Timeline for Self-adjointness of fractional laplacian
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 5, 2023 at 12:38 | comment | added | Christian Remling | $T_1$ is the first operator you define (= closure of the fractional Laplacian on $C_0^{\infty}$), and $T_2$ is the second one (via the third displayed equation). | |
| Oct 5, 2023 at 6:30 | comment | added | B.Hueber | Can you maybe specify what $T_{1,2}$ in your comment are? | |
| Oct 4, 2023 at 20:40 | comment | added | Christian Remling | I don't think that's a good summary of my comment. My hope is that I said exactly what I meant to say in a clear manner. The description of the domain you give follows from my comment when combined with material in your post that I didn't check. | |
| Oct 4, 2023 at 14:34 | comment | added | B.Hueber | @ChristianRemling thanks for the comment. So you are basically saying that $\mathcal{D}(\overline{\Delta_{g}}^{s})=H^{2s}(M)$? | |
| Oct 4, 2023 at 14:30 | comment | added | Christian Remling | If we denote your first and second version of the operator by $T_{1,2}$, then the second definition really just says that $T_2=T_1^*$. ($C_0^{\infty}$ is a core of $T_1$, so you get the third displayed equation for all $\varphi\in D(T_1)$.) Since $T_1$ was self-adjoint, this gives $T_2=T_1$. | |
| Oct 4, 2023 at 13:32 | history | asked | B.Hueber | CC BY-SA 4.0 |