solved, thank you all

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- November 25th 2008, 05:46 PMfrankmelody[Solved]limit of ansequence
solved, thank you all

- November 26th 2008, 07:47 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 uniformly. Let g be the function that is equal to f on A and 0 outside A. Then .

On the other hand, it follows from Parseval's equality that as n→∞. But since g is zero off A. This converges to as n→∞, by the uniform convergence. That gives your contradiction. - November 26th 2008, 09:11 AMfrankmelody
thank you for your help!