Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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.

Liouville's theorem: every bounded entire function is constant

Statement

Every bounded entire function is constant.

More explicitly, if f:C→C is holomorphic and there is a real M≥0 such that ∣f(z)∣≤M for every z∈C, then f is constant.

Facts & Assumptions

Given: An entire function f:C→C and a real M≥0 with ∣f(z)∣≤M for every z; the Euclidean identification of C=R[x]/(x2+1) as the Euclidean plane and as a normed real algebra: what the identification preserves and the definition of complex domain in A complex domain is a nonempty connected open subset of C.

[L1]

If f is holomorphic on D(a,R), 0<r<R, and ∣f∣≤M on the radius-r circle, then ∣f(n)(a)∣≤n!M/rn for every natural n (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

[L2]

A holomorphic function on a complex domain whose derivative vanishes everywhere is constant (A holomorphic function with zero derivative on a domain is constant).

[L3]

The Euclidean plane R2 is polygonally connected and connected (Rn is polygonally connected, connected, locally path-connected and locally connected).

Proof

technique · direct
1.1givenL1

Fix a∈C and r>0. Since f is holomorphic on D(a,r+1) and its modulus is at most M on the radius-r circle, [L1] with derivative order one gives ∣f′(a)∣≤M/r.

1.2givenL3

Under the identification in the given data, [L3] makes C connected; it is also nonempty and open in itself, so it is a complex domain.

2.1step 1.1choosealgebra

If ∣f′(a)∣>0, choose r=M/∣f′(a)∣+1; then M/r<∣f′(a)∣, contradicting step 1.1, so f′(a)=0.

3.1step 2.1step 1.2L2∎

Since a was arbitrary, step 2.1 gives f′=0 throughout the domain of step 1.2, and [L2] makes f constant; this also covers M=0 and every constant entire function.

Depends on

Used by

Dependency tree · two levels

27 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