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Recall if <math>f\in L^1_{loc}, </math>  the result for #3 a follows from Lebesgue differentiation theorem.   
 
Recall if <math>f\in L^1_{loc}, </math>  the result for #3 a follows from Lebesgue differentiation theorem.   
  
Next if <math>f\notin L^1_{loc}</math> consider the following:
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Next if <math>f\notin L^1_{loc}</math> consider the following proof:
 
WLOG <math>f\geq 0 </math> by replacing <math> f </math> with <math> |f|.</math>
 
WLOG <math>f\geq 0 </math> by replacing <math> f </math> with <math> |f|.</math>
  
Let <math>x\in \mathbb{R}^n</math>.   
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Since <math>f\notin L^1_{loc}, \exists x\in \mathhbb{R}^n  and \exists K\subset \mathbb{R}^n, K </math> compact,s.t. <math>\int_Kf=\infty</math>.   
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Choose any <math>y\in\mathbb{R}^n </math>.  Then choose a cube <math>Q\supseteq K</math> centered at <math>y </math> which is possible since <math>K</math> compact implies <math>K </math> bounded. 
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Then <math> f^*(y)\geq \dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty </math>, so we have the result, namely <math>\{f*\geq f} =\mathbb{R}^n </math>
  
'''Case 1''', <math>\exists K\subset \mathbb{R}^n, K </math> compact, and<math>\int_Kf=\infty</math>. 
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QED
Choose a cube <math>Q\supseteq K</math> with <math>|Q|<\infty </math> which is possible since <math>K</math> compact implies <math>K </math> bounded.  WLOG <math>x</math> is the center of <math>Q.</math>
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Then <math>f^*(x)\geq\dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty</math>, so our result holds on <math>  \{x|\exists K s.t. \intKf=\infty \} </math>.
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Revision as of 15:06, 9 July 2008

Recall if $ f\in L^1_{loc}, $ the result for #3 a follows from Lebesgue differentiation theorem.

Next if $ f\notin L^1_{loc} $ consider the following proof: WLOG $ f\geq 0 $ by replacing $ f $ with $ |f|. $

Since $ f\notin L^1_{loc}, \exists x\in \mathhbb{R}^n and \exists K\subset \mathbb{R}^n, K $ compact,s.t. $ \int_Kf=\infty $. Choose any $ y\in\mathbb{R}^n $. Then choose a cube $ Q\supseteq K $ centered at $ y $ which is possible since $ K $ compact implies $ K $ bounded. Then $ f^*(y)\geq \dfrac{\int_Qf}{|Q|} \geq \dfrac{\int_Kf}{|Q|}=\infty $, so we have the result, namely $ \{f*\geq f} =\mathbb{R}^n $

QED

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010