Timeline for A question on characterizing a Banach space containing no copy of $l_{1}$
Current License: CC BY-SA 3.0
7 events
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| Sep 29, 2015 at 14:12 | comment | added | Dongyang Chen | @BillJohnson I am able to prove your characterization. I show that if there is a non-compact operator from $X$ into $l_{1}$, then $X^{*}$ contains a copy of $c_{0}$. | |
| Sep 29, 2015 at 12:23 | comment | added | Dongyang Chen | @BillJohnson Could you give a sketch of the proof of your characterization? I am not sure. | |
| Sep 28, 2015 at 23:33 | comment | added | Dongyang Chen | Thanks, Philip. Your counterexample shows that the condition "$X$ contains no copy of $l_{1}$" does not characterize that any operator from $X$ into $l_{1}$ is compact. | |
| Sep 28, 2015 at 23:08 | vote | accept | Dongyang Chen | ||
| Sep 28, 2015 at 23:06 | comment | added | Dongyang Chen | Could you give a sketch of the proof of the necessary part? I am not sure. | |
| Sep 28, 2015 at 22:31 | comment | added | Bill Johnson | There is a non compact operator from $X$ into $\ell_1$ iff and only if $\ell_1$ is isomorphic to a complemented subspace of $X$. For that you need one additional comment; namely, that every subspace of $\ell_1$ contains a small complemented subspace that is isomorphic to $\ell_1$. | |
| Sep 28, 2015 at 22:27 | history | answered | Philip Brooker | CC BY-SA 3.0 |