| Line 1: | Line 1: | ||
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> | ||
| + | |||
| + | Taking the limit of both side as <math>n</math> go to infinity, we get | ||
| + | |||
| + | <math>\lim_{n\to \infty}||f||_{n} = ||f||_{\infty}</math> | ||
Revision as of 14:33, 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} $
