Timeline for Existence of an extension operator $E: W_0^{s,p}(\Omega)\rightarrow W^{s,p}(\mathbb{R}^d)$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9 at 2:26 | comment | added | Liding Yao | @AymanMoussa I see, that makes sense! I guess there is a more constructive way for that but I don't get a clue. | |
| Apr 8 at 5:54 | comment | added | Ayman Moussa | @LidingYao why do you say the result is false with $s=1$ ? We're only speaking of $W^{s,p}_0$- extension, so the functions at stake cannot " see " the boundary. | |
| Apr 8 at 0:22 | comment | added | Liding Yao | I don't think this is right. Did you use E being extension? If so why the argument fail if s=1? Notice that not all open sets are $W^{1,p}$-extension domain. | |
| Apr 7 at 21:18 | comment | added | Perelman | @AymannMoussa I think this works i will try it out. thank you | |
| Apr 7 at 21:18 | vote | accept | Perelman | ||
| Apr 7 at 21:14 | comment | added | Ayman Moussa | Well, if you want to prove that the graph of $E$ is closed, you can use the sequential characterization : you assume that $u_n$ converges to $u$ and $E(u_n)$ converges to some $v$ and then you just have to prove that $v=E(u)$ (that is : the graph of $E$ is closed sequentially). | |
| Apr 7 at 20:21 | comment | added | Perelman | Interesting approach, but why can we say that $E(u_n)$ converges? | |
| Apr 7 at 19:39 | history | answered | Ayman Moussa | CC BY-SA 4.0 |