Let $C$ be a closed convex set of $\mathbb{R}^n$ $(n\geq 1)$, with asymptotic cone $C^{as}$ having for interior $\text{Int}\big(C^{as}\big)$. Let $u\in\mathbb{R}^n\setminus\{0\}$ such that


\begin{align}
u \notin C^{as} \text{ and }(-u) & \in \text{Int}\big(C^{as}\big).
\end{align}


For $x\in C$, let $p(x)$ be the projection of $x$ on $\partial C$ toward the direction $u$:


$$p(x)=x+\inf\{\lambda\geq 0;\,x+\lambda u\notin C\}\,u.$$


**Question:** is $p$ Lipschtiz?


This question is motivated by the answers given for the close question: [Is this projection on the boundary of a convex Lipschitz?](https://mathoverflow.net/questions/426915/is-this-projection-on-the-boundary-of-a-convex-lipschitz/426918?noredirect=1#comment1097752_426918)



Lipschitz aspect of a projection on the boundary of a convex
$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$Yes, $p$ is Lipschitz.


Indeed,
\begin{equation\*}
 p(x)=x+f(x)u
\end{equation\*}
for $x\in C$, where
\begin{equation\*}
 f(x):=\inf E\_x,\quad E\_x:=\{t\ge0\colon x+tu\notin C\}. 
\end{equation\*}


So, it suffices to show that the function $f$ is Lipschitz on $C$.


Let us use the [Rockafellar, p. 62](https://www.google.com/books/edition/Convex_Analysis/1TiOka9bx3sC?hl=en&gbpv=1) notation $0^+C$ for $C^{as}$ and then $(0^+C)^\circ$ for the interior of $0^+C$.


For each $x\in C$, we have
\begin{equation\*}
 f(x)=\infty\iff E\_x=\emptyset\iff (\forall t\ge0\ x+tu\in C)\iff u\in0^+C; 
\end{equation\*}
the latter $\iff$ follows by [Rockafellar, Theorem 8.3, p. 63](https://www.google.com/books/edition/Convex_Analysis/1TiOka9bx3sC?hl=en&gbpv=1). So, in view of the condition $u\notin0^+C$, we see that


\begin{equation}
 \text{$0\le f<\infty$ on $C$. } \tag{0}\label{0}
\end{equation}


Without loss of generality (wlog), the vector $u$ is of unit length.
Since
\begin{equation\*}
 v:=-u\in(0^+C)^\circ,
\end{equation\*}
there is some real $\ep>0$ such that
\begin{equation\*}
 v+\ep B\subseteq 0^+C, \tag{1}\label{1}
\end{equation\*}
where $B$ is the closed unit ball centered at the origin.


It is enough to show that the restriction of $f$ to the intersection of $C$ with any 2-D affine plane parallel to the vector $u$ is $L$-Lipschtiz, with the same Lipschitz constant
\begin{equation\*}
 L:=1+l,\quad l:=1/\ep \tag{2}\label{2}
\end{equation\*}
for all such 2-D affine planes.


Therefore, wlog the dimension $n$ is $2$. Also then, wlog $-u=v=(0,1)$. It then follows that
\begin{equation\*}
 f(a,b):=f((a,b))=b-g(a) \tag{3}\label{3}
\end{equation\*}
for $(a,b)\in C\subseteq\R^2$, where
\begin{equation\*}
 g(a):=\inf\{y\in\R\colon(a,y)\in C\}
\end{equation\*}
for all real $a$, so that $C$ is the epigraph $\epi(g):=\{(a,y)\in\R^2\colon y\ge g(a)\}$ of the (necessarily convex) function $g$. Here, the standard convention $\inf\emptyset=\infty$ is used (even though we will see in a moment that in our case in fact we have $g(a)<\infty$ for all real $a$).


By \eqref{1} and \eqref{2}, $\epi(g)$ contains $\epi(h)$ for a function $h$ given by a formula of the form $h(a)=B+l|a-A|$ for some real $A,B$ and all real $a$. Then $g\le h$; in particular, $g<\infty$ on $\R$. Also, by \eqref{3} and \eqref{0}, $g(a)>-\infty$ for some real $a$. So, for the convex function $g$ we have $g>-\infty$ on $\R$. So, $-\infty<g<\infty$ on $\R$.


Now it follows from $g\le h$ that the limit $g'(\infty)$ of the (say) right derivative $g'(a)$ of the (finite convex) function $g$ as $a\to\infty$ is $\le h'(\infty)=l$. Similarly, the limit $g'(-\infty)$ of the right derivative $g'(a)$ of the convex function $g$ as $a\to-\infty$ is $\ge h'(-\infty)=-l$. Since the right derivative $g$ of the convex function $g$ is nondecreasing, we see that $-l\le g'\le l$. So, $g$ is $l$-Lipschitz.


Thus, by \eqref{3} and \eqref{2}, $f$ is indeed $L$-Lipschtiz. $\quad\Box$


