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<math>\int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq \int_{0}^{1}w(y)dy + \int_{1}^{\infty}w(y)dy \leq \mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}}</math> | <math>\int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq \int_{0}^{1}w(y)dy + \int_{1}^{\infty}w(y)dy \leq \mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}}</math> | ||
| − | Let <math>c=\mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}}</math>. But then, <math>c</math> will also depends on | + | Let <math>c=\mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}}</math>. But then, <math>c</math> will also depends on X. |
Latest revision as of 16:28, 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{'}}} $
Let $ c=c_{0}\int_{0}^{\infty}\frac{dy}{y^{p/p{'}}} $. DONE!!!
Posted after 5pm Friday:
The above answer is wrong. It should be:
$ \int_{X}g d\mu = \int_{0}^{\infty}w(y)dy \leq \int_{0}^{1}w(y)dy + \int_{1}^{\infty}w(y)dy \leq \mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}} $
Let $ c=\mu(X) + c_{0}\int_{1}^{\infty}\frac{dy}{y^{p/p{'}}} $. But then, $ c $ will also depends on X.
