Let $C$ be a nonempty closed subset of $\mathbb R^n$. It is known that any such set satisfies the following condition



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> **(Unique CPP a.e).** *For **almost every** $x \in \mathbb R^n$, there exists a unique point in $c(x) \in C$ such that $\|x-c(x)\| = \mbox{dist}(x,C) := \inf\_{c \in C}\|x-c\|$.*
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For example, see **Theorem 1** of <http://tzamfirescu.tricube.de/TZamfirescu-110.pdf>.


In particular, if $C$ is a closed convex subset of $\mathbb R^n$, then it is a folklore fact that $C$ verifies he unique **CPP** **everywhere**.


For any $\epsilon > 0$, let $C^\epsilon := \{x \in \mathbb R^n \mid d(x,A) \le \epsilon\}$ be the $\epsilon$-expansion of $C$.



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> **Question.** *Under what minimalistic conditions on $C$ does there exist a function $\epsilon:(0,\infty) \to [0,\infty)$, such that $\limsup\_{L \to \infty} \epsilon(L) = 0$ and the mapping $u\_C:x \mapsto (x-c(x))/\|x-c(x)\|$ is $L$-Lipschitz a.e on $\mathbb R^n \setminus C^{\epsilon(L)}$ for any sufficiently large $L>0$ ?*
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* I'm particularly interested in the case of sets of the form $C := \cap\_{i \in I} C\_i$ with $C\_i := \{x \in \mathbb R^n \mid f\_i(x) \le 0\}$, where $f\_i:\mathbb R^n \to \mathbb R$ are sufficiently smooth functions. Even the case where $I$ is a singleton $C$ is the sublevel set of a sufficiently smooth function is already interesting.
* The case where $C$ is a closed convex subset of $\mathbb R^n$ has been solved here <https://mathoverflow.net/a/412676/78539>: it suffices to take $\varepsilon(L) = 2/L$.



Conditions for Lipschitzness of boundary normal vector, almost everywhere
Given a nonempty closed set $C\subseteq\mathbb{R}^n$, let $S\_C\subseteq\mathbb{R}^n$ denote the set of points for which the closest point in $C$ is not unique. Suppose $C$ satisfies each of the following:


* $S\_C$ is nonempty (i.e., $C$ is not convex),
* there exists $\rho>0$ such that $C^\rho\cap S\_C=\emptyset$, and
* the closest point mapping $c\colon(\mathbb{R}^n\setminus S\_C)\to C$ is continuous.


(Every nonempty closed nonconvex subset $C\subseteq\mathbb{R}^n$ with smooth boundary that I can think of satisfies these property.)


Let $r\_C<\infty$ denote the supremum of such $\rho$. Below, we show that for every $\epsilon\in(0,r\_C)$, it holds that $u\_C$ is **not** Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$ for any null set $Z$.


Select $p\in\mathbb{R}^n$ of distance $r\_C$ from $C$ for which there exist $x,y\in C$ with $x\neq y$ such that
$$
\|x-p\|=\|y-p\|=r.
$$
Then for each $t\in(0,1)$, the points $x(t):=tx+(1-t)p$ and $y(t):=ty+(1-t)p$ are of distance less than $r\_C$ from $C$, and so $c(x(t))=x$ and $c(y(t))=y$. Furthermore,
$$
u\_C(x(t))=\frac{x(t)-c(x(t))}{\|x(t)-c(x(t))\|}=\frac{x(t)-x}{\|x(t)-x\|}=\frac{p-x}{\|p-x\|},
$$
and similarly
$$
u\_C(y(t))=\frac{p-y}{\|p-y\|}.
$$
Notably, for small $t$, $x(t)$ and $y(t)$ are arbitrarily close to each other, while $u\_C(x(t))$ and $u\_C(y(t))$ remain a fixed distance apart.


Given $\epsilon\in(0,r\_C)$, a null set $Z$, and a constant $L>0$, we now show that $u\_C$ is not $L$-Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$. Fix $\delta>0$ to be selected later. Select $t\_0\in(0,1/2)$ small enough so that $x(t\_0)$ and $y(t\_0)$ avoid $C^\epsilon$ and $\|x(t\_0)-y(t\_0)\|<\delta$. Select a continuous path $q\colon[0,1]\to \mathbb{R}^n\setminus S\_C$ such that $q(0)=x(t\_0)$, $q(1)=y(t\_0)$, and $q(s)\not\in Z$ for almost every $s\in(0,1)$. By the continuity of $c$, there exist $s\_0,s\_1\in[0,1]$ such that $q(s\_0),q(s\_1)\not\in Z$ and
$$
\max\Big\{\|q(s\_0)-x(t\_0)\|,\|q(s\_1)-y(t\_0)\|,\|c(q(s\_0))-x\|,\|c(q(s\_1))-y\|\Big\}<\delta.
$$
As we will see, $q(s\_0)$ and $q(s\_1)$ witness that $u\_C$ is not $L$-Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$.


After some straightforward manipulations, we have
\begin{align\*}
\|u\_C(q(s\_0))-u\_C(x(t\_0))\|
&=\bigg\|\frac{q(s\_0)-c(q(s\_0))}{\|q(s\_0)-c(q(s\_0))\|}-\frac{x(t\_0)-x}{\|x(t\_0)-x\|}\bigg\|\\
&\leq\frac{4\delta}{\|x(t\_0)-x\|}=\frac{4\delta}{(1-t\_0)r\_C}\leq\frac{8\delta}{r\_C},
\end{align\*}
and similarly
$$
\|u\_C(q(s\_1))-u\_C(y(t\_0))\|\leq\frac{8\delta}{r\_C}.
$$
Then
$$
\|u\_C(q(s\_0))-u\_C(q(s\_1))\|
\geq\|u\_C(x(t\_0))-u\_C(y(t\_0))\|-\frac{16\delta}{r\_C}
=\frac{\|x-y\|-16\delta}{r\_C},
$$
but
$$
\|q(s\_0)-q(s\_1)\|
\leq\|x(t\_0)-y(t\_0)\|+2\delta
\leq 3\delta.
$$
Selecting $\delta<\min\{\frac{1}{32},\frac{1}{6Lr\_c}\}\cdot\|x-y\|$ then gives
$$
\|u\_C(q(s\_0))-u\_C(q(s\_1))\|>L\|q(s\_0)-q(s\_1)\|,
$$
as claimed.


