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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. | ||
| − | Now, let <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} | + | <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 16:37, 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{'}}} $
