| Line 1: | Line 1: | ||
| − | + | 2a) | |
| + | |||
| + | First, subdivide <math> [0,1] </math> into tenths and remove the interior of the seventh. And let <math> E1 = [0,1]\(\frac{7}{10},\frac{8}{10}) </math>. | ||
| − | |||
<math>\int_X |f|^p = \int_{\{x:|f|\leq 1 \}} |f|^p+\int_{\{x:|f|> 1 \}} |f|^p \leq \int_{\{x:|f|\leq 1 \}} |f|^r+\int_{\{x:|f|> 1 \}} |f|^s <\infty</math> | <math>\int_X |f|^p = \int_{\{x:|f|\leq 1 \}} |f|^p+\int_{\{x:|f|> 1 \}} |f|^p \leq \int_{\{x:|f|\leq 1 \}} |f|^r+\int_{\{x:|f|> 1 \}} |f|^s <\infty</math> | ||
Latest revision as of 08:32, 16 July 2008
2a)
First, subdivide $ [0,1] $ into tenths and remove the interior of the seventh. And let $ E1 = [0,1]\(\frac{7}{10},\frac{8}{10}) $.
$ \int_X |f|^p = \int_{\{x:|f|\leq 1 \}} |f|^p+\int_{\{x:|f|> 1 \}} |f|^p \leq \int_{\{x:|f|\leq 1 \}} |f|^r+\int_{\{x:|f|> 1 \}} |f|^s <\infty $
b) Recall that a function $ \sigma (x):(r,s)\rightarrow \mathbb{R} $ is convex iff $ \sigma(tp_1+(1-t)p_2) \leq t\sigma (p_1)+(1-t)\sigma( p_2) $ for any $ t\in [0,1] $ and any $ r<p_1<p_2<s $.
Now, by Holder,
$ \int|f|^{tp_1}|f|^{(1-t)p_2} \leq \|f^{tp_1}\|_{\frac{1}{t}}\|f^{tp_2}\|_{\frac{1}{1-t}} $ since f is in $ L^{p_1} $ and $ L^{p_2} $ and since $ t+(1-t)=1 $.
Since log is increasing we may take log of both sides while preserving the inequality:
$ log[\int|f|^{tp_1}|f|^{(1-t)p_2}] \leq log[\|f^{tp_1}\|_{\frac{1}{t}}\|f^{tp_2}\|_{\frac{1}{1-t}}] =log[(\int|f|^{p_1})^t(\int|f|^{p_2})^{1-t}]= log[(\int|f|^{p_1})^t]+log[(\int|f|^{p_2})^{1-t}] $ $ = tlog[\int|f|^{p_1}]+(1-t)log[\int|f|^{p_2}]= tlog(\phi (p_1))+(1-t)log(\phi (p_2)) $
So $ tlog(\phi (p_1))+(1-t)log(\phi (p_2)) \geq log(\int|f|^{tp_1}|f|^{(1-t)p_2}) = log(\int|f|^{tp_1+(1-t)p_2})=log(\phi (tp_1+(1-t)p_2)). $
Thus $ log(\phi) $ is convex on $ (r,s). $
--Wardbc 19:57, 14 July 2008 (EDT)
