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| − | a | + | '''a)''' |
| + | Notice that <math>\mu(\{|f|>0\})>0</math>, so we have | ||
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| + | <math>(\int_{X}|f|^{n})^{1/n} \leq (\mu(X)||f||_{\infty})^{1/n}</math> | ||
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| + | Taking the limit of both side as <math>n</math> go to infinity, we get | ||
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| + | <math>\lim_{n\to \infty}||f||_{n} \leq ||f||_{\infty}</math> | ||
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| + | Let <math>M<||f||_{\infty} </math>, and <math>E=\{|f|>M\}</math>, then | ||
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| + | <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> | ||
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| + | So, <math>(\int_{X}|f|^{n})^{1/n} \geq (\mu(X)||f||_{\infty})^{1/n}</math> | ||
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| + | and we have the identity. | ||
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| + | Notice that it is true for true for finite measure space. | ||
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| + | '''b)''' | ||
| + | <math>\lim_{n\to\infty}\frac{\int_{X}|f|^{n+1}}{\int_{X}|f|^{n}} = \lim_{n\to\infty}\frac{(||f||_{n+1})^{n+1}}{(||f||_{n})^{n}} = \frac{(||f||_{\infty})^{n+1}}{(||f||_{\infty})^{n}}=||f||_{\infty}</math> | ||
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| + | '''c)''' | ||
| + | If the space is of infinite measure, it is not true. Let <math>f(x)=1</math> for all real <math>x</math>, we have a counter example. | ||
Latest revision as of 14:50, 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}} = \lim_{n\to\infty}\frac{(||f||_{n+1})^{n+1}}{(||f||_{n})^{n}} = \frac{(||f||_{\infty})^{n+1}}{(||f||_{\infty})^{n}}=||f||_{\infty} $
c) If the space is of infinite measure, it is not true. Let $ f(x)=1 $ for all real $ x $, we have a counter example.
