Timeline for On Riesz decomposition of Volterra operator
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2022 at 16:13 | comment | added | Yonah Borns-Weil | In fairness, the Volterra operator can be unsettling at first, because it's very easy to implicitly assume that every compact operator has eigenvalues (in particular, that's true for self-adjoint operators and for finite-rank operators, for different reasons.) | |
| Sep 26, 2022 at 14:21 | comment | added | Christian Remling | In other words, $n=1$ works, since $\lambda\notin\sigma(T)$, so $R(T-\lambda)=L^2$, $N(T-\lambda)=0$. | |
| Sep 26, 2022 at 8:20 | comment | added | Jochen Glueck | I agree with @YonahBorns-Weil's comment, and I'd add that even the entire spectrum of $T$ consists of $0$ only. | |
| Sep 26, 2022 at 3:51 | comment | added | Yonah Borns-Weil | I am a bit puzzled by this question. $T$ has no point spectrum away from $0$, so Isn't the kernel of $(T-\lambda)^n$ always just the zero function? | |
| Sep 25, 2022 at 21:46 | history | asked | Ali | CC BY-SA 4.0 |