---
tag: B51J3BC610
kind: statement
role: theorem
name: Exterior cone condition
msc: [31B25, 31B20, 35J25]
uses: [B7948M365K, 02BH3MCBQB, 05D5EGMFYV]
about: [B7948M365K]
sources:
  - url: https://en.wikipedia.org/wiki/Dirichlet_problem
---

## Statement

Let $n \ge 2$, let $\Omega \subseteq \R^n$ be a bounded [[open set|02BH3MCBQB]], and let $\xi \in \partial\Omega$ be a [[boundary|05D5EGMFYV]] point.
Suppose there is a truncated closed circular cone $C$ with vertex $\xi$ and nonempty interior such that $C \cap \Omega = \emptyset$.
Then $\xi$ is a [[regular boundary point|B7948M365K]] of $\Omega$.
Suppose every boundary point of $\Omega$ satisfies this condition.
Then the Dirichlet problem on $\Omega$ has a solution continuous on $\overline{\Omega}$ for every continuous boundary function.

## 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|STMT1K1865]].
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.

## Explanations

### L1 human

If you can place a small solid ice-cream cone outside the set with its tip at a boundary point, then that point is good for the Dirichlet problem.
The solution then takes the prescribed value there.
Points on smooth boundaries and corners of polygons and polyhedra all pass this test.

### L2 human

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.