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Jul
2
awarded  Curious
Jun
10
accepted surface area of the graph of a convex function
Jun
10
comment surface area of the graph of a convex function
Of course my definition for "area" is the $n-1$ dimensional Hausdorff measure. I suppose the last part of your answer doesn't restrict itself to 3 dimension. The argument works for any dimension with no modification. Thanks for this viewpoint. I was tempted to ask about the proof when $f$ is not differentiable at all.
Jun
10
asked surface area of the graph of a convex function
May
13
comment Min-Max Principle and Harnack's inequality
Thanks for this very well explained answer.
May
13
accepted Min-Max Principle and Harnack's inequality
May
10
comment Min-Max Principle and Harnack's inequality
@GiuseppeNegro just trying to see if somebody might have an idea. I am not "requiring" people to know or answer. That's the purpose of this website, right? People raise questions and people discuss.
May
10
comment Min-Max Principle and Harnack's inequality
@GiuseppeNegro It was merely a sentence saying that. I can't find it again... sorry. But as for my doubt, there is at least no easy way to prove maximum principle with harnack, right?
May
10
asked Min-Max Principle and Harnack's inequality
May
3
asked Upper semi-continuity of log-concave functions
Apr
17
asked reducibility of an operator
Apr
6
revised proof on Poincare's inequality.
added 2 characters in body
Apr
6
comment proof on Poincare's inequality.
which notation do you need me to explain?
Apr
6
asked proof on Poincare's inequality.
Apr
5
accepted Minimizing sequence that has certain property
Apr
3
asked Minimizing sequence that has certain property
Mar
30
accepted Limiting argument when proving inequality in Sobolev space
Mar
29
comment Limiting argument when proving inequality in Sobolev space
this makes a hell lot more sense!! Thanks!
Mar
29
comment Limiting argument when proving inequality in Sobolev space
So I should interpret this inequality as: For any $f$ in $W^{1,1}$, there is a $\tilde{f}$, s.t. $\tilde{f}=f$ in the distributional sense with $\tilde{f}\in L^{\infty}$ and that inequality holds. Am I right?
Mar
29
comment Limiting argument when proving inequality in Sobolev space
first thank you for pointing out the Cauchy sequence part. I was stupid to not have seen this. But why does $f_n$ necessarily converge to this same $f$ in $L^{\infty}$?