Timeline for Are smooth functions with compact support a core for the Laplacian on compact manifolds with boundary?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2021 at 17:04 | comment | added | Giorgio Metafune | In that case it is true, provided $\partial M$ is smoooth. The usual way is to prove first for the half space and then use local coordinates to flatten the boundary. Basically one needs a $C^k$, $ k \geq 2$, boundary to approximate with $C^k$ functions vanishing at the boundary. | |
| Oct 28, 2021 at 13:58 | comment | added | Alex M. | @GiorgioMetafune: Does the situation change if I ask not about $C_0 ^\infty (M)$, but about the smooth functions that vanish on $\partial M$? | |
| Oct 28, 2021 at 11:14 | comment | added | Alex M. | @GiorgioMetafune: You guessed it correctly, I meant "core", not "form core". Thank you. | |
| Oct 28, 2021 at 11:08 | comment | added | Giorgio Metafune | I have only one doubt about the definition. For "core" I mean a dense subset of the domain of the operator with respect to the graph norm (which is a Sobolev case in the smooth case). Sometimes one means "form core", that is a dense subset for the domain of the associated quadratic form (which is usually $H^1$ for $p=2$). In the first case the assertion is clearly false in the interval: smooth functions with compact support are not dense in the domain of the operator, which is $H^2\cap H^1_0$ (that is only $u(0)=0$). In the closure one has also $u'(0)=0$.) | |
| Oct 28, 2021 at 8:49 | comment | added | Alex M. | @GiorgioMetafune: Funny, I just talked to somebody who told me the exact opposite of what you say. In particular, I was told that what I ask is true on the interval $[0,1]$, while you claim that it is false... I imagined that this couldn't be possible in mathematics. | |
| Oct 28, 2021 at 7:23 | comment | added | Giorgio Metafune | If $C_0^\infty(M)$ stands for smooth functions with support far away from the booundary, then they are not a core for example when $M$ is a bounded open subset of $R^n$. In fact you can have two (actually many) maximal acceretive extensions corresponding to Dirichlet and Neumann boundary conditions. | |
| Oct 28, 2021 at 5:57 | history | asked | Alex M. | CC BY-SA 4.0 |