Continuity of distance function without triangle inequality We say that a continuous function $\rho : \mathbb{C}^n \to \mathbb{R}$ is a distance function if the following three conditions hold:
1.)  $\rho \geq 0$
2.) $\rho (z) =0$ iff $z=0$
3.) $\rho(cz)= |c| \cdot \rho(z)$ for all $c \in \mathbb{C}$ and $z \in \mathbb{C}^n$
Let $C \subset \mathbb{C}^n $ be closed. Define $d: \mathbb{C}^n \to \mathbb{R}$ so that $d(z) = \inf_{w \in C} \rho(z -w)$. Show that $d$ is a continuous function.
 A: It follows from your conditions that $\rho(z)\geq a|z|$ with $a>0$. Indeed, $\rho$ is continuous and positive away from $0$, so it must reach a positive minimum on the unit sphere of $\mathbb{C}^n$ by compactness, say $a$. Then by the last condition for $z\neq0$ we have $\rho(z)=|z|\rho(z/|z|)\geq a|z|$.
Having this we can replace $\inf_{w \in C}\rho(z-w)$ in the definition of $d(z)$ by $\inf_{w \in C_R}\rho(z-w)$, where $C_R$ is the intersection of $C$ with a closed ball of appropriately chosen radius $R$, without changing its values on $z$ from some ball $B_r(z_0)$ around any $z_0$. Indeed, if the value of $|w|$ is large, so is the value of $|z-w|\geq|w|-|z|\geq|w|-r$, and hence so is the value of $\rho(z-w)\geq a(|w|-r)$. Therefore, the values of $\rho(z-w)$ for large $w$ are uniformly large as long as $z\in B_r(z_0)$, and do not affect the infimum over $C$.
The point of this is that $C_R$ is closed and bounded now, i.e. compact, and $\rho(z-w)$ is jointly continuous on $B_r(z_0)\times C_R$. It is a general fact that the infimum of continuous family of continuous functions over a compact set is continuous, see a proof here or here. Thus, $d$ is continuous at $z_0$, and $z_0$ was arbitrary.
