| Line 9: | Line 9: | ||
Now, let <math>g=|f|^{p{'}}</math>, then <math>w(y)=\mu(\{g>y\} \leq \frac{c_{0}}{y^{p/p{'}}}</math> | Now, let <math>g=|f|^{p{'}}</math>, then <math>w(y)=\mu(\{g>y\} \leq \frac{c_{0}}{y^{p/p{'}}}</math> | ||
| − | <math>\int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq c_{0}\int_{0}^{\infty}\frac{dy}{y^{p | + | <math>\int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq c_{0}\int_{0}^{\infty}\frac{dy}{y^{p/p{'}}}</math> |
Revision as of 15:38, 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{'}} $, then $ w(y)=\mu(\{g>y\} \leq \frac{c_{0}}{y^{p/p{'}}} $
$ \int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq c_{0}\int_{0}^{\infty}\frac{dy}{y^{p/p{'}}} $
