CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Strong maximum principle needs connectedness

Statement refuted

For $n \geq 2$ , the connectedness hypothesis cannot be omitted from the strong harmonic maximum principle. Let $Ω = B_{1} (0) \cup B_{1} (3 e_{1})$ , and set $u = 1$ on the first ball and $u = 0$ on the second. It attains an interior global maximum without being globally constant.

Facts & Assumptions

Given: The objects and hypotheses in the statement refuted.

[F1]

A harmonic function on a connected nonempty open set attaining an interior global extremum is constant. (Strong maximum principle for harmonic functions).

Counterexample

technique · direct

1.1

The two open balls are disjoint, since their centers are distance three apart. Every point has a neighborhood on which $u$ is constant; consequently $u$ is smooth with $Δ u = 0$ everywhere in $Ω$ .

givenalgebra

2.1

Every point of the first ball attains the global maximum $1$ , whereas every point of the second has value $0$ . Thus $u$ is not constant on $Ω$ , which is disconnected. This is precisely the hypothesis absent from the strong theorem.

F1step 1.1∎

Depends on

Strong maximum principle for harmonic functions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Hunter, Notes on Partial Differential Equations (standard reference, not scraped)