I have thought about the following question for several years. This question may be stupid or not interesting. My question is: Is there a subspace $U$ of $l{_1}$ such that the quotient $l_{1}/U$ is isomorphic to $L_{1}(\mu)$ for some measure $\mu$, but $U$ is not Lipschitz complemented in $l_{1}$.
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asked Jul 6, 2014 at 16:41
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$\begingroup$ The question is very interesting IMO. $\endgroup$– Bill JohnsonCommented Jul 6, 2014 at 18:02
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$\begingroup$ @BillJohnson I must be missing something -- isn't this handled by Godefroy--Kalton? (If there were a Lipschitz complementation, then surely there'd be a Lipschitz right inverse to the quotient map, hence a bounded linear right inverse by the GK results; and we know that $L_1[0,1]$ is not isomorphic to any complemented subspace of $\ell_1$... $\endgroup$– Yemon ChoiCommented Jul 6, 2014 at 19:28
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$\begingroup$ @Yemon Choi: A Lipschitz right inverse to the quotient gives a Lipschitz projection, but the converse is false. for example, $C[0,1]$ is Lipschitz complemented in every superspace by Lindenstrauss' 1964 paper, but, as you know, it is not complemented in every separable superspace; in particular the functions that are right continuous, left continuous except at the rationals, and have left limits at the rationals. $\endgroup$– Bill JohnsonCommented Jul 6, 2014 at 21:38
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$\begingroup$ @BillJohnson Thanks - I knew I was missunderstanding something basic $\endgroup$– Yemon ChoiCommented Jul 6, 2014 at 22:08
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1$\begingroup$ Dear Dongyang Chen, please follow this link - mathoverflow.net/contact - to arrange for your two accounts to be merged: #41619 #54727. $\endgroup$– François G. DoraisCommented Jul 7, 2014 at 2:18
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This is not an answer, just a related question: In the particular case when $U$ is complemented in its bidual, you have that $U$ is Lipschitz complemented in $\ell_1$ if and only if $U$ is complemented in $\ell_1$ (by a classical result of Lindenstrauss) if and only if $U$ is isomorphic to $\ell_1$ (Pelczynski). It means that if you find $U$ non-isomorphic to $\ell_1$, complemented in its bidual and such that $\ell_1/U$ is isomorphic to $L_1(\mu)$, you will get the Lipschitz non-complementability for free. Now the question is: is there such a space $U$? Can you have for example for a dual $U\subset \ell_1$ ($U$ not isomorphic to $\ell_1$) that $\ell_1/U$ is isomorphic to some $L_1(\mu)$?
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$\begingroup$ 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 $l_{1}$. $\endgroup$ Commented Jul 7, 2014 at 1:02
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1$\begingroup$ @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! $\endgroup$ Commented Jul 7, 2014 at 23:24