Timeline for Smoothing operator raising the smoothness exactly by one
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 5, 2015 at 20:29 | comment | added | Igor Belegradek | I have no idea. Since none of the alternatives works, it puzzles me why you even posted this. | |
| Apr 5, 2015 at 19:33 | comment | added | Anton Petrunin | @IgorBelegradek, I think you know what I mean, do not you? | |
| Apr 5, 2015 at 12:33 | comment | added | Igor Belegradek | Still does not work. You cannot really mean $C^k\setminus C^{k-1}$ because this is the empty set. If you mean $C^k\setminus C^{k+1}$, then it is unclear how $S$ is defined on $C^{k+1}$. If you mean to use the formula separately for each $k$, then it is unclear why $S$ is continuous. | |
| Apr 5, 2015 at 3:03 | comment | added | Anton Petrunin | @IgorBelegradek corrected. | |
| Apr 5, 2015 at 3:02 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
added 89 characters in body
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| Apr 5, 2015 at 1:21 | comment | added | Igor Belegradek | This does not work. I need the operator for each $k$ while in your constriction $k$ varies. If we fix $k$, and let $S(f)=\sigma(f)(1+h_i)$ with $h_i\in C^{k+1}-C^k$ and $h_i\to 0$ as $i\to\infty$, then the operator does not have property (2). | |
| Apr 4, 2015 at 19:45 | history | answered | Anton Petrunin | CC BY-SA 3.0 |