关于一个算子的收敛性

On Convergence of An Operator

本文在文[2][3]的基础上,进一步讨论了((?)x)的收敛性问题,得到若f(t)=0(t^(2at^2))(t→∞),并满足在有限区间上有界及一定的连续条件,则有Bn(f,X)→f(x)(n→∞)且给出了一个使Bn(f,x)不存在的函数数类。

Suppose that f(t) is defined in [0, ∞),let Bn be a generalization of S. Bernstein's Polynomial to infinite interval, i. e Bn(f,x)=e^(-(nx)2) sum from K=0 to ∞ f(K^(1/2)/n)(nx)^(2k)/K! x∈[0, ∞) The following theorem is obtained Theorem 1 Let f(t) be a function in [0, ∞) (Ⅰ) If f(t)=0(t^(2λt^2)), r∈[0,∞) f(t) be continuous in x, then lim (?)(f,x)=f(x) (Ⅱ) If f(t)≥t^(2λ(t^2)t^2) when λ(t) is any monotonically increasing function so that lim (?), (t)=∞ ,then Bn(f,x) does not exist.