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Now, let <math>g=|f|^{p{'}}</math> | Now, let <math>g=|f|^{p{'}}</math> | ||
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Revision as of 16:31, 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.
Now, let $ g=|f|^{p{'}} $
$ \int_{X}gd\mu $
