Exterior cone condition

Remarks

Proof idea: regularity at $\xi$ only gets easier when $\Omega$ shrinks, so it suffices to treat the complement of the cone near $\xi$. For that set one can build a barrier of the form $\abs{x - \xi}^{\alpha} \phi(\theta)$, where $\theta$ is the angle from the axis of the cone. Here $\phi$ solves an eigenvalue problem on the sphere. For $n \ge 3$ one can instead use Wiener's criterion. The cone puts in each dyadic shell a ball of radius comparable to the shell, so each term of Wiener's series is bounded below. Example: every bounded open set with Lipschitz boundary, such as a cube or a convex polytope, satisfies the condition at each boundary point. The condition is sufficient but not necessary: in $\R^3$ a tip whose cross-sections shrink like a power of the distance is still regular, although no cone fits. Without such a condition regularity can fail: the centre of a punctured ball has no exterior cone and is irregular.

Going deeper

For a student

The exterior cone condition is the usual practical way to check that a domain is regular. It generalizes the exterior sphere condition, where a ball outside the set touches the boundary at the point, and it covers all Lipschitz domains. It is a consequence of Wiener's criterion, and its barrier gives quantitative control of how fast the solution approaches its boundary values.