(New page: The case <math>\mu(X)=\infty</math> the inequality is true. Suppose <math>\mu(X)</math> is finite, we have Given <math>p^{'}=\frac{p+r}{2}</math>, <math>\int_{X}|f|^{r}d\mu = \int_{X}|f...) |
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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 | + | <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> |
Revision as of 15:13, 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} $
