Timeline for Subspaces of $l_{1}$ are not Lipschitz complemented in $l_{1}$
Current License: CC BY-SA 3.0
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| Jul 7, 2014 at 23:24 | comment | added | Tony Prochazka | @DongyangChen Aha, so solving your problem would in particular give you a separable Banach space which is not a Lipschitz retract of its bidual. Nice! | |
| Jul 7, 2014 at 1:02 | comment | added | Dongyang Chen | Thank you for your suggestion. But, if a subspace $U$ of $l_{1}$ is such that $l_{1}/U$ is isomorphic to $L_{1}(\mu)$, then $U^{**}$ is complemented in $l_{1}^{**}$. This is due to J.Lindenstrauss(J.Lindenstrauss, On a certain subspace of $l_{1}$,1964.).Thus $U$ is Lipschitz complemented in its bidual if and only if $U$ is Lipschitz complemented in $l_{1}$. | |
| Jul 6, 2014 at 19:37 | history | answered | Tony Prochazka | CC BY-SA 3.0 |