 x n+1 p 2 + ε n. x n+1 p 2 x n p 2 2 ψ( x n p ) x n p +α 2 n(ψ( x n p )) 2 + 2 ε n x n p 2 2 ψ( x n p ) x n p +α 2 n(ψ(m)) 2 + 2 ε n, for some M > 0. Since { x n p } is bounded, thus, for λ n = x n p 2, we obtain the following recursive inequality: where λ n+1 λ n ϕ(λ n ) + ω n, ω n = [ (ψ(m)) 2 + 2ε n ] ϕ(t) = 2 tψ( t). So {x n } converges strongly to p by Lemma 8. If γ n = 1, the following result is clearly gained. Corollary 12 Let C be a nonempty closed convex subset of a real reflexive strictly convex Banach space E with a uniformly Gâteaux differentiable norm, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a weakly contractive mapping with function ψ. Suppose lim t n =. Let { }, {β n } be two sequences in (0,1) with + β n 1. The sequence {x n } is given by x n+1 = f(x n ) + β n x n + (1 β n )T (t n )x n. Let x 1 C assume that { }, {β n }, ψ satisfy the following conditions: (i) lim = 0; (ii) = ; E-ISSN: 2224-2880 92 Issue 1, Volume 12, January 2013

(iii) 0 < lim inf β n lim sup β n < 1; (iv) inf{ψ( x n q )/ x n q : x n q, n N} = δ > 0, q F. Then as n, {x n } converges strongly to some common fixed point p of F such that p is the unique solution in F to the variational inequality (11). Corollary 13 Let C be a nonempty closed convex subset of a real reflexive strictly convex Banach space E with a uniformly Gâteaux differentiable norm, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a fixed contractive mapping with contractive coefficient β [0, 1). Suppose lim t n =. Let { }, {β n } be two sequences in (0,1) with + β n 1(n 1), {γ n } a sequence in [0,1]. The sequence {x n } is given by (10). Let x 1 C assume that { }, {β n }, {γ n }, ψ satisfy the following conditions: (i) lim = 0; (ii) = ; (iii) 0 < lim inf β n lim sup β n < 1; (iv) inf{ψ( x n q )/ x n q : x n q, n N} = δ > 0, forq F ; (v) lim γ n+1 γ n = 0 lim inf γ n > 0. Then as n, {x n } converges strongly to some common fixed point p of F such that p is the unique solution in F to the variational inequality (11). Corollary 14 Let C be a nonempty closed convex subset of a real uniformly convex Banach space E with a uniformly Gâteaux differentiable norm, {T (t)} a u.a.r nonexpansive semigroup from C into itself such that F := F ix(f) = t>0 F ix(t (t)), f : C C a weakly contractive mapping with function ψ. Suppose lim t n =. Let { }, {β n } be two sequences in (0,1) with + β n 1(n 1), {γ n } a sequence in [0,1]. The sequence {x n } is given by (10). Let x 1 C assume that { }, {β n }, {γ n }, ψ satisfy the following conditions: (i) lim = 0; (ii) = ; (iii) 0 < lim inf β n lim sup β n < 1; (iv) inf{ψ( x n q )/ x n q : x n q, n N} = δ > 0, forq F ; (v) lim γ n+1 γ n = 0 lim inf γ n > 0. Then as n, {x n } converges strongly to some common fixed point p of F such that p is the unique solution in F to the variational inequality (11). Remark 15 When { }, {β n }, {γ n }, ψ satisfy different conditions, the results in this paper extend improve some related results considered by Song Xu [8], Xu [9], Wu, Chang Yuan[10] the other authors. Acknowledgements: The authors would like to thank editors referees for many useful comments suggestions for the improvement of the article. The research was partially supported by the Natural Science Foundation of Zhejiang Province (Y6100696) the National Natural Science Foundation of China (11071169). References: [1] B. Halpern, Fixed points of nonexpansive maps, Bull. Amer. Math. Soc., 73, 1967, pp. 957 961. [2] R. Wittmann, Approximation of fixed points of nonexpansive mappings, Arch. Math., 59, 1992, pp. 486 491. [3] N. Shioji W. Takahashi, Strong convergence of approximated sequences for nonexpansive mappings in Banach spaces, Proc. Amer. Math. Soc., 125, 1997, pp. 3641 3645. [4] H. K. Xu, Viscosity approximation methods for nonexpansive mappings, J. Math. Anal. Appl., 298, 2004, pp. 279 291. [5] R. Chen Y. Song, Convergence to common fixed point of nonexpansive semigroup, J. Comput. Appl. Math., 200, 2007, pp. 566 575. [6] T. Suzuki, On strong convergence to common fixed points of nonexpansive semigroup in Hilbert spaces, Proc. Amer. Math. Soc., 131, 2002, pp. 2133 2136. [7] T. D. Benavides, G. L. Acedo H. K. Xu, Construction of sunny nonexpansive retraction in Banach space, Bull. Austral. Math. Soc., 66(1), 2002, pp. 9 16. [8] Y. Song S. Xu, Strong convergence theorems for nonexpansive semigroup in Banach spaces, J. Math. Anal. Appl., 338, 2008, pp. 152 161. E-ISSN: 2224-2880 93 Issue 1, Volume 12, January 2013

[9] H. K. Xu, A strong convergence theorem for contraction semigroup in Banach spaces, Bull. Austral. Math. Soc., 72, 2005, pp. 371 379. [10] D. P. Wu, S. S. Chang G. X. Yuan, Approximation of common fixed points for a family of finite nonexpansive mappings in Banach space, Nonlinear Anal., 63, 2005, pp. 987 999. [11] Y. H. Wang Y. H. Xia, Strong convergence for asymptotically pseudo-contractions with the demi-closedness principle in Banach spaces, Fixed Point Theory Appl., 45, 2012, pp. 1 8. [12] Ya. I. Alber S. Guerre-Delabriere, Principles of weakly contractive maps in Hilbert Spaces, in:i.gohberg,yu.lyubich(eds.), New Results in operator Theory, Advances appl., 98, Birkhauser,Basel, 1997, pp. 7 22. [13] R. E. Megginson, An introduction to Banach space Theory, Spring Verlag, New York 1998. [14] W. Takahashi Y. Ueda, On Reich s strong convergence for resolvents of accretive operators, J. Math. Anal. Appl., 104, 1984, pp. 546 553. [15] T. Suzuki, Strong convergence of Krasnoselskii Mann s type sequences for one-parameter nonexpansive semigroup without Bochner integrals, J. Math. Anal. Appl. 305, 2005, pp. 227 239. E-ISSN: 2224-2880 94 Issue 1, Volume 12, January 2013