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| − | a/<math>\mu({|f|>0})>0</math>, so we have | + | a/<math>\mu(\{|f|>0\})>0</math>, so we have |
<math>(\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n}</math> | <math>(\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n}</math> | ||
Revision as of 13:33, 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} $
