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Dixon's glued function is entire and vanishes at infinity

Statement

Let Ω⊆C be open, let f:Ω→C be holomorphic and let Γ be a complex chain which is a cycle, with trace in Ω and null-homologous in Ω (Null-homologous cycles and homologous cycles in an open set). Let g be the filled difference quotient of f on Ω×Ω (The filled difference quotient of a holomorphic function is jointly continuous) and put

h0(z)=12πi∫Γg(ζ,z) dζ  (z∈Ω),Ω0={z∈C∖Γ∗:n(Γ,z)=0},h1(z)=12πi∫Γf(ζ)ζ−z dζ  (z∈Ω0).

Then Ω0 is open, Ω∪Ω0=C, h0 is holomorphic on Ω, h1 is holomorphic on Ω0, and h0=h1 on Ω∩Ω0. Consequently the function h equal to h0 on Ω and to h1 on Ω0 is a well-defined entire function; it is bounded, and for every ε>0 there is R>0 with ∣h(z)∣<ε whenever ∣z∣>R.

Facts & Assumptions

Given: An open Ω, a holomorphic f:Ω→C, and a cycle Γ=∑k<rmkγk with Γ∗⊆Ω which is null-homologous in Ω; the plane carries the Euclidean metric of C=R[x]/(x2+1) as the Euclidean plane and as a normed real algebra: what the identification preserves.

[L1]

If γ is a rectifiable contour, W is open and φ is continuous on γ∗×W with φ(w,⋅) holomorphic on W for each w∈γ∗, then z↦∫γφ(ζ,z) dζ is holomorphic on W (A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic).

[L2]

The filled difference quotient g of a holomorphic f on Ω equals (f(ζ)−f(z))/(ζ−z) off the diagonal and f′(z) on it, and is continuous on Ω×Ω (The filled difference quotient of a holomorphic function is jointly continuous).

[L3]

For each fixed z∈Ω the map ζ↦g(ζ,z) is holomorphic on Ω, and for each fixed ζ∈Ω the map z↦g(ζ,z) is holomorphic on Ω (The filled difference quotient is holomorphic in each variable separately).

[L4]

For a complex chain Γ and φ continuous on Γ∗, the function z↦(2πi)−1∫Γφ(ζ)(ζ−z)−1 dζ is holomorphic on C∖Γ∗ (The Cauchy transform of a cycle is holomorphic off its trace, with the expected derivatives).

[L5]

For a cycle Γ the trace Γ∗ is compact, the index is locally constant on C∖Γ∗, the set Ω0 of points off the trace where the index vanishes is open, and there is R1>0 with n(Γ,p)=0 whenever ∣p∣>R1 (The index of a cycle is locally constant off its trace and vanishes far from it).

[L6]

A cycle Γ with trace in Ω is null-homologous in Ω when n(Γ,p)=0 for every p∈C∖Ω (Null-homologous cycles and homologous cycles in an open set).

[L7]

∫Γf dz=∑k<r, mk≠0mk∫γkf dz, and n(Γ,p)=(2πi)−1∫Γdz/(z−p) for p∉Γ∗ (Integration over a complex chain and the index of a chain); a chain is a finite list of integer-weighted contours whose trace is the union of the γk∗ with mk≠0 (Complex chains, their traces, and cycles).

[L8]

If ∣f(z)∣≤M on the trace of a rectifiable contour γ, with M≥0, then ∣∫γf dz∣≤M L(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length); complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand).

[L10]

A function holomorphic on all of C is entire, and holomorphy on an open set is complex differentiability at each of its points (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L11]

A set is closed exactly when its complement is open, and a set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space); a set is bounded when it is empty or lies inside a ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L12]

∣zw∣=∣z∣∣w∣ and ∣z+w∣≤∣z∣+∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[L14]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L15]

Finite linear combinations of holomorphic functions are holomorphic (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · direct
1.1givenL5L9L11

By [L5] the trace Γ∗ is compact, the set Ω0 is open, and there is R1>0 with n(Γ,p)=0 for ∣p∣>R1; by [L9] and [L11] there is also R0>0 with Γ∗⊆{w:∣w∣≤R0}.

1.2givenL6L7

If p∉Ω then p∉Γ∗, because Γ∗⊆Ω, and n(Γ,p)=0 by [L6]; so p∈Ω0. Hence Ω∪Ω0=C.

1.3givenL1L2L3L7L13L15

For each k<r with mk≠0 the trace γk∗ lies in Γ∗⊆Ω, and g is continuous on γk∗×Ω by [L2] with z↦g(w,z) holomorphic on Ω for each fixed w by [L3]; so [L1] makes z↦∫γkg(ζ,z) dζ holomorphic on Ω, and [L15] therefore makes the finite linear combination h0 holomorphic on Ω.

2.1step 1.1L4L14

The restriction of f to Γ∗ is continuous by [L14], so [L4] makes z↦(2πi)−1∫Γf(ζ)(ζ−z)−1dζ holomorphic on C∖Γ∗; since Ω0⊆C∖Γ∗ is open by step 1.1, h1 is holomorphic on Ω0.

2.2step 1.1L2L7L8L13

Let z∈Ω∩Ω0. Then z∉Γ∗, so ζ≠z for every ζ∈Γ∗ and [L2] gives g(ζ,z)=(f(ζ)−f(z))/(ζ−z) there; splitting the integral by [L8] and [L13] gives h0(z)=h1(z)−f(z) n(Γ,z) through [L7], and n(Γ,z)=0 because z∈Ω0, so h0(z)=h1(z).

2.3step 1.1L7L8L9L12L13L14

Let Mf=0 when Γ∗=∅, and otherwise choose a real Mf≥0 with ∣f(w)∣≤Mf for every w∈Γ∗; such a bound exists because f is continuous on the compact trace by [L14], so its image is compact and therefore bounded by [L9]. Put M(Γ)=∑k<r, mk≠0∣mk∣L(γk). For z∈Ω0 with ∣z∣>R0 and ζ∈Γ∗, [L12] gives ∣ζ−z∣≥∣z∣−R0>0, so [L7], [L8] and [L13] give ∣h1(z)∣≤MfM(Γ)/(2π(∣z∣−R0)).

3.1step 1.2step 1.3step 2.1step 2.2L10

By steps 1.2, 1.3, 2.1 and 2.2 the assignment h=h0 on Ω and h=h1 on Ω0 is a well-defined function on C, and it is complex differentiable at every point because each point lies in one of the two open sets on which the corresponding piece is holomorphic; so h is entire by [L10].

4.1step 3.1step 2.3L7L9L11L14∎

Let ε>0 and take R=max⁡{R0,R1}+MfM(Γ)/(2πε)+1. For ∣z∣>R step 1.1 puts z in Ω0, so h(z)=h1(z) by step 3.1 and step 2.3 gives ∣h(z)∣<ε. Taking ε=1 produces one such R, and h is continuous on the compact disc {∣z∣≤R} by [L9] and [L14], hence bounded there by [L9]; so h is bounded on C. If Γ∗=∅ then every integral over Γ is 0 by [L7] and h is identically 0, which satisfies both conclusions.

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