Let $C$ be a closed convex set of $\mathbb{R}^n$ $(n\geq 1)$, and $u\in\mathbb{R}^n\setminus\{0\}$ such that $u$ does not belong to the asymptotic cone of $C$ and is nowhere tangent to $\partial C$. For $x\in C$, let $p(x)$ be the projection of $x$ on $\partial C$ toward the direction $u$:
$$p(x)=x+\inf\{\lambda\geq 0;\,x+\lambda u\notin C\}\,u$$
Hence $p$ is defined from $C$ to $\partial C$.
Question: Is $p$ Lipschitz?
The answers given lead to asking the following close question: Lipschitz aspect of a projection on the boundary of a convex
Best Answer
No. E.g., let $n=2$ and $$C=\{(s,t)\in\mathbb R^2\colon t\ge s^2\}.$$ Then the asymptotic cone of $C$ is $K:=\{(0,t)\colon t\ge0\}$. Let $u=(1,0)\notin K$. Then $$p((0,t))=(\sqrt t,t)$$ for real $t\ge0$. So, $p$ is not Lipschitz.
Somehow I missed the condition that the direction $u$ be nowhere tangent to $\partial C$ -- my apologies for that. Yet, for this $C$ still there is a truly "bad" admissible direction $u$, such that the projection along $u$ is not Lipschitz.
Indeed, let $u=(0,-1)$. Then $u\not\in K$ and $u$ is nowhere tangent to $\partial C$. Take any $s$ and $h$ such that $0<h<s<\infty$. Let $x=(s,s^2)$ and $y=(s-h,s^2)$. Let $d$ denote the Euclidean distance. Then $p(x)=(s,s^2)$, $p(y)=(s-h,(s-h)^2)$, $d(x,y)=h$, and $d(p(x),p(y))\ge s^2-(s-h)^2>sh$. So, letting $s\to\infty$, we have $\dfrac{d(p(x),p(y))}{d(x,y)}\to\infty$. So, $p$ is not Lipschitz.