Timeline for Does $\pm A \leq B$ imply that $B^{-1} A$ is bounded?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2021 at 13:52 | comment | added | MaoWao | Not necessarily. But in this case, $D(B^{-1}A)=D(A)$ is dense. | |
| Apr 13, 2021 at 13:46 | comment | added | Mikael de la Salle | @MaoWao Thanks. But is $T^* S^*$ densely defined? | |
| Apr 13, 2021 at 9:42 | comment | added | MaoWao | If $ST$ is densely defined, then $(ST)^\ast\supset T^\ast S^\ast$. Thus $B^{-1}A\subset (AB^{-1})^\ast$ is bounded as well. | |
| Apr 13, 2021 at 9:19 | comment | added | Severin Schraven | That is super neat, thanks! Indeed, I assume that the image of $B$ is the full space (was a bit sloppy there). | |
| Apr 13, 2021 at 8:28 | history | answered | Mikael de la Salle | CC BY-SA 4.0 |