Given a (closed) separable subspace $M$ of $\ell_\infty$, I am interested in conditions implying that the quotient $\ell_\infty/M$ is isomorphic to a subspace of $\ell_\infty$.

It is not difficult to see that being $M$ reflexive is sufficient, and Bourgain proved that $\ell_\infty/c_0$ does not admit an equivalent strictly convex norm (while $\ell_\infty$ does). So $\ell_\infty/c_0$ is not isomorphic to a subspace of $\ell_\infty$.

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    $\begingroup$ Nice question. The only immediate comment I have is that $\ell_\infty/M$ does not embed into $\ell_\infty$ if $M$ is separable and $c_0$ embeds into $M$. WLOG [Lindenstrauss-Rosenthal] by the subspace homogeneity property of $\ell_\infty$ you can assume $c_0 \subset M$, and $c_0(R)$ embeds into $\ell_\infty/c_0$, and every operator from $c_0(R)$ into $\ell_\infty$ has separable range because weakly compact subsets of $\ell_\infty$ are separable since $\ell_\infty$ is the dual of a separable space. $\endgroup$ – Bill Johnson Nov 15 '13 at 16:16
  • $\begingroup$ A second comment is that if $M$ is isomorphic to a separable conjugate space, then $\ell_\infty/M$ does embed into $\ell_\infty$. The reason is that then $M$ is isomorphic to a weak$^*$ closed subspace of $\ell_\infty$, so by the subspace homogeneity of $\ell_\infty$ you can assume that $M$ is weak$^*$ closed. $\endgroup$ – Bill Johnson Nov 15 '13 at 19:18
  • $\begingroup$ Continuing the 2nd comment, it is then enough to assume that $M$ embeds into a separable conjugate space, for again by subspace homogeneity you can assume WLOG that $M$ is contained in a weak$^*$ closed separable subspace of $\ell_\infty$ and use the fact that if both $Y$ and $X/Y$ embed into $\ell_\infty$, then so does $X$. $\endgroup$ – Bill Johnson Nov 15 '13 at 19:35
  • $\begingroup$ @Bill Johnson: Thank you very much for your useful comments. $\endgroup$ – M.González Nov 16 '13 at 11:44

While I do not have a complete answer to the OP’s question, I made enough observations that I think it is worthwhile to record them as an answer.

(1) If $X$ and $Y$ are isomorphic (closed) subspaces of $\ell_\infty$, then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $\ell_\infty/Y$ embeds into $\ell_\infty$.

Indeed, it is clear that $\ell_\infty/X$ embeds into $\ell_\infty$ iff $(\ell_\infty \oplus \ell_\infty)/(X \oplus \{0\})$ embeds into $\ell_\infty$, so without loss of generality we can assume that $\ell_\infty$ embeds into both $\ell_\infty/X$ and into $\ell_\infty/Y$. We then get from [LR, [Theorem 3(i)] that every isomorphism from $X$ onto $Y$ extends to an automorphism of $\ell_\infty$.

(2) Suppose that $Y$ is a subspace of $\ell_\infty$ and $X$ is a subspace of $Y$. Assume that $\ell_\infty/Y$ embeds into $\ell_\infty$. Then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $Y/X$ embeds into $\ell_\infty$.

The “only if” part is clear because $Y/X$ embeds into $\ell_\infty/X$. The other direction is an easy consequence of Lindentrauss’ observation that if $U\subset V$ and both $U$ and $V/U$ embed into $\ell_\infty$, then $V$ embeds into $\ell_\infty$. To prove this observation, let $T$ be an isomorphism from $U$ into $\ell_\infty$ and $S$ an isomorphism from $V/U$ into $\ell_\infty$. The space $\ell_\infty$ is $1$-injective (this is immediate from the Hahn-Banach theorem), so $T$ extends to a bounded linear mapping (which we also denote by $T$) from $V$ into $\ell_\infty$. Denoting the quotient map from $V$ to $V/U$ by $Q$, we see that $T \oplus SQ: x\mapsto (Tx, SQx)$ defines an isomorphism from $V$ into $\ell_\infty \oplus_\infty \ell_\infty \equiv \ell_\infty$.

(3) Suppose that $X$ is a subspace of of $\ell_\infty$ and $X$ is isomorphic to $Y^*$ for some separable $Y$. Then $\ell_\infty/X$ embeds into $\ell_\infty$.

Indeed, $Y$, being separable, is isomorphic to the quotient space $\ell_1/W$ for some subspace $W$ of $\ell_1$, and $W^\perp$ in $\ell_\infty$ is isomorphic to $X$. But $\ell_\infty/{W^\perp}$ is isometric to $W^*$, which embeds into $\ell_\infty$ because $W$, being separable, is a quotient of $\ell_1$. Hence by (1), the space $\ell_\infty/X$ also embeds into $\ell_\infty$.

(4) Suppose that $X$ a subspace of of $\ell_\infty$ and $X$ embeds into a separable conjugate space. Then $\ell_\infty/X$ embeds into $\ell_\infty$.

This follows from (3), (2), and the fact that every separable space embeds into $\ell_\infty$.

(5) $\ell_\infty/{c_0}$ does not embed into $\ell_\infty$. Hence if $X$ is a separable subspace of of $\ell_\infty$ and $X$ contains a subspace isomorphic to $c_0$, then $\ell_\infty/X$ does not embed into $\ell_\infty$.

The second statement follows from the first and (1), (2). The first statement is another observation that I think is due to Lindenstrauss; namely, that $\ell_\infty/{c_0}$ contains an isometric copy of $c_0(\Gamma)$ with $|\Gamma|=2^{\aleph_0}$ (consider the image in $\ell_\infty/{c_0}$ of characteristic functions of $2^{\aleph_0}$ infinite subsets of the natural numbers all of whose pairwise intersections are finite). The space $c_0(\Gamma)$ does not embed into $\ell_\infty$ because the unit vector basis for the space is a non separable set that has weakly compact closure, and every weakly compact subset of the dual to a separable space is separable.

(6) If $X$ is a subspace of $\ell_\infty$ that is isomorphic to $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$. Hence if $X$ is a subspace of $\ell_\infty$ that embeds into $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$.

The second statement follows from the first by what is now standard reasoning. For the first statement, note that $L_1(0,1)$ embeds into $C[0,1]^*$ as a complemented subspace, and by (3), if $Y$ is a subspace of $\ell_\infty$ isomorphic to $C[0,1]^*$, then $\ell_\infty/Y$ embeds into $\ell_\infty$. Now use (2) and (1).

[LR] J. Lindenstrauss and H. P. Rosenthal, Automorphisms in $c_0$, $\ell_1$, and $m$, Israel J. Math. 7 (1969), 227–239.

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