henryforever14
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 Nov 7 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. Nov 7 asked minimize p norm of f+c Sep 21 revised determinant in terms of quadratic form evaluated at a point deleted 605 characters in body; edited tags; edited title Sep 21 awarded Citizen Patrol Sep 21 comment determinant in terms of quadratic form evaluated at a point @copper.hat I sincerely apologize for that. I am deleting this post. Sep 21 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. Sep 21 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" . Sep 21 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. Sep 21 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? Sep 21 asked determinant in terms of quadratic form evaluated at a point Sep 6 awarded Tumbleweed Aug 30 asked Does the symmetric decreasing rearrangement of a smooth function preserve smoothness? Aug 29 awarded Popular Question Aug 16 awarded Nice Question Aug 7 answered How to prove: “Every subspace of $V$ invariant under $T$ is also invariant under $T^*$ if and only if $T$ is normal.”? 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.