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Nov
18 |
comment |
Ways to prove the fundamental theorem of algebra
It is open on the set A by the Open Mapping Theorem of Complex Analysis. It is closed because proper maps are closed (see math.stackexchange.com/questions/501510/…). Non-constant complex polynomials are proper because they diverge at infinity. |
Jun
29 |
awarded | Great Answer |
Jun
2 |
awarded | Notable Question |
May
13 |
awarded | Nice Answer |
Apr
27 |
awarded | Favorite Question |
Feb
10 |
awarded | Nice Question |
Feb
4 |
awarded | Good Question |
Nov
17 |
awarded | Popular Question |
Oct
23 |
awarded | Yearling |
Sep
24 |
awarded | Autobiographer |
Jul
2 |
awarded | Curious |
Mar
19 |
asked | Invertibility of Hankel operators? |
Mar
6 |
comment |
An inequality for Fourier transform
It seems to me that the inequality follows from Plancherel inequality and the identity $\widehat{(-\Delta)^{-\alpha}f}=|x|^{-2\alpha}\widehat{f}$. |
Jan
8 |
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Better bound for Hardy-Littlewood maximal function
Testing your estimate on the characteristic of a ball, you see that when $\alpha\rightarrow+\infty$ you get a contradiction. |
Oct
23 |
awarded | Yearling |
Oct
15 |
comment |
Elliptic Harnack inequality for 1D Schrodinger operator?
A Harnack inequality with constants depending on the radius may be found in Aizenman, M.; Simon, B. Brownian motion and Harnack inequality for Schrödinger operators. |
Oct
8 |
awarded | Favorite Question |
Oct
1 |
awarded | Caucus |
Aug
5 |
comment |
Converse to Chow's theorem in sub-riemannian geometry
Thank you, Dario! |
Jul
18 |
accepted | Converse to Chow's theorem in sub-riemannian geometry |