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

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.
