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 Mar25 comment borel measurable functions and measurable functions Thanks! Very neat construction! Mar25 comment borel measurable functions and measurable functions @PhoemueX Thanks, that solves the problem! Feb7 comment boundary of the support of a continuous function Thanks. That's a nice example! Nov7 comment minimize p norm of f+c Never mind. I've already found a counter example. $e^x$ in $L^4$ doesn't follow this rule. Sep21 comment determinant in terms of quadratic form evaluated at a point @copper.hat I sincerely apologize for that. I am deleting this post. Sep21 comment determinant in terms of quadratic form evaluated at a point @WillJagy I am not trying to offend any of you. Perhaps what I said is not so polite. I apologize for that. Also to copper.hat. I can't at two users in one comment. Sep21 comment determinant in terms of quadratic form evaluated at a point This is not what I asked anyway. Towards the end, if you note that $A^*=det(A)A^{-1}$ you just got what I said in my post "A simple calculation reveals" . Sep21 comment determinant in terms of quadratic form evaluated at a point @WillJagy $n=1$ doesn't yield anything interesting. By positive definite I meant symmetric positive definite. Personally, I don't believe there is a relationship. If one just take $b$ to be an eigenvector, then the quantity given above would just be the product of the $n-1$ remaining eigenvalues. Since determinant is the product of $n$ eigenvalues, there is no guarantee that one might be bigger than the other. It all depends on whether the last eigenvalue is bigger or smaller than 1. The person who told me this seemed very certain. So I decided to ask in case people here have seen this before. Sep21 comment determinant in terms of quadratic form evaluated at a point @WillJagy Do you know that such a formula exists or are you just making a general comment? Jun10 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. May13 comment Min-Max Principle and Harnack's inequality Thanks for this very well explained answer. May10 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. May10 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? Apr6 comment proof on Poincare's inequality. which notation do you need me to explain? Mar29 comment Limiting argument when proving inequality in Sobolev space this makes a hell lot more sense!! Thanks! Mar29 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? Mar29 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 $L^{\infty}$? Mar21 comment What does it mean for a distribution to be in $L_2$? I am trying to get a little intuition here. So based on what you said, a lot of distributions are not in $L_2$ since for a general distribution, you are not even lucky enough to get that kind of representation. Am I right? Feb10 comment What is the characteristics for the wave equation with space dimension more than 1? What do you mean by there are no characteristics in higher spatial dimensions? I thought we need to just find a function $\phi(t,x,y)$, s.t. $\phi_t^2=\phi_x^2+\phi_y^2$. And the set$\{\phi=const\}$ would be the family of characteristic surfaces. When I tried to solve the equation, I got planes. But my intuition tells me that it should be something like the cone you mentioned. I don't know where I went wrong. Apr8 comment Approximating measurable function by continuous ones @GiuseppeNegro I did check on that one. But isn't regularity of the measure needed there?