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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Derivative estimate proof of one sided harmonic liouville

Statement

Assume the Axiom of Countable Choice ACω. Let n≥2, R>0, and u be real harmonic on BR(a). If S=sup⁡BR(a)∣u∣<∞, then ∣∇u(a)∣≤nS/R. If u≥0, the stronger estimate ∣∇u(a)∣≤nu(a)/R holds without assuming a finite global supremum. Consequently, a nonnegative entire harmonic function has gradient zero everywhere.

Facts & Assumptions

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

[F1]

Smooth sphere data have a unique smooth harmonic replacement given by the displayed sphere kernel, continuous with those data on the boundary. (Smooth sphere data have a harmonic replacement under Countable Choice).

[F2]

Every partial derivative of a smooth harmonic function is smooth harmonic. (Derivatives of harmonic functions are harmonic).

[F3]

A classical harmonic function has the ball mean-value property. (Ball mean-value property for harmonic functions under Countable Choice).

[F4]

A continuous function with the ball mean-value property is smooth and harmonic. (Continuous ball-mean-value functions are harmonic).

Proof

technique · direct
1.1F2F3F4given

The ball mean property and the continuous mean-value theorem make u smooth. Its partial derivatives are therefore harmonic by the smooth derivative lemma, whose smoothness hypothesis is now satisfied.

2.1F1step 1.1algebra

On every sphere of radius 0<s<R, the trace of u is smooth. Harmonic replacement and uniqueness represent u inside the smaller ball by this trace. Differentiate the kernel at its center and write y=a+sθ. It gives ∇u(a)=ns1∣∂Bs∣∫∂Bs(a)u(y)θ dS(y), while the undifferentiated center formula gives u(a)=∣∂Bs∣−1∫∂Bs(a)u. Derivative passage is justified by the same compact-sphere bounds as in replacement.

3.1step 2.1algebra

Taking vector norms and using ∣θ∣=1 bounds the gradient by (n/s) times the average of ∣u∣. This is at most nS/s in the bounded signed case, and exactly nu(a)/s as an upper bound in the nonnegative case. Let s↑R to obtain both estimates.

4.1step 3.1algebra∎

For a nonnegative entire harmonic function, the positive estimate holds at each fixed a for every R>0. Let R→∞ to obtain ∇u(a)=0. This is also a second route to one-sided Liouville after shifting and, if necessary, negating the function.

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