(New page: # <math>\log ||f||_p=\log \left(\int |f|^p\right)^{1/p}=\frac{1}{p}\log\left(\int|f|^p\right)\geq\frac{1}{p}\int\log|f|^p=\int\log|f|d\mu</math> The last but two inequality is due to the...) |
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First inequality is by <math>log(x)\leq x-1</math> from hint; the second equality is due to the property of probability space<math>\int d\mu=1</math> | First inequality is by <math>log(x)\leq x-1</math> from hint; the second equality is due to the property of probability space<math>\int d\mu=1</math> | ||
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| + | By L' Hospital Rule,<math>\phi(p)=\frac{|f|^p-1}{p}\to log|f|,(p\to0)</math> | ||
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| + | Since <math>\phi(p)</math> is monotone increasing, by Monotone Convergence Theorem, <math>\int\frac{|f|^p-1}{p}\to\int\log|f|d\mu</math> | ||
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| + | Therefore,<math>\lim\limits_{p\to0}\log||f||_p=\int_X\log|f|d\mu</math>, and the result follows by using Exp on the both sides. | ||
Latest revision as of 10:15, 10 July 2008
$ \log ||f||_p=\log \left(\int |f|^p\right)^{1/p}=\frac{1}{p}\log\left(\int|f|^p\right)\geq\frac{1}{p}\int\log|f|^p=\int\log|f|d\mu $
The last but two inequality is due to the integral form of Jensen's inequality.
$ \log||f||_p=\frac{1}{p}\log\left(\int|f|^p\right)\leq\frac{1}{p}\left(\int|f|^p-1\right)=\frac{1}{p}\int(|f|^p-1)=\int\frac{|f|^p-1}{p} $
First inequality is by $ log(x)\leq x-1 $ from hint; the second equality is due to the property of probability space$ \int d\mu=1 $
By L' Hospital Rule,$ \phi(p)=\frac{|f|^p-1}{p}\to log|f|,(p\to0) $
Since $ \phi(p) $ is monotone increasing, by Monotone Convergence Theorem, $ \int\frac{|f|^p-1}{p}\to\int\log|f|d\mu $
Therefore,$ \lim\limits_{p\to0}\log||f||_p=\int_X\log|f|d\mu $, and the result follows by using Exp on the both sides.
