Timeline for When is $CAC^{-1}$ bounded for $C$ Hilbert-Schmidt?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2014 at 14:10 | answer | added | holloway | timeline score: 1 | |
| Sep 22, 2014 at 13:27 | comment | added | Simon Henry | Because the question was "Which requirements on A..." and I have no idea on if and how my counterexample really allow to says something about that ? (and because I think that the answer is "this is true for almost no $A$" and I was hoping that someone would prove this) | |
| Sep 22, 2014 at 1:28 | comment | added | Yemon Choi | @SimonHenry Why not leave this as an answer? | |
| Sep 9, 2014 at 9:02 | comment | added | Simon Henry | Here is a counterexample: take $H$ generated by two orthogonal sequence $(a_n)$ and $(b_n)$, $A$ the operator $2*Id$ + the operator which switch $a_n$ and $b_n$ for all $n$, and $C$ the operator which acts on $a_n$ by multiplication by $1/n$ and on $b_n$ by multiplication by $1/n^2$. then $(C A C^{-1}) (b_n)=n.b_n$ Generalizing this idea you can construct counterexample for a very large class of $A$. (for example for any operator which have two infinite dimensional disjoint spectral projection...) | |
| Sep 9, 2014 at 7:00 | review | First posts | |||
| Sep 9, 2014 at 8:35 | |||||
| Sep 9, 2014 at 6:57 | history | asked | BSP | CC BY-SA 3.0 |