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.)



weak*-closed convex = closed convex?
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.


