Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Maximum principle with limsup control at infinity

Statement

Let n≥2, let Ω⊆Rn be unbounded and open, and let u∈C2(Ω)∩C(Ω‾) be subharmonic. Suppose M∈R, u≤M on ∂Ω, and lim sup⁡x∈Ω∣x∣→∞u(x)≤M. Here the last condition means that for every ε>0 there is R0 with u(x)≤M+ε whenever x∈Ω and ∣x∣>R0. Then u≤M on Ω‾, even if ∂Ω is empty.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

A subharmonic C2 function continuous on the closure of a bounded nonempty open set is bounded above by its boundary maximum. (Weak maximum principle for the laplacian).

Proof

technique · direct
1.1given

Fix x∈Ω and ε>0, choose R0 from the infinity hypothesis, and choose R>max⁡(R0,∣x∣). The set D=Ω∩BR is bounded nonempty open; its boundary lies in (∂Ω∩B‾R)∪(Ω‾∩∂BR).

2.1F1step 1.1algebra∎

On the first part u≤M. On the second part the infinity bound, extended from Ω by continuity when necessary, gives u≤M+ε. The weak maximum principle on D yields u(x)≤M+ε. Let ε↓0; the original boundary inequality then also includes all closure points.

Depends on

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources