Timeline for Is the set of compact operators closed with the strong topology?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2021 at 15:44 | comment | added | Robert Furber | The in the two previously mentioned examples one has that every operator is the limit of a sequence of operators with finite-dimensional range (called finite-rank operators), which are a fortiori compact operators. In fact the set of finite-rank operators is dense $\mathcal{L}(X)$ in the strong topology, but there cannot be a sequence of finite-rank operators approximating the identity in the strong topology unless the space $E$ has the bounded approximation property, which not all separable Banach spaces have, but is implied by the existence of a Schauder basis. | |
| Jul 10, 2021 at 20:08 | comment | added | Nik Weaver | On Hilbert space every bounded operator is a strong limit of a sequence of compact operators. | |
| Jul 10, 2021 at 17:41 | comment | added | DCM | Have you thought about the case where the $T_n$ are 'approximation' operators? (e.g. $T_nx = (x_1,\dots,x_n,0,0,\dots)$ on $l^p$) | |
| Jul 10, 2021 at 16:59 | history | asked | Malik Amine | CC BY-SA 4.0 |