Báo cáo hóa học: "VISCOSITY APPROXIMATION FIXED POINTS FOR NONEXPANSIVE AND m-ACCRETIVE OPERATORS" ppt

VISCOSITY APPROXIMATION FIXED POINTS FOR NONEXPANSIVE AND m-ACCRETIVE OPERATORS RUDONG CHEN AND ZHICHUAN ZHU Received 10 June 2006; Accepted 22 July 2006 Let X be a real reflexive Banach space, let C be a closed convex subset of X,andletA be an m-accretive operator with a zero. Consider the iterative method that generates the sequence {x n } by the algorithm x n+1 = α n f (x n )+(1− α n )J r n x n ,whereα n and γ n are two sequences satisfying certain conditions, J r denotes the resolvent (I + rA) −1 for r>0, and let f : C → C be a fixed contractive mapping. The strong convergence of the algorithm {x n } is proved assuming that X has a weakly continuous duality map. Copyright © 2006 R. Chen and Z. Zhu. 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 X bearealBanachspace,letC be a nonempty closed convex subset of X,andlet T : C → C be a nonexpansive mapping if for all x, y ∈ C,suchthat Tx− Ty≤x − y. (1.1) We use F(T) to denote the set of fixed points of T, that is, F(T) ={x ∈ C : x = Tx}.And  denotes weak convergence, → denotes strong convergence. Recall that a self-mapping f : C → C is a contraction on C if there exists a constant β ∈ (0,1) such that   f (x) − f (y)   ≤ βx − y, x, y ∈ C. (1.2) Brow der [2] considered an iteration in a Hilbert space as follows. Fix u ∈ C and define a contraction T t : C → C by T t x = tu+(1− t)Tx, x ∈ C, (1.3) where t ∈ (0,1). Banach’s contraction mapping principle guarantees that T t has a unique fixed point x t in C. Xu [7] defined the following one viscosity iteration for nonexpansive mappings in uniformly smooth Banach space. Hindawi Publishing Corporation Fixed Point Theory and Applications Volume 2006, Article ID 81325, Pages 1–10 DOI 10.1155/FPTA/2006/81325 2 Nonexpansive mappings Theorem 1.1 [2, Theorem 4.1, page 287]. Let X be a uniformly smooth Banach space, let C be a closed convex subset of X, T : C → C is a nonexpansive mapping with F(T) = φ, and f ∈ Π C ,whereΠ C denotes the set of all contractions on C. Then {x t } defined by the following: x t = tf  x t  +(1− t)Tx t , x ∈ C, (1.4) convergesstronglytoapointinF(T).DefineQ : Π C → F(T) by Q( f ): = lim t→0 x t , f ∈  C , (1.5) then Q(f) solves the variational inequality  (I − f )Q( f ), J  Q( f ) − p  ≤ 0, f ∈  C , p ∈ F(T). (1.6) Xu [8] proved the strong convergence of {x t } defined by (1.3)inareflexiveBanach space with a weakly continuous duality map J ϕ with gauge ϕ. Recall that an operator A with D(A) and range R(A)inX is said to be accretive, if for each x i ∈ D(A)andy i ∈ Ax i ,(i = 1, 2) such that  y 2 − y 1 ,J  x 2 − x 1  ≥ 0, (1.7) where J is the duality map from X to the dual space X ∗ given by J(x) =  f ∈ X ∗ :  x, f  = x 2 =f  2  , x ∈ X. (1.8) An accretive operator A is m-accretive if R(I + λA) = X for all λ>0. Denote by J r the resolvent of A for r>0, J r = (I + rA) −1 . (1.9) It is known that J r is a nonexpansive mapping from X to C := D(A) which wil l be assumed convex. Also in [8], Xu considered the following algorithm: x n+1 = α n u +  1 − α n  J r n x n , n ≥ 0, (1.10) where u ∈ C is arbitrarily fixed, {α n } is a sequence in (0,1), and {r n } is a sequence of positive numbers. Xu proved that if X is a reflexive Banach space with weakly continuous duality mapping, then the sequence {x n } given by (1.10) converges strongly to a point in F provided the sequences {α n } and {r n } satisfy cer tain conditions. The main purpose of this paper is to consider the following two iterations both in a refle xive Banach space X which has a weakly continuous duality mapping: x t = tf  x t  +(1− t)Tx t , t ∈ (0,1), (1.11) x n+1 = α n f  x n  +  1 − α n  J r n x n , n ≥ 0. (1.12) R. Chen and Z. Zhu 3 2. Preliminaries In order to prove our main results, we need the following lemmas. The proof of Lemma 2.1 can be found in [5, 6]. Lemma 2.2 is an immediate consequence of the subdifferential inequality of the function (1/2) · 2 . Lemma 2.1. Let {a n } n be a sequence of nonnegat ive real numbers such that a n+1 ≤  1 − α n  a n + α n β n , n ≥ 0, (2.1) where {α n } n ⊂ (0,1),andβ n satisfy the conditions: (i) lim n→∞ α n = 0, (ii)  ∞ n=1 α n =∞, (iii) limsup n→∞ β n ≤ 0. Then lim n→∞ a n = 0. Lemma 2.2. Let X be an arbit rary real Banach space. Then x + y 2 ≤x 2 +2  y,J(x + y)  , x, y ∈ X. (2.2) Recall that a gauge is a continuous strictly increasing function ϕ :[0,+ ∞) → [0, +∞) such that ϕ(0) = 0andϕ(t) →∞. Associated to a gauge ϕ is the duality map J ϕ : X → X ∗ defined by J ϕ (x) =  f ∈ X ∗ :  x, f  = xϕ   x  ,  f =ϕ   x  , x ∈ X. (2.3) Following Browder [3], we say that a Banach space X has a weakly continuous duality map if there exists a gauge ϕ for which t he duality map J ϕ is single valued and weak-to- weak ∗ sequentially continuous, that is, if {x n } is a sequence in X weakly convergent to a point x, then the sequence {J ϕ (x n )} converges weakly ∗ to J ϕ (x). It is known that  p has a weakly continuous duality map for all 1 <p< ∞.Set Φ(t) =  t 0 ϕ(τ)dτ, τ ≥ 0. (2.4) Then J ϕ (x) = ∂Φ   x  , x ∈ X, (2.5) where ∂ denotes the subdifferential in the sense of convex analysis. We also need the next lemma, and the first part of Lemma 2.3 is an immediate conse- quence of the subdifferential inequality and the proof of the second part can be found in [4]. Lemma 2.3. Assume that X has a weakly continuous duality map J ϕ with gauge ϕ. (i) For all x, y ∈ X, there holds the inequality Φ   x + y  ≤ Φ   x  +  y,J ϕ (x + y)  . (2.6) 4 Nonexpansive mappings (ii) Assumeasequence {x n } in X is weakly convergent to a point x. Then there holds the identity limsup n→∞ Φ    x n − y    = limsup n→∞ Φ    x n − x    + Φ   y − x  , x, y ∈ X. (2.7) Lemma 2.4 is the resolvent identity which can be found in [1]. Lemma 2.4. For λ,μ>0, there holds the identity J λ x = J μ  μ λ x +  1 − μ λ  J λ x  , x ∈ X. (2.8) 3. Main results Theorem 3.1. Let X be a real reflexive Banach space and have a weakly continuous duality mapping J ϕ with ϕ.SupposeC is a closed convex subset of X,andT : C → C is a nonexpansive mapping, let f : C → C be a fixed contractive mapping. For t ∈ (0,1), {x t } is defined by (1.11). Then T has a fixed point if and only if {x t } remains bounded as t → 0 + , and in this case, {x t } converges strongly to a fixed point of T as t → 0 + . Proof. Assume first that F(T) = φ.Takeu ∈ F(T), it follows that   x t − u   =   tf  x t  +(1− t)Tx t − u   ≤ t   f  x t  − u   +(1− t)   Tx t − u   ≤ tβ   x t − u   + t   f (u) − u   +(1− t)   x t − u   =  1 − (1 − β)t    x t − u   + t   f (u) − u   . (3.1) Hence   x t − u   ≤ 1 1 − β   f (u) − u   . (3.2) Therefore, {x t } is bounded, so are {Tx t } and { f (x t )}. Next assume that {x t } is bounded as t → 0 + .Assumet n → 0 + and {x t n } is bounded. Since X is reflexive, we may assume that x t n  p for some p ∈ C.SinceJ ϕ is weakly con- tinuous, we have by Lemma 2.3, limsup n→∞ Φ    x t n − x    = limsup n→∞ Φ    x t n − p    + Φ    x − p    , ∀x ∈ X. (3.3) Put g(x) = limsup n→∞ Φ    x t n − x    , x ∈ X. (3.4) It follows that g(x) = g(p)+Φ   x − p  , x ∈ X. (3.5) R. Chen and Z. Zhu 5 Since   x t n − Tx t n   = t n 1 − t n   f (x t n ) − x t n   −→ 0, (3.6) we obtain g(Tp) = limsup n→∞ Φ    x t n − Tp    ≤ limsup n→∞ Φ    Tx t n − Tp    ≤ limsup n→∞ Φ    x t n − p    = g(p). (3.7) On the other hand, however, g(Tp) = g(p)+Φ   Tp− p  . (3.8) From (3.7)and(3.8), we get Φ   Tp− p  ≤ 0. (3.9) Hence Tp = p and p ∈ F(T). Now we prove that {x t } converges strongly to a fixed point of T provided it remains bounded when t → 0. Let {t n } be a sequence in (0,1) such that t n → 0andx t n  p as n →∞. Then the argu- ment above shows that p ∈ F(T). We next show that x t n → p.Asamatteroffact,wehave by Lemma 2.3, Φ    x t n − p    = Φ    t n  Tx t n − p  +  1 − t n  f (x t n  − p     ≤ Φ  t n   Tx t n − p    +  1 − t n  f  x t n  − p, J ϕ  x t n − p  ≤ t n Φ    x t n − p    +  1 − t n  f  x t n  − f (p),J ϕ  x t n − p  +  1 − t n  f (p) − p,J ϕ  x t n − p  . (3.10) This implies that Φ    x t n − p    ≤  f  x t n  − f (p),J ϕ  x t n − p  +  f (p) − p,J ϕ  x t n − p  ≤ β   x t n − p     J ϕ  x t n − p    +  f (p) − p,J ϕ  x t n − p  = βΦ    x t n − p    +  f (p) − p,J ϕ  x t n − p  , (3.11) that is, Φ    x t n − p    ≤ 1 1 − β  f (p) − p,J ϕ  x t n − p  . (3.12) Now noting that x t n  p implies J ϕ (x t n − p)  0, we get Φ    x t n − p    −→ 0. (3.13) Hence x t n → p. 6 Nonexpansive mappings We have proved for any sequence {x t n } in {x t : t ∈ (0,1)} that there exists a subse- quencewhichisstilldenotedby {x t n } thatconvergestosomefixedpointp of T.To prove that the entire net {x t } converges strongly to p, we assume there exists another sequence {s n }∈(0,1) such that x s n → q,thenq ∈ F(T). We have to show p = q. Indeed, for u ∈ F(T), it is easy to see that  x t − Tx t ,J ϕ  x t − u  = Φ    x t − u    +  u − Tx t ,J ϕ  x t − u  ≥ Φ    x t − u    −   u − Tx t   ·   J ϕ  x t − u    ≥ Φ    x t − u    − Φ    x t − u    = 0. (3.14) On the other hand, since x t − Tx t = t 1 − t  f  x t  − x t  , (3.15) we get for t ∈ (0,1) and u ∈ F(T),  x t − f  x t  ,J ϕ  x t − u  ≤ 0. (3.16) Since the sets {x t − u} and {x t } are bounded and a Banach space X has a weakly contin- uous duality map J ϕ ,thenJ ϕ is single valued and weak-to-weak ∗ sequentially continuous, for an y u ∈ F(T), by x s n → q(s n → 0), we have   x s n − f  x s n  −  q − f (q)    −→ 0  s n −→ 0  ,    x s n − f  x s n  ,J ϕ  x s n − u  −  q − f (q),J ϕ (q − u)    =    x s n − f  x s n  −  q − f (q)  ,J ϕ  x s n − u  +  q − f (q),J ϕ  x s n − u  − J ϕ (q − u)    ≤   x s n − f  x s n  −  q − f (q)      J ϕ  x s n − u    +    q − f (q),J ϕ  x s n − u  − J ϕ (q − u)    as s n −→ 0. (3.17) Therefore, we get  q − f (q),J ϕ (q − u) = lim s n →0  x s n − f  x s n  ,J ϕ  x s n − u  ≤ 0. (3.18) Interchange p and u to obtain  q − f (q),J ϕ (q − p)  ≤ 0. (3.19) Interchange q and u to obtai n  p − f (p),J ϕ (p − q)  ≤ 0. (3.20) This implies that  (p − q) −  f (p) − f (q)  ,J ϕ (p − q)  ≤ 0. (3.21) R. Chen and Z. Zhu 7 That is, p − qϕ   p − q  ≤ βp − qϕ   p − q  . (3.22) This is a contradiction, so we have p = q. The proof is complete.  Remark 3.2. Theorem 3.1 is proved in a weaker condition than [7, Theorem 4.1], and the method of proof is different from [7], we introduce a continuous strict increasing function. Next two main results are about accretive operators, we consider the problem of find- ing a zero of an m-accretive operator A in a reflexive Banach space X,0 ∈ Ax. Denote by F(A) the zero set of A, that is, F(A):=  x ∈ D(A):0∈ Ax  = A −1 (0). (3.23) Theorem 3.3. Suppose that X is reflexive and has a weakly continuous duality map J ϕ with gauge ϕ.SupposethatA is an m-accretive operator in X such that C = D(A) is convex with F(A) = φ,and f : C → C is a fixed contractive map. Assume (i) α n → 0 and  ∞ n=0 α n =∞, (ii) γ n →∞. Then the sequence {x n } defined by (1.12) converges strongly to a point in F(A). Proof. Firstweprove {x t } is bounded. Indeed, take u ∈ F(A), then   x n+1 − u   ≤ α n   f  x n  − u   +  1 − α n    J r n x n − u   ≤ α n β   x n − u   + α n   f (u) − u   +  1 − α n    x n − u   ≤  1 − (1 − β)α n    x n − u   + α n   f (u) − u   . (3.24) By induction, we get   x n − u   ≤ max    x 0 − u   , 1 1 − β   f (u) − u    ∀ n ≥ 0. (3.25) This implies that {x n } is bounded, so are { f (x n )} and {J r n x n }, and hence   x n+1 − J r n x n   = α n   f  x n  − J r n x n   −→ 0. (3.26) We next prove limsup n→∞  f (p) − p,J ϕ  x n − p  ≤ 0, p ∈ F(A). (3.27) By Theorem 3.1,putp = lim t→0 x t ,wetakeasubsequence{x n k } of {x n } such that limsup n→∞  f (p) − p,J ϕ  x n − p  = limsup n→∞  f (p) − p,J ϕ  x n k − p  . (3.28) 8 Nonexpansive mappings Since X is reflexive, we may further assume that x n k  x. Moreover, since   x n+1 − J r n x n   −→ 0, (3.29) we obtain J r n k −1 x r n k −1 x. (3.30) Taking the limit as k →∞in the relation  J r n k −1 x r n k −1 ,A r n k −1 x r n k −1  ∈ A, (3.31) we get [ x,0] ∈ A, that is, x ∈ F(A). Hence by (3.28)and(3.18), we have limsup n→∞  f (p) − p,J ϕ  x n − p  =  f (p) − p,J ϕ (x − p)  ≤ 0. (3.32) That is, (3.27)holds. Finally, we prove that x n → p. We apply Lemma 2.3 to get Φ    x n+1 − p    = Φ     1 − α n  J r n x n − p  + α n  f  x n  − p     = Φ     1 − α n  J r n x n − p  + α n  f  x n  − f (p)  + α n  f (p) − p    ) ≤ Φ     1 − α n  J r n x n − p  + α n  f  x n  − f (p)     + α n  f (p) − p,J ϕ  x n+1 − p  ≤  1 − (1 − β)α n  Φ    x n − p   )+α n  f (p) − p,J ϕ  x n+1 − p  . (3.33) Applying Lemma 2.1,weget Φ    x n − p    −→ 0. (3.34) That is, x n − p→0, that is, x n → p. The proof is complete.  Theorem 3.4. Suppose that X is reflexive and has a weakly continuous duality map J ϕ with gauge ϕ.SupposethatA is an m-accretive operator in X such that C = D(A) is convex with F(A) = φ,and f : C → C is a fixed contractive map. Assume (i) α n −→ 0, ∞  n=0 α n =∞, ∞  n=1   α n+1 − α n   < ∞, (ii) γ n ≥ ε ∀n, ∞  n=1   γ n+1 − γ n   < ∞. (3.35) Then {x n } defined by (1.12) converges strongly to a point in F(A). R. Chen and Z. Zhu 9 Proof. We only include the differences. We have x n+1 = α n f  x n  +  1 − α n  J γ n x n , x n = α n−1 f  x n−1  +  1 − α n−1  J γ n−1 x n−1 . (3.36) Thus, x n+1 − x n = α n f  x n  +  1 − α n  J γ n x n −  α n−1 f  x n−1  +  1 − α n−1  J γ n−1 x n−1  = α n  f  x n  − f  x n−1  +  α n − α n−1  f  x n−1  − J γ n−1 x n−1  +  1 − α n  J γ n x n − J γ n−1 x n−1  . (3.37) If γ r n−1 ≤ γ n ,byLemma 2.4,weget J γ n x n = J γ n−1  γ n−1 γ n x n +  1 − γ n−1 γ n  J γ n x n  , (3.38) we have   J γ n x n − J γ n−1 x n−1   ≤ γ n−1 γ n   x n − x n−1   +  1 − γ n−1 γ n    J γ n x n − x n−1   ≤   x n − x n−1   +  γ n − γ n−1 γ n    J γ n x n − x n−1   ≤   x n − x n−1   + 1 ε   γ n−1 − γ n     J γ n x n − x n−1   . (3.39) It follows from the above results that   x n+1 − x n   ≤   α n − α n−1     f  x n−1  − J γ n−1 x n−1   +  1 − α n    x n − x n−1   + 1 ε  1 − α n    γ n−1 − γ n     J γ n x n − x n−1   + α n β   x n − x n−1   ≤ M    α n − α n−1   +   γ n−1 − γ n    +  1 − (1 − β)α n    x n − x n−1   , (3.40) where M>0 is some appropriate constant. Similarly, we can prove the last inequality if γ n−1 ≥ γ n . By assumptions (i) and (ii) and Lemma 2.1,wehave   x n+1 − x n   −→ 0. (3.41) This implies that   x n − J γ n x n   ≤   x n+1 − x n   +   x n+1 − J γ n x n   . (3.42) Since x n+1 − J γ n x n =α n  f (x n ) − J γ n x n →0. It follows from (3.42)that   A γ n x n   = 1 γ n   x n − J γ n x n   ≤ 1 ε   x n − J γ n x n   −→ 0. (3.43) Now if {x n k } is a subsequence of {x n } converging weakly to a point x, then taking the limit as k →∞in the relation  J γ n k x n k ,A γ n k x n k  ∈ A, (3.44) 10 Nonexpansive mappings we ge t [ x,0] ∈ A, that is, x ∈ F(A). We therefore conclude that all weak limit points of {x n } are zeros of A. TherestoftheprooffollowsfromTheorem 3.3. The proof is complete.  Acknowledgment This work is supported by the National Science Foundation of China, Grants 10471033 and 10271011. References [1] V. Barbu, Nonlinear Semigroups and Differential Equations in Banach Spaces, Editura Academiei Republicii Socialiste Rom ˆ ania, Bucharest; Noordhoff, Leiden, 1976. [2] F.E.Browder,Convergence of approximants to fixed points of nonexpansive non-linear mappings in Banach spaces, Archive for Rational Mechanics and Analysis 24 (1967), no. 1, 82–90. [3] , Convergence theorems for sequences of nonlinear operators in Banach spaces, Mathema- tische Zeitschrift 100 (1967), no. 3, 201–225. [4] T C. Lim and H K. Xu, Fixed point theorems for asymptotically nonexpansive mappings,Non- linear Analysis 22 (1994), no. 11, 1345–1355. [5] H K. Xu, Iterative algorithms for nonlinear operators, Journal of the London Mathematical Soci- ety. Second Series 66 (2002), no. 1, 240–256. [6] , An iterative approach to quadratic optimization, Journal of Optimization Theory and Applications 116 (2003), no. 3, 659–678. [7] , Viscosity approximation methods for nonexpansive mappings, Journal of Mathematical Analysis and Applications 298 (2004), no. 1, 279–291. [8] , Strong convergence of an iterative method for nonexpansive and accretive operators,Jour- nal of Mathematical Analysis and Applications 314 (2006), no. 2, 631–643. Rudong Chen: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China E-mail address: chenrd@tjpu.edu.cn Zhichuan Zhu: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China E-mail address: zhuzcnh@yahoo.com.cn . VISCOSITY APPROXIMATION FIXED POINTS FOR NONEXPANSIVE AND m-ACCRETIVE OPERATORS RUDONG CHEN AND ZHICHUAN ZHU Received 10 June 2006; Accepted 22 July. subset of X,andlet T : C → C be a nonexpansive mapping if for all x, y ∈ C,suchthat Tx− Ty≤x − y. (1.1) We use F(T) to denote the set of fixed points of T, that is, F(T) ={x ∈ C : x = Tx} .And . defined the following one viscosity iteration for nonexpansive mappings in uniformly smooth Banach space. Hindawi Publishing Corporation Fixed Point Theory and Applications Volume 2006, Article ID