Timeline for Subspaces of $l_p$ and Banach-Mazur distance
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2013 at 23:38 | comment | added | Bill Johnson | Yes, it is not obvious. What is clear is that $(\sum_{n=1}^\infty E_p)_p$ cannot have the uniform approximation property, while $\ell_p$ does. This does not require the specifics of Szankowski's construction but uses another concept, the uniform approximation property. AFAIK, there is no really simple proof using only elementary facts about $\ell_p$. $$ $$ Another approach is to show that there are finite dimensional subspaces $F_n$ of $\ell_p$ s.t. the GL-constants of $F_n$ tend to infinity, but that also uses a non elementary concept and requires more arguments. | |
| Jan 13, 2013 at 22:17 | comment | added | Theo | Thank you, I am modestly familiar with Szankowski's construction, so I think I understand what the choice of $(E_n)$ can be. But it is not clear to me why $\gamma_p(E_n)\to\infty$, when $X$ fails AP. I can see that by looking at Szankowski's concrete definition (LT(II), Theorem 1.g.4 ) and taking $E_n=\span\{z_1, z_2,\dots z_n\}$ as in the proof, but I suppose there exists a simpler, abstract argument why any $(E_n)$ as above have the factorization constant going to infinity. | |
| Jan 13, 2013 at 21:12 | vote | accept | Theo | ||
| Jan 13, 2013 at 18:10 | vote | accept | Theo | ||
| Jan 13, 2013 at 21:12 | |||||
| Jan 13, 2013 at 18:05 | history | answered | Bill Johnson | CC BY-SA 3.0 |