Let $M$ be a closed subspace of $l^\infty$. Suppose that the quotient $l^{\infty}/M$ is isomorphic to $l^\infty$. Is it true that $M$ is complemented in $l^\infty$?
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4$\begingroup$ The "predual" version of this question -- namely, "if a subspace of $\ell^1$ is isomorphic to $\ell^1$, is it complemented in $\ell^1$? -- has a negative answer, see J. Bourgain, A counterexample to a complementation problem, numdam.org/item?id=CM_1981__43_1_133_0 I therefore suspect that the answer to your question is negative but I'm not sure. $\endgroup$– Yemon ChoiCommented Aug 27, 2013 at 16:25
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2$\begingroup$ By a result of Lindenstrauss, the OP really asks whether $M$ is isomorphic to $\ell_\infty$. $\endgroup$– Tomasz KaniaCommented Aug 27, 2013 at 16:28
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1$\begingroup$ If you are happy with @BillJohnson's answer then you should click the tick mark to "accept it". $\endgroup$– Yemon ChoiCommented Sep 3, 2013 at 4:00
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Notice that the comment by Yemon gives a negative answer to the question. Bourgain constructs a short exact sequence
$0 \to \ell_1 \to \ell_1 \to X \to 0$
that does not split. The space $X$ is not $\mathcal{L}_1$ for then (since the kernel of the quotient map is a dual space and the quotient mapping onto any $\mathcal{L}_1$ space locally lifts) the quotient map would lift by a classical result of J. Lindenstrauss. Dualizing the diagram, you get a quotient mapping from $\ell_\infty$ onto $\ell_\infty$ whose kernel is not $\mathcal{L}_\infty$ and thus is not complemented in $\ell_\infty$.
Incidentally, from such an $\ell_\infty$ example you can deduce the $\ell_1$ result by localization; that is, modulo standard localization arguments, your question is equivalent to the question Bourgain answered. The moral is "don't look for an easy proof".
answered Aug 27, 2013 at 18:06
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1$\begingroup$ Thanks Bill: I'd forgotten the Lindenstrauss lifting result, which is why I couldn't see that X was not ${\mathcal L}^1$ $\endgroup$ Commented Aug 27, 2013 at 18:22
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$\begingroup$ 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. $\endgroup$– MartinCommented Aug 27, 2013 at 23:51
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1$\begingroup$ 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 $\ell_1$. $\endgroup$ Commented Aug 28, 2013 at 0:18
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1$\begingroup$ 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 $E$. $\endgroup$ Commented Aug 28, 2013 at 0:26
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1$\begingroup$ 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. $\endgroup$ Commented Aug 28, 2013 at 0:32