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Jan 9, 2011 at 13:41 vote accept Italo
Jan 7, 2011 at 16:40 answer added Deane Yang timeline score: 1
Jan 7, 2011 at 16:13 comment added Italo it is correct $F_{\epsilon}(x)$ because $dist(F_{\epsilon}(x),N)\leq |F(y)-F_{\epsilon}(x)|$ for all $y\in B(x,\epsilon)$ so i should use that kind of estimate proceed
Jan 7, 2011 at 16:05 comment added Deane Yang On the left, is the "$F_\epsilon(x)$" correct or should it be $F_\epsilon(y)$?
Jan 7, 2011 at 15:59 comment added Italo I'm really sorry i made a copy and paste error, now i fixed it
Jan 7, 2011 at 15:58 history edited Italo CC BY-SA 2.5
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Jan 7, 2011 at 15:43 comment added Denis Serre The inequality is false twice. First, it is not homogeneous in $\epsilon$. The fraction is the right-hand side should be deleted. Now the second integral must be taken on a larger ball $B(x;2\epsilon)$, assuming that the support of $\phi_\epsilon$ is $B(x;\epsilon)$.
Jan 7, 2011 at 15:41 comment added Deane Yang And have you tried to prove this inequality directly yourself using the fundamental theorem of calculus and the definition of $F_\epsilon$? What goes wrong?
Jan 7, 2011 at 15:40 comment added Deane Yang I agree that the inequality does not look right. The two sides do not scale the same, do they? Are you supposed to have the $1/\mu(B(x,\epsilon))$ factor on both sides or just on the right?
Jan 7, 2011 at 15:38 history edited Denis Serre CC BY-SA 2.5
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Jan 7, 2011 at 15:31 history edited Andrey Rekalo CC BY-SA 2.5
TeX corrected
Jan 7, 2011 at 15:26 history edited Denis Serre CC BY-SA 2.5
added 4 characters in body; deleted 16 characters in body
Jan 7, 2011 at 15:18 history asked Italo CC BY-SA 2.5