Banach 空间中强增生映象方程的迭代解(英文)  

Iterative Solution of Strongly Accretive Mapping Equations in Banach Space

作　　者：何晓林[1] 

机构地区：[1]泸州医学院基础部

出　　处：《应用基础与工程科学学报》1998年第3期12-19,共8页Journal of Basic Science and Engineering

摘　　要：设Ｅ为一实Ｂａｎａｃｈ空间，映象Ｔ：Ｅ→Ｅ一致连续、强增生．设映象Ｓ：ｘ→ｆ－Ｔｘ＋ｘ，ｘ∈Ｅ的值域有界且实序列｛αｎ｝∞ｎ＝０，｛βｎ｝∞ｎ＝０［０，１］满足条件αｎ→０，βｎ→０（ｎ→∞）和∑∞ｎ＝０αｎ＝∞，则Ｉｓｈｉｋａｗａ迭代序列｛ｘｎ｝∞ｎ＝０：ｘ０∈Ｅｘｎ＋１＝（１－αｎ）ｘｎ＋αｎＳｙｎｙｎ＝（１－βｎ）ｘｎ＋βｎＳｘｎ强收敛于方程Ｔｘ＝ｆ的唯一解．若Ｅ的对偶空间Ｅ＊是一致凸的且Ｔｘ＝ｆ的解存在，则上述结论在不假定Ｔ连续的情形下仍然成立．Let E be a real Banach space and T:E→E be a uniformly continuous strongly accretive mapping. Define S:E→E by Sx=f-Tx+x, x∈E.  Suppose that the range of S is bounded and that {α n} ∞ n=0 , {β n} ∞ n=0  0,1  satisfy α n→0, β n→0(n→∞) and ∑∞n=0α n =∞. Then the Ishikawa Iteration sequence {x n} ∞ n=0 :∈E x n+1 =(1-α n)x n+α nSy n y n=(1-β n)x n+β nSx nstrongly converges to the unique solution of the equation Tx=f. If the dual space E * of E is uniformly convex and the solution of Tx=f exists, then the above conclusion still holds without assuming the continuity of T.

关 键 词：ISHIKAWA迭代 增生映象 BANACH空间 

分 类 号：O177.2[理学—数学]