Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adapted
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every holomorphic function on a star-shaped domain has a primitive

Statement

Let U⊆C be open and star-shaped with respect to a∈U. Every holomorphic function f:U→C has the primitive

F(z)=∫ℓazf(ζ) dζ.

Facts & Assumptions

Given: A set U open for dC and star-shaped with respect to a∈U, meaning that Φ[U] is open and star-shaped at Φ(a) in Star-shaped open subsets of Euclidean space. This is equivalent to containment of every segment from a to a point of U, by The complex coordinate isometry transports topology and affine segments. We have and a holomorphic f:U→C.

[L1]
[L2]

Goursat's theorem makes the integral of a holomorphic function around every filled triangle in its domain zero (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L3]

Vanishing triangle integrals construct the displayed primitive for a continuous function on a star-shaped domain (Vanishing integrals around triangles construct a primitive for a continuous function on a star-shaped domain).

Proof

technique · direct
1.1L1L2

By [L1], f is continuous, and by [L2] its integral around every filled triangle contained in U is zero.

2.1step 1.1L3∎

The hypotheses of [L3] now hold, so the displayed function is holomorphic on U and has derivative f.

Depends on

Dependency tree · 2 levels

Sources