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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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A locally bounded holomorphic function extends across a coordinate hyperplane

Statement

Let m2, let ΩCm be a domain, and let

H:=Ω{zm=0}.

If f:ΩHC is holomorphic and locally bounded near H, then there exists a unique holomorphic F:ΩC such that F=f on ΩH.

Facts & Assumptions

Given: A domain ΩCm, a holomorphic function f:ΩHC, and local boundedness near the coordinate hyperplane H.

[L1]

A bounded punctured slice extends holomorphically, and the missing value depends holomorphically on the remaining parameters (A locally bounded punctured slice has a holomorphic parameter extension).

[L2]

A holomorphic function on a connected open set is determined by its values on a nonempty open subset (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[L3]

Holomorphic extension means agreement on some nonempty open overlap (Holomorphic extension and domains of holomorphy in several variables).

Proof

technique · direct
1.1

Let p=(p,0)H. By local boundedness, choose a product neighborhood U×{w<R}Ω of p on which f is bounded whenever 0<w<R, after translating coordinates so that pm=0. Applying [L1] on this product neighborhood gives a holomorphic function Fp:U×{w<R}C extending f across the slice U×{0}.

givenL1
2.1

The functions Fp and f agree on the nonempty open overlap U×{0<w<R}, so each Fp is a local holomorphic extension in the sense of [L3].

step 1.1L3
3.1

If two such neighborhoods overlap, their local extensions agree on the nonempty open subset of the overlap where w0, because both equal f there. The overlap is connected after shrinking if necessary, so [L2] makes the two local extensions equal on the whole overlap.

step 2.1L2
4.1

The local extensions therefore glue to a single holomorphic function on Ω that agrees with f off H. Uniqueness follows from [L2], since two global extensions agree on the nonempty open set ΩH.

step 3.1L2

Depends on

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