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

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.