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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27
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Positive linear combinations and finite maxima preserve subharmonicity

Statement

Let Ω⊆C be a complex domain.

  1. If u1,…,um are subharmonic on Ω and α1,…,αm≥0, then α1u1+⋯+αmum is subharmonic on Ω, where terms with αj=0 are omitted (so an all-zero combination is the zero function).
  2. If u1,…,um are subharmonic on Ω, then u(z)=max⁡{u1(z),…,um(z)} is subharmonic on Ω.

Facts & Assumptions

Given: Subharmonic functions u1,…,um on a complex domain Ω.

[L1]

Subharmonicity is equivalent to harmonic comparison on compactly contained discs (Subharmonicity is equivalent to harmonic comparison on compactly contained discs).

[L2]

The defining submean inequality for subharmonicity is linear in the function being averaged (Subharmonic functions on plane domains).

[L3]

A subharmonic function is finite almost everywhere (Plane subharmonic functions are locally integrable).

Proof

technique · direct
1.1givenL2L3algebra

Let I={j:αj>0}. If I is empty, the combination is the harmonic zero function. Otherwise it means the well-defined extended-real sum ∑j∈Iαjuj; no product 0⋅(−∞) occurs. Finite sums with positive coefficients preserve upper semicontinuity, and [L3] shows that all summands are finite simultaneously almost everywhere, so their sum is not identically −∞. On any closed disc, multiply the submean inequality for uj by αj>0 and sum over I to obtain the submean inequality for the combination.

1.2L1given

For the finite maximum, upper semicontinuity is preserved by finite maxima. Let D(a,r)‾⊆Ω and let h be continuous on the closure, harmonic on the disc, and satisfy h≥max⁡(u1,…,um) on the boundary. Then h≥uj on the boundary for every j, so [L1] gives h≥uj throughout the disc for every j. Therefore h≥max⁡(u1,…,um) on the disc. Another use of [L1] shows that the maximum is subharmonic.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the two closure properties.

Depends on

Used by

Dependency tree · two levels

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Sources