Timeline for Smooth functions on subsets of $\mathbb{R}^n$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 28, 2018 at 11:41 | vote | accept | Roberto Frigerio | ||
| Sep 27, 2018 at 15:13 | comment | added | Piotr Hajlasz | You are in Pisa so there are plenty of people around you who would know the answer right away (Alberti, Ambrosio, Magnani...) | |
| Sep 27, 2018 at 15:03 | answer | added | Piotr Hajlasz | timeline score: 6 | |
| Sep 27, 2018 at 14:53 | comment | added | Igor Khavkine | Beaten to the punch by Christian Remling. :-) The current Wikipedia page does not explicitly say that the extension can be $C^m$ for $m=\infty$, but the EoM page does. In any case, Whitney's original article does cover the $m=\infty$ case. | |
| Sep 27, 2018 at 14:46 | comment | added | Christian Remling | I think this will follow from the Whitney extension theorem, after having extended $f$ to $\overline{X}$: en.wikipedia.org/wiki/Whitney_extension_theorem Certainly in one dimension, it's straightforward to deduce the claim from Borel's theorem, and I think a more elaborate version of this argument should work in general. | |
| Sep 27, 2018 at 14:05 | comment | added | Roberto Frigerio | Of course the answer is yes if $X$ is an open subset of the Euclidean space. My question may be reformulated as follows: is it true that if $f\colon X\to\mathbb{R}$ admits a (local) $C^k$ extension $F_k$ for every $k$, then it admits a $C^\infty$ extension? If $F_k\neq F_h$ for every h\neq k$, this does not seem obvious to me. | |
| Sep 27, 2018 at 13:59 | history | asked | Roberto Frigerio | CC BY-SA 4.0 |