For $E$ a Banach space and $E'$ (the continuous dual) a separable space, show that the closed unit ball $\overline{B}_{E} \subset E$ is weakly metrisable.

My work so far: As $E'$ is separable, we know that the closed unit ball $\overline{B}_{E''} \subset E''$ in the weak-$*$ topology is metrisable. Then Goldstine's lemma says that $\Phi(\overline{B}_E)$ is weak-$*$ dense in $\overline{B}_{E''}$ and we know $\Phi$ is an isometry, so $\overline{B}_E$ is metrisable. Finally, $\Phi^{-1}: (E'', \text{weak-}*) \to (E, \text{weak})$ is continuous (proved earlier), so the induced topology on $\overline{B}_E$ agrees with the weak topology.

($\Phi$ refers to the evaluation map, $E \to E''$ defined as $f \mapsto \varphi_f$ with $\varphi_f(F) = F(f)$ for $F \in E'$).

My question: Is this correct? It doesn't seem to me like I used the full power of Goldstine's lemma. Also, is there a more direct proof using $E'$ separable $\Rightarrow E$ separable?


1 Answer 1


Your logic is correct, but I don't think you even need Goldstine's lemma here. You already know that weak* topology is metrizable on bounded subsets of $E''$. The restriction of this topology to $E\subset E''$ agrees with the weak topology on $E$, by definition (both come from the same set of linear functionals). And any subspace of a metrizable topological space is metrizable, it need not be dense.

That said, a typical direct proof of the stated result is to pick a countable dense subset $\{\phi_n\}$ of the unit ball of $E'$ and verify that the metric $$ d(x, y) = \sum_{n=1}^\infty 2^{-n} |\phi_n(x-y)| $$ metrizes the weak topology on the unit ball of $E$.

  • $\begingroup$ Thank you for the reply! I felt like a direct proof as you say would be slightly wasteful, since that method was used to show the quoted result that the closed unit ball in $E''$ is weak-$*$ metrisable, so something higher level should be doable. I see how Goldstine isn't necessary though! $\endgroup$
    – B. Mehta
    Apr 10, 2018 at 1:54

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.