Timeline for Quotients of l^infty
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2013 at 0:32 | comment | added | Martin | Now this is crystal clear. Thank you very much! | |
| Aug 28, 2013 at 0:32 | comment | added | Bill Johnson | Bourgain did not give this argument because it was well known to experts and he hates (or at least hated at that time) to write more than is absolutely necessary. | |
| Aug 28, 2013 at 0:26 | comment | added | Bill Johnson | Here is how you get the $E_n$. Take $E$ an uncomplemented copy of $\ell_1$ in $L_1$ and let $G_n$ be the span in $E$ of the first $n$ unit vector basis elements. Suppose you have uniformly bounded projections $P_n$ from $L_1$ onto $G_n$. Endow $E=\ell_1$ with its weak$^*$ topology, which makes the unit ball compact and pass to an ultra limit of $P_n$. This gives a projection from $L_1$ onto $E$. | |
| Aug 28, 2013 at 0:18 | comment | added | Bill Johnson | Martin, you have to localize Bourgain's theorem. You get from it that for every $n$ there is a finite dimensional subspace $E_n$ of $L_1$ with $E_n$ $C$-isomorphic to $\ell_1^{m_n}$ ($m_n$ is the dimension of $E_n$) and any projection from $L_1$ onto $E_n$ has norm at least $n$ (see next comment). Here $C$ is a constant independent of $n$. You get a superspace $F_n$ of $E_n$ in $L_1$ which is, say, $2$-isomorphic to $\ell_1^{k_n}$. Now take the $\ell_1$ sum of $E_n$ in the $\ell_1$ sum of $F_n$ to get an uncomplemented copy of $\ell_1$ in $\ell_1$. | |
| Aug 27, 2013 at 23:51 | comment | added | Martin | I'm a bit confused: doesn't Bourgain "only" construct a short exact sequence $0 \to \ell_1 \to L_1 \to X \to 0$? [This also yields the answer to the question by taking duals, using that $L_\infty$ and $\ell_\infty$ are isomorphic.] If I understand your last paragraph correctly, one can infer a short exact sequence $0 \to \ell_1 \to \ell_1 \to Y \to 0$ from this, but Bourgain doesn't seem to spell that out in his paper. | |
| Aug 27, 2013 at 18:22 | comment | added | Yemon Choi | Thanks Bill: I'd forgotten the Lindenstrauss lifting result, which is why I couldn't see that X was not ${\mathcal L}^1$ | |
| Aug 27, 2013 at 18:06 | history | answered | Bill Johnson | CC BY-SA 3.0 |