I have an open bounded connected domain $\Omega \subseteq \mathbb{R}^n$ and I would like to say that for every two $x,y \in \Omega$ there is a path $\gamma$ from $x$ to $y$ of length at most $C|x-y|$ where $C$ does not depend on $x,y$. (This is my definition of quasiconvexity)
Would any of the following properties imply this?
- $\Omega$ has the exterior sphere property
- $\Omega$ has $C^1$ boundary
If not, are there any simple conditions on $\Omega$ which can guarantee quasiconvexity?
Exterior sphere property is not enough. For example, this domain, bounded by internally tangent circles, has the exterior sphere property but is not quasiconvex.
Having $C^1$ boundary is enough. Indeed, suppose quasiconvexity fails: i.e., the intrinsic distance $d(x,y)$ is not bounded by $C|x-y|$. Then there are sequences $x_n$ and $y_n$ in $\Omega$ such that $d(x_n,y_n)/|x_n-y_n|\to\infty$. Since $\Omega$ is bounded, this is possible only when $|x_n-y_n|\to 0$. By picking a subsequence, we can assume $x_n,y_n\to p\in \overline{\Omega}$. If $p$ is an interior point of $\Omega$, then $d(x_n,y_n) =|x_n-y_n|$ for all large $n$, contrary to the above. So, $p\in \partial\Omega$. In an appropriate coordinate system, $p=0$ and the piece of $ \Omega$ near $p$ is of the form $\{v:v_n>f(v_1,\dots,v_{n-1})\}$ where $f$ is smooth, with gradient $0$ at the origin. But this piece is bi-Lipschitz equivalent to a piece of half-space via $v\mapsto v-f(v_1,\dots,v_{n-1})\,e_n $. Therefore, $d(x_n,y_n)/|x_n-y_n|$ stays bounded, a contradiction.
You may notice that it suffices for $\Omega$ to have Lipschitz boundary.