Timeline for Are "most" operators on an infinite-dimensional complex Banach space "diagonalizable"?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2015 at 20:15 | vote | accept | Tim Campion | ||
| Apr 9, 2015 at 13:34 | comment | added | Bill Johnson | The Argyros-Haydon spaces I mentioned have Schauder bases, so the finite rank operators are dense in the compact operators--I neglected to mention that. I don't know whether on every Banach space the closure of the diagonal operators contains the compact operators. | |
| Apr 8, 2015 at 23:46 | comment | added | Tim Campion | Thanks, that was silly of me. I still don't understand why every operator being of the form $\lambda I + K$ with $K$ compact implies that the diagonal operators are dense, although I believe I see how this would follow if finite-rank operators are dense in compact operators. | |
| Apr 8, 2015 at 23:11 | comment | added | Robert Israel | No, that's not even true in finite dimensions (where every operator is compact). | |
| Apr 8, 2015 at 22:00 | comment | added | Tim Campion | @Bill As might be clear from the question, I know very little functional analysis. I believe that the spectral theory of compact operators shows that every compact operator on a Hilbert space is diagonalizable: is this true for all Banach spaces? If so, then in the spaces you're talking about, every operator is digaonalizable, right? | |
| Apr 8, 2015 at 21:31 | comment | added | Bill Johnson | There are infinite dimensional Banach spaces on which every bounded linear operator is of the form $\lambda I + K$ with $K$ a compact operator. On such a space the diagonal operators are dense. | |
| Apr 8, 2015 at 20:53 | comment | added | Tim Campion | Very nice. I'll refrain from accepting this answer for a little while only because it's not immediately obvious how to extend the argument to more general Banach spaces. | |
| Apr 8, 2015 at 20:20 | history | answered | Robert Israel | CC BY-SA 3.0 |