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<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. | <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. | ||
| − | + | <math>\int_{X}|f|^{p^{'}} = \int_{0}^{\infty}\mu(\{|f|>y^{\frac{p}{p^{'}}}\})</math> | |
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Revision as of 15:22, 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.
$ \int_{X}|f|^{p^{'}} = \int_{0}^{\infty}\mu(\{|f|>y^{\frac{p}{p^{'}}}\}) $
