Suppose that $\Omega$ is bounded domain in $\mathbb{R}^n$ with $C^k$ boundary. Why is it that for points sufficiently close to the boundary the distance function $d = d( \cdot, \partial \Omega)$ is $C^k$? I can see it is enough to take a small ball around the boundary, and take a $C^k$ function $\psi$ defining the boundary and show that this give me a $C^k$ distance function, then use compactness. But, I don't see how to do this.
Thoughts?
Thanks
Let's suppose the boundary of $C$ is a $C^k$ $(n-1)$-manifold, let's call it $M$. Focus on a small neighbourhood of a point $P$ in $M$. If this is small enough then the part of $M$ in this neighbourhood has a vector field of unit normals pointing into the inside of $\Omega$. Call the part of the neighbourhood $M'$ and this vector field $v$. Then there's a map $\Phi:M'\times\Bbb R\to\Bbb R^n$ with $\Phi(m,t)=m+tv(m)$. This is $C^k$ and in a neighbourhood of $(P,0)$ it is a diffeomorphism. For a point $Q$ in $\Omega$ near enough $P$ we can write $Q=\Phi(m,t)$ with $t>0$, the distance of $Q$ to $M$, and with $m$ the closest point in $M$ to $Q$. The map $Q\mapsto t$ will then be $C^k$.
Of course, I am sweeping a lot of finicky details under the carpet here....
