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Let $E$ be a Banach space.

It is known that if for any equivalent norm on $E^*$ the closed unit ball of $E^*$ is weakly* closed, then $E$ is reflexive (a very short proof is in the book by Fabian, Habala, Hajek, Montesinos and Zizler).

It is easier to show that if the closed unit ball of any closed subspace of $E^*$ is weak* closed, then $E$ is reflexive (but now the norm is fixed).

Hence, for any non-reflexive space $E$ there is an equivalent norm on $E^*$ such that the unit ball is NOT weak* closed, and a closed subspace of $E^*$ such that the unit ball is NOT weak* closed.

I wonder if these criteria can be joined, i.e. is is true that for any non-reflexive space $E$ there is an equivalent norm on $E^*$, such that the unit balls of all infinite-dimensional closed subspaces are not weak* closed, and a closed subspace, whose unit ball in any equivalent norm is not closed?

Equivalently:

  1. If for every closed subspace $F$ of $E^*$ there is an equivalent norm on $E$ (or, equivalently, on $F$), such that $\overline{B}_{F}$ is weak* closed, does it follow that $E$ is reflexive?

  2. If for every equivalent norm on $E^*$ there is an infinite-dimensional closed subspace $F$ of $E^*$ such that $\overline{B}_{F}$ is weak* closed, does it follow that $E$ is reflexive?

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Question 2 has a negative answer if $E$ contains a reflexive subspace $F$ because the closed unit ball of $F$ under any equivalent norm is weakly compact and hence weak$^*$ compact.

Question 1 has a positive answer. If $E$ is not reflexive, let $F$ be the kernel of some functional in $E^{**}\sim E$.Then $F$ is norming for $E$, which implies that the bipolar of any neighborhood of $0$ in $F$ contains a neighborhood of $0$ in $E^*$. But the bipolar of a convex balanced set is the weak$^*$ closure of the set.

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  • $\begingroup$ Thank you for your answer, but by "weak* closed" I meant "relatively weak* closed in $F$". Perhaps, that was indeed not clear from the context. However, as it turns out, your version is sufficient for my specific purposes. Note sure what to do with the question now. Thank you again. $\endgroup$ – erz Sep 8 '17 at 0:46

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