CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-09-23 (gpt-6-sol)

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Unbounded punctured harmonic singularity is not removable

Statement refuted

A harmonic function on a punctured ball need not have a harmonic extension at the puncture without growth control. On $0 < ∣ x ∣ < 1$ , use $u (x) = \log ∣ x ∣$ for $n = 2$ and $u (x) = ∣ x ∣^{2 - n}$ for $n \geq 3$ . Neither has even a continuous extension at zero.

Facts & Assumptions

Given: The objects and hypotheses in the statement refuted.

Counterexample

technique · direct

1.1givenalgebra

For $r = ∣ x ∣ > 0$ , differentiating $f (r)$ gives $Δ f (r) = f'' (r) + (n - 1) f' (r) / r$ . For $n = 2$ , $f' = 1 / r$ and $f'' = - 1 / r^{2}$ cancel. For $n \geq 3$ , $f' = (2 - n) r^{1 - n}$ and $f'' = (2 - n) (1 - n) r^{- n}$ also cancel. Thus both profiles are harmonic away from zero.

2.1step 1.1algebra∎

The logarithm tends to $- \infty$ as $r ↓ 0$ , and the negative power tends to $+ \infty$ . Continuity at zero would require a finite limit, so no harmonic extension exists. These profiles also violate boundedness near the point. Their signed ratios to the critical growth profiles are $- 1$ for $n = 2$ and $1$ for $n \geq 3$ , so in both cases the absolute ratio is one rather than tending to zero.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Gantumur, Harmonic functions (standard reference, not scraped)