5
$\begingroup$

The weak $\sigma$-$(V,V')$ topology on a normed space V is Hausdorff?

So it is claimed that The weak $\sigma$-$(V,V')$ topology on a BANACH space V is Hausdorff?

In my proof I used the Hahn-Banach theorem to find the existence of linear function that separates any two arbitrarily chosen points. Then it easily follows that V is Hausdorff w.r.t the weak topology.

Nowhere can I see the necessity of V being a Banach space. So it holds true when V is just a normed space?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Tomás's answer is correct. You can also argue as follows: If $W$ is the completion of $V$ then $V' = W'$ and $V$ with the $\sigma(V,V')$-topology is a subspace of $W$ with the $\sigma(W,W')$-topology. Subspaces of Hausdorff spaces are Hausdorff. Megginson's book on Banach spaces carefully distinguishes between results that hold for normed spaces and those that only hold for Banach spaces. $\endgroup$ – Martin Mar 20 '13 at 23:28
2
$\begingroup$

You can assume even less from the space, for example, if $V$ is a topological vector space (I assume that it's topology is Hausdorff) with a convex base, then it's weak topology is Hausdorff. To understand more about it, you can consult this book in the part of weak topologies.

Let me add a litlle bit more. If $V$ is a locally convex space, then it's topological dual $V'$ is not degenerate, hence, it can be prove that the bilinear form $\langle\cdot,\cdot\rangle:V\times V'\rightarrow\mathbb{R}$ separate points, i.e. if $\langle v,v'\rangle =0$ for all $v'\in V'$ then $v=0$ and if $\langle v,v'\rangle =0$ for all $v\in V$ then $v'=0$. Denote by $(V')^\#$ the algebraic dual of $V'$ and define $\phi:V\rightarrow (V')^\#$ , by $$\phi(v)(v')=\langle v,v'\rangle$$

Because the bilinear form separate points, we can prove that $\phi$ is one-to-one, hence we can identify $V$ with the subspace $\phi(V)\subset (V')^\#$. On the other hand $(V')^\#\subset \mathbb{R}^{V'}$, where $\mathbb{R}^{V'}$ is the set of all functions from $V'$ to $\mathbb{R}$. Now we consider the product topology on $\mathbb{R}^{V'}$ and we define the weak topology $\sigma (V,V')$ in $V$ as the restriction of the product topology of $\mathbb{R}^{V'}$ to $\phi(V)$. From this definition you can see that the weak topology is in fact Hausforff.

| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ It is not enough just to assume that $V$ is a topological vector space; look at $L^p$ for $0<p<1$. However, local convexity is sufficient. $\endgroup$ – Mateusz Wasilewski Mar 20 '13 at 18:20
  • $\begingroup$ Hence, you may call $ V $ a locally convex topological vector space. $\endgroup$ – Leonard Huang Mar 20 '13 at 19:01

Your Answer

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