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seen Nov 8 at 20:09

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 symmetric decreasing rearrangement of a smooth function preserves 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.