Let $X$ be a normed space, $X^*$ its dual space, $(X^*, w^*)$ is completely Hausdorff.

Proof: Let $f, g \in X^*$, $f\neq g$ then $\exists x\in X$ such that $f(x)\neq g(x)$ i.e. $\hat{x}(f)\neq \hat{x}(g)$ (here $\hat{x}\in X^{**}$, $\hat{x}(h)=h(x)$ for $h\in X^*$) so $(X^*, w^*)$ is completely Hausdorff.

Note: A top space is Completely Hausdorff when for any given pair of points in it there exists a real valued continuous function that separates them.

The whole thing seems pretty straight forward but it is always good to check.

EDIT: I'm thinking, if the scalar field is complex, we can take either $\text{Re} \hat{x}$ or $\text{Im} \hat{x}$ as the separating continuous function. Also, this function is usually required to map into $[0,1]$. It is sufficient that it maps into $\mathbb{R}$ if the image is bounded though. The whole thing does not bother me because I'm only interested is proving that the unit closed ball $(B_{X^*}, w^*)$ is Hausdorff, which by Banach-Alaoglu is compact (so its image by a real cont function will be bounded). However I asked the question in a more general sense and I am wondering now.

  • 1
    $\begingroup$ Looks okay to me. $\endgroup$ – Brian M. Scott Dec 16 '14 at 2:41
  • $\begingroup$ I'd remark that "every pair of points can be separated by a continuous function" isn't quite the same as "there is a continuous function that separates every pair of points". $\endgroup$ – Jonas Dec 16 '14 at 2:54
  • 1
    $\begingroup$ In general, the weak-star topology will be Tychonoff by virtue of being a $T_0$ vector space topology. $\endgroup$ – Henno Brandsma Dec 16 '14 at 20:23

The proof is mostly correct, except the step where you note that $\hat x\in X^{**}$: this makes it seem like you are trying to show that the weak topology is completely Hausdorff on a dual space.

That $\hat x\in X^{**}$ is immaterial, what matters here is that $x\in X$.

A different way to see that weak$^{*}$ topology is completely Hausdorff (and actually Tychonoff) - and as a consequence, so is the weak topology on any dual space, because it is finer) is to note that the weak$^{*}$ topology is just the topology of pointwise convergence on $X$.

Since complete regularity is preserved by arbitrary products and by taking a subspace, it follows that any subspace of ${\bf R}^X$ (or ${\bf C}^X$), including $X^*$, is completely regular (Hausdorff).


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.