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My first question is the following:

Q1: Let $X$ be a Banach space. If its dual $X^\*$ is weak* separable, does $X$ admit an infinite-dimensional and separable quotient $X/M$?

To the best of my knowledge, the dual $X^\*$ is weak* separable, when $X$ satisfies one of the following:

(i) $X$ is separable;

(ii) $X$ is the dual of a separable Banach space;

(iii) $X$ is Hereditary Indecomposable Banach space. That is, every infinite-dimensional closed subspace of $X$ can not be written as a direct sum of two infinite-dimensional closed subspaces.

And, I see that if $X$ satisfies (i) or (ii), $X$ admit an infinite-dimensional and separable quotient.

Q2: Is it true that $X$ admit an infinite-dimensional and separable quotient, if $X$ satisfies (iii) ?

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BTW: If $X^*$ is not HI, then $X$ has a separable quotient. Indeed, then by Gowers $X^*$ contains a subspace with an unconditional basis and hence, by James, a copy of $c_0$, $\ell_1$, or an infinite dimensional reflexive space. In the last case, $X$ has a reflexive quotient. In the first case, $X$ contains a complemented subspace isomorphic to $\ell_1$ by Bessaga-Pelczynski. In the middle case, $X$ has a quotient isomorphic to either $c_0$ or $\ell_1$ by combining results of Rosenthal and mine and Hagler and mine. –  Bill Johnson Oct 25 '11 at 16:36
    
In the previous comment, "If $X^∗$ is not HI" should be "If $X^∗$ has no HI subspace". –  Bill Johnson Oct 25 '11 at 20:53
    
Thanks,Bill. Great "BTW". Your answer is a good application of Gowers' dichotomy. –  Qingping Zeng Oct 26 '11 at 4:27

1 Answer 1

Q1 is equivalent to the separable quotient problem. Indeed, given $Y$ infinite dimensional, let $W$ be any separable infinite dimensional subspace of $X^*$ and let $Y$ be the annihilator of $W$ in $X$. Then the dual of $X/Y$ is the weak$^*$ closure of $W$ in $X^*$.

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