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'''a)''' | '''a)''' | ||
Notice that <math>\mu(\{|f|>0\})>0</math>, so we have | Notice that <math>\mu(\{|f|>0\})>0</math>, so we have | ||
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and we have the identity. | and we have the identity. | ||
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| + | Notice that it is true for true for finite measure space. | ||
'''b)''' | '''b)''' | ||
| + | <math>\lim_{n\to\infty}\frac{\int_{X}|f|^{n+1}}{\int_{X}|f|^{n}}</math> | ||
Revision as of 13:46, 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.
Notice that it is true for true for finite measure space.
b) $ \lim_{n\to\infty}\frac{\int_{X}|f|^{n+1}}{\int_{X}|f|^{n}} $
