Let $V$ be a topological vector space and $V^*$ be the space of linear functionals induced with the weak-* topology. Can we say that $V^*$ is Hausdorff? Here is my attempt:

Let $\lambda\ne\lambda'\in V^*$. Then, there exists $v\in V$ such that $\lambda(v)\ne\lambda'(v)$. Choose $|\lambda(v)-\lambda'(v)|/2>\epsilon>0$. Then, the open sets $$\{\psi\in V^*: \|\psi-\lambda\|_v<\epsilon\}\qquad \{\psi\in V^*:\|\psi-\lambda'\|_v<\epsilon\}$$ are disjoint. So, every distinct point can be seperated by disjoint open sets, thus $V^*$ is Hausdorff.


This is indeed correct. You might want to expand on why these (basic) open sets are indeed disjoint (triangle inequality argument).

Abstractly, this argument works because the point-evaluations $e_v$, where $v \in V$, form a family of functions (to a Hausdorff space, namely the underlying field) that separates points, in this case basically by definition (different functionals are distinct functions so must be different for some point of the domain), while for the weak topology on $V$ we need some assumption on $V$ to ensure that all linear functionals can separate points, but then also the weak topology is Hausdorff.

  • $\begingroup$ So, I used the fact that the topological scalar field $K$ is Hausdorff. $\endgroup$ – Emre May 22 '16 at 12:45
  • $\begingroup$ @E.Girgin because the norm of distinct points is positive. So you use it implicitly. $\endgroup$ – Henno Brandsma May 22 '16 at 12:59

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