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10h
comment Are convex functions enough to determine a measure?
@ByronSchmuland Thanks! It's a nice argument. Please change the very last measure to $\nu(dx)$. I will then accept your answer. Thanks a lot!
22h
comment Are convex functions enough to determine a measure?
@ByronSchmuland I am sorry, but is that function still convex? For example, the function $f(x,y)=xy$ should not be convex.
22h
comment Are convex functions enough to determine a measure?
@ByronSchmuland, what is the analogous functions for $f$ and $g$ in higher dimensions?
22h
comment Are convex functions enough to determine a measure?
So it does work if the measures are compactly supported.
1d
comment Are convex functions enough to determine a measure?
didn't realize compact domain would make such a huge difference. Thank you. For your last remark, suppose we are dealing with measures that have compact support, how may one prove convex functions are enough? I suppose it would require some approximation of continuous functions by convex functions that I am not aware of.
Jul
25
comment Confusion about the definition of reflexive relation
@BrianM.Scott Thanks! Edited.
Jul
24
comment Is the boundedness necessary to extend harmonically?
Why are the conditions different between harmonic functions and holomorphic function?
Jul
24
comment Is the boundedness necessary to extend harmonically?
So a priori, you don't need to know $v$ is bounded. As long as $v(z) \leq o(\log |z|)$, it will be still true that $v_\epsilon(z)$ will go to $-\infty$. It is the analysis that forced $v$ to be bounded. That means, the statement can be weakened as: "if $u$ is harmonic in the punctured disk and $u(z) \leq o(\log |z|)$, then $u$ can be extended harmonically at the origin."
Jul
20
comment Define a relation — with functions and derivatives
Let me just remark that in your case, a relation $D$ on $F$ means $D$ is a relation from $F$ to $F$, or, $D\subset F\times F$. Also, personally I would prefer not to use the words "domain" and "range". "domain" gives people a sense that you need to associate everything in it with something in the range, which is not the case in relation. It also might help if you just run some tests before you jump into the problem. For example, is $x^2 Dx$? Is $e^xDe^x+1$?
Jul
20
comment How do I get a English version of an article in French?
Try google translate. Works for me when I try to read in German.
Jun
30
comment second fundamental form and connection forms
I guess what the author meant by "orthonormal frame" is orthonormal list of vector fields induced from local parametrization.
Jun
24
comment why does Lie bracket of two coordinate vector fields always vanish?
@JamesS.Cook, sure, that I agree. Thanks with the help.
Jun
24
comment why does Lie bracket of two coordinate vector fields always vanish?
@JamesS.Cook, "Let's see, if the commutator is nontrivial then I don't think that means it is not possible. I certainly can find vector fields on the plane which have nontrivial Lie Bracket." I meant vector fields with nontrivial commutator can't be coordinate derivations.
Jun
24
comment why does Lie bracket of two coordinate vector fields always vanish?
@JamesS.Cook, so this in a sense tells me if I start out with two vector fields such that the Lie bracket doesn't vanish, then there is no way I could find a coordinate system such that they happen to be coordinate derivations. Now, is the reverse true? That is, if I start out with two vector fields such that the Lie bracket DOES vanish, is it true that I can find a coordinate system with them being coordinate derivations? Thanks!
Mar
25
comment borel measurable functions and measurable functions
Thanks! Very neat construction!
Mar
25
comment borel measurable functions and measurable functions
@PhoemueX Thanks, that solves the problem!
Feb
7
comment boundary of the support of a continuous function
Thanks. That's a nice example!
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.
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.