Regular boundary point

A boundary point is regular for the Dirichlet problem when the Perron solution approaches the prescribed continuous boundary value there. The existence of a barrier implies regularity.

An open set satisfies an exterior cone condition at $z$ if a solid cone with vertex $z$ lies outside the set in some neighbourhood of $z$. The stronger global version requires the closures of the cone and the set to meet only at $z$. In two dimensions, a power barrier for an exterior cone makes such a point a regular boundary point for the Dirichlet problem. Local cone conditions suffice for local boundary regularity; a global cone is an additional geometric assumption.

For a planar set outside a global solid cone of half-angle $\alpha$, choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$, $\beta =\pi /[2(\pi -{\alpha }_0)]$, and $0<\nu <\beta$. If $\psi$ is the angle from the opposite axis, the Laplacian in polar coordinates givesThe cosine has a positive uniform lower bound outside the original cone. On a bounded set, $\Delta g$ therefore has a positive uniform lower bound as well. The positive function $-g$ is a barrier for the Dirichlet problem. An unbounded set does not receive the same global lower bound from this construction.

A domain satisfies the exterior sphere condition at a boundary point when a ball contained in the complement is tangent to the boundary only at that point. Fundamental solutions of the Laplace equation centred at the ball's centre provide a harmonic barrier, so such a point is regular.

If $B(y,R)$ is an exterior ball tangent to a planar domain at $x_0$, thenis a harmonic function in the domain, vanishes at $x_0$, and is positive at every other point of its closure. It is therefore a barrier for the Dirichlet problem and proves that $x_0$ is a regular boundary point.