Let $A\subset \mathbb R^n$ be a compact convex set. For $x\in \mathbb R^n$ denote $p(x)$ a unique nearest point to $x$ in $K$, i.e. $|x-p(x)|= \mathrm{dist}(x,K)$. As shown in this post function $p$ is continuous. I'd like to have some properties of this function as $x\to \infty$. Namely, denote $\bar l_x =x/|x|$, so $x=r\bar l_x$, $r>0$. Is it true that
- For all $\bar l_x\in S^{n-1}$ there exists a point $p(\infty\bar l_x):=\lim_{r\to+\infty}p(r\bar l_x)$?
- Does there holds the relation $$ p(x)=p(\infty\bar l_x)+O\left(\frac{1}{|x|}\right),\quad x\to \infty? $$ Meaning, for any $\varepsilon>0$ there exist $C, R>0$ s.t. $ |p(x)-p(\infty\bar l_x)|\le \frac{C}{|x|} $ if $|x|>R$.
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It seems to be plausible if the boundary $\partial A\in C^1$ and for a ball the asymptotics of $p(x)$ can be written out explicitly: $p(x)=p(\infty\bar l_x)+O\left(\frac{1}{|x|}\right)$. But generally $\partial K$ is just Lipschitz. Secondly, $A$ can have dimention less that $n$, say, to be a segment. The last case can be checked directly though and I cannot find a counterexample.
