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1、Journal of Mathematical Research )AbstractLet E be a real Banach space and let A be an m-accretive operator with a zero.Define a sequence xn as follows: xn+1= nf(xn) + (1 n)Jrnxn, where n, rn are sequences satisfying certain conditions, and Jrdenotes the resolvent (I+rA)1for r 1. Strong convergence
2、of the algorithm xn is obtained provided that E either has a weakly continuous duality map or is uniformly smooth.Keywordsfixed point; nonexpansive mapping; m-accretive operator; viscosity approximation; weakly continuous duality map; uniformly smooth Banach space.Document code AMR(2000) Subject Cla
3、ssification 47H06; 47H10Chinese Library Classification O1771. IntroductionIn the sequel, we assume that E is a real Banach space with norm kk, denote the fixed pointset by F(T) = x E;Tx = x, the weak convergence by , the strong convergence by .A mapping T with its domain D(T) and range R(T) in E is
4、called nonexpansive (respectivelycontractive) if for all x,y D(T) such that kTx Tyk kx yk (respectively kTx Tyk kx yk for some 0 0.Denote by F the zero set of A; i.e.,F := A1(0) = x D(A) : 0 Ax.Denote by Jrthe resolvent of A for r 0: Jr= (I + rA)1.It is known that Jris a nonexpansive mapping from E
5、to C := D(A). For the proof of our main results, we shall need the following lemmas.Lemma 2.12Let E be a uniformly smooth Banach space and let T : C C be a nonexpansivemapping with a fixed point. For each fixed u C and every t (0,1), the unique fixed point xt C of the contraction x 7 tu + (1 t)Tx co
6、nverges strongly as t 0 to fixed point of T.Define Q : C F(T) by Qu = s limt0xt. Then Q is the unique sunny nonexpansive retractfrom C onto F(T); that is, Q satisfies the property:hu Qu,J(z Qu)i 0, u C, z F(T).Lemma 2.24Let n be a nonnegative real sequence that satisfies the condition: n+1(1n)n+nnfo
7、r all n n0, where the sequence n 0,1, and n satisfies the conditions: (i) limnn= 0; (ii) n=0n= ; (iii) limnn= 0, then limnn= 0.Lemma 2.3 (The Sub-differential Inequality) Let E be a Banach space, J the normalized dualitymapping from E into 2E, x,y E, kx + yk2 kxk2+ 2hy,j(x + y)i,j(x + y) J(x + y).Le
8、mma 2.45(The Resolvent Identity) For , 0, there holds the identity: Jx = J(x + (1 )Jx), x E.Lemma 2.56Assume that c2 c1 0. Then kJc1x xk kJc2x xk for all x E.Lemma 2.67Assume that E has a weakly continuous duality map Jwith gauge .(i) For all x,y E, there holds the inequality(kx + yk) (kxk) + hy,J(x
9、 + y)i.(ii) Assume a sequence xn in E is weakly convergent to a point x. Then there holds theidentity limsup n(kxn yk) = limsup n(kxn xk) + (ky xk).3. Main resultsLet E be a real Banach space, C a nonempty closed convex subset of E and T be a nonex- pansive mapping from C into itself with F(T) 6= .
10、For t (0,1) and f C, let xt C be theunique fixed point of the contraction x 7 tf(x) + (1 t)Tx on C; that isxt= tf(xt) + (1 t)Txt.(1)582YAN L X and ZHOU H YTheorem 3.1 Let E be a reflexive Banach space and has a weakly continuous duality map Jwith a gauge . Let C be a closed convex subset of E and T
11、be a nonexpansive mapping from C into itself with F(T) 6= , f C. For t (0,1), xt C is the unique solution in C to Eq.(3.1).Then T has a fixed point if and only if xt remains bounded as t 0+, and in this case, xtconverges strongly to a fixed point of T. If we define Q : C F(T) byQ(f) := limt0xt,f C,(
12、2)then Q(f) solves the variational inequalityh(I f)Q(f),J(Q(f) p)i 0,f C,p F(T).(3)In particular, if f = u C is a constant, then (2) reduces to the sunny nonexpansive retractionfrom C onto F(T), hQ(u) u,J(Q(u) p)i 0,u C,p F(T).Proof Necessity. Assume that F(T) 6= . Take p F(T), for t (0,1).kxt pk =
13、kt(f(xt) p) + (1 t)(Txt p)k tkf(xt) pk + (1 t)kTxt pk tkf(xt) f(p)k + tkf(p) pk + (1 t)kTxt pk (1 t + t)kxt pk + tkf(p) pk.We obtain kxt pk 1 1kf(p) pk. Therefore xt is bounded.Sufficiency. Assume that xt is bounded as t 0+. Assume that tn 0+and xtnis bounded. Since E is reflexive, we may assume tha
14、t xtn z for some z C. Since Jisweakly continuous, by Lemma 2.6, we have limsupn(kxtn xk) = limsupn(kxtnzk) + (kx zk),x E. Put g(x) = limsupn(kxtn xk),x E. It follows that g(x) = g(z) + (kx zk),x E. xn is bounded, so are f(xtn) and Txtn, we getkxtn Txtnk =tn 1 tnkf(xtn) xtnk 0.g(Tz) = limsup n(kxtn T
15、zk) = limsup n(kTxtn Tzk) limsup n(kxtn zk) = g(z).(4)On the other hand, g(Tz) = g(z) + (kTz zk).(5)Combining (4) and (5) yields that (kTz zk) 0. Hence Tz = z, and z F(T).Next we claim that xt converges strongly to a fixed point of T provided that it remains bounded as t 0+. Let tn be a sequence in (0,1) such that tn 0 and xtn z, as n .Then z F(T) by the above arguments. We show that xtn z. In fact, by Lemma 2.6,(kxtn
