solved, thank you all

Printable View

- Nov 25th 2008, 04:46 PMfrankmelody[Solved]limit of ansequence
solved, thank you all

- Nov 26th 2008, 06:47 AMOpalg
Suppose, for a contradiction, that there is a set of positive measure on which f is not zero. Then by Egorov's theorem there is a (maybe smaller) set A on which $\displaystyle f_n(x)\to f(x)$ uniformly. Let g be the function that is equal to f on A and 0 outside A. Then $\displaystyle \int_A|g|^2d\mu>0$.

On the other hand, it follows from Parseval's equality that $\displaystyle \langle g,f_n\rangle\to0$ as n→∞. But $\displaystyle \langle g,f_n\rangle = \int_X g\bar{f_n}\,d\mu = \int_A g\bar{f_n}\,d\mu$ since g is zero off A. This converges to $\displaystyle \int_A |g|^2d\mu$ as n→∞, by the uniform convergence. That gives your contradiction. - Nov 26th 2008, 08:11 AMfrankmelody
thank you for your help!