Timeline for Is the restriction map $C^1\ni f\mapsto\left.f\right|_K$ a continuous map?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2020 at 17:36 | answer | added | Alexander Schmeding | timeline score: 0 | |
| Oct 10, 2020 at 16:09 | answer | added | Jochen Wengenroth | timeline score: 2 | |
| Oct 9, 2020 at 23:45 | comment | added | Pietro Majer | Isn't already continuous $C^1(E,E)\ni g\mapsto g_{|K}\in C^1(K,E)$, wrto this topology on $C^1(E,E)$? | |
| Oct 9, 2020 at 19:41 | comment | added | 0xbadf00d | @DCM I'm interested in the case $E=\mathbb R^d$ as well. But if I'm not missing anything, even in the infinite-dimensional case, $C^1(E,E)$ endowed with the topology induced by compact convergence of the functions and their Fréchet derivatives should always be a Locally convex topological vector space. | |
| Oct 9, 2020 at 18:40 | comment | added | DCM | Also... what topology do you give to $C^1(E,E)$? Are you mainly interested in the finite dimensional case or do you need to allow $E$ infinite dimensional? | |
| Oct 9, 2020 at 18:20 | comment | added | DCM | Re. $C^1(K,E)$ always being complete with the norm you suggest - I might be wrong, but I'm not sure that's even true for all compact subsets of $\mathbb{R}^d$, never mind when $E$ is something more exotic (it is true for all 'nice' compact subsets of $\mathbb{R}^d$ of course). | |
| Oct 9, 2020 at 17:43 | history | asked | 0xbadf00d | CC BY-SA 4.0 |