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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

1.1

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.

givenL1

2.1

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 $f (z) = (2 π i)^{- 1} \int_{γ} f (ζ) / (ζ - z) d ζ$ .

step 1.1L2choose

3.1

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 $q = 0$ .

step 2.1L3L4L7L8L9algebra

4.1

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

step 2.1step 3.1L5algebra

5.1

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 $0! = 1$ .

step 1.1step 2.1step 4.1L6choose

6.1

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 $Ω$ .

step 1.1step 5.1∎

Depends on

Used by

Dependency tree · two levels

82 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

Lars Ahlfors, Complex Analysis, 3rd ed., Ch. 5 §1.2 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Theorem 4.4 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Theorem 2.2.3 (standard reference, not scraped)
Steven G. Krantz, A Guide to Complex Variables, §3.1.6 (standard reference, not scraped)