Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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  • 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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A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically

Statement

Let m≥1, let U⊆Cm be a nonempty connected open set, and let f:U→C be holomorphic (Holomorphic functions on an open subset of Cm). If there is a nonempty open set W⊆U such that f(z)=0 for every z∈W, then f≡0 on U.

This is the several-variable identity theorem at the strength the page supports: the hypothesis is a nonempty open set of zeros. An accumulation point of the zero set is neither assumed nor sufficient in several variables; the companion page records that stronger one-variable statement as false here.

Facts & Assumptions

Given: A nonempty connected open set U⊆Cm, a holomorphic function f:U→C, and a nonempty open set W⊆U on which f=0.

[L1]

Holomorphic functions of several variables are smooth; every point a∈U has a polydisc Δr(a)⊆U on which f(z)=∑αcα(z−a)α,cα=∂zαf(a)α!; and all mixed complex derivatives are holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic).

[L3]

The mixed complex derivative notation ∂zα and the zero-order identity ∂z0f=f are those in the power-series and smoothness statement [L1]; Ck maps and multi-index derivative notation in Euclidean space supplies the underlying multi-index arithmetic.

[L4]

Polydiscs in Cm are the coordinatewise discs of Balls, polydiscs and the distinguished boundary in Cm.

Proof

technique · direct
1.1givenL1L3

For every multi-index α, the derivative ∂zαf is holomorphic and therefore continuous on U by [L1]; since f=0 on the open set W, every derivative of f is also 0 on W, so the set A:={z∈U:∂zαf(z)=0 for every α} contains W and is therefore nonempty.

2.1step 1.1

The set A is closed in U, because it is the intersection over all multi-indices α of the closed zero sets of the continuous functions ∂zαf.

2.2step 1.1L1L4

The set A is open in U: if a∈A, choose a smaller polydisc Δ⊆U centred at a; then every coefficient ∂αf(a)/α! in the power-series expansion of f on Δ is 0, so [L1] gives f=0 on Δ, and hence every derivative vanishes on Δ as well, which means Δ⊆A.

3.1step 2.1step 2.2L2L3∎

The set A is a nonempty subset of U that is both open in U and closed in U, so connectedness and [L2] force A=U; in particular f=∂z0f vanishes at every point of U, hence f≡0 on U.

Remarks

  • Why the hypothesis is open-set vanishing and not an accumulation point. In one complex variable, accumulation of zeros implies equality by local factorisation and isolated zeros. In several variables the zero set of a nonzero holomorphic function can contain whole complex hypersurfaces, so the open-set hypothesis is the honest form at this stage.

  • What the proof really uses. The proof needs only two page-level tools: holomorphic smoothness and the local power-series expansion. Once every derivative at one point vanishes, the power series on a smaller polydisc is identically zero, and connectedness propagates that local vanishing to the whole set.

Depends on

Used by

Dependency tree · two levels

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Sources