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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?

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I will give a proof of 2) and leave 3) to you.

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 $0<r<1-p(x)$ 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 $y=t(\frac y t)+(1-t)0 \in C$.

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