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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

  1. 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

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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