Timeline for Classification of subsymmetric basic sequences
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2016 at 11:30 | answer | added | Anso | timeline score: 0 | |
| Mar 19, 2016 at 20:22 | comment | added | Tomasz Kania | Cross-posted at MSe: math.stackexchange.com/questions/1699572/… | |
| Mar 16, 2016 at 19:32 | vote | accept | ragrigg | ||
| Mar 16, 2016 at 11:55 | comment | added | Ben W | Not every basic sequence is bounded or bounded away from zero. In fact, for any basic sequence $(x_n)_{n=1}^\infty$ and any sequence of nonzero scalars $(a_n)_{n=1}^\infty$, the sequence $(a_nx_n)_{n=1}^\infty$ is basic. However, if $(x_n)_{n=1}^\infty$ is subsymmetric then it must be seminormalized, i.e. admit $M\in(0,\infty)$ such that $M^{-1}\leq\|x_n\|\leq M$ for every $n\in\mathbb{N}$. | |
| Mar 16, 2016 at 11:49 | answer | added | Ben W | timeline score: 1 | |
| Mar 16, 2016 at 4:41 | answer | added | Bill Johnson | timeline score: 2 | |
| Mar 16, 2016 at 3:53 | comment | added | ragrigg | My questions about boundedness came to mind while remembering Rosenthal's $\ell_1$ Theorem. If every subsymmetric basic sequence is bounded, then that theorem would allow you to conclude that a subsequence is either weakly Cauchy or equivalent to the canonical basis of $\ell_1$. But a weakly Cauchy sequence need not be weakly null, right? Anyway, the conclusion would not be the same as the conclusion of the statement that originated my question. | |
| Mar 16, 2016 at 2:50 | history | asked | ragrigg | CC BY-SA 3.0 |