About the regularity of the boundary of a set

Question

Consider $\Omega_1, \Omega_2$ two convex, bounded domains in $R^n$ with $\Omega_1 \supset \overline{\Omega_2}$ and suppose that $\operatorname{dist}(x, \partial \Omega_2) =\min_{y \in \partial \Omega_2} |x - y| = \lambda>0$ for all $x \in \partial \Omega_1$ ($\lambda$ independent of $x$).

Someone know if in this case the boundary $\partial \Omega_1$will be $C^1$ ? If you dont know a proof, a reference it is ok.

I am asking this because if is true it will help me understand some things that I am studying.

I searched a lot for a reference , but I didn't find anything. Appears that it is true...

Thanks for your attention

Answer 1

So I think the answer is yes. We can suppose without loss of generality that $\Omega_1$ and $\Omega_2$ are closed.

We first claim that you have $\Omega_1=\{x\colon d(x,\Omega_2)\le \lambda\}$. To see this, note if $x\in\Omega_1$, we must have $d(x,\Omega_2)\le \lambda$ - if $d(x,\Omega_2)>\lambda$, then you can use convexity to find a boundary point with this property. Conversely, if $d(x,\Omega_2)\le\lambda$ and $x\not\in\Omega_1$, let $y$ be the nearest point of $\Omega_2$ to $x$ and let $z$ be the first point on the line segment $[x,y]$ belonging to $\Omega_1$. Then $z\in\partial\Omega_1$ and $d(z,\Omega_2)<\lambda$ which is a contradiction.

Now for $x\in \mathbb R^n$, let $\Phi(x)$ be the closest point of $\Omega_2$. A standard Hilbert space argument shows that this is unique, as if there were two closest points, then their midpoint would be 1) in $\Omega_2$ and 2) closer. You can extend this argument to show that $\Phi$ is a continuous map.

Given this, you get the required differentiability, as the normal at the boundary points of $\Omega_1$, given by $(x-\Phi(x))/\lambda$, varies continuously with $x$.