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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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A conjugate difference quotient characterizes antiholomorphic maps

Statement

Let U⊆C be open, let a∈U, and let f:U→C. The limit

G=lim⁡h→0h≠0f(a+h)−f(a)hˉ

exists if and only if f is real totally differentiable at a and fz(a)=0. In that case fzˉ(a)=G. Consequently, the conjugate quotient exists at every point of U exactly for real-differentiable antiholomorphic maps, and its value is fzˉ.

Facts & Assumptions

Given: An open set U⊆C, a point a∈U, and a map f:U→C.

[F1]

Total differentiability at a means f(a+h)=f(a)+Df(a)h+r(h) with ∣r(h)∣/∣h∣→0 (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder).

[F2]

For a real-differentiable map, Df(a)h=fz(a)h+fzˉ(a)hˉ (The Wirtinger derivatives ∂zf and ∂zˉf, and antiholomorphic functions).

[L1]

Every real-linear map between Euclidean spaces has a matrix and is bounded by a constant times the Euclidean norm (Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0).

[L2]

Conjugation is a real-field automorphism with z‾‾=z, the modulus is multiplicative, and zz‾=∣z∣2 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive). Since z‾‾=z and zz‾=∣z∣2, one has ∣z‾∣2=z‾ z‾‾=z‾z=∣z∣2, and both moduli are nonnegative, so ∣z‾∣=∣z∣, so in particular ∣hˉ∣=∣h∣.

Proof

technique · direct
1.1

Suppose the conjugate quotient tends to G, and put r(h)=f(a+h)−f(a)−Ghˉ. Then ∣r(h)∣/∣h∣=∣(f(a+h)−f(a))/hˉ−G∣→0.

givenL2algebra
1.2

Conversely, suppose f is real totally differentiable and fz(a)=0. By [F1] and [F2], f(a+h)−f(a)=fzˉ(a)hˉ+r(h) with ∣r(h)∣/∣h∣→0.

givenF1F2
1.3

For the identity map, [F2] gives fz=1 and fzˉ=0, while its conjugate quotient is h/hˉ and has incompatible values 1 and −1 on real and imaginary increments. For conjugation, [F2] gives fz=0 and fzˉ=1, and its conjugate quotient is identically 1. These two tests confirm the placement of the conjugates and the barred coefficient.

F2L2algebra
2.1

The map h↦Ghˉ is real-linear and bounded by ∣G∣∣h∣, so [F1] and step 1.1 show that f is real totally differentiable with Df(a)h=Ghˉ.

step 1.1F1L1L2
3.1

Comparing this differential with [F2] gives fz(a)=0 and fzˉ(a)=G.

step 2.1F2algebra
4.1

Dividing by hˉ and using ∣hˉ∣=∣h∣ gives (f(a+h)−f(a))/hˉ=fzˉ(a)+r(h)/hˉ→fzˉ(a). This proves the reverse implication and the value of the limit.

step 1.2L2algebra∎

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