Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Maximum principle for plurisubharmonic functions

Statement

Let u be plurisubharmonic on a domain Ω⊆Cm. If u attains its finite global maximum at a point of Ω, then u is constant on Ω.

Facts & Assumptions

Given: A plurisubharmonic function u on a domain Ω⊆Cm and a point a∈Ω with u(a)=sup⁡Ωu<+∞.

[L1]

Plurisubharmonicity is tested by subharmonicity on every affine complex line (Plurisubharmonic functions).

[L2]

A subharmonic function of one complex variable that attains a finite interior maximum is constant on its connected component (A plane subharmonic function with an interior maximum is constant on its component).

Proof

technique · direct
1.1L1L2given

Choose a small Euclidean ball B(a,r)⊆Ω. Fix z∈B(a,r), and restrict u to the affine complex line through a and z. By [L1], that restriction is subharmonic on the connected line-domain component containing the segment from a to z, and the global bound makes its value at a a finite maximum. Hence [L2] makes it constant on that component, so u(z)=u(a). Thus u is constant on B(a,r).

2.1step 1.1given∎

Put S:={z∈Ω:u(z)=u(a)}. It is nonempty. Every point of S is another point where the finite global maximum is attained, so the argument of step 1.1 makes S open. Since u≤u(a) everywhere and upper semicontinuity makes {u≥u(a)} closed, S={u≥u(a)} is closed. Connectedness of Ω gives S=Ω, so u is constant.

Depends on

Used by

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Dependency tree · two levels

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Sources