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<math>\lim_{n\to \infty}||f||_{n} = ||f||_{\infty}</math> | <math>\lim_{n\to \infty}||f||_{n} = ||f||_{\infty}</math> | ||
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| + | Let <math>M<||f||_{\infty} </math> | ||
Revision as of 14:35, 11 July 2008
a/$ \mu(\{|f|>0\})>0 $, so we have
$ (\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n} $
Taking the limit of both side as $ n $ go to infinity, we get
$ \lim_{n\to \infty}||f||_{n} = ||f||_{\infty} $
Let $ M<||f||_{\infty} $
