Timeline for Unbounded linear operator defined on $l^2$
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2012 at 10:37 | comment | added | Martin Sleziak | Almost identical question has been posted also at MSE: Discontinuous linear functional | |
| Jul 9, 2010 at 13:12 | comment | added | Pietro Majer | Actually, you immediately have unbounded linear operators on a normed spaces as soon as you have a Hamel basis, and as you know, in general the existence of a Hamel basis on a linear space is ensured by the Zorn lemma. Then, if $(x_i)$ is any Hamel basis and $(y_i)$ is any family of vectors indicized on the same set, there is a unique linear map sending $x_i$ to $y_i$, and it is certainly unbounded if e.g. the $y_i$ are chosen in such a way that $|y_i|/|x_i|$ is unbounded. | |
| Jul 9, 2010 at 12:33 | vote | accept | falagar | ||
| Jul 9, 2010 at 10:49 | comment | added | Willie Wong | You probably know this already, but $T$ of course cannot be symmetric by the Hellinger-Toeplitz theorem. en.wikipedia.org/wiki/Hellinger%E2%80%93Toeplitz_theorem | |
| Jul 9, 2010 at 10:27 | comment | added | Matthew Daws | My understanding is that you can't do this. But, it would be interesting to know how tight this is logically: are there models of ZF where every linear map $\ell^2\rightarrow\ell^2$ is bounded? | |
| Jul 9, 2010 at 10:15 | comment | added | Yemon Choi | Good point, Rasmus. For some reason when I write that I thought there was a distinction, but a quick check in Rudin tells me I was mistaken. (I think I was thinking of closed operators, in which case every closed operator with full domain is necessarily bounded by the Closed Graph Theorem.) | |
| Jul 9, 2010 at 10:00 | comment | added | Rasmus | Dear Yemon Choi, I'm confused: What's the difference between the two things you mention? | |
| Jul 9, 2010 at 9:45 | answer | added | Jeff Schenker | timeline score: 19 | |
| Jul 9, 2010 at 9:36 | comment | added | Yemon Choi | By $D(T)$, do you mean the domain of $T$ in the usual sense for unbounded operators; or are you just looking for an everywhere-defined, unbounded linear map from $\ell^2$ to itself? | |
| Jul 9, 2010 at 9:08 | history | asked | falagar | CC BY-SA 2.5 |