13
$\begingroup$

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 $\ell_\infty$.

$\endgroup$
  • 1
    $\begingroup$ 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
13
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.