# Closed and convex subset of a locally convex space

For the following problem concerning functional analysis I cannot seem to find a solution. It states as follows.

Let $(E, \tau)$ be a locally convex vector space and $S \subseteq E$ with $S \ni 0$. Show that $S$ is closed and convex iff there is a subset $T \subseteq E^*$ such that $$S = \bigcap_{f \in T}\{\text{Re}\, f(x) \leq 1\}.$$

The "$\Rightarrow$"-direction seems to somehow follow from the separation theorem version of the Hahn-Banach theorem as I read it in Rudin's "Functional Analysis", p. 59, Theorem 3.4. Although I do not really know how to apply it here?

For the "$\Leftarrow$"-direction, the closedness of $S$ as intersection of closed sets seems clear to me. For the convexity again I am not so sure but I suppose this is due to the space being locally convex. Since $0 \in S$ we can find some absolutely convex base around 0.

What is the correct line of argument here?

• The $\Leftarrow$ direction doesn't need any particular properties of $E$ such as local convexity. In every topological space, the intersection of an arbitrary family of closed sets is closed. And in every real (or complex) vector space, the intersection of an arbitrary family of convex sets is convex. – Daniel Fischer Nov 22 '17 at 16:24
• But how would I argue then that the sets are convex? – Taufi Nov 23 '17 at 9:00
• They are preimages of a a convex set ($\{ z : \operatorname{Re} z \leqslant 1\}$) under linear maps. – Daniel Fischer Nov 23 '17 at 12:43