Hereditarily indecomposable Banach spaces and Separable Quotient problem

Question

A Banach space $X$ is called indecomposable if there exists no infinite-dimensional subspaces $M$ and $N$ such that $X = M \bigoplus N$. If every infinite-dimensional closed subspace of $X$ is indecomposable, then $X$ is called Hereditarily indecomposable.

The well known "Separable Quotient problem" asks that does every Banach space has an infinite-dimensional separable quotient space？

My question is that: Does there exist an inseparable Banach space $X$ such that its dual space $X^*$ is Hereditarily indecomposable?

PS: It is easy to see that if such Banach space $X$ exists, then $X$ has no infinite-dimensional separable quotient space. Thus the well known "Separable Quotient problem" is sloved in the negative.

Answer 1

The dual to any non separable Banach space is decomposable. I mentioned this in my post

Decomposable Banach Spaces

EDIT 9/1/11. Lindenstrauss' method is described in his paper

MR0205040 (34 #4875) Lindenstrauss, Joram On nonseparable reflexive Banach spaces. Bull. Amer. Math. Soc. 72 1966 967–970. (Reviewer: R. C. James).

You can also read about it in Zizler's article in volume II of the Handbook of the Geometry of Banach Spaces.

Given a separable subspace $X$ of the Banach space $Y$, there is a separable superspace $Z$ of $X$ in $Y$ s.t. for every finite dimensional subspace $E$ of $Y$, there is a linear operator $T_E$ from $E$ to $Z$ so that $\|T_E\| < 1+ 1/\dim(E)$ and $T_E$ is the identity of the intersection of $E$ with $Z$. Extend $T_E$ to a (discontinuous, non linear) map from $Y$ to $Z$ by letting $T_E$ be zero on the complement of $Z$. The finite dimensional subspaces of $Y$ are directed by inclusion--this turns $(T_E)$ into a net. You get a subnet s.t. for each $f$ in $Y^*$ and $y$ in $Y$, $f(T_E)(y)$ converges pointwise to, say, $S(f)(y)$. You can check that $S$ is in fact a norm on projection on $Y^*$ with kernel $Z^\perp$.