Timeline for Orthogonal projection
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2014 at 7:41 | comment | added | Pietro Majer | OK, by invertible operator I mean a linear homeomorphism $ A:H\to H$ , thus more than just $0\in\rho(A)$ . So yes, the answer was somehow misleading. | |
| Nov 15, 2014 at 7:38 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 14, 2014 at 22:44 | comment | added | Christian Remling | Yes, I also interpret $\rho(G_0)$ as the resolvent set (= complement of the spectrum), so $0\in\rho(G_0)$ precisely if $G_0$ is injective and $G_0^{-1}\in B(H)$ (but $G_0$ itself need not be bounded, of course). | |
| Nov 14, 2014 at 22:15 | comment | added | Pietro Majer | But I understand $0\in\rho(G_0)$ as: $G_0$ being an invertible operator; therefore: $0\notin\sigma(G_0)$ and $\sigma(G_0)$ bounded... Why we don't need the latter? | |
| Nov 14, 2014 at 22:08 | comment | added | Pietro Majer | Yes, but I preferred to write it independently from the assumption "compact resolvent", to make a precise statement. Yes, "compact resolvent" implies 0 is automatically isolated, and $\sigma(G)$ unbounded. | |
| Nov 14, 2014 at 20:54 | comment | added | Christian Remling | In other words, the answer to the second question is just "yes." | |
| Nov 14, 2014 at 20:52 | comment | added | Christian Remling | The last statement is misleading: $0$ is automatically isolated in the spectrum because $G$ was assumed to have compact resolvent; $\sigma(G)$ will not be bounded, but we don't need this either. | |
| Nov 14, 2014 at 18:54 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 14, 2014 at 18:32 | history | answered | Pietro Majer | CC BY-SA 3.0 |