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Dixon's glued function is entire and vanishes at infinity

Statement

Let ΩC be open, let f:ΩC 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)=12πiΓg(ζ,z)dζ  (zΩ),Ω0={zCΓ:n(Γ,z)=0},h1(z)=12πiΓf(ζ)ζzdζ  (zΩ0).

Then Ω0 is open, ΩΩ0=C, h0 is holomorphic on Ω, h1 is holomorphic on Ω0, and h0=h1 on ΩΩ0. 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:ΩC, and a cycle Γ=k<rmkγk with ΓΩ 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 wγ, then zγφ(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))/(ζz) 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 zΩ the map ζg(ζ,z) is holomorphic on Ω, and for each fixed ζΩ the map zg(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)1Γφ(ζ)(ζz)1dζ is holomorphic on CΓ (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 CΓ, 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 pCΩ (Null-homologous cycles and homologous cycles in an open set).

[L7]

Γfdz=k<r,mk0mkγkfdz, and n(Γ,p)=(2πi)1Γdz/(zp) for pΓ (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 mk0 (Complex chains, their traces, and cycles).

[L8]

If f(z)M on the trace of a rectifiable contour γ, with M0, then γfdzML(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=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, 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
1.1

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:wR0}.

givenL5L9L11
1.2

If pΩ then pΓ, because ΓΩ, and n(Γ,p)=0 by [L6]; so pΩ0. Hence ΩΩ0=C.

givenL6L7
1.3

For each k<r with mk0 the trace γk lies in ΓΩ, and g is continuous on γk×Ω by [L2] with zg(w,z) holomorphic on Ω for each fixed w by [L3]; so [L1] makes zγkg(ζ,z)dζ holomorphic on Ω, and [L15] therefore makes the finite linear combination h0 holomorphic on Ω.

givenL1L2L3L7L13L15
2.1

The restriction of f to Γ is continuous by [L14], so [L4] makes z(2πi)1Γf(ζ)(ζz)1dζ holomorphic on CΓ; since Ω0CΓ is open by step 1.1, h1 is holomorphic on Ω0.

step 1.1L4L14
2.2

Let zΩΩ0. Then zΓ, so ζz for every ζΓ and [L2] gives g(ζ,z)=(f(ζ)f(z))/(ζz) 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 zΩ0, so h0(z)=h1(z).

step 1.1L2L7L8L13
2.3

Let Mf=0 when Γ=, and otherwise choose a real Mf0 with f(w)Mf for every wΓ; 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,mk0mkL(γk). For zΩ0 with z>R0 and ζΓ, [L12] gives ζzzR0>0, so [L7], [L8] and [L13] give h1(z)MfM(Γ)/(2π(zR0)).

step 1.1L7L8L9L12L13L14
3.1

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].

step 1.2step 1.3step 2.1step 2.2L10
4.1

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 {zR} 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.

step 3.1step 2.3L7L9L11L14

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