| Line 1: | Line 1: | ||
| − | a | + | |
| + | '''a)''' | ||
| + | Notice that <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> | ||
| Line 9: | Line 11: | ||
Let <math>M<||f||_{\infty} </math>, and <math>E=\{|f|>M\}</math>, then | 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} \geq (\mu(E)M^{n})^{1/n}</math> | + | <math>\lim_{n\to \infty}||f||_{n} \geq \lim_{n\to \infty}(\int_{E}|f|^{n})^{1/n} \geq (\mu(E)M^{n})^{1/n} = M</math> |
| + | |||
| + | So, <math>(\int_{X}|f|^{n})^{1/n} \geq (\mu(X)||f||_{\infty})^{1/n}</math> | ||
| + | |||
| + | and we have the identity. | ||
| + | |||
| + | '''b)''' | ||
Revision as of 13:43, 11 July 2008
a) Notice that $ \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} \geq (\mu(E)M^{n})^{1/n} = M $
So, $ (\int_{X}|f|^{n})^{1/n} \geq (\mu(X)||f||_{\infty})^{1/n} $
and we have the identity.
b)
