Let $C$ be a (nonempty) compact subset of euclidean $\mathbb R^n$, and consider the set-valued map $p\_C:\mathbb R^n \to 2^C$ defined by
$$
p\_C(x) = \{c \in C \mid \|x-c\| = \mbox{dist}(x,C)\},
$$
where $\mbox{dist}(x,C) := \inf\_{c \in C}\|x-c\|$ is the distance of $x$ from $C$.



> 
> **Question.** *Under what minimal conditions on $C$ is $p\_C$ Lipschitz-continuous w.r.t Hausdorff distance ?*
> 
> 
> 


**Note.** The case where $C$ is closed and convex is fully solved. Indeed, in such a case, $p\_C$ is single-valued and non-expansive (and therefore $1$-Lipschitz).



On the Lipschitz continuity of $x \mapsto \arg\min_{c \in C}d(x,c)$ w.r.t Hausdorff distance
Seems like in fact convex sets are the only ones for which $p\_C$ is continuous. To prove this, we can begin by noticing that for a set $C$ with the property, $p\_C$ can only take one point sets as values. If not, let $x,a\_1,a\_2$ be different points with $a\_1,a\_2\in p\_C(x)$. Then any point $y$ in the open segment from $x$ to $a\_i$ has $p\_C(y)=\{a\_i\}$, so $p\_C$ cannot be continuous at $x$.


So, if for a compact $C$, $p\_C$ is continuous, then it is single valued. So [this](https://math.stackexchange.com/questions/4344363/if-a-compact-set-is-not-convex-some-ball-is-tangent-to-it-at-several-points) shows that $C$ has to be convex.


Edit: From Dustin G. Mixon's comment above, it seems like the nearest point mapping is well studied. A reference that if a compact set $C$ is not convex, then the nearest point mapping is not single valued would be welcome.


