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