Is the level set $\{f=1\}$ of the Minkowski functional of $C$ equal to the boundary of $C$?

Question

Let $C$ be a convex and compact subset of $\mathbb{R}^d$. Assume that $\boldsymbol{0}$ belongs to the interior of $C$.

The Minkowski functional of $C$ is \begin{align*} f \colon \mathbb{R}^d & \to [0,+\infty)\\ \boldsymbol{x} &\mapsto \min\{\tau\ge 0 : \boldsymbol{x} \in \tau C\} \end{align*}

I know from the general theory that $$ \mathrm{int}(C)\subset\{f<1\}\subset C=\{f\le 1\} $$ where $\mathrm{int}(C)$ is the topological interior of $C$.

This in principle would allow for points $\boldsymbol{x}\in \partial C$ in the boundary of $C$ to belong to $\{f<1\}$.

However, I have a feeling that this cannot happen, i.e., that $\{f<1\} = \mathrm{int}(C)$ or, equivalently said, that $\{f=1\} = \partial C$.

Is that the case?

Answer 1

Your gut feeling seems correct to me.

Let $\varepsilon>0$ such that $B(0,\varepsilon)\subseteq C$ and assume $x\in \tau C$ for some $\tau\in (0,1)$. Then, $\frac{1}{\tau}x\in C$ and accordingly, we get that

$$ B(x,(1-\tau)\varepsilon)=\tau \left\{\frac{1}{\tau}x\right\}+(1-\tau)B(0,\varepsilon)\subseteq C, $$ which proves that $x$ is an interior point.