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Given <math>p^{'}=\frac{p+r}{2}</math>, | Given <math>p^{'}=\frac{p+r}{2}</math>, | ||
| − | <math>\int_{X}|f|^{r}d\mu \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p^{'}} \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p}</math> | + | <math>\int_{X}|f|^{r}d\mu \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p^{'}} \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p}</math> by Holder. |
| + | |||
| + | Let <math>g=\f}^{p{'}}</math> | ||
| + | |||
| + | <math>\int_{X}|f|^{p^{'}} = </math> | ||
Revision as of 15:19, 11 July 2008
The case $ \mu(X)=\infty $ the inequality is true.
Suppose $ \mu(X) $ is finite, we have
Given $ p^{'}=\frac{p+r}{2} $,
$ \int_{X}|f|^{r}d\mu \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p^{'}} \leq \int_{X}|f|^{p^{'}}(\mu(X))^{1-r/p} $ by Holder.
Let $ g=\f}^{p{'}} $
$ \int_{X}|f|^{p^{'}} = $
