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?

  • 2
    $\begingroup$ This is still Alaoglu's theorem at play here, just set $Y = X'$, so then $Y' = X''$. $\endgroup$ – Christopher A. Wong Mar 28 '13 at 19:57
  • $\begingroup$ Of course! What is wrong with me! Thanks! $\endgroup$ – 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.

Maybe you meant compact instead of closed ?

  • $\begingroup$ Thanks for your comment. But now I realise (due to Christopher A. Wong comment) that this does hold for a vector space X. When I mentioned X being reflexive I was of course generalising massively. $\endgroup$ – user58514 Mar 28 '13 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy