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If $X'$ is the topological dual of a Banach space, then is that true that a convex set is closed (for the norm on $X'$ given by $\lVert f \rVert_{X'} := \sup \frac{\langle f , x \rangle}{\lVert x \rVert_{X}}$) if and only if it is weak*-closed?

If it is false, is there a class of infinite-dimensional spaces on which it is true?

(I know that a convex set is weakly closed if and only if it is closed but I could not find such a result (or its negation) for the weak*-topology.)

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    $\begingroup$ Would the closed halfspace generated by a continuous linear functional that is not in the predual work as a counterexample? $\endgroup$
    – Michael Greinecker
    Commented Aug 31, 2016 at 8:59
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    $\begingroup$ As Michael Greinecker says, the closed convex sets differ in general. We could also consider the kernel of a continuous linear functional. Weak-* closed convex and closed convex are the same if and only if the space is reflexive, i.e. the weak and weak-* topologies coincide. $\endgroup$
    – Robert Furber
    Commented Aug 31, 2016 at 10:23

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No it is false in general.

Yes, there is a class of spaces where it is true: these are exactly the reflexive spaces.

Suppose $X$ is not reflexive. Then considering $X$ embedded into $X''$, we have $X \subsetneq X''$. If $\alpha \in X'' \setminus X$, then its kernel is norm closed, weakly closed, convex, but not weak* closed.

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