Timeline for Finite dimensional subspaces of $L^1.$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2012 at 20:59 | comment | added | Bill Johnson | Yes, Greg, and you can then prove Lindenstrauss' result by using ultraproducts. | |
| Apr 20, 2012 at 19:18 | history | edited | Bill Johnson | CC BY-SA 3.0 |
Corrected typo.
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| Apr 20, 2012 at 19:18 | comment | added | Bill Johnson | Right; the $\ell_1$ is a typo. I'll change it. | |
| Apr 20, 2012 at 19:16 | comment | added | Igor Rivin | @Greg. That would make sense... | |
| Apr 20, 2012 at 19:14 | comment | added | Greg Kuperberg | @Igor $L^1$ can only give you more, because $\ell^1$ embeds isometrically into $L^1$. I think that $\ell^1$ only gives you the duals of those limits of zonotopes whose sphere measure is pure point. | |
| Apr 20, 2012 at 19:11 | comment | added | Greg Kuperberg | The polygon case of Lindenstrauss' 2D result is easy to visualize: Every centrally symmetric polygon is a zonogon (and therefore also a dual zonogon). | |
| Apr 20, 2012 at 19:10 | comment | added | Igor Rivin | I assume that the distinction between $L^1$ and $\ell^1$ is intentional. Are you, in fact, saying that Greg's answer works for the latter and not for the former? | |
| Apr 20, 2012 at 19:05 | history | answered | Bill Johnson | CC BY-SA 3.0 |