Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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.

Bounded holomorphic disc functions have Poisson boundary data and Fatou limits under countable choice

Statement

Assume countable choice. Let f be bounded and holomorphic on D, and set M=sup⁡z∈D∣f(z)∣. Then there is a unique φ∈L∞(T,m) with f=P[φ]. It satisfies ∥φ∥∞=M, and f has nontangential limit φ(ζ) for almost every ζ within every cone ΓA(ζ), A>1. The zero function is allowed.

Facts & Assumptions

Given: Countable choice, bounded holomorphic f, and its finite bound M.

[F1]
[F2]

Under countable choice, Parseval identifies the squared L2 norm with the sum of the squared Fourier coefficients, and every square-summable bilateral coefficient sequence comes from a unique L2 class. (The Parseval identity for Fourier series, Riesz–Fischer: the Fourier coefficient map is onto the space of square-summable families, The Axiom of Countable Choice (ACω))

[F3]

Haar measure is a probability measure; Holder gives L2⊆L1. The Poisson kernel is (1−∣z∣2)/∣ζ−z∣2, and P[φ](z)=∫P(z,ζ)φ(ζ) dm(ζ). For any L1 datum its Poisson integral converges nontangentially almost everywhere to that datum under countable choice. (The one-dimensional torus and its normalized Haar integral, Complex Holder, Minkowski, and the quotient norm, The Poisson kernel on the unit disc, The Poisson kernel is positive, has total mass one, and concentrates at a boundary point, The Poisson integral of a finite complex boundary measure, Fatou limits for Poisson extensions of L1 boundary data)

Proof

1.1F1F2F3givenconstructalgebra

For 0<r<1, [F1] identifies the Fourier coefficients of fr as cnrn for n≥0 and zero for n<0, by uniform termwise integration and character orthogonality. By [F2], for each N, ∑n=0N∣cn∣2r2n≤∫T∣f(rζ)∣2 dm≤M2. Let r↑1 at this fixed finite N, then take the supremum in N: ∑n≥0∣cn∣2≤M2. Riesz-Fischer in [F2] gives a unique φ∈L2 with coefficients cn for n≥0 and zero for n<0.

2.1F1F3step 1.1algebra

This datum is in L1 by [F3]. The geometric identity, for ∣z∣<1 and ∣ζ∣=1, gives P(z,ζ)=1+∑n≥1(znζ−n+z‾nζn). The series converges absolutely uniformly in ζ at each fixed z, with total absolute bound 1+2∑n≥1∣z∣n<∞. Its integral against φ therefore converges termwise, since the error is bounded by its uniform norm times ∥φ∥1. Using its prescribed Fourier coefficients yields P[φ](z)=∑n≥0cnzn=f(z).

3.1F3step 2.1algebra∎

By [F3], f=P[φ] has the asserted nontangential limits. Since ∣f(z)∣≤M, passing to these limits gives ∣φ∣≤M almost everywhere, so φ∈L∞. Conversely positivity and unit mass of the kernel give ∣f(z)∣≤∥φ∥∞, hence M=∥φ∥∞. If another bounded datum ψ has P[ψ]=f, its nontangential limits are ψ by [F3]; the same limits give ψ=φ almost everywhere. If M=0 then f=0 and every coefficient and the unique datum are zero, so all claims remain valid. Only countable choice, in [F2] and [F3], has been used.

Depends on

Used by

Dependency tree · two levels

116 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