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Exterior disc points and exterior cone points are regular

Statement

Let ΩC be a bounded complex domain and let ζΩ.

  1. If there is a closed disc B(c,R)CΩ with ζB(c,R), then ζ is regular.
  2. If, after a rigid motion sending ζ to 0, the domain lies locally in a sector of opening angle <π, then ζ is regular.

Facts & Assumptions

Given: A bounded complex domain Ω and a boundary point ζΩ.

[L1]

A boundary point is regular exactly when it admits a barrier (A boundary point is regular exactly when it admits a barrier).

[L2]

A negative local subharmonic peak with a strictly negative bound on a smaller seam globalizes to a barrier (A local strict subharmonic peak function globalizes).

Proof

technique · direct
1.1

In the exterior-disc case put Φ(z)=ζczc,b(z)=ReΦ(z)1. The center c is outside Ω, so Φ and b are holomorphic and harmonic respectively on Ω. Moreover Φ(z)=R/zc<1 for zΩ, hence b<0, while b(z)0 as zζ. On the compact set ΩV, for any neighbourhood V of ζ, the continuous function ReΦ1 has a strictly negative maximum: equality could hold only when Φ=1, namely at z=ζ. Thus b is a global barrier and [L1] makes ζ regular.

L1givenconstruct
1.2

In the exterior-cone case, after translation and rotation take ζ=0 and suppose that near 0 the domain lies in S={reit:0<r<r0, t<θ},0<θ<π/2. Choose θ with θ<θ<π/2, put λ=π/(2θ), and use the branch of zλ on the larger sector t<θ. Then q(z)=Re(zλ) is harmonic and negative on ΩS, and tends to 0 at the origin. On a sufficiently small circle z=ρ, the angular margin gives q(z)ρλcos(λθ)<0(zΩD(0,ρ)). Thus [L2] globalizes q to a barrier, and [L1] gives regularity.

L1L2givenconstruct
2.1

The two barrier constructions prove the two regularity criteria.

step 1.1step 1.2

Depends on

Used by

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Sources