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seen Jul 26 at 19:42

Jul
14
comment What dimension bound is known on the singular set of a linear combination of eigenfunctions of Laplacian?
I am not bothered about making any stipulations about multiplicity or normalization. I guess I could really ask it for a general linear combination $f = a_1\phi_1 + \dots + a_m\phi_m$; $a_j \in \mathbb{R}$.
Jul
14
revised What dimension bound is known on the singular set of a linear combination of eigenfunctions of Laplacian?
added some relevant context
Jul
14
asked What dimension bound is known on the singular set of a linear combination of eigenfunctions of Laplacian?
Jun
28
awarded  Necromancer
May
1
awarded  Notable Question
Apr
8
asked Closed geodesics in free smooth loop space?
Mar
18
revised Reference for higher order Campanato Lemmas, e.g. `Sufficiently fast L^2 decay on balls to affine functions implies C^{1,\alpha}'
Edited after Willie Wong's answer to expand title
Mar
18
comment Reference for higher order Campanato Lemmas, e.g. `Sufficiently fast L^2 decay on balls to affine functions implies C^{1,\alpha}'
Thanks Willie, this is exactly it.
Mar
18
accepted Reference for higher order Campanato Lemmas, e.g. `Sufficiently fast L^2 decay on balls to affine functions implies C^{1,\alpha}'
Mar
18
asked Reference for higher order Campanato Lemmas, e.g. `Sufficiently fast L^2 decay on balls to affine functions implies C^{1,\alpha}'
Feb
13
comment Must the Lebesgue measure of a $\rho$ - neighbourhood of an $(n-2)$ - dimensional set be at least $c\rho^2$?
Thanks for the comment fedja, I see your point. I don't know what made me think it was true now.
Feb
13
asked Must the Lebesgue measure of a $\rho$ - neighbourhood of an $(n-2)$ - dimensional set be at least $c\rho^2$?
Oct
7
awarded  Nice Question
Oct
6
comment Why do roots of polynomials tend to have absolute value close to 1?
I know very little about the math going on in the background but a colleague demystified this for me by saying that basically it is a special feature of the basis you have chosen, namely the monomials. If you choose a different basis or even just weight each monomial by a factor, the roots will tend to congregate on a different set. Basically although a) any polynomial can arise and b) you chose them randomly... They aren't as generic as as you think; they've in fact been chosen in a special way.
Aug
31
awarded  Yearling
Aug
31
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Aug
3
awarded  Necromancer
Jul
2
awarded  Curious
Jan
24
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Jul
15
awarded  Nice Answer