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Old November 25th, 2008, 06:46 PM
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Default [Solved]limit of ansequence

solved, thank you all

Last edited by frankmelody; November 28th, 2008 at 10:37 AM. Reason: solved
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Old November 26th, 2008, 08:47 AM
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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 f_n(x)\to f(x) uniformly. Let g be the function that is equal to f on A and 0 outside A. Then \int_A|g|^2d\mu>0.

On the other hand, it follows from Parseval's equality that \langle g,f_n\rangle\to0 as n→∞. But \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 \int_A |g|^2d\mu as n→∞, by the uniform convergence. That gives your contradiction.
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Old November 26th, 2008, 10:11 AM
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thank you for your help!
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