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

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Weak maximum principle for the laplacian

Statement

Let $n \geq 2$ and let $Ω \subseteq R^{n}$ be bounded, nonempty and open. If $u \in C^{2} (Ω) \cap C (\overline{Ω})$ and $Δ u \geq 0$ , then $\max_{\overline{Ω}} u = \max_{\partial Ω} u .$ No connectedness or boundary smoothness is required.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

Adding $ε ∣ x ∣^{2}$ to a subharmonic function, with $ε > 0$ , excludes interior local maxima. (Strict subharmonic perturbation).

[F2]

A Euclidean subset is compact if and only if it is closed and bounded. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F3]

A continuous real function on a nonempty compact metric space attains its maximum and minimum. (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct

1.1F2F3given

Choose $R > 0$ with $\overline{Ω} \subseteq {\overline{B}}_{R} (0)$ . The closure is nonempty compact. For each $ε > 0$ , continuity makes $u_{ε} = u + ε ∣ x ∣^{2}$ attain its maximum on that closure.

2.1F1F2F3step 1.1

The maximizer cannot lie in $Ω$ , so it lies in $\partial Ω$ . In particular the boundary is nonempty; it is closed and bounded, hence compact, and $m = \max_{\partial Ω} u$ exists.

3.1step 2.1algebra∎

For every $x \in \overline{Ω}$ , $u (x) \leq u_{ε} (x) \leq m + ε R^{2}$ . Letting $ε ↓ 0$ yields $u (x) \leq m$ . Since a boundary maximizer belongs to the closure, equality of the maxima follows.

Depends on

Dependency tree · 2 levels

Sources

Hunter, Notes on Partial Differential Equations