Timeline for Schauder basis of $L^1_{\mathrm{loc}}(\mathbb{R}^n,H)$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2022 at 21:56 | vote | accept | ABIM | ||
| Jan 27, 2022 at 12:55 | comment | added | ABIM | @JohanessHahn I thought it wasn't true (but maybe there is some assumption implicitly working where which I'm not aware of). | |
| Jan 26, 2022 at 22:00 | comment | added | Johannes Hahn | Honestly, no. I found a flaw in my original argument and haven't had the time to try to fix it yet. At least I found some articles that claim your more general conjecture is true that the tensor product of two Schauder basis is a Schauder basis of the (projective) tensor product, but alas it was stated without a reference. Is that because it is a well known fact to people in functional analysis? I'm hopeful. | |
| Jan 25, 2022 at 23:53 | comment | added | ABIM | @JohanessHahn any luck on the full detailed version? | |
| Jan 25, 2022 at 23:52 | vote | accept | ABIM | ||
| Jan 26, 2022 at 14:50 | |||||
| Jan 24, 2022 at 13:50 | comment | added | ABIM | I guess, since $L^1(Q_k,H)\cong L^1(Q_k)\hat{\otimes} H \cong L^1([0,1]^n)\otimes H$ and then, the density of $\{\psi_{\cdot}\}$ in $L^1([0,1]^n)$ implies that $\{\psi_{\cdot}\cdot h_{\cdot}\}$ is dense in $L^1(Q_k,H)$. Thus, by your semi-norm formulation of the topology in $L^1_{loc}(\mathbb{R},H)$ we would have our conclusion; no? | |
| Jan 24, 2022 at 12:52 | vote | accept | ABIM | ||
| Jan 24, 2022 at 12:52 | |||||
| Jan 24, 2022 at 11:45 | history | answered | Johannes Hahn | CC BY-SA 4.0 |