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Removable singularity for bounded harmonic functions under Countable Choice

Statement

Assume the Axiom of Countable Choice. Let n≥2, let Ω⊆Rn be open and p∈Ω. If u∈C2(Ω∖{p}) is harmonic and bounded in some punctured neighborhood of p, it has a unique harmonic extension to Ω. More generally the same conclusion holds under u(x)=o(∣x−p∣2−n) for n≥3, or u(x)=o(log⁡(1/∣x−p∣)) for n=2, as x→p.

Facts & Assumptions

Given: Countable Choice and the objects and hypotheses in the statement (The Axiom of Countable Choice (ACω)).

[F1]

Smooth real sphere data have a smooth harmonic replacement continuous on the closed ball with the prescribed boundary values. (Smooth sphere data have a harmonic replacement under Countable Choice).

[F2]

Classical comparison applies to bounded nonempty open sets and C2 functions continuous on the closure when Laplacian and boundary inequalities hold. (Comparison principle for classical subharmonic functions).

[F3]

Harmonic functions have the ball mean-value property. (Ball mean-value property for harmonic functions under Countable Choice).

[F4]

Continuous ball-mean-value functions are smooth harmonic. (Continuous ball-mean-value functions are harmonic).

Proof

technique · direct
1.1F1F3F4given

Translate p=0 and choose 0<R<1 so that B‾R⊂Ω. Under the given Countable Choice premise, ball mean values and the smoothness theorem make u smooth on the punctured domain; its trace on ∂BR is smooth. Let h be its harmonic replacement on BR, and w=u−h. Then w is harmonic off zero and vanishes on the outer sphere. The replacement is bounded on the closed ball.

2.1step 1.1algebra

For 0<δ<R, put Aδ=max⁡∣x∣=δ∣w(x)∣, a finite nonnegative number. On δ<r=∣x∣<R take ϕδ(x)=Aδ(δ/r)n−2 if n≥3, and ϕδ(x)=Aδlog⁡(R/r)/log⁡(R/δ) if n=2. For radial f, differentiation of ∂if(r)=f′(r)xi/r gives Δf=f′′+(n−1)f′/r. Substitution proves Δϕδ=0 in both dimensions. Each barrier equals Aδ on the inner sphere and is nonnegative on the outer sphere.

3.1F2step 2.1algebra

Comparison on the bounded annulus applied separately to w and −w against ϕδ yields ∣w(x)∣≤ϕδ(x). In the bounded case Aδ remains uniformly bounded for small δ. At fixed x≠0, the factor δn−2 tends to zero for n≥3, and the denominator log⁡(R/δ) tends to infinity for n=2. Thus w(x)=0.

4.1step 1.1step 3.1algebra∎

Under the stated little-o hypotheses, Aδ=o(δ2−n) or Aδ=o(log⁡(1/δ)), because h is bounded. The same barrier bound again tends to zero at each fixed nonzero x. Hence u=h on the punctured ball in either case. Define u(0)=h(0) and retain the original u outside the ball; equality on their overlap proves harmonicity locally everywhere. Any continuous extension must have this value at zero, proving uniqueness.

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