# Minkowski functional of a convex neighbourhood

I want to prove the following theorem:

Theorem

Let $$X$$ be a topological vector space. If $$C \subseteq X$$ is a convex subset such that $$0 \in C^\circ$$, then

1. the Minkowski functional $$p_C$$ is sublinear
2. $$C^\circ = \{ x \in X : p(x) < 1 \}$$
3. $$\overline{C} = \{ x \in X : p(x) \leq 1 \}$$

What I know:

The assumption $$0 \in C^\circ$$ implies that $$C$$ is absorbing. Since $$C$$ is also convex, the function $$p_C$$ is sublinear. Now, let $$A = \{ x \in X : p_C(x) < 1 \}$$ and $$B = \{ x \in X : p_C(x) \leq 1 \}$$. Since $$p_C$$ is sublinear, we have $$p_A = p_B = p_C$$.

The problem is that $$p_A = p_B = p_C$$ does not directly imply $$C^\circ = A$$ and $$\overline{C} = B$$. How should I proceed?

If $$x \in C^{0}$$ then $$(1+\frac 1 n) x \in C$$ for $$n$$ sufficiently large and this implies $$p((1+\frac 1 n) x) \leq 1$$. It follows that $$p(x) \leq \frac 1 {1+\frac 1 n}<1$$.
Now let $$p(x) <1$$. Let $$0 and consider $$x+ rC^{0}$$. This is an open set containing $$x$$. let us show that it is contained in $$C$$. For $$c \in C^{0}$$ we have $$p(x+rc) \leq p(x)+rp(c)\leq p(x)+r <1$$ To finish the proof we need the fact that $$p(y) <1$$ implies $$y \in C$$. By definition of $$p$$ there exists $$t<1$$ such that $$y \in tC$$. But then $$y=t(\frac y t)+(1-t)0 \in C$$.