Minkowski Functional Evaluated at a Point Outside the Convex Set

Question

Let $K$ be a convex set in a vector space such that $0$ is an internal point of $K$, let $p_K$ be the associated Minkowski functional, and let $z \notin K$. Why must $p_K(z) \ge 1$ hold? If it were true that $z \in p_K(z) \cdot K$, then if $p_K(z) < 1$ were the case, I could write $z = p_K(z) \cdot \frac{z}{p_K(z)} + (1 - p_K(z)) \cdot 0 \in K$, which would be a contradiction... However, I don't know if it's true that $z \in p_K(z) \cdot K$.

Answer 1

It's not necessarily true that $z\in p_K(z)\cdot K$; consider the open ball of unit radius for $K$ and $z=\langle 1,0,0,\ldots\rangle$. However, since $K$ is a convex body containing $\mathbf{0}$, then $p\leq q\implies p\cdot K\subseteq q\cdot K$ (can you see why?), and so $p\gt p_K(z)\implies z\in p\cdot K$ (likewise, make sure you can understand why this is so!). So if $p_K(z)\lt 1$, then...