PDE: Properties of boundaries

Question

I lecture, we had a remark that polygonal domains in $\mathbb R^2$ have Lipschitz boundary, see: https://www.mat.tuhh.de/veranstaltungen/isem18/pdf/Lecture07.pdf

My question is simple and maybe elementary: Why this remark holds true?

Answer 1

This is "obvious" if we draw a picture (and you definitely should do that), but let me try to answer in text.

Let $P$ be the polygon. We need to show that for every $p\in \partial P$, there exists a $r>0$ and a rotation $R$ such that $P\cap B_r(p)$ is the same as $p+R(V)$, where $V=\{(x,y): y>f(x)\} \cap B_r(0)$ for some Lipschitz function $f$.

If $p$ is not a corner point, then $B_r(p)\cap P$ is a half-disk for $r$ small enough. To see how this formally fits the definition, we can choose $f=0$ and a rotation $R$ that maps $e_1$ into the tangent at $p$ and $e_2$ into the inner normal.

If $p$ is a corner point of the polygon, then for small $r>0$ the set $B_r(p)\cap P$ is a piece of the disk cut out by two half-lines emanating from $p$. We need to find a Lipschitz function such that we can write this as a super level set again. Choose a "normal" direction $\nu$ such that $-\nu$ points out of the polygon and $\nu$ points into the polygon. Choose $R$ that maps $e_1$ into $-i\nu$ and $e_2$ into $\nu$. We can again write $P\cap B_r(p)=p+R(V)$, where now $V=B_r(0)\cap \{(x,y): y>f(x)\}$ with $f(x)=ax$ for $x<0$ and $f(x)=bx$ for $x\ge 0$. You can choose the numbers $a$ and $b$ such that $p+R(x,ax)$ an $p+R(x,bx)$ parametrise the sides of your polygon that emanate at $p$.