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?

  • $\begingroup$ 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, 2013 at 23:28
  • $\begingroup$ According to Theorem 3.10 at Rudin's Functional Analysis: let $X$ be a vector space and $Y$ be a separating vector space of linear functionals on $X$. Let $\tau$ be the weak topology on $X$ induced by $Y$. In this context, $(X,\tau)$ is a locally convex Hausdorff topological vector space. On the other hand, this is not so straightforward. In order to understand this, one needs to study the construction of locally convex topologies on vector spaces from families of seminorms. $\endgroup$
    – gpr1
    Jun 5, 2021 at 12:32

1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$ Mar 20, 2013 at 18:20
  • $\begingroup$ Hence, you may call $ V $ a locally convex topological vector space. $\endgroup$ Mar 20, 2013 at 19:01

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