This is followup to the following question: [On the Lipschitz continuity of the unit-normal vector field of a polytope](https://mathoverflow.net/q/412663/78539)




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Let $C$ be a (nonempty) closed convex subset of $\mathbb R^n$. Note that to every $x \in \mathbb R^n$ corresponds a point $c(x) \in C$ which is closest to $x$. Let $C' := \mathbb R^n \setminus C$ and define $u:C' \to \mathbb R^n$ by $u(x) := (x-c(x))/\|x-c(x)\|$.



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> **Question.** *What are generic conditions on $C$ under which $u\_C$ is Lipschitz-continuous on $C'$ ?*
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Are there any such conditions linked with a "classical" notion of smoothness or curvature of (the boundary of) a closed convex set ?


Examples
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* If $C := \{0\}$, then $u$ is the function $x \mapsto x/\|x\|$, which is definitely not Lipschitz on its domain $C'=\mathbb R^n\setminus\{0\}$.
* If $C = B\_n$, the unit-ball in $\mathbb R^n$, then $u$ is the map $x \mapsto x/\|x\|$, which is $1$-Lipschitz on $C'$.
* If $A$ is a positive-definite matrix and $C = \{x \in \mathbb R^n \mid x^\top A x \le 1\}$, then $u$ is the map $x \mapsto Ax/\|Ax\|$, which is $\mbox{cond}(A)$-Lipschitz on $C'$, where $\mbox{cond}(A) := \|A\|\_{op}\|A^{-1}\|\_{op}$ is the condition number of $A$. The previous example is a particular case with $\mbox{cond}(A) = 1$.



Sufficient conditions for the boundary unit-normal vector field of a closed convex set to be Lipschitz continuous
(1) A necessary condition: For every $x\in\partial C$, it holds that the [polar cone](https://en.wikipedia.org/wiki/Dual_cone_and_polar_cone) of $C-x$ is one-dimensional.


(The polar cone is always at least one-dimensional by part 4 of Prop 4 of [these lecture notes](https://people.orie.cornell.edu/dsd95/teaching/orie6300/lec05.pdf).)


Suppose there exists $x\in\partial C$ such that the polar cone of $C-x$ has dimension at least $2$, and select distinct unit vectors $v$ and $w$ from this polar cone. Then $u(x+\epsilon v)=v$ and $u(x+\epsilon w)=w$ for every $\epsilon>0$. (This is part 2 of the same Prop 4.) Sending $\epsilon\to0$ gives arbitrarily close inputs whose outputs are a fixed distance apart, thereby breaking Lipschitz.


(2) A sufficient condition: $C$ is a compact sublevel set $\{x\in\mathbb{R}^n:f(x)\leq y\_0\}$ of some twice continuously differentiable convex function $f\colon\mathbb{R}^n\to\mathbb{R}$ and $y\_0>\min(f)$.


Indeed, since $y\_0>\min(f)$, it holds that $\nabla f(x)\neq0$ for every $x\in\partial C$, and by compactness, there exists $\delta>0$ such that $\|\nabla f(x)\|>\delta$ for all $x\in\partial C$. Since $x\mapsto \nabla f(x)$ is Lipschitz on $\partial C$ (again, by compactness), it follows that the mapping $n\colon\partial C\to S^{n-1}$ defined by $n(x):=\nabla f(x)/\|\nabla f(x)\|$ is also Lipschitz. Finally, the nearest point mapping $c\colon(\mathbb{R}^n\setminus C)\to\partial C$ is $1$-Lipschitz, and so the composition $u=n\circ c$ is Lipschitz.


