# Regularity of the distance function

Let $\Omega \subseteq \mathbb{R}^N$ be open and bounded, and set:

$$d(x):=\text{dist} (x,\partial \Omega) =\inf_{y\in \partial \Omega} |x-y|\;.$$ I would appreciate if somebody could verify my proof. I tried to show that it is Lipschitz continuous:

Let $\forall x,y \in \Omega$, and WLOG assume that $d(x)\geq d(y)$. Let $\forall \varepsilon >0$. By definition of infimum, $\exists z \in \partial \Omega$ such that $d(y)+\varepsilon > |y-z|$ and $|x-z| \geq d(x)$. Putting everything together, we obtain that $0 \leq d(x)-d(y) \leq |x-z| - |y-z| +\varepsilon \leq |x-y| + \varepsilon$. Since $\varepsilon$ was arbitrary, done.

• ‘Let $\forall x,y\in\Omega$’ and ‘Let $\forall\varepsilon>0$’ are meaningless. You mean ‘Let $x,y\in\Omega$ be arbitrary’ and ‘Let $\varepsilon>0$ be arbitrary’. – Brian M. Scott Feb 20 '12 at 21:27

Yeah, you proof is correct. Whatever you are proved more, you did not use the assumption that $\partial \Omega$ is closed and bounded since $\Omega$ is too.
So the result you proved is $d_{A}:\mathbb{R}^n\rightarrow \mathbb{R}$ given by
$$d_{A}(x)=\inf_{A}|x-y|$$
PS: of course $A$ is non-empty!