# Constructing a closed, convex subset of $X^{\ast}$ that is not weakly-* closed

I'm asked to show that if $X$ is a non-reflexive Banach space, there exists (norm) closed and convex subsets of $X^\ast$ that are not $w^{\ast}$-closed. In other words, there's no analogue of Mazur's Lemma for the $w^{\ast}$-topology. As a hint, I'm given that a functional in a TVS is continuous iff it's kernel is closed.

I thought about two main approaches to tackle the problem:

(1st) $X$ non-reflexive implies at least one $\psi \in X^{\ast \ast}$ s.t. $\psi \notin i[X]$, $i[X]$ being the canonical embbeding of $X$ in it's second dual. Therefore, $\ker \psi$ is a closed, convex subset of $X^{\ast}$, and I'd have to show that it's not $w^\ast$-closed. The problem here is that I just can't find any direct connection between $w^\ast$-closedness and $\psi \notin i[X]$, and also, I don't see where I could use $X$ Banach.

(2nd) $X$ Banach non-reflexive implies that $i[X]$ as above is a proper, closed subspace of $X^{\ast \ast}$. Therefore, I could use a separation theorem to find $\Gamma$ in $X^{\ast \ast \ast}$ s.t. $\text{Re } \Gamma (\psi_{0}) > \sup_{x \in X} \Gamma(i(x))$ for some $\psi_{0} \notin i[X]$, and then try to get some useful result from embedding $X^{\ast}$ in $X^{\ast \ast \ast}$, maybe by using Goldstine's theorem about $w^{\ast}$-density. The problem here is that it seems too confusing, and I just couldn't get very far with so many embeddings. Specially considering that embedding is an $w-w^{\ast}$ homeomorphism, so if I get results about $w^{\ast}$ in the embedding of the first dual in the third dual, it translates as a result about $w$ in the first dual.

Anyone could give me hints at least about which direction follow?

• For 1, if $\psi$ is weak* continuous, then $\{x^*\in X^* : |\psi x^*|<1\}$ is weak* open. Show then that there are $x_1 ,\ldots, x_n$ in $X$ with ${\rm ker} \,\psi\supset \cap{\rm ker}\, x_i$. But then, $\psi$ is a linear combination of the $x_i$. Oct 6, 2013 at 22:17

Your 1st approach is good. You need to notice that your $\psi$ is not $w^*$-continuous, and that a functional is continuous if and only if its kernel is closed.