# Is the unit ball in $X''$ weak-star closed?

Is the unit ball in $X''$ weak-star closed?

I reached a point in my argument where it would suffice to show that the unit ball in $X''$ is weak-star closed. (Where $X$ is just some topological space and $X'$ is the continuous dual of $X$, etc.)

If $X$ was reflexive then of course this result would hold due to the unit ball in $X$ being weakly closed, therefore the unit ball in $X''$ would be weak-star closed.

But does this result hold in general?

• This is still Alaoglu's theorem at play here, just set $Y = X'$, so then $Y' = X''$. – Christopher A. Wong Mar 28 '13 at 19:57
• Of course! What is wrong with me! Thanks! – user58514 Mar 28 '13 at 20:09

I presume you intend for $X$ to be a normed space (you just wrote "topological space"), otherwise there is no obvious notion of a unit ball in $X''$. I'll assume for simplicity we are dealing with real normed spaces (the complex case is analogous).
This is true, and it doesn't need Alaoglu's theorem either. For $x \in X'$, let $T_x : X'' \to \mathbb{R}$ be the evaluation map $T_x(y) = y(x)$. By definition of the weak-* topology, all the maps $T_x$ are weak-* continuous. Now by definition of the norm on $X''$, $y$ is in the closed unit ball $B$ of $X''$ iff we have $|y(x)| \le 1$ for all $\|x\|_{X'} \le 1$. Thus we can write $$B = \bigcap_{\|x\|_{X'} \le 1} T_x^{-1}([-1,1]).$$ This is an intersection of weak-* closed sets, hence is weak-* closed.
First of all I assume $X$ is a Banach space (it has to be a normed vector space, or else I don't know what you mean by unit ball) and if it is not complete then it is not true.
Second, you seem to know that the unit ball of $X$ (with no other assumption that it's a Banach space) is weakly closed, so of course the same is true for its bidual, which is another Banach space in its own right.