Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: 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.
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Weak maximum principle needs boundedness or control at infinity

Statement refuted

Without boundedness or control at infinity, a harmonic function continuous on the closure of an open set and zero on its nonempty boundary need not be nonpositive inside. For n2, take Ω={xRn:xn>0} and u(x)=xn.

Facts & Assumptions

Given: The objects and hypotheses in the statement refuted.

[F1]

The weak maximum principle assumes a bounded nonempty open set and a subharmonic C2 function continuous on the closure. (Weak maximum principle for the laplacian).

Counterexample

technique · direct
1.1

The half-space is open, connected and unbounded, with nonempty boundary {xn=0}. The affine function is smooth on all of Rn, has Laplacian zero, and vanishes on that boundary.

givenalgebra
2.1

For every t>0, u(ten)=t>0, and these values tend to infinity as t. Hence the proposed interior bound fails. The bounded-open-set hypothesis of the weak maximum principle does not hold for this example.

F1step 1.1algebra

Depends on

Used by

Nothing in the library uses this result yet.

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