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visits | member for | 5 years, 1 month |
seen | yesterday | |
stats | profile views | 1,183 |
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 |
awarded | Disciplined |
Aug 3 |
awarded | Necromancer |
Jul 2 |
awarded | Curious |
Jan 24 |
awarded | Popular Question |
Jul 15 |
awarded | Nice Answer |
May 24 |
awarded | Good Question |
Mar 30 |
comment |
What's a mathematician to do?
Probably my favourite answer on MO. |
Mar 21 |
answered | Applications of Rademacher's Theorem |
Mar 11 |
comment |
What goes wrong for the Sobolev embeddings at $k=n/p$?
I'm not sure I quite appreciate exactly what is being asked. You lay out two cases and then say that what is essentially just a third case is an example of the other two not working. You say that the Sobolev embedding theorem "fails" or "goes wrong" when $k=n/p$, but one might say that it is simply neither of the two cases you lay out at the start. Nothing "fails", it just happens to be its own special case. |
Feb 27 |
awarded | Yearling |