| Line 5: | Line 5: | ||
Taking the limit of both side as <math>n</math> go to infinity, we get | Taking the limit of both side as <math>n</math> go to infinity, we get | ||
| − | <math>\lim_{n\to \infty}||f||_{n} | + | <math>\lim_{n\to \infty}||f||_{n} \leq ||f||_{\infty}</math> |
| − | Let <math>M<||f||_{\infty} </math> | + | Let <math>M<||f||_{\infty} </math>, and <math>E=\{|f|>M\}</math>, then |
| + | |||
| + | <math>\lim_{n\to \infty}||f||_{n} \geq \lim_{n\to \infty}(\int_{E}|f|^{n})^{1/n}</math> | ||
Revision as of 13:40, 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} \leq ||f||_{\infty} $
Let $ M<||f||_{\infty} $, and $ E=\{|f|>M\} $, then
$ \lim_{n\to \infty}||f||_{n} \geq \lim_{n\to \infty}(\int_{E}|f|^{n})^{1/n} $
