Continuity of Minkowski functional in locally convex topological vector space

Question

Let $X$ be a locally convex topological vector space over $\mathbb{R}$ or $\mathbb{C}$ and let $p_C(x)=\inf (\lbrace t>0 \mid t^{-1}x \in C\rbrace)$ be the Minkowski functional for an arbitrary open convex neighbourhood $C$ of $0$ in $X$. I need to show that for any $C$ as above that the Minkowski functional is continuous, but I have a hard time getting started on the proof, since I can't figure out what the pre-image of the Minkowski functional is for an arbitrary open subset of the positive real line $[0,\infty)$. Can I show it without assuming that $C$ is balanced?

Answer 1

It can be easily shown that every Minkowski functional functional satisfy in$$p_c(x-y)\leq p_c(x)-p_c(y)$$

Answer 2

Here is a try for how to solve it.

Since the topology on $\mathbb{R}$ is generated by the open intervals $(a,b)$ with $a<b$, the subspace topology on $[0, \infty)$ is generated by the sets of the form $(a,b)\cap [0, \infty)$ it suffices to prove that $p_C^{-1}((a,b)\cap [0, \infty))$ is open in $X$. Now, since the topology on $X$ is generated by sets of the form $p+C$ for $C \in \gamma$ and $p \in X$, where $\gamma$ is local base of convex sets $C$, it is enough to show that for any $x \in p_C^{-1}((a,b)\cap [0, \infty))$, there exists a $C' \in \gamma$ such that $x + C' \subseteq p_C^{-1}((a,b)\cap [0, \infty))$.

Now, there are four possibilities for what the set $(a,b)\cap [0, \infty)$ is:

$(a,b)\cap [0, \infty)=(a,b)$ if $b>a>0$

$(a,b)\cap [0, \infty)=(0,b)$ if $a=0$

$(a,b)\cap [0, \infty)=[0,b)$ if $a<0$ and $b>0$

$(a,b)\cap [0, \infty)=\emptyset$ if $b\leq0$

Take for example the first case $p_C^{-1}((a,b))$:

Let $x_0 \in p_C^{-1}((a,b))$, then $a<p_C(x_0)<b$, since $x_0+C$ is an open neighborhood of $x_0$ from the base for the topology on $X$, we have the required open set. Now here's the tricky part: I want to show that $x_0+C \subseteq p_C^{-1}((a,b))$, but I run into trouble because of the following analysis ($c_0 \in C$):

$p_C(x_0+c_0)\leq p_C(x_0)+p_C(c_0) < b+1$

But since an element in $p_C^{-1}((a,b))$ has to satisfy "$<b$", this does not work in this case (or the others?). Any help would be much appreciated.