Timeline for Is this operator invertible?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 5, 2019 at 16:19 | answer | added | Jochen Glueck | timeline score: 5 | |
| Apr 5, 2019 at 15:12 | comment | added | Nik Weaver | Of course not. You can find an easy counterexample to that statement. | |
| Apr 5, 2019 at 14:59 | comment | added | Saj_Eda | What if $\|B\|>1$, then it's never invertible? | |
| Apr 5, 2019 at 14:52 | comment | added | Nik Weaver | Any operator of the form $I - B$ with $\|B\| < 1$ is invertible. So $G$ will be invertible if $\int_0^t \|T(t-s)A(s)\|\, ds \leq \alpha < 1$ for all $t$. I don't think the fact that $T$ is a semigroup has any particular bearing on the question. | |
| Apr 5, 2019 at 14:37 | history | asked | Saj_Eda | CC BY-SA 4.0 |