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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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The logarithm of the modulus of a holomorphic function is plurisubharmonic

Statement

Let ΩCm be a domain and let f be holomorphic on Ω, not identically zero on any connected component. Define

u(z)=logf(z),

with the convention u(z)= at the zeros of f. Then u is plurisubharmonic on Ω.

Facts & Assumptions

Given: A holomorphic function f on a domain ΩCm, not identically zero on any connected component.

[L1]

Plurisubharmonicity is tested on affine complex lines (Plurisubharmonic functions).

[L2]

For a one-variable holomorphic function, the logarithm of the modulus is subharmonic with the value at its zeros (The logarithm of the modulus of a holomorphic function is subharmonic).

Proof

technique · direct
1.1

Fix an affine complex line in Ω. The restriction of f to that line is a one-variable holomorphic function, and by the componentwise hypothesis it is not identically zero on the connected component under consideration. Therefore [L2] makes the restriction of u=logf subharmonic or identically there.

L2given
2.1

The function u is upper semicontinuous because it is a logarithm of a continuous modulus away from the zero set and has value on the zero set. Step 1.1 is exactly the line test from [L1], so u is plurisubharmonic on Ω.

L1step 1.1

Depends on

Used by

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Dependency tree · two levels

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Sources