TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Harnack convergence principle

Statement

Let $n \geq 2$ , let $Ω \subseteq R^{n}$ be a domain, and let $u_{1} \leq u_{2} \leq \dots$ be real harmonic functions on $Ω$ . Either $u_{j} (x) \to + \infty$ for every $x \in Ω$ , or the sequence converges uniformly on each compact subset of $Ω$ to a harmonic function.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

On any nonempty compact subset of a domain, nonnegative harmonic functions have supremum bounded by a fixed constant times their infimum. (Harnack inequality on compact subsets).

[F2]

A locally uniform limit of harmonic functions on an open set is harmonic. (Locally uniform limits of harmonic functions are harmonic).

Proof

technique · direct

1.1

If no point has a bounded-above scalar sequence $u_{j} (x)$ , each scalar sequence is increasing and unbounded above, hence tends to $+ \infty$ . Otherwise fix $a \in Ω$ at which it is bounded above; there $u_{j} (a)$ converges to a finite real number.

given

2.1

For any nonempty compact $K \subset Ω$ , apply compact Harnack on $K \cup {a}$ to the harmonic nonnegative difference $u_{j} - u_{i}$ , $j \geq i$ . It yields $\sup_{K} ∣ u_{j} - u_{i} ∣ \leq C (u_{j} (a) - u_{i} (a))$ . The right side tends to zero as $i, j \to \infty$ , uniformly in those indices.

F1step 1.1

3.1

The real completeness property gives a pointwise limit, and the uniform Cauchy bound proves convergence uniformly on $K$ . Empty compact sets impose no condition. Since $K$ was arbitrary, convergence is locally uniform and the harmonic-limit theorem makes the limit harmonic.

F2step 2.1∎

Depends on

Used by

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Sources

Gantumur, Harmonic functions (standard reference, not scraped)