ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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.

A function whose modulus attains its maximum only on the distinguished boundary of a bidisc

Example

Let $f (z) = z_{0} z_{1}$ on the closed unit bidisc ${\overline{Δ}}_{(1, 1)} (0) \subseteq C^{2}$ . Then $∣ f ∣ \leq 1$ on the whole closed bidisc, and equality holds exactly on the distinguished boundary

$Γ_{(1, 1)} (0) = {(z_{0}, z_{1}) : ∣ z_{0} ∣ = ∣ z_{1} ∣ = 1} .$

By contrast, the topological boundary also contains points such as $(1, 0)$ , where $∣ f ∣ = 0$ . So the maximum-modulus information here is carried by the distinguished boundary and not by the whole topological boundary.

Facts & Assumptions

Given: The function $f (z) = z_{0} z_{1}$ on the closed unit bidisc.

[L1]

If $f$ is continuous on a closed polydisc and holomorphic on its interior, its modulus there is bounded by, and attains the same supremum as on, the distinguished boundary (The modulus of a holomorphic function on a closed polydisc is bounded by its supremum on the distinguished boundary).

[L2]

The distinguished boundary of the unit bidisc is $Γ_{(1, 1)} (0) = {(z_{0}, z_{1}) : ∣ z_{0} ∣ = ∣ z_{1} ∣ = 1}$ (Balls, polydiscs and the distinguished boundary in $C^{m}$ ).

Verification

technique · direct

1.1

For every $z$ in the closed unit bidisc, $∣ f (z) ∣ = ∣ z_{0} z_{1} ∣ = ∣ z_{0} ∣ ∣ z_{1} ∣ \leq 1$ , and equality holds if and only if $∣ z_{0} ∣ = ∣ z_{1} ∣ = 1$ , that is, exactly on the distinguished boundary of [L2].

givenL2

2.1

The point $(1, 0)$ lies on the topological boundary of the closed unit bidisc but not on the distinguished boundary: every ball about $(1, 0)$ meets the bidisc interior, while points with first coordinate of modulus $> 1$ lie arbitrarily close outside it. At that point $∣ f (1, 0) ∣ = 0$ . So the whole topological boundary does not by itself identify where the maximum is attained.

step 1.1

3.1

This is exactly the concrete content of [L1] for the function $f (z) = z_{0} z_{1}$ : the boundary points that matter are the distinguished ones.

step 1.1step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

30 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

J. Lebl, Tasty Bits of Several Complex Variables, v4.4, §1.2 (standard reference, not scraped)