If $E\subset X^{*}$ is bounded, then so is its weak* closure

Question

If $X$ is a Banach space and $E\subset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?

(I've shown that if $F\subset X$ is norm-bounded, then its weak closure is also norm-bounded...)

Answer 1

I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R=\{\phi\in X^*: \|\phi\|\le R\}$ is weak*-closed; indeed, any bounded set is contained in such a ball.

Take any $\phi \in X^*\setminus B_R$. By definition of the norm, there is a unit vector $u\in X$ such that $|\phi(u)|>R$. Let $\epsilon = |\phi(u)|-R$. The set $$U = \{\psi\in X^* : |\psi(u)-\phi(u)|<\epsilon\}$$ is weak* open, contains $\phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.

Answer 2

I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded.