Timeline for Does $\pm A \leq B$ imply that $B^{-1} A$ is bounded?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2021 at 8:29 | comment | added | Mikael de la Salle | I added an answer. | |
| Apr 13, 2021 at 8:28 | answer | added | Mikael de la Salle | timeline score: 8 | |
| Apr 13, 2021 at 8:20 | comment | added | Severin Schraven | @MikaeldelaSalle Thank you for your comment. Is equivalence you mention easy to see? | |
| Apr 13, 2021 at 8:15 | vote | accept | Severin Schraven | ||
| Apr 13, 2021 at 8:13 | comment | added | Mikael de la Salle | For bounded positive operators, the condition $A^2 \leq B^2$ is equivalent to saying that $B^{-1} A$ has norm $\leq 1$. So your second question should have a positive answer. (But I am not very used to unbounded operators; what you get for free is that $A B^{-1}$ has dense domain and is bounded of norm $\leq 1$, I am not even sure that $B^{-1} A$ has dense domain). | |
| Apr 13, 2021 at 2:41 | answer | added | Michael Renardy | timeline score: 12 | |
| Apr 12, 2021 at 23:21 | history | asked | Severin Schraven | CC BY-SA 4.0 |