Faithfullness of the Minkowski functional

Question

Let $X$ be a locally convex topological vector space. I need to show that the Minkowski functional $p_C$ for $C$ a convex open neighborhood of $0$ coming from the local base of convex sets, is faithful, that is, I must prove that for any $x \in X$, $x=0$ if an only if $p_C(x)=0$ for all $C$ as defined. The first way ($\Rightarrow$) is easy enough, but the other way ($\Leftarrow$) I am getting stuck. I have the feeling that the only way is through an indirect proof, but I can't figure out what the best strategy is.

Answer 1

Suppose $x=\mathbf{0}$. Let $C \in \gamma$ be arbitrary , where $\gamma$ is a local base of convex sets. Let $p_C$ be the Minkowski-functional associated with $C$. If $t=0$ it follows that: \begin{equation} p_C(\mathbf{0})=p_C(0\mathbf{0})=0p_C(\mathbf{0})=0 \end{equation} Which was to be proved. Conversely, suppose $p_C(x)=0$ for all $C \in \gamma$ and assume for contradiction that $x \neq \mathbf{0}$. Since $X$ is Hausdorff, there exists open neighborhoods $U_x$ and $U_\mathbf{0}$ of respectively $x$ and $\mathbf{0}$ in $X$ , such that $U_x \cap U_\mathbf{0} = \emptyset$. Since $X$ is locally convex the open neighborhood $U_\mathbf{0}$ contains an open, convex neighborhood $C' \in \gamma$. Since $C' \subseteq U_\mathbf{0}$ it follows that also $C' \cap U_x = \emptyset$. Hence $x \not\in C'$ and so the Minkowski functional $p_{C'}(x) \geq 1$. This contradicts that $p_C(x)=0$ for all $C \in \gamma$. Therefore, $x = \mathbf{0}$, as was to be proved.

This does not use the Hahn-Banach Theorem, I think.