| Line 5: | Line 5: | ||
Assuming <math>x>0 </math> we have | Assuming <math>x>0 </math> we have | ||
| − | <math>\dfrac{F(x)}{x^{\frac{1}{2}} \leq ||f\chi_{[0,x]}||_2 \to 0 </math> as <math>x\to 0^+ </math> | + | <math>\dfrac{F(x)}{x^{\frac{1}{2}}} \leq ||f\chi_{[0,x]}||_2 \to 0 </math> as <math>x\to 0^+ </math> |
by absoulute continuity, since <math>f^2 \in L^1. </math> | by absoulute continuity, since <math>f^2 \in L^1. </math> | ||
-Matty | -Matty | ||
Latest revision as of 10:39, 18 July 2008
- 4. Use Ho(e)lder's Inequality to obtain
$ 0\leq \int_0^xf\leq \big[ \int_0^x(f^2) ] ^\frac{1}{2} [ \int_0^x 1^2]^\frac{1}{2}= ||f\chi_{[0,x]}||_2 x^\frac{1}{2} $
Assuming $ x>0 $ we have
$ \dfrac{F(x)}{x^{\frac{1}{2}}} \leq ||f\chi_{[0,x]}||_2 \to 0 $ as $ x\to 0^+ $
by absoulute continuity, since $ f^2 \in L^1. $
-Matty
