TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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.

Vitali-Porter convergence theorem for holomorphic functions

Statement

Assume the Axiom of Choice. Let $Ω$ be a plane domain and let $(f_{n})$ be a locally bounded sequence in $H (Ω)$ . Suppose there is a set $E \subseteq Ω$ with an accumulation point in $Ω$ such that $f_{n} (z)$ converges for every $z \in E$ . Then $(f_{n})$ converges locally uniformly on $Ω$ to a holomorphic function.

Facts & Assumptions

Given: Choice, a plane domain $Ω$ , a locally bounded holomorphic sequence $(f_{n})$ , and a set $E \subseteq Ω$ on which the pointwise limit exists with an accumulation point in $Ω$ .

[L1]

Locally bounded holomorphic families are normal (Montel's theorem: every locally bounded holomorphic family is normal).

[L2]

Two holomorphic functions that agree on a set with an accumulation point in the domain agree everywhere (Identity theorem for holomorphic functions).

Proof

technique · direct

1.1

By [L1], every subsequence of $(f_{n})$ has a further subsequence converging locally uniformly to a holomorphic limit. Any two such subsequential limits agree on $E$ , because the scalar sequence $f_{n} (z)$ has a fixed pointwise limit there, so [L2] makes them equal on all of $Ω$ .

L1L2given

2.1

If the whole sequence did not converge locally uniformly to that common limit, then some subsequence would stay a definite distance away on a compact set. Applying [L1] again to that subsequence would produce a further subsequence converging locally uniformly to the same limit from step 1.1, which is impossible.

L1givenassume-contradischarge-contradiction

3.1

Therefore the full sequence converges locally uniformly on $Ω$ to a holomorphic function.

given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Ch. 5 §§5.1-5.2 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 2 §5.2 and Ch. 8 §3.2 (standard reference, not scraped)
Sheldon Axler, Paul Bourdon, and Wade Ramey, Harmonic Function Theory, Ch. 2 (standard reference, not scraped)