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I have the following question related to the previous posts Hereditarily indecomposable Banach spaces and Separable Quotient problem and Weak star separable and separable quotient problem

Question. Does every infinite dimensional Banach space $X$ have an infinite dimensional quotient $Y$ which is isomorphic to a subspace of $\ell^ \infty $.

This is equivalent to the property that the dual $X^*$ has a w* closed subspace $Z$ with its unit ball w* separable. Note that this is not equivalent to the property that $Z$ is w* separable https://arxiv.org/abs/1112.5710 .

A possible positive answer to the question will reduce the general SQP to the subspaces of $\ell^ \infty $.

In the case of the subspaces of $\ell^ \infty $ we know that the infinite dimensional representable subspaces (i.e. subspaces which are analytic as a subset of $\ R^N$ ) have an infinite dimensional separable quotient https://arxiv.org/abs/0805.2032 (Thm. 50) . This class includes the duals of separable Banach spaces.

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  • $\begingroup$ @YCor I think that now is clear. Thanks! $\endgroup$
    – S Argyros
    Commented Jan 6, 2022 at 13:45
  • $\begingroup$ Here are some thoughts concerning the general SQP. Assume that the initial space $X$ does not have a separable quotient. Then $X^*$ does not contain neither an unconditional basic sequence nor a reflexive subspace. By Gowers dichotomy $X^*$ is saturated by separable HI subspaces not containing reflexive subspace.Each one of them is either like Gowers Tree space or its predual. (A terrified structure for anyone thinking for a counterexample to SQP). In the case where $X$ fails SQP consider a separable W of $X^*$ which is HI and let $Z^*$ its w* closure. $\endgroup$
    – S Argyros
    Commented Jan 7, 2022 at 20:52
  • $\begingroup$ Assume that $Z^*$ remains HI . Denote by $Z$ the quotient of $X$ with its dual the space $Z^*$. Choose $(z_n)_n$ a sequence in the unit ball of $Z$ that norms the space $W$. The sequence $(z_n )_n$ defines a w* continuous operator from $Z^*$ to $l^\infty$ which is isometric on the subspace $W$. Since $Z^*$ is HI this operator is an isomorphism on a subspace of $Z^*$ of finite codimension . Assume that it is isomorphism on the whole $Z^*$. then the unit ball of $Z^*$ is w* metrisable which yields that $Z$ is separable. $\endgroup$
    – S Argyros
    Commented Jan 7, 2022 at 21:06
  • $\begingroup$ I do not understand "Since $𝑍^∗$ is HI this operator is an isomorphism on a subspace of $𝑍^∗$ of finite codimension". The operator does not range in $𝑍^∗$. What am I missing? $\endgroup$
    – Bill Johnson
    Commented Jan 8, 2022 at 0:07
  • $\begingroup$ @Bill The general statement is the following. If $X$ is HI and T a bounded linear operator from $X$ to $Y$ then either T is strictly singular or it is isomorphism on a finite codimensional subspace of $X$. The result is included in the Memoirs AMS with Tolias. However the statement if $Z^*$ is HI then Z is separable follows immediately from your result that non separable spaces have decomposable dual. $\endgroup$
    – S Argyros
    Commented Jan 8, 2022 at 8:41

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