Timeline for What is $\left\| u \right\|_{ H_{0}^{k}, H_{0}^{k}}$ norm when $H_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int_{M} u \operatorname{vol}_{g}=0\right\}$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 28, 2022 at 13:13 | vote | accept | Elio Li | ||
| Jun 27, 2022 at 19:53 | answer | added | username | timeline score: 2 | |
| Jun 27, 2022 at 19:13 | comment | added | Willie Wong | Oof, I don't like this notation. Normally for operator norms I would write $\| - \|_{X\to Y}$ because the one they chose, $\| - \|_{X,Y}$, does not make clear whether they want it to be the operator norm of a mapping from $X\to Y$ or from $Y\to X$. This becomes a bit of an issue later on when they write $\| e^{-t\Delta} \|_{H^k_0, H^0_0}$. At a quick glance it is not obvious to me which direction they meant. | |
| Jun 27, 2022 at 14:58 | comment | added | Elio Li | It seems that the $\Delta$ in this paper is $-\Delta$. I add the link in the question. | |
| Jun 27, 2022 at 14:58 | history | edited | Elio Li | CC BY-SA 4.0 |
added 72 characters in body
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| Jun 27, 2022 at 14:48 | comment | added | Willie Wong | @username given the context, your interpretation is almost certainly the correct one (that $\| e^{-t\Delta}\|_{H_0^k,H_0^k}$ is the operator norm of $e^{-t\Delta}$ as a mapping from $H^k_0$ to itself). Can you post it as an answer? | |
| Jun 27, 2022 at 7:16 | comment | added | username | Taking $K:f\to (\Delta)^{-1}f$ as a (compact) operator between $H^k_0$ and $H^k_0$, it is the usual linear operator norm, perhaps? | |
| Jun 27, 2022 at 4:22 | history | asked | Elio Li | CC BY-SA 4.0 |