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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.

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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 $Z^\perp$.

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Thanks,Bill. Your answer is very useful to me. As you mentioned in your post…, it is an easy consequence of lemmas of J. Lindenstrauss that the dual to any non separable Banach space is decomposable. Could you please give me a link to Lindenstrauss's paper and, give me some hints if you feel convenient. A future question: For a given non separable Banach space X, whether every infinite-dimensional quotient space X/M is decomposable. – Qingping Zeng Sep 1 '11 at 4:44

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