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

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A holomorphic function equals its Taylor series throughout the largest centred disc in its domain

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $a \in Ω$ . Put

$ρ_{a} : = {\begin{cases} d (a, C ∖ Ω), & C ∖ Ω \neq \emptyset, \\ + \infty, & Ω = C . \end{cases}$

Then $ρ_{a} > 0$ , the disc $D (a, ρ_{a})$ is the largest centred open disc contained in $Ω$ , with $D (a, + \infty) = C$ , and

$f (z) = \sum_{n \geq 0} \frac{f^{(n)} (a)}{n!} (z - a)^{n} (∣ z - a ∣ < ρ_{a}) .$

Every holomorphic function equals its Taylor series throughout the largest centred open disc contained in its domain.

Facts & Assumptions

Given: An open set $Ω \subseteq C$ , a holomorphic function $f : Ω \to C$ , and a point $a \in Ω$ ; the Taylor series convention of The Taylor series of a holomorphic function at a point and the whole-plane element $+ \infty$ of The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined.

[L1]

For a point $x$ and a nonempty subset $A$ of a metric space, the distance $d (x, A)$ is the greatest lower bound of ${d (x, y) : y \in A}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L2]

If $f$ is holomorphic on $D (a, R)$ , $0 < r < R$ , $∣ z - a ∣ < r$ , and $γ (t) = a + r \exp (i t)$ , then $f (z) = (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - z) d ζ$ (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

[L3]

Complex modulus is multiplicative, vanishes exactly at zero, and satisfies $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L4]

If a real $q$ satisfies $∣ q ∣ < 1$ , then $q^{k} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L5]

Uniform convergence of continuous integrands on the trace of a fixed rectifiable contour permits passage of the limit through the complex line integral (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L6]

For every natural $n$ , Cauchy's higher-derivative formula gives $f^{(n)} (a) = n! (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - a)^{n + 1} d ζ$ on a compactly contained circle (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L7]

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

[L8]

Closed bounded subsets of the Euclidean plane, and in particular circles of positive radius, are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L9]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Proof

technique · direct

1.1givenL1

If $Ω \neq C$ , openness gives $s > 0$ with $D (a, s) \subseteq Ω$ , so every $w \in C ∖ Ω$ has $∣ w - a ∣ \geq s$ and [L1] gives $ρ_{a} \geq s > 0$ ; moreover $∣ z - a ∣ < ρ_{a}$ forces $z \in Ω$ , while every $R > ρ_{a}$ contains a point of the complement by the defining greatest-lower-bound property. If $Ω = C$ , the stated $+ \infty$ convention gives the same largest-disc conclusion.

2.1step 1.1L2choose

Fix $z$ with $∣ z - a ∣ < ρ_{a}$ , and choose $r = (∣ z - a ∣ + ρ_{a}) / 2$ when $ρ_{a}$ is finite and $r = ∣ z - a ∣ + 1$ otherwise; then $∣ z - a ∣ < r < ρ_{a}$ , the radius- $r$ circle and its interior lie in $Ω$ by step 1.1, and [L2] gives $f (z) = (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - z) d ζ$ .

3.1step 2.1L3L4L7L8L9algebra

On $∣ ζ - a ∣ = r$ , the finite geometric identity gives $1 / (ζ - z) = \sum_{n = 0}^{N} (z - a)^{n} / (ζ - a)^{n + 1} + E_{N} (ζ)$ with $∣ E_{N} (ζ) ∣ \leq q^{N + 1} / (r - ∣ z - a ∣)$ for $q = ∣ z - a ∣ / r < 1$ ; [L7], [L8], and [L9] bound $∣ f ∣$ on the circle, so [L3] and [L4] make $f E_{N} \to 0$ uniformly there, including the case $z = a$ where $q = 0$ .

4.1step 2.1step 3.1L5algebra

By [L5], step 3.1 may be integrated term by term in the limit, and step 2.1 becomes $f (z) = \sum_{n \geq 0} (z - a)^{n} (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - a)^{n + 1} d ζ$ .

5.1step 1.1step 2.1step 4.1L6choose

Choose a radius $R$ with $r < R < ρ_{a}$ when $ρ_{a}$ is finite, and take $R = r + 1$ in the whole-plane case. Then $f$ is holomorphic on $D (a, R)$ and the radius- $r$ circle is compactly contained there, so for every natural $n$ , [L6] identifies the integral coefficient in step 4.1 with $f^{(n)} (a) / n!$ , including $n = 0$ and $0! = 1$ .

6.1step 1.1step 5.1∎

Since the point $z$ was arbitrary in the disc identified in step 1.1, step 5.1 proves the displayed Taylor equality throughout the largest centred open disc contained in $Ω$ .

A holomorphic function equals its Taylor series throughout the largest centred disc in its domain

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources