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| − | For any <math> x\in[0,1], |x^{\frac{1}{n}}f(x)| \leq |f(x)| </math> | + | For any <math> x\in[0,1], |x^{\frac{1}{n}}f(x)| \leq |f(x)| </math>. |
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| + | <math>\lim_{n \rightarrow \infty} x^{\frac{1}{n}} = 1</math> if <math>x > 0</math>, then <math> lim_{n \rightarrow \infty}|x^{\frac{1}{n}}f(x)| = |f(x)|</math> a.e. on <math>[0,1]</math>. Since <math> f \in L^1</math>, then by DCT <math>\lim_{n \rightarrow \infty}\int_0^1x^{\frac{1}{n}}|f(x)|\mbox{d}x = \int_0^1|f(x)|\mbox{d}x < \infty. | ||
Latest revision as of 03:14, 10 July 2008
For any $ x\in[0,1], |x^{\frac{1}{n}}f(x)| \leq |f(x)| $.
$ \lim_{n \rightarrow \infty} x^{\frac{1}{n}} = 1 $ if $ x > 0 $, then $ lim_{n \rightarrow \infty}|x^{\frac{1}{n}}f(x)| = |f(x)| $ a.e. on $ [0,1] $. Since $ f \in L^1 $, then by DCT $ \lim_{n \rightarrow \infty}\int_0^1x^{\frac{1}{n}}|f(x)|\mbox{d}x = \int_0^1|f(x)|\mbox{d}x < \infty. $
