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I asked this question on Math StackExchange first, but it was not answered.

If $X$ is a Banach space and $Z$ is a subset of $X^*$, consider the annihilator of $Z$ in $X^{**}$:

$$ Z^{\perp}=\{x^{**}\in X^{**} : x^{**}(Z)=0\} $$

and the pre-anihilator of $Z$ in $X$:

$$ Z^{\top}=\{x\in X : y^*(x)=0, \forall y^*\in Z\} $$

It is easy to see that $Z^\top\subseteq Z^{\perp}$ when the elements of $X$ are viewed as functionals on $X^*$ via the canonical embedding. It is also clear that when $X$ is reflexive we have equality. My question: is there a characterization of those subsets $Z$ of $X^*$ that give equality in the non-reflexive case as well?

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As I understand the conditions are: (1) The closed linear span of $Z$ in $X^*$ is weak$^*$ closed; (2) The subspace $Z^\top$ in $X$ is reflexive.

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  • $\begingroup$ I see it, thank you. I like the first condition, but I was wondering whether the second could be rephrased as a property intrinsic to $Z$ and $X^*$, rather than going to $X$. Not sure if what I just said makes sense :). $\endgroup$
    – Markus
    Commented Apr 7, 2017 at 17:16
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    $\begingroup$ The following condition is equivalent condition to (2): The quotient space $X^*/W$ is reflexive, where $W$ is the weak$^*$-closure of the linear span of $Z$. $\endgroup$
    – Mikhail Ostrovskii
    Commented Apr 7, 2017 at 17:45

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